jan wiegerinck version march 1, 2011 · 9.4. green’s function 103 9.5. beurling-nevanlinna...

Advanced Function Theory

Jan Wiegerinck

version March 1, 2011

Faculty of Mathematics, University of Amsterdam, Science

Park 904 Amsterdam, The Netherlands

E-mail address: [email protected]

Contents

Preface vii

Chapter 1. Holomorphic functions 11.1. Holomorphic functions revisited 11.2. Approximation theory 31.3. Exercises 6

Chapter 2. Normal families 92.1. Introduction 92.2. Spaces of continuous functions 92.3. Normal families of holomorphic functions 112.4. Normal families of meromorphic functions. 112.5. Exercises. 14

Chapter 3. Conformal Mappings 173.1. Introduction 173.2. Fractional linear transformations revisited 173.3. The hyperbolic metric on B 193.4. The Riemann mapping theorem 203.5. Boundary behavior of the Riemann map 223.6. Exercises. 26

Chapter 4. Complex Dynamics 294.1. Iteration of holomorphic maps 294.2. Classification of fixed points 314.3. Julia and Fatou sets 354.4. Polynomials of degree 2 364.5. Exercises 37

Chapter 5. Product representations 395.1. Products 395.2. Products of functions 415.3. Weierstrass theory 435.4. Mittag-Leffler’s principal part representation 465.5. Exercises 47

Chapter 6. Growth and zeros of entire functions of finite order 496.1. Introduction 496.2. The growth scale 496.3. The Borel-Caratheodory inequality 506.4. Blaschke factors, Jensen’s formula, and counting functions 516.5. Canonical products 54

iii

iv CONTENTS

6.6. Hadamard’s product theorem 566.7. Exercises 58

Chapter 7. Some special functions 617.1. Elementary properties of the Gamma-function 617.2. Product representations for 1/Γ 637.3. The Hankel integral for the Gamma-function 647.4. Asymptotics of Γ 657.5. The zeta-function 677.6. Hankel representation and functional equation for the

zeta-function 697.7. Zeros of the Riemann-zeta function 717.8. Elliptic functions 727.9. Exercises 75

Chapter 8. Elements of potential theory 818.1. Harmonic functions 818.2. The Poisson integral 828.3. The Dirichlet problem on the disc 848.4. Subharmonic functions and maximum principles 868.5. Harmonic measure 908.6. Phragmen-Lindelof Theorems 928.7. Exercises 94

Chapter 9. The Dirichlet problem 979.1. Preliminaries 979.2. Perron’s method 979.3. Boundary behavior of the Perron function 1009.4. Green’s function 1039.5. Beurling-Nevanlinna Theorem 1059.6. Green’s function and dynamics 1059.7. Exercises 105

Chapter 10. Some real analysis 10710.1. The maximal function 10710.2. Lebesgue’s differentiation theorem 11010.3. Boundary behavior of Poisson integrals 11310.4. Positive and h1 harmonic functions on the disc 11510.5. Exercises 117

Chapter 11. Hardy spaces 11911.1. The definition of Hp(B) 11911.2. Two words on Fourier series 12011.3. The Hardy space H2(B) 12011.4. Blaschke products 12211.5. The Hardy space H1(B) 12411.6. Additional results 12611.7. Exercises 126

Chapter 12. Applications of Algebraic Topology 12712.1. Results from Algebraic Topology 127

CONTENTS v

12.2. Weierstrass theory of analytic continuation, Riemann surfaces12812.3. Examples in Function Theory, Picard’s theorem 13212.4. Sheaves 13512.5. Exercises 13612.6. Appendix: proofs of the results of section 12.1 137

Bibliography 141

Index 143

Preface

These notes are designed for a second course in function theory, given atthe master level. It is a topic course; many topics are treated, but only to alimited depth. The idea is that a student who has completed the course hasa fairly broad view of function theory, and will be able to decide in whichdirection to proceed. In particular, it should be relatively easy to figure outa topic for a master’s thesis.

Even in a topics course one has to make a selection. Many things remainuntouched. We will treat a few topics that were hinted at in the elementarycourse, like Riemann Mapping Theorem and related results, product rep-resentations, and Picard’s theorem. Normal families are then unavoidable.My personal interests, in particular in potential theory and in the theory ofHardy spaces, motivated the choice for the rest of the material.

Every chapter has its own short introduction and ends with an exerciseset.

The notes owe much, if not all, to the (Dutch) notes of Prof. J. Kore-vaar’s course “voortgezette functietheorie”. Students who can read Dutchmay want to obtain Korevaar’s notes too.

Jan Wiegerinck

vii

CHAPTER 1

Holomorphic functions

We review in section 1 some results from the introductory course. Weemphasize the Cauchy-Pompeiu formula, theorem 1.1.2. In section 2 weprove the theorems of Runge and Hartogs-Rosenthal. The proofs of theseapproximation theorems are based on the Cauchy-Pompeiu representationformula.

1.1. Holomorphic functions revisited

Holomorphic functions can be defined in three ways, that all give thesame class of functions. Namely, as complex analytic functions, i.e. locallythe sum of a convergent power series, as complex differentiable (holomor-phic) functions, i.e. f(z + h) = f(z) + Ah + o(h) as h → 0, and finally as

C1-functions that satisfy the Cauchy-Riemann equations ∂f∂z = 0, (see be-

low). The latter will be our starting point and we commence with Green’sformula

Theorem 1.1.1. Let G be a bounded domain in R2 with piecewise C1-boundary ∂G. Suppose f and g are continuously differentiable on G. Then

∫

G

(

∂f

∂x− ∂g

∂y

)

dx ∧ dy =

∫

∂Gg dx+ f dy.

We recall the notation

∂

∂z= 1

2

(

∂

∂x+

1

i

∂

∂y

)

,(1.1.1)

∂

∂z= 1

2

(

∂

∂x− 1

i

∂

∂y

)

,(1.1.2)

dz = dx+ idy,(1.1.3)

dz = dx− idy.(1.1.4)

Then 2idx ∧ dy = dz ∧ dz and dfdef= ∂f

∂xdx + ∂f∂ydy = ∂f

∂z dz + ∂f∂z dz. The

complex form of Green’s theorem for C1-smooth functions defined on theclosure of a domain in C, becomes

(1.1.5)

∫

G

∂f

∂zdz ∧ dz =

∫

∂Gf dz.

For functions satisfying the Cauchy-Riemann equations ∂f∂z = 0 we see that

(1.1.5) is just Cauchy’s first theorem. The Cauchy-Pompeiu formula follows

when we apply (1.1.5) to f(ζ)ζ−z on G \B(z, ε) and letting ε→ 0. It reads

1

2 1. HOLOMORPHIC FUNCTIONS

Theorem 1.1.2 (Cauchy-Pompeiu). Let G be a domain in C with piece-wise C1-boundary, and let f be a C1-function on G. Then

f(z) =1

2πi

∫

∂G

f(ζ)

ζ − zdζ −

∫

G

∂f∂ζ

(ζ)

ζ − zdζ ∧ dζ

.

For holomorphic functions f this yields Cauchy’s formula

f(z) =1

2πi

∫

∂G

f(ζ)

ζ − zdζ.

The integrand may be developed into a power series, resulting in the powerseries representation for holomorphic functions. Indeed, for a ∈ G and|z − a| < d(a, ∂G) we have

(1.1.6) f(z) =1

2πi

∫

∂G

f(ζ)

(ζ − a) − (z − a)dζ =

1

2πi

∫

∂G

f(ζ)

ζ − a

1

1 − z−aζ−a

dζ

=

∞∑

n=0

(

1

2πi

∫

∂G

f(ζ)

(ζ − a)n+1dζ

)

(z − a)n.

Easy consequences are

• Non constant holomorphic functions have isolated zeros.• The uniqueness theorem: Two holomorphic functions on a domainG that coincide on a set E with density point inside G are equal.

• Weierstrass’ theorem: If fn is a sequence of holomorphic func-tions that converges uniformly on compact sets in a domain G, thenlimn→∞ fn = f exists and is a holomorphic function. Moreover, thederivatives of fn converge uniformly to those of f .

• The residue theorem. If f is holomorphic on G except for isolatedsingularities a1, . . . an then

1

2πi

∫

∂Gf(ζ) dζ =

n∑

1

Res (f, aj).

• The argument principle. For a function f that is holomorphic on asmooth domain G except maybe for poles, and, moreover, is con-tinuous and zero-free on the closure G, one has

Z(f) − P (f) =1

2πi

∫

∂G

f ′(z)f(z)

dz =1

2π

∫

∂Gd arg(f(z)).

Here Z(f) is the number of zeros, while P (f) is the number of polesof f on G, all counted with multiplicity.

• Rouche’s theorem. If f and g are holomorphic on a bounded domainG and continuous up to the boundary of G and if |f | > |g| on ∂G,then Z(f + g) = Z(f)

• Open mapping theorem. A non-constant holomorphic function isan open map. [The image of an open set is always open]

• Maximum principle: Let f be a non-constant holomorphic functionon a domain G. Then |f | has no (weak) local maximum on G.

In a later chapter we will need the following theorem of Hurwitz.

1.2. APPROXIMATION THEORY 3

Theorem 1.1.3. Let D ⊂ C be a bounded domain and let fn be asequence of holomorphic functions that converges to f 6≡ 0 uniformly oncompact sets in D. Let D′ ⊂ D have smooth boundary and be such that|f | > 0 on ∂D′. Then there exists n0 such that for n > n0 the functions fand fn have the same number of zeros (counted with multiplicity) on D′. Ifthe functions fn are injective then f is injective or f is a constant.

Proof. We apply Rouche’s theorem. For sufficiently large n0 and n >n0, we have

max∂D′

|fn(z) − f(z)| < min∂D′

|f(z)|.So f and f + (fn − f) = fn have the same number of zeros on D′.

Now suppose that all fn are injective and that f is non-constant. Letz1 ∈ D. Then fn − fn(z1) has exactly one zero on D. The functions fn −fn(z1) converge uniformly on compact sets in D to f − f(z1). If f(z2) =f(z1), (z1 6= z2), then f − f(z1) would have at least two zeros in a smoothlybounded domain D′ that is compactly contained in D. Changing D′ slightlyif necessary, we can assume that f − f(z1) has no zeros on the boundary ofD′. Because f − f(z1) is not identically 0, this contradicts the first part ofthe theorem.

Noteworthy is the following converse of Cauchy’s theorem. It will be ofuse later on.

Theorem 1.1.4 (Morera). Suppose that f is continuous on a domain Dand has the property that for every closed path of integration T in D that

(1.1.7)

∫

Tf(z) dz = 0,

then f is holomorphic on D.

In fact it is sufficient if (1.1.7) holds for paths without self intersectionsthat consist of finitely many line segments parallel to the coordinate axes

1.2. Approximation theory

The most famous approximation theorem is due to Weierstrass. We willquote it below. Approximation theory is an important subject in complexanalysis. References are firstly Gaier’s book, [17]. The books [39, 4] havesome interesting sections on approximation theory too.

Theorem 1.2.1 (Weierstrass). Let K be compact in Rn. Every contin-uous function f on K can be approximated uniformly by polynomials.

The modern way to prove this theorem and the ones that will follow, is byfunctional analysis. (See exercise 1.3.3 for a proof of Weierstrass theorem).The third year course functional analysis [48] will do for our purposes. Wewish to show that a subspace S (the polynomials) is dense (in the uniformnorm) in a larger space X (some subspace of C(K), the continuous functionson a compact set K). The Hahn Banach theorem says that ifX is a completemetric topological vector space and x ∈ X \ S, then there exists a boundedlinear functional L on X such that

L|S = 0, L(x) 6= 0.


Thus, to prove that S = X it is sufficient to show that for continuouslinear functionals on X

L|S = 0 ⇒ L ≡ 0 on X.

Now, by another application of Hahn-Banach and because X is a subspaceof C(K), any functional on X can be extended to all of C(K).

The continuous linear functionals on C(K) are well known: accordingto the Riesz representation theorem they are represented by the complexregular Borel measures on K. These form a normed vector space that isdenoted by M(K). The norm ‖µ‖ of µ ∈ M(K) is the norm as a linearfunctional on C(K).

‖µ‖ = supf∈C(K),|f |≤1

∣

∣

∣

∣

∫

Kf(x) dµ(x)

∣

∣

∣

∣

.

Thus with this norm, M(K) is isometrically isomorphic to C(K)∗. Soour strategy for proving S = X comes down to proving the following. Everyµ ∈M(K) that annihilates S must annihilate X.

Definition 1.2.2. Let µ be a finite complex Borel measure on C withcompact support K. The Cauchy Transform of µ is the function

µ(z) =

∫

K

dµ(ζ)

ζ − z.

Lemma 1.2.3. The function µ is locally in L1(C). It is holomorphic onC \K with derivative

µ(m)(z) = m!

∫

K

dµ(ζ)

(ζ − z)m+1, z ∈ C \K.

Proof. As a convolution of a compactly supported measure and a lo-cally integrable function, µ is locally in L1, i.e. µ ∈ L1(B(a, r)) for everyball B(a, r). Indeed, applying Fubini’s theorem

∫

B|µ| dx dy ≤

∫

K|dµ(ζ)|

∫

B

1

|z − ζ|dx dy ≤ C‖µ‖.

The fact that µ is holomorphic off K and the formula for the derivatives ata ∈ C \ K follow in the usual way by developing 1

z−ζ into a power series

about a.

Theorem 1.2.4 (Runge). Let f be holomorphic on a neighborhood of acompact set K ⊂ C. For each component Cj of C \ K a point pj ∈ Cj isgiven. Then f is on K the uniform limit of a sequence of rational functionswith, atmost, poles in pj.

Proof. Let R denote the rational functions in C with poles only inpj and R the closure of R in C(K). We want to show f ∈ R. By theHahn-Banach theorem and Riesz representation theorem, this is equivalentwith

for every µ ∈M(K)

(

∀Q ∈ R

∫

Qdµ = 0 =⇒∫

f dµ = 0

)

.

Because f is defined in a neighborhood D of K, there exist finitely manysmooth Jordan curves γj ⊂ D \ K with pj in the component Ωj of C \ γj

1.2. APPROXIMATION THEORY 5

that does not contain K. We choose the orientation on γj in such a way

that γ = ∪γj is the boundary of a domain D, K ⊂ D ⊂ D, and

f(z) =

∫

γ

f(ζ)

ζ − z

dζ

2πi, z ∈ D.

Let µ ∈M(K). Then∫

Kf(z) dµ(z) =

∫

K

(∫

γ

f(ζ)

ζ − z

dζ

2πi

)

dµ(z)

=∑

j

∫

γj

−f(ζ)µ(ζ)dζ

2πi.

(1.2.1)

Now µ is holomorphic on Ωj. If µ annihilates all functions in R, then

µ(m)(pj) = 0 for all m ≥ 0 and all j. It follows that µ ≡ 0 on Ωj, andtherefore on γj . Then (1.2.1) equals 0 and µ annihilates f .

Corollary 1.2.5. If C \ K consists of one component, then f is auniform limit of polynomials on K.

Proof. Let B(0, R) be a large disc that contains K and choose p1 out-side B(0, R). Runge’s theorem shows that we can approximate f uniformlyon K with rational functions that have only one pole, namely at p1. Theserational functions are on compact sets in B(0, r) uniform limit of their Taylorpolynomials.

For the proof of an other approximation theorem, we need to know alittle more about Cauchy Transforms of measures.

Lemma 1.2.6 (Bishop). Let µ ∈M(K) be a compactly supported complexmeasure. Suppose that µ = 0 a.e. dx ∧ dy. Then µ = 0.

Proof. Choose R so large that K ⊂ B(0, R). Let f be a C1-functionthat is supported in B(0, R). We apply the Cauchy-Pompeiu formula and(1.1.2) and Fubini’s theorem.

∫

Kf(z) dµ(z) =

∫

K

(

−∫

B

∂f∂ζ

ζ − z

dζ ∧ dζ2πi

)

dµ(z) =

∫

B

∂f

∂ζµ(ζ)

dζ ∧ dζ2πi

= 0.

We see that µ annihilates all C1 functions in B(0, R). These form a densesubspace of C(K), hence µ annihilates C(K), and therefore µ = 0.

Theorem 1.2.7 (Hartogs-Rosenthal). Suppose that K is a compact inC of 2-dimensional Lebesgue measure 0. Then the rational functions withpoles outside K are dense in C(K).

Proof. Let µ ∈M(K) and suppose that µ annihilates all rational func-tions with poles off K. We wish to show that µ = 0. The Cauchy transformµ is analytic on C \K and is formed by integrating µ against a special ra-tional function 1/(ζ − z). Hence µ = 0 on C \K. From Bishop’s Lemma weconclude that µ = 0.

We have only seen a glimpse of complex approximation theory. See e.g.Gaier’s book [17] for much more. We only quote the famous Mergelyanapproximation theorem.


Theorem 1.2.8 (Mergelyan). Let K be a compact set in C such thatC\K is connected. Let f be a continuous function on K that is holomorphicon the interior Ko of K. Then f is the uniform limit of a sequence ofholomorphic polynomials.

1.3. Exercises

1.3.1. What can you say about the location of the zeros of

pn(z) = 1 +z

1!+z2

2!+ · · · + zn

n!, n→ ∞?

1.3.2. Show that on the annulus A(0, 1, 2) the function 1/z can not beapproximated uniformly by polynomials in z. In fact, it may be easiest toshow that ‖P − 1/z‖∞ ≥ 1 for every holomorphic polynomial P .

1.3.3. Prove Weierstrass’ theorem 1.2.1 along the following lines, (thisproof was found by de Branges, [10], cp. [4]). Let X denote the set of Borelmeasures µ on C(K) that annihilate the polynomials and have ‖µ‖ ≤ 1.Then X is a compact convex set in the dual space M(K) = C(K)∗. Applythe Krein-Millman theorem that asserts that X has an extreme point µ0. IfX 6= 0, we can assume ‖µ0‖ = 1.

(1) Show that µ0 has a support that contains at least two distinctpoints, say x1 x2.

(2) Show that there is a polynomial g with 0 < g < 1 on K andg(x1) 6= g(x2).

(3) Now show that gµ0

‖gµ0‖ and (1−g)µ0

‖(1−g)µ0‖ are in X and that

µ0 = ‖gµ0‖gµ0

‖gµ0‖+ ‖(1 − g)µ0‖

(1 − g)µ0

‖(1 − g)µ0‖,

a convex combination of elements in X.

1.3.4. Suppose that K is a compact in C of 2-dimensional Lebesguemeasure 0. Fix a point pj in every component Ωj of C \ K. Prove thatevery continuous function on K can be uniformly approximated by rationalfunctions with poles only in the pj .

1.3.5. Let f be defined on B(0, 1) by

f(z) =

1 for Re z ≥ 0,

0 for Re z < 0.

Prove that there exists a sequence of holomorphic polynomials that con-verges pointwise to f on B(0, 1).

1.3.6. Let f be defined on B(0, 1) by

f(z) =

1 for z = 0,

0 for z 6= 0.

Prove that there exists a sequence of holomorphic polynomials that con-verges pointwise to f on B(0, 1).

1.3.7. Prove that there exists a sequence of holomorphic polynomialsthat converges pointwise to 0 on B(0, 1), but not uniformly.

1.3. EXERCISES 7

1.3.8. Prove that every continuous function on B(0, R) that is holomor-

phic on B(0, R), is a uniform limit of holomorphic polynomials on B(0, R)(without using Mergelyan’s theorem!).

1.3.9. Let K = B(0, R)∪ [−R−M,R+M ]. Let A(K) be the subspaceof C(K) that consists of continuous functions on K that are holomorphicon B(0, R). Show that f ∈ A(K) is a uniform limit of holomorphic polyno-mials on K without using Mergelyan’s theorem. (Hint: Use the functions

z2n

(R+ε)2n+z2n to show that every measure µ on K that annihilates the ratio-

nal functions with poles off K, can be written as µ1 + µ2 such that µ1 issupported on B(0, R) and µ2 supported on [−R −M,R +M ], while µ1, µ2

annihilate these rational functions too).

1.3.10. Prove Carleman’s theorem: Let f be continuous on R and εa positive continuous function on R. Prove that there exists a holomorphicfunction h on C such that for t ∈ R

|f(t) − h(t)| < ε(t).

(Hint: Use the previous exercise (or Mergelyan) on an increasing set of K’s,and control the polynomials so that they converge uniformly on compactsets.)

1.3.11. Try to prove Runge’s theorem in the original way. Let K becompact in C and let f be holomorphic in a neighborhood of K.

(1) Use Riemann sums as an approximation for the Cauchy integral inorder to prove that holomorphic functions can be approximated byrational ones (with location of the poles unprescribed).

(2) Show that for a /∈ K the function 1(z−a)n can be uniformly approx-

imated on K by polynomials in 1z−b if d(b,K) > |b− a|.

(3) Use part 2 to shift the poles in the approximating functions to pj.

1.3.12. Let Dn = B(0, 1 − 1/(n + 1)), (n ≥ 1), and D0 = ∅. Let Kn

be compact sets in Dn \ Dn−1 and suppose that the complement of Kn isconnected. Show that every function that is holomorphic on a neighborhoodof ∪nKn is a uniform limit of holomorphic functions on B(0, 1).

1.3.13. Use the previous exercise to show that there exist holomorphicfunctions f on B(0, 1) such that for no θ ∈ [0, 2π) the limit limr→1 f(reiθ)exists.

1.3.14. Let K be compact subset of the unit circle without (relative)

interior. Let ϕ be continuous on K = z = reiθ : eiθ ∈ K, 0 < r < 1. Showthat there exists a holomorphic function f on B(0, 1) such that

limr→1

f(reiθ) − ϕ(reiθ) = 0 for eiθ ∈ K.

1.3.15. Use the previous exercise to show that there exists a non con-stant holomorphic function on the unit disc, such that limr→1 f(reiθ) = 0for almost every 0 ≤ θ < 2π. (Hint: ϕ(z) = 1/(|z| − 1) what does this do toRe f , so?)

CHAPTER 2

Normal families

2.1. Introduction

In this chapter we will study the set of holomorphic functions on adomain in C as a topological vector space. It will turn out that Hilbertor Banach structure is too much to ask. However, our functions form acomplete metric space, and this metric space shares one important propertywith finite dimensional Euclidean space: Every bounded sequence has aconvergent subsequence. This important property makes it easy to decidethat certain limits exist and this will allow us to prove the Riemann mappingtheorem in a subsequent chapter.

2.2. Spaces of continuous functions

As the holomorphic functions on an open set Ω ⊂ C form a subset ofthe set of continuous functions C(Ω) on Ω, we start by studying spaces ofcontinuous functions.

Recall that if K is a compact set in a metric space M then the complexvalued continuous functions on K form a Banach space C(K) with supre-mum norm ‖f‖K = supx∈K |f(x)|. The associated metric is dK(f, g) =‖f − g‖K . Also recall that a sequence of functions is Cauchy in C(K) ifand only if it is a uniformly convergent sequence. Now let Ω be open ina compact metric space M with metric ρ. We can write Ω as a countableunion of compact sets:

Ω = ∪∞1 Kn,

where

Kn = x ∈ Ω : |x| ≤ n, ρ(x, ∂Ω) ≥ 1/n.We turnC(Ω) into a metric space as follows: Denote the sup-norm metric

on Kn by dn. Then dn/(1 + dn) is an equivalent metric that is bounded by1.

Proposition 2.2.1. The function d defined on C(Ω) × C(Ω) by

d(f, g) =∞∑

n=1

1

2ndn(f, g)

1 + dn(f, g)

is a metric on C(Ω). The space C(Ω) is complete under d and convergencein d is equivalent with uniform convergence on compact subsets.

Proof. Exercise 2.5.1.

Definition 2.2.2. Let Mi be metric spaces with metric di. A family ofcontinuous functions F from M1 to M2 is called equicontinuous on E ⊂ M1,

9

10 2. NORMAL FAMILIES

if for each ε > 0 there is a δ = δ(ε,E) such that for all z,w ∈ E and allf ∈ F

d1(z,w) < δ =⇒ d2(f(z), f(w)) < ε.

The family F is called locally equicontinuous on M1 if each point p ∈ M1

has a neighborhood E such that F is equicontinuous on E.

Equicontinuity means that the continuity is uniform for each functionin F , but also uniform over the functions in F . Usually M2 is the real orcomplex numbers, so that F consists of continuous functions.

Recall that a topological space is called separable if it has a countabledense subset.

Theorem 2.2.3 (Arzela-Ascoli). Let M be a separable metric space andlet F ⊂ C(M) be a locally equicontinuous family of continuous functions onM that is bounded in each point of M. Then each sequence fn ⊂ F hasa subsequence that converges uniformly on compact sets in M. Moreover, ifgn is any sequence in F that converges pointwise on a dense subset of M,then the sequence converges on all of M and uniformly on compact subsetsof M.

Proof. Let A = an be a dense sequence in M. Let fn be a sequencein F . We will first construct a subsequence gk that converges in points ofA. Secondly we will show that this sequence is in fact uniformly convergenton compact sets in M.

Step 1. Since F is bounded in each point of M we can find a subsequencef1,j such that f1,j(a1) converges. From the sequence f1,j we extracta subsequence f2,j such that f2,j(a2) converges. Continuing in thisfashion, we find a sequence of subsequences, each one containing the nextones as subsequences,

f1,j ⊃ f2,j ⊃ f3,j ⊃ . . . .

We now form the diagonal sequence

gi, gi = fi,i.

Then limi→∞ gi(ak) exists for every k, because the tail of this sequence is asubsequence of fk,j(ak)j .

Step 2. Next we show that any sequence in F that converges on a denseset A is in fact uniformly convergent on any compact subset K of M.

Let ε > 0. Let p ∈ M. Then there is a δ = δp such that |f(z)− f(w)| <ε for all z,w ∈ B(p, δ) and all f ∈ F . Now K ⊂ ∪p∈KB(p, δp) and Kis compact, Thus we can select a finite cover B1, B2, . . . , Bn of K. Weselect in each ball Bi a point wi ∈ Bi ∩ A and choose k0 so large that|gj(wi) − gk(wi)| < ε if j, k > k0. Let z ∈ K, then z ∈ Bi for some i. Wehave for k, j > k0

|gj(z)− gk(z)| ≤ |gj(z)− gj(wi)|+ |gj(wi)− gk(wi)|+ |gk(wi)− gk(z)| ≤ 3ε.

Definition 2.2.4. Let Ω be open in a metric space M. A familyF ⊂ C(Ω) is called normal if every sequence in F has a subsequence thatconverges uniformly on compact sets in Ω

2.4. NORMAL FAMILIES OF MEROMORPHIC FUNCTIONS. 11

2.3. Normal families of holomorphic functions

Let H(Ω) denote the collection of holomorphic functions on an open setΩ ⊂ C. As a subset of C(Ω) it inherits a metric structure. Weierstrass’well-known result from elementary function theory states that a sequence ofholomorphic functions that converges uniformly on compact sets of Ω has alimit that is holomorphic on Ω. In other words, H(Ω) is a closed subspaceof C(Ω) and therefore a complete metric space.

Definition 2.3.1. A family F ∈ C(Ω) is called locally bounded if forevery a ∈ Ω there exists a neighborhood Ua of a and a constant M such that

|f | ≤M on Ua for all f ∈ F .

Theorem 2.3.2 (Montel). Let F be a locally bounded family of holo-morphic functions on an open set Ω. Then F is normal. Moreover, if asequence in F is pointwise convergent, then it is uniformly convergent oncompact subsets.

Proof. In view of the theorem of Arzela-Ascoli 2.2.3 it suffices to showthat F is locally equicontinuous. Let a ∈ Ω. Choose M and r > 0 suchthat B(a, r) ⊂ Ω and |f | < M on B(a, r) for all f ∈ F . Then we find forz,w ∈ B(a, r/2) and f ∈ F

|f(z) − f(w)| =

∣

∣

∣

∣

∣

1

2πi

∫

C(a,r)f(ζ)

(

1

ζ − z− 1

ζ − w

)

dζ

∣

∣

∣

∣

∣

≤ M |z − w|(r/2)2

r.

The last expression will be less than ǫ > 0 independently of f ∈ F if |z−w|is sufficiently small.

Remark 2.3.3. Montel’s theorem says that every locally bounded familyof holomorphic functions in H(Ω) has a limit point in H(Ω).

Corollary 2.3.4 (Stieltjes-Vitali). Let Ω be a domain in C and fna locally bounded sequence of holomorphic functions on Ω. Suppose thatE ⊂ Ω has a limit point in Ω and that fn(x) converges for every x ∈ E.Then the sequence fn is uniformly convergent on compact sets in Ω.

Proof. By Montel’s theorem the sequence fn has a limit point, sayf ,in H(Ω). It suffices to show that the sequence has exactly one limit point.

Let f and f be limit points of fn. Then f = f on E. By the uniqueness

theorem, f ≡ f .

2.4. Normal families of meromorphic functions.

Normality is more naturally treated in the setting of meromorphic func-tions. For a proper understanding of these we recall some facts about theRiemann sphere C∗ = C ∪∞. We identify the Riemann sphere C∗ with theball

S = (p, q, t) : p2 + q2 + (t− 12)

2 = ( 12)

2 ⊂ R3,


through

Φ : x+ iy 7→(

x

1 + x2 + y2,

y

1 + x2 + y2,

x2 + y2

1 + x2 + y2

)

; ∞ 7→ (0, 0, 1).

(2.4.1)

The standard metric on R3 is induced by the norm ‖(p, q, t)‖ =√

p2 + q2 + t2.We restrict this metric to S and transfer it by (2.4.1) to C∗. Thus we arriveat the chordal metric χ:

χ(z,w) =|z − w|

√

(|z|2 + 1)(|w|2 + 1), z, w,∈ C;

χ(∞, z) =1

√

(|z|2 + 1), z ∈ C.

(2.4.2)

Definition 2.4.1. Let γ : [0, 1] → C be a C1 curve in C. The chordallength of L(γ) is the length of the image of γ as a subset of S ⊂ R3.

Lemma 2.4.2. Let γ : [0, 1] → C be a C1 curve in C. Then

(2.4.3) L(γ) =

∫

γ

|dz|1 + |z|2 =

∫ 1

0

|γ′(t)|1 + |γ(t)|2 dt.

Proof. The chordal length L(γ) is the supremum of the sums of thelengths of the inscribed polygons of Φ(γ). That is,

(2.4.4) supn−1∑

j=1

χ(γ(tj+1), γ(tj)),

where the sup is taken over all n and all possible partitions 0 = t1 < t2 <. . . < tn = 1. Let ε > 0. There exist δ > 0 such that

∣

∣

∣

∣

∣

1√

(|z|2 + 1)(|w|2 + 1)− 1

1 + |z|2

∣

∣

∣

∣

∣

< ε, if |z − w| < δ.

This is ensured if the stepsize max |tj − tj−1| is sufficiently small. Then∣

∣

∣

∣

∣

∣

n−1∑

j=1

χ(γ(tj+1), γ(tj)) −n−1∑

j=1

|γ(tj+1) − γ(tj)|1 + |γ(tj)|2

∣

∣

∣

∣

∣

∣

< ε

n−1∑

j=1

|γ(tj+1)−γ(tj)| ≤ ε|γ|.

This formula expresses that a Riemann sum for the first integral in (2.4.3) isarbitrarily close the supremum in (2.4.4) and the first equality follows; thesecond follows by substitution of z = γ(t).

Alternatively, one can compute

L(γ)def=

∫ 1

0

√

((Φ γ)′(t)))τ ((Φ γ)′(t)))dt

directly but tediously. The result is again (2.4.3). (Here we have denotedby Aτ the transposed of a matrix A).

We also can identify C∗ with the projective space P1(C) via

z ↔ [z, 1]; ∞ ↔ [1, 0].

2.4. NORMAL FAMILIES OF MEROMORPHIC FUNCTIONS. 13

Definition 2.4.3. A function f on a domain Ω ⊂ C∗ is called meromor-phic if f is holomorphic except for poles, or if f is identically equal to ∞.Equivalently, f can be written as a quotient of two holomorphic functionsf1, f2 without common zeros; equivalently f is a holomorphic (and thereforecontinuous) map from Ω to P1(C), (the map is given by [f1, f2], with f1, f2

as before).

Remark 2.4.4. Notice that normality of meromorphic functions is de-fined by Definition 2.2.4, as soon as we understand what uniform convergenceof a sequence of meromorphic functions means. With the chordal metric athand this is easy. A sequence of meromorphic functions fn defined ona neighborhood of a set E ⊂ C, converges uniformly to f (in the chordalmetric, as continuous functions with values in S) on E if

∀ε∃n0 such that ∀n > n0 : χ(fn(z), f(z)) < ε, z ∈ E.

Definition 2.4.5. For a meromorphic function f on a domain Ω wedefine the spherical derivative

(2.4.5) µ(f) : Ω 7→ R≥0, µ(f)(z) = limh→0

χ(f(z + h), f(z))

|h| .

Lemma 2.4.6. Let f be meromorphic on a domain Ω. Then

(2.4.6) µ(f) =

|f ′(z)|1 + |f(z)|2 if z is not a pole;

1

Aif z is a pole of order 1 and A = |Res (f, z)| > 0;

0 else.

Proof. Suppose that z is not a pole of f . Then f is holomorphic onsome ball B(z, r). Taking |h| < r we have

(2.4.7) µ(f)(z) = limh→0

χ(f(z + h), f(z))

|h| =

limh→0

|f(z + h) − f(z)||h|

1√

1 + |f(z + h)|2√

1 + |f(z)|2=

|f ′(z)|1 + |f(z)|2 .

If z is a pole, we have

(2.4.8) µ(f)(z) = limh→0

χ(f(z + h), f(z))

|h| =

limh→0

1

|h|√

1 + |f(z + h)|2= lim

h→0

1

|hf(z + h)|

=

1

Aif z is a pole of order 1 and A = |Res (f, z)| > 0;

0 if z is a pole of higher order.

Theorem 2.4.7 (Marty). A family of meromorphic functions F on adomain Ω is normal if and only if µ(F) := µ(f) : f ∈ F is locally bounded.


Proof. The only if part is left as an exercise, cp. 2.5.9. It suffices toprove that F is locally equicontinuous (in the chordal metric). For eachpoint z ∈ Ω we find a ball Bz ⊂ Ω and a constant M such that µ(f) < Mon B independent of f . Let z,w ∈ B and let γ be the C1-curve (in fact, itis an interval) in Bz given by γ(t) = z+ t(w− z). Then χ(f(z), f(w)) is lessthan the length of the curve f γ in S. Hence

(2.4.9) χ(f(z), f(w)) ≤∫ 1

0

|f γ′(t)|1 + |f γ(t)|2 dt =

∫

γµ(f)(z)|dz|

≤ supζ∈γ

µ(f)(ζ)|γ| ≤M |z − w|.

This estimate is independent of f and shows equicontinuity of F .

Remark 2.4.8. It is not easy to show that a given particular family ofholomorphic functions is normal. E.g., we shall see in a later chapter thatgiven 3 different points in C∗, the family of meromorphic functions that omitthese three values is in fact normal. Compare this to the Picard theoremthat an entire meromorphic function that omit 3 values is constant! Bothstatements are non-trivial and clearly related. Bloch formulated a principlethat may paraphrased as follows “If a family of holomorphic functions hasa property that would guarantee an entire function to be constant, thenthe family is normal.” Surprizingly, something like this can be proved! SeeZalcman, [50].

2.5. Exercises.

2.5.1. Complete the proof of Proposition 2.2.1.

2.5.2. Let f = f(x, y) be a function on R2 and let fy(x) = f(x, y). Showthat if f is continuous on −R ≤ x ≤ R, then fyy is an equicontinuousfamily on −R ≤ x ≤ R. Show that the converse is not necessarily trueand formulate an extra condition on the family fyy that guarantees theconverse.

2.5.3. Let F = e−λz, λ > 0, on H = z : Re z > 0. Show that F isa normal family; what are the possible limits?

2.5.4. Let F be a family of holomorphic functions on a domain G.Assume that no function in F assumes values in the disc B(a, r), or even inthe interval [0, 1]; prove that F is normal.

2.5.5. Let F be a normal family in H(G). Show that F ′ = f ′|f ∈ Fis normal. What about the converse? What if we look at this from the pointof view of meromorphic functions?

2.5.6. Let G be a domain in C, a ∈ G, M > 0. Let F = f ∈H(G)|f(a) = 0,D(f) < M, where

D(f) =

∫

G|f ′(z)|2 dxdy.

Show that F is normal. [D(f) is called the Dirichlet integral and equals thearea of f(G) counted with multiplicity.]

2.5. EXERCISES. 15

2.5.7. Let F be a family of holomorphic functions on a domain G thatis normal as a family of meromorphic functions. Show that a limit of asequence in F is either holomorphic or identically ∞.

2.5.8. Prove that for z,w ∈ C∗, the Riemann sphere, we have χ(z,w) =χ(1/z, 1/w).

2.5.9. Prove that a normal family of holomorphic functions on a domainΩ is locally bounded. Prove the inverse direction of Marty’s theorem: anormal family of meromorphic functions has locally bounded µf .

2.5.10. Let f be a meromorfic function on a domain Ω ⊂ C and letg = 1/f . Prove that µ(f) = µ(g).

2.5.11. Let f be a bounded holomorphic function on the strip S = z :−1 < y < 1. Suppose limx→∞ f(x) = A. Show that

limx→∞

f(x+ iy) = A

uniformly on −1 + δ < y < 1 − δ, (δ > 0). [Consider the functions fn(z) =f(n+ z), n = 1, 2, . . ..]

CHAPTER 3

Conformal Mappings

3.1. Introduction

From elementary function theory we recall that injective holomorphicmaps are conformal, that is, angles between curves as well as their orien-tation are preserved under such a map. In the first section we will reviewsome facts about the simplest conformal maps, the fractional linear trans-formations.

We will next make a small excursion to hyperbolic geometry on the unitdisc. Our excursion will pay off in the next sections.

The main theorem will be the famous Riemann Mapping theorem, whichstates that every simply connected domain in C, but not equal to C, isconformally equivalent to the unit disc.

A second important theorem is Caratheodory’s. It states that the Rie-mann map extends to a homeomorphism of the closure of the domains ifand only if the boundaries of the domains are Jordan curves.

We will also discuss Schwarz-reflection, which enables us to “compute”certain Riemann maps for polygon-like domains.

References for this chapter are the books by Nehari, [30] and Pom-merenke, [34, 35]. Also [1, 12, 13, 15, 27, 38, 39] have the Riemannmapping theorem and (much) more.

The unit disc B(0, 1) will throughout this chapter be denoted by B, itsboundary by T.

3.2. Fractional linear transformations revisited

A fractional linear transformation or Mobius transformation is a map

w = f(z) =az + b

cz + d, ad− bc 6= 0.

Such an f is a 1-1 holomorphic homeomorphism of the Riemann sphere.Hence it is conformal, meaning that it preserves angles between curves.Generalized circles, i.e. circles and straight lines, are mapped onto gener-alized circles. Mobius transforms are completely determined by prescribingthe (different) images of three different points.

Example 3.2.1. Let z1, z2, z3 be different points in C. Then

f(z) =z3 − z2z3 − z1

z − z1z − z2

is a Mobius transform that maps z1 to 0, z2 to ∞ and z3 to 1.

Example 3.2.2. Similarly we send three different points z1, z2, z3 todifferent points w1, w2, w3 by composing the map from the previous example

17

18 3. CONFORMAL MAPPINGS

with the inverse of a similar one.

f = g−1 h, or g(f(z)) = h(z)

with

h(z) =z3 − z2z3 − z1

z − z1z − z2

, g(w) =w3 −w2

w3 −w1

w − w1

w − w2.

We now look at conformal self mappings of the unit disc B. We needSchwarz’ Lemma.

Lemma 3.2.3 (Schwarz). Let f be a holomorphic map from B to B.Suppose that f(0) = 0. Then

|f(z)| ≤ |z|, z ∈ B,(3.2.1)

and therefore

|f ′(0)| ≤ 1.(3.2.2)

If equality occurs in (3.2.1) for some z, |z| > 0, or in (3.2.2), then f(z) =eiθz for some θ ∈ R.

Proof. The function f is bounded by 1. The function g(z) = f(z)/z isholomorphic with a removable singularity at 0 on B; note that g(0) = f ′(0).On C(0, r) we find |g(z)| < 1/r, hence by the maximum principle, we have|g(z)| < 1/r on B(0, r). Letting r → 1 we find |g| ≤ 1 on B. This proves(3.2.1) and (3.2.2). If equality occurs in (3.2.1) in a non-zero point of B, themaximum principle asserts that g be constant; similarly for (3.2.2). Thisproves the last statement of the lemma.

Theorem 3.2.4. The conformal mappings of the unit disc B are preciselythe fractional linear transformations of the form

(3.2.3) w = f(z) = eiθz − a

1 − az, a ∈ B, θ ∈ R.

Proof. Let f be as in (3.2.3). Then f(a) = 0, so it suffices to showthat f(T) ⊂ T, since f is a homeomorphism of the Riemann sphere (andmaps generalized circles onto generalized circles). Let z ∈ T. Then

∣

∣

∣

∣

eiθz − a

1 − az

∣

∣

∣

∣

=

∣

∣

∣

∣

z − a

zz − az

∣

∣

∣

∣

=

∣

∣

∣

∣

1

z

∣

∣

∣

∣

∣

∣

∣

∣

z − a

z − a

∣

∣

∣

∣

= 1.

Now let g be a 1-1 conformal self mapping of B. Composing g withsuitable f of the form (3.2.3), we can obtain that g f(0) = 0. From theSchwarz lemma we find |(g f)′(0)| ≤ 1. Applying Schwarz’ lemma to theinverse map we find that |(g f)′(0)| ≥ 1. A third application shows thatg f(z) = eiθz, which implies that g is a Mobius tranformation.

Definition 3.2.5. The conformal self mappings of a domainG are calledautomorphisms of G. This set is denoted by Aut(G).

Remark 3.2.6. Aut(G) is a group; Aut(B) is apparently the group ofMobius tranformations (3.2.3).

3.3. THE HYPERBOLIC METRIC ON B 19

3.3. The hyperbolic metric on B

One of the pleasant properties of the standard metric on C is that itis invariant under a large group of linear transformations. The hyperbolic,or Poincare metric on B will have the property that it is invariant underAut(B).

We will first compute the metric in infinitesimal form. Suppose thatw = f(z), f ∈ Aut(B), with f(z0) = w0. Let T, S ∈ Aut(B) be such thatT (w0) = 0, S(z0) = 0, given by (3.2.3). Then T f = eiθS, in other words,for some θ ∈ [0, 2π),

w − w0

1 − w0w= eiθ

z − z01 − z0z

, θ ∈ R.

We take the modulus on both sides and let z → z0. This gives a relation forthe line element or infinitesimal metric at w0 and at z0.

|dw|1 − |w0|2

=|dz|

1 − |z0|2.

Observe that this relation is valid for every pair (z0, f(z0)). Apparently theexpression

(3.3.1)|dz|

1 − |z|2is invariant under Aut(B).

Definition 3.3.1. The line element of the hyperbolic metric of B is theexpression (3.3.1). The hyperbolic length of a C1-curve γ ⊂ B is

(3.3.2) LH(γ) =

∫

γ

|dz|1 − |z|2 .

The hyperbolic distance between z1, z2 ∈ B is

(3.3.3) ρ(z1, z2) = minγ(0)=z0,γ(1)=z1

LH(γ).

Theorem 3.3.2. Hyperbolic length and hyperbolic distance are invariantunder Aut(B). For z ∈ B, |z| = r we have

(3.3.4) ρ(0, z) = ρ(0, r) =1

2log

1 + r

1 − r.

The geodesics in the hyperbolic metric are straight lines through 0 and arcsof circles that are orthogonal to T.

Proof. Let γ : [0, 1] → B be a curve in B and let T ∈ Aut(B). Then∫

Tγ

|dz|1 − |z|2 =

∫

γ

|dT (z)|1 − |T (z)|2 =

∫

γ

|dz|1 − |z|2 .

The last equality follows from (3.3.1); the (sceptical) reader is invited toperform the computation from classical calculus principles. It follows thatLH and ρ are invariant under Aut(B). For z ∈ B we find that ρ(0, z) =ρ(0, |z|). Next, note that |dz| = |dx + idy| ≥ dx, so that for 0 < r < 1 anda curve γ with γ(0) = 0 and γ(1) = r,

(3.3.5)

∫

γ

|dz|1 − |z|2 ≥

∫

γ

dx

1 − x2 − y2≥∫ r

0

dx

1 − x2.


Hence

(3.3.6) ρ(0, r) = minγ(0)=0,γ(1)=r

∫

γ

|dz|1 − |z|2 =

∫ r

0

dx

1 − x2=

1

2log

1 + r

1 − r.

This proves (3.3.4) and also shows that the geodesic connecting 0 and r isthe segment [0, r]. It follows that the geodesics passing through 0 are thestraight lines passing through 0. Applying an element of Aut(B), we seethat a geodesic passing through z ∈ B will be (an arc of) the circle passingthrough z and meeting T orthogonally.

Proposition 3.3.3. Let f : B → B be a holomorphic function. Thenρ(f(z1), f(z2)) ≤ ρ(z1, z2). If z1 6= z2 equality occurs if and only if f ∈Aut(B).

Proof. This is a consequence of the Schwarz Lemma. Details are leftas exercise 3.6.1.

Example 3.3.4. If we do not require that f is defined on all of B, f mayincrease hyperbolic length on its domain of definition. Let g(z) = z2 and leth be the principal value of

√z on G = B \ (−1, 0]. Let γ be a C1-curve in

G, γ′ = h(γ), so that γ = g(γ′). Then by Proposition 3.3.3

(3.3.7) LH(γ′) > LH(g(γ′)) = LH(γ).

This example is the crux in the following Lemma, that will be used inthe proof of the Riemann mapping theorem.

Lemma 3.3.5 (Koebe). Let G ⊂ B be simply connected and not equal toB. Suppose that 0 ∈ G. Then there exists an injective holomorphic mappingϕ from G into B such that

ϕ(0) = 0, ϕ′(0) > 1.

Proof. There exists b ∈ B \ G and B(0, δ) ⊂ G. Let T1 ∈ Aut(B) besuch that T1(b) = 0. Let ω be a holomorphic root of z on T1(G) and letT2 ∈ Aut(B) have the property that T2ωT1(0) = 0. We set ϕ = T2ωT1.Then ϕ increases all hyperbolic distances, because the Ti preserve hyperbolicdistances and ω increases them. In particular we find for z ∈ B(0, δ) ⊂ Gthat

ρ(0, ϕ(z)) =1

2log

1 + |ϕ(z)|1 − |ϕ(z)| > ρ(0, z) =

1

2log

1 + |z|1 − |z| .

This implies that for all z ∈ B(0, δ)

|ϕ(z)| > |z|, |z| < δ.

Since |ϕ(z)/z| cannot have a minimum at 0, it follows that |ϕ′(z)| > 1.Multiplying with a suitable eiθ gives the result.

3.4. The Riemann mapping theorem

Riemann formulated his mapping theorem in 1851. Complete proofswere given much later, by Osgood (1900) and Caratheodory and Koebe(1912). Our proof is closest to the one by Caratheodory and Koebe.

3.4. THE RIEMANN MAPPING THEOREM 21

Theorem 3.4.1 (Riemann mapping Theorem). Let G be a simply con-nected domain in C, not equal to C. Let B be the open unit disc. Then foreach point a ∈ G there is precisely one biholomorphic map from G onto B,such that g(a) = 0, g′(a) > 0.

Proof. Step 1. First we will prove uniqueness. Suppose that g is as inthe theorem and let f be any holomorphic map from G to B with f(a) = 0,f ′(a) > 0. Then h = f g−1 maps B to B, and has h(0) = 0, h′(0) > 0. BySchwarz’ lemma |h(w)| ≤ |w| and h′(0) = f ′(a)/g′(a) ≤ 1. We find

(3.4.1) f ′(a) ≤ g′(a).

Moreover, equality can only occur if h(z) = z, that is, if f ≡ g. If g is anotherfunction with the properties of the theorem, we would find g′(a) ≤ g′(a) andg′(a) ≤ g′(a), so g ≡ g.

Thus, it is natural to study the family F that consists of all injectiveholomorphic maps from G to B with the property that f(a) = 0 and f ′(a) >0.

Step 2. The family F is non-empty.If G omits disc B(c, r) then z 7→ r/(z− c) maps G injectively into B and

composition with a suitable T ∈ Aut(B) gives an element of F .For arbitrary domains we proceed as follows. Because G is simply con-

nected and is not equal to C, there exist a b ∈ C \ G and a holomorphicbranch of log(z − b) on G. Then we can define a holomorphic branch ω of√z − b on G by

ω(z) = e12

log(z−b).

Suppose ω(z1) = ±ω(z2). Then z1 − b = ω(z1)2 = ω(z2)

2 = z2 − b. Thus ωis injective and ω(z1) 6= −ω(z2), because ω has no zeros on G. Now ω(G) isopen and contains a ball B(d, r). Then values in −B(d, r) = B(−d, r) areomitted by ω, and the first part of this step shows

r

ω(z) + d∈ F .

Step 3. The family F is a normal family, and the set f ′(a) : f ∈ F isbounded.

Indeed, all functions in F are bounded by 1, so Montel’s theorem 2.3.2applies. Now let B(a, δ) ⊂ G. Then for all f ∈ F

|f ′(a)| =1

2π

∣

∣

∣

∣

∣

∫

C(a,δ)

f(ζ)

(ζ − a)2dζ

∣

∣

∣

∣

∣

≤ 1

δ.

Step 4. There exists a g ∈ F such that g′(a) is maximal.Choose a sequence fj ⊂ F with

f ′j(a) → µ = supf∈F

f ′(a), as j → ∞.

Because F is normal, there exists a subsequence that converges uniformlyon compact subsets to a holomorphic function g on G. Clearly,

g(a) = 0, g′(a) = µ, |g(z)| ≤ 1 on B.

Since g′(a) > 0, we see that g is not constant; Hurwitz’ theorem 1.1.3 showsthat g is injective and the maximum principle shows that |g| < 1 on G.


Step 5. The function g is the Riemann map.We have only to prove that g is surjective. If not, then g(G) is a simply

connected subdomain of B not equal to B and 0 ∈ g(G). By Koebe’s Lemma3.3.5, there exists an injective holomorphic map ϕ : g(G) → B with theproperty that ϕ(0) = 0, ϕ′(0) > 1. Then ϕ g is in F , but (ϕ g)′(a) =ϕ′(0)g′(a) > g′(a). This contradicts the maximality of g′(a). Therefore, g issurjective and the proof is complete.

Corollary 3.4.2 (Riemann mapping theorem). Let G1, G2 be domainsas in the theorem. Let a ∈ G1, b ∈ G2. Then there exists precisly onebijective conformal map (biholomorphic map) f : G1 → G2 with f(a) = b,f ′(a) > 0.

Proof. Combine the Riemann map from G1 to B with the inverse ofthe one from G2 to B.

Domains G1 and G2 that admit a map as in the Corollary are calledbiholomorphically, or conformally, equivalent.

3.5. Boundary behavior of the Riemann map

We will adapt Schwarz’ reflection principle to show that biholomorphicmaps between domains with real analytic boundary extend analytically overthe boundary. Schwarz’ method also gives us some explicit maps from a halfspace to a polygon. We start with Caratheodory’s theorem, which deals withcontinuous extensions of biholomorphic maps.

To fix conventions we recall the following. Let I = [a, b] be a compactinterval in R. A continuous map γ : I → C is called a curve in C. A curveis called a closed if γ(a) = γ(b). A curve (closed) curve is called simple ifγ(x) = γ(y) ⇒ x = y ∨ x, y = a, b. A simple curve is called an arc, asimple closed curve is called a Jordan curve.

Often curves will be identified with their image. A closed curve may berepresented as a continuous map γ : T → C.

Definition 3.5.1. A compact set K in a metric space is called locallyconnected at a ∈ K if for every sequence zn ⊂ K that converges to a thereexist Ln ⊂ K such that a, zn ∈ Ln and if n tends to ∞

• diam(Ln) → 0;• Ln is a connected set.

Theorem 3.5.2 (Caratheodory). Let D be a simply connected domainin C∗ with more than one boundary point. Then the boundary ∂D is lo-cally connected if and only if the inverse Riemann map f : B → D extendscontinuously to B.

Proof. First suppose that f extends continuously to ∂B. Then f(B)is compact as the image of a compact set under f . Hence D ⊂ f(B). Alsof(B) ⊂ D, so f(B) = D. Let a ∈ ∂D and let zn ⊂ ∂D converge to a.Choose ζn ∈ f−1(zn). Let X ⊂ T be the set of limit points of ζn. Thenf(X) = a.

The smallest of the two arcs in T that connect two points α, β ∈ T

will be denoted by [α, β]. The point in X closest to ζn is denoted by ηn.

3.5. BOUNDARY BEHAVIOR OF THE RIEMANN MAP 23

Then Ln = f([ζn, ηn]) is a connected subset of ∂D. Because f is uniformlycontinuous on T, we see that diam(Ln) → 0 as n tends to ∞.

Next suppose that ∂D is locally connected. We can assume that ∞ =f(0) ∈ D. Fix ζ0 ∈ T and let γρ = B ∩ C(ζ0, ρ), (ρ > 0). The length Lρ ofthe curve Γρ = f(γρ) is given by

Lρ =

∫

γρ

|f ′(ζ)| |dζ|.

By the Cauchy Schwarz inequality we obtain

L2ρ ≤

∫

γρ

12 |dζ|∫

γρ

|f ′(ζ)|2 |dζ| ≤ πρ

∫

γρ

|f ′(ζ)|2 |dζ|.

Therefore, using that on γρ we have |dζ| = ρdθ, (ζ = ρeiθ),∫ δ

0

L2ρ

ρdρ < π

∫∫

B∩B(ζ0,δ)|f ′(ζ)|2 dξdη = π area[f(B ∩B(ζ0, δ))] <∞.

Hence there is a sequence ρn → 0 such that Lρn → 0. The (open) curvesΓn = Γρn have finite length. Thus if ζ ∈ γn tends to an endpoint of γn in T,f(ζ) can have at most one limit point. In other words, f can be continuouslyextended to γn and Γn has endpoints an and bn on ∂D. Because Lρn → 0we have |an − bn| → 0. Taking a subsequence if necessary, we can assumethat an, hence also bn, tend to w0 ∈ ∂D.

By local connectivity there exist connected sets Ln ⊂ ∂Dn with an, bn,and w0 ∈ Ln, and diam(Ln) → 0. Because f is a homeomorphism on B andγn separates B into two connected components, the curve Γn separates Dinto two connected components. One of them contains ∞ = f(0), the otheris bounded and is denoted by Dn. In fact, Dn is contained in a boundedcomponent of C∗ \ (Γn ∪ Ln). Otherwise there would be a simple arc fromz0 ∈ Dn to ∞ in C∗ \(Γn∪Ln), followed by an arc from ∞ to z0 in C∗ \(Ln)that crosses Γn exactly once, together forming a simply closed Jordan curvein C∗ \ Ln that would separate an from bn. This would contradict theconnectivity of Ln. Now diam(Γn ∪Ln) → 0 and hence also diam(Dn) → 0.The compacts Dn are nested, and have w0 as intersection. With f(ζ0) = w0,this implies that f is continuous at ζ0.

Observe that we did not use much of the properties of B. In fact, we onlyused that at a point ζ0 ∈ T the set C(ζ0, ρ)∩B is an arc for 0 < ρ < ρ0. Thisis clearly the case for domains bounded by a Jordan curve that is piecewiseC1. Hence we have

Corollary 3.5.3 (of the proof). Let G be a simply connected domainbounded by a piecewise C1 Jordan curve, and D a simply connected domainwith at least two boundary points. Then the Riemann map f : G → Dextends continuously to G if and only if ∂D is locally connected.

Definition 3.5.4. A Jordan domain is a domain bounded by a Jordancurve.

A Jordan domain is necessarily simply connected. The following resultwill proved in full generality in a later chapter. Now we have to restrictourselves to the case of piecewise C1-boundary.


Theorem 3.5.5. Let G1 and G2 be simply connected Jordan domains.Then the Riemann map f : G1 → G2 extends to a homeomorphism f : G1 →G2.

Proof. (For piecewise C1-boundary.) From Corollary 3.5.3 we see thatboth f and f−1 extend continuously to the boundary of their domain. Itfollows that f f−1 is continuous up to the boundary of G2. Since this mapis the identity on G2, it is the identity on G2.

A weak converse is

Proposition 3.5.6. Let G1, G2 be bounded Jordan domains and let fbe holomorphic on G1 that extends continuously to a homeomorphism from∂G1 to ∂G2. Then f is a biholomorphism from G1 to G2.

Proof. Let w ∈ G2. The increase in arg(f(ζ) − w) when ζ traverses∂G1 equals 2π. The Argument Principle yields that f assumes the valuew exactly once. If w /∈ G2 then the increase in the argument of f(ζ) − wequals 0, and the value w is not assumed.

Next we will study Schwarz’ reflection.

Definition 3.5.7. A curve γ : I → C is called real analytic if γ is therestriction of a holomorphic function f , defined in a neighborhood of I withthe property that γ′ 6= 0 on I.

Theorem 3.5.8 (Schwarz’ reflection). Let D be a simply connected Jor-dan domain in C that is bounded by a simple closed real analytic curve J ,(so that J is given by as an injective image of the circle under a holomor-phic map γ defined in a neighborhood of the circle). Then the Riemann map

f : D → B admits an analytic extension f defined on a neighborhood of J .

Proof. The complement of D has at least two points, and so the Rie-mann map exists. By Theorem 3.5.5 the Riemann map extends continuouslyto J , and f(J) ⊂ T. Applying a Mobius transformation, we can replace B

by the upper half plane H, so that the image of J under the new map isR ∪ ∞. Moreover we can also replace the circle on which γ is defined byR ∪ ∞.

Now let g : U → C be an analytic extension of γ to a neighborhood ofR. Since g is locally a homeomorphism, g−1(D) will be either in the upperhalf plane or in the lower half plane. We assume that it is in the upper halfplane. Then h = f g is a holomorphic map from a domain U ⊂ H that is aone sided neighborhood of R, to H; h is continuous up to R and h(R) ⊂ R.

We extend h to U = z : z ∈ U by h(z) = h(z). This is holomorphicon U ∪ U and continuous on U ∪ ((−ε, ε) + I)∪ U . Morera’s theorem showsthat h is holomorphic on this set. Since g is locally biholomorphic, we canextend f by setting f = h g−1 in a neighborhood of J .

Remark 3.5.9. We have phrased this theorem in the setting of conformalmappings. But in fact we have shown

Let Ji be real analytic curves in the boundary of domains Di (i = 1, 2).If f : D1 → D2 is analytic and admits a continuous extension f that mapsJ1 into J2 then f extends analytically over J1.

3.5. BOUNDARY BEHAVIOR OF THE RIEMANN MAP 25

Theorem 3.5.10 (Schwarz-Christoffel). Let a1 < a2 < . . . < an be real,0 < βk < 1 and

∑nk=1 βk = 2. Then the map

(3.5.1) w = f(z) = A

∫ z

0

dζ

(ζ − a1)β1 · · · (ζ − an)βn+B

is an injective conformal mapping from the upper half plane H onto an n-gonG with outer angles βkπ.

The functions (z − ak)βk denote on H the principal values of the corre-

sponding multiple valued functions. However, it is most convenient to viewthem as defined and holomorphic on C\ak−is, s > 0. Then the integrandextends holomorphically over every segment in R \ a1, . . . ak. Integrationmay be performed over segments parallel to the coordinate axes.

G

β3π

b3β4π

b4

β1πb1

β2π

b2

a1 a2 a3 a4

H

Figure 1. Schwarz-Christoffel map

Proof. Because all βk < 1, the integrand in (3.5.1) is in L1(R) for realζ and limz→t f(z) exists for all t ∈ R, and the limit equals the value of(3.5.1) at t ∈ R viewed as a real integral. In particular it follows that fextends continuously to R. Because the integrand extends holomorphicallyover every segment in R \ a1, . . . ak, the same is true for f .

On each of the intervals (−∞, a1), (a1, a2), . . . the argument of (z−ak)βk

is constant, hence so is the argument of

f ′(z) = A1

(z − a1)β1 · · · (z − an)βn.

These intervals are therefore mapped onto straight line segments in C. Ifz passes through ak from left to right only the argument of (z − ak)

−βk

increases from −πβk to 0. Hence the argument of f ′ increases at ak withπβk. The image of (ak, ak+1) is rotated over πβk compared to the image of(ak−1, ak). Because

∑

βk = 2 the image of (−∞, a1) and of (an,∞) havethe same direction.

At this point we conclude that the image of R is a piecewise linear arcwith an increasing change of direction βkπ at the nodes f(ak). Either it isa simple closed curve, so that it bounds an n-gon; or it may have one point


of self intersection, or it may be an arc. The latter two possibilities implythat f(−∞) 6= f(∞). However, this is ruled out because

f(∞) − f(−∞) = limA→∞

∫ A

−Af ′(ζ) dζ = 0,

which is seen by replacing the contour in the upper half space with half acircle with radius A and observing that the integral over this path tends to0 as A→ ∞.

Hence the boundary map is a homeomorphism from R∪∞ to the n-gonand Proposition 3.5.6 shows that f is conformal.

3.6. Exercises.

3.6.1. Complete the proof of Proposition 3.3.3.

3.6.2. Prove that there does not exist an injective conformal mappingfrom B onto C.

3.6.3. Show that B is not conformally equivalent with the annulusA(0, 1, 2). Show that neither of these is conformally equivalent with B\0.

3.6.4. Determine Aut(H), where H = z : Im z > 0.

3.6.5. Show that

w = 12(z +

1

z)

defines a Riemann map from C∗ \B (and also from B) to C∗ \ [−1, 1]. Whatis the image of circles C(0, r) and half-lines arg z = θ?

3.6.6. Let 0 < a < 1 and let G(r) (0 < r < 1 − a) be the domain

B \B(a, r).

(1) Find the Riemann map G(1 − a) → B.(2) Find the Riemann map G(r) → A(0, ρ, 1), where ρ has to be deter-

mined, depending on r.

3.6.7. Consider a triangle in the hyperbolic geometry (i.e., its edgesare hyperbolic geodesics). Show that the sum of its angles is < π.

3.6.8. Let 0 < k < 1. Show that

f(z) =

∫ z

0

dζ

[(1 − ζ2)(1 − k2ζ2)]1/2

defines a Riemann map from H to a rectangle with vertices f(±1), f(±1/k).[The function f is called an elliptic integral.] What can you say about theposition of the rectangle?

Let g = f−1 be the inverse map. Apply Schwarz reflection to extendg analytically to neighboring rectangles. Continue this reflection proces.Observe that in this way g can be extended to a meromorphic function onthe C. Show that g is doubly periodic and compute the periods. We call gan elliptic function .

3.6. EXERCISES. 27

3.6.9. Treat the integral

f(z) =

∫ z

0

dζ

(1 − ζ2)1/2

and its inverse similarly as f and g in the previous exercise.

3.6.10. Suppose that f maps the rectangle R1 with vertices 0, a, a+ib, ib(a, b > 0) biholomorphically to the rectangle R2 with vertices 0, a′, a′ +ib′, ib′ (a′, b′ > 0) in such a way that the vertices of R1 are mapped to thecorresponding ones of R2. Show that b/a = b′/a′. [ Extend f to a conformalmap of C to itself.]

3.6.11. Prove that A(0, r1, R1) is conformally equivalent withA(0, r2, R2)if and only if R1/r1 = R2/r2. [Use Remark 3.5.9.]

C∗

CnC

C3

C4

C1

C2

Figure 2. Exercise 3.6.12

3.6.12. Let D be a domain obtained by deleting a (small) closed ballwith boundary C∗ from a bigger ball with boundary C. The boundary ofD consists of the two disjoint circles C, C∗ Suppose that there exists afinite sequence of circles C1, . . . , Cn with the following property. Each Cj istangent to C and C∗, as well as to Cj−1 and Cj+1 (with the convention thatC0 = Cn and Cn+1 = C1). Show that every circle in D that is tangent to Cand C∗ is the starting point of a sequence of circles with this property.

CHAPTER 4

Complex Dynamics

Complex dynamics deals with the dynamical systems that occur whenholomorphic functions are iterated. One studies the possible orbits of apoint, invariant domains, fixed points, and much more. It is a subjectin itself, even if one would restrict oneself to very simple maps, such aspolynomials of degree 2.

In the present chapter we will only see a glimpse of the subject, but itwill be enough to understand that much from classical function theory isindispensible to complex dynamics, Moreover, function theory allows us togive the proper definition of famous concepts like Mandelbrot set or Juliaset.

Excellent general introductions to complex dynamics in one variable arethe books [28, 11].

4.1. Iteration of holomorphic maps

Let f : Ω → Ω be a holomorphic function that maps a domain Ω ⊂ C∗

to itself. Write fn = f · · · f for the n times iteration of f with theunderstanding that f0(z) = z. We wish to study the iterated system fn :n ∈ N and its orbits fn(z). To find out what one can expect, we presentsome examples.

Example 4.1.1. For Ω we take the Riemann sphere C∗. Let f(z) = αz.(α ∈ C). Then fn(z) = αnz. We observe

(1) If α = 1 all maps are the identity, an uninteresting case.(2) If |α| < 1 there are two fixed points 0 and ∞. If z ∈ C \ 0 we see

that limn→∞ fn(z) = 0. Observe that the family fn is normalon C, . The point 0 is an attracting fixed point and C is its basin ofattraction. On the other hand, ∞ is a repelling fixed point; pointsclose to ∞ will move away from ∞ if f is iterated.

(3) If |α| > 1 then 0 is repelling, ∞ attracting and the basin of attrac-tion is ∞ ∪ |z| > 0.

(4) If |α| = 1 and αn 6= 1 for every n, then f is an irrational rotation.0 and ∞ are fixed points, which in this case are called irrationallyneutral.

(5) If αn = 1 for some n, then f is a periodic map. 0 and ∞ are againfixed points, which in this case are called rationally neutral.

Example 4.1.2. Again Ω = C∗. Let f(z) = z2. Then fn(z) = z2n. If

z ∈ D then limn→∞ fn(z) = 0. 0 is an attracting fixed point with basinof attraction D. If z ∈ C∗ \ D, then limn→∞ fn(z) = ∞. Thus ∞ isan attracting fixed point with basin of attraction C∗ \ D. On C(0, 1) all

29

30 4. COMPLEX DYNAMICS

kind of things happen: 1 is a fixed point; ei2π/3 is a periodic point withorbit e2πi/3, e4πi/3 of period 2. Other points on the unit circle can bepre-periodic. (like i 7→ −1 7→ 1 7→ 1 . . . ) or may have a dense orbit.

The definition that goes with these examples is

Definition 4.1.3. Let f : Ω → Ω be a holomorphic function that mapsa domain Ω ⊂ C∗ to itself.

(1) a ∈ Ω is called a fixed point for f if f(a) = a; λ = f ′(a) is calledthe multiplier of a

(2) A fixed point is• Attracting if λ satisfies 0 < |λ| < 1; Superattracting if λ = 0.• Repelling if |λ| > 1• Rationally Neutral if for some n ∈ N λn = 1.• Irrationally Neutral if |λ| = 1 and for no n ∈ N λn = 1.

(3) a ∈ Ω is called a periodic point for f if there exists n ∈ N so that ais a fixed point for fn. The smallest possible n is called the period.Periodic points can be attracting, repelling, etc. according to thebehavior of the fixed point for fn. Fixed points are periodic withperiod 1.

(4) a ∈ Ω is called a preperiodic point for f if for some j ∈ N f j(a) isa periodic point.

(5) The basin of attraction of a fixed point a isAa = z : limn→∞ fn(z) =a

(6) The Fatou set F of f is the open subset of Ω on which the familyfn is normal.

(7) The Julia set J of f is the complement of the Fatou set.

In Example 4.1.2 the Julia set is C(0, 1) and the Fatou set consists ofA0 ∪ A∞, the two basins of attraction of the superattracting fixed points 0and ∞ respectively; 1 is a repelling fixed point. In this example the dynamicson F is very simple. All points in the basin of attraction tend to the fixedpoint. But on J the dynamics is more complicated: Periodic points aredense in J , non periodic points are also dense in J . This difference inbehavior of the dynamics on J and F occurs always.

It is fairly straightforward that the Fatou set of R and of Rk (k ∈ N

fixed) are equal, and hence so are their Julia sets. Cf. Exercise 4.5.2.

To study the behavior of the dynamical system generated by f at somefixed point, it is natural to try to find a favourable coordinate system, inwhich the behavior is easily understood.

Definition 4.1.4. Let U, V be open in C; f : U → U , g : V → Vholomorphic. Then f is called conformally conjugate to g if there exists abiholomorphic map ϕ : U → V such that

f = ϕ−1 g ϕ.Remark 4.1.5. It is easily seen that if f is conformally conjugate to g,

then the image under ϕ of fixed points, periodic points, basins of attraction,Fatou - and Julia sets for f are likewise for g. Moreover, multipliers andperiods are kept, cf. Exercise 4.5.3

4.2. CLASSIFICATION OF FIXED POINTS 31

Example 4.1.6. Consider the map f : z → z2 − 2. It has a superat-tracting fixed point at ∞. The map ϕ : z 7→ z + 1/z is biholomorphic fromU = |z| > 1 to V = C∗ \ [−2, 2], and satisfies ϕ2(z) − 2 = ϕ(z2), hencez2 = ϕ−1 f ϕ(z), i.e. f is conformally conjugate to z2. As ϕ is defined onthe whole basin of attraction of ∞ for z2, its image C∗ \ [−2, 2] will be thebasin of attraction of ∞ for f .

4.2. Classification of fixed points

Let f be holomorphic from Ω → Ω and let a be a fixed point of fwith multiplier λ. In this section we will see that if |λ| 6= 1 the map f isconformally conjugate to λ(z − a) in a neighborhood of a. The first resultin this direction is due to Schroder, [43], who introduced the functionalequation (4.2.1). It dates back to 1871 and is the oldest result in complexdynamics.

Theorem 4.2.1 (Kœnigs, [21]). Let U be a domain in C∗, a ∈ U andf : U → U holomorphic with f(a) = a. Suppose that λ = f ′(a) satisfies0 < |λ| < 1. Then f is in a neighborhood of a conformally conjugate to thelinear map z 7→ λz.

Proof. We can assume a = 0. Then we have to find a biholomorphicmap ϕ defined on a neighborhood V of 0 that satisfies the functional equation

(4.2.1) ϕ f = λϕ,

We will try to solve this by iteration via

(4.2.2)1

λϕn f = ϕn+1, ϕ0(z) = z,

and wish to prove that the sequence (ϕn) converges. The limit ϕ = limϕnwill necessarily solve (4.2.1). Observe that from (4.2.2) we find

(4.2.3) ϕn =1

λnfn.

We expand f at 0:

f(z) = λz + a2z2 + · · ·

and find |f(z) − λz| < C|z|2 for |z| < δ sufficiently small. Also, taking δeven smaller if necessary,

(4.2.4) |f(z)| < |λz| + C|z|2 ≤ (|λ| + Cδ)|z| < |z| for |z| < δ.

Hence

(4.2.5) |fn(z)| < (|λ| + Cδ)n|z| for |z| < δ.

Choosing δ even smaller, if necessary, we can achieve that (|λ|+Cδ)2|λ| = ρ < 1.

Now for |z| < δ

|ϕn+1(z) − ϕn(z)| =

∣

∣

∣

∣

(1/λ)f(fn(z)) − fn(z)

λn

∣

∣

∣

∣

≤M1

|λn| |fn(z)|2 ≤M

(|λ| + Cδ)2n|z||λ|n ≤ ρn

(4.2.6)


for some constant M > 0. We conclude from (4.2.6) that (with m < n)

|ϕm(z) − ϕn(z)| < Cn∑

j=m+1

ρj ≤ Cρm

1 − ρ< ε

if m,n are sufficiently large. The sequence (ϕn) thus converges uniformly on|z| < δ to a holomorphic function ϕ. Because ϕ′(0) = 1, ϕ is biholomorphicin a neighborhood of 0.

Corollary 4.2.2. Suppose now that f has a repelling fixed point at awith multiplier |λ| > 1. Then f is conformally conjugate to z 7→ λz.

Proof. Again we can assume that a = 0. As |λ| > 1 the map f islocally invertible at 0 with inverse g, and g′(0) = 1/λ. So g is conformallyconjugate to z/λ, i.e. ∃ϕ with

ϕ g =1

λϕ.

Hence

ϕ f = λϕ.

For superattracting fixed points Bottcher [9] found the following result.

Theorem 4.2.3. Let f(z) = z0+ap(z−z0)p+H.O.T be holomorphic in aneighborhood of z0. Then f is conformally conjugate to zp. The conjugatingmap ϕ is unique up to multiplication by a p− 1-th root of unity.

Proof. We can assume z0 = 0. Next, changing coordinates z = a1/pp z,

where any p-th root of ap may be choosen, we can assume that ap = 1. Thedevelopment of f about the fixed point 0 is now

(4.2.7) f(z) = zp + cp+1zp+1 + · · · .

Notice that a p-th root of f exists in a neighborhood of 0, and can be choosenof the form z(1 +O(z)).

We are looking for a solution ϕ of the functional equation

(4.2.8) ϕ f(z) = (ϕ(z))p.

Again we try to find ϕ via iteration as follows. Put ϕ0(z) = z and define forn ≥ 1

(4.2.9) ϕn−1 f(z) = (ϕn(z))p.

We see that ϕ1(z) = f1/p = z(1 +O(z)). More generally

(4.2.10) ϕn(z) = fn(z))1

pn .

We will show that the sequence (ϕn) converges in a neighborhood of 0, Thelimit function ϕ solves (4.2.8). Indeed,

ϕn+1(z)

ϕn(z)=f fn(z)

1pn+1

fn(z)1

pn=

(fn(z)p(1 +O(fn(z))))1

pn+1

fn(z)1

pn

= (1 +O(fn(z)))1

pn+1 = 1 +1

pn+1O(zp

n).

(4.2.11)

4.2. CLASSIFICATION OF FIXED POINTS 33

It follows that

ϕ = limϕn = ϕ1 limN→∞

N∏

1

ϕn+1

ϕn

exists and is a solution of (4.2.8).For the uniqueness up to roots of unity, it suffices to see how zp can be

conjugate to itself, i.e. ϕ(zp) = ϕp(z) for some holomorphic ϕ. This is leftas Exercise 4.5.4.

The situation for neutral fixed points is much less clear. If f has arationally neutral fixed point at a, then for some p ∈ N the map g = fp

will have multiplier 1 at a. Take a = 0 as we may. If the identity wereconjugated to g by ϕ, we would have g = ϕ−1 id ϕ and g would bethe identity! There are, however, conjugations to standard forms, of whichTheorem 4.2.4 below is an example. For irrationally neutral fixed pointswith multiplier λ the situation is even more complicated. Depending on thenumber theoretic properties of λ conjugation to an irrational rotation mayor may not be possible.

Theorem 4.2.4 (Fatou). Let g be holomorphic and given in a neighbor-hood of 0 by

(4.2.12) g(z) = z + az2 + · · · , a 6= 0,

Then g is conformally conjugate to z 7→ z+ 1 (at ∞) and to z 7→ z + z2 (at0).

Proof. Conjugation with z 7→ az reduces the situation to the casea = 1. Next we conjugate with z 7→ −1/z, which moves the fixed point to∞. The expansion of the conjugated function at ∞ is

(4.2.13) f(z) = z + 1 +b

z+ · · · .

We make the following preliminary observations. There exists a constantC > 0 so that Re z > C =⇒ Re f(z) > C, and for Re z > C

(4.2.14) Re fn(z) ≥ Re z + n/2,

and

(4.2.15) n/2 < |fn(z)| < |z| + 2n.

In fact, a more precise estimate is possible and will be needed. One mighthope for an estimate like fn(z) = z + n+O(nε). We have for Re z > C

|fn+1(z) − fn(z) − 1| = |f(fn(z)) − fn(z) − 1|

=

∣

∣

∣

∣

b

fn(z)

∣

∣

∣

∣

+O((fn)−2) =2b

n+O(1/n2).

(4.2.16)

From this we find

fn(z) − n− z = (fn − fn−1) + (fn−1 − fn−2) · · · + (f(z) − z) + z − n− z

=

n∑

j=1

2b

j+O(1/j2) = O(log n) +O(1) = O(log n).

(4.2.17)


A first setup for an iterative process leading to a conjugation ϕ that conju-gates f to z + 1, i.e.

(4.2.18) ϕ f = ϕ+ 1

would be

(4.2.19) ϕn+1 = ϕn f − 1.

With ϕ0(z) = z this leads to

(4.2.20) ϕn = fn − n,

and hence by (4.2.16)

(4.2.21) ϕn+1(z) − ϕn(z) = f(fn(z)) − fn(z) − 1 =2b

n+O(1/n2).

This would at best give ϕj − ϕk ∼ 2b| log k − log j| which is not enough tomake the series uniform convergent. To compensate for b/n (∼ b log(n +1) − b log n) we modify ϕn as follows

(4.2.22) ϕn = fn − n− b log n.

This then leads to the following estimate valid for Re z > C

ϕn+1(z) − ϕn(z) = f(fn(z)) − fn(z) − 1 − b log(1 + 1/n)

=b

fn(z)− b

n+O(

1

n2) = b(

1

n+ z +O(log n) +O(1/n)− 1

n) +O(

1

n2)

= −b2b log n+ z

n2+O(

1

n2).

(4.2.23)

Because∑ logn

n2 converges, the sequence (ϕn)n is uniform Cauchy on com-pact sets. The limit function ϕ = limϕn will satisfy (4.2.18). We have nowconjugated g to z+1, which in particular can be done in case g(z) = z+ z2,and the theorem is proven.

Remark 4.2.5. A similar result holds for maps of the form

f(z) = z + a1z1+p + a2z

2+p + h.o.t.

Again one can assume a1 = 1. Introducing g(z) = f(z)p and w = zp onsuitable sectors, we find

g(w) = w + pw2 +O(w2+1/p).

The method and the estimates as in the proof of Theorem 4.2.4 are possible,but much more complicated because of the fractional power. See [11], alsofor different ways of proving conjugacies.

In all cases the conjugations ϕ were defined in a small neighborhood ofthe fixed point. How big is the domain where conjugation is possible?

Theorem 4.2.6. Let a be an attractive fixed point with multiplier λ,0 < |λ| < 1. of f , and let ϕ be the conjugation to λz. The Schroder equation(4.2.1) defines an analytic continuation of ϕ to the basin of attraction Aa.

4.3. JULIA AND FATOU SETS 35

Proof. We can assume that a = 0. The conjugation ϕ then satisfiesfor every n the functional equation

(4.2.24) ϕ(z) =1

λnϕ(fn(z)),

by repeated application of (4.2.1). Every compact K ⊂ Aa is mappedinto arbitrarily small neighborhoods of a by fn if n is choosen large enough.Hence (4.2.24) defines an extension of ϕ toK and we conclude that ϕ extendsto Aa.

4.3. Julia and Fatou sets

We will discuss some properties of the Julia and Fatou sets of rationalfunctions and polynomials, i.e. of holomorphic maps from C∗ to itself. Thedegree of a rational function R = P/Q with P,Q polynomials without com-mon factor, extends the notion of degree of a polynomial. It is defined byd(R) = maxd(P ), d(Q). A rational function of degree d gives a d-foldbranched cover of C∗. Every point in C∗ has d-preimages under R countedwith multiplicity. The multiplicity at z is defined as the smallest k ≥ 1 suchthat R(k)(z) 6= 0. A points with multiplicity > 1 is called a critical point.In this section R will always be a rational functions of degree > 1. Observethat d(Rn) = d(R)n.

Theorem 4.3.1. The Julia set of a rational function R of degree > 1 isnon-empty.

Proof. Suppose that R has Fatou set C∗, then (Rn)n has a subsequence(Rnj)j that is uniformly convergent in the chordal metric χ to a holomorphicfunction f : C∗ → C∗. Then f is a rational function or a constant. Aconstant c is impossible, as this would mean that χ(Rnj (z), c) < 1/2 for all

z ∈ C∗ if j is sufficiently large, which contradicts the fact that Rnj(C∗) =C∗. If f is rational it has a finite zeroset Zf . Consider U = z ∈ C∗ :χ(f(z), 0) < 1/2. For j > j0 sufficiently large χ(Rnj (z), f(z) < 1/2 so thatχ(Rnj(z), 0) < 3/4 if z ∈ U , and Rnj(z) 6= 0 outside U . It follows that Rnj

and f will be holomorphic on U .By application of Theorem 1.1.3 Rnj will have the same number of zeros

as f , for j > j1 sufficiently large, contradicting #ZRnj tends to ∞.

A set E is completely invariant under f if f(E) ⊂ E and f(C∗ \ E) ⊂C∗ \ E. Because R is surjective, E is completely invariant under R if andonly if R−1(E) = E.

Theorem 4.3.2. The Julia set J of R is completely invariant.

Proof. We show that F is completely invariant. Clearly (Rn)n is nor-mal onR−1(F), henceR−1(F) ⊂ F . Also R(F) is open and the family (Rn)nis normal on R(F), since (Rn+1)n is normal on F , therefore R(F) ⊂ F . Nowif z ∈ F , then R(z) ∈ F , so that z ∈ R−1(F). In other words, F ⊂ R−1(F),and F , as well as J are completely invariant.

Montel’s Theorem, cf. Exercise 12.5.7 later on, asserts that a family ofmeromorphic functions on an open set is normal if it omits 3 or more values.


Hence it follows that for z ∈ J and U any open neighborhood of z the familyRn cannot omit a set Ez consisting of more than two values.

Theorem 4.3.3. The set Ez is independent of z. (We will denote it byE) If it consists of one point, then R is conformally conjugate to a polyno-mial (with exceptional set ∞). If it consists of two points, R is conjugate tozm for some m ≥ 2. In both cases E ⊂ F .

Proof. We have C∗ \E ⊂ ∪nRn(U) for every open neighborhood U ofz. Suppose that w ∈ ∪∞

n=1Rn(U) for every U containing z, then we have

R(w) ∈ ∪∞n=1R

n+1(U). Thus w 6∈ E implies R(w) 6∈ E hence R−1(E) ⊂ E.If Ez contains 2 points, we conjugate these to 0 and ∞ and after conjugationwith 1/z we can assume 0 7→ 0 and ∞ 7→ ∞. It follows that R is conjugateto zk for some k ≥ 2. If Ez consists of 1 point only, we can conjugate it to∞ and so R is conjugate to a polynomial.

In both cases it is clear that Ez is independent of z and consists ofsuperattracting fixed points, so Ez belongs to F .

The set E of values that are omitted is called the exceptional set. Thepoints in ∪∞

1 R−n(z) are called the backward iterates of z (under R)

This has interesting consequences for the Julia set.

Theorem 4.3.4. Let R be a rational function. The backward iteratesunder R of a point z ∈ J are dense in J . Any completely invariant subsetof J is dense in J .

Proof. If z 6∈ E, then for every neighborhood Uw of a point w ∈ Jthere exists n such that R−n(z) ∩ Uw 6= ∅. In other words w ∈ ∪nR−n(z).Hence J ⊂ ∪nR−n(z). If z ∈ J the inverse iterates will belong to J and so

J = ∪nR−n(z).In case E is a completely invariant it contains all the backward iterates

of some point in J .

Remark 4.3.5. This theorem provides a method for drawing the Juliaset. Start with a point on J (e.g. a repelling periodic point, which one canshow that always exists) and plot the backward iterates.

Theorem 4.3.6. Let R be a rational function. Then J is perfect, i.e.J is closed and does not contain isolated points.

Proof. Let z0 ∈ J . If z0 is not a periodic point we consider its back-ward orbit X = ∪∞

1 R−j(z0). Because X is dense in J , every neighborhood

of z0 will contain points of X, and these points will not coincide with z0because z0 is not periodic.

If z0 is periodic, we consider an iterate Rm for which z0 is a fixed point.Because the degree of Rm is larger than 1 and z0 /∈ E, there is a pointz1 6= z0 such that Rm(z1) = z0. The inverse orbit of z1 will not contain z0,and the argument can be completed as in the non-periodic case.

4.4. Polynomials of degree 2

Let P denote a polynomial of degree 2. One can show that P is conjugateto z2 + c for some c, cf. Exercise 4.5.9, Thus we will consider polynomials of

4.5. EXERCISES 37

the form Pc(z) = z2 + c only. They all have a superattracting point at ∞and a critical point at 0.

In Section 9.6 we will prove the following result.

Theorem 4.4.1. Let P be a polynomial. The Julia set J of P is con-nected if and only if the forward orbit of every critical point of P is bounded.

This gives rise to the following definition.

Definition 4.4.2. The Mandelbrot set M is the set of all c ∈ C suchthat Pc has a connected Julia set.

To find out if c belongs to M we iterate the finite critical point (0) andsee if the iterates tend to ∞. We leave the proof of the following theoremas an exercise (4.5.10)

Theorem 4.4.3. The Mandelbrot set M is closed in c : |c| ≤ 2. Itconsists of precisely those c with the property that for every n |Pnc (z)| ≤ 2.The intersection of M with the real axis is the interval [−2, 1/4].

The Mandelbrot set is connected and simply connected.

Remark 4.4.4. One can show that for quadratic polynomials the Juliaset is either connected, or totally disconnected.

4.5. Exercises

4.5.1. Show that the connected components of a basin of attractionare simply connected.

4.5.2. Show that for a rational map R the Julia set of R and of Rk areequal.

4.5.3. Prove the claims of Remark 4.1.5.

4.5.4. Provide the proof of the uniqueness up to roots of unity inTheorem 4.2.3. Why is it sufficient to understand conjugacy of zp to itself?

4.5.5. Let 0 be a superattracting fixed point of R and let ϕ conjugateR to zd. Show that ϕ can be analytically extended to (atleast) A0 \X whereX consists of the points in A0 where R′ = 0 and their backward orbits.

4.5.6. Consider the map z + z2 with neutral fixed point 0. The realaxis is mapped into itself. Show that point x > 0 are repelled and point−1 < x < 0 are attracted to 0. We call (the vector) 1 a repelling directionand −1 an attracting direction. Use Theorem 4.2.4 to show that orbits of apoint in A0 will be tangent to the attracting direction. A0 is an attractingpetal.

4.5.7. (Continued) Now consider z + zk+1 at 0. Extend the notion ofattracting and repelling directions. There should be k of both Determinethem and estimate the attracting petals that make up A0.

4.5.8. Let P (z) = −z + z5, This map is best studied via P 2. Oneshould find that P 2 has 8 attracting petals. What does this mean for P .Compare this to a perturbation P (z) = −z + z5 + εz6.


4.5.9. Show that every polynomial of degree 2 is conjugate to a poly-nomial of the form z2 + c.

4.5.10. Prove Theorem 4.4.3 by completing the following outline.

(1) Show that |c| > 2 implies limn→∞ Pnc (0) = ∞.(2) Show that if |c| <≤ 2 and |Pmc (0)| > 2 then limn→∞ Pnc (0) = ∞.(3) Conclude that M = c : ∀m |Pmc (0)| ≤ 2(4) View Pmc as a holomorphic function in the c-variable and use this

to show that M = is simply connected. (Cf. Exercise 4.5.1).(5) Prove the claim of the theorem for real c.

CHAPTER 5

Product representations

A polynomial can be represented in two ways, as a sum of monomials oras a product. Holomorphic functions have representations as power series.This corresponds with writing a polynomial as a sum. In this chapter we willmeet product representations for holomorphic functions. We will first studyinfinite products of complex numbers and of holomorphic functions. We willprove Weierstrass’ theorem stating that every sequence in C without limitpoint is the set of zeros of an entire function. The theorem holds true muchmore generally, though. The chapter ends with the Mittag Leffler theoremin C.

5.1. Products

Let un ∈ C and pn =∏nj=1 un. In analogy with the definition of infinite

sums we could define the infinite product by

(5.1.1) p = limn→∞

pn =∞∏

j=1

un,

if the limit of the pn exists. However, this is not entirely satisfactory. Wewould like to keep the property that a product equals 0 if and only if one ofits factors equals 0, but with un = 1/2 we get lim pn = 0.

We therefore give the following definition.

Definition 5.1.1. Let un ∈ C. The product (5.1.1) is called convergentif there exists k0 such that

• uk 6= 0 for k ≥ k0,• limn→∞ uk0uk0+1 · · · un = p∗ exists and is not equal to 0.

Lemma 5.1.2. The product (5.1.1) converges if and only if for everyε > 0 there exists k0 such that for k0 ≤ m < n

(5.1.2) |um+1 · · · un − 1| < ε.

In particular if (5.1.1) converges, then limm→∞ um = 1.

Proof. If the product (5.1.1) converges, then

|um+1 · · · un − 1| =

∣

∣

∣

∣

uk0 · · · unuk0 · · · um

− 1

∣

∣

∣

∣

→∣

∣

∣

∣

p∗

p∗− 1

∣

∣

∣

∣

= 0 as m,n→ ∞.

For the other direction we can assume all uj 6= 0. Now first apply (5.1.2)with ε = 1/2, fixed m, and n variable and set M = |u0 . . . um−1|. Then forall n

|pn| = M |um0 . . . un| ∈ [M

2, 2M ].

39

40 5. PRODUCT REPRESENTATIONS

Another application of (5.1.2) gives for k < l sufficiently large

|pk − pl| ≤ |pk||1 − uk−1 . . . ul| < 2Mε,

and pn converges. The remaining part is immediate.

Because their factors tend to 1, we will write our products in the form

(5.1.3)

∞∏

1

(1 + vk).

Convergence will depend on how rapidly vk tend to 0.

Lemma 5.1.3. Let vk > 0. Then

(5.1.4)

∞∏

1

(1 + vk) converges ⇐⇒∞∑

1

vk converges.

If 0 < vk < 1 then

(5.1.5)

∞∏

1

(1 − vk) converges ⇐⇒∞∑

1

vk converges.

Proof. We have 1 + v < ev. HenceN∏

1

(1 + vk) ≤ ePN

1 vk

and convergence of∑∞

1 vk implies converges of the product in (5.1.4). Onthe other hand, convergence of the product in (5.1.4) implies that vk → 0,hence, assuming now that v is positive, there exists k0 such that 1 + vk >evk/2, for k > k0. We can assume k0 = 1. Then

N∏

1

(1 + vk) ≥ ePN

1 vk/2

and (5.1.4) has been proved. We leave (5.1.5) to the reader.

Example 5.1.4.∞∏

1

(1 +1

k2) converges,

∞∏

1

(1 +1

k) diverges to ∞,

∞∏

1

(1 − 1

k) diverges to 0.

Definition 5.1.5. The product (5.1.3) is called absolutely convergent if∞∏

1

(1 + |vk|)

converges. (Or by Lemma 5.1.3, if∑ |vk| converges.)

Lemma 5.1.6. An absolutely convergent product is convergent; conver-gence and limit are independent of the order of the factors.

Proof. Exercise 5.5.1

5.2. PRODUCTS OF FUNCTIONS 41

5.2. Products of functions

Definition 5.2.1. Let vk be a sequence of bounded functions on a setE ⊂ C. The product

(5.2.1)∞∏

1

(1 + vk(z))

converges uniformly on E if the product (5.2.1) converges for each z ∈ Eand the sequence of partial products converges uniformly on E.

Theorem 5.2.2. Suppose that∑∞

1 |vk(z)| converges uniformly on E ⊂C, then (5.2.1) converges uniformly on E.

Proof. Because vk(z) tends to 0 uniformly and are bounded, they arein fact uniformly bounded on E. Moreover, there exists k0 such that fork ≥ k0 the factors 1 + vk(z) 6= 0. We may assume that k0 = 1. Now weestimate, by expanding the product,

(5.2.2)

∣

∣

∣

∣

pnpm

− 1

∣

∣

∣

∣

= |(1 + vm+1) · · · (1 + vn) − 1|

≤ (1 + |vm+1|) · · · (1 + |vn|) − 1 ≤ ePn

m+1 |vk| − 1.

Here vk = vk(z) and pk = pk(z). For m,n→ ∞ we see that (5.2.2) tends to0. Taking m = 0 in (5.2.2) we see that for a suitable constant M

|pn(z)| < M on E,

independently of z ∈ E and n = 1, 2, . . ..Let ε > 0. Multiplying (5.2.2) through with pm we obtain

|pn − pm| < ε|pm| < εM,

uniformly on E, for n and m sufficiently large.

Theorem 5.2.3. Let vk be holomorphic on an open set D ⊂ C. Supposethat

∑ |vk(z)| converges uniformly on compact sets in D. Then the product(5.2.1) represents a holomorphic function on D. Its zeros are precisely thezeros of the factors of (5.2.1), counting multiplicities. Outside the zeros ofthe product, the following formula for logarithmic differentiation holds.

(5.2.3) (log p)′ =p′(z)p(z)

=∞∑

k=1

v′k(z)

1 + vk(z).

Proof. The partial products pn of (5.2.1) converge uniformly on com-pact sets in D, hence the limit function p is holomorphic. Morover, p hasa zero of order l if and only if the partial products have such a zero from

a large enough index on. By Weierstrass’ theorem the derivatives p(k)n also

converge uniformly on compact sets to p(k). For the logarithmic derivativewe observe that

p′

p= lim

n→∞p′npn

= limn→∞

n∑

k=1

v′k(z)

1 + vk(z),

uniformly on compact sets.


Example 5.2.4. The product

p(z) =

∞∏

k=1

(

1 − z2

k2

)

converges uniformly on compact sets in C. Thus p(z) represents an entirefunction (holomorphic function on all of C). Its zeros are the non-zerointegers. The function

(5.2.4) zp(z) = z∞∏

k=1

(

1 − z2

k2

)

is also entire and Z is its zero set. We compare (5.2.4) with sin(πz) which hasthe same zeros with the same multiplicity. The quotient q(z) = sin(πz)/(zp(z))

is a zero free entire function, hence it is of the form eg(z). We have found

(5.2.5) sin(πz) = eg(z)zp(z).

We next try to determine the entire function g. Logarithmic differentiationof (5.2.5) yields

(5.2.6) π cot(πz) = g′(z) +1

z+p′(z)p(z)

= g′(z) +1

z+

∞∑

k=1

(

1

z − k+

1

z + k

)

,

another ordinary differentiation leads to

(5.2.7)−π2

sin2(πz)= g′′(z) − 1

z2−

∞∑

k=1

(

1

(z − k)2+

1

(z + k)2

)

=

g′′(z) −∞∑

j=−∞

1

(z − j)2.

The final equality (5.2.7) is justified by uniform convergence. As sin2(πz)has period 1 and the final sum in (5.2.7) has also period 1, the same is truefor g′′. From (5.2.7) we also see that for 0 ≤ x ≤ 1 lim|y|→∞ g′′(x+ iy) = 0.

Thus g′′ is a bounded entire function, and hence a constant that must beequal to 0. It follows that g′ is constant. But from (5.2.6) we see thatg′(z) = −g′(−z), so g′ ≡ 0 and g itself is a constant g(0). Finally

eg(0) = limz→0

sin(πz)

zp(z)= π.

We have found the product representation for the sine:

sin(πz) = πz∞∏

k=1

(

1 − z2

k2

)

,(5.2.8)

and the principal part expansions

(5.2.9)

π cot(πz) =1

z+

∞∑

k=1

(

1

z − k+

1

z + k

)

,−π2

sin2(πz)=

∞∑

j=−∞

1

(z − j)2.

(5.2.10)

5.3. WEIERSTRASS THEORY 43

5.3. Weierstrass theory

We now plan to prove that there exist entire functions with a prescribedset of zeros in C. Let ak be a sequence in C \ 0 without finite limitpoint, so ak → ∞. There may be finite repetitions among the ak.

If there are only finitely many ak then both

n∏

k=1

(z − ak) and

n∏

k=1

(1 − z

ak)

will be polynomials that vanish exactly in ak. If there are infinitely many,the first products will not converge but the second may yield an entirefunction, namely if

∑ 1|ak| converges.

Example 5.3.1. The product

∞∏

k=1

(1 − z

k2)

converges uniformly on compact sets in C and represents an entire function

that vanishes exactly if z = k2, k = 1, 2, . . . (in fact it equals sinπ√z

π√z

). The

product∞∏

k=1

(1 − z

k)

does not converge to a holomorphic function. The factors tend to 1, but tooslowly.

Weierstrass’ idea is to modify the factors with a non-zero factor, to havethem approach 1 fast enough. For |z| < k we can expand

1 − z

k= expp.v. log(1 − z/k) = exp−z

k− 1

2

z2

k2− . . .,(5.3.1)

so that for |z| ≤ R and k → ∞

(1 − z

k)ez/k = exp− 1

2

z2

k2− . . . ≈ 1 − 1

2

z2

k2.(5.3.2)

Apparently the factors in the adapted product

∞∏

k=1

(1 − z

k)ez/k

converge rapidly to 1 and the product represents an entire function withzero set 1, 2, 3, . . .. Similarly, the product

z∞∏

k=1

(1 − z

k)ez/k

∞∏

k=1

(1 +z

k)e−z/k

represents an entire function with zero set equal to Z. Because

(1 − z

k)ez/k(1 +

z

k)e−z/k = 1 − z2

k2,

this function equals sin(πz)/π.


Definition 5.3.2. The k-th elementary factor of Weierstrass Ek(u) isdefined by

E0(u) = 1 − u,(5.3.3)

Ek(u) = (1 − u) exp

u+1

2u2 + · · · + 1

kuk

, k = 1, 2, . . . .(5.3.4)

For small |u| < 1 we can write

(5.3.5) Ek(u) = exp(log(1 − u)) exp

u+1

2u2 + · · · + 1

kuk

= 1 − 1

k + 1uk+1 + H.O.T. .

In fact we have

Lemma 5.3.3. For |u| < 1/2 and k ≥ 0 the following holds

|Ek(u) − 1| ≤ 2|u|k+1.

Proof. We deal with the case k ≥ 1. Then for |u| < 1 we find from(5.3.5) that Ek(u) = exp(gk(u)), where gk is the k-tail of the series forp.v. log(1 − u). For |u| < 1/2 we have

(5.3.6) |gk(u)| ≤∞∑

j=k+1

|u|jj

≤ |u|k+1

k + 1

(

1

1 − |u|

)

≤ 2|u|k+1

k + 1.

We use |ez − 1| ≤ 2|z| if |z| < 1/2. Then

|Ek(u) − 1| ≤ 4|u|k+1

k + 1≤ 2|u|k+1.

Theorem 5.3.4 (Weierstrass on C). Let f be an entire function with azero at 0 of multiplicity m ≥ 0 and further zeros a1, a2, . . . (finite multiplic-ities may occur) then there exists an entire function g such that f has theproduct representation

(5.3.7) f(z) = eg(z)zm∞∏

k=1

Ek

(

z

ak

)

.

If the product is infinite it converges uniformly on compact sets in C. More-over for every sequence ak, ak 6= 0 without finite limit point, the productin (5.3.7) represents an entire function with a zero at 0 of order m and zerosexactly in ak (counting the multiplicities).

Proof. We will prove that for every sequence ak ⊂ C \ 0 withoutlimit point the product in (5.3.7) converges uniformly on B(0, R), R > 0.Let

vk(z) = Ek

(

z

ak

)

− 1.

By Lemma 5.3.3

(5.3.8) |vk(z)| ≤ 2

∣

∣

∣

∣

z

ak

∣

∣

∣

∣

k+1

.

5.3. WEIERSTRASS THEORY 45

Now we have |z| < R and there exists k0 with |ak| > 2R if k > k0. Then|vk(z)| < ( 1

2)k and

∑ |vk(z)| is uniformly convergent on B(0, R). By Theo-rem 5.2.2 the product

p(z) =∞∏

k=1

Ek

(

z

ak

)

is uniformly convergent on every disc B(0, R). According to Theorem 5.2.3p(z) represents an entire function with zeros ak. Thus the second partof the theorem has been proved. For the first part, we have that zmp(z)is an entire function with the same zeros as f (counting multiplicities).Then f(z)/(zmp(z)) = H(z) is a zero-free entire function, hence a branchof log(H(z)) can be holomorphically defined on C by

g(z) =

∫ z

0

H ′(ζ)H(ζ)

dζ + p.v. logH(0).

This proves (5.3.7).

Remark 5.3.5. It may not be necessary to have the order of the Weier-strass factors increase; this depends on the rate of growth of the zeros ak.Example 5.3.1 shows that sometimes order 0 or 1 suffices. In fact we have

Proposition 5.3.6. Suppose that

(5.3.9)

∞∑

1

1

|ak|s+1

converges. Then∞∏

k=1

Es

(

z

ak

)

converges to an entire function with zeros ak.Proof. Apply Lemma 5.3.3 and Theorem 5.2.3.

The assumptions of this proposition are fulfilled for interesting classesof functions.

The drawback of the Weierstrass product representation is that the func-tion g is difficult to determine in general. Compare the work done in theexample of the sine in 5.2.4. Hadamard’s theorem 6.6.1 will solve this prob-lem for a large class of functions.

Theorem 5.3.7 (Weierstrass on arbitrary domains). Let D be a domainin C and let an be a sequence in D without limit point in D. Then thereexists a holomorphic function f on D that is 0 precisely on an (countingmultiplicities).

For the proof we refer to the exercises 5.5.9.

Example 5.3.8. We construct a function with zeros in the points of thelattice

(5.3.10) Λ = k1ω1 + k2ω2 : k1, k2 ∈ Z.


Here ω1, ω2 ∈ C \ 0, ω1/ω2 /∈ R. We order the non-zero lattice pointsapproximately by their size. For every n ∈ N there are 4n lattice points

ω = k1ω1 + k2ω2, |k1| + |k2| = n,

positioned on a parallellogram Vn. Notice that Vn = nV1. The sequence akis now formed by the lattice point on V1, then on V2 et cetera. The order ofthe ak on any particular Vn is immaterial. Let

d = min|z| : z ∈ V1, D = max|z| : z ∈ V1.Then if ak ∈ Vn we have nd ≤ |ak| ≤ nD. We will estimate the sumsinvolved in the convergence of the Weierstrass product.

∞∑

k=1

1

|ak|≥

∞∑

n=1

4n

nD= ∞,

∞∑

k=1

1

|ak|2≥

∞∑

n=1

4n

(nD)2= ∞,(5.3.11)

∞∑

k=1

1

|ak|3≤

∞∑

n=1

4n

(nd)3<∞.(5.3.12)

We conclude from Theorem 5.3.4 that

(5.3.13) σ(z) = z∏

(

1 − z

ak

)

exp

(

z

ak+

z2

2a2k

)

is an entire function with zeros of order 1 in the lattice points. The func-tion σ is the σ-function of Weierstrass; despite the periodicity of its zerosand its resemblance of the sine, cf. 5.3.1, it is not periodic. Logarithmicdifferentiation yields Weierstrass’ ζ-function

(5.3.14) ζ(z) =σ′(z)σ(z)

=1

z+

∞∑

k=1

(

1

z − ak+

1

ak+

z

a2k

)

.

One further differentiation leads to the famous Weierstrass P-function.

(5.3.15) P(z) = − d

dz

σ′(z)σ(z)

=1

z2+

∞∑

k=1

(

1

(z − ak)2− 1

a2k

)

.

This is an elliptic function, i.e., an even, doubly periodic meromorphic func-tion on C; its periods are ω1 and ω2. Double periodicity of its derivative∑

ω∈Λ−2

(z−ω)3is immediate. This leads to periodicity of P as follows.

(5.3.16) P(z + ωj) − P(z) = constant = P( 12ωj) − P(− 1

2ωj) = 0.

We will return to the P-function in a later chapter.

5.4. Mittag-Leffler’s principal part representation

Let f be a meromorphic function on C. At any point a ∈ C the functionf admits a Laurent expansion

(5.4.1) f(z) =−1∑

k=−mck(z − a)k +

∞∑

k=0

ck(z − a)k.

The first sum (with negative exponents k) is called the principal part off at a. It will be 0 for all a except for a countable set a1, a2, . . . without

5.5. EXERCISES 47

limit point in C. In analogy with Weierstrass’ theorem, one may look for ameromorphic function with prescribed principal parts

hk(z) =

mk∑

j=1

ckj(z − ak)j

in ak. If there are only finitely many k, then∑

k

hk(z)

is a solution to this problem, but in general there may be problems with theconvergence of this sum. The idea is again to compensate hk, now by addingholomorphic functions, which will not affect the principal parts, in such away that uniform convergence on compact sets occurs. We met alreadyexamples in (5.3.14), (5.3.15) where hk = 1

z−ak, respectively hk = 1

(z−ak)2.

The compensating functions are in fact the first few terms of the Taylorexpansion of hk about 0. Subtracting these will make the series converge.

We have all but proved

Theorem 5.4.1 (Mittag-Leffler). Let f be meromorphic on C and letak be the sequence of different poles of f . Then there exist polynomials Pkand a holomorphic function g on C such that

(5.4.2) f(z) =∞∑

k=1

(hk(z) − Pk(z)) + g(z).

Here hk is the principal part of f at ak and for Pk one may take a Taylorpolynomial of hk of sufficiently high order.

In fact, just as there is a version of Weierstrass’ theorem for arbitrarydomains in C, there is also a version of Mittag-Leffler’s theorem for arbitrarydomains in C. See Ex. 5.5.13.

5.5. Exercises

5.5.1. Prove Lemma 5.1.6.

5.5.2. Let vk be in C. Show that convergence of both the series∑

vkand

∑ |vk|2 implies convergence of the product∏

(1 + vk).

5.5.3. Show that∏∞k=1(1+ i

k ) diverges, whereas∏∞k=1 |1+ i

k | converges.

5.5.4. Compute the value of the Wallis-product 21

23

43

45

65

67 · · · .

5.5.5. Compute∏∞k=1(1 − z

k2 ).

5.5.6. Prove that∏∞k=1(1 − kzk) defines a holomorphic function f on

the unit disc B(0, 1). What are the zeros of f ; show that f has no analyticcontinuation to any domain D that contains B(0, 1) as a proper subset.

5.5.7. Show that the product∏

p prime(1− p−z) converges on H = z :

Re z > 1. Show that on H

∏

pprime

(1 − p−z)−1 =∞∑

n=1

1

nzdef= ζ(z).


[Recognize the sum of the geometric series!] The function ζ is the famousRiemann-zeta function. Notice that this formula serves to show that thereare infinitely many primes! See also Exercise 7.9.24.

5.5.8. Let Ij = (aj , bj), j = 1, 2, . . ., be open intervals in [−1, 1] thatare mutually disjoint. Let

gn(z) =z + 1

z − 1

n∏

j=1

z − bjz − aj

.

Prove that the sequence gn converges uniformly on compact sets on thedomain C∗ \ [−1, 1]∪⋃∞

j=1 Ij to a function g. Show that g is constant if and

only if∑∞

j=1(bj − aj) = 2.

5.5.9. Prove the Weierstrass theorem 5.3.4 for arbitrary simply con-nected domains. One can copy the proof for C, but instead of using ap-proximation of log(1−u) by a Taylor series, approximate by polynomials asprovided by Runge’s theorem.

5.5.10. Determine the Mittag-Leffler representation or at least theprincipal part of

1

sinπz, cot z.

5.5.11. Prove that

1

ez − 1= −1

2+

1

z+

∞∑

k=1

2z

z2 + 4k2π2.

5.5.12. Let ak be a sequence in C \ 0. Write down a meromorphicfunction with poles at ak and corresponding principal part 1/(z−ak) in case(i)∑

1/|ak| converges, (ii)∑

1/|ak|2 converges, (iii)∑

1/|ak|3 converges.Same question with principal parts 1/(z − ak)

2.

5.5.13. Prove Mittag-Leffler’s theorem. Let f be meromorphic on adomain D and let ak be the sequence of different poles of f . Then thereexist holomorphic functions Pk and a holomorphic function g on D such that

(5.5.1) f(z) =

∞∑

k=1

(hk(z) − Pk(z)) + g(z).

Moreover given a sequence ak without limit point in D and principal partshk at ak, there exist holomorphic functions Pk on D such that the series

(5.5.2)∞∑

k=1

(hk(z) − Pk(z))

represents a meromorphic function on D with principal parts hk.

CHAPTER 6

Growth and zeros of entire functions of finite order

6.1. Introduction

A precise relation between growth order and the size of the zero-set ofan entire function is provided by Hadamard’s theorem, which is the maintheorem of the present chapter. The voyage to this theorem starts withdefining a growth scale for entire functions. Next we will prove two resultsthat are of interest by themselves: The Borel-Caratheodory inequality andJensen’s theorem. Eventually all this is wrapped together with the theoryof the previous chapter and some clever estimates to produce the desiredresult.

References for this chapter are [8, 26]. All holomorphic functions in thischapter are assumed to be non-constant.

6.2. The growth scale

Let f be a holomorphic function on B(0, R). The maximum modulus off is the function M = Mf defined by

M(r) = Mf (r) = max|z|=r

|f(z)|, 0 ≤ r < R.

The maximum principle implies that M is a strictly increasing function ofr.

Definition 6.2.1. Let f be an entire function. The order of f is

(6.2.1) ρ = ρ(f) = lim supr→∞

log logM(r)

log r.

If f has finite order ρ > 0, the ρ-type, or shortly type of f is

(6.2.2) τ = τρ(f) = lim supr→∞

logM(r)

rρ.

Remarks 6.2.2. A little longwinded, but perhaps more instructive arethe following two alternatives

• ρ may also be defined as

ρ = mins ≥ 0 : ∀ε > 0∃rε : logM(r) ≤ rs+ε, r > rε.• τ may be defined as

τ = mins ≥ 0 : ∀ε > 0∃rε : M(r) ≤ e(s+ε)rρ, r > rε.

The order of f may be infinite; the ρ-type of f of finite order ρ > 0 maybe infinite too. An entire function of order 1 and finite 1-type is called afunction of exponential type.

49

50 6. GROWTH AND ZEROS OF ENTIRE FUNCTIONS OF FINITE ORDER

Examples 6.2.3. Most of the functions we would ordinarily meet havefinite order, but it is not hard to write down functions of infinite order.

(1) ez2

has order 2 and 2-type 1.(2) sin(πz) has order 1 and type π.

(3) e(ez) has infinite order.

(4) sin(z1/2)

z1/2 has order 1/2 and type 1.

(5) Polynomials have order 0.

Eventually Hadamard’s theorem will provide a means to construct func-tions of any order. In fact functions of prescribed order and type can beconstructed.

Remark 6.2.4. We compare M(r) to functions like earb. Other growth-

scales are useful in more advanced theory. See [26].

6.3. The Borel-Caratheodory inequality

Let f be holomorphic on B(0, R). Define

(6.3.1) Af (r) = A(r) = max|z|=r

Re f(z), r < R.

A(r) may well be negative, but as M(r) it is a strictly increasing functionof r (consider F = ef ). Clearly |A(r)| ≤ M(r). The Borel-Caratheodoryinequality goes in the opposite direction.

Lemma 6.3.1 (Borel-Caratheodory). Let f be holomorphic in a neigh-

borhood of B(0, R). Then

(6.3.2) M(r) ≤ 2r

R− rA(R) +

R+ r

R− r|f(0)|, 0 ≤ r ≤ R.

Proof. First we assume f(0) = 0, so that A(r) > 0 if r > 0. Consider

(6.3.3) F (z) =f(z)

2A(R) − f(z)=N

D.

Note that |N |2 = (Re f)2+(Im f)2 and |D|2 = (2A(R)−Re f)2+(Im f)2 ≥(Re f)2+(Im f)2. It follows that F is holomorphic on B(0, R) with F (0) = 0and |F | ≤ 1. We can apply the Schwarz Lemma and find

(6.3.4) |F (z)| ≤ |z|/R,or, with |z| = r,

(6.3.5) |f(z)| ≤ r

R|2A(R) − f(z)| ≤ r

R(2A(R) + |f(z)|).

Rewriting this gives (6.3.2), (with f(0) = 0).For the general case we consider

g(z) = f(z) − f(0).

Applying the special case and using Ag(R) ≤ Af (R) + |f(0)|, we find

Mf (r) ≤Mg(r)+|f(0)| ≤ 2r

R− rAg(R)+|f(0)| ≤ 2r

R− rAf (R)+

R+ r

R− r|f(0)|.

6.4. BLASCHKE FACTORS, JENSEN’S FORMULA, AND COUNTING FUNCTIONS 51

Corollary 6.3.2. Let f be an entire function of order ≤ ρ and zero-free. Then there exists a polynomial of degree ≤ ρ such that

f(z) = eP (z), z ∈ C.

Proof. The function g(z) = p.v. log f(z) is well-defined and holomor-phic on C, because f is zero-free and C is simply connected. We have forevery ε > 0 that

Re g(z) ≤ |z|ρ+ε, |z| = r > rε.

Apply the Borel-Caratheodory lemma with R = 2r sufficiently large andfind

(6.3.6) Mg(r) ≤ 2Ag(R) + 3|g(0| ≤ 2(2r)ρ+ε + 3|g(0)| ≤ Crρ+ε.

Now the Cauchy inequalities for the Taylor coefficients of g at 0, imply thatg is a polynomial of degree at most ρ+ε. As ε is arbitrary, g is a polynomialof degree ρ at most.

A refinement of this idea will turn out useful.

Corollary 6.3.3. Let g be an entire function such that for a sequenceRk → ∞ we have A(Rk, g) ≤ CεR

ρ+εk (ε > 0). Then g is a polynomial of

degree ≤ ρ.

Proof. We find M(g,Rk/2) ≤ C(Rk/2)ρ+ε by another application of

the Borel-Caratheodory lemma. The Cauchy estimates on the C(0, Rk/2)yield again that g is a polynomial of degree ρ at most.

6.4. Blaschke factors, Jensen’s formula, and counting functions

Let f be holomorphic on B(0, R) and let 0 < r ≤ R. We can factor outthe zeros of f inside B(0, R) without changing the norm of f on C(0, r).This is done with the help of Blaschke factors:

Ba(z) =

a−|a|

r(z−a)r2−za , if a ∈ B(0, r) \ 0,

zr , if a = 0.

The rational function Ba is holomorphic outside z = r2/a while on C(0, r)we have |Ba| = 1. The factor a

−|a| turns out to be useful when considering

convergence properties of infinite Blaschke products, cf. exercise 6.7.14. Fora1, . . . , ak ∈ B(0, r) we introduce the (finite) Blaschke product

Ba1,...,ak=

k∏

j=1

Baj .

It follows that if f has zeros precisly in a1, . . . , ak (counted with multiplicity),then

g(z) = f(z)/Ba1,...,ak

is holomorphic and zero free on B(0, r).Hence a holomorphic branch of log g can be defined; we will just denote

it by log g too. The mean value equality will apply to log g and taking realparts will give us:


Theorem 6.4.1 (Jensen’s formula). Let f be holomorphic on the discB(0, R) such that f(0) 6= 0. Assume f has zeros a1, a2, . . . with |a1| ≤ |a2| ≤. . . < R. Then for 0 < r < R we have the equality

(6.4.1) log |f(0)| =1

2π

∫ π

−πlog |f(reiθ)| dθ −

∑

|ak|<rlog

r

|ak|.

Proof. First assume as a special case that f has no zeros on C(0, r).Let a1, . . . , am be the zeros inside C(0, r). Again with g(z) = f(z)/Ba1,...,am

we have

(6.4.2) log(g(0)) =1

2π

∫ 2π

0log g(reiθ) dθ.

Hence, taking real parts,

(6.4.3) log |g(0)| =

∫ 2π

0log |g(reiθ)| dθ

2π.

Therefore

(6.4.4)

log |f(0)| −∑

|ak |<rlog

|ak|r

=

∫ 2π

0log |g(reiθ)| dθ

2π=

∫ 2π

0log |f(reiθ)| dθ

2π

This proves the theorem in the special case. Now assume that f has zeros onC(0, r0). Then, because the zeros form a discrete set, there are no zeros onC(0, r) for r close to r0. We will show that each side of (6.4.4) is continuousas a function of r at r0. For the lefthand side this is obvious. For therighthand side we write

f(z) = f(z)∏

|ak |=r(z − ak),

with f zero free on C(0, r) and the product consisting of only finitely manyfactors. We substitute this in the righthand side of (6.4.4). Then it remainsto show that for every ak on C(0, r0) the integral

∫ π−π log |reiθ − ak|dθ is a

continuous function of r at r0. It is no loss of generality to take r0 = 1 andak = 1. But then,

(6.4.5)

∫ π

−πlog |reiθ − 1|2 dθ =

∫ π

−πlog((r − 1)2 + 4r sin2(θ/2)) dθ

The integrand is majorized by the integrable function |2 log | sin θ2 || + log 5

and continuity of (6.4.5) follows from Lebesgue’s theorem.

Corollary 6.4.2.∫ π

−πlog |f(reiθ)| dθ

is an increasing function of r

Definition 6.4.3. The counting function of the zeros of an entire func-tion is defined as

(6.4.6) n(t) = n(t, f) = the number of zeros of f on B(0, t)

6.4. BLASCHKE FACTORS, JENSEN’S FORMULA, AND COUNTING FUNCTIONS 53

If f(0) 6= 0, then the integrated counting function or Nevanlinna countingfunction is defined as

(6.4.7) N(r) = N(r, f) =

∫ r

0

n(t)

tdt.

Theorem 6.4.4. Let f be holomorphic on B(0, R), f(0) 6= 0. Then

N(r) =1

2π

∫ π

−πlog |f(reiθ)| dθ − log |f(0)|.

Proof. We apply summation by parts to the sum of logarithms inJensen’s formula. Keep in mind that n(t) = j on [|aj |, |aj+1|).

(6.4.8)∑

|ak|≤rlog

r

|ak|= n(r)(log r) −

∑

|ak|≤rlog |ak|

= (log |a2| − log |a1|) + 2(log |a3| − log |a2|) + . . . + n(r)(log r − log |an(r)|)

=

n(r)−1∑

j=1

∫ |aj+1|

|aj |

n(t)

tdt+

∫ r

|an(r)|

n(t)

tdt =

∫ r

0

n(t)

tdt.

Now the theorem follows from Jensen’s formula.

Corollary 6.4.5. Suppose that f is an entire function and f(0) 6= 0and λ > 1. Then

n(r) log λ ≤∫ λr

r

n(t)

tdt = N(λr)−N(r) ≤ N(λr) ≤ log(Mf (λr))−log |f(0)|.

Apparently the growth of f puts a bound on the amount of zeros in thedisc B(0, r), (r → ∞). In particular

Corollary 6.4.6. Suppose f is of finite exponential type τ . Then forsome C > 0

n(r) ≤ Cr, |ak| ≥ k/C, if |ak| > 1.

Proof. The previous corollary gives the estimate for n(r), with C =(τ + ε)λ/ log λ and r sufficiently large. Next, observe that n(ak) ≥ k (withequality if the zero ak is simple), therefore the first equality gives k ≤ C|ak|.

Similarly one shows

Corollary 6.4.7. Let f be of finite order ρ, f(0) 6= 0, and zerosa1, a2, . . . ordered by size. For every ε > 0 there exist constants Cε, cε > 0such that

n(r, f) ≤ Cεrρ+ε, |ak| ≥ cεk

1/(ρ+ε).

In particular

(6.4.9)∑ 1

|ak|βconverges for β > ρ.

Notice that convergence in (6.4.9) is absolute and (hence) independentof the order of summation.


6.5. Canonical products

The convergence result for the zeros of a function f of order ρ that wasobtained in Corollary 6.4.7 can be combined with Proposition 5.3.6 to obtaina canonical product of the zeros of f . We will show that such a product hasthe same (or smaller) order as f . First we introduce some terminology.

Definition 6.5.1. Let ak be a sequence in C \ 0 that diverges to∞. The exponent of convergence of ak is

σ = infβ :∑

k

|ak|−β converges.

The genus of ak is the least integer such that∑

k

|ak|−s−1

converges.

Note that s ≤ σ ≤ s+ 1.

Examples 6.5.2. If σ is entire, then each inequality can occur.

• The sequence defined by ak = k2 has σ = 1/2 and s = 0.• The sequence defined by ak = k has σ = 1 and s = 1.• The sequence defined by ak = k log2 k has σ = 1 and s = 0.

Recall from Definition 5.3.2 the elementary factors of Weierstrass Ek.

Definition 6.5.3. The canonical product of a sequences ak ⊂ C \ 0of finite genus s is the function defined by

p(z) =∞∏

k=1

Es

(

z

ak

)

.

We will call s the genus of this product.

Lemma 6.5.4. Suppose that f is an entire function of order ρ, f(0) 6= 0.Then the exponent of convergence σ and the genus s of the zero-set of fhave the following properties:

σ ≤ ρ, s ≤ ρ.

The canonical product p of genus s of the zeros of f converges uniformly oncompact sets in C. The function p is entire and has the same zeros (countingmultiplicity) as f .

Proof. Corollary 6.4.7 states that∑

k |ak|β converges for β > ρ. Henceσ ≤ ρ and s ≤ ρ. Proposition 5.3.6 yields the other statements of thelemma.

It is possible to make good estimates on the size of the function p inLemma 6.5.4. To do so we need a lemma.

Lemma 6.5.5. For every nonnegative integer s there exist positive con-stants C1, . . . , C4 (that depend on s) such that the elementary factors of

6.5. CANONICAL PRODUCTS 55

Weierstrass Es(u) satisfy the following estimates

−C1|u|s+1 ≤ log |Es(u)| ≤ C2|u|s+1, |u| ≤ 1

2, s = 0, 1, . . . ;(6.5.1)

log |1 − u| − C3|u|s ≤ log |Es(u)| ≤ C4|u|s, |u| ≥ 1

2, s = 1, 2, . . . ;

(6.5.2)

log |1 − u| = log |E0(u)| ≤ C4|u|ε, |u| ≥ 1

2, (0 < ε < 1).(6.5.3)

Proof. The proof consists of estimates similar to the ones in the proofof Lemma 5.3.3. We will only indicate the idea, and leave details to thereader.

log |Es(u)| ∗= log |1 − u| + Re

(

u+1

2u2 + · · · + 1

kuk)

∗∗= Re

∑

j>s

uj

j.

Now∗∗= will lead to (6.5.1), and

∗= to (6.5.2), while (6.5.3) is elementary.

Theorem 6.5.6. Let f be an entire function of finite order ρ, f(0) 6= 0,and p the canonical product associated to the zeros ak of f . Then p hasorder ρ1 equal to the exponent of convergence σ of ak. That is, for everyε > 0 there is a constant Aε such that

(6.5.4) log |p(z)| ≤ Aε|z|σ+ε.

Next, for α > 0 put

Sα = C \(

B(0, 1) ∪k B(ak, |ak|−α))

.

For every ε > 0 there exists a constant Bε = Bε(α) such that

(6.5.5) log |p(z)| ≥ −Bε|z|σ+ε, z ∈ Sα.

Proof. We will treat the case that σ is not entire, so that s < σ < s+1in detail. First the inequality (6.5.4). Let rk = |ak|, r = |z|. In view ofLemma 6.5.5 it is natural to split log |p| as follows:

(6.5.6) log |p(z)| =∑

rk<2r

log

∣

∣

∣

∣

Es

(

z

ak

)∣

∣

∣

∣

+∑

rk≥2r

log

∣

∣

∣

∣

Es

(

z

ak

)∣

∣

∣

∣

= Σ1 + Σ2.

To Σ1 we apply (6.5.2) and to Σ2 (6.5.1). Then

(6.5.7) log |p(z)| ≤∑

rk<2r

C4

(

r

rk

)s

+∑

rk≥2r

C2

(

r

rk

)s+1

.

Now if 2r/rk > 1, then (2r/rk)s < (2r/rk)

σ+ε and (r/rk)s < 2(r/rk)

σ+ε, so

∑

rk<2r

(

r

rk

)s

< 2rσ+ε∑

rk<2r

rk−σ−ε.

Similarly, if 2r/rk < 1, then for ε > 0 sufficiently small σ+ ε < s+ 1, hence(r/rk)

s+1 ≤ (r/rk)σ+ε and

∑

rk≥2r

(

r

rk

)s+1

≤ rσ+ε∑

rk≥2r

rk−σ−ε.


Now∑

r−σ−εk is convergent, so we conclude from the above that

log |p(z)| < Cεrσ+ε.

In other words, p has order ρ1 ≤ σ. If ρ1 would be strictly less than σ, thezeros of p would have convergence exponent at most ρ1. Therefore, ρ1 = σ.

Next we will prove (6.5.5). Again we split the sum and observe that theimportant difference with the previous part is the presence of the logarithmin (6.5.2). Thus the following estimate, which is valid on Sα, suffices toprove (6.5.5)

(6.5.8)∑

rk<2r

log |1 − z/ak| =∑

rk<2r

log |ak − z| −∑

rk<2r

log |ak|

≥ −(1 + α)∑

rk<2r

log |ak| ≥ −(1 + α)n(2r) log 2r ≥ −Crσ+2ε.

We used that on Sα we have |ak − z| ≥ |ak|−α and, in the last line, thatn(r) ≤ Crσ+ε, cf. Corollary 6.4.7.

The case where σ is integer or where s = 0 is left as an exercise, cf. Ex6.7.13.

6.6. Hadamard’s product theorem

All our efforts now culminate in the following fundamental theorem.

Theorem 6.6.1 (Hadamard). Let f be a non-constant entire functionof finite order ρ. Then f admits a canonical product representation:

(6.6.1) f(z) = zmeq(z)p(z),

where m ≥ 0 is the multiplicity of the zero at the origin, q is a polynomial ofdegree ν, ν ≤ ρ, and p is the canonical product associated to the zeros akof f in C \ 0. We have ρ = maxσ, ν.

We call the integer maxs, ν the genus of f .

Proof. By considering f(z)/zm, which is a function of order ρ too, wecan assume f(0) 6= 0. Let p denote the canonical product associated to thezero-set ak of f . By Lemma 6.5.4, both the genus s and the exponentof convergence σ of ak are ≤ ρ. Thus f/p is zero-free on C and we have(6.6.1), but with q an entire function.

To show that q is a polynomial we will estimate f/p on suitable circlesC(0, R). We will use Theorem 6.5.6 and its notation. Fix α > σ suchthat

∑ |ak|−α is convergent. With this choice of α, the total length of thediameters of the discs B(ak, |ak|−α) is finite and thus there exists a sequenceof radii Rj → ∞ such that C(0, Rj) ⊂ Sα. Now it follows from Theorem6.5.6 that on C(0, Rk)

Re q(z) = log |f(z)| − log |p(z)| ≤ C1Rρ+εk + C2R

σ+εk ≤ CRρk, |z| = Rk.

Application of Corollary 6.3.3 gives that q is a polynomial of degree ν ≤ ρ.We now have proven (6.6.1) and shown that maxσ, ν ≤ ρ. But it

follows from (6.6.1) that ρ ≤ maxσ, ν, and we are done.

6.6. HADAMARD’S PRODUCT THEOREM 57

Example 6.6.2. The sine product revisited. As sinπ√z

π√z

is an entire func-

tion of order 1/2 with zeros ak = k2, Hadamard’s theorem gives immediately

sinπ√z

πz= ec

∞∏

k=1

(1 − z

k2).

(Here c is a polynomial of degree ≤ 1/2, hence a constant). Substitutingz = 0 we find c = 0. Replacing z by z2, the usual product formula is found.

Starting out with sinπz without the square root trick, we find thatσ = s = 1, so that Hadamard leads to

sinπz = zea+bz∏

n∈Z

(1 − z/n)ez/n.

We compute ea = limz→0sinπzz = π. Moreover, sinπz

πz and the canonicalproduct are both even functions, so b = 0.

In a later chapter we will prove the famous Picard’s theorem, which statesthat an entire function takes on all values in C with at most one exception.For functions of finite order this follows from the following corollaries.

Corollary 6.6.3. Suppose a function f has finite non-integer order ρ.Then f takes on all values in C.

Proof. Because ρ is non-integer, ρ = σ. Therefore f has infinitelymany zeros. By considering f(z) − w, which has the same order, we seethat f assumes every value w infinitely often, and as often as it assumes thevalue zero, in the sense that the exponent of convergence of the zeros of fand of f − w equal ρ.

Corollary 6.6.4. If f has integer order and is non-constant, then fassumes all values with at most one exception.

Proof. If f − a is zerofree, it is of the form eP (z) for some polynomialP and eP assumes all values except 0.

Hadamard’s theorem is only the beginning of a refined analysis of therelation between growth and zero-distribution of entire functions. The booksby Boas [8] and Levin [26] are excellent for further study. As an examplewe treat functions of finite exponential type.

Example 6.6.5. Let f be an entire function of exponential type, f(0) 6=0. Let ak denote the zeros of f then by Corollary 6.4.6

(6.6.2) ak ≥ ck, C > 0,

So the genus of ak is either 0 or 1. Apply Hadamard’s theorem. If s = 0then

f(z) = ea+bz∞∏

k=1

(1 − z

ak).

If s = 1 then

(6.6.3) f(z) = ea+bz∞∏

k=1

(1 − z

ak)ez/ak .


In this case one can show that

(6.6.4) S(r) =∑

|ak|<r

1

ak≤ C, (r → ∞)

It is a theorem of Lindelof that (6.6.2) and (6.6.4) characterize the zero setof functions of exponential type, i.e., if (6.6.2) and (6.6.4) are fulfilled, thenthe canonical product will be of exponential type.

6.7. Exercises

6.7.1. Compute (or estimate) M(r) for the following functions

sin z, cos√z,

∫ c

−cg(t)e−iztdt, g ∈ L1(−c, c).

6.7.2. Let f(z) =∑∞

k=0 ckzk be an entire function. Show that for all

r > 0, n 6= 0

|cn|rn ≤M(r) ≤∞∑

k=0

|ck|rk.

6.7.3. Determine the order of the following entire functions

(i) sin z2, (ii) ecos z, (iii)

∞∏

k=1

(1 − z/k!),

(iv) σ(z) (Weierstrass-σ), cf. 5.3.8, (v) P (z), a polynomial of degree ν,

(vi) eP (z), (P as in (v)), (vii)

∞∑

k=0

zk

kαk, α > 0.

6.7.4. Let fj be entire of order ρj , j = 1, 2. What can you say about theorder of f1 + f2 and about the order of f1f2? [Distinguish between ρ1 = ρ2

and ρ1 6= ρ2].

6.7.5. let f be entire and non-constant. Show that f and f ′ have thesame order and type. [Estimate Mf (r) in terms of Mf ′(r) and Mf ′(r) interms of Mf (r + 1)].

6.7.6. The Bessel function is defined by

J0 = 1 − z2

22+

z4

2242− · · · + (−1)k

z2k

2242 . . . (2k)2+ · · · .

Show that it has order one and determine its exponential type.

6.7.7. Let f(z) =∑∞

k=0 ckzk be an entire function and suppose that

|f(z)| ≤ eA|z|B

for some A,B and r sufficiently large. Show that for k > k0

and r > r0, with r0, k0 sufficiently large

|ck| ≤eAr

B

rk,

and hence

|ck| ≤(

eAB

k

)kB

,

6.7. EXERCISES 59

6.7.8. Let f(z) =∑∞

k=0 ckzk be an entire function. Using the previous

exercise, try prove the following theorem, cf. [26].The order of f equals

(6.7.1) ρ = lim supn→∞

n log n

− log |cn|.

If ρ is finite then the ρ-type σ of f satisfies

(6.7.2) (σeρ)1/ρ = lim supn→∞

(

n1ρ n√

|cn|)

.

6.7.9. Check how good the estimates for the counting functions are forthe case f(z) = sinπz

z .

6.7.10. Derive the fundamental theorem of Algebra (“A polynomial ofdegree n has n zeros”) from Jensen’s formula.

6.7.11. Prove Corollary 6.4.7.

6.7.12. Consider the function

f(z) =

∫ ∞

0eztt−t dt.

a: Show that f is entire.b: Show that f is bounded on the half plane Re z < 0.c: By a suitable change of the path of integration, show that for τ ∈

(−π/2, π/2) the function f is bounded on the half plane Re (zeiτ ) <0.

d: Compute the order of f .e: Modify f to a function that is bounded on every line through 0.f: Adding a constant, one can assume that f has zeros and modify f

to an entire function that has the property that limr→∞ g(reiθ) = 0.

6.7.13. Complete the proof of Theorem 6.5.6.

6.7.14. Let f be a bounded holomorphic function on the unit disc withzeros ak 6= 0. Use Jensen’s formula to prove that

∑∞k=1 log |ak| converges.

Show that this is equivalent to convergence of

(6.7.3)

∞∑

k=1

(1 − |ak|).

If a sequence of points ak in B(0, 1) has the property that (6.7.3) con-verges, we say that it satisfies the Blaschke condition.

Show that if ak in B(0, 1) satisfies the Blaschke condition, then theBlaschke product

∞∏

1

|ak|−ak

z − ak1 − akz

converges uniformly on compact sets in B(0, 1).

CHAPTER 7

Some special functions

In this chapter we will apply the theory of product representations,growth and zero sets to some classical functions. We will study Euler’sGamma-function cf. [47, 3], the Riemann zeta-function, cf. [16, 24], andelliptic functions and their relatives, cf. [2] [46] for some older and extensivereferences.

7.1. Elementary properties of the Gamma-function

The Gamma-function was introduced by Euler, as a means to interpolatefactorials. A convenient definition is

Definition 7.1.1 (Euler 1764).

(7.1.1) Γ(z) =

∫ ∞

0tz−1e−tdt, Re z > 0.

Here tz−1 = e(z−1) log t as usual (with log t real valued on R≥0).

One easily shows that Γ is holomorphic on Re z > 0. Integration byparts gives the

First functional equation

(7.1.2) Γ(z + 1) = zΓ(z), Re z > 0.

As Γ(1) = 1 = 0! (by convention) we have in particular Γ(n + 1) = n! forpositive integers.

We can write (7.1.2) also in the form

(7.1.3) Γ(z) =Γ(z + 1)

z=

Γ(z + 2)

(z + 1)z= . . . =

Γ(z + n+ 1)

(z + n)(z + n− 1) · · · z .

The righthand side is clearly defined as a meromorphic function on the(increasing set of) halfplanes Re z > −n− 1, and thus we see that Γ admitsa meromorphic extension to C with poles in 0,−1,−2, . . .. The nature ofthe poles becomes apparent from the following splitting of the integral in(7.1.1).

(7.1.4) Γ(z) =

∫ 1

0tz−1e−tdt+

∫ ∞

1tz−1e−tdt.

The last integral represents an entire function, say g(z), cf. Exercise 7.9.7.The first integral can be written as follows.

(7.1.5)

∫ 1

0tz−1e−tdt =

∫ 1

0

∞∑

n=0

(−1)ntz+n−1

n!dt =

∞∑

n=0

(−1)n

n!(z + n).

61

62 7. SOME SPECIAL FUNCTIONS

We have found the Mittag-Leffler representation:

(7.1.6) Γ(z) =∞∑

n=0

(−1)n

n!(z + n)+ g(z).

Cf. Theorem 5.4.2.Euler also found the famous second functional equation. We will derive

it now, starting from the observation that Γ(z)Γ(1− z) is holomorphic on C

except for poles of order 1 in Z. This suggests multiplying with sinπz. Thefunction

(7.1.7) f(z) := Γ(z)Γ(1 − z) sin πz

is an entire function. By the first functional equation and elementary prop-erties of the sine it is periodic with period 1:

f(z+ 1) = Γ(z+ 1)Γ(−z) sin π(z+ 1) =z

−zΓ(z)Γ(1− z)(− sin π(z)) = f(z).

We estimate f on the strip 1 ≤ z ≤ 2. There we find, with z = x+ iy,

(7.1.8) |Γ(z)| ≤∫ ∞

0|tz−1e−t|dt ≤

∫ ∞

0|te−t|dt = Γ(2) = 1.

Next

|Γ(1 − z)| =

∣

∣

∣

∣

Γ(3 − z)

(2 − z)(1 − z)

∣

∣

∣

∣

≤∣

∣

∣

∣

1

(2 − z)(1 − z)

∣

∣

∣

∣

,

and finally,

| sinπz| < e|πy|.

Combining these estimates gives

(7.1.9) |f(z)| < Ce|πy|, z = x+ iy ∈ C.

The periodicity of f allows us to view f as a function of w = e2πiz. Moreprecisely, let h be any holomorphic branch of 1

2πi logw on a simply connecteddomain D in C\0. Then f h is holomorphic on D. The value f h(w) isindependent of the choice of the branch and the domain D ∋ w. It followsthat f h defines a holomorphic function g on C\0. Using (7.1.9) we find

|g(w)| ≤ Ce|Re12

logw| ≤ |w| 12 + |w|− 12 , w 6= 0.

Developing g in a Laurent series and applying the Cauchy estimates yieldsthat g equals a constant, hence so does f . We compute it as follows:

f(0) = limz→0

f(z) = limz→0

Γ(z) sinπz = π,

because the residue of Γ at 0 equals 1. We have found the second functionalequation

(7.1.10) Γ(z)Γ(1 − z) =π

sinπz.

Corollary 7.1.2. The function 1/Γ(z) = 1πΓ(1 − z) sin πz is an entire

function. All its zero’s are of order one. They are located in 0,−1,−2, . . ..

7.2. PRODUCT REPRESENTATIONS FOR 1/Γ 63

7.2. Product representations for 1/Γ

We wish to apply Hadamard’s theorem 6.6.1 to find a product repre-sentation for 1/Γ. We need its order. From (7.1.2), (7.1.3), and (7.1.8) weobtain on x = Re z > 0

log |Γ(z)| ≤ C

[x]−1∑

1

log |z − j| ≤ C|z| log |z|,

where [x] = maxn ∈ Z : n ≤ x, while if Re z ≤ 2, |z + k| ≥ δ > 0

log |Γ(z)| < D.

Here C, D are suitible positive constants. Since sinπz has order 1, weconclude that 1/Γ has order 1, see also Ex.7.9.11. Hadamard’s theoremgives the canonical product representation:

(7.2.1)1

Γ(z)= zeγz+b

∞∏

k=1

(

1 +z

k

)

e−zk .

Using zΓ(z) = Γ(z + 1) and substituting z = 0 we find b = 0. The constantγ is the Euler-Mascheroni constant. It is found in its usual form by lettingz = 1 in (7.2.1) and expanding the product:

1 = eγ limn→∞

n∏

k=1

(k + 1

k)e−1/k = eγ lim

n→∞elog(n+1)−(1+1/2+1/3+···1/n),

or, because log(n+ 1) − log n→ 0 as n→ ∞,

(7.2.2) γ = limn→∞

(1 + 1/2 + 1/3 + · · · 1/n − log n).

Plugging in this value of γ we obtain

Theorem 7.2.1 (Euler product definition).

Γ(z) = limn→∞

nzn!

z(z + 1) · · · (z + n), z 6= 0,−1,−2, . . . .

The convergence is uniform on compact sets that do not meet Z≤0.

Proof. From (7.2.1) and (7.2.2), for z 6= 0,−1, · · · ,

(7.2.3) Γ(z)

= limn→∞

n!ez(1+1/2+···+1/n)

z(z + 1) · · · (z + n)ez logn−z(1+1/2+···+1/n)

= limn→∞

nzn!

z(z + 1) · · · (z + n).


7.3. The Hankel integral for the Gamma-function

The second functional equation (7.1.10) gives rise to an integral repre-sentation for 1/Γ that is valid in the whole complex plane. This is the socalled Hankel representation . Starting with the substitution s = −t we havefor Re z < 0

Γ(1 − z) =

∫ ∞

0t−ze−tdt =

∫ 0

−∞(−s)−zesds.

Now for s < 0 we have, taking the limit of the principal value coming fromthe upper halfplane,

(−s)−z = e−z p.v. log(−s) = e−z(p.v. log s−πi) = eπizp.v. s−z.

Thus

Γ(1 − z) = eπiz∫ 0

−∞p.v. s−zesds.

In this integral it should be understood that the argument of s equals π.If we take the limit from the lower halfplane, the argument of s equals

−π and we we find similarly

Γ(1 − z) = e−πiz∫ 0

−∞p.v. s−zesds.

We add these integrals and obtain

2Γ(1−z) =

(

e−πiz∫ 0

−∞(from below) − eπiz

∫ −∞

0(from above)

)

p.v. s−zesds.

Combining these two path of integrations into one path L, rearranging andapplying (7.1.10) we obtain

1

2πi

∫

Lp.v. s−zesds =

(eπiz − e−πiz)2πi

Γ(1 − z) =sin(πz)

πΓ(1 − z) =

1

Γ(z).

0L′ 0

L′′

Figure 1. Paths of the Hankel integral

Using Cauchy’s theorem, we now change the path of integration to apath L′ as indicated in the picture. At the origin this is possible since

7.4. ASYMPTOTICS OF Γ 65

Re z < 0. The new integral represents an entire function. It equals 1/Γ forRe z < 0; by the unicity theorem, the equality is valid over all of C.

Take any L′′ equivalent to one of the path’s in the picture, i.e. startingin the third quadrant, ending in the second, not meeting (−∞, 0] and havingthe Re s → −∞ at the ends of the path. For such L′′ we have found theHankel representation

(7.3.1)1

2πi

∫

L′′

p.v. s−zesds =1

Γ(z), z ∈ C.

7.4. Asymptotics of Γ

What is the asymptotic behavior of n! ? A first estimate is obtained byconsidering log n! =

∑nj=1 log j as a Riemann sum:

∫ n

1log t dt ≤ log n! ≤ log n+

∫ n

1log t dt, n ≥ 1.

Thus

log n! = n log n− n+ O(log n), as n→ ∞.

We will obtain a sharper estimate for the Gamma-function using themethod of Laplace. Fix x > 0 and start with

Γ(x+ 1) =

∫ ∞

0txe−tdt.

The main contribution in the integral will come from integration over thepart were the integrand is large. We find by elementary calculus that theintegrand has a maximum where its t- derivative vanishes, that is at t = x.It is now convenient to rescale (i.e. make a substitution) in such a way thatthe new integrand will have a max at the same fixed point for every x. Thisis done by substituting t = xs, and yields

Γ(x+ 1) = xx+1

∫ ∞

0sxe−sxds.

The maximum is now at s = 1. We prefer Taylor developments at the originand make a further substitution u = s− 1, which will put the maximum att = 0. Hence,

(7.4.1) Γ(x+ 1)/xx+1e−x =

∫ ∞

−1exp[x log(1 + u) − xu] du = I(x).

At 0 we develop the exponent in a Taylor series with respect to u (with x aparameter).

(7.4.2) x log(1 + u) − xu = −xu2

2+xu3

3− xu4

4+ H.O.T..

As this is only precise on a small interval about 0, we split the integral asfollows

I(x) =

∫ −δ

−1+

∫ δ

−δ+

∫ ∞

δ= I−1 + I0 + I1.


The value of δ will be chosen depending on x later on. From (7.4.1) and(7.4.2) one may expect that I0 behaves as

∫ δ

−δe−xu

2/2du = x−1/2

∫

√xδ

−√xδe−u

2/2du.

If δ(x)√x→ ∞ this will converge to (2π/x)1/2, while otherwise the limit is

untractable, or 0, which would mean that we don’t count the contributionfrom the max at all! Therefore, it is wise to choose δ > 1/

√x. However,

additional restrictions are needed to control the other terms in the Taylordevelopment. We proceed as follows.

I0(x) =

∫ δ

−δe

−xu2

2 exp

(

xu3

3− xu4

4+ H.O.T.

)

du(7.4.3)

=

∫ δ

−δe

−xu2

2

(

1 +xu3

3+ O(x2u6 + xu4)

)

du(7.4.4)

=

∫ δ

−δe

−xu2

2

(

1 + O(x2u6 + xu4))

du(7.4.5)

=

∫ ∞

−∞e

−xu2

2 + O(

∫ ∞

−∞e

−xu2

2 (x2u6 + xu4) du)

(7.4.6)

= (2π/x)1/2 + O(x−3/2)(7.4.7)

Some remarks are in order: The higher order terms in (7.4.3) are O(xu4), as

long as δ < x−1/4; To pass to (7.4.4), observe that all higher order terms thatcome from the expansion of exp(xu3/3+Cxu4) in (7.4.4) can be absorbed inO(x2u6 + xu4) when δ = x−1/3. Next, (7.4.5) is valid because the function

e−xu2

2 xu3/3 is odd, hence it does not contribute to the integral; (7.4.6) is

valid if we choose δ = x−1/3, so that x2u6 ≥ 1 for |u| ≥ δ; (7.4.7) followsfrom the substitution s = u

√x. The integrals I−1 and I1 are much smaller,

because the integrand is very small if |u| > δ. We can estimate I−1 by

I−1 ≤ max−1≤u≤−δ

exp(x(log(1 + u) − u)) = exp(−xδ2/2) = e−x3/2.

In I1 we have an infinite interval; the trick is to make the integral look likean integral of a small function against a finite measure as follows

(7.4.8) I1 =

∫ ∞

δ(1 + u)2 exp[x log(1 + u) − xu]

du

1 + u2

≤ maxδ≤u≤∞

exp(x+ 2) log(1 + u) − xu)∫ ∞

0

du

(1 + u)2

≤ Cexp(x+ 2) log(1 + δ) − xδ) ≤ C exp(−xδ2/2) = Ce−x3/2.

We used that for x > 23/2 the maximum is attained at u = δ and concludethat

I(x) = (2π/x)1/2 [1 + O(1/x)] .

We have shown

7.5. THE ZETA-FUNCTION 67

Theorem 7.4.1.

(7.4.9) Γ(x+ 1) = xx+1/2e−x(2π)1/2[

1 + O(1

x)

]

.

The expression between [ ] can be expanded further. It is clear from theabove computation that the contribution of I1 and I−1 is minute comparedto that of I0. Looking at the expansion of I0 we observe that a better resultmay be obtained by further expansion of the Taylor series in (7.4.4). Weleave this as exercise 7.9.16. Another way to asymptotic development of theGamma-function is described in Exercise 7.9.19.

Next we wish to study asymptotics of Γ(z) for complex values of z. Wewill estimate

(7.4.10)

∫ ∞

0tze−tdt

for z → ∞ in the halfplane Re z > 0. We will use the saddle point method,also known as method of steepest decent. As in the case of real z > 0, itseems natural to find the point were the integrand is maximal, and followthe previous method. However, as Im z 6= 0, there may (and will) occurcancellation in the integral, which makes good estimates hard, if not impos-sible.

The way out is as follows. First observe that by Cauchy’s theorem wemay replace the path of integration [0,∞) by any halfline Λθ = reiθ :r > 0 with |θ| < π/2 − ε. Next look at the graph of | exp(z log t − t)| =exp(Re gz(t)) above such a halfline. It will be best to pick the halflinewere the maximum of exp(Re gz(t)) is minimal. Why? Well, for the valueof the integral it does not matter which Λθ is chosen, but it is clear thatcancellation is minimal if

∫

Λθ

exp(Re gz(t))dt

is minimal. In view of our experience with real x, this is likely to happenfor θ0 with maxexp(Re gz(t)), z ∈ Λθ0 minimal. This means that weare looking for a saddlepoint of the graph of exp(Re gz(t)), which occurs if∇Re gz = 0 or g′z(t) = 0. We find that this happens if t = z.

With θ = arg z, our starting point will thus be(7.4.11)∫

λθ

exp[zp.v. log t− t] dtt=z(1+v)

= p.v. zz+1e−z∫ ∞

−1exp[z log(1 + v) − zv]dv.

For |θ| < π/2− ε the method of Laplace may be applied again. The result is

Theorem 7.4.2.

(7.4.12) Γ(z + 1) = p.v. zz+1/2e−z(2π)1/2[

1 + O(1

z)

]

.

7.5. The zeta-function

The definition of the Riemann zeta-function is due to Euler, who alsowas aware of the functional equation that we will meet in the next section.However, a proof and thorough study had to wait until Riemann (1859).


The classical definition is

(7.5.1) ζ(z) =

∞∑

n=1

1

nz, Re z > 1.

One establishes easily that the series in (7.5.1) converges uniformly on thehalfplanes Re z > 1 + δ, (δ > 0), and represents a holomorphic functionon Re z > 1. The connection to number theory becomes already apparentfrom the following product representation due to Euler, cf. Exercise 5.5.7.

(7.5.2) ζ(z) =

∞∏

j=1

1

1 − p−zj, Re z > 1.

Here and in the next display, the summation is over all primes pj . Indeed,

(7.5.3)∞∏

j=1

1

1 − p−zj=

∞∏

j=1

( ∞∑

k=0

p−kzj

)

=∑

α

(pα11 pα2

2 · · · )−z =∑

n−z.

The manipulation with sums and products is justified by uniform conver-gence. Observe that the unique prime factorization of integers has beenused!

An integral representation can be obtained via multiplication by theGamma-function. Observe that

(7.5.4) Γ(z)n−z =

∫ ∞

0

(

t

n

)z

e−tdt

t

t=sn=

∫ ∞

0sz−1e−snds.

For the zeta-function this gives

(7.5.5) Γ(z)ζ(z) =

∫ ∞

0

∞∑

n=1

(e−s)nsz−1ds

=

∫ ∞

0sz−1e−s

1

1 − e−sds =

∫ ∞

0sz−1 1

es − 1ds.

Interchanging sum and integral may be justified by uniform convergence,or by summing explicitly the first N terms of the geometric series that isinvolved, and letting N → ∞. In other words, we have obtained the firstintegral representation for ζ .

(7.5.6) ζ(z) =1

Γ(z)

∫ ∞

0

tz−1

et − 1dt, Re z > 1.

We will derive a kind of Mittag-Leffler expansion for the integral figuringin (7.5.6) as we did for Γ in (7.1.6). It will exhibit ζ as a meromorphicfunction with only a single pole.

Theorem 7.5.1. There exists an entire function g and real numbers Bnsuch that

(7.5.7) ζ(z) =1

Γ(z)

1

z − 1+

∞∑

n=1

Bnn!

1

z + n− 1+ g(z)

.

The righthand side of (7.5.7) is a meromorphic function with a single pole,of order 1 at z = 1.

7.6. HANKEL REPRESENTATION AND FUNCTIONAL EQUATION FOR ζ 69

The numbers Bn are called the Bernoulli numbers.

Proof. We split the integral in (7.5.6) into∫ 10 and g(z) =

∫∞1 . The

function g is entire, cf. Section 7.1. For∫ 10 we develop the integrand. Write

(7.5.8)t

et − 1=

∞∑

0

Bnn!tn = 1 − t

2+t2

12− . . . .

(Observe that the singularity at 0 is removable!) The Bernoulli numbers Bnare defined by the identity (7.5.8). They are in fact rational, cf. Exercise7.9.21. It is easy to see that t/(et− 1)− 1+ t/2 is even, hence B2k+1 = 0 fork ≥ 1. The radius of convergence of this series equals the distance to thenearest singularity, which is 2π. Hence for every ε > 0

∣

∣

∣

∣

Bnn!

∣

∣

∣

∣

≤ 1

(2π − ε)nas n→ ∞.

It follows that∫ 1

0

tz−1

et − 1dt =

∞∑

n=0

∫ 1

0

Bnn!tntz−2 =

∞∑

n=0

Bnn!

1

z + n− 1.

(The series for tz−1/(et − 1) converges uniformly on [0, 1] if Re z − 2 ≥−1 + ε and interchanging of sum and integral is permitted.) This provesformula (7.5.7) for Re z > 1. The right-hand side of (7.5.7) is a meromorphicfunction, which extends ζ. It has only one pole at 1, because the zeros of1/Γ remove the possible poles at 0,−1, . . ..

Corollary 7.5.2. The functions

(z − 1)ζ(z), ζ(z) − 1

z − 1

are entire.

7.6. Hankel representation and functional equation for the

zeta-function

Using the multiple valuedness of tz we will now derive a Hankel repre-sentation for the zeta-function, similar to what we did in Section 7.3. Weintroduce a cut in the complex plane from −∞ to 0 over the reals. Considerthe integral

I(z) =

∫

Lδ

p.v. sz−1

e−s − 1ds.

The path of integration Lδ is indicated in the picture below. It consists ofan interval (−∞,−δ] supposed to be part of the lower halfplane, so thatp.v. arg s = −π there and p.v. (s)z−1 = p.v. (−s)z−1e−iπz−iπ, a circle |s| =δ and an interval [−δ,∞) in the upper halfplane, so that p.v. arg s = π andp.v. (s)z−1 = p.v. (−s)z−1eiπz+iπ. I(z) is defined for all z. We proceed asin Section 7.3, and contract the circle |s| = δ to the origin, assuming now


that Re z > 1, so that the integrals are convergent. Taking into accountone sign change because of reversing the direction of integration, we find

(7.6.1) I(z) =(

eπiz − e−iπz)

∫ 0

−∞

p.v. (−s)z−1

e−s − 1ds

= 2i sinπzΓ(z)ζ(z) = 2πiζ(z)

Γ(1 − z), Re z > 1.

As first and last term in this equality are entire functions, these are equalfor all z. We have found the Hankel representation

(7.6.2) ζ(z) =1

2πiΓ(1 − z)

∫

Lδ

p.v. sz−1

e−s − 1ds, z 6= 1.

Lδ

0 δ N

−(N + 1/2)2πi

(N + 1/2)2πi×

×

××

2πi

WN

Figure 2. Paths of the Hankel integral for ζ

Let g denote the integrand in (7.6.2). We will manipulate the path ofintegration in (7.6.2). Let N ∈ N, and let WN be the path of integrationconsisting of the complex intervals [−N,−N − (N + 1/2)i], [−N + −(N +1/2)i,N−(N+1/2)i], [N−(N+1/2)i,N+(N+1/2)i], [N+(N+1/2)i,−N+(N+1/2)i] and [−N+(N+1/2)i,−N ]. Together with (Lδ)N = Lδ∩Re s ≥−N we obtain a closed path of integration in the cut plane. The residuetheorem gives us

1

2πi

∫

(Lδ)N

g(s) ds =1

2πi

∫

WN

g(s) ds −∑

interior residues.

The residues are located at the simple poles s = 2nπi, 1 ≤ |n| ≤ N , and

(7.6.3) Res (g, 2πin) = e(z−1)p.v. log 2nπi lims→2nπi

s− 2nπi

e−s − 1

= −e(z−1)p.v. log 2nπi = −(2|n|π)z−1e(z−1)i argni.

Keeping in mind that argni = (Signn)π/2 we find that the sum of theresidues equals

−(2π)z−1N∑

n=1

nz−12 cos(z − 1)π/2 = 2(2π)z−1 sin(πz/2)

N∑

n=1

1

n1−z .

7.7. ZEROS OF THE RIEMANN-ZETA FUNCTION 71

If Re z < 0 the limit for N → ∞ exists and equals

2(2π)z−1 sin(πz/2)ζ(1 − z).

It remains to show that when Re z < 0 the integral of g over WN tends to0; it is clear that the integral over (Lδ)N then tends to the one over Lδ. Fors = N + it and z = x+ iy we have

|g(s)| ≤ exp[(x− 1) log |s| + yp.v. arg s]

1 − e−N≤ 2 exp[(x− 1) logN + π|y|].

so that the integral over [N − (N + 1/2)i,N + (N + 1/2)i] can be estimatedby (2N+1)2Nx−1eπ|y| = O(Nx). This tends to 0 as N → ∞, because x < 0.On the intervals I± = s = σ ± (N + 1/2)2πi, −N < σ < N, we find

∣

∣e−s − 1∣

∣ = e−σ + 1 > 1.

Hence

|∫

I±

g(s) ds| ≤ eπ|y|∫ N

−N|s|x−1dσ = O(Nx),

again tending to 0 as N → ∞. On the intervals [−N,−N − (N + 1/2)i] and[−N + (N + 1/2)i,−N ] we have |g(s)| ≤ CNx−1e−N , hence the integrals ofg over these intervals vanish exponentially fast as N → ∞.

We have arrived at the Functional equation for the zeta-function

Theorem 7.6.1.

(7.6.4) ζ(z) = 2(2π)z−1 sinπz

2Γ(1 − z)ζ(1 − z), z 6= 1.

Proof. Indeed, we have just derived (7.6.4) for Re z < 0. However,both sides of (7.6.4) extend analytically to z 6= 1 and the unicity theoremcompletes the proof.

7.7. Zeros of the Riemann-zeta function

The Euler product formula (7.5.2) shows that ζ has no zeros on Re z >1. Let us consider (7.6.4) and take Re z < 0. The right-hand side has zerosprecisely at the zeros of sin(πz/2); these are called the trivial zeros of thezeta-function.

If there are any other zeros, they have to be in the critical strip 0 ≤Re z ≤ 1. For such z we have from (7.6.4): ζ(z) = 0 ⇔ ζ(1 − z) = 0. Also,

from (7.5.1) ζ(z) = ζ(z), (first for Re z > 1 and then everywhere by analyticcontinuation), so that ζ(z) = 0 ⇔ ζ(z) = 0. Apparently the zeros come inquadruples z, z, 1 − z, 1 − z, unless Re z = 1/2.

The famous Riemann hypothesis states

All non trivial zeros of ζ are on Re z = 1/2.

Are there zeros in the critical strip?

Theorem 7.7.1. The zeta-function has infinitely many zeros inside thecritical strip 0 ≤ Re z ≤ 1.


Proof. Formϕ(z) = z(z − 1)Γ(z/2)ζ(z);

this is an entire function; it has its zero’s in the critical strip. Its order isatmost one. That it is ≤ 1 follows from the fact that (z− 1)ζ(z) and 1/Γ(z)have both order one, and so invoking Hadamard’s theorem (or rather its

proof), Γ grows slower than e|z|1+ε

on appropriate circles.In fact it has order 1 and type ∞, because of the growth of Γ(z/2) on

the positive axis, and the fact that ζ(x) → 1 as x→ ∞.So the exponent of convergence of the zeros ak is 1. If it were σ < 1

then we could write

ϕ(z) = eaz+b∏

(

1 − z

ak

)

, a, b ∈ C

The product would have order less then 1 and the exponential has order oneand type |a|, which is impossible.

See [16] for the more precise result n(r) ∼ ( rπ ) log r.

Remark 7.7.2. Riemann’s hypothesis remains open until today. It hasbecome one of the “Clay-foundation problems”, meaning that it has a prizeof a million dollar on its head. Partial results to the effect that a positivefraction of number of zeros are indeed located on Re z = 1/2 have beenestablished. With the use of computers, it has been shown that the first few1010 zeros are indeed on the critical line.

How can a numerical computation “prove” that a zero is precisely onthe critical line? Well, if a zero is just very close to Re z = 1/2 there willbe a companion mirror image on the other side of the line. The argumentprinciple, even in a numerical form will detect two zeros close to each other.But this has never happened so far!

Remark 7.7.3. Euler’s product representation reveals a connection be-tween prime numbers and the zeta-function. The prime number theoremstates that π(N), the number of primes less than N , behaves as follows.

π(N)def=∑

pj≤N1 ∼ N

logN, if N → ∞.

The prime number theorem can be deduced fairly easily from the not sodifficult fact that ζ has no zeros on the line Re z = 1. See [33, 23];Exercise 7.9.24 goes a little in this direction.

7.8. Elliptic functions

We say that a function f on C has period ω ∈ C\0 if f(z+ω) = f(z).It is called doubly periodic if it has periods ω1, ω2 with ω1/ω2 /∈ R

If ω1 and ω2 are periods, then so is k1ω1 + k2ω2 k1, k2 ∈ Z, (k1, k2) 6=(0, 0). The periods of a doubly periodic function form a period lattice Λ, thatis, a free abelian subgroup of C that spans C over R. If Λ is generated by ω1

and ω2, we call ω1, ω2 primitive periods. Primitive periods ω1, ω2 determinea parallellogram E with vertices 0, ω1, ω2, ω1 + ω2. It is convenient todemand, as we shall do, that the point 0 as well as the open edges (0, ω1)and (0, ω2) belong to E.

7.8. ELLIPTIC FUNCTIONS 73

For every w ∈ C, any translate Π = E + w = e + w, e ∈ E is calleda fundamental parallellogram or fundamental domain for the lattice Λ, andalso for any doubly periodic function with period lattice Λ.

Definition 7.8.1. An elliptic function (with respect to some periodlattice Λ) is a non-constant meromorphic function f on the complex planethat is doubly periodic with respect to Λ. The order of f is the number ofpoles (counted with multiplicity) in any fundamental domain of f withoutpoles of f on its boundary.

Theorem 7.8.2. Let f be an elliptic function with period lattice L andfundamental domain Π.

• f cannot be holomorphic• The sum of the residues of f on Π equals 0.• The order N of f is at least 2.• On Π the function f assumes every complex value w precisely N

times. (counting multiplicities)

Proof. Exercise 7.9.28.

Next we will construct a basic elliptic function with period lattice Λ andfundamental periods ω1, ω2. By theorem 7.8.2, if it has only one pole in Π,it must have order 2 at least. We may put such a pole at the origin; ourfunction, if it exists, will have poles of order 2 at all points of Λ and nowhereelse. We arrive at the Weierstrass P-function.

Theorem 7.8.3. For every period lattice Λ and fundamental periodsω1, ω2 there exists precisely one elliptic function P of order 2 that has Lau-rent expansion

P(z) =1

z2+ c1z + . . . .

In fact,

(7.8.1) P(z) =1

z2+∑

ω∈Λ′

(

1

(z − ω)2− 1

ω2

)

where Λ′ = Λ \ 0. The Laurent expansion of P at 0 is given by

(7.8.2) P(z) =1

z2+

∞∑

j=1

(2j + 1)z2j

(

∑

ω∈Λ′

ω−2j−2

)

.

Proof. Unicity: Suppose P1 and P2 are elliptic functions that satisfythe requirements of the theorem. Then P1 − P2 is a holomorphic ellipticfunction, hence a constant. The constant equals 0 because the Pi have noconstant term.

Next we recall, cf. (5.3.15), that (7.8.1) represents a meromorphic func-tion. This can be shown as follows. Take K a compact in C that does notmeet Λ. Then there is a constant CK such that

∣

∣

∣

∣

1

(z − ω)2− 1

ω2

∣

∣

∣

∣

=

∣

∣

∣

∣

2zω − z2

(z − ω)2ω2

∣

∣

∣

∣

≤ CK |ω|−3

and inequality (5.3.11) shows that the series (7.8.1) converges uniformly onK. It was shown in Example 5.3.8 that P is doubly periodic.


The Laurent expansion of 1(z−ω)2 − 1

ω2 is

1

(z − ω)2− 1

ω2=

1

ω2

∞∑

n=1

(−2

n

)

(

− z

ω

)n=

∞∑

n=1

(n+ 1)zn

ωn.

From here on we will write

sj =∑

ω∈Λ′

ω−j.

Observing that s2j+1 = 0 because ω ∈ Λ′ ⇒ −ω ∈ Λ′, we find at 0

(7.8.3) P(z) =1

z2+

∞∑

n=1

∑

ω∈Λ′

(n+ 1)zn

ωn=

1

z2+

∞∑

j=1

(2j + 1)s2j+2z2j .

Corollary 7.8.4. The derivative of P can be written (on a neighbor-hood of 0) as

(7.8.4) P ′(z) = −∑

ω∈Λ

−21

(z − ω)3=

−2

z3+

∞∑

j=1

(2j + 1)2js2j+2z2j−1.

Proof. The first equality follows by differentiation of the Mittag-Lefflerseries (7.8.1), the second by differentiation of the Laurent expansion (7.8.3).

Definition 7.8.5. Given a period lattice Λ we define

g2 = 60s4 = 60∑

Λ′

ω−4(7.8.5)

g3 = 140s6 = 140∑

Λ′

ω−6(7.8.6)

Theorem 7.8.6. The Weierstrass P-function is a solution of the differ-ential equation

(7.8.7)

(

dw

dz

)2

= 4w3 − g2w − g3 = (w − e1)(w − e2)(w − e3).

Here e1 = P(ω1/2), e2 = P(ω2/2), e3 = P((ω1 +ω2)/2). Note that e1 +e2 +e3 = 0.

Proof. From (7.8.3) and (7.8.4) we can compute the Laurent expansionof

(7.8.8) h = (P ′)2 − 4P3 + g2P + g3.

The function h is an elliptic function with poles (if any) on Λ. A computationshows that there are only powers of z with positive exponent left in theexpansion of (7.8.8) at 0. It follows that h ≡ 0 and we have the firstequality in (7.8.7)

From equation (7.8.4) we infer that P ′ is odd, (f(z) = −f(−z)), first ina neighborhood of 0 and by the unicity theorem on all of C. If ω is one ofthe numbers ω1, ω2, ω3 = ω1 + ω2, then

P ′(ω/2) = P ′(−ω/2 + ω) = P ′(−ω/2) = −P ′(ω/2) = 0 or ∞.

7.9. EXERCISES 75

However, on Π the function P ′ assumes the value ∞ only at 0, and the orderof P ′ equals 3. Hence ω1/2, ω2/2, ω3/2 are the only zeros of P ′ on Π andthey are simple zeros. It follows that P assumes the values ei = P(ωi/2)with multiplicity 2 at ωi/2. Because the order of P equals 2, the values eidiffer from each other and are on Π assumed only at ωi/2. We are done.

Observe that we could have derived (7.8.7) from scratch as follows. Startwith (P ′)2 it has a pole of order 6 at 0. Subtract a suitable multiple of P3 sothat the term 4z−6 vanishes. Look at the Laurent development. It has againa pole at 0, but of lower order (in fact its order is 2). Subtracting a suitable(and computable!) multiple of P, we find that the remaining function iselliptic and holomorphic, hence it is a (computable) constant.

Corollary 7.8.7. The map P = (P,P ′) : Π \ 0 → C2 is one to oneonto the complex curve E = t2 = 4s3 − g2s− g3 ⊂ C2.

Proof. We have seen that P maps into E. Because P is even and hasorder two, we find for z,w ∈ Π that P(z) = P(w) implies z = w or z = −wmod Λ; in the second case P ′(z) = −P ′(w). We conclude that P is injective.

Now let (s, t) ∈ E As P assumes all values on Π twice (counting mul-tiplicities), there exist z ∈ Π with P(z) = P(−z + ω3). Again becauseP ′(z) = −P ′(z + ω3) and (7.8.7) we find that (s, t) is assumed.

Remark 7.8.8. It is convenient to extend the map P to all of Π, byadding a point at ∞ on the curve E, that is, consider E, the closure of Ein P2, with homogeneous coordinates [X0,X1,X2], so that C2 corresponds

to X0 6= 0. In homogeneous coordinates E would be given by

X22X0 = X3

1 − g2X3X20 − g3X

30 .

Written in these coordinates P becomes

P (z) = [1,P(z),P ′(z)] = [z3, z3P(z), z3P ′(z)].

Now let z → 0 and find P (0) = [0, 0, 1]. The map P extends as a covering

map to all of C. It then induces a map P : C/Λ → E, which is a homeomor-phism. We may identify C/Λ with a torus as follows; start with Π and gluethe opposite sides of the Π. We conclude that E is topologically a torus too!

7.9. Exercises

7.9.1. Compute Γ(n+ 1/2) and Γ′(1).

7.9.2. Compute (in terms of values of Γ)

a.

∫ ∞

0xαe−x

2dx, b.

∫ 1

0

(

log1

x

)α

dx, (α > −1),

∫ ∞

0(log t)e−tdt.

7.9.3. Show that∫ ∞

0tz−1e−stdt = Γ(z)p.v. s−z, Re z > 0, Re s > 0.

7.9.4. Compute (as improper Riemann integrals)∫ ∞

0tx−1e−itdt, 0 < x < 1,

∫ ∞

0

sin(αt)

tβ, α > 0, 0 < β < 2.


7.9.5. Compute the Fresnel-integrals∫ ∞

0sin(x2) dx

∫ ∞

0cos(x2). (Improper)

7.9.6. Prove that for real values of y

|Γ(iy)|2 =π

y sinhπy|Γ(1/2 + iy)|2 =

π

y coshπy.

7.9.7. Prove that the Gamma-function (7.1.1) is holomorphic on H =

Re z > 0. [Show that the integrals∫ A0 are holomorphic and converge

uniformly on compact set in H as A → ∞.] Observe that the same argu-ment shows that the function g in the Mittag-Leffler representation (7.1.9)is entire.

7.9.8. Derive the first functional equation for the Gamma-function(7.1.2)

7.9.9. Let 0 < a < 1. Deduce the second functional equation for Γ asfollows. By a suitable transformation (u = tv), show that

Γ(1 − a)Γ(a) =

∫∫

u,t>0t−aua−1e−t−udt du

=

∫ ∞

0

va−1

v + 1dv =

1

a

∫ ∞

0

1

w1/a + 1dw.

(7.9.1)

Compute the last integral (apply the residue theorem and pick a contourthat contains the segment re2πa, 0 < r < R).

7.9.10. Prove that

Γ(a)Γ(b) = Γ(a+ b)

∫ ∞

0

va−1

(v + 1)a+bdv = Γ(a+ b)

∫ 1

0xa−1(1 − x)b−1dx.

Derive the formula for Euler’s Beta-function

(7.9.2)

B(z,w)def=

∫ 1

0tz−1(1−t)w−1dt = 2

∫ π/2

0sin2z−1 θ cos2w−1 θ dθ =

Γ(z)Γ(w)

Γ(z + w),

Re z,Rew > 0.

7.9.11. Fill in the details of the computation of the order of 1/Γ andshow that it is a function of order 1 and type ∞.

7.9.12. Compute the volume ωn of the unit ball in Rn. Show

ωn = 2ωn−1

∫ 1

0(1 − x2

n)(n−1)/2dxn = πn/2/Γ( 1

2n+ 1).

Also, show that the volume of the sphere in Rn equals σn = nωn.

7.9.13. Prove the duplication formula for the Gamma-function

Γ(2z) = 22z−1π−1/2Γ(z)Γ(z + 12).

7.9.14. Find a formula for Γ(z)Γ(z + 13)Γ(z + 2

3).

7.9. EXERCISES 77

7.9.15. Determine the order of the Mittag-Leffler function

Eα(z) =∞∑

n=0

zn

Γ(nα+ 1), α > 0.

For 0 < α < 2, prove that

Eα(z) =1

α

1

2πi

∫

L

exp(p.v. ζ1/α)

ζ − zdζ, πα/2 ≤ arg z ≤ 2π − πα/2,

where L is the path Re (p.v. ζ1/α) = 0. Conclude that Eα is bounded onsectors πα/2 + ε ≤ arg z ≤ 2π − πα/2 − ε.

7.9.16. Prove the asymptotic development

(7.9.3) Γ(x+ 1) = xx+1/2e−x(2π)1/2[

1 +1

12x+ O(

1

x

2

)

]

7.9.17. Prove Gauss’ formula:

(7.9.4)d

dxlog Γ(x+ 1) = lim

n→∞

(

log n−n∑

k=1

1

x+ k

)

= limn→∞

(∫ ∞

0

e−t − e−nt

tdt −

∫ ∞

0

1 − e−nt

et − 1e−xtdt

)

=

∫ ∞

0

(

e−t

t− e−xt

et − 1

)

dt, x > 0.

7.9.18. (continued) Prove Binet’s formulas

d

dxlog Γ(x+ 1) =

1

2x+ log x−

∫ ∞

0

(

12 −

1

t+

1

et − 1

)

e−xtdt,

log Γ(x+ 1) = (x+ 12) log x− x+ c+ ρ(x),

ρ(x) =

∫ ∞

0

(

1

et − 1− 1

t+ 1

2

)

e−xtdt

t, x > 0.

Show in addition that c = 12 log 2π.

7.9.19. (continued) Use the Mittag Leffler representation of 1/et − 1(Exercise 5.5.11) to prove that

0 < ρ(x) =1

π

∫ ∞

0

x

x2 + v2log

1

1 − e−2πvdv <

1

12x.

7.9.20. Prove that the holomorphic branch of log Γ(z+1) that is definedon C \ (−∞, 0] and positive on the positive real axis can be represented as

log Γ(z + 1) = (z +1

2) log z − z +

1

2log 2π + ρ(z),

with

ρ(z) =1

π

∫ eiα∞

0

z

z2 + v2log

1

1 − e−2πvdv, |α| < π/2,

for | arg z − α| < π/2. (Note that the integral is over the halfline of pointswith argument α).


7.9.21. Find a recurrency relation for the Bernoulli numbersBn startingfrom the relation

t =

( ∞∑

0

Bnn!tn

)( ∞∑

1

tn

n!

)

.

Conclude that Bn are rational.

7.9.22. Prove that the functions in Corollary 7.5.2 have order 1.

7.9.23. Prove that

1−z − 2−z + 3−z − . . . = (1 − 21−z)ζ(z),Re z > 0.

Conclude that (i) ζ has a pole at 1 with principal part 1/(z − 1). (ii) ζ hasno zeros on the interval (0, 1).

7.9.24. Let pj denote the j-th prime. Use the Euler- product represen-tation to prove that

∞∑

j=1

p−1j = ∞.

7.9.25. Prove that

ζ ′(z)ζ(z)

=∞∑

n=1

Λ(n)

nz, Re z > 1,

with Λ the Von Mangoldt function

Λ(n) =

log p if n = pk, p prime, k ∈ N,

0 in all other cases.

7.9.26. Prove that ζ(2n) = 12(−1)n−1(2π)2kB2n/(2n!), n = 1, 2, 3, . . ..

7.9.27. Prove that ζ(1 − n) = (−1)n−1Bn/n, n = 1, 2, 3, . . ..

7.9.28. Complete the proof of Theorem 7.8.2. [What can you say for

suitably chosen fundamental domains Π about∫

bΠ f(z) dz,∫

bΠf ′(z)f(z) dz, and

∫

bΠf ′(z)f(z)−w dz?]

7.9.29. Suppose that f is elliptic of order n with respect to Λ. Showthat the order of f ′ is m with n+ 1 ≤ m ≤ 2n.

7.9.30. Show that P ′′ is a polynomial in P and determine this polyno-mial.

7.9.31. Let f be an elliptic function with respect to the lattice Λ.Suppose that the zeros of f in Π are aj and a′j = −aj mod Λ, while its

poles are bj and b′j = −bj mod Λ; j = 1, . . . , n. Prove that for suitable C

f(z) = C(P(z) −P(a1)) · · · (P(z) − P(an))

(P(z) − P(b1)) · · · (P(z) − P(bn)).

If ak = 0 or bk = 0, we have to omit the corresponding factor.

7.9.32. (Continued) Working with the lattice Λ fixed, prove that forevery elliptic function f , there exist rational functions R1, R2 such that

f(z) = R1(P(z)) + P ′(z)R2(P(z)).

7.9. EXERCISES 79

7.9.33. The Lattes example.[25, 11]

(1) Apply the previous exercise to compute the rational function Rsuch that

P(2z) = R(P(z)).

(2) Observe that P defines a conjugation (at least where P is injective)between z 7→ 2z and z 7→ R(z).

(3) Show that ∞ is a repelling fixed point of R.(4) Show that R has the property that all points of the form 2−mk +

i2−nl, (k, l ∈ Z, m,n ∈ N) belong to the inverse orbit of ∞. (usethe conjugation with z 7→ 2z.)

(5) Show that for R J is dense in C∗ and hence J = C∗.

CHAPTER 8

Elements of potential theory

8.1. Harmonic functions

Definition 8.1.1. A function h on a domain D in C is called harmonicif it is real valued, C2, and satisfies Laplace’s equation

(8.1.1) ∆h =∂2h

∂x2+∂2h

∂y2= 4

∂2h

∂z∂z= 0.

Occasionally a complex valued function that satisfies (8.1.1) is also calledharmonic.

Theorem 8.1.2. Let f be holomorphic on a domain D. Then Re f andIm f are harmonic functions on D. If h is a harmonic function on a simplyconnected domain D, then h = Re f for some holomorphic function f onD.

Proof. If f is holomorphic, then ∂f∂z = 0 hence 4 ∂2f

∂z∂z = ∆f = 0. But ∆respects real and imaginary part, that is, for real valued functions p and q,∆(p+ iq) = ∆p+ i∆q is the splitting in real and imaginairy part. It followsthat ∆Re f = 0 and ∆Im f = 0.

Now let h be harmonic on a simply connected domain D. Then ∂h∂z is

holomorphic. Indeed,∂(∂h∂z )

∂z=

∂2h

∂z∂z= 0.

Since D is simply connected, the function ∂h∂z has a holomorphic primitive f

on D. We compute

∂(f + f)

∂z=∂f

∂z=∂h

∂z,

∂(f + f)

∂z=∂f

∂z=∂h

∂z=∂h

∂z.

We see that f + f and h are two real valued functions that have the samepartial derivatives, hence they differ by a real constant:

h = f + f + c,

orh = Re (2f + c).

If harmonic functions h1 and h2 have the property that h1 + ih2 isholomorphic, then h2 is called a harmonic conjugate of h1.

Example 8.1.3. If D is not simply connected, there exist harmonicfunctions that are not the real part of a holomorphic function. On theannulus A(0, 1, 2) the function log |z| is harmonic, but it is not the real partof a holomorphic function on A(0, 1, 2).

81

82 8. ELEMENTS OF POTENTIAL THEORY

Corollary 8.1.4. We have the following elementary properties:

(1) Harmonic functions are infinitely often differentiable.(2) Partial derivatives of harmonic functions are harmonic.(3) If f : D1 → D2 is a holomorphic map and h is harmonic on D2,

then h f is harmonic on D1.(4) A harmonic function on a domain D does not assume a local max-

imum or a local minimum on D.(5) Harmonic functions satisfy the Mean Value Equality: If h is har-

monic on D and B(a,R) ⊂ D, then

(8.1.2) h(a) =1

2π

∫ 2π

0h(a+ reit) dt, 0 < r < R.

Proof. We use the fact that we can write h = Re f for some holomor-phic function f on a suitable disc B(a, r). Then (1) and (2) and (3) areimmediate. Next, (4) follows from the fact that such an f is either an openmap or constant (Open Mapping Theorem). Finally, (5), i.e. (8.1.2) followsfrom the Cauchy integral formula for f , the substitution ζ = a + reit, andtaking the real part in the resulting formula.

f(a) =1

2πi

∫

C(a,r)

f(ζ)

ζ − adζ =

1

2π

∫ 2π

0f(a+ reit) dt.

8.2. The Poisson integral

Let h be harmonic on B = B(0, 1) and continuous on B. The mean valueequality represents h(0) in terms of the boundary values of h. We now wishto represent the value h(z0) at an arbitrary point z0 ∈ B. This is done witha suitable automorphism of B. Let

(8.2.1) w = Tz0(z) =z − z01 − z0z

and h = h T−1z0 .

Then Tz0 ∈ Aut(B) maps C(0, 1) to itself and h is harmonic on B andcontinuous on B. We apply the Mean Value Equality (8.1.2) to obtain

(8.2.2) h(z0) = h(0) =1

2π

∫ π

−πh(eiψ) dψ =

1

2πi

∫

C(0,1)h(w)

dw

w.

Logarithmic differentiation of Tz0 in (8.2.1) and keeping in mind that 1/z = zon C(0, 1), leads to

dw

w= d logw = d log Tz0(z) =

(

1

z − z0+

z01 − z0z

)

dz

=

(

z

z − z0+

z0z − z0

)

dz

z=

1 − |z0|21 − 2Re (z0z) + |z0|2

dz

z.

(8.2.3)

Substitute w = Tz0(z) in (8.2.2), combine with (8.2.3), and put z0 = reiθ,z = eiϕ. This finally yields

(8.2.4) h(reiθ) =1

2π

∫ π

−πh(eiϕ)

1 − r2

1 − 2r cos(θ − ϕ) + r2dϕ, reiθ ∈ B.

We have proved

8.2. THE POISSON INTEGRAL 83

Theorem 8.2.1. Let h be harmonic on B and continuous on B. Then hadmits the Poisson integral representation

h(reiθ) =

∫ π

−πPr(θ − ϕ)h(eiϕ) dϕ, 0 ≤ r < 1, θ ∈ R,

where Pr is the Poisson kernel

(8.2.5) Pr(t) =1

2π

1 − r2

1 − 2r cos t+ r2.

Remark 8.2.2. The Poisson kernel Pr, 0 ≤ r < 1, has the followingproperties:

(1) Pr is a positive continuous function with period 2π.(2)

1 =

∫ π

−πPr(θ − ϕ) 1 dϕ =

∫ π

−πPr(ϕ) dϕ.

(3) Scaling and translation leads to a Poisson integral for harmonicfunctions h on B(a,R) that are continuous up to the boundary

h(a+ reiθ) =

∫ π

−πPr/R(θ − ϕ)h(a+Reiϕ) dϕ

=1

2π

∫ π

−πh(a+Reiϕ)

R2 − r2

R2 − 2rR cos(θ − ϕ) + r2dϕ.

(8.2.6)

Here 0 ≤ r < R, θ ∈ R.

Of course (8.2.4) still makes sense when we replace h by an arbitrarycontinuous function, or even an integrable one, that is defined on C(0, 1).We are led to

Definition 8.2.3. Let µ be a finite Borel measure on C(0, 1). ThePoisson integral of µ is defined as

(8.2.7) P [µ](reiθ) =

∫ π

−πPr(θ − ϕ) dµ(ϕ), reiθ ∈ B.

The Poisson integral of a continuous or an L1 function f on C(0, 1) is definedas the Poisson integral of the associated measure f dϕ,

(8.2.8) P [f ](reiθ) =

∫ π

−πPr(θ − ϕ)f(eiϕ) dϕ, reiθ ∈ B.

Example 8.2.4. For the harmonic functions

rn cos(nθ) = Re zn, rn sin(nθ) = Im zn, n = 0, 1, 2, . . .

Theorem 8.2.1 reads

(8.2.9) P [cos(n·)](reiθ) = rn cos(nθ), P [sin(n·)](reiθ) = rn sin(nθ).

As a consequence we obtain the Fourier series of Pr

(8.2.10) 2πPr(eiθ) = 1 + 2

∞∑

n=1

rn cos(nθ) =

∞∑

n=−∞r|n|einθ.

See exercise 8.7.3.

Theorem 8.2.5. Let µ be a finite (real) Borel measure on C(0, 1). ThenP [µ] is harmonic on B.


Proof. Because Pr is a continuous function on C(0, 1), the integral(8.2.7) is well defined and finite. We will write the Poisson kernel as the realpart of a holomorphic function. With z0 = reiθ, z = eiϕ we have in view of(8.2.3)

2πPr(θ − ϕ) =

(

z

z − z0+

z0z − z0

)

= 2Rez

z − z0− 1.

It follows that P [µ](z0) is the real part of

f(z0) =1

2π

∫

C(0,1)

(

2z

z − z0− 1

)

dµ(z)

iz=

1

πiµ(z0) + C.

Thus, up to constants the function f equals the Cauchy transform of µ andtherefore f is holomorphic.

The Poisson integral of a measure on C(a,R) is of course harmonic onB(a,R).

Corollary 8.2.6. If hn is a sequence of harmonic functions on a do-main D and limn→∞ hn = h uniformly on compact sets in D, then h isharmonic.

Proof. We only have to prove that h is harmonic on balls B(a, r) with

B(a, r) ⊂ D. Uniform convergence of hn to h on C(a,R) implies that thePoisson representation P [hn] on B(a, r) of hn will converge to P [h] which isharmonic.

8.3. The Dirichlet problem on the disc

The Dirichlet problem for the Laplace operator ∆ on a domain D in C

reads as follows.

“Given a continuous function g on ∂D, find a continuous function F onD such that F is harmonic on D and F |∂D = g.”

We can solve the Dirichlet problem for the disc.

Theorem 8.3.1. Let g be continuous on C(0, 1). The Poisson integralP [g] is a solution to the Dirichlet problem on B.

This theorem will be a consequence of

Theorem 8.3.2. Let g be an integrable function on C(0, 1) and let h =P [g]. At every point ζ0 = eiϕ0

lim infζ→ζ0,|ζ|=1

g(ζ) ≤ lim infz→ζ0,|z|<1

h(z)

≤ lim supz→ζ0,|z|<1

h(z) ≤ lim supζ→ζ0,|ζ|=1

g(ζ).(8.3.1)

Proof of Theorem 8.3.1. Note that (8.3.1) implies that if g is con-tinuous at ζ0, then

limz→ζ0,|z|<1

h(z) = g(ζ0).

This proves Theorem 8.3.1.

8.3. THE DIRICHLET PROBLEM ON THE DISC 85

Proof of Theorem 8.3.2. By considering P [−g] = −h, it followsthat the first and third inequality in (8.3.1) are equivalent. The secondinequality is trivial, so it suffices to prove the third. In the third inequal-ity, we can assume that lim supζ→ζ0,|ζ|=1 g(ζ) = M < ∞; otherwise there isnothing to prove. If M is finite, then we may replace g by g −M ; this willhave the effect that h is replaced by h−M , therefore assuming that M = 0is no loss of generality. We thus only have to prove

(8.3.2) lim supζ→ζ0,|ζ|=1

g(ζ) ≤ 0 =⇒ lim supz→ζ0,|ζ|<1

h(z) ≤ 0.

Suppose for a moment that (8.3.2) has been proven. If now M = −∞,then for every positive constant A lim supζ→ζ0,|ζ|=1 g + A ≤ 0, therefore

lim supz→ζ0,|z|<1 h(z) ≤ −A, so that (8.3.1) follows in this case too.

It remains to prove (8.3.2). After a rotation we can in assume thatζ0 = 1.

Divide the circle C(0, 1) in two arcs

γ1 = eiϕ : −δ ≤ ϕ ≤ δ;(8.3.3)

γ2 = eiϕ : δ < ϕ < 2π − δ.(8.3.4)

For ε > 0 we choose δ so small that on γ1 we have g < ε. Because thePoisson kernel is positive and its integral over the circle equals 1, we obtain

(8.3.5) I1def=

∫ δ

−δPr(θ − ϕ)g(eiϕ) dϕ ≤ ε

∫ π

−πPr(θ − ϕ) dϕ.

Now z = reiθ will be close to 1, so we can assume that − 12δ < θ < 1

2δ. Forζ = eiϕ ∈ γ2 we find 1

2δ < ϕ−θ < 2π− 12δ. This implies cos(ϕ−θ) < cos( 1

2δ),therefore

0 <1 − r2

1 − 2r cos(ϕ− θ) + r2<

1 − r2

1 − 2r cos( 12δ) + r2(cos2( 1

2δ) + sin2( 12δ))

<1 − r2

r2 sin2( 12δ)

< ε,

if r is sufficiently close to 1.It now follows that

I2def=

∫ 2π−δ

δPr(θ − ϕ)g(eiϕ) dϕ ≤ ε

2π

∫ π

−π|g(eiϕ)| dϕ = ε‖g‖1.

The conclusion is that

h(reiθ) = P [g] = I1 + I2 < (1 + ‖g‖1)ε,

for − 12δ < θ < 1

2δ and r sufficiently close to 1. This implies (8.3.2).

Remark 8.3.3. Let F : B → D be the Riemann map from B to a simplyconnected domain D. If F extends to a homeomorphism from B to D, thenone can easily solve the Dirichlet problem on D. Let g be continuous on ∂D,then g1 = g F is continuous on ∂B. A solution h1 of the Dirichlet problemfor g1 on B leads to the solution h = h1 F−1 on D for g. See exercise 8.7.7.


8.4. Subharmonic functions and maximum principles

Definition 8.4.1. Let D be a domain in C. A function v : D →R∪−∞ is called upper semi-continuous (USC) if z ∈ D : v(z) < M isopen. Equivalently, for every z ∈ D

lim supw→z

v(w) ≤ v(z).

Equivalently, v is the pointwise limit of a decreasing sequence of continuousfunctions on D.

Observe that this definition is purely topological; it can be made on anytopological space.

Definition 8.4.2. Let D be a domain in C. A function

v : D → R ∪ −∞that is not identically −∞ is called subharmonic if it is upper semi-continuousand satisfies the Mean Value Inequality: for every a ∈ D there exists R = Rasuch that

(8.4.1) v(a) ≤ v(a, r)def=

1

2π

∫ π

−πv(a+ reiϕ) dϕ, (r < Ra).

A function v : D → R∪∞ that is not identically ∞ is called superharmonicif −v is subharmonic. Some authors call the function identically −∞ alsosubharmonic. This is a matter of taste.

Example 8.4.3. Let f be holomorphic and not identically 0 on a domainD in C. Then v = log |f | is a subharmonic function. (v is a continuousfunction to the extended reals; it satisfies the Mean Value Equality at pointswhere f 6= 0 and the Mean Value Inequality at points where f = 0.) Also|f | and |f |p (p > 0) are subharmonic, cf. Exercise 8.7.12.

Let µ be a finite, positive Borel measure on C that has compact support.The function

(8.4.2) u(z) =

∫

C

log1

|z − ζ| dµ(ζ)

can be defined as the increasing limit of the functions

(8.4.3) un(z) =

∫

C

minn, log 1

|z − ζ| dµ(ζ).

It is called the (logarithmic) potential of µ . Potentials are superharmonic,see Exercise 8.7.13, and one can show that locally all subharmonic functionscan be written as the difference of a harmonic function and a potential. SeeExercise 8.7.14.

8.4.1. Maximum principles. Subharmonic functions share an impor-tant property with harmonic ones. They cannot assume a (local) maximumon their domain, unless they are constant. Loosely speaking “subharmonicfunctions assume their max at the boundary of their domain”.

We will denote the boundary of a domain D in C by ∂D. In particularfor unbounded domains D we have opportunity to consider the boundary asa subset of the Riemann sphere C ∪∞; we denote this boundary as ∂∗D.

8.4. SUBHARMONIC FUNCTIONS AND MAXIMUM PRINCIPLES 87

Theorem 8.4.4. Let v be subharmonic on a domain D ⊂ C. Supposethat a ∈ D is such that v(a) ≥ v(z) for all z ∈ D. Then v is a constant.

Proof. We introduce the set

D1 = b ∈ D : v(b) = v(a) = maxz∈D

v(z).

The mean value inequality implies for b ∈ D∫ π

−π

(

v(b) − v(b+ reiθ))

dθ ≤ 0, 0 ≤ r < Rb.

Now if b ∈ D1, then the integrand is a non-negative lower semi-continuousfunction. It follows that the integrand is identically equal to 0 and thenB(b,Rb) ⊂ D1. Therefore D1 is open and a ∈ D1. However,

D2 = D \D1 = z ∈ D : v(z) − v(a) < 0is open because v(z) − v(a) is upper semi-continuous. Hence D2 must beempty by connectivity of D.

Theorem 8.4.5. Let v be subharmonic on a domain D. If

(8.4.4) lim supz→ζz∈D

v(z) ≤ 0

for every ζ ∈ ∂∗D, thenv(z) ≤ 0 on D.

Proof. Let M = supz∈D v(z). There exists a sequence ak in D suchthat ak converges to a limit a ∈ D ∪ ∂∗D, while v(ak) →M . If M > 0 thena ∈ D. On the compact set ak ∪ a the upper semi-continuous functionv assumes its maximum. This must be M . It follows from Theorem 8.4.4that v ≡M but this contradicts (8.4.4).

Corollary 8.4.6. Let v be subharmonic on D and let D1 be a subdo-main of D. If u is harmonic on D1 and for all ζ ∈ ∂∗D1,

lim supz→ζ

(v(z) − u(z)) ≤ 0,

then v − u ≤ 0 on D1.

Corollary 8.4.7. Let v be continuous and subharmonic on D. Then vsatisfies the mean value inequality on every disc B(a, r) that is contained inD.

Proof. Take u = P [v]B(a,r) in the previous corollary, Then

v(a) ≤ u(a) =1

2π

∫ π

−πv(a+ reit)dϕ.

Corollary 8.4.8. If v : D → R∪ −∞ is continuous, 6≡ −∞ and hasthe property that for every subdomain D1 ⊂ D

[

u harmonic on D1 and lim supz→∂∗D1

(v(z) − u(z)) ≤ 0

]

⇒ v ≤ u on D,

then v is subharmonic.


Proof. Let B(a, r) ⊂ D. Consider the harmonic function

u = P [v]B(a,r).

At ζ ∈ C(a, r) we have by Theorem 8.3.2 that

lim infz→ζ,|z−a|<r

u(z) ≥ lim infw→ζ,w∈C(a,r)

v(w) = v(ζ).

Hence u ≥ v on B(a, r). But

u(a) =1

2π

∫ π

−πv(a+ reiϕ)dϕ,

and because v(a) ≤ u(a), the mean value inequality for v at a has beenproved.

8.4.2. Further properties of subharmonic functions.

Theorem 8.4.9. Let vn be a sequence of subharmonic functions on adomain D in C.

(1) The maximum of finitely many vn is subharmonic.(2) The supremum

h(z) = supnvn(z)

is subharmonic if h is upper semi-continuous.(3) If the sequence vn is pointwise monotonically decreasing to a func-

tion h that is not identically −∞, then

h(z) = limn→∞

vn(z)

is a subharmonic function.

Proof. For (1) and (2) we have to check the mean value inequality. Forevery j = 1, 2, . . . we have

(8.4.5) vj(a) ≤ vj(a, r) ≤1

2π

∫ π

−πsupkvk(a+ reiϕ) dϕ = h(a, r).

Hence also h(a) ≤ h(a, r).For (3) we first note that the limit of a decreasing sequence of upper semi-

continuous functions is again upper semi-continuous. Indeed, z : h(z) <a = ∪jz : vj(z) < a is open. Observe that by Corollary 8.4.7 all vj satisfythe Mean Value Inequality on every disc in D. The Mean Value Inequalityfor h will be a consequence of Fatou’s Lemma from integration theory, seee.g. [39]. The usual form of this lemma reads

∫

Xlim inf undµ ≤ lim inf

∫

Xun dµ,

for a sequence of positive functions un on a measure space (X,µ).Our functions vj are bounded from above; adding a negative constant,

we can assume that the vj are negative on C(a, r). Fatou’s Lemma is then

8.4. SUBHARMONIC FUNCTIONS AND MAXIMUM PRINCIPLES 89

applied to the sequence −vj. This leads to

(8.4.6) h(a) = lim supj→∞

vj(a) ≤ lim supj→∞

1

2π

∫ π

−πvj(a+ reiϕ) dϕ

≤ 1

2π

∫ π

−πlim supj→∞

vj(a+ reiϕ) dϕ = h(a, r).

Corollary 8.4.10. The Corollaries 8.4.7 and 8.4.8 are also valid ifcontinuity is replaced by upper semi-continuity.

Proof. For Corollary 8.4.7 Let un be continuous on C(a, r) and suchthat un ↓ v on C(a, r). Denote the harmonic extension of un to B(a, r) alsoby un. Then by Corollary 8.4.6 and Fatou’s lemma

v(a) ≤ lim supn→∞

un(a) ≤ lim supn→∞

1

2π

∫ π

−πun(a+re

iϕ)dϕ =1

2π

∫ π

−πv(a+reiϕ)dϕ.

The argument for Corollary 8.4.8 is similar.

Lemma 8.4.11. Let D be a domain in C and let v ∈ C2(D). The functionv is subharmonic if and only if ∆v ≥ 0.

Proof. We pick a point a ∈ D, which we can assume to be 0, anddevelop v in a Taylor series about a = 0.

(8.4.7) v(x+ iy) = v(0) + vx(0)x+ vy(0)y+

+ 12vxx(0)x

2 + vxy(0)xy + 12vyy(0)y

2 + o(r2).

Here r2 = x2 + y2 and vx, vxy denote a partial derivative with respect to x,respectively, a second partial derivative with respect to x and y, etc. Averag-ing over C(0, r) and keeping in mind that

∫

C(0,r) x dθ = 0,∫

C(0,r) x2 dθ = πr2

leads to

(8.4.8) v(0, r) = v(0) +1

4∆v(0)r2 + o(r2).

Suppose that the mean value inequality holds for small r. Then we seefrom (8.4.8) that ∆v(0) ≥ 0 by letting r go to 0. In the other direction, if∆v(0) > 0 then (8.4.8) yields that for small r the Mean Value Inequalityholds. Finally, suppose that ∆v ≥ 0 on D. Let

vn(z) = v(z) +|z|2n.

Because ∆vn ≥ 1/n, vn is subharmonic. Also vn ↓ v, and Theorem 8.4.9, no3, shows that v is subharmonic.

The next theorem shows that for bounded subharmonic functions Theo-rem 8.4.5 may be strengthened in the sense that small exceptional sets canbe allowed. Cf. Exercise 8.7.18.

Theorem 8.4.12. Let v be a subharmonic function, bounded by M on abounded domain D. Assume

lim supz→ζ,z∈D

v(z) ≤ 0, ζ ∈ ∂D \ ζ1, . . . , ζn.


Then v ≤ 0 on D.

Proof. Let d = diamD <∞. The function

vε(z) = v(z) + εn∑

k=1

log |z − ζk|d

, ε > 0

is subharmonic and has lim supz→ζ vε(z) ≤ 0 for all ζ ∈ ∂D. By Theorem8.4.5 vε(z) ≤ 0 on D for every ε > 0. Letting ε ↓ 0 with z fixed, we obtainv(z) ≤ 0.

Proposition 8.4.13. Harmonic functions are completely determined bytheir boundary values. More precisely, let u be harmonic on a domain D. If

limz→ζ

u(z) = 0, for all ζ ∈ ∂D,

then u ≡ 0. The Dirichlet problem on D with finite continuous boundaryvalues has at most one solution.

Proof. Applying Theorem 8.4.5 to u as well as to −u gives the firststatement. If v1 and v2 are two solutions for the Dirichlet problem, thenv1 − v2 has boundary values 0 and hence is identically equal to 0.

8.5. Harmonic measure

We begin with a preliminary definition of harmonic measure; later onwe will give a more comprehensive definition.

Definition 8.5.1. Let D be a domain with boundary ∂∗D consisting offinitely many Jordan curves. Let E be a subarc of ∂∗D. Harmonic measureω(E, z0,D) at z0 ∈ D of E relative to D is the value at z0 of the boundedharmonic function ω(E, z,D) on D that has boundary values 1 at points ofEo and 0 at points of ∂∗D \ E; it is the solution of the Dirichlet problemwith boundary values 1 at E and 0 at ∂D \E.

Remark 8.5.2. The interior and closure of E may be defined via theparametrizing interval of a Jordan curve in the boundary (or the relativetopology) of the boundary. At endpoints of E the boundary values are notprescribed; by Theorem 8.4.12 these points are of no consequence for thevalues of ω(E, z,D), but see Exercise 8.7.20 for what happens if we leaveout boundedness.

At this point we only know that the Dirichlet problem can be solved forthe ball and for domains that are conformal to the ball via a Riemann mapthat extends as a homeomorphism of the boundary. (This explains partlythe restrictions that we have put on D).

From the maximum principle, Theorem 8.4.12, we obtain immediately

Proposition 8.5.3. Let D be as above and let E, Ei be proper, nonemptysubarcs of ∂∗D. Then for z ∈ D

0 < ω(E, z,D) < 1.

Also,

ω(∂∗D, z,D) ≡ 1.

8.5. HARMONIC MEASURE 91

Harmonic measure is monotonic:

E1 ⊂ E2 ⇒ ω(E1, z,D) ≤ ω(E2, z,D),

and harmonic measure is addditive: If E1 ∩ E2 = ∅, then

ω(E1, z,D) + ω(E2, z,D) = ω(E1 ∪ E2, z,D).

Combined with some elementary measure theory this proposition impliesthat for fixed z ∈ D the function ω(E, z,D) extends to a probability measureon the Borel measurable subsets E of ∂∗D.

Example 8.5.4. (A) On B = B(0, 1) we compute harmonic measure ofan arc

γ = ζ = eiϕ : α ≤ ϕ ≤ β ⊂ C(0, 1).

This is just the Poisson integral.

ω(γ, z,B) =

∫ β

αPr(θ − ϕ) dϕ, z = reiθ.

In particular

ω(γ, 0,B) =β − α

2π.

(B) Next we study the half disc

(8.5.1) SR = z : 0 < |z| < R, 0 < arg z < π.Let CR = Reiϕ, 0 ≤ ϕ ≤ π. The harmonic measure is given by

(8.5.2) ω(Cr, z, SR) =2

πp.v. arg

R+ z

R− z.

This formula is most easily obtained with some Fourier analysis. Let f bedefined on C(0, R) by

f(R(eiθ)) = 1 if 0 ≤ θ ≤ π;

f(R(eiθ)) = −1 if −π < θ < 0.

Let u(z) = P [f ](z), the Poisson integral of f . Then u(z)+u(z) is a boundedharmonic function with boundary values 0, so u(z) = −u(z) and it followsthat u(x) = 0, (−1 < x < 1). We conclude that ω(CR, z, SR) = u(z).To compute u, recall that P [f ](z) can be written as a Fourier-type series.Because f is odd, using a sine-series is most natural.

P [f ](z) =∞∑

n=0

an[f ]rn

Rnsin(nϕ), z = reiϕ,

where

an =1

π

∫ π

−πf(θ) sin(nθ) dθ =

2

π

∫ π

0sin(nθ) dθ =

4

nπ, for n odd,

0, for n even.

We infer that

P [f ](z) =∑

n odd

4

nπ

rn

Rnsin(nϕ) =

∑

n odd

4

nπ

Im zn

Rn

=2

πIm p.v. log

1 + z/R

1 − z/R=

2

πp.v. arg

R+ z

R− z.


(C) For a third example, let H be the upper half space, and Lt be thehalfline Lt = y = 0, x ≤ t. Then

ω(Lt, z,H) =1

πIm p.v. log(z − t) =

1

πp.v. arg(z − t).

The harmonic measure of an interval [a, b] easily follows:

ω([a, b], z,H) = p.v. argz − b

z − a.

Compare Exercise 8.7.9.

The next Two-Constants Theorem is a bit awkwardly formulated, dueto our limited definition of harmonic measure. It goes back to Carleman,the brothers Nevanlinna and Ostrowski. See e.g. [31].

Theorem 8.5.5 (Two-Constants Theorem). Let v be a bounded subhar-monic function on a domain D, the boundary ∂∗D of which consists of afinite union of Jordan curves in C∗. Let E be a subset of ∂∗D consisting offinitely many Jordan arcs. Suppose that

(8.5.3) lim supz→ζ

v(z) ≤

m1, for ζ ∈ E, ζ not an end point,

m2, for ζ ∈ ∂∗D \ E.

Then

(8.5.4) v(z) ≤ m1ω(E, z,D) +m2(1 − ω(E, z,D))def= u(z).

Proof. The righthand side u(z) of (8.5.4) represents a bounded har-monic function. The function v(z)− u(z) is subharmonic, bounded and haslim supz→∂D∗ v(z) − u(z) ≤ 0, except maybe at the end points of E. ThusTheorem 8.4.12 applies and the conclusion is that v ≤ u on D.

8.6. Phragmen-Lindelof Theorems

Harmonic measure is a fine tool to prove maximum principles for holo-morphic functions on unbounded domains. Results in this section are dueto Phragmen and Lindelof around 1906.

Theorem 8.6.1. Let f be holomorphic on the upper half-space H =Im z > 0. Suppose that

lim supz→ξ

|f(z)| ≤ B for all ξ ∈ R.

If f is of exponential type 0, that is, for every ε > 0 there exists an Rε suchthat

(8.6.1) |f(z)| < eε|z|, |z| > Rε;

or if, with M(R)def= sup0<ϕ<π |f(R(eiϕ))|, more weakly

(8.6.2) lim infR→∞

logM(R)

R≤ 0.

Then

|f(z)| ≤ B on H.

8.6. PHRAGMEN-LINDELOF THEOREMS 93

Proof. We may assume f 6≡ 0 and B = 1. (In case B = 0 we addε > 0, show that |f | < ε, and let ε→ 0.) We will apply the Two-ConstantsTheorem 8.5.5 to the bounded subharmonic function v = log |f | on the halfdisc SR defined in (8.5.1). We fix z = reiθ and take E = CR, m1 = logM(R),m2 = logB = 0 and obtain

(8.6.3) log |f(z)| ≤ ω(CR, z, SR) logM(R).

By (8.5.2)

ω(CR, z, SR) < 2r/R, for R > 2r.

The weakest condition (8.6.2) in the theorem gives already that for everyε > 0 there exists a sequence Rk tending to ∞ with logM(Rk) < εRk.The conclusion is that

log |f(z)| ≤ 2εr.

Letting ε→ 0 we find that log |f(z)| ≤ 0, or |f(z)| ≤ 1.

Corollary 8.6.2. Let f be holomorphic in a sector Sα = 0 < arg z <α and continuous on Sα and such that

|f(z)| ≤ B on ∂Sα.

If f satisfies

lim infR→∞

logM(R)

Rπ/α≤ 0, with M(R) = sup

0<ϕ<α|f(R(eiϕ))|,

then

|f(z)| ≤ B on Sα.

Proof. The conformal mapping z 7→ p.v. zα/π sends H to Sα. Thefunction F (z) = f(zα/π) is defined on H and satisfies the conditions ofTheorem 8.6.1, therefore it is bounded by B and so is f .

We can now refine Theorem 8.6.1 somewhat.

Corollary 8.6.3. Let f be holomorphic on H continuous on H andsuch that

|f(x)| ≤ B on R.

Suppose that f is of bounded exponential type on H,

lim supR→∞

logM(R)

R= τ.

Then

|f(x+ iy)| ≤ Beτy on H.

Proof. Even careful analysis of (8.6.3) gives only log |f |/B ≤ 4πτy. We

need a trick. Apply Corollary 8.6.2 to the function f(z) = f(z)eiτz on both

of the sectors z : arg z ∈ (0, π/2) and z : arg z ∈ (π/2, π). Since f is

bounded on the positive imaginary axis, we find that f is bounded on eachof these sectors. By Theorem 8.6.1 f is bounded by B. We conclude that

|f(x+ iy)| ≤ B|e−iτ(x+iy)| = Beτy on H.


8.7. Exercises

8.7.1. Show that a bounded harmonic function on C is constant.

8.7.2. Determine the harmonic functions that only depend on Re zand the ones that only depend on |z|.

8.7.3. Derive the Fourier representation for Pr given by (8.2.10).

8.7.4. Let µ be a Borel measure on T. Show that P [µ] ≡ 0 impliesthat µ = 0.

8.7.5. Show that a bounded subharmonic function u on B \ 0 can beextended to a bounded subharmonic function on B. (Consider u+ ε log |z|.)Can you generalize this result?

8.7.6. Prove that the Dirichlet problem on B \ 0 with boundaryfunction f(eiθ) = 0, f(0) = 1 has no solution.

8.7.7. Let f be a conformal map on the unit disc B that extends to ahomeomorphism B → f(B). Solve the Dirichlet problem on a f(B).

8.7.8. Show that a bounded subharmonic function on C is constant.(Think Riemann sphere!)

8.7.9. Complete the computation of ω([a, b], z,H) in example 8.5.4.

8.7.10. Prove that

u(z) =

∞∑

n=1

2−n log |z − 1/n|

is a subharmonic function on the disc B(0, 1/2) that is not continuous at 0.Can you show that u is subharmonic on C?

8.7.11. Show that a radial function f , (i.e. f(z) = f(|z|)), is subhar-monic if and only if it is a convex function of log r.

8.7.12. Let u be subharmonic on a domainD and let ϕ be an increasing,convex function. Show that ϕ f is subharmonic. [One can try the smoothcase first, i.e u and ϕ are C2 and ϕ′, ϕ′′ ≥ 0.] For the general case one mayuse Jensen’s inequality, see [39]. It reads Let µ be a probability measure onX and let f ∈ L1(dµ) be positive. If ϕ is convex on the image of f , then

ϕ(

∫

Xf dµ) ≤

∫

Xϕ f dµ.

Show that for p > 0 and f holomorphic on D the function |f |p is sub-harmonic.

8.7.13. Show that the potential u(z) defined in 8.4.2 is indeed super-harmonic. Compute ∆u in distributional sense.

8.7.14 (Green’s identities). Let G be a domain as in Theorem 1.1.1.We use the notation of Lemma 8.4.11, and the standard notation for thegradient; ∇u = (ux, uy). Apply Theorem 1.1.1 to f = uvx, g = −uvy to

obtain Green’s second formula for u, v ∈ C2(G)

(8.7.1)

∫

∂Guvx dy − uvy dy =

∫

G(u∆v + ∇u∇v) dxdy.

8.7. EXERCISES 95

Next show that

(8.7.2)

∫

∂G(uvx − vux) dy − (uvy − vuy) dx =

∫

G(u∆v − v∆u) dxdy.

Now let ∂∂N be the directional derivative in the direction of the unit outward

normal in a point of ∂G. Derive Green’s third formula

(8.7.3)

∫

∂Gu∂v

∂N− v

∂u

∂N=

∫

G(u∆v − v∆u) dxdy.

8.7.15. (Continued.) Show that if v is subharmonic on G, then

(8.7.4)

∫

∂G

∂v

∂Nds ≥ 0,

and that equality occurs for harmonic functions.

8.7.16. Use Green’s formula (8.7.3) to prove a representation formulafor C2-functions v. (Apply (8.7.3) with u = log(1/|z − a| on the domainsGδ = G \B(a, δ), let δ → 0 and study what happens on C(a, δ)).

v(a) =1

2π

∫

∂G

∂v

∂Nlog(1/|z − a|) ds +

1

2π

∫

∂G(−v)∂ log 1/|z − a|

∂Nds

+1

2π

∫

G(−∆v) log

1

|z − a|dxdy.

(8.7.5)

8.7.17. Show that v ∈ C2(B(0, R) has the representation

(8.7.6) v(0) =1

2π

∫ π

−πv(Reiϕ) dϕ− 1

2π

∫

B(0,R)∆v(z) log

R

|z|dxdy.

If v is a subharmonic function of the form log |f |, compare this to Jensen’sformula (Theorem 6.4.1). Green’s function that will be introduced in thenext chapter, Definition 9.4.1, will allow such a representation for manydomains.

8.7.18. Call a set X ⊂ C polar if there exists a subharmonic funtion onC such that u|X = −∞. Formulate and prove a version of Theorem 8.4.12for functions that are bounded off a polar subset of the circle.

8.7.19. Show that a countable union of polar sets is polar.

8.7.20. Show that the Poisson kernel had boundary values 0 on C(0, 1)\1. Make a connection to Theorem 8.4.12.

8.7.21. Analyze the boundary behavior of the function

Im

(

1 + z

1 − z

)2

, (z ∈ B)

8.7.22. Consider the function

f(z) =

∫ ∞

0eztt−t.

a: Show that f is entire.b: Show that f is bounded on the half plane Re z < 0.


c: By a suitable change of the path of integration, show that for τ ∈(−π/2, π/2) the function f is bounded on the half plane Re (zeiτ ) <0.

d: Compute the order of f . [This can be done similarly to the waythe refinements of Stirling’s formula are proven, or by using thePhragmen-Lindelof theory.]

e: Modify f to a function that is bounded on every line through 0.f: Adding a constant, make sure that f has zeros and modify f to

an entire function that for every θ ∈ [0, 2π) has the property thatlimr→∞ g(reiθ) = 0.

CHAPTER 9

The Dirichlet problem

In this chapter we will continue our study of subharmonic functions andtry to solve the Dirichlet problem in quite general situations. This will bedone by means of the Perron method. It is noteworthy that the ideas ofPerron can be used for solving Dirichlet type problems for large classes ofpartial differential equations, cf. [20].

9.1. Preliminaries

Let D be a simply connected domain in C. In Remark 8.3.3 we havenoticed that if f : B → D is the Riemann map, and D has nice boundary,then we can solve the Dirichlet problem on D by pulling back the data toB, solve the Dirichlet problem on B and pushing everything back to D. Infact one obtains the following formula.

u(w) = P [ϕ f ](f−1(w)),

where ϕ is the boundary function and u the solution to the Dirichlet problem.But we have also seen, see Exercise 8.7.6 that the punctured ball B\0

is an example of a domain for which one can not solve the Dirichlet problem,and in any case if the domain is not simply connected, a method based onthe Riemann mapping theorem cannot work.

Let us start with a slightly more precise definition of the Dirichlet prob-lem and then define Dirichlet domains.

Definition 9.1.1. The Dirichlet problem (for the Laplacian) for a realvalued boundary function g defined on the boundary ∂∗D of a domain D isto find a function u on D such that

(9.1.1) ∆u = 0, limz→ζ

u(z) = g(ζ), ζ ∈ ∂∗D.

This is tricky where g is not continuous. A domain D is called a Dirichletdomain if the Dirichlet problem on D can be solved for every continuousboundary function g.

9.2. Perron’s method

If u is harmonic on a domain D and has boundary function g, then,by the maximum principle 8.4.5 u majorizes every subharmonic functionwith boundary function g, and a fortiori every subharmonic function withboundary function smaller or equal to g. In other words, if u is the solutionof the Dirichlet problem with boundary function g, then u is the sup of allsubharmonic functions with boundary values less than g. Perron had thestriking idea to use this property to define u.

97

98 9. THE DIRICHLET PROBLEM

Definition 9.2.1. Suppose that D is a domain in C∗ with (extended)boundary ∂∗D and suppose that g is a function on ∂∗D with values in theextended reals. The Perron family P(g) = P(g,D) is the set consisting ofsubharmonic functions v on D with

(9.2.1) lim supz→ζ

v(z) ≤ g(ζ), for all ζ ∈ ∂∗D.

The Perron function U is the supremum of the Perron family:

(9.2.2) U(z) = supv∈P

v(z).

We observed already that if the Dirichlet problem has a solution, thenU = u. Our programme will be to show first that U is harmonic if the Perronfamily is non-empty, and second to formulate conditions that guarantee thatthe Perron function is the solution of the Dirichlet problem. Notice that P(g)contains some constant functions if g is bounded away from −∞.

9.2.1. Harmonicity of the Perron function. As auxiliary resultswe need two propositions of Harnack.

Proposition 9.2.2 (Harnack’s inequalities). Let u be a non-negative

harmonic function on B(a,R). Then for 0 ≤ r < R we have

R− r

R+ ru(a) ≤ u(a+ reit) ≤ R+ r

R− ru(a).

On B(a,R/2) this gives

1

3u(a) ≤ u(a+ reit) ≤ 3u(a), |z − a| < R/2.

Proof. For 0 ≤ r < R

(R− r)2 ≤ R2 + 2rR cos(θ − ϕ) + r2 ≤ (R+ r)2,

hence for the Poisson kernel P

R− r

R+ r≤ 2πPr/R(θ − ϕ) =

R2 − r2

R2 + 2rR cos(θ − ϕ) + r2≤ R+ r

R− r.

Now u is continuous and positive on C(a,R), so that we can estimate itsPoisson integral as follows.

R− r

R+ r

∫ π

−πu(a+Reiϕ)

dϕ

2π≤∫ π

−πPr/R(θ − ϕ)u(a+Reiϕ) dϕ

≤ R+ r

R− r

∫ π

−πu(a+ reiϕ)

dϕ

2π.

(9.2.3)

Now u is harmonic so the first and last integral equal u(a) and we aredone.

Proposition 9.2.3 (Harnack’s convergence principle). Let unn be anon-decreasing sequence of harmonic functions on a domain D. u1(z) ≤u2(z) ≤ . . . , and let

u(z) = limn→∞

un(z), z ∈ D.

Then either u is harmonic on D or u is identically equal to +∞. Theconvergence is in both cases uniform on compact subsets of D. (In the latter

9.2. PERRON’S METHOD 99

case this means for every M there exists n0 such that un > M on ourcompact set if n > n0).

Proof. Observe that u(z) is well defined (possibly +∞) for every z ∈D.

Suppose that a ∈ D and u(a) <∞. Pick R > 0 such that B(a, 2R) ⊂ D.Let ε > 0. For n sufficiently large and all k > 0 we have

0 < (un+k − un) (a) < ε.

Hence by Harnack’s inequalities applied on B(a, 2R)

0 < (un+k − un) (z) < 3ε, z ∈ B(a,R).

The sequence un converges thus uniformly on B(a,R) and the limit func-tion U is harmonic on B(a,R). It follows that the set D1 where u is finiteand harmonic is open and the convergence in D1 is uniform on compactsubsets.

Next, suppose that u(a) = +∞. Again, pick R > 0 such that B(a, 2R) ⊂D. Let M > 0. For all n sufficiently large we find un(a)− u1(a) > M . NowHarnack’s inequalities imply that

un(z) − u1(z) > M/3, z ∈ B(a,R).

In other words, unn converges uniformly to +∞ on B(a,R) and it followsthat the set D2 where u = +∞ is open and that on D2 the sequence unnconverges uniformly to +∞ on compact subsets.

Because D = D1 ∪D2 is connected, either D1 or D2 must be empty.

Definition 9.2.4 (Poisson modification). Suppose that v is subhar-monic on a domain D and B(a,R) ⊂ D. The Poisson modification of vrelative to B(a,R) is the function v(= vB(a,R)) obtained by replacing v byits Poisson integral on B(a,R), that is,

v(z) =

P [v]B(a,R) for z ∈ B(a,R);

v(z) for z ∈ D \B(a,R).

Lemma 9.2.5. The Poisson modification v of a subharmonic functionv on D is subharmonic on D and harmonic on B(a,R). In particular, itdoesn’t assume the value −∞ on B(a,R).

Proof. We only have to check subharmonicity at points ζ ∈ C(a,R).Because v ≥ v on B(a,R) and v(ζ) = v(ζ) and v satisfies the mean valueinequality, the mean value inequality holds at ζ. Since

lim supz∈B(a,R)→ζ

P [v]B(a,r) ≤ lim supξ∈C(a,r)→ζ

v(ξ) and lim supz→ζ

v(z) ≤ v(ζ),

the function v is upper semi-continuous.

Theorem 9.2.6 (Harmonicity of the Perron function). Let D be a do-main in C∗ and g a function on ∂∗D with values in the extended reals. ThePerron function U is harmonic on D or constant ∞ or −∞. The latter onlyoccurs if P(g) = ∅.


Proof. Suppose that P(g) 6= ∅. Fix a ∈ D and a ball B(a,R) ⊂ D.Let vnn be a sequence in P(g) such that limn→∞ vn(a) = U(a). Formun = maxv1, . . . , vn ∈ P(g) and the Poisson modifications un with respectto B(a,R). Observe that un ∈ P(g), the sequence unn is increasing and unis harmonic on B(a,R). From Harnack’s convergence principle we concludethat U0 = limn→∞ un is either identically +∞ or harmonic on B(a,R).

We claim that U = U0 on B(a,R). If U0 = +∞ on B(a,R) we have noth-ing left to prove. We know already that U0 ≤ U on B(a,R). Let b ∈ B(a, r)and let wnn be a sequence in P(g) such that limn→∞wn(b) = U(b). Formw′n = maxw1, . . . , wn, vn and the corresponding Poisson modifications wn.

As above W0 = limn→∞ wn is either identically +∞ or harmonic on B(a, r)Observe that W0(a) = U0(a) = U(a) < ∞. Thus W0 is harmonic, and byconstruction W0 ≥ U0 on B(a,R). But then the maximum principle appliedto W0 − U0 yields W0 = U0 on B(a,R), so that U(b) = W0(b) = U0(b).

The conclusion is that the set where U is harmonic is open and the setwhere U = +∞ is also open. Hence one of them is empty and the proof isfinished.

9.3. Boundary behavior of the Perron function

The behavior at the boundary of the Perron function P (g,D) dependsof course both on g as well as on D, but the main influence comes fromD and more precisely we shall see that the boundary behaviour at a pointζ ∈ ∂∗D is mostly determined by local properties of ∂∗D at ζ. We need thenotion of barrier.

Definition 9.3.1. Let D be a domain and ζ ∈ ∂∗D. A barrier or barrierfamily at a is a collection of positive superharmonic functions

βδ(z) = β(z, a, δ), 0 < δ < δ0,

which are defined on D and have the property that

limz→a

βδ(z) = 0, βδ(z) ≥ 1 for z ∈ D \B(a, δ).

Theorem 9.3.2. Let D be a domain in C∗, a ∈ ∂∗D and g a boundedreal valued function on ∂∗D. Suppose that there exists a barrier βδ at a,and suppose that the Perron function U for g is not equal to −∞. Then

lim infζ∈∂∗D→a

g(ζ) ≤ lim infz∈D→a

U(z) ≤ lim supz∈D→a

U(z) ≤ lim supζ∈∂∗D→a

g(ζ).

Proof. As in the proof of Theorem 8.3.2 it suffices to show that the lastinequality holds under the additional assumption that lim supζ→a g(ζ) = 0.Let ε > 0, then there is a neighborhood T of a in ∂∗D such that g < ε onT . Since g is bounded, there exists a barrier βδ and a constant M such thatMβ + ε > g on T . Now if v ∈ P(g), then v −Mβ − ε is subharmonic andless than 0 at ∂∗D, hence less then 0 on D. This holds for all v ∈ P(g),hence also for U . It follows that

limz∈D→a

U(z) ≤ limz∈D→a

Mβ(z) + ε = ε.

Since ε can be chosen arbitrarily small, the statement is proved.

9.3. BOUNDARY BEHAVIOR OF THE PERRON FUNCTION 101

Remark 9.3.3. To derive the first inequality from the third, one usesthat if u is harmonic so is −u. However, the Perron method also worksfor classes that are not closed under multiplying by -1. So a direct proofof the first inequality can be useful. Here it is. We can assume thatlim infζ→a g(ζ) = 0. Let ε > 0. Then g(ζ) > −ε on a neighborhood ofa. Therefore, with M sufficiently large and δ sufficiently small, the functionv(z) = −Mβδ(z) − ε will belong to P(g), and we have

lim infz→a

U(z) ≥ limz→a

v(z) = −ε.

Corollary 9.3.4 (Dirichlet problem). Suppose that D is a domain withthe property that each of its boundary points admits a barrier, then D isa Dirichlet domain. I.e. the classical Dirichlet problem (with continuousboundary function g) can be solved on D.

Remark 9.3.5. In fact if g is bounded and continuous except for a finiteset of discontinuities, then U has the right boundary values at the pointsof continuity, and is bounded. The Perron function U is often called thePerron solution of the Dirichlet problem, whether or not it has the rightboundary values. It is also convenient to call U the Perron subsolution, incontrast to the completely analogously defined function

H(z) = infv∈G

v(z),

where G consists of the superharmonic functions with boundary values ≥ g.H is called the Perron supersolution. It is also harmonic if g is bounded.

Example 9.3.6. Let D be a domain in C and let a ∈ ∂∗D have theproperty that there exists a C1 arc γ that ends in a. Then there exists abarrier at a. Indeed, after translation and rotation, we can assume thata = 0 and that, for δ0 sufficiently small, γ ∩B(0, δ) is contained in a closedsector S = z = reit : 0 ≤ r < δ0, |t − π| ≤ π/4. Then there exists a

holomorphic branch f of4√z on B(0, δ) \ γ that has positive real part (and

is real on the positive real axis). In particular Re f > 0 on D ∩ B(0, δ0).We define for 0 < δ < δ0 a barrier as follows.

(9.3.1) βδ(z) = min 1

δ1/4Re f(z), 1, z ∈ D ∩B(0, δ0),

and extend it by 1 on the rest of D.

With a little more effort we can prove the following result.

Theorem 9.3.7. Let D be a domain in C∗, a ∈ ∂∗D a generalized bound-ary point. Suppose that a belongs to a continuum K in C∗\D, i.e. a compactconnected set K ⊂ C∗ \D. Then D admits a barrier at a.

Proof. The components of C∗ \ K are simply connected open sets.The domain D is contained in one of these, and enlarging it if necessarywe can assume that D is simply connected. By a suitable fractional lineartransformation we can achieve that a = 0 and ∞ ∈ K, moreover it sufficesto construct β1. It then suffices to construct a harmonic function β0 onB \K that has values in (0, 1), satisfies β0(z) → 0 as z ∈ D → 0, and hasboundary values 1 at D ∩ C(0, 1). The function β0 can be extended to allof D by 1 and the resulting function will be superharmonic.


It is convenient to transfer our problem to the upper halfplane H =w : Imw > 0 via a holomorphic branch f of 1

i log z. Such a branch iswell-defined on D because D is simply connected and 0 /∈ D. In fact f isconformal and injective. Note that for every 0 < r ≤ 1 D ∩ C(0, r) is acountable disjoint union of arcs of total length ≤ 2πr. The image underf of D ∩ C(0, r) consists of countably many disjoint segments in the liney = − log r with total length equal to the increase of the argument of zover these arcs, which is less than 2π. In particular D ∩ C(0, 1) is mappedonto a countable union E of disjoint intervals Ej of the reals with totallength less than 2π. If z → 0 in D, then Im f(z) → ∞ in H.

Let E = [a, b] an interval in R. We computed in Example 8.5.4 C theharmonic measure

ω(E,w,H) =1

πarg

w − b

w − a=

1

π

∫ b

a

v

(u− t)2 + v2dt.

Thus

h(w) =∑

j

ω(Ej , w,H) =1

π

∫

E

v

(u− t)2 + v2dt

is a harmonic function (use Harnack’s convergence principle). In fact itfollows just as in the case of the ball, Theorem 8.3.2 that h has boundaryvalues 1 on E and 0 on R \ E. We have the easy estimate

h(w) ≤ 1

π

∫

E

1

vdt ≤ 2

v→ 0, as v → ∞.

It follows that the function

β0(z) = h(f(z))

has all the properties that we demand.

Corollary 9.3.8. If the extended boundary of a domain D is a unionof continua, then D is a Dirichlet domain. If D is simply connected and notequal to C, then D is a Dirichlet domain. (The boundary is a continuum).

We can now give a general definition of harmonic measure on Dirichletdomains.

Theorem 9.3.9. Let D be a Dirichlet domain in C∗. Then for everyz ∈ D there exists a measure ωz on ∂∗D such that for all g ∈ C(∂∗D)

(9.3.2) Ug(z) =

∫

∂∗Dg(ζ) dωz(ζ).

Here Ug is the (Perron-)solution for g of the Dirichlet problem on D. Forevery bounded Borel measurable function g on ∂∗D the integral (9.3.2) isharmonic. In particular, for a Borel subset E in ∂∗D, the function

ω(E, z,D) = ωz(E) =

∫

Edωz(ζ)

is harmonic in z.

9.4. GREEN’S FUNCTION 103

Proof. The linearity of the operator g 7→ Ug(z) follows from the uniquesolvability of the Dirichlet problem in Dirichlet domains. By the Riesz repre-sentation theorem, there exists a measure ωz that represents this functional.Of course the integral (9.3.2) is harmonic if g is continuous.

The set L of bounded measurable functions g with the property that(9.3.2) is harmonic is a linear subspace of L∞(∂∗D). By Harnack’s monotoneconvergence theorem, it is also closed under monotone pointwise limits. Itfollows that L contains the indicator functions of (relatively) open subsets of∂∗D. The Monotone Class Theorem, cf. [5] now states that L = L∞(∂∗D).As a consequence, the indicator functions of measurable set E ⊂ ∂∗D are inL. This proves the last statement.

Remark 9.3.10. The theorem goes through for domains that are notDirichlet, but it is more difficult to prove the linearity (in g) of the Perronsolution Ug. See Exercises 9.7.1 and its successors.

9.4. Green’s function

From the theory of partial differential equations the reader may knowGreen’s function for a linear partial differential operator P on a domainΩ with respect to x ∈ Ω as the fundamental solution at x of P that hasboundary values 0. A fundamental solution at x is a function Fx such that

PFx = δx,

the Dirac distribution at x. Our Green’s function will actually be thesame thing, with the differential operator being the Laplacian, but we willavoid the theory of partial differential equations and distributions as muchas possible.

Definition 9.4.1. Let D ⊂ C∗ be a subdomain of C∗, and a a point inD. A function G(z, a) = Ga(z) is called a Green’s function for D with poleat a if it satisfies the following conditions

(1) Ga is harmonic on D \ a.(2) There exists a harmonic function ha in a neighborhood U of a such

that

Ga(z) = log1

|z − a| + ha(z) on U a 6= ∞.

If a = ∞ the condition is

G∞(z) = log |z| + h(z),

with h harmonic in a neighborhood of ∞.(3)

limz→ζ

Ga(z) = 0, for all ζ ∈ ∂∗D.

Remark 9.4.2. Green’s function need not exist. If the boundary istoo small, the boundary behavior makes no sense (D = C∗) or can not beachieved (D = C \ a1, a2, . . . , an). See Exercise 9.7.7.

Simple properties of Green’s function are gathered in the following propo-sitions

Proposition 9.4.3. Let D be a domain in C∗, a ∈ D.


(1) There is at most one Green’s function g(z, a) for D with pole at a.(2) Green’s function is positive and superharmonic on D.

Proof. For 1. observe that the difference of two Green functions isharmonic and has boundary values identically 0, hence the difference is thezero function. Next, 2. follows from the minimal principal for superharmonicfunctions.

Proposition 9.4.4. Suppose that f : z → f(z) is a conformal map froma domain D1 to D2, a ∈ D1, b = f(a) ∈ D2. Assume that Green’s functionfor D2 at b exists and denote it by G2(w, b). Then

G1(z) = G1(z, a) = G2(f(z), b) = G2(f(z), f(a))

is Green’s function for D1 at a.

Proof. It is clear that G1 is a positive superharmonic function that isharmonic on D1 \ a. At a we have for some harmonic function h, definedon a neighborhood of b in D2

G1(z, a) = G2(f(z), f(a)) = − log |f(z) − f(a)| + h(f(z))

= − log |z − a| − log

∣

∣

∣

∣

f(z) − f(a)

z − a

∣

∣

∣

∣

+ h(f(z)) = − log |z − a| + h(z).

(9.4.1)

Here h is harmonic in a neighborhood of a because f(z)−f(a)z−a is holomorphic

on D1 and h f is harmonic in a neighborhood of a. Finally, suppose thatzn → ζ ∈ ∂∗D1, and that a subsequence, which we may also denote by zn,has the property that G2(f(zn), b) → c > 0. Taking another subsequence ifnecessary, we can assume that limn→∞ f(zn) = w0 exists. Then w0 ∈ D2

but not in ∂∗D2, but then zn = f−1(f(zn)) → f−1(w0) ∈ D1, which is acontradiction.

Example 9.4.5. Green’s function for the unit disc B with pole at 0is obviously G(z, 0) = − log |z|. Applying an automorphism f(z) = (z −a)/(1 − az) of B with f(a) = 0 we obtain

G(z, a) = G(f(z), f(a)) = − log

∣

∣

∣

∣

z − a

1 − az

∣

∣

∣

∣

.

A slight variation gives us Green’s function for the upper halfspace H.

G(z, a) = − log |z − a| + log |z − a|.Indeed (z−a)/(z− a) maps H to B and a to 0. We can also observe directlythat for x ∈ R x− a = x− a.

The important existence result is as follows.

Theorem 9.4.6. Let D be a bounded Dirichlet domain in C∗, a ∈ D.Then D admits a Green function G with pole in a. In fact

G(z, a) = − log |z − a| +∫

∂Dlog |ζ − a| dωz(ζ).

9.7. EXERCISES 105

Proof. Let h be a solution of the Dirichlet problem with boundaryvalues log |z−a|. ThenG(z, a) = − log |z−a|+h(z) satisfies all the conditionsof Green’s function. Application of formula 9.3.2 for h completes the proof.

We now give an application to bounded holomorphic function.

Theorem 9.4.7. Suppose that D is a domain that admits Green func-tions G(z, a). Let f be a bounded holomorphic function on D with zeros ina1, a2, . . .. If f(z0) 6= 0 for z0 ∈ D, then

(9.4.2)

∞∑

k=1

G(z0, ak) <∞.

If the sum in (9.4.2) equals ∞, then any bounded holomorphic function withzeros in an is identically zero.

Proof. We can assume that |f | < 1 on D. Assume that f 6≡ 0. Let vnbe the function on D defined by

vn(z) = log |f(z)| +n∑

k=1

G(z, ak).

Then vn is subharmonic, because each pole of a Green function is compen-sated by the corresponding zero of f . Moreover, vn has boundary valuesless than 0. The maximum principle shows that vn(z0) ≤ 0. This implies(9.4.2).

9.5. Beurling-Nevanlinna Theorem

A very useful application of Green’s function is the Beurling-NevanlinnaTheorem. It gives a very precize estimate for subharmonic functions on adisc, if we only have limited knowledge about its size. The problem of findingsuch estimates is known as the Carleman Milloux problem.

Theorem 9.5.1. Let u be subharmonic on the unit disc. Suppose thatU satisfies the following estimates

(9.5.1) sup|z|=r

u(z) ≤ 0 and inf|z|=r

u(z) ≤ −1 0 ≤ r < 1.

Then

(9.5.2) u(z) ≤ − 2

πarcsin

(

1 − |z|1 + |z|

)

, |z| < 1.

Proof.

9.6. Green’s function and dynamics

9.7. Exercises

9.7.1. Prove that a C∞-smooth function on a closed disc in C may bewritten as the sum of a smooth subharmonic and a smooth superharmonicfunction.

Let f be continuous on a compact set K ⊂ C. Show that for every discD containing K and for every ε > 0 there exist on D a subharmonic functionu and a superharmonic function v such that |f − u− v| < ε on K.


9.7.2. Let D be a subdomain of C. Show that for every boundedfunction g on ∂∗D we have the inequality Ug ≤ Hg.

9.7.3 (Existence of harmonic measure for arbitrary domains D). Provethat if g the restriction to ∂∗D of a function that is continuous and subhar-monic on a neighborhood of D, then Ug = Hg.

9.7.4. Use the previous exercises to show that for every continuousg on ∂∗D we have Ug = Hg. Conclude that the operator g 7→ Ug(z) iswell defined linear and continuous for continuous g. Complete the proof ofTheorem 9.3.9 for general domains in C.

9.7.5 (relative extremal function). Let D be a (bounded) domain in C

and let E ⊂ D. The relative extremal function of E with respect to D is thefunction

ω(E, z,D) = supu∈F

u(z), z ∈ D,

where F = u : u subharmonic on D,u ≤ 0, u|E ≤ −1. Show the following:The upper semi-continuous regularization ω∗ of ω is subharmonic on D. IfE is compact, then ω is harmonic on D \ E.

Give examples where ω is not subharmonic and also where ω∗ > −1 onE.

9.7.6. Let u be the function defined in Exercise 8.7.10 and observe thatu(0) > −4. Let D be the domain that is the interior of

D = z : |z| ≤ 10, u(z) ≥ −10.Show that D is not a Dirichlet domain. (In fact there is no barrier functionfor 0.)

9.7.7. Show that there every positive superharmonic function on C isconstant. Show that for a finite set E, the domain C \ E does not admit aGreen’s function.

9.7.8. let D be a domain with smooth boundary ∂D and let v ∈ C2(D).Use Exercise 8.7.17 to infer the following representation formula:

(9.7.1) v(a) =1

2π

∫

∂D(−v) ∂

∂NGa ds+

1

2π

∫

D(−∆v)Ga dxdy, a ∈ D.

Give an interpretation of this formula for harmonic v.

CHAPTER 10

Some real analysis

It turns out that the boundary behavior of Poisson integrals and offunctions in so called Hardy spaces is a matter that is better suited fora treatment in real analysis. Before we can study boundary behavior ofPoisson integrals, we have to introduce the maximal function of Hardy andLittlewood, which we will use to prove Lebesgue’s famous differentiationtheorem. Hardy spaces will be the subject of the next chapter. We willwork in the half space H = z : Im z > 0 with boundary R. Similar resultshold true for the disc, but more importantly, with slightly more effort onecan extend “everything” to Hn = (x, y) : x ∈ Rn, y > 0. All of this sectionand much more can be found in [44] and its companion [45]. See also [39].

10.1. The maximal function

A locally integrable function f on R is a measurable function f such thatf is integrable over every bounded interval in R. The set of locally integrablefunctions on R is denoted by L1

loc(R). From integration theory we know thatit is convenient to work with equivalence classes of integrable functions thatare equal almost everywhere, which can be assembled into the nice Banachspaces Lp (1 ≤ p ≤ ∞). We will, however, adopt the convention thatf ∈ Lp means that “f is a measurable function that represents an elementof the space of equivalence classes Lp” The reader should easily recognize thesituations below where a statement is valid for an equivalence class, ratherthan one particular function in the class.

Definition 10.1.1. Let f ∈ L1loc(R). The maximal function of f is

defined by

Mf(x) = supr>0

1

2r

∫ r+x

−r+x|f(y)| dy.

The maximal function may well be infinite. Let µ be a complex Borelmeasure, with total variation |µ|. The maximal function of µ is defined by

Mµ(x) = supr>0

1

2r

∫ r+x

−r+x|dµ|.

The distribution function λ = λg of a measurable function g on R is definedby

λ(α) = λg(α) = |x : |g(x)| > α|, α ≥ 0.

We have written |A| for the Lebesgue measure of a measurable A ⊂ R.

Proposition 10.1.2. Let f ∈ L1loc

(R), then Mf has the following prop-erties

(1) Mf is semi-continuous from below;

107

108 10. SOME REAL ANALYSIS

(2) Mf is subadditive: M(f1 + f2) ≤Mf1 +Mf2;(3) For f ∈ L∞(R) we have Mf ≤ ‖f‖∞.

Proof. For (1) we observe that

(10.1.1) Mf(x) > λ⇒ ∃ r0, ε > 0 such that1

2r0

∫ r0+x

−r0+x|f(y)| dy > λ+ ε

Then for δ > 0 sufficiently small

1

2r0

∫ −r0+δ+x

−r0+x|f(y)| dy < ε/2

and

1

2r0

∫ r0+x

r0−δ+x|f(y)| dy < ε/2,

hence for |ξ| < δ we obtain

Mf(x+ ξ) ≥ 1

2r0

∫ r0+ξ+x

−r0+ξ+x|f(y)| dy > λ.

For (2) we compute

M(f + g) = supr>0

1

2r

∫ r+x

−r+x|f(y) + g(y)| dy

≤ supr>0

1

2r

∫ r+x

−r+x|f(y)| dy + sup

r>0

1

2r

∫ r+x

−r+x|g(y)| dy.

Finally, 3 is obvious.

Proposition 10.1.3. Let f ∈ Lp(R), p ≥ 1. Then we have

(1) λf (α) ≤ 1αp ‖f‖pp;

(2) ‖f‖pp = −∫∞0 αp dλ(α) = p

∫∞0 λ(α)αp−1dα.

The integral in (2) is a Riemann-Stieltjes integral, which is well definedbecause λ is monotonic.

Proof. For (1):

αpλf (α) =

∫

|f |>ααp dx ≤

∫

|f |>α|f |p dx ≤ ‖f‖pp.

For (2) we remark that it should be intuitively clear; −dλ(α) “is” the mea-sure of the set |f | = α. This can be turned into a proof via step functions.Alternatively, an application of Fubini’s theorem gives

‖f‖pp =

∫

R

∫ |f(y)|p

01 dx dy

Fubini=

∫ ∞

0

∫

|f(y)|>x(1/p)

1 dy dx

=

∫ ∞

0λ(x(1/p)) dx =

∫ ∞

0λ(x) dxp = −

∫ ∞

0xp dλ(x).

(10.1.2)

The last equality is by integration by parts over compact intervals, assumingthat in the limit the stock term vanishes. It remains to prove just that.

From (1) it follows that the stock term is bounded by 2‖f‖pp. Moreover,it is clear that the stock term vanishes if f is bounded and has compact

10.1. THE MAXIMAL FUNCTION 109

support. We now write f = f1 + f2 where f1 is bounded and has compactsupport, while ‖f2‖pp < ε. Because

λf (a) ≤ λf1(a/2) + λf2(a/2)

we obtain

(10.1.3) lim supδ→0,M→∞

|Mpλf (M) − δpλf (δ)|

= lim supδ→0,M→∞

|Mpλf2(M/2) − δpλf2(δ/2)| < 2p+1ǫ.

In the following lemma, known as Vitali’s covering lemma, we constructa special sequence of intervals. Such a sequence is known as a Vitali sequence.

Lemma 10.1.4 (Vitali). Let E be a measurable subset of R and supposethat E ⊂ ⋃j∈J Ij , where Ijj∈J is a family of intervals, the lenght of which

is uniformly bounded. Then there exists a subsequence Ijk : k ∈ N thatconsists of mutually disjoint intervals such that

∑

|Ijk | > |E|/5.

Proof. We will choose the sequence Ikdef= Ijk inductively, by choosing

essentially the largest interval that the conditions allow at each step. ChooseI1 such that |I1| > 1

2 supj∈J |Ij|.If I1, . . . , In are chosen, then we determine In+1 such that

|In+1| > 12 sup|Ij | : j ∈ J, Ij ∩ Ik = ∅, k = 1, . . . , n.

Continuing in this manner we obtain a sequence of intervals Ik. Fromthe construction it is clear that these intervals are mutually disjoint. If now∑ |Ik| = ∞, there is nothing left to prove. Otherwise |Ik| → 0, which weassume from now on.

Let x ∈ E. Then x ∈ Ij for some j. Suppose that Ij was not chosen.Let Ik be the first interval that was chosen with |Ik| < 1

2Ij . Then we hadchosen earlier an In, n < k, with In ∩ Ij 6= ∅ and |In| ≥ |Ij |/2. (Otherwisewe would have chosen Ik!) It follows that Ij ⊂ 5In and hence E ⊂ ⋃

5Ik,We conclude that |E| < 5

∑ |Ik|.

Theorem 10.1.5. If f ∈ L1(R), then

λMf (α) ≤ 5

α‖f‖1.

If f ∈ Lp(R), p ≥ 1, then Mf is finite a.e..

Proof. Let Eα = x : Mf(x) > α. For x ∈ Eα there exists an intervalIx of the form Ix = (−r + x, r + x) such that

1

2r

∫ r+x

−r+x|f(y)| dy > α.

Observe that independently of x ∈ Eα

|Ix| = 2r <1

α

∫

|f(y)| dy.


It follows thatEα ⊂

⋃

x∈Eα

Ix

and we can choose a Vitali subsequence Ik. Then the following holds

λMf (α) = |Eα| ≤ 5∑

|Ik| ≤ 5∑ 1

α

∫

Ik

|f | ≤ 5

α

∫

|f |,

because the intervals Ik are disjoint.Now suppose the f ∈ Lp. Define

f1(x) =

f(x) if |f(x)| ≤ 1,

0 otherwise(10.1.4)

and

f2(x) = f(x) − f1(x).(10.1.5)

Then f1 ∈ L∞ and f2 6= 0 only on a set E with |E| <∞. This implies thatf2 ∈ L1 by Holder’s inequality.

‖f2‖1 ≤∫

E|f2| ≤

(∫

E1q)1/q (∫

E|f2|p

)1/p

<∞.

We have Mf1 < 1 and

(10.1.6) Mf2 = ∞ ⊂⋂

α>0

Eα

Because |Eα| ≤ 5‖f2‖/α, (10.1.6) implies thatMf2 is finite a.e. The theoremnow follows from the inequality

Mf ≤Mf1 +Mf2.

Remark 10.1.6. The theorem is also valid for Borel measures.

λMµ(α) ≤ 5

α

∫

|dµ|

and Mµ is finite a.e.. The proof is the same.

10.2. Lebesgue’s differentiation theorem

For a continuous function f on R we know

(10.2.1) limr→0

1

2r

∫ r+x

−r+xf(y) dy = f(x);

this is just a reformulation of the main theorem of integral calculus. If fis only in L1

loc we cannot expect (10.2.1) to remain valid for all x; think ofthe characteristic function of an interval. However, surprizingly, Lebesgue’stheorem asserts that if f ∈ L1

loc then (10.2.1) is true for almost all x! Weneed some preparation.

Let χ ∈ L1(R) such that∫

χdx = 1. Form

χǫ(x) = χ(ǫx)/ǫ, ε > 0.

The family χε is an example of an approximate identity for convolution.In other words, it satisfies

10.2. LEBESGUE’S DIFFERENTIATION THEOREM 111

(1)∫

R|χǫ| dx = C <∞;

(2)∫

Rχǫ dx = 1;

(3)∫

R\V |χǫ| dx→ 0 if ε→ 0 for every neighborhood V of 0.

Lemma 10.2.1. Let χ ∈ L1(R) be as above and let f ∈ Lp(R), p ≥ 1.Then the convolutions f ∗ χǫ approach f in Lp sense:

limǫ→0

‖f − f ∗ χǫ‖p = 0.

Proof. The continuous functions with compact support are dense inLp. Hence we can find a continuous function g supported on an interval Iof finite length l such that ‖g − f‖p < δ. Then

‖f − f ∗ χǫ‖p =

∫

R

∣

∣

∣

∣

∫

R

(f(x− t) − f(x))χǫ(t) dt

∣

∣

∣

∣

p

dx

≤ Cp∫

R

(∫

R

|f(x− t) − f(x)| |χǫ(t)| dtC

)p

dx

≤ Cp∫

R

∫

R

|f(x− t) − f(x)|p |χǫ(t)| dtC

dx

≤ Cp∫

R

∫

R

|f(x− t) − g(x− t)|p + |f(x) − g(x)|p |χǫ(t)|C

dt dx

+ Cp∫

R

∫

R

|g(x− t) − g(x)|p |χǫ(t)|C

dt dx.

(10.2.2)

Indeed, the second inequality is an instance of Jensen’s inequality, cf. [39],which reads ϕ

∫

f dµ ≤∫

ϕ f dµ for a convex function ϕ (in our case ·p),and a probability measure µ (in our case |χε(t)| dt/C). The third inequalityis the triangle inequality for Lp.

Now we estimate the last expression in (10.2.2). We apply Fubini’stheorem to the first integral; the result is less than 2δ. For the secondintegral we notice that g is uniformly continuous. We choose an intervalV = [−r, r], r < 1

2 so small that |g(x−t)−g(x)| < δ/l and we choose ε small.With property 3 of an approximate identity we find

∫

R\V χǫ ≤ δ/(lM), for

ε sufficiently small. Thus we can estimate

∫

R

∫

R

|g(x− t) − g(x)|p |χǫ(t)|C

dt dx =

∫

I+V

∫

V+

∫

R

∫

R\V≤ (l + 1)(δ/l)p + 2lMpδ/(lM),

(10.2.3)

and the proof is complete.

The following lemma contains the crux of the argument.

Lemma 10.2.2. For f ∈ L1loc

limr→0

1

2r

∫ r+x

−r+xf(y) dy = f(x) a.e.(x)


Proof. Without loss of generality f is real valued and has compactsupport. Then f ∈ L1(R). Put

fr(x) =1

2r

∫ r+x

−r+xf(y) dy.

In view of Lemma 10.2.1 we obtain ‖fr − f‖1 → 0, hence there exists asequence rk tending to 0 such that frk → f a.e..

Define for real valued functions g ∈ L1(R) an operator Ω as follows.

Ωg(x) = lim supr→0

gr(x) − lim infr→0

gr(x).

We observe that Ωg(x) ≤ 2Mg(x) and that for a continuous function h,Ωh = 0 and Ω(g + h) = Ωg.

Let ε > 0. There is a continuous function h with ‖h − f‖1 < ε. Thenfor a suitable constant C > 0

λΩf (α) = λΩ(f−h)(α) ≤ C

α‖f − h‖1 ≤ C

αǫ.

Because ε is arbitrarily small, λΩf (α) = 0. This true for every α, henceΩ(f) = 0 a.e.. In other words, limr→0 fr(x) exists for almost every x. Butthe sequence frk converges to f for almost every x. We conclude thatlimr→0 fr(x) = f(x) a.e..

We have finally arrived at Lebesgue’s theorem.

Theorem 10.2.3 (Lebesgue). Let f ∈ L1loc

. Then

limr→0

1

2r

∫ r+x

−r+x|f(y) − f(x)| dy = 0,

for almost all x ∈ R.

Proof. Let c ∈ R. Lemma 10.2.2 applied to the function |f(x) − c|,shows that

limr→0

1

2r

∫ r+x

−r+x|f(y) − c| dy = |f(x) − c|

for x outside a set Ec of measure 0. Let E =⋃

c∈QEc, then |E| = 0 and for

x /∈ E and ε > 0 we find

limr→0

1

2r

∫ r+x

−r+x|f(y) − f(x)| dy ≤ lim

r→0

1

2r

∫ r+x

−r+x|f(y) − c| dy + |f(x) − c|

= 2|f(x) − c| < ε,

if c ∈ Q is chosen close enough to f(x).

Definition 10.2.4. Let f be a locally integrable function (NOT anequivalence class). The set X ⊃ R \ E for which the limit in the theoremexists is called the Lebesgue set of f .

Now we need the theorem of Radon-Nikodym, but only in the settingof Lebesgue measure, to be able to deal with measures. Let (X,A) be ameasurable space, see e.g. [39] for a proof and for the general formulation.A measure ν on X is called singular with respect to a measure µ on X,if there exists a set A ∈ A with µ(A) = 0 such that ν(L) = 0 for everymeasurable L ⊂ X \ A. We write ν ⊥ µ.

10.3. BOUNDARY BEHAVIOR OF POISSON INTEGRALS 113

Theorem 10.2.5 (Radon-Nikodym). Let µ be a Borel measure that isfinite on compact sets in R. Then there exist a function f ∈ L1

locand a

singular measure (with respect to dt) ν such that

µ = f dt+ ν.

This splitting of µ in absolute continuous part f dt and singular part νis unique and is called the Radon-Nikodym decomposition of µ with respectto dt.

Proposition 10.2.6. Let ν be a singular Borel measure that is finite oncompact sets in R. Then

(10.2.4) limr→0

1

2r

∫ r+x

−r+x|dν| = 0 a.e. .

Proof. We can assume that ν is finite and vanishes outside a finiteinterval I. Let ε > 0. There exists a compact set K of measure 0, with‖ν|K‖ > |ν| − ε. We put νε = ν − ν|K , then ‖vε‖ < ε. Introduce for Borelmeasures µ

Wµ(x) = lim supr→0

1

2r

∫ r+x

−r+x|dµ|,

So that Wµ ≤ Mµ. Now clearly Wν(x) ≤ Wνε(x) + Wν|K (x), WνK(x) = 0

for x /∈ K. Because K has measure 0, we have, using Remark 10.1.6, for asuitable constant C

λWν (α) = λWνε(α) ≤ λMνε

(α) ≤ Cε

α.

Letting ε → 0, we conclude that λWν (α) = 0 for every positive α, henceWν(x) = 0 a.e., which finishes the proof.

Definition 10.2.7. Let f ∈ L1loc(R). A Lebesgue point of f is a point

x0 ∈ R such that

lim supr→0

1

2r

∫ r+x

−r+x|f(t) − f(x)| dt = 0.

Let µ = f dt + ν be the Radon-Nikodym decomposition of µ. A Lebesguepoint of µ is a point x0 ∈ R such that

lim supr→0

1

2r

∫ r+x

−r+x|f(t) − f(x)| dt + |dν| = 0.

10.3. Boundary behavior of Poisson integrals

We study Poisson integrals on the upper halfspace H = y > 0. Thishas the small advantage that formulas are a little simpler than in the disc, aslight disadvantage is that R is not compact. Recall that the Poisson kernelfor H is given by

(10.3.1) Py(t) =1

2π

y

t2 + y2.

For a σ-finite measure on R that satisfies

(10.3.2)

∫

R

dµ(t)

1 + t2<∞,


the Poisson integral is

(10.3.3) P [µ](z) =

∫

R

Py(x− t) dµ(t) z = x+ iy;

for f ∈ L1loc such that f dt satisfies (10.3.2) one defines

(10.3.4) P [f ](z) =

∫

R

Py(x− t)f(t) dt z = x+ iy.

Lemma 10.3.1. Let f ∈ Lp(R). Then

(10.3.5) ‖f(x) − P [f ](x+ iy)‖p → 0 if y → 0.

Proof. The functions Py(t) form an approximate identity. As P [f ] isthe convolution of this approximate identity with f , Lemma 10.2.1 applies.

How about the boundary behavior? We will study the radial limitlimy→0 P [f ](x0 + iy) and the non-tangential limit lim

zn.t.→ x0

P [f ](z). Here

limz

n.t.→ x0means that z → x0 within some cone |x− x0| < Cy, where C is

an arbitrary, positive constant.

Lemma 10.3.2. For every C > 0 there are A,B > 0 such that

y

(x− t)2 + y2≤ A

By

t2 + (By)2.

Proof. The function ht(x + iy) = y(x−t)2+y2

is positive and harmonic

on H. If |x| < Cy then z ∈ B((1+C2)iy, C√

1 + C2y), while the larger discB((1 +C2)iy, (1 +C2)y) belongs to H. The ratio of the radii of these discsis only dependent on C. We apply Harnack’s inequality and find a constantA > 0 such that h(x+ iy) < Ah((1 +C2)iy).

Theorem 10.3.3. Let µ be a σ-finite measure on R that satisfies (10.3.2).Let µ = f dt + ν be the Radon-Nikodym decomposition of µ, with f ∈ L1

loc

a locally integrable function, and ν ⊥ dt a singular measure. For everyLebesgue point x0 of µ, hence for almost every point in R

limz

n.t.→ x0

P [µ] = f(x0).

Proof. If µ = 0 on an interval about x0, then (the proof of) theorem8.3.2 shows that limz→x0 P [dµ](z) = 0 for arbitrary approach to x0. We cantherefore assume that µ = χIdµ, where χI is the characteristic function ofa bounded interval I about 0. In particular µ is finite. We set

(10.3.6) µr(x) =1

2r

∫ x+r

x−r(|f(t)| dt + |dν|).

Observe that

(10.3.7) µr <|µ|2r.

Now assume that x0 is a Lebesgue point of f . In particular the valuef(x0) is well defined. We can assume after translations that x0 = 0 andf(x0) = 0. Thus

(10.3.8) limr→0

µr(0) = 0.

10.4. POSITIVE AND h1 HARMONIC FUNCTIONS ON THE DISC 115

We have to prove that

limz

n.t.→ 0

P [µ](z) = 0.

We have the following relation

|P [µ](z)| ≤ P [|µ|](z) =1

2π

∫

y

(x− t)2 + y2| dµ(t)|.

By Lemma 10.3.2 the final integral can be estimated by

(10.3.9) A

∫

By

t2 + (By)2|dµ(t)|.

We will show that 10.3.9 tends to 0 as y → 0. We may take B = 1 and nowintegrate by parts.∫

y

t2 + y2|dµ(t)| =

∫

2ty

(t2 + y2)2

∫ t

0|dµ|

=

∫

2t22y

(t2 + y2)2

(

1

2t

∫ t

0|dµ|

)

dt ≤∫

2t22y

(t2 + y2)2µt(0) dt.

In the last inequality we substituted (10.3.6). By (10.3.8) there is a neigh-borhood V of 0 in R such that µt(0) < ε on V and by (10.3.7) there existsa constant C such that µt(0) < C for t /∈ V . Hence

∫

R

2t22y

(t2 + y2)2µt(0) dt =

∫

V+

∫

R\V

≤ ǫ

∫

V

2t22y

(t2 + y2)2dt+ C

∫

R\V

2t22y

(t2 + y2)2dt.

In the last expression the first term is O(ε) and the last term tends to 0 asy ↓ 0.

Remark 10.3.4. It is Lebesgue’s theorem that gives this theorem itsstrength.

The proof is fast, but may look a bit tricky. One can base a proof on thetechniques of Section 2, Either by estimating the Poisson Kernel in termsof the maximal function, see Exercise 10.5.2 and [45], or by introducing amaximal function for the Poisson Kernel and estimating that in terms of themaximal function, cf. [39]. The slight drawback of such an approach is thatthe connection with Lebesgue points remains hidden.

10.4. Positive and h1 harmonic functions on the disc

The results of the previous section can be transferred to the unit disc B

without difficulties. We will not carry out the details.Let µ be a positive (finite) Borel measure on T = ∂B. From Theorem

8.2.5 we see immediately that P [µ] is a positive harmonic function on B. Itis remarkable that every positive harmonic function is of this form.

Theorem 10.4.1 (Herglotz representation). The following are equivalent

a: The function u is positive and harmonic on B.b: There exists a unique positive Borel measure µ on T such that

u = P [µ].


Proof. We only have to prove a.⇒b. Let u be a positive harmonicfunction on B. For every 0 < r < 1 the function

ur(z) = u(rz) = P [ur](z), z ∈ B

is positive and harmonic and ur|T is a positive continuous function that maybe identified with the positive measure µr = ur(e

iθ) dθ2π on T. The mass ofµr equals

∫

T

ur(eiθ)dθ

2π= u(0).

Hence all µr belong to the closed ball with radius u(0) in the dual spaceC(T)∗. The Banach-Alaoglu theorem, see [40], states that such a ball iscompact in the weak∗ topology. This means that for every sequence (µrj )jwe can find a subsequence (µrk)k that converges to a measure µ in the sensethat for every function f ∈ C(T)

limk→∞

∫

T

f µrk =

∫

T

f µ.

Recalling that a measure µ is positive if and only if∫

fµ ≥ 0 for everycontinuous f ≥ 0, we observe that a weak∗ limit of positive measures ispositive.

For f we take the Poisson kernel and obtain

u(Reiθ) = limk→∞

urk(Reiθ) = limk→∞

∫

T

PR(θ − ϕ) dµrk

=

∫

T

PR(θ − ϕ) dµ(ϕ).

(10.4.1)

Equation (10.4.1) shows that µ is a positive measure as we are looking for.Any other sequence rk would lead to a positive measure with the samePoisson integral, but then by exercise 8.7.4 this measure has to equal µ.

Herglotz type representations are not possible for arbitrary harmonicfunctions. Some limitations on the growth of the function at the boundaryare necessary.

Definition 10.4.2. Let u be harmonic on B. If there exists M > 0 suchthat

sup0≤r<1

1

2π

∫ π

−π|u(r(eiθ)| dθ ≤M,

then we say that u is in the class h1(B). (“little h-1”).

Theorem 10.4.3 (Herglotz representation). For a real valued harmonicfunction the following are equivalent

c: The function u is in h1(B).d: There exists a unique signed Borel measure µ on T such that

u = P [µ].

Proof. If µ = µ1 − µ2 with µi positive, then P [µ] = P [µ1] − P [dµ2],hence

1

2π

∫ π

−π|u(r(eiθ)| dθ ≤ P [dµ1](0) + P [dµ2](0),

independently of r, so d. implies c.

10.5. EXERCISES 117

In the other direction, c. implies that the measures µr = u(reiθ) dθ forma bounded sequence in C(T)∗ so there exists a weak limit point µ ∈ C(T∗)by the Banach-Alaoglu Theorem. The proof can now be completed as thatof the previous theorem.

In the sequel we will wish to view holomorphic functions as a subclassof the complex valued harmonic functions. A complex valued function iscalled harmonic if its real and imaginary part are both harmonic. harmonicfunction!complex valued Definition 10.4.2 serves also to define the class h1

for complex valued functions. The complex version of the Herglotz theoremis as follows.

Theorem 10.4.4. For a complex valued harmonic function the followingare equivalent

c: The function u is in h1(B).d: There exists a unique complex Borel measure µ on T such that

u = P [µ].

The proof goes just like the proof in the real case.

Corollary 10.4.5. Let u ∈ h1 so that P [µ] = u and let µ = f∗dθ + νbe the Radon-Nikodym decomposition of µ. Then u∗ = f a.e.

Proof. This follows immediately from Theorem 10.4.3 and Theorem10.3.3.

10.5. Exercises

10.5.1. Let E be a measurable set in R. Show that

limh→0

1

2h

∫

−h<t−x<ht∈E

dt = 1, a.e. x ∈ E.

10.5.2. Give a proof of Theorem 10.3.3 (but without referral to theLebesgue set) following the lines of thought of Section 2 as follows. Introduceagain an operator Ω but now a little differently:

Ωf(x) = lim supy

n.t.→ 0

P [f ](x+ iy) − lim infy

n.t.→ 0

P [f ](x+ iy).

If we can show that Ωf(x) ≤ CMf (x) we can copy Section 2. For this itsuffices to show that for positive f , P [f ](x+iy) ≤ CMf (x). One can assumex = 0. Prove this statement by finding functions ϕy such that

∫

y

y2 + t2f(t) dt ≤

∫

ϕy(r)

∫

R

χr(t)f(t) dt dr,

where χr =I[−r,r]

2r .

10.5.3. Find the Herglotz representation for the Poisson kernel Pr(θ),(z = reiθ).

10.5.4. Let u be a positive harmonic function on B. Show that foralmost every θ the limit limr→1 u(re

iθ) = u∗(eiθ) exists. Show that u∗ isin L1(T). Show that it is in general not true that u = P [u]. What can gowrong?


10.5.5. Give a description of the non-negative harmonic functions withboundary values equal to 0 almost everywhere.

10.5.6 (More Herglotz). Let f be holomorphic on B and assume thatRe f > 0 and that f(0) = 1. Prove that there exists a probability measureµ on T such that

f(z) =1

2π

∫

T

ζ + z

ζ − zdµ(ζ).

10.5.7. Let f be holomorphic on B and assume that Re f > 0 and thatf(0) = 1. Prove Caratheodory’s inequality expressing that

f(z) = 1 +

∞∑

n=1

anzn.

with |an| ≤ 2. Show that

1 − |z|1 + |z| ≤ |f(z)| ≤ 1 + |z|

1 − |z| .

10.5.8. Prove Landau’s inequality If f =∑∞

n=0 anzn is holomorphic on

B and |f | < 1, then |an| < 2(1 − |a0|).10.5.9. The Bohr radius is defined as the supremum of all r such that

f(z) =∑∞

n=0 anzn is holomorphic on B and |f | < 1 implies

∑

n=0∞ |an|rn <1. Use Landau’s inequality to prove that the Bohr radius is larger or equalthan 1/3.

This use of Landau’s inequality is due to L. Aizenberg. Show that theBohr radius equals 1/3 by studying the fractional linear transformations

z − a

1 − az, 0 < a < 1.

H. Bohr showed that the radius was at least 1/6. The sharp 1/3 is due toM. Riesz, I. Schur and N. Wiener.

10.5.10. Give an example of a harmonic function on B that is not inh1.

CHAPTER 11

Hardy spaces

Hardy spaces are intersections of spaces of holomorphic functions andcertain Lp spaces. We will study Hardy spaces on the unit disc B = B(0, 1).But it makes sense to study HP spaces on other domains. The theoryis virtually the same for the half space, but becomes more complicated ifthe underlying domain is no longer simply connected or has bad boundary.We will be using a little Fourier analysis, integration theory and functionalanalysis. One question we will address is that of existence of boundaryvalues of Hp and related functions. Here, the theory of the previous chapterwill be used fruitfully. A second question deals with the possible zero sets ofHp functions. Surprisingly, the zero sets of an H1 function is also the zeroset of an H∞ function, although the class of H1 functions is much larger.We note that in case of functions of several variables such a statement iscompletely false, cf. [41, 42].

11.1. The definition of Hp(B)

The vector space Hp = Hp(B) consists of the holomorphic functions onB, (with usual addition) for which the Lp means

(11.1.1) Mp(r, f) =

(

1

2π

∫ π

−π|f(reiθ)|pdθ

)1/p

remain finite as r ↑ 1. H∞(B) consists of the bounded holomorphic functionson B. In other words,

M∞(r, f) = maxθ∈[0,2π)

|f(reiθ)|

remains finite as r ↑ 1.Observe that because |f |p is subharmonic, the Lp means Mp(r, f) in-

crease with r. By definition then, for f ∈ Hp the limit

‖f‖p = limr→1

Mp(r, f)

exists and is finite. One shows as for the classical Lp spaces that for p ≥ 1‖f‖p is a norm on Hp. But the relation with Lp is more than superficial.Let T = C(0, 1) and Lp(T) = Lp(T, dλ) the usual space of p-integrablefunctions with respect to normalized Lebesgue measure (λ(T) = 1). Thespace Hp can be viewed as an isometrically embedded closed subspace ofLp(T). More precisely every f ∈ Hp admits a boundary function f∗ ∈ Lp(T)and ‖f‖p = ‖f∗‖p. (The first norm is in Hp, the second in Lp(T) but thevery equality shows that there will be no risk in confusing them).

Finally, Holder’s inequality shows that for p > 1, Hp ⊂ H1.

119

120 11. HARDY SPACES

11.2. Two words on Fourier series

The functions eint : n ∈ Z form a complete orthonormal system, orbasis of L2(T). As a consequence, every f ∈ L2(T) can be written as

f(eit) =∑

n∈Z

(f, ein·)eint,

where the sum is a limit in the L2 metric and the inner product is

(f, g) =

∫ π

−πf(eit)g(eit)

dt

2π.

We often denote the n-th Fourier coefficient by

(11.2.1) cn[f ] = (f, ein·)

(

=

∫ π

−πf(eit)e−int

dt

2π

)

.

The space

l2(Z) = (cn)n∈Z :∑

n∈Z

|cj |2 <∞

is a Hilbert space with the inner product

(c, d) =∑

Z

cj dj .

The map f 7→ (cn[f ])n is an isometry from L2(T) onto l2(Z). This is theso-called Riesz-Fischer theorem .

If f ∈ L1, then the Fourier coefficients are also defined, but the inter-pretation of the series is much more difficult.

11.3. The Hardy space H2(B)

Recall the Hilbert space of square summable sequences

l2 = l2(N) = a = (an)n :

∞∑

j=0

|aj |2 <∞

with inner product

(a, b) =∑

ajbj .

Let f =∑∞

n=0 anzn be a holomorphic function on B. Then

M2(r)2 = M2(f, r)

2 =1

2π

∫ π

−π|f(reiθ)|2dθ

=1

2π

∫ π

−π

∞∑

n=0

∞∑

m=0

anamrn+mei(n−m)θdθ =

∞∑

n=0

|an|2r2n.(11.3.1)

Clearly M2(r) is increasing with r; If∑∞

n=0 |an|2 is bounded, i.e. an ∈ l2,then limr↑1M2(r) = ‖f‖2

2 =∑∞

n=0 |an|2. If limr↑1M2(r) = M exists, thenfor each N

N∑

n=0

|an|2 ≤ limr↑1

M2(r) = M,

therefore the partial sums of∑∞

n=0 |an|2 are bounded and so is the infinitesum.

We have proved

11.3. THE HARDY SPACE H2(B) 121

Proposition 11.3.1. Let f(z) =∑∞

n=0 anzn. Then f ∈ H2(B) if and

only if the sequence of Taylor coefficients an of f is in l2. The norm on l2

can be transferred to H2(B), and H2(B) becomes a Hilbert space with norm‖f‖2. In fact

‖f‖22 = (

∞∑

n=0

|an|2),(11.3.2)

and inner product

(f, g) =

∞∑

n=0

an[f ]an[g].(11.3.3)

We now bring in some Fourier analysis. Let f(z) =∑∞

n=0 anzn ∈ H2.

We set z = reiθ and let r tend to 1. One obtains formally

(11.3.4) limr↑1

f(reiθ) =

∞∑

n=0

aneinθ.

The Riesz-Fischer theorem says that there exists a function f∗ ∈ L2(T)such that (11.3.4) is the Fourier series of f∗. Vice versa, each functiong∗(eiθ) =

∑∞n=0 bne

inθ ∈ L2(T) has a sequence of Fourier coefficients (bn)n ∈l2.

In fact, (11.3.4) is valid in L2 sense:

‖f∗ − fr‖22 = ‖

∞∑

n=0

(1 − rn)aneinθ‖2

2 =

∞∑

n=0

(1 − rn)2|an|2 → 0, as r → 1.

What is the precise relation between f and f∗? Because f∗ ∈ L2(T) ⊂L1(T), the Cauchy integral is wel defined.

f =1

2πi

∫

T

f∗(ζ)ζ − z

dζ.

The n-th Taylor coefficients of f at 0

an[f ] =1

2πi

∫

T

f∗(ζ)ζn+1

dζ =1

2π

∫ π

−πf∗(eiθ)e−inθ dθ = an.

We have proved the first and second statement of the following theorem.

Theorem 11.3.2. Let f ∈ H2(B) then

a: f∗ ∈ L2(T), and fr → f∗ in L2(T).b: f equals the Cauchy integral of f∗,

f(z) =1

2πi

∫

T

f∗(ζ)ζ − z

dζ,

c: f equals the Poisson integral of f∗,

f(z) = P [f∗](reiθ), z = reiθ.

d: limr→1 f(reiθ) = f∗(eiθ) for almost all θ


Proof. For c. we proceed as follows. By (8.2.10) we have

Pr(t) =1

2π

∞∑

n=−∞r|n|eint,

where for each 0 ≤ r < 1 the convergence is uniform on T. Hence

P [f∗](reiθ) =

∫ π

−πPr(θ − ϕ)f∗(eiϕ)dϕ

=

∞∑

n=−∞r|n|

(

1

2π

∫ π

−πein(θ−ϕ)f∗(eiϕ)dϕ

)

=

∞∑

n=−∞r|n|an[f

∗]einθ = f(reiθ).

for d. we note that f is in h1, hence Corollary 10.4.5 gives that

f(z) = P [h+ dν](z),

where h is in L1 and ν is singular, and this representation is unique. Fromunicity we conclude that f∗ = h and ν = 0. Next we apply Theorem 10.3.3and d. is proved.

Remark 11.3.3. The possible conflict caused by the notation f∗ bothfor L2 boundary values and for the non tangential boundary limit a.e. isresolved by this theorem.

Theorem 11.3.4. Let u =∑

n∈Z cneint ∈ L2(T). Then u = f∗ for some

f ∈ H2(B) if and only if cn = 0 for n < 0.

Proof. If cn = 0 for all n < 0 then by Proposition 11.3.1 f(z) =∑∞

n=0 cnzn is in H2(B) and by Theorem 11.3.2 a. f∗ = u. Next, if f∈H2(B),

then∫ 2π

0f∗(eit)eintdt− lim

r→1

∫

T

f(rz)zndz

iz= 0

by Cauchy’s theorem.

11.4. Blaschke products

We cannot immediately extend the results from the previous sectionto H1 because for h1 some singular measures may occur in the Poissonrepresentation. If f ∈ H1 we could try to divide out the zeros and then takea square root and apply the H2-theory. To make this idea work, we need tostudy the zeros of H1 functions.

Without loss of generality we will assume in this section that f 6≡ 0 andf(0) 6= 0. In view of Jensen’s formula Theorem 6.4.1 a restriction on thezero set of Hp-functions is not a big surprize. In fact if ak (0 < |ak| < 1)are the zeros of a holomorphic function f on the unit disc B, then Jensen’sformula states for r < 1

(11.4.1)∑

|ak|<rlog

r

|ak|=

1

2π

∫ π

−πlog |f(reiθ)| dθ − log |f(0)|.

11.4. BLASCHKE PRODUCTS 123

For positive a and p we have p log a < ap, hence

(11.4.2)1

2π

∫ π

−πlog |f(reiθ)| dθ < 1

pMp(r, f).

Therefore, if f ∈ Hp, then

(11.4.3)∑

|ak |<rlog

r

|ak|≤ 1

p‖f‖pp − log |f(0)|,

independently of r, i.e.∑

k log 1|ak| is bounded. We have proved

Theorem 11.4.1. The zeros of a function f ∈ Hp, with f(0) 6= 0, satisfythe Blaschke condition which may be stated as one of the following equivalentconditions.

(1)∑

log 1|ak| <∞;

(2)∏ |ak| converges;

(3)∑

(1 − |ak|) converges.

In fact, more generally, the zeros of f satisfy the Blaschke condition if andonly if

(11.4.4) lim supr→1

∫ π

−πlog |f(reiθ)| dθ <∞,

Proof. After all we have done in the previous paragraph, we just haveto check that the three versions of the Blaschke condition are indeed equiv-alent. This is left Exercise 11.7.1. It remains to observe that the Blaschkecondition implies (11.4.4). This is again a direct consequence of Jensens’sformula (11.4.1).

Theorem 11.4.2. Suppose 0 < |a1| ≤ |a2| ≤ . . . < 1 are points in B thatsatisfy the Blaschke condition. Then the Blaschke product

(11.4.5)

∞∏

k=1

|ak|−ak

z − ak1 − akz

converges uniformly on compact sets in B to a holomorphic function B(z).The function B satisfies |B(z)| < 1 and for almost all θ the limit B∗(eiθ) =limr→1B(reiθ) exists and has absolute value 1.

Proof. Let 1 + vk be the k-th factor in (11.4.5). Then for |z| ≤ R < 1

|vk| =

∣

∣

∣

∣

ak + |ak|z(1 − akz)ak

∣

∣

∣

∣

(1 − |ak|) ≤1 +R

1 −R(1 − |ak|).

The Blaschke condition in combination with Theorem 5.2.2 now implies thatthe product (11.4.5) converges uniformly on B(0, R).

Since all factors in (11.4.5) are absolutely bounded by 1, so is B. Itfollows that B ∈ H2, hence B∗(eiθ) exists for almost all θ, and |B∗| ≤ 1almost everywhere.

To prove that |B∗| = 1 a.e., we first observe that every factor, hence alsoevery finite partial product

(11.4.6) Bn(z) =

n∏

k=1

z − ak1 − akz

|ak|−ak


has absolute value 1 on T. Therefore ‖B∗‖2 = ‖(B/Bn)∗‖2. Next we notethat for all 0 < r < 1

(11.4.7)B

Bn(0) = M2(B/Bn, 0) ≤M2(B/Bn, r) ≤ ‖(B/Bn)∗‖2 = ‖B∗‖2.

Because limn→∞B(0)/Bn(0) = 1, we have ‖B∗‖2 = 1. Hence∫ π

−2π(1 − |B∗(eit)|2) dt = 0,

But the integrand is a non-negative L1-function, hence it must be 0 a.e., inother words, |B∗| = 1 a.e.

11.5. The Hardy space H1(B)

We can now execute our plan for dealing with H1. We fix f ∈ H1 andassume without loss of generality that f(0) 6= 0. Let B be the Blaschkeproduct formed by the zeros of f , so that f/B is a zero free holomorphicfunction on the disc. We claim

Lemma 11.5.1 (F. Riesz factorization theorem). ‖f/B‖1 = ‖f‖1.

Proof. Indeed, because |B| < 1 on B, ‖f/B‖1 ≥ ‖B‖1. Next if B isa finite Blaschke product, there is nothing to prove. Otherwise, with Bn apartial Blaschke product as before, we have for fixed 0 < r < 1

(11.5.1) M1(r,f

Bn) < ‖ f

Bn‖1 = ‖f‖1.

Letting n → ∞ we obtain that for every r < 1, M1(r,fB ) ≤ ‖f‖1, and the

lemma is proved.

Next let ϕ be a holomorphic branch of√

f/B and ϕ1 = Bϕ. The lemmashows that ϕ and ϕ1 are in H2, and ‖ϕ1‖2

2 = ‖ϕ‖22 = ‖f‖1. We now state

our main result

Theorem 11.5.2. Let f(z) =∑∞

n=0 anzn be in H1. Then limr→1 f(reit) =

f∗(eit exists for almost all t ∈ [0, 2π). The boundary function f∗ is in L1(T).Its Fourier coefficients cn[f

∗] are

(11.5.2) cn[f∗] = 0, if n < 0 cn[f

∗] = an, if n ≥ 0.

The boundary function f∗ is L1-limit of the functions fr(eit) = f(reit):

(11.5.3) ‖f∗ − fr‖1 =1

2π

∫ π

−π|f∗(eit) − f(reit)| dt → 0, r ↑ 1.

In particular

‖f‖H1 = ‖f∗‖L1 .

The function f equals the Poisson integral, as well as the Cauchy integralof f∗.

If f 6≡ 0 then log |f∗| is in L1(T). As a consequence, f∗ cannot vanishon a set of positive measure.

Finally, every g ∈ L1(T) which has cn[g] = 0 for n < 0 is the boundaryfunction of an element of H1.

11.5. THE HARDY SPACE H1(B) 125

Proof. With the notation as before we can write f(z) = ϕ(z)ϕ1(z),with ϕ,ϕ1 in H2. As the radial limits of ϕ and ϕ1 exist a.e. we can definef∗(eit) = ϕ∗(eit)ϕ1(e

it) for almost all t. Because ϕ∗ and ϕ∗1 are in L2 we

find with Cauchy-Schwarz

‖f∗‖1 =

∫ π

−πϕ∗(eit)ϕ1(e

it)dt

2π≤ ‖ϕ∗‖2‖ϕ∗

1‖2.

So f∗ is in L1. We now prove (11.5.3) using the triangle equation, Cauchy-Schwarz and theorem 11.3.2:

‖f∗ − fr‖1 = ‖ϕ∗ϕ∗1 − ϕ∗(ϕ1)r + ϕ∗(ϕ1)r − ϕr(ϕ1)r‖1

≤ ‖ϕ∗‖2‖ϕ∗1 − (ϕ1)r‖2 + ‖(ϕ1)r‖2‖ϕ∗ − ϕr‖2 → 0, as r ↑ 1.

It follows that ‖f∗‖1 = ‖f‖, because ‖fr‖1 = M1(f, r).From (11.5.3) and (11.2.1) we also obtain the following relation for the

Fourier coefficients

cn[f∗] = lim

r↑1cn[fr].

Now (11.5.2) follows immediately.The Cauchy and Poisson representation can be proved similarly. E.g.

f(z) = limr↑1

fr(z) = limr↑1

P [fr](z) = P [f∗](z).

The last equality follows from (11.5.3).If g ∈ L1(T) has the property that cn[g] = 0 for n < 0, then P [g] ∈ h1.

Expanding P [g] gives

P [g](z) =∑

n∈Z

cn[g]r|n|eint =

∞∑

0

cn[g]zn, (z = reit),

so P [g] ∈ H1 and p[g]∗ = g. In the final part of the proof we will deal withthe integrability of log |f∗|. For x > 0 we introduce log+ x = maxlog x, 0,log− x = max− log x, 0. So log = log+ − log− and | log | = log+ + log−.We can assume that f ∈ H1(B) is zero-free, in view of Lemma 11.5.1.Because log x < x we have

1

2π

∫ π

−πlog+ |f∗(eit)| dt ≤ ‖f∗‖1,

which is finite. Now observe that log |f | is harmonic. By the mean valueproperty

1

2π

∫ π

−πlog− |f∗(reit)| dt = − log |f(0)| + 1

2π

∫ π

−πlog+ |f∗(reit)| dt

≤ − log |f(0)| + ‖f∗‖1.

(11.5.4)

Now Fatou’s lemma can be applied on any sequence rnn increasing to 1.The conclusion is that

1

2π

∫ π

−πlog− |f∗(eit)| dt =

1

2π

∫ π

−πlim infn→∞

log− |f∗(rneit)| dt

≤ lim infn→∞

1

2π

∫ π

−πlog− |f∗(rneit)| dt ≤ − log |f(0)| + ‖f∗‖1.

(11.5.5)


Integrability of log− |f∗| implies that f∗ cannot equal 0 on a set of positivemeasure.

11.6. Additional results

11.7. Exercises

11.7.1. prove that the different versions of the blaschke condition areequivalent.

11.7.2. Let f be holomorphic on B, and such that f assumes no valuesin some disc B(a, r). Prove that f has finite radial limits almost everywhereon T.

11.7.3. Consider the space H2(B). Prove that for a ∈ B, the mapla : f 7→ f(a) defines a continuous linear functional on H2(B). By RieszRepresentation Theorem, la can be represented by taking the inner productwith a function ga ∈ H2(B). Determine ga.

11.7.4. Let g(eit) =∑∞

n=−∞ cneint be a continuous function on T.

Define

(11.7.1)1

2πi

∫

T

g(ζ)

ζ − zdζ =

f1(z), |z| < 1

f2(z), |z| > 1.

Compute the Laurent series of f1 and f2. Show that f1 and f2(z) belong toH2(B). Show that the same holds if g ∈ L2(T).

Next show that on T one has the Plemelj formula

(11.7.2) g(eit) = f∗1 (eit) − f∗2 (eit) a.e.

11.7.5. Let g ∈ L2[a, b], a < b ∈ R and let

f(z) =1

2πi

∫ b

a

g(t)

t− zdt.

Computelimy↓0

f(x+ iy) − f(x− iy),

in L2-sense.

11.7.6. Prove the following theorem, which is due to F. &M. Riesz

Theorem 11.7.7. Suppose that µ is a complex measure on T and theFourier coefficients

cn[µ] =

∫ π

−πe−int

dµ(t)

2π

are zero for n < 0. Then µ is absolutely continuous with respect to Lebesguemeasure on T.

CHAPTER 12

Applications of Algebraic Topology

Parts of algebraic topology, such as the theory of covering spaces andfundamental groups, as well as the theory of sheaves, have there roots infunction theory. In this chapter we will discuss some of the results in functiontheory that have a distinct algebraic topology flavour. In the first sectionwe will collect the results of algebraic topology that we will use withoutproof. The appendix to this chapter is a copy of the first section, but nowwith proofs included. Highlights of this chapter are: a proof of the Picardtheorem to the effect that a non-constant meromorphic function on C omitsat most two values of the Riemann sphere and a topological characterizationof the maximal analytic extension of a (multiple valued) analytic function.

12.1. Results from Algebraic Topology

Definition 12.1.1. Let Y andX be topological spaces and π a surjectivelocal homeomorphism from Y to X. Then we call Y a space over X. (Moreformally, the triple (Y, π,X) is a space over X). The French espace etale iscommonly used in non-french texts too. The map π is called the projection.For x ∈ X we call π−1(x) the fiber above x.

Definition 12.1.2. Let (Y, π,X) be a space over X and let f : Z → Xbe a continuous map from a topological space Z to X. A continuous mapf : Z → Y is called a lifting of f if π f = f .

In the rest of this section we assume that all topological spaces will beHausdorff and connected.

In fact, some of the results below remain valid without assuming thatall spaces involved have these nice properties. This is pointed out in theappendix.

Theorem 12.1.3. Let (Y, π,X) be a space over X and let f : Z → X becontinuous and suppose that Z is connected. Suppose g1 and g2 are liftingsof f and that g1(z) = g2(z) at some z ∈ Z. Then g1 = g2.

Definition 12.1.4. Let X be a topological space and let I denote theinterval [0, 1]. Two curves γ0, γ1 : I → X with γi(0) = a and γi(1) = b, arecalled homotopic if there exists a continuous map F : I × I :→ X such that

F (0, u) = a, F (1, u) = b, F (t, 0) = γ0(t), F (t, 1) = γ1(t).

The map F is called a homotopy between γ0 and γ1. The curves γu : t 7→F (t, u) are called the homotopic family associated to F .

Definition 12.1.5. A topological space X is called simply connected ifevery closed curve γ in X is homotopic to the constant curve t 7→ γ(0),

127

128 12. APPLICATIONS OF ALGEBRAIC TOPOLOGY

i.e., there exists a continuous map F : I2 → X such that F (t, 0) = γ(t),F (0, u) = F (1, u) = F (t, 1) = γ(0).

Theorem 12.1.6 (Monodromy theorem). Let (Y, π,X) be a space overX and let a, b ∈ X, a′ ∈ π−1(a). Let F : I2 → X be continuous andF (0, u) = a, F (1, u) = b (0 ≤ u ≤ 1). Suppose that for every u ∈ I the curveγu : t → F (t, u) admits lifting to γu : I → Y such that γu(0) = a′. Then

F : (t, u) 7→ γu(t) is continuous and is a lifting of F to Y . In particular,

F (1, u) = γu(1) is independent of u ∈ I.

Definition 12.1.7. Let (Y, π,X) be a space over X. We call (Y, π,X)a covering space if for every x ∈ X there exists a fundamental neighborhoodUx of x. This means that

π−1(Ux) = ∪α∈AVα,with Vα ∩ Vβ = ∅, (α 6= β ∈ A) and π|Vα is a homeomorphism onto Ux.

Proposition 12.1.8. Let I = [a, b] be a closed interval and let (Y, π,X)be a covering space. Suppose that f : I → X is continuous and let x0 ∈π−1(f(a)). Then there exists precisely one lifting g : I → X of f such thatπ g = f and g(a) = x0.

Recall that a topological space is locally path connected if for every pointx and every neighborhood U of x there exists a neighborhood V of x con-tained in U with the property that for every a, b ∈ V there exists a curveγ : [0, 1] → V such that γ(0) = a and γ(1) = b.

Theorem 12.1.9. Let (Y, π,X) be a covering space and let Z be a con-nected, locally path connected and simply connected topological space. Letf : Z → X be continuous and suppose that f(z0) = x0. Suppose π(y0) = x0.Then there exists a lifting g : Z → Y such that π g = f and g(z0) = y0.

12.2. Weierstrass theory of analytic continuation, Riemann

surfaces

In the present section we discuss classical methods of analytic continua-tion – techniques based on power series, the relation with Riemann surfacesand the emerging of the sheaf concept.

12.2.1. Weierstrass theory of analytic continuation. We will con-sider tripels (a,U, f), where a ∈ C, U is a connected open neighborhood ofa and f is a function on U into some non specified, but fixed set X. Twotripels (a,U, f), (a′, U ′, f ′) are called equivalent is a = a′ and f = f ′ on aneighborhood U ′′ of a contained in U ∩ U ′. This is indeed an equivalencerelation, as is easily seen. The equivalence class of (a,U, f) is called thegerm of f at a and we will write [(a,U, f)] of [f ]a. We will meet germs ofcontinuous and of smooth functions, with values in R, C or worse, but forus the most prominent case is that f is holomorphic. In the setting of thischapter, the holomorphic functions on an open set U ⊂ C are denoted byO(U). The tripel (a,U, f), f ∈ O(U) is then called a function element at apoint a ∈ C.

Using the uniqueness theorem for power series expansions of holomorphicfunctions, one sees that for a ∈ U ∩ U , elements (a,U, f) and (a, U , f) at

12.2. ANALYTIC CONTINUATION 129

the same point a are equivalent if and only if f and f have the same powerseries expansion at a: [f ]a = [f ]a.

Thus germs of holomorphic functions can be identified with convergentpower series. If no confusion is possible we may occasionally identify germsof holomorphic functions with their representatives.

Definition 12.2.1. A function element (b, V, f) is called a direct analyticcontinuation of the element (a,U, f) if V ∩ U is nonempty and g = f on acomponent of V ∩ U. [Some authors require that g be equal to f on everycomponent of V ∩ U ]. More generally, an element (b, V, g) at b is calledan analytic continuation of (a,U, f) if there is a finite chain of elements(ak, Uk, fk), k = 0, 1, . . . , p which links (a,U, f) to (b, V, g) by successivedirect continuations:

(a0, U0, f0) = (a,U, f), (ap, Up, fp) = (b, V, g)

and

(ak, Uk, fk) is a direct analytic continuation of (ak−1, Uk−1, fk−1)

for k = 1, . . . , p.

One loosely speaks of an analytic continuation of f ∈ O(U) to V . IfV ∩ U is nonempty, the uniqueness theorem shows that (a,U, f) has atmost one direct analytic continuation (b, V, g) for given b ∈ V and a givencomponent of V ∩U. [On a different component of V ∩U , g may be differentfrom f .] In the case of a chain as above, one may insert additional elementsto ensure that ak belongs to Uk ∩ Uk−1 for k = 1, . . . , p. Such a chainmay be augmented further to obtain analytic continuation along an arcγ : [0, 1] → C from a to b, namely, if γ is chosen as follows: γ(0) = a,γ(1) = b and there is a partitioning 0 = t0 < t1 < . . . < tp = 1 suchthat γ(tk) = ak and the subarc of γ corresponding to the interval [tk, tk+1]belongs to Uk, k = 0, . . . , p− 1. One can then define a continuous chain ofelements (at, Ut, ft) = (γ(t), Uk, fk), tk ≤ t ≤ tk+1, k = 0, . . . , p − 1 whichlinks (a,U, f) to (b, V, g).

Given an element (a,U, f) at a and a point b, different chains startingwith (a,U, f) may lead to different [more precisely, inequivalent] elementsat b.

Example 12.2.2. Start with the function element

(12.2.1) (1, Re z > 0, p.v. log z)

at the point z = 1 of C. Here the principal value of

log z = log |z| + i arg z, z 6= 0

denotes the value with imaginary part > −π but ≤ +π. Hence in our initialelement, log z has imaginary part between −π/2 and π/2. One may continuethis element analytically to the point z = −1 along the upper half of the unitcircle. At any point eit, for the moment only for 0 ≤ t ≤ π, one may use thehalf-plane Ht = t − π/2 < arg z < t + π/2 as basic domain and on thishalf-plane one will obtain the holomorphic branch of log z with imaginarypart between t − π/2 and t + π/2. On the half-plane Re z < 0 as basic


domain around z = −1, our analytic continuation will thus give the branchof log z with imaginary part between π/2 and 3π/2.

We continue this process and arrive at H2π (which as a subset of C isjust the right half-plane) with a branch of the logarithm with imaginarypart in (3π/2, 5π/2). Continuing in this way and observing that the basicdomains are related by Ht+2π = Ht, we will find function elements with thevalue of the imaginary part increased by 2π, each time we return to a basicdomain.

However, one may continue the original element (12.2.1) in the oppositedirection, along the lower half of the unit circle. The proces will be similarto the one described above, but this time the imaginary part of our functionelement decreases by 2π for each time we return to a basic domain.

We have constructed the complete analytic function log z.

Definition 12.2.3 (Weierstrass). The totality of all equivalence classesof function elements (b, V, g) (or of all convergent power series gb) at pointsb ∈ C, which may be obtained from a given element (a,U, f) by unlimitedanalytic continuation, is called the complete analytic function F generatedby (a,U, f).

Example 12.2.4 (√z). In a completely analogous fashion we may con-

struct the complete analytic function√z. We keep the notation of the

previous example. Recall that a value of√z is just a value of e

12 log z.

We start with the function element (1,H0,p.v.√z) = (1,H0, e

12p.v. log z);

on Ht we obtain (with log the complete analytic function!) (eit,Ht,√z) =

(eit,Ht, e12 log z). Now we find that the sign of the square root changes each

time we return to a basic domain Ht. Hence, (eit,Ht,√z) = (eit,Ht+4π,

√z).

There are only two function elements associated to a basic domain!

12.2.2. Riemann surface for F. As the examples show, a completeanalytic function F may be multivalued over C. In order to get a betterunderstanding of such a function, we will introduce a multisheeted Riemannsurface R for F over (an open subset of) C on which F may be interpretedas a single-valued function. These Riemann surfaces will be special cases ofa space over a domain in C (Definition 12.1.1

Definition 12.2.5. The Riemann surface associated to a complete an-alytic function F is a space (R, π,G) over a domain G as follows. Thepoints of the Riemann surface R for F as in Definition 12.2.3 are thegerms p = [(b, V, g)] = [g]b where (b, V,G) is a function element of F .G = b : ∃[g]b ∈ R and the projection π is given by π : p = [g]b 7→ b.The topology on R is defined by declaring a basis for it as follows. A basicneighborhood of p = [g]b will be a sets of the form

(12.2.2) N = N (p, V, g) = [g]z : z ∈ V ,

where (p, V, g) is a function element of F . Thus N consists of the germs ofg at points of V . The basic neighborhoods indeed form a basis for a topol-ogy on R, as one checks rapidly, cp. Exercise 12.5.1. Small enough basicneighborhoods will separate the points of R, (Exercise 12.5.2, so we have

12.2. ANALYTIC CONTINUATION 131

a Hausdorff space. The restriction π | N (b, V,G) establishes a homeomor-phism of N ⊂ R onto V in C. I.e., (R, π, π(R)) is a space over the open setπ(R) ⊂ C.

Over each point b ∈ C, the Riemann surface R for F will have as manylayers as there are different equivalence classes [(b, V, g)] in F at b. (it maybe that there is no layer at all)

Smaller Riemann surfaces can be defined in this way too. Instead ofconsidering analytic extensions along arbitrary curves in C, we may restrictourselves to curves inside a prescribed (connected) open set U in C. Weobtain a space over (a subset of) U . Notice that it may be smaller thenπ−1(U) in R, e.g. If we take for U the right halfplane and extend theprincipal value of the logarithm along curves in U , then only one sheet isobtained.

Theorem 12.2.6 ([12]). Let R be the Riemann surface associated toa function element (a,U, f). An element (b, V, g) is obtained by analyticcontinuation of (a,U, f) along an arc γ in C if and only if R contains anarc σ above γ which connects [f ]a and [g]b

Proof. Suppose that (b, V, g) is obtained by analytic continuation of(a,U, f) along γ. Then there is a chain of function elements (γ(t), Ut, ft)that connect (a,U, f) and b(V, g). Define σ by σ(t) = [ft]γ(t). It is clearthat π σ = γ. Since π is a local homeomorphism, σ is continuous. On theother hand if σ connects [f ]a and [f ]b in R, then γ = π σ is a curve inC that connects a to b. For every t we choose a representative (γ(t), Ut, ft)of σ(t). The open sets Ut cover γ. We take a finite subcover Utj, then(γ(tj), Utj , ftj ), j = 1, . . . , n, is an analytic continuation of (a,U, f) to(b, V, g).

Theorem 12.2.7 (Classical monodromy). Let G be a simply connecteddomain in C. Let U be a connected open set in G, let a ∈ U , and let (a,U, f)be a function element. Suppose that (a,U, f) can be extended along everycurve γ in G that starts in a. Then the possible multiple valued functionobtained by unlimited analytic continuation along these curves is in factsingle valued.

Proof. Let R denote the Riemann surface over G obtained by analyticextension of (a,U, f) along curves in G. Let γ1, γ2 be two curves that bothconnect a to b ∈ G. We have to show that the elements [g1]b and [g2]b inR obtained by analytic continuation along γ1 and γ2 respectively, are thesame.

Let F be a homotopy between γ1 and γ2. By assumption, analyticcontinuation of (a,U, f) along each curve γu = F (u, ·) is possible, that is thecurves γu can be lifted to curves γu in R. But now the monodromy theorem12.2.6 shows that γu(1) is independent of u, so that [g1]b = γ1(1) = γ2(1) =[g2]b.

The complete analytic function F is made into a single-valued functionon its Riemann surface through the simple definition F([g]b) = g(b). Wenow let q = [h]z run over the neighborhood N (p, V, g) in R. The result is

F(q) = F([h]z) = h(z) = g(z), ∀q = [h]z ∈ N (p, V, g).


Thus on the Riemann surface, F is locally given by an ordinary holomorphicfunction g on a domain V ⊂ C “under” R. Taking this state of affairs as anatural definition of holomorphy on R, the function F will be holomorphic.Setting [(a,U, f)] = p0 and identifying N (p0, U, f) with U , one will haveF = f on U . In that way the Riemann surface R will provide a maximalcontinuation or existence domain for the function f ∈ O(U): every germ ofevery analytic continuation is represented by a point of R.

There are also more geometric theories of Riemann surfaces, not directlytied to functions F . Riemann surfaces are examples of so-called domainsX = (X,π) over C. The latter are Hausdorff spaces X with an associatedprojection π to C. Every point of X must have a neighborhood on which πestablishes a homeomorphism onto a domain in C. The C coordinates zj canserve as local coordinates on X; different points of X over the same pointz ∈ C may be distinguished by means of an additional coordinate, cf. [29].

Given a function element (a,U, f) and a boundary point b of U , theremay or may not exist a direct analytic continuation (b, V, g) at b.

There always exist functions f ∈ O(U) that can not be continued analyt-ically across any boundary point of U . This is easily seen: using Weierstraßtheorem mentioned at the end of 1.10 one constructs a holomorphic functionf on U such that the boundary of U is in the closure of the zeroes of f , cf.Chapter 6. However, this latter result is typical for functions of one complexvariable, and does not hold for functions of several complex variables.

12.3. Examples in Function Theory, Picard’s theorem

Example 12.3.1. The map z 7→ zn is a covering map from C \ 0 toC \ 0.

Example 12.3.2 (Continuous branch of the logarithm). First we observethat (C, exp,C \ 0), with exp(z) = ez the exponential map, is a coveringspace. Indeed, let z ∈ C \ 0. Then Bz = B(z, |z|) is a fundamentalneighborhood. Choose a holomorphic branch of the logarithm hz on Bz,which is possible because arg can be chosen continuous with values in aninterval of length π, and

exp−1(B(z, |z|)) = ∪j∈Z(h(B(z, |z|)) + 2jπi).

Now let Z be a connected, locally path connected and simply connectedtopological space and f : Z → C a continuous zero free function. Applicationof theorem 12.1.9 then shows that there exists a continuous function F : Z →C such that eF (x) = f(x), in other words, F = log f i.e., we have defined acontinuous branch of log f may be defined on Z. Moreover, for z0 ∈ Z andgiven y0 ∈ exp−1(f(z0)), i.e., y0 is a value of log(f(z0)), the theorem givesthat we may demand F (z0) = y0 and that then F is unique.

If Z is an open, simply connected set in C and f is holomorphic, thenso is F , because locally we can write F = hz f . In this situation we mayalso define the logarithm as an integral

F (z) =

∫ z

z0

f ′(ζ)f(ζ)

dζ + y0.

12.3. EXAMPLES IN FUNCTION THEORY, PICARD’S THEOREM 133

12.3.1. The modular function and Picard’s theorem. Let H bethe upper halfspace.

Lemma 12.3.3. Aut(H) consists of the fractional linear transformationsof the form

F (z) =az + b

cz + d, a, b, c, d ∈ R, ad− bc > 0.

If ad− bc = 1 then F−1(z) = dz−b−cz+a .

The proof is left as Exercise 12.5.3.

Definition 12.3.4. The modular group PGL(2,Z) is the group of frac-tional linear transformations of the form

az + b

cz + d, a, b, c, d ∈ Z, ad− bc = 1.

Proposition 12.3.5. The modular group PGL(2,Z) is a subgroup ofAut(H).

Proof. This is immediate from Lemma 12.3.3.

We will need the subgroup M of PGL(2,Z) generated by the two ele-ments

T : z 7→ z + 2, S : z 7→ z

2z + 1.

Let G0 be the domain given by 0 < Re z < 1, |z − 1/2|2 > 1/4. Itsboundary consists of three arcs: the halflines Γ1 = z = iy, y ≥ 0, Γ2 =z = 1 + iy, y ≥ 0, and the halfcircle Γ0 = |z − 1/2|2 = 1/4, y ≥ 0.

Let λ be the conformal map from G0 to H, such that λ(0) = 0, λ(1) = 1and λ(∞) = ∞. Such a map exists according to the Riemann mappingtheorem and the Caratheodory theorem. By prescribing the values at 0, 1,∞ we made sure that there is only one map λ!

Next we will extend λ to all of H by continued application of the Schwarzreflection principle.

Theorem 12.3.6 (Modular Function). The map λ may be continuedanalytically to H. The continuation is called the modular function. Themodular function is invariant under the transformations T and S:

λ(T (z)) = λ(z) = λ(S(z)).

The triple (H, λ,C \ 0, 1) is a covering space over C \ 0, 1.Proof. We apply the Schwarz reflection principle and reflect in Γ1,

yielding an analytic extension of λ to the mirror image G1 of G0 containedin −1 < x < 0; λ(−x + iy) = λ(x+ iy). Similarly, reflection in Γ2 is

performed, yielding λ(2 − x + iy) = λ(x+ iy). We can continue in thisfashion and obtain that λ is extended to z : z = w+ j, w ∈ G0, j ∈ Z andmoreover

(12.3.1) λ(T (z)) = λ(z).

Indeed, T = R2 R1 with Rj the reflection in Γj .Next we reflect in Γ0. The reflection is given by z 7→ z

2z−1 The imageof G0 will be the domain G2 in 0 < x < 1 bounded by the circular arcs Γ0,


and the images of Γ1 and Γ2, which are circular arcs γ1 ⊂ |z − 1/4| = 1/4passing through 0 and 1/2, and γ2 ⊂ |z − 3/4| = 1/4 passing through 1/2and 1, respectively. Again γ1 and γ2 intersect the real axis perpendicularly.Apparently G0 ∪ G2 can be written as 1/2G0 ∪ (1/2G0 + 1/2) that is, theunion of two scaled copies of G0. We may continue as before, performingreflections in γ1. The image of the domain 0 < x < 1/2, |z − 1/4| > 1/4will be a scaled copy of G2 contained in H and bounded by γ1 and subarcs of|z − 1/8| = 1/8 and |z − 3/8| = 1/8. Reflection in γ2 leads to a similarresult. We end up with a union of 4 copies of G0 scaled by 1/4.

We conclude that a series of suitable reflections will reach every point inH.

Composition of reflection in Γ0 with the reflection in Γ1 yields the holo-morphic map S, it follows that

(12.3.2) λ(S(z)) = λ(z)

when left– and righthand side are defined.Schwarz’ reflection associated to the reflections described above (or the

relations (12.3.1) and (12.3.2)) extend λ to all of H. Every possible sequenceof reflections gives an analytic extension of λ along a suitable curve (deter-mined by the reflections) and along every curve in H analytic extensionis possible by choosing suitable reflections. From the classical monodromytheorem 12.2.7 we obtain that λ is single valued on H. We also concludethat λ(z) = λ(w) implies ∃m ∈ M such that w = m(z). Indeed, there existm1,m2 ∈ M such that m1(z),m2(w) ∈ G0 ∪ G1 \ 0, 1, but λ is injectiveon this set so m1(z) = m2(w).

We have proved that λ is an analytic function on H that is invariantunder the group M :

λ(m(z)) = λ(z), z ∈ H, m ∈M.

We now claim that (H, λ,C \ 0, 1) is a covering space. Firstly, λ issurjective: points in the upper halfplane are values of λ on G0, the real axis–except for 0 and 1– is the image of (∂G0)∩H and the lower halfplane is theimage of G1. Secondly, we have to show that every z ∈ C \ 0, 1 admits afundamental neighborhood. If Im z > 0 we can take the upper half plane Uas a fundamental neighborhood. Indeed, λ is a homeomorphism for m(G0)onto U for all m ∈M , λ−1(U) = ∪m∈Mm(G0) and m1(G0)∩m2(G0) = ∅ ifm1 6= m2.

The lower halfplane is treated similarly; for points z 6= 0, 1 on the realaxis we can take a disk about z that does not meet 0 or 1.

Theorem 12.3.7 (Picard). A meromorphic function on C that omitsthree (or more) values in C∗ is constant.

Proof. Suppose that f is meromorphic and omits at least three val-ues. By composition with a suitable linear fractional transformation, wecan assume that the values 0, 1,∞ are not in the image of f . We considerthe covering map (H, λ,C \ 0, 1) and the map f : C → C \ 0, 1 and arenow in the situation of theorem 12.2.6. Thus there exists a continuous mapf : C → H such that λ f = f . Observe that f is holomorphic, because

12.4. SHEAVES 135

f and λ are holomorphic and λ is locally invertible. But then, because H

and B are biholomorphically equivalent, and the theorem of Liouville, f isconstant! Hence so is f .

Remark 12.3.8. This theorem of Picard is often called Picard’s littletheorem. A stronger result is Picard’s big theorem, which states that a holo-morphic function in an arbitrary punctuated neighborhood of an essentialsingularity assumes all values except possibly two. The little theorem is thespecial case where the singularity is at ∞ and the neighborhood is all of C.

12.4. Sheaves

Sheaves were introduced and studied by Cartan, Leray and Serre. Theywere used by Cartan and Grauert in connection with the solution of theLevi-problem. Sheaves have been a highly successful tool in several partsof mathematics, particularly in algebraic geometry. This section is meantas an introduction to the concept only. The reason to discuss sheaves atall in this course, is that the setting of complex analysis is a very naturalenvironment to introduce them.

Often a sheave is defined as a special kind of presheaf (that is a functorfrom the category of open subsets in a topological space to the category ofsets (or groups, rings, etc.) with some additional properties). Then, surpris-ingly, it turns out that there is a one-one correspondence between sheavesand spaces X over a base space Y . As this definition is a bit complicated,we will use the one-one correspondence; this means that we only give anew name to “espace etale”. Nevertheless, we look at sheaves from a quitedifferent perspective.

Definition 12.4.1. A sheaf F over a space X with projection π is atriple (F , π,X) where F and X are topological spaces and π is a surjectivelocal homeomorphism. In other words a sheaf F over X is a space over X

A section of (F , π,X) over an open U ⊂ X is a continuous map σ : U →F such that σ π is the identity mapping on U . The sections over U aredenoted by F(U) or Γ(U) = Γ(U,F). A stalk of (F , π,X) is a subset of Fof the form π−1(x) where x ∈ X.

A sheaf of rings, (abelian) groups, etc. is a sheaf F with the propertythat the stalks F(x) have the structure of a ring, respectively, an (abelian)group, etc. of which the algebraic operations like addition or multiplicationare continuous.

Continuity means the following: form the product space F × F withproduct topology and consider the subset

F · F = (f1, f2) ∈ F × F : π(f1) = π(f2).Now addition (for example) in the stalks of F gives rise to a map

+ : F · F → F , (f1, f2) 7→ f1 + f2,

which has to be continuous.

Example 12.4.2. Let D be a domain in C.


(1) The constant sheaves C × D, Z × D, etc. over D. Projection isordinary projection on D. Observe that C (and Z etc.) need beequipped with the discrete topology.

(2) The Riemann surfaces (R, π,D) of Definition 12.2.5 equipped withthe usual topology, that is, defined by the basic neighborhoodsN (p, V, g).

(3) The sheaf of (all) germs of holomorphic functions on U , denoted byOU , with projection π : [f ]a 7→ a. Here [f ]a denotes the germ of ananalytic function f at a point a ∈ U . For OU to become a sheaf wehave to give it a topology that makes π a local homeomorphism.This can be done in a way similar to Example 2. or Definition12.2.5: A base for the topology is given by the sets

(12.4.1) N (V, f) = [f ]a : a ∈ V where f ∈ O(V ).

Sections over V can be identified with holomorphic functions on V :To a holomorphic function f on V we associate the section

σf : a 7→ [f ]a.

It is an easy exercise to check that σ is continuous. It should also beclear that a continuous section f ∈ O(V ) leads in a unique way to aholomorphic function defined by z 7→ fz(z), where fz(z) = g(z) forany representant g of fz. The fact that O(V ) indicates both sectionsover V and holomorphic functions on V reflects this association.

(4) Let K denote an algebra of functions on U . Thus K could be C∞(U)or ∧p,q(U) the (p, q)-forms on U (our functions may well be vectorvalued!) A germ of a function in K was defined in Section 12.2.1.As in the previous example these germs together form a sheaf; abase for its topology is given similar to (12.2.2). We thus obtainthe sheaf C∞

U of germs of smooth functions on U , the sheaf ∧p,qUof germs of smooth p, q forms on U , the sheaf O∗

U of holomorphiczero free functions on U , the sheaf MU of germs of meromorphicfunctions on U , etc. Again sections and functions can be identified.

It is easily seen that the examples 1,3,4, have the property that the stalksare abelian groups or have even more algebraic structure. We leave it to thereader to check that the algebraic operations are continuous.

The language of sheaves allows a slick description of complete analyticfunction:

Theorem 12.4.3. Let (z, U, f) be a function element. The Riemann sur-face of the complete analytic function associated to (z, U, f) is the connectedcomponent of fz in OC.

Proof. This follows immediately from theorem 12.2.6.

12.5. Exercises

12.5.1. Show that the sets N in Definition 12.2.5 form a basis for atopology on R.

12.6. APPENDIX: PROOFS OF THE RESULTS OF SECTION 12.1 137

12.5.2. Show that the topological space R of definition 12.2.5 is Haus-dorff. Consider the case of germs [f ]a, [g]b with a 6= b and with a = bseparately.

12.5.3. Prove Lemma 12.3.3

12.5.4. Show that every pair of entire functions (f, g) that solves theequation

ef + eg ≡ 1

consists of constant functions.

12.5.5. Show that the Riemann surface of the complete analytic func-tion associated to

1

sin(

log z2πi

)

− 1

is not a covering space over its projection.

12.5.6. Prove the Poincare-Volterra Theorem . Any fiber of a completeanalytic function associated to a function element of a holomorphic functionis atmost countably infinite. The fact that the Euclidean topology on C hasa countable base may come in handy.

12.5.7. Show that Montel’s theorem applies to normal families as fol-lows. If F is a family of holomorphic (meromorphic) functions on a domainD and there are three values in C∗ that are not assumed by members of F ,then F is normal.

12.5.8. Let R denote a finite Blaschke product with a simple zero at0. Prove the following

(1) 0 and ∞ are attracting fixed points and A0 = B, A∞ = C∗ \ B.(2) Let ϕ be holomorphic on B and conjugate R to λz at 0. Then ϕ

has infinitely many zeros as well as critical points in B and C(0, 1)is in the closure.of the zeros

(3) ϕ−1 is a multivalued function. On its Riemann surface the actionof R is “translated” to multiplying by λ and shifting to anothersheet.

12.6. Appendix: proofs of the results of section 12.1

Theorem 12.6.1. Let (Y, π,X) be a space over X and let f : Z → X becontinuous and suppose that Z is connected. Suppose g1 and g2 are liftingsof f and that g1(z) = g2(z) at some z ∈ Z. Then g1 = g2.

Proof. Let E = z : g1(z) = g2(z). Then E is non empty by assump-tion.

E is open: Let z ∈ E; take a neighborhood U of w = g1(z) = g2(z), suchthat π|U is a homeomorphism with inverse σ. Let W = g−1

1 (U) ∩ g−12 (U), a

neighborhood of z. Now π gi = f on W so that gi = σ f , i = 1, 2, on W .In other words, g1 = g2 on W .

E is closed because Y is a Hausdorff space and g1 and g2 are continuous.Because Z is connected, E = Z.


Remark 12.6.2. Apparently, only Z need be connected; Y must beHausdorff, and so will be X via π. Local path connectedness does not playa role here.

Theorem 12.6.3 (Monodromy theorem). 12.1.6 Let (Y, π,X) be a spaceover X and let a, b ∈ X, a′ ∈ π−1(a). Let F : I2 → X be continuousand F (0, u) = a, F (1, u) = b. Suppose that for every u ∈ I the curveγu : t → F (t, u) admits lifting to γu : I → Y such that γu(0) = a′. Then

F : (t, u) 7→ γu(t) is continuous and is a lifting of F to Y . In particular,

F (1, u) = γu(1) is independent of u ∈ I.

Proof. For every (t, u) ∈ I2 we find a neighborhood U = Ut,u of F (t, u)such that π|U is a homeomorphism; denote its local inverse by σt,u. Thereare (relatively) open intervals It about t, and Iu about u such that F (It ×Iu) ⊂ π(Ut,u). By compactness we can cover I2 with finitely many Itj × Iuj ,(j = 1, . . . , n). Let

T = t ∈ I : F is continuous on I × [0, t],and let s = supT ≤ 1. We will show that s = 1 and that s ∈ T .

Let j be such that s ∈ Itj . The map G : (t, u) 7→ σtj ,uj F (t, u) is

continuous and a lifting of F on Itj × Iuj . Since G(tj , uj) = F (tj, uj) and

since F (·, uj) is continuous, it follows from Theorem 12.1.6 that F (·, uj) =

G(·, uj). Take t0 in Itj such that F (t0, u) is continuous. (If s = 0 use

F (0, u) = a′, and choose t0 < s otherwise.) Again by Theorem 12.1.6

F (t0, ·) = G(t0, ·) on Iuj . A final application of Theorem 12.1.6 shows now

that F (·, u) = G(·, u) for every u, or F = G on Itj × Iuj . It follows that F iscontinuous on every relative open square Itj × Iuj that meets t = s. Hences = 1 and s ∈ T , which proves the theorem.

Remark 12.6.4. The theorem hinges on Theorem 12.1.6, and so it needsthe same assumptions.

Proposition 12.6.5. Let I = [a, b] be a closed interval and let (Y, π,X)be a covering space. Suppose that f : I → X is continuous and let x0 ∈π−1(f(a)). Then there exists precisely one lifting g : I → X of f such thatπ g = f and g(a) = x0.

Proof. Observe that by Theorem 12.1.6 there can at most be one liftingg with g(a) = x0. Let f : I → X be continuous. For every t ∈ I we choose afundamental neighborhood Ut of f(t). f(I) is compact and can be coveredby finitely many of the Ut, say U1, . . . Un. Let Wn = f−1(Un). We can nowassume a = 0, b = 1. Let

T = t ∈ I : on [0, t] there exists a continuous lifting g of f with f(0) = a′.(Any two liftings will coincide on the intersection of their domains by The-orem 12.1.6) Clearly 0 ∈ T . Let s = supT . Then s ≤ 1 and we shall seeagain that s ∈ T and s = 1. Suppose s ∈ Wn. Choose t0 ∈ Wn such that gis continuous at t0. Now g(t0) ⊂ V with V some homeomorphic preimageof Un. Let σUn → V be the local inverse of π. Then σ f is a continuouslifting of f on Wn which extends g to Wn. We conclude that s = 1 ands ∈ T , which proves the proposition.

12.6. APPENDIX: PROOFS OF THE RESULTS OF SECTION 12.1 139

Theorem 12.6.6. 12.1.9 Let (Y, π,X) be a covering space and let Z be aconnected, locally path connected and simply connected topological space. Letf : Z → X be continuous and suppose that f(z0) = x0. Suppose π(y0) = x0

then there exists a lifting g : Z → Y such that π g = f and g(z0) = y0.

Proof. We define g as follows. From the assumptions it follows that forevery z ∈ Z there exists a curve γz : I → Z such that γz(0) = z0, γz(1) = z.Then f γ is a curve in X with f γ(0) = x0, hence it may be lifted to acurve Γ : I → Y with Γ(0) = y0. We define

g(z) = Γ(1), z ∈ Z.

Because any two curves that connect z0 to z are homotopic, it follows fromTheorem 12.1.6 that Γ(1) is independent of the choice of γ. Hence g is welldefined.

Next we wish to prove that g is continuous. Let z ∈ Z. Take a funda-mental neighborhood U of f(z), a local inverse σ of π with σ(f(z) = g(z)and a path connected neighborhood W of z contained in f−1(U). On W wehave a continuous lifting of f defined by σ f . It coincides with g, becauseit coincides with g on curves in W starting at z.

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Index

C(Ω), 9Lp means, 119ρ-type, 49

Beurling-Nevanlinna Theorem, 105Dirac distribution, 103fixed point, 30multisheeted Riemann surface, 130

analytic continuation, 129Weierstrass theory, 128chain of elements, 129direct, 129

analytic continuation along an arc, 129analytic function

multivalued, 130approximate identity, 110, 114arc, 22argument principle, 2asymptotic development, 67, 77automorphisms, 18

b, 100backward iterates, 36Banach-Alaoglu theorem, 116barrier family, 100basin of attraction, 29, 30basis, 120Bernoulli numbers, 69Bessel function, 58Beta-function, 76biholomorphically equivalent, 22Binet’s formulas, 77Bishop’s lemma, 5Blaschke condition, 59, 123Blaschke factor, 51Blaschke product, 51, 59, 123Bohr radius, 118Borel-Caratheodory inequality, 50

canonical product, 54Caratheodory’s inequality, 118Carleman, 92Carleman Milloux problem, 105Cauchy, 1

Cauchy integral, 121Cauchy integral formula, 82Cauchy Transform, 4Cauchy’s formula, 2Cauchy-Pompeiu formula, 1Cauchy-Riemann equations, 1chordal length, 12chordal metric, 12, 35complete analytic function, 130completely invariant, 35conformally equivalent, 22Continuous branch of the logarithm,

132continuum, 101counting function, 52covering space, 128critical point, 35critical strip, 71curve, 22

homotopic, 127Jordan, 22real analytic, 24simple, 22

degree of a rational function, 35direct analytic continuation, 129Dirichlet domain, 97, 101Dirichlet problem, 84, 97

Perron solution, 101uniqueness, 90

distribution function, 107doubly periodic, 73duplication formula, 76

elementary factor of Weierstrass, 44, 54,55

elliptic function, 26, 46, 61, 73elliptic integral, 26equicontinuous, 9espace etale, 127Euler, 68

Beta-function, 76Euler-Mascheroni constant, 63exceptional set, 36exponent of convergence, 54

143

144 INDEX

exponential type, 49, 57

F. Riesz factorization theorem, 124Fatou set, 30fiber, 127fixed point, 29

attracting, 29repelling, 29irrationally neutral, 29rationally neutral, 29

Fourier coefficient, 120Fourier coefficients, 124fractional linear transformation, 17Fresnel-integrals, 76function

Hp, 119Bessel, 58complex analytic, 1complex differentiable, 1distribution, 107elliptic, 26, 46, 73elliptic- (see elliptic function), 61Gamma- (see Gamma-function), 61harmonic, 81holomorphic, 1locally integrable, 107maximal, 107meromorphic, 12of exponential type, 49, 57of finite order, 49relative extremal, 106subharmonic, 86superharmonic, 86upper semi-continuous, 86Von Mangoldt, 78Weierstrass-P , 46, 73Weierstrass-σ, 46Weierstrass-ζ, 46zeta, 67zeta- (see zeta-function), 61

function element, 128Functional equation

zeta-function, 71Functional equation for the

zeta-function, 71fundamental domain, 73fundamental neighborhood, 128fundamental parallellogram, 73fundamental solution, 103

Gamma-function, 61asymptotic development, 67, 77duplication formula, 76Euler-product, 63first functional equation, 61second functional equation, 62

Gauss’ formula, 77Generalized circles, 17

genus, 54, 56of a product, 54

geodesic, 19germ, 128Green’s formula, 1Green’s function, 95, 103Green’s second formula, 94Green’s third formula, 95

Hadamard’s product theorem, 56Hankel representation

for 1/Γ, 64, 65for zeta-function, 70

Hardy space, 119Hardy spaces, 107harmonic conjugate, 81harmonic function, 81

complex valued, 117harmonic measure, 90, 102Harnack, 98Harnack’s convergence principle, 98Harnack’s inequalities, 98Harnack’s inequality, 114Herglotz representation, 115, 116homotopic curves, 127homotopic family, 127homotopy, 127Hurwitz, 2hyperbolic distance, 19hyperbolic length, 19hyperbolic metric, 19

inequalityCaratheodory, 118Harnack, 98Landau, 118

infinitesimal metric, 19integrated counting function, 53

Jensen’s formula, 52, 95, 122Jensen’s inequality, 94, 111Jordan curve, 22Jordan domain, 23Julia set, 29, 30

Landau’s inequality, 118Lattes example, 79lattice, 45, 72Lebesgue point, 113, 114Lebesgue set, 112Lebesgue’s differentiation theorem, 110,

112lemma

Bishop, 5Schwarz, 18Vitali, 109

lifting, 127limit

INDEX 145

non-tangential, 114radial, 114

line element, 19line element of the hyperbolic metric, 19locally connected set, 22locally equicontinuous, 10locally path connected, 128logarithmic differentiation, 41

Mobius transformation, 17Mandelbrot set, 29, 37maximal function, 107maximum modulus, 49maximum principle, 86, 92

for holomorphic functions, 2mean value equality, 82Mean Value Inequality, 86meromorphic function, 12method of Laplace, 65, 67method of steepest decent, 67metric

chordal, 12, 35hyperbolic, 19Poincare, 19

metric space, 9Mittag-Leffler expansion, 68Mittag-Leffler function, 77Mittag-Leffler representation, 48, 62Mittag-Leffler theorem, 46modular function, 133modular group, 133monodromy theorem, 128, 138

classical, 131Montel’s theorem, 137multiplicity, 35multiplier, 30

Nevanlinna, 92Nevanlinna counting function, 53normal, see also subnormalnormal family, 29, 30number theory, 68

open mapping theorem, 2order, 49orthogonal system

complete, 120Ostrowski, 92

period, 30, 72periodic point, 30Perron family, 98Perron function, 98Perron method, 97Perron solution, 101Perron subsolution, 101Perron supersolution, 101Picard’s big theorem, 135

Picard’s little theorem, 135Picard’s theorem, 57, 133, 134Plemelj formula, 126Poincare metric, 19Poincare-Volterra Theorem, 137Poisson integral, 83, 121

boundary behavior, 113Poisson kernel, 83

Fourier series of, 83Poisson modification, 99polar set, 95potential, 86preperiodic point, 30presheaf, 135prime number theorem, 72primes, 68primitive periods, 72principal part, 46, 48projection, 135

Radon-Nikodym decomposition, 113relative extremal function, 106repelling fixed point, 29Riemann

zeta-function, 48Riemann hypothesis, 71Riemann mapping theorem, 20Riemann sphere, 11, 17, 29Riemann surface, 128, 130

associated to a complete analyticfunction, 130

Riemann zeta-function, 67first integral representation, 68product representation, 68trivial zeros, 71

Riemann-Stieltjes integral, 108Riemann-zeta function, 48Riesz-Fischer theorem, 120, 121

saddle point method, 67Schwarz reflection principle, 133Schwarz’ Lemma, 18Schwarz’ reflection, 24section, 135separable, 10sheaf , 135simply connected, 127singular measure, 113space over, 127spherical derivative, 13stalk, 135subharmonic function, 86superharmonic function, 86supremum norm, 9

theoremTwo-Constants, 92Arzela-Ascoli, 10

146 INDEX

Banach-Alaoglu, 116Caratheodory, 133F. and M. Riesz, 126F. Riesz factorization, 124Hadamard, 56, 63Hahn Banach, 3Hartogs-Rosenthal, 5Hurwitz, 2Lebesgue

differentiation, 110, 112Lindelof, 58Marty, 13Mergelyan, 6Mittag-Leffler, 46Montel, 35Montel for normal families, 137Morera, 3, 24open mapping, 2Phragmen-Lindelof, 92Picard, 57, 133, 134Poincare-Volterra, 137prime number, 72Radon-Nikodym, 113residue, 2Riemann mapping, 20, 133Riesz representation, 4Riesz-Fischer, 120, 121Rouche, 2Runge, 4Schwartz-Christoffel, 25uniqueness, 2Weierstrass, 6

approximation, 3uniform convergence, 2

Weierstrass product, 45total variation, 107transformation

fractional linear, 17Mobius, 17

Two-Constants Theorem, 92type, 49

upper semi-continuous function, 86

Vitali, 109Vitali sequence, 109Vitali’s covering lemma, 109Von Mangoldt function, 78

Wallis-product, 47Weierstrass, 3, 128P-function, 46, 73σ-function, 46ζ-function, 46elementary factor, 44, 54, 55product-theorem, 44, 45, 48

Weierstrass’ theorem, 2

zeta-function, 61functional equation, 71Riemann, 67

jan wiegerinck version march 1, 2011 · 9.4. green’s function 103 9.5. beurling-nevanlinna...

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