Hardy Spaces, Singular Integrals and the Geometry of EuclideanDomains of Locally Finite Perimeter !

Steve Hofmann, Emilio Marmolejo-Olea, Marius Mitrea,

Salvador Perez-Esteva and Michael Taylor

Abstract

We study the interplay between the geometry of Hardy spaces and functional analytic proper-ties of singular integral operators (SIO’s), such as the Riesz transforms as well as Cauchy-Cli!ordand harmonic double layer operator, on the one hand and, on the other hand, the regularityand geometric properties of domains of locally finite perimeter. Among other things, we giveseveral characterizations of Euclidean balls, their complements, and half-spaces, in terms of theaforementioned SIO’s.

1 Introduction

Hardy spaces (of holomorphic functions) have originally been considered in domains in the plane,a context in which the two-dimensional Euclidean space is identified with the field of complexnumbers. Defining Hardy spaces in the higher dimensional setting presupposes the existence of anadditional structure, playing a role in relation to Rn somewhat analogous to the role played by Cin relation to R2. A natural choice for us is to consider the embedding

Rn !" C"n (1.1)

where C"n is the Cli!ord algebra with n generators. That is, C"n is the minimal enlargement of Rn

to an associative, unitary algebra with the property that X #X = $|X|2 for every X % Rn (whenn is odd, it is also understood that that C"n is not generated, as an algebra, by any proper subspaceof Rn). In this context, the role of the Cauchy-Riemann operator is played by the Dirac operatorD, defined as Df =

!nj=1 ej # (#jf) for C"n-valued functions f . Null-solutions of D, referred to as

monogenic functions, play the role of holomorphic functions in C.Given a domain " in Rn, and p % (1,&), let H p

± (#") be the boundary Hardy (or Smirnov)spaces obtained by taking (pointwise) traces of inner and outer monogenics (having p-th powerintegrable nontangential maximal functions) on #". When " is a two-sided NTA domain (cf. ap-pendix) with an Ahlfors regular boundary, we show that H p

± (#") are well-defined, closed subspacesof Lp(#", d$)' C"n and the following direct sum decomposition holds:

Lp(#", d$)' C"n = H p+ (#")(H p

! (#"). (1.2)!2000 Math Subject Classification. Primary: 49Q15, 42B20 Secondary 26B15, 30G35

Key words: Hardy spaces, double layer potential, Riesz transforms, Cli!ord algebras, Cauchy-Cli!ord operator,domains of locally finite perimeter, SKT domains, Cli!ord-Szego projections, characterizations of balls, half-spaces

1

Here, the boundary measure $ is the restriction of the (n$1)-dimensional Hausdor! measure Hn!1

to #". Also, Lp(#", d$) ' C"n simply denotes the Lp-space of functions on #" with values in theHilbert space C"n. This extends previous work in [2], [26], where the case of Lipschitz domains wastreated, and in [3], [4], [35], [20] where the authors have dealt with the two-dimensional case.

One significant feature of these considerations emphasized in this paper is that the geometryof Hardy spaces in Calderon’s decomposition (1.2) contains a remarkable amount of informationabout the regularity and shape of the domain " itself. A concrete result substantiating this claimis as follows. Let

<)"H 2

+ (#") , H 2! (#")

#:= arccos

"sup

f±"H 2±(!!)

)f+, f!*+f+++f!+

#(1.3)

be the angle between the closed subspaces H 2+ (#") and H 2

! (#") of L2(#", d$) ' C"n (with )·, ·*and + · + denoting the natural inner product and norm in L2(#", d$)' C"n, respectively).

Theorem 1.1. In the class of bounded, two-sided NTA domains with Ahlfors regular boundaries,the following holds. If % :=<)

$H 2

+ (#") , H 2! (#")

%is su!ciently close to &/2 (relative to the NTA

and Ahlfors constants of ") then " is a '-SKT domain, for some ' = '(%) > 0 such that ' , 0 as% - &/2 (cf. Appendix for definitions pertaining to terminology). In addition,

<)"H 2

+ (#") , H 2! (#")

#=

&

2./ " is a ball. (1.4)

This is remarkable, in that the specification % = &/2 not only implies that " is smooth, but actuallydetermines the shape of ".

A condition that is equivalent to % = &/2 and, at the same time, does not employ the Cli!ordalgebra formalism nor does it involve the outward unit normal to ", is as follows. For 1 0 k 0 n,recall that the Riesz transform Rk associated with " is the formal convolution operator on #" withthe kernel 2

"n"1

xk|X|n , where (n!1 stands for the surface area of Sn!1. The most familiar setting

is when " = Rn+, in which case it is well-known that

!nk=1 R2

k = $I and RjRk = RkRj for allj, k % {1, ..., n} (called URTI, i.e., the usual Riesz transform identities). It is perhaps less knownthat the URTI are valid when " is a ball in Rn. What is, however, most remarkable about theseconsiderations is that, under some mild measure theoretic background assumptions, the converseof the latter statement is also true. Specifically, we have the following.

Theorem 1.2. If " 1 Rn is a two-sided NTA domain with an Ahlfors regular boundary, then

#" is a sphere, or a (n$ 1)-plane ./n&

k=1

R2k = $I and RjRk = RkRj 2 j, k % {1, ..., n}. (1.5)

Thus, in the setting of the above theorem, if the URTI hold then " is a ball if it is unbounded, thecomplement of a ball if it is unbounded and has compact boundary, and a half-space if it has anunbounded boundary.

Another characterization of balls, complements of balls, and half-spaces which avoids involvingthe outward unit normal reads as follows:

for every X± % "± there hold'

!!

#Y!X+,Y!X"$|Y!X+|n|Y!X"|n dHn!1(Y ) = 0 and

'

!!

(Y!X+)j(Y!X")k!(Y!X+)k(Y!X")j

|Y!X+|n|Y!X"|n dHn!1(Y ) = 0 2 j, k % {1, ..., n},(1.6)

2

where we have set "+ := " and "! := Rn \ ". See Theorem 2.12. Hence, in the class of domainsdescribed in Theorem 1.1,

#" is a sphere, or a (n$ 1)-plane ./ (1.6) holds. (1.7)

The theory of the Hardy spaces H 2± (#") is also strongly intertwined with that of a cer-

tain principal-value singular integral operator C on #", called the Cauchy-Cli"ord operator (cf.(2.19) for a formal definition). One basic aspect of this relationship is that the proximity of<)

"H 2

+ (#") , H 2! (#")

#to &/2 turns out to be closely related to the degree of failure for C to be

be self-adjoint, as an operator on L2(#", d$)' C"n. Somewhat more precisely,

<)"H 2

+ (#") , H 2! (#")

#is close to

&

2./ +C $ C %+L(L2(!!,d#)&C$n) is small (1.8)

and

<)"H 2

+ (#") , H 2! (#")

#=

&

2./ C = C %. (1.9)

As with the Hardy spaces themselves, the Cauchy-Cli!ord operator C also encodes a remarkableamount of information about the geometry of ". In the final section of [17], N. Kerzman is askingwhether the spectral properties of C $ C % can be related in a significant way to the geometry ofthe underlying domain ". In a slightly more general context than that of Theorem 1.1, it has beenestablished in [10] that

C $ C % is compact on L2(#", d$)' C"n ./ " is a regular SKT domain. (1.10)

Here we augment this with the following:

Theorem 1.3. Let " 1 Rn be a UR (uniformly rectifiable) domain which satisfies #" = #". Then

C $ C % = 0 ./ " is either a ball, the complement of a (closed) ball, or a half-space. (1.11)

In fact, instead of C = C %, i.e. C is self-adjoint on L2(#", d$)'C"n, we could assume the seeminglyconsiderably weaker condition that C is normal, i.e. C C % = C %C . The class of UR domains isdefined in the appendix. Here we only wish to mention that every two-sided NTA domain withAhlfors regular boundary is a UR domain.

In the two dimensional setting, when the Cli!ord algebra is replaced by the field of complexnumbers and C is the classical Cauchy operator in complex analysis, we succeed in refining theabove theorem by proving the following perturbation result:

Theorem 1.4. Assume that " 1 R2 is a bounded, connected UR domain satisfying #" = #".Then for every ' > 0 there exists ) > 0, which depends on ' > 0 and the geometrical characteristicsof ", with the property that

+C %1$ 12+L2(!!,d#)&C < ) =/ inf

D disk+*! $ *D+L2(R2) < '. (1.12)

As a corollary, for every ' > 0 there exists ) > 0 such that

+C $ C %+L$L2(!!,d#)&C

% < ) =/(

" is a '-SKT domain and

infD disk +*! $ *D+L2(R2) < '(1.13)

in the class of all bounded, connected domains " 1 R2 which satisfy a two-sided local John con-dition and have Ahlfors regular boundaries, with a certain fixed common bound on the geometricalconstants involved in describing these characteristics.

3

As is well-known, the scalar part (i.e., the coe#cient of the multiplicative unit e0 of the Cli!ordalgebra C"n) of C is K, the harmonic double layer (cf. (2.25) for a formal definition) when bothoperators act on scalar-valued functions. More specifically, in the matrix representation of C inthe standard basis in C"n in which all entries are operators mapping scalar-valued functions toscalar-valued functions, the entry (1, 1) (corresponding to the scalars in C"n) is exactly the doublelayer potential. In particular, C = C % =/ K = K%, with K% the adjoint of K. It is thereforesurprising that the seemingly much weaker condition K = K% contains, at least for boundeddomains, essentially the same amount of information as C = C %. Concretely, we have:

Theorem 1.5. Assume that " 1 Rn, n 3 3, is a bounded UR domain with #" = #" and for whichK = K%. Then " is a ball.

As a corollary, retaining the same background hypotheses as above,

C = C % ./ K = K%. (1.14)

It is unclear to us whether perturbation results similar to those discussed in Theorem 1.4 hold indimension n 3 3. See the discussion at the end of § 4 for more in this regard.

As already alluded to above, C contains significant information regarding the geometry of ".In the class of two-sided NTA domains with Ahlfors regular boundaries, we prove the estimate+C +

L$L2(!!,d#)&C$n

% 3 12 and show that the extremal case +C +

L$L2(!!,d#)&C$n

% = 12 occurs if and

only if #" is a sphere, or a (n $ 1)-plane. Likewise, +C + C %+L$L2(!!,d#)&C$n

% 3 1 with equality

precisely when #" is a sphere, or a (n $ 1)-plane. These, along with several other closely relatedresults, are described in Theorem 4.21.

Let us consider for a moment Calderon’s decomposition (1.2) with p = 2 and denote by P±the Cli!ord-Szego projections of L2(#", d$)'C"n onto H 2

+ (#") and H 2! (#"), respectively. In this

paper we also show that for every ' > 0 there exist ) > 0 and R > 0 with the property that

+I $P+ $P!+L$L2(!!,d#)&C$n

% 0 ) =/ supX"!!, 0<r<R

))))Hn!1(B(X, r) 4 #")

(n!1rn!1$ 1

)))) 0 ', (1.15)

in the class of all bounded domains " 1 Rn that are two-sided NTA and have Ahlfors regularboundaries, with a certain fixed common bound on the diameter, NTA and Ahlfors constants. Thiscan be viewed as the higher dimensional analogue of a result by G.David [4] for chord-arc domainsin the plane.

The organization of the paper is as follows. In § 2 we extend the scope of the Cli!ord analysisin Lipschitz domains from [26] to a much larger class of sets, namely two-sided NTA domains withAhlfors regular boundaries. This section is divided into five subsections. In § 2.1 we review theCli!ord algebra formalism. In § 2.2 we introduce the Cauchy-Cli!ord operator and recall severalrecent results from [10], which are of basic importance for the current work. In the Cli!ord algebrasetting, we define in § 2.3 Hardy spaces and establish the validity of Calderon’s decomposition(1.2). In § 2.4 we introduce the Cli!ord-Szego projections P± and prove a number of very usefulidentities involving these operators. Finally, in § 2.5, we compute and estimate the norms of variousoperators related to C and P± in terms of the angle % between H 2

+ (#") and H 2! (#").

The aim of § 3 is to relate % (defined by (1.3)) to the regularity of ". Here we prove the firstpart of Theorem 1.1 and establish (1.15). In § 4 we discuss several characterizations of Euclideanballs and half-spaces in terms of Hardy spaces and singular integral operators. This section is

4

partitioned into three subsection. In § 4.1 we motivate our enterprise by recalling several knownresults scattered in the literature, and then go on and forge further links between them. We thenproceed to study the two dimensional case in detail in § 4.2. Higher dimensional results pertainingto characterizations of balls and half-spaces are contained in § 4.3. Finally, § 5 is an appendixwhere, for the convenience of the reader, we collect a number of definitions from geometric measuretheory.

Acknowledgments. Part of this research was carried out while some of the authors were visitingInstituto de Matematicas Unidad Cuernavaca, University of Missouri-Columbia and the Ban! In-ternational Research Station. We thank these institutions for their hospitality. The work of authorswas also supported in part by US NSF grants DMS-0245401, DMS-0653180, DMS-FRG0456306,DMS-0456861, as well as by DAIC-CONACyT # 48633. Last, but not least, we are particularlygrateful to the referee for his/her very careful reading of the manuscript and for making numeroussuggestions which have led to the present version.

2 Cli!ord analysis in NTA Ahlfors regular domains

2.1 The Cli!ord algebra formalism

We shall make use of the Cli!ord algebra formalism, which we now briefly review. As shownin such standard texts as [21], elements in the Cli!ord algebra with n generators, C"n, can beuniquely written in the form u =

!nl=0

!'|I|=luI eI with uI % R, where eI stands for the product

ei1 # ei2 # · · · # eil if I = (i1, i2, . . . , il) with 1 0 i1 < i2 < · · · < il 0 n, e0 := e( := 1, and!'

Iindicates that the sum is performed over (strictly) increasingly ordered n-tuples I. In particular,we have the embedding

Rn !" C"n, Rn 5 X = (xj)1)j)n 6n&

j=1

xjej % C"n. (2.1)

The Cli!ord conjugation on C"n, denoted by ‘bar’, is defined as the unique real-linear involutionon C"n for which eI#eI = eI#eI = 1 for any multi-index I. Define the scalar part of u =

!'I uIeI %

C"n as u0 := u(, and endow C"n with the natural Hilbert space structure

)u, v* :=&

I

'uIvI , if u =

&

I

'uIeI , v =

&

I

'vIeI % C"n. (2.2)

It follows that

|u|2 = (u# u)0 = (u# u)0, )u, v* = (u# v)0 = (u# v)0, 2u, v % C"n, (2.3)

X = $X and X #X = $|X|2 for any X % Rn, (2.4)

u + u = 2 u0 for any u % Rn # Rn, (2.5)

u = u and u# v = v # u, for any u, v % C"n, (2.6)

|u# v| 0 cn|u||v|, for any u, v % C"n, (2.7)

X # Y + Y #X = $2)X,Y *, for any X,Y % Rn. (2.8)

5

Furthermore, if Ma denote the operator of multiplication by a (from the left) then, in the aboveCli!ord algebra setting,

(Ma)% = Ma, 2 a % C"n. (2.9)

where star denotes adjunction with respect to the inner product in C"n. As a corollary of (2.9) and(2.6), we have

)a# b, c* = )b, a# c* and )a# b, c* = )a, c# b*, 2 a, b, c % C"n. (2.10)

Let us also recall here the Dirac operator associated with C"n, namely

D :=n&

j=1

Mej#j . (2.11)

As is well-known,

$D2 = $, (2.12)

the Laplacian in Rn. In particular, if

E(X) :=

*+,

+-

1(n!1(2$ n)

1|X|n!2

, if n 3 3,

12&

log |X|, if n = 2,

X % Rn \ {0}, (2.13)

is the fundamental solution for the Laplacian in Rn, then

$(DE)(X) = $ 1(n!1

X

|X|n , X % Rn \ {0}, (2.14)

becomes, thanks to (2.12), a fundamental solution for the Dirac operator D in Rn.

2.2 Regular SKT domains and the Cauchy-Cli!ord operator

Let " 1 Rn be a fixed domain of locally finite perimeter. Essentially, this is the largest classof domains for which some version of the classical Gauss-Green formula continues to hold. Inparticular, there exists a suitably-defined concept of outward unit normal to #", which we denoteby + = (+1, ..., +n), and the role of the surface measure on #" is played by $ := Hn!17 #". Hereand elsewhere, Hk denotes k-dimensional Hausdor! measure. There are several excellent accountson these topics, including the monographs by H. Federer [9], W. Ziemer [34], and L.Evans andR.Gariepy [8].

Recall that the measure-theoretic boundary #%" of a domain " 8 Rn is defined by

#%" :=.

X % #" : lim supr*0

|Br(x) 4 "|rn

> 0, lim supr*0

|Br(x) \ "|rn

> 0/

, (2.15)

where |E| stands for the Lebesgue measure of E 8 Rn. As is well-known, if " has locally finiteperimeter, then the outward unit normal is defined $-a.e. on #%". In particular, if

Hn!1(#" \ #%") = 0, (2.16)

6

then + is defined $-a.e. on #".Following [10], we shall call a nonempty open set " 1 Rn a UR (uniformly rectifiable) domain

provided #" is uniformly rectifiable (in the sense of Definition 5.8) and (2.16) holds. Let usemphasize that, by definition, a UR domain " has locally finite perimeter as well as an Ahlforsregular boundary. We remind the reader that a closed set % 1 Rn is called Ahlfors regular providedthere exist 0 < C1 0 C2 < & such that

C1 Rn!1 0 Hn!1$B(X, R) 4 %

%0 C2 Rn!1, (2.17)

for each X % % and R % (0,&) (if % is compact, we require (2.17) only for R % (0, 1]). Theconstants C1, C2 intervening in (2.17) will be referred to as the Ahlfors constants of #".

For further use, let us point out that, as is apparent from definitions,

" 1 Rn is an UR domain with #" = #"

=/ Rn \ " is an UR domain, with the same boundary as ". (2.18)

It is also relevant to recall that a two-sided NTA domain with an Ahlfors regular boundary is a URdomain; cf. [[10], § 3].

For a UR domain " 1 Rn, the (principal value) Cauchy-Cli!ord operator C associated with #"is given by

C f(X) := lim%*0+

1(n!1

'

Y #!!|X!Y |>%

X $ Y

|X $ Y |n # +(Y )# f(Y ) d$(Y ), X % #", (2.19)

where (n!1 is the surface area of the unit sphere Sn!1 in Rn, and f is a C"n-valued function on#" (the reader is referred to [10, Proposition 3.24] for a discussion pertaining to the existence ofprincipal value singular integral operators on the boundaries of UR domains). Its formal adjoint isdefined as

C %f(X) = $ lim%*0+

1(n!1

'

Y #!!|X!Y |>%

+(X)# X $ Y

|X $ Y |n # f(Y ) d$(Y ), X % #", (2.20)

i.e.,

C % = M&C M& , (2.21)

where M& denotes the operator of Cli!ord multiplication from the left by + 6!n

j=1 +jej .Next, for each p % (1,&) we let Lp(#", d$) denote the Lebesgue space of $-measurable, p-th

power integrable functions on #", and denote by Lp(#", d$)'C"n the space of C"n-valued functionswith components in Lp(#", d$). In the case when " 1 Rn is an UR domain, the operator

C : Lp(#", d$)' C"n $" Lp(#", d$)' C"n (2.22)

is well-defined and bounded for any p % (1,&) (cf. [10]). In this setting, the operator (2.25) is thefunctional analytic adjoint of K in (2.25). Likewise, the operator (2.20) is the functional analyticadjoint of C in (2.19). Furthermore, since M& is a unitary operator on L2(#", d$)' C"n and (2.21)gives that C M& = $M&C %, it follows that in this setting C and $C % are unitarily equivalent.

7

Given a domain " 1 Rn of locally finite perimeter and for which (2.16) holds, define the Riesztransforms (Rk)1)k)n by

Rkf(X) := lim%*0+

2(n!1

'

Y #!!|X!Y |>%

xk $ yk

|X $ Y |n f(Y ) d$(Y ), X % #", (2.23)

where k % {1, ..., n}, and f is a (possibly Cli!ord algebra-valued) function on #". It is then clearfrom (2.19)-(2.23) that

C = 12

n&

k=1

RkMekM& . (2.24)

Let also recall here the harmonic double layer and its formal adjoint

Kf(X) := lim%*0+

1(n!1

'

Y #!!|X!Y |>%

)+(Y ), Y $X*|X $ Y |n f(Y ) d$(Y ), X % #", (2.25)

K%f(X) := lim%*0+

1(n!1

'

Y #!!|X!Y |>%

)+(X),X $ Y *|X $ Y |n f(Y ) d$(Y ), X % #", (2.26)

considered in the same geometric measure theoretic setting as before.Next, call " 8 Rn a regular Semmes-Kenig-Toro (SKT) domain if it is a two-sided NTA domain

with an Ahlfors-regular boundary and for which the unit normal is in VMO(#", d$), Sarason’sspace of functions of vanishing mean oscillation. This piece of terminology, which originated in[10], replaces what was previously called chord-arc domains with vanishing constant. A somewhatdi!erent, albeit equivalent, point of view on this topic can be found in the Appendix.

We now record a series of results recently established in [10] (the reader is again advised toconsult the Appendix for the relevant definitions). To state the first theorem, recall that thecommutator of two operators A,B is defined as [A, B] := AB $BA.

Theorem 2.1. Assume that " 1 Rn is an open set satisfying a two-sided local John conditionand such that #" is Ahlfors regular and compact. Also, denote by + = (+1, ..., +n) the outward unitnormal to ". Then the following statements are equivalent:

(i) the harmonic double layer K and the commutators [M&j , Rk], between the Riesz transformsand multiplication by the components of the unit normal, are compact operators on Lp(#", d$)for some (and, hence, all) p % (1,&);

(ii) C $ C % is a compact operator on Lp(#", d$)' C"n for some (hence all) p % (1,&);

(iii) " is a regular SKT domain.

The result below, which has recently been obtained in [10], involves two new concepts, bothdefined in the Appendix, namely the '-SKT domain and the John condition. Heuristically, thelatter condition can be thought of as a curvilinear, scale-invariant version of starlikeness. Asfor the former, this means that the boundary of the domain in question is Ahlfors regular, well-approximated by planes at scales 9 ', and the mean-oscillations of the unit normal are 0 '; see(5.5) in connection with the definition of VMO(#", d$).

8

Theorem 2.2. Let " 8 Rn be a bounded open set that satisfies a two-sided John condition andwhose boundary is Ahlfors regular.

Then there exists a constant Co > 1, depending only on the John and Ahlfors constants on ",with the following significance. Assume that there exists ' > 0, su!ciently small relative to theJohn and Ahlfors constants of ", with the property that

dist (+ , VMO(#", d$)) < ', (2.27)

where the distance is taken in the John-Nirenberg space BMO(#", d$), of functions of boundedmean oscillations. Then " is a 'o-SKT domain, with 'o = Co'. In addition, there exists R > 0with the property that for every X % #" and r % (0, R]

))))$($(X, r))(n!1rn!1

$ 1)))) 0 Co', (2.28)

where $(X, r) := B(X, r) 4 #".

In order to continue, we make one more definition. Specifically, given a Banach space (X , + · +), set

L(X ) := the space of all bounded linear operators on X ,

Cp (X ) := the space of all linear compact operators on X .(2.29)

As is well-known, L(X ) becomes a Banach space when equipped with the natural norm +T+L(X ) :=sup {+Tx+ : x % X , +x+ 0 1}, and Cp (X ) is a closed subspace of L(X ). The theorem belowappears in [10].

Theorem 2.3. Let " 1 Rn be an UR domain and assume that p % (1,&). Then there existsC > 0, depending only on n, p, and the Ahlfors constants of #", such that

dist"+ , VMO(#", d$)

#0 C

0dist

"C $ C % , Cp (Lp(#", d$)' C"n)

#11/n. (2.30)

The results in this subsection will play a prominent role in subsequent considerations. Although,strictly speaking, the Cli!ord algebra setting from §2.1 does not contain the standard field ofcomplex numbers, very similar considerations apply to this latter context, leading to analogousresults to the ones discussed above. In this scenario, the Cli!ord-Cauchy operator takes the familiarform

C f(z) := lim%*0+

12&i

'

"#!!

|z!'|>%

f(,), $ z

d,, z % #", " 1 R2 6 C. (2.31)

This remark will be tacitly assumed in what follows. In fact, it is possible to slightly alter theCli!ord algebra formalism discussed in §2.1 in order to ensure a more uniform theory, in which(2.19) becomes precisely (2.31) when the Cli!ord algebra in question has precisely one imaginaryunit. See, e.g., [26] for the details of this construction.

2.3 Hardy spaces and Calderon’s decomposition

Assume that " 1 Rn be an open set and abbreviate

"+ := ", "! := Rn \ ". (2.32)

9

In analogy with the classical setting of functions of one complex variable, for each p % (1,&) definethe Hardy spaces H p("±) by

H p("+) := {u : "+ " C"n : N (u) % Lp(#", d$), Du = 0 in "+}, (2.33)

with the convention that if "+ is unbounded and #" is bounded, the decay condition

u(x) = O(|x|1!n) as |x|"& (2.34)

is also included. The space H p("!) is defined analogously. It is useful to observe that, thanks to(2.12), functions in H p("±) are harmonic. Next, the boundary Hardy spaces are defined as

H p± (#") := {u|!! : u % H p("±)}. (2.35)

Above, for u : "± " C"n, we define the nontangential maximal function of u by

Nu(Z) := N(u(Z) := sup {|u(X)| : X % "±, |X $ Z| < (1 + -) dist (X, #")}, Z % #", (2.36)

where - > 0 is a fixed parameter, and we make the convention that Nu(Z) = 0 whenever thesupremum is taken over an empty set.

In addition to the principal value Cauchy-Cli!ord operator C from (2.19) let us also recall herethe Cauchy-Cli!ord operator

Cf(X) :=1

(n!1

'

!!

X $ Y

|X $ Y |n # +(Y )# f(Y ) d$(Y ), X % ", (2.37)

mapping a C"n-valued function f defined on #" into a C"n-valued function defined in ".Up to this point, the above considerations are purely formal, as the class of domains to which "

belongs. From [10, Section 3], it follows that if " is a UR domain then for every f % Lp(#", d$)'C"n,p % (1,&), C f is meaningfully defined a.e. on #",

+N (Cf)+Lp(!!,d#) 0 C+f+Lp(!!,d#)&C$n, +C f+Lp(!!,d#) 0 C+f+Lp(!!,d#)&C$n

, (2.38)

D(Cf) = 0 in ", and Cf)))!!

= (12I + C )f a.e. on #", (2.39)

where the boundary trace is taken in a nontangential pointwise sense.Let us also note here that if " 1 Rn is an NTA domain with an Ahlfors regular boundary and

p % (1,&), the following Fatou type theorem holds

2u % H p("+) =/ u)))!!

exists a.e. in the nontangential pointwise sense, (2.40)

and that the following Cauchy’s reproducing formula is valid

2u % H p("+) =/ u = C(u|!!) in ". (2.41)

Indeed, as has been noted in [10], Theorem 6.4 on p. 112 of [14] gives that any function u whichis harmonic in an NTA domain " and nontangentially bounded from below on E 1 #" has anontangential limit (Xo-a.e. on E (where (Xo is the harmonic measure with pole at Xo % ").Now, the fact that u|!! exists for every u % H p("+), p % (1,&), is a consequence of (2.12), theabove local Fatou theorem applied to Ek := {X % #" : Nu(X) < 2k}, k = 1, 2..., and the mutualabsolute continuity between the surface and harmonic measures proved in [5].

The key to establishing (2.41) is the following version of Green’s formula from [10]:

10

Theorem 2.4. Let " 1 Rn be an open set which is either bounded or has an unbounded boundary.Assume that #" is Ahlfors regular and satisfies (2.16). Denote by + the outward unit normal to#" and set $ := Hn!17#". Then

'

!

div.v dX ='

!!

)+,.v))!!* d$ (2.42)

holds for each vector field .v % C0(") that satisfies

|.v|, div.v % L1("), N.v % L1(#", d$) 4 Lploc(#", d$) for some p % (1,&),

and the pointwise nontangential trace .v))!!

exists $-a.e. on #".(2.43)

For " 1 Rn NTA domain with an Ahlfors regular boundary, we deduce from (2.38), (2.39) and(2.40) that the operator

C : Lp(#", d$)' C"n $" H p("+) (2.44)

is well-defined, bounded and onto for each p % (1,&). From this and (2.38)-(2.40) we then obtain(12I + C )2 = 1

2I + C so that, ultimately,

C 2 = 14I on Lp(#", d$)' C"n, 1 < p < &. (2.45)

As a consequence, if " is an NTA domain with an Ahlfors regular boundary, and p % (1,&), then

Im"

12I + C : Lp(#", d$)' C"n

#= H p

+ (#") = Ker"$1

2I + C : Lp(#", d$)' C"n

#. (2.46)

Likewise, if "! is an NTA domain with an Ahlfors regular boundary and p % (1,&), then also

Im"$1

2I + C : Lp(#", d$)' C"n

#= H p

! (#") = Ker"

12I + C : Lp(#", d$)' C"n

#. (2.47)

Proposition 2.5. Suppose " 1 Rn is a two-sided NTA domain with an Ahlfors regular boundary(which makes it a UR domain), and fix p % (1,&). Then the spectrum of the operator C onLp(#", d$)'C"n is {$1

2 ,+12}, the numbers ±1

2 are eigenvalues and H p± (#") are the corresponding

eigenspaces.

Proof. It su#ces to observe that, thanks to (2.45),

(/I $ C )!1 =1

/2 $ 14

(/I + C ), / % R \ {±12}, (2.48)

as operators in Lp(#", d$)' C"n. Then everything follows from this and (2.46)-(2.47). !We conclude this subsection with a result which is going to play a basic role for the goals we

have in mind. As a preamble, we remind the reader a piece of terminology. Given a two closedsubspaces X0,X1 of a Banach space X, we say that X is the direct sum of X0 and X1 (and writeX = X0 (X1) provided any x % X can be uniquely written as x = x0 + x1 with xj % Xj , j = 0, 1,Note that, as a consequence of the Open Mapping Theorem, the assignments X 5 x :" xj % Xj ,j = 0, 1, are continuous (i.e., the direct sum decomposition X = X0 (X1 is ‘topological’).

11

Theorem 2.6. Assume that " 1 Rn is a two-sided NTA domain with an Ahlfors regular boundary(making it a UR domain). Then the following decomposition is valid for each p % (1,&):

Lp(#", d$)' C"n = H p+ (#")(H p

! (#"). (2.49)

Proof. Any f % Lp(#", d$)'C"n can be written as f+$f! where f± := (±12I +C )f % H p

± (#"), by(2.46)-(2.47). Furthermore, thanks to the second estimate in (2.38) (recall that a two-sided NTAdomain with an Ahlfors regular boundary is a UR domain), there exists C = C(", p) > 0 suchthat +f±+Lp(!!,d#)&C$n

0 C+f+Lp(!!,d#)&C$n. To see that the sum in (2.49) is direct, assume that

f % H p+ (#") 4H p

! (#"). Then (12I + C )f vanishes, by (2.47) and is equal to f by (2.46). Thus,

H p+ (#") 4H p

! (#") = 0, finishing the proof of the theorem. !

2.4 Cli!ord-Szego projections and the Kerzman-Stein operator

Fix a two-sided NTA domain " 1 Rn with an Ahlfors regular boundary, and define the Cli"ord-Szego projections

P± : L2(#", d$)' C"n $" H 2± (#") !" L2(#", d$)' C"n (2.50)

as the orthogonal projections of L2(#", d$) ' C"n onto the closed subspace H 2± (#"). It has been

shown in [10] that there exists ) > 0 with the property that P± in (2.50) extend to

P± : Lp(#", d$)' C"n $" H p± (#") (2.51)

in a continuous and onto fashion for each p % (2 $ ), 2 + )). Furthermore, when " is a boundedregular SKT domain, then this is true for each p % (1,&).

We now discuss a version of Kerzman-Stein’s formula (cf. [18]) in the Cli!ord algebra settingfrom [10]. Specifically, we note that formulas (2.46)-(2.47) and the definition of P± readily implythat

P±(±12I + C ) = ±1

2I + C , (;12I + C )P± = 0. (2.52)

From the second formula above and duality we also obtain

P±(;12I + C %) = 0. (2.53)

Subtracting this from the first formula in (2.52) then gives P±(±I + C $C %) = ±12I + C . Now, if

A := C $ C % : Lp(#", d$)' C"n $" Lp(#", d$)' C"n, 1 < p < &, (2.54)

it has been observed in [10] that there exists ) > 0 so that

I ±A : Lp(#", d$)' C"n $" Lp(#", d$)' C"n are invertible for 2$ ) < p < 2 + ). (2.55)

The above considerations then justify the following Kerzman-Stein type formula

P± = (12I ± C ) < (I ±A )!1, (2.56)

valid in Lp(#", d$)' C"n, if 2$ ) < p < 2 + ). This, (2.55) and (2.22) then show that P± extendas bounded operators on Lp(#", d$)' C"n for 2$ ) < p < 2 + ) and, in fact,

P± : Lp(#", d$)' C"n $" H p± (#") are onto for 2$ ) < p < 2 + ). (2.57)

Several other properties of interest are collected in the proposition below.

12

Proposition 2.7. Let " 1 Rn be a two-sided NTA domain with an Ahlfors regular boundary. Thenthere exists ) > 0 such that if 2$ ) < p < 2 + ) then the following identities are valid:

(P+ $P!)!1 = C + C %, (2.58)

C $ C % = P+P!(C + C %)$ (C + C %)P!P+, (2.59)

I $P+ $P! = (P+P! $P!P+)(C + C %), (2.60)

(I + 4C C %)!1 = (P+ $P!)C , (2.61)

(I + 4C %C )!1 = (P+ $P!)C %, (2.62)

4(I + 4C %C )!1[C ,C %](I + 4C C %)!1 = $(P+ $P!)(C $ C %), (2.63)

as operators in Lp(#", d$)' C"n.

Proof. As is apparent from Theorem 2.6, (2.38)-(2.39) and (2.41), the operators P± := 12I ± C

are, respectively, the skew projection of L2(#", d$)' C"n onto H 2(#"±) parallel to (with kernel)H 2(#"+). Then, from an abstract point of view, (2.58)-(2.63) considered as operator identitieson L2(#", d$) ' C"n, are formulas relating the skew projections P± to the orthogonal projectionsP±. Given that there exists ) > 0 with the property that all operators involved are boundedLp(#", d$) ' C"n if 2 $ ) < p < 2 + ), the same identities are then valid on Lp(#", d$) ' C"n,2$ ) < p < 2 + ), by density.

For the convenience of the reader we, nonetheless, include here a direct argument. To getstarted, we note that (2.52)-(2.53) imply

(P+ $P!)(C + C %) = P+

0(12I + C ) + ($1

2I + C %)1$P!

0($1

2I + C ) + (12I + C %)

1

= P+(12I + C ) + P+($1

2I + C %)$P!($12I + C )$P+(1

2I + C %)

= (12I + C ) + 0$ ($1

2I + C )$ 0 = I. (2.64)

Hence also (C + C %)(P+ $P!) = I, by duality. From this, (2.58) follows. To continue, note thatobserve that (2.52) implies

P+($12I + C ) = P+

"(12I + C )$ I

#= 1

2I + C $P+, (2.65)

and, further, that

P+P!($12I + C ) = 1

2I + C $P+. (2.66)

From the second formula in (2.52) and duality we also obtain P!(12I + C %) = 0, so that

P+P!(12I + C %) = 0. (2.67)

Adding (2.67) and (2.66) then yields

12I + C $P+ = P+P!(C + C %), (2.68)

so by subtracting (2.68) from its dual version we arrive at (2.59). Going further, analogously to(2.68) we have

P!P+(C + C %) = $12I + C + P!. (2.69)

13

By subtracting this from (2.68) we arrive at (2.60). Turning our attention to (2.61), we write

(14I + C C %)!1 =

0C (C + C %)

1!1= (C + C %)!1C!1 = 4(P+ $P!)C (2.70)

by (2.45) and (2.58). Finally, (2.62) is proved similarly and (2.63) follows by subtracting (2.62)from (2.61). This concludes the proof of the proposition. !

2.5 The geometry of Hardy spaces

Recall that given two closed subspaces H1, H2 of a Hilbert space H with inner product )·, ·* andnorm + · +, the angle between H1 and H2, denoted by <) (H1,H2), is the unique number % % [0,&/2]for which

cos % = sup.)f1, f2*/+f1++f2+ : f1 % H1, f2 % H2

/. (2.71)

In particular,

H1=H2 ./ <) (H1,H2) =&

2, (2.72)

and it is straightforward to check that if Pj : H " H, j = 1, 2, are the orthogonal projections ontoH1, H2, then

+P1P2+L(H) = +P2P1+L(H) = cos (<) (H1,H2)). (2.73)

Theorem 2.8. Let " 1 Rn be a two-sided NTA domain with an Ahlfors regular boundary, and set

% :=<)"H 2

+ (#") , H 2! (#")

#. (2.74)

Then

+P+P!+L$L2(!!,d#)&C$n

% = +P!P++L$L2(!!,d#)&C$n

% = cos %, (2.75)

+C +L$L2(!!,d#)&C$n

% = +C %+L$L2(!!,d#)&C$n

% =12

cot"%

2

#, (2.76)

+± 12I + C +L

$L2(!!,d#)&C$n

% =1

sin %, (2.77)

+(P+ $P!)!1+L$L2(!!,d#)&C$n

% = +C + C %+L$L2(!!,d#)&C$n

% =1

sin %, (2.78)

+/I + C %C +L$L2(!!,d#)&C$n

% = +/I + C %C +L$L2(!!,d#)&C$n

% = / +14

cot2"%

2

#, (2.79)

for every / 3 0, and

cos % 0 +C $ C %+L$L2(!!,d#)&C$n

% 0 2 cot %, (2.80)

cos % 0 +I $P+ $P!+L$L2(!!,d#)&C$n

% 0 2 cot %, (2.81)

cos %

20 +[C , C %]+

L$L2(!!,d#)&C$n

% 0 2cos %

sin2 %. (2.82)

14

Proof. As was the case with Proposition 2.7 (cf. the comment at the beginning of its proof), theabove formulas follow from general Hilbert space geometrical considerations. A direct argumentgoes as follows. Clearly, (2.75) is a direct consequence of (2.73). Also, standard functional analysisgives +C %+

L$L2(!!,d#)&C$n

% = +C +L$L2(!!,d#)&C$n

%. Thus, as far as (2.76) is concerned, there remains

to prove that the first operator norm is 12 cot (%/2). A more general result of this type is proved

in (2.98) below. Here we only wish to comment that, in the language of the skew projections P±,we have C = 1

2(P+ $ P!), so matters can be easily reduced to computing the norm of a 2 > 2matrix by considering all possible restrictions of C onto two dimensional subspaces span (h+, h!),h± % H 2

± (#"), then taking the supremum of the norm.Moving on, formula (2.77) simply expresses the fact that the operator norm of the skew projec-

tions P± := ±12I + C is 1/ sin %. Next, we write

)(C + C %)f, f* = )(12I + C )f, f*+ )($1

2I + C %)f, f*

= )f+, f+ $ f!*+ )f+ $ f!, f!* = +f++2 $ +f!+2, (2.83)

for every f % L2(#", d$)' C"n. Here, as usual, f± := (±12I + C )f % H 2

± (#"). Since C + C % is anormal operator (in fact, self-adjoint), we may then write

+C + C %+L$L2(!!,d#)&C$n

% = sup.|)(C + C %)f, f*| / +f+2 : f % L2(#", d$)' C"n

/

= sup. | +f++2 $ +f!+2|

+f+ $ f!+2: f± % H 2

± (#")/

= sup2

| +f++2 $ +f!+2|+f++2 $ 2)f+, f!*+ +f!+2

: f± % H 2± (#")

3. (2.84)

Elementary calculus shows that for every - % [0,&/2]

max)1,)2>0

|/21 $ /2

2|/2

1 + /22 $ 2/1/2 cos -

=1

sin -. (2.85)

Thus, for any two vectors v1, v2 in a Hilbert space

max)1,)2>0

| +/1v1+2 $ +/2v2+2|+/1v1+2 $ 2)/1v1,/2v2*+ +/2v2+2

=1

sin (<) (v1, v2)). (2.86)

Returning with this in (2.84) then yields

+C + C %+L$L2(!!,d#)&C$n

% =1

sin"<)

$H 2

+ (#") , H 2! (#")

%# , (2.87)

finishing the justification of (2.78). As far as (2.79) is concerned, this is a consequence of (2.76)and elementary functional analysis.

Next, observe that when (2.75), (2.78) and (2.59) are used in concert, they entail

+C $ C %+L$L2(!!,d#)&C$n

% 0 2+P+P!+L$L2(!!,d#)&C$n

%+C + C %+L$L2(!!,d#)&C$n

%

0 2 (cos %)(sin %)!1 = 2 cot %. (2.88)

15

This justifies the upper bound in (2.80). As for the lower bound in (2.80), note that for eachf± % H 2

± (#") we have

)(C $ C %)f+, f!* = )C f+, f!* $ )f+, C f!*

= 12)f+, f!*+ 1

2)f+, f!* = )f+, f!*. (2.89)

Consequently,

+C $ C %+L$L2(!!,d#)&C$n

% = sup.)(C $ C %)g, h*/+g++h+ : g, h % L2(#", d$)' C"n

%/

3 sup.)(C $ C %)f+, f!*/+f+++f!+ : f± % H 2

± (#")/

= sup.)f+, f!*/+f+++f!+ : f± % H 2

± (#")/

= cos %, (2.90)

as desired.Turning our attention to (2.81), we first note that, (2.60), (2.75) and (2.78) give

+I $P+ $P!+L$L2(!!,d#)&C$n

% 0 2 (cos %)(sin %)!1 = 2 cot %, (2.91)

which is the upper bound in (2.81). Since for each f± % H 2± (#") we have

)(I $P+ $P!)f+, f!* = $)f+, f!*, (2.92)

the estimate cos % 0 +I$P+$P!+L$L2(!!,d#)&C$n

% follows much as in (2.89)-(2.90). This justifies

(2.81).As for (2.82), we first note that

[C , C %] = (C $ C %)(C + C %) (2.93)

by (2.45) and its dual version. In turn, this and (2.58) also give

[C , C %](P+ $P!) = C $ C %. (2.94)

Consequently,

+[C , C %]+L$L2(!!,d#)&C$n

% = +(C $ C %)(C + C %)+L$L2(!!,d#)&C$n

%

0 +C + C %+L$L2(!!,d#)&C$n

%+C % $ C +L$L2(!!,d#)&C$n

%

0 2cot %

sin %= 2

cos %

sin2 %, (2.95)

by (2.93) and (2.78)-(2.80), and

cos % 0 +C $ C %+L$L2(!!,d#)&C$n

%

0 +[C , C %]+L$L2(!!,d#)&C$n

%+P+ $P!+L$L2(!!,d#)&C$n

%

0 2+[C , C %]+L$L2(!!,d#)&C$n

%, (2.96)

16

by (2.94) and (2.80). In concert, (2.95)-(2.96) prove (2.82), thus concluding the proof of thetheorem. !

A similar type of argument as in the proof of (2.78) can be used to obtain resolvent normformulas for the Cauchy-Cli!ord operator.

Proposition 2.9. Suppose that " 1 Rn is a two-sided NTA domain with an Ahlfors regularboundary, and let % be as in (2.74). Then

+(/I $ C )!1+L$L2(!!,d#)&C$n

% =1)))/ + 1

2

)))

41 +

2(sec %)

?4/2 tan2% + 1 + 2/ tan2% $ 1

(2.97)

for every / % R \ {±12}.

Proof. Thanks to (2.48), it su#ces to show that

+/I + C +L$L2(!!,d#)&C$n

% =)))/ + 1

2

)))

41 +

2(sec %)

?4/2 tan2% + 1 + 2/ tan2% $ 1

(2.98)

for every / % R, if / @= $12 (incidentally, the case / = $1

2 is covered by (2.77)).With this goal in mind, we observe that for each number / % R and each function f %

L2(#", d$) ' C"n, we may write (/I + C )f = (/ $ 12)f + (1

2I + C )f = (/ $ 12)(f+ $ f+) + f+ =

(/ + 12)f+ $ (/$ 1

2)f! where, as before, f± := (±12I + C )f . Thus,

+/I + C +2L$L2(!!,d#)&C$n

% = sup.+(/I + C )f+2/+f+2 : f % L2(#", d$)' C"n

/

= sup.+(/ + 1

2)f+ $ (/$ 12)f!+2/+f+ $ f!+2 : f± % H 2

± (#")/

(2.99)

= sup

((/ + 1

2)2+f++2 + 2(/2 $ 14))f+, f!*+ (/$ 1

2)2+f!+2

+f++2 $ 2)f+, f!*+ +f!+2: f± % H 2

± (#")

5.

For f± % H 2± (#"), nonzero, let - be such that cos - = )f+, f!*/+f+++f!+ and set t := +f++/+f!+.

An elementary (yet tedious) analysis shows that / @= $12 and - % (0,&/2)

maxt>0

(/ + 12)2t2 + 2(/2 $ 1

4)(cos-)t + (/$ 12)2

t2 $ 2 (cos -)t + 1

="/ + 1

2

#2.1 +

2(sec -)

?4/2 tan2- + 1 + 2/ tan2-$ 1

/. (2.100)

Then formula (2.98) follows from (2.99) and (2.100). !Next, we record an useful consequence of Theorem 2.8.

Corollary 2.10. Assume that " 1 Rn is a two-sided NTA domain with an Ahlfors regular bound-ary. The following are equivalent:

(i) There holds

L2(#", d$)' C"n = H 2+ (#")(H 2

! (#"), orthogonal sum. (2.101)

17

(ii) The Cauchy-Cli"ord operator satisfies either of the following six conditions

C $ C % = 0, C C % $ C %C = 0,

+12I + C +

L$L2(!!,d#)&C$n

% = 1, + $ 12I + C +

L$L2(!!,d#)&C$n

% = 1,

+C +L$L2(!!,d#)&C$n

% = 12 , +C + C %+L

$L2(!!,d#)&C$n

% = 1.

(2.102)

(iii) The Cli"ord-Szego projections satisfy either of the following four conditions

P+P! = 0, P!P+ = 0,

P+ + P! = I, +(P+ $P!)!1+L$L2(!!,d#)&C$n

% = 1.(2.103)

Proof. This is a consequence of Theorem 2.8, which shows that each of the conditions (i) $ (iii)above is equivalent to having % = &/2. !Remark. As is apparent from Theorem 2.8, in general we have

+± 12I + C +L

$L2(!!,d#)&C$n

% 3 1, +C +L$L2(!!,d#)&C$n

% 3 12 ,

+C + C %+L$L2(!!,d#)&C$n

% 3 1, +(P+ $P!)!1+L$L2(!!,d#)&C$n

% 3 1.(2.104)

What the above corollary shows is that extremal case in each of these four inequalities occursprecisely when <)

$H 2

+ (#") , H 2! (#")

%= *

2 . Later on, we shall show that, in turn, this lattercondition actually determines the shape of ".

Next, consider

R±(#") :=

*,

-

N&

j=1

X $Xj

|X $Xj |n# aj : N % N, Xj % "+, aj % C"n

67

8 . (2.105)

That is, R±(#") are the right Cli!ord module spanned by rational functions of the form X!Z|X!Z|n ,

X % #", when Z % "+.

Theorem 2.11. Suppose that " 1 Rn is a two-sided NTA domain whose boundary is Ahlforsregular and, as usual, set "+ := ", "! := Rn \ ", $ := Hn!17 #". Then the spaces R±(#") aredense in H 2

± (#"), respectively. Hence, as a consequence of this and (2.71), one has

cos"<)

$H 2

+ (#") , H 2! (#")

%#(2.106)

= sup

*++,

++-

'

!!)P+(X), P!(X)* d$(X)

"9!! |P+(X)|2 d$(X)

#1/2"9!! |P!(X)|2 d$(X)

#1/2: P± % R±(#")

6++7

++8.

Proof. We proceed in a series of steps, starting with

Step I. For every & % (H 2+ (#"))% there exists g % L2(#", d$) ' C"n satisfying (1

2I + C %)g = gand for which

&(f) ='

!!)f, g* d$, 2 f % H 2

+ (#"). (2.107)

18

Indeed, as a consequence of Riesz’s representation theorem, there exists h % H 2+ (#") with the

property that

&(f) ='

!!)f, h* d$, 2 f % H 2

+ (#"). (2.108)

Since every f % H 2+ (#") can be expressed as f = (1

2I + C )f , it follows that (2.107) holds withg := (1

2I + C %)h.

Step II. A functional & % (H 2+ (#"))% vanishes identically if and only if

&" X $ ·|X $ ·|n # a

#= 0, 2 a % C"n, 2X % "!. (2.109)

To prove this claim, we first observe that X!·|X!·|n # a % H 2

+ (#") for every a % C"n and everyX % "!. We may therefore use the representation formula (2.107), which holds for some functiong % L2(#", d$)' C"n with (1

2I + C %)g = g, in order to conclude that

0 ='

!!

: X $ Y

|X $ Y |n # a , g(Y );

d$(Y ) =:'

!!

X $ Y

|X $ Y |n # g(Y ) d$(Y ) , a;, (2.110)

for every X % "! (here, (2.9) is also used). In turn, this implies that C(+ # g) = 0 in "! and,further, that ($1

2I + C )(+ # g) = 0 on #" by going nontangentially to the boundary. By (2.21), itfollows that (1

2I + C %)g = 0 on #", hence ultimately g = 0 given that (12I + C %)g = g. This and

(2.107) then prove that & 6 0.

Step III. The spaces R±(#") are dense in H 2± (#"), respectively.

For the choice plus of the sign, this is a direct consequence of Hahn-Banach’s extension theoremand the result proved in Step II. The choice minus of the sign is handled similarly. !

To state our next result, for any two vectors X = (xj)j , Y = (yj)j in Rn we define

X > Y :="xjyk $ xkyj

#

1)j,k)n. (2.111)

Theorem 2.12. Assume that " 1 Rn is a two-sided NTA domain whose boundary is Ahlforsregular and, as before, set "+ := ", "! := Rn \ ", $ := Hn!17 #". Then

<)$H 2

+ (#") , H 2! (#")

%=

&

2./

'

!!

(Y $X+)# (Y $X!)|Y $X+|n|Y $X!|n

d$(Y ) = 0 2X± % "±. (2.112)

Coordinate-wise,

<)$H 2

+ (#") , H 2! (#")

%=

&

2./

*++,

++-

9

!!

#Y!X+,Y!X"$|Y!X+|n|Y!X"|n d$(Y ) = 0 and

9

!!

(Y!X+),(Y!X")|Y!X+|n|Y!X"|n d$(Y ) = 0 2X± % "±.

(2.113)

Proof. Note that (2.10) implies that, for every X± % "± and every a± % C"n,'

!!

: Y $X+

|Y $X+|n# a+ ,

Y $X!|Y $X!|n

# a!;

d$(Y )

= $:'

!!

(Y $X+)# (Y $X!)|Y $X+|n|Y $X!|n

d$(Y ) , a! # a+

;. (2.114)

19

This identity shows that if the integral in the right hand-side of (2.114) vanishes for every X± % "±then

9!!)P+, P!* d$ = 0 for every P± % R±(#"). Hence, in this case, <)

$H 2

+ (#") , H 2! (#")

%= *

2by (2.106).

Conversely, if <)$H 2

+ (#") , H 2! (#")

%= *

2 then H 2+ (#")=H 2

! (#") by (2.72). In particular, forevery X± % "± and a± % C"n, the Cli!ord algebra-valued functions Y!X+

|Y!X+|n # a+ % H 2+ (#") and

Y!X"|Y!X"|n # a! % H 2

! (#") are orthogonal. Thus, by (2.114), the integral in the right hand-side of(2.112) necessarily vanishes for every X± % "±.

Finally, (2.113) is a consequence of (2.112) and the fact that the components of X # Y arerecovered in )X, Y * and X > Y . !

3 Hardy spaces and SKT domains

We start with some motivation for the material in this section. Assume that " is a simply connected,open subset of R2 (naturally identified with the complex field C) and whose boundary #" is arectifiable (orientable) curve. Substituting the Cauchy-Riemann operator # in place of the Diracoperator D in the considerations in § 2.3, it is then possible to define Hardy spaces H 2

± (#") in asimilar fashion as before. In this setting, the Cli!ord-Szego projections P± map from L2(#", d$)'Conto H 2

± (#"). G.David has established the following result (which appears as Theoreme 2 on p. 236in [4]):

Theorem 3.1. Given ' > 0 there exists ) > 0 with the property that

+I $P+ $P!+L$L2(!!,d#)&C

% 0 ) =/ #" is chord-arc with constant 0 1 + ', (3.1)

in the class of chord-arc domains " 8 R2.

Recall that a curve % is called chord-arc, with constant 0 C, if

length of the arc in between z1 and z2 is 0 C|z1 $ z2|, 2 z1, z2 % %. (3.2)

An equivalent way of describing (3.2) is to demand that |s$ t| 0 C|z(s)$ z(t)| for s, t % R, wherez(s) is a parametrization of #" with |z'(s)| = 1.

One of our goals in this section is to prove a higher dimensional extension of Theorem 3.1.David’s proof of this result uses the arc-length parametrization of #" in order to reduce mattersto an estimate from below of the norm of the commutator between the Hilbert transform (on R)and a unitary operator (on L2(R)) associated with a certain change of variables. This strategymakes essential use of the two-dimensional setting, so a di!erent approach is required in higherdimensions. To handle this problem, we shall make use of the recent results from [10].

First we establish the following result, which has independent interest.

Theorem 3.2. Assume that " 1 Rn is a bounded, two-sided NTA domain whose boundary isAhlfors regular. Then there exists C > 0, depending only on the NTA and Ahlfors constants of ",such that

dist"+ , VMO(#", d$)

#0 C

n?

cot % (3.3)

where % :=<)$H 2

+ (#") , H 2! (#")

%.

Furthermore, if % is su!ciently close to &/2 (relative to the NTA and Ahlfors constants of ")then " is a '-SKT domain, for some ' = '(%) > 0 such that ' , 0 as % - &/2.

20

Proof. By (2.80),

dist"C $ C % , Cp (L2(#", d$)' C"n)

#0 +C $ C %+L

$L2(!!,d#)&C$n

% 0 2 cot %, (3.4)

so (3.3) follows from this and (2.30). Finally, the last part in the statement of the theorem is aconsequence of (3.3) and Theorem 2.2. !

The above theorem shows that % :=<)$H 2

+ (#") , H 2! (#")

%encodes significant information about

the regularity of ". In fact, the closer % is to &/2, the more regular " becomes. As we shall see inSection 4, the limit case % = &/2 is special in that not only does this ensure that " is smooth, butsuch a specification actually determines the shape of ".

Our next result can be viewed as the higher dimensional analogue of Theorem 3.1. To state it,recall that $(X, r) := B(X, r)4 #" and that (n!1 stands for the surface area of the unit sphere inRn.

Theorem 3.3. For every ' > 0 there exist ) > 0 and R > 0 with the property that

+I $P+ $P!+L$L2(!!,d#)&C$n

% 0 ) =/ supX"!!, 0<r<R

))))Hn!1($(X, r))

(n!1rn!1$ 1

)))) 0 ', (3.5)

in the class of all bounded domains " 1 Rn which are two-sided NTA and have Ahlfors regularboundaries, with a certain fixed common bound on the diameter, NTA and Ahlfors constants.

Proof. From Theorem 2.2 it follows that for every ' > 0 there exist ) > 0 and R > 0 such that

dist"+ , VMO(#", d$)

#0 ) =/ sup

X"!!, 0<r<R

))))$($(X, r))(n!1rn!1

$ 1)))) 0 '. (3.6)

Then (3.5) is a consequence of this, (3.3) and (2.81). !

4 Characterizations of balls and half-spaces

4.1 Motivation

The following characterization of disks in the plane has been noted by N. Kerzman and E.M. Steinin [18].

Theorem 4.1. Let " 1 R2 be a simply-connected, bounded, C- domain. Then the Szego projectioncoincides with the (trace of the) Cauchy operator on #" if and only if " is a disk.

In general, the Szego projection coincides with the (trace of the) Cauchy operator on #" if andonly if the latter is a self-adjoint operator. In this formulation, and with Cli!ord algebras playingthe role of the complex structure in R2, the result in Theorem 4.1 has been extended to smoothsubdomains of the sphere Sn!1, n 3 2, by P.Van Lancker in [32].

The discussion above suggests making the following conjecture, which can be thought of as ahigher-dimensional, geometrically sharp version of the Kerzman-Stein result in Theorem 4.1:

Conjecture 4.2. Let " be a nonempty, open subset of Rn, of finite perimeter, and for which

#" = #". (4.1)

Recall the Cauchy-Cli"ord operator from (2.19) and its formal adjoint introduced in (2.20). Thenthe kernels of C and C % coincide a.e. if and only if " is a ball, the complement of a ball, or ahalf-space.

21

A few comments are in order here. First, the class of domains of locally finite perimeter is thelargest one within which one can still talk about a reasonable concept of surface measure and unitnormal, which are the main ingredients entering the definition on the Cauchy-Cli!ord operator C .Second, the adjoint C % is taken in a formal sense. As already pointed out, when " happens to bea UR domain then C turns out to be bounded on, say, L2(#", d$) ' C"n, and C % is in fact thefunctional analytic adjoint of C in this case. Finally, condition (4.1) is necessary if the conclusion inConjecture 4.2 is to hold, for otherwise an open ball without its center would be a counterexampleto the left-to-right implication.

Recall next the harmonic double layer potential operator K and its adjoint K% from (2.25)-(2.26).

Conjecture 4.3. Assume that " is a nonempty, bounded, open subset of Rn, of finite perimeterand for which (4.1) holds. Then the kernels of the integral operators K and K% coincide a.e. ifand only if " is a ball.

Clearly, Conjecture 4.3 implies Conjecture 4.2 since K, K% are the scalar components of C andC %, respectively. If " is su#ciently smooth, then the conclusion in Conjecture 4.3 is known tohold (see [22] for the class of Lipschitz domains). The goal here is to explore the extent to whichConjecture 4.3 is true for larger classes of domains, for which substantially less regularity is assumedto begin with.

To continue, let " 8 Rn be a domain of locally finite perimeter, with surface measure $ andunit normal +. In this setting, recall the harmonic single layer potential

S f(X) :='

!!

E(X $ Y )f(Y ) d$(Y ), X /% #", (4.2)

and its boundary version

Sf(X) :='

!!

E(X $ Y )f(Y ) d$(Y ), X % #", (4.3)

where E is as in (2.13). Let us also define

Lp0(#", d$) :=

.f % Lp(#", d$) :

9!!f d$ = 0

/, 1 < p < &. (4.4)

In [19], J. Kral and D.Medkova prove the following result.

Theorem 4.4. Let " 8 R2, a bounded, connected open set whose boundary is a rectifiable, Ahlforsregular Jordan curve. Then

S1 = constant on #" ./ K"L2

0(#", d$)#8 L2

0(#", d$) ./ " is a disk. (4.5)

While the aforementioned result was established using conformal mapping techniques in [19], herewe shall prove an extension of Theorem 4.4 via a conceptually new proof, of purely real-variablemethods, using results on stability of isoperimetric problems, and combining ideas and techniquesfrom [25] and [10]. See Theorem 4.13 in § 4.2 for details. Here we only wish to make a couple ofcomments pertaining to (4.5). First, in the context of Theorem 4.4, the equivalences in (4.5) canbe augmented by

K"L2

0(#", d$)#

= 0 ./ " is a disk. (4.6)

22

Indeed, the left-to-right implication in (4.6) is covered by the second equivalence in (4.5), whereasthe right-to-left implication in (4.6) is a direct consequence of the fact that

" 1 R2 is a disk of radius R =/ )Y $X, +(Y )*|X $ Y |2 =

12R

, 2X, Y % #". (4.7)

This obviously implies that Kf = 0 on #" for any f % L20(#", d$).

Our second comment is that the first equivalence in (4.5) also holds in the higher dimensionalsetting as indicated in the lemma below.

Lemma 4.5. Let " 1 Rn be a bounded, connected UR domain. Then

S 1 = constant in " ./ S1 = constant on #"

./ K%1 = constant on #" ./ K%1 = 12 on #"

./ K"L2

0(#", d$)#8 L2

0(#", d$). (4.8)

Proof. If S 1 is constant in " then, going nontangentially to the boundary, shows that S1 isconstant on #". On the other hand, if S1 6 c on #", then u := S 1 is a harmonic functionin ", with the property that the nontangential maximal function of its gradient is in L2(#", d$)and u|!! 6 c on #". Using Green’s formula from Theorem 2.4 and the jump-relations for layerpotentials established in [10], we may then write

'

!|Au|2 dX =

'

!!u #&u d$ = c

'

!!#&u d$ = c

'

!div (Au) dX = 0. (4.9)

Hence, u is constant in " which further forces ($12I + K%)1 = #&u = 0 on #". Thus, S1 6 c on

#" implies K%1 = 12 on #".

Next, if K%1 is constant on #" then so is ($12I + K%)1. On the other hand, the integral of

($12I + K%)1 = #&S 1 on #" should vanish, by the Divergence Theorem (in the form presented

in Theorem 2.4), so necessarily K%1 6 12 on #". Going further, if K%1 6 1

2 on #" then u := S 1satisfies #&u = ($1

2I + K%)1 = 0 on #", so that u is constant in " by the first equality in(4.9). Finally,

9!! Kf d$ = 0 for every f % L2

0(#", d$) if and only if9!! fK%1 d$ = 0 for every

f % L20(#", d$) if and only if K%1 is a constant. !

Moving on, for a Borel measure µ in Rn define the (possibly infinite) potential

Uµ(X) := $'

RnE(X $ Y ) dµ(Y ), X % Rn. (4.10)

Fix now a bounded Borel set B in Rn (no extra regularity assumed), and denote by P(B) the setof all Borel probability measures on B. Notice that when B = #", for some bounded open set "in Rn, and µ % P(#"), then Uµ can be identified (up to a sign) with the single layer potential

S µ(X) :='

!!E(X $ Y ) dµ(Y ), X % Rn. (4.11)

For further reference, let us also recall here that the capacity of a bounded Borel set B is definedas

Cap (B) := sup.

µ(B) : µ % P(B), Uµ 0 1 on B/

. (4.12)

23

As is well-known, Cap (B) = 0 implies that B has zero (n-dimensional) Lebesgue measure. Also,the capacitary function is subadditive and nondecreasing, i.e.

Cap (B1) 0 Cap (B2) if B1 8 B2, Cap (B1 BB2) 0 Cap (B1) + Cap (B2). (4.13)

For later use, let us also point out that

Cap (B(X, r)) = cnrn!2, n 3 3, X % Rn, r > 0. (4.14)

Next, with B as above, define the potential energy of µ % P(B) as

E(µ) := $'

B

'

BE(X $ Y ) dµ(X)dµ(Y ), (4.15)

where E is as in (2.13). According to Theorem 6.3 in [33], a minimizer µ of E , called equilibriumdistribution, always exists. Specifically, we have:

Proposition 4.6. If B is compact, there exists a probability measure µ % P(B) that minimizes theenergy. That is, µ solves the minimal-energy problem E(µ) = 0, where

0 := min.E(µ) : µ % P(B)

/. (4.16)

We shall also need the following uniqueness result (cf. Theorem 9.1. in [33]):

Proposition 4.7. Let µ' be a probability measure on a compact set B such that Uµ$ = 0', constant,µ'-a.e. on B and Uµ$ 0 0' everywhere in Rn. Then

µ = µ'. (4.17)

The equilibrium distribution µ of a compact set B has a number of distinguished properties.First,

µ is carried by #B, (4.18)

(in the sense that µ(B \ #B) = 0). This implies that 0 can, in fact, be obtained by minimizing

Eb(µ) := $'

!B

'

!BE(X $ Y ) dµ(X)dµ(Y ), (4.19)

over all Borel probability measures µ on #B. Second, there exists an exceptional set <B 1 B suchthat

Uµ(X) 0 0, 2X % Rn, (4.20)

Uµ(X) = 0, 2X % B\ <B, (4.21)

where 0 is as in (4.16). To describe the properties of the exceptional set <B 1 B, recall the notionof capacity introduced earlier. We have:

<B 8 #B and Cap ( <B) = 0, (4.22)

and

if B 8 Rn, n 3 3, is compact and X0 % B is such that CC,R > 0

with Cap ({X % B : |X $X0| 3 r} 3 Crn!2, 2 r % (0, R)

5=/ X0 % B\ <B. (4.23)

See Theorem 7.1 and Theorem 10.1 in [33].

24

Proposition 4.8. Let B := ", where " 8 Rn, n 3 3, is a bounded, connected open set whichsatisfies an interior corkscrew condition. Then Uµ(X) = 0 for every X % B.

Proof. We claim that <B is empty. Indeed, if X0 % <B, then X0 % #" by (4.22) in which casethe interior corkscrew condition implies that there exist X% % " and R > 0, M > 1 such thatB(X%, r/M) 8 {X % B : |X $X0| 3 r} for every r % (0, R). In concert with (4.13)-(4.14) thisthen implies Cap ({X % B : |X $X0| 3 r} 3 Crn!2 for every r % (0, R) which, in turn, forcesX0 % B\ <B by (4.23). This contradicts the fact that X0 % <B and shows that <B is empty. With thisin had, the desired conclusion follows from (4.21). !

We are now in a position to state and prove the potential theoretic result which is most signif-icant for the goals we have in mind.

Theorem 4.9. Let " 8 Rn, n 3 3, be a bounded, connected open set satisfying a two-sidedcorkscrew condition and whose boundary is Ahlfors regular. Then the equilibrium distribution on "is constant, in the sense that

µ = /Hn!17 #" for some / > 0, (4.24)

if and only of S 1 is constant on ".

Proof. Set $ := Hn!17 #". In one direction, if dµ = / d$ for some constant / > 0, then we maywrite / (S 1)(X) = Uµ(X) = 0 for every X % ", by (4.11) and Proposition 4.8. This shows thatS 1 is constant in ". In the opposite direction, if S 1 is constant on ", take / := 1/$(#") > 0 sothat / $ % P(#"). For the type of domains specified in the statement of the theorem, it has beenproved in [10] that the operator

S : Lp(#", d$) $" W 1,pn/(n!1)loc (Rn), (4.25)

is well-defined and bounded for every p % (1,&), where W 1,qloc (Rn), 1 < q < &, denotes the local

version of the usual scale of Lp-based Sobolev spaces of order one in Rn. From this and standardembedding results it follows that

S 1 is nonnegative and continuous in Rn, harmonic in Rn \ #"

and, given that n 3 3, decays at infinity.(4.26)

Thus, the maximum principle applies and gives that S/ 0 0 in Rn. With the help of Proposition 4.7we may then conclude that µ = /$, finishing the proof. !

Recall the boundary Hardy spaces H 2± (#") and note that 1 % H 2

+ (#"), where 1 denotes theconstant function on #". Below we show that whether the equilibrium distribution of " is aconstant hinges on whether H 2

! (#") is contained in L20(#", d$)' C"n.

Proposition 4.10. Let " 8 Rn be a bounded, connected, two-sided NTA domains with an Ahlforsregular boundary. Then

1 %0H 2! (#")

1../ the equilibrium distribution of " is a constant, (4.27)

in the sense of (4.24).

25

Proof. If 1=H 2! (#") = Im

"$1

2I + C : L2(#", d$) ' C"n

#, it follows that ($1

2I + C %)1 = 0 and,

further, (12I + C )+ = 0 by (2.21). Next, recall (2.37) and set u := C+ % H 2("). It follows that

u|!! = (12I + C )+ = 0 hence u = C(u|!!) = 0 in ", by (2.41). In summary, the above reasoning

shows that DS 1 = C+ = 0 in ", where D is as in (2.11). As a consequence, |AS 1| = |DS 1| = 0 in" which proves that S 1 is constant in ". At this stage, the fact that the equilibrium distributionof " is a constant is a direct consequence of Theorem 4.9. This justifies the left-to-right implicationin (4.27). In fact, the converse implication is also valid since all steps above are reversible. !

Let " be a bounded, open subset of Rn. The following conjecture has been formulated byP.Gruber:

the equilibrium distribution of " is a constant

(in the sense of (4.24)) if and only if " is a ball.(4.28)

Let us highlight a connection between Conjecture 4.3 and Gruber’s conjecture (4.28). Specifically,we have:

Proposition 4.11. In the class of bounded, connected open sets in Rn, n 3 3, satisfying a two-sidedcorkscrew condition and whose boundaries are Ahlfors regular,

Gruber’s conjecture =/ Conjecture 4.3. (4.29)

Proof. As already noted earlier, it is clear that if " is a ball then K = K%. In view of Theorem 4.9it su#ces to show that if " 1 Rn is as in the statement of the proposition and has the propertythat K = K%, then S 1 = constant in ". This, however, follows from Lemma 4.5 and the fact thatK1 = 1

2 . !Work of L.E. Payne and G.A.Philippin in [28], [27] shows that Gruber’s conjecture has a positive

answer in the class of starlike C2,%-domains in Rn, n 3 3. Subsequent work by W.Reichel in [29],[30], has shown that Gruber’s conjecture holds in the class of C2,%-domains. On the other hand,E.Martensen has shown in [24] that this is also the case for piecewise smooth domains in R2.Finally, in [25], O.Mendez and W.Reichel have shown that Gruber’s conjecture is valid both in theclass of convex domains in Rn, n 3 3, as well as the class of two-dimensional Lipschitz domains.We shall further strengthen the latter result in Corollary 4.14 in § 4.2. For future purposes, herewe only wish to record the higher-dimensional version of the main result in [25] as follows:

Theorem 4.12. Let " 1 Rn, n 3 3, be a bounded convex domain with the property that

S1 = constant on #". (4.30)

Then " is a ball.

4.2 The two dimensional case

Here we continue the discussion initiated in the previous subsection, by focusing on the two di-mensional setting. The main theorem in this subsection is contained in the following perturbationresult, involving the classical Cauchy operator (2.31).

Theorem 4.13. Assume that " 1 R2 is a bounded, connected UR domain satisfying #" = #".Then for every ' > 0 there exists ) > 0, which depends on ' and the geometrical characteristics

26

of " (more specifically, the diameter and the constants implicit in the definition of UR domains),with the property that

+C %1$ 12+L2(!!,d#)&C < ) =/ inf

D disk+*! $ *D+L2(R2) < '. (4.31)

Furthermore, in the class of domains as in the first part of the statement, the following holds:

K%1 = constant on #" ./ " is a disk. (4.32)

Proof. Inspection of (4.31) reveals that

C = K + iQ (4.33)

where, with 1 := $i+ denoting the unit tangent along #",

Qf(z) := lim%*0+

12&

'

"#!!

|z!'|>%

), $ z, 1(,)*|, $ z|2 f(,) d$(,), z % #". (4.34)

Let us also note that

Q% = #+S, (4.35)

with S as in (4.3), and #+ the directional derivative along 1 . As a consequence,

C % = K% + i#+S (4.36)

and

+($12I + K%)1+2L2(!!,d#) + +#+S1+2L2(!!,d#) = +C %1$ 1

2+2L2(!!,d#)&C. (4.37)

Next, a direct calculation shows that for every harmonic function u in R2 \ " there holds

div"

12X|Au(X)|2 $ )X,Au(X)*Au(X)

#= 0, X % R2 \ ". (4.38)

Hence, assuming that N (Au) % L2(#", d$), Theorem 2.4 and (2.18) give

0 ='

!(BR\!)

"12)X, +(X)*|Au(X)|2 $ )X,Au(X)*#&u(X)

#d$(X)

='

!BR

"12R|Au(X)|2 $R!1|)X,Au(X)*|2

#d$(X)

$'

!!

"12)X, +(X)*|Au(X)|2 $ )X,Au(X)*#&u(X)

#d$(X), (4.39)

where BR is the ball of radius R (assumed to be su#ciently large) with center at the origin.Let us specialize the above to the case when u := S 1. From the definition of the single layer

we then have the following asymptotic expansion

(Au)(X) = 12*H

1(#")X/|X|2 + O(|X|!2) as |X|"&. (4.40)

27

Using (4.40) in (4.39) and letting R "& then yields'

!!

"12)X, +(X)*|Au(X)|2 $ )X,Au(X)*#&u(X)

#d$(X) = $ 1

4*

"H1(#")

#2. (4.41)

Next, consider the tangential derivative #+ along #". Two properties of this operator, estab-lished in [10], are going to be of importance for us here. First, #+w is well defined for any functionw with N (Aw) % L2(#", d$) and for which the pointwise nontangential traces (Aw)|!!, w|!! exist$-a.e. on #". Second, if w is as above then

(Aw)|!! = [1 · (Aw)|!!]1 + [+ · (Aw)|!!]+. (4.42)

Returning to the mainstream discussion, these considerations ensure that

Au = (#+u)1 + (#&u)+ and |Au|2 = |#+u|2 + |#&u|2 $-a.e. on #", (4.43)

+#+u+2L2(!!,d#) + +#&u$ 1+2L2(!!,d#) = +C %1$ 12+

2L2(!!,d#). (4.44)

In particular, assuming that +C %1$ 12+

2L2(!!,d#) 0 ), we obtain

+#+u+L2(!!,d#) + +#&u$ 1+L2(!!,d#) 0 C). (4.45)

To continue, for $-a.e. X % #" we write

12)X, +(X)*|Au(X)|2 $ )X,Au(X)*#&u(X) = $1

2)X, +(X)*+ 2(X) (4.46)

where the error term 2(X) is given by

2(X) := $12)X, +(X)*

0|#&u(X)|2 $ 1

1$ )X, 1(X)*#+u(X)#&u(X).

Note that, by (4.45)'

!!|2(X)| d$(X) 0 C), (4.47)

where C > 0 depends only on the geometrical characteristics of ". In conjunction with (4.41), thisimplies

14*

"H1(#")

#2= $

'

!!

"12)X, +(X)*|Au(X)|2 $ )X,Au(X)*#&u(X)

#d$(X)

= 12

'

!!)X, +(X)* d$(X) + O()) = H2(") + O()), (4.48)

where the last equality follows from (2.42) with .v(X) := X. In turn, this implies"H1(#")

#2$ 4&H2(") 0 C)H2("). (4.49)

We can now appeal to the following isoperimetric stability result, given as Theorem 1.1 in [23].Given (4.49), there exists a disk D% such that

+*! $ *D#+L2(R2) < C)1/4, (4.50)

28

where C depends only on the geometrical characteristics of ". From this, (4.31) readily follows,completing the proof of the first part of the theorem.

As for (4.32), first note that if K%1 is a constant function in #", then Lemma 4.5 gives thatK%1 = 1

2 and S 1 is constant in ". As shown in [10], the latter condition forces #+S1 = 0 on #".In concert with (4.36), this gives that C %1 = 1

2 on #". Having shown this, (4.31) then gives that" coincides a.e. with a disk, proving (4.32). Since we are also assuming that #" = #", elementarytopological considerations then show that " must actually be a disk. !

We conclude this section with a few corollaries of Theorem 4.13 (and its proof), starting with:

Corollary 4.14. Assume that " 1 R2 is a connected, bounded UR domain satisfying #" = #".Then

S 1 = constant in " ./ S1 = constant on #" ./ K = K%

./ K%1 = constant on #" ./ K%1 = 12 on #" ./ " is a disk. (4.51)

In particular, Conjecture 4.2 and Conjecture 4.3 are both valid in the class of connected, boundedUR domains " 1 R2 for which #" = #".

Proof. If " is a disk, it is easy to see that K = K%, so in this case K%1 = K1 = 12 . Everything else

now follows from Theorem 4.13 and Lemma 4.5. !Next we show that the equivalence AS 1 = 0 in " D " is a disk is stable under L- perturba-tions.

Corollary 4.15. Suppose " 1 R2 is a bounded, connected UR domain with the property that#" = #". Then for every ' > 0 there exists ) > 0, which depends on ' and the geometricalcharacteristics of ", for which

+AS 1+L%(!) < ) =/ infD disk

+*! $ *D+L2(R2) < '. (4.52)

Proof. Note that (4.52) entails +($12I + K%)1+L%(!!,d#) = +#&S 1+L%(!!,d#) 0 +AS 1+L%(!) and

+#+S1+L%(!!,d#) = +)1,AS 1*+L%(!!,d#) 0 +AS 1+L%(!). In combination with (4.37) these implythat +C %1 $ 1

2+L2(!!,d#) is small if +AS 1+L%(!) is small, so the desired conclusion follows fromTheorem 4.13. !

Corollary 4.16. For every ' > 0 there exists ) > 0 for which

+C $ C %+L$L2(!!,d#)&C

% < ) =/(

" is a '-SKT domain and

infD disk +*! $ *D+L2(R2) < '.(4.53)

in the class of all connected, bounded domains " 1 R2 which satisfy a two-sided local John con-dition and have Ahlfors regular boundaries, with a certain fixed common bound on the geometricalconstants involved in describing these characteristics.

Proof. Since C 1 = 12 , +C $C %+

L$L2(!!,d#)&C

% < ) implies that +C %1$ 12+L2(!!,d#)&C < ). Granted

this, Theorem 2.2 and Theorem 2.3 give that " is a '-SKT domain for ' = '()) > 0 with '()) " 0as ) " 0+. Now, (4.53) follows from this and Theorem 4.13. !

29

4.3 Higher dimensional results

The nature of the results in this subsection is that certain analytical conditions (involving singularintegral operators) do in fact determine the shape of the underlying domain. Our first major resultin this regard is as follows.

Theorem 4.17. Let " 1 Rn be a UR domain which satisfies #" = #". Then

C = C % ./ " is either a ball, the complement of a (closed) ball, or a half-space. (4.54)

Proof. To get started, (2.20) gives

C = C % ./ +(X)# (X $ Y ) = $(X $ Y )# +(Y ) for $-a.e. X, Y % #". (4.55)

Now, if " = B(Z0, R), for each X,Y % #" we have +(X) = (X $ Z0)/R, +(Y ) = (Y $ Z0)/R, andsince

(X $ Z0)# (X $ Y ) = (X $ Z0)# [(X $ Z0)$ (Y $ Z0)]

= $R2 $ (X $ Z0)# (Y $ Z0) = $(X $ Y )# (Y $X0), (4.56)

we obtain +(X) # (X $ Y ) = $(X $ Y ) # +(Y ) for every X, Y % #". Hence, C = C %. Likewise,the last equality in (4.55) checks out if " is the complement of a (closed) ball, or a half-space (inwhich case #" is a (n$ 1)-plane). This proves the left-to-right implication in (4.54). The crux ofthe matter is, of course, the opposite implication, to which we now turn.

To this end, assume that the domain " is as in the statement of the theorem and, in addition,satisfies +(X)# (X $ Y ) = $(X $ Y )# +(Y ) for $-a.e. X, Y % #". From this and (2.4), it followsthat

+(X) =X $ Y

|X $ Y | # +(Y )# X $ Y

|X $ Y | , for $-a.e. X, Y % #". (4.57)

Fixing Y , this shows that, after eventually modifying + on a set of $-measure zero, + % C0(#").With this in hand and relying on the results proved in [11], we may deduce that

" is a strongly C1-domain. (4.58)

Next, we record a general identity to the e!ect that

a# b# a = |a|2b$ 2)a, b*a, 2 a, b % Rn. (4.59)

Indeed, for every a, b % Rn formula (2.8) allows us to write a# b = $b# a$ 2)a, b* which furtherimplies that a # b # a = $b # a2 $ )a, b*a = |a|2b $ 2)a, b*a, proving (4.59). When used in thecontext of (4.57), formula (4.59) yields

+(X) = +(Y )$ 2: X $ Y

|X $ Y | , +(Y ); X $ Y

|X $ Y | , (4.60)

for all X,Y % #".To continue, after a translation and a rotation, it can be assumed that

0 % #" and +(0) = $en = (0, ..., 0,$1) % Rn. (4.61)

30

In this setting, let

3 : {x' % Rn!1 : |x'| < r} $" R, 3(0') = 0, A'3(0') = 0' (4.62)

(where A' denotes the gradient with respect to x' % Rn!1) be a C1 function with whose graphcoincides with #" in a neighborhood of 0 % Rn. More precisely, there exists an open cylinderC := {(x', xn) : |x'| < r, |xn| < a} such that

C 4 " = C 4 {(x', xn) : xn > 3(x')}, (4.63)

C 4 #" = C 4 {(x', xn) : xn = 3(x')}. (4.64)

Hence,

+(x',3(x')) =(A'3(x'),$1)=1 + |A'3(x')|2

, if x' is near 0' % Rn!1. (4.65)

By making Y = 0 in (4.60) and using the fact that +(0) = $en we arrive at

+(X) = $en + 2xn

|X|2 X, for X = (x', xn) % #" near 0 % Rn. (4.66)

In particular, for X = (x1, ..., xn) % #" near 0 % Rn,

+j(X) = 2xjxn

|X|2 for 1 0 j 0 n$ 1, and +n(X) = $1 + 2x2

n

|X|2 , (4.67)

so that further, on account of (4.65) and (4.66),

#j3(x') = $ +j(X)+n(X)

=2xjxn

|X|2 $ 2x2n

=2xj3(x')

|x'|2 $ 32(x'), 1 0 j 0 n$ 1, (4.68)

for each x' near 0' % Rn!1. Thus, in a neighborhood of 0' % Rn!1,

#j3(x') =2xj3(x')

|x'|2 $ 32(x'), 1 0 j 0 n$ 1. (4.69)

Consider now the expression

3(x')|x'|2 + 32(x')

for x' @= 0. (4.70)

Note that, for j % {1, . . . , n $ 1}, the partial derivative of (4.70) with respect to xj is a fractionwith the numerator

#j3(x')(|x'|2 + 32(x'))$ 3(x')(2xj + 23(x')#j3(x')). (4.71)

This, however, vanishes identically, thanks to (4.69). Hence, the expression in (4.70) is constantfor x' near 0'. Denoting the value of this constant by ± 1

2R , for some 0 < R 0 &, and solving theensuing equation in the unknown 3(x'), we obtain the following possible expressions for 3(x') forx' near 0' % Rn!1:

3+(x') = R$=|x'|2 + R2, 3!(x') = $R +

=|x'|2 + R2, 3o(x') 6 0. (4.72)

31

The solutions 3±(x') = ±R ;=|x'|2 + R2 correspond, respectively, to the situation when the

boundary of " agrees with that of B(±Ren, R), near 0 % Rn. On the other hand, the solution3o(x') 6 0 corresponds to the case when #" is flat (i.e., is contained in a (n $ 1)-plane) near0 % #".

In summary, the argument above shows that near each boundary point, #" is either flat, oragrees with a sphere. Thus, #" = As BAf , where As is the subset of #" consisting of points nearwhich #" agrees with some sphere, and Af is the subset of #" consisting of points near which #"is flat. We continue the proof, under the hypothesis that As is nonempty. To fix ideas, assume thatX% % #" is such that there exists a ball B 1 Rn with the property that #" and #B coincide nearX%. In this case, consider % := #" 4 #B which is therefore a closed subset of #B, with nonemptyinterior %/ relative to #B (viewed as a topological space, with the structure inherited from Rn).The claim we make at this stage is that #(%/), considered in #B, is empty. Indeed, if X % #(%/)then X % % and X /% %/. In particular, X % #" and we can run the same argument as above (thistime, for X % #" in place of 0 % #") in order to conclude that there exist an open neighborhoodU of X along with either an Euclidean ball B', or a (n$ 1)-plane &, such that U 4 B' = U 4 #",or U 4 & = U 4 #". Hence, W := (U 4 #B) 4 %/ is a nonempty open set (relative to the topologyof #B) with the property that either W 8 U 4 #" = U 4 #B', or W 8 U 4 #" = U 4 &. As aconsequence, either #B 4 &, or #B 4 #B', has nonempty interior in #B. This rules out the formereventuality and, further, forces B = B'. In turn, this implies that U 4 #B = U 4 #" which showsthat X % U 4 #B 8 %/ and, further, that X % %/, which is a contradiction. Thus, %/ is an open,nonempty, boundaryless subset of the connected topological space #B. We can therefore concludethat % = #B, i.e. #B 8 #".

As a consequence of the reasoning above, #" is a union of spheres if As is nonempty. Since,nonetheless, the equality +(X)# (X $ Y ) = $(X $ Y )# +(Y ) may fail if X,Y belong to di!erentspheres, we may ultimately conclude that #" is a sphere itself, say #" = #B, for some ball B 1 Rn.That actually " = B, or " = Rn \B then follows from (4.63) and the following general topologicalresult

Let O1, O2 be two open subsets of Rn with the property that #O1 = #O2 @= E. Then

2x % #O1 C r > 0 such that B(x, r) 4O1 = B(x, r) 4O2 / O1 = O2. (4.73)

See [11] for a proof. Hence, " is a ball, or the complement of a closed ball if As is nonempty. Avery similar reasoning shows that " is a half-space when Af is nonempty, thus finishing the proofof the theorem. !

To state our next result, recall the Riesz transforms Rk, 1 0 k 0 n, from (2.23). In the case when" 1 Rn is a half-space (so that #" is a (n$ 1)-plane in Rn), it is well-known that

!nk=1 R2

k = $Iand RjRk = RkRj for all j, k % {1, ..., n}. It turns out that the same identities are also valid when" is a ball in Rn. Somewhat surprisingly, the validity of the aforementioned identities actuallydetermines the shape of the domain in question. Concretely, we have the following.

Theorem 4.18. Assume that " 1 Rn is a two-sided NTA domain with an Ahlfors regular boundary.Then

" is a ball, the complement of a (closed) ball, or a half-space

./n&

k=1

R2k = $I and RjRk = RkRj 2 j, k % {1, ..., n}. (4.74)

32

Proof. From (2.24) it follows that

C M& = $12

n&

k=1

RkMek . (4.75)

Squaring both sides of this identity then yields

C M&C M& = 14

n&

j,k=1

RjRkMejMek .

= $14

n&

k=1

R2k + 1

4

&

1)j 0=k)n

RjRkMejMek

= $14

n&

k=1

R2k + 1

8

&

1)j<k)n

(RjRk $RkRj)Mej1ek . (4.76)

Thus, based on (4.76) and (2.21), we may conclude that

C C % = $14

n&

k=1

R2k + 1

8

&

1)j<k)n

(RjRk $RkRj)Mej1ek . (4.77)

In turn, this implies

C C % = 14I ./

n&

k=1

R2k = $I and RjRk = RkRj 2 j, k % {1, ..., n}. (4.78)

On the other hand, (2.45) and Proposition 2.5 give that

C C % = 14I ./ C C % = C 2 ./ C = C %. (4.79)

At this point, the equivalence in (4.74) becomes a consequence of (4.78)-(4.79) and Theorem 4.17.The proof is therefore complete. !

The aim is now to show that the Hardy spaces H 2± (#") are orthogonal (if and) only if " is a

ball, its complement, or a half-space.

Theorem 4.19. Let " 1 Rn be a two-sided NTA domain with an Ahlfors regular boundary. Then

<)"H 2

+ (#") , H 2! (#")

#=

&

2./ #" is a sphere, or a (n$ 1)-plane. (4.80)

In other words,

L2(#", d$)' C"n = H 2+ (#")(H 2

! (#") orthogonal sum ./

" is a ball, the complement of a (closed) ball, or a half-space. (4.81)

Proof. If " 1 Rn is as in the statement of the theorem then necessarily " has finite perimeter andsatisfies #" = #". Then the desired results follow from Corollary 2.10 and Theorem 4.17. !

Recall that a linear bounded operator on a Hilbert space is called normal if it commutes withits adjoint. While the class of normal operators is strictly larger than that of self-adjoint operators,below we show that in (4.54) of Theorem 4.17 the demand that C is self-adjoint can be relaxed toasking that C is normal.

33

Corollary 4.20. For a two-sided NTA domain " 1 Rn with an Ahlfors regular boundary, thefollowing conditions are equivalent:

(i) C % L$L2(#", d$)' C"n

%is a normal operator, i.e.

C C % = C %C . (4.82)

(ii) #" is a sphere, or a (n$ 1)-plane.

Proof. This is a consequence of (4.80) and (2.82). Parenthetically, we note that the more deli-cate implication (i) / (ii) can also be justified by observing that if C is a normal operator onL2(#", d$)' C"n then, as is well-known,

+C +2L$L2(!!,d#)&C$n

% = +C 2+L$L2(!!,d#)&C$n

%. (4.83)

On account of (2.45) and (2.76), we obtain from this cot2 (%/2) = 1 where % % (0,&/2] is as in(2.74). This forces % = &/2 and the desired conclusion now follows from (4.80). !

In concert, Corollary 2.10 and Theorem 4.17 also give the following result.

Theorem 4.21. Assume " 1 Rn is a two-sided NTA domain with an Ahlfors regular boundary.Then the following conditions are equivalent:

+C +L$L2(!!,d#)&C$n

% = 12 , +C + C %+L

$L2(!!,d#)&C$n

% = 1,

+($12I + C )+L

$L2(!!,d#)&C$n

% = 1, +(12I + C )+L

$L2(!!,d#)&C$n

% = 1,

P+ + P! = I, +(P+ $P!)!1+L$L2(!!,d#)&C$n

% = 1,

#" is a sphere, or a (n$ 1)-plane.

(4.84)

From Theorem 2.12 and Theorem 4.19 we also have:

Theorem 4.22. Let " 1 Rn be a two-sided NTA domain whose boundary is Ahlfors regular. Set"+ := ", "! := Rn \ ", and $ := Hn!17 #". Then

#" is a sphere, or a (n$ 1)-plane

./'

!!

(Y $X+)# (Y $X!)|Y $X+|n|Y $X!|n

d$(Y ) = 0 2X± % "±. (4.85)

Moving on, we note that it is trivial to check that if " 1 Rn is a ball then K = K%. Results inthe converse direction when n = 2 are in Corollary 4.14. Below we identify a geometrical frameworkwithin which the claim in Conjecture 4.3 holds in the higher dimensional setting.

Theorem 4.23. Assume that " 1 Rn, n 3 3, is a UR domain satisfying #" = #" and for whichK = K%. Then " is a ball.

Proof. At the level of kernels, the operator identity K = K% tells us that $-a.e. X % #" theidentity

)X $ Y, +(Y )* = $)X $ Y, +(X)* (4.86)

holds for $-a.e. Y % #". Thus, for $-a.e. Y % #" the identity (4.86) holds for $-a.e. X % #".

34

The first order of business is to show that (4.86) implies that there exists a subset L 1 #" suchthat $(#" \ L) = 0 and for which " is a strongly Lipschitz domain near every point in L. To seethis fix X0 % #%", the reduced boundary of ", with the property that there exists E0 8 #" with$(E0) = 0 and such that

)X0 $ Y, +(Y )* = $)X0 $ Y, +(X0)*, 2Y % #" \ E0. (4.87)

Let us alo denote by E1 a subset of #" of zero $-measure such that for each Y % #" \ E1 formula(4.86) holds for $-a.e. X % #". If )X0$Y, +(X0)* = 0 for every Y % #" \ (E0 BE1), it follows thatthere exists a (n$ 1)-plane H0 in Rn such that

#" \ (E0 B E1) 8 H0. (4.88)

On the other hand, we claim that the inclusion

#" \ (E0 B E1) !" #" (4.89)

has dense range. Indeed, the failure of this claim would imply the existence of some point X# % #"and number R > 0 such that $(#" 4 B(X#, R)) = 0. However, for any open set of locally finiteperimeter, it has been established in [11] that

#" = #" =/ $(B(X, r) 4 #") > 0 2X % #", 2 r > 0. (4.90)

This contradiction proves the that the inclusion in (4.89) has dense range. From this and (4.88),we may then conclude that #" 8 H0. With this in hand, and given that #" = #", it is not toodi#cult to show that #" is a strongly Lipschitz domain near X0.

Next, we deal with the remaining case, i.e., when there exists Y0 % #" \ (E0 BE1) such that

)X0 $ Y0, +(X0)* @= 0. (4.91)

In particular, we also have

)X0 $ Y0, +(Y0)* @= 0, (4.92)

by (4.87) and (4.89). By continuity, it follows that )X $ Y0, +(Y0)* @= 0 for every X % #" near X0.In turn, since Y0 /% E1, this and (4.86) entail )X $ Y0, +(X)* @= 0 for $-a.e. X % #" near X0. Inother words, the continuous vector field X :" X $ Y0 is transversal to #" near X0. According to acriterion proved in [11], " is a strongly Lipschitz domain near X0.

Moving on, with ")(a, b) denoting the angle between the vectors a, b % Rn, formula (4.86) gives

")"+(X), Y!X

|Y!X|

#= ")

"+(Y ), X!Y

|X!Y |

#(4.93)

for $ ' $-a.e. (X,Y ) % #"> #". We now claim that an open set " 1 Rn which is of locally finiteperimeter and satisfies (4.86) is necessarily convex. Our strategy for proving this is to show that" satisfies the following geometric property:

X, Y % #", I(X, Y ) 4 " @= E =/ I(X,Y ) 8 ", (4.94)

where I(X, Y ) denotes the open line segment with endpoints X and Y . At the end of the currentproof, we will show that if " is a bounded, open set with #" = #", and such that (4.94) holds,then is in fact " is convex. For the time being, the goal is to check that (4.94) holds.

35

To facilitate the understanding of subsequent considerations, let us first explain the main ideain the argument in the case when " is a smooth domain. In this scenario, if X, Y % #" are suchthat I(X,Y ) 4 " @= E and yet I(X, Y ) 8 ", then there exists a point Z % I(X, Y ) 4 #". If wedenote by % the angle between +(X) and Y $X, then (4.94) tells us that the angle between +(Y )and X $ Y is also %. Employing (4.94) two more times, we see that, in the one hand, the anglebetween +(Z) and X $ Z is % and, on the other hand, the angle between +(Z) and Y $ Z is thesame as the angle between +(Y ) and X $ Y , i.e. once again %. Altogether, % satisfies % = & $ %,i.e., % = &/2. Cf. picture below:

Ωθ

θ

Y

Z

X

θ

θ

This, however, entails that +(X) is perpendicular to X$Y for each Y % #", an impossibility giventhat " is bounded.

We now turn to the actual task of checking that (4.94) holds in the more general geometricalcontext described in the statement of the theorem. By continuity, it su#ces to actually prove theclaim in (4.94) when X % % for some dense subset % of #", and when Y % %(X) where %(X) is adense subsets of #", possibly depending on X.

Concretely, let us first take % := {X % #" : (4.93) holds for (this X and) $-a.e. Y % #"}, thenfix a point X % %. Denote by F 8 #" a set of $-measure zero with the property that (4.93) holdsfor (the chosen X and) every Y % #"\F . Then recall the set L of points near which " is a stronglyLipschitz domain (introduced earlier in the proof), define %(X) := L \ F and fix Y % %(X). Ourgoal is to show that

I(X, Y ) 4 " @= E =/ I(X, Y ) 8 ". (4.95)

Seeking a contradiction, assume that I(X,Y ) 4 " @= E and yet I(X, Y ) @8 ". Then a simpleargument based on the connectivity of the segment I(X, Y ) shows that, necessarily, there existsa point Z % I(X,Y ) 4 #". By hypothesis, there exists a sequence of points {Zj}j"N with theproperty that (4.93) holds both for the pair X, Zj as well as the pair Y, Zj (in place of X, Y ), andsuch that Zj " Z as j "&.

To continue, set % := ")"+(X), Y!X

|Y!X|

#. From (4.93), used three times: first for X, Y , then for

Y, Zj and, finally, for X, Zj , it then follows that

% = ")"+(X), Y!X

|Y!X|

#= ")

"+(Y ), X!Y

|X!Y |

#= ")

"+(Y ), Z!Y

|Z!Y |

#

= limj*-

")"+(Y ), Zj!Y

|Zj!Y |

#= lim

j*-")"+(Zj),

Y!Zj

|Y!Zj |

#

= & $ limj*-

")"+(Zj),

X!Zj

|X!Zj |

#= & $ lim

j*-")"+(X), Zj!X

|Zj!X|

#

= & $ ")"+(X), Z!X

|Z!X|

#= & $ ")

"+(X), Y!X

|Y!X|

#

= & $ %. (4.96)

36

Above, we have used the fact that

limj*-

0")"+(Zj),

Y!Zj

|Y!Zj |

#+ ")

"+(Zj),

X!Zj

|X!Zj |

#1= &. (4.97)

To see this, by eventually passing to a subsequence we can assume that +(Zj) converges to some( % Sn!1. Then

limj*-

0")"+(Zj),

Y!Zj

|Y!Zj |

#+ ")

"+(Zj),

X!Zj

|X!Zj |

#1

= limj*-

0arccos

"&(Zj)·(Y!Zj)

|Y!Zj |

#+ arccos

"&(Zj)·(X!Zj)

|X!Zj |

#1

= arccos"

"·(Y!Z)|Y!Z|

#+ arccos

""·(X!Z)|X!Z|

#

= &, (4.98)

since the sums of the angles made by ( with X $ Z and X $ Z, respectively, is &.The above calculation forces % = & $ %, hence ultimately % = &/2. In other words,

+(X) is perpendicular on I(X, Y ). (4.99)

The idea now is that since #" looks like a Lipschitz domain near Y (recall that Y % L), twoscenarios are possible. Namely, either

CYk % %(X) near Y , 1 0 k 0 n, with {!*

XYk : 1 0 k 0 n} linearly independent

and such that I(X, Yk) 4 " @= E as well as I(X,Yk) 4 #" @= E for every k,(4.100)

or

I(X, Y#) 4 #" = E 2Y# % %(X) near Y , with Y# @= Y . (4.101)

When (4.100) happens, we run the same argument as above with Yk in place of Y and, much as in(4.99), arrive at the conclusion that +(X) is perpendicular on I(X, Yk) as well. Since the vectors!*

XYk, 1 0 k 0 n} are linearly independent this is, however, impossible.In the second case, i.e., when (4.101) happens, we may conclude that there exists a (one-

component, open, circular) infinite cone ' with vertex at X and symmetry axis along!*XY such

that if 'o := ' 4 " then

I(X, Y#) 1 'o 2Y# % %(X) near Y , with Y# @= Y . (4.102)

Given that, as a topological space, #" is simply connected near Y , by density, we then obtain'o 8 ", hence

'o 8 (")/. (4.103)

Note that

#" = #" = " \ (")/ =/ #" 4 (")/ = E. (4.104)

In concert, (4.103)-(4.104) can be used to conclude that 'o 4 #" = E. However, recall that #" 5Z % I(X, Y ) 1 ' 4 " = 'o, which contradicts the conclusion just derived. This shows that the

37

implication (4.95) holds and, further, that (4.94) holds. As claimed there, this proves that " is abounded convex domain (hence, in particular, strongly Lipschitz).

The next step in the proof is to observe that K = K% also forces K%1 = K1 = 12 , i.e. that

($12I + K%)1 = 0. From Lemma 4.5 we may then deduce that S 1 is constant in " (since a

convex domain is connected). Ultimately, by taking boundary traces, we arrive at the conclusionthat S1 = constant on #". At this stage, given that we have already proved that " is convex,Theorem 4.12 applies and gives that " is a ball.

There remains to prove that if " is a bounded, open set with #" = #", and such that (4.94)holds, then " is convex. To see this, fix two arbitrary points Z1, Z2 % " and denote by I(Z1,$&),I(Z1, Z2), I(Z2, +&), respectively, the three components in which the points Z1, Z2 divide the(infinite) line L passing through them. Since " is bounded, a simple connectivity argument showsthat the intersection of both I(Z1,$&) and I(Z2,+&) with #" is a nonempty (closed) subset ofL. In particular, there exist X, Y % L 4 #" with

I(X, Z1), I(Y, Z2) 1 ". (4.105)

In particular, I(X,Y )4" @= E. Thus, by (4.94), I(X, Y )4" = "B#". Hence, as soon as we provethat

I(X,Y ) 4 #" = E (4.106)

we may conclude that I(Z1, Z2) 1 I(X, Y ) 1 ", which proves that " is convex.As regards (4.106), we reason by contradiction and assume that there exists Z % I(X, Y )4 #".

To continue, let r > 0 be su#ciently small so that B(Z1, r)B(Z2, r) 8 ". By taking arbitraryZ#

1 % B(Z1, r), Z#2 % B(Z2, r) and running a similar argument for these points in place of Z1, Z2,

we arrive at the conclusion that there exists a circular, open cylinder C, of radius r, whose symmetryaxis is I(Z1, Z2) which is contained in ". In particular, B(Z, r) 8 " and, hence, B(Z, r) 8 (")/.From this and (4.104) we then conclude that Z % B(Z, r) 4 #" = E, a contradiction. This justifiesthe claim made about the role of the implication (4.94), and finishes the proof of the theorem. !

5 Appendix

A word of explanation regarding the current terminology is in order. What we here call SKT do-mains have been previously called in the literature chord arc domains. The latter notion originatedin the two dimensional setting, where the defining condition is that the length of a boundary arcbetween two points does not exceed a fixed multiple of the length of a chord between these points.In higher dimensions, where this phenomenon becomes somewhat more sophisticated, this notionoriginated in S. Semmes [31] and was further developed in [15]–[16]. In the higher dimensionalsetting, this “chord arc” designation no longer adequately captures the essential features of suchdomains, and in [10] we have proposed to call them SKT (Semmes-Kenig-Toro) domains. Likewise,we have relabeled what was previously called in these papers chord arc domains with vanishingconstant, calling them regular SKT domains.

For the convenience of the reader, here we summarize the definitions of Reifenberg flat, NTAand SKT domains, and recall the John-type condition from [10]. To get started, we remind thereader that for each A,B 1 Rn,

D[A,B] := max.

sup{dist (a,B) : a % A} , sup{dist (b, A) : b % B}/

(5.1)

is the Hausdor! distance between the sets A, B.

38

Definition 5.1. Let % 1 Rn be a compact set and let ' % (0, 142

2). We say that % is '-Reifenberg

flat if there exists R > 0 such that for every x % % and every r % (0, R] there exists a (n $ 1)-dimensional plane L(x, r) which contains x and such that

1r

D[% 4B(x, r) , L(x, r) 4B(x, r)] 0 ', (5.2)

where D[A,B] denotes the Hausdor" distance between the sets A and B.

Definition 5.2. We say that a bounded open set " 1 Rn has the separation property if thereexists R > 0 such that for every x % #" and r % (0, R] there exists an (n $ 1)-dimensional planeL(x, r) containing x and a choice of unit normal vector to L(x, r), .nx,r, satisfying

{y + t.nx,r % B(x, r) : y % L(x, r), t < $ r4} 1 ",

{y + t.nx,r % B(x, r) : y % L(x, r), t > r4} 1 Rn \ ".

(5.3)

Moreover, if " is unbounded, we also require that #" divides Rn into two distinct connected com-ponents and that Rn \ " has a non-empty interior.

Definition 5.3. Let " 1 Rn be a bounded open set and ' % (0, 'n). Call " a '-Reifenberg flatdomain if " has the separation property and #" is '-Reifenberg flat.

Definition 5.4. A (bounded) open set " 1 Rn is called an NTA domain (in the sense of Jerison andKenig) provided " satisfies a two-sided corkscrew condition, along with a Harnack chain condition.

We say that " 1 Rn satisfies the interior corkscrew condition if there are constants M > 1 andR > 0 such that for each x % #" and r % (0, R) there exists y = y(x, r) % ", called corkscrew pointrelative to x, such that |x $ y| < r and dist(y, #") > M!1r. Also, " 1 Rn satisfies the exteriorcorkscrew condition if "c := Rn \" satisfy the interior corkscrew condition. Finally, " satisfies thetwo sided corkscrew condition if it satisfies both the interior and exterior corkscrew conditions.

The Harnack chain condition is defined as follows (with reference to M and R as above). First,given x1, x2 % ", a Harnack chain from x1 to x2 in " is a sequence of balls B1, . . . , BK 1 " suchthat x1 % B1, x2 % BK and Bj 4Bj+1 @= E for 1 0 j 0 K $ 1, and such that each Bj has a radiusrj satisfying M!1rj < dist(Bj , #") < Mrj. The length of the chain is K. Then the Harnackchain condition on " is that if ) > 0 and x1, x2 % " 4 Br/4(z) for some z % #", r % (0, R),and if dist(xj , #") > ) and |x1 $ x2| < 2k), then there exists a Harnack chain B1, . . . , BK fromx1 to x2, of length K 0 Mk, having the further property that the diameter of each ball Bj is3 M!1 min

$dist(x1, #"), dist(x2, #")

%.

An open set " 1 Rn is said to be a two-sided NTA domain if both " and Rn \ " are NTAdomains.

Definition 5.5. Let ' % (0, 'n), where 'n is a fixed, su!ciently small number. A bounded set" 1 Rn of finite perimeter is said to be a '-SKT domain if " is a '-Reifenberg flat domain, #" isAhlfors regular and there exists r > 0 such that

supx"!!

>sup

"3"(x,r)

"'$

"|+ $ +"|2 d$

#1/2?

< ', (5.4)

with the supremum taken over all surface balls $ contained in $(x, r) := #" 4 B(x, r). Here, asbefore, + is the measure-theoretic outward unit normal to #" and +" :=

9$"+ d$.

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Definition 5.6. Call a bounded open set " 1 Rn a regular SKT domain if " is a '-SKT domainfor some ' % (0, 'n) and, in addition, + % VMO(#", d$). The last condition means that

lim supr*0+

>supx"!!

"'$

"(x,r)|+ $ +"(x,r)|2 d$

#1/2?

= 0. (5.5)

We also recall the following definition from [10].

Definition 5.7. Let " 1 Rn be an open set. This is said to satisfy a local John condition ifthere exist % % (0, 1) and R > 0 (required to be & if #" is unbounded), called the John constants of", with the following significance. For every p % #" and r % (0, R) one can find pr % B(p, r) 4 ",called John center relative to $(p, r) := B(p, r)4#", such that B(pr, %r) 1 " and with the propertythat for each x % $(Q, r) one can find a rectifiable path 0x : [0, 1] " ", whose length is 0 %!1r andsuch that

0x(0) = x, 0x(1) = pr, and dist (0x(t), #") > % |0x(t)$ x| 2 t % (0, 1]. (5.6)

Finally, " is said to satisfy a two-sided local John condition if both " and Rn \ " satisfy a localJohn condition.

Finally, following G. David and S. Semmes [6] we make the following.

Definition 5.8. Call % 1 Rn uniformly rectifiable provided it is Ahlfors regular and thefollowing holds. There exist ), M % (0,&) (called the UR constants of %) such that for each x % %,R > 0, there is a Lipschitz map 3 : Bn!1

R " Rn (where Bn!1R is a ball of radius R in Rn!1) with

Lipschitz constant 0 M , such that

Hn!1$% 4BR(x) 4 3(Bn!1

R )%3 )Rn!1. (5.7)

If % is compact, this is required only for R % (0, 1].

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Steve HofmannDepartment of MathematicsUniversity of MissouriColumbia, MO 65211, USAe-mail: hofmann@math.missouri.edu

Marius MitreaDepartment of MathematicsUniversity of MissouriColumbia, MO 65211, USAe-mail: marius@math.missouri.edu

Emilio Marmolejo-OleaInstituto de MatematicasUnidad CuernavacaUniversidad Nacional Autonoma de MexicoA.P. 273-3 ADMON 362251 Cuernavaca, Mor.Mexicoe-mail: emilio@matcuer.unam.mx

Salvador Perez-EstevaInstituto de MatematicasUnidad CuernavacaUniversidad Nacional Autonoma de MexicoA.P. 273-3 ADMON 362251 Cuernavaca, Mor.Mexicoe-mail: salvador@matcuer.unam.mx

Michael TaylorMathematics DepartmentUniversity of North CarolinaChapel Hill, NC 27599, USAe-mail: met@email.unc.edu

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