Analisis de Medida Tesis de Grado

CY535/Garnett 0 521 47018 8 January 27, 2005 15:41

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Harmonic Measure

During the last two decades several remarkable new results were discovered aboutharmonic measure in the complex plane. This book provides a survey of these resultsand an introduction to the branch of analysis that contains them. Many of these results,due to Bishop, Carleson, Jones, Makarov, Wolff, and others, appear here in book formfor the first time.The book is accessible to students who have completed standard graduate courses inreal and complex analysis. The first four chapters provide the needed backgroundmaterial on univalent functions, potential theory, and extremal length, and each chapterhas many exercises to further inform and teach the reader.

J O H N B . G A R N E T T is Professor of Mathematics at the University of California,Los Angeles.

D O N A L D E . M A R S H A L L is Professor of Mathematics at the University ofWashington.

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ii


NEW MATH MONOGRAPHS

Editorial Board

Bla BollobsWilliam FultonFrances KirwanPeter SarnakBarry Simon

For information about Cambridge University Press mathematics publicationsvisit http://publishing.cambridge.org/stm/mathematics

iii


iv


Harmonic Measure

JOHN B. GARNETT

University of California, Los Angeles

DONALD E. MARSHALL

University of Washington

v

CAMBRIDGE UNIVERSITY PRESS

Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, So Paulo

Cambridge University PressThe Edinburgh Building, Cambridge CB2 8RU, UK

First published in print format

ISBN-13 978-0-521-47018-6

ISBN-13 978-0-511-34894-5

Cambridge University Press 2005

2005

Information on this title: www.cambridge.org/9780521470186

This publication is in copyright. Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

ISBN-10 0-511-34894-0

ISBN-10 0-521-47018-8

Cambridge University Press has no responsibility for the persistence or accuracy of urls for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.

Published in the United States of America by Cambridge University Press, New York

www.cambridge.org

hardback

eBook (EBL)

eBook (EBL)

hardback


To Dolores and Marianne

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viii


Contents

Preface page xiii

I. Jordan Domains 11. The Half-Plane and the Disc 12. Fatous Theorem and Maximal Functions 63. Carathodorys Theorem 134. Distortion and the Hyperbolic Metric 165. The HaymanWu Theorem 23

Notes 25Exercises and Further Results 26

II. Finitely Connected Domains 371. The Schwarz Alternating Method 372. Greens Functions and Poisson Kernels 413. Conjugate Functions 504. Boundary Smoothness 59


III. Potential Theory 731. Capacity and Greens Functions 742. The Logarithmic Potential 773. The Energy Integral 794. The Equilibrium Distribution 825. Wieners Solution to the Dirichlet Problem 896. Regular Points 937. Wiener Series 978. Polar Sets and Sets of Harmonic Measure Zero 102

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x Contents

9. Estimates for Harmonic Measure 104Notes 112Exercises and Further Results 112

IV. Extremal Distance 1291. Definitions and Examples 1292. Uniqueness of Extremal Metrics 1333. Four Rules for Extremal Length 1344. Extremal Metrics for Extremal Distance 1395. Extremal Distance and Harmonic Measure 1436. The

dx(x)

Estimate 146Notes 149Exercises and Further Results 150

V. Applications and Reverse Inequalities 1571. Asymptotic Values of Entire Functions 1572. Lower Bounds 1593. Reduced Extremal Distance 1624. Teichmllers Modulsatz 1665. Boundary Conformality and Angular Derivatives 1736. Conditions More Geometric 184


VI. Simply Connected Domains, Part One 2001. The F. and M. Riesz Theorem 2002. Privalovs Theorem and Plessners Theorem 2033. Accessible Points 2054. Cone Points and McMillans Theorem 2075. Compression and Expansion 2126. Pommerenkes Extension 216


VII. Bloch Functions and Quasicircles 2291. Bloch Functions 2292. Bloch Functions and Univalent Functions 2323. Quasicircles 2414. Chord-Arc Curves and the A Condition 2465. BMO Domains 253



Contents xi

VIII. Simply Connected Domains, Part Two 2691. The Law of the Iterated Logarithm for Bloch Functions 2692. Harmonic Measure and Hausdorff Measure 2723. The Number of Bad Discs 2814. Brennans Conjecture and Integral Means Spectra 2865. Numbers and Polygonal Trees 2896. The Dandelion Construction and (c) (a) 2967. Baernsteins Example on the HaymanWu Theorem 302


IX. Infinitely Connected Domains 3151. Cantor Sets 3152. For Certain , dim < 1. 3243. For All , dim 1. 331


X. Rectifiability and Quadratic Expressions 3471. The Lusin Area Function 3482. Square Sums and Rectifiability 3613. A Decomposition Theorem 3724. Schwarzian Derivatives 3805. Geometric Estimates of Schwarzian Derivatives 3846. Schwarzian Derivatives and Rectifiable Quasicircles 3937. The BishopJones H

12 Theorem 397

8. Schwarzian Derivatives and BMO Domains 4089. Angular Derivatives 411

10. A Local F. and M. Riesz Theorem 41511. Ahlfors Regular Sets and the HaymanWu Theorem 420


Appendices 435A. Hardy Spaces 435B. Mixed Boundary Value Problems 441C. The Dirichlet Principle 447D. Hausdorff Measure 456E. Transfinite Diameter and Evans Functions 466F. Martingales, Brownian Motion, and Kakutanis Theorem 470G. Carlemans Method 480


xii Contents

H. Extremal Distance in Finitely Connected Domains 484I. McMillans Twist Point Theorem 497J. Bloch Martingales and the Law of the Iterated Logarithm 503

K. A Dichotomy Theorem 512L. Two Estimates on Integral Means 518

M. Calderns Theorem and Chord-Arc Domains 520

Bibliography 531Author Index 555Symbol Index 559Subject Index 561


Preface

Several surprising new results about harmonic measure on plane domains havebeen proved during the last two decades. The most famous of these results areMakarovs theorems that harmonic measure on any simply connected domainis singular to Hausdorff measure for all > 1 but absolutely continuousto for all < 1. Also surprising was the extension by Jones and Wolff ofMakarovs > 1 theorem to all plane domains. Further important new resultsinclude the work of Carleson, Jones and Wolff, and others on harmonic measurefor complements of Cantor sets; the work by Carleson and Makarov, Bertilsson,Pommerenke, and others on Brennans tantalizing conjecture that for univalentfunctions

||2pdxdy < if 43 < p < 4; several new geometric condi-tions that guarantee the existence of angular derivatives; and the Jones squaresum characterization of subsets of rectifiable curves and its applications byBishop and Jones to a variety of problems in function theory.

We wrote this book to explain these exciting new results and to providebeginning students with an introduction to this part of mathematics. We havetried to make the subject accessible to students who have completed graduatecourses in real analysis from Folland [1984] or Wheeden and Zygmund [1977],for example, and in complex analysis from Ahlfors [1979] or Gamelin [2001],for example.

The first four chapters, along with the appendices on Hardy spaces, Hausdorffmeasures and martingales, provide a foundation that every student of functiontheory will need. In Chapter I we solve the Dirichlet problem on the half-planeand the disc and then on any simply connected Jordan domain by using theCarathodory theorem on boundary continuity. Chapter I also includes briefintroductions to hyperbolic geometry and univalent function theory. In Chap-ter II we solve the Dirichlet problem on domains bounded by finitely manyJordan curves and study the connection between the smoothness of a domainsboundary and the smoothness of its Poisson kernel. Here the main tools are

xiii


xiv Preface

two classical theorems about conjugate functions. Chapter II and the discus-sion in Chapter III of Wieners solution of the Dirichlet problem on arbitrarydomains follow the 1985 UCLA lecture course by Carleson. The introductionto extremal length in Chapter IV is based on the Institut MittagLeffler lec-tures of Beurling [1989]. Chapter V contains some applications of extremallength, such as Teichmllers Modulsatz and some newer theorems about an-gular derivatives, that are not found in other books. Chapter VI is a blend ofthe classical theorems of F. and M. Riesz, Privalov, and Plessner and the morerecent theorems of McMillan, Makarov, and Pommerenke on the comparisonof harmonic measure and one dimensional Hausdorff measure for simply con-nected domains. Chapter VII surveys the beautiful circle of ideas around Blochfunctions, univalent functions, quasicircles, and Ap weights. Chapter VIII isan exposition of Makarovs deeper results on the relations between harmonicmeasure in simply connected domains and Hausdorff measures and the work ofCarleson and Makarov concerning Brennans conjecture. Chapter IX discussesharmonic measure on infinitely connected plane domains. Chapter X begins byintroducing the Lusin area function, the Schwarzian derivative, and the Jonessquare sums, and then applies these ideas to several problems about univalentfunctions and harmonic measures. The thirteen appendices at the end of the textprovide further related material.

For space reasons we have not treated some important related topics. Theseinclude the connections between Chapters VIII and IX and thermodynamicalformalism and several other connections between complex dynamics and har-monic measure. We have emphasized Wieners solution of the Dirichlet prob-lem instead of the Perron method. The beautiful Perron method can be foundin Ahlfors [1973] and Tsuji [1959]. We also taken a few detours around thetheory of prime ends. There are excellent discussions of prime ends in Ahlfors[1973], Pommerenke [1975], and Tsuji [1959]. Finally, the theory of harmonicmeasure in higher dimensions has a different character, and we have omitted itentirely.

At the end of each chapter there is a brief section of biographical notesand a section called Exercises and Further Results". An exercise consistingof a stated result without a reference is meant to be homework for the reader.Further results" are outlines, with detailed references, of theorems not in thetext.

Results are numbered lexicographically within each chapter, so that Theo-rem 2.4 is the fourth item in Section 2 of the same chapter, while Theorem III.2.4is from Section 2 of Chapter III. The same convention is used for formulas, sothat (3.2) is in the same chapter, while (IV.6.4) refers to (6.4) from Chapter IV.

Many of the results that inspired us to write this book are also covered in


Preface xv

Pommerenkes excellent book [1991]. However, our emphasis differs from theone in Pommerenke [1991] and we hope the two books will complement eachother.

Some unpublished lecture notes from a 1986 Nachdiplom Lecture courseat Eidgenssische Technische Hochschule Zurich by the first listed author andthe out-of-print monograph Garnett [1986] were preliminary versions of thepresent book.

The web pagehttp://www.math.washington.edu/ marshall/HMcorrections.html

will list corrections to the book. Though we have tried to avoid errors, theobservant reader will no doubt find some. We would appreciate receiving emailat [email protected] about any errors you come across.

Many colleagues, friends, and students have helped with their comments andsuggestions. Among these, we particularly thank A. Baernstein, M. Benedicks,D. Bertilsson, C. J. Bishop, K. Burdzy, L. Carleson, S. Choi, R. Chow, M.Essn, R. Gundy, P. Haissinski, J. Handy, P. Jones, P. Koosis, N. Makarov, P.Mateos, M. ONeill, K. yma, R. Prez-Marco, P. Poggi-Corridini, S. Rohde,I. Uriarte-Tuero, J. Verdera, S. Yang and S. Yoshinobu.

We gratefully acknowledge support during the writing of this book by theRoyalty Research Fund of the University of Washington, the University ofWashingtonUniversity of Bergen Faculty Exchange Program, the Institut desHautes Etudes Scientifiques, the Centre de Recerca Matemtica, Barcelona, andthe National Science Foundation.

Los Angeles and Seattle John B. GarnettSeattle and Bergen Donald E. Marshall


xvi


IJordan Domains

To begin we construct harmonic measure and solve the Dirichlet problem inthe upper half-plane and the unit disc. We next prove the Fatou theorem onnontangential limits. Then we construct harmonic measure on domains boundedby Jordan curves, via the Riemann mapping theorem and the Carathodorytheorem on boundary correspondence. We review two topics from classicalcomplex analysis, the hyperbolic metric and the elementary distortion theory forunivalent functions. We conclude the chapter with the theorem of Hayman andWu on lengths of level sets. Its proof is an elementary application of harmonicmeasure and the hyperbolic metric.

1. The Half-Plane and the Disc

Write H = {z : Imz > 0} for the upper half-plane and R for the real line. Sup-pose a < b are real. Then the function

= (z) = arg(

z bz a

)= Im log

(z bz a

)is harmonic on H, and = on (a, b) and = 0 on R \ [a, b].

a b

z

Figure I.1 The harmonic function (z).

1


2 I. Jordan Domains

Viewed geometrically, (z) = Re(z) where (z) is any conformal mappingfrom H to the strip {0 < Rez < } which maps (a, b) onto {Rez = } andR \ [a, b] into {Rez = 0}. Let E R be a finite union of open intervals andwrite E =nj=1(aj , bj ) with bj1 < aj < bj . Set

j = j (z) = arg(

z bjz aj

)and define the harmonic measure of E at z H to be

(z, E,H) =n

j=1

j

. (1.1)

Then

(i) 0 < (z, E,H) < 1 for z H,(ii) (z, E,H) 1 as z E, and

(iii) (z, E,H) 0 as z R \ E .The function (z, E,H) is the unique harmonic function on H that satisfies (i),(ii), and (iii). The uniqueness of (z, E,H) is a consequence of the followinglemma, known as Lindelfs maximum principle.

Lemma 1.1 (Lindelf). Suppose the function u(z) is harmonic and boundedabove on a region such that = C. Let F be a finite subset of andsuppose

lim supz

u(z) 0 (1.2)

for all \ F. Then u(z) 0 on .Proof. Fix z0 / . Then the map 1/(z z0) transforms into a boundedregion, and thus we may assume is bounded. If (1.2) holds for all ,then the lemma is the ordinary maximum principle. Write F = {1, . . . , N },let > 0, and set

u(z) = u(z) N

j=1log(

diam()|z j |

).

Then u is harmonic on and lim supz u(z) 0 for all . Thereforeu 0 for all , and

u(z) lim0

Nj=1

log(

diam()|z j |

)= 0.


1. The Half-Plane and the Disc 3

Lindelf [1915] proved Lemma 1.1 under the weaker hypothesis that isinfinite. See also Ahlfors [1973]. Exercise 3 and Exercise II.3 tell more aboutLindelfs maximum principle.

Given a domain and a function f C(), the Dirichlet problem forf on is to find a function u C() such that u = 0 on and u| = f .Theorem 1.2 treats the Dirichlet problem on the upper half-plane H .

Theorem 1.2. Suppose f C(R {}). Then there exists a unique functionu = u f C(H {}) such that u is harmonic on H and u|H = f .Proof. We can assume f is real valued and f () = 0. For > 0, take disjointopen intervals Ij = (tj , tj+1) and real constants cj , j = 1, . . . , n, so that thestep function

f(t) =n

j=1cj Ij

satisfies f f L(R) < . (1.3)Set

u(z) =n

j=1cj(z, Ij ,H).

If t R \ Ij , thenlim

Hztu(z) = f(t)

by (ii) and (iii). Therefore by (1.3) and Lemma 1.1,supH

u1(z) u2(z) < 1 + 2.Consequently the limit

u(z) lim0

u(z)

exists, and the limit u(z) is harmonic on H and satisfies

supH

|u(z) u(z)| 2.

We claim that

lim supzt

|u(z) f (t)| (1.4)


4 I. Jordan Domains

for all t R. It is clear that (1.4) holds when t / Ij . To verify (1.4) at theendpoint tj+1 Ij Ij+1, notice that by (ii), (iii), and Lemma 1.1,

supH

cj(z, Ij ,H) + cj+1(z, Ij+1,H) (cj + cj+12)

(z, Ij Ij+1,H)

cj cj+12

,while

limztj+1

(cj + cj+1

2

)(z, Ij Ij+1,H) = cj + cj+12 .

Hence all limit values of u(z) at tj+1 lie in the closed interval with endpointscj and cj+1, and then (1.3) yields (1.4) for the endpoint tj+1.

Now let t R. By (1.4)lim sup

zt|u(z) f (t)| sup

zH|u(z) u(z)| + lim sup

zt|u(z) f (t)| 3.

The same estimate holds if t = . Therefore u extends to be continuous on Hand u|H = f . The uniqueness of u follows immediately from the maximumprinciple.

For a < b, elementary calculus gives

(x + iy, (a, b),H) = 1

(tan1

( x ay

) tan1( x by

))= b

a

y(t x)2 + y2

dt

.

If E R is measurable, we define the harmonic measure of E at z H to be

(z, E,H) =

E

y(t x)2 + y2

dt

. (1.5)

When E is a finite union of open intervals this definition (1.5) is the same asdefinition (1.1). For z = x + iy H, the density

Pz(t) = 1

y(x t)2 + y2

is called the Poisson kernel for H . If f C(R {}), the proof of Theo-rem 1.2 shows that

u f (z) =R

f (t)Pz(t)dt,

and for this reason u f is also called the Poisson integral of f .


1. The Half-Plane and the Disc 5

Note that the harmonic measure (z, E,) is a harmonic function in its firstvariable z and a probability measure in its second variable E . If z1, z2 H then

0 < C1 (z1, E,H)(z2, E,H)

C < ,

where C depends on z1 and z2 but not on E . This inequality, known as Har-nacks inequality, is easily proved by comparing the kernels in (1.5).

Now let D be the unit disc {z : |z| < 1} and let E be a finite union of openarcs on D. Then we define the harmonic measure of E at z in D to be

(z, E,D) ((z), (E),H), (1.6)where is any conformal map of D onto H. This harmonic function satisfiesconditions analogous to (i), (ii), and (iii), so that by Lemma 1.1 the definition(1.6) does not depend on the choice of . It follows by the change of variables(z) = i(1 + z)/(1 z) that

(z, E,D) =

E

1 |z|2|ei z|2

d2

.

An equivalent way to find this function is by a construction similar to (1.1).This construction is outlined in Exercise 1.

Theorem 1.3. Let f (ei ) be an integrable function on D and set

u(z) = u f (z) = 2

0f (ei ) 1 |z|

2

|ei z|2d2

. (1.7)

Then u(z) is harmonic on D. If f is continuous at ei0 D, thenlim

Dzei0u(z) = f (ei0). (1.8)

Clearly (1.8) also holds if the integrable function f is changed on a measurezero subset of D \ {ei0}. The function u = u f is called the Poisson integralof f and the kernel

Pz() = 121 |z|2|ei z|2

is the Poisson kernel for the disc. If f C(D) then

U (z) ={

u f (z), z Df (z), z D

is the solution of the Dirichlet problem for f on D.In the special case when f (ei ) is continuous, Theorem 1.3 follows from

Theorem 1.2 and a change of variables. Conversely, Theorem 1.3 shows that


6 I. Jordan Domains

Theorem 1.2 can be extended to f L1(dt/(1 + t2)), again by changing vari-ables.

Proof of Theorem 1.3. We may suppose f is real valued. From the identity

Re(ei + z

ei z)

= 2 Pz(),

we see that u is the real part of the analytic function 20

f (ei )ei + z

ei zd2

,

and therefore that u is a harmonic function. One can also see u is harmonic bydifferentiating the integral (1.7).

Suppose f is continuous at ei0 and let > 0. Then| f (ei ) f (ei0)| <

on an interval I = (1, 2) containing 0. Setting

u(z) =

[0,2]\I1 |z|2|ei z|2 f (e

i )d2

+ f (ei0)(z, I,D),

we have

|u(z) u(z)| =

I

1 |z|2|ei z|2 ( f (e

i ) f (ei0)) d2

(z, I,D) .However, limzei0 u(z) = f (ei0) by the definition of u. Therefore

lim supzei0

|u(z) f (ei0)| < ,

and (1.8) holds when f is continuous at ei0 .

2. Fatous Theorem and Maximal Functions

When f L1(D) the limit (1.8) can fail to exist at every D; see Ex-ercise 7. However, there is a substitute result known as Fatous theorem, inwhich the approach z is restricted to cones. For D and > 1, wedefine the cone

( ) ={z : |z | < (1 |z|)}.

The cone ( ) is asymptotic to a sector with vertex and angle 2 sec1() thatis symmetric about the radius [0, ]. The cones ( ) expand as increases.


2. Fatous Theorem and Maximal Functions 7

( )0

z

2 sec1

Figure I.2 The cone ( ).

A function u(z) on D has nontangential limit A at D iflim

( )zu(z) = A (2.1)

for every > 1.A good example is the function u(z) = e z+1z1 . This function u(z)is continuous on D \ {1}, and |u( )| = 1 on D \ {1}, but u(z) has nontangen-tial limit 0 at = 1. With fixed > 1, the nontangential maximal functionof u at is

u( ) = sup( )

|u(z)|.

If u has a finite nontangential limit at , then u( ) < for every > 1.We write |E | for the Lebesgue measure of E D.

Theorem 2.1 (Fatous theorem). Let f (ei ) L1(D) and let u(z) be thePoisson integral of f . Then at almost every = ei D,

lim( )z

u(z) = f ( ) (2.2)

for all > 1. Moreover, for each > 1{ D : u( ) > } 3 + 6 || f ||1 . (2.3)When u(z) is the Poisson integral of f L1(D) the function u = u f is also

called the solution to the Dirichlet problem for f , even though u convergesto f on D only nontangentially and only almost everywhere.

Inequality (2.3) says the operator L1(D) f u is weak-type 1-1. Itfollows from (2.2) that u( ) < almost everywhere, but (2.3) is a sharper,


8 I. Jordan Domains

quantitative result. In the proof of the theorem we derive (2.2) from the estimate(2.3).

The proof of Fatous theorem is a standard approximate identity argu-ment from real analysis that derives almost everywhere convergence for allf L1(D) from(a) an estimate such as (2.3) for the maximal function, and(b) the almost everywhere convergence (2.2) for all functions in a dense subset

of L1(D), such as C(D).

See Stein [1970]. We will use this approximate identity argument again later.

Proof. As promised, we first assume (2.3) and show (2.3) implies (2.2). Fix temporarily. We may assume f is real valued. Set

Wf ( ) = lim supz

|u f (z) f ( )|.

Then Wf ( ) u( ) + | f ( )|. Chebyshevs inequality gives{ : | f ( )| > } || f ||1

,

so that by (2.3),{ : Wf ( ) > } { : u( ) > /2}+ { : | f ( )| > /2} 8 + 12

|| f ||1 .

(2.4)

Fix > 0 and let g C(D) be such that || f g||1 2. Now Wg( ) = 0 byTheorem 1.3, and hence

Wf ( ) = Wf g( ).Applying (2.4) to f g then gives{ : Wf ( ) > } (8 + 12)2

= (8 + 12).

Therefore, for any fixed , (2.2) holds almost everywhere. Because the cones increase with , it follows that (2.2) holds for every > 1, except for ina set of measure zero.

To prove (2.3) we will dominate the nontangential maximal function with asecond, simpler maximal function. Let f L1(D) and write

M f ( ) = supI

1|I |

I| f |d



for the maximal average of | f | over subarcs I D that contain . The functionM f is called the HardyLittlewood maximal function of f . The function M fis simpler than u because it features characteristic functions of intervals insteadof Poisson kernels.

Lemma 2.2. Let u(z) be the Poisson integral of f L1(D) and let > 1.Then

u( ) (1 + 2)M f ( ). (2.5)Proof. Assume = 1. Fix z so that 0 = arg z has |0| . Define

Pz () = sup{

Pz() : | | || }

=

1

21 + |z|1 |z| , | | |0|

max(Pz(), Pz()), |0| < | | .

Pz()

Pz ()

00Figure I.3 The function Pz .

The function Pz satisfies

(i) Pz () is an even function of [, ],(ii) Pz () is decreasing on [0, ], and

(iii) Pz () Pz().The even function Pz is the smallest decreasing majorant of Pz on [0, ]. Wemay assume f (ei ) 0, so that

f (ei )Pz()d

f (ei )Pz ()d.


10 I. Jordan Domains

Then properties (i) and (ii) implyf (ei )Pz ()d ||Pz ||1 M f (1) (2.6)

because Pz is the increasing limit of a sequence of functions of the formcj(

12j

(j ,j )()

)with cj 0 and

cj ||Pz ||1 .

Pz ()

0 n0Figure I.4 Approximating Pz by a step function.

Now we claim that when z (1),||Pz ||1 (1 + 2). (2.7)

Note that (iii), (2.6), and (2.7) imply (2.5). To prove (2.7) we first assume/2 0 = arg z /2. Then by the law of sines,

|0|1 |z|

|0||1 z|

2| sin 0||1 z| =

2|sin |

1

2,

where = arg(z 1)/z is explained by Figure I.2. If /2 |0| andz (1), then |1 z| 1 and

|0|1 |z|

|0||1 z| .

Hence (see Figure I.3)

Pz 1 = 2 |0|

Pz()d + 2|0|21 + |z|1 |z| (1 + 2).

That proves (2.7) and therefore Lemma 2.2.



By Lemma 2.2, the inequality (2.3) will follow from the simpler inequality{ D : M f ( ) > } 3 f 1

, (2.8)

which says that the operator L1 f M f is also weak-type 1-1.To prove the inequality (2.8), we use a covering lemma.

Lemma 2.3. Let be a positive Borel measure on D and let {Ij } be a fi-nite sequence of open intervals in D. Then {Ij } contains a pairwise disjointsubfamily {Jk} such that

(Jk) 13(

Ij). (2.9)

Proof. Because the family {Ij } is finite, we may assume that no Ij is containedin the union of the others. Writing Ij = {ei : (aj , bj )},we may also assumethat

0 a1 < a2 < . . . . < an < 2.Then bj+1 > bj , because otherwise Ij+1 Ij , and bj1 < aj+1, because oth-erwise Ij Ij1 Ij+1. If n > 1, then bn < b1 + 2 and bn1 < a1 + 2 .Consequently, the family of even-numbered intervals Ij is pairwise disjoint.The family of odd-numbered intervals Ij is almost pairwise disjoint; only thefirst and last intervals can intersect. If

j even(Ij ) 13

(Ij),

we take the even-numbered intervals to be the subfamily {Jk}. Otherwisej odd

(Ij ) 23(

Ij).

In that case, if

(I1) 12j odd

(Ij ),

we take for {Jk} the family of odd-numbered intervals, omitting the first intervalI1, while if

(I1) >12j odd

(Ij ),

we take {Jk} = {I1}. Then in each case (2.9) holds for the subfamily {Jk}.



Lemma 2.4. The operator f M( f ) is weak-type 1-1: If f L1, then{ D : M f ( ) > } 3 f 1

. (2.8)

Proof. Let K be a compact subset of E = { D : M f ( ) > }. For each E there is an open interval I such that I and

1|I |

I| f |d > ,

so that

|I | < 1

I| f |d.

Cover K by finitely many such intervals {Ij : 1 j n}, and let {Jk} be thepairwise disjoint subfamily given by Lemma 2.3. Then

|K | Ij 3 |Jk | 3 Jk | f |d 3 || f ||1,

and letting |K | tend to |E| establishes Lemma 2.4.

By (2.5) and (2.8), inequality (2.3) holds with constant 3 + 6, and the proofof Fatous theorem is complete.

Corollary 2.5. If u is a bounded harmonic function on D, then for every > 0and for almost every = ei D,

f ( ) = lim( )z

u(z)

exists, u(z) is the Poisson integral of f , and || f || = supzD |u(z)|.Proof. Let rn 1 and let fn(ei ) = u(rnei ). By the BanachAlaoglu theo-rem, the sequence { fn} has a weak-star cluster point f L(D) satisfying|| f || lim sup || fn|| supD|u(z)|. Since u(rnz) is the Poisson integral offn , and Poisson kernels are in L1, u must be the Poisson integral of f . Butthen |u(z)| || f ||. The corollary now follows from Fatous theorem.

In particular, the corollary implies that for any measurable E D, thereexists a unique bounded harmonic function u(z) on D such that u(z) has non-tangential limit E almost everywhere. It is the function u(z) = (z, E,D).


3. Carathodorys Theorem 13

3. Carathodorys TheoremLet be a simply connected domain in the extended plane C. We say is aJordan domain if = is a Jordan curve in C.Theorem 3.1 (Carathodory). Let be a conformal mapping from the unitdisc D onto a Jordan domain . Then has a continuous extension to D, andthe extension is a one-to-one map from D onto .

Because mapsD onto, the continuous extension (also denoted by) mustmap Donto = , and because is one-to-one on D, (ei )parameterizesthe Jordan curve .

Before we prove Carathodorys theorem, we use it to solve the Dirichletproblem on a Jordan domain . Let f be Borel function on such that f is integrable on D . If w = 1(z), then

u(z) u f (z) = 2

0f (ei ) 1 |w|

2

|ei w|2d2

(3.1)

is harmonic on , and by Theorem 3.1 and Theorem 1.3,

limz

u(z) = f ( ) (3.2)

whenever 1( ) D is a point of continuity of f . In particular, if f iscontinuous on then (3.2) holds for all and u(z) = u f (z) solves theDirichlet problem for f on .

If f is a bounded Borel function on , then f is Borel and the integral(3.1) is defined. For any Borel set E we use (3.1) with f = E to definethe harmonic measure of E relative to by:

(z, E,) = (w, 1(E),D) =1(E)

1 |w|2|ei w|2

d2

. (3.3)

Then E (z, E) is a Borel measure on , and (3.1) can be rewritten as

u(z) =

f ( )d(z, ). (3.4)

Equations (3.3) and (3.4) do not depend on the choice of , because for everyconformal self map T of D,

(T (w), T (1(E)),D) = (w, 1(E),D).When f is a bounded Borel function on , (3.4) and Fatous theorem give

supz

|u(z)| = f L().



Moreover, Corollary 2.5 shows that every bounded harmonic function on canbe expressed in the form (3.4).

The principal goal of this book is to find geometric properties of the harmonicmeasure (z, E) more explicit than the definition (3.3). But (3.3) already pointsout the key issue: for a Jordan domain questions about harmonic measure areequivalent to questions about the boundary behavior of conformal mappings.

Proof of Theorem 3.1. We may assume is bounded. Fix D. First weshow has a continuous extension at . Let 0 < < 1, write

B(, ) = {z : |z | < }and set = D B(, ). Then () is a Jordan arc having length

L() =

|(z)|ds.

By the CauchySchwarz inequality

L2()

|(z)|2ds,

so that for < 1, 0

L2()

d

DB(,)|(z)|2dx dy

= Area((D B(, ))) < . (3.5)

n(n )

n

Un

n

n

Figure I.5 The crosscuts n and (n ).

Therefore there is a sequence n 0 such that L(n) 0. When L(n) < ,the curve (n ) has endpoints n , n , and both of these endpoints mustlie on = . Indeed, if n , then some point near n has two distinct


3. Carathodorys Theorem 15

preimages in D because maps D onto , and that is impossible because isone-to-one. Furthermore,

|n n| L(n) 0. (3.6)Let n be that closed subarc of having endpoints n and n and having smallerdiameter. Then (3.6) implies diam(n) 0, because is homeomorphic tothe circle. By the Jordan curve theorem the curve n (n ) divides the planeinto two regions, and one of these regions, say Un , is bounded. Then Un ,because C \ is arcwise connected. Since

diam(Un) = diam(n (n )

) 0,we conclude that

diam(Un) 0. (3.7)Set Dn = D {z : |z | < n}. We claim that for large n, (Dn) = Un . Ifnot, then by connectedness (D \ Dn) = Un and

diam(Un) diam((B(0, 1/2))

)> 0,

which contradicts (3.7). Therefore diam((Dn)) 0 and

(Dn) consistsof a single point, because (Dn+1) (Dn). That means has a continuousextension to { } D. It is an exercise to show that the union over of theseextensions defines a continuous map on D .

Let also denote the extension : D . Since (D) = , maps Donto . To show is one-to-one, suppose (1) = (2) but 1 = 2. Theargument used to show n also shows that (D) = , and so we canassume j D, j = 1, 2. The Jordan curve{

(r1) : 0 r 1} {(r2) : 0 r 1}

bounds a domain W , and then 1(W ) is one of the two components ofD \

({r1 : 0 r 1} {r2 : 0 r 1}

).

But since (D) ,(D 1(W )) W = {(1)}

and is constant on an arc of D. It follows that is constant, either bySchwarz reflection principle or by the Jensen formula, and this contradictionshows (1) = (2).



One can also prove is one-to-one by repeating for 1 the proof that iscontinuous. See Exercise 13(b). Exercise 12 gives the necessary and sufficientcondition that have a continuous extension to D.

The CauchySchwarz trick used to prove (3.5) is known as a lengthareaargument. The lengtharea method is the cornerstone of the theory of extremallength. See Chapter IV.

4. Distortion and the Hyperbolic Metric

Let D be the open unit disc. The hyperbolic distance from z1 D to z2 Dis

(z1, z2) = D(z1, z2) = inf z2

z1

|dz|1 |z|2 , (4.1)

where the infimum is taken over all arcs in D connecting z1 to z2. LetM denotethe set of conformal self maps of D :

T (z) = z a1 az , a D, || = 1.

When T M, we have|T (z)|

1 |T (z)|2 =1

1 |z|2 ,

and thus the hyperbolic distance is conformally invariant,

(T (z1), T (z2)) = (z1, z2), T M. (4.2)This conformal invariance is the main reason we are interested in the hyperbolicdistance.

The hyperbolic metric is the infinitesimal form |dz|/(1 |z|2) of the hy-perbolic distance. Taking

T (z) = z z11 z1z

gives

(z1, z2) = (0, T (z2)) = T (z2)

0

|dz|1 |z|2 .

Therefore the hyperbolically shortest arc from 0 to T (z2) is the radius[0, T (z2)], and its hyperbolic length is

(0, T (z2)) = 12 log(1 + |T (z2)|

1 |T (z2)|).


4. Distortion and the Hyperbolic Metric 17

In general

(z1, z2) = 12 log

1 + z2 z11 z1z2

1

z2 z11 z1z2 , (4.3)

and the hyperbolically shortest curve, or the geodesic, from z1 to z2 is a segmentof a diameter of D or an arc of a circle in D orthogonal to D.

By (4.3) we have z2 z11 z1z2 = e2(z1,z2) 1

e2(z1,z2) + 1 = tanh (z1, z2).

Write

t = t (d) = tanh(d) = e2d 1

e2d + 1 .

Then the hyperbolic ball B = {z : (z, a) < d} is the euclidean disc

{z : z a1 az

< t},and a calculation shows that B has euclidean radius

r(a, d) = t (1 |a|2)

1 t2|a|2 (4.4)

and euclidean distance to D

dist(B, D) =( 1 t

1 + |a|t)(1 |a|). (4.5)

Therefore, if d is fixed, the euclidean distance dist(B, D) and the euclideandiameter of B are both comparable to dist(a, D). However, for a = 0 theeuclidean center of B is not a. Figure I.6 shows two hyperbolic balls with thesame hyperbolic radius and two geodesics with the same hyperbolic length.

z1z2

a1 a2

x10

Figure I.6 Hyperbolic balls and geodesics.



Now assume (z) is a univalent function in D, that is, assume is analyticand one-to-one on D. After dilating, translating, and rotating the domain (D), is normalized by (0) = 0 and (0) = 1, so that

(z) = z + a2z2 + . (4.6)Important examples are the Koebe functions

(z) = (z) = z(1 z)2 , || = 1. (4.7)

Note that

(z) =

n=1nn1zn

maps D to the complement of the radial slit [/4,].Theorem 4.1 (Koebe one-quarter theorem). Assume(z) is a univalent func-tion on D. If (z) has the form (4.6) then

|a2| 2 (4.8)and

dist(0, (D)) 14. (4.9)

Equality holds in (4.8) or (4.9) if and only if is a Koebe function (4.7).Proof. First note that (4.8) implies (4.9). Indeed, suppose w / (D). Then(4.8) holds for the univalent function

g(z) = w(z)w (z) = z + (a2 +

1w

)z2 + ,

so that a2 + 1w

2, (4.10)and together (4.10) and (4.8) give

|w| 14.

To prove (4.8) we form the odd function

f (z) = z

(z2)

z2= z + a2

2z3 + .



Then f is univalent because is univalent, and the C = C {} valuedfunction

F(z) = 1f (z) =1z a2

2z + = 1

z+

n=1

bnzn (4.11)

is also univalent in D. To complete the proof we use the following lemma, calledthe area theorem, which we will prove after we establish (4.8).Lemma 4.2 (area theorem). If the univalent function F(z) satisfies (4.11),then

n=1

n|bn|2 1. (4.12)

To establish (4.8) we apply (4.12) to F = 1/ f . Since b1 = a2/2 and |b1| 1,we have |a2| 2.

The reader can verify that equality in either of (4.8) or (4.9) implies is aKoebe function.

Proof of Lemma 4.2. The lemma is called the area theorem because of itsproof. For r < 1, the Jordan curve r = {F(rei ) : 0 2} encloses anarea A(r), and by Greens theorem,

A(r) = i2

r

wdw = i2

20

F(rei )F

(rei )d.

Therefore by (4.11) and Fourier series,

A(r) = (

1r2

n=1n|bn|2r2n

)and

1

n=1n|bn|2 = lim

r1A(r)

0,

which yields (4.12).

Theorem 4.3 (Koebes estimate). Let (z) be a conformal mapping from theunit disc D onto a simply connected domain . Then for all z D

14|(z)|(1 |z|2) dist((z), ) |(z)|(1 |z|2). (4.13)



Proof. Fix z0 D. Then the univalent function

(z) =(

z+z01+z0z

) (z0)

(z0)(1 |z0|2

)satisfies (0) = 0 and (0) = 1. Hence if w / (D), then w (z0)(z0)(1 |z0|2)

14by (4.9), and this gives the left-hand inequality in (4.13).

To prove the right-hand inequality, fix z D, takef (w) = 1

((z) + dist((z), )w),

and apply the Schwarz lemma at w = 0 to the function

g(w) = f (w) z1 z f (w) .

We will often use the invariant form of (4.13):

Corollary 4.4. Let be a conformal mapping from a simply connected domain1 onto a simply connected domain 2, and let (z0) = w0. Then

| (z0)|4

dist(w0, 2)dist(z0, 1)

4| (z0)| (4.14)

Proof. Applying (4.13) to (z) = (z0 + dist(z0, 1)z) gives the left-handinequality and the same argument with 1 gives the right-hand inequality.

In a simply connected domain = C, the hyperbolic distance is definedby moving back to D via a conformal map : D . We write

(w1, w2) = D(z1, z2)when wj = (zj ). By (4.2), (w1, w2) does not depend on the choice of theconformal map . The quasihyperbolic distance from w1 to w2 is

Q(w1, w2) = inf w2w1

|dw|dist(w, )

,

in which the infimum is taken over all arcs in joining w1 to w2. Since (4.13)can be written as

|dz|1 |z|2

|dw|dist(w, )

4|dz|1 |z|2 ,



where w = (z), we have(w1, w2) Q(w1, w2) 4(w1, w2). (4.15)

Consequently the geometric statement following (4.4) and (4.5) about hyper-bolic distances near D remains approximately true in every simply connecteddomain with nontrivial boundary.

Let be any proper open subset of C. Then there exist closed squares {Sj },having pairwise disjoint interiors and sides parallel to the axes, such that

(i) Sj has sidelength (Sj ) = 2nj(ii) = Sj , and

(iii) diam(Sj ) dist(Sj , ) < 4 diam(Sj ).The squares {Sj } are called Whitney squares. Here is one way to constructWhitney squares in the case diam() < : Let 2N+1 diam() < 2N+2and partition the plane into squares having sides parallel to the axes and side-length 2N . We call these squares 2N squares. Include in the family {Sj }any 2N square S which satisfies (iii), and divide each of the remain-ing 2N squares into four squares of side 2N1. Next include in {Sj } any ofthese new 2N1 squares contained in and satisfying (iii), and continue. (SeeFigure I.7). Finding Whitney squares in the case diam() = is an exercisefor the reader. Whitney squares can be viewed as replacements for hyperbolicballs since there are universal constants c1 < c2 such that each Tj contains ahyperbolic ball of radius c1 and is contained in a hyperbolic ball of radius c2.

{Tk} {Sj }Figure I.7 Whitney squares in D and .

Whitney squares are almost conformally invariant in the following sense:Assume is simply connected, let : D be a conformal mapping, let{Sj } be the Whitney squares for and let {Tk} be the Whitney squares for D.Then by (4.15) there is a constant M , not depending on the map , such that for



each k, (Tk) is contained in at most M squares Sj , and 1(Sk) is containedin at most M squares Tj . In particular, for each d > 0, there is M(d) suchthat every hyperbolic ball {z : (z, a) < d} in is covered by M(d) Whitneysquares.

Theorem 4.5. Let (z) be a univalent function satisfying (0) = 0 and (0) = 1. Then

|z|(1 + |z|)2 |(z)|

|z|(1 |z|)2 , (4.16)

and1 |z|

(1 + |z|)3 |(z)| 1 + |z|

(1 |z|)3 . (4.17)

Moreover,

1 |z||z|(1 + |z|)

| (z)||(z)|

1 + |z||z|(1 |z|) . (4.18)

Inequality (4.16) is known as the growth theorem, while (4.17) is called thedistortion theorem.

Proof. The critical inequalities are (4.17) and we prove them first. Fix z0 Dand take

f (z) =(

z+z01+z0z

) (z0)

(z0)(1 |z0|2

) . (4.19)Then f is univalent on D, f (0) = 0 and f (0) = 1. By (4.8),

| f (0)| = (z0)1 |z0|2 (z0) 2z0

4,so that when z0 = rei ,ei (z0) (z0) 2|z0|1 |z0|2

41 |z0|2 . (4.20)Applying the general formula

Rezg(z)|z| =

Regr

to g = log , we then obtain2r 41 r2

rlog | (z0)| 2r + 41 r2 . (4.21)

Integrating (4.21) along the radius [0, z] gives both inequalities in (4.17).


5. The HaymanWu Theorem 23

To prove the upper bound in (4.16), integrate the upper bound in (4.17)along [0, z]. To prove the lower bound, we can assume that |(z)| < 14 , because|z|(1+|z|)2 0 there exists I I such that Iand |I | < . Prove the Vitali covering lemma: If I is a Vitali cover of Eand if > 0 there exists a pairwise disjoint sequence {Ij } I such that|E \ Ij | < .

10. Assume u(z) is the Poisson integral of a finite positive measure on D,and let d = f ()d + ds be the Lebesgue decomposition of . Provethat almost everywhere d , u has nontangential limit f (), and that almosteverywhere ds , u has nontangential limit . Hint: Use Lemma 2.3.

11. Let u(z) be real and harmonic in D.(a) If 1 < < , then

u u C, M(u).Moreover, if 0 < p < , then

||u ||L p(D) C , ||u||L p(D).


Exercises and Further Results 29

(b) If 1 p < and if u is the Poisson integral of f L p, then

limr1

|u(rei ) f (ei )|d = 0.

(c) If p > 1 show u is the Poisson integral of f L p(D) if and only if

supr 0 and u(0) = 1.(e) Show u is the Poisson integral of a finite (signed) Borel measure on Dif and only if

supr



n such that length(n) 0. Let Un be the component of \ n suchthat z / Un . Prove that (Un) D.

14. Let f (z) be an analytic function on D and assume | f (z)| 1.(a) Prove Picks theorem: f (z) f (w)1 f (w) f (z)

z w1 wz.

Equivalently, prove

( f (z), f (w)) (z, w).In other words, analytic maps from D to D are Lipschitz with respect to thehyperbolic metric.Hint: Repeat the proof of the Schwarz lemma, using (S f )/T for suitableS and T in M.(b) Deduce that

| f (z)|1 | f (z)|2

11 |z|2 .

(c) If equality holds in (a) at any two points z and w, or if equality holds in(b) at any point, then f M.

15. (a) Prove (4.4) and (4.5).(b) Find the euclidean center of the hyperbolic ball {z : (z, a) < d}.

16. Let (z) be univalent on D and assume (z) = 0 for every z.(a) Prove

|(0)| 4|(0)|,using (4.9).(b) Prove (

1 |z|1 + |z|

)2 |(z)||(0)|

(1 + |z|1 |z|

)2.

Hint: Fix z and apply (a) to

(w) = ( w + z

1 + zw)

to get (z)(z)

41 |z|2 ,and integrate along [0, z].



17. (a) Show that if equality holds anywhere in (4.8), (4.9), (4.16), (4.17), (4.18),or (4.21) then is a Koebe function, defined by (4.7).(b) On the other hand, equality in (4.12) holds if and only if C \ F(D) hasarea zero.

18. (a) Find an upper bound for the constant M mentioned in the remarks pre-ceding Figure I.7.(b) Give a construction of Whitney squares in the case diam() = .

19. A region of the form

Q = {z = rei : 0 < < 0 + (Q), 1 (Q) r < 1}is called a box or a Carleson box, and the arc

{ei : 0 < < 0 + (Q)}is the base of Q. See Figure I.11. A finite measure on D is a Carlesonmeasure if there is a constant C such that

(Q) C(Q)for every Carleson box.

aQ

Q

(Q)

(Q)

ei0

Figure I.11 A Carleson box.

(a) Let > 1. Prove is a Carleson measure if and only if there is C suchthat

({z D : u(z) > }) C{ D : u > }

whenever u is the Poisson integral of f L1(D). Hint: Write the set{u > } =

Ij where the Ij are pairwise disjoint open arcs on D, and

define Tj = D \

/Ij ( ). The set Tj looks like a tent with base Ij .



Tj

Tk

Ij

Ik

Figure I.12 Tents.

Then (Tj ) C(Ij ) and |u| on D \

Tj . Conversely, let Q be thebox with base I, and let u(z) = 4(z, I,D). Then u(z) > on Q whileby (2.3), {u > } C|I |.(b) If is a finite measure, then is a Carleson measure if and only if

supaD

1 |a|2|1 az|2 d(z) < .

Hint: Let a be the center of Q. See page 239 of Garnett [1981].(c) Let be a countable union of rectifiable curves in D. Then

supTM

length(T ()) <

if and only if arc length on is a Carleson measure.(d) A region in the upper half-plane H of the form

Q = {(x, y) : a x a + (Q), 0 < y (Q)}is also called a box or Carleson box and a finite measure on H is called aCarleson measure if there is a constant C such that

(Q) C(Q)for every Carleson box in H. A box Q is called dyadic if (Q) = 2k anda = j2k for some integers j and k. If (Q) C(Q) for all dyadic boxes,prove that is a Carleson measure on H.(e) If is a Carleson measure on D and is a conformal map of H onto D,then prove is a Carleson measure on H. Conversely, if is a Carlesonmeasure on H supported in the unit box [0, 1] [0, 1], then 1 is aCarleson measure on D.

Part (a) is due to E. M. Stein (unpublished).



20. Let be a conformal mapping from D to a simply connected domain , letL be any line and set = 1( L). Then(a) Arc length on is a Carleson measure; in other words,

length( Q) C(Q)for any box Q = {rei : 0 < < 0 + (Q), 1 (Q) r < 1}.(b) Moreover, for any disc B(z, r),

length( B(z, r)) C r. (E.3)Hint: If r < 1 |z| or r 0.4 use a linear map and the Hayman-Wutheorem. If 1 |z| < r < 0.4 then there is a Carleson box Q D B(z, r)such that (Q) Cr, and part (a) gives (E.3).

21. Let (z) be univalent on D and assume (z) = 0 for every z.(a) There is C1 < , independent of , such that

D

(z)(z)

dxdy C1.Hint: Write w = u + iv = (z), w . Then(

D

(z)(z)

dxdy)2 D

( 11 |z|

) 23 dxdy

D

(z)(z)

2(1 |z|) 23 dxdy

C2

(1 |1(w)|) 23 dudv|w|2 .

Now integrate over {|w| > 1} and {|w| 1} separately, and useExercise 16.(b) Let f = log . There is C2 < , independent of such that

Q| f |dxdy C2(Q) (E.4)

for any Carleson box

Q = {z = rei : 0 0 + (Q), 1 (Q) r < 1}.In other words, | f (z)|dxdy is a Carleson measure. For the proof we mayassume (Q) is small. Let a = aQ = (1 2(Q))ei(0+

(Q)2 ) and use (a) with

w = (z a)/(1 az). See Figure I.11.



(c) It follows from (b) that almost everywhere and in L1 norm,limr1

f (rei ) = f (ei )

exists and f (z) is the Poisson integral of f (ei ). In fact, f (ei ) BMO(for bounded mean oscillation), which by definition means there exists aconstant C3 < , independent of , such that for every arc I D,

infC

1|I |

I| f |d C3. (E.5)

Indeed, if Q has base I, then one can show1|I |

I| f f (aQ)|d C

Q

| f |dxdy.

See Jones [1980]. Baernstein [1976] gave the first proof of (E.5). Cima andSchober [1976] and Pommerenke [1976] have related results. When e f (z) isnot univalent, (E.4) is not equivalent to (E.5). However, in general (E.5) isequivalent to

Q| f |2(1 |z|)dxdy C(Q).

See Fefferman and Stein [1972], Baernstein [1980], Garnett [1981], or Ap-pendix F.

22. Again let be univalent and assume(z) = 0 for all z. Prove that for r 1/2

I (r) = (rei )

(rei )

2d C41 r log( 21 r ), (E.6)where C4 does not depend on . Hayman [1980] attributes the followingargument to P. L. Duren: By Exercise 16,

{|z|



Now take

1 R2 = 12(1 r2).

Hayman [1980] shows that the logarithm in (E.6) is necessary.23. Let 0 < p < . An analytic function f (z) on the disc D is in the Hardy

space H p if

sup0



(c) If p > 12 , a similar argument gives1

2

20

|(rei )|pd Cp 1(1 r)2p1 .

(d) If p > 12 , (c) and (4.18) imply1

2

20

| (rei )|pd Cp 1(1 r)3p1 .

For p = 1, (c) implies (d) if is analytic but not necessarily univalent.(e) The Koebe function (z) = z/(1 z)2 is not in H 12 .In a deep paper [1974] Baernstein proved that the extremal functions forparts (b) and (c) are the rotates of the Koebe function. See Duren [1983] forthese results and more, and see Appendix A for the basic theory of Hardyspaces.

24. (a) Show that the best constant in the HaymanWu theorem is smaller than4 (Rohde [2002]). Hint: By (5.3) if |()| > 2 then () lies ina cone at with opening < C and || > 1 C. Moreover the curve = (D) makes an angle with the radius through () approximatelytan1(1/2). Then use (4.20) to show that if makes a larger angle with theradius at a point z then || < 2 on an interval centered at z/|z| oflength (1 |z|). Use this to show that there is a > 0, independent of sothat || < 2 on a subset of the circle of length at least .(b) Use (5.3) to show that if we also require (0) L in the HaymanWutheorem, then the best constant is less than 4 2.

25. (a) Let 1 2 be Jordan domains, and let E = 1 2. Then on 1,(z, E,1) (z, E,2).

(b) Let 0 < < 1 and let be a Jordan domain such that{|z| < 1 } {|z| < 1 + }.

Let E = {1 arg z 2}, where 2 < 1 + 2. Show(0, E,) (2 1)2

C log(1).

Hint: By (a) the worst case in (b) is for regions of the following form. LetI = [ei1 , ei2 ]. Let = D \ {r z : z I, r 1 } where < |I |, and letE = {|z| = 1 }. Show that if |z| = 1 , and dist(z, E) , then

(z, E,) 2(z, I, D).(c) Show the estimate in (b) is sharp, except for the value of C .


IIFinitely Connected Domains

In this chapter we solve the Dirichlet problem on a domain bounded by afinite number of Jordan curves. For a simply connected Jordan domain theproblem was solved in Chapter I via the theorem of Carathodory. For a multiplyconnected domain the problem will be reduced to the simply connected caseusing the Schwarz alternating method.

Solving the Dirichlet problem on a domain is equivalent to constructingharmonic measure on . In Section 2 we describe harmonic measure in termsof the normal derivative of Greens function in the case when consists ofanalytic curves. In Section 4 we study the relation between the smoothness of and the smoothness of the Poisson kernel (the RadonNikodym derivativeof harmonic measure against arc length). This relation hinges on two classicalestimates for conjugate functions which we prove in Section 3.

1. The Schwarz Alternating MethodLet be a plane domain such that is a finite union of pairwise disjointJordan curves

= 1 2 . . . p.We say is a finitely connected Jordan domain. A bounded function f on is piecewise continuous if there is a finite set E such that f is continuouson \ E and f has left and right limits at each point of E . In this sectionwe solve the Dirichlet problem for piecewise continuous boundary data on afinitely connected Jordan domain.

Theorem 1.1. Let be a finitely connected Jordan domain, and let f be abounded piecewise continuous function on . Then there is a unique function

37


38 II. Finitely Connected Domains

u(z) = u f (z), bounded and harmonic on , such thatlimz

u(z) = f ( ) (1.1)

at every point of continuity of f . Moreover,sup

|u| sup

| f |.

Proof. We may assume is bounded. The uniqueness of u f is immediate fromLindelfs maximum principle, Lemma I.1.1. The existence of u f in case p = 1was treated in Section I.3, and so we assume p > 1. Take a Jordan arc withendpoints a, b / such that 1 = \ is simply connected and such that is a finite set. See Figure II.1. Then (z) =

zazb has a single valued

analytic branch defined on 1, and we can solve the Dirichlet problem on 1by transplanting it to the Jordan region (1). Take a second Jordan arc suchthat 1 = \ is simply connected, is a finite set, and = .We can also solve the Dirichlet problem on 1.

a

a

b

b

Figure II.1 The proof of Theorem 1.1.

Let E be a finite set and set F = E ( ) ( ). Supposewithout loss of generality that f C( \ E) is positive and bounded. To start,let u1 be the solution to the Dirichlet problem on 1 with boundary data

u1( ) ={ f ( ),

max f, .Then u1 is harmonic on 1 and continuous on \ F and u1 matches its bound-ary data on 1 \ F . Next let u1 be the solution to the Dirichlet problem on 1with boundary data u1( ), 1.Then u1 is harmonic on1 and continuouson \ F and u1 matches its boundary data on 1 \ F . In particular, u1 = u1on \ F. By the Lindelf maximum principle we have u1 max f = u1on , and therefore u1 u1 on \ F . Now let u2 be the solution to theDirichlet problem on 1 with boundary data u1( ), 1. On we


1. The Schwarz Alternating Method 39

have u2 = u1 u1,while on \ F we have u2 = f = u1. Therefore u2 u1on 1, again by Lemma I.1.1. Consequently u2 u1 = u1 on , and bythe maximum principle, u2 u1 on \ F .

Continuing in this way we obtain a decreasing sequence

u1 u1 u2 u2 u3 . . .of positive functions, alternately harmonic on 1 and 1. By Harnacks prin-ciple, Exercise I.5, the limit

u(z) = limn un(z) = limn u

n(z)

is a bounded harmonic function on such that u u1 max f .To complete the proof we show

limz

u(z) = f ( ) (1.1)

whenever is a point of continuity of f . We may assume / E . Takea neighborhood V of such that W = V is a Jordan domain and such thatW (E ) = . Let be a conformal map of D onto W. By Carathodorystheorem (D) = W, and for w = (z) W

un(w) =1()

1 |z|2|ei z|2 f (e

i )d2

+D\1()

1 |z|2|ei z|2 un (e

i )d2

.

Because 1() is a neighborhood of 1( ) in D, the first integral ap-proaches f ( ) as w . Because |un| sup | f |, the second integral tendsto zero, uniformly in n, as w . Therefore (1.1) holds and the theorem isproved.

If is a finitely connected Jordan domain, Theorem 1.1 shows that the mapf u f (z) is a bounded linear functional on C(). Define the harmonicmeasure of a relatively open subset U by

(z,U ) (z,U,) = sup{u f (z) : f C(), 0 f U },and for an arbitrary subset E of , set

(z, E) (z, E,) = inf{(z,U ) : U open in , U E}.



This procedure, which mimics the usual proof of the Riesz representation the-orem, shows (z, E) is a Borel measure on such that

u f (z) =

f ( )d(z, ) (1.2)

for continuous f . When is simply connected, this definition of harmonicmeasure agrees with the earlier definition (I.3.3).

For every z1, z2 there exists, by virtue of Harnacks inequality, a con-stant c = c(z1, z2) such that

1c(z1, E) (z2, E) c(z1, E), (1.3)

and the constants c(z1, z2) remain uniformly bounded if z1 and z2 remain ina compact subset of . If f L1(, d), and in particular if f is boundedand Borel, there is a sequence { fn} in C() such that for some fixed z0 ,

| fn( ) f ( )| d(z0, ) 0.

Write

un(z) =

fn( )d(z, )

and

u(z) = u f (z) =

f ( )d(z, ). (1.4)

Then by (1.3)un(z) u(z)

for all z . By (1.3) we also see that the harmonic functions un(z) are uni-formly bounded on compact subsets of . Thus by Harnacks principle thelimit function u(z) is harmonic on . We call u the solution to the Dirichletproblem for f on . If f is bounded, then we also have

supz

|u f (z)| || f ||L(,d).

Moreover, if f is bounded and continuous at , then (1.1) also holds at . See Theorem 2.7 or Exercise 1(b) for a proof.

Note however that the Schwarz alternating method cannot be applied directlyto a bounded Borel function because the hypothesis of Lindelfs maximumprinciple holds only for piecewise continuous functions.

The next section will give a much more explicit description of the measure(z, E) when has some additional smoothness.


2. Greens Functions and Poisson Kernels 41

2. Greens Functions and Poisson KernelsAgain let be a finitely connected Jordan domain, and assume is bounded.For fixed w , let h(z, w) be the solution to the Dirichlet problem for theboundary value f ( ) = log | w| C(), and define Greens functionwith pole w to be

g(z, w) = log 1|z w| + h(z, w). (2.1)

Then g(z, w) is continuous in z \ {w}, andg(z, w) > 0 on , (2.2)g(, w) = 0 on , (2.3)z g(z, w) is harmonic on \ {w}, (2.4)z g(z, w) log 1|w z| is harmonic at w. (2.5)

These properties are easily derived from Theorem 1.1 and the definition (2.1).By the maximum principle, properties (2.3), (2.4), and (2.5) determine g(z, w)uniquely.

When is unbounded we fix a / . For w = , we let h(z, w) solve theDirichlet problem on for f ( ) = log w

a, and define

g(z, w) = log z az w

+ h(z, w).For w = , we instead use f ( ) = log 1

a to define h(z,) and set

g(z,) = log |z a| + h(z,). These definitions are independent of thechoice of a, and with them (2.2)(2.5) still hold and determine g(z, w).

Suppose is a conformal mapping from one finitely connected Jordan do-main onto another finitely connected Jordan domain . Then (z) whenever z , because : is a homeomorphism. It follows that

g((z), (w)) = g(z, w) (2.6)because Greens functions are characterized by (2.3), (2.4), and (2.5).

If is the unit disc D then clearly

g(z, w) = log1 zwz w

. (2.7)Consequently Greens function for any simply connected Jordan domain canbe expressed in terms of the conformal mapping : D.



Theorem 2.1. Let be a simply connected domain bounded by a Jordan curve,let w and let : D be a conformal map with (w) = 0. Then

g(z, w) = log |(z)|.Proof. Immediate from (2.6) and (2.7).

By definition, an analytic arc is the image ((1, 1)) of the open interval(1, 1) under a map one-to-one and analytic on a neighborhood in C of (1, 1)to C. An analytic Jordan curve is a Jordan curve that is a finite union of (open)analytic arcs.

Lemma 2.2. Let be a finitely connected Jordan domain. Then there exists afinitely connected Jordan domain such that consists of finitely manypairwise disjoint analytic Jordan curves and there exists a conformal map from onto which extends to be a homeomorphism from to .

1

2

3

1

2

3

3 2 1

1(2)

1(3)

Figure II.2 The proof of Lemma 2.2.

Proof. Write = 1 . . . p, where each j is a Jordan curve. Let 1 bethe component of C \ 1 containing , where C is the extended plane, and let1 be a conformal map of 1 onto D. Let 2 be the component of C \ 1(2)containing 1(), and let 2 be a conformal map of 2 onto D. Repeating thisprocess for each boundary curve, we obtain a conformal map p from toa region such that consists of finitely many pairwise disjoint analytic



Jordan curves. Applying Carathodorys theorem to each k , we see that pextends to be a homeomorphism from to .

In Exercise 5 the proof of Lemma 2.2 is used to define Greens function innon-Jordan, finitely connected domains.

Theorem 2.3. Let be a finitely connected Jordan domain and let z1, z2 .Then

g(z1, z2) = g(z2, z1). (2.8)

Proof. By Lemma 2.2 we may assume consists of analytic Jordan curves.When consists of analytic curves, an argument with Schwarz reflection,which we will use many times and prove in Lemma 2.4, shows there is aneighborhood V of to which z g(z, w) has a harmonic extension. Henceg(z, w) is C on some neighborhood V of and we can use Greens theoremin the form

U(uv vu)dxdy =

U

(uv

n v u

n

)ds,

where n is the unit normal vector pointing out from the domain U .Fix distinct z1, z2 . We apply Greens theorem on the domain

= \({|z z1| } {|z z2| }

),

when is small, with u(z) = g(z, z1) and v(z) = g(z, z2). Because u = v = 0on ,

(uv

n v u

n

)ds = 0,

and because u and v are harmonic on , the area integral in Greens theoremvanishes. We conclude that

20

g(z1 + ei , z1)gr

(z1 + ei , z2) d2

20

g(z1 + ei , z2)gr

(z1 + ei , z1) d2=

20

(g(z2 + ei , z2)gr

(z2 + ei , z1) d2

20

g(z2 + ei , z1)gr

(z2 + ei , z2) d2 .

(2.9)



For small, g(zj + ei , zj ) 2 log (1/) and for k = j , g(z, zj ) has boundedderivatives near zk . That means the first and third integrals in (2.9) approach 0as 0. By (2.1),

gr

(zj + ei , zj ) = log

+ O(1),so that, as 0, the second integral tends to g(z1, z2) and the fourth integraltends to g(z2, z1). Therefore (2.8) holds. Lemma 2.4. Suppose is a finitely connected Jordan domain and suppose is an analytic arc. Let u(z) be a harmonic function in . If

limz

u(z) = 0

for all , then there is an open set W such that u extends to beharmonic on W . If also u(z) > 0 in , then

u

n( ) < 0 (2.10)

for all .

VD

v = 0

v > 0

u = 0

u > 0

Figure II.3 Straightening an analytic arc in .

Proof. Let . Because is an analytic arc, there is a neighborhood V of and a conformal map : V D such that ( ) = 0,

(V ) = D = D {Imw < 0},and

( V ) = (1, 1).



Set

v(w) =

u(1(w)), w D = D {Imw < 0},

u(1(w)), w D+ = D {Imw > 0},

0, w (1, 1).Then v is continuous in D and v has the mean value property over sufficientlysmall circles centered at any w D. Hence v is harmonic in D and U = v defines a harmonic extension of u to V .

If U1 and U2 are extensions of u to neighborhoods V1 and V2 such thatV1 V2 is connected, then U1 = U2 in the component of V1 V2 thatcontains V1 V2 . It follows that u has a harmonic extension to some openset W .

If u > 0 in , then clearly u/n 0 on . The strict inequality (2.10) willhold at if and only if

(v/y

)(0) < 0. On D, there is an analytic function

h = v iv with Imh = v and h(0) = 0. The Taylor expansion of h at 0 ish(w) = anwn + O(|wn+1|),

with an = 0. But if n 2, then h(D) D+ = , which is a contradiction.Hence a1 = h(0) =

(v/y

)(0) = 0 and (2.10) holds.

When consists of analytic curves, Greens function provides a formulafor harmonic measure that generalizes the Poisson integral formula for D.

Theorem 2.5. Assume consists of finitely many pairwise disjoint analyticJordan curves, and let z . Then Greens function g(, z) extends to beharmonic (and hence real analytic) on a neighborhood of and

g(, z)n

> 0 (2.11)

on , where n is the unit outer normal at . If u C() is harmonicon , then

u(z) =

g(, z)n

u( )ds( )

2. (2.12)

Because of (2.12),

Pz( ) = 12g(, z)

n

is called the Poisson kernel for .



Proof. Fix z . By Lemma 2.4, g(, z) extends to be harmonic (and realanalytic) on some neighborhood of and then (2.11) follows from (2.10).

To prove (2.12) we first assume that u is C on a neighborhood of . Wethen apply Greens theorem on = \ {w : |w z| < } with small andv(w) = g(w, z). Because w g(w, z) = u = 0 on and because g = 0 on, Greens theorem collapses into

g(, z)n

u( )ds( )

2=

20

g(z + ei , z)u(z + ei )

r

d2

2

0u(z + ei )g(z + e

i , z)

r

d2

.

Since g(z + ei , z) 2 log 1/ for small , and since u is C, we have

lim0

20

g(z + ei , z)u(z + ei )

r

d2

= 0.

As we have seen, (2.1) yieldsg(z + ei , z)

r= log

+ O(1),

while u(z + ei ) = u(z) + O(). Hence

2

0u(z + ei )g

r(z + ei , z) d

2= u(z) + O(),

and that gives (2.12) when u is C on a neighborhood of .To prove (2.12) in general, set = {w : g(w, z) > }where is small.

By uniqueness, has Greens function g(w, z) = g(w, z) . Thereforegn

(, z) = gn

(, z)

on . In a neighborhood of a point 0 , the function = g + i g is aconformal map and

12

N

g(, z)n

u( )ds ={Rez=}(N )

u 1ds,

which converges to{Rez=0}(N )

u 1ds = 12

g(, z)n

u( )ds.

Therefore1

2

g(, z)n

u( )ds 12

g(, z)n

u( )ds (2.13)



as 0. But (2.12) holds for u on because u is C on a neighborhood of , and hence (2.13) yields (2.12) for . Corollary 2.6. If consists of finitely many pairwise disjoint analytic Jordancurves and if z , then

d(z, ) = g(z, )n

ds( )2

. (2.14)

In other words, harmonic measure for z is absolutely continuous to arclength on , the density

dds

= 12

g(z, )n

= Pz( )

is real analytic on , and

c1 1 define

( ) ={z : |z | < dist(z, )}.

If u(z) is a bounded harmonic function on , then for ds almost all thelimit

lim( )z

u(z) = f ( ) (2.16)

exists,

u(z) =

Pz( ) f ( )ds( ), (2.17)

and

sup

|u(z)| = || f ||L . (2.18)



Conversely, if f is a bounded Borel function on , then (2.17) defines abounded harmonic function u(z) on such that (2.18) holds and (2.16) holdsds almost everywhere. Moreover, if f is continuous at 0 , then

limz0

u(z) = f (0). (2.19)

By (2.18), (2.16) and (2.17) establish an isometry between the space ofbounded harmonic functions on and L(, ds) when consists of ana-lytic curves.

Proof. A simple localization argument gives the existence of the nontangentiallimit f . If I is an open arc on , there exists a neighborhood V I suchthat V = I and V is simply connected, and there exists a conformalmapping defined on V such that (V ) = D and (I ) is an arc on D.It follows that maps conical approach regions at V into cones at( ):

(V ( ) B( )

) ()(( )),where B( ) = {z : |z | < = ( )}. Then if u(z) is a bounded harmonicfunction on , we can apply Fatous theorem to u (1) to obtain (2.16) dsalmost everywhere on V .

The proof of (2.17) is exactly the same as the proof of (2.12) exceptthat the dominated convergence theorem is applied in (2.13). By (2.16) wehave || f || sup |u(z)|, and because Pz 0 and

Pzds = 1, we havesup |u(z)| || f ||. Hence (2.16) and (2.17) imply (2.18).

To prove the converse, let f L(, ds). Then the discussion follow-ing (1.4) shows (2.17) defines a bounded harmonic function u(z) on andsup |u(z)| || f ||. Therefore by (2.16), u has almost everywhere a nontan-gential limit, which we will temporarily call F , and u is the Poisson integralof F . What we must prove is that F = f almost everywhere. Again let V be aneighborhood of an open arc I = V such that V is simply connected.For h L(, ds), define

vh(z) =

Pz( )h( )ds( )

IPz(, V )h( )ds( ),

where Pz(, V ) is the Poisson kernel for z V . If h C() then by(1.2), Theorem 1.1, (2.12), and (3.2) of Chapter I, limz vh(z) = 0 for all I , and hence by Lemma 2.4, vh extends to be harmonic in a neighborhoodW of I which does not depend on h. Thus by Exercise I.5(e) or (3.5) below, ifJ is a compact subset of I and if > 0 then there is a neighborhood N of J ,depending only on ||h|| and , so that |vh | < in N . Now take hn C()



so that hn converges to f in L1, and ||hn|| || f ||. For each z V N ,vhn (z) converges to v f (z) and so |v f (z)| < . Because > 0 was arbitrary, weconclude that v f (z) 0 as z J . But by Theorem I.1.3,

F( ) f ( ) = lim( )z

v f (z) = 0

almost everywhere in J . Consequently F = f almost everywhere and (2.16)and (2.18) hold for all f L(, ds).

Finally, if f is continuous at 0 I then

I Pz(, V ) f ( )ds( ) is continuousat 0 by (3.2) of Chapter I and so by the continuity of v f , (2.19) holds.

The converse can be proved another way. Using the real analyticity ofg(w, z), one can refine the proof of Lemma I.2.2 and show that

sup( )

|u(z)| C(,)Ms f ( ), (2.20)

where u is the Poisson integral (2.17) and where the maximal function Ms f ( )is the supremum of the averages of f over arcs with :

Ms f ( ) = sup

1( )

| f |ds.

A variation on the covering lemma shows that Ms is weak-type 1-1, and anapproximation, as in the proof of Theorem I.2.1, then yields (2.16) for thePoisson integral of f . This is the argument that must be used in the Euclideanspaces Rd , with d 3, and in other situations.

With some care, the conformal mapping proof of (2.16) in the text can alsobe parlayed into a proof of the maximal estimate (2.20). See Exercise 9.

Let be any finitely connected Jordan domain and let be the conformalmap, given in Lemma 2.2, of onto a domain , where consists ofanalytic Jordan curves. Since is a homeomorphism of onto , harmonicmeasure can be transplanted from to via , just as it was in Section I.3for simply connected Jordan domains. This gives an alternate but equivalentdefinition of harmonic measure for .

In Section 4 we will consider two questions. Let be a finitely connectedJordan domain.

Question 1. If has some degree of differentiability and if f C() alsohas some degree of differentiability along , how smooth is the solution u f (z),as z approaches ?

Question 2. What smoothness conditions on , weaker than real analyticity,will ensure that g(z, )/n exists on and that (2.14) and (2.15) still hold?



These two questions are equivalent. Their answers will depend on Kelloggstheorem about the boundary behavior of conformal mappings. The proof ofKelloggs theorem in turn depends on the estimates for conjugate functions inthe next section.

3. Conjugate FunctionsLet f L1(D) be real. For convenience we write f () for f (ei ). If u(z) isthe Poisson integral of f on D, then u(z) is harmonic and real and there existsa unique harmonic function u(z) such that u(0) = 0 and

F = u + i uis analytic on D. The function u is called the conjugate function or harmonicconjugate of u. The nontangential limit

f () = lim(ei )zei

u(z) (3.1)

exists almost everywhere, and this has an easy proof from Fatous theorem: Wemay assume f 0, so that

G(z) = e(u(z)+i u(z))

is bounded and analytic on D. By Corollary I.2.5, G has a nontangential limitG(ei ) almost everywhere. Since |G(ei )| = e f () and f L1, |G(ei )| > 0almost everywhere. At such ei , G is continuous and non-zero on the coneK = (ei ). Consequently log G = (u + i u) has a continuous extension toK {z : |G(z) G(ei )| < 12 |G(ei |} and the limit (3.1) exists at ei .

There is a close connection between conjugate functions and conformalmappings. If u is harmonic and if |u| < /2, then

(z) = z

0ei(u+i u)( )d

is a conformal map from D to a simply connected domain and u = arg .Indeed, if a = b D, then

(b) (a) = (b a) 1

0(a + t[b a])dt = 0,

because Re() > 0.When f is bounded, or even continuous, it can happen that f is not bounded.

For an example, let u + i u be the conformal map of D onto the region{0 < x < 1/(1 + |y|)}.


3. Conjugate Functions 51

Then u is continuous on D by Carathodorys theorem, but u is not bounded.For a second example see Exercise 11. The next two theorems get around theobstruction that f may be unbounded even when f is continuous.Theorem 3.1 (Zygmund). Let f L(D) be real with || f || 1.(a) For 0 < < /2 there is a constant C, depending only on , such that

e| f ()| d2

C.

(b) If f C(D), then for all <

sup0



such that f p < /(2). Then

p(rei ) =N

n=0rn(an sin n bn cos n)

is bounded, while part (a) gives

B = sup0



CauchyRiemann equations,

|u(z)| = F (z) =

eit f (t)(eit z)2

dt

. (3.4)Therefore (as in Exercise I.5(e))

|u(z)| (

1

|eit z|2dt

) f 2(1 |z|)1 f . (3.5)

We will often use the following consequence of (3.5): If u(z) is harmonicand |u(z)| M on B(z, R), then

supB(z, R2 )

|u| 4MR

. (3.6)

To prove (3.6) simply apply (3.5) to U (w) = u(z + Rw).The next theorem shows that f C if and only if the estimate (3.5) can be

upgraded to

|u| = O((1 |z|)1).Theorem 3.2. Let 0 < < 1, let f L(D) be real, and let u(z) be thePoisson integral of f . Then the following conditions are equivalent:(a) f C.(b) f C.(c) |u(z)| = O((1 |z|)1).(d) u C(D); that is, for all z1, z2 D,

|u(z1) u(z2)| = O(|z1 z2|).

Moreover there is a constant C1, independent of , such that

|| f ||C C1(1 ) || f ||C , (3.7)

|u(z)| C1(1 )(1 |z|)

1|| f ||C , (3.8)

and

supz1 =z2

|u(z1) u(z2)||z1 z2|

C1

sup|z|



Proof. Clearly (d) (a) because if (d) holds, u is uniformly continuous onD. We first show (a) (c) and establish (3.8) and then show (c) (d)and establish (3.9). Then (a) (b) and inequality (3.7) will follow because|u| = |u| by the CauchyRiemann equations.

Assume (a) holds, that is, assume u is the Poisson integral of f C .We prove (3.8). By (3.5) we may assume |z| 1/2. Let eit0 = z/|z|. Because

eit/(eit z)2dt = 0, (3.4) yields

|u(z)| =

eit ( f (t) f (t0))

(eit z)2dt

,so that

|u(z)|

| f (t) f (t0)||eit z|2

dt

=|tt0|



However,

|u(zj ) u(wj )| = rj

(1)rj

u

t(teij )dt

11

(1 t)1dt

,

while

|u(w1) u(w2)| |w1 w2|Maxj(1 |wj |

)1 .Therefore (d) and (3.9) hold.

ei1

ei2

z1

z2

w1 = (1 )z1w2 = (1 )z2

Figure II.4

Now suppose (a) holds. Then (c) holds and |u| = O((1 |z|)1). There-fore (d) and (3.9) hold for u, and u extends continuously to D where u hasboundary value f . It follows that f C because (d) implies (a). Finally sincef = f + u(0), it follows that (a) (b) and that (3.7) holds.

Let k be a non-negative integer, let 0 < 1 and let f C(D). We sayf Ck+ if f is k times continuously differentiable on D and (d/d)k f Cif > 0. If F(z) is analytic on D we say F Ck+(D) if F and its first kderivatives F , F , , F (k) extend continuously to D and if there is C suchthat

|F (k)(z1) F (k)(z2)| C |z1 z2|

for all z1, z2 D.Corollary 3.3. Assume k is a non-negative integer and assume 0 < < 1. Letf C(D) be real and let F(z) = u(z) + i u(z) be the Herglotz integral of f .Then F Ck+(D) if and only if f Ck+(D).Proof. Because f = ReF, it is clear that f Ck+(D) if F Ck+(D).



Assume f Ck+ . If f () has Fourier series

f ()

anein ,

then F has Taylor series

F(z) = a0 + 2

n=1anz

n .

Therefore d f/d, which has Fourier seriesd fd

inanein ,

has Herglotz integral i zF (z) and the corollary follows from the special casek = 0, which is Theorem 3.2.

Theorem 3.2 fails when = 1. The harmonic conjugate of a continuouslydifferentiable function on D need not have a continuous derivative, and theconjugate of a Lipschitz function, that is, a function satisfying f ( + t) f () M |t |,need not be a Lipschitz function. See Exercise 13. For the same reason, Corol-lary 3.3 is also false when = 0 and k is a non-negative integer. However,conjugation does preserve the Zygmund class Z of continuous f on D suchthat

supt>0

f ( + t) + f ( t) 2 f ()t

< .

The Zygmund class has norm

f Z = f + supt>0

f ( + t) + f ( t) 2 f ()t

.

When f Z, we say f is a Zygmund function.Define 2u(z) = ux = (|uxx (z)|2 + |uyx (z)|2) 12

=F (z) = 2

eit f (t)(eit z)3 dt

,



where F = u + i u is the Herglotz integral of f L1(D). If f C , then(3.8) and (3.6), applied to ux in the disc B(z, 1|z|2 ), give us2u(z) = O((1 |z|)2). (3.10)Conversely, if (3.10) holds, then integrating F along radii shows thatu(z) = O((1 |z|)1), and f C. Thus (3.10) provides yet anothercharacterization of C functions, in terms of second derivatives. Zygmund func-tions have a similar characterization.

Theorem 3.4. Let f L(D) be real and let u(z) be the Poisson integral off . Then the following are equivalent:(a) f Z.(b) f Z.(c) 2u(z) = O((1 |z|)1).There is a constant C such that f Z C f Z (3.11)and

1C f Z || f || + sup

zD

{(1 |z|)2u(z)} C f Z . (3.12)

Notice that by (3.12) and (3.10), Z C for all < 1. In particular, iff Z, then f and f are continuous. On the other hand, if f is Lipschitz, thenclearly f Z. Thus we have the increasing scale of spaces

C1 Lipschitz Z C C, < 1, where C1 denotes the space of continuously differentiable functions andC is the space of continuous functions.

Proof. The logic is the same as in the proof of Theorem 3.2. First assume (a)holds. Fix z D . We may assume z = |z| = Rez and |z| > 12 . Because

ei

(ei z)3 d = 0

and f () has Herglotz integral F(z), we have2u

x2(z) = ReF (z) = 1

eit( f (t) + f (t) 2 f (0))

(eit |z|)3 dt



and 2ux2

(z) = ReF (z) 1

|t |1|z|

f Z |t |(1 |z|)3 dt +

c

1|z|


4. Boundary Smoothness 59

Becauseu(rei(+t)) + u(rei(t)) 2u(rei ) |t |2 sup|w|=r

2u2

(w),

the last two terms on the right side of (3.13) are O(|t |) by (c). Then becauselims1(1 s) us (sei) = 0 by (c), an integration by parts shows

f () u(rei) = 1

r

(1 s)2u

s2(sei )ds + (1 r)u

r(rei).

Therefore when (c) holds, the sum of the first three lines of the right side of(3.13) is

t

(ur (re

i(+t)) + ur (rei(t)) 2ur (rei ))+ O(t)

t

ur (rei(+t)) ur (rei )+ t

ur (rei(t)) ur (rei )+ O(t) 2t

2

sup|w|=r

2ur

(w)+ O(t) C t.

That proves (a), and the left-hand inequality in (3.12) can be seen by followingthe constants in the previous argument.

Since2u = 2u, it follows that (a) (b) and that (3.11) holds.

4. Boundary SmoothnessLet be a Jordan domain with boundary and let be a conformal map-ping from D onto , so that extends to a homeomorphism from D to theJordan curve = . In this section we examine the connection between thesmoothness of and the differentiability of on D. When has some de-gree of smoothness, we also study the relation between the differentiability off C() and the differentiability of its solution u f to the Dirichlet problemat points of .

The results do not depend on the choice of the mapping : D becauseany other such map has the form T, with T M. We first show that thesmoothness of in a neighborhood of 1( ) depends only on the smoothnessof in a neighborhood of .



Theorem 4.1. Let 1 and 2 be Jordan domains such that 1 2 and let 1 2 be an open subarc. Let j be a conformal map of D onto j .Then = 12 1 has an analytic continuation across 11 ( ), and = 0on 11 ( ).

Proof. The analytic function = 12 1 from D into D has a continuousand unimodular extension to the arc 11 ( ). By Schwarz reflection has ananalytic and one-to-one extension to a neighborhood of 11 ( ) in C and hence = 0 on 11( ).

Let be an arc parameterized as {(t) : a < t < b}. We say has a tangentat 0 = (t0) if

limtt0

(t) 0|(t) 0| = e

i (4.1)

and

limtt0

(t) 0|(t) 0| = e

i . (4.2)

If (4.1) and (4.2) hold then has unit tangent vector ei at 0. Except forreversals of orientation, the existence of a tangent at 0 and its value ei donot depend on the choice of the parameterization t (t). We say has acontinuous tangent if has a tangent at each and if ei( ) is continuouson .

Theorem 4.2. The curve has a tangent at = (ei ) if and only if the limit

limDzei

arg(

(z) z ei

)(4.3)

exists and is finite. In that case,

limDzei

arg(

(z) z ei

)= ( )

2(modulo 2). (4.4)

The curve has a continuous tangent if and only if arg(z) has a continuousextension to D. If has a continuous tangent, then C(D) for all < 1and is rectifiable.

There exist Jordan domains and conformal maps : D such that has a continuous tangent but C1(D). An example can be built from theconnection in Section 3 between conjugate functions and conformal mappings:If u(ei ) is continuous and |u| < 2 , then u(ei ) = arg (ei ), where is aconformal mapping onto a Jordan domain, but u(ei ) = log |(ei )| maynot be bounded above or below.


4. Boundary Smoothness 61

Proof. Set

v(z) = arg(

(z) z ei

).

Then v(z) is harmonic on D and v is continuous on D \ {ei }. If has a tangentat = (ei ) or if the limit (4.3) exists at ei , then v(z) is bounded on D. Thusin either case v is the Poisson integral of v(ei ). Therefore v has a continuousextension to ei if and only if v|D has a continuous extension to ei , and if andonly if has a tangent at . Clearly, (4.4) holds when v is continuous at ei .

Now suppose has a continuous tangent. Then is continuous on D.If h = 0, then

Ah(z) = arg((zeih) (z)

z(eih 1))

is continuous on D and harmonic on D. For |z| < 1,limh0

Ah(z) = arg(z).

For |z| = 1 there exists k, 0 < |k| < |h|, such that

arg((zeih) (z)

h

)= arg

((zeih) (z)

)= ((zeik)),

by the proof of the

Analisis de Medida Tesis de Grado

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