analytic capacity, the cauchy transform, and non-homogeneous calder³nâ€“zygmund theory

Progress in Mathematics307

Xavier Tolsa

Analytic Capacity, the Cauchy Transform, and Non-homogeneous Calderón–Zygmund Theory

Ferran Sunyer i Balaguer Award winning monograph

Progress in MathematicsVolume 307

For further volumes:http://www.springer.com/series/4848

Series EditorsHyman Bass, University of Michigan, Ann Arbor, USA Jiang-Hua Lu, The University of Hong Kong, Hong Kong SAR, China

Yuri Tschinkel, Courant Institute of Mathematical Sciences, New York, USAJoseph Oesterlé, Université Pierre et Marie Curie, Paris, France

http://www.springer.com/series/4848

Xavier Tolsa

Analytic Capacity,the Cauchy Transform,and Non-homogeneousCalderón–Zygmund Theory

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DOI 10.1007/978-3- - ISBN 978-3-3 - - ISBN 978-3-3 - - (eBook)

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International Publishing Switzerland 4

part ment de Mat tiqueUniversit

Xavier Tolsaa e emà s

at Autònoma de BarcelonaBellaterra (Cerdanyola)

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Catalonia

Library of Congress Control Number: 2013956424

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Ferran Sunyer i Balaguer (1912–1967) was a self-taught Catalan mathematician who, in spite of aserious physical disability, was very active in researchin classical mathematical analysis, an area in whichhe acquired international recognition. His heirs cre-ated the Fundacio Ferran Sunyer i Balaguer insidethe Institut d’Estudis Catalans to honor the memoryof Ferran Sunyer i Balaguer and to promote mathe-matical research.

Each year, the Fundacio Ferran Sunyer i Balaguerand the Institut d’Estudis Catalans award an in-ternational research prize for a mathematical mono-graph of expository nature. The prize-winning mono-graphs are published in this series. Details about theprize and the Fundacio Ferran Sunyer i Balaguer canbe found at

http://ffsb.iec.cat/EN

This book has been awarded theFerran Sunyer i Balaguer 2013 prize.

The members of the scientific commiteeof the 2013 prize were:

Alejandro AdemUniversity of British Columbia

Nuria FagellaUniversitat de Barcelona

Joseph OesterleUniversite de Paris VI

Eero SaksmanUniversity of Helsinki

Alan WeinsteinUniversity of California at Berkeley

http://ffsb.iec.cat/EN

Ferran Sunyer i Balaguer Prize winners since 2003:

2003 Fuensanta Andreu-Vaillo, Vincent Casellesand Jose M. MazonParabolic Quasilinear Equations MinimizingLinear Growth Functionals, PM 223

2004 Guy DavidSingular Sets of Minimizers for theMumford-Shah Functional, PM 233

2005 Antonio Ambrosetti and Andrea MalchiodiPerturbation Methods and SemilinearElliptic Problems on Rn, PM 240

Jose SeadeOn the Topology of Isolated Singularities inAnalytic Spaces, PM 241

2006 Xiaonan Ma and George MarinescuHolomorphic Morse Inequalities andBergman Kernels, PM 254

2007 Rosa Miro-RoigDeterminantal Ideals, PM 264

2008 Luis BarreiraDimension and Recurrence in HyperbolicDynamics, PM 272

2009 Timothy D. BrowningQuantitative Arithmetic of Projective Vari-eties, PM 277

2010 Carlo MantegazzaLecture Notes on Mean Curvature Flow,PM 290

2011 Jayce Getz and Mark GoreskyHilbert Modular Forms with Coefficients inIntersection Homology and Quadratic BaseChange, PM 298

2012 Angel Cano, Juan Pablo Navarrete andJose SeadeComplex Kleinian Groups, PM 303

To Montse, Laura, and Sılvia

Contents

Introduction 1

Basic notation 11

1 Analytic capacity 15

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.2 Definition and basic properties of analytic capacity . . . . . . . . . 15

1.3 Removable sets and the Painleve problem . . . . . . . . . . . . . . 20

1.4 Two technical lemmas . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.5 The Cauchy transform and Vitushkin’s localization operator . . . . 22

1.6 Relationship with Hausdorff measures . . . . . . . . . . . . . . . . 28

1.7 Rectifiable sets and Vitushkin’s conjecture . . . . . . . . . . . . . . 34

1.8 Historical remarks, the semiadditivity of γ, etc. . . . . . . . . . . . 37

1.8.1 Some historical remarks . . . . . . . . . . . . . . . . . . . . 37

1.8.2 Uniform approximation by rational functions and the semi-additivity of analytic capacity . . . . . . . . . . . . . . . . . 37

1.8.3 The approach by duality to analytic capacity . . . . . . . . 39

1.8.4 Other spaces and capacities . . . . . . . . . . . . . . . . . . 40

1.8.5 Hausdorff measures . . . . . . . . . . . . . . . . . . . . . . . 41

2 Calderon-Zygmund theory with non-doubling measures 45

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

2.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

2.3 Covering theorems and maximal operators . . . . . . . . . . . . . . 49

2.4 Doubling cubes and balls . . . . . . . . . . . . . . . . . . . . . . . 53

2.5 Some standard estimates . . . . . . . . . . . . . . . . . . . . . . . . 55

2.6 Calderon-Zygmund decomposition . . . . . . . . . . . . . . . . . . 57

2.7 Weak (1,1) boundedness of Calderon-Zygmund operators . . . . . . 60

2.8 Cotlar’s inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

2.9 The good lambda method . . . . . . . . . . . . . . . . . . . . . . . 68

2.10 Historical remarks and further results . . . . . . . . . . . . . . . . 73

ix

x Contents

3 The Cauchy transform and Menger curvature 753.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 753.2 The curvature of a measure . . . . . . . . . . . . . . . . . . . . . . 763.3 The T 1 theorem for the Cauchy singular integral operator . . . . . 813.4 The Cauchy transform on Lipschitz graphs . . . . . . . . . . . . . 843.5 The Cauchy transform on AD regular curves . . . . . . . . . . . . 883.6 Curvature and Jones’ β’s . . . . . . . . . . . . . . . . . . . . . . . 913.7 Historical remarks and further results . . . . . . . . . . . . . . . . 98

3.7.1 The Cauchy transform and curvature . . . . . . . . . . . . . 983.7.2 The T 1 theorem . . . . . . . . . . . . . . . . . . . . . . . . 983.7.3 The Cauchy transform on Lipschitz graphs and AD regular

curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 993.7.4 Application of the curvature method to other kernels . . . . 99

4 The capacity γ+ 1034.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1034.2 Smoothing of the Cauchy kernel by mollification . . . . . . . . . . 1044.3 Dualization of the weak (1,1) inequality . . . . . . . . . . . . . . . 1074.4 The Denjoy conjecture . . . . . . . . . . . . . . . . . . . . . . . . . 1104.5 Semiadditivity of γ+ . . . . . . . . . . . . . . . . . . . . . . . . . . 1124.6 Some potential theory for γ+ . . . . . . . . . . . . . . . . . . . . . 115

4.6.1 A couple of technical lemmas about curvature . . . . . . . . 1154.6.2 A maximum principle for curvature and for Uμ . . . . . . . 1174.6.3 A dual version for the capacity γ+ . . . . . . . . . . . . . . 120

4.7 The capacity γ+ of some Cantor sets . . . . . . . . . . . . . . . . . 1254.8 A quantitative version of Denjoy’s conjecture . . . . . . . . . . . . 1274.9 Relationship between γ+ and Wolff’s potentials . . . . . . . . . . . 1294.10 Historical remarks and further results . . . . . . . . . . . . . . . . 132

4.10.2 Analytic capacity and curvature . . . . . . . . . . . . . . . 1324.10.3 The analytic capacity of the Cantor sets E(λ) and Wolff’s

potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1334.10.4 Capacities associated with the signed Riesz kernels . . . . . 134

5 A Tb theorem of Nazarov, Treil and Volberg 1375.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1375.2 The exceptional set S . . . . . . . . . . . . . . . . . . . . . . . . . 1385.3 The suppressed operators KΘ . . . . . . . . . . . . . . . . . . . . . 1405.4 Dyadic lattices and the martingale decomposition . . . . . . . . . . 143

5.4.1 The dyadic Carleson embedding theorem . . . . . . . . . . 1435.4.2 Random dyadic lattices . . . . . . . . . . . . . . . . . . . . 1455.4.3 Transit and terminal squares . . . . . . . . . . . . . . . . . 1465.4.4 The martingale decomposition . . . . . . . . . . . . . . . . 146

5.5 Good and bad squares and functions . . . . . . . . . . . . . . . . . 152

4.10.1 Denjoy’s conjecture, theDavie-Øksendal dualization, and γ+ 132..

Contents xi

5.6 Estimates for good functions . . . . . . . . . . . . . . . . . . . . . 1535.6.1 The main lemma for good functions . . . . . . . . . . . . . 1535.6.2 Beginning of the proof of Lemma 5.13 . . . . . . . . . . . . 154

5.7 Estimate of the sum S1: distant squares . . . . . . . . . . . . . . . 1565.8 Estimate of the sum S4 . . . . . . . . . . . . . . . . . . . . . . . . 160

5.8.1 Splitting of S4 . . . . . . . . . . . . . . . . . . . . . . . . . 1605.8.2 Estimate of Str

4,1 via a paraproduct . . . . . . . . . . . . . . 1615.8.3 Estimate of Sterm

4,1 . . . . . . . . . . . . . . . . . . . . . . . 1685.9 Estimate of S2 + S3: the diagonal term . . . . . . . . . . . . . . . . 1725.10 Cotlar’s inequality revisited . . . . . . . . . . . . . . . . . . . . . . 1765.11 The final probabilistic argument . . . . . . . . . . . . . . . . . . . 181

5.11.1 The low probability of bad squares and functions . . . . . . 1815.11.2 The nice set G . . . . . . . . . . . . . . . . . . . . . . . . . 1855.11.3 The functions Φ, Ψw, and the operators K and C . . . . . . 1875.11.4 The key estimates involving C and K . . . . . . . . . . . . . 1885.11.5 The final step . . . . . . . . . . . . . . . . . . . . . . . . . . 191

5.12 Historical remarks and further results . . . . . . . . . . . . . . . . 1935.12.1 The Tb theorem and analytic capacity . . . . . . . . . . . . 1935.12.2 Other Tb-like theorems . . . . . . . . . . . . . . . . . . . . 194

6 The comparability between γ and γ+ 1956.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1956.2 An argument for sets of finite length . . . . . . . . . . . . . . . . . 1966.3 Outline of the argument for proving that γ ≈ γ+ . . . . . . . . . . 2006.4 An intermediate approximation in terms of γ+ . . . . . . . . . . . 2026.5 Construction of the good measures μ and ν . . . . . . . . . . . . . 206

6.5.1 The construction of μ and ν and the proof of (a)–(c) . . . . 2076.5.2 The exceptional set H and the proof of (e), (f) . . . . . . . 2086.5.3 Proof of (d) . . . . . . . . . . . . . . . . . . . . . . . . . . . 2096.5.4 Proof of (g) . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

6.6 Dyadic lattices and the exceptional sets HD and TD . . . . . . . . 2166.6.1 The construction of HD . . . . . . . . . . . . . . . . . . . . 2166.6.2 The accretivity condition and the exceptional set TD . . . . 2176.6.3 The size of HD ∪ TD . . . . . . . . . . . . . . . . . . . . . . 217

6.7 Application of a Tb theorem . . . . . . . . . . . . . . . . . . . . . . 2186.8 The final induction argument . . . . . . . . . . . . . . . . . . . . . 2206.9 Estimate of the Cauchy integral and AD regular curves . . . . . . 221

6.9.1 Estimate of the Cauchy integral . . . . . . . . . . . . . . . . 2216.9.2 Another proof of the L2 boundedness of the Cauchy trans-

forms on AD regular curves . . . . . . . . . . . . . . . . . . 2256.10 Historical remarks and further results . . . . . . . . . . . . . . . . 226

6.10.1 Analytic capacity . . . . . . . . . . . . . . . . . . . . . . . . 2266.10.2 Other capacities . . . . . . . . . . . . . . . . . . . . . . . . 228

xii Contents

7 Curvature and rectifiability 2317.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2317.2 The quantitative version of the David-Leger theorem . . . . . . . . 2337.3 The two squares condition, Jones’ β’s, and curvature . . . . . . . . 2357.4 The sets Z, E1 and E2 . . . . . . . . . . . . . . . . . . . . . . . . . 2417.5 Construction of the Lipschitz graph . . . . . . . . . . . . . . . . . . 2437.6 E and Γ are close to each other . . . . . . . . . . . . . . . . . . . . 2537.7 The set E1 is small . . . . . . . . . . . . . . . . . . . . . . . . . . . 2567.8 The set E2 is small . . . . . . . . . . . . . . . . . . . . . . . . . . . 258

7.8.1 The implications E2 big ⇒ ‖A′‖2 big ⇒ c2(H1�Γ) big . . . 2587.8.2 The implication c2(H1�Γ) big ⇒ c2(μ) big . . . . . . . . . . 259

7.9 Three applications . . . . . . . . . . . . . . . . . . . . . . . . . . . 2767.9.1 A characterization of rectifiable sets in terms of pointwise

curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2767.9.2 A quantitative version of David’s theorem . . . . . . . . . . 2777.9.3 Characterization of γ in terms of measures with bounded

upper density . . . . . . . . . . . . . . . . . . . . . . . . . . 2797.10 Curvature, β’s, and rectifiability in the AD regular case . . . . . . 2817.11 Historical remarks and further results . . . . . . . . . . . . . . . . 286

7.11.1 About the results in this chapter . . . . . . . . . . . . . . . 2867.11.2 Rectifiability, uniform rectifiability and Riesz transforms in

arbitrary dimensions . . . . . . . . . . . . . . . . . . . . . . 2867.11.3 Other results in connection with curvature . . . . . . . . . . 288

8 Principal values for the Cauchy transform 2898.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2898.2 L2 boundedness implies existence of principal values . . . . . . . . 290

8.2.1 Reduction to a class of dense functions . . . . . . . . . . . . 2908.2.2 Existence of principal values on Lipschitz graphs and recti-

fiable sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2938.2.3 Plemelj formulas on Lipschitz graphs . . . . . . . . . . . . . 2968.2.4 Weak convergence and existence of principal values for mea-

sures with zero density . . . . . . . . . . . . . . . . . . . . . 3008.2.5 The final argument for the proof of Theorem 8.1 . . . . . . 307

8.3 Finiteness of the maximal Cauchy transform . . . . . . . . . . . . . 3098.4 Some consequences . . . . . . . . . . . . . . . . . . . . . . . . . . . 3128.5 Historical remarks and further results . . . . . . . . . . . . . . . . 314

8.5.1 Existence of principal values on rectifiable sets . . . . . . . 3148.5.2 From the analytic information to rectifiability . . . . . . . . 315

9 RBMO(μ) and H1atb(μ) 319

9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3199.2 The space RBMO(μ) . . . . . . . . . . . . . . . . . . . . . . . . . . 321

9.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 321

Contents xiii

9.2.2 The coefficients KQ,R . . . . . . . . . . . . . . . . . . . . . 3219.2.3 The definition of RBMO(μ) . . . . . . . . . . . . . . . . . . 3239.2.4 Equivalent definitions . . . . . . . . . . . . . . . . . . . . . 3269.2.5 Boundedness of CZO’s from L∞(μ) to RBMO(μ) . . . . . . 3309.2.6 Some examples . . . . . . . . . . . . . . . . . . . . . . . . . 334

9.3 The John-Nirenberg inequality . . . . . . . . . . . . . . . . . . . . 3369.4 The Hardy spaces H1,p

atb(μ) . . . . . . . . . . . . . . . . . . . . . . . 340

9.5 The duality H1,patb(μ) - RBMO(μ) . . . . . . . . . . . . . . . . . . . 347

9.6 Another maximal operator and another covering theorem . . . . . 3539.7 The sharp maximal operator . . . . . . . . . . . . . . . . . . . . . 3569.8 Two interpolation theorems . . . . . . . . . . . . . . . . . . . . . . 3609.9 The T 1 theorem for the Cauchy transform again . . . . . . . . . . 3659.10 The T 1 theorem in terms of RBMO(μ) and BMOρ(μ) . . . . . . . 3699.11 Historical remarks and further results . . . . . . . . . . . . . . . . 376

9.11.1 About RBMO(μ) and BMO(μ) . . . . . . . . . . . . . . . . 3769.11.2 Other related results . . . . . . . . . . . . . . . . . . . . . . 3769.11.3 Tb theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . 378

Bibliography 381

Index 393

Introduction

This book studies some of the striking advances that have occurred regarding ana-lytic capacity and its relationship with rectifiability in the decade 1995-2005. TheCauchy transform plays a fundamental role in this area and it is one of the mainthemes of the book too. Another important topic is the so-called non-homogeneousCalderon-Zygmund theory, whose development has been largely motivated by theproblems arising in connection with analytic capacity.

The analytic capacity of a compact set E ⊂ C is defined as

γ(E) = sup |f ′(∞)|,where the supremum is taken over all analytic functions f : C \ E → C whosemodulus is bounded by 1, and f ′(∞) = limz→∞ z

(f(z)−f(∞)

). Analytic capacity

was introduced by Ahlfors in 1947 in order to study the Painleve problem. Thisproblem consists in characterizing removable singularities for bounded analyticfunctions (removable, for short) in metric and/or geometric terms. Ahlfors showedthat a compact set in the complex plane is removable if and only if its analyticcapacity vanishes. This characterization is neither geometric nor metric, and thusis far from being a solution of Painleve’s problem.

In the 1950’s analytic capacity was rediscovered by Vitushkin while he wasstudying uniform approximation of analytic functions by rational functions incompact subsets of the plane. Indeed, some of the criteria he gave for uniformrational approximation were stated in terms of analytic capacity and a variant ofthis notion, the continuous analytic capacity. Further, because of the applicationto this type of problems, he raised the question on the semiadditivity of analyticcapacity. Namely, does there exist an absolute constant c such that all compactsets E,F ⊂ C satisfy

γ(E ∪ F ) ≤ c(γ(E) + γ(F )

)?

Another relevant contribution of Vitushkin is connected to the Painleve prob-lem. He showed that there are compact sets with positive length (or 1-dimensionalHausdorff measure) which are removable. The example that he constructed turnedout to have orthogonal projections of zero length in almost all directions. So heconjectured that all removable sets are of this form. This became known as Vi-tushkin’s conjecture and remained open for a long time. Let us remark that, by

, , 1 OI 10.1007/978-3- - -6_1,

© Springer

X. Tolsa Analytic Capacity, the Cauchy Transform, and Non-homogeneous Calderón–Zygmund TheoryProgress in Mathematics 307, D 319 00596


2 Introduction

a well-known theorem of Besicovitch, for a compact set E ⊂ C with finite length,Vitushkin’s conjecture is equivalent to saying that E is removable if and only ifit is purely unrectifiable (i.e. its intersection with any rectifiable curve has zerolength).

Given a measure μ on C and a function f ∈ L1loc(μ), the Cauchy transform

(or Cauchy singular integral operator) is defined by

Cμf(z) =∫

1

w − zf(w) dμ(w),

whenever the integral makes sense. In 1977 Calderon [5] proved that the Cauchytransform is bounded in L2(μ) for μ equal to the arc length on a Lipschitz graphwith sufficiently small slope. As a corollary of this result, it follows that anyrectifiable set with positive length has positive analytic capacity and thus it isnon-removable. This result, previously known as Denjoy’s conjecture, is equivalentto saying that any removable set with finite length is purely unrectifiable. So thisis the solution of one of the implications in Vitushkin’s conjecture for sets withfinite length.

In 1986 Mattila [98] showed that Vitushkin’s conjecture in full generalityis false. He proved this in a quite astonishing way: he showed that for sets withnon-σ-finite length the property of having projections of zero length in almost alldirections is not invariant by conformal mappings, unlike removability. Later on,Jones and Murai [76] showed an explicit example of a non-removable set (withnon-σ-finite length) with projections of zero length in almost all directions. So theconjecture fails for sets with non-σ-finite length.

The discovery of the relationship between the Cauchy kernel and Mengercurvature by Melnikov in 1995 is a factor of major importance in the subsequentadvances concerning analytic capacity and the Painleve problem. While studyinga discrete version of analytic capacity in [113] he found the nice identity

∑σ∈S3

1

zσ1 − zσ2

1

zσ1 − zσ3

=1

R(z1, z2, z3)2, z1, z2, z3 ∈ C, (0.1)

where S3 is the group of permutations of three elements and R(z1, z2, z3) standsfor the radius of the circumference through z1, z2, z3. The reciprocal of R(z1, z2, z3)is called Menger curvature of z1, z2, z3. The preceding formula is so surprising andhas so many relevant consequences that is quite often called “Melnikov’s miracle”.

Given a finite measure with linear growth μ, that is,

μ(B(x, r)) ≤ c0 r for all x ∈ Rd and all r > 0, (0.2)

by integrating the identity (0.1) with respect to μ three times one easily deducesthat, for all ε > 0,

Introduction 3

∫ ∣∣∣∣∫|z−w|>ε

1

w − zdμ(w)

∣∣∣∣2 dμ(z)=

1

6

∫∫∫Tε

1

R(z1, z2, z3)2dμ(z1) dμ(z2) dμ(z3) +O(μ(C)), (0.3)

where |O(μ(C))| ≤ c μ(C) and Tε is the set of triples (z1, z2, z3) such that |zi−zj| >ε for i �= j. This identity was proved by Melnikov and Verdera [114] and it is veryimportant because it relates the L2(μ) norm of the ε-truncated Cauchy transformof μ on the left side with something metric/geometric, namely the triple integralof squared Menger curvature with respect to μ. Following Melnikov, this tripleintegral will be called curvature of μ in this book. In a sense, the equation (0.3)provides a bridge between the L2(μ) boundedness of the Cauchy singular integraland rectifiability.

After Melnikov’s discovery, to complete the solution of Vitushkin’s conjecturefor sets of finite length, it remained to show that if E ⊂ C is a purely unrectifiablecompact set with finite length, then it is removable. This was accomplished byMattila, Melnikov and Verdera [103] in 1996 under the additional assumptionthat E is Ahlfors-David regular, that is,

c−1r ≤ H1(E ∩B(z, r)) ≤ c r for 0 < r ≤ diam(E), z ∈ E, (0.4)

where H1 stands for the Hausdorff 1-dimensional (or length) measure. Besidesthe relationship between the Cauchy kernel and curvature described above, theirarguments used deep results from harmonic analysis and geometric measure theory.

The proof of Vitushkin’s conjecture in full generality for sets of finite lengthwas finally obtained by David [23] in 1998. To this end, he had to develop suitabletools from geometric measure theory and harmonic analysis, analogous to theones used by Mattila, Melnikov and Verdera [103], now without any regularityassumption on the set E. It is also worth mentioning that very shortly after David’scelebrated completion of the proof of Vitushkin’s conjecture, Nazarov, Treil andVolberg [126] obtained a very powerful Tb theorem which provides an alternativeargument for the proof of the last step of the conjecture.

For sets with infinite length, Painleve’s problem does not have a solution sonice as David’s one for sets with finite length. We have to content ourselves witha characterization of removability in terms of the notion of curvature of measures.This characterization is a corollary of the comparability (proved by Tolsa [160]in 2003) between analytic capacity and another capacity denoted by γ+ whichhas a useful quantitative description in terms of measures with linear growth andfinite curvature. Another important consequence of the comparability between γand γ+ deals with the semiadditivity question. Indeed, from the aforementionedquantitative description of γ+, one easily deduces that γ+ is semiadditive. Thenthe fact that γ and γ+ are comparable implies that γ is semiadditive too.

The advances regarding analytic capacity that culminated in the proof ofVitushkin’s conjecture for sets with finite length and the proof of the semiadditivity

4 Introduction

of analytic capacity were possible because of the spectacular progress that occurredpreviously in the areas of harmonic analysis (more precisely, Calderon-Zygmundtheory) and geometric measure theory. This is an example where some problemsfrom complex analysis are solved using real variable techniques.

The connection between analytic capacity and Calderon-Zygmund theorystems from the essential role that the Cauchy transform plays in many questionsarising in complex analysis. For example, concerning the Painleve problem, wehave already mentioned above the relevance of Calderon’s theorem on the L2

boundedness of the Cauchy transform on Lipschitz graphs with small slope for theproof of Denjoy’s conjecture. The smallness assumption on the slope was removedin 1982 by Coifman, McIntosh and Meyer in their outstanding work [16].

To study the L2(μ) boundedness of singular integral operators such as theone given by the Cauchy transform, some powerful tools were developed. This isthe case, for instance, of the so-called T 1 theorem of David and Journe [25] andof other various Tb theorems like the ones in David, Journe and Semmes [26] orChrist [14].

In almost all the classical Calderon-Zygmund theory, and in particular, in theT 1 and Tb theorems just mentioned, an essential assumption is that the underlyingmeasure μ in the space is doubling, that is,

μ(B(x, 2r)) ≤ c μ(B(x, r)) for all x ∈ supp(μ) and all r > 0.

By the early 1990’s it was a general belief in the area that the framework ofhomogeneous spaces, where the underlying measure is assumed to be doubling,was the most natural context where one could obtain reasonable results. However,often to deal with questions on analytic capacity one is faced with problems inCalderon-Zygmund theory where the doubling assumption fails. For example, typ-ically to estimate the analytic capacity of an arbitrary compact set E ⊂ C onehas to work with measures such as the Hausdorff 1-dimensional measure H1�E,which may not satisfy the above doubling condition. Instead of this assumption,one is led naturally to consider measures μ which satisfy the linear growth con-dition (0.2). Under this assumption on the growth of the μ (or an analogous n-dimensional version), in the decade 1995-2005 it was shown that large parts of theclassical Calderon-Zygmund theory remain valid. For example, a non-doublingversion of the T 1 theorem for the Cauchy transform was proved by Tolsa [154]and for general singular integral operators by Nazarov, Treil and Volberg [123].Also, in addition to the Tb theorems suitable to study analytic capacity provedby David [23] and by Nazarov, Treil and Volberg [126], other Tb-like theoremsfor non-homogeneous spaces appeared in Nazarov, Treil and Volberg [127] and[125]. To summarize, Calderon-Zygmund theory was developed in the setting ofnon-homogeneous spaces.

Concerning the aforementioned advances on geometric measure theory, thedevelopment of multiscale techniques for rectifiability played a fundamental role inthe solution of Vitushkin’s conjecture and in the study of the relationship between

Introduction 5

the L2 boundedness of the Cauchy transform and rectifiability. The developmentof these techniques started in 1990 with the celebrated “traveling salesman theo-rem” of Peter Jones [75], where he obtained a characterization of rectifiable setsin terms of the so-called β coefficients. This characterization is reminiscent of themultiscale methods from Littlewood-Paley theory. David and Semmes ([28], [29])made further progress in these techniques, even in higher dimensions, and intro-duced the notion of “uniform rectifiability”. It turns out that the n-dimensionaluniformly rectifiable sets are a natural framework to study the L2 boundedness ofn-dimensional singular integral operators with an odd kernel.

This book contains complete proofs of most of the results explained above.In particular, the reader will find the detailed arguments for Vitushkin’s conjec-ture and for the semiadditivity of analytic capacity. In the text, the relationshipbetween the L2 boundedness of the Cauchy integral operator, the existence of prin-cipal values, and rectifiability is also discussed. The necessary non-homogeneousCalderon-Zygmund theory to prove these results is studied too, together with ad-ditional topics, such as the theory about BMO and Hardy spaces.

There are other books which study questions about analytic capacity. Someclassical references are the books by Gamelin [47] or Garnett [54], which werewritten more than thirty years ago, and so they do not discuss the exceptionaladvances from the last decade. Mattila’s text [100] from 1995 is more modern, al-though not enough to include these advances either. It contains a chapter which isdevoted to analytic capacity and the Painleve problem and another to singular in-tegrals and rectifiability. There is more overlap between our text and the books byPajot [132] and Dudziak [35], which are more recent. The book by Pajot containsthe proof of Vitushkin’s conjecture for sets satisfying the Ahlfors-David regularitycondition (0.4). On the other hand, Dudziak’s book includes detailed argumentsof the full conjecture. However, the comparability γ ≈ γ+ and the semiadditivityof analytic capacity are not proved, neither in Pajot [132] nor in Dudziak [35].Moreover, the non-homogeneous Calderon-Zygmund theory is discussed in muchless detail in these books than in the present one. It is also worth mentioningVolberg’s book [180]. Although, strictly speaking it does not deal with analyticcapacity but with its higher dimensional version, the so-called Lipschitz harmoniccapacity κ, the arguments in the book to show the comparability between κ andκ+ also apply to analytic capacity, perhaps with minor modifications.

I will now describe briefly the content of each chapter of this book. Thefirst one summarizes some classical results on analytic capacity and the Painleveproblem. The relationship of these notions with Hausdorff measures is discussed.Also, the important localization operator of Vitushkin is introduced. Most of thematerial from this chapter is well known and can be found in references such asGamelin [47], Garnett [54], or Mattila [100]

In Chapter 2 some basic results on non-homogeneous Calderon-Zygmundtheory are obtained. In particular, by means of a suitable Calderon-Zygmund de-composition, it is shown that the L2(μ) boundedness of singular integral operators

6 Introduction

implies that they are of weak type (1, 1) (and also bounded from the space of fi-nite real measures M(Rn) to L1,∞(μ)). Also, it is proved that Cotlar’s inequalityholds in this context. Finally, a localization result that follows from a good lambdainequality is discussed.

Chapter 3 is concerned with the Cauchy integral operator. Its relationshipwith Menger curvature is explained, and the identities (0.1) and (0.3) are proved.Using this relationship and the localization result obtained at the end of Chapter2, we prove the T 1 theorem for the Cauchy transform. The L2 boundedness ofthe Cauchy transform on Lipschitz graphs is deduced from this T 1 theorem and aFourier type estimate for the curvature of the arc length measure. A new applica-tion of the good lambda type theorem from Chapter 2 yields the L2 boundedness ofthe Cauchy transform with respect to arc length on Ahlfors-David regular curves.At the end of the chapter, a different approach to estimating curvature and theCauchy transform on Lipschitz graphs and Ahlfors-David regular curves which isbased on the β numbers of Peter Jones is described.

In Chapter 4 we introduce the capacity γ+. We show how this can be esti-mated from below by means of a suitable dualization of the weak (1, 1) estimatefor the Cauchy transform (or more precisely, of the boundedness from the spaceM(C) to L1,∞(μ)). At this point, by using the L2(μ) boundedness of the Cauchytransform on Lipschitz graphs, we are ready to prove Denjoy’s conjecture. Recallthat this asserts that any subset of positive length of a rectifiable curve is non-removable. In this chapter we also characterize γ+ in terms of measures μ withlinear growth and finite curvature, or equivalently, in terms of the L2(μ) norm ofthe Cauchy singular integral operator Cμ. An immediate corollary of this charac-terization is the semiadditivity of γ+. Next, by variational arguments, it is provedthat this capacity admits a dual formulation, analogous to the one of many clas-sical capacities originated by positive kernels. This new characterization will playan important role in the proof of the comparability between γ and γ+. Finally,the behavior of γ+ on some Cantor sets and its connection with another capacityfrom non-linear potential theory is discussed.

Chapter 5 is devoted to proving the Tb theorem for non-doubling measuresfrom Nazarov, Treil and Volberg [126] mentioned above , which will be used toshow later the comparability between γ and γ+. This will be also a necessarytool for Vitushkin’s conjecture. The proof of this Tb theorem is one of the mosttechnical and delicate parts of the book. Nevertheless, some of the techniquesinvolved are very powerful and original, like the “suppressed operators” or therandom dyadic lattices, and they have been applied to other different situations(for instance, to the solution of the A2 conjecture by Hytonen [68]). So surely thereader will benefit a lot from a careful reading of the chapter.

The main result of Chapter 6 is the comparability between γ and γ+, andas a corollary one gets the semiadditivity of analytic capacity. The argument forcomparability involves an induction on scales technique and a careful applicationof the Tb theorem proved in the previous chapter. At the end of the chapter wesee how the comparability can be applied to deduce an estimate of the Cauchy

Introduction 7

integral on Ahlfors-David regular Jordan curves which generalizes an old result ofMelnikov valid for analytic curves [111].

In Chapter 7 we prove that if a set has finite length and its arc lengthmeasure has finite curvature, then it is rectifiable. This is the so-called “David-Leger curvature theorem”. This result was first obtained by David. However, hisarguments have remained unpublished and instead a different proof by Leger [83]appeared later. The book follows the arguments of Leger, although some steps ofthe proof have been modified substantially. The David-Leger theorem, togetherwith the comparability of γ+ and γ, yields the proof of Vitushkin’s conjecture.Let us remark that the original proof of Vitushkin’s conjecture by Guy David wassomewhat different. Moreover, for the sake of exposition, we have reversed thehistorical order of the two main ingredients of the proof. Indeed, the first stepof David’s proof consisted of the curvature theorem, while the last one was anappropriate Tb-like theorem (which relied on a previous joint result with Mattila[27]). In our book, the Tb part of the proof of Vitushkin’s conjecture is subsumedin the comparability γ ≈ γ+, which in turn uses the Tb theorem of Nazarov, Treiland Volberg explained in Chapter 5 instead of the original one of David.

Chapter 8 discusses the relationship among existence of principal values forthe Cauchy transform, finiteness of the maximal Cauchy transform, and rectifia-bility. It is shown that if Cμ is bounded in L2(μ) and μ has no point masses, thenthe principal values of Cμf(z) exist μ-a.e. for any f ∈ Lp(μ), with 1 ≤ p < ∞.Also, the sets E ⊂ C with finite length which are rectifiable are characterized interms of the principal values of CH1�E1 and in terms of the corresponding max-imal Cauchy transform. The David-Leger curvature theorem is an essential toolfor many results in this chapter,

The last chapter of the book, Chapter 9, goes back to the subject of Calderon-Zygmund theory in non-homogeneous spaces. It deals with some questions concern-ing the space RBMO(μ) and the Hardy space H1,∞

atb (μ), which are some variants

of the classical spaces BMO(μ) and the atomic Hardy space H1,∞at (μ). In the usual

homogeneous setting, the space BMO(μ) plays an important role in the study ofCalderon-Zygmund operators. For instance, Calderon-Zygmund operators whichare bounded in L2(μ) are also bounded from L∞(μ) into BMO(μ). This turns outto be false when the underlying measure μ is non-doubling. As a consequence, thespace BMO(μ) is not suitable to deal with Calderon-Zygmund operators whenthe underlying measure μ is non-doubling. We will see that RBMO(μ) (whichstands for “regular bounded mean oscillations”) is an appropriate substitute. Inparticular, Calderon-Zygmund operators are bounded from L∞(μ) to RBMO(μ).Also, some variant of the classical John-Nirenberg inequality holds for RBMO(μ).Further, its predual is a Hardy type space, denoted by H1,∞

atb (μ). By interpolating

between the pairs (L∞(μ),RBMO(μ)) and (H1,∞atb (μ), L1(μ)) we will obtain a new

proof of the T 1 theorem for the Cauchy integral operator. Moreover, we will showthat the T 1 theorem admits a formulation analogous to the classical one in termsof the condition of Cμ1 ∈ RBMO(μ). The same holds for any singular integral

8 Introduction

operator with a Calderon-Zygmund kernel.

There are some parts of the book that can be read independently of theothers. For example, a reader who is only interested in Calderon-Zygmund theoryin non-homogeneous spaces might read only Chapter 2, Chapter 3 (if he is inter-ested on the Cauchy transform), and Chapter 9. On the other hand, to understandthe semiadditivity of analytic capacity Chapters 1-6 suffice, and for Vitushkin’sconjecture also Chapter 7. For most of the results that deal with the relationshipbetween the Cauchy transform and rectifiability it is enough to read Chapters 2,3, 7 and 8.

A sufficient prerequisite to read this book is a basic knowledge of measuretheory, integration, Lp spaces, and complex analysis. Knowing the classical homo-geneous Calderon-Zygmund theory is not necessary to understand the content ofthe book, although the reader might find it interesting to compare the classical“homogeneous” techniques with the ones explained in this book. No knowledge ofgeometric measure theory (Hausdorff measures and rectifiability) is needed either.In fact, no profound result from geometric measure theory is used in the book.Most of the required results from this area are proved in the text, and otherwisean appropriate reference is provided.

There are some topics on analytic capacity that are not covered in the text.For instance, the theory about uniform rational approximation developed by Vi-tushkin and his school is not studied (see Gamelin [47], for example), and the usefulcontinuous analytic capacity is not discussed either. The book does not discussthe approach to analytic capacity by duality using the Garabedian function ei-ther. This approach is studied in detail in the books by Garnett [54] and Dudziak[35]. Indeed, in the latter reference, this is a crucial ingredient for the proof ofVitushkin’s conjecture. In our book we have preferred to follow a real variableapproach, which relies more on the use of non-homogeneous Calderon-Zygmundtheory and is more appropriate to study the capacity γ+. Another result that isout the scope of this text is the proof (from Tolsa [163]) of the fact that bilipschitzmaps keep analytic capacity invariant modulo multiplicative estimates. That is tosay, if f : C → C is bilipschitz, then γ(E) ≈ γ(f(E)), where the constant involvedin the comparability only depends on the bilipschitz constant of f .

I have tried to give appropriate references of all the results proved in thetext. Moreover, at the end of each chapter there is a section where additionalrelated results and references are mentioned. However, there is no attempt atcompleteness. The selection of references only reflects my personal interests and/orknowledge. I apologize in advance for any possible (undesired) inaccuracy or lackof recognition.

Acknowledgements. I would like to thank the following people for reading someparts of this book, finding misprints, and making useful suggestions: Vasilis Chou-sionis, John Garnett, Daniel Girela Sarrion, Joan Mateu, Pertti Mattila, Mark

Introduction 9

Melnikov, Joan Orobitg, Mari Carmen Reguera, and Joan Verdera. Of course, Iam solely responsible for any error or lack of recognition.

While I was writing this book I was partially supported by the grantsMTM2010-16232 (Spain) and 2009SGR-000420 (Catalonia).

Basic notation

Most of the notation used in this book is standard. Otherwise, it will be introducedwhen necessary. However, for the reader’s convenience, in this section we recordsome basic definitions and useful notation that we will use.

Given A ⊂ Rd, the characteristic function of A is defined by χA(x) = 1 ifx ∈ A and 0 otherwise. The complementary of A is denoted either by Rd \ A orAc. The symmetric difference of two sets A,B equals AΔB = (A \ B) ∪ (B \ A).The closure of A is denoted by A, its interior by

◦A, and its boundary by ∂A. The

distance from x ∈ Rd to A is denoted by dist(x,A), and the diameter of A bydiam(A). Also, for a given δ > 0, Uδ(A) stands for the δ-neighborhood of A. Thatis, Uδ(A) = {x ∈ Rd : dist(x,A) < δ}.

An open ball centered at x with radius r is denoted by B(x, r), and a closedball by B(x, r). The radius of a ball B sometimes will be written as r(B). By acube Q ⊂ Rd (or a square, in the planar case) we mean a cube (or square) withsides parallel to the axes. We denote by �(Q) its side length. The notation Q(x, r)stands for an open cube with center x and side length 2r. The analogous notationQ(x, r) is used for closed cubes. In general, cubes are assumed to be closed unlessthey are dyadic, or written in the form Q(x, r), or stated otherwise.

We recall now the definition of dyadic cubes in Rd. For m ∈ Z, Dm is thefamily of cubes of the form{

x ∈ Rd : ki 2−m ≤ xi < (ki + 1) 2−m for 1 ≤ i ≤ d

},

where ki are arbitrary integers and x = (x1, . . . , xd). Cubes of this form are calleddyadic. The family of all dyadic cubes (the so-called dyadic lattice) is written asD =⋃

m∈ZDm. Observe that, for each m ∈ Z, the cubes from Dm form a partition

of Rd. Given Q ∈ Dm, there are 2d cubes from Dm+1 which are contained in Q.They are the so-called children or sons of Q. Also, given j ≥ 0, Dj(Q) denotes thefamily of the dyadic cubes contained in Q with side length 2−j�(Q).

The line that passes through two points x, y ∈ Rd is denoted by Lx,y. Giventwo lines L1, L2, the notation �L1, L2 stands for the angle they form (the smallestone, say). Also, for x, y, z ∈ Rd, x, y, z is the angle with vertex y and sides y, xand y, z.

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12 Basic notation

As usual in the field of harmonic analysis, the letters c and C stand forpositive constants (quite often absolute constants) which may change their valuesat different occurrences. On the other hand, constants with subscripts, such asc1, retain their value at different occurrences in a same chapter (and they changeamong different chapters). The notation A � B means that there is a positiveconstant c such that A ≤ cB. Also, A ≈ B is equivalent to A � B � A.

For a function f : C → C, ∂f and ∂f stand for the complex derivatives

∂f(z) =1

2

(∂x f(z)− i ∂y f(z)

),

∂f(z) =1

2

(∂x f(z) + i ∂y f(z)

).

All measures in this book are assumed to be Borel measures. Recall that aRadon measure μ in a metric space X is a Borel measure such that

(i) μ(K) < ∞ for all compact sets K ⊂ X ,

(ii) μ(V ) = sup{μ(K) : K ⊂ V is compact} for all open sets V ⊂ X ,

(iii) μ(A) = inf{μ(V ) : A ⊂ V, V is open} for all A ⊂ X .

If A is μ-measurable and μ is Radon, then for all ε > 0 there exists a closed setC ⊂ A such that μ(A \ C) < ε.

It turns out that, in Rd, any Borel measure which is locally finite is a Radonmeasure.

The closed support (or just, support) of a measure μ is denoted by supp(μ).The restriction of μ to a set A ⊂ Rd is written as μ�A. That is, for B ⊂ Rd, onesets μ�A(B) = μ(A ∩B). The image measure (or push forward) of μ on X undera mapping f : X → Y is a measure on Y defined by f#μ(A) = μ(f−1(A)).

Real and complex measures are also assumed to be Borel in the book. Thevariation of a real or complex measure ν is denoted by |ν|. Its total variation is‖ν‖ = |ν|(Rd). The vector space of all Borel real (or complex if we are in thecomplex plane) finite measures is denoted by M(Rd). This is a Banach space withthe total variation norm. On the other hand, M+(R

d) stands for the subset ofpositive measures from M(Rd).

The Lebesgue measure in Rd is written as Ld, or just by dx (or dy or dt. . . )inside some integral if the meaning is clear from the context. On the other hand,for a rectifiable curve Γ ⊂ C with a parameterization γ : [a, b] → Γ, dz (or dzΓ, ordw. . . ) stands for the usual complex measure on Γ given by the image measure ofγ′(t) dt, with the usual orientation when Γ is closed.

For 1 ≤ p ≤ ∞, we denote by Lp(μ) the Banach spaces of the μ-measurablefunctions f : Rd → R (or C) such that the following norm is finite:

‖f‖Lp(μ) =

(∫|f |p dμ

)1/p.

Basic notation 13

For p = ∞, L∞(μ) is also a Banach space with the norm ‖ · ‖L∞(μ) equal tothe μ-essential supremum. As usual, p′ will denote the conjugate index of p, i.e.p′ = p/(p− 1), so that Lp′

(μ) is the dual of Lp′(μ) for 1 ≤ p < ∞.

We also consider the weak Lebesgue spaces Lp,∞(μ), for 1 ≤ p < ∞. Recallthat f ∈ Lp,∞(μ) if

‖f‖Lp,∞(μ) = supλ>0

λμ({x ∈ Rd : |f(x)| > λ})1/p

is finite. When μ coincides with the Lebesgue measure on Rd, for short we write‖ · ‖p = ‖ · ‖Lp(μ) and ‖ · ‖p,∞ = ‖ · ‖Lp,∞(μ)

The identity operator in a vector space such as Lp(μ) is denoted by Id.

Chapter 1

Analytic capacity

1.1 Introduction

In this chapter we will introduce the notion of analytic capacity and we will studysome of its basic properties, its connection with the Painleve problem, and itsrelationship with Hausdorff measures. We will also review some results on theCauchy transform and Vitushkin’s localization operator that are useful for thestudy of analytic capacity.

Most of the results in this chapter are rather classical and have alreadyappeared in several books or monographs, such as Gamelin [47], Garnett [54],Pajot [132], or Verdera [174], for example.

1.2 Definition and basic properties of analytic capacity

The analytic capacity of a compact set E ⊂ C is

γ(E) := sup |f ′(∞)|, (1.1)

where the supremum is taken over all analytic functions f : C\E−→C with |f | ≤ 1in C \ E, and f ′(∞) = limz→∞ z(f(z)− f(∞)).

One says that a function f : C \ E−→C is admissible for E if it is analyticand bounded by 1 in modulus in C \ E. So the supremum that defines γ(E) istaken over all admissible functions for E.

Let us consider a couple of easy examples: the capacity of a point is zero,because the only functions that are analytic and bounded in the complementaryare constants. On the other hand the capacity of a ball B(0, r) is positive, as canbe seen by taking the function f(z) = r/z.

Consider the Laurent expansion of f near ∞:

f(z) = a0 +a1z

+a2z2

+ · · · .

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16 Chapter 1. Analytic capacity

Clearly, f ′(∞) = a1, and also

f ′(∞) =1

2πi

∫Γ

f(z) dz, (1.2)

where Γ is any rectifiable curve surrounding E with a suitable orientation. Letus remark that, in general, f ′(∞) �= limz→∞ f ′(z). Instead, f ′(∞) = g′(0) forg(z) = f(1z ).

If A ⊂ C is an arbitrary set, then we define

γ(A) = supE⊂A,E compact

γ(E).

The outer boundary of a compact set E is the boundary of the unboundedcomponent of C \ E. It is denoted by ∂oE. Obviously, ∂oE ⊂ ∂E.

Proposition 1.1. The following properties hold:

(a) If E ⊂ F , then γ(E) ≤ γ(F ).

(b) For all z, λ ∈ C,γ(z + λE) = |λ|γ(E).

(c) For every compact set E ⊂ C,

γ(E) = γ(∂oE).

Proof. The statements (a) and (c) are straightforward consequences of the defini-tion, while (b) follows by an easy change of variables. �Proposition 1.2. Let E ⊂ C be compact. The supremum in (1.1) is attained, andany admissible function f which attains it satisfies f(∞) = 0 if γ(E) > 0.

As a consequence, of this proposition, one obtains an equivalent definitionfor γ(E) if in the supremum (1.1) one asks f(∞) = 0 too.

Proof. To see that the supremum is attained one just has to notice that the familyof admissible functions for E is a normal family, and so one can consider a sequenceof admissible functions {fk}k such that f ′

k(∞) → γ(E), and take f to be the limitof a convergent subsequence.

To check that f(∞) = 0 if f is admissible and attains the supremum whenγ(E) > 0, consider the function

g(z) =f(z)− f(∞)

1− f(∞) f(z).

It is clear |g(z)| ≤ 1 for all z ∈ C \ E, g(∞) = 0, and moreover,

g′(∞) = limz→∞

z(f(z)− f(∞)

)1− f(∞) f(z)

=f ′(∞)

1− |f(∞)|2 ,

and thus, |g′(∞)| > |f ′(∞)| if γ(E) = f ′(∞) �= 0. �

1.2. Definition and basic properties of analytic capacity 17

Proposition 1.3. Let E ⊂ C be a compact connected set different from a singlepoint, and let f be a conformal map of the unbounded connected component ofC∞ \ E to the unit disk satisfying f(∞) = 0. Then γ(E) = |f ′(∞)|.Proof. Since f is admissible, it is obvious that γ(E) ≥ |f ′(∞)|. On the other hand,if g is also admissible and g(∞) = 0, then g ◦ f−1 : B(0, 1) → B(0, 1) is analyticand fixes the origin. Thus, by the Schwarz lemma, |g ◦ f−1(z)| ≤ |z| for all z inthe unit disk. Consequently, |g(z)| ≤ |f(z)| for all z ∈ C \ E , which implies that|g′(∞)| ≤ |f ′(∞)|. �

Proposition 1.4. The analytic capacity of a disk is its radius. The analytic capacityof a segment of length � equals �/4.

Proof. The first statement follows by taking into account that, for a disk B(z0, r),the function

f(z) =r

z − z0

maps conformally C∞ \ B(z0, r) to the unit disk and f ′(∞) = r.

For the second one, notice that the function

f(z) =

(z +

1

z

)�

4

maps conformally the unit disk to C∞ \ [−�/2, �/2] and satisfies f(0) = ∞. There-fore,

γ([−�/2, �/2]

)= lim

z→∞ |z f−1(z)| = limz→0

|f(z) z| = �

4. (1.3)

�

Recall that the 1/4 Koebe theorem asserts that if f : B(0, 1) → C is analytic,univalent, and f(0) = 0, f ′(0) = 1, then B(0, 1/4) ⊂ f

(B(0, 1)

). In the next

proposition we show an important application of this result to analytic capacity.

Proposition 1.5. If E ⊂ C is compact and connected, then

diam(E)/4 ≤ γ(E) ≤ diam(E).

Proof. The second inequality follows from the preceding proposition and the factthat E is contained in a closed disk with radius diam(E).

For the first one, let U ⊂ C∞ be the unbounded component of C∞ \ E andconsider the conformal mapping f : U → B(0, 1), with f(∞) = 0. Take z1, z2 ∈ Esuch that |z1 − z2| = diam(E) and consider the function

g(z) =γ(E)

f−1(z)− z1.


This is a univalent map in the unit disk because f−1 is so. Further, g(0) = 0 andarguing as in (1.3), |z f−1(z)| → γ(E) as z → 0, and thus

|g′(0)| = limz→0

∣∣∣∣ γ(E)

z(f−1(z)− z1

) ∣∣∣∣ = 1.

As z2 is not in the range of f−1, γ(E)/(z2 − z1) is not either in the range of g.Therefore, by Koebe’s 1/4 theorem,

γ(E)

|z2 − z1| ≥1

4,

and the proposition follows. �Corollary 1.6. If γ(E) = 0, then E is totally disconnected.

As shown above, for a continuum E ⊂ C, the analytic capacity can becomputed easily if one knows the conformal mapping between the unboundedcomponent of C∞ \ E and the unit disk, and moreover one has the estimatediam(E)/4 ≤ γ(E) ≤ diam(E). For disconnected sets the situation is much moredifficult. Obtaining precise identities in simple cases is already difficult and esti-mates analogous to the preceding one are missing. See Murai [120], for example,for some results involving the analytic capacity of two parallel segments of equallength.

Proposition 1.7 (Outer regularity of γ). Let {En}n≥0 be a sequence of compactsets in C such that En+1 ⊂ En for each n. Then

γ

( ⋂n≥0

En

)= lim

n→∞ γ(En).

Proof. Let us write E =⋂

n≥0 En. Since {γ(En)}n is a non-increasing sequence,it is clear that the limit on the right-hand side above exists. Also, taking intoaccount that E ⊂ En for all n, we get γ(E) ≤ limn→∞ γ(En).

For the converse inequality, for each n, take a function fn admissible for En

such that f ′n(∞) = γ(En). Since {fn}n is a normal family on C \ E, there exists

some subsequence {fnk}k which is uniformly convergent on compact subsets of

C \ E to some function f . It turns out that f is admissible for E and moreover,using (1.2),

f ′(∞) = limk→∞

f ′nk(∞) = lim

k→∞γ(Enk

) = limn→∞ γ(En).

Therefore, γ(E) ≥ |f ′(∞)| = limn→∞ γ(En). �Corollary 1.8. If E ⊂ C is compact, then

γ(E) = inf{γ(U) : U open, U ⊃ E}.

1.2. Definition and basic properties of analytic capacity 19

Proposition 1.9. For every compact set E, there is an admissible function f for Esuch that f ′(∞) = γ(E). Such a function is unique in the unbounded componentof C∞ \ E if γ(E) > 0.

The function f in the proposition is called the Ahlfors function of E.

Proof. The existence of f has already been shown in Proposition 1.2. So let usturn to the uniqueness. Suppose that there are two admissible functions f1, f2,

such that f ′1(∞) = f ′

2(∞) = γ(E) and f1(∞) = f2(∞) = 0. Then f =f1 + f2

2is

also admissible and f ′(∞) = γ(E), f(∞) = 0. Let g =f2 − f1

2, so that f1 = f − g

and f2 = f + g. From the inequalities

|f ± g|2 = |f |2 + |g|2 ± 2 Re(fg) ≤ 1,

we infer that |f |2 + |g|2 ≤ 1. Using also that 1 + |f | ≤ 2, we deduce

|g|22

≤ 1− |f |22

=(1 − |f |) (1 + |f |)

2≤ 1− |f |.

Thus,

|f |+ |g|22

≤ 1.

If f1 �= f2 near ∞, then g �= 0 in the unbounded component of C \E, and we may

consider the Laurent series ofg2

2for z near ∞:

g(z)2

2=

anzn

+an+1

zn+1+ · · · , an �= 0.

Since g(∞) = 0, we have n ≥ 2.For ε > 0, take the function

f(z) = f(z) + ε an zn−1 g(z)2

2.

If ε is small enough so that |ε an zn−1| ≤ 1 in some bounded neighborhood of E,then we deduce that

|f(z)| ≤ |f(z)|+∣∣∣ε an zn−1 g(z)2

2

∣∣∣ ≤ |f(z)|+ |g(z)|22

≤ 1

in that neighborhood, and thus in the whole unbounded component of C \ E bythe maximum principle. On the other hand,

f ′(∞) = f ′(∞) + ε |an|2 > γ(E),

which is a contradiction. �Let us remark that a slight modification in the arguments for Proposition 1.7

shows that the Ahlfors functions of En, n ≥ 0, converge on compact subsets tothe Ahlfors function of

⋂n≥0 En.


1.3 Removable sets and the Painleve problem

A compact set E ⊂ C is said to be removable for bounded analytic functions(or just, removable) if for every open set Ω containing E, each bounded functionanalytic on Ω \ E has an analytic extension to Ω. For example, by a well-knowntheorem of Riemann, if E is a finite collection of points, then it is removable. Also,using Baire’s category theorem, one can extend this result to the case where Eis countable and compact. On the other hand, a disk B(z0, r) is not removable:just consider the function f(z) = 1/(z − z0), which is analytic and bounded inC \ B(z0, r) and cannot be extended analytically to the whole B(z0, r).

Painleve’s problem consists in characterizing removable sets for bounded an-alytic functions in a metric/geometric way. Because of the next result, essentiallydue to Ahlfors [2], this turns out to be equivalent to describing compact sets withzero analytic capacity in metric/geometric terms.

Proposition 1.10. Let E ⊂ C be compact. The following are equivalent:

(i) E is removable for bounded analytic functions.

(ii) There exists an open set Ω ⊃ E such that every bounded function analytic inΩ \ E has an analytic extension to Ω.

(iii) Every function analytic and bounded in C \ E is constant.

(iv) γ(E) = 0.

Proof. That (i) ⇒ (ii) is trivial. The implication (ii) ⇒ (iii) is also immediate:given f : C \E → C analytic and bounded, considering its restriction to Ω \E, itturns out that f can be extended to an entire bounded function, and thus it mustbe constant by Liouville’s theorem.

(iii) ⇒ (iv) is a direct consequence of the definition of γ. To prove the converseimplication, let E ⊂ C be compact with γ(E) = 0. Suppose that there exists anon-constant bounded analytic function f : C \ E → C, so that f(∞) �= f(z0) forsome z0 ∈ C \ E. Consider the function

g(z) =f(z)− f(z0)

z − z0for z �= z0,

and g(z) = f ′(z0). It is easy to check that g is bounded in C \ E, g(∞) = 0 andg′(∞) = f(∞)− f(z0) �= 0, and thus γ(E) > 0, which is a contradiction.

To show that (iii) ⇒ (ii), just take Ω = C. Finally, let us see that (iii) ⇒(i). To this end, consider an arbitrary open set Ω ⊃ E and take f : Ω \ E → C

analytic and bounded. We have to show that f can be extended analytically tothe whole of Ω. We may assume that Ω is connected (otherwise, we consider eachcomponent separately). Then Ω\E is connected because E is totally disconnected(by Corollary 1.6, since (iii) implies that γ(E) = 0). Take smooth curves Γ1,Γ2 ⊂ Ωsurrounding E, with Γ2 very close to E, and consider a point z inside Γ1 and

1.4. Two technical lemmas 21

outside Γ2. Then we have

f(z) =1

2πi

∫Γ1

f(w)

w − zdw − 1

2πi

∫Γ2

f(w)

w − zdw = f1(z) + f2(z).

It is easy to check that f1(z) and f2(z) do not depend on the precise curves Γ1,Γ2,as long as z is inside Γ1 and outside Γ2. Then f1 is analytic in Ω and f2 in C \E.Moreover, since f1 is bounded near ∂E, it turns out that f2 is also bounded near∂E, and so in the whole C \ E, by the maximum principle. Since f2 vanishes at∞, (iii) implies that f2 = 0, and thus f = f1 is analytic in Ω. �

Let us insist on the fact that saying that a compact set is removable is thesame as saying that it has zero analytic capacity. In a sense, the reader should thinkthat analytic capacity measures the size of a set as a non-removable singularityfor bounded analytic functions.

A stronger version of the implication (iv) ⇒ (i) will be proved in Proposition1.18 below. Moreover, that proof avoids the technical problem of the constructionof the curves Γ1 and Γ2 in the preceding proposition.

1.4 Two technical lemmas

Lemma 1.11. Let E ⊂ C be compact. Suppose that E ⊂ B(z0, δ). Let f be admis-sible for E and consider the Laurent expansion of f centered at z0:

f(z) =

∞∑n=1

an(z − z0)n

, |z − z0| > δ.

We have |an| ≤ nδn−1γ(E).

Proof. Suppose for simplicity that z0 = 0. The function

g(z) = zn−1

(f(z)−

n−1∑j=1

ajzj

)

is analytic outside E and satisfies g(∞) = 0. Moreover, since |aj| ≤ δj , we inferthat for z ∈ B(0, δ),

|g(z)| ≤ |zn−1 f(z)|+n−1∑j=1

δj |zn−1−j| ≤ nδn−1.

By the maximum principle, ‖g‖∞ ≤ nδn−1. Hence, g/nδn−1 is admissible for Eand thus

|an| = |g′(∞)| ≤ nδn−1γ(E). �


From the preceding lemma we deduce the following property of γ, which willbe very useful for the proof of the semiadditivity of analytic capacity in Chapter6.

Lemma 1.12. Let E ⊂ C be compact. Suppose that f is admissible for E andsatisfies f(∞) = 0. Then, for all z ∈ C such that dist(z, E) ≥ 3

2 diam(E), we have

|f(z)| ≤ cγ(E)

dist(z, E).

If moreover f ′(∞) = 0, then

|f(z)| ≤ cdiam(E) γ(E)

dist(z, E)2.

Proof. Set δ = diam(E). Take z0 ∈ E. Then E ⊂ B(z0, δ). We consider theLaurent expansion of f centered at z0:

f(z) =∞∑

n=1

an(z − z0)n

, |z − z0| > δ.

Notice that the condition dist(z, E) ≥ 32 δ implies that |z − z0| ≥ 3

2 δ. From thepreceding lemma we deduce that

|f(z)| ≤∞∑n=1

n δn−1γ(E)

|z − z0|n ≤ c γ(E)

|z − z0| .

If moreover f ′(∞) = 0, then

|f(z)| ≤∞∑n=2

n δn−1γ(E)

|z − z0|n ≤ c δ γ(E)

|z − z0|2 . �

1.5 The Cauchy transform and Vitushkin’s localizationoperator

The Cauchy transform of a (possibly complex) finite measure ν on C is defined by

Cν(z) =∫

1

ξ − zdν(ξ). (1.4)

The integral is absolutely convergent for a.e. z ∈ C, with respect to Lebesguemeasure. This follows easily from Fubini’s theorem, taking into account that∫K

1|z| dL2(z) < ∞ on compact sets K.

The Cauchy transform appears naturally in complex analysis. For instance,by Cauchy’s integral formula, if f is a function analytic in a simply connected

1.5. The Cauchy transform and Vitushkin’s localization operator 23

open set Ω ⊂ C, Γ ⊂ Ω is a closed rectifiable Jordan curve, and z is a point whichbelongs to the bounded component of C \ Γ (and so z ∈ Ω), then f(z) = Cν(z),where ν is the complex measure

ν =1

2πif(z) dzΓ.

Proposition 1.13. If ν is a complex measure, then Cν is locally integrable in C

(with respect to Lebesgue measure). Moreover, the integral that defines the Cauchytransform in (1.4) is absolutely convergent for a.e. z ∈ C, with respect to Lebesguemeasure. Further, Cν is analytic in C\suppν, Cν(∞) = 0 and (Cν)′(∞) = −ν(C).

The easy proof is left for the reader.A straightforward consequence of the proposition is that, if a set E supports

a measure ν such that ν(C) �= 0 and Cν is bounded in C \ E, then γ(E) �= 0and so E is not removable. One should view the Cauchy transform as a tool forconstructing analytic functions. Usually, the difficulties arise when one tries tocheck that the constructed functions are bounded in modulus.

In the remainder of this section we assume that the reader is familiar with thevery basics of the theory of distributions. By using distributions, the properties ofthe Cauchy transform and the localization operator of Vitushkin that we will seebelow become much more natural. In fact, it is possible to talk about Vitushkin’slocalization operator without appealing to distributions, like in Gamelin [47], butthen the results look less transparent. The current section is the only place in thebook where distributions are used. The reader will find all the necessary results,and much more, in Rudin [141, Chapter 6], for example.

The definition of Cauchy transform also makes sense if ν is a compactlysupported (complex) distribution. In this case we set

Cν = −1

z∗ ν.

A key fact that explains why the Cauchy transform is so important in complexanalysis is the following:

Theorem 1.14. The kernel1

πzis the fundamental solution of the ∂ operator. That

is,

∂1

πz= δ0,

where δ0 is the Dirac delta at the origin. As a consequence, if ν is a compactlysupported distribution on C,

∂(Cν) = −π ν. (1.5)

Also, if f ∈ L1loc(C) (or, more generally, f ∈ D′) is analytic in a neighborhood of

∞ and f(∞) = 0, thenC(∂f) = −π f.


All the identities in the preceding theorem must be understood in the senseof distributions. For the proof, see Conway [19, p. 195], for instance. Notice that,as a consequence of the last statement in the theorem, if a distribution ν satisfies∂ν = 0 (and, in particular, it is analytic in a neighborhood of ∞) and vanishes at∞, then ν = 0. Further, any function f ∈ L1

loc(C) or distribution f ∈ D′ which isanalytic in a neighborhood of ∞ and vanishes at ∞ is the Cauchy transform of aunique compactly supported distribution, namely 1

π ∂f .

Proposition 1.15. If ν is a compactly supported distribution, then Cν is analyticin C \ supp(ν), and moreover,

Cν(∞) = 0 and (Cν)′(∞) = −〈ν, 1〉 = −ν(C).

Here, 〈·, ·〉 stands for the pairing between distributions of compact supportand the corresponding test functions (i.e. C∞ functions).

Proof. By the preceding theorem, ∂(Cν) = −π ν, and so Cν is analytic out ofsupp(ν). To show that Cν(∞) = 0, take r > 0 such that supp(ν) ⊂ B(0, r), andlet ϕ : C → R be a C∞ radial function such that 0 ≤ ϕ ≤ 1, which vanishes on

B(0, r/2), and equals 1 on C \B(0, r). Write kr(z) = ϕ(z)1

z, for z ∈ C. It is easy

to check that

ν ∗ 1

z= ν ∗ kr in C \ B(0, 2r),

in the sense of distributions. Moreover, since kr(z) is a C∞ radial function,

ν ∗ kr(z) = 〈ν, τzkr〉,

where τzkr(w) = kr(w − z). It is easy to check that τzkr → 0 as z → ∞ in C∞

with the topology of test functions, and so

〈ν, τzkr〉 → 0 as z → ∞.

That is, Cν(∞) = −ν ∗ 1

z(∞) = −ν ∗ kr(∞) = 0.

To prove that (Cν)′(∞) = 〈ν, −1〉, consider a radial C∞ approximation ofthe identity {ψε}ε>0, so that suppψε ⊂ B(0, ε) and

∫ψε dL2 = 1 for all ε > 0.

Then we have

ψε ∗ Cν = ψε ∗(−1

z∗ ν)= −1

z∗ (ψε ∗ ν

)= C(ψε ∗ ν).

Thus we deduce that C(ψε ∗ ν) is analytic in C \ Uε(supp(ν)), converges locallyuniformly to Cν on compact subsets of C \ Uε(supp(ν)), and vanishes at ∞. Fromequation (1.2) one infers easily that

limε→0

(C(ψε ∗ ν))′(∞) = (Cν)′(∞).


On the other hand, by Proposition 1.13, since ψε ∗ ν is a C∞ function (and thusa measure),

(C(ψε ∗ ν))′(∞) = −

∫1 d(ψε ∗ ν) = −〈(ψε ∗ ν), 1〉 = −〈ν, 1 ∗ ψε〉 = −〈ν, 1〉,

and thus the last claim in the proposition follows. �Given f ∈ L1

loc(C) and ϕ ∈ C∞ compactly supported, we define

Vϕf := ϕf +1

πC(f ∂ϕ). (1.6)

For a fixed function ϕ, one calls Vϕ Vitushkin’s localization operator (associatedwith ϕ). The same definition (1.6) makes sense if f is a distribution from D′.

Proposition 1.16. Let f ∈ L1loc(C) (or more generally, f ∈ D′) and ϕ ∈ C∞ be

compactly supported. Then we have

Vϕf =−1

πC(ϕ ∂f),

in the sense of distributions.

In the identity above, ∂f should be understood in the sense of distributions.

Proof. By (1.6) and (1.5),

∂(Vϕf) = f ∂ϕ+ ϕ ∂f +1

π∂ C(f ∂ϕ) = ϕ ∂f = ∂

(−1

πC(ϕ ∂f)

).

Moreover, since ϕf , f ∂ϕ and ϕ ∂f are all compactly supported, it follows thatboth Vϕf and −1

π C(ϕ ∂f) are analytic in a neighborhood of∞ and vanish at∞. Bythe remark after Theorem 1.14, it turns out that both distributions are equal. �

Observe that if f = Cν, where ν is a compactly supported complex measureor distribution, then

Vϕ(Cν) = C(ϕν). (1.7)

This fundamental identity justifies why Vϕ is called (Vitushkin’s) localization op-erator: Cν is a analytic in C\supp(ν), and so supp(ν) can be understood as the setof singularities of Cν. By (1.7), it turns out that Vϕ(Cν) is analytic in the largerset C \ supp(ϕν). So the singularities are now localized to supp(ν) ∩ supp(ϕ).

In the next proposition we show that the operator Vϕ enjoys other nice prop-erties, besides the one about localization of singularities.

Proposition 1.17. Let ϕ ∈ C∞ be supported in a ball Br of radius r, with ‖ϕ‖∞ ≤c4 and ‖∇ϕ‖∞ ≤ c4/r. For any function f ∈ L1

loc(C), the following propertieshold:

(i) ‖Vϕf‖∞ ≤ c5‖f χBr‖∞,


(ii) ‖Vϕf‖∞ ≤ c6 ωf (r), where ωf stands for the modulus of continuity of f ,

(iii) Vϕf is holomorphic outside supp(∂f) ∩ supp(ϕ),

(iv) if f is bounded in Br, then Vϕf is continuous where f is.

The constants c5 and c6 depend only on c4.

Proof. By (1.6),

‖Vϕf‖∞ ≤ ‖ϕf‖∞ +1

π‖C(f ∂ϕ)‖∞.

Clearly, ‖ϕf‖∞ ≤ c4 ‖χBr f‖∞. Also, for any z ∈ C,

|C(f ∂ϕ)(z)| ≤∫Br

1

|w − z| |f(w)| |∂ϕ(w)| dL2(w)

≤ ‖fχBr‖∞ ‖∇ϕ‖∞∫Br

1

|w − z| dL2(w) ≤ c ‖fχBr‖∞,

and thus (i) follows.By Proposition 1.16, Vϕ vanishes on constants. Then replacing f by f−f(z0)

in (i), where z0 is the center of Br, we get (ii).The third statement is a direct consequence of the identity Vϕf = −1

π C(ϕ ∂f).The fourth one follows from (1.6), taking into account that the integral∫

f(w)

w − zdL2(w)

depends continuously on z when f is bounded on Br. �Proposition 1.18. Let Ω ⊂ C be open and E ⊂ C compact with γ(E) = 0. Thenevery function analytic and bounded in Ω \ E can be extended analytically to thewhole set Ω.

Notice that it is not assumed that E ⊂ Ω. In particular, it may happen thatE ∩ ∂Ω �= ∅. This is the main difference with Proposition 1.10 above.

Proof. We may assume Ω to be bounded. Consider a grid of squares {Qi}i∈I in C

with side length �(Qi) = � for every i ∈ I. Let {ϕi}i∈I be a partition of unity ofC∞ functions subordinated to the squares {2Qi}i∈I , so that supp(ϕi) ⊂ 2Qi foreach i ∈ I and

∑i∈I ϕi ≡ 1 on C.

Extend f by zero to C \ (Ω \ E). Since f vanishes out of a bounded set,f = −1

π C(∂f) and Vϕf is identically zero except for finitely many indices i ∈ I.So we have

f =−1

π

∑i∈I

C(ϕi ∂f) =∑i∈I

Vϕif.

Notice that, since supp(∂f) ⊂ E ∪ ∂Ω, then for each i ∈ I,

supp(∂Vϕif) ⊂ 2Qi ∩ (E ∪ ∂Ω).


As a consequence, if 2Qi ∩ ∂Ω = ∅, then Vϕif is analytic out of 2Qi ∩ E. SinceVϕif vanishes at ∞, ‖Vϕf‖∞ < ∞ (by (i) in the preceding proposition), and

γ(2Qi ∩E) ≤ γ(E) = 0,

we infer that

Vϕif ≡ 0 if 2Qi ∩ ∂Ω = ∅.

Therefore,

f =∑

i∈I:2Qi∩∂Ω�=∅

Vϕif.

Thus, f is analytic in Ω \ U4(∂Ω). Since � is arbitrarily small, it turns out that fis analytic in the whole of Ω. �

In the next proposition we show how Vitushkin’s localization operator can beused to prove the semiadditivity of analytic capacity in two very particular cases,which will be useful for future purposes in this book.

Proposition 1.19. (a) For a compact set E ⊂ C and a closed disk or squareD ⊂ C, we have

γ(E ∪D) ≤ c (γ(E) + γ(D)).

(b) Given two closed rectangles R,S ⊂ C (we do not assume their sides to beparallel to the axes),

γ(R ∪ S) ≤ c (γ(R) + γ(S)).

Proof. Let f be the Ahlfors function of E ∪ D, and extend it by zero on E ∪ D.Let ϕ ∈ C∞ be supported on 2D such that ϕ = 1 in a neighborhood of D, with‖∇ϕ‖∞ ≤ c r−1, where r is the radius (or side length) of D. Now set f1 = Vϕfand f2 = f − f1. By Proposition 1.17, f1 is bounded by some constant c′, and sothe same holds for f2 (for another constant c′′). Therefore, since f1 is holomorphicoff 2D and f2 off E,

γ(E ∪D) ≤ (|f ′1(∞)|+ |f ′

2(∞)|) ≤ c(γ(2D) + γ(E)

).

Then (a) follows from the fact that γ(2D) = 2 γ(D).

Consider now the rectangles R and S. Let LR, �R be the side lengths of R,and LS , �S the side lengths of S. Suppose �R ≤ LR and �S ≤ LS . Assume firstthat LR = M�R and LS = N�S with M,N ∈ Z. So we can write R =

⋃Mi=1 Pi and

S =⋃N

j=1 Qj , where Pi and Qj are squares of side lengths �R and �S respectively.Take positive functions ϕi ∈ C∞, 1 ≤ i ≤ M , with supp(ϕi) ⊂ 2Pi for each i,

satisfying∑M

i=1 ϕi ≤ 1 on C,∑M

i=1 ϕi = 1 on R, and ‖∇ϕi‖∞ ≤ c�−1R . Let ψj ,


1 ≤ j ≤ N , be the analogous functions for S. Suppose that �R ≥ �S . Settingψj = ψj

(1−∑M

i=1 ϕi

), we have

M∑i=1

ϕi +

N∑j=1

ψj = 1

on R ∪ S and so, if f is the Ahlfors function of R ∪ S extended by zero to R ∪ S,we have

f =M∑i=1

Vϕif +N∑j=1

Vψjf.

Moreover, for some c > 0, cVϕif is admissible for 2Pi, and also cVψjf for 2Qj

(because it is easy to check that ‖ψj‖∞ ≤ 1 and ‖∇ψj‖∞ ≤ c �−1S ). Thus,

|f ′(∞)| ≤M∑i=1

γ(2Pi) +

N∑j=1

γ(2Qj)

≤ c (M�R +N�S) ≤ c(γ(R) + γ(S)

),

where we used that for any connected set (in particular, any square), its analyticcapacity is comparable to its diameter.

Suppose now that R and S are arbitrary rectangles. Let M be the leastinteger such that M ≥ LR/�R, and N the least integer such that N ≥ LS/�S. LetR0 be a rectangle containing R with side lengths �R and M�R, and S0 anotherrectangle containing S with side lengths �S and N�S. Then

γ(R ∪ S) ≤ γ(R0 ∪ S0) ≤ c(γ(R0) + γ(S0)

) ≈ γ(R) + γ(S),

where in the last inequality we took into account that γ(R) ≈ diam(R) ≈ diam(R0)≈ γ(R0), and analogously for S and S0. �

1.6 Relationship with Hausdorff measures

We start by recalling the definition of Hausdorff measure and Hausdorff dimension(in Rd). Given s ≥ 0 and 0 < ε ≤ ∞, for A ⊂ Rd, we write

Hsε(A) = inf

{∑i

diam(Ai)s : A ⊂

⋃i

Ai, diam(Ai) ≤ ε}. (1.8)

The s-dimensional Hausdorff measure of A is:

Hs(A) = supε>0

Hsε(A) = lim

ε→0Hs

ε(A).

It is well known that Hs is a Borel regular measure (see Mattila [100, Chapter 4],for instance), although it is not a Radon measure because it is not locally finite,

1.6. Relationship with Hausdorff measures 29

except in the case s = d. It is also immediate to check that H0 coincides with thecounting measure. On the other hand, for s = d one has

Hd = 2d α(d)−1Ld,

where α(d) stands for the Lebesgue measure of the d-dimensional unit ball. In the1-dimensional case, which is the most relevant for the study of analytic capacity,the measure H1, is also called length. It coincides with other classical definitionsof length in sets such as rectifiable curves.

When ε = ∞ (and so there is no restriction on the diameters of the setsAi in (1.8)), the quantity Hs∞(A) is called s-dimensional Hausdorff content of A.Although this is not a Borel measure, it is subadditive and it is straightforwardto check that

Hs(A) = 0 ⇔ Hs∞(A) = 0.

Moreover, the Hausdorff content has some advantages over the Hausdorff measure.For instance, it is always finite on bounded sets and quite usually it is easier toestimate than the Hausdorff measure. On the other hand, we will see below thatanalytic capacity is very connected to the 1-dimensional Hausdorff content.

Next, we wish to introduce the notion of Hausdorff dimension. To this end,we need first the following auxiliary result.

Lemma 1.20. For 0 ≤ s < t and A ⊂ Rd, we have

(i) If Hs(A) < ∞, then Ht(A) = 0.

(ii) If Ht(A) > 0, then Hs(A) = ∞.

Proof. Clearly, both statements are equivalent. To prove (i), consider a coveringA ⊂ ⋃iAi with diam(Ai) ≤ ε and

∑i diam(Ai)

s ≤ Hsε(A) + 1. Then

Htε(A) ≤

∑i

diam(Ai)t ≤ εt−s

∑i

diam(Ai)s ≤ εt−s

(Hsε(A) + 1

).

Letting ε → 0, the right side tends to 0 if Hs(A) < ∞, and so we deduce thatHt(A) = 0. �

The Hausdorff dimension of a non-empty set A ⊂ Rd is defined as

dimH(A) = sup{s : Hs(A) > 0}.

We also set dim(∅) = 0. It is easy to realize that the Hausdorff dimension is amonotone set function:

E ⊂ F ⇒ dimH(E) ≤ dimH(F ).

Since Hd(Rd) is σ-finite, from the preceding lemma it turns out that dimH(A) ≤ dfor all sets A ⊂ Rd. One also deduces that Hs(A) = ∞ if s < dimH(A), and


Ht(A) = 0 if t > dimH(A). In the borderline case s = dimH(A), Hs(A) can attainany value from [0,+∞].

For much more information on Hausdorff measures, we suggest that thereader have a look at any of the books Mattila [100], Falconer [40], Federer [42],Carleson [7], or Rogers [140], for example.

The first result that connects Hausdorff measures and contents to analyticcapacity is the following old theorem, essentially due to Painleve.

Theorem 1.21 (Painleve). For every compact set E ⊂ C, we have

γ(E) ≤ H1∞(E).

In particular, if H1(E) = 0, then E is removable.

Proof. Given ε > 0, cover E with a family of sets Ai such that∑

i diam(Ai) ≤H1

∞(E) + ε. One can replace each Ai by an open ball Bi of radius ri slightlybigger than diam(Ai) if necessary so that E ⊂ ⋃iBi and

∑i ri ≤ H1

∞(E) + 2ε.Since E is compact we may assume that this is a finite family. Consider the curveΓ = ∂o

(⋃i Bi

). By (1.2), if f is the Ahlfors function of E, we have

|f ′(∞)| =∣∣∣∣ 12πi∫Γ

f(z) dz

∣∣∣∣ ≤∑i

1

2π

∫Γ∩∂Bi

|f(z)| dH1(z)

≤∑i

ri ≤ H1∞(E) + 2ε.

Since ε > 0 is arbitrary, the theorem follows. �

The inequality in the preceding theorem can be sharpened to γ(E) ≤12 H1∞(E), which is optimal, as shown if one chooses E to be a ball (see Dudziak[35, p. 25] for some details, for example).

For subsets of the real line we have additional information: it turns out that

γ(E) =1

4H1(E) if E ⊂ R. (1.9)

Notice that for this type of sets, the Hausdorff measure H1 coincides with thecontentH1∞. We will not prove the identity (1.9). Instead, we will content ourselveswith the following estimate.

Proposition 1.22. If E ⊂ C is a compact set contained in R, then

1

4H1(E) ≤ γ(E) ≤ 1

πH1(E).

Let us remark that the precise and more difficult estimate γ(E) ≤ 14 H1(E)

in (1.9) (which we will not prove) is due to Pommerenke [135].


Proof. By the outer regularity of analytic capacity and by compactness, we mayassume that E is a finite collection of disjoint intervals. In this case, for anyε > 0, one can easily construct a curve (or a family of curves) Γ surrounding Ewith H1(Γ) ≤ 2H1(E) + ε. Arguing as in the proof of Theorem 1.21, if f is anadmissible function for E, we obtain

|f ′(∞)| =∣∣∣∣ 12πi∫Γ

f(z) dz

∣∣∣∣ ≤ 1

2πH1

∞(Γ) ≤ 1

π

(H1∞(E) + ε

).

Consequently, γ(E) ≤ H1∞(E)/π.For the other inequality, consider the function f = 1

2C(H1�E). Observe thatif z = x+ i y �∈ E, then

Im f(z) =1

2

∫E

Im1

t− zdH1(t) =

1

2

∫E

y

(t− x)2 + y2dH1(t).

Therefore, Im f(z) = 0 if y = 0, and otherwise,

| Im f(z)| < 1

2

∫R

|y|t2 + y2

dH1(t) =π

2.

So f maps C \ E to the strip | Im(z)| < π/2.Consider the map

ϕ(z) =ez − 1

ez + 1.

Notice that ϕ maps conformally the strip | Im(z)| < π/2 to the unit disk. Theng = ϕ ◦ f is admissible for E, and since f(∞) = 0 and g(∞) = ϕ(0) = 0, we have

γ(E) ≥ |g′(∞)| = limz→∞ |z g(z)| = lim

z→∞ |z f(z)|∣∣∣ϕ(f(z))

f(z)

∣∣∣ = |f ′(∞)ϕ′(0)|.

Using that f ′(∞) = −H1(E)/2 and ϕ′(0) = 1/2, we get γ(E) ≥ H1(E)/4. �Analogously to (1.9), when E ⊂ ∂B(0, 1), one has

γ(E) = sinH1(E)

4.

We will not prove this estimate. See Murai [118] for further details.

The proposition just proved shows that the compact subsets of lines areremovable if and only if they have zero length. Denjoy tried to extend this resultto subsets of rectifiable curves and he believed he had managed in [33]. Howeverthere was an important gap in his proof, and this question remained open fora long time. This became known as Denjoy’s conjecture, and it was only solvedin the affirmative, by Calderon [5], after the proof of the L2 boundedness of theCauchy transform on Lipschitz graphs with small enough slope. In this book, wewill prove Denjoy’s conjecture in Chapter 4.


Now we turn to sets of Hausdorff dimension bigger than one. First we needto review a key tool: Frostman’s Lemma. The proof below is somewhat differentfrom the classical one that can be found in the literature (for example, in Mattila[100, Chapter 4] or Carleson [7, Chapter 2]).

Theorem 1.23 (Frostman Lemma). Let 0 < s ≤ d and consider a compact setE ⊂ Rd. Then Hs∞(E) > 0 if and only if there exists a non-trivial Borel measureμ supported on E such that μ(B) ≤ r(B)s for any ball B. Furthermore, we canfind μ so that μ(E) ≥ c−1Hs

∞(E), with the constant c depending only on d.

Proof. Suppose first that such a measure μ exists, and let us see that Hs∞(E) > 0.Indeed, consider a covering

⋃iAi ⊃ E, and take for each i a point xi ∈ Ai. Since

the union of the balls B(xi, diam(Ai)) covers E, we get∑i

diam(Ai)s ≥ c−1

∑i

μ(B(xi, diam(Ai))

) ≥ c−1μ(E).

Taking the infimum over all possible coverings of E, we obtainHs∞(E) ≥ c−1 μ(E).For the converse implication of the theorem, assume that E is contained in

a dyadic cube Q0. The measure μ will be constructed as a weak limit of measuresμn, n ≥ 0. The first measure is

μ0 = Hs∞(E)

Ld�Q0

Ld(Q0).

For n ≥ 1, each measure μn vanishes in Rd \Q0, it is absolutely continuous withrespect to Lebesgue measure, and in each cube from Dn(Q0), it has constantdensity. It is defined from μn−1 as follows. If P ∈ Dn(Q0) and P is a dyadic childof Q ∈ Dn−1(Q0) (then we write P ∈ Ch(Q)), we set

μn(P ) =Hs

∞(P ∩ E)∑R∈Ch(Q) Hs∞(R ∩ E)

μn−1(Q). (1.10)

Observe that ∑P∈Ch(Q)

μn(P ) = μn−1(Q) for all Q ∈ Dn−1(Q0),

and thus μn(Rd) = μn−1(R

d).As said above, μ is just a weak limit of the measures μn. The fact that μ is

supported on E is easy to check: from the definition of μn in (1.10), μn(P ) = 0 ifP ∈ Dn(Q0) does not intersect E. As a consequence, μk(P ) = 0 for all k ≥ n too,and thus,

supp(μk) ⊂ U2−n+1 diam(Q0)(E) for all k ≥ n.

From this condition, one gets that supp(μ) ⊂ U2−n+1 diam(Q0)(E), for all n ≥ 0,which proves the claim.


Next we will show that

μn(P ) ≤ Hs∞(P ∩E) for all P ∈ Dn(Q0).

This follows easily by induction: it is clear for n = 0, and if it holds for n− 1 andQ is the dyadic parent of P , then

μn−1(Q) ≤ Hs∞(Q ∩ E) ≤

∑R∈Ch(Q)

Hs∞(R ∩ E).

Thus, from (1.10), we infer that μn(P ) ≤ Hs∞(P ∩ E), as claimed. As a conse-

quence, for all j ≥ n,

μj(P ) ≤ Hs∞(P ∩ E) for all P ∈ Dn(Q0).

Moreover, by construction, all the dyadic cubes which do not intersect Q0 havezero measure μj .

Since every open ball Br of radius r with 2−n−1�(Q0) ≤ r < 2−n�(Q0) iscontained in a union of at most 2d dyadic cubes Pk with side length 2−n�(Q0), weget

μj(Br) ≤2d∑k=1

μj(Pk) ≤2d∑k=1

Hs∞(Pk ∩ E) ≤ 2d diam(Pk)

s ≤ c rs,

for all j ≥ n. Letting j → ∞, we infer that μ(Br) ≤ c rs. �Remark 1.24. In fact, in the preceding proof we have shown that for any compactset E ⊂ Rd,

Hs∞(E) ≈ sup

{μ(E) : μ(B(x, r)) ≤ rs ∀x ∈ Rd, r > 0

}.

Theorem 1.25. Let s > 1. For every compact set E ⊂ C, we have

γ(E) ≥ c(s)Hs∞(E)1/s,

where c(s) = c′ s−1s > 0 and c′ is an absolute constant.

Observe that c(s) → 0 as s → 1.

Proof. Let μ be a Borel measure supported on E such that μ(E) ≥ c−1Hs∞(E)

satisfying μ(B(z, r)) ≤ rs for all z ∈ C, r > 0. Consider the function f = Cμ.Observe that f(∞) = 0 and |f ′(∞)| = μ(E). To estimate ‖f‖∞, for any z ∈ C wewrite

|Cμ(z)| ≤∫

1

|w − z| dμ(w) =∫ ∞

0

μ({w : 1/|w − z| > t}) dt

=

∫ ∞

0

μ(B(z, 1/t)) dt ≤∫ ∞

0

min(μ(E), 1/ts

)dt.


The last integral equals∫ 1/μ(E)1/s

0

μ(E) dt+

∫ ∞

1/μ(E)1/s1/ts dt = μ(E)1−

1s +

1

s− 1

(μ(E)

1s

)s−1

=s

s− 1μ(E)1−

1s .

Therefore, ‖f‖∞ ≤ ss−1 μ(E)1−

1s , and so

γ(E) ≥ |f ′(∞)|‖f‖∞ ≥ s− 1

sμ(E)1/s ≥ c

s− 1

sHs

∞(E)1/s. �

In the case s = 2, in the plane, H2∞ coincides with the area modulo a scaling

factor, and so we have γ(E) ≥ cL2(E)1/2, for some absolute constant c > 0.Arguing as in Theorem 1.25, a more careful estimate of the integral that definesC(L2�E) shows that γ(E) ≥ √L2(E)/π. This inequality is sharp since equalityholds when E is a disk. See Garnett [54, p.79] or Dudziak [35, p.16].

1.7 Rectifiable and purely unrectifiable sets and

Vitushkin’s conjecture

To summarize, in the preceding section we have seen that the relationship betweenHausdorff measure and analytic capacity is the following:

• If dimH(E) > 1, then γ(E) > 0.

• γ(E) ≤ H1∞(E) ≤ H1(E). In particular, if dimH(E) < 1, then γ(E) = 0.

By the statements above, it turns out that dimension 1 is the critical dimension inconnection with analytic capacity. Moreover, a natural question arises: is it truethat γ(E) > 0 if and only if H1(E) > 0?

It took some time to realize that the answer is no. This was first shown byVitushkin in the 1960’s, by constructing a compact set with positive length andvanishing analytic capacity. A typical example of such a set is the so-called cornerquarters Cantor set. It is constructed in the following way: consider a squareQ0 with side length 1. Now replace Q0 by four squares Q1

i , i = 1, . . . , 4, withside length 1/4 contained in Q0, so that each Q1

i contains a different vertex of Q0.Analogously, in the next stage each Q1

i is replaced by four squares with side length1/16 contained in Q1

i so that each one contains a different vertex of Q1i . So we will

have 16 squares Q2k of side length 1/16. We proceed inductively (see Figure 1.1),

and we set En =⋃4n

i=1 Qni and E =

⋂∞n=1 En. This is the corner quarters Cantor

set. It is not difficult to check that 0 < H1(E) < ∞. Indeed, from the fact that

4n∑i=1

�(Qni ) = 1 for all n ≥ 1,

1.7. Rectifiable sets and Vitushkin’s conjecture 35

Q0 E1 E2 E3

Figure 1.1: The square Q0 and the sets E1, E2 and E3, which appear in the firststages of the construction of the corner quarters Cantor set.

one deduces that H12−n+1/2(E) ≤ H1

2−n+1/2(En) ≤ 21/2, and thus letting n → ∞,

it follows that H1(E) ≤ 21/2. To show that H1(E) > 0 we will use Frostman’slemma: consider a probability measure μ supported on E such that μ(Qn

i ) = 4−n

for all i, n. It follows easily that μ has linear growth. That is, μ(B(z, r)) ≤ c r forall z ∈ C and all r > 0. This implies that

H1(E) ≥ H1∞(E) ≥ c−1μ(E) = c−1.

The proof of the fact that γ(E) = 0 is more difficult, and it was obtained indepen-dently by Garnett [53] and Ivanov [69], [70], shortly after Vitushkin’s example. Inthis book we will obtain a different proof of this result, as a corollary of the maintheorem of Chapter 6, namely the comparability between analytic capacity andthe so-called capacity γ+.

In [179], Vitushkin suggested that sets of zero analytic capacity can be charac-terized in terms of the Favard length. To define this notion, we need some notation:for θ ∈ [0, π), let pθ denote the orthogonal projection onto the line through theorigin and direction (cos θ, sin θ). Given a Borel set E ⊂ C, its Favard length is

Fav(E) =

∫ π

0

H1(pθ(E)) dθ. (1.11)

Vitushkin conjectured that γ(E) > 0 if and only if Fav(E) > 0. In 1986 Mattila[98] showed that this conjecture is false. He proved this in a quite astonishing way:he showed that the property of having positive Favard length is not invariant underconformal mappings while removability for bounded analytic functions is. Observethat Mattila’s argument does not tell which implication in the above conjecture isfalse.

In [76] Jones and Murai constructed a set with zero Favard length and positiveanalytic capacity. An easier example using curvature was obtained more recentlyby Joyce and Morters [79]. It is not known yet if the converse implication inVitushkin’s conjecture holds. That is, it is not known if there exist sets withpositive Favard length and zero analytic capacity.


Although Vitushkin’s conjecture is not true in full generality, it turns outthat it holds in the particular case of sets with finite length. This was proved byG. David [23] in 1998. In the case of finite length, the notion of Favard lengthis closely related to the one of rectifiability. A set E ⊂ Rd is called countablyrectifiable (or just, abusing language, rectifiable) if it is H1-almost all containedin a countable union of rectifiable curves (a rectifiable curve is a curve of finitelength). That is to say, there are curves Γi, i ≥ 1, with H1(Γi) < ∞ such that

H1(E \⋃i

Γi

)= 0.

On the other hand, E ⊂ Rd is called purely unrectifiable if it intersects anyrectifiable curve at most in a set of zero length.

Theorem 1.26. Let E ⊂ Rd with H1(E) < ∞. Then there exists a rectifiableBorel set B such that E \ B is purely unrectifiable. Thus, E can be decomposedas E = Er ∪ Eu, where Er is rectifiable and Eu is purely unrectifiable, and thedecomposition is unique up to sets of zero length.

Proof. Let s = supB H1(E ∩ B), where the sup is taken over all rectifiable Borelsets B ⊂ Rd. For every j ≥ 1, take a Borel set Bj such that H1(E ∩Bj) ≥ s−1/j.Consider the set B =

⋃j Bj. Then E \ B is purely unrectifiable, since otherwise

there would be a Borel set A such that A ∩B = ∅ and H1(A ∩E) > 0, and so

H1((A ∪B) ∩ E) = H1(E ∩A) +H1(E ∩B) > H1(E ∩B).

For the decomposition above, set Er = E∩B and Eu = E\B. The uniquenessup to sets of zero length is straightforward. �

In his work on Vitushkin’s conjecture, David proved the following.

Let E ⊂ C be compact with H1(E) < ∞. Then γ(E) = 0 if and only if E is purelyunrectifiable.

To be precise, let us remark that the “if” part is not due to David. In fact, itis an easy consequence of the solution of Denjoy’s conjecture. The “only if” partof the theorem, which is more difficult, is the one proved by David. This resultcan be considered as the solution of Painleve’s problem for sets with finite length.We will see a complete proof along this book. The main steps are contained inChapters 5, 6, and 7.

A well-known theorem of Besicovitch asserts that, for sets E of finite length,Fav(E) = 0 if and only if E is purely unrectifiable. In particular, the corner quar-ters Cantor set described above is purely rectifiable and has zero Favard length.Because of the Besicovitch theorem, David’s result is equivalent to saying thatγ(E) = 0 if and only Fav(E) = 0.

We will not prove the Besicovitch theorem on projections in this book. SeeMattila [100, Chapter 18] or Falconer [40, Chapter 6] for detailed arguments, forexample.

1.8. Historical remarks, the semiadditivity of γ, etc. 37

1.8 Historical remarks, uniform approximation byrational functions, the semiadditivity of analytic

capacity, and further results

1.8.1 Some historical remarks

Painleve [131] proved in 1888 that if a set has zero 1-dimensional Hausdorff mea-sure, then it is removable. He did not state this result exactly in this way becauseat that time the notion of Hausdorff measure had not been introduced. Althoughhe studied the problem of the removability by bounded analytic functions, it isnot clear if he formulated explicitly what now is known as the Painleve problem.See Pajot [132, p.78] for more information on this question.

Analytic capacity was introduced by Ahlfors in 1947 in [2]. In this work heshowed that E ⊂ C is removable if and only if γ(E) = 0. Also, he deduced severalproperties of the Ahlfors function. For example, assuming that E is a finite union ofn disjoint analytic arcs, by using variational arguments he proved that the Ahlforsfunction is unique and that it has n + 1 zeros in the unbounded component ofC∞ \ E (on the other hand, the proof of the uniqueness of the Ahlfors functionfor arbitrary domains is due to Havinson, and the proof that we showed in thischapter is from Fisher [46]).

1.8.2 Uniform approximation by rational functions and thesemiadditivity of analytic capacity

Vitushkin in the 1950’s and 1960’s studied problems on uniform rational approx-imation on compact sets of the complex plane. He showed that analytic capacityplays an essential role in this type of problems.

Let A(E) be the algebra of the complex continuous functions on E which

are analytic in◦E. Also, denote by R(E) the algebra formed by uniform limits

of rational functions with poles off E, which by Runge’s theorem coincide withuniform limits of functions analytic in neighborhoods of E. Recall that if C \E isconnected, the classical theorem of Mergelyan asserts that all functions from A(E)are uniform limits (in E) of analytic polynomials, and thus A(E) = R(E). For ageneral compact set E, Vitushkin proved the next result (see Vitushkin [179]).

Theorem 1.27. Let E ⊂ C be compact and let f ∈ C(C). The following are equiv-alent:

(a) f ∈ R(E).

(b) For any open disk D ⊂ C with radius δ and all ϕ ∈ C∞c (D),∣∣∣∣∫ f ∂ϕ dL2

∣∣∣∣ ≤ ε(δ) δ ‖∇ϕ‖∞ γ(D \ E),


where ε(·) is some function such that ε(δ) → 0 as δ → 0.

(c) For any open square Q ⊂ C with side length δ,∣∣∣∣∫∂Q

f(z) dz

∣∣∣∣ ≤ c ε(δ) γ(Q \ E),


Let us remark that, by a result of Paramonov [134], the squares Q in (c) canbe replaced by disks. That is, the conditions above are equivalent to the following:

(c′) For any open disk D ⊂ C with radius δ,∣∣∣∣∫∂D

f(z) dz

∣∣∣∣ ≤ c ε(δ) γ(D \ E),


To state another of the main theorems of Vitushkin in this area we need firstto define the so-called continuous analytic capacity. Given a compact set E ⊂ C,its continuous analytic capacity is

α(E) = sup |f ′(∞)|,

where the supremum is taken over all continuous functions f : C−→C with ‖f‖∞ ≤1 which are analytic in C \ E. If A ⊂ C is an arbitrary set, then one defines

α(A) = supE⊂A,E compact

α(E).

We have:

Theorem 1.28 (Vitushkin [179]). For a compact set E ⊂ C the following are equiv-alent:

(a) R(E) = A(E).

(b) α(D \ ◦E) = α(D \ E) for all open disks D.

(c) For some constant c, α(D \ ◦E) ≤ c α(D \ E) for all open disks D.

Because of the applications to this type of problems, Vitushkin raised thequestion of the semiadditivity of γ and α. Namely, does there exist an absoluteconstant c such that

γ(E ∪ F ) ≤ c (γ(E) + γ(F )) ? (1.12)

Also, does this hold for α instead of γ?A first result concerning these questions was obtained in the 1960’s by Mel-

nikov [111]. He showed that the semiadditivity of γ and α holds for subsets E


and F which are separated by an analytic curve Γ, with the constant c in (1.12)depending on Γ. Vitushkin [179] extended the result to the case where the curvewhich separates E and F is piecewise C1+ε. Davie [30] showed that it is enoughto assume that the curve is C1 and its derivative is Dini continuous.

On the other hand, for disjoint connected compact sets E,F , Suita [152]proved in 1984 the subadditivity of γ, that is, (1.12) holds with c = 1 in thisparticular case. It is not known if the subadditivity holds for arbitrary compactsets E,F ⊂ C.

For arbitrary compact sets E,F , the semiadditivity of γ was proved in 2003in Tolsa [160], by showing its comparability with the so-called capacity γ+, byreal analytic methods, unlike the preceding partial results of Melnikov, Vitushkin,Davie, and Suita. In this book, we will see the proof of this result in Chapter 6.

The semiadditivity of α also holds, as shown in Tolsa [162]. The arguments forα follow, up to some point, the scheme of the ones for γ, although the adaptationis non-trivial. The proof of the semiadditivity of α is not included in this bookbecause the main interest of this capacity lies in the application to the theory ofuniform rational approximation, which will not be studied here. An interestingconsequence of this result is the so-called inner boundary conjecture. The innerboundary of a compact set E ⊂ C, is the subset of the boundary points of Ewhich do not belong to the boundary of any component of C \ E.

Theorem 1.29 (Inner boundary conjecture, Tolsa [162]). Let E ⊂ C be compactand let ∂iE be its inner boundary. If α(∂iE) = 0, then R(E) = A(E).

For more information on the relationship between analytic capacity and thetheory of uniform approximation by rational functions, see the books by Gamelin[47] and Zalcman [183], or the more modern survey by Verdera [174], for example.

1.8.3 The approach by duality to analytic capacity

This is another topic that is not covered in this book. Given a domain Ω ⊂ C and0 < p < ∞, let Hp(Ω) be the Hardy space of the analytic functions h : Ω → C

such that |h(z)|p has a harmonic majorant. Garabedian [48] proved that if Econsists of a finite union of disjoint analytic arcs and Ω is the unbounded connectedcomponent of C∞ \ E, then

γ(E) = inf1

2π

∫E

|h| dH1, (1.13)

where the infimum is taken over all functions inH1(Ω) such that h(∞) = 1. It turnsout that there is a unique function h ∈ H1(Ω) which attains the infimum above.This is the so-called Garabedian function, which has a close relationship withthe Szego kernel. From (1.13) and the study of the properties of the Garabedianfunction it follows that

γ(E) = sup |f ′(∞)|2, (1.14)


where the supremum is taken over all functions f ∈ H2(Ω) such that f(∞) = 0and

‖f‖H2(Ω) := lim supε→0

(1

2π

∫∂(Ω\Uε(E))

|f(z)|2 dH1(z)

)1/2≤ 1.

The preceding results are explained in detail in Garnett [54, Chapter 1]. The bookby Carleson [7] also discusses this topic in some detail in Chapter 6, by some-what different arguments. For other questions and other approaches by duality toanalytic capacity, see Khavinson [80] and the references therein.

1.8.4 Other spaces and capacities

The problem of removability for bounded analytic functions and the theory ofuniform rational approximation also make sense for other spaces of functions. Forexample, one can think of the removability of compact sets for BMO or Lips

analytic functions. One says that f : Rd → R belongs to BMO(Rd) if

supQ

1

Ld(Q)

∫Q

|f −mQf | dLd < ∞, (1.15)

where the supremum is taken over all the cubes Q in Rd and mQf is the mean off over Q with respect to the Lebesgue measure. It turns out that E is removablefor BMO(C) analytic functions if and only if E has zero length. This was shownby Kral [82].

Recall that a function f : Ω → C is from Lips(Ω) if there exists a constant csuch that

|f(z)− f(w)| ≤ c |z − w|s for all z, w ∈ Ω.

For 0 < s < 1, Carleson [6] proved that a compact set E ⊂ C with H1+s(E) = 0is removable for Lips analytic functions. Later on, Dolzhenko [34] showed that theLips removability of E is equivalent to the condition H1+s(E) = 0. Further, Mel-nikov [112] proved that the corresponding capacity associated with this problemis comparable to the Hausdorff content Hs+1∞ . The case s = 1 is more difficult andwas solved by Uy [171]. Other results in connection with removability for someclasses of analytic and harmonic functions are discussed in Carleson [7, Chapters6,7].

Vitushkin’s theorems on uniform rational approximation have been extendedto the approximation in BMO (Verdera [172]), Lips (O’Farrell [128]), and othernorms. See the survey by Verdera [174] for more information and additional ref-erences to related works. See also the remarkable works by Mazalov [106] and[107] on approximation by bianalytic functions and by solutions of elliptic partialdifferential equations.

The natural generalization of analytic capacity to higher dimensions is theso-called Lipschitz harmonic capacity κ. This is defined as follows. For a compact


set E ⊂ Rd, one setsκ(E) = sup

∣∣〈Δϕ, 1〉∣∣where the supremum is taken over all Lipschitz functions ϕ : Rd → R which areharmonic in Rd \ E with ‖∇ϕ‖∞ ≤ 1. Also 〈Δϕ, 1〉 denotes the action of thecompactly supported distributional Laplacian Δϕ on the function 1. It turns outthat E ⊂ Rd is removable for Lipschitz harmonic functions if and only if κ(E) = 0.Moreover, κ and its C1 version κc are the relevant capacities to study the problemof C1 harmonic approximation in compact subsets of Rd, as shown by Paramonov[133], who has extended Vitushkin’s approximation theorems to this context.

As in the case of bounded analytic functions, the description of removablesets for Lipschitz harmonic functions in metric and geometric terms is a diffi-cult problem, because Lipschitz harmonic capacity is difficult to describe in suchterms. See Section 6.10 of Chapter 6 and Section 7.11 of Chapter 7 for additionalinformation on this topic.

1.8.5 Hausdorff measures

First we will recall some results on Hausdorff densities. To this end, we need todefine the upper and lower s-dimensional densities of a Borel measure μ at x.Given s > 0, we set

Θ∗s(x, μ) := lim supr→0

μ(B(x, r))

(2r)s, Θs

∗(x, μ) := lim infr→0

μ(B(x, r))

(2r)s. (1.16)

These are the upper and lower s-dimensional densities, respectively. If they agree,then Θs(x, μ) := Θ∗s(x, μ) = Θs

∗(x, μ) is the so-called s-dimensional density ofμ at x. For μ = Hs�A, we write Θ∗s(x,A) = Θ∗s(x,Hs�A), and analogously forΘs

∗(x,A) and Θs(x,A).The following result is rather elementary:

Theorem 1.30. Let A ⊂ Rd with Hs(A) < ∞.

(a) 2−s ≤ Θ∗s(x,A) ≤ 1 for Hs-almost all x ∈ A.

(b) If A is Hs-measurable, then Θ∗s(x,A) = 0 for Hs-almost all x ∈ Rd \A.For the proof see Mattila [100, Chapter 6], for example.

In the preceding section we introduced the notion of rectifiability. There isan analogous notion in higher dimensions: a set A ⊂ Rd is called n-rectifiable ifthere are Lipschitz functions fi : R

n → Rd, i = 1, 2, . . . such that

Hn(A \⋃i

fi(Rn))= 0.

On the other hand, A is called purely n-unrectifiable if Hn(A ∩ F ) = 0 for everyn-rectifiable set A. Any set with finite Hn-measure can be decomposed into an


n-rectifiable set and a purely n-unrectifiable set, such as in Theorem 1.26. Theproof is very similar.

Rectifiable sets can be characterized in terms of densities. The followingtheorem is due to Besicovitch in the case n = 1, d = 2, to Marstrand [88] forn = 2, d = 3, and to Mattila [97] in full generality.

Theorem 1.31. Let n be a positive integer and let A be an Hn-measurable subsetof Rd with Hn(A) < ∞.

(a) A is n-rectifiable if and only if the density Θn(x,A) exists and equals 1 forHn-almost all x ∈ A.

(b) A is purely unrectifiable if and only if Θn∗ (x,A) < 1 for Hn-almost all x ∈ A.

Let us remark that in the case n = 1, d = 2, Besicovitch proved that ifA is purely unrectifiable, then Θ1∗(x,A) ≤ 3/4 for Hn-almost all x ∈ A. Preissand Tiser [139] have improved this result by showing that 3/4 can be replaced by(2+

√46)/12 � 0.7319. The (still open) conjecture is that 1/2 is the right constant.

Another important result regarding densities is the following.

Theorem 1.32. Let μ be a finite Radon measure in Rd and s > 0. Suppose that thedensity Θs(x, μ) exists and is positive and finite for μ-almost all x ∈ Rd. Then sis an integer and μ is rectifiable, that is, μ is of the form μ = gHn�A, for somerectifiable Borel set A and some non-negative function g ∈ L1(Hn�A).

The fact that s must be integer in the preceding result is due to Marstrand[89], while the more difficult fact that μ is rectifiable is a celebrated theorem ofPreiss [138].

Rectifiable and purely unrectifiable sets can also be characterized in termsof their orthogonal projections. This has already been mentioned in Section 1.7.We recall the precise result below. For simplicity we only state the 1-dimensionalresult in R2, which is due to Besicovitch. However, there is a natural generalizationto n-dimensional Hausdorff measures in Rd by Federer, which we skip.

Theorem 1.33. Let E ⊂ C be an H1-measurable set with 0 < H1(E) < ∞. Thefollowing are equivalent:

(a) E is purely unrectifiable.

(b) H1(pθ(E)) = 0 for almost all θ ∈ [0, π).

(c) There are two different angles θ1, θ2 ∈ [0, π) such that

H1(pθ1(E)) = H1(pθ2(E)) = 0.

The criterion in (c) is a very useful tool to show that some Cantor type setsare purely unrectifiable. For example, the corner quarters Cantor set from Section


1.7 has zero length projections on the horizontal and vertical axes (in fact, theyhave Hausdorff dimension 1/2). This implies that the set is purely unrectifiable.

For most of the results stated above some good references are the books ofFalconer [40], Federer [42], and Mattila [100], for example. Falconer’s text dealsmainly with the 1-dimensional case in the plane, while the ones of Federer andMattila consider the general n-dimensional case in Rd. On the other hand, theaforementioned theorem of Preiss is explained in detail in the nice book [32] byDe Lellis.

Chapter 2

Basic Calderon-Zygmund theorywith non-doubling measures

2.1 Introduction

In this chapter we will obtain some basic results of Calderon-Zygmund theory inRd. In the classical literature, the underlying measure μ is assumed to be doubling.That is, there exists some constant c > 0 such that

μ(B(x, 2r)) ≤ c μ(B(x, r)) for all x ∈ supp(μ), r > 0.

Notice that this condition is asked only for x ∈ supp(μ).However, for many applications to geometric analysis, in particular for ques-

tions related to analytic capacity, the preceding doubling assumption is too restric-tive. Instead, it is more natural to ask for an upper growth condition for μ. So, tostudy n-dimensional singular integral operators (SIO’s) in Rd, with 0 < n ≤ d, wewill consider measures μ satisfying the following growth condition of degree n:

μ(B(x, r)) ≤ c0 rn for all x ∈ Rd, r > 0. (2.1)

(when n = 1, we say that μ has linear growth). Sometimes we will also talkabout c0-growth of degree n or about c0-linear growth if we are interested in thedependence of some estimates on precise value of the constant c0. It turns out thatif μ has no point masses, the growth of degree n for μ is necessary for the L2(μ)boundedness of any SIO whose kernel K(x, y) satisfies |K(x, y)| ≈ |x − y|−n (seeDavid [22, p. 56]).

One of the main difficulties that arises when one deals with a non-doublingmeasure μ is due to the fact that the non-centered maximal Hardy-Littlewoodoperator

Mncμ f(x) := sup

{1

μ(B)

∫B

|f | dμ : B closed ball, x ∈ B

}

, , OI 10.1007/978-3- - -6_ ,

© Springer



454

46 Chapter 2. Calderon-Zygmund theory with non-doubling measures

may fail to be of weak type (1, 1) (the superindex “nc” stands for non-centered).Sometimes the centered version of the operator, that is,

Mμf(x) = supr>0

1

μ(B(x, r))

∫B(x,r)

|f | dμ,

is a good substitute ofMncμ f , because using Besicovitch’s covering theorem one can

show that Mμ is bounded from L1(μ) into L1,∞(μ), and in Lp(μ), for 1 < p ≤ ∞.However, one cannot always use the centered maximal Hardy-Littlewood operatorinstead of the non-centered one. In these cases other arguments are required.

Let us see some examples of non-doubling measures in the plane. Considerfirst the function f : R → R given by

f(x) =1

x2e−1/x if x �= 0, and f(0) = 0.

Let μ be the planar Lebesgue measure which is located between the graph of fand the horizontal axis, for 0 ≤ x ≤ 1. Let us see that μ fails to be doubling atthe origin. Indeed, for r > 0 small, we have

μ(Q(0, r)) = μ ({(x, y) : 0 ≤ x ≤ r}) =∫ r

0

f(x) dx = e−1/r.

Also, μ(Q(0, 2r)) = e−1/2r. Therefore,

μ(Q(0, 2r))

μ(Q(0, r))=

e−1/2r

e−1/r= e1/2r → ∞ as r → 0.

Of course, this implies that μ(B(0,2r))μ(B(0,r)) also tends to ∞ as r → 0.

In our second example (also in the plane), we set

μ = H2�[−1, 1]2 +H1�([−1, 1]× {0}).That is, μ is the addition of area on a square and length on a segment that bisectsthe square. Let Qn = Q(xn, �n) be a sequence of squares contained in [−1, 1]2 with�n → 0 such that each Qn does not intersect the segment [−1, 1]×{0}, while 2Qn

does, so that μ(Qn) = �2n and μ(2Qn) = 2�n + 4�2n. We have

μ(2Qn)

μ(Qn)>

2�n�2n

=2

�n→ ∞ as n → ∞.

Our third example is more sophisticated and was explained to me by Mattila.Given a positive sequence of integers {Nm}m≥1 tending to ∞, we construct aplanar Cantor set E of the form

E =∞⋂

m=0

Em, Em =⋃

j∈Jm

Dmj ,

2.2. Preliminaries 47

where the sets Dmj of the generation m are closed disks of equal radius rm =

1/#Jm for all j ∈ Jm. For m = 0, the family J0 is made up of a unique element,corresponding to D0

1 = B(0, 1). For m ≥ 1,

#Jm = Nm ·#Jm−1.

To construct the disks Dmj , j ∈ Jm, we perform the following procedure in each

disk Dm−1i : we choose Nm disks of the family {Dm

j }j∈Jm to be contained in Dm−1i ,

so that one of them is concentric with Dm−1i , while the other Nm − 1 disks are

internally tangent to ∂Dm−1i and they are uniformly separated (see Figure 2.1).

Since Em ⊂ Em−1 for all m ≥ 1, the Cantor set E is non-empty. In fact, onecan even check that it has positive and finite length, taking into account that the∑

j∈Jmdiam(Dm

j ) = 2 for any fixed generation m.We choose μ to be the natural associated probability measure, so that

μ(Dmj ) =

1

#Jmfor all j ∈ Jm.

It is easy to check that μ has linear growth. Let us see that it is not doubling. Form ≥ 1, j ∈ Jm−1, consider a disk Dm−1

j . Notice that its center belongs to E, and

that inside 12 D

m−1j there is only one disk of the family {Dm

k }k∈Jm , call it Dms(j).

Therefore, μ(12 D

m−1j

)= μ(Dm

s(j)) = 1/#Jm. Thus,

μ(Dm−1

j

)μ(12 D

m−1j

) = #Jm#Jm−1

= Nm → ∞ as m → ∞,

which implies that μ is not doubling. Moreover, if Nm grows slow enough so that∑∞m=1

1Nm

= ∞, then μ-almost every x ∈ E belongs to infinitely many disks12 D

m−1j . This follows easily from the second Borel-Cantelli lemma (we leave it for

the reader to check the details). It can be shown that this implies that there doesnot exist any measurable subset A ⊂ E of positive μ-measure such that μ�A isdoubling.

2.2 Preliminaries

Recall that a Radon measure on Rd has growth of degree n if there exists someconstant c0 such that μ(B(x, r)) ≤ c0r

n for all x ∈ Rd, r > 0. Sometimes forshort, we will just say that μ is of degree n. When n = 1, we say that μ has lineargrowth. If there exists some constant c such that

c−1rn ≤ μ(B(x, r)) ≤ c rn for all x ∈ supp(μ), 0 < r ≤ diam(supp(μ)), (2.2)

then we say that μ is n-dimensional AD regular or, simply, AD regular. On theother hand, a set E ⊂ Rd is called n-dimensional AD regular if Hn�E is n-dimensional AD regular. Clearly, if a measure is AD regular, then it is doubling.


Figure 2.1: The disk D01 and the families of disks {D1

j}j∈J1 and {D2j}j∈J2 for the

construction of the Cantor set E, with N1 = 5, N2 = 13.

The space of finite complex Radon measures in Rd is denoted by M(Rd).This is a Banach space with the norm of the total variation: ‖ν‖ = |ν|(Rd).

We say that K(·, ·) : Rd × Rd \ {(x, y) ∈ Rd × Rd : x = y} → C is an n-dimensional Calderon-Zygmund kernel if there exist constants c > 0 and η, with0 < η ≤ 1, such that the following inequalities hold for all x, y ∈ Rd, x �= y:

|K(x, y)| ≤ c

|x− y|n , and, if |x− x′| ≤ |x− y|/2,

|K(x, y)−K(x′, y)|+ |K(y, x)−K(y, x′)| ≤ c |x− x′|η|x− y|n+η

.(2.3)

Notice that, without the condition |x − x′| ≤ |x − y|/2, the last inequality alsoholds if |x− y| ≈ |x′ − y|, possibly with a different constant c. This follows easilyusing the first inequality in (2.3).

Given a Radon measure ν on Rd, possibly complex, we define

Tν(x) :=

∫K(x, y) dν(y), for x ∈ Rd \ supp(ν). (2.4)

We also define Tν(x) as above for any x ∈ Rd where the integral on the right sidemakes sense. We say that T is an n-dimensional singular integral operator (SIO)with kernel K(·, ·). The integral in the definition may not be absolutely convergentif x ∈ supp(ν). For this reason, we consider the following ε-truncated operatorsTε, ε > 0:

Tεν(x) :=

∫|x−y|>ε

K(x, y) dν(y), x ∈ Rd.

2.3. Covering theorems and maximal operators 49

Observe that now the integral on the right-hand side converges absolutely if, forinstance, |ν|(Rd) < ∞.

Given a fixed positive Radon measure μ on Rd and f ∈ L1loc(μ), we write

Tμf(x) := T (f μ)(x) x ∈ Rd \ supp(f μ),

andTμ,εf(x) := Tε(f μ)(x).

The integral that defines Tε(f μ)(x) is absolutely convergent for all x ∈ Rd if, forexample, f ∈ Lp(μ) for some 1 ≤ p < ∞ and μ is of degree n. Indeed, from thepolynomial growth of degree n of μ, one easily deduces that(∫

|x−y|>ε

|K(x, y)|p′dμ(y)

)1/p′

< ∞.

We say that Tμ is bounded in Lp(μ) if the operators Tμ,ε are bounded inLp(μ) uniformly on ε > 0. Analogously, with respect to the boundedness fromL1(μ) into L1,∞(μ). We also say that T is bounded from M(Rd) to L1,∞(μ) ifthere exists some constant c such that for all ν ∈ M(Rd) and all λ > 0,

μ({x ∈ Rd : |Tεν| > λ}) ≤ c‖ν‖

λ

uniformly on ε > 0.Singular integral operators which are bounded in L2(μ) are called Calderon-

Zygmund operators (CZO’s).The Cauchy transform is the SIO in C originated by the kernel

K(x, y) :=1

y − x, x, y ∈ C.

It is denoted by C. That is to say,

Cν(x) :=∫

1

y − xdν(y), x ∈ C \ supp(ν).

2.3 Covering theorems and maximal operators

First we recall the classical 3r-covering theorem.

Theorem 2.1. Let B be a finite family either of closed balls or of open balls from ametric space. Then there exists a subfamily {Bi}i∈I ⊂ B of pairwise disjoint ballssuch that ⋃

B∈BB ⊂⋃i∈I

3Bi.


Proof. Let B = {Bj}Nj=1 and assume that r(Bj) ≥ r(Bj+1) for j = 1, . . . , N − 1.Set Bj1 := B1. Let Bj2 be a biggest ball from B which is not contained in 3Bi1

(if it exists). Let Bj3 be a biggest ball which is not contained in 3Bj1 ∪ 3Bj2 (ifit exists). We continue this selection process as long as possible. Since the totalnumber of balls is finite, the process finishes after a finite number of steps. We setI = {j1, j2, . . .}. It is clear that, by construction,⋃

B∈BB ⊂⋃i∈I

3Bi.

Moreover, the balls from B are pairwise disjoint. Indeed, consider Bjh , Bjk ∈ Bwith jh < jk, so that r(Bjh ) ≥ r(Bjk ). If these balls intersect, then 3Bjh ⊃ Bjk ,which is not possible by the construction above. �

The following is a variant of the result above for infinite families.

Theorem 2.2. Let B be a family either of closed balls or of open balls from a metricspace such that

sup{diam(B) : B ∈ B} < ∞.

Then there exists a finite or countable sequence {Bi}i∈I of pairwise disjoint ballssuch that ⋃

B∈BB ⊂⋃i∈I

5Bi.

The proof is quite similar to the one of Theorem 2.1. For the detailed argu-ments, see Mattila [100, Chapter 2], for example.

Notice that Theorems 2.1 and 2.2 are valid in the general setting of metricspaces. In particular, they hold for families of cubes in Rd.

Now we recall the Besicovitch covering theorem.

Theorem 2.3. There are integer constants pd and qd depending only on d suchthat, if A ⊂ Rd is bounded and each x ∈ A is the center of some closed ball Bx,then:

(a) There exists a subfamily {Bi}i∈I ⊂ {Bx}x∈A such that

χA ≤∑i∈I

χBi ≤ pd.

In other words, the balls Bi cover A and have bounded overlap.

(b) There are subfamilies B1, . . . ,Bqd ⊂ {Bx}x∈A such that the balls from eachfamily Bi are pairwise disjoint and

A ⊂qd⋃i=1

⋃B∈Bi

B.

2.3. Covering theorems and maximal operators 51

For the proof, we refer the reader to Mattila [100, Chapter 2] again.A family of cubes or balls with finite overlap, such as the one in the statement

(a), is said to be almost disjoint.

Remark 2.4. The preceding theorem is not valid in general metric spaces. However,it also holds with cubes instead of balls.

Recall that the centered maximal Hardy-Littlewood operator Mμ appliedto a complex Radon measure ν is defined by

Mμν(x) = supr>0

|ν|(B(x, r))

μ(B(x, r)),

so that for f ∈ L1loc(μ) we have Mμf = Mμ(f μ). It is straightforward to check

that in the supremum defining Mμ it does not matter if we take open or closedballs.

Using Besicovitch’s covering theorem we will prove the following result:

Theorem 2.5. Let μ be a Radon measure in Rd. The centered maximal Hardy-Littlewood operator Mμ is bounded from M(Rd) to L1,∞(μ) and in Lp(μ), for1 < p ≤ ∞.

Let us remark that this theorem holds for an arbitrary Radon measure μ inRd. Neither doubling conditions nor growth conditions on μ are necessary.

Proof. It is immediate to check that Mμ is bounded in L∞(μ), and thus, by theMarcinkiewicz interpolation theorem, it is enough to show that it is bounded fromM(Rd) to L1,∞(μ) (which implies that Mμ is of weak type (1, 1) with respect toμ). To this end, fix R > 0, take ν ∈ M(Rd) and λ > 0, and write

ΩR ={x ∈ supp(μ) ∩B(0, R) : Mμν(x) > λ

}.

For each x ∈ ΩR, let Bx be a ball centered at x such that

|ν|(Bx)

μ(Bx)> λ.

By the Besicovitch covering theorem, there exists some subfamily {Bi}i∈I ⊂ {Bx}xwhich covers ΩR with bounded overlapping, that is,

∑i∈I χBi ≤ pd. Then

μ(ΩR) ≤∑i∈I

μ(Bi) ≤∑i∈I

|ν|(Bi)

λ≤ pd

λ‖ν‖.

Since this holds for all R > 0, we get

μ({

x ∈ Rd : Mμν(x) > λ}) ≤ pd

λ‖ν‖. �


Another maximal operator that we will use below is the following:

Nμν(x) = sup

{ |ν|(B)

μ(B): B open ball, x ∈ B, μ(3B) ≤ 3d+1μ(B)

}. (2.5)

For f ∈ L1loc(μ) we set Nμf ≡ Nμ(f μ).

Using now the 3r-covering theorem we will prove the following:

Theorem 2.6. Let μ be a Radon measure in Rd. The maximal operator Nμ isbounded from M(Rd) to L1,∞(μ) and in Lp(μ), for 1 < p ≤ ∞. Moreover, for anycomplex Radon measure ν and any λ > 0, the following holds:

μ({x : Nμν(x) > λ}) ≤ c

1

λ|ν|({x : Nμν(x) > λ}). (2.6)

Proof. The Lp(μ) boundedness, for 1 < p < ∞ follows by interpolation from theboundedness from M(Rd) to L1,∞(μ) and in L∞(μ). In turn, the boundednessfrom M(Rd) to L1,∞(μ) is a straightforward consequence of (2.6).

Given ν ∈ M(Rd) and λ > 0, let

Ω ={x ∈ supp(μ) : Nμν(x) > λ

}.

Let K ⊂ Ω be an arbitrary compact subset. For each x ∈ K, let Bx be an openball containing x such that μ(3Bx) ≤ 3d+1μ(Bx) and

|ν|(Bx)

μ(Bx)> λ.

Consider a finite subfamily Bx1 , . . . , Bxm such that K ⊂ ⋃mj=1 Bxj . By the 3r-

covering Theorem 2.1, there exists a disjoint subfamily {Bi}i∈I ⊂ {Bxj}1≤j≤m

such that the balls 3Bi, i ∈ I, cover K. Then we have

μ(K) ≤∑i∈I

μ(3Bi) ≤ 3d+1∑i∈I

μ(Bi) ≤ 3d+1∑i∈I

|ν|(Bi)

λ.

Since Nμν(x) > λ for all x ∈ ⋃i∈I Bi,∑i∈I

|ν|(Bi) ≤ |ν|({x : Nμν(x) > λ}).As the preceding estimates are independent of K, we get

μ({

x ∈ Rd : Nμν(x) > λ}) ≤ 3d+1

λ|ν|({x : Nμν(x) > λ}). �

Remark 2.7. If in the definition of Nμ one asks the balls B to satisfy

μ(αB) ≤ β μ(B)

2.4. Doubling cubes and balls 53

for fixed constants α, β > 1 instead of μ(3B) ≤ 3d+1μ(B), the resulting operatoris also bounded from M(Rd) to L1,∞(μ) and in Lp(μ), for 1 < p ≤ ∞. This is astraightforward consequence of the results from Section 9.6 of Chapter 9, whichrely on a slightly more difficult covering theorem.

Finally, we introduce the n-dimensional radial maximal operator:

MR,nν(x) = supr>0

|ν|(B(x, r))

rn. (2.7)

Notice that if μ has growth of degree n, then

MR,nν(x) ≤ c0 Mμν(x), (2.8)

and thus MR,n is bounded in Lp(μ), 1 < p ≤ ∞, and from M(Rd) to L1,∞(μ) (infact, this can also be deduced by a direct application of the previous 3r-coveringtheorem, arguing as in the proof of Theorem 2.6).

2.4 Doubling cubes and balls

Given α, β > 1, we say that a cube (or ball) Q is (α, β)-doubling if μ(αQ) ≤β μ(Q), where αQ is the cube (ball) concentric with Q with diameter α diam(Q).For definiteness, if α and β are not specified, by a doubling cube (ball) we meana (2, 2d+1)-doubling cube (ball).

For β > αn, as μ satisfies the growth condition (2.1), there are a lot of “big”doubling cubes and balls. To be precise, given any point x ∈ supp(μ) and c > 0,there exists some (α, β)-doubling cube (or ball) Q centered at x with �(Q) ≥ c(respectively, with r(Q) > c). Indeed, suppose that there are no doubling cubescentered at x with side length greater than c. Then we take Q = Q(x, c), and itfollows that μ(αmQ) > βmμ(Q) for each m ≥ 1, and so

μ(αmQ)

�(αmQ)n>

βm−1μ(αQ)

�(αmQ)n=

(β

αn

)m−1μ(αQ)

�(αQ)n→ ∞ as m → ∞,

since β > αn and μ(αQ) > 0. It is clear then that (2.1) is violated for m bigenough. A similar argument works with balls instead of cubes.

The next lemma states that there are a lot of “small” doubling cubes too:

Lemma 2.8. Let β > αd. If μ is a Radon measure in Rd, then for μ-a.e. x ∈ Rd

there exists a sequence of (α, β)-doubling cubes (or balls) {Qk}k centered at x with�(Qk) → 0 as k → ∞ (respectively, with r(Qk) → 0).

Notice that the preceding statement is valid for any Radon measure on Rd.In particular, it is not necessary to assume the growth condition (2.1).


Proof. We show the detailed arguments for cubes. The proof for balls is almostidentical.

Let Z ⊂ Rd be the set of points x such that there does not exist a sequenceof (α, β)-doubling cubes {Qk}k≥0 centered at x with side length decreasing to 0;and let Zj ⊂ Rd be the set of points x such that there does not exist any (α, β)-doubling cube Q centered at x with �(Q) ≤ 2−j. Clearly, Z =

⋃j≥0 Zj , and thus

proving the lemma is equivalent to showing that μ(Zj) = 0 for every j ≥ 0.

Let Q0 be a fixed cube with side length 2−j and let k ≥ 1 be some integer.For each z ∈ Q0 ∩ Zj, let Qz be a cube centered at z with side length α−k�(Q0).Since the cubes αhQz are not (α, β)-doubling for h = 0, . . . , k−1 and αkQz ⊂ 2Q0,we have

μ(Qz) ≤ β−1μ(αQz) ≤ · · · ≤ β−kμ(αkQz) ≤ β−kμ(2Q0). (2.9)

By the Besicovitch theorem, there exists a subfamily {zm} ⊂ Q0 ∩ Zj suchthat Q0 ∩Zj ⊂

⋃m Qzm and moreover

∑m χQzm

≤ pd. This is a finite family andthe number N of points zm can be easily estimated as follows. If Ld stands for theLebesgue measure on Rd,

N (α−k�(Q0))d =

N∑m=1

Ld(Qzm) ≤ pdLd(2Q0) = pd (2�(Q0))d.

Thus,

N ≤ pd2dαkd.

As a consequence, since the family {Qzm}1≤m≤N covers Q0 ∩ Zj , using (2.9) weget

μ(Q0 ∩ Zj) ≤N∑

m=1

μ(Qzm) ≤ Nβ−kμ(2Q0) ≤ pd2dαkdβ−kμ(2Q0).

Since β > αd, the right-hand side tends to 0 as k → ∞. Therefore μ(Q0∩Zj) = 0,and since the cube Q0 is arbitrary, we are done. �

Remark 2.9. The preceding proof shows that the sequence of cubes (or balls) {Qk}in Lemma 2.8 can be taken so that �(Qk) = αnk , for some integer nk such thatnk → −∞ as k → ∞. Also, it is immediate to check that we may assume thecubes or balls to be either open or closed.

Remark 2.10. Given f ∈ L1loc(μ) and β > αd, by the Lebesgue differentiation

theorem, for μ-almost all x ∈ Rd, every sequence of cubes {Qk}k centered at xwith �(Qk) → 0 satisfies

limk→∞

1

μ(Qk)

∫Qk

f dμ = f(x). (2.10)

2.5. Some standard estimates 55

By the preceding lemma, for μ-a.e. x ∈ Rd, there exists a sequence of (α, β)-doubling cubes {Qk}k with �(Qk) → 0 satisfying (2.10). As a consequence,

|f(x)| ≤ lim supr→0

β

μ(Q(x, αr))

∣∣∣∣∫Q(x,r)

f dμ

∣∣∣∣ for μ-a.e. x ∈ Rd.

Since this holds for all β > αd, we infer that

|f(x)| ≤ lim supr→0

αd

μ(Q(x, αr))

∣∣∣∣∫Q(x,r)

f dμ

∣∣∣∣ for μ-a.e. x ∈ Rd.

The analogous statements hold for balls instead of cubes.

2.5 Some standard estimates

In this section we collect some easy standard estimates which will be used later.

Lemma 2.11. Let μ be a positive measure on Rd, x ∈ Rd, and ρ, η > 0, such that

μ(B(x, r)) ≤ c0rn

for all r ≥ ρ. Then ∫|y−x|≥ρ

1

|y − x|n+ηdμ(y) ≤ c(n, η)

c0ρη

.

Proof. We split the domain of integration into annuli centered at x:∫|y−x|≥ρ

1

|y − x|n+ηdμ(y) =

∞∑k=0

∫2kρ≤|y−x|<2k+1ρ

1

|y − x|n+ηdμ(y)

≤∞∑k=0

μ(B(x, 2k+1ρ))

2(n+η)kρn+η

≤∞∑k=0

c0 (2k+1ρ)n

2(n+η)kρn+η=

c(n, η)c0ρη

. �

Lemma 2.12. Let ν be a complex measure on Rd. Let x, x′ ∈ Rd such that |x−x′| ≤12 dist(x, supp ν) =: ρ. If T is an n-dimensional SIO, we have

|Tν(x)− Tν(x′)| ≤ c|x− x′|η

ρηMR,n ν(x). (2.11)

Proof. Using the second inequality in (2.3), we get

|Tν(x)− Tν(x′)| ≤∫

|K(x, y)−K(x′, y)| d|ν|(y)

≤∫|x−y|>ρ

c|x− x′|η|x− y|n+η

d|ν|(y).

2.6. Calderon-Zygmund decomposition 57

2.6 Calderon-Zygmund decomposition

Lemma 2.14 (Calderon-Zygmund decomposition). Let μ be a Radon measure onRd. For every ν ∈ M(Rd) with compact support and every λ > 2d+1 ‖ν‖/‖μ‖, wehave:

(a) There exists a family of almost disjoint cubes {Qi}i and a function f ∈ L1(μ)such that

|ν|(Qi) >λ

2d+1μ(2Qi), (2.13)

|ν|(ηQi) ≤ λ

2d+1μ(2ηQi) for η > 2, (2.14)

ν = f μ in Rd \⋃i Qi, with |f | ≤ λ μ-a.e. (2.15)

(b) For each i, let Ri be a (6, 6n+1)-doubling cube concentric with Qi, with�(Ri) > 4�(Qi) and let wi =

χQi∑k χQk

. Then there exists a family of func-

tions ϕi with supp(ϕi) ⊂ Ri, each ϕi with constant sign, satisfying∫ϕi dμ =

∫Qi

wi dν, (2.16)

∑i

|ϕi| ≤ B λ (2.17)

(where B is some fixed constant depending only on d and n), and

‖ϕi‖L∞(μ) μ(Ri) ≤ c |ν|(Qi). (2.18)

Let us remark that if the measure ν above is complex, then instead of realfunctions ϕi with constant sign, in (b) we get functions of the form ϕi = αi hi,where αi ∈ C is constant and hi is a non-negative function.

Proof. (a) Let H be the set of those points from supp(ν) such that there existssome cube Q centered at x satisfying

|ν|(Q) >λ

2d+1μ(2Q).

For each x ∈ H , let Qx be a cube centered at x such that the preceding inequalityholds for Qx but fails for the cubes Q centered at x with �(Q) > 2�(Qx). Noticethat the condition λ > 2d+1 ‖ν‖/‖μ‖ guaranties the existence of Qx.

Now we can apply Besicovitch’s covering theorem to get an almost disjointsubfamily of cubes {Qi}i ⊂ {Qx}x which cover H and satisfy (2.13) and (2.14) byconstruction.

To prove (2.15), denote by Z the set of points y ∈ supp(ν) such that theredoes not exist a sequence of (2, 2d+1)-|ν|-doubling cubes centered at y with side


length tending to 0, so that |ν|(Z) = 0, by Lemma 2.8. By the definitions of H andZ, for every x ∈ supp(ν)\(H ∪Z), there exists a sequence of (2, 2d+1)-|ν|-doublingcubes Pk centered at x, with �(Pk) → 0, such that

|ν|(Pk) ≤ λ

2d+1μ(2Pk),

and thus

|ν|(2Pk) ≤ 2d+1|ν|(Pk) ≤ λμ(2Pk).

This implies that ν�(H ∪ Z)c is absolutely continuous with respect to μ and thatν�Hc = ν�(H ∪Z)c = fμ with |f | ≤ λ, by the Lebesgue-Radon-Nikodym theorem(see Mattila [100, p. 36-39], for instance).

(b) Assume first that the family of cubes {Qi}i is finite. Then we may supposethat this family is ordered in such a way that the sizes of the cubes Ri are non-decreasing (i.e. �(Ri+1) ≥ �(Ri)). The functions ϕi that we will construct will beof the form ϕi = αi χAi , with αi ∈ R (or αi ∈ C) and Ai ⊂ Ri. We set A1 = R1

and ϕ1 = α1 χR1 , where the constant α1 is chosen so that∫Q1

w1 dν =∫ϕ1 dμ.

Suppose that ϕ1, . . . , ϕk−1 have been constructed, satisfy (2.16) and

k−1∑i=1

|ϕi| ≤ B λ,

where B is some constant which will be fixed below.

Let Rs1 , . . . , Rsm be the subfamily of R1, . . . , Rk−1 such that Rsj ∩Rk �= ∅.As �(Rsj ) ≤ �(Rk) (because of the non-decreasing sizes of Ri), we have Rsj ⊂ 3Rk.Taking into account that for i = 1, . . . , k − 1∫

|ϕi| dμ ≤ |ν|(Qi)

by (2.16), and since Rk is (6, 6n+1)-doubling and (2.14), we get

∑j

∫|ϕsj | dμ ≤

∑j

|ν|(Qsj ) ≤ c|ν|(3Rk) ≤ cλμ(6Rk) ≤ c2λμ(Rk).

Therefore,

μ({∑

j |ϕsj | > 2c2λ})

≤ μ(Rk)

2.

So we set

Ak = Rk ∩{∑

j |ϕsj | ≤ 2c2λ},

and then μ(Ak) ≥ μ(Rk)/2.

2.6. Calderon-Zygmund decomposition 59

The constant αk is chosen so that for ϕk = αk χAkwe have

∫ϕk dμ =∫

Qkwk dν. Then we obtain

|αk| ≤ |ν|(Qk)

μ(Ak)≤ 2|ν|(12Rk)

μ(Rk)≤ c3λ

(this calculation also applies to k = 1). Thus,

|ϕk|+∑j

|ϕsj | ≤ (2c2 + c3)λ in Ak.

If we choose B = 2c2 + c3, (2.17) follows, by induction.

Now it is easy to check that (2.18) also holds. Indeed we have

‖ϕi‖L∞(μ) μ(Ri) ≤ c |αi|μ(Ai) = c

∣∣∣∣∫Qi

wi dν

∣∣∣∣ ≤ c |ν|(Qi).

Suppose now that the collection of cubes {Qi}i is not finite. For each fixed Nwe consider the family of cubes {Qi}1≤i≤N . Then, as above, we construct functionsϕN1 , . . . , ϕN

N with supp(ϕNi ) ⊂ Ri satisfying∫

ϕNi dμ =

∫Qi

wi dν,

N∑i=1

|ϕNi | ≤ B λ,

and

‖ϕNi ‖L∞(μ) μ(Ri) ≤ c |ν|(Qi).

Notice that the sign of ϕNi equals the sign of

∫wi dν and so it does not depend

on N .

Then there is a subsequence {ϕk1}k∈I1 which is convergent in the weak ∗

topology of L∞(μ) to some function ϕ1 ∈ L∞(μ). Now we can consider a subse-quence {ϕk

2}k∈I2 with I2 ⊂ I1 which is also convergent in the weak ∗ topology ofL∞(μ) to some function ϕ2 ∈ L∞(μ). In general, for each j we consider a subse-quence {ϕk

j }k∈Ij with Ij ⊂ Ij−1 that converges in the weak ∗ topology of L∞(μ)to some function ϕj ∈ L∞(μ). It is easily checked that the functions ϕj satisfy therequired properties. �

In the next lemma we prove a very useful estimate involving non-doublingcubes which relies on the idea that the mass μ which lives on non-doubling cubesmust be small.


Lemma 2.15. Let μ be a Radon measure on Rd. If Q ⊂ R are concentric cubes suchthat there are no (α, β)-doubling cubes (with β > αn) of the form αkQ, k ≥ 0,with Q ⊂ αkQ ⊂ R, and xQ stands for the center of Q, then∫

R\Q

1

|x− xQ|n dμ(x) ≤ c1μ(R)

�(R)n,

where c1 depends only on α, β, n, and d.

Proof. Let N be the least integer such that αNQ ⊂ R. For 0 ≤ k ≤ N we haveμ(αkQ) ≤ μ(αNQ)/βN−k. Then∫

αNQ\Q

1

|x− xQ|n dμ(x) ≤N∑

k=1

∫αkQ\αk−1Q

1

|x− xQ|n dμ(x)

≤ c

N∑k=1

μ(αkQ)

�(αkQ)n

≤ c

N∑k=1

βk−N μ(αNQ)

α(k−N)n �(αNQ)n

≤ cμ(αNQ)

�(αNQ)n

∞∑j=0

(αn

β

)j≤ c

μ(R)

�(R)n.

Also, since �(αNQ) ≈ �(R),∫R\αNQ

1

|x− xQ|n dμ(x) ≤ cμ(R)

�(R)n. �

2.7 Weak (1,1) boundedness of Calderon-Zygmundoperators

Theorem 2.16. Let μ be a Radon measure on Rd satisfying the growth condition(2.1). If Tμ is an n-dimensional singular integral operator which is bounded inL2(μ), then T is bounded from M(Rd) to L1,∞(μ). In particular, Tμ is of weaktype (1, 1).

The preceding result was first obtained by Nazarov, Treil and Volberg [124],although a previous proof valid only for the Cauchy transform appeared in Tolsa[154]. The proof below is from Tolsa [158].

Proof. We have to show that for any ν ∈ M(Rd) and λ > 0,

μ({x ∈ Rd : |Tεν(x)| > λ}) ≤ c

‖ν‖λ

. (2.19)

2.7. Weak (1,1) boundedness of Calderon-Zygmund operators 61

Suppose first that ν has compact support. Clearly, we may assume that λ >2d+1‖ν‖/‖μ‖.

Let {Qi}i be the almost disjoint family of cubes of Lemma 2.14. Let Ri bethe smallest (6, 6n+1)-doubling cube of the form 6kQi, k ≥ 1. Then we can writeν = g μ+ β, with

gμ = χRd\⋃i Qiν +∑i

ϕiμ

and

β =∑i

βi :=∑i

(wi ν − ϕiμ) ,

where the functions ϕi satisfy (2.16), (2.17), (2.18) and wi =χQi∑k χQk

.

By (2.13) we have

μ(⋃

i

2Qi

)≤ c

λ

∑i

|ν|(Qi) ≤ c

λ‖ν‖.

So we have to show that

μ({

x ∈ Rd \⋃i

2Qi : |Tεν(x)| > λ})

≤ c

λ‖ν‖. (2.20)

Since βi(Ri) = 0 and supp(βi) ⊂ Ri, using also that ‖βi‖ ≤ c |ν|(Qi), by Lemma2.13 we deduce that ∫

Rd\2Ri

|Tεβi| dμ ≤ c ‖βi‖ ≤ c |ν|(Qi).

Let us see now that ∫2Ri\2Qi

|Tεβi| dμ ≤ c |ν|(Qi) (2.21)

too. On the one hand, by (2.18) and using the L2(μ) boundedness of Tμ and thatRi is (6, 6

n+1)-doubling we get∫2Ri

|Tμ,εϕi| dμ ≤(∫

2Ri

|Tμ,εϕi|2 dμ)1/2

μ(2Ri)1/2

≤ c

(∫|ϕi|2 dμ

)1/2μ(Ri)

1/2 ≤ c |ν|(Qi).

On the other hand, since supp(wiν) ⊂ Qi, if x ∈ 2Ri \ 2Qi, then |Tε(wiν)(x)| ≤c |ν|(Qi)/|x− xQi |n, and so∫

2Ri\2Qi

|Tε(wi ν)| dμ ≤ c |ν|(Qi)

∫2Ri\2Qi

1

|x− xQi |ndμ(x).


By Lemma 2.15, the integral on the right-hand side is bounded by some constantindependent of Qi and Ri, since there are no (6, 6n+1)-doubling cubes of the form6kQi between 6Qi and Ri. Therefore, (2.21) holds.

Then we have∫Rd\⋃k 2Qk

|Tεβ| dμ ≤∑i

∫Rd\2Qi

|Tεβi| dμ

≤ c∑i

|ν|(Qi) ≤ c ‖ν‖.

Therefore,

μ({

x ∈ Rd \⋃i

2Qi : |Tεβ(x)| > λ/2})

≤ c

λ‖ν‖. (2.22)

The corresponding integral for the function g is easier to estimate. Takinginto account that |g| ≤ c λ, we get

μ({

x ∈ Rd \⋃i

2Qi : |Tμ,εg(x)| > λ/2})

≤ c

λ2

∫|g|2 dμ ≤ c

λ

∫|g| dμ. (2.23)

Also, we have ∫|g| dμ ≤ |ν|

(Rd \⋃i

Qi

)+∑i

∫|ϕi| dμ

≤ ‖ν‖+∑i

|ν|(Qi) ≤ c ‖ν‖.

Now, from (2.22) and (2.23) we get (2.20).

Suppose now that ν is not compactly supported while μ has compact support.Let N0 be such that suppμ ⊂ B(0, N0), and for some N > N0, let νN = χB(0,N) ν.Then, for x ∈ supp(μ),

|Tε(ν − νN )(x)| ≤ c|ν|(Rd \B(0, N))

N −N0≤ c

‖ν‖N −N0

≤ λ

2

if N − N0 > 2c‖ν‖/λ. Thus, for such N , since both μ and νN have compactsupport,

μ({x ∈ Rd : |Tεν(x)| > λ}) ≤ μ

({x ∈ Rd : |TενN (x)| > λ/2})≤ c

‖νN‖λ

≤ c‖ν‖λ

.

On the other hand, if μ is not compactly supported, then we take μN =μ�B(0, N). Clearly, if Tμ is bounded in L2(μ), then TμN is bounded in L2(μN ),and so

μN

({x ∈ Rd : |Tεν(x)| > λ}) ≤ c‖ν‖λ

uniformly on N , and then (2.19) follows in full generality. �

2.8. Cotlar’s inequality 63

Remark 2.17. In fact, in the preceding theorem we showed that, for any givenε > 0, if Tμ,ε is bounded in L2(μ), then Tε is bounded from M(Rd) to L1,∞(μ),with a bound depending only on the growth constant c0 and on ‖Tμ,ε‖L2(μ)→L2(μ).

2.8 Cotlar’s inequality

This inequality involves a new maximal operator which we proceed to define. Givenν ∈ M(Rd), we set

Mμν(x) = sup

{ |ν|(B(x, r))

μ(B(x, r)): r > 0, μ(B(x, 3r)) ≤ 3d+1μ(B(x, r))

}.

As usual, for f ∈ L1loc(μ) we set Mμf ≡ Mμ(f μ). Notice that Mμν ≤ Mμν, and

so Mμ is bounded in Lp(μ), and from M(Rd) to L1,∞(μ) (this can be deduced

from the fact that Mμν ≤ Nμ too).If T is a SIO, the maximal operator T∗ is

T∗ν(x) = supε>0

|Tεν(x)| for ν ∈ M(Rd), x ∈ Rd,

and the δ-truncated maximal operator T∗,δ is

T∗,δν(x) = supε>δ

|Tεν(x)| for ν ∈ M(Rd), x ∈ Rd.

We also set Tμ,∗ f ≡ T∗ (f μ) and Tμ,∗,δ f ≡ T∗,δ (f μ) for f ∈ L1loc(μ).

Theorem 2.18 (Cotlar’s inequality). Let μ be a positive Radon measure on Rd andlet T be an n-dimensional SIO. Let 0 < s ≤ 1, and 0 < δ ≤ ε. Suppose that forsome fixed x ∈ Rd,

μ(B(x, r)) ≤ c0 rn for r ≥ ε (2.24)

and that Tδ is bounded from M(Rd) to L1,∞(μ). Then we have

|Tε ν(x)| ≤ csMμ(|Tδν|s)(x)1/s + cs,T Mμν(x), for ν ∈ M(Rd). (2.25)

Thus, if μ has growth of degree n, then for all x ∈ Rd,

T∗,δ ν(x) ≤ csMμ(|Tδν|s)(x)1/s + cs,T Mμν(x), for ν ∈ M(Rd). (2.26)

The constant cs depends only on the constant c0 in (2.24), s, n, and d, and cs,T =c(1 + ‖Tδ‖M(Rd)→L1,∞(μ)), with c depending only on c0, s, n and d.

If we assume that there exists Tν defined in a reasonable sense (for instance,as a principal value, or as some kind of a weak limit), then we get the more classicalversion of Cotlar’s inequality

T∗ ν(x) ≤ cs

(Mμ(|Tν|s)(x)1/s +Mμν(x)

).


Cotlar’s inequality with non-doubling measures is due to Nazarov, Treil andVolberg [124], although not in the form stated above, which is from Tolsa [153].

To prove Theorem 2.18 we will need some lemmas. In the first one we reviewthe well-known Kolmogorov’s inequality:

Lemma 2.19. Let μ be a positive Radon measure on Rd and f : Rd −→ C a Borelfunction from L1,∞(μ). Then for 0 < s < 1 and for any μ-measurable set A ⊂ Rd

with μ(A) < ∞, (1

μ(A)

∫A

|f |sdμ)1/s

≤ (1 − s)−1/s ‖f‖L1,∞(μ)

μ(A).

Proof. We have∫A

|f |sdμ =

∫ ∞

0

s λs−1 μ({x ∈ A : |f(x)| > λ}) dλ

≤∫ ∞

0

s λs−1 min(μ(A), λ−1 ‖f‖L1,∞(μ)

)dλ.

Letting a = ‖f‖L1,∞(μ)/μ(A), the last integral equals∫ a

0

s λs−1 μ(A) dλ +

∫ ∞

a

s λs−2 ‖f‖L1,∞(μ) dλ = μ(A) as +s ‖f‖L1,∞(μ) a

s−1

1− s

=1

1− sμ(A)1−s ‖f‖sL1,∞(μ). �

Also, we will need the following result (notice the resemblances to Lemma2.15).

Lemma 2.20. Let x ∈ Rd, 0 < r < R, with R = 3Nr, and take β > 3n. If

β μ(B(x, 3kr)) ≤ μ(B(x, 3k+1r)) for k = 1, . . . , N − 1,

then we have

|TRν(x) − Trν(x)| ≤ c(β)μ(B(x,R))

RnMμν(x),

for all ν ∈ M(Rd).

Proof. We set Bk = B(x, 3kr). Then

|TRν(x)− Trν(x)| =∣∣∣∣∣∫r<|y−x|≤3Nr

K(x, y) dν(y)

∣∣∣∣∣�

N∑k=1

∫3k−1r<|y−x|≤3kr

1

|y − x|n d|ν|(y)

�N∑

k=1

|ν|(Bk)

(3kr)n�

N∑k=1

μ(Bk)

(3kr)nMμν(x).


We have

μ(B1) ≤ 1

βμ(B2) ≤ · · · ≤ 1

βN−1μ(BN ).

Thusμ(B1)

(3r)n≤ 3n

β

μ(B2)

(32r)n≤ · · · ≤

(3n

β

)N−1μ(BN )

(3Nr)n.

Since3n

β< 1, we get

N∑k=1

μ(Bk)

(3kr)n≤ c

μ(BN )

(3Nr)n,

and the lemma follows. �We are ready now for the proof of Cotlar’s inequality. This will follow by

combining Lemma 2.20 with the usual arguments for doubling measures.

Proof of Theorem 2.18. Clearly, it is enough to prove the estimate (2.25). Let ε > δand x ∈ Rd. From the growth condition (2.24) we infer that there exists somem ≥ 1 such that

0 < μ(B(x, 3mε)) ≤ 3d+1μ(B(x, 3m−1ε)) (2.27)

(see Section 2.4). We assume that m is the least integer ≥ 1 such that (2.27) holds.Set ε′ = 3mε. By Lemma 2.20,

|Tεν(x) − Tε′/3ν(x)| ≤ cMμν(x).

Also, it is straightforward to check that |Tε′/3ν(x) − Tε′ν(x)| ≤ cMμν(x). There-fore,

|Tεν(x) − Tε′ν(x)| ≤ cMμν(x).

Thus it only remains to show that

|Tε′ν(x)| ≤ csMμ(|Tδν|s)(x)1/s + c′s,TMμν(x). (2.28)

Sinceμ(B(x, ε′)) ≤ 3d+1μ(B(x, ε′/3)), (2.29)

we can apply the usual “doubling argument” to prove (2.28). To this end, we set

ν1 = χB(x,ε′)ν, dν2 = ν − dν1.

For y ∈ B(x, ε′/3), since ε′ > 3δ we have Tε′ν2(x) = Tδν2(x) = Tν2(x) andTδν2(y) = Tν2(y). Using the growth condition (2.24) and the gradient conditionfor the kernel of T , arguing as in Lemma 2.12, it follows that

|Tδν2(y)− Tδν2(x)| = |Tν2(y)− Tν2(x)| ≤ c supr≥ε

|ν2|(B(x, r))

rn≤ c2Mμν(x).


Therefore,

|Tε′ν(x)| = |Tδν2(x)| ≤ |Tδν2(y)|+ c2Mμν(x)

≤ |Tδν1(y)|+ |Tδν(y)|+ c2Mμν(x). (2.30)

Assume first s = 1. If Tε′ν(x) �= 0, let 0 < λ < |Tε′ν(x)|. For y ∈ B(x, ε′/3),by (2.30) either c2Mμν(x) > λ/3 or |Tδν(y)| > λ/3 or |Tδν1(y)| > λ/3. Therefore,either

λ < 3c2Mμν(x),

or

B(x, ε′/3) = {y ∈ B(x, ε′/3) : |Tδν(y)| > λ/3}∪ {y ∈ B(x, ε′/3) : |Tδν1(y)| > λ/3}.

We have

μ({y ∈ B(x, ε′/3) : |Tδν(y)| > λ/3}) ≤ 3

λ

∫B(x,ε′/3)

|Tδν|dμ

≤ 3

λμ(B(x, ε′/3)) Mμ(Tδν)(x),

and by the boundedness of Tδ from M(Rd) to L1,∞(μ) and (2.29), setting cT =‖Tδ‖M(Rd)→L1,∞(μ),

μ({y ∈ B(x, ε′/3) : |Tδν1(y)| > λ/3}) ≤ cT

‖ν1‖λ

= cT|ν|(B(x, ε′))

λ

≤ c cTμ(B(x, ε′/3))

λMμν(x).

Thus,

μ(B(x, ε′/3)) ≤ μ(B(x, ε′/3))λ

(3 Mμ(Tδν)(x) + c cT Mμν(x)

).

In any case we obtain λ < 3Mμ(Tδν)(x) + c(1 + cT )Mμν(x). Since this holdsfor 0 < λ < |Tε′ν(x)|, (2.28) follows when s = 1.

Assume now 0 < s < 1. From (2.30) we get

|Tε′ν(x)|s ≤ |Tδν1(y)|s + |Tδν(y)|s + cMμν(x)s.

Integrating with respect to μ for y ∈ B(x, ε′/3), dividing by μ(B(x, ε′/3)) andraising to the power 1/s we obtain

|Tε′ν(x)| ≤ cs

⎡⎣( 1

μ(B(x, ε′/3))

∫B(x,ε′/3)

|Tδν1|sdμ)1/s

+

(1

μ(B(x, ε′/3))

∫B(x,ε′/3)

|Tδν|sdμ)1/s

+Mμν(x)

⎤⎦ . (2.31)


Recalling (2.29), the second term on the right-hand side of (2.31) can be esti-

mated by Mμ(|Tδν|s)(x)1/s. On the other hand, the first term is estimated usingKolmogorov’s inequality, the boundedness of Tδ from M(Rd) to L1,∞(μ), and(2.29):(

1

μ(B(x, ε′/3))

∫B(x,ε′/3)

|Tδν1|sdμ)1/s

�‖Tδν1‖L1,∞(μ)

μ(B(x, ε′/3))

≤ cT ‖ν1‖μ(B(x, ε′/3))

≤ c cTMμν(x).

Now (2.28) follows. �

As in the classical doubling case, a direct consequence of Cotlar’s inequalityand Theorem 2.16 is the following result.

Theorem 2.21. Let μ be a Radon measure on Rd of degree n. If Tμ is an n-dimensional SIO bounded in L2(μ), then Tμ,∗ is bounded in Lp(μ), for p ∈ (1,∞),and T∗ is bounded from M(Rd) to L1,∞(μ).

Proof. By Theorem 2.16, interpolation, and duality, Tμ is bounded in Lp(μ), p ∈(1,∞), and from M(Rd) to L1,∞(μ). Then by Cotlar’s inequality (2.26) withs = 1 it is clear that Tμ,∗,δ is bounded in Lp(μ), p ∈ (1,∞), uniformly on δ > 0.Hence, by monotone convergence, Tμ,∗ is also bounded in Lp(μ), p ∈ (1,∞). Theboundedness of T∗ from M(Rd) to L1,∞(μ), as in the classical doubling case,requires some additional work. By monotone convergence, it is enough to showthat

μ({x : T∗,δ ν(x) > λ}) ≤ ‖ν‖

λ(2.32)

uniformly on λ, δ > 0. By Cotlar’s inequality (2.26) with s = 1/2, for λ > 0 wehave

μ({x : T∗,δν(x) > λ}) ≤ μ

({x : Mμν(x) >

λ

2c1/2,T

})+ μ({

x : Mμ(|Tδν|1/2)(x)2 >λ

2c1/2

}). (2.33)

The first term on the right-hand side of (2.33) satisfies

μ({

x : Mμν(x) >λ

2c1/2,T

})≤ c

‖ν‖λ

,

by the boundedness of Mμ from M(Rd) to L1,∞(μ). To estimate the second termon the right-hand side of (2.33) we will use the non-centered restricted maximal


Hardy-Littlewood operator Nμ introduced above. Obviously,

μ({

x : Mμ(|Tδν|1/2)(x)2 >λ

2c1/2

})≤ μ({

x : Nμ(|Tδν|1/2)(x) > λ1/2

(2c1/2)1/2

}). (2.34)

Recall that, by the doubling condition on the balls B(y, r), the operator Nμ isbounded from M(Rd) to L1,∞(μ). In fact, for any σ ∈ M(Rd), the followingsharper condition holds:

μ({x : Nμσ(x) > λ}) ≤ c

1

λ|σ|({x : Nμσ(x) > λ}). (2.35)

Let us remark that an estimate such as this one does not hold for the centeredoperator Mμ, in general.

Now take

A ={x : Nμ(|Tδν|1/2)(x) > λ1/2

(2c1/2)1/2

}.

Applying (2.35) to σ = |Tδν|1/2μ and using Kolmogorov’s inequality, we obtain

μ(A) ≤ c1

λ1/2

∫A

|Tδν|1/2dμ

≤ c1

λ1/2μ(A)1/2 ‖Tδν‖1/2L1,∞(μ) ≤ c

1

λ1/2μ(A)1/2 ‖ν‖1/2.

Therefore, μ(A) ≤ c ‖ν‖/λ, and by (2.34),

μ({

x : Mμ(|Tδν|1/2)(x)2 >λ

2c1/2

})≤ c

‖ν‖λ

,

and so (2.32) holds. �

2.9 The good lambda method

In this section we will prove the following theorem, which will be an essentialingredient for the proof of the T 1 theorem for the Cauchy transform in nextchapter.

Theorem 2.22. Let μ be a Radon measure on Rd of degree n and Tμ an n-dimensional SIO. Let β > 0 be big enough (depending only on d) and let θ > 0.Suppose that for every (2, β)-doubling cube Q there exists some subset GQ ⊂ Q,with μ(GQ) ≥ θ μ(Q), such that T∗ is bounded from M(Rd) to L1,∞(μ�GQ), withnorm bounded uniformly on Q. Then Tμ is bounded in Lp(μ), for 1 < p < ∞, withits norm depending on p and on the preceding constants.

2.9. The good lambda method 69

Roughly speaking, this result asserts that if for every doubling square thereexists a big piece (in terms of μ) where T∗ is bounded from M(Rd) to L1,∞,then Tμ is bounded in Lp(μ), for 1 < p < ∞. This will be proved by a means of agood lambda inequality, which is a well-known technique in the classical Calderon-Zygmund theory that extends quite well to the non-doubling setting (although onehas to take care of some additional subtleties).

To prove Theorem 2.22 we will use a Whitney’s decomposition of some openset. In the next lemma we show the precise version of the required decomposition.

Lemma 2.23. If Ω ⊂ Rd is open, Ω �= Rd, then Ω can be decomposed as

Ω =⋃i∈I

Qi,

where Qi, i ∈ I, are closed dyadic cubes with disjoint interiors such that for someconstants R > 20 and D0 ≥ 1 the following holds:

(i) 10Qi ⊂ Ω for each i ∈ I.

(ii) RQi ∩ Ωc �= ∅ for each i ∈ I.

(iii) For each cube Qi, there are at most D0 cubes Qj such that 10Qi∩10Qj �= ∅.Further, for such cubes Qi, Qj, we have �(Qi) ≈ �(Qj).

Moreover, if μ is a positive Radon measure on Rd and μ(Ω) < +∞, the subfamilyof (10, 2D0)-doubling cubes {Qj}j∈S from {Qi}i∈I satisfies

μ

( ⋃j∈S

Qj

)≥ 1

2μ(Ω). (2.36)

The cubes Qi from the above decomposition are usually calledWhitney cubes.

Proof. Whitney’s decomposition in dyadic cubes satisfying (i), (ii) and (iii) is awell-known result. See for example Stein [148, pp. 167-169].

To prove that the subfamily of (10, 2D0)-doubling cubes {Qj}i∈S from{Qi}i∈I satisfies (2.36), observe that

μ(Qj) <1

2D0μ(10Qj) if j ∈ I \ S.

Since ∑j∈I

χ10Qj ≤ D0χΩ,

we deduce that ∑j∈I\S

μ(Qj) ≤ 1

2D0

∑j∈I

μ(10Qj) ≤ 1

2μ(Ω).

Thus,

μ

( ⋃j∈S

Qj

)= μ(Ω)− μ

( ⋃j∈I\S

Qj

)≥ 1

2μ(Ω). �


For technical reasons, in the proof of Theorem 2.22 it will be more convenientto use the centered maximal Hardy-Littlewood operator associated with cubesinstead of balls. For ν ∈ M(Rd), we define

MQμ ν(x) = sup

r>0

|ν|(Q(x, r))

μ(Q(x, r)).

As usual, we set MQμ (f) ≡ MQ

μ (f μ). From the Besicovitch covering theorem(which also holds for cubes), arguing as in the proof of Theorem 2.5, it followsthat MQ

μ is bounded from M(Rd) to L1,∞(μ), and in Lp(μ), for 1 < p ≤ ∞. If μhas growth of degree n, then it turns out that

MR,nν(x) ≤ cMQμ ν(x),

analogously to (2.8).

Proof of Theorem 2.22. We will assume that for every (10, 2D0)-doubling cube Q(where D0 is the constant in Lemma 2.23) there exists some subset GQ ⊂ Q, withμ(GQ) ≥ θ μ(Q), such that T∗ is bounded from M(Rd) into L1,∞(μ�GQ), withnorm bounded uniformly on Q. This is a weaker assumption than the one in thetheorem, since (10, 2D0)-doubling cubes are (2, D0)-doubling.

As mentioned above, our arguments rely on a good λ inequality. More pre-cisely, we will prove that for all ε > 0 there exists δ = δ(ε) > 0 such that

μ({

x : Tμ,∗f(x) > (1 + ε)λ, MQμ f(x) ≤ δλ

}) ≤ (1− θ

4

)μ({

x : Tμ,∗f(x) > λ})

(2.37)for every compactly supported f ∈ L1(μ). We will show below that the Lp(μ)boundedness of Tμ,∗ follows from this estimate.

To prove (2.37) consider a Whitney decomposition of the open set

Ωλ ={x : Tμ,∗f(x) > λ

}into a family of dyadic cubes {Qi}i∈I with disjoint interior such that, for eachi ∈ I, 10Qi ⊂ Ωλ and RQi ∩ Ωc

λ �= ∅, as in Lemma 2.23. To this end, notice thatthe fact that f ∈ L1(μ) is compactly supported implies that Ωλ is a bounded setand thus has finite measure.

Consider a cubeQi, i ∈ I. We claim that if x ∈ Qi satisfies Tμ,∗f(x) > (1+ε)λand MQ

μ f(x) ≤ δλ, thenTμ,∗(χ2Qif)(x) > ελ/2, (2.38)

assuming that δ is small enough. To show this, take z ∈ RQi \ Ωλ, so thatTμ,∗f(z) ≤ λ. Since x, z ∈ RQi ⊂ B(z, d1/2R�(Qi)) =: Bi, using the linear growthof μ, by standard arguments (such as the ones in Lemma 2.12) it follows easilythat ∣∣Tμ,∗(χRd\2Bi

f)(x) − Tμ,∗(χRd\2Bif)(z)∣∣ ≤ cMQ

μ f(x).

2.9. The good lambda method 71

Since 2Bi is a ball centered at z �∈ Ωλ, by definition, we have

Tμ,∗(χRd\2Bif)(z) ≤ Tμ,∗f(z) ≤ λ.

Also, since x ∈ Qi, it is easy to check that Tμ,∗(χ2Bi\2Qif)(x) ≤ cMQ

μ f(x).Therefore,

Tμ,∗(χRd\2Qif)(x) ≤ Tμ,∗(χ2Bi\2Qi

f)(x) + Tμ,∗(χRd\2Bif)(x)

≤ Tμ,∗(χ2Bi\2Qif)(x) + Tμ,∗(χRd\2Bi

f)(z)

+∣∣Tμ,∗(χRd\2Bi

f)(x)− Tμ,∗(χRd\2Bif)(z)∣∣

≤ cMQμ f(x) + λ+ cMQ

μ f(x) ≤ (1 + c10δ)λ,

and thus

Tμ,∗(χ2Qif)(x) ≥ Tμ,∗f(x)− Tμ,∗(χRd\2Qif)(x) > (1 + ε)λ− (1 + c10δ)λ.

So the claim follows if we choose δ such that c10δ ≤ ε/2.Let {Qi}i∈S ⊂ {Qi}i∈I be the subfamily of those cubes such that

μ(10Qi) ≤ 2D0μ(Qi), if i ∈ S.

Notice that, by Lemma 2.23,

μ

( ⋃i∈S

Qi

)≥ 1

2μ(Ωλ). (2.39)

Recall now that, by the assumptions in the theorem, taking β ≥ 2D0, for eachi ∈ S there exists a subset Gi ⊂ Qi with μ(Gi) ≥ θ μ(Qi) such that T∗ is boundedfrom M(Rd) to L1,∞(μ�Gi). Together with (2.38) this implies that

μ({x ∈ Gi : Tμ,∗f(x) > (1 + ε)λ, MQ

μ f(x) ≤ δλ})≤ μ({x ∈ Gi : Tμ,∗(χ2Qif)(x) > ελ/2}) ≤ c

ελ‖χ2Qif‖L1(μ).

Notice that, if Qi contains some point x such that MQμ f(x) ≤ δλ, then

‖χ2Qif‖L1(μ) ≤∫Q(x,2(Qi))

|f | dμ ≤ μ(Q(x, 2�(Qi)))MQμ f(x)

≤ μ(5Qi)MQμ f(x) ≤ 2D0δλμ(Qi).

Thus,

μ({x ∈ Gi : Tμ,∗f(x) > (1 + ε)λ, MQ

μ f(x) ≤ δλ}) ≤ c11δ

εμ(Qi).


Using the preceding estimate for all the cubes Qi, i ∈ S ⊂ I, and the factthat μ(Gi) ≥ θ μ(Qi), we get

μ({

x : Tμ,∗f(x) > (1 + ε)λ, MQμ f(x) ≤ δλ

})≤∑

i∈I\Sμ(Qi) +

∑i∈S

μ(Qi \Gi)

+∑i∈S

μ({

x ∈ Gi : Tμ,∗f(x) > (1 + ε)λ, MQμ f(x) ≤ δλ

})≤∑

i∈I\Sμ(Qi) +

∑i∈S

(1− θ)μ(Qi) +∑i∈S

c11δ

εμ(Qi)

= μ(Ωλ)− θ∑i∈S

μ(Qi) +∑i∈S

c11δ

εμ(Qi).

By (2.39), we obtain

μ({

x : Tμ,∗f(x) > (1 + ε)λ, MQμ f(x) ≤ δλ

}) ≤ (1− θ

2

)μ(Ωλ) +

c11δ

εμ(Ωλ).

Therefore, if we choose δ = δ(ε) small enough, the right-hand side does not exceed(1− θ

4

)μ(Ωλ), and so (2.37) follows.

It remains to check that (2.37) implies the Lp(μ) boundedness of Tμ,∗. Thearguments are standard. Take f ∈ Lp(μ) with compact support. Then we have

μ({

x : Tμ,∗f(x) > (1 + ε)λ}) ≤ μ

({x : Tμ,∗f(x) > (1 + ε)λ, MQ

μ f(x) ≤ δλ})

+ μ({

x : MQμ f(x) > δλ

})≤(1− θ

4

)μ({

x : Tμ,∗f(x) > λ})

+ μ({

x : MQμ f(x) > δλ

}).

Multiplying by pλp−1 this inequality and integrating with respect to λ in (0,∞),we obtain

1

(1 + ε)p‖Tμ,∗f‖pLp(μ) ≤

(1− θ

4

)‖Tμ,∗f‖pLp(μ) +

1

δp‖MQ

μ f‖pLp(μ). (2.40)

Thus, assuming that ‖Tμ,∗f‖Lp(μ) < ∞ and taking ε > 0 so that

(1 + ε)p(1− θ

4

)=(1− θ

8

)(recall that θ does not depend on ε, unlike δ), we deduce that

‖Tμ,∗f‖Lp(μ) ≤ c(θ, δ) ‖MQμ f‖Lp(μ) ≤ c(θ, δ) ‖f‖Lp(μ).

2.10. Historical remarks and further results 73

If we do not assume that ‖Tμ,∗f‖Lp(μ) < ∞, then notice that, for any m ≥ 1, the

function gm = inf(m,Tμ,∗f

)also satisfies the inequality (2.37). Moreover, since

f ∈ Lp(μ) has compact support, it is easy to check that gm ∈ Lp(μ) (since gm isbounded and |gm(x)| ≤ c(gm)/|x|n for x → ∞). Thus arguing as above, we get‖gm‖Lp(μ) ≤ c(θ, δ) ‖f‖Lp(μ), and letting m → ∞, we are done. �

Remark 2.24. With minor modifications in the arguments above, one can askthe assumption in Theorem 2.22 to hold only for (2, β′)-doubling cubes or ballscentered on supp(μ), instead of all the (2, β)-doubling cubes.

Observe also that from Theorems 2.21 and 2.22 one easily infers that theboundedness of T from M(Rd) to L1,∞(μ) implies the L2(μ) boundedness of Tμ.This has the following, a priori non-trivial, consequence.

Proposition 2.25. Let μ and σ be Borel measures of degree n in Rd, and let Tbe an n-dimensional SIO such that Tμ is bounded in L2(μ) and Tσ is bounded inL2(σ). Then Tμ+σ is bounded in L2(μ+ σ).

Proof. It is clear that if both μ and σ are measures with growth of degree n, thenso is μ + σ (possibly with a worse constant). By Theorem 2.21, T∗ is boundedfrom M(Rd) to L1,∞(μ) and from M(Rd) to L1,∞(σ). Trivially, this implies thatT∗ is bounded from M(Rd) to L1,∞(μ + σ), which is equivalent to the L2(μ+ σ)boundedness of Tμ+σ, as explained above. �

2.10 Historical remarks and further results

As mentioned in the Introduction, for a long time it was believed that the naturalframework to study singular integral operators is the one of homogenous spaces,where the underlying measure μ is assumed to be doubling. The doubling as-sumption is basic in many parts of the classical Calderon-Zygmund theory. This“classical doubling theory” is covered in many books on harmonic analysis. SeeCoifman and Weiss [18], Duoandikoetxea [36], Grafakos [57], Journe [78], Stein[148], Stein [149], or Torchinsky [170], for example. A main motivation to developthe Calderon-Zygmund theory for non-doubling measures in Rd arises from theapplication to the study of analytic capacity.

There are other important works in the area of non-homogeneous Calderon-Zygmund theory which appeared more or less simultaneously to the contributionsof Nazarov, Treil, Volberg and Tolsa explained in this chapter. In particular, it isworth mentioning that in 1998 David solved Vitushkin’s conjecture for sets withfinite length by proving a suitable Tb theorem which did not assume any doublingcondition. This result relied on a previous joint work with Mattila [27] (althoughit was published later).

Concerning the good lambda theorem explained in Section 2.9, in some earlierworks Sawyer ([144], [145]) already noticed that the good lambda inequalities aregood tools to deal with non-doubling weights and fractional integrals, by using


the finite superposition of the Whitney cubes. In the Ahlfors-David regular case,David [20] used a good lambda inequality to show that for the Lp(μ) boundednessof a singular integral operator Tμ it suffices that each cube Q centered at supp(μ)contains a big piece GQ ⊂ Q where Tμ�GQ

is bounded in L2(μ�GQ). Theorem2.22 is a variant of this fact, adapted to the non-doubling situation. On the otherhand, as far as I know, the use of the boundedness from M(Rd) to L1,∞(μ�GQ) toprove the good lambda inequality is rather new. It has the advantage of avoidingduality (that is used in David [20]) and thus it can be extended more easily tooperators which may be non-linear. Other related good lambda inequalities usefulfor non-doubling measures appeared in the works by Tolsa [154] and Verdera [175],in connection with the T 1 theorem for the Cauchy transform.

Most of the results from the non-homogeneous Calderon-Zygmund theoryfrom this chapter also hold in more general settings. Indeed, Nazarov, Treil andVolberg proved in [124] that if μ is a measure with growth of degree n in a gen-eral separable metric space, the L2(μ) boundedness of n-dimensional Calderon-Zygmund operators implies their boundedness from L1(μ) to L1,∞(μ). Further,they showed that Cotlar’s inequality remains valid in this situation too.

Other results on non-homogeneous Calderon-Zygmund theory have been ex-tended to the so-called geometrically doubling spaces. A metric space X is calledgeometrically doubling if every ball B(x, r) ⊂ X does not contain more than Ndisjoint balls of radius r/2, with N independent of x,R. The polynomial growthof the measure μ in Rd is replaced by an upper doubling condition for μ in X :one assumes that there exists a function λ(x, r), non-decreasing in r, such thatμ(B(x, r)) ≤ cλ(x, r) and λ(x, 2r) ≤ cλ(x, r) for any point x ∈ X and all r > 0.In this context, one considers singular integral operators whose kernels satisfysome size and smoothness conditions involving the function λ. It turns out thatthe L2(μ) boundedness of these operators implies that they are also bounded inLp(μ), for 1 < p < ∞, and of weak (1, 1) type. Analogously, by a Cotlar typeinequality it follows that the associated maximal operators are bounded in thesame range of spaces. Some of the additional difficulties that appear in the set-ting of the geometrically doubling spaces are due to the fact that the Besicovitchcovering theorem may fail. These difficulties can be overcome by using alternativecovering arguments. See the works of Anh and Duong [3] and Hytonen, Liu, Yangand Yang [65] for the detailed results and techniques.

Chapter 3

The Cauchy transform andMenger curvature

3.1 Introduction

Recall that the Cauchy transform is the 1-dimensional singular integral in C as-sociated with the kernel

K(x, y) :=1

y − x, x, y ∈ C.

So the Cauchy transform of a Radon measure μ in C equals

Cμ(x) :=∫

1

y − xdμ(y), x ∈ C \ supp(μ).

When μ is fixed, we set Cμ(f) = C(f μ)).

Of course, the results and techniques of Calderon-Zygmund theory explainedin Chapter 2 apply to the Cauchy transform (with the parameters n = 1, d = 2).Moreover, because of the relationship of the Cauchy kernel with Menger curvature,discovered by Melnikov [113], some results are easier to prove for the Cauchytransform than for general SIO’s.

In this chapter we will explain in detail the relationship between the Cauchytransform and Menger curvature, and we will exploit this connection to give arather simple proof of the T 1 theorem for the Cauchy singular integral operator,i.e. for Cμ. As a byproduct, we will obtain a proof of the L2 boundedness of theCauchy transform on Lipschitz graphs and on regular curves, with respect to arclength.

, , OI 10.1007/978-3- - -6_ ,

© Springer



755

76 Chapter 3. The Cauchy transform and Menger curvature

3.2 The curvature of a measure

Given three pairwise different points x, y, z ∈ C, their Menger curvature is

c(x, y, z) =1

R(x, y, z),

where R(x, y, z) is the radius of the circumference passing through x, y, z (withR(x, y, z) = ∞, c(x, y, z) = 0 if x, y, z lie on a same line). If two among the pointsx, y, z coincide, for convenience, we set c(x, y, z) = 0.

Proposition 3.1. Let x, y, z ∈ C be pairwise different. Then

c(x, y, z) =2 dist(x, Ly,z)

|x− y| |x− z| =4S(x, y, z)

|x− y| |x− z| |y − z| =2 sin xyz

|x− z| , (3.1)

where Ly,z is the line passing through y, z, S(x, y, z) stands for the area of thetriangle of vertices x, y, z, and xyz is the angle in the same triangle opposite tothe side xz.

The preceding identities follow by elementary geometry. However, for thereader’s convenience we show the detailed arguments.

�

x z

y

y0 Γ

Figure 3.1: The circumference Γ and the points x, y, y0, z.

Proof. Denote by Γ the circumference passing through x, y, z, and by R its radius.Let y0 ∈ Γ be the point opposite to z in Γ, so that the segment with end pointsy0, z is a diameter of Γ. See Figure 3.1. In this situation, if y and y0 are in the samehalf plane determined by the line Lx,z, then xyz = xy0z. If they are in differenthalf planes, then xyz = π − xy0z. So, in any case,

sin xyz = sin xy0z.

On the other hand, y0xz is a right angle because y0, z are the end points of adiameter of Γ. Thus

sin xy0z =|x− z||y0 − z| =

|x− z|2R

.

3.2. The curvature of a measure 77

So

c(x, y, z) =1

R=

2 sin xy0z

|x− z| =2 sin xyz

|x− z| .

Since dist(x, Ly,z) = |x− y| sin xyz, we infer that


|x− y| |x− z| .

Finally, from the fact that S(x, y, z) = 12 |y − z| dist(x, Ly,z), we get

c(x, y, z) =4S(x, y, z)

|x− y| |x− z| |y − z| . �

The relationship between Menger curvature, the Cauchy transform and an-alytic capacity originates from the following striking identity, discovered by Mel-nikov:

Proposition 3.2. Let z1, z2, z3 ∈ C be pairwise different. We have

c(z1, z2, z3)2 =∑s∈S3

1

(zs2 − zs1)(zs3 − zs1), (3.2)

where S3 is the group of permutations of three elements.

Proof. Setting a = z2 − z1, b = z3 − z1, the sum on the right-hand side of (3.2)equals

2 Re

(1

ab+

1

(a− b)a+

1

b(b− a)

)= 4

|a|2|b|2 − Re(ab)2

|a|2|b|2|b− a|2 .

Since

|a|2|b|2 − Re(ab)2 = |a|2|b|2(1− cos2 z2z1z3)

= |a|2|b|2 sin2 z2z1z3 = 4S(z1, z2, z3)2,

the lemma follows. �For a positive Radon measure μ, we write

c2μ(x) =

∫ ∫c(x, y, z)2 dμ(y)dμ(z),

and we define the curvature of μ as

c2(μ) =

∫c2μ(x) dμ(x) =

∫ ∫ ∫c(x, y, z)2 dμ(x)dμ(y)dμ(z). (3.3)

The notion of curvature of a measure was introduced by Melnikov [113]. In thispaper he studied a discrete version of analytic capacity and he proved a lower


bound for analytic capacity involving the curvature of measures. See Section 4.10from the next chapter for more details.

Let us see some examples. Suppose first that μ is the arc length measure ona circumference Γ of radius r. Then R(x, y, z) = r for all x, y, z ∈ Γ, and thus

c2μ(x) =1

r2μ(Γ)2 = 4π2

and c2(μ) = 8π3r.Consider now the case where μ is the arc length on a segment L. Clearly,

c2μ(x) = 0 for all x ∈ L, and so c2(μ) = 0. Let us calculate c2μ(x) for x �∈ L. Bytranslation and rotation invariance, we may assume that L is contained in the realaxis and x belongs to the imaginary upper half axis. So L = [a, b], with a, b ∈ R,and x = (0, h). Then

c2μ(x) =

∫∫y,z∈L

(2 d(x, L)

|x− y| |x− z|)2

dH1L(y) dH1

L(z)

= 4h2

(∫L

1

|x− y|2 dH1L(y)

)2.

The last integral equals∫ b

a

1

t2 + h2dt =

1

h

(arctan

b

h− arctan

a

h

).

Thus,

c2μ(x) = 4

(arctan

b

h− arctan

a

h

)2.

Notice that if a < 0 < b, then c2μ(x) tends to 4π2 as h → 0. So the function c2μ(·)is not continuous in C. On the other hand, the preceding identity also shows thatc2μ(x) ≤ 4π2 for all x ∈ C.

In our third example we consider the corner quarters Cantor set E describedin Section 1.7 of Chapter 1, and we let μ be the Hausdorff 1-dimensional measureon E. We are going to check that c2μ(x) = ∞ for all x ∈ E, and so c2(μ) = ∞.Recall that E =

⋂∞n=1 En, where En is made up of a disjoint union of 4n squares

Qni with side length 4−n. For a fixed n ≥ 0, denote by Qn(x) the square Qn

i thatcontains x. We write

c2μ(x) ≥∞∑k=1

∫y∈Qk−1(x)\Qk(x)

∫z∈Qk−1(x)\[Qk(x)∪Qk(y)]

c(x, y, z)2dμ(y)dμ(z). (3.4)

By the geometry of the set E it is easy to check that, given y ∈ Qk−1(x) \Qk(x),we have

c(x, y, z) ≥ C1

�(Qk(x))for all z ∈ Qk−1(x) \ (Qk(x) ∪Qk(y)),

3.2. The curvature of a measure 79

for some constant C > 0. Moreover, by the self-similarity of E,

μ(Qki ) = 4−k μ(E) ≈ 4−k for 1 ≤ i ≤ 4k,

and so μ(Qk−1(x) \ [Qk(x) ∪Qk(y)]) ≈ μ(Qk−1(x) \Qk(x)) ≈ 4−k. Hence,

c2μ(x) ≥ C

∞∑k=1


4−k

�(Qk(x))2dμ(y) ≈

∞∑k=1

(4−k

4−k

)2= ∞.

From the above examples the reader may suspect that, given a set E ⊂ C withH1(E) < ∞ and μ = H1�E, the finiteness of c2(μ) is related to the rectifiabilityproperties of E. This is indeed the case: the David-Leger curvature theorem assertsthat if c2(μ) < ∞, then E is rectifiable. This is a deep and difficult result that willbe proved in Chapter 7.

On the other hand, one may consider the curvature of measures different fromthe arc length on some subset. For example, take a set E ⊂ C with finite area andset μ = L2�E. Then using symmetry and the estimate R(x, y, z) ≥ |x− y|/2, onegets

c2μ(x) ≤ 2

∫∫|x−y|≥|x−z|

1

R(x, y, z)2dμ(y) dμ(z)

≤ 8

∫∫|x−y|≥|x−z|

1

|x− y|2 dμ(y) dμ(z)

≤ 8

∫π|x− y|2|x− y|2 dμ(y) = 8πL2(E).

Thus, c2(μ) ≤ 8πL2(E)2.

When one integrates three times with respect to μ the identity (3.2), onegets another nice identity, which relates the Cauchy transform of a measure toits curvature. The precise result, from Melnikov and Verdera [114], is in the nextproposition. The statement involves the ε-truncated curvature

c2ε(μ) =

∫∫∫|x−y|>ε|y−z|>ε|x−z|>ε

c(x, y, z)2 dμ(x)dμ(y)dμ(z).

Proposition 3.3. Let μ be a finite Radon measure on C with c0-linear growth. Wehave

‖Cεμ‖2L2(μ) =1

6c2ε(μ) +O(μ(C)), (3.5)

with

|O(μ(C))| ≤ c c20 μ(C),

where c is some absolute constant.


The identity (3.5) is remarkable because it relates an analytic notion (theCauchy transform of a measure) with a metric-geometric one (curvature). It hasplayed a key role in many of the recent results on analytic capacity.

Proof. We have

‖Cεμ‖2L2(μ) =

∫∫∫|x−y|>ε|x−z|>ε

1

(y − x)(z − x)dμ(y)dμ(z)dμ(x)

=

∫∫∫|x−y|>ε|x−z|>ε|y−z|>ε

1


+

∫∫∫|x−y|>ε|x−z|>ε|y−z|≤ε

1


=: I1 + I2. (3.6)

Consider first the integral I1. Permuting x, y, z and using Fubini and Mel-nikov’s identity (3.2), by the symmetry of the domain of integration we get

I1 =1

6

∫∫∫|z1−z2|>ε|z1−z3|>ε|z2−z3|>ε

∑s∈S3

1

(zs2 − zs1)(zs3 − zs1)dμ(z1)dμ(z2)dμ(z3)

=1

6

∫∫∫|z1−z2|>ε|z1−z3|>ε|z2−z3|>ε

c(z1, z2, z3)2 dμ(z1)dμ(z2)dμ(z3) =

1

6c2ε(μ).

To estimate the integral I2 in (3.6), notice that, by the conditions in thedomain of integration, the side with vertices y, z is the shortest one in the trianglewith vertices x, y, z. Thus, |x− y| ≈ |x− z|, and so

|I2| ≤ c

∫∫∫|x−y|>ε|x−z|>ε|y−z|≤ε

1

|y − x|2 dμ(y)dμ(z)dμ(x).

Integrating with respect to z, by the linear growth of μ, we derive

|I2| ≤ c c0 ε

∫∫|x−y|>ε

1

|y − x|2 dμ(y)dμ(x).

Using again the linear growth of μ, splitting the domain {y : |x − y| > ε} intoannuli, it follows that ∫

x:|x−y|>ε

1

|y − x|2 dμ(x) ≤ cc0ε.

Thus we get,|I2| ≤ c c20μ(C),

and the proposition follows. �

3.3. The T 1 theorem for the Cauchy singular integral operator 81

In the next lemma we generalize the identity (3.5) to the case involvingthree possibly different measures. To simplify notation, we will denote by MR the1-dimensional radial maximal operator, instead of MR,1.

Lemma 3.4. Let νj, j = 1, 2, 3, be finite real Radon measures on C. Then we have∑s∈S3

∫Cε(νs2) Cε(νs3) dνs1 =

∫∫∫|x−y|>ε|x−z|>ε|y−z|>ε

c(x, y, z)2 dν1(x)dν2(y)dν3(z) +R,

where

|R| ≤ C∑s∈S3

∫MRνs2 MRνs3 dνs1 ,

with C being an absolute constant.

Proof. Arguing as in the preceding proposition, we get∫Cε(ν2) Cε(ν3) dν1 =

∫∫∫|x−y|>ε|x−z|>ε

dν1(x) dν2(y) dν3(z)

(y − x)(z − x)

=

∫∫∫|x−y|>ε|x−z|>ε|y−z|>ε

· · ·+∫∫∫

|x−y|>ε|x−z|>ε|y−z|≤ε

· · · =: I1 + I2. (3.7)

As above, in the domain of integration I2, we have |x− y| ≈ |x− z|, and thus

I2 �∫∫∫

|x−y|>ε|y−z|≤ε

dν1(x) dν2(y) dν3(z)

|y − x|2 =

∫∫|x−y|>ε

ν3(B(y, ε)) dν1(x) dν2(y)

|y − x|2 .

Since ∫|x−y|>ε

1

|y − x|2 dν1(x) �1

εMRν1(y),

we deduce

I2 �∫

ν3(B(y, ε))

εMRν1(y)dν2(y) �

∫MRν1(y)MRν3(y) dν2(y).

Permuting the indices 1, 2, 3 in (3.7) and summing over all possible permutations,using the identity (3.2), we are done. �

3.3 The T1 theorem for the Cauchy singular integral

operator

We will prove the following version of the T 1 theorem for the Cauchy singularintegral operator.


Theorem 3.5. Let μ be a Radon measure on C with linear growth. The followingconditions are equivalent:

(a) Cμ is bounded in L2(μ).

(b) For all ε > 0 and all the squares Q ⊂ C,

‖Cμ,εχQ‖L2(μ�Q) ≤ c μ(2Q)1/2,

with c independent of ε.

(c) For all the squares Q ⊂ C,

c2(μ�Q) ≤ c μ(2Q).

To prove this theorem we will exploit the relationship between the Cauchytransform andMenger curvature and we will use Theorem 2.22, which was obtainedby means of a good λ inequality. Our arguments are new, although they have someresemblances to the ones in Tolsa [154] and, especially, in Verdera [175].

Notice that the equivalence (b) ⇔ (c) follows by a direct application of (3.5)to the measure μ�Q, for all the squares Q ⊂ C. So we only have to prove (b) ⇒(a). To this end, we need the following key lemma:

Lemma 3.6. Let μ be a finite measure with linear growth on C, that is,

μ(B(x, r)) ≤ c0r for all x ∈ C, r > 0.

Suppose that ‖Cεμ‖2L2(μ) ≤ c1μ(C) for all ε > 0. Then there exists a subset G ⊂ C

with μ(G) ≥ μ(C)/4 such that Cμ�G : L2(μ�G) → L2(μ�G) is bounded with normbounded above by some constant depending only on c0 and c1.

Proof. From the assumptions in the lemma and Proposition 3.3, we deduce

c2(μ) ≤ c2 μ(C).

Given c3 > 0, let

Aε :={x ∈ C : |Cεμ(x)| ≤ c3 and c2μ(x) ≤ c23

}.

Since∫c2μ(x) dμ(x) = c2(μ) ≤ c2μ(C) and

∫ |Cεμ|2 dμ ≤ cμ(C), we infer thatμ(Aε) ≥ μ(C)/2 if c3 is chosen big enough, by Chebyshev.

We want to show that the Cauchy integral operator Cμ�Aε,ε is bounded inL2(μ�Aε). To this end we introduce an auxiliary “curvature operator”: for x, y ∈Aε, consider the kernel K(x, y) :=

∫c(x, y, z)2 dμ(z), and let T be the operator

Tf(x) =

∫K(x, y)f(y) dμ(y).

3.3. The T 1 theorem for the Cauchy singular integral operator 83

By Schur’s lemma, T is bounded in Lp(μ�Aε) for all p ∈ [1,∞], because for allx ∈ Aε, ∫

Aε

K(x, y) dμ(y) =

∫Aε

K(y, x) dμ(y)

=

∫y∈Aε

c(x, y, z)2 dμ(y)dμ(z) ≤ c2μ(x) ≤ c23.

From Lemma 3.4 we deduce that, given a non-negative (real) function fsupported on Aε, we have

4

∫|Cε(f μ)|2 dμ =

∫∫∫|x−y|>ε|x−z|>ε|y−z|>ε

c(x, y, z)2f(x)f(y) dμ(x)dμ(y)dμ(z)

− 2Re

∫(Cεμ) Cε(f μ) f dμ+O(‖f‖2L2(μ)).

Thus,∫|Cε(f μ)|2 dμ ≤ 1

4

∣∣〈Tf, f〉∣∣+ 1

2

∫ ∣∣(Cεμ) Cε(f μ) f∣∣ dμ+ c‖f‖2L2(μ). (3.8)

To estimate the first term on the right-hand side we use the L2(μ�Aε) boundednessof T (recall that supp(f) ⊂ Aε):∣∣〈Tf, f〉∣∣ ≤ ‖Tf‖L2(μ) ‖f‖L2(μ) ≤ c‖f‖2L2(μ).

To deal with the second integral on the right-hand side of (3.8), notice that |Cεμ| ≤c3 on the support of f , and so∫ ∣∣(Cεμ) Cε(f μ) f

∣∣ dμ ≤ c3

∫ ∣∣Cε(f μ) f∣∣ dμ ≤ c3‖Cε(f μ)‖L2(μ)‖f‖L2(μ).

By (3.8) we get

‖Cε(f μ)‖2L2(μ) ≤ c‖f‖2L2(μ) +c32‖Cε(f μ)‖L2(μ)‖f‖L2(μ),

which implies that ‖Cε(f μ)‖L2(μ) ≤ c‖f‖L2(μ).So far we have proved the L2(μ�Aε) boundedness of Cμ�Aε,ε. If Aε were

independent of ε, we would set G := Aε and we would be done. Unfortunately thisis not the case and we have to work a little more. We set

Gε :={x ∈ C : C∗,εμ(x) ≤ c4 and c2μ(x) ≤ c24

},

where c4 is some constant big enough (with c4 > c3) to be chosen below. By Re-mark 2.17, Theorem 2.21, and the discussion above, we know that C∗,ε is boundedfrom M(C) into L1,∞(μ�Aε) (with constants independent of ε). Thus,

μ({

x ∈ Aε : C∗,εμ(x) > c4}) ≤ cμ(C)

c4.


If c4 is big enough, the right-hand side of the preceding inequality does not exceedμ(C)/4 ≤ μ(Aε)/2. Thus,

μ(Gε) ≥ μ({

x ∈ Aε : C∗,εμ(x) ≤ c4}) ≥ 1

2μ(Aε) ≥ 1

4μ(C).

We set

G :=⋂ε>0

Gε.

Notice that, by definition, Gε ⊂ Gδ if ε > δ and so we have

μ(G) = limε→0

μ(Gε) ≥ 1

4μ(C).

By the same argument used for Aε, it follows that Cμ�Gε,ε is bounded on L2(μ�Gε)(with constant independent of ε), and thus Cμ�G is bounded on L2(μ�G). �

Proof of Theorem 3.5. As remarked above, we only have to prove (b) ⇒ (a). Thisfollows by a direct application of the preceding lemma and Theorem 2.22. Indeed,for any given β > 0, consider a (2, β)-doubling square Q. By the assumption (b),

‖Cμ,εχQ‖L2(μ�Q) ≤ cμ(2Q)1/2 ≤ cβ1/2 μ(Q)1/2

uniformly on ε > 0. By Lemma 3.6 applied to the measure μ�Q, we infer that thereexists some subset GQ ⊂ Q with μ(GQ) ≥ μ(Q)/4 such that the Cauchy transformis bounded in L2(μ�GQ), and thus C∗ is bounded from M(C) into L1,∞(μ). ThenTheorem 2.22 ensures that Cμ is bounded in L2(μ). �

Remark 3.7. By the Remark 2.24 just after the proof of Theorem 2.22, one can askconditions (a) and (b) in Theorem 3.5 to hold only for squares or balls centeredon supp(μ), instead of all the squares.

3.4 The Cauchy transform on Lipschitz graphs

In this section we will prove that the Cauchy transform is bounded on Lipschitzgraphs. This important result was first proved by Calderon [5] assuming the slopeof the Lipschitz graph small enough. The proof in full generality was obtainedlater by Coifman, McIntosh, and Meyer [16].

The arguments below are not the original ones. They are due to Melnikovand Verdera [114]. They rely on the use of curvature by means of the identity (3.5)and the T 1 theorem.

For x, y, z ∈ C, we have

S(x, y, z) =1

2

∣∣(y1 − x1)(z2 − x2)− (y2 − x2)(z1 − x1)∣∣,

3.4. The Cauchy transform on Lipschitz graphs 85

where the subindices 1, 2 stand for the horizontal and vertical coordinates, respec-tively. Therefore, we have

c(x, y, z) = 2

∣∣(y1 − x1)(z2 − x2)− (y2 − x2)(z1 − x1)∣∣

|x− y| |x− z| |y − z|

= 2|x1 − y1| |x1 − z1| |y1 − z1|

|x− y| |x− z| |y − z|

∣∣∣∣∣∣∣z2 − x2

z1 − x1− y2 − x2

y1 − x1

z1 − y1

∣∣∣∣∣∣∣ . (3.9)

Remark 3.8. Let I ⊂ R be an interval and A : I → R a Lipschitz function.Consider three points X = (x,A(x)), Y = (y,A(y)), and Z = (z, A(z)), withx, y, z ∈ I. By the preceding identity

2(1 + ‖A′‖2∞

)3/2∣∣∣∣∣∣∣∣A(z)−A(x)

z − x− A(y)−A(x)

y − x

z − y

∣∣∣∣∣∣∣∣≤ c(X,Y, Z

) ≤ 2

∣∣∣∣∣∣∣∣A(z)−A(x)

z − x− A(y)−A(x)

y − x

z − y

∣∣∣∣∣∣∣∣ . (3.10)

Let us see how the last term on the right-hand side can be estimated in terms of‖A′′‖∞, assuming A′ to be Lipschitz. To simplify notation, for a < b, write

m[a,b]A′ =

1

b − a

∫ b

a

A′(t) dt.

That is, m[a,b]A′ is the mean of A′ over the interval [a, b]. Suppose that x < y < z.

By the fundamental theorem of calculus, we have

A(z)−A(x)

z − x− A(y)−A(x)

y − x= m[x,z]A

′ −m[x,y]A′ (3.11)

=(y − x

z − xm[x,y]A

′ +z − y

z − xm[y,z]A

′)−m[x,y]A

′

=z − y

z − x

(m[y,z]A

′ −m[x,y]A′).

Observe now that∣∣m[y,z]A′ −A′(y)

∣∣ ≤ ‖A′′‖∞ (z − y) and∣∣m[x,y]A

′ −A′(y)∣∣ ≤ ‖A′′‖∞ (y − x).

Therefore,∣∣m[y,z]A′ −m[x,y]A

′∣∣ ≤ ‖A′′‖∞ (z − y) + ‖A′′‖∞ (y − x) = ‖A′′‖∞ (z − x).


From this estimate and (3.11), we derive∣∣∣∣∣∣∣∣A(z)−A(x)

z − x− A(y)−A(x)

y − x

z − y

∣∣∣∣∣∣∣∣ ≤ ‖A′′‖∞ (3.12)

and soc(X,Y, Z) ≤ 2‖A′′‖∞. (3.13)

In the preceding remark we have estimated the Menger curvature of threepoints in terms of ‖A′′‖∞. Integrating with respect to arc length, one can boundthe curvature of the arc length measure on the graph of A in terms of ‖A′′‖∞.Nevertheless, using the Plancherel identity, it turns out that the curvature of arclength measure can be estimated in terms of the first derivative of A, instead ofthe second one.

Lemma 3.9. Let A : R → R be an absolutely continuous function such that A′ ∈L2(R). We have

∫∫∫R3

∣∣∣∣∣∣∣∣A(z)−A(x)

z − x− A(y)− A(x)

y − x

z − y

∣∣∣∣∣∣∣∣2

dx dy dz = 2π2‖A′‖22, (3.14)

and

(3.15)

8π2‖A′‖22(1 + ‖A′‖2∞

)3 ≤∫∫∫

R3

c((x,A(x)), (y,A(y)), (z, A(z))

)2dx dy dz ≤ 8π2‖A′‖22.

Proof. Let us consider new variables h = y−x and k = z−x. Applying Plancherelin x, the triple integral in (3.14) equals

∫∫∫R3

∣∣∣∣∣∣∣∣e2πikξ − 1

k− e2πihξ − 1

hk − h

∣∣∣∣∣∣∣∣2

|A(ξ)|2 dξ dh dk

=

∫∫∫R3

∣∣∣∣∣∣∣e2πiu − 1

u− e2πiv − 1

vu− v

∣∣∣∣∣∣∣2

|ξA(ξ)|2 dξ du dv

=1

4π2‖A′‖2∫∫

R2

∣∣∣∣∣∣∣e2πiu − 1

u− e2πiv − 1

vu− v

∣∣∣∣∣∣∣2

du dv.

3.4. The Cauchy transform on Lipschitz graphs 87

Let E(u) = (e2πiu − 1)/u, so that the last double integral equals

I =

∫∫R2

∣∣∣∣E(u + t)− E(u)

t

∣∣∣∣2 du dt.

Since the Fourier transform of E(u) is 2πi χ[0,1](ξ), we get

I = 4π2

∫R

∫ 1

0

∣∣∣∣e2πiξt − 1

t

∣∣∣∣2 dξ dt = 4π2

∫R

∫ 1

0

∣∣∣∣e2πiu − 1

u

∣∣∣∣2 |ξ| dξ du.By Plancherel in u:

I = 4π2 ‖E‖22∫ 1

0

ξ dξ = 8π4,

and so we are done with (3.14).Finally, the inequality (3.15) is a straightforward consequence of (3.10) and

(3.14). �From the last lemma we get:

Lemma 3.10. Let A : R → R a Lipschitz function and let Γ ⊂ R2 be its graph. LetΠ be the orthogonal projection onto the horizontal axis, and let μ be the measureon Γ given by

dμ(x) =1√

1 + |A′(Π(x))|2 dH1�Γ(x).

Then, for any interval I ⊂ R, we have

c2(μ�Π−1(I)) ≤ 16π2 ‖A′‖2∞ �(I).

Proof. The lemma follows by localizing the estimates obtained above. Since thecurvature is invariant by translation, we may assume that I = [0, b] and that

A(b) = 0. Consider the auxiliary Lipschitz function A given by:

A(x) =

⎧⎪⎪⎨⎪⎪⎩0 if x ≤ −b,A(0) b−1 x+A(0) if −b < x ≤ 0,A(x) if 0 < x ≤ b,0 if x > b.

Then, since the image measure of μ by Π is the Lebesgue measure on the x axis,we have

c2(μ�Π−1(I)) ≤∫∫∫

R3

c((x, A(x)), (y, A(y)), (z, A(z))

)2dx dy dz

≤ 8π2‖A′‖22.

Since ‖A′‖∞ ≤ ‖A′‖∞, we have ‖A′‖22 ≤ 2�(I)‖A′‖2∞. �


Theorem 3.11. Let A : R → R be a Lipschitz function and Γ ⊂ R2 its Lipschitzgraph. Consider the measure μ = H1�Γ. Then the Cauchy transform Cμ is boundedin L2(μ), and the norm does not exceed some constant depending only on ‖A′‖∞.

Proof. We will use the T 1 theorem proved in the previous section. So it is enoughto show that there exists some constant c depending only on ‖A′‖∞ such that forevery square Q,

‖Cμ,ε(χQ)‖2L2(μ) ≤ c μ(2Q),

uniformly on ε > 0, or equivalently, that

c2(μ�Q) ≤ c μ(2Q).

Obviously, we may assume that Q ∩ Γ �= ∅. By the previous lemma,

c2(μ�Q) ≤ c(‖A′‖∞)�(Q) ≤ c(‖A′‖∞)μ(2Q),

and so we are done. Notice that the last inequality holds because Q ∩ Γ �= ∅.Another possible approach towards the application of the T 1 theorem consists

in considering only squares Q centered in Γ, recalling Remark 3.7. In this case, wehave

c2(μ�Q) ≤ c(‖A′‖∞)�(Q) ≤ c(‖A′‖∞)μ(Q). �Let us remark that, in order to prove the preceding theorem, it is not neces-

sary to apply a T 1 theorem for non-doubling measures, since the measure H1�Γ isclearly doubling. So the usual “doubling” T 1 theorem of David and Journe suffices.Moreover, changing slightly the arguments above, one can also prove the theoremby interpolation between the pairs (H1(R), L1(R)) and (L∞(R), BMO(R)). SeeMelnikov and Verdera [114] for more details.

3.5 The Cauchy transform on AD regular curves

In this section we will show that the Cauchy transform is bounded in L2 withrespect to arc length on AD regular curves. This was first proved by David in [20].Recall that an AD regular curve is a curve Γ satisfying

H1(Γ ∩B(x, r)) ≤ c0 r for all x ∈ C, r > 0.

Notice that since Γ is a connected set, it automatically satisfies the lower estimateH1(Γ ∩ B(x, r)) ≥ r/2 for all x ∈ Γ and 0 < r < diam(Γ). For the proof we willfollow the original argument of David, which consists in showing that Γ has bigpieces of Lipschitz graphs. That is, for every ball B centered at Γ with radiusr(B) ≤ diam(Γ) there exists a possibly rotated Lipschitz graph GB with its slopebounded above uniformly such that H1(Γ ∩ GB ∩ B) ≥ c−1 r. Applying then thegood lambda Theorem 2.22 and the fact that the Cauchy transform is boundedon Lipschitz graphs, one derives the L2(H1�Γ) boundedness of CH1�Γ.

3.5. The Cauchy transform on AD regular curves 89

Let us remark that, since H1�Γ is a doubling measure, the full strength ofTheorem 2.22 is not needed. A suitable doubling version would suffice, as in David[20].

Recall also that the L2 boundedness of the Cauchy transform implies thelinear growth condition. Together with David’s theorem, this characterizes theAD regular curves as those curves such that the Cauchy transform is bounded inL2 with respect to arc length.

Theorem 3.12. Let Γ ⊂ C be an AD regular curve. Then CH1�Γ is bounded inL2(H1�Γ), with its norm bounded above by some constant depending only on theAD regularity constant of Γ.

As remarked above, it is enough to prove the following.

Lemma 3.13. Let Γ ⊂ C be an AD regular curve. For every ball B centered at Γwith radius r(B) ≤ diam(Γ) there exists a (possibly rotated) Lipschitz graph GB

with slope bounded above uniformly such that

H1(Γ ∩GB ∩B) ≥ c−1 r.

Proof. Set B = B(x0, r) and let g = (g1, g2) : [0, s0]−→C be an arc lengthparametrization of the arc of Γ between x0 and the first time that Γ hits ∂B(x0, r),so that g(0) = x0 and g(s0) ∈ ∂B(x0, r). After, a change of coordinates we mayassume that x0 = 0 and g(s0) = (r, 0).

We will find the required subset Γ ∩ GB in g([0, s0]). To this end, we willprove the following.

Claim. There exist a set E ⊂ [0, s0] and a Lipschitz function h : [0, s0] → R suchthat L1(E) ≥ r/2,

h(s) = g1(s) for all s ∈ E,

and1

2c0≤ h′(s) ≤ 1 for all s ∈ [0, s0].

To check that the lemma follows from this claim, notice that h is invertibleand h−1 is Lipschitz, and then consider the curve

Γ ={(h(s), g2(s)) : s ∈ [0, s0]

}={(t, g2 ◦ h−1(t)) : t ∈ h([0, s0])

}.

Clearly, Γ is the graph of the Lipschitz function g2 ◦ h−1 and, setting g(t) =(t, g2 ◦ h−1(t)), we get

Γ ∩ Γ ⊃ g(E) = g ◦ h(E).

So choosing GB = Γ, we have H1(GB ∩ Γ) ≥ H1(g(h(E))) ≥ L1(h(E)) ≥ r/2c0,and we are done.

To prove the claim, we apply the rising sun lemma. Notice first that

s0 = H1(g([0, s0])) ≤ c0 r = c0 g1(s0). (3.16)


r

s0

g1

r

s0

��

��

��

��

��

��

h

Figure 3.2: On the left, the function g1. On the right, the function h is drawn withthick lines.

Now we take the function

h(s) = sup0≤t≤s

[g1(t) +

1

2c0(s− t)].

See Figure 3.2. This is a Lipschitz function, and from the fact that |g′| = 1 a.e.we infer that ‖g′1‖∞ ≤ 1, and then it easily follows that

‖h′‖∞ ≤ 1.

Moreover, by construction

h′(s) ≥ 1

2c0for a.e. s ∈ [0, s0].

The set E of those s ∈ [0, s0] where h(s) = g1(s) is closed. Thus, its com-plementary in (0, s0) is an, at most, countable union of open intervals Ik. In eachinterval Ik, by construction we have h′ = 1/(2c0). Thus, using also (3.16),∫

[0,s0]\Eh′(s) ds =

∑k

1

2c0�(Ik) ≤ s0

2c0≤ r

2.

On the other hand,∫ s0

0

h′(s) ds = h(s0)− h(0) ≥ g1(s0)− g1(0) = r.

Therefore,

L1(E) ≥ L1(E) ‖h′‖∞ ≥∫E

h′(s) ds ≥ r

2,

and so the claim and the lemma hold. �

3.6. Curvature and Jones’ β’s 91

3Q

Q

E ∩ 3Q

VQ

Figure 3.3: The squares Q, 3Q, and the strip VQ in the definition of βE(Q).

3.6 Curvature and Jones’ β’s

In this section we investigate the relationship between curvature of measures andthe Jones β numbers. From this relationship, in combination with the Jones’ trav-eling salesman theorem and the T 1 theorem, we will get a new a proof of theL2 boundedness of the Cauchy transform on Lipschitz graphs and on AD regu-lar curves. Moreover, by means of this new approach we will get a sharp boundfor the L2 norm of the Cauchy transform on Lipschitz graphs, as well as sharpquantitative estimates for the analytic capacity of subsets of rectifiable curves inChapter 4. However, the results in this section are not necessary either for theproof of Vitushkin’s conjecture or for the semiadditivity of analytic capacity, andthus can be skipped without much harm if preferred.

To state the traveling salesman theorem of Peter Jones [75], first we introducesome notation. Given E ⊂ C and a square Q, let VQ be an infinite closed strip (orline in the degenerate case) of smallest possible width which contains E ∩ 3Q, andlet w(VQ) denote the width of VQ. Then we set

βE(Q) =w(VQ)

�(Q). (3.17)

See Figure 3.3. So βE(Q) measures in a scale invariant way how well E can beapproximated in 3Q by some line.

Theorem 3.14 (Jones’ traveling salesman theorem). A set E ⊂ C is contained ina rectifiable curve Γ (with finite length) if and only if∑

Q∈DβE(Q)2�(Q) < ∞. (3.18)


Moreover, the length of the shortest curve Γ containing E satisfies

H1(Γ) ≈ diam(E) +∑Q∈D

βE(Q)2�(Q),

with absolute constants.

The theorem also holds for sets E contained in Rd. The proof of the “if” partof the theorem in Jones [75] is also valid in this case. The “only if” part for Rd

was obtained by Okikiolu [129]. We will not prove Jones’ theorem in this book.We refer the reader either to the original proofs in Jones [75] and Okikiolu [129],or to the more friendly arguments from Garnett and Marshall [56, Chapter X].An alternative proof based on the use of Crofton’s formula is given in Bishop andPeres [4].

The next theorem shows how the curvature of a measure with linear growthsupported on a curve or, more generally, a connected set can be estimated in termsof Jones β’s.

Theorem 3.15. If μ is a measure with c0-linear growth supported on a connectedset Γ ⊂ C, then

c2(μ) ≤ c c20∑Q∈D

βΓ(Q)2μ(Q),

where c is an absolute constant.

For the proof we need a couple of auxiliary results. The first one is thefollowing.

Lemma 3.16. Let E ⊂ C and consider a square Q ⊂ C. Let x, y, z ∈ 3Q ∩ E besuch that

|x− y| ≈ |x− z| ≈ |y − z| ≈ �(Q). (3.19)

Then

c(x, y, z) ≤ cβE(Q)

�(Q),

with c depending on the comparability constants in (3.19).

Proof. Let V be a closed strip that contains E∩3Q, and thus x, y, z. Since V ∩3Qis convex, the triangle with vertices x, y, z is contained in V ∩ 3Q. Thus, the areaof V ∩3Q is greater than or equal to the area S(x, y, z) of the triangle with verticesx, y, z. On the other hand,

Area of V ∩ 3Q ≤ w(V ) diam(Q)

(recall that w(V ) is the width of V ). Thus, S(x, y, z) ≤ 21/2w(V ) �(Q). Sincethis holds for all the strips V containing E ∩ 3Q, we deduce that S(x, y, z) ≤21/2 β(Q) �(Q)2, and so

c(x, y, z) =4S(x, y, z)

|x− y| |x− z| |y − z| ≤ cβ(Q) �(Q)2

�(Q)3= c

β(Q)

�(Q). �


In the preceding lemma we have seen how the Menger curvature of the triple(x, y, z) can be estimated in terms of the β’s when the sides of the triangle withvertices x, y, z have comparable lengths. To deal with the triples of points whichoriginate triangles whose sides lengths are not comparable we will use the followinglemma.

Lemma 3.17. Let x, y ∈ C, and for i = 1, . . . , N , let zi ∈ C be such that

|x− zi| ≤ 2−i|x− y| for i = 1, . . . , N .

Then

c(x, y, zN ) ≤N∑i=1

2−i+3 c(x, zi−1, zi),

where z0 = y.

Proof. Denote by ΠL the orthogonal projection onto a line L. For i = 1, . . . , N ,define inductively wN = zN and wi−1 = ΠLx,zi−1

(wi). See Figure 3.4. Then wehave

dist(zN , Lx,y) ≤N∑i=1

|wi−1 − wi|.

By Thales theorem,

|wi−1 − wi| = dist(wi, Lx,zi−1) = dist(zi, Lx,zi−1)|wi − x||zi − x| .

Therefore, we get

c(x, y, zN ) =2 dist(zN , Lx,y)

|zN − x| |zN − y| ≤N∑i=1

2 dist(zi, Lx,zi−1)

|zN − x| |zN − y||wi − x||zi − x| .

Observe now that, for all i,

|wi − x| ≤ |wi+1 − x| ≤ · · · ≤ |wN − x| = |zN − x|.So we have

c(x, y, zN) ≤N∑i=1

2 dist(zi, Lx,zi−1)

|zi − x| |zN − y| .

Now we take into account that

|y − zN | ≥ |x− y| − |x− zN | ≥ 1

2|x− y|,

and

|zi − zi−1| ≤ |zi − x|+ |zi−1 − x| ≤ 2−i+2 |x− y| ≤ 2−i+3 |y − zN |.


z3 = w3

z2 z1

w1

w2

�

�

�

��

�

�

�

x w0 y = z0

Figure 3.4: The points x, y, z0, . . . , z3, and w0, . . . , w3 in the proof of Lemma 3.17,with N = 3.

Then we obtain

c(x, y, zN ) ≤N∑i=1

2−i+3 2 dist(zi, Lx,zi−1)

|zi − x| |zi − zi−1| =N∑i=1

2−i+3 c(x, zi−1, zi). �

Proof of Theorem 3.15. We have

c2(μ) ≤ 6

∫∫∫|x−y|≥|y−z|≥|x−z|

c(x, y, z)2 dμ(x) dμ(y) dμ(z).

Given a square Q ⊂ C, we write (x, y) ∈ Q if x ∈ Q and moreover 12�(Q) ≤

|x − y| ≤ �(Q). For every pair of points x, z ∈ C it is easy to check that there

exists a square Q ∈ D such that (x, z) ∈ Q. Also, if |x − y| ≥ |x − z|, then there

exists some dyadic square R ⊃ Q such that (x, y) ∈ R. Then we have

c2(μ) ≤ 6∑

Q,R∈DR⊃Q

∫∫(x,z)∈Q

(∫y:(x,y)∈R|x−y|≥|y−z|≥|x−z|

c(x, y, z)2 dμ(y)

)dμ(x) dμ(z).

(3.20)To estimate c(x, y, z) for x, y, z ∈ Γ as in the integrals above we wish to apply

Lemma 3.17. Notice that in the domain of integration, |x−z| ≤ |x−y|/2, so thereexists N ≥ 1 such that

2−N−1|x− y| < |x− z| ≤ 2−N |x− y|.

Since Γ is connected, for i = 1, . . . , N − 1 there are points zi ∈ Γ such that|x− zi| = 2−i|x− y|. Then we have

c(x, y, z) ≤ 8

N∑i=1

2−i c(x, zi−1, zi),


with z0 = y and zN = z. Observe that from the condition |x− zi| = 2−i|x− y| ≤2−i�(R) for i = 0, . . . , N , we infer that zi−1, zi ∈ 3S, where S is the (i − 1)-thdescendant of R which contains Q. Moreover,

|x− zi| ≈ |x− zi−1| ≈ |zi − zi−1| ≈ �(S).

Then, by Lemma 3.16, we have c(x, zi−1, zi) ≤ c βΓ(S)/�(S) and thus

c(x, y, z) ≤ c∑

S∈D:Q⊂S⊂R

�(S)

�(R)

βΓ(S)

�(S)= c

∑S∈D:Q⊂S⊂R

βΓ(S)

�(R).

Now, by the Cauchy-Schwarz inequality,( ∑S∈D:Q⊂S⊂R

βΓ(S)

)2≤( ∑

S∈D:Q⊂S⊂R

βΓ(S)2 �(R)1/2

�(S)1/2

)( ∑S∈D:Q⊂S⊂R

�(S)1/2

�(R)1/2

)

≤ c∑

S∈D:Q⊂S⊂R

βΓ(S)2 �(R)1/2

�(S)1/2.

As a consequence,

c(x, y, z)2 ≤ c∑

S∈D:Q⊂S⊂R

βΓ(S)2

�(R)3/2�(S)1/2.

We plug the preceding estimate into (3.20), and we take into account that

(x, z) ∈ Q implies that z ∈ 3Q, and (x, y) ∈ R that y ∈ 3R, and then we obtain

c2(μ) ≤ c∑

Q,R∈DR⊃Q

∫∫(x,z)∈Q

(∫y∈3R

∑S∈D:

Q⊂S⊂R

βΓ(S)2

�(R)3/2�(S)1/2dμ(y)

)dμ(x) dμ(z)

≤ c∑

Q,R∈DR⊃Q

∑S∈D:

Q⊂S⊂R

βΓ(S)2

�(R)3/2�(S)1/2μ(3R)μ(Q)μ(3Q).

By the linear growth of μ and Fubini, we get

c2(μ) ≤ c c20∑S∈D

βΓ(S)2∑

Q∈D:Q⊂S

μ(Q)�(Q)

�(S)1/2

∑R∈D:R⊃S

1

�(R)1/2

≤ c c20∑S∈D

βΓ(S)2∑

Q∈D:Q⊂S

μ(Q)�(Q)

�(S).

Finally, notice that∑Q∈D:Q⊂S

μ(Q)�(Q)

�(S)=∑k≥0

2−k∑

Q∈D:Q⊂S

(Q)=2−k(S)

μ(Q) ≤ 2μ(S),

and so the theorem follows. �


Remark 3.18. By analogous arguments to the ones used above, one can show thatif μ is an AD regular measure with AD regularity constant c0 (i.e., the constant cin (2.2) equals c0), then

c2(μ) ≤ c5∑Q∈D

βE(Q)2μ(Q),

where E = suppμ and c5 depends on c0.

In the next theorem we apply the preceding results to prove a sharp estimatefor the L2 norm of the Cauchy transform on Lipschitz graphs which is due toMurai [115].

Theorem 3.19. Let A : R → R be a Lipschitz function and Γ ⊂ C its graph.Consider the measure

dμ(x) =1√

1 + |A′(Π(x))|2 dH1�Γ(x).

Then the Cauchy transform Cμ is bounded in L2(μ), and

‖Cμ‖L2(μ)→L2(μ) ≤ c (1 + ‖A′‖∞)1/2, (3.21)


Observe that the statement in the theorem is equivalent to saying that thesingular integral operator

TAf(x) =

∫x− y + i

(A(x) −A(y)

)|x− y|2 + |A(x) −A(y)|2 f(y) dy

is bounded in L2(R) with ‖TA‖L2(R)→L2(R) ≤ c (1 + ‖A′‖∞)1/2.

Proof. Write

S = supQ

‖Cμ(χQ)‖L2(μ�Q)

μ(Q)1/2,

where the supremum is taken over all the squares Q centered at Γ. We have

‖Cμ‖L2(μ)→L2(μ) ≤ c (c0 + S), (3.22)

where c0 is the linear growth constant of μ. This follows easily by applying theT 1 theorem to the measure σ = 1

c0+S μ. Observe further that in our case c0 ≤ 2,since for any interval I ⊂ R, μ(I × R) = �(I).

By (3.22), to prove the theorem it suffices to show that

‖Cμ(χQ)‖2L2(μ�Q) ≤ c (1 + ‖A′‖∞)μ(Q)


for any square Q centered at Γ. Consider such a square Q, and let ΓQ = ∂Q ∪(Γ∩Q). This is a connected set (this may fail for Γ∩Q when ‖A′‖∞ is big) whichcontains supp(μ�Q). From Theorems 3.15 and 3.14, we deduce

c2(μ�Q) ≤ c∑P∈D

βΓQ(P )2μ(P ) ≤ c∑P∈D

βΓQ(P )2�(P ) ≤ cH1(ΓQ).

Notice now that

H1(ΓQ) = 4�(Q) +H1(Γ ∩Q) ≤ 5H1(Γ ∩Q),

because H1(Γ ∩Q) ≥ �(Q) (recall that Q is centered on Γ). Therefore,

c2(μ�Q) ≤ cH1(Γ ∩Q) ≤ c (1 + ‖A′‖∞)μ(Q).

By the Melnikov-Verdera identity from Proposition 3.3, we deduce that

‖Cμ(χQ)‖2L2(μ�Q) ≤1

6c2(μ�Q) + c μ(Q) ≤ c (1 + ‖A′‖∞)μ(Q),

and so we are done. �Theorem 3.20. Let Γ ⊂ C be an AD regular curve with constant c0. That is, H1�Γhas c0-linear growth. Then CH1�Γ is bounded in L2(H1�Γ) and

‖CH1�Γ‖L2(H1�Γ)→L2(H1�Γ) ≤ c c3/20 ,


Proof. Write

S = supB

‖CH1�Γ(χB)‖L2(H1�Γ∩B)

H1(Γ ∩B)1/2,

where the supremum is taken over all the balls B ⊂ C centered at Γ. Since

‖CH1�Γ‖L2(H1�Γ)→L2(H1�Γ) ≤ c (c0 + S),

to prove the theorem it suffices to show that

‖CH1�Γ(χB)‖2L2(H1�Γ∩B) ≤ c c30H1(Γ ∩B)

for any ball B centered at Γ. To prove this estimate, reducing B if necessary, wemay assume that ∂B∩Γ �= ∅. Let ΓB = ∂B∪(Γ∩B). Notice that this a connectedAD regular set which contains Γ∩B with an AD regularity constant not exceedingc c0. Then, from the fact that c2(H1�Γ∩B) ≤ c2(H1�ΓB) together with Theorems3.15 and 3.14, we derive

c2(H1�Γ∩B) ≤ c c20∑Q∈D

βΓB (Q)2H1(ΓB ∩Q) ≤ c c30H1(ΓB) ≤ c c30H1(Γ ∩B).

From Proposition 3.3 we deduce that

‖CH1�Γ(χB)‖2L2(H1�Γ∩B) ≤ c (c30 + c20)H1(Γ ∩B) ≤ c c30 H1(Γ ∩B). �



3.7.1 The Cauchy transform and curvature

As mentioned above, the relationship between the Cauchy kernel and Mengercurvature was discovered by Melnikov [113] while studying a discrete version ofanalytic capacity. The nice identity (3.5) was proved by Melnikov and Verdera in[114] in order to obtain a new proof of the L2 boundedness of the Cauchy transformon Lipschitz graphs, similar to the first one explained in this chapter.

The limit of the “error term” O(μ(C)) in (3.5) as ε → 0 can be computedprecisely. In fact, in Tolsa and Verdera [169], assuming that MRμ ∈ L2(μ) andc2(μ) < ∞, the following identity is proved:

‖p.v.Cμ‖2L2(μ) =1

6c2(μ) +

π2

3

∫Θ1(z, μ)2dμ(z),

where p.v.Cμ stands for the principal value of the Cauchy transform of μ:

p.v.Cμ(z) = limε→0

Cεμ(z).

Let us remark that this principal value exists μ-a.e. under the above assumptions(see Corollary 8.2 in this book). Recall also that

Θ1(z, μ) = limr→0

μ(B(z, r))

2r.

From the conditions MRμ ∈ L2(μ) and c2(μ) < ∞ and the David-Leger curvatureTheorem 7.1, one can easily show that Θ1(z, μ) exists for μ-a.e. z ∈ C.

3.7.2 The T1 theorem

The idea of using a good lambda inequality to prove the T 1 theorem for the Cauchytransform stems from Tolsa [154], and it was further exploited in Verdera [175].For general singular integral operators, this technique is not so useful (as far as Iknow). The T 1 theorem for n-dimensional singular integral operators with respectto a measure of degree n which may be non-doubling was proved by Nazarov,Treil and Volberg [123], almost at the same time that the proof for the Cauchytransform was obtained in Tolsa [154]. The result from Nazarov, Teil and Volberg[123] reads as follows.

Theorem 3.21 (T 1 theorem). Let μ be a Borel measure on Rd with growth of degreen, with 0 < n ≤ d. Let Tμ be an n-dimensional singular integral operator. ThenTμ is bounded in L2(μ) if and only if

‖Tμ,εχQ‖L2(μ) ≤ c μ(Q)1/2 and ‖T ∗μ,εχQ‖L2(μ) ≤ c μ(Q)1/2 (3.23)

uniformly on ε > 0, for any cube Q ⊂ Rd.


The proof of the preceding result in Nazarov, Treil and Volberg [123] is basedon the study of the matrix of Tμ with respect to the Haar basis associated withrandom dyadic lattices. The techniques are similar (although considerably simpler)to the ones that will be used in Chapter 5 to prove a Tb theorem suitable for thestudy of analytic capacity.

A different proof of the above T 1 theorem appears in a subsequent paper fromTolsa [157]. Instead of the Haar basis, it relies on some kind of smooth wavelets,following some ideas from Coifman’s proof of the T 1 theorem on homogeneousspaces and from David, Journe and Semmes [26] for the Tb theorem.

Let us remark that the conditions (3.23) can be weakened. For instance, theycan be replaced by

‖Tμ,εχQ‖L2(μ�Q) ≤ c μ(2Q)1/2 and ‖T ∗μ,εχQ‖L2(μ�Q) ≤ c μ(2Q)1/2

uniformly on ε > 0, for any cube Q ⊂ Rd. This follows from the results in Tolsa[157] or Nazarov, Treil and Volberg [127]. Moreover, the T 1 theorem can also bestated in terms of BMO type conditions, as in the classical doubling setting. Thiswill be explained in more detail in Section 9.10 of Chapter 9.

3.7.3 The Cauchy transform on Lipschitz graphs and AD regularcurves

By now there are many different proofs of the L2 boundedness of the Cauchytransform on Lipschitz graphs with respect to arc length. In addition to the originalone by Coifman, McIntosh and Meyer [16] and to the ones in this chapter, one canfind others in David [20], Coifman, Journe, and Semmes [15], Jones [74], Journe[78], or Murai [117], for example.

The proof of the L2 boundedness of the Cauchy transform on AD regularcurves in Section 3.5 is based on the original argument of David [20] and onhis exposition in [22, Section III.4]. On the other hand, Theorem 3.15 on therelationship between curvature and β’s is due to Peter Jones, although it was firstpublished in Pajot [132, Chapter 3]. The fact that this result in combination withJones’ traveling salesman theorem and the T 1 theorem yields the boundedness ofthe Cauchy transform on Lipschitz graphs with the sharp power 1/2 was explainedto me by Verdera. The sharpness of the power 1/2 in (3.21) was shown by David[21] by constructing a Lipschitz graph which approximates the corner quartersCantor set.

3.7.4 Application of the curvature method to other kernels

The curvature method (or symmetrization method) applied in this chapter to theCauchy kernel does not work for most of the other Calderon-Zygmund kernels.This is due to the fact that, for most of the kernels K, the sum

S(z1, z2, z3) =∑s∈S3

K(zs2 − zs1)K(zs3 − zs1) (3.24)


may be negative for some triples of pairwise different points z1, z2, z3 ∈ C.One of the Calderon-Zygmund kernels for which the sum above is always

non-negative is the real part of the Cauchy kernel:

K(z) =−Re z

|z|2 .

In this case one gets, for z1, z2, z3 ∈ C:

1

2c(z1, z2, z3)

2 =∑s∈S3

K(zs2 − zs1)K(zs3 − zs1).

So if μ is a compactly supported measure with linear growth, one deduces that

‖Re Cεμ‖2L2(μ) =1

12c2ε(μ) +O(μ(C)).

By using the T 1 theorem, it turns out that, somewhat surprisingly, the L2(μ)boundedness of Re Cμ is equivalent to the L2(μ) boundedness of Cμ. Of course, ananalogous phenomenon occurs with the imaginary part of the Cauchy kernel.

For each n ≥ 1, consider the 1-dimensional kernel

Kn(z) =(Re z)2n−1

|z|2n . (3.25)

Notice thatK1 coincides with the real part of the Cauchy kernel (modulo the minussign). It turns out that the symmetrization method also works for each kernel Kn,n ≥ 1. This happens because the sum S(z1, z2, z3) in (3.24) with K replaced byKn is non-negative for any triple of pairwise different points z1, z2, z3 ∈ C. SeeChousionis, Mateu, Prat and Tolsa [10].

Other interesting kernels for which the curvature method is useful are theones of the Calderon commutators. Recall that they are the Calderon-Zygmundoperators in R associated with the kernels

Cn(x, y) =1

x− y

(A(x) −A(y)

x− y

)n, x, y ∈ R, x �= y,

where A : R → R is a Lipschitz function and n ≥ 1. Mateu and Verdera (see [176])discovered that the sum analogous to the one in (3.24) with Cn(xi, xj) instead ofK(zi− zj) is always non-negative. Moreover, in the case of C1, this is comparableto the square of the Menger curvature of the points {(xi, A(xi))}i=1,2,3.

For some 1-dimensional kernels in Rd we have analogous results. For instance,for the 1-dimensional Riesz kernelK(x) = x/|x|2, with x ∈ Rd\{0}, we always haveS(x1, x2, x3) ≥ 0 too [in this case the product in (3.24) should be replaced by thescalar productK(zs2−zs1)·K(zs3−zs1)]. However, for n-dimensional kernels with ngreater than 1 the situation is much worse. It seems that there are no interesting


kernels in this case for which the symmetrization method can be applied. Forexample, for the n-dimensional Riesz kernels x/|x|n+1, x ∈ Rd \ {0}, n > 1, thesum S(x1, x2, x3) may change sign (see Farag [41]). This causes great difficultiesfor the extension of some of the results in this book, such as Vitushkin’s conjecture,to higher dimensions (where analytic functions are replaced by Lipschitz harmonicfunctions).

On the other hand, for the fractional signed Riesz kernels with homogeneity−α, with 0 < α < 1, that is for Kα(x) = x/|x|α+1, it turns out that

0 ≤ S(x1, x2, x3) ≈ 1

max1≤i<j≤3(|xi − xj |)2α . (3.26)

This was noticed by Prat [136], and she used it to prove there are no sets E ⊂ Rd

with 0 < Hα(E) < ∞ such that the α-dimensional Riesz transform

Rαμf(x) =

∫x− y

|x− y|α+1f(y) dμ(y) (3.27)

is bounded in L2(μ), for μ = Hα�E. See also Mateu, Prat and Verdera [93] for analternative argument.

Chapter 4

The capacity γ+

4.1 Introduction

The capacity γ+ of a compact set E ⊂ C is

γ+(E) := sup{μ(E) : supp(μ) ⊂ E, ‖Cμ‖L∞(C) ≤ 1}, (4.1)

and the capacity γ+ of an arbitrary set A ⊂ C is defined as

γ+(A) = sup{γ+(E) : E ⊂ A, E compact}.

Notice that γ+ is defined like γ in (1.1) with the additional constraint thatf has to coincide with Cμ, where μ is some positive Radon measure supportedon E (observe that (Cμ)′(∞) = −μ(C) for any Radon measure μ). To be precise,there is another little difference: in (1.1) we asked ‖f‖L∞(C\E) ≤ 1, while in (4.1)‖f‖L∞(C) ≤ 1 (for f = Cμ). Trivially, we have γ+(E) ≤ γ(E). In Chapter 6 wewill show that the converse inequality γ(E) ≤ c γ+(E) also holds. For the momentin this chapter we will study the capacity γ+. We will characterize it in terms ofthe Cauchy transform and also in terms of curvature, and finally we will exploitsome (basic) techniques of potential theory to get further information on γ+.

In the next proposition we state some easy properties of γ+, whose proof isleft for the reader.

Proposition 4.1. Let E,F ⊂ C be compact. The following properties hold:

(a) If E ⊂ F , then γ+(E) ≤ γ+(F ).

(b) For all z, λ ∈ C, γ+(z + λE) = |λ| γ+(E).

From the definition it is not clear if γ+(E) = γ+(∂oE) or γ+(E) ≈ γ+(∂oE).The latter assertion is true, but its proof is not elementary: this follows from thefact that γ(E) = γ(∂oE) and the theorem stating that γ ≈ γ+.

, , OI 10.1007/978-3- - -6_ ,

© Springer



1036

104 Chapter 4. The capacity γ+

4.2 Smoothing of the Cauchy kernel by mollificationand outer regularity of γ+

For some arguments, the fact that the truncated Cauchy kernel χ|z|>ε−1z is not

smooth may cause troubles. Usually, a good solution in these situations consistsin considering a smooth version of χ|z|>ε

−1z . The smoothing by mollification of

−1z is particularly convenient. In this section we deal with this question.

Lemma 4.2. Let ϕ ∈ L∞(C) be a non-negative radial function supported in aball B(0, r) with L1(C) norm (with respect to Lebesgue measure) equal to 1. The

function K(z) =−1

z∗ ϕ is continuous and satisfies

K(z) =−1

zif |z| ≥ r,

and‖K‖∞ ≤ 2πr ‖ϕ‖∞.

Proof. The function K is continuous because the Cauchy kernel is in Lploc for

1 ≤ p < 2 and ϕ is compactly supported and in L∞. Also, K(z) = −1/z for|z| ≥ r because the Cauchy kernel is analytic (and thus harmonic) in the interiorof the ball B(z, r), ϕ is radial, and

∫ϕ(z) dL2(z) = 1.

Also, for any z ∈ C,

|K(z)| ≤ ‖ϕ‖∞∫|z−ξ|≤r

1

|z − ξ| dL2(ξ) ≤ 2πr ‖ϕ‖∞. �

As in the preceding chapter, to simplify notation we will use the notationMRν instead of MR,1ν. That is, for any complex measure ν and x ∈ C, we set

MRν(x) = supr>0

|ν|(B(x, r))

r.

Lemma 4.3. Let ϕ ∈ L∞(C) be a non-negative radial function supported in the ballB(0, 1) with L1(C) norm (with respect to Lebesgue measure) equal to 1. Write

ϕε(x) =1

ε2ϕ(xε

), for x ∈ C,

and let Cε the integral operator associated with the kernel−1

z∗ϕε. If ν is a complex

measure, then Cεν is a continuous function satisfying

|Cεν(x) − Cεν(x)| ≤ Cϕ MRν(x), (4.2)

where Cϕ depends only on ϕ. As a consequence, if μ is some measure with linear

growth, the Cauchy transform is bounded in Lp(μ) if and only if Cμ,ε is boundedin Lp(μ) uniformly in ε > 0, and it is bounded from M(C) to L1,∞(μ) if and only

if the same holds for Cμ,ε uniformly in ε.

4.2. Smoothing of the Cauchy kernel by mollification 105

Proof. By Lemma 4.2, the kernel Kε =−1

z∗ϕε is uniformly continuous and coin-

cides with the Cauchy kernel on C\B(0, ε). Then it follows that Cεν is continuousand satisfies

Cεν(x) − Cεν(x) =∫|x−y|≤ε

Kε(x− y) dν(y).

Thus,|Cεν(x) − Cεν(x)| ≤ ‖Kε‖∞ |ν|(B(x, ε)) ≤ ‖Kε‖∞εMRν(x).

By Lemma 4.2,

‖Kε‖∞ ≤ 2πε ‖ϕε‖∞ =2π

ε‖ϕ‖∞,

and so (4.2) follows with Cϕ = 2π ‖ϕ‖∞.The last assertions in the lemma follow from (4.2) applied to f μ, because

the maximal operator MR is bounded in Lp(μ) and from M(C) to L1,∞(μ) whenμ has linear growth. �Example. Let ϕ = π−1χB(0,1). Then one can check that

−1

z∗ ϕε =

⎧⎪⎪⎨⎪⎪⎩−1

zif |z| > ε,

− z

ε2if |z| ≤ ε.

(4.3)

We will need the following technical lemma below.

Lemma 4.4. Let ϕ and Cε be as in Lemma 4.3 and let ν be a complex measure.For 0 < ε ≤ δ, we have

‖Cδν‖∞ ≤ ‖Cεν‖∞ + C′ϕ‖MRν‖∞, (4.4)

with C′ϕ = 12π‖ϕ‖∞. Also,

‖Cδν‖∞ ≤ ‖Cεν‖∞ + C′′ϕ‖MRν‖∞. (4.5)

Proof. Write Kε =−1

z∗ ϕε and Kε,δ =

−1

z∗ ϕε ∗ ϕδ. Notice that

Kε,δ ∗ ν = (Kε ∗ ν) ∗ ϕδ = Cεν ∗ ϕδ,

and thus‖Kε,δ ∗ ν‖∞ ≤ ‖Cεν‖∞.

Recall that Kδ(z) = −1/z if z ≥ δ, and since supp(ϕδ ∗ ϕε) ⊂ B(0, ε + δ), byLemma 4.2, Kε,δ(z) = −1/z if z ≥ ε+ δ. Therefore,

|Kδ ∗ ν(x) −Kε,δ ∗ ν(x)| ≤∫|x−y|≤ε+δ

|Kδ(x− y)−Kε,δ(x− y)| d|ν|(y)

≤ (‖Kδ‖∞ + ‖Kε,δ‖∞) |ν|(B(0, ε+ δ)). (4.6)


By Lemma 4.2 again,

‖Kδ‖∞ ≤ 2πδ ‖ϕδ‖∞ = 2πδ−1 ‖ϕ‖∞and

‖Kε,δ‖∞ ≤ 2π(ε+ δ) ‖ϕε ∗ ϕδ‖∞ ≤ 4πδ‖ϕδ‖∞ = 4πδ−1 ‖ϕ‖∞.

Thus, by (4.6),

|Kδ ∗ ν(x) −Kε,δ ∗ ν(x)| ≤ 6πδ−1 ‖ϕ‖∞ |ν|(B(0, 2δ)) ≤ 12π‖ϕ‖∞ ‖MRν‖∞.

Then (4.4) is deduced from this estimate and the fact that ‖Kε,δ ∗ν‖∞ ≤ ‖Cεν‖∞.

Now, (4.5) follows easily from (4.4) and the estimate ‖Cλν − Cλν‖∞ ≤Cϕ‖MRν‖∞ for every λ > 0, by (4.2). �

Now we turn to the outer regularity of γ+ The arguments are quite standard.

Proposition 4.5. Let En ⊂ C, n ≥ 1, be a sequence of compact sets such thatEn+1 ⊂ En for all n. For E =

⋂n En, we have

γ+(E) = limn→∞ γ+(En).

Proof. Since {γ+(En)}n is a non-increasing sequence, it is convergent. Moreover

γ+(E) ≤ limn→∞ γ+(En),

because E ⊂ En for each n. To prove the converse inequality, given ε > 0, for eachn let μn be a (positive) measure supported on En such that γ+(En) ≤ μ(En) + εand ‖Cμn‖L∞(C) ≤ 1. Let μ be a weak ∗ limit of a subsequence {μnk

}k. Then μ issupported on E. To check that ‖Cμ‖L∞(C) ≤ 1 we will use the mollification trick:we consider the function ϕ = π−1χB(0,1), and for each ε > 0 we set

Cεμnk:= ϕε ∗ (Cμnk

) =

(−1

z∗ ϕε

)∗ μnk

.

Since ‖Cμnk‖L∞(C) ≤ 1, it turns out that

‖Cεμnk)‖L∞(C) = ‖ϕε ∗ (Cμnk

)‖L∞(C) ≤ 1.

Moreover, by the weak ∗ convergence of μnkto μ and the continuity of the kernel

−1z ∗ ϕε, (−1

z∗ ϕε

)∗ μnk

(z) →(−1

z∗ ϕε

)∗ μ(z) for all z ∈ C.

As a consequence, ‖Cεμ‖L∞(C) ≤ 1. Since by the Lebesgue differentiation theorem

Cεμ(z) → Cμ(z) for a.e. z ∈ C as ε → 0,

we infer that ‖Cμ‖L∞(C) ≤ 1.Finally we obtain

γ+(E) ≥ μ(E) = limk→∞

μnk(Enk

) ≥ lim supk→∞

γ+(Enk)− ε = lim

n→∞ γ+(En)− ε. �

4.3. Dualization of the weak (1,1) inequality 107

4.3 Dualization of the weak (1,1) inequality

In this section we show how the weak (1, 1) inequality for the Cauchy transformcan be dualized. This is a key step which will allow us to connect analytic capacityand the capacity γ+ to the L2 boundedness of the Cauchy transform.

Given a locally compact Hausdorff space X , let C0(X) be the set of C-valuedcontinuous functions vanishing at ∞. This is a Banach space when equipped withthe sup norm. Let M(X) be the space of complex Radon measures, which is aBanach space dual from C0(X) when it is equipped with the total variation norm.

Let X and Y be locally compact Hausdorff spaces, and T : M(X) → C0(Y )a bounded linear operator. We say that T t : M(Y ) → C0(X) is the transpose ofT if ∫

T (ν) dσ =

∫T tσ dν for all ν ∈ M(X) and σ ∈ M(Y ).

The next theorem is due to Davie and Øksendal [31], although for the proofwe follow the approach from Christ [13, p. 107] quite closely.

Theorem 4.6. Let X,Y be locally compact Hausdorff spaces, μ a Radon measure onX, and T : M(X) → C0(X) a bounded linear operator. Suppose that the transposeoperator T t : M(Y ) → C0(X) is bounded from M(Y ) to L1,∞(μ), that is, there issome constant A such that

μ({x ∈ X : |T tν(x)| > λ}) ≤ A

‖ν‖λ

∀λ > 0

for all ν ∈ M(X). Then for each Borel set E ⊂ X with μ(E) < ∞ there existssome Borel function h supported on E, with 0 ≤ h ≤ 1, such that

μ(E) ≤ 2

∫h dμ and ‖T (hμ)‖∞ ≤ 3A.

Proof. Suppose that this fails for some Borel set E ⊂ X . Write

B0 = {f : X → [0, 1] : f is a Borel function such that

f = 0 in X \ E and∫f dμ ≥ μ(E)/2},

B1 = {T (fμ) : f ∈ B0},B2 = {g ∈ C0(Y ) : ‖g‖∞ < 3A}.

Since B1 and B2 are disjoint convex sets from C0(Y ) and B2 is open, by the Hahn-Banach theorem (see Rudin [142, Theorem 4.3]) there exists some ν ∈ C0(Y )∗ =M(Y ) such that Re ν(h) > Re ν(g) for all h ∈ B1 and g ∈ B2. This is equivalentto saying that

Re

(∫T (f μ) dν

)> Re

(∫g dν

)for all f ∈ B0, g ∈ B2.


Since

supg∈B2

Re

(∫g dν

)= 3A‖ν‖,

we infer that

Re

(∫f T tν dμ

)≥ 3A‖ν‖ for all f ∈ B0. (4.7)

By the weak (1, 1) inequality above with λ = 2A‖ν‖/μ(E), we deduce

μ({x ∈ X : |T tν(x)| > λ}) ≤ A

‖ν‖λ

=μ(E)

2,

and thus

μ({x ∈ E : |T tν(x)| ≤ λ}) ≥ μ(E)

2.

Let f be the characteristic function of the set in the left-hand side above, so thatf ∈ B0. Then we have ∣∣∣∣∫ f T tν dμ

∣∣∣∣ ≤ λμ(E) = 2A‖ν‖,

which contradicts (4.7). �

The following lemma relates weak (1, 1) estimates for the Cauchy integraloperator to L∞ estimates (which are then connected to γ+ and γ).

Lemma 4.7. let μ be a Radon measure with linear growth on C. The followingstatements are equivalent:

(a) The Cauchy transform is bounded from M(C) to L1,∞(μ).

(b) For any Borel E ⊂ C with μ(E) < ∞ there exists some function h supportedon E, with 0 ≤ h ≤ 1, such that

μ(E) ≤ 2

∫h dμ, and ‖Cε(hμ)‖L∞(C) ≤ c ∀ε > 0.

The constant c in (b) depends only on the norm of the Cauchy transform fromM(C) to L1,∞(μ), and conversely.

Proof. First we show (b)⇒ (a). It is enough to prove that for any complex measureν ∈ M(C) and any λ > 0,

μ({x ∈ C : Re(Cεν(x)) > λ}) ≤ c‖ν‖

λ.

4.3. Dualization of the weak (1,1) inequality 109

To this end, let E be a Borel subset of the set on left-hand side above, withμ(E) < ∞, and let h be a function supported on E fulfilling the properties in thestatement (b) of the lemma. Then we have

μ(E) ≤ 2

∫h dμ ≤ 2

λRe

(∫(Cεν)h dμ

)=

−2

λRe

(∫Cε(hμ) dν

)≤ c‖ν‖

λ.

Let us turn our attention to the implication (a) ⇒ (b). We may assume Eto be compact, by the inner regularity of μ. Consider the function ϕ = π−1χB(0,1)

and, for each ε > 0, the integral operator Cε associated with the smooth kernel−1z ∗ε, as in Section 4.2. If the Cauchy transform is bounded fromM(C) to L1,∞(μ),

then the operators Cε are also bounded from M(C) to L1,∞(μ) uniformly on ε.They are also bounded from M(C) to C0(C) with norm depending on ε. On the

other hand, for each ε, the transpose of Cε is −Cε, and so, by Theorem 4.6, thereexists some function hε supported on E, with 0 ≤ hε ≤ 1, such that

μ(E) ≤ 2

∫hε dμ and ‖Cε(hε μ)‖∞ ≤ c.

Since ‖MR(hε μ)‖∞ ≤ ‖MRμ‖∞ ≤ c, from Lemma 4.4 we deduce that

‖Cδ(hε μ)‖∞ ≤ c′ for δ ≥ ε.

Let h : C → [0, 1] be a weak-� limit in L∞(μ) of some subsequence {hεk}k≥0, withεk ↘ 0, so that h is also supported on E and

μ(E) ≤ 2

∫h dμ.

Since ‖Cδ(hεk μ)‖∞ ≤ c′ for δ ≥ εk, and the kernel of Cδ is in L1loc(μ), we infer

that‖Cδ(hμ)‖∞ ≤ c′ for every δ > 0,

and thus, by (4.2),

‖Cδ(hμ)‖∞ ≤ c′′ for every δ > 0. �

Remark 4.8. Notice that if E supports a non-zero Radon measure μ with lineargrowth such that the Cauchy integral operator Cμ is bounded in L2(μ), then thereexists some non-zero function h with 0 ≤ h ≤ χE such that ‖Cε(hμ)‖L∞(C) ≤ cuniformly on ε, by Theorem 2.16 and the preceding lemma. Letting ε → 0, we inferthat |C(hμ)(z)| ≤ c for all z �∈ E also for a.e. z ∈ C (with respect to Lebesguemeasure), and so γ(E) ≥ γ+(E) > 0. A more precise result will be proved inTheorem 4.14 below.


4.4 The Denjoy conjecture

In this section we will prove the so-called Denjoy conjecture:

Theorem 4.9 (The Denjoy conjecture). Let Γ ⊂ C be a rectifiable curve and E ⊂ Γcompact. Then γ(E) > 0 if and only if and H1(E) > 0.

Recall that a set E ⊂ C is called rectifiable (or countably rectifiable) if itis H1-almost all contained in a countable union of curves of finite length, andit is purely unrectifiable it does not have rectifiable subset with positive length.An equivalent way of stating Denjoy’s conjecture consists in saying that, for arectifiable set E ⊂ C, γ(E) > 0 if and only if H1(E) > 0.

To prove Theorem 4.9 we need first a couple of lemmas.

Lemma 4.10. Let F ⊂ C satisfy the following property:∣∣∣∣ Im z − Imw

Re z − Rew

∣∣∣∣ ≤ cF for all z, w ∈ F , (4.8)

for some fixed constant cF . Then F is contained in the graph of a Lipschitz functionA : R → R.

Proof. From (4.8), it turns out that if z, w ∈ F and Re z = Rew, then z = w.Thus we can define the map

A : ReF −→ ImF ⊂ R

Re z �−→ Im z.

By (4.8), it is clear that A is Lipschitz with constant ≤ cF . Moreover, A can beextended to a Lipschitz map A : R → R by the formula:

A(x) = inf{A(x) + cF |x− y| : y ∈ ReF

}.

It is straightforward to check that A(x) = A(x) if x ∈ ReF , and that, moreover,A is Lipschitz, with Lipschitz constant ≤ cF . �Lemma 4.11. Let Γ be a rectifiable curve and E ⊂ Γ with H1(E) > 0. Then thereexists a compact subset F ⊂ E with H1(F ) > 0 which is contained in the (possiblyrotated) graph of a Lipschitz function A : R → R.

Proof. Consider an arc length parametrization of Γ, g : [a, b] → C, so that Γ =g([a, b]), g is differentiable a.e. in [a, b], and |g′(t)| = 1 a.e. t ∈ [a, b]. Let t0 ∈(a, b) be a Lebesgue point for g′ and suppose (after a rotation if necessary), thatg′(t0) = 1 (i.e. the tangent at g(t0) is horizontal). Let I0 ⊂ [a, b] be an intervalcontaining t0 such that ∫

I0

|g′(t)− g′(t0)| dt ≤ 1

20L1(I0). (4.9)

4.4. The Denjoy conjecture 111

Let G = {t ∈ I0 : ∃ g′(t) and |g′(t) − 1| ≤ 1/10}. From (4.9), it follows thatL1(G) > 0. For each m ≥ 1, consider the set

Gm =

{t ∈ G :

∣∣∣∣g(s)− g(t)

s− t− g′(t)

∣∣∣∣ ≤ 1

10if |s− t| ≤ 1

m

}.

Since⋃

m≥1 Gm = G, we infer that L1(Gm) > 0 for m big enough. For such an m,

consider an interval J ⊂ I0 of length 1/2m such that L1(J ∩ Gm) > 0. Observethen that if s, t ∈ J ∩Gm, then∣∣∣∣g(s)− g(t)

s− t− 1

∣∣∣∣ ≤ ∣∣∣∣g(s)− g(t)

s− t− g′(t)

∣∣∣∣+ |g′(t)− 1| ≤ 1

10+

1

10=

1

5.

This implies that ∣∣∣∣Re g(s)− Re g(t)

s− t

∣∣∣∣ ≥ 1− 1

5=

4

5

and ∣∣∣∣ Im g(s)− Im g(t)

s− t

∣∣∣∣ ≤ 1

5.

Thus, ∣∣∣∣ Im g(s)− Im g(t)

Re g(s)− Re g(t)

∣∣∣∣ ≤ 1

4for all s, t ∈ J ∩Gm.

Therefore, by the preceding lemma, g(J ∩ Gm) is contained in the graph of aLipschitz function y = A(x). Thus, by taking a compact subset F0 ⊂ J ∩Gm withL1(F0) > 0 and letting F = g(F0), we are done. �Proof of Denjoy’s conjecture. Recall that, by Theorem 1.21, γ(E) ≤ H1∞(E) ≤H1(E). Thus we only have to show that, for E ⊂ Γ with H1(E) > 0, we haveγ(E) > 0. By Lemma 4.11, there exists a compact subset F ⊂ E with H1(F ) > 0contained in a (possibly rotated) Lipschitz graph. Since the Cauchy transform isbounded in L2 with respect to arc length on this Lipschitz graph, it is also boundedin L2(H1�F ). As shown in Remark 4.8, γ+(F ) > 0, and thus γ(E) ≥ γ+(E) ≥γ+(F ) > 0. �

Notice that, in fact, the preceding argument shows that γ+(E) > 0 for E ⊂ Γwith H1(E) > 0. So what we have proved can be restated as follows.

Corollary 4.12. Let E ⊂ C be a subset with finite length, that is, H1(E) < ∞. Ifγ+(E) = 0, then E is purely unrectifiable.

Finally in this section, using Lemma 4.11, we prove the following character-ization of rectifiable sets with finite length, which we will need later.

Proposition 4.13. Let E ⊂ C. Then E is (countably) rectifiable if and only if thereexist Z ⊂ C with H1(Z) = 0 and a family {Γi}i≥0 of possibly rotated Lipschitzgraphs such that

E ⊂ Z ∪⋃i

Γi.


Proof. It is clear that if E ⊂ Z ∪ ⋃i Γi, with Z and Γi as above, then E isrectifiable.

To prove the converse implication it is enough to show that if G ⊂ C is arectifiable curve with finite length, then there exist Z and Γi as above such that

G ⊂ Z ∪⋃i

Γi.

To this end, write s = supB H1(G ∩ B), where the sup is taken over all sets Bwhich are given by a countable union of possibly rotated Lipschitz graphs. Thenfrom Lemma 4.11 it follows easily that s = H1(Γ).

For every j ≥ 1, take a set Bj which coincides with a countable union ofpossibly rotated Lipschitz graphs such that H1(G ∩ Bj) ≥ s − 1/j. Consider theset B =

⋃j Bj . Then Z = G \ B has zero length and B can be written as a

countable union of possibly rotated Lipschitz graphs, and thus the propositionfollows. �

4.5 Semiadditivity of γ+ and its characterization in

terms of curvature

We denote by Σ(E) the set of Radon measures supported onE such that μ(B(x, r))≤ r for all x ∈ C, r > 0.

Theorem 4.14. For any compact set E ⊂ C we have

γ+(E) ≈ sup{μ(E) : μ ∈ Σ(E), ‖Cεμ‖L∞(μ) ≤ 1 ∀ε > 0

}≈ sup{μ(E) : μ ∈ Σ(E), ‖Cεμ‖2L2(μ) ≤ μ(E)∀ε > 0

}≈ sup{μ(E) : μ ∈ Σ(E), c2(μ) ≤ μ(E)

}≈ sup{μ(E) : μ ∈ Σ(E), ‖Cμ‖L2(μ)→L2(μ) ≤ 1

}.

In the statement above, ‖Cμ‖L2(μ)→L2(μ) stands for the operator norm of Cμon L2(μ). That is, ‖Cμ‖L2(μ)→L2(μ) = supε>0 ‖Cμ,ε‖L2(μ)→L2(μ).

Proof. We write

S1 := sup{μ(E) : μ ∈ Σ(E), ‖Cεμ‖L∞(μ) ≤ 1 ∀ε > 0

},

S2 := sup{μ(E) : μ ∈ Σ(E), ‖Cεμ‖2L2(μ) ≤ μ(E)∀ε > 0

},

S3 := sup{μ(E) : μ ∈ Σ(E), c2(μ) ≤ μ(E)

},

S4 := sup{μ(E) : μ ∈ Σ(E), ‖Cμ‖L2(μ)→L2(μ) ≤ 1

}.

We will show that γ+(E) � S1 � S2 ≈ S3 � S4 � γ+(E). The inequality S3 � S4

requires more work than the others. We will prove this in two different ways: oneuses the T 1 theorem and the other not (and so it is more elementary).

4.5. Semiadditivity of γ+ 113

Proof of γ+(E) � S1. Let μ be supported on E such that ‖Cμ‖L∞(C) ≤ 1 withγ+(E) ≤ 2μ(E). It is enough to show that μ has linear growth and ‖Cεμ‖L∞(μ) ≤ cuniformly on ε > 0.

First we will prove the linear growth of μ. For any x ∈ C and R > 0, we have∫∫z:|z−x|≤R

1

|z − ξ| dL2(z) dμ(ξ) ≤ 2πRμ(E).

Then by Fubini it turns out that for almost all r > 0,∫∫z:|z−x|=r

1

|z − ξ| dH1(z) dμ(ξ) < ∞.

For such r, by Fubini again, we have

μ(B(x, r)) = −∫|z−x|=r

Cμ(z) dz

2πi≤ r.

By approximation the same estimate holds for every r > 0, and thus the lineargrowth of μ follows.

To deal with the L∞(μ) norm of Cε, we use the smoothed operator Cε associ-ated with the kernel −1

z ∗ (πε2)−1χB(0,ε). Since μ is compactly supported, we havethe following identity:

Cεμ =1

z∗ χε

πε2∗ μ =

χε

πε2∗ Cμ.

This equality must be understood in the sense of distributions and Cμ as a functionfrom L1

loc(C). As a consequence, if ‖Cμ‖L∞(C) ≤ 1, we infer that ‖Cεμ‖L∞(C) ≤ 1

and so ‖Cεμ‖L∞(μ) ≤ 1 by the continuity of Cεμ, for all ε > 0. Since μ has lineargrowth, we have

|Cεμ(x) − Cεμ(x)| ≤ c for all x ∈ C.,

and so ‖Cεμ‖L∞(μ) ≤ c uniformly on ε > 0.

Proof of S1 � S2. Trivial.

Proof of S2 ≈ S3. This is a direct consequence of Proposition 3.3.

Proof of S3 � S4 using the T 1 theorem. Let μ be supported on E with lineargrowth such that c2(μ) ≤ μ(E) and S3 ≤ 2μ(E). We set

A :={x ∈ E : c2μ(x) ≤ 2

}.

By Chebyshev μ(A) ≥ μ(E)/2. Moreover, for any set B ⊂ C,

c2(μ�B ∩ A) ≤∫∫∫

x∈B∩A

c(x, y, z)2 dμ(x)dμ(y)dμ(z)

=

∫x∈B∩A

c2μ(x) dμ(x) ≤ 2μ(B).


In particular, this estimate holds when B is any square in C, and so Cμ�A isbounded on L2(μ�A), by Theorem 3.5. Thus S4 � μ(A) ≈ S3.

Proof of S3 � S4 using Lemma 3.6. Take μ supported on E with linear growthsuch that c2(μ) ≤ μ(E) and S3 ≤ 2μ(E). By Proposition 3.3, we deduce that‖Cεμ‖2L2(μ) ≤ cμ(E) uniformly on ε > 0. By Lemma 3.6, there exists G ⊂ E with

μ(G) ≥ μ(E)/4 such that Cμ�G : L2(μ�G) → L2(μ�G) is bounded with normbounded above by some absolute constant. Thus, the measure ν = μ�G is sup-ported on E, has linear growth, and satisfies ν(E) ≥ μ(E)/4 and ‖Cν‖L2(ν),L2(ν) ≤c.

Proof of S4 � γ+(E). This is a direct consequence of Lemma 4.7, the subsequentRemark 4.8, and the fact that the L2(μ) boundedness of Cμ implies its boundednessfrom M(C) to L1,∞(μ), as shown in Theorem 2.16. �

Recall that for an arbitrary set A ⊂ C, γ+(A) = sup{γ+(E) : E ⊂ Acompact}. It is straightforward to check that the characterizations on γ+ obtainedin Theorem 4.14 also hold for arbitrary sets A ⊂ C.

From the preceding theorem, since the term

sup{μ(E) : μ ∈ Σ(E), ‖Cμ‖L2(μ)→L2(μ) ≤ 1

}is countably semiadditive, we deduce that γ+ is also countably semiadditive.

Corollary 4.15. The capacity γ+ is countably semiadditive on Borel sets. That is,if Ei, i = 1, 2, . . ., is a countable (or finite) family of Borel sets, we have

γ+

( ∞⋃i=1

Ei

)≤ C

∞∑i=1

γ+(Ei).

Proposition 4.16. If ν is any complex measure on C, then

γ+({x : C∗ν(x) > λ}) ≤ C‖ν‖λ

. (4.10)

In [173] Verdera asked if the estimate (4.10) is true with γ+ replaced by γ.Obviously, one obtains a positive answer after showing that γ ≈ γ+ holds.

Proof. Let G = {x : C∗ν(x) > λ}. Take μ ∈ Σ(G) such that γ+(G) ≈ μ(G) so thatthe Cauchy transform is bounded in L2(μ) with norm ≤ 1. Then the maximalCauchy transform is bounded from M(C) to L1,∞(μ) and so

γ+(G) � μ(G) = μ({C∗ν(x) > λ}) � ‖ν‖

λ. �

4.6. Some potential theory for γ+ 115

4.6 Some potential theory for γ+

Another consequence of Theorem 4.14 is that the capacity γ+ can be characterizedin terms of the following potential, introduced by Verdera [175]:

Uμ(x) = MRμ(x) + c2μ(x)1/2. (4.11)

The precise result is the following.

Corollary 4.17. For any compact set E ⊂ C we have

γ+(E) ≈ sup{μ(E) : supp(μ) ⊂ E, Uμ(x) ≤ 1 ∀x ∈ E

}. (4.12)

The proof of this corollary follows easily from the fact that

γ+(E) ≈ sup{μ(E) : μ ∈ Σ(E), c2(μ) ≤ μ(E)

},

using Chebyshev.The preceding characterization of γ+ in terms of Uμ is interesting because it

suggests that some techniques of potential theory could be useful to study γ+. Inthis section we will obtain some results which support this assertion.

4.6.1 A couple of technical lemmas about curvature

Next we show how the Menger curvature of three points changes as one of thesepoints moves.

Lemma 4.18. Let x, y, z ∈ C be three pairwise different points, and let x′ ∈ C besuch that

A−1|x− y| ≤ |x′ − y| ≤ A|x− y|, (4.13)

where A > 0 is some constant. Then

|c(x, y, z)− c(x′, y, z)| ≤ (4 + 2A)|x− x′|

|x− y||x− z| . (4.14)

Proof. Since x �= y, we have x′ �= y by (4.13). If x′ = z, then c(x′, y, z) = 0. Inthis case, (4.14) is straightforward:

|c(x, y, z)− c(x′, y, z)| = c(x, y, z) ≤ 2

|x− y| = 2|x− x′|

|x− y||x− z| .

For x′ �= y and x′ �= z we have

|c(x, y, z)− c(x′, y, z)| =∣∣∣∣2 dist(x, Lyz)

|x− y||x− z| −2 dist(x′, Lyz)

|x′ − y||x′ − z|∣∣∣∣

≤ 2| dist(x, Lyz)− dist(x′, Lyz)|

|x− y||x− z|+ 2 dist(x′, Lyz)

∣∣∣∣ 1

|x− y||x− z| −1

|x′ − y||x′ − z|∣∣∣∣

= I + II. (4.15)


To estimate the term I, notice that | dist(x, Lyz)− dist(x′, Lyz)| ≤ |x− x′| and so

I ≤ 2|x− x′|

|x− y||x− z| .

We turn now to the term II in (4.15). We have

II = 2 dist(x′, Lyz)

∣∣∣∣ |x′ − y||x′ − z| − |x− y||x− z||x− y||x− z||x′ − y||x′ − z|

∣∣∣∣ .Now we write

| |x′ − y||x′ − z| − |x− y||x− z| |= | (|x′ − y| − |x− y|) |x′ − z|+ (|x′ − z| − |x− z|) |x− y| |≤ |x′ − x||x′ − z|+ |x′ − x||x − y|.

Thus, using that dist(x′, Lyz) ≤ |x′ − y| and dist(x′, Lyz) ≤ |x′ − z|, we obtain

II ≤ 2|x− x′|(

dist(x′, Lyz)|x′ − z||x− y||x− z||x′ − y||x′ − z| +

dist(x′, Lyz)|x− y||x− y||x− z||x′ − y||x′ − z|

)≤ 2

|x− x′||x− y||x− z| + 2

|x− x′||x′ − y||x− z| ≤ (2 + 2A)

|x− x′||x− y||x− z| .

Now, adding the inequalities obtained for I and II, we get (4.14). �

Given two Radon measures ν and μ (with no point masses, say) and x ∈ C,we write

c2(x, μ, ν) =

∫∫1

R(x, y, z)2dμ(y)dν(z).

We also set

c(x, μ, ν) =(c2(x, μ, ν)

)1/2.

Our next objective consists in estimating the difference between c(x, μ, ν)and c(x, μ, ν) when x and x′ are far from the supports of μ and ν.

Lemma 4.19. Let x, x′ ∈ C and μ and ν Radon measures on C. Let r > 0 andsuppose that x′ ∈ B(x, r) and supp(μ) ∪ supp(ν) ⊂ B(x, 2r)c. Then

|c(x, μ, ν)− c(x′, μ, ν)| ≤ C(MRμ(x)MRν(x)

)1/2 ≈ (MRμ(x′)MRν(x

′))1/2

.(4.16)

Proof. It is immediate to check thatMRμ(x) ≈ MRμ(x′) andMRν(x) ≈ MRν(x

′),using that μ and ν are supported far away from x and x′. Thus,

MRμ(x)MRν(x) ≈ MRμ(x′)MRν(x

′).


Also, by Lemma 4.18,∣∣∣∣(∫∫ c(x, y, z)2 dμ(y)dν(z)

)1/2−(∫∫

c(x′, y, z)2 dμ(y)dν(z))1/2∣∣∣∣

≤(∫∫

|c(x, y, z)− c(x′, y, z)|2 dμ(y)dν(z))1/2

≤ c

(∫∫ |x− x′|2|x− y|2|x− z|2 dμ(y)dν(z)

)1/2≤ cr

(∫1

|x− y|2 dμ(y)

∫1

|x− z|2 dν(z)

)1/2.

Since |x − y| > r for y ∈ supp(μ), it follows by standard estimates (splitting thedomain of integration into annuli, for example) that∫

1

|x− y|2 dμ(y) ≤ c

rMRμ(x),

and analogously, ∫1

|x− z|2 dν(z) ≤ c

rMRν(x),

and then (4.16) follows. �

4.6.2 A maximum principle for curvature and for Uμ

In the next lemma we prove a kind of maximum principle for c2(x, μ, ν) which willbe needed to obtain a new characterization for γ+, dual to the one in Corollary4.17.

Lemma 4.20. Let μ and ν be positive Radon measures such that

c2(x, μ, ν) ≤ β for all x ∈ supp(ν).

Then

c2(x, μ, ν) ≤ 2β + C1 MRν(x) min(MRμ(x), sup

z∈supp(ν)

MRμ(z))

for all x ∈ C,

(4.17)where C1 is some absolute constant.

Proof. We only have to prove (4.17) when x �∈ supp(ν). Let r = dist(x, supp(ν)) >0, and let x′ ∈ supp(ν) be such that |x−x′| = r. Then we split c2(x, μ, ν) as follows:

c2(x, μ, ν) = c2(x, μ�B(x, 2r), ν�B(x, 2r)) + c2(x, μ�B(x, 2r), ν�B(x, 2r)c)

+ c2(x, μ�B(x, 2r)c, ν�B(x, 2r)) + c2(x, μ�B(x, 2r)c, ν�B(x, 2r)c).(4.18)


We shall estimate each one of the terms on the right-hand side above separately.For the first one, using c(x, y, z) ≤ 2|x− z|−1 ≤ 2r−1 for z ∈ supp(ν), we obtain

c2(x, μ�B(x, 2r), ν�B(x, 2r)) =4

r2

∫|y−x|<2r

∫|x−z|<2r

dμ(y)dν(z)

≤ 4μ(B(x, 2r)) ν(B(x, 2r))

r2.

We have

μ(B(x, 2r))

r≤ 2MRμ(x) and

ν(B(x, 2r))

r≤ 2MRν(x). (4.19)

For the first quotient, we can also use the estimate

μ(B(x, 2r))

r≤ μ(B(x′, 3r))

r≤ 3MRμ(x

′). (4.20)

Thus we get

c2(x, μ�B(x, 2r), ν�B(x, 2r)) ≤ cMRν(x) min(MRμ(x), MRμ(x

′)).

For the second term in (4.18) we have:

c2(x, μ�B(x, 2r), ν�B(x, 2r)c) ≤∫y∈B(x,2r)

∫|x−z|≥2r

4

|x− z|2 dμ(y)dν(z)

= μ(B(x, 2r))

∫|x−z|≥2r

4

|x− z|2 dν(z).

Using the estimates (4.19) and (4.20) for μ(B(y, 2r)) and the fact that the lastintegral above does not exceed cMRν(x), we get again

c2(x, μ�B(x, 2r), ν�B(x, 2r)c) ≤ cMRν(x) min(MRμ(x), MRμ(x

′)).

For the third term we argue analogously:

c2(x, μ�B(x, 2r)c, ν�B(x, 2r)) ≤∫|x−y|≥2r

∫z∈B(x,2r)

4

|x− y|2 dμ(y)dν(z)

= ν(B(x, 2r))

∫|x−y|≥2r

4

|x− y|2 dμ(y)

≤ cMRν(x)MRμ(x).

Taking into account that |x− y| ≈ |x′ − y| in the integral above, we also have∫|x−y|≥2r

4

|x− y|2 dμ(y) ≤ c

∫|x′−y|>c−1r

1

|x′ − y|2 dμ(y) ≤ cMRμ(x′),


and thus

c2(x, μ�B(x, 2r)c, ν�B(x, 2r)) ≤ cMRν(x) min(MRμ(x), MRμ(x

′)).

Finally, we consider the last term in (4.18). Applying Lemma 4.19 toμ�B(x, 2r)c and ν�B(x, 2r))c, we get

|c(x, μ�B(x, 2r)c, ν�B(x, 2r)c)− c(x′, μ�B(x, 2r)c, ν�B(x, 2r)c)|

�(MR (μ�B(x, 2r)c) (x)MR (ν�B(x, 2r)c) (x)

)1/2.

Therefore,

c2(x, μ�B(x, 2r)c, ν�B(x, 2r)c) ≤ 2c2(x′, μ, ν) + cMR (μ�B(x, 2r)c) (x)MRν(x).

Since MR (μ�B(x, 2r)c) (x) ≈ MR (μ�B(x, 2r)c) (x′), we have

MR (μ�B(x, 2r)c) (x) � min(MRμ(x), MRμ(x

′)),

and thus

c2(x, μ�B(x, 2r)c, ν�B(x, 2r)c)

≤ 2c2(x′, μ, ν) + cMRν(x) min(MRμ(x), MRμ(x

′)).

Adding all the estimates obtained for the terms in (4.18), recalling that x′ ∈supp(ν), we derive (4.17). �

Let us remark that, by slightly more accurate estimates, one can prove that

c2(x, μ, ν)1/2 ≤ β1/2 + C1 MRν(x)1/2 ·min

(MRμ(x), sup

z∈supp(ν)

MRμ(z))1/2

.

Also, 2β can be replaced by αβ in (4.17), for arbitrary 0 < α < 1, at the price ofincreasing C1.

Corollary 4.21. If μ is a Radon measure such that Uμ(x) ≤ A for all x ∈ supp(μ),then

Uμ(x) ≤ cA for all x ∈ C,


Proof. Since MRμ(x) ≤ A for all x ∈ supp(μ), we infer that for y �∈ supp(μ),

μ(B(y, r)) ≤ μ(B(y′, 2r)) for all r > 0,

where y′ is a closest point to y in supp(μ). Thus, MRμ(y) ≤ 2A for all x ∈ C.From the preceding lemma with μ = ν, since c2(x, μ, μ) ≤ A2 for all x ∈ supp(μ),we deduce that

c2(y, μ, μ) ≤ cA2. �Corollary 4.22. For any compact set E ⊂ C we have

γ+(E) ≈ sup{μ(E) : supp(μ) ⊂ E, Uμ(x) ≤ 1 ∀x ∈ C

}. (4.21)


4.6.3 A dual version for the capacity γ+

In this section we will obtain a characterization of γ+ in terms of the potential Uwhich can be understood as a dual version of (4.12) or (4.21).

Theorem 4.23. If μ is any positive Radon measure on C, then

γ+({x ∈ C : Uμ(x) > λ}) ≤ c‖μ‖λ

. (4.22)

Proof. Consider the Borel sets

E = {x : Uμ(x) > λ},E1 = {x : MRμ(x) > λ/2}

andE2 = {x : MRμ(x) ≤ λ/2 and cμ(x) > λ/2}.

Hence, E ⊂ E1 ∪ E2, and γ+(E) ≤ c(γ+(E1) + γ+(E2)).It is not difficult to check that

γ+(E1) ≤ c‖μ‖λ

. (4.23)

Indeed, we only have to take ν ∈ Σ(E1) such that γ+(E1) ≈ ‖ν‖ and c2ν(x) ≤ 1for all x ∈ C. By the linear growth of ν, the maximal operator MR is boundedfrom M(C) to L1,∞(ν), and thus we get

γ+(E1) ≤ c ν(E1) = ν({MRμ(x) > λ/2}) ≤ c

‖μ‖λ

.

Now we will show that

γ+(E2) ≤ c‖μ‖λ

. (4.24)

Let us take ν ∈ Σ(E2), such that γ+(E2) ≈ ‖ν‖ and c2ν(x) ≤ 1 for all x ∈ C.Observe that

c2(μ, ν, ν) ≤ ‖μ‖ = β‖ν‖,where β = ‖μ‖/‖ν‖. Then there exists a closed subset F ⊂ supp(ν) such thatν(F ) ≥ ‖ν‖/4 and c2(x, μ, ν) ≤ 2β for all x ∈ F . Therefore, c2(x, μ, ν�F ) ≤ 2β forall x ∈ F , and by Lemma 4.20,

c2(x, μ, ν�F ) ≤ c λ+ 4β

for all x ∈ C, since MRμ(z) ≤ λ/2 for all z ∈ supp(ν). Thus, using Fubini,

γ+(E2) ≤ c

∫dν�F ≤ c

λ2

∫c2(x, μ, μ) dν�F (x)

=c

λ2

∫c2(y, μ, ν�F ) dμ(y) ≤ c

λ2

∫(c λ+ 4β) dμ

≤ c

λ2

(λ ‖μ‖+ ‖μ‖2

‖ν‖).


So we obtain

γ+(E2) ≤ c2

(‖μ‖λ

+‖μ‖2

λ2 γ+(E2)

). (4.25)

From this inequality we derive (4.24). Indeed, if ‖μ‖/λ ≥ ‖μ‖2/(λ2γ+(E2)), then(4.25) implies γ+(E2) ≤ 2c2

1λ‖μ‖, and if ‖μ‖/λ ≤ ‖μ‖2/(λ2γ+(E2)) we also get

(4.24). �Theorem 4.24. For E ⊂ C compact, we have

γ+(E) ≈ inf{‖μ‖ : μ ∈ M+(C), Uμ(x) ≥ 1 ∀x ∈ E}. (4.26)

Before proving Theorem 4.24 we will state and prove the following corollary.

Corollary 4.25. Let E ⊂ C be compact. Then γ+(E) = 0 if and only if there existssome measure μ ∈ M+(C) such that Uμ(x) = ∞ for all x ∈ E.

Proof. If Uμ(x) = ∞ for all x ∈ E, then γ+(E) = 0. This follows easily fromTheorem 4.24 or directly from (4.22).

Suppose that γ+(E) = 0. By Theorem 4.24, for each n there exists somemeasure μn ∈ M+(C) with ‖μn‖ ≤ 2−n and such that MRμn(x) + cμn(x) ≥1 for all x ∈ E. We set μ =

∑∞n=1 nμn. Then μ ∈ M+(C) and, for each n,

MRμ(x) + cμ(x) ≥ n (MRμn(x) + cμn(x)) ≥ n for x ∈ E. �Let us remark that, with a little of additional effort, the measure μ in the

proof of the corollary can be constructed so that it also satisfies Uμ(y) < ∞ forall y ∈ Ec.

In Section 3.2 of Chapter 3 we saw that for the corner quarters set E andthe measure μ = H1�E, c2μ(x) = ∞ for all x ∈ E. From the above corollary weinfer that γ+(E) = 0.

We will need the following estimate for the proof of Theorem 4.24:

Lemma 4.26. Let ν ∈ M+(C). Let Q ⊂ C be a closed square with side length δ. LetL be a closed segment of length δ/2 parallel to one side of Q and concentric with

Q. Suppose that ν�Q = a · H1�L and ν

(3

2Q \Q)

= 0. Then there exists some

absolute constant c3 such that for all x, y ∈ Q,

c2ν(x) ≤10

9c2ν(y) + c3 min

(MRν(x), MRν(y)

)2.

Proof. Notice first that, by Lemma 4.19,

|cν�C\ 32Q

(x)− cν�C\ 32Q

(y)| � MR

(ν�C \ 3

2Q)(x) � min

(MRν(x), MRν(y)

).

Therefore,

c2ν�C\ 32Q

(x) ≤ 10

9c2ν�C\ 3

2Q(y) + cmin

(MRν(x), MRν(y)

)2.


On the other, by an application of Lemma 4.20 (or by a direct calculation), iteasily follows that

c2ν� 32Q

(x) = c2ν�L(x) � MR(ν�L)(x)2 ≈ a2 ≈ MR(ν�L)(y)2.

Thus,

c2ν� 32Q

(x) � min(MRν(x), MRν(y)

)2.

Finally, denoting by cQ the center of Q, we have

c2(x, ν� 32Q, ν�C \ 3

2Q) ≤ c

∫z∈L

∫t�∈ 3

2Q

1

|cQ − t|2 dν(z)dν(t),

since |z − t| ≈ |cQ − t| for z, t in the domain of integration. Therefore,

c2(x, ν� 32Q, ν�C \ 3

2Q) � ν(Q)

∫t�∈ 3

2Q

1

|cQ − t|2 dν(t)

� ν(Q)

�(Q)MR(ν�C \ 3

2Q)(cQ)

� min(MRν(x), MRν(y)

)2.

Adding the different estimates for the curvature, our claim follows. �

To prove Theorem 4.24 we will need to apply a variational argument on some‘nice’ approximation of E by another compact E. The following lemma deals withthe variational argument on E.

Lemma 4.27. Consider a grid of squares of side length δ > 0 in C with sides parallelto the axes. Take a finite collection of closed squares {Qi}i∈I of the grid. For eachi ∈ I, let Li be the closed segment of length δ/2 concentric with Qi and parallel

to the x axis. Set E =⋃

i∈I Li. Let Σ0(E) be the subset of Σ(E) of measures μ of

the form μ =∑

i∈I ai H1�Li. There exists a measure ν ∈ Σ0(E) such that

‖ν‖2‖ν‖+ c2(ν)

= supμ∈Σ0(E)

‖μ‖2‖μ‖+ c2(μ)

. (4.27)

Any maximal measure ν satisfies

c2(ν) ≤ 2‖ν‖ (4.28)

and

Uν(x) ≥ c4 (4.29)

for all x ∈ E, where c4 > 0 is some absolute constant.


Proof. The existence of ν follows easily by a compactness argument. We leave thedetails for the reader.

Let us see that (4.28) holds. Since ν is maximal and ν/2 is also in Σ0(E), wehave

‖ν‖2‖ν‖+ c2(ν)

≥ ‖ν‖2/4‖ν‖/2 + c2(ν)/8

.

Therefore,‖ν‖2

+c2(ν)

8≥ ‖ν‖

4+

c2(ν)

4.

That is, c2(ν) ≤ 2‖ν‖.Now we turn our attention to (4.29). Let x ∈ E and let Li be such that

x ∈ Li. Suppose that MRν(x) ≤ 1/20. For each λ > 0, we define the measureνλ = ν + λH1�Li. We claim that if λ is small enough, then

MRνλ(y) ≤ 1 for all y ∈ E. (4.30)

This is easy to check for y ∈ Li. Indeed, taking into account that ν is a constantmultiple of H1�Li on Li, it follows easily that MRν(y) ≤ 1/5 for all y ∈ Li. So,for λ small enough,

MRνλ(y) ≤ 1/4 for all y ∈ Li. (4.31)

Consider now the case y ∈ E\Li. IfB(y, r)∩Li = ∅, then νλ(B(y, r)) = ν(B(y, r)).Otherwise, B(y, r) contains some point y′ ∈ Li and so B(y, r) ⊂ B(y′, 2r). From(4.31) we deduce that

νλ(B(y, r))

r≤ νλ(B(y′, 2r))

r≤ 2Mνλ(y

′) ≤ 1

2.

Thus in any case our claim (4.30) holds. That is to say, νλ ∈ Σ0(E) for λ > 0small enough.

Since ν satisfies (4.27), the function

f(λ) =‖νλ‖2

‖νλ‖+ c2(νλ),

satisfies f ′(0) ≤ 0. Observe that f(λ) equals

(‖ν‖+ λ δ/2)2

‖ν‖+ λ δ/2 + c2(ν) + 3λ c2(H1�Li, ν, ν) + 3λ2c2(ν,H1�Li,H1�Li).

So,

f ′(0) =‖ν‖ δ (‖ν‖+ c2(ν)) − ‖ν‖2 (δ/2 + 3 c2(H1�Li, ν, ν))

(‖ν‖+ c2(ν))2.

Therefore, f ′(0) ≤ 0 if and only if

δ (‖ν‖+ c2(ν)) ≤ ‖ν‖ (δ/2 + 3 c2(H1�Li, ν, ν)).


That is,

‖ν‖+ 2c2(ν)

‖ν‖ ≤ 6 c2(H1�Li, ν, ν)

δ.

Therefore, c2(H1�Li, ν, ν)/δ ≥ 16 . So there exists some x′ ∈ Li such that

c2(x′, ν, ν) ≥ 13 . By Lemma 4.26

c2ν(x) ≥9

10

(1

3− c3 MRν(x)

2

).

If c3 ·MRν(x)2 ≤ 1/6, then

c2ν(x) ≥3

20. (4.32)

So we have proved that if MRν(x) ≤ min(1/20, 1/(6c3)1/2), then (4.32) holds, and

the lemma follows. �

Proof of Theorem 4.24. Let μ ∈ M+(C) be such that MRμ(x) + cμ(x) ≥ 1 for allx ∈ E. By Theorem 1, γ+(E) ≤ c ‖μ‖.

It remains to show that for each ε > 0 there exists some measure μ ∈ M+(C)such that γ+(E) ≥ c ‖μ‖ − ε, where c > 0 is some absolute constant, with μsatisfying MRμ(x) + cμ(x) ≥ 1 on E. From the outer regularity of γ+ on com-pact sets proved in Proposition 4.5, there exists some δ > 0 such that γ+(E) ≥γ+(U2δ(E)) − ε, where U2δ(E) stands for the 2δ-neighborhood of E. Consider agrid of squares of side length δ > 0 with sides parallel to the axes. Let {Qi}i∈I

the closed squares of the grid that intersect E. For each i ∈ I, let Li be the closedsegment of length δ/2 centered in Qi and parallel to the x axis. Let E =

⋃i∈I Li.

Notice that E ⊂ U2δ(E). Thus γ+(E) ≥ γ+(E)− ε.

Let ν ∈ Σ0(E) be the maximal measure of Lemma 4.27 for the compact set

E. Since c2(ν) ≤ 2‖ν‖, we have γ+(E) ≥ c ‖ν‖. We know that MRν(x)+cν(x) ≥ c4for all x ∈ E, with c4 > 0. We are going to show that this also holds for x ∈ E,changing the constant c4 by another constant c′4 smaller enough. Clearly, this willfinish the proof of the theorem: just take μ = ν/c′4.

Let x ∈ E. Suppose that MRν(x) ≤ c5, where c5 > 0 is some constant muchsmaller than c4, which will be fixed below. Let Qi be a square of the grid suchthat x ∈ Qi. It is not difficult to check that for y ∈ Li, MRν(y) ≤ 4 c5. Thus fory ∈ Li, cν(y) ≥ c4 − 4 c5 ≥ c4/2, assuming c5 ≤ c4/8. Now, by Lemma 4.26

c2ν(x) ≥9

10

(c244

− c3 MRν(x)2

)≥ 9

10

(c244

− c3 · c25)

≥ c2416

,

if c5 is chosen small enough. �

4.7. The capacity γ+ of some Cantor sets 125

4.7 The capacity γ+ of some Cantor sets

We construct a Cantor set E(λ) ⊂ C as follows. Let λ := {λn}n ⊂ R be asequence satisfying 0 < λn ≤ 1/3 for all n. In R, we consider the sets K0 = [0, 1],K1 = [0, λ1] ∪ [1 − λ1, 1], and for each n, Kn made up of a finite union of closedintervals which have been obtained from Kn−1 in the following way. We replaceeach connected component Kn−1,j of Kn−1 by the two endmost closed intervalscontained in this same component, each one with length equal to λn times thelength of connected component Kn−1,j. We set

K(λ) =∞⋂

n=1

Kn,

and then

E(λ) = K(λ)×K(λ).

Let us observe that Kn is the union of 2n closed intervals of length �n =λ1 · · ·λn. So, if we set En = Kn ×Kn, we have

E(λ) =∞⋂n=0

En,

with

En =

4n⋃j=1

Qn,j, (4.33)

where Qn,j is a square with side length �n for every j = 1, ..., 4n.

If λn = 1/4 for all n, then E(λ) coincides with the corner quarters Cantorset from Section 1.7 in Chapter 1. Recall that this is a self similar set with positiveand finite 1-dimensional Hausdorff measure which is purely unrectifiable.

The Cantor sets E(λ) were introduced by Garnett in [54], where he asked forwhich sequences λ the set E(λ) has non-zero analytic capacity.

Let μ be the probability measure supported on E(λ) which assigns equalmeasure to each square Qn,j from the same generation n, so that μ(Qn,j) = 4−n.

Notice that the quotient 4−n

ncoincides with the linear density μ(Qn,j)/�(Qn,j) =:

θn for 1 ≤ j ≤ 4n.

Theorem 4.28. Let E(λ) be as above. There exists an absolute constant c such that

c−1( ∞∑k=0

θ2k

)−1/2

≤ γ+(E(λ)) ≤ c( ∞∑k=0

θ2k

)−1/2

. (4.34)

We need the following lemma.


Lemma 4.29. Let μ be the probability measure on E(λ) defined above. There existssome constant C > 0 such that for all x ∈ E(λ),

C−1 c2μ(x) ≤∞∑k=0

θ2k ≤ C c2μ(x). (4.35)

Proof. Let us see the left inequality of (4.35) first. Given x ∈ E(λ) and k ≥ 0 wedenote by Qk(x) the square of generation k that contains x. We have

c2μ(x) ≤ 2

∞∑k=1


∫z∈Qk−1(x)

c(x, y, z)2 dμ(y)dμ(z)

≤ C

∞∑k=1


∫z∈Qk−1(x)

1

�2kdμ(y)dμ(z)

= C

∞∑k=1


4−k

�2kdμ(y) = C

∞∑k=1

(4−k

�k

)2.

Now, we are going to prove the right-hand inequality in (4.35). As in (3.4),we write

c2μ(x) ≥∞∑k=1


∫z∈Qk−1(x)\[Qk(x)∪Qk(y)]

c(x, y, z)2dμ(y)dμ(z).

By the geometry of the set E(λ), given x ∈ E(λ) and y ∈ Qk−1(x) \ Qk(x) wehave

c(x, y, z) ≥ C1

�kfor all z ∈ Qk−1(x) \ (Qk(x) ∪Qk(y)),

for some constant C > 0. Hence,

c2μ(x) ≥ C

∞∑k=1


4−k

�2kdμ(y) ≥ C

∞∑k=1

(4−k

�k

)2,

with C > 0. �

Proof of Theorem 4.28. Let μ the probability measure on E(λ) defined above. Itis straightforward to check that for all x ∈ E(λ), MRμ(x) � supk θk and so

MRμ(x) �(∑

k

θ2k

)1/2.

Thus,

Uμ(x) = MRμ(x) + cμ(x) ≈(∑

k

θ2k

)1/2.

4.8. A quantitative version of Denjoy’s conjecture 127

Consider the measure

μ = μ ·( ∞∑

k=0

θ2k

)−1/2

.

Since Uμ(x) ≈ 1 for all x ∈ E(λ), from Corollary 4.17 we deduce that

γ+(E(λ)) �(∑

k

θ2k

)−1/2

,

and from Theorem 4.24,

γ+(E(λ)) �(∑

k

θ2k

)−1/2

. �

Remark 4.30. Arguing as above, we also get that for En as in (4.33),

γ+(En) ≈( n∑k=0

θ2k

)−1/2

. (4.36)

In particular, for the n-generation of the corner quarters Cantor set, we obtain

γ+(En) ≈ 1

n1/2.

4.8 A quantitative version of Denjoy’s conjecture

In this section we go back to Denjoy’s conjecture. By using the Jones’ travelingsalesman Theorem 3.14 and the relation between β’s and curvature from Theorem3.15, we will prove the following result, which can be understood as a quantitativeversion of Denjoy’s conjecture.

Theorem 4.31. Let Γ ⊂ C be a rectifiable curve and let E ⊂ Γ be compact. Then

γ+(E) ≥ c−1H1∞(E)3/2

H1(Γ)1/2. (4.37)

This was first proved by Murai [116] in the particular case where Γ is arectifiable graph. Verdera realized that the more general statement in Theorem4.31 follows from Jones’ theorems and the relationship between analytic capacityand curvature discovered by Melnikov.

Proof. By Frostman’s lemma, there exists a measure μ supported on E such thatμ(E) ≈ H1

∞(E) and μ(B(x, r)) ≤ r for all x ∈ C and r > 0. By Theorems 3.15and 3.14,

c2(μ) ≤ c∑Q∈D

βΓ(Q)2μ(Q) ≤ c∑Q∈D

βΓ(Q)2�(Q) ≤ c6 H1(Γ),


where c6 is some absolute constant.

Consider now the measure

σ =

(μ(E)

c6H1(Γ)

)1/2μ.

Since μ(E) � H1∞(E) ≤ H1(Γ), it follows that σ has c-linear growth, for some

absolute constant c. On the other hand,

c2(σ) =

(μ(E)

c6H1(Γ)

)3/2c2(μ) ≤ μ(E)3/2

[c6H1(Γ)]1/2= σ(E).

Thus, by Theorem 4.14,

γ+(E) � σ(E) =μ(E)3/2

[c6H1(Γ)]1/2. �

Murai [119] also proved that the power 3/2 in the (4.37) is optimal in thefollowing sense: if H1(Γ) = 1 and E ⊂ Γ, then the estimate γ+(E) � H1

∞(E)α

fails for α < 3/2. We are going to prove this by using a variation of Murai’s ideas.

Consider the n-th generation En of the corner quarters Cantor set. Then itis easy to check that ∂En is contained in a curve Γn of length at most c n, wherec is some absolute constant. Indeed, we may take

Γn =n⋃

j=0

∂Ej .

Note that Γn is a curve with self intersections, by taking a suitable parameteriza-tion. See Figure 4.1. Contracting Γn and ∂En by a factor of c/n, we get a curveGn with H1(Gn) = 1 and a set Fn ⊂ Gn with

H1∞(Fn) ≈ 1

nH1

∞(∂En) ≈ 1

n

and

γ+(Fn) ≈ 1

nγ+(∂En) ≈ 1

n3/2,

since γ(∂En) ≈ 1/n1/2 (see Remark 4.30). As H1∞(Fn)

3/2 ≈ 1/n3/2 too, we have

γ+(Fn) ≈ H1∞(Fn)

3/2,

which shows the sharpness of the power 3/2.

4.9. Relationship between γ+ and Wolff’s potentials 129

Figure 4.1: The union of the sets ∂Ej , 0 ≤ j ≤ 3, which form the curve Γ3.

4.9 Relationship between γ+ and Wolff’s potentials

For α > 0, 1 < p < ∞ with 0 < αp < d, the Riesz capacity Cα,p of a compact setE ⊂ Rd is defined as

Cα,p(E) = supμ

μ(E)p,

where the supremum runs over all positive measures μ supported on E such that

Iα(μ)(x) =

∫1

|x− y|d−αdμ(x)

satisfies ‖Iα(μ)‖p′ ≤ 1, where as usual p′ = p/(p− 1). The capacities Cα,p arise inthe so-called non-linear potential theory

It is easy to check that Cα,p is a homogeneous capacity of degree d−αp, thatis,

Cα,p(λE) = |λ|d−αp Cα,p(E)

for any compact set E ⊂ Rd and λ ∈ R. It is well known that sets with positivecapacity Cα,p have non-σ-finite Hausdorff measure Hd−αp. On the other hand,

not all sets with non-σ-finite Hd−αp measure have positive capacity Cα,p. SeeAdams and Hedberg [1, Chapter 5] for these and many other results on non-linearpotential theory.

For our purposes, the description of Riesz capacities in terms of Wolff poten-tials is more useful than the above definition of Cα,p. Consider

Wμα,p(x) =

∫ ∞

0

(μ(B(x, r))

rd−αp

)p′−1dr

r.

A well-known theorem of Wolff asserts that

Cα,p(E) ≈ supμ

μ(E), (4.38)


where the supremum is taken over all measures μ supported on E such thatWμ

α,p(x) ≤ 1 for all x ∈ E. See Adams and Hedberg [1, Chapter 4], for instance.

Notice that for the indices α = 23 , p = 3

2 , in R2, we have

Wμ2/3,3/2(x) =

∫ ∞

0

(μ(B(x, r))

r

)2dr

r.

Observe also that C2/3,3/2 is homogeneous of degree 1.The next proposition shows the relationship between the potential Uμ(x) =

MRμ(x) + c2μ(x)1/2 introduced in (4.11) and the Wolff potential Wμ

2/3,3/2(x).

Proposition 4.32. Let μ be a finite Radon measure on C. Then there exists anabsolute constant c such that

Uμ(x) ≤ c Wμ2/3,3/2(x)

1/2 for all x ∈ C.

In particular, we have

c2(μ) ≤ c

∫Wμ

2/3,3/2 dμ.

Proof. Clearly, the second statement is a consequence of the first one. To provethe first statement, observe that, for every x ∈ C and r > 0,(

μ(B(x, r))

r

)2≤ c

∫ 2r

r

(μ(B(x, t))

t

)2dt

t≤ c Wμ

2/3,3/2(x),

and thus it follows that MRμ(x) ≤ c Wμ2/3,3/2(x)

1/2.

On the other hand,

c2μ(x) ≤ 2

∫∫|x−z|≤|x−y|

c(x, y, z)2 dμ(y) dμ(z)

≤ 8

∫∫|x−z|≤|x−y|

1

|x− y|2 dμ(y) dμ(z) = 8

∫μ(B(x, |x − y|))

|x− y|2 dμ(y).

Now we write1

|x− y|2 = 2

∫ ∞

|x−y|

1

r3dr,

and by Fubini we get

c2μ(x) ≤ 16

∫∫y,r: |x−y|≤r

μ(B(x, |x − y|))r3

dr dμ(y)

≤ 16

∫ ∞

0

μ(B(x, r))

r3

∫|x−y|≤r

dμ(y) dr

= 16

∫ ∞

0

μ(B(x, r))2

r3dr = 16 Wμ

2/3,3/2(x),

which yields the proposition. �

4.9. Relationship between γ+ and Wolff’s potentials 131

Corollary 4.33. There exists an absolute constant c such that

γ+(E) ≥ c−1 C2/3,3/2(E),

for any compact set E ⊂ C.

Proof. This is a direct consequence of the preceding proposition and the char-acterization of the capacities γ+ and C2/3,3/2 in terms of the potentials Uμ and

Wμ2/3,3/2, respectively. �

The comparability between γ and C2/3,3/2 is false (for instance, if E is a seg-

ment, γ(E) > 0, while C2/3,3/2(E) = 0 because it has finite length). Nevertheless,for Cantor type sets E(λ) such as the ones considered in the previous section itturns out that γ+(E(λ)) ≈ C2/3,3/2(E(λ)):

Proposition 4.34. The Cantor sets E(λ) introduced in (4.33) satisfy

γ+(E(λ)) ≈ C2/3,3/2(E(λ)),

with constants independent of λ.

Proof. Let μ be the probability measure supported on E(λ) which assigns equalmeasure to each square Qn,j from the same generation n, so that μ(Qn,j) = 4−n.Recall that, letting θn := μ(Qn,j)/�(Qn,j), we have

C2/3,3/2(E(λ)) � γ+(E(λ)) ≈( ∞∑n=0

θ2n

)−1/2

.

To prove that C2/3,3/2(E(λ)) �(∑∞

n=0 θ2n

)−1/2

, denote by Qn(x) the square

Qn,j from the n-th generation in the construction of E(λ) that contains x. It isstraightforward to check that for all x ∈ E(λ),

Wμ2/3,3/2(x) ≈

∞∑n=0

(μ(Qn(x))

�(Qn(x))

)2=

∞∑n=0

θ2n.

Thus, if we consider the measure

ν =

( ∞∑n=0

θ2n

)−1/2

μ,

we have W ν2/3,3/2(x) � 1 for all x ∈ E(λ). From (4.38) we infer that

C2/3,3/2(E(λ)) � ν(E(λ)) =

( ∞∑n=0

θ2n

)−1/2

. �



4.10.1 Denjoy’s conjecture, the Davie-Øksendal dualization, andγ+

The theorem stating that rectifiable sets with positive length are non-removableis called Denjoy’s conjecture because Denjoy [33] claimed in 1909 to have provedthis result. However, there was a serious gap in his arguments and the conjectureremained open until the proof of the L2 boundedness of the Cauchy transformwith respect to arc length on Lipschitz graphs with small slope by Calderon [5]in 1977. It seems that, at the beginning, Calderon was not aware that Denjoy’sconjecture was an important corollary of his theorem, using the characterizationof analytic capacity in (1.14) that follows via the Garabedian duality.

In this chapter, instead of using the Garabedian duality to derive Denjoy’sconjecture from Calderon’s theorem, our arguments rely on the dualization of theweak (1, 1) inequality for the Cauchy transform by means of Theorem 4.6, fromDavie and Øksendal. This approach yields more information about the rectifiablesets with positive length: it shows that they also have positive capacity γ+.

A related previous argument to dualize the weak (1, 1) inequality for anothersingular integral, namely the Beurling transform, appears in the work of Uy [171]about the removable singularities for Lipschitz analytic functions in the plane.However, his argument does not ensure the positivity of the required functionh ∈ L∞ whose Beurling transform is in L∞ too.

An open problem posed by Peter Jones consists in finding a constructive ar-gument to show the existence of a bounded non-constant function which is analyticout of an arbitrary compact subset E of a Lipschitz graph, assuming H1(E) > 0.Up to now, the known arguments rely on the use of the Hahn-Banach theorem.

After Davie and Øksendal showed how to dualize the weak (1, 1) inequalityfor the Cauchy transform in [31], it was natural to introduce the notion of γ+.To the best of my knowledge, this capacity first appeared in the book of Murai[117, Chapter III], where the connection between γ+ and the L2 estimates for theCauchy transform was further investigated.

4.10.2 Analytic capacity and curvature

The relationship between the analytic capacity, curvature, and the Cauchy kernelwas found by Melnikov in his seminal paper [113]. By using the Garabedian dualityand a discrete version of analytic capacity he proved that for any compact setE ⊂ C,

γ(E) ≥ c−1 supμ∈Σ(E)

μ(E)2

μ(E) + c2(μ), (4.39)

where c is some absolute constant and Σ(E) is the set of Radon measures supportedon E such that μ(B(x, r)) ≤ r for all x ∈ C, r > 0. One can easily check that the


right-hand side above is comparable to

sup{μ(E) : μ ∈ Σ(E), c2(μ) ≤ μ(E)

},

and thus also comparable to γ+(E), by Theorem 4.14.It is also worth mentioning that a sharper version of the inequality (4.39)

was obtained in Tolsa [164]. In this work it was shown that

γ(E) ≥ supμ

μ(E)2

4π3 μ(E) + 1

6π c2(μ)

, (4.40)

where the supremum is taken over all Radon measures μ supported on E suchthat Θ∗1(x, μ) < ∞ and Θ1∗(x, μ) ≤ 1 at μ-a.e. x ∈ E. Recall that the densitiesΘ∗1(x, μ) and Θ1

∗(x, μ) were defined in (1.16). So the preceding inequality can beapplied to μ = H1�E if H1(E) < ∞, say. The constants 1/6π and 4π/3 are sharpin the sense that the equality is attained when E is a circumference and μ = H1�E.On the other hand, it is not clear if the precise estimate (4.40) holds with γ+(E)instead of γ(E) on the left side (in my opinion, this is quite unlikely).

The characterization of the capacity γ+ in terms of curvature and the L2

boundedness of the Cauchy transform in Theorem 4.14 was essentially proved inTolsa [154]. On the other hand, most of the results from Section 4.6 are fromTolsa [159]. Nevertheless, the idea of introducing the potential Uμ is from Verdera[175]. In this work he showed that if one defines the energy associated to μ asE(μ) =

∫Uμ dμ, then

γ(E) ≥ c−1 sup{E(μ)−1 : supp(μ) ⊂ E, ‖μ‖ = 1},where c is some positive absolute constant. Again, by Theorem 4.14 and Chebyshevit is not difficult to see that the right-hand side above is comparable to γ+(E).

In the work [80], Khavinson obtained a characterization by duality for γ+different from the one in Theorem 4.24. Indeed, given a compact set E ⊂ C, let

γ+(E) = sup{μ(E) : μ ∈ M+(C), suppμ ⊂ E, |Cμ(z)| ≤ 1 ∀z ∈ Ec}.Notice that this capacity is slightly different from γ+(E) if L2(E) > 0, since inthis definition the condition |Cμ(z)| ≤ 1 is only asked for z ∈ Ec, and not forL2-a.e. z ∈ C. Khavinson proved that

γ+(E) = inf{‖ν‖ : ν ∈ M(C), supp ν ⊂ Ec, Re(Cν(z)) ≥ 1 ∀z ∈ E}.Other approaches by duality to the capacities γ, γ+, and the so-called Cauchycapacity γc are also discussed in Khavinson [80].

4.10.3 The analytic capacity of the Cantor sets E(λ) and Wolff’spotentials

In [54] Garnett considered the problem about the analytic capacity of the Cantorsets E(λ) from Section 4.7. He claimed that γ(E(λ)) > 0 if and only if

∑∞k=0 θk <


∞. However, later Eiderman found a mistake in his proof. The inequality

γ+(E(λ)) ≥ c−1( ∞∑k=0

θ2k

)−1/2

from Theorem 4.28, which is due to Mattila [101], disproves Garnett’s claim. Theconverse inequality was obtained by Eiderman [39], by using different argumentsfrom the ones in Section 4.7.

The relationship between the capacity γ+ and the capacity C2/3,3/2 wasfirst noticed by Mateu, Prat and Verdera in [93]. However, the estimate Uμ(x) ≤c Wμ

2/3,3/2(x)1/2 from Proposition 4.32 already appears in Mattila [101], although

not stated in these terms, perhaps because the connection with non-linear poten-tial theory was not noticed by Mattila.

4.10.4 Capacities associated with the signed Riesz kernels

In Rd, for 0 < α < d, one can consider capacities γα associated with the α-dimensional signed vector-valued Riesz kernels Kα(x) = x/|x|1+α. Given a com-pact set E ⊂ Rd, one sets

γα(E) = sup |〈T, 1〉| ,where the supremum is taken over all the distributions T supported on E suchthat Kα ∗ T is an L∞ function with norm ‖Kα ∗ T ‖L∞ ≤ 1. For α = d − 1, γαcoincides with the Lipschitz harmonic capacity modulo a constant factor.

The capacities γα,+ are defined in an analogous way, restricting the supremumabove to the positive Radon measures T supported on E. By arguments quitesimilar to the ones in Theorem 4.14, relying on the non-homogenous T 1 theoremof Nazarov, Treil and Volberg [123] which avoids the curvature method, one canshow that

γα,+(E) ≈ sup{μ(E) : μ ∈ Σα(E), ‖Rα

μ‖L2(μ)→L2(μ) ≤ 1},

where Σα(E) stands for the subset of those positive measures μ supported on Esuch that μ(B(x, r)) ≤ rα for all x, r and Rα

μ is the α-dimensional Riesz transformwith respect to μ. The preceding characterization of γα,+ shows that it is countablysemiadditive. For the details, see Prat [136], [137].

The capacities γα of some higher dimensional versions of the sets E(λ) fromSection 4.7 were studied in Mateu and Tolsa [95], Tolsa [168] and Eiderman andVolberg [38].

For 0 < α < 1, it was shown by Mateu, Prat and Verdera [93] that γα,+ ≈C 2

3 (d−α), 32. In the proof, the “curvature type” estimate (3.26) is essential. On the

other hand, following ideas similar to the ones for the comparability between γand γ+ (which will be explained in Chapter 6), they also proved in the same workthat γα,+ ≈ γα, and thus one gets

γα ≈ C 23 (d−α), 32

for 0 < α < 1.


This comparability fails for the integer values of α ∈ (0, d). On the other hand, itis natural to conjecture that it holds for every non-integer α with 0 < α < d. Forthe moment, the problem is open.

Chapter 5

A Tb theorem of Nazarov, Treiland Volberg

5.1 Introduction

Our main objective in this chapter consists in obtaining a Tb-like theorem suitablefor the proof of the comparability between γ and γ+ in the next chapter. Thetheorem that we will prove is from Nazarov, Treil and Volberg [126]. To state theresult, first we need some notation concerning dyadic lattices.

The usual lattice of dyadic squares from C will be denoted by D0 in thischapter, For every w ∈ C, we consider the translated dyadic lattice

D(w) = w +D0. (5.1)

The aforementioned Tb theorem is the following.

Theorem 5.1. Let μ be a finite measure supported on a compact set F ⊂ C.Suppose that there exist a complex measure ν and, for each w ∈ C, two subsetsHD(w), TD(w) ⊂ C made up of dyadic squares from D(w) such that:

(a) Every ball Br of radius r such that μ(Br) > c0r is contained in⋂

w∈CHD(w).

(b) ν = b μ, where b is some function such that ‖b‖L∞(μ) ≤ cb.

(c)∫C\HD(w)

C∗ν dμ ≤ c∗ μ(F ), for all w ∈ C.

(d) If Q ∈ D(w) is such that Q �⊂ TD(w), then μ(Q) ≤ cacc|ν(Q)| (i.e. Q is anaccretive square).

(e) μ(HD(w) ∪ TD(w)) ≤ δ0 μ(F ), for all w ∈ C and some δ0 < 1.

, , OI 10.1007/978-3- - -6_ ,

© Springer



1377

138 Chapter 5. A Tb theorem of Nazarov, Treil and Volberg

Then there exists a subset G ⊂ F \⋂w∈C

(HD(w) ∪ TD(w)

)such that

(i) μ(G) ≥ c−11 μ(F ),

(ii) μ�G has c0-linear growth,

(iii) the Cauchy transform is bounded in L2(μ�G).

The constant c1 and the bound for the L2(μ�G) norm depend only on c0, cb, c∗,cacc and δ0.

Observe that from the condition (a) in the theorem one infers that μ�C \⋂w∈CHD(w) has linear growth. On the other hand, the estimate in (c) ensures

that C∗ν is not too big in C \ HD(w). The statement (d) is a kind of accretivitycondition for ν, or equivalently, for the bounded function b such that ν = b μ. Thelast condition (e) asserts, in a sense, that the bad sets HD(w), TD(w), are not toobig.

The theorem above is also valid for more general singular integral operators.However, for simplicity we only consider the case of the Cauchy transform. Thearguments below for the Cauchy transform can be adapted easily to more generalsingular integrals. The relationship between the Cauchy kernel and curvature isnot used in this chapter. For more details, see Section 5.12.

The remainder of this chapter is devoted to the proof of Theorem 5.1. Weassume that F , μ, ν, HD(w) and TD(w) satisfy the hypotheses of the theorem.

5.2 The exceptional set S

We setS0 = {x ∈ F : C∗ν(x) > α},

where α is some big constant which will be chosen below. For the moment, let ussay that α � c0cb. For x ∈ S0, let

e(x) = sup{ε : ε > 0, |Cεν(x)| > α}. (5.2)

For x ∈ F \ S0, we set e(x) = 0. We define the exceptional set S as

S =⋃

x∈S0

B(x, e(x)).

The next lemma shows that μ(S \HD(w)) is small if α is taken big enough.

Lemma 5.2. Let w ∈ C. If y ∈ S \ HD(w), then C∗ν(y) > α − 8c0cb. Thus, ifα ≥ 16c0cb, then

μ(S \HD(w)) ≤ 2c∗α

μ(F ). (5.3)

5.2. The exceptional set S 139

Proof. Observe that if y ∈ S \ HD(w), then y ∈ B(x, e(x)) for some x ∈ S0.Let ε0(x) be such that |Cε0(x)ν(x)| > α and y ∈ B(x, ε0(x)). To prove the firststatement in the lemma we will show that

|Cε0(x)ν(x) − Cε0(x)ν(y)| ≤ 8c0cb, (5.4)

and we will be done. We have

|Cε0(x)ν(x)− Cε0(x)ν(y)|≤ |Cε0(x)(ν�B(y, 2ε0(x)))(x)| + |Cε0(x)(ν�B(y, 2ε0(x)))(y)|+ |Cε0(x)(ν�B(y, 2ε0(x))

c)(x)− Cε0(x)(ν�B(y, 2ε0(x))c)(y)|. (5.5)

Notice that the first two terms on the right-hand side are bounded above by

|ν|(B(y, 2ε0(x)))

ε0(x)≤ cbμ(B(y, 2ε0(x)))

ε0(x)≤ 2c0cb,

since y �∈ HD(w). The last term on the right-hand side of (5.5) does not exceed∫C\B(y,2ε0(x))

∣∣∣∣ 1

z − x− 1

z − y

∣∣∣∣ d|ν|(z) = ∫C\B(y,2ε0(x))

|x− y||z − x||z − y| d|ν|(z)

≤ 2cbε0(x)

∫C\B(y,2ε0(x))

1

|z − y|2 dμ(z),

where we have applied that |x − y| ≤ ε0(x) and |z − x| ≥ |z − y|/2 in the lastinequality. As y �∈ HD(w), we get the following standard estimate:

2cbε0(x)

∫C\B(y,2ε0(x))

1

|z − y|2 dμ(z)

= 2cbε0(x)

∞∑k=1

∫2kε0(x)≤|z−y|<2k+1ε0(x)

1

|z − y|2 dμ(z)

≤ 2cbε0(x)

∞∑k=1

μ(B(y, 2k+1ε0(x)))

22kε0(x)2≤ 4c0cb.

So we get

|Cε0(x)(ν�B(y, 2ε0(x))c)(x) − Cε0(x)(ν�B(y, 2ε0(x))

c)(y)| ≤ 4c0cb,

and (5.4) holds.

To prove (5.3), note that C∗ν(y) > α/2 for all y ∈ S \HD(w) if α ≥ 16c0cb,and then by the Chebyshev inequality,

μ(S \HD(w)) ≤ 2

α

∫F\HD(w)

C∗ν dμ ≤ 2c∗α

μ(F ). �


We set δ1 = (δ0 + 1)/2 (so that δ0 < δ1 < 1). To prove Theorem 5.1 we willchoose α = max(16c0cb, 4c∗/(1− δ0)), so that

μ(HD(w) ∪ TD(w)) + μ(S \HD(w)) ≤(δ0 +

2c∗α

)μ(F ) ≤ δ1μ(F ) (5.6)

for all w ∈ C.

5.3 The suppressed operators KΘ

Let Θ : C−→[0,∞) be a Lipschitz function with Lipschitz constant ≤ 1. Forx, y ∈ C, we write

kΘ(x, y) =y − x

|y − x|2 +Θ(x)Θ(y).

Following Nazarov, Treil and Volberg [126], we call kΘ a “suppressed kernel”.For ε ≥ 0, we set

KΘ,εν(x) =

∫C\B(x,ε)

kΘ(x, y) dν(y).

The operator KΘ,ε is the (ε-truncated) Θ-suppressed Cauchy transform. We alsoset

KΘ,∗ν(x) = supε>0

|KΘ,εν(x)|

and

KΘν(x) = KΘ,0ν(x),

assuming that the integral that defines KΘ,0ν(x) exists (this is the case, if forexample Θ(x) > 0 for all x ∈ C).

Given f ∈ L1(μ), to simplify notation, we write

KΘ,εf := KΘ,ε(fμ), KΘ,∗f := KΘ,∗(fμ), KΘf := KΘ(fμ).

Lemma 5.3. Let Θ : C−→[0,∞) be a Lipschitz function with Lipschitz constant≤ 1. The kernel kΘ is an antisymmetric Calderon-Zygmund kernel. In fact, itsatisfies the following estimates:

|kΘ(x, y)| ≤ 3(|x− y|2 +Θ(x)2 +Θ(y)2)1/2 , (5.7)

and

|∇xkΘ(x, y)| + |∇ykΘ(x, y)| ≤ 135

|x− y|2 +Θ(x)2 +Θ(y)2. (5.8)

5.3. The suppressed operators KΘ 141

Notice that the constants involved in these estimates do not depend on Θ.Possibly, they are not optimal. Observe also that, when Θ(x) = Θ(y) = 0, theestimates (5.7) and (5.8) are the typical ones from Calderon-Zygmund kernels.

Proof. Clearly, the estimates (5.7) and (5.8) imply that kΘ is a Calderon-Zygmundkernel. To prove (5.7) and (5.8), first we show that

|x− y|2 +Θ(x)Θ(y) ≥ 1

9

(|x− y|2 +Θ(x)2 +Φ(y)2). (5.9)

To this end, we distinguish several cases. If |x− y| ≤ Θ(x)/2, then

|Θ(x)−Θ(y)| ≤ |x− y| ≤ 1

2Θ(x),

and thus 12 Θ(x) ≤ Θ(y) ≤ 2Θ(x). So we have

Θ(x)Θ(y) ≥ 1

2Θ(x)2 and Θ(x)Θ(y) ≥ 1

2Θ(y)2.

We deduce that

|x− y|2 +Θ(x)Θ(y) ≥ |x− y|2 + 1

4Θ(x)2 +

1

4Θ(y)2,

and so (5.9) holds in this case. By symmetry, the same estimate holds if |x− y| ≤Θ(y)/2.

If |x− y| > 12 max(Θ(x),Θ(y)), then

|x− y|2 ≥ 1

8

(Θ(x)2 +Θ(y)2

),

and thus9

8|x− y|2 ≥ 1

8

(|x− y|2 +Θ(x)2 +Θ(y)2),

and so (5.9) also holds.The estimate (5.7) is a direct consequence of (5.9) and the definition of

kΘ(x, y). Concerning the gradient condition (5.8), we have

|∇xkΘ(x, y)| ≤ 1

|x− y|2 +Θ(x)Θ(y)+

|x− y|(2|x− y|+Θ(y))(|x− y|2 +Θ(x)Θ(y))2

≤ 3

|x− y|2 +Θ(x)Θ(y)+

|x− y|Θ(y)(|x− y|2 +Θ(x)Θ(y))2 .

Using (5.9) and the estimate |x− y|Θ(y) ≤ 12

(|x− y|2 +Θ(y)2), we obtain

|∇xkΘ(x, y)| ≤27 +

81

2|x− y|2 +Θ(x)2 +Θ(y)2

.

The analogous inequality for |∇ykΘ(x, y)| is proved in a similar way, and then(5.8) follows. �


Lemma 5.4. Let x ∈ C and suppose that ε ≥ Θ(x). Let σ be a complex Radonmeasure on C. Then∣∣Cεσ(x) −KΘ,εσ(x)

∣∣ ≤ c supr≥ε

|σ|(B(x, r))

r.

Proof. We have∣∣∣∣kΘ(x, y)− 1

y − x

∣∣∣∣ = ∣∣∣∣ y − x

|y − x|2 +Θ(x)Θ(y)− y − x

|y − x|2∣∣∣∣

=|x− y|Θ(x)Θ(y)

|x− y|2 (|x− y|2 +Θ(x)Θ(y))≤ Θ(x)Θ(y)

|x− y|3

≤ Θ(x)(Θ(x) + |x− y|)|x− y|3 =

Θ(x)2

|x− y|3 +Θ(x)

|x− y|2 .

Thus,

|KΘ,εσ(x) − Cεσ(x)| ≤∫|x−y|>ε

(Θ(x)2

|x− y|3 +Θ(x)

|x− y|2)d|σ|(y).

From the estimates∫|x−y|>ε

1

|x− y|3 d|σ|(y) � 1

ε2supr≥ε

|σ|(B(x, r))

r

and ∫|x−y|>ε

1

|x− y|2 d|σ|(y) � 1

εsupr≥ε

|σ|(B(x, r))

r,

the claim in the lemma follows. �Lemma 5.5. Let x ∈ C and r0 ≥ 0 be such that μ(B(x, r)) ≤ c0 r for r ≥ r0 and|Cεν(x)| ≤ α for ε ≥ r0. If Θ(x) ≥ r0, then

|KΘ,εν(x)| ≤ α+ c cb c0 (5.10)

for all ε > 0.

Proof. From the preceding lemma, we infer that if ε ≥ Θ(x) (and so ε ≥ r0), then

|KΘ,εν(x)| ≤ |Cεν(x)|+ c supr≥ε

|ν|(B(x, r))

r

≤ α+ c cb supr≥ε

μ(B(x, r))

r≤ α+ c cb c0.

On the other hand, for ε < Θ(x),

|KΘ,εν(x)| ≤ cb

∫B(x,Θ(x))

|kΘ(x, y)|dμ(y) +∣∣∣∣∫

C\B(x,Θ(x))

kΘ(x, y)dν(y)

∣∣∣∣. (5.11)

5.4. Dyadic lattices and the martingale decomposition 143

To estimate the first integral on the right-hand side we use the fact that

|kΘ(x, y)| ≤ c

Θ(x)and μ(B(x,Θ(x))) ≤ c0Θ(x).

The last estimate holds because Θ(x) ≥ r0. The second integral on the right-handside of (5.11) equals KΘ,Θ(x)ν(x). This term is bounded by α + c cb c0, as shownabove. Then we infer that (5.10) also holds in this case. �

We writeR(x) = sup

{r > 0 : μ(B(x, r)) > c0 r

}.

Let Θ : C → [0,∞) be a 1-Lipschitz function such that, for some w ∈ C,

Θ(x) ≥ dist(x,C \ (HD(w) ∪ S))

(in fact, we might choose Θ(x) = dist(x,C\(HD(w)∪S)), for example). Obviously,kΘ(x, y) = 1/(y−x) for x, y ∈ {z ∈ C : Θ(z) = 0}. Observe thatHD(w)∪S containsall non-Ahlfors disks and all the disks B(x, e(x)), x ∈ F , and so

Θ(x) ≥ max(R(x), e(x)

).

From the definition of S and the preceding lemma with r0 = max(R(x), e(x)

), we

deduce:

Lemma 5.6. Let w ∈ C and let Θ : C−→[0,+∞) be a Lipschitz function withLipschitz constant 1 such that Θ(x) ≥ dist(x,C \ (HD(w)∪S)) for all x ∈ C. ThenKΘ,∗ν(x) ≤ c4 for all x ∈ F , with c4 depending on c0, cb, c∗, and δ0.

5.4 Dyadic lattices and the martingale decomposition

5.4.1 The dyadic Carleson embedding theorem

In this subsection, our ambient space is Rd. Given a Radon measure σ ∈ Rd andsome dyadic lattice D from Rd, the associated dyadic maximal Hardy-Littlewoodoperator (with respect to σ and D) is defined by

Mσ,dν(x) = supQ∈D:Q�x

|ν|(Q)

σ(Q),

for any complex measure ν ∈ M(Rd). As usual, for f ∈ L1loc(σ), we set Mσ,df(x) =

Mσ,d(f σ)(x). The following result is well known, but for completeness we will showall the details.

Theorem 5.7. The operator Mσ,d is bounded from M(Rd) to L1,∞(Rd) and also inLp(σ), for 1 < p ≤ ∞.

Observe that no doubling conditions are assumed on the measure σ.


Proof. It is clear that Mσ,d is bounded in L∞(σ), and thus by interpolation it isenough to prove the boundedness from M(Rd) to L1,∞(Rd).

Given ν ∈ M(Rd) and λ > 0, let

Ωλ = {x ∈ Rd : Mσ,dν(x) > λ}.Let Iλ ⊂ D be the subfamily of those dyadic cubes such that |ν|(Q)/σ(Q) > λ.Note that

Ωλ =⋃

Q∈Iλ

Q.

Consider an arbitrary finite subfamily Iλ,0 ⊂ Iλ, and let Imaxλ,0 ⊂ Iλ,0 be the

subfamily of the maximal, and thus disjoint, cubes from Iλ,0. Then we have

σ

( ⋃Q∈Iλ,0

Q

)=∑

Q∈Imaxλ,0

σ(Q) ≤ 1

λ

∑Q∈Imax

λ,0

|ν|(Q) ≤ ‖ν‖λ

.

Since this holds for any finite subfamily Iλ,0 ⊂ Iλ, we infer that

σ(Ωλ) ≤ ‖ν‖λ

. �

Theorem 5.8 (Dyadic Carleson embedding theorem). Let σ be a Radon measureon Rd. Let D be some dyadic lattice from Rd and let {aQ}Q∈D be a family ofnon-negative numbers. Suppose that for every cube R ∈ D we have∑

Q∈D:Q⊂R

aQ ≤ c2 σ(R). (5.12)

Then every family of non-negative numbers {wQ}Q∈D satisfies∑Q∈D

wQ aQ ≤ c2

∫supQ�x

wQ dσ(x). (5.13)

Also, if f ∈ L2(σ), ∑Q∈D

|〈f〉σ,Q|2 aQ ≤ c c2‖f‖2L2(σ), (5.14)

where 〈f〉σ,Q =∫Qf dσ/σ(Q) and c is an absolute constant.

Proof. To prove (5.13), notice that by Fubini,∑Q∈D

wQ aQ =∑Q∈D

∫ wQ

0

aQ dλ =

∫ ∞

0

∑Q∈D:wQ>λ

aQ dλ. (5.15)

For each λ > 0, let

Ωλ =⋃

Q∈D:wQ>λ

Q.


Let Iλ ⊂ D be the family of the dyadic cubes such that wQ > λ. Consider anarbitrary finite subfamily Iλ,0 ⊂ Iλ, and let Imax

λ,0 ⊂ Iλ,0 be the subfamily of themaximal, and thus disjoint, cubes from Iλ,0. From the condition (5.12) and thedisjointness of the cubes from Imax

λ,0 we get∑Q∈Iλ,0

aQ ≤∑

R∈Imaxλ,0

∑Q⊂R

aQ ≤ c2∑

R∈Imaxλ,0

σ(R) ≤ c2 σ(Ωλ).

Since this holds for any finite subfamily Iλ,0 ⊂ Iλ, we deduce that∑Q∈Iλ

aQ ≤ c2 σ(Ωλ).

Recalling (5.15), we get ∑Q∈D

wQ aQ ≤ c2

∫ ∞

0

σ(Ωλ) dλ.

Since Ωλ coincides with {x ∈ Rd : w∗(x) > λ}, where w∗(x) = supQ�x wQ, theintegral on the right-hand side above equals∫ ∞

0

σ({x ∈ Rd : w∗(x) > λ}) dλ =

∫w∗(x) dσ(x),

which yields (5.13).To prove the estimate (5.14), we take wQ = |〈f〉σ,Q|2, and then from (5.13)

and the L2(σ) boundedness of Mσ,d we deduce∑Q∈D

|〈f〉σ,Q|2 aQ ≤ c2‖Mσ,df‖2L2(σ) ≤ c c2‖f‖2L2(σ) �

5.4.2 Random dyadic lattices

In the rest of the chapter, we assume that

F ⊂ 1

8S0, where S0 =

[0, 2N]2,

with N big enough. Notice that S0 is a square which belongs to the usual dyadiclattice D0 of C ≡ R2. We set

Ω =[−2N−4, 2N−4

]2 ⊂ R2 ≡ C,

and we consider dyadic lattices D(w) as in (5.1), with w ∈ Ω. We write

Q0(w) ≡ w + S0,


so that Q0(w) ∈ D(w). Notice that for each w ∈ Ω, we have

F ⊂ 1

4Q0(w).

Let PΩ be the uniform probability on Ω, that is, PΩ coincides with thenormalized Lebesgue measure on the square Ω.

5.4.3 Transit and terminal squares

For a fixed w ∈ Ω, consider the dyadic lattice D ≡ D(w). Let Q ∈ D be containedin Q0

D = Q0(w) with μ(Q) �= 0. The square Q is called terminal if Q ⊂ HD ∪ TD.Otherwise, it is called transit. The set of terminal squares is denoted by Dterm,and the set of transit squares by Dtr. From the assumption (e) in Theorem 5.1 itfollows that Q0

D is always transit. Notice that we assume all transit and terminalsquares to be contained in Q0

D.

Lemma 5.9. Let Q ∈ Dtr. Then

μ(λQ) ≤ c0 �(λQ) for all λ ≥ 1,

and

μ(Q) ≤ cacc |ν(Q)|.Proof. By the definition of transit squares, since Q �⊂ HD, any ball B(x, r) con-taining Q satisfies μ(B(x, r)) ≤ c0 r, and so if wQ is the center of Q,

μ(λQ) ≤ μ(B(wQ, λ�(Q))

) ≤ c0�(λQ).

The condition μ(Q) ≤ cacc |ν(Q)| holds just because Q �⊂ TD. �

5.4.4 The martingale decomposition

For f ∈ L1loc(μ) (we assume always f to be real, for simplicity) and any square Q

with μ(Q) �= 0, we set

〈f〉Q =1

μ(Q)

∫Q

f dμ.

We define the operator Ξ as

Ξf =〈f〉Q0

D〈b〉Q0

D

b,

where b is the complex function from Theorem 5.1. It follows easily that Ξf ∈L2(μ) if f ∈ L2(μ), and Ξ2 = Ξ. Moreover, the definition of Ξ does not depend onthe choice of the lattice D(w), w ∈ Ω. The transpose of Ξ is

Ξtf =〈fb〉Q0

D〈b〉Q0

D

.


For a fixed dyadic square Q ∈ D, the set of, at most four, children of Q,whose μ-measure is not zero is denoted by CH(Q).

For any square Q ∈ Dtr and any f ∈ L1loc(μ), we define the function ΔQf as

follows:

ΔQf =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩0 in C \⋃P∈CH(Q) P ,( 〈f〉P

〈b〉P − 〈f〉Q〈b〉Q

)b in P if P ∈ CH(Q) ∩ Dtr,

f − 〈f〉Q〈b〉Q b in P if P ∈ CH(Q) ∩ Dterm.

The operators ΔQ satisfy the following properties.

Lemma 5.10. For all f ∈ L2(μ) and all Q ∈ Dtr,

(a) ΔQf ∈ L2(μ),

(b)∫ΔQf dμ = 0,

(c) ΔQ is a projection, i.e. Δ2Q = ΔQ,

(d) ΔQ Ξ = ΞΔQ = 0,

(e) If R ∈ Dtr and R �= Q, then ΔQΔR = 0.

(f) The transpose of ΔQ is

ΔtQf =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩0 in C \⋃P∈CH(Q) P ,

〈fb 〉P〈b〉P − 〈fb〉Q

〈b〉Q in P if P ∈ CH(Q) ∩ Dtr,

f − 〈fb〉Q〈b〉Q in P if P ∈ CH(Q) ∩ Dterm.

All the properties of the lemma are easily checked. We leave the details forthe reader.

Lemma 5.11. For any f ∈ L2(μ), we have the decomposition

f = Ξf +∑

Q∈Dtr

ΔQf, (5.16)

where the sum is unconditionally convergent in L2(μ). Moreover, there exists someconstant c3 depending on cb and cacc such that

c−13 ‖f‖2L2(μ) ≤ ‖Ξf‖2L2(μ) +

∑Q∈Dtr

‖ΔQf‖2L2(μ) ≤ c3‖f‖2L2(μ). (5.17)


Proof. The second inequality in (5.17). We need to introduce the following oper-ators DQ, for Q ∈ D:

DQf =

{0 in C \Q,

〈f〉P − 〈f〉Q in P ∈ CH(Q).

We also define Ef = 〈f〉Q0D. Then it is well known that, because of orthogonality,

‖Ef‖2L2(μ) +∑Q∈D

‖DQf‖2L2(μ) = ‖f‖2L2(μ). (5.18)

Consider first a square P ∈ CH(Q) ∩ Dtr. In this case we have

ΔQf |P =〈f 〉P b〈b〉P − 〈f〉Qb

〈b〉Q=

(1

〈b〉P − 1

〈b〉Q

)〈f 〉P b+ 1

〈b〉Q(〈f〉P − 〈f〉Q

)b.

Since∣∣〈b〉P 〈b〉Q

∣∣ ≥ c−2acc and b is bounded, we get(ΔQf |P

)2 �(DQb|P

)2〈f 〉2P + |DQ(f)|2.Therefore,∑

Q∈Dtr

∑P∈CH(Q)∩Dtr

‖ΔQf‖2L2(μ�P ) �∑Q∈D

∑P∈CH(Q)

‖χPDQb‖2L2(μ) 〈f 〉2P

+∑Q∈D

‖DQf‖2L2(μ) =: I + II.

From (5.18), we infer that II ≤ ‖f‖2L2(μ).Concerning the term I, note that it can be rewritten as

I =∑Q∈D

‖χQDQb‖2L2(μ) 〈f 〉2Q,

where Q is the father of Q. To estimate it we will apply the dyadic Carleson em-bedding Theorem 5.8. Let us check that the numbers aQ := ‖χQDQb‖2L2(μ) satisfy

the packing condition (5.12). By (5.18), taking into account that b is bounded, foreach square R ∈ D we have∑

Q⊂R

‖χQDQb‖2L2(μ) = ‖DRb‖2L2(μ�R) +∑Q�R

‖DQb‖2L2(μ)

≤ cμ(R) +∑Q⊂R

‖DQ(bχR)‖2L2(μ) ≤ cμ(R). (5.19)


So (5.12) holds and then I ≤ c‖f‖2L2(μ).

Now we have to deal with the terminal squares. If Q ∈ Dtr and P ∈ Dterm,then we have

ΔQf |P =

(f − 〈f〉P b

〈b〉Q

)+

( 〈f〉P〈b〉Q − 〈f〉Q

〈b〉Q

)b.

Since b is bounded and |〈b〉Q| ≥ c−1acc, we get

|ΔQf |P | �(|f |+ 〈|f |〉P

)+∣∣〈f〉P − 〈f〉Q

∣∣=(|f |+ 〈|f |〉P

)+∣∣DQf |P

∣∣.Therefore,

∑Q∈Dtr

∑P∈CH(Q)∩Dterm

‖ΔQf‖2L2(μ�P )

≤ c∑

Q∈Dtr


∫P

|f |2 dμ+ c∑Q∈D

‖DQf‖2L2(μ). (5.20)

For the first sum on the right-hand side above, note that the squares P ∈ Dterm

whose father is a transit square are pairwise disjoint. For the last sum, we onlyhave to use (5.18). Then we obtain∑

Q∈Dtr


‖ΔQf‖2L2(μ�P ) ≤ c‖f‖2L2(μ),

which finishes the proof of the second inequality in (5.17).

Proof of the first inequality in (5.17) for finite sums. Given a finite subset F ⊂ Dtr

and the fixed function f in the statement of the lemma, we will prove the firstinequality in (5.17) for the function

g = Ξf +∑Q∈F

ΔQf.

Notice that, from the properties (d) and (e) of Lemma 5.10, we infer that Ξg = Ξf ,ΔQg = ΔQf for Q ∈ F , and ΔQg = 0 for Q �∈ F , and thus the identity (5.16)holds for g.

By (5.16) and the fact that Ξ and ΔQ are projections, we have

g = Ξg +∑

Q∈Dtr

ΔQg = Ξ2g +∑

Q∈Dtr

Δ2Qg. (5.21)


Therefore, ∫|g|2 dμ =

∫ (Ξ2g +

∑Q∈Dtr

Δ2Qg)g dμ

=

∫(Ξg)(Ξtg) dμ+

∑Q∈Dtr

∫(ΔQg)(Δ

tQg) dμ

≤(‖Ξg‖2L2(μ) +

∑Q∈Dtr

‖ΔQg‖2L2(μ)

)1/2×(‖Ξtg‖2L2(μ) +

∑Q∈Dtr

‖ΔtQg‖2L2(μ)

)1/2. (5.22)

So if we show that

‖Ξtg‖2L2(μ) +∑

Q∈Dtr

‖ΔtQg‖2L2(μ) ≤ c‖g‖2L2(μ), (5.23)

we will be done. The arguments to prove this are very similar to the ones to provethe second inequality in (5.17). For completeness, we show the details.

It is straightforward to check that

‖Ξtg‖L2(μ) ≤ c‖g‖L2(μ) = ‖g‖L2(μ),

and so we only have to estimate∑

Q∈Dtr ‖ΔtQg‖2L2(μ). Again, our strategy consists

in relating ΔtQg to DQg. If P ∈ CH(Q) is a transit square, then we have (using

(f) from Lemma 5.10),

ΔtQg|P =

〈gb〉P − 〈gb〉Q〈b〉Q + 〈gb〉P

(1

〈b〉P − 1

〈b〉Q

)=

1

〈b〉QDQ(gb)|P − 〈gb〉P〈b〉P 〈b〉QDQb|P .

Since |〈b〉Q|, |〈b〉P | ≥ c−1acc, we obtain∑

Q∈Dtr

∑P∈CH(Q)∩Dtr

‖ΔtQg‖2L2(μ�P )

≤ c∑Q∈D

‖DQ(gb)‖2L2(μ) + c∑Q∈D

∑P∈CH(Q)

‖〈gb〉PDQb|P ‖2L2(μ�P ). (5.24)

From (5.18) we deduce∑Q∈D

‖DQ(gb)‖2L2(μ) ≤ ‖gb‖2L2(μ) ≤ c‖g‖2L2(μ).


Now observe that the last term in (5.24) can be rewritten as∑Q∈D

|〈gb〉Q|2‖χQDQb‖2L2(μ).

To estimate this term we apply again the dyadic Carleson embedding theorem,recalling the packing condition (5.19). We obtain∑

Q∈D

∑P∈CH(Q)

‖〈gb〉PDQb|P ‖2L2(μ�P ) ≤ c‖gb‖2L2(μ) ≤ c‖g‖2L2(μ).

Finally we deal with the terminal squares. If Q ∈ Dtr and P ∈ Dterm, then

ΔtQg|P =

(g − 〈gb〉P

〈b〉Q

)+

( 〈gb〉P〈b〉Q − 〈gb〉Q

〈b〉Q

).

Since b is bounded and |〈b〉Q| ≥ c−1acc, we get

|ΔtQg|P | �

(|g|+ 〈|g|〉P)+∣∣〈gb〉P − 〈gb〉Q

∣∣=(|g|+ 〈|g|〉P

)+∣∣DQ(gb)|P

∣∣.Therefore,∑

Q∈Dtr


‖ΔtQg‖2L2(μ�P )

≤ c∑

Q∈Dtr


∫P

|g|2 dμ+ c∑Q∈D

‖DQ(gb)‖2L2(μ).

Arguing as in (5.20), we obtain∑Q∈Dtr


‖ΔtQg‖2L2(μ�P ) ≤ c‖g‖2L2(μ),

and so (5.23) follows.

Proof of (5.16). First we show that the series∑

Q∈Dtr ΔQf converges uncondi-

tionally in L2(μ). Indeed, let F ⊂ Dtr be a finite subset and let g =∑

Q∈F ΔQf .Then by (5.17),∥∥∥∥∑

Q∈FΔQf

∥∥∥∥2L2(μ)

= ‖g‖2L2(μ) ≤ c∑

Q∈Dtr

∥∥ΔQg∥∥2L2(μ)

= c∑Q∈F

∥∥ΔQf∥∥2L2(μ)

.

From this condition, recalling that∑Q∈Dtr

∥∥ΔQf∥∥2L2(μ)

≤ c‖f‖2L2(μ) < ∞,


one deduces easily that the series∑

Q∈Dtr ΔQf converges unconditionally in L2(μ).Now, to prove (5.16), it is enough to show that

Ξf(x) + limn→∞

∑Q∈Dtr:(Q)≥2−n

ΔQf(x) = f(x) μ-a.e. x ∈ C.

This is clear from the definitions of Ξf and ΔQf if x is contained in some terminalsquare. If x ∈ supp(μ) does not belong to any terminal square, then the suminside the limit above equals 〈f〉P b(x)/〈b〉P , which converges to f(x) μ-a.e. by the(dyadic) Lebesgue differentiation theorem.

Proof of the first inequality in (5.17) in full generality. This is proved as above inthe special case where the sum over Q ∈ Dtr is a finite sum, replacing g by thefunction f itself, and using the identity (5.16) in (5.21). �

Remark 5.12. Lemma 5.11 also holds by replacing Ξ and ΔQ with Ξt and ΔtQ,

respectively. In fact, from the inequality (5.23), it turns out that for all f ∈ L2(μ),

‖Ξtf‖2L2(μ) +∑

Q∈Dtr

‖ΔtQf‖2L2(μ) ≤ c ‖f‖2L2(μ). (5.25)

The converse inequality is derived from (5.22):

‖f‖2L2(μ) ≤ c(‖Ξtf‖2L2(μ) +

∑Q∈Dtr

‖ΔtQf‖2L2(μ)

).

We leave the details about the L2(μ) unconditional convergence for the reader.

5.5 Good and bad squares and functions

We say that a subset V ⊂ C has M -thin boundary if

μ({x ∈ C : dist(x, ∂V ) ≤ r}) ≤ Mr

for all r ≥ 0.We now define the bad squares. Let D1 = D(w1) and D2 = D(w2), with

w1, w2 ∈ Ω, be two dyadic lattices. We say that a transit square Q ∈ Dtr1 is bad

(with respect to D2) if either

(a) there exists a transit square R ∈ Dtr2 such that dist(Q, ∂R) ≤ �(Q)1/4�(R)3/4

and �(R) ≥ 2m�(Q) (where m is some positive integer to be fixed below), or

(b) there exists a transit square R ∈ Dtr2 such that 2−m�(Q) ≤ �(R) ≤ 2m�(Q),

dist(Q,R) ≤ 2m�(Q), and at least one of the children P ∈ CH(R) does nothave M -thin boundary.

5.6. Estimates for good functions 153

Of course, if Q is not bad, then we say that it is good.Notice that we declare the square Q ∈ D1 to be good or bad depending

on another lattice D2. This makes a big difference with respect to the previousdefinition of transit and terminal squares, which depends on the lattice to whichthe squares belong. Observe also that the definition of bad squares depends on theconstantsm and M . So strictly speaking, bad squares should be called (m,M)-badsquares.

Bad squares do not appear very often in dyadic lattices. To be precise, wewill see in Lemma 5.28 that, given εb > 0 arbitrarily small, if the constants mand M are chosen big enough in the definition of bad squares, then for each fixedQ ∈ D1 = D(w1), the probability that it is bad with respect to a dyadic latticeD2 = D(w2), w2 ∈ Ω, does not exceed εb. That is,

PΩ({w2 ∈ Ω : Q ∈ D1 is bad with respect to D(w2)}

) ≤ εb.

The notion of good and bad squares allows us now to introduce the definitionof good functions. Remember that given any fixed dyadic latticeD1 = D(w1), everyfunction f ∈ L2(μ) can be written as

f = Ξf +∑

Q∈Dtr1

ΔQf.

We say that f is D1-good with respect to D2 (or simply, good) if ΔQf = 0 for allbad squares Q ∈ Dtr

1 (with respect to D2).

5.6 Estimates for good functions

5.6.1 The main lemma for good functions

We consider the set WD = HD ∪ S ∪ TD, and we call it the total exceptional set.We define the function ΦD as

ΦD(x) = dist(x,C \WD).

Notice that ΦD is a Lipschitz function with Lipschitz constant 1 which vanisheson C \WD. Observe also that ΦD ≥ dist(x,C \ (HD ∪S)), so that Lemma 5.6 canbe applied to ΦD.

Given f ∈ Lp(μ) and g ∈ Lp′(μ), we set

〈f, g〉 :=∫

f g dμ.

Notice the absence of complex conjugation on g. By duality, to show that KΘ isbounded in L2(μ), it is enough to show that

|〈KΘf, g〉| ≤ c5‖f‖L2(μ)‖g‖L2(μ)


for all f, g ∈ L2(μ). In the next lemma we show that this holds for good functionsf, g ∈ L2(μ) and an appropriate 1-Lipschitz function Θ.

Lemma 5.13. Let D1 = D(w1) and D2 = D(w2), with w1, w2 ∈ Ω, be two dyadiclattices. Given ε > 0, let Θ : C−→[ε,+∞) be a Lipschitz function with Lipschitzconstant 1 such that Θ(x) ≥ max(ΦD1(x), ΦD2(x)) for all x ∈ C. If f is D1-goodwith respect to D2, and g is D2-good with respect to D1, then

|〈KΘf, g〉| ≤ c5‖f‖L2(μ)‖g‖L2(μ), (5.26)

where c5 depends on c0, cb, c∗, δ0 and εb, but not on ε.

The assumption that Θ(x) ≥ ε > 0 for all x ∈ C is taken for technicalreasons. For instance, this ensures that the operator KΘ is bounded in Lp(μ), forall 1 ≤ p ≤ ∞, with the norm possibly depending on ε, and so, in particular, KΘfis well defined for all f ∈ L2(μ). The important point of the theorem is that theestimate (5.26) does not depend on ε.

5.6.2 Beginning of the proof of Lemma 5.13

Let f, g be functions as in the lemma. Then

f = Ξf +∑

Q∈Dtr1

ΔQf, g = Ξg +∑

Q∈Dtr2

ΔQg. (5.27)

Moreover, we may (and will) assume that the squares Q that appear in the firstsum are good with respect to D2, while the ones in the sum for g are good withrespect to D1. As Θ(x) ≥ ε for all x ∈ C, ‖KΘ‖L2(μ)→L2(μ) < ∞ (with normdepending in ε), and thus

KΘf = KΘ(Ξf) +∑

Q∈Dtr1

KΘ(ΔQf), (5.28)

with the sum converging in L2(μ) (unconditionally). The analogous identity alsoholds for g.

To prove the lemma we will assume that ΔQf = 0 except for a finite family ofsquares Q ∈ Dtr

1 and ΔRg = 0 except for another finite family of squares R ∈ Dtr2 .

Of course, the estimates that we will obtain below will not depend on the numberof non-zero components ΔQf and ΔRg. This assumption will avoid any problemabout convergence which might arise in the arguments below and will allow usto change the order of summation in the sums involving the squares Q,R. Thelemma in full generality follows by taking L2(μ) limits, using the finiteness of theL2(μ) norm of KΘ and the identity (5.28).

We say that f is very D1-good with respect to D2 (or abusing language,just very good) if it is D1-good with respect to D2 and moreover ΔQf = 0 exceptfor a finite family of squares Q ∈ Dtr

1 . We also consider the analogous notion

5.6. Estimates for good functions 155

interchanging D1 by D2. So as remarked above, in the rest of the proof of Lemma5.13 we assume that f and g are very good functions.

To simplify notation we write Q0 = Q0D1

and R0 = Q0D2

. Notice that, sinceF = supp(μ) ⊂ S0 ∩ Q0 ∩ R0, he have 〈b〉S0 = 〈b〉Q0 = 〈b〉R0 . Then, since KΘ isantisymmetric,

〈KΘ(Ξf), Ξg〉 = 〈f〉Q0〈g〉R0

〈b〉2Q0

〈KΘb, b〉 = 0.

So we get

〈KΘf, g〉 = 〈KΘ(Ξf), g〉+ 〈KΘf, Ξg〉+∑

Q∈Dtr1 , R∈Dtr

2

〈KΘ(ΔQf), ΔRg〉. (5.29)

The first term on right-hand side is easy to handle. Indeed, recall that, by Lemma5.6, KΘ,∗ν(x) ≤ c4 for all x ∈ F , with c4 depending on c0, cb, c∗, and δ0, and so

|KΘb(x)| = |KΘν(x)| ≤ c4.

Therefore, since Q0 is a transit square,

‖KΘ(Ξf)‖L2(μ) =|〈f〉Q0 ||〈b〉Q0 | ‖KΘb‖L2(μ) ≤ c |〈f〉Q0 |μ(Q0)1/2 ≤ c ‖f‖L2(μ). (5.30)

Consequently, ∣∣〈KΘ(Ξf), g〉∣∣ ≤ c ‖f‖L2(μ) ‖g‖L2(μ).

We deal with the second term on the right-hand side of (5.29) in an analogousway, by duality. Indeed,

〈KΘf, Ξg〉 = −〈f, KΘ(Ξg)〉.As in (5.30), we have

‖KΘ(Ξg)‖L2(μ) ≤ c ‖g‖L2(μ),

and thus ∣∣〈KΘf, Ξg〉∣∣ ≤ c ‖f‖L2(μ) ‖g‖L2(μ).

Therefore, to prove the lemma it is enough to estimate∑Q∈Dtr

1 , R∈Dtr2

〈KΘ(ΔQf), ΔRg〉. (5.31)

We say that two squares Q ∈ D1, R ∈ D2 are distant if

dist(Q,R) ≥ min(�(Q), �(R)

)1/4 ·max(�(Q), �(R)

)3/4.

So if Q and R are disjoint squares which are not distant and both are good, fromthe definition in Section 5.5, it turns out that 2−m �(R) < �(Q) < 2m �(R).


We split the sum (5.31) as follows:∑Q∈Dtr

1 , R∈Dtr2

〈KΘ(ΔQf), ΔRg〉 =∑

Q∈Dtr1 , R∈Dtr

2Q∩R=∅

· · · +∑

Q∈Dtr1 , R∈Dtr

2Q∩R �=∅

· · · (5.32)

=∑

Q∈Dtr1 , R∈Dtr

2Q,R distant

+∑

Q∈Dtr1 , R∈Dtr

2Q∩R=∅

Q,R not distant

+∑

Q∈Dtr1 , R∈Dtr

2Q∩R �=∅

2−m(R)≤(Q)≤2m(R)

+∑

Q∈Dtr1 , R∈Dtr

2Q∩R �=∅

(Q)<2−m(R) or (R)<2−m(Q)

=: S1 + S2 + S3 + S4.

To prove Lemma 5.13, we will show below that each sum on the right-hand sideabove is bounded by c ‖f‖L2(μ) ‖g‖L2(μ). Section 5.7 deals with the sum S1; whileSection 5.8 is devoted to S4. Finally, in Section 5.9, S2 + S3 is estimated.

5.7 Estimate of the sum S1: distant squares

We set∣∣∣∣∣ ∑Q∈Dtr

1 , R∈Dtr2

Q,R distant

〈KΘ(ΔQf), ΔRg〉∣∣∣∣∣ ≤ ∑

Q∈Dtr1 , R∈Dtr

2Q,R distant(Q)≤(R)

|〈KΘ(ΔQf), ΔRg〉|

+∑

Q∈Dtr1 , R∈Dtr

2Q,R distant(Q)>(R)

|〈KΘ(ΔQf), ΔRg〉|

=: S1,1 + S1,2.

By the antisymmetry of KΘ, it is enough to deal with the sum S1,1. The sum S1,2

is estimated analogously, by duality.We write

D(Q,R) := dist(Q,R) + �(Q) + �(R).

We call D(Q,R) the “long distance” between Q and R.

Lemma 5.14. Let Q,R ⊂ C be disjoint squares and let ϕQ, ψR ∈ L2(μ) be functionssupported on Q and R, respectively. If dist(Q, supp(ψR)) ≥ �(Q) and

∫ϕQ dμ = 0,

then

|〈KΘϕQ, ψR〉| ≤ c�(Q)

dist(Q, supp(ψR))2‖ϕQ‖L1(μ)‖ψR‖L1(μ). (5.33)

If moreover �(Q) ≤ �(R) and for some constant c6 > 0,

dist(Q, supp(ψR)) ≥ c6 �(Q)1/4 �(R)3/4, (5.34)

5.7. Estimate of the sum S1: distant squares 157

then

|〈KΘϕQ, ψR〉| ≤ c�(Q)1/2 �(R)1/2

D(Q,R)2‖ϕQ‖L1(μ)‖ψR‖L1(μ), (5.35)

with c depending on c6.

Proof. Let zQ be the center of Q. From the assumptions of the lemma we inferthat if s ∈ Q and t ∈ supp(ψR), then

|zQ − t| ≈ |s− t| ≥ �(Q) ≥ |s− zQ|.Thus,

|〈KΘϕQ, ψR〉| =∣∣∣∣∫ ∫ kΘ(t, s)ϕQ(s)ψR(t) dμ(s) dμ(t)

∣∣∣∣=

∣∣∣∣∫ ∫ (kΘ(t, s)− kΘ(t, zQ))ϕQ(s)ψR(t) dμ(s) dμ(t)

∣∣∣∣≤ c

∫ ∫ |s− zQ||t− zQ|2 ϕQ(s)ψR(t) dμ(s) dμ(t)

≤ c�(Q)

dist(Q, supp(ψR))2‖ϕQ‖L1(μ)‖ψR‖L1(μ).

Suppose now that �(Q) ≤ �(R) and that (5.34) holds. From the precedingestimate, if dist(Q,ψR) ≥ �(R), then (5.35) follows easily: we only have to plugthe inequality �(Q) ≤ �(Q)1/2�(R)1/2 in the numerator, and to take into accountthat

dist(Q, supp(ψR)) ≈ dist(Q, supp(ψR)) + �(Q) + �(R) ≈ D(Q,R).

If dist(Q,ψR) ≤ �(R), then

�(Q)

dist(Q, supp(ψR))2≤ �(Q)(

c6 �(Q)1/4�(R)3/4)2 =

c−26 �(Q)1/2�(R)1/2

�(R)2,

which also gives (5.35), since in this case D(Q,R) ≈ �(R). �The main ingredients to deal with the sum S1,1 are the estimates in the

preceding lemma together with a well-known version of Schur’s lemma, which werecall next.

Lemma 5.15 (Schur’s lemma). Let I be some set of indices, and for each i ∈ Ia number wi > 0. Suppose that, for some constant a ≥ 0, the matrix {Ti,j}i,j∈I

satisfies ∑j

|Ti,j |wj ≤ awi for each i

and ∑i

|Ti,j |wi ≤ awj for each j.

Then the matrix {Ti,j}i,j∈I defines a bounded operator in �2 with norm ≤ a.


Proof. Let {xQi}i∈I ∈ �2(I), and set yi =∑

j∈I Ti,jxj . We intend to show that∑i |yi|2 ≤ a2

∑i |xQi |2. So we write

|Ti,j ||xj | =(|Ti,j |1/2 w1/2

j

) (|Ti,j |1/2 w−1/2j |xj |

),

and then by Cauchy-Schwarz,

|yi|2 ≤(∑

j∈I

|Ti,j|wj

)(∑j∈I

|Ti,j |w−1j |xj |2

)≤ awi

∑j∈I

|Ti,j|w−1j |xj |2.

Summing on i and interchanging the sums, we get∑i

|yi|2 ≤ a∑j

∑i

|Ti,j|wi w−1j |xj |2 ≤ a2

∑j

|xj |2. �

Lemma 5.16. If f, g ∈ L2(μ) are very good functions, then we have

S1,1 ≤ c∑

Q∈Dtr1 , R∈Dtr

2

(Q)≤(R)

�(Q)1/2 �(R)1/2

D(Q,R)2μ(Q)1/2 μ(R)1/2 ‖ΔQf‖L2(μ) ‖ΔRg‖L2(μ),

(5.36)and

∑Q∈Dtr

1 , R∈Dtr2

(Q)≤(R)

�(Q)1/2 �(R)1/2

D(Q,R)2μ(Q)1/2 μ(R)1/2 ‖ΔQf‖L2(μ) ‖ΔRg‖L2(μ)

≤ c ‖f‖L2(μ) ‖g‖L2(μ). (5.37)

Therefore,S1,1 ≤ c ‖f‖L2(μ) ‖g‖L2(μ).

Proof. By Lemma 5.14, for Q and R such that Q ∩ R = ∅, �(Q) ≤ �(R), anddist(Q,R) > �(Q)1/4�(R)3/4, we have

|〈TΔQf,ΔRg〉| ≤ c�(Q)1/2 �(R)1/2

D(Q,R)2μ(Q)1/2 μ(R)1/2 ‖ΔQf‖L2(μ) ‖ΔRg‖L2(μ),

and thus (5.36) follows.To prove (5.37), we consider the matrix {TQ,R}Q∈Dtr

1 ,R∈Dtr2

defined by

TQ,R =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩�(Q)1/2 �(R)1/2

D(Q,R)2μ(Q)1/2 μ(R)1/2, if (Q,R) ∈ Dtr

1 ×Dtr2

and �(Q) ≤ �(R),

0, otherwise.

5.7. Estimate of the sum S1: distant squares 159

Since ∑Q∈D1

‖ΔQf‖2L2(μ) ≤ c ‖f‖2L2(μ),

and analogously with g instead of f , we only have to show that the matrix{TQ,R}Q∈D1,R∈D2 generates a bounded operator in �2, in the sense that

∑Q∈Dtr

1 ,R∈Dtr2

TQ,R aQ bR ≤ c

( ∑Q∈Dtr

1

a2Q

)1/2( ∑R∈Dtr

2

b2Q

)1/2(5.38)

for all �2 sequences {aQ}Q∈D1 , {bR}R∈D2 .By Schur’s lemma it suffices to prove that∑

Q∈Dtr1 : (Q)≤(R)

TQ,R μ(Q)1/2 ≤ c μ(R)1/2 for all R ∈ Dtr2 , (5.39)

and ∑R∈Dtr

2 : (Q)≤(R)

TQ,R μ(R)1/2 ≤ c μ(Q)1/2 for all Q ∈ Dtr1 . (5.40)

First we consider (5.39). For a fixed R ∈ Dtr2 , we get

∑Q∈Dtr

1 : (Q)≤(R)

TQ,R μ(Q)1/2 = μ(R)1/2∑

Q∈D1: (Q)≤(R)

�(Q)1/2 �(R)1/2

D(Q,R)2μ(Q)

= μ(R)1/2∑k≥0

2−k/2∑

Q∈D1: (Q)=2−k(R)

�(R)

D(Q,R)2μ(Q). (5.41)

We estimate the last sum as follows:∑Q∈D1: (Q)=2−k(R)

�(R)

D(Q,R)2μ(Q) ≤ c

∑j≥0

∑Q∈D1:Q∩2jR �=∅

(Q)=2−k(R)

�(R)(�(2jR)

)2 μ(Q) (5.42)

≤ c∑j≥0

μ(2j+2R)

2j�(2jR).

Since R is a transit square, we have

μ(2j+2R) ≤ 21/2c0 �(2j+2R) (5.43)

(recall Lemma 5.9). Therefore, the last sum in (5.42) does not exceed some constantdepending on c0, and consequently, from (5.41) we deduce (5.39).


Let us consider (5.40). The arguments are quite similar to the preceding ones.For all Q ∈ Dtr

1 ,

∑R∈Dtr

2 : (Q)≤(R)

TQ,R μ(R)1/2 = μ(Q)1/2∑

R∈D2: (Q)≤(R)

�(Q)1/2 �(R)1/2

D(Q,R)2μ(R)

= μ(Q)1/2∑k≥0

2−k/2∑

R∈D2: (Q)=2−k(R)

�(R)

D(Q,R)2μ(R). (5.44)

For each k ≥ 0, denote by Qk a square concentric with Q with side length 2k�(Q) =�(R). We have

∑R∈D2: (R)=2k(Q)

�(R)

D(Q,R)2μ(R) ≤ c

∑j≥0

∑R∈D2:R∩2jQk �=∅

(R)=2k(Q)

�(Qk)(�(2jQk)

)2 μ(R)

≤ c∑j≥0

μ(2j+2Qk)

2j�(2jQk)≤ c c0,

since Q is a transit square and so (5.43) holds with Qk instead of R. Then (5.40)follows from the preceding estimate and (5.44). �

The fact that the only squares that appear in the martingale decompositionof f and g are transit squares plays a key role in the preceding proof. In fact, weused that any ball Br containing some square from the martingale decompositionsatisfies μ(Br) ≤ c0r.

5.8 Estimate of the sum S4

5.8.1 Splitting of S4

Recall that the sum S4 was defined in (5.32) as

S4 =∑

Q∈Dtr1 , R∈Dtr

2Q∩R �=∅

(Q)<2−m(R) or (R)<2−m(Q)


Q∈Dtr1 , R∈Dtr

2Q∩R �=∅

(Q)<2−m(R)

+∑

Q∈Dtr1 , R∈Dtr

2Q∩R �=∅

(R)<2−m(Q)

=: S4,1 + S4,2.

Because of the antisymmetry of KΘ, it is enough to estimate only one of the sumson the right-hand side, say S4,1.

Recall that the functions f and g are supposed to be good. So we may andwill assume the squares Q ∈ D1 in the sum to be good with respect to D2, and

5.8. Estimate of the sum S4 161

analogously for the squares R ∈ D2 with respect to D1. In particular, everyQ ∈ D1

in the sum S4,1 fulfills

dist(Q, ∂S) > �(Q)1/4�(S)3/4 for every S ∈ D2 such that �(S) ≥ 2m�(Q).

As every R in S4,1 satisfies �(R) ≥ 2m+1�(Q), we can choose S to be any of thefour children of R. Since Q∩R �= ∅, it turns out that Q must be contained in onethese four children, which we denote by RQ.

Now we distinguish two cases, according to whether RQ is a transit or aterminal square:

S4,1 =∑

Q∈Dtr1 , R∈Dtr

2Q⊂R

(Q)<2−m(R)RQ transit

〈KΘ(ΔQf), ΔRg〉+∑

Q∈Dtr1 , R∈Dtr

2Q⊂R

(Q)<2−m(R)RQ terminal

〈KΘ(ΔQf), ΔRg〉 (5.45)

=: Str4,1 + Sterm

4,1 .

5.8.2 Estimate of Str4,1 via a paraproduct

By duality, we have

Str4,1 = −

∑Q∈Dtr

1 , R∈Dtr2

Q⊂R(Q)<2−m(R)

RQ transit

〈ΔQf, KΘ(ΔRg)〉 (5.46)

= −∑

Q∈Dtr1 , R∈Dtr

2Q⊂R


〈ΔQf, KΘ(χRQ ΔRg)〉 −∑

Q∈Dtr1 , R∈Dtr

2Q⊂R


〈ΔQf, KΘ(χR\RQΔRg)〉.

We estimate the last sum in the next lemma.

Lemma 5.17. Let f, g ∈ L2(μ) be very good functions. We have,∑Q∈Dtr

1 , R∈Dtr2

Q⊂R(Q)<2−m(R)

RQ transit

∣∣〈ΔQf, KΘ(χR\RQΔRg)〉∣∣ ≤ c ‖f‖L2(μ) ‖g‖L2(μ). (5.47)

Proof. Let Q,R be as in the sum above and assume that they are good. Since�(Q) < 2−m�(R), we have

dist(Q, supp(χR\RQ

ΔRg)) ≥ dist(Q, ∂RQ)

≥ �(Q)1/4�(RQ)3/4 = 2−3/4�(Q)1/4�(R)3/4.


From Lemma 5.14 (choosing c6 = 2−3/4 in (5.34)), we deduce that∣∣〈ΔQf, KΘ(χR\RQΔRg)〉∣∣

≤ c�(Q)1/2 �(R)1/2

D(Q,R)2μ(Q)1/2 μ(R)1/2 ‖ΔQf‖L2(μ)‖ΔRf‖L2(μ).

As a consequence, the sum in (5.47) is bounded by some constant times the sumin (5.37), and so by c ‖f‖L2(μ) ‖g‖L2(μ), by Lemma 5.16. �

For g ∈ L2(μ) as above, consider the complex number

cR,Q(g) =〈g〉RQ

〈b〉RQ

− 〈g〉R〈b〉R ,

so that χRQ ΔRg = cR,Q(g)χRQ b. The first sum on the right-hand side of (5.46)equals ∑

Q∈Dtr1 , R∈Dtr

2Q⊂R


cR,Q(g) 〈ΔQf, KΘ(χRQ b)〉.

Our next objective consists in replacing KΘ(χRQ b) by KΘb in this term, in orderto be able to use our main assumption on KΘb (namely, that it is a boundedfunction). Finally, we will estimate the resulting sum by a suitable paraproductinvolving KΘb.

Lemma 5.18. If f, g ∈ L2(μ) are very good functions, then we have∑Q∈Dtr

1 , R∈Dtr2

Q⊂R


∣∣cR,Q(g) 〈ΔQf, KΘ(χC\RQb)〉∣∣ ≤ c ‖f‖L2(μ) ‖g‖L2(μ).

Proof. To estimate∣∣〈ΔQf, KΘ(χC\RQ

b)〉∣∣ we argue as in Lemma 5.14. Indeed,recall that

dist(Q,C \RQ) ≥ �(Q)1/4�(RQ)3/4 = 2−3/4�(Q)1/4�(R)3/4 ≥ 2−3/4�(Q). (5.48)

Let zQ be the center of Q. From the assumptions of the lemma, we infer that fory ∈ Q and x ∈ C \RQ,

|zQ − x| ≈ |y − x| � �(Q) ≥ 21/2|y − zQ|.


Thus,

|KΘ(ΔQf)(x)| =∣∣∣∣∫ kΘ(x, y)ΔQf(y) dμ(y)

∣∣∣∣=

∣∣∣∣∫ (kΘ(x, y)− kΘ(x, zQ))ΔQf(y) dμ(y)

∣∣∣∣≤ c

∫ |y − zQ||x− zQ|2 ΔQf(y) dμ(y)

≤ c�(Q)

|x− zQ|2 ‖ΔQf‖L1(μ).

So we get∣∣〈ΔQf, KΘ(χC\RQb)〉∣∣ = ∣∣〈KΘ(ΔQf), χC\RQ

b〉∣∣ (5.49)

≤ c

∫x∈C\RQ

�(Q)

|x− zQ|2 |b(x)| dμ(x) ‖ΔQf‖L1(μ).

Taking into account that ‖b‖∞ ≤ cb, we get∫x∈C\RQ

�(Q)

|x− zQ|2 |b(x)| dμ(x) ≤ cb∑k≥0

∫x∈2k+1RQ\2kRQ

�(Q)

|x− zQ|2 dμ(x) (5.50)

≤ c�(Q)μ(2RQ)

dist(zQ, ∂RQ)2+ c∑k≥1

�(Q)μ(2k+1RQ)

�(2kRQ)2.

Recall that, since RQ is a transit square, μ(λRQ) ≤ 21/2 c0 �(λRQ) for all λ ≥ 1.Using also the estimate (5.48), it follows that the right-hand side of (5.50) doesnot exceed

c�(Q) �(RQ)(

�(Q)1/4�(RQ)3/4)2 + c∑k≥1

�(Q) �(2k+1RQ)

�(2kRQ)2≤ c

�(Q)1/2

�(RQ)1/2+

�(Q)

�(RQ)

≤ c�(Q)1/2

�(RQ)1/2.

Therefore, by (5.49) and Cauchy-Schwarz,

∣∣〈ΔQf, KΘ(χC\RQb)〉∣∣ ≤ c

�(Q)1/2

�(RQ)1/2μ(Q)1/2 ‖ΔQf‖L2(μ).

Concerning the coefficient cR,Q(g), we have

‖ΔRg‖2L2(μ) ≥∫RQ

|ΔRg|2 dμ = |cR,Q(g)|2∫RQ

|b|2 dμ

≥ |cR,Q(g)|2 |〈b〉RQ |2 μ(RQ) ≥ c−2acc |cR,Q(g)|2 μ(RQ).


We obtain

∣∣cR,Q(g) 〈ΔQf, KΘ(χC\RQb)〉∣∣ ≤ c

�(Q)1/2

�(RQ)1/2μ(Q)1/2

μ(RQ)1/2‖ΔQf‖L2(μ)‖ΔRg‖L2(μ),

and thus∑Q∈Dtr

1 , R∈Dtr2

Q⊂R(Q)<2−m(R)

RQ transit

∣∣cR,Q(g) 〈ΔQf, KΘ(χC\RQb)〉∣∣

≤ c∑

Q∈Dtr1 , R∈Dtr

2

Q∈I(R)

�(Q)1/2

�(R)1/2μ(Q)1/2

μ(RQ)1/2‖ΔQf‖L2(μ)‖ΔRg‖L2(μ), (5.51)

where the notation Q ∈ I(R) means that Q is contained in some of the sons of R.For each R ∈ Dtr

2 , by Cauchy-Schwarz, we have

∑Q∈Dtr

1

Q∈I(R)

�(Q)1/2

�(R)1/2μ(Q)1/2

μ(RQ)1/2‖ΔQf‖L2(μ)

≤( ∑

Q∈Dtr1

Q∈I(R)

�(Q)1/2

�(R)1/2‖ΔQf‖2L2(μ)

)1/2( ∑Q∈Dtr

1

Q∈I(R)

�(Q)1/2

�(R)1/2μ(Q)

μ(RQ)

)1/2.

It is easy to check that the last factor is bounded by some absolute constant, bysplitting the sum according to the size of the squares Q. Therefore, the sum onthe left-hand side of (5.51) is bounded by

c∑

R∈Dtr2

‖ΔRg‖L2(μ)

( ∑Q∈Dtr

1 :Q⊂R

�(Q)1/2

�(R)1/2‖ΔQf‖2L2(μ)

)1/2

≤ c

( ∑R∈Dtr

2

‖ΔRg‖2L2(μ)

)1/2( ∑R∈Dtr

2

∑Q∈Dtr

1 :Q⊂R

�(Q)1/2

�(R)1/2‖ΔQf‖2L2(μ)

)1/2,

by Cauchy-Schwarz again. The last double sum equals

∑Q∈Dtr

1

‖ΔQf‖2L2(μ)

∑R∈Dtr

2 :R⊃Q

�(Q)1/2

�(R)1/2≤ c∑

Q∈Dtr1

‖ΔQf‖2L2(μ).


So we get∑Q∈Dtr

1 , R∈Dtr2

Q⊂R


∣∣cR,Q(g) 〈ΔQf, KΘ(χC\RQb)〉∣∣

≤ c

( ∑R∈Dtr

2

‖ΔRg‖2L2(μ)

)1/2( ∑Q∈Dtr

1

‖ΔQf‖2L2(μ)

)1/2,

and thus the lemma follows. �By Lemmas 5.17 and 5.18, we know that∣∣Str

4,1

∣∣ ≤ c ‖f‖L2(μ) ‖g‖L2(μ)

is equivalent to proving that∣∣∣∣∣ ∑Q∈Dtr

1 , R∈Dtr2

Q⊂R(Q)<2−m(R)

RQ transit

cR,Q(g) 〈ΔQf, KΘb〉∣∣∣∣∣ ≤ c ‖f‖L2(μ) ‖g‖L2(μ). (5.52)

Observe that the term 〈ΔQf, KΘb〉 does not depend on R, and so the sum on Ronly affects the coefficients cR,Q.

Let us consider the function

g = g − Ξg.

For a good square Q ∈ Dtr1 with �(Q) < 2−m�(R0), we have

cR,Q(g) =〈g〉RQ

〈b〉RQ

− 〈g〉R〈b〉R =

〈g〉RQ

〈b〉RQ

− 〈g〉R〈b〉R = cR,Q(g).

Since 〈g〉R0 = 0, we get∑R∈Dtr

2R⊃Q

(R)>2m(Q)RQ transit

cR,Q(g) =∑

R∈Dtr2

R⊃Q(R)>2m(Q)RQ transit

cR,Q(g) =〈g〉R(Q)

〈b〉R(Q), (5.53)

where R(Q) is the smallest transit square R ∈ Dtr2 containing Q such that �(R) ≥

2m�(Q). Notice that such a square always exists, since R0 is transit and, Q beinggood with F ∩ Q �= ∅, it does not intersect the boundary of R0, and so it is


contained in R0. Observe also that in the special case where no square RQ asin the sums above exists, the last identity in (5.53) is still valid because thenR(Q) = R0 and so 〈g〉R(Q) = 0.

From the preceding argument we infer that the sum on the left-hand side of(5.52) can be rewritten as follows:

∑Q∈Dtr

1

(Q)<2−m(R0)

〈g〉R(Q)

〈b〉R(Q)〈ΔQf, KΘb〉 =

∑Q∈Dtr

1Q good

(Q)<2−m(R0)

〈g〉R(Q)

〈b〉R(Q)〈f, Δt

QKΘb〉.

Given a function ψ ∈ L2(μ) such that ΔRψ = 0 except for a finite numberof squares R ∈ Dtr

2 , we consider the following paraproduct:

ΠKΘb(ψ) =∑

Q∈Dtr1

Q good(Q)<2−m(R0)

〈ψ〉R(Q)

〈b〉R(Q)Δt

QKΘb,

so that ∑Q∈Dtr

1 , R∈Dtr2

Q⊂R(Q)<2−m(R)

RQ transit

cR,Q(g) 〈ΔQf, KΘb〉 = 〈f, ΠKΘb(g)〉. (5.54)

In the following lemma we show that the paraproduct ΠKΘb extends to a boundedoperator in L2(μ).

Lemma 5.19. The paraproduct ΠKΘb extends to a bounded operator in L2(μ).

Proof. Take ϕ, ψ ∈ L2(μ) so that only a finite number of components ΔQϕ andΔRψ are non-zero. Of course, these functions are dense in L2(μ). Then we have

〈ϕ, ΠKΘb(ψ)〉 =∑

Q∈Dtr1

Q good

(Q)<2−m(R0)

〈ψ〉R(Q)

〈b〉R(Q)〈ΔQϕ,KΘb〉.

Clearly, we may restrict the sum above to the Q such ΔQϕ �≡ 0. Doing so, wewrite


∣∣〈ϕ, ΠKΘb(ψ)〉∣∣ ≤ c−1

acc

∑Q∈Dtr

1Q good

(Q)<2−m(R0)ΔQϕ �≡0

|〈ψ〉R(Q)|∣∣〈ΔQϕ,KΘb〉

∣∣‖ΔQϕ‖L2(μ)

‖ΔQϕ‖L2(μ)

≤ c−1acc

( ∑Q∈Dtr

1Q good

(Q)<2−m(R0)ΔQϕ �≡0

|〈ψ〉R(Q)|2∣∣〈ΔQϕ,KΘb〉

∣∣2‖ΔQϕ‖2L2(μ)

)1/2( ∑Q∈Dtr

1

‖ΔQϕ‖2L2(μ)

)1/2.

Since the last factor on the right-hand side does not exceed c ‖ϕ‖L2(μ), to provethe lemma it is enough to show that the first one is bounded by c ‖ψ‖L2(μ). Thesum corresponding to this factor can be rewritten as

∑R∈Dtr

2

|〈ψ〉R|2∑

Q∈F (R)

∣∣〈ΔQϕ,KΘb〉∣∣2

‖ΔQϕ‖2L2(μ)

,

whereF (R) = {Q ∈ Dtr

1 : Q good, ΔQϕ �≡ 0, R(Q) = R}.By Carleson’s Theorem 5.8, it suffices to prove that the coefficients

aR :=∑

Q∈F (R)

∣∣〈ΔQϕ,KΘb〉∣∣2

‖ΔQϕ‖2L2(μ)

satisfy the packing condition (5.12). Observe that, since ΔQ = Δ2Q,∣∣〈ΔQϕ,KΘb〉

∣∣2‖ΔQϕ‖2L2(μ)

=

∣∣〈ΔQϕ,ΔtQKΘb〉

∣∣2‖ΔQϕ‖2L2(μ)

≤ ‖ΔtQKΘb‖2L2(μ).

Therefore, taking into account that the families F (R) are not overlapping, andthat every square from F (R) is contained in R, for every S ∈ D2,∑

R∈D2:R⊂S

aR ≤∑

R∈D2:R⊂S

∑Q∈F (R)

‖ΔtQKΘb‖2L2(μ)

≤∑

Q∈D1:Q⊂S

‖ΔtQ(χSKΘb)‖2L2(μ).

By (5.25), the last sum does not exceed c ‖χSKΘb‖2L2(μ), and then recalling that

‖KΘb‖∞ ≤ c4 we deduce ∑R∈D2:R⊂S

aR ≤ c μ(S),

as wished. �


Gathering the previous results, we get:

Lemma 5.20. If f, g ∈ L2(μ) are very good functions, we have∣∣Str4,1

∣∣ ≤ c ‖f‖L2(μ) ‖g‖L2(μ).

Proof. As remarked above, from (5.46) and Lemmas 5.17 and 5.18, the estimateclaimed by the lemma is equivalent to proving∣∣∣∣∣ ∑

Q∈Dtr1 , R∈Dtr

2Q⊂R


cR,Q(g) 〈ΔQf, KΘb〉∣∣∣∣∣ ≤ c ‖f‖L2(μ) ‖g‖L2(μ).

As noticed in (5.54), the left-hand side of the previous inequality equals|〈f, ΠKΘb(g)〉|, and then we have

|〈f, ΠKΘb(g)〉| ≤ c ‖f‖L2(μ) ‖g‖L2(μ) ≤ c ‖f‖L2(μ) ‖g‖L2(μ),

because of the L2(μ) boundedness of ΠKΘb. �

5.8.3 Estimate of Sterm4,1

Observe that the sum Sterm4,1 defined in (5.45) equals

Sterm4,1 =

∑Q∈Dtr

1 , R∈Dterm2

(Q)≤2−m(R)Q⊂R

〈KΘ(ΔQf), ΔRg〉,

where R ∈ Dtr2 is the father of R. Notice that, to ease the notation, we have not

written above the condition that R is a transit square (on the other hand, if R isnot a transit square, the operator ΔR has not been defined).

Given R as in the preceding sum, we consider a Whitney decomposition ofR into a family W (R) of squares from D2 satisfying:

• the squares from W (R) are pairwise disjoint and cover the interior of R,

• dist(S, ∂R) = �(S) for every S ∈ W (R),

• 3S ⊂ R for every S ∈ W (R), and moreover the family of dilated squares{2S}S∈W (R) has bounded overlap. That is,∑

S∈W (R)

χ2S ≤ c,

for some absolute constant c.


For S ∈ W (R), we write

gR,S = χ2S ΔRg, gR,S = χR\2S ΔRg.

Recall also that zQ stands for the center of the square Q. Then, for every R ∈Dterm

2 , we write∑Q∈Dtr

1

(Q)≤2−m(R)Q⊂R


S∈W (R)

∑Q∈Dtr

1

(Q)≤2−m(R)Q⊂R, zQ∈S

〈KΘ(ΔQf), ΔRg〉 (5.55)

=∑

S∈W (R)

∑Q∈Dtr

1


〈KΘ(ΔQf), gR,S〉+∑

S∈W (R)

∑Q∈Dtr

1


〈KΘ(ΔQf), gR,S〉.

In the next lemma we deal with the last sum.

Lemma 5.21. If f, g ∈ L2(μ) are very good functions and m is chosen big enough,we have∑

R∈Dterm2

∑S∈W (R)

∑Q∈Dtr

1


|〈KΘ(ΔQf), gR,S〉| ≤ c ‖f‖L2(μ) ‖g‖L2(μ).

Proof. Observe that if Q ∈ Dtr1 is a good square from the sum above and m is big

enough, then

8�(Q) ≤ �(Q)1/4 �(R)3/4 ≤ dist(Q, ∂R) ≤ dist(zQ, ∂R) ≤ �(S). (5.56)

Then we deduce

dist(Q, supp gR,S) ≥ dist(Q, ∂2S) ≥ �(S)− �(Q)

2≥ �(S)

4≥ �(Q)1/4 �(R)3/4

4.

In this situation, since ΔQf has zero μ-mean, we can apply the estimate (5.35)from Lemma 5.14, and so

|〈KΘ(ΔQf), gR,S〉| ≤ c�(Q)1/2 �(R)1/2

D(Q, R)2‖ΔQf‖L1(μ)‖gR,S‖L1(μ).

Since ‖gR,S‖L1(μ) ≤ ‖ΔRg‖L1(μ), by Cauchy-Schwarz we get∑R∈Dterm

2

∑S∈W (R)

∑Q∈Dtr

1


|〈KΘ(ΔQf), gR,S〉|

≤ c∑

Q∈Dtr1 ,R∈Dterm

2

(Q)≤2−m(R)Q⊂R

�(Q)1/2 �(R)1/2

D(Q, R)2μ(Q)1/2μ(R)1/2‖ΔQf‖L2(μ)‖ΔRg‖L2(μ).


Recalling that R ∈ Dtr2 , it is straightforward to check that the right-hand side of

the last inequality is bounded by the left-hand side of (5.37) in Lemma 5.16 (timessome constant c), and thus by c ‖f‖L2(μ) ‖g‖L2(μ). �

It remains to estimate the first sum on the right-hand side of (5.55) (aftersumming on R ∈ Dterm

2 ). This will finish our task with Sterm4,1 , and also with the

term S4 from (5.32).

Lemma 5.22. If f, g ∈ L2(μ) are very good functions and m is chosen big enough,then ∑

R∈Dterm2

∑S∈W (R)

∑Q∈Dtr

1


|〈KΘ(ΔQf), gR,S〉| ≤ c ‖f‖L2(μ) ‖g‖L2(μ),

and thus

|Sterm4,1 | ≤ c ‖f‖L2(μ) ‖g‖L2(μ).

Proof. The second statement follows from Lemma 5.21 and the first inequality,which we proceed to prove.

As shown in (5.56), if Q and S ⊂ R are squares from the sum in the lemmaand zQ ∈ S, �(Q) ≤ �(S)/8. So it turns out that Q ⊂ 2S. Since R is terminal and2S ⊂ R, for every x ∈ 2S we have

Θ(x) ≥ dist(x, ∂R) ≥ �(2S)

4.

Then we infer that for all x ∈ R, y ∈ C,

|kΘ(x, y)| ≤ c

Θ(x)≤ c

�(S).

Hence, ∣∣KΘgR,S(x)∣∣ ≤ c

‖gR,S‖L1(μ)

�(S)≤ c

‖gR,S‖L2(μ) μ(2S)1/2

�(S). (5.57)

Since Δ2Q = ΔQ, we obtain∑

Q∈Dtr1


|〈KΘ(ΔQf), gR,S〉| =∑

Q∈Dtr1


|〈ΔQf, ΔtQKΘ(gR,S)〉| (5.58)

≤( ∑

Q∈Dtr1 :Q⊂2S

‖ΔQf‖2L2(μ)

)1/2( ∑Q∈Dtr

1 :Q⊂2S

‖ΔtQKΘ(gR,S)‖2L2(μ)

)1/2.


By Remark 5.12 and (5.57) we have

∑Q∈Dtr

1 :Q⊂2S

‖ΔtQKΘ(gR,S)‖2L2(μ) =

∑Q∈Dtr

1 :Q⊂2S

‖ΔtQ(χ2SKΘ(gR,S)‖2L2(μ)

≤ ‖χ2SKΘ(gR,S)‖2L2(μ) ≤ c‖gR,S‖2L2(μ) μ(2S)

2

�(S)2.

We may assume that any square 2S appearing in the preceding inequality containsa transit square Q ∈ Dtr

1 . Otherwise the corresponding summand is zero. ByLemma 5.9 we know that μ(λQ) ≤ 21/2 c0 �(λQ) for all λ ≥ 1. This implies thatμ(2S) ≤ c c0�(2S). Therefore,

∑Q∈Dtr

1 :Q⊂2S

‖ΔtQKΘ(gR,S)‖2L2(μ) ≤ c ‖gR,S‖2L2(μ).

From (5.58), summing over S ∈ W (R),

∑S∈W (R)

∑Q∈Dtr

1


|〈KΘ(ΔQf), gR,S〉|

≤ c∑

S∈W (R)

( ∑Q∈Dtr

1 :Q⊂2S

‖ΔQf‖2L2(μ)

)1/2‖gR,S‖L2(μ)

≤ c

( ∑S∈W (R)

∑Q∈Dtr

1 :Q⊂2S

‖ΔQf‖2L2(μ)

)1/2( ∑S∈W (R)

‖gR,S‖2L2(μ)

)1/2.

By the bounded overlapping of the squares 2S,

∑S∈W (R)

∑Q∈Dtr

1 :Q⊂2S

‖ΔQf‖2L2(μ) ≤ c∑

Q∈Dtr1 :Q⊂R

‖ΔQf‖2L2(μ),

and also,

∑S∈W (R)

‖gR,S‖2L2(μ) =∑

S∈W (R)

∫2S

|ΔRg|2 dμ ≤ c ‖ΔRg‖2L2(μ).


Therefore,∑R∈Dterm

2

∑S∈W (R)

∑Q∈Dtr

1


|〈KΘ(ΔQf), gR,S〉| (5.59)

≤ c∑

R∈Dterm2

( ∑Q∈Dtr

1 :Q⊂R

‖ΔQf‖2L2(μ)

)1/2‖ΔRg‖L2(μ)

≤ c

( ∑R∈Dterm

2

∑Q∈Dtr

1 :Q⊂R

‖ΔQf‖2L2(μ)

)1/2( ∑R∈Dterm

2

‖ΔRg‖2L2(μ)

)1/2.

Since the terminal squares R from Dterm2 with R ∈ Dtr

2 are pairwise disjoint, wehave ∑

R∈Dterm2

∑Q∈Dtr

1 :Q⊂R

‖ΔQf‖2L2(μ) ≤∑

Q∈Dtr1

‖ΔQf‖2L2(μ).

Also, since each square has four children, and R ∈ Dtr2 ,∑

R∈Dterm2

‖ΔRg‖2L2(μ) ≤ 4∑

R∈Dtr2

‖ΔRg‖2L2(μ).

So the lemma follows from (5.59). �

5.9 Estimate of S2 + S3: the diagonal term

Recall that

S2 + S3 =∑

Q∈Dtr1 , R∈Dtr

2Q∩R=∅

Q,R not distant

〈KΘ(ΔQf), ΔRg〉 +∑

Q∈Dtr1 , R∈Dtr

2Q∩R �=∅

2−m (R)≤(Q)≤2m (R)

〈KΘ(ΔQf), ΔRg〉.

Recall that, since f and g are good functions, we may and do assume that thesquares Q and R in the range of summation of S2 and S3 are good. Then theconditions “Q ∩ R = ∅, Q,R not distant”, in S2, imply that 2−m�(R) < �(Q) <2m �(R). Thus we have

|S2 + S3| ≤∑

Q∈Dtr1 , R∈Dtr

2Q,R not distant

2−m (R)≤(Q)≤2m (R)

∣∣〈KΘ(ΔQf), ΔRg〉∣∣.

Recall also that Q,R not being distant means that

dist(Q,R) < min(�(Q), �(R)

)1/4 ·max(�(Q), �(R)

)3/4.

5.9. Estimate of S2 + S3: the diagonal term 173

In particular, dist(Q,R) < max(�(Q), �(R)

). So we have

|S2 + S3| ≤∑

Q∈Dtr1 , R∈Dtr

2

dist(Q,R)<max((Q),(R))

2−m (R)≤(Q)≤2m (R)

∣∣〈KΘ(ΔQf), ΔRg〉∣∣. (5.60)

To estimate this sum notice that each square Q can interact with a numberof squares R bounded by some constant depending only on m, and vice versa.Therefore, it is enough to show that∣∣〈KΘ(ΔQf), ΔRg〉

∣∣ ≤ c ‖ΔQf‖L2(μ) ‖ΔRg‖L2(μ)

for all Q and R in the sum above. We will prove this by using the assumption thatthe good squares have M -thin boundaries. Recall that a square Q is said to havethin boundaries if

μ({x ∈ C : dist(x, ∂Q) ≤ r}) ≤ Mr for all r ≥ 0.

Lemma 5.23. Let Ω1 ⊂ C be open and Ω2 = C \ Ω1. Suppose that the boundaryΓ = ∂Ω1 = ∂Ω2 is M -thin. That is,

μ({x ∈ C : dist(x,Γ) ≤ r}) ≤ Mr for all r ≥ 0.

If ϕ1, ϕ2 ∈ L2(μ) are supported on Ω1 and Ω2 respectively, then∣∣〈KΘϕ1, ϕ2〉∣∣ ≤ c ‖ϕ1‖L2(μ) ‖ϕ2‖L2(μ).

Proof. From the assumption that Γ is M -thin, it follows that μ(Γ) = 0. Thus,

∣∣〈KΘϕ1, ϕ2〉∣∣ ≤ ∫

x∈Ω1

∫y∈Ω2

|kΘ(x, y)| |ϕ1(x)| |ϕ2(y)| dμ(x) dμ(y).

To prove the lemma it is enough to find a function λ : C \ Γ → (0,∞) such that∫Ω1

|kΘ(x, y)|λ(x) dμ(x) ≤ cλ(y) for all y ∈ Ω2 (5.61)

and ∫Ω2

|kΘ(x, y)|λ(y) dμ(y) ≤ cλ(x) for all x ∈ Ω1. (5.62)

This is just the continuous version of the Schur Lemma 5.15. Its proof is similar:roughly speaking, sums must be replaced by integrals. We leave the details for thereader.


We write d(x) = dist(x,Γ) and we set

λ(x) =1

d(x)1/2.

For all x ∈ Ω1, y ∈ Ω2, we have

|kΘ(x, y)| ≤ 1

|x− y| ≤1

max(dist(x,Γ), dist(y,Γ)

) = 1

max(d(x), d(y)).

Therefore, for every y ∈ Ω2,∫Ω1

|kΘ(x, y)|λ(x) dμ(x) ≤∫Ω1

1

max(d(x), d(y))

1

d(x)1/2dμ(x)

≤∫x∈Ω1:d(x)≤d(y)

1

d(y)

1

d(x)1/2dμ(x)

+

∫x∈Ω1:d(x)>d(y)

1

d(x)

1

d(x)1/2dμ(x) = I1 + I2.

Concerning I1, using that Γ is M -thin,

I1 ≤ 1

d(y)

∑k≥0

μ({

x ∈ Ω1 : 2−k−1d(y) < d(x) ≤ 2−kd(y)}) 1(

2−k−1d(y))1/2

≤ 1

d(y)

∑k≥0

M2−kd(y)1(

2−k−1d(y))1/2 =

cM

d(y)1/2.

For I2 we obtain an analogous estimate

I2 ≤∑k≥0

μ({

x ∈ Ω1 : 2kd(y) < d(x) ≤ 2k+1d(y)}) 1(

2k−1d(y))3/2

≤∑k≥0

M2k+1d(y)1(

2k−1d(y))3/2 =

cM

d(y)1/2,

and thus (5.61) is proved. The estimate (5.62) follows analogously. �Lemma 5.24. If all the children of Q ∈ Dtr

1 and R ∈ Dtr2 have M -thin boundaries,

then ∣∣〈KΘ(ΔQf), ΔRg〉∣∣ ≤ c ‖ΔQf‖L2(μ) ‖ΔRg‖L2(μ).

As a consequence,|S2 + S3| ≤ c ‖f‖L2(μ) ‖g‖L2(μ).

Proof. The last assertion follows from the first one and the fact ifQ is D1-good withrespect to D2, then the children of all the squares R ∈ Dtr

2 such that 2−m�(Q) ≤�(R) ≤ 2m�(Q), dist(Q,R) ≤ 2m�(Q) have M -thin boundaries. In particular, this

5.9. Estimate of S2 + S3: the diagonal term 175

applies to the squares R that appear in the sum in (5.60). The analogous statementholds interchanging the squares Q ∈ Dtr

1 by R ∈ Dtr2 .

Let us turn our attention to the first assertion of the lemma, and consider Qand R such that their children have M -thin boundaries. We have


P∈CH(Q),S∈CH(R)

〈KΘ(χPΔQf), χSΔRg〉.

For each P ∈ CH(Q) and S ∈ CH(R) we write fQ,P = χPΔQf and gR,S = χSΔRg.Then

〈KΘfQ,P , gR,S〉 = 〈KΘ(χP\S fQ,P ), gR,S〉+ 〈KΘ(χP∩S fQ,P ), χS\P gR,S〉 (5.63)

+ 〈KΘ(χP∩S fQ,P ), χS∩P gR,S〉.

The two first terms on the right-hand side are bounded in absolute value byc ‖ΔQf‖L2(μ)‖ΔRg‖L2(μ). In the case of 〈KΘ(χP\S fQ,P ), gR,S〉, this is due to thefact that

supp(χP\S fQ,P ) ⊂ C \ S, supp(gR,S) ⊂ S,

and the boundary ∂S is M -thin. So we can apply Lemma 5.23. Analogously con-cerning the second term on the right-hand side of (5.63):

supp(χP∩S fQ,P ) ⊂ P, supp(χS\P gR,S) ⊂ C \ P,

and the boundary ∂P is M -thin.To deal with the last term on the right-hand side of (5.63), we distinguish

two cases. Suppose first that one of the squares P, S, say P , is terminal. ThenΘ(z) ≥ dist(z, ∂P ) for all z ∈ P . Therefore,

|kΘ(x, y)| ≤ 1

max(Θ(x),Θ(y)

) ≤ 1

max(dist(x, ∂P ), dist(y, ∂P )

)for all x, y ∈ P ∩ S. Then almost the same arguments as the ones in the proof ofLemma 5.23 give that

|〈KΘ(χP∩S fQ,P ), χS∩P gR,S〉| ≤ c ‖ΔQf‖L2(μ)‖ΔRg‖L2(μ).

Indeed, in the arguments above involving Schur’s test, one only has to replaceboth Ω1 and Ω2 by P .

Finally suppose that both squares P, S are transit. In this case, from thedefinition of ΔQf and ΔRg it turns out that χP∩S fQ,P and χP∩S gR,S are bothconstant multiples of χP∩Sb, and then by the antisymmetry of kΘ,

〈KΘ(χP∩S fQ,P ), χS∩P gR,S〉 = 0.

This finishes the proof of Lemma 5.24 and, thus, of Lemma 5.13. �


5.10 Cotlar’s inequality revisited

Recall that, for x ∈ C, we write

R(x) = sup{r > 0 : μ(B(x, r)) > c0 r}.We write R(x) = 0 if the set on the right-hand side is empty. We also set

H =⋃

x∈C:R(x)>0

B(x,R(x)). (5.64)

Recall that the assumption (a) of Theorem 5.1 ensures that H is contained in theexceptional set HD(w) for each w ∈ Ω.

Our goal in this section is to prove the following result, which will be neededin the next section in order to complete the proof of Theorem 5.1.

Lemma 5.25. Let Θ : C−→[0,+∞) be a Lipschitz function with Lipschitz constant1 such that Θ(x) ≥ R(x) for all x ∈ C. Let T be a one-dimensional Calderon-Zygmund operator with kernel t(x, y) satisfying

|t(x, y)| ≤ 1

max(ε, dist(x,C \H), dist(y,C \H))for all x, y ∈ C. (5.65)

Then we have‖T∗‖L2(μ)→L2(μ) ≤ c

(1 + ‖T ‖L2(μ)→L2(μ)

),

where c depends only on c0 and on the Calderon-Zygmund constants of the kernelt(x, y) (i.e. the constants in (2.3) in Chapter 2).

Observe that if t(x, y) satisfies (5.65), then

|t(x, y)| ≤ 1

max(ε,R(x),R(y)).

The kernel 19 kΘ satisfies (5.65), because the assumption

Θ(x) ≥ max(ΦD1(x), ΦD2(x))

ensures thatΘ(x) ≥ dist(x,C \HD1) ≥ dist(x,C \H),

and then, by Lemma 5.3, |kΘ(x, y)| ≤ 9/max(Θ(x),Θ(y)).We will prove Lemma 5.25 by means of a Cotlar type inequality analogous to

the one proved in Section 2.8 of Chapter 2. The precise statement is the following.

Lemma 5.26. Let T be a singular integral operator satisfying the assumptions ofLemma 5.25. Suppose that T is bounded from L1(μ) into L1,∞(μ). Then for everyf ∈ L2(μ) and x ∈ C we have

T∗ f(x) ≤ cMμ(Tf)(x) + c(‖T ‖L1(μ)→L1,∞(μ) + 1

)Mμf(x), (5.66)

where c depends only on c0 and on the Calderon-Zygmund constants of the kernelt(x, y).

5.10. Cotlar’s inequality revisited 177

Since Mμ is bounded in L2(μ) with absolute constants, it is clear that Lemma5.25 follows from the preceding result, after showing that

‖T ‖L1(μ)→L1,∞(μ) ≤ c(‖T ‖L2(μ)→L2(μ) + 1

). (5.67)

Proof of Lemma 5.26. By Theorem 2.18 in Chapter 2, we know that if T isbounded from M(C) to L1,∞(μ), and

μ(B(x, r)) ≤ c0 r for r ≥ ε, (5.68)

then

|Tε ν(x)| ≤ cMμ(Tν)(x) + c(‖T ‖M(C)→L1,∞(μ) + 1

)Mμν(x),

for ν ∈ M(C). It is straightforward to check that if instead of the boundedness ofT from M(C) to L1,∞(μ) one assumes that T is of weak type (1, 1) with respectto μ in Theorem 2.18, then one gets

|Tε f(x)| ≤ cMμ(Tf)(x) + c(‖T ‖L1(μ)→L1,∞(μ) + 1

)Mμf(x), (5.69)

for f ∈ L1(μ).So we only have to prove that the preceding estimate also holds without the

assumption (5.68), for our particular operator T . That is to say, it remains to showthat if R(x) > 0, then (5.69) holds for 0 < ε < R(x). To this end, we set

|Tε f(x)| ≤ |TR(x) f(x)|+∫0<|x−y|<R(x)

|t(x, y)| |f(y)| dμ(y).

The term |TR(x) f(x)| is bounded by the right-hand side of (5.69), while for thelast integral we use the fact that |t(x, y)| ≤ 1/R(x), and so∫

0<|x−y|<R(x)

|t(x, y)| |f(y)| dμ(y) ≤ 1

R(x)

∫B(x,R(x))

|f | dμ

≤ c0μ(B(x,R(x))

∫B(x,R(x))

|f | dμ ≤ c0 Mμf(x).

Thus, (5.69) holds for all ε > 0. �

To complete the proof of Lemma 5.25, it remains to prove (5.67). We carryout this task in the following lemma.

Lemma 5.27. Let T be a singular integral operator satisfying the assumptions ofLemma 5.25. If T is bounded in L2(μ), then it is also bounded from L1(μ) toL1,∞(μ). Moreover, we have

‖T ‖L1(μ)→L1,∞(μ) ≤ c(‖T ‖L2(μ)→L2(μ) + 1

).


Proof. The arguments are just an adaptation of the ones of Theorem 2.16 inChapter 2. We have to show that

μ({x ∈ C : |Tf(x)| > λ}) ≤ c

(‖T ‖L2(μ)→L2(μ) + 1) ‖f‖L1(μ)

λ. (5.70)

Clearly, we may assume that λ > 8‖f‖L1(μ)/‖μ‖.Let {Qi}i be the almost disjoint family of squares of Lemma 2.14 of Chapter

2 (notice that no assumption on the linear growth of μ is needed). Let Ri bethe smallest (6, 36)-doubling square of the form 6kQi, k ≥ 1. Then we can writef = g + b, with

g = χC\⋃i Qif +∑i

ϕi

andf =∑i

bi :=∑i

(wi f − ϕi) ,

where the functions ϕi satisfy (2.16), (2.17) and (2.18) of Lemma 2.14 from Chap-ter 2 and wi =

χQi∑k χQk

.

By (2.13) we have

μ(⋃

i

2Qi

)≤ c

λ

∑i

∫Qi

|f | dμ ≤ c

λ‖f‖L1(μ).

So we have to show that

μ({

x ∈ C \⋃i

2Qi : |Tf(x)| > λ})

≤ c(‖T ‖L2(μ)→L2(μ) + 1

) ‖f‖L1(μ)

λ. (5.71)

The left-hand side of (5.71) does not exceed

μ({

x ∈ C : |Tg(x)| > λ/2})

+ μ({

x ∈ C \⋃i

2Qi : |Tb(x)| > λ/2})

. (5.72)

To estimate the first term we use the L2(μ) boundedness of T , and we take intoaccount that |g| ≤ c λ, as usual:

μ({

x ∈ C : |Tg(x)| > λ/2})

≤ 4

λ2

∫|Tg|2 dμ ≤ 4 ‖T ‖L2(μ)→L2(μ)

λ2

∫|g|2 dμ

≤ c ‖T ‖L2(μ)→L2(μ)

λ

∫|g| dμ.

Also, we have ∫|g| dμ ≤

∫C\⋃i Qi

|f | dμ+∑i

∫|ϕi| dμ

≤ ‖f‖L1(μ) +∑i

‖fχQi‖L1(μ) ≤ c ‖f‖L1(μ).

5.10. Cotlar’s inequality revisited 179

Thus,

μ({

x ∈ C : |Tg(x)| > λ/2})

≤ c ‖T ‖L2(μ)→L2(μ)

λ‖f‖L1(μ).

Let us turn our attention to the second term in (5.72). By Chebyshev,

μ({

x ∈ C \⋃i

2Qi : |Tb(x)| > λ/2})

≤ 2

λ

∫C\⋃k 2Qk

|Tb| dμ (5.73)

≤ 2

λ

∑i

∫C\2Qi

|Tbi| dμ.

We claim that for each i,∫C\2Qi

|Tbi| dμ ≤ c‖fχQi‖L1(μ). (5.74)

Notice that from this claim and (5.73) one deduces

μ({

x ∈ C \⋃i

2Qi : |Tb(x)| > λ/2})

≤ c∑i

‖fχQi‖L1(μ) ≤ c‖f‖L1(μ),

which proves the lemma.

It only remains to prove (5.74). Let Bi = B(xQi ,R(xQi )/2), where xQi isthe center of Qi, and set∫

C\2Qi

|Tbi| dμ ≤∫Bi

|Tbi| dμ+

∫2Ri\(2Qi∪Bi)

|Tbi| dμ+

∫C\(2Ri∪Bi)

|Tbi| dμ.(5.75)

To estimate the first integral on the right-hand side, observe that if x ∈ Bi, thendist(x,C \H) ≥ R(xQi )/2, and so

|t(x, y)| ≤ 2

R(xQi ).

Therefore, using that ‖bi‖L1(μ) ≤ c ‖fχQi‖L1(μ) (which follows from the definitionof bi and the condition (2.18) on ϕi), we get

|Tbi(x)| ≤ 2

R(xQi )‖bi‖L1(μ) ≤ c

R(xQi)‖fχQi‖L1(μ).

Thus, ∫Bi

|Tbi| dμ ≤ c

R(xQi)‖fχQi‖L1(μ) μ(Bi) ≤ c ‖fχQi‖L1(μ),

since μ(Bi) ≤ μ(B(xQi ,R(xQi)) ≤ c0 R(xQi ).

5.11. The final probabilistic argument 181

Splitting the domain of integration in the last integral into annuli, taking intoaccount that μ(B(x, r)) ≤ 2c0 r for r ≥ max(�(Ri),

12R(xQi)), by standard argu-

ments we obtain∫|x−xQi

|>max((Ri),12R(xQi

))

�(Ri)

|x− xQi |2dμ(x) ≤ c �(Ri)

max(�(Ri),12R(xQi ))

≤ c.

Thus, the last integral on the right-hand side of (5.75) is bounded by c ‖fχQi‖L1(μ)

too. This proves our claim (5.74), and so we are done. �

5.11 The final probabilistic argument

5.11.1 The low probability of bad squares and functions

Recall that in Section 5.5 we declared a transit square Q ∈ Dtr1 to be bad with

respect to D2 if either

(a) there exists a transit square R ∈ Dtr2 such that dist(Q, ∂R) ≤ �(Q)1/4�(R)3/4

and �(R) ≥ 2m�(Q) (where m is some positive integer to be fixed below), or

(b) there exists a transit square R ∈ Dtr2 such that 2−m�(Q) ≤ �(R) ≤ 2m�(Q),

dist(Q,R) ≤ 2m�(Q), and at least one of the children P ∈ CH(R) does nothave M -thin boundary.

In the next lemma we show that bad squares happen with quite small prob-ability, if the parameters m,M are chosen appropriately.

Lemma 5.28. Let 0 < εb < 1 be any fixed (small) number. Suppose that the con-stants m and M are big enough in the definition of bad squares (depending onlyon εb). Let D1 = D(w1) be any fixed dyadic lattice. For each fixed Q ∈ D1, theprobability that it is bad with respect to a dyadic lattice D2 = D(w2), w2 ∈ Ω, doesnot exceed εb. That is,

PΩ({w2 ∈ Ω : Q ∈ D1 is bad with respect to D(w2)}

) ≤ εb.

Proof. Recall that F ⊂ 18S

0, where S0 =[0, 2N]2, with N big enough. We defined

Ω =[−2N−4, 2N−4

]2 ⊂ R2,

and for w ∈ Ω we set Q0(w) ≡ w + S0, so that Q0(w) ∈ D(w), and for eachw ∈ Ω then we have F ⊂ 1

4Q0(w). The probability measure PΩ is the normalized

Lebesgue measure on the square Ω.

Choice of m. Let D1 = D(w1), with w1 ∈ Ω, and consider Q ∈ Dtr1 . First we will

estimate the probability PΩ such that Q is bad because of the condition (a) above.That is, we deal with

PΩ({w2 ∈ Ω : ∃R ∈ D(w2)

tr such that dist(Q, ∂R) ≤�(Q)1/4�(R)3/4, (5.76)

and �(R) ≥ 2m�(Q)}).


To estimate this probability, for Q ∈ Dtr1 and k ≥ m fixed, let R(w2) ∈ D(w2) be

the square with side length 2k�(Q) containing the center zQ. Suppose moreover

that 2k�(Q) ≤ �(S0). Notice that zR(w2), the center of R(w2), belongs to 2kQ, and

moreover for any subset A ⊂ 2kQ,

PΩ({w2 ∈ Ω : zR(w2) ∈ A}) = L2(A)

L2(2kQ).

It is easy to check that there exists some R′ ∈ D(w2) with side length 2k�(Q) suchthat dist(Q, ∂R′) ≤ �(Q)1/4�(R′)3/4 if and only if

zQ ∈ Ud(R(w2)) ∩R(w2),

where d = �(Q)1/4 �(R(w2))3/4 + 1

2�(Q) and Ud(A) stands for d-neighborhood ofA. This is equivalent to saying that

zR(w2) ∈ Ud(2kQ) ∩ 2kQ.

(this can be checked by taking into account that both conditions are equivalentto 2k−1�(Q)− d ≤ |zQ − zR(w2)|∞ ≤ 2k−1�(Q), where | · |∞ is the ∞-norm in R2).Therefore,

PΩ({

w2∈ Ω : ∃R ∈ D(w2) such that dist(Q, ∂R) ≤ �(Q)1/4�(R)3/4, (5.77)

and �(R) = 2k�(Q)})

=L2(2kQ ∩ Ud(∂(2kQ))

)L2(2kQ)

≤ 4d 2k�(Q)

(2k�(Q))2.

Using thatd ≤ 2�(Q)1/4 �(R(w2))

3/4 = 2 · 23k/4 �(Q),

we infer that the probability on the left-hand side of (5.77) does not exceed

8 · 23k/4 2k�(Q)2

(2k�(Q))2= 8 · 2−k/4.

Recalling that all transit squares from D(w2) have side length at most equal to�(S0), we deduce that the probability in (5.76) is not greater than∑

k: k≥m, (2kQ)≤(S0)

8 · 2−k/4 ≤ 8 · 2−m/4

1− 2−1/4≤ 60 · 2−m/4.

We choose m so that 60 · 2−m/4 ≤ εb/2.

Choice of M . For the condition (b) in the definition of bad squares, we have toestimate the probability

PΩ({w2 :∃R ∈ D(w2)

tr with 2−m�(Q)≤ �(R)≤ 2m�(Q), dist(Q,R)≤ 2m�(Q),(5.78)

and some P ∈ CH(R) does not have M -thin boundary}).


Clearly, this probability is not greater than

PΩ({w2 :∃R ∈ D(w2) such that 2−m−1�(Q)≤ �(R)≤ min(2m�(Q), �(S0)),

(5.79)

dist(Q,R)≤ 2m�(Q), and ∂R is not M -thin}).Let Q and R be as in (5.79). Recall that ∂R is called M -thin if

μ({x ∈ C : dist(x, ∂R) ≤ r}) ≤ Mr for all r ≥ 0. (5.80)

If r ≥ �(Q), this condition is satisfied for all the squares R as in (5.79) for someM big enough due to the fact that Q is a transit square. Indeed, since

{x ∈ C : dist(x, ∂R) ≤ r} ⊂(1 +

2r

�(R)

)R (5.81)

⊂ Q(zQ,

1

2�(Q) + 2m�(Q) + �(R) + r

)⊂ Q(zQ,

1

2r + 2mr + 2mr + r

)⊂ Q(zQ, 4 · 2mr

),

we deduce that (5.80) holds for r ≥ �(Q) with M = 21/2 · 2m+3c0, by Lemma 5.9.Consider now the case r < �(Q). As shown in (5.81),

{x ∈ C : dist(x, ∂R) ≤ r} ⊂ Q(zQ, 4 · 2m�(Q)

)= 2m+3Q.

Let us write σ = μ�2m+3Q, so that

μ({x ∈ C : dist(x, ∂R) ≤ r}) = σ

({x ∈ C : dist(x, ∂R) ≤ r}) for r < �(Q).

Consider the families V(w2) and H(w2) consisting of the vertical and horizontallines, respectively, which are boundaries of the squares fromD(w2) with side length2−m−1�(Q). From the preceding arguments, if we choose M ≥ 21/2 · 2m+3c0 andw2 is as in the subset in (5.79), then there exists some line from V(w2) ∪ H(w2)which intersects 2m+3Q and is not σ-M -thin (a line L is called σ-M -thin if (5.80)holds with σ instead of μ and L instead of ∂R). Therefore, the probability (5.79)does not exceed

PΩ({w2 : ∃L∈ V(w2) such that L ∩ 2m+3Q �= ∅ and L is not σ-M -thin})

+ PΩ({w2 : ∃L∈ H(w2) such that L ∩ 2m+3Q �= ∅ and L is not σ-M -thin})

=: pV + pH.

We will use the fact that σ(2m+3Q) ≤ 21/2 2m+3c0 �(Q) to estimate pV . The argu-ments for pH are analogous.

Given t ∈ R, we denote by Lt the vertical line in C containing t. Let B ⊂ R

the subset of those t such that Lt is not σ-M -thin. Let I ⊂ R be an interval of


length 2−m−1�(Q). Since the grid V(w2) is (2−m−1�(Q))-periodic, it follows easily

that

PΩ({w2 : ∃L ∈ V(w2), L ⊂ I × R, such that L is not σ-M -thin})

=L1(B ∩ I)

L1(I). (5.82)

To estimate L1(B ∩ I), consider the vertical projection π : C → R, i.e. π(z) =Re(z), and let τ be the measure on R given by the image measure of σ by π.Observe that, for any t ∈ R, Lt is not σ-M -thin if and only if

MRτ(t) := supr>0

τ(B(t, r))

r> M.

Recall that the maximal operator MR is bounded from M(C) to L1,∞(R), andthus

L1(B ∩ I) ≤ L1(B) ≤ c

M‖τ‖ =

c

Mσ(2m+3Q) (5.83)

≤ c c0 2m+3

M�(Q) =

c c0 22m+4

ML1(I).

Let J ⊂ R be the interval π(2m+3Q) = Re(2m+3Q), and split it into 22m+4 disjointintervals Ik of length 2−m−1�(Q) (open in one extreme and closed in the other).Let Gk ⊂ Ω be the subset of those w2 such that the unique line from V(w2)intersecting Ik is σ-M -thin. From (5.82) and (5.83) we infer that

PΩ(Gk) = 1− L1(B ∩ Ik)

L1(Ik)≥ 1− c c0 2

2m+4

M.

If all the lines from V(w2) which intersect 2m+3Q are σ-M -thin, this means that

w2 ∈ ⋂22m+4

k=1 Gk, and therefore,

pV = 1− PΩ

( 22m+4⋂k=1

Gk

)= PΩ

( 22m+4⋃k=1

Gck

)

≤22m+4∑k=1

PΩ(Gck) ≤

22m+4∑k=1

c c0 22m+4

M=

c c0 24m+8

M.

The same estimate holds for pH, and thus the probability (5.78) is bounded byc c0 2

4m+9 M−1 = εb/2, assuming that we choose M = c c0 24m+10 ε−1

b .If we add the estimates obtained for the probabilities (5.76) and (5.78), the

lemma follows. �The preceding result has the following consequence regarding good and bad

functions.


Lemma 5.29. Let f ∈ L2(μ) and for w = (w1, w2) set

fbad(w) =∑

Q∈D(w1)∩ bad(w2)

ΔQf,

where “Q ∈ bad(w2)” means “Q is bad with respect to D(w2)”. Then we have

EΩ2(‖fbad(w)‖2L2(μ)

) ≤ c23εb‖f‖2L2(μ),

where EΩ2

stands for the expectation with respect to PΩ2

= PΩ × PΩ.

Proof. From Lemma 5.11 we deduce that

‖fbad(w)‖2L2(μ) ≤ c3∑


‖ΔQf‖2L2(μ).

Then, by Lemma 5.28,

EΩ2(‖fbad(w)‖2L2(μ)

) ≤ c3

∫∫ ∑Q∈D(w1)∩ bad(w2)

‖ΔQf‖2L2(μ) dPΩ(w1) dP

Ω(w2)

= c3

∫ ∑Q∈D(w1)

‖ΔQf‖2L2(μ)PΩ({

w2 : Q ∈ bad(w2)})

dPΩ(w1)

≤ c3εb

∫ ∑Q∈D(w1)

‖ΔQf‖2L2(μ) dPΩ(w1) ≤ c23εb‖f‖2L2(μ). �

5.11.2 The nice set G

Let WD1 , WD2 be the total exceptional sets corresponding to two dyadic latticesD1 = D(w1), D2 = D(w2). We showed in (5.6) that

μ(F \WD(w)) ≥ (1− δ1)μ(F ),

with 0 < δ1 < 1 for all w ∈ Ω. For each x ∈ F we consider the probabilities

p0(x) = PΩ({w ∈ Ω : x ∈ F \WD(w)}

),

andp(x) = PΩ2({(w1, w2) ∈ Ω× Ω : x ∈ F \ (WD(w1) ∪WD(w2)) }

).

Recall that PΩ is the normalized Lebesgue measure on Ω and PΩ2

= PΩ × PΩ.Since the sets F \WD(w1) and F \WD(w2) are independent, we have p(x) = p0(x)

2.In other words, by Fubini:

p(x) =

∫∫χF\D(w1)(x)χF\D(w2)(x) dP

Ω(w1) dPΩ(w2) = p0(x)

2.


Notice that∫F

p0(x) dμ(x) =

∫w∈Ω

∫χF\WD(w)

(x) dμ(x) dPΩ(w)

=

∫w∈Ω

μ(F \WD(w)) dPΩ(w) ≥ (1− δ1)μ(F ).

LetG = {x ∈ F : p0(x) > (1− δ1)/2}, (5.84)

and B = F \G. We have

μ(B) ≤ 2

1 + δ1

∫F

(1− p0(x)

)dμ(x)

=2

1 + δ1

(μ(F )−

∫F

p0(x) dμ(x)

)≤ 2δ1

1 + δ1μ(F ).

Thus,

μ(G) ≥ 1− δ11 + δ1

μ(F ).

Observe, moreover, that for every x ∈ G we have p(x) = p0(x)2 > (1−δ1)

2/4 =: β.Roughly speaking, this means that for a big proportion of dyadic lattices D(w1),D(w2), the points from G are in F \ (WD(w1) ∪WD(w2)).

Notice also that if x ∈ ⋂w∈C

(HD(w) ∪ TD(w)

), then x �∈ F \WD(w) for every

w ∈ Ω and so p0(x) = 0. From the definition (5.84), we infer that x �∈ G. In otherwords,

G ⊂ F \⋂w∈C

(HD(w) ∪ TD(w)

),

as wished.Now we define

Φ(w1,w2)(x) = dist(x, F \ (WD(w1) ∪WD(w2))

).

Observe that if x ∈ G, then

PΩ2({(w1, w2) : Φ(w1,w2)(x) = 0}) ≥ p(x) > β.

We deduce that

μ({x ∈ F : p(x) > β}) ≥ μ(G) ≥ 1− δ1

1 + δ1μ(F )

and

μ({

x ∈ F : PΩ2({(w1, w2) : Φ(w1,w2)(x) = 0}) > β})

≥ 1− δ11 + δ1

μ(F ).

In the remaining part of this chapter we will prove that the Cauchy transformis bounded in G.


5.11.3 The functions Φ, Ψw, and the operators K and CFor a fixed ε0 > 0, we define

Φ(x) = ε0 + infB⊂Ω2,PΩ2(B)=β

sup(w1,w2)∈B

Φ(w1,w2)(x). (5.85)

Notice that Φ is a 1-Lipschitz function such that Φ(x) = ε0 for all x ∈ G. Moreover,

Φ(x) ≥ max(dist(x,C \H), e(x), ε0

) ≥ max(R(x), e(x), ε0

)for all x ∈ F ,

where e(x) was defined in (5.2) and H and R(x) in (5.64). This is due to the factthat Φ(w1,w2)(x) ≥ max

(dist(x,C \H), e(x)

)for all x ∈ F, (w1, w2) ∈ Ω×Ω, since

all non-Ahlfors disks are contained in HD for any choice of the lattice D, and Sdoes not depend on D.

Given w = (w1, w2) ∈ Ω2, we also set

Ψw(x) = max(Φ(x),Φw(x)

).

We write

Kf(x) =

∫KΨwf(x) dP

Ω2

(w).

So the kernel of K is:

k(x, y) =

∫kΨw(x, y) dP

Ω2

(w).

The analogous definition for truncated operators is the following:

Cf(x) =∫

CΨw(x)f(x) dPΩ2

(w),

so that the kernel of C is

c(x, y) =

∫cΨw(x, y) dP

Ω2

(w),

with cΨw(x, y) =1

x−y χ|x−y|≥Ψw(x). Observe that k(x, y) is a Calderon-Zygmund

kernel (with absolute constants), because it is the average of the Calderon-Zyg-mund kernels kΨw(x, y). On the contrary, this is not the case for c(x, y), in general.

We will show below that the operator KΦ is bounded in L2(μ), with normindependent of ε0. Since Φ equals ε0 on G, this will yield the boundedness of theCauchy transform in L2(μ�G) (notice ε0 is arbitrary and that G does not depend

on ε0). In the proof of the L2(μ) boundedness of KΦ, the operators K and C willplay a key role.

Let us insist on the fact that none of the constants below will depend on thechoice of ε0.


5.11.4 The key estimates involving C and K

Lemma 5.30. Consider the truncated Cauchy kernels

cΦ(x, y) =1

x− yχ|x−y|≥Φ(x), cΨw(x, y) =

1

x− yχ|x−y|≥Ψw(x),

and set also c(x, y) = EΩ2

cΨw(x, y). The associated operators are denoted by CΦ,CΨw , and C, respectively. For f ∈ L2(μ) and x ∈ C, the following estimates hold:

|CΦf(x)| ≤ 2β−1C∗f(x), (5.86)

and

|CΨwf(x)| ≤ 2β−1C∗f(x) for each w ∈ Ω2, (5.87)

where C∗ is the maximal operator associated with the kernel c(x, y):

C∗f(x) = supt>0

|Ctf(x)| = supt>0

∣∣∣∣∫|x−y|≥t

c(x, y) f(y) dμ(y)

∣∣∣∣.Notice that C is the average of the operators CΨw over w ∈ Ω2. So the

inequality (5.87) asserts that each operator CΨw is controlled by the maximal

version of the average operator C, which might look surprising at first sight.

Proof. First we will prove (5.86). Observe that

c(x, y) =1

x− yPΩ2({w ∈ Ω2 : |x− y| ≥ Ψw(x)}

)=

1

x− yvx(|x− y|), (5.88)

wherevx(t) = PΩ2({w ∈ Ω2 : t ≥ max(Φ(x), Φw(x)) = Ψw(x)}

).

This function satisfies the following properties:

1. t < Φ(x) ⇒ vx(t) = 0.

2. vx is non-decreasing in t and right-continuous.

3. t ≥ Φ(x) ⇒ vx(t) ≥ β.

The first two properties are clear. The last one follows from the fact that if t ≥Φ(x), then

vx(t) = PΩ2({w ∈ Ω2 : t ≥ Φw(x))}), (5.89)

and moreover, recalling the definition of Φ(x) in (5.85),

t ≥ infB⊂Ω2,PΩ2 (B)≥β

supw′∈B

Φw′(x),


which means that there exists some subset B ⊂ Ω2 such that P (B) ≥ β andsupw∈B Φw(x) ≤ t. So B is contained in the subset on the right-hand side of

(5.89), and then vx(t) ≥ PΩ2

(B) ≥ β.From (5.88) and the fact that vx(t) vanishes for t < Φ(x), we infer that

c(x, y) = c(x, y)χ|x−y|≥Φ(x)

=1

x− yvx(|x− y|)χ|x−y|≥Φ(x) = vx(|x− y|) cΦ(x, y).

Thus,cΦ(x, y) = vx(|x− y|)−1 c(x, y)χ|x−y|≥Φ(x). (5.90)

To prove (5.86) we will show that, using the preceding identity and the semi-continuity, the boundedness and the monotonicity of 1/vx, we can write the kernelcΦ(x, y) as a convex combination of the kernels c(x, y)χ|x−y|≥t, for t ≥ Φ(x). In-deed, since vx(t)

−1 is continuous from the right and non-increasing, we can definethe Lebesgue-Stieljes measure σ on R given by⎧⎨⎩

σ(a, b] = vx(a)−1 − vx(b)

−1 if (a, b] ⊂ (Φ(x),∞),

σ(−∞,Φ(x)] = 0.

Observe that σ is a finite positive measure. Indeed, from the definition above it isclear that σ(R) ≤ supR

1vx

≤ 1β . Now we write

∫t>Φ(x)

c(x, y)χ|x−y|≥t dσ(t) = c(x, y)σ(Φ(x), |x− y|]

= c(x, y)χ|x−y|≥Φ(x)

(vx(Φ(x))

−1 − vx(|x− y|)−1). (5.91)

Therefore, by (5.90),

cΦ(x, y) = vx(Φ(x))−1 c(x, y)χ|x−y|≥Φ(x) −

∫t>Φ(x)

c(x, y)χ|x−y|≥t dσ(t).

Multiplying this identity by f(y) and integrating with respect to μ on y, we get

CΦf(x) = vx(Φ(x))−1 CΦ(x)f(x)−

∫t>Φ(x)

Ctf(x) dσ(t), (5.92)

where Ct and CΦ(x) are the t-truncated and Φ(x)-truncated versions of the op-

erator C defined above, respectively. From the preceding identity, since σ(R) =vx(Φ(x))

−1 ≤ β−1, we infer that

|CΦf(x)| ≤ β−1C∗f(x) +∫t>Φ(x)

C∗f(x) dσ(t) ≤ 2β−1C∗f(x),


and so (5.86) follows.Let us turn our attention to (5.87) now. Notice that by the definition of

cΨw(x, y), we have

cΨw(x, y) = cΦ(x, y)χ|x−y|≥Ψw(x) = vx(|x− y|)−1 c(x, y)χ|x−y|≥Ψw(x). (5.93)

This is the same identity as (5.90), replacing Φ by Ψw. Arguing as in (5.91),integrating with respect to σ for t > Ψw(x), we obtain the same identity as (5.92),with Ψw instead of Ψ, namely,

CΨwf(x) = vx(Ψw(x))−1 CΨw(x)f(x)−

∫t>Ψw(x)

Ctf(x) dσ(t),

and so, as above, we get

|CΨwf(x)| ≤ β−1C∗f(x) +∫t>Φ(x)

C∗f(x) dσ(t) ≤ 2β−1C∗f(x). �

Lemma 5.31. We have

‖KΦ‖L2(μ)→L2(μ) ≤ cβ−1‖K‖L2(μ)→L2(μ) + cβ−1, (5.94)

and

‖KΨw‖L2(μ)→L2(μ) ≤ cβ−1‖K‖L2(μ)→L2(μ) + cβ−1 for each w ∈ Ω2. (5.95)

Proof. Recall that, by Lemma 5.4, if Θ : C−→(0,+∞) is a Lipschitz function withLipschitz constant 1, for ε ≥ Θ(x), we have

|KΘ,εf(x)− Cεf(x)| ≤ c supr≥ε

1

r

∫B(x,r)

|f | dμ. (5.96)

For an arbitrary ε ≥ 0, write

CΘ,εf(x) = Cmax(Θ(x),ε)f(x).

Then we claim that

|KΘ,εf(x)− CΘ,εf(x)| ≤ c supr≥Θ(x)

1

r

∫B(x,r)

|f | dμ.

Indeed, if ε ≥ Θ(x), this follows from (5.96), while for ε < Θ(x), using that|kΘ(x, y)| ≤ c/Θ(x) and (5.96) with ε = Θ(x), we get

|KΘ,εf(x)− CΘ,εf(x)| ≤ |KΘ,Θ(x)f(x)− CΘf(x)|+

∫B(x,Θ(x))

|kΘ(x, y)| |f(y)| dμ(y)

≤ c supr≥Θ(x)

1

r

∫B(x,r)

|f | dμ.


Suppose now that Θ is one of the Lipschitz functions Φ or Ψw introducedabove. Since μ(B(x, r)) ≤ c0 r for all r ≥ Θ(x), the last supremum above isbounded by c0 Mμf(x). That is,

|KΘ,εf(x)− CΘ,εf(x)| ≤ c c0 Mμf(x) for all ε ≥ 0. (5.97)

Choosing Θ = Φ and ε = 0, from Lemma 5.30, we deduce that

|KΦf(x)| ≤ |CΦf(x)|+ c c0 Mμf(x) ≤ 2β−1C∗f(x) + c c0 Mμf(x). (5.98)

Thus, ‖KΦ‖L2(μ)→L2(μ) ≤ 2β−1‖C∗‖L2(μ)→L2(μ) + c. On the other hand, settingΘ = Ψw in (5.97) and taking the mean on w ∈ Ω2, we get

|Kεf(x)− Cεf(x)| ≤ c c0 Mμf(x) for all ε ≥ 0.

Taking supremum in ε, we deduce that

C∗f(x) ≤ K∗f(x) + c c0 Mμf(x),

where K∗f(x) = supε>0 |Kεf(x)|. Thus, by (5.98),

|KΦf(x)| ≤ |CΦf(x)| ≤ 2β−1K∗f(x) + c c0 β−1Mμf(x),

where we took also into account that 0 < β < 1. By Lemma 5.25, we know that‖K∗‖L2(μ)→L2(μ) ≤ c ‖K‖L2(μ)→L2(μ) + c. Therefore,

‖KΦ‖L2(μ)→L2(μ) ≤ cβ−1‖K‖L2(μ)→L2(μ) + cβ−1. (5.99)

The proof of (5.95) is analogous. �

5.11.5 The final step

Lemma 5.32. The operator K is bounded in L2(μ).

Proof. Given w = (w1, w2), that is, given two dyadic lattices D(w1) and D(w2),and two functions f, g ∈ L2(μ), we write

〈KΨwf, g〉 = 〈KΨwfgood(w), ggood(w)〉+ 〈KΨwfgood(w), gbad(w)〉 (5.100)

+ 〈KΨwfbad(w), g〉,where

fgood(w) = Ξf +∑

Q∈D(w1)∩ good(w2)

ΔQf, fbad(w) =∑


ΔQf,

and

ggood(w) = Ξg +∑

Q∈D(w2)∩ good(w1)

ΔQg, gbad(w) =∑


ΔQg.


Of course, “good(wi)” means “good with respect to D(wi)”, and analogously for“bad(wi)”.

We will average equation (5.100) over w ∈ Ω2. Concerning the left-hand side,notice that

EΩ2 〈KΨwf, g〉 = 〈Kf, g〉.Now we turn our attention to the right-hand side of (5.100). First, by Lemma5.13, for each w ∈ Ω2, we have

|〈KΨwfgood(w), ggood(w)〉| ≤ c5‖f‖2 ‖g‖2.On the other hand,

|〈KΨwfgood(w), gbad(w)〉| ≤ ‖KΨw‖L2(μ)→L2(μ)‖f‖L2(μ)‖gbad(w)‖L2(μ).

Recall that by Lemma 5.29,

EΩ2(‖gbad(w)‖2L2(μ)

) ≤ c23εb‖g‖2L2(μ).

Therefore,

|EΩ2〈KΨwfgood(w), gbad(w)〉| ≤ EΩ2(‖KΨwf‖2L2(μ)

)1/2EΩ2(‖gbad(w)‖2L2(μ)

)1/2≤ c3ε

1/2b sup

w∈Ω2

‖KΨw‖L2(μ)→L2(μ)‖f‖L2(μ)‖g‖L2(μ).

Arguing analogously, we get

|EΩ2〈KΨwfbad(w), g〉| ≤ c3ε1/2b sup

w∈Ω2

‖KΨw‖L2(μ)→L2(μ)‖f‖L2(μ)‖g‖L2(μ).

Since, by (5.95),

supw∈Ω2

‖KΨw‖L2(μ)→L2(μ) ≤ cβ−1‖K‖L2(μ)→L2(μ) + cβ−1,

from the estimates above, averaging (5.100) over w ∈ Ω2, we deduce∣∣〈Kf, g〉∣∣ ≤ (c5+c3ε1/2b

[cβ−1‖K‖L2(μ)→L2(μ)+cβ−1

])‖f‖L2(μ)‖g‖L2(μ). (5.101)

Taking suprema in f, g ∈ L2(μ), we derive

‖K‖L2(μ)→L2(μ) ≤ c5 + c3ε1/2b

[cβ−1‖K‖L2(μ)→L2(μ) + cβ−1

].

Recall that β = (1− δ1)2/4, and δ1 is independent of εb (it was defined just before

Section 5.3). As a consequence, if εb is chosen small enough, then we infer that

‖K‖L2(μ)→L2(μ) ≤ 2c5. �


Now we conclude easily the (hard) proof of Theorem 5.1:

Corollary 5.33. The operator KΦ is bounded in L2(μ) with a bound independentof ε0, and thus the Cauchy transform is bounded in L2(μ�G).

Proof. The first assertion is a direct consequence of Lemma 5.32 and (5.94). Thesecond one is due to the fact that Φ(x) = ε0 on G and so the operatorKΦ coincideswith the Cauchy ε0-regularized transform on L2(μ�G), with kernel

x− y

|x− y|2 + ε20.

From Lemma 5.4 we infer that the Cε0 is bounded in L2(μ) uniformly on ε0. �


5.12.1 The Tb theorem and analytic capacity

Most of this chapter is based on the arguments from Nazarov, Treil and Volberg[126]. Another reference where one can find the proof of the Tb theorem above isVolberg’s book [180], where it is used to study the Lipschitz harmonic capacity inRd. Moreover, in Volberg’s monograph, Theorem 5.1 is proved in a greater gener-ality, namely, for general n-dimensional antisymmetric singular integral operatorsin Rd (replacing the linear growth condition by growth of the degree n out of theexceptional set). With minor modifications, the proof we saw in this chapter forthe Cauchy kernel also works for k(x, y). Let us mention that, if k(x, y) is thekernel of such a singular integral operator, a possible way to define the suppressedkernel kΘ(x, y) is the following, as shown in Volberg [180]:

kΘ(x, y) = k(x, y)1

1 + |k(x, y)|2Θ(x)nΘ(y)n.

On the other hand, if χ : [0,∞) → [0, 1] is a smooth cut-off function which vanishesidentically in [0, 1/2] and equals 1 in [1,+∞), one can also consider the kernel

kΘ(x, y) = χ

( |x− y|2Θ(x)Θ(y)

)k(x, y).

This also coincides with k(x, y) if Θ(x) or Θ(y) vanish. Moreover, it has the ad-

vantage that k(x, y) = 0 if |x− y| ≤ cmax(Θ(x),Θ(y)), for some suitable absoluteconstant c.

The Tb theorem that David proved in [23] to solve Vitushkin’s conjecture (forsets with finite length) is of a qualitative nature and it seems difficult to derivea quantitative version from it suitable to prove later the comparability betweenγ and γ+. Moreover, David’s Tb theorem relies on a previous joint result withMattila [27] (where the case where b is a real-valued function is studied), and the


sum of the papers by David and Mattila [27] and David [23] is longer than theone by Nazarov, Treil and Volberg [126]. For these reasons, in this book we havefollowed the approach in the latter paper.

5.12.2 Other Tb-like theorems

The classical Tb theorem in homogeneous spaces was proved by David, Journeand Semmes [26] in 1985. A previous result for the particular case Tb = 0 wasobtained by McIntosh and Meyer [108]. Most of the usual Tb-like theorems requirestronger conditions than the ones in Theorem 5.1, and then they guaranty theL2(μ) boundedness on the whole suppμ. For example, instead of the L1(μ) typecondition (c) on C∗(b μ) in Theorem 5.1, in the Tb theorem of David, Journe andSemmes [26] (for an antisymmetric kernel, say) one requires the condition Tb ∈BMO(μ). Also, an appropriate accretivity condition is necessary for all the dyadiccubes Q ⊂ Rd. Namely, for any such cube, one requires the existence of another

cube Q′ with �(Q) = �(Q′), dist(Q,Q′) ≤ c�(Q), such that∣∣∣∫Q′ b dμ

∣∣∣ ≥ c−1μ(Q′).In [15], Coifman, Journe and Semmes obtained a simpler proof of the Tb theoremusing a martingale decomposition with respect to the function b. Another relatedand more flexible result is the so-called local Tb theorem from Christ [14], wherethe existence of a unique function b good for all the dyadic squares is replaced bythe existence of a collection of functions bQ, where each bQ is associated with adyadic cube Q and is required to be good only for Q, roughly speaking.

The Tb theorem of David, Journe and Semmes has been extended to the caseof non-homogeneous spaces by Nazarov, Treil and Volberg in [127], and a version ofthe local Tb theorem of Christ has been obtained by Nazarov, Treil and Volbergin [125]. The arguments in these papers use dyadic martingale decompositions,like in Coifman, Journe and Semmes [15]. The averaging with respect to randomdyadic lattices is one of the key ingredients in the non-homogeneous situation, likein this chapter. Let us also mention that Hytonen and Martikainen have provedsome extensions of these Tb theorems to the more general case of upper doublingmeasures in [67] and [66]. See Section 9.11 of Chapter 9 for some precise statementsregarding Tb theorems on non-homogeneous spaces.

Chapter 6

The comparability between γand γ+, and the semiadditivityof analytic capacity

6.1 Introduction

The main result of this chapter is the following.

Theorem 6.1. There exists an absolute constant c such that

γ(E) ≤ c γ+(E)

for any compact set E ⊂ C. As a consequence, γ(E) ≈ γ+(E).

As a corollary, using the characterization of γ+ in terms of curvature, we get:

Corollary 6.2. A compact set E ⊂ C is non-removable for bounded analytic func-tions if and only if it supports a non-zero Radon measure with linear growth andfinite curvature.

Also, since γ+ is countably semiadditive, we deduce:

Corollary 6.3. Let E ⊂ C be compact. Let Ei, i ≥ 1, be Borel sets such thatE =⋃∞

i=1 Ei. Then

γ(E) ≤ c∞∑i=1

γ(Ei),


, , OI 10.1007/978-3- - -6_ ,

© Springer



1958

196 Chapter 6. The comparability between γ and γ+

6.2 An argument for sets of finite length

In this section we will prove the following theorem, due to David [23].

Theorem 6.4. Let E ⊂ C be compact with H1(E) < ∞ and γ(E) > 0. Thenγ+(E) > 0.

Clearly, this result can be considered as a particular case of Theorem 6.1.So the proof of Theorem 6.4 can be skipped by the reader without much harmas it will come for free later. On the other hand, to prove Vitushkin’s conjecture,the full strength of the comparability between γ and γ+ is not needed, and thearguments in this section suffice (together with the curvature theorem of Davidand Leger from the next chapter).

We have decided to include the proof of Theorem 6.4 for two reasons: firstbecause it follows by a quite direct application of the Tb theorem of Nazarov, Treiland Volberg; and second because, by examining its proof, one can understandbetter why a more or less direct application of a Tb-like theorem does not sufficeto prove Theorem 6.1.

We need first the following proposition (that will also be used below in theproof of Theorem 6.1).

Proposition 6.5. Let E ⊂ C be compact with H1(E) < ∞, and let f : C \ E → C

be analytic such that ‖f‖∞ ≤ 1 and f(∞) = 0. Then there is a complex measureν supported on E such that f(z) = Cν(z) for all z �∈ E. Moreover, this measuresatisfies

|ν(B(z, r))| ≤ r for all z ∈ C, r > 0,

and it can be written as ν = bH1�E, where b is a measurable function such that|b(z)| ≤ 1 for all z ∈ E.

Proof. Given ε > 0, we cover E with a family of sets Aε,i with diam(Aε,i) < εsuch that

∑i diam(Aε,i) ≤ H1(E) + ε. We can replace each Aε,i by an open ball

Bε,i centered in E ∩ Aε,i, with radius rε,i < ε slightly bigger than diam(Aε,i) ifnecessary so that E ⊂ ⋃i Bε,i and

∑i rε,i ≤ H1(E) + 2ε. Since E is compact we

may assume that this is a finite family. Consider the curve (or family of curves)Γε = ∂

(⋃iBε,i

), oriented so that the winding number is 1 about each z ∈ ⋃iBε,i.

Then, for z ∈ C \⋃iBε,i, we have

f(z) = − 1

2πi

∫Γε

f(w)

w − zdw.

We take the complex measure

dνε(w) = −f(w)

2πidw�Γε,

so that f(z) = Cνε(z) for z ∈ C \⋃iBε,i. Observe that ‖νε‖ ≤∑i rε,i ≤ H1(E) +2ε. Thus, there exists a sequence εk → 0 such that the complex measures νεk

6.2. An argument for sets of finite length 197

converge weakly � to another complex measure ν. It is clear that supp ν ⊂ E andf(z) = Cν(z) for all z �∈ E.

The arguments to show that |ν(B(z, r))| ≤ r for all z ∈ C, r > 0, are verysimilar to the ones for the particular case where ν is positive, used in Theorem4.14. For the reader’s convenience, we repeat the details. Since H1(E) < ∞, wehave |Cν(w)| ≤ 1 a.e. in C, with respect to Lebesgue measure. For every z ∈ C,by Fubini it turns out that for almost all r > 0,∫∫

|ξ−z|=r

1

|ξ − w| dH1(ξ) d|ν|(w) < ∞.

For such r, by Fubini again, we have

|ν(B(z, r))| =∣∣∣∣∫|ξ−z|=r

Cν(ξ) dξ

2πi

∣∣∣∣ ≤ r. (6.1)

By approximation the same estimate holds for every r > 0.To show that ν = bH1�E, where b is some function such that |b(z)| ≤ 1 for all

z ∈ E, it is enough to show that |ν(A)| ≤ H1(A) for any measurable subset A ⊂ E,by the Radon-Nikodym theorem. Since ν and H1�E are Radon measures, if weshow that |ν(U)| ≤ H1(U∩E) for any open set U , we are done. To this end, let ϕ bea continuous function such that 0 ≤ ϕ ≤ χU , and let δ = dist(supp(ϕ), C\U) > 0.Then we have ∫

ϕdν =−1

2πilimk→0

∫Γεk

ϕf(w) dw.

Thus, if we denote by Ik the family of those indices i such that Bεk,i∩supp(ϕ) �= ∅,we have∣∣∣∣∫ ϕdν

∣∣∣∣ ≤∑i∈Ik

rεk,i =∑i

rεk,i −∑i�∈Ik

rεk,i ≤ H1(E) + 2εk −∑i�∈Ik

rεk,i.

To deal with the last sum, we claim that if εk < δ/2, then E \ U ⊂ ⋃i�∈IkAεk,i.

Indeed, if x ∈ E \U , then there exists some i such that x ∈ Aεk,i \U ⊂ Bεk,i \U .Since diam(Bεk,i) < δ, Bεk,i ∩ suppϕ = ∅ and so i �∈ Ik and the claim follows.Then we deduce that H1

εk(E \ U) ≤∑i�∈Ik

rεk,i, and thus∣∣∣∣∫ ϕdν

∣∣∣∣ ≤ H1(E) + 2εk −H1εk(E \ U).

Letting k → ∞, we obtain∣∣∣∣∫ ϕdν

∣∣∣∣ ≤ H1(E)−H1(E \ U) = H1(U ∩ E).

Since this holds for any arbitrary function ϕ such that 0 ≤ ϕ ≤ χU , we get|ν(U)| ≤ H1(U ∩E), as wished. �


In the proposition we have just proved, the assumption that E has finitelength is essential. In fact, given an arbitrary compact set E ⊂ C and f analyticand bounded in C \ E, there may not exist a measure ν such that f = Cν. Onthe other hand, by Theorem 1.14, recall that, in the sense of distributions, f =C(π−1 ∂f), assuming f to be defined appropriately on E (setting f = 0 on E, forexample). So f is always the Cauchy transform of a distribution, namely π−1 ∂f .

Now we are ready to prove Theorem 6.4.

Proof of Theorem 6.4. Consider the measure μ = H1�E. To prove the theoremit is enough to show that there exists some subset G ⊂ E with μ(G) > 0 suchthat μ�G has linear growth and Cμ�G is bounded in L2(μ�G), by Theorem 4.14.To this end we wish to apply the Tb theorem of Nazarov, Treil and Volberg inthe preceding chapter. So, using the notation in Theorem 5.1, for each w ∈ C,we have to construct a complex measure ν supported on E and some subsetsHD(w), TD(w) ⊂ C made up of dyadic squares from D(w) = w + D0 (where D0 isthe usual dyadic lattice) such that


w∈C HD(w).

(b) dν = b dμ, with ‖b‖L∞(μ) ≤ cb.

(c)∫C\HD(w)

C∗ν dμ ≤ c∗ μ(E), for all w ∈ C.

(d) If Q ∈ D(w) is such that Q �⊂ TD(w), then μ(Q) ≤ cacc|ν(Q)| (i.e. Q is anaccretive square).

(e) μ(HD(w) ∪ TD(w)) ≤ δ0 μ(E), for all w ∈ C and some δ0 < 1.

If these assumptions hold, then there exists a subset G ⊂ F such that μ(G) ≥c−11 μ(F ), μ�G has c0-linear growth, and the Cauchy transform is bounded onL2(μ�G). The constant c1 and the bound for the L2 norm depend only on theconstants above.

We take the complex measure ν given by Proposition 6.5, which clearly sat-isfies the assumption (b) above, with cb = 1.

Let us see how we construct HD(w) for any given w ∈ C. Since H1(E) <∞, as explained in Theorem 1.30, the following estimate holds for the upper 1-dimensional density of μ:

lim supr→0

μ(B(x, r))

2r≤ 1 for μ-a.e. x ∈ C.

Thus, for all ε > 0 there exists some δ > 0 such that the set

Eδ :={x ∈ C : μ(B(x, r)) ≤ 3r for 0 < r ≤ δ

}satisfies μ(Eδ) ≥ (1 − ε)μ(E). Take such δ. Notice that for each x ∈ Eδ, we have

μ(B(x, r)) ≤ max(3,

μ(E)

δ

)r, for all r > 0.

6.2. An argument for sets of finite length 199

Let M = max(3, μ(E)

δ

)and consider the set

H =⋃

B ball:μ(B)>10Mr(B)

B. (6.2)

Each ball B which appears in the definition of H is covered by at most foursquares Q from D(w) such that 2r(B) < �(Q) ≤ 4r(B). We denote by DH(w) thecollection of all these dyadic squares covering all the balls in the union in (6.2),and we define

HD(w) =⋃

Q∈DH(w)

Q.

By definition we have H ⊂ HD(w), and thus the assumption (a) above, for theapplication of the Nazarov-Treil-Volberg theorem, is satisfied with c0 = 10M .

It remains to define the exceptional set TD(w) and to check that the assump-tions (c), (d), and (e) also hold.

First we estimate μ(HD(w)). To this end, observe that if x ∈ HD(w), then

there exists a ball B0 such that μ(B0) > 10Mr(B0) and dist(x,B0) ≤ 4√2 r(B0),

and thus B0 ⊂ B(x, 10r(B0)). This implies that μ(B(x, 10r(B0)) ≥ μ(B0) >10Mr(B0) and so x �∈ Eδ. That is, HD(w) ⊂ E \Eδ and then

μ(HD(w)) ≤ εμ(E). (6.3)

Let us turn our attention to the exceptional set TD(w) now. Write cacc =4μ(E)/γ(E) = 4|ν(E)|/γ(E). If a dyadic square Q ∈ D(w) satisfies

μ(Q) ≥ cacc|ν(Q)|, (6.4)

we write Q ∈ DT (w). The exceptional set TD(w) is just defined by

TD(w) =⋃

Q∈DT (w)

Q.

Clearly, the assumption (d) holds by construction.To prove (e), first we will estimate |ν(HD∪TD)|. LetDHT (w) be the subfamily

of maximal (and thus disjoint) squares from DH(w) ∪ DT (w). Using that ν = b μwith ‖b‖L∞(μ) ≤ 1 and the condition (6.4) for the squares in DT (w), we get

|ν(HD(w) ∪ TD(w))| ≤∑

Q∈DHT (w)

|ν(Q)|

≤∑

Q∈DHT (w)∩DH(w)

|ν(Q)|+∑

Q∈DHT (w)∩DT (w)

|ν(Q)|

≤∑

Q∈DHT (w)∩DH(w)

μ(Q) + c−1acc

∑Q∈DHT (w)∩DT (w)

μ(Q).


The squares in the first sum are contained in HD(w) and then, by (6.3), this sumis bounded above by εμ(E). Thus

|ν(HD(w) ∪ TD(w))| ≤ εμ(E) + c−1acc μ(E) = εμ(E) +

1

4|ν(E)|.

Therefore, if we choose ε = γ(E)/4μ(E), we have |ν(HD(w) ∪ TD(w))| ≤ |ν(E)|/2.As a consequence,

|ν(E \ (HD(w) ∪ TD(w)))| ≥ 1

2|ν(E)|,

which implies that

μ(E \ (HD(w) ∪ TD(w))) ≥ |ν(E \ (HD(w) ∪ TD(w)))| ≥ 1

2|ν(E)| = 1

2γ(E),

and thus,

μ(HD(w) ∪ TD(w)) ≤ μ(E)− 1

2γ(E) =

(1− γ(E)

2H1(E)

)μ(E).

So (e) holds with δ0 = 1− γ(E)/2H1(E).

It only remains to prove (c). We will use the regularized operators Cε intro-duced in Section 4.2 of Chapter 4. So take a C∞ non-negative radial function ψsupported on B(0, 1) with L1 norm equal to 1, and let Cε be the operator associ-ated with the kernel 1

z ∗ ψε. Since |Cν(z)| ≤ 1 for all z �∈ E and L2(∂E) = 0, then|Cν(z)| ≤ 1 L2-a.e. z ∈ C. Thus

‖Cεν‖L∞ = ‖ψε ∗ Cν‖L∞ ≤ ‖Cν‖L∞ ≤ 1.

Since Cεν is continuous in C, we deduce that |Cεν(z)| ≤ 1 for all z ∈ C, ε > 0. By

(4.2) in Lemma 4.3 we have |Cεν(z) − Cεν(z)| ≤ cψ MRν(z). For z ∈ E \ HD(w),from the linear growth condition (i.e. the condition (a) above) and (b), we deducethat MRν(z) ≤ MRμ(z) ≤ c0, and so |Cεν(z)| ≤ 1 + cψ c0 for all ε > 0, whichyields (c). �

6.3 Outline of the argument for proving that γ ≈ γ+

To prove Theorem 6.1, we have to show that there exists an absolute constant csuch that

γ(E) ≤ c γ+(E)

for every compact set E ⊂ C. By the characterization of γ+ obtained in Theorem4.14, one way to do this consists in proving that there exists some measure μsupported on E, with μ(E) ≈ γ(E), such that μ(B(x, r)) ≤ r, and such that theCauchy transform Cμ is bounded in L2(μ) with absolute constants.

6.3. Outline of the argument for proving that γ ≈ γ+ 201

In the preceding section, in the case where E has finite length we have shownthat there exists some subset G ⊂ E with positive length such that Cμ is boundedin L2(μ), for μ = H1�G. However, the constants that we obtained involve the termγ(E)/H1(E) and so explode when H1(E) tends to infinite or γ(E) to 0.

Recall that, for sets with finite length, the proof of Theorem 6.4 stems fromthe existence of a complex measure ν0 (which would be a distribution in the generalcase of infinite length) supported on E such that

‖Cν0‖∞ ≤ 1, (6.5)

|ν0(E)| = γ(E), (6.6)

dν0 = b0 dH1�E, with ‖b0‖∞ ≤ 1. (6.7)

By a rather direct application of a Tb theorem such as the one of Nazarov,Treil and Volberg we cannot expect to prove that the Cauchy transform is boundedwith respect to a measure μ with total mass comparable to γ(E), with absoluteconstants. Let us explain the reasons in some detail. Suppose for example thatthere exists some function b such that dν0 = b dμ and we use the information aboutν0 given by (6.5), (6.6) and (6.7). The estimate (6.5) is quite good for our purposes(recall that in the assumption (c) in Theorem 5.1 we asked

∫C\HD(w)

C∗ν0 dμ ≤c∗ μ(E)). On the other hand, assuming that μ(E) ≈ γ(E), from (6.6) we deducethat |ν0(E)| ≈ μ(E), which is a global accretivity condition that can be used toconstruct a small exceptional set TD(w), so that the dyadic squares from D(w) notcontained in TD(w) satisfy the condition μ(Q) ≤ cacc|ν(Q)|.

The main obstacle comes from (6.7). Notice that this implies that |ν0|(E) ≤H1(E), where |ν0| stands for the variation of ν0. This estimate is completely uselesssince we do not have any control on H1(E) (it may be infinite in the general case).Instead, we would need |ν0|(E) ≤ c μ(E), because of condition (b) in Theorem 5.1.As far as we know, all Tb-like theorems in the literature require estimates suchthis one, and often stronger assumptions involving the L∞ norm of b.

To prove Theorem 6.1, we need to work with a measure “better” than ν0,which we call ν. This new measure will be a suitable modification of ν0 with therequired estimate for its variation. To construct ν, we consider a set F containingE made up of a finite disjoint union of squares: F =

⋃i∈I Qi. One should think

that the squares Qi approximate E at some “intermediate scale”. For each squareQi, we take a complex measure νi supported on Qi such that νi(Qi) = ν0(Qi) and|νi|(Qi) = |νi(Qi)| (that is, νi will be a constant multiple of a positive measure).Then we set ν =

∑i νi. So, if the squares Qi are big enough, the variation |ν|

will be sufficiently small. On the other hand, the squares Qi cannot be too big,because we will need

γ+(F ) ≤ c γ+(E). (6.8)

In this way, we will construct a complex measure ν supported on F satisfying

|ν|(F ) ≈ |ν(F )| = γ(E). (6.9)


Taking a suitable measure μ such that supp(μ) ⊃ supp(ν) and μ(F ) ≈ γ(E),we will be ready for the application of the Tb theorem. Indeed, notice that (6.9)implies that ν satisfies a global accretivity condition and that also the variation|ν| is controlled. On the other hand, if we have been careful enough, we will havealso some useful estimates on |Cν|, since ν is an approximation of ν0 (in fact, whendefining ν in the paragraph above, the measures νi should have been constructedin a smoother way). Using the Tb Theorem 5.1, we will get

γ+(F ) ≥ c−1μ(E),

and then γ+(E) ≥ c−1γ(E) by (6.8), and we will be done.In order to obtain the right estimates on the measures ν and μ we will need

to assume that γ(E ∩ Qi) ≈ γ+(E ∩ Qi) for each square Qi. For this reason, wewill use an induction argument: we will prove that γ(E ∩R) ≈ γ+(E ∩R) for allthe rectangles R, by induction on the size of R.

6.4 An intermediate approximation in terms of γ+

This section deals with the construction of the intermediate squares Qi that formthe set F described in the preceding section.

Lemma 6.6. Let E ⊂ C be compact. There exists a bounded open set Ω containingE and a Whitney decomposition Ω =

⋃i∈J Qi, where {Qi}i∈J are Whitney squares,

with∑

i∈J χ14Qi ≤ c χΩ, such that:

(a) γ+(Ω) ≈ γ+(E).

(b)∑i∈J

γ+(E ∩ 2Qi) ≤ c γ+(E).

(c) If γ+(E) ≤ c2 diam(E), with c2 > 0 small enough, then

diam(Qi) ≤ 1

10diam(E) for every i ∈ J .

The constants involved in (a) and (b) are absolute.

Notice that (a) implies that the squares Qi in the lemma are not too big and(b) that they are not too small. That is, they belong to some intermediate scale.The construction of these squares is one of the key steps for the proof of Theorem6.1.

Proof. By Theorem 4.24 there exists some measure σ ∈ M+(C) satisfying σ(E) ≈γ+(E) and Uσ(x) ≥ 1 for all x ∈ E. Let λ be some constant with 0 < λ ≤ 1/100which will be fixed below. Let Ωλ ⊂ C be the open set

Ωλ := {x ∈ C : Uσ(x) > λ}.

6.4. An intermediate approximation in terms of γ+ 203

Notice that E ⊂ Ωλ, and by Theorem 4.23 we have

γ+(Ωλ) ≤ Cλ−1σ(E) ≤ Cλ−1γ+(E), (6.10)

and, obviously γ+(E) ≤ γ+(Ωλ), because E ⊂ Ωλ.Let Ωλ =

⋃i∈J Qi be a Whitney decomposition of Ωλ, where {Qi}i∈J is the

usual family of Whitney squares with disjoint interiors, satisfying 20Qi ⊂ Ωλ, ρQi∩(C \ Ωλ) �= ∅ (where ρ > 20 is some fixed absolute constant), and

∑i∈J χ14Qi ≤

c χΩλ. Recall that in Lemma (2.23) we already described the properties that the

Whitney squares satisfy. Notice that now some of the constants in those propertiesare different. By replacing the squares described in Lemma 2.23 by their sons, say,we obtain a new Whitney decomposition with the constants 14 and 20 that we areasking for now.

Proof of (b). Let us see now that (b) holds if λ has been chosen small enough. Wewill show below that if x ∈ E ∩ 2Qi �= ∅ for some i ∈ J , then

Uσ�4Qi(x) > 1/4, (6.11)

assuming that λ is small enough. This implies E ∩ 2Qi ⊂ {Uσ�4Qi> 1/4} and

then, by Theorem 4.23, we have

γ+(E ∩ 2Qi) ≤ c σ(4Qi).

Using the finite overlap of the squares 4Qi, we deduce∑i∈J

γ+(E ∩ 2Qi) ≤ c∑i∈J

σ(4Qi) ≤ c σ(E) ≤ c γ+(E),

and so (b) follows.Now we have to show that (6.11) holds for x ∈ E ∩ 2Qi. Let z ∈ ρQi \ Ω,

so that dist(z,Qi) ≈ dist(∂Ω, Qi) ≈ �(Qi). Since MRσ(z) ≤ Uσ(z) ≤ λ, it easilyfollows that for any ball B with radius r(B) ≥ �(Qi)/4 and B ∩ 2Qi �= ∅, we have

σ(B)

�(B)≤ c3λ � 1

1000, (6.12)

where the constant c3 depends on the Whitney decomposition (in particular, onthe constant ρ) and we assume that λ is very small.

Remember that Uσ(x) ≥ 1. If MRσ(x) > 1/2, then

σ(B(x, r))

r> 1/2

for some r < �(Qi)/4 because the balls centered at x with radius greater than�(Qi)/4 satisfy (6.12). Since B(x, r) ⊂ 4Qi, we infer that Uσ�4Qi

(x)≥M(σ�4Qi)(x)> 1/2.


Assume now that MRσ(x) ≤ 1/2. In this case, cσ(x) > 1/2. We decomposec2σ(x) =: c2(x, σ, σ) as follows:

c2(x, σ, σ) = c2(x, σ�4Qi, σ�4Qi) + 2c2(x, σ�4Qi, σ�C \ 4Qi)

+ c2(x, σ�C \ 4Qi, σ�C \ 4Qi).

We wish to see thatcσ�4Qi

(x) > 1/4. (6.13)

So it is enough to show that the last two terms in the equation above are sufficientlysmall. First we deal with c2(x, σ�4Qi, σ�C \ 4Qi):

c2(x, σ�4Qi, σ�C \ 4Qi) ≤∫y∈4Qi

∫t∈C\4Qi

4

|t− x|2 dσ(y)dσ(t)

= 4 σ(4Qi)

∫t∈C\4Qi

1

|t− x|2 dσ(t)

≤ c σ(4Qi)MRσ(z)

�(4Qi)≤ cMRσ(z)

2 ≤ cλ2, (6.14)

where we used Lemma 2.11 in the second inequality.For the term c2(x, σ�C \ 4Qi, σ�C \ 4Qi) we write

c2(x, σ�C \ 4Qi, σ�C \ 4Qi) = c2(x, σ�C \ 2ρQi, σ�C \ 2ρQi)

+ 2c2(x, σ�C \ 2ρQi, σ�2ρQi \ 4Qi)

+ c2(x, σ�2ρQi \ 4Qi, σ�2ρQi \ 4Qi).

Arguing as in (6.14), it easily follows that the last two terms are bounded aboveby cMRσ(z)

2 ≤ cλ2 again. So we get,

c2σ�4Qi(x) ≥ c2σ(x)− c2σ�C\2ρQi

(x) − cλ2. (6.15)

We are left with the term c2σ�C\2ρQi(x). Since x, z ∈ ρQi, from Lemma 4.19 we

infer that|cσ�C\2ρQi

(x) − cσ�C\2ρQi(z)| ≤ cMRσ(z) ≤ cλ.

Taking into account that cσ(z) ≤ λ, we get

cσ�C\2ρQi(x) ≤ (1 + c)λ.

Thus, by (6.15), we obtain

c2σ�4Qi(x) ≥ 1

4− cλ2 ≥ 1

16,

if λ is small enough. That is, we have proved (6.13), and so in this case (6.11)holds too.

6.4. An intermediate approximation in terms of γ+ 205

Proof of (c). Now we have to show that diam(Qi) ≤ 110 diam(E).

It is immediate to check that

Uσ(x) ≤ 100σ(E)

dist(x,E)for all x �∈ E

(of course, 100 is not the best constant here). Thus, for x ∈ Ω \ E we have

λ < Uσ(x) ≤ 100 σ(E)

dist(x,E).

Therefore,

dist(x,E) ≤ 100λ−1σ(E) ≤ cλ−1γ+(E) ≤ 1

20diam(E),

taking the constant c2 in the lemma small enough. As a consequence, diam(Ω) ≤1110 diam(E). Since 20Qi ⊂ Ω for each i ∈ I, we have

20 diam(Qi) ≤ diam(Ω) ≤ 11

10diam(E),

which implies (c). �Example 6.7. Let EN be the N -th generation of the corner quarters Cantor set.In this case the squares Qi, i ∈ J , from the preceding lemma can be constructedin a different way. Indeed, suppose for simplicity that N is even, and consider theset EN/2 of the N/2-th generation. Then we let {Qi}i∈J be the squares of side

length 4−N/2 that form EN/2. We claim that these squares and Ω := EN/2 satisfythe properties (a), (b) and (c) above, for E = EN . To check this, recall that byTheorem 4.28 and the subsequent remark we know that γ+(EN ) ≈ N−1/2. Then

γ+(EN/2) ≈(N

2

)−1/2

≈ N−1/2 ≈ γ+(EN ),

and thus the property (a) holds. Concerning (b), observe that EN ∩2Qi = EN ∩Qi

coincides with a translation of 4−N/2EN/2, since Qi belongs to the N/2 generation,

EN ∩ Qi is made up of squares of the N -th generation and �(Qi) = 4−N/2 andEN ∩Qi. Thus,

γ+(EN ∩ 2Qi) ≈ 4−N/2 γ+(EN/2) ≈ 4−N/2 γ+(EN ),

and therefore, using that there are 4N/2 squares in the family J ,∑i∈J

γ+(E ∩ 2Qi) ≈ 4N/24−N/2 γ+(EN ) = γ+(EN ).

Finally, (c) holds if N ≥ 2, say.

The intermediate set F described in Section 6.3 consists of the union of thosesquares from the family {Qi}i∈J from Lemma 6.6 such that γ(E ∩ 2Qi) �= 0. Soin the particular case of EN , we may take F = EN/2 (for N even).


6.5 Construction of the good measures μ and ν forTheorem 6.1

For a given compact set E ⊂ C, in the next lemma, we construct some measuresμ (positive) and ν (complex) suitable for the application of a Tb theorem. Thesemeasures will not be supported on E, but on the intermediate set F made up ofsome of the intermediate squares from Lemma 6.6.

Lemma 6.8 (Main Lemma). Let E ⊂ C be a compact set with H1(∂E) < ∞ andγ(E) > 0. Let {Qi}i∈I be the finite subfamily of those intermediate squares Qi,i ∈ J , from Lemma 6.6 such that γ(E ∩ 2Qi) �= 0. Let F =

⋃i∈I Qi. Suppose that

γ+(E)

γ(E)≤ γ+(E ∩ 2Qi)

γ(E ∩ 2Qi)�= 0 for all i ∈ I. (6.16)

Then there exist a positive Radon measure μ and a complex Radon measure ν,both supported on F , and a subset H ⊂ F , such that:

(a) c−1a γ(E) ≤ μ(F ) ≤ ca γ(E).

(b) ν = b μ, with ‖b‖L∞(μ) ≤ cb.

(c) |ν(F )| = γ(E).

(d)∫F\H C∗ν dμ ≤ cd μ(F ).

(e) If μ(B(x, r)) > c0r (for some big constant c0), then B(x, r) ⊂ H. In partic-ular, μ(B(x, r)) ≤ c0r for all x ∈ F \H, r > 0.

(f) H is of the form H =⋃

k∈IHB(xk, rk), with

∑k∈IH

rk ≤ 5c−10 μ(F ) =:

εμ(F ).

(g) All the squares Q ⊂ C satisfy

|ν(Q)| ≤ cg�(Q).

The constants ca, cb, cd, and cg are absolute, while c0 can be chosen arbitrarilylarge (and so ε > 0 arbitrarily small).

We have preferred to use the notation ca, cb, cd, cg instead of c4, c5, c6, c7, say,because these constants will play an important role in the proof of Theorem 6.1.Of course, the constant cb does not depend on b (it is an absolute constant). Theset H depends on the choice of c0.

The properties (a), (b), (c) and (d) from the lemma are proved in Subsections6.5.1-6.5.3. These are the basic properties which must satisfy μ and ν in order toapply the Tb theorem of Nazarov, Treil and Volberg with absolute constants. In(d) notice that instead of the L∞(μ) or BMO(μ) norm of Cν, we estimate theL1(μ) norm of C∗ν out of the set H .

6.5. Construction of the good measures μ and ν 207

Roughly speaking, the exceptional set H contains the part of μ without lineargrowth. The properties (e) and (f) describe H and are proved in Subsection 6.5.2.Observe that (f) means that H is a rather small set, in a sense. This is deducedfrom the key property (g) satisfied by ν, which can be considered as a kind oflinear growth.

6.5.1 The construction of μ and ν and the proof of (a)–(c)

It is easily seen that there exists a family of C∞ functions {gi}i∈J such that,for each i ∈ J , supp(gi) ⊂ 2Qi, 0 ≤ gi ≤ 1, and ‖∇gi‖∞ ≤ c/�(Qi), so that∑

i∈J gi = 1 on Ω, where Ω is the open set constructed in Lemma 6.6.

Let f(z) be the Ahlfors function of E, with f(z) = 0 for z ∈ ◦E, and consider

a complex measure ν0 supported on ∂E such that Cν0(z) = f(z) for z �∈ ∂E and‖Cν0‖L∞(C) ≤ 1. Such a measure exists, by Proposition 6.5, since H1(∂E) < ∞.Moreover, ∣∣ν0(E)

∣∣ = γ(E).

The measure ν that we will construct in this subsection will be a suitable mod-ification of ν0. As explained in Section 6.3, the main drawback of ν0 is that theonly information that we have about its variation |ν0| is that |ν0| = b0 dH1

E , with‖b0‖∞ ≤ 1. This is a very bad estimate if we try to apply some kind of Tb theoremin order to show that the Cauchy transform is bounded (with absolute constants).For the variation |ν| we will have a much better control. This will the main ad-vantage of ν over ν0.

First we define the measure μ. For each i ∈ I, let Γi be a circumferenceconcentric with Qi and radius γ(E ∩ 2Qi)/10. Observe that Γi ⊂ 1

2Qi for each i.We set

μ =∑i∈I

H1�Γi.

Let us define ν now:

ν =∑i∈I

1

H1(Γi)

∫gi dν0 · H1�Γi.

Notice that supp(ν) ⊂ supp(μ) ⊂ F .

We have dν = b dμ, with b =∫gi dν0

H1(Γi)on Γi. To estimate ‖b‖L∞(μ) notice that,

for each i ∈ J ,

|C(giν0)(z)| ≤ c for all z �∈ E ∩ 2Qi. (6.17)

Indeed, recall that C(giν0) = Vgi(Cν0) (where Vgi is the Vitushkin localizationoperator), and then (6.17) follows from Proposition 1.17. Inequality (6.17) impliesthat ∣∣∣∣∫ gi dν0

∣∣∣∣ = ∣∣(C(gi ν0))′ (∞)∣∣ ≤ cγ(E ∩ 2Qi) = cH1(Γi). (6.18)


As a consequence, ‖b‖L∞(μ) ≤ c, and (b) is proved.On the other hand, from the last estimates we also deduce that

∫gi dν0 = 0

if i ∈ J \ I, and so

γ(E) =∣∣ν0(E)∣∣ = ∣∣∣∣∑

i∈I

∫gi dν0

∣∣∣∣ = ∣∣∣∣∑i∈I

ν(Qi)

∣∣∣∣ = ∣∣ν(F )∣∣,

which yields (c).It remains to check that (a) also holds. From (6.18), the assumption that

γ+(E)

γ(E)≤ γ+(E ∩ 2Qi)

γ(E ∩ 2Qi)�= 0 for all i ∈ I,

and the property (b) in Lemma 6.6, we obtain the following:

γ(E) = |ν0(E)| =∣∣∣∣∑i∈I

∫gi dν0

∣∣∣∣ ≤∑i∈I

∣∣∣∣∫ gi dν0

∣∣∣∣≤ c∑i∈I

γ(E ∩ 2Qi) = c μ(F )

≤ cγ(E)

γ+(E)

∑i∈I

γ+(E ∩ 2Qi) ≤ cγ(E),

which gives (d) (notice that γ+(E) �= 0 because γ+(E ∩ 2Qi) �= 0).

6.5.2 The exceptional set H and the proof of (e), (f)

Let c0 ≥ 100ca be some fixed constant. Given x ∈ F , r > 0, we say that B(x, r)is a non-Ahlfors disk if μ(B(x, r)) > c0r. For a fixed x ∈ F , if there exists somer > 0 such that B(x, r) is a non-Ahlfors disk, then we say that x is a non-Ahlforspoint. For any x ∈ F , we write

R(x) = sup{r > 0 : B(x, r) is a non-Ahlfors disk}.

If x ∈ F is an Ahlfors point, we set R(x) = 0. We say that R(x) is the Ahlforsradius of x.

Observe that (a) implies that μ(F ) ≤ caγ(E) ≤ caγ(F ) ≤ ca diam(F ). There-fore,

μ(B(x, r)) ≤ μ(F ) ≤ ca diam(F ) ≤ 100car

for r ≥ diam(F )/100. Thus R(x) ≤ diam(F )/100 for all x ∈ F .

Consider the set

H0 =⋃

x∈F,R(x)>0

B(x,R(x)).


By Vitali’s 5r-covering theorem there is a disjoint family {B(xh,R(xh))}h suchthat H0 ⊂ ⋃hB(xh, 5R(xh)). Let

H =⋃h

B(xh, 5R(xh)). (6.19)

Since H0 ⊂ H , all non-Ahlfors disks are contained in H and then

dist(x, F \H) ≥ R(x) (6.20)

for all x ∈ F .Since μ(B(xh,R(xh))) ≥ c0R(xh) for every h, we get∑

h

R(xh) ≤ 1

c0

∑h

μ(B(xh,R(xh))) ≤ 1

c0μ(F ). (6.21)

6.5.3 Proof of (d)

In this subsection we will show that∫F\H

C∗ν dμ ≤ cd μ(F ). (6.22)

We will work with the regularized operators Cε introduced in Section 4.2 ofChapter 4. So consider a C∞ non-negative radial function ψ supported on B(0, 1)

with L1 norm equal to 1, and let Cε be the operator associated with the kernel1z ∗ ψε. Write also

C∗ν(x) = supε>0

|Cεν(x)|.

Remember that |Cν0(z)| ≤ 1 for all z �∈ ∂E. Since L2(∂E) = 0, this is alsotrue L2-a.e. z ∈ C. Then we have

‖Cεν0‖L∞ = ‖ψε ∗ Cν0‖L∞ ≤ ‖Cν0‖L∞ ≤ 1.

Since Cεν0 is continuous in C, we deduce that |Cεν0(z)| ≤ 1 and C∗ν0(z) ≤ 1 forall z ∈ C, ε > 0.

To estimate C∗ν, we will study the function C∗(ν− ν0). This will be the mainpoint for the proof of (6.22).

We set νi := ν�Qi.

Lemma 6.9. We have

C∗(νi − giν0)(z) ≤ c for every z ∈ C \ supp(νi − giν0) (6.23)

and

C∗(νi − giν0)(z) ≤ c �(Qi)μ(Qi)

dist(z, 2Qi)2for every z ∈ C \ 14Qi. (6.24)


Notice that∫(dνi − gi dν0) = 0. By the smoothness of the kernels of the

operators Cε, ε > 0, and by standard estimates, then it easily follows that

C∗(νi − giν0)(z) ≤ c �(Qi) (|ν|(Qi) + |ν0|(2Qi))

dist(z, 2Qi)2.

This inequality is not useful for our purposes because to estimate |ν0|(2Qi) weonly can use that |ν0|(2Qi) ≤ H1(∂E∩2Qi). However, we do not have any controlover H1(∂E ∩ 2Qi) (we only know that it is finite, by our assumptions on E). Theestimate (6.24) is much sharper.

Proof of the lemma. We set αi = νi−giν0. The first estimate in the lemma followseasily. Indeed, we already saw that |C(giν0)(z)| � 1 for z �∈ supp(giν0). Also,|Cνi(z)| � 1 for z �∈ Γi since

νi =1

H1(Γi)

∫gi dν0 · H1�Γi,

with1

H1(Γi)

∫gi dν0 � 1.

Thus,|Cαi(z)| ≤ |C(giν0)(z)|+ |Cνi(z)| � 1

for z �∈ supp(αi). In particular, |Cαi(z)| � 1 for L2-a.e. z ∈ C. Therefore, for everyε > 0,

‖Cεαi‖L∞ = ‖ψε ∗ Cαi‖L∞ ≤ ‖Cαi‖L∞ � 1,

which yields (6.23).To prove the second estimate in the lemma we have to show that

|Cεαi(z)| ≤ c �(Qi)μ(Qi)

dist(z, 2Qi)2(6.25)

for all ε > 0.Assume first ε ≤ dist(z, 2Qi)/2. Remember that

supp(αi) ⊂ Γi ∪ (E ∩ 2Qi) ⊂ 2Qi.

Since |Cαi(w)| ≤ c for all w �∈ supp(αi) and αi(C) = 0, by Lemma 1.12 we have

|Cαi(w)| ≤ c diam(supp(αi)) γ(supp(αi))

dist(w, supp(αi))2if dist(w, 2Qi) ≥ 5�(Qi).

Therefore,

|Cαi(w)| ≤ c �(Qi)γ(Γi ∪ (E ∩ 2Qi))

dist(w, 2Qi)2if dist(w, 2Qi) ≥ 5�(Qi). (6.26)


Moreover, by Proposition 1.19 we have

γ(Γi ∪ (E ∩ 2Qi)) ≤ c(γ(Γi) + γ(E ∩ 2Qi)).

Therefore, by the definition of Γi, we get

γ(Γi ∪ (E ∩ 2Qi)) ≤ c γ(E ∩ 2Qi) = cμ(Qi). (6.27)

If w ∈ B(z, ε) (for z ∈ C \ 14Qi), then dist(w, 2Qi) ≥ 5�(Qi) and dist(w, 2Qi) ≈dist(z, 2Qi). By (6.26) and (6.27) we obtain

|Cαi(w)| ≤ c �(Qi)μ(Qi)

dist(z, 2Qi)2.

Making the convolution with ψε, (6.25) follows for ε ≤ dist(z, 2Qi)/2.Suppose now that ε > dist(z, 2Qi)/2. Write h = ψε ∗ αi, so that

Cεαi = ψε ∗ −1

z∗ αi = C(h dL2).

Therefore,

|Cεαi(z)| ≤∫ |h(ξ)|

|ξ − z| dL2(ξ) ≤ c ‖h‖∞ [L2(supp(h))]1/2. (6.28)

We have to estimate ‖h‖∞ and L2(supp(h)). Observe that, if we write �i = �(Qi)and we denote by zi the center of Qi, we have

supp(h) ⊂ supp(ψε) + supp(αi) ⊂ B(0, ε) +B(zi, 2�i) = B(zi, ε+ 2�i).

Thus, L2(supp(h)) ≤ c ε2, since �i ≤ ε.Let us deal with ‖h‖∞ now. Let ηi be a C∞ function supported on 3Qi which

is identically 1 on 2Qi and such that ‖∇ηi‖∞ ≤ c/�i. Taking into account thatαi(2Qi) = 0, we have

h(w) =

∫ψε(ξ − w) dαi(ξ) =

∫ (ψε(ξ − w) − ψε(zi − w)

)dαi(ξ)

=�iε3

∫ε3

�i

(ψε(ξ − w)− ψε(zi − w)

)ηi(ξ) dαi(ξ) =:

�iε3

∫ϕw(ξ)ηi(ξ) dαi(ξ).

We will prove below that

‖C(ϕwηi αi)‖L∞(C) ≤ c. (6.29)

Let us assume this estimate for the moment. Since C(ϕwηi αi) is analytic in C \supp(αi), taking into account (6.27) we get∣∣∣∣ �iε3

∫ϕw(ξ)ηi(ξ) dαi(ξ)

∣∣∣∣ ≤ c �iε3

γ(Γi ∪ (E ∩ 2Qi)) ≤ c �iμ(Qi)

ε3.


Therefore,

‖h‖∞ ≤ c �iμ(Qi)

ε3.

From (6.28) and the estimates on ‖h‖∞ and L2(supp(h)), we deduce that

|Cεαi(z)| ≤ c �(Qi)μ(Qi)

ε2≤ c �(Qi)μ(Qi)

dist(z, 2Qi)2.

It remains to prove (6.29). Observe that C(ϕwηi αi) = Vϕwηi(Cαi) and recallthat Cαi is a bounded function. By Proposition 1.17, since supp(ϕwηi) ⊂ 3Qi, itis enough to show that

‖ϕwηi‖∞ ≤ c (6.30)

and

‖∇(ϕwηi)‖∞ ≤ c

�i. (6.31)

For ξ ∈ 3Qi, we have

|ϕw(ξ)| = ε3

�i

∣∣ψε(ξ − w)− ψε(zi − w)∣∣ ≤ ε3 ‖∇ψε‖∞ ≤ c,

which yields (6.30). Finally, (6.31) follows easily too:

‖∇(ϕwηi)‖∞ ≤ ‖∇ϕw‖∞ + ‖ϕw χ3Qi‖∞‖∇ηi‖∞ ≤ c

�i.

We are done. �

Now we are ready to prove (6.22). We write∫F\H

C∗ν dμ ≤∫F\H

C∗ν0 dμ+

∫F\H

C∗(ν − ν0) dμ

≤ μ(F \H) +∑i∈I

∫F\H

C∗(νi − giν0) dμ. (6.32)

To estimate the last integral we use Lemma 6.9 and recall that ‖C∗(νi−giν0)‖L∞(μ)

≤ c:∫F\H

C∗(νi − giν0) dμ ≤ cμ(14Qi) +

∫F\(14Qi∪H)

c �(Qi)μ(Qi)

dist(z, 2Qi)2dμ(z). (6.33)

Let N ≥ 1 be the least integer such that (14N+1Qi \ 14NQi) \H �= ∅, and take


some fixed z0 ∈ (14N+1Qi \ 14NQi) \H . We have∫F\(14Qi∪H)

1

dist(z, 2Qi)2dμ(z) =

∞∑k=N

∫(14k+1Qi\14kQi)\H

1

dist(z, 2Qi)2dμ(z)

≤ c

∞∑k=N

μ(14k+1Qi)

�(14k+1Qi)2

≤ c

∞∑k=N

μ(B(z0, 2�(14k+1Qi)))

�(14k+1Qi)2

≤ c

∞∑k=N

c0�(14k+1Qi)

�(14k+1Qi)2

≤ c c01

�(14NQi)≤ c c0

1

�(Qi).

Notice that in the second inequality we have used that z0 ∈ F \ H , and soμ(B(z0, r)) ≤ c0 r for all r. Recalling (6.33), we obtain∫

F\HC∗(νi − giν0) dμ ≤ c μ(14Qi).

Thus, by the finite overlap of the squares 14Qi, i ∈ I, and (6.32), we get∫F\H

C∗ν dμ ≤ c μ(F \H) + c∑i∈I

μ(14Qi) ≤ c μ(F ). (6.34)

Now, recall that|C∗ν(z)− C∗ν(z)| ≤ cMRν(z) (6.35)

(see Lemma 4.3). By (b), if z ∈ F \H , we have MRν(z) ≤ cMRμ(z) ≤ c. Thus(6.34) and (6.35) imply ∫

F\HC∗ν(z) dμ(z) ≤ c μ(F ).

6.5.4 Proof of (g)

We must show that for all the squares R ⊂ C, we have |ν(R)| ≤ Cg�(R). To thisend we will need a couple of technical lemmas.

Lemma 6.10. Let R ⊂ C be a square such that there exists some Whitney squareQi, i ∈ J , satisfying R ∩ 2Qi �= ∅ and �(R) ≤ 4�(Qi). Then

μ(R) ≤ c0 �(R),

where c0 is some big enough absolute constant.


Proof. If R intersects a unique circumference Γj , j ∈ I, then

μ(R) = H1(Γj ∩R) ≤ c�(R),

and we are done.Suppose now that R intersects at least two circumferences Γj and Γj′ , with

j, j′ ∈ I. Since Γj ⊂ 12Qj, we deduce that

�(R) ≥ 1

4�(Qj).

The assumption R ∩ 2Qi �= ∅ with �(R) ≤ 4�(Qi) implies that R ⊂ 9Qi. Observenow that if Qk is a Whitney square such that Qk ∩ 9Qi �= ∅, then �(Qi) ≈ �(Qk)(recall the property (iii) of Lemma 2.23). In particular, this happens for Qk = Qj .Then we get

μ(R) ≤∑

k:Qk∩9Qi �=∅

μ(Qk) ≤ c∑

k:Qk∩9Qi �=∅

�(Qk) ≤ c �(Qi) ≈ �(Qj) ≤ 4�(R).

�Lemma 6.11. Let R ⊂ C be a square such that �(R) ≥ 4 �(Qi) for each Whitneysquare Qi such that 2Qi ∩R �= ∅. Let LR = {h ∈ I : 2Qh ∩ ∂R �= ∅}. Then∑

h∈LR

�(Qh) ≤ c �(R).

Proof. For a square Qh, h ∈ LR, we have �(Qh) ≤ �(R)/4, and then it follows thatH1(4Qh ∩ ∂R) ≥ c−1�(Qh). Then, by the bounded overlap of the squares 4Qh, weobtain ∑

h∈LR

�(Qh) ≤ c∑h∈LR

H1(4Qh ∩ ∂R) ≤ c �(R). (6.36)

�Now we are ready to prove (g). By Lemma 6.10 we may assume �(Qi) ≤

�(R)/4 if 2Qi ∩R �= ∅. Otherwise, |ν(R)| ≤ cbμ(R) ≤ cbc0�(R).From the fact that ‖Cν0‖L∞(C) ≤ 1, we deduce |ν0(R)| ≤ c �(R), arguing as

in (6.1). So we only have to estimate the difference |ν(R)− ν0(R)|.Let {Qi}i∈IR , IR ⊂ I, be the subfamily of Whitney squares such that Qi∩R �=

∅, and let {Qi}i∈JR , JR ⊂ I, be the Whitney squares such that Qi ⊂ R. We write

|ν(R)− ν0(R)| =∣∣∣ν( ⋃

i∈IR

(Qi ∩R))− ν0

( ⋃i∈IR

(Qi ∩R))∣∣∣

≤∣∣∣ν( ⋃

i∈IR\JR

(Qi ∩R))− ν0

( ⋃i∈IR\JR

(Qi ∩R))∣∣∣

+∣∣∣ν( ⋃

i∈JR

Qi

)− ν0

( ⋃i∈JR

Qi

)∣∣∣=: A+B.


First we deal with the term A:

A ≤∑

i∈IR\JR

|ν|(Γi) +∑

i∈IR\JR

|ν0(Qi ∩R)|.

We take into account now that |ν|(Γi) ≤ cbH1(Γi). Also, from the fact that |Cν0| ≤1, arguing as in (6.1), we get |ν0(Qi ∩R)| ≤ cH1(∂(Qi ∩R)). Thus,

A ≤ c∑

i∈IR\JR

[H1(Γi) +H1(∂(Qi ∩R))] ≤ c

∑i∈IR\JR

�(Qi).

Notice now that if i ∈ IR \JR, then Qi∩R �= ∅ and Qi �⊂ R. Therefore, Qi∩∂R �=∅. From Lemma 6.11 we deduce A ≤ c �(R).

Let us turn our attention to B:

B =

∣∣∣∣∑i∈JR

∫gi dν0 −

∫⋃

i∈JRQi

dν0

∣∣∣∣=

∣∣∣∣∫ (∑i∈JR

gi − χ⋃i∈JR

Qi

)dν0

∣∣∣∣≤∑j∈JR

∣∣∣∣∫Qj

(∑i∈JR

gi − 1)dν0

∣∣∣∣+ ∣∣∣∣∫C\⋃j∈JR

Qj

∑i∈JR

gi dν0

∣∣∣∣=: B1 +B2.

We consider first the term B1. If∑

i∈JRgi �≡ 1 on Qj, we write j ∈ MR. In

this case there exists some h ∈ I \ JR such that gh �≡ 0 on Qj. So 2Qh ∩Qj �= ∅,with Qh �⊂ R. Thus, 2Qh ∩ ∂R �= ∅. That is, h ∈ LR.

For each h ∈ LR there are at most c4 squares Qj such that 2Qh ∩ Qj �= ∅.Moreover, for these squares Qj we have �(Qj) ≤ c �(Qh). Then, by Lemma 6.11,we get ∑

j∈MR

�(Qj) ≤ c c4∑h∈LR

�(Qh) ≤ c �(R). (6.37)

Now we set

B1 =∑

j∈MR

∣∣∣∣∫Qj

(∑i∈JR

gi − 1)dν0

∣∣∣∣≤∑

j∈MR

(∣∣∣∣∫Qj

∑i∈JR

gi dν0

∣∣∣∣+ |ν0(Qj)|).

We have |ν0(Qj)| ≤ c �(Qj) and also∣∣∣∣∫Qj

∑i∈JR

gi dν0

∣∣∣∣ ≤ ∑i∈JR

∣∣∣∣∫Qj

gi dν0

∣∣∣∣ ≤ c �(Qj),


because #{i ∈ JR : supp(gi) ∩Qj �= ∅} ≤ c and |C(giν0)| ≤ c for each i. Thus, by(6.37) we get

B1 ≤ c �(R).

Finally we have to estimate B2. We have

B2 ≤∑i∈JR

∣∣∣∣∫C\⋃j∈JR

Qj

gi dν0

∣∣∣∣ = ∑i∈JR

B2,i.

Observe that if B2,i �= 0, then supp(gi) ∩ supp(ν0) \⋃

j∈JRQj �= ∅. As a conse-

quence, 2Qi ∩Qh �= ∅ for some h ∈ I \ JR. Since Qi ⊂ R and Qh �⊂ R, we deducethat either 2Qi ∩ ∂R �= ∅ or Qh ∩ ∂R �= ∅. So either i ∈ LR or h ∈ LR. Takinginto account that �(Qi) ≈ �(Qh), arguing as above we get

B2 ≤ c∑i∈LR

�(Qi) + c∑i∈JR

∑h∈LR:Qh∩2Qi �=∅

�(Qi)

≤ c �(R) + c∑h∈LR

∑i∈I:Qh∩2Qi �=∅

�(Qh) ≤ c �(R).

6.6 Dyadic lattices and the exceptional sets HD and TDLet D0 be the usual dyadic lattice. For w ∈ C, we write

D(w) = w +D0.

That is, D(w) is a translation of D0 by the vector w. In the remaining part of thischapter, by “a dyadic lattice” we mean a lattice D(w), for some w ∈ C.

6.6.1 The construction of HDLet F , μ, ν and H be as in the Main Lemma 6.8, and let D be a fixed dyadic latticefrom the collection of lattices {D(w)}w∈C. Remember that H =

⋃k∈IH

B(xk, rk)∩F , with ∑

k∈IH

rk ≤ 5

c0μ(F ) = ε μ(F ).

Consider the family of dyadic squares DH ⊂ D such that R ∈ DH if there existssome ball B(xk, rk), k ∈ IH , satisfying

B(xk, rk) ∩R �= ∅ (6.38)

and2rk < �(R) ≤ 4rk. (6.39)

Notice that ⋃k

B(xk, rk) ⊂⋃

R∈DH

R. (6.40)

6.6. Dyadic lattices and the exceptional sets HD and TD 217

We take a subfamily of disjoint maximal squares {Rk}k∈IH from DH such that⋃R∈DH

R =⋃

k∈IH

Rk,

and we define the dyadic exceptional set HD as

HD =⋃

k∈IH

Rk.

Observe that (6.40) implies H ⊂ HD and, since for each ball B(xk, rk) there areat most four squares R ∈ DH satisfying (6.38) and (6.39), we obtain∑

k∈IH

�(Rk) ≤ 16∑k

rk ≤ 80

c0μ(F ) = 16ε μ(F ). (6.41)

6.6.2 The accretivity condition and the exceptional set TDLooking at conditions (a), (b) and (c) from Main Lemma 6.8 one can guess thatthe function b will behave as an accretive function on many squares from thedyadic lattice D. We deal with this question in this subsection.

Let us define the exceptional set TD. If a dyadic square R ∈ D satisfies

μ(R) ≥ cT |ν(R)|, (6.42)

where cT is some big constant which will be chosen below, we set R ∈ DT . Let{Rk}k∈IT ⊂ DT be the subfamily of maximal, and thus disjoint, dyadic squaresfrom DT . The exceptional set TD is

TD =⋃

k∈IT

Rk.

6.6.3 The size of HD ∪ TDIn the next lemma we show μ(F \ (HD ∪TD)) is big. That is, that it is comparableto μ(F ).

Lemma 6.12. If c0 and cT are chosen big enough, then

|ν(HD ∪ TD)| ≤ 1

2|ν(F )|

andμ(HD ∪ TD) ≤ δ1μ(F ),

with δ1 = 1− 1

2ca cb< 1.


Proof. Let {Rk}k∈IHT be the subfamily of maximal (and thus disjoint) squaresfrom

{Rk}k∈IH ∪ {Rk}k∈IT ,

so thatHD ∪ TD =

⋃k∈IHT

Rk.

From the property (g) of the Main Lemma 6.8, (6.42) and (6.41), we get

|ν(HD ∪ TD)| ≤∑

k∈IHT

|ν(Rk)|

≤∑k∈IH

|ν(Rk)|+∑k∈IT

|ν(Rk)|

≤ cg∑k∈IH

�(Rk) + c−1T

∑k∈IT

μ(Rk)

≤ 16cgε μ(F ) + c−1T μ(F )

≤ ca(16cgε+ c−1T ) |ν(F )|.

So if we choose ε small enough and cT big enough, we obtain

|ν(HD ∪ TD)| ≤ 1

2|ν(F )|.

Hence,

|ν(F \ (HD ∪ TD))| ≥ |ν(F )| − |ν(HD ∪ TD)| ≥ 1

2|ν(F )|.

Therefore,

μ(F ) ≤ ca|ν(F )| ≤ 2ca |ν(F \ (HD ∪ TD))| ≤ 2cacb μ(F \ (HD ∪ TD)).

Thus, μ(HD ∪ TD) ≤ δ1μ(F ), with δ1 = 1− 12cacb

. �

6.7 Application of a Tb theorem for the proof ofTheorem 6.1

Let us recall the Tb theorem of Nazarov, Treil and Volberg proved in Chapter 5:

Theorem 6.13. Let μ be a measure supported on a compact set F ⊂ C. Suppose thatthere exist a complex measure ν and, for each w ∈ C, two sets HD(w), TD(w) ⊂ C

made up of dyadic squares from D(w) such that


w∈CHD(w).

(b) ν = b μ, where b is some function such that ‖b‖L∞(μ) ≤ cb.

6.7. Application of a Tb theorem 219

(c)∫C\HD(w)

C∗ν dμ ≤ c μ(F ), for all w ∈ C.

(d) If Q ∈ D(w) is such that Q �⊂ TD(w), then μ(Q) ≤ cd|ν(Q)| (i.e. Q is anaccretive square).

(e) μ(HD(w) ∪ TD(w)) ≤ δ0 μ(F ), for all w ∈ C and some δ0 < 1.

Then there exists a subset G ⊂ F such that

(i) μ(G) ≥ c−1μ(F ),


(iii) the Cauchy transform is bounded on L2(μ�G).

The constant c and the bound for the L2 norm depend only on the constants above.

If we apply this Tb theorem to the set F and to the measures μ and νconstructed in Main Lemma 6.8, we get:

Lemma 6.14. Let E ⊂ C be a compact set such that H1(∂E) < ∞ and γ(E) > 0.Let {Qi}i∈J be the finite subfamily of those intermediate squares Qi, i ∈ I, fromLemma 6.6 such that γ(E ∩ 2Qi) �= 0. Let F =

⋃i∈I Qi. Suppose that

γ+(E)

γ(E)≤ γ+(E ∩ 2Qi)

γ(E ∩ 2Qi)�= 0 for all i ∈ I. (6.43)

Then there exists a measure μ supported on F and a subset G ⊂ F such that

• c−1a γ(E) ≤ μ(F ) ≤ ca γ(E) and μ(F ) ≤ c5μ(G),

• μ(G ∩B(x, r)) ≤ c0r for all x ∈ G, r > 0, and

• the Cauchy transform is bounded on L2(μ�G) with

‖Cμ�G‖L2(μ�G)→L2(μ�G) ≤ c6.

The constants ca, c0, c5, c6 are absolute constants.

Proof. This is a direct consequence of Main Lemma 6.8, Lemma 6.12, and Theorem6.13. Let us notice that the set G obtained in Theorem 6.13 fulfills

G ⊂ C \⋂w∈Ω

HD ⊂ C \H,

and so μ(B(x, r)) ≤ c0r for all x ∈ G. �


6.8 The final induction argument for the proof ofTheorem 6.1

Lemma 6.15. Let E ⊂ C be a compact set such that H1(∂E) < ∞ and γ(E) > 0.There exists some absolute constant B such that if

(a) γ+(E) ≤ c2 diam(E), and

(b)γ+(E)

γ(E)≤ γ+(E ∩Q)

γ(E ∩Q)�= 0 for all the squares Q such that γ(E ∩Q) �= 0 and

diam(Q) ≤ diam(E)/10,

then γ(E) ≤ Bγ+(E).

Proof. By Lemmas 6.6 and 6.14, there are sets F, G and a measure μ supportedon F such that

(i) E ⊂ F and γ+(E) ≈ γ+(F ),

(ii) μ(F ) ≈ γ(E),

(iii) G ⊂ F and μ(G) ≈ μ(F ),

(iv) μ(G ∩B(x, r)) ≤ c0r for all x ∈ G, r > 0, and ‖C‖L2(μ�G)→L2(μ�G) ≤ c6.

From (iv) and (iii), we get

γ+(F ) ≥ c−1μ(G) ≥ c−1μ(F ).

Then, by (ii), the preceding inequality, and (i),

γ(E) ≤ c μ(F ) ≤ c γ+(F ) ≤ B γ+(E). �

As a corollary we deduce:

Lemma 6.16. Let E ⊂ C be a compact set such that H1(∂E) < ∞. There existssome absolute constant A0 such that if γ(E∩Q) ≤ A0 γ+(E∩Q) for all the squaresQ with diam(Q) ≤ diam(E)/10, then γ(E) ≤ A0 γ+(E).

Proof. We may assume that γ(E) > 0, otherwise the statement in the propositionis trivial.

If γ+(E) > c2 diam(E), then we get γ+(E) > c2 γ(E) by Proposition 1.5, andso we are done provided that A0 ≥ c−1

2 . We set A0 = max(1, c−12 , B). If γ+(E) ≤

c2 diam(E), then we also have γ(E) ≤ A0γ+(E). Otherwise, the assumption (b)in Lemma 6.15 holds, as well as (a), and we deduce γ(E) ≤ Bγ+(E) ≤ A0γ+(E),which is a contradiction. �

Notice that any constant A0 ≥ max(1, c−12 , B) works in the argument above.

So Lemma 6.16 holds for any constant A0 sufficiently big.

6.9. Estimate of the Cauchy integral and AD regular curves 221

Proof of Theorem 6.1. By the outer regularity of γ+ on compact sets, if d > 0 is

small enough, then γ+

(Ud(E))≤ γ+(E)+ε, where ε > 0 is some arbitrarily small

constant. Consider the compact set E0 made up of all the closed dyadic squares(from the usual standard dyadic lattice, say) of side length d/2 that intersect E(assuming d to be of the form 2−k). We will prove below that γ(E0) ≤ c γ+(E0).From this inequality, since E ⊂ E0 ⊂ Ud(E), we get

γ(E) ≤ γ(E0) ≤ c γ+(E0) ≤ c (γ+(E) + ε),

and then we are done.To prove that γ(E0) ≤ cγ+(E0), we are going to show by induction on n that

if R is a closed rectangle with sides parallel to the axes and diameter ≤ 4n−1d,n ≥ 0, then

γ(R ∩E0) ≤ A0γ+(R ∩ E0). (6.44)

Notice that if diam(R) ≤ d/4, then R can intersect at most four of the dyadicsquares of side length d that form E0. In any case it is easy to check that R ∩E0

is either a rectangle R1 or a polygon which can be decomposed as the union oftwo closed rectangles R1, R2 (with R1 ∩R2 �= ∅). In the first case, (6.44) followsfrom the fact that for every rectangle R1 we have γ+(R1) ≈ diam(R1) (usingthe characterization of γ+ in terms of curvature, for instance), which implies thatγ(R1) ≤ A0γ+(R1), assuming A0 sufficiently big. In the second case, by Lemma1.19, we have

γ(R1 ∪R2) ≤ c (γ(R1) + γ(R2)) ≤ c (γ+(R1) + γ+(R2)) ≤ 2c γ+(R1 ∪R2).

Let us see now that if (6.44) holds for all rectangles R with diameter ≤ 4nd,

then it also holds for a rectangle R with diameter ≤ 4n+1d. We only have toapply Lemma 6.16 to the set R ∩ E0. Indeed, take a square Q with diameter≤ diam(R ∩ E0)/10. By the induction hypothesis we have

γ(Q ∩ R ∩ E0) ≤ A0 γ+(Q ∩ R ∩ E0),

because Q ∩ R is a rectangle with diameter ≤ 4nd. Therefore,

γ(R ∩ E0) ≤ A0 γ+(R ∩ E0)

by Lemma 6.16. �

6.9 Estimate of the Cauchy integral and AD regularcurves

6.9.1 Estimate of the Cauchy integral

In this subsection we will show a nice application of the comparability betweenγ and γ+ to the so-called estimate of the Cauchy integral. The arguments will


involve some of the techniques developed above for the proof of γ ≈ γ+, such asthe approximation by means of the intermediate squares from Section 6.4.

The problem of estimating the Cauchy integral over the boundary of a do-main was posed by Vitushkin in connection with the theory of uniform rationalapproximation on compact subsets of the complex plane. The problem consists incharacterizing those bounded domains G ⊂ C with rectifiable boundary for whichthere exists a constant c(G) such that for any compact set E ⊂ G and any functionf bounded and holomorphic on G \E the following estimate holds:∣∣∣∣∫

∂G

f(z) dz

∣∣∣∣ ≤ c(G) ‖f‖∞γ(E).

Vitushkin also raised the analogous question for functions f continuous on G andholomorphic on G \ E, interchanging γ(E) by the continuous analytic capacityα(E).

Theorem 6.17. Let G be a bounded open set in C whose boundary ∂G is a finite dis-joint union of Jordan rectifiable (closed) curves such that the measure H1�∂G haslinear growth and the Cauchy integral operator CH1�∂G is bounded in L2(H1�∂G).Then there exists some constant c(G) such that for any compact set E ⊂ G andany function f ∈ H∞(G \E), we have∣∣∣∣∫

∂G

f(z) dz

∣∣∣∣ ≤ c(G) ‖f‖∞γ(E). (6.45)

The constant c(G) depends only on the linear growth constant of H1�∂G and onthe L2(H1�∂G) norm of CH1�∂G.

For example, if ∂G is a single Jordan curve satisfying H1(B(z, r)) ≤ c0 r forall z ∈ ∂G and all r > 0, then ∂G is an AD regular curve and thus CH1�∂G isbounded on L2(H1�∂G), and (6.45) holds.

Let us remark that the linear growth ofH1�∂G in the theorem is a superfluousassumption, because it follows from the L2(H1�∂G) boundedness of CH1�∂G. In thetheorem we assume f to be defined in ∂G by its boundary values, which are knownto exist.

The motivation of Vitushkin for the question above is due to the fact thatfrom (6.45) one can deduce the semiadditivity of γ for disjoint compact sets whichare separated by an AD regular curve. Indeed, let E,F ⊂ C be compact andconsider an AD regular Jordan curve Γ such that E lies in the bounded componentof C \ Γ and F in the unbounded one. Let f be the Ahlfors function of E ∪ F .Denote by G the bounded component of C \ Γ and let B(0, R) be an open ballcontaining G ∪ F . Then

γ(E ∪ F ) = f ′(∞) =1

2πi

∫∂B(0,R)

f(z) dz

=1

2πi

∫∂(B(0,R)\G)

f(z) dz +1

2πi

∫∂G

f(z) dz.


By Theorem 6.17, we have∣∣∣∣∫∂G

f(z) dz

∣∣∣∣ ≤ c(G)‖f‖∞γ(E).

The same theorem can also be applied to the domain B(0, R) \ G because thearc length measure on ∂(B(0, R) \ G) = ∂B(0, r) ∪ Γ has linear growth and theCauchy transform is bounded in L2 with respect to this measure. In fact, the L2

boundedness can be derived from the fact that ∂B(0, r) ∪ Γ is a subset of an ADregular curve: just connect ∂B(0, r) to Γ by means of a segment. So we have∣∣∣∣∣

∫∂(B(0,R)\G)

f(z) dz

∣∣∣∣∣ ≤ c′(G)‖f‖∞γ(F ).

Then we getγ(E ∪ F ) ≤ c′′(G)

(γ(E) + γ(F )

).

Proof of Theorem 6.17. We assume that f vanishes on C \ G and also, by homo-geneity, that ‖f‖∞ ≤ 1.

Let Ω be the open set containing E defined in Lemma 6.6, and Ω =⋃

i∈I Qi

a Whitney decomposition into squares satisfying the conditions mentioned in thesame lemma. Now we consider a partition of unity: let {ϕi}i∈I be a family of C∞

functions such that 0 ≤ ϕi ≤ 1, ‖∇ϕi‖∞ ≤ c/�(Qi) and supp(ϕi) ⊂ 32Qi for each

i ∈ I, so that∑

i∈I ϕi = 1 in Ω.Consider the (finite) subfamily of squares {Qj}j∈J , J ⊂ I, such that 2Qj ∩

E �= ∅. Notice that ψ :=∑

j∈J ϕj = 1 in a neighborhood of E. Moreover, ψ is acompactly supported C∞ function because J is finite. Then we have

f =∑j∈J

Vϕjf + V1−ψf,

where V··· stands for Vitushkin’s localization operator. Although 1 − ψ is notcompactly supported, V1−ψf also makes sense: we just set V1−ψf = − 1

πC((1 −ψ)∂f). Observe that V1−ψf is bounded in G, although its L∞ norm may dependon #J . Also, it is holomorphic in G because

∂(V1−ψf) = (1− ψ) ∂f = 0 in G,

since 1−ψ vanishes in a neighborhood of E and f is holomorphic in G \E. Thus,∫∂G

f(z) dz =∑j∈J

∫∂G

Vϕjf(z) dz. (6.46)

Let J1 ⊂ J be the set of indices j such that 2Qj ∩ ∂G = ∅, and J2 = J \ J1.By the definition of γ, for j ∈ J1 we have∣∣∣∣∫

∂G

Vϕjf(z) dz

∣∣∣∣ ≤ c γ(E ∩ 2Qj),


because E ∩ 2Qj ⊂ G and ‖Vϕjf‖∞ ≤ c. Therefore, from Lemma 6.6 and the factthat γ ≈ γ+, we derive∣∣∣∣∑

j∈J1

∫∂G

Vϕjf(z) dz

∣∣∣∣ ≤ c∑j∈J1

γ(E ∩ 2Qj) ≤ c γ(E). (6.47)

For j ∈ J2 we will show below that∣∣∣∣∫∂G

Vϕjf(z) dz

∣∣∣∣ ≤ c(γ(E ∩ 2Qj) +H1(∂G ∩ 5Qj)

). (6.48)

Before proving this estimate, let us see that (6.45) follows from (6.46), (6.47), and(6.48). Indeed, appealing to Lemma 6.6 again, by the finite overlap of the squares5Qj, we get∣∣∣∣∑

j∈J2

∫∂G

Vϕjf(z) dz

∣∣∣∣ ≤ c(∑j∈J

γ(E ∩ 2Qj) +∑j∈J

H1(∂G ∩ 5Qj))

≤ c(γ(E) +H1(∂G ∩ Ω)

).

Because of the L2(H1�∂G) boundedness of the Cauchy transform, we have

H1(∂G ∩ Ω) ≤ c γ+(∂G ∩Ω) ≤ c γ+(Ω) ≤ c γ+(E).

Thus, ∣∣∣∣∑j∈J2

∫∂G

Vϕjf(z) dz

∣∣∣∣ ≤ c γ(E),

which, together with (6.47), yields (6.45).It only remains to prove (6.48) for j ∈ J2. Let Dj be a disk with radius

�(Qj)/4 whose center zj coincides with the center of Qj . Consider the followingmeasure:

νj :=−1

πL2(Dj)

(∫f ∂ϕj dL2

)L2�Dj .

We want to compare Vϕjf to the function gj := Cνj. Notice that

1

π

∫f ∂ϕj dL2 = (Vϕjf)

′(∞) = g′j(∞).

Then, from the definition of γ, the fact that ‖Vϕjf‖∞ ≤ c, and the semiadditivityof γ, we get∣∣∣∣ 1π

∫f ∂ϕj dL2

∣∣∣∣ ≤ c γ((E ∪ ∂G) ∩ 2Qj

) ≤ c(γ(E ∩ 2Qj) + γ(∂G ∩ 2Qj)

)≤ c(γ(E ∩ 2Qj) +H1(∂G ∩ 2Qj)

). (6.49)


Hence,

‖gj‖∞ ≤∣∣∣∣ c

�(Qj)

∫f ∂ϕj dL2

∣∣∣∣ ≤ cγ(E ∩ 2Qj) +H1(∂G ∩ 2Qj)

�(Qj)≤ c. (6.50)

In the last inequality we used the linear growth of H1�∂G. Since gj and Vϕjfare bounded in modulus by some absolute constant and (Vϕjf)

′(∞) = g′j(∞), thefollowing estimate holds for z �∈ 5Qj, by Lemma 1.12:

|Vϕjf(z)− gj(z)| ≤c �(2Qj)γ

((E ∪ ∂G) ∩ 2Qj

)dist(z, 2Qj)2

≤ c �(Qj)[γ(E ∩ 2Qj) +H1(∂G ∩ 2Qj)

]|z − zj|2 . (6.51)

Let us estimate the integral∫∂G

|Vϕjf(z)− gj(z)| dH1(z) =

∫∂G∩5Qj

+

∫∂G\5Qj

=: I1 + I2.

From the uniform boundedness of gj and Vϕjf , it follows immediately that I1 ≤cH1(∂G ∩ 5Qj). To deal with I2 we use (6.51) and the fact that the arc lengthmeasure on ∂G has linear growth. Then we get

I2 ≤∫∂G\5Qj

c �(Qj)[γ(E ∩ 2Qj) +H1(∂G ∩ 2Qj)

]|z − zj |2 dH1(z)

≤ c[γ(E ∩ 2Qj) +H1(∂G ∩ 2Qj)

].

Thus ∣∣∣∣∫∂G

Vϕjf(z) dz

∣∣∣∣ ≤ c[γ(E ∩ 2Qj) +H1(∂G ∩ 5Qj)

]+

∣∣∣∣∫∂G

gj(z) dz

∣∣∣∣.By Fubini and (6.49) we obtain∣∣∣∣∫

∂G

gj(z) dz

∣∣∣∣ = ∣∣∣∣∫∂G

Cνj(z) dz∣∣∣∣ = c |νj(G)| ≤ c

[γ(E ∩ 2Qj) +H1(∂G ∩ 2Qj)

].

So (6.48) holds, and the theorem follows. �

6.9.2 Another proof of the L2 boundedness of the Cauchytransforms on AD regular curves

We will show now how the comparability γ ≈ γ+ together with the good lambdaTheorem 2.22 gives a straightforward proof of the L2 boundedness of the Cauchytransform on AD regular curves. Recall that two more direct (and natural) proofs


have already been obtained in Chapter 3. One is based on the fact that AD regularcurves have big pieces of Lipschitz graphs and the other relies on Jones’ travelingsalesman theorem and the relationship between curvature and Jones’ β’s.

So consider an AD regular curve Γ ⊂ C and assume for simplicity thatdiam(Γ) = ∞. By Theorem 2.22 and Remark 2.24, it is enough to show that foreach ball B centered at some point in Γ there exists some subset GB ⊂ Γ∩B suchthat H1(GB) ≥ η r(B) and CH1�GB

is bounded in L2(H1�GB), with norm boundedabove independently of B. To find GB , notice that Γ∩B contains a connected setof diameter ≥ r(B). Therefore,

γ+(Γ ∩B) ≈ γ(Γ ∩B) ≈ r(B).

Thus there exists a measure μ supported on Γ∩B, with μ(Γ∩B) ≈ r(B) such thatμ(B(x, r)) ≤ r for all x, r, and Cμ is bounded in L2(μ), with absolute constants.From the linear growth condition on μ and the Radon-Nikodym theorem it followseasily that μ = gH1�Γ∩B, for some measurable function g such that 0 ≤ g(x) ≤ 1for H1-a.e. x ∈ Γ ∩B. Since∫

Γ∩B

g dH1 � r(B) ≥ c−10 H1(Γ ∩B),

one easily deduces that there exists a subset GB ⊂ B with H1(GB) � r(B) suchthat g(x) � 1 on GB , with constants depending on c0. Then CH1�GB

is boundedin L2(H1�GB), and we are done.


6.10.1 Analytic capacity

To my knowledge, the first one to apply a Tb-like theorem in combination withstopping time arguments to the study of analytic capacity was Michael Christ.Using these techniques, in [14] he proved that if E ⊂ C is 1-dimensional ADregular and γ(E) > 0, then there exists a subset F ⊂ E with H1(F ) > 0 such thatthe Cauchy transform is bounded in L2 with respect H1�F . This result and theideas used in the proof were very influential in the subsequent research on analyticcapacity, in particular, in the solution of Vitushkin’s conjecture and in the proofof the comparability between γ and γ+.

Let us mention that, prior to the proof of this comparability, Peter Jonesshowed around 1999 that any continuum Γ ⊂ C supports a measure μ satisfyingμ(Γ) ≈ diam(Γ), μ(B(x, r)) ≤ r for all x, r and c2(μ) � diam(Γ). In other words,γ+(Γ) � diam(Γ), which is equivalent to saying that

γ+(Γ) ≈ γ(Γ)


for any continuum Γ ⊂ C, in view of Proposition 1.5. The proof can be found inPajot [132]. The arguments used by Jones are of geometric type, very differentfrom the ones in the present chapter.

Another relevant particular case is the one of the Cantor sets E(λ) introducedin Section 4.7 of Chapter 4. Mateu, Tolsa and Verdera [96] proved that γ(E(λ)) ≈γ+(E(λ)) before the obtention of the analogous result for arbitrary compact setsin Tolsa [160]. The arguments for the sets E(λ) also use an induction on scales anda Tb-like theorem, like the proof for general compact sets in this chapter. For theCantor sets E(λ), the induction on scales and the rest of the proof are particularlytransparent (recall the Example 6.7). The arguments in Tolsa [160] were inspiredin part by the ideas from this particular case.

The estimate of the Cauchy integral in Theorem 6.17 is from Melnikov andTolsa [110]. The same estimate had been proved in 1966 by Melnikov [111] for adomain G with analytic boundary. Later on, Vitushkin [178] extended it to thecase where the boundary of G is made up of piecewise C1+ε curves. In [30], Daviegeneralized the result to the case of C1 curves whose tangent satisfies a Dini typecondition. Arguing as in Section 6.9.1, from the results of Melnikov, Vitushkin andDavie one derives the semiadditivity of γ for disjoint compact sets separated bycurves which are analytic, piecewise C1+ε, or C1 with Dini continuous tangent,respectively. All these curves are AD regular, and thus the Cauchy transform isbounded with respect to arc length on them. So the results of Melnikov, Vitushkinand Davie can be considered as particular cases of Theorem 6.17.

Let us remark that in Melnikov and Tolsa [110] it is also shown that if theestimate (6.45) holds for any subset set E ⊂ G and any function f ∈ H∞(G \E),then the Cauchy transform CH1�∂G is bounded in L2(H1�∂G). Further, in thiswork there are analogous results which involve the continuous analytic capacity αin lieu of γ.

The comparability γ ≈ γ+ has other interesting consequences besides thesemiadditivity of γ. For example, before proving γ ≈ γ+, it was not even known ifthe class of sets with vanishing analytic capacity is invariant under an affine mapsuch as x+ iy �→ x+ i 2y, x, y ∈ R (this question was raised by O’Farrell, as far asI know). However, this is true for γ+ (and so for γ), because its characterizationin terms of curvature of measures. Similarly, it is quite easy to check that theclass of sets with vanishing capacity γ+ is invariant under C1+ε diffeomorphisms.The analogous fact for bilipschitz diffeomorphisms also holds, although it is moredifficult. This is a consequence of the following result, proved in Tolsa [163].

Theorem 6.18. Let μ be a Radon measure supported on a compact E ⊂ C, suchthat μ(B(x, r)) ≤ r for all x ∈ E, r > 0 and c2(μ) < ∞. Let ϕ : C → C be abilipschitz mapping. Then there exists a positive constant C depending only on ϕsuch that

c2(ϕ#μ) ≤ C(μ(E) + c2(μ)

).


Recall that ϕ#μ stands for image measure of μ by ϕ. See also Garnett andVerdera [55] for a preliminary result on Cantor sets. From the characterization ofγ+ in terms of curvature and the comparability γ ≈ γ+, one deduces:

Corollary 6.19. Let E ⊂ C be a compact set and ϕ : C → C a bilipschitz map.There exists a positive constants c depending only on ϕ such that

c−1γ(E) ≤ γ(ϕ(E)) ≤ c γ(E).

From Theorem 6.18 and the T 1 theorem for the Cauchy transform one alsogets the following.

Corollary 6.20. Let ϕ : C−→C be a bilipschitz map and μ a Radon measure on C

with linear growth. Set σ = ϕ#μ. If Cμ is bounded in L2(μ), then Cσ is boundedin L2(σ).

The results analogous to the ones in the Corollaries 6.19 and 6.20 for theLipschitz harmonic capacity and for the n-dimensional Riesz transform in Rn+1

are open at the time of this writing.

Observe, by the way, that the question whether positive Favard length for acompact set E ⊂ C implies positive analytic capacity γ(E), which was mentionedin Section 1.7 of Chapter 1, can be restated in a more geometric way thanks toTheorem 6.1. Indeed, this is equivalent to asking if any compact set with positiveFavard length supports a positive measure with linear growth and finite curvature.

Now we turn our attention to another open problem, which concerns theso-called Cauchy capacity γc. This is defined like the usual analytic capacity,although the supremum in (1.1) is taken over all the functions f : C \E → C with‖f‖∞ ≤ 1 which are the Cauchy transform of some complex measure supportedon E. Clearly, γ+ ≤ γc ≤ γ and thus γ ≈ γc. However, it is not known if γ = γc.In fact, this is true for sets with finite length, by Proposition 6.5, but unknown forsets with infinite length. For more information about γc and this open problemsee Khavinson [80].

6.10.2 Other capacities

The arguments to prove that γ ≈ γ+ have been adapted successfully to the studyof other related capacities. For example, in Tolsa [162] it has been shown that thecontinuous analytic capacity satisfies

α(E) ≈ α+(E)

≈ sup{μ(E) : supp(μ) ⊂ E, Θ1μ(x) = 0 ∀x ∈ E, c2(μ) ≤ μ(E)}.

Recall that Θ1μ(x) is the 1-dimensional density of μ at x (see (1.16)). From this

characterization of α it follows that α is countably semiadditive


Volberg [180] has also been able to adapt the scheme from Theorem 6.1 forthe Lipschitz harmonic capacity κ. He has shown that

κ(E) ≈ κ+(E) ≈ sup{μ(E) : μ ∈ Σn(E), ‖Rn

μ‖L2(μ)→L2(μ) ≤ 1}, (6.52)

where Σn(E) stands for the subset of the positive measures μ supported on Esuch that μ(B(x, r)) ≤ rn for all x, r and Rn is the n-dimensional Riesz transformwith respect to μ in Rn+1 (which is associated with the vectorial kernel x/|x|n+1).Since curvature and permutations are not useful in higher dimensions, one of themain difficulties consists in finding a potential which plays the same role as Uμ forthe construction of the intermediate approximating squares. The potential takenby Volberg is the following:

MR,nμ(x) +Rn∗μ(x) +

n+1∑i=1

Rn,(i)μ,∗(Rn,(i)μ

)(x), (6.53)

where Rn,(i) is i-th scalar component of Rn, and the subindex ∗ stands for themaximal version of the corresponding operator, as usual.

By combining the techniques for the continuous analytic capacity and theLipschitz harmonic capacity, it also follows that the C1 harmonic capacity κc iscomparable to its positive version κc,+, as shown by Ruiz de Villa and Tolsa [143].

For 0 < α < d, the capacity γα in Rd associated with the α-dimensionalRiesz transform (see Section 4.10.4 of Chapter 4) is comparable to γα,+. This wasproved by Mateu, Prat and Verdera [93] in the case 0 < α < 1 and by Prat [137]for 1 < α < d (except in the case α = d − 1, which corresponds to the Lipschitzharmonic capacity). For 0 < α < 1 a potential of “curvature type” works, becauseof (3.26), while for α > 1 a potential similar to (6.53) is required.

There are works which deal with other more “exotic” capacities associatedwith Calderon-Zygmund kernels, which have been shown to be semiadditive byproving their comparability with their corresponding positive versions too. Forexample, Mateu, Prat and Verdera [94] have studied the capacities in Rd associatedwith the scalar signed Riesz kernel xi/|x|n+1, where xi is the i-th component ofx. On the other hand, the paper by Chousionis, Mateu, Prat and Tolsa [11] dealswith the capacities in the plane associated with the vectorial kernels(

x2j−11

|x|2j ,x2j−12

|x|2j),

for j ≥ 1.

Chapter 7

Curvature and rectifiability

7.1 Introduction

In this chapter we will prove the following.

Theorem 7.1. Let E ⊂ C be an H1-measurable set with H1(E) < ∞. If c2(H1�E) <∞, then E is rectifiable.

This result was first proved by David. However, his arguments have remainedunpublished and instead a different proof by Leger [83] appeared later. Moreover,the arguments of Leger have the advantage of generalizing more easily to higherdimensions than the original ones of David. So quite often the theorem above isreferred to as the “David-Leger curvature theorem”, for short.

From Theorem 7.1, the characterization of γ+ in terms of curvature, and thecomparability1 between γ and γ+, one easily gets a positive answer to Vitushkin’sconjecture for sets with finite length. This is David’s theorem [23]:

Theorem 7.2. Let E ⊂ C be compact with H1(E) < ∞. Then γ(E) = 0 if and onlyif E is purely unrectifiable.

Proof. Under the assumption that H1(E) < ∞, the fact that γ(E) = 0 forces E tobe purely unrectifiable is just a restatement of the solution of Denjoy’s conjecture(see Theorem 4.9).

Suppose now that γ(E) > 0. Then γ+(E) > 0, either by Theorem 6.1 orTheorem 6.4, and thus E supports a non-zero measure μ with linear growth andfinite curvature. Such a measure μ must be of the form μ = hH1�E, for someh ∈ L1(H1�E). Indeed, if F ⊂ C satisfies H1�E (F ) = 0, then, for any ε > 0,E ∩ F can be covered by a family of balls B(xi, ri) such that

∑i ri < ε. We

1In this situation, we might appeal to David’s Theorem 6.4, instead of using the full strength

of the comparability between γ and γ+.

, , OI 10.1007/978-3- - -6_ ,

© Springer



231

232 Chapter 7. Curvature and rectifiability

deduce thatμ(F ) = μ(E ∩ F ) ≤ c

∑i

ri < c ε,

and thus μ(F ) = 0. Thus, μ = hH1�E by the Radon-Nikodym theorem.Let δ > 0 be such that the set Gδ := {x ∈ E : h(x) > δ} has positive

μ-measure. Then we have

c2(H1�Gδ) ≤ δ−3c2(μ�Gδ) < ∞,

since H1�Gδ = h−1μ�Gδ, with h−1 ≤ δ−1 on Gδ. Thus, by the David-Legertheorem, we infer that Gδ is rectifiable. So E is not purely unrectifiable. �

Using the countable semiadditivity of γ, David’s theorem can be extendedto the case of σ-finite length:

Corollary 7.3. Let E ⊂ C be compact with σ-finite length. Then γ(E) = 0 if andonly if E is purely unrectifiable.

Proof. We already know that if E is rectifiable, then γ(E) > 0, by the solution ofDenjoy’s conjecture. On the other hand, if E is purely unrectifiable, then we writeE =⋃∞

n=1 En, whereEn ⊂ C are purely unrectifiable Borel sets withH1(En) < ∞.Then γ(En) = 0 for every n, by David’s theorem, and thus

γ(E) ≤ c

∞∑n=1

γ(En) = 0. �

For sets with non-σ-finite length, the characterization of removability in termsof pure unrectifiability fails. Indeed, let F ⊂ R be the classical 1/3 middle Cantorset, and let E(1/3) = F × F ⊂ C. This set is not removable because, by Theorem4.28,

γ(E(1/3)) ≈( ∞∑

n=0

(4−n

3−n

)2)−1/2

=

( ∞∑n=0

(9

16

)2)−1/2

> 0.

Alternatively, it is not difficult to check that dimH(E(1/3)) > 1 (in fact,dimH(E(1/3)) = log 4/ log 3, see Mattila [100, Sections 4.10-4-13]), which impliesthat γ(E) > 0, by Theorem 1.25.

On the other hand, E(1/3) is purely unrectifiable. That is, H1(E(1/3)∩Γ) =0 for any rectifiable curve. This can be deduced from the fact that the orthogonalprojections of E(1/3) both on the horizontal and vertical axes equal the 1/3 Cantorset and thus have zero length; recalling also that the length of the orthogonalprojections of any subset of positive length of a rectifiable curve Γ vanishes atmost in one direction (see Falconer [40, Theorem 6.10]).

By choosing an appropriate sequence λ = {λn}n in the construction of thesets E(λ) from Section 4.7 of Chapter 4, we can get an example of a set withHausdorff dimension 1 which is non-removable and purely unrectifable (of course,

7.2. The quantitative version of the David-Leger theorem 233

in this case E(λ) will have non-σ-finite length). Indeed, it suffices to take λ suchthat λn ↘ 1/4 with

∑∞n=0 θ

2n < ∞, for θn = (4nλ1 · · ·λn)

−1. For example, thechoice λn = n+3

4(n+2) , which yields θn = 3n+3 , does the job.

7.2 The quantitative version of the David-Leger

theorem

Theorem 7.1 follows as a consequence of the following more quantitative result.

Theorem 7.4. For any constant c0 ≥ 1, there exists a number η > 0 such thatif μ is a positive Radon measure supported on a ball B(x0, R) which satisfiesμ(B(x0, R)) ≥ R, μ(B(x, r)) ≤ c0 r for all x ∈ C, r > 0, and c2μ(x) ≤ η forμ-a.e. x ∈ C, then there exists a Lipschitz graph Γ with slope cΓ ≤ 1/10 such thatμ(Γ) ≥ 99

100 μ(C). Moreover, the slope cΓ can be taken arbitrarily small, assumingη small enough.

Let us remark that the above condition

c2μ(x) ≤ η for μ-a.e. x ∈ C,

is equivalent toc2μ(x) ≤ η for all x ∈ supp(μ),

since c2μ(x) ≤ lim supy→x c2μ(y), as the reader can easily check. On the other hand,

in general,c2μ(x) �= lim

y→xc2μ(y).

To see this, one can take μ equal to the arc length measure on an interval I ⊂ C,and a sequence of points yn approaching to some point in x ∈ I orthogonally toI, say.

The preceding theorem will be proved in Sections 7.3-7.8 of this chapter. Toshow how this yields Theorem 7.1, we first need the following auxiliary result.

Lemma 7.5. Let μ be some positive finite Radon measure without atoms on C. Ifc2(μ) < ∞, then

limr→0

c2(μ�B(x, r))

μ(B(x, r))= 0 for μ-almost all x ∈ C.

Proof. For each m ≥ 1, let

Am ={x ∈ C : lim supr→0 c

2(μ�B(x, r))[μ(B(x, r))]−1 > 1/m}.

For r > 0, write

c2(r)(μ) =

∫∫∫|x−y|≤r



Notice that limr→0 c2(r)(μ) = 0, because c2(μ) < ∞ and μ has no atoms.

Given any r > 0 and m ≥ 1, for each x ∈ Am there exists some ballB(x, s) with s < r/2 such that μ(B(x, s)) ≤ mc2(μ�B(x, s)). With this typeof balls, we consider a Besicovitch covering of Am. That is, Am ⊂ ⋃iB(xi, si),with∑

i χB(xi,si) ≤ c. Let us remark that, for the application of this theorem, wemay assume supp(μ) to be bounded if necessary. Then we get

μ(Am) ≤∑i

μ(B(xi, si)) ≤ m∑i

c2(μ�B(xi, si)) ≤ cm c2(r)(μ),

which tends to 0 as r → 0. Thus μ(Am) = 0 for each m. �Let A ⊂ Rd and s ≥ 0. Recall that if Hs(A) < ∞, as explained in Theorem

1.30, the upper 1-dimensional density satisfies

2−s ≤ Θ∗s(x,A) ≤ 1 for Hs-a.e. x ∈ A. (7.1)

We will appeal to this result in the argument below.

Proof of Theorem 7.1 using Theorem 7.4. Let E ⊂ C be an H1-measurable setsatisfying H1(E) < ∞ and c2(H1�E) < ∞. By Theorem 1.26 it follows that,in order to prove that E is rectifiable, it is enough to show that for every H1-measurable subset F ⊂ E with H1(F ) > 0 there exists a measurable rectifiablesubset A ⊂ F with H1(A) > 0.

For each m ≥ 1, let

Fm ={x ∈ F : H1(F ∩B(x, r)) ≤ 3r for 0 < r ≤ 1/m

}.

From the statement in (7.1), it is easy to check that H1(F \⋃m Fm) = 0, and thusthere exists some subset Fm ⊂ F with H1(Fm) > 0.

Let σ = H1�Fm and notice that c2(σ) ≤ c2(H1�E) < ∞. Given any η > 0,by (7.1) and Lemma 7.5, for σ-a.e. x ∈ Fm there exists a ball B(x, rx) with0 < rx < 1/2m such that

rx2

≤ σ(B(x, rx)) ≤ 3 rx and c2(σ�B(x, rx)) ≤ η σ(B(x, rx)).

For such x ∈ Fm, by Chebyshev, this implies that

G :={y ∈ B(x, rx) : c

2σ�B(x,rx)

(y) ≤ 2η}

satisfies σ(G) ≥ σ(B(x, rx))/2 ≥ rx/4. Then we can apply Theorem 7.4 to μ =4 σ�G, with B(x0, R) = B(x, rx). Indeed, μ(B(x0, R)) = 4 σ(G) ≥ rx = R, μ haslinear growth (with c0 = 12) by the definition of Fm and σ, and

c2μ(y) = 16 c2σ�G(y) ≤ 32 η for μ-a.e. y ∈ C.

Thus, if η is taken small enough, there exists a subset A ⊂ Fm ∩ B(x, rx) withpositive length which is contained in a Lipchitz graph, and so A is rectifiable. �

7.3. The two squares condition, Jones’ β’s, and curvature 235

7.3 The two squares condition, Jones’ β’s, andcurvature

In this section we will introduce some notions and prove some preliminary resultsfor the proof of Theorem 7.4.

Given a ball B = B(x, r), we write

Θ(B) =μ(B)

r.

So Θ(B) is the average 1-dimensional density of μ on B. The following lemma willbe used many times in the proof of Theorem 7.4, in the next sections.

Lemma 7.6. Let μ be a measure with c0-linear growth on C and let δ > 0. Thereexists some constant c1 ≥ 1 depending only on c0 and δ such that if B is someball satisfying Θ(B) ≥ δ > 0, then there are two balls B1 and B2 satisfying

• dist(B1, B2) ≥ 10c−11 r(B), and

• μ(B ∩Bi) ≥ c−11 r(B).

Above, by c0-linear growth, we mean that μ(B(x, r)) ≤ c0 r for all x ∈ C,r > 0.

To prove the lemma we will show a slightly more general result, and forconvenience we will work with squares instead of balls. Given a, b > 0, we saythat a square Q ⊂ C satisfies the two squares condition with parameters a, b withrespect to μ if there exist two squares Q1, Q2 ⊂ Q such that

• dist(Q1, Q2) ≥ a �(Q),

• �(Qi) ≤ a

10�(Q) for i = 1, 2, and

• μ(Qi) ≥ bμ(Q) for i = 1, 2.

We write Q ∈ TSCμ(a, b).

Lemma 7.7. Let μ be an arbitrary Radon measure on C. For some positive integerm, let a = 1/m. Also, let b be an arbitrary positive constant. If Q �∈ TSCμ(a, b),then there exists a square P ⊂ Q with �(P ) = 3a�(Q) such that

μ(P ) ≥(1− 100b

a2

)μ(Q).

Although the above b is an arbitrary positive constant, the last estimate isonly non-trivial for b small enough, that is, for b such that 1− 100b

a2 > 0.

Proof. We set N := 10m. We split Q into N2 squares Qk, k = 1, . . . , N2, withdisjoint interiors, of side length �(Q)/N = a�(Q)/10. We put

G = {Qk : 1 ≤ k ≤ N2, μ(Qk) ≥ bμ(Q)}.


Suppose that G is non-empty, and take Qj ∈ G. Since Q �∈ TSCμ(a, b), any othersquare Qk ∈ G satisfies dist(Qj , Qk) < a �(Q). Therefore, all the squares from Gare contained in a square P concentric with Qj of side length 3a�(Q), since

2(2a10

+ a)�(Q) ≤ 3a�(Q).

Thus, ∑k:Qk∈G

μ(Qk) ≤ μ(P ∩Q).

Of course this estimate also holds if G is empty, choosing P as any square of sidelength 3a�(Q) centered at Q ∩ suppμ, say.

On the other hand,∑k:Qk �∈G

μ(Qk) ≤ b∑

k:Qk �∈Gμ(Q) ≤ bN2μ(Q) =

100b

a2μ(Q).

Then we get

μ(Q) ≤ μ(P ∩Q) +100b

a2μ(Q),

and so

μ(P ∩Q) ≥(1− 100b

a2

)μ(Q). �

Now to prove Lemma 7.6, we just consider the measure μ�B and a squareQ concentric with B and side length 2r(B) and choose b = a2/200. For theseparameters, if a is small enough, then Q ∈ TSCμ�B(a, b). Otherwise the squareP found in Lemma 7.7 satisfies μ(P ) ≥ μ(B ∩ Q)/2 ≥ δ r(B)/2, which violatesthe c0-linear growth for a small enough, recalling that �(P ) = 3a�(Q). It is clearthat the two squares condition for μ�B implies the existence of B1 and B2 as inLemma 7.6.

Given E = supp(μ) and a ball B = B(x, r), we consider the Jones’ β coeffi-cient of B (with respect to E):

β(B) = infL

(sup

x∈E∩3B

dist(x, L)

r

),

where the infimum is taken over all lines L ⊂ C. So β(B) is the width of thethinnest strip VB containing E∩3B divided by diam(B). Note that β(B) measureshow close E ∩ 3B is to some L line in a scale invariant way. See Figure 7.1. Recallthat in (3.17) we introduced an analogous notion for squares instead of balls.

In the following lemmas we will show how the β coefficients can be controlledin terms of curvature.


3B

B

E ∩ 3B

L

VB

Figure 7.1: The balls B, 3B, a minimizing line L, and the strip VB in the definitionof β(B).

Lemma 7.8. Let B ⊂ C be a ball. Let x0 ∈ B be such that c2μ�B(x0) < ∞. Thenthere exists a line L passing through x0 such that, for any ε0 > 0,

μ({y ∈ B : dist(y, L) > ε0 r(B)}) ≤ 2 r(B) c2μ(x0)

1/2

ε0.

Proof. Since c2μ�B(x0) < ∞, there exists some z0 ∈ B such that

∫B

c(x0, z0, y)2 dμ(y) ≤

c2μ�B(x0)

μ(B).

Let L be the line through x0 and z0. Recall that

c(x0, z0, y) =2 dist(y, L)

|x0 − y| |z0 − y| .

Then we have

μ({y ∈ B : dist(y, L) > ε0 r(B)}) ≤ ∫

B

dist(y, L)

ε0 r(B)dμ(y)

≤ 2r(B)

ε0

∫B

2 dist(y, L)

|x0 − y| |z0 − y| dμ(y)


≤ 2r(B)μ(B)1/2

ε0

(∫B

c(x0, z0, y)2dμ(y)

)1/2=

2r(B) c2μ�B(x0)1/2

ε0. �

Lemma 7.9. Let μ be a measure with c0-linear growth and let δ > 0. There existsa constant c2 depending only on c0 and δ such that if B is some ball satisfyingΘ(3B) ≥ δ > 0, then

β(B)2 ≤ c2(c0, δ) supx∈supp(μ)∩3B

c2μ�3B(x). (7.2)

Proof. Let

ε0 =4

δsup

x∈supp(μ)∩3B

(c2μ�3B(x)

)1/2.

By the preceding lemma, since μ(3B) ≥ δr(3B), we deduce that there exists a lineL such that

μ({y ∈ 3B : dist(y, L) > ε0 r(3B)}) ≤ 2r(3B)

ε0sup

x∈supp(μ)∩3B

(c2μ�3B(x)

)1/2≤ μ(3B)

2.

Thus, if we consider the strip V = {y ∈ C : dist(y, L) ≤ ε0 r(3B)}, then μ(V ∩3B) ≥ μ(3B)/2. By Lemma 7.6 applied to μ�V and 3B, we infer that there are twoballs B1, B2 such that dist(B1, B2) ≥ 10c−1

1 r(3B), and μ(3B∩V ∩Bi) ≥ c−11 r(3B)

with c1 = c1(c0, δ/2). Notice that, for all y ∈ 3B ∩ V ∩B1 and z ∈ 3B ∩ V ∩B2,

�(L,Ly,z) ≤ c · width of V

r(B),

with c depending on c1, and so c = c(c0, δ). As a consequence, there exists someconstant c3 = c3(c0, δ) > 1 such that

Ly,z ∩ 9B ⊂ V ′ := {y ∈ C : dist(y, L) ≤ c3 ε0 r(3B)}. (7.3)

We distinguish now two cases. Suppose first that

supp(μ) ∩ 3B ⊂ 2V ′ := {y ∈ C : dist(y, L) ≤ 2c3 ε0 r(3B)}.Then by definition we have β(B) ≤ 2c3 ε0 and we are done.

In the other case, consider x ∈ supp(μ)∩3B\2V ′. Then for all y ∈ 3B∩V ∩B1,z ∈ 3B ∩ V ∩ B2 we claim that dist(x, L) ≤ 2 dist(x, Ly,z). Indeed, let x′ ∈ Ly,z

be such that dist(x, Ly,z) = |x− x′|. Sincedist(x, Ly,z) ≤ diam(3B) = 6r(B),


we deduce that x′ ∈ B(x, 6r(B)) ⊂ 9B. From (7.3) we infer that dist(x′, L) ≤c3 ε0 r(3B). So,

dist(x, L) ≤ |x− x′|+ dist(x′, L)

≤ dist(x, Ly,z) + c3 ε0 r(3B) ≤ dist(x, Ly,z) +1

2dist(x, L),

which proves the claim. As a consequence,


|x− y| |x− z| ≥dist(x, L)

4 r(3B)2.

Therefore,

c2μ(x) ≥∫∫

y∈3B∩V∩B1

z∈3B∩V ∩B2

c(x, y, z)2 dμ(y) dμ(z)

≥ dist(x, L)2

16 r(3B)4μ(3B ∩ V ∩B1)μ(3B ∩ V ∩B2) ≥ c(c0, δ)

dist(x, L)2

r(B)2,

and so (7.2) holds in this case too. �Remark 7.10. We recall that, for t > 0, the t-truncated pointwise curvature of μequals

c2t,μ(x) =

∫∫∫|x−y|>t|x−z|>t|y−z|>t

c(x, y, z)2 dμ(y) dμ(z).

With some additional effort one can show that the preceding lemma also holds ifone replaces c2μ�3B(x) by c2kr(B),μ�3B(x), for some 0 < k < 1 depending only on c0and δ. However, we will not need this for the arguments below and so we skip thedetails. For a related argument, we refer the reader to Leger [83, Lemma 2.5].

The following auxiliary lemma will be used below too.

Lemma 7.11. Let μ be a measure with c0-linear growth. Let B1 = B(x, r), B2 =B(y, r) be balls such that Θ(B1) ≥ δ, Θ(B2) ≥ δ, |x − y| ≤ 3

2 r, and let L1, L2 belines such that

supx∈supp(μ)∩3Bi

dist(x, Li)

r≤ ε0 for i = 1, 2, (7.4)

with 0 ≤ ε0 ≤ 3. Then there is some constant c4 depending only on c0 and δ suchthat

(a) for all t ∈ L1, dist(t, L2) ≤ c4 ε0(r+ |t−x|), and for all t ∈ L2, dist(t, L1) ≤

c4 ε0(r + |t− y|),

(b) �(L1, L2) ≤ c4 ε0.


Proof. We may assume that ε0 is small enough, since otherwise the statementsabove are trivial.

Observe that B1 ⊂ 3B2 and B2 ⊂ 3B1. By Lemma 7.6 there are two pointsz, w ∈ B1 ∩ supp(μ) such that |z − w| ≥ 10c−1

1 r, with c1 depending on c0 andδ. Let z1 and w1 be the orthogonal projections of z and w onto L1, respectively.Also, let z2 and w2 be the orthogonal projections of z1 and w1 onto L2. Then|z1 − z| ≤ ε0 r and |w1 − w| ≤ ε0 r, by (7.4). In particular, z1 ∈ B(x, (1 + ε0)r).By (7.4) again, notice that dist(z, L2) ≤ ε0 r because z ∈ 3B2 ∩ suppμ, and thus

|z1 − z2| = dist(z1, L2) ≤ |z1 − z|+ dist(z, L2) ≤ 2ε0 r,

and analogously |w1 − w2| ≤ 2ε0 r. Moreover,

|z1 − w1| ≥ |z − w| − 2ε0 r ≥ 1

2c−11 r,

if ε0 is small enough (depending on c0 and δ). So we have found two points z1, w1 inL1 which are separated by a distance c−1

1 r/2, so that both are at a small distance(≤ 2ε0 r) from L2, and moreover z1 ∈ B(x, (1 + ε0)r). By elementary geometry,the first statement in (a) follows. Indeed, let L′

2 be the line parallel to L2 passingthrough z1. Since dist(L2, L

′2) = |z1 − z2|, for every t ∈ L1 we get

dist(t, L2) ≤ dist(t, L′2) + |z1 − z2| ≤ dist(t, L′

2) + 2ε0 r.

By Thales theorem, now we have

dist(t, L′2) ≤ dist(w1, L

′2)

|z1 − t||z1 − w1| ≤ 2c1 dist(w1, L

′2)

|z1 − t|r

.

Using that dist(w1, L′2) ≤ dist(w1, L2) + |z1 − z2| ≤ 4ε0 r, we obtain

dist(t, L2) ≤ 2ε0 r + 8c1ε0 |z1 − t|,

which implies the first statement in (a). The second one is proved analogously.Concerning (b), we have

sin�(L1, L2) = sin�(L1, L′2) =

dist(w1, L′2)

|z1 − w1| ≤ 4ε0 r

c−11 r/2

= 8c1ε0. �

It is easy to check that for any ball B there exists a minimizing line for β(B).Although this line may not be unique, by the preceding lemma, it turns out thattwo minimizing lines LB and L′

B must be close to each other. More precisely, theymust satisfy the statements (a) and (b) from Lemma 7.11, with ε0 = β(B). Inparticular, one deduces easily that

distH(LB ∩ 4B,L′B ∩ 4B) ≤ c β(B),

7.4. The sets Z, E1 and E2 241

where distH stands for the Hausdorff distance. Recall that the Hausdorff distancebetween two sets E1, E2 ⊂ Rd is defined by

distH(E1, E2) = max(supx∈E1

dist(x,E2), supy∈E2

dist(y, E1)).

Alternatively,

distH(E1, E2) = inf{ε > 0 : E1 ⊂ Uε(E2), E2 ⊂ Uε(E1)

}.

Below we will denote by LB some (fixed) minimizing line for β(B). Of course,we have 3B ∩ LB = ∅.

7.4 Stopping conditions for the proof of Theorem 7.4and the sets Z, E1, E2

In this section we start the construction of the Lipschitz graph Γ for the proof ofTheorem 7.4.

Here and in the subsequent Sections 7.5-7.8 we suppose that μ is a measuresatisfying the assumptions of Theorem 7.4. We write E = suppμ. We may as-sume that the line L0 that minimizes β(B(x0, R)) passes through x0, by replacingB(x0, R) by another ball centered at L0 ∩ B(0, R) with at most double radius ifnecessary. Then the condition μ(B(x0, R)) ≥ R will be replaced by

μ(B(x0, R)) ≥ R

2.

Further, by translating and rotating μ, we may and will also assume that B(x0, R)= B(0, R), and that L0 coincides with the horizontal axis R× {0}.

We consider some constants ε, δ, α such that

0 < ε δ, α 1

whose precise values will be fixed along the proof of Theorem 7.4.

Remark 7.12. The constant δ will be chosen in Lemma 7.33. It depends only on c0and on the number 99/100 appearing in Theorem 7.4, and its choice is independentof the other parameters α and ε. For instance, one can take δ = 10−10, say.

Concerning α, any value such that 0 < α ≤ 1/100 works. One should thinkthat cα is the maximal slope of the Lipschitz graph that we will construct, wherec is some absolute constant.

On the other hand, ε depends on the parameters c0, δ, α. Moreover, a keypoint in the proof is that ε α, δ.

Some of the constants appearing in the construction below are absolute,while others depend on c0, or both on c0 and δ. For the latter case, we will usethe notation c(δ) (possibly with a subindex). Any constant denoted by c or c(δ),with or without a subindex, may also depend on c0.


Let x ∈ E and 0 < r ≤ 50R. We say that a ball B = B(x, r) is good, and wewrite B ∈ G, if(a) Θ(B) ≥ δ, and

(b) �(LB, L0) ≤ α, for some best approximating line LB.

Observe that if B satisfies the condition (a) then Θ(3B) ≥ δ/3, and recalling thatc2μ(x) ≤ η for all x ∈ E, we deduce that

β(B) ≤ c2(c0, δ/3) η1/2,

by Lemma 7.9. We will assume that η has been chosen small enough in Theorem7.4 so that c2(c0, δ/3) η

1/2 ≤ ε, for fixed c0, δ, ε, and so β(B) ≤ ε.We say that B is very good, and we write B ∈ VG, if B(x, s) is good for

r ≤ s ≤ 50R (the radius of B is still called r and r ≤ 50R). Notice that all theballs from G and VG are assumed to be centered at points from E. Observe alsothat any ball B centered at E with radius not greater than 50R which containsB(0, R) is very good. However, we cannot ensure that B(0, R) ∈ VG because itmay happen 0 ∈ E.

For x ∈ E, we set

h(x) = inf{r : 0 < r ≤ 50R, B(x, r) ∈ VG}.One should think that, in fact, h(x) ≤ 2R for all x ∈ E, because B(0, R) ∩ E =B(x, 2R) ∩ E for all x ∈ E and δ, ε 1.

We denote by Z the subset of those x ∈ E such that h(x) = 0 (i.e. B(x, r) isgood for all 0 < r ≤ 50R). We also define:

• E1 is the set of those x ∈ E \ Z such that Θ(B(x, h(x))) ≤ δ.

• E2 = E \ (E1 ∪ Z).

We clearly haveE = Z ∪ E1 ∪ E2.

Lemma 7.13. Let x ∈ E and r such that h(x) < r ≤ 50R. Then Θ(B(x, r))) ≥ δ,β(B(x, r)) ≤ ε, and �(LB(x,r), L0) ≤ α. If x ∈ E2, then �(LB(x,2h(x)), L0) ≥ α/2,for any best approximating line LB(x,2h(x)), assuming ε small enough.

Proof. The statements for h(x) < r ≤ 50R are just a consequence of the definitionsof Z, E1, and E2, and Lemma 7.9, as mentioned above.

If x ∈ E2, then h(x) > 0 and Θ(B(x, h(x))) > δ by definition. Also, byconstruction there exists a sequence rn ↗ h(x) such that either Θ(B(x, rn)) ≤ δfor all n or �(LB(x,rn), L0) > α for all n. The first assertion cannot hold, becausethis would imply that Θ(B(x, h(x)) ≤ δ (recall that we are assuming the ballsB(x, rn) to be open). Take rn such that 9

10h(x) ≤ rn < h(x). From the fact that�(LB(x,rn), L0) > α and Lemma 7.11 applied to B1 = B2 = B(x, rn) and the linesL1 = LB(x,rn), L2 = LB(x,2h(x)), we infer that �(LB(x,2h(x)), L0) ≥ α− c(δ)ε. If εis small enough, then �(LB(x,2h(x)), L0) ≥ α/2. �

7.5. Construction of the Lipschitz graph 243

7.5 Construction of the Lipschitz graph

Lemma 7.14. Let B1, B2 ∈ VG be such that B1 ⊂ B2. We have

x ∈ LB1 ∩ 2B1 =⇒ dist(x, LB2) ≤ c(δ) ε r(B2). (7.5)

Notice that the condition B1 ∈ VG ensures that β(B1) ≤ ε, and thus thecenter of B1 is very close to LB1 , which implies that LB1 ∩ 2B1 = ∅ (and alsoLB1 ∩ 1

2B1 = ∅, say) if ε is taken small enough.

Proof. This is a direct consequence of the fact that β(B) ≤ ε for any ball B ∈ VG.Indeed, by Lemma 7.6, we find two points y, z ∈ B1 ∩ E such that |y − z| ≥c1(δ)

−1 r(B1). Let y′ and z′ be the orthogonal projections of y and z onto LB1 ,respectively. If ε is assumed to be small enough, we obtain

|y′ − z′| ≥ 1

2c1(δ)

−1 r(B1), (7.6)

and also y′, z′ ∈ 2B1.Since β(B1) ≤ ε and β(B2) ≤ ε, we have dist(y, LB1) ≤ ε r(B1) and

dist(y, LB2) ≤ ε r(B2), and analogously replacing y by z. Then we deduce

dist(y′, LB2) ≤ ε r(B1) + ε r(B2) ≤ 2ε r(B2),

and analogously for z′. So we have found two points y′, z′ ∈ 2B1 ∩ LB1 whichsatisfy (7.6). Then the lemma follows easily from Thales theorem. �

We denote by Π the orthogonal projection onto L0, and by Π⊥ the orthogonalprojection onto L⊥

0 , the line which is orthogonal to L0 and passes through theorigin. That is to say, for x ∈ C, Π(x) is the horizontal coordinate of x, and Π⊥(x)the vertical one.

Lemma 7.15. Let B1, B2 ∈ VG and x ∈ LB1 ∩ 2B1, y ∈ LB2 ∩ 2B2. Then

|Π⊥(x)−Π⊥(y)| ≤ 6α(|Π(x) −Π(y)|+ 2r(B1) + 2r(B2)

). (7.7)

Proof. Suppose first that r(B1), r(B2) ≤ 6R. Let B0 be a ball concentric with B1

with r(B0) = |x − y| + 2r(B1) + 2r(B2), so that B1, B2 ⊂ B0. Since the centersof B1 and B2 are inside B(0, R), we have |x− y| ≤ 2R+ 2r(B1) + 2r(B2) ≤ 26R.Therefore

r(B0) ≤ 26R+ 12R+ 12R = 50R,

and so B0 ∈ VG.Denote by x′ and y′ the orthogonal projections of x and y onto LB0 , respec-

tively. By Lemma 7.14, we have

|x− x′| ≤ c ε r(B0) and |y − y′| ≤ c ε r(B0).


Therefore,

|Π⊥(x)−Π⊥(y)| ≤ |x− x′|+ |Π⊥(x′)−Π⊥(y′)|+ |y′ − y|≤ c ε r(B0) + |Π⊥(x′)−Π⊥(y′)|.

Since x′, y′ ∈ 2B0 ∩ LB0 and the slope of LB0 is smaller that tanα ≤ 1.1α (for αsmall) we have |Π⊥(x′)−Π⊥(y′)| ≤ 4.4α r(B0). Thus, for ε small enough, we get

|Π⊥(x)−Π⊥(y)| ≤ 5αr(B0)

≤ 5α(|Π(x)− Π(y)|+ |Π⊥(x)−Π⊥(y)|+ 2r(B1) + 2r(B2)

).

Assuming that α is small enough (α ≤ 1/50, say), we get

|Π⊥(x)−Π⊥(y)| ≤ 6α(|Π(x) −Π(y)|+ 2r(B1) + 2r(B2)

).

Suppose now that r(B1), r(B2) ≥ 6R. Since B1∩E = B2∩E = B(0, R)∩E =E the minimizing lines for β(B1) and β(B2) coincide with the ones for β(B(0, R)).Although these lines may not be unique, for simplicity we will assume that for anyball B that contains B(0, R) we have chosen LB = L0. So we have LB1 = LB2 = L0

and thus Π⊥(x) = Π⊥(y) = 0 and (7.7) is trivial.If r(B1) ≥ 6R but r(B2) < 6R, then LB1 = L0, Π

⊥(x) = 0, and so (7.7)follows if we show that

|Π⊥(y)| ≤ 6α 12R = 72αR. (7.8)

This is an easy consequence of Lemma 7.14. Indeed, let B0 be a ball centered atsome point in E with radius 50R, so that B0 ∈ VG, LB0 = L0, and moreoverB2 ⊂ B0. Then Lemma 7.14 tells us that

|Π⊥(y)| = dist(x, LB0) ≤ c(δ) ε 50R,

and thus (7.8) holds for ε α sufficiently small.The case where r(B2) ≥ 6R but r(B1) < 6R is analogous. �

Remark 7.16. By the preceding lemma and the triangle inequality, it follows easilythat if B1, B2 ∈ VG, and for some c ≥ 1, x ∈ LB1 ∩ cB1, y ∈ LB2 ∩ cB2, then

|Π⊥(x) −Π⊥(y)| ≤ c′α(|Π(x) −Π(y)|+ r(B1) + r(B2)

),

with c′ depending only on c.On the other hand, if B1 ⊂ B2 are as in Lemma 7.14, then

x ∈ LB1 ∩ cB1 =⇒ dist(x, LB2) ≤ c′ ε r(B2), (7.9)

with c′ depending on c, c0, δ.


Remark 7.17. Another straightforward consequence of the preceding lemma isthat, for any two ball B1, B2 ∈ VG,

dist(B1, B2

) ≤ c[dist(Π(B1),Π(B2)

)+ r(B1) + r(B2)

].

Now we consider functions d and D defined by

d(x) = infB(z,r)∈VG

[|x− z|+ r], x ∈ C,

andD(p) = inf

x∈Π−1(p)d(x) = inf

B(z,r)∈VG[|p−Π(z)|+ r

], p ∈ L0.

Recall that the balls from VG are centered at E, by construction, and thus in theinfimums that define the functions d and D, we have z ∈ E.

The functions d and D are 1-Lipschitz. This can be deduced from the factthat the infimum of any family of non-negative 1-Lipschitz functions is another1-Lipschitz function. Also, it is easy to check that d(x) ≤ h(x) for x ∈ E. So if weset

Z0 = {x ∈ C : d(x) = 0},we have Z ⊂ Z0, and Z0 is closed. One should think of d as a regularized versionof the function h, which moreover has the advantage of being defined on the wholeC, unlike h.

Lemma 7.18. We have H1(Z0) < ∞ and

μ�Z0 = ρZ0 H1�Z0,

where ρZ0 is some function such that c−1δ ≤ ρZ0 ≤ c0 H1-a.e. on Z0.

Proof. Let us check that any compact set K ⊂ Z0 satisfies

H1(K) ≤ c

δμ(K).

Given 0 < τ R, by Besicovitch’s theorem, there exists a covering K ⊂⋃iB(zi, ri), with zi ∈ K, 0 < ri < τ , and finite overlap. Since zi ∈ Z0, there exists

z′i ∈ Z such that |zi−z′i| ≤ ri/2. Since B(z′i, ri/2) ∈ VG and B(z′i, ri/2) ⊂ B(zi, ri),we get

μ(B(zi, ri)) ≥ μ(B(z′i, ri/2)) ≥δ

2ri

(assuming τ ≤ R, say), and thus

H1τ (K) ≤

∑i

ri ≤ 2

δ

∑i

μ(B(zi, ri)) ≤ c

δμ(Uτ (K)),

where in the last inequality we took into account that the balls B(zi, ri) havebounded overlap and are contained in the τ -neighborhood of K. As τ → 0, by


definition H1τ (K) → H1(K) and, since μ is a Radon measure, μ(Uτ (K)) → μ(K),

which proves our claim. In particular, this tells us that H1(Z0) < ∞. Using nowthat both H1�Z0 and μ are Radon measures, we deduce that H1(A) ≤ c μ(A)/δfor any Borel subset A ⊂ Z0. By the Lebesgue-Radon-Nikodym theorem, we inferthat

H1�Z0 = gZ0 μ�Z0, (7.10)

with 0 ≤ gZ0(x) ≤ c/δ for μ-a.e. x ∈ Z0.Now given any Borel setA ⊂ Z0 and any τ > 0, consider a coveringA ⊂ ⋃i Ai

by Borel sets such that

H1τ (A) ≥

∑i

diam(Ai)− τ. (7.11)

By the c0-linear growth of μ, μ(Ai) ≤ c0 diam(Ai), and thusH1τ (A) ≥ c−1

0 μ(A)−τ.So letting τ → 0, we infer that μ(A) ≤ c0 H1(A). The Lebesgue-Radon-Nikodymtheorem ensures that

μ�Z0 = ρZ0 H1�Z0,

with 0 ≤ ρZ0(x) ≤ c0 for H1-a.e. x ∈ Z0. Together with (7.10) and the fact thatgZ0(x) ≤ c/δ, this proves the lemma. �

The following is a straightforward consequence of Lemma 7.15.

Lemma 7.19. For all x, y ∈ C, we have

|Π⊥(x) −Π⊥(y)| ≤ 6α |Π(x) −Π(y)|+ 4d(x) + 4d(y). (7.12)

Proof. Given x, y ∈ C and any τ > 0, consider balls B1 = B(z1, r1) and B2 =B(z2, r2) from VG such that

|x− z1|+ r1 ≤ d(x) + τ and |y − z2|+ r2 ≤ d(y) + τ.

Then the balls B′1 = B(z1, d(x) + τ) and B′

2 = B(z2, d(y) + τ) contain x and yrespectively.

Suppose first that d(x), d(y) ≤ 5R. In this case, if τ ≤ R, then B′1, B

′2 ∈ VG

and their radius is at most 6R. By Lemma 7.15, all w1 ∈ LB′1∩ 2B′

1 and w2 ∈LB′

2∩ 2B′

2 satisfy

|Π⊥(w1)−Π⊥(w2)| ≤ 6α(|Π(w1)−Π(w2)|+ 2d(x) + 2d(y) + 4τ

). (7.13)

Notice that

|Π⊥(x) −Π⊥(y)| (7.14)

≤ |x− z1|+ 2r(B′1) + |Π⊥(w1)−Π⊥(w2)|+ 2r(B′

2) + |z2 − y|≤ |Π⊥(w1)−Π⊥(w2)|+ 3d(x) + 3d(y) + 6τ,


and analogously,

|Π(w1)−Π(w2)| ≤ |Π(x) −Π(y)|+ 3d(x) + 3d(y) + 6τ. (7.15)

From (7.13) and (7.15) we derive

|Π⊥(w1)−Π⊥(w2)| ≤ 6α(|Π(x) −Π(y)|+ 5d(x) + 5d(y) + 10τ

).

Taking into account that α ≤ 1/100, (7.14) and the last estimate yield

|Π⊥(x)−Π⊥(y)|≤ 6α(|Π(x) −Π(y)|+ 5d(x) + 5d(y) + 10τ

)+ 3d(x) + 3d(y) + 6τ

≤ 6α |Π(x) −Π(y)|+ 4d(x) + 4d(y) + 7τ.

Recalling that τ > 0 is arbitrarily small, the lemma follows in this case.Suppose now that d(x) > 5R. Since B(z0, 2R) ∈ VG for every z0 ∈ B(0, R)∩

E, we have d(x) ≤ |x− z0|+ 2R ≤ |x|+ 3R. From the estimates

5R < d(x) ≤ |x|+ 3R

we infer that |x| ≥ 2R. Since dist(x,E) ≥ dist(x,B(0, R)

) ≥ |x| − R, it turns outthat

d(x) ≥ |x| −R ≥ 1

2|x|. (7.16)

If |y| ≤ |x|, we deduce that

|x− y| ≤ |x|+ |y| ≤ 2|x| ≤ 4d(x),

which implies (7.12). If |y| > |x|, then from the fact that |x| ≥ 2R, we infer that|y| ≥ 2R too. Arguing as above we deduce that (7.16) holds replacing x by y, andthus

|x− y| ≤ |x|+ |y| ≤ 2|y| ≤ 4d(y).

So (7.12) also holds in this situation.The arguments for the case d(y) > 5R are analogous. �As a consequence of the preceding lemma, if x, y ∈ C are such that d(x) =

d(y) = 0, then|Π⊥(x)−Π⊥(y)| ≤ 6α |Π(x) −Π(y)|.

Therefore, the map Π : Z0 → L0 is injective and one can define a function A onΠ(Z0) by setting

A(Π(x)) = Π⊥(x) for x ∈ Z0, (7.17)

which is Lipschitz with constant ≤ 6α.To extend the function A to the whole line L0 we will use a variant of the

Whitney’s extension theorem. For each p ∈ L0 such that D(p) > 0, i.e. p ∈


L0 \Π(Z0), we call Rp the largest dyadic interval from L0 (which we identify withR) containing p such that

(Rp) ≤ 1

20inf

q∈Rp

D(q).

We consider the collection of the different intervals Rp, p ∈ L0 \Π(Z0), and relabelit as {Ri}i∈I .

Lemma 7.20. The intervals {Ri}i∈I have disjoint interiors in L0 and satisfy thefollowing properties:

(a) If p ∈ 15Ri, then 5(Ri) ≤ D(p) ≤ 50(Ri).

(b) There exists an absolute constant c such that if 15Ri ∩ 15Rj = ∅, then

c−1(Rj) ≤ (Ri) ≤ c (Rj).

(c) For each i ∈ I, there are at most N intervals Rj such that 15Ri∩ 15Rj = ∅,where N is some absolute constant.

(d) L0 \Π(Z0) =⋃

i∈I Ri =⋃

i∈I 15Ri.

Proof. The intervals {Ri}i∈I have disjoint interiors by construction. To see (a) fora given p ∈ 15Ri, take an arbitrary u ∈ Ri, and notice that since D is 1-Lipschitz,then

D(p) ≥ D(u)− 15(Ri) ≥ 20(Ri)− 15(Ri) = 5(Ri).

On the other hand, by the definition of Ri, there exists some u ∈ Ri, the fatherof Ri, such that D(u) ≤ 20(Ri) = 40(Ri), and thus

D(p) ≤ D(u) + 9(Ri) ≤ 49(Ri).

The statements (b) and (c) are an immediate consequence of (a). Regarding(d), by definition we have L0 \Π(Z0) =

⋃i∈I Ri. On the other hand, from (a) we

infer that if p ∈ 15Ri, then D(p) = 0 and thus p ∈ Π(Z0). So

Π(Z0) ∩⋃i∈I

15Ri = ∅,

which implies that⋃

i∈I Ri =⋃

i∈I 15Ri. �

Write now

I0 = {i ∈ I : Ri ∩B(0, 10R) = ∅}.The following holds.

Lemma 7.21. (a) If i ∈ I0, then (Ri) ≤ R and 3Ri ⊂ [−12R, 12R].


(b) If Ri ∩ [−2R, 2R] = ∅ (in particular if i ∈ I0), then

(Ri) ≈ dist(0, Ri) ≈ |p| for all p ∈ Ri.

Proof. For (a), take Ri with i ∈ I0. That is, Ri ∩ B(0, 10R) = ∅. Then we have3Ri ⊂ [−10R− 2(Ri), 10R+ 2(Ri)]. Recall now that

(Ri) ≤ 1

20infq∈Ri

D(q),

and let z0 ∈ E ⊂ B(0, R). Since B(z0, 2R) ∈ VG, for all x ∈ B(0, 10R) we have

d(x) ≤ |x− z0|+ 2R ≤ 13R,

and so D(p) ≤ 13R for every p ∈ [−10R, 10R]. Therefore, as Ri ∩B(0, 10R) = ∅,(Ri) ≤ 13R/20 ≤ R, and so 3Ri ⊂ [−12R, 12R].

To prove (b), suppose now that Ri ∩ [−2R, 2R] = ∅. In this case, for everyp ∈ Ri we have dist(Π−1({p}), E) ≈ |p| ≥ R, and then D(p) ≈ |p|. Together withLemma 7.20 (a) this yields

|p| ≈ D(p) ≈ (Ri),

which also implies that (Ri) ≈ dist(0, Ri). �

Lemma 7.22. There exists some absolute constant c ≥ 1 such that for any i ∈ I0there exists a ball Bi ∈ VG satisfying

(a) (Ri) ≤ diam(Bi) ≤ c (Ri),

(b) dist(Π(Bi), Ri) ≤ c (Ri).

Proof. Fix p ∈ Ri. From the definition of D(·) we infer there exists some ballB = B(x, r) ∈ VG such that |p − Π(x)| + r ≤ 1.1D(p) ≈ (Ri). Clearly, the ballB satisfies the condition (b), as well as the right-hand inequality in (a). If theleft-hand inequality does not hold, we just replace B by a concentric ball withradius (Ri)/2, which does the job. �

For each i ∈ I0, we denote by Ai the affine function Ai : L0 → L⊥0 whose

graph is LBi , where Bi is the ball associated with Ri given by the preceding lemma.Since Bi ∈ VG, it turns out that Ai is Lipschitz with constant not exceedingtanα ≤ 1.1α, by the property (b) defining the good balls. On the other hand, fori ∈ I0, we set Ai ≡ 0.

Lemma 7.23. There exists an absolute constant c such that if 10Ri∩10Rj = ∅ forsome i, j ∈ I, then

(a) dist(Bi, Bj) ≤ c (Ri) if moreover i, j ∈ I0,

(b) |Ai(p)−Aj(p)| ≤ c(δ) ε (Ri) for p ∈ 100Ri,


(c) |A′i −A′

j | ≤ c(δ) ε.

Proof. For i, j ∈ I0, Lemmas 7.20 and 7.22 ensure that r(Bi) ≈ r(Bj) and thatdist(Π(Bi),Π(Bj)) ≤ c (Ri). From Remark 7.17 applied to the balls Bi, Bj, weinfer that

dist(Π⊥(Bi),Π⊥(Bj)) � dist(Π(Bi),Π(Bj)) + r(Bi) + r(Bj) � (Ri),

and thus (a) follows.For i, j ∈ I0, the properties (b) and (c) follow from (a) and Lemma 7.11,

considering an auxiliary ball B0 concentric with Bi containing both Bi and Bj ,say. The details are left for the reader.

For i, j ∈ I0, Ai ≡ Aj ≡ 0, and thus (b) and (c) are trivial.Finally, if i ∈ I0 and j ∈ I0, then Ri ∩ B(0, 10R) = ∅. From the condition

10Ri ∩ 10Rj = ∅, we know that (Ri) ≈ (Rj). By Lemma 7.21 (b) we deducethat

(Ri) ≈ (Rj) ≈ dist(0, Rj) ≈ R.

Then, by Lemma 7.11 again, LBi is very close to L0, i.e. dist(x, L0) ≤ c(δ)ε(Ri)for all x ∈ LBi ∩ c′Bi, with c(δ) depending on c′, which yields (b) and (c). �

Recall that the function A has already been defined in Π(Z0) (see (7.17)). Soit remains to define it only in L0\Π(Z0). To this end, we first introduce a partitionof unity in L0 \ Π(Z0). For each i ∈ I, we can find a function ϕi ∈ C∞(L0) suchthat χ2Ri ≤ ϕi ≤ χ3Ri , with

|ϕi′| ≤ c

(Ri)and |ϕi

′′| ≤ c

(Ri)2.

Then, for each i ∈ I, we set

ϕi =ϕi∑j∈I ϕj

.

Clearly, the family {ϕi}i∈I is a partition of unity subordinated to the sets {3Ri}i∈I ,and each function ϕi satisfies

|ϕi′| ≤ c

(Ri)and |ϕi

′′| ≤ c

(Ri)2,

taking into account Lemma 7.20.Recall that L0 \Π(Z0) =

⋃i∈I Ri =

⋃i∈I 3Ri. For p ∈ L0 \Π(Z0), we define

A(p) =∑i∈I0

ϕi(p)Ai(p).

Observe that in the preceding sum we can replace I0 by I because Ai ≡ 0 fori ∈ I \ I0.

We denote by Γ the graph {(p,A(p)) : p ∈ L0}.In the next lemma we show that A is compactly supported and globally

Lipschitz in L0.


Lemma 7.24. The function A : L0 → L⊥0 is supported in [−12R, 12R] and is

cα-Lipschitz, where c is an absolute constant.

Proof. For the support of A, just notice that, by the definition,

supp(A) ⊂ Π(Z0) ∪⋃i∈I0

supp(ϕi) ⊂ Π(Z0) ∪⋃i∈I0

3Ri.

It is clear that Π(Z0) ⊂ [−12R, 12R]. On the other hand, by Lemma 7.21 (a),3Ri ⊂ [−12R, 12R] for all i ∈ I0, and so it follows that supp(A) ⊂ [−12R, 12R].

Concerning the Lipschitz character of A, we have to show that for everyp, q ∈ L0,

|A(p)−A(q)| ≤ c α |p− q|. (7.18)

If p, q ∈ Π(Z0), this has already been proved. So we assume that p ∈ Π(Z0) andwe denote by Ri the interval such that p ∈ Ri.

We consider first the case q ∈ 3Ri. Recall that 3Ri ⊂ L0 \ Π(Z0), by (d)from Lemma 7.20. We denote by I(Ri) and I(3Ri) the sets of indices k ∈ I suchthat supp(ϕk) ∩ Ri = ∅ and such that supp(ϕk) ∩ 3Ri = ∅, respectively (noticethat, by construction, these sets are finite). Since A is a C∞ function on 3Ii it isclear that (7.18) follows if we show that the derivative of A does not exceed c α,in modulus. To this end, set

A′ =∑

k∈I(3Ri)

ϕk A′k +∑

k∈I(3Ri)

ϕ′k Ak. (7.19)

To estimate the first sum on the right-hand side we just take into account that‖A′

k‖∞ ≤ tanα ≤ 1.1α for all k, and thus∑k∈I(3Ri)

ϕk |A′k| ≤ 1.1α

∑k∈I(3Ri)

ϕk = 1.1α. (7.20)

For the last sum in (7.19), note that∑

k∈I(3Ri)ϕ′k ≡ 0 since

∑k∈I(3Ri)

ϕk ≡ 1 on3Ri. As a consequence, in 3Ri we have∣∣∣∣ ∑

k∈I(3Ri)

ϕ′k Ak

∣∣∣∣ = ∣∣∣∣ ∑k∈I(3Ri)

ϕ′k (Ak −Ai)

∣∣∣∣ ≤ cN

(Ri)sup

k∈I(3Ri),s∈3Ri

|Ak(s)−Ai(s)|,

(7.21)where N is the absolute constant in (c) of Lemma 7.20. From (b) in Lemma 7.23,we derive |Ak(s) − Ai(s)| ≤ c(δ) ε (Ri) α (Ri) for s ∈ 3Ri. Plugging thisestimate into (7.21) gives∣∣∣∣ ∑

k∈I(3Ri)

ϕ′k Ak

∣∣∣∣ ≤ c(δ)N ε α in 3Ri, (7.22)

which together with (7.20) implies that |A′| ≤ c α in 3Ri.


Suppose now that q ∈ 3Ri∪Π(Z0) in (7.18), and let Rj the interval such thatq ∈ Rj . Since |p − q| ≥ (Ri), using the property (b) in Lemma 7.20 one easilydeduces that |p− q| ≥ c−1(Rj) too. For all k ∈ I(Rj) and h ∈ I(Rk), by Remark7.16 applied to x = (p,Ak(p)), y = (q, Ah(q)), Bk, and Bh, we get

|Ak(p)−Ah(q)| ≤ c α(|p− q|+ r(Bk) + r(Bh)

)≤ c α(|p− q|+ (Ri) + (Rj)

)≤ c α |p− q|.

Therefore,

|A(p) −A(q)| =∣∣∣∣ ∑k∈I(Ri)

ϕk(p)Ak(p)−∑

h∈I(Rj)

ϕh(q)Ah(q)

∣∣∣∣≤∑

k∈I(Ri)

∑h∈I(Rj)

ϕk(p)ϕh(q) |Ak(p)−Ah(q)|

≤ c α |p− q|∑

k∈I(Ri)

∑h∈I(Rj)

ϕk(p)ϕh(q) = c α |p− q|.

The arguments for the case q ∈ Π(Z0) (and so q ∈ 3Ri) are analogous to theones for q ∈ 3Ri ∪Π(Z0): loosely speaking, we may consider that q coincides withthe interval Rj in the preceding estimates, and set Ah = A, etc. �

Remark 7.25. We assume that α is chosen small enough so that

‖A′‖∞ ≤ cα ≤ 1

10.

Remark 7.26. From the arguments in the proof of Lemma 7.24 it follows easilythat for p ∈ 3Ri, i ∈ I,

|A′(p)−A′i(p)| ≤ c(δ) ε.

Indeed, from (7.19) we get

A′ −A′i =∑

k∈I(3Ri)

ϕk (A′k −A′

i) +∑

k∈I(3Ri)

ϕ′k Ak. (7.23)

From (c) in Lemma 7.23, we have |A′k − A′

i| ≤ c(δ) ε in the equation above, andthus ∑

k∈I(3Ri)

ϕk |A′k −A′

i| ≤ c(δ) ε∑

k∈I(3Ri)

ϕk = c(δ) ε.

On the other hand, it has already been shown in (7.22) that the last sum in (7.23)is bounded by c(δ)N ε, in modulus in 3Ri.

Next we will estimate the second derivative of A out of Π(Z0).

7.6. E and Γ are close to each other 253

Lemma 7.27. If p ∈ 15Ri, i ∈ I, then

|A′′(p)| ≤ c(δ) ε

(Ri). (7.24)

Proof. We consider first the case when p ∈ Ri. We have

A′′ =(∑

j∈I

ϕj Aj

)′′=∑j∈I

ϕ′′j Aj + 2

∑j∈I

ϕ′j A

′j ,

because A′′j ≡ 0 for all j ∈ I. Notice that in each of these sums there is a bounded

number of non-zero terms for a fixed point p. Since∑

j∈I ϕj ≡ 1, then∑

j∈I ϕ′j =

0. So, for each i ∈ I we get

A′′ =∑j∈I

ϕ′′j (Aj −Ai) + 2

∑j∈I

ϕ′j (A

′j −A′

i).

If p ∈ Ri is in the support of some ϕj , then 3Ri∩3Rj = ∅, and thus (Ri) ≈ (Rj).Then (b) and (c) from Lemma 7.23 tell us that |Ai(p)−Aj(p)| ≤ c(δ) ε (Ri) and|A′

i−A′j | ≤ c(δ) ε for i, j ∈ I. Using that |ϕ′

j | ≤ c/(Rj) and that |ϕ′′j | ≤ c/(Rj)

2,(7.24) follows.

Suppose now that p ∈ 15Ri. By (d) from Lemma 7.20 there exists someinterval Rj , j ∈ I, that contains p. It has already been shown above that

|A′′(p)| ≤ c(δ) ε

(Rj).

Since (Rj) ≈ (Ri) by (b) from Lemma 7.20, the estimate (7.24) also holds inthis case. �

7.6 E and Γ are close to each other

Lemma 7.28. There exists an absolute constant c5 such that every x ∈ B(0, 10R)satisfies

dist(x,Γ) ≤ c5 d(x).

Proof. Let y =(Π(x), A(Π(x))

). By Lemma 7.19, we have

dist(x,Γ) ≤ |x− y| = |Π⊥(x) −Π⊥(y)| ≤ 4d(x) + 4d(y). (7.25)

If Π(x) ∈ Π(Z0), then y ∈ Z0, and thus d(y) = 0 and (7.25) proves the lemma.If Π(x) ∈ Π(Z0), let Ri, i ∈ I, be such that Π(x) = Π(y) ∈ Ri. From Π(x) ∈Ri ∩ B(0, 10R) = ∅, we derive that i ∈ I0 and so there is an associated ballBi ∈ VG fulfilling the properties of Lemma 7.22. Then d(y) ≤ dist(y,Bi) + r(Bi).By the definition of A, it follows that

dist(y,Bi) ≤ c r(Bi) ≈ (Ri) ≤ cD(Π(x)) ≤ c d(x).

Thus, d(y) ≤ c d(x), which together with (7.25) proves the lemma. �


Lemma 7.29. Let B ∈ VG and x ∈ Γ ∩ 3B. Then

dist(x, LB) ≤ c(δ) ε r(B). (7.26)

Proof. If d(x) = 0, then x ∈ Z0 ⊂ E and thus the estimate follows from β(B) ≤ ε.Otherwise, let p = Π(x), and recall that A(p) =

∑j∈I0

ϕj(p)Aj(p).We distinguish two cases. Suppose first that∑

j∈I0

ϕj(p) = 1.

In this situation, since (p,A(p)) is a convex combination of the points (p,Aj(p))such that ϕj(p) = 0, (7.26) follows if for all j such that ϕj(p) = 0 we have

dist((p,Aj(p)), LB) ≤ c ε r(B), (7.27)

with c depending on δ. To prove this estimate, notice that since x ∈ 3B,

D(p) ≤ d(x) ≤ 4 r(B).

Thus, if Ri is the interval that contains p,

(Ri) ≤ 1

20D(p) ≤ 1

5r(B). (7.28)

As a consequence, if 3Rj ∩ Ri = ∅, by Lemma 7.20 (b), (Rj) ≤ c r(B). Inparticular, this holds for j such that ϕj(p) = 0. Thus, for such j’s, r(Bj) ≤ c r(B)and then

dist(Π(Bj),Π(B)

) ≤ dist(Π(Bj),Π(Bi)

)+ dist((Π(Bi),Π(B)

)≤ c((Ri) + (Rj) + r(B)

) ≤ c r(B).

From Remark 7.17 we deduce that

dist(Bj , B) ≤ c[dist(Π(Bj),Π(B)

)+ r(Bj) + r(B)

] ≤ c r(B),

which implies that Bj ⊂ c′B for some absolute constant c′ ≥ 1. Then, by Lemma7.14 and Remark 7.16,

dist((p,Aj(p)), LcB

) ≤ c ε r(B),

which together with Lemma 7.11 implies (7.27).Suppose now that ∑

j∈I0

ϕj(p) < 1.

In this case, there exists some Rh with h ∈ I0 such that p ∈ 3Rh. By Lemma 7.21,

R � (Rh) ≈ dist(0, Rh).

7.6. E and Γ are close to each other 255

Moreover, if Ri is the interval that contains p, then

(Ri) � |p| � R.

Indeed, the first estimate follows from Lemma 7.20 (a), and the second one becausep = Π(x) ∈ Π(B) with B ∈ VG.

As (Ri) ≈ (Rh), we infer that (Rh) ≈ R. Also, this implies that (Rj) ≈ Rfor any j ∈ I0 such that p ∈ 3Rj , and so LBj is very close to L0. In fact, fromLemma 7.11 applied to the measure μ�3Bj and to B1 = B2 = B(0, cR), with c > 1such that Bj ⊂ B(0, cR), if follows that

distH(LBj ∩B(0, cR), L0 ∩B(0, cR)

) ≤ c εR. (7.29)

On the other hand, arguing as in (7.28), one deduces that (Rj) � r(B), and sor(B) ≈ R too (because we assume r(B) ≤ 50R). By (7.29) then we get

|Aj(p)| = dist((p,Aj(p)), L0

) ≤ c εR ≤ c ε r(B)

for all these j’s. Thus

dist((p,A(p)), L0

) ≤∑j

ϕj(p) |Aj(p)| ≤ c ε r(B).

Due to the fact that r(B) ≈ R, L0 is also very close to LB (by Lemma 7.11), andthus,

dist((p,A(p)), LB

) ≤ c ε r(B). �Lemma 7.30. We have

dist(x,Γ) ≤ c(δ) ε d(x) for all x ∈ E. (7.30)

Proof. Recall first that if d(x) = 0, then x ∈ Γ, and so (7.30) holds. Given x ∈ Ewith d(x) > 0, take a ball B0 = B(z, r) ∈ VG such that |x − z| + r ≤ 1.1d(x).Consider the ball B′

0 = B(z, r′), with

r′ = min(50R, 2(c5 + 2)d(x)

),

where c5 first appeared in Lemma 7.28. It is easy to check that B′0 ∈ VG and

x ∈ 13B

′0. Also,

12B

′0 ∩ Γ = ∅. This is clear if r′ = 50R, and otherwise, if r′ =

2(c5 + 2)d(x), this follows from

dist(z,Γ) ≤ |x− z|+ dist(x,Γ) ≤ 1.1.d(x) + c5 d(x) = (c5 + 1.1)d(x).

The preceding lemma tells us that any y ∈ Γ ∩B′0 satisfies

dist(y, LB′0) ≤ c(δ) ε r(B′

0).

In other words,Γ ∩B′

0 ⊂ Uc(δ) ε r(B′0)(LB′

0).


From this fact, taking into account that x ∈ 13B

′0 and 1

2B′0 ∩ Γ = ∅, we infer that

dist(x,Γ) ≤ dist(x, LB′0) + c(δ) ε r(B′

0).

Since β(B′0) ≤ c(δ)ε, we get

dist(x,Γ) ≤ c′(δ) ε r(B′0) ≤ c′′(δ) ε d(x). �

Remark 7.31. From the preceding lemma it follows easily that for each i ∈ I0,

dist(Bi,Γ ∩ Π−1(Ri)) ≤ c (Ri). (7.31)

Indeed, if Bi = B(zi, ri), then

dist(zi,Γ) ≤ c(δ)εd(zi) ≤ c(δ)ε ri ≤ c(δ)ε (Ri) ≤ (Ri).

Also, recall that dist(Π(Bi), Ri) ≤ c(Ri). From these two inequalities, (7.31)follows.

Lemma 7.32. We have

dist(x, L0) ≤ c(δ)εR for all x ∈ Γ. (7.32)

Proof. Since suppA ⊂ [−12R, 12R], (7.32) clearly holds for x ∈ Γ such that|Π(x)| > 12R.

To deal with the case when |Π(x)| ≤ 12R, first we claim that

Γ ∩ Π−1[−12R, 12R] ⊂ B(0, 30R)

(of course, the number 30 is not optimal here). Indeed, this follows from the factthat A(−12R) = A(12R) = 0 and ‖A′‖∞ ≤ c α ≤ 1/10, for α small enough.

So if x ∈ Γ and |Π(x)| ≤ 12R, then x ∈ B(0, 30R). Thus, by Lemma 7.29applied to B = B(z0, 50R) for some z0 ∈ E,

dist(x, L0) ≤ c(δ) εR. �

7.7 The set E1 is small

In the previous section we constructed a Lipschitz function A with slope ≤ cαwhose graph Γ contains the set Z defined above. So to prove Theorem 7.4 it isenough to show that μ(Z) ≥ 99

100 μ(E) or, equivalently, that

μ(E1) + μ(E2) ≤ 1

100μ(E).

In the current section we show that μ(E1) is small.

7.7. The set E1 is small 257

Lemma 7.33. Suppose that δ is taken small enough (for instance, δ = 10−10) andthat ε and α are chosen small enough too (as explained in Remark 7.12), then

μ(E1) ≤ 1

1000μ(E).

Proof. Loosely speaking, to prove the lemma we will take into account that eachpoint of E1 is the center of a ball with very small density and that this family ofballs is almost aligned along the Lipschitz graph Γ.

Let us go through the details. For each x ∈ E1, we have

μ(B(x, h(x))

) ≤ δ h(x).

On the other hand, B(x, λh(x)) ∈ VG for all λ > 1 such that λh(x) ≤ 50R, andso d(x) ≤ h(x). Then, by Lemma 7.30,

dist(x,Γ) ≤ c(δ)ε d(x) ≤ c(δ)ε h(x) ≤ 1

4h(x).

As a consequence, Γ ∩B(x, h(x)/2) = ∅ and since Γ is a connected set,

H1(Γ ∩B(x, h(x))

) ≥ h(x).

By the Besicovitch covering theorem, there exists a subfamily of points {xj}jsuch that the balls {B(xj , h(xj))}j are pairwise disjoint and, moreover,

μ(E1) ≤ q2∑j

μ(B(xj , h(xj))

),

where q2 is an absolute constant (see Theorem 2.3). Then we have

μ(E1) ≤ q2δ∑j

h(xj) ≤ q2δ∑j

H1(Γ ∩B(xj , h(xj))

)= q2δH1

(Γ ∩⋃j

B(xj , h(xj))).

Since h(x) ≤ 2R for all x ∈ E, we deduce that B(xj , h(xj)) ⊂ B(0, 3R), and so

μ(E1) ≤ q2 δH1(Γ ∩B(0, 3R)

) ≤ q2δ 10R,

taking into account that ‖A′‖∞ ≤ 1/10 by Remark 7.25. Thus,

μ(E1) ≤ q2δ 20μ(E),

and the lemma follows. �


7.8 The set E2 is small

To show that μ(E2) is small will require a somewhat bigger effort than in the caseof E1. Roughly speaking, the strategy will be the following:

E2 big ⇒ ‖A′‖2 big ⇒ c2(H1�Γ) big ⇒ c2(μ) big,

which contradicts the fact that c2(μ) is small.

7.8.1 The implications E2 big ⇒ ‖A′‖2 big ⇒ c2(H1�Γ) bigWe need the following auxiliary result.

Lemma 7.34. Let B1 = B(x1, r1), B2 = B(x2, r2) ⊂ C be pairwise disjoint balls.Suppose that �(Lx1,x2 , L0) ≤ 1

100 . Then Π(12 B1

) ∩ Π(12 B2

)= ∅.

Proof. This follows by elementary geometry. �Lemma 7.35. If ε is taken small enough, then

μ(E2) ≤ c α−2‖A′‖22.Proof. For every x ∈ E2, take the ball B(x, h(x)). By the 5r-covering theorem,there exists a family of balls B(xi, 20h(xi)), i ∈ J , which covers E2 and such that

the balls B(xi, 4h(xi)) are pairwise disjoint. Write Bi := B(xi, 2h(xi)) and recallthat B(xi, 2h(xi)) ∈ VG. Then, by Lemma 7.30,

dist(xi,Γ) ≤ c(δ) ε d(xi) ≤ c(δ) ε h(xi) <1

2h(xi).

Thus Γ ∩ 14 Bi = ∅. So there exist y1, y2 ∈ Bi ∩ Γ such that |Π(y1) − Π(y2)| ≥

c−1h(xi). By Lemma 7.29,

dist(y, LBi) ≤ c(δ) ε r(Bi) for all y ∈ Γ ∩ Bi.

In particular, this holds for y1 and y2. So if we denote by y′1, y′2 the orthogonalprojection onto LBi

of y1, y2, respectively, we have

|y1 − y′1|+ |y2 − y′2| ≤ c(δ) ε r(Bi) = c(δ) ε h(xi).

Recall that, by Lemma 7.13, �(LBi, L0) ≥ α/2. Assuming ε small enough

this implies that

|A(Π(y1))−A(Π(y2))| = |Π⊥(y1)−Π⊥(y2)|≥ |Π⊥(y′1)−Π⊥(y′2)| − |y1 − y′1| − |y2 − y′2|≥ 1

2α |Π(y′1)−Π(y′2)| − |y1 − y′1| − |y2 − y′2|

≥ 1

2α |Π(y1)−Π(y2)| − 2|y1 − y′1| − 2|y2 − y′2|

≥ c−1αh(xi)− c(δ) ε h(xi) ≥ c−1αh(xi).


Thus, ∫Π(Bi)

|A′(p)|dp ≥∣∣∣∣∣∫ Π(y2)

Π(y1)

A′(p) dp

∣∣∣∣∣= |A(Π(y1))−A(Π(y2))| ≥ c−1αh(xi).

Therefore,

‖A′‖22,Π(Bi)

≥ c−1α2 h(xi).

As the balls 2Bi, i ∈ J , are pairwise disjoint, by Lemma 7.34 we deduce that theintervals Π(Bi) ⊂ L0, i ∈ J , are also pairwise disjoint. Then we have

μ(E2) ≤∑i∈J

μ(B(xi, 20h(xi))) ≤ 20c0∑i∈J

h(xi)

≤ c α−2∑i∈J

‖A′‖22,Π(Bi)

≤ c α−2‖A′‖22. �

Recall now that by Lemma 3.9 we have c2(H1�Γ) ≈ ‖A′‖22, with absoluteconstants, taking into account that ‖A′‖∞ ≤ c α ≤ 1/10. So we have

μ(E2) ≤ c α−2c2(H1�Γ). (7.33)

7.8.2 The implication c2(H1�Γ) big ⇒ c2(μ) big

This is the most laborious part of the argument to show that μ(E2) is small.For x ∈ C such that Π(x) ∈ Z0, we denote by Rx the interval Ri, i ∈ I, which

contains Π(x), and we set x = (Rx). If Π(x) ∈ Π(Z0), we write Rx = {Π(x)}and we set x = 0. In a sense, one should think that the latter case correspondsto the one where the intervals Rx have side length equal to zero.

To simplify notation, in this subsection we write Π(x) = x1 and Π⊥(x) = x2.

Localization

Recall that the intervals Ri, i ∈ I0, are the ones from Ri, i ∈ I, that intersectB(0, 10R). Observe that if p ∈ L0 ∩ B(0, 10R), then D(p) � R. Thus, (Ri) � Rfor all i ∈ I0. Thus, setting

Γ0 := Z0 ∪⋃i∈I0

Γ ∩ Π−1(Ri),

we deduce that Γ0 ⊂ B(0, cR), for some absolute constant c. Moreover, Γ0 is aconnected subset of Γ. That is,

Γ0 = Γ ∩Π−1(R0),


where R0 ⊂ L0 is some interval. We also set

Γext := Γ \ Γ0.

First we will show that the part of the curvature of c2(H1�Γ) that involvesΓext is very small.

Lemma 7.36. We have

c2(H1�Γext,H1�Γ,H1�Γ) ≤ c(δ)ε1/2μ(E).

To simplify notation, below sometimes we will write H1F instead of H1�F ,

for F ⊂ C.

Proof. Since |x1| ≥ 10R for all x ∈ Γext, we have

c2(H1�Γext,H1�Γ,H1�Γ) ≤∫∫∫

|x1|≥10R

c(x, y, z)2 dH1Γ(x) dH1

Γ(y) dH1Γ(z)

(7.34)

≤ 6

∫∫∫|x1|≥10R|x1|≥|y1|≥|z1|


Γ(y) dH1Γ(z),

where the last inequality follows easily by symmetry. Indeed, writing “. . . ” insteadof “c(x, y, z)2 dH1

Γ(x) dH1Γ(y) dH1

Γ(z)”, we have∫∫∫|x1|≥10R

. . . =

∫∫∫|x1|≥10R|y1|≥10R|z1|≥10R

. . .+ 2

∫∫∫|x1|≥10R|y1|≥10R|z1|<10R

. . .+

∫∫∫|x1|≥10R|y1|<10R|z1|<10R

. . .

≤ 6

∫∫∫|x1|≥|y1|≥|z1|≥10R

. . .+ 4

∫∫∫|x1|≥|y1|≥10R>|z1|

. . .+ 2

∫∫∫|x1|≥10R>|y1|≥|z1|

. . .

Clearly, this yields (7.34).Notice now that if |x1| ≥ 10R, since dist(Π−1({x1}), E) ≈ |x1|, we get

(Rx) ≈ D(x1) ≈ |x1|, by Lemma 7.20 (a). This will be used below.In order to simplify notation, we denote by D the domain of integration of

the integral on the right-hand side of (7.34). That is,

D ={(x, y, z) ∈ Γ3 : |x1| ≥ 10R and |x1| ≥ |y1| ≥ |z1|

}.

To estimate that integral, we distinguish several cases for (x, y, z) ∈ D:

1) Suppose first that |y1| > 14R.This implies that |x1| ≥ 14R in the domain of integration. Since supp(A) ⊂[−12R, 12R], we deduce that x, y ∈ L0 and thus Lx,y = L0. If |z1| ≥ 12R, thenz ∈ L0 too, and c(x, y, z) = 0. If |z1| ≤ 12R, taking into account that dist(z, L0) ≤c(δ)εR because of Lemma 7.32, we have

c(x, y, z) =2 dist(z, Lx,y)

|z − x| |z − y| ≤c(δ)εR

|z − x| |z − y| ≤c(δ)εR

|x| |y| .


Thus, by Lemma 2.11, using the linear growth of H1Γ,∫∫∫

(x,y,z)∈D|y1|>14R


Γ(y) dH1Γ(z) (7.35)

≤∫∫∫

|x1|≥|y1|≥14R|z1|≤12R

c(δ)ε2R2

|x|2 |y|2 dH1Γ(x) dH1

Γ(y) dH1Γ(z)

≤ c(δ)ε2R3

∫∫|x1|≥14R|y1|≥14R

1

|x|2 |y|2 dH1Γ(x) dH1

Γ(y) ≤ c(δ)ε2R.

2) Suppose that |y1| ≤ 14R, z1 ∈ 3Rx, and y1 ∈ 3Rx.In this case, we have |x1|+ |z1| ≤ cR because, using Lemma 7.20 (a),

|x1| ≈ D(x1) ≈ (Rx) ≈ D(y1) � |y1| ≤ 14R. (7.36)

Also, we deduce that (Rx) ≈ R. Recall that in Remark 3.8 it was shown that ifJ ⊂ L0 is an interval and x, y, z lie in the graph of J � t �→ A(t), then∣∣∣∣∣∣∣∣

A(z1)−A(x1)

z1 − x1− A(y1)−A(x1)

y1 − x1

z1 − y1

∣∣∣∣∣∣∣∣ ≤ ‖A′′‖∞,J , (7.37)

and alsoc(x, y, z) ≤ 2 ‖A′′‖∞,J .

By Lemma 7.27 we have

‖A′′‖∞,3Rx ≤ c(δ) ε

(Rx)≈ c(δ) ε

R. (7.38)

Then we infer that

c(x, y, z) ≤ c(δ)ε

R, (7.39)

and so ∫∫∫(x,y,z)∈D|y1|≤14Ry1∈3Rxz1∈3Rx


Γ(y) dH1Γ(z) (7.40)

≤∫∫∫

|x1|≤cR|y1|≤cR|z1|≤cR

c(δ)ε2

R2dH1

Γ(x) dH1Γ(y) dH1

Γ(z) ≤ c(δ)ε2R.

3) Suppose that |y1| ≤ 14R, z1 ∈ 3Rx, and y1 ∈ 3Rx.Arguing as in (7.36), the fact that z1 ∈ 3Rx and |z1| ≤ 14R, implies that |x1| ≈ R,which also yields |z1| ≈ R. Moreover, the sign of x1 equals the one of z1 and is


different from the one of y1, because otherwise the inequalities |x1| ≥ |y1| ≥ |z1|would imply that y1 ∈ 3Rx. Therefore, |y − x| ≈ |y − z| ≈ R.

By Lemma 7.32, dist(x, L0)+dist(y, L0)≤c(δ)εR, and so we have �(L0, Lx,y)≤ c(δ)ε. We claim that

�(L0, Lx,z) ≤ c(δ)ε (7.41)

too. This follows easily if |x − z| ≈ (Rx), taking into account that dist(x, L0) +dist(z, L0) ≤ c(δ)εR and that (Rx) � R. Otherwise, consider z′ ∈ Γ such thatz′1 ∈ Rx and |x− z′| = (Rx)/4, so that �(L0, Lx,z′) ≤ c(δ)ε. From (7.37) appliedto x, z′, z, we infer that∣∣tan�(L0, Lx,z)− tan�(L0, Lx,z′)

∣∣ = ∣∣∣∣A(z1)−A(x1)

z1 − x1− A(z′1)−A(x1)

z′1 − x1

∣∣∣∣≤ |z1 − z′1| ‖A′′‖∞,3Rx

≤ c (Rx) ‖A′′‖∞,3Rx .

Thus, taking into account that ‖A′′‖∞,3Rx ≤ c(δ) ε(Rx)

, we deduce that

�(L0, Lx,z) ≤∣∣tan�(L0, Lx,z)

∣∣≤ ∣∣tan�(L0, Lx,z′)

∣∣+ c (Rx) ‖A′′‖∞,3Rx

≤ ∣∣tan�(L0, Lx,z′)∣∣+ c(δ) ε ≤ c′(δ) ε,

which proves (7.41).We deduce that

sin yxz ≈ �(Lx,y, Lx,z) ≤ �(L0, Lx,y) + �(L0, Lx,z) ≤ c(δ)ε,

and so

c(x, y, z) =2 sin yxz

|y − z| � c(δ)ε

R. (7.42)

Now an estimate such as (7.40) also holds and thus∫∫∫(x,y,z)∈D|y1|≤14Ry1 �∈3Rxz1∈3Rx


Γ(y) dH1Γ(z) ≤ c(δ)ε2R. (7.43)

4) Suppose that |y1| ≤ 14R, z1 ∈ 3Rx, and y1 ∈ 2Rx.Recalling that |z1| ≤ |y1|, in this case we also have |x1|+ |z1| ≤ cR, and moreover|y − z| ≈ R, as (Rx) ≈ R.

In this situation, the estimates of the preceding case interchanging y by zhold, and so one also gets c(x, y, z) ≤ c(δ)ε/R. Indeed, since |x − z| � R anddist(x, L0) + dist(z, L0) ≤ c(δ)εR, we deduce that �(L0, Lx,z) ≤ c(δ)ε. Arguingas above, we also obtain �(L0, Lx,y) ≤ c(δ)ε, because y ∈ 2Rx. So the estimate(7.42) holds. Finally, one obtains an inequality analogous to (7.43).


5) Suppose that |y1| ≤ 14R, z1 ∈ 3Rx, and y1 ∈ 2Rx.We have |y1|+ |z1| ≤ cR, although x1 may be arbitrarily big. In this situation,

|x− y| ≈ |x− z| ≈ |x|. (7.44)

Let us estimate the Menger curvature c(x, y, z) assuming first that |y− z| >ε1/2R. From (7.44) we deduce that

�(L0, Lx,y) ≤ c(δ)εR

|x| and �(L0, Lx,z) ≤ c(δ)εR

|x| .

Thus, yxz ≤ �(L0, Lx,y) + �(L0, Lx,z) ≤ 2 c(δ)εR/|x|, and

c(x, y, z) =2 sin yxz

|y − z| ≤ c(δ)εR

|x| |y − z| ≤c(δ)ε1/2

|x| .

In the case |y − z| ≤ ε1/2R, we use that c(x, y, z) ≤ 2/|x− y| ≈ 1/|x|. Thenwe get∫∫∫

(x,y,z)∈D|y1|≤14Ry1 �∈2Rx

z1 �∈3Rx


Γ(y) dH1Γ(z)

≤∫∫∫

|x|≥10R|y|≤cR|z|≤cR

|y−z|>ε1/2R

c(δ)ε

|x|2 dH1Γ(x) dH1

Γ(y) dH1Γ(z)

+

∫∫∫|x|≥10R|y|≤cR

|y−z|≤ε1/2R

c

|x|2 dH1Γ(x) dH1

Γ(y) dH1Γ(z)

≤∫|x|≥10R

c(δ)εR2

|x|2 dH1Γ(x) +

∫|x|≥10R

c(δ)ε1/2R2

|x|2 dH1Γ(x) ≤ c(δ)ε1/2R.

Finally, gathering all the results obtained for the different cases 1),. . . ,5), weobtain

c2(H1�Γext,H1�Γ,H1�Γ) ≤ c(δ)ε1/2R ≤ c(δ)ε1/2μ(E). �

Very close, close, and far pairs of points

In the next subsections we will estimate the curvature of H1�Γ0. To simplifynotation, we set σ := H1�Γ0.

Recall that, for x ∈ C, we set x = (Rx). We say that two points x, y ∈ Γare very close, and we set (x, y) ∈ VC if |x1 − y1| ≤ x + y. We say that they areclose, and we set (x, y) ∈ Cl, if |x1 − y1| ≤ ε−1/10(x + y). Finally, they are far,and we write (x, y) ∈ Far, if |x1 − y1| > ε−1/10(x + y). Notice that the relationsof very close, close, and far points are symmetric with respect to x and y.


Given (x, y, z) ∈ Γ3, there are three possibilities: either two of the points inthe triple are very close, or no pair of points is very close but there is at least onepair that is close, or all the pairs of points are far. So we can split the curvatureof σ as follows:

c2(σ) ≤ 3

∫∫∫(x,y)∈VC

c(x, y, z)2 dσ(x) dσ(y) dσ(z)

+ 3

∫∫∫(x,y)∈Cl\VC(x,z) �∈VC(y,z) �∈VC


+

∫∫∫(x,y)∈Far(x,z)∈Far(y,z)∈Far


=: c2VC(σ) + c2Cl\VC(σ) + c2Far(σ).

Now we need to prove a technical lemma. Abusing notation, below we writep := (0,p) for all p ∈ L0 ≡ R.

Lemma 7.37. For each p, q ∈ L0, we have

q ≤ 5

2p +

1

20|p− q|.

If x, y ∈ Γ are very close, that is, (x, y) ∈ VC, then

|x1 − y1| ≤ 4 min(x, y).

Proof. Suppose first that p > 0 and let Ri be such that p ∈ Ri. Recall that, bydefinition,

1

40infs∈Ri

D(s) < (Ri) ≤ 1

20infs∈Ri

D(s),

where Ri is the father of Ri. Since D : L0 → R is a 1-Lipschitz function, we inferthat

D(p) ≤ (Ri) + infs∈Ri

D(s) ≤ 42 (Ri).

Using again that D is 1-Lipschitz,

D(q) ≤ D(p) + |p− q| ≤ 42 (Ri) + |p− q|.

Therefore, by the definition of q,

q ≤ 1

20D(q) ≤ 42

20(Ri) +

1

20|p− q|,

and so the first claim in the lemma follows in this case.


If p = 0, the arguments are analogous and even easier:

q ≤ 1

20D(q) ≤ 1

20D(p) +

1

20|p− q| = 1

20|p− q|.

The second statement in the lemma is a straightforward consequence of thefirst one: if x, y ∈ Γ are very close, then

|x1 − y1| ≤ x + y ≤ x +5

2x +

1

20|x1 − y1| = 7

2x +

1

20|x1 − y1|,

which implies |x1 − y1| ≤ 4x. Analogously, one gets |x1 − y1| ≤ 4y. �

The curvature c2VC(σ) is very small

Lemma 7.38. We havec2VC(σ) ≤ c(δ)ε1/2 μ(E).

Proof. Recall that

c2VC(σ) = 3

∫∫∫|x1−y1|≤x+y

c(x, y, z)2 dσ(x) dσ(y) dσ(z).

By Lemma 7.37, for x, y, z in the domain of integration, we have |x1 − y1| ≤ 4x,and so we can assume that x > 0. We distinguish several cases:

1) Suppose first that |x1 − z1| ≤ 5x.In this case x1, y1, z1 ∈ 11Rx. By Lemma 7.27,

‖A′′‖∞,11Rx ≤ c(δ) ε

x, (7.45)

and then from Remark 3.8 we infer that c(x, y, z) ≤ c(δ) ε/x,. Thus∫∫∫|x1−y1|≤x+y|x1−z1|≤5x


≤∫∫∫

|x1−y1|≤4x|x1−z1|≤5x

c(δ)ε2

2xdσ(x) dσ(y) dσ(z) ≤ c(δ)ε2σ(C).

2) Suppose now that 5x < |x1 − z1| ≤ ε−1/2x.For x ∈ suppσ = Γ0, let Ri, i ∈ I0, be such that Π(x) ∈ Ri. Consider theassociated ball Bi ∈ VG given by Lemma 7.22. Let us check that �(Lx,y, LBi) ≤c(δ)ε. This follows easily arguing as in (7.41). Indeed, let x′ ∈ Γ be a point veryclose to x and different from x, so that Π(x′) = x′

1 ∈ Ri. From (7.37) applied tox, y, x′ and (7.45) we infer that∣∣tan�(L0, Lx,x′)− tan�(L0, Lx,y)

∣∣ = ∣∣∣∣A(x′1)−A(x1)

x′1 − x1

− A(y1)−A(x1)

y1 − x1

∣∣∣∣≤ |x′

1 − y1| ‖A′′‖∞,9Rx ≤ c(δ) ε.


Letting x′ → x, we obtain∣∣A′(x1)− tan�(L0, Lx,y)∣∣ ≤ c(δ) ε.

Since |A′(x1)−A′i| ≤ c(δ) ε (see Remark 7.26 and recall that the parametrization

of LBi is given by x2 = Ai(x1)), we derive∣∣A′i − tan�(L0, Lx,y)

∣∣ ≤ c(δ) ε,

which implies that �(Lx,y, LBi) ≤ c(δ)ε, as wished.Now let B′

i be the ball concentric with Bi with radius

r(B′i) = min(15R, c6|x− z|),

where c6 ≥ 1 is an absolute constant big enough to guarantee that Bi ⊂ B′i and

x, y, z ∈ B′i (to find c6, recall that |x1 − z1| > 5x). We claim that

�(LBi , LB′i) ≤ c(δ)ε r(B′

i)

r(Bi)≤ c(δ)ε1/2.

The first inequality follows easily from Lemma 7.14, by taking two points fromLBi ∩ 2Bi with mutual distance comparable to r(Bi). The second one comes fromthe fact that r(B′

i) ≤ cε−1/2r(Bi).On the other hand, recalling that r(B′

i) = min(15R, c6|x − z|) and that, byLemma 7.29,

dist(x, LB′

i

)+ dist(z, LB′

i

) ≤ c(δ)ε r(B′i),

we deduce that �(LB′i, Lx,z) ≤ c(δ)ε. Therefore,

�(Lx,y, Lx,z) ≤ �(Lx,y, LBi) + �(LBi , LB′i) + �(LB′

i, Lx,z)

≤ c(δ)ε+ c(δ)ε1/2 + c(δ)ε ≤ c(δ)ε1/2.

So we have

c(x, y, z) =2 sin(yxz)

|y − z| ≤ c(δ)ε1/2

x,

because |y1 − z1| ≥ |x1 − z1| − |x1 − y1| ≥ 5x − 4x = x. So we obtain∫∫∫|x1−y1|≤x+y

5x<|x1−z1|≤ε−1/2x


≤∫∫∫

|x1−y1|≤4x|x1−z1|≤ε−1/2x

c(δ)ε

2xdσ(x) dσ(y) dσ(z).

The last integral is bounded by c(δ)ε1/2σ(C), as

σ({z : |x1 − z1| ≤ ε−1/2x}

)� ε−1/2x.


3) In the last case, we assume that |x1 − z1| > ε−1/2x.We just use the trivial inequality c(x, y, z) ≤ 2/|x−z|, so that by the linear growthof σ, ∫

|x1−z1|>ε−1/2x

4

|x− z|2 dσ(z) ≤ c ε1/2

x.

Then we get∫∫∫|x1−y1|≤x+y

|x1−z1|>ε−1/2x


≤∫∫∫

|x1−y1|≤4x|x1−z1|>ε−1/2x

4

|x− z|2 dσ(x) dσ(y) dσ(z) ≤ c ε1/2σ(C).

Finally, gathering all the results obtained for the cases 1), 2), 3), and noticingthat σ(C) = H1(Γ0) ≈ μ(E), we are done. �

The curvature c2Cl\VC(σ) is very small

Lemma 7.39. We have

c2Cl\VC(σ) ≤ c(δ)ε1/10 μ(E).

Proof. Recall that

c2Cl\VC(σ) = 3

∫∫∫(x,y)∈Cl\VC(x,z) �∈VC(y,z) �∈VC

c(x, y, z)2 dσ(x) dσ(y) dσ(z),

where (x, y) ∈ Cl means that |x1 − y1| ≤ ε−1/10(x + y), and (x, y) ∈ VC that|x1 − y1| > x + y. By the symmetry on x, y of the preceding triple integral, it isenough to consider the case x ≥ y. For such x, y, z in the domain of integration,we consider several cases:

1) Suppose first that |y1 − z1| ≤ ε1/5x.In this case we appeal to the trivial estimate c(x, y, z) ≤ 2/|x − y|, and then weobtain ∫∫∫

(x,y)∈Cl\VC(x,z),(y,z) �∈VC

|y1−z1|≤ε1/5xx≥y


≤∫∫∫

|x1−y1|>x|y1−z1|≤ε1/5x

4

|x− y|2 dσ(x) dσ(y) dσ(z)

≤∫∫

|x1−y1|>x

c ε1/5x|x− y|2 dσ(x) dσ(y)

≤∫

c ε1/5 dσ(x) = c ε1/5 σ(C).


2) Suppose that |x1 − z1| > ε−1/5x.Now we just use the inequality c(x, y, z) ≤ 2/|x−z|. Observe also that the condition|x1 − y1| ≤ ε−1/10(x + y) together with the fact that x ≥ y implies that|x1 − y1| ≤ 2 ε−1/10x. Applying Lemma 2.11 we derive∫∫∫

(x,y)∈Cl\VC(x,z),(y,z) �∈VC

|x1−z1|>ε−1/5xx≥y


≤∫∫∫

|x1−y1|≤2ε−1/10x|x1−z1|>ε−1/5x

4

|x− z|2 dσ(x) dσ(y) dσ(z)

≤∫∫

|x1−y1|≤2ε−1/10x

c ε1/5

xdσ(y) dσ(x)

≤∫

c ε1/5ε−1/10 dσ(x) = c ε1/10 σ(C).

3) Suppose finally that |y1 − z1| > ε1/5x and |x1 − z1| ≤ ε−1/5x.We claim that

c(x, y, z) ≤ c(δ)ε3/5

|x− y| . (7.46)

To prove this estimate, recall that there exists a ball Bi ∈ VG, i ∈ I0, such thatr(Bi)+dist(x,Bi) ≤ c x. Let B be a concentric ball which contains Bi and x, y, z,with radius

r(B) ≈ max(|x− z|, |x− y|),so that moreover, r(B) ≤ 50R.

Assume first that |x− z| ≥ |x− y|. By Lemma 7.29, dist(x, LB) ≤ c(δ)ε r(B)and dist(z, LB) ≤ c(δ)ε r(B), and thus

�(Lx,z, LB) ≤ c(δ)ε r(B)

|x− z| ≤ c(δ)ε.

Taking into account that dist(y, LB) ≤ c(δ)εr(B) too, we get

�(Ly,z, LB) ≤ c(δ)ε r(B)

|y − z| ≈ c(δ)ε |x− z||y − z| ≤ c(δ)ε ε−1/5x

ε1/5x= c(δ)ε3/5.

So, sin xzy ≤ c(δ)ε3/5 and

c(x, y, z) =2 sin(xzy)

|x− y| ≤ c(δ)ε3/5

|x− y| .

Suppose now that |x−z| < |x−y|, so that r(B) ≈ |x−y| ≤ 2 ε−1/10 x. Fromthe inequalities dist(x, LB) ≤ c(δ)ε r(B) and dist(y, LB) ≤ c(δ)ε r(B), we deducethat

�(Lx,y, LB) ≤ c(δ)ε r(B)

|x− y| ≤ c(δ)ε.


Recalling also that dist(z, LB) ≤ c(δ)ε r(B) and that |x−z| ≥ x (because (x, z) ∈VC), we get

�(Lx,z, LB) ≤ c(δ)ε r(B)

|x− z| ≤ c(δ)ε ε−1/10xx

= c(δ)ε9/10.

Therefore, sin yxz ≤ c(δ)ε9/10. Since

|y − z| > ε1/5x ≥ c−1ε1/5 ε1/10 |x− y| = c−1ε3/10 |x− y|,

we get

c(x, y, z) =2 sin(yxz)

|y − z| ≤ c(δ)ε9/10

ε3/10|x− y| =c(δ)ε3/5

|x− y| .

So (7.46) holds in the present situation too.

With (7.46) at hand, we obtain∫∫∫(x,y)∈Cl\VC(x,z),(y,z) �∈VC

|y1−z1|>ε1/5x|x1−z1|≤ε−1/5xx≥y


≤∫∫∫

|x1−y1|>x|x1−z1|≤ε−1/5x

c(δ)ε6/5

|x− y|2 dσ(x) dσ(y) dσ(z)

≤∫∫

|x1−y1|>x

c(δ)ε x|x− y|2 dσ(y) dσ(x) ≤ c(δ) ε σ(C).

The lemma follows from the estimates of the cases 1), 2), 3) above, sinceσ(C) ≈ μ(E). �

Estimation of c2Far(σ)

In this subsection we will obtain the following.

Lemma 7.40. We have

c2Far(σ) ≤ c(δ) c2(μ) + c(δ)ε1/5 μ(E).

To prove this lemma we will compare the curvature c2Far(σ) to (part of) thecurvature of μ. To this end, first we need to approximate the measure σ = H1�Γ0

by another measure absolutely continuous with respect to μ, of the form g μ, withg ∈ L∞(μ). This task is carried out in the next lemma.


Lemma 7.41. For each i ∈ I0 there exists some function gi ∈ L∞(μ), gi ≥ 0supported on Bi such that∫

gi dμ = H1(Γ ∩ Π−1(Ri)) = σ(Π−1(Ri)), (7.47)

and ∑i∈I0

gi � c(δ), (7.48)

Proof. The arguments are somewhat similar to the ones for the Calderon-Zygmunddecomposition of Lemma 2.14.

We assume first that the family {Ri}i∈I0 is finite. We also suppose that(Ri) ≤ (Ri+1) for all i ∈ I0. The functions gi that we will construct will be of theform gi = αiχAi , with αi ≥ 0 and Ai ⊂ Bi (recall the construction of the balls Bi

in Lemma 7.22). We set α1 := σ(Π−1(R1))/μ(B1) and A1 := suppμ∩B1, so that∫g1dμ = σ(Π−1(R1)). Notice by the way that ‖g1‖∞ = σ(Π−1(R1))/μ(B1) ≤ c(δ),

since B1 ∈ VG.To define gk, k ≥ 2, we argue by induction. Suppose that g1, . . . , gk−1 have

been constructed, satisfy (7.47), and∑k−1

i=1 gi ≤ b, where b is some constant de-pending on δ which will be chosen below. Let Bs1 , . . . , Bsm be the subfamily ofB1, . . . , Bk−1 such that Bsj ∩Bk = ∅. Since (Rsj ) ≤ (Rk) (because of the non-decreasing sizes of (Ri), i ∈ I0), we deduce that dist(Rsj , Rk) ≤ c (Rk), and thuswe have Rsj ⊂ c7Rk, for some absolute constant c7. Using (7.47) for i = sj , weget∑

j

∫gsj dμ =

∑j

σ(Π−1(Rsj )) ≤ σ(Π−1(c7Rk)) ≤ c r(Bk) ≤ c8(δ)μ(Bk).

Therefore,

μ({∑

j

gsj > 2c8(δ)})

≤ 1

2μ(Bk).

So we setAk := suppμ ∩Bk ∩

{∑j

gsj ≤ 2 c8(δ)},

and then μ(Ak) ≥ μ(Bk)/2. Also, we put αk :=σ(Π−1(Rk))

μ(Ak), so that gk = αkχAk

satisfies∫gk dμ = σ(Π−1(Rk)). Then we have

αk ≤ 2σ(Π−1(Rk))

μ(Bk)≤ c r(Bk)

μ(Bk)≤ c9(δ).

Thus,

gk +∑j

gsj ≤ 2c8(δ) + c9(δ).


We choose b := 2c8(δ) + c9(δ) and (7.48) follows. Of course, this bound is inde-pendent of the number of functions.

Suppose now that {Ri}i∈I0 is not finite. For each fixed N we consider afamily of intervals {Ri}1≤i≤N . As above, we construct functions gN1 , . . . , gNN withsupp(gNi ) ⊂ Bi satisfying∫

gNi dμ = σ(Π−1(Ri)) and

N∑i=1

gNi ≤ b.

Then there is a subsequence {gk1}k∈I1 which is convergent in the weak ∗ topology ofL∞(μ) to some function g1 ∈ L∞(μ). Now we take another convergent subsequence{gk2}k∈I2 , I2 ⊂ I1, in the weak ∗ topology of L∞(μ) to another function g2 ∈L∞(μ), etc. We have supp(gi) ⊂ Bi. Further, (7.47) and (7.48) also hold, becauseof the weak ∗ convergence. �

We recall now the following technical result, which was proved in Subsection4.6.1 of Chapter 4.

Lemma 7.42. Let x, y, z ∈ C be three pairwise different points, and let x′ ∈ C besuch that

a−1|x− y| ≤ |x′ − y| ≤ a|x− y|, (7.49)

where a > 0 is some constant. Then

|c(x, y, z)− c(x′, y, z)| ≤ (4 + 2a)|x− x′|

|x− y||x− z| . (7.50)

Proof of Lemma 7.40. To simplify notation, we write Ri = Γ ∩ Π−1(Ri). Recallthat

Γ0 = Z0 ∪⋃i∈I0

Ri.

In Lemma 7.41 we showed how H1�Ri can be approximated by a measure sup-ported on Bi, for each i ∈ I0. Notice that, by Remark 7.31,

dist(Bi, Ri) ≤ c(Ri). (7.51)

Concerning H1�Z0, recall that Z0 ⊂ Γ, and moreover by Lemma 7.18,

μ�Z0 = ρZ0 H1�Z0,

where ρZ0 is some function such that c−1δ ≤ ρZ0 ≤ c0 H1-a.e. on Z0. So,

H1�Z0 =1

ρZ0

μ�Z0,

and the function gZ0 := 1/ρZ0 is bounded uniformly by some constant dependingon δ.


Now we consider the measures νi = gi μ, for i ∈ I0, with gi as in Lemma7.41, and we set

ν = gZ0 μ�Z0 +∑i∈I0

νi.

This new measure should be understood as an approximation of σ = H1�Γ0, whichcoincides with σ�Z0 = H1�Z0 on Z0. We will show below that

c2Far(σ) ≤ 4c2(ν) + c(δ)ε1/5μ(E). (7.52)

Observe that, by Lemma 7.41 and the discussion above on gZ0 , ν is of the formν = g μ, where g is some function uniformly bounded by c(δ). As a consequence,

c2(ν) ≤ c(δ) c2(μ),

and so the lemma follows from (7.52).To prove (7.52), first we need to introduce some additional notation. Given

Borel measures τ1, τ2, τ3, we set

c2(τ1, τ2, τ3) =

∫∫∫c2(x, y, z) dτ1(x) dτ2(y) dτ3(z).

On the other hand, this triple integral restricted to those (x, y, z) such that

|x1 − y1| ≥ ε−1/10(x + y), |x1 − z1| ≥ ε−1/10(x + z),

and |y1 − z1| ≥ ε−1/10(y + z), (7.53)

is denoted by c2Far(τ1, τ2, τ3). So we have

c2Far(σ) = c2Far(σ�Z0) + c2Far(σ�Γ0 \ Z0) (7.54)

+ 3 c2Far(σ�Z0, σ�Γ0 \ Z0, σ�Γ0 \ Z0) + 3 c2Far(σ�Z0, σ�Z0, σ�Γ0 \ Z0).

The first term on the right-hand side is very easy to deal with. Indeed,

c2Far(σ�Z0) = c2Far(H1�Z0) ≤ c(δ) c2(μ�Z0).

To study c2Far(σ�Γ0 \ Z0) we set

c2Far(σ�Γ0 \ Z0) =∑

i,j,k∈I0

c2Far(σ�Ri, σ�Rj , σ�Rk

).

By definition, if c2Far(σ�Ri, σ�Rj , σ�Rk

) = 0, then there exists x ∈ Ri, y ∈ Rj and

z ∈ Rk satisfying (7.53). Then it follows easily that

dist(Ri, Rj) ≥ε−1/10

2

((Ri) + (Rj)

), dist(Ri, Rk) ≥ ε−1/10

2

((Ri) + (Rk)

),

(7.55)

and dist(Rj , Rk) ≥ ε−1/10

2

((Rj) + (Rk)

).


We denote by JFar the set of those indices (i, j, k) ∈ I30 such that these inequalitieshold, so that

c2Far(σ�Γ0 \ Z0) ≤∑

(i,j,k)∈JFar

c2(σ�Ri, σ�Rj , σ�Rk

).

Consider (i, j, k) ∈ JFar and x, x′ ∈ Ri ∪Bi, y, y′ ∈ Rj ∪Bj, and z, z′ ∈ Rk ∪

Bk. Because of (7.55) and (7.51), taking into account that (Rh) ≈ diam(Rh) ≈diam(Bh) for each h ∈ I, the sets Ri ∪ Bi, Rj ∪ Bj , and Rk ∪Bk are far to eachother for ε small enough, and then by applying Lemma 7.42 three times, we deduce

c(x, y, z)2 ≤ 2 c(x′, y′, z′)2

+c 2x

|x− y|2 |x− z|2 +c 2y

|y − x|2 |y − z|2 +c 2z

|z − x|2 |z − y|2 .

We set

Tx(y, z) =2x

|x− y|2 |x− z|2 ,

and analogously, Ty(x, z), Tz(x, y). Then, integrating on x ∈ Ri, y ∈ Rj , and

z ∈ Rk with respect to σ, we get

c2(σ�Ri,σ�Rj , σ�Rk

) ≤ 2 c(x′, y′, z′)2 σ(Ri)σ(Rj)σ(Rk)

+ c

∫∫∫x∈Ri

y∈Rj

z∈Rk

[Tx(y, z) + Ty(x, z) + Tz(x, y)

]dσ(x) dσ(y) dσ(z).

On the other hand, by analogous arguments, we have

c(x′, y′, z′)2 νi(Ri) νj(Rj) νk(Rk) ≤ 2 c2(νi, νj , νk)

+ c

∫∫∫ [Tx(y, z) + Ty(x, z) + Tz(x, y)

]dνi(x) dνj(y) dνk(z).

Taking into account that νh(Rh) = σ(Rh) for all h, from the preceding two in-equalities, summing on (i, j, k) ∈ JFar, we deduce

c2Far(σ�Γ0 \ Z0) ≤ 4 c2(ν) (7.56)

+ c

∫∫∫|x−y|≥ 1

2 ε−1/10(x+y)

|x−z|≥ 12 ε

−1/10(x+z)

|y−z|≥ 12 ε

−1/10(y+z)

[Tx(y, z) + Ty(x, z) + Tz(x, y)

]dσ(x) dσ(y) dσ(z)

+ c

∫∫∫|x−y|≥ 1

2 ε−1/10(x+y)

|x−z|≥ 12 ε

−1/10(x+z)

|y−z|≥ 12 ε

−1/10(y+z)

[Tx(y, z) + Ty(x, z) + Tz(x, y)

]dν(x) dν(y) dν(z).


To estimate the first triple integral on the right-hand side, notice that∫∫|x−y|≥ 1

2 ε−1/10(x+y)

|x−z|≥ 12 ε

−1/10(x+z)

Tx(y, z) dσ(y) dσ(z) (7.57)

≤(∫

|x−y|≥ 12 ε

−1/10x

x|x− y|2 dσ(y)

)(∫|x−z|≥1

2 ε−1/10x

x|x− z|2 dσ(z)

)

=

(∫|x−y|≥ 1

2 ε−1/10x

x|x− y|2 dσ(y)

)2

≤ c ε1/5,

where the last inequality follows from Lemma 2.11, using the linear growth ofσ. Analogous estimates hold permuting x, y, z, and also interchanging σ by ν,recalling that ν has linear growth with some constant depending on δ. Pluggingthem into (7.56) we get

c2Far(σ�Γ0 \ Z0) ≤ 4 c2(ν) + c(δ) ε1/5 μ(E). (7.58)

Now it remains to estimate the last two terms of (7.54). The arguments aresimilar to the preceding ones. Since σ�Z0 = ν�Z0, we have

c2Far(σ�Z0, σ�Γ0 \ Z0, σ�Γ0 \ Z0) = c2Far(ν�Z0, σ�Γ0 \ Z0, σ�Γ0 \ Z0)

andc2Far(σ�Z0, σ�Z0, σ�Γ0 \ Z0) = c2Far(ν�Z0, ν�Z0, σ�Γ0 \ Z0).

Concerning the term c2Far(σ�Z0, σ�Γ0 \ Z0, σ�Γ0 \ Z0), the main difference withrespect to the estimates above for c2Far(σ�Γ0 \ Z0) is that Tx(y, z) equals zero in

this case (since x = x′), and instead of integrating over σ�Ri and νi and thensumming on i, one integrates over σ�Z0. Then one obtains

c2Far(σ�Z0, σ�Γ0 \ Z0, σ�Γ0 \ Z0) ≤ 4 c2(ν) (7.59)

+ c

∫∫∫|x−y|≥1

2 ε−1/10y

|x−z|≥ 12 ε

−1/10z

|y−z|≥ 12 ε

−1/10(y+z)

[Ty(x, z) + Tz(x, y)

]dσ(x) dσ(y) dσ(z)

+ c

∫∫∫|x−y|≥1

2 ε−1/10y

|x−z|≥ 12 ε

−1/10z

|y−z|≥ 12 ε

−1/10(y+z)

[Ty(x, z) + Tz(x, y)

]dν(x) dν(y) dν(z).

The last two triple integrals are estimated as in (7.57), and then it follows that

c2Far(σ�Z0, σ�Γ0 \ Z0, σ�Γ0 \ Z0) ≤ 4 c2(ν) + c(δ) ε1/5 μ(E). (7.60)

Finally, the arguments for c2Far(σ�Z0, σ�Z0, σ�Γ0 \ Z0) are very similar. Inthis case, both terms Tx(y, z) and Ty(x, z) vanish, and analogously we also get

c2Far(σ�Z0, σ�Z0, σ�Γ0 \ Z0) ≤ 4 c2(ν) + c(δ) ε1/5 μ(E). (7.61)


The details are left for the reader.Adding the estimates obtained in (7.58), (7.60), and (7.61), the lemma fol-

lows. �

Summary

Gathering the previous results, we obtain:

Lemma 7.43. We have

c2(H1�Γ) ≤ c(δ) c2(μ) + c(δ)ε1/10 μ(E).

Proof. We have

c2(H1�Γ) ≤ c2(H1�Γ0) + 3 c2(H1�Γext,H1�Γ,H1�Γ)≤ c2VC(σ) + c2Cl\VC(σ) + c2Far(σ) + 3 c2(H1�Γext,H1�Γ,H1�Γ). (7.62)

Now we just recall that, by Lemma 7.36,

c2(H1�Γext,H1�Γ,H1�Γ) ≤ c(δ)ε1/2μ(E);

by Lemma 7.38,c2VC(σ) ≤ c(δ)ε1/2 μ(E);

by Lemma 7.39,c2Cl\VC(σ) ≤ c(δ)ε1/10 μ(E);

and by Lemma 7.40,

c2Far(σ) ≤ c(δ) c2(μ) + c(δ)ε1/5 μ(E).

Plugging these estimates into (7.62), the lemma follows. �We can now prove that μ(E2) is small and conclude the proof of Theorem 7.4:

Lemma 7.44. If ε and η are chosen small enough, then

μ(E2) ≤ 1

1000μ(E).

Proof. Recalling that, by (7.33),

μ(E2) ≤ c α−2c2(H1�Γ),

from Lemma 7.43 we get

μ(E2) ≤ c(δ)α−2 c2(μ) + c(δ)α−2ε1/10 μ(E).

Thus we are done, assuming ε and η small enough. �


7.9 Three applications

7.9.1 A characterization of rectifiable sets in terms of pointwisecurvature

Theorem 7.45. Let E ⊂ C be an H1-measurable set with H1(E) < ∞. Then E isrectifiable if and only if

c2H1�E(x) < ∞ for H1-a.e. x ∈ E.

Proof. Suppose that c2H1�E(x) < ∞ for H1-a.e. x ∈ E. For every positive integerm, let

Em = {x ∈ E : c2H1�E(x) ≤ m}.Then c2(H1�Em) ≤ mH1(E), and so by Theorem 7.1, Em is rectifiable. SinceE coincides with

⋃m Em up to a set of H1 measure zero, it turns out that E is

rectifiable.Let now E be an arbitrary H1-measurable rectifiable set with finite length.

Then E ⊂ Z ∪⋃i Γi, where H1(Z) = 0 and Γi, i ≥ 1, are pieces of finite lengthof possibly rotated Lipschitz graphs, as shown in Proposition 4.13. It is enough toshow that c2H1�E(x) < ∞ for H1-a.e. x ∈ Γi.

Since c2(H1�Γi) < ∞, for every ε > 0 there exist M > 0 and a compactsubset F ⊂ Γi such that H1(Γi \ F ) ≤ ε and c2H1�E(x) ≤ M for all x ∈ F . Thus,

c2H1�F (x) ≤ M for all x ∈ F .

From the maximum principle in Lemma 4.20, we infer that

c2H1�F (x) ≤ M ′ for all x ∈ C,

for some constant M ′ depending on M and the linear growth of H1�F . Integratingon E, we deduce that

c2(H1�F,H1�F,H1�E) ≤ M ′ H1(E).

Now, it follows that there exist some M ′′ > 0 and a compact subset F0 ⊂ F suchthat H1(F \ F0) ≤ ε and c2(x,H1�F,H1�E) ≤ M ′′ for all x ∈ F0, and thus

c2(x,H1�F0,H1�E) ≤ M ′′ for all x ∈ F0.

Again by the maximum principle in Lemma 4.20,

c2(x,H1�F0,H1�E) ≤ M ′′′ for all x ∈ C,

with M ′′′ depending on M ′′ and the linear growth of H1�F0. Integrating on E,

c2(H1�F0,H1�E,H1�E) ≤ M ′′′ H1(E),

7.9. Three applications 277

which implies that

c2(x,H1�E,H1�E) < ∞ for H1-a.e. x ∈ F0.

Since H1(Γi \ F0) ≤ 2ε and ε > 0 is arbitrary, we get the desired conclusion. �An alternative argument to show that if E is an H1-measurable rectifiable

set with finite length, then c2H1�E(x) < ∞ for H1-a.e. x ∈ C is the following. Recallthat, by Theorem 4.23,

γ+({x ∈ C : Uμ(x) > λ}) ≤ c‖μ‖λ

,

for any arbitrary Radon measure μ and for λ > 0. In particular,

γ+({x ∈ C : c2H1�E(x) = ∞}) = 0.

Since any subset of positive length of E has positive capacity γ+, it follows that

H1({x ∈ E : c2H1�E(x) = ∞}) = 0.

7.9.2 A quantitative version of David’s theorem

We will prove the following result.

Theorem 7.46. Consider a Borel set E ⊂ C with H1(E) < ∞, and write E =Erectif∪Eunrectif, where Erectif is a Borel rectifiable set and Eunrectif a Borel purelyunrectifiable set. Then

γ(E) ≤ H1(Erectif).

Notice first that a straightforward application of David’s theorem and thesemiadditivity of analytic capacity yields

γ(E) ≤ c[γ(Erectif) + γ(Eunrectif)

]= c γ(Erectif) ≤ cH1(Erectif),

for some absolute constant c. Proving the result with c = 1 requires more carefularguments. We need the following auxiliary lemma.

Lemma 7.47. Consider E ⊂ C such that H1(E) < ∞, with E = Erectif ∪Eunrectif,where Erectif is a Borel rectifiable set and Eunrectif is Borel and purely unrectifiable.Let b ∈ L∞(H1�E) be a complex function such that |C(bH1�E)(z)| ≤ 1 for allz ∈ E. Then b(z) = 0 for H1-a.e. z ∈ Eunrectif.

Proof. Consider an arbitrary λ > 0. Let z ∈ Eunrectif be a density point both ofχEunrectif

and b, with respect to the measure H1�E, and suppose that |b(z)| ≥ λ.Let ϕ be a C∞ radial decreasing function supported on B(0, 2), which equals 1 onB(0, 1). For r > 0, let

ϕz,r(x) = ϕ

(x− z

r

).


Then|C(ϕz,r bH1�E)(x)| ≤ c for all x ∈ E,

by Proposition 1.17. Therefore,

γ(E ∩B(z, 2r)) ≥ c−1

∣∣∣∣∫ ϕz,r b dH1�E∣∣∣∣ .

So from the semiadditivity of analytic capacity and David’s theorem, we get∣∣∣∣∫ ϕz,r b dH1�E∣∣∣∣ ≤ c[γ(Erectif ∩B(z, 2r)) + γ(Eunrectif ∩B(z, 2r))

](7.63)

= c γ(Erectif ∩B(z, 2r)) ≤ cH1(Erectif ∩B(z, 2r)).

Since z is a density point of χEunrectif, for any ε > 0,

H1(Erectif ∩B(z, 2r)) ≤ εH1(E ∩B(z, 2r)) (7.64)

for every r > 0 small enough.On the other hand, since z is a density point for b and ϕ is a radially de-

creasing nice function,

limr→0

(∫ϕz,r dH1�E

)−1 ∫ϕz,r b dH1�E = b(z).

Therefore, for r > 0 small enough,∣∣∣∣∫ ϕz,r b dH1�E∣∣∣∣ > 1

2|b(z)|∫

ϕz,r dH1�E ≥ λ

2H1(E ∩B(z, r)). (7.65)

From (7.63), (7.64) and (7.65), we deduce that

λ

2H1(E ∩B(z, r)) < εH1(E ∩B(z, 2r))

for every r small enough. In other words, the ball B(z, r) is not (2, λ/(2ε))- dou-bling. However, if ε is chosen big enough (depending on λ), this may happen onlyfor a set of zero H1�E-measure, by Lemma 2.8. Therefore,

H1({z ∈ Eunrectif : |b(z)| ≥ λ}) = 0 for all λ > 0,

which proves the lemma. �To prove Theorem 7.46 we invoke Proposition 6.5, which tells us that if E ⊂ C

is compact with H1(E) < ∞, and f : C \ E → C is analytic with ‖f‖∞ ≤ 1 andf(∞) = 0, then there is a function b ∈ L∞(H1�E), with |b(z)| ≤ 1 for all z ∈ E,such that f(z) = C(bH1�E)(z) for all z ∈ E. By the preceding lemma, b vanisheson Eunrectif and thus

|f ′(∞)| =∣∣∣∣∫ b dH1�E

∣∣∣∣ ≤ ∫Erectif

|b| dH1 ≤ H1(Erectif).

7.9. Three applications 279

7.9.3 Characterization of γ in terms of measures with boundedupper density

Recall that for a compact set E ⊂ C,

γ(E) ≈ sup{μ(E) : μ(B(x, r)) ≤ r ∀x ∈ C, ∀r > 0, and c2(μ) ≤ μ(E)

}. (7.66)

It is clear that the linear growth condition cannot be completely dispensed with.Otherwise, this would imply that the analytic capacity of a point is positive.However, the linear growth condition can be weakened, as the following theoremshows.

Theorem 7.48. Let E ⊂ C be compact. Then

γ(E) ≈ supμ

μ(E),

where the supremum is taken over all the Radon measures μ such that c2(μ) ≤μ(E), and whose upper 1-dimensional upper density satisfies:

Θ∗1(x, μ) = lim supr→0

μ(B(x, r))

2r≤ 1 for μ-a.e. x ∈ E.

The main tool to prove this theorem is the next proposition, which may havesome independent interest.

Proposition 7.49. Let μ be some Radon measure supported on B(x0, R) such thatΘ∗1(x, μ) ≤ 1 for μ-a.e. x ∈ C. If c2(μ) ≤ a μ(B(x0, R)), then μ(B(x0, R)) ≤ AR,where A is some constant depending only on a.

Proof. Suppose that μ(B(x0, R)) = AR for some big constant A > 0. Let

E = {x ∈ B(x0, R) : c2μ(x) ≤ 2a}.By Chebyshev, μ(E) ≥ μ(B(x0, R))/2. We set

En = {x ∈ E : μ(B(x, r)) ≤ 3r if 0 < r ≤ 1/n}.Since μ(E \ ⋃∞

n=1 En) = 0, we can choose n big enough so that μ(En) ≥ ‖μ‖/3.We write ν = μ|En

.We claim that there exists some ball B(x1, R1), with x1 ∈ B(x0, R), such

that

ν(B(x1, R1)) ≥ A

100R1 (7.67)

andν(B(x, r)) ≤ 100Ar for all x ∈ C and 0 < r ≤ R1. (7.68)

Indeed, let Q0 be the square concentric with B(x0, R) and side length 4R. LetD(Q0) be the collection of dyadic squares generated by Q0 which are contained


in Q0. That is, D(Q0) =⋃∞

k=0 Dk(Q0), where Dk(Q0) stands for the family of 22k

squares with disjoint interiors and side length 2−k(Q0) contained in Q0.For each square Q ∈ D(Q0) with side length ≤ 1/10n we have ν(Q) ≤

10(Q) < A20 (Q). On the other hand, ν(Q0) ≥ A

10 (Q0). So there exists some

square Q1 ∈ D(Q0) such that ν(Q1) ≥ A20 ((Q1) whose side length is minimal. If

we choose x1 as the center of Q1 and we set R1 = (Q1), then B(x1, R1) fulfillsthe required properties.

Now we consider the measure σ = 200A ν�B(x1, R1). Notice that suppσ ⊂

B(x1, 2R1), ‖σ‖ ≥ 2R1, σ(B(x, r)) ≤ 20000 r for all x ∈ C, and

c2σ(x) ≤40000

A2c2ν(x) ≤

40000

A2c2μ(x) ≤

80000 a

A2

for σ-a.e. x ∈ C. By Theorem 7.4, if

A � a1/2, (7.69)

then there exists a Lipschitz graph Γ with slope smaller than 1/10 such thatσ(Γ) ≥ 99

100 ‖σ‖. Therefore,

μ(Γ ∩B(x1, R1)) ≥ 99

100μ(B(x1, R1) ∩ En) ≥ 99

10000AR1.

Since Θ∗1(x, μ) ≤ 1 μ-a.e., arguing as in (7.11) we obtain

μ(Γ ∩B(x1, R1)) ≤ H1(Γ ∩B(x1, R1)) ≤ 10R1.

Thus, A � 1.So we have shown that either (7.69) fails or A � a1/2. In other words,

A � max(1, a1/2). �

Proof of Theorem 7.48. Let

S = sup{μ(E) : Θ∗1(x, μ) ≤ 1 for μ-a.e. x ∈ E, and c2(μ) ≤ μ(E)

}.

By the characterization (7.66) of γ(E), it is clear that S � γ(E). To prove theconverse inequality, consider a measure μ supported on E such that Θ∗1(x, μ) ≤ 1for μ-a.e. x ∈ E, and c2(μ) ≤ μ(E). Let

F = {x ∈ E : c2μ(x) ≤ 2},so that μ(F ) ≥ μ(E)/2. Then the measure σ = μ�F satisfies

c2(σ�A) ≤ 2σ(A),

for any measurable set A ⊂ C. From Lemma 7.49 we infer that σ has c0-lineargrowth for some absolute constant c0. Thus, by (7.66) we have

γ(E) � σ(E) ≥ 1

2μ(E).

Taking the supremum on μ, we get γ(E) � S. �

7.10. Curvature, β’s, and rectifiability in the AD regular case 281

7.10 Curvature, β’s, and rectifiability in the AD regularcase

In this section we will show how in the particular case that E ⊂ C is AD regular(1-dimensional) with c2(H1�E) < ∞ the rectifiability of E can be deduced fromthe Jones’ traveling salesman Theorem 3.14. To this end, we will need to estimatethe β coefficients of E in terms of the curvature c2(H1�E).

Recall that

βE(Q) = infV

w(V )

(Q),

where the infimum is taken over all infinite strips V that contain E ∩ 3Q andw(V ) denotes the width of V . Recall also that in Theorem 3.15 and Remark 3.18we showed that if μ is a 1-dimensional AD regular measure, then

c2(μ) ≤ c∑Q∈D

βE(Q)2μ(Q),

where E = suppμ. In the next theorem we prove the converse estimate.

Theorem 7.50. Let μ be a 1-dimensional AD regular measure and set E = suppμ.There exist some constant C depending only on the AD regularity constant of μsuch that, for any square R ∈ D with (R) ≤ 2 diamE, we have∑

Q∈D:Q⊂R

βE(Q)2(Q) ≤ C c2(μ�30R). (7.70)

Also, ∑Q∈D

βE(Q)2(Q) ≤ C c2(μ). (7.71)

To prove the preceding result it is convenient to work with the βE,p coeffi-cients. Given 1 ≤ p < ∞ and a square Q ⊂ C, we define

βE,p(Q) = infL

(1

(Q)

∫E∩3Q

(dist(x, L)

(Q)

)pdμ(x)

)1/p,

where the infimum is taken over all the lines L ⊂ C. We also write βE,∞ ≡ βE .The coefficients βE,p, for 1 ≤ p < ∞, were introduced by David and Semmes [28]in their pioneering study of the so-called uniform rectifiability.

In the next lemma we collect some easy properties of the βp’s.

Lemma 7.51. Let μ be a 1-dimensional AD regular measure and let E = suppμ.The following properties hold:


(a) If Q ⊂ R are squares such that (Q) ≈ (R), then

βE,p(Q) ≤ c βE,p(R), (7.72)

with c depending on the comparability constant. Moreover, if Q ∩ E = ∅,(Q) ≤ 2 diam(E), and LQ and LR are lines that minimize βE,p(Q) andβE,p(R), respectively, then

distH(LQ ∩ 5R, LR ∩ 5R) ≤ c βE,p(R) (R). (7.73)

(b) Let Q ⊂ C be a square. If 1 ≤ p1 ≤ p2 ≤ ∞, then

βE,p1(Q) ≤ c βE,p2(Q). (7.74)

Moreover, if Q ∩ E = ∅, (Q) ≤ 2 diam(E), and and LQ,p1 and LQ,p2 arelines that minimize βE,p1(Q) and βE,p2(Q), respectively, then

distH(LQ,p1 ∩ 5Q, LQ,p2 ∩ 5Q) ≤ c βE,p2(Q) (Q). (7.75)

(c) Let DE be the family of dyadic squares that intersect E with side length notexceeding 2 diam(E). Then, for all R ∈ D,∑Q∈D:Q⊂R

βE,p(Q)2(Q) ≤ c(p)∑

Q∈DE :Q⊂3R

βE,p(Q)2(Q) for all 1 ≤ p ≤ ∞.

All the statements in this lemma are elementary and we leave the proof forthe reader.

The converse inequality to the one stated in (7.74) fails. Nevertheless, thefollowing weaker statement holds.

Theorem 7.52. Let μ be a 1-dimensional AD regular measure and let E = suppμ.Also, let DE be the family of dyadic squares that intersect E with side length notexceeding 2 diam(E). Let 1 ≤ p ≤ ∞. For any square R ∈ DE, the following holds:∑

Q∈DE :Q⊂R

βE,∞(Q)2 (Q) ≤ c∑

Q∈DE :Q⊂3R

βE,p(Q)2(Q).

For the proof we need the next key lemma.

Lemma 7.53. Let μ be a 1-dimensional AD regular measure and let E = suppμ.Consider a square R ∈ DE and let LR be a line which minimizes βE,1(R). For anyx ∈ 3R ∩ E, the following holds:

dist(x, LR) ≤ c∑

Q∈D:z∈Q⊂3R

βE,1(Q) (Q), (7.76)

where c depends on the AD regularity constants of μ.


Proof. Let n0 ≥ 1 be some integer to be fixed below, and consider the sequenceof dyadic squares R = Q0 ⊃ Q1 ⊃ Q2 . . . such that x ∈ Qm for each m ≥ 1 and(Qm) = 2−mn0(R). Let ε0 be some (small) constant that will be fixed belowtoo. Let N ≥ 0 be the least integer such that βE,1(QN ) ≥ ε0. If N does notexist because βE,1(Qm) < ε0 for all m ≥ 0, we let N be an arbitrary positiveinteger. Let aN be any point from QN and for m = N − 1, N − 2, . . . , 0 let am bethe orthogonal projection of am+1 onto a line that minimizes βE,1(Qm), that wedenote by LQm . Then we have

dist(aN , LR) ≤ dist(aN , LQN ) +

N−1∑m=0

|am − am+1|. (7.77)

Our next objective consists in showing that

|am − am+1| � βE,1(Qm)(Qm) for m = 0, 1, . . . , N − 1. (7.78)

Let us see first that (7.76) follows from this estimate. Indeed, from (7.77) and(7.78) we infer that

dist(aN , LR) � dist(aN , LQN ) +N−1∑m=0

βE,1(Qm)(Qm).

If βE,1(Qm) < ε0 for all m ≥ 0 (in this case N was chosen as an arbitrarypositive integer), we let N → ∞ in the preceding inequality, and then aN → xand dist(aN , LQN ) → 0, and so (7.76) follows. If βE,1(QN ) ≥ ε0, then

|x− aN |+ dist(aN , LQN ) � (QN ) ≤ ε−10 βE,1(QN )(QN).

Thus, by (7.77) and (7.78),

dist(x, LR) ≤ |x− aN |+ dist(aN , LQN ) +N−1∑m=0

|am − am+1|

� ε−10

N∑m=0

βE,1(Qm)(Qm).

To prove (7.78) we wish to use the estimate (7.73) in (a) from Lemma 7.51.Then we need to show first that am ∈ 5Qm for m = N, N − 1, . . . , 1. We argueby backward induction. Indeed, for m = N , this holds by the definition of aN .Assume now that am+1 ∈ 5Qm+1 and let us see that am ∈ 5Qm. Remember thatfor m = N − 1, N − 2, . . . , 1, we have βE,1(Qm) ≤ ε0. By the AD regularity of μ,all points y ∈ Qm+1 ⊂ Qm satisfy

dist(y, LQm) ≤ C(ε0)(Qm) ≤ (Qm+1)/2,


assuming that ε0 has been taken small enough (depending also on the choice ofn0). So we infer that LQm ∩5Qm+1 = ∅. Recall that, by the induction hypothesis,am+1 ∈ 5Qm+1. If n0 has been chosen big enough we deduce that

|am+1 − am| = dist(am+1, LQm) ≤ diam(5Qm+1) ≤ (Qm). (7.79)

Also, for n0 big enough again, 5Qm+1 ⊂ 2Qm, and thus am+1 ∈ 2Qm. This factand (7.79) imply that am ∈ 5Qm.

The estimate (7.78) follows now easily from (7.73) using the fact that am,am+1 ∈ 5Qm:

|am − am+1| ≤ distH(LQm ∩ 5Qm, LQm+1 ∩ 5Qm) � βE,1(Qm) (Qm). �Proof of Theorem 7.52. Since βE,1(Q) ≤ c βE,p(E) for any p > 1, it suffices toprove the theorem for p = 1.

Consider a square Q ∈ DE . Using the estimate (7.75) in Lemma 7.51 andapplying Lemma 7.53 to a point x = uQ ∈ 3Q such that dist(uQ, LQ,∞) ≥12 βE,∞(Q) (Q), where LQ,∞ minimizes βE,∞(Q), we deduce that

βE,∞(Q) (Q) ≤ c∑

P∈D:uQ∈P⊂3Q

βE,1(P ) (P ).

Therefore, by the Cauchy-Schwarz inequality,

βE,∞(Q)2 (Q)2 ≤ c

( ∑P∈D:uQ∈P⊂3Q

βE,1(P )2 (P )3/2

)( ∑P∈D:uQ∈P⊂3Q

(P )1/2

)≤ c

∑P∈D:uQ∈P⊂3Q

βE,1(P )2 (P )3/2 (Q)1/2.

As a consequence,∑Q∈DE :Q⊂R

βE,∞(Q)2 (Q) ≤ c∑

Q∈DE:Q⊂R

∑P∈DE :P⊂3Q

βE,1(P )2(P )3/2

(Q)1/2

≤ c∑

P∈DE:P⊂3R

βE,1(P )2 (P )3/2∑

Q∈DE :3Q⊃P

1

(Q)1/2

≤ c∑

P∈DE:P⊂3R

βE,1(P )2 (P ). �

Proof of Theorem 7.50. The estimate (7.71) in the theorem is a consequence of(7.70). To prove (7.70) for any R ∈ D with (R) ≤ 2 diam(E) note that, byTheorem 7.52 and (b),(c) in Lemma 7.51,∑

Q∈D:Q⊂R

βE,∞(Q)2(Q) ≤c∑

Q∈DE :Q⊂3R

βE,∞(Q)2(Q)

≤c∑

Q∈DE :Q⊂7R

βE,2(Q)2(Q).


To estimate the last sum consider a square Q ∈ DE contained in 7R. Note that(Q) ≤ 4(R) ≤ 4 diam(E) and 3Q ⊂ 15R and 5Q ⊂ 30R. Also, for all x, y ∈E ∩ 5Q,

βE,2(Q)2(Q)3 ≤∫z∈3Q

dist(z, Lx,y)2 dμ(z).

Thus,

βE,2(Q)2(Q) ≤ c (Q)2∫z∈3Q

dist(z, Lx,y)2

|z − x|2 |z − y|2 dμ(z)

= c (Q)2∫z∈3Q

c(x, y, z)2 dμ(z).

Denote by c0 the AD regularity constant of μ, and for x ∈ E ∩ 3Q consider theopen annulus

Ax,Q = A(x,

1

8c20(Q),

1

4(Q)).

Notice that it is contained in 5Q and moreover

μ(Ax,Q

) ≥ 1

4c0(Q)− 1

8c0(Q) =

1

8c0(Q).

Taking also into account that μ(3Q) ≥ c−1 (Q) (because Q ∩ E = ∅), we obtain

βE,2(Q)2(Q) ≤ C(Q)2

μ(3Q)

∫∫x∈3Qz∈3Q

1

μ(Ax,Q)

∫y∈Ax,Q

c(x, y, z)2 dμ(x) dμ(y) dμ(z)

≤ C

∫∫∫x∈3Qy∈Ax,Q

z∈3Q


Therefore,∑Q∈DE :Q⊂7R

βE,2(Q)2(Q)

≤ C∑k≥−2

∑Q∈DE :Q⊂7R

(Q)=2−k(R)

∫∫∫x∈3Qy∈Ax,Q

z∈3Q


≤ C∑k≥−2

∫∫∫x∈15Rc−20 2−k−3(R)<|x−y|<2−k−2(R)

z∈15R


≤ C c2(μ�30R),

as wished. Note that in the second inequality we used the finite superposition ofthe squares 3Q for Q such that (Q) = 2−k(R), while in the third one the finitesuperposition of the sets

{y ∈ C : c−20 2−k−3(R) < |x− y| < 2−k−2(R)},


for each x ∈ 15R. �

From Jones’ traveling salesman Theorem 3.14 and Theorem 7.50 we deducethe following.

Corollary 7.54. Let μ be a 1-dimensional AD regular measure. If c2(μ) < ∞, thensupμ is rectifiable.

Remark 7.55. From the estimate (7.70), we deduce that if the Cauchy transformis bounded in L2(μ) and μ is 1-dimensional AD regular, then∑

Q∈D:Q⊂R

βE(Q)2(Q) ≤ C(R)

for any dyadic square R. Jones [75] showed that this implies that μ is supportedon an AD regular curve. That is, in the language of David and Semmes [28],μ is uniformly rectifiable. This nice result was proved by Mattila, Melnikov andVerdera [103], relying on some techniques from David and Semmes [28] instead ofJones’ traveling salesman theorem.


7.11.1 About the results in this chapter

The proof above of the David-Leger theorem is based on Leger’s paper [83]. How-ever, the arguments in Section 7.8 for the proof of the smallness of the key set E2

are different from the ones by Leger [83]. In turn, the arguments in this paper canbe considered as a (highly non-trivial) adaptation to the non-homogeneous settingof the David and Semmes [28] arguments for the uniformly rectifiable sets in theAD regular situation.

I learned the quantitative version of David’s theorem stated in Theorem 7.46from Dudziak’s book [35]. However, the arguments in the proof above are differentfrom Dudziak’s. On the other hand, Theorem 7.48 is from Tolsa [159]. Theorem7.50 was proved by Jones, although he did not publish it and instead it appearedin Pajot’s book [132]. The analogous estimate to (7.70) with the coefficients βE

replaced by βE,2 had been proved previously by Mattila, Melnikov and Verdera[103]. The arguments to prove Theorem 7.50 in Section 7.10 are based on therelationship between the coefficients βE and βE,2 in Theorem 7.52 and on ideasfrom Mattila, Melnikov and Verdera [103].

7.11.2 Rectifiability, uniform rectifiability and Riesz transforms inarbitrary dimensions

David and Semmes extended in [28] and [29] the techniques of quantitative rectifi-ability introduced by Jones in [74] and [75] to AD regular measures in dimensions


greater than 1 and developed a deep theory of the so-called uniform rectifiability.An n-dimensional AD regular measure μ in Rd is said to be uniformly n-rectifiable(or simply, uniformly rectifiable) if there exist θ,M > 0 so that, for each x ∈ suppμand 0 < r ≤ diam(suppμ), there is a Lipschitz mapping g from the n-dimensionalball Bn(0, r) ⊂ Rn to Rd such that g has Lipschitz norm ≤ M and

μ(B(x, r) ∩ g(Bn(0, r))

) ≥ θ rn.

In the language of David and Semmes [29], this means that suppμ has big piecesof Lipschitz images of Rn. In the case n = 1, this turns out to be equivalent tosaying that μ is supported on an AD regular curve.

The setting of uniformly rectifiable measures is a natural framework to studythe action of n-dimensional singular integrals with odd kernels. Indeed, David andSemmes [28] showed that an n-dimensional regular measure in Rd is uniformly n-rectifiable if and only if all n-dimensional singular integral operators Tμ associatedwith an odd kernel smooth enough are bounded in L2(μ). They asked if the L2(μ)boundedness of a single operator, namely the n-dimensional Riesz transform, orthe Cauchy transform for n = 1, suffices for the uniform n-rectifiability of μ.This question is now called the David-Semmes problem. As mentioned in Remark7.55, for n = 1 this was answered in the affirmative by Mattila, Melnikov andVerdera [103]. The problem for n > 1 is more difficult because there is not a niceformula such as the one of Melnikov and Verdera (3.5) which connects the geometryof the measure to the kernel of the n-dimensional Riesz transform, and so the“curvature method” does not work. Nevertheless, in codimension 1, i.e. n = d− 1,quite recently the solution of the problem has been completed by Nazarov, Tolsaand Volberg [121]. Their methods rely on some quasiorthogonality techniques,variational estimates, and a deep characterization of uniform rectifiability by Davidand Semmes [29]. For n different from 1 and d− 1 the problem is still open.

By combining the results from Nazarov, Tolsa and Volberg [121] with othersby Eiderman, Nazarov and Volberg [37] for measures with vanishing lower n-dimensional density, Nazarov, Tolsa and Volberg [122] have proved the following.

Theorem 7.56. Let E ⊂ Rn+1 be a set such that Hn(E) < ∞. If the Riesz trans-form Rn

Hn�E is bounded in L2(Hn�E), then E is n-rectifiable.

A corollary of this result and Volberg’s theorem about the comparabilityκ ≈ κ+ (see (6.52)) concerns the generalization of Vitushkin’s conjecture to higherdimensions, for sets with finite n-dimensional Hausdorff measure:

Theorem 7.57. Let E ⊂ Rn+1 be a compact set with Hn(E) < ∞. Then E isremovable for Lipschitz harmonic functions in Rn+1 if and only if E is purelyn-unrectifiable.

Recall that a subset E ⊂ Rn+1 is said to be removable for Lipschitz harmonicfunctions if, for each open set Ω ⊂ Rn+1, every Lipschitz function f : Ω → R thatis harmonic in Ω \ E is harmonic in the whole Ω.


7.11.3 Other results in connection with curvature

The results above dealing with rectifiability, β’s, and curvature have been extendedin different directions. For example, in Tolsa [159] a very different proof of The-orem 7.1 has been obtained, relying on the use of the sharp inequality (4.40). In[84], Lerman obtained a version of Jones’ traveling salesman Theorem 3.14 whichinvolves very general Borel measures μ on Rd and some appropriate L2(μ) versionsof Jones’ β’s. Ferrari, Franchi and Pajot [45] have extended the “if” part of theJones’ Theorem 3.14 to the Heisenberg group. Schul [147] proved a version of thesame theorem which is valid for Hilbert spaces. On the other hand, Hahlomaa [59]has obtained a version of the David-Leger Theorem 7.1 suitable for metric spaces.In the case of AD regular metric spaces, the relationship between curvature ofmeasures and rectifiability has been explored by Hahlomaa [58] and Schul [146].Lerman and Whitehouse have studied some analogs of curvature in higher dimen-sions in [86] and [85] in connection with rectifiability. Other energy type integralsinvolving curvature have been investigated by Strzelecki, Szumanska and von derMosel [150] and by Strzelecki and von der Mosel [151]. A different proof of the factthat the L2 boundedness of the Cauchy transform implies uniform rectifiability,in the setting of AD regular measures, has been obtained recently by Jaye andNazarov [71] relying the study of the reflectionless measures for the Cauchy kernel.See Section 8.5.2 for the notion of reflectionless measures.

It is also worth mentioning that Chousionis, Mateu, Prat and Tolsa [10] haveproved that, for μ = H1�E with H1(E) < ∞, the L2(μ) boundedness of thesingular integral operator associated with any kernel Kn as in (3.25), with respectto the measure μ, implies the rectifiability of E. This follows from a suitableadaptation of the arguments used above in the proof of Theorem 7.1. It would beinteresting to find other new convolution kernels (in the plane and homogeneousof degree −1, say) such that the L2(μ) boundedness of the associated singularintegral operator implies the rectifiability of E. A much more ambitious objectiveconsists in characterizing the class of such kernels.

Chapter 8

Principal values for the Cauchytransform and rectifiability

8.1 Introduction

Given a complex measure ν ∈ M(Rd) and a Calderon-Zygmund operator T , wewrite

p.v.T ν(x) = limε→0

Tεν(x),

whenever the limit exists. This is the so-called principal value of the singularintegral. Analogously, for a fixed positive Radon measure μ and f ∈ Lp(μ),

p.v.Tμf(x) = limε→0

Tμ,εf(x).

The existence of principal values for Calderon-Zygmund operators is a quitedelicate question. This is one of the reasons why in Chapter 2 we defined theL2(μ) boundedness of Tμ in terms of the uniform boundedness of the truncatedoperators Tμ,ε.

Let us remark that, in general, the L2(μ) boundedness of Tμ does not implythe μ-almost everywhere existence of principal values. This is the case for instancefor the singular integral operator associated with the kernelK(x, y) = 1/|x−y|n+i t

in Rn, where t > 0 is some fixed constant, with μ equal to the Lebesgue measure.This is shown in Duoandikoetxea [36, Chapter 5]. Other examples on some Cantorsets are studied in Chousionis [9].

However, it turns out that, given any Radon measure μ with no point masses,the L2(μ) boundedness of the Cauchy transform Cμ implies the μ-a.e. existence ofp.v.Cμf(x) for all f ∈ Lp(μ), 1 ≤ p < ∞, and also of p.v.Cν(x) for all ν ∈ M(C).This was first proved in Tolsa [153] and it is one of the main results that we willsee in this chapter. Our arguments will combine some ideas from Tolsa [153] withothers from Mattila and Verdera [105].

, , OI 10.1007/978-3- - -6_ ,

© Springer



28910

290 Chapter 8. Principal values for the Cauchy transform

In the second part of this chapter we will prove other results in the conversedirection. In particular, we will show that the μ-a.e. existence of p.v.Cμ(x) (andmore generally, the μ-a.e. finiteness of C∗μ(x)) implies the L2(μ) boundedness ofthe Cauchy transform in some subsets of positive μ-measure. Combining this factwith the David-Leger theorem, we will deduce a nice characterization of rectifiablesets both in terms of existence principal values for the Cauchy transform andfiniteness of the maximal Cauchy transform.

8.2 L2 boundedness implies existence of principal values

The main result from this section is the following.

Theorem 8.1. Let μ be a Radon measure on C with linear growth such that theCauchy transform Cμ is bounded in L2(μ). Then

(a) for 1 ≤ p < ∞ and all f ∈ Lp(μ), p.v.Cμf(x) exists for μ-a.e. x ∈ C;

(b) for all ν ∈ M(C), p.v.Cν(x) exists for μ-a.e. x ∈ C.

Before turning to the proof of the theorem, we show an interesting conse-quence.

Corollary 8.2. Let μ be a finite Radon measure in C such that Uμ(x) = MRμ(x)+c2μ(x)

1/2 < ∞ for μ-a.e. x ∈ C. Then p.v.Cμ(x) exists for μ-a.e. x ∈ C.

Proof. For m ≥ 1, let Am = {x ∈ C : Uμ(x) ≤ m}, so that

μ(C \⋃m≥1

Am

)= 0.

By the T 1 theorem for the Cauchy transform, it follows that Cμ�Amis bounded in

L2(μ�Am). Then, from (b) in the preceding theorem, p.v.Cμ(x) exists for μ-a.e.x ∈ Am. �

To prove the theorem above, first we will review some rather classical tech-niques in Section 8.2.1, and subsequently we will distinguish two particular casesof measures μ. First we will consider the case where μ is the arc length on a Lips-chitz graph (or more generally, on a rectifiable set), and second we will study thecase where μ has zero density μ-a.e. In Subsection 8.2.5 we will complete the proofof the theorem.

8.2.1 Reduction to a class of dense functions

Theorem 8.3. Let μ be a Radon measure on Rd with growth of degree n and letTμ be an n-dimensional Calderon-Zygmund operator (bounded in L2(μ)). Supposethat, for some 1 ≤ p0 < ∞, there exists a class of functions A ⊂ Lp0(μ) which isdense in Lp0(μ) such that, for all f ∈ A, the principal value p.v.Tμf(x) exists forμ-a.e. x ∈ Rd. Then

8.2. L2 boundedness implies existence of principal values 291

(a) for 1 ≤ p < ∞ and all f ∈ Lp(μ), p.v.Tμf(x) exists for μ-a.e. x ∈ Rd;

(b) for all ν ∈ M(Rd), p.v.T ν(x) exists for μ-a.e. x ∈ Rd.

Proof. (a) First we will show that for all f ∈ Lp0(μ), the principal value p.v.Tμf(x)exists for μ-a.e. x ∈ Rd.

For each m ≥ 1, take gm ∈ A such that ‖f−gm‖Lp0(μ) ≤ 1/m. Let Fm be the

subset of those points x ∈ Rd such that p.v.Tμgm(x) exists, and let F =⋂

m≥1 Fm,so that F has full μ-measure.

To check that p.v.Tμf(x) exists for μ-almost all x ∈ F , it is enough to provethat

μ

({x ∈ F : lim sup

ε1,ε2→0|Tμ,ε1f(x)− Tμ,ε2f(x)| > λ

})= 0 for all λ > 0. (8.1)

For x ∈ F and every m ≥ 1, we write

|Tμ,ε1f(x)−Tμ,ε2f(x)| ≤ |Tμ,ε1gm(x)−Tμ,ε2gm(x)| + |Tμ,ε1f(x)−Tμ,ε1gm(x)|+ |Tμ,ε2f(x)− Tμ,ε2gm(x)|

≤ |Tμ,ε1gm(x) − Tμ,ε2gm(x)| + 2Tμ,∗(f − gm)(x).

Since |Tμ,ε1gm(x)− Tμ,ε2gm(x)| → 0 as ε1, ε2 → 0, we infer that

lim supε1,ε2→0

|Tμ,ε1f(x)− Tμ,ε2f(x)| ≤ 2Tμ,∗(f − gm)(x).

Thus, recalling that Tμ,∗ is bounded from Lp0(μ) to Lp0,∞(μ),

μ

({x ∈ F : lim sup

ε1,ε2→0|Tμ,ε1f(x)− Tμ,ε2f(x)| > λ

})≤ μ({

x ∈ F : 2Tμ,∗(f − gm)(x) > λ}) ≤ c ‖f − gm‖p0

Lp0(μ)

λp0≤ c

(mλ)p0.

Letting m → ∞, (8.1) follows.Consider now the case of an arbitrary 1 ≤ p < ∞. Since Lp0(μ) ∩ Lp(μ) is

dense in Lp(μ) and by the preceding argument we know that for all f in this classthe principal value p.v.Tμf exists μ-a.e., by applying what we have just provedabove (with p replacing p0), we infer that for any arbitrary f ∈ Lp(μ), p.v.Tμfexists μ-a.e.

(b) We have now to prove that for a given ν ∈ M(Rd), p.v.T ν exists μ-a.e. By theRadon-Nikodym theorem, we can decompose ν as

ν = f μ+ νs,

where f ∈ L1(μ) and νs is mutually singular with μ. From (a) we already knowthat p.v.Tμf exists μ-a.e., and thus we only have to deal with p.v.T νs. So it is


enough to show that

μ

({x ∈ Rd : lim sup

ε1,ε2→0|Tε1νs(x) − Tε2νs(x)| > λ

})= 0 for all λ > 0. (8.2)

Since μ and νs are mutually singular, there are disjoint Borel sets A and B suchthat μ = μ�A and νs = νs�B. For any δ > 0, let G ⊂ B be a closed subset suchthat

|νs|(B \G) ≤ δ.

Since G is closed, for all x ∈ Rd \G, Tε1(χG νs)(x) = Tε2(χG νs)(x) for ε1, ε2 smallenough. Therefore,

lim supε1,ε2→0

|Tε1νs(x) − Tε2νs(x)| = lim supε1,ε2→0

∣∣Tε1(χB\G νs)(x) − Tε2(χB\G νs)(x)∣∣

≤ 2T∗(χB\G νs)(x).

Thus, by the boundedness of T∗ from M(Rd) to L1,∞(μ),

μ

({x ∈ Rd \G : lim sup

ε1,ε2→0|Tε1νs(x) − Tε2νs(x)| > λ

})≤ μ({

x ∈ Rd \G : 2T∗(χB\G νs)(x) > λ}) ≤ c |νs|(B \G)

λ≤ c δ

λ,

which yields (8.2) because μ(G) = 0 and δ can be chosen arbitrarily small. �

In the next proposition we show that to find a dense class as in the precedingtheorem it is enough to study constant functions on open sets.

Proposition 8.4. Let μ be a Radon measure on Rd with growth of degree n andlet Tμ be an n-dimensional Calderon-Zygmund operator in Rd. Let U ⊂ Rd withμ(U) < ∞ such that p.v.TμχU (x) exists for μ-a.e. x ∈ U . Then for any f ∈ C1(Rd)with compact support p.v.Tμf(x) exists for μ-a.e. x ∈ U .

In particular, notice that if μ(Rd) < ∞, the existence of p.v.Tμ(x) for μ-a.e.x ∈ Rd implies that, for all f ∈ C1(Rd) compactly supported, p.v.Tμf(x) existsfor μ-a.e. x ∈ Rd.

Proof. Consider x ∈ U such that p.v.TμχU (x) exists. We claim that for f ∈ C1(Rd)with compact support p.v.Tμf(x) exists. Indeed, we write

p.v.Tμf(x) = p.v.Tμ

(f − f(x)χU

)(x) + f(x) p.v.TμχU (x).

Clearly, it is enough to check that p.v.Tμ

(f − f(x)χU

)(x) exists. To this end, for


0 < ε < δ < dist(x,Rd \ U), we set

Tμ,ε

(f − f(x)χU

)(x) =

∫|x−y|>ε

K(x, y)(f(y)− f(x)χU (y)

)dμ(y) (8.3)

=

∫ε<|x−y|≤δ

K(x, y)(f(y)− f(x)

)dμ(y)

+

∫|x−y|>δ

K(x, y)(f(y)− f(x)χU (y)

)dμ(y).

Using the estimate |K(x, y)| ≤ c/|x−y|n, the mean value theorem, and the growthof degree n of μ, we get∫

|x−y|≤δ

∣∣K(x, y)(f(y)− f(x)

)∣∣ dμ(y) ≤ c ‖∇f‖∞∫|x−y|≤δ

1

|x− y|n−1dμ(y)

≤ c ‖∇f‖∞ δ.

From the dominated convergence theorem, we infer that

limε→0

∫ε<|x−y|≤δ

K(x, y)(f(y)− f(x)

)dμ(y)

exists. Then by (8.3) we deduce that p.v.Tμ

(f−f(x)χU

)(x) exists, as claimed. �

8.2.2 Existence of principal values on Lipschitz graphs andrectifiable sets

Theorem 8.5. Let A : R → R be a Lipschitz function and Γ = {(x, y) ∈ C : y =A(x)} its graph. Let μ = H1�Γ. Then, for all f ∈ Lp(μ), 1 ≤ p < ∞, p.v.Cμf(z)exists for μ-almost every z ∈ Γ. Also, for all ν ∈ M(C), p.v.Cν(z) exists forμ-almost every z ∈ Γ.

Proof. By Theorem 8.3, it is enough to prove that p.v.Cμf(z) exists μ-a.e. fora class of functions f dense in L1(μ), say. In our case, this class consists of thefunctions of the form

f(z) =1 + i A′(z1)

(1 +A′(z1)2)1/2g(z),

where z1 = Re z and g is C1 and compactly supported. In this way, it turns outthat

Cμ,εf(w) =∫|z−w|>ε

1

z − wf(z) dH1

Γ(z) =

∫|z−w|>ε

1

z − wg(z) dzΓ,

with

dzΓ =1 + i A′(z1)

(1 +A′(z1)2)1/2dH1�Γ(z).


We will show that p.v.C(g dzΓ)(w) exists whenever w is a point of differentiabilityof Γ, that is, whenever the function A is differentiable at w1 = Re w. Since A isdifferentiable a.e., we will be done.

First we will study the principal value associated with

Tεg(w) =

∫|z1−w1|>ε

1

z − wg(z) dzΓ,

for w ∈ Γ. Observe that the truncation in the integral that defines Tε is differentfrom the one for Cμ,εf . By translating Γ, we may assume that w = 0. Then for0 < ε < 1 we have

Tεg(0) =

∫ε<|z1|≤1

g(z)− g(0)

zdzΓ + g(0)

∫ε<|z1|≤1

1

zdzΓ +

∫|z1|>1

g(z)

zdzΓ.

As in the proof of Proposition 8.4, for the first integral we take into account that|g(z)− g(0)|/|z| ≤ ‖g′‖∞ and then it follows easily that

limε→0

∫ε<|z1|≤1

g(z)− g(0)

zdzΓ

exists. Thus, to prove the existence of p.v.T g(0) it suffices to show the existenceof

limε→0

∫ε<|z1|≤1

1

zdzΓ. (8.4)

To this end we consider the branch of the complex logarithm log z = log |z|+i arg zdefined in C minus the negative imaginary axis, with the argument such that−π/2 < arg z < 3π/2. Then, since log z is analytic in a neighborhood of Γ ∩ {z :ε < |z1| ≤ 1}, denoting γ(t) = t+ i A(t) for t ∈ R, we have∫

ε<|z1|≤1

1

zdzΓ = log(γ(1))− log(γ(ε)) + log(γ(−ε))− log(γ(−1)). (8.5)

Since γ is differentiable at 0, we may assume that

log(γ(−ε)) = log(γ(ε)) + π i+ τ,

where τ ∈ C is very small in modulus. In this situation, it follows easily thatlog(γ(−ε))− log(γ(ε)) = log

(γ(−ε)/γ(ε)

), and then

limε→0

[log(γ(−ε))− log(γ(ε))

]= lim

ε→0log

γ(−ε)− γ(0)

εγ(ε)− γ(0)

ε

= log(−1) = π i.

Therefore, the limit in (8.4), and thus p.v.T g(0), exists.


To prove the existence of p.v.C(g dzΓ)(w) it is enough to check thatCε(g dzΓ)(0)− Tε0g(0) tends to 0 as ε → 0, with ε0 = ε (1 +A′(0)2)−1/2. Clearly,∣∣Cε(g dzΓ)(0)− Tε0g(0)

∣∣ ≤ ∫F

1

|z| |g(z)| dH1(z), (8.6)

whereF =(Γ \ B(0, ε)

)Δ {z ∈ Γ : |z1| > ε0}

(recall that Δ stands for the symmetric difference). Observe that z ∈ Γ \ B(0, ε)is equivalent to √

1 +

(A(z1)−A(0)

z1

)2|z1| > ε,

while |z1| > ε0 is equivalent to√1 +A′(0)2 |z1| > ε.

It follows easily that∣∣∣∣∣∣√1 +

(A(z1)−A(0)

z1

)2−√1 +A′(0)2

∣∣∣∣∣∣ ≤ c|A(z1)−A(0)−A′(0) z1|

|z1| ,

with c possibly depending on ‖A′‖∞. Given any arbitrary δ > 0, if |z1| is smallenough, the right-hand side above is smaller than δ. Then we deduce that

F ⊂ {z ∈ Γ : (1− c δ)ε ≤ |z1| ≤ (1 + c δ)ε},

for |z1| small enough. From (8.6), recalling that |g| = |f |,∣∣Cε(g dzΓ)(0)− Tε0g(0)∣∣ ≤ ∫

(1−c δ)ε≤|z1|≤(1+c δ)ε

|f(z)||z| dH1�Γ(z) ≤ c δ ‖f‖∞,

assuming δ small enough for the last inequality. Thus the difference on the left-hand side above converges to 0 as ε → 0, as wished. �Remark 8.6. For the record, observe that in the last part of the proof we showedthat if g is a compactly supported C1 function and w ∈ Γ is a point of differen-tiability of Γ,

p.v.C(g dzΓ)(w) = limε→0

∫|z1−w1|>ε

1

z − wg(z) dzΓ.

Although this will not be needed below, let us remark that the preceding identityalso holds H1-a.e. for g ∈ Lp(μ), or even for a general measure ν ∈ M(C) insteadof g dzΓ. This follows from the density of C1 functions in Lp(μ) and using the weak(p, p) boundedness of CH1�Γ,∗, by techniques analogous to the ones in Theorem 8.3.


As a corollary of Theorem 8.5, we obtain the following result from Mattilaand Melnikov [102].

Corollary 8.7. Let E ⊂ C be rectifiable and let ν ∈ M(C). Then p.v.Cν(z) existsfor H1-almost every z ∈ E.

Proof. Recall that H1-almost all E is contained in a countable union of possiblyrotated Lipschitz graphs Γj , j ≥ 1 (see Proposition 4.13). Since p.v.Cν(z) existsfor H1-almost every z ∈ Γj , the corollary follows. �

In particular, the preceding result implies that if E ⊂ C is rectifiable and hasfinite length, then p.v.C(H1�E) exists for H1-a.e. x ∈ E.

8.2.3 Plemelj formulas on Lipschitz graphs

Although it is not necessary for the proof of Theorem 8.1, at this point it is worthstudying the classical Plemelj’s formulas, in the particular case of Lipschitz graphs.First we have to introduce some notation. Recall that for z ∈ C, we set z1 = Re(z)and z2 = Im(z). Given α > 0 and w ∈ C, we consider the open cone (with verticalaxis)

X(w,α) ={z ∈ C : |z1 − w1| < α |z2 − w2|

},

and the half open cones

X+(w,α) ={z ∈ X(w,α) : z2 > w2

},

X−(w,α) ={z ∈ X(w,α) : z2 < w2

}.

Let A : R → R be Lipschitz and let Γ be its graph. Observe that if α <1/‖A′‖∞, then, for every w ∈ Γ,

X+(w,α) ⊂ {z ∈ C : z2 > A(z1)},

and analogously,X−(w,α) ⊂ {z ∈ C : z2 < A(z1)}.

On the graph Γ of A, we consider the usual complex measure

dzΓ =1 + i A′(z1)

(1 +A′(z1)2)1/2dH1�Γ(z).

The next theorem deals with the non-tangential limits

C+(f dzΓ)(w) := limX+(α,w)�ξ→w

C(f dzΓ)(ξ),

C−(f dzΓ)(w) := limX−(α,w)�ξ→w

C(f dzΓ)(ξ),

for f ∈ Lp(H1�Γ), 1 ≤ p < ∞.


Theorem 8.8. Let A : R → R and let Γ ⊂ C be its Lipschitz graph. Let 0 < α <1/‖A′‖∞. Then, for all f ∈ Lp(H1�Γ), 1 ≤ p < ∞, the non-tangential limitsthat define C+(f dzΓ)(w) and C−(f dzΓ)(w) exist for H1-a.e. w ∈ Γ. Moreover, thefollowing identities hold H1-a.e. in Γ:

1

2πiC+(f dzΓ)(w) =

1

2πip.v.C(f dzΓ)(w) +

1

2f(w), (8.7)

and1

2πiC−(f dzΓ)(w) =

1

2πip.v.C(f dzΓ)(w) − 1

2f(w). (8.8)

The identities (8.7) and (8.8) are the so-called Plemelj formulas.

Proof. First we assume that f is compactly supported and of class C1 and weshow that the limit that defines C+(f dzΓ)(w) exists for H1-a.e. w ∈ Γ and that(8.7) holds.

Let w ∈ Γ be a point of differentiability of Γ and take ξ ∈ X+(w,α). For afixed ε > 0, we write

C(f dzΓ)(ξ) =

∫|z1−w1|>ε

f(z)

z − ξdzΓ +

∫|z1−w1|≤ε

f(z)− f(w)

z − ξdzΓ (8.9)

+ f(w)

∫|z1−w1|≤ε

1

z − ξdzΓ.

To deal with the last integral, we consider the branch of the complex logarithmlog(z) = ln |z|+ i arg(z) defined in C minus the positive imaginary axis, with theargument such that π/2 < arg(z) < 5π/2. With this choice, observe that sinceξ is above Γ, log(z − ξ) is well defined and analytic for z belonging to a thinneighborhood of Γ. Then, denoting γ(t) = t+ i A(t) for t ∈ R, since log(z − ξ) isa primitive of 1/(z − ξ) (with respect to z), the last integral in (8.9) equals

log[γ(w1 + ε)− ξ

]− log[γ(w1 − ε)− ξ

]. (8.10)

Moreover, we assume that |ξ − w| ε and ε small, so that the argument ofγ(w1 + ε) − γ(w1 − ε) is then close to π and thus it is the interval (π/2, 5π/2),recalling that w is a point of differentiability of Γ. Then the expression in (8.10)equals

ln

∣∣∣∣γ(w1 + ε)− ξ

γ(w1 − ε)− ξ

∣∣∣∣+ i arg(γ(w1 + ε)− ξ)− i arg(γ(w1 − ε)− ξ)

= ln

∣∣∣∣γ(w1 + ε)− ξ

γ(w1 − ε)− ξ

∣∣∣∣+ i argγ(w1 + ε)− ξ)

γ(w1 − ε)− ξ)

= logγ(w1 + ε)− ξ

γ(w1 − ε)− ξ.


Then, letting ξ → w with ξ ∈ X+(α,w) in (8.9), we get

C+(f dzΓ)(w) =

∫|z1−w1|>ε

f(z)

z − wdzΓ +

∫|z1−w1|≤ε

f(z)− f(w)

z − wdzΓ

+ f(w) logγ(w1 + ε)− w

γ(w1 − ε)− w.

Observe that the second integral on the right-hand side is convergent becausef ∈ C1. Now, we let ε → 0 in this equation. The first integral on the right-handside tends to p.v.C(f dzΓ)(w) (recall Remark 8.6), the second one tends to 0, while

limε→0

logγ(w1 + ε)− w

γ(w1 − ε)− w= lim

ε→0log

γ(w1 + ε)− w

εγ(w1 − ε)− w

ε

= log(−1) = π i, (8.11)

since γ is differentiable at w1. So we obtain (8.7).

The identity (8.8) for a compactly supported function f of class C1 is provedin an analogous way. In this case, for ξ ∈ X−(w,α), one has to take a determinationof log(z− ξ) such that −5π/2 < arg(z− ξ) < −π/2, so that the right-hand side of(8.11) equals −π i.

To prove that (8.7) holds H1-a.e. for a general function f ∈ Lp(H1�Γ), for1 ≤ p < ∞, we will show that the H1 measure of the set{

w ∈ Γ: lim supX+(α,w)�ξ→w

∣∣∣∣C(f dzΓπi )(ξ)− p.v.C(fdzΓπi

)(w) − f(w)

∣∣∣∣ >λ

}(8.12)

is zero for all λ > 0, and we will be done. Given δ > 0, let g ∈ C1(C) such that‖f − g‖Lp(H1�Γ) ≤ δ. Since (8.7) holds for g,

lim supX+(α,w)�ξ→w

∣∣∣∣C(f dzΓπi

)(ξ)− p.v.C

(fdzΓπi

)(w) − f(w)

∣∣∣∣= lim sup

X+(α,w)�ξ→w

∣∣∣∣C((f − g)dzΓπi

)(ξ)− p.v.C

((f − g)

dzΓπi

)(w) − (f − g)(w)

∣∣∣∣≤ lim sup

X+(α,w)�ξ→w


)(ξ)

∣∣∣∣+ C∗((f − g)

dzΓπi

)(w) + |f − g|(w).

We claim that

lim supX+(α,w)�ξ→w


)(ξ)

∣∣∣∣ ≤ C∗((f−g)

dzΓπi

)(w)+cMR(f−g)(w). (8.13)


Assume this for the moment. Then the H1 measure of the set (8.12) does notexceed

H1

({w ∈ Γ : 2 C∗

((f − g)

dzΓπi

)(w) + |f − g|(w) + cMR(f − g)(w) >λ

})≤

c ‖f − g‖pLp(H1�Γ)λp

≤ c δp

λp,

where we took into account that C∗ and MR are of weak type (p, p) with respectto H1�Γ, and we used Chebyshev. Letting δ → 0, it follows that the set (8.12) haszero H1-measure.

It only remains to prove (8.13) now. From the fact that ξ ∈ X+(α,w), itfollows that dist(ξ,Γ) ≈ |ξ − w| =: t, and then one gets∣∣C((f − g)(χB(w,2t)dzΓ

)(ξ)∣∣ ≤ ∫

B(w,2t)∩Γ

1

|z − ξ| |f(z)− g(z)| dH1(z) (8.14)

≤ c

t

∫B(w,2t)∩Γ

|f − g| dH1 ≤ cMR(f − g)(w).

Similarly, it is easy to check that∣∣C((f − g)χB(w,2t)cdzΓ)(ξ)− C((f − g)χB(w,2t)cdzΓ)(w)

∣∣ ≤ cMR(f − g)(w),

and so ∣∣C((f − g)χB(w,2t)cdzΓ)(ξ)∣∣ ≤ C∗((f − g)dzΓ)(w) + cMR(f − g)(w).

From this estimate and (8.14), the claim (8.13) follows.The proof that (8.8) holds H1-a.e. for a general function f ∈ Lp(H1�Γ), for

1 ≤ p < ∞, is analogous. �The following is a nice consequence of the Plemelj formulas.

Corollary 8.9. Let Γ ⊂ C be a Lipschitz graph in C. For f ∈ Lp(H1�Γ), with1 < p < ∞, let

Tf(z) =1

πip.v.C(f dzΓ)(z).

Then T is a bounded operator in Lp(H1�Γ) such that

T ◦ T = Id

in Lp(H1�Γ).Proof. The boundedness of T in Lp(H1�Γ) for 1 < p < ∞ has already been proved.For f ∈ Lp(H1�Γ) and z ∈ Γ, write T+f(z) =

1πi C+(f dzΓ)(z). From Theorem 8.8,

we haveT+f = Tf + f.


In other words, T+ = T + Id. As a consequence,

T+ ◦ T+ = T ◦ T + 2T + Id.

By Cauchy’s integral formula, it easily follows that T+ ◦ T+(f) = 2T+(f). Thatis, T+ ◦ T+ = 2T+. Thus,

2T + 2 Id = T ◦ T + 2T + Id,

which yields T ◦ T = Id. �When Γ is the real line, T = −iH , where H stands for the Hilbert transform.

So the identity T ◦ T = Id generalizes the well-known identity H ◦H = −Id.

8.2.4 Weak convergence and existence of principal values formeasures with zero density

In this section we consider a Radon measure μ in Rd that has growth of degree nand zero n-dimensional density μ-a.e., that is,

Θn(x, μ) := limr→0

μ(B(x, r))

(2r)n= 0 for μ-a.e. x ∈ Rd.

We will show that if Tμ is an n-dimensional Calderon-Zygmund operator boundedin L2(μ) and antisymmetric, i.e. its kernel verifies

K(x, y) = −K(y, x) for all x = y,

then p.v.Tμf(x) exists for all f ∈ Lp(μ) and μ-a.e. x ∈ Rd. This was first provedfor the Cauchy transform by Tolsa [153], and later on it was extended to arbitraryantisymmetric Calderon-Zygmund operators by Mattila and Verdera [105]. Belowwe will follow their approach. First we will prove that Tμ,εf converges weakly inLp(μ) as ε → 0, for all f ∈ Lp(μ) with 1 < p < ∞ (without assuming zero n-dimensional density) and subsequently we will turn to the existence of principalvalues.

Theorem 8.10. Let μ be a Radon measure in Rd that has growth of degree n. Let Tμ

be an n-dimensional Calderon-Zygmund operator which is antisymmetric. Then,for all 1 < p < ∞ and f ∈ Lp(μ), Tμ,εf has a weak limit in Lp(μ) as ε → 0.Further, denoting by Tw

μ f such a weak limit, we have

Twμ f(x) = lim

r→0

1

μ(B(x, r))

∫B(x,r)

Tμ(f χB(x,r)c) dμ for μ-a.e. x ∈ Rd. (8.15)

Recall that Calderon-Zygmund operators are assumed to be bounded inL2(μ). Also, saying Tμ,εf has a weak limit Tw

μ f in Lp(μ) as ε → 0 means that for

all g ∈ Lp′(μ),

limε→0

∫Tμ,εf g dμ =

∫Twμ f g dμ.


From the theorem, it follows that Twμ is an operator bounded in Lp(μ) for all

1 < p < ∞.

In the identity (8.15), notice that Tμ(f χB(x,r)c) = p.v.Tμ(f χB(x,r)c) is de-fined for all y ∈ B(x, r), since (recalling that B(x, r) is open)∫

B(x,r)c

1

|y − z|n |f(z)| dμ(z)

≤(∫

|y−z|>dist(y,B(x,r)c)

1

|y − z|np′ dμ(z)

)1/p′

‖f‖Lp(μ) < ∞.

Finally, notice also that from (8.15) it follows that for f ∈ Lp(μ), 1 < p < ∞,

Twμ f(x) = Tμf(x) for μ-a.e. x ∈ supp(f). (8.16)

Proof of Theorem 8.10. Suppose first that μ(Rd) < ∞. To begin with, we willshow that Tμ,ε1 has a weak limit in Lp(μ) as ε → 0. By antisymmetry, for allε > 0 and any open ball B we have∫

B

Tμ,εχB dμ = 0.

Therefore, ∫B

Tμ,ε1 dμ =

∫B

Tμ,εχBc dμ.

Since B is open,

TμχBc(x) = limε→0

Tμ,εχBc(x) for all x ∈ B.

As |Tμ,εχBc(x)| ≤ Tμ,∗χBc(x) and Tμ,∗χBc ∈ L1(μ), from the dominated conver-gence theorem we deduce that

limε→0

∫B

Tμ,ε1 dμ =

∫B

TμχBc dμ.

To prove the weak convergence in Lp(μ) of Tμ,ε1 as ε → 0, we have to showthe existence of

limε→0

∫Tμ,ε1 g dμ (8.17)

for all g ∈ Lp′(μ). By the previous argument and linearity, we already know that

the limit exists if g belongs to the subspace S ⊂ Lp′(μ) of finite linear combinations

of characteristic functions of open balls. For an arbitrary function g ∈ Lp′(μ) we

argue by density: from Vitali’s covering theorem (see Mattila [100, Chapter 2], for


example), it is easy to verify that S is dense in Lp′(μ). For any δ > 0, let b ∈ S

satisfy ‖g − b‖Lp′(μ) ≤ δ. Then, for all ε1, ε2 > 0,

∫ (Tμ,ε11− Tμ,ε21

)g dμ

=

∫ (Tμ,ε11− Tμ,ε21

)b dμ+

∫ (Tμ,ε11− Tμ,ε21

)(g − b) dμ.

The second integral is bounded in modulus by 2‖Tμ,∗1‖Lp(μ) ‖g − b‖Lp′(μ) ≤2δ ‖Tμ,∗1‖Lp(μ), and thus

lim supε1,ε2→0

∣∣∣∣∫ (Tμ,ε11− Tμ,ε21)g dμ

∣∣∣∣ ≤ 2δ ‖Tμ,∗1‖Lp(μ),

since the limit (8.17) exists with g replaced by b. As δ is arbitrarily small, theleft-hand side above vanishes, which means that Tμ,ε1 converges weakly in Lp(μ)as ε → 0, as claimed.

Take now some function f ∈ L∞(μ), and set f = f+2‖f‖L∞(μ) and σ = f μ.

Notice that f is positive and that f ≈ ‖f‖L∞(μ), and thus Tσ is bounded inL2(σ). As a consequence, applying what we have proved above to σ, we infer that

Tσ,ε1 = Tμ,εf converges weakly in Lp(σ) as ε → 0. This is equivalent to saying

that it converges weakly in Lp(μ) as ε → 0 (since f is bounded above and awayfrom 0). By linearity, this implies that Tμ,εf converges weakly in Lp(μ) too.

To prove the weak convergence of an arbitrary function f ∈ Lp(μ) we argueby density again. For any δ > 0, consider h ∈ L∞(μ) such that ‖f − h‖Lp(μ) ≤ δ.

Then for any g ∈ Lp′(μ) we write∫ (

Tμ,ε1f − Tμ,ε2f)g dμ

=

∫ (Tμ,ε1h− Tμ,ε2h

)g dμ+

∫ (Tμ,ε1(f − h)− Tμ,ε2(f − h)

)g dμ.

The first integral on the right-hand side tends to 0 as ε1, ε2 → 0, while the lastone is bounded in modulus by 2‖T∗(f − h)‖Lp(μ)‖g‖Lp′(μ) ≤ cδ ‖g‖Lp′(μ). Since δis arbitrarily small, we get

limε1,ε2→0

∫ (Tμ,ε1f − Tμ,ε2f

)g dμ = 0,

which yields the weak convergence of Tμ,εf in Lp(μ).Consider now the case when μ is not a finite measure. In this situation, we

already know that the limit

limε→0

∫Tμ,εf g dμ


exists for all f ∈ Lp(μ) and g ∈ Lp′(μ) which are compactly supported. By density

arguments analogous to the ones above, we infer that the limit also exists withoutthe assumption on the compact support of f and g.

It remains to prove (8.15). First we will show that, for any open ball B,∫B

Tμ(f χBc) dμ =

∫B

Twμ (f χBc) dμ. (8.18)

To this end, notice that since Twμ (f χBc) is the weak limit of the functions

Tμ,ε(f χBc) in Lp(μ),

limε→0

∫B

Tμ,ε(f χBc) dμ =

∫B

Twμ (f χBc) dμ.

On the other hand, since Tμ,ε(f χBc)(x) → Tμ(f χBc)(x) for all x ∈ B (recallthat B is open), by the dominated convergence theorem, taking into account that|Tμ,ε(f χBc)(x)| ≤ Tμ,∗(f χBc)(x) and that Tμ,∗(f χBc) ∈ L1(μ�B), we also have

limε→0

∫B

Tμ,ε(f χBc) dμ =

∫B

Tμ(f χBc) dμ,

and thus (8.18) holds.By the Lebesgue differentiation theorem, for f ∈ Lp(μ) and μ-a.e. x ∈ Rd,

Twμ f(x) = lim

r→0

1

μ(B(x, r))

∫B(x,r)

Twμ f(x) dμ

= limr→0

1

μ(B(x, r))

(∫B(x,r)

Twμ (f χB(x,r)c) dμ+

∫B(x,r)

Twμ (f χB(x,r)) dμ

).

By (8.18), to prove (8.15) it suffices to show that

limr→0

1

μ(B(x, r))

∫B(x,r)

Twμ (f χB(x,r)) dμ = 0 for μ-a.e. x ∈ Rd.

In fact, from the antisymmetry of Twμ (which follows from that of Tμ,ε, ε > 0), we

deduce that∫B(x,r)

Twμ χB(x,r) dμ = 0. So we have∫

B(x,r)

Twμ (f χB(x,r)) dμ =

∫B(x,r)

Twμ

((f −mB(x,r)f)χB(x,r)

)dμ

= −∫B(x,r)

(f −mB(x,r)f)Twμ χB(x,r) dμ,

where mB(x,r)f stands for the μ-mean of f over B(x, r). Then the Schwarz in-

equality, the Lebesgue differentiation theorem, and the Lp′(μ) boundedness of Tw

μ

give

limr→0

1

μ(B(x, r))

∫B(x,r)

(f −mB(x,r)f)Twμ χB(x,r) dμ = 0 for μ-a.e. x ∈ Rd,


and so we are done. �Theorem 8.11. Let μ be a Radon measure in Rd that has growth of degree n and zeron-dimensional density μ-a.e. Let Tμ be an n-dimensional antisymmetric Calderon-Zygmund operator. Then, for all 1 < p < ∞ and f ∈ Lp(μ), p.v.Tμf(x) exists forμ a.e. x ∈ Rd and coincides with Tw

μ f(x). Also, for all ν ∈ M(C), p.v.T ν(x) exists

for μ a.e. x ∈ Rd.

Proof. For f ∈ Lp(μ), 1 < p < ∞, the coincidence of p.v.Tμf(x) with Twμ f(x) for

μ a.e. x ∈ Rd follows from the fact that∫Twμ f g dμ = lim

ε→0

∫Tμ,εf g dμ =

∫p.v.Tμf g dμ for all g ∈ Lp′

(μ).

The first identity is a direct consequence of the definition of weak convergencein Lp(μ), and the second one follows from the dominated convergence theorem,taking into account that Tμ,∗f ∈ Lp(μ).

Let us turn our attention to the existence of principal values. From the anti-symmetry of Tμ,ε we deduce that Tw

μ is also antisymmetric, and thus for any ballB, ∫

B

Twμ χB dμ = 0. (8.19)

As in the previous theorem, this identity will play an important role here.From Theorem 8.3 and Proposition 8.4 we infer that it suffices to prove the

existence of p.v.Tμf(x) for f = χU and μ-a.e. x ∈ U , where U ⊂ Rd is open andμ(U) < ∞. By this observation, it turns out that it is enough to show that p.v.Tμexists μ-a.e. in Rd, assuming that μ(Rd) < ∞. Let x ∈ suppμ be a Lebesgue pointfor Twμ with zero n-dimensional μ-density, such that (8.15) holds with f = 1,and such that there exists a sequence of balls B(x, 4ki), ki → −∞, which are(4, 4d+1) doubling. Recall that Lemma 2.8 and Remark 2.9 ensure that the set ofsuch points x has full μ-measure. We claim that for such x, we have

limε→0

Tεμ(x) = Twμ(x).

To prove this, for any given δ > 0, let r0 > 0 be such that

μ(B(x, r))

rn≤ δ and

∣∣∣∣Twμ(x) − 1

μ(B(x, r))

∫B(x,r)

Twμ dμ

∣∣∣∣ ≤ δ (8.20)

for 0 < r ≤ r0. Let p > 10 be some big integer to be fixed later and supposethat 2pε ≤ r0. Let ε0 be the maximum of the numbers 4j , with j ∈ Z, suchthat 4j ≤ ε and B(x, 4j) is (4, 4d+1)-doubling. From Lemma 2.20 (in fact, from asuitable analogous version) and the first estimate in (8.20) together with the factthat ε0 ≤ r0, we infer that

|Tεμ(x)− Tε0μ(x)| ≤ cδ.


We claim now that there exists some ε1 ∈ [ε0, 2ε0] such that

1

μ(B(x, ε1))

∫∫y∈B(x,ε1)z∈B(x,2ε1)\B(x,ε1)

1

|y − z|n dμ(y) dμ(z) ≤ c δ1/n. (8.21)

Assume for the moment that such ε1 exists. Then

|Tε0μ(x)− Tε1μ(x)| ≤∫B(x,ε1)\B(x,ε0)

c

|x− y|n dμ(y) ≤ μ(B(x, 2ε0))

εn0≤ cδ

since 2ε0 ≤ r0. Thus|Tεμ(x)− Tε1μ(x)| ≤ cδ. (8.22)

Now we write

Tε1μ(x)−1

μ(B(x, ε1))

∫B(x,ε1)

Twμ dμ

= TμχB(x,pε1)\B(x,ε1)(x)

+

(TμχB(x,pε1)c(x)−

1

μ(B(x, ε1))

∫B(x,ε1)

Twμ χB(x,pε1)c dμ

)

− 1

μ(B(x, ε1))

∫B(x,ε1)

Twμ χB(x,pε1) dμ

=: I1 + I2 − I3.

We will estimate each term Ii separately.To deal with I1 we set

|I1| =∣∣TμχB(x,pε1)\B(x,ε1)(x)

∣∣ ≤ ∫B(x,pε1)\B(x,ε1)

c

|x− y|n dμ(y)

≤ cμ(B(x, pε1))

εn1≤ c pnδ,

by (8.20), because pε1 ≤ 2pε0 ≤ 2pε ≤ r0.Let us consider I2 now. By (8.16), we know that for μ-a.e. y ∈ B(x, ε1), we

have Twμ χB(x,pε1)c(y) = TμχB(x,pε1)c(y). Then, from Lemma 2.12, it follows that

∣∣TμχB(x,pε1)c(x)− TμχB(x,pε1)c(y)∣∣ ≤ c

|x− y|η[(p− 1)ε1]η

MR,n μ(x) ≤ c p−η,

where η is the positive constant in the definition (2.3). Averaging with respect toy ∈ B(x, ε1), we derive

|I2| =∣∣∣∣∣TμχB(x,pε1)c(x)−

1

μ(B(x, ε1))

∫B(x,ε1)

Twμ χB(x,pε1)c dμ

∣∣∣∣∣ ≤ c p−η.


Finally we turn our attention to I3. From (8.19), we derive

I3 =1

μ(B(x, ε1))

∫B(x,ε1)

Twμ χB(x,pε1)\B(x,ε1) dμ.

By (8.16) again, it turns out that Twμ χB(x,pε1)\B(x,ε1)(y) = TμχB(x,pε1)\B(x,ε1)(y)

for μ-a.e. y ∈ B(x, ε1). As a consequence, for such y, denoting by K(·, ·) the kernelof Tμ,

|Twμ χB(x,pε1)\B(x,ε1)(y)| ≤

∫B(x,pε1)\B(x,ε1)

|K(y, z)| dμ(z).

For y ∈ B(x, ε1) and z ∈ B(x, 2ε1), we have |K(y, z)| ≤ c/εn1 , and thus∫B(x,pε1)\B(x,2ε1)

|K(y, z)| dμ(z) ≤ cμ(B(x, pε1))

εn1≤ c pnδ.

Therefore, appealing to (8.21), we infer that

|I3| ≤ c pnδ +1

μ(B(x, ε1))

∫∫y∈B(x,ε1)z∈B(x,2ε1)\B(x,ε1)

c

|y − z|n dμ(y) dμ(z)

≤ c pnδ + c δ1/n.

Gathering the estimates obtained for I1, I2 and I3, and recalling (8.22), weget ∣∣∣∣Tεμ(x)− 1

μ(B(x, ε1))

∫B(x,ε1)

Twμ dμ

∣∣∣∣ � δ + δ1/n + pnδ + p−η.

Now it is clear that, by an appropriate choice of δ > 0 (small) and p > 10 (big),the term on the left-hand side above will be arbitrarily small, which proves theassertion that limε→0 Tεμ(x) = Twμ(x).

It remains to prove the claim (8.21). For t ∈ [ε0, 2ε0], we set

J(t) =1

μ(B(x, t))

∫∫y∈B(x,t)z∈B(x,2t)\B(x,t)

1

|y − z|n dμ(y) dμ(z).

It is enough to show that ∫ 2ε0

ε0

J(t) dt ≤ c δ1/nε0. (8.23)

We have∫ 2ε0

ε0

J(t) dt ≤ 1

μ(B(x, ε0))

∫∫∫ε0≤t≤2ε0|x−y|<tt≤|x−z|<2t

1

|y − z|n dμ(y) dμ(z) dt.


Observe now that the triples (y, z, t) in the domain of integration of the last integralsatisfy y, z ∈ B(x, 4ε0), and |x− y| < t ≤ |x− z|. The latter condition implies thatt is contained in an interval of length |x− z| − |x− y| ≤ |y − z|. Then we obtain∫ 2ε0

ε0

J(t) dt ≤ 1

μ(B(x, ε0))

∫∫y∈B(x,4ε0)z∈B(x,4ε0)

(∫ |x−z|

|x−y|

1

|y − z|n dt

)dμ(y) dμ(z)

(8.24)

≤ 1

μ(B(x, ε0))

∫∫y∈B(x,4ε0)z∈B(x,4ε0)

1

|y − z|n−1dμ(y) dμ(z).

To estimate the last integral, we fix some q > 1 and, for each y ∈ B(x, 4ε0) wewrite ∫

z∈B(x,4ε0)

1

|y − z|n−1dμ(z) ≤

∫z∈B(x,4ε0):|y−z|>q−1ε0

1

|y − z|n−1dμ(z)

+

∫|y−z|≤q−1ε0

1

|y − z|n−1dμ(z).

We have∫z∈B(x,4ε0):|y−z|>q−1ε0

1

|y − z|n−1dμ(z) ≤ μ(B(x, 4ε0))

[q−1ε0]n−1≤ c δ qn−1ε0.

On the other hand, using the growth of degree n of μ, it follows easily that∫|y−z|≤q−1ε0

1

|y − z|n−1dμ(z) ≤ c q−1ε0.

Now, taking into account that B(x, ε0) is (4, 4d+1)-doubling, from the last esti-mates and (8.24) we derive∫ 2ε0

ε0

J(t) dt ≤ μ(B(x, 4ε0))

μ(B(x, ε0))(c δ qn−1ε0 + c q−1ε0) ≤ c (δ qn−1 + q−1)ε0.

Choosing q = δ−1/n, (8.23) follows, and so the claim (8.21) is proved. �

8.2.5 The final argument for the proof of Theorem 8.1

We need the following auxiliary result.

Lemma 8.12. Let μ be a finite Radon measure on Rd. For given λ2 ≥ λ1 > 0, letE,F be bounded Borel sets such that

E ⊂ {x ∈ Rd : Θ∗n(x, μ) ≥ λ1},


andF ⊂ {x ∈ Rd : λ1 ≤ Θ∗n(x, μ) ≤ λ2}.

There exist some absolute constants c1, c2, and c3 such that Hn(E) ≤ c1μ(E)

λ1and μ�F = gHn�F, where g is some Borel function such that c2 λ1 ≤ g(x) ≤ c3 λ2

for Hn-a.e. x ∈ F .

Proof. First we deal with Hn(E). Let ε > 0 be small (in particular, ε < λ1/2).Let U be an open set containing E such that μ(U) ≤ μ(E) + ε. For every x ∈ E,there exists a ball B(x, r) contained in U with radius 0 < r < ε/2 such thatμ(B(x, r)) ≥ (λ1 − ε)(2r)n. By the Besicovitch covering theorem, there existsa covering E ⊂ ⋃∞

i=1 B(xi, ri), with xi ∈ E, μ(B(xi, ri)) ≥ (λ1 − ε)(2ri)n and∑

i χB(xi,ri) ≤ M χU , where M is a universal constant. Then

Hnε (E) ≤

∑i

(2ri)n ≤∑i

μ(B(xi, ri))

λ1 − ε≤ M

λ1 − εμ(U) ≤ M

λ1 − ε

(μ(E) + ε

).

Therefore, letting ε → 0,

Hn(E) ≤ M

λ1μ(E). (8.25)

Now we turn our attention to F . From the first statement in the lemma, weknow that Hn(F ) < ∞. For ε > 0 and m ≥ 1, let

Fm ={x ∈ F : μ(B(x, r)) ≤ (λ2 + ε) (2r)n for 0 < r ≤ 1/m

}.

For any Borel set G ⊂ F and for any ε > 0, there is a covering⋃

i B(xi, ri) ⊃G ∩ Fm, where xi ∈ G ∩ Fm, 0 < ri ≤ 1/m, and μ(B(xi, ri)) ≤ (λ2 + ε) (2ri)

n,with∑

i rni ≤ Hn(G ∩ Fm) + ε. Then we get

μ(G ∩ Fm) ≤∑i

μ(B(xi, ri)) ≤ (λ2 + ε)∑i

(2ri)n ≤ (λ2 + ε)2n(H1(G) + ε).

Since this holds for all ε > 0 and all m ≥ 1, we deduce μ(G) ≤ 2nλ2 H1(G).Together with (8.25) applied to G, this yields

M−1λ1 H1(G) ≤ μ(G) ≤ 2nλ2 H1(G).

Therefore, μ�F and Hn�F are mutually absolutely continuous and there is a func-tion g, with M−1λ1 ≤ g ≤ 2n λ2, such that μ�F = gHn�F . �Proof of Theorem 8.1. Let f : C → C be an arbitrary C1 function with compactsupport. By Theorem 8.3, it is enough to show that p.v.Cμf(x) exists μ-a.e. Clearlythis exists for all x ∈ supp(fμ).

To deal with the case where x ∈ supp(fμ), for each m ≥ 1 let

Em ={x ∈ supp(fμ) : Θ∗1(x, μ) > 1/m

}

8.3. Finiteness of the maximal Cauchy transform 309

and setF ={x ∈ supp(fμ) : Θ∗1(x, μ) = 0

},

so that supp(fμ) = F ∪⋃m≥1 Em.

From the preceding lemma, it follows that H1(Em) < ∞ and, recalling thatμ has linear growth, we infer that μ�Em and H1�Em are mutually absolutelycontinuous, and there exists some Borel function gm such that μ�Em = gmH1�Em,with gm satisfying c2 m

−1 ≤ gm(x) ≤ c for H1-a.e. x ∈ Em. Since the Cauchytransform is bounded in L2(μ), it turns out that c2(μ�Em ∩D) < ∞ for any diskD ⊂ C (as μ(Em ∩D) < ∞). By the David-Leger theorem, it follows that Em ∩Dis rectifiable, and then, from Corollary 8.7, we deduce that p.v.Cμf(x) exists forμ-a.e. x ∈ Em ∩D. Thus p.v.Cμf exists μ-a.e. in

⋃m Em.

From Theorem 8.11 we know that p.v.Cμf exists μ-a.e. in F too, and so weare done. �

8.3 Finiteness of the maximal Cauchy transform impliesL2 boundedness in some subset

Recall that for a finite Borel measure μ and x ∈ C, we set

MRμ(x) = supr>0

μ(B(x, r))

r

andC∗μ(x) = sup

ε>0|Cεμ(x)|.

Theorem 8.13. Let μ be a Radon measure with compact support on C and considera μ-measurable set E with μ(E) > 0 such that

E ⊂ {x ∈ C : MRμ(x) < ∞ and C∗μ(x) < ∞}.Then there exists a Borel subset G ⊂ E with μ(G) > 0 such that μ�G has lineargrowth and the Cauchy integral operator Cμ�G is bounded in L2(μ�G).

Our proof of this result, which is based on Tolsa [155], relies on the use of theTb theorem of Nazarov, Treil and Volberg from Chapter 5 applied to the particularcase b ≡ 1. In this situation, this result reads as follows.

Theorem 8.14. Let μ be a measure supported on a compact set F ⊂ C. Supposethat, for each w ∈ C, there exists a set HD(w) made up of dyadic squares fromD(w) = w +D0, where D0 is the usual dyadic lattice of C, such that


w∈CHD(w).

(c)∫C\HD(w)

C∗μ dμ ≤ c∗ μ(F ), for all w ∈ C.

(e) μ(HD(w)) ≤ δ0 μ(F ), for all w ∈ C and some δ0 < 1.


Then there exists a Borel subset G ⊂ F \⋂w∈CHD(w) such that

(i) μ(G) ≥ c−11 μ(F ),


(iii) the Cauchy transform is bounded in L2(μ�G).

The constant c1 and the bound for the L2(μ�G) norm depend only on the otherconstants above.

Proof of Theorem 8.13. We intend to apply Theorem 8.14, for F = suppμ. So, foreach w ∈ C, we have to construct a set HD(w) made up of dyadic squares fromD(w) = w + D0 which satisfies the assumptions (a), (c), and (e). We will definean open set H (independent of w), and then we just will take HD(w) = H for allw ∈ C. Indeed, notice that any open set can be decomposed into dyadic squaresfrom any given dyadic lattice.

First, by the outer regularity of μ, there exists an open set V ⊃ C \ E suchthat

μ(V ∩ E) ≤ μ(E)

4. (8.26)

We will take a set of the form H = V ∪ H1 ∪ H2 where, loosely speaking, H1

will take care of the points where MRμ(x) is too big and H2 of the points whereC∗μ(x) is too big.

We define H1 as follows. Given a big enough constant M0 > 0 which will befixed below, we set H0 = {x ∈ C : MRμ(x) > M0}. For every x ∈ H0, we set

r(x) = sup{r > 0 : μ(B(x, r)) > M0r},so that μ(B(x, r(x))) ≥ M0r(x). We define

H1 =⋃

x∈H0

B(x, r(x)).

Obviously, H1 is open and H0 ⊂ H1. Let us check that

μ(E ∩H1) ≤ μ(E)

4if M0 is big enough. (8.27)

We setEn = {x ∈ E : MRμ(x) < n}.

Since MRμ(x) < ∞ for all x ∈ E, μ(E \ ⋃n En) = 0. We take m such thatμ(E \ Em) < μ(E)/10. We claim that if y ∈ H1, then MRμ(y) ≥ M0/2. Indeed,suppose that y ∈ B(x, r(x)), for some x ∈ H0. Then

μ(B(y, 2r(x))) ≥ μ(B(x, r(x))) ≥ M0r(x),

and the claim follows. As a consequence, E ∩ H1 ⊂ E \ EM0/2 (we assume thatM0 is an even integer), and thus μ(E ∩H1) → 0 as M0 → ∞, which proves (8.27).

8.3. Finiteness of the maximal Cauchy transform 311

Next we turn our attention to H2. For some fixed big constant k � M0, wedefine

L = {x ∈ C \H1 : C∗μ(x) > k}.For every x ∈ L, consider ε0(x) > 0 such that |Cε0(x)μ(x)| > k and |Cεμ(x)| ≤ kfor all ε > 2ε0(x). Then we set

H2 =⋃x∈L

B(x, ε0(x)).

Notice that H2 is open and C∗μ(x) ≤ k if x ∈ H2, because L ⊂ H2.Let us check that

μ(E ∩H2 \H1) ≤ μ(E)

4if k is chosen big enough. (8.28)

By construction, if y ∈ H2 \ H1, then y ∈ B(x, ε0(x)) for some x ∈ L. First wewill show that

|Cε0(x)μ(x)− Cε0(x)μ(y)| ≤ cM0, (8.29)

where c is some absolute constant. In fact, we have

|Cε0(x)μ(x)− Cε0(x)μ(y)| (8.30)

≤ |Cε0(x)(χB(y,2ε0(x))μ)(x)| + |Cε0(x)(χB(y,2ε0(x))μ)(y)|+ |Cε0(x)(χC\B(y,2ε0(x))μ)(x) − Cε0(x)(χC\B(y,2ε0(x))μ)(y)|.

Notice that the first two terms on the right-hand side are bounded by

μ(B(y, 2ε0(x)))

ε0(x)≤ 2M0,

since y ∈ H1. The last term on the right-hand side of (8.30) is estimated as follows:∫C\B(y,2ε0(x))

∣∣∣∣ 1

z − x− 1

z − y

∣∣∣∣ dμ(z) = ∫C\B(y,2ε0(x))

|x− y||z − x||z − y| dμ(z)

≤ c ε0(x)

∫C\B(y,2ε0(x))

1

|z − y|2 dμ(z),

where we have applied that |x − y| < ε0(x) and |z − x| ≈ |z − y| in the lastinequality. As y ∈ H1, by Lemma 2.11 we obtain

|Cε0(x)(χC\B(y,2ε0(x))μ)(x) − Cε0(x)(χC\B(y,2ε0(x))μ)(y)|= |C(χC\B(y,2ε0(x))μ)(x) − C(χC\B(y,2ε0(x))μ)(y)| ≤ cM0,

and (8.29) holds. As a consequence,

H2 \H1 ⊂ {y ∈ C : C∗μ(y) > k − cM0}.


Since C∗μ(y) < ∞ for all y ∈ E, we deduce that μ({

y ∈ E : C∗μ(y) > k − cM0

})tends to 0 as k → ∞, and thus

μ(E ∩H2 \H1) → 0 as k → ∞,

and so (8.28) follows.Setting HD(w) = H = V ∪H1∪H2, it is clear that then the property (a) from

Theorem 8.14 holds with c0 = M0. Moreover, since C∗μ(x) ≤ k for every x ∈ H2,we have ∫

C\HD(w)

C∗μ dμ ≤ k μ(F ).

On the other hand,

μ(E ∩HD(w)

) ≤ μ(E ∩ V ) + μ(E ∩H1) + μ(E ∩H2 \H1) ≤ 3

4μ(E),

by (8.26), (8.27) and (8.28). Thus, μ(HD(w)

) ≤ μ(F \E)+ 34 μ(E) < μ(F ). So the

assumptions of Theorem 8.14 are fulfilled and then we deduce that there exists aBorel subset G ⊂ F \ H ⊂ F \ V ⊂ E with μ(G) > 0 such that μ�G has lineargrowth and the Cauchy integral operator Cμ�G is bounded in L2(μ�G). �

8.4 Some consequences

Corollary 8.15. Let μ be a positive finite Radon measure on C satisfying MRμ(x) <∞ and C∗μ(x) < ∞ for μ-a.e. x ∈ C. Then there is a countable collection of Borelsets En with μ(C\⋃n En) = 0 such that, for each n, μ(En) < ∞, μ�En has lineargrowth, and Cμ�En

is bounded in L2(μ�En) (and thus c2(μ�En) < ∞).

Proof. Write E = suppμ and let

s = sup μ(⋃

n

Fn

),

where the supremum is taken over all the countable unions of Borel sets Fn ⊂ Esuch that μ(Fn) < ∞, μ�Fn has linear growth, and Cμ�Fn

is bounded in L2(μ�Fn).It is easy to check that there exists a countable family of Borel sets {Em}m≥1 forwhich the preceding supremum is attained. Indeed, for each j ≥ 1, take a familyof Borel sets F j

n suitable for the supremum above such that μ(⋃

n Fjn

) ≥ s− 1/j.Then we just put {Em} = {F j

n}n,j≥1.So the corollary is proved if we show that s = μ(E). This follows easily from

Theorem 8.13. Indeed, suppose that s < μ(E) and take a sequence {Em} thatattains the supremum s. Then Theorem 8.13 ensures that there exists a Borel setG ⊂ E \⋃m Em, with μ(G) > 0, such that μ�G has linear growth and the Cauchyintegral operator Cμ�G is bounded in L2(μ�G). Then μ (G ∪⋃m Em) > s, whichis a contradiction. �

8.4. Some consequences 313

Also we have:

Corollary 8.16. Let μ be a finite Borel measure on measure on C satisfying 0 <Θ∗1(x, μ) < ∞ and C∗μ(x) < ∞ for μ-a.e. x ∈ C. Then there exist some rectifiableset F ⊂ C and some non-negative functions g ∈ L1(H1�F ) such that μ = gH1�F.Proof. By the preceding corollary, there are subsets En ⊂ C such that μ(C \⋃

n En) = 0 and, for each n, μ(En) < ∞, μ�En has linear growth, and c2(μ�En) <∞. Since 0 < Θ∗1(x, μ�En) < ∞ for μ-a.e. x ∈ En, arguing as in the proof ofTheorem 8.1 we find subsets En,i and non-negative functions gn,i such that

μ�En =∑i

gn,iH1�En,i.

Since c2(gn,iH1�En,i) ≤ c2(μ�En) < ∞, by the David-Leger theorem it followsthat En,i is rectifiable. �

Corollary 8.17. Let μ be a positive finite Radon measure on C satisfying Θ∗1(x, μ)< ∞ for μ-a.e. x ∈ C. Then C∗μ(x) < ∞ for μ-a.e. x ∈ C if and only if theprincipal value limε→0 Cεμ(x) exists for μ-a.e. x ∈ C.

Proof. It is obvious that if the principal value limε→0 Cεμ(x) exists, then C∗μ(x) <∞.

If we assume C∗μ(x) < ∞ for μ-a.e. x ∈ C, then by Corollary 8.15 there is acountable collection of compact sets En such that μ(C \⋃n En) = 0 and, for eachn μ�En has linear growth and Cμ�En

is bounded in L2(μ�En). Finally, by Theorem8.1, the principal value limε→0 Cεμ(x) exists for μ-a.e. x ∈ En. �

Consider the case where μ is a finite 1-dimensional Hausdorff measure on anH1-measurable set E ⊂ C with H1(E) < ∞. Recall that a fundamental result ofBesicovitch asserts that E is rectifiable if and only if

Θ1(x,E) = limε→0

H1(E ∩B(x, ε))

2εexists for H1-a.e. x ∈ E,

and in this case Θ1(x,E) = 1 H1-a.e. in E. This characterization should be com-pared with the following.

Corollary 8.18. Let E ⊂ C be H1-measurable with H1(E) < ∞. The following areequivalent:

(a) E is rectifiable.

(b) c2H1�E(x) < ∞ for H1-a.e. x ∈ E.

(c) C∗(H1�E)(x) < ∞ for H1-a.e. x ∈ E.

(d) limε→0 Cε(H1�E)(x) exists for H1-a.e. x ∈ E.


Proof. Notice that MR(H1�E)(x) < ∞ for H1�E-a.e. x ∈ E, since Θ∗1(x,E) ≤ 1for H1-a.e. x ∈ E (see Theorem 1.30).

The equivalence (a) ⇔ (b) was proved in Theorem 7.45, and (c) ⇔ (d) hasbeen shown in the preceding corollary. The implication (a) ⇒ (d) was establishedin Corollary 8.7, while (c) ⇒ (a) follows from Corollary 8.15 and the David-Legertheorem. �

Let us remark that an analogous result holds for a measure μ in C such that0 < Θ∗1(x, μ) < ∞ and C∗μ(x) < ∞ for μ-a.e. x ∈ C. This follows easily fromLemma 8.12, arguing as above.


8.5.1 Existence of principal values on rectifiable sets

The existence of principal values on rectifiable sets holds for other singular integralsbesides the Cauchy transform. Indeed, Albert Mas [90] has shown that if T is ann-dimensional singular integral operator in Rd associated with an odd kernel K(x)which satisfies

|∇K(x)| ≤ c

|x|n+jfor 0 ≤ j ≤ 2 and all x ∈ Rd \ {0},

then the principal value limε→0 Tεν(x) exists Hn-a.e. on n-rectifiable sets, for anycomplex measure ν. Mas obtained this result by studying the ρ-variation of thefamily T ν := {Tεν}ε>0. This is defined by

Vρ(T ν)(x) := sup{εm}

(∑m∈Z

|Tεm+1ν(x) − Tεmν(x)|ρ)1/ρ

,

where the pointwise supremum is taken over all the decreasing sequences {εm}m∈Z

⊂ (0,∞). Mas has shown that, for ρ > 2, Vρ ◦ T is bounded from M(Rd) toL1,∞(Hn�Γ) for any n-dimensional Lipschitz graph Γ with small enough slope.It is easy to check that Vρ(T ν)(x) < ∞ implies the existence of limε→0 Tεν(x).In fact, one can think of the finiteness of the ρ-variation as quantification of theexistence of the principal value.

In the work [91], Mas and Tolsa have studied the ρ-variation of the familyof the truncated Riesz transforms Rn

μ := {Rnμ,ε}ε>0 on uniformly n-rectifiable

measures μ. They have shown that, for an n-dimensional AD regular measure μin Rd and ρ > 2, the operator Vρ ◦ Rn

μ is bounded in L2(μ) if and only if μ isuniformly n-rectifiable.

The Plemelj formulas also hold for more general rectifiable curves than justthe Lipschitz graphs studied in this chapter. The fact that Corollary 8.9 can beobtained from the Plemelj formulas was explained to me by Luis Escauriaza.


8.5.2 From the analytic information to rectifiability

In Corollary 8.16, if one replaces the assumption 0 < Θ∗1(x, μ) < ∞ by Θ∗1(x, μ) >0 μ-a.e., then one deduces that there exists some rectifiable set F ⊂ C, some non-negative function g ∈ L1(H1�F ), and some measure μ0 made up of countablemany point masses such that

μ = μ0 + gH1�F.This has been shown by Tolsa [165]. Under the stronger assumptions that Θ1∗(x, μ)> 0 and that p.v.Cμ(x) exists μ-a.e., this had been proved previously by Mattila[99] using tangent measures (see Mattila’s book [100, Chapter 14] for the notionof tangent measure), before the discovery of the relationship between the Cauchykernel and Menger curvature. See also the work of Jones and Poltoratski [77] fora related result.

Theorem 8.13 holds in more generality than just for the Cauchy transformbecause its main ingredient, the Nazarov-Treil-Volberg Theorem from Chapter 5, isvalid for other singular integral operators, such as the n-dimensional antisymmetricsingular integral operators in Rd. In particular it holds for the n-dimensional Riesztransforms in Rn+1. Then, from Theorem 7.56, one deduces easily that Corollaries8.16, 8.17, and 8.18 (except the statement in (b)) are also valid with the Cauchytransform replaced by the Riesz transform of codimension 1.

For other kernels only some partial results are known. For example, for then-dimensional Riesz transform in Rd, assuming n to be integer, given E ⊂ Rd

with Hn(E) < ∞, the existence of p.v.Rn(Hn�E)(x) Hn-a.e. on E implies thatE is n-rectifiable. This was proved by Tolsa [167], and previously by Mattila andPreiss [104] under the additional assumption that Θn

∗ (x,E) > 0 Hn-a.e. in E.For s ∈ [0, d] \ Z Vihtila [177] proved that there are no sets E ⊂ Rd with

0 < Hs(E) < ∞ such that Θs∗(x,E) > 0 and Rs

∗(Hs�E)(x) < ∞ Hs-a.e. inE. From Prat’s work [136], it follows that the assumption Θs

∗(x,E) > 0 can beremoved in the case 0 < s < 1. This also happens for d − 1 < s < d, as shownby Eiderman, Nazarov and Volberg [37]. For the other non-integers s, this is anopen problem. Nevertheless, Ruiz de Villa and Tolsa [143] proved that, under thestronger condition of the existence of p.v.Rs(Hs�E) Hs-a.e. on E, the assumptionΘs

∗(x,E) > 0 is superfluous too.Other kernels in the plane were studied by Huovinen in his PhD thesis [62].

For j ≥ 1, set

Kj(z) =z2j−1

|z|2j , z ∈ C \ {0},and let Tj be the singular integral associated with Kj . By the method of tangentmeasures, Huovinen proved that, for E ⊂ C with H1(E) < ∞, the H1-a.e. exis-tence in E of principal values of Tj(H1�E)(z) (or of a non-trivial finite linear com-bination of the Tj(H1�E)’s) implies the rectifiability of E, assuming Θ1

∗(z, E) > 0for H1-a.e. z ∈ E. By now it is not known if the same result holds without thelatter assumption on the lower 1-dimensional density of H1�E.


On the other hand, Jaye and Nazarov [73] have shown that there exists apurely unrectifiable set E ⊂ C, with 0 < H1(E) < ∞, such that the singular

integral operator T2,H1�E associated with K2(z) =z3

|z|4 is bounded in L2(H1�E).

The set E that they have constructed turns out to have Θ1∗(z, E) = 0 at allpoints z ∈ E. By Cotlar’s lemma, the L2(H1�E) boundedness of T2,H1�E impliesthe H1�E-a.e. finiteness of T2,∗(H1�E)(z). However, in this example the principalvalues p.v.T2H1�E(z) fail to exist at H1-a.e. z ∈ E. It is not known if one canconstruct a similar counterexample such that Θ1

∗(z, E) > 0 for H1-a.e. z ∈ E, orwith the additional property of being self-similar. See Chousionis and Mattila [12]for a partial result in this direction.

The aforementioned result of Jaye and Nazarov was preceded by anothercounterexample due to Huovinen [63]. He constructed a purely unrectifiable setE ⊂ C with 0 < H1(E) < ∞ such that the singular integral operator associatedwith the kernel

K(z) =x y2

|z|4 , 0 = z = x+ i y

is bounded in L2(H1�E) and, moreover, the corresponding principal values existsH1-a.e. in E. In fact, he showed that the same phenomenon happens for any oddCalderon-Zygmund kernel which vanishes on the real line and such that K(z) =K(z). See also David [24] for related counterexamples.

Finally, it is worth mentioning a work concerning the so-called reflectionlessmeasures. A complex measure with no point masses ν ∈ M(C) is called reflec-tionless if p.v.Cν(z) exists and vanishes |ν|-a.e. Melnikov, Poltoratski and Volberg[109] proved that if ν is reflectionless and C∗ν ∈ L1(|ν|), then ν ≡ 0. For the proofthey used the identity

2 C((p.v.Cν)ν)(z) = Cν(z)2 for L2-a.e. z ∈ C.

Then from the fact that p.v.Cν vanishes |ν|-a.e. one infers that Cν vanishes a.e. inthe plane with respect to planar Lebesgue measure and, as a consequence, ν ≡ 0.The preceding identity follows from the fact that∑

s∈S3

1

(zs2 − zs1)(zs3 − zs1)= 0,

where S3 is the group of permutations of three elements. Notice that the sum abovelooks very much like the sum involved in Melnikov’s identity (3.2) (although herethere is no conjugation).

For the n-dimensional Riesz transforms in Rd one cannot argue as above.However, one can still show that, assuming ν to be a positive finite measure withno point masses, if Rn

∗ν ∈ L1(ν) and p.v.Rnν(x) exists and vanishes ν-a.e., thenν ≡ 0 too. However, it is an open problem to prove (or disprove) the existence ofnon-trivial reflectionless signed measures for the Riesz transforms.


The recent works [71] and [72] by Jaye and Nazarov show that the problemof the characterization of reflectionless measures for Calderon-Zygmund operatorsis strongly related to the David-Semmes problem and other connected questions.

Chapter 9

RBMO(μ) and H1atb(μ)

9.1 Introduction

Let μ be a doubling Radon measure in Rd such that suppμ = Rd. A functionf ∈ L1

loc(μ) is said to belong to BMO(μ) (the space of functions with boundedmean oscillation with respect to μ) if there exists some constant c1 such that

supQ

1

μ(Q)

∫Q

|f(x) −mQ(f)| dμ(x) ≤ c1, (9.1)

where the supremum is taken over all the cubes Q ⊂ Rd and mQ(f) stands forthe mean of f over Q with respect to μ, i.e. mQ(f) =

∫Qf dμ/μ(Q). The optimal

constant c1 is the BMO(μ) norm of f . If suppμ = Rd, then (usually) one definesBMO(μ) by asking the supremum (9.1) to be bounded only for the cubes whichare centered at suppμ.

From now on, throughout all this chapter we will assume that μ is a Radonmeasure on Rd with growth of degree n. That is,

μ(B(x, r)) ≤ c0 rn for all x ∈ Rd, r > 0. (9.2)

Under this assumption, it is well known that if μ is doubling, then any n-dimen-sional singular integral operator Tμ which is bounded in L2(μ) is also boundedfrom L∞(μ) to BMO(μ). In a sense, this property replaces the boundedness inL∞(μ) of Tμ, which fails in general.

However, when μ is non-doubling, the boundedness from L∞(μ) to BMO(μ)may fail too. Hence if one wants to work with a BMO type space which fulfills someof the fundamental properties related with singular integral operators, a differentspace is required. The first candidate is the so-called BMOρ(μ) space. For ρ > 1,one says that a function f ∈ L1

loc(μ) belongs to BMOρ(μ) if for some constant c2

supQ

1

μ(ρQ)

∫Q

|f(x)−mQ(f)| dμ(x) ≤ c2, (9.3)

, , OI 10.1007/978-3- - -6_ ,

© Springer



319

320 Chapter 9. RBMO(μ) and H1atb(μ)

with the supremum taken now over all the cubes Q such that μ(Q) = 0.It is not difficult to check that if Tμ is bounded in L2(μ), then it is also

bounded from L∞(μ) to BMOρ(μ) (this will be proved in Section 9.2.5 below).This is the main advantage of BMOρ(μ) over BMO(μ). Nevertheless, the newspace BMOρ(μ) does not have all the nice properties that one would like. Firstof all, it happens that the definition of BMOρ(μ) depends on the constant ρ > 1that we choose. Not only does the BMOρ(μ) norm of f (i.e. the optimal constantc2 in (9.3)) depend on ρ, but the space itself depends on ρ.

On the other hand, given p ∈ [1,∞), one says that f ∈ BMOpρ(μ) if

supQ

1

μ(ρQ)

∫Q

|f −mQ(f)|p dμ < ∞.

In case μ is doubling measure, by the John-Nirenberg inequality, all the spacesBMOp(μ) ≡ BMOp

ρ=1(μ) coincide. This is not the case if μ is non-doubling.In this chapter, in Section 9.2, we will introduce a new BMO type space

suitable for non-doubling measures which will satisfy some of the properties of theusual “doubling” BMO(μ), such as for example the John-Nirenberg inequality.This space will be a (proper, in general) subspace of the spaces BMOp

ρ(μ). It willbe small enough to fulfill the properties that we have mentioned and big enoughin order that singular integral operators which are bounded in L2(μ) are alsobounded from L∞(μ) to our new space of BMO type.

We will show that if Tμ is bounded in L2(μ) and g ∈ L∞(μ), then the os-cillations of f = Tμ(g) satisfy not only the condition given by (9.3), but anotherregularity condition. Then the functions of our new space will be the functionssatisfying (9.3) together with this additional regularity condition about their oscil-lations. This space will be denoted by RBMO(μ) (this stands for ‘regular boundedmean oscillations’). Notice that we have not written RBMOρ(μ). This is because,as we will see, the definition will not depend on ρ, for ρ > 1.

Because of all the good properties that RBMO(μ) enjoys, this seems to bea good substitute of the classical space BMO(μ) for non-doubling measures ofdegree n.

In this chapter we will also introduce a Hardy space of atomic type: H1,∞atb (μ)

(the subindex ‘atb’ stands for ‘atomic block’). This space will be made up offunctions of the form f =

∑i bi, where the bi’s will be some elementary functions,

which we will call atomic blocks.We will see that this new atomic space enjoys some very interesting prop-

erties. For instance, singular integral operators which are bounded in L2(μ) arealso bounded from H1,∞

atb (μ) to L1(μ). Moreover, we will show that H1,∞atb (μ) is the

predual of RBMO(μ), and that, as in the doubling case, there is a collection ofspaces H1,p

atb(μ), p > 1, that coincide with H1,∞atb (μ).

In Section 9.8, as an application of all the results proved for RBMO(μ) andH1,∞

atb (μ), we will obtain an interpolation theorem: we will prove that if a linear

9.2. The space RBMO(μ) 321

operator is bounded from L∞(μ) to RBMO(μ) and from H1,∞atb (μ) to L1(μ), then

it is bounded in Lp(μ), 1 < p < ∞. By means of this interpolation theorem, we willobtain a new proof of the T 1 theorem for the Cauchy transform for non-doublingmeasures.

9.2 The space RBMO(μ)

9.2.1 Introduction

If μ is a doubling measure and f is a function belonging to BMO(μ), it is easilychecked that if Q,R are two cubes of comparable size with Q ⊂ R, both centeredat suppμ, then

|mQ(f)−mR(f)| ≤ c ‖f‖BMO(μ) (9.4)

In case μ is not doubling and f ∈ BMOρ(μ), it is easily seen that

|mQ(f)−mR(f)| ≤ μ(ρR)

μ(Q)‖f‖BMOρ(μ), (9.5)

and that is all one can get. So if μ(Q) is much smaller than μ(R), then mQ(f)may be very different from mR(f), and one does not have any useful information.However, to prove most results dealing with functions in BMO(μ), some kind ofcontrol in the changes of the mean values of f , such as the one in (9.4), appearsto be essential.

We will see that if Tμ is a singular integral operator that is bounded in L2(μ)and g ∈ L∞(μ), then the oscillations of Tμ(g) satisfy some properties which willbe stronger than (9.5). Some of these properties will be stated in terms of somecoefficients KQ,R, where Q ⊂ R are cubes in Rd. Next we will introduce and studythese coefficients.

9.2.2 The coefficients KQ,R

Throughout the rest of the paper, unless otherwise stated, any cube will be aclosed cube in Rd with sides parallel to the axes such that μ(Q) = 0, unless statedotherwise.

Given two cubes Q ⊂ R in Rd, we set

KQ,R = 1 +

NQ,R∑k=1

μ(2kQ)

(2kQ)n, (9.6)

where NQ,R is the first integer k such that (2kQ) ≥ (R). The coefficient KQ,R

measures how close Q is to R, in some sense. For example, if Q and R havecomparable sizes, then KQ,R is bounded above by some constant which dependson the ratio (R)/(Q) (and on the constant c0 of (9.2)).


Recall that given α, β > 1, we say that some cube Q ⊂ Rd is (α, β)-doublingif μ(αQ) ≤ β μ(Q). Due to the fact that μ satisfies the growth condition (9.2),there are a lot of “big” doubling cubes if we choose β > αn. To be precise, givenany point x ∈ suppμ and r > 0, there exists some (α, β)-doubling cube Q centeredat x with (Q) ≥ r.

Recall also that in Lemma 2.8 we showed that there exist a lot of smalldoubling cubes too. That is, if we choose β > αd, then for μ-a.e. x ∈ Rd thereexists a sequence of (α, β)-doubling cubes {Qk}k centered at x with (Qk) → 0 ask → ∞.

If β is not specified, by an α-doubling cube we will mean a (α, αd+1)-doublingcube. If neither α nor β are specified, we assume that α = 2 and β = 2d+1.

In the following lemma we show some of the properties of the coefficientsKQ,R.

Lemma 9.1. We have:

(a) If Q ⊂ R ⊂ S are cubes in Rd, then KQ,R ≤ KQ,S, KR,S ≤ cKQ,S andKQ,S ≤ c (KQ,R +KR,S).

(b) If Q ⊂ R have comparable sizes, KQ,R ≤ c.

(c) If N is some positive integer and the cubes 2Q, 22Q, . . . 2N−1 are non-(2, β)-doubling (with β > 2n), then KQ,2NQ ≤ c, with c depending on β and c0.

(d) If N is some positive integer and for some β < 2n,

μ(2NQ) ≤ βμ(2N−1Q) ≤ β2μ(2N−2Q) ≤ · · · ≤ βNμ(Q),

then KQ,2NQ ≤ c, with c depending on β and c0.

Proof. The properties (a) and (b) are easy and are left for the reader. Let us see(c). For β > 2n, we have μ(2k+1Q) > β μ(2kQ) for k = 1, . . . , N − 1. Thus

μ(2kQ) <μ(2NQ)

βN−k

for k = 1, . . . , N − 1. Therefore,

KQ,2NQ ≤ 1 +

N−1∑k=1

μ(2NQ)

βN−k (2kQ)n+

μ(2NQ)

(2NQ)n

≤ 1 + c0 +μ(2NQ)

(2NQ)n

N−1∑k=1

1

βN−k 2(k−N)n

≤ 1 + c0 + c0

∞∑k=1

(2n

β

)k≤ c.


Let us check the fourth property. For β < 2n, we have

KQ,2NQ ≤ 1 +N∑

k=1

βkμ(Q)

(2kQ)n

≤ 1 +μ(Q)

(Q)n

N∑k=1

βk

2kn

≤ 1 + c0

∞∑k=1

(β

2n

)k≤ c. �

Note that, in a sense, the property (c) of Lemma 9.1 says that if the densityof the measure μ in the concentric cubes grows much faster than the size of cubes,then the coefficients KQ,2NQ remain bounded, while the fourth property says thatif the measure grows too slowly, then they also remain bounded.

Remark 9.2. If we replace the numbers 2k in the definition (9.6) by αk, for someα > 1, we will obtain a coefficient Kα

Q,R. It is easy to check that KQ,R ≈ KαQ,R

(with constants that may depend on α and c0).Also, if we set

K ′Q,R = 1 +

∫ (R)

(Q)

μ(B(xQ, r))

rn−1dr,

or

K ′′Q,R = 1 +

∫(Q)≤|y−xQ|≤(R)

1

|y − xQ|n dμ(y),

where xQ is the center of Q, then it is easily seen that KQ,R ≈ K ′Q,R ≈ K ′′

Q,R. Thedefinitions of K ′

Q,R and K ′′Q,R have the advantage of not depending on the grid of

cubes, unlike the one of KQ,R.

9.2.3 The definition of RBMO(μ)

Let ρ > 1 be some fixed constant. We say that f ∈ L1loc(μ) is in RBMO(μ) if there

exists a collection of numbers {fQ}Q (i.e. for each cube Q, there exists some realor complex number fQ) and some constant c3 such that

1

μ(ρQ)

∫Q

|f(x)− fQ| dμ(x) ≤ c3 for any cube Q (9.7)

and,|fQ − fR| ≤ c3 KQ,R for any two cubes Q ⊂ R. (9.8)

Recall that we assume all the cubes Q,R to have non-zero μ-measure. Then wewrite ‖f‖∗ = inf c3, where the infimum is taken over all the constants c3 and allthe numbers {fQ}Q satisfying (9.7) and (9.8). It is easily checked that ‖ · ‖∗ is a


norm in the quotient space of RBMO(μ) modulo constants. Abusing notation, wewill identify RBMO(μ) with its quotient space modulo constants.

Let us remark that the space RBMO(μ) depends on the integer n becauseof the definition of the coefficients KQ,R. Also, the definition of the norm ‖ · ‖∗depends on the constant ρ chosen in (9.7). However, if we write ‖ · ‖∗,(ρ) insteadof ‖ · ‖∗, we have

Lemma 9.3. The norms ‖ · ‖∗,(ρ), ρ > 1, are equivalent.

Proof. Let ρ > η > 1 be some fixed constants. Obviously, ‖f‖∗,(ρ) ≤ ‖f‖∗,(η). Sowe only have to show ‖f‖∗,(η) ≤ c ‖f‖∗,(ρ). To this end, it is enough to prove thatfor a fixed collection of numbers {fQ}Q satisfying

supQ

1

μ(ρQ)

∫Q

|f − fQ| dμ ≤ 2 ‖f‖∗,(ρ)

and

|fQ − fR| ≤ 2KQ,R ‖f‖∗,(ρ) for any two cubes Q ⊂ R,

we have

1

μ(ηQ0)

∫Q0

|f − fQ0 | dμ ≤ c ‖f‖∗,(ρ) for any fixed cube Q0. (9.9)

For any x ∈ Q0 ∩ suppμ, let Qx be a cube centered at x with side lengthη−110ρ (Q0). Then (ρQx) = η−1

10 (Q0), and so ρQx ⊂ ηQ0. By Besicovitch’s cov-

ering theorem, there exists a family of points {xi}i ⊂ Q0 ∩ suppμ such that thecubes {Qxi}i form an almost disjoint covering of Q0 ∩ suppμ. Since Qxi and Q0

have comparable sizes,

|fQxi− fQ0 | ≤ c ‖f‖∗,(ρ),

with c depending on η and ρ. Therefore,∫Qxi

|f − fQ0 | dμ ≤∫Qxi

|f − fQxi| dμ+ |fQ0 − fQxi

|μ(Qxi)

≤ c ‖f‖∗,(ρ) μ(ρQxi).

Then we get∫Q0

|f − fQ0 | dμ ≤∑i

∫Qxi

|f − fQ0 | dμ ≤ c ‖f‖∗,(ρ)∑i

μ(ρQxi).

Since ρQxi ⊂ ηQ0 for all i, we obtain∫Q0

|f − fQ0 | dμ ≤ c ‖f‖∗,(ρ) μ(ηQ0)N,


where N is the number of cubes of the Besicovitch covering. Now it is easy tocheck that N is bounded by some constant depending only on η, ρ and d: if Ld isthe Lebesgue measure on Rd and pd is the Besicovitch constant in Rd, we have

N Ld(Qxi) =∑i

Ld(Qxi) ≤ pd Ld(ηQ0).

Thus

N ≤ pd Ld(ηQ0)

Ld(Qxi)= pd

(10ηρ

η − 1

)d,

and (9.9) holds. �Throughout the rest of the chapter we will assume that the constant ρ in the

definition of RBMO(μ) is 2.

Remark 9.4. In fact, in the preceding lemma we have seen that if cf is someconstant and {fQ}Q is some fixed collection of numbers satisfying

supQ

1

μ(ρQ)

∫Q

|f − fQ| dμ ≤ cf

and|fQ − fR| ≤ cf KQ,R for any two cubes Q ⊂ R,

then for the same numbers {fQ}Q, for any η > 1 we have

supQ

1

μ(ηQ)

∫Q

|f − fQ| dμ ≤ c cf ,

with c depending on η.Also, arguing as above, it can be shown that we also obtain equivalent defi-

nitions for the space RBMO(μ) if, in the conditions above, instead of cubes withnon-zero μ-measure we consider only cubes centered at points in suppμ. Further-more, it does not matter if we take balls instead of cubes either.

Notice that if (9.7) is satisfied, then (9.3) also holds. Indeed, for any cube Qand any a ∈ R one has ∫

Q

|f −mQf | dμ ≤ 2

∫Q

|f − a| dμ. (9.10)

In particular this holds for a = fQ. So the conditions (9.7), (9.8) are stronger than(9.3).

Observe also that, as a consequence of (9.8), if Q ⊂ R are cubes with com-parable size, then

|fQ − fR| ≤ c ‖f‖∗, (9.11)

taking into account the property (b) of Lemma 9.1.


Proposition 9.5. (a) RBMO(μ) is a Banach space of functions (modulo additiveconstants).

(b) L∞(μ) ⊂ RBMO(μ), with ‖f‖∗ ≤ 2‖f‖L∞(μ).

(c) If f ∈ RBMO(μ), then |f | ∈ RBMO(μ) and ‖ |f | ‖∗ ≤ ‖f‖∗.(d) If f, g ∈ RBMO(μ), then min(f, g), max(f, g) ∈ RBMO(μ) and

‖min(f, g)‖∗, ‖max(f, g)‖∗ ≤ ‖f‖∗ + ‖g‖∗.Proof. The property (a) was already mentioned above. For (b), we just choosefQ = mQ(f). The third property is also easy to prove: given the family {fQ}Qassociated to f , we take

(|f |)Q := |fQ|.Then we have

1

μ(ρQ)

∫Q

∣∣|f | − |fQ|∣∣ dμ ≤ 1

μ(ρQ)

∫Q

|f − fQ| dμ

and ∣∣|fQ| − |fR|∣∣ ≤ |fQ − fR|,

which implies (c). The fourth property follows from the third one, putting

max(f, g) =f + g + |f − g|

2, min(f, g) =

f + g − |f − g|2

. �

9.2.4 Equivalent definitions

Given a cube Q ⊂ Rd, let N be the smallest integer ≥ 0 such that 2NQ is doubling.We denote by Q this cube (recall that Q exists because otherwise the growthcondition (9.2) on μ would fail). This notation will be used in the whole chapter.

To check if some function f belongs to RBMO(μ) does not look easy fromthe definition in (9.7) and (9.8) because it is not clear how to choose a suitablecollection of numbers {fQ}Q. In the next lemma we show that we may take fQ =mQf .

Proposition 9.6. Let ρ > 1 be fixed. If f ∈ RBMO(μ), then

1

μ(ρQ)

∫Q

|f −mQf | dμ ≤ c ‖f‖∗ for any cube Q ⊂ Rd (9.12)

and|mQf −mRf | ≤ c ‖f‖∗KQ,R for any two cubes Q ⊂ R. (9.13)

Conversely, if f ∈ L1loc(μ) and

1

μ(ρQ)

∫Q

|f −mQf | dμ ≤ c4 for any cube Q ⊂ Rd (9.14)


and

|mQf −mRf | ≤ c4 KQ,R for any two doubling cubes Q ⊂ R, (9.15)

then f ∈ RBMO(μ), with ‖f‖∗ ≤ c c4.

Notice that in (9.15) one needs the condition (9.15) only for doubling cubesQ and R.

For the proof we need the following auxiliary result.

Lemma 9.7. Let {fQ}Q be a collection of numbers and some constant A such that

|fQ − fR| ≤ AKQ,R for any two doubling cubes Q ⊂ R.

Then

|fQ − fR| ≤ cAKQ,R for any two cubes Q ⊂ R. (9.16)

Proof. Notice that if Q ⊂ R, then

|mQf −mRf | ≤ KQ,R A,

because Q, R are doubling. So (9.16) follows if KQ,R ≤ cKQ,R. However, in gen-

eral, Q ⊂ R does not imply Q ⊂ R, and so we have to modify the argument.

Suppose first that (R) ≥ (Q). Then Q ⊂ 4R. We write R0 = 4R. Then wehave

|fQ − fR| ≤ |fQ − fR0 |+ |fR0 − fR|. (9.17)

Applying the properties of Lemma 9.1 repeatedly, we get

KQ,R0≤ cKQ,R0 ≤ c (KQ,R +KR,R0)

≤ c (KQ,R +KR,R +KR,4R +K4R,R0) ≤ cKQ,R.

Since Q ⊂ R0 and they are doubling cubes, we have

|fQ − fR0 | ≤ AKQ,R0≤ cAKQ,R.

Now we are left with the second term on the right-hand side of (9.17). We have

KR,R0≤ c(KR,4R +K4R,R0

) ≤ c ≤ cKQ,R.

Due to the fact that R ⊂ R0 are doubling cubes,

|fR0 − fR| ≤ AKR,R0≤ cAKQ,R,

and, by (9.17), we infer that (9.16) holds in this case.


Assume now (R) < (Q). Then R ⊂ 4Q. There exists some m ≥ 1 such that

(R) ≥ (2mQ)/10 and R ⊂ 2mQ ⊂ 4Q. Since R and 2mQ have comparable sizes,

we have KR,2mQ ≤ c. Then, if we write Q0 = 4Q, we get

KR,Q0≤ c (KR,2mQ +K2mQ,4Q +K4Q,Q0

) ≤ c.

Also,KQ,Q0

≤ c (KQ,4Q +K4Q,Q0) ≤ c.

Therefore,

|fQ − fR| ≤ |fQ − fQ0 |+ |fQ0 − fR|≤ AKQ,Q0

+AKR,Q0≤ cA ≤ cAKQ,R. �

Proof of Proposition 9.6. First we show that if f ∈ RBMO(μ), then (9.12) and(9.13) hold. If Q is a doubling cube, since (9.7) holds with ρ = 2 (by Lemma 9.3),we have

|mQf − fQ| =∣∣∣∣ 1

μ(Q)

∫Q

(f − fQ) dμ

∣∣∣∣ ≤ ‖f‖∗ μ(2Q)

μ(Q)≤ c ‖f‖∗.

Therefore, for any cube Q (non-doubling, in general), taking into account thatKQ,Q ≤ c we get

|fQ −mQf | ≤ |fQ − fQ|+ |fQ −mQf | ≤ c ‖f‖∗.Thus

1

μ(ρQ)

∫Q

|f −mQf | dμ ≤ 1

μ(ρQ)

∫Q

|f − fQ| dμ

+1

μ(ρQ)

∫Q

|fQ −mQf | dμ ≤ c ‖f‖∗.

It only remains to show (9.13). This follows easily. Indeed, suppose first that Q ⊂ Rare doubling cubes. Then we have

|mQf −mRf | ≤ |mQf − fQ|+ |fQ − fR|+ |fR −mRf |≤ c ‖f‖∗ +KQ,R ‖f‖∗ + c ‖f‖∗ ≤ cKQ,R ‖f‖∗.

If Q ⊂ R are arbitrary cubes which may be non-doubling, then (9.13) follows fromthe preceding estimate and Lemma 9.7 applied to {fQ}Q = {mQf}Q.

We turn now to the second statement in the lemma. From (9.15) and Lemma9.7, we infer that

|mQf −mRf | ≤ c c4KQ,R for any two cubes Q ⊂ R.

Then, setting fQ = mQf for all the cubes Q, we deduce that f ∈ RBMO(μ) and

‖f‖∗ ≤ c c4. �


Remark 9.8. Arguing as above, it is easily seen that one obtains equivalent normsand the same space RBMO(μ) if we replace (2, 2d+1)-doubling cubes in (9.14) and(9.15) of the space RBMO(μ) by (α, β)-doubling cubes, for any choice of α > 1and β > αn.

Observe also that, as a consequence of (9.13), if Q ⊂ R are doubling cubeswith comparable sizes, then

|mQf −mRf | ≤ c ‖f‖∗, (9.18)

taking into account the property (b) of Lemma 9.1.

Remark 9.9. In fact, (9.11) also holds for any two doubling cubes with comparablesizes such that dist(Q,R) � (Q). To see this, let Q0 be a cube concentric withQ, containing Q and R, and such that (Q0) ≈ (Q). Then K

Q0,Q0≤ c, and thus

we have KQ,Q0

≤ c and KR,Q0

≤ c (we have used the properties (a), (b) and (c)

of Lemma 9.1). Then |mQf −mQ0

f | ≤ c ‖f‖∗ and |mRf −mQ0

f | ≤ c ‖f‖∗, whichyield (9.11).

In the next lemma we still show more equivalent ways of defining RBMO(μ).

Proposition 9.10. Let ρ > 1 be fixed. For a function f ∈ L1loc(μ), the following are

equivalent:

(a) f ∈ RBMO(μ).

(b) There exists some constant cb such that for any cube Q∫Q

|f −mQf | dμ ≤ cb μ(ρQ) (9.19)

and

|mQf −mRf | ≤ cb KQ,R

(μ(ρQ)

μ(Q)+

μ(ρR)

μ(R)

)for any two cubes Q ⊂ R.

(9.20)

(c) There exists some constant cc such that for any doubling cube Q∫Q

|f −mQf | dμ ≤ cc μ(Q) (9.21)

and

|mQf −mRf | ≤ cc KQ,R for any two doubling cubes Q ⊂ R. (9.22)

Moreover, the best constants cb and cc are comparable to the RBMO(μ) norm off .


Proof. Assume ρ = 2 for simplicity. First we show (a) ⇒ (b). If f ∈ RBMO(μ),then (9.19) holds with cb = 2‖f‖∗, as remarked in (9.10). Also, we write

|mQf −mRf | ≤ |mQf −mQf |+ |mQf −mRf |+ |mRf −mRf |.

By (9.13), we know that |mQf −mRf | ≤ c ‖f‖∗KQ,R. Also, from (9.12), for anycube Q we have

|mQf −mQf | ≤ mQ(|f −mQf |) ≤ c ‖f‖∗μ(2Q)

μ(Q), (9.23)

and analogously with respect to R. So we get,

|mQf −mRf | ≤ c

(KQ,R +

μ(2Q)

μ(Q)+

μ(2R)

μ(R)

)‖f‖∗

≤ cKQ,R

(μ(2Q)

μ(Q)+

μ(2R)

μ(R)

)‖f‖∗.

Thus f satisfies (9.20) too.The implication (b) ⇒ c) is easier: one only has to consider doubling cubes

in (b).Let us see now (c)⇒ (a). By Proposition 9.6, we only have to show that

(9.14) holds for any cube Q, non-doubling in general. We know that for μ-almostall x ∈ Q there exists some doubling cube centered at x with side length 2−k (Q),for some k ≥ 1. We denote by Qx the biggest cube satisfying these properties.Observe that KQx,Q

≤ c, and then

|mQxf −mQf | ≤ c cc. (9.24)

By Besicovitch’s covering theorem, there are points xi ∈ Q such that μ-almost all Q is covered by a family of cubes {Qxi}i with bounded overlap. By(9.24), using that Qxi ⊂ 2Q, we get∫

Q

|f −mQf | dμ ≤∑i

∫Qxi

|f −mQf | dμ

≤∑i

∫Qxi

|f −mQxif | dμ+

∑i

|mQf −mQxif |μ(Qxi)

≤ c cc μ(2Q). �

9.2.5 Boundedness of CZO’s from L∞(μ) to RBMO(μ)

Now we are going to see that if a singular integral operator is bounded in L2(μ),then it is bounded from L∞(μ) to RBMO(μ). First of all, notice that for f ∈L∞(μ), the integral

∫|x−y|>ε

K(x, y) f(y) dμ(y) may not be absolutely convergent.


This problem can be solved in the following way. Given a cube Q0 centered at theorigin with side length > 3ε, we write f = f1 + f2, with f1 = f χ2Q0 . For x ∈ Q0,we define

Tμ,εf(x) = Tμ,εf1(x) +

∫(K(x, y)−K(0, y)) f2(y) dμ(y). (9.25)

Now it easily seen that both integrals in this equation are convergent, and moreoverone can check that the definition above of Tμ,εf as an RBMO(μ) function doesnot depend on the particular chosen cube Q0.

The main result that we will prove in this section is the following.

Theorem 9.11. Let T be an n-dimensional singular integral operator. If for anycube Q and any function a supported on Q∫

Q

|Tμ,εa| dμ ≤ c ‖a‖L∞ μ(ρQ) (9.26)

uniformly on ε > 0 (in particular, this holds if Tμ is bounded in L2(μ)), then Tμ

is bounded from L∞(μ) to RBMO(μ).

We need the following auxiliary estimate.

Lemma 9.12. Let T be an n-dimensional singular integral operator, let Q ⊂ Rd besome cube, and let f ∈ Lp(μ), for some p ∈ [1,∞), with supp f ⊂ Rd \ 2Q. Forx, y ∈ Q we have

|Tμ,εf(x)− Tμ,εf(y)| ≤ c ‖f‖L∞(μ),

for all ε > 0.

The proof is very standard and it is similar to the one of Lemma 2.12. How-ever, for the reader’s convenience we will show the detailed arguments.

Proof. Let K(·, ·) be the kernel of T . Writing

χ|x−z|>εK(x, z)− χ|y−z|>εK(y, z)

= χ|x−z|>ε

(K(x, z)−K(y, z)

)+ (χ|x−z|>ε − χ|y−z|>ε)K(y, z),

we get

|Tμ,εf(x)− Tμ,εf(y)| ≤ ‖f‖L∞(μ)

∫z �∈2Q|x−z|>ε

∣∣K(x, z)−K(y, z)∣∣dμ(z) (9.27)

+ ‖f‖L∞(μ)

∫D∩(2Q)c

|K(y, z)| dμ(z),

where

D = {z : |x− z| > ε}Δ {z : |y − z| > ε}.


On the other hand, since (QR)≈(R), we havemR|Tμ,ε(f χQR\2R)| ≤ c ‖f‖L∞(μ).Therefore,

|mQ(Tμ,εf)−mR(Tμ,εf)| ≤ c

NQ,R∑k=1

μ(2k+1Q)

(2k+1Q)n‖f‖L∞(μ)

+ c

(μ(2ρQ)

μ(Q)+

μ(2ρR)

μ(R)

)‖f‖L∞(μ)

≤ cKQ,R

(μ(2ρQ)

μ(Q)+

μ(2ρR)

μ(R)

)‖f‖L∞(μ).

So we have proved that (9.28) holds for f ∈ L∞(μ) ∩ Lp0(μ).

If f ∈ Lp(μ) for all p ∈ [1,∞), then the integral∫|x−y|>εK(x, y) f(y) dμ(y)

may be not convergent. In this case, Tμ,εf has to be defined as in (9.25). Theproof follows by estimates analogous to the ones above. The details are left for thereader. �

Let us remark that in Theorem 9.38 we will show that the condition (9.26)holds if and only if Tμ is bounded from L∞(μ) to RBMO(μ).

9.2.6 Some examples

Example 9.13. Assume d = 2 and n = 1. Let E ⊂ C be a 1-dimensional ADregular set. Recall that this means that

c−1 r ≤ H1(E ∩B(x, r)) ≤ c r for all x ∈ E, 0 < r ≤ diam(E).

We set μ = H1�E.For any Q centered at some point of suppμ, we have

μ(2kQ) ≈ (2kQ)

if (2kQ) ≤ diam(E). GivenQ ⊂ R, then it is easy to check that if (R) ≤ diam(E),

KQ,R ≈ 1 + log(R)

(Q), (9.30)

and if (R) > diam(E),

KQ,R ≈ 1 + logdiam(E)

(Q). (9.31)

So in this case we have RBMO(μ) = BMO(μ), since any function f ∈ BMO(μ)satisfies (9.7) and (9.8), with fQ = mQf for all the cubes Q centered at suppμ,and with KQ,R satisfying (9.30) and (9.31). Indeed, recall that in Remark 9.4 wesaid that it is enough to deal with the centered cubes at suppμ to check thatf ∈ RBMO(μ).


Example 9.14. We assume again d = 2 and n = 1. Let μ be the planar Lebesguemeasure restricted to the unit square [0, 1]× [0, 1]. This measure is doubling, butnot AD regular (for n = 1). Now one can check that the coefficients KQ,R areuniformly bounded. That is, for any two squares Q ⊂ R,

KQ,R ≈ 1.

Let us take R0 = [0, 1]2 and Q ⊂ R0. Then, if f ∈ RBMO(μ),

|mQ(f −mR0f)| = |mQf −mR0f | ≤ KQ,R0 ‖f‖∗ ≤ c ‖f‖∗.Since this holds for any square Q ⊂ R0, by the Lebesgue differentiation theoremf −mR0f is a bounded function, with

‖f −mR0f‖L∞(μ) ≤ ‖f‖∗.Therefore, now RBMO(μ) coincides with L∞(μ) modulo constants functions,

which is strictly smaller than BMO(μ).

Example 9.15. Suppose d = 2 and n = 1 again. Let μ be a measure on the realaxis such that in the intervals [−2,−1] and [1, 2] is the linear Lebesgue measure,on the intervals [−1/2,−1/4], [1/4, 1/2] is the linear Lebesgue measure times ε,with ε > 0 very small, and μ = 0 elsewhere. We consider the function f =ε−1 (χ[1/4,1/2] − χ[−1/2,−1/4]). It is easily checked that for ρ ≤ 2,

‖f‖BMOρ(μ) ≈ ε−1.

Indeed, ‖f‖BMOρ(μ) ≤ 2‖f‖L∞(μ) ≤ 2ε−1. On the other hand, consider the squareQ = [−1/2, 1/2]2. Then mQf = 0 and so∫

Q

|f −mQf | dμ =

∫Q

|f | dμ ≈ ε−1 μ(Q) = ε−1 μ(ρQ),

taking into account that ρQ does not intersect R\ [−1, 1] in the last identity. Thus,‖f‖BMOρ(μ) � ε−1.

On the other hand, we have

‖f‖BMO5(μ) ≈ 1.

Indeed, if Q intersects only one component of suppμ, then∫Q|f −mQf | dμ = 0

because f is constant in each component. On the other hand, if Q intersects both[1/4, 1/2] and [−1/2,−1/4], then ρQ ⊃ [−5/2, 5/2] and so μ(ρQ) ≈ (Q). Thenwe deduce ∫

Q

|f −mQf | dμ ≤ c

∫Q

|f | dμ ≤ c(Q) ≈ μ(ρQ).

An analogous estimate holds for any square which intersects at least two differentcomponents of suppμ. So we infer that ‖f‖BMO5(μ) � 1. The converse inequalitycan be checked by considering the square Q = [−1/2, 1/2]2, say.


Finally, concerning the RBMO(μ) norm of f we have

‖f‖∗ ≈ ε−1,

since for the 2-doubling squares [−1/2, 1/2]2, [1/4, 1/2]2 we have

c ‖f‖∗ ≥ |m[−1/2,1/2]2f −m[1/4,1/2]2f | = ε−1

and ‖f‖∗ ≤ c ‖f‖L∞(μ) ≤ c ε−1.

9.3 The John-Nirenberg inequality

The following result is a version of the classical John-Nirenberg inequality (see[36, Chapter 6], for example) suitable for the space RBMO(μ).

Theorem 9.16. Let f ∈ RBMO(μ) and let {fQ}Q be a collection of numbers sat-isfying

supQ

1

μ(2Q)

∫Q

|f − fQ| dμ ≤ c5 ‖f‖∗ (9.32)

and

|fQ − fR| ≤ c5 KQ,R ‖f‖∗ for any two cubes Q ⊂ R. (9.33)

Then for any cube Q and any λ > 0 we have

μ({x ∈ Q : |f(x)− fQ| > λ}) ≤ c6 μ(ρQ) exp

(−c7 λ

‖f‖∗

), (9.34)

with c6 and c7 depending on c5 and ρ > 1 (but not on λ).

To prove this theorem we will use the following straightforward result:

Lemma 9.17. Let f ∈ RBMO(μ) and let {fQ}Q be a collection of numbers sat-isfying (9.32) and (9.33). If Q and R are cubes such that (Q) ≈ (R) anddist(Q,R) � (Q), then

|fQ − fR| ≤ c ‖f‖∗.Proof. Let R′ be the smallest cube concentric with R containing Q and R. Since(Q) ≈ (R′) ≈ (R), we have KQ,R′ ≤ c and KR,R′ ≤ c. Then

|fQ − fR| ≤ |fQ − fR′ |+ |fR − fR′ | ≤ c (KQ,R +KR,R′) ‖f‖∗ ≤ c ‖f‖∗. �

Proof of Theorem 9.16. We will prove (9.34) for ρ = 2. The proof for other valuesof ρ is similar. Recall that if (9.32) and (9.33) are fulfilled, then (9.32) is alsosatisfied substituting “μ(2Q)” by “μ(ρQ)”, for any ρ > 1 (see Remark 9.4).

Let f ∈ RBMO(μ). Assume first that f is bounded. Let Q0 be some fixedcube in Rd. We write Q′

0 = 32Q0.

9.3. The John-Nirenberg inequality 337

Let B be some positive constant which will be fixed later. By Remark 2.10,for μ-almost any x ∈ Q0 such that |f(x) − fQ0 | > B ‖f‖∗, there exists somedoubling cube Qx centered at x satisfying

mQx(|f − fQ0 |) > B ‖f‖∗ (9.35)

with side length 2−k (Q0) for some integer k ≥ 0, and such that

(Qx) ≤ 1

10(Q0).

Moreover, we assume that Qx is the biggest doubling cube of this form. By Besi-covitch’s covering theorem, there exists an almost disjoint subfamily {Q1

i }i of thecubes {Qx}x such that μ-a.e. x ∈ Q0 such that |f(x)− fQ0 | > B ‖f‖∗ is containedin⋃

iQ1i . Then, since Q1

i ⊂ Q′0 and |fQ0 − fQ′

0| ≤ c ‖f‖∗, we have

∑i

μ(Q1i ) ≤∑i

1

B ‖f‖∗

∫Q1

i

|f − fQ0 | dμ (9.36)

≤ c

B ‖f‖∗

∫Q′

0

|f − fQ0 | dμ

≤ c

B ‖f‖∗ |fQ0 − fQ′0|μ(Q′

0) +c

B ‖f‖∗

∫Q′

0

|f − fQ′0| dμ.

Since (9.32) holds if we change “μ(2Q)” by “μ(43Q)”, we have∫Q′

0

|f − fQ′0| dμ ≤ c μ(2Q0) ‖f‖∗,

and, by (9.36),∑

i μ(Q1i ) ≤

c μ(2Q0)

B. So if we choose B big enough,

∑i

μ(Q1i ) ≤

μ(2Q0)

2d+2. (9.37)

Now we want to see that for each i we have

|fQ1i− fQ0 | ≤ 2B ‖f‖∗, (9.38)

for B big enough. We consider the cube 2Q1i . If (2Q

1i ) > 10(Q0), then there exists

some cube 2mQ1i , m ≥ 1, containing Q0 and such that (Q0) ≈ (2mQ1

i ) ≤ (2Q1i ).

Thus|fQ1

i− fQ0 | ≤ |fQ1

i− f2Q1

i|+ |f2Q1

i− f2mQ1

i|+ |f2mQ1

i− fQ0 |.

The first and third summands on the right-hand side are bounded by c ‖f‖∗ be-cause Q1

i and 2Q1i on the one hand and 2mQ1

i and Q0 on the other hand have


comparable sizes. The second sum is also bounded by c ‖f‖∗ due to the fact thatthere are no doubling cubes of the form 2kQ1

i between Q1i and 2mQ1

i , and thenKQ1

i ,2mQ1

i≤ c.

Assume now 110 (Q0) < (2Q1

i ) ≤ 10(Q0). Then

|fQ1i− fQ0 | ≤ |fQ1

i− f

2Q1i

|+ |f2Q1

i

− fQ0 |.

Since 2Q1i and Q0 have comparable sizes, by Lemma 9.17 we have |f

2Q1i

− fQ0 | ≤c ‖f‖∗. And since K

Q1i ,2Q

1i

≤ c (KQ1i ,2Q

1i+K

2Q1i ,2Q

1i

), we also have |fQ1i− f

2Q1i

| ≤c ‖f‖∗. So (9.38) holds in this case too.

If (2Q1i ) ≤ 1

10 (Q0), then by the choice of Q1i , we have m

2Q1i

(|f − fQ0 |) ≤B ‖f‖∗, which implies

|m2Q1

i

(f − fQ0)| ≤ B ‖f‖∗. (9.39)

Thus

|fQ1i− fQ0 | ≤ |fQ1

i− f

2Q1i

|+ |f2Q1

i

−m2Q1

i

f |+ |m2Q1

i

f − fQ0 |.

As above, the term |fQ1i−f

2Q1i

| is bounded by c ‖f‖∗. The last one equals |m2Q1i

(f−fQ0)|, which is estimated in (9.39). For the second one, since 2Q1

i is doubling, wehave

|f2Q1

i

−m2Q1

i

f | ≤ 1

μ(2Q1i )

∫2Q1

i

|f − f2Q1

i

| dμ ≤ cμ(2 · 2Q1

i )

μ(2Q1i )

‖f‖∗ ≤ c ‖f‖∗.

So (9.38) holds in any case, for B big enough.Now we operate with each cube Q1

i as we did with Q0. Then we obtain afamily of cubes Q2

i,j with finite overlap which satisfy

|f(x)− fQ1i| ≤ B ‖f‖∗ for μ-a.e. x ∈ Q1

i \⋃

j Q2i,j .,∑

j

μ(Q2i,j) ≤

μ(2Q1i )

2d+2≤ μ(Q1

i )

2

(using that Q1i is (2, 2d+1) doubling in the last inequality), and

|fQ2i,j

− fQ1i| ≤ 2B ‖f‖∗ for all j.

We gather the cubes Q2i,j corresponding to all the Q1

i ’s in a unique family

{Q2j}, and then we will have

∑j

μ(Q2j) ≤∑i

μ(Q1i )

2≤ μ(2Q0)

2 · 2d+2,

9.3. The John-Nirenberg inequality 339

and moreover, for μ-a.e. x ∈ Q0 \⋃

j Q2j ,

|f(x)− fQ0 | ≤ |f(x)− fQ1i|+ |fQ1

i− fQ0 | ≤ 4B ‖f‖∗.

Repeating this process infinitely many times, for each N ≥ 1 we get a familyof cubes {QN

i } such that∑i

μ(QNi ) ≤ μ(2Q0)

2N−1 · 2d+2=

μ(2Q0)

2N+d+1,

and, for μ-a.e. x ∈ Q0 \⋃

i QNi ,

|f(x) − fQ0 | ≤ 2BN ‖f‖∗.

To prove (9.34), suppose first that λ ≥ 2B ‖f‖∗, and take N ≥ 1 such that2BN ≤ λ/ ‖f‖∗ < 2B(N + 1). Then we have

μ({x ∈ Q0 : |f(x)− fQ0 | > λ}) ≤∑

i

μ(QNi ) ≤ μ(2Q0)

2N+d+1

=μ(2Q0)

2d+1e−N log 2 ≤ μ(2Q0)

2d+1e−c7λ/ ‖f‖∗ ,

with c7 = log 2/4B.If λ < 2B ‖f‖∗, then c72B − c7λ/‖f‖∗ > 0, and thus

μ({x ∈ Q0 : |f(x)− fQ0 | > λ}) ≤ μ(2Q0) ≤ μ(2Q0) e

c72B−c7λ/‖f‖∗

= c μ(2Q0) e−c7λ/‖f‖∗ ,

and so (9.34) also holds. �From Theorem 9.16 we will deduce easily that the following spaces RBMOp(μ)

coincide for all p ∈ [1,∞). Given ρ > 1 and p ∈ [1,∞), we say that f ∈ L1loc(μ) is

in RBMOp(μ) if there exists a collection of numbers {fQ}Q and some constant c9such that (

1

μ(ρQ)

∫Q

|f − fQ|p dμ)1/p

≤ c9 for any cube Q (9.40)

and|fQ − fR| ≤ c9 KQ,R for any two cubes Q ⊂ R. (9.41)

Recall that we assume all the cubes to have non-zero μ-measure. The minimalconstant c9 is the RBMOp(μ) norm of f , denoted by ‖ · ‖∗,p. Arguing as forp = 1, one can show that the definition of the space does not depend on theconstant ρ > 1, and one obtains other equivalent definitions, setting fQ = mQf ,for example.

We have the following corollary of the John-Nirenberg inequality:


Corollary 9.18. For p ∈ [1,∞), the spaces RBMOp(μ) coincide, and the norms‖ · ‖∗,p are equivalent.

Proof. The conditions (9.8) and (9.41) are the same. So we only have to compare(9.7) and (9.40).

For any f ∈ L1loc(μ), the inequality ‖f‖∗ ≤ ‖f‖∗,p follows from Holder’s

inequality. To obtain the converse one we will apply John-Nirenberg. If f ∈RBMO(μ), then

1

μ(ρQ)

∫Q

|f −mQf |p dμ =1

μ(ρQ)

∫ ∞

0

p λp−1 μ({

x : |f(x)−mQf | > λ})

dλ

≤ c p

∫ ∞

0

λp−1 exp

(−c λ

‖f‖∗

)dλ ≤ c ‖f‖p∗,

and so ‖f‖∗,p ≤ c ‖f‖∗. �

9.4 The Hardy spaces H1,patb(μ)

If we say that f is in BMOρ(μ) when it satisfies (9.3), it seems that we have to

consider the atomic space H1,∞at,ρ (μ) consisting of functions of the form

f =∑i

λi ai, (9.42)

where λi ∈ R (or C),∑

i |λi| < ∞ and, for each i, ai is a function supported ona cube Qi, with ‖ai‖L∞(μ) ≤ μ(ρQi)

−1, and∫ai dμ = 0 (that is, ai is an atom).

Obviously, H1,∞at,ρ (μ) is the usual atomic space H1,∞

at (μ) ≡ H1,∞at,ρ=1(μ) when μ is

a doubling measure (see Duoandikoetxea [36, Chapter 6]). With this definition,a singular integral operator which is bounded in L2(μ), is also bounded fromH1,∞

at,ρ (μ) to L1(μ) (taking ρ > 1). However, analogously to BMOρ(μ), this is notthe right space to deal with.

We are going to introduce another space of atomic type, H1,∞atb (μ), which has

better properties than H1,∞at,ρ (μ). For a fixed ρ > 1, a function b ∈ L1

loc(μ) is calledan atomic block if

1. there exists some cube R such that supp(b) ⊂ R,

2.

∫b dμ = 0,

3. there are functions aj supported on cubes Qj ⊂ R and numbers λj ∈ R (orC) such that b =

∑∞j=1 λjaj , and

‖aj‖L∞(μ) ≤(μ(ρQj)KQj ,R

)−1.

9.4. The Hardy spaces H1,patb(μ) 341

Then we write|b|H1,∞

atb (μ) =∑j

|λj |

(to be rigorous, we should think that b is not only a function, but a structureformed by the function b, the cubes R and Qj , the functions aj , etc.)

We say that f ∈ H1,∞atb (μ) if there are atomic blocks bi such that

f =∞∑i=1

bi,

with∑

i |bi|H1,∞atb (μ) < ∞. The H1,∞

atb (μ) norm of f is

‖f‖H1,∞atb (μ) = inf

∑i

|bi|H1,∞atb (μ),

where the infimum is taken over all the possible decompositions of f in atomicblocks.

Observe the difference with the atomic space H1,∞at,ρ (μ). The size condition on

the functions aj is similar (if we do not worry about the coefficient KQ,R), but thecancellation property

∫aj dμ = 0 is substituted by a requirement which offers a

higher degree of freedom: we gather some terms λjaj into an atomic block b, andthen we must have

∫b dμ = 0.

Notice also that if we take atomic blocks bi made up of a unique function ai,we derive H1,∞

at,ρ (μ) ⊂ H1,∞atb (μ).

We will now introduce the atomic spaces H1,patb(μ) (and later we will prove

that they coincide with H1,∞atb (μ)). For a fixed ρ > 1 and p ∈ (1,∞), a function

b ∈ L1loc(μ) is called a p-atomic block if

1. there exists some cube R such that supp(b) ⊂ R,

2.

∫b dμ = 0,

3. there are functions aj supported in cubes Qj ⊂ R and numbers λj ∈ R (orC) such that b =

∑∞j=1 λjaj , and

‖aj‖Lp(μ) ≤ μ(ρQj)1/p−1 K−1

Qj ,R.

We write |b|H1,patb(μ)

=∑

j |λj | (as in the case of H1,∞atb (μ), to be rigorous we should

think that b is not only a function, but a structure formed by the function b, thecubes R and Qj , the functions aj , etc.). Then we say that f ∈ H1,p

atb(μ) if there are

p-atomic blocks bi such that f =∑∞

i=1 bi, with∑

i |bi|H1,patb(μ)

< ∞. The H1,patb(μ)

norm of f is

‖f‖H1,patb(μ)

= inf∑i

|bi|H1,patb(μ)

,


where the infimum is taken over all the possible decompositions of f in p-atomicblocks.

From the definitions above, it is easily checked that for 1 < p ≤ q ≤ ∞,H1,∞

atb (μ) ⊂ H1,qatb(μ) ⊂ H1,p

atb(μ), and moreover,

‖f‖H1,patb(μ)

≤ ‖f‖H1,qatb(μ)

≤ ‖f‖H1,∞atb (μ).

In the next proposition we show some elementary properties of the spacesH1,p

atb(μ).

Proposition 9.19. For 1 < p ≤ ∞, we have:

(a) H1,patb(μ) is a Banach space.

(b) H1,patb(μ) ⊂ L1(μ), with ‖f‖L1(μ) ≤ ‖f‖H1,p

atb(μ).

(c) The definition of H1,patb(μ) does not depend on the constant ρ > 1.

Proof. The proofs of the properties (a) and (b) are very standard and we skipthem.

To prove (c) we use an argument similar to the one of Lemma 9.3. Givenρ > η > 1, it is obvious that H1,p

atb,ρ(μ) ⊂ H1,patb,η(μ) and ‖f‖H1,p

atb,η(μ)≤ ‖f‖H1,p

atb,ρ(μ).

For the converse inequality, given a p-atomic block b =∑

j λjaj with supp(aj) ⊂Qj ⊂ R, it is not difficult to see that each function aj can be written as

aj =

N∑k=1

aj,k,

where the aj,k’s are functions such that ‖aj,k‖Lp(μ) ≤ ‖aj‖Lp(μ) for all k, withsupp(aj,k) ⊂ Qj,k, so that Qj,k are cubes such that (Qj,k) ≈ (Qj) and ρQj,k ⊂ηQj. Moreover, N is a fixed number depending on ρ and η.

Then for each j, k we have

‖aj,k‖Lp(μ) ≤ ‖aj‖Lp(μ) ≤(μ(ρQj)

1/p′KQj ,R

)−1

≤ c(μ(ηQj,k)

1/p′KQj,k,R

)−1

,

taking into account that KQj,k,R ≈ KQj ,R. As a consequence, we infer that|b|H1,p

atb,ρ(μ)≤ c |b|H1,p

atb,η(μ), which yields ‖f‖H1,p

atb,ρ(μ)≤ c ‖f‖H1,p

atb,η(μ). �

Unless otherwise stated, we will assume that the constant ρ in the definitionof H1,p

atb(μ) is equal to 2.

We introduce now a distinguished type of p-atomic blocks. For 1 < p ≤ ∞,we say that a function b ∈ L1

loc(μ) is a special p-atomic block if

1. there exists some doubling cube R such that supp(b) ⊂ R,

2.

∫b dμ = 0,


3. there is a function a supported on a doubling cube Q ⊂ R and numbersλ, cR ∈ R (or C) such that b = λa+ cRχR, and

‖a‖Lp(μ) ≤ μ(Q)1/p−1 K−1Q,R.

Observe that, from the condition∫b dμ = 0 we get |cR|μ(R) ≤ ‖a‖L1(μ) and thus

|b|H1,patb(μ)

≈ |λ|.From the next proposition it follows that if in the definition of H1,p

atb(μ) onereplaces p-atomic blocks by special p-atomic blocks, then one gets an equivalentdefinition of H1,p

atb(μ).

Proposition 9.20. Let 1 < p ≤ ∞ and let b be a p-atomic block. Then there is afamily of special p-atomic blocks {bj}j such that b =

∑j bj and∑

j

|bj|H1,patb(μ)

≤ c |b|H1,patb(μ)

.

Proof. Let R be a cube that supports b and suppose that b =∑

j λjaj , where eachaj is supported on some cube Qj ⊂ R, as in the definition of p-atomic blocks.

Let R0 be a doubling cube containing 2R such that K2R,R0 ≤ c. For a fixedcube Qj , since μ-almost every x ∈ Rd is the center of arbitrarily small doublingcubes, by the Besicovitch covering theorem we deduce that μ-almost all of Qj

can be covered by a family of doubling cubes Qj,k with finite overlap which arecontained in 2Qj so that KQj,k,2Qj ≤ c. It follows that

KQj,k,R0 ≤ c(KQj,k,2Qj +K2Qj,2R +K2R,R0) ≤ cK2Qj ,2R ≤ cKQj ,R.

Letwj,k =

χQj,k∑k χQj,k

,

so that∑

k wj,k = 1 μ-a.e. on Qj . Observe that∑j,k

λj wj,k aj =∑j

λj aj = b.

Let us definebj,k = λjwj,k aj + cj,k χR0 ,

with the constant cj,k chosen so that∫bj,k dμ = 0, which guaranties that each bj,k

is a special p-atomic block. Since∫b dμ = 0, we have

∑j,k cj,k = 0, and thus∑

j,k

bj,k = b.

It remains to estimate∑

j,k |bj,k|H1,patb(μ)

. Clearly,

|bj,k|H1,patb(μ)

≤ c |λj | ‖aj wj,k‖Lp(μ) μ(Qj,k)1/p′

KQj,k,R0 .


Therefore, summing on k and recalling that KQj,k,R0 ≤ cKQj ,R,∑k

|bj,k|H1,patb(μ)

≤ c |λj |KQj ,R

∑k

‖aj wj,k‖Lp(μ) μ(Qj,k)1/p′

≤ c |λj |KQj ,R

(∑k

‖aj χQj,k‖pLp(μ)

)1/p (∑k

μ(Qj,k))1/p′

.

By the finite overlap of the family of cubes {Qj,k}k,∑

k ‖aj χQj,k‖pLp(μ) ≤

c ‖aj‖pLp(μ), and also∑

k μ(Qj,k) ≤ c μ(2Qj). Therefore, summing also on j we

obtain ∑j,k

|bj,k|H1,patb(μ)

≤ c∑j

|λj |KQj ,R ‖aj‖Lp(μ) μ(2Qj)1/p′

≤ c∑j

|λj | = c |b|H1,patb(μ)

. �

Now we are going to see that if a singular integral operator is bounded inL2(μ), then it is also bounded from H1,∞

atb (μ) to L1(μ). In fact, we will replace theassumption of L2(μ) boundedness by the same weaker assumption as in Theorem9.11. By the way, observe that for any f ∈ H1,∞

atb (μ) the integral∫|x−y|>ε

K(x, y) f(y) dμ(y)

makes sense for any x ∈ Rd since |χ|x−y|>εK(x, y)| ≤ c/εn and f is integrable.

Theorem 9.21. Let T be an n-dimensional singular integral operator. If for anycube Q and any function a supported on Q∫

Q

|Tμ,εa| dμ ≤ c ‖a‖L∞(μ) μ(ρQ) (9.43)

uniformly on ε > 0, then T is bounded from H1,∞atb (μ) to L1(μ).

An analogous result can be stated with H1,∞atb (μ) replaced by H1,p

atb(μ). How-

ever, we skip it, since it will come for free later, after showing that H1,patb(μ) =

H1,∞atb (μ).

Proof. First we will show that

‖Tμ,εb‖L1(μ) ≤ c |b|H1,∞atb,2ρ(μ)

(9.44)

for any atomic block b with supp(b) ⊂ R, b =∑

j λj aj , where the aj ’s are functionssatisfying the property 3 of the definition of atomic block.


consider a decomposition f =∑

i bi such that∑

i |bi|H1,∞atb,2ρ(μ)

≈ ‖f‖H1,∞atb (μ) and

then by the continuity of Tμ,ε we get Tμ,ε

(∑i bi)=∑

i Tμ,εbi. Thus, using theestimate (9.44), we obtain

‖Tμ,εf‖L1(μ) ≤∑i

‖Tbi‖L1(μ) ≤ c∑i

|bi|H1,∞atb,2ρ(μ)

≈ ‖f‖H1,∞atb (μ), (9.48)

as wished.In the general situation we reduce to the previous case by considering a

double truncation of the kernel K(x, y). Indeed, let ϕ(x) be a radial C∞ functionsuch that χB(0,1) ≤ ϕ ≤ χB(0,2) and for δ > ε > 0 write

Kδ(x, y) = ϕ(x− y

δ

)K(x, y).

It is easily checked that Kδ(x, y) is a Calderon-Zygmund kernel with constantsuniform on δ. For f ∈ L1(μ), set

Tμ,ε,δf(x) =

∫|x−y|>ε

Kδ(x, y) f(y) dμ(y).

Since χ|x−y|>εKδ(x, y) coincides with χε<|x−y|≤δ K(x, y) outside the annulus δ ≤|x− y| ≤ 2δ, we infer that∣∣Tμ,ε,δf(x)− (Tεf(x)− Tδf(x))

∣∣ ≤ ∫δ<|x−y|≤2δ

|K(x, y) f(y)| dμ(y)

≤ c

δn

∫|x−y|≤2δ

|f(y)| dμ(y).

From this estimate it is straightforward to check that the operator Tμ,ε,δf−(Tεf−Tδf) is bounded in L1(μ) uniformly on ε and δ. As a consequence, applying (9.43)to Tμ,ε and Tμ,δ, we deduce that for any function a supported on Q,∫

Q

|Tμ,ε,δa| dμ ≤ c ‖a‖L∞(μ) μ(ρQ) (9.49)

uniformly on ε and δ. On the other hand, Tμ,ε,δ is bounded in L1(μ), with normpossibly depending on ε, δ. Indeed, for any x ∈ Rd,∫

|Kε,δ(x, y)| dμ(y) ≤ cμ(B(x, 2δ))

εn≤ c

δn

εn.

Also, an analogous estimate holds for all y ∈ Rd integrating with respect to x, andso by appealing to Schur’s lemma we derive the L1(μ) boundedness of Tμ,ε,δ. Thisimplies that

‖Tμ,ε,δ‖H1,∞atb (μ)→L1(μ) < ∞.

9.5. The duality H1,patb(μ) - RBMO(μ) 347

Therefore, as in (9.48), from (9.49) we deduce

‖Tμ,ε,δf‖L1(μ) ≤ c ‖f‖H1,∞atb

(μ),

uniformly on ε and δ.It is immediate to check that Tε,δf(x) → Tεf(x) as δ → ∞. Moreover,

|Tε,δf(x)| ≤ c ε−n ‖f‖L1(μ) for any x ∈ Rd. Then by the dominated convergence

theorem, given an arbitrary bounded set D ⊂ Rd,∫D

|Tεf | dμ = limδ→∞

∫D

|Tε,δf | dμ ≤ supδ>0

‖Tμ,ε,δf‖L1(μ) ≤ c ‖f‖H1,∞atb (μ).

Since this estimate holds with a constant c independent of the bounded set D, weget ‖Tμ,εf‖L1(μ) ≤ c ‖f‖H1,∞

atb (μ). �

We will see below, in Theorem 9.38, that the condition (9.43) holds if andonly if T is bounded from H1,∞

atb (μ) to L1(μ).

9.5 The duality H1,patb(μ) - RBMO(μ)

Our objective in this section consists in showing that the dual of H1,patb(μ) is

RBMO(μ), for 1 < p ≤ ∞, and that the spaces H1,patb(μ) coincide. For simplic-

ity, we will consider only the case of real-valued functions. The arguments forcomplex-valued functions are very similar and are left for the reader. First weneed an auxiliary technical result.

Lemma 9.22. Let f ∈ RBMO(μ) be real-valued. Given q > 0, we set

fq(x) =

⎧⎪⎪⎨⎪⎪⎩f(x) if |f(x)| ≤ q,

qf(x)

|f(x)| if |f(x)| > q.

Then fq ∈ RBMO(μ), with ‖fq‖∗ ≤ c ‖f‖∗.Proof. For any function g, we set g = g+ − g−, with g+ = max(g, 0) and g− =−min(g, 0).

By Proposition 9.5, ‖f+‖∗, ‖f−‖∗ ≤ c ‖f‖∗. Since fq,+ = min(f+, q) andfq,− = min(f−, q), we have ‖fq,+‖∗, ‖fq,−‖∗ ≤ c ‖f‖∗. Thus ‖fq‖∗ ≤ ‖fq,+‖∗ +‖fq,−‖∗ ≤ c ‖f‖∗. �Remark 9.23. Let f ∈ RBMO(μ) be real-valued and let {fQ}Q be a collectionof numbers satisfying (9.32) and (9.33). We set fQ,+ = max(fQ, 0) and fQ,− =−min(fQ, 0) and we take

fq,Q = min(fQ,+, q)−min(fQ,−, q).


It is easily seen that

supQ

1

μ(2Q)

∫Q

|fq(x)− fq,Q| dμ(x) ≤ c ‖f‖∗

and

|fq,Q − fq,R| ≤ cKQ,R ‖f‖∗ for any two cubes Q ⊂ R.

From now on, in this section all functions are assumed to be real-valued.

The proof about the duality between H1,patb(μ) and RBMO(μ) and the coin-

cidence of the spaces H1,patb(μ) for 1 < p ≤ ∞ has been split into several lemmas.

The first one is the following.

Lemma 9.24. For 1 < p ≤ ∞,

RBMO(μ) ⊂ H1,patb(μ)

∗.

That is, for g ∈ RBMO(μ), the linear functional

Lg(f) =

∫f g dμ

defined over functions f ∈ Lp(μ) with compact support extends to a unique con-tinuous linear functional Lg over H1,p

atb(μ). Moreover,

‖Lg‖H1,patb(μ)

∗ ≤ c ‖g‖∗.

Proof. Notice that, by Proposition 9.20, it follows easily that finite sums of specialp-atomic blocks are dense in H1,∞

atb (μ). Therefore, Lp(μ) integrable functions with

compact support are dense in H1,∞atb (μ), and so by the Hahn-Banach theorem, if

the functional Lg is bounded over this type of functions, then it extends uniquely

to the whole space H1,patb(μ).

To deal with the boundedness of Lg, first we will prove that if b is a p-atomicblock and g ∈ RBMO(μ), then∣∣∣∣∫ b g dμ

∣∣∣∣ ≤ c |b|H1,patb(μ)

‖g‖∗.

Suppose that supp(b) ⊂ R, b =∑

j λj aj , where the aj ’s are functions satisfying

property 3 of the definition of p-atomic block. Since∫b dμ = 0,

∣∣∣∣∫ b g dμ

∣∣∣∣ = ∣∣∣∣∫R

b (g − gR) dμ

∣∣∣∣ ≤ ∑j

|λj | ‖aj‖Lp(μ)

(∫Qj

|g − gR|p′dμ

)1/p′

,

(9.50)


where p′ = p/(p− 1), as usual. As g ∈ RBMO(μ) = RBMOp′(μ), we have(∫

Qj

|g − gR|p′dμ

)1/p′

≤(∫

Qj

|g − gQj |p′dμ

)1/p′

+ |gR − gQj |μ(Qj)1/p′

≤ c ‖g‖∗ μ(2Qj)1/p′

+ cKQj,R ‖g‖∗ μ(Qj)1/p′

≤ cKQj ,R ‖g‖∗ μ(2Qj)1/p′

.

Plugging this estimate into (9.50) we get∣∣∣∣∫ b g dμ

∣∣∣∣ ≤ c∑j

|λj | ‖g‖∗ = c |b|H1,patb(μ)

‖g‖∗.

Let f ∈ H1,patb(μ) be Lp(μ) integrable and compactly supported, and suppose

first that g ∈ RBMO(μ) is bounded. Let f =∑

j bj be a decomposition of finto atomic blocks such that

∑j |bj|H1,p

atb(μ)≤ 2‖f‖H1,p

atb(μ). Since the sum

∑j bj

converges in L1(μ) and g is bounded,∣∣∣∣∫ f g dμ

∣∣∣∣ ≤∑j

∣∣∣∣∫ bj g dμ

∣∣∣∣ ≤ c∑j

|bj|H1,patb(μ)

‖g‖∗ ≤ c ‖f‖H1,patb(μ)

‖g‖∗.

For a general function g ∈ RBMO(μ), consider a sequence of bounded func-tions {gq}q≥1 as in Lemma 9.22. As ‖gq‖∗ ≤ c ‖g‖∗,∣∣∣∣∫ f gq dμ

∣∣∣∣ ≤ c ‖f‖H1,patb(μ)

‖gq‖∗ ≤ c ‖f‖H1,patb(μ)

‖g‖∗.

Since |gq| ≤ |g| and gq converges to g pointwise, by the dominated convergence

theorem (recalling that f is from Lp(μ) and compactly supported and g ∈ Lp′loc(μ),

by John-Nirenberg), ∣∣∣∣∫ f g dμ

∣∣∣∣ = limq→∞

∣∣∣∣∫ f gq dμ

∣∣∣∣and the lemma follows. �

The following lemma is a kind of partial converse to the preceding one.

Lemma 9.25. For 1 < p ≤ ∞,

H1,patb(μ)

∗ ∩ Lp′loc(μ) ⊂ RBMO(μ).

That is, if a functional from H1,patb(μ)

∗is of the form Lg for some g ∈ Lp′

loc(μ),

with Lg(f) =∫f g dμ for f ∈ H1,p

atb(μ) ∩ Lp(μ) with compact support, then g ∈RBMO(μ).

Moreover, we have‖Lg‖H1,p

atb(μ)

∗ ≈ ‖g‖∗.


To prove this result we need first to define another equivalent norm forRBMO(μ). First we introduce some notation. Given a cube Q ⊂ Rd and f ∈L1loc(μ), let αQ(f) be the constant for which infα∈R mQ(|f − α|) is attained. By a

variational argument, it follows easily that the constant αQ(f) (which may be notunique) satisfies

μ({x ∈ Q : f(x) > αQ(f)}

) ≤ 1

2μ(Q) (9.51)

and

μ({x ∈ Q : f(x) < αQ(f)}

) ≤ 1

2μ(Q). (9.52)

Given f ∈ L1loc(μ), we denote by ‖f‖◦ the minimal constant c10 such that

1

μ(Q)

∫Q

|f − αQ(f)| dμ ≤ c10 for any doubling cube Q (9.53)

and

|αQ(f)− αR(f)| ≤ c10 KQ,R for any two doubling cubes Q ⊂ R. (9.54)

Lemma 9.26. ‖ · ‖◦ is a norm which is equivalent with ‖ · ‖∗.Proof. To show that ‖f‖∗ ≤ c ‖f‖◦ for any f ∈ L1

loc(μ) we use the characterizationof RBMO(μ) in Proposition 9.10 (c). For a doubling cube Q, from the definitionof ‖ · ‖◦ we deduce

|mQ(f)− αQ(f)| = |mQ(f − αQ(f))| ≤ mQ|f − αQ(f)| ≤ c ‖f‖◦.Then from (9.53) and (9.54) (with c10 = ‖f‖◦), we obtain

1

μ(Q)

∫Q

|f −mQf | dμ ≤ c ‖f‖◦ for any doubling cube Q

and

|mQf −mRf | ≤ c ‖f‖◦KQ,R for any two doubling cubes Q ⊂ R.

Thus, ‖f‖∗ ≤ c ‖f‖◦.To prove the converse inequality, consider a doubling cube Q. From the def-

inition of αQ(f) we get

|mQ(f)− αQ(f)| = |mQ(f − αQ(f))|≤ mQ|f − αQ(f)| ≤ mQ|f −mQ(f)| ≤ c ‖f‖∗.

As a consequence, from this estimate and Proposition 9.10 (c) again, it followseasily that

1

μ(Q)

∫Q

|f − αQf | dμ ≤ c ‖f‖∗ for any doubling cube Q ⊂ Rd

and

|αQf − αRf | ≤ c ‖f‖∗KQ,R for any two doubling cubes Q ⊂ R. �


Proof of Lemma 9.25. We have to prove that ‖g‖◦ ≤ c ‖Lg‖H1,patb(μ)

∗ , since the con-

verse inequality has already been proved. First we will check that

1

μ(Q)

∫Q

|g − αQ(g)| dμ ≤ c ‖Lg‖H1,patb(μ)

∗ for any doubling cube Q. (9.55)

Let f be such that it vanishes out of Q and, for x ∈ Q, f(x) = 1 if g(x) > αQ(g),f(x) = −1 if g(x) < αQ(g), and f(x) = ±1 if g(x) = αQ(g), so that

∫f dμ = 0

(this is possible because of (9.51) and (9.52)). Then∫Q

|g − αQ(g)| dμ =

∣∣∣∣∫Q

(g − αQ(g)) f dμ

∣∣∣∣=

∣∣∣∣∫ f g dμ

∣∣∣∣ ≤ ‖Lg‖H1,patb(μ)

∗ ‖f‖H1,patb(μ)

.

Since f is a p-atomic block and Q is doubling, ‖f‖H1,patb(μ)

≤ |f |H1,patb(μ)

≤ c μ(Q).

Therefore ∫Q

|g − αQ(g)| dμ ≤ ‖Lg‖H1,patb(μ)

∗ μ(Q),

as wished.To estimate |αQ(g)− αR(g)| for doubling cubes Q ⊂ R we take

f =1

μ(R)χR − 1

μ(Q)χQ.

So∫f dμ = 0, and f is an atomic block. Since Q and R are doubling, ‖f‖H1,p

atb(μ)≤

‖f‖H1,∞atb (μ) ≤ cKQ,R. Now we write

Lg(f) =

∫f g dμ

=1

μ(R)

∫R

(g − αR(g)) dμ− 1

μ(Q)

∫Q

(g − αQ(g)) dμ+ (αR(g)− αQ(g)).

Therefore, using (9.55) for Q and R,

|αQ(g)−αR(g)|≤ |Lg(f)|+ 1

μ(R)

∫R

|g − αR(g)| dμ+1

μ(Q)

∫Q

|g − αQ(g)| dμ

≤ ‖Lg‖H1,patb(μ)

∗‖f‖H1,patb(μ)

+ c ‖Lg‖H1,patb(μ)

∗

≤ cKQ,R ‖Lg‖H1,patb(μ)

∗ .

Thus, ‖g‖◦ ≤ c ‖Lg‖H1,patb

(μ)∗ . �


Lemma 9.27. For 1 < p < ∞,

H1,patb(μ)

∗ ⊂ Lp′loc(μ) modulo constants.

Remark 9.28. The statement in the lemma means that there exists some functiong ∈ Lp′

loc(μ) such that for any function f ∈ H1,patb(μ) ∩ Lp(μ) which is compactly

supported we have L(f) =∫f g dμ.

Proof of Lemma 9.27. For N ≥ 1, let QN = [−N,N ]d and let μN = μ�QN . LetLp0(μN ) be the subspace of the functions from Lp(μN ) with zero mean value, and

let CN be the space of functions that are constant on supp(μN ). Lp0(μN ) can be

considered as a subspace of H1,patb(μ) (understanding that the functions of Lp

0(μN )vanish out of the support of μN ). Observe that L|Lp

0(μ)belongs to Lp

0(μN )∗ (taking

the Lp norm for Lp0(μN )), because

|L(f)| ≤ ‖L‖H1,patb(μ)

∗ ‖f‖H1,patb(μ)

≤ ‖L‖H1,patb(μ)

∗ μ([−N), N ]d)1/p′ ‖f‖Lp(μN ).

Since Lp0(μN )∗ = Lp′

(μN )/CN , there exists some function gN ∈ Lp′(μN ), which is

unique μN -a.e. up to constants, such that

L(f) =

∫f gN dμ for any function f ∈ Lp

0(μN ).

Taking into account that Lp0(μN ) ⊂ Lp

0(μN+1), we have∫f gN+1 dμ =

∫f gN dμ for any function f ∈ Lp

0(μN ),

which implies that gN = gN+1+ c μN -a.e. for some constant c. So redefining gN+1

if necessary, we may assume that gN and gN+1 coincide μN -a.e.As a consequence, there exists some function g such that for every f ∈

H1,patb(μ) which is compactly supported and Lp(μ) integrable (and so with has zero

mean),

L(f) =

∫f g dμ. �

Finally, we have:

Theorem 9.29. For 1 < p < ∞, H1,patb(μ) = H1,∞

atb (μ). Also, H1,∞atb (μ)

∗= RBMO(μ).

As above, the last statement must be understood in the sense of Remark9.28.

In the proof below we will appeal to Banach’s closed range theorem. Recallthat this asserts that if X,Y are Banach spaces, X∗, Y ∗ stand for their duals, andT : X → Y is a bounded linear operator, then T (X) is closed in Y if and only ifT ∗(Y ∗) is closed in X∗, where T ∗ is the adjoint of T . See Yosida [182, p.205], forexample.

9.6. Another maximal operator and another covering theorem 353

Proof. By Lemmas 9.24, 9.25, and 9.27, we know that H1,patb(μ)

∗= RBMO(μ) for

1 < p < ∞. Thus, to prove the theorem it is enough to show that H1,∞atb (μ) =

H1,patb(μ) for 1 < p < ∞.

We consider the maps

i : H1,∞atb (μ) −→ H1,p

atb(μ)

i∗ : RBMO(μ) = H1,patb(μ)

∗ −→ H1,∞atb (μ)

∗.

The map i is an inclusion and i∗ is the canonical injection of RBMO(μ) into

H1,∞atb (μ)

∗(with the identification g ≡ Lg for g ∈ RBMO(μ)). By Lemma 9.25, we

have‖Lg‖H1,∞

atb (μ)∗ ≈ ‖g‖∗ ≈ ‖Lg‖H1,p

atb(μ)∗ ,

and thus i∗(RBMO(μ)) is a closed subspace ofH1,∞atb (μ)

∗. By Banach’s closed range

theorem, H1,∞atb (μ) is also closed in H1,p

atb(μ).

To prove that H1,∞atb (μ) = H1,p

atb(μ), by the Hahn-Banach theorem, we have

to show that any functional from H1,patb(μ)

∗which vanishes on H1,∞

atb (μ) vanishes

identically on the whole H1,patb(μ). This functional is of the form Lg for some g ∈

RBMO(μ) (because of the identification RBMO(μ) = H1,patb(μ)

∗). Since Lg vanishes

on H1,∞atb (μ), ‖g‖∗ = 0 by Lemma 9.25 applied with p = ∞, and thus Lg vanishes

identically on H1,patb(μ), as wished. �

Example 9.30. By the previous theorem and the fact that for an AD regularmeasure μ the space RBMO(μ) coincides with BMO(μ), we derive that in this casewe have H1,∞

atb (μ) = H1,∞(μ) too (using the same sort of uniqueness argument asabove). However, this does not hold for all doubling measures μ. For instance, inExample 9.14 (μ equal to the Lebesgue measure on [0, 1]2, with n = 1, d = 2)since RBMO(μ) = L∞(μ) modulo constants, we have H1,∞

atb (μ) = {f ∈ L1(μ) :∫f dμ = 0}.

9.6 Another maximal operator and another covering

theorem

For a given ρ > 1 and f ∈ L1loc(μ), we consider the non-centered maximal operator

M (ρ)μ f(x) = sup

Q�x

1

μ(ρQ)

∫Q

|f | dμ.

One can assume either that the cubes in the supremum are closed, or that theyare open, as it easily follows that the value of the sup is the same in both cases.

We wish to prove that, for any ρ > 1, M(ρ)μ is of weak type (1, 1) and bounded in

Lp(μ), for p > 1. To this end we need the following covering lemma.


Theorem 9.31. Let δ be a distance in Rd which is equivalent with the Euclideandistance. That is,

δ(x, y) ≈ |x− y| for all x, y ∈ Rd. (9.56)

Let ρ > 1, and let {Bi}i∈I be a finite family of open δ-balls (i.e., balls with respectto the metric δ). Then there exists a subfamily {Bi}i∈H ⊂ {Bi}i∈I such that

(a)⋃

i∈I Bi ⊂ ⋃i∈H ρBi, where ρBi stands for the δ-ball with radius ρ r(Bi)which is concentric with Bi, and

(b)∑

i∈H χBi ≤ c, where c depends only on d, ρ, and the constant involved in(9.56).

For ρ ≥ 3 this theorem is a direct consequence of the 3r-covering Theorem2.1 from Chapter 2. In this case the balls from the family H turn out to be disjoint.So the main novelty in the preceding result is that the constant ρ is allowed to beany value strictly bigger than 1.

Proof. We choose the family of indices H by the following algorithm: Let Bs1 ,s1 ∈ I, a δ-ball with maximal radius. We set s1 ∈ H . Next, let s2 ∈ I be such thatBs2 has maximal radius among the δ-balls Bi, i ∈ I, which are not contained inρBs1 , and set s2 ∈ H . In general, if s1, . . . , sj have already been chosen as indicesfrom H , we take sj+1 such that r(Bsj+1 ) is maximal among the δ-balls Bi not

contained in⋃j

k=1 ρBsk , and we set sj+1 ∈ H . If there is no index sj+1 satisfyingthe preceding requirements, then the construction of H is finished.

The above algorithm readily yields⋃

i∈I Bi ⊂⋃

i∈H ρBi. So we only have toprove the finite superposition claimed in (b). First, consider two δ-balls Bi, Bj withr(Bi) ≥ r(Bj) such that Bi ∩ Bj = ∅ and Bj ⊂ ρBi. From the latter conditionswe infer that

2 r(Bj) ≥ diam(Bj) ≥ (ρ− 1) r(Bi). (9.57)

Of course, both the diameter and the radius of the balls Bi, Bj are taken withrespect to the distance δ. So the radius of the balls Bi and Bj in the abovesituation are comparable. On the other hand, denoting by xi, xj the centers ofBi, Bj , we also have

δ(xi, xj) ≥ (ρ− 1) r(Bi). (9.58)

Otherwise, every z ∈ Bj satisfies

δ(xi, z) ≤ δ(xi, xj) + δ(xj , z) < (ρ− 1)r(Bi) + r(Bj) ≤ ρ r(Bi),

and thus Bj ⊂ ρBi.Let Bt1 , . . . , Btm be δ-balls from {Bi}i∈H such that

m⋂k=1

Btk = ∅.

9.6. Another maximal operator and another covering theorem 355

Suppose that Bt1 is the one with maximal radius. Since each δ-ball Btk intersectsBt1 , we infer that

δ(xt1 , xtk) ≤ r(Bt1) + r(Btk ) ≤ 2r(Bt1 ).

So it turns out that 2Bt1 contains points xt1 , . . . , xtm such that, by (9.58) and(9.57),

δ(xtj , xtk) ≥ (ρ− 1) max(r(Btj ), r(Btk )

) ≥ 1

2(ρ− 1)2 r(Bt1 ).

Since δ is equivalent with the Euclidean distance, we infer that m is bounded aboveby some constant depending only on d, ρ, and the constant involved in (9.56). �

Let us remark that the above covering theorem is valid for arbitrary metricspaces which are geometrically doubling. The following is an easy consequence.

Theorem 9.32. Let μ be a Radon measure in Rd, and let ρ > 1. The operator M(ρ)μ

is bounded from M(Rd) to L1,∞(μ) and in Lp(μ) for every p ∈ (1,∞].

Proof. The boundedness in L∞(μ) is trivial. Thus, by the Marcinkiewicz interpo-

lation theorem it is enough to show that M(ρ)μ is bounded from M(Rd) to L1,∞(μ).

So let ν ∈ M(Rd) and λ > 0, and set

Ω ={x ∈ Rd : M (ρ)

μ ν(x) > λ}.

Take a compact set K ⊂ Ω, and for each x ∈ K, let Qx be an open cube containingx such that

|ν|(Qx)

μ(ρQx)> λ.

Let {Qi}i∈I ⊂ {Qx}x be a finite subfamily that covers K. By the precedingcovering theorem (with the distance δ induced by the sup-norm in Rd), thereexists a subfamily {Qi}i∈H ⊂ {Qi}i∈I with bounded overlapping such that K ⊂⋃

i∈H ρQi. Then

μ(K) ≤∑i∈H

μ(ρQi) ≤∑i∈H

|ν|(Qi)

λ≤ c

‖ν‖λ

.

Since this holds uniformly for any compact K ⊂ Ω, we get

μ({

x ∈ Rd : M (ρ)μ ν(x) > λ

}) ≤ c‖ν‖λ

. �


9.7 The sharp maximal operator

The classical (centered) sharp maximal operator when μ is doubling is defined as

M �doubf(x) = sup

Q

1

μ(Q)

∫Q

|f −mQf | dμ,

where the supremum is taken over the cubes Q centered at x and f ∈ L1loc(μ).

Then one has f ∈ BMO(μ) if and only if M �doubf ∈ L∞(μ). For f ∈ Lp(μ) (with∫

f dμ = 0 when μ is finite), it turns out that ‖f‖Lp(μ) ≈ ‖M �doubf‖Lp(μ). This

operator plays an important role in the proof of interpolation results involvingBMO(μ), or in the study of commutators, for example.

We will now introduce an analogous sharp maximal operator suitable forour space RBMO(μ) enjoying properties similar to the ones of the classical sharp

operator, denoted above by M �doub. We define

M �f(x) = supQ�x

1

μ(32Q)

∫Q

|f −mQf | dμ+ supQ ⊂ R : x ∈ Q,Q, R doubling

|mQf −mRf |KQ,R

. (9.59)

Notice that the cubes that appear in these supremums may be non-centered at x.It is clear that f ∈ RBMO(μ) if and only if M �f ∈ L∞(μ) (recall Remark 9.8).

We consider the non-centered doubling maximal operator Nμ:

Nμf(x) = supQ � x,

Q doubling

1

μ(Q)

∫Q

|f | dμ.

Observe that |f(x)| ≤ Nμf(x) for μ-a.e. x ∈ Rd, by Remark 2.10. Moreover, theoperator Nμ is of weak type (1, 1) and bounded in Lp(μ), p ∈ (1,∞]. Indeed, if Qis doubling and x ∈ Q, we can write

1

μ(Q)

∫Q

|f | dμ ≤ 2d+1

μ(2Q)

∫Q

|f | dμ ≤ 2d+1M (2)μ f(x).

So Nμf(x) ≤ 2d+1M(2)μ f(x). As a consequence, from Theorem 9.32 we infer that

Nμ is bounded from M(Rd) to L1,∞(μ) and also in Lp(μ), for 1 < p ≤ ∞.

Note that M � is also bounded from M(Rd) to L1,∞(μ), and in Lp(μ), for1 < p ≤ ∞. This follows from the fact that the first supremum in the definition

of M �f is bounded by M(3/2)μ f(x) +Nμf(x), and the second one by 2Nμf(x).

Remark 9.33. We have

M �|f |(x) ≤ 2d+4M �f(x).

9.7. The sharp maximal operator 357

This is easy to check: Assume that x ∈ Q and Q is doubling. Then we have∣∣mQ|f | − |mQf |∣∣ = ∣∣∣∣ 1

μ(Q)

∫Q

(|f(x)| − |mQf |) dμ(x)∣∣∣∣

≤ 1

μ(Q)

∫Q

|f(x)−mQf | dμ(x)

≤ 2d+1M �f(x). (9.60)

Therefore, if Q ⊂ R are doubling,∣∣mQ|f | −mR|f |∣∣ ≤ |mQf −mRf |+ 2d+2M �f(x)

≤ (KQ,R + 2d+2)M �f(x) ≤ 3 · 2d+1KQ,RM �f(x).

Thus

supQ ⊂ R : x ∈ Q,Q, R doubling

|mQ|f | −mR|f | |KQ,R

≤ 3 · 2d+1M �f(x).

For the other supremum, by (9.60) we have

supQ�x

1

μ(32Q)

∫Q

∣∣ |f(x)| −mQ|f |∣∣ dμ(x)

≤ supQ�x

1

μ(32Q)

∫Q

∣∣ |f(x)| − |mQf |∣∣ dμ(x) + 2d+1 M �f(x)

≤ supQ�x

1

μ(32Q)

∫Q

|f(x) −mQf | dμ(x) + 2d+1M �f(x)

≤ 2d+2M �f(x).

Theorem 9.34. Let f ∈ L1loc(μ), with

∫f dμ = 0 if ‖μ‖ < ∞. For 1 < p < ∞, if

inf(1,Nμf) ∈ Lp(μ), then we have

‖Nμf‖Lp(μ) ≤ c ‖M �f‖Lp(μ). (9.61)

Proof. We assume ‖μ‖ = ∞. The proof for ‖μ‖ < ∞ is similar. For some fixedη < 1 and all ε > 0, we will prove that there exists some δ > 0 such that for anyλ > 0 the following good λ inequality holds:

μ({x : Nμf(x) > (1 + ε)λ, M �f(x) ≤ δλ}) ≤ η μ

({x : Nμf(x) > λ}). (9.62)

Consider the sets Ωλ = {x : Nμf(x) > λ} and

Eλ = {x : Nμf(x) > (1 + ε)λ, M �f(x) ≤ δλ}.For the moment we assume f ∈ Lp(μ) for some 1 < p < ∞, so that Nμf ∈ Lp(μ)too. Consider an arbitrary compact subset Fλ ⊂ Eλ. For each x ∈ Fλ, among the


doubling cubes Q that contain x and such that mQ|f | > (1 + ε/2)λ, we considerone cube Qx which has ‘almost maximal’ side length, in the sense that if somedoubling cube Q′ with side length ≥ 2(Qx) contains x, then mQ′ |f | ≤ (1+ε/2)λ.It is easy to check that this maximal cube Qx exists, because f ∈ Lp(μ).

Let Rx be the cube centered at x with side length 3(Qx). We write Sx = Rx.So assuming δ small enough we have mSx |f | > λ, and then Sx ⊂ Ωλ. Indeed,by construction, we have KQx,Sx ≤ c11. Then, as Qx ⊂ Sx are doubling cubescontaining x, ∣∣mQx |f | −mSx |f |

∣∣ ≤ KQx,Sx M�|f |(x) ≤ c112

d+4δλ.

Thus, for δ < c−111 2

−d−5ε,

mSx |f | > (1 + ε/2)λ− c112d+4δλ > λ.

By the Besicovitch covering theorem (since Fλ is bounded) there are qd (de-pending only on d) subfamilies Dk = {Sk

i }i, k = 1, . . . , qd, of cubes Sx such thatthe cubes from

⋃qdk=1 Dk cover Fλ, they are centered at points xk

i ∈ Fλ, and eachsubfamily Dk is formed by pairwise disjoint cubes. Note that at least one of thesubfamilies Dk, say D1, satisfies

μ

(⋃i

S1i

)≥ 1

qdμ

(⋃i,k

Ski

).

We will prove that for each cube S1i ,

μ(S1i ∩ Fλ) ≤ 1

2qdμ(S1

i ) (9.63)

if δ is assumed small enough. From this inequality one gets

μ

(Fλ ∩⋃i

S1i

)≤ 1

2qd

∑i

μ(S1i ) ≤

1

2qdμ(Ωλ).

Then

μ(Fλ) ≤ μ

(⋃i,k

Ski \⋃i

S1i

)+ μ

(Fλ ∩⋃i

S1i

)(9.64)

≤(1− 1

qd

)μ

(⋃i

S1i

)+

1

2qdμ(Ωλ)

≤(1− 1

2qd

)μ(Ωλ).

Let us prove (9.63). Let y ∈ S1i ∩ Fλ. If Q � y is doubling and such that

mQ|f | > (1+ ε)λ, then (Q) ≤ (S1i )/8. Otherwise, 30Q ⊃ S1

i ⊃ Qx1i, and since Q

9.7. The sharp maximal operator 359

and 30Q are doubling, we have∣∣∣mQ|f | −m30Q

|f |∣∣∣ ≤ K

Q,30QM �|f |(y) ≤ c12 δλ ≤ ε

2λ,

assuming c12δ < ε/2, and so

m30Q

|f | > (1 + ε/2)λ,

which contradicts the choice of Qx1ibecause 30Q ⊃ Qx1

iand (30Q) > 2(Qx1

i).

So Nμf(y) > (1 + ε)λ implies

Nμ(χ 54S

1if)(y) > (1 + ε)λ.

On the other hand, we also have

m 54S

1i

|f | ≤ (1 + ε/2)λ,

since 54S

1i is doubling and its side length is > 2(Qx1

i). Therefore, we get

Nμ

(χ 5

4S1i|f | −m 5

4S1i

|f |)(y) > ε

2λ,

and then, by the weak (1, 1) boundedness of Nμ and the fact that S1i is doubling,

μ(S1i ∩ Fλ) ≤ μ

({y : Nμ

(χ 5

4S1i(|f | −m 5

4S1i

|f |))(y) > ε

2λ})

≤ c

ελ

∫54S

1i

(|f | −m 54S

1i

|f |) dμ≤ c

ελμ(2S1

i )M�|f |(x1

i )

≤ c13 δ

εμ(S1

i ).

Thus (9.63) follows by choosing δ < ε/2qdc13. This finishes the proof of (9.64),and since Fλ is an arbitrary compact subset of Eλ, one gets (9.62).

Under the assumption f ∈ Lp(μ), if we multiply both sides of (9.62) by pλp−1

and we integrate with respect to λ ∈ (0,∞), (9.61) follows arguing as in (2.40).

Suppose now that f ∈ Lp(μ). We consider the functions fq, q ≥ 1, introducedin Lemma 9.22. Since for all functions g, h ∈ L1

loc(μ) and all x we have M �(g +h)(x) ≤ M �g(x) + M �h(x) and M �|g|(x) ≤ cM �g(x), operating as in Lemma9.22 we get M �fq(x) ≤ cM �f(x). On the other hand, |fq(x)| ≤ q inf(1, |f |)(x) ≤q inf(1,Nμf)(x) and so fq ∈ Lp(μ). Therefore,

‖Nμfq‖Lp(μ) ≤ c ‖M �fq‖Lp(μ) ≤ c ‖M �f‖Lp(μ).

Taking the limit as q → ∞, (9.61) follows. �


9.8 Two interpolation theorems

The following result is an immediate consequence of the properties of the sharpmaximal operator.

Theorem 9.35. Let 1 < p < ∞ and T be a linear operator bounded in Lp(μ) andfrom L∞(μ) to RBMO(μ). Then T extends continuously to Lr(μ), p < r < ∞.

Proof. Assume ‖μ‖ = ∞. The operator M � ◦ T is sublinear and it is bounded inLp(μ) and in L∞(μ). By the Marcinkiewicz interpolation theorem, it is boundedin Lr(μ), for p < r < ∞. That is,

‖M �Tf‖Lr(μ) ≤ c ‖f‖Lr(μ).

Take f ∈ Lr(μ) with compact support. Then f ∈ Lp(μ) and so Tf ∈ Lp(μ).Thus Nμ(Tf) ∈ Lp(μ), and so inf(1,Nμ(Tf)) ∈ Lr(μ). From Theorem 9.34 wededuce that

‖Tf‖Lr(μ) ≤ c ‖M �Tf‖Lr(μ) ≤ c ‖f‖Lr(μ).

The proof for ‖μ‖ < ∞ is similar: Given f ∈ Lr(μ), we write f =(f − ∫ f dμ

)+∫f dμ. It easily seen that the same argument as for ‖μ‖ = ∞ can be applied to

the function f − ∫ f dμ. On the other hand, T is bounded in Lr(μ) over constantfunctions. Indeed, since T is bounded from L∞(μ) to RBMO(μ), we get

‖T 1‖Lr(μ) ≤ ‖T 1−mRd(T 1)‖Lr(μ) + ‖mRd(T 1)‖Lr(μ) (9.65)

≤ c μ(Rd)1/r + ‖mRd(T 1)‖Lr(μ).

The second inequality follows by taking a cube centered at some point from suppμand letting its side length tend to ∞, for instance. Also, we have

‖mRd(T 1)‖Lr(μ) = |mRd(T 1)|μ(Rd)1r

≤ ‖T 1‖Lp(μ) μ(Rd)

1r− 1

p ≤ c μ(Rd)1/r ≡ c ‖1‖Lr(μ),

and so ‖T 1‖Lr(μ) ≤ c ‖1‖Lr(μ). �

The main theorem of this section is another interpolation result which is notas immediate as the previous one. Using this result, we will be able to obtain anew proof of the T 1 theorem for the Cauchy integral in the next section. Thestatement is the following.

Theorem 9.36. Let T be a linear operator which is bounded from H1,∞atb (μ) to L1(μ)

and from L∞(μ) to RBMO(μ). In the case ‖μ‖ < ∞, suppose also that T 1 ∈L1(μ). Then T extends continuously to Lp(μ), 1 < p < ∞.

To prove the theorem we need the following variant of the Calderon-Zygmunddecomposition obtained in Lemma 2.14.

9.8. Two interpolation theorems 361

Lemma 9.37 (Variant of the Calderon-Zygmund decomposition). For 1 ≤ p <∞,consider f ∈ Lp(μ) with compact support. For any λ > 0 (with λ > 2d+1 ‖f‖L1(μ)/‖μ‖ if ‖μ‖ < ∞), we have:

(a) There exists a family of almost disjoint (i.e. with a bounded overlap) cubes{Qi}i such that

1

μ(2Qi)

∫Qi

|f |p dμ >λp

2d+1, (9.66)

1

μ(2ηQi)

∫ηQi

|f |p dμ ≤ λp

2d+1for all η > 2, (9.67)

|f | ≤ λ μ-a.e. on Rd \⋃iQi. (9.68)

(b) For each i, let wi =χQi∑k χQk

and let Ri be a (6, 6n+1)-doubling cube concentric

with Qi, with (Ri) > 4(Qi). Then there exists a family of functions ϕi, eachwith constant sign, with supp(ϕi) ⊂ Ri, satisfying∫

ϕi dμ =

∫Qi

f wi dμ, (9.69)

∑i

|ϕi| ≤ B λ (9.70)

(where B is some constant), and if 1 < p < ∞,(∫Ri

|ϕi|p dμ)1/p

μ(Ri)1/p′ ≤ c

λp−1

∫Qi

|f |p dμ. (9.71)

(c) For 1 < p < ∞, if Ri is the smallest (6, 6n+1)-doubling cube of the family{6kQi}k≥1 and we set b =

∑i(f wi − ϕi), then

‖b‖H1,patb

(μ) ≤c

λp−1‖f‖pLp(μ). (9.72)

If the function f above is complex, then instead of real functions ϕi withconstant sign, in (b) we get functions of the form ϕi = αi hi, where αi ∈ C isconstant and hi is a non-negative function.

Although the arguments for the proof of the preceding lemma are very similarto the ones of Lemma 2.14, for the reader’s convenience we will show the details.

Proof. We will assume ‖μ‖ = ∞.

(a) Taking into account Remark 2.10, for μ-almost all x ∈ Rd such that |f(x)|p >λp, there exists a cube Qx satisfying

1

μ(2Qx)

∫Qx

|f |p dμ >λp

2d+1


and such that if Q′x is centered at x with (Q′

x) > 2(Qx), then

1

μ(2Q′x)

∫Q′

x

|f |p dμ ≤ λp

2d+1.

Now we can apply Besicovitch’s covering theorem to get an almost disjoint sub-family of cubes {Qi}i ⊂ {Qx}x satisfying (9.66), (9.67) and (9.68).

(b) Assume first that the family of cubes {Qi}i is finite. Then we may supposethat this family is ordered in such a way that the sizes of the cubes Ri are non-decreasing (i.e. (Ri+1) ≥ (Ri)). The functions ϕi that we will construct will beof the form ϕi = αi χAi , with αi ∈ R (or C) and Ai ⊂ Ri. We set A1 = R1 and

ϕ1 = α1 χR1 ,

where the constant α1 is chosen so that∫Q1

f w1 dμ =∫ϕ1 dμ.

Suppose that ϕ1, . . . , ϕk−1 have been constructed, satisfy (9.69) and∑k−1i=1 |ϕi| ≤ B λ, where B is some constant (which will be fixed below).

Let Rs1 , . . . , Rsm be the subfamily of R1, . . . , Rk−1 such that Rsj ∩Rk = ∅.As (Rsj ) ≤ (Rk) (because of the non-decreasing sizes of Ri), we have Rsj ⊂ 3Rk.Taking into account that for i = 1, . . . , k − 1∫

|ϕi| dμ ≤∫Qi

|f | dμ

by (9.69), and using that Rk is (6, 6n+1)-doubling and (9.67), we get

∑j

∫|ϕsj | dμ ≤

∑j

∫Qsj

|f | dμ ≤ c

∫3Rk

|f | dμ

≤ c

(∫3Rk

|f |p dμ)1/p

μ(3Rk)1/p′

≤ cλμ(6Rk)1/p μ(3Rk)

1/p′ ≤ c14λμ(Rk).

Therefore,

μ({∑

j |ϕsj | > 2c14λ})

≤ μ(Rk)

2.

So we set

Ak = Rk ∩{∑

j |ϕsj | ≤ 2c14λ},

and then

μ(Ak) ≥ μ(Rk)/2.

9.8. Two interpolation theorems 363

The constant αk is chosen so that for ϕk = αk χAkwe have

∫ϕk dμ =∫

Qkf wk dμ. Then we obtain

|αk| ≤ 1

μ(Ak)

∫Qk

|f | dμ ≤ 2

μ(Rk)

∫Qk

|f | dμ

≤ 2

μ(Rk)

∫12Rk

|f | dμ ≤(

2

μ(Rk)

∫12Rk

|f |p dμ)1/p

≤ c15λ

(this calculation also applies to k = 1). Thus,

|ϕk|+∑j

|ϕsj | ≤ (2c14 + c15)λ.

If we choose B = 2c14 + c15, (9.70) follows.Now it is easy to check that (9.71) also holds. Indeed we have(∫

Ri

|ϕi|p dμ)1/p

μ(Ri)1/p′

= |αi|μ(Ai)1/p μ(Ri)

1/p′

≤ c |αi|μ(Ai) = c

∣∣∣∣∫Qi

f wi dμ

∣∣∣∣≤ c

(∫Qi

|f |p dμ)1/p

μ(Qi)1/p′

,

and by (9.66), (∫Qi

|f |p dμ)1/p

μ(2Qi)1/p′ ≤ c

λp−1

∫Qi

|f |p dμ. (9.73)

Thus we get (9.71).

Suppose now that the collection of cubes {Qi}i is not finite. For each fixed Nwe consider the family of cubes {Qi}1≤i≤N . Then, as above, we construct functionsϕN1 , . . . , ϕN

N with supp(ϕNi ) ⊂ Ri satisfying∫

ϕNi dμ =

∫Qi

f wi dμ,

N∑i=1

|ϕNi | ≤ B λ (9.74)

and, if 1 < p < ∞,(∫Ri

|ϕNi |p dμ)1/p

μ(Ri)1/p′ ≤ c

λp−1

∫Qi

|f |p dμ. (9.75)


By (9.74) and (9.75) there is a subsequence {ϕk1}k∈I1 which is convergent in the

weak ∗ topology of L∞(μ) and in the weak ∗ topology of Lp(μ) to some functionϕ1 ∈ L∞(μ) ∩ Lp(μ). Now we can consider a subsequence {ϕk

2}k∈I2 with I2 ⊂ I1which is convergent also in the weak ∗ topologies of L∞(μ) and Lp(μ) to somefunction ϕ2 ∈ L∞(μ) ∩ Lp(μ). In general, for each j we consider a subsequence{ϕk

j }k∈Ij with Ij ⊂ Ij−1 that converges in the weak ∗ topologies of L∞(μ) andLp(μ) to some function ϕj ∈ L∞(μ) ∩ Lp(μ).

We have supp(ϕi) ⊂ Ri and, by the weak ∗ convergence in L∞(μ) and Lp(μ),the functions ϕi also satisfy (9.69) and (9.71). To get (9.70), notice that for eachfixed m, by the weak ∗ convergence in L∞(μ),

m∑i=1

|ϕi| ≤ Bλ,

and so (9.70) follows.

(c) For each i, we consider the atomic block bi = f wi−ϕi, supported on the cubeRi. Since KQi,Ri ≤ c, by (9.73) and (9.71) we have

|bi|H1,patb(μ)

≤ c

λp−1

∫Qi

|f |p dμ,

which implies (9.72). �Proof of Theorem 9.36. Suppose first that ‖μ‖ = ∞. In this case, the functionsf ∈ L∞(μ) having compact support with

∫f dμ = 0 are dense in Lp(μ), 1 < p <

∞. For such functions we will show that

‖M �Tf‖Lp(μ) ≤ c ‖f‖Lp(μ) 1 < p < ∞. (9.76)

By Theorem 9.34, this implies

‖Tf‖Lp(μ) ≤ c ‖f‖Lp(μ).

Indeed, note that if f ∈ L∞(μ) has compact support and∫f dμ = 0, then f ∈

H1,∞atb (μ) and Tf ∈ L1(μ). Thus Nμ(Tf) ∈ L1,∞(μ), and then inf(1,Nμ(Tf)) ∈

Lp(μ). So the hypotheses of Theorem 9.34 are satisfied.

Given any function f ∈ Lp(μ), 1 < p < ∞, for λ > 0 we take a family ofalmost disjoint cubes {Qi}i as in the previous lemma, and a collection cubes {Ri}ias in (c) from the same lemma. Then we can write

f = b+ g =∑i

(χQi∑k χQk

f − ϕi

)+ g.

By (9.68) and (9.70), we have ‖g‖L∞(μ) ≤ c λ, and by (9.72),

‖b‖H1,patb(μ)

≤ c

λp−1‖f‖pLp(μ).

9.9. The T 1 theorem for the Cauchy transform again 365

Due to the boundedness of T from L∞(μ) to RBMO(μ), we have

‖M �Tg‖L∞(μ) ≤ c16 λ.

Therefore,

{M �Tf > (c16 + 1)λ} ⊂ {M �Tb > λ}.Since M � is of weak type (1, 1), we get

μ({M �Tb > λ}) ≤ c

‖Tb‖L1(μ)

λ.

On the other hand, as T is bounded from H1,∞atb (μ) to L1(μ),

‖Tb‖L1(μ) ≤ c ‖b‖H1,∞atb (μ) ≤

c

λp−1‖f‖pLp(μ).

Thus

μ({x ∈ Rd : M �(Tf) > λ}) ≤ c

‖f‖pLp(μ)

λp.

So the sublinear operatorM �T is of weak type (p, p) for all p ∈ (1,∞). By theMarcinkiewicz interpolation theorem we infer that M �T is bounded in Lp(μ) forall p ∈ (1,∞). In particular, (9.76) holds for a bounded function f with compactsupport such that

∫f dμ = 0.

In the case ‖μ‖ < ∞, the arguments are similar to the ones in (9.65) in theproof of Theorem 9.35, using now that ‖T 1‖L1(μ) < ∞. The details are left for thereader. �

9.9 The T1 theorem for the Cauchy transform again

Before studying the particular case of the Cauchy transform, we prove the followingresult which has its own interest.

Theorem 9.38. Let T be an n-dimensional singular integral operator and ρ > 1some fixed constant. The following conditions are equivalent:

(a) For any cube Q and any function a supported on Q∫Q

|Tμ,εa| dμ ≤ c ‖a‖L∞(μ) μ(ρQ) (9.77)

uniformly on ε > 0.

(b) Tμ is bounded from L∞(μ) to RBMO(μ).

(c) Tμ is bounded from H1,∞atb (μ) to L1(μ).


Proof. We have already seen (a)=⇒(b) in Theorem 9.11 and (a)=⇒(c) in Theorem9.21.

Let us prove (b)=⇒(a). Suppose that ρ = 2, for example. Let a ∈ L∞(μ) bea function supported on some cube Q. Assume first that μ(Rd \ 5Q) > 0 (this isalways the case if ‖μ‖ = ∞). We have∫

Q

|Tμ,εa−mQ(Tμ,εa)| dμ ≤ c ‖a‖L∞(μ) μ(2Q). (9.78)

So it is enough to show that

|mQ(Tμ,εa)| ≤ c ‖a‖L∞(μ). (9.79)

Let x0 ∈ suppμ be the point (or one of the points) in Rd \ (5Q)◦ which isclosest to Q. We set d0 = dist(x0, Q). We assume that x0 is a point such thatsome cube with side length 2−kd0, k ≥ 2, is doubling. Otherwise, we take y0 insuppμ∩B(x0, (Q)/100) satisfying this condition, and we interchange x0 with y0.

We denote by R a cube concentric with Q with side length max(10d0, (Q)).So KQ,R ≤ c. Let Q0 be the biggest doubling cube centered at x0 with side length

2−k d0, k ≥ 2. Then Q0 ⊂ R, with KQ0,R ≤ c, and thus

|mQ0(Tμ,εa)−mQ(Tμ,εa)| ≤ |mQ0(Tμ,εa)−mR(Tμ,εa)| (9.80)

+ |mQ(Tμ,εa)−mR(Tμ,εa)|≤ c(KQ0,R +KQ,R

)‖Tμ,εa‖RBMO(μ)

≤ c ‖a‖L∞(μ).

Moreover, dist(Q0, Q) ≈ d0 and so, for y ∈ Q0,

|Tμ,εa(y)| ≤ cμ(Q)

dn0‖a‖L∞(μ) ≤ c ‖a‖L∞(μ),

because (Q) < d0. Then we get |mQ0(Tμ,εa)| ≤ c ‖a‖L∞(μ), and from (9.80), weobtain (9.79).

Suppose now that μ(Rd \ 5Q) = 0. Obviously, this implies that 5 diam(Q) ≥diam(suppμ). Let m be the least integer such that

10 diam(Q)

m<

1

2diam(suppμ). (9.81)

We split Q into cubes Qj, j = 1, . . . ,m2, of side length (Q)/m with disjointinteriors. Since (Qj) ≈ diam(suppμ), the number of cubes Qj which intersectsuppμ does not exceed some absolute constant N . We denote by {Qj}j∈J thisfamily of cubes, so that #J ≤ N . We set

aj =χQj∑k χQk

a.

9.9. The T 1 theorem for the Cauchy transform again 367

Notice that aj coincides with a in the interior of each cube Qj but it may differfrom a on the boundary. Now we set∫

Q

|Tμ,εa| dμ ≤∑j∈J

∫Q\2Qj

|Tμ,εaj | dμ+∑j∈J

∫2Qj

|Tμ,εaj | dμ.

To estimate the first sum on the right-hand side we take into account that

|Tμ,εaj(x)| ≤ cμ(Qj)

(Qj)n‖aj‖∞ ≤ c ‖aj‖∞ for x ∈ Q \ 2Qj,

while for the second one we use the fact that (a) holds for the cubes 2Qj and thefunctions aj (since (9.81) ensures that μ(Rd \ 10Qj) > 0). Then we get∫

Q

|Tμ,εa| dμ ≤ c∑j∈J

‖aj‖L∞(μ) μ(Q) + c∑j∈J

‖aj‖L∞(μ) μ(4Qj)

≤ cN ‖a‖L∞(μ) μ(2Q).

Now we are going to prove (c)=⇒(a). Let a ∈ L∞(μ) be supported on a cubeQ. Assume ρ = 2 and suppose first that μ(Rd \ 5Q) > 0. We consider the sameconstruction as in (b)=⇒(a). The cubes Q, Q0 and R are taken as above, and theysatisfy Q,Q0 ⊂ R, KQ,R ≤ c, KQ0,R ≤ c and dist(Q0, Q) ≥ (Q). Recall also thatQ0 is doubling.

We take the atomic block (supported on R)

b = a+ cQ0 χQ0 ,

where cQ0 is the constant that ensures that∫b dμ = 0. For y ∈ Q we have

|Tμ,ε(cQ0 χQ0)(y)| ≤ c|cQ0 |μ(Q0)

dist(Q,Q0)n≤ c

‖a‖L1(μ)

dist(Q,Q0)n

≤ cμ(Q)

(Q)n‖a‖L∞(μ) ≤ c ‖a‖L∞(μ).

Then we have∫Q

|Tμ,εa| dμ ≤∫Q

|Tμ,εb| dμ+ c ‖a‖L∞(μ) μ(Q)

≤ c ‖b‖H1,∞atb (μ) + c ‖a‖L∞(μ) μ(Q)

≤ cKQ,R ‖a‖L∞(μ) μ(2Q) + cKQ0,R |cQ0 |μ(2Q0)

+ c ‖a‖L∞(μ) μ(Q).

Since Q0 is doubling, we have

|cQ0 |μ(2Q0) ≤ c ‖a‖L1(μ) ≤ c ‖a‖L∞(μ) μ(Q).


Therefore, ∫Q

|Tμ,εa| dμ ≤ c ‖a‖L∞(μ) μ(2Q).

If μ(Rd \ 5Q) = 0, operating as in the implication (b)=⇒(a), we get that (a)also holds. �

Next we intend to obtain another proof of the T 1 theorem for the Cauchyintegral operator. A basic tool for the proof will be the interpolation theorembetween (H1,∞

atb (μ), L1(μ)) and (L∞(μ),RBMO(μ)). Also, the relationship betweenMenger curvature and the Cauchy transform found by Melnikov and Verdera willbe fundamental. Recall that if μ is finite measure with linear growth in the plane,then

‖Cεμ‖2L2(μ) =1

6c2ε(μ) +O(μ(C)),

where c2ε(μ) is the ε-truncated curvature of μ and |O(μ(C)| � μ(C). From thepreceding identity we get the following.

Lemma 9.39. Let μ be some measure on C with linear growth and let Q ⊂ C besome square. If

‖Cμ,εχQ‖L2(μ�Q) ≤ c μ(2Q)1/2 uniformly on ε > 0,

then for any bounded function a supported on Q,

‖Cμ,εa‖L2(μ�Q) ≤ c ‖a‖L∞(μ) μ(2Q)1/2 uniformly on ε > 0.

Proof. Clearly, it is enough to prove the estimate assuming that ‖a‖L∞(μ) = 1.Consider the function g = a + 2χQ. So g is non-negative, and g ≈ 1 on Q. Byapplying the Melnikov-Verdera identity first to the measure gμ and then to χQμ(notice that both have linear growth), we get∫

Q

|Cμ,εg|2 dμ ≤∫

|Cμ,εg|2 g dμ ≤ 1

6c2ε(gμ) + c

∫g dμ

� c2ε(μ) + μ(Q) �∫Q

|Cμ,εχQ|2 dμ+ μ(Q) � μ(2Q).

Therefore,

‖Cμ,εa‖L2(μ�Q) ≤ ‖Cμ,εg‖L2(μ�Q) + 2‖Cμ,εχQ‖L2(μ�Q) ≤ c μ(2Q)1/2. �

Theorem 9.40 (T 1 theorem for the Cauchy transform). The Cauchy transform isbounded in L2(μ) if and only if∫

Q

|Cμ,εχQ|2 dμ ≤ c μ(2Q) for any square Q ⊂ C (9.82)

uniformly on ε > 0.

9.10. The T 1 theorem in terms of RBMO(μ) and BMOρ(μ) 369

Proof. If (9.82) holds, from the preceding lemma and Theorem 9.38 we infer thatthe Cauchy transform is bounded from H1,∞

atb (μ) to L1(μ) and from L∞(μ) toRBMO(μ). Moreover, the assumption (9.82) implies that Cμ,ε1 ∈ L1(μ) uniformlyon ε > 0 if ‖μ‖ < ∞. So by the interpolation Theorem 9.36 we deduce that Cμ isbounded in Lp(μ) for 1 < p < ∞. �

9.10 The T1 theorem in terms of RBMO(μ) andBMOρ(μ)

In this section we will prove a version of the T 1 theorem involving RBMO(μ)and BMOρ(μ) and the notion of weak boundedness. This version is closer to theclassical way of stating the T 1 theorem than the one in terms of characteristicfunctions of cubes.

We say that Tμ is weakly bounded if

|〈Tμ,εχQ, χQ〉| ≤ cμ(2Q) uniformly on ε > 0 for every cube Q ⊂ Rd.

Observe that if Tμ is antisymmetric, like the Cauchy transform, then Tμ is weaklybounded.

The main result from this section is the following.

Theorem 9.41 (T 1 theorem in terms of RBMO and BMOρ). Let μ be a Borelmeasure on Rd with growth of degree n, with 0 < n ≤ d. Let ρ > 1 and let Tμ bean n-dimensional singular integral operator. The following are equivalent:

(a) Tμ is bounded in L2(μ).

(b) Tμ,ε1, T∗μ,ε1 ∈ RBMO(μ) uniformly on ε > 0 and Tμ is weakly bounded.

(c) Tμ,ε1, T∗μ,ε1 ∈ BMOρ(μ) uniformly on ε > 0 and Tμ is weakly bounded.

To prove this theorem, we will relate the conditions (b) and (c) to the L2(μ)boundedness of Tμ and T ∗

μ over characteristic functions of cubes. This will allowus to reduce the result above to a suitable version of the T 1 theorem involvingcubes with small boundaries.

Given t > 0, a cube Q ⊂ Rd is said to have t-small boundary if

μ({x ∈ 2Q : dist(x, ∂Q) ≤ λ (Q)}) ≤ t λ μ(2Q)

for all λ > 0. Below we state a T 1 theorem for general Calderon-Zygmund op-erators which is due to Nazarov, Treil and Volberg. Although the result is notstated explicitly in any paper by them (as far as I know), it follows easily fromthe arguments of Nazarov, Treil and Volberg [127].

Theorem 9.42 (T 1 theorem involving cubes with small boundaries). Let μ be aBorel measure on Rd with growth of degree n, with 0 < n ≤ d. Let t > 0 be big


enough, depending on n, d, and c0. Let Tμ be an n-dimensional singular integraloperator. Then Tμ is bounded in L2(μ) if and only if


uniformly on ε > 0, for any cube Q ⊂ Rd which has t-small boundary.

We will prove this theorem only in the particular case of the Cauchy trans-form. For general singular integral operators we refer the reader to the papermentioned above. We need the following auxiliary result.

Lemma 9.43. Let μ be a Radon measure on Rd. Let t be some constant big enough(depending only on d). Let Q ⊂ Rd be any fixed cube. Then there exists a concentriccube Q′ with Q ⊂ Q′ ⊂ 1.1Q which has t-small boundary.

Proof. Suppose for simplicity that Q is centered at the origin. Let σ = μ�2Q and,for a ∈ R, let Hj(a) be the hyperplane

Hj(a) = {x ∈ Rd : xj = a},

where xj is the j-th coordinate of x. To prove the existence of Q′ it is enough toshow that there exists some a ∈ [(Q), 1.05(Q)] such that for all j = 1, . . . , d andλ > 0,

σ(Uλ(Q)(Hj(a))

)λ (Q)

≤ t‖σ‖(Q)

andσ(Uλ(Q)(Hj(−a))

)λ (Q)

≤ t‖σ‖(Q)

. (9.83)

Recall that Uδ(A) stands for the δ-neighborhood of A. To prove the existence of a,

consider the image measures νj = Πj�σ and νj = Πj#σ, where Πj , Πj : Rd → R

are defined by Πj(x) = xj and Πj(x) = −xj . Observe that (9.83) is equivalent tosaying that for all j = 1, . . . , d and λ > 0,

νj (I(a, λ(Q)))

λ (Q)≤ t

‖σ‖(Q)

andνj (I(a, λ(Q)))

λ (Q)≤ t

‖σ‖(Q)

,

where I(a, λ(Q)) ⊂ R is the interval centered at a with length 2λ(Q). In otherwords,

Mνj(a) ≤ t‖σ‖(Q)

and Mνj(a) ≤ t‖σ‖(Q)

, (9.84)

where M ≡ ML1 is the centered maximal Hardy-Littlewood operator in R.Take the measure ν =

∑dj=1(νj + νj). Since ‖νj‖ = ‖νj‖ = ‖σ‖ for all j, we

have ‖ν‖ = 2d‖σ‖. Observe that for (9.84) to hold for every j, it suffices that

Mν(a) ≤ t‖σ‖(Q)

= t‖ν‖

2d (Q).


From the boundedness of M from M(R) to L1,∞(R), we deduce that

L1

({a ∈ R : Mν(a) > t

‖ν‖2d (Q)

})≤ c

2d (Q)

t.

Thus, if t is chosen big enough, then there exists some a ∈ [(Q), 1.05(Q)] such

that Mν(a) ≤ t ‖ν‖2d (Q) , as wished. �

Proof of Theorem 9.42 for the Cauchy transform. By Theorem 3.5 (or Theorem9.40), it is enough to show that∫

Q

|Cμ,εχQ|2 dμ ≤ c μ(2Q) for any square Q ⊂ C.

Our assumption is that this holds for the squares Q with t-small boundary.Consider an arbitrary square Q ⊂ C. Split it into four half open-closed dis-

joint squares Q1, . . . , Q4 with (Qi) = (Q)/2 for each i. Take now four squaresQ′

1, . . . , Q′4 with t-small boundaries such that Qi ⊂ Q′

i ⊂ 1.1Qi for i = 1, . . . , 4.By our assumption,

‖Cμ,εχQ′i‖2L2(μ�Q′

i)≤ c μ(2Q′

i) ≤ c μ(2Q),

since 2Q′i ⊂ 2Q. Using now the relationship between the Cauchy transform and

curvature,

‖Cμ,εχQi‖2L2(μ�Qi)≤ 1

6c2ε(μ�Qi) + c μ(Qi)

≤ 1

6c2ε(μ�Q′

i) + c μ(Q′i)

≤ ‖Cμ,εχQ′i‖2L2(μ�Q′

i)+ c μ(Q′

i) ≤ c μ(2Q).

Thus,

‖Cμ,εχQ‖L2(μ�Qi) ≤4∑

i=1

‖Cμ,εχQi‖L2(μ�Qi) ≤ c μ(2Q)1/2. �

A key estimate satisfied by the cubes with small boundary, which we willneed below, is the following:

Lemma 9.44. Let μ be a Borel measure on Rd with growth of degree n. Let Q ⊂ Rd

be a cube with t-small boundary. For 1 ≤ p < ∞, we have∫x∈Q

(∫y∈2Q\Q

1

|x− y|n dμ(y)

)pdμ(x) ≤ c t μ(2Q), (9.85)

for 1 ≤ p < ∞, with c depending on d, c0 and p. As a consequence,

‖Tμ,εχQ‖L2(μ�Qc) ≤ c μ(2Q)1/2 uniformly on ε > 0. (9.86)


Proof. Take x ∈ Q and y ∈ 2Q \Q. Then |x− y| ≥ dist(x, ∂Q) and thus∫y∈2Q\Q

1

|x− y|n dμ(y) ≤∫dist(x,∂Q)≤|x−y|≤c (Q)

1

|x− y|n dμ(y).

Using the growth of degree n of μ, it is easy to check that the last integral is

bounded above by c log

(2 +

(Q)

dist(x, ∂Q)

). Hence we get

∫x∈Q

(∫y∈2Q\Q

1

|x− y|n dμ(y)

)p

dμ(x) (9.87)

≤ c

∫x∈Q

(log

(2 +

(Q)

dist(x, ∂Q)

))pdμ(x).

For j ≥ 0, let

Vj ={x ∈ Q : 2−j−1(Q) < dist(x, ∂Q) ≤ 2−j(Q)

}.

Using that μ(Vj) ≤ t 2−jμ(2Q), from (9.87) we infer that

∫x∈Q

(∫y∈2Q\Q

1

|x− y|n dμ(y)

)pdμ(x)

≤ c∑j≥0

∫Vj

(log(2 + 2j+1

))pdμ(x) ≤ c t

∑j≥0

jp2−jμ(2Q) ≤ c t μ(2Q). �

The next proposition concentrates the main technical difficulties for the re-duction of Theorem 9.41 to the T 1 Theorem 9.42.

Proposition 9.45. Let μ be a Borel measure on Rd with growth of degree n, with0 < n ≤ d. Let t > 0 be big enough (as in Lemma 9.43) and let ρ > 1. Let Tμ bean n-dimensional singular integral operator. The following are equivalent:

(a) For any cube Q ⊂ Rd with t-small boundary,

‖Tμ,εχQ‖L2(μ�Q) ≤ c μ(2Q)1/2 uniformly on ε > 0.

(b) Tμ,ε1 ∈ RBMO(μ) uniformly on ε > 0 and Tμ is weakly bounded.

(c) Tμ,ε1 ∈ BMOρ(μ) uniformly on ε > 0 and Tμ is weakly bounded.

Proof. We will show the equivalences (a)⇔(b) and (b)⇔(c). We start with thelatter one.

(b)⇒(c). This is just a consequence of the inclusion BMOρ(μ) ⊂ RBMO(μ).


(c)⇒(b). By Proposition 9.10 (c), it is enough to show that for any (3ρ, (3ρ)d+1)-doubling cube Q, ∫

Q

|Tμ,ε1−mQ(Tμ,ε1)| dμ ≤ c μ(Q) (9.88)

and, for any pair of (3ρ, (3ρ)d+1)-doubling cubes Q ⊂ R,

|mQ(Tμ,ε1)−mR(Tμ,ε1)| ≤ KQ,R. (9.89)

Clearly, the condition (9.88) follows from the fact that Tμ,ε1 ∈ BMOρ(μ) uniformlyon ε > 0 and so we only have to deal with (9.89). To this end, take a cube Q′ witht-small boundary concentric with Q such that Q ⊂ Q′ ⊂ 1.1Q, and take anothercube R′ with t-small boundary such that 1.1R ⊂ R′ ⊂ 1.12R. In this way, weensure that Q′ ⊂ R′. Moreover, notice that Q′ and R′ are (2, (3ρ)d+1)-doubling,since 2Q′ ⊂ 3ρQ and 2R′ ⊂ 3ρR. Write Q′

R′ = 2NQ′,R′+1Q′, and set

|mQ′(Tμ,ε1)−mR′(Tμ,ε1)|≤ |mQ′(Tμ,εχQ′)|+ |mQ′(Tμ,εχ2Q′\Q′)|+ |mQ′(Tμ,εχQ′

R′\2Q′)|+ |mQ′(Tμ,εχRd\Q′

R′ )−mR′(Tμ,εχRd\Q′R′ )|

+ |mR′(Tμ,εχR′)|+ |mR′(Tμ,εχQ′R′\2R′)|+ |mR′(Tμ,εχQ′

R′∩2R′\R′)|= M1 +M2 +M3 +M4 +M5 +M6 +M7.

From the weak boundedness of Tμ and the fact that Q′ and R′ are (2, (3ρ)d+1)-doubling, we get M1+M5 ≤ c. On the other hand, by applying the estimate (9.85)to Q′ (which has t-small boundary), we obtain

M2 = |mQ′(Tμ,εχ2Q′\Q′)|

≤ 1

μ(Q′)

∫∫x∈Q′

y∈2Q′\Q′

1

|x− y|n dμ(x) dμ(y) ≤ cμ(2Q′)μ(Q′)

≤ c.

By the same argument, we get

M7 = |mR′(Tμ,εχQ′R′∩2R′\R′)| ≤ μ(2R′)

μ(R′)≤ c.

Also, it is easily seen that |Tμ,εχQ′R′\2Q′(x)| ≤ cKQ′,R′ for all x ∈ Q′, and so

M3 = |mQ′(Tμ,εχQ′R′\2Q′)| ≤ cKQ′,R′ .

On the other hand, recall that by Lemma 9.12 if x ∈ Q′ and y ∈ R′, we have

|Tμ,εχRd\Q′R′ (x) − Tμ,εχRd\Q′

R′ (y)| ≤ c,


and so M4 ≤ c. Finally, since (Q′R′) ≈ (R′), |Tμ,εχQ′

R′\2R′(x)| ≤ c for x ∈ R′,and thus M6 ≤ c′. Therefore,

|mQ′(Tμ,ε1)−mR′(Tμ,ε1)| ≤ cKQ′,R′ ≤ cKQ,R.

To complete the proof of (9.89) it is enough to show that

|mQ(Tμ,ε1)−mQ′(Tμ,ε1)|+ |mR(Tμ,ε1)−mR′(Tμ,ε1)| ≤ c.

This follows easily. Indeed,

|mQ(Tμ,ε1)−mQ′(Tμ,ε1)| =∣∣mQ

(Tμ,ε1−mQ′(Tμ,ε1)

)∣∣≤ mQ

(∣∣Tμ,ε1−mQ′(Tμ,ε1)∣∣)

≤ cmQ′(∣∣Tμ,ε1−mQ′(Tμ,ε1)

∣∣) ≤ c,

where we took into account that μ(Q′) ≤ μ(3ρQ) ≤ c μ(Q) in the second inequalityand and, in the very last one, that

mQ′(∣∣Tμ,ε1−mQ′(Tμ,ε1)

∣∣) ≤ ‖Tμ,ε1‖BMOρ(μ)μ(ρQ′)μ(Q′)

≤ cμ(3ρQ)

μ(Q)≤ c.

An analogous argument shows that |mR(Tμ,ε1)−mR′(Tμ,ε1)| ≤ c. Thus the proofthat Tμ,ε1 belongs to RBMO(μ) is completed.

(a)⇔(b). Consider a cube Q with t-small boundary, and let Q0 be a (3, 3d+1)-doubling cube containing Q and concentric with Q such that KQ,Q0 ≤ c. Take now

a cube Q with t-small boundary concentric with Q0 such that Q0 ⊂ Q ⊂ 1.1Q0.Since 2Q ⊂ 3Q0, it turns out that Q is (2, 3d+1)-doubling.

To prove (a)⇔(b), we claim that it suffices to prove that for any cube Q witht-small boundary,∣∣∣∣∣(∫

Q

|Tμ,ε1−mQ(Tμ,ε1)|2 dμ)1/2

−(∫

Q

|Tμ,εχQ|2 dμ)1/2∣∣∣∣∣ ≤ c μ(2Q)1/2,

(9.90)under the assumption that either (a) or (b) holds. In fact, if Tμ,ε1 ∈ RBMO(μ),then the first integral in the inequality above is bounded above by μ(2Q)1/2, sinceRBMO2(μ) = RBMO(μ) and |mQ(Tμ,ε1) − mQ(Tμ,ε1)| ≤ c, because KQ,Q ≤ c.

Thus, ‖Tμ,εχQ‖L2(μ�Q) ≤ c μ(2Q)1/2.

On the other hand, if ‖Tμ,εχQ‖L2(μ�Q) ≤ c μ(2Q)1/2, then we infer that∫Q

|Tμ,ε1−mQ(Tμ,ε1)| dμ ≤ cμ(2Q).

From the fact that this holds for any cube Q with t-small boundary one easilydeduces that Tμ,ε1 ∈ BMO3(μ). Indeed, given an arbitrary cube P , let Q be with


t-small boundary concentric with P and such that P ⊂ Q ⊂ 1.1P . Then we have∫P

|Tμ,ε1−mP (Tμ,ε1)| dμ

≤∫P

|Tμ,ε1−mQ(Tμ,ε1)| dμ +∣∣mP (Tμ,ε1)−mQ(Tμ,ε1)

∣∣μ(P )

≤∫Q

|Tμ,ε1−mQ(Tμ,ε1)| dμ + mP

(∣∣Tμ,ε1−mQ(Tμ,ε1)∣∣)μ(P )

≤ 2

∫Q

|Tμ,ε1−mQ(Tμ,ε1)| dμ

≤ c μ(2Q) ≤ c μ(3P ).

Thus, Tμ,ε1 ∈ BMO3(μ), as claimed. From the equivalence (b)⇔(c) with ρ = 3,we get Tμ,ε1 ∈ RBMO(μ).

Let us turn our attention to (9.90) now. The left-hand side of the inequalityis bounded above by(∫

Q

|Tμ,ε1−mQ(Tμ,ε1)− Tμ,εχQ|2 dμ)1/2

=

(∫Q

|Tμ,εχRd\Q −mQ(Tμ,ε1)|2 dμ)1/2

.

For each x ∈ Q, we write the function inside the last integral as follows

Tμ,εχRd\Q(x)−mQ(Tμ,ε1) = Tμ,εχ2Q\Q(x) + Tμ,εχ2Q\2Q(x) (9.91)

−mQ(Tμ,εχQ)−mQ(Tμ,εχ2Q\Q)

+[Tμ,εχRd\2Q(x) −mQ(Tμ,εχRd\2Q)

].

The second term on the right-hand side is bounded by some constant becauseKQ,Q ≤ c. Also, |mQ(Tμ,εχQ)| ≤ c, either by the boundedness of Tμ,ε over char-

acteristic functions of cubes with small t-boundaries under the assumption (a), orby the weak boundedness if (b) holds, taking also into account in both cases that

Q is (2, 3d+1)-doubling.The fourth term on the right-hand side of (9.91) is also bounded by some

constant, by (9.85), since Q has t-small boundary and is doubling. On the otherhand, from Lemma 9.12 it easily follows that∣∣Tμ,εχRd\2Q(x)−mQ(Tμ,εχRd\2Q)

∣∣ ≤ c.

Therefore, the left-hand side of (9.90) does not exceed(∫Q

|Tμ,εχ2Q\Q|2 dμ)1/2

+ c μ(Q)1/2.


By (9.85), since Q has t-small boundary, the integral above is bounded by μ(2Q),and thus (9.90) follows. �Proof of Theorem 9.41. By Theorem 9.42, the L2(μ) boundedness of Tμ is equiv-alent to the fact that


uniformly on ε > 0, for any cube Q ⊂ Rd with t-small boundary (for some fixed bigconstant t). Theorem 9.41 is a direct consequence of this result and Proposition9.45. �


9.11.1 About RBMO(μ) and BMO(μ)

Most of the results explained in this chapter are from Tolsa [156]. In turn thearguments in this work are an adaptation of varying difficulty of the techniquesfrom the classical homogeneous case. For example, the proof of the duality betweenRBMO(μ) and H1,∞

atb (μ) (obtained in the classical setting by Fefferman [43]) isbased on the approach by Journe [78]. On the other hand, the Example 9.15 isessentially from Nazarov, Treil and Volberg [127]. It is also worth mentioning thatmany of the results regarding RBMO(μ) and H1,∞

atb (μ) from this chapter havebeen extended to the more general setting of geometrically doubling spaces inworks such as Hytonen [64], Hytonen, Liu, Yang and Yang [65], or Liu, Yang andYang [87].

A different BMO space has been studied by Mateu, Mattila, Nicolau andOrobitg in [92]. Given an arbitrary Radon measure μ in Rd, let BMOnc(μ) bethe space of the functions f ∈ L1

loc(μ) such that the supremum in (9.1) takenover all the cubes Q ∈ Rd with sides parallel to the coordinate axes is finite.The notation “nc” refers to the fact that the cubes considered in the supremummay be non-centered at suppμ. It is shown in [92] that if μ puts no mass in anyhyperplane orthogonal to one of the coordinate axes, then BMOnc(μ) satisfiesthe usual John-Nirenberg inequality, and its predual can be identified with theatomic Hardy space H1

at(μ) = H1at,ρ=1(μ) defined in (9.42). Also, an interpolation

theorem between the pairs (H1at(μ), L

1(μ)) and (L∞(μ),BMOnc(μ)) is proved.However, in general, singular integral operators which are bounded in L2(μ) mayfail to be bounded from L∞(μ) to BMOnc(μ) and from H1

at(μ) to L1(μ), as shownby Verdera [175].

9.11.2 Other related results

A result that generalizes a classical theorem of Coifman, Rochberg and Weiss [17]on commutators has been obtained in Tolsa [156]. Namely, it has been proved


that if Tμ is a singular integral bounded in L2(μ) and b ∈ RBMO(μ), then thecommutator

[Tμ, b](f) = b Tμf − Tμ(b f)

is bounded in Lp(μ), for 1 < p < ∞. This follows from a pointwise estimate ofM �([Tμ, b](f)) in terms of other maximal bounded operators in Lp(μ).

The space H1,∞atb (μ) can be characterized in terms of a grand maximal func-

tion. Given f ∈ L1loc(μ), let

MΦf(x) = supϕ∼x

∣∣∣∣∫ f ϕ dμ

∣∣∣∣ ,where the notation ϕ ∼ x means that ϕ ∈ L1(μ) ∩ C1(Rd) and satisfies

1. ‖ϕ‖L1(μ) ≤ 1,

2. 0 ≤ ϕ(y) ≤ 1

|y − x|n for all y ∈ Rd, and

3. |ϕ′(y)| ≤ 1

|y − x|n+1for all y ∈ Rd.

It is shown in Tolsa [161] that a function f ∈ L1loc(μ) belongs to H1,∞

atb (μ) if andonly if f ∈ L1(μ),

∫f dμ = 0, and MΦf ∈ L1(μ). Moreover, in this case

‖f‖H1,∞atb (μ) ≈ ‖f‖L1(μ) + ‖MΦf‖L1(μ).

This can be considered as the extension of an analogous result by Fefferman andStein [44] for the usual Hardy space in Rd, with respect to the Lebesgue measure.Let us also remark that some of the techniques from [161] have been used in Tolsa[157] to develop a Littlewood-Paley type theory useful to obtain a new proof ofthe T 1 theorem, and also in Tolsa [166] to study the boundedness of Calderon-Zygmund operators Tμ in the weighted space Lp(wμ), where w is a function suchthat 0 < w(x) < ∞ for μ-a.e. x ∈ Rd.

Many additional results on Calderon-Zygmund theory with non-doublingmeasures have been proved recently. For example, there are additional results onweights by Garcıa-Cuerva and Martell [51], Orobitg and Perez [130], Komori [81];on commutators by Chen and Sawyer [8], Hu, Meng and Yang [61]; on multilinearcommutators by Hu, Meng and Yang [60]; on fractional integrals by Garcıa-Cuervaand Martell [52], Garcıa-Cuerva and Gatto [49]; on Lipschitz spaces by Garcıa-Cuerva and Gatto [50]; etc. See also the very recent book by Yang, Yang andHu [181] for more information about the Hardy space H1,∞

atb (μ) and other relatedtopics.


9.11.3 Tb theorems

We are going to state one of the versions of the Tb theorem devised by Nazarov,Treil and Volberg in [127] for non-homogeneous spaces. To this end, we needto introduce some additional terminology. We say that an operator S is weaklybounded if ∣∣〈SχQ, χQ〉

∣∣ ≤ c μ(2Q) for every cube Q ⊂ Rd.

We say that b ∈ L1loc(μ) is weakly accretive if there exists some positive constant

δ such that ∣∣∣∣∫Q

b dμ

∣∣∣∣ ≥ δ μ(Q) for every cube Q ⊂ Rd.

The aforementioned theorem of Nazarov, Treil and Volberg is the following.

Theorem 9.46 (Tb theorem). Let μ be a Borel measure on Rd with growth ofdegree n, and let Tμ be an n-dimensional singular integral operator. Let b1, b2 betwo bounded and weakly accretive functions. Then Tμ is bounded in L2(μ) if andonly if the operator b2Tμ,εb1 is weakly bounded uniformly in ε > 0, and Tμ,εb1,T ∗μ,εb2 belong to BMOρ(μ) uniformly in ε > 0, for some ρ > 1.

Above, abusing notation we have denoted by b2Tμ,εb1 the operator

b2Tμ,εb1(f) := b2 · Tμ,ε(b1f).

Notice that if Tμ is antisymmetric and b1 = b2 =: b, then bTμ,εb is always weaklybounded. The assumption that Tμ,εb1, T

∗μ,εb2 belong to BMOρ(μ) uniformly in ε >

0 is equivalent to the condition that Tμ,εb1, T∗μ,εb2 belong to RBMO(μ) uniformly

in ε > 0.

Nazarov, Treil and Volberg have also proved a “local Tb theorem” in [125],which generalizes a result by Christ [14] for homogeneous spaces:

Theorem 9.47 (local Tb theorem). Let μ be a Borel measure on Rd with growthof degree n, and let Tμ be an n-dimensional singular integral operator. Supposethat for every cube Q ⊂ Rd there exist functions b1Q and b2Q such that, for someconstants 0 < δ,B < ∞,

(a) supp biQ ⊂ Q for i = 1, 2,

(b) ‖biQ‖L∞(μ) ≤ 1 for i = 1, 2,

(c)∣∣∣∫Q biQ dμ

∣∣∣ ≥ δ μ(Q) for i = 1, 2, and

(d) ‖Tμ,εb1Q‖L∞(μ) + ‖T ∗

μ,εb2Q‖L∞(μ) ≤ B for all ε > 0.

Then Tμ is bounded in L2(μ).

In the case when Tμ is antisymmetric and b1Q = b2Q =: bQ for every cube Q,Nazarov, Treil and Volberg have shown that bQ need not be supported on Q, i.e.the assumption (a) can be eliminated. Further the condition (d) can be replacedby


(d’) ‖Tμ,εbQ‖BMO2ρ(μ)

≤ B, for some fixed ρ > 1.

See also the work of Hytonen and Martikainen [66] for a further generalization ofthe theorem above to the setting of the so-called upper doubling measures and forthe correction of a gap in the arguments in [125].

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Index

ΔQ, 147Ξf , 147γ, 15γ+, 103

AD regular curve, 88, 97AD regular measure, 47AD regular set, 47admissible function, 15A(E), 37Ahlfors function, 19Ahlfors-David regular, 3almost disjoint, 51(α, β)-doubling cube or ball, 53analytic capacity, 15atomic block, 320atomic Hardy space, 320

bad square, 153Banach’s closed range theorem, 352Besicovitch covering theorem, 50Besicovitch theorem on projections,

42β, 236βE(Q), 91, 281βp, 281βE,p(Q), 281big piece, 69big pieces of Lipschitz images, 287bilipschitz maps and γ, 227BMO(μ), 319BMO(Rd), 40BMOρ(μ), 319BMOp

ρ(μ), 320bounded overlap, 50

B(x, r), 11B(x, r), 11

C1 harmonic capacity, 41Calderon commutators, 100Calderon-Zygmund decomposition,

57Calderon-Zygmund kernel, 48Calderon-Zygmund operator, 49Cantor sets, 125capacity γ+, 103Carleson’s embedding theorem, 143,

144Cauchy capacity, 228Cauchy transform, 22Cν, 22centered maximal Hardy-Littlewood

operator, 51characterization of γ+ in terms of

curvature, 112CH(Q), 147closed range theorem, 352commutator, 376comparability between γ and γ+, 195construction of the Lipschitz graph,

243continuous analytic capacity, 38corner quarters Cantor set, 34Cotlar’s inequality, 63, 176countably rectifiable, 36covering theorem, 49, 3533r-covering theorem, 495r-covering theorem, 50curvature and β’s, 92, 281curvature of a measure, 77

, , OI 10.1007/978-3- - -6,

© Springer



393

394 Index

CZO, 49

David’s theorem, 231David-Leger curvature theorem, 231,

233David-Semmes βp, 281David-Semmes problem, 287Davie and Øksendal duality, 107Denjoy’s conjecture, 110density, 41dimH , 29distant squares, 155distH , 241doubling cube or ball, 53doubling measure, 4dual version of γ+, 120, 121duality to study analytic capacity, 39dualization of the weak (1,1) inequal-

ity, 107dyadic Carleson embedding theorem,

143, 144dyadic cube, 11dyadic maximal Hardy-Littlewood

operator, 143

estimate of the Cauchy integral, 221exceptional set H , 208exceptional set S, 138exceptional sets HD and TD, 216existence of principal values, 290

Fav, 35Favard length, 35, 228f ′(∞), 15Frostman’s Lemma, 32

Garabedian function, 39geometrically doubling metric space,

74good λ, 68good ball, 242good function, 153good lambda, 68good square, 153

grand maximal, 377growth of degree n, 45

Hardy space, 320H1,∞

atb (μ), 341

H1,patb(μ), 341

Hausdorff content, 29Hausdorff dimension, 29Hausdorff distance, 241Hausdorff measure, 28Hs∞, 29Hs, 28Huovinen counterexample, 316

image measure, 12induction on scales, 220inner boundary conjecture, 39intermediate approximation by

squares, 202interpolation theorems, 360

Jaye-Nazarov counterexample, 316John-Nirenberg inequality, 336Jones’ β’s, 236Jones’ β’s, 91Jones’ traveling salesman theorem,

91

κ, 41κc, 41Koebe’s theorem, 17Kolmogorov’s inequality, 64KQ,R, 321

Lebesgue differentiation theorem, 54length, 29linear growth, 2, 45Lipschitz graphs and curvature, 86Lipschitz graphs and the Cauchy

transform, 84Lipschitz harmonic capacity , 40local Tb theorem, 378long distance, 156lower s-dimensional density, 41Lp,∞(μ), 13

Index 395

Marstrand, 42martingale decomposition, 146maximal Hardy-Littlewood operator,

51maximal Hardy-Littlewood operator

associated with cubes, 70maximal operator, 63maximal radial operator, 53maximum principle, 117maximum principle for Uμ, 119maximum principle for curvature,

117Melnikov inequality, 132Melnikov’s identity, 77Melnikov-Verdera identity, 79Menger curvature, 76Mμ, 51

Mμ, 63MQ

μ , 70

M(ρ)μ , 353

mollification of the Cauchy kernel,104

MΦ, 377mQ(f), 319M(Rd), 12M+(R

d), 12MR,n, 53M �, 356Murai, 96, 127Murai’s quantitative version of Den-

joy’s conjecture, 127

Nazarov-Treil-Volberg theorem, 137Nμ, 52Nμ, 356non-tangential limits, 296NQ,R, 321

orthogonal projection, 35outer regularity of γ, 18outer regularity of γ+, 106

Painleve’s theorem, 30Painleve’s problem, 20

paraproduct, 166Plemelj formulas, 296pointwise curvature, 276Pommerenke, 30potential, 115, 229Preiss theorem, 42principal value, 289probability of bad squares, 181pθ, 35purely unrectifiable, 36p.v.T ν(x), 289

Q, 326quantitative David’s theorem, 277Q(x, r), 11Q(x, r), 11

radial maximal operator, 53Radon measure, 12random dyadic lattice, 145RBMO(μ), 320R(E), 37rectifiable, 36reflectionless measure, 316removable for Lipschitz harmonic

functions, 287removable set, 20Riesz capacity, 129Riesz transform, 101rising sun lemma, 89

Sawyer, 73Schur’s lemma, 157semiadditivity of γ, 195semiadditivity of γ+, 112sharp estimate for the Cauchy trans-

form on Lipschitz graphs,96

singular integral operator, 48SIO, 48small boundary, 369smoothing of the Cauchy kernel, 104special atomic block, 342support of a measure, 12

396 Index

suppressed operator, 140symmetric difference, 11

T 1 theorem, 81, 98, 365, 369tangent measure, 315T∗, 63Tb theorem, 137, 378terminal square, 146thin boundary, 152total exceptional set, 153totally disconnected set, 18transit square, 146traveling salesman theorem, 91two squares condition, 235

Uδ(A), 11uniform rectifiability, 286, 287upper s-dimensional density, 41

variation, 314variation of a singular integral, 314very good ball, 242very good function, 154VG, 242Vitushkin’s conjecture, 231Vitushkin’s localization operator, 25Vitushkin’s theorems, 37Vϕ, 25

weak (1,1) boundedness of Calderon-Zygmund operators, 60

weak convergence, 300weak Lebesgue spaces, 13weak limit, 300weakly bounded, 369, 378Whitney cubes, 69Whitney’s decomposition, 69Wolff potential, 129

zero density measure, 300

analytic capacity, the cauchy transform, and non-homogeneous calder³nâ€“zygmund theory

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