The Cauchy transform, potential theory, and conformal mapping

The Cauchy Transform, Potential Theory and Conformal Mapping explores the most central result in all of classical function theory, the Cauchy integral formula, in a new and novel way based on an advance made by Kerzman and Stein in 1976.

The book provides a fast track to understanding the Riemann Mapping Theorem. The Dirichlet and Neumann problems for the Laplace opera-tor are solved, the Poisson kernel is constructed, and the inhomog-enous Cauchy-Reimann equations are solved concretely and efficiently using formulas stemming from the Kerzman-Stein result.

These explicit formulas yield new numerical methods for computing the classical objects of potential theory and conformal mapping, and the book provides succinct, complete explanations of these methods.

Four new chapters have been added to this second edition: two on quadrature domains and another two on complexity of the objects of complex analysis and improved Riemann mapping theorems.

The book is suitable for pure and applied math students taking a begin-ning graduate-level topics course on aspects of complex analysis as well as physicists and engineers interested in a clear exposition on a fundamental topic of complex analysis, methods, and their application.

K25868

w w w . c r c p r e s s . c o m

The Cauchy Transform, Potential Theory and Conformal Mapping 2nd Edition


The C

auchy Transform, Potential T

heory and Conform

al Mapping

Steven R. Bell

Bell

2nd Edition

Mathematics

K25868_cover.indd 1 10/9/15 11:27 AM


Steven R. BellPurdue UniversityWest Lafayette, Indiana, USA

CRC PressTaylor & Francis Group6000 Broken Sound Parkway NW, Suite 300Boca Raton, FL 33487-2742

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Contents

Preface vii

Table of symbols xi

1 Introduction 1

2 The improved Cauchy integral formula 3

3 The Cauchy transform 9

4 The Hardy space, Szego projection, and Kerzman-Stein

formula 13

5 The Kerzman-Stein operator and kernel 17

6 The classical definition of the Hardy space 21

7 The Szego kernel function 27

8 The Riemann mapping function 33

9 A density lemma and consequences 35

10 Solution of the Dirichlet problem in simply connected

domains 43

11 The case of real analytic boundary 47

12 The transformation law for the Szego kernel 51

13 The Ahlfors map of a multiply connected domain 59

14 The Dirichlet problem in multiply connected domains 65

15 The Bergman space 69

v

vi Contents

16 Proper holomorphic mappings and the Bergman

projection 77

17 The Solid Cauchy transform 87

18 The classical Neumann problem 93

19 Harmonic measure and the Szego kernel 97

20 The Neumann problem in multiply connected domains 107

21 The Dirichlet problem again 111

22 Area quadrature domains 113

23 Arc length quadrature domains 123

24 The Hilbert transform 131

25 The Bergman kernel and the Szego kernel 135

26 Pseudo-local property of the Cauchy transform 141

27 Zeroes of the Szego kernel 149

28 The Kerzman-Stein integral equation 153

29 Local boundary behavior of holomorphic mappings 159

30 The dual space of A∞(Ω) 165

31 The Green’s function and the Bergman kernel 177

32 Zeroes of the Bergman kernel 183

33 Complexity in complex analysis 187

34 Area quadrature domains and the double 191

A The Cauchy-Kovalevski theorem for the

Cauchy-Riemann operator 197

Bibliographic Notes 199

Bibliography 203

Index 207

Preface

The Cauchy integral formula is an old and beautiful result. Anybodywith an undergraduate mathematics education knows something aboutthis formula. Since the Cauchy integral has been studied for so long byso many people, it is tempting to believe that everything there is toknow about the integral is known. Such beliefs are always wrong. In1978, Norberto Kerzman and Elias M. Stein [K-S] discovered a verybasic, yet previously unknown, property of the Cauchy integral. Theydiscovered that the Cauchy integral is nearly self adjoint. This discoveryconstituted a shift in the bedrock of complex analysis. The vibrationswere felt immediately in the areas of boundary behavior of holomorphicfunctions, Hardy spaces, and mapping problems. I hope this book willhelp dissipate the shock wave through some areas that were thought tobe peacefully settled forever.

Students encountering complex analysis for the first time are de-lighted to see how many of the basic theorems follow directly from theCauchy integral formula. In the first course on complex variables, theCauchy integral is applied only to holomorphic functions. This book isintended to be part of a second course on complex variables. In it, theCauchy integral will be applied to nonholomorphic functions. Since theCauchy integral of a nonholomorphic function does not reproduce thatfunction, this process gives rise to a transform. The Cauchy transformmaps functions defined on the boundary of a domain to holomorphicfunctions on the domain.

The Kerzman-Stein breakthrough allows many of the basic objectsof a second course in complex analysis to be described in terms of theCauchy transform. More theorems than ever before can be seen to bedirect corollaries of simple facts about the Cauchy integral. In this book,we will deduce the Riemann mapping theorem, we will solve the Dirichletand Neumann problems for the Laplace operator, we will construct thePoisson kernel, and we will solve the inhomogeneous Cauchy-Riemannequations. We will do all this in a very constructive way using formulasstemming from the Kerzman-Stein theorem.

Kerzman and Stein made their discovery while studying the Szegokernel in several complex variables. Discussion of the Szego projectionand kernel have always been considered to be too advanced to include in

vii

viii Preface

a second course in complex analysis. However, because of the Kerzman-Stein result, we will be able to discuss these objects in very simple andconcrete terms. We will also be able to study the Bergman kernel. Theseclassical objects are very useful in the study of conformal mappings andwe explore many of their applications.

Another motivation for writing this book has been that the new for-mulas expressing the classical objects of potential theory and conformalmapping in terms of Cauchy transforms and Szego projections have ledto new numerical methods for computing these objects. In the past, ithas been difficult for nonspecialists to find accessible descriptions of thebackground material needed to understand the new formulas. A readerwho is interested only in understanding the Kerzman-Stein numericalmethod for computing conformal mappings may read Chapters 1–8, thenskip to Chapter 28.

Many classical problems in conformal mapping and potential theoryon finitely connected domains in the plane can be reduced by means ofthe Riemann mapping theorem to problems on domains whose bound-aries consist of finitely many real analytic curves. In this book, we studymany problems for which this standard reduction is applicable, however,the full force of the real analyticity of the boundary curves is only rarelyneeded. In most situations we will encounter, it will suffice to studyproblems on a domain whose boundary curves are merely C∞ smooth.This is why we fixate on problems of C∞ behavior of objects on domainswith C∞ smooth boundaries. Although many complex analysts may lookupon such a fixation as being unwholesome, I believe it is justified in thecontext of this book. I should also mention that many results that weprove in the C∞ category can be quite routinely generalized to applyunder milder smoothness hypotheses. I did not feel compelled to explainthese more general results because they are irrelevant to the main themeof the book.

I have written this book thinking that my readers have completeda one-semester course on complex analysis and that they know somerudimentary facts about real analysis and Hilbert space. The facts fromHilbert space and measure theory are so few and so basic that theycould easily be learned along the way. I also assume that my readerknows about C∞ partitions of unity and a little about differential forms,enough to understand Green’s Theorem in planar domains. Any voids inthe reader’s background can be filled by consulting Ahlfors [Ah], Rudin[Ru1, Ru2], and Spivak [Sp].

I wrote the first edition of this book thinking of it as a tour of abeautiful region of classical analysis viewed through the new lenses of theKerzman-Stein theorem. Writing the first edition propelled me to studyquadrature domains, complexity in complex analysis, and improvements

Preface ix

upon the Riemann mapping theorem. I have added four new chaptersto this second edition: Chapters 22 and 23 on quadrature domains andChapters 33 and 34 on complexity of the objects of complex analysisand improved Riemann mapping theorems. This new material bringsthe tour to a destination where I currently wander.

I would like to thank David Barrett, Harold Boas, Young-Bok Chung,Anthony Thomas, and Alan Legg for reading preliminary drafts of thisbook and for making many constructive suggestions. I’d like to thankSteve Krantz for planting the idea of writing this book in my head.

West Lafayette, April 2015

Table of symbols

(∂/∂z) the ∂-operator 4(∂/∂z) the ∂-operator 4ϕz = ∂ϕ/∂z 6ϕz = ∂ϕ/∂z 6〈u, v〉b inner product on L2(bΩ) 11〈u, v〉Ω inner product on L2(Ω) 69〈h, g〉 extension of L2(Ω) inner product to

A∞(Ω)×A−∞(Ω) 166‖u‖ the norm on L2(bΩ) (or L2(Ω)) 11 (69)‖u‖s the norm on Cs(bΩ) (or Cs(Ω)) 35 (165)‖u‖−s the norm on A−s(Ω) 165‖u‖As the Cs norm on an arc A ⊂ bΩ 141∫

integral on a curve in the plane 5∫ ∫integral on an open set in the plane 5

P.V.∫

Principal Value integral 17∆arg h increase in argument of h around bΩ 40, 60ρ boundary defining function 6ωk harmonic measure function 97[Ajk ] matrix of periods 100A Kerzman-Stein operator 13A(z, w) Kerzman-Stein kernel 14, 19A∞(Ω) holomorphic functions in C∞(Ω) 9A∞(bΩ) ≡ A∞(Ω) 13A−∞(Ω) the dual of A∞(Ω) 165A−s(Ω) a subspace of A−∞(Ω) 165bΩ boundary of Ω 3B the Bergman projection 70C Cauchy transform 9C∗ adjoint of Cauchy transform 12Ca(z) kernel of Cauchy transform 27Ck(Ω) functions with k continuous derivatives

on Ω 3C∞(Ω) = ∩∞

k=0Ck(Ω) 3

C∞(bΩ) C∞ smooth functions on bΩ 3

xi

xii Table of symbols

Cω(bΩ) real analytic functions on real analytic bΩ 48Cs(bΩ) functions on bΩ with s continuous

derivatives 35Dr(a) disc of radius r about a 5E Poisson extension operator 45F ′k = 2∂ωk/∂z 97

F ′ the linear span of the functions F ′k 98

G the classical Green’s operator 73G(z, a) the Green’s function 177Ga(z) = (1/2π)(z − a)−1 30H the Hilbert transform 131 (17)H2(bΩ) Hardy space 13H2(bΩ)⊥ orthogonal complement of H2(bΩ)

in L2(bΩ) 15H2(Ω) the Bergman space 69H2(Ω)⊥ the orthogonal complement of H2(Ω)

in L2(Ω) 74Ka(z) = K(z, a) the Bergman kernel function 70L2(bΩ) square integrable functions on bΩ 11L2(Ω) square integrable functions on Ω 69La(z) = L(z, a) Garabedian’s kernel 29ℓ(z, w) = L(z, w)− (2π)−1(z − w)−1 145P the Szego projection 13P⊥ the projection onto H2(bΩ)⊥ 16P.V. Principal Value 17S(z) the Schwarz function 118Sa(z) = S(z, a) the Szego kernel 27T the complex unit tangent vector function 3

1

Introduction

One reason that the unit disc D1(0) in the complex plane is the mostpleasant place on earth to do function theory is that the monomialszn are orthogonal in two basic inner product spaces associated to thedisc; they are orthogonal with respect to the boundary arc length innerproduct

〈u, v〉b =∫u(eiθ)v(eiθ) dθ

and with respect to the area measure inner product

〈u, v〉 =∫∫

D1(0)

u v dx dy.

Therefore, after the theory of Hilbert space had been developed, it wasinevitable that the spaces obtained by forming the Hilbert space comple-tions of the polynomials with respect to the boundary and the area innerproducts would be studied. These spaces turned out to be as heavenlyas the unit disc itself. On these two spaces, classical complex analysis,measure theory, and functional analysis blend together naturally to yieldscores of theorems that any mathematician would find appealing.

The completion of the polynomials with respect to the boundary in-ner product is the Hardy space, and the completion with respect to thearea inner product is the Bergman space. These spaces have assumedprominent spots in the literature of graduate mathematics. The purposeof this book is to define and study the analogous spaces for domains inthe plane more general than the disc and to reap some of the remark-able applications that these spaces have found in potential theory andconformal mapping. Although the subject matter of this book has notbeen considered elementary enough to become a standard part of thegraduate mathematics curriculum, I hope to present it in such a waythat it will be seen that it could easily be included in a second courseon complex variables. Because this is part of my mission, and because itrequired very little extra effort and ink, I have tried to make this bookcomprehensible to a first year graduate student in mathematics.

1

2 The Cauchy Transform, Potential Theory, and Conformal Mapping

The paper of Kerzman and Stein [K-S] is at the foundation of thepresent work; it is the Kerzman-Stein viewpoint that allows the clas-sical results I will describe to be simplified to the point that they canbe understood by beginners in complex analysis. Although this book isprimarily expository, it does have a research component. It is researchto look at old results from a new point of view, one that allows proofsto be streamlined and simplified. New outlooks always give rise to newtheorems. I have added the last two chapters of this new second editionof the book to illustrate this point.

2

The improved Cauchy integral formula

In this book, we will study functions defined inside and on the boundaryof a bounded domain Ω in the plane with C∞ smooth boundary. Such adomain is finitely connected and its boundary consists of finitely manyC∞ smooth simple closed curves. We let bΩ denote the boundary ofΩ. If Ω is n-connected, there are C∞ complex valued functions zj(t) oft ∈ [0, 1] ⊂ R, j = 1, . . . n, that parameterize the n boundary curves ofΩ in the standard sense. This means that, if zj(t) parameterizes the j-thboundary curve of Ω, it is understood that zj(t) and all its derivativesagree at the endpoints t = 0 and t = 1, that z′j(t) is nowhere vanishing,and that zj(t) traces out the curve exactly once. Furthermore, −iz′j(t)is a complex number representing the direction of the outward pointingnormal vector to the boundary at the point zj(t). We say that a functiong defined on the boundary of Ω is C∞ smooth on bΩ if, for each j,g(zj(t)) is a C

∞ function of t on [0, 1], all of whose derivatives agree atthe endpoints 0 and 1. This definition seems to depend on the choiceof the parameterization functions zj(t), but it is an easy exercise tosee that it in fact does not. We let C∞(bΩ) denote the space of C∞

functions on bΩ. One of the most important C∞ smooth functions onthe boundary of Ω is the complex unit tangent function T . If z ∈ bΩ, thenT (z) is equal to the complex number of unit modulus that represents thedirection of the tangent vector to bΩ at z pointing in the direction of thestandard orientation of the boundary. To be precise, T is characterizedby the formula T (zj(t)) = z′j(t)/|z′j(t)|. Since the differential dz is givenby dz = z′j(t) dt and the differential ds of arc length is given by ds =|z′j(t)| dt, we see that dz = T ds.

If k is a positive integer, Ck(Ω) denotes the space of continuous com-plex valued functions on Ω whose partial derivatives up to and includingorder k exist and are continuous on Ω and extend continuously to Ω.The space C∞(Ω) is the set of functions in Ck(Ω) for all k.

Everyone learns that a holomorphic function can be represented byits Cauchy integral. Everyone should also learn that even nonholomor-phic functions have Cauchy integral representations. In this book, unlessexplicitly stated otherwise, Ω will denote a bounded domain in the planewith C∞ smooth boundary.

3


Theorem 2.1. If u is a function in C1(Ω), then

u(z) =1

2πi

∫

bΩ

u(ζ)

ζ − zdζ +

1

2πi

∫∫

Ω

∂u/∂ζ

ζ − zdζ ∧ dζ

for all z ∈ Ω.

To make this book self-contained, we will spend a moment to de-fine the notation used in this theorem before starting its proof. Thedifferential of a complex valued function u(z), considered as a functionof (x, y) ∈ R2 (via z = x + iy), is a differential one-form given bydu = (∂u/∂x)dx + (∂u/∂y)dy. If we define complex valued one-formsdz = dx + idy and dz = dx − i dy, then, after some linear algebra, wemay express du as F dz+Gdz where F is defined to be ∂u/∂z and G isdefined to be ∂u/∂z. The result of this linear algebra gives

∂u

∂z=

1

2

(∂u

∂x− i

∂u

∂y

)

∂u

∂z=

1

2

(∂u

∂x+ i

∂u

∂y

).

Note that, by the Cauchy-Riemann equations, u is holomorphic if andonly if (∂u/∂z) = 0; this is why the ∂-operator ∂/∂z is so important incomplex analysis. Also note that if u is holomorphic, then the formulain Theorem 2.1 reduces to the classical Cauchy integral formula.

The operators ∂/∂z and ∂/∂z will be used on almost every pageof this book. The following properties of these operators will be usedroutinely. The complex conjugate of ∂u/∂z is ∂u/∂z and the complexconjugate of ∂u/∂z is ∂u/∂z. Suppose that u(z) is defined as a compo-sition of functions via u(z) = g(w) where w = f(z). The complex chainrule can be written

∂u

∂z=∂u

∂w

∂w

∂z+∂u

∂w

∂w

∂z, and

∂u

∂z=∂u

∂w

∂w

∂z+∂u

∂w

∂w

∂z.

These formulas can be verified by simply writing out both sides in termsof real derivatives, or preferably, they can be deduced by using the chainrule for differential forms and the definition of dz and dz given above.

Before we can begin the proof of Theorem 2.1, we need to explain acomplex version of integration by parts that is analogous to the Green’sidentities of classical analysis. Let ζ denote a complex variable. Sinced[u dζ] = du∧dζ, we may use elementary properties of differential formsto see that,

d[u dζ] =∂u

∂ζdζ ∧ dζ + ∂u

∂ζdζ ∧ dζ =

∂u

∂ζdζ ∧ dζ.

The improved Cauchy integral formula 5

We may use this fact in an application of Stokes’ Theorem to deducethat ∫

bΩ

u dζ =

∫∫

Ω

∂u

∂ζ∂ζ ∧ ∂ζ.

Although it may be old fashioned, we like to write∫∫

for the integralover Ω and

∫for the integral over the lower dimensional bΩ. This last

result will be used repeatedly in this book. The complex conjugate ofthe identity is also important:

∫

bΩ

u dζ =

∫∫

Ω

∂u

∂ζ∂ζ ∧ ∂ζ.

We will call these two identities complex Green’s identities. Notice that,because dζ = dx + i dy and dζ = dx − i dy, it follows that ∂ζ ∧ ∂ζ =−2i dx dy, and hence the double integral in the formula represents anintegral with respect to Lebesgue area measure.

Proof of Theorem 2.1. Let Dǫ(z) denote the disc of radius ǫ aboutthe point z. To prove the Cauchy formula, let Ωǫ denote the domainΩ − Dǫ(z) for small ǫ > 0. Apply the complex Green’s identity onthe domain Ωǫ using the function U(ζ) = u(ζ)/(ζ − z). Note that∂U/∂ζ = (∂u/∂ζ)/(ζ − z). Thus, if we parameterize the boundary ofDǫ(z) in the counterclockwise sense, we may write

∫

bΩ

u(ζ)

ζ − zdζ −

∫

bDǫ(z)

u(ζ)

ζ − zdζ =

∫∫

Ωǫ

∂u/∂ζ

ζ − zdζ ∧ dζ.

We now let ǫ → 0. Because u is continuous, the integral over theboundary of Dǫ(z) tends to 2πi u(z). We now claim that the inte-grand in the double integral is in L1. To see this, use polar coordi-nates centered at z in order to write dζ ∧ dζ = 2i dx dy = 2ir dr dθ and|ζ − z|−1dζ ∧ dζ = 2i dr dθ. Finally, because the integrand is in L1(Ω),the Cauchy formula follows.

The Dirichlet problem for the Laplace operator is a very importanttool in the study of harmonic functions. To study holomorphic functions,we will need to understand solutions to the ∂-equation, that is, solutionsto ∂u/∂z = v.

Theorem 2.2. Suppose that v ∈ C∞(Ω). Then the function u definedvia

u(z) =1

2πi

∫∫

Ω

v(ζ)

ζ − zdζ ∧ dζ

satisfies ∂u/∂z = v and u ∈ C∞(Ω).


Since the complex conjugate of ∂u/∂z is equal to ∂u/∂z, Theorem 2.2implies that the equation ∂u/∂z = v can also be solved with u ∈ C∞(Ω)via an integral formula similar to the one in the statement of the theorem.

To prove Theorem 2.2, we will need the following lemma. The lemmais somewhat technical in nature. The reader can be assured that the effortspent on its proof will be fully rewarded later when we will use it to giveshort and easy proofs of some classical theorems in analysis.

Before we can state the lemma, we need to define some terminologyand mention some elementary facts. A function will be said to vanish toorder m on the boundary of Ω if it, together with its partial derivatives(with respect to x and y) up to and including order m, vanish on bΩ.Two functions are said to agree to order m on the boundary of Ω if theirdifference vanishes to orderm on bΩ. Note that to say that two functionsagree to order zero on the boundary simply means that they are equalon the boundary.

Because the boundary of Ω is C∞ smooth, there is a real valuedfunction ρ that is C∞ smooth on a neighborhood of Ω with the propertythat Ω = ρ < 0, bΩ = ρ = 0, and dρ 6= 0 on bΩ. Such a function iscalled a defining function for Ω. We will indicate briefly how to constructsuch a ρ, leaving the details to the reader. The implicit function theoremimplies that, given a sufficiently small arc Γ in the boundary of Ω, it ispossible to find a C∞ diffeomorphism defined on a neighborhood of Γthat maps Γ one-to-one onto a segment in the real axis of the complexplane. A suitable local defining function for Ω near Γ can be obtainedby pulling back the function ±Im z via the diffeomorphism. A globalρ can be constructed from a finite number of local ones by means of apartition of unity.

We will need to know that a function ψ in C∞(Ω) vanishes to ordermon bΩ if and only if ψ = θρm+1 for some function θ ∈ C∞(Ω). This fact iseasy to understand in case Ω is the upper half plane and ρ(z) = −Im z. Inthis case, the fact follows from Taylor’s formula with an explicit integralas the remainder term. To understand the general case, use the localdiffeomorphisms described above to reduce the question to the easy case.We can now state and prove the lemma.

Lemma 2.1. Suppose that v ∈ C∞(Ω). Then, for each positive integerm, there exists a function Φm ∈ C∞(Ω) that vanishes on the boundaryof Ω such that ∂Φm/∂z and v agree to order m on bΩ.

Proof of the Lemma. Let ρ be a defining function for Ω. We will con-struct the functions Φm inductively. A subscript z will denote differen-tiation with respect to z. Let χ be a C∞ function on C that is equal toone on a small neighborhood of bΩ and that vanishes on a neighborhoodof the zero set of ρz (that agrees with the set where dρ = 0). First, we

The improved Cauchy integral formula 7

will construct a function Φ0 that satisfies the conclusion of the lemmafor m = 0 by setting Φ0 = θ0ρ and then choosing θ0 appropriately. Since

(θ0ρ)z = (θ0)zρ+ θ0ρz,

it is clear that it would be a good idea to set θ0 = v/ρz. This wouldyield that (θ0ρ)z = (θ0)zρ + v and hence, that (θ0ρ)z − v vanishes onbΩ. The only problem with doing this is that ρz vanishes at some pointsin Ω. This is where the function χ comes in. By setting θ0 = χv/ρz,we obtain a function Φ0 such that (Φ0)z is equal to v on bΩ. In fact,(Φ0)z − v = Ψ0ρ where

Ψ0 = (χ− 1)vρ−1 +∂

∂z(χv/ρz) ,

and it is clear that Ψ0 is in C∞(Ω).Now suppose that we have constructed functions Φk for k < m that

satisfy the conclusion of the lemma. Then (Φm−1)z − v = Ψm−1ρm for

some Ψm−1 ∈ C∞(Ω). We will complete the induction process by settingΦm = Φm−1 − θmρ

m+1 and by choosing θm appropriately. Now

(Φm)z − v = (Φm−1)z − v − (θmρm+1)z

= Ψm−1ρm − (θm)zρ

m+1 − θm(m+ 1)ρmρz.

The term in this expression containing ρm+1 vanishes to order m onbΩ and can be ignored. We would like to choose θm so that the termscontaining ρm cancel out near bΩ. This can easily be done by takingθm = χΨm−1/((m + 1)ρz). With this choice, we have established that(Φm)z − v = Ψmρ

m+1 for a function Ψm ∈ C∞(Ω). This completes theinduction.

Proof of Theorem 2.2. We will prove the theorem by showing that u isin Cm(Ω) for each positive integer m. Given a positive integer m, let Φdenote the function Φm furnished by the lemma and let Ψ = v−Φz be thecorresponding function that vanishes to orderm on bΩ. By Theorem 2.1,

Φ(z) =1

2πi

∫∫

Ω

∂Φ/∂ζ


Subtracting this equation from the definition of u yields that

u(z)− Φ(z) =1

2πi

∫∫

Ω

Ψ(ζ)


Now, we may consider Ψ to be a Cm function on all of C by extending


it to be equal to zero on C − Ω. We may perform a simple change ofvariables in the last integral to obtain

u(z)− Φ(z) = − 1

2πi

∫∫

C

Ψ(z − ζ)

ζdζ ∧ dζ. (2.1)

Because 1/ζ is locally in L1 and because Ψ has compact support, it ispermissible to differentiate m times under the integral sign to see thatu − Φ is in Cm(Ω); hence, so is u. Since this is true for each positiveinteger m, it follows that u is in C∞(Ω).

Finally, we must show that ∂u/∂z = v. Suppose z0 ∈ Ω and let χ bea function in C∞

0 (Ω) that is one on a neighborhood of z0. Now

u(z) =1

2πi

∫∫

Ω

χv

ζ − zdζ ∧ dζ + 1

2πi

∫∫

Ω

(1− χ)v


Since (1 − χ) vanishes near z0, we may differentiate under the secondintegral when z is close to z0 to see that the second integral is holomor-phic in z near z0. Hence, ∂u/∂z is equal to ∂/∂z of the first integral for znear z0. Since χv has compact support, we may treat the first integral asan integral over C and we may change variables and differentiate underthe integral as we did above to obtain

∂u

∂z(z) = − 1

2πi

∫∫

C

(∂(χv)/∂ζ

)(z − ζ)

ζdζ ∧ dζ

for z near z0. We may further manipulate this integral by reversing thechange of variables and by using Theorem 2.1. We obtain

∂u

∂z=

1

2πi

∫∫

Ω

(∂(χv)/∂ζ

)

ζ − zdζ ∧ dζ = χv

for z near z0. Hence, ∂u/∂z = v near z0 and the proof is complete.

3

The Cauchy transform

If u is a C∞ function defined on the boundary of a bounded domain Ωin the plane with C∞ smooth boundary, then the Cauchy transform ofu is a holomorphic function Cu on Ω given by

(Cu)(z) = 1

2πi

∫

bΩ

u(ζ)

ζ − zdζ.

In the study of harmonic functions, the Poisson integral plays a veryimportant role; the Poisson integral establishes a one-to-one correspon-dence between continuous functions on the boundary and harmonic func-tions on the interior that assume those functions as boundary values. Incomplex analysis, the Cauchy transform plays a similar part; however,the interaction is more subtle because not all functions on the boundarycan be the boundary values of a holomorphic function. In this chapter,we will spell out some basic properties of the Cauchy transform.

Let A∞(Ω) denote the space of holomorphic functions on Ω that arein C∞(Ω).

Theorem 3.1. The Cauchy transform maps C∞(bΩ) into A∞(Ω).

Proof. Let u ∈ C∞(bΩ) and let U be a function in C∞(Ω) that is equalto u on bΩ. Theorem 2.1 allows us to write

U = Cu+ 1

2πi

∫∫

Ω

Uζ(ζ)

ζ − zdζ ∧ dζ,

and it follows from Theorem 2.2 that the function defined by the doubleintegral is in C∞(Ω). The theorem is proved.

Theorem 3.1 allows us to view the Cauchy transform as an operatorthat maps the space C∞(bΩ) into C∞(Ω), or even as an operator fromC∞(bΩ) into itself. Theorem 3.1 also implies the following result.

Theorem 3.2. Suppose that h is a holomorphic function on Ω thatextends to be a continuous function on Ω. If the boundary values of hare in C∞(bΩ), then h ∈ C∞(Ω).

9


The analogue of this theorem for harmonic functions is also true, butit is considerably more difficult to prove. It is remarkable that Theo-rem 3.2 can be proved so easily.

It is quite easy to deduce a local version of Theorem 3.2. Suppose Γ isan open arc in the boundary of Ω, and suppose that h is a holomorphicfunction on Ω that extends continuously to Ω such that the boundaryvalues of h on Γ are C∞ smooth. Let z0 be a point in Γ and let χ be areal valued function in C∞(bΩ) with compact support in Γ that is equalto one on a small neighborhood of z0. Now Ch = C(χh) + C[(1 − χ)h].Since χh is in C∞(bΩ), the term C(χh) is in A∞(Ω). Since (1 − χ)his zero in a neighborhood of z0, a quick glance at the formula for theCauchy transform reveals that, in fact, the term C[(1 − χ)h] extendsholomorphically past the boundary near z0. Thus, we have proved thefollowing theorem.

Theorem 3.3. Suppose that h is a holomorphic function on Ω thatextends to be a continuous function on Ω. If the boundary values of hare C∞ smooth on an open arc Γ in bΩ, then all the derivatives of hextend continuously from Ω to Ω ∪ Γ.

The proofs of Theorems 2.2 and 3.1 contain a constructive methodfor computing the boundary values of the Cauchy transform of a smoothfunction. Since we will continue to use the notation set up in thoseproofs in the remainder of this chapter, it is worth summarizing thekey ingredients in the construction in the form of the next theorem. Letu ∈ C∞(bΩ) and let U be a function in C∞(Ω) that is equal to u onbΩ. For a given positive integer m, let Φ be a function furnished byLemma 2.1 such that Uz −Φz vanishes to order m on bΩ and Φ vanisheson bΩ. We may now apply the Cauchy integral formula of Theorem 2.1to obtain

U − Φ = Cu+ 1

2πi

∫∫

Ω

Ψ(ζ)

ζ − zdζ ∧ dζ

where Ψ = Uz − Φz. Since Ψ can be viewed as a function in Cm(C) viaextension by zero, we may change variables and differentiate under theintegral m times as we did in the proof of Theorem 2.2 to see that Cuand its derivatives up to order m extend continuously to the boundary.Since Φ vanishes on the boundary and since U = u on the boundary, wemay express the boundary values of the Cauchy integral as follows.

Theorem 3.4. Suppose that u ∈ C∞(bΩ). If m is a positive integer,there is a function Ψ ∈ C∞(Ω) that vanishes to order m on the boundarysuch that the boundary values of Cu are expressed via

(Cu)(z) = u(z)− 1

2πi

∫∫

Ω

Ψ(ζ)

ζ − zdζ ∧ dζ, z ∈ bΩ.

The Cauchy transform 11

One of the goals of this book is to define and study the space offunctions on the boundary of Ω that arise as the L2 boundary values ofholomorphic functions on Ω. To do this, we will need to study the L2

adjoint of the Cauchy transform.Let ds denote the differential element of arc length on the boundary

of Ω and let zj(t), j = 1, . . . , n, denote functions that parameterize the nboundary curves of Ω. For u and v in C∞(bΩ), the L2 inner product on bΩof u and v is defined via 〈u, v〉b =

∫bΩu v ds. The space L2(bΩ) is defined

to be the Hilbert space obtained by completing the space C∞(bΩ) withrespect to this inner product. It is not hard to see that L2(bΩ) is equal tothe set of complex valued functions u on bΩ such that u(zj(t)) is a mea-

surable function of t for each j and ‖u‖2 =∑n

j=1

∫ 1

0|u(zj(t))|2|z′j(t)| dt

is finite, and that this definition is independent of the choice of the pa-rameterization of the boundary.

Suppose that u and v are in C∞(bΩ). We know that Cu is also inC∞(bΩ). We wish to construct a function C∗v in C∞(bΩ) that satisfies〈Cu, v〉b = 〈u, C∗v〉b for all u ∈ C∞(bΩ). Although we have used thenotation C∗, and we think of C∗ as being the adjoint of C, we mustemphasize that, for the time being, we must refer to C∗ as the formaladjoint of C; this is because the crucial property 〈Cu, v〉b = 〈u, C∗v〉bwill only be shown to hold for u and v in C∞(bΩ), a space that is not aHilbert space. Later, we will see that C∗ agrees with the L2 adjoint of C.

Given a positive integer m, we may express the boundary values ofCu as described in Theorem 3.4. For z ∈ bΩ, let us write (Cu)(z) =u(z)− I(z) where we define

I(z) = 1

2πi

∫∫

Ω

Ψ(ζ)

ζ − zdζ ∧ dζ. (3.1)

Now, since m ≥ 1, the function Ψ vanishes to at least order one onthe boundary, and hence Ψ = θρ2 for some defining function ρ andθ ∈ C∞(Ω). It follows that the integrand in the double integral I is acontinuous function of (ζ, z) on Ω × bΩ. This will allow us to changethe order of integration (via Fubini’s theorem) when we write out theexpression for 〈Cu, v〉b.

Note that since the unit tangent vector function T is unimodular, itfollows that T = 1/T . Since dz = T ds, it also follows that ds = T dz.Now,

〈Cu, v〉b = 〈u − I, v〉b,


and writing out the term 〈I, v〉b and using Fubini’s theorem, we obtain

∫

z∈bΩ

(1

2πi

∫∫

ζ∈Ω

Ψ(ζ)

ζ − zdζ ∧ dζ

)v(z) ds

=

∫∫

ζ∈Ω

Ψ(ζ)

(1

2πi

∫

z∈bΩ

v(z)

ζ − zds

)dζ ∧ dζ

=

∫∫

Ω

Ψ C(−v T ) dζ ∧ dζ.

The function C(−v T ) is in A∞(Ω). Remember that Ψ was constructedto be equal to (∂/∂ζ)(U −Φ) where Φ = 0 on bΩ. Thus, we may furthermanipulate the last integral by applying the complex Green’s formula toget

∫∫

Ω

∂

∂ζ

[(U − Φ)C(−v T )

](−dζ ∧dζ) =

∫

bΩ

(U −Φ)C(v T ) dζ = 〈u, V 〉b

where V is the complex conjugate of TC(v T ). If we now insert this ex-pression for 〈I, v〉b into 〈Cu, v〉b = 〈u−I, v〉b, we get 〈Cu, v〉b = 〈u, C∗v〉bwhere

C∗v = v − T C(v T ). (3.2)

For now, C∗v is only defined when v ∈ C∞(bΩ). In the next chapterwe will extend the definition to L2(bΩ). Notice that whereas C couldbe viewed either as an operator that maps C∞(bΩ) into C∞(Ω) or intoC∞(bΩ), C∗ can only be viewed as an operator from C∞(bΩ) to itself.

4

The Hardy space, the Szego projection,and the Kerzman-Stein formula

Let A∞(bΩ) denote the set of functions on bΩ that are the boundaryvalues of functions in A∞(Ω). The Hardy space, H2(bΩ), is defined to bethe closure in L2(bΩ) of A∞(bΩ). (This is not the standard definitionof the Hardy space given in, say Stein [St], however, we will explainlater why it is equivalent to any of the more standard definitions.) Tostreamline our notation, we will identify the space A∞(Ω) with A∞(bΩ).

Because H2(bΩ) is a closed subspace of L2(bΩ), we may considerthe orthogonal projection P of L2(bΩ) onto H2(bΩ). This projection iscalled the Szego projection. At the moment, we know that P is a boundedoperator on L2(bΩ) and that the Cauchy transform maps C∞(bΩ) intoitself. We would also like to know that the Cauchy transform satisfies anL2 estimate and that the Szego projection maps C∞(bΩ) into itself. Thekey to deducing what we would like to know from what we now knowis the Kerzman-Stein formula, which relates the Szego projection to theCauchy transform.

The Kerzman-Stein formula is

P (I +A) = C (4.1)

where I denotes the identity operator and A is the Kerzman-Steinoperator defined to be equal to (C − C∗). Since we have only defined theoperator C∗ on C∞(bΩ), we will first establish the truth of this formulawhen it acts on functions in C∞(bΩ). Later, we will see that it is validon L2(bΩ). Because this formula plays a central role in this book, we willprove it in two different ways. The first proof rests on the observationthat functions of the form HT where H ∈ H2(bΩ) are orthogonal toH2(bΩ). To see this, let h and H be functions in H2(bΩ), and let hi andHi be sequences of functions in A

∞(Ω) that converge in L2(bΩ) to h andH , respectively. Now

〈h,HT 〉b = limi→∞

〈hi, HiT 〉b = limi→∞

∫

bΩ

hiHi dz = 0

13


by Cauchy’s theorem. Hence HT ⊥ H2(bΩ). Now if u ∈ C∞(bΩ), then

[I + (C − C∗)]u = u+ Cu− (u− TC(uT )) = Cu+ TC(uT ).

The term Cu is in A∞(Ω) and the term TC(uT ) is of the form THwith H ∈ A∞(Ω), and is therefore orthogonal to H2(bΩ). It follows thatP [I + (C − C∗)]u = Cu, and this is the Kerzman-Stein formula.

The second proof uses the following general fact about projections:If 〈u,H〉b = 〈v,H〉b for all functions H in a dense subspace of H2(bΩ),then Pu = Pv. Let u ∈ C∞(bΩ) and H ∈ A∞(Ω). Now, using the factthat CH = H , it follows that

〈[I + (C − C∗)]u,H〉b = 〈u, [I − (C − C∗)]H〉b = 〈u, C∗H〉b = 〈Cu,H〉b.

Since this identity holds for all H ∈ A∞(Ω) (which is a dense subspaceof H2(bΩ)), we conclude that P [I + (C − C∗)]u = P (Cu) = Cu.

At first glance, it is not apparent why the Kerzman-Stein formulashould be useful. Its utility stems from the fact that the Kerzman-Steinoperator A = C − C∗ is a much better operator than either C or C∗. Wewill prove that A can be written

(Au)(z) =∫

ζ∈bΩ

A(z, ζ)u(ζ) ds

where A(z, ζ) is infinitely differentiable as a function of (z, ζ) ∈ bΩ× bΩ.What this means is that the Cauchy transform is very close to beingself-adjoint. Let us assume this point for the moment and deduce someof its consequences. The most important consequence is that A is anoperator that maps L2(bΩ) into C∞(bΩ) and that satisfies an L2 estimate‖Au‖ ≤ c‖u‖. Hence, because ‖Pu‖ ≤ ‖u‖ for any u ∈ L2(bΩ), theKerzman-Stein formula yields that the Cauchy transform satisfies theL2 estimate ‖Cu‖ ≤ (1 + c)‖u‖ for u ∈ C∞(bΩ). Since C∞(bΩ) is densein L2(bΩ), and since C maps C∞(bΩ) into A∞(Ω) ⊂ H2(bΩ), we haveproved the following theorem.

Theorem 4.1. The Cauchy transform extends to be a bounded operatorfrom L2(bΩ) into H2(bΩ).

We use the same symbol C to denote the extension of C to L2(bΩ)as defined above. In fact, from this point on, unless stated otherwise, wewill consider C to be defined as this operator on L2(bΩ).

It follows from Theorem 4.1 that formula (3.2)

C∗v = v − T C(v T )

expressing C∗ in terms of C extends by the density of C∞(bΩ) in L2(bΩ)

Hardy space, Szego projection, and Kerzman-Stein formula 15

to define a bounded operator (which we also denote by C∗) from L2(bΩ)to itself. Since the identity 〈Cu, v〉b = 〈u, C∗v〉b holds when u and v arein C∞(bΩ), and since C∞(bΩ) is dense in L2(bΩ), this identity holds foru and v in L2(bΩ). This shows that C∗ is the L2 adjoint of C and thatwe are no longer abusing notation.

We proved the Kerzman-Stein formula P [I + (C − C∗)] = C when itoperates on C∞(bΩ). We now see that this same formula is also validon L2(bΩ) when the operators C and C∗ are understood as operators onL2(bΩ).

We are now in a position to take the L2 adjoint of the Kerzman-Steinidentity. Using the facts thatA∗ = (C−C∗)∗ = −A, that (PA)∗ = A∗P ∗,and that P ∗ = P (because P is a projection), we obtain

(I −A)P = C∗. (4.2)

If we subtract this formula from the Kerzman-Stein formula, we get

PA+AP = A.

Thus, PA = A(I − P ). Now, because P = C − PA, we may write

P = C − A(I − P ). (4.3)

Since A maps L2(bΩ) into C∞(bΩ), we see that A(I − P ) also has thisproperty. Because both A(I − P ) and C preserve the space C∞(bΩ) wemay say the same about the Szego projection.

Theorem 4.2. The Szego projection maps C∞(bΩ) into itself.

Let H2(bΩ)⊥ denote the orthogonal complement of the Hardy spacein L2(bΩ). Suppose that v ∈ H2(bΩ)⊥. Then it follows that 0 =〈Cu, v〉b = 〈u, C∗v〉b for all u ∈ L2(bΩ), and therefore, that C∗v = 0.Hence, formula (3.2) for C∗ reveals that v = T H where H = C(Tv) isan element of the Hardy space.

Conversely, we showed in the first proof of the Kerzman-Stein formulathat any function of the form v = T H , whereH ∈ H2(bΩ), is orthogonalto the Hardy space. We have almost proved the following theorem.

Theorem 4.3. A function u ∈ L2(bΩ) has an orthogonal decomposition

u = h+ T H

where h = Pu ∈ H2(bΩ) and T H ∈ H2(bΩ)⊥. Furthermore H =P (uT ). If u is in C∞(bΩ), so are h and H. Functions of the form TGwhere G ∈ A∞(Ω) form a dense subspace of H2(bΩ)⊥.


The only part of Theorem 4.3 that we have not yet proved is thatH =P (uT ). This can be seen by multiplying the orthogonal decompositionu = h + HT by T and then taking the complex conjugate. This givesuT = H + Th, that is an orthogonal decomposition for uT in which Happears as the holomorphic part.

Let P⊥ = I − P denote the orthogonal projection of L2(bΩ) ontoH2(bΩ)⊥. Theorem 4.3 reveals that

P⊥u = T P (uT ). (4.4)

We close this chapter by mentioning that Theorem 4.2 can be lo-calized in the same way that we localized Theorem 3.2 to yield Theo-rem 3.3. Indeed, since P = C − A(I − P ), and since A maps L2(bΩ)into C∞(bΩ), any local regularity property that C has is passed on toP . Hence, when the localization argument used to deduce Theorem 3.3is applied in L2(bΩ) to the Cauchy transform, the following theorem isobtained. We will say that a function v in L2(bΩ) is C∞ smooth on anopen arc Γ ⊂ bΩ if there is a function v on Γ that is C∞ smooth suchthat v is equal almost everywhere on Γ to v.

Theorem 4.4. Suppose that u is a function in L2(bΩ) that, on an openarc Γ in bΩ, is C∞ smooth. Then Cu and Pu are C∞ smooth on Γ.

5

The Kerzman-Stein operator and kernel

The missing ingredient in the proofs of the theorems in Chapter 4 isthe proof that the Kerzman-Stein operator A is represented by a kernelfunction A(z, ζ) that is C∞ as a function of (z, ζ) ∈ bΩ × bΩ. To provethis result, we need formula (3.2) which expresses C∗ in terms of C andwe need the classical Plemelj formula which gives an alternate methodto that of Chapter 3 for expressing the boundary values of a Cauchytransform.

For a point z0 ∈ bΩ and for small ǫ > 0, let γǫ(z0) denote the partof the boundary of Ω that is not contained in the disc of radius ǫ aboutz0, parameterized in the same sense as bΩ. If u ∈ C∞(bΩ), we will letHu denote the function on bΩ given by the following principal valueintegral. (The H stands for “Hilbert.” This operator is closely related tothe classical Hilbert transform.) If z0 ∈ bΩ, then

(Hu)(z0) = P.V.1

2πi

∫

bΩ

u(ζ)

ζ − z0dζ

which is defined via

limǫ→0

1

2πi

∫

γǫ(z0)

u(ζ)

ζ − z0dζ.

We will now prove the Plemelj Theorem, which says that this limit existsand that

(Cu)(z0) =1

2u(z0) + (Hu)(z0). (5.1)

The Plemelj formula is easy to verify if u(z) is a constant function.Indeed, let Cǫ denote the curve that traces out the part of the circle ofradius ǫ about z0 that lies inside Ω so that γǫ(z0) ∪ Cǫ is a curve thatrepresents the boundary of Ω − Dǫ(z0) parameterized in the positivesense. The contour integral of 1/(ζ − z0) about γǫ ∪ Cǫ is zero and it iseasy to check that the part of this integral about Cǫ tends to −πi as ǫtends to zero. Now that the formula is established for constant functions,we may consider the Cauchy transform of u− u(z0). Since this functionvanishes at z0, the integrand in the Cauchy transform is not singular

17


at z0. By letting a point z in Ω approach z0 along the inward pointingnormal to bΩ at z0, we see that [C(u− u(z0))](z) tends to the integral

1

2πi

∫

bΩ

u(ζ)− u(z0)

ζ − z0dζ.

But we also know that [C(u−u(z0))](z) tends to (Cu)(z0)−u(z0). Hence,

(Cu)(z0) = u(z0) + limǫ→0

1

2πi

∫

γǫ(z0)

u(ζ)− u(z0)

ζ − z0dζ

= u(z0) +P.V.1

2πi

∫

bΩ

u

ζ − z0dζ −P.V.

1

2πi

∫

bΩ

u(z0)

ζ − z0dζ

= u(z0) + (Hu)(z0)−1

2u(z0)

and the proof of (5.1) is complete.Let

A(z, ζ) =1

2πi

(T (ζ)

ζ − z− T (z)

ζ − z

)for z, ζ ∈ bΩ.

If we combine the Plemelj formula with formula (3.2) for C∗, we obtain

Au = Cu− C∗u = Hu+ T H(uT )

= P.V.1

2πi

∫

ζ∈bΩ

A(z, ζ)u(ζ) ds.

We will now prove that A(z, ζ) is infinitely differentiable as a function of(z, ζ). It then will follow that the principal value integral for Au above isa standard integral and the proof will be finished. To simplify the com-putations, let us suppose that the boundary curves of Ω have been pa-rameterized with respect to arc length. We wish to see that A(z(t), z(s))is C∞ smooth as a function of (t, s) as t and s range over the various pa-rameter intervals. We should also check that the values of these functionsand their derivatives agree at the endpoints of the parameter intervals.However, since nothing special is really happening at the endpoints ofthe parameter intervals, we may assume that t and s stay in the interiorof the intervals. Since A(z(t), z(s)) is clearly C∞ smooth when t and sare in different parameter intervals, and when t and s belong to the sameinterval and t 6= s, we need only show that A(z(t), z(s)) is C∞ in (t, s)when t and s belong to the interior of the same parameter interval and|t− s| < ǫ for some small ǫ > 0.

The Kerzman-Stein operator and kernel 19

The proof rests on the elementary fact that, if Y (s, t) is C∞ in (s, t)and Y (t, t) = 0, then Y (s, t) = (s − t)W (s, t) where W (s, t) is also C∞

in (s, t). To understand this fact, write

Y (s, t) =

∫ s

t

∂Y

∂x1(x1, t) dx1

and make the change of variables x1 = t+ u(s− t) to get

Y (s, t) =

(∫ 1

0

∂Y

∂x1(t+ u(s− t), t) du

)(s− t).

The function represented by the integral is easily seen to be C∞ smoothin both variables s and t.

Since z(s) − z(t) is a C∞ function of (s, t) that vanishes at (t, t),we may use the fact above to see that the difference quotient Q(s, t) =(z(s)− z(t))/(s− t) is a C∞ function of (s, t). Note that Q(t, t) = z′(t),and therefore, that Q(s, t) is nonvanishing when t and s are close to-gether. Also, since s represents arc length, it follows that T (z(s)) = z′(s).We may now write

2πiA(z(t), z(s)) =1

s− t

(z′(s)

Q(s, t)− z′(t)

Q(s, t)

).

Thus, we see that 2πi(s − t)A(z(t), z(s)) is equal to a C∞ functionR(s, t). Furthermore, R(t, t) = 1−1 = 0; so R(s, t) = (s−t)X(s, t) whereX(s, t) is also C∞ smooth. After dividing out by (s − t), we concludethat 2πiA(z(t), z(s)) = X(s, t) is C∞ in (s, t), and this is precisely whatit means to say that A(z, ζ) is C∞ smooth on bΩ × bΩ. The proof iscomplete.

Notice that A(z, w) = −A(w, z). It follows that A(z, z) is apure imaginary number. In fact, by studying the Taylor expansion ofA(z(t), z(s)) in the t variable at t = s, it is not hard to show thatA(z, z) = 0. Since we will not need this fact, we will not prove it.

It is worth pointing out that in the course of our work above, wehave proved the following classical theorem.

Theorem 5.1. The transform H maps C∞(bΩ) into itself and extendsto be a bounded operator on L2(bΩ).

6

The classical definition of the Hardyspace

We have defined the Hardy space as a subspace of L2(bΩ). We will nowidentify the Hardy space with a space of holomorphic functions on Ω.If u ∈ L2(bΩ), then Cu has been defined to be the limit in L2(bΩ) ofCuj where uj is any sequence of functions in C∞(bΩ) converging to u inL2(bΩ). The functions Cuj are in A∞(Ω) and it is easy to see that theyconverge uniformly on compact subsets of Ω to a holomorphic functionH . Although we have been thinking of the Cauchy transform C as anoperator on L2(bΩ), let us agree to abuse our notation and also use thesymbol C to represent the classical Cauchy integral,

(Cu)(z) = 1

2πi

∫

ζ∈bΩ

u(ζ)

ζ − zdζ

for z ∈ Ω. The holomorphic function H is given by H(z) = (Cu)(z). Thepurpose of this chapter is to show that this dual use of the symbol Cis not an abuse. In fact, we will show that H has L2 boundary valuesgiven by Cu. Furthermore, there is a one-to-one correspondence betweenelements h of H2(bΩ) and holomorphic functions H on Ω arising as theirCauchy integrals. In this chapter, we will use lowercase letters to denotefunctions on the boundary in L2(bΩ) and we will let uppercase lettersdenote holomorphic functions on Ω. In particular, if h and H are used inthe same paragraph, they will be related via the Cauchy integral formulaH(z) = (Ch)(z). When this chapter is finished, we will be justified touse the same symbol for h and H .

For ǫ > 0, let Ωǫ denote the set of points in Ω that are a distance ofmore than ǫ from the boundary of Ω. For small ǫ > 0, Ωǫ is a boundeddomain with C∞ smooth boundary. Let z(t) denote a parameterizationof the boundary of Ω in the standard sense. (We are dropping the sub-script j’s to streamline the notation. It is understood that z(t) representsparameterizations of all the boundary components of Ω.) The functionzǫ(t) = z(t) + iǫT (z(t)) parameterizes the curve obtained by allowing apoint at a distance ǫ along the inward pointing normal to z(t) ∈ bΩ totrace out a curve as z(t) ranges over the boundary. It is a standard fact

21


about curves in the plane that, if ǫ is sufficiently small, zǫ(t) parame-terizes the boundary of Ωǫ in the standard sense. Let δ > 0 be a smallpositive number such that Ωǫ is a C∞ domain parameterized by zǫ(t)when 0 < ǫ < δ.

Classically, the Hardy space was defined to be the space of holomor-phic functions H on Ω such that

sup0<ǫ<δ

(∫|H(zǫ(t))|2|z′ǫ(t)| dt

)1/2

is finite. We will prove that the space of holomorphic functions on Ωsatisfying this classical Hardy condition is equal to the space of functionsgiven by Cauchy integrals of functions in the Hardy space as we havedefined it.

Theorem 6.1. Functions that satisfy the classical Hardy condition areCauchy integrals of functions in H2(bΩ). In particular, bounded holo-morphic functions are Cauchy integrals of functions in H2(bΩ).

Proof. Suppose H is a holomorphic function that satisfies the classi-cal Hardy condition. The family of functions uǫǫ>0 on bΩ defined viauǫ(z(t)) = H(zǫ(t)) is easily seen to be bounded in L2(bΩ) because|z′(t)| < c|z′ǫ(t)| with c independent of ǫ. We may therefore find a se-quence ǫj tending to zero such that uǫj converges weakly to a functionu in L2(bΩ). We now claim that H = Cu. To see this, let Cǫ denote theCauchy transform associated to the domain Ωǫ. If a ∈ Ω and ǫ is small,we may write

(Cu)(a)−H(a) = (Cu− CǫH)(a) = C(u− uǫ)(a) + (Cuǫ − CǫH)(a).

Now, if we let ǫ → 0 through values yielding terms in our weakly con-vergent subsequence, it is clear that C(u − uǫ)(a) tends to zero. Next,we may write out the integral defining (Cuǫ − CǫH)(a) in terms of t toobtain

1

2πi

∫H(zǫ(t))

(z′(t)

z(t)− a− z′ǫ(t)

zǫ(t)− a

)dt.

This quantity tends to zero because H(zǫ(t)) is bounded in L2 (as afunction of t) and the other term in the integral tends uniformly tozero as ǫ tends to zero. Hence H = Cu. We have shown that H is theCauchy integral of a function in L2(bΩ). We will finish the proof byshowing that u ∈ H2(bΩ), i.e., that Cu = u. To see this, we will provethat u is orthogonal to H2(bΩ)⊥. In fact, it will be enough to show that〈u, TG〉bΩ = 0 for all G ∈ A∞(Ω) since, by Theorem 4.3, functions ofthis form are dense in H2(bΩ)⊥. Let Tǫ denote the unit tangent vector

The classical definition of the Hardy space 23

function associated to bΩǫ. By writing out the integrals and using theweak convergence of uǫ to u as we did above, we will see that 〈u, TG〉bΩ−〈H,TǫG〉bΩǫ

tends to zero as ǫ tends to zero through values indexingthe weakly convergent subsequence. But, 〈H,TǫG〉bΩǫ

= 0 by Cauchy’stheorem. Hence, it will follow that u ∈ H2(bΩ), and the proof will befinished. Now,

〈u, TG〉bΩ − 〈H,TǫG〉bΩǫ= 〈u − uǫ, TG〉bΩ + I

where

I =

∫H(zǫ(t)) (z

′(t)G(z(t)) − z′ǫ(t)G(zǫ(t))) dt.

The term 〈u − uǫ, TG〉bΩ tends to zero as ǫ tends to zero because ofthe weak convergence of uǫ to u, and the term I tends to zero becauseH(zǫ(t)) is bounded in L2 (with respect to dt) and the other functionin the integrand tends uniformly to zero as ǫ tends to zero. The proof isfinished.

The function u in the proof above can be thought of as the L2(bΩ)boundary values of H . We will make this association more clear in amoment. First, however, to round out the picture, we must show that afunction that is the Cauchy integral of a function in H2(bΩ) satisfies theclassical Hardy condition.

Theorem 6.2. Cauchy integrals of functions in H2(bΩ) are holomorphicfunctions on Ω that satisfy the classical Hardy condition.

Proof. Suppose that H ∈ A∞(Ω). Define

N(ǫ) =

∫|H(zǫ(t))|2|z′ǫ(t)| dt.

Notice that N(0) = ‖H‖2. We want to show that N(ǫ) satisfies an esti-mate of the form N(ǫ) ≤ CN(0) where C > 0 is a constant that does notdepend on H . Having established this inequality for functions in A∞(Ω),we can approximate a function g ∈ H2(bΩ) by functions Gj ∈ A∞(Ω).Letting G denote the Cauchy integral of g, it then follows from theL2(bΩ) convergence of Gj to g on bΩ that the Gj converge uniformly oncompact subsets of Ω to G. Hence, for a fixed ǫ, the Gj converge to Gin the L2 norm on bΩǫ. Now, by writing the uniform estimate appliedto the Gj and letting j → ∞, we see that the L2(bΩǫ) norm of G isbounded by the same constant C times the H2(bΩ) norm of g.

We now return to proving the estimate N(ǫ) ≤ CN(0) when H ∈A∞(Ω). Note that N(ǫ) is a C∞ function of ǫ on 0 ≤ ǫ < δ and thatN(ǫ) ≥ 0 if ǫ > 0. We now wish to compute the derivative N ′(ǫ). In


order to do this, we will need to know an elementary property of thecurves that bound Ωǫ. It is that, at a point zǫ(t) = z(t) + iǫT (z(t)), theinward pointing unit normal vector to bΩǫ is the same as the inwardpointing unit normal vector to bΩ at z(t). From this, it follows thatTǫ(zǫ(t)) = T (z(t)). Let Dǫ denote the differentiation with respect to ǫoperator. Note that

Dǫ (zǫ(t)) = iT (z(t)) = iTǫ(zǫ(t)). (6.1)

Now, by differentiating under the integral, we obtain

N ′(ǫ) =

∫H ′(zǫ(t))H(zǫ(t))Dǫ (zǫ(t)) |z′ǫ(t)| dt

+

∫H(zǫ(t))H ′(zǫ(t))Dǫ (zǫ(t)) |z′ǫ(t)| dt

+

∫|H(zǫ(t))|2Dǫ (|z′ǫ(t)|) dt.

Using formula (6.1) and the complex Green’s formula, the first of thesethree integrals is seen to be equal to

i

∫

bΩǫ

H ′ H dz = i

∫∫

Ωǫ

H ′H ′ dz ∧ dz,

which is less than or equal to zero because i dz ∧ dz = −2 dx ∧ dy. Thesecond of the three integrals is just the complex conjugate of the first.Since the first integral is a negative real number, it follows that so is thesecond integral. Hence, N ′(ǫ) is seen to be less than or equal to the thirdintegral. Let M denote an upper bound for the quantity

Dǫ|z′ǫ(t)||z′ǫ(t)|

as ǫ ranges over 0 ≤ ǫ < δ and t ranges over its domain. We may nowestimate the third integral to obtain

N ′(ǫ) ≤MN(ǫ) (6.2)

for 0 ≤ ǫ < δ.The next argument should be familiar to anyone who has been in

the same room with an ODE. Multiplying inequality (6.2) by e−Mǫ andsubtracting, we obtain

e−MǫN ′(ǫ)−Me−MǫN(ǫ) ≤ 0.

Integrating this inequality between 0 and ǫ yields

e−MǫN(ǫ)−N(0) ≤ 0,

The classical definition of the Hardy space 25

and the proof of the desired inequality is complete, and hence, so is theproof of the theorem.

Although we will not need to know it, the following fact is interesting.It is not too hard to show that, if z(t) is the parameterization of bΩ withrespect to arc length, then

Dǫ (|z′ǫ(t)|) = iz′′(t)

z′(t).

(Some of the tricks used in this computation include

|z′ǫ(t)| =|z′(t)|z′(t)

z′ǫ(t)

and |z′(t)| ≡ 1.) Incorporating this identity in the expression above forN ′ yields that

N ′(ǫ) = −4

∫∫

Ωǫ

|H ′|2 dx ∧ dy + i

∫|H(zǫ(t))|2

z′′(t)

z′(t)dt.

It is interesting that, if Ω is the unit disc, then z′′(t)/z′(t) = i, andtherefore, N is a strictly decreasing function of ǫ.

To really understand the Hardy space, we need to prove one moretheorem. This theorem will allow us to say, in a strong sense, thatH2(bΩ)is equal to the space of holomorphic functions on Ω with L2 boundaryvalues.

Given h ∈ H2(bΩ), let H(z) = (Ch)(z) be the holomorphic functionon Ω given by the Cauchy integral of h and let uǫ denote the familyof functions on bΩ defined via uǫ(z(t)) = H(zǫ(t)) as above.

Theorem 6.3. If h ∈ H2(bΩ), then uǫ → h in L2(bΩ) as ǫ→ 0.

Proof. We have shown that there is a constant C (which is independentof h and ǫ) such that ‖uǫ‖ ≤ C‖h‖. Let λ > 0 and let G ∈ A∞(Ω) besuch that ‖h−G‖ < λ. Let Uǫ denote the family of functions defined onbΩ corresponding to G, i.e., Uǫ(z(t)) = G(zǫ(t)). Now,

‖uǫ − h‖ ≤ ‖uǫ − Uǫ‖+ ‖Uǫ −G‖+ ‖G− h‖.

The first term is less than or equal to Cλ, the second term tends to zeroas ǫ → 0 because G is continuous on Ω, and the third term is less thanλ. The proof is complete.

The association of a classical Hardy function H to its L2 boundaryvalues sets up a one-to-one correspondence between the space of classical


Hardy functions and H2(bΩ). Hence, we are justified in using the samesymbol h to denote a function in H2(bΩ) and the holomorphic functionon Ω that is its Cauchy integral.

We will need the following fact later, which is a direct consequenceof the work above. (Recall that if a sequence converges in L2, then thereis a subsequence converging pointwise almost everywhere.)

Theorem 6.4. If h is a holomorphic function on Ω such that |h| < 1on Ω, then |h| ≤ 1 on bΩ as a function in H2(bΩ).

This is also an opportune moment to state another theorem that wewill need later when we study the Bergman space in Chapter 15. LetAδ denote the annular region consisting of points in Ω that are withina distance of δ to the boundary of Ω. Using the notation set up in theproof of Theorem 6.2, notice that

∫ δ

0

N(ǫ) dǫ =

∫ δ

0

(∫|H(zǫ(t))|2|z′ǫ(t)| dt

)dǫ =

∫∫

Aδ

|H |2 dx dy

if H ∈ A∞(Ω). Hence, the estimate N(ǫ) ≤ C N(0) derived in the courseof the proof of Theorem 6.2 reveals that there is a constant C suchthat the L2 norm of H with respect to Lebesgue area measure on Aδ

is bounded by a constant times the H2(bΩ) norm of H . Since the valueof H(z) at a point z in Ω that is at a distance greater than δ fromthe boundary is easily bounded by a uniform constant times the H2(bΩ)norm ofH via the Cauchy integral formula, we have proved the followingtheorem.

Theorem 6.5. If h is in H2(bΩ), then |h|2 is integrable over Ω withrespect to Lebesgue area measure, and there is a constant C which doesnot depend on h such that

∫∫

Ω

|h|2 dx dy ≤ C‖h‖2.

7

The Szego kernel function

For a ∈ Ω and z ∈ bΩ, let Ca(z) denote the complex conjugate of

1

2πi

T (z)

z − a.

With this notation, Ca is the kernel that defines the Cauchy integral inthe sense that (Cu)(a) = 〈u,Ca〉b. If h is a function in the Hardy space,we have identified h with the holomorphic function on Ω given by Ch.The value of h at a point a ∈ Ω can be computed via

h(a) = (Ch)(a) = 〈h,Ca 〉b = 〈h, P (Ca)〉b = 〈h, Sa〉b

where Sa = PCa. The function S(z, a) defined to be S(z, a) = Sa(z) iscalled the Szego kernel function.

Another way to think of the Szego kernel is via the Riesz representa-tion theorem. The formula h(a) = 〈h,Ca〉b shows that |h(a)| ≤ ‖h‖‖Ca‖,and therefore, that evaluation at a ∈ Ω is a continuous linear func-tional on H2(bΩ). Hence, there is a unique function Ea ∈ H2(bΩ)that represents this functional in the sense that h(a) = 〈h,Ea〉b for allh ∈ H2(bΩ). We showed above that PCa also has this property. BecauseEa is uniquely determined, it follows that the function Ea defined by therepresenting property is the same as the function Sa we defined aboveas PCa.

Because P maps C∞(bΩ) into itself, it follows that Sa(z) is in A∞(Ω)

as a function of z when a ∈ Ω is fixed. We will now show that S(z, a) iscontinuous as a function of both variables (z, a) on Ω×Ω and that S(z, a)is antiholomorphic in a. Notice that if a, z ∈ Ω, then Sa(z) = 〈PCa, Cz〉b,and, using the identity

Sa1(z)− Sa2

(z) = 〈P (Ca1− Ca2

), Cz〉b

and the estimate ‖Pu‖ ≤ ‖u‖, we may therefore estimate

|Sa1(z)− Sa2

(z)| ≤ ‖Ca1− Ca2

‖‖Cz‖.

If a1 and a2 are restricted to be in a compact subset of Ω, then ‖Ca1−

27


Ca2‖ can be made uniformly small by demanding that |a1−a2| be small.

Furthermore, ‖Cz‖ is uniformly bounded when z is restricted to be in acompact set. Using these facts and the basic estimates, it is easy to showthat |S(z1, a1)− S(z2, a2)| is small when (z1, a1) is close to (z2, a2) in acompact subset of Ω× Ω. Hence S(z, a) is continuous on Ω× Ω. To seethat S(z, a) is antiholomorphic in a, we compute a difference quotientusing the ideas above to obtain

Sa(z)− Sa0(z)

a− a0= 〈P

(Ca − Ca0

a− a0

), Cz〉b.

Since (a − a0)−1(Ca − Ca0

) tends in L2(bΩ) to C′(a0) where [C′(a)](ζ)is defined to be equal to

∂

∂aCa(ζ) = − 1

2πi

T (ζ)

(ζ − a)2,

it follows that the difference quotient tends to 〈P (C′(a0)), Cz〉b, and thisshows that S(z, a) is antiholomorphic in a. By repeating this argument,it can be shown that S(z, a) is in fact C∞ smooth as a function of bothvariables (z, a) on Ω × Ω. The proof boils down to showing that it ispermissible to differentiate under the operators and inner product in theexpression Sa(z) = 〈PCa, Cz〉b, and this follows from the L2 estimatefor P and the uniform differentiability of the Cauchy kernel. Later in thebook (Chapter 26), we will prove a much stronger result. We will provethat S(z, a) is in C∞((Ω × Ω) − D) where D = (z, z) : z ∈ bΩ is theboundary diagonal.

It is not hard to see that the Szego kernel is hermitian symmetric,i.e., that S(a, b) = S(b, a) for all a, b ∈ Ω. Indeed, using the representingproperty of Sb, we obtain

Sa(b) = 〈Sa, Sb〉b,

and this is clearly the complex conjugate of

Sb(a) = 〈Sb, Sa〉b,

which follows from the representing property of Sa. Note that, if we seta = b in the work above, we obtain

S(a, a) = Sa(a) =

∫

bΩ

|Sa|2 ds.

This identity shows that S(a, a) > 0. Indeed, since the L2 pairing ofSa with the function that is identically one gives the value one, it fol-lows that Sa cannot be identically zero as a function in L2(bΩ). Hence,

The Szego kernel function 29

S(a, a) > 0 for each a ∈ Ω. We will need this fact later when we studythe relationship between the Szego kernel and the Riemann mappingfunction.

For h ∈ H2(bΩ), using the hermitian symmetry of S(z, a), we maywrite

h(a) =

∫

z∈bΩ

S(a, z)h(z) ds.

The Szego kernel is the kernel for the Szego projection in the classicalsense of integral operators because

(Pu)(a) = 〈Pu, Sa〉b = 〈u, Sa〉b =∫

z∈bΩ

S(a, z)u(z) ds.

Theorem 4.3 allows us to define another important kernel function.Consider the orthogonal decomposition of the Cauchy kernel Ca(ζ). Weknow that (PCa)(ζ) = S(ζ, a). Therefore, the orthogonal decompositionfor Ca given by Theorem 4.3 is Ca = Sa +HaT , where Ha is in A∞(Ω).Conjugating the orthogonal decomposition of Ca and using the hermitiansymmetry of S(z, a) yields

1

2πi

T (ζ)

ζ − a= S(a, ζ) +Ha(ζ)T (ζ).

The function Ha(ζ) is in A∞(Ω) as a function of ζ for fixed a ∈ Ω. Also,

it can be seen from the decomposition formula that Ha is holomorphicin a ∈ Ω for fixed ζ ∈ bΩ because the same is true of the other terms inthe formula. Garabedian’s L kernel is defined via

1

iL(ζ, a) =

1

2πi

1

ζ − a−Ha(ζ).

When the function Ha is expressed in terms of the Garabedian kerneland when this formula is plugged back into the orthogonal decompositionof the Cauchy kernel, the following identity is obtained.

The Szego and Garabedian kernels are related via

S(a, ζ) =1

iL(ζ, a)T (ζ) for a ∈ Ω, ζ ∈ bΩ. (7.1)

This identity is very important and is at the heart of many of the ap-plications of the Szego kernel to problems in conformal mapping. Wewill use (7.1) in various forms many times in later chapters and so wetake this opportunity to list several different ways to write this identity.Since |T (ζ)| = 1, it follows that 1/T (ζ) = T (ζ). Using this fact and theshorthand notation Sa(ζ) = S(ζ, a) and La(ζ) = L(ζ, a), (7.1) can be


written in any of the following equivalent ways. Thinking of a as beingfixed in Ω and Sa and La as functions of ζ ∈ bΩ, we have

Sa = −iLaT,

Sa = iLaT ,

La = iSaT ,

La = −iSaT.

Notice that when we write identity (7.1) in the form L(ζ, a) =i Sa(ζ)T (ζ), Theorem 4.3 shows that L(ζ, a) is orthogonal to H2(bΩ)as a function of ζ. Thus,

L(ζ, a) = P⊥(L(·, a))(ζ).

But L(ζ, a) = 12π (ζ − a)−1 − Ha(ζ) where Ha ∈ A∞(Ω), and since

P⊥Ha = 0, it follows that

L(ζ, a) = (P⊥Ga)(ζ)

where Ga is defined to be

Ga(ζ) =1

2π(ζ − a).

It can be read off from the formula defining the Garabedian kernelthat, for fixed a ∈ Ω, L(ζ, a) is a meromorphic function of ζ on Ω with asingle simple pole at ζ = a having residue 1/(2π). Also, L(ζ, a) extendsC∞ smoothly up to the boundary as a function of ζ. Because Ha(ζ) isholomorphic in a, it follows that, for fixed ζ, L(ζ, a) is holomorphic in aon Ω− ζ.

Alternatively, L(ζ, a) can be seen to be holomorphic in a by differen-tiating under the operator in the identity L(ζ, a) = (P⊥Ga)(ζ) and byusing reasoning analogous to that which we used to study the smooth-ness properties of the Szego kernel above.

We now show that identity (7.1) characterizes the Szego and Garabe-dian kernels.

Theorem 7.1. Suppose that σ(z) is a holomorphic function on Ω thatextends continuously to Ω and suppose λ(z) is holomorphic on Ω− a,extends continuously to bΩ, and has a simple pole with residue 1/(2π)at z = a. If

σ(z) =1

iλ(z)T (z) for z ∈ bΩ, (7.2)

then σ(z) = S(z, a) and λ(z) = L(z, a).

The Szego kernel function 31

Proof. We may write λ(z) = 12π (z − a)−1 + h(z) where h is a holo-

morphic function on Ω that extends continuously to Ω. But λ = iσT ,and therefore λ is orthogonal to holomorphic functions. Thus λ(z) =(P⊥λ)(z) = (P⊥(Ga + h))(z) = (P⊥Ga)(z) = L(z, a), and (7.1) yieldsthat σ(z) = S(z, a). The proof is finished.

Another way to think of Garabedian’s kernel is as the kernel for theorthogonal projection P⊥. Indeed, P⊥u = HT where H = P (T u). Wemay evaluate H at a point a ∈ Ω by using identity (7.1) in the followingcomputation:

H(a) = P (T u)(a) =

∫

ζ∈bΩ

S(a, ζ)T (ζ)u(ζ) ds

=1

i

∫

ζ∈bΩ

L(ζ, a)u(ζ) ds.

It is possible to manipulate the identities of this chapter to deducethat L(a, b) = −L(b, a) for a 6= b in Ω. Indeed, L(a, b) is equal to theresidue of 2πL(z, a)L(z, b) at the point a because L(z, a) has a simplepole at a with residue (2π)−1. The same reasoning shows that L(b, a) isequal to the residue of 2πL(z, a)L(z, b) at the point b. Hence, we mayuse the residue theorem to compute,

L(a, b) + L(b, a) = 2π∑

Res L(z, a)L(z, b) =1

i

∫

bΩ

L(z, a)L(z, b) dz

=

∫

bΩ

L(z, a)1

iL(z, b)T (z) ds =

∫

bΩ

L(z, a)S(b, z) ds = 0

because, as mentioned above, L(z, a) is orthogonal to H2(bΩ).All the facts and formulas we have derived for a general domain

become particularly simple when the domain is the unit disc.

Theorem 7.2. The Szego kernel of the unit disc U is given by

S(z, a) =1

2π(1− az)

and the Garabedian kernel is given by

L(z, a) =1

2π(z − a).

A function u in L2(bU) has an orthogonal decomposition

u = h+H

where h = Pu and H = z P (zu). Notice that H is holomorphic on Uand vanishes at the origin.


Proof. On the unit disc, the complex unit tangent at a point z in theboundary is given by T (z) = iz. Hence the orthogonal decompositionfollows immediately from the formulas in Theorem 4.3.

The Szego kernel is equal to the Szego projection of the Cauchykernel. On the unit disc, the Cauchy kernel turns out to be a holomorphicfunction. Indeed, we may write

Ca(z) = − 1

2πi

T (z)

z − a=

1

2π

z

z − a.

Now, using the fact that z = 1/z when |z| = 1, we obtain

Ca(z) =1

2π

1

1− az

which is a holomorphic function in A∞(U). Hence, it follows that Sa =PCa = Ca and the formula for the Szego kernel is proved.

Since Sa = Ca, the function Ha in the orthogonal expansion of Ca

must be zero. Since

L(z, a) =1

2π

1

z − a− iHa(z),

the formula for the Garabedian kernel is proved.

The equality Sa = Ca that we just saw is valid on the unit disc isvery special. In fact, Kerzman and Stein [K-S] proved that this identityholds for the Szego kernel of a domain Ω if and only if Ω is equal to adisc.

There are some interesting theorems lurking in the background ofTheorem 7.2. For example, if u ∈ C∞(bU), the orthogonal decompo-sition u = h + H gives a harmonic extension of u to the closed discand this extension must agree with the one given by the classical Pois-son integral formula (see Ahlfors [Ah, p. 166]). Since P maps C∞(bU)into A∞(U), Theorem 7.2 reveals that the harmonic extension of u is inC∞(U) whenever u ∈ C∞(bU) and it is given by Pu+ z P (zu). We willreturn to the problem of finding the harmonic extension of a functiondefined on the boundary of a more general domain in Chapter 10 wherewe will relate the solution of this problem to the Szego projections ofsimple functions.

8

The Riemann mapping function

In this chapter, we will assume that Ω is a bounded simply connecteddomain in C with C∞ smooth boundary. We wish to illustrate as quicklyand easily as possible that the Szego and Garabedian kernels are inti-mately tied to questions in conformal mapping.

The Riemann mapping theorem asserts that, for a point a ∈ Ω,there is a one-to-one holomorphic mapping f of Ω onto D1(0) such thatf(a) = 0. If we require that f ′(a) be real and positive, then f is uniquelydetermined. We will call this mapping the Riemann mapping functionassociated to a.

In order to make the proof of the next theorem as short and easyas possible, we will assume Caratheodory’s theorem, which states thatRiemann maps of the types of domains we are studying extend con-tinuously to the boundary. Later we will prove the Riemann mappingtheorem and the theorem below from first principles without relying onCaratheodory’s result.

Theorem 8.1. The Riemann mapping function f associated to a pointa in a bounded simply connected domain Ω with C∞ smooth boundary isgiven by

f(z) =S(z, a)

L(z, a)

where S(z, a) is the Szego kernel and L(z, a) is the Garabedian kernelassociated to Ω.

Proof. Since S(a, a) is a positive real number, the function

λ(z) = S(z, a)/f(z)

is a meromorphic function on Ω which extends continuously to theboundary and which has a single simple pole at the point a. Further-more, the function σ(z) = f(z)L(z, a) is holomorphic on Ω and extendscontinuously to the boundary. Since f maps the boundary of Ω into theunit circle, it follows that 1/f = f on the boundary of Ω, and sinceS(z, a) and L(z, a) satisfy formula (7.1), we see that σ and λ satisfyidentity (7.2). Thus Theorem 7.1 implies that λ(z) = cL(z, a) for some

33


constant c, i.e., that cf(z) = S(z, a)/L(z, a). Now, because formula (7.1)implies that |S(z, a)| = |L(z, a)| for z ∈ bΩ, we conclude that |c| = 1.Finally, because f ′(a) and S(a, a) are real and positive, and because theresidue of L(z, a) at a is 1/(2π), it follows that c is real and positive.Thus, c = 1, and the proof is complete.

We have assumed Caratheodory’s theorem about continuous exten-sion of the Riemann map in order to prove Theorem 8.1. Soon, we willprove from first principles a much stronger result than Caratheodory’stheorem about the boundary behavior of the Riemann map. We willprove the following theorem of Painleve. (It is interesting to note thatPainleve proved his theorem before Caratheodory proved his. See [Be-K]for the history of this question.)

Theorem 8.2. The Riemann mapping function f , mapping a boundedsimply connected domain Ω with C∞ smooth boundary onto the unit disc,is C∞ smooth up to the boundary. Furthermore, f ′ is nonvanishing onΩ. Hence, f−1 is C∞ smooth up to the boundary of the unit disc.

9

A density lemma and consequences

We now go back to studying the Szego kernel function on a generalbounded domain Ω with C∞ smooth boundary. In particular, Ω is againallowed to be multiply connected.

For fixed a ∈ Ω, we will let Sa(z) denote the function of z given bySa(z) = S(z, a). Let Σ denote the (complex) linear span of the set offunctions Sa(z) : a ∈ Ω. It is easy to see that Σ is a dense subspace ofH2(bΩ). Indeed, if h ∈ H2(bΩ) is orthogonal to Σ, then h(a) = 〈h, Sa〉b =0 for each a ∈ Ω; thus h ≡ 0. In this chapter, we will prove that

Theorem 9.1. Σ is dense in A∞(Ω).

To say that Σ is dense in A∞(Ω) means that, given a function h ∈A∞(Ω), there is a sequence Hj ∈ Σ such that Hj(z) tends uniformlyon Ω to h(z), and each derivative of Hj(z) tends uniformly on Ω to thecorresponding derivative of h(z). To prove that Σ is dense in A∞(Ω), wemust show that, given h ∈ A∞(Ω), ǫ > 0, and a positive integer s, thereis a function H ∈ Σ such that |h−H | < ǫ on Ω, and |h(m) −H(m)| < ǫon Ω for each derivative of order m ≤ s.

If s is a positive integer, let us define the Cs(bΩ) norm ‖u‖s of afunction u defined on the boundary of Ω to be equal to the supremumof the derivatives |(dm/dtm)u(z(t))| as the parameter t ranges over itsdomain and m ranges from zero to s. Although this norm depends onthe parameterization z(t) of the boundary, the corresponding space offunctions with finite Cs(bΩ) norm does not.

Although the proof is somewhat technical, the idea is quite simple. InTheorem 3.4, we expressed the boundary values of the Cauchy transformof a smooth function u as an integral,

Cu = u− 1

2πi

∫∫

ζ∈Ω

Ψ(ζ)

ζ − zdζ ∧ dζ

where Ψ is a function in C∞(Ω) that vanishes to high order on bΩ. Theidea of the proof is to approximate the integral in this expression bya Riemann sum

∑Ni=1 ci

1ai−z . We will see that this can be done in a

uniform way as z ranges over the boundary of Ω. Having done this, we

35


take the complex conjugate of the equation and multiply through byT (z) to obtain

(Cu)T ≈ uT −∑

ciCai

where Ca denotes the Cauchy kernel. Now, since (Cu)T is orthogonal tothe Hardy space, it follows that P ((Cu)T ) = 0, and therefore, we maytake the Szego projection of this last formula to obtain

P (uT ) ≈∑

ciP (Cai).

Now, since P (Ca) = Sa, we have P (uT ) ≈ ∑ciSai

. If we wanted toapproximate a function H in A∞(Ω) by a function in Σ, we would simplychoose u so that H = uT , in which case, the reasoning above wouldproduce an element in Σ that is close to H = PH = P (uT ). That is theidea. Now, here are the details.

If you go back to the proofs of Theorems 3.1 and 4.2 and countderivatives, you will see that we also proved the following result.

Theorem 9.2. Given a positive integer s, there is a positive integern = n(s) and a constant K = K(s) such that

‖Pu‖s ≤ K‖u‖n and ‖Cu‖s ≤ K‖u‖n

for all u ∈ C∞(bΩ). Consequently, since C∞(bΩ) is dense in Cn(bΩ),the same inequalities hold for all u ∈ Cn(bΩ). In particular, it followsthat Pu and Cu are in Cs(bΩ) whenever u ∈ Cn(bΩ).

With some additional effort, the estimates in Theorem 9.2 can besharpened to allow n = s + 1. We will not prove this fact because, forour purposes, it will suffice to know that the estimate holds for somen. With no additional effort, Theorem 9.2 can be sharpened to read asfollows.

Given a positive integer s, there is a positive integer n = n(s) and aconstant K = K(s) such that

sup∣∣∣∣dm

dmz(Cu)(z)

∣∣∣∣ : z ∈ Ω, 0 ≤ m ≤ s ≤ K‖u‖n,

for all u ∈ Cn(bΩ). The analogous estimate also holds for the Szegoprojection.

To see that this apparently stronger statement is a direct consequenceof Theorem 9.2, take a sequence of functions in C∞(bΩ) tending to u inthe Cn(bΩ) norm. Use the maximum principle to see that the boundaryestimate in Theorem 9.2 implies the uniform estimate for derivative on

A density lemma and consequences 37

Ω when applied to elements in the sequence. Finally, a simple limitingargument implies the estimate for u.

Theorem 9.2 is the main ingredient in the proof of Theorem 9.1, towhich we now return. Let n and ǫ > 0 be given. By Theorem 3.4, we maywrite the Cauchy transform of a function u ∈ C∞(bΩ) as Cu = u − Iwhere, for z ∈ bΩ,

(I)(z) = 1

2πi

∫∫

ζ∈Ω

Ψ(ζ)

ζ − zdζ ∧ dζ

and Ψ is a function in C∞(Ω) that vanishes to order n on the boundaryof Ω. Since Ψ vanishes to order n on the boundary, we may think of Ψ asa function in Cn(C) by extending it to be zero outside Ω. We now claimthat, given δ > 0, we can find a function Ψδ in Cn(C) that has compactsupport in Ω such that for any derivative Dα of order n or less, we have|Dα(Ψδ − Ψ)| < δ on Ω. To construct such a Ψδ, we take a partition ofunity subordinate to a covering of Ω by small discs, thereby enabling usto assume that Ψ is a function supported in a small disc Dr(z0) wherez0 ∈ bΩ. The complex number w0 = −iT (z0) represents the outwardpointing normal vector to bΩ at z0. Now, if the disc is small enough, thetranslated function Ψδ(z) = Ψ(z + λw0) will be well defined on Ω andwill have the properties we seek provided λ is chosen small enough. Theclaim is proved.

Define

(Iδ)(z) =1

2πi

∫∫

ζ∈Ω

Ψδ(ζ)


We next claim that it is possible to choose δ so small that the modulusof I−Iδ and the moduli of the derivatives of I−Iδ of order n or less areless than ǫ/2 on Ω. In fact, to see this, we make the change of variablesin the integral as we did in the proof of Theorem 2.2. This allows us towrite

(I)(z)− (Iδ)(z) = − 1

2πi

∫∫

ζ∈C

Ψ(z − ζ)− Ψδ(z − ζ)

ζdζ ∧ dζ,

and by differentiating under the integral, and using the fact that thekernel 1/ζ is locally integrable, our claim follows.

Now, because Ψδ has compact support, for z ∈ bΩ, we may approxi-mate the integral defining Iδ by a (finite) Riemann sum

S(z) = 1

2πi

∑ci

1

ai − z

in such a way that ‖S − Iδ‖n < ǫ/2.


We have now shown that ‖u−Cu−S‖n < ǫ where ǫ can be taken tobe arbitrarily small. If we now multiply u − Cu − S by T and take thecomplex conjugate, we see that the Cn(bΩ) norm of

Tu− TCu−∑

ciCai

can be made arbitrarily small. Next, we take the Szego projection ofthis function. Using the facts that P (TCu ) = 0 and S(z, a) = (PCa)(z),we see that by choosing n sufficiently large and ǫ sufficiently small, theCs(bΩ) norm of

P (Tu)−∑

ciS(·, ai)can be made arbitrarily small by virtue of Theorem 9.2. To finish theproof of Theorem 9.1, we need only note that a function h in A∞(Ω)can be written as Tu where u = Th. Hence h = Ph = P (Tu) can beapproximated in Cs(bΩ) norm by functions in Σ.

The following corollaries will be seen to be simple consequences ofTheorem 9.1.

Corollary 9.1. Given a point w0 in the boundary of Ω, the functionh(z) = S(z, w0) cannot be identically zero as a function of z on Ω. Also,the function H(z) = L(z, w0) cannot be identically zero as a function ofz on Ω.

For a ∈ Ω, let La(z) = L(z, a), and let Λ denote the linear span ofLa(z) : a ∈ Ω.

Corollary 9.2. Λ is dense in C∞(bΩ) ∩ H2(bΩ)⊥ in the sense of ap-proximation in Cs(bΩ) norms for arbitrarily high s.

To prove Corollary 9.1, note that, according to Theorem 9.1, thefunction h(z) ≡ 1 can be approximated uniformly on bΩ by functionsin Σ. If S(z, w0) were identically zero in z, then every function in Σwould vanish at w0. Thus, it would be impossible to approximate h ≡ 1uniformly near w0, contradicting Theorem 9.1. The statement aboutthe nonvanishing of L(z, w0) now follows from that of S(z, w0) and theidentity iS(z, w0) = L(w0, z)T (w0), which holds for all z ∈ Ω.

Corollary 9.2 follows from the density of Σ, identity (7.1), and theorthogonal decomposition of L2(bΩ). Indeed, if HT is in C∞(bΩ) ∩H2(bΩ)⊥, we may approximate H by elements sj of Σ. Now HT isapproximated by the sequence of functions sjT which belong to Λ by

virtue of identity (7.1) rewritten in the form S(z, a)T (z) = −iL(z, a).Recall that L(z, a) = −L(a, z) for a and z in Ω. Hence, if L(b, a) were

to vanish for some values of a and b in Ω, a 6= b, it would follow thatL(a, b) = 0 too. Consequently, L(z, a)L(z, b) would be a holomorphic


function of z on all of Ω that is in C∞(Ω). To be precise, the zero ofL(z, a) at z = b would cancel out the pole of L(z, b) at z = b and thezero of L(z, b) at z = a would cancel out the pole of L(z, a) at z = a.This observation allows us to give a simple proof of the nonvanishing ofthe Garabedian kernel in simply connected domains.

Theorem 9.3. If Ω is a bounded simply connected domain with C∞

smooth boundary, then L(z, a) 6= 0 for all z ∈ Ω− a.

It is a general fact that L(a, b) 6= 0 for all a, b ∈ Ω, a 6= b, even if Ωis a smooth multiply connected domain. We will prove this more generalresult in Chapter 13.

Proof. Suppose L(a, b) = 0 for some a and b in Ω, a 6= b. Let L(z) =L(z, a)L(z, b) and let S(z) = S(z, a)S(z, b). Identity (7.1) yields that,

−L(z)T (z) = S(z)T (z), z ∈ bΩ. (9.1)

As remarked above, L is in A∞(Ω). So is S. Hence, by Theorem 4.3, for-mula (9.1) implies that LT is orthogonal to H2(bΩ) and also orthogonalto the space of conjugates of functions in H2(bΩ). In a simply connecteddomain, this forces us to conclude that L ≡ 0. Indeed, it follows fromthis orthogonality that, for any h ∈ A∞(Ω), we have

0 =

∫

z∈bΩ

LT h ds =

∫

z∈bΩ

L h dz.

Now, since dz ∧ dz = 2i dx ∧ dy, it follows from the complex Green’sformula that ∫∫

Ω

L h′ dx ∧ dy = 0

for all h ∈ A∞(Ω). Since L is in A∞(Ω), and since Ω is simply connected,there is a function h(z) in A∞(Ω) such that h′(z) = L(z). Hence, itfollows that

∫∫Ω|L|2 dx ∧ dy = 0 and therefore, that L ≡ 0, i.e., that

L(z, a)L(z, b) is identically zero. This implies that, either L(z, a) ≡ 0, orL(z, b) ≡ 0, and it is impossible for either of these functions to vanishidentically in z ∈ bΩ because they have poles at a and b, respectively.

We have shown that L(z, a) is nonvanishing for all z ∈ Ω−a. To seethat L(z0, a) 6= 0 for z0 ∈ bΩ, let zj be a sequence of points in Ω that tendto z0. Let ǫ be small enough so that the disc D2ǫ(a) about a is compactlycontained in Ω, and let Ωǫ = Ω−D2ǫ(a). The functions Hj(w) = L(zj, w)are holomorphic and nonvanishing on Dǫ(a) as functions of w if j islarge. We claim that these functions converge uniformly on Dǫ(a) toH0(w) = L(z0, w) as j → ∞. Assuming this claim, it then follows byHurwitz’s Theorem, that either the limit function is identically zero, or


never zero. By Corollary 9.1, L(z0, w) cannot be identically zero. Thus,L(z0, w) is never zero, and we conclude that L(z0, a) 6= 0. Therefore,to finish the proof, it will suffice to verify the claim by showing thatL(z, w) is continuous for (z, w) in Ωǫ ×Dǫ(a). Recall that Lw = P⊥Gw

where Gw(z) = [2π(z − w)]−1. Hence Lw = Gw − P (Gw). Theorem 9.2shows that P (Gw)(z) is continuous on (z, w) ∈ (Ω ×Dǫ(a)). It is clearthat Gw(z) is continuous on (z, w) ∈ (Ωǫ × Dǫ(a)). Hence, L(z, w) iscontinuous on (z, w) ∈ (Ωǫ × Dǫ(a)), and the proof of Theorem 9.3 iscomplete.

We are now in a position to prove Theorem 8.2. In fact, the proof weare about to give includes an existence proof for the Riemann mappingfunction associated to a point a in a bounded simply connected domainΩ with C∞ smooth boundary.

Proof of Theorem 8.2. Theorem 9.3 says that La is nonvanishing on bΩ.It follows from (7.1) that Sa is nonvanishing on bΩ too. Hence, we arejustified in using the argument principle to count the zeroes of Sa in Ω,and to count the pole of La in Ω. The main tool we will use to relatethese numbers is identity (7.1). Let ∆arg h denote the net increase in theargument of a nonvanishing function h(z) defined on bΩ as z traces outthe boundary of Ω in the standard sense. Note that ∆arg h = −∆arg hand that ∆arg T = 2π.

It follows from (7.1) that −∆arg Sa = (∆arg La) + 2π. The increaseof arg La around bΩ is −2π because La has a single simple pole at a,and no zeroes in Ω. Hence, ∆arg Sa = 0 and we conclude that Sa hasno zeroes in Ω. In fact, Sa is nonvanishing on Ω because, as mentionedabove, Sa does not vanish on bΩ.

We can now say that the map given by f = Sa/La is in A∞(Ω), andthat |f | = 1 on bΩ. Note that f has a simple zero at a due to the poleof La at a and the fact that S(a, a) 6= 0. Since f(a) = 0, the maximumprinciple yields that |f | < 1 on Ω, i.e., that f maps Ω into the unit disc.For w in the unit disc, consider the integral

N(w) =1

2πi

∫

bΩ

f ′(z)

f(z)− wdz.

The argument principle says that N(w) is equal to the number of pointsz ∈ Ω satisfying f(z) = w. Since N(w) is integer valued and holomorphicon the unit disc, and since N(0) = 1, it follows that N(w) ≡ 1. Weconclude that f is a one-to-one map of Ω onto the unit disc.

To see that f ′ cannot vanish on bΩ, we use the Harnack inequality([Ah, p. 243]). Suppose z0 ∈ bΩ. We may assume that f(z0) = 1. Leth(z) = 1 − Re z. This function is harmonic and assumes its minimum


value of zero on the closed unit disc at z = 1. The function h f isharmonic on Ω, is in C∞(Ω), and assumes its minimum value of zero onΩ at z0. Let Dr(w0) be a small disc of radius r contained in Ω that isinternally tangent to the boundary of Ω at z0. If we now apply Harnack’sinequality to the positive harmonic function h f on Dr(w0), we obtain

r − |z − w0|r + |z − w0|

h(f(w0)) ≤ h(f(z)).

But h(f(w0)) = c is a positive constant. Hence,

h(f(z))− h(f(z0))

r − |z − w0|≥ c

r + |z − w0|≥ c

2r.

By letting z approach z0 along the inward pointing normal to bΩ at z0,we deduce that the normal derivative of h f at z0 is nonzero. Hence,f ′(z0) cannot vanish. The proof of Theorem 8.2 is complete.

We remark that the argument just given to show that f ′ is nonvan-ishing on the boundary contains a proof of the following classical resultknown as the Hopf lemma.

Theorem 9.4. Suppose that Ω is a bounded domain with C∞ smoothboundary and that φ is a harmonic function on Ω that is in C1(Ω). Ifthe maximum value of φ on Ω is attained at a boundary point z0, thenthe normal derivative of φ at z0 is strictly positive.

We mention one last result. We proved it above in the existence proofof the Riemann map.


smooth boundary and a ∈ Ω, then S(z, a) 6= 0 for all z ∈ Ω.

10

Solution of the Dirichlet problem insimply connected domains

Given a continuous function ϕ on bΩ, the classical Dirichlet problem isto find a harmonic function u on Ω that extends continuously to bΩ andthat agrees with ϕ on bΩ. In this chapter, we will relate the solutionof this problem to the Szego projection. In fact, we will prove that thesolution to the analogous problem in the C∞ setting exists and is wellbehaved. We will then use the C∞ result to solve the classical problem.

Theorem 10.1. Suppose Ω is a bounded simply connected domain withC∞ smooth boundary and suppose ϕ is a function in C∞(bΩ). Let a ∈ Ωand let Sa(z) = S(z, a) and La(z) = L(z, a). Then, the function u =h+H, where h and H are holomorphic functions in A∞(Ω) given by

h =P (Sa ϕ)

Saand H =

P (La ϕ)

La,

solves the Dirichlet problem for ϕ. Note that it follows that u ∈ C∞(Ω).

We remark that, since every harmonic function on a simply connecteddomain can be written as the real part of a holomorphic function, it ispossible to decompose any harmonic function as g + G. If a harmonicfunction u is decomposed in two ways as u = g1 + G1 = g2 + G2, theng1 − g2 = G2 − G1 is both holomorphic and antiholomorphic, hence,constant. Thus g and G in the decomposition u = g + G are uniquelydetermined up to additive constants. The functions h and H in thetheorem are therefore uniquely determined by the condition that H(a) =0, which follows from the fact that La has a pole at a. The maximumprinciple implies that the solution to the Dirichlet problem is unique.Hence, although the functions h and H depend on the choice of a, thesolution to the Dirichlet problem does not.

Note that it follows from Theorem 10.1 and the preceding remarksthat if a harmonic function on Ω that is in C∞(Ω) is decomposed asg +G where g and G are holomorphic on Ω, then it must be that g andG are in A∞(Ω).

It is easy to deduce from Theorem 10.1 that the solution to the clas-sical Dirichlet problem exists. Indeed, if ϕ is a continuous function on

43


bΩ, let ϕj be a sequence of functions in C∞(bΩ) that converge uniformlyon bΩ to ϕ. The Maximum Principle can now be used to see that thesolutions uj to the Dirichlet problems corresponding to ϕj converge uni-formly on Ω to a function u that is harmonic on Ω, continuous on Ω,and that assumes ϕ as its boundary values. This same limiting argumentreveals that the formula in Theorem 10.1 expresses the solution to theDirichlet problem even when ϕ is merely assumed to be continuous onbΩ. It must be pointed out, however, that in this case, the functions hand H need not extend continuously to the boundary, even though thesolution h+H does.

Proof of Theorem 10.1. The function Saϕ has an orthogonal decompo-sition Saϕ = g + GT where g = P (Saϕ) and GT = P⊥(Saϕ). Iden-tity (7.1) yields that T = −iSa/La. Hence, Saϕ = g − i GSa/La, andupon dividing this equation by Sa, we obtain

ϕ =g

Sa− i

G

La

.

By Theorem 9.5, Sa does not vanish on Ω, and by Theorem 9.3, La

does not vanish on Ω−a. Hence, the first term in this decomposition isseen to be in A∞(Ω) and the second term is the conjugate of a functionin A∞(Ω). It follows that the sum is a harmonic function in C∞(Ω)that agrees with ϕ on bΩ and we have found a solution to the Dirichletproblem.

To finish the proof of the theorem, we must show that iG = P (La ϕ ).

Formula (4.4) says that P⊥v = TP (vT ). Thus, G = P (SaϕT ). But,identity (7.1) implies that SaT = −iLa. Therefore, G = −iP (Laϕ), andthe proof is finished.

The proof just given is somewhat sneaky. However, there is an easyway to see that the formulas should be true. Suppose u is a harmonicfunction on Ω in C∞(Ω) and u = h+H where h and H are in A∞(Ω),and H(a) = 0. Then Sau = Sah+H Sa. This is actually an orthogonaldecomposition of Sau because, by (7.1), Sa = i LaT , and so H Sa =iHLaT ; the zero of H at a cancels the pole of La at a, and hencethis term is in H2(bΩ)⊥. Thus, P (Sau) = Sah. A similar argumentyields the formula for H . We could not use this more straightforwardapproach in our proof because we could not say in advance that, for agiven ϕ ∈ C∞(bΩ), a solution u exists to the Dirichlet problem. Evenif we knew a solution existed, we could not say that h and H in thedecomposition for u must be in A∞(Ω). In fact, we could not even saythat h and H must be in H2(bΩ). However, these facts were byproductsof the sneaky proof above.

Solution of the Dirichlet problem in simply connected domains 45

Theorem 10.1 can be localized. The explicit form of the solutionoperator to the Dirichlet problem given in Theorem 10.1 together withTheorem 4.4 yield the following result.

Theorem 10.2. Suppose u is a harmonic function on Ω that extendscontinuously to Ω. If the boundary values of u on an open arc Γ ⊂ bΩ areC∞ smooth there, then all partial derivatives of u extend continuouslyto Ω ∪ Γ.

The formulas in Theorem 10.1 yield interesting results if we think of ϕas merely being in L2(bΩ). Indeed, if ϕ ∈ L2(bΩ), then the decompositionϕ = h + H, where h = P (Saϕ)/Sa and H = P (Laϕ )/La, is valid onthe boundary of Ω. Consider the harmonic function u on Ω given byu = h + H . If z is a boundary point of Ω, let uǫ(z) = u(z + iǫT (z)).Because of Theorem 6.3, we may assert that uǫ tends to u in L2(bΩ) as ǫtends to zero. Hence, it is possible to solve the Dirichlet problem startingwith L2(bΩ) boundary data, thereby obtaining a harmonic function thathas the data as its boundary values in an L2 sense. Another interestingconsequence of the kind of reasoning above is an L2 maximum principle.To be specific, if a harmonic function u has vanishing boundary valuesin the sense that uǫ, as defined above, tends to zero in L2(bΩ), then umust be zero in the interior.

If ϕ is a continuous function on bΩ, the Poisson extension of ϕ,Eϕ, is defined to be equal to the harmonic function u on Ω that solvesthe Dirichlet problem with boundary data ϕ. We have shown that thePoisson extension operator is related to the Szego projection in simplyconnected domains via the identity in Theorem 10.1, and that E mapsC∞(bΩ) into C∞(Ω). When the integrals for the Szego projections arewritten out in that identity, we obtain the Poisson kernel for a simplyconnected domain. Indeed,

(Eϕ)(z) =∫

w∈bΩ

p(z, w)ϕ(w) ds

where, for z ∈ Ω and w ∈ bΩ, the Poisson kernel p(z, w) is given by

p(z, w) =S(z, w)S(w, a)

S(z, a)+S(z, w)L(w, a)

L(z, a). (10.1)

This formula for the Poisson kernel seems to depend on the choice of a ∈Ω. However, if p1(z, w) and p2(z, w) were the Poisson kernels correspond-ing to choosing a = a1 and a = a2, respectively, then, by the unique-ness of the solution to the Dirichlet problem, q(w) = p1(z, w)− p2(z, w)would be orthogonal to every continuous function on bΩ, and conse-quently would have to be identically zero. Hence, the kernel p(z, w) does


not depend on the point a. It is interesting to note that we can let a = zto see that

p(z, w) =|S(w, z)|2S(z, z)

. (10.2)

The function on the right hand side of this equality is known as thePoisson-Szego kernel. We have shown that, in a simply connected do-main, the Poisson kernel and the Poisson-Szego kernel coincide.

The Poisson kernel of a domain Ω has many of the properties of theclassical Poisson kernel on the unit disc. Most of these key propertiescan be read off from formulas (10.1) and (10.2). For example, p(z, w)is strictly positive on Ω × bΩ by (10.2) and the fact that the Szegokernel is nonvanishing in a simply connected domain. It is also clearthat

∫w∈bΩ

p(z, w) ds = 1 for all z ∈ Ω because the function u ≡ 1 is theharmonic extension of 1. Formula (10.1) shows that p(z, w) is harmonicin z on Ω for fixed w ∈ bΩ.

There is one other well known property of the Poisson kernel for theunit disc that we would like to prove for p(z, w), but we will have to savethe proof for later. It will be proved in Chapter 26 that, given w0 ∈ bΩand δ > 0, p(z, w) tends to zero uniformly in w on the set bΩ−Dδ(w0)as z tends to the boundary while staying in the set Ω ∩Dδ/2(w0).

11

The case of real analytic boundary

A domain in the plane is said to have real analytic boundary if its bound-ary can be (locally) parameterized by a function z(t) = x(t) + iy(t)where the real valued functions x(t) and y(t) are equal to their (real)Taylor series expansions in (t − t0) in a neighborhood of each point t0in the parameter space. A function v(x, y) will be said to be real ana-lytic on an open set in the plane if it can be expanded in a power seriesv(x, y) =

∑anm(x−x0)n(y− y0)m that converges on a neighborhood of

each point (x0, y0) in the open set. Note that harmonic functions are realanalytic because they are locally the real part of holomorphic functions.

In this chapter, we will indicate how the arguments we have given inthe C∞ setting can be modified to give analogous results for domainswith real analytic boundary.

Theorem 11.1. Suppose Ω is a bounded domain with real analyticboundary. Suppose that v is a function in C∞(Ω) that extends to bedefined on a neighborhood of Ω in such a way that the extension is realanalytic on a neighborhood of bΩ. Then the solution u,

u(z) =1

2πi

∫∫

Ω

v(ζ)

ζ − zdζ ∧ dζ,

to the equation, ∂u/∂z = v, also extends to be real analytic in a neigh-borhood of bΩ.

Since the complex conjugate of ∂u/∂z is equal to ∂u/∂z, Theo-rem 11.1 implies a similar statement about the existence of nice solutionsto the equation ∂u/∂z = v.

As in the C∞ case, the theorem is a simple consequence of the CauchyIntegral formula and a lemma.

Lemma 11.1. Suppose that Ω is a bounded domain with real analyticboundary and that v ∈ C∞(Ω) extends to be real analytic in a neigh-borhood of bΩ. Then, there exists a function Φ ∈ C∞(Ω) that vanisheson the boundary of Ω such that ∂Φ/∂z = v near bΩ, i.e., such that∂Φ/∂z − v is in C∞

0 (Ω).

47


Proof of the lemma. We give a complete, elementary, and self containedproof of this lemma in Appendix A. Here, we show how the lemma canbe seen to be a direct consequence of the Cauchy-Kovalevski theorem.(see Folland [Fo] for a proof of this famous theorem). Indeed, every curvein the plane is noncharacteristic for the Laplace operator ∆. Hence, wemay solve the Cauchy problem, ∆ψ = v near the boundary of Ω withCauchy conditions ψ = 0 on bΩ and ∇ψ = 0 on bΩ. The solution ψ willbe defined and real analytic on a neighborhood of bΩ. We now claim thatthe function ϕ = 4(∂/∂z)ψ solves our Cauchy problem near bΩ. Indeed,since ∆ = 4(∂/∂z)(∂/∂z), and since ∆ψ = v, it follows that ∂ϕ/∂z = v.Furthermore, because ψ and ∇ψ vanish on bΩ, it follows that ϕ = 0 onbΩ. To extend our solution ϕ to all of Ω, we simply use a C∞ cutofffunction χ that is compactly supported inside the set where ψ is definedand real analytic, and that is equal to one on a small neighborhood ofbΩ. Now the function Φ, which is defined to be zero where χ is zero andwhich is defined to be equal to χϕ on the support of χ, is a functionwith the properties we seek.

When the boundary of a domain is real analytic, we may define whatit means for a function to be real analytic on the boundary. A functionu defined on bΩ is said to be real analytic if u(z(t)) is a real analyticfunction of t when z(t) is a real analytic parameterizing function forbΩ. It is an easy exercise to see that u is real analytic on bΩ if andonly if u(z(t)) is real analytic for a single real analytic parameterizationz(t). We will use the symbol Cω(bΩ) to denote the space of real analyticfunctions on bΩ. It is an elementary fact that Cω(bΩ) coincides withthe space of continuous functions on bΩ that are restrictions to bΩ offunctions that are defined and real analytic in a neighborhood of bΩ.In fact, we will now show that Cω(bΩ) is equal to the set of functionson bΩ that are restrictions to bΩ of functions that are holomorphic ina neighborhood of bΩ. To see this, suppose that z(t) =

∑cn(t − t0)

n

is a real analytic parameterization of bΩ with t near t0. Consider thefunction f(ζ) =

∑cn(ζ − t0)

n which is defined and holomorphic nearζ = t0. Since z

′(t0) 6= 0, it follows that f ′(ζ) 6= 0 near ζ = t0. Hence,we may define a holomorphic inverse F (z) to f near z(t0). This functionhas the property that it maps the boundary curve of Ω near z(t0) one-to-one onto a segment in the real axis of the complex plane. If u isin Cω(bΩ), then u(z(t)) is real analytic near t = t0. Let U(ζ) be theholomorphic function defined on a neighborhood of ζ = t0 obtained byreplacing the real variable t in the power series expansion for u(z(t)) bya complex variable ζ. Now U(F (z)) is a holomorphic function definedin a neighborhood of z(t0) whose restriction to the boundary of Ω nearz(t0) agrees with u.

The case of real analytic boundary 49

This is an opportune moment to prove the general version of theSchwarz reflection principle because the argument in the last paragraphis at the heart of its proof. Suppose f is a holomorphic function definedon one side of a real analytic curve γ1 that extends continuously up toγ1 and that maps γ1 into another real analytic curve γ2. The Schwarzreflection principle asserts that f must extend holomorphically past γ1.To prove this, we restrict our attention to a small neighborhood of a pointin γ1. As in the last paragraph, we may construct holomorphic functionsF1 and F2 such that Fj is holomorphic in a neighborhood of γj andmaps γj into a segment of the real axis in the complex plane, j = 1, 2.By restricting to a small enough neighborhood, we may guarantee thatthe functions Fj have nonvanishing derivatives near γj . We may alsosuppose that F1 maps the side of γ1 on which f is defined into the upperhalf plane. Now consider the map H = F2 f F−1

1 . It is defined on asubregion of the upper half plane, it extends continuously to a segmentin the real axis, and it maps this segment in the real axis into the realaxis. Hence, we may apply the classical Schwarz reflection principle (seeAhlfors [Ah, p. 172]) to see that H extends holomorphically across thereal axis. Now f = F−1

2 H F1 is seen to extend holomorphicallythrough γ1.

Now that we have proved the reflection principle, it is an opportunemoment to mention reflection functions. Later in this book, when weconsider the boundary behavior of the Szego and Garabedian kernels ofdomains with real analytic boundaries in detail, we will need to knowof the existence of antiholomorphic reflection functions. Suppose thatγ is a real analytic curve. An antiholomorphic reflection function R(z)for γ is a function satisfying the following properties. It is defined andantiholomorphic on a neighborhood of γ, R(R(z)) = z for all z, R locallymaps one side of γ to the other side, and R(w) = w for w on γ. To seethat such functions exist, it is enough to check that they exist locallybecause analytic continuation may be used to obtain global reflectionsfrom local ones. To do the local construction, let F (z) be a holomorphicfunction as we produced above which is defined on a neighborhood ofa point in γ and which maps γ into the real axis. The function z 7→ zis an antiholomorphic reflection for the real axis. Hence, it follows thatR(z) = F−1(F (z) ) defines an antiholomorphic reflection for γ.

We will use reflection functions later when we show that kernel func-tions extend past the boundary in both variables simultaneously. At themoment, we will be satisfied to show only that the kernels extend pastthe boundary in one variable when the other variable is held fixed at apoint in the interior. This result will follow easily from the next theorem.

Theorem 11.2. Suppose Ω is a bounded domain with real analyticboundary. The Cauchy transform maps Cω(bΩ) into itself. So does the


Szego projection. Hence, for a fixed point a ∈ Ω, the kernel functionsS(z, a) and L(z, a) both extend holomorphically past the boundary of Ωin the z variable.

The proof of Theorem 11.2 mirrors the proofs of Theorems 3.1and 4.2. One of the key steps is to prove that the Kerzman-Stein kernelA(z, w) is in Cω(bΩ× bΩ), but this is routine.

The Szego kernel is the Szego projection of the Cauchy kernel, whichis obviously a function in Cω(bΩ). The Garabedian kernel is equal toP⊥Ga where Ga(z) = (2π)−1(z − a)−1. Hence, it follows from Theo-rem 11.2 that the kernels S(z, a) and L(z, a) associated to a domainwith real analytic boundary extend holomorphically past the boundaryas functions of z when a ∈ Ω is fixed.

Another interesting consequence of Theorem 11.2 is that if a holo-morphic function that extends continuously to the boundary of a domainwith real analytic boundary has real analytic boundary values near agiven boundary point, then that function must extend holomorphicallypast the boundary near the point.

All of the rest of the theorems we have proved in the C∞ setting cannow be routinely generalized to the Cω case. We mention one of themthat is particularly interesting and useful. We note that we are using theelementary fact that a real analytic function on a connected open setthat is harmonic on a small open subset must be harmonic on the wholeset.

Theorem 11.3. Suppose Ω is a bounded simply connected domain withreal analytic boundary and suppose ϕ is a function in Cω(bΩ). The solu-tion to the Dirichlet problem with boundary data ϕ extends to be definedon a neighborhood of Ω in such a way as to be harmonic there.

12

The transformation law for the Szegokernel under conformal mappings

To deduce the transformation laws for the Szego projection and kernelunder conformal mappings, we will require the following result.

Theorem 12.1. Suppose that f : Ω1 → Ω2 is a biholomorphic map-ping between bounded domains with C∞ smooth boundaries. Then f ∈C∞(Ω1) and f ′ is nonvanishing on Ω1. Consequently, f

−1 ∈ C∞(Ω2).Furthermore, f ′ is equal to the square of a function in A∞(Ω1).

The term biholomorphic means that f is a one-to-one holomorphicmap of Ω1 onto Ω2 (and consequently f−1 is holomorphic on Ω2). Toprove this theorem, we will need the following lemma which will alsoprove to be useful later.

Lemma 12.1. Suppose Ω is a bounded domain with C∞ smooth bound-ary. Then Ω is biholomorphic to a bounded domain with real analyticboundary, i.e., there exists a bounded domain Ω2 with real analyticboundary and a biholomorphic map of Ω onto Ω2. The biholomorphicmap extends C∞ smoothly to the boundary and its derivative is nonva-nishing on Ω.

Proof of the lemma. This is a standard construction in conformal map-ping. We proceed by induction on the connectivity of Ω. If Ω is 1-connected, i.e., simply connected, we use a Riemann mapping functionto map Ω onto the unit disc (which has real analytic boundary). Inthis case, the lemma reduces to Theorem 8.2. Suppose the lemma hasbeen established for (n − 1)-connected domains, and suppose that Ω isn-connected. Let γ denote one of the inner boundary curves of Ω, i.e.,one that bounds a bounded component of the complement of Ω in C.Let Ωn−1 denote the (n − 1)-connected domain obtained by filling inthe hole in Ω bounded by γ, and let Dγ denote the domain enclosedby γ. By our induction hypothesis, there is a biholomorphic map G ofΩn−1 onto a domain G(Ωn−1) with real analytic boundary. Now, whenG is restricted to Ω, we obtain a biholomorphic map of Ω onto the do-main G(Ω) which is an n-connected domain such that all of its boundary

51


curves but possibly one are real analytic. The one boundary curve thatmight not be real analytic is G(γ), the image of γ under G. Let z0 be apoint in G(Dγ). By composing G with the mapping F (z) = 1/(z − z0),we map Ω to a domain F (G(Ω)) such that the inner boundaries are realanalytic, and only the outer boundary might not be. Finally, to completethe induction, we use a Riemann map H that maps the simply connecteddomain obtained by filling in the holes of F (G(Ω)) onto the unit disc.The mapping H F G maps Ω into a subdomain of the unit disc. Theinner boundaries are real analytic. The outer boundary is the unit cir-cle, which is also real analytic. The statement about the smoothness upto the boundary of this map follows from the smoothness of the mapsused in its construction (see Theorem 8.2). Similarly, the nonvanishingof the derivative of this map on Ω follows from the nonvanishing of thederivatives of the maps used in its construction. The proof of the lemmais complete.

Proof of Theorem 12.1. Using the lemma, let h1 : Ω1 → G1 be a bihol-omorphic mapping of Ω1 onto a domain G1 with real analytic boundaryand let h2 : Ω2 → G2 be a biholomorphic mapping of Ω2 onto a do-main G2 with real analytic boundary. Consider the biholomorphic mapH = h2 f h−1

1 of G1 onto G2. Using Lemma 11.1, let Φ ∈ C∞(G2) bea function that vanishes on bG2 such that ∂Φ/∂z = 1 near bG2, i.e., suchthat 1 − ∂Φ/∂z has compact support in G2. We now claim that Φ Hextends continuously to G1 and that Φ H is harmonic on G1 near theboundary. Let d1(z) and d2(z) denote the distances from a point z tothe boundaries of G1 and G2, respectively. It is easy to see that, givenan ǫ > 0, there is a δ > 0 such that if a point z ∈ G1 satisfies d1(z) < δ,then d2(H(z)) < ǫ. Indeed, if this were not the case, we could constructa sequence of points zn in G1 such that d1(zn) → 0 as n → ∞, butd2(H(zn)) remains bounded away from zero. By taking a subsequence,it would be possible to find a sequence of points zn in G1 converging toa boundary point z0 ∈ bG1 such that H(zn) converges to a point w0 inG2, implying that H−1(w0) = z0, which is absurd. Since Φ = 0 on bG2,it follows that ΦH is continuous up to bG1 and is zero on the boundary.Since ∂Φ/∂z = 1 near bG2, it follows that (∂Φ/∂z) H = 1 near bG1.Because ∆ = 4(∂/∂z)(∂/∂z), we can show that Φ H is harmonic nearbG1 by showing that (∂/∂z)(Φ H) is holomorphic near bG1. Let us usesubscript z’s to indicate differentiation with respect to z. The chain ruleyields that (Φ H)z = H ′[Φz H ]. But, as mentioned above, Φz His equal to one near bG1, and hence, (Φ H)z = H ′ near bG1, whichis holomorphic. Since Φ H is continuous up to bG1, zero on bG1, andharmonic near bG1, we may apply the classical Schwarz Reflection Prin-ciple to see that Φ H extends past the boundary of G1 as a harmonic

The transformation law for the Szego kernel 53

function. Now, since (∂/∂z)(Φ H) = H ′ near bG1, we deduce that H ′

extends holomorphically past bG1. Thus, it follows that H ∈ C∞(G1)and we can say that f = h2 H h−1

1 is in C∞(Ω1). Since the sameargument will show that F = f−1 is in C∞(Ω2), it follows from theidentity, F ′(f(z)) = 1/f ′(z), that f ′ cannot vanish at any point in Ω1.

To finish the proof, we must see that f ′ is the square of a function inA∞(Ω1). Let Ti(z) denote the unit tangent vector functions associatedto bΩi, i = 1, 2. Observe that, if z(t) locally parameterizes the boundaryof Ω1 in the standard sense, then ζ(t) = f(z(t)) locally parameterizes theboundary of Ω2. As z(t) traces out the boundary of Ω1 in the standardsense, f(z(t)) traces out the boundary of Ω2 in the standard sense. Tosee this, consider two unit vectors originating from a point z ∈ bΩ1, onepointing in the direction of T1(z), the other pointing in the direction ofthe inward normal vector. Since f is conformal in the interior, and since fis smooth up to the boundary, f satisfies the Cauchy-Riemann equationsat z. The identity, ζ′(t) = f ′(z(t))z′(t), yields that, as a linear map, f ′(z)maps T1(z) to a tangent vector at f(z). Furthermore, since f maps Ω1

into Ω2, the inward pointing normal vector gets mapped to an inwardpointing normal vector at f(z). The fact that f preserves the sense ofangles at z forces us to conclude that ζ′(t) points in the direction of thestandard orientation. This shows that f(z(t)) traces out the boundaryof Ω2 in the standard sense. Hence, upon dividing ζ′(t) = f ′(z(t))z′(t)by its modulus, we obtain the identity,

T2(f(z)) = T1(z)f ′(z)

|f ′(z)| . (12.1)

Now, as any particular boundary component of Ωi (i = 1 or 2) is tracedout exactly once in the standard sense, the argument of Ti varies atotal of ±2π. Hence, (12.1) reveals that the argument of f ′(z) varies atotal of either zero or ±4π on each boundary component of Ω1. Sincethese numbers are all even multiples of 2π, we claim that it follows thatf ′(z) has a single valued square root on Ω1. Indeed, to see this, considerthe variation of arg f ′ around any closed curve γ in Ω1. The variation is

given by the integral∫γ

1if ′′

f ′dz. Because the boundary curves of Ω1 form

a homology basis for Ω1, it follows that the integral is an even multipleof ±2π. This means that, if we were to analytically continue the germ ofa square root for f ′ around γ, we would come back to our starting pointwith the original germ. Hence, f ′ has a single valued square root on Ω1.Let

√f ′(z) denote one of the square roots of f ′. Note that equation

(12.1) can now be rewritten

√f ′(z)T2(f(z)) =

√f ′(z)T1(z). (12.2)


We may now state the transformation rule for the Szego projection,the Szego kernel, and the Garabedian kernel under biholomorphic maps.Let a subscript one or two indicate that the function or projection underdiscussion is associated to Ω1 or Ω2, respectively.

Theorem 12.2. Suppose that f : Ω1 → Ω2 is a biholomorphic map-ping between bounded domains with C∞ smooth boundaries. The Szegoprojections transform according to the formula

P1

(√f ′ (ϕ f)

)=√f ′ ((P2ϕ) f)

for all ϕ ∈ L2(bΩ2). The Szego kernels transform according to

S1(z, w) =√f ′(z) S2(f(z), f(w))

√f ′(w).

The Garabedian kernels transform according to

L1(z, w) =√f ′(z) L2(f(z), f(w))

√f ′(w).

The notation√f ′ (ϕ f) stands for: √f ′ times the quantity, ϕ com-

posed with f .

Proof. Define an operator Λ1 that maps C∞(bΩ2) into C∞(bΩ1) via

Λ1ϕ =√f ′ (ϕ f).

It is clear that Λ1 maps A∞(Ω2) into A∞(Ω1). It is also easy to verifythat Λ1 is isometric in L2 norms, i.e., that

‖Λ1ϕ‖L2(bΩ1) = ‖ϕ‖L2(bΩ2).

Indeed, because f(z(t)) parameterizes the boundary of Ω2 in the stan-dard sense when z(t) parameterizes the boundary of Ω1 in the standardsense, it follows that ds2 = |f ′(z(t))z′(t)| dt = |f ′| ds1 where dsi denotesthe differential element of arc length on bΩi, i = 1, 2. Thus,

∫

bΩ2

|ϕ|2 ds =∫

bΩ1

|f ′||ϕ f |2 ds. (12.3)

The inverse of Λ1 can be written down in terms of the inverse of f . Ifwe write F = f−1, and if we choose the square root of F ′ that satisfies√F ′(f(z)) = 1/

√f ′(z), then it is easy to check that the operator Λ2

given byΛ2ψ =

√F ′ (ψ F )


is the inverse of Λ1. Hence, it follows that Λ1 extends uniquely to aHilbert space isomorphism of L2(bΩ2) onto L

2(bΩ1) which also restrictsto be an isomorphism between the subspaces H2(bΩ2) and H2(bΩ1).We will now show that this fact alone implies the transformation rule,P1Λ1 = Λ1P2, for the Szego projections. Before we can begin, we recallthe standard fact from Hilbert space theory that norm preserving oper-ators must also preserve inner products. That this is so is easily seen bymeans of the polarization identity,

〈u, v〉 = 1

4(‖u+ v‖2 − ‖u− v‖2)− i

4(‖u+ iv‖2 − ‖u− iv‖2).

Thus, we may assert that 〈Λ1u,Λ1v〉bΩ1= 〈u, v〉bΩ2

and similarly for Λ2.An important consequence of the fact that the operators Λ1 and Λ2

preserve the inner products is that

〈Λ1ϕ , ψ〉bΩ1= 〈ϕ , Λ2ψ〉bΩ2

(12.4)

for all ϕ ∈ L2(bΩ2) and ψ ∈ L2(bΩ1). Indeed, since Λ1Λ2 is the identity,we may write

〈Λ1ϕ , ψ〉bΩ1= 〈Λ1ϕ , Λ1Λ2ψ〉bΩ1

= 〈ϕ , Λ2ψ〉bΩ2.

We now claim that if u ∈ H2(bΩ2)⊥, then Λ1u ∈ H2(bΩ1)

⊥. To seethis, suppose that h ∈ H2(bΩ1) and u ∈ H2(bΩ2)

⊥. Then (12.4) yields

〈Λ1u, h〉bΩ1= 〈u,Λ2h〉bΩ2

= 0

because Λ2h ∈ H2(bΩ2). This proves our claim. Now the transformationrule follows by decomposing a function ϕ ∈ L2(bΩ2) as ϕ = h+ u whereh = P2ϕ and u = P⊥

2 ϕ. Then Λ1ϕ = Λ1h+Λ1u is an orthogonal sum andthe transformation formula rule follows because P1Λ1ϕ = Λ1h = Λ1P2ϕ.

We now turn to the proof of the transformation formulas for thekernels. Let Sa(z) = S2(z, a). If h ∈ H2(bΩ1), then using (12.4), we maywrite

〈h,Λ1Sa〉bΩ1= 〈Λ2h, Sa〉bΩ2

= (Λ2h)(a).

Furthermore, if G(z) =√F ′(a)S1(z, F (a)), then

〈h,G〉bΩ1= (Λ2h)(a).

Thus 〈h,Λ1Sa〉bΩ1= 〈h,G〉bΩ1

for all h ∈ H2(bΩ1), and we must con-clude that Λ1Sa = G, i.e., that

√f ′(z)S2(f(z), a) = S1(z, F (a))

√F ′(a),

which is equivalent to the transformation rule for the Szego kernel statedin the theorem. The formula for the Garabedian kernel follows fromthe Szego kernel formula and identities (7.1) and (12.2). The proof isfinished.


The transformation formula for the Szego kernels under a biholomor-phic map gives rise to another nice formula for the Riemann mappingfunction.

Theorem 12.3. Suppose that Ω is a bounded simply connected domainwith C∞ smooth boundary and suppose that f is a biholomorphic map-ping from Ω onto the unit disc U such that for a certain point a in Ω,we have f(a) = 0 and f ′(a) > 0. Then

f ′(z) = 2πS(z, a)2

S(a, a).

Proof. The proof of this identity rests on the simple fact that the Szegokernel SU (z, w) for the unit disc satisfies SU (z, 0) ≡ 1/(2π), which canbe read off from Theorem 7.2. Hence, the transformation formula yields

1

2π

√f ′(z) =

√f ′(z)SU (f(z), 0) = cS(z, a)

where c = [√f ′(a)]−1. Since f ′(a) > 0, we may assume that we are

dealing with a square root function that makes c a positive real number.Plugging z = a into this formula yields that f ′(a) = 2πS(a, a). Now, wemay square the formula and replace f ′(a) by its expression in terms ofthe Szego kernel to obtain the formula for the derivative of the Riemannmap.

We remark that another, and more classical, way to think about theSzego kernel is in terms of an orthonormal basis for H2(bΩ). Supposethat hi∞i=1 is such a basis. We will prove that

S(z, a) =

∞∑

i=1

hi(z)hi(a), (12.5)

with absolute and uniform convergence in z on compact subsets of Ω.Granted this fact, the transformation formula for the Szego kernel isseen to be a direct consequence of the fact that the operator Λ1 sendsan orthonormal basis for H2(bΩ2) to an orthonormal basis for H2(bΩ1).

To prove (12.5), notice that the coefficients ci in the orthogonal ex-pansion Sa =

∑cihi for the Szego kernel are given by ci = 〈Sa, hi〉b =

hi(a). The Cauchy integral formula h(w) = 〈h,Cw〉b gives rise to thebasic estimate |h(w)| ≤ ‖h‖‖Cw‖ which shows that convergence inH2(bΩ) implies uniform convergence on compact subsets of Ω. Hence,∑∞

i=1 hi(z)hi(a) converges in z uniformly on compact subsets of Ω toSa(z). The absolute convergence of the series follows from the observa-tion that

S(a, a) =

∞∑

i=1

|hi(a)|2.


Hence, the Cauchy-Schwarz inequality yields

∞∑

i=1

|hi(z)hi(a)| ≤(

∞∑

i=1

|hi(z)|2)1/2( ∞∑

i=1

|hi(a)|2)1/2

=√S(z, z)

√S(a, a)

and absolute convergence is proved.

13

The Ahlfors map of a multiplyconnected domain

Suppose Ω is a bounded simply connected domain. Everyone knows that,among all holomorphic functions h that map Ω into the unit disc, theRiemann mapping function associated to a point a ∈ Ω is the uniquefunction in this class making h′(a) real and as large as possible. Hence,finding the Riemann map is equivalent to solving an extremal problem.In Chapter 8, we showed that the solution to this extremal problemcan also be expressed as the quotient of the Szego and the Garabediankernels, f(z) = S(z, a)/L(z, a). In this chapter, we will consider thisquotient when Ω is a multiply connected domain. We will show that itis a mapping of the domain onto the unit disc, that it solves the sameextremal problem, and that it has many of the geometric features onewould expect of a “Riemann mapping function” of a multiply connecteddomain. The map is known as the Ahlfors mapping.

Because we will be studying the extremal problem mentioned above,let us spell it out. Let F denote the set of holomorphic functions on Ωmapping Ω into the unit disc.The Extremal Problem. Given a ∈ Ω, find all the functions h ∈ Fthat maximize |h′(a)|.

To study the Ahlfors map, we will need to know a generalized versionof the argument principle that allows zeroes to occur on the boundary.In the discussion of the argument principle that follows, we will allowfunctions to have zeroes and poles on the boundary because we willneed this version later in the book. Let Ω be a bounded domain withC∞ smooth boundary. Suppose that h is meromorphic in a neighborhoodof Ω, and that h is not identically zero. Then the zeroes and poles of hare isolated. Let ziNi=1 denote the zeroes of h that lie in Ω, let piQi=1

denote the poles of h that lie in Ω, let biMi=1 denote the zeroes of hthat lie on bΩ, and let BiRi=1 denote the poles of h that lie on bΩ. Forsmall ǫ > 0, let γǫ = bΩ−

(∪Mi=1Dǫ(bi)

)−(∪Ri=1Dǫ(Bi)

). We assume that

ǫ is small enough so that the closures of all the discs Dǫ(bi) and Dǫ(Bj)are mutually disjoint and the set γǫ consists of finitely many connectedsmooth arcs. On each of these arcs, the increment of the argument of

59


h(z) as z moves along the arc in the positive sense is well defined. Wewill prove that as ǫ → 0, the sum of the increments of the argumentof h(z) along all these arcs tends to an angle ∆arg h, and this angle isrelated to the number of zeroes and poles of h according to the followingformula (we let mh(z) denote the multiplicity of a zero or pole of h atz).The generalized argument principle

N∑

i=1

mh(zi) +1

2

M∑

i=1

mh(bi)−Q∑

i=1

mh(pi)−1

2

R∑

i=1

mh(Bi) =1

2π∆arg h

In words, this generalized argument principle says that zeroes orpoles in the interior of Ω contribute to the increment of the argumentof h as usual, whereas zeroes and poles on the boundary contributehalf the normal amount to the increment. To prove this fact, let Cǫ(bi)denote the arc of the circle that bounds Dǫ(bi) lying outside of Ω andlet Cǫ(Bi) denote the arc of the circle that bounds Dǫ(Bi) lying outsideof Ω. We assume that these circular arcs are parameterized so that γǫ ∪(∪Mi=1Cǫ(bi)

)∪(∪Ri=1Cǫ(Bi)

)represents the boundary of

Ωǫ = Ω ∪(∪Mi=1Dǫ(bi)

)∪(∪Ri=1Dǫ(Bi)

),

parameterized in the standard sense. We may now apply the classicalargument principle to h on Ωǫ. As we let ǫ tend to zero, the incrementof arg h on Cǫ(bi) is easily seen to approach πmh(bi), and the incrementof arg h on Cǫ(Bi) is seen to approach −πmh(Bi). This completes theproof. We may now prove the following theorem.

Theorem 13.1. Suppose Ω is a bounded n-connected domain with C∞

smooth boundary and let a ∈ Ω be given. Then La(z) is nonvanishing forz ∈ Ω− a. The function Sa(z) is nonvanishing on bΩ and has exactlyn − 1 zeroes in Ω. The function f(z) = Sa(z)/La(z) maps Ω onto theunit disc and is an n-to-one map (counting multiplicities). Among allholomorphic functions h that map Ω into the unit disc, the functionsthat maximize the quantity |h′(a)| are given by eiθf(z) for some realconstant θ. Furthermore, f is uniquely characterized as the solution tothis extremal problem such that f ′(a) > 0. Also, f extends to be inC∞(Ω), f ′ is nonvanishing on the boundary, and f maps each boundarycurve one-to-one onto the boundary of the unit disc.

Proof. Proving this theorem is easiest when Ω is a bounded finitely con-nected domain with real analytic boundary; so that is what we nowassume. Later, we will relax this hypothesis. With the real analytic as-sumption, we know by Theorem 11.2 that Sa(z) and La(z) extend holo-morphically past the boundary. This fact, together with identity (7.1),

The Ahlfors map of a multiply connected domain 61

is ninety percent of the proof. The other ten percent is contained in thegeneralized argument principle proved above.

The function f has a simple zero at z = a because S(a, a) > 0 andLa has a simple pole at z = a. Besides this fact, at the moment, we knowonly that f is a meromorphic function on a neighborhood of Ω. Let λ be acomplex number of unit modulus. We want to consider how many timesf(z) assumes the value λ on Ω. To do this, let G(z) = Sa(z) − λLa(z).We will first prove that G(z) has exactly n zeroes on Ω, one on eachboundary component of Ω. Observe that formula (7.1) implies that, onbΩ,

GT = SaT − λLaT = i La − λi Sa = −iλ(Sa − λLa) = −iλG.

Thus, we have the identity,

G2 = −iλ|G2|T . (13.1)

Let γini=1 denote the simple closed real analytic curves that representthe n boundary components of Ω. We will now show that G has at leastone zero on each γi. Indeed, if G has no zero on γi, identity (13.1) showsthat the increment of arg G2 around γi is the same as the incrementof arg T . But the increment of the argument of G2, the square of aholomorphic function, around γi is either zero or an even multiple of±2π, and the increment of arg T is ±2π. Hence, equality is out of thequestion. Thus, G must have at least one zero on γi.

Let ziNi=1 denote the zeroes of G that lie in Ω and let biMi=1 denotethe zeroes of G that lie on bΩ. Observe that G has a single simple poleat a. Thus, the generalized argument principle yields

−1 +

N∑

i=1

mG(zi) +1

2

M∑

i=1

mG(bi) =1

2π∆arg G.

Now, identity (13.1) reveals that

1

2π∆arg G2 =

1

2π∆arg T = −1 + (n− 1) = n− 2.

Therefore, 12π∆arg G = 1

2 (n− 2). But the term∑M

i=1mG(bi) is at leastn because G has at least one zero on each boundary component of Ω.When these numbers are plugged into the argument principle, we areforced to conclude that G has no zeroes in Ω, and exactly one zero oneach component bΩ.

Using what we know about G, we will now show that f extends to beholomorphic on a neighborhood of Ω, that |f | < 1 on Ω, and that |f | = 1on bΩ. Indeed, if f had a pole at some point in Ω, then there would exist


a point p ∈ Ω near the pole such that |f(p)| > 1 and La(p) 6= 0. Considerthe function |f(z)| along a curve in Ω joining p to a that does not passthrough a zero of La. Since |f(p)| > 1 and |f(a)| = 0, the intermediatevalue theorem implies that there is a point z0 ∈ Ω (which is not a zero ofLa) such that f(z0) = λ, a complex number of unit modulus. However,at such a point z0, the function G associated to this λ that we studiedabove would have to vanish, and we have shown that it cannot vanishat an interior point. Hence, f has no poles in Ω. Now, if there is a pointp in Ω with |f(p)| > 1, we may apply the same reasoning to obtain thesame contradiction. Hence, |f | ≤ 1 on Ω. Since f(a) = 0, the maximumprinciple implies that |f | < 1 on Ω. Identity (7.1) yields that |f | = 1 onthe dense subset of bΩ where La is nonvanishing. By continuity, |f | = 1on all of bΩ.

Next, we show that La is nonvanishing on Ω − a. We have shownthat if |λ| = 1, then Gλ = Sa − λLa has exactly one zero on eachboundary component of Ω and no zeroes inside Ω. Also, because f =Sa/La is holomorphic in a neighborhood of Ω, it follows that if La(z0) =0 with z0 ∈ Ω, then Sa(z0) = 0 too. Hence, Gλ and Sa must vanishwherever La does. Because Gλ cannot vanish in Ω, this yields that La

cannot vanish in Ω. To see that La cannot vanish on bΩ either, supposeLa(z0) = 0, z0 ∈ bΩ. Then Sa(z0) = 0, and Gλ(z0) = 0 for any λ of unitmodulus. We have shown that Sa cannot be identically zero on Ω; thusthere is a point ξ0 in the same boundary component of bΩ that z0 is insuch that Sa(ξ0) 6= 0. Since Sa(z0) = 0, it follows that ξ0 6= z0. Formula(7.1) shows that |Sa(ξ0)| = |La(ξ0)|. Hence, there is a λ with |λ| = 1such that Gλ(ξ0) = 0. We have now shown that, for this particularchoice of λ, Gλ has zeroes at two points, z0 and ξ0, in a single boundarycurve. This is a contradiction. Therefore, La is nonvanishing on Ω−a.Formula (7.1) now shows that Sa is nonvanishing on bΩ too. We can alsoread off from (7.1) that

−∆arg Sa = ∆arg La +∆arg T = 2π[−1 + 1− (n− 1)].

Hence, ∆arg Sa = 2π(n− 1), and we deduce that Sa has exactly n− 1zeroes in Ω, and, because of the pole of La at a, that f has exactly nzeroes in Ω.

Since La is nonvanishing, and since Gλ vanishes exactly once on eachboundary component of Ω when |λ| = 1, it follows that f maps eachboundary component one-to-one onto the unit circle. This fact alonecould be used to see that f is an n-to-one map (counting multiplicities)of Ω onto the unit disc. We will use the following simple argument toprove this instead. Given w in the unit disc, the number of times thatf assumes the value w (counting multiplicities) on Ω is given by the

The Ahlfors map of a multiply connected domain 63

integral

M(w) =1

2πi

∫

bΩ

f ′(z)

f(z)− wdz.

We know that M(0) = n. Since M(w) is a holomorphic function of won the unit disc taking values in the integers, M must be constant, i.e.,M(w) ≡ n.

Finally, we must see that f is extremal, i.e., that, among all holo-morphic functions on Ω that map into the unit disc, f has the propertythat |f ′(a)| is as large as possible. First, note that f ′(a) = 2πS(a, a)because L(z, a) has a simple pole at z = a with residue 1/2π. Since mul-tiplication by eiθ preserves the class of functions that map into the unitdisc, and since f ′(a) > 0, we may restrict our attention to the class offunctions h that map into the disc such that h′(a) > 0. Since this classis a normal family, we know extremal functions exist. Furthermore, if his extremal, it must be that h(a) = 0. Indeed, if this is not the case, byforming the composition M h where M(z) is the Mobius transforma-tion M(z) = (z − h(a))/(1 − h(a) z), we obtain a map in the class withstrictly larger derivative at a, which contradicts the extremal assump-tion. Hence, we may restrict our attention to the class F+ of functionsh that are holomorphic on Ω, mapping Ω into the unit disc, such thath(a) = 0 and h′(a) > 0. Note that, by Theorem 6.4, F+ may be viewedas a subset of H2(bΩ).

Consider the function L2a. It is meromorphic on a neighborhood of Ω

and has a single pole at z = a. We now claim that the residue of L2a at

a is zero. Indeed, by (7.1), L2aT = iLaSa. Hence, the residue of L2

a at acan be computed via

2πiResaL2a =

∫

bΩ

L2aT ds =

∫

bΩ

iLaSa ds = 〈iLa, Sa〉b,

and this last quantity is zero because La = i SaT is orthogonal toH2(bΩ). Hence,

L2a =

1

4π2

1

(z − a)2+Ha

where Ha is holomorphic on a neighborhood of Ω. Hence, by the residuetheorem, if h ∈ F+, then

h′(a) = 4π2Resa(L2ah) =

2π

i

∫

bΩ

L2ahT ds ≤ 2π

∫

bΩ

|L2a| ds (13.2)

since |h| ≤ 1 on bΩ. But, by (7.1), the L2(bΩ) norm of La is equal to thatof Sa, which we know is equal to S(a, a)1/2. Hence, we have shown thath′(a) ≤ 2πS(a, a) = f ′(a), and therefore, that f is an extremal function.


We next show that f is the unique extremal function in the class F+.This turns out to be an easy consequence of a measure theory exercisewhich asserts that, if |v| ≤ |u| and

∫v =

∫|u|, then v = |u|. Suppose

that h ∈ F+ is extremal, and let v = −iL2ahT and u = |La|2. Because

h maps into the unit disc, it follows from Theorem 6.4 that |v| ≤ |u| onbΩ. Now the reasoning used to deduce (13.2) shows that

∫v =

∫|u| and

it follows that v = |u|, i.e., that −iL2ahT = |La|2. Solving this equation

for h shows that h = i LaT/La and using (7.1) yields that h = Sa/La asdesired.

To finish the proof in the case of real analytic boundary, we mustshow that f ′ is nonvanishing on the boundary. Suppose z0 ∈ bΩ andlet u denote the harmonic function on Ω given by u(z) = Re f(z0) f(z).This function is in C∞(Ω) and assumes its maximum value of one on Ωat the boundary point z0. The Hopf lemma (Theorem 9.4) states thatthe normal derivative of u at z0 is nonzero. The chain rule now impliesthat f ′(z0) cannot vanish either. The proof is complete for domains withreal analytic boundaries.

To prove the theorem in the general case, we use Lemma 12.1. LetG : Ω → Ω2 be a biholomorphic map of our C∞ smooth domain Ω onto adomain Ω2 with real analytic boundary. We know that G ∈ C∞(Ω) andthat G′ is nonvanishing on Ω. Given a ∈ Ω, it is clear that the solutionto the extremal problem for Ω at a is given by f = eiθf2 G where f2 isthe solution to the extremal problem for Ω2 at G(a) with f ′

2(G(a)) > 0and eiθ is a complex number of unit modulus chosen so that f ′(a) > 0.To be precise,

eiθ = G′(a)/|G′(a)| =√G′(a)

/√G′(a).

We know that f2(z) = S2(z,G(a))/L2(z,G(a)). Now, using the trans-formation formulas for the Szego and Garabedian kernels under G, weobtain f = S(z, a)/L(z, a) and the proof is finished.

14

The Dirichlet problem in multiplyconnected domains

We have postponed the study of the Dirichlet problem in multiply con-nected domains until now because, before Theorem 13.1, we did notknow that the Garabedian kernel was nonvanishing on a multiply con-nected domain, and this is an important ingredient in our approach tothe problem.

Life in a multiply connected domain is complicated by the fact thatnot every harmonic function can be globally written as the sum of aholomorphic and an antiholomorphic function. In order to obtain a the-orem analogous to Theorem 10.1 in a multiply connected domain, wewill have to add some terms to put the functions involved in the spaceof functions h+H : h,H ∈ H2(bΩ).

Theorem 4.3 implies that the operator that sends a function ϕ ∈C∞(bΩ) to the function in C∞(bΩ) given by

h+ T H

where h = Pϕ and H = P (Tϕ) is the identity operator. Thus, we maydecompose Sa ϕ on bΩ as

Sa ϕ = h+ T H

where h = P (Saϕ) and H = P (TSaϕ). Next, using (7.1), we substi-tute −iSa/La for T and divide the identity by Sa. What we get is adecomposition of ϕ on the boundary of Ω as h+H where

h =P (Sa ϕ)

Saand H = i

P (SaϕT )

La.

Using (7.1) again, we may replace SaT in the expression for H by −iLa.We have proved most of the statements in the following theorem.

Theorem 14.1. Suppose Ω is a bounded finitely connected domain withC∞ smooth boundary and suppose ϕ is a function in C∞(bΩ). Let a ∈ Ωbe given. Then, on the boundary, the function ϕ can be decomposed as

ϕ = h+H,

65


where h is a meromorphic function on Ω that extends C∞ smoothly upto bΩ given by

h =P (Sa ϕ)

Sa

and H is a holomorphic function in A∞(Ω) given by

H =P (La ϕ)

La.

Furthermore, if ϕ is equal to g + G for some g and G in H2(bΩ), thenthe function h has no poles and is in A∞(Ω). Thus, in this case, h+His the harmonic extension of ϕ to Ω.

Proof. The decomposition of ϕ on the boundary as h + H was provedabove. To see that h is meromorphic, we use the fact that Szego kernelSa(z) has exactly n − 1 zeroes in Ω when Ω is an n-connected domain,and Sa(z) does not vanish for any z ∈ bΩ (see Theorem 13.1). Thesmoothness up to the boundary follows from the fact that P preservesC∞(bΩ) and the fact that Sa(z) is in A∞(Ω). The function H is inA∞(Ω) because La(z) has a single simple pole at a and does not vanishfor any z ∈ Ω − a, and furthermore, La extends C∞ smoothly up tothe boundary.

Next, we show that if ϕ ∈ C∞(bΩ) can be decomposed as ϕ = g+Gwhere g and G are in H2(bΩ), then g and G must be in A∞(Ω). Notethat if ϕ = g + G, then, using (7.1), we see that Saϕ = Sag + i GLaT .By subtracting G(a) from G and adding G(a) to g, we may assumethat G(a) = 0. In this case, it follows that GLa is in H2(bΩ). Hence,Sag + i GLaT is an orthogonal decomposition of Saϕ. It follows thatSag = P (Saϕ) and this shows that g ∈ C∞(bΩ). Consequently, g ∈A∞(Ω). In fact, we have shown that g = P (Saϕ)/Sa on the boundary,i.e., that g = h on bΩ. Since a holomorphic function that is continuousup to the boundary cannot vanish on an open set in the boundary, itfollows that g must be equal to h on Ω, too. Hence, h has no poles in Ω.Since g = h, it follows that G = H , and the proof is complete.

If Ω is n-connected, let bin−1i=1 be a set of points comprised of one

point from each of the bounded connected components of the comple-ment of Ω in C, and let γi denote the boundary curve of Ω bounding thecomponent of C− Ω containing bi. The function

ψi = log |z − bi|

is a harmonic function on a neighborhood of Ω. We claim that ψi cannotbe expressed as the real part of a single valued holomorphic function on

The Dirichlet problem in multiply connected domains 67

Ω. Indeed, locally, ψi has a harmonic conjugate given by a continuouschoice of an argument of z− bi plus a constant. If such a local conjugateis continued around the curve γi in the standard sense, a simple com-putation based on the argument principle shows that the ending valuediffers from the starting value by −2π. Hence, ψi cannot have a singlevalued harmonic conjugate on Ω. Also note that the variation of a localharmonic conjugate for ψi around γj where j 6= i is zero. We may repeat

the argument above to deduce that if ψ, given by ψ =∑n−1

i=1 ciψi, is thereal part of a holomorphic function on Ω, then all the c’s must be zero.

We have shown that ψi cannot be written on bΩ as gi + Gi for anygi, Gi ∈ H2(bΩ). It follows that if we express ψi on bΩ as hi +Hi wherehi and Hi are given by the formulas in Theorem 14.1, then hi musthave a pole at one of the zeroes of the Szego kernel. In fact, the samereasoning shows that if cin−1

i=1 are constants that are not all zero, then

ψ =∑n−1

i=1 ciψi cannot be written as the real part of a holomorphicfunction on Ω, and therefore cannot be written on bΩ as g + G whereg,G ∈ H2(bΩ). This means that

∑cihi has at least one pole at a zero

of the Szego kernel.Let ajn−1

j=1 denote the n − 1 zeroes of Sa(z) in Ω. For simplicity,let us suppose, for the moment, that each zero has multiplicity one. (Infact, we will prove in Chapter 27 that the zeroes of Sa become simplezeroes as a tends to the boundary of Ω. Hence, it is actually possible tochoose a so that this condition is met.) The fact that

∑cihi has at least

one pole at a zero of the Szego kernel if not all the ci’s are zero meansthat if

n−1∑

i=1

ci (P (ψiSa)) (aj) = 0 for j = 1, . . . , n− 1,

then ci = 0 for all i. This implies that the determinant of this linearsystem is nonzero. Hence, given ϕ ∈ C∞(bΩ), we may solve the linearsystem,

n−1∑

i=1

ci (P (ψiSa)) (aj) = P (ϕSa)(aj),

j = 1, . . . , n − 1, for ci, i = 1, . . . , n − 1. Having solved the system, wededuce that, when ϕ−∑ ciψi is expressed as h+H via the formulas inTheorem 14.1, the function h has no poles at the zeroes of Sa. Hence,the harmonic extension of ϕ to Ω is given by

h+H +∑

ci log |z − bi|.

Now, in the case that Sa(z) has zeroes of multiplicity greater thanone, we must solve a linear system analogous to the one above. However,


corresponding to each zero of multiplicity m, there must be m linearequations, one stemming from point evaluation at the zero, and m − 1additional equations arising from point evaluation of the first m − 1derivatives at the zero. The details are not hard. We leave them to thereader.

We have proved the following theorem.

Theorem 14.2. Suppose Ω is a bounded domain with C∞ smoothboundary and suppose ϕ is a function in C∞(bΩ). Then the solutionto the Dirichlet problem with boundary data ϕ exists and is in C∞(Ω).

As in the simply connected case, this theorem, which solves theDirichlet problem in C∞, can be used to prove that the classical Dirichletproblem is solvable for continuous functions.

Note that by revisiting Theorem 11.3 in the multiply connected set-ting, and noting that the functions log |z − bi| are harmonic on an openset containing the closure of the domain, we obtain the following theo-rem.

Theorem 14.3. Suppose Ω is a bounded domain with smooth real anal-tyic boundary and suppose ϕ is a function in Cω(bΩ). The solution tothe Dirichlet problem with boundary data ϕ exists and extends to be har-monic on a neighborhood of Ω.

15

The Bergman space

To begin, we suppose that Ω is merely a domain in the plane of finite areaand we do not make any assumptions about the nature of the boundary.The Bergman space, denoted H2(Ω), is the space of holomorphic func-tions on Ω that are square integrable on Ω with respect to area measuredA = dx ∧ dy = i

2dz ∧ dz, i.e., h in H2(Ω) are holomorphic functionssuch that

∫∫Ω|h|2 dA < ∞. We may think of H2(Ω) as being a subset

of L2(Ω) by adopting the standard convention that two functions thatagree almost everywhere are the same function. Theorem 6.5 shows thatH2(bΩ) is a subset of H2(Ω) when the domain is bounded and smooth.

We wish to define the orthogonal projection of L2(Ω) onto its sub-space of holomorphic functions similarly to the way we defined the Szegoprojection of L2(bΩ) onto H2(bΩ). To do this, we need to see that H2(Ω)is a closed subspace of L2(Ω). To prove this, we will use the fact that thevalue of a holomorphic function at the center of a disc is given by theaverage of the function over the disc. Thus, if z0 ∈ Ω and h ∈ H2(Ω),we may estimate the value of h(z0) by averaging h over the disc Dr(z0)where r is any radius that is less than the distance d from z0 to theboundary of Ω. Using Holder’s inequality, we obtain

|h(z0)| ≤1

πr2

∫∫

Dr(z0)

|h| dA ≤ 1√π r

(∫∫

Dr(z0)

|h|2 dA)1/2

,

and this last expression is less than or equal to (√π r)−1‖h‖ where ‖h‖

denotes the L2(Ω) norm of h. By letting r tend to d, we obtain the im-proved estimate |h(z0)| ≤ C‖h‖ where C = 1/(

√π d). Thus, if hj is a

sequence of holomorphic functions on Ω that converge in L2(Ω) to u,then hj converges uniformly on compact subsets of Ω to a holomorphicfunction h. Since convergence in L2(Ω) implies pointwise convergencealmost everywhere of a subsequence, it follows that u = h almost every-where, i.e., that u ∈ H2(Ω).

The space L2(Ω) is a Hilbert space with inner product defined via〈u, v〉Ω =

∫∫Ωuv dA. Since H2(Ω) is a closed subspace of a Hilbert space,

it too is a Hilbert space.

69


The estimate |h(a)| ≤ C‖h‖ that we proved above for h ∈ H2(Ω)and a ∈ Ω shows that evaluating a function h ∈ H2(Ω) at a pointa ∈ Ω is a continuous linear functional on the Hilbert spaceH2(Ω). Thus,the Riesz representation theorem implies that there is a function Ka inH2(Ω) that represents this functional. The function Ka is called theBergman kernel function and it is standard to write K(z, a) = Ka(z).Because Ka(b) = 〈Ka,Kb〉Ω and Kb(a) = 〈Kb,Ka〉Ω it follows thatK(a, b) = K(b, a). It also follows thatK(a, a) = 〈Ka,Ka〉Ω = ‖Ka‖2 andthis quantity must be positive because not every holomorphic functionin H2(Ω) vanishes at a (for example, f(z) ≡ 1 is such a nonvanishingfunction).

Since H2(Ω) is a closed subspace of L2(Ω), we may consider theorthogonal projection B of L2(Ω) ontoH2(Ω). This operator is called theBergman projection. The Bergman kernel is the kernel for the Bergmanprojection in the sense that

(Bu)(a) = 〈Bu,Ka〉Ω = 〈u,Ka〉Ω =

∫∫

z∈Ω

K(a, z)u(z) dA.

The Bergman space is an important tool in the study of conformalmappings. That this should be so can be understood from the simplefact that if f is holomorphic, then |f ′|2 is equal to the real Jacobiandeterminant of f viewed as a mapping from R2 into itself. Thus, if f :Ω1 → Ω2 is a biholomorphic mapping between bounded domains, then,without worrying about such things as convergence, the classical changeof variables formula reads

∫∫

Ω2

ϕ dA =

∫∫

Ω1

|f ′|2(ϕ f) dA.

Taking ϕ = |h|2 in this formula leads one to guess that the transforma-tion h 7→ f ′(h f) is a norm preserving operator between the Bergmanspaces associated to Ω2 and Ω1. The notation f ′(ϕ f) stands for f ′

times the quantity, ϕ f . This transformation is important, and so wewill give it a name; let Λ1ϕ = f ′(ϕf). In fact, this transformation is soimportant that it is worthwhile to be more careful in defining it. If ϕ isin C∞

0 (Ω2), then Λ1ϕ is in C∞0 (Ω1) and the change of variables formula

above is valid. Replacing ϕ by |ϕ|2 in the change of variables formulashows that ‖Λ1ϕ‖Ω1

= ‖ϕ‖Ω2. Now C∞

0 (Ω2) is dense in L2(Ω2), andtherefore, this estimate shows that Λ1 extends uniquely as a boundedoperator from L2(Ω2) to L2(Ω1). Furthermore, it is clear that Λ1 pre-serves holomorphic functions. Let F = f−1 and, for ψ ∈ L2(Ω1), defineΛ2ψ = F ′(ψ F ) analogous to the way we defined Λ1φ above. It is easyto check, via the identity f ′(z) = 1/F ′(f(z)), that Λ2 is the inverse toΛ1. Thus, Λ1 is an isometry between L2(Ω2) and L2(Ω1) that restricts

The Bergman space 71

to be an isometry between H2(Ω2) and H2(Ω1). These facts, togetherwith a little standard Hilbert space theory, can be used to prove anotheruseful identity,

〈Λ1u, v〉Ω1= 〈u,Λ2v〉Ω2

, (15.1)

that holds for all u ∈ L2(Ω2) and v ∈ L2(Ω1). Indeed, we proved aresult identical to this when we proved identity (12.4). Recall that thepolarization identity implies that an isometry also preserves the innerproduct, and so 〈Λ1u1,Λ1u2〉Ω1

= 〈u1, u2〉Ω2. Hence, since Λ1Λ2 is the

identity operator, we may write

〈Λ1u, v〉Ω1= 〈Λ1u,Λ1(Λ2v)〉Ω1

= 〈u,Λ2v〉Ω2

and the identity is proved.In this chapter, we want to prove some facts about the Bergman

projection and the Bergman kernel. To motivate why such a programmight be fruitful, let us take a moment to show that the Bergman kernelfunction of a bounded simply connected domain is related to Riemannmapping functions. Let f : Ω → D1(0) be a Riemann mapping functionof a bounded simply connected domain Ω onto the unit disc such thatf(a) = 0 and f ′(a) > 0. Let F = f−1. The averaging property ofholomorphic functions implies that 〈H, 1〉D1(0) = πH(0) for all H ∈H2(D1(0)). Now, if we apply the conjugate of the identity we provedabove, using the functions u ≡ 1 and v = h ∈ H2(Ω), we obtain

〈h, f ′〉Ω = 〈F ′(h F ) , 1〉D1(0) = πF ′(0)h(F (0)) = ch(a)

where c = πF ′(0) = π/f ′(a). It follows that the function k(z) = c−1f ′(z)has the property that 〈h, k〉Ω = h(a) for all h ∈ H2(Ω). This propertyis shared by the Bergman kernel Ka. Hence, by the uniqueness clausein the Riesz representation theorem, it follows that k(z) ≡ Ka(z), i.e.,that f ′(z) = cK(z, a) where c = π/f ′(a). Letting z = a in this formulareveals that πK(a, a) = f ′(a)2. Since f ′(a) > 0, we may deduce thatf ′(a) =

√πK(a, a). Now we can write down the classical formula

f ′(z) = CK(z, a)

where C =√π/K(a, a). Thus, learning things about the Bergman kernel

is going to yield information about conformal mappings.Incidentally, we may use the formula for the Riemann map proved

above to determine the Bergman kernel for the unit disc U . The averag-ing property for holomorphic functions implies that πh(0) = 〈h, 1〉U forevery h ∈ H2(U). Hence, it follows that the Bergman kernel KU (z, w)for the disc satisfies KU (z, 0) ≡ 1

π for z ∈ U . Fix a point a ∈ U . Letf(z) = (z − a)/(1− az). This map is the Riemann map of the unit disc


onto the unit disc mapping a to the origin. Hence f ′(z) = C KU (z, a).It follows that KU (z, a) = c(1− az)−2 where c = (1− |a|2)/C. But sinceKU (z, 0) =

1π , we deduce that c = 1/π and therefore that

KU (z, w) =1

π(1− wz)2.

We now assume that Ω is a bounded domain in the plane with C∞

smooth boundary. In this setting we wish to relate the Bergman projec-tion to the Dirichlet problem, and hence, to the Szego projection. To dothis, we require the following lemma.

Lemma 15.1. Suppose that Ω is a bounded domain with C∞ smoothboundary. If ϕ is a function in C∞(Ω) that vanishes on bΩ, then ∂ϕ/∂zis orthogonal to H2(Ω).

Proof. Suppose ϕ is as in the statement of the lemma. If h is in A∞(Ω),then a simple application of the complex Green’s identity reveals that

〈h, ∂ϕ/∂z〉Ω =

∫∫

Ω

∂

∂z(hϕ)

(− i

2dz ∧ dz

)= − i

2

∫

bΩ

h ϕ dz = 0.

Since we do not yet know that A∞(Ω) is dense in H2(Ω), we cannot usea density argument to deduce the lemma from this simple computation.Instead, we must resort to the following machinations.

Consider a finite covering of Ω by small discs and suppose that χjis a C∞ partition of unity that is subordinate to the cover. If we canprove that ∂(χjϕ)/∂z is orthogonal to H2(Ω) for each j, then it followsthat ∂ϕ/∂z =

∑∂(χjϕ)/∂z is also orthogonal to H2(Ω). Hence, we

may reduce our problem to proving the lemma for ϕ ∈ C∞(Ω) that issupported on a small set of the form Dǫ(z0) ∩ Ω where z0 is a fixedpoint in bΩ. Of course, we continue to assume that ϕ = 0 on bΩ. Letξ0 = iT (z0), i.e., let ξ0 denote the complex number that represents theinward pointing unit normal vector to bΩ at z0. If ǫ > 0 is sufficientlysmall, the translation of the region Dǫ(z0) ∩ Ω by a small distance δ inthe ξ0 direction will be compactly contained in Ω. Given h ∈ H2(Ω), forz ∈ Ω, define hδ(z) = h(z + δξ0) if z + δξ0 is in Ω and define hδ(z) = 0otherwise. It is a standard fact in measure theory that hδ tends to h inL2(Ω) as δ → 0. If h ∈ H2(Ω), then

〈h, ∂ϕ/∂z〉Ω = limδ→0

〈hδ, ∂ϕ/∂z〉Dǫ(z0)∩Ω

and this last quantity is zero because hδ is holomorphic on a neighbor-hood of the support of ϕ and we may use the complex Green’s identityon Dǫ(z0) ∩ Ω as we did in the simple case above.


We need to define an operator related to the Dirichlet problem. Theclassical Green’s operator G is the solution operator to the followingproblem. Given v ∈ C∞(Ω), then Gv is equal to the function u satisfying

∆u = v on Ω

u = 0 on bΩ. (15.2)

Theorem 15.1. Suppose that Ω is a bounded domain with C∞ smoothboundary. Then the classical Green’s operator for Ω is well defined andmaps C∞(Ω) into itself.

Proof. By Theorem 2.2, we can solve the equation ∂u/∂z = v in theC∞(Ω) category, and, since ∂u/∂z = ∂u/∂z, we can also solve the equa-tion ∂u/∂z = v in the C∞(Ω) category. Let u1 be a function in C∞(Ω)that satisfies ∂u1/∂z =

14v and let u2 be a function in C∞(Ω) that satis-

fies ∂u2/∂z = u1. Because (∂/∂z)(∂/∂z) =14∆, it follows that ∆u2 = v.

Finally, let u3 be the harmonic function on Ω that has u2 as its bound-ary values (u3 is in C∞(Ω) by Theorem 14.2). Now u = u2 − u3 solvesproblem (15.2).

This solution u is the unique solution to problem (15.2) in the follow-ing strong sense. Assume that U is a function that is merely continuouson Ω and C∞ smooth on Ω. Assume further that ∆U = v on Ω andU = 0 on bΩ. Then u − U is a harmonic function on Ω that is contin-uous on Ω and that vanishes on bΩ, and so the maximum principle forharmonic functions implies that u−U ≡ 0. Hence, the Green’s operatoris well defined and the proof is finished.

The next theorem shows how the Bergman projection is related tothe Dirichlet problem via the Green’s operator.

Theorem 15.2. Suppose that Ω is a bounded domain with C∞ smoothboundary. For u ∈ C∞(Ω), the Bergman projection Bu of u is given by

Bu = u− 4∂

∂zG∂u

∂z.

It follows that B maps C∞(Ω) into itself. It also follows that the Bergmankernel Ka(z) = K(z, a) is in A∞(Ω) as a function of z for each fixeda ∈ Ω.

Proof. Let ϕ = 4G∂u∂z . Since ϕ vanishes on bΩ, it follows that (∂ϕ/∂z)

is orthogonal to H2(Ω). Hence B(∂ϕ/∂z) = 0 and it follows that Bu =B(u− ∂

∂zϕ). But u− ∂

∂zϕ is holomorphic because the computation,

∂

∂z

(∂

∂z4G

∂u

∂z

)=

1

4∆

(4G

∂u

∂z

)=∂u

∂z,


shows that ∂∂z (u− ∂

∂zϕ) is zero, i.e., that u− ∂∂zϕ satisfies the Cauchy-

Riemann equations. Hence, Bu = B(u − ∂∂zϕ) = u − ∂

∂zϕ, and this iswhat we wanted to see.

To finish the proof, we need to express the Bergman kernel as theBergman projection of a smooth function. The easiest way to do this isto let ϕa be a function in C∞

0 (Ω) that is radially symmetric about thepoint a such that

∫∫Ωϕa dA = 1. Because the value of a holomorphic

function h ∈ H2(Ω) is equal to (2π)−1∫ 2π

0 h(a + reiθ) dθ when r is lessthan the distance from a to bΩ, it can be verified by integrating inpolar coordinates centered at a that 〈h, ϕa〉Ω = h(a) for all h ∈ H2(Ω).Hence, Ka = Bϕa and the smoothness of Ka follows from the fact thatthe Bergman projection preserves C∞(Ω) because the Green’s operatordoes.

The fact that B preserves the space C∞(Ω) has an important corol-lary.

Corollary 15.1. If Ω is a bounded domain with C∞ smooth boundary,then A∞(Ω) is dense in H2(Ω).

Proof. Given h ∈ H2(Ω), let uj be a sequence in C∞(Ω) that convergesto h in L2(Ω). Now Buj is a sequence in A∞(Ω) that converges in L2(Ω)to Bh = h.

The formula in Theorem 15.2 is called Spencer’s formula. It is worthremarking that the proof of Spencer’s formula contained a theorem aboutsolving a ∂-problem. The problem is, given v ∈ C∞(Ω), find u ∈ C∞(Ω)such that ∂u/∂z = v with u orthogonal to the Bergman space. Scrutinyof the proof of Spencer’s formula reveals that a solution to this problemis given by u = 4 ∂

∂zGv. Uniqueness is easy because if u1 and u2 bothsolve the problem, then u1 − u2 would be a holomorphic function thatis orthogonal to holomorphic functions, and hence u1 − u2 ≡ 0.

It is useful to know that A∞(Ω) is dense in the Bergman space. Next,we seek a nice dense subspace of the orthogonal complement H2(Ω)⊥

of H2(Ω) in L2(Ω). The next few results do not require us to assumethat the boundary of our domain is smooth, and therefore we drop thisassumption. We will only assume that the domain Ω under study isbounded. A function g in L2(Ω) is called a weak solution to the Cauchy-Riemann equations (or a distributional solution to the Cauchy-Riemannequations) if ∫∫

Ω

g∂ϕ

∂zdA = 0

for every ϕ ∈ C∞0 (Ω). Since ∂ϕ/∂z = ∂ϕ/∂z, an equivalent way of


stating this condition is to say

〈g, ∂ϕ∂z

〉Ω = 0

for every ϕ ∈ C∞0 (Ω). To make sense of this definition, consider what it

implies about a function g with continuous first partial derivatives. Inthis setting, we may integrate by parts via the complex Green’s identityto obtain ∫∫

Ω

g∂ϕ

∂zdA = −

∫∫

Ω

∂g

∂zϕ dA.

If this integral vanishes for all smooth compactly supported ϕ, then ∂g∂z

must be zero and therefore g satisfies the Cauchy-Riemann equations.Hence, a weak solution to the Cauchy-Riemann equations that has con-tinuous first partial derivatives is a strong solution.

We will need the following classical result, known as Weyl’s lemma,that says that a locally integrable function that is a weak solution tothe Cauchy-Riemann equations is holomorphic (in the sense that thereexists a holomorphic function with which it agrees almost everywhere).

Lemma 15.2. Suppose Ω is a bounded domain and suppose g ∈ L2(Ω)is a weak solution to the Cauchy-Riemann equations. Then g ∈ H2(Ω).Consequently, the set of functions F of the form ∂ϕ

∂z , where ϕ ∈ C∞0 (Ω),

is a dense subspace of H2(Ω)⊥.

Proof. Let θ denote a function in C∞0 (D1(0)) that is radially symmetric

such that∫∫

D1(0)θ dA = 1, and let θǫ(z) = ǫ−2θ(z/ǫ). Consider the

convolution gǫ = θǫ ∗ g which is well defined on the set Ωǫ of points in Ωthat are greater than a distance of ǫ from the boundary. First, we showthat gǫ is holomorphic on Ωǫ. Indeed, it is a standard exercise in realanalysis to see that gǫ is C

∞ on Ωǫ. Furthermore, on Ωǫ,

∂

∂zgǫ =

(∂θǫ∂z

)∗ g =

∫∫

w∈Ω

∂

∂z[θǫ(z − w)] g(w)dA.

But ∂∂z [θǫ(z − w)] = − ∂

∂w [θǫ(z − w)]. Therefore, the last integral aboveis equal to

−∫∫

w∈Ω

∂

∂w[θǫ(z − w)] g(w)dA,

and this quantity is zero by the definition of weak solutions to theCauchy-Riemann equations. Hence gǫ is holomorphic on Ωǫ. Let ǫj be asequence of positive numbers that tend to zero and let gj = gǫj on Ωǫj

and gj = 0 on Ω − Ωǫj . It is another standard exercise in real analysisto see that gj converges in L2(Ω) to g. As in the proof that H2(Ω) is a


closed subspace of L2(Ω), we see that g is locally the L2 limit of holomor-phic functions and that, therefore, g must be equal almost everywhereto a function holomorphic on Ω.

Finally, we must check the density statement. If F is not dense inH2(Ω)⊥, then there would exist a nonzero element g of H2(Ω)⊥ that isorthogonal to F . But this orthogonality condition is equivalent to sayingthat g is a weak solution to the Cauchy-Riemann equations. Thus, bythe lemma, g ∈ H2(Ω), and we must conclude that g ≡ 0, contrary tohypothesis. The proof is finished.

16

Proper holomorphic mappings and theBergman projection

We showed in the last chapter that the Bergman kernel is related toconformal mappings. We now wish to study a more general class ofholomorphic mappings between domains than conformal mappings. Wewill show that the Bergman projection and kernel are also useful in thestudy of these mappings.

A continuous mapping is called proper if the inverse image of anycompact set is compact. It is easy to check that biholomorphic mappingsare proper. Finite Blaschke products are examples of proper holomor-phic maps of the unit disc into itself, and it is a standard exercise incomplex analysis to show that these maps constitute all possible properholomorphic self maps of the unit disc. The Ahlfors map studied in Chap-ter 13 is another example of a proper holomorphic mapping. Indeed, iff : Ω1 → Ω2 is a holomorphic mapping between bounded domains thatextends continuously up to the boundary, then it is an easy exerciseto check that the condition that f be proper is equivalent to the condi-tion that f(bΩ1) ⊂ bΩ2. Proper holomorphic mappings between domainsshare many of the nice qualities held by conformal mappings.

Suppose f : Ω1 → Ω2 is a proper holomorphic map between boundeddomains. We wish to define the operator Λ1 for f as we did for bi-holomorphic maps in Chapter 15. Given a function φ on Ω2, we defineΛ1φ = f ′(φ f), which is defined on Ω1. We wish to show that Λ1 mapsL2(Ω2) into L

2(Ω1). As in the biholomorphic case, a key element of theproof will be that |f ′|2 is equal to the real Jacobian determinant of f ,viewed as a mapping of R2 into itself. However, before we can use theclassical change of variables formula, we will need to know some elemen-tary properties of proper holomorphic maps. We need to know that fmaps Ω1 onto Ω2, and that there is a positive integer m, known as themultiplicity of f , such that (counting multiplicities), f is an m-to-onemap of Ω1 onto Ω2.

Since clearly f cannot be constant, the open mapping theorem saysthat f(Ω1) is an open subset of Ω2. We claim that f(Ω1) is also a rel-atively closed subset of Ω2, and that therefore, since Ω2 is connected,

77


f(Ω1) = Ω2. That f(Ω1) is closed in Ω2 follows from the definitionof a proper map. Indeed, if wi is a sequence in f(Ω1) converging to apoint w0 ∈ Ω2, then the set K = ∪∞

i=0wi is compact. Hence, f−1(K)is compact. We may choose a sequence zi of points in Ω1 such thatf(zi) = wi. Since f

−1(K) is compact, there is a subsequence of zi con-verging to a point z0 ∈ f−1(K) ⊂ Ω1. Now the continuity of f impliesthat f(z0) = w0, and hence, that w0 ∈ f(Ω1).

We have shown that f maps Ω1 onto Ω2. Let w0 ∈ Ω2. Since f isnot constant, the set f−1(w0) is discrete in Ω1. Since f is proper, thisset is also compact, and it must therefore be a finite set in Ω1. Writef−1(w0) = a1, a2, . . . , an where the ai are distinct and let mi denotethe multiplicity of the zero of f(z) − w0 at z = ai. Let m =

∑ni=1mi.

We now claim that this number does not depend on the choice of w0. Tosee this, we use the fact proved in most elementary books on complexvariables (see Ahlfors [Ah, p. 132]) that, for sufficiently small ǫ > 0,there are neighborhoods Ui of each ai such that f : Ui → Dǫ(w0) is anmi-to-one covering map of Ui − ai onto Dǫ(w0) − w0. In fact, themap is given as a composition of the map w0 + (z − ai)

mi with a one-to-one holomorphic map. By shrinking ǫ if necessary, we may assumethat the Ui are disjoint. We will now prove there is a ρ < ǫ such that if|w − w0| < ρ, then f−1(w) ⊂ ∪n

i=1Ui. If this were not true, there wouldexist a sequence wi tending to w0 and points zi ∈ Ω1 − (∪n

i=1Ui) suchthat f(zi) = wi. By the same argument we used to show that f(Ω1) isclosed in Ω2, it would follow that there is a convergent subsequence of ziconverging to a point z0 ∈ Ω1. By continuity, f(z0) = w0, and therefore,z0 = ak for some k. But this forces us to conclude that some of the zilie in Uk, contrary to assumption. Hence, the existence of ρ is assured,and it follows that if |w − w0| < ρ and w 6= w0, then f

−1(w) consists ofexactly m distinct points, mi of them falling in Ui. Hence, the numberm associated to w is the same as that associated to w0, and we haveshown that m(w) is a locally constant function of w on Ω2. A locallyconstant function on a connected set is constant.

In fact, the proof above gives more than the existence of m. Let Ldenote the set of points z ∈ Ω1 such that f ′(z) = 0. This set is knownas the branch locus of f and the set f(L) is called the image of thebranch locus. The proof above shows that f(L) is a discrete subset ofΩ2 and that for each w not in f(L), the set f−1(w) consists of exactlym distinct points. Furthermore, if w ∈ f(L), the set f−1(w) consists ofstrictly fewer than m points. Let V2 = f(L) and let V1 = f−1(V2). Thesets Vi are discrete subsets of Ωi, i = 1, 2, and we have shown that f isan m-to-one (unbranched) covering map of Ω1 − V1 onto Ω2 − V2. Notethat discrete subsets are sets of measure zero with respect to Lebesguearea measure. Thus, integrals over Ωi are equal to integrals over Ωi−Vi.

Proper holomorphic mappings and the Bergman projection 79

We are now prepared to study the operator Λ1. Let ϕ ∈ C∞0 (Ω2−V2).

Since V2 is discrete, it is easy to check that such functions are dense inL2(Ω2). Because f is an m-to-one unbranched covering map of Ω1 − V1onto Ω2 − V2, we may apply the classical change of variables formula towrite

∫∫

Ω1

|f ′|2|φ f |2 dA =

∫∫

Ω1−V1

|f ′|2|φ f |2 dA

= m

∫∫

Ω2−V2

|φ|2 dA = m

∫∫

Ω2

|φ|2 dA.

This formula is easy to understand when ϕ is supported in a very smalldisc. Hence, we can verify the formula by using a partition of unity χjand by summing the integrals

∫∫Ω2χj |ϕ|2 dA. Our calculation shows that

‖Λ1ϕ‖2 = m‖ϕ‖2. By density, Λ1 extends uniquely to all of L2(Ω2) as abounded operator to L2(Ω1) with operator norm

√m. Since convergence

in L2 implies almost everywhere convergence of a subsequence, it followsthat the extension of Λ1 to L2(Ω2) can be expressed via the formulaΛ1φ = f ′(φ f), and so our original definition of Λ1 agrees with theoperator constructed by extension.

Let a subscript i on a Bergman projection indicate that it is associ-ated to the domain Ωi. Lemma 15.2 will allow us to give a simple proof ofthe following transformation formula for the Bergman projections underproper holomorphic mappings.

Theorem 16.1. Suppose f : Ω1 → Ω2 is a proper holomorphic mapbetween bounded domains. Then, the operator Λ1 commutes with theBergman projection in the sense that B1Λ1 = Λ1B2, i.e.,

B1 (f′(ϕ f)) = f ′ ((B2ϕ) f)

for all ϕ ∈ L2(Ω2).

Proof. The formula is clearly true when ϕ = h ∈ H2(Ω2). In this case,B2h = h. Now Λ1h is holomorphic and in the Bergman space of Ω1

because Λ1 is a bounded operator. Hence, B1Λ1h = Λ1h. Thus B1Λ1h =Λ1h = Λ1B2h.

Because the formula is true on H2(Ω2), and because Λ1 and theBergman projections are bounded linear operators, we may reduce ourtask to showing that the transformation formula holds for ϕ in F2, thedense subspace of H2(Ω2)

⊥ mentioned in Lemma 15.2. If ψ has compactsupport in Ω2, then ψ f has compact support in Ω1. Furthermore, thecomplex chain rule yields (∂/∂z)(ψ f) = f ′[(∂ψ/∂z) f)]. Hence, ifϕ ∈ F2, then Λ1ϕ ∈ F1, the dense subspace of H

2(Ω1)⊥ of Lemma 15.2.

Therefore, for such a ϕ, it follows that B2ϕ = 0, and B1Λ1ϕ = 0, and so


B1Λ1ϕ = Λ1B2ϕ because both of these expressions are zero. The proofof the transformation formula is complete.

Next, we prove a general theorem about the boundary behavior ofproper maps using the transformation formula for the Bergman projec-tions.

Theorem 16.2. Suppose f : Ω1 → Ω2 is a proper holomorphic map be-tween bounded domains with C∞ smooth boundaries. Then f ∈ C∞(Ω1)and f ′ is nonvanishing on bΩ1. It follows that f maps the boundary ofΩ1 into the boundary of Ω2.

Proof. Because of Lemma 12.1, we may assume that Ω1 and Ω2 havereal analytic boundaries. Lemma 11.1 implies that there is a functionΦ ∈ C∞(Ω2) such that Φ = 0 on bΩ2 and such that 1 − (∂Φ/∂z) hascompact support in Ω2. Remember that the conjugate of ∂u/∂z is ∂u/∂z.Since Φ vanishes on bΩ2, ∂Φ/∂z is orthogonal to holomorphic functionson Ω2 by Lemma 15.1. Now, the function ψ given by

ψ = 1− ∂Φ

∂z

is such that B2ψ ≡ 1 and ψ ∈ C∞0 (Ω2). Let 1 denote the function that

is identically one on Ω2. The transformation formula for the Bergmanprojections yields that

f ′ = f ′(1 f) = f ′ ((B2ψ) f) = B1 (f′(ψ f)) .

Since f is proper and ψ ∈ C∞0 (Ω2), it follows that f

′(ψf) is in C∞0 (Ω1).

Hence, f ′ is equal to the projection of a function in C∞(Ω1), and by The-orem 15.2, f ′ ∈ C∞(Ω1); so f ∈ C∞(Ω1). We have mentioned before thata proper map that extends continuously to the boundary must map theboundary to the boundary; so f(bΩ1) ⊂ bΩ2. Now the Schwarz ReflectionPrinciple implies that f extends to be holomorphic in a neighborhoodof Ω1.

Finally, we must show that f ′ is nonvanishing on bΩ1. Let z0 ∈ bΩ1.We may make local holomorphic changes of variables near z0 and nearf(z0) so that we are dealing with a map F that is holomorphic nearthe origin that maps the real line into itself, and such that F (0) = 0.We may further assume that F maps the upper half plane near theorigin into the upper half plane. In the changed variables, F can bewritten F (z) = zmh(z) where h is holomorphic near zero, h(0) 6= 0,and m ≥ 1. Note that, because F maps R into R, h(z) is real valuedon the real axis. Since F maps the upper half plane into itself nearthe origin, it follows that h(0) > 0. Let H(z) be a holomorphic m-th root of h(z) such that H(0) is a positive real number. Note that


zH(z) is one-to-one near the origin and that zH(z) maps R into R.Furthermore, zH(z) maps the upper half plane near the origin into theupper half plane because H(0) > 0. Hence, there is a small disc Dǫ(0)such that z : Im z > 0 ∩Dǫ(0) is contained in the image under zH(z)of a neighborhood of the origin. Now, if m > 1, it is easy to see thatF (z) = [zH(z)]m maps the upper half plane near the origin onto a fullneighborhood of the origin. This contradiction forces us to conclude thatm = 1, and consequently f ′(z0) 6= 0. The proof is complete.

Alternatively, we could have used a Harnack inequality argument likethe one used in the proof of Theorem 8.2 to prove that f ′(z0) 6= 0.

We now wish to show that the Bergman kernel transforms underproper holomorphic mappings. Suppose that f : Ω1 → Ω2 is a properholomorphic mapping between bounded domains. Again, we drop anyassumptions about the smoothness of the boundaries of Ωi. Let Ki(z, w)denote the Bergman kernel function associated to Ωi, i = 1, 2. We haveshown that there are discrete sets Vi ⊂ Ωi, i = 1, 2, such that f is anm-to-one covering map of Ω1 − V1 onto Ω2 − V2. If Dǫ(w0) ⊂ Ω2 − V2,then there are holomorphic functions Fk(w), k = 1, 2, . . . ,m, mappingDǫ(w0) into Ω1 − V1 such that f(Fk(w)) = w there. The functions Fk

are called the local inverses to f .We need to construct an operator like the Λ2 operator we used in the

discussion of biholomorphic maps. Given ϕ ∈ C∞(Ω1), we define Λ2ϕon Ω2 − V2 by setting

Λ2ϕ =

m∑

k=1

F ′k(ϕ Fk)

on small discs contained in Ω2 − V2 as above. To see that this gives riseto a well defined C∞ function on Ω2 − V2, notice that on a small disc,the definition is clearly C∞ and is independent of the order in whichwe choose to label the local inverses. Furthermore, the functions definedon two small discs agree on the intersection of the discs if they overlap.Hence, Λ2ϕ is a globally well defined C∞ function on Ω2−V2. We wish todefine this operator on L2(Ω1). Suppose that ϕ ∈ C∞(Ω1). To see thatΛ2ϕ is square integrable, we will use the simple inequality |∑m

k=1 ck|2 ≤m∑m

k=1 |ck|2 which can be deduced by applying the Schwarz inequalityfor the inner product of two vectors in Cm to the vectors (c1, . . . , cm)and (1, . . . , 1). Now

∫

Ω2−V2

∣∣∣∣∣

m∑

k=1

F ′k(ϕ Fk)

∣∣∣∣∣

2

dA ≤ m

∫

Ω2−V2

m∑

k=1

|F ′k(ϕ Fk)|2 dA,

and, using the facts that |F ′k|2 represents a real Jacobian and that f is


an m-sheeted covering map of Ω1 − V1 onto Ω2 − V2, it follows that thelast quantity is equal to

m

∫

Ω1−V1

|ϕ|2 dA.

As before, the calculation above is more transparent if ϕ is assumed tohave very small support in Ω1−V1. The general case can be deduced byusing a partition of unity. Since V2 is a set of measure zero in Ω2, it followsthat Λ2ϕ can be viewed as an element of L2(Ω2), and our calculationshows that ‖Λ2ϕ‖L2(Ω2) ≤

√m‖ϕ‖L2(Ω1). Hence, since C

∞(Ω1) is densein L2(Ω1), it follows from this estimate that Λ2 has a unique extensionas a bounded operator from L2(Ω1) to L

2(Ω2).We next claim that Λ2 mapsH2(Ω1) intoH

2(Ω2). Given h ∈ H2(Ω1),we know that Λ2h is holomorphic on Ω2 − V2 and that Λ2h is squareintegrable over Ω2 − V2. The following generalization of the Riemannremovable singularity theorem will show that Λ2h ∈ H2(Ω2).

Theorem 16.3. Suppose that h is holomorphic on Ω− V where Ω is abounded domain and V is a discrete subset of Ω. If

∫Ω−V |h|2 dA < ∞,

then h has removable singularities at each point in V .

Proof. The theorem is purely local, so we might as well assume thatΩ is the unit disc and that V = 0. The monomials zn∞n=−∞ areorthogonal on each annulus Aρ

ǫ = z : ǫ < |z| < ρ where 0 < ǫ < ρ < 1.Furthermore, h has a Laurent expansion

∑∞n=−∞ anz

n that convergesuniformly on each of these annuli. Hence

∫∫

Aρǫ

|h|2 dA =

∞∑

n=−∞

|an|2∫∫

Aρǫ

|zn|2 dA

is bounded independent of ǫ. But a simple computation in polar coordi-nates shows that

∫∫Aρ

ǫ|zn|2 dA tends to infinity as ǫ tends to zero when

n < 0. Hence, we must conclude that an = 0 if n < 0, i.e., that h has aremovable singularity at the origin. The proof is finished.

Next, we show that Λ1 and Λ2 are adjoint operators.

Theorem 16.4. If f : Ω1 → Ω2 is a proper holomorphic mapping be-tween bounded domains, then

〈Λ1u, v〉Ω1= 〈u,Λ2v〉Ω2

for all u ∈ L2(Ω2) and v ∈ L2(Ω1).


Proof. This is easy to prove if u is supported in a small disc Dǫ(w0) ⊂Ω2 − V2 such that the local inverse maps Fk(w) define biholomorphicmaps of Dǫ(w0) onto disjoint domains Wk in Ω1 − V1. In this case, wecan use the analogous result for biholomorphic maps to see that

〈Λ1u, v〉Ω1=

∫∫

Ω1

f ′(u f) v dA =

m∑

k=1

∫∫

Wk

f ′(u f) v dA

=

m∑

k=1

∫∫

Dǫ(w0)

u F ′k(v Fk) dA =

∫∫

Dǫ(w0)

u Λ2v dA

= 〈u,Λ2v〉Ω2.

The theorem now follows because the linear span of the set of functionsin L2(Ω2) that are supported in small discs of this type forms a densesubset of L2(Ω2), and a standard limiting argument can be used.

Theorem 16.4 will allow us to deduce the transformation formula forthe Bergman kernel functions under proper holomorphic mappings.

Theorem 16.5. If f : Ω1 → Ω2 is a proper holomorphic mapping be-tween bounded domains, then the Bergman kernels Kj(z, w) associatedto Ωj, j = 1, 2, transform according to

f ′(z)K2(f(z), w) =

m∑

k=1

K1(z, Fk(w))F ′k(w)

where the multiplicity of the mapping is m and the functions Fk(w)denote the local inverses to f .

Notice that when f is a biholomorphic mapping, the transformationrule in the statement of the theorem reduces to

f ′(z)K2(f(z), w) = K1(z, F (w))F ′(w)

where F = f−1. We may let w = f(ζ) in this formula and use the factthat f ′(ζ) = 1/F ′(f(ζ)) to write the last formula in the equivalent form

f ′(z)K2(f(z), f(ζ))f ′(ζ) = K1(z, ζ). (16.1)

This result is known as the transformation formula for the Bergmankernel under biholomorphic mappings, and Theorem 16.5 is its general-ization to the case of proper holomorphic mappings.

Proof of Theorem 16.5. Fix a point w in Ω2 − V2 and let g1(z) =K2(z, w). Notice that if h ∈ H2(Ω1), then

〈h,Λ1g1〉Ω1= 〈Λ2h, g1〉Ω2

= (Λ2h)(w)


because of the reproducing property of the Bergman kernel on Ω2. Next,define G2(z) =

∑mk=1K1(z, Fk(w))F ′

k(w). Now

〈h,G2〉Ω1=

m∑

k=1

F ′k(w)〈h,K1(·, Fk(w))〉Ω1

=

m∑

k=1

F ′k(w)h(Fk(w)) = (Λ2h)(w)

by the reproducing property of the Bergman kernel on Ω1. We haveshown that 〈h,Λ1g1〉Ω1

= 〈h,G2〉Ω1for any h ∈ H2(Ω1). It follows that

Λ1g1 = G2, and this is precisely the transformation formula we areseeking to prove. Hence, the formula holds when w ∈ Ω2 − V2. Theformula is true for w in the discrete set V2 in the sense that those pointsare removable singularities of the anti-holomorphic function on the righthand side of the formula, and therefore, when the function on the righthand side is given the proper value at these points, the transformationformula remains valid by continuity.

In defining the operator Λ2, we showed that the operator norm of Λ2

was less than or equal to√m, where m is the multiplicity of the proper

map f . Actually, the operator norm of Λ2 is equal to√m, which is the

same as the operator norm of Λ1. We will now prove this. Observe that,since f(Fk(w)) = w, it follows that F ′

k(w)f′(Fk(w)) = 1, and therefore,

that Λ2Λ1u = mu for all u in L2(Ω2). This fact allows us to deducethat the range R of Λ1 is closed. Indeed, if Λ1ui tends to U , then ui =1mΛ2Λ1ui tends to

1mΛ2U , and this shows that U is equal to Λ1(

1mΛ2U).

Since R is closed, L2(Ω1) has an orthogonal decomposition as R⊕R⊥.The adjoint law shows that R⊥ is equal to the kernel of the operator Λ2.Hence, given U ∈ L2(Ω1), there exists a u ∈ L2(Ω2) and a V ∈ L2(Ω1)with Λ2V ≡ 0 such that U = f ′(uf)+V . This decomposition allows usto determine the operator norm of Λ2. Indeed, the decomposition showsthat Λ2U = Λ2Λ1u = mu. But ‖U‖2 = ‖Λ1u‖2+‖V ‖2 and we know fromabove that ‖Λ1u‖ =

√m‖u‖. Hence, ‖Λ2U‖ = m‖u‖ =

√m‖Λ1u‖ ≤√

m‖U‖. It is clear that equality holds in this last inequality wheneverV = 0, and it follows that the operator norm of Λ2 is

√m.

The same reasoning shows that the image of H2(Ω2) under Λ1 is aclosed subspace of H2(Ω1), and that therefore any G ∈ H2(Ω1) can bedecomposed as G = f ′(g f) +H where g ∈ H2(Ω2) and H ∈ H2(Ω1)with Λ2H ≡ 0. Furthermore, the operator norm of Λ2 as an operatorfrom H2(Ω1) to H

2(Ω2) is equal to√m.

In case f is a biholomorphic map, the operator Λ1 is an isometryand Λ2 is its inverse. In case f is a proper holomorphic mapping ofmultiplicity m > 1, the operator m−1/2Λ1 is isometric and m−1/2Λ2 is


an isometric right inverse. If f does not have a holomorphic inverse, Λ1

does not map onto H2(Ω1).

17

The Solid Cauchy transform

We now wish to express the Bergman projection in terms of the Szegoprojection. Suppose Ω is a bounded domain with C∞ smooth bound-ary and suppose v ∈ C∞(Ω). We want to compute Bv, the Bergmanprojection of v. Define

(Λv)(z) =1

2πi

∫∫

w∈Ω

v(w)

w − zdw ∧ dw

for z ∈ Ω. The operator Λ maps C∞(Ω) into itself and (∂Λv/∂z) = v onΩ (see the remark after Theorem 2.2). Let E denote the Poisson extensionoperator mapping a function u ∈ C∞(Ω) to the harmonic function on Ωthat has the same boundary values as u. In this book, we have expressedE in terms of the Szego projection (see Theorem 10.1 and Chapter 14).

We now claim that

Bv =∂

∂z(EΛv). (17.1)

To see this, suppose that h ∈ A∞(Ω). Writing u = Λv, we may computeas follows.

∫∫

Ω

v(z) h(z) dx ∧ dy = − 1

2i

∫∫

Ω

∂u

∂zh dz ∧ dz

= − 1

2i

∫

bΩ

u h dz = − 1

2i

∫

bΩ

(Eu) h dz

= − 1

2i

∫∫

Ω

∂Eu∂z

h dz ∧ dz

=

∫∫

Ω

G(z) h(z) dx ∧ dy

where G = (∂/∂z)(EΛv) is in A∞(Ω). We have shown that

〈v, h〉Ω = 〈G, h〉Ωfor all h ∈ A∞(Ω). But A∞(Ω) is dense in the Bergman space (Corol-lary 15.1). Hence, it follows that this inner product identity holds for allh in the Bergman space and it follows that G = Bv. The proof of theidentity is complete.

87


If we combine formula (17.1) with Theorem 14.3, we obtain the fol-lowing result, that we will need in Chapter 22.

Theorem 17.1. The Bergman projection associated to a bounded do-main with smooth real analytic boundary maps compactly supported func-tions in L2 to holomorphic functions that extend to be holomorphic ona neighborhood of the closure.

Indeed, the operator Λ maps compactly supported functions to func-tions that are holomorphic in a neighborhood of the boundary, then Emaps such functions to functions that are are harmonic on a neighbor-hood of the closure of the domain, and finally, ∂/∂z of the end resultproduces a function holomorphic on a neighborhood of the closure.

In the simply connected case, the holomorphic part of Eu is explicitvia Theorem 10.1 and we obtain

Bv =∂

∂z

(P (SaΛv)

Sa

).

In the n-connected case, the formulas developed in Chapter 14 showthat linear combinations of the functions log |z − bi| enter the picture.The formulas are uglier, but not too difficult to write out. We omit thedetails.

The Cauchy transform was very important in the study of the Hardyspace where it served to expose properties of the Szego projection. Therelationship between the operator Λ and the Bergman projection leadsus to define a kind of Cauchy transform that is relevant to the Bergmanspace and projection. The improved Cauchy integral formula in Theo-rem 2.1 decomposes a function into two pieces. The first piece, that isan integral over the boundary, gives rise to the Cauchy transform. Wenow define an operator that comes from the second piece, the integralover the interior of Ω in that formula. If U ∈ L2(Ω), we define the solidCauchy transform CsU of U to be the antiholomorphic function on thecomplement of Ω given by

(CsU)(z) =1

2πi

∫∫

ζ∈Ω

U(ζ)

ζ − zdζ ∧ dζ for z 6∈ Ω.

In Chapter 3, we saw that the Cauchy integral had well defined boundaryvalues. A similar result is true for the solid Cauchy transform.

Theorem 17.2. If Ω is a bounded domain with C∞ smooth boundary,then the solid Cauchy transform maps C∞(Ω) into the space of antiholo-morphic functions on C− Ω that vanish at infinity and that extend C∞

smoothly up to the boundary of Ω.

The Solid Cauchy transform 89

Proof. It is clear that the solid Cauchy transform of a bounded functionvanishes at infinity. The hard part of the theorem is the smoothnessassertion.

Given U ∈ C∞(Ω), Lemma 2.1 implies that there is a function ϕ ∈C∞(Ω) that vanishes on bΩ such that U −∂ϕ/∂z vanishes to order m onthe boundary. Let Ψ = U−∂ϕ/∂z. The complex Green’s identity revealsthat Cs(∂ϕ/∂z) ≡ 0, and therefore, it follows that CsU = CsΨ. Now itis a simple matter to extend Ψ to all of C by setting it equal to zerooutside of Ω, to change variables, and to differentiate under the integralas we did way back in the proof of Theorem 2.2, to see that CsU extendsto be m times differentiable on C. Since m can be taken to be arbitrarilylarge, it follows that CsU extends C∞ smoothly to the boundary.

A byproduct of the proof above is that the boundary values of CsUagree with the boundary values of the function ΛU defined above. Con-sequently, it follows that the Bergman projection can be expressed viaBv = (∂/∂z)(ECsv). To better understand this identity, we will examinethe solid Cauchy transform in more detail.

The solid Cauchy transform and the classical Cauchy transform areadjoint operators in a sense that we will now explain. We will prove thatif U ∈ C∞(Ω) and v ∈ C∞(bΩ), then

〈CsU, v〉bΩ = 〈U, 2iC(vT )〉Ω.

Indeed, using the same notation that we used in the proof of Theo-rem 17.2, we may write CsU = CsΨ and

〈CsU, v〉bΩ =

∫

z∈bΩ

1

2πi

(∫∫

ζ∈Ω

Ψ(ζ)

ζ − zdζ ∧ dζ

)v(z) ds.

Since Ψ can be assumed to vanish to high order on the boundary, wemay use Fubini’s theorem as we did in the proof of the adjoint identity(3.2) for the Cauchy transform to see that the last integral is equal to

∫∫

ζ∈Ω

Ψ(ζ)

(1

2πi

∫

z∈bΩ

v(z)

ζ − zds

)dζ ∧ dζ = 〈Ψ, 2iC(vT )〉Ω.

Finally, since Ψ = U − ∂ϕ/∂z and since, by Lemma 15.1, functions ofthe form ∂ϕ/∂z with ϕ vanishing on the boundary are orthogonal toH2(Ω), we may reduce this last inner product to 〈U, 2iC(vT )〉Ω, and wehave proved that 〈CsU, v〉bΩ = 〈U, 2iC(vT )〉Ω.

This adjoint property will allow us to extend the solid Cauchy trans-form to be an operator from L2(Ω) to L2(bΩ). Recall that Theorem 6.5


implies that there is a constant C such that ‖C(vT )‖Ω ≤ C‖vT‖bΩ =C‖v‖bΩ. Hence, if U ∈ C∞(Ω) and v ∈ C∞(bΩ), then

|〈CsU, v〉bΩ| = |〈U, 2iC(vT )〉Ω| ≤ 2C‖U‖Ω‖v‖bΩ.

Since C∞(bΩ) is dense in L2(bΩ), it follows that

‖CsU‖bΩ = sup|〈CsU, v〉bΩ| : v ∈ C∞(bΩ) and ‖v‖bΩ = 1,

and therefore, we see that ‖CsU‖bΩ ≤ 2C‖U‖Ω. Now, since C∞(Ω) isdense in L2(Ω), this estimate shows that Cs extends uniquely as an op-erator from L2(Ω) to L2(bΩ). Let us state this as a theorem.

Theorem 17.3. If Ω is a bounded domain with C∞ smooth boundary,then the solid Cauchy transform extends to be a bounded operator fromL2(Ω) to L2(bΩ).

The solid Cauchy transform gives rise to a method for measuringsmoothness of holomorphic functions. We know that if h ∈ A∞(Ω),then Csh is an antiholomorphic function on the complement of Ω thatextends C∞ smoothly to the boundary of Ω. We will now show that ifh ∈ H2(Ω) is such that Csh extends smoothly to bΩ, then it must bethat h ∈ A∞(Ω).

Theorem 17.4. Suppose that Ω is a bounded domain with C∞ smoothboundary. A function h in the Bergman space is also in A∞(Ω) if andonly if

supw 6∈Ω

∣∣∣∣∫∫

z∈Ω

h(z)

(z − w)ndA

∣∣∣∣ <∞

for each positive integer n, i.e., if and only if the solid Cauchy transformof h extends C∞ smoothly to bΩ.

The reader might imagine that the problem of testing the finitenessof the integrals in Theorem 17.4, if not impossible, should be at least asdifficult as estimating bounds for the derivatives of h directly. To showthat this may not be the case, we will give a simple application of thetheorem before proving it. Given a point a ∈ Ω, let h(z) = Ka(z) whereKa denotes the Bergman kernel function. The reproducing property ofthe kernel yields that the integral in the statement of the theorem issimply (a − w)−n, which is clearly bounded as a function of w on thecomplement of Ω for each positive integer n. Hence, Theorem 17.4 leadsto a two line proof that the Bergman kernel K(z, w) is in A∞(Ω) as afunction of z for each fixed w ∈ Ω. Theorem 17.4 can also be used togive short proofs that conformal and proper holomorphic maps betweensmooth domains must extend smoothly to the boundary.

We now turn to the proof of Theorem 17.4.

The Solid Cauchy transform 91

Proof. The idea of the proof is simple. If h ∈ A∞(Ω), we know that

h = Bh =∂

∂zECsh,

and this formula should extend to remain valid if h is merely assumedto be in the Bergman space. Hence, it should follow that if Csh hassmooth boundary values, then h must be in A∞(Ω) because the otheroperators in the formula for the Bergman projection behave well whenacting on C∞ smooth functions. The idea is simple, and it is going towork. However, some of the points in the argument that seem obviousrequire proof.

Suppose that h is a function in the Bergman space such that all thesuprema in the hypothesis of the theorem are finite. This means thatCsh defines an antiholomorphic function on the complement of Ω thatextends C∞ smoothly to bΩ. The first obvious sounding step in theproof requiring justification concerns the boundary values of Csh. Whenh is merely in the Bergman space, the definition of Csh as an elementof L2(bΩ) was made by means of a sequence of functions hi ∈ A∞(Ω)such that hi → h in L2(Ω). Then Cshi → Csh in L2(bΩ). We must showthat Csh as defined as a limit of a sequence agrees with the actual C∞

boundary values of Csh, treated as a function on the complement of Ω.Fortunately, the groundwork has been laid in Chapter 6. Let F (z) = 1/z

and consider the set Ω given as the image of the complement of Ω underthe map F . We also assume that the origin is in Ω so that the set Ω is abounded domain with C∞ smooth boundary. By composing with F−1,we may consider Cshi and Csh to be antiholomorphic functions on Ωsuch that the boundary values of Cshi converge in L2(bΩ) to a functionH in that space. Note that we are using the fact that Cs maps L2(Ω)

to functions vanishing at ∞ and hence 0 ∈ Ω, is a removable singularityfor the antiholomorphic functions on Ω under study. Now, the functionsCshi converge uniformly on compact subsets to Csh and it follows fromTheorem 6.3 that H is equal to the smooth boundary values of Csh.Thus, the first sticky point in the proof is past; we have shown that Csh,as an element of L2(bΩ), is in C∞(bΩ).

Next, we must show that the operator (∂/∂z)E can be defined tomake sense as an operator on L2(bΩ). Let a be a point in Ω, and supposethat u is a harmonic function on Ω that is in C∞(Ω). Now ∂u/∂z is inA∞(Ω) and

∂u

∂z(a) =

∫∫

Ω

∂u

∂zKa dA

= − 1

2i

∫∫

Ω

∂u

∂zKa dz ∧ dz = − 1

2i

∫

bΩ

uKa dz,


where we have used the reproducing property of the Bergman kernel Ka

and the complex Green’s identity. Hence, it follows that

∣∣∣∣∂u

∂z(a)

∣∣∣∣ ≤1

2‖u‖‖Ka‖

where the norms denote L2(bΩ) norms. We know that Ka is in A∞(Ω);hence ‖Ka‖ is well defined and finite. This shows that the formula h(a) =(∂/∂z)(ECsh)(a) makes sense and is stable under the limiting process oftaking hi tending to h in L2(Ω). Finally, we may state, therefore, that his in A∞(Ω).

Now that we have studied the boundary behavior of the Cauchytransform and the solid Cauchy transform, we can look at the improvedCauchy integral formula in Theorem 2.1 in a new light. Given a functionu in C∞(bΩ), we can find a function U ∈ C∞(Ω) such that U = u on theboundary. If we write out the improved Cauchy integral formula for Uand restrict to the boundary, we obtain the following classical theorem.

Theorem 17.5. If Ω is a bounded domain with C∞ smooth boundary,then every function u in C∞(bΩ) can be decomposed uniquely as a sumh+H where h ∈ A∞(Ω) and H is a holomorphic function on the com-plement of Ω that vanishes at infinity and that extends C∞ smoothly tobΩ.

We have proved that there is a decomposition u = h + H . To seethat the decomposition is unique, suppose that h1+H1 and h2+H2 aretwo such decompositions. Then h1 − h2 = H2 − H1 on the boundary,and Morera’s Theorem can be used to see that the continuous functionon the whole complex plane that is defined to be equal to h1 − h2 in Ωand equal to H2 −H1 outside Ω is an entire function. Since this entirefunction vanishes at infinity, it must be the zero function, and uniquenessis proved.

18

The classical Neumann problem

The Dirichlet problem is very important in the study of harmonic func-tions and conformal mapping. Equally important is the classical Neu-mann problem. Suppose Ω is a bounded domain with C∞ smooth bound-ary. Given a function ψ in C∞(bΩ), the Neumann problem is to find afunction ϕ ∈ C∞(Ω) that is harmonic on Ω such that the normal deriva-tive of ϕ on the boundary is equal to ψ. We will use the notation ∂ϕ/∂nto denote the normal derivative of ϕ. The first observation to be madeis that not every ψ can be equal to the normal derivative of a harmonicfunction. Indeed, if ϕ is harmonic, then Gauss’ theorem yields

∫

bΩ

(∂ϕ/∂n) ds =

∫∫

Ω

∆ϕ dA = 0.

Thus, it is necessary to stipulate that∫bΩψ ds = 0.

We will prove that the Neumann problem can be solved by expressingthe solution in terms of the Szego projections of explicit functions. Inthe case that Ω is simply connected, the result can be stated as follows.

Theorem 18.1. Suppose Ω is a bounded simply connected domain withC∞ smooth boundary, and let a ∈ Ω be given. Suppose ψ is a functionin C∞(bΩ) satisfying

∫bΩψ ds = 0. Then, a solution to the Neumann

problem

∆ϕ = 0 on Ω

∂ϕ

∂n= ψ on bΩ

is given by ϕ = h+H where h and H are holomorphic functions on Ωdefined via

h′ = Sa P(ψ/La

), and

H ′ = La P(ψ/Sa

).

Furthermore, the solution ϕ is in C∞(Ω).

We remark that, in the setting of Theorem 18.1, Sa is nonvanishing on

93


Ω. It is also worth pointing out that, even though La has a simple pole ata, the expression in the statement of the theorem for H ′ is holomorphicnear a because, using the fact that S(a, z) = S(z, a), we have

P(ψ/Sa

)(a) =

∫

z∈bΩ

S(a, z)ψ

S(z, a)ds =

∫

bΩ

ψ ds = 0.

We will state a version of this theorem for multiply connected do-mains in Chapter 20. It is somewhat more involved because not everyharmonic function on a multiply connected domain can be expressed ash +H and not every holomorphic function is equal to the derivative ofa holomorphic function.

It is not hard to show that the solution to the Neumann problemis unique up to the addition of an arbitrary constant. Indeed, if ϕ is aharmonic function in C∞(Ω) such that ∂ϕ/∂n ≡ 0, then Green’s identityimplies that

0 =

∫

bΩ

ϕ∂ϕ

∂nds =

∫∫

Ω

|∇ϕ|2 dA

and this shows that ϕ must be constant on Ω. Consequently, the deriva-tives h′ and H ′ in the decomposition of the solution in Theorem 18.1are uniquely determined, and therefore, the expressions for h′ and H ′ donot depend on the choice of a.

It can be shown that the functions defined in the statement of The-orem 18.1 make sense even when ψ is merely assumed to be continuouson bΩ and that the normal derivative of ϕ = h + H exists in a weaksense and is equal to ψ. We will not study the more subtle problemsthat arise in the study of harmonic functions with continuous boundarynormal derivatives in this book.

Proof of Theorem 18.1. To motivate the proof, let us suppose for themoment that we know that a solution ϕ exists that is in C∞(Ω). Anyharmonic function on a simply connected domain can be expressed ash + H where h and H are holomorphic functions that are determinedup to additive constants. Furthermore, since ϕ ∈ C∞(Ω), it follows fromdifferentiating ϕ = h + H with respect to z and z that h and H mustbe in A∞(Ω). If z(s) parameterizes the boundary of Ω in the standardsense with respect to arc length s, then (d/ds)h(z(s)) = h′(z(s))z′(s) =h′(z(s))T (z(s)). Since h satisfies the Cauchy-Riemann equations on theboundary, the normal derivative ∂h/∂n is equal to −i times the deriva-tive of h in the tangential direction pointing in the standard sense. Thus,∂h/∂n = −iT (z)h′(z). Similarly, ∂H/∂n = i T (z)H ′(z). Hence,

ψ =∂ϕ

∂n= −iT (z)h′(z) + i T (z)H ′(z). (18.1)

The classical Neumann problem 95

At this point, the orthogonal decomposition of Theorem 4.3 shouldspring to mind. We may use formula (7.1) to express T as

T = iLa

Sa,

and we may plug this expression into (18.1) and divide by La to obtain

ψ

La

=h′

Sa+ i T

H ′

La

. (18.2)

Now, because of the nonvanishing of Sa and La on Ω, it follows thath′/Sa and H ′/La are both in A∞(Ω) (the pole of La implies that thesecond function vanishes at a, but this is of no consequence to us). Hence,(18.2) is an orthogonal sum, and we obtain

h′

Sa= P

(ψ/La

).

Furthermore,

−i H′

La= P

(Tψ

La

),

and, using the fact that T/La = 1/(i Sa ) by (7.1), we see that

H ′

La= P

(ψ/Sa

).

To prove Theorem 18.1, we need only trace this argument backwards.Indeed, define ϕ = h+H where h and H are holomorphic functions onΩ whose derivatives satisfy

h′ = Sa P(ψ/La

), and

H ′ = La P(ψ/Sa

).

Then ϕ is in C∞(Ω) and

∂ϕ

∂n= −iT (z)h′(z) + i T (z)H ′(z) = −iSaP (ψ/La )T + i LaP (ψ/Sa )T .

The proof will be finished after we prove the following lemma whichholds even when Ω is not assumed to be simply connected.

Lemma 18.1. If Ω is a bounded finitely connected domain with C∞

smooth boundary and if ψ ∈ L2(bΩ), then

ψ = −iSaP (ψ/La )T + i LaP (ψ/Sa )T .


Proof of the lemma. Divide the equation in the statement of the lemmaby La, and use identity (7.1) to see that our problem is equivalent toproving that

ψ

La

= P (ψ/La ) + i P (ψ/Sa )T .

This is an orthogonal sum. The holomorphic part is obviously what itshould be. The proof of the lemma will be finished if we show that

P⊥

(ψ

La

)= i P (ψ/Sa )T .

Since P⊥u = P (uT )T , this is equivalent to showing that

P

(ψ

LaT

)= −iP (ψ/Sa ),

and this follows directly from identity (7.1). The proof of the lemma isdone and therefore, so is the proof of the theorem.

19

Harmonic measure and the Szego kernel

The expression −ih′T + iH ′T appears as the normal derivative of h+Hin the proof of Theorem 18.1. In a simply connected domain, we willsee that the condition −ih′T + iH ′T = 0 forces h′ ≡ 0 and H ′ ≡ 0.Before we treat the Neumann problem in a multiply connected domain,we must determine which functions h and H in H2(bΩ) satisfy the con-dition −ihT + iHT = 0, i.e., satisfy hT = HT . By Theorem 4.3, afunction u that is expressible as both hT and HT would be orthogo-nal to H2(bΩ) and orthogonal to conjugates of functions in H2(bΩ). Wewill see that such functions are closely related to the classical harmonicmeasure functions (which we define below).

In this chapter, we will be dealing with a bounded n-connected do-main Ω with C∞ smooth boundary curves. Let γjnj=1 denote then boundary curves of Ω. For convenience, we may assume that γn isthe outer boundary curve, i.e., the one that also bounds the unboundedcomponent of the complement of Ω in C. The harmonic measure func-tions ωjnj=1 associated to Ω are defined as follows. The function ωj isequal to the harmonic function on Ω that solves the Dirichlet problemwith boundary data equal to one on γj and equal to zero on the otherboundary curves. Note that we know that these harmonic measure func-tions are in C∞(Ω) by Theorem 14.2.

An important holomorphic function associated to ωj is the functionF ′j = 2∂ωj/∂z. The prime in the notation F ′

j is traditional; F ′j is the

derivative of a multi-valued holomorphic function. To see this, supposethat v is a harmonic conjugate for ωj on a small disc contained in Ωand define Fj = ωj + iv there. By the Cauchy-Riemann equations, F ′

j =(∂/∂x)ωj−i(∂/∂y)ωj = 2∂ωj/∂z. Thus, F

′j is the derivative of the multi-

valued function obtained by analytically continuing around Ω a germ ofωj + iv where v is a local harmonic conjugate for ωj .

Our next theorem will show how the functions F ′j are related to the

Szego kernel and its zeroes. Let a ∈ Ω be a given point. By Theorem 13.1,the function Sa(z) = S(z, a) has exactly n− 1 zeroes in Ω (counted withmultiplicity). We will prove in Chapter 27 that the zeroes of S(z, a)become simple zeroes as a approaches the boundary of Ω. We assume

97


now that a has been chosen so that the zeroes of S(z, a) are all simplezeroes. Let ajn−1

j=1 denote these distinct n− 1 zeroes of Sa.

Let Q denote the space of functions in L2(bΩ) that are orthogonalto both the Hardy space H2(bΩ) and to the space of functions that arecomplex conjugates of functions in H2(bΩ).

Let F ′ denote the (complex) linear span of F ′j : j = 1, 2, . . . , n−1.

Although the function F ′n is not included in the spanning set for F ′, this

function is in F ′. Indeed, the maximum principle shows that∑n

j=1 ωj ≡1 and consequently,

∑nj=1 F

′j ≡ 0.

Theorem 19.1. The space Q of functions in L2(bΩ) orthogonal to bothH2(bΩ) and the space of functions that are conjugates of H2(bΩ) func-tions is equal to hT : h ∈ F ′. Furthermore, F ′ is equal to the complexlinear span of

L(z, aj)S(z, a) : j = 1, 2, . . . , n− 1.

F ′ is also equal to the complex linear span of

L(z, a)S(z, aj) : j = 1, 2, . . . , n− 1.

The part of this theorem that states that F ′ is equal to the span ofL(z, aj)S(z, a) was proved by Schiffer in [Sch].

Proof. Notice that the functions L(z, aj)S(z, a) are in A∞(Ω) becausethe pole of L(z, aj) at aj is cancelled by the zero of S(z, a) at aj . Sim-ilarly, the functions L(z, a)S(z, aj) are in A∞(Ω) because the pole ofL(z, a) at a is cancelled by the zero of S(z, aj) at a. Let L denote thecomplex linear span of L(z, aj)S(z, a) : j = 1, 2, . . . , n− 1. We makethe convention that, if G is a subset of H2(bΩ), then GT will denote theset of functions of the form gT , where g ∈ G. We will prove the theoremby first showing that F ′T ⊂ Q ⊂ LT . The dimension of LT as a vectorspace over the complex numbers is less than or equal to (n − 1). If weshow that F ′T is a vector space of dimension (n − 1), it follows thatF ′T = Q = LT .

That F ′T ⊂ Q follows from Theorem 4.3 and the identity,

F ′jT = −F ′

jT on bΩ, (19.1)

which we now prove. This is an important fact. We will also need itin the next chapter when we study the Neumann problem in multiplyconnected domains. Let z(t) parameterize a boundary curve of Ω. Sinceωj is constant on boundary curves, we have

0 =d

dtωj(z(t)) =

∂ωj

∂zz′(t) +

∂ωj

∂zz′(t).

Harmonic measure and the Szego kernel 99

Since (∂ωj/∂z) is equal to the complex conjugate of (∂ωj/∂z), identity(19.1) is proved.

We will next prove that Q ⊂ LT . If u ∈ Q, then by Theorem 4.3, ucan be written

u = hT = HT,

where h and H are elements of H2(bΩ). Using identity (7.1) in the formT = i La/Sa, we obtain

ihLa

Sa= HT,

and therefore,

ih

Sa=H

La

T .

Now, H/La is a holomorphic function in H2(bΩ). Hence, the functionon the right hand side of the last equation is orthogonal to H2(bΩ).Therefore,

iP⊥

(h

Sa

)=H

La

T .

But

ih(z)

Sa(z)= G(z) +

n−1∑

j=1

cj1

z − aj,

where G(z) is in H2(bΩ) and cj = ih(aj)/S′(aj , a), the prime denoting

differentiation in the holomorphic variable. Therefore, since P⊥G = 0,we have

H

La

T =

n−1∑

j=1

cjP⊥

(1

z − aj

).

It is a consequence of the orthogonal decomposition of the Cauchy kernelthat we used to define the Garabedian kernel that

P⊥

(1

z − aj

)= 2πL(z, aj).

Hence,

H

La

T =n−1∑

j=1

cj2πL(z, aj).

Finally, we may multiply through by La and use identity (7.1) in theform La = −iSaT to obtain

hT = HT =

n−1∑

j=1

−icj2πL(z, aj)S(z, a)T (z).

100The Cauchy Transform, Potential Theory, and Conformal Mapping

Hence h =∑n−1

j=1 −icj2πL(z, aj)S(z, a) and we have proved that Q ⊂LT .

Next, we show that the dimension of F ′ is (n−1) by proving that thefunctions F ′

j , j = 1, . . . , n − 1 are linearly independent. To see this, wewill prove the classical fact that the (n− 1)× (n− 1) matrix of periods[Ajk ] defined via

Ajk =

∫

γj

F ′k dz, j = 1, . . . , n− 1,

is nonsingular. It is proved in most books on elementary complex analy-sis that a function g ∈ A∞(Ω) is the derivative of a holomorphic functionon the multiply connected domain Ω if and only if the periods

∫

γj

g dz

vanish for j = 1, 2, . . . n− 1 (see [Ah, p. 146]). Suppose that there exist

constants ck, not all zero, so that∑n−1

k=1 Ajkck = 0 for each j. Then it

follows that all the periods of the function∑n−1

k=1 ckF′k vanish, and so

G′ =∑n−1

k=1 ckF′k for some G ∈ A∞(Ω). Now, G′ = 2(∂ω/∂z) where

ω =∑n−1

k=1 ckωk. Since not all the ck are zero, and since ω = 0 on γn,it is clear that the function ω cannot be an anti-holomorphic function.Hence, G′ is not identically zero and therefore

∫∫Ω|G′|2 dA 6= 0. But

1

4

∫∫

Ω

|G′|2 dA =

∫∫

Ω

∂ω

∂z

∂ω

∂zdA =

1

2i

∫

bΩ

ω∂ω

∂zdz

by the complex Green’s identity. If we expand this last integral, we obtain

n−1∑

j=1

n−1∑

k=1

cj ck

∫

bΩ

ωj∂ωk

∂zdz

and this last sum is a linear combination of the numbers∑n−1

k=1 Ajkck,which we have assumed to be zero. This contradiction shows that thematrix of periods is nonsingular, and therefore, the proof that F ′T =Q = LT is complete.

To finish the proof, we must see that L is equal to the complex linearspan L of L(z, a)S(z, aj) : j = 1, 2, . . . , n−1. The spaceQ of functionsin L2(bΩ) orthogonal to H2(bΩ) and conjugates of functions in H2(bΩ)is, by its definition, invariant under complex conjugation. By (7.1), wemay write

L(z, aj)S(z, a)T (z) = −S(z, aj)L(z, a)T (z). (19.2)


Hence, Q is equal to the space of conjugates of functions in LT , whichby (19.2) is equal to LT . Since, Q = LT , it follows that L = L, and theproof is complete.

When we study the Neumann problem in multiply connected do-mains, we will need to know the following classical fact about the func-tions F ′

j . It is a direct consequence of the result proved above that thematrix of periods [Ajk] is nonsingular.

Lemma 19.1. Given a holomorphic function h ∈ A∞(Ω), there existconstants cj and another holomorphic function H ∈ A∞(Ω) such that

h = H ′ +∑n−1

k=1 ckF′k.

Thus every function in A∞(Ω) is the derivative of a function inA∞(Ω) modulo the space of linear combinations of the F ′

j .Since [Ajk] is nonsingular, there exist complex constants cj , j =

1, . . . , n− 1 satisfying

∫

γj

h dz =

n−1∑

k=1

Ajkck, j = 1, . . . , n− 1.

Now, all the periods of h −∑n−1k=1 ckF

′k vanish, and so this function is

equal to the derivative of a holomorphic function H on Ω. It is clear thatH ∈ A∞(Ω) and the lemma is proved.

We close this chapter by studying a classical application of the har-monic measure functions to the conformal classification of multiply con-nected domains. Given an n-connected domain Ω with C∞ smoothboundary, we describe the construction of a biholomorphic map of Ωonto a canonical n-connected domain that is an annulus minus n − 2disjoint closed circular arcs. The annulus and each of the closed circulararcs are centered at the origin and the n−2 disjoint arcs each span anglesthat are less than 2π. Lemma 12.1 allows us to reduce the problem ofconstructing such a map to the case in which Ω has real analytic bound-ary. One of the virtues of such a domain is that its harmonic measurefunctions ωk extend to be harmonic on a neighborhood of Ω. Indeed,Theorem 14.2 shows that these functions extend continuously to theboundary, and the Schwarz reflection principle applies to yield the har-monic extensions. Harmonic extendibility of the functions ωk, in turn,implies that the functions F ′

k extend holomorphically to a neighborhoodof Ω.

The function f(z) we will construct mapping Ω onto an annular do-main of the type described above will extend holomorphically past theboundary of Ω, and it will map the outer boundary γn of Ω onto the unitcircle. To motivate the construction, let us take a moment to see why


such an f should be related to the harmonic measure functions. Con-sider the function ln |f(z)|. It is harmonic on a neighborhood of Ω andconstant on each boundary curve of Ω. Hence, there exist real constantsr1, r2, . . . , rn such that

ln |f(z)| =n∑

k=1

rkωk.

Since |f | = 1 on γn, it follows that rn = 0.We now differentiate this equa-tion with respect to z. Notice that ln |f(z)| = 1

2 ln |f(z)|2 = 12 ln f(z)f(z)

and so

∂

∂zln |f(z)| = 1

2

1

f(z)f(z)

∂

∂z

(f(z)f(z)

)=f ′(z)

2f(z).

Also recall that (∂/∂z)ωk = 12F

′k. Hence, we obtain

f ′(z)

f(z)=

n−1∑

k=1

rkF′k. (19.3)

Let us now restrict our attention to a simply connected subdomain Ω0

of Ω. Let G(z) be a holomorphic antiderivative of∑n−1

k=1 rkF′k on Ω0.

Notice that f satisfies the ODE, f ′ −G′f = 0, and hence, (fe−G)′ ≡ 0.It follows that f = ceG on Ω0 for some choice of a constant c. The key tothe construction of f will be to use (19.3) to determine an appropriatechoice of the constants rk, and then to find a multi-valued antiderivativeG of

∑n−1k=1 rkF

′k that will make f = eG a single valued holomorphic

function with the desired properties.To put (19.3) in a more useful form, we integrate it around the bound-

ary curve γj . The left hand side of the equation represents 1i times the

increase in the argument of f(z) as z traces out γj in the standard sense.Let ∆jarg f represent this increase and, as before, let Ajk denote theperiod,

∫γjF ′k dz. We may rewrite the integrated equation in the form

1

i∆jarg f =

n−1∑

k=1

Ajkrk, j = 1, . . . , n, (19.4)

which is an n × (n − 1) linear system. Actually, the n-th equation inthe system must be dependent upon the first (n− 1) equations because∑n

j=1 Ajk =∫bΩ F

′k dz = 0 by Cauchy’s theorem. Keep in mind that we

have shown that the submatrix [Ajk]j,k=1,...,n−1 is nonsingular.We now describe how to construct from scratch a mapping f like the

one we assumed to exist above. The key to the construction is the system


(19.4). We seek a function f that is holomorphic on a neighborhood ofΩ mapping the outer boundary of Ω onto the unit circle, mapping γ1onto a circle of radius R1 < 1 centered at the origin, and mappingγ2, . . . , γn−1 to circular arcs of radii between R1 and 1. As z traces outγn in the standard sense, we want f(z) to trace out the unit circle inthe standard sense, and so we need ∆narg f = 2π. Similarly, as z tracesout γ1 in the standard sense (clockwise), we want f(z) to trace out acircle of radius R1 < 1. For f to preserve the sense of angles, it mustbe that f(z) traces out this circle of radius R1 in the clockwise sense,and so we need ∆1arg f = −2π. As z traces out any of the other n− 1boundary curves, we expect f(z) to move back and forth along a circulararc spanning an angle of less than 2π, and so we need ∆jarg f = 0 forj = 2, . . . , n−1. Now we may determine constants rkn−1

k=1 using the first(n − 1) of the equations in system (19.4) using the increments of arg faround the boundary curves that we just specified. Fortunately, because2π + 0 + · · · + 0 − 2π = 0, the n-th equation in that system is satisfiedbecause it represents the sum of the first (n− 1) equations. Let rn = 0.Identity (19.1) shows that Ajk = −Ajk, and so the matrix of periods isa matrix of pure imaginary numbers. Since the inhomogeneous term inthe system (19.4) is also pure imaginary, it follows that the constants rkare real numbers.

Pick a point z0 ∈ Ω and, given z ∈ Ω, let Γ(z0, z) denote a curve in Ωstarting at z0 and ending at z ∈ Ω. Let H(z) =

∑nk=1 rkF

′k(z) and define

G(z) =∫Γ(z0,z)

H(ζ) dζ. Since the periods of H are all integer multiples

of 2πi, it follows that, although G may not be uniquely defined on Ω,the real part of G and eG are. Furthermore, by restricting attention tosimply connected subdomains of Ω containing z0, where we may think ofG as being a well defined holomorphic function, it is easily seen that eG

defines a a single valued holomorphic function on Ω. In fact, eG extendsto be holomorphic on a neighborhood of Ω.

The mapping we desire is now given as f = eG. To see that the mod-ulus of f is constant on each boundary curve of Ω, let ω =

∑n−1k=1 rkωk.

Restrict attention to a simply connected subdomain Ω0 of Ω containingz0. We may think of G as being a well defined single valued holomorphicfunction on Ω0. We will see that Re G−ω is a constant on this domain.Since (∂/∂z)Re h = 1

2h′ for any holomorphic function h, we see that

(∂/∂z)(Re G − ω) ≡ 0 on Ω0, and so Re G − ω is a real valued anti-holomorphic function there. Consequently, Re G − ω is equal to a realconstant λ on Ω0. Since Re G is well defined on Ω, this constant doesnot depend on the choice of the subdomain Ω0. Hence |eG| = eλeω on Ω.Let us now redefine f to be equal to ceG where c = e−λ. We may nowstate that |f | = erk on γk for each k = 1, . . . , n.

The next part of the proof follows directly from the classical argument


principle. If w does not lie on any of the circles of radius erk , k = 1, . . . , n,then the number of solutions z of the equation f(z) = w is given byN (w)where

N (w) =1

2πi

∫

bΩ

f ′(ζ)

f(ζ)− wdζ =

n∑

k=1

Nk(w),

where

Nk(w) =1

2πi

∫

γk

f ′(ζ)

f(ζ)− wdζ.

Our construction guarantees that N (0) is given by the sum −1 + 0 +· · · + 0 + 1, and so N (0) = 0. Also note that N1(w) is equal to −1 if|w| < er1 and equal to zero if |w| > er1 . Furthermore,Nn(w) is equal to 1if |w| < 1 and equal to zero if |w| > 1. If k = 2, . . . , n−1, then Nk(w) = 0if |w| 6= erk . Suppose w is in the image of Ω under f . Because f is anopen mapping, we may find a point w0 in f(Ω) that is arbitrarily closeto w and that does not lie on any of the circles of radii erk , k = 1, . . . , n.In order for the numbers Nk(w0) to add up to something greater thanor equal to one, it must be that er1 < |w0| < 1. This shows that f(Ω)is contained in the annulus A = w : er1 < |w| < 1. Because f iscontinuous up to the boundary of Ω, it follows that er1 ≤ erk ≤ 1 fork = 2, . . . , n − 1. Because the argument of f has been constructed tohave a net change of 2π on γn and γ1, we know that f maps γ1 onto thecircle of radius er1 and f maps γn onto the unit circle. We also knowthat there is exactly one point z0 in Ω satisfying f(z0) = w0 when w0 isa point in the annulus A that does not lie on any of the circles |w| = erk .

To finish the proof, we must use the generalized version of the argu-ment principle that we described in Chapter 13. If h is a holomorphicfunction defined on γk, let z1, z2, . . . , zM denote the finite set of zerosof h on γk. Let γk(ǫ) denote the set of curves formed by removing thesegments γk ∩Dǫ(zi), i = 1, . . . ,M , each segment parameterized in thesame sense as γk. We define ∆karg h to be the limit as ǫ tends to zeroof the sum of the increases in arg h over the segments in γk(ǫ). Theproof that this limit is well defined is contained in the proof given inChapter 13 of the generalized argument principle.

Consider the triangle with vertices at the origin, the point R, and apoint Reiθ, 0 ≤ θ < 2π. It is a simple exercise in high school geometry tosee that θ is equal to −π plus two times the angle from the real axis tothe line segment joining R to Reiθ. Suppose |w0| = erk . The geometricfact just mentioned shows that

d arg (f(z)− w0) =1

2d arg f(z) on γk


near points z where f(z) 6= w0, and this shows that

1

2π∆karg (f(z)− w0) =

1

2Nk(0).

If |w0| 6= erj , then f(z)−w0 has no zeroes on γj and so 12π∆jarg (f(z)−

w0) = Nj(w0). The generalized argument principle states that the num-ber of solutions to f(z) = w0 with z ∈ Ω plus one half times the numberof solutions to f(z) = w0 with z ∈ bΩ is equal to the sum

1

2π

n∑

j=1

∆jarg (f(z)− w0),

which we know is equal to the sum

∑ 1

2Nk(0) +

∑Nj(w0)

where the first sum ranges over all indices k such that |w0| = erk , andthe second sum ranges over indices j such that |w0| 6= erj . If |w0| = 1,the only nonzero term in the sum is 1

2Nn(0) = 12 and this shows that

there is exactly one z0 in bΩ with f(z0) = w0 and no solutions in Ω.Hence, f maps γn one-to-one onto the unit circle. Similarly, if |w0| = er1

we deduce that f maps γ1 one-to-one onto the circle of radius er1 . Wemay now also assert that er1 < erk < 1 for k = 2, . . . , n − 1. Suppose2 ≤ k ≤ n − 1. If |w0| = erk , then the counting equation yields thefollowing information about the number of solutions to f(z) = w0 withz ∈ Ω. There is either exactly one solution z0 with z0 ∈ Ω, or thereare exactly two distinct solutions to f(z) = w0 lying in bΩ, or there isexactly one solution to f(z) = w0 of multiplicity two in bΩ. Becausethe net increase of arg f(z) is zero around γk, a single valued branch ofarg f(z) can be defined on neighborhood of γk. At the points on γk wherearg f(z) assumes local maxima or minima, the mapping f must take onits values with multiplicity two. Furthermore, the counting argumentabove shows that there must be exactly one maximum and exactly oneminimum, otherwise f would trace over sections of the circle |w| = erk

more than twice. The same reasoning shows that the difference betweenthe maximum and the minimum must be less than 2π. Finally, we mayalso conclude that the arc f(γk) is disjoint from all the others. The proofis finished.

20

The Neumann problem in multiplyconnected domains

The procedure we used in Chapter 18 to solve the Neumann problemin a simply connected domain must be altered to apply in a multiplyconnected domain because not every harmonic function can be expressedglobally as a sum h+H. The modification involves the harmonic measurefunctions and the functions F ′

j studied in Chapter 19.

Suppose ϕ is a harmonic function on Ω in C∞(Ω). We claim thatthere exist functions h and H in A∞(Ω) and constants cj such that

ϕ = h+H +

n−1∑

j=1

cjωj .

To see this, note that, by Lemma 19.1, it is possible to choose constants cjso that the holomorphic function ∂ϕ/∂z is equal to the derivative h′ of afunction h ∈ A∞(Ω) plus 1

2

∑cjF

′j . Now ∂/∂z annihilates ϕ−h−∑ cjωj ;

so this function is an antiholomorphic function H where H ∈ A∞(Ω).This proves our claim.

The normal derivative of ωj is easily computed. If z0 is a point inthe boundary, we may choose ǫ > 0 so small that Dǫ(z0)∩Ω is a simplyconnected domain. On this set, we may find a harmonic conjugate forωj , and therefore, we may write ωj = Re Fj , where Fj is a holomor-phic antiderivative of F ′

j . Hence, locally, ωj = 12Fj +

12Fj , and we may

compute,∂ωj

∂n= − i

2F ′jT +

i

2F ′jT .

But recall from Chapter 19 that F ′jT = −F ′

jT on bΩ. Hence, in fact,

∂ωj

∂n= −iF ′

jT.

Using this fact, the normal derivative of ϕ on bΩ is seen to be equalto

∂ϕ

∂n= −ih′T + iH ′T − i

∑cjF

′jT. (20.1)

107


The next theorem shows how to solve the Neumann problem on amultiply connected domain by relating the functions h′ and H ′ and thecoefficients cj to Szego projections of known functions.

Theorem 20.1. Suppose Ω is a bounded n-connected domain with C∞

smooth boundary, and let a ∈ Ω be given. Suppose ψ ∈ C∞(bΩ) satisfies∫bΩψ ds = 0. Then, a solution to the Neumann problem

∆ϕ = 0 on Ω

∂ϕ

∂n= ψ on bΩ

is given by

h+H +

n−1∑

j=1

cjωj

where h and H are holomorphic functions on Ω defined via

h′ = SaP (ψ/La ) +

n−1∑

j=1

AjF′j , and

H ′ = LaP (ψ/Sa ) +

n−1∑

j=1

BjF′j ,

where the constants Aj and Bj are determined by the condition thatthe functions on the right hand side be the derivative of a holomorphicfunction (i.e., by the condition that the periods of the functions on theright vanish). The constants cj in the solution are then given by

cj = −Bj −Aj .

The key element in the proof of Theorem 20.1 is Lemma 18.1.

Proof. Given ψ ∈ C∞(bΩ), Lemma 18.1 yields a decomposition of ψ as−igT + i GT where g is in A∞(Ω), and G is a meromorphic functionon Ω with (possibly) a simple pole at a that extends smoothly to bΩ.If it is further assumed that

∫bΩψ ds = 0, then the argument after the

statement of Theorem 18.1 shows that G has no pole at a, and it followsthat G ∈ A∞(Ω). In the simply connected case, we could antidifferentiateg and G to obtain the solution to the Neumann problem associated toψ. In the multiply connected case we are studying now, although we canno longer antidifferentiate every holomorphic function, we can still usethis decomposition to solve the Neumann problem. Here are the details.

The Neumann problem in multiply connected domains 109

The function Sa P(ψ/La

)may not have a holomorphic antideriva-

tive on Ω, but there do exist a function h ∈ A∞(Ω) and constants Aj

such that

h′ = SaP (ψ/La ) +

n−1∑

j=1

AjF′j .

Similarly, there exist a function H ∈ A∞(Ω) and constants Bj such that

H ′ = LaP (ψ/Sa ) +

n−1∑

j=1

BjF′j .

Let cj = −Bj−Aj, and let ϕ = h+H+∑n−1

j=1 cjωj . The normal derivativeof ϕ is given by (20.1). Now, using Lemma 18.1, the definitions of h andH , and the fact that F ′

jT = −F ′jT , we see that

∂ϕ

∂n= ψ − i

n−1∑

j=1

(Aj +Bj + cj )F′jT = ψ,

and Theorem 20.1 is proved.

21

The Dirichlet problem again

In this chapter, we reconsider the Dirichlet problem in a multiply con-nected domain. This time, we have the machinery of the harmonic mea-sure functions developed in Chapter 19 at our disposal.

Suppose Ω is a bounded n-connected domain with C∞ smoothboundary and suppose ϕ is a function in C∞(bΩ). Let a ∈ Ω be chosenso that the n − 1 zeroes of Sa are distinct and simple. (That this canbe done will be proved in Chapter 27.) Theorem 14.1 allows us to writeϕ = h +H where h is a meromorphic function on Ω that extends C∞

smoothly up to bΩ given by h = P (Sa ϕ)/Sa and H is in A∞(Ω) and isgiven by H = P (La ϕ )/La. Notice that the set of points at which h mayhave poles is a subset of the set of zeroes ajn−1

j=1 of Sa.First, we will prove that the following system can be solved for any

choice of coefficients Bkn−1k=1 :

n−1∑

j=1

cjP (Saωj)(ak) = Bk, k = 1, 2, . . . , n− 1.

Let Ajk = P (Saωj)(ak). We may use identity (7.1) to compute

Ajk =

∫

bΩ

S(ak, z)S(z, a)ωj ds = −i∫

γj

L(z, ak)S(z, a) dz, (21.1)

and this shows that Ajk is the classical period of −iL(z, ak)S(z, a)around γj . Now, by Theorem 19.1, the functions L(z, ak)S(z, a) and thefunctions F ′

k span the same space. We showed in Chapter 19 that thematrix of periods associated to the F ′

k is nonsingular. Hence, it followsthat det [Ajk] 6= 0 and our system can be solved.

Let ϕ ∈ C∞(bΩ) be given and set Bk = P (Saϕ)(ak). Define ψ =

ϕ−∑n−1j=1 cjωj where the cj satisfy the linear system,

n−1∑

j=1

Ajkcj = Bk, k = 1, 2, . . . , n− 1.

Now define

h =P (Sa ψ)

Sa

111


and

H =P (La ψ)

La.

The linear system was conceived so that P (Saψ) has a zero at each ofthe zeroes of Sa. Since the zeroes of Sa are simple zeroes, this meansthat h, which is the quotient of these two functions, has no poles in Ω.Thus h and H are both in A∞(Ω) and it follows from Theorem 14.1that h+H gives the harmonic extension of ψ to Ω. Hence, the harmonicextension of ϕ is given by h+H+

∑n−1j=1 cjωj. We summarize this result

in the following theorem.

Theorem 21.1. Let Ω be a bounded n-connected domain with C∞

smooth boundary and let Ajk denote the matrix of periods for the func-tions −iL(z, ak)S(z, a) as given by (21.1). Given ϕ ∈ C∞(bΩ), let cjsolve the system

n−1∑

j=1

Ajkcj = P (Saϕ)(ak), k = 1, 2, . . . , n− 1.

The harmonic extension of ϕ to Ω is given by

h+H +

n−1∑

j=1

cjωj,

where, using the notation ψ = ϕ −∑ cjωj, h and H are functions inA∞(Ω) given by

h =P (Sa ψ)

Saand

H =P (La ψ)

La.

22

Area quadrature domains

In the next two chapters, we hope to convey the beauty of the subjectof quadrature domains in the plane, and the utility of the tools andtechniques described in the previous chapters to understand quadraturedomains. To make these chapters self-contained, we do not go into theutmost generality. For the whole story, see the Bibliographic Notes onpage 200.

The unit disc is the most famous of area quadrature domains. Whena holomorphic function is averaged over the disc with respect to areameasure, the value of the function at the origin is obtained. More gener-ally, a domain Ω in the plane of finite area is called an area quadraturedomain if the average of a holomorphic function in the Bergman spaceover the domain with respect to area measure is a finite linear combina-tion of values of the function and its derivatives at finitely many pointsin Ω. The points and the coefficients are fixed in this “quadrature iden-tity” and the identity holds for all holomorphic functions in the Bergmanspace. In other words, there exist points zjNj=1 in Ω, complex constantscjk, and nonnegative integers mj such that

∫∫

Ω

h dA =

N∑

j=1

mj∑

k=0

cjkh(k)(zj) (22.1)

for all h in H2(Ω). In this chapter, we will show that area quadraturedomains share many of the properties of the unit disc and that they turnout to be very abundant. We will also explain how to view our resultsas an improvement upon the Riemann mapping theorem.

Before we begin, we must define higher order versions of the Bergmankernel. In Chapter 15, we showed that the averaging property of holo-morphic functions allowed us to the write the Bergman kernel Ka as theBergman projection of a function ϕa in C∞

0 (Ω). We now take a closerlook at this fact with an eye to differentiating it with respect to a. Let θbe a real valued radially symmetric C∞ function compactly supportedin the unit disc such that

∫∫θ dA = 1. Fix a point a0 in Ω and let ǫ > 0

be less than the distance from a0 to the boundary of Ω. For a near a0,

113


the function ϕa that we described in Chapter 15, may be written

ϕa(z) =1

ǫ2θ

(z − a

ǫ

).

Recall thath(a) = 〈h, ϕa〉Ω (22.2)

for all h ∈ H2(Ω) and that is why we were able to conclude that Ka =Bϕa. Let ϕ

ma be defined to be (∂m/∂am)ϕa. Note that this function is

in C∞0 (Ω) and that it is the conjugate of (∂m/∂am)ϕa because ϕa is

real valued. If a is close to a0, we may differentiate (22.2) m times withrespect to a to obtain

h(m)(a) = 〈h, ϕma 〉Ω.

Let Kma = Bϕm

a . This function represents the functional h 7→ h(m)(a) inthe sense that

h(m)(a) = 〈h,Kma 〉Ω.

LetK0a denote the Bergman kernelKa. The Bergman span of Ω is defined

to be the space of all linear combinations of functions of the form Kma as

a ranges over Ω and m ranges over nonnegative integers. We may equate〈Km

a ,Kz〉Ω with the conjugate of 〈Kz,Kma 〉Ω to deduce that

Kma (z) =

∂m

∂amK(z, a).

We remark here that equating 〈Knb ,K

ma 〉Ω with the conjugate of

〈Kma ,K

nb 〉Ω allows one to express all the derivatives of the Bergman

kernel in both variables in terms of integrals that are easily bounded oncompact sets via the basic estimate proved in Chapter 15. In this way,it can be deduced that K(z, w) is C∞ smooth in z and w in Ω × Ω,holomorphic in z, and antiholomorphic in w.

We now turn to the study of area quadrature domains. Assume thatΩ is an area quadrature domain. Note that the function that is identicallyone is in the Bergman space because of the finite area assumption, andif h is another function in the Bergman space, then

∫∫

Ω

h dA = 〈h, 1〉Ω.

Thus, we see that h is in L1(Ω) by the Cauchy-Schwarz inequality, andso the integral in the quadrature identity for Ω is well defined. Let Qdenote the following linear combination of functions Kk

zj in the Bergmanspan inspired by the quadrature identity (22.1) that we assume holds:

Q(z) =

N∑

j=1

mj∑

k=0

cjkKkzj . (22.3)

Area quadrature domains 115

Notice that〈h, 1〉Ω = 〈h,Q〉Ω

for h ∈ H2(Ω) because both sides yield the quadrature identity in h.Since the functions 1 and Q are both in the Bergman space, and sinceh is arbitrary, if must be that Q ≡ 1. The converse of this fact is alsoclearly true, namely that if the function 1 is in the Bergman span, thenΩ is an area quadrature domain.

Theorem 22.1. A domain of finite area is an area quadrature domain ifand only if the function that is identically equal to one is in the Bergmanspan.

We will now combine this theorem with the transformation proper-ties of the Bergman kernel functions under biholomorphic mappings toobtain the following theorem.

Theorem 22.2. Suppose that f : Ω1 → Ω2 is a biholomorphic mappingbetween domains of finite area. Then Ω2 is an area quadrature domainif and only if f ′ belongs to the Bergman span of Ω1.

Proof. Suppose f is a biholomorphic mapping as in the statement of thetheorem. Because |f ′|2 is equal to the absolute value of the real Jacobiandeterminate of the mapping as a mapping from R2 into itself, it followsfrom the classical change of variables formula (see Chapter 15) that

∫∫

Ω2

h dA = 〈h, 1〉Ω2=

∫∫

Ω1

|f ′|2(h f) dA = 〈f ′(h f), f ′〉Ω1.

If f ′ is in the Bergman span, then the last inner product yields a linearcombination of values and derivatives of f ′(hf) at finitely many pointsin Ω1, which is a fixed finite linear combination of values and derivativesof h at finitely many points in Ω2. Consequently, Ω2 is an area quadraturedomain.

Conversely, suppose that Ω2 is an area quadrature domain. Let F =f−1. If h is a function in H2(Ω1), then identity (15.1) yields

〈h, f ′〉Ω1= 〈F ′(h F ), 1〉Ω2

=

∫∫

Ω2

F ′(h F ) dA,

and this last integral is equal to a linear combination of values andderivatives of F ′(hF ) by virtue of the quadrature identity for Ω2. Sucha linear combination is a linear combination of values and derivatives ofh at finitely many points in Ω1. There is a function Q in the Bergmanspan of Ω1 that yields the same linear combination when paired withh. Since 〈h, f ′〉Ω1

= 〈h,Q〉Ω1holds for all h in the Bergman space, it

follows that f ′ = Q, i.e., that f ′ is in the Bergman span.


When Theorem 22.2 is combined with the Riemann mapping theoremand the known formula for the Bergman kernel on the disc, the followinglovely result is obtained.

Theorem 22.3. A simply connected domain of finite area is an areaquadrature domain if and only if the inverse of a Riemann mappingfunction is rational and without poles on the boundary.

Proof. Suppose that Ω is a simply connected domain of finite area and letf : Ω → D1(0) be a Riemann mapping function associated to a point inΩ. Recall that the Bergman kernel for the unit disc is π−1/(1−zw)2. LetF = f−1 be the inverse of the Riemann map. Theorem 22.2 yields that Ωis an area quadrature domain if and only if F ′ is in the Bergman span ofthe disc. It is easy to verify that the Bergman span of the disc is exactlythe set of rational functions with residue free poles outside the closedunit disc. (Note that the complex polynomials belong to this space.)Such rational functions have rational antiderivatives and the set of allsuch antiderivatives is the set of rational functions with poles outsidethe closed unit disc.

We remark that the reasoning in the proof of Theorem 22.3 can beused to give a quick proof that discs are the only “one point” simplyconnected quadrature domains. Indeed, if Ω is a simply connected areaquadrature domain satisfying

∫∫

Ω

h dA = ch(a)

for all h in the Bergman space, let f be a Riemann map that takes a tothe origin and let K(z, w) denote the Bergman kernel associated to Ω.Putting the function h ≡ 1 in the quadrature identity shows that c isequal to the area of Ω. The proof of Theorem 22.1 yields that Ka is equalto the constant C = 1/c. The transformation formula for the Bergmankernels yields

F ′(z)K(F (z), a) =1

π(1− z f(a) )2f ′(a), (22.4)

and since f(a) = 0 and K(·, a) is constant, we conclude that F ′(z) is aconstant. Consequently F (z) is complex linear and Ω is a disc.

It is interesting to next consider simply connected domains in theplane with two point quadrature identities. Similar reasoning can beused, mapping one of the points a1 in the quadrature identity to theorigin via a Riemann map and using the transformation formula, toshow that such domains must be the famous Neumann ovals. Indeed,


the two point quadrature identity yields constants c1 and c2 such thatc1K(z, a1)+ c2K(z, a2) ≡ 1. Adding up the appropriate linear combina-tions of equation (22.4) yields

F ′(z) = A1 +A2

(1− z f(a2) )2,

where A1 and A2 are constants. It follows that F (z) = C0 + A1z +B2/(z − b), where C0, A1, B2 are complex constants and b = 1/ f(a2)is a point outside the closed unit disc. If F is one-to-one on the unitdisc, then F (D1(0)) is indeed a two point area quadrature domain. SeeShapiro [Sh] for more about these special domains. We will soon see thatcontinuing this line of thought leads to a class of quadrature domainsthat are “dense” in the realm of simply connected domains with smoothboundaries.

Theorem 22.3 reveals that simply connected area quadrature domainshave algebraic Riemann mapping functions and particularly nice bound-aries. In particular, the function F in the proof is holomorphic in aneighborhood of the closed unit disc. If F ′ is nonvanishing on the closedunit disc, then it follows that the boundary of the simply connected areaquadrature domain must be C∞ smooth and real analytic (and, in fact,real algebraic). If F ′ has a zero on the unit circle, then near such a zero,for F to map one-to-one onto a domain, the zero can be at most a simplezero and F maps the zero to a boundary point of Ω that is the terminalend of a cusp pointing into the domain. (Otherwise, the local mappingtheorem would show that F is not one-to-one on a small disc aboutthe boundary point intersected with the unit disc.) Thus, Ω either hasC∞ smooth boundary, or is C∞ smooth except at finitely many inwardpointing cusps.

We will now show that simply connected area quadrature domainswithout cusps in the boundary are “dense” in the realm of simply con-nected domains with smooth boundary. Suppose that Ω is a simply con-nected domain bounded by a C∞ smooth curve. Let f denote a Riemannmap associated to a point in the domain. We know that f is in C∞(Ω)and that the inverse F of f is in C∞(D1(0)). If ρ < 1, then F (ρz) isholomorphic in a neighborhood of the closed unit disc and we may ap-proximate F (ρz) in C∞(D1(0)) by a Taylor polynomial P (z). By takingρ to be sufficiently close to one and taking the polynomial to be suf-ficiently close to F (ρz) in C∞(D1(0)), we obtain a domain P (D1(0))that is an area quadrature domain without cusps that is as close toΩ in C∞ as desired in the sense that the biholomorphic map P f isas close to the identity in C∞(Ω) as desired. This is the improvementupon the Riemann mapping theorem alluded to above. Instead of map-ping the domain to the unit disc, which might be a far away quadrature


domain, we have mapped to a nearby quadrature domain. Since quadra-ture domains have many properties in common with the unit disc, thisnearby quadrature domain might be more useful than a far away disc innumerical computations.

The argument above can be reworked on a domain bounded by aJordan curve to obtain a mapping that is uniformly close to the iden-tity map and that maps to an area quadrature domain without cuspsin the boundary. The key to the argument is Caratheodory’s theoremabout continuous extension to the boundary of Riemann maps in thissetting. This yields Gustafsson’s theorem about the uniform density ofarea quadrature domains among Jordan domains.

We now reveal one of the very special properties of area quadraturedomains that make them akin to the unit disc. Simply connected areaquadrature domains without cusps in the boundary have a well behaved“Schwarz function” that is quite useful. The Schwarz function for theunit disc is S(z) = 1/z. Notice that it is meromorphic on the disc, itis holomorphic on a neighborhood of the unit circle, and S(z) = z onthe unit circle. We remark here that the antiholomorphic reflection func-tion studied in Chapter 11 is given by R(z) = S(z) = 1/z. To motivatewhat we are about to do next, keep in mind that we would expect an-tiholomorphic reflection functions to be invariant under biholomorphicmappings that extend holomorphically past the boundaries.

Suppose that Ω is a simply connected area quadrature without cuspsand let f be a Riemann map with inverse F as above. Since F is rational,we may think of F as defined on the whole complex plane minus finitelymany poles. The Schwarz function S(z) for Ω is defined via

S(z) = F (1/ f(z) ).

Note that S is holomorphic where it is well defined because a compositionof antiholomorphic functions is holomorphic. Also note that S has nosingularities on the unit circle and at worst poles where f has zeroes orwhere 1/ f maps to a pole of F . We claim that S(z) is such that S(z) = zon the boundary of Ω and S extends meromorphically to Ω. Indeed, ifz is a boundary point of Ω, then f(z) is a point on the unit circle, andso f(z) = 1/ f(z). Since F is the inverse of f , it follows that S(z) = z.Also, 1/f maps Ω onto-to-one onto the outside of the closed unit discunion the point at infinity. Since F is rational, we conclude that S(z)is meromorphic on Ω. Since S(z) has no singularities on the boundary,S(z) extends to be holomorphic on a neighborhood of the boundary.

We now turn to the study of area quadrature in the multiply con-nected category. Since the Riemann mapping theorem is no longer atour disposal, we will have to come up with new techniques, but we willobtain many similar results to those in the simply connected setting.


Suppose Ω1 is a finitely connected domain of finite area. The same ar-gument used in the proof of Lemma 12.1 (see also Ahlfors [Ah, p. 252])yields that there exists a biholomorphic mapping f : Ω1 → Ω2 to abounded domain Ω2 with smooth real analytic boundary. Let F = f−1

denote the inverse of f . Theorem 22.2 states that Ω1 is an area quadra-ture domain if and only if F ′ is in the Bergman span of Ω2. Note thatTheorem 17.1 together with the fact that Km

a is the Bergman projec-tion of a function in C∞

0 (Ω) shows that the Bergman span of a boundeddomain with smooth real analytic boundary consists of functions thatextend holomorphically past the boundary. Thus, we have proved thefollowing theorem.

Theorem 22.4. If Ω is a finitely connected area quadrature domain,then Ω is a bounded domain bounded by finitely many nonintersectingreal analytic curves that are either C∞ smooth or C∞ smooth except atfinitely many inward pointing cusps.

Suppose that Ω is a finitely connected area quadrature domain with-out cusps in the boundary. We will now show that Ω has a Schwarzfunction that extends meromorphically to Ω exactly as in the simplyconnected case. Since Ω has C∞ smooth boundary, we may apply the im-proved Cauchy integral formula of Theorem 2.1 to the function u(z) = zto obtain

z = Cz + 1

2πi

∫∫

w∈Ω

1

w − zdw ∧ dw

for z ∈ Ω. Recall that the boundary values of the integral over Ω can beevaluated by letting z approach the boundary from the outside of thedomain (see Theorem 17.2). If z is outside the closure of the domain,then the quadrature identity for Ω yields a rational function R(z) in zwith no poles outside Ω. Thus, we conclude that z = h(z)+R(z) on theboundary, where h = Cz is a holomorphic function in C∞(Ω) and R(z)is a rational function without poles outside Ω. The Schwarz function istherefore given by h+R and is seen to extend meromorphically to Ω.

To illustrate the utility of the Schwarz function, we will use it toprove the following result.

Theorem 22.5. Suppose that Ω is a finitely connected area quadraturedomain without cusps in the boundary. Then the complex polynomialsbelong to the Bergman span of Ω.

Proof. Suppose that Ω is a domain as in the statement of the theo-rem. Let S(z) denote the Schwarz function for Ω. Given a function


h ∈ A∞(Ω), observe that the complex Green’s identity yields

〈h, zn〉Ω =(−i/2)n+ 1

∫∫

Ω

∂

∂z

[h(z)zn+1

]dz ∧ dz

= − (i/2)

n+ 1

∫

bΩ

h(z)zn+1 dz

= − i/2

n+ 1

∫

bΩ

h(z)S(z)n+1 dz. (22.5)

This last integral is equal to a finite linear combination of values andderivatives of h at finitely many points via the residue theorem. Thereis a function q in the Bergman span that produces this same linearcombination when paired with such an h. Since A∞(Ω) is dense inH2(Ω),we conclude that zn = q(z), i.e., that zn is in the Bergman span. Sincethis argument works for any nonnegative integer n, we conclude that allcomplex polynomials are in the Bergman span.

Since the function 1 being in the Bergman span is equivalent to thearea quadrature condition, one can say that a smooth finitely connecteddomain of finite area is an area quadrature domain if and only if theBergman span contains all complex polynomials.

We remark that letting n = 0 in equation (22.5) shows that if abounded domain Ω with C∞ smooth boundary has a Schwarz functionS(z) that is meromorphic on Ω, extends continuously to the boundary,and satisfies S(z) = z on the boundary, then Ω is a smooth area quadra-ture domain. Thus, having a Schwarz function like this is equivalent tobeing an area quadrature in the realm of smooth domains. Note that theSchwarz function has poles at the points that appear in the quadratureidentity and the numbers mj + 1 are the orders of the poles. It can beshown that the existence of a Schwarz function with the properties wehave been using is equivalent to the area quadrature condition in muchmore generality and we direct the reader to the Bibliographic Notes foravenues to explore.

A theorem that is closely related to the fact that one-point quadra-ture domains are discs states that the only domains with complex ra-tional Schwarz functions are discs. Indeed, if S(z) is a rational Schwarzfunction for a domain, then R(z) = S(z) is the antiholomorphic reflec-tion function studied in Chapter 11. The fact that S(z) = z on theboundary yields that R(z) = z on the boundary. Consequently, R R isholomorphic near the boundary and equal to z on the boundary. There-fore, R(R(z)) = z globally, i.e., R is equal to its own inverse. The onlyrational functions of z with this property are the linear fractional trans-formations. We conclude that S(z) has at most one pole (which must


fall inside the domain), and therefore equation (22.5) with n = 0 showsthat the domain is a one-point quadrature domain, and hence, a disc.

We remark that the Bergman kernel and the Schwarz function arehighly related in area quadrature domains, and that will be the subjectof the last chapter of this book.

We will close this chapter by sketching a proof that any boundeddomain with C∞ smooth boundary can be approximated by finitelyconnected area quadrature domains without cusps via a biholomorphicmapping that is as C∞-close to the identity as desired.

We have already handled the simply connected case, so suppose thatΩ is a bounded multiply connected domain with C∞ smooth boundary.We will prove in Chapter 30 that the space of all linear combinationsof functions of the form Ka as a ranges over Ω is dense in A∞(Ω), andhence the Bergman span of Ω is dense in A∞(Ω). We now approximatethe function H(z) ≡ 1 in A∞(Ω) by a function Q in the Bergman span.Suppose that Ω has n boundary curves γj , j = 1, . . . , n. Let the first n−1of those curves denote inner boundary curves and let bj , j = 1, . . . , n−1be points inside the bounded domains determined by the inner boundarycurves, one per curve. We may also approximate the functions hj(z) =1/(z− bj) in A∞(Ω) by functions qj in the Bergman span. Note that theperiod matrix

∫γihj dz as i and j range from 1 to n− 1 is nonsingular.

Hence, by making our approximations close enough, we may assume thatthe period matrix of the functions qj is also nonsingular. Since Q is closeto one, the periods of Q are small. Hence, by taking Q sufficiently closeto one, we may assume that there are small constants ǫj such that the

periods of Q−∑n−1j=1 ǫjqj are zero. Because the periods are zero, we know

there is an antiderivative f of this last function on Ω. If we make all ourapproximations sufficiently close, we see that f can be made arbitrarilyclose to the identity map (since f ′ is close to one). In particular, wemay assume that f is one-to-one on Ω and close to the function z inC∞(Ω). Since f ′ is in the Bergman span, the image of Ω under f isan n-connected area quadrature domain without cusps in the boundarythat is as C∞ close to Ω as desired. This result can clearly be viewed asan improvement upon the Riemann mapping theorem, which does nothold in the multiply connected domain setting.

To show that a domain bounded by finitely many nonintersectingJordan curves can be approximated by a smooth area quadrature do-main (Gustafsson’s theorem [Gu1]), first map the domain to a domaincontained in the unit disc via a Riemann map f associated to the simplyconnected domain given by the inside of the outer boundary curve. If Fis the inverse of this map, we may use the dilated map Fρ(z) = F (ρz)as we did in the simply connected case to obtain a map Fρ f that isclose to the identity map that maps the domain to a nearby domain


bounded by Jordan curves, where the outer boundary is now real ana-lytic. If b is a point inside one of the bounded domains determined byan inner boundary curve of this new domain, we may invert the domainusing the map 1/(z − b) to send that inner boundary curve to the outerboundary. We now repeat the process for this new outer boundary, not-ing that the boundary curve represented by the old outer boundary isnow a real analytic curve, too. Hence, two of the boundary curves arereal analytic curves at the end of this next iteration. We now composewith the inverse of 1/(z−b) (which is b+(1/w)) to obtain a domain closeto the original domain where two of the boundary curves are real ana-lytic. It is now clear how to continue this process until a nearby domainwith real analytic boundary is obtained. Finally, we map this domainwith real analytic boundary to a quadrature domain via a map that isC∞ close to the identity and compose to obtain a map to a quadraturedomain that is uniformly close to the identity.

23

Arc length quadrature domains

Not only is the unit disc the most famous of area quadrature domains,it is also the most famous of boundary arc length quadrature domains.When a holomorphic function that extends continuously to the bound-ary is averaged over the unit circle with respect to arc length measure,the value of the function at the origin is obtained. More generally, afinitely connected domain in the plane bounded by n nonintersecting C1

smooth curves is called an arc length quadrature domain if the averageof a holomorphic function that extends continuously to the closure ofthe domain over the boundary with respect to arc length measure is afinite linear combination of values of the function and its derivatives atfinitely many points in the domain. The points and the coefficients arefixed in this quadrature identity even though the holomorphic functionis allowed to vary. In this chapter, we will restrict our attention to arclength quadrature domains with C∞ smooth boundaries, and we will callsuch domains smooth arc length quadrature domains. In this context, abounded domain Ω with C∞ smooth boundary is called a smooth arclength quadrature domain if there exist points zjNj=1 in Ω, complexconstants cjk, and nonnegative integers mj such that

∫

bΩ

h ds =

N∑

j=1

mj∑

k=0

cjkh(k)(zj) (23.1)

for all h in the Hardy spaceH2(bΩ). In this chapter, we will show that arclength quadrature domains are to the Szego kernel as area quadraturedomains are to the Bergman kernel. We will also consider domains thatare like the unit disc in that they are both area and arc length quadraturedomains. Once again, these results can be viewed as improvements uponthe Riemann mapping theorem.

The reader will notice that this chapter is highly parallel to the pre-vious one.

To begin, we must define higher order versions of the Szego kernelmuch like we did in Chapter 22 for the Bergman kernel. In Chapter 11,we showed that the Szego kernel Sa was the Szego projection of the

123


Cauchy kernel Ca. We can differentiate the identity

h(z) = 〈h,Ca〉b = 〈h, Sa〉b

exactly as we did for the Bergman kernel to obtain higher order functionsSma which are Szego projections of (∂m/∂am)Ca. Analogous arguments

show that Sma (z) = (∂m/∂am)S(z, a). Note that Sm

a is in A∞(Ω) and

h(m)(a) = 〈h, Sma 〉b.

Let S0a denote Sa.

The Szego span of Ω is defined to be the space of all linear combina-tions of functions of the form Sm

a as a ranges over Ω and m ranges overnonnegative integers. Note that Theorem 9.1 yields that the Szego spanis dense in A∞(Ω) in smooth domains.

We now show that there is a theorem analogous to Theorem 22.1 forsmooth arc length quadrature domains.

Assume that Ω is a smooth arc length quadrature domain. Note thatif h is in the Hardy space, then

∫

Ω

h ds = 〈h, 1〉b.

We assume that the quadrature identity (23.1) holds. Let Q denote thefollowing linear combination of functions Sk

zj in the Szego span:

Q(z) =

N∑

j=1

mj∑

k=0

cjkSkzj . (23.2)

Notice that〈h, 1〉b = 〈h,Q〉b

for h ∈ H2(bΩ) because both sides yield the quadrature identity in h.Since the functions 1 and Q are both in the Hardy space, and since h isarbitrary, if must be that Q ≡ 1. The converse of this fact is also clearlytrue, namely that if Ω is a bounded domain with C∞ smooth boundaryand the function 1 is in the Szego span, then Ω is a smooth arc lengthquadrature domain. Thus, we have proved the following theorem.

Theorem 23.1. A bounded domain with C∞ smooth boundary is asmooth arc length quadrature domain if and only if the function thatis identically equal to one is in the Szego span.

We can combine this theorem with transformation properties of theSzego kernel functions under biholomorphic mappings to obtain the fol-lowing theorem analogous to Theorem 22.2.

Arc length quadrature domains 125

Theorem 23.2. Suppose that f : Ω1 → Ω2 is a biholomorphic mappingbetween bounded domains with C∞ smooth boundaries. Then Ω2 is asmooth arc length quadrature domain if and only if

√f ′ belongs to the

Szego span of Ω1.

Proof. Suppose f is a biholomorphic mapping as in the statement of thetheorem. Theorem 12.1 informs us that f extends nicely to the boundaryand there is a holomorphic square root of f ′ that also extends nicely.Let

√f ′ denote one of the two possible square root functions. If h is in

H2(bΩ2), it follows from the discussion before equation (12.3) that∫

bΩ2

h ds = 〈h, 1〉bΩ2=

∫∫

bΩ1

|f ′|(h f) dA = 〈√f ′(h f),

√f ′〉bΩ1

.

If√f ′ is in the Szego span, then the last inner product yields a linear

combination of values and derivatives of√f ′(h f) at finitely many

points in Ω1, which is a fixed finite linear combination of values andderivatives of h at finitely many points in Ω2. Consequently, Ω2 is asmooth arc length quadrature domain.

Conversely, suppose that Ω2 is an arc length quadrature domain. LetF = f−1. If h is a function in H2(bΩ1), then identity (12.4) yields

〈h,√f ′〉bΩ1

= 〈√F ′(h F ), 1〉bΩ2

=

∫

bΩ2

√F ′(h F ) ds,

and this last integral is equal to a linear combination of values andderivatives of

√F ′(hF ) by virtue of the quadrature identity for Ω2. Such

a linear combination is a linear combination of values and derivatives ofh at finitely many points in Ω1. There is a function Q in the Szego spanof Ω1 that yields the same linear combination when paired with h. Since〈h,√f ′〉bΩ1

= 〈h,Q〉bΩ1holds for all h in the Hardy space, it follows that√

f ′ = Q, i.e., that√f ′ is in the Szego span.

When Theorem 23.2 is combined with the Riemann mapping theoremand the known formula for the Szego kernel on the disc, the followinganalogue to Theorem 22.3 is obtained.

Theorem 23.3. A bounded simply connected domain with C∞ smoothboundary is an arc length quadrature domain if and only if the derivativeof the inverse of a Riemann mapping function is the square of a rationalfunction without poles on the closed unit disc.

Proof. Suppose that Ω is a bounded simply connected domain with C∞

smooth boundary and let f : Ω → D1(0) be a Riemann mapping functionassociated to a point in Ω. Recall that the Szego kernel for the unit discis (2π)−1/(1 − zw). Let F = f−1 be the inverse of the Riemann map.


Theorem 23.2 yields that Ω is an area quadrature domain if and only if√F ′ is in the Szego span of the disc. It is easy to verify that the Szego

span of the disc is exactly the set of rational functions with poles outsidethe closed unit disc. (Note that the complex polynomials belong to thisspace.) Hence,

√F ′ is in the Szego span if and only if F ′ is the square

of such a function and the proof is complete.

We remark here that if Ω is a one-point simply connected smooth arclength quadrature domain, then we can adapt the proof of Theorem 23.3just as we did after the proof of Theorem 22.3 to deduce that Ω must bea disc. Indeed, the argument shows that the square root of the derivativeof the inverse of a Riemann map that maps the one point to the origin isconstant, and so the inverse is complex linear and the domain is a disc.

Theorem 23.3 reveals that simply connected smooth arc lengthquadrature domains have very nice boundaries. Indeed, the function Fin the proof is holomorphic in a neighborhood of the closed unit disc andF ′ must be nonvanishing on the closed unit disc for it to have a squareroot that is smooth up to the boundary. It therefore follows that theboundary of the simply connected arc length quadrature domain mustbe C∞ smooth and real analytic.

We will now show that smooth simply connected arc length quadra-ture domains are dense in the realm of simply connected domainswith smooth boundary. Suppose that Ω is a simply connected domainbounded by a C∞ smooth curve. Let f denote a Riemann map associ-ated to a point in the domain, which we know is in C∞(Ω). The inverseF of f is in C∞(D1(0)). If ρ < 1, let Fρ(z) = F (ρz). Notice that Fρ isholomorphic in a neighborhood of the closed unit disc. There is a holo-morphic square root of

√F ′ρ that also extends smoothly to the closure

of the unit disc. We may approximate√F ′ρ in C∞(D1(0)) by a Taylor

polynomial q(z). Let P (z) be a polynomial that is an antiderivative ofq(z)2. We may choose the constant of integration so that P ′(z) is closeto F ′

ρ in C∞(D1(0)). By taking ρ to be sufficiently close to one and tak-ing the polynomial approximates sufficiently close to their targets, weobtain a domain P (D1(0)) that is both a smooth arc length quadratureand an area quadrature domain without cusps that is as close to Ω inC∞ as desired in the sense that the biholomorphic map P f is as closeto the identity in C∞(Ω) as desired. This is an improvement upon theimprovement upon the Riemann mapping theorem of the last chapter.We call smooth domains that are both area and arc length quadraturedomains double quadrature domains. Instead of mapping the domain tothe unit disc, which might be a far away double quadrature domain, wehave mapped to a nearby double quadrature domain. Double quadraturedomains have even more properties in common with the unit disc.


We now turn to the study of arc length quadrature domains in themore general finitely connected setting.

First, we show that smooth arc length quadrature domains have thevery special property that their complex unit tangent vector functionsT (z) extend meromorphically to the domain. Indeed, if we differenti-ate identity 7.1 with respect to a and add up a linear combination ofconjugates of functions in the Szego span that sum to one (given byTheorem 23.1), we obtain that 1 = HT on the boundary, where H(z)is a meromorphic function on Ω that extends smoothly to the boundaryand is given by a linear combination of the Garabedian kernel La(z)and its derivatives in a at finitely many points a in Ω. Hence 1/H isthe meromorphic extension we seek. Note that, since T = 1/T , it fol-lows that T extends as the meromorphic function H to Ω. The followingcomputation shows that this condition of extendibility of T is equivalentto the arc length quadrature condition in the realm of smooth domains.Indeed, if Ω is a bounded C∞ smooth domain and h is in A∞(Ω), then

∫

bΩ

h ds =

∫

bΩ

h T T ds =

∫

bΩ

h H dz,

where H is the meromorphic extension of T , and this last integral is alinear combination of values and derivatives of h at finitely many pointsin Ω via the residue theorem. Since A∞(Ω) is dense in the Hardy space,this identity extends to H2(bΩ) and we conclude that Ω is an arc lengthquadrature domain.

It can be shown that the extension property of T is equivalent tothe arc length quadrature condition in much more generality and, onceagain, we direct the curious reader to the Bibliographic Notes.

We next show that the boundaries of smooth arc length quadraturedomains are particularly nice. Suppose Ω1 is a bounded domain withC∞ smooth boundary. We know that there exists a biholomorphic map-ping f : Ω1 → Ω2 to a bounded domain Ω2 with smooth real analyticboundary. Let F = f−1 denote the inverse of f . Theorem 23.2 statesthat Ω1 is an area quadrature domain if and only if

√F ′ is in the Szego

span of Ω2. Theorem 11.2 can be used to show that the Szego span ofa bounded domain with smooth real analytic boundary consists of func-tions that extend to be holomorphic in a neighborhood of the closureof the domain. Thus,

√F ′ extends holomorphically past the boundary,

and consequently, so does F ′ and F . As in the area quadrature domaincase, F ′ can vanish to at most first order on the boundary in order tomap one-to-one to a domain, but since

√F ′ extends, F ′ cannot vanish

on the boundary to first order. We have proved the following theorem.

Theorem 23.4. If Ω is a finitely connected smooth arc length quadrature


domain, then Ω is a bounded domain bounded by finitely many nonin-tersecting C∞ smooth real analytic curves.

In the last chapter, we saw that being an area quadrature domainwas essentially equivalent to the complex polynomials belonging to theBergman span. The analogous theorem for arc length quadrature do-mains is as follows.

Theorem 23.5. Suppose that Ω is a bounded domain with C∞ smoothboundary. Then the complex polynomials belong to the Szego span of Ωif and only if Ω is a smooth double quadrature domain.

Proof. Suppose that Ω is a smooth double quadrature domain. Let S(z)denote the Schwarz function for Ω and let H denote the meromorphicextension of T to Ω. Given a function h ∈ A∞(Ω), observe that

〈h, zn〉b =∫

bΩ

h zn T T ds =

∫

bΩ

hS(z)nH(z) dz,

which is a finite linear combination of values and derivatives of h atfinitely many points via the residue theorem. There is a function q inthe Szego span that produces this same linear combination when pairedwith such an h. Since A∞(Ω) is dense in H2(bΩ), we conclude thatzn = q(z), i.e., that zn is in the Szego span. Since this argument worksfor any nonnegative integer n, we conclude that all complex polynomialsare in the Szego span.

To prove the converse, if the complex polynomials are in the Szegospan, then z and 1 are. The fact that 1 is in the span implies that thedomain is an arc length quadrature domain. The function z is equalto z/1, which in turn is equal to a quotient of functions in the Szegospan. If we apply versions of identity (7.1) differentiated with respect toa to these functions, we see that the quotient is equal to the conjugateof a quotient of linear combinations of the Garabedian kernel and itsderivatives in the second variable. Thus, z is equal to the boundaryvalues of a meromorphic function, namely, the Schwarz function S(z),and we conclude that Ω is an area quadrature domain, too.

We close this chapter by remarking that it is possible to prove thatany bounded domain with C∞ smooth boundary can be approximatedby a smooth arc length quadrature domain via a biholomorphic mappingthat is as C∞-close to the identity as desired. The arguments closely par-allel those used in the last chapter, but are somewhat more complicatedbecause of the square roots. A considerably more difficult result is toprove that any bounded domain with C∞ smooth boundary can be ap-proximated by a smooth double quadrature domain via a biholomorphic


mapping that is as C∞-close to the identity as desired. This result is aneven better improvement upon the Riemann mapping theorem than ourprevious ones because double quadrature domains are even more like theunit disc. We refer the interested reader to the Bibliographic Notes forreferences to these results.

24

The Hilbert transform

Suppose that Ω is a bounded simply connected domain with C∞ smoothboundary. Let a ∈ Ω be a fixed point. Given a real valued function uin C∞(bΩ), we may identify u as the boundary values of a real valuedharmonic function U on Ω that is in C∞(Ω) (see Theorem 10.1). We willshow that there is a real valued harmonic conjugate function V (meaningthat U + iV is holomorphic on Ω) such that V is also in C∞(Ω). Wecan make V uniquely determined by specifying that V (a) = 0. Underthese conditions, let v denote the restriction to the boundary of V . Thefunction v is called the Hilbert transform of u and we write Hu = v. Inthis chapter, we will prove that the Hilbert transform is a well definedlinear operator and we will prove the classical facts that H maps C∞(bΩ)into itself and that H extends uniquely to be a bounded operator onL2(bΩ).

Given a real valued function u in C∞(bΩ), we have expressed theharmonic extension U of u as a sum h+H where h and H are in A∞(Ω)and are given by h = P (Sau)/Sa and H = P (Lau)/La (Theorem 10.1).Since u is real valued, the maximum principle shows that its harmonicextension U is also real valued, and it follows that the imaginary partsof h and H must be equal. Hence, h and H differ by a real constant onΩ. In fact, since H(a) = 0, we deduce that h(z) = H(z) + h(a) for allz ∈ Ω, and therefore, that the harmonic extension of u is given by

U(z) = h(a) + 2Re H(z).

We have expressed the harmonic extension of u as the real part of the hol-omorphic function h(a)+2H(z), and we may write U + iV = h(a)+2H .The real valued function V in this formula is a harmonic conjugatefunction for U and it is uniquely determined by the condition thatV (a) = 2H(a) = 0. The Hilbert transform Hu of u is defined to bethe function v on bΩ given by the boundary values of V . We have shownthat Hu = 2Im H .

Since Im h = Im H, we may summarize our work above in the fol-lowing theorem.

Theorem 24.1. The Hilbert transform on a bounded simply connected

131


domain Ω with C∞ smooth boundary is given by

Hu = 2Im

(P (La u)

La

),

or

Hu = 2Im

(P (Sa u)

Sa

)(24.1)

where P denotes the Szego projection associated to Ω. It follows that theHilbert transform is a linear operator mapping C∞(bΩ) into itself thatextends uniquely to be a bounded operator from L2(bΩ) to itself.

Indeed, since Sa is a nonvanishing function in C∞(bΩ) and since Pis a bounded operator on L2(bΩ), the fact that the Hilbert transformextends to be a bounded operator on L2(bΩ) can be read off from for-mula (24.1).

Let f denote the Riemann mapping function that maps Ω one-to-one onto the unit disc with f(a) = 0 and f ′(a) > 0. It was shown inChapter 12 that the Szego kernel transforms under the map f via

S(z, a) =√f ′(z)SU (f(z), f(a))

√f ′(a), (24.2)

where SU (z, w) denotes the Szego kernel of the unit disc. Now, SU (z, 0) ≡(2π)−1. Hence, it follows from (24.2) that

S(z, a) = (2π)−1√f ′(z)

√f ′(a). (24.3)

When this formula is plugged into (24.1), we obtain the next theorem.

Theorem 24.2. The Hilbert transform can be written

Hu = 2Im

(P(u√f ′)

√f ′

)

where f is the Riemann mapping function that maps Ω one-to-one ontothe unit disc with f(a) = 0 and f ′(a) > 0.

Formula (24.1) can be used to estimate the constant C in the L2

estimate ‖Hu‖ ≤ C‖u‖. Indeed, we have

‖Hu‖ ≤ 2‖S−1a P (Sau)‖ ≤ 2

(Maxw∈bΩ

|Sa(w)−1|)‖P (Sau)‖,

and

‖P (Sau)‖ ≤ ‖Sau‖ ≤(Maxw∈bΩ

|Sa(w)|)‖u‖.

The Hilbert transform 133

Let

M = Maxw∈bΩ

|S(w, a)|, and

m = minw∈bΩ

|S(w, a)|.

We have proved that the smallest possible constant C in the L2 estimate‖Hu‖ ≤ C‖u‖ for the Hilbert transform satisfies

C ≤ 2M

m.

The fraction M/m can also be described in terms of the Riemann map-ping function f that maps Ω one-to-one onto the unit disc with f(a) = 0and f ′(a) > 0. Let

Λ = Maxw∈bΩ

|f ′(w)|, and

λ = minw∈bΩ

|f ′(w)|.

Then, using (24.3), we see that M/m = Λ1/2/λ1/2.We mention one last formula. The Szego projection transforms under

the Riemann map f according to the identity

P(√

f ′(ϕ f))=√f ′ ((PUϕ) f) ,

where PU denotes the Szego projection on the unit disc U and ϕ ∈L2(bU). Apply this identity to the function ϕ = u f−1 and plug theresult into the formula of Theorem 24.2 to obtain

Hu = 2Im(PU (u f−1)

) f.

Actually, this last formula is not hard to derive from first principles andit is possible to make this result the starting point of the theory as analternative to basing the reasoning on Theorem 4.3 and identity (7.1).

25

The Bergman kernel and the Szegokernel

The reader should suspect that, because boundary integrals can read-ily be turned into solid integrals by means of the Green’s formula, theBergman kernel and the Szego kernel of a domain should be closely re-lated. In this chapter, we show that they are very closely related indeed.


smooth boundary, then the Bergman kernel K(z, a) associated to Ω isrelated to the Szego kernel via the identity

K(z, a) = 4πS(z, a)2.

To prove this identity, we use the relationships that exist between thederivative of a Riemann map and the two kernels. Let f be a biholomor-phic map of Ω onto the unit disc such that f(a) = 0 and f ′(a) > 0. It wasshown in Chapter 15 that f ′(z) = C K(z, a) where C =

√π/K(a, a).

Theorem 12.3 states that f ′(z) = cS(z, a)2 where c = 2π/S(a, a). Henceit follows that K(z, a) = (c/C)S(z, a)2. By plugging in z = a into thislast formula and by using the expressions for c and C, it can be deducedthat c/C = 4π.

If Ω is multiply connected, then the relationship between the ker-nels is not as direct. In the next theorem, the functions F ′

j(z) denotethe derivatives of the classical harmonic measure functions that wereintroduced in Chapter 19.

Theorem 25.2. Suppose that Ω is a bounded n-connected domain withC∞ smooth boundary. Then the Bergman kernel and Szego kernel arerelated via

K(z, a) = 4πS(z, a)2 +

n−1∑

j=1

λjF′j(z),

where the coefficients λj are constants in z which depend on a.

Proof. Define G(z) = K(z, a)−4πS(z, a)2. To prove the theorem, we willshow that GT is orthogonal in L2(bΩ) to H2(bΩ) and to the space of

135


conjugates of functions in H2(bΩ). Then Theorem 19.1 implies that G =∑λjF

′j and this is what we want to see. Since K(z, a) and S(z, a) are in

A∞(Ω) as functions of z for fixed a ∈ Ω, Theorem 4.3 yields immediatelythat GT is orthogonal to the space of conjugates of functions in H2(bΩ).To see that GT is orthogonal to H2(bΩ), let h ∈ A∞(Ω) and compute

〈KaT, h〉bΩ =

∫

bΩ

KaT hds =

∫

bΩ

Ka h dz

=

∫∫

Ω

Ka h′dz ∧ dz = 2i 〈Ka, h′〉Ω = 2i h′(a)

(by the reproducing property of the Bergman kernel). Next, observe that,by identity (7.1),

(Sa)2T = −(La)

2T .

By integrating this identity around the boundary with respect to ds, wesee that the residue of (La)

2 at z = a is zero by applying the residuetheorem on the right hand side and Cauchy’s theorem on the left. Nowwe may compute

〈(Sa)2T, h〉bΩ =

∫

bΩ

(Sa)2T hds = −

∫

bΩ

(La)2T hds

= −∫

bΩ

(La)2 h dz = −(−2πi)1

(2π)2h′(a)

because (2πLa)2 = (z − a)−2 +H where H ∈ A∞(Ω). Hence, we have

shown that∫bΩGT h ds = 0. Since A∞(Ω) is dense in H2(bΩ), it follows

that GT is orthogonal to H2(bΩ), and the proof is complete.

The numbers λj in Theorem 25.2 are functions of a. In fact, it is nothard to show that λj(a) is an antiholomorphic function of a given by

λj(a) =

n−1∑

k=1

CjkF ′k(a)

for some constants Cjk . To see this, we integrate the formula in Theo-rem 25.2 around one of the n− 1 inner boundary curves γk to obtain

∫

γk

(K(z, a)− 4πS(z, a)2

)dz =

n−1∑

j=1

λj

∫

γk

F ′j(z) dz =

n−1∑

j=1

Akjλj

where [Akj ] denotes the nonsingular matrix of periods that we discussedin Chapter 19. Let γǫk represent a curve that is homotopic to γk, butthat is inside Ω (such as the curve traced out by a point at a distance of

The Bergman kernel and the Szego kernel 137

ǫ along the inward pointing unit normal vector to γk for small ǫ.) Wemay now write

∫

γǫk

(K(z, a)− 4πS(z, a)2

)dz =

n−1∑

j=1

Akjλj(a).

This shows that the λj(a) are antiholomorphic functions of a becausethe kernels are antiholomorphic in a. Now consider the idea of approxi-mating the integral in the identity above by a finite Riemann sum. SinceK(z, a)− 4πS(z, a)2 is the conjugate of K(a, z)− 4πS(a, z)2, it followsfrom Theorem 25.2 that K(z, a) − 4πS(z, a)2 is in the linear span ofF ′

k(a)n−1k=1 for each fixed z. Hence, as a function of a, the Riemann sum

represents a function in the linear span of the F ′k(a). A simple limiting

argument now shows that a limit of such Riemann sums must also lie inthis span and the proof is complete.

Therefore, the Bergman and Szego kernels are actually related via

K(z, a) = 4πS(z, a)2 +

n−1∑

j,k=1

CjkF′j(z)F

′k(a) (25.1)

for some constants Cjk.We may use the results of Chapter 19 to write the formula relating

the Szego kernel to the Bergman kernel in a different form. Let us usethe notation S′(z, w) to denote the function ∂

∂zS(z, w), i.e., the primedenotes differentiation in the holomorphic variable.

Theorem 25.3. Suppose that Ω is a bounded n-connected domain withC∞ smooth boundary. For a point a ∈ Ω, suppose that the zeroes of theSzego kernel S(z, a) are given as the set ajn−1

j=1 of n− 1 distinct pointsin Ω. The Bergman kernel associated to Ω is related to the Szego kernelvia the identity

K(z, a) = 4πS(z, a)2 + 2πn−1∑

j=1

K(aj , a)

S′(aj , a)L(z, aj)S(z, a)

Proof. Since, by Theorem 19.1, the linear span of the functions F ′j is the

same as the linear span of the functions L(z, aj)S(z, a), it is clear fromTheorem 25.2 that there are constants cj such that

K(z, a) = 4πS(z, a)2 + 2π

n−1∑

j=1

cjL(z, aj)S(z, a).


The values of the constants cj are easily determined because the func-tions Gj(z) = L(z, aj)S(z, a) are such that

Gj(ak) = 0 if k 6= j, and

Gj(aj) =S′(aj , a)

2π

since S(ak, a) = 0 for each k and L(z, aj) has a single simple pole atz = aj with residue 1/(2π). The proof is finished.

Another way to relate the Bergman kernel to the Szego kernel is bymeans of the connection of both of these objects to the Dirichlet problem.We will now show that the Bergman kernel associated to Ω is directlyrelated to the Poisson extension of the boundary values of the function1/(z − a) to Ω.

Theorem 25.4. Let Ω denote a bounded domain with C∞ smoothboundary. Let a ∈ Ω and let u denote the Poisson extension to Ω ofϕ(z) = (2πi)−1/(z − a). The Bergman kernel function K(z, a) is givenby

K(z, a) = −2i∂u

∂z.

Proof. If h ∈ A∞(Ω), then

h(a) =

∫

bΩ

hϕ dz =

∫∫

Ω

h∂u

∂zdz ∧ dz

=

∫∫

Ω

h(z) G(z) dx ∧ dy

whereG(z) is the holomorphic function on Ω given by−2i(∂u/∂z). Thus,the inner product of a function h ∈ A∞(Ω) with G is equal to the valueof h at a. Since A∞(Ω) is dense in the Bergman space, this holds truefor all h in the Bergman space. This reproducing property characterizesthe Bergman kernel, and therefore K(z, a) = G(z) and the theorem isproved.

The formula in Theorem 25.4 is most interesting on a multiply con-nected domain. On a simply connected domain, the Bergman kernelK(z, a) is a constant times the derivative of the Riemann mapping func-tion mapping a to the origin given by Sa/La. From this, it is not hardto deduce that

K(z, a) = 2S(a, a)∂

∂z

(S(z, a)

L(z, a)

).

We proved in Chapter 16 that the Bergman projection and kernel

The Bergman kernel and the Szego kernel 139

transform under proper holomorphic mappings. We close this chapterby showing that the Szego kernel transforms under certain proper hol-omorphic maps. In order to prove this result, we will need a fact aboutproper holomorphic maps that follows from the material in Chapter 16.

Suppose g is a proper holomorphic mapping of a bounded multiplyconnected domain Ω1 with C∞ smooth boundary onto a bounded sim-ply connected domain Ω2 with C∞ smooth boundary. Suppose that themultiplicity of g is m. Let V2 denote the discrete subset of Ω2 that isthe image of the branch locus of g. Let Gk, k = 1, . . . ,m, denote thelocal inverses to g. We proved in Chapter 16 that the Gk are defined onΩ2 − V2, that they extend smoothly to the boundary of Ω2, and thatthey map the boundary of Ω2 into the boundary of Ω1.

If ωj is the harmonic measure function equal to one on the j-thboundary curve of Ω1 and equal to zero on the others, then

m∑

k=1

ωj(Gk(w)) (25.2)

is equal to a constant (which is a positive integer) for w ∈ Ω2 − V2. Toprove this fact, notice that the sum defines a positive harmonic functionon Ω2 − V2. Since this harmonic function is bounded above by m andbelow by zero, and since the set V2 is discrete, the possible singularitiesat points in V2 are removable. Thus, we may think of the sum as defin-ing a harmonic function u on all of Ω2. Since the functions Gk extendsmoothly to bΩ2 and since they map bΩ2 into bΩ1, it follows that u ex-tends smoothly to Ω2 and that it is positive integer valued on bΩ2. Theboundary of Ω2 is connected since Ω2 is simply connected. A continuousinteger valued function on a connected set must be constant. Hence, uis constant on bΩ2 and the maximum principle shows that u is constanton Ω2.

By differentiating equation (25.2) with respect to z and using thecomplex chain rule, we see that

0 ≡m∑

k=1

F ′j(Gk(w))G

′k(w). (25.3)

Recall that the operator Λ2 that we used in Chapter 16 was defined asΛ2u =

∑mk=1G

′k(uGk), and thus, identity (25.3) can be expressed very

simply in terms of the operator Λ2 as Λ2F′j = 0. Now, apply Λ2 in the

z variable to the identity in Theorem 25.2 to obtain

Λ2K1(·, a) = 4πΛ2S1(·, a)2.Note that the F ′

j terms drop out because Λ2F′j = 0. The trans-

formation formula for the Bergman kernels yields that Λ2K1(·, a) =


K2(·, g(a))g′(a), which in a simply connected domain, is equal to4πS2(·, g(a))2g′(a). Hence, we have proved that, under these conditionsthe Szego kernel transforms under g via

g′(z)S2(g(z), w)2 =

m∑

k=1

S1(z,Gk(w))2G′

k(w).

It is worth pointing out that an Ahlfors map associated to a boundedmultiply connected domain with C∞ smooth boundary satisfies the con-ditions under which this transformation formula has been proved to bevalid.

26

Pseudo-local property of the Cauchytransform and consequences

Let Ω denote a bounded domain with C∞ smooth boundary. The Cauchytransform associated to Ω is an example of an operator that is not local.A local operator Q on L2(bΩ) would have the property that, given afunction u ∈ L2(bΩ) that vanishes on an open arc A in the boundary,then Qu also vanishes on this arc. It is easy to see that the Cauchy trans-form does not satisfy this property because the transform of a functionvanishing on an arc extends holomorphically past that arc. However,the Cauchy transform is an example of a pseudo-local operator. Thismeans that, given an open connected arc contained in the boundary of abounded domain Ω with C∞ smooth boundary, if a function in L2(bΩ)is C∞ smooth on this arc, then so is its Cauchy transform. In this chap-ter, we will study this property in more detail and deduce some of itsconsequences.

If A is an open connected arc in the boundary of Ω, we will let ‖u‖Asdenote the Cs norm of a function u on A. To be precise, fix a C∞

parameterization z(t) of the arc A such that z(t) traces out the arc as t

ranges from a to b. We define ‖u‖As to be the supremum of | ∂k

∂tku(z(t))|

over a < t < b and 0 ≤ k ≤ s. We will continue to use the unadornedsymbol ‖u‖ to denote the L2(bΩ) norm of u.

Let γ be an open connected arc contained in the boundary of Ω andlet Γ be another such arc that compactly contains γ. The Cauchy trans-form satisfies the following property known as a pseudo-local estimate.


‖Cu‖γs ≤ K(‖u‖Γn + ‖u‖

)

for all u ∈ L2(bΩ).

It should be remarked that part of the conclusion of this theorem isthat if the norms on the right hand side of the inequality make sense,then the norm on the left hand side makes sense too.

It is possible to prove an improved version of Theorem 26.1 in which

141


n is taken to be equal to s+ 1 but we will not prove this version. It willsuffice for us to know only that the estimate holds for some value of n.

Proof. Theorem 26.1 is a straightforward consequence of Theorem 9.2.To see this, let χ be a function in C∞(bΩ) that is equal to one on aneighborhood of the closure of γ in bΩ and equal to zero on bΩ−Γ. Now,we may write Cu = C(χu) + C((1− χ)u). Because (1− χ)u is supportedaway from γ, it is clear that we may differentiate under the integral signdefining the Cauchy transform and apply Holder’s inequality to see that

‖C((1− χ)u)‖γs ≤ (constant)

(∫

bΩ−γ

|u|2ds)1/2

≤ (constant)‖u‖.

To analyze the other part of Cu, we use Theorem 9.2 to obtain

‖C(χu)‖γs ≤ K‖χu‖bΩn ≤ (constant)‖u‖Γn.

Combining these two inequalities now yields the theorem.

Recall that, as a consequence of the Kerzman-Stein identity, theSzego projection is related to the Cauchy transform via identity (4.3),

P = C − A(I − P ).

Because the Kerzman-Stein operator A is given by integration against akernel in C∞(bΩ× bΩ), it follows that

‖A(I − P )u‖γs ≤ C‖(I − P )u‖ ≤ C‖u‖.

Thus, the pseudo-local estimate for the Cauchy transform implies thesame kind of estimate for the Szego projection.


‖Pu‖γs ≤ K(‖u‖Γn + ‖u‖

)

for all u ∈ L2(bΩ).

As was the case for Theorem 26.1, Theorem 26.2 can be proved usingn = s+ 1, but we will not need this refinement.

One of the most interesting consequences of this pseudo-local esti-mate is the following result.

Theorem 26.3. The Szego kernel function is in C∞((Ω×Ω)−∆) where∆ = (z, z) : z ∈ bΩ denotes the boundary diagonal set.

Pseudo-local property of the Cauchy transform and consequences 143

In view of identity (25.1), it follows from Theorem 26.3 and the factthat the functions F ′

j are all in A∞(Ω) that Bergman kernel K(z, w) is

also in C∞((Ω× Ω)−∆).

Proof. Let z0 and w0 be distinct points in the boundary of Ω and letǫ > 0 be small enough that the closures of the balls of radius ǫ about z0and w0 do not intersect. As in Chapter 7, let Cw(ζ) denote the complexconjugate of

1

2πi

T (ζ)

ζ − w.

For z and w in Ω, we know that

S(z, w) = (PCw)(z) = (CCw)(z)− (PACw)(z)

where the last equality follows from the Kerzman-Stein identity P (I +A) = C. Let us define H1(z, w) = (CCw)(z) and H2(z, w) = (PACw)(z).The term H1(z, w) is the interesting part of S(z, w); the term H2(z, w)turns out to be very well behaved.

To analyze H2(z, w), notice that, for ζ ∈ bΩ,

(ACw)(ζ) =

∫

ξ∈bΩ

A(ζ, ξ)Cw(ξ) ds,

and, because Cw(ξ) is the conjugate of the kernel function of the Cauchytransform, this is equal to the complex conjugate of the Cauchy trans-form of the function ψζ(ξ) = A(ζ, ξ) evaluated at w. Now, because thesefunctions and their derivatives in ξ are bounded on bΩ as ζ ranges overthe boundary, and because the Cauchy transform satisfies the estimatein Theorem 9.2, it follows that (ACw)(ζ) and all its derivatives in ware bounded on Ω as ζ ranges over bΩ. We will now repeat this argu-ment for derivatives of (ACw)(ζ) with respect to ζ. Let ζ(t) representa parameterization of the boundary and, when w is taken to be a fixedpoint in Ω, consider derivatives of (ACw)(ζ(t)) with respect to t. Thek-th derivative is given by

∫

ξ∈bΩ

∂k

∂tkA(ζ(t), ξ)Cw(ξ) ds,

which is equal to the complex conjugate of the Cauchy transform of thefunction ψk

ζ (ξ) defined to be the complex conjugate of (∂k/∂tk)A(ζ(t), ξ).Since this function and its derivatives with respect to ξ are boundedon bΩ, we may repeat the reasoning above to see that the function(∂k/∂tk)(ACw)(ζ(t)) and all its derivatives in w are bounded on Ω asζ ranges over bΩ. We have shown that (ACw)(ζ) is in C∞(bΩ × Ω) as


a function of (ζ, w). If we now apply the Szego projection in ζ, we canuse the uniform estimate of Theorem 9.2 for P to deduce that H2(z, w)is in C∞(Ω× Ω).

We now assume that z ∈ Dǫ(z0)∩Ω and w ∈ Dǫ(w0)∩Ω. To analyzeH1(z, w), we let χ be a function in C∞(bΩ) such that χ = 1 inDǫ(z0)∩bΩand χ = 0 in Dǫ(w0) ∩ bΩ. We now split H1(z, w) into two pieces via

H1(z, w) = (C(χCw)) (z) + (C((1− χ)Cw)) (z).

Consider the first term in this sum. Notice that (χCw)(ζ) and its deriva-tives in ζ are bounded on bΩ as w ranges over Dǫ(w0)∩Ω. Furthermore,derivatives of (χCw)(ζ) with respect to w give rise to functions of ζ withthe same property. Thus, it follows that the first term in the sum for H1

is in C∞((Dǫ(z0) ∩ Ω)× (Dǫ(w0) ∩Ω)).The second term can also be seen to belong to this class by observing

that(C((1− χ)Cw)) (z) = (C((1− χ)Cz)) (w),

and the same reasoning can be applied to this term.

In the course of the proof of Theorem 26.3, we proved that

S(z, w) =1

4π2

∫

ζ∈bΩ

1

(ζ − z)(ζ − w)ds+H(z, w)

where H(z, w) is in C∞(Ω × Ω) and where the integral is equal toH1(z, w) = (CCw)(z). By letting z = a and w = a in this formula,it becomes clear that S(a, a) tends to +∞ as a tends to a point in theboundary at the same rate as

1

4π2

∫

ζ∈bΩ

1

|ζ − a|2 ds.

This last integral can easily be seen to blow up like a constant times theinverse of the distance from a to the boundary.

We proved in Chapter 10 that the Poisson kernel on a bounded simplyconnected domain with C∞ smooth boundary is given by

p(z, w) =S(z, w)S(w, a)

S(z, a)+S(z, w)L(w, a)

L(z, a)=

|S(z, w)|2S(z, z)

.

We can now assert that the Poisson kernel has all the familiar propertiesthat we admire about the Poisson kernel on the unit disc. We can nowadd to the properties proved in Chapter 10 that p(z, w) is in C∞((Ω ×bΩ)−∆) where ∆ = (z, z) : z ∈ bΩ. Also, for a fixed point w0 ∈ bΩ andδ > 0, p(z, w) tends to zero uniformly in w on the set bΩ−Dδ(w0) as z


tends to the boundary while staying in the set Ω∩Dδ/2(w0). When thesetwo properties are added to the ones proved in Chapter 10, we know allthe properties of the Poisson kernel needed to prove Schwarz’s theoremabout the solution of the Dirichlet problem with continuous boundarydata.

We now turn to the study of the behavior of the Garabedian kernelL(z, a) when z and a are both close to the boundary.

Theorem 26.4. If Ω is a bounded domain with C∞ smooth boundary,then the function ℓ(z, w) defined via

L(z, w) =1

2π(z − w)+ ℓ(z, w)

is a function on Ω×Ω that is holomorphic in z and w and that extendsto be in C∞(Ω× Ω).

Proof. We claim that, as a function of z, ℓ(z, a) is the Szego projectionof the function Ga defined to be (2π)−1(z − a)−1. To see this, note thatℓ(z, a) is holomorphic in z on Ω. Hence

ℓ(z, a) = P (ℓ(·, a)) = PLa − PGa = −PGa

because La = i SaT is orthogonal to holomorphic functions.Now the Kerzman-Stein identity P (I +A) = C allows us to write

PGa = CGa − PAGa.

A simple calculation using the residue theorem reveals that (CGa)(z) iszero for all z ∈ Ω and a ∈ Ω. Hence, the proof will be finished if we provethat (AGa) is in C

∞(Ω× Ω). But

(AGa)(z) =1

2π

∫

ξ∈bΩ

A(z, ξ)1

ξ − ads,

and this integral represents the Cauchy transform of the function ψz(ξ) =iA(z, ξ)T (ξ) evaluated at a. Hence, we may reason exactly as we didin the proof of Theorem 26.3 to see that all the mixed derivatives of(AGa)(z) in a and z are bounded on bΩ×Ω. This completes the proof.

There are even stronger theorems about the boundary behavior of theSzego and Garabedian kernels in domains with real analytic boundaries.

Theorem 26.5. On a bounded domain with real analytic boundary,the Szego kernel S(z, w) extends to be defined on a neighborhood of


(Ω × Ω) − (z, z) : z ∈ bΩ as a function that is holomorphic in zand antiholomorphic in w, and the Garabedian kernel is given by

L(z, w) =1

2π(z − w)+ ℓ(z, w),

where ℓ(z, w) extends to be holomorphic in z and w on a neighborhoodof (Ω× Ω).

Note that, because the functions F ′j all extend holomorphically to

a neighborhood of Ω, it follows from Theorem 26.5 and identity (25.1),that the Bergman kernel K(z, w) enjoys the same extension propertiesas the Szego kernel.

Proof. Suppose that Ω is a bounded domain with real analytic bound-ary, and let γ denote one of the boundary curves of Ω. We will needto use a reflection function for this real analytic curve as described inChapter 11. Let R(z) denote an antiholomorphic reflection function forγ. For example, 1/z is an antiholomorphic reflection function for the unitcircle. Recall that such a function is defined and antiholomorphic in aneighborhood of γ, fixes γ, and is locally diffeomorphic near γ. Near γ,R(z) maps the outside of Ω to the inside, and the inside to the outside,and R(R(z)) = z. We may define a reflection like this for each of theboundary curves of Ω; we will use the same symbol R to denote eachof them. In this way, we may view R as an antiholomorphic functiondefined on a small neighborhood of bΩ.

Fix a point a in Ω and let w be another point in Ω that we will allowto vary. By (7.1), we may write S(a, z) = −iL(z, a)T (z) and S(w, z) =−iL(z, w)T (z) when z ∈ bΩ. After dividing the second equation by thefirst, and using the fact that R(z) = z on the boundary, we may write

S(w,R(z))

S(a,R(z))=L(z, w)

L(z, a)for z ∈ bΩ. (26.1)

The function on the left hand side of this equality is defined and holo-morphic for z near γ on the outside of Ω. We know by Theorem 11.2 thatL(z, w) extends holomorphically past the boundary as a function of z foreach fixed w ∈ Ω. Hence the function on the left hand side agrees withthe holomorphic extension of L(z, w)/L(z, a) outside of Ω. A particularlyinteresting consequence of this formula is that L(z, w)/L(z, a) is seen toextend holomorphically to a neighborhood of γ that is independent of w.Now consider what (26.1) implies as w is allowed to tend to a boundarypoint w0 ∈ bΩ. Let wk be a sequence in Ω tending to w0. We deducethat there is a neighborhood O of Ω such that, as a function of z, eachL(z, wk) extends holomorphically to O − wk, and as wk tends to w0,


these functions converge uniformly on compact subsets of O − w0 toa function L0(z) that is holomorphic on O − w0. By writing contourintegrals about a small fixed circle around w0 that give the coefficientsof the Laurent expansion of L(z, wk) about the point wk, and by takinguniform limits under integral signs, it is seen that L0(z) has a simple poleat w0 with residue 1/(2π). Since L0(z) agrees with L(z, w0) inside Ω, wehave produced a meromorphic extension of L(z, w0). We now abandonour L0(z) notation and allow L(z, w0) to denote the meromorphic func-tion defined on O. Note that since L(z, w) = −L(w, z), we may deducethe same extension property in the w variable.

We have now defined L(z, w) for (z, w) in (O×Ω)∪ (Ω×O). We willcomplete the proof by defining L(z, w) on the rest of O ×O. It followsfrom Theorems 26.3 and 26.4 that identity (7.1) holds even when bothvariables are on the boundary, that is

S(w, z) = −iL(z, w)T (z) for z, w ∈ bΩ, z 6= w.

Hence, if z, w ∈ bΩ and z 6= w, we may use (7.1) in both variables towrite

−iT (w)L(z, w)T (z) = T (w)S(w, z) = −iL(w, z).Hence, if z, w ∈ bΩ, and z 6= w, we have

T (z)L(z, w)T (w) = L(z, w). (26.2)

Fix a point a in Ω and notice that T (ζ) = i Sa(ζ)/La(ζ). Plugging thisinto the last identity yields

− L(z, w)

La(z)La(w)=

L(z, w)

Sa(z)Sa(w)

when z, w ∈ bΩ, z 6= w. Using the fact that ζ = R(ζ) when ζ ∈ bΩ, wemay write

− L(z, w)

La(z)La(w)=

L(R(z), R(w))

Sa(R(z))Sa(R(w)).

This identity holds when z and w are on the boundary. The function onthe left extends to be holomorphic for z, w inside Ω, z 6= w, the functionon the right extends to be holomorphic for z, w outside Ω, z 6= w. Wemay reason as we did above to see that the function on the right handside of this identity defines the meromorphic extension of the functionon the left to z and w that are outside of Ω. It also follows that theextension is holomorphic in both variables z and w when z 6= w andthe singularity at z = w is a simple pole with residue 1/(2π). We havenow produced a meromorphic extension of L(z, w) to O×O. When the


principal part (2π)−1(z − w)−1 is subtracted off, we see that ℓ(z, w)extends holomorphically to O×O. Finally the extendibility of the Szegokernel follows from that of L(z, w) via identity (7.1) and the fact thatholomorphic functions with real analytic boundary values must extendpast the boundary.

The function ℓ(z, w) pops up in many places in the study of conformalmapping. We close this chapter by deriving Burbea’s formula [Bu1] whichrelates ℓ(z, w) to the Kerzman-Stein kernel. Identity (26.2) allows us towrite

L(z, w)T (z)− L(z, w)T (w) = 0

for z, w ∈ bΩ, z 6= w. Since L(z, w) = ℓ(z, w) + (2π)−1(z − w)−1, thismeans that

ℓ(z, w)T (z)− ℓ(z, w)T (w) = − T (z)

(2π)(z − w)+

T (w)

(2π)(z − w).

The right hand side of this last formula is −iA(w, z), and we have de-duced Burbea’s formula,

ℓ(z, w)T (z)− ℓ(z, w)T (w) = −iA(w, z).

27

Zeroes of the Szego kernel

In this chapter, we determine the behavior of the zeroes of the functionSa(z) as a tends to a point in the boundary. We suppose that Ω isa bounded n-connected domain with C∞ smooth boundary. We knowthat, for a ∈ Ω the function

f(a)(z) =S(z, a)

L(z, a)(27.1)

is the Ahlfors mapping associated to Ω, which is a branched n-to-onecovering map of Ω onto the unit disc (see Chapter 13). Notice thatf(a)(a) = 0 because of the pole of L(z, a) at z = a and that f ′

(a)(a) is

equal to 2πS(a, a). The n-to-one map f(a) must have n− 1 other zeroesbesides the one at a; these zeroes coincide with the zeroes of S(z, a)since L(z, a) is nonvanishing. As we have before, we list these zeroes(with multiplicity) a1, a2, . . . , an−1. When we want to emphasize thedependence of these zeroes on a we will write aj = Zj(a). As before, letγj , j = 1, . . . , n, denote the boundary curves of Ω.

Theorem 27.1. Let wk be a sequence in Ω that tends to a point a inthe boundary curve γm of Ω. As wk tends to a, the zeroes Zj(wk) ofS(z, wk) become simple zeroes, and it is possible to order them so thatfor each j, j 6= m, there is a point aj ∈ γj such that Zj(wk) tends to aj.

If a is a point in the boundary of Ω, then S(z, a) is nonvanishing onΩ as a function of z and has exactly n − 1 zeroes on the boundary ofΩ, one on each boundary component not containing a. Furthermore, thezeroes are simple in the sense that S′(aj , a) 6= 0 for each zero aj. Thesame is true of L(z, a). In fact, the zeroes of L(z, a) coincide with thoseof S(z, a).

Proof. If f : Ω1 → Ω2 is a biholomorphic map between bounded domainswith C∞ smooth boundaries, then the Szego kernels associated to Ω1

and Ω2 transform according to the formula

S1(z, w)2 = f ′(z)S2(f(z), f(w))

2f ′(w). (27.2)

Since Ω is biholomorphically equivalent to a domain whose boundary

149


curves are real analytic (via a conformal map that extends to be a C∞

diffeomorphism of the closures of the domains), and since (27.2) showsthat the boundary behavior of the zeroes of the Szego kernel is confor-mally invariant, we may assume that Ω is a domain whose boundarycurves are real analytic. By Theorem 26.5, such a domain has the virtuethat its Szego kernel extends to be real analytic on a neighborhood of(Ω× Ω)− (z, z) : z ∈ bΩ and its Garabedian kernel is given by

L(z, a) =1

2π

1

z − a+ ℓ(z, a),

where ℓ(z, a) extends to be holomorphic in a neighborhood of (Ω× Ω).We first prove that the Ahlfors maps f(wk) tend to a constant of unit

modulus as k → ∞. Using formula (7.1), thinking of a as being the pointin the boundary and z as being the “other point,” we have

S(z, a) = −iL(a, z)T (a) = iL(z, a)T (a).

We know that neither S(z, a) nor L(z, a) can vanish for z in Ω whena ∈ bΩ. Consequently, S(z, a)/L(z, a) = iT (a) for z ∈ Ω. We thereforesee that f(wk) tends uniformly on compact subsets of Ω to the constantfunction iT (a) as k → ∞. Hence, it follows that the zeroes of S(z, wk)must tend to the boundary as k → ∞.

Next, we show that L(z, a) has at least one zero on each of the curvesγj , j 6= m. Suppose that L(z, a) is nonvanishing for z ∈ γj . When bothpoints a and z are in the boundary and a 6= z, identity (26.2) yields

T (z)L(z, a)T (a) = L(z, a).

Let ∆jL denote the increase in arg L(z, a) as z traces out γj in the stan-

dard sense. The last identity reveals that

±2π +∆jL = −∆j

L,

and, therefore, that ∆jL = ±π. But this is impossible; ∆j

L must be aninteger multiple of 2π. Hence, L(z, a) has at least one zero on γj . Thesame is true of S(z, a) because, by virtue of identity (7.1), the zeroes inthe z variable of S(z, a) and L(z, a) coincide when a ∈ bΩ.

Fix a positive integer j, j 6= m. Let R(z) denote an antiholomorphicreflection function for γj (like the function 1/z for the unit circle; suchfunctions were constructed in Chapter 11). This reflection function fixesγj , is locally diffeomorphic near γj , locally maps the outside of Ω to theinside, and R(R(z)) = z. Fix a point A in Ω. By (7.1), we may writeS(A, z) = −iL(z, A)T (z) and S(wk, z) = −iL(z, wk)T (z) for z ∈ bΩ.

Zeroes of the Szego kernel 151

After dividing the second equation by the first, and using the fact thatR(z) = z on the boundary, we may write

S(wk, z)

S(A, z)=L(R(z), wk)

L(R(z), A)for z ∈ bΩ.

The function on the left hand side of this equality is antiholomorphic in aneighborhood of γj ; so is the function on the right. These two functionsagree on γj . Hence, they must agree on a neighborhood of γj . Fromthis, we deduce that, if S(z, wk) has a zero Z outside of Ω near γj , thenL(z, wk) must have a zero inside Ω at the reflected point R(Z), whichwe know is impossible. Hence, zeroes of S(z, wk) in the z variable thatoccur near the boundary must be inside Ω.

Suppose now that S(aj , a) = 0 where aj ∈ γj . We know that S(z, w)extends holomorphically in z and antiholomorphically in w to a neigh-borhood of (aj , a). Let Cǫ denote a circle of radius ǫ centered at aj . SinceS(z, a) cannot be identically zero in z (Corollary 9.1), the zero of S(z, a)at z = aj is isolated, and hence, there is a small ǫ > 0 such that S(z, wk)is nonvanishing on Cǫ if k is large. Now, the argument principle can beused to see that the integral

1

2πi

∫

Cǫ

(∂/∂z)S(z, wk)

S(z, wk)dz

is equal to the number of zeroes of S(z, wk) for z inside Cǫ. As wk tendsto a, the integers given by these integrals tend to the number of zeroes ofS(z, a) inside Cǫ which is a positive integer greater than or equal to one.As shown above, the zeroes of S(z, wk) must lie inside Ω. Hence, thereis a subsequence of zeroes of S(z, wk) in Ω that approach aj ; denote itZj(wk). We now know that the n − 1 zeroes of S(z, wk) approach theboundary, that there is a least one point on each of the n− 1 boundarycurves γj , j 6= m, where S(z, a) vanishes, and that each zero of S(z, a)on the boundary is approached by a sequence of zeroes of S(z, wk) insideΩ. The pigeon hole principle now yields that the zeroes of S(z, wk) mustseparate into n − 1 simple zeroes, that S(z, a) has exactly one zero oneach boundary curve γj , j 6= m, and that the zeroes of S(z, wk) canbe numbered as claimed so that Zj(wk) tends to the unique zero aj ofS(z, a) on γj . Now, because none of the zeroes of S(z, wk) migrate fromthe outside of Ω to the boundary and since the zeroes of S(z, wk) insideare simple, an application of Hurwitz’s theorem yields that the zeroesaj must also be simple zeroes.

Let us call the set (z, w) ∈ Ω× Ω : S(z, w) = 0 the Szego variety.It is a natural object associated to a domain. Identity (27.2) shows thatthe Szego variety is invariant under conformal changes of variable, i.e.,


if f : Ω1 → Ω2 is a biholomorphic map, then the mapping (z, w) 7→(f(z), f(w)) maps the Szego variety associated to Ω1 onto the Szegovariety associated to Ω2; the zeroes of S1(z, a) get mapped under f tothe zeroes of S2(z, f(a)).

28

The Kerzman-Stein integral equation

The constructive methods described in this book for studying the Szegokernel function, the Dirichlet problem, and conformal mapping give riseto reasonable schemes for numerically computing such objects. In thischapter, we describe the Kerzman-Stein integral equation and how it canbe used in a simply connected domain to compute the Szego kernel, theRiemann mapping function, and solutions to the Dirichlet problem.

Assume that Ω is a bounded simply connected domain with C∞

smooth boundary that is parameterized in the standard sense by a com-plex function z(t) in C∞[a, b]. We emphasize that the parameter t neednot be related to the arc length. For z and w in bΩ, the Kerzman-Steinkernel A(z, w) is given by

A(z, w) =1

2πi

(T (w)

w − z− T (z)

w − z

)

where T (z(t)) = z′(t)/|z′(t)| is the unit tangent vector function. Thekernel A(z, w) is a skew-hermitian function, i.e. A(z, w) = −A(w, z).Also, A(z, w) is in C∞(bΩ × bΩ) as a function of (z, w). A Kerzman-Stein integral equation is one of the form

u(z)±∫

w∈bΩ

A(z, w)u(w) ds = v(z)

where, given a function v defined for z ∈ bΩ, the problem is to find afunction u on bΩ satisfying this integral equation. It is remarkable thatall the domain functions described in this book are solutions to Kerzman-Stein integral equations with fairly simple functions v on the right handside. Let us begin by showing that the Szego kernel Sa(z) = S(z, a) isthe unique solution to the Kerzman-Stein integral equation,

Sa(z)−∫

w∈bΩ

A(z, w)Sa(w) ds = Ca(z)

where

Ca(z) =1

2πi

T (z)

a− z

153


is the Cauchy kernel. Indeed, the adjoint of the Kerzman-Stein identity(see formula (4.2)) is (I−A)P = C∗, and the Szego kernel was defined asthe projection of the Cauchy kernel, i.e., Sa = PCa. Hence, (I−A)Sa =

C∗Ca. But, by formula (3.2), C∗Ca = Ca − TC(CaT ), and an explicitcomputation using the residue theorem shows that C(CaT ) = 0. Hence,C∗Ca = Ca and the integral equation is proved. That Sa is the uniquesolution will be explained momentarily.

We will now show that the Kerzman-Stein integral equation for theSzego kernel can be seen to be an integral equation of a very classical typeknown as a Fredholm integral equation of the second kind. FollowingKerzman and Trummer [K-T] it is convenient to let z = z(s) wheres ∈ [a, b], to write out the integral in terms of the parameter t, and tomultiply the whole equation by |z′(s)|1/2. To make the end result moretransparent, let us define

k(s, t) = |z′(s)|1/2A(z(s), z(t))|z′(t)|1/2,c(s) = |z′(s)|1/2Ca(z(s)),

σ(s) = |z′(s)|1/2Sa(z(s)).

The kernel k(s, t) is skew-hermitian and in C∞([a, b]× [a, b]). Now, theintegral equation becomes

σ(s) −∫ b

a

k(s, t)σ(t) dt = c(s),

which is a Fredholm integral equation of the second kind. This is a verywell understood equation. Since the kernel is skew-hermitian and C∞

smooth, it follows from the standard theory of integral equations (seeYosida [Yo1]) that, given any function c(s) in C∞[a, b], there exists aunique solution σ(s) to the integral equation that is also in C∞[a, b].Hence the Szego kernel is uniquely determined as the solution to theKerzman-Stein integral equation above.

We should also mention that there are many efficient methods fornumerically computing solutions to this kind of equation (see Trummer[Tr]). The easiest method is to partition the interval [a, b] as a = t0 <t1 < · · · < tn = b and to write down an approximation for the integralequation in the form of a Riemann sum

σ(s)−n∑

i=1

k(s, ti)σ(ti)∆ti = c(s).

Writing down this equation for s = ti, i = 1, 2, . . . , n, gives n linearequations in the n unknowns σ(ti), i = 1, 2, . . . , n. The fact that k(s, t)is skew-hermitian guarantees that this linear system can be solved.

The Kerzman-Stein integral equation 155

Having computed the boundary values of Sa(z), we may use identity(7.1) to obtain the boundary values of La(z), and we may evaluate theSzego and Garabedian kernels at points in the interior of Ω by meansof the Cauchy integral formula. Note that, because La(z) has a singlesimple pole at a with residue 1/(2π), we must use the Cauchy integralin the form

La(z) =1

2π

1

z − a+ C

(La −

1

2π

1

z − a

).

Now that we have computed the boundary values of Sa and La,we also know the boundary values of the Riemann mapping functionf = Sa/La that maps Ω one-to-one onto the unit disc with f(a) = 0 andf ′(a) > 0. The Riemann map can be evaluated at points in the interiorby means of the Cauchy integral formula.

Next, we wish to show how to compute the Szego projection of a givenfunction v defined on bΩ. The Kerzman-Stein identity is P (I + A) =C. Hence, if we solve the Kerzman-Stein integral equation (I + A)u =v for u, then Pv = Cu and we have expressed the Szego projectionof v as an explicit Cauchy integral. This gives a reasonable methodfor computing Pv at points in Ω. If the boundary values of Pv aredesired, the constructive methods used in the proof of Theorem 3.1 canbe adapted to give a numerical method for computing the boundaryvalues of Cu, assuming that u is a sufficiently smooth function.

Now that we know how to compute Szego projections, we can solvethe Dirichlet problem via Theorem 10.1. Suppose we are given a bound-ary data function ϕ. The first step in the procedure is to compute thesolution u to the Kerzman-Stein integral equation

(I +A)u = Saϕ.

To actually do this, it is convenient to transform this equation by writingthe integral in terms of t and by multiplying the equation by |z′(s)|1/2.Define k(s, t) as before, and set

λ(s) = |z′(s)|1/2Sa(z(s))ϕ(z(s)),

µ(s) = |z′(s)|1/2u(z(s)).

The equation becomes

µ(s) +

∫ b

a

k(s, t)µ(t) dt = λ(s).

Solving this equation for µ yields the function u on the boundary of Ω.


We now claim that the solution to the Dirichlet problem with bound-ary data ϕ is given by

CuSa

− iC(uT )La

.

The first fraction in this term is equal to P (Saϕ)/Sa. Our claim willfollow from Theorem 10.1 if we show that P (Laϕ) = iC(uT ). We know

that (I +A)u = Saϕ. But, using (3.2), (I +A)u = Cu+TC(uT). Hence,after taking conjugates and multiplying through by T , we obtain

TCu+ C(uT ) = SaTϕ

and, by Theorem 4.3, it follows that C(uT ) = P (SaTϕ). But SaT = −iLa

by (7.1) and the proof of our claim is complete.Hence, once the solution u to the Kerzman-Stein integral equation

has been computed, the solution to the Dirichlet problem can be evalu-ated at an interior point w ∈ Ω by summing

1

2πi

1

Sa(w)

∫ b

a

u(z(t))

z(t)− wz′(t) dt

and −i times the complex conjugate of

1

2πi

1

La(w)

∫ b

a

u(z(t))

z(t)− w|z′(t)| dt.

Because the Kerzman-Stein kernel can be written down explicitly interms of a parameterization of the boundary of a domain, certain resultsof classical analysis become particularly easy to prove. One exampleof such a result is the stability of the Riemann mapping function undersmooth perturbations of a domain. Suppose that Ω is a simply connecteddomain whose boundary curve is parameterized by a C∞ function z(t),a ≤ t ≤ b. For small ǫ > 0, suppose that wǫ(t) is a complex valuedC∞ function on [a, b] such that all the t derivatives of wǫ(t) at t = aand t = b match. Suppose further that wǫ(t) and each of its derivativeswith respect to t tends uniformly to zero on [a, b] as ǫ → 0. For smallǫ > 0, let Ωǫ denote the domain whose boundary is parameterized byzǫ(t) = z(t) +wǫ(t). Notice that Ωǫ “tends” to Ω as ǫ tends to zero. If ais a point in Ω, let F (ǫ, a, z) denote the inverse f−1(z) of the Riemannmapping function f that maps Ωǫ onto the unit disc with f(a) = 0 andf ′(a) > 0.

Theorem 28.1. There is an ǫ0 > 0 such that F (ǫ, a, z) is C∞ as afunction of (ǫ, a, z) on [0, ǫ0]× Ω×D1(0).

The Kerzman-Stein integral equation 157

The proof of this theorem is quite routine. Let Aǫ(z, w) denotethe Kerzman-Stein kernel for Ωǫ. It is not difficult to show thatAǫ(zǫ(t), zǫ(s)) and any of its derivatives in s and t tend uniformly toA(z(t), w(t)) and its corresponding derivatives as ǫ → 0. The Riemannmaps can be written explicitly in terms of the Szego and Garabediankernels which are solutions to Kerzman-Stein integral equations. Sincethe kernels and the inhomogeneous terms in these integral equations varysmoothly in all the variables, including ǫ, the result follows from classicaltheorems about stability of solutions to Fredholm integral equations ofthe second kind with parameters ǫ and a (see Yosida [Yo1]).

29

Local boundary behavior ofholomorphic mappings

The classical Schwarz Reflection Principle of one complex variable is atheorem about boundary behavior of holomorphic mappings. It is easy tostate and easy to prove. The purpose of this chapter is to state and provethe following analogue of the reflection principle in the C∞ category.

Theorem 29.1. Suppose that γ1 and γ2 are C∞ smooth curves in thecomplex plane and suppose there are a point z0 ∈ γ1 and a disc D cen-tered at z0 such that D − γ1 consists of exactly two simply connectedcomponents, which we denote by D+ and D−. Suppose that there isa holomorphic function f defined on D+ that extends continuously toγ1 and such that the extension maps γ1 into γ2. Then f extends C∞

smoothly up to γ1 near z0. Furthermore, if f is not a constant function,there is a positive integer n such that f (n)(z0) 6= 0.

If the two curves in the statement of the theorem were real analytic,the classical Schwarz Reflection Principle states that f would extendholomorphically past the point z0, and hence, the statement about anonvanishing derivative of f at z0 is obvious because holomorphic func-tions are constant if and only if all their derivatives vanish at a point.In the C∞ case, this result is not obvious.

To prove Theorem 29.1 we will first show that, as in the real analyticcase, we may assume that γ1 is the real axis in the complex plane, thatz0 = 0, and that D+ is the upper half of the unit disc. To make thisreduction, let Ω be a small simply connected domain with C∞ smoothboundary contained in D+ whose boundary includes an open arc of thecurve γ1 containing z0. We use the Riemann mapping theorem to map Ωto the upper half plane via the unit disc in such a way that z0 is mappedto the origin. We have shown that, in this setting, the Riemann mappingfunction is a local C∞ diffeomorphism up to the boundary. We may alsoassume that f(0) = 0.

The proof of the theorem rests on the following lemmas. We will usethe shorthand notation φz and φz to denote derivatives of φ with respectto z and z, respectively. Let U+ denote the upper half of the unit disc

159


and let U− denote the lower half. The first lemma can be thought ofas a C∞ reflection principle for functions that are almost holomorphicnear the real axis and that satisfy the other hypotheses of the classicalreflection principle.

Lemma 29.1. Suppose that v is a complex valued C∞ function on U+

that extends continuously to the real axis and that is real valued on thereal axis. If vz extends C∞ smoothly to the real axis and vanishes toinfinite order along the real axis, then the function that is defined to beequal to v(z) for z in U+ ∪ R, and equal to v(z) for z in U−, is C∞

smooth on the unit disc.

Let d1(z) = |Im z| denote the distance from a point z to the real axis,and let d2(w) denote the distance from a point w to the curve γ2. LetU+(r) denote the upper half of the disc of radius r centered at the origin.Remember that we are now studying a holomorphic map f defined onU+ that extends continuously to the real axis R and that maps R intothe C∞ curve γ2.

Lemma 29.2. There is a constant C > 0 and a radius r > 0 such thatd2(f(z)) ≤ Cd1(z) for z ∈ U+(r).

In the language of partial differential equations, the next lemmawould be called a unique continuation theorem for the ∂/∂z operator.We will give a very elementary proof of this lemma below using only theCauchy integral formula.

Lemma 29.3. Suppose v is a C∞ function on the unit disc such that|vz | ≤ C|v| for some positive constant C. If v vanishes to infinite orderat the origin, then v is identically zero.

Let us now show how the lemmas imply the theorem. Afterwards,we will prove the lemmas. We require a function Φ that is an almostholomorphic mapping of γ2 into the real axis. To be precise, we needΦ to be C∞ smooth on a neighborhood of f(z0) = 0 such that Φz isnonvanishing on γ2, such that Φz vanishes to infinite order along γ2, andsuch that Φ is real valued along γ2. Note that the fact that Φz(0) 6= 0,Φz(0) = 0 implies that Φ is a local diffeomorphism near the origin. Toobtain such a function Φ, we will use the Riemann mapping theorem aswe did above to map a small one sided neighborhood of γ2 near f(z0) = 0onto the upper half plane in such a way that 0 gets mapped to 0; call thisRiemann map F . Now F is C∞ smooth up to γ2 near 0 and therefore canbe extended to be C∞ smooth in a neighborhood of 0. Since F ′ cannotvanish along γ2, we may take Φ to be equal to the extension of F on asmall neighborhood of 0.

Consider the function v = Φ f . It is defined and C∞ on U+(r)

Local boundary behavior of holomorphic mappings 161

for some r > 0. For convenience, we may assume that r = 1. Notethat Φ f extends continuously to the real axis and is real there. Wewish to prove that vz = (Φz f)f ′ extends C∞ smoothly to the realaxis so that Lemma 29.1 can be applied. Because f is a bounded hol-omorphic function on U+, the classical Cauchy estimates yield that|f (n)(z)| ≤ cn(Im z)−n near the origin. Let Dm denote the differen-tial operator ∂m

∂zi∂zj of order m = i + j. Since Φz vanishes to infiniteorder on γ2, we have an estimate of the form |DmΦz(w)| ≤ K d2(w)

N

near f(0) for each positive integer N (where K > 0 is a constant thatdepends on N and m). A typical term in a derivative of vz can be writ-ten as (DmΦz) f times a product of derivatives of f and f . Hence, ifwe choose N in the estimate for DmΦz to be sufficiently large, we mayuse Lemma 29.2 and the Cauchy estimates for f to deduce that all thederivatives of (Φ f)z tend to zero as Im z tends to zero. Thus, (Φ f)zextends smoothly to the real axis and vanishes to infinite order there.Lemma 29.1 yields that Φ f extends C∞ smoothly up to the real axisnear the origin. Now it follows that f extends C∞ smoothly up to thereal axis near the origin because Φ is a local C∞ diffeomorphism nearthe origin.

Finally, we must prove the finite vanishing condition usingLemma 29.3. Assume that f vanishes to infinite order at the origin.We know that v = Φ f extends to be a C∞ function on the unit discvia reflection. We need to see that the extension satisfies the hypothesesof Lemma 29.3. The infinite order vanishing of Φz along γ2 allows us toconclude that, for any positive integer N , there is a positive constantCN such that for z in U+ near 0,

|vz(z)| = |Φz(f(z))| |f ′(z)| ≤ CNd2(f(z))N .

Now, since Φz(0) 6= 0, it follows that |Φ(w)| ≥ |Im Φ(w)| ≥ cd2(w) forw in a neighborhood of 0. Thus, |v(z)| ≥ cd2(f(z)) ≥ (constant)|vz(z)|if z is restricted to be in a small enough neighborhood of the origin.The same inequality holds in the lower half disc by reflection (since∂∂z v(z) = vz(z)). Hence, Lemma 29.3 implies that v is identically zeronear the origin; therefore f is constant. The proof of Theorem 29.1 willbe complete after we have proved the lemmas.

Proof of Lemma 29.1. Define a function λ(z) on the unit disc to be equalto vz(z) for z ∈ U+, equal to zero for real z, and equal to vz(z) for z ∈ U−.Since vz vanishes to infinite order along the real axis, the function λ isseen to be C∞ on the unit disc U . Now let u be the solution to the∂-problem uz = λ given by

u(z) =1

2πi

∫∫

U

λ(ζ)

ζ − zdζ ∧ dζ


(Theorem 2.2). The function u is C∞ smooth on the unit disc and it iseasy to check that u satisfies the reflection property u(z) = u(z). Nowu−v is holomorphic on U+ and real on the real axis. Hence, the classicalreflection principle implies that u − v extends holomorphically to thewhole unit disc via reflection. Thus, v extends to be C∞ smooth on thewhole unit disc via reflection and the proof is complete.

Proof of Lemma 29.2. Choose a normal direction ν to the curve γ2 nearf(0) = 0. We wish to construct two C∞ subharmonic functions ρ+ and

ρ− in a neighborhood of 0 that both vanish on γ2 such that ∂ρ+

∂ν >

0 and ∂ρ−

∂ν < 0. To construct ρ+, let Ω be a small domain with C∞

smooth boundary whose boundary coincides with γ2 near the origin suchthat ν is an outward pointing normal to Ω at 0. Let φ be a solution tothe Dirichlet problem: ∆φ = 1 on Ω with φ = 0 on the boundary ofΩ. Since φ is C∞ smooth up to the boundary, we may extend φ as aC∞ subharmonic function to a neighborhood of the origin. The classicalHopf lemma (Theorem 9.4) implies that ∂φ

∂ν (z) > 0 for z ∈ γ2 near 0.Thus, we may define ρ+ to be equal to the extension of φ restricted to asmall neighborhood of 0. To construct ρ−, we repeat the argument aboveusing a domain Ω whose boundary agrees with γ2 near 0 such that ν isan inward pointing normal to the boundary of Ω near 0. By shrinkingthe neighborhood of 0 under consideration, we may assume that thefunctions ρ± are nonzero off of γ2 in their domain of definition. Nowdefine ρ = supρ+, ρ−. This function is subharmonic in a neighborhoodof the origin, is zero on γ2 and positive off of γ2, and there are positiveconstants c1 and c2 such that c1d2(w) ≤ ρ(w) ≤ c2d2(w) for w near 0.

We will restrict our attention to a small enough half disc U+(r) sothat the composition ρ f is defined. For convenience, we may assumethat r = 1. Note that ρ f is a nonnegative subharmonic function onU+ that is continuous up to the real axis and that vanishes there. Bycomposing with a conformal map of the unit disc onto U+ that mapsthe boundary point 1 to the origin, we may reduce our task to provingthe following proposition.

Proposition. Suppose that λ is a nonnegative subharmonic function onthe unit disc that is continuous up to the boundary such that λ(eiθ) = 0for θ in the range −δ < θ < δ for some δ > 0. Then there is a positiveconstant C such that

λ(ζ) ≤ C (1 − |ζ|)for all ζ in the sector |arg ζ| < δ/2.

Proof of the Proposition. Let P (ζ, θ) denote the Poisson kernel for the

Local boundary behavior of holomorphic mappings 163

unit disc. Now, we may write

λ(ζ) ≤∫ π

−π

P (ζ, θ)λ(θ) dθ.

But if |arg ζ| < δ/2 and |θ| > δ, we may estimate the Poisson kernel

P (ζ, θ) ≤ 2

π

1− |ζ||eiδ/2 − eiδ|2 .

If we use this inequality in the preceding inequality, we obtain the desiredestimate. The proof of Lemma 29.2 is now complete.

Proof of Lemma 29.3. Define a function λ on the unit disc to be equalto vz/v where v is nonzero and equal to zero where v is zero. Note thatλ is a bounded measurable function. Define a function u on the unit discU via

u(z) =1

2πi

∫∫

U

λ(ζ)


This function u is continuous on U because 1/(ζ− z) tends to 1/(ζ− z0)in L1(U) as z tends to a point z0 ∈ U . We now claim that u also hasthe property that it is C∞ where v is nonzero and uz = vz/v there. Tosee this, suppose that v(z0) 6= 0, and let χ be a function in C∞

0 (U) thatis equal to one on a neighborhood of z0 and that is supported in the setwhere v is nonvanishing. Next, split the integral defining u into the twopieces,

1

2πi

∫∫

U

χλ

ζ − zdζ ∧ dζ and

1

2πi

∫∫

U

(1 − χ)λ


Because χλ is in C∞(U), Theorem 2.2 yields that the first integral definesa function that is C∞ smooth on U whose derivative with respect to zis equal to χλ, which near z0 is equal to vz/v. Because (1−χλ) vanishesnear z0, it is permissible to differentiate under the second integral whenz is near z0 to see that the second integral defines a holomorphic functionthere, and so its derivative with respect to z is zero. The proof of theclaim is complete.

To finish the proof, consider the function h = ve−u. This function iscontinuous on U and is holomorphic where it is not zero. Thus, Rado’sTheorem implies that h is holomorphic on all of U . Now suppose thatv vanishes to infinite order at the origin. Then h must also vanish toinfinite order at the origin, and this implies that h is identically zero.Consequently, v too is identically zero and Lemma 29.3 is proved.

30

The dual space of A∞(Ω)

Throughout this book, when Ω is a bounded domain with C∞ boundary,we have singled out the space A∞(Ω) as a particularly nice subspace ofwhatever space of holomorphic functions on Ω we were studying. In thischapter, we will study A∞(Ω) for its own sake, and we will prove arepresentation theorem for linear functionals on this space. An exampleof a typical linear functional on A∞(Ω) is given by h 7→ h(n)(a) where ais any point in Ω and n is a positive integer.

We assume that Ω is a bounded domain with C∞ smooth boundary.The topology on A∞(Ω) is the Frechet space topology defined by thecountable family of norms given by

‖h‖s = sup|h(n)(a)| : 0 ≤ n ≤ s, a ∈ Ω.

We will also use the notation ‖u‖s when u is nonholomorphic to de-note the supremum over Ω of |Dαu| as Dα ranges over all real partialderivatives of order between zero and s. A linear functional on A∞(Ω)is continuous if and only if there is a positive integer N such that itis continuous in the s-norm for each positive integer s ≥ N . Let d(z)denote the distance from a point z in Ω to the boundary of Ω. The dualspace of A∞(Ω) will be identified with the space A−∞(Ω) that is definedto be the space of holomorphic functions g on Ω such that

sup|g(z)|d(z)s : z ∈ Ω

is finite for some positive integer s. Hence, the space A−∞(Ω) consistsof those holomorphic functions g on Ω that satisfy a growth estimate ofthe form |g(z)| ≤ c d(z)−s for some constant c and positive integer s.

For a positive s, we define the space A−s(Ω) to be the set of holo-morphic functions g on Ω such that the −s norm ‖g‖−s, defined by thesupremum above, is finite. The spaces A−s(Ω) are Banach spaces andA−∞(Ω) is their union, and as such, inherits a natural topology knownas the inductive limit topology (See [Yo2, p. 28]). We will not concernourselves with what that topology might be. Suffice it to say that a se-quence of functions gj in A−∞(Ω) converges to g if and only if gj → gin some A−s(Ω). Also, a linear functional on A−∞(Ω) is continuous ifand only if it is continuous as a functional on each subspace A−s(Ω).

165


Note that the estimate

|h(z)| ≤ 1√π d(z)

‖h‖L2(Ω)

that we proved early in Chapter 15 for functions h in H2(Ω) yields thatH2(Ω) ⊂ A−1(Ω).

The following result is the main ingredient in the proof that the dualspace of A∞(Ω) is equal to A−∞(Ω).

Lemma 30.1. Suppose that Ω is a bounded domain with C∞ smoothboundary. Given a positive integer s, there is a constant C such that

∣∣∣∣∫∫

Ω

h g dA

∣∣∣∣ ≤ C ‖h‖s‖g‖−s

for all h in A∞(Ω) and g in H2(Ω).

It will be a consequence of this lemma that we can define an innerproduct 〈h, g〉 that agrees with the standard L2 inner product

〈h, g〉 =∫∫

Ω

h g dA

when h and g are in L2(Ω), but that also makes sense wheneverh ∈ A∞(Ω) and g ∈ A−∞(Ω), even though |h g| may be far from be-ing integrable on Ω. The duality between A∞(Ω) and A−∞(Ω) will beexhibited via this extended inner product. This means that, given a lin-ear functional λ on A∞(Ω), there will exist a g ∈ A−∞(Ω) such thatλ(h) = 〈h, g〉 for all h ∈ A∞(Ω). Actually, it can also be shown thatA∞(Ω) is the dual space of A−∞(Ω) via this pairing, and we will provethis later.

Proof of Lemma 30.1. To prove the lemma, we must go back and ex-amine the proof of Lemma 2.1. At this time, the reader would benefitfrom going back and rereading the proof of Lemma 2.1, replacing alldifferentiations with respect to z by differentiations with respect to z.

Given a positive integer s and a function v ∈ C∞(Ω), we know thatthere exists a function Φs ∈ C∞(Ω) vanishing on the boundary of Ωsuch that v and ∂Φs/∂z agree to order s on the boundary. If we agreeto construct Φs by the procedure used in the proof of Lemma 2.1, wemay define an operator that maps a function v to the constructed Φs.Actually, we will be interested in an operator Qs defined via

Qsv = v − ∂Φs

∂z.

The dual space of A∞(Ω) 167

We will see that this operator is a linear differential operator of orders + 1 of a very nice kind. To gain an understanding of this operator,we must reconsider the procedure used in the process of constructingΦs from v. Let ρ denote the C∞ defining function for Ω used in theconstruction of Φs. (This means that Ω = ρ < 0, that ρ = 0 on bΩ,and that dρ 6= 0 on bΩ.) We claim that Qsv can be written as

Qsv = ρs+1Lsv

where Ls is a linear partial differential operator of order s + 1 of theform

Ls =∑

k≤s+1

Ak∂k

∂zk

where the coefficients Ak are functions in C∞(Ω). To see this, we willrepeat the inductive procedure used in the proof of Lemma 2.1, payingcloser attention to the functions defined along the way.

We will use a subscript z to indicate differentiation with respect toz. The starting point of the induction is easy. Indeed,

Q0v = v − ∂

∂z(θ0ρ)

where θ0 is chosen to make Q0v vanish on the boundary. As in the proofof Lemma 2.1, after expanding the derivative, it becomes clear that anideal choice for θ0 would be θ0 = χv/ρz, where χ is a fixed functionthat is equal to one near the boundary of Ω and that vanishes in aneighborhood of the zero set of ρz . If we plug in this expression for θ0in the definition of Q0v and expand the derivatives, we obtain

Q0v = v − χv − ρ (χv/ρz)z .

But v − χv is in C∞0 (Ω) because χ is equal to one in a neighborhood of

the boundary. In fact, (1− χ)/ρ is in C∞0 (Ω), and we may finally write

Q0v = ρ[(1− χ)ρ−1v − (χv/ρz)z

],

and this shows that the differential operator L0 exists and has the desiredproperties. Notice that the order of L0 is one.

Now suppose that we have proved the existence of the operators Lj

for j = 1, 2, . . . , s − 1. Remember that Φs = Φs−1 − θsρs+1 where θs

is an explicit function designed to make Qsv vanish to order s on theboundary. Now,

Qsv = v − (Φs)z = [v − (Φs−1)z ] + (θsρs+1)z

= Qs−1v + (θsρs+1)z = ρs(Ls−1v) + (θsρ

s+1)z .


When we expand the derivative in the last expression, we obtain

Qsv = ρs [(Ls−1v) + (s+ 1)θsρz] + (θs)zρs+1.

The function θs is chosen precisely to make the expression inside the lastpair of square brackets zero near the boundary. To imitate the proof ofLemma 2.1, we set

θs = −χ(Ls−1v)

(s+ 1)ρz.

Plugging this expression for θs back into the formula for Qsv, we obtain

Qsv = ρs+1

[(1 − χ)ρ−1(Ls−1v)−

∂

∂z

(χ(Ls−1v)

(s+ 1)ρz

)],

and the existence of Ls can be read off. Notice that the order of Ls isone greater than the order of Ls−1. This completes the induction.

The operator Qs is relevant to the proof of Lemma 30.1 because, byLemma 15.1, functions of the form (∂/∂z)Φs are orthogonal to H2(Ω).Hence, if v ∈ C∞(Ω) and g ∈ H2(Ω), it follows that

〈v, g〉 = 〈Qsv, g〉 =∫∫

Ω

(Lsv)ρs+1 g dA.

Since there are positive constants c1 and c2 such that c1d(z) ≤ −ρ(z) ≤c2d(z) for z ∈ Ω, we conclude that

|〈h, g〉| ≤ C‖h‖s+1‖g‖−(s+1),

where the constant C is equal to a constant times the area of Ω timesthe sum of the suprema of the moduli of the coefficients |Ak| appearingin the expansion of Ls as a differential operator. The proof of the lemmais complete.

Before we can use Lemma 30.1 to define the inner product 〈h, g〉 ofa function h ∈ A∞(Ω) with a function g ∈ A−∞(Ω), we must proveanother lemma. To say that A∞(Ω) is dense in A−∞(Ω) means thatgiven g ∈ A−∞(Ω), there is a positive integer s and a sequence gj inA∞(Ω) such that gj tends to g in A−s(Ω). We remark that it may bethat g belongs to A−m(Ω), but the sequence gj converging to g can onlybe produced in A−s(Ω) for s larger than m.

Lemma 30.2. If Ω is a bounded domain with C∞ smooth boundary,then A∞(Ω) is dense in A−∞(Ω). In fact, given g ∈ A−s(Ω), there isa sequence gj in A∞(Ω) such that gj tends to g in A−(s+1)(Ω). Conse-quently, H2(Ω) is dense in A−∞(Ω).


Proof. This lemma is easy to prove if Ω is the unit disc. In this case,we will show that, given g ∈ A−s(D1(0)), the sequence gj(z) = g(rjz)tends to g in the −(s + 1) norm whenever 0 < rj < 1 is a sequence ofreal numbers tending to one. Indeed, d(z) = 1− |z| on the unit disc andd(rz) > d(z) when 0 < r < 1. Hence, |g(rz)|d(z)s ≤ |g(rz)|d(rz)s ≤ cwhere c = ‖g‖−s. Now |g(z)− g(rz)|d(z)s+1 can be made small on anycompact subset of D1(0) by choosing r close to one because g(rz) tendsuniformly on compact subsets of D1(0) to g as r → 1. Furthermore,

|g(z)− g(rz)|d(z)s+1 = [(|g(z)|+ |g(rz)|)d(z)s] d(z) ≤ 2cd(z),

and this quantity is small near the boundary. It follows that ‖g −gj‖−(s+1) → 0 as j → ∞.

Using the result on the unit disc, we may prove the lemma for abounded simply connected domain Ω with C∞ smooth boundary via aRiemann mapping. In this case, let f : Ω → D1(0) denote a Riemannmapping of Ω onto the disc, and let F denote the inverse of this map.Let d1(z) = 1 − |z| denote the boundary distance function on the unitdisc and let dΩ(z) denote the boundary distance function on Ω. Becausef and F extend C∞ smoothly to the boundary, and because f ′ and F ′

are nonvanishing on the boundary, there are constants c1 and c2 suchthat

c1dΩ(F (z)) ≤ d1(z) ≤ c2dΩ(F (z))

for all z ∈ D1(0). Another way to write the same inequality is

c1dΩ(w) ≤ d1(f(w)) ≤ c2dΩ(w) for all w ∈ Ω.

Now, given g ∈ A−s(Ω), it follows from the distance estimate that g Fis in A−s(D1(0)). We may now take a sequence hj in A

∞(D1(0)) tendingto g F in A−(s+1)(D1(0)). The sequence hj f is a sequence in A∞(Ω)and it follows from the distance estimates that this sequence tends to gin A−(s+1)(Ω). The lemma is proved for simply connected domains.

We now assume that Ω is a bounded 2-connected domain with C∞

smooth boundary. Let γ1 denote the outer boundary curve of Ω and letγ2 denote the inner boundary curve. Let Ω1 denote the simply connecteddomain bounded by γ1. Note that Ω1 is obtained from Ω by filling inthe hole cut out by the inner curve γ2. Given a function g ∈ A−s(Ω),we will construct a function g1 ∈ A−s(Ω1) that has the same boundarybehavior as g near γ1. Let χ be a function in C∞(C) such that χ ≡ 1 ina neighborhood of γ1 and χ ≡ 0 in a neighborhood of γ2 and the domainenclosed by γ2. Since χ is zero near γ2 and inside γ2, we may think of χgas being a function defined on Ω1 by extending it to be zero inside thehole cut out by γ2. Let v1 = (∂/∂z)(χg). Since g is holomorphic and sinceχ is one near γ1 it follows that v1 is zero near γ1. Since χ is zero near


γ2 and inside γ2, it is clear that v1 ∈ C∞(Ω1). (In fact, v1 ∈ C∞0 (Ω1).)

Let u1 be a function in C∞(Ω1) that satisfies ∂u1/∂z = v1. Now thefunction

g1 = (χg)− u1

is in A−s(Ω1) and has the same boundary behavior as g near γ1 in thesense that, near γ1, it differs from g by a function that is C∞ smoothup to that curve. Since Ω1 is simply connected, there is a sequence h1jof functions in A∞(Ω1) that tend to g1 in A−(s+1)(Ω1). Notice that byrestricting all functions to Ω, we may think of h1j as a sequence in A∞(Ω)

that tends to g1 in A−(s+1)(Ω).Next, we turn our domain inside out and repeat the argument above.

Let Ω be the domain that is the image of Ω under the map F (z) =1/(z− a) where a is a fixed point in the interior of the domain bounded

by γ2. We may repeat the argument above on Ω, using g F−1 in placeof g and (1−χ) F−1 in place of χ. When we compose again with F tohave functions defined on Ω, we obtain a function

g2 = (1 − χ)g − u2

in A−s(Ω) that has the same boundary behavior as g near γ2, but thatis C∞ smooth up to γ1. The function u2 is in C∞(Ω). We also obtain asequence h2j of functions in A∞(Ω) that tend to g2 in A−(s+1)(Ω). Now

h1j + h2j is a sequence in A∞(Ω) that tends to g1 + g2 = g − u1 − u2 in

A−(s+1)(Ω). But u1+u2 is in A∞(Ω). Thus h1j+h

2j+u1+u2 is a sequence

in A∞(Ω) converging to g in A−(s+1)(Ω) and the lemma is proved for2-connected domains.

The method used to prove the lemma in 2-connected domains easilygeneralizes to n-connected domains. We leave it to the reader to completethe proof.

With Lemmas 30.1 and 30.2 in place, we may finally define the ex-tended inner product. Suppose that h ∈ A∞(Ω) and g ∈ A−∞(Ω). Thatg is in A−∞(Ω) means that g ∈ A−s(Ω) for some positive integer s. ByLemma 30.2, there is a sequence gj ∈ H2(Ω) tending to g in A−(s+1)(Ω).Lemma 30.1 implies that 〈h, gj〉 is a Cauchy sequence converging to anumber that we define to be 〈h, g〉. (That this number does not dependon the choice of the approximating sequence follows easily from the es-timate in Lemma 30.1.) For convenience, we define 〈g, h〉 to be thecomplex conjugate of 〈h, g〉.

Theorem 30.1. Suppose that Ω is a bounded domain with C∞ smoothboundary. If λ is a continuous linear functional on A∞(Ω), then thereis a unique g in A−∞(Ω) such that λ(h) = 〈h, g〉 for all h ∈ A∞(Ω).


Proof. It follows from Lemma 30.1 that a function g in A−s(Ω) defines acontinuous linear functional λ on A∞(Ω) via λ(h) = 〈h, g〉. To motivatethe proof, we now consider the problem of expressing this g in termsof the linear functional it defines. The Bergman kernel should come tomind. Using the notation Ka(z) = K(z, a) to write the Bergman kernelas we did in Chapter 15, we should expect that g(a) = 〈g,Ka〉, i.e., thatg(a) = λ(Ka). Since Ka ∈ A∞(Ω) by Theorem 15.2, the inner productmakes sense. We must see that the Bergman kernel has the reproducingproperty even when applied to functions in A−s(Ω). To do this, let gj bea sequence in A∞(Ω) that converges to g in A−s(Ω). It is easy to see thatconvergence in A−s(Ω) implies uniform convergence on compact subsetsof Ω. But gj(a) = 〈gj ,Ka〉, and so

g(a) = limj→∞

gj(a) = limj→∞

〈gj ,Ka〉 = 〈g,Ka〉,

as desired, and the strategy we should employ to prove the theorem isclear. Incidentally, the argument just given shows that if a functional isrepresented by g ∈ A−∞(Ω), then g is uniquely determined.

Now suppose that λ is an arbitrary continuous linear functional onA∞(Ω). Define a function on Ω via

g(a) = λ(Ka) for a ∈ Ω.

We must show that g is holomorphic, that g ∈ A−s(Ω) for some positiveinteger s, and that λ(h) = 〈h, g〉 for all h ∈ A∞(Ω).

To see that g is holomorphic, we would like to differentiate underthe operator, noting that Ka(z) is antiholomorphic in a. That this canbe done will follow from the fact that Ka(z) is in C∞(Ω × Ω) as afunction of (z, a), a property of Ka(z) that we deduced from formula(25.1) and Theorem 26.3. However, rather than quote this result, it willbe worthwhile to prove it from scratch here because we will need to useelements of the following more direct proof later in this chapter. Thestarting point is a fact that was proved in Chapter 15, namely, thatthe Bergman kernel is given as the Bergman projection of an explicitfunction in C∞

0 (Ω) as follows. Let θ denote a C∞ function on C that iscompactly supported in the unit disc such that θ is radially symmetricand

∫∫θ dA = 1. Let a0 be a point in Ω and let ǫ > 0 be a number that

is smaller than half the distance d(a0). For a in Dǫ(a0), let

ϕa(z) =1

ǫ2θ

(z − a

ǫ

).

The proof of Theorem 15.2 contained a proof that Ka = Bϕa. Note thatϕa(z) is in C

∞0 (Ω×D2ǫ(a0)) as a function of (z, a). Hence, to see that


Ka(z) is in C∞(Ω×Ω), we must differentiate under the operator in the

formula Ka(z) = (Bϕa)(z), making sure that the Bergman projection Bacts uniformly in z as a varies.

We know that B maps C∞(Ω) into itself (Theorem 15.2). We claimthat it follows that, given a positive integer s, there is a positive integerN and a constant C such that

‖Bu‖s ≤ C‖u‖N (30.1)

for all u ∈ C∞(Ω). To see this, we will first use the closed graph theoremto deduce that B is a continuous operator from C∞(Ω) into itself withrespect to the standard Frechet space topology on C∞(Ω). To see thatthe graph of B is closed in C∞(Ω)×C∞(Ω), notice that if (uj , Buj) con-verges in the graph topology to (u, v), then the uniform convergence of ujto u implies that uj → u in L2(Ω), and hence, that Buj → Bu in L2(Ω).But holomorphic functions converging in L2 also converge uniformly oncompact subsets, and we deduce that Bu = v. Thus the graph is closedand the closed graph theorem yields that B is continuous. Suppose thats is a positive integer and suppose that our claim about the norm es-timate (30.1) is not true. Then there exists a sequence un of functionsin C∞(Ω) such that ‖Bun‖s > n‖un‖n. By dividing the functions unby ‖Bun‖s, we may assume further that ‖Bun‖s = 1 for each n. Underthese conditions, the norm inequality implies that ‖un‖n < 1/n. Hence,the sequence un tends to the zero function in C∞(Ω), yet ‖Bun‖s = 1for all n. This violates the continuity of B in the topology of C∞(Ω) andthe norm estimate is proved.

We may now write a difference quotient for a partial derivative in thea variable in the formula Ka(z) = (Bϕa)(z), and use the norm estimates(30.1) to deduce that it is permissible to differentiate under the operatorsign in this formula. We may repeat this argument for partial derivativesof order two and up. Let Dk

a denote a partial derivative in the a variableof order k. The norm estimates imply that

∂m

∂zmDk

aKa(z) =∂m

∂zm(B(Dk

aϕa))(z),

and it can be read off that Ka(z) is in C∞(Ω×Dǫ(a0)) as a function of

(z, a). Since a0 was arbitrary, it follows that Ka(z) is in C∞(Ω× Ω).

Knowing this fact about Ka(z), the continuity of λ allows us todifferentiate with respect to a under the operator sign in the formulag(a) = λ(Ka). Since Ka(z) is antiholomorphic in a, it follows that g(a)is holomorphic in a on Ω.

Next, we will prove that g ∈ A−∞(Ω). Let us redefine the symbol


ϕa(z) to denote the function

ϕa(z) =1

ǫ2θ

(z − a

ǫ

)

where now ǫ = 12d(a). It is easy to see that the norm ‖ϕa‖n grows as a

constant times d(a)−n−2 as a tends to the boundary of Ω. Hence, sinceKa = Bϕa, estimate (30.1) implies that, given a positive integer s, thereis a positive integer N and a constant C such

‖Ka‖s ≤ C d(a)−N for all a ∈ Ω.

Now, since λ is a continuous linear functional on A∞(Ω), there mustexist a positive integer s and a constant c such that

|λ(h)| ≤ c‖h‖s for all h ∈ A∞(Ω).

To see this, assume the contrary and derive a contradiction in the sameway that we deduced estimate (30.1) for the Bergman projection. It nowfollows that there exists a positive integer N and a constant C such that

|λ(Ka)| ≤ C d(a)−N ,

and this is exactly what it means for g to be in A−N (Ω), and so g ∈A−∞(Ω).

We may now assert that

g(a) = 〈g,Ka〉 = λ(Ka),

and hence, thatλ(Ka) = 〈Ka, g〉.

Thus, g represents the linear functional λ on the subset X of A∞(Ω)given by the linear span of the set of functions Ka : a ∈ Ω. We willcomplete the proof of the theorem by showing that X is dense in A∞(Ω).

Let s be a positive integer, and let h be a given function in A∞(Ω).To prove the density statement, it will suffice to prove that ‖h − κ‖scan be made as small as desired by choosing κ ∈ X appropriately. LetN be a positive integer for which there exists a constant C such that‖Bu‖s ≤ C‖u‖N for all u ∈ C∞(Ω). From the proof of Lemma 30.1, weknow that h = B(QNh) where QNh is a function in C∞(Ω) vanishing toorder N on bΩ. Because of this vanishing, it is possible to find a functionψ in C∞

0 (Ω) such that ‖QNh−ψ‖N is as small as desired. As before, letθ denote a C∞ function on C that is compactly supported in the unitdisc such that θ is radially symmetric and

∫∫θ dA = 1. Given ǫ > 0, let

θǫ =1

ǫ2θ(zǫ

).


It is a simple exercise to show that the convolution θǫ ∗ ψ tends to ψ inC∞(Ω) as ǫ→ 0. By choosing ǫ sufficiently small, and by approximatingthe integral in the convolution by a Riemann sum, it is possible to finda (finite) sum

∑ciθǫ(z − ai) that is as close to ψ in the N -norm as we

please. (In this sum, the points ai may be assumed to be in the supportof ψ and ǫ may be assumed to be much smaller than the distance fromthe support of ψ to the boundary of Ω.) We now know that we mayapproximate QNh in the N -norm by sums of the form

∑ciθǫ(z − ai).

But the Bergman projection of such a sum is equal to∑ciKai

, whichis an element κ of X . Hence,

h−∑

ciKai= B

(QNh−

∑ciθǫ(z − ai)

)

and by arranging for the N -norm of the function acted upon by B to besmall, the norm estimate for the Bergman projection can be used to seethat the s-norm of h− κ can be made arbitrarily small. This completesthe proof of the theorem.

We now prove the companion theorem to Theorem 30.1.

Theorem 30.2. Suppose that Ω is a bounded domain with C∞ smoothboundary. If λ is a continuous linear functional on A−∞(Ω), then thereis a unique h in A∞(Ω) such that λ(g) = 〈g, h〉 for all g ∈ A−∞(Ω).

Proof. The proof of this theorem will be somewhat easier than thepreceding proof. Suppose that λ is a continuous linear functional onA−∞(Ω). The continuity of λ means that there are constants Cs foreach positive integer s such that

|λ(g)| ≤ Cs‖g‖−s.

For w 6∈ Ω, let

Hw(z) =1

2πi

1

(w − z).

DefineG(w) = λ(Hw) for w 6∈ Ω.

The functions Hw(z) are uniformly bounded in A−1(Ω) as w ranges overthe complement of Ω. Hence G is a bounded function on C − Ω. Thecontinuity of λ can be used to see that it is permissible to differentiatewith respect to w under the operator sign; hence, G(w) is a holomorphicfunction onC−Ω. Furthermore, since (∂/∂w)Hw(z) = −(2πi)−1/(w−z)2is uniformly bounded in A−2(Ω) as w ranges over C−Ω, it follows thatG′(w) is bounded there. Similar reasoning shows that all the derivativesof G are bounded on C− Ω, and hence G is a holomorphic function on


C − Ω that extends C∞ smoothly to bΩ. Also, given a point w0 ∈ bΩ,since Hw tends to Hw0

in A−2(Ω) as w tends to w0 through values ofw outside Ω, it follows that G(w0) = λ(Hw0

), and hence, that the C∞

extension of G to bΩ agrees with the actual values of G on bΩ. Let h(z)be the holomorphic function on Ω defined via

h = −2i∂

∂zE G

where E G denotes the Poisson extension to Ω of the boundary valuesof G. Since the boundary values of G are C∞ smooth, we know thath ∈ A∞(Ω). We now claim that h represents the linear functional λ inthe sense that λ(g) = 〈g, h〉 for all g ∈ A−∞(Ω). Actually, it will beenough to prove this equation for g ∈ A∞(Ω) because such functionsform a dense subspace of A−∞(Ω). Now, given g ∈ A∞(Ω), we mayapproximate the Cauchy integral formula

g(z) =1

2πi

∫

w∈bΩ

g(w)

w − zT (w) ds

by a Riemann sum

R(z) =1

2πi

∑ g(wk)

wk − zT (wk) ∆sk.

It is a simple matter to check that, by choosing a sufficiently fine parti-tion, we may approximate g by R in A−2(Ω). Hence, we may approximateλ(g) by

λ(R) =1

2πi

∑g(wk)T (wk)∆sk λ

(1

wk − z

).

This last sum is an approximation to the integral

∫

w∈bΩ

g(w)G(w) dw =

∫∫

w∈Ω

g(w)∂

∂w(EG)(w) dw ∧ dw

which is equal to ∫∫

Ω

g h dA

where h = −2i(∂/∂z)EG. The proof is completed by taking limits asthe partition width in the Riemann sum is allowed to tend to zero.

Finally, that h is uniquely determined follows from the identity

h(a) = 〈h,Ka〉 = λ(Ka),

and the proof is complete.


We close this chapter by making some remarks about the extendedinner product. This inner product is nondegenerate, meaning that, ifh ∈ A∞(Ω) is such that 〈h, g〉 = 0 for all g ∈ A−∞(Ω), then h ≡ 0, andif g ∈ A−∞(Ω) is such that 〈h, g〉 = 0 for all h ∈ A∞(Ω), then g ≡ 0.This fact can be deduced by noting that a function that is orthogonalto Ka in the extended inner product for a ∈ Ω must vanish at a.

Although we have defined the extended inner product rather ab-stractly in terms of the limit of a sequence, it is possible to define it inmore concrete terms by studying more carefully the operator Qs that wedefined in the course of the proof of Lemma 30.1. For example, it can beshown that if h ∈ A∞(Ω) and g ∈ A−∞(Ω), then

〈h, g〉 = limǫ→0

∫∫

Ωǫ

h g dA

where Ωǫ denotes the domain consisting of points in Ω that are a distancegreater than ǫ from the boundary of Ω. Another way to express theextended inner product is by means of the operator Qs directly. If h isin A∞(Ω), then Qsh is a function in C∞(Ω) that vanishes to order s onthe boundary, i.e., |(Qsh)(z)| ≤ c d(z)s+1 for some constant c. It followsthat (Qsh)(z)d(z)

−s is bounded on Ω. Consequently, if g ∈ A−s(Ω), then(Qsh) g is bounded on Ω. It can be shown that

〈h, g〉 =∫∫

Ω

(Qsh) g dA.

Proofs of these facts use the following tool. It can be shown that, nearthe boundary, the function −d(z) defines a C∞ smooth function that hasthe properties of a defining function for Ω. Thus, we may use a definingfunction ρ for Ω in the construction of Qs such that ρ(z) = −d(z) for znear the boundary of Ω. If ǫ > 0 is small, the function ρǫ(z) = ǫ+ ρ is adefining function for Ωǫ. If we use this defining function to construct anoperator Qǫ

s for the domain Ωǫ, it can easily be shown that Qǫsh tends

to Qsh in the s-norm as ǫ tends to zero. This allows us to write

〈h, g〉 =∫∫

Ω

Qsh g dA = limǫ→0

∫∫

Ωǫ

Qǫsh g dA = lim

ǫ→0

∫∫

Ωǫ

h g dA.

The details of the argument are somewhat technical, but not difficult.

31

The Green’s function and the Bergmankernel

In this chapter, we define the Green’s function G(z, a) associated to abounded domain Ω with C∞ smooth boundary, and we show how thisfunction is related to the Bergman kernel of the domain. For a fixedpoint a ∈ Ω, the Green’s function is defined via G(z, a) = − ln |z −a| + ua(z) where ua(z) is the harmonic function of z on Ω that solvesthe Dirichlet problem with boundary data equal to ln |z − a|. In view ofTheorem 14.2,G(z, a) is a harmonic function of z on Ω−a that extendsC∞ smoothly up to bΩ, that vanishes on bΩ, and that has the propertythat G(z, a) + ln |z − a| is bounded near a (and so it has a removablesingularity at a). It is a consequence of the maximum principle that theseproperties characterizeG(z, a). In fact, the condition that G(z, a) extendC∞ smoothly to the boundary is much stronger than necessary. It isenough to know only that G(z, a) extends continuously to the boundary.The maximum principle also yields that G(z, a) > 0 if z, a ∈ Ω andz 6= a.

If f : Ω1 → Ω2 is a biholomorphic mapping between bounded do-mains with C∞ smooth boundaries, then the Green’s functions G1 andG2 associated to Ω1 and Ω2, respectively, satisfy the identity

G1(z, a) = G2(f(z), f(a)).

To understand this identity, note that f(z) is in C∞(Ω1) by The-orem 16.2. Furthermore, we may write f(z) − f(a) = (z − a)H(z)where H is a nonvanishing holomorphic function in A∞(Ω1). Henceln |f(z) − f(a)| = ln |z − a| plus a harmonic function in C∞(Ω1) andthe rest of the proof becomes clear.

On the unit disc U , it is apparent that GU (z, 0) = − ln |z|, andthe transformation property applied to the biholomorphic mapping (z−a)/(1− az) of U onto itself reveals that

GU (z, a) = − ln

∣∣∣∣z − a

1− az

∣∣∣∣ . (31.1)

Notice that GU (z, a) = GU (a, z), and that, as a function of (z, a) ∈

177


U × U , G extends to be in C∞((U × U)−D

)where D denotes the

diagonal set (z, z) : z ∈ U. We will now see that these properties holdtrue for Green’s functions associated to any bounded domain with C∞

smooth boundary.

Theorem 31.1. The Green’s function G(z, a) associated to a boundeddomain Ω with C∞ smooth boundary is such that G(z, a) = G(a, z) and,consequently, G is harmonic in each variable separately when z 6= a.Furthermore, G extends to be in C∞

((Ω× Ω)−D

)where D denotes

the diagonal set (z, z) : z ∈ Ω. Also, the function G(z, a) + ln |z − a|has removable singularities at z = a ∈ Ω and can viewed as an elementof C∞

((Ω× Ω)− (z, z) : z ∈ bΩ

).

Proof. If Ω is simply connected, we may use a Riemann map to relatethe Green’s function of Ω to the Green’s function of the unit disc andthe truth of the theorem follows from the transformation formula for theGreen’s functions, formula (31.1), and the fact that Riemann mappingfunctions extend C∞ smoothly to the boundary.

To prove the theorem in case Ω is n-connected, n > 1, we will needto use the following enhanced version of Theorem 21.1 which allows thedata in the Dirichlet problem to depend smoothly on a parameter. Thistheorem is proved by differentiating under the operators in the formulasin the statement of Theorem 21.1. The argument is routine, and so weomit it here.

Theorem 31.2. Suppose that Ω and O are bounded domains with C∞

smooth boundaries. Suppose that ϕa(z) is a function of (z, a) ∈ bΩ×Oin C∞(bΩ×O). The solution ua(z) to the Dirichlet problem on Ω withboundary data ϕa is in C∞(Ω×O) as a function of (z, a).

We now return to the proof of Theorem 31.1. Let γn denote the outerboundary of Ω and let Ω0 denote the inside of γn, i.e., Ω0 is the simplyconnected domain obtained from Ω by filling in all its holes. Let G0(z, a)denote the Green’s function associated to Ω0 and consider the functionϕa(z) = G0(z, a) as a C∞ function on bΩ × Ω. Let ua(z) denote thesolution on Ω to the Dirichlet problem with boundary data equal to ϕa.Now it is easy to see that G(z, a) = G0(z, a)− ua(z) for (z, a) ∈ Ω× Ω.Let O denote a small disc that is compactly contained in Ω. Since ϕa(z)is C∞ smooth on bΩ×O for any suchO, Theorem 31.2 yields that G(z, a)is in C∞((Ω×Ω)−(ζ, ζ) : ζ ∈ Ω). Now, for a small ǫ > 0, let O denotethe domain consisting of the set of points in Ω that are within a distanceǫ of γn. Notice that, becauseG0(z, a) vanishes on γn, it follows that ϕa(z)is in C∞(bΩ×O). Theorem 31.2 implies that ua(z) is in C

∞(Ω×O) as afunction of (z, a). It follows that G(z, a) extends C∞ smoothly in (z, a) to(bΩ×γn)−(ζ, ζ) : ζ ∈ γn, and G is seen to have the desired smoothness

The Green’s function and the Bergman kernel 179

property near the outer boundary. To see that G(z, a) also extends C∞

smoothly in (z, a) to (bΩ×γk)−(ζ, ζ) : ζ ∈ γk for the other boundarycurves γk, k = 1, . . . , n− 1, simply use a biholomorphic map of the form1/(z−z0) to map Ω to a domain whose outer boundary corresponds to γkand repeat the argument just given. Invoke the transformation propertyof Green’s functions under biholomorphic maps to finish the argument.

Finally, we must prove the symmetry property G(z1, z2) = G(z2, z1).Given two points z1 and z2 in Ω, let Ωǫ denote the domain obtained fromΩ by removing the closed discs of radius ǫ about z1 and z2. We will usethe real Green’s identity on Ωǫ, which says that

∫∫

Ωǫ

(v∆u− u∆v) dA =

∫

bΩǫ

(v∂u

∂n− u

∂v

∂n

)ds

for any u, v ∈ C∞(Ωǫ), where dA denotes area measure, ds denotes arclength measure, and ∂/∂n represents the normal derivative operator.Apply this identity using u(ζ) = G(ζ, z1) and v(ζ) = G(ζ, z2) to obtain

0 =

∫

bΩǫ

(v∂u

∂n− u

∂v

∂n

)ds.

Let Cǫ(zi) denote the circle of radius ǫ about zi. Since u and v vanishon bΩ, this last equation becomes

0 =

2∑

i=1

∫

Cǫ(zi)

(v∂u

∂n− u

∂v

∂n

)ds.

But u(ζ) = − ln |ζ − z1|+ u(ζ) and v(ζ) = − ln |ζ − z2|+ v(ζ), where uand v are harmonic functions on Ω in C∞(Ω). It is now an elementaryexercise in analysis to see that the first integral in the sum tends to−G(z1, z2) as ǫ→ 0 and the second one tends to G(z2, z1). The proof iscomplete.

We studied the Poisson kernel associated to a simply connected do-main in Chapter 10. Having defined the Green’s function on a boundedmultiply connected domain with C∞ smooth boundary, we may nowdescribe the Poisson kernel on such a domain. Suppose that u(z) is aharmonic function in C∞(Ω). Let Ωǫ denote the domain obtained fromΩ by removing the closed disc of radius ǫ about z0 and let Cǫ denote thecircle of radius ǫ about z0. We will now use the real Green’s identity onΩǫ, taking u to be the present u and v(ζ) = G(ζ, z0). We obtain

0 =

∫∫

Ωǫ

(v∆u − u∆v) dA =

∫

bΩǫ

(v∂u

∂n− u

∂v

∂n

)ds,


and similar reasoning to that which we used to prove the symmetryproperty of the Green’s functions shows that this integral is equal to

∫

bΩ

−u∂v∂n

ds+

∫

Cǫ

(v∂u

∂n− u

∂v

∂n

)ds.

The second integral in this sum tends to u(z0) as ǫ → 0, and we obtainthe Poisson formula

u(z0) =

∫

bΩ

p(z0, ζ)u(ζ) ds, (31.2)

where the Poisson kernel is given by

p(z0, ζ) =∂G(ζ, z0)

∂n,

the normal derivative acting in the ζ variable.The next theorem shows how the Green’s function and the Bergman

kernel are connected.

Theorem 31.3. Suppose that Ω is a bounded domain with C∞ smoothboundary. The Green’s function G(z, w) and the Bergman kernel K(z, w)associated to Ω are related via

K(z, w) = − 2

π

∂2G(z, w)

∂z∂w.

Proof. Let us agree to use subscripts to denote differentiation so that

Gzw(z, w) =∂2G(z, w)

∂z∂w.

Notice that∂

∂zln |z − w| = 1

2(z − w),

and so∂2

∂z∂wln |z − w| ≡ 0.

Hence, it follows from Theorem 31.1 that Gzw(z, w) is a function inC∞

((Ω× Ω)− (z, z) : z ∈ bΩ

)that is holomorphic in z and antiholo-

morphic in w. It also follows that we may write

Gz(z, w) = − 1

2(z − w)+H(z, w) (31.3)

where, for any fixed point z0 ∈ Ω, H(z0, w) is a bounded function in wnear z0. Fix a point z0 ∈ Ω, and let Ωǫ denote the domain obtained from

The Green’s function and the Bergman kernel 181

Ω by removing the closed disc of radius ǫ about z0 and let Cǫ denotethe circle of radius ǫ about z0 parameterized in the clockwise sense, i.e.,in the proper sense when viewed as a boundary curve of Ωǫ. We wish toshow that − 2

πGzw(z0, w) has the reproducing property,

h(z0) = − 2

π

∫∫

Ω

Gzw(z0, w)h(w) dA

for all h ∈ A∞(Ω). Since K(z0, w) is also an antiholomorphic functionof w satisfying this property, and since A∞(Ω) is dense in the Bergmanspace (Corollary 15.1), the equality of the two kernels will follow. Now,given h ∈ A∞(Ω),

∫∫

Ω

Gzw(z0, w)h(w) dA = limǫ→0

∫∫

Ωǫ

Gzw(z0, w)h(w) dA

= limǫ→0

∫∫

Ωǫ

∂

∂w[Gz(z0, w)h(w)]

1

2idw ∧ dw

= limǫ→0

1

2i

∫

bΩǫ

Gz(z0, w)h(w) dw.

Since G(z, w) = 0 when z ∈ Ω and w ∈ bΩ, it follows that Gz(z0, w) = 0when w ∈ bΩ, and so this last integral reduces to be an integral over justthe portion of bΩǫ consisting of the circle Cǫ. Using (31.3), we obtain

∫∫

Ω

Gzw(z0, w)h(w) dA

= limǫ→0

1

2i

∫

Cǫ

(− 1

2(z0 − w)+H(z0, w)

)h(w) dw

= (−2πi)(1

2i)(1

2)h(z0),

and the theorem is proved.

32

Zeroes of the Bergman kernel

In this chapter, we will prove that the Bergman kernelK(z, a) associatedto a bounded n-connected domain Ω with C∞ smooth boundary hasexactly n − 1 zeroes in Ω as a function of z whenever a ∈ Ω is nearenough to the boundary. The transformation rule (16.1) for the Bergmankernels under biholomorphic mappings shows that the number of zeroesof the Bergman kernel is invariant under such mappings. Hence, in viewof Lemma 12.1, we may study the problem on domains Ω that have realanalytic boundary.

We have repeatedly seen that C∞ properties of the classical functionsand operators in C∞ smooth domains translate to extension properties incase the domains under study are assumed to have real analytic bound-aries. We will now see that the Green’s function and Bergman kerneladhere to this rule.

We now suppose that Ω is a bounded domain with real analyticboundary and we let R(z) denote an antiholomorphic reflection functiondefined on a neighborhood U of bΩ (see Chapter 11). Let U1 = Ω∩U andlet U2 = U − Ω. The definition of an antiholomorphic reflection allowsus to assume that R is an antiholomorphic one-to-one map of U1 ontoU2 such that R−1 = R. Thus U2 = R(U1). Let Ω = Ω ∪ U2. Note that Ωis an open set containing Ω. We will use the following Schwarz reflectionprinciple for harmonic functions on Ω. Suppose that a is a point in Ω andthat u(z) is harmonic in Ω−a. Suppose further that u is continuous upto bΩ and that u vanishes on bΩ. Then the function defined to be equal tou(z) when z ∈ Ω and −u(R(z)) when z ∈ U2 defines an extension of u. If

a ∈ (Ω−U1), then this extension is harmonic on Ω−a. If a ∈ U1, then

the extension is harmonic on Ω − a,R(a). This reflection principle isproved by reducing the question to the classical reflection principle [Ah,p. 172] by means of the holomorphic maps described in Chapter 11 thatlocally map a real analytic curve into the real axis. From this reflectionprinciple, we read off that the Green’s function G(z, a) extends to be

harmonic in z on Ω− a if a ∈ Ω− U1 and to Ω− a,R(a) if a ∈ U1.In the second case, we may write G(z, a) = ln |R(z) − a| + u(z, a) forz near R(a) where u(z, a) is harmonic in both variables near (R(a), a).We may now read off the extension properties of the Bergman kernel.

183


Notice that for z near R(a),

∂G(z, a)

∂z=

∂∂zR(z)

2(R(z)− a)+∂u(z, a)

∂z,

and so∂2G(z, a)

∂z∂a=

∂∂zR(z)

2(R(z)− a)2+∂2u(z, a)

∂z∂a,

and it follows that the Bergman kernel K(z, a) extends to Ω× Ω mero-morphically in z and antimeromorphically in a in such a way thatK(z, a)

is holomorphic in z on all of Ω when a ∈ Ω − U1 and holomorphic onΩ− R(a) with a double pole at z = R(a) when a ∈ U1.

It is natural to also consider the function

Λ(z, w) = − 2

π

∂2G(z, w)

∂z∂w.

In fact, this function will play an important role in our study of thezeroes of the Bergman kernel. Since

∂2

∂z∂wln |z − w| = 1

2(z − w)2,

it follows that Λ(z, w) is a function in C∞((Ω× Ω)−D

)where D de-

notes the diagonal set (z, z) : z ∈ Ω. Furthermore, Λ(z, w) is holo-morphic in z and in w, Λ(z, w) = Λ(w, z), and for a fixed point w ∈ Ω,Λ(z, w) (as a function of z) has a double pole at w with principal part− 1

π (z −w)−2. When we apply the Schwarz reflection argument to Λ we

deduce that Λ(z, a) extends to Ω× Ω meromorphically in both variables

z and a in such a way that Λ(z, a) is holomorphic in z on Ω− a witha double pole at z = a. (Note that, since (∂2/∂z∂a) ln |R(z)− a| ≡ 0 forz near R(a), it follows that Λ(z, a) has no singularity at R(a).)

If z0 ∈ Ω and if ζ(t) parameterizes a boundary curve of Ω, thenG(z0, ζ(t)) ≡ 0, and differentiating this formula with respect to t yields

∂G

∂ζ(z0, ζ(t))ζ

′(t) +∂G

∂ζ(z0, ζ(t))ζ′(t) ≡ 0,

and we deduce that

∂G

∂ζ(z0, ζ)T (ζ) = −∂G

∂ζ(z0, ζ)T (ζ) for ζ ∈ bΩ. (32.1)

Since G is a real valued function, it follows that Gζ and Gζ are complex

Zeroes of the Bergman kernel 185

conjugates of each other, and so we deduce that ∂G∂ζ (z0, ζ)T (ζ) is a pure

imaginary number when ζ ∈ bΩ.We next consider the normal derivative of the Green’s function. Re-

call formula (18.1) which states that the normal derivative of a functionϕ given as a sum h+H where h and H are holomorphic functions thatare C∞ smooth up to the boundary can be expressed

∂ϕ

∂n= −iT (z)h′(z) + i T (z)H ′(z).

Suppose that w0 is a boundary point of Ω and that z ∈ Ω. Near w0, wemay find a harmonic conjugate function V (w) for the harmonic functionG(z, w) making G(z, w) + iV (w) a holomorphic function of w that ex-tends C∞ smoothly up to the boundary of Ω near w0. Hence, near w0,we may express G as G(z, w) = h(w)+h(w) where h = 1

2 (G+iV ). Using

the fact that g′ = 2 ∂∂zRe g when g is holomorphic, we obtain

∂

∂nwG(z, w) = −iT (w)∂G

∂w(z, w) + i T (w)

∂G

∂w(z, w), (32.2)

the subscript w on n indicating that the normal derivative acts in the wvariable. We may now combine (32.1) and (32.2) to obtain

∂

∂nwG(z, w) = −2iT (w)

∂G

∂w(z, w). (32.3)

Since G(z, w) is a positive harmonic function of w ∈ Ω−z that is C∞

smooth up to bΩ and that vanishes on the boundary, the Hopf lemma(Theorem 9.4) informs us that the normal derivative in (32.3) is strictlynegative for z ∈ Ω. Now, let w be a fixed point in bΩ and consider thefunction on the left hand side of (32.3) as a function of z on Ω − w.It is a strictly negative harmonic function of z on Ω that extends C∞

smoothly to Ω− w and that vanishes on bΩ− w. The Hopf lemmacan be used again to see that the normal derivative in the z variable of∂

∂nwG(z, w) is strictly positive. This implies that

∂

∂z

∂

∂nwG(z, w) 6= 0 if z, w ∈ bΩ, z 6= w.

If we differentiate (32.3) with respect to z and use this fact, we obtain

∂2G

∂z∂w(z, w) 6= 0 if z, w ∈ bΩ, z 6= w,

i.e., Λ(z, w) 6= 0 if z, w ∈ bΩ, z 6= w. If we differentiate (32.1) withrespect to z, we obtain

∂2G

∂z∂w(z, w)T (w) = − ∂2G

∂z∂w(z, w)T (w) for z ∈ Ω and w ∈ bΩ,


i.e.,

Λ(z, w)T (w) = −K(z, w)T (w) for z ∈ Ω and w ∈ bΩ. (32.4)

Hence, it also follows that K(z, w) 6= 0 if z, w ∈ bΩ, z 6= w. Furthermore,(32.4) implies that, as functions of z, K(z, w) and Λ(z, w) have the samezeroes in Ω when w ∈ bΩ.

By allowing z to approach a boundary point, formula (32.4) is seento remain true even when z is in the boundary provided that z 6= w. Ifwe multiply (32.4) by K(w, z) = K(z, w), we obtain

K(w, z)Λ(z, w)T (w) = −|K(z, w)|2T (w).

Consider z to be a fixed point in bΩ. This last identity allows us todetermine the increase in the argument of the holomorphic function ofw given by K(w, z)Λ(z, w) as w traces out bΩ − z in the standardsense. Indeed, since |K(z, w)|2 is nonvanishing, we obtain

∆arg K(w, z)Λ(z, w) + ∆arg T (w) = ∆arg T (w).

Since ∆arg T (w) = (2 − n)(2π) and since ∆arg T (w) = −(2 − n)(2π),we deduce that

∆arg K(w, z)Λ(z, w) = 2(n− 2)(2π).

For our fixed z ∈ bΩ, both K(w, z) and Λ(z, w) (as functions of w)have double poles at w = z and no zeroes on bΩ− z. The generalizedargument principle (see Chapter 13) therefore yields that the number ofzeroes of K(w, z)Λ(z, w) in Ω is equal to 2(n− 2)+ 1

2 (2+ 2) = 2(n− 1),and since K(w, z) and Λ(z, w) have the same zeroes, it follows that theyboth have (n− 1) zeroes.

It is now an easy exercise using continuity and the argument principleto see that K(z, w) has exactly n−1 zeroes in Ω as a function of z whenw ∈ Ω is sufficiently close to bΩ.

We remark that if w ∈ Ω is not close to the boundary, then K(z, w)may have fewer than n− 1 zeroes. Indeed, the Hartog’s Extension The-orem of several complex variables implies that any zero of K(z, w) mustpropagate to the boundary of Ω×Ω. Since K(z, w) is nonvanishing nearbΩ× bΩ, it follows that the zero variety of K(z, w) exits Ω×Ω throughthe portion of the boundary given by (Ω × bΩ) ∪ (bΩ × Ω). Hence, theBergman kernel must vanish at certain points K(z0, w0) with z0 ∈ bΩand w0 ∈ Ω. This is a good place to terminate this chapter, lest we betempted to start studying functions of several complex variables.

33

Complexity in complex analysis

Throughout this book, we have seen that there are beautiful formu-las relating almost any two objects we have defined. In particular, theBergman, Szego, and Poisson kernels are all closely connected. In thischapter, we address the question of just how complicated these kernelsare. Are they true functions of two complex variables, or might they becomprised of more elementary functions of one complex variable? Webegin by showing that the Szego kernel associated to a smooth domainis a rational combination of more elementary functions. In the simplyconnected setting, we may use a Riemann map to map to the unit discand use the transformation formula for the Szego kernels to see that theSzego kernel is a simple rational combination of the Riemann map anda square root of its derivative. Thus, this question is interesting only inthe multiply connected setting.

Suppose that Ω is a bounded n-connected domain with C∞ smoothboundary. We saw in Chapter 27 that a point a in Ω can be chosen sothat the n− 1 zeroes of the Szego kernel S(z, a) are distinct and simple.Let a1, . . . , an−1 denote these simple zeroes in Ω, and let a0 = a. Let fdenote the Ahlfors map fa associated to a. Recall that the zeroes of fare given by aj , j = 0, . . . , n− 1.

We will now show that the functions him(z) = S(z, ai)f(z)m, where

i = 0, . . . , n − 1 and m = 0, 1, 2, · · · , are nearly an orthogonal basis forthe Hardy space. Indeed, if m > k, then, using the fact that f = 1/f onthe boundary,

〈him, hjk〉b =∫

bΩ

Saifm Saj

fk ds =

∫

bΩ

Saifm−k Saj

ds

= S(aj , ai)f(aj)m−k = 0.

Hence, him and hjk are orthogonal if m 6= k. If m = k, notice that

〈him, hjk〉b =∫

bΩ

SaiSaj

ds = S(aj , ai).

The functions Saiare clearly linearly independent. Indeed, if

∑ciSai

≡0, then every function g in the Hardy space would satisfy

∑cig(ai) = 0,

187


which is absurd. Thus, if we perform a Gram-Schmidt orthogonalizationof the functions Sai

to get G0, . . . , Gn−1 we obtain an orthonormal setof functions Him = Gif

m (since |f | = 1 on the boundary). We nowclaim that this orthonormal set is complete. Indeed, if g is a functionin the Hardy space that is orthogonal to all of these functions, then gis orthogonal to all the original him too. The case m = 0 shows thatg(ai) = 0 for each i. Thus g is divisible by f (which has simple zeroes ata0, a1, . . . , an−1). Now, the case m = 1 yields

〈g, him〉b =∫

bΩ

g Saif ds =

∫

bΩ

(g/f)Saids = 0.

and this last integral is equal to the value of g/f at the removable sin-gularity, ai. Hence g vanishes to order 2 at each ai. We may continue inthis manner to see that g must vanish to infinite order at each ai. Thisshows, of course, that g must be identically zero, and we conclude thatthe orthogonal system is complete.

We next expand the Szego kernel Sw = S(z, w) in terms of our new-found orthonormal basis to see that

Sw =

∞∑

m=1

n−1∑

i=0

cimGifm,

wherecim = 〈Sw, Gif

m〉b,which is equal to the conjugate of Gi(w)f(w)

m. Thus, we have shownthat

S(z, w) =

∞∑

m=1

n−1∑

i=0

Gi(z)Gi(w)(f(z) f(w) )m

=1

(1− f(z) f(w) )

n−1∑

i=0

Gi(z)Gi(w).

This shows that the Szego kernel is a rational combination of finitelymany functions of one complex variable, just as in the simply connectedcase. Formula (25.1) now yields that the Bergman kernel is comprised offinitely many functions of one complex variable. Because it is possible toexpress the Poisson kernel in terms of the Szego kernel, it is also possibleto see that the Poisson kernel can be written in terms of finitely manyholomorphic functions and harmonic functions of one variable, but wewill not pursue this further here.

We close this chapter by showing that the formula for the Szego

Complexity in complex analysis 189

kernel can be further manipulated to yield

S(z, w) =

∑n−1i,j=0 cijSai

(z)Saj(w)

(1 − f(z) f(w) ), (33.1)

where [cij ] is the inverse matrix to the n×nmatrix S(ai, aj) as i and j runfrom 0 to n− 1. Indeed, since the functions Gj are linear combinationsof the functions Sai

, it follows that there are such constants cij . Todetermine the constants, write out the identity at the point (z, ak) usingthe fact that f(ak) = 0 to obtain

Sak=

n−1∑

i,j=0

cijS(aj , ak)Sai.

We showed above that the Sakare linearly independent, so it must be

thatn−1∑

j=0

cijS(aj , ak)

is equal to 1 if i = k and equal to zero otherwise, i.e., that [cij ] is thepromised inverse matrix. It is not hard to show that c00 = 1/S(a, a) andc0j = cj0 = 0 for all j 6= 0 because S(a, aj) = 0 for j = 1, . . . , n− 1, andconsequently, the (n− 1)× (n− 1) blocks of the matrices, letting i andj range from 1 to n − 1 are inverse to each other. We leave this to thereader.

34

Area quadrature domains and thedouble

Bjorn Gustafsson [Gu1] discovered the connection between the Schottkydouble of a domain and the property of being an area quadrature do-main. In this last chapter, we will describe just enough of this theoryto be able to prove that the Bergman kernel K(z, w) associated to anarea quadrature domain without cusps in the boundary is particularlysimple. It is a rational combination of z and the Schwarz function S(z)in the sense that K(z, w) is a rational function of z, S(z), w, and S(w).Consequently, since S(z) = z on the boundary, the Bergman kernel isrational in z, z, w, and w when z and w are boundary points, z 6= w.This is another instance of our claim that area quadrature domains sharemany properties with the unit disc. We will later combine this result withthe improved Riemann mapping theorem that says that bounded finitelyconnected domains are biholomorphic to area quadrature domains, anduse the transformation formula for the Bergman kernels to see that theBergman kernel is a rational combination of three functions of one com-plex variable, in general.

A rudimentary understanding of Riemann surfaces would make thischapter easier to digest for the reader, but it is not required. We willdescribe the basic notions that we will need.

Suppose that Ω is a bounded n-connected domain with C∞ smoothboundary. We now describe a compact Riemann surface Ω associated toΩ known as the double of Ω. The surface will be described by an atlasconsisting of n + 2 open connected sets. The first set is the domain Ωitself with the coordinate chart z mapping Ω into the complex plane.The second set is a copy of Ω that we will call Ω with the coordinatechart z. The double Ω is obtained by gluing Ω to Ω along correspondingboundary curves via coordinate charts described as follows. Let f denotean Ahlfors map associated to a point a in Ω. If γj is one of the boundary

curves of Ω, let Uj denote a thin collared neighborhood of γj in Ω with

the property that Uj ∩ Ω and Uj ∩ Ω correspond to the same set in Ω

under the obvious identification of Ω and Ω. We know that f extends C∞

smoothly up to γj and that f ′ is nonvanishing on γj . Hence, by taking

191


0 < ρ < 1 close enough to one we may assume that f ′ is nonvanishingon Uj and f maps Uj ∩Ω one-to-one onto the annulus z : ρ < |z| < 1.The coordinate chart on Uj is defined to be f(z) on Uj ∩ Ω and equal

to 1/ f(z) on Uj ∩ Ω. This chart maps Uj one-to-one onto the annulusz : ρ < |z| < 1/ρ. We obtain n open sets in our atlas this way, onefor each boundary curve. Think of the open sets Uj as lines of tape

gluing the edges of Ω to the corresponding edges of Ω. Think of Ω asthe “bottom half” of Ω and Ω as the “top half.” Since the transitionfunctions between overlapping open sets in the atlas are holomorphic,a compact one-dimensional complex manifold, i.e., a compact Riemannsurface, is obtained.

We remark that another more classic way to do the gluing above usesmaps from small open connected sets about points in the boundary of Ωto discs that are symmetric about the real axis. The side of U in Ω getsmapped to the upper half of the disc via a map F and the side of U in Ωgets mapped via the “reflected” function F on the Ω side. More chartsare needed in this method to cover the whole boundary. The Schwarzreflection theorem shows that the transition functions are holomorphic.However the gluing is done, it is the Schwarz reflection theorem that isat the heart of the double.

The maximum modulus theorem yields that a holomorphic functionon a compact Riemann surface must be constant; meromorphic func-tions are much more interesting, but still drastically restricted. Indeed,a meromorphic function h on Ω that extends smoothly to the bound-ary further extends meromorphically to the double if and only if thereis an antimeromorphic function H on Ω that extends smoothly to theboundary in such a way that h = H on the boundary. The meromorphicfunction on Ω is given by h on Ω and H on Ω. It is meromorphic on thetape strips Uj because of the following fact: If a function is continuouson a disc and analytic away from a C1 smooth curve, then it is analyticon the disc. This follows from Morera’s theorem because it is easy toshow that the contour integral around any triangle in the disc must bezero.

The Ahlfors map satisfies f(z) = 1/ f(z) on the boundary, and con-sequently, Ahlfors maps extend meromorphically to the double.

From this point on, we assume that Ω is an area quadrature domainwithout cusps in the boundary. We know that the Schwarz function S(z)associated to Ω extends meromorphically to Ω and satisfies S(z) = z onthe boundary. This fact gives rise to two meromorphic functions G1 andG2 on Ω defined as follows. The function G1 is equal to z on Ω andequal to S(z) on Ω. The function G2 is equal to S(z) on Ω and equal to

z on Ω. We will show soon that these two functions form a primitive pair,meaning that they generate all the meromorphic functions on the double,

Area quadrature domains and the double 193

but first, we show that the Bergman kernel function Ka associated to apoint a in Ω extends to the double of Ω as a meromorphic function.

Recall from Chapter 22 that, because Ω is an area quadrature do-main, there is an element in the Bergman span identically equal to one.Rewrite identity (32.4) by taking the conjugate and using subscripts aswe have throughout the book to obtain

KaT = −ΛaT (34.1)

on the boundary. This identity can be differentiated with respect to a toobtain

Kma T = −Λm

a T ,

where Λma (z) = (∂m/∂am)Λ(z, a). Let Λ0

a(z) = Λ(z, a) and call the com-plex linear span of functions of the form Λm

a as a ranges over Ω and mranges over nonnegative integers the Λ-span. The fact that one is in theBergman span is now seen to imply that

T = λT ,

on the boundary, where λ is in the Λ-span. If we now divide iden-tity (34.1) by this last formula, we obtain

Ka = Λa/λ

on the boundary, and this shows thatKa extends meromorphically to thedouble. This same argument shows that any holomorphic or meromor-phic function that satisfies an identity like (34.1) on an area quadraturedomain extends meromorphically to the double. To be precise, if H is ameromorphic function on an area quadrature domain Ω without cuspsthat extends smoothly to the boundary and satisfies an identity

H(z)T (z) = G(z)T (z) (34.2)

on the boundary, where G is meromorphic on Ω and extends smoothlyto the boundary, then H extends meromorphically to the double.

There are quite a few functions that satisfy an identity like (34.2).For example, the functions F ′

j given as derivatives 2∂ωj/∂z of the har-monic measure functions studied in Chapter 19 do (see identity (19.1)).Conjugating products of identity (7.1) and using the fact that 1/T = Tshow that functions of the form H(z) = S(z, a)S(z, b) do. When we com-bine the fact that these functions extend meromorphically to the doublewith identities (33.1) and (25.1), we see that the Bergman kernel is givenby K(z, w) =

∑ni,j=0 cijkmSai

(z)Saj(z)Sak

(w)Sam(w)

(1− f(z) f(w) )2+

n−1∑

j,k=1

CjkF′j(z)F

′k(w),


which is a rational combination of functions of one variable that extendmeromorphically to the double.

If we now give the promised proof that G1 and G2 generate all themeromorphic functions on the double, we will have completed the mis-sion of this chapter.

Let p1, . . . , pk denote the poles of S(z) in Ω. Note that the function

G1 has no poles on the Ω side of Ω, no poles on the bΩ part, and poles atthe points p1, . . . , pk on the Ω side of Ω corresponding to p1, . . . , pk. Letmj denote the order of the pole of S(z) at pj , which is the same as theorder of the pole of G1 at pj. We may choose a small ǫ > 0 so that there

is a small open set Wj about pj in Ω such that Wj − pj gets mappedby G1 in an mj-to-one manner onto w : |w| > 1/ǫ. Pick a point w0

in the complex plane with |w0| > 1/ǫ. If we choose ǫ small enough, theclosure of the sets Wj will be disjoint, Ω will be contained in D1/ǫ(0),

and we obtain N = m1 +m2 + · · ·+mk distinct points in G−11 (w0) that

fall on the Ω side of Ω. Denote these points by q1, . . . , qN . Since thesepoints are distinct, and since G2 is equal to z on the Ω side of Ω, wesee that G2 separates these points. We have shown that G2 separatesthe points of G−1

1 (w0), which are all points of multiplicity one for G1.This is a basic property of a primitive pair (see Farkas and Kra [F-K,p. 249]), and we will now use it to show that G1 and G2 generate the

meromorphic functions on Ω.It is a standard result in Riemann surface theory to show that the

number N is locally constant on C, and so it is also globally constant.It is the generic number of preimages of w in Ω under G1, counted withmultiplicity. For all but finitely many points w, these preimages are Ndistinct points.

Suppose that h is a meromorphic function on Ω. We will now showthat there are rational functions Ri such that

h(z) =N−1∑

i=0

Ri(G1(z))G2(z)i (34.3)

on Ω. Let w = G1(z) and let z1, . . . , zN denote the points in G−11 (w)

listed according to multiplicity, but in no particular order. For all butfinitely many points in the complex plane, these points will be distinct,and away from this finite set, we may think of zj as locally being aholomorphic function of w. When w is close to w0, we consider the Ninstances of equation (34.3) that would have to hold in order for thefunctions Ri to be well defined near w0, namely

h(zj) =

N−1∑

i=0

Ri(w)G2(zj)i

Area quadrature domains and the double 195

for j = 1, . . . , N . The zj are distinct points on the Ω side, and in fact,G2(zj)

i = zij is just a Vandermonde style matrix. Hence, the values ofRi(w) are uniquely determined via Cramer’s Rule. We remark that, al-though the order of the zj were not specified, the symmetric functionsinvolved in Cramer’s Rule make the order of no consequence. The defini-tion of the functions Ri near w0 can be uniquely analytically continuedto the whole complex plane, minus the finitely many points where somepoints in the list z1, . . . , zN occur with multiplicities greater than one.However, even at these isolated singularities, the formulas obtained fromCramer’s Rule show that the singularities are either removable or poles,even at the point at infinity. Hence the Ri are holomorphic functions onthe extended complex plane minus finitely many points where they havepoles, i.e., the Ri are rational. Since (34.3) holds near the points qj , itholds in general by the uniqueness of analytic continuation.

A simply connected domain that is not equal to the whole com-plex plane can be mapped to the unit disc via a Riemann map f . Thetransformation formula for the Bergman kernel shows that the Bergmankernel K(z, w) associated to the domain is a rational combination off ′(z), f(z), f ′(w), and f(w). A finitely multiply connected domain canbe mapped to a bounded domain with real analytic boundary, whichcan be mapped to an area quadrature domain without cusps. Let S(z)denote the Schwarz function for this area quadrature domain and letf denote the biholomorphic map from the multiply connected domainto the quadrature domain. Such a map f has been called a Gustafssonmap (see [Be8]). Perhaps a better name for it would be a Gustafsson-Riemann map, as we now explain. If we combine the results above withthe transformation formula for the Bergman kernel, we deduce that theBergman kernel K(z, w) for the original multiply connected domain isf ′(z) times f ′(w) times a rational function of f(z), S(f(z)), f(w), andS(f(w)). This is why we think of the mapping f as being a general-ized “Riemann” map in the multiply connected setting. It allows us topull back facts and formulas on quadrature domains to more generaldomains, just as the Riemann map allows pull backs of formulas on thedouble quadrature domain, the unit disc.

We end by mentioning that the generalized Riemann mapping men-tioned above can be improved. The target domain can be taken to be adouble quadrature domain. The complex unit tangent vector function Tassociated to a double quadrature domain extends meromorphically tothe double. Hence, it must be a rational function of z and the Schwarzfunction, i.e., a rational function of z and z. Double quadrature domainsare even more like the unit disc than area quadrature domains. On adouble quadrature domain, it can also be shown that the Szego kernelis like the Bergman kernel in that it is a rational function of z and the


Schwarz function, and thus, it too can be pulled back to the originaldomain in a concrete way. Proofs of these facts currently depend on adeeper understanding of Riemann surface theory (see [B-G-S]), so we donot pursue them further here. We close by delivering the reader to thedoorstep of a very interesting new subject in complex analysis, built onvery classical foundations using new tools given by the Kerzman-Steintheorem.

A

The Cauchy-Kovalevski theorem for theCauchy-Riemann operator

The purpose of this appendix is to give an elementary proof (shown tome by David Barrett) of the following theorem, which is a version ofthe Cauchy-Kovalevski theorem for the Cauchy problem for the ∂/∂zoperator.

Theorem A.1. Suppose that γ is a real analytic curve and that v isa complex valued real analytic function on a neighborhood of γ. Thereexists a real analytic function Φ on a neighborhood of γ such that

∂

∂zΦ = v

on the neighborhood and such that Φ vanishes on γ.

Proof. It is clear that the properties satisfied by the function Φ makeit uniquely determined near γ. Indeed, if Φ1 and Φ2 both satisfy theconditions of the theorem, then Φ1 − Φ2 is a holomorphic function ona neighborhood of γ vanishing along γ, and so Φ1 ≡ Φ2 near γ. Thisobservation allows us to restrict our attention to a small neighborhoodof a small segment in γ because we may use analytic continuation tocreate global solutions from local solutions.

The first step in the proof is to reduce the problem to the case inwhich γ is an open line segment L in the real axis and v is real analyticon a neighborhood of the closure of L. As we did in the proof of theSchwarz reflection principle, we may write down a holomorphic functionF on a neighborhood of the origin with F ′(0) 6= 0 and F mapping thereal axis onto a segment of γ containing a specified point. Let f denotethe inverse to F . If (∂/∂z)Φ = v near γ, then

∂

∂z(Φ F ) = F ′ (v F ).

It follows that if we can solve

∂

∂z(φ) = F ′ (v F )

197


with φ defined near the origin and vanishing along the real axis, thenΦ = φ f solves our original problem. Since F ′ (v F ) is real analyticnear the origin, we have reduced our problem to the case where γ is aline segment in the real axis containing the origin.

Suppose that v is real analytic near the origin. Then v can be ex-pressed as a convergent power series via

v(x+ iy) =

∞∑

n,m=0

cnmxnym.

The complex extension of v to C×C is the holomorphic function v(τ, ζ)of two complex variables (τ, ζ) ∈ C2 defined near (0, 0) given by

v(τ, ζ) =

∞∑

n,m=0

cnmτnζm.

For x and y near zero, we define

φ(x + iy) =2y

i

∫ 1

0

v(x+ i(1− t)y , ty ) dt.

It is easy to check that φ is real analytic near the origin and that φvanishes along the real axis. We claim that φ also satisfies ∂φ/∂z = vnear the origin. To see this, we compute

∂

∂zφ(x + iy) =

1

2

(∂

∂x− i

∂

∂y

)φ(x + iy)

=y

i

∫ 1

0

∂

∂τv(x + i(1− t)y, ty) dt+

∫ 1

0

v(x+ i(1− t)y, ty) dt

+ y

∫ 1

0

∂

∂τv(x+ i(1− t)y, ty)i(1− t) dt

+ y

∫ 1

0

∂

∂ζv(x+ i(1− t)y, ty)t dt

=

∫ 1

0

d

dt[tv(x+ i(1− t)y, ty)] dt = v(x, y),

and the proof is complete.

Another way to construct the function φ in the last paragraph of theproof is to express v as a power series in z and z. Let u =

∑cnmz

nzm

denote an antiderivative of this function with respect to z obtained bytermwise integration. The function φ is then given by u minus u whereu is the holomorphic function

∑cnmz

nzm in z, which agrees with u onthe real line.

Bibliographic Notes

Most of the theorems in this book are very old and well known results.What is new in this book is the way in which the old theorems arededuced from properties of the Szego projection and Cauchy transformwhich follow from the Kerzman-Stein theorem. In these notes, I willattempt to point out the parts of this book which differ from the standardapproach to the subject.

Most of the material in Chapter 2 is standard (see for example [Ho],Chapter 1). The proof of Lemma 2.1 and its application to proving The-orem 2.2 are based on a lemma proved in [Be3] which was used laterto study the boundary behavior of biholomorphic mappings in severalcomplex variables in [Be-Li] and [Be4].

Formula (3.2) relating the adjoint of the Cauchy transform to theCauchy transform was used implicitly by Kerzman and Stein in [K-S]where they showed that the Cauchy kernel is nearly self adjoint and theyproved the Kerzman-Stein formula. However, they deduced (3.2) fromthe boundedness of the Cauchy transform as an operator on L2(bΩ) andthe Plemelj formula. In this book, we were able to deduce (3.2) fromfirst principles, and later use it to deduce the Kerzman-Stein identity,and from that, the boundedness of the Cauchy transform in L2(bΩ).

It is traditional to define the Hardy space according to the classicaldefinition in Chapter 6. It is then a rather difficult task to prove thatA∞(Ω) is dense in H2(bΩ). The Kerzman-Stein identity allowed me toprove the basic facts about the Cauchy transform, which in turn allowedme to define H2(bΩ) as the closure in L2(bΩ) of A∞(Ω), and to deducethe standard facts about the Hardy space with a minimum of effort.

The orthogonal decomposition of L2(bΩ) given in Theorem 4.3 isdue to Schiffer [Sch]. Also, Theorem 19.1 relating the linear span of thederivatives of the harmonic measure functions to the Szego kernel wasproved by Schiffer in the same work.

In [Bu1, Bu2], J. Burbea set down much of the groundwork for study-ing the Kerzman-Stein theorem, the Szego kernel, and the Garabediankernel in multiply connected domains. It is in [Bu1] that Burbea provedthe formula described in Chapter 25 that I called Burbea’s formula.

The Garabedian kernel was defined by Garabedian in [Ga] where heused it to show that the Ahlfors map associated to a multiply connected

199

200 Bibliographic Notes

domain is given as the quotient of the Szego kernel and the Garabediankernel. See Bergman [Ber], Nehari [Ne], Grunsky [Gr], and Fisher [Fi]for alternate proofs of the existence of the Ahlfors map and of Garabe-dian’s formula. In the literature, the Garabedian kernel is constructed asthe unique function satisfying identity (7.1). The method of proof usesa delicate extremal problem involving Green’s functions and the Pois-son kernel. The characterization of the Garabedian kernel in this bookas falling out of the orthogonal decomposition of the Cauchy kernel ismuch easier than the classical approach. This allows the proof of theexistence of the Ahlfors map given in Chapter 13 (which otherwise fol-lows the description of Nehari’s proof given by Bergman in [Ber]) to bestreamlined.

The density lemmas and their consequences described in Chapter 9are motivated by similar results proved for the Bergman kernel in severalcomplex variables in [Be3, Be6, Be-Li].

The approach to studying the classical Dirichlet and Neumann prob-lems by means of the Szego projection is based on [Be1, Be2]. However,in these papers, the formulas were deduced from known properties ofthe solution operators to the classical PDE problems. The novelty of theapproach of this book is that the formulas relating the solution opera-tors to the Szego projection can be deduced from first principles, therebyallowing the properties of the solution operators to the classical prob-lems to be read off from properties of the Szego projection. An alternateapproach to the study of the Dirichlet problem by means of the Szegoprojection can be found in [Sh].

The transformation formulas for the Bergman projection and kernelunder proper holomorphic maps were proved in [Be5] where they wereused to study boundary regularity of proper holomorphic mappings be-tween domains in several complex variables.

The transformation formula for the Szego kernel under certain properholomorphic mappings in Chapter 25 was proved by Moonja Jeong [Je]in her Purdue University PhD thesis.

The study of area quadrature domains was pioneered by Shapiro andAharonov [A-S] in the simply connected case and by Gustafsson [Gu1] inthe multiply connected case. Chapter 22 follows the approach of Y. Avci[Av] and [Be8, Be9, Be10]. More about the history of this subject andits usefulness can be found in [Cr], [EGKP], [G-S], and [Sh].

Arc length quadrature domains were studied by Shapiro and Ullemar[S-U] in the simply connected case and by Gustafsson [Gu2] in the mul-tiply connected case. Chapter 23 follows the approach in [B-G-S]. SeeShapiro’s book [Sh] for more about the history and applications of theseideas.

Theorem 25.4 turned out to be useful in the study of domains with the

Bibliographic Notes 201

property that solutions to the Dirichlet problem with rational boundarydata must be rational (see [BEKS]).

Theorem 27.1 about the zeroes of the Szego kernel was proved in[Be2]. Results in Chapter 32 about the zeroes of the Bergman kernelwere proved by Suita and Yamada in [Su-Ya].

The description of the numerical methods for computing the Szegokernel in Chapter 28 follows Kerzman and Trummer [K-T]. Treatmentsof this subject can also be found in [He] and [Tr].

Theorem 29.1 was proved by Alinhac, Baouendi, and Rothschild in[A-B-R]. The proof given in Chapter 29 follows the alternate proof givenin [Be-Le].

The results about the duality between A∞(Ω) and A−∞(Ω) were firststudied in the case of the unit disc in [Ko1, Ko2, T-W], and were gener-alized to several complex variables in [Be6, Be7, Be-Bo]. The one variableversions of these results presented here are based on the approach usedin several variables.

Chapters 33 and 34 follow [Be11] and rely heavily on Gustafsson’swork about the connections between quadrature domains and meromor-phic functions on the double [Gu1].

Bibliography

[A-S] D. Aharonov, and H. S. Shapiro, Domains on which ana-lytic functions satisfy quadrature identities, Journal d’AnalyseMathematique 30 (1976), 39–73.

[Ah] L. Ahlfors, Complex analysis, McGraw Hill, New York, 1979.

[A-B-R] S. Alinhac, M. S. Baouendi, and L. P. Rothschild, Uniquecontinuation and regularity for holomorphic functions at theboundary, Duke Math. J. 61 (1990), 635–653.

[Av] Y. Avci, Quadrature identities and the Schwarz function, Stan-ford University PhD thesis, 1977.

[Be1] S. Bell, Solving the Dirichlet problem in the plane by means ofthe Cauchy integral, Indiana Math. J. 39 (1990), 1355–1371.

[Be2] , The Szego projection and the classical objects of po-tential theory in the plane, Duke Math. J. 64 (1991), 1–26.

[Be3] , Non-vanishing of the Bergman kernel function atboundary points of certain domains in Cn, Math. Ann. 244

(1979), 69–74.

[Be4] , Biholomorphic mappings and the ∂-problem, Ann.Math. 114 (1981), 103–113.

[Be5] , Proper holomorphic mappings and the Bergman pro-jection, Duke Math. J. 48 (1981), 167–175.

[Be6] , A representation theorem in strictly pseudoconvex do-mains, Illinois J. Math. 26 (1982), 19–26.

[Be7] , A duality theorem for harmonic functions, MichiganMath. J. 29 (1982), 123–128.

[Be8] , The Bergman kernel and quadrature domains in theplane, Operator Theory: Advances and Applications 156 (2005),35–52.

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[Be9] , Quadrature domains and kernel function zipping, Arkivfor Matematik 43 (2005), 271–287.

[Be10] , Density of quadrature domains in one and several com-plex variables, Complex Variables and Elliptic Equations 54

(2009), 165–171.

[Be11] , An improved Riemann mapping theorem and complex-ity in potential theory, Arkiv for Matematik 51 (2013), 223–249.

[BEKS] S. Bell, P. Ebenfelt, D. Khavinson, and H. Shapiro, On the clas-sical Dirichlet problem in the plane with rational data, Journald’Analyse Mathematique 100 (2006), 157–190.

[Be-Bo] S. Bell and H. P. Boas Regularity of the Bergman projectionand duality of holomorphic function spaces, Math. Ann. 267(1984), 473–478.

[B-G-S] S. Bell, B. Gustafsson, and Z. Sylvan, Szego coordinates,quadrature domains, and double quadrature domains, Compu-tational Methods and Function Theory 11 (2011), No. 1, 25–44.

[Be-K] S. Bell and S. Krantz Smoothness to the boundary of conformalmaps, Rocky Mountain Math. J. 17 (1987), 23–40.

[Be-Le] S. Bell and L. Lempert A C∞ Schwarz Reflection Principle inone and several complex variables, J. of Differential Geometry32 (1990), 899–915.

[Be-Li] S. Bell and E. Ligocka A simplification and extension of Feffer-man’s theorem on biholomorphic mappings, Invent. Math. 57(1980), 283–289.

[Ber] S. Bergman The kernel function and conformal mapping, Math.Surveys 5, AMS, Providence, 1950.

[Bu1] J. Burbea The Cauchy and the Szego kernels on multiply con-nected regions, Rendiconti del Circolo Math. di Palermo 31

(1982), 105–118.

[Bu2] , The Riesz projection theorem in multiply connectedregions, Bollettino della Unione Matematica Italiana 14 (1977),143–147.

[Cr] D. Crowdy, Quadrature domains and fluid dynamics, OperatorTheory: Advances and Applications 156 (2005), 113–129.

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[Eb] P. Ebenfelt, Singularities encountered by the analytic continu-ation of solutions to Dirichlet’s problem, Complex Variables 20

(1992), 75–91.

[EGKP] P. Ebenfelt, B. Gustafsson, D. Khavinson, and M. Putinar,Quadrature domains and their applications, Operator Theory:Advances and Applications 156, Birkhauser, Basel, 2005.

[F-K] H. Farkas, and I. Kra, Riemann Surfaces, Springer-Verlag, NewYork, 1980.

[Fi] Stephen D. Fisher Function theory on planar domains, Wiley,New York, 1983.

[Fo] G. Folland Introduction to partial differential equations, Prince-ton U. Press, Princeton, 1976.

[Ga] P. Garabedian Schwarz’s lemma and the Szego kernel function,Trans. Amer. Math. Soc. 67 (1949), 1–35.

[Gr] H. Grunsky Lectures on theory of functions in multiply con-nected domains, Vandenhoeck and Ruprecht, Gottingen, 1978.

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[Gu2] , Applications of half-order differentials on Riemann sur-faces to quadrature identities for arc-length, Journal d’AnalyseMath. 49 (1987), 54–89.

[G-S] B. Gustafsson, and H. Shapiro, What is a quadrature domain?Operator Theory: Advances and Applications 156 (2005), 1–25.

[He] P. Henrici Applied and computational complex analysis, Vol. 3,John Wiley, New York, 1986.

[Ho] L. Hormander An introduction to complex analysis in severalvariables, North Holland, Amsterdam, 1973.

[Je] Moonja Jeong Approximation theorems on mapping propertiesof the classical kernel functions of complex analysis, PurdueUniversity PhD Thesis, 1991.

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[K-T] N. Kerzman and M. Trummer Numerical conformal mappingvia the Szego kernel, in Numerical conformal mapping, L. N.Trefethen, ed., North Holland, Amsterdam, 1986, 111–123.

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[Ru2] , Real and complex analysis, McGraw Hill, New York,1966.

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[Sh] H. S. Shapiro The Schwarz function and its generalization tohigher dimensions, Univ. of Arkansas Lecture Notes in theMathematical Sciences, Wiley, New York, 1992.

[S-U] H. Shapiro, and C. Ullemar, Conformal mappings satisfying cer-tain extremal properties and associated quadrature identities,Research Report TRITA-MAT-1986-6, Royal Inst. of Technol-ogy, 40 pp., 1981.

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[St] E. M. Stein Boundary behavior of holomorphic functions,Princeton University Press, Princeton, 1972.

[Su-Ya] N. Suita and A. Yamada On the Lu Qi-Keng conjecture, Proc.Amer. Math. Soc. 59 (1976), 222–224.

[T-W] B. A. Taylor and D. L. Williams Ideals in rings of analyticfunctions with smooth boundary values, Canad. J. Math. 22

(1970), 1266–1283.

[Tr] M. Trummer An efficient implementation of a conformal map-ping method based on the Szego kernel, SIAM J. of Numer.Anal. 23 (1986), 853–872.

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[Yo2] , Functional Analysis, Springer-Verlag, New York, 1980.

The Cauchy Transform, Potential Theory and Conformal Mapping explores the most central result in all of classical function theory, the Cauchy integral formula, in a new and novel way based on an advance made by Kerzman and Stein in 1976.

The book provides a fast track to understanding the Riemann Mapping Theorem. The Dirichlet and Neumann problems for the Laplace opera-tor are solved, the Poisson kernel is constructed, and the inhomog-enous Cauchy-Reimann equations are solved concretely and efficiently using formulas stemming from the Kerzman-Stein result.

These explicit formulas yield new numerical methods for computing the classical objects of potential theory and conformal mapping, and the book provides succinct, complete explanations of these methods.

Four new chapters have been added to this second edition: two on quadrature domains and another two on complexity of the objects of complex analysis and improved Riemann mapping theorems.

The book is suitable for pure and applied math students taking a begin-ning graduate-level topics course on aspects of complex analysis as well as physicists and engineers interested in a clear exposition on a fundamental topic of complex analysis, methods, and their application.

K25868

w w w . c r c p r e s s . c o m



The C

auchy Transform, Potential T

heory and Conform

al Mapping

Steven R. Bell

Bell

2nd Edition

Mathematics

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The Cauchy transform, potential theory, and conformal mapping

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