Harmonic Measure (New Mathematical Monographs)

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Harmonic Measure During the last two decades several remarkable new results were discovered about harmonic measure in the complex plane. This book provides a survey of these results and 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 form for the ﬁrst time. The book is accessible to students who have completed standard graduate courses in real and complex analysis. The ﬁrst four chapters provide the needed background material on univalent functions, potential theory, and extremal length, and each chapter has 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 of

Washington.

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NEW MATH MONOGRAPHS Editorial Board Béla Bollobás William Fulton Frances Kirwan Peter Sarnak Barry Simon For information about Cambridge University Press mathematics publications visit http://publishing.cambridge.org/stm/mathematics

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Harmonic Measure JOHN B. GARNETT University of California, Los Angeles DONALD E. MARSHALL University of Washington

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CAMBRIDGE UNIVERSITY PRESS

Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521720601 © Cambridge University Press 2005 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. First published in print format 2008

ISBN-13 978-0-511-41011-6

eBook (NetLibrary)

ISBN-13 978-0-521-72060-1

paperback

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.

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To Dolores and Marianne

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Contents

page xiii

Preface I. 1. 2. 3. 4. 5.

Jordan Domains The Half-Plane and the Disc Fatou’s Theorem and Maximal Functions Carathéodory’s Theorem Distortion and the Hyperbolic Metric The Hayman–Wu Theorem Notes Exercises and Further Results

1 1 6 13 16 23 25 26

II. 1. 2. 3. 4.

Finitely Connected Domains The Schwarz Alternating Method Green’s Functions and Poisson Kernels Conjugate Functions Boundary Smoothness Notes Exercises and Further Results

37 37 41 50 59 66 66

III. 1. 2. 3. 4. 5. 6. 7. 8.

Potential Theory Capacity and Green’s Functions The Logarithmic Potential The Energy Integral The Equilibrium Distribution Wiener’s Solution to the Dirichlet Problem Regular Points Wiener Series Polar Sets and Sets of Harmonic Measure Zero

ix

73 74 77 79 82 89 93 97 102

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Contents

9. Estimates for Harmonic Measure Notes Exercises and Further Results IV. 1. 2. 3. 4. 5. 6.

V. 1. 2. 3. 4. 5. 6.

VI. 1. 2. 3. 4. 5. 6.

VII. 1. 2. 3. 4. 5.

104 112 112

Extremal Distance Deﬁnitions and Examples Uniqueness of Extremal Metrics Four Rules for Extremal Length Extremal Metrics for Extremal Distance Extremal d x Distance and Harmonic Measure The θ(x) Estimate Notes Exercises and Further Results

129 129 133 134 139 143 146 149 150

Applications and Reverse Inequalities Asymptotic Values of Entire Functions Lower Bounds Reduced Extremal Distance Teichmüller’s Modulsatz Boundary Conformality and Angular Derivatives Conditions More Geometric Notes Exercises and Further Results

157 157 159 162 166 173 184 193 194

Simply Connected Domains, Part One The F. and M. Riesz Theorem Privalov’s Theorem and Plessner’s Theorem Accessible Points Cone Points and McMillan’s Theorem Compression and Expansion Pommerenke’s Extension Notes Exercises and Further Results

200 200 203 205 207 212 216 221 221

Bloch Functions and Quasicircles Bloch Functions Bloch Functions and Univalent Functions Quasicircles Chord-Arc Curves and the A∞ Condition BMO Domains Notes Exercises and Further Results

229 229 232 241 246 253 257 258

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Simply Connected Domains, Part Two The Law of the Iterated Logarithm for Bloch Functions Harmonic Measure and Hausdorff Measure The Number of Bad Discs Brennan’s Conjecture and Integral Means Spectra β Numbers and Polygonal Trees The Dandelion Construction and (c) ⇒ (a) Baernstein’s Example on the Hayman–Wu Theorem Notes Exercises and Further Results

269 269 272 281 286 289 296 302 305 307

IX. 1. 2. 3.

Inﬁnitely Connected Domains Cantor Sets For Certain , dim ω < 1. For All , dim ω ≤ 1. Notes Exercises and Further Results

315 315 324 331 341 342

X. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

Rectiﬁability and Quadratic Expressions The Lusin Area Function Square Sums and Rectiﬁability A Decomposition Theorem Schwarzian Derivatives Geometric Estimates of Schwarzian Derivatives Schwarzian Derivatives and Rectiﬁable Quasicircles 1 The Bishop–Jones H 2 −η Theorem Schwarzian Derivatives and BMO Domains Angular Derivatives A Local F. and M. Riesz Theorem Ahlfors Regular Sets and the Hayman–Wu Theorem Notes Exercises and Further Results

347 348 361 372 380 384 393 397 410 412 416 421 426 427

Appendices Hardy Spaces Mixed Boundary Value Problems The Dirichlet Principle Hausdorff Measure Transﬁnite Diameter and Evans Functions Martingales, Brownian Motion, and Kakutani’s Theorem Carleman’s Method

435 435 441 447 456 466 470 480

VIII. 1. 2. 3. 4. 5. 6. 7.

A. B. C. D. E. F. G.

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Contents

H. I. J. K. L. M.

Extremal Distance in Finitely Connected Domains McMillan’s Twist Point Theorem Bloch Martingales and the Law of the Iterated Logarithm A Dichotomy Theorem Two Estimates on Integral Means Calderón’s Theorem and Chord-Arc Domains

484 497 503 512 518 520

Bibliography Author Index Symbol Index Subject Index

531 555 559 561

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Preface

Several surprising new results about harmonic measure on plane domains have been proved during the last two decades. The most famous of these results are Makarov’s theorems that harmonic measure on any simply connected domain is singular to Hausdorff measure α for all α > 1 but absolutely continuous to α for all α < 1. Also surprising was the extension by Jones and Wolff of Makarov’s α > 1 theorem to all plane domains. Further important new results include the work of Carleson, Jones and Wolff, and others on harmonic measure for complements of Cantor sets; the work by Carleson and Makarov, Bertilsson, Pommerenke, and others on Brennan’s tantalizing conjecture that for univalent functions |ϕ |2− p d xd y < ∞ if 43 < p < 4; several new geometric conditions that guarantee the existence of angular derivatives; and the Jones square sum characterization of subsets of rectiﬁable curves and its applications by Bishop and Jones to a variety of problems in function theory. We wrote this book to explain these exciting new results and to provide beginning students with an introduction to this part of mathematics. We have tried to make the subject accessible to students who have completed graduate courses 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, Hausdorff measures and martingales, provide a foundation that every student of function theory will need. In Chapter I we solve the Dirichlet problem on the half-plane and the disc and then on any simply connected Jordan domain by using the Carathéodory theorem on boundary continuity. Chapter I also includes brief introductions to hyperbolic geometry and univalent function theory. In Chapter II we solve the Dirichlet problem on domains bounded by ﬁnitely many Jordan curves and study the connection between the smoothness of a domain’s boundary and the smoothness of its Poisson kernel. Here the main tools are xiii

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Preface

two classical theorems about conjugate functions. Chapter II and the discussion in Chapter III of Wiener’s solution of the Dirichlet problem on arbitrary domains follow the 1985 UCLA lecture course by Carleson. The introduction to extremal length in Chapter IV is based on the Institut Mittag–Lefﬂer lectures of Beurling [1989]. Chapter V contains some applications of extremal length, such as Teichmüller’s Modulsatz and some newer theorems about angular derivatives, that are not found in other books. Chapter VI is a blend of the classical theorems of F. and M. Riesz, Privalov, and Plessner and the more recent theorems of McMillan, Makarov, and Pommerenke on the comparison of harmonic measure and one dimensional Hausdorff measure for simply connected domains. Chapter VII surveys the beautiful circle of ideas around Bloch functions, univalent functions, quasicircles, and A p weights. Chapter VIII is an exposition of Makarov’s deeper results on the relations between harmonic measure in simply connected domains and Hausdorff measures and the work of Carleson and Makarov concerning Brennan’s conjecture. Chapter IX discusses harmonic measure on inﬁnitely connected plane domains. Chapter X begins by introducing the Lusin area function, the Schwarzian derivative, and the Jones square sums, and then applies these ideas to several problems about univalent functions and harmonic measures. The thirteen appendices at the end of the text provide further related material. For space reasons we have not treated some important related topics. These include the connections between Chapters VIII and IX and thermodynamical formalism and several other connections between complex dynamics and harmonic measure. We have emphasized Wiener’s solution of the Dirichlet problem instead of the Perron method. The beautiful Perron method can be found in Ahlfors [1973] and Tsuji [1959]. We also taken a few detours around the theory of prime ends. There are excellent discussions of prime ends in Ahlfors [1973], Pommerenke [1975], and Tsuji [1959]. Finally, the theory of harmonic measure in higher dimensions has a different character, and we have omitted it entirely. At the end of each chapter there is a brief section of biographical notes and a section called “Exercises and Further Results". An exercise consisting of 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 the text. Results are numbered lexicographically within each chapter, so that Theorem 2.4 is the fourth item in Section 2 of the same chapter, while Theorem III.2.4 is from Section 2 of Chapter III. The same convention is used for formulas, so that (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

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Pommerenke’s excellent book [1991]. However, our emphasis differs from the one in Pommerenke [1991] and we hope the two books will complement each other. Some unpublished lecture notes from a 1986 Nachdiplom Lecture course at Eidgen¨ssische Technische Hochschule Zurich by the ﬁrst listed author and the out-of-print monograph Garnett [1986] were preliminary versions of the present book. The web page http://www.math.washington.edu/ marshall/HMcorrections.html will list corrections to the book. Though we have tried to avoid errors, the observant reader will no doubt ﬁnd some. We would appreciate receiving email at [email protected] about any errors you come across. Many colleagues, friends, and students have helped with their comments and suggestions. Among these, we particularly thank A. Baernstein, M. Benedicks, D. Bertilsson, C. J. Bishop, K. Burdzy, L. Carleson, S. Choi, R. Chow, M. Essèn, R. Gundy, P. Haissinski, J. Handy, P. Jones, P. Koosis, N. Makarov, P. Mateos, M. O’Neill, K. ∅yma, R. Pérez-Marco, P. Poggi-Corradini, S. Rohde, I. Uriarte-Tuero, J. Verdera, S. Yang and S. Yoshinobu. We gratefully acknowledge support during the writing of this book by the Royalty Research Fund of the University of Washington, the University of Washington–University of Bergen Faculty Exchange Program, the Institut des Hautes Etudes Scientiﬁques, the Centre de Recerca Matemàtica, Barcelona, and the National Science Foundation.

Los Angeles and Seattle Seattle and Bergen

John B. Garnett Donald E. Marshall

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I Jordan Domains

To begin we construct harmonic measure and solve the Dirichlet problem in the upper half-plane and the unit disc. We next prove the Fatou theorem on nontangential limits. Then we construct harmonic measure on domains bounded by Jordan curves, via the Riemann mapping theorem and the Carathéodory theorem on boundary correspondence. We review two topics from classical complex analysis, the hyperbolic metric and the elementary distortion theory for univalent functions. We conclude the chapter with the theorem of Hayman and Wu on lengths of level sets. Its proof is an elementary application of harmonic measure 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. Suppose a < b are real. Then the function z−b z−b = Im log θ = θ(z) = arg z−a z−a is harmonic on H, and θ = π on (a, b) and θ = 0 on R \ [a, b]. z θ

a b Figure I.1 The harmonic function θ (z). 1

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Viewed geometrically, θ (z) = Reϕ(z) where ϕ(z) is any conformal mapping from H to the strip {0 < Rez < π } which maps (a, b) onto {Rez = π} and R \ [a, b] into {Rez = 0}. Let E ⊂ R be a ﬁnite union of open intervals and write E = nj=1 (a j , b j ) with b j−1 < a j < b j . Set z − bj θ j = θ j (z) = arg z − aj and deﬁne the harmonic measure of E at z ∈ H to be ω(z, E, H) =

n θj j=1

π

.

(1.1)

Then (i) 0 < ω(z, E, H) < 1 for z ∈ H, (ii) ω(z, E, H) → 1 as z → E, and (iii) ω(z, E, H) → 0 as z → R \ E. The function ω(z, E, H) is the unique harmonic function on H that satisﬁes (i), (ii), and (iii). The uniqueness of ω(z, E, H) is a consequence of the following lemma, known as Lindelöf’s maximum principle. Lemma 1.1 (Lindelöf). Suppose the function u(z) is harmonic and bounded above on a region such that = C. Let F be a ﬁnite subset of ∂ and suppose lim sup u(z) ≤ 0

(1.2)

z→ζ

for all ζ ∈ ∂ \ F. Then u(z) ≤ 0 on . Proof. Fix z 0 ∈ / . Then the map 1/(z − z 0 ) transforms into a bounded region, and thus we may assume is bounded. If (1.2) holds for all ζ ∈ ∂, then the lemma is the ordinary maximum principle. Write F = {ζ1 , . . . , ζ N }, let ε > 0, and set N diam() u ε (z) = u(z) − ε log . |z − ζ j | j=1

Then u ε is harmonic on and lim supz→ζ u ε (z) ≤ 0 for all ζ ∈ ∂. Therefore u ε ≤ 0 for all ε, and N diam() u(z) ≤ lim ε log = 0. ε→0 |z − ζ j | j=1

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Lindelöf [1915] proved Lemma 1.1 under the weaker hypothesis that ∂ is inﬁnite. See also Ahlfors [1973]. Exercise 3 and Exercise II.3 tell more about Lindelöf’s maximum principle. Given a domain and a function f ∈ C(∂), the Dirichlet problem for f 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 function u = 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 ε > 0, take disjoint open intervals I j = (t j , t j+1 ) and real constants c j , j = 1, . . . , n, so that the step function f ε (t) =

n

cj χ Ij

j=1

satisﬁes

fε − f ∞ < ε. L (R )

(1.3)

Set u ε (z) = If t ∈ R \

n

c j ω(z, I j , H).

j=1

∂ I j , then lim u ε (z) = f ε (t)

Hz→t

by (ii) and (iii). Therefore by (1.3) and Lemma 1.1, sup u ε1 (z) − u ε2 (z) < ε1 + ε2 . H

Consequently the limit u(z) ≡ lim u ε (z) ε→0

exists, and the limit u(z) is harmonic on H and satisﬁes sup |u(z) − u ε (z)| ≤ 2ε. H

We claim that lim sup |u ε (z) − f (t)| ≤ ε z→t

(1.4)

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for all t ∈ R. It is clear that (1.4) holds when t ∈ / ∂ I j . To verify (1.4) at the endpoint t j+1 ∈ ∂ I j ∩ ∂ I j+1 , notice that by (ii), (iii), and Lemma 1.1, c j + c j+1 sup c j ω(z, I j , H) + c j+1 ω(z, I j+1 , H) − ω(z, I j ∪ I j+1 , H) 2 H c j − c j+1 , ≤ 2 while

lim

z→t j+1

c j + c j+1 2

ω(z, I j ∪ I j+1 , H) =

c j + c j+1 . 2

Hence all limit values of u ε (z) at t j+1 lie in the closed interval with endpoints c j and c j+1 , and then (1.3) yields (1.4) for the endpoint t j+1 . Now let t ∈ R. By (1.4) lim sup |u(z) − f (t)| ≤ sup |u(z) − u ε (z)| + lim sup |u ε (z) − f (t)| ≤ 3ε. z→t

z∈H

z→t

The same estimate holds if t = ∞. Therefore u extends to be continuous on H and u|∂ H = f . The uniqueness of u follows immediately from the maximum principle. For a < b, elementary calculus gives

x − a

x − b 1 ω(x + i y, (a, b), H) = tan−1 − tan−1 π y y b y dt . = 2 + y2 π (t − x) a If E ⊂ R is measurable, we deﬁne the harmonic measure of E at z ∈ H to be y dt . (1.5) ω(z, E, H) = 2 2 E (t − x) + y π When E is a ﬁnite union of open intervals this deﬁnition (1.5) is the same as deﬁnition (1.1). For z = x + i y ∈ H, the density Pz (t) =

1 y π (x − t)2 + y 2

is called the Poisson kernel for H . If f ∈ C(R ∪ {∞}), the proof of Theorem 1.2 shows that u f (z) = f (t)Pz (t)dt, R

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

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Note that the harmonic measure ω(z, E, ) is a harmonic function in its ﬁrst variable z and a probability measure in its second variable E. If z 1 , z 2 ∈ H then 0 < C −1 ≤

ω(z 1 , E, H) ≤ C < ∞, ω(z 2 , E, H)

where C depends on z 1 and z 2 but not on E. This inequality, known as Harnack’s 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 open arcs on ∂D. Then we deﬁne the harmonic measure of E at z in D to be ω(z, E, D) ≡ ω(ϕ(z), ϕ(E), H),

(1.6)

where ϕ is any conformal map of D onto H. This harmonic function satisﬁes conditions analogous to (i), (ii), and (iii), so that by Lemma 1.1 the deﬁnition (1.6) does not depend on the choice of ϕ. It follows by the change of variables ϕ(z) = i(1 + z)/(1 − z) that 1 − |z|2 dθ . ω(z, E, D) = iθ 2 E |e − z| 2π 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 (eiθ ) be an integrable function on ∂D and set 2π 1 − |z|2 dθ f (eiθ ) iθ . u(z) = u f (z) = |e − z|2 2π 0

(1.7)

Then u(z) is harmonic on D. If f is continuous at eiθ0 ∈ ∂D, then lim

Dz→eiθ0

u(z) = f (eiθ0 ).

(1.8)

Clearly (1.8) also holds if the integrable function f is changed on a measure zero subset of ∂D \ {eiθ0 }. The function u = u f is called the Poisson integral of f and the kernel Pz (θ ) =

1 1 − |z|2 2π |eiθ − z|2

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

u f (z), z ∈ D U (z) = f (z), z ∈ ∂D is the solution of the Dirichlet problem for f on D. In the special case when f (eiθ ) is continuous, Theorem 1.3 follows from Theorem 1.2 and a change of variables. Conversely, Theorem 1.3 shows that

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Theorem 1.2 can be extended to f ∈ L 1 (dt/(1 + t 2 )), again by changing variables. Proof of Theorem 1.3. We may suppose f is real valued. From the identity eiθ + z = 2π Pz (θ ), Re iθ e −z we see that u is the real part of the analytic function 2π eiθ + z dθ f (eiθ ) iθ , e − z 2π 0 and therefore that u is a harmonic function. One can also see u is harmonic by differentiating the integral (1.7). Suppose f is continuous at eiθ0 and let ε > 0. Then | f (eiθ ) − f (eiθ0 )| < ε on an interval I = (θ1 , θ2 ) containing θ0 . Setting 1 − |z|2 dθ f (eiθ ) u ε (z) = + f (eiθ0 )ω(z, I, D), iθ 2 2π [0,2π]\I |e − z| we have

|u(z) − u ε (z)| =

I

1 − |z|2 iθ iθ0 dθ ( f (e ) − f (e )) ≤ εω(z, I, D) ≤ ε. iθ 2 |e − z| 2π

However, lim z→eiθ0 u ε (z) = f (eiθ0 ) by the deﬁnition of u ε . Therefore lim sup |u(z) − f (eiθ0 )| < ε, z→eiθ0

and (1.8) holds when f is continuous at eiθ0 .

2. Fatou’s Theorem and Maximal Functions When f ∈ L 1 (∂D) the limit (1.8) can fail to exist at every ζ ∈ ∂D; see Exercise 7. However, there is a substitute result known as Fatou’s theorem, in which the approach z → ζ is restricted to cones. For ζ ∈ ∂D and α > 1, we deﬁne the cone α (ζ ) = z : |z − ζ | < α(1 − |z|) . The cone α (ζ ) is asymptotic to a sector with vertex ζ and angle 2 sec−1 (α) that is symmetric about the radius [0, ζ ]. The cones α (ζ ) expand as α increases.

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2. Fatou’s Theorem and Maximal Functions

β ζ z 0

α (ζ )

2 sec−1 α

Figure I.2 The cone α (ζ ). A function u(z) on D has nontangential limit A at ζ ∈ ∂D if lim

α (ζ )z→ζ

u(z) = A

(2.1) z+1

for every α > 1. A good example is the function u(z) = e z−1 . This function u(z) is continuous on ∂D \ {1}, and |u(ζ )| = 1 on ∂D \ {1}, but u(z) has nontangential limit 0 at ζ = 1. With ﬁxed α > 1, the nontangential maximal function of u at ζ is u ∗α (ζ ) = sup |u(z)|. α (ζ )

If u has a ﬁnite nontangential limit at ζ , then u ∗α (ζ ) < ∞ for every α > 1. We write |E| for the Lebesgue measure of E ⊂ ∂D. Theorem 2.1 (Fatou’s theorem). Let f (eiθ ) ∈ L 1 (∂D) and let u(z) be the Poisson integral of f . Then at almost every ζ = eiθ ∈ ∂D, lim

α (ζ )z→ζ

u(z) = f (ζ )

for all α > 1. Moreover, for each α > 1 3 + 6α || f ||1 . {ζ ∈ ∂D : u ∗α (ζ ) > λ} ≤ λ

(2.2)

(2.3)

When u(z) is the Poisson integral of f ∈ L 1 (∂D) the function u = u f is also called the solution to the Dirichlet problem for f , even though u converges to f on ∂D only nontangentially and only almost everywhere. Inequality (2.3) says the operator L 1 (∂D) f → u ∗α is weak-type 1-1. It follows from (2.2) that u ∗α (ζ ) < ∞ almost everywhere, but (2.3) is a sharper,

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quantitative result. In the proof of the theorem we derive (2.2) from the estimate (2.3). The proof of Fatou’s theorem is a standard approximate identity argument from real analysis that derives almost everywhere convergence for all f ∈ L 1 (∂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 L 1 (∂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 α temporarily. We may assume f is real valued. Set W f (ζ ) = lim sup |u f (z) − f (ζ )|. α z→ζ

Then W f (ζ ) ≤

u ∗α (ζ ) + | f (ζ )|.

Chebyshev’s inequality gives

{ζ : | f (ζ )| > λ} ≤ || f ||1 , λ so that by (2.3), {ζ : W f (ζ ) > λ} ≤ {ζ : u ∗ (ζ ) > λ/2} + {ζ : | f (ζ )| > λ/2} α 8 + 12α || f ||1 . ≤ λ

(2.4)

Fix ε > 0 and let g ∈ C(∂D) be such that || f − g||1 ≤ ε2 . Now Wg (ζ ) = 0 by Theorem 1.3, and hence W f (ζ ) = W f −g (ζ ). Applying (2.4) to f − g then gives 2 {ζ : W f (ζ ) > ε} ≤ (8 + 12α)ε = (8 + 12α)ε. ε Therefore, for any ﬁxed α, (2.2) holds almost everywhere. Because the cones α increase with α, it follows that (2.2) holds for every α > 1, except for ζ in a set of measure zero. To prove (2.3) we will dominate the nontangential maximal function with a second, simpler maximal function. Let f ∈ L 1 (∂D) and write 1 | f |dθ M f (ζ ) = sup I ζ |I | I

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9

for the maximal average of | f | over subarcs I ⊂ ∂D that contain ζ. The function M f is called the Hardy–Littlewood maximal function of f . The function M f is simpler than u ∗α because it features characteristic functions of intervals instead of Poisson kernels. Lemma 2.2. Let u(z) be the Poisson integral of f ∈ L 1 (∂D) and let α > 1. Then u ∗α (ζ ) ≤ (1 + 2α)M f (ζ ).

(2.5)

Proof. Assume ζ = 1. Fix z so that θ0 = arg z has |θ0 | ≤ π. Deﬁne Pz∗ (θ ) = sup Pz (ϕ) : |θ | ≤ |ϕ| ≤ π ⎧ 1 1 + |z| ⎪ ⎨ , |θ| ≤ |θ0 | = 2π 1 − |z| ⎪ ⎩ max(Pz (θ ), Pz (−θ )), |θ0 | < |θ | ≤ π.

Pz∗ (θ )

Pz (θ )

−π

−θ0

θ0

π

Figure I.3 The function Pz∗ . The function Pz∗ satisﬁes (i) Pz∗ (θ ) is an even function of θ ∈ [−π, π], (ii) Pz∗ (θ ) is decreasing on [0, π ], and (iii) Pz∗ (θ ) ≥ Pz (θ ). The even function Pz∗ is the smallest decreasing majorant of Pz on [0, π ]. We may assume f (eiθ ) ≥ 0, so that f (eiθ )Pz (θ )dθ ≤ f (eiθ )Pz∗ (θ )dθ.

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Then properties (i) and (ii) imply f (eiθ )Pz∗ (θ )dθ ≤ ||Pz∗ ||1 M f (1)

(2.6)

because Pz∗ is the increasing limit of a sequence of functions of the form 1 χ (−θ j ,θ j ) (θ ) cj 2θ j with c j ≥ 0 and c j ≤ ||Pz∗ ||1 .

Pz∗ (θ )

0 θ0

−π

Figure I.4 Approximating

Pz∗

θn

π

by a step function.

Now we claim that when z ∈ α (1), ||Pz∗ ||1 ≤ (1 + 2α).

(2.7)

Note that (iii), (2.6), and (2.7) imply (2.5). To prove (2.7) we ﬁrst assume −π/2 ≤ θ0 = arg z ≤ π/2. Then by the law of sines, |θ0 | |θ0 | π α | sin θ0 | π α |sin β| πα ≤α ≤ = ≤ , 1 − |z| |1 − z| 2 |1 − z| 2 1 2 where β = arg(z − 1)/z is explained by Figure I.2. If π/2 ≤ |θ0 | ≤ π and z ∈ α (1), then |1 − z| ≥ 1 and |θ0 | |θ0 | ≤α ≤ π α. 1 − |z| |1 − z| Hence (see Figure I.3) Pz∗ 1 = 2

π |θ0 |

Pz (θ )dθ +

2|θ0 | 1 + |z| ≤ (1 + 2α). 2π 1 − |z|

That proves (2.7) and therefore Lemma 2.2.

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11

By Lemma 2.2, the inequality (2.3) will follow from the simpler inequality 3 f 1 , (2.8) {ζ ∈ ∂D : M f (ζ ) > λ} ≤ λ which says that the operator L 1 f → M f is also weak-type 1-1. To prove the inequality (2.8), we use a covering lemma. Lemma 2.3. Let μ be a positive Borel measure on ∂D and let {I j } be a ﬁnite sequence of open intervals in ∂D. Then {I j } contains a pairwise disjoint subfamily {Jk } such that 1 μ(Jk ) ≥ μ Ij . (2.9) 3 Proof. Because the family {I j } is ﬁnite, we may assume that no I j is contained in the union of the others. Writing I j = {eiθ : θ ∈ (a j , b j )}, we may also assume that 0 ≤ a1 < a2 < . . . . < an < 2π. Then b j+1 > b j , because otherwise I j+1 ⊂ I j , and b j−1 < a j+1 , because otherwise I j ⊂ I j−1 ∪ I j+1 . If n > 1, then bn < b1 + 2π and bn−1 < a1 + 2π . Consequently, the family of even-numbered intervals I j is pairwise disjoint. The family of odd-numbered intervals I j is almost pairwise disjoint; only the ﬁrst and last intervals can intersect. If 1 μ(I j ) ≥ μ Ij , 3 j even

we take the even-numbered intervals to be the subfamily {Jk }. Otherwise 2 μ(I j ) ≥ μ Ij . 3 j odd

In that case, if μ(I1 ) ≤

1 μ(I j ), 2 j odd

we take for {Jk } the family of odd-numbered intervals, omitting the ﬁrst interval I1 , while if 1 μ(I j ), μ(I1 ) > 2 j odd

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

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Lemma 2.4. The operator f → M( f ) is weak-type 1-1: If f ∈ L 1 , then {ζ ∈ ∂D : M f (ζ ) > λ} ≤ 3 f 1 . (2.8) λ Proof. Let K be a compact subset of E λ = {ζ ∈ ∂D : M f (ζ ) > λ}. For each ζ ∈ E λ there is an open interval I such that ζ ∈ I and 1 | f |dθ > λ, |I | I so that |I | <

1 λ

| f |dθ. I

Cover K by ﬁnitely many such intervals {I j : 1 ≤ j ≤ n}, and let {Jk } be the pairwise disjoint subfamily given by Lemma 2.3. Then 3 |K | ≤ I j ≤ 3 |Jk | ≤ λ3 Jk | f |dθ ≤ λ || f ||1 , and letting |K | tend to |E λ | establishes Lemma 2.4.

By (2.5) and (2.8), inequality (2.3) holds with constant 3 + 6α, and the proof of Fatou’s theorem is complete.

Corollary 2.5. If u is a bounded harmonic function on D, then for every α > 1 and for almost every ζ = eiθ ∈ ∂D, f (ζ ) =

lim

α (ζ )z→ζ

u(z)

exists, u(z) is the Poisson integral of f , and || f ||∞ = supz∈D |u(z)|. Proof. Let rn → 1 and let f n (eiθ ) = u(rn eiθ ). By the Banach–Alaoglu theorem, the sequence { f n } has a weak-star cluster point f ∈ L ∞ (∂D) satisfying || f ||∞ ≤ lim sup || f n ||∞ ≤ supD |u(z)|. Since u(rn z) is the Poisson integral of f n , and Poisson kernels are in L 1 , u must be the Poisson integral of f . But then |u(z)| ≤ || f ||∞ . The corollary now follows from Fatou’s theorem. In particular, the corollary implies that for any measurable E ⊂ ∂D, there exists a unique bounded harmonic function u(z) on D such that u(z) has nontangential limit χ E almost everywhere. It is the function u(z) = ω(z, E, D).

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3. Carathéodory’s Theorem Let be a simply connected domain in the extended plane C∗ . We say is a Jordan domain if = ∂ is a Jordan curve in C∗ . Theorem 3.1 (Carathéodory). Let ϕ be a conformal mapping from the unit disc D onto a Jordan domain . Then ϕ has a continuous extension to D, and the extension is a one-to-one map from D onto . Because ϕ maps D onto , the continuous extension (also denoted by ϕ) must map ∂D onto = ∂, and because ϕ is one-to-one on ∂D, ϕ(eiθ ) parameterizes the Jordan curve . Before we prove Carathéodory’s theorem, we use it to solve the Dirichlet problem on a Jordan domain . Let f be Borel function on such that f ◦ ϕ is integrable on ∂D . If w = ϕ −1 (z), then 2π 1 − |w|2 dθ f ◦ ϕ(eiθ ) iθ (3.1) u(z) ≡ u f (z) = |e − w|2 2π 0 is harmonic on , and by Theorem 3.1 and Theorem 1.3, lim u(z) = f (ζ )

z→ζ

(3.2)

whenever ϕ −1 (ζ ) ∈ ∂D is a point of continuity of f ◦ ϕ. In particular, if f is continuous on then (3.2) holds for all ζ ∈ and u(z) = u f (z) solves the Dirichlet problem for f on . If f is a bounded Borel function on , then f ◦ ϕ is Borel and the integral (3.1) is deﬁned. For any Borel set E ⊂ we use (3.1) with f = χ E to deﬁne the harmonic measure of E relative to by: 1 − |w|2 dθ . (3.3) ω(z, E, ) = ω(w, ϕ −1 (E), D) = iθ 2 ϕ −1 (E) |e − w| 2π Then E → ω(z, E) is a Borel measure on ∂, and (3.1) can be rewritten as f (ζ )dω(z, ζ ). (3.4) u(z) = ∂

Equations (3.3) and (3.4) do not depend on the choice of ϕ, because for every conformal self map T of D, ω(T (w), T (ϕ −1 (E)), D) = ω(w, ϕ −1 (E), D). When f is a bounded Borel function on ∂, (3.4) and Fatou’s theorem give sup |u(z)| = f L ∞ (ω) . z∈

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Moreover, Corollary 2.5 shows that every bounded harmonic function on can be expressed in the form (3.4). The principal goal of this book is to ﬁnd geometric properties of the harmonic measure ω(z, E) more explicit than the deﬁnition (3.3). But (3.3) already points out the key issue: for a Jordan domain questions about harmonic measure are equivalent to questions about the boundary behavior of conformal mappings. Proof of Theorem 3.1. We may assume is bounded. Fix ζ ∈ ∂D. First we show ϕ has a continuous extension at ζ . Let 0 < δ < 1, write B(ζ, δ) = {z : |z − ζ | < δ} and set γδ = D ∩ ∂ B(ζ, δ). Then ϕ(γδ ) is a Jordan arc having length L(δ) = |ϕ (z)|ds. γδ

By the Cauchy–Schwarz inequality

L 2 (δ) ≤ π δ so that for ρ < 1, ρ 0

L 2 (δ) dδ ≤ π δ

γδ

|ϕ (z)|2 ds,

D∩B(ζ,ρ)

|ϕ (z)|2 d x d y

(3.5)

= π Area ϕ(D ∩ B(ζ, ρ)) < ∞.

ζ γ δn

αn

σn Un

βn

ϕ(γδn )

Figure I.5 The crosscuts γδn and ϕ(γδn ). Therefore there is a sequence δn ↓ 0 such that L(δn ) → 0. When L(δn ) < ∞, the curve ϕ(γδn ) has endpoints αn , βn ∈ , and both of these endpoints must lie on = ∂. Indeed, if αn ∈ , then some point near αn has two distinct

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preimages in D because ϕ maps D onto , and that is impossible because ϕ is one-to-one. Furthermore, |αn − βn | ≤ L(δn ) → 0.

(3.6)

Let σn be that closed subarc of having endpoints αn and βn and having smaller diameter. Then (3.6) implies diam(σn ) → 0, because is homeomorphic to the circle. By the Jordan curve theorem the curve σn ∪ ϕ(γδn ) divides the plane into two regions, and one of these regions, say Un , is bounded. Then Un ⊂ , because C∗ \ is arcwise connected. Since

diam(∂Un ) = diam σn ∪ ϕ(γδn ) → 0, we conclude that diam(Un ) → 0.

(3.7)

Set Dn = D ∩ {z : |z − ζ | < δn }. We claim that for large n, ϕ(Dn ) = Un . If not, then by connectedness ϕ(D \ D n ) = Un and

diam(Un ) ≥ diam ϕ(B(0, 1/2)) > 0, which contradicts (3.7). Therefore diam(ϕ(Dn )) → 0 and ϕ(Dn ) consists of a single point, because ϕ(Dn+1 ) ⊂ ϕ(Dn ). That means ϕ has a continuous extension to {ζ } ∪ D. It is an exercise to show that the union over ζ of these extensions deﬁnes a continuous map on D . Let ϕ also denote the extension ϕ : D → . Since ϕ(D) = , ϕ maps D onto . To show ϕ is one-to-one, suppose ϕ(ζ1 ) = ϕ(ζ2 ) but ζ1 = ζ2 . The argument used to show αn ∈ also shows that ϕ(∂D) = , and so we can assume ζ j ∈ ∂D, j = 1, 2. The Jordan curve ϕ(r ζ1 ) : 0 ≤ r ≤ 1 ∪ ϕ(r ζ2 ) : 0 ≤ r ≤ 1 bounds a domain W ⊂ , and then ϕ −1 (W ) is one of the two components of D \ {r ζ1 : 0 ≤ r ≤ 1} ∪ {r ζ2 : 0 ≤ r ≤ 1} . But since ϕ(∂D) ⊂ , ϕ(∂D ∩ ∂ϕ −1 (W )) ⊂ ∂ W ∩ ∂ = {ϕ(ζ1 )} and ϕ is constant on an arc of ∂D. It follows that ϕ is constant, either by Schwarz reﬂection principle or by the Jensen formula, and this contradiction shows ϕ(ζ1 ) = ϕ(ζ2 ).

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One can also prove ϕ is one-to-one by repeating for ϕ −1 the proof that ϕ is continuous. See Exercise 13(b). Exercise 12 gives the necessary and sufﬁcient condition that ϕ have a continuous extension to D. The Cauchy–Schwarz trick used to prove (3.5) is known as a length–area argument. The length–area method is the cornerstone of the theory of extremal length. See Chapter IV.

4. Distortion and the Hyperbolic Metric Let D be the open unit disc. The hyperbolic distance from z 1 ∈ D to z 2 ∈ D is z2 |dz| , (4.1) ρ(z 1 , z 2 ) = ρD (z 1 , z 2 ) = inf 1 − |z|2 z1 where the inﬁmum is taken over all arcs in D connecting z 1 to z 2 . Let M denote the set of conformal self maps of D : T (z) = λ

z−a , a ∈ D, |λ| = 1. 1 − az

When T ∈ M, we have 1 |T (z)| = , 2 1 − |T (z)| 1 − |z|2 and thus the hyperbolic distance is conformally invariant, ρ(T (z 1 ), T (z 2 )) = ρ(z 1 , z 2 ), T ∈ M.

(4.2)

This conformal invariance is the main reason we are interested in the hyperbolic distance. The hyperbolic metric is the inﬁnitesimal form |dz|/(1 − |z|2 ) of the hyperbolic distance. Taking T (z) =

z − z1 1 − z1 z

gives

T (z 2 )

ρ(z 1 , z 2 ) = ρ(0, T (z 2 )) = 0

|dz| . 1 − |z|2

Therefore the hyperbolically shortest arc from 0 to T (z 2 ) is the radius [0, T (z 2 )], and its hyperbolic length is 1 + |T (z )| 1 2 ρ(0, T (z 2 )) = log . 2 1 − |T (z 2 )|

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z − z ⎞ 1 2 1+ 1 1 − z1 z2 ⎟ ⎜ ρ(z 1 , z 2 ) = log ⎝ z − z ⎠ , 1 2 2 1− 1 − z1 z2

(4.3)

⎛

and the hyperbolically shortest curve, or the geodesic, from z 1 to z 2 is a segment of a diameter of D or an arc of a circle in D orthogonal to ∂D. By (4.3) we have z − z e2ρ(z 1 ,z 2 ) − 1 1 2 = tanh ρ(z 1 , z 2 ). = 2ρ(z ,z ) 1 2 +1 1 − z1 z2 e Write t = t (d) = tanh(d) =

e2d − 1 . e2d + 1

Then the hyperbolic ball B = {z : ρ(z, a) < d} is the euclidean disc z−a {z : < t}, 1 − az and a calculation shows that B has euclidean radius r (a, d) = and euclidean distance to ∂D dist(B, ∂D) =

t (1 − |a|2 ) 1 − t 2 |a|2

1−t (1 − |a|). 1 + |a|t

(4.4)

(4.5)

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

z1 0 a1

z2 x1 a2

Figure I.6 Hyperbolic balls and geodesics.

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Now assume ψ(z) is a univalent function in D, that is, assume ψ is analytic and one-to-one on D. After dilating, translating, and rotating the domain ψ(D), ψ is normalized by ψ(0) = 0 and ψ (0) = 1, so that ψ(z) = z + a2 z 2 + · · · . Important examples are the Koebe functions z , |λ| = 1. ψ(z) = ψλ (z) = (1 − λz)2

(4.6)

(4.7)

Note that ψλ (z) =

∞

nλn−1 z n

n=1

maps D to the complement of the radial slit [−λ/4, ∞]. Theorem 4.1 (Koebe one-quarter theorem). Assume ψ(z) is a univalent function on D. If ψ(z) has the form (4.6) then |a2 | ≤ 2

(4.8)

and 1 . (4.9) 4 Equality holds in (4.8) or (4.9) if and only if ψ is a Koebe function (4.7). dist(0, ∂ψ(D)) ≥

Proof. First note that (4.8) implies (4.9). Indeed, suppose w ∈ / ψ(D). Then (4.8) holds for the univalent function g(z) =

1 wψ(z) = z + (a2 + )z 2 + · · · , w − ψ(z) w

so that a2 + 1 ≤ 2, w and together (4.10) and (4.8) give 1 . 4 To prove (4.8) we form the odd function ψ(z 2 ) a2 = z + z3 + · · · . f (z) = z z2 2 |w| ≥

(4.10)

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Then f is univalent because ψ is univalent, and the C∗ = C ∪ {∞} valued function ∞

F(z) =

1 1 a2 1 bn z n = − z + ··· = + f (z) z 2 z

(4.11)

n=1

is also univalent in D. To complete the proof we use the following lemma, called the 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|bn |2 ≤ 1.

(4.12)

n=1

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 ψ is a Koebe function. Proof of Lemma 4.2. The lemma is called the area theorem because of its proof. For r < 1, the Jordan curve r = {F(r eiθ ) : 0 ≤ θ ≤ 2π } encloses an area A(r ), and by Green’s theorem, −i −i 2π ∂F A(r ) = wdw = F(r eiθ ) (r eiθ )dθ. 2 r 2 0 ∂θ Therefore by (4.11) and Fourier series, ∞ 1 n|bn |2 r 2n A(r ) = π 2 − r n=1

and 1−

∞ n=1

n|bn |2 = lim

r →1

A(r ) ≥ 0, π

which yields (4.12).

Theorem 4.3 (Koebe’s estimate). Let ϕ(z) be a conformal mapping from the unit disc D onto a simply connected domain . Then for all z ∈ D 1 |ϕ (z)|(1 − |z|2 ) ≤ dist(ϕ(z), ∂) ≤ |ϕ (z)|(1 − |z|2 ). 4

(4.13)

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Proof. Fix z 0 ∈ D. Then the univalent function z+z 0 ϕ 1+z − ϕ(z 0 ) z 0

ψ(z) = ϕ (z 0 ) 1 − |z 0 |2 satisﬁes ψ(0) = 0 and ψ (0) = 1. Hence if w ∈ / ϕ(D), then w − ϕ(z ) 1 0 ≥ ϕ (z 0 )(1 − |z 0 |2 ) 4 by (4.9), and this gives the left-hand inequality in (4.13). To prove the right-hand inequality, ﬁx z ∈ D, take

f (w) = ϕ −1 ϕ(z) + dist ϕ(z), ∂ w , and apply the Schwarz lemma at w = 0 to the function g(w) =

f (w) − z . 1 − z f (w)

We will often use the invariant form of (4.13): Corollary 4.4. Let ψ be a conformal mapping from a simply connected domain 1 onto a simply connected domain 2 , and let ψ(z 0 ) = w0 . Then dist(w0 , ∂2 ) |ψ (z 0 )| ≤ ≤ 4|ψ (z 0 )| 4 dist(z 0 , ∂1 )

(4.14)

Proof. Applying (4.13) to ϕ(z) = ψ(z 0 + dist(z 0 , ∂1 )z) gives the left-hand inequality and the same argument with ψ −1 gives the right-hand inequality. In a simply connected domain = C, the hyperbolic distance is deﬁned by moving back to D via a conformal map ϕ : D → . We write ρ (w1 , w2 ) = ρD (z 1 , z 2 ) when w j = ϕ(z j ). By (4.2), ρ (w1 , w2 ) does not depend on the choice of the conformal map ϕ. The quasihyperbolic distance from w1 ∈ to w2 ∈ is w2 |dw| , Q (w1 , w2 ) = inf w1 dist(w, ∂) in which the inﬁmum is taken over all arcs in joining w1 to w2 . Since (4.13) can be written as |dz| |dw| 4|dz| ≤ , ≤ 1 − |z|2 dist(w, ∂) 1 − |z|2

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where w = ϕ(z), we have ρ (w1 , w2 ) ≤ Q (w1 , w2 ) ≤ 4ρ (w1 , w2 ).

(4.15)

Consequently the geometric statement following (4.4) and (4.5) about hyperbolic distances near ∂D remains approximately true in every simply connected domain 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 ) = 2−n j (ii) = Sj , and (iii) diam(Sj ) ≤ dist(Sj , ∂) < 4 diam(Sj ). The squares {Sj } are called Whitney squares. Here is one way to construct Whitney squares in the case diam() < ∞: Let 2−N +1 ≤ diam() < 2−N +2 and partition the plane into squares having sides parallel to the axes and sidelength 2−N . We call these squares “2−N squares”. Include in the family {Sj } any 2−N square S ⊂ which satisﬁes (iii), and divide each of the remaining 2−N squares into four squares of side 2−N −1 . Next include in {Sj } any of these new 2−N −1 squares contained in and satisfying (iii), and continue. (See Figure I.7). Finding Whitney squares in the case diam() = ∞ is an exercise for the reader. Whitney squares can be viewed as replacements for hyperbolic balls since there are universal constants c1 < c2 such that each Tj contains a hyperbolic 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 ϕ : 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 ϕ, such that for

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each k, ϕ(Tk ) is contained in at most M squares Sj , and ϕ −1 (Sk ) is contained in at most M squares Tj . In particular, for each d > 0, there is M(d) such that every hyperbolic ball {z : ρ (z, a) < d} in is covered by M(d) Whitney squares. Theorem 4.5. Let ψ(z) be a univalent function satisfying ψ(0) = 0 and ψ (0) = 1. Then |z| |z| ≤ |ψ(z)| ≤ , (1 + |z|)2 (1 − |z|)2

(4.16)

1 − |z| 1 + |z| ≤ |ψ (z)| ≤ . (1 + |z|)3 (1 − |z|)3

(4.17)

|ψ (z)| 1 + |z| 1 − |z| ≤ ≤ . |z|(1 + |z|) |ψ(z)| |z|(1 − |z|)

(4.18)

and

Moreover,

Inequality (4.16) is known as the growth theorem, while (4.17) is called the distortion theorem. Proof. The critical inequalities are (4.17) and we prove them ﬁrst. Fix z 0 ∈ D and take z+z 0 ψ 1+z − ψ(z 0 ) 0z .

(4.19) f (z) = ψ (z 0 ) 1 − |z 0 |2 Then f is univalent on D, f (0) = 0 and f (0) = 1. By (4.8), 1 − |z 0 |2 | f (0)| = ψ (z 0 ) − 2z 0 ≤ 4, ψ (z 0 ) so that when z 0 = r eiθ , iθ e ψ (z 0 ) 2|z 0 | 4 ψ (z ) − 1 − |z |2 ≤ 1 − |z |2 . 0 0 0

(4.20)

Applying the general formula Re

zg (z) ∂Reg = |z| ∂r

to g = log ψ , we then obtain ∂ 2r + 4 2r − 4 ≤ . (4.21) log |ψ (z 0 )| ≤ 1 − r2 ∂r 1 − r2 Integrating (4.21) along the radius [0, z] gives both inequalities in (4.17).

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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 |ψ(z)| < 41 , because |z| < 41 . Then by (4.9) there exists an arc γ ∈ D with ψ(γ ) = [0, ψ(z)], (1+|z|)2 and integrating |ψ (z)||dz| along γ then gives the lower bound in (4.16). To prove (4.18), apply (4.16) at −z 0 to the function f deﬁned by (4.19). Again, equality any place in (4.16), (4.17), (4.18), or (4.21) implies that ψ is a Koebe function.

5. The Hayman–Wu Theorem We give a very elementary proof, based on an idea of the late K. Øyma [1992], of the theorem of Hayman and Wu. The Hayman–Wu theorem will be a recurrent topic throughout this book. Theorem 5.1 (Hayman–Wu). Let ϕ be a conformal mapping from D to a simply connected domain and let L be any line. Then (5.1) length ϕ −1 (L ∩ ) ≤ 4π. Hayman and Wu [1981] gave the ﬁrst proof of (5.1) with 4π replaced by some large unknown constant. Øyma [1992] obtained the constant 4π , Rohde [2002] proved that the best constant in (5.1) is strictly smaller than 4π , and Øyma [1993] proved that the best constant is at least π 2 . The sharp constant in (5.1) is not known. See Exercises 24 and VI.3. We present Øyma’s elementary proof, as modiﬁed by Rohde. For the proof it will be convenient to replace the hyperbolic metric ρ(z 1 , z 2 ) by the pseudohyperbolic metric, deﬁned in D by z1 − z2 = tanhρ(z 1 , z 2 ), δD (z 1 , z 2 ) = 1 − z1 z2 and in by δ (w1 , w2 ) = δD (ϕ −1 (w1 ), ϕ −1 (w2 )). Proof. We can assume that ϕ is analytic and one-to-one in a neighborhood of D and that L = R . Let L k denote the components of ∩ L and let k be that component of ∩ {z : z ∈ } such that L k ⊂ k . Then k is a Jordan domain symmetric about R. When k = j, ∂k ∩ ∂ j ⊂ R and ∂k ∩ ∂ j contains at most one point, because

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is simply connected.

k

j

Figure I.8 The domains k are shaded. By symmetry there is a conformal mapping ψk : k → −iH such that ψk (L k ) = R+ and ψk extends continuously to k . For ζ ∈ ∂ϕ −1 (k ) ∩ ∂D, set α = ϕ(ζ ), x = |ψk (α)|, β = ψk−1 (x), and z = ϕ −1 (β). Then the composition ≡ ϕ −1 ◦ ψk−1 (|ψk ◦ ϕ|) is a smooth map of ϕ −1 ( ∂k ∩ ∂ \ P) ⊂ ∂D onto ϕ −1 ( L k ) \ P where P and P are ﬁnite sets.

ϕ

ψk β k

z ζ

x α

ψk (α) Figure I.9 The map z = (ζ )

To prove Theorem 5.1, it sufﬁces to show that |∇| ≤ 2.

(5.2)

To prove (5.2), suppose that I = (ζ, ζ ) is an open interval contained in ϕ −1 (∂k ) ∩ ∂D. Set α = ϕ(ζ ), x = |ψk (α )|, β = ψk−1 (x ), and z = ϕ −1 (β ).

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5. The Hayman–Wu Theorem

Then by Pick’s theorem (see Exercise 14(a))

x − x , δD ((ζ ), (ζ )) = δ (β, β ) ≤ δk (β, β ) = δ−i H (x, x ) = x + x

and by Lindelöf’s maximum principle ω(x, ψk (ϕ(I )), −iH) = ω(z, I, ϕ −1 (k )) ≤ ω((ζ ), I, D). Letting ζ → ζ we obtain the inequality 1 1 − |(ζ )|2 |∇| ≤π· , 2 1 − |(ζ )| 2π |ζ − (ζ )|2

(5.3)

which easily implies (5.2). This completes the proof of the Hayman–Wu theo rem. Because it does not depend on the mapping ϕ, the conclusion of Theorem 5.1 is actually stronger than (5.1). See Exercise 20. In Chapter VII we will see that for some 1 < p < 2, independent of ϕ p −1 (5.4) (ϕ ) (w) |dw| < C p , L∩

but the largest permissible p is unknown. A slit disc shows (5.4) fails at p = 2, and a counterexample for some p < 2, due to Baernstein, is given in Chapter VIII. In Chapter X we will determine the class of curves L for which (5.1) holds.

Notes This chapter is a survey of elementary material which can be found in many other sources. We learned Lemma 2.3 from Garsia [1970], who attributed it to W. H. Young. The proof of Lemma 2.3 also shows that every open cover has χ I ≤ 2. In Wheeden and Zygmund [1977] the a subcover {Ik } such that k higher dimensional formulation of this statement is called the “simple Vitali lemma”. The classical Vitali covering lemma is an easy consequence of this or Lemma 2.3. See Exercise 9 or Mattila’s book [1995], which has much about covering lemmas. Stein’s classic book [1970] also gives an excellent discussion of the relation between covering lemmas, maximal functions, and almost everywhere convergence and a clear introduction to Whitney squares. Section 4 follows Duren [1983], Hayman [1958], and Goluzin [1952]. We call (4.13) Koebe’s estimate and we will refer to it as such many times, but in

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[1907] Koebe only proved the left half of (4.13) with a constant ρ ≤ 41 . Later, in [1916] Bieberbach showed ρ = 41 . See Duren [1983].

Exercises and Further Results 1. (a) If E is the arc (eiθ1 , eiθ2 ) of ∂D in Figure I.10, then iθ2 e −z 2π ω(z, E, D) = 2 arg iθ − (θ2 − θ1 ) = 2ϕ − (θ2 − θ1 ), (E.1) e 1 −z where ϕ is the angle subtended at z by the arc (eiθ1 , eiθ2 ). Hint: Verify (i), (ii), and (iii) from Section 1 via elementary geometry and use Lemma 1.1. An alternative approach from the Poisson integral formula is to write 2eiθ d 2 eiθ + z = − 1 = log(eiθ − z) − θ eiθ − z eiθ − z dθ i and integrate the real part of this expression over the interval (θ1 , θ2 ). (b) We also have ω(z, E) =

1 (ϕ2 − ϕ1 ), 2π

where eiϕ j is the other endpoint of the chord extending from eiθ j through z. This is easy from (E.1) and elementary geometry. eiϕ1

2π ω

eiθ2 ϕ E 0

z

eiθ1

eiϕ2 Figure I.10 Harmonic measure of an arc. (c) Using ϕ(z) = i(1 − z)/(1 + z), and a change of variables, derive the Poisson integral formula for D from the Poisson integral formula for H. Hints:

eiθ + z 1 − |z|2 1 y = Re iθ = Im and . iθ 2 |e − z| e −z (x − t)2 + y 2 t−z

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27

2. If eiθ ∈ ∂D is ﬁxed, the level curves of z → Pz (eiθ ) are circles in D tangent to ∂D at eiθ . If E is an arc of ∂D, the level set {ω(z, E) = t} is the arc of a circle cutting ∂D at the endpoints of E in the angle π t. 3. (a) The Lindelöf maximum principle, Lemma 1.1, is false if u(z) is not assumed to be bounded above. (b) But Lindelöf’s maximum principle is also true for subharmonic functions: If u(z) is a subharmonic function on a domain such that = C, if u(z) is bounded above, and if lim sup u(z) ≤ 0, z→ζ

for all ζ ∈ ∂ \ F, where F is a ﬁnite subset of ∂, then u(z) ≤ 0 on . See Ahlfors [1973]. (c) Show that the conclusion in 3(b) still holds if the hypothesis = C is replaced by the weaker hypothesis ∂ \ F = ∅. 4. Let u be a real valued continuous function on a region . Prove that the following three conditions are equivalent: (a) For each sufﬁciently small disc D ⊂ and each v harmonic on D, u − v and v − u satisfy the maximum principle on D. (b) u satisﬁes the mean value property for each sufﬁciently small disc D contained in . (c) u is harmonic on . Hint: Use the Poisson integral on D. 5. (a) Harnack’s inequality. Let p and q be points of an arbitrary plane domain . Prove there is a constant C p,q such that whenever u(z) is positive and harmonic on , 1 u( p) ≤ ≤ C p,q . C p,q u(q) (b) If p and q remain in a compact subset K ⊂ , then there is a constant C K < ∞ such that C p,q ≤ C K . (c) Harnack’s principle. Let {n } be a sequence of domains and let u n be a harmonic function on n . Assume is a domain such that every z ∈ has a neighborhood Vz such that for n sufﬁciently large, Vz ⊂ n , and u n ≤ u n+1 on Vz . Then either u n tends to ∞ uniformly on every compact subset of , or there is a harmonic function u on and u n converges to u uniformly on every compact subset of . (d) Let n and be as in (c), and let u n be a harmonic function on n . Now assume there is Mz < ∞ such that for n sufﬁciently large, supVz |u n | ≤ Mz . Prove {u n } has a subsequence converging uniformly on all compact subsets

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I. Jordan Domains of to a function harmonic on . (e) If u is harmonic on D and |u| ≤ M then |∇u| ≤

2M . 1 − |z|2

it 1 e +z Re it . 2π e −z 6. The boundary Harnack inequality. Let u and v be positive harmonic functions on D with u(0) = v(0), let I ⊂ ∂D be an open arc and assume Hint: Pz (t) =

lim u(z) = lim v(z) = 0

z→ζ

z→ζ

for all ζ ∈ I. Prove that for every compact K ⊂ D ∪ I there is a constant C(K ) independent of u and v such that on K ∩ D 1 u(z) ≤ ≤ C(K ). C(K ) v(z)

7.

8.

9.

10.

11.

Hint: Using a conformal map of a subregion of D onto D, you may assume u and v are continuous on D. Estimate the Poisson kernel on ∂D \ I . See Wu [1978] for a more general result. Show there is a Borel set A ⊂ ∂D such that any function almost everywhere equal to χ A is not continuous at any point ζ ∈ ∂D. Then show that for f = χ A , (1.8) fails for all ζ ∈ ∂D. Suppose G(z) is continuous on the closed disc and analytic on the open disc. Use the Poisson kernel to show that if dG(eiθ )/dθ exists at θ0 , then G (z) has nontangential limit at eiθ0 . A Vitali cover of a measurable set E ⊂ ∂D is a family I of arcs I such that for every ζ ∈ E and every δ > 0 there exists I ∈ I such that ζ ∈ I and |I | < δ. Prove the Vitali covering lemma: If I is a Vitali cover of E and if ε > 0 there exists a pairwise disjoint sequence {I j } ⊂ I such that |E \ I j | < ε. Assume u(z) is the Poisson integral of a ﬁnite positive measure μ on ∂D, and let dμ = f (θ )dθ + dμs be the Lebesgue decomposition of μ. Prove that almost everywhere dθ, u has nontangential limit f (θ ), and that almost everywhere dμs , u has nontangential limit ∞. Hint: Use Lemma 2.3. Let u(z) be real and harmonic in D. (a) If 1 < α < β, then u ∗α ≤ u ∗β ≤ Cα,β M(u ∗α ). Moreover, if 0 < p < ∞, then ||u ∗α || L p (∂ D) . ||u ∗β || L p (∂ D) ≤ Cα,β

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29

(b) If 1 ≤ p < ∞ and if u is the Poisson integral of f ∈ L p , then lim |u(r eiθ ) − f (eiθ )|dθ = 0. r →1

(c) If p > 1 show u is the Poisson integral of f ∈ L p (∂D) if and only if dθ < ∞. sup |u(r eiθ )| p 2π r <1 (d) Show u is the Poisson integral of a probability measure μ on ∂D u(z) = Pz (θ )dμ(θ), if and only if u > 0 and u(0) = 1. (e) Show u is the Poisson integral of a ﬁnite (signed) Borel measure on ∂D if and only if dθ sup |u(r eiθ )| < ∞. (E.2) 2π r <1 If (E.2) holds, then u has an almost everywhere nontangential limit f (θ ), but u(z) need not be the Poisson integral of f. See Exercise 10. (f) If u is the Poisson integral of a ﬁnite signed measure ν on ∂D, then the dθ measures u(r eiθ ) 2π converge weak-star to ν as r → 1. It follows that u determines the measure ν uniquely. (g) Show u is the Poisson integral of f ∈ L 1 (∂D) if and only if the family {u(r eiθ ) : 0 < r < 1} is uniformly integrable in θ. (h) Let 1 ≤ p ≤ ∞. If u ∗α ∈ L p (∂D), then u is the Poisson integral of some f ∈ L p (∂D). See Appendix A for the corresponding results when p < 1. 12. A compact set K is locally connected if whenever U is a relatively open subset of K and z ∈ U ⊂ K there is a relatively open subset V of K such that V is connected and z ∈ V ⊂ U. Let be a simply connected domain such that ∂ contains at least two points. Prove ∂ is locally connected if and only if the Riemann map ϕ : D → extends continuously to D. Hint: For one direction, use the uniform continuity of ϕ. For the other direction follow the proof of Carathéodory’s theorem. 13. (a) Let be a simply connected domain and let σ ⊂ be a crosscut, that is, a Jordan arc in having distinct endpoints in ∂. Prove that \ σ has two components 1 and 2 , each simply connected, and β j = ∂ j \ σ is connected. (b) Let be simply connected, let ψ : → D be conformal and ﬁx z ∈ and ζ ∈ ∂. Assume ζ and z can be separated by a sequence of crosscuts

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γn ⊂ such that length(γn ) → 0. Let Un be the component of \ γn such that z ∈ / Un . Prove that ψ(Un ) → α ∈ ∂D. 14. Let f (z) be an analytic function on D and assume | f (z)| ≤ 1. (a) Prove Pick’s theorem: f (z) − f (w) z − w ≤ 1 − f (w) f (z) 1 − wz . Equivalently, prove ρ( f (z), f (w)) ≤ ρ(z, w). In other words, analytic maps from D to D are Lipschitz with respect to the hyperbolic metric. Hint: Repeat the proof of the Schwarz lemma, using (S ◦ f )/T for suitable S and T in M. (b) Deduce that 1 | f (z)| ≤ . 1 − | f (z)|2 1 − |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 : ρ(z, a) < d}. 16. Let ϕ(z) be univalent on D and assume ϕ(z) = 0 for every z. (a) Prove |ϕ (0)| ≤ 4|ϕ(0)|, using (4.9). (b) Prove

1 − |z| 1 + |z|

2

|ϕ(z)| ≤ ≤ |ϕ(0)|

Hint: Fix z and apply (a) to ψ(w) = ϕ

1 + |z| 1 − |z|

2 .

w+z 1 + zw

to get ϕ (z) 4 , ≤ ϕ(z) 1 − |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 ψ is a Koebe function, deﬁned by (4.7).

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(b) On the other hand, equality in (4.12) holds if and only if C \ F(D) has area zero. 18. (a) Find an upper bound for the constant M mentioned in the remarks preceding Figure I.7. (b) Give a construction of Whitney squares in the case diam() = ∞. 19. A region of the form Q = {z = r eiθ : θ0 < θ < θ0 + (Q), 1 − (Q) ≤ r < 1} is called a box or a Carleson box, and the arc {eiθ : θ0 < θ < θ0 + (Q)} is the base of Q. See Figure I.11. A ﬁnite measure on D is a Carleson measure if there is a constant C such that μ(Q) ≤ C(Q) for every Carleson box.

(Q) Q aQ

eiθ0 (Q)

Figure I.11 A Carleson box.

(a) Let α > 1. Prove μ is a Carleson measure if and only if there is Cα such that μ({z ∈ D : u(z) > λ}) ≤ Cα {ζ ∈ ∂D : u ∗ > λ} α

L 1 (∂D).

Hint: Write the set whenever u is the Poisson integral of f ∈ {u ∗α > λ} = I j where the I j are pairwise disjoint open arcs on ∂D, and deﬁne Tj = D \ ζ ∈I / j α (ζ ). The set Tj looks like a tent with base I j .

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Ij Tj Ik Tk Figure I.12 Tents. Then μ(Tj ) ≤ Cα (I j ) and |u| ≤ λ on D \ Tj . Conversely, let Q be the box with base I, and let u(z) = 4λω(z, I, D). Then u(z) > λ on Q while by (2.3), ∗ {u > λ} ≤ Cα |I |. α (b) If μ is a ﬁnite measure, then μ is a Carleson measure if and only if sup

a∈D

1 − |a|2 dμ(z) < ∞. |1 − az|2

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 sup length(T ()) < ∞

T ∈M

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 μ on H is called a Carleson measure if there is a constant C such that μ(Q) ≤ C(Q) for every Carleson box in H. A box Q is called dyadic if (Q) = 2−k and a = j2−k for some integers j and k. If μ(Q) ≤ C(Q) for all dyadic boxes, prove that μ is a Carleson measure on H. (e) If μ is a Carleson measure on D and τ is a conformal map of H onto D, then prove μ ◦ τ is a Carleson measure on H. Conversely, if μ is a Carleson measure on H supported in the unit box [0, 1] × [0, 1], then μ ◦ τ −1 is a Carleson measure on D. Part (a) is due to E. M. Stein (unpublished).

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20. Let ϕ be a conformal mapping from D to a simply connected domain , let L be any line and set = ϕ −1 ( ∩ L). Then (a) Arc length on is a Carleson measure; in other words, length( ∩ Q) ≤ C(Q) for any box Q = {r eiθ : θ0 < θ < θ0 + (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-Wu theorem. 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 ϕ(z) be univalent on D and assume ϕ(z) = 0 for every z. (a) There is C1 < ∞, independent of ϕ, such that ϕ (z) d xd y ≤ C1 . ϕ(z) D

Hint: Write w = u + iv = ϕ(z), w ∈ . Then ϕ (z) 2 1 2 2 3 ϕ (z) 2 d xd y d xd y ≤ (1 − |z|) 3 d xd y ϕ(z) 1 − |z| ϕ(z) D

D

D

≤ C2

1 − |ϕ −1 (w)|

2 dudv 3

|w|2

.

Now integrate over ∩ {|w| > 1} and ∩ {|w| ≤ 1} separately, and use Exercise 16. (b) Let f = log ϕ. There is C2 < ∞, independent of ϕ such that | f |d xd y ≤ C2 (Q) (E.4) Q

for any Carleson box Q = {z = r eiθ : θ0 ≤ θ ≤ θ0 + (Q), 1 − (Q) ≤ r < 1}. In other words, | f (z)|d xd y is a Carleson measure. For the proof we may (Q) assume (Q) is small. Let a = a Q = (1 − 2(Q))ei(θ0 + 2 ) and use (a) with w = (z − a)/(1 − az). See Figure I.11.

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(c) It follows from (b) that almost everywhere and in L 1 norm, lim f (r eiθ ) = f (eiθ )

r →1

exists and f (z) is the Poisson integral of f (eiθ ). In fact, f (eiθ ) ∈ BMO (for bounded mean oscillation), which by deﬁnition means there exists a constant C3 < ∞, independent of ϕ, such that for every arc I ⊂ ∂D, 1 | f − α|dθ ≤ C3 . (E.5) inf α∈C |I | I Indeed, if Q has base I, then one can show 1 | f − f (a Q )|dθ ≤ C | f |d xd y. |I | I Q

See Jones [1980]. Baernstein [1976] gave the ﬁrst proof of (E.5). Cima and Schober [1976] and Pommerenke [1976] have related results. When e f (z) is not univalent, (E.4) is not equivalent to (E.5). However, in general (E.5) is equivalent to | f |2 (1 − |z|)d xd y ≤ C(Q). Q

See Fefferman and Stein [1972], Baernstein [1980], Garnett [1981], or Appendix F. 22. Again let ϕ be univalent and assume ϕ(z) = 0 for all z. Prove that for r ≥ 1/2 iθ 2 C4 ϕ (r e ) 2 I (r ) = dθ ≤ log , (E.6) ϕ(r eiθ ) 1−r 1−r where C4 does not depend on ϕ. Hayman [1980] attributes the following argument to P. L. Duren: By Exercise 16, iθ 1 ϕ (r e ) 2 d xd y ≤ 8π log + 8π log 2. ϕ(r eiθ ) 1−r {|z|
Take r < R < 1. Since I (r ) increases with r , iθ 2 ϕ (r e ) 2 I (r ) ≤ 2 d xd y 2 R −r ϕ(r eiθ ) {r <|z|

1 2 + 8π log 2 . 8π log ≤ 2 R − r2 1− R

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Now take 1 (1 − r 2 ). 2 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 2π p sup | f (r eiθ )| p dθ = || f || Hp < ∞. 1 − R2 =

0
Let ψ(z) be a univalent function on D, normalized by ψ(z) = z + a2 z 2 + . . . . (a) For 0 < p < ∞, verify the identity

|ψ| p = p 2 |ψ| p−2 |ψ |2 . (b) If 0 < p <

1 2

then ψ ∈ H p and p

||ψ|| H p ≤

C , 1 − 2p

where C is independent of p. This result is due to Prawitz [1927]. Green’s theorem and (a) give 2π d 1 p2 iθ p |ψ(r e )| dθ = |ψ(z)| p−2 |ψ |2 d xd y dr 2π 0 2πr {|z|
p2 = 2πr Write M(r ) = sup|z|≤r |ψ(z)|. Then 2π p2 d 1 |ψ| p dθ ≤ dr 2π 0 2πr

|w| p−2 d A(w). ψ({|z|
|w| p−2 d A(w) =

|w|<M(r )

p p

M(r ) , r

while by the distortion theorem, M(r ) ≤ Therefore 1 2π

2π

r

|ψ| dθ ≤ p p

0

≤p

0 r 0

r . (1 − r )2 p

M(t) t

dt

t p−1 (1 − t)−2 p dt = O

1 . 1 − 2p

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(c) If p > 21 , a similar argument gives 2π 1 1 |ψ(r eiθ )| p dθ ≤ C p . 2π 0 (1 − r )2 p−1 (d) If p > 21 , (c) and (4.18) imply 2π 1 1 |ψ (r eiθ )| p dθ ≤ C p . 2π 0 (1 − r )3 p−1 For p = 1, (c) implies (d) if ψ is analytic but not necessarily univalent. 1 (e) The Koebe function ψ(z) = z/(1 − z)2 is not in H 2 . In a deep paper [1974] Baernstein proved that the extremal functions for parts (b) and (c) are the rotates of the Koebe function. See Duren [1983] for these results and more, and see Appendix A for the basic theory of Hardy spaces. 24. (a) Show that the best constant in the Hayman–Wu theorem is smaller than 4π (Rohde [2002]). Hint: By (5.3) if |∇(ζ )| > 2 − ε then (ζ ) lies in a cone at ζ with opening < Cε and || > 1 − Cε. Moreover the curve α = (∂D) makes an angle with the radius through (ζ ) approximately tan−1 (1/2). Then use (4.20) to show that if α makes a larger angle with the radius at a point z then |∇| < 2 − ε on an interval centered at z/|z| of length δ(1 − |z|). Use this to show that there is a δ > 0, independent of ϕ so that |∇| < 2 − ε on a subset of the circle of length at least δ. (b) Use (5.3) to show that if we also require ϕ(0) ∈ L in the Hayman–Wu theorem, then the best constant is less than 4π − 2. 25. (a) Let 1 ⊂ 2 be Jordan domains, and let E = ∂1 ∩ ∂2 . Then on 1 , ω(z, E, 1 ) ≤ ω(z, E, 2 ). (b) Let 0 < δ < 1 and let be a Jordan domain such that {|z| < 1 − δ} ⊂ ∂ ⊂ {|z| < 1 + δ}. Let E = ∂ ∩ {θ1 ≤ arg z ≤ θ2 }, where θ2 < θ1 + 2π. Show ω(0, E, ) − (θ2 − θ1 ) ≤ Cδ log( 1 ). 2π δ Hint: By (a) the worst case in (b) is for regions of the following form. Let I = [eiθ1 , eiθ2 ], let = D \ {r z : z ∈ I, r ≥ 1 − δ} where δ < |I |, and let E = ∂ ∩ {|z| = 1 − δ}. Show that if |z| = 1 − δ, and dist(z, E) ≥ δ, then ω(z, E, ) ≤ 2ω(z, I, ∂ D). (c) Show the estimate in (b) is sharp, except for the value of C.

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In this chapter we solve the Dirichlet problem on a domain bounded by a ﬁnite number of Jordan curves. For a simply connected Jordan domain the problem was solved in Chapter I via the theorem of Carathéodory. For a multiply connected domain the problem will be reduced to the simply connected case using the Schwarz alternating method. Solving the Dirichlet problem on a domain is equivalent to constructing harmonic measure on ∂. In Section 2 we describe harmonic measure in terms of the normal derivative of Green’s function in the case when ∂ consists of analytic curves. In Section 4 we study the relation between the smoothness of ∂ and the smoothness of the Poisson kernel (the Radon–Nikodym derivative of harmonic measure against arc length). This relation hinges on two classical estimates for conjugate functions which we prove in Section 3.

1. The Schwarz Alternating Method Let be a plane domain such that ∂ is a ﬁnite union of pairwise disjoint Jordan 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 continuous on ∂ \ E and f has left and right limits at each point of E. In this section we solve the Dirichlet problem for piecewise continuous boundary data on a ﬁnitely connected Jordan domain. Theorem 1.1. Let be a ﬁnitely connected Jordan domain, and let f be a bounded piecewise continuous function on ∂. Then there is a unique function 37

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u(z) = u f (z), bounded and harmonic on , such that lim u(z) = f (ζ )

(1.1)

z→ζ

at every point of continuity ζ of f . Moreover, sup |u| ≤ sup | f |.

∂

Proof. We may assume is bounded. The uniqueness of u f is immediate from Lindelöf’s maximum principle, Lemma I.1.1. The existence of u f in case p = 1 was treated in Section I.3, and so we assume p > 1. Take a Jordan arc σ with endpoints a, b ∈ / such that 1 = \ σ is simply!connected and such that σ ∩ ∂ is a ﬁnite set. See Figure II.1. Then ϕ(z) = z−a z−b has a single valued analytic branch deﬁned on 1 , and we can solve the Dirichlet problem on 1 by transplanting it to the Jordan region ϕ(1 ). Take a second Jordan arc σ such that 1 = \ σ is simply connected, σ ∩ ∂ is a ﬁnite set, and σ ∩ σ = ∅. We can also solve the Dirichlet problem on 1 .

a

a

σ

b

σ

b

Figure II.1 The proof of Theorem 1.1. Let E ⊂ ∂ be a ﬁnite set and set F = E ∪ (σ ∩ ∂) ∪ (σ ∩ ∂). Suppose without loss of generality that f ∈ C(∂ \ E) is positive and bounded. To start, let u 1 be the solution to the Dirichlet problem on 1 with boundary data

f (ζ ), ζ ∈ ∂ u 1 (ζ ) = max∂ f, ζ ∈ σ . Then u 1 is harmonic on 1 and continuous on \ F and u 1 matches its boundary data on ∂1 \ F. Next let u 1 be the solution to the Dirichlet problem on 1 with boundary data u 1 (ζ ), ζ ∈ ∂1 . Then u 1 is harmonic on 1 and continuous on \ F and u 1 matches its boundary data on ∂1 \ F. In particular, u 1 = u 1 on ∂ \ F. By the Lindelöf maximum principle we have u 1 ≤ max∂ f = u 1 on σ ∩ , and therefore u 1 ≤ u 1 on \ F. Now let u 2 be the solution to the Dirichlet problem on 1 with boundary data u 1 (ζ ), ζ ∈ ∂1 . On σ ∩ we

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have u 2 = u 1 ≤ u 1 , while on ∂ \ F we have u 2 = f = u 1 . Therefore u 2 ≤ u 1 on 1 , again by Lemma I.1.1. Consequently u 2 ≤ u 1 = u 1 on σ ∩ , and by the maximum principle, u 2 ≤ u 1 on \ F. Continuing in this way we obtain a decreasing sequence u 1 ≥ u 1 ≥ u 2 ≥ u 2 ≥ u 3 . . . of positive functions, alternately harmonic on 1 and 1 . By Harnack’s principle, Exercise I.5, the limit u(z) = lim u n (z) = lim u n (z) n→∞

n→∞

is a bounded harmonic function on such that u ≤ u 1 ≤ max∂ f . To complete the proof we show lim u(z) = f (ζ )

z→ζ

(1.1)

whenever ζ ∈ ∂ is a point of continuity of f . We may assume ζ ∈ / σ ∪ E. Take a neighborhood V of ζ such that W = V ∩ is a Jordan domain and such that W ∩ (E ∪ σ ) = ∅. Let ϕ be a conformal map of D onto W. By Carathéodory’s theorem ϕ(∂D) = ∂ W, and for w = ϕ(z) ∈ W 1 − |z|2 dθ u n (w) = f ◦ ϕ(eiθ ) iθ − z|2 −1 |e 2π ϕ (∂) +

∂ D\ϕ −1 (∂)

1 − |z|2 dθ u n ◦ ϕ(eiθ ) . iθ 2 |e − z| 2π

Because ϕ −1 (∂) is a neighborhood of ϕ −1 (ζ ) in ∂D, the ﬁrst integral approaches f (ζ ) as w → ζ. Because |u n | ≤ sup∂ | f |, the second integral tends to zero, uniformly in n, as w → ζ. Therefore (1.1) holds and the theorem is proved. If is a ﬁnitely connected Jordan domain, Theorem 1.1 shows that the map f → u f (z) is a bounded linear functional on C(∂). Deﬁne the harmonic measure of a relatively open subset U ⊂ ∂ by ω(z, U ) ≡ ω(z, U, ) = sup{u f (z) : f ∈ C(∂), 0 ≤ f ≤ χ U }, and for an arbitrary subset E of ∂, set ω(z, E) ≡ ω(z, E, ) = inf{ω(z, U ) : U open in ∂, U ⊃ E}.

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This procedure, which mimics the usual proof of the Riesz representation theorem, shows ω(z, E) is a Borel measure on ∂ such that f (ζ )dω(z, ζ ) (1.2) u f (z) = ∂

for continuous f . When is simply connected, this deﬁnition of harmonic measure agrees with the earlier deﬁnition (I.3.3). For every z 1 , z 2 ∈ there exists, by virtue of Harnack’s inequality, a constant c = c(z 1 , z 2 ) such that 1 (1.3) ω(z 1 , E) ≤ ω(z 2 , E) ≤ cω(z 1 , E), c and the constants c(z 1 , z 2 ) remain uniformly bounded if z 1 and z 2 remain in a compact subset of . If f ∈ L 1 (∂, dω), and in particular if f is bounded and Borel, there is a sequence { f n } in C(∂) such that for some ﬁxed z 0 ∈ , | f n (ζ ) − f (ζ )| dω(z 0 , ζ ) → 0. Write

u n (z) =

and

f n (ζ )dω(z, ζ )

u(z) = u f (z) =

f (ζ )dω(z, ζ ).

(1.4)

Then by (1.3) u n (z) → u(z) for all z ∈ . By (1.3) we also see that the harmonic functions u n (z) are uniformly bounded on compact subsets of . Thus by Harnack’s principle the limit function u(z) is harmonic on . We call u the solution to the Dirichlet problem for f on . If f is bounded, then we also have sup |u f (z)| ≤ || f || L ∞ (,dω) . z∈

Moreover, if f is bounded and continuous at ζ ∈ ∂, then (1.1) also holds at ζ . See Theorem 2.7 or Exercise 1(b) for a proof. Note however that the Schwarz alternating method cannot be applied directly to a bounded Borel function because the hypothesis of Lindelöf’s maximum principle holds only for piecewise continuous functions. The next section will give a much more explicit description of the measure ω(z, E) when ∂ has some additional smoothness.

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2. Green’s Functions and Poisson Kernels Again 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 the boundary value f (ζ ) = log |ζ − w| ∈ C(∂), and deﬁne Green’s function with pole w to be g(z, w) = log

1 + h(z, w). |z − w|

(2.1)

Then g(z, w) is continuous in z ∈ \ {w}, and g(z, w) > 0

on ,

(2.2)

g(ζ, w) = 0

on ∂,

(2.3)

z → g(z, w) is harmonic on \ {w}, 1 z → g(z, w) − log is harmonic at w. |w − z|

(2.4) (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 the Dirichlet problem on for f (ζ ) = log ζζ−w −a , and deﬁne z−a + h(z, w). g(z, w) = log z − w 1 to deﬁne h(z, ∞) and set For w = ∞, we instead use f (ζ ) = log ζ −a g(z, ∞) = log |z − a| + h(z, ∞). These deﬁnitions are independent of the choice of a, and with them (2.2)–(2.5) still hold and determine g(z, w). Suppose ϕ is a conformal mapping from one ﬁnitely connected Jordan domain onto another ﬁnitely connected Jordan domain . Then ϕ(z) → ∂ whenever z → ∂, because ϕ : → is a homeomorphism. It follows that g (ϕ(z), ϕ(w)) = g (z, w) because Green’s functions are characterized by (2.3), (2.4), and (2.5). If is the unit disc D then clearly 1 − zw . g(z, w) = log z−w

(2.6)

(2.7)

Consequently Green’s function for any simply connected Jordan domain can be expressed in terms of the conformal mapping ψ : → D.

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Theorem 2.1. Let be a simply connected domain bounded by a Jordan curve, let w ∈ and let ψ : → D be a conformal map with ψ(w) = 0. Then g(z, w) = − log |ψ(z)|.

Proof. Immediate from (2.6) and (2.7).

By deﬁnition, an analytic arc is the image ψ((−1, 1)) of the open interval (−1, 1) under a one-to-one and analytic map ψ deﬁned on a neighborhood of (−1, 1). 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 a ﬁnitely connected Jordan domain ∗ such that ∂∗ consists of ﬁnitely many pairwise disjoint analytic Jordan curves and there exists a conformal map from onto ∗ which extends to be a homeomorphism from to ∗ .

2 ψ3 ◦ ψ2 ◦ ψ1

3

∗

1 ψ1

ψ3

ψ1 (2 ) ψ1 (3 )

ψ2

Figure II.2 The proof of Lemma 2.2. Proof. Write ∂ = 1 ∪ . . . ∪ p , where each j is a Jordan curve. Let 1 be the component of C∗ \ 1 containing , where C∗ is the extended plane, and let ψ1 be a conformal map of 1 onto D. Let 2 be the component of C∗ \ ψ1 (2 ) containing ψ1 (), and let ψ2 be a conformal map of 2 onto D. Repeating this process for each boundary curve, we obtain a conformal map ψ p from to a region ∗ such that ∂∗ consists of ﬁnitely many pairwise disjoint analytic

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Jordan curves. Applying Carathéodory’s theorem to each ψk , we see that ψ p extends to be a homeomorphism from to ∗ . In Exercise 5 the proof of Lemma 2.2 is used to deﬁne Green’s function in non-Jordan, ﬁnitely connected domains. Theorem 2.3. Let be a ﬁnitely connected Jordan domain and let z 1 , z 2 ∈ . Then g(z 1 , z 2 ) = g(z 2 , z 1 ).

(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 a neighborhood V of ∂ to which z → g(z, w) has a harmonic extension. Hence g(z, w) is C ∞ on some neighborhood V of ∂ and we can use Green’s theorem in the form ∂u ∂v (uv − vu)d xd y = −v ds, u ∂n ∂n ∂U U

where n is the unit normal vector pointing out from the domain U. Fix distinct z 1 , z 2 ∈ . We apply Green’s theorem on the domain ε = \ {|z − z 1 | ≤ ε} ∪ {|z − z 2 | ≤ ε} , when ε is small, with u(z) = g(z, z 1 ) and v(z) = g(z, z 2 ). Because u = v = 0 on ∂, ∂u ∂v −v ds = 0, u ∂n ∂n ∂ and because u and v are harmonic on ε , the area integral in Green’s theorem vanishes. We conclude that 2π ∂g dθ g(z 1 + εeiθ , z 1 ) (z 1 + εeiθ , z 2 ) ε ∂r 2π 0 2π ∂g dθ g(z 1 + εeiθ , z 2 ) (z 1 + εeiθ , z 1 ) −ε ∂r 2π 0 (2.9) 2π ∂g dθ iθ iθ =ε (g(z 2 + εe , z 2 ) (z 2 + εe , z 1 ) ∂r 2π 0 2π ∂g dθ g(z 2 + εeiθ , z 1 ) (z 2 + εeiθ , z 2 ) . −ε ∂r 2π 0

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For ε small, g(z j + εeiθ , z j ) ≤ 2 log (1/ε) and for k = j, g(z, z j ) has bounded derivatives near z k . That means the ﬁrst and third integrals in (2.9) approach 0 as ε → 0. By (2.1), ∂ log ε ∂g (z j + εeiθ , z j ) = + O(1), ∂r ∂ε so that, as ε → 0, the second integral tends to g(z 1 , z 2 ) and the fourth integral tends to g(z 2 , z 1 ). Therefore (2.8) holds. −

Lemma 2.4. Suppose is a ﬁnitely connected Jordan domain and suppose γ ⊂ ∂ is an analytic arc. Let u(z) be a harmonic function in . If lim u(z) = 0

z→ζ

for all ζ ∈ γ , then there is an open set W ⊃ γ ∪ such that u extends to be harmonic on W . If also u(z) > 0 in , then ∂u (ζ ) < 0 ∂n

(2.10)

for all ζ ∈ γ . u=0

D V

u>0

ζ

γ

v=0

ψ

v>0 Figure II.3 Straightening an analytic arc in ∂.

Proof. Let ζ ∈ . Because γ is an analytic arc, there is a neighborhood V of ζ and a conformal map ψ : V → D such that ψ(ζ ) = 0, ψ(V ∩ ) = D− = D ∩ {Imw < 0}, and ψ(γ ∩ V ) = (−1, 1).

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Set

⎧ u(ψ −1 (w)), ⎪ ⎪ ⎪ ⎪ ⎨ v(w) = −u(ψ −1 (w)), ⎪ ⎪ ⎪ ⎪ ⎩ 0,

45

w ∈ D− = D ∩ {Imw < 0}, w ∈ D+ = D ∩ {Imw > 0}, w ∈ (−1, 1).

Then v is continuous in D and v has the mean value property over sufﬁciently small circles centered at any w ∈ D. Hence v is harmonic in D and U = v ◦ ψ deﬁnes a harmonic extension of u to V . If U1 and U2 are extensions of u to neighborhoods V1 and V2 such that V1 ∩ V2 ∩ γ is connected, then U1 = U2 in the component of V1 ∩ V2 that contains V1 ∩ V2 ∩ γ . It follows that u has a harmonic extension to some open set W ⊃ γ ∪ . If u > 0 in , then clearly ∂u/∂n ≤ 0 on γ . The strict inequality (2.10) will

hold at ζ 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 is h(w) = an w n + O(|w n+1 |), 2, then h(D− ) ∩ D+ = ∅, which is a contradiction. with an = 0. But if n ≥

Hence a1 = h (0) = − ∂v/∂ y (0) = 0 and (2.10) holds. When ∂ consists of analytic curves, Green’s function provides a formula for harmonic measure that generalizes the Poisson integral formula for D. Theorem 2.5. Assume ∂ consists of ﬁnitely many pairwise disjoint analytic Jordan curves, and let z ∈ . Then Green’s function g(ζ, z) extends to be harmonic (and hence real analytic) on a neighborhood of ∂ and −∂g(ζ, z) >0 ∂nζ

(2.11)

on ∂, where nζ is the unit outer normal at ζ ∈ ∂. If u ∈ C() is harmonic on , then −∂g(ζ, z) ds(ζ ) u(z) = u(ζ ) . (2.12) ∂nζ 2π ∂ Because of (2.12), Pz (ζ ) = − is called the Poisson kernel for .

1 ∂g(ζ, z) 2π ∂nζ

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Proof. Fix z ∈ . By Lemma 2.4, g(ζ, z) extends to be harmonic (and real analytic) 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 ∂. We then apply Green’s theorem on ε = \ {w : |w − z| < ε} with ε small and v(w) = g(w, z). Because w g(w, z) = u = 0 on ε and because g = 0 on ∂, Green’s theorem collapses into 2π ∂g(ζ, z) ds(ζ ) ∂u(z + εeiθ ) dθ − u(ζ ) g(z + εeiθ , z) =ε ∂nζ 2π ∂r 2π ∂ 0 2π iθ ∂g(z + εe , z) dθ u(z + εeiθ ) . −ε ∂r 2π 0 Since g(z + εeiθ , z) ≤ 2 log 1/ε for small ε, and since u is C ∞ , we have 2π ∂u(z + εeiθ ) dθ lim ε g(z + εeiθ , z) = 0. ε→0 ∂r 2π 0 As we have seen, (2.1) yields ∂g(z + εeiθ , z) ∂ log ε =− + O(1), ∂r ∂ε while u(z + εeiθ ) = u(z) + O(ε). Hence 2π ∂g dθ −ε u(z + εeiθ ) (z + εeiθ , z) = u(z) + O(ε), ∂r 2π 0 and that gives (2.12) when u is C ∞ on a neighborhood of ∂. To prove (2.12) in general, set δ = {w ∈ : g(w, z) > δ} where δ is small. By uniqueness, δ has Green’s function gδ (w, z) = g(w, z) − δ. Therefore ∂g ∂gδ (ζ, z) = (ζ, z) ∂nζ ∂nζ g is a on ∂δ . In a neighborhood of a point ζ0 ∈ ∂, the function ϕ = g + i conformal map and 1 −∂g(ζ, z) u(ζ )ds = u ◦ ϕ −1 ds, 2π N ∩∂δ ∂nζ {Rez=δ}∩ϕ(N ) which converges to {Rez=0}∩ϕ(N )

u◦ϕ

−1

1 ds = 2π

∂

−∂g(ζ, z) u(ζ )ds. ∂nζ

Therefore 1 −∂g(ζ, z) −∂g(ζ, z) 1 u(ζ )ds → u(ζ )ds 2π ∂δ ∂nζ 2π ∂ ∂nζ

(2.13)

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as δ → 0. But (2.12) holds for u on δ because u is C ∞ on a neighborhood of ∂δ , and hence (2.13) yields (2.12) for . Corollary 2.6. If ∂ consists of ﬁnitely many pairwise disjoint analytic Jordan curves and if z ∈ , then dω(z, ζ ) = −

∂g(z, ζ ) ds(ζ ) . ∂nζ 2π

(2.14)

In other words, harmonic measure for z ∈ is absolutely continuous to arc length on ∂, the density dω 1 ∂g(z, ζ ) = Pz (ζ ) =− ds 2π ∂nζ is real analytic on ∂, and dω < c2 ds

c1 <

(2.15)

for positive constants c1 and c2 . Proof. The identity (2.14) follows from (1.2) and (2.12) and the fact that Borel measures are determined by their actions on continuous functions. The inequalities (2.15) are immediate from (2.11). One objective of this book is to compare harmonic measure for general domains to more geometrical measures such as arc length, and Corollary 2.6 is the ﬁrst result of this kind. Theorem 2.7. Assume ∂ consists of ﬁnitely many pairwise disjoint analytic Jordan curves and for ζ ∈ ∂ and α > 1 deﬁne α (ζ ) = z ∈ : |z − ζ | < α dist(z, ∂) . If u(z) is a bounded harmonic function on , then for ds almost all ζ ∈ ∂ the limit lim

α (ζ )z→ζ

exists,

u(z) = f (ζ )

(2.16)

u(z) =

∂

Pz (ζ ) f (ζ )ds(ζ ),

(2.17)

and sup |u(z)| = || f || L ∞ .

(2.18)

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Conversely, if f is a bounded Borel function on ∂, then (2.17) deﬁnes a bounded harmonic function u(z) on such that (2.18) holds and (2.16) holds ds almost everywhere. Moreover, if f is continuous at ζ0 ∈ ∂, then lim u(z) = f (ζ0 ).

z→ζ0

(2.19)

By (2.18), (2.16) and (2.17) establish an isometry between the space of bounded harmonic functions on and L ∞ (∂, ds) when ∂ consists of analytic curves. Proof. A simple localization argument gives the existence of the nontangential limit f . If I is an open arc on ∂, there exists a neighborhood V ⊃ I such that V ∩ ∂ = I and V ∩ is simply connected, and there exists a conformal mapping ψ deﬁned on V such that ψ(V ∩ ) = D and ψ(I ) is an arc on ∂D. It follows that ψ maps conical approach regions at ζ ∈ V ∩ ∂ into cones at ψ(ζ ):

ψ V ∩ α (ζ ) ∩ Bδ (ζ ) ⊂ β(α) (ψ(ζ )), where Bδ (ζ ) = {z : |z − ζ | < δ = δ(ζ )}. Then if u(z) is a bounded harmonic function on , we can apply Fatou’s theorem to u ◦ (ψ −1 ) to obtain (2.16) ds almost everywhere on V ∩ ∂. The proof of (2.17) is exactly the same as the proof of (2.12) except that the dominated convergence theorem is applied in (2.13). By (2.16) we have || f ||∞ ≤ sup |u(z)|, and because Pz ≥ 0 and ∂ Pz ds = 1, we have sup |u(z)| ≤ || f ||∞ . Hence (2.16) and (2.17) imply (2.18). To prove the converse, let f ∈ L ∞ (∂, ds). Then the discussion following (1.4) shows (2.17) deﬁnes a bounded harmonic function u(z) on and sup |u(z)| ≤ || f ||∞ . Therefore by (2.16), u has almost everywhere a nontangential limit, which we will temporarily call F, and u is the Poisson integral of F. What we must prove is that F = f almost everywhere. Again let V be a neighborhood of an open arc I = V ∩ ∂ such that V ∩ is simply connected. For h ∈ L ∞ (∂, ds), deﬁne Pz (ζ )h(ζ )ds(ζ ) − Pz (ζ, V )h(ζ )ds(ζ ), vh (z) = ∂

I

where Pz (ζ, 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, lim z→ζ vh (z) = 0 for all ζ ∈ I , and hence by Lemma 2.4, vh extends to be harmonic in a neighborhood W of I which does not depend on h. Thus by Exercise I.5(e) or (3.5) below, if J is a compact subset of I and if ε > 0 then there is a neighborhood N of J , depending only on ||h||∞ and ε, so that |vh | < ε in N . Now take h n ∈ C(∂)

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so that h n converges to f in L 1 , and ||h n ||∞ ≤ || f ||∞ . For each z ∈ V ∩ N , vh n (z) converges to v f (z) and so |v f (z)| < ε. Because ε > 0 was arbitrary, we conclude that v f (z) → 0 as z → ζ ∈ J . But by Theorem I.1.3, F(ζ ) − f (ζ ) =

lim

α(ζ ) z→ζ

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 ζ0 ∈ I then I Pz (ζ, V ) f (ζ )ds(ζ ) is continuous at ζ0 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 of g(w, z), one can reﬁne the proof of Lemma I.2.2 and show that sup |u(z)| ≤ C(α, )Ms f (ζ ),

α (ζ )

(2.20)

where u is the Poisson integral (2.17) and where the maximal function Ms f (ζ ) is the supremum of the averages of f over arcs γ ⊂ ∂ with ζ ∈ γ : 1 | f |ds. Ms f (ζ ) = sup γ ζ (γ ) γ A variation on the covering lemma shows that Ms is weak-type 1-1, and an approximation, as in the proof of Theorem I.2.1, then yields (2.16) for the Poisson integral of f . This is the argument that must be used in the Euclidean spaces Rd , with d ≥ 3, and in other situations. With some care, the conformal mapping proof of (2.16) in the text can also be parlayed into a proof of the maximal estimate (2.20). See Exercise 9. Let be any ﬁnitely connected Jordan domain and let ϕ be the conformal map, given in Lemma 2.2, of onto a domain ∗ , where ∂∗ consists of analytic Jordan curves. Since ϕ is a homeomorphism of onto ∗ , harmonic measure can be transplanted from ∗ to via ϕ, just as it was in Section I.3 for simply connected Jordan domains. This gives an alternate but equivalent deﬁnition of harmonic measure for . In Section 4 we will consider two questions. Let be a ﬁnitely connected Jordan domain. Question 1. If ∂ has some degree of differentiability and if f ∈ C(∂) also has 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, ζ )/∂nζ exists on ∂ and that (2.14) and (2.15) still hold?

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

3. Conjugate Functions Let f ∈ L 1 (∂D) be real. For convenience we write f (θ ) for f (eiθ ). If u(z) is the Poisson integral of f on D, then u(z) is harmonic and real and there exists a unique harmonic function u (z) such that u (0) = 0 and F = u + i u is analytic on D. The function u is called the conjugate function or harmonic conjugate of u. The nontangential limit f (θ ) =

lim

α (eiθ )z→eiθ

u (z)

(3.1)

exists almost everywhere, and this has an easy proof from Fatou’s theorem: We may assume f ≥ 0, so that u (z)) G(z) = e−(u(z)+i

is bounded and analytic on D. By Corollary I.2.5, G has a nontangential limit G(eiθ ) almost everywhere. Since |G(eiθ )| = e− f (θ) and f ∈ L 1 , |G(eiθ )| > 0 almost everywhere. At such eiθ , G is continuous and non-zero on the cone u ) has a continuous extension to K = α (eiθ ). Consequently log G = −(u + i K ∩ {z : |G(z) − G(eiθ )| < 21 |G(eiθ |} and the limit (3.1) exists at eiθ . There is a close connection between conjugate functions and conformal mappings. If u is harmonic and if |u| < π/2, then z u )(ζ ) ei(u+i dζ ϕ(z) = 0

is a conformal map from D to a simply connected domain and u = arg ϕ . Indeed, if a = b ∈ D, then 1 ϕ (a + t[b − a])dt = 0, ϕ(b) − ϕ(a) = (b − a) 0

because Re(ϕ ) > 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|)}.

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51

u is not bounded. Then u is continuous on D by Carathéodory’s theorem, but For a second example see Exercise 11. The next two theorems get around the obstruction 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 < λ < π/2 there is a constant Cλ , depending only on λ, such that dθ eλ| f (θ)| ≤ Cλ . 2π (b) If f ∈ C(∂D), then for all λ < ∞ u (r eiθ ) dθ eλ sup < ∞. 2π 0

2π

u (r e eλ

0

iθ )

dθ ≤ sec λ. 2π

Then by (3.1) and Fatou’s lemma 2π dθ eλ f (θ) ≤ sec λ. 2π 0 By repeating this argument with − f (θ ), we then obtain part (a) with constant Cλ = 2 sec λ. To prove (b), ﬁx λ < ∞ and take a trigonometric polynomial p(θ ) =

N (an cos nθ + bn sin nθ ) n=0

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such that f − p∞ < π/(2λ). Then p (r eiθ ) =

N

r n (an sin nθ − bn cos nθ )

n=0

is bounded, while part (a) gives Bλ = sup

0

u − p)(r e eλ|(

iθ )|

dθ < ∞. 2π

Therefore, since | u | ≤ | p | + |(u − p)|, we have u (r eiθ )| dθ p ∞ eλ| < ∞, ≤ Bλ eλ sup 2π 0
and (b) is proved. Let 0 < α < 1. The Lipschitz class C α is

" f (θ + t) − f (θ )∞ α ∞ <∞ . C = f ∈ L (∂D) : sup tα t>0

Every f ∈ C α agrees almost everywhere with a function continuous on ∂D. The class C α is given the Lipschitz norm f C α = f ∞ + sup t>0

f (θ + t) − f (θ )∞ . tα

(3.2)

In a moment we shall prove Privalov’s theorem that f ∈ C α whenever f ∈ C α . In the Poisson integral formula π it e +z dt u(z) = Re f (t) , it 2π −π e − z the kernel (eit + z)/(eit − z) is analytic in z ∈ D and real at z = 0. Then the uniqueness of u shows π it e +z dt u(z) + i u (z) = F(z) = f (t) . (3.3) it 2π −π e − z The analytic function F(z) in (3.3) is called the Herglotz integral of f . Write ∂u ∂u ∇u = , ∂x ∂y when u is differentiable on an open plane set. If u is harmonic and bounded on D then u is the Poisson integral of some f ∈ L ∞ (∂D) and by (3.3) and the

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53

Cauchy–Riemann equations, |∇u(z)| = F (z) =

π

−π

eit f (t) dt . (eit − z)2 π

Therefore (as in Exercise I.5(e)) π dt 1 f ∞ ≤ 2(1 − |z|)−1 f ∞ . |∇u(z)| ≤ it − z|2 π |e −π

(3.4)

(3.5)

We will often use the following consequence of (3.5): If u(z) is harmonic and |u(z)| ≤ M on B(z, R), then sup |∇u| ≤ B(z, R2 )

4M . R

(3.6)

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

|∇u| = O (1 − |z|)α−1 . Theorem 3.2. Let 0 < α < 1, let f ∈ L ∞ (∂D) be real, and let u(z) be the Poisson integral of f . Then the following conditions are equivalent: (a) f ∈ C α . (b) f ∈ Cα.

(c) |∇u(z)| = O (1 − |z|)α−1 . (d) u ∈ C α (D); that is, for all z 1 , z 2 ∈ D,

|u(z 1 ) − u(z 2 )| = O |z 1 − z 2 |α . Moreover there is a constant C1 , independent of α, such that C1 || f ||C α , α(1 − α)

(3.7)

C1 (1 − |z|)α−1 || f ||C α , (1 − α)

(3.8)

|| f ||C α ≤ |∇u(z)| ≤ and sup

z 1 =z 2

$ # |u(z 1 ) − u(z 2 )| C1 1−α ≤ |∇u(z)| . sup (1 − |z|) |z 1 − z 2 |α α |z|<1

(3.9)

The equivalence (a) ⇐⇒ (b) was ﬁrst proved by Privalov [1916] who worked directly with the imaginary part of the integral (3.3); (a) ⇐⇒ (c) was proved by Hardy and Littlewood in [1931].

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Proof. Clearly (d) ⇒ (a) because if (d) holds, u is uniformly continuous on D. 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 Cauchy–Riemann equations. Assume (a) holds, that is, assume u is the Poisson integral of f ∈ C α . (3.8). By (3.5) we may assume |z| ≥ 1/2. Let eit0 = z/|z|. Because Weitprove it e /(e − z)2 dt = 0, (3.4) yields π eit ( f (t) − f (t )) dt 0 |∇u(z)| = , it − z)2 (e π −π so that

|∇u(z)| ≤ =

π

π

| f (t) − f (t0 )| dt |eit − z|2 π +

|t−t0 |<1−|z|

1−|z|≤|t−t0 |<π

.

The inequality 1 − |z| ≤ |eit − z|, valid for all t, gives 1−|z| | f (t) − f (t0 )| dt dt 2 f α tα ≤ it 2 2 |e − z| π (1 − |z|) 0 π |t−t0 |<1−|z| 2 f α = (1 − |z|)α−1 . π(1 + α) When 1 − |z| ≤ |t − t0 | < π , the inequality |t − t0 |2 ≤ c|eit − z|2 , valid with c independent of z, gives π | f (t) − f (t0 )| dt dt t α−2 ≤ 2c f α it − z|2 |e π π 1−|z|<|t−t0 |<π 1−|z| 2c f α ≤ (1 − |z|)α−1 . π(1 − α) Therefore we obtain (3.8) with 2 π 0<α<1

C1 = sup

c+

(1 − α) (1 + α)

=

2(c + 1) . π

If (c) holds, we may assume (1 − |z|)1−α |∇u(z)| ≤ 1. Let z j = r j eiθ j ∈ D, as in Figure II.4. We may assume |z j | ≥ 1/2 and δ = |z 1 − z 2 | ≤ 1/2. Set w j = (1 − δ)z j . Then |u(z 1 ) − u(z 2 )| ≤ |u(z 1 ) − u(w1 )| + |u(z 2 ) − u(w2 )| + |u(w1 ) − u(w2 )|.

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3. Conjugate Functions

However,

|u(z j ) − u(w j )| =

while

rj (1−δ)r j

1 ∂u iθ j δα (1 − t)α−1 dt ≤ (te )dt ≤ , ∂t α 1−δ

α−1 ≤ δα . |u(w1 ) − u(w2 )| ≤ |w1 − w2 |Max j 1 − |w j |

Therefore (d) and (3.9) hold.

z1 w1 = (1 − δ)z 1 w2 = (1 − δ)z 2

δ z2

eiθ1 eiθ2

Figure II.4 Now suppose (a) holds. Then (c) holds and |∇ u | = O((1 − |z|)α−1 ). Therefore (d) and (3.9) hold for u , and u extends continuously to ∂D where u has boundary value f . It follows that f ∈ C α because (d) implies (a). Finally since f = − f + u(0), it follows that (a) ⇐⇒ (b) and that (3.7) holds. Let k be a non-negative integer, let 0 ≤ α < 1 and let f ∈ C(∂D). We say f ∈ C k+α if f is k times continuously differentiable on ∂D and (d/dθ )k f ∈ C α if α > 0. If F(z) is analytic on D we say F ∈ C k+α (D) if F and its ﬁrst k derivatives F , F , · · · , F (k) extend continuously to D and if there is C such that |F (k) (z 1 ) − F (k) (z 2 )| ≤ C|z 1 − z 2 |α for all z 1 , z 2 ∈ D. Corollary 3.3. Assume k is a non-negative integer and assume 0 < α < 1. Let f ∈ C(∂D) be real and let F(z) = u(z) + i u (z) be the Herglotz integral of f . Then F ∈ C k+α (D) if and only if f ∈ C k+α (∂D). Proof. Because f = ReF, it is clear that f ∈ C k+α (∂D) if F ∈ C k+α (D). As-

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sume f ∈ C k+α . If f (θ ) has Fourier series f (θ ) ∼

∞

an einθ ,

−∞

then F has Taylor series F(z) = a0 + 2

∞

an z n .

n=1

Therefore d f /dθ , which has Fourier series ∞

df inan einθ , ∼ dθ −∞ has Herglotz integral i z F (z) and the corollary follows from the special case k = 0, which is Theorem 3.2. Theorem 3.2 fails when α = 1. The harmonic conjugate of a continuously differentiable function on ∂D need not have a continuous derivative, and the conjugate of a Lipschitz function, that is, a function satisfying f (θ + t) − f (θ ) ≤ M|t|, need not be a Lipschitz function. See Exercise 13. For the same reason, Corollary 3.3 is also false when α = 0 and k is a non-negative integer. However, conjugation does preserve the Zygmund class Z ∗ of continuous f on ∂D such that f (θ + t) + f (θ − t) − 2 f (θ )∞ sup < ∞. t t>0 The Zygmund class has norm f Z ∗ = f ∞ + sup t>0

When f ∈ Deﬁne

Z ∗,

f (θ + t) + f (θ − t) − 2 f (θ )∞ . t

we say f is a Zygmund function. 1 ∇2 u(z) = ∇u x = |u x x (z)|2 + |u yx (z)|2 2 π eit f (t) 2 = F (z) = dt , π −π (eit − z)3

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57

where F = u + i u is the Herglotz integral of f ∈ L 1 (∂D). If f ∈ C α , then (3.8) and (3.6), applied to u x in the disc B(z, 1−|z| 2 ), give us

∇2 u(z) = O (1 − |z|)α−2 . (3.10) Conversely, then integrating F along radii shows that

if (3.10) holds, ∇u(z) = O (1 − |z|)α−1 , and f ∈ C α . Thus (3.10) provides yet another characterization of C α functions, in terms of second derivatives. Zygmund functions have a similar characterization.

Theorem 3.4. Let f ∈ L ∞ (∂D) be real and let u(z) be the Poisson integral of f . Then the following are equivalent: (a) f ∈ Z ∗ . (b) f ∈ Z ∗ .

(c) ∇2 u(z) = O (1 − |z|)−1 . There is a constant C such that (3.11) f Z ∗ ≤ C f Z ∗ and 1 f ∗ ≤ || f ||∞ + sup (1 − |z|) ∇2 u(z) ≤ C f ∗ . Z Z C z∈D

(3.12)

Notice that by (3.12) and (3.10), Z ∗ ⊂ C α for all α < 1. In particular, if f are continuous. On the other hand, if f is Lipschitz, then f ∈ Z ∗ , then f and clearly f ∈ Z ∗ . Thus we have the increasing scale of spaces C 1 ⊂ Lipschitz ⊂ Z ∗ ⊂ C α ⊂ C, α < 1, where C 1 denotes the space of continuously differentiable functions and C 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| > 21 . Because eiθ dθ = 0 (eiθ − z)3 and f (−θ ) has Herglotz integral F(z), we have it

e f (t) + f (−t) − 2 f (0) 1 ∂ 2u (z) = ReF (z) = dt ∂x2 π (eit − |z|)3

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and ∂ 2u 2 (z) = ReF (z) ∂x 1 ≤ π

|t|≤1−|z|

f Z ∗ |t| c dt + (1 − |z|)3 π

1−|z|<|t|≤π

f Z ∗ dt |t|2

≤ C(1 − |z|)−1 f Z ∗ . 2 Unfortunately this trick does not us help with ImF (z) = ∂∂x∂uy . Instead,

we apply (3.6) to U = ∂ 2 u/∂ x 2 on B(z, (1−|z|) 2 ), getting ∂ 3u C f Z ∗ . ≤ 2 ∂ y∂ x (1 − |z|)2

An integration then yields |z| ∂ 2u f Z ∗ C f Z ∗ ds ≤ C (z) ≤ F (0) + . 2 ∂ y∂ x (1 − s) 1 − |z| 0 That proves (c), and the right inequality in (3.12). Now assume (c) holds. Then by (3.10), f ∈ C α and f and f are continuous. Fix θ and t with 0 < t ≤ π and set r = 1 − (t/π). Then f (θ + t) + f (θ − t) − 2 f (θ ) = f (θ + t) − u(r ei(θ+t) ) + f (θ − t) − u(r ei(θ−t) ) (3.13)

+ 2u(r eiθ ) − 2 f (θ ) + u(r e

i(θ+t)

) − u(r e ) iθ

+ u(r ei(θ−t) ) − u(r eiθ ).

ei(θ+t) eiθ r ei(θ+t) r eiθ r ei(θ−t) Figure II.5

ei(θ−t)

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59

Because 2 u(r ei(θ+t) ) + u(r ei(θ−t) ) − 2u(r eiθ ) ≤ |t|2 sup ∂ u (w) , 2 |w|=r ∂θ the last two terms on the right side of (3.13) are O(|t|) by (c). Then because iα lims→1 (1 − s) ∂u ∂s (se ) = 0 by (c), an integration by parts shows 1 ∂ 2u ∂u iα (1 − s) 2 (seiθ )ds + (1 − r ) (r eiα ). f (α) − u(r e ) = ∂s ∂r r Therefore when (c) holds, the sum of the ﬁrst three lines of the right side of (3.13) is t

u r (r ei(θ+t) ) + u r (r ei(θ−t) ) − 2u r (r eiθ ) + O(t) π ≤

t t u r (r ei(θ+t) ) − u r (r eiθ ) + u r (r ei(θ−t) ) − u r (r eiθ ) + O(t) π π

≤

∂ 2u 2t 2 sup (w) + O(t) ≤ C t. π |w|=r ∂r ∂θ

That proves (a), and the left-hand inequality in (3.12) can be seen by following the constants argument. in the previous Since ∇2 u = ∇2 u , it follows that (a) ⇐⇒ (b) and that (3.11) holds.

4. Boundary Smoothness Let be a Jordan domain with boundary and let ϕ be a conformal mapping from D onto , so that ϕ extends to a homeomorphism from ∂D to the Jordan curve = ∂. In this section we examine the connection between the smoothness of and the differentiability of ϕ on ∂D. When has some degree of smoothness, we also study the relation between the differentiability of f ∈ C() and the differentiability of its solution u f to the Dirichlet problem at points of . The results do not depend on the choice of the mapping ϕ : D → because any other such map has the form ϕ ◦ T, with T ∈ M. We ﬁrst show that the smoothness of ϕ in a neighborhood of ϕ −1 (ζ ) depends only on the smoothness of in a neighborhood of ζ .

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Theorem 4.1. Let 1 and 2 be Jordan domains such that 1 ⊂ 2 and let γ ⊂ ∂1 ∩ ∂2 be an open subarc. Let ϕ j be a conformal map of D onto j . Then ψ = ϕ2−1 ◦ ϕ1 has an analytic continuation across ϕ1−1 (γ ), and ψ = 0 on ϕ1−1 (γ ). Proof. The analytic function ψ = ϕ2−1 ◦ ϕ1 from D into D has a continuous and unimodular extension to the arc ϕ1−1 (γ ). By Schwarz reﬂection ψ has an analytic and one-to-one extension to a neighborhood of ϕ1−1 (γ ) in C and hence ϕ = 0 on ϕ1 −1 (γ ). Let be an arc parameterized as {ζ (t) : a < t < b}. We say has a tangent at ζ0 = ζ (t0 ) if ζ (t) − ζ0 = eiτ |ζ (t) − ζ0 |

(4.1)

ζ (t) − ζ0 = −eiτ . |ζ (t) − ζ0 |

(4.2)

lim t↓t0

and lim t↑t0

If (4.1) and (4.2) hold then has unit tangent vector eiτ at ζ0 . Except for reversals of orientation, the existence of a tangent at ζ0 and its value eiτ do not depend on the choice of the parameterization t → ζ (t). We say has a continuous tangent if has a tangent at each ζ ∈ and if eiτ (ζ ) is continuous on . Theorem 4.2. The curve has a tangent at ζ = ϕ(eiθ ) if and only if the limit ϕ(z) − ζ (4.3) lim arg z − eiθ Dz→eiθ exists and is ﬁnite. In that case, ϕ(z) − ζ π = τ (ζ ) − θ − lim arg iθ iθ z−e 2 Dz→e

(modulo 2π ).

(4.4)

The curve has a continuous tangent if and only if argϕ (z) has a continuous extension to D. If has a continuous tangent, then ϕ ∈ C α (∂D) for all α < 1 and is rectiﬁable. There exist Jordan domains and conformal maps ϕ : D → such that ∂ has a continuous tangent but ϕ ∈ C 1 (∂D). An example can be built from the connection in Section 3 between conjugate functions and conformal mappings: If u(eiθ ) is continuous and |u| < π2 , then u(eiθ ) = arg ϕ (eiθ ), where ϕ is a conformal mapping onto a Jordan domain, but u (eiθ ) = − log |ϕ (eiθ )| may not be bounded above or below.

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Proof. Set

61

ϕ(z) − ζ . v(z) = arg z − eiθ

Then v(z) is harmonic on D and v is continuous on D \ {eiθ }. If has a tangent at ζ = ϕ(eiθ ) or if the limit (4.3) exists at eiθ , then v(z) is bounded on D. Thus in either case v is the Poisson integral of v(eiθ ). Therefore v has a continuous extension to eiθ if and only if v|∂ D has a continuous extension to eiθ , and if and only if has a tangent at ζ. Clearly, (4.4) holds when v is continuous at eiθ . Now suppose has a continuous tangent. Then τ ◦ ϕ is continuous on ∂D. If h = 0, then ϕ(zei h ) − ϕ(z) Ah (z) = arg z(ei h − 1) is continuous on D and harmonic on D. For |z| < 1, lim Ah (z) = argϕ (z).

h→0

For |z| = 1 there exists k, 0 < |k| < |h|, such that ϕ(zei h ) − ϕ(z) = arg ϕ(zei h ) − ϕ(z) = τ (ϕ(zeik )), arg h by the proof of the mean value theorem from calculus. Consequently π lim Ah (z) = τ (ϕ(z)) − argz − , h→0 2 with the convergence uniform on ∂D, and hence arg(ϕ ) is the Poisson integral of the continuous function π τ (ϕ(eiθ )) − θ − . 2 Now suppose argϕ (z) has a continuous extension to D. For r < 1, the curve iθ r = {ϕ(r eiθ ) : 0 ≤ θ ≤ 2π } has tangent eiτ (ϕ(r e )) satisfying eiτ (ϕ(r e

iθ ))

= ieiθ eiargϕ (r e ) , iθ

and eiτ (ϕ(z)) has a continuous extension to D \ {0}. Then for h > 0 arg ϕ(ei(θ+h) ) − ϕ(eiθ ) = lim arg ϕ(r ei(θ+h) ) − ϕ(r eiθ ) r →1

= lim τ (ϕ(r ei(θ+kr ) )), r →1

where 0 < kr < h, again by the proof of the mean value theorem. Therefore lim arg ϕ(ei(θ+h) ) − ϕ(eiθ ) = lim τ (ϕ(z)), h↓0

z→eiθ

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and a similar argument for h < 0 yields lim arg ϕ(ei(θ+h) ) − ϕ(eiθ ) = lim τ (ϕ(z)) + π. h↑0

z→eiθ

Thus has a continuous tangent. If has a continuous tangent, then by part (b) of Theorem 3.1, iθ iθ λ sup |ϕ (r e )| dθ = sup e−λ(arg ϕ )(r e ) dθ = Bλ < ∞ r <1

r <1

for all λ < ∞, where Bλ depends on λ and ϕ. Take λ = 1/(1 − α), where 0 < α < 1. Let a < b < a + π . Then for any r < 1, Hölder’s inequality gives 1−α b b |ϕ (r eiθ )|dθ ≤ |b − a|α |ϕ (r eiθ )|λ dθ ≤ |b − a|α Bλ1−α . a

a

Therefore if a = θ0 < θ1 < . . . < θn = b, then b n |ϕ(r eiθ j ) − ϕ(r eiθ j−1 )| ≤ |ϕ (r eiθ )|dθ ≤ |b − a|α Bλ 1−α , j=1

(4.5)

a

and letting r tend to 1, we see that is rectiﬁable and that ϕ ∈ C α .

Let k be a non-negative integer and let 0 ≤ α < 1. We say the curve is of class C k+α if is rectiﬁable and if, in the arc length parameterization = {γ (s) : 0 ≤ s ≤ () = length ()}, the function γ is k times continuously differentiable and d k γ /ds k ∈ C α if α > 0. Theorem 4.3 (Kellogg). Let k ≥ 1 and 0 < α < 1. Then the following conditions are equivalent: (a) is of class C k+α . (b) arg ϕ ∈ C k−1+α (∂D). (c) ϕ ∈ C k+α (D) and ϕ = 0 on D. Note that if α = 0 and k ≥ 1 then (a) does not imply (c) but (a) is still equivalent to (b). We need the elementary lemma. Lemma 4.4. Let k be a positive integer, and let 0 ≤ α < 1. Let f ∈ C 1 ([0, 1]) satisfy f > 0 and let g ≡ f −1 . Then g ∈ C k+α if and only if f ∈ C k+α . Proof. The case α = 0 and k = 1 is clear since g (y) = 1/ f (g(y)) > 0. If α = 0 and k ≥ 2, the proof is by induction: if f ∈ C k and g ∈ C k−1 , then f ◦ g ∈ C k−1 and because f > 0, g = 1/( f ◦ g) ∈ C k−1 . Hence g ∈ C k . Now suppose α > 0 and k = 1. If f ∈ C α then

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63

1 1 − |g (y1 ) − g (y2 )| = f (g(y1 )) f (g(y2 ))

≤

| f (g(y2 )) − f (g(y1 ))| min | f |2

≤ C|g(y2 ) − g(y1 )|α ≤ C |y2 − y1 |α , and so g ∈ C α . Finally, suppose α > 0 and k ≥ 2. If f ∈ C k+α , then g ∈ C k and g (k) can be written as a sum of products of the functions g (1) , . . . , g (k−1) , f (2) ◦ g, . . . , f (k) ◦ g. All these functions are C 1 , except perhaps f (k) ◦ g. But f (k) ◦ g ∈ C α , and thus g (k) ∈ C α . The converse follows from interchanging g and f . Proof of Theorem 4.3. If arg ϕ ∈ C k−1+α , 0 < α < 1, then by Corollary 3.3, log |ϕ | ∈ C k−1+α , and by taking exponentials ϕ ∈ C k−1+α (D) and ϕ = 0. Conversely, if ϕ ∈ C k−1+α (D) and ϕ = 0, then ei arg ϕ = ϕ /|ϕ | ∈ C k−1+α , and arg ϕ ∈ C k−1+α . Hence (b) and (c) are equivalent. Assume (c) holds. If θ |ϕ (eit )|dt, s(θ ) = 0

s

s (θ )

then > 0 and 4.4, and by (4.4)

=

arg

|ϕ (eiθ )|

∈ C k−1+α . Thus θ (s) ∈ C k−1+α by Lemma

π dγ = arg ϕ (eiθ(s) ) + + θ (s). ds 2

(4.6)

Since arg ϕ ∈ C k−1+α and θ ∈ C k−1+α , we conclude from Corollary 3.3 that dγ /ds ∈ C k−1+α and that is of class C k+α . Therefore (a) holds. Now assume (a) holds, that is, assume is of class C k+α . Then by Theorem 4.2, arg ϕ ∈ C and by (4.6), dγ (s(θ ))/ds ∈ C. If k = 1, so that dγ /ds ∈ C α , then dγ (s(θ1 )) − dγ (s(θ2 )) ≤ C|s(θ1 ) − s(θ2 )|α (4.7) ds ds so that by (4.5)

dγ (s(θ1 )) − dγ (s(θ2 )) ≤ C|θ1 − θ2 |(1−ε)α ds ds

for any ε > 0. Thus arg ϕ ∈ C (1−ε)α and by Corollary 3.3, ϕ ∈ C (1−ε)α . Thus s (θ ) = |ϕ (eiθ )| ∈ C (1−ε)α and |s(θ1 ) − s(θ2 )| ≤ K |θ1 − θ2 |. But then

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by (4.6) and (4.7), arg ϕ ∈ C α . That proves (b) when k = 1. Moreover, by Theorem 3.2 we have s (θ ) = |ϕ (eiθ )| ∈ C α , and s ∈ C 1+α . If k ≥ 2, we use induction. If dγ /ds ∈ C k−1+α and if s(θ ) ∈ C k−1+α , then by (4.6), arg ϕ ∈ C k−1+α , so that s = |ϕ | ∈ C k−1+α , again by Theorem 3.2, and s ∈ C k+α . That gives (b) in general. Let l be a non-negative integer and let 0 ≤ β < 1. If is of class C k+α and if f ∈ C() we say f ∈ C l+β () if f (γ (s)) ∈ C l+β , when viewed as a function of arc length on . Corollary 4.5. Suppose is of class C k+α , where k + α > 1, and suppose f ∈ C l+β (). Set n + σ = min(k + α, l + β), where 0 < σ < 1 and n is a non-negative integer. Let ϕ be a conformal map of D onto , let G be the Herglotz integral of f ◦ ϕ, and let F = G ◦ ϕ −1 . Then F ∈ C n+σ (). Proof. Use Corollary 3.3, Theorem 4.3, and Lemma 4.4.

The same result holds for ﬁnitely connected Jordan domains whose boundary curves are of class C k+α , except that conjugate functions and Herglotz integrals cannot be deﬁned in multiply connected domains. Corollary 4.6. Let ∂ be a ﬁnite union of pairwise disjoint C k+α Jordan curves, where k + α > 1, and let f ∈ C(∂) be a C l+β function of arc length on each component of ∂. Set n + σ = min(k + α, l + β), where 0 < σ < 1 and n is a non-negative integer. Then u(z) = u f (z) and its ﬁrst n partial derivatives extend continuously to and |D n u(z 1 ) − D n u(z 2 )| ≤ K |z 1 − z 2 |σ , for all z 1 , z 2 ∈ , where D n denotes any n-th partial derivative. Proof. By Theorem 4.1 it is enough to work in some neighborhood of ζ ∈ ∂. Let J be the component of ∂ such that ζ ∈ J . Let 1 be that component of C∗ \ J such that ⊂ 1 , and let u 1 be the solution to the Dirichlet problem on 1 with boundary value f . Near J , u 1 has the required smoothness by Corollary 4.5. If ϕ1 is a conformal map of D on 1 , then v = (u − u 1 ) ◦ ϕ1 is harmonic on A = {r < |z| < 1}, for some r < 1, v is continuous on A, and v = 0 on {|z| = 1}. By reﬂection v extends to be harmonic across ∂ D. Hence v ∈ C ∞ and u = u 1 + v ◦ ϕ1−1 has as much smoothness as u 1 and ϕ1 both have.

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Corollary 4.6 answers Question 1 from the end of Section 2. The next corollary answers Question 2. Corollary 4.7. If ∂ consists of ﬁnitely many pairwise disjoint Jordan curves of class C 1+α , where α > 0, then dω(z, ζ ) = −

∂g(z, ζ ) ds(ζ ) . ∂nζ 2π

(4.8)

In other words, harmonic measure for z ∈ is absolutely continuous to arc length on ∂, and the density dω 1 ∂g(ζ, z) =− ds 2π ∂nζ is of class C α (∂) and satisﬁes c1 <

dω < c2 ds

(4.9)

for positive constants c1 and c2 . Proof. Let ϕ be a conformal map from ∗ onto , where ∂∗ consists of analytic Jordan curves. If ζ = ϕ(ζ ∗ ) and z = ϕ(z ∗ ), then 1 ∂g∗ (ζ ∗ , z ∗ ) ∂g (ζ, z) = ∂nζ ∂nζ ∗ |ϕ (ζ ∗ )| and |ϕ (ζ ∗ )| =

ds(ζ ) , ds(ζ ∗ )

by Corollary 4.6 and the uniqueness of Green’s function. Since harmonic measure is conformally invariant, (4.8) now follows from the case when ∂ is analytic, as in Corollary 2.6. By Corollary 4.6, ∂g/∂nζ ∈ C α (∂). And (4.9) holds because |ϕ | > 0 on ∂∗ , by Theorem 4.3. This answers Question 2 from Section 2. When ∂ is of class C 1 , harmonic measure is absolutely continuous to arc length, but the density may not be continuous, bounded or bounded below. See Exercise 14. Exercise 20 shows how Green’s theorem can be applied when ∂ is of class 1+α , without ﬁrst mapping to a domain with real analytic boundary. C

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Notes This chapter is based on Lennart Carleson’s 1985 lectures at UCLA. The conformal maps in Figure II.2 were computed using the Zipper algorithm from Marshall [1993]. Stein [1970] gives a good introduction to the scale of Lipschitz spaces beyond C α and Z ∗ . For more complete discussions see also Adams [1975], Adams and Hedberg [1996], Bennett and Sharpley [1988], Stein [1993], and Triebel [1983] and [1992]. See Zygmund [1959] for the classical picture. Theorem 4.3 and its corollaries are due to Kellogg [1929]. Other boundary problems are discussed in Appendix B and a different approach, through the Dirichlet principle, is presented in Appendix C.

Exercises and Further Results 1. Suppose f is real valued, bounded, and Borel measurable on ∂, where is a ﬁnitely connected Jordan domain and let u = u f be the solution to the Dirichlet problem on for f given by (1.4). (a) Prove lim inf f (ζ ) ≤ lim inf u(z) ≤ lim sup u(z) ≤ lim sup f (ζ ).

∂ζ →ζ0

z→ζ0

z→ζ0

∂ζ →ζ0

(b) If f is continuous at ζ ∈ ∂, prove lim z→ζ u(z) = f (ζ ). Hint: Use Theorem I.1.3, Carathéodory’s theorem, and Theorem 1.1 applied to a piecewise constant function. 2. A numerical routine for solving the Dirichlet problem in a ﬁnitely connected Jordan domain bounded by C 1+α curves can be based on the following construction: With u n and u n+1 as the proof of Theorem 1.1, ﬁnd functions h and k, explicitly in terms of the data f and Green’s functions g and g of the simply connected domains 1 and 1 , such that ku n ds on σ . u n+1 = h + σ

It follows that u satisﬁes the integral equation kuds on . u=h+ σ

3. (a) Let be a ﬁnitely connected Jordan domain, let E be a ﬁnite union of open arcs in ∂ and let f (z) be a bounded analytic function on . If | f (z)| ≤ M,

z ∈ ,

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and lim sup | f (z)| ≤ m,

ζ ∈ E,

z→ζ

then | f (z)| ≤ m ω M 1−ω ,

z ∈ ,

where ω = ω(z, E, ). Hint: Apply Exercise I.3(b) with log | f (z)|. (b) On the annulus {r1 < |z| < r2 } show the circle {|z| = r2 } has harmonic measure log |z| − log r1 . log r2 − log r1 (c) (three circles theorem) Suppose that f (z) is analytic on the annulus {r1 < |z| < r2 }, and suppose that m j = sup lim sup | f (z)| |ζ |=r j

z→ζ

is ﬁnite for j = 1, 2. Then λ(r ) 1−λ(r ) , sup | f (z)| ≤ m 1 m 2

|z|=r

where λ(r ) =

log r2 − log r . log r2 − log r1

(d) (Lindelöf) Suppose f (z) is bounded and analytic on the disc D. Let γ be an arc in D terminating at a point ζ ∈ ∂D, and assume lim

γ z→ζ

f (z) = a.

Then f (z) has nontangential limit a at ζ. Hint: Take a = 0. If is a nontangential cone terminating at ζ , then lim inf z∈∂ ω(z, γ ) > 0. Then (a) can be used. (e) (Lindelöf) Suppose f (z) is bounded and analytic on D. If the nontangential limit f (eiθ ) is continuous in some deleted arc (α − δ, α) ∪ (α, α + δ) and if the two limits lim f (ei(α+θ) ) θ↑0

and lim f (ei(α+θ) ) θ↓0

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both exist, then these two limits are equal. (f) Let ϕ be a univalent function on D. If σ is an arc in D with endpoint ζ ∈ ∂D and if lim ϕ(z) = w,

σ z→ζ

then ϕ has nontangential limit w at ζ. 4. Suppose is a ﬁnitely connected Jordan domain, and suppose n are ﬁnitely connected Jordan domains bounded by disjoint C 1+α Jordan curves such that n ⊂ n+1 ⊂ and n = . Fix z ∈ and let dωn denote harmonic measure for z with respect to the region n . Prove that dωn converges weakstar to dω where dω is harmonic measure for z with respect to . In other words, if f ∈ C(), then prove lim f dωn = f dω. n

∂n

∂

Hint: If u f is the harmonic extension of f |∂ to , lim (u f − f )dωn = 0 and thus any weak-star cluster point dμ of dωn satisﬁes ∂ f dμ = u f (z). So by the Riesz representation theorem and (1.2), dμ = dω. 5. Let be a ﬁnitely connected domain such that ∂ has no isolated points. (a) Prove there is a conformal map ϕ of onto a domain ∗ bounded by ﬁnitely many pairwise disjoint analytic Jordan curves. (b) Deﬁne g (z, w) = g∗ (ϕ(z), ϕ(w)), and verify (2.2), (2.4), (2.5), and lim

z→ζ ∈∂

g (z, w) = 0.

(2.3 )

(c) Then (2.3 ), (2.4), and (2.5) uniquely determine g and this deﬁnition of g is independent of the choice of ∗ and ϕ. (d) Using the preceding results and the Riemann mapping theorem, derive the conclusion of Theorem 2.1 for any simply connected domain not bounded by a Jordan curve. 6. Let be a simply connected plane domain and assume that there exists a (Green’s) function g (z, w) on × \ {z = w} satisfying (2.3 ), (2.4), and (2.5). Let u(z) = g(z, w) + log |z − w|. Prove the Riemann mapping theorem by showing that u (z)) ϕ(z) = ϕ(z, w) = (z − w)e−(u(z)+i

deﬁnes a conformal mapping from onto D. Hint: To show ϕ is one-to-one, compare − log |T ◦ ϕ(z, w)| to g(z, w ) for some T ∈ M. 7. (a) If is a doubly connected Jordan domain, if E is one of the boundary

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components, and if ω = ω(z, E, ), show that for some choice of a > 0, ω) ϕ = ea(ω+i

deﬁnes a single valued conformal map of onto an annulus. (b) Let be a domain such that C∗ \ has exactly two components. Prove there is a conformal mapping from onto a ring {z : r1 < |z| < r2 } where 0 ≤ r1 < r2 ≤ ∞. (c) Let be a ﬁnitely connected Jordan domain and suppose z 0 ∈ . Set u ε (z) = log(1/ε)ω(z, Bε , \ Bε ), where Bε = {z : |z − z 0 | ≤ ε}. Then |u ε − g (z, z 0 )| ≤ Cω(z, Bε , \ Bε ), and hence u ε → g (z, z 0 ), uniformly on compact subsets of \ {z 0 }. (d) Let be a ﬁnitely connected Jordan domain bounded by C 1+α curves. The period of the conjugate g (z, z 0 ) of Green’s function around a component j of ∂ is equal to the harmonic measure of j at z 0 . 8. (a) Let u(z) be the Poisson integral of f ∈ L 2 . Use Green’s theorem to show 2 1 1 | f − f (0)|2 dθ = |∇u(z)|2 log d xd y 2π π |z| D 2 ≥ |∇u(z)|2 (1 − |z|)d xd y π D 4 1 ≥ |∇u(z)|2 log d xd y. 3π |z| D

Hints: Replace f by u(r eiθ ), use the identity (u 2 ) = 2|∇u|2 and Green’s theorem, and send r → 1. The ﬁnal inequality is a direct calculation using polar coordinates and Fourier series. The identity on the left can also be proved with Fourier series. (b) If f ∈ H 2 (see Appendix A), then by (a) 2 1 1 | f − f (0)|2 dθ = | f (z)|2 log d xd y 2π π |z| D 2 ≥ | f (z)|2 (1 − |z|)d xd y π D 4 1 | f (z)|2 log d xd y. ≥ 3π |z|

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In particular, the two area integrals are comparable. 9. (a) In Theorem 2.7, verify (2.16) when f ∈ L 1 (ds). (b) Prove (2.20) by following the conformal mapping proof of (2.16) given in the text. 10. (a) Prove that every analytic Jordan curve is the image of ∂D under an analytic map that is one-to-one on a neighborhood of ∂D. (b) If is a domain in C∗ then we can ﬁnd subdomains n , each bounded by ﬁnitely many pairwise disjoint analytic Jordan curves so that n ⊂ n+1 and = n . Hint: Use the proof of Lemma 2.2. 11. (a) If f ∈ L 1 (∂D), and if u is the Poisson integral of f , then π 2r sin(ϕ − θ ) dθ f (eiθ ) . u (r eiϕ ) = 2 1 − 2r cos(ϕ − θ ) + r 2π −π (b) Almost everywhere on ∂D, f (ϕ) = lim

ε→0 |θ −ϕ|>ε

cot

ϕ−θ 2

f (eiθ )

dθ . 2π

(c) Deﬁne the modulus of continuity of the function f as ω f (δ) = sup{| f (θ ) − f (ϕ)| : |θ − ϕ| < δ}. The function f is called Dini continuous if ω f (t) dt < ∞. t 0 Prove that f is continuous if f is Dini continuous, in fact, δ π ω(t) ω(t) dt. dt + δ ω f (δ) ≤ t t2 0 δ h(r ) exists and (d) Let h(θ ) be odd and increasing on (−π, π]. Then limr →1 is ﬁnite if and only if h(θ ) dθ < ∞. 0 θ (e) Use (d) to show (c) is sharp. 12. Let be a simply connected domain. Write d(w) = dist(w, ∂) and let ρ(w, w0 ) denote the hyperbolic distance in . Let ϕ : D → be conformal. Then ϕ ∈ C α (∂D) if and only if 1 lim sup 2ρ(w, w0 ) + log d(w) < ∞ α d(w)→0 for any w0 ∈ . Hint: Use Theorem 3.2 and Theorem I.4.3. See Becker and Pommerenke [1982a].

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13. Corollary 3.3 is false when k = 1 and α = 0. Worse yet, there exists f ∈ C 1 (∂D) such that f is not a Lipschitz function. 14. There is a simply connected domain such that ∂ ∈ C 1 but such that no conformal map ϕ : D → is class C 1 on D. Worse yet, |ϕ | can have inﬁnite nontangential limit at some point on ∂D. 15. In Theorem 4.2, if has a continuous tangent, then ϕ ∈ H p for all ﬁnite p. In other words, |ϕ (r eiθ )| p dθ < ∞. sup 0
See Appendix A. 16. Let be a Jordan domain with C 1 boundary parameterized by arc length = {γ (s) : 0 < s < (γ )},

17.

18. 19.

20.

and let ϕ be a conformal map from D onto . (a) If γ is Dini continuous, then ϕ extends continuously to D and ϕ = 0. See Warschawski [1932a] and [1961]. (b) If γ ∈ C 1 and if γ is Dini continuous, then ϕ extends continuously to D. See Warschawski [1932a]. Let be a ﬁnitely connected Jordan domain with n + 1 boundary components 0 , 1 , . . . , n and let u(z) = ω(z, j , ). Prove that u(z) has n − 1 critical points in . In other words, prove the analytic function f (z) = u x (z) − iu y (z) has n − 1 zeros in , when counted with multiplicities. Hint: We may assume each k is an analytic curve. Then f = 0 on ∂ and since u is constant on k , the vector ∇u = f is parallel to the normal vector on k . Hence arg f decreases by 2π as k is traversed counterclockwise, and the argument principle then shows that f has n − 1 zeros in . Prove that conditions (a) and (b) of Theorem 4.3 are equivalent even when α = 0. (a) Formulate a notion of nontangential convergence for a ﬁnitely connected Jordan domain with C 1 boundary. Use it to show that ω(z, E, ) is the unique bounded harmonic function on with nontangential limits 1 on E and 0 on ∂ \ E, almost everywhere with respect to arc length on ∂. (b) If ∂ is C 1+ε for some ε > 0, then under the conformal mapping cones α (ζ ) correspond to cones α (ϕ(ζ )), and then (2.16) holds for f ∈ L 1 (ds). Green’s theorem can be used on a ﬁnitely connected domain bounded by a ﬁnite number of pairwise disjoint C 1+α curves with functions u and v in C 1 (∂) ∩ C 2 (). Hint: Use a ﬁnite number of Riemann maps and Theorem 4.3.

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21. More generally, Green’s theorem also holds for Lipschitz domains, which will be deﬁned in this exercise. A K -Lipschitz graph is a set of the form = {(x, β(x)) : |x| ≤ M}, where β is a real valued Lipschitz continuous function: |β(x 2 ) − β(x1 )| ≤ K |x2 − x1 |, for some constant K . A Lipschitz graph is rectiﬁable, and since β is absolutely continuous, arc length on has the form ! ds = 1 + (β (x))2 d x, and almost everywhere, (β (x), −1) n= % 1 + (β (x))2 is a vector normal to . For B > sup[−M,M] |β(x)|, set = β = {(x, y) : |x| < M, β(x) < y < B}. (a) Prove that Green’s theorem holds on in this form: Suppose u ∈ C 1 () and v ∈ C 2 (), suppose u, ∇u, v, ∇v, and v are bounded on , and suppose u and ∇v extend continuously to . Then

∂v uv + ∇u · ∇v d xd y = u ds. (E.1) ∂ ∂n Hint: Approximate by C ∞ subdomains as follows: Let χ ∈ C ∞ (−1, 1) satisfy χ ≥ 0 and 1 χ d x = 1. −1

1χ x (δ) ∞ χ δ ( δ ) and β (x) = δ ∗ β(x) ∈ C , ∇β (δ) (x) → ∇β(x) almost everywhere.

β (δ) (x) → β(x) uniIf χ δ (x) = Show there are δn ↓ 0 formly, and such that βn = δn + β (δn ) satisﬁes βn > βn+1 > β. Then derive (E.1) from Green’s theorem for βn . (b) A ﬁnitely connected Jordan domain is a Lipschitz domain if each p ∈ ∂ has a neighborhood V so that V ∩ ∂ is the image of a Lipschitz graph under a linear mapping. Prove Green’s theorem (E.1) holds for the Lipschitz domain under the same smoothness assumptions on u and v that we made in (a). That is, u ∈ C 1 ();and v ∈ C 2 (); u, ∇u, v, ∇v, and v are bounded on ; and u and ∇v extend continuously to .

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III Potential Theory

The goal of this chapter is to solve the Dirichlet problem on an arbitrary plane domain . There are three traditional ways to solve this problem: (i) The Wiener method is to approximate from inside by subdomains n of the type studied in Chapter II and to show that the harmonic measures ω(z, E, n ) converge weak-star to a limit measure on ∂. With Wiener’s method one must prove that the limit measure ω(z, E, ) does not depend on the approximating sequence n . (ii) The Perron method associates to any bounded function f on ∂ a harmonic function P f on . The function P f is the upper envelop of a family of subharmonic functions constrained by f on ∂. Perron’s method is elegant and general. With Perron’s method the difﬁculty is linearity; one must prove that P− f = −P f , at least for f continuous. (iii) The Brownian motion approach, originally from Kakutani [1944a], identiﬁes ω(z, E, ) with the probability that a randomly moving particle, starting at z, ﬁrst hits ∂ in the set E. This method has considerable intuitive appeal, but it leaves many theorems hard to reach. We follow Wiener and use the energy integral to prove that the limit ω(z, E, ) is unique. This leads to the notions of capacity, equilibrium distribution, and regular point and to the characterization of regular points by Wiener series. For the Perron method see Ahlfors [1979] or Tsuji [1959]. Appendix F below includes Kakutani’s theorem for the discrete version of Brownian motion. We conclude the chapter with some potential theoretic estimates for harmonic measure.

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1. Capacity and Green’s Functions Let E be a compact plane set such that = C∗ \ E is a ﬁnitely connected Jordan domain. By Chapter II and a conformal mapping, we see that has / Green’s function g (z, ∞) with pole at ∞, and if a ∈ g (z, ∞) = log |z − a| + h(z, ∞), where h(z, ∞) is harmonic on and continuous on ∂ and h(ζ, ∞) = − log |ζ − a|

ζ ∈ ∂.

(Recall that u(z) is harmonic at ∞ if u(1/z) is harmonic on a neighborhood of 0.) The quantity γ = γ (E) = h(∞, ∞), is called Robin’s constant for E, and we have g (z, ∞) = log |z| + γ + o(1),

as z → ∞.

(1.1)

Deﬁne the logarithmic capacity of E to be Cap(E) = e−γ (E) . Thus Cap(E) > 0 in the case at hand. Suppose 1 and 2 are ﬁnitely connected Jordan domains such that ∞ ∈ j and set E j = C∗ \ j . Assume there is a conformal map ψ of 1 onto 2 , such that for |z| large ψ(z) = az + b0 +

b1 + ... z

with a > 0. Then g1 (z, ∞) = g2 (ψ(z), ∞), so that by (1.1) γ (E 1 ) = γ (E 2 ) + log a and Cap(E 2 ) = aCap(E 1 ).

(1.2)

In particular, the capacity of a closed disc is the radius of the disc, because gC∗ \D (z, ∞) = log |z|. Now let E be any compact plane set and write for the component of C∗ \ E such that ∞ ∈ . Fix a sequence {n } of ﬁnitely connected domains such that ∞ ∈ n ⊂ n ⊂ n+1 ⊂ ,

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such that =

&

75

n ,

and such that ∂n consists of C 1+α Jordan curves for some α > 0. See Exercise II.10(b). Deﬁne E n = C \ n . As we shall soon see, all our results will be independent of the sequence {n }. Because n ⊂ n+1 , it follows from from the maximum principle that gn+1 (z, ∞) > gn (z, ∞) on n , and hence that γ (E n+1 ) > γ (E n ) and Cap(E n+1 ) < Cap(E n ). Now deﬁne Cap(E) = lim Cap(E n ). n

(1.3)

Because g (z, w) is an increasing function of , an interlacing of the domains n shows that the deﬁnition (1.3) does not depend on the choice of the sequence ' = C \ , then by deﬁnition, {n }. Note that if E ' = Cap(∂ E). ' Cap(∂ E) = Cap(E) = Cap( E)

(1.4)

By deﬁnition, the quantity Robin’s constant 1 = lim γ (E n ) γ (E) ≡ log n Cap(E) is Robin’s constant for the arbitrary compact set E. If E ⊂ F then by (1.3) Cap(E) ≤ Cap(F) and γ (E) ≥ γ (F).

(1.5)

If Cap(E) > 0, then limn γ (E n ) = γ (E) < ∞, and by Harnack’s principle g (z, ∞) = lim gn (z, ∞) n

deﬁnes a positive harmonic function on having expansion g (z, ∞) = log |z| + γ (E) + o(1)

(1.6)

at ∞. When z = ∞, the symmetry (2.8) of Chapter II shows that lim gn (∞, w) = lim gn (w, ∞) n

n

exists for all w ∈ \ {∞}. Thus by Harnack’s principle the limit g (z, w) = lim gn (z, w) n

(1.7)

exists for all z, w ∈ with z = w, and g (z, w) satisﬁes conditions (2.2), (2.4), and (2.5) of Chapter II. The function g (z, w) is Green’s function for with pole at w.

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Example 1.1. Suppose E is compact and connected. Let be the component of C∗ \ E such that ∞ ∈ and let ψ : → C∗ \ D be the conformal mapping such that for |z| large, b1 + ..., z

ψ(z) = az + b0 + with a > 0. For r > 1,

r = {z : |ψ(z)| > r } is bounded by an analytic Jordan curve and gr (z, ∞) = log |ψ(z)/r |. Then by (1.7), g (z, ∞) = log |ψ(z)| and by (1.2), 1 = Cap(D) = aCap(E). For example, the normalized conformal map ϕ = ψ −1 of C∗ \ D onto the complement of the interval [−2, 2] is ϕ(z) = z +

1 z

and so Cap([−2, 2]) = 1. Consequently Cap([α, β]) =

β −α 4

for every interval [α, β] ⊂ R, by (1.2).

Example 1.2. If E ⊂ ∂D, then Cap(E) ≥ sin |E|/4 . Proof. If E is an arc on ∂D, then after a conformal mapping, Example 1.1 gives Cap(E) = sin(|E|/4). Because of (1.3) we may assume E is a ﬁnite union of arcs. Deﬁne iθ e +z 1 dθ, F(z) = 4 E eiθ − z

ω on D. and let ω(z) = ω(z, E, D). Then F(1/z) = −F(z) and F = π2 ω + i Therefore π π − ≤ ReF ≤ 2 2 on = C∗ \ E and F(0) = −F(∞) = |E|/4. Hence H (z) = ei F(z) maps

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77

into the right half-plane and H (∞) = H (0). Now H (z) + H (0) G(z) = log z H (z) − H (0) is superharmonic on and lim inf G(z) ≥ 0.

z→∂

By the maximum principle, G(z) > 0 in so that G(z) ≥ gn (z, ∞) for all n. For |z| large G(z) = log |z| − log sin

|E| 4

+ ...,

so that by (1.1) − log sin

|E| 4

≥ γ (∂n ) → γ (E)

and Cap(E) ≥ sin(|E|/4).

2. The Logarithmic Potential Let μ be a ﬁnite, compactly supported signed (Borel) measure. The logarithmic potential of μ is the function 1 Uμ (z) = log dμ(ζ ). |ζ − z| By Fubini’s theorem, the integral Uμ is absolutely convergent for area almost every z. Lemma 2.1. If μ > 0, the potential Uμ is lower semicontinuous and superharmonic. Proof. The lemma holds because for ζ ﬁxed, the function z → log

1 |ζ − z|

is lower semicontinuous and superharmonic.

The next theorem connects the notions of logarithmic potential, Green’s function, capacity, and harmonic measure. Suppose = C∗ \ E is bounded by a ﬁnite family of disjoint C 1+α Jordan curves, α > 0, write μ E for the harmonic

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measure of ∞ relative to , dμ E = dω(∞, · , ) = −

1 ∂g(ζ, ∞) ds = P∞ (ζ )ds. 2π ∂nζ

Theorem 2.2. If = C∗ \ E is connected and bounded by ﬁnitely many pairwise disjoint C 1+α Jordan curves, then the integral Uμ E (z) is absolutely convergent at every z ∈ C. The potential Uμ E is continuous on C and g(z, ∞) = γ (E) − Uμ E (z), Uμ E (z) < γ (E),

z ∈ ,

(2.1)

z ∈ ,

(2.2)

z ∈ E = C∗ \ .

(2.3)

and Uμ E (z) = γ (E),

Later we shall see that μ E is the unique probability measure μ on E such that Uμ is constant on E. For this reason μ E is called the equilibrium distribution of E. Proof. We can assume 0 ∈ / . Clearly, the integral Uμ (z) is absolutely convergent at all z ∈ / ∂. Since log |ζ | dω(z, ζ ), g(z, ∞) = log |z| − ∂

we have

γ (E) = −

∂

log |ζ | dμ E (ζ ).

On the other hand, for ﬁxed z 0 ∈ , z − g(z, z 0 ) = log z−z 0

ζ dω(z, ζ ), log ζ − z0 ∂

(2.4)

because the right side of (2.4) satisﬁes the conditions (2.3), (2.4), and (2.5) from Chapter II and those conditions determine Green’s function. Then sending z to ∞ yields 1 dμ E (ζ ) g(∞, z 0 ) = γ (E) − log |ζ − z 0 | = γ (E) − Uμ E (z 0 ). For z ∈ , (2.1) is then a consequence of the symmetry of Green’s function, g(∞, z) = g(z, ∞). Then because g(z, ∞) > 0 on , (2.1) implies (2.2).

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For z ∈ / , v(ζ ) = log (|ζ |/|ζ − z|) is harmonic on a neighborhood of and 1 0 = v(∞) = log |ζ |dμ E (ζ ) + log dμ E (ζ ) |ζ − z| = −γ (E) + Uμ E (z). Therefore (2.3) holds for z ∈ / . If z ∈ ∂, the lower semicontinuity of Uμ gives Uμ (z) ≤ lim inf Uμ (w) ≤ γ (E) < ∞. w→z

Since the integrand is bounded below, that means the integral Uμ (z) converges absolutely. Because ∂ consists of C 1+α curves, Area(∂) = 0. Then by the superharmonicity of Uμ and the continuity of g(z, ∞), ( ) dξ dη dξ dη Uμ (z) ≥ lim sup Uμ (ζ ) + γ = γ. π δ2 π δ2 δ→0 ∩B(z,δ) B(z,δ)\ Consequently (2.1) and (2.3) hold at all z ∈ ∂, and it follows that Uμ is continuous in C . Let E be a compact set with Cap(E) > 0, and let E n = C∗ \ n be as in Section 1. Then by (1.7) and Theorem 2.2 any weak-star cluster point μ E of the sequence {μ E n } satisﬁes both (2.1) and (2.2) on . In Section 4 we will use the energy integral to show that there is a unique weak-star limit μ E independent of the sequence E n and to establish a version of (2.3) for Uμ E on E. A different proof of the uniqueness of the weak-star limit {μ E n } for a bounded domain is given in Exercise 4.

3. The Energy Integral Let ν be a signed measure with compact support. If log 1 d|ν|(ζ ) d|ν|(z) < ∞, |z − ζ |

(3.1)

we say ν has ﬁnite energy and deﬁne the energy integral I (ν) by 1 dν(ζ )dν(z) = Uν (z)dν(z). I (ν) = log |z − ζ | The energy integral has a very important property: It is positive deﬁnite on the space of signed measures with ﬁnite energy and zero integral.

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Theorem 3.1. If (3.1) holds and if I (ν) = 0, then ν ≡ 0.

dν = 0, then I (ν) ≥ 0. Moreover, if

1 . By Green’s theorem Proof. Write L(z) = log |z| 1 L(z − w) f (w)dudv f (z) = − 2π

(3.2)

whenever f ∈ C ∞ has compact support. First consider the special case of an absolutely continuous signed measure dν = h(z)d xd y where h ∈ C ∞ has compact support and satisﬁes h(z)d xd y = 0. (3.3) Then the convolution Uν = L ∗ h is also C ∞ and by (3.3), |Uν (z)| ≤

C |z|

(3.4)

C |z|2

(3.5)

and |∇Uν (z)| ≤

for |z| large. For any f ∈ C ∞ with compact support, Green’s theorem (3.2) and Fubini’s theorem give Uν f d xdy = −2π h f d xdy, Uν f d xdy = because h and f have compact support. Therefore Uν = −2π h. Now by (3.4), (3.5), and Green’s theorem again 1 1 Uν Uν d xd y = |∇Uν |2 d xd y. I (ν) = Uν hd xdy = − 2π 2π That shows I (ν) ≥ 0 in this special case. Moreover, if I (ν) = 0 then ∇Uν = 0 1 Uν = 0. and h = − 2π To derive the full Theorem 3.1 from the special case we employ a standard molliﬁcation argument. Let K ∈ Cc∞ (C) be a compactly supported inﬁnitely differentiable function such that (i) K is radial, K (z) = K (|z|), (ii) K ≥ 0, and (iii) K d xdy = 1.

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Set K ε (z) = ε −2 K (z/ε) and let νε be the absolutely continuous measure having density h ε (z) = K ε ∗ ν(z) = K ε (z − w)dν(w). Then

f dνε =

lim

ε→0

f dν,

(3.6)

for all continuous f . Furthermore, h ε ∈ C ∞ has compact support and satisﬁes (3.3). By Fubini’s theorem and (i) K ε ∗ K ε ∗ L(z − ζ )dν(z)dν(ζ ), I (νε ) = where

K ε ∗ K ε (z) =

and

K ε (z − w)K ε (w)dudv

K ε ∗ K ε ∗ L(w) =

K ε ∗ K ε (z) L(w − z)d xd y.

Now because K ε is a radial function, K ε ∗ K ε is also a radial function, and therefore since L(z) is superharmonic, K ε ∗ K ε ∗ L(z) ≤ L(z).

(3.7)

Also, the continuity of L(z) viewed as a map to (−∞, ∞], gives

Since

K ε ∗ K ε ∗ L(z) → L(z), dν = 0, log

1 dν(ζ )dν(z) = |αz − αζ |

as ε → 0.

log

1 dν(ζ )dν(z), |z − ζ |

for any α > 0, and we can assume ν has support in {z : |z| < 1/2}. Therefore by (3.1), (3.7) and dominated convergence, lim I (νε ) = I (ν),

ε→0

and we see that I (ν) ≥ 0. Now assume I (ν) = 0. Write Uε = Uνε . Then |∇Uε |2 d xd y = I (νε ) → 0.

(3.8)

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1 We also have Uε (z) = O |z| uniformly in ε, because all h ε , ε < 1, satisfy (3.3) and vanish outside a common compact set. Then by (3.8) and Lemma 3.2 below, |Uε (z)|2 d xd y = 0. lim ε→0

Let f ∈ C ∞ have compact support. Then by (3.6) and Green’s theorem once more, −1 f Uε d xd y = 0. f dν = lim f dνε = lim ε→0 ε→0 2π Therefore ν = 0.

Lemma 3.2. Assume Un (z) ∈ C ∞ (C) satisfy |Un (z)| ≤ C/|z|,

(3.9)

and ||∇Un ||2 → 0. Then

|Un |2 d xd y → 0,

(3.10)

K

for every compact set K . Proof. Assume K ⊂ [−L , L] × [−R, R]. Then L |Un (x, y)|2 d x ≤ 2 |Un (x, y) − Un (x, −L)|2 d x + 2C 2 /L [−L ,L]

≤2

−L L −L

y −R

2 ∂Un 2 ∂ y dy d x + 2C /L

≤ 4L||∇Un ||22 + 2C 2 /L , and a second integration gives (3.10) when L is large compared to R.

4. The Equilibrium Distribution As promised at the end of Section 2, we use the energy integral to show that the sequence {μ E n } has a unique limit μ E and that Uμ E = γ (E) on E \ A, where Cap(A) = 0. Write P(E) for the set of all Borel probability measures

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83

on the compact set E. Let be the component of C∗ \ E such that ∞ ∈ , and recall the sequences {n } and E n = C∗ \ n deﬁned in Section 1. Also deﬁne ' = C \ . E Theorem 4.1. Assume Cap(E) > 0. Then sequence {μn } = {μ E n } converges weak-star to some μ = μ E ∈ P(E), and the limit does not depend on the choice of the sequence {E n }. Moreover, γ (E) = I (μ E ) = inf{I (σ ) : σ ∈ P(E)}

(4.1)

and I (σ ) > γ (E) for all σ ∈ P(E) with σ = μ E . In other words, μ E is the unique probability measure on E satisfying γ (E) = I (μ E ). It follows that the measure μ E is the unique μ ∈ P(E) satisfying Uμ = γ (E) almost everywhere dμ. By deﬁnition μ E is the equilibrium distribution of E, and Uμ E is called the equilibrium potential. Lemma 4.2. Let E be a compact set and let {μn } be a sequence in P(E). If μn converges weak-star to μ ∈ P(E) then Uμ (z) ≤ lim inf Uμn (z),

z ∈ C,

(4.2)

and I (μ) ≤ lim inf I (μn ).

(4.3)

Proof of Lemma 4.2. Write h(z, ζ ) = log 1/|z − ζ | and for N > 0 deﬁne h N (z, ζ ) = min(h(z, ζ ), N ). Then each h N is continuous, h = lim h N , and h N ≤ h. Therefore h N (z, ζ )dμ(ζ ) = lim h N (z, ζ )dμn (ζ ) n ≤ lim inf h(z, ζ )dμn (ζ ) n

= lim inf Uμn . n

Because h N increases to h, this gives (4.2). Because μn × μn converges to μ × μ weak-star, a similar argument gives (4.3). Proof of Theorem 4.1. We ﬁrst assume that = C∗ \ E is bounded by ﬁnitely many pairwise disjoint C 1+α Jordan curves and in that case we prove μ E is the unique minimizer for (4.1). For such E, Uμ E = γ (E) on E by (2.3), so that I (μ E ) = Uμ E dμ E = γ (E). Thus if σ ∈ P(E), then γ (E) = Uμ E dσ = Uσ dμ E

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by Fubini’s theorem. But if σ = μ E , then by Theorem 3.1, 0 < I (σ − μ E )

= I (σ ) + I (μ E ) −

Uμ E dσ −

Uσ dμ E

= I (σ ) − γ (E). Therefore (4.1) holds, and I (σ ) = γ (E) if and only if σ = μ E . Now suppose E is any compact set. By the Banach–Alaoglu theorem there is a subsequence {μn j } of {μ E n } that converges weak-star to some μ ∈ P(E 1 ). By (4.3), I (μ) ≤ lim inf I (μn j ) = lim inf γ (E n j ) = γ (E). Moreover, any such μ has support contained in * ∂

+

, En

' ⊂ E, = ∂E

(4.4)

n

because μ E n has support ∂ E n . If σ ∈ P(E) ⊂ P(E n ) then, because (4.1) holds for the sets E n , I (σ ) ≥ lim γ (E n ) = γ (E). n

Therefore I (σ ) ≥ I (μ) = γ (E). μ. By Fubini’s theorem Suppose σ ∈ P(E) satisﬁes I (σ ) = γ (E) and σ = Uμ dσ = Uσ dμ, and by Theorem 3.1 (4.5) 0 < I (σ − μ) = I (σ ) + I (μ) − 2 Uμ dσ. Therefore

Uμ dσ < γ (E).

(4.6)

But if σt = tσ + (1 − t)μ,

0 < t < 1,

then σt ∈ P(E) so that

γ (E) ≤ I (σt ) = t 2 I (σ ) + (1 − t)2 I (μ) + 2t (1 − t) ≤ γ (E) + 2

Uμ dσ − γ (E) t + O(t 2 ).

Uμ dσ (4.7)

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85

For t small (4.6) and (4.7) are in contradiction and hence μ is the unique extremal for the minimizing problem (4.1). The uniqueness implies {μn } converges weakstar, because the sequence {μn } has a unique weak-star accumulation point. The uniqueness also shows that the limit μ cannot depend on the choice of {E n }. For any compact set E, ' ⊂ ∂ E ⊂ E ⊂ E, ' ∂E ' By (1.4) and the uniqueness in Theoand by (4.4), μ E is supported on ∂ E. rem 4.1, μ E = μ∂ E = μ E' = μ∂ E'. The capacity of any Borel set A is deﬁned as Cap(A) = sup{Cap(E) : E compact, E ⊂ A}.

(4.8)

Corollary 4.3. If E is compact and Cap(E) = 0, then I (μ) = +∞ for all μ ∈ P(E). Consequently, if A is Borel with Cap(A) = 0 and if μ is a positive measure with I (μ) < ∞, then μ(A) = 0. Proof. Suppose E is compact and Cap(E) = 0 and suppose μ ∈ P(E). Then P(E) ⊂ P(E n ) and I (μ) ≥ γ (E n ) → ∞ by Theorem 4.1. If A is a Borel set with Cap(A) = 0 and if σ is a positive measure χwith σ (A) > 0, then take E ⊂ A E compact, such that σ (E) > 0. Then μ = σ (E) σ ∈ P(E) and I (μ) = ∞. Thus I (σ ) = ∞. In particular, (4.8) and Corollary 4.3 imply that Cap(A) = 0 ⇒ Area(A) = 0. Sets of capacity zero are also called polar sets. Theorem 4.4. Let E be a compact set such that Cap(E) > 0 and let μ E be the equilibrium distribution for E. Then Uμ E (z) ≤ γ (E),

z∈C

' Uμ E (z) = γ (E) − g (z, ∞) < γ (E), z ∈ = C∗ \ E, and

' : Uμ E (z) = γ (E)} = 0. Cap {z ∈ E

(4.9) (4.10)

(4.11)

Proof. Since Uμ En ≤ γ (E n ) and γ (E n ) → γ (E), (4.9) follows from Lemma 4.2. By (1.7), (2.1), and the weak-star convergence of μ E n , we conclude that

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Uμ E (z) = γ (E) − g (z, ∞) for z ∈ . If Uμ E (z) = γ (E) at some z ∈ , then by the maximum principle, Uμ E is constant on , but lim Uμ E (z) = −∞.

z→∞

Therefore (4.10) holds. ' such that Cap(K ) > 0 To prove (4.11), let K be any compact subset of E and take σ ∈ P(K ) with I (σ ) < ∞. For 0 < t < 1, the inequality (4.7) still obtains, so that the coefﬁcient of t in (4.7) is non-negative: Uμ E dσ − γ (E) ≥ 0. But then by (4.9)

Uμ E dσ = γ (E)

and Uμ E (z) = γ (E) almost everywhere with respect to σ . Thus Cap(K ) = 0 ' : Uμ E (z) = γ (E)}. whenever K ⊂ {z ∈ E Theorem 4.5. Suppose ψ is a contraction: |ψ(z) − ψ(w)| ≤ |z − w| for all z, w ∈ E. Then Cap(ψ(E)) ≤ Cap(E). Consequently, if E ∗ = {|z| : z ∈ E} is the circular projection of E onto R, then Cap(E) ≥ Cap(E ∗ ) ≥

|E ∗ | , 4

(4.12)

where |E ∗ | denotes the length of E ∗ .

E

E

0

E∗

E∗

E∗

R

E Figure III.1 Circular projection. In Theorem 4.5, circular projections can also be replaced by vertical projec = {x : x + i y ∈ E}. The proof is the same. tions: E

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Proof. We may assume that Cap(ψ(E)) > 0. Let μ ∈ P(ψ(E)) be such that I (μ) < ∞. By Lemma 4.6 below there is ν ∈ P(E) such that ψ∗ (ν) = μ, where ψ∗ (ν) is deﬁned by ψ∗ (ν)(A) = ν(ψ −1 (A))

(4.13)

for all Borel A ⊂ ψ(E). Then 1 log dν(z)dν(w) ≥ I (ν), I (μ) = |ψ(z) − ψ(w)| E E and by Theorem 4.1, γ (E) ≤ γ (ψ(E)) and Cap(E) ≥ Cap(ψ(E)). In proving (4.12) we may assume that E is a ﬁnite union of smooth curves and ∗ E is a ﬁnite union of intervals in R . Let F be the interval obtained by sliding the intervals of E ∗ together without altering their lengths. Then F is obtained from E by applying two contractions, and hence Cap(E) ≥ Cap(E ∗ ) ≥ Cap(F). But then by Example 1.1, Cap(F) = |F|/4 = |E ∗ |/4. Lemma 4.6. Let E be a compact set and let ψ be a continuous map from E onto ψ(E). For ν ∈ P(E) deﬁne ψ∗ (ν) by (4.13) above. Then ψ∗ maps P(E) onto P(ψ(E)). Proof. By its deﬁnition, the map ψ∗ : P(E) → P(ψ(E)) is continuous, weakstar to weak-star. Given μ ∈ P(ψ(E)), let μn be a sequence of atomic measures μn =

Nn

(n)

a j δz (n) , j

j=1

(n)

(n)

where δz denotes the point-mass at z, such that z j ∈ ψ(E), such that a j > 0 (n) (n) (n) and a j = 1, and such that μn −→ μ weak-star. Pick any w j ∈ ψ −1 (z j ), and set νn =

Nn j=1

(n)

a j δw(n) . j

Then ψ∗ (νn ) = μn , and if a subsequence of {νn } converges weak-star to ν, then ν ∈ P(E) and ψ∗ (ν) = μ. In contrast to Theorem 4.5, there exist sets E ⊂ R with |E| = 0 but Cap(E) > 0. Example 4.7. The Cantor set is obtained from the unit interval K 0 = [0, 1] as follows. The (closed) set K n consists of 2n intervals of length 3−n obtained by removing the middle (open) one-third of each interval in K n−1 . The Cantor

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set is K =

∞ +

Kn ,

n=0

and Cap(K ) ≥

1 . 9

(4.14)

K0 K1 K2 K3 K4 K5 Figure III.2 The Cantor set. To prove (4.14), note that by (1.3) Cap(K ) = limn Cap(K n ). The set K n consists of two subsets K n1 and K n2 , with dist(K n1 , K n2 ) = 13 such that Cap(K n1 ) = Cap(K n2 ) = Cap(K n−1 )/3. In other words, γ (K n1 ) = γ (K n2 ) = γ (K n−1 ) + log 3. j

Let μ = (μ1 + μ2 )/2 where μ j is the equilibrium distribution for K n , j = 1, 2. Then by Theorem 4.1, 1 γ (K n1 ) γ (K n2 ) 1 μ1 + μ2 log = + + dμ1 (ζ )dμ2 (z), I 2 4 4 2 |z − ζ | so that by (4.1), γ (K n ) ≤ 2

γ (K n−1 ) + log 3 4

+

log 3 γ (K n−1 ) + 2 log 3 = . 2 2

Thus γ (K ) = lim γ (K n ) ≤ 2 log 3 and Cap(K ) ≥ 1/9.

Theorem 4.5 provides another proof of the Koebe one-quarter theorem I.4.1. Corollary 4.8. Let ϕ(z) be a univalent function on the unit disc D. If ϕ is normalized by ϕ(0) = 0 and ϕ (0) = 1, then ϕ(D) ⊃ {w : |w| < 1/4}.

(4.15)

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Proof. Let a = inf{|w| : w ∈ / ϕ(D)}. Then ψ(z) =

1 ϕ( 1z )

= z + b0 +

b1 + ... z

is a conformal map from C∗ \ D to a domain and E = C∗ \ is a connected set with circular projection E ∗ = [0, 1/a]. By Theorem 4.5 and Example 1.1, 1 = Cap(E) ≥ Cap(E ∗ ) =

1 , 4a

which gives (4.15).

5. Wiener’s Solution to the Dirichlet Problem Let be a domain in C∗ such that Cap(C \ ) > 0, let f ∈ C(∂), and let {n } be the subdomains introduced in Section 1. Extend f to a continuous function on C∗ , still called f , and set f (ζ )dω(z, ζ, n ), z ∈ n . u n (z) = ∂n

Then Theorem 4.1 yields the following fundamental result from Wiener [1924a]. Theorem 5.1 (Wiener). If Cap(C \ ) > 0, the sequence {u n (z)} converges uniformly on all compact subsets of to a harmonic function u(z), and u(z) does not depend on the choice of the sequence {n } nor on the choice of the extension of f ∈ C(∂). We write u(z) = u f (z) and we call u f (z) the Wiener solution to the Dirichlet problem with boundary values f (ζ ). Proof. Suppose ∞ ∈ . As in Section 4, let μn = μ E n = ω(∞, ·, n ) where E n = C \ n . By Theorem 4.1, the sequence {μn } converges weak-star to μ E , where E = C \ . Hence f (ζ )dμ E n (ζ ) = f (ζ )dμ E (ζ ), (5.1) lim u n (∞) = lim n→∞

n→∞ ∂

∂

and the latter integral does not depend on the sequence {n }. It also does not depend on the extension of f because the limit measure μ E is supported on ∂.

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For z ∈ and z = ∞, let T be a Möbius transformation such that T z = ∞ and note that Cap(∂) > 0 ⇐⇒ Cap(T (∂)) > 0.

(5.2)

Indeed, if Cap(∂) > 0, then by (1.7), g (w, T −1 (∞)) exists and g (w, T −1 (∞)) = lim gn (w, T −1 (∞)) n

= lim gT (n ) (T w, ∞) = gT () (T w, ∞), n

and that implies (5.2). The argument for z = ∞ shows limn→∞ u n (z) = u(z) exists for all z ∈ and the limit u(z) is independent of {n } and the extension of f . Because the functions {u n (z)} are uniformly bounded and harmonic, the convergence u n (z) → u(z) is uniform on compact subsets of and the limit u(z) is harmonic in . Theorem 5.1 enables us to deﬁne harmonic measure for any domain with Cap(C∗ \ ) > 0, because it says the weak-star limit ω(z, · , ) = lim ω(z, · , n ) n→∞

(5.3)

exists and is independent of the choice of {n }. Then for any f ∈ C(∂), u f (z) = f (ζ )dω(z, ζ, ), ∂

where u f (z) is the Wiener solution to the Dirichlet problem for f (ζ ). By deﬁnition, the limit measure ω(z, ·, ) is the harmonic measure for z relative to . By Exercise 4 of Chapter II, this deﬁnition of harmonic measure agrees with the deﬁnition in Chapter II for ﬁnitely connected Jordan domains. As in Chapter I, ω(z, A, ) is a harmonic function of z ∈ and 0 ≤ ω(z, A, ) ≤ 1. By Harnack’s inequality, ω(z 1 , E, ) ≤ C z 1 ,z 2 ω(z 2 , E, ) for every pair z 1 , z 2 of points in , where the constant C z 1 ,z 2 > 0 depends only on z 1 , z 2 , and . Thus the measures ω(z, ·, ), z ∈ , are mutually absolutely continuous. If z = ∞, then (5.1) shows that ω(∞, · , ) = μ∂ . If z ∈ and if T is a Möbius transformation with T (z) = ∞, then for every Borel set A, ω(z, A, ) = μT (∂) (T (A)). By (5.2) and Corollary 4.3 Cap(A) = 0 ⇒ ω(z, A, ) = 0

(5.4)

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for every with Cap(C \ ) > 0 and every z ∈ . If Cap(C∗ \ ) = 0 then, as we shall see in Section 8, each bounded or positive harmonic function on is constant. Therefore we cannot deﬁne harmonic measure or solve the Dirichlet problem when Cap(C∗ \ ) = 0. A major goal of this book is to understand the relation between harmonic measure and other measures. Every ﬁnite F ⊂ ∂ has harmonic measure zero. (One proof: Cap(F) = 0 by (4.1), so that ω(z, F, ) = 0 for all z ∈ by (5.4).) We shall need the following quantitative form of this fact. Let {n } be the increasing sequence of domains introduced in Section 1. Proposition 5.2. If E = C \ satisﬁes Cap(E) > 0, then for any ζ ∈ ∂, and any ε > 0, lim

sup

δ→0 z∈n \Bε (ζ )

ω(z, ∂n ∩ Bδ (ζ ), n ) = 0

(5.5)

uniformly in n.

Jn ε δ

1 M

M n ∂n Figure III.3 Proof of Proposition 5.2. Proof. Suppose ∞ ∈ and ζ = 0. Let J = E ∩ {z : 1/M ≤ |z| ≤ M} with M so that Cap(J ) > 0. For n as before, set Jn = {z : 1/M ≤ |z| ≤ M} \ n . See Figure III.3. Let μn be the equilibrium distribution of Jn and let γn be Robin’s constant for Jn . Then z − ζ 1 dμn (ζ ) Un (z) ≡ log log − Uμn (z) = |z| z Jn

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is harmonic on C∗ \ (Jn ∪ {0}). For δ < 1/M we have 1 Un (z) ≥ log − 1 , z ∈ ∂ Bδ (0), Mδ and 1 − γn , z ∈ C. M For δ < ε, these inequalities and the maximum principle give Un (z) ≥ log

ω(z, ∂n ∩ Bδ (0), n ) ≤

Un (z) + log M + γn log( 1δ − M) + γn

on {|z| > δ} ∩ n . Hence if |z| ≥ ε ≥ δ,

log 1 + Mε + log M + γn . ω(z, ∂n ∩ Bδ (0), n ) ≤

log 1δ − M + γn

But because M is ﬁxed and J ⊂ Jn ⊂ B M (0), the quantities γn are bounded above and below, and therefore (5.5) holds. Suppose 0 ∈ / . Then by Theorem 5.1 and the deﬁnition of g(z, ∞) for ﬁnitely connected domains, h(z, ∞) = g(z, ∞) − log |z| coincides with the solution to the Dirichlet problem in with boundary value − log |z|. The following proposition gives a similar result when = C∗ . Proposition 5.3. Let E ⊂ {z : |z| < R} be a compact set such that Cap(E) > 0 and let = C∗ \ E. Then Uμ E is the solution to the Dirichlet problem on R = ∩ B(0, R) with boundary data

γ (E) for ζ ∈ ∂, f (ζ ) = Uμ E (ζ ) for |ζ | = R, and g (·, ∞) is the solution to the Dirichlet problem on R with boundary data γ (E) − f (ζ ). Proof. Because Uμ (z) = γ (E) − g (z, ∞), it is enough to show that Uμ E is the solution of a Dirichlet problem. We may suppose the domains {n } satisfy {z : |z| ≥ R} ⊂ 1 . Extend f to C(C) so that ⎧ on E, ⎨ γ (E) on ∂n , f = γ (E n ) ⎩ on |z| = R, Uμ E

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and let u n be the Wiener solution of Dirichlet’s problem on nR = n ∩ B R (0) for the boundary data f |∂nR . Then u n − Uμ En is continuous on nR and harmonic on n , and hence u n (z) − Uμ En (z) = Uμ E (w) − Uμ En (w) dω(z, w, nR ) |w|=R

on

nR .

Because Uμ En → Uμ E uniformly on {|w| = R}, it follows that lim u n (z) = lim Uμ En (z) = Uμ E (z) n

n

for z ∈ R . Therefore Uμ E is the Wiener solution to the Dirichlet problem in R for the boundary data f |∂ R . With the same proof, Proposition 5.3 is also valid if B(0, R) is replaced by any bounded Jordan region such that E ∩ ∂ = ∅.

6. Regular Points Assume Cap(C∗ \ ) > 0, let ζ ∈ ∂ and let f ∈ C(∂). In this section we determine when the Wiener solution u f has boundary value f (ζ ) at ζ. The point ζ ∈ ∂ is a regular point for if lim u f (z) = f (ζ )

z→ζ

(6.1)

for all f ∈ C(∂). Equivalently, ζ ∈ ∂ is a regular point if lim ω(z, ·, ) = δζ

z→ζ

in the weak-star topology, where δζ is the point mass at ζ. A point ζ ∈ ∂ not regular is called an irregular point. If ∂ consists of ﬁnitely many Jordan curves, then by Chapter II every boundary point is regular. However, if = {z : |z| < 1} \ {0}, then every bounded harmonic function on extends to be harmonic at 0. Therefore (6.1) fails at ζ = 0 except in the rare cases where 2π dθ f (eiθ ) , f (0) = 2π 0 and therefore 0 is an irregular point of ∂. Theorem 6.1. The point ζ ∈ ∂ is a regular point for if and only if there is ρ > 0 such that for every ε > 0, there exists a function Vε , harmonic in

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∩ {z : |z − ζ | < ρ}, satisfying Vε (z) ≥ 0,

(6.2)

lim sup Vε (z) ≤ ε,

(6.3)

Vε (z) > 1/2 on {z : ε < |z − ζ | ≤ ρ} ∩ .

(6.4)

z→ζ

and By deﬁnition a barrier is a harmonic function V on such that lim V (z) = 0,

z→ζ

and lim inf V (z) > 0 z→α

for α ∈ ∂ \ {ζ }. If V is a barrier, then there are constants Cε > 0 such that the functions Vε = Cε V satisfy (6.2), (6.3), and (6.4). Proof. Let ζ be a regular point. Fix ε > 0. For 0 < δ < ε choose f ∈ C(∂) such that 0 ≤ f ≤ 1, {z ∈ ∂ : f (z) = 0} = {ζ }, and f (z) = 1, when |z − ζ | > δ. Then clearly Vε (z) = u f (z) satisﬁes (6.2) and (6.3), and when δ is small enough (6.4) also holds by Proposition 5.2. Now suppose the functions Vε (z) exist. Let f ∈ C(C∗ ) satisfy 0 ≤ f ≤ 1/2 and f −1 (0) = ζ

(6.5)

and let n be as above. If δ > 0, then for ε sufﬁciently small, f (z) ≤ Vε (z) + δ on {z : |z − ζ | < ρ} ∩ ∂n . Let u n solve the Dirichlet problem on the subdomain n for the boundary value f . Then u n (z) ≤ sup f ≤ 1/2 ≤ Vε (z) on n ∩ {z : |z − ζ | = ρ}, and by the maximum principle u n (z) ≤ Vε (z) + δ

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on n ∩ {|z − ζ | < ρ}. Thus 0 ≤ u f (z) ≤ Vε (z) + δ on ∩ {|z − ζ | < ρ}. Because δ is arbitrary, (6.3) now yields lim u f (z) = 0.

z→ζ

Because every real g ∈ C(C∗ ) is a constant plus a linear combination of two functions satisfying (6.5), it follows that ζ is a regular point. Theorem 6.1 implies that the regularity of ζ is a local question; it depends only on the behavior of in some neighborhood of ζ . Corollary 6.2. Let ζ ∈ ∂. If the connected component of ∂ containing ζ consists of more than one point, then ζ is a regular point for . In particular, if is simply connected, then every point of ∂ is a regular point. Proof. For ε > 0, let E ε be the component of ∂ ∩ B(ζ, ε/2) containing ζ . Then E ε = {ζ }. Let ψε be a Riemann map of C∗ \ E ε onto D with ψε (∞) = 0. Then |ψε (z)| → 1 as C∗ \ E ε z → E ε and log ψε1(z) Vε (z) ≡ inf log ψε1(z) ε<|z|<1

satisﬁes (6.2), (6.3), and (6.4) with ρ = 1.

Theorem 6.3. Assume E = C∗ \ is a compact plane set with Cap(E) > 0 and let μ E be the equilibrium distribution for E. Then for ζ ∈ ∂ the following conditions are equivalent: ζ is a regular point, lim g (z, ∞) = 0,

z→ζ

Uμ E (ζ ) = γ (E).

(6.6) (6.7) (6.8)

Proof. Take R > 0 such that E ⊂ B(0, R). Then by Proposition 5.3, g (·, ∞) is the solution to the Dirichlet problem in R = ∩ B(0, R), for the boundary data

0 for z ∈ ∂, for |z| = R. g (z, ∞) If ζ ∈ ∂ is a regular point, then by Theorem 6.1, ζ is also a regular point for R and hence (6.7) holds.

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Now suppose (6.7) holds. Then by (4.10) Uμ E (z) = γ (E) − g (z, ∞) → γ (E),

(6.9)

as z ∈ → E. Because log 1/|z| is superharmonic, we have 1 Uμ E (z)d xd y, Uμ E (ζ ) ≥ πr 2 B(ζ,r )

while by Theorem 4.4, Uμ E (z) ≤ γ (E) for all z. Therefore 0 ≤ γ (E) − Uμ E (ζ ) ≤ lim sup r →0

1 πr 2

(6.10)

(γ (E) − Uμ E (z))d xd y B(ζ,r )∩

1 + lim sup 2 r →0 πr

(γ (E) − Uμ E (z))d xd y. B(ζ,r )\

By (6.9) and (6.10) the ﬁrst limit is 0, while by the remark after Corollary 4.3 and Theorem 4.4, Area({z ∈ / : Uμ E (z) = γ (E)}) = 0, so that the second limit is also 0. Hence (6.8) holds.

To show that (6.8) implies ζ is regular, set E ε = E ∩ B ζ, ε2 /4 . If ζ and {z : |z − ζ | = ε2 /4} lie in the same component of E ε ∪ {z : |z − ζ | = ε2 /4}, ∗ then the complement

2 in C of that component is a simply connected domain ε ⊃ ∩ B ζ, ε /4 , and ζ ∈ ∂ε . By Corollary 6.2, ζ is regular for ε , and then by Theorem 6.1, ζ is regular

for . Thus we may assume there is a (closed) C 2 Jordan curve σε ⊂ ∩ B ζ, ε2 /4 such that ζ is in the bounded component Wε of C \ σε . Let με be the equilibrium distribution for E ε , and let Uε be its logarithmic potential. We will verify conditions (6.2), (6.3), and (6.4) for the functions Vε (z) = 1 −

Uε (z) . γ (E ε )

and this will show that ζ is regular. First note that e−γ (E ε ) = Cap(E ε ) > 0, because if not, then by Corollary 4.3, μ E (B(ζ, ε2 /4)) = 0. But then ζ would be in the unbounded component of V = C \ supp(μ E ), where supp(μ E ) is the closed support of μ E . Then (4.10) and the maximum principle would give Uμ E (ζ ) < γ (E), contrary to the assumption (6.8).

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By Theorem 4.4, (6.2) holds for Vε . Note that if |z − ζ | > ε, then 1 2 log dμ E ε (w) ≤ log . Uε (z) = |z − w| ε Eε Because γ (E ε ) ≥ log ε42 , we have Vε (z) > 1/2 in |z − ζ | ≥ ε and (6.4) holds for Vε . Choose Mε > 0 so that when z ∈ σε ,

(6.11) γ (E ε ) − Uε (z) ≤ Mε γ (E) − Uμ E (z) . By Proposition 5.3, each side of (6.11) is the solution to a Dirichlet problem in Wε ∩ with boundary data 0 on ∂ ∩ Wε , and hence the inequality (6.11) continues to hold in Wε ∩ . Sending z → ζ then yields (6.3). Then ζ is a regular point by Theorem 6.1. Corollary 6.4 (Kellogg’s theorem). Let be any domain in C∗ . Then

Cap {ζ ∈ ∂ : ζ is irregular} = 0.

Proof. We may assume ∞ ∈ / . Then the corollary is immediate from Theorem 4.4 and Theorem 6.3. Corollary 6.5. Let be any domain in C∗ with Cap(C∗ \ ) > 0. Then ω(z, {ζ ∈ ∂ : ζ is irregular}, ) = 0 for all z ∈ . Proof. By (6.8), the set of irregular points is a Borel set, so the corollary follows from (5.4) and Corollary 6.4.

7. Wiener Series In a second paper [1924b], Wiener gave a geometric condition necessary and sufﬁcient for ζ ∈ ∂ to be a regular point for . Let An = An (ζ ) be the ring An = {z : 2−n ≤ |z − ζ | ≤ 2−n+1 }, and recall the notation

γ (E) = log 1/Cap(E) .

Theorem 7.1 (Wiener). Let be a domain in C∗ and let ζ ∈ ∂. Then the following are equivalent:

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ζ is a regular point, ∞

n

n=1

γ (An \ )

∞ n=1

= ∞,

n = ∞. γ (An ∩ ∂)

(7.1) (7.2)

(7.3)

Proof. We can assume that ζ = 0, and ⊂ B(0, 1/2). It is clear that (7.3) implies (7.2) because the function E → γ (E) is decreasing. Now assume (7.2). By the deﬁnitions of regular point and capacity, we can replace by a subdomain so that, for each n, An \ is bounded by ﬁnitely many C 2 Jordan curves and so that the series (7.2) still diverges. Write γn = γ (An \ ), let μn be the equilibrium distribution for An \ , and let u n (z) = Uμn (z). Fix ε > 0. Then because γn ≥ n log 2 there are integers N1 < N2 < . . . Nk → ∞ such that N j+1 −1

1≤ε

n log 2 ≤ 1 + ε. γn

(7.4)

Nj

Set N j+1 −1

Wj = ε

un . γn Nj

By our smoothness assumption on An \ , W j is continuous on B(0, 1), and W j is harmonic and positive on ∩ B(0, 1). Also, by (7.4) N j+1 −1

W j (0) = ε

N j+1 −1 u n (0) n log 2 ≥ε ≥ 1. γn γn Nj

(7.5)

Nj

On the other hand, if z ∈ Ak , k ≥ 1, then ⎧ 1, if |n − k| ≤ 1 ⎪ ⎪ ⎪ ⎪ u n (z) ⎨ (k+1) log 2 , if n > k + 1 ≤ γn ⎪ γn ⎪ ⎪ ⎪ ⎩ (n+1) log 2 , if n < k − 1. γn Hence by (7.4), W j (z) ≤ 3ε +

Nj + 1 (1 + ε) ≤ 1 + 5ε Nj

(7.6)

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on B(0, 1) provided N j is sufﬁciently large. Also, for any δ > 0, sup W j (z) ≤ ε

(7.7)

|z|≥δ

if N j is sufﬁciently large. Set Vj = 1 + 5ε − W j . Then by (7.5), (7.6), and (7.7) the functions Vj satisfy the hypotheses of Theorem 6.1 (with 5ε), and ζ = 0 is a regular point of ∂. Now assume ζ = 0 is a regular point of ∂ and assume the series (7.3) converges. Again we may replace by a subdomain so that, for each n, An ∩ ∂ consists of ﬁnitely many C 2 Jordan curves and so that (7.3) still converges. We may also assume ⊂ B(0, 1/2). Write γn = γ (An ∩ ∂), let μn be the equilibrium distribution for An ∩ ∂, and let u n = Uμn be its potential. Fix ε, 0 < ε < 13 and take N ≥ 3 so that ∞ n < ε. γn

(7.8)

n=N

Let F(z) ∈ C() be harmonic on with 0 ≤ F ≤ 1, F(0) = 1, and F = 0 on ∂ \ B(0, 2−N +1 ). Since ζ = 0 is a regular point there exists s < 2−N such that F(z) > 1 − ε on ∩ B(0, s). Let 0 < r < s be so small that if Ur (z) is the equilibrium potential for ∂ ∩ B(0, r ), then Vr (z) =

Ur (z) γ (∂ ∩ B(0, r ))

≤ε

on {|z| ≥ s}. Let A be the annulus {r ≤ |z| ≤ 2−N +1 }. Let U denote the equilibrium potential for ∂ ∩ A and take V (z) =

U (z) . γ (∂ ∩ A)

Then V + Vr ≥ F by Proposition 5.3. Consequently V ≥ 1 − 2ε on

{|z| = s} ∪ B(0, s) ∩ ∂ ∩ A . Hence by the maximum principle, V (0) ≥ 1 − 2ε. On the other hand, V (z) ≤ principle also gives

2 N

on |z| = 21 , so that when 2−M < r the maximum

V (z) ≤

M u n (z) N

Hence by (7.8) V (0) ≤ ε +

(7.9)

2 N,

γn

+

2 . N

which contradicts (7.9).

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Theorem 7.1 has some important applications. The ﬁrst shows there exist totally disconnected sets K such that each ζ ∈ K is a regular boundary point for = C∗ \ K . Example 7.2. The Cantor set K = ∞ n=1 K n of Example 4.7 has Cap(K ) > 0. Suppose ζ ∈ K and suppose α ∈ K ∩ A j (ζ ). Then α belongs to an interval I ⊂ K j of length 3− j and there is a second interval J ⊂ K j of length 3− j with dist(I, J ) = 3− j . See Figure III.4. If j is such that 2− j−1 > 3− j+1 then one of the three rings A j−1 (ζ ), A j (ζ ), or A j+1 (ζ ) contains either I or J . Hence for some n ∈ { j − 1, j, j + 1} Cap(An ∩ K ) ≥ Cap(I ∩ K ) = Cap(J ∩ K ) = 3− j Cap(K ). Therefore n ≥C >0 γ (An ∩ K ) for some C > 0 and for inﬁnitely many n and by (7.3) every ζ is a regular point for = C ∗ \ K .

A j+1

Aj

A j−1 α J

2− j

I 2−( j−1)

2−( j−2)

Figure III.4 Annuli and the Cantor set. See Corollary 7.4 and Corollary D.3 of Appendix D for other proofs that every point of the Cantor set K is regular. Corollary 7.3 (Beurling). Let E be a compact set. Assume 0 ∈ E and set E ∗ = {|z| : z ∈ E}. If dr = ∞, (7.9) ∗ E r then 0 is a regular point for = C∗ \ E. Proof. Set E n = E ∩ An and E n ∗ = {|z| : z ∈ E n }. Then Cap(E n ) ≥ |E n∗ |/4 n by Theorem 4.5, and consequently we have 2 Cap(E n ) = ∞ by (7.9). n But 2 Cap(E n ) ≤ Cn/γ (E n ), because Cap(E n ) ≤ 2−n+1 and the function

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x log 1/x is increasing on [0, 1/e]. Therefore the series (7.3) also diverges and 0 is a regular point. Corollary 7.4. Assume ζ ∈ E = ∂. If

Cap E ∩ B(ζ, r ) > 0, lim sup r r →0

(7.11)

or if lim sup r →0

log(1/r ) > 0, γ (E ∩ B(ζ, r ))

(7.12)

then ζ is a regular point of ∂. Proof. It is clear that (7.11) implies (7.12) and that (7.12) implies n lim sup > 0. −n n→∞ γ (E ∩ B(ζ, 2 )) Then by Lemma 7.5 below lim sup n→∞

∞ n

k > 0, γ (E ∩ Ak )

and (7.3) again diverges.

Lemma 7.5. Suppose E 1 , . . . , E n are disjoint Borel subsets of B(0, 1/2). Then 1 1 ≤ . γ ( Ek ) γ (E k ) n

k=1

Proof. By the deﬁnition of capacity, we can suppose each E k is bounded by ﬁnitely many pairwise disjoint C 2 Jordan curves. Let μ be the equilibrium distribution for E = E k , let μk be the equilibrium distribution for E k , let σ = μk /γ (E k ) and let Uμ , Uμk , and Uσ be their potentials. Then because Uμ = γ (E) on E and supp(σ ) ⊂ E, we have 1 γ (E) = Uμ dσ = Uσ dμ, γ (E k ) and because Uμk ≥ 0 on E and Uμk = γ (E k ) on E k , we have Uσ ≥ 1 on supp(μ). Consequently Uσ dμ ≥ 1 and the lemma follows.

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8. Polar Sets and Sets of Harmonic Measure Zero In this section we prove a strong form of the maximum principle. It implies that if a compact set E is a polar set, i.e. Cap(E) = 0, and if u > 0 is a harmonic function on C∗ \ E, then u is constant. Theorem 8.1. Suppose is a domain such that Cap(C∗ \ ) > 0, and suppose A is a Borel subset of ∂ such that ω(z 0 , A, ) = 0,

(8.1)

holds for some z 0 ∈ . Let u(z) be a subharmonic function in . If u is bounded above in and if lim sup u(z) ≤ m

(8.2)

z→ζ

for all ζ ∈ ∂ \ A, then u(z) ≤ m for all z ∈ . The condition (8.1) is essential for Theorem 8.1. For example, if K is a compact subset of ∂ such that 0 < ω(z 0 , K ) < 1 for some z 0 ∈ , let u(z) = ω(z, K , ). If A = (∂ \ K ) ∩ {regular points} then ω(z 0 , A, ) > 0 by Corollary 6.5. But lim z→ζ u(z) = 0 on A, while u(z 0 ) > 0. Because a ﬁnite set has zero harmonic measure, Theorem 8.1 is a generalization of Lindelöf’s maximum principle, Lemma I.1.1. Proof. On , deﬁne V (ζ ) =

⎧ u(ζ ), ⎪ ⎨

ζ ∈ ,

⎪ ⎩ lim sup u(z),

ζ ∈ ∂.

z→ζ

Then by hypothesis, V is bounded above, and V is upper semicontinuous on because subharmonic functions are upper semicontinuous. By (8.1) and (8.2), V (ζ )dω(z, ζ, ) ≤ m (8.3) ∂

for every z ∈ . Let {n } be the subdomains from before. Then for any z ∈ , ω(z, ·, ) = lim ω(z, ·, n ), n→∞

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103

with respect to weak-star convergence of measures on . Let Vk (ζ ) be a sequence of continuous functions that decreases to V (ζ ) on , and let u k,n (z) be the harmonic extension of Vk to n . Because u is upper semicontinuous on n ⊂ ,

lim sup u(z) − u k,n (z) = u(ζ ) − Vk (ζ ) ≤ 0 n z→ζ

at every α ∈ ∂n . Hence by the ordinary maximum principle, Vk (ζ )dω(z, ζ, n ) u(z) ≤ u k,n (z) = ∂n

for z ∈ n . Then by the weak-star convergence Vk (ζ )dω(z, ζ, ), u(z) ≤

(8.4)

so that by monotone convergence u(z) ≤

(8.5)

∂

∂

V (ζ )dω(z, ζ, ),

z ∈ . Then (8.3) and (8.5) imply that u(z) ≤ m in =

n .

Theorem 8.2. Suppose A is a Borel set. Then A is a polar set if and only if ω(z, A ∩ ∂, ) = 0

(8.6)

for every domain such that Cap(C∗ \ ) > 0 and for every z ∈ . Proof. By (4.8) and the regularity of harmonic measure, we may assume that the set A is compact. If Cap(A) = 0, then (8.6) follows from (5.4). Conversely, if Cap(A) > 0 let μ A be the equilibrium distribution for A and let be the unbounded component of C \ A. Then Cap(C∗ \ ) = Cap(A) > 0 and ω(∞, A ∩ ∂, ) = μ A (A) = 1. More quantitative forms of Theorem 8.2 will be given in the next section. Because of Theorem 8.2, Nevanlinna [1953] called a polar set a set of absolute harmonic measure 0. Corollary 8.3. Suppose is a domain such that Cap(C∗ \ ) > 0, and suppose u is subharmonic and bounded above in . If lim sup u(z) ≤ m z→ζ

for all ζ ∈ ∂ \ A, where A is a Borel set and Cap(A) = 0, then sup u(z) ≤ m.

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Corollary 8.3 is immediate from Theorems 8.1 and 8.2. Corollary 8.4. Suppose E is a compact polar set. If u(z) is subharmonic and bounded above in C∗ \ E, then u is constant. If ⊃ E is a Jordan domain then every bounded harmonic function on \ E extends to be harmonic on E. Conversely, if Cap(E) > 0, if is the unbounded component of C∗ \ E, and if A is a compact subset of E such that 0 < μ E (A) < 1, then u(z) = ω(z, A, ) is a non-constant bounded harmonic function on . If E is a compact polar set then by Corollary 8.4 every positive harmonic function in C∗ \ E is constant and every bounded harmonic function in C∗ \ E is constant. Proof. Let z 0 ∈ C∗ \ E, δ < dist(z 0 , E), and = C∗ \ (E ∪ B(z 0 , δ)). Then Cap(C∗ \ ) > 0 and by Corollary 8.3, sup u(z) ≤ sup u(z).

∂ B(z 0 ,δ)

Letting δ → 0 we obtain sup u(z) ≤ u(z 0 ).

C∗ \E

Because z 0 is arbitrary in C∗ \ E, u is constant. To prove the second statement, we shrink so that u is continous on ∂. Let v be the solution to the Dirichlet problem on for the function u. Then by Corollary 8.3, u = v and v is the harmonic extension of u to . The converse assertion is clear.

9. Estimates for Harmonic Measure We use capacity and potentials to estimate ω(z, E, D \ E) where E is a closed subset of the unit disc D. By (5.3) we may assume ∂ E ⊂ D is compact and consists of ﬁnitely many analytic curves. First consider the case when dist(E, ∂D) is large. Theorem 9.1. Suppose E ⊂ {z : 0 < δ < |z| < r < 1} and set γ (E) = − log (Cap(E)) . Then there are constants c1 (r ) and c2 (δ, r ) such that c2 (δ, r ) c1 (r ) ≤ ω(0, E, D \ E) ≤ . γ (E) γ (E)

(9.1)

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Proof. Let

105

1 − zζ g(z, ζ ) = log z−ζ

be Green’s function on D and let μ E be the equilibrium distribution for E. Then the Green potential V (z) = g(z, ζ )dμ E (ζ ) ∂E

is continuous on D \ E and harmonic on D \ E. Also, log r1 ≤ V (0) ≤ log 1δ , V = 0 on ∂D, and for z ∈ E, γ (E) + log(1 − r 2 ) ≤ V (z) ≤ γ (E) + log(1 + r 2 ) by (2.3). Hence

log 1δ log r1 ≤ ω(0, E, D \ E) ≤ , γ (E) + log(1 + r 2 ) γ (E) + log(1 − r 2 )

(9.2)

provided that γ (E) + log(1 − r 2 ) > 0. This gives the left-hand inequality in (9.1) with c1 (r ) = (log r )2 / log(r + 1/r ), because γ (E) ≥ log(1/r ). If γ (E) ≥ 2 log 1/(1 − r 2 ) the upper bound in (9.2) gives the upper bound in (9.1) with constant 2 log(1/δ), and if γ (E) ≤ 2 log 1/(1 − r 2 ), the upper bound in (9.1) follows from the trivial estimate ω(0, E, D \ E) ≤ 1 ≤

1 2 log 1−r 2

γ (E)

.

The next result is called the Beurling projection theorem. Theorem 9.2 (Beurling). If E ⊂ D \ {0}, and E ∗ = {|z| : z ∈ E} is the circular projection of E, then ω(z, E, D \ E) ≥ ω(−|z|, E ∗ , D \ E ∗ ). Proof. By Green’s theorem 1 ω(z, E, D \ E) = ω(z) = 2π

∂E

g(z, ζ )

∂ω(ζ ) ds ∂nζ (9.3)

=

∂E

g(z, ζ )dσ (ζ ),

in which the normal nζ points out of D \ E, so that σ ≥ 0. Consider the circular projection σ ∗ of σ , σ ∗ (A) = σ ({z ∈ E : |z| ∈ A}),

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and its Green potential

V (z) =

g(z, |ζ |)dσ ∗ .

(∂ E)∗

z E E

−|z| 0

E∗ E∗

E∗ E

D

Figure III.5 Circular Projection in D. Because g(−|z|, |ζ |) ≤ g(z, ζ ) ≤ g(|z|, |ζ |), we have V (−|z|) ≤ ω(z, E, D \ E) ≤ V (|z|). Let z n → z ∈ E. By the right-hand inequality, lim inf V (|z n |) ≥ 1, so that by a comparison, ω(z, E ∗ , D \ E ∗ ) ≤ V (z). With the left-hand inequality that gives ω(−|z|, E ∗ , D \ E ∗ ) ≤ V (−|z|) ≤ ω(z),

which proves the theorem.

Corollary 9.3. Let be a simply connected domain, let z ∈ and let ζ ∈ ∂. Then for r sufﬁciently small, 1

ω(z, B(ζ, r ) ∩ ∂, ) ≤ Cr 2 .

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107

Proof. We may assume ζ = 0 and z = −1. We project ∂ \ B(0, r ) to the positive real axis and apply Beurling’s theorem in C∗ \ B(0, r ) to get ω(z, B(0, r ) ∩ ∂, ) ≤ ω (−1, B(0, r ), C \ (B(0, r ) ∪ [r, ∞))) √ √ i− r 4 2 = tan−1 r . = arg √ π π i+ r

∂

B(0, r )

−1 Figure III.6 The proof of Corollary 9.3. The case when dist(E, ∂D) is small leads to the connection between harmonic measure and Carleson measures. We work in the upper half-plane H and with dyadic Carleson boxes of the form Q = { j2−k ≤ x ≤ ( j + 1)2−k ; 0 < y ≤ 2−k }, and write (Q) = 2−k . Fix z 0 = i and for convenience assume E is compact and E ⊂ 0 ≤ x ≤ 1, 0 < y ≤ 1/2 . (9.4) Let A = A E be the set of positive measures σ on E such that (i) σ is a Carleson measure with constant 1, σ (Q) ≤ (Q),

(9.5)

for every dyadic Carleson box Q (see Exercise I.19), and (ii) the top half T (Q) = Q ∩ {y ≥ (Q)/2} of every dyadic Carleson box Q satisﬁes 1 σ (T (Q)) ≤ (Q) 1 + γ (E ∩ T (Q)) − γ (T (Q)) (9.6) =

1 + log

1 Cap T (Q) Cap E∩T (Q)

.

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Theorem 9.4. Assume E satisﬁes (9.4). Then there are constants C1 and C2 such that C1 ω(i, E, H \ E) ≤ sup{σ (E) : σ ∈ A E } ≤ C2 ω(i, E, H \ E).

(9.7)

In the special case z = 0 and E ⊂ {|ζ | > 1/2}, Beurling’s projection theorem (with a larger constant) is a consequence of Theorem 9.4. Suppose σ0 (A) = ω(0, A ∩ E ∗ , D \ E ∗ ) and let σ be any positive measure on E that projects circularly onto σ0 . Then an elementary comparison gives σ (Q) ≤ σ0 (Q ∗ ) ≤ C(Q ∗ ) for some constant C, so that C −1 σ satisﬁes (9.5). To check (9.6) for Cσ notice that by (4.12) the right side of (9.6) becomes smaller when E is replaced by E ∗ and then apply the right-hand inequality in (9.2) with δ = 1 − (Q) and r = (1 + δ)/2. When E ⊂ H we still write E ∗ = {|z| : z ∈ E} for the circular projection of E, but for z = x + i y ∈ H we deﬁne z ∗ = −|x| + i y. The estimate ω(z, E, H \ E) ≥

2 ω(z ∗ , E ∗ , H) 3

from Hall [1937] is known as Hall’s lemma. With a different constant, the most important special case, E ⊂ {arg z < π/4} and z = z ∗ of Hall’s lemma also follows easily Theorem 9.4. See Exercise 20. Proof of Theorem 9.4. Write ζ = ξ + iη ∈ E. Green’s function for the halfplane H is z − ζ g(z, ζ ) = log , z − ζ and g(z, ζ ) satisﬁes g(z, ζ ) = log and

η + O(1), |z − ζ |

if

z − ζ 2 4yη , g(z, ζ ) ∼ −1= z − ζ |z − ζ |2

|z − ζ | |z − ζ |

if |z − ζ | >

The Green potential of a measure μ on H is G μ (z) = g(z, ζ )dμ(ζ ). H

≤ c < 1,

η . 4

(9.8)

(9.9)

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Let μ Q denote the (logarithmic) equilibrium distribution of E ∩ T (Q). Then for z ∈ E ∩ T (Q) we have G μ Q (z) = g(z, ζ )dμ Q (ζ ) ∼ 1 + γ (E ∩ T (Q)) − γ (T (Q)) by (9.8). Thus if we deﬁne νQ =

μQ , 1 + γ (E ∩ T (Q)) − γ (T (Q))

then for some constant C C −1 ≤

E∩T (Q)

g(z, ζ )dν Q ≤ C

(9.10)

on E ∩ T (Q). To prove the left half of (9.7) we may assume E ⊂ {y ≥ 2−N }. We inductively deﬁne a measure ν = ν N on E as follows: Set ν0 = 0. Assume that νk is deﬁned and has support in E ∩ {ζ : η ≤ 2−N +k }. Let Q ∈ Dk = {Q dyadic : (Q) = 2−N +k+1 and Cap(E ∩ T (Q)) > 0}. If inf E∩T (Q) G νk < 1, take β Q > 0 such that inf (G νk (z) + β Q G ν Q (z)) = 1.

E∩T (Q)

(9.11)

If inf E∩T (Q) G νk ≥ 1, take β Q = 0. Then deﬁne βQ νQ . νk+1 = νk + Dk+1

Finally take ν = ν N , and σ = ην. Then by construction G ν ≥ 1 on E and G ν = 0 on R. But by (9.4) and (9.9), G ν (i) ≤ C ηdν(ζ ) = Cσ (E). E

Therefore by the maximum principle ω(i, E, H \ E) ≤ G ν (i) ≤ Cσ (E), and if we can show that cσ ∈ A E for some constant c, then the left-hand inequality in (9.7) will hold with C1 = (cC)−1 .

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To verify (9.5) for cσ let Q be a dyadic box of side 2−N +k+1 and ﬁrst assume β Q > 0. Then by (9.11) there exists z ∈ E ∩ T (Q) such that 1 = G νk (z) + β Q G ν Q (z) g(z, ζ )

≥ Q

dσ η

≥C

σ (T (Q)) +C (Q)

≥C

σ (Q) , (Q)

Q\T (Q)

y dσ |z − ζ |2

and therefore (9.5) holds for Q. But if β Q = 0 then σ (T (Q)) = 0 and σ (Q) = {σ (Q ) : Q ⊂ Q, β Q > 0, Q maximal}, so that (9.5) also holds for Q. To verify (9.6) for the dyadic box Q observe that σ (T (Q)) ≤ ν(T (Q)) (Q) ≤

βQ 1 + γ (E ∩ T (Q)) − γ (T (Q))

≤C by (9.10) and (9.11). To prove the right-hand part of (9.7) let σ ∈ A E and set σ σ = σ (T (Q))μ Q and ν = . η Q dyadic

Then G ν is harmonic on H \ E and Gν = 0 on R since E is compact. By (9.4) and (9.9) ηdν = σ (E). G ν (i) ∼ E

(9.12)

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Now let z ∈ E and ﬁx a dyadic box Q with z ∈ T (Q). Then by (9.5) and (9.9) dσ (ζ ) g(z, ζ )dν(ζ ) ≤ C y 2 |ζ −z|≥(Q)/4 |ζ −z|≥(Q)/4 |z − ζ | ∞ σ ({|ζ − z| ≤ 2n−2 (Q)}) ≤ Cy 22n (Q)2 n=0

≤ C, and by (9.6) and (9.8)

|ζ −z|<(Q)/4

g(z, ζ )dν(ζ ) ≤ C

for some constant C. Therefore G ν (z) ≤ C on E and by (9.12) ω(i, E, H \ E) ≥ C −1 G ν (i) ≥ Cσ (E).

That proves the right-hand inequality in (9.7).

Every theorem in this section was proved using Green potentials and Green potentials have a theory parallel to the potential theory in Sections 2, 3, and 4. The Green potential of a signed measure σ supported on E ⊂ D is the function g(z, ζ )dσ (ζ ). G σ (z) = E

Assume E is compact and Cap(E) > 0. Then by (9.3) and a limiting process there is a positive measure ν on E such that ω(z, E, D \ E) = G ν (z).

(9.13)

Moreover, there exists unique μ ∈ P(E) having minimal Green energy:

" G τ dτ : τ ∈ P(E) ≡ γG (E). G μ dμ = inf The minimum energy γG (E) is called the Green capacity of E, and except for a set of capacity zero, G μ = γG (E) on E. Then (9.13) becomes ω(z, E, D \ E) =

1 G μ (z). γG (E)

See Exercise 18 for the proofs of these results and for an application.

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Notes Sections 1 through 6 follow a UCLA lecture course given by Lennart Carleson. Most of Section 7 is in Wiener’s original paper [1924b]. Actually, Wiener did not prove Theorem 7.1 in [1924b]. He proved the Rd , d ≥ 3, version of Theorem 7.1, but for C = R2 he proved ζ ∈ ∂ is regular if and only if ∞ n=1

2n = ∞, γ (Jn \ )

(N.1)

where Jn =

&

Ak : 2n−1 < k ≤ 2n = z : 2n−1 ≤ log2

1 ≤ 2n . |z − ζ |

In Rd Wiener’s proof hinged on an elegant estimate of Green potentials (see [1924b], p. 133). In R2 the analogous estimate holds for the big rings Jn , but it is 1 decays more slowly in R2 than |X1 | does in false for the rings An because log |z| Rd . The ﬁrst proof of Theorem 7.1 was published by de la Vallée Poussin [1949]. See also Tsuji [1959]. Our proof follows Tsuji [1959] when the series diverges and copies Wiener [1924b] when it converges. See Exercise 9 for Wiener’s proof of (N.1) and for the equivalence of (N.1) and (7.2). Corollary 7.3 is from Beurling’s thesis [1933]. Theorem 9.4 is from Carleson [1982]. Potential theory is a vast subject and we have only touched on the two-dimensional theory. For the broader picture we refer to the books of Brelot [1965], Carleson [1967a], Landkof [1972], Ransford [1995], Tsuji [1959], and Wermer [1974]. See Doob [1984] for the connections with probability, and see the books of Heinonen, Kilpeläinen, and Martio [1993] and Adams and Hedberg [1996] for nonlinear potential theory. Appendix E gives a geometric description of capacity by transﬁnite diameter and constructs Evans functions for sets of capacity zero.

Exercises and Further Results 1. Suppose p(z) = ak z k + . . . + a0 is a polynomial. Set = {| p| > ε} and ω = ω(∞, ., ). Then p(z) 1 . (i) g (z, ∞) = log k ε a 1 k (ii) γ (∂) = log . k ε 1 p (z) dz ∂g ds = . (iii) dω = − ∂n 2π k p(z) 2πi

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(iv) The harmonic measure at ∞ of a component J of ∂ is the number of zeros of p in the bounded component of C \ J , divided by k. (v) Formulate and prove similar results for rational functions. 2. Let be a ﬁnitely connected Jordan domain such that ∞ ∈ and every component of ∂ is a C 2 curve. Let G(z) = g (z, ∞) and let {z j } be the (ﬁnite) set of critical points of G in . Prove ∂G ∂G 1 log G(z j ), ds = γ (∂) + 2π ∂ ∂n ∂n where γ (∂) is Robin’s constant for ∂. This identity from Ahlfors [1947] will play a critical role in Chapter IX. Hint: Use Green’s theorem and the Taylor expansion of G x − i G y near each critical point z j . 3. Let be a bounded plane domain and let v(z) be subharmonic in a neighborhood of . Then there is a unique positive ﬁnite measure μ supported on and a unique function u(z) harmonic on such that v(z) = u(z) − Uμ (z) on . Hint: If v ∈ C 2 (), take dμ = v 2π d xd y. In the general case, replace v by v ∗ K ε as in the proof of Theorem 3.1 and take a limit. This result is known as the F. Riesz decomposition theorem for subharmonic functions. 4. (a) Let g be a compactly supported real valued C 2 function on the plane. Then g = ϕ1 − ϕ2 where ϕ1 and ϕ2 are C 2 and subharmonic on an open set containing {z : g(z) = 0}. (b) Let be a bounded plane domain and let n ⊃ n−1 . . . be an increasing sequence of ﬁnitely connected subdomains of such that = n n and ∂n consists of C 2 Jordan curves. Fix z ∈ 1 and consider the sequence μn of probability measures on deﬁned by μn (E) = ω(z, E ∩ ∂n , n ). Show that if ϕ is continuous on and subharmonic on , then ϕdμn ≤ ϕdμn+1 . Using this result, part (a), and the boundedness of {μn }, prove that for all f ∈ C(), f dμn exists. lim n

It follows that for bounded the Wiener deﬁnition of harmonic measure does not depend on the approximating sequence n . (c) In Theorem 4.1, if γ = lim γn = ∞, then for any μ ∈ P(E) the sequence {E n } can be chosen so that μ is the weak-star limit of the sequence {μn }.

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5. A property is said to hold p.p. (for presque partout) on a set E if the subset of E where the property fails has capacity zero. Suppose E is compact and γ (E) > 0. Let M(E) = {ν : ν is a signed measure supported on E}. Then 1 = inf{ν(E) : ν ∈ M(E) and Uν ≥ 1 p.p. on E} γ (E) = sup{ν(E) : ν ∈ M(E) and Uν ≤ 1 p.p. on E}. 6. Let = D\ {0} and let f ∈ C(∂). Prove lim z→0 u f (z) = f (0) if and only dθ . if f (0) = ∂ D f (eiθ ) 2π 7. Let p(z) = z d + ad−1 z d−1 + . . . + a0 be a monic polynomial of degree d ≥ 2 and let p (n) be the n-th iterate p (n) (z) = p ◦ p ◦ . . . ◦ p(z) and write

−1

p (−n) = p (n) .

The Julia set of p(z) is E = z : p (n) is not a normal family in any neighborhood of z . Here “normal” includes convergence to the constant f = ∞. Thus, for example, if p(z) = z 2 then E is the unit circle {|z| = 1}. The Fatou set of p(z) is the complement C ∗ \ E. The basin of attraction of ∞ for p(z) is = z : lim p (n) (z) = ∞ n→∞

C∗

and ⊂ \ E. It is known that the Julia set E is a non-empty, compact, perfect set, that E is completely invariant, E = p(E) = p−1 (E), and that E = ∂. See Chapter III of Carleson and Gamelin [1993]. This exercise gives Brolin’s beautiful [1965] description of the harmonic measure ω(∞, ·, ). (a) Let U0 be a disc such that U0 ⊂ p(U0 ), so that C∗ \ U0 ⊂ . The disc U0 = {|z| < 1 +

d−1

|a j |}

j=0

is an example. If Un = p (−n) (U0 ) and E n = Un then p(E n+1 ) = E n , E n = E. Let gn (z) denote Green’s function for p −1 (E n ) = E n+1 , and C ∗ \ E n with pole at ∞. Then gn (z) = log |z| + γn + o(1),

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where Cap(E n ) = e−γn . Show that gn ( p(z)) , d by uniqueness. Then conclude that γn+1 = γn /d and Cap(E) = 1. (b) Let μ E be the equilibrium distribution for E. It is known that if V is open and V ∩ E = ∅, then for some n, p (n) (V ∩ E) = E. Again see Carleson and Gamelin [1993]. Use this fact and Theorem 4.5 to show that μ E has closed support E. (c) If V is a neighborhood of z ∈ E then by Montel’s theorem n p (n) (V ) can omit at most two values ζ1 and ζ2 , and ζ1 and ζ2 do not depend on / {ζ1 , ζ2 } and take the point mass μ0 = δz 0 . the point z. Fix any point z 0 ∈ (−n) (z 0 ) = {z 1 , . . . , z d n }, counting points with their multiplicities, Write p and form the discrete measure 1 δz j . μn = n d gn+1 (z) =

j

Show μn ( p(A)) = μn+1 (A) for any Borel set A. Show μn converges weakstar to μ E . Hint: Use the identity | p (n) (z) − z 0 | = |z − z j | j

to estimate Uμn , and use Lemma 4.2 to show that any weak-star convergent subsequence of {μn } has limit μ E . (d) If A ⊂ E, then μ E ( p −1 (A)) = μ E (A). In other words, p is a measure preserving transformation on the measure space (E, μ E ). Moreover, p is strongly mixing: If f, g ∈ L 2 (E, μ E ), then (n) f ( p (z))g(z)dμ E (z) = lim f dμ E gdμ E . n→∞

8. (a) Prove ζ is a regular point for if and only if lim z→ζ g(z, z 0 ) = 0. Give a proof with Wiener series and give a proof without Wiener series. (b) Prove ζ is a regular point for if and only if ζ is a regular point for ∩ B(ζ, r ), r > 0. Give a proof with Wiener series and give a proof without Wiener series. (c) Let j = C \ E j , where E 1 and E 2 are disjoint compact sets and let = 1 ∩ 2 . Prove ζ ∈ ∂ is regular for if and only if ζ is regular for 1 or ζ is regular for 2 . 9. Let ζ ∈ ∂. (a) Let α < 1 and set G n = {z : α n ≤ |z − ζ | ≤ α n−1 }. Prove

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ζ is a regular point if and only if ∞

n

n=1

γ (G n \ )

= ∞,

and if and only if ∞ n=1

n = ∞. γ (G n ∩ ∂)

(b) Lemma 7.5 has a converse. Let Jn = {Ak : 2n−1 < k ≤ 2n }. There is a universal constant C < ∞ so that if γ (Jn \ ) > C2n and if E k ⊂ Ak for 2n−1 < k ≤ 2n then n

2 2n−1 +1

1 4 ≤ . γ (E k ) γ (∪E k )

Hint: If μ = μk /γ (E k ) where μk is the equilibrium distribution for E k , then ||μ|| = 1/γ (E k ). By estimating Uμk dμm as in the proof of Theorem 7.1, show that I (μ) ≤ 3||μ|| + C1 2n ||μ||2 . Now apply Theorem 4.1 and the hypothesis on γ (Jn \ ) above. (c) Set Br = {z : |z − ζ | ≤ r }. Prove that the following conditions are equivalent to the regularity of ζ : (i)

∞

(ii) 0

1

(iii) 0

(iv) (v)

2n = ∞, γ (Jn \ )

1 ds = ∞, γ (B2−s \ )

1 dr = ∞, γ (Br \ ) r 1 γ (B2−n \ )

= ∞,

n = ∞. γ (An \ )

Hint: Note that the integrand in (ii) is decreasing. Bound the left side of each displayed formula by a multiple of the left side of the subsequent formula. If the ﬁrst sum is inﬁnite, then apply Theorem 7.1 to conclude ζ is regular, and if the ﬁrst sum is ﬁnite, use part (b) to show the sum in (v) is ﬁnite.

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(d) Wiener’s paper [1924b] proved that (i) implies ζ is a regular point. Here is his argument. Suppose ζ = 0 and suppose ∞ 2n 2 n=1

γ2n

= ∞,

(E.1)

where γn = γ (Jn \ ). Set rn = 2−2 . Then Jn = {z : rn ≤ |z| ≤ rn−1 } and Jn ⊂ Bn−1 = {z : |z| ≤ rn−1 }. Deﬁne rn−2 − z w/r ¯ n−2 dμn (w) log , Wn (z) = γ z−w n Jn \ n

where μn is the equilibrium distribution for Jn \ . The function Wn is a Green potential because the kernel in its deﬁnition is Green’s function for the disc Bn−2 . Then Wn (z) = 0

on |z| = rn−2 ,

2 rn−1 1 Wn (z) ≤ 1 + log rn−2 + ≤1 γn rn−2

(E.2)

on Jn ∩ ,

for n ≥ 2, and the critical rn−2 − rn+1rn−1 /rn−2 C2n 1 ≥ Wn (z) ≥ log γn rn+1 + rn−1 γn

(E.3)

on |z| ≤ rn+1 . (E.4)

Let Vn,m be the Wiener solution to the Dirichlet problem on the domain B0 \ ( m p=0 J2n+2 p \ ) for the boundary data

0, if z ∈ m p=0 J2n+2 p \ , Vn,m (z) = 1, if |z| = r0 . We show by induction on m that m C22n+2 p 1− Vn,m ≤ γ2n+2 p

on {z : |z| < r2n+2m+1 }.

(E.5)

p=0

2 To prove (E.5), we may assume m p=0 J2n+2 p \ is bounded by disjoint C Jordan curves. Then by the maximum principle, (E.2), (E.3), and (E.4), Vn,0 ≤ 1 − W2n ≤ 1 −

C22n . γ2n

on B2n+1 . If (E.5) holds for m = k − 1, then by the maximum principle k−1 - C22n+2 p Vn,k (z) ≤ 1 − W2n+2k (z) 1− γ2n+2 p p=0

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on B2n+2k+1 , so that by (E.4), we obtain (E.5) when m = k. Now by (E.1) and (E.5) lim Vn,m (0) = 0.

m→∞

For |z| = r2n−2 and all m, Vn,m (z) ≥ ω(z, B0 , B0 \ Dn ) ≥ εn , where Dn = {z : |z| ≤ r2n−1 } and it follows via Theorem 6.1 that 0 is a 22n+1 regular point. A similar argument works if γ2n+1 = ∞.

10. (a) Let E ⊂ { 21 ≤ |z| ≤ 1} be a continuum with E ∩ {z : |z| = 21 } = ∅ and E ∩ {|z| = 1} = ∅. Then 2 1 −1 ω(0, E, D \ E) ≥ tan √ . π 8 This follows from Beurling’s projection theorem, but it also has an elementary proof, albeit with a smaller constant: We can assume 21 ∈ E. Set E = {z : z ∈ E}. Since E is a continuum, the component U of D \ (E ∪ E) satisfying 0 ∈ U has [ 21 , 1] ∩ U = ∅ and ∂U ⊂ ∂D ∪ E ∪ E.

E 1 2

E U

Figure III.7 Lower bound via reﬂection. Then by the maximum principle, ω(z, E, D \ E) + ω(z, E, D \ E) ≥ ω(z, (E ∪ E) ∩ ∂U, U )

1 1 ≥ ω z, [ , 1], D \ [ , 1] . 2 2 Hence

1 2 1 1 2ω(0, E, D \ E) ≥ ω(0, [ , 1], D \ [ , 1]) = tan−1 √ . 2 2 π 8

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See Milloux [1924] and Garnett [1986]. (b) Let ϕ : D → be conformal, let I be an arc of ∂D with length |I | < 1 and center ζ I and let z I = (1 − |I |)ζ I . Let ω denote the harmonic measure on for the point ϕ(0). Show that if C > 1, then

ω B(ϕ(z I ), Cdist(ϕ(z I ), ∂)) ≥ c|I |, where c is a constant depending only on C. 11. Let K be a compact set and set = C∗ \ K . The set K is called uniformly perfect if there is a constant A such that Cap(K ∩ B(z, δ)) > Aδ

(E.6)

for all z ∈ K and all δ < diam(K ). (a) Condition (E.6) holds if and only if there is a constant η > 0 such that ω(z, B(z, 2d(z)), ) ≥ η for all z ∈ , where d(z) = dist(z, K ). Exercise IX.3 will give several other conditions equivalent to (E.6). (b) Assume K satisﬁes (E.6), let U be an open subset of ∂ and let u(z) be continuous on and harmonic on . If u = 0 on U and if J ⊂ U is compact, then u(z) ≤ C(dist(z, J ))α for constants C = C(J ) and α = α(A) satisfying 0 < α ≤ 1. See Jerison and Kenig [1982a], Lemma 4.1. In particular, if K is uniformly perfect and g(z) = g (z, ∞), then on , g(z) ≤ C(dist(z, K ))α

(E.7)

and |g(z 1 ) − g(z 2 )| ≤ C |z 1 − z 2 |α . (c) However, (E.7) does not imply (E.6). If K = {0} ∪

∞ &

{|z| = 2−n , |argz| ≥ εn } 2

n=1

for small εn , then (E.7) holds but (E.6) fails. 12. Øksendal [1972] showed that for any plane domain, harmonic measure is singular to area measure (although ∂ may have positive area). The following proof is from a lecture by Carleson: Let K be a compact subset of ∂ with Area(K ) > 0 and suppose ζ is a point of area density of K : Area(K ∩ B(ζ, δ)) >1−ε π δ2

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for small δ. Then there is

δ 2

< r (δ) <

3δ 4

so that

|{|z − ζ | = r (δ)} \ K | < 6ε, 2πr (δ) and by a comparison, δ ω(z, B(ζ, ) ∩ ∂, ) < 42ε 2 on ∩ {|z − ζ | = δ}. Taking 42ε < 1/4 yields 22n ω(z, B(ζ, 2−n ) ∩ ∂, ) −→ 0, and by the Radon–Nikodym theorem ω is singular to area measure. 13. Let be a domain such that Cap(∂) > 0 and let E ⊂ ∂ be a Borel set such that ω(z, E, ) > 0 for some z ∈ . Prove sup ω(z, E, ) = 1.

Find a proof without using the Perron solution. 14. Let be a simply connected domain and let E ⊂ R ∩ ∂ have |E| = 0. Prove ω(z, E, ) = 0 for all z ∈ . Hint: The result is easy if ⊂ H or if = {z : z ∈ }. In general, we can assume 0 = ∩ H = ∅. Then 0 is simply connected. Write R ∩ = Jj where Jj is an open interval and let j be the component of ∩ {z : z ∈ } with Jj ⊂ j . Then j is also simply connected. Now if z 0 ∈ 0 , then ω(z 0 , E, ) = ω(ζ, E, )dω0 (z 0 , ζ ) j

=

Jj

j

Jj

∂ j \∂

ω(z, E, )dω j (ζ, z)dω0 (z 0 , ζ ),

but by symmetry ω j (ζ, ∂ j \ , j ) = 1/2 for ζ ∈ Jj . The result now follows from Exercise 13. See Øksendal [1980]. 15. (a) Let 0 ∈ , let ζ ∈ ∂ satisfy |ζ | > 1, and let 0 < r < 21 . Then ω(0, ∂ ∩ B(ζ, r ), ) ≤ C

sup |z−ζ |< 3r 2

g (z, 0)

with constant C independent of . Hint: Assume is a ﬁnitely connected Jordan domain with smooth boundary. Take χ ∈ C ∞ (B(ζ, 3r2 )) such that

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121

0 ≤ χ ≤ 1, χ (z) = 1 on B(ζ, r ), and |χ | ≤ rC2 . Write −1 ∂g(0, w) χ ds(w) ω(0, ∂ ∩ B(ζ, r ), ) ≤ 2π ∂ ∂nw and apply Green’s theorem. See Jerison and Kenig [1982a]. (b) Let be a simply connected domain containing the disc D, let z 1 ∈ satisfy |z 1 | > 1, and set g(z) = g (z, z 1 ). Let be the hyperbolic geodesic of joining 0 to z 1 and for 0 ≤ t ≤ 1 deﬁne δ(t) = Then

sup

∩{|z|=t}

dist(z, ∂).

1 1 dt . g(0) ≤ C0 exp − 2 0 δ(t)

Hint: Let ρ(z, w) be the hyperbolic distance on and for 0 ≤ t ≤ 1 pick z t ∈ with |z t | = t. Then g(z) ∼ e−2ρ(z,z 1 ) and by Koebe’s estimate 1 ds dt ≥ . ρ(0, z 1 ) ≥ 4 dist(z, ∂) 0 4δ(t) Carleson and Jones [1992]. (c) Let be a simply connected domain containing the disc{|z| < 21 }, let ζ ∈ ∂ satisfy |ζ | > 1, and let δ(t) = sup{dist(z, ∂) : z ∈ , |z − ζ | = t}. Then for r < 14 ,

1 1 dt ω(0, ∂ ∩ B(ζ, r ), ) ≤ C exp − . 2 3r2 δ(t)

Hint: Use (a) and (b). See Jones and Makarov [1995]. (d) Part (c) is false with 23 r replaced by r . Hint: ζ = 2, = 2D \ {|z − 2| ≤ r, |y| ≥ ε}. (e) Let = {z : | arg z| < θ } and Br = {|z| ≤ r }. Use part (c) to show ω(1, Br , \ Br ) ≤ Cr 1/(2 sin θ) . Also show by an explicit conformal map that ω(1, Br , \ Br ) = Cr π/θ . Theorem IV.6.2 gives the accurate upper bound Cr π/θ .

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For related results see also Beurling [1933], Exercise 24, Theorem G.1, and the comments following the proof of Theorem G.1. 16. (a) The following result is from Gehring and Hayman [1962]. Let ϕ : D → be univalent and assume ϕ extends continuously to the open arc σ = ∂D ∩ {Imz > 0}. Then

ϕ((−1, 1)) ≤ K ϕ(σ ) , with constant K not depending on ϕ. Set I j = [1 − 2− j , 1 − 2− j−1 ] and σ j = σ ∩ (B(1, 2− j ) \ B(1, 2− j−1 ), for j = 0, 1, . . . . Then clearly inf ω(z, σ j , D) ≥ c1 > 0.

(E.8)

z∈I j

We may assume j = (ϕ(σ j )) < ∞. Suppose we have sup |ϕ (z)| ≤ c2 2 j j

(E.9)

Ij

for all j. Then summing over j, and doing the same for −I j we get 1 |ϕ (x)|d x ≤ c2 (ϕ(σ )), −1

which gives the Gehring–Hayman theorem. Now by Theorem I.4.3, (E.9) holds if and only if dist(ϕ(z), ∂) ≤ c3 j

(E.10)

for some constant c3 and for all z ∈ I j . Let c3 be large and let β j be the arc length midpoint of the curve ϕ(σ j ). Then B j = B(β j , c3 j ) ⊃ ϕ(σ j ).

If (E.10) fails at z j , then ϕ(z j ) ∈ / B j . But using the inversion T z = z−βj j and the continuum E = D ∩ T (∂), we see by (E.8) and the Beurling projection theorem that (E.10) will hold at z j if c3 is sufﬁciently large. (b) Here are two consequences. If there is an arc γ ⊂ D, joining 0 to 1 such that (ϕ(γ )) = |ϕ (z)|ds = < ∞, γ

then

ϕ((0, 1)) =

0

1

|ϕ (x)|d x ≤ K .

(E.11)

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123

Write Iz = ∂D ∩ {|ζ − z| < 2(1 − |z|)}, for z ∈ D. If ϕ extends continuously to Iz then (ϕ(Iz )) ≥ K dist(ϕ(z), ∂). Inequality (E.11) will be used in the next chapter. See Heinonen and Näkki [1994], Heinonen and Rohde [1993], and Bonk, Koskela, and Rohde [1998] for extensions. 17. Suppose E ⊂ {ζ : |ζ | < δ} and suppose δ < r < 1. Then there are constants c1 (δ, r ) and c2 (δ, r ) such that if |z| = r , then c1 (δ, r ) c2 (δ, r ) ≤ ω(z, E, D \ E) ≤ . γ (E) γ (E) This can be proved directly by mimicking the proof of Theorem 9.1, or derived from the statement of Theorem 9.1 by relating γ (E) to γ (T E) where T is a Möbius of the disc. transformation z be Green’s function for D and let σ be a compactly 18. Let g(z, ζ ) = log 1−ζ z−ζ supported signed measure on D. (a) Prove that the Green potential G σ (z) = g(z, ζ )dσ (ζ ) satisﬁes J (σ ) = G σ dσ ≥ 0 and J (σ ) = 0 ⇐⇒ σ ≡ 0. Hint: Follow the argument in Section 3. (b) Let E ⊂ D be compact. Prove Cap(E) > 0 if and only if there exists μ ∈ P(E) such that J (μ) < ∞. Assuming Cap(E) > 0, prove there is a unique μ ∈ P(E) such that

" G μ dμ = inf G τ dτ : τ ∈ P(E) ≡ γG (E). Also show that G μ (z) = γG (E) for all z ∈ E except for set of capacity zero and that on D \ E ω(z, E, D \ E) =

1 G μ (z). γG (E)

(c) Let E ⊂ [−1, 0) be compact and let E ∗ = [−1, a] be a closed interval dt dt with the same logarithmic measure as E, = . Prove ∗ t E E t ω(0, E, D \ E) ≥ ω(0, E ∗ , D \ E ∗ ). This result, called the “Beurling shove theorem”, is also from Beurling’s thesis [1933]. See Essèn and Haliste [1989], Pruss [1999], and Betsakos and Solynin [2000] for reﬁnements. = { 19. Write z = x + i|y| when z = x + i y and write F z : z ∈ F}.

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(a) Let F be a compact subset of D, and let z ∈ D \ F. Prove D \ F). ω(z, F, D \ F) ≥ ω( z, F, Hint: The left side is superharmonic in D and the right side is harmonic in D \ F. Consequently we may assume z ∈ D ∩ R. In fact, via a Möbius transformation of the disc, we can assume z = 0. Represent the right side as the Green potential of a positive measure μ on F. Write F + = F ∩ {Imz ≥ 0} deﬁne and F − = F ∩ {Imz < 0}. When A ⊂ F, ν(A) = μ(F + ∩ A) + μ(F − ∩ A) D \ F). and let v(z) be the Green potential of ν. Show v(z) ≥ ω(z, F, (b) Let E be a compact plane set of positive capacity, let = C \ E, and let Prove ∗ = C \ E. ) g (x, w) ≤ g∗ (x, w for x ∈ R ∪ {∞} and w ∈ C. Hint: We may take x = ∞. Now use the general and trivial fact

g (∞, w) = lim log Rω w, ∂ B(0, R), B(0, R) \ E . R→∞

(c) Let be a simply connected domain with 0 ∈ , let a = inf ∩ R, b = sup ∩ R, and E = (−∞, a) ∪ ∂ ∪ (b, ∞), containing the lower half-plane. Then and let 1 be the component of C \ E 1 is simply connected. Let ψ : → D and ψ1 : 1 → D be conformal mappings with ψ(0) = ψ1 (0) = 0. Prove |ψ (x)| ≤ 4|ψ1 (x)|,

x ∈ R.

Consequently the Hayman–Wu theorem, Theorem I.5.1, can be reduced to the special case of L ⊂ . That case has been proved by B. Flinn [1983]. Parts (a), (b), and (c) are attributed to Baernstein in Fernández, Heinonen, and Martio [1989]. 20. Hall’s lemma asserts that if E ⊂ H is compact, E ∗ = {|z| : z ∈ E}, and z ∗ = −|Rez| + iImz, then ω(z, E, H \ E) ≥

2 ω(z ∗ , E ∗ , H). 3

(E.12)

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z∗ E

E

125

z E

E∗ E∗ E∗ Figure III.8 Hall’s lemma. (a) Derive the special case z = z ∗ and E ⊂ {arg z < π4 } of Hall’s lemma from Theorem 9.4. Hint: For 0 < x < 1, x ≤ 1/(1 + log(1/x)). Split E into E ∩ D and E ∩ C \ D, using a reﬂection to estimate the harmonic measure of the latter. (b) The general case is also proved using Green potentials. See Hall [1937] or Duren [1970]. (c) Contrary to intuition, (E.12) is not true with the constant 23 replaced by 1. See Hayman [1974] and Marshall and Sundberg [1989]. (d) (radial Hall’s lemma) If E ⊂ D and if δ = dist(0, E) > 0 then ω(0, E, D \ E) ≥ Cω(0, E ∗ , D), where E ∗ is the radial projection of E onto ∂D and C is independent of E. Hint: Assume that E is the union of circular slits contained in {|z| > 21 } such that no radius meets E more than once. Let V be the Green potential of |dζ |/(|ζ | log(1/|ζ |). Then V (z) ≤ cω(z, E, D \ E) and V (0) = |E ∗ | (Øksendal, private communication). 21. Let Q = {0 < x < 1, 0 < y < 1} and let {Q j } be a sequence of pairwise disjoint dyadic squares Q j = {k j 2−n j < x < (k j + 1)2−n j , 0 < y < 2−n j } with sidelength (Q j ) = 2n j ≤ 41 . Set = Q \ Q j and z 0 = 21 + i 34 . Prove that given ε > 0 there is δ > 0 such that if ω(z 0 , ∂ Q j , ) < δ then (Q j ) < ε. 22. The following results are from Beurling’s elegant paper [1940]. Also see Beurling’s thesis [1933]. Let u(z) be the Poisson integral of a real function f ∈ L 2 (∂D). (a) If f has Fourier series ∞

a0 an cos nθ + bn sin nθ , + 2 n=1

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then

S( f ) ≡ D(u) = D

1

2π ∂u

2 0

|∇u|2 d xd y =

0

∂r

2

r dr dθ = π

n(an 2 + bn 2 ).

(b) Assume a0 = u(0) = 0, and assume D(u) < ∞. Set R ∂u iθ dr. U (Re ) = ∂r 0 Then (i) |u| ≤ U, (ii) U is subharmonic, (iii) F(θ ) = U (eiθ ) ∈ L 2 (∂D), and 2 1 1 2π F(θ ) dθ ≤ n(an 2 + bn 2 ). 2π 0 2 (c) Prove

0

2π ∂U

∂θ

2 (r eiθ ) dθ ≤ r 2

2π ∂U 0

∂r

2 (r eiθ ) dθ.

Hint: First suppose U ∈ C 2 . Then by a partial integration (with respect to θ ), R 2π 2π

∂U 2 ∂U 2 ∂U 2 r2 − U dθ dr = r2 dθ. ∂r ∂r ∂θ 0 0 0 (d) Because D(U ) =

∂U 2 ∂r

+

1 ∂U 2 dθ, r 2 ∂θ

part (c) implies that D(U ) ≤ D(u) and by the Dirichlet principle (see Appendix C), S(F) ≤ S( f ). In fact, S(F) < S( f ) because U is not harmonic. (e) Let μ be a measure on ∂D and deﬁne h n = cos nθdμ(θ), kn =

sin nθdμ(θ).

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Then

Uμ (r e ) = iθ

log

∞

rn 1 dμ(t) = (h n cos nθ + kn sin nθ ). r eiθ − eit n 0

(f) If E ⊂ ∂D is compact, if Cap(E) < ∞, and if μ = μ E , then h n 2 + kn 2 . I (μ) = lim Uμ (r eiθ )dμ(θ) = r →1 n If S( f ) < ∞, then . % 1 iθ (an h n + bn kn ) ≤ S( f ) I (μ). lim u(r e )dμ(θ) = r →1 π (g) If S( f ) ≤ π, then E λ = {eiθ : lim sup |u(r eiθ )| > λ} r →1

satisﬁes Cap(E λ ) ≤ e−λ . 2

(h) Therefore, if S( f ) < ∞, then on a set of capacity zero, 1 except ∂u iθ iθ limr →1 u(r e ) exists, and in fact, 0 | ∂r (r e )|dr < ∞. n n|cn |2 < ∞, and (i) Prove the theorem of Fejér that if f (z) = ∞ 0 cn z , iθ inθ cn e = α. Thus if S( f ) < ∞, the Fourier limr →1 f (r e ) = α, then series an cos nθ + bn sin nθ converges except on a set of capacity zero. 23. (a) Let ϕ(z) be a univalent function on the unit disc. Prove that ϕ has a 1 nontangential limit and that 0 |ϕ (r ζ )|dr < ∞ for all ζ ∈ ∂D \ E, where Cap(E) = 0. This is also from Beurling [1940]. (b) Let f (z) = cn z n be analytic on D, and assume n|cn |2 ≤ 1. Then 1 E λ = {ζ ∈ ∂D : | f (r ζ )|dr > λ} 0

2 e−λ .

2

1 Consequently, 2π |E λ | < e1−λ . The second estimate has Cap(E λ ) < is from Beurling [1933]. See also Marshall [1989]. 24. (a) Now let be a simply connected domain, and let z, z 0 ∈ . Consider the family of rectiﬁable curves γ in connecting z to z 0 . With Beurling, deﬁne λ(z 0 , z) = sup{ inf |ψ (z)|ds : ψ univalent on , Area(ψ()) ≤ π }. γ ∈ γ

Prove e−2g(z,z 0 ) + e−λ = 1. 2

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(b) Now let E ⊂ ∂. Consider the family of rectiﬁable curves γ in having one endpoint z 0 and the other endpoint in E. With Beurling, deﬁne L(z 0 , E) = sup{ inf |ψ (z)|ds : ψ univalent on , Area(ψ()) ≤ π }. γ ∈ γ

Prove ω(z 0 , E, ) ≤ e1−L

2 (z

0 ,E)

.

This estimate, from Beurling’s thesis [1933], is a precursor of extremal length. Note that (b) follows from 23(b).

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IV Extremal Distance

Extremal distance is a conformally invariant version of distance. As such it is a powerful tool for estimating conformal invariants like harmonic measure in terms of more geometric quantities. We begin the chapter with the basic properties of extremal length and of extremal distance, which is the most important application of extremal length. We then characterize certain extremal distances in Jordan domains by means of conformal mappings from the domains to slit rectangles. This characterization leads to fundamental estimates for harmonic measure by extremal distance and the famous integral dx . θ (x)

1. Deﬁnitions and Examples By deﬁnition, a path family in a domain is a non-empty set of countable unions of rectiﬁable arcs in . An element γ ∈ is called a curve even though γ may not be connected and may have many self-intersections. The euclidean length of the path family is ds, (1.1) inf γ ∈ γ

where ds denotes arc length. The quantity (1.1) is not conformally invariant, but a conformally invariant version of (1.1) is sup inf |ϕ (z)|ds , (1.2) ϕ

γ ∈ γ

where the supremum is taken over all conformal maps ϕ : → ϕ() such that Area ϕ() = 1. See Exercise III.24 for an inequality about (1.2). However, 129

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the supremum in (1.2) is difﬁcult to estimate geometrically and (1.2) becomes more manageable if we use functions more general than |ϕ (z)| in the arc length formula. By deﬁnition, a metric is a non-negative Borel measurable function ρ on such that the area, A(, ρ) ≡ ρ 2 d xd y,

satisﬁes 0 < A(, ρ) < ∞. When ρ is a metric and is a path family, we deﬁne the ρ-length of by L(, ρ) = inf ρ|dz|, γ ∈ γ

and the extremal length of by λ () = sup ρ

L(, ρ)2 . A(, ρ)

(1.3)

A conformal mapping ϕ transforms the metric ρ on into the metric ρ(ϕ −1 (z))|(ϕ −1 ) (z)| on ϕ(). Therefore λϕ() (ϕ()) = λ () and the extremal length (1.3) is conformally invariant. An extremal metric for is a metric ρ which attains the supremum (1.3). In many important cases there exists an extremal metric of the form |ϕ (z)| for some conformal mapping ϕ, and in those cases (1.3) is just the square of (1.2). The advantage of the more general expression (1.3) is that every metric ρ provides a lower bound for λ (). Indeed, one of the main results of the theory, Theorem 6.1 below, is proved using metrics not of the form |ϕ (z)|. Because of its homogeneity, the ratio (1.3) is unchanged if the metric ρ is multiplied by a positive constant. Thus we can normalize the metric to satisfy A(, ρ) = 1 so that λ () = sup{L 2 (, ρ) : A(, ρ) = 1}, or we can normalize the metric by L(, ρ) = 1 so that λ ()−1 = inf{A(, ρ) : L(, ρ) = 1}. The quantity λ ()−1 is called the modulus or module of . Some authors emphasize moduli instead of extremal lengths. Extremal distance is the most useful instance of extremal length. When E ⊂ and F ⊂ the extremal distance d (E, F) from E to F is deﬁned by d (E, F) = λ (),

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where is the family of connected arcs in that join E and F. The conjugate ∗ (E, F), is deﬁned to be the extremal length of the family extremal distance, d ∗ (E, F) we allow a curve ∗ of curves that separate E from F. To compute d ∗ ∗ γ ∈ to be any ﬁnite union (not necessarily connected) of arcs and closed curves in such that E and F lie in the boundaries of distinct components of \ γ . See Figure IV.1.

γ ∗ ∈ ∗

E

γ ∈ E

F

γ ∗ ∈ ∗

Figure IV.1 Connecting and separating curves. Example 1.1 (The rectangle). If R = {(x, y) : 0 < x < and 0 < y < h} is a rectangle of length and height h, then the extremal distance between the vertical sides E and F is (1.4) dR (E, F) = . h The rectangle example is the building block for many results about extremal length. Proof. Let be the family of connected arcs in R joining E and F and let ρ be a metric on R. Then 2 L 2 (, ρ) ≤ ρ(x + i y)d x ≤ ρ 2 (x + i y)d x (1.5) 0

0

by the Cauchy–Schwarz inequality. Integrating (1.5) with respect to y gives L 2 (, ρ)h ≤ A(R, ρ) and hence dR (E, F) = sup ρ

L 2 (, ρ) ≤ . A(R, ρ) h

Equality holds when ρ = 1, and that proves (1.4).

(1.6)

By the condition for equality in the Cauchy–Schwarz inequality, (1.5) is an equality if and only if ρ is constant almost everywhere d xd y. Therefore

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every extremal metric is constant almost everywhere. The conjugate extremal distance dR∗ (E, F) is the extremal distance between the two horizontal sides of R, and thus dR∗ (E, F) = h . More generally, let be a Jordan domain, let E = [ζ1 , ζ2 ] and F = [ζ3 , ζ4 ] be two disjoint subarcs of ∂, where ζ1 , ζ2 , ζ3 , and ζ4 are listed in the counterclockwise ordering of ∂. There exist a unique > 0 and a conformal mapping ϕ from to a rectangle R = {(x, y) : 0 < x < , 0 < y < 1} so that ζ1 , ζ2 , ζ3 , and ζ4 are mapped respectively to the corners i, 0, , and + i. To prove the rectangle exists, map to a half-plane and apply a Schwarz– Christoffel map or use Theorem 4.1. To prove is unique use equality (1.4) and the conformal invariance of extremal lengths or apply Schwarz reﬂection to the map between two rectangles. By conformal invariance = d (E, F) and the extremal metrics for d (E, F) are the constant multiples of |ϕ |. Thus (1.3) is the square of (1.2) in this case. For the rectangle we also have ∗ (E, F) = · d (E, F) d

ζ1

ζ4

F

E ζ2

ϕ

1 = 1.

ϕ(E)

(1.7)

ϕ(F)

ζ3 Figure IV.2 Extremal distance in a quadrilateral.

Recall that every metric gives a lower bound for d (E, F). Equality (1.7) ∗ (E, F), and is signiﬁcant because every metric also gives a lower bound for d ∗ when (1.7) holds every lower bound for d (E, F) provides an upper bound for d (E, F). We shall see in Theorem 4.1 and Theorem H.1 in Appendix H that (1.7) holds in very general circumstances. Example 1.2 (The annulus). The extremal distance between the two boundary circles Cr and C R of the annulus A = {z : r < |z| < R} is 1 R log . dA (Cr , C R ) = 2π r

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133

Proof. The reasoning is similar to the rectangle case. Since (1.3) is a supremum, we can assume 0 < r < R < ∞. Let be the family of connected arcs in A joining Cr to C R and let ρ be a metric on A. Then again by Cauchy–Schwarz, R 2

R R 2 iθ ρ(te )dt ≤ log ρ(teiθ )2 tdt. (1.8) L (, ρ) ≤ r r r Integrating (1.8) with respect to θ gives 2π L 2 (, ρ) ≤ log

R r

A(A, ρ)

and consequently dA (Cr , C R ) = sup ρ

R L 2 (, ρ) 1 ≤ log . A(A, ρ) 2π r

(1.9)

We will get equality holds in (1.8) if and only if ρ(z) = c/|z| a.e., where c is a positive constant,

R and for such ρ we also get equality in (1.9). Thus 1 dA (Cr , C R ) = 2π log r . The proof shows that the extremal metrics for dA (Cr , C R ) have the form c/|z|. The conjugate extremal distance, dA∗ (Cr , C R ), is equal to the extremal length of the family of closed curves in A separating the two boundary circles Cr and C R . By a similar argument we also have dA∗ (Cr , C R ) =

1 2π .

R = dA (Cr , C R ) log r

More generally, a ring domain is a doubly connected plane . By Exercise 7 of Chapter II every ring domain is conformally equivalent to an annulus of the form A = {z : r < |z| < R}. Therefore the module of the family of curves 1 in separating the two components of ∂ is λ ()−1 = 2π log Rr , and an

(z)| extremal metric is |ψ |ψ(z)| where ψ : → A is a conformal map. The quantity λ ()−1 is also called the module of the ring domain and denoted by

λ ()−1 = mod() =

R 1 log . 2π r

2. Uniqueness of Extremal Metrics It is an open problem to determine when a path family has an extremal metric. However, if an extremal metric does exist, then up to a multiplicative constant it is unique area almost everywhere.

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Theorem 2.1. Let be a path family on and let ρ1 and ρ2 be metrics on satisfying L 2 (, ρ1 ) L 2 (, ρ2 ) = . A(, ρ1 ) A(, ρ2 ) Let ρ3 = c1 ρ1 + c2 ρ2 , where c j =

1 − 21 . 2 A(, ρ j )

Then

L 2 (, ρ3 ) L 2 (, ρ1 ) ≥ . A(, ρ3 ) A(, ρ1 )

(2.1)

If equality holds in (2.1) then c1 ρ1 = c2 ρ2 a.e. d xdy. In particular, if ρ1 and ρ2 are extremal metrics for , normalized by A(, ρ1 ) = A(, ρ2 ), then ρ1 = ρ2 a.e. d xd y. Proof. Without loss of generality we assume A(, ρ1 ) = A(, ρ2 ) = 1. Then L(, ρ1 ) = L(, ρ2 ) and by the deﬁnition of L(, ρ), 1 L(, ρ1 ) + L(, ρ2 ) = L(, ρ1 ). L(, ρ3 ) ≥ 2 On the other hand, by the Cauchy–Schwarz inequality, 1 1 1 ρ1 ρ2 d xd y ≤ 1. (2.2) A(, ρ3 ) = A(, ρ1 ) + A(, ρ2 ) + 4 4 2 Therefore (2.1) holds. If equality occurs in (2.1), then A(, ρ3 ) = 1 and equality occurs in (2.2), so that ρ1 = ρ2 almost everywhere d xd y, again by the condition for equality in the Cauchy–Schwarz inequality. In particular, if ρ1 and ρ2 are extremal metrics, then equality holds in (2.1) and ρ1 = ρ2 almost everywhere.

3. Four Rules for Extremal Length Extremal lengths obey four basic rules. 1. The extension rule. Let ⊂ be domains and let be a path family in . Then λ () = λ ().

(3.1)

Moreover, if is a path family in such that every γ ∈ contains some γ ∈ , then λ ( ) ≥ λ ().

(3.2)

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Equality (3.1) says that the extremal length depends on the path family and not the domain , and for this reason we often write λ() for λ (). The extension rule shows that d (E, F) is decreased if any of the sets E, F, or is increased and d (E, F) = d\(E∪F) (E, F) when E and F are closed. = ∩ E

F F

E

Figure IV.3 Extension rule. In Figure IV.3, two applications of the extension rule give d (E , F ) ≥ d (E, F) ≥ d (E, F). Proof. The curves in the family are longer and fewer than the curves in the family . Therefore λ ( ) ≥ λ () and inequality (3.2) will follow from equality (3.1). Let ρ be any metric on and set ρ = ρ | . Then L(, ρ) = L(, ρ ) and A(, ρ) ≤ A( , ρ ). Taking the supremum over ρ yields λ () ≤ λ (). Now let ρ be any metric on and deﬁne the metric ρ on by ρ = ρ χ . Then L(, ρ) = L(, ρ ) and A(, ρ) = A( , ρ ). Taking the supremum over ρ, we get λ () ≥ λ (), which proves (3.1). 2. The serial rule. Let 1 and 2 be path families contained in disjoint open sets 1 and 2 respectively, and let be a path family contained in a domain ⊃ 1 ∪ 2 . If each γ ∈ contains some γ1 ∈ 1 and some γ2 ∈ 2 , then λ() ≥ λ(1 ) + λ(2 ).

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E

γ

F

1

2

G

Figure IV.4 Serial rule. In Figure IV.4, the crosscut F divides into regions 1 and 2 with E ⊂ ∂1 and G ⊂ ∂2 , and by the serial rule, d (E, G) ≥ d1 (E, F) + d2 (F, G). Proof. If either λ(1 ) or λ(2 ) is 0 or ∞, the result follows from the extension rule. Otherwise, choose metrics ρ1 in 1 and ρ2 in 2 , normalized by the conditions L(1 , ρ1 ) = A(1 , ρ1 ) and L(2 , ρ2 ) = A(2 , ρ2 ). Then the metric ρ = ρ1 χ 1 + ρ2 χ 2 on satisﬁes L(, ρ) ≥ L(1 , ρ1 ) + L(2 , ρ2 ) and A(, ρ) = A(1 , ρ1 ) + A(2 , ρ2 ) = L(1 , ρ1 ) + L(2 , ρ2 ). Thus L 2 (1 , ρ1 ) L 2 (, ρ) L 2 (2 , ρ2 ) ≥ L(1 , ρ1 ) + L(2 , ρ2 ) = + . A(, ρ) A(1 , ρ1 ) A(2 , ρ2 ) Taking the supremum over all ρ1 and ρ2 , we obtain λ() ≥ λ(1 ) + λ(2 ). 3. The parallel rule. Let 1 and 2 be path families contained in disjoint open sets 1 and 2 respectively. If is a path family in ⊃ 1 ∪ 2 such that every γ ∈ 1 ∪ 2 contains some γ ∈ , then 1 1 1 ≥ + . λ() λ(1 ) λ(2 )

(3.3)

E2

E1 1

F

G2 G1 Figure IV.5 Parallel rule.

2

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In Figure IV.5, F is a crosscut dividing into regions 1 and 2 and E j = E ∩ ∂ j and G j = G ∩ ∂ j j = 1, 2. By the parallel rule, 1 1 1 ≥ + . d (E, G) d1 (E 1 , G 1 ) d2 (E 2 , G 2 ) In situations where (1.7) holds, this inequality can also be obtained by applying the serial rule to the conjugate extremal distances. Proof. If λ() > 0, let ρ be a metric on , normalized by L(, ρ) = 1. Then L(1 , ρ) ≥ 1 and L(2 , ρ) ≥ 1 and A(, ρ) ≥ A(1 , ρ) + A(2 , ρ) ≥

1 1 + . λ(1 ) λ(2 )

Taking the supremum (1.3) over all ρ, we obtain (2.3).

4. The symmetry rule. Let T : → satisfy T ◦ T (z) = z and suppose that either T (z) or T (z) is analytic. If is a path family in such that T () = , then

2 " L (, ρ) λ() = sup (3.4) : ρ = (ρ ◦ T ) |JT | , A(, ρ)

where |JT | = |T | when T is analytic and |JT | = |T | when T (z) is analytic. Proof. If ρ1 = (ρ ◦ T )|JT | then ρ|dz| = γ

T −1 (γ )

ρ1 |dz|.

Because T maps onto , L(, ρ) = L(, ρ1 ), and because T is one-to-one, A(, ρ) = A(, ρ1 ). Then by Theorem 2.1, ρ2 = 21 (ρ + ρ1 ) satisﬁes L 2 (, ρ2 ) L 2 (, ρ) ≥ . A(, ρ2 ) A(, ρ) But since T ◦ T (z) = z, we have ρ2 ◦ T |JT | = ρ2 , and hence (3.4) holds.

For example, let T (z) = z, let ⊂ {z : Imz > 0}, and let 1 ⊃ ∪ T () be a domain such that T (1 ) = 1 . Let E, F ⊂ ∂ and set E 1 = E ∪ T (E) and F1 = F ∪ T (F). See Figure IV.6. Then 1 (3.5) d (E, F) 2 because to compute d1 (E 1 , F1 ) it sufﬁces by the symmetry rule to consider only metrics ρ satisfying ρ(z) = ρ(z) and ρ = ∞ on R ∩ 1 . If γ is a curve d1 (E 1 , F1 ) =

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connecting E 1 to F1 , then reﬂecting γ ∩ T () yields a curve γ ⊂ ∪ R that joins E to F and ρ|dz| = ρ|dz|. γ

γ

Because ρ is a metric, we can ﬁnd curves ν ⊂ with ρ-length arbitrarily close to the length of γ . Then (3.5) follows since A(1 , ρ) = 2 A(, ρ). We will use this idea several times.

γ

F

γ

E R T (E) T ()

γ

T (F)

Figure IV.6 Symmetry rule. There is an analogy between extremal distance and electrical resistance or resistivity. When the opposite ends of an electrical conductor are joined to the two terminals of a battery and the remaining sides of the conductor are insulated, there is a drop in potential energy from one end of the conductor to the other, and this drop is equal to the voltage of the battery. Suppose the conductor is a uniformly thin sheet of material in the shape of a Jordan domain . Put a copper coating on two arcs of ∂ and attach each copper arc to a terminal of the battery. Assume the atmosphere acts as an insulator on the rest of the surface of the region. Then the electrical resistance of the conductor is the extremal distance between the two intervals, the magnitude of the electric ﬁeld is the extremal metric, the “ﬁeld lines” are the shortest curves in the extremal metric, and the “equipotential lines” are the shortest curves for the conjugate extremal distance. The extension rule says that the resistance d (E, F) increases if or E or F is decreased. The serial and parallel rules correspond to the familiar rules for calculating resistance when circuits are attached in series or in parallel.

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4. Extremal Metrics for Extremal Distance Let be a Jordan domain and let E and F be disjoint closed subsets of ∂, each consisting of a ﬁnite union of arcs. Recall that the extremal distance between E and F is the extremal length of the family of curves in connecting E to F, and the conjugate extremal distance is the extremal length of the family of curves separating E from F. In this section everything will hinge on the following observation: Let R = {(x, y) : 0 < x < 1, 0 < y < h}

be a rectangle with sides parallel to the axes, let = R \ L j where {L j } is a ﬁnite family of horizontal line segments in R, and let E and F be the vertical sides of the rectangle R. Then d (E, F) = dR (E, F) by the argument used in Example 1.1. Moreover, every extremal metric for d (E, F) is constant almost everywhere. The same holds for the conjugate extremal distance ∗ d (E, F) = dR∗ (E, F)

because by deﬁnition the “curves” that separate E from F need not be connected.

L1 L2

E

F

L3 Figure IV.7 Slit rectangle. Now suppose there exists a conformal mapping ϕ : → R \ L j , where R and L j are as in the preceding paragraph, such that ϕ is continuous on , ϕ(E) is the left vertical side of R, and ϕ(F) is the right vertical side of R. Then by the conformal invariance of extremal length and the preceding observation, ∗ (E, F), d (E, F) = dϕ() (ϕ(E), ϕ(F)) = 1/ h = 1/d

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and an extremal metric for d (E, F) is ρ0 (z) = |ϕ (z)|. In the next theorem, we determine when can be mapped to a slit rectangle R so that E and F correspond to the vertical edges of R. Theorem 4.1. Let be a Jordan domain and let E and F be ﬁnite unions of closed subarcs of ∂. Assume E ∩ F = ∅. Then there is a rectangle R having sides parallel to the axes and a conformal map ϕ of onto the rectangle R with a ﬁnite number of horizontal line segments removed such that ϕ ∈ C() and ϕ(E) and ϕ(F) are the vertical sides of the rectangle if and only if there is an arc σ ⊂ ∂ such that E ⊂ σ and F ∩ σ = ∅.

(4.1)

In this case, the extremal distance from E to F is the ratio of the length to height of this rectangle, the conjugate extremal distance satisﬁes ∗ (E, F) = 1/d (E, F), d

and the extremal metrics on for d (E, F) are the positive constant multiples of ρ0 (z) = |ϕ (z)|. E

F

ϕ ϕ(F)

E ϕ(E) F E Figure IV.8 Extremal distance and slit rectangles.

Proof. Assume ϕ is a conformal map of onto the rectangle R with a ﬁnite number of horizontal line segments removed such that ϕ(E) and ϕ(F) are the vertical sides of R. Let γ be a curve in ϕ() connecting the top and bottom edges of the rectangle. The curve γ divides ϕ() into two regions U1 and U2 such that ϕ(E) ⊂ ∂U1 and ϕ(F) ⊂ ∂U2 . Then ϕ −1 (γ ) is a curve in with two endpoints on ∂. Because is a Jordan curve, these endpoints divide ∂ into two arcs σ1 and σ2 with E ⊂ σ1 and F ⊂ σ2 . Conversely, assume there is an arc σ such that (4.1) holds. By the conformal invariance of extremal distance we may suppose that is the unit disc D and that E ∪ F ⊂ ∂D. Let d = C∗ \ (E ∪ F) and let ω(z) = ω(z, F, d ) be the ω(z) be the harmonic conjugate of ω in D. harmonic measure of F in d . Let We claim that ϕ = ω + i ω

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141

is the conformal map promised by the theorem. Because ω is harmonic in d , ϕ extends to be analytic across ∂D \ (E ∪ F). By the symmetry of d and ω, ∂ω ∂ ω = =0 ∂θ ∂r

(4.2)

on ∂D \ (E ∪ F). Let ζ ∈ E ◦ , where E ◦ is the relative interior of E in ∂D. Because ω = 0 on E, we can by the Schwarz reﬂection principle extend ϕ to be analytic in a neighborhood W of ζ . The extension satisﬁes Reϕ > 0 on W ∩ D and Reϕ < 0 on W \ D. This implies ϕ (ζ ) = 0 and ∂ω/∂r ≤ 0 by Lemma II.2.4. Then because ∂ω/∂θ = 0, we obtain ∂ω ∂ ω = <0 (4.3) ∂θ ∂r at ζ , and hence on all of E ◦ . Applying the same argument to 1 − ϕ on F ◦ we obtain ∂ ω ∂ω = >0 (4.4) ∂θ ∂r on F ◦ . Let ζ be an endpoint of E ∪ F and let W be a neighborhood of ζ so that U = W \ γ is simply connected, where γ is the component of E ∪ F containing ζ . Then ω is continuous on W and ϕ = ω + i ω is analytic in U . By mapping C∗ \ γ to the upper half-plane and using Schwarz reﬂection again, we see that ϕ extends to be continuous on D ∪ {ζ }. We conclude that ϕ is continuous on D. Now let us follow ϕ(z) as z traverses ∂D counterclockwise. On each component of E, ω = 0 and ω is strictly decreasing, by (4.3). On each component of F, ω = 1 and ω is strictly increasing, by (4.4). On any component ω is constant, by (4.2). If both endγk of ∂D \ (E ∪ F), 0 < ω < 1 and points of γk are in E, then the curve ϕ(γk ) traces a horizontal line segment L k ⊂ {z : 0 ≤ Rez ≤ 1}, beginning and ending on {z : Rez = 0}. If both endpoints are in F, then ϕ(γk ) traces a horizontal segment L k ⊂ {z : 0 ≤ Rez ≤ 1}, beginning and ending on {z : Rez = 1}. There are only two components γk having endpoints on both E and F; on these ω is constant and ω = Reϕ goes from 0 to 1 or from 1 to 0. Thus ϕ(∂D) contains the boundary of a rectangle R and the contour ϕ(∂D) has winding number 1 about each point in R \ L k and winding number 0 about each point in C \ R. The argument principle then shows that & ϕ(D) = R \ Lk

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and that ϕ is one-to-one. The statements about the extremal distance and the extremal metrics now follow from Theorem 2.1 and the observation made at the beginning of this section. By the proof of Example 1.2, the extremal distance and conjugate extremal distance between the boundary circles of an annulus will also not change if radial slits are removed from the annulus. Thus we can replace the rectangle R by an annulus and obtain a ring domain version of Theorem 4.1. Theorem 4.2. Suppose is bounded by two disjoint Jordan curves 1 , 2 and suppose E and F are ﬁnite unions of closed arcs with E ⊂ 1 and F ⊂ 2 . Then there is a conformal map ψ of onto an annulus with ﬁnitely many radial slits removed such that ψ is continuous on and the images of E and F are the two boundary circles. The extremal metrics for the extremal distance between E and F are constant multiples of |ψ (z)|/|ψ(z)| and d (E, F) =

1 1 ∗ (E, F) = 2π log R, d

where R > 1 is the ratio of the radii of the boundary circles.

E E

F

ψ F

ψ(F)

E ψ(E) Figure IV.9 Extremal distance and slit annuli. Proof. By conformal invariance, we may suppose that 2 = ∂D and that 1 ⊂ D is an analytic Jordan curve. If ∂ \ (E ∪ F) = ∅, take two copies of and form the doubled Riemann surface d by attaching them along ∂ \ (E ∪ F). Then d is a torus with ﬁnitely many arcs removed, where the arcs correspond to the intervals in E ∪ F. If E ∪ F = ∂, then set d = . Let ω(z) = ω(z, Fd , d ) be the harmonic measure of the double Fd of F in d . (The proof of Theorem B.4 in Appendix B explains the construction of d and the solution of the Dirichlet problem on d .) Let ω be the (locally deﬁned)

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harmonic conjugate of ω in . Choose a Jordan curve γ ⊂ homologous to ∂D. If ∂ ω ds = 0, (4.5) γ ∂s then ϕ = ω + i ω is single valued and analytic in and by the proof of Theorem 4.1 ϕ extends to be continuous on . But then by (4.2) and (4.4), ω increases as z traces ∂D, in contradiction to (4.5). Thus there exists a non-zero constant c so that ∂ ω ds = 2π. c γ ∂s Then ψ = ecϕ is single valued and analytic on , and ψ is continuous on ∂. As ω decreases by 2π . Hence z traces ∂D, c ω increases by 2π and, as z traces 1 , c ψ(∂) contains the boundary of an annulus, and the remaining assertions of the theorem follow just as in Theorem 4.1.

5. Extremal Distance and Harmonic Measure We begin with the special case of a rectangle. Set R L = {z : |Rez| < L and |Imz| < 1}. Lemma 5.1. If E L = {z ∈ ∂ R L : |Rez| = L} is the union of the vertical edges of R L , then π

e− 2 L ≤ ω(0, E L , R L ) ≤

8 −π L e 2 . π

(5.1)

The estimates in (5.1) are sharp, because π

lim ω(0, E L , R L )e 2 L =

L→∞

8 π

and

π

lim ω(0, E L , R L )e 2 L = 1.

L→0

(5.2)

The lemma connects the harmonic measure of the two ends of R L at its center point to the extremal distance L between the two ends. When L ≥ 1, the lower π bound in (5.1) can be improved to (.93)(8/π)e− 2 L . When L < 1, better bounds can be obtained by applying (5.1) to ω(0, E 1/L , R1/L ) = 1 − ω(0, E L , R L ). See Exercise 10. Proof. Let SL be the inﬁnite strip {z : Rez > −L and |Imz| < 1} and let ω L (z) be the harmonic measure of the left edge ∂ SL ∩ {Rez = −L} in SL . Using the

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conformal map z → eπ z/2 and some elementary geometry, we obtain * π , π 2 e 2 z − (−ie− 2 L ) ω L (z) = arg . π π π e 2 z − ie− 2 L In particular, ω L (0) = Thus for z ∈ ∂ R L ,

4 π

(5.3)

π

tan−1 (e− 2 L ). By (5.3) ω L (z) ≤ ω L (L) on Rez = L.

ω(z, E, R L ) ≤ ω L (z) + ω L (−z) ≤ (1 + ω L (L))ω(z, E, R L ). By the maximum principle, this inequality persists at z = 0, and we conclude that 2ω L (0) ≤ ω(0, E, R L ) ≤ 2ω L (0). 1 + ω L (L) Because ω L (L) = ω2L (0) =

4 tan−1 (e−π L ), π

we obtain 8 π

1+

π

tan−1 (e− 2 L ) 4 π

tan−1 (e−π L )

≤ ω(0, E, R L ) ≤

π 8 tan−1 (e− 2 L ). π

(5.4)

The estimates (5.1) now follow from the elementary inequalities

π π t ≤ tan−1 (t) ≤ min t, , 4 4 valid for 0 ≤ t ≤ 1, and both equalities in (5.2) are immediate from (5.4).

Now let be a Jordan domain, let E be an arc on ∂ and let z 0 ∈ . Consider all Jordan arcs σ ⊂ that join z 0 to ∂ \ E, and deﬁne λ(z 0 , E) = sup d\σ (σ, E), σ

where the supremum is taken over all such Jordan arcs. E

z0 σ

Figure IV.10 Distance from a point to an arc.

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Both quantities λ(z 0 , E) and ω(z 0 , E, ) are conformally invariant but ω(z, E, ) is strictly increasing in E, while λ(z 0 , E) is strictly decreasing in E. This means there is a function connecting λ to ω, and the next result examines that function. Theorem 5.2. Let be a Jordan domain, let E be a subarc of ∂ and let z 0 ∈ . Then e−π λ(z 0 ,E) ≤ ω(z 0 , E, ) ≤

8 −π λ(z 0 ,E) . e π

Moreover lim ω(z 0 , E, )eπ λ(z 0 ,E) =

λ→∞

8 π

and

lim ω(z 0 , E, )eπ λ(z 0 ,E) = 1.

λ→0

By Theorem 5.2 every choice of the arc σ and every choice of the metric ρ give an upper bound for ω(z 0 , E, ), because λ(z 0 , E) and the extremal distance d\σ (σ, E) are both suprema. This idea of varying σ is from Beurling [1989]. Proof. By conformal invariance we may suppose that = D, that z 0 = 0 and that E is an arc on ∂D. Let E 1 and E 2 be the two disjoint arcs on ∂D given by E 1 ∪ E 2 = {eiθ : e2iθ ∈ E}. Because the arcs E 1 and E 2 are symmetric about 0, there is a conformal map f of D onto a rectangle R such that R has center j = f (E j ), 0 and sides parallel to the axes, such that f (0) = 0 and such that E j = 1, 2, are the vertical sides of R. Let σ be any arc in D connecting 0 to √ ∂D \ E and let σ = {z ∈ R : ( f −1 (z))2 ∈ σ }. The map z is conformal on √ D \ σ and the two branches of f ( z) map D \ σ conformally onto the two components of R \ σ . Hence by the serial rule and by conformal invariance, 1 , E 2 ) ≥ d( E 1 , 2 ) = 2dD (σ, E). σ ) + d( σ, E dR (E Equality holds if σ0 is the vertical line segment in R through 0, so that 1 , E 2 ). 2λ(0, E) = d R ( E

√

σ 0

E

z

E1

0 E2

f

Figure IV.11 Proof of Theorem 5.2.

1 E

0

σ

2 E

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Since ω(z 2 , E, D) = ω(z, E 1 ∪ E 2 , D), we have 1 ∪ E 2 , R). ω(0, E, D) = ω(0, E 1 ∪ E 2 , D) = ω(0, E Then by Lemma 5.1, e−π λ(0,E) ≤ ω(0, E, D) ≤

8 −πdD (σ,E) . e π

The upper estimate is also valid when E is a ﬁnite union of arcs on ∂. In that case we consider all Jordan arcs σ in connecting z 0 to ∂ \ E, and deﬁne λ(z 0 , E) = sup d\σ (σ, E). σ

Theorem 5.3. Let be a Jordan domain, and let E be a ﬁnite union of arcs contained in ∂. Then 8 (5.5) ω(z 0 , E, ) ≤ e−π λ(z 0 ,E) . π Proof. We can assume = D and z 0 = 0. Let σ be an arc from 0 to ∂D \ E and set λ = d\σ (σ, E). Write σ1 = {z : z 2 ∈ σ } and {eiθ : e2iθ ∈ E} = E 1 ∪ E 2 where σ1 separates E 1 from E 2 and z 2 (E 1 ) = z 2 (E 2 ) = E. By Theorem 4.1, there is a conformal map ϕ of D onto a rectangle R with horizontal slits removed j = ϕ(E j ) are the vertical ends of R. Then as in Theorem 5.2 we obtain so that E 2 ) ≥ 2dD (σ, E) and consequently d R ( E1, E 8 −π λ . e π Taking the supremum over σ then gives (5.5). ω(0, E, D) ≤

There is no lower bound in Theorem 5.3 because the slits can consume nearly all of the harmonic measure. See Exercise 12.

6. The

dx θ(x)

Estimate

Every lower bound for extremal distance yields an upper bound for harmonic measure, and by (1.3) every metric ρ yields a lower bound for extremal distance. Constructing good metrics is therefore an important method. Consider the special case of a strip domain: = {(x, y) : |y − m(x)| < θ(x)/2, a < x < b},

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having varying width θ (x) and mid-line y = m(x).

θ (x)

a

y = m(x)

x Figure IV.12 A strip domain.

b

Let a = x0 < x1 < . . . < xn = b and set σ j = ∩ {z : Rez = x j }, where j = 0, . . . , n. By the serial rule, d (σ0 , σn ) ≥

n

d(σ j−1 , σ j ).

j=1

If x = x j − x j−1 is small, the region between σ j−1 and σ j is approximately a thin rectangle having x as base, θ (x j ) as height, and x j + im(x j ) as the midpoint of its right vertical side. Under the linear map z − im(x j ) , θ (x j ) this rectangle is sent to a rectangle centered on R with height 1 and width x/θ(x j ), so that d(σ j−1 , σ j ) ≈ x/θ(x j ). Consider the (non-analytic) map x dt y − m(x) : (x, y) −→ , (6.1) θ (x) a θ (t) b 1 dt. The extremal metric from onto a rectangle of height 1 and length a θ(t) for d (σ0 , σn ) is given by ρ0 (z) = |ϕ (z)| = |∇Reϕ(z)|, where ϕ is a conformal map of onto some rectangle, but the key idea here is to replace the analytic ϕ by and use as metric ρ(z) = |∇Re| =

1 , θ (x)

z = x + i y.

If γ is any curve connecting σ0 to σn in , then b 1 ρ(z)|dz| ≥ d x. γ a θ (x)

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Moreover

b

ρ 2 d yd x = a

so that

d (σ0 , σn ) ≥

a

b

1 d x, θ (x)

1 d x. θ (x)

More generally, let be a ﬁnitely connected Jordan domain and assume E ⊂ {z : Rez ≤ a} ∩ ∂ and F ⊂ {z : Rez ≥ b} ∩ ∂ are ﬁnite unions of arcs. Suppose I x ⊂ {z : Rez = x} separates E from F and let θ (x) be the length of I x , a < x < b. We assume that θ (x) is measurable but we do not assume that I x = ∩ {z : Rez = x} or that I x is connected. Deﬁne the metric ρ A by ⎧ 1 ⎨ θ(x) if (x, y) ∈ I x , ρ A (x, y) = ⎩ 0 elsewhere in . Then by the above argument b 1 d x. (6.2) d (E, F) ≥ a θ (x) Theorem 6.1. Let be a Jordan domain and let z 0 ∈ . Let b > x0 = Rez 0 and suppose F ⊂ {z ∈ ∂ : Rez ≥ b}. Assume that for x0 < x < b, there exists I x ⊂ ∩ {Rez = x} separating z 0 from F and set θ (x) ≡ (I x ). Then b dx 8 . (6.3) ω(z 0 , F) ≤ exp −π π x0 θ (x)

I x2

I x3

F

\

F

z0 I x1

x0

I x3

\ F

R x1 x2 x3 b Figure IV.13 Crosscut estimate of harmonic measure.

F

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For example, if I x = {z ∈ : Rez = x}, then θ (x) is lower semicontinuous and measurable. denote the points of which are not separated from z 0 by some Proof. Let which are separated from z 0 by all I x , I x and let F denote the points of ∂ by replacing x0 < x < b. We can suppose that F is a ﬁnite union of arcs on ∂ −1 with ϕ (|z| < r ) where ϕ is a conformal map of onto D. Let σ ⊂ {Rez = x 0 } . By the maximum principle, be a curve connecting z 0 to ∂ ), ω(z 0 , F, ) ≤ ω(z 0 , F, and then Theorem 5.3 and inequality (6.2) give (6.3).

Theorem 6.1 remains true if θ (x) is replaced by ({z ∈ : Rez = x}), because that change only increases the right side of (6.3). When {z ∈ : Rez = x} contains several crosscuts that separate z 0 from F, sharper versions of (6.3) are available. See Appendix G. Theorem H.8 in Appendix H extends Theorem 6.1 to ﬁnitely connected domains. Theorem 6.1 is often used in polar coordinates, as in the following theorem. Theorem 6.2. Let be a Jordan domain and let E ⊂ ∩ {|z| ≥ R}. If z 0 ∈ satisﬁes r0 = max(|z 0 |, dist(0, ∂)) < R. Suppose that Jr ⊂ {z ∈ : |z| = r } separates z 0 from E, r0 < r < R, and let r (r ) be the length of Jr . Then R dr 8 ω(z 0 , E) ≤ exp −π , (6.4) π r0 r (r ) if (r ) is measurable. Proof. Deﬁne the metric ⎧ 1 ⎨ r (r ) in Jr , ρ A (z) = ⎩ 0 in \ {Jr : r0 < r < R}, for z ∈ and r = |z| and repeat the proof of Theorem 6.1.

Notes Much of this chapter is based on Beurling’s Mittag–Lefﬂer Institut lectures, published in [1989], and on Ahlfors [1973]. In [1973] Ahlfors says that Beurling invented extremal length in 1943 or 1944, but waited until 1946 to announce his results. The application of the Cauchy–Schwarz inequality in the proof of (1.4) is often referred to as the length–area principle; it goes back to early

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work by Grötzsch [1928] and others. Both Beurling [1989] and Ahlfors [1973] obtained the extremal metric |ϕ (z)| = |∇Reϕ(z)| of Section 4 by solving a mixed Dirichlet–Neumann problem for Reϕ(z). Lemma 5.1 is from Marshall and Sundberg [1989]. Theorem 5.3 is in Beurling [1989], with different constants. The ﬁrst form of Theorem 6.1 was proved by Ahlfors [1930], but we have followed Beurling [1989]. Fuchs [1967], Hersch [1955], and Ohtsuka [1970] have other presentations. Lord Rayleigh [1871], [1876] also derived the estimate (6.2) in 3 dimensions for the resistance of a conductor using the serial rule and proved that the conjugate resistance (extremal distance) is the reciprocal of the resistance for a quadrilateral. [1988] says Carleman was Baernstein dx to measure a domain and calls it the ﬁrst to use integrals of the form θ(x) “surely one of the most brilliant ideas in the history of complex function theory”. Carleman’s original [1933] estimate of harmonic measure by the above integral, proved with a differential inequality, will be given in Appendix G. In Appendix H, extremal distances on ﬁnitely connected domains will be studied by embedding the domains in their Riemann surface doubles.

Exercises and Further Results 1. Let be a Jordan domain and let A1 , B1 , A2 , and B2 be arcs of ∂ listed in counterclockwise order. Assume these arcs have disjoint interiors and assume A1 ∪ A2 ∪ B1 ∪ B2 = ∂. Prove Area() ≥ distC (A1 , A2 )distC (B1 , B2 ), where distC denotes euclidean distance. Prove equality holds only if A1 , B1 , A2 , and B2 are the four sides of a rectangle. 2. Suppose two annuli A1 and A2 are each bounded by concentric circles. Prove that if A1 and A2 are conformally equivalent, then the ratios of the radii of their bounding circles are the same. Find one proof with extremal lengths and ﬁnd one proof without extremal lengths. 3. Let S be a one-to-one analytic or conjugate analytic map from a domain 1 onto a domain 2 such that 1 ∩ 2 = ∅. Let 1 be a path family in 1 , set 2 = S(1 ) and suppose is a path family in 1 ∪ 2 . (a) If γ1 ∪ S(γ1 ) ∈ for every γ1 ∈ 1 , and if every γ ∈ contains some γ1 ∈ 1 and some γ2 ∈ 2 , then λ() = λ(1 ) + λ(2 ) = 2λ(1 ). In other words, equality holds in the serial rule. Moreover, λ( ∗ ) = λ() where ∗ = {γ1 ∪ S(γ1 ) : γ1 ∈ 1 }. For example, if S(z) = z and S() = , set

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Exercises and Further Results 1 = ∩ {z : Imz > 0}. Then if E ⊂ ∂ ∩ {z : Imz > 0}, we obtain d (E, S(E)) = 2d1 (E, R ∩ ).

(b) If every γ1 ∈ 1 and every γ2 ∈ 2 contains some γ ∈ and if whenever γ ∈ , γ ∪ (S −1 γ ∩ 1 ) contains some γ1 ∈ 1 , then 1 1 1 2 = + = . λ() λ(1 ) λ(2 ) λ(1 ) In other words, equality holds in the parallel rule. For example, if S(z) = z and S() = , and if E 1 , F1 ⊂ ∂ ∩ {z : Imz > 0}, then for E 2 = S(E 1 ), F2 = S(F1 ), and 1 = ∩ {z : Imz > 0}, 1 1 2 1 = + = . d (E 1 ∪ E 2 , F1 ∪ F2 ) d1 (E 1 , F1 ) d2 (E 2 , F2 ) d1 (E 1 , F1 ) 4. Let a < b < c be real, let be the family of arcs in C joining [−∞, a] to [b, c], and let + be the family of arcs in the upper half-plane joining [−∞, a] to [b, c]. (a) Show that b − a 2λ() = λ( + ) = f , c−b where f (t) is strictly increasing, f (0) = 0, and f (∞) = ∞. (b) Write the function f as an elliptic integral, using the Schwarz–Christoffel formula. (c) If C is the circle centered on R through a and c, let ζ be the intersection of {Rez = b} and C in {Imz > 0}. Show that . 2 c−b 8 + ω(ζ, (−∞, a] ∪ [b, c], H) = tan−1 ∼ e−π λ( )/2 . π b−a π 5. A ring domain separates the two sets E and F if E and F are contained in different components of C \ . (a) (Grötzsch [1928]) If is a ring domain separating {0, r } from the unit circle {ζ : |ζ | = 1}, then mod() ≤ mod(D \ [0, r ]),

(E.1)

where mod() denotes the modulus of . Hint: Apply the extension rule for the (larger) family of curves in D separating {0, r } from ∂D. Then use the symmetry rule, and consider curves in D ∩ {Imz > 0} connecting [−1, 0] to [r, 1].

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(b) (Teichmüller [1938]) If z 1 , z 2 , and 0 are three distinct points and if is a ring domain separating {0, z 1 } from {z 2 , ∞}, then

mod() ≤ mod C∗ \ [−|z 1 |, 0] ∪ [|z 2 |, ∞] . (E.2) Many authors call (E.2) Teichmüller’s module theorem, but in [1938] Teichmüller called a weak

version of Theorem 4.1 of Chapter V “Der Modulsatz”. (c) If = C∗ \ [0, 1 − ε] ∪ [1, ∞] then mod() ≥

C log 1ε

.

Hint: See Exercise 4 and (1.7) or Lehto and Virtanen [1973], page 61. (d) If ∂ = 1 ∪ 2 where 1 is a rectiﬁable Jordan curve with (1 ) ≤ 1 and dist(1 , 2 ) > ε, then mod() ≥ Cε.

Hint: If E = {z : dist(z, 1 ) < ε} and ρ = χ E , then ρ 2 d xd y ≤ Cε. (e) Given R > 0, there exists A > 0 such that every ring domain with mod() > A contains an annulus {r1 < |z − z 0 | < r2 } with r2 /r1 > R. 6. (a) Let A and B be disjoint arcs in C. Then 2 1 dist(A, B) dC (A, B) ≥ . π diam(A) + dist(A, B) Hint: Take z 0 ∈ A such that dist(z 0 , B) = dist(A, B) and deﬁne ⎧ 1 if |z − z 0 | < dist(A, B) + diam(A), ⎨ dist(A,B) ρ(z) = ⎩ 0 if |z − z 0 | ≥ dist(A, B) + diam(A). (b) Suppose is a ring domain having boundary components 1 and 2 such that with respect to the spherical metric ρ(z) = (1 + |z|2 )−1 , diam( j ) ≥ δ. Prove π mod() ≤ 2 . 4δ If also dist(1 , 2 ) ≤ ε ≤ δ, prove tan( 2δ ) 2 1 ≥ log . mod() π tan( 2ε ) See Lehto and Virtanen [1973], page 34. 7. (Beurling’s criterion for extremal metrics) Let be a path family in a domain and let ρ0 be a metric on . Suppose contains a subfamily 0 such that (i) γ ρ0 |dz| = L(, ρ0 ) for all γ ∈ 0 , and

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(ii) If h is real valued and Borel measurable in and if γ h|dz| ≥ 0 for all γ ∈ 0 , then hρ0 d xd y ≥ 0. Then ρ0 is an extremal metric for . It follows that all other extremal metrics are of the form cρ0 , where c > 0 is a constant. See Ahlfors [1973]. 8. Let ϕ : D → be conformal, let I be an arc of ∂D with length |I | < π and center ζ I and let z I = (1 − |I |)ζ I and let σ be any arc in D that connects the endpoints of I . Prove dist(ϕ(z I ), ∂) ≤ Cdiamϕ(σ ).

I ζI σ zI

Figure IV.14 9. Let = (0, 1) × (0, 1) be the unit square, let F = {(1, y) : 0 ≤ y ≤ 1}, and let E = {(0, y) : 0 ≤ y ≤ 1} ∪ {(1/2, 1)}. (a) Prove d (E, F) = 1. Hint: let χ Bε ∩ (z) , ρε (z) = 1 + |z − ζ | log 1/|z − ζ | where Bε is a ball of radius ε centered at ζ = (1/2, 1). (b) Let E δ consists of the set E together with an interval centered at (1/2, 1) of length δ, and let ρδ = |ϕδ | be the extremal metric for d (E δ , F). Prove limδ→0 ρδ exists and is the constant metric, but the limit is not an extremal metric for d (E 0 , F). (c) Prove there is no extremal metric for d (E 0 , F). See Ohtsuka [1970]. 10. Let R1 = {z : |Rez| < L , |Imz| < 1}, R2 = {z : |Imz| ≤ 1}, where L > 2. Let be a Jordan domain with R1 ⊂ ⊂ R2 . Let J = {(0, y) : |y| ≤ 1}, E ⊂ {z ∈ R2 : Rez ≤ −L} and F ⊂ {z ∈ R2 : Rez ≥ L} be such that E and F are ﬁnite unions of closed arcs in ∂. Then there is a constant C such that d (E, F) ≤ d (E, J ) + d (J, F) + ce−π L/2 .

(E.3)

In other words, if L is large then equality almost holds in the serial rule. This result will be used in Chapter VIII. The proof of Theorem V.5.7 will include a simple special case of (E.3).

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Hints: By the extension rule, d (E, F) ≥ L. By Theorem 4.1 there exists a conformal map ϕ from to a rectangle R with a ﬁnite number of horizontal slits removed. We may assume R = {z : |Rez| < d (E, F), |Imz| < 1}, Im(ϕ(z) − z) = 0 on the top and bottom edges of R1 , Reϕ = −d (E, F) on E, and Reϕ = d (E, F) on F. Since |Im(ϕ(z) − z)| ≤ 2 on the vertical edges of R1 , |Im(ϕ(z) − z)| ≤ C1 e−π L/2 on R1 ∩ {|Rez| ≤ 2} by Lemma 5.1. By Schwarz reﬂection, ϕ extends to a conformal mapping from some domain containing the rectangle {z : |Rez| < L , |Imz| < 3} into the rectangle {z : |Rez| < d (E, F), |Imz| < 3}, and |Im(ϕ(z) − z)| ≤ C1 e−π L/2 on D2 = {|z| < 2}. Differentiating the Herglotz integral formula on ∂D2 shows that there is a constant C such that |ϕ (z) − 1| < Ce−π L/2 , for z ∈ J . It follows that sup Reϕ − inf Reϕ ≤ 2 sup |Imϕ (z)| ≤ 2Ce−π L/2 . J

J

J

Therefore (E.3) holds because d (E, J ) ≥ 21 (d (E, F) + inf J Reϕ) and d (J, F) ≥ 21 (d (E, F) − sup J Reϕ). 11. Let R L = {|Rez| < L , |Imz| < 1} and let E L = ∂ R L ∩ {|Rez| = L} be as in Lemma 5.1. π (a) Prove that ω(0, E L , R L ) ≥ (.93) π8 e− 2 L when L ≥ 1. See Marshall and Sundberg [1989]. π π (b) If L < 1, prove that 1 − π8 e− 2L ≤ ω(0, E L , R L ) ≤ 1 − (.93) π8 e− 2L . (c) Let S = {|Imz| < 1} and let FL = ∂ S ∩ {|Rez| < L}. Prove ω(z, E L , R L ) =

+∞

2ω(z + (4n + 2)L , FL , S).

n=−∞

(d) Using a conformal map, ﬁnd explicit formulae for ω(z, FL , S), and ω(0, E L , R L ). 12. Let R L and E L be as in Exercise 10. (a) Set π − 1 2 . xn = sin ω(0, E 2n L , R2n L ) 2 2 = 21 (xn + 1/xn ). Hint: Use the Schwarz reﬂection principle to Prove xn−1 relate the conformal map of D onto R L to the conformal map of D onto R2L .

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(b) Given n and L, set 1 π L2n−1 , e 2 . 1 yk = (yk−1 + 1/yk−1 ), 1 ≤ k ≤ n, 2 2 1 . ω = tan−1 π yn − 1/yn

y0 =

Prove |ω(0, E L , R L ) − ω| ≤ Ce−π L2

n−1

.

When n = 3 and L ≥ 1 the computation of ω is accurate to 16 decimal places. In general this method gives quadratic convergence to ω(0, E L , R L ), which is considerably faster than other methods such as the one given in Exercise 11. (c) Use (a) to ﬁnd a quadratically convergent algorithm to compute the conformal map of D onto R L . Hint: For large N , approximate R2n L by a semiinﬁnite strip. See Marshall and Sundberg [1989]. 13. (a) Given ε > 0, construct a set E ⊂ ∂D consisting of ﬁnitely many closed arcs so that ω(0, E, D) < ε but λ(0, E) = supσ dD\σ (σ, E) = 1 where the supremum is taken over all arcs σ connecting 0 to ∂D. Hint: Remove horizontal slits from a rectangle. (b) Show that for each set E ⊂ ∂D that consists of n arcs, there are n possible rectangles that occur in the proof of Theorem 5.3. Also show that the vertical line through 0 can meet the excised slits in the proof of Theorem 5.3. Is there a curve σ so that λ(z 0 , E) = dD (σ, E)? 14. Let be simply connected, let z 0 , z 1 ∈ , and let γ j be an arc in joining z j to ∂. Then

Min g(z 0 , z 1 ), 1 ≤ C exp −π d\γ1 ∪γ2 (γ1 , γ2 ) for some absolute constant C. See Beurling [1989]. 15. Let L n = {r eiθn : 0 ≤ r ≤ 1/2, θn = 2−n π } and = {z : |z| < 1, Imz > 0, |z − i/4| > 1/4} \

∞ &

Ln.

n=1

Let E = {eiθ : 0 ≤ θ ≤ π/6} and F = {−r : 0 ≤ r ≤ 1}. If denotes the curves in connecting E to F and n denotes the curves in connecting E to Fn = F ∪ B(0, 2−n ), then for ρ ≡ 1, inf ρ|dz| ≤ 1 γ ∈n

γ

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and

inf

γ ∈ γ

√ 3 π 1 ρ|dz| = + + > 1. 2 8 4

The approximation does not work here because the prime end corresponding to 0 includes the interval [0, 1/2]. See Ohtsuka [1970]. 16. (a) Let γn ⊂ have endpoints γn (0) → a ∈ ∂ and γn (1) → b ∈ ∂. Moreover suppose that there are arcs Ik ⊂ ∩ {|z − a| = δk } with δk → 0 so that Ik separates all but ﬁnitely many {γn (0)} from some ﬁxed point z 0 ∈ . Similarly suppose there are arcs Jk ⊂ {|z − b| = δk } so that Jk separates all but ﬁnitely many {γn (1)} from z 0 . Let ρ be a metric on with A(, ρ) < ∞. Prove that for all ε > 0 there exists a curve σε ⊂ connecting a to b so that ρ|dz| ≤ lim inf ρ|dz| + ε. σε

γn

Hint: Use the length–area principle. See Ohtsuka [1970]. (b) Suppose is a ﬁnitely connected Jordan domain and suppose E n and Fn are each ﬁnite unions of pairwise disjoint closed arcs in ∂ with E n ⊃ E n+1 and Fn ⊃ Fn+1 . Let E = E n and F = Fn . Use (a) to prove that lim d (E n , Fn ) = d (E, F).

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V Applications and Reverse Inequalities

In Section 1 we present Ahlfors’ solution of the Denjoy problem, historically the ﬁrst application of an estimate like (IV.6.4). In Section 2 we give Beurling’s converse to Theorem IV.6.1. It is proved by constructing a new metric. In Section 3 we discuss reduced extremal distance and derive a recent sharp theorem of Balogh and Bonk about the lengths of conformal images of radii. It is also proved by constructing a new metric. In Section 4 we prove Teichmüller’s theorem on the addition of moduli of rings. In Sections 5 and 6 we use extremal distances to to obtain two different, deﬁnitive results on the existence of angular derivatives.

1. Asymptotic Values of Entire Functions An entire function f (z) has asymptotic value a if there is a Jordan arc tending to ∞ such that f (z) = a.

lim

z→∞

For example

z

f (z) =

e−w dw n

0

has n distinct asymptotic values, one along each of the curves arg z = 2π j/n, j = 1, . . . , n. Write M(r ) = sup | f (z)| |z|=r

for the maximum of | f (z)| on {|z| = r }. Theorem 1.1. If the non-constant entire function f (z) has n distinct ﬁnite

157

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asymptotic values, then

lim inf r →∞

logM(r ) r n/2

> 0.

(1.1)

Proof. Assume f has n distinct asymptotic values a j taken along n distinct curves j . Moving each j slightly, we may assume that j is a simple polygonal arc whose vertices tend only to ∞. Because the asymptotic values a j are distinct, we may further move the j so that j ∩ k = {0} for j = k, and so that 0 ∈ 1 when n = 1. Then the j divide the plane into n Jordan domains G 1 , G 2 , . . . , G n , and when n > 1 each G j is bounded by two distinct k . On each domain G j the function f (z) must be unbounded. When n = 1, this holds because f is not constant. When n > 1, it follows from Lindelöf’s theorem, Exercise II.3(d), applied to f ◦ ϕ where ϕ is a conformal map from D onto G j . We can assume | f (z)| ≤ 1 on j , and by the maximum principle we can choose R0 large so that for each j = 1, . . . , n there is a z j ∈ G j with |z j | = R0 and | f (z j )| > 1. Let R > R0 . Write ω j (R) = ω(z j , {|z| = R}, G j ∩ {|z| < R}), and write j (r ) for the angular measure of G j ∩ {|z| = r }. Then by the maximum principle log| f (z j )| ≤ ω j (R)logM(R) and hence by (IV.6.4),

R dr 8 logM(R). log | f (z j )| ≤ exp −π π R0 r j (r ) √ (When n = 1, G 1 can be mapped to a Jordan domain using z and (IV.6.4) can then be applied.) By our assumptions on the j , n

j (r ) = 2π,

j=1

and by the Cauchy–Schwarz inequality, ⎞ ⎞⎛ ⎛ n n 1 ⎠⎝ ⎝ j (r )⎠ ≥ n 2 , j (r ) j=1

j=1

so that n j=1

1 n2 ≥ . j (r ) 2π

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Hence

n 1 R dr n R ≥ log , n r (r ) 2π R j 0 R0 j=1

and there is j = j (R) such that exp −π

R

R0

dr r j (r )

≤

We conclude that 8 n 0 < min log| f (z j )| ≤ R02 j π

R0 R

n/2 .

logM(R) n

R2

,

from which (1.1) follows.

2. Lower Bounds For a domain of the form = {(x, y) : |y − m(x)| < θ(x)/2, a < x < b}, the inequality (IV.6.3) of Theorem IV.6.1 is not sharp, because the mapping deﬁned in (IV.6.1) is not conformal. However, minor smoothness assumptions on m(x) and θ (x) will yield a lower bound for ω(z 0 , F) that complements (IV.6.3). Lower bounds on harmonic measure come from upper bounds on extremal distances or, equivalently, from lower bounds on conjugate extremal distances. In view of (IV.6.1) a natural metric for studying the conjugate extremal distance in is 1 y − m(x) (y − m)θ + θm 2 ρ B (x, y) = ∇ . (2.1) = θ 2 (x) + θ (x) θ2 This metric was invented by Beurling [1989]. Let E = ∂ ∩ {x = a} and F = ∂ ∩ {x = b}. If γ is any curve in connecting the curve y = m(x) + θ (x)/2 to the curve y = m(x) − θ (x)/2, then y−m ρ B |dz| ≥ ∇ · dz = 1. θ γ γ

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Furthermore, A(, ρ B ) =

b

a

=

a

b

) (y − m)θ + θm 2 1 + d yd x θ2 θ2 m− θ2 b 2 1 m (x) + 12 θ (x)2 dx + d x. θ (x) θ (x) a

m+ θ2

Thus d (E, F) =

1 ≤ ∗ d (E, F)

a

(

b

dx + θ (x)

a

b

1 m (x)2 + 12 θ (x)2 d x. θ (x)

(2.2)

Theorem 2.1. Suppose = {(x, y) : |y − m(x)| < θ(x)/2, a < x < b} is a Jordan domain and suppose θ and m are absolutely continuous on (a, b). Suppose further that {(x, y) : |x − x 0 | < δ, |y − y0 | < δ} ⊂ , and set z 0 = x0 + i y0 . If F = ∂ ∩ {(x, y) : x = b}, then ( ) b 1 1 + m (x)2 + 12 θ (x)2 dx , ω(z 0 , F) ≥ C(δ) exp −π θ (x) x0 where

(2.3)

π x0 +δ θ (x)d x C(δ) = exp − 2 δ x0 −δ

depends only on δ and θ (x) for x ∈ (x0 − δ, x0 + δ). Proof. Deﬁne the metric ρ for (x, y) ∈ by ⎧ ρ B (x, y) ⎪ ⎪ ⎪ ⎪ ! ⎪ ⎪ ⎪ 1 ⎪ ⎨ ρ B2 (x, y) + δ 2 ρ (x, y) = ⎪ ⎪ ⎪ 1δ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 0

if x > x0 + δ, if x0 < x < x0 + δ, if x0 − δ < x < x0 , elsewhere in ,

where ρ B is the Beurling metric (2.1). Let σ be any curve connecting z 0 to ∂. If γ is any curve in \ σ that sep|dz| ≥ 1. Indeed suppose γ ⊂ {z : Rez > x0 }. Then arates σ from F, then γ ρ ρ |dz| ≥ ρ |dz| ≥ 1. On the other hand, if γ ⊂ {z : Rez > x0 }, then the B γ γ

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euclidean length of γδ = γ ∩ {z : x0 − δ < Rez < x0 + δ} is at least δ, because |dz| ≥ γδ 1δ |dz| ≥ 1. γ separates σ from F. Hence γ ρ Also, b m+θ/2 1 x0 +δ ρ 2 d xd y = ρ B2 (x, y)d yd x + 2 θ (x)d x. δ x0 −δ x0 m−θ/2

Thus,

b 1 1 ≤ dx ∗ (σ, F) d θ (x) x0 b 2 1 m (x) + 12 θ (x)2 1 x0 +δ + θ (x)d x dx + 2 θ (x) δ x0 −δ x0

d (σ, F) =

by (2.2). Because the crosscut σ is arbitrary, the result now follows from Theorem IV.5.2. Theorem 2.1 implies that Theorem IV.6.1 is the best result possible for domains with smooth boundaries. Corollary 2.2. Suppose = {(x, y) : |y − m(x)| < domain and suppose

θ(x) 2 ,

x > a} is a Jordan

{(x, y) : |x − x0 | < δ, |y − y0 | < δ} ⊂ . Set z 0 = x0 + i y0 . If

∞ m (x)2

+ θ (x)2 θ (x)

a

d x = A < ∞,

then whenever b > x0 and F = {z ∈ : Rez ≥ b}, b b dx dx 8 ≤ ω(z 0 , F) ≤ exp −π , C exp −π π x0 θ (x) x0 θ (x) where C is a constant depending only on A and θ (x) for x ∈ (x 0 − δ, x0 + δ). The lower bound (2.3) and the upper bound (IV.6.3) were both obtained by studying metrics of the form ρ = |∇u|. Metrics of this form have the obvious advantage that lengths can be estimated by |∇u||dz| ≥ ∇u · dz , γ

γ

which is the change in u along γ . Other metrics of this form will appear in Section 6.

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3. Reduced Extremal Distance Let be a ﬁnitely connected Jordan domain, let z 0 ∈ , and let E be a ﬁnite union of closed subarcs of ∂. The reduced extremal distance δ (z 0 , E) is a conformally invariant version of the distance ds : γ ⊂ , γ joins z 0 to E inf γ

in from z 0 to E. Let us try to measure the distance from z 0 to E by deleting a small disc, Bε = Bε (z 0 ) = {z : |z − z 0 | < ε} from and computing the extremal distance d\Bε (∂ Bε , E). If β < ε, then by the serial rule and Example IV.1.2, d\Bβ (∂ Bβ , E) ≥ d\Bε (∂ Bε , E) + d Bε \Bβ (∂ Bβ , ∂ Bε ) 1 log(ε/β). = d\Bε (∂ Bε , E) + 2π

(3.1)

Hence 1 logε 2π is a decreasing function of ε. We will show in a moment that h(ε) is bounded above. It will then follow that the three limits 1 log ε, lim d\Bε (∂ Bε , E) + ε→0 2π h(ε) = d\Bε (∂ Bε , E) +

lim d\Bε (∂ Bε , ∂) +

ε→0

and

1 log ε, 2π

δ(z 0 , E) = δ (z 0 , E) = lim d\Bε (∂ Bε , E) − d\Bε (∂ Bε , ∂) ε→0

all exist and are ﬁnite. The reduced extremal distance is deﬁned to be the third limit δ(z 0 , E). To see that h(ε) is bounded above, choose a component E 1 of E and let ψ be a conformal map of C∗ \ E 1 to D such that ψ(z 0 ) = 0. Then ψ(Bε ) ⊃ B(0, aε) for some a > 0, so that by conformal invariance and the extension rule, h(ε) ≤ d\Bε (∂ Bε , E 1 ) +

1 logε 2π

≤ dD\B(0,aε) (∂ B(0, aε), ∂D) + and h(ε) is bounded.

1 1 1 logε = log , 2π 2π a

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Lemma 3.1. The reduced extremal distance δ(z 0 , E) is conformally invariant. Proof. Suppose ϕ is a conformal map deﬁned on . We may suppose that z 0 = 0 and ϕ(z 0 ) = 0. Because ϕ is conformal at 0, there exist ε0 > 0 and K > 0 such that for α = ε|ϕ (0)| − K ε2 and β = ε|ϕ (0)| + K ε2 , Bα ⊂ ϕ(Bε ) ⊂ Bβ 1 when ε < ε0 . Because k(α) = dϕ()\Bα (∂ Bα , ϕ(E)) + 2π log α is decreasing and bounded above and because limε→0 log(β/α) = 0, we have

lim dϕ()\Bα (∂ Bα , ϕ(E)) − dϕ()\Bβ (∂ Bβ , ϕ(E)) = lim (k(α) − k(β)) = 0. ε→0

ε→0

We conclude that

δ(0, E) = lim d (∂ Bε , E) − d (∂ Bε , ∂) ε→0

= lim dϕ() (ϕ(∂ Bε ), ϕ(E)) − dϕ() (ϕ(∂ Bε ), ∂ϕ()) ε→0

= lim dϕ() (∂ Bα , ϕ(E)) − dϕ() (∂ Bα , ∂ϕ()) ε→0

= δ(0, ϕ(E)), and δ(z 0 , E) is a conformal invariant.

Recall that when E is a compact plane set, the logarithmic capacity of E is e−γ E , where g(z, ∞) = log |z| + γ E + o(1) (as z → ∞), is Green’s function for C∗ \ E, and γ E is Robin’s constant for E In [1950] and [1952] Ahlfors and Beurling obtained the following connection between capacity and reduced extremal distance. Theorem 3.2 (Ahlfors–Beurling). If E is a ﬁnite union of closed arcs in ∂D, then δD (0, E) = δDc (∞, E) = γ E /π, where γ E is Robin’s constant for E. Proof. Set d = C∗ \ E and G(z) = g(z) + g(1/z), where g(z) is Green’s function in d with pole at ∞. Note that g(z) − g(1/z) − log |z| is harmonic in d and zero on ∂d and hence g(1/z) = g(z) − log |z| in d . Thus G(z) = 2g(z) − log |z| and g(0) = lim|z|→∞ g(z) − log |z| = γ E . Since G is symmetric about the unit circle, ∂G/∂r = 0 on ∂D \ E and G = 0 on E. If g ) is analytic in D g is a harmonic conjugate of g in D, then f (z) = ze−2(g+i and − log | f | = G. By repeating the proof of Theorem IV.4.1, but working with the symmetric function G instead of with ω, we see that f extends to be

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continuous on D, that | f | = 1 on E, and that ∂(arg f )/∂θ = 0 on ∂D \ E. The argument principle shows that f is a conformal map of D onto the disc with radial slits removed, because f has exactly one zero in D. Furthermore, ∂D = f (E) and f (0) = 0. The extremal distance from Fε ≡ {z : | f (z)| = ε} to 1 log(1/ε). Because g(0) = γ E , we have ε = |z|e−2g(z) ∼ |z|e−2γ E E is then 2π on Fε and hence Fε is close to the circle of radius εe2γ E . But then the proof of Lemma 3.1 yields δD (0, E) = lim dD (Fε , E) − dD (Fε , ∂D) ε→0 1 1 1 1 log − log 2γ = lim ε→0 2π ε 2π εe E = γ E /π.

The proof of Theorem 3.2 was a reprise of the proof of Theorem IV.4.1, with harmonic measure replaced by Green’s function and for that reason reduced extremal distance leads to estimates in which harmonic measures are replaced by capacities. Theorem 3.2 includes an estimate for the harmonic measure of E ⊂ ∂D, because by Example III.1.2, |E| |E| ≥ = ω(0, E). (3.2) 4 2π When E is a single arc, equality holds in the ﬁrst inequality in (3.2). Thus we have the following corollary. e−γ E ≥ sin

Corollary 3.3. If is a Jordan domain and if E is a ﬁnite union of closed arcs on ∂, then ω(z 0 , E) ≤ e−π δ(z 0 ,E) . If E is a single arc, then 2 2 sin−1 (e−π δ(z 0 ,E) ) ≥ e−π δ(z 0 ,E) . π π Pﬂuger [1955] gives a quantitative version of Theorem 3.2. ω(z 0 , E) =

Corollary 3.4 (Pﬂuger). If = D, if E is a ﬁnite union of closed arcs in ∂D, and if ∂ Bε is the circle centered at 0 of radius ε, then γE γE log 1/ε log(1 − ε) log 1/ε + + ≤ dD (∂ Bε , E) ≤ + . π 2π π π 2π Proof. By (3.1) the function h(ε) = dD (∂ Bε , E) − dD (∂ Bε , ∂D) = dD (∂ Bε , E) +

1 log ε 2π

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increases to γ E /π as ε → 0. This proves the right-hand inequality. By Theorem III.2.2 , we have 1 log dμ(ζ ), γ E − g(z) = Uμ (z) = |z − ζ | E g ) as where μ is a probability measure and g(z) = g (z, ∞). If f = ze−2(g+i in the proof of Theorem 3.2, then for z ∈ ∂ Bε

log | f | = −2g(z) + log ε ≤ −2γ E − 2 log(1 − ε) + log ε. Thus if log 1/r = 2γ E + 2 log(1 − ε) − log ε, dD (∂ Bε , E) ≥ d f (D) (∂ Br , ∂D) = dD (∂ Br , ∂D) 1 (2γ E + 2 log(1 − ε) − log ε). = 2π That proves the left-hand inequality.

See Exercise 2 for a proof of Theorem 3.2 using logarithmic potentials only. The next result introduces a metric found in Bonk, Koskela, and Rohde [1998] and Balogh and Bonk [1999]. Fix z 0 ∈ and let

" ds : γ is a curve in and z 0 , w ∈ γ dist (z 0 , w) = inf γ

be the euclidean distance in from z 0 to w. Theorem 3.5. Let ψ : D → be a normalized univalent function, ψ(0) = 0 and ψ (0) = 1, and let E ⊂ {ζ ∈ ∂ : dist (0, ζ ) ≥ R}. Then 1 log R. (3.3) 2π Equality holds in (3.3) if is a disc of radius R, with radial slits removed. δ (0, E) ≥

Proof. We may assume that ∂ is an analytic curve. Since ψ(0) = 0 and |ψ (0)| = 1, 1 1 log . δ (0, E) = lim d (∂ Bε , E) − ε→0 2π ε By the extension rule, we may suppose that E = {z ∈ : dist (0, z) ≥ R} and by another application of the extension rule, we may suppose that dist (0, ζ ) ≤ R for all ζ ∈ ∂ and E = {ζ ∈ ∂ : dist (0, ζ ) = R}. Deﬁne a metric ρ on by ρ(z) =

1 . dist (0, z)

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If γ : [0, L) → is a curve parameterized by arc length with γ (0) = 0, then |γ (s)| ≤ dist (0, γ (s)) ≤ s. Any curve σ connecting ∂ Bε to E in \ Bε can be extended to a curve γ connecting 0 to E by adding a line segment of length ε. Thus R 1 R ρ(z)ds = ρ(z)ds ≥ ds = log s ε σ |γ |≥ε ε and

ρ 2 (z)d xd y ≤

\Bε

ε<|w|
1 R d A(w) = 2π log , 2 |w| ε

where Bε = {z : |z| < ε}. Therefore d (∂ Bε , E) = d\Bε (∂ Bε , E) ≥ and

1 R log 2π ε

1 1 1 δ(0, E) = lim d (∂ Bε , E) − log ≥ log R. ε→0 2π ε 2π

(3.4)

By Theorem IV.4.2, equality holds in (3.4) when E is the circle of radius R and is the disc of radius R with radial slits removed. Corollary 3.6 (Balogh–Bonk). Let ψ(z) be a normalized univalent function, ψ(0) = 0 and ψ (0) = 1, in D and let R > 0. Then there is a constant C, independent of ψ and R, such that 1 $ # |ψ (r ζ )|dr > λ ≤ Cλ−1/2 . Cap ζ ∈ ∂D : 0

Proof. Set = ψ(D). By an approximation, we may suppose that ∂ is an analytic Jordan curve. By the Gehring–Hayman inequality, Exercise III.16, 1 |ψ (r ζ )|dr ≤ K dist (0, ψ(ζ )) 0

for some constant K independent of ψ and ζ . Corollary 3.6 now follows from √ Theorems 3.2 and 3.5 with R = λ/K and C = K . For a set E ⊂ ∂D, (3.2) shows that capacity can be replaced by length in the statement of Corollary 3.6. See Exercises III.22 and III.23 for precursors of Corollary 3.6.

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4. Teichmüller’s Modulsatz

4. Teichmüller’s Modulsatz Let = {R1 < |z| < R2 } and let J ⊂ be a continuum that separates into disjoint ring domains 1 , 2 with {|z| = R1 } ⊂ ∂1 and {|z| = R2 } ⊂ ∂2 .

J

R1

R2

1 2 Figure V.1 Teichmüller’s Modulsatz. By the serial rule mod(1 ) + mod(2 ) ≤ mod() =

R 1 2 log , 2π R1

(4.1)

and equality holds in (4.1) if J is a circle {|z| = r }. Theorem 4.1 says that if equality approximately holds in (4.1) then J is approximately a circle centered at 0. Theorem 4.1 (Teichmüller). If mod(1 ) + mod(2 ) ≥

R 1 2 log −δ 2π R1

(4.2)

and if δ > 0 is sufﬁciently small, then there is a C < ∞ independent of and δ such that . sup J |z| 1 ≤ 1 + C δ log . (4.3) inf J |z| δ In [1938] Teichmüller derived his Modulsatz, Theorem 4.1, from his special Modulsatz, Theorem 4.2 below, and we do the same. Let G be a simply connected domain such that 0 ∈ G. For small ρ deﬁne Mρ = Mρ (G) = mod(G \ B(0, ρ)).

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As in (3.1), Mρ +

log ρ log ρ ≤ Mρ + , 2π 2π

(4.4)

if ρ < ρ. Let ϕ : D → G be a conformal mapping with ϕ(0) = 0. Then for small ρ

Mρ (G) = mod D \ ϕ −1 (B(0, ρ)) , so that by the proof of Lemma 3.1, lim Mρ +

ρ→0

log ρ 1 = log |ϕ (0)|. 2π 2π

1 log |ϕ (0)| the reduced modulus of G. If G ⊂ C ∗ is simWe call M(G) = 2π ply connected and ∞ ∈ G, the reduced modulus of G is deﬁned by

M(G) =

1 1 (0)|, log | 2π ψ

where ψ : D → G is a conformal map with ψ(0) = ∞. It follows from Example III.1.1 that 2π M(G) = γ∂G when ∞ ∈ G. If J is a continuum and if G 1 and G 2 are disjoint components of C∗ \ J such that 0 ∈ G 1 and ∞ ∈ G 2 , then by (4.1) M(G 1 ) + M(G 2 ) ≤ 0, and equality holds if J is a circle {|z| = r }. Theorem 4.2. If M(G 1 ) + M(G 2 ) ≥ −δ

(4.5)

and if δ > 0 is sufﬁciently small, then there is a C < ∞ independent of J and δ such that . sup J |z| 1 ≤ 1 + C δ log . (4.6) inf J |z| δ By inequality (4.4), Theorem 4.1 is a corollary of Theorem 4.2. Proof. We follow Pommerenke [1991]. We assume J is not a circle. Let ϕ(z) =

∞ n=1

an z n

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4. Teichmüller’s Modulsatz be a conformal map of D onto G 1 . Also let ∞

ψ(z) =

b−1 bn z n + z n=0

be a conformal map of D onto G 2 . Then by (4.5) b−1 ≤ 2π δ. log a1

(4.7)

Inequality (4.7) is the main character in the proof of Theorem 4.2. By the proof of the area theorem, Lemma I.4.2, we have π

∞

∞

n|an |2 = Area(G 1 ) ≤ π |b−1 |2 − n|bn |2 .

n=1

n=1

Therefore |b−1 |2 − |a1 |2 ≥ |b1 |2 +

∞ n |bn |2 + |an |2 . n=2

In particular, |a1 | < |b−1 |, because J is not a circle. Set |z| = r = |a1 /b−1 |. Then by (4.7), e−2π δ ≤ r < 1 and (1 − r 2 ) log(1/(1 − r 2 )) ≤ C1 δ log(1/δ), so that by the Cauchy–Schwarz inequality *∞ , ∞ r 2n |ϕ(z) − a1 z|2 ≤ n|an |2 n n=2

n=1

≤ |b−1 |2 − |a1 |2 log Thus

1 1 ≤ C1 |b−1 |2 δ log . 1 − r2 δ

* inf |z| ≥ inf |ϕ(z)| ≥ |b−1 | 1 − C2

z∈J

|z|=r

.

1 δ log δ

, ≡ A(δ)

(4.8)

when δ is small, because by (4.7) |a1 | ≥ 1 − 2π δ. |b−1 | Similarly, on |z| = r , 2 * ∞ b 1 1 −1 2 ψ(z) − n|bn | log ≤ C1 |b−1 |2 δ log . + b0 ≤ z 1 − r2 δ n=1

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But then by (4.7), supz∈J |z − b0 | ≤ sup|z|=r |ψ(z) − b0 | √ ! ≤ |b−1 | 1 + C1 δ log 1δ ≡ B(δ),

(4.9)

when δ is small. Therefore J lies inside the circle centered at b0 of radius B(δ) and outside the circle centered at 0 of radius A(δ). Consequently . 1 |b0 | ≤ B(δ) − A(δ) = C3 |b−1 | δ log , δ and with (4.8) and (4.9) this gives (4.6). Corollary 4.3. Let R > 1, let = {1 < |z| < R} be the annulus with boundary curves 1 = {|z| = 1} and 2 = {|z| = R}, let σ be a curve in joining 1 ∈ 1 to 2 , and assume sup | arg z| = ε.

(4.10)

σ

If ε < ε0 < π , there is a constant C = C(ε0 ) such that if 1 < R ≤ e then d\σ (1 , 2 ) ≥

1 Cε2 , log R + 2π log 1ε

(4.11)

while if R > e then d\σ (1 , 2 ) ≥

1 C ε2 log R +

. 2π log R log log R + log 1ε

2 1

1 σ

R

R2 2ε

Figure V.2 , σ and their reﬂection about |z| = R.

(4.12)

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By the extension rule, 1 log R (4.13) 2π and equality holds in (4.13) if σ is the radial slit [1, R]. Corollary 4.3 says that if equality almost holds in (4.13) then σ is almost a radial segment. d\σ (1 , 2 ) ≥

Proof. We ﬁrst prove (4.11) in the case R ≤ e and then derive (4.12) for R > e from (4.11). Write λ = d\σ (1 , 2 ) and assume λ<

log R + δ. 2π

(4.14)

and Let σ be the extensions of and σ by reﬂection through {|z| = R}, so ) = 2mod(). Then by the symmetry that σ joins z = 1 to z = R 2 and mod( \ rule (see Exercise IV.3(a)) the family of symmetric curves in σ joining a satisﬁes λ( ) = 2λ. Set point eiθ ∈ 1 to R 2 eiθ ∈ ∂ B = e2π

2 /log

R

.

The map z → exp

−πi log z log R

\ sends σ to the slit ring 1 \ [1, B] bounded by and B = {Bz : z ∈ } and cut along the positive real axis. The map also sends 1 and {|z| = R 2 } to the slit [1, B] from to B and sends to the family of closed curves in 1 separating from B. Therefore for δ sufﬁciently small, π 2 1 π mod(1 ) ≥ δ + ... ≥ −2 2λ log R log R log B − Cδ. 2π Let ϕ(z) be a conformal map from D to the domain inside of such that ϕ(0) = 0. Let r > 0 be so small that 2 = Bϕ({r < |z| < 1}) ⊃ 1 and set 3 = 2 \ (1 ∪ ). Then

1 1 log , mod(2 ) = 2π r while

1 1 log − O(r ), mod(3 ) ≥ 2π Br ≥

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because

ϕ −1 (3 ) = D \ ϕ −1 Bϕ({|z| ≤ r }) ⊃ D \ {|z| ≤ Br − O(r 2 )}.

Therefore for r small, mod(1 ) + mod(3 ) ≥ mod(2 ) − Cδ.

Let J = (Bϕ)−1 = ϕ −1 ( B1 ). Then by (4.15) and Theorem 4.1 . 1 sup |z|/inf |z| ≤ 1 + C δ log . J δ J

(4.15)

(4.16)

By (4.10) we have −πε0

πε0

⊂ {e log R ≤ |z| ≤ e log R }, −πε0

πε0

B ⊂ {Be log R ≤ |z| ≤ Be log R }, and πε0

−πε0

e log R ≤ c(ε0 )Be log R

with c(ε0 ) < 1 because R ≤ e. By Koebe’s theorem J ⊂ {|z| ≤ c (ε0 ) < 1} so that by (4.16) and the Schwarz lemma . 1 sup |z|/inf |z| ≤ 1 + C δ log . δ Therefore by deﬁnition of

. ε ≤ C δ log

1 δ

and δ≥C

ε2 log 1ε

.

Now assume R > e. The map z → log z sends to a region log bounded by {Rez = 0}, {Rez = log R}, a curve γ from 0 to {Rez = log R}, and γ + 2πi. Let n be the integer such that n − 1 < log R ≤ n and let n be the region bounded by {Rez = 0}, {Rez = log R}, γ , and γ + 2π ni. Let E n = ∂n ∩ {Rez = 0} and let Fn = ∂n ∩ {Rez = log R}. Then by the parallel rule and conformal invariance 1 n n ≥ = . dn (E n , Fn ) d1 (E 1 , F1 ) d\σ (1 , 2 )

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By the R ≤ e case of (4.11) and a rescaling, we have dn (E n , Fn ) ≥

1 (ε/n)2 log R + C 2π n (log n/ε)3

and so d\σ (1 , 2 ) ≥

1 C ε2 log R +

. 2π log R log log R + log 1ε

Slightly weaker forms of (4.3) and (4.11) are easier to prove and sufﬁcient for most applications. See Exercise 4. Teichmüller’s [1938] paper actually shows that the inequality (4.3) is sharp, except for the choice of constant. For his example Teichmüller formed a Riemann surface (which is C∗ ) by ﬁxing q > 1, taking two domains D1 = D and D2 = {1 < |z| ≤ ∞}, and making the identiﬁcations: (i) eiθ ∈ ∂D1 with eiqθ ∈ ∂D2 , for 0 ≤ |θ | < π/q, (ii) eiθ ∈ ∂D1 with e−iθ ∈ ∂D1 , for π/q ≤ |θ | < π , and (iii) eiπ/q with e−iπ/q and with eiπ . By the uniformization theorem this gives two domains G 1 and G 2 such that J = C∗ \ (G 1 ∪ G 2 ) is a slit cardioid. Teichmüller showed (4.6) is sharp by varying the parameter q. It follows that (4.3) is also sharp.

5. Boundary Conformality and Angular Derivatives Let be a simply connected domain, let ζ ∈ ∂, and let F : → be a conformal mapping. In this section we study two kinds of local boundary behavior: (a) The conformality of F at ζ , and (b) The existence of a non-zero angular derivative F (ζ ). We seek geometric conditions on that are necessary and sufﬁcient for either (a) or (b) to hold. Let ζ ∈ C. For θ ∈ [0, 2π ), β ∈ (0, π/2), and ε > 0, deﬁne the truncated cone βε (ζ, θ ) = {z : | arg(z − ζ ) − θ| < β, 0 < |z − ζ | < ε}. Deﬁnition 5.1. We say that ∂ has an inner tangent with inner normal eiθ at ζ ∈ ∂ if for every β ∈ (0, π/2) there is an ε = ε(β) > 0 so that βε (ζ, θ ) ⊂ .

(5.1)

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When θ = π/2 we say ∂ has [vertical] vertical inner normal.

βε (ζ ) ζ

β ε

Figure V.3 Truncated cone at ζ . When ∂ has an inner tangent, the inner normal eiθ may not be unique. For example, at each point on a slit there are two inner tangents with opposite normal directions and the Jordan domain D ∩ {0 < arg z < 3π/2} has inner tangent at 0 with inner normal eiθ for all θ ∈ [π/2, π ]. Nevertheless, when ∂ has an inner tangent, we ﬁx one eiθ for which (5.1) holds and write βε (ζ ) = βε (ζ, θ ). If ∂ has an inner tangent at ζ and if lim

βε (ζ )z→ζ

f (z) = A,

we say the function f has nontangential limit A at ζ. Deﬁnition 5.2. Suppose F : → is a conformal mapping and suppose ∂ has an inner tangent at ζ ∈ ∂. We say F is conformal at ζ if F has a nontangential limit F(ζ ) ≡

lim

F(z)

arg

F(z) − F(ζ ) z−ζ

βε (ζ )z→ζ

and if the limit Aζ =

lim

βε (ζ )z→ζ

(5.2)

exists for every β ∈ (0, π/2). If F is conformal at ζ and if |α − θ| < π/2, then as z → ζ the image arc F({arg(z − ζ ) = α}) is asymptotic to the ray {arg(w − F(ζ )) = α + Aζ }. In other words, F preserves angles between nontangential rays. If F is conformal

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at ζ , then ∂ F() has an inner tangent at F(ζ ) and F −1 is conformal at F(ζ ). If is a Jordan domain, if F : D → is a conformal mapping, and if ∂ has a tangent at F(ζ ), then by Theorem II.4.2, F is conformal at ζ . Moreover, in this case the convergence z → ζ need not be restricted to cones β (ζ ). In general, F can be conformal at ζ ∈ ∂D even when does not have a tangent at F(ζ ). See Exercise VII.12 or Theorem 5.5 below. Deﬁnition 5.3. Suppose F : → is a conformal map deﬁned on and suppose ∂ has an inner tangent at ζ ∈ ∂. We say F has angular derivative F (ζ ) if for all β ∈ (0, π/2) the two limits F(ζ ) ≡

lim

βε (ζ )z→ζ

F(z)

and F (ζ ) ≡

lim

βε (ζ )z→ζ

F(z) − F(ζ ) z−ζ

(5.3)

exist and are ﬁnite. Lemma 5.4. Suppose F : → is a conformal map deﬁned on and suppose ∂ has an inner tangent at ζ ∈ ∂. Then F has an angular derivative at ζ if and only if F has a ﬁnite nontangential limit at ζ . Proof. If F has nontangential limit L, then by integration F has ﬁnite nontangential limit F(ζ ) and by the fundamental theorem of calculus (5.3) holds with F (ζ ) = L . Conversely, if F has an angular derivative at ζ and if z n → ζ nontangentially, then uniformly in some disc {w : |w| < δ}, gn (w) =

F(z n + (ζ − z n )w) − F(ζ ) → (w − 1)F (ζ ). ζ − zn

Therefore lim F (z n ) = lim gn (0) = F (ζ ) and F has nontangential limit F (ζ ) at ζ .

It is clear that if F has a non-zero angular derivative at ζ then F is conformal at ζ and Aζ = arg F (ζ ). Furthermore, F −1 also has a non-zero angular derivative at F(ζ ). On the other hand, F can be conformal at ζ ∈ ∂ without having an angular derivative at ζ . See Exercise VI.8. However, the conformality of F at ζ does yield some information about the difference quotients (5.3). Suppose ∂ has an inner tangent with vertical inner

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normal at 0 ∈ ∂ and suppose F is conformal at 0. Form the functions * , F(εz) − F(0) εi f ε (z) = log · . εz F(εi) − F(0) Then Im f ε converges to 0, as ε → 0, uniformly on compact subsets of the half annulus A = {z : 1/2 < |z| < 2} ∩ H. Since Re f ε (i) = 0, it follows that Re f ε converges to 0 uniformly on compact subsets of A. Therefore F(εz) − F(0) εi lim · = lim e fε = 1 ε→0 εz F(εi) − F(0) ε→0 uniformly on compact subsets of A. Consequently F has an angular derivative at 0 if and only if the vertical limit lim

ε→0

F(εi) − F(0) εi

exists. This fact also follows from Lindelöf’s theorem, Exercise II.3(d). For the rest of this section we assume ϕ:H→ is a conformal mapping from upper half-plane H onto a domain such that 0 ∈ ∂ and we assume ϕ has nontangential limit lim

α (0)z→0

ϕ(z) = 0.

(5.4)

It will often be convenient to transform both H and ≡ ϕ(H) to “strips.” The “standard strip” S = {z : |Imz| < π/2} is mapped by the function τ (z) = ie−z onto H so that τ (+∞) = 0 and τ (−∞) = ∞. Then ψ(z) = τ −1 ◦ ϕ ◦ τ (z) = iπ/2 − log ϕ(ie−z )

(5.5)

deﬁned by is a conformal map of S onto the region = {iπ/2 − log w : w ∈ }

or

}. = {ie−z : z ∈

With this correspondence the cones β1 (0) in H or in are transformed to half-strips Sδ = {z : |Imz| < π/2 − δ and Rez > 0},

δ ∈ (0, π/2)

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5. Boundary Conformality and Angular Derivatives . With this transformation (5.4) holds if and only if in S or in lim

Sδ z→+∞

Reψ(z) = +∞

for all δ ∈ (0, π/2), and condition (5.2) holds if and only if lim Imψ(z) − Imz = −Aζ Sδ z→+∞

for all δ ∈ (0, π/2). Similarly, ∂ has an inner tangent at 0 with a vertical normal if and only if for each δ ∈ (0, π/2) there is xδ > 0 such that . xδ + Sδ ⊂

α (0)

i π2

(5.6)

i π2

− log(w)

R xδ

xδ + Sδ

0 Figure V.4 Transforming the half-plane version to the strip version.

−i π2

The ﬁrst theorem gives local geometric conditions on a region = ϕ(H) that are necessary and sufﬁcient for ϕ to be conformal ζ ∈ R. The theorem is from Ostrowski [1937]. For convenience we treat only the special case ζ = 0, ϕ(0) = 0, and Aζ = 0 in (5.2); the other cases reduce to this case by translations and rotations. Theorem 5.5 (Ostrowski). Suppose is a simply connected domain in C and suppose 0 ∈ ∂. Let ϕ : H → be a conformal map having nontangential limit ϕ(0) = 0. If ϕ is conformal at 0 and if (5.2) holds with A0 = 0, then ∂ has an inner tangent at 0 with a vertical inner normal and dist(x, ∂) = 0. x Rx→0 lim

(5.7)

Conversely, if ∂ has an inner tangent at 0 with a vertical inner normal and if (5.7) holds, then there exists a conformal map ϕ of H onto having nontangential limit ϕ(0) = 0 such that ϕ is conformal at 0 and the limit A0 in (5.2) is 0.

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The comb in Figure V.5 has tines [xn − i, xn ] with xn = −x−n → 0. If ϕ is a conformal map of the upper half-plane onto the complement of the comb (including ∞) and if ϕ(0) = 0, then by Ostrowski’s theorem ϕ is conformal at 0 if and only if limn→∞ xn+1 /xn = 1. 0

xn

xn − i Figure V.5 A comb region. Proof. If ϕ(0) = 0 and if the limit in (5.2) is 0, then for all θ ∈ (0, π) the ray {r eiθ : r > 0} and its image are asymptotic and therefore ∂ has an inner tangent at 0 with a vertical inner normal. If (5.7) fails, there is ε > 0 and a sequence xn → 0 such that dist(xn , ∂) ≥ εxn for all n. We may suppose xn > 0 and xn ∈ by (5.1). The sequence of harmonic functions xn z u n (z) = arg −1 ϕ (xn z) is then a normal family on Aε = B(1, ε) ∪ {z ∈ H : 1 − ε < |z| < 1 + ε}, and for 0 < δ < π/2, u n (z) converges to 0 on Aε ∩ {z : δ < arg z < π − δ}. Fix z = 1 − i 2ε ∈ Aε . Then u n (z) converges to 0, so that arg ϕ −1 (xn z) < 0 for large n. Because xn z ∈ and ϕ −1 () = H, this is a contradiction and (5.7) is established. Conversely, assume ∂ has an inner tangent at 0 with a vertical inner normal and assume (5.7) holds. Let Er be the component of ∩ {z : |z| = r } such that ir ∈ Er and let be a conformal map of onto D. By the proof of Carathéodory’s Theorem I.3.1, applied to (instead of −1 ), we conclude that lim (z)

z∈Er r →0

exists in ∂D. Composing −1 with a linear fractional transformation, we obtain a conformal map ϕ of H onto such that lim ϕ −1 (z) = 0.

z∈Er r →0

(5.8)

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179

Because conformality is a local property of ∂, we may suppose contains iR+ = {i y : y > 0} and limz→∞ |ϕ −1 (z)| = +∞. In terms of the and that if ψ is given by (5.5) then strip regions this means that R ⊂ −1 ψ (R) is a curve in S along which Reψ −1 (x) → +∞ as x → +∞ and Reψ −1 (x) → −∞ as x → −∞. Condition (5.7) becomes ) = 0, lim dist(z, ∂

(5.9)

z∈∂S x→+∞

such that the set Er is replaced by the component Fx of {z : Rez = x} ∩ Fx ∩ R = ∅, and (5.8) translates to lim Reψ −1 (z) = +∞.

(5.10)

z∈Fx x→+∞

Take ζ ∈ ∂S with Imζ = π/2. For δ < 1, set Bδ = {z : |z − ζ | < δ}, and let satisfying Uδ ∩ FReζ = ∅. We claim there is Uδ be the component of Bδ ∩ an absolute constant C so that when Uδ = ∅ 1 1 −2 −1 diam ψ (Uδ ) ≤ C log . (5.11) δ

∂S

ζ

∂S

Uδ FReζ

ψ −1 (Uδ )

ψ

Figure V.6 Distortion near the ends of vertical crosscuts. Indeed, by the extension rule and the annulus example d (Uδ , R) ≥ dC (Bδ , R) ≥ C +

1 1 log . 2π δ

(5.12)

By conformal invariance and the extension rule, −1 −1 −1 d (Uδ , R) = dS (ψ (Uδ ), ψ (R)) ≤ dS (ψ (Uδ ), L),

(5.13)

where L = −πi/2 + R is the component of ∂S such that ζ ∈ / L. Let be the family of curves in S separating the connected set ψ −1 (Uδ ) from L. Using the spherical metric, as in Exercise IV.6, we obtain 1 = λS () ≥ C(diam ψ −1 (Uδ ))2 . dS (ψ −1 (Uδ ), L)

(5.14)

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Now (5.11) follows from (5.12), (5.13), and (5.14). The same estimate (5.11) also holds if ζ ∈ ∂S and Imζ = −πi/2. Finally, if δ > 0 and if z ∈ ∂ Sδ then dist(z, ∂) < 2δ for Rez is sufﬁciently large by (5.9). Thus by (5.11) 1 1 dist(ψ −1 (z), ∂S) < C(log )− 2 δ and 1 1 |Imz − Imψ −1 (z)| ≤ C(log )− 2 . δ

(5.15)

In other words γ = ψ −1 (∂ Sδ ) ∩ {Rez > x(δ)} consists of two curves situated 1 outside Sε , where ε = 2C(log 1/δ)− 2 . Moreover, by (5.10) γ → +∞. By the maximum principle and (5.15), 1 1 |Imψ(w) − Imw| ≤ C(log )− 2 δ when w ∈ M + Sε if M > 0 is sufﬁciently large. Because δ > 0 was arbitrary, this proves that ϕ has nontangential limit 0 at 0 and that (5.2) holds with limit A0 = 0. Ostrowski’s theorem gave a geometric condition at p ∈ ∂ which is necessary and sufﬁcient for the conformal map ϕ : H → from the disc or halfplane to to be conformal at ϕ −1 ( p). The next theorem, due to Rodin and Warschawski [1976] and independently to Jenkins and Oikawa [1977], gives an extremal length condition that is necessary and sufﬁcient for the conformal map to have non-zero angular derivative at ϕ −1 ( p). In Section 6 we will reinterpret this extremal length condition more geometrically. Because the existence of a non-zero angular derivative depends on the domain and not on the choice of the conformal map ϕ, we make the following deﬁnition: Deﬁnition 5.6. If is simply connected and if 0 ∈ ∂, we say has a positive angular derivative at 0 ∈ ∂ if there is a conformal map ϕ of H onto which has nontangential limit ϕ(0) = 0 and which has angular derivative ϕ (0) such that 0 < ϕ (0) < ∞. Theorem 5.7 (Jenkins–Oikawa, Rodin–Warschawski). Suppose is a simply connected domain such that 0 ∈ ∂ and iR+ = {i y : y > 0} ⊂ . For r > 0, let Er denote the component of ∩ {z : |z| = r } containing the point ir . Then has positive angular derivative at 0 if and only if (a) has an inner tangent at 0 with a vertical inner normal, and (5.16) (b) lims
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If the Ostrowski condition (5.7) holds, then Theorem 5.7 is still true if condition (b) is replaced by the weaker (b ) limn>m→+∞ d (E 2−m , E 2−n ) −

1 π

log 2n−m = 0.

See Exercise 5. Er

Es 0

Figure V.7 Positive angular derivative at a tine. Proof. First assume that has a positive angular derivative at 0. Then (a) holds by Theorem 5.5. To establish (b), we transform the problem using (5.5). The = {iπ/2 − log w : w ∈ } conformal map ψ of the standard strip S onto satisﬁes lim ψ(z) − z = C

Rez→+∞ z∈Sδ

for each δ ∈ (0, π/2), where C is a (ﬁnite) real number; or equivalently, lim

Rew→+∞ w∈Sδ

w − ψ −1 (w) = C

(5.17)

∩ {z : Rez = x} such for each δ ∈ (0, π/2). Let Fx denote the component of that x ∈ Fx . Then (b) will hold if and only if π d (Fs , Ft ) − |s − t| → 0

(5.18)

, we may suppose that the constant C in (5.17) as s, t → +∞. By translating is zero. That will not affect (5.18). With ε > 0 and C as in (5.11) choose δ so that C(log 1/δ)−1/2 < ε. By Theorem 5.5, we can choose xε so that xε + Sδ ⊂ and B(ζ, δ) ∩ ∂ = ∅ if ζ ∈ ∂S satisﬁes Reζ > xε . Let σζ be the component of {z : |z − ζ | = δ} ∩ such that σζ ∩ FReζ = ∅. Then if Uζ is the component of \ σζ that meets

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B(ζ, δ), we have by (5.11) 1 1 diamψ −1 (Uζ ) = diamψ −1 (σζ ) ≤ C(log )− 2 < ε. δ By (5.17), we may also suppose that |Reψ −1 (z) − x| < ε for all z ∈ Sδ ∩ Fx and all x > xε . Consequently, sup |Reψ −1 (z) − x| < ε, x > xε .

z∈Fx

Thus for t > s > xε , {z : s − ε < Rez < t + ε, |Imz| < π/2} ⊃ ψ −1 (Fs ) ∪ ψ −1 (Ft ),

(5.19)

whereas {z : s + ε < Rez < t − ε, |Imz| < π/2} ∩ (ψ −1 (Fs ) ∪ ψ −1 (Ft )) = ∅.

(5.20)

Then by the extension rule and conformal invariance t − s − 2ε < π dS (ψ −1 (Fs ), ψ −1 (Ft )) = π d (Fs , Ft ) < t − s + 2ε and (5.18) follows. Conversely, assume (a) and (b) hold. We ﬁrst establish (5.7). If (5.7) fails, then as in the proof of Theorem 5.5 we may suppose there is a region A = {z : Imz > 0, 1 − ε < |z| < 1 + ε} ∪ B(1, ε) and a sequence xn → 0, xn > 0 such that xn A ⊂ . Let rn = (1 + ε)xn and sn = (1 − ε)xn . Then by the extension rule and conformal invariance, d (Ern , E sn ) ≤ d A ({|z| = 1 + ε}, {|z| = 1 − ε}) 1+ε 1 rn 1 = −C(ε) + log , = −C(ε) + log π 1−ε π sn for some C(ε) > 0. Letting xn → 0 then contradicts (b), and therefore (5.7) holds. Now choose ϕ as in Theorem 5.5, deﬁne ψ by (5.5), and recall that Fx is ∩ {Rez = x} containing x. We claim that the variation of that component of Rez along the image of Fx satisﬁes Vx = sup |Reψ −1 (u) − Reψ −1 (v)| → 0

(5.21)

u,v∈Fx

as x → +∞. If (5.21) is true, then by a comparison with rectangles as in (5.19) and (5.20) we obtain via (5.18) that lim

Rew→+∞ w∈Sδ

Rew − Reψ −1 (w)

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5. Boundary Conformality and Angular Derivatives exists for all δ ∈ (0, π/2). Then by Theorem 5.5 lim

Rew→+∞ w∈Sδ

Imw − Imψ −1 (w) = 0,

and (5.17) follows. But (5.17) clearly implies that ϕ has a positive angular derivative. Thus to complete the proof of Theorem 5.7, we need only prove (5.21). To prove (5.21) note that by (5.18) there exists xδ > 0 so that for all r, s, t with xδ < r < s < t < ∞ we have approximate equality in the serial rule: d (Fr , Fs ) + d (Fs , Ft ) ≥ d (Fr , Ft ) − δ. Let f be the conformal map of the subregion of S between ψ −1 (Fr ) and ψ −1 (Ft ) onto the rectangle R = {z : 0 < Imz < π, 0 < Rez < π d (Fr , Ft )}, −1 −1 so that f ◦ ψ (Fr ) is the left vertical end Ir of R and f ◦ ψ (Ft ) is the right vertical end It of R. Then Is = f ◦ ψ −1 (Fs ) divides R into two regions U1 and U2 , and dU1 (Ir , Is ) + dU2 (Is , It ) ≥ d R (Ir , It ) − δ.

(5.22)

Is Ir

U1

U2

It

ε Figure V.8 Image of a vertical crosscut. with ends Reﬂect R through its top edge, obtaining a rectangle R Ir and It and R contains regions U1 and U2 separated by a curve Is which connects as to as + 2πi. By the symmetry rule, dU1 ( Ir , Is ) + dU2 ( Is , I t ) ≥ d R ( Ir , It ) − δ/2.

(5.23)

s connects as ∈ R to as + 2πi, Theorem 4.1 can now be applied to Because T z to derive (5.21) from (5.23). {e : z ∈ R} Exercises 4 and 5 include proofs of (5.21) without Teichmüller’s Modulsatz. When ∂ is smooth, Theorem 5.7 and the estimates in Sections 2 and IV.6 yield necessary and sufﬁcient geometric conditions for the existence of a positive angular derivative.

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Corollary 5.8. Suppose is a simply connected domain given by = {r eiα : α1 (r ) < α < α2 (r ); r > 0}, where α j are absolutely continuous in (0, 1), α1 (r ) < π/2 < α2 (r ), α1 (0) = 0, α2 (0) = π, and 1 α j (r )2 r dr < ∞, 0

for j = 1, 2. Let (r ) = α2 (r ) − α1 (r ). Then has a positive angular derivative at 0 if and only if 1 1 1 − dr (5.24) r (r ) πr 0 exists. A related result can be found in Rodin and Warschawski [1976]. Proof. Since α1 (0) = 0 and α2 (0) = π , has an inner tangent with a vertical inner normal. Let α1 (e−x ) + α2 (e−x ) π − , 2 2

m(x) =

θ (x) = α2 (e−x ) − α1 (e−x ), and = {x + i y : m(x) − θ (x)/2 < y < m(x) + θ (x)/2}. onto . Let Fx = ∩ {z : Rez = x}. Then ϕ(z) = ie−z is one-to-one from For s, t sufﬁciently large, by (IV.6.2) and (2.2) we have the estimates t t t 2 1 m (x) + 12 θ (x)2 1 1 d x ≤ d dx + d x. (Fr , Fs ) ≤ θ (x) s θ (x) s θ (x) s By our assumptions on α j , lim

s,t→+∞ s

t

1 m (x)2 + 12 θ (x)2 d x = 0. θ (x)

(5.25)

Thus by Theorem 5.7, has a positive angular derivative at 0 if and only if t 1 1 lim − d x = 0, (5.26) s,t→+∞ s θ (x) π and by a change of variables, (5.26) is equivalent to (5.24).

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185

6. Conditions More Geometric In this section we seek more geometric conditions on a simply connected domain that are equivalent to the existence of a non-zero angular derivative. We concentrate on the normalized domain = {iπ/2 − log w : w ∈ } . From this perspective, has an inner tangent and on the point z = +∞ ∈ ∂ , and has a positive and vertical inner normal if and only if (5.6) holds for . If is a simply angular derivative if and only if (5.6) and (5.18) hold for connected domain satisfying (5.6) and (5.18), we say that has an angular derivative at +∞. We do not require that the map z → e−z be one-to-one on . We begin by observing that under mild assumptions, condition (5.26) ∩ {z : s < Rez < t} and equivalent to a statement about areas. Set s,t = Ss,t = {z : s < Rez < t, |Imz| < π/2} and note that then t (π − θ )d x = Area(Ss,t \ s,t ) − Area(s,t \ Ss,t ). s

Now if ∂ has an inner tangent at 0 with a vertical inner normal, then for x sufﬁciently large π ≤ θ (x) ≤ 2π. 2 Hence if π − θ has constant sign on (s, t), then t 1 1 s θ − π dx 1 2 ≤ ≤ . (6.1) 2 2π π2 Area(Ss,t \ s,t ) − Area(s,t \ Ss,t ) In particular, if ) + Area( \ S) < ∞ Area(S \ then

t 1 1 lim − dx = 0 s,t→∞ s θ (x) π

and (5.26) holds. When (5.25) fails, upper and lower estimates can still be found in terms of is replaced by a smaller region with a Lipschitz boundary. Let M > 0. areas if Recall that an M-Lipschitz graph is the graph of a function h for which |h(s) − h(t)| ≤ M|s − t|

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is a region such that R ⊂ . Let T M denote the for all s, t ∈ R. Suppose collection of isosceles triangles T such that (i) T ⊂ , (ii) T has base on R, and (iii) T has sides with slope ±M, and deﬁne M =

&

{T : T ∈ T M }.

(6.2)

M ⊂ and ∂ M is the largest domain such that R ⊂ Then M consists of two M-Lipschitz graphs. ∂

\ M ∂S

M ∂

∂S

Figure V.9 M-Lipschitz subregion. ⊃ R is a simply connected domain such that Theorem 6.1. Suppose ∩ {z : Rez < 0} = ∂S ∩ {z : Rez < 0}. ∂

(6.3)

be the region deﬁned by (6.2) above. M ⊂ Let has an angular derivative at +∞ M ) < ∞. Then (a) Suppose Area(S \ if and only if M \ S) < ∞. Area( has an angular derivative at +∞ M \ S) < ∞. Then (b) Suppose Area( if and only if M ) < ∞, Area(S \ provided that M > 8π . The hypothesis (6.3) is not crucial because the existence of an angular derivative at +∞ is a local property of ∂ near +∞. See Marshall [1995] for an example where M is small and (b) fails.

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M ) < ∞ and Area( M \ S) < ∞, then Proof. If Area(S \ lim dist(z, ∂S) = 0

Rez→+∞ z∈∂ M

(6.4)

has an angular derivative at +∞, then by and (5.9) holds. Conversely, if Ostrowski’s theorem, (6.4) holds. Thus we may assume (6.4) during the re ∩ {z : Rez = s} such mainder of the proof. If Fs denotes the component of has an angular derivative at +∞ if and only if that s ∈ Fs , then lim d (Fs , Ft ) −

s,t→∞ s
t −s = 0. π

(6.5)

To prove Theorem 6.1 we will obtain upper and lower bounds for the left side of (6.5). The upper bound is given by Lemma 6.2, and the lower bound by Lemma 6.4. M is given by the two curves y = h 1 (x) + π/2 and Now suppose that ∂ y = −h 2 (x) − π/2, −∞ < x < +∞, where h 1 > −π/2 and h 2 > −π/2. By (6.4), lim x→+∞ h j (x) = 0. As before, for any region U , let Us,t = U ∩ {z : s < Rez < t}. Lemma 6.2 (Sastry). Suppose |h j (x)| ≤ π M 2 for s < x < t. Then d (Fs , Ft ) −

M )s,t M \ S)s,t t −s Area( (4M 2 + 2)Area(S \ − . ≤ 2 π π (4M 2 + 2)π 2

Sastry [1995] proved this lemma with different constants and used a piecewise constant metric. We will use a metric of the form ρ = |∇u|, where it is easier to estimate lengths. Proof. To give an upper bound for extremal distance, we construct a metric that gives a lower bound for the conjugate extremal distance. The idea is that the onto S is nearly the identity in Sδ and it compresses the conformal map of \ S and expands near ∂ ∩ S. The expansion and compression regions near ∂ will be accomplished by imitating Beurling’s metric from Section 2 in those regions. Because d M )s,t (Fs , Ft ), (Fs , Ft ) ≤ d M (Fs , Ft ) = d(

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without loss of generality, we may suppose (for notational convenience) that = ( M )s,t . Deﬁne : h 1 (x) > 0 and − h 1 /(2M 2 ) < y < h 1 }, U1 = {(x, y + π/2) ∈ : h 1 (x) < 0 and 2h 1 < y < h 1 }, U2 = {(x, y + π/2) ∈ : h 2 (x) > 0 and − h 2 < y < h 2 /(2M 2 )}, U3 = {(x, y − π/2) ∈ : h 2 (x) < 0 and − h 2 < y < −2h 2 }, U4 = {(x, y − π/2) ∈ and \ U∗ =

4

j=1 U j .

∂ ∂S

U1

Fs

U1 U2

U∗ U4

∂S

Ft U4

U3 ∂ Figure V.10 Proof of Sastry’s lemma.

onto S which ﬁxes Consider the continuous map (x, y) → (x, u(x, y)) of ∗ and is linear in y on \ U . Thus ⎧ y for (x, y) ∈ U ∗ , ⎪ ⎪ ⎪ 2 ⎪ (y − (h 1 + π/2))/(2M + 1) + π/2 for (x, y) ∈ U1 , ⎨ u(x, y) = 2(y − (h 1 + π/2)) + π/2 for (x, y) ∈ U2 , ⎪ ⎪ ⎪ for (x, y) ∈ U3 , (y + h 2 + π/2)/(2M 2 + 1) − π/2 ⎪ ⎩ 2(y + h 2 + π/2) − π/2 for (x, y) ∈ U4 .

U∗

. Note that Set ρ = |∇u| on ⎧ ⎨1 2 |∇u| ≤ 4(1 + M 2 ) ⎩ (1 + M 2 )/(2M 2 + 1)2

on U ∗ , on U2 ∪ U4 , on U1 ∪ U3 .

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Thus,

189

|∇u|2 d yd x −

Ss,t

d yd x

≤ 4(1 + M 2 )Area(U2 ∪ U4 ) +

1 + M2 Area(U1 ∪ U3 ) (2M 2 + 1)2

1 \ Ss,t ) − 2Area(Ss,t \ ) Area( 2M 2 1 ) − \ Ss,t ). = (4M 2 + 2)Area(Ss,t \ Area( 4M 2 + 2 −

If γ is a curve connecting {y = h 1 (x) − π/2} to {y = −h 2 (x) − π/2} and , then contained in |∇u||dz| ≥ ∇u · dz = π. γ

Thus

γ

d (Fs , Ft ) ≤

|∇u|

2 d yd x

(inf γ γ |∇u||dz|)2 1 ) ≤ 2 π(t − s) + (4M 2 + 2)Area(Ss,t \ π 1 \ Ss,t ) . Area( − 4M 2 + 2

The next step is to derive a lower bound for the extremal distance in = ∪ j σ j where each σ j is an open arc with endpoints M \ ∂ (6.5). Write ∂ where Rez l < Rez r . For v ∈ C \ R, let Tv denote the triangle z lj , z rj ∈ ∂ j j in T M with vertex v. There is a unique v j with Rez lj ≤ Rev j ≤ Rez rj so that z lj , z rj ∈ ∂ Tv j . Then σ j ⊂ ∂ Tv j and σ j consists of at most two line segments. If |Imz lj | ≤ |Imz rj |, let B j = {z : |z − z lj | < Rez rj − Rez lj }. Otherwise let B j = {z : |z − z rj | < Rez rj − Rez lj }. Lemma 6.3. Area(B j ) ≤

8π M

z∈σ j

π |Imz| − d x, 2

z = x + i y.

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σj

vj

∂ z rj ∂S

z lj

Bj Figure V.11 Proof of Lemma 6.3. Proof. Suppose f is deﬁned and continuous on [0, 1] with

1 0 ≤ x ≤ a, f (x) = −1 a ≤ x ≤ 1. Then by elementary geometry or calculus 1 | f (x)|d x 0

is minimal when a =

1 2

and f (0) = − 41 . Thus 1 1 | f (x)|d x ≥ . 8 0

If Imz > 0 on σ j , the map * (x, y) →

x − Rez lj Rez rj − Rez lj

,

y − π/2

,

M(Rez rj − Rez lj )

transforms σ j into the graph of one such f . Thus M(Rez rj − Rez lj )2 M |Imz| − π d x ≥ = Area(B j ). 2 8 8π z∈σ j The inequality of the lemma for Imz < 0 follows by a reﬂection about R. The quantity

z∈σ j

π |Imz| − d x 2

is the total area “between” σ j and ∂S. using the The next lemma gives a lower bound for the extremal distance in geometry of M . When = M , it follows immediately from the discussion at the end of Section 5.

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6. Conditions More Geometric Lemma 6.4. If ε > 0 there is an s0 < ∞ so that for s0 < s < t < ∞,

M )s,t t −s (M − 8π )Area(S \ ≥ π Mπ 2 M \ S)s,t (M + 8π )Area( − − ε. 2 Mπ M \ S)s,t → 0 and Proof. Suppose 0 ≤ t − s ≤ ε. Then as s, t → ∞, Area( M )s,t → 0 as s, t → ∞ by (6.4) and the inequality follows. Now Area(S \ given by suppose that t − s ≥ ε. Let ρ be the metric on

ρ = 1 for z ∈ M ∪ j B j , . 0 elsewhere on d (Fs , Ft ) −

The metric ρ will provide the lower bound for the extremal distance. First we s,t . By Lemma 6.3 compute the ρ-area of ρ 2 d xd y − d xd y s,t

Ss,t

M )s,t + M \ S)s,t − Area(S \ ≤ Area(

Area(B j )

(6.6)

j

8π 8π M \ S)s,t − (1 − M )s,t . )Area( )Area(S \ M M connecting Fs to Ft , with s < t. Note that Now suppose γ is a curve in ≤ (1 +

Rez rj − Rez lj ≤

4 max ||Imz| − π/2| → 0 M z∈σ j

(6.7)

as σ j → +∞, by (6.4). Thus by deleting an initial and terminal portion of γ , M ∪ j B j and that if necessary, we may suppose that γ begins and ends in if γ meets σ j , then s < Rez lj < Rez rj < t. M ∂

∂ γ2

Fs γ1

Ft γ3

Figure V.12 Proof of Lemma 6.4. Thus we can write γ = either

k

γk where {γk } are disjoint subarcs of γ such that

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M ∪ j B j , or (i) γk ⊂ (ii) γk meets some σ j and connects FRezl to FRezrj with s < Rez lj < Rez rj < t. j If (i) holds, then γk ρ|dz| is at least equal to the change in Rez along γk . If (ii) holds, then M ∪ B j )) ≥ Rez rj − Rez lj . length(γk ∩ ( Indeed, if z lj is the center of B j and if ζ ∈ γk ∩ σ j , then the shortest curve from is the straight line from z l to ζ , since |Imζ | ≥ |Imz l |. Since FRezl to ζ in j j j γk must intersect ∂ B j , it must have length at least the radius of B j , namely Rez rj − Rez lj . A similar argument works if z rj is the center of B j . Together with (6.7) this implies, for s, t sufﬁciently large, that ρ|dz| ≥ t − s − ε. γ

We conclude by (6.6) that for t − s ≥ ε, d (Fs , Ft ) ≥

(t − s − ε)2 π(t − s) + (1 + 8π M )Area( M \ S)s,t − (1 −

8π M )Area(S \ M )s,t

.

M is Lipschitz and (6.4) holds, we have, as s, t → ∞, both Because ∂ M )s,t /(t − s) → 0. Hence for Area( M \ S)s,t /(t − s) → 0 and Area(S \ s, t sufﬁciently large with t − s ≥ ε, M )s,t t −s (M − 8π )Area(S \ ≥ 2 π Mπ M \ S)s,t (M + 8π )Area( − − ε. 2 Mπ concluding the proof of Lemma 6.4. d (Fs , Ft ) −

M ) < ∞ We now complete the proof of Theorem 6.1. We have Area(S \ M )s,t → 0 as s, t → +∞. Likewise, the condition if and only if Area(S \ M \ S)s,t → 0 as s, t → +∞. M \ S) < ∞ is equivalent to Area( Area( Thus if Area(S \ M ) < ∞ and Area( M \ S) < ∞, by Lemmas 6.2 and 6.4, lim d (Fs , s,t→∞

Ft ) − (t − s)/π = 0,

and hence by the discussion at the beginning of the proof of the Theorem 6.1, has an angular derivative at +∞.

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M ) < ∞ and if has an angular derivative at +∞, then by If Area(S \ (6.5) and Lemma 6.2, M \ S)s,t → 0 Area( as s, t → +∞. That proves part (a) of Theorem 6.1. If M > 8π and if M \ S) < ∞ then by (6.5) and Lemma 6.4, Area( M )s,t → 0 Area(S \ as s, t → +∞, and that proves part (b) of Theorem 6.1.

There exist regions that do not satisfy the hypotheses of Theorem 6.1, but have angular derivative at +∞. See Exercise 8.

Notes Theorem 1.1 was conjectured by Denjoy in [1907] and proved by Ahlfors in [1930]. Beurling [1933] and Carleman [1933] give different proofs. Ahlfors’ paper, which predated the theory of extremal length, uses the length–area method. He obtained a version of (IV.6.4), with a slightly different (r ) and a constant larger than π8 , but his version was good enough for (1.1). See also Warschawski [1942] for another variant. We have followed Ahlfors’ original argument, but starting from (IV.6.4). Carleman’s proof is in Appendix G, and Beurling’s proof is in Exercise 1. Lord Rayleigh [1871] gave the three-dimensional version of Beurling’s upper estimate (2.2) for the resistance in a conductor using the parallel rule, assuming the conductor is a solid of revolution about an axis (m (x) ≡ 0 in the two dimensional version) with smooth boundary. Maxwell [1891] gave an account of Rayleigh’s method with a slight improvement of the upper bound. Reduced extremal distance is close to a method used by Beurling in his thesis and explained in Exercise III.23. Theorem 3.2 is from Ahlfors and Beurling [1952], but Corollary 3.6 is a recent result by Balogh and Bonk [1999]. An extension of Theorem 3.2 to ﬁnitely connected domains is given in Theorem H.9 of Appendix H. Jenkins [1970] has a slightly weaker but easier analogue of Theorem 4.1. See Exercise 4. Bertilsson [1999] proved (4.11) directly and derived Teichmüller’s Modulsatz as a corollary. He also showed that the estimate ε 2 /(log 1/ε) in (4.11) is sharp. See Exercise 10. The proof of Ostrowski’s theorem given here is based on Warschawski [1967]. The estimate (5.11) is due to J. Wolff [1935]. The problem of ﬁnding

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geometric conditions equivalent to the existence of a non-zero angular derivative is very old, dating back at least to Ahlfors [1930]. Warschawski [1967], Rodin and Warschawski [1977], Baernstein [1988], and their references provide a history of the problem. The proof of Theorem 6.1 is from Marshall [1995]. Theorem 6.1(a) was proved by Burdzy [1986], using Brownian excursions, in the half-plane form given in Exercise 9, which is the context of the original angular derivative problem. Half of Theorem 6.1(a), in the “strip” form given in the text, was proved by Rodin-Warschawski [1986] and the other half was proved by Sastry [1995]. For other proofs of the half-plane version of Theorem 6.1(a), see Carroll [1988] and Gardiner [1991]. Theorem 6.1(b) was proved by Marshall [1995] and answers a question of Burdzy [1986].

Exercises and Further Results 1. In [1933] Beurling derived Theorem 1.1 from his own inequality, proved in Exercise III.22 above. With the notation of the proof of Theorem 1.1, take D j = G j ∩ {|z| > 1}, assume | f | ≤ 1 on ∂ D j , but take z j ∈ D j such that log | f (z j )| ≥ e. For R > Max|z j | set M j (R) =

sup

D j ∩{|z|=R}

Beurling shows there is R0 such that if log M j (R) =

| f (z)|.

R σ j (R) R0

then for all R > R0 , n j=1

1 < 2. σ j (R)

(E.1)

and (E.1) clearly implies (1.1). To prove (E.1), write ω j (R) = ω(z j , |z| = R, D j ∩ {|z| < R}). Then e ≤ ω j (R) log M j (R). Now let C j be the family of curves in D j connecting z j to ∂ D j ∩ {|z| = R} and deﬁne L j = sup inf (ψ(γ )) : ψ is conformal on D j and Areaψ(D j ) = π . γ ∈C j

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Beurling’s inequality from Exercise III.24(b) gives 2

ω j ≤ e1−L j , so that 2

log M j (R) ≥ e L j .

(E.2)

Under the logarithm D j is mapped to a region D ∗j with area A j and A j ≤ 2π log R. (E.3) Using ψ(z) =

π 1 2 log z Aj

gives L 2j ≥

R π π R 2 ≥ log R log log Aj |z j | Aj R0

(E.4)

if R0 ≥ Max|z j |2 , and together (E.2), (E.3), and (E.4) imply (E.1). 2. This exercise outlines a proof of Theorem 3.2 that does not use Green’s functions. (a) Let μ be the equilibrium distribution for E ⊂ ∂D and let Uμ be its logarithmic potential. Set V (z) = log

1 2

Uμ (z) + Uμ (1/z) and prove

1 1 1 1 ≤ V (z) + log ≤ log . 1 + |z| 2 |z| 1 − |z|

(b) Use Green’s theorem to prove 1 1 2 |∇V | d xd y = π γ E + log , Dε (V ) ≡ 2 ε ε where ε = {z ∈ D : V (z) > (log ε)/2}. (c) Set uε =

V + 21 log 1ε γ E + 21 log 1ε

.

Then ρ = |∇u ε | as the extremal metric for d({z : V = (log ε)/2}, E). (d) Deduce Theorem 3.2 from (c) and (a). 3. Compare the estimates of harmonic measure given by Theorem IV.5.2, Corollary 3.3, and Exercise III.24(b) when E is an arc on the boundary of a Jordan domain.

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4. The following variations on Theorem 4.1 and Corollary 4.3 are from Jenkins [1970]. (a) Let be the annulus bounded by 1 = {|z| = 1} and 2 = {|z| = R} where R > 1, and let σ be a curve joining 1 to 2 such that 0 = inf arg z < sup arg z = ε. σ

σ

Prove d\σ (1 , 2 ) ≥

log R 1 ε3 + + O(ε5 ). 2π 108 π 2 log R

Hint: If a curve γ ⊂ \ σ joins 1 to 2 and enters { 3ε < arg z < ds ε2 ≥ (log R)2 + . 9 γ |z|

2ε 3 }, then

(b) Let R be the rectangle {0 < x < L; 0 < y < π } with left vertical side E 1 and right vertical side E 2 . Suppose that T is a curve connecting the horizontal sides of R such that a = inf Rez < sup Rez = a + ε. T

T

Then T divides R into two components 1 and 2 such that E j ⊂ ∂ j . Prove d1 (E 1 , T ) + d2 (T, E 2 ) ≤

L ε3 + O(ε5 ). − π 27π 3

Hint: Bound d j (E j , T )−1 from below using the metric ⎧ 1 ⎨ √ 2 2 , if a + ε/3 < x < a + 2ε/3; π +ε /9 ρ(z) = ⎩1 otherwise. π, (c) Use (b) and the symmetry rule to prove Theorem 4.1 with the weaker 1 estimate δ 3 in (4.3). 5. Suppose is a simply connected domain such that {i y : y > 0} ⊂ and 0 ∈ ∂. For r > 0 let Er be the arc of ∩ {|z| = r } such that ir ∈ Er . Assume has inner tangent at 0 with vertical inner normal and satisﬁes the Ostrowski condition dist(x, ∂) lim = 0. x Rx→0

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197

(a) Prove that 1 r log = 0, π s where the limit is restricted to 1 < r/s < C and r, s → 0. Hint: Use estimate (5.11) together with a normal families argument on {z : 1 < |z| < C} ∩ H} as given in the comments after Deﬁnition 5.3. (b) Prove lim d (Er , E s ) −

lim d (Er , E s ) −

s
1 r log = 0 π s

if and only if lim

n>m→+∞

d (E 2−m , E 2−n ) −

(c) Use part (a) to prove (5.21). 6. (a) Show that if ∞ x0

and

θ (x)2 dx < ∞ θ (x)

t lim

s,t→+∞ s

1 log 2n−m = 0. π

1 1 − θ (x) π

dx = 0

then lim θ (x) = π.

x→+∞

⊂ S then (b) Show that if lim x→+∞ θ (x) = π and if

t π 2 s θ1 − π1 d x = 1, lim s,t→∞ Area(Ss,t \ s,t ) ⊃ S using Area(s,t \ Ss,t ). and prove a similar limit holds if 7. Let = {(x, y) : y > min (|x|/ log(1/|x|), 1)}. Then = ∂ has a tangent at 0. (a) If ϕ+ is a conformal map of H onto with ϕ(0) = 0, then ϕ+ has inﬁnite angular derivative at 0. (b) If ϕ− is a conformal map of H onto C \ with ϕ(0) = 0, then ϕ− has zero angular derivative at 0.

(c) Prove that if is replaced by 1 = {(x, y) : y > min |x|/(log |x|)2 , 1 }, then the conformal maps ϕ+ and ϕ− have non-zero angular derivatives at 1. = {(x, y) : |y − m(x)| < 1 }. 8. Set 2

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∞ ∞ (a) If |m (x)| ≤ 1, 0 |m (x)|2 d x < ∞, and 0 |m(x)|d x = ∞, then by has an angular derivative at +∞, but Theorem 6.1 (2.2) and Theorem 5.7 does not apply. ∞ ∞ (b) If |m (x)| ≤ 1, 0 |m(x)|d x < ∞, and 0 |m (x)|2 d x = ∞, then by has an angular derivative at +∞, but this fact can not be Theorem 6.1 deduced from (2.2). 9. Let be a simply connected domain containing the positive imaginary axis and suppose 0 ∈ ∂. Let h M denote the smallest M-Lipschitz function whose graph is contained in . (a) (Burdzy [1986]) Suppose 1 h (x) χ h M >0 M d x < ∞. (E.5) x2 −1 Then has a positive angular derivative at 0 if and only if 1 h (x) χ h M <0 M d x > −∞. x2 −1

(E.6)

For example if H ⊂ then has a positive angular derivative at 0 if and only if (E.6) holds. (b) (Marshall [1995]) There is an M0 < ∞ so that if M > M0 and if h M satisﬁes (E.6) then has a positive angular derivative at 0 if and only if h M satisﬁes (E.5). For example, if ⊂ H, then has a positive angular derivative at 0 if and only if (E.5) holds. show that Theorem 6.1 can be derived (c) If τ (z) = ie−z is one-to-one on from parts (a) and (b), but with 8π replaced by a larger constant. See Marshall [1995]. 10. Let = {(x, y) : 0 < x < 1, h(x) < y < h(x) + 1}, where h(x) = ε log(1 + x/ε)/ log(1 + 1/ε) and set E = {Rez = 0} ∩ ∂ and F = {Rez = 1} ∩ ∂. Let ϕ be a conformal map of onto a rectangle R = {(x, y) : 0 < x < 1, 0 < y < H }, with Reϕ = 0 on E and Reϕ = 1 on F. Then d (E, F) =

1 1 = . 2 H |∇Reϕ| d A

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Show that

|∇Reϕ|

2d A

199

≥ 1 − C logε1/ε and hence 2

d (E, F) ≤ 1 + C

ε2 log 1ε

.

Hint: Write Reϕ = x + u(x, y). By Green’s theorem, it is enough to show 1 2 ε 2 udy ≤ C ! |∇u| d xd y . ∂ log 1ε Since |h | is small, it is enough to prove this for v(x, y) ≡ u(x, h(x) + y) deﬁned on the unit square S. Relate v on ∂ S to |∇v| on S by integrating the gradient along lines with slope −1. See Bertilsson [1999].

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VI Simply Connected Domains, Part One

The results of this chapter are largely elementary and independent of the three preceding chapters. First we prove the classical [1916] F. and M. Riesz theorem. Next we prove three “almost everywhere” theorems, due to Privalov, Plessner, and McMillan. Their proofs all rely on elementary measure theory and a cone construction (see Figure VI.2) which is a geometric variation on the proof of Lusin’s theorem. We then give Makarov’s elegant proof that on every simply connected domain ω ⊥ α , α > 1, and the extension by Pommerenke that almost everywhere identiﬁes a non-zero angular derivative at ζ with the existence of a cone at ϕ(ζ ).

1. The F. and M. Riesz Theorem By deﬁnition an analytic function f (z) on D is in the Hardy space H p for 0 < p < ∞, if p | f (r eiθ )| p dθ = || f || H p < ∞. sup 0
See Appendix A for a brief introduction to the Hardy spaces H p . Theorem 1.1. Let be a domain such that ∂ = is a Jordan curve and let ϕ be a conformal map from D onto . Then the curve is rectiﬁable if and only if ϕ ∈ H 1 . If ϕ ∈ H 1 , then ||ϕ || H 1 = () = 1 ().

(1.1)

Proof. By Carathéodory’s theorem, ϕ extends to a homeomorphism from D to . First assume ϕ ∈ H 1 . Let {0 = θ0 < θ1 < . . . . < θn = 2π } be any partition 200

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201

of [0, 2π]. Then n n iθ j ϕ(r eiθ j ) − ϕ(r eiθ j−1 ) ϕ(e ) − ϕ(eiθ j−1 ) = lim r →1

j=1

= lim

r →1

j=1 n θ j

j=1

θ j−1

ϕ (r eiθ )ir eiθ dθ

(1.2)

≤ ||ϕ || H 1 . But () is the supremum, over all partitions, of the left side of (1.2). Therefore is rectiﬁable and () ≤ ||ϕ || H 1 . Conversely, assume is rectiﬁable. Then given r < 1, choose a partition {θ0 < θ1 < . . . . < θn } of [0, 2π ], so that n ϕ(r eiθ j ) − ϕ(r eiθ j−1 ) ≥ (r ) − ε, j=1

where r = ϕ({|z| = r }). Write ψ(z) =

n ϕ(zeiθ j ) − ϕ(zeiθ j−1 ) . j=1

Then ψ is subharmonic on D and by Carathéodory’s theorem ψ is continuous on D, so that sup ψ(z) = sup ψ(eiθ ) ≤ (). θ

D

Thus

|ϕ (r eiθ )|dθ = (r ) ≤ ψ(r ) + ε ≤ () + ε.

Therefore ϕ ∈ H 1 and the equality (1.1) holds.

Theorem 1.1 is equivalent to a theorem about harmonic measure. Assume that = ∂ is a rectiﬁable Jordan curve and let A ⊂ be a Borel set. Write A = ϕ(E) and z 0 = ϕ(0). Then by Carathéodory’s theorem 1 |E|. 2π When A is an arc, Theorem A.1 from Appendix A and the proof of (1.1) yield |ϕ (r eiθ )|dθ = |ϕ (eiθ )|dθ, (1.3) 1 (A) = 1 (ϕ(E)) = lim ω(z 0 , A, ) =

r →1 E

E

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because ϕ ∈ H 1 , and if (1.3) holds for arcs then (1.3) also holds for all Borel sets A ⊂ ∂. Consequently ω(A) = 0 ⇒ 1 (A) = 0. Conversely, 1 (A) = 0 ⇒ ω(A) = 0, because by Corollary A.2,

{θ : |ϕ (θ )| = 0} = 0.

Thus when is rectiﬁable, harmonic measure for and linear measure on are mutually absolutely continuous, ω 1 ω. This argument proves the following: Theorem 1.2 (F. and M. Riesz). Let be a simply connected plane domain such that = ∂ is a rectiﬁable Jordan curve and let ϕ : D → be conformal. Then ϕ ∈ L 1 (∂D). For any Borel set E ⊂ ∂D, |ϕ |dθ, 1 (ϕ(E)) = E

and for any Borel set A ⊂ ∂, ω(A) = 0 ⇐⇒ 1 (A) = 0. Theorem 1.2 is equivalent to Theorem 1.1 because (1.3) holds only if ϕ

(1.4) ∈ H 1.

In [1936] Lavrentiev gave a quantitative version of (1.4). See Exercise 2 and (5.1) below. See Exercise 4 for an example of a Jordan domain and a set E ⊂ R ∩ ∂ such that 1 (E) > 0 but ω(E) = 0. Theorem 1.1 has another important geometric consequence. Let ϕ be a conformal map from D to a domain such that ∂ = is a rectiﬁable Jordan curve. Then by Fatou’s theorem, Theorem 1.1, and Corollary A.2, ϕ has a non-zero nontangential limit ϕ (ζ ) at almost all ζ ∈ ∂D. But by calculus, ϕ has a non-zero angular derivative at ζ ∈ ∂D whenever ϕ has a non-zero nontangential limit at ζ . Therefore ϕ has a non-zero angular derivative ϕ (ζ ) at 1 almost every ϕ(ζ ) ∈ . In this way Theorem 1.1 is an improvement, almost everywhere, on the pointwise theorems II.4.2 and V.5.7. Corollary 1.3. Let be a domain such that ∂ = is a rectiﬁable Jordan curve and let ϕ be a conformal map from D onto . Then ϕ has a non-zero

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angular derivative at ζ and has a tangent at ϕ(ζ ), both for Lebesgue almost every ζ ∈ ∂D and for 1 almost every ϕ(ζ ) ∈ . In general, ϕ can have a non-zero angular derivative at a point ζ , so that ∂ then has an inner tangent at ϕ(ζ ) even though ∂ does not have a tangent at ϕ(ζ ). Bishop [1987] exhibited a domain bounded by a non-rectiﬁable Jordan curve such that for almost every ζ ∈ ∂D, ϕ is conformal at ζ but has no tangent at ϕ(ζ ). The domain is {− f (x) < y < f (x), 0 < x < 1} where f (x) is continuous, f (0) = f (1), f (x) > 0 on (0, 1), and f has no ﬁnite or inﬁnite derivative at any point. See Exercise 9. If ∂ is not a Jordan curve, then ϕ will not be one-to-one on ∂D. However, the nontangential limit ϕ(ζ ) can still exist, and in that case the existence of an inner tangent at ϕ(ζ ) depends on the point ζ ∈ ∂D and not just on the point ϕ(ζ ) ∈ ∂. But since ϕ is one-to-one on D, there can be at most two points ζ ∈ ϕ −1 (w) at which ϕ is conformal.

2. Privalov’s Theorem and Plessner’s Theorem Let ζ ∈ ∂D, let α > 1 and let 0 < h < 1. The truncated cone αh is αh (ζ ) = {z : |z − ζ | < α(1 − |z|) < αh}.

αh 2 sec−1 (α)

0

ζ h

Figure VI.1 The truncated cone αh . Let u(z) be a harmonic or meromorphic function on D. Deﬁnition 2.1. The function u is nontangentially bounded at ζ if there exist M < ∞, α > 1, and 0 < h < 1 such that |u(z)| ≤ M on αh (ζ ). Notice that the deﬁnition of nontangential boundedness only requires that u(z) be bounded on a single truncated cone, whereas the deﬁnition of nontangential convergence requires that u(z) has a limit through every cone α (ζ ).

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Theorem 2.2 (Privalov). Suppose E ⊂ ∂D and suppose u(z) is nontangentially bounded at each ζ ∈ E. Then u(z) has a nontangential limit at almost every ζ ∈ E. Proof. Let ε > 0. By a little measure theory there exists a compact set K ⊂ E with |E \ K | < ε and there exist constants α, h, and M such that |u| < M on & U= αh (ζ ). ζ ∈K

To prove the theorem it is enough to show u has a nontangential limit almost everywhere on K . Let Ui be one component of U and let ϕi be a conformal map of D onto Ui . Then u ◦ ϕi is a bounded harmonic function on D and by Fatou’s theorem, u ◦ ϕi has a ﬁnite nontangential limit almost everywhere on ∂D. Because α and h are ﬁxed, ∂Ui is a rectiﬁable Jordan curve. See Figure VI.2. For Lebesgue almost every ζ ∈ K ∩ ∂Ui , ϕi is conformal at ϕi −1 (ζ ) and u ◦ ϕi has a ﬁnite nontangential limit at ϕi −1 (ζ ), both by the F. and M. Riesz theorem. But when ϕi is conformal at ϕi −1 (ζ ), u ◦ ϕi has a nontangential limit at ϕi −1 (ζ ) if and only if u has a nontangential limit at ζ . Since K = i K ∩ ∂Ui , we conclude that u has a ﬁnite nontangential limit at almost every ζ ∈ K .

ϕi K Ui

Uj

K

D

K h Figure VI.2 Cone domains Ui . Many proofs in this book will be simple variations on Figure VI.2 and the previous argument. Deﬁne a cone domain to be a domain of the form & & α (ζ ) or U = αh (ζ ), U= ζ ∈K

ζ ∈K

where K ⊂ ∂D is compact, α > 1, and 0 < h < 1. Cone domains will be recurrent in this chapter and again in Chapter X. Theorem 2.3. Let f (z) be a meromorphic function on D and let E ⊂ ∂D have |E| > 0. If for each ζ ∈ E there is α = α(ζ ) > 1 such that lim

α (ζ )z→ζ

f (z) = 0,

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3. Accessible Points then f (z) ≡ 0 in D.

Proof. Let K ⊂ E be compact with |K | > 0 and take the cone domains Ui and the maps ϕi as in the proof of Privalov’s theorem. Then f ◦ ϕi is bounded and analytic on D, and by the F. and M. Riesz theorem, f ◦ ϕi has nontangential limit 0 on a set of positive measure. Then by (1.4), f ◦ ϕi ≡ 0 so that f ≡ 0. Nontangential approach regions are necessary in Theorems 2.2 and 2.3. Bagemihl and Seidel [1954] showed that there is a non-constant analytic function on D having radial limit 0 almost everywhere on ∂D. See Exercise 5. Deﬁnition 2.4. Let f (z) be a meromorphic function on D.

A point ζ ∈ ∂D is a Plessner point for f if for all α > 1 and all 0 < h < 1, f αh (ζ ) is dense in C. Theorem 2.5 (Plessner). Let f (z) be a nonconstant meromorphic function on D. Then there are pairwise disjoint Borel subsets N , G, and P of ∂D such that ∂D = N ∪ G ∪ P and (a) |N | = 0, (b) At each ζ ∈ G, f has a ﬁnite nontangential limit f (ζ ) and f (ζ ) = 0, and (c) Each ζ ∈ P is a Plessner point for f . Proof. Let P be the set of Plessner points for f and let E = ∂D \ P. Then P and E are Borel sets, and the theorem is equivalent to the assertion that f has a ﬁnite non-zero nontangential limit at almost every ζ ∈ E. Let {wn } be a countable dense subset of C and set # E n = ζ : there exist α > 1, 0 < h < 1, $ and ε > 0 such that f − wn | ≥ ε on h (ζ ) . α

Then E = n E n . By Privalov’s theorem 1/( f (z) − wn ) has a nontangential limit almost everywhere on E n and hence f has a nontangential limit at almost every ζ ∈ E. By Theorem 2.3, applied to f and to 1/ f , the limit is ﬁnite and non-zero almost everywhere.

3. Accessible Points Now let be a simply connected domain and let ϕ : D → be a conformal mapping. We do not assume ∂ is a Jordan curve. Nevertheless, ϕ has a nontangential limit ϕ(ζ ) at almost every ζ ∈ ∂D, because the function z ϕ(z) − ϕ(0)

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is bounded and analytic on D. When ϕ has a ﬁnite nontangential limit at ζ ∈ ∂D we deﬁne ϕ(ζ ) to be that nontangential limit. Thus we have deﬁned ϕ(ζ ) at almost every ζ ∈ ∂D. Better yet, by Corollary V.3.6, 1 ϕ (r ζ ) dr < ∞ (3.1) 0

except on a capacity zero subset of ∂D. If (3.1) holds then ϕ has a radial limit at ζ , and it follows from Lindelöf’s theorem, Exercise 3(d) of Chapter II, that ϕ has a nontangential limit at ζ. Therefore the set where ϕ(ζ ) is not deﬁned is smaller than many sets of linear measure zero; it has capacity zero and Hausdorff dimension zero. We say w ∈ ∂ is an accessible point of ∂ if w is the endpoint of an open arc σ ⊂ . Lemma 3.1. If ϕ has a nontangential limit ϕ(ζ ) at ζ ∈ ∂D, then ϕ(ζ ) ∈ ∂ and ϕ(ζ ) is an accessible point of ∂. Conversely, every accessible w ∈ ∂ is a nontangential limit w = ϕ(ζ ). Proof. Suppose ϕ has nontangential limit at ζ ∈ ∂D. Then clearly ϕ(ζ ) ∈ . But if ϕ(ζ ) ∈ then there is z ∈ D with ϕ(ζ ) = ϕ(z), and that means that ϕ −1 is not single valued in some neighborhood of ϕ(z). Therefore ϕ(ζ ) ∈ ∂ and clearly ϕ(ζ ) is an accessible point. For the converse, suppose w ∈ ∂ is the endpoint of an arc σ ⊂ . The preceding argument above shows that limσ z→w |ϕ −1 (z)| = 1. Because ϕ has nontangential limits a.e. and ϕ −1 (σ ) is an arc, Theorem 2.3, applied to ϕ(z) − w, shows that limσ z→w ϕ −1 (z) = ζ ∈ ∂D exists. It then follows from Lindelöf’s theorem, Exercise II.3(d), that ϕ has nontangential limit w at ζ. Write A = {accessible points of ∂}. Then A is the range of the boundary function ϕ(ζ ), deﬁned except on a set of capacity zero, and A is dense in ∂. Theorem 3.2. Let be a simply connected domain. Then the set A of accessible points is an ω-measurable subset of ∂ and for all z ∈ ,

If E ⊂ A, then

ω(z, A, ) = 1.

(3.2)

ω z, E, = ω ϕ −1 (z), ϕ −1 (E), D .

(3.3)

In particular, ω(z, E, ) = 0 ⇐⇒ |ϕ −1 (E)| = 0.

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Proof. By Egoroff’s theorem and Lemma 3.1, there are closed sets K n ⊂ ∂D such that |∂D \ K n | < 1/n and such that ϕ is continuous on the compact set ∪ K n π/2 (eiθ ). Hence ϕ(K n ) ⊂ A is compact and ω(∪n ϕ(K n )) = 1. Thus A is ω-measurable and (3.2) holds. We may assume z = ϕ(0). Let rn ↑ 1, and set n = ϕ({z : |z| < rn }), and let f ∈ C(). Then by the deﬁnition of harmonic measure, by Lemma 3.1, and by dominated convergence, 1 1 f (ϕ(rn eiθ ))dθ = f (ϕ(eiθ ))dθ. (3.4) f (ζ )dω(ζ ) = lim n→∞ 2π 2π ∂ Each side of (3.3) deﬁnes a Borel measure and since Borel measures are determined by their actions on continuous functions, (3.4) also implies (3.3).

4. Cone Points and McMillan’s Theorem Recall from Chapter V that if ϕ has a non-zero angular derivative at ζ , then ϕ is conformal at ζ . Using conjugate functions or Ostrowski’s theorem and Corollary V.5.8 it is easy to make an example where ϕ has no non-zero angular derivative at ζ but ϕ is conformal at ζ. See Exercise 8 and Exercise V.7. On the other hand, Theorem 6.1 below will show that if ϕ is conformal at all ζ ∈ E ⊂ ∂D, then ϕ has a non-zero angular derivative at almost every ζ ∈ E. We permanently write # $ G = ζ ∈ ∂D : ϕ has non-zero angular derivative at ζ and

# B = ζ ∈ ∂D : ϕ has a nontangential limit at ζ, and $ lim inf |ϕ (z)| = 0, for all α > 1 .

(4.1)

α (ζ )z→ζ

Theorem 4.1. Let ϕ be a univalent function on D. Then G ∩ B = ∅ and |G ∪ B| = 2π. Proof. Because ϕ has a nontangential limit almost everywhere, this theorem is nothing but Theorem 2.3 and Plessner’s theorem, applied to the function ϕ . We call a point w ∈ ∂ a cone point of ∂ if w is the vertex of an open isosceles triangle T ⊂ . Every cone point is an accessible boundary point. In

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review, we know that the following implications hold pointwise: ϕ has non-zero angular derivative at ζ ⇓ ϕ is conformal at ζ ⇓ ∂ has an inner tangent at ϕ(ζ ) ⇓ ϕ(ζ ) is a cone point of ∂, but that all three converse implications fail pointwise. However, Theorem 6.1 will show that these four conditions are almost everywhere equivalent. Write K = K () = cone points for . Then it is clear that ϕ(G) ⊂ K . Theorem 4.2 (McMillan). Let be a bounded simply connected domain. Then K is a Borel set, with σ -ﬁnite 1 measure, and when E ⊂ K , ω(E) = 0 ⇐⇒ 1 (E) = 0.

(4.2)

Moreover, at almost every w ∈ K , ∂ has an inner tangent. Proof. Let {L n } be the countable set of lines having rational slope and rational y intercept. For each n, put w ∈ K n if w ∈ ∂ \ L n and w is the vertex of an open isosceles triangle Tn (w) ⊂ such that (a) Tn (w) has base on the line segment L n ∩ {|z| ≤ n}, (b) Tn (w) has vertex angle πn , and (c) Tn (w) has height h n (w) satisfying n1 ≤ h n (w) ≤ n. Then K = n K n , where K n is compact, so that K is an Fσ set. Note that dist(K n , L n ) ≥ n1 . Let n = K n Tn (w). Then n ⊂ , and by (a), (b), and (c), n has ﬁnitely many components n, j . Each component n, j is bounded by a Jordan curve n, j , and except for endpoints on L n , these Jordan curves are pairwise disjoint. We have K ⊂ (n, j \ L n ) and K has σ -ﬁnite 1 measure. If we assume the line L n is horizontal, then n, j is the union of an arc of L n and the graph of a Lipschitz function deﬁned on that arc. In particular n, j is rectiﬁable, and has an inner tangent at 1 almost every point of n, j , and thus at 1 almost every point of K .

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w U

Tn (w) Ln Figure VI.3 n-Lipschitz subregion. Then to prove (4.2) we can assume E ⊂ n, j \ L n . Suppose 1 (E) > 0. Then by the F. and M. Riesz theorem, ω(z, E, n, j ) > 0 for all z ∈ n, j , and therefore ω(z, E, ) > 0. Conversely, suppose E ⊂ n, j \ L n is a compact set such that 1 (E) = 0. We may suppose L n is the real axis. Set & Tn (w). U= E

units in the direction of n, j to a parallel line L n . When Translate L n by w ∈ E, let Tn (w) ⊂ Tn (w) be the isosceles triangle having vertex w, base on π . Set L n , and vertex angle n+1 & Tn (w). V = 1 10n

E 1 units in the direction of n, j to a parallel line L n . Repeat Translate L n by 5n this construction with L n and with triangles Tn (w) now having bases on L n π and vertex angle n+2 . Set & Tn (w). W = E

Then W has ﬁnitely many components. We may assume E has no isolated points so that W is not a triangle. We can also assume W , and hence U and V , is connected by replacing E by a subset if necessary. Then W ⊂ V ⊂ U ⊂ n, j ⊂ and like n, j , ∂U, ∂ V , and ∂ W consist of a segment of L n , L n , L n , respectively, and a Lipschitz graph over that segment. Each component τk of ∂ V \ E is an arc having endpoints in E, and these endpoints are also the endpoints of a component σk of ∂ W \ E and a component γk of ∂U \ E. Each of the arcs

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τk , σk , γk is a polygonal arc having at most three sides and angles bounded below. ∂

γk C∗

\ Uk

τk ⊂ ∂ V \ E σk ⊂ ∂ W \ E

L n

L n L n Figure VI.4 Nested regions in McMillan’s theorem. Let Uk be the unbounded component of C \ (τk ∪ σk ). Only one σk satisﬁes σk ∩ L n = ∅. Then for each E inf ω(z, σk , Uk ) = α > 0, γk

uniformly in k, by a normal families argument for instance. Therefore inf ω(z, ∂ \ E, \ V ) ≥ α > 0

∂U \E

(4.3)

and by the maximum principle inf ω(z, ∂ \ E, ) ≥ α > 0.

∂U \E

But ∂U is rectiﬁable and 1 (E) = 0, so by the F. and M. Riesz theorem, ω(z, E, U ) = 0. Thus ω(z, ∂ \ E, ) ≥ α on U ⊃ ∂ V \ E and β = inf ω(z, ∂ \ E, ) ≥ α. ∂ V \E

Therefore

ω(z, ∂ \ E, ) = ω(z, ∂ \ E, \ V ) +

∂V

ω(ζ, ∂ \ E, )dω\V (z, ζ )

≥ ω(z, ∂ \ E, \ V ) + βω(z, ∂ V, \ V ) = (1 − β)ω(z, ∂ \ E, \ V ) + β, for all z ∈ \ V . But taking z ∈ ∂U \ E then gives β ≥ α ≥ (1 − β)α + β, so that β = 1 and ω(E) = 0.

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211

In [1969] McMillan proved Theorem 4.2 and also the following striking result, known as the McMillan twist point theorem: At almost every ζ ∈ ∂D \ G, arg(ϕ(z) − ϕ(ζ )) is unbounded above and unbounded below on every curve γ ⊂ D having one endpoint ζ. When arg(ϕ(z) − ϕ(ζ )) is so unbounded, ϕ(ζ ) is called a twist point of ∂. In particular, for almost every ζ ∈ ∂D \ G, the radial image {ϕ(r ζ ) : 0 < r < 1} is a rectiﬁable arc in with endpoint ϕ(ζ ), but arg(w − ϕ(ζ )) has no upper or lower bound along this arc. McMillan’s twist point theorem will be proved in Appendix I. Example 4.3 below will show that we can have |∂D \ G| = 2π. When ∂ is a Jordan curve, ϕ is one-to-one and the notions of twist point or inner tangent depend only on ∂ near the point w = ϕ(ζ ) ∈ ∂. But suppose ∂ is not a Jordan curve and let w ∈ ∂. If there exists ζ0 ∈ ϕ −1 (w) such that on some curve γ ⊂ D with endpoint ζ0 , arg(ϕ(z) − ϕ(ζ0 )) is unbounded above and below,

(4.4)

then (4.4) holds for every ζ ∈ ϕ −1 (w) and for every curve γ ∈ D ending at ζ0 . Now assume is a Jordan curve and let 1 and 2 be the two simply connected components of C∗ \ . Fix p j ∈ j , let ϕ j : D → j be a conformal mapping with ϕ j (0) = p j and write ω j = ω( p j , ·, j ). Set Tn() = w ∈ : has a tangent at w . If w ∈ Tn() then by Theorem II.4.2, ϕ j is conformal at ϕ j −1 (w) for j = 1, 2. Consequently, by Theorem 4.2 the three measures ω1 , ω2 , and 1 are in the same measure class on Tn = Tn(), χ Tn ω1 χ Tn 1 χ Tn ω2 χ Tn ω1 . Theorem 6.3 below will show, without using the twist point theorem, that on \ Tn(), ω1 ⊥ ω2 . Write Twi() for the set of twist points of . In [1987], C. J. Bishop also made an example of a Jordan curve for which ω1 (Twi()) = 1, but

ω2 Tn() ∪ Twi() = 0.

Bishop’s construction is outlined in Exercise 10. Example 4.3 (The von Koch snowﬂake). Let 0 ⊂ 1 . . . be the increasing sequence of Jordan domains deﬁned as follows: ∂0 = 0 is the equilateral √ triangle with vertices {0, 1, 21 + i 23 }; n = ∂n consists of 3 · 4n segments Jn, j of length 3−n ; and n+1 is obtained from n by replacing the middle third

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In, j of each Jn, j by the two remaining sides of the unique equilateral triangle Tn, j satisfying In, j = Jn, j ∩ Tn, j and Tn, j ∩ n = ∅. Clearly n+1 ⊃ n .

0

1 2 Figure VI.5 The von Koch snowﬂake.

5

Then = n is a Jordan domain and ∂ = = lim n , where the limit is taken in the Hausdorff metric. The Jordan curve = ∂ is called the von Koch log 4 snowﬂake. The Hausdorff dimension of is log 3 and 0 < log 4 < ∞. See log 3

Appendix D. Let ϕ be the conformal map of D onto . If has an inner tangent at ϕ(ζ ), then for any ε > 0, contains a truncated cone with vertex ϕ(ζ ) and with aperture π − ε. Consequently ϕ(ζ ) ∈ n ( ∩ n ). But E = n ( ∩ n ) 2 is the union of a sequence of linear images of the Cantor set and dimE = log log 3 . Therefore the set of points of with an inner tangent in has linear measure 0 and by Theorem 4.2, harmonic measure√0 with respect to . If is a copy of rotated by 90 degrees and scaled by 1/ 3, then six copies of in a hexagonal pattern ﬁt exactly along the outer boundary of . Thus the conformal maps ϕ respectively have angular derivatives only on a set and ϕ of D onto and of measure 0 by Theorem 4.2. The theorems in Section 6 below imply that the three measures 1 , ω1 = ω , and ω2 = ω are pairwise mutually singular on .

5. Compression and Expansion Let be a simply connected domain, let ω be harmonic measure for some ﬁxed z 0 ∈ and let ϕ : D → be a conformal mapping. If α (∂) > 0 and ω ⊥ α , then the conformal map ϕ compresses a set E ⊂ ∂D having full harmonic measure into a set ϕ(E) having α (ϕ(E)) = 0, and ϕ expands a set E ⊂ ∂ with ω(E) = 0 into a set ϕ(E) with α (ϕ(E)) > 0. On the other hand, if ω α , then ϕ cannot compress a set E with |E| > 0 into a set with α (ϕ(E)) = 0.

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If ∂ is a rectiﬁable Jordan curve then by the F. and M. Riesz theorem, ω 1 ω. In [1936] Lavrentiev made the more precise lower estimate ω(E) ≤

C log 1 (∂) , 1 + | log 1 (E)|

(5.1)

when ∂ is rectiﬁable, E ⊂ ∂, and dist(z 0 , ∂) ≥ 1. See Exercise 2 for the proof. If ∂ is not rectiﬁable, no inequality like (5.1) can hold. Lavrentiev [1936] exhibited a Jordan domain for which ω 1 . McMillan and Piranian [1973] constructed a Jordan domain such that ω ⊥ 1 . Then Kaufman and Wu [1982] built, for any measure function of the form 1 , 2 a Jordan domain for which ω ⊥ h . An even sharper result will be proved in Chapter VIII. These examples show ϕ can compress a set E with |E| = 2π into ϕ(E) where h (ϕ(E)) = 0. On the other hand, ϕ cannot compress harmonic measure very much. By Beurling’s projection theorem (see Corollary III.9.3) ω 1 . In the very in2 ﬂuential paper [1973], Carleson improved Beurling’s estimate to ω β for some unknown β, 21 < β < 1. Then, in the paper that inspired much of this book, Makarov [1985] proved the following theorem. h(t) = t exp(| log t|α ), 0 < α <

Theorem 5.1 (Makarov). Let ω be harmonic measure for a simply connected domain and let 0 < α < 1. Then ω α . However, ϕ cannot expand harmonic measure very much either. In [1972] and [1981] Øksendal proved ω ⊥ 2 for plane domains of any connectivity, and he conjectured that ω ⊥ α for all α > 1 and all domains. See Exercise III.12 for a proof of Øksendal’s result. Makarov, also in [1985], established the Øksendal conjecture for simply connected domains: Theorem 5.2 (Makarov). Let ω be harmonic measure for a simply connected domain and let h(t) be a measure function such that lim

t→0

Then ω is singular to h , ω ⊥ h .

h(t) = 0. t

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In particular, ω ⊥ α if α > 1. Together Theorems 5.1 and 5.2 say that on a simply connected domain, harmonic measure has dimension 1 and ϕ can neither expand nor compress harmonic measure very much. Later Jones and Wolff [1988] proved that for every plane domain, ω ⊥ α for all α > 1. The Jones–Wolff theorem will be proved in Chapter IX. Theorem 5.1 depends on Makarov’s [1985] law of the iterated logarithm. It will be proved in Chapter VIII. Here we give Makarov’s elegant proof of Theorem 5.2. We need a simple lemma. Let E ⊂ ∂D and let S = {z k } be a sequence in D. We say S is nontangentially dense a.e. on E if there is α > 1 such that for almost every ζ ∈ E, ζ ∈ S ∩ α (ζ ). Lemma 5.3. Assume {z k } is nontangentially dense a.e. on E ⊂ ∂D and let ϕ be the conformal map from D onto a simply connected domain . Then set wk = ϕ(z k ), rk = dist(wk , ∂), Bk = B(wk , 2rk ), and

& V = ∂ ∩ Bk ∩ A, where A is the set of accessible boundary points. Then |E \ ϕ −1 (V )| = 0.

(5.2)

Proof. Suppose (5.2) is false and let Wk be the component of Bk ∩ such that wk ∈ Wk .

Wk

wk

Bk

Figure VI.6 Proof of Lemma 5.3. Since ∂ is connected, ω(wk , V, ) ≥ ω(wk , ∂ ∩ Bk , Wk ) ≥ c, by the Beurling projection theorem; see Exercise 10 of Chapter III. Consequently u(z) = ω(ϕ(z), V, ) satisﬁes u(z k ) ≥ c. But u has nontangential limits a.e. and almost every ζ ∈ E is the nontangential limit of a subsequence of

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215

{z k }, so that lim

α (ζ )z→ζ

u(z) ≥ c

almost everywhere on E. On the other hand, by Theorem 3.2 u(z) = ω(z, ϕ −1 (V ), D), so that u(z) has nontangential limit χ ϕ −1 (V ) almost everywhere on ∂D. Consequently χ ϕ −1 (V ) ≥ c

almost everywhere on E, and (5.2) follows. Proof of Theorem 5.2. Fix ε > 0. When z ∈ D and |z| > 21 , deﬁne I (z) = ζ ∈ ∂D : |z − ζ | < 2(1 − |z|) . Then 1 − |z|2 ≤ c|I (z)| and ζ ∈ I (z) if and only if z ∈ 2 (ζ ).

z

I (z)

Figure VI.7 I (z) consists of the base points of all cones 2 containing z. Let ϕ be the conformal mapping from D to , and set α = 2. By Theorem 2.3, lim inf ϕ (z) < ∞ α (ζ )z→ζ

for almost every ζ ∈ ∂D, and the sets E n = ζ ∈ ∂D : lim inf ϕ (z) < n α (ζ )z→ζ

satisfy E n ⊂ E n+1 and E n = 2π. Every ζ ∈ E n is covered by arbitrarily small arcs I (z) such that ϕ (z) < n, (5.3) and 1 − |z|2 < δn ,

(5.4)

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where δn < ε/n is so small that ε h(t) (5.5) < n+2 t n2 whenever t < 4nδn . By the Vitali covering lemma, Exercise I.9, there is a sequence {z n, j } satisfying (5.3) and (5.4) such that & En \ I (z n, j ) = 0 (5.6) j

and

|I (z n, j )| ≤ 2π.

j

Then by (5.3), (5.4), and (5.5),

Cε h 2|ϕ (z n, j )|(1 − |z n, j |2 ) ≤ n . 2 j

With ε ﬁxed, we take {z k } = Bk = B(wk , 2rk ), and

n {z n, j },

(5.7)

wk = ϕ(z k ), rk = dist(wk , ∂),

Vε = ∂ ∩ (∪Bk )). Then by (5.4) and Theorem I.4.3, 2rk ≤ 2|ϕ (z k )|(1 − |z k |2 ) ≤ 4ε, so that (5.7) yields

h(2rk ) ≤ Cε.

k

Consequently, V = ∞ m V m1 satisﬁes h (V ) = 0. But by (5.6), {z k } is nontangentially dense a.e. in ∂D. Therefore ω( · , V 1 , ) = 1 m

by Lemma 5.3 and ω( · , V, ) = 1.

6. Pommerenke’s Extension In [1986a], Pommerenke extended Makarov’s argument to obtain the following improvement of Theorem 5.2. We recall the sets # $ G = ζ ∈ ∂D : ϕ has non − zero angular derivative at ζ ,

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# B = ζ ∈ ∂D : ϕ has a nontangential limit at ζ, but $ lim inf |ϕ (z)| = 0, for all α > 1 , α (ζ )z→ζ

and

# $ K = cone points for .

Then we know G ∩ B = ∅, |G ∪ B| = 2π and ϕ(G) ⊂ K . Theorem 6.1 (Pommerenke). Let be a simply connected domain and let ϕ : D → . Then there is a subset S ⊂ ϕ(B) \ K such that 1 (S) = 0.

(6.1)

ω(S ∪ ϕ(G)) = 1.

(6.2)

and

In particular, the set P = ϕ(G) ∪ S has σ −ﬁnite 1 measure and ω(P) = 1. Consequently, 1 (K \ ϕ(G)) = ω(K \ ϕ(G)) = 0. Proof. We follow the proof of Theorem 5.2. Again by the Vitali covering lemma, there are {z n, j } such that |ϕ (z n, j )| < 2−n−3 ,

(6.3)

& B \ I (z n, j ) = 0, j

and

|I (z n, j )| ≤ 2π.

j

Now take wn, j = ϕ(z n, j ), rn, j = dist(wn, j , ∂), Bn, j = B(wn, j , 2rn, j ), and & Vn = ∂ ∩ ( Bn, j ). j

Then as before, (6.3) yields 2rn, j ≤ 2|ϕ (z n, j )|(1 − |z n, j |2 ) ≤ cπ 2−n , j

j

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and V =

+&

Vn

k n≥k

has 1 (V ) = 0. Set S = V ∩ ϕ(B). Then (6.1) holds for S. By Beurling’s projection theorem, ω(wn, j , Vn , ) ≥ c, and & {z n, j } n≥N

is nontangentially dense on B. Therefore by Lemma 5.3 & B \ ϕ −1 (A ∩ Vn ) = 0 n≥k

for any k, where A is the set of accessible points. Hence B \ ϕ −1 (S) = 0, and (6.2) then follows from Theorem 4.1. Since 1 (S) = 0, if we replace S by S \ K then (6.1) and (6.2) still hold by Theorem 4.2 and ω(K \ ϕ(G)) = 0. By Theorem 4.2, 1 (K \ ϕ(G)) = 0. By McMillian’s twist point theorem, Appendix I, ω almost every w ∈ S is a twist point of ∂. From Theorems 6.1 and 4.2 we have the following corollary. Corollary 6.2. Let be a simply connected domain and let S ⊂ ∂ be the set given in Theorem 6.1. Then ω 1 ⇐⇒ ω(S) = 0 ⇐⇒ |G| = 2π, and ω ⊥ 1 ⇐⇒ |G| = 0. Theorem 6.3. Suppose is a Jordan curve, and let 1 and 2 be the two components of the complement C∗ \ . Let E be a Borel subset of such that ω(z j , E, ) > 0 for z j ∈ j , j = 1, 2, and let ω j | E be the restriction of ω j = ω(z j , ·, j ) to E. Then ω1 | E ⊥ ω2 | E if and only if 1 (Tn() ∩ E) = 0,

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219

6. Pommerenke’s Extension where Tn() is the set of tangent points of .

Browder and Wermer [1963] gave the ﬁrst example of a Jordan domain for which 1 (Tn()) = 0. See also Gamelin and Garnett [1971] and McMillan and Piranian [1973]. Proof. Assume 1 (Tn() ∩ E) > 0. On Tn (), ω j 1 ω j by Theorem 4.2, so that for j = 1, 2, ω j | E (Tn()) > 0 and ω1 | E ⊥ ω2 | E . Now assume ω1 | E ⊥ ω2 | E . The argument is exactly like the proof of Theorem V.1.1. We may assume dist(z j , ) ≥ 1. Let w ∈ . For 0 < t < 1, let Jj (t) be any arc in j ∩ {z : |z − w| = t} such that Jj separates z j from w, and write tθ j (t) = (Jj (t)). Then by Theorem IV.6.2,

Because have

1

θ1

+

1 θ2

1 dt 8 . ω j (B(w, r )) ≤ exp −π π r tθ j (t)

(θ1 + θ2 ) ≥ 4, (by Cauchy–Schwarz) and θ1 + θ2 ≤ 2π, we 1 2 1 + ≥ . θ1 θ2 π

Therefore ω1 (B(w, r ))ω2 (B(w, r )) ≤

1 2dt 64 64 exp − = 2 r 2. 2 π t π r

(6.4)

By assumption, there is a compact set E N ⊂ E such that ω j (E N ) > 0 and ω1 (S) ≤ ω2 (S) ≤ N ω1 (S) N for all Borel S ⊂ E N . Cover S ⊂ E N by balls B(wk , rk ), wk ∈ , and rk < 1. Then by (6.4), π rk ≥ √ ω j (S), 8 N and hence π 1 (S) ≥ √ ω j (S). 4 N

(6.5)

Let ϕ j be a conformal map of D onto j , let G j ⊂ ∂D be the set where ϕ j has non-zero angular derivative and let S = Sj be the set given by Theorem 6.1

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that satisﬁes (6.1) and (6.2) for j . Then ω1 (S ∩ E N ) = 0 by (6.1) and (6.5). Therefore

ω1 ϕ1 (G 1 ) ∩ ϕ2 (G 2 ) ∩ E N > 0 so that by Theorem 4.2,

1 ϕ1 (G 1 ) ∩ ϕ2 (G 2 ) ∩ E N > 0,

which proves the theorem because ϕ1 (G 1 ) ∩ ϕ2 (G 2 ) ⊂ Tn().

The proof of Theorem 6.3 yields a purely geometric result about the tangent points of a Jordan curve . Let w ∈ , take θ j (t) as in that proof, and set ε(w, t) = max{|π − θ j (t)| : j = 1, 2}. Then by Taylor series 1 1 2 2 ε(w, t) 2 + ≥ + , θ1 θ2 π π π and the proof of (6.4) yields 2 64 ω1 (B(w, r )) ω2 (B(w, r )) ≤ 2 exp − 2 r r π π

r

1

ε2 (w, t)

dt . t

(6.6)

But by Theorem 4.2 the left side of (6.6) is bounded below at 1 (or ω j ) almost every point of Tn(). Consequently we have the following corollary. Corollary 6.4. If is a Jordan curve, then at 1 almost every tangent point w of , 1 dt ε2 (w, t) < ∞. (6.7) t 0 Carleson has conjectured that conversely, has a tangent at almost every point where (6.7) holds. Bishop and Jones [1994] has a result slightly weaker than this conjecture. See the remarks following Theorem X.2.5 below. We continue to assume 1 and 2 are the two complementary components of a Jordan curve . Let K j be the cone points for j , let G j be the set where the conformal map ϕ j of D onto j has non-zero angular derivative, and let Sj ⊂ ∂ j \ K j be a set satisfying (6.1) and (6.2) with respect to j . Note that

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221

ϕ j (G j ) ∩ Sj = ∅. Consider the set

E = ϕ1 (G 1 ) ∪ S1 ∪ ϕ2 (G 2 ) ∪ S2 = ϕ1 (G 1 ) ∩ ϕ2 (G 2 )

∪ S1 ∩ S2

∪ ϕ1 (G 1 ) \ ϕ2 (G 2 ) ∪ S2 ∪ ϕ2 (G 2 ) \ ϕ1 (G 1 ) ∪ S1

∪ S1 \ S2 ) ∪ S2 \ S1 ).

By (6.2), ω1 (E) = ω2 (E) = 1. By Theorems 6.3 and 6.1 we have the following table. Subsets of

Measures

ϕ1 (G 1 ) ∩ ϕ2 (G 2 ) ⊂ Tn()

ω1 1 ω2 ω1

S1 ∩ S2

ω1 ⊥ ω2 , while 1 = 0

ϕ j (G j ) \ ϕk (G k ) ∪ Sk Sj \ Sk

ωk = 0, but ω j 1 ω j

K 1 ∪ K 2 \ ϕ1 (G 1 ) ∪ ϕ2 (G 2 )

ωk = 0 and 1 = 0 ω1 = ω2 = 1 = 0

Examples that limit these results can be found in Exercises 8, 9 and 10.

Notes Sections 1 through 4 are classical. Sections 5 is from Makarov [1985]. See Pommerenke [1986a] and Bishop, Carleson, Garnett and Jones [1989] for the results of Section 6.

Exercises and Further Results 1. (a) Let ϕ be the conformal mapping from D to a simple connected domain , and assume ϕ ∈ H 1 . Prove that ∂ is a rectiﬁable curve, although not necessarily a Jordan curve. (b) Let K be a compact connected subset of C such that 1 (K ) < ∞. Let j be the components of C∗ \ K , and let ϕ j be a conformal mapping from D onto j . If j is bounded, then ϕ j ∈ H 1 , while if ∞ ∈ j and ϕ(0) = ∞,

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then z 2 ϕ j (z) ∈ H 1 . Furthermore, 2π ϕ j dθ = 21 (K ). j

0

See Alexander [1989]. 2. Suppose is a bounded Jordan domain with rectiﬁable boundary, and suppose z 0 ∈ is such that dist(z 0 , ∂) ≥ 1. Then for E ⊂ ∂, ω(z 0 , E) ≤

C log 1 (∂) . 1 + log 1 (E)

See Lavrentiev [1936]. Hint: Take z 0 = ϕ(0). Then (∂) = |ϕ (0)| ≥ 1. By subharmonicity, log |ϕ |dθ ≥ 0

|ϕ |ds and

so that by Jensen’s inequality log |ϕ | ≤ 2 log+ |ϕ | 1 1 ≤ 4π log ϕ 1

= 4π log 1 (∂). Write E = ϕ(S). We may assume 2π ω(E) = |S| and 1 (E) = are small. Again by Jensen’s inequality, 1 1 (E) log |ϕ |dθ ≤ log , |S| S |S| so that |S| log

|S| ≤ 1 (E)

∂ D\S

S

|ϕ |dθ

log |ϕ |dθ ≤ 4π log 1 (∂).

3. We recall the Hayman–Wu theorem: There exists a constant C such that if ϕ : D → is univalent and if L is a line, then length ϕ −1 (L) ≤ C. In Section 5 of Chapter I, we showed that C ≤ 4π. Here we give in outline Øyma’s [1993] example that C ≥ π 2 . (a) If I is an interval in R and if 0 < ε < 1, let S I,ε be the longer circular arc in H having the same endpoints as I and meeting

R at angles ε and π − ε. Let I,ε be the unbounded component of C∗ \ S I,ε ∪ S I,ε/2 . Then

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for z ∈ I ω(z, S I,ε , I,ε ) =

(π + ε/2) 1 < + ε. (2π − ε/2) 2

(b) Given δ > 0, there is b > 0 and ε > 0 such that if I is a subarc of ∂D with |I | (1 − δ) |I |. 2 Hint: The level set = {z : ω(z, I ) = 21 + ε} is a circular arc, and if b and ε are small enough, then length(J ) ≥ length() > (1 − δ) π2 |I |. (c) Fix δ > 0, and let ε and b be from part (b). Construct open intervals In ⊂ R of bounded length such that for n = m, In ∩ Im = ∅, S In ,ε/2 ∩ S Im ,ε/2 = ∅, and

& In = 0. R \

Write Sn = S In ,ε , let Vn be the Jordan domain with ∂ Vn = In ∪ Sn , and set & = {y < 0} ∪ (Vn ∪ In ) . Then ∂ is locally rectiﬁable, so that if ϕ : D → is conformal, then by the F. and M. Riesz theorem, ϕ −1 (R \ In ) = 0. Hence ϕ −1 (Sn ) = 2π. Taking ϕ with |Imϕ(0)| large gives |ϕ −1 (Sn )| (1 − δ) |ϕ −1 (Sn )| 2 and thus length ϕ −1 (R) = length ϕ −1 (In ) ≥ (1 − δ)π 2 . 4. There exists a simply connected Jordan domain and a set E ⊂ R ∩ ∂ such that 1 (E) = |E| > 0 but ω(z, E, ) = 0. Let 0 = (0, 1) × (0, 1),

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and E 0 = [0, 1] and form closed sets E n+1 ⊂ E n and domains n+1 ⊂ n thusly: E n consists of 2n intervals In, j with |In, j | = an , 2n+1 an+1 < 2n a n , but lim 2n an > 0. Let In, j \ E n+1 be the middle open subinterval Jn, j ⊂ In, j bn

with |Jn, j | = an − 2an+1 . Let bn ↓ 0 satisfy lim 2n e−π an = 0, and take &

J n, j × [0, bn+1 ] . n+1 = n \ j

Then E = E n has |E| > 0 and = E ⊂ ∂, and for z ∈ ,

n is a Jordan domain such that

ω(z, E, ) ≤ lim ω(z, E n , n ) = 0 n

by Theorem IV.6.1 or Theorem VI.6.1. See Lohwater and Seidel [1948]. 5. (a) Let γ ⊂ ∂D be an arc with non-empty interior and let f (z) be a meromorphic function on D such that lim f (r ζ ) = 0

r →1

for all ζ ∈ γ . Prove f ≡ 0. Hint: Use Baire category and the proof of Privalov’s theorem. (b) If E ⊂ ∂D is an Fσ set of the ﬁrst Baire category in ∂D, and if g(z) is any continuous function on the open disc D, then there is a function f (z) analytic on D such that

lim f (r ζ ) − g(r ζ ) = 0 r →1

for all ζ ∈ E. (c) There is a non-constant analytic function f (z) on D having radial limit 0 almost everywhere on ∂D. See Bagemihl and Seidel [1954]. 6. Let f (z) be an analytic function on D and let E ⊂ ∂D. If Re f is nontangentially bounded on E, then f has a nontangential limit almost everywhere on E. 7. Let be a simply connected domain, let α ⊂ ∂ be a compact set, let σ be an arc in connecting z 0 ∈ to some ζ ∈ ∂ \ α, and let be the set of curves in joining σ to α. Prove ω(z 0 , α, ) ≤ Hint: Use Theorems 3.2 and IV.5.3.

8 −π λ() . e π

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Exercises and Further Results α

α

z0

σ α

α Figure VI.8 Extremal distance estimate of harmonic measure.

8. (a) Let u(z) be real valued and continuous on D, and harmonic on D, and u (r ) = −∞. Then assume supz∈D |u(z)| < 1 but limr →1 z u )(ζ ) ei(u+i dζ ϕ(z) = 0

is a conformal mapping from D to a simply connected Jordan domain, ϕ is conformal at ζ = 1, but ϕ does not have a non-zero angular derivative at ζ = 1. (b) Find a simply connected domain such that for some ζ ∈ ∂D, the conformal map ϕ : D → is not conformal at ζ even though ∂ has an inner tangent at ϕ(ζ ). 9. Suppose the Jordan curve is a ﬁnite union of linear images w = Az + B of graphs k = {y = f k (x)}. Let 1 and 2 be the components of C∗ \ and let ω j be harmonic measure for some point in j . (a) Prove that ω j 1 on . Hint: Use the twist point theorem from Appendix I or use Lemmas I.2 and I.3 directly. (b) Suppose f k has no ﬁnite or inﬁnite derivative at any point. For examples, see Hobson [1926]. Then ω j 1 but ω1 ⊥ ω2 . Consequently if ϕ j is the conformal map from D to j then at almost every ζ ∈ ∂D, ϕ is conformal but has no tangent at ϕ(ζ ). (c) Suppose f k (x) = 2−n cos 2n x. Then f k is nowhere differentiable, but f k ∈ Z ∗ . See Zygmund [1959], page 47. It follows that has tangents ω j almost everywhere. (d) If f k is nowhere differentiable, but f k ∈ Z ∗ , then the graph {y = f k (x)} has a vertical tangent on a set of positive linear measure. See Bishop [1987] for (a)–(d).

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10. From the interval E 0 = [0, 1] remove the middle open interval I1,1 of length |I1,1 | = 1/8, obtaining a set E 1 . Inductively, obtain E n by removing the middle interval In, j of length |In, j | = 2−2n−1 from each of the 2n−1 components of E n−1 . That leaves E = E n with |E| = 3/4. Each deleted interval In, j is the base of a rectangle Rn, j having dimensions 2−2n−1 × 2−n−2 . Let the “tower” Tn, j consist of the three sides of ∂ Rn, j \ In, j . Then & Tn, j σ1 = E ∪ n, j

is a Jordan arc connection 0 to 1. Note that σ1 falls inside the triangle with vertices {0, 1, 1/2 + i/6}. On each side of the unit square [0, 1] × [0, 1] put an isometric copy of σ , with the towers pointing out from [0, 1] × [0, 1], to obtain a Jordan curve 1 . Let 1 denote the bounded domain with ∂1 = 1 .

Tj3 Jj3 I j3 Jj3 Figure VI.9 First generation towers. Partition each tower Tn, j into 2n + 1 segments of length |In, j | = 2−2n−1 , and repeat the tower construction on each segment using the same length ratios as before. That gives a second Jordan curve 2 bounding a domain 2 ⊃ 1 . Continuing, we get a Jordan curve = lim n bounding a domain be the exterior C∗ \ ( ∪ ). Write ω(E) = ω(E, ) = n n . Let ). Prove ω 1 ω but ω ⊥ 1 (Bishop [1987]). and ω = ω(E,

Figure VI.10 Second generation towers.

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11. Let be the von Koch snowﬂake curve. Prove 0 < log 4 () < ∞, log 3

log 4 log 3 .

(See Appendix D for the deﬁnition and consequently that dim() = of α .) 12. Let S = {z k } be a discrete sequence in D. Then the following are equivalent: (i) S is a nontangentially dense a.e. on ∂D. (ii) For a.e. ζ ∈ ∂D there is α > 0 such that ζ ∈ S ∩ α (ζ ). (iii) For every bounded harmonic function u(z) on D, sup |u(z)| = sup |u(z k )|. D

S

(iv) There is M < ∞ such that for every bounded analytic function f (z) on D, sup | f (z)| ≤ M sup | f (z k )|. D

S

(v) For each z ∈ D, there exist positive weights λk such that for every bounded harmonic function u on D, u(z) = λk u(z k ). (vi) For some z 0 ∈ D \ S, there are complex weights μk with |μk | < ∞ such that for every bounded analytic function f on D, f (z) = μk f (z k ). See Brown, Shields, and Zeller [1960], and Hoffman and Rossi [1967]. 13. Let be a simply connected domain and ﬁx a point z 0 ∈ . For w ∈ ∂, set ω(δ) = ω(w, δ) = ω(z 0 , B(w, δ) ∩ ∂, ). Then at ω almost every cone point, the limit lim

δ→0

ω(δ) =L δ

exists and 0 < L < ∞. Let S be the set given by Theorem 6.1. Then ω almost everywhere on S lim sup δ→0

ω(δ) = ∞. δ

Hint: Use the results or proofs in Section 5. Bishop [1994] conjectured that lim inf δ→0

ω(δ) =0 δ

almost everywhere on S. This has been proved recently by S. Choi [2004].

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14. Let be a simply connected domain, and let ϕ : D → be conformal. Fix w0 ∈ . For accessible ζ ∈ ∂D and a = ϕ(ζ ), let γ (a, r ) be an arc of the circle {|w − a| = r } such that γ (a, r ) is a crosscut of separating w0 from {ϕ(r ζ ) : r0 < r < 1} for some r0 < 1. (Thus γ (a, r ) actually depends on ζ.) Write L(a, r ) = (γ (a, r )), and

r

A(a, r ) =

L(a, s)ds. 0

McMillan [1970] proved these results: (a) Almost everywhere with respect to ω, lim sup r →0

A(a, r ) 1 ≥ . πr 2 2

(b) Consequently, ω almost everywhere, lim sup r →0

L(a, r ) 1 ≥ . 2πr 2

(c) There is an example for which, ω almost everywhere, lim inf r →0

A(a, r ) A(a, r ) = 0, and lim sup = 1. 2 πr πr 2 r →0

See McMillan [1970] and [1969] or O’Neill [1999] for the proof of (a), and note that (b) also follows from Theorems 5.2 and IV.6.2. In [1970] had McMillan conjectured that ω almost everywhere, lim inf r →0

A(a, r ) 1 ≤ πr 2 2

and L(a, r ) 1 ≤ . 2πr 2 These conjectures were recently established by O’Neill and Thurman [2000] and [2001]. lim inf r →0

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VII Bloch Functions and Quasicircles

1. Bloch Functions An analytic function g(z) on D is a Bloch function if ||g||B = sup |g (z)|(1 − |z|2 ) < ∞. z∈D

(1.1)

We write B for the space of Bloch functions and we call ||g||B the Bloch norm1 of g. Because d z + z 0 (1.2) g = g (z 0 )(1 − |z 0 |2 ), dz 1 + z 0 z z=0

we have ||g||B = sup |(g ◦ T ) (0)|, T ∈M

(1.3)

where M is the set of conformal self maps of D: T (z) = λ

z+a , 1 + az

with a ∈ D and |λ| = 1. Thus the Bloch space B is invariant under M: ||g ◦ T ||B = ||g||B , T ∈ M.

(1.4)

If g is bounded, then g ∈ B by (1.2). In fact by (II.3.4) and (II.3.5), g ∈ B if only Reg is bounded. The function g(z) =

∞

z2

n

n=1 1 Actually ||g|| is only a semi-norm, but ||g|| = 0 only if g is constant. B B

229

(1.5)

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is a good example of a Bloch function in no H p space. See Theorem 1.3 below. A function g(z) on D is normal if {g ◦ T : T ∈ M} is a normal family, in the extended sense that f ≡ ∞ is a permitted limit. By Marty’s theorem (Ahlfors [1979]) g(z) is normal if and only if |(g ◦ T ) (z)| 2 T ∈M 1 + |g ◦ T (z)| sup

is bounded on each compact subset of D. Since M is transitive, it follows from (1.2) that g is normal if and only if |g (z)|(1 − |z|2 ) < ∞. 1 + |g(z)|2 D

sup Hence (1.1) yields:

Theorem 1.1. Every Bloch function is a normal function. Recall that the hyperbolic distance from z 1 ∈ D to z 2 ∈ D is z2 dz ρ(z 1 , z 2 ) = inf , 1 − |z|2 z1 where the inﬁmum is taken over all arcs in D connecting z 1 to z 2 . Theorem 1.2. Let g(z) be analytic on D. Then g ∈ B if and only if g is Lipschitz continuous as a map from the hyperbolic metric on D to the euclidian metric on C. Furthermore, |g(z) − g(w)| . ρ(z, w) z,w∈D

||g||B = sup Proof. If

|g(z 1 ) − g(z 2 )| ≤ B

z2 z1

1 |dz|, 1 − |z|2

then a differentiation shows ||g||B ≤ B. Conversely, if g ∈ B then we get |g(z 1 ) − g(z 2 )| ≤ ||g||B ρ(z 1 , z 2 ) by integrating along a hyperbolic geodesic. Theorem 1.3. Let g(z) be analytic on D and let G(z) be the primitive z g(w)dw. G(z) = 0

(1.6)

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1. Bloch Functions

Then g ∈ B if and only if G is continuous on D and G(eiθ ) is a Zygmund function, G(eiθ ) ∈ Z ∗ . When g ∈ B, there is a constant C independent of g such that 1 (1.7) (gB + |g(0)|) ≤ G Z ∗ ≤ C(gB + |g(0)|). C The Bloch function g has nontangential limit at eiθ0 ∈ ∂D if and only if its primitive G(eiθ ) is differentiable at θ0 . Proof. Suppose g ∈ B. Then (1.6) gives, for 0 < r < s < 1, s G(seiθ ) − G(r eiθ ) = g(teiθ )eiθ dt ≤ |g(r eiθ )|(s − r ) r 1 + t gB s log dt, + 2 1−t r

(1.8)

and hence G extends continuously to D. Then (1.1) and Theorem II.3.4 give (1.7). The proof of (1.8) also gives the estimate 1 1 |g(seiθ ) − g(r eiθ )|ds ≤ gB ρ(r, s)ds r

r

2 1+r ≤ gB (1 − r ).

= gB log

(1.9)

Now suppose gB ≤ 1 and suppose g has nontangential limit 0 at eiθ . Then (1.9) implies that dG dθ = 0. Indeed, let ε be small and take α > 1 so that (1 − εt)ei(θ+t) ∈ α (eiθ ) for all small t.

ei(θ+t) (1 − εt)ei(θ+t) eiθ α (ζ )

Figure VII.1

(1.10)

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Set r = 1 − εt. Then G(e

12:30

i(θ+t)

) − G(e ) = lim iθ

=

sei(θ+t)

s→1 seiθ r ei(θ+t) r eiθ

g(z)dz g(z)dz − eiθ

1

g(seiθ )ds

(1.11)

r i(θ+t)

g(r e ) + (1 − r )e 1 + ei(θ+t) g(sei(θ+t) ) − g(r ei(θ+t) ) ds. i(θ+t)

r

By (1.10) the ﬁrst three terms on the far right-hand side of (1.11) are each o(t), because g(z) → 0 when α (eiθ ) z → eiθ , and by (1.9) the last term in (1.11) is bounded by 2 = O(εt). gB log 1+r Hence G(ei(θ+t) ) − G(eiθ ) = o(t) and G (eiθ ) = 0. The converse assertion, that G has a nontangential limit wherever is a simple property of the Poisson kernel. See Exercise I.8.

dG dθ

exists,

Theorem 1.3 shows that the Bloch function (1.5) has a nontangential limit at no point, because the Weierstrass function Re G(z) = 2−n cos(2n θ ) is differentiable at no point. See Katznelson [1968], page 106.

2. Bloch Functions and Univalent Functions We ﬁx a pair of functions ϕ(z) and g(z) analytic on D such that

g(z) = log ϕ (z) , ϕ(0) = 0, or equivalently

ϕ(z) =

z

(2.1)

e g(w) dw.

0

Theorem 2.1. If ϕ is univalent on D, then g ∈ B and ||g||B ≤ 6. Conversely, there is β < 1 such that if ||g||B < β, then ϕ is univalent and ϕ(D) is bounded by a Jordan curve.

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2. Bloch Functions and Univalent Functions Proof. Fix z ∈ D and write ψ(w) =

ϕ

w+z 1+zw

− ϕ(z)

ϕ (z)(1 − |z|2 )

.

Then ψ is a normalized univalent function, ψ(0) = 0 and ψ (0) = 1, and by Theorem I.4.1, |ψ (0)| ≤ 4. Then because ψ (0) =

ϕ (z) (1 − |z|2 ) − 2z = g (z)(1 − |z|2 ) − 2z, ϕ (z)

it follows that |g (z)|(1 − |z|2 ) ≤ 6. To prove the converse, let I be an arc on ∂D with |I | ≤ π . Let C I be the circle that intersects ∂D orthogonally at the endpoints of I , and let z I be the point on C I closest to 0. Lemma 2.2. If gB ≤ β < 2, then at each ζ ∈ ∂D, ϕ(ζ ) = limDz→ζ ϕ(z) exists and ϕ(ζ ) − ϕ(z I ) − (ζ − z I )ϕ (z I ) ≤ 2β ϕ (z I ) |I | (2.2) 2−β for all ζ ∈ I. Furthermore, ϕ extends continuously to D and ϕ ∈ C 1−β/2 (D). Proof. Let J = ∂D ∩ {Rew ≥ 0} and let T ∈ M satisfy T (J ) = I. Then T (w) =

z I w + |z I | . |z I | 1 + |z I |w

Note that when Rew ≥ 0, |T (w)| ≤ 1 − |z I |2 ≤ |I |

(2.3)

T

I zI

0

J

Figure VII.2

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Suppose for a moment that the limit ϕ(ζ ) exists. Then ζ

g(z)−g(z ) I − 1 dz e ϕ(ζ ) − ϕ(z I ) − ϕ (z I )(ζ − z I )= ϕ (z I ) = ϕ (z I )

zI T −1 (ζ )

e g(T (w))−g(T (0)) −1 T (w)dw.

0

e|z|

− 1| ≤ − 1 is trivial by Taylor series. With (2.3) and the hypothesis gB ≤ β, it yields T −1 (ζ ) β 1 + |w| 2 ϕ(ζ ) − ϕ(z I ) − ϕ (z I )(ζ −z I ) ≤ ϕ (z I ) |I | −1 d|w| 1 − |w| 0 2β ϕ (z I ) |I |, ≤ 2−β The inequality |e z

which is (2.2). Since (1.6) and (2.1) give |g(0)| |ϕ (z I )| ≤ e

2 1 − |z I |

β/2

,

Theorem II.3.2 shows that ϕ extends continuously to D, ϕ ∈ C 1−β/2 (D), and |ϕ(ζ1 ) − ϕ(ζ2 )| ≤

C |z 1 − z 2 |1−β/2 . 2−β

To conclude the proof of Theorem 2.1, let ζ1 , ζ2 ∈ ∂D and let I be the shorter arc of ∂ D having endpoints ζ1 and ζ2 . Then by (2.2) and triangle inequalities, |ϕ(ζ1 ) − ϕ(ζ2 )| ≥ |ϕ (z I )||ζ1 − ζ2 | −

4β |ϕ (z I )||I | 2−β

2πβ ≥ 1− |ϕ (z I )||ζ1 − ζ2 |, 2−β

(2.4)

2πβ < 1, ϕ is a one-to-one map from because |I | ≤ π2 |ζ1 − ζ2 |. Thus when 2−β ∂D onto a Jordan curve, and it then follows by the argument principle that ϕ is univalent on D.

Lemma 2.2 proves more than it says. A Jordan curve is a quasicircle if there is a constant A such that |w1 − w| + |w − w2 | ≤ A, (2.5) |w1 − w2 | whenever w ∈ lies on that subarc of having endpoints w1 and w2 and smaller diameter. See Ahlfors [1963] and [1966]. We call (2.5) the Ahlfors condition. Any C 1 curve or Lipschitz curve is a quasicircle and the von Koch

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snowﬂake is a nonrectiﬁable quasicircle. A quasidisc is a domain bounded by a quasicircle. w1

w w1 w2

w2

Quasicircle

w

Not Quasicircle Figure VII.3

Theorem 2.3. There is β0 < 1 such that if gB < β < β0 , then ϕ(∂D) is a quasicircle with constant √ A ≤ A(β) ≤ 2(1 + C1 β). (2.6) Proof. Take w j = ϕ(ζ j ) and let w = ϕ(ζ ) lie on the smaller diameter subarc ϕ(I ) of ∂D with endpoints ζ1 and ζ2 . Replacing ϕ by ϕ ◦ T for some T ∈ M if necessary, we can assume that |I | ≤ π. Then by (2.2), ϕ(ζ j ) − ϕ(ζ ) − (ζ j − ζ )ϕ (z I ) ≤ 4β ϕ (z I ) |I |, 2−β so that 8β |ϕ (z I )||I | |w1 − w| + |w − w2 | ≤ ϕ (z I )|(|ζ1 − ζ | + |ζ − ζ2 | + 2−β

4πβ ≤ 1+ |ϕ (z I )| |ζ1 − ζ | + |ζ − ζ2 | . 2−β Then by (2.4) we obtain (2.5) with the estimate 2 + (4π − 1)β √ 2. A ≤ A(β) = 2 − (2π + 1)β Therefore ϕ(∂D) is a quasicircle and (2.6) holds provided β<

2 . (2π + 1)

The constants β and β0 in Theorems 2.1 and 2.3 are not sharp. Becker [1972] showed that ϕ is univalent if gB ≤ 1, and that ϕ(∂D) is a quasicircle

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if gB < 1. The constant 1 is sharp in both of Becker’s results, but when gB = 1, ϕ(∂D) √ and Pommerenke [1984]. √ is still a Jordan curve. See Becker The constant 2 in (2.6) is needed because A = 2 for a circle. Later we will need the following geometric result, which can be viewed as a converse of Theorem 2.3. For δ > 0 write 1 and Imz > δ}, Dδ = {z : |z| < δ and L δ = {x ∈ R : x + iδ ∈ ∂ Dδ }. When w ∈ choose w0 ∈ ∂ to satisfy |w − w0 | = dist(w, ∂), and write τ (z) = τw (z) = w0 − i(w − w0 )z, Dδ (w) = τ (Dδ ), and L δ (w) = τ (L δ ). Theorem 2.4. Let ϕ be a conformal map from D onto a simply connected domain , and let g(z) = log(ϕ (z)). Then the following conditions are equivalent: (a) There are η > 0 and M < ∞ such that

sup 1 − |z|2 g (z) ≥ η, (2.7) {z:ρ(z,z 0 )<M}

for all z 0 ∈ D. (b) The family of maps ψζ (z) =

ϕ

z+ζ 1−ζ z

− ϕ(ζ ) , ζ ∈ D,

ϕ (ζ ) 1 − |ζ |2

contains no sequence converging, uniformly on compact subsets of D, to a map of the form z ψ (λ) (z) = , |λ| = 1. 1 + λz (c) There is δ > 0 such that for all w ∈ , either Dδ (w) ∩ ∂ = ∅, or there exists w1 ∈ L δ (w) so that B(w1 , δ|w − w0 |) ∩ ∂ = ∅.

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Dδ (w) w w0

w w−

0|

δ|

Dδ (w) w w0 w1

L δ (w)

w w−

0|

δ|

L δ (w)

Figure VII.4 The two possibilities in Condition (c). Proof. Suppose (a) is false. Then there exists {z n } ⊂ D, with z n → λ ∈ ∂D such that lim

sup

n→∞ {z:ρ(z,z )≤M} n

(1 − |z|2 )|g (z)| = 0,

for all M < ∞. Then, uniformly on compact subsets of D, z+z n g − g(z n ) → 0, 1 + zn z and ϕ

z+z n 1+z n z ϕ (z n )

so that ψz n (z) =

ϕ

z+z n 1+z n z

→ 1,

ϕ (z n )(1 + z n z)2

→

1 (1 + λz)2

.

(2.8)

Because ψz n (0) = ψ (λ) (0) = 0, it follows that ψz n (z) → ψ (λ) (z) and that (b) is false. Now suppose (b) fails. Then there is a sequence {z n } → μ ∈ ∂D such that ψz n (z) → ψ (λ) (z), uniformly on compact subsets of D. If μ = λ, then we can reverse the previous argument to conclude that (a) does not hold. If μ = λ then by (2.8) z+z n 1 + μz 2 ϕ 1+z nz . (2.9) → ϕ (z n ) 1 + λz

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Fix |ζ | < 1 and |z| < 1. Set αn =

ζ + zn 1 + zn ζ βn z + ζ 1 + μζ , βn = , β= , and wn = . 1 + zn ζ 1 + zn ζ 1 + μζ 1 + ζ βn z

Then αn → μ and wn + z n z + αn = . 1 + z n wn 1 + αn z Because wn → (βz + ζ )/(1 + ζ βz) ∈ D and (2.9) holds uniformly on compact subsets of D

z+αn

wn +z n ϕ 1+α ϕ 1+z ϕ (z n ) nz n wn = ·

ζ +z n ϕ (αn ) ϕ (z n ) ϕ 1+z nζ

βz+ζ ,2 * ,2 * 1 + μ 1+ζ 1 + λζ βz →

βz+ζ 1 + μζ 1 + λ 1+ζ βz * ,2 1 + μz = , 1 + σz where σ =λ

1 + λζ 1 + μζ · . 1 + μζ 1 + λζ

(2.10)

When |σ | = 1 but σ = μ, we can solve (2.10) for ζ with |ζ | < 1. Take σk → μ and αn,k so that limn→∞ αn,k = μ and limn→∞ ψαn,k = ψ (σk ) . Then a diagonalization argument will give (2.9) with λ = μ. Hence (a) implies (b). The equivalence of (b) and (c) is nothing but the Carathéodory convergence theorem; see Tsuji [1959], page 381. We remark that if (2.7) fails for M = 1 then it fails for all M < ∞ by Theorems 1.6 and 2.1. Theorem 2.4 is from Jones [1989]. Exercise 2 gives a result from Pommerenke [1978] that complements Theorems 2.3 and 2.4. Together Theorems 1.3 and 2.1 set up a one-to-one correspondence u h ) , log ϕ = (u h + i

(2.11)

between univalent functions ϕ and small norm Zygmund functions h. This correspondence was ﬁrst noticed by Duren, Shapiro, and Shields [1966]. We give two applications. Example 2.5. There exists a Jordan domain such that the conformal mapping ϕ : D → has a non-zero angular derivative at no point.

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Proof. Actually we will give two examples. For the ﬁrst example, let n log(ϕ ) = ε z2 . If ε is small, then ϕ maps D to a quasidisc; and by the remark following Theorem 1.3, ϕ has an angular derivative at no ζ ∈ ∂D. The second example is the von Koch snowﬂake, Example VI.4.3. The conformal map from D to the interior domain has a non-zero angular derivative at no point of ∂. Indeed if there is an inner tangent at ζ , then ζ belongs to n for some n, as shown in Example VI.4.3. If ζ is a vertex of n , then either contains a truncated cone with opening 4π/3 or contains no truncated cone with opening greater than π/3 at ζ . If ζ is not a vertex then contains the union of a half disc and an equilateral triangle where the disc, centered at ζ , can be arbitrarily small and the size of the equilateral triangle is comparable to the diameter of the disc. See Figure VII.5. By the easy half of Ostrowski’s Theorem V.5.5, does not have a nonzero angular derivative at ζ . In fact the conformal map of the unit disc onto is not conformal at any point of ∂D. ζ

Figure VII.5 Union of a half disc and triangle. Duren, Shapiro, and Shields [1966] introduced (2.11) in order to construct a Jordan domain that is not a Smirnov domain. If is a Jordan domain with rectiﬁable boundary the conformal map ϕ : D → satisﬁes ϕ ∈ H 1 and iθ iθ log |ϕ (z)| = Pz (e ) log |ϕ (e )|dθ − Pz (eiθ )dμs (θ ), ∂D

∂D

where the singular measure μs is the weak-star limit μs = lim log |ϕ (eiθ )|dθ − log |ϕ (r eiθ )|dθ . r →1

Because ϕ ∈ H 1 , log |ϕ (z)| ≤

Pz (eiθ ) log |ϕ (eiθ )|dθ

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by Theorem A.1 and it follows that μs ≥ 0. The domain is called a Smirnov domain if μs = 0, or equivalently if log |ϕ (z)| = log |ϕ (eiθ )|Pz (eiθ )dθ (2.12) for all (or one) z ∈ D. Smirnov domains are important in approximation theory because is a Smirnov domain if and only if, for 1 ≤ p < ∞, the analytic polynomials a0 + a1 z + a 2 x 2 + . . . are L p (, ds) dense in the space, called H p (), of analytic functions satisfying ( f ◦ ϕ)(ϕ )1/ p ∈ H p (D). See Duren [1970], p. 173. Example 2.6. There exists a Jordan domain with rectiﬁable boundary such that is not a Smirnov domain. Proof. Kahane [1969] exhibited an increasing function h ∈ Z ∗ such that h = 0 almost everywhere. A different construction appears in Piranian [1966]. We may assume h Z ∗ is small. Then the map ϕ deﬁned by (2.11) has image ϕ(D) bounded by a rectiﬁable Jordan curve and Pz (θ )dμ(θ), log |ϕ (z)| = ∂D

where μ is the singular measure having primitive h. Hence ϕ(D) is not a Smirnov domain. To construct Kahane’s function we partition the unit interval into 4n subintervals, In, j = [ j4−n , ( j + 1)4−n ), and deﬁne a sequence μn of probability measures. Let dμ0 = d x. When n ≥ 0, assume dμn is absolutely continuous with constant density dn, j ≥ 0 on each In, j . Write In, j = In+1,4 j ∪ In+1,4 j+1 ∪ In+1,4 j+2 ∪ In+1,4 j+3 . If dn, j > 0, deﬁne dn+1,k =

dn, j − 1, if k = 4 j, 4 j + 3; dn, j + 1, if k = 4 j + 1, 4 j + 2.

If dn, j = 0, deﬁne dn+1,k = 0 for all In+1,k ⊂ In, j . Then the sequence μn converges weak-star, because μn+1 (In, j ) = μn (In ). Take μ = lim μn and h(θ ) = μ([0,

θ )). 2π

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If Sn is the closed support of μn , then Sn+1 ⊂ Sn and Sn = 0 because an unbiased random walk on the integers which begins at 1 almost surely hits 0. See for example Feller [1968]. Therefore h is singular. To see that h ∈ Z ∗ , note that whenever I and J are adjacent intervals of the same length, |μ(I ) − μ(J )| ≤ C|I |.

(2.13)

By construction, (2.13) holds whenever I = In, j and J = In, j+1 , even though In, j and In, j+1 may fall in different In−1,k . In general, let 4−n ≤ |I |/2 < 4−n+1 , and take In, j ⊂ I and In,k ⊂ J . Then by the 4-adic case of (2.13), |μ(In, j ) − μ(In,k )| ≤ 2C4−n , μ(I ) − μ(In, j ) ≤ 4C4−n , and μ(J ) − μ(In,k ) ≤ 4C4−n . That establishes (2.13) for general I and J and hence h ∈ Z ∗ .

3. Quasicircles We ﬁrst describe without proof some of the magical properties of quasiconformal mappings and quasicircles. The proofs can be found in the delightful books of Ahlfors [1966] and Lehto and Virtanen [1973], and some are outlined in the exercises. We end the section with the Jerison and Kenig characterization of quasicircles in terms of harmonic measure. Let and be domains in the extended plane C∗ , let f : → be an orientation preserving homeomorphism, and let K ≥ 1. Then we say f is a K-quasiconformal mapping if (a) f is absolutely continuous on almost every horizontal or vertical line segment in , and (b) the (area almost everywhere deﬁned) derivatives fz =

fx − i f y 2

and

fz =

fx + i f y 2

satisfy | fz | ≤

K −1 | fz | K +1

(3.1)

almost everywhere on . When condition (a) holds we say f is AC L, for absolutely continuous on lines. The smallest constant K for which (3.1) holds is called the dilatation of f on .

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Let f : → be K -quasiconformal, let z ∈ \ {∞, f −1 (∞)}, and suppose B(z, r ) ⊂ . Deﬁne m(z, r ) = inf f (w) − f (z) , |w−z|=r

M(z, r ) =

sup f (w) − f (z) ,

|w−z|=r

and H (z) = lim sup

r →0

M(z, r ) . m(z, r )

f

z

m

r

f (z) M

Figure VII.6 The quasiconformal image of a small circle. When f is AC L it follows from (3.1) that H (z) ≤ K

(3.2)

almost everywhere on . Conversely, f is K -quasiconformal if f is an orientation preserving ACL homeomorphism for which (3.2) holds almost everywhere. The equivalence of (3.1) and (3.2) is easy to prove when f is an orientation preserving diffeomorphism, but the general case requires the differentiation theorem of Gehring and Lehto [1959]. Surprisingly, Heinonen and Koskela [1995] derived (3.1) from a weaker form of (3.2). If f is a K -quasiconformal, then f −1 is also K -quasiconformal, and f is a 1-quasiconformal mapping if and only if f is a conformal mapping. Let f be a K -quasiconformal on a domain , let be a path family in , and write f () = { f (γ ) : γ ∈ }. Then (3.1) implies 1 λ () ≤ λ f () ( f ()) ≤ K λ (). K

(3.3)

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Therefore a K -quasiconformal mapping can only increase or decrease an extremal distance by the factor K , 1 d (E, F) ≤ d f () ( f (E), f (F)) ≤ K d (E, F), K whenever f is K -quasiconformal on a neighborhood of . It follows that a K -quasiconformal mapping can only increase or decrease the module of a ring domain W ⊂ by the same factor K , 1 mod( f (W )) ≤ mod(W ) ≤ K mod( f (W )). (3.4) K Conversely, if f is an orientation preserving homeomorphism deﬁned on an open set U and if (3.4) holds for every ring W ⊂ U , then f is K -quasiconformal on U . Again it is not hard to prove that an orientation preserving diffeomorphism is quasiconformal if and only if it satisﬁes (3.4), but in the general case the argument is much deeper. One consequence of (3.3) is the large-scale version of (3.2): Proposition 3.1. If f : C → C is K -quasiconformal, then for all z ∈ C and all r > 0, M(z, r ) ≤ e8K m(z, r ). In other words, if |z 1 − z| ≤ |z 2 − z|, then | f (z 1 ) − f (z)| ≤ e8K | f (z 2 ) − f (z)|. The proof of Proposition 3.1 is outlined in Exercise 3. If f : C → C is K -quasiconformal and if f (∂D) = ∂D, then (3.4) implies there is a constant M = M(K ) such that | f (ei(θ+t) ) − f (eiθ )| 1 ≤ M, ≤ M | f (eiθ ) − f (ei(θ−t) )|

(3.5)

for all θ and all t < π. The converse is also true. See Beurling and Ahlfors [1956], Ahlfors [1966], or the sketch in Exercise 4. Homeomorphisms of ∂D satisfying (3.5) are called quasisymmetric functions. Theorem 3.2. Let f be an orientation preserving homeomorphism of ∂D to ∂D. Then f is the restriction to ∂D of a quasiconformal homeomorphism of C to C if and only if (3.5) holds. If (3.5) holds with constant M then the extension satisﬁes (3.1) with K ≤ K (M), and if f is K -quasiconformal and f (∂D) = ∂D, then (3.5) holds with M ≤ M(K ). Let f : C → C be K -quasiconformal. It then follows easily from (3.4) that = f (∂D) is a quasicircle with constant A ≤ A(K ). The converse holds also.

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Theorem 3.3. Let be a Jordan curve in the plane. Then satisﬁes the Ahlfors condition (2.5) if and only if = f (∂D), where f : C → C is K -quasiconformal .

(3.6)

If (2.5) holds with constant A, then (3.6) holds with K ≤ K (A), and if (3.6) holds with constant K , then (2.5) holds with A ≤ A(K ). See Exercise 5 for a proof of Theorem 3.3. Corollary 3.4. Let be a quasidisc and let ϕ : D → be a conformal mapping. Then ϕ has an extension to a quasiconformal mapping from C onto C. Proof. Let f be a quasiconformal mapping with = f (D) and f (0) = ϕ(0). Set F = f −1 ◦ ϕ. Then F is a homeomorphism of D and F is quasiconformal on D. Extend F to C \ D by reﬂection F(z) =

1 F( 1z )

.

Then F is a quasiconformal self map of C and f ◦ F is the desired quasicon formal extension of ϕ. A positive measure μ on a Jordan curve is a doubling measure if there is a constant C such that μ(I ) ≤ Cμ(J ) whenever I and J are adjacent subarcs of with diam(I ) ≤ diam(J ). Interchanging I and J , we see that μ is doubling if and only if diam(I ) ≤ C diam(J ) whenever I and J are adjacent arcs with μ(I ) ≤ μ(J ). If f : ∂D → ∂D is a homeomorphism, then f satisﬁes the Beurling–Ahlfors condition (3.5) if and only if μ(E) = | f (E)| is a doubling measure on ∂D. Kahane’s measure in Example 2.6 is a doubling measure on R singular to Lebesgue measure. From Jerison and Kenig [1982a] we have the following theorem. Theorem 3.5. Suppose is a Jordan curve in the plane and suppose 1 and 2 are the two components of the complement C∗ \ . Let z j ∈ j and let ω j (E) = ω(z j , E, j ). Then is a quasicircle if and only if both ω1 and ω2 are doubling measures on . Proof. Assume ω1 and ω2 are doubling measures and assume z 2 = ∞ ∈ 2 . Let ϕ1 : D → 1 and ϕ2 : C∗ \ D → 2 be conformal mappings such that

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ϕ2 (∞) = ∞ and ϕ1 (0) = z 1 , and deﬁne the welding map h : ∂D → ∂D by h = ϕ2−1 ◦ ϕ1 . Let I and J be adjacent subarcs of ∂D such that |I | = |J | ≤ π. Because ω1 is doubling, diam(ϕ1 (J )) ≤ C1 diam(ϕ1 (I )). Then because ω2 is doubling, |ϕ −1 (ϕ1 (I ))| |ϕ2−1 (ϕ1 (J ))| = ω2 (ϕ1 (J )) ≤ C2 C1 ω2 (ϕ1 (I )) = 2 . 2π 2π Consequently |h(J )| ≤ C1 C2 , |h(I )| and after interchanging I and J we see that h satisﬁes (3.5). Now let H : C → C denote the quasiconformal extension of h given by Corollary 3.4 and deﬁne

ϕ1 (z), if |z| ≤ 1, 1 (z) = ϕ2 (H (z)), if |z| > 1. Then 1 is a quasiconformal map such that 1 (∂D) = , and is a quasicircle. Conversely, assume is a quasicircle and let ϕ : D → 1 . Let I1 and I2 be adjacent arcs of with diam(I1 ) = diam(I2 ) ≤ diam()/4. Let I3 be an arc adjacent to I2 with diam(I3 ) = diam(I2 ) and I3 ∩ I1 = ∅, and let w j , w j+1 be the endpoints of I j , j = 1, 2, 3. Then by the Ahlfors condition (2.5), dist(I1 , I3 ) ≤ |w2 − w3 | ≤ Adiam(I2 ) = Adiam(I1 ) = Adiam(I3 ), so that by Exercise 6(b) of Chapter IV and a rescaling, dC (I1 , I3 ) ≤ A . Set Jk = ϕ −1 (Ik ). We may suppose that ω(0, ∪k Jk , D) ≤ 1/2. Then by Corollary 3.4 and (3.4), dC (J1 , J3 ) ≤ K A . Hence diam(J2 ) ≤ C diam(J1 ), for otherwise an annulus of large modulus separates J1 and J3 . Thus ω1 is doubling. The proof for ω2 is of course the same. The proof of doubling in Jerison and Kenig [1982a] is a different argument that applies to higher dimensional generalizations of quasidiscs known as nontangentially accessible domains. See Exercise 9 for a Jordan curve on which ω2 is doubling but ω1 is not doubling. A one-sided version of the deﬁnition of quasicircle yields domains called John domains, which are discussed in Exercise 13.

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4. Chord-Arc Curves and the A∞ Condition A compact set K is called Ahlfors regular if there is a constant A such that 1 (K ∩ B(z, r )) ≤ Ar for all z ∈ K . A Jordan curve is called a chord-arc curve or a Lavrentiev curve if is an Ahlfors regular quasicircle. Thus is a chord-arc curve if and only if is rectiﬁable and there is a constant A such that for all w1 , w2 ∈ , (w1 , w2 ) ≤ A , |w1 − w2 |

(4.1)

where (w1 , w2 ) is the length of the shorter arc of joining w1 to w2 .

Figure VII.7 Chord-arc curve. Note how (4.1) is the rectiﬁable analogue of (2.5). Curves satisfying (4.1) were introduced by Lavrentiev in [1936]. A Jordan curve is called a Lipschitz curve if = {e F(θ)+iθ : 0 ≤ θ ≤ 2π }, where F(θ + 2π ) = F(θ ) and |F(θ + t) − F(θ )| ≤ A|t| for all t. The smallest constant A is called the Lipschitz constant for . It is clear that every Lipschitz curve is a chord-arc curve. The object of this section is to prove a result, parallel to Theorem 3.5, that characterizes chord-arc curves via harmonic measure. Let μ and ν be continuous positive measures on a curve . We say μ and ν are A∞ -equivalent if there exists ε < 1 and η < 1 such that whenever I is a subarc of and E ⊂ I , μ(E) ν(E) ≤ ε ⇒ ≤ η. μ(I ) ν(I )

(4.2)

To see that (4.2) deﬁnes an equivalence relation, simply replace E by I \ E. It is not hard to show that ν μ ν if μ and ν are A∞ -equivalent. If is a chord-arc curve and if μ is A∞ -equivalent to arc length on , then μ is a doubling measure. However, not every doubling measure on the unit circle

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is A∞ -equivalent to arc length and Example 2.6 exhibits a doubling measure singular to Lebesgue measure. A stopping time argument given in Exercise 14 shows that A∞ is equivalent to this stronger looking condition: There exist α > 0 and C > 0 such that for all arcs I ⊂ and all E ⊂ I , ν(E) α ν(E) 1 μ(E) α ≤ . ≤C C −1 ν(I ) μ(I ) ν(I ) When μ = k(θ )dθ and ν = dθ are A∞ -equivalent, we say k is an A∞ -weight and write k(θ ) ∈ A∞ . Now let be a rectiﬁable curve and let μ = k(z)ds be a positive ﬁnite measure on . We say k(z) satisﬁes a reverse Hölder condition if there exist δ > 0 and C > 0 such that whenever I ⊂ is a subarc having length (I ), 1 1 1 1+δ k(z)1+δ ds ≤C k(z)ds. (I ) I (I ) I When p < ∞ we say k(z) is an A p -weight if there is C > 0 such that for all arcs I ⊂ , * , p−1 1 1 1 1 p−1 k(z)ds ds ≤ C. (4.3) (I ) I (I ) I k(z) The A p -condition (4.3) is important because of the famous theorems of Muckenhoupt and of Hunt, Muckenhoupt, and Wheeden telling us that when μ is a measure on the unit circle and 1 < p < ∞, the Hardy–Littlewood maximal operator f → M f or the conjugation operator f → f is bounded on L p (μ) if p and only if μ = kd x where k is an A -weight. See Muckenhoupt [1972], Hunt, Muckenhoupt and Wheeden [1973], or Garnett [1981]. Proposition 4.1. Let be a rectiﬁable curve and let μ = k(z)ds be a ﬁnite positive measure on . Then the following conditions are equivalent: (a) μ and arc length measure ds are A∞ -equivalent, (b) k(z) satisﬁes a reverse Hölder condition, and (c) There is p, 1 < p < ∞, such that k(z) is an A p -weight. There exist δ(ε, η), p(ε, η), and C = C(ε, η) such that when (a) holds with constants ε and η, (b) holds for C and all δ < δ(ε, η) and (c) holds for C and all p ≥ p(ε, η). Some parts of the proof of Proposition 4.1 are easy applications of Hölder’s inequality but at one point a stopping time is needed. See Exercise 14, Coifman and Fefferman [1974], Garnett [1981], or Torchinsky [1986]. Theorem 4.3 will show that every chord-arc curve satisﬁes the conditions of Proposition 4.1. The easier Theorem 4.2 shows that Lipschitz curves satisfy a

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strong form of condition (c). Theorem 4.2. Let be a domain bounded by a Lipschitz curve (as above), let ϕ : D → be a conformal mapping, and let ω(E) = ω(0, E, ). Then there is a p < 2, depending only on the Lipschitz constant of , such that |ϕ | and 1 dω is an A 2− p -weight on . 1/|ϕ | are A p -weights on ∂D and k = ds Proof. Let ϕ : D → with ϕ(0) = 0. Then almost everywhere the magnitude of the angle between the tangent vector to and the radius vector is eiθ ϕ (eiθ ) π ≤ arg(1 + i A) = α < , arg iθ ϕ(e ) 2 and so by Zygmund’s theorem, Theorem II.3.1, , 1 * , * λ zϕ λ ϕ sup ϕ Pa dθ zϕ Pa dθ a∈D * λ , λ1 * zϕ ϕ = sup ϕ ◦ T dt zϕ ◦ T dt < ∞, T ∈M for all λ < π/(2α). If I is an interval in ∂D, take a ∈ D such that a/|a| is the center of I and 1 − |a| = |I |/2. Then Pa ≥ C/|I | on I and since |ϕ(z)/z| is bounded above and below on ∂D, ,1 * * , λ 1 1 1 λ dt < ∞. |ϕ | dt sup |I | I |I | I ϕ I This shows that |1/ϕ | ∈ A p with p = 1 + 1/λ > 1 + 2α/π and a similar argument shows that |ϕ | ∈ A p . Because k = 1/|ϕ ◦ ϕ −1 |, we obtain for J = ϕ(I ) 1 q−1 q−1 1 1 1 kds ds (J ) J (J ) J k ,q−1 * q 1 q−1 dθ |I | I |ϕ | =

* 1 |I |

/* ≤

1 |I |

,q

I

I

|ϕ |dθ 1 dθ |ϕ |

,

and k ∈ Aq for q > π/(π − 2α).

1 |I |

,

|ϕ |

q q−1

dθ

q q−1

0q < ∞,

I

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Theorem 4.3 (Lavrentiev). Let be the bounded domain bounded by a chordarc curve , let z 0 ∈ , and let ω(E) = ω(z 0 , E, ). (a) Then ω and arc length on are A∞ -equivalent, with constants depending only on the chord-arc constant of and dist(z 0 , ∂)/diam(). (b) Moreover, if ϕ : D → is a conformal mapping with ϕ(0) = z 0 , then arc length and harmonic measure for z 0 on the curves ϕ({|z| = r }), 0 < r < 1 are A∞ -equivalent with constants independent of r . Proof. We follow Jerison and Kenig [1982b]. To prove (a), let I be a small arc on ∂D, form the box Q = {z = tζ : 1 − |I | < t < 1; ζ ∈ I }, and let z Q be its center. Given that ϕ has a quasiconformal extension to C, Proposition 3.1 tells us there is r > 0 such that B(ϕ(z Q ), r ) ⊂ ϕ(Q) ⊂ B(ϕ(z Q ), Cr ),

(4.4)

where C depends only on the quasiconformality of the extension of ϕ. Since is a chord-arc curve, (ϕ(I )) ≤ Adiam(ϕ(I )) ≤ Ar. Let J = (1 − |I |)I be the top edge of Q. By the Koebe estimates (I.4.13), (ϕ(J )) = |ϕ (z)|ds ≤ 4dist(ϕ(J ), ) ≤ Ar. J

Now let L = (ζ, ζ0 ), ζ0 = (1 − |I |)ζ, be one of the edges in ∂ Q \ (I ∪ J ) of Q. Let c = e8K , where K is the dilatation of the quasiconformal extension of ϕ from C to C, and deﬁne ζk ∈ L to be the point of smallest modulus such that |ϕ(ζk ) − ϕ(ζ )| = c−k |ϕ(ζ0 ) − ϕ(ζ )|. If 2 N ≥ c, then |ϕ(ζk ) − ϕ(ζ )| ≤ 2 N |ϕ(ζk+1 ) − ϕ(ζ )|, and because ϕ −1 also has a K -quasiconformal extension, Proposition 3.1 and induction yield |ζk − ζ | ≤ (2c) N |ζk+1 − ζ |.

(4.5)

Write L k = [ζk+1 , ζk ]. Then by Proposition 3.1 and the deﬁnition of ζk+1 , dist(ϕ(L k ), ) ≤ c|ϕ(ζk ) − ϕ(ζ )| = c−k+1 |ϕ(ζ0 ) − ϕ(ζ )|,

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so that by the Koebe estimates (ϕ(L k )) ≤ 4c−k+1 |ϕ(ζ0 ) − ϕ(ζ )| log With (4.5) we then obtain (ϕ(L)) ≤ |ϕ(ζ0 ) − ϕ(ζ )|

|ζ − ζ | k . |ζk+1 − ζ |

4c−k+1 N log(2c) ≤ Cr.

k

Thus we have established (∂ϕ(Q)) ≤ C r.

(4.6)

With (4.6) and (4.4) we can apply Lavrentiev’s estimate (VI.5.1) to the domain ϕ(Q) and the point ϕ(z Q ), and the bounds will not depend on Q. (E) Let E ⊂ ϕ(I ). Given ε > 0 there exists δ > 0 such that if (ϕ(I )) < δ then by (VI.5.1) ω(ϕ(z Q ), E, ϕ(Q)) ω(z Q , ϕ −1 (E), Q) = < ε. ω(z Q , I, Q) ω(ϕ(z Q ), ϕ(I ), ϕ(Q))

(4.7)

But if (4.7) holds with suitably small ε, then |ϕ −1 (E)| ω(z 0 , E, ) = < 1/2, ω(z 0 , ϕ(I ), ) |I | and we conclude that ω and ds are A∞ -equivalent on ∂. (b) By Lemma 4.4 below, the function β |ϕ (zeiθ )|dθ + log I (z) = log α

|z|/A |ϕ(zeiβ ) − ϕ(zeiα )|

is subharmonic on D for A > 0. Because is chord-arc, we can choose A so that I ≤ 0 on ∂D. Then by the maximum principle, I (r ) ≤ 0, and therefore ϕ({|z| = r }) is a chord-arc curve with a constant independent of r . Lemma 4.4. If f (z) is analytic and nowhere zero on D and if 0 ≤ α < β < 2π then β | f (zeiθ )|dθ J (z) = log α

is subharmonic on D. Proof. Write f = g 2 . Uniformly on compact sets, the function J (z) is a limit of functions of the form n |g j |2 , K (z) = log j=1

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and a calculation with Cauchy–Schwarz shows that ∂2 K (z) ≥ 0. ∂z∂z

Lemma 4.4 is also true without the assumption that f is nowhere zero, but the calculation is a bit more tedious. Corollary 4.5. Let be a bounded domain bounded by a chord-arc curve and let ϕ : D → be a conformal map. Then (a) is a Smirnov domain, and (b) There is q > 0 such that ϕ1 ∈ H q . The index q in (b) depends only on the chord-arc constant of ∂ and on the dilatation of the quasiconformal extension of ϕ. Proof. We ﬁrst prove (b). By Theorem 4.3, the measures dθ and |ϕ (r eiθ )|dθ are A∞ -equivalent, with constants independent of r . Thus by part (c) of Propo1 , depending only on the constants of ∂, such that sition 4.1, there is q = p−1 q dθ 1 <∞ sup iθ ϕ (r e ) 2π r and (b) holds. Recall that is a Smirnov domain if and only if dθ log |ϕ (eiθ )| . log |ϕ (0)| = 2π ∂D Because ϕ ∈ H 1 , we have

log |ϕ (0)| ≤

∂D

log |ϕ (eiθ )|

dθ . 2π

−q ∈ H 1 by (b), so that we also have But ϕ dθ log |ϕ (eiθ )| . log |ϕ (0)| ≥ 2π ∂D Therefore is a Smirnov domain and (a) holds.

Theorem 4.6. Let be a Jordan curve and let be the bounded component of C \ . Then is a chord-arc curve if and only if the following three conditions all hold: (i) is a quasicircle,

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(ii) Arc length on is A∞ -equivalent to harmonic measure for any z ∈ , and (iii) is a Smirnov domain. Examples showing that Theorem 4.6 is sharp are given in Exercise 16. Proof. We have already seen that (i), (ii), and (iii) hold when is a chord-arc curve. Now assume (i), (ii), and (iii) hold. Let ϕ : D → , let I ⊂ ∂D be an arc such that diam(ϕ(I )) ≤ diam()/2, let Q = {z = tζ : 1 − |I | < t < 1; ζ ∈ I }, and let z Q be its center. Because ∂ Q is a quasicircle and ϕ has a quasiconformal extension, there exists r > 0 and C > 0 such that B(ϕ(z Q ), r ) ⊂ ϕ(Q) ⊂ B(ϕ(z Q ), Cr ) and the endpoints w1 and w2 of ϕ(I ) satisfy |w1 − w2 | ≥

r . C

By (ii) and Proposition 4.1, |ϕ | satisﬁes the A p -condition 1 1 1 1− p 1 p−1 dθ |ϕ |dθ ≤ C |I | I |I | I ϕ for some p > 1. But by Jensen’s inequality, 1− p 1 1 1 1 p−1 ≤ exp log |ϕ |dθ dθ |I | I ϕ |I | I so that

1 |ϕ |dθ ≤ C exp log |ϕ |dθ |I | I I ≤ C exp log |ϕ (eiθ )|Pz Q (θ )dθ .

1 |I |

∂D

But by (iii) and (2.12), log |ϕ (z Q )| = so that 1 |I |

∂D

log |ϕ (eiθ )|Pz Q (θ )dθ,

|ϕ |dθ ≤ C|ϕ (z Q )| ≤ I

2Cr , |I |

by Koebe’s estimate. Therefore (ϕ(I )) = |ϕ |dθ ≤ 2C 2 |w1 − w2 | I

(4.8)

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The left inequality in (4.8) is a “reverse Jensen’s inequality” for A p -weights. Pommerenke [1991] gives a more geometric proof of Theorem 4.6. Theorem 4.7. Let be a quasicircle, let be the bounded component of C \ , and let ϕ : D → be a conformal mapping. Then is a chord-arc curve if and only if arc length and harmonic measure for ϕ(0) are A∞ -equivalent on ϕ({|z| = r }), 0 < r < 1, with constants that are independent of r . Proof. If is a chord-arc curve then by Theorem 4.3 arc length and harmonic measure on ϕ({|z| = r }) are A∞ -equivalent uniformly in r . Conversely, the A∞ -equivalence implies conditions (i), (ii), and (iii) of Theorem 4.6 hold on ϕ({|z| = r }), uniformly in r . Consequently the curves ϕ({|z| = r }) are chord arc uniformly in r < 1, and is chord-arc.

5. BMO Domains Let f ∈ L 1 (∂D). Recall from Appendix F that f has bounded mean oscillation, f ∈ BMO, if 1 sup | f − f I |dθ = || f ||BMO < ∞, (5.1) I |I | I where I is a subarc of ∂D and fI =

1 |I |

f dθ I

is the average of f over I . Condition (5.1) is very strong. By the John–Nirenberg theorem (see Exercise F.2), it implies that f ∈ L p (∂D) for all p < ∞. Condition (5.1) also implies that if the Poisson integral f (z) is analytic in D, then | f (z)|(1 − |z|2 ) ≤ C|| f ||BMO . See Exercise F.3. In other words, || f ||B ≤ C|| f ||BMO if f (z) is analytic and f (eiθ ) ∈ BMO. In this section we describe the simply connected domains for which the Bloch function g = log(ϕ (z)) ∈ BMO. A second characterization will be given in Chapter X. Let k(θ ) ≥ 0 be a function on the unit circle ∂D and write g = log k.

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Lemma 5.1. Suppose k(θ )dθ is A∞ -equivalent to dθ . Then g = log k ∈ BMO. Conversely, if g ∈ BMO then there is a constant b > 0 such that ebg dθ and dθ are A∞ -equivalent. Proof. Suppose k(θ )dθ and dθ are A∞ -equivalent. Let I ⊂ ∂D be a subarc. By Proposition 4.1, k satisﬁes the A p -condition (4.3), 1 1

g I − g p−1 exp(g − g I )dθ exp ≤C (5.2) dθ |I | I |I | I p−1 By Jensen’s inequality 1

g I − g p−1 1 exp(g − g I )dθ ≥ 1 and exp ≥ 1. dθ |I | I |I | I p−1 Thus (5.2) holds if and only if each factor is bounded and we have 1 1 exp (g − g I )dθ ≤ exp(g − g I )dθ ≤ C |I | I |I | I and

1 g −g 1

g − g I p−1 I exp exp ≤ C . dθ ≤ dθ |I | I p − 1 |I | I p−1

Therefore 1 sup I |I |

g − g I dθ < ∞ I

and g ∈ BMO. Conversely if g ∈ BMO then by the John–Nirenberg theorem (Exercise F.2) there is b > 0 such that 1 eb|g−g I | dθ < ∞. sup I |I | I Then ebg ∈ A2 and ebg dθ and dθ are A∞ -equivalent by Proposition 4.1.

Now let ϕ : D → be a conformal mapping and write g = log(ϕ ). We say is a BMO domain if g = log(ϕ ) ∈ BMO. We say is a chord-arc domain if is bounded by a chord-arc curve. If is a chord-arc domain, then by Lemma 5.1 and Theorem 4.3, Reg(eiθ ) = log |ϕ (eiθ )| ∈ BMO, so that g ∈ BMO by Exercise F.3. More generally, Zinsmeister [1984] proved that every domain bounded by an Ahlfors regular Jordan curve is a BMO domain. However, not every Jordan BMO domain has an Ahlfors regular boundary. See Exercise 17.

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Theorem 5.2. Every chord-arc domain is a BMO domain. Conversely, there is β > 0 such that if g ∈ BMO and ||g||BMO < β, then is bounded by a chord-arc curve. Proof. We have just seen that g = log(ϕ ) ∈ BMO when is a chord-arc domain. Conversely, since ||g||B ≤ C||g||BMO , Theorem 2.3 gives us a β0 such that ∂ is a quasicircle if ||g||BMO < β0 . By Lemma 5.1 there is β1 such that if ||g||BMO < β1 , then |ϕ (r eiθ )| ∈ A∞ , uniformly in r < 1. Thus by Theorem 4.7, is a chord-arc domain if ||g||BMO ≤ min(β0 , β1 ). Theorem 5.3. The following are equivalent. (a) is a BMO domain. (b) There exist δ > 0 and C > 0 such that if z 0 ∈ there is a subdomain U ⊂ such that (i) z 0 ∈ U, (ii) ∂U is rectiﬁable and (∂U) ≤ C dist(z 0 , ∂), and (iii) ω(z 0 , ∂ ∩ ∂U, U) ≥ δ. (c) There exist δ > 0 and C > 0 such that if z 0 ∈ there is a subdomain U ⊂ such that (i) z 0 ∈ U and dist(z 0 , ∂) ≤ C dist(z 0 , ∂U), (ii) ∂U is chord-arc with constant at most C and (∂U) ≤ C dist(z 0 , ∂), and (iii) (∂ ∩ ∂U) ≥ δ dist(z 0 , ∂).

z0 U

Figure VII.8 Chord-arc subdomain. A map G : C → C is bilipschitz if there is a constant A such that A−1 |z 1 − z 2 | ≤ |G(z 1 ) − G(z 2 )| ≤ A|z 1 − z 2 | whenever z 1 = z 2 .

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Corollary 5.4. The family of BMO domains is invariant under bilipschitz homeomorphisms of the plane. Proof. Condition (c) is invariant under bilipschitz mappings.

Corollary 5.5. Let be a quasicircle, and let 1 and 2 be the components of C ∗ \ . Then 1 is a BMO domain if and only if 2 is a BMO domain. Proof. Let F : C → C be a quasiconformal map such that 1 = F(D). The quasiconformal reﬂection G : 1 → 2 is deﬁned by * , 1 G(z) = F

, F −1 (z) and Ahlfors [1963] shows that G can be chosen to be bilipschitz.

Proof of Theorem 5.3. We show (a) ⇒ (b) ⇒ (c) ⇒ (a). (a) ⇒ (b): By (a) we may assume ϕ −1 (z 0 ) = 0. Indeed, if T ∈ M has T (0) = ϕ −1 (z 0 ) then || log T ||BMO < Const, so that by the conformal invariance of BMO (see Exercise F.3),

gT (z) = log (ϕ ◦ T ) (z) = log ϕ (T (z)) + log T (z), has ||gT ||BMO ≤ ||g||BMO + Const. Fix α > 1 and recall the nontangential maximal function G(ζ ) = (g − g(0))∗α (ζ ) = sup |g(z) − g(0)|. α (ζ )

Let λ > 0 be large and form E = {ζ : G(ζ ) ≤ λ} and V = E α (ζ ). If α is sufﬁciently large, then ω(z, E, D) ≤ 1/2 on ∂V \ E. It follows that ω(0, E, V) ≥ |E| π − 1 ≥ δ if λ is sufﬁciently large. On V, |ϕ (z)| ≤ eλ |ϕ (0))| ≤ 4eλ dist(z 0 , ∂), and thus (b) holds for U = ϕ(V). (b) ⇒ (c): This is another cone domain construction. See Exercise 19. (c) ⇒ (a): Set ϕ(0) = z 0 . Since the domain U in (c) has chord-arc boundary, ϕ has a non-zero angular derivative at almost all points of ϕ −1 (∂U ∩ ∂) and by Theorem 4.3, ω(z 0 , ∂U ∩ ∂, ) ≥ ω(z 0 , ∂U ∩ ∂, U) > ε = ε(C, δ) > 0. Consequently we can construct a cone domain & α (ζ ) ⊂ ϕ −1 (U) V= E

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such that |g(w) − g(0)| ≤ M on V and ω(0, E, D) ≥ ε, where E ⊂ ∂D is compact. In particular, |g(ζ ) − g(0)| ≤ M on E and |E| ≥ 2π ε. Let I1 be a component of ∂D \ E with center c I and take w1 = (1 − ε|I |)c I and z 1 = ϕ(w1 ). Then w1 has bounded hyperbolic distance to V, so that |g(w1 ) − g(0)| ≤ C M since ||g||B ≤ 6. Repeating this construction with z 1 in place of z 0 , we obtain E 1 ⊂ I1 with ω(w1 , E 1 , D) ≥ ε and |g(ζ ) − g(w0 )| ≤ 2C M on E 1 . Since ω(w1 , I1 ∩ E 1 , D) ≥ ε − ω(w1 , ∂D \ I1 , D) ≥ ε −

2 ε ε≥ , π 3

we obtain |I1 ∩ E 1 | ≥ c2 ε2 |I1 |. We repeat this construction on all complementary intervals of E and of the newly constructed sets E j ∩ I j . After n generations we arrive at a set Fn such that |g(ζ ) − g(0)| ≤ nC M on Fn and such that |∂D \ Fn | ≤ |∂D \ E| (1 − c2 ε2 )n . Thus ω(0, {ζ : |g(ζ ) − g(0)| > λ}, D) ≤ c3 e−c4 λ , where c3 and c4 depend only on the constants in (c). Applying this argument to z+w0 ), we obtain ψ(z) = ϕ( 1+w 0z ω(w0 , {ζ ∈ ∂D : |g(ζ ) − g(w0 )| > λ and |ζ − w0 | ≤ 2(1 − |w0 |)}, D) ≤ c5 e−c4 λ . That means that g ∈ BMO.

Notes Section 1 follows Makarov [1990a]. Anderson, Clunie, and Pommerenke [1974] and Anderson and Pitt [1988] and [1989] have more on Bloch functions. The ﬁrst Jordan domain not a Smirnov domain was constructed by Keldysh and Lavrentiev [1937]. We have only touched the beautiful theory of quasicircles. Gehring [1982] has much more, including seventeen different characterizations of quasicircles. Chord-arc curves originated in Lavrentiev [1936] and resurfaced in Pommerenke [1977], Coifman and Meyer [1979], and Jerison and Kenig [1982a] and [1982b]. It is an open problem to show that {log ϕ : ϕ : D → is conformal and ∂ is chord-arc} is a connected subset of BMO. See Semmes [1986b], [1988a] and [1988b], Zinsmeister [1984], [1985], and [1989], Astala and Zinsmeister [1991], MacManus [1994], and Bishop and Jones [1994] for more about chord-arc curves and BMO domains. Theorem 5.3 is from Bishop and Jones [1994].

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Exercises and Further Results 1. Let ϕ be a univalent function in D and suppose 0 < r < R < 1. (a) Use Theorem 2.1 and Exercise I.23(d) to show there is C independent of r and R such that (1 − r )3 it |ϕ (Rei(θ+t) )|dθ ≤ C |ϕ (r e )|. (1 − R)3 |θ|<1−r (b) Use (a) to show (1 − r )2 |ϕ (Reit )|dt ≤ C |ϕ (r eit )|dt. (1 − R)2 2. Let g ∈ B. We say g ∈ B0 the little Bloch space if lim |g (z)|(1 − |z|2 ) = 0.

|z|→1

A Jordan curve is asymptotically conformal if |w − w1 | + |w − w2 | → 0 as |w1 − w2 | → 0, |w1 − w2 | w∈(w1 ,w2 ) sup

where (w1 , w2 ) is the subarc of having endpoints w1 and w2 and smaller diameter. Let g = log(ϕ ) where ϕ is the conformal map from D to the domain bounded by a Jordan curve . Prove is asymptotically conformal if and only if g ∈ B0 . (Pommerenke [1978]). 3. (a) The spherical distance between z ∈ C ∗ and w ∈ C ∗ is 2ds S(z, w) = inf 2 γ γ 1 + |z| with the inﬁmum taken over all rectiﬁable arcs joining z to w. If |z| = 1 then S(0, z) = S(z, ∞) = π2 . (b) Let |z 1 | = |z 2 | = 1 and let be the family of curves that separate {0, z 1 } 1 from {z 2 , ∞}. Using the metric ρ(z) = π2 1+|z| 2 , prove π . 4 (c) Use (b) to prove Proposition 3.1. Hint: We may assume z = f (z) = 0, |z 1 | = |z 2 | = 1, and | f (z 1 )| < | f (z 2 )|. Let be the set of circles of radius | f (z 1 | < r < | f (z 2 )|. Then λ() ≥

λ() ≤

| f (z 2 )| 1 log , 2π | f (z 1 )|

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while by (b) λ( f −1 ()) ≥

π . 4

This proof is from Gehring [1982]. 4. (a) Let f : C → C be quasiconformal and suppose f (∂D) = ∂D. Use (3.4) and Exercise IV.5 to show that (3.5) holds for f . (b) The converse, that f has a quasiconformal extension whenever (3.5) holds, is easier to prove on the line. Let f : R → R be an increasing homeomorphism such that for all x ∈ R and all h > 0, f (x + h) − f (x) 1 ≤ ≤ M. M f (x) − f (x − h) Extend F to the plane by deﬁning 1 1 ( f (x + t y) + f (x − t y))dt F(x + i y) = 2 0 i 1 + ( f (x + t y) − f (x − t y))dt 2 0 when y = 0 and F(x) = f (x). Prove F : C → C is a homeomorphism. (c) Prove F is ACL and satisﬁes (3.1) almost everywhere. (d) Prove F is bilipschitz, C −1 |z − w| ≤ |F(z) − F(w)| ≤ C|z − w|, where C = C(M). See Beurling and Ahlfors [1956] or Ahlfors [1966]. 5. Let be a Jordan curve that satisﬁes the quasicircle condition (2.5) and let I and J be disjoint subarcs of . Let 1 and 2 be the components of C ∗ \ with ∞ ∈ 2 . (a) Prove C −1 d1 (I, J ) ≤ d2 (I, J ) ≤ Cd1 (I, J ), where the constant C depends only on the constant in (2.5). Elementary proofs can be found in Ahlfors [1963] and Gehring [1982]. (b) Let ϕ1 : D → 1 and ϕ2 : C∗ \ D → 2 be conformal mappings such that ϕ2 (∞) = ∞, and let h : ∂D → ∂D be the welding map h = ϕ2−1 ◦ ϕ1 . Using part (a) and Exercise IV.5 shows that h satisﬁes the Beurling–Ahlfors M-condition (3.5). (c) Let H : C → C be the quasiconformal extension of h guaranteed by (b)

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and Theorem 3.2. Deﬁne

1 (z) =

ϕ1 (z), if |z| ≤ 1; ϕ2 (H (z)), if |z| > 1.

Then 1 is a quasiconformal map such that 1 (∂D) = , and is a quasicircle. (d) Let 1 < α < 2. Construct a quasicircle α such that 0 < α (α ) < ∞. See Gehring and Väisälä [1973]. (e) Exhibit a Jordan curve of positive area. It is known that every quasicircle has area zero. 6. (a) A domain is an M-extremal distance domain (or M-QED domain) if whenever E and F are disjoint compact sets in , d (E, F) ≤ MdC (E, F). Note that the reverse inequality with M = 1 is trivial. Show the unit disc D is a 2-QED domain and show the constant 2 is sharp. It then follows by (3.3) that a K −quasidisc is an M(K )-QED domain. Now suppose is a Jordan curve and deﬁne 1 and 2 as in Exercise 5. Assume that 1 is a QED domain, and let I and J be disjoint arcs on . Then M −1 d1 (I, J ) ≤ d2 (I, J ) ≤ Md1 (I, J ), in which the second inequality is obtained from taking conjugate extremal distances. From this it follows as in Exercise 5 that is a quasicircle. See Gehring and Martio [1985]and Yang [1992] and [1994] for further results. (b) Let ϕ be a conformal mapping from a quasidisc to a simply connected domain W and let U ⊂ W be a quasidisc. Then ϕ −1 (U ) ⊂ is also a quasidisc. This is easy from (a). See Fernández, Heinonen, and Martio [1989]. 7. Let ϕ : D → be conformal and let L be a line. Then there are p > 1 and C > 0, independent of and ϕ, such that |(ϕ −1 ) (w)| p |dw| ≤ C. L∩

This result is due to Fernández, Heinonen, and Martio [1989]. By Exercise III.16 we may assume L = ∂U where U ⊂ is a closed disc. Then by Exercise 6, ϕ −1 (L) is a quasicircle. On the other hand, by (I.4.15) ϕ −1 (L) is an Ahlfors regular curve. Therefore ϕ −1 (L) is a chord-arc curve and |(ϕ −1 ) | is an A∞ -weight, which means that (ϕ −1 ) ∈ L p (∂U ). Moreover, a check of the constants will show p and C are independent of and ϕ.

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8. Let be a simply connected domain and let ϕ : D → be conformal. For z ∈ D and |z| > 1/2, write I (z) = {ζ ∈ ∂D : |ζ − z| < 2(1 − |z|)}. (a) If is a quasidisc, then C(K )(1 − |z|)|ϕ (z)| ≤ diamϕ(I (z)) ≤ C(K )−1 (1 − |z|)|ϕ (z)| and

(E.1)

dist ϕ(z), ϕ(I (z)) ≤ C(K )(1 − |z|)|ϕ (z)|,

where K is the smallest dilatation of a quasiconformal extension of ϕ to a quasiconformal : C∗ → C ∗ with (∞) = ∞. Hint: Use (3.4) and estimate the modulus of C ∗ \ I (z) ∪ B(z, (1 − |z|)/2) . (b) Conversely, if (E.1) holds whenever |z| < 1/2, then is a quasidisc. (c) For z ∈ write d(z) = dist(z, ∂) and for α > 1 and w ∈ ∂ write α (w) = {z ∈ : |z − w| < αd(z)}. Then there are β = β(α, K ) > 1 and γ (α, K ) > 1 such that when ζ ∈ ∂D, ϕ(β) (ζ ) ⊂ α (ϕ(ζ )) ⊂ ϕ(γ (ζ )). Hint: Use (a). (d) If ∂ is a chord-arc curve, then (E.1) can be improved to C(, K )(1 − |z|)|ϕ (z)| ≤ |ϕ (ζ )|ds ≤ C(, K )−1 (1 − |z|)|ϕ (z)|. I (z)

9. Let be a Jordan curve in the plane, let 1 and 2 be the two components of C∗ \ , so that ∞ ∈ 2 , let ϕ1 : D → 1 and ϕ2 : C∗ \ D → 2 be conformal mappings, and let h = ϕ2−1 ◦ ϕ1 be the welding map. (a) Let ω j be harmonic measure for some point in j . Prove h = 0 if and only if ω1 ⊥ ω2 . (b) Prove there exists such that ω1 is doubling but ω2 is not doubling. Hint: See Bishop’s towers, Figure VI.10. 10. A Jordan curve in ⊂ C ∗ such that ∞ ∈ is called a quasicircle if is the quasiconformal image of R, or equivalently, is a Möbius image of a bounded quasicircle. Prove that a graph {y = f (x) : −∞ < x < ∞} is a quasicircle if and only if f is a Zygmund function, f ∈ Z ∗ (Jerison and Kenig [1982a]). 11. Let be a quasicircle in C ∗ and assume ∞ ∈ and write 1 and 2 for the complementary components of . Ahlfors [1963] proved that in this case there exists a quasiconformal G : 1 → 2 which is bilipschitz, C −1 |z − w| ≤ |G(z) − G(w)| ≤ C|z − w|.

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(a) If 1 ⊂ 1 is a rectiﬁable Jordan curve, then 2 = G(1 ) ⊂ 2 is rectiﬁable and 1 ∩ 2 = 1 ∩ . (b) Let ω j be harmonic measure for some point z j ∈ j , and let K j be the set of cone points for j . If ω1 (K 1 ) > 0, then ω1 (K 1 ∩ K 2 ) = ω1 (K 1 ) and χK

1 ∩K 2

ω1 χ K 1 ∩K 2 1 χ K 1 ∩K 2 ω2 .

These results are from Bishop’s thesis [1987]. 12. Suppose the Jordan curve is a ﬁnite union of linear images w = Az + B of graphs k = {y = f k (x)}. Let 1 and 2 be the components of C∗ \ and let ω j be harmonic measure for some point in j . If f k ∈ Z ∗ , then by Exercises 10 and 11 is a quasicircle and ω1 χ Tn 1 ω2 ω1 . See Exercise VI.9 and Bishop [1987]. 13. Let be a simply connected domain such that ∞ ∈ / ∂. Then is called a John domain if there is z 0 ∈ and c > 0 such that for every z 1 ∈ there is an arc σ ∈ joining z 0 to z 1 such that dist(z, ∂) ≥ c|z − z 1 | for all z ∈ σ. Thus in Figure VII.9 a slit disc is a John domain but the teardrop is not a John domain. This exercise will survey some properties of John domains. See John [1961], Martio and Sarvas [1979] and especially Näkki and Väisälä [1991] for much more.

z0

John

Not John Figure VII.9

(a) Recall that a crosscut of is a Jordan arc γ ⊂ having both its endpoints in ∂. Every crosscut γ of divides \ γ into two components 1 and 2 . Prove is a John domain if and only if there is a constant C such that

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263

whenever γ is a crosscut of , min diam( j ) ≤ C diam(γ ). j

Figure VII.10 Both sides John. (b) The boundary of a John domain is locally connected. Consequently, if ϕ is a conformal map from D onto a John domain, then ϕ has a continuous extension to ∂D. However, a John domain need not be a Jordan domain. (c) A Jordan curve is a quasicircle if and only if both of its complementary components are John domains. (d) Let be a bounded simply connected domain and let ϕ : D → be a conformal mapping. Recall I (z) = {ζ ∈ ∂D : |ζ − z| < 2(1 − |z|)} and write Bz = D ∩ B(z, 2(1 − |z|)). The next theorem is from Pommerenke [1982a]. See also Pommerenke [1964] and [1991]. Theorem. The following are equivalent: (1) is a John domain. (2) There is C1 such that for whenever 1/2 < |z| < 1, diamϕ(Bz ) ≤ C1 dist(ϕ(z), ∂). (3) There exist α, 0 < α ≤ 1, and C2 such that if 0 < r < s < 1 and ζ ∈ ∂D, 1 − s α−1 |ϕ (sζ )| . ≤ C2 |ϕ (r ζ )| 1−r (4) There exist α, 0 < α ≤ 1, and C3 such that if I ⊂ J are subarcs of ∂D |I | α diamϕ(I ) . ≤ C3 diamϕ(J ) |J | To prove (1) ⇒ (2), use Beurling’s projection theorem to show that dist(ϕ(z), ϕ(Jj )) ≤ K dist(ϕ(z), ∂),

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where Jj , j = 1, 2, are the two components of 2I (z) \ 23 I (z). Then by Exercise III.16(b) obtain two geodesic arcs γ j with initial point z and terminal points in Jj , j = 1, 2, such that (ϕ(γ j )) ≤ Cdist(ϕ(z), ϕ(Jj )). Write Uz for the domain bounded by γ1 , γ2 and the arc of ∂D containing I (z). When is a John domain (a) shows diam(ϕ(Uz )) ≤ Cdist(ϕ(z), ∂). But because Bz \ Uz has bounded hyperbolic diameter, Koebe’s theorem gives diamϕ(Bz \ Uz ) ≤ Cdist(ϕ(z), ∂) even if is not a John domain. To show (2) ⇒ (3), we suppose ζ = 1 and deﬁne 1 |ϕ (x + i y)|2 d xd y. g(s) = s

|y|≤1−x

By (2) g(s) ≤ Area(ϕ(Bs )) ≤ C12 (1 − s)2 |ϕ (s)|2 . By (I.4.14) there is an absolute constant a > 0 such that g(s) ≥ a(1 − s)2 |ϕ (s)|2 and g (s) ≤ −a(1 − s)|ϕ (s)|2 , since the sets {x + i y : |z| ≤ 1 − x} have bounded hyperbolic diameter. Therefore d (1 − s)−α g(s) ≤ 0, ds with α = a/2C12 and we obtain (3) with this α and C2 = a −1/2 C1 . Note that when (2) holds, ϕ ∈ C α . If (3) holds then by Koebe’s estimate (1) holds with z 0 = ϕ(0) and with σ ϕ([0, ϕ −1 (z)]). Thus (3) ⇒ (1) and a hyperbolic geodesic can be taken for the curve σ . To show (3) ⇒ (4), let I ⊂ J be subarcs of D. We can assume J = I (z) and I = I (w). Then by (2) diamϕ(I ) ≤ C 1 (1 − |w|)|ϕ (w)|.

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Exercises and Further Results

Let z =

|z| |w| w.

Then by (3) (1 − |w|)|ϕ (w)| ≤ C2

1 − |w| α 1 − |z |

|ϕ (z )|.

But |ϕ (z )| ≤ a|ϕ (z)| again via (I.4.14) and for arbitrary simply connected domains 1 − |z| diamϕ(I (z)) ≥ a diamϕ B z, . 2 Therefore (4) holds with the exponent α from (3). Now assume (4) and let 1/2 < |z| < 1. Given ε > 0 there is A > 0 such that by Theorem V.3.5 |ϕ(ζ ) − ϕ(z)| ≤ A dist(ϕ(z), ∂) on 2I (z) \ I j where | I j |/|2I (z)| ≤ ε. Then if ε is sufﬁciently small (4) implies that max j diam(ϕ(I j )) ≤ 21 diamϕ(2I (z)) and hence

1 diamϕ(2I (z)). 2 It then follows from V.3.5 exactly as in the proof of (1) ⇒ (2) that diamϕ(Bz ) ≤ C dist(ϕ(z), ∂). Therefore (2) holds. (e) If is a John domain then harmonic measure is doubling. 14. (a) If two measures μ and ν are A∞ -equivalent, then ν μ ν. (b) If μ and ν are A∞ -equivalent and if μ is a doubling measure, then ν is a doubling measure. (c) There is a doubling measure on the unit interval singular to Lebesgue measure. Hint: Start with a quasicircle for which ω2 ⊥ ω1 . (d) Assume (4.2) holds for μ = d x and ν on [0, 1]. Let N ≥ 1 and suppose N E ⊂ I ⊂ [0, 1] satisfy μ(E) < 2ε . We may suppose I is a dyadic interval. Let {I j } be the maximal dyadic subintervals of I such that diamϕ(2I (z)) ≤ 2 A dist(ϕ(z), ∂) +

μ(E ∩ I j ) ≥ ε. μ(I j )

By Lebesgue’s theorem, μ I j \ E = 0. Let I j∗ denote the dyadic interval containing I j with |I j∗ | = 2|I j |. Since I j is maximal, ν(E ∩ I j∗ ) ≤ ην(I j∗ ) by (4.2). Set E 1 = I j∗ . Then E ⊂ E 1 , ν(E) ≤ ν(E 1 ), and μ(I j )

ε N −1 2 μ(E 1 ) . ≤2 ≤ μ(E) < μ(I ) μ(I ) ε 2

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Continuing by induction we obtain μ(E) ε N ν(E) ⇒ ≤ ≤ ηN , μ(I ) 2 ν(I ) which means 1 μ(E) log η1 /log 2ε ν(E) . ≤ ν(I ) η μ(I ) (e) Adapt the proof outlined in (d) to the case when μ and ν are continuous positive measures. (f) If k is an A∞ -weight, then k satisﬁes the reverse Hölder condition. Fix I and λ large, and assume I kd x = 1. Let {I j,1 } be the maximal intervals in the dyadic decomposition of I such that

and set E 1 =

λ < f I j,1 ≤ 2λ, I j,1 . Then k ≤ λ on I \ E 1 and |E 1 | C 1 ≤ . |I | λ 1−λ

Now let {I j,2 } be the maximal intervals in the dyadic decomposition of I such that

and set E 2 =

λ2 < f I j,1 ≤ 2λ2 , I j,2 ⊂ E 1 . Then 1 |E 2 | 2n−1 C n 1−λ . ≤ |E 1 | λn

Continuing, we obtain the reverse Hölder condition 1 1 1 1+δ 1+δ k dx ≤C kd x |I | I |I | I 1 . whenever 1 + δ < 1−α (g) Conversely, the reverse Hölder condition and Hölder’s inequality yield δ . A∞ for α = 1+δ ∞ (h) By Hölder’s inequality, A p implies A∞ with α = p−1 p . Conversely, A and its symmetry imply 1 1 1 C 1+δ δ dx ≤ |I |, k kd x I I I kd x

from which A p follows with p = 1 + 1δ .

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15. Let be a Jordan domain, let ϕ1 : D → and ϕ2 : C ∗ \ D → C∗ \ be conformal mappings. (a) Prove that the welding map ϕ2 −1 ◦ ϕ1 is in A∞ if and only if is a BMO domain and a quasidisc. (Astala and Zinsmeister [1991]). (b) Prove |ϕ | ∈ A∞ and is a Smirnov domain if and only if there is a constant K such that |ϕ |ds ≤ K dist(ϕ(z), ∂) Iz

|eiθ

z |z| |

− < 1 − |z|}. where Iz = {θ : (c) Prove that is an Ahlfors regular John domain if and only if is a Smirnov domain and |ϕ | is an A∞ -weight. See Pommerenke [1982a] and Zinsmeister [1984]. 16. (a) Example 2.6 is a quasicircle for which |ϕ | is an A∞ -weight, but it is not a Smirnov domain. (b) Construct a Zygmund function f (θ ) such that f ≥ 1, f ∈ L 1 but / BMO. Then = {r eiθ : 0 ≤ r < f (θ )} is a quasidisc and a Smirnov f ∈ domain, but |ϕ | is not in A∞ . (c) Construct a Jordan Smirnov domain such that |ϕ | ∈ A∞ but is not a quasidisc. 17. (a) Let ϕ map D to a domain bounded by an Ahlfors regular Jordan curve. Then for p < 1/5 there is C such that |ϕ | p ds ≤ C(1 − |z|)|ϕ (z)| p . I (z)

Hint: By Hölder, p |ϕ | ds ≤ I (z)

I (z)

p |ϕ (ζ )| ds |ϕ(ζ ) − ϕ(z)|2

2p

I (z)

|ϕ(ζ ) − ϕ(z)| 1− p ds

1− p .

The ﬁrst factor can be estimated directly, using the regularity of ∂D, and the second factor can be handled with Prawitz’s theorem. (b) Under the assumptions in (a), g = log(ϕ ) ∈ BMO. Hint: If p is small, then (ϕ ) p = ψ where ψ is a conformal map of D to a quasidisc. But by (a), |ψ | ∈ A∞ , so that log(ψ ) ∈ BMO. (c) Let ϕ be a conformal map from the half-plane onto the region above the curve = {y = sin(x 2 ), x ∈ R}. Then log(ϕ ) ∈ BMO, but is not Ahlfors regular. See Zinsmeister [1984] and [1985]. 18. The space VMO, for vanishing mean oscillation, consists of all g ∈ BMO

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such that 1 lim sup δ→0 |I |<δ |I |

|g − g I |ds = 0, I

and VMOA = VMO ∩ BMOA. See Sarason [1975] or Garnett [1981]. A Jordan curve is said to be asymptotically smooth if is rectiﬁable and if lim

sup

δ→0 |w1 −w2 |<δ

(w1 , w2 ) = 1, |w1 − w2 |

whenever w1 , w2 ∈ and (w1 , w2 ) is the length of the shorter subarc joining w1 to w2 . Let g = log(ϕ ) where ϕ maps D to a Jordan domain. Prove g ∈ VMOA if and only if ∂ is asymptotically smooth. See Pommerenke [1978]. 19. Let be a Jordan domain such that B(0, 1) ⊂ . Assume (∂) ≤ M and let E ⊂ ∂ satisfy ω(0, E, ) ≥ ε. Prove there is δ = δ(ε, M) such that there is a chord-arc domain U ⊂ with B(0, 1) ⊂ U, ω(0, E ∩ ∂U, U) ≥ δ and (E ∩ ∂U) ≥ δ. Hint: Take a union of ﬁxed angle cones over a subset of E. 20. Let be a Jordan curve bounding domains 1 and 2 and let ϕ j : D → j be conformal. Suppose, as happens in Example 2.6, that log |ϕ1 | is the Poisson integral of a negative singular measure on ∂D, so that 1 is badly not a Smirnov domain. Prove there is c > 0 such that |ϕ2 | ≥ c on ∂D, and that consequently 2 is a Smirnov domain. Hint: Use the estimate (6.4). This result is due to Jones and Smirnov [1999].

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VIII Simply Connected Domains, Part Two

We begin the chapter with the famous theorems from Makarov [1985], [1987], and [1990a] on Hausdorff measures and harmonic measure in simply connected domains. Next we discuss the Brennan conjecture on the L p integrability of derivatives of univalent functions and prove some related theorems from Carleson and Makarov [1994]. Then we give Baernstein’s counterexample to the L p version of the Hayman–Wu theorem.

1. The Law of the Iterated Logarithm for Bloch Functions Theorem 1.1 (Makarov). There is a constant C > 0 such that if g(z) is a Bloch function on D, then for a.e. ζ ∈ ∂D |g(r ζ )|

lim sup ! r →1

log

1 1−r

1 log log log 1−r

≤ CgB .

(1.1)

Inequality (1.1) is called Makarov’s law of the iterated logarithm because it was ﬁrst proved by Makarov in [1985] and because it resembles the classical Khinchin law of the iterated logarithm in Feller [1968] or Chung [1974]. Consider for example the bad Bloch function g(z) =

∞

n

z2 .

(1.2)

n=1

The series (1.2) behaves much like a sum of independent identically distributed random variables, and Salem and Zygmund proved in [1950] that its partial

269

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sums Sn (z) =

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k

z 2 satisfy lim sup √

|Sn (ζ )| =1 n log log n

(1.3)

for almost every ζ ∈ ∂D. But if g(z) is deﬁned by (1.2), then (1.3) is equivalent to |g(r ζ )| lim sup ! = 1, 1 1 r →1 log 1−r log log log 1−r which is an instance of (1.1). Theorem 1.1 has a converse, proved by Jones in [1989]. Theorem 1.2. If ||g||B ≤ 1 and if there exist β > 0 and M < ∞ such that for all z 0 ∈ D,

1 − |z|2 g (z) ≥ β, (1.4) sup {z:ρ(z,z 0 )<M}

then almost everywhere on ∂D, Reg(r ζ )

lim sup ! r →1

log

1 1−r

1 log log log 1−r

≥ c(β, M) > 0.

(1.5)

When ϕ is a conformal mapping from D onto a domain , Theorem VII.2.4 gave a geometric condition on that is necessary and sufﬁcient for the Bloch function g = log(ϕ ) to satisfy condition (1.4). For example, (1.4) holds when ϕ maps D to the domain inside the snowﬂake curve. Proof of Theorem 1.1. This proof is due to Carleson (unpublished) and independently to Pommerenke [1986b]. We give Pommerenke’s version, which yields the best known constant C = 1 in (1.1). A different proof, from Makarov [1990a], is given in Appendix J. We may assume g(0) = 0 and ||g||B = 1. Let p ≥ 0 be an integer and consider the integral means 2π 1 |g(r eiθ )|2 p dθ. I p (r ) = 2π 0 The proof begins with the identity d 4 p 2 r 2π |g(r eiθ )|2 p−2 |g (r eiθ )|2 dθ, (r I p (r )) = dr 2π 0

(1.6)

which is known as Hardy’s identity. Hardy’s identity can be proved by examining the Fourier series of |g| p or by differentiating I p (r ) directly. A corollary

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of Hardy’s identity is the inequality 1 p 1 p ≤ p! log , I p (r ) ≤ p! log 1 − r2 1−r

271

(1.7)

valid for ||g||B = 1 and for integral p ≥ 0. Indeed, (1.7) is trivial when p = 0, and to prove (1.7) when p ≥ 1 we use (1.6) and induction to get 4 p2r d I p−1 (r ) (r I p (r )) ≤ dr (1 − r 2 )2 1 p−1 4 pp!r log ≤ (1 − r 2 )2 1 − r2 d d 1 p ≤ p! r log , dr dr 1 − r2 and since I p (0) = 0, an integration then yields (1.7). Now let us apply the Hardy–Littlewood maximal theorem to the function |g(r eiθ )| p ∈ L 2 . By inequality (1.7), gr∗ (eiθ ) = supρ 1 and set A p (r ) = Then

1

1 1−r

1 log 1−r

A p (s)ds ≥

r

while by (1.8) 1

≤ C p! r

p+1

1 1 log log 1−r

α .

1 C 1 , p log 1 p log log 1 α 1−r 1−r

|gs∗ (eiθ )|2 p dθ ds

A p (s)

r

1

1

1

log log

1 1−s

α

1 log

1 1−s

ds ≤ Cα p!. 1−s

It follows, via Fubini’s theorem and Chebychev’s inequality, that the set ( ) 1 ∗ iθ 2 p 2 A p (s)|gs (e )| ds > Cα p p! Ep = θ : r

(1.9)

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satisﬁes |E p | ≤ of gs∗ ,

1 . Therefore if θ p2

∈ /

p> p0

E p , then by (1.9) and the deﬁnition

α 1 −1 3

1 2p 1 C 2 p Cα2 p p 2 p p! 2 p log log 1−r |g(r eiθ )| ! ≤ . (1.10) . 1 1 1 log log log log 1−r log log log 1−r 1−r 1 in (1.10) and using Stirling’s formula, we Finally setting p = log log log 1−r obtain (1.1) almost everywhere with constant C = 1.

Jones’ [1989] proof of Theorem 1.2 is elementary and similar to the classical proof of the law of the iterated logarithm for Bernoulli trials in Feller [1968]. We give his proof in martingale language in Appendix J.

2. Harmonic Measure and Hausdorff Measure A real valued function h(t) on an interval (0, t0 ) is called a logarithmicoexponential function or an L-function if h(t) is deﬁned by a ﬁnite algebraic combination of exponential functions and logarithm functions. The main theorem about L-functions is from Hardy [1954]: If h and g are L-functions, then the limit h(t) lim t↓0 g(t) exists in the extended interval [0, ∞]. Because the derivative of an L-function is again an L-function, it follows that every L-function is monotone on some interval (0, t1 ). Every measure function in this section will be assumed to be a positive, increasing and continuous L-function. When we are comparing harmonic measure ω to a Hausdorff measure h , we may also assume its measure function h(t) satisﬁes h(t) = ∞. t→0 t lim

(2.1)

Indeed if limt→0 h(t) t = 0 then we have ω ⊥ h by Theorem VI.5.2, and if h(t) 0 < limt→0 t < ∞ then the situation is explained by Corollary VI.6.2. We write h −1 for the inverse function of h, so that h −1 (h(t)) = t. Theorem 2.1 (Makarov). Let h be a positive increasing L-function satisfying (2.1). Let ϕ be a conformal mapping from D onto a simply connected domain and let ω denote harmonic measure for some point w0 ∈ . Then

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(a) ω h ⇐⇒ lim inf

(1 − r 2 )|ϕ (r ζ )| > 0 a.e. on ∂D, h −1 (1 − r )

(b) ω ⊥ h ⇐⇒ lim inf

(1 − r 2 )|ϕ (r ζ )| = 0 a.e. on ∂D, and h −1 (1 − r )

r →1

r →1

273

(c) there is a set A ⊂ ∂ of σ -ﬁnite h measure such that ω(A) = 1 if and only if lim inf r →1

(1 − r 2 )|ϕ (r ζ )| < ∞ a.e. on ∂D. h −1 (1 − r )

Theorem 2.2 (Makarov). There is an absolute constant C > 0 such that if h(t) = te

! C log

1 t

log log log

1 t

,

then for every simply connected domain , ω h .

(2.2)

Conversely, there is c < C such that if h(t) = te

! c log

1 t

log log log

1 t

,

(2.3)

then there exists a Jordan domain for which ω ⊥ h .

(2.4)

Note that Theorem 2.2 contains Theorem VI.5.1: If α < 1, then w α . Let us derive Theorem 2.2 from Theorem 2.1 before we prove Theorem 2.1. Proof of Theorem 2.2. Let be any simply connected domain, let ϕ be a conformal map from D onto and let g = log(ϕ ). Then by Theorem VII.2.1, ||g||B ≤ 6, and by Theorem 1.1 there is a constant C > 0 such that lim sup ! r →1

|Reg(r ζ )|

1 ≤C 1 log 1−r log log log 1−r

almost everywhere on ∂D. Take h −1 1 (t) = te

! −C log

1 t

log log log

1 t

Then by (2.5), lim inf r →1

(1 − r )|ϕ (r ζ )| h −1 1 (1 − r )

≥1

.

(2.5)

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almost everywhere, and by Theorem 2.1(a), ω h 1 . But then a little calculus shows that ! 1 C log t log log log 1t h 1 (t) = o te as t → 0, and (2.2) follows. Conversely, let be any Jordan domain that satisﬁes condition (c) of Theorem VII.2.4. The snowﬂake is one example. Let ϕ be the conformal mapping from D to and let g = log(ϕ ). Then by Theorem VII.2.4 and Theorem 1.2, there is c3 > 0 such that lim sup . r →1

−Reg(r ζ ) ≥ c3 > 0, 1 1 log 1−r log log log 1−r

almost everywhere on ∂D. If c < c3 , take c with c < c < c3 and set h −1 2 (t)

= te

! −c log

1 t

log log log

1 t

.

Then lim inf r →1

(1 − r )|ϕ (r ζ )| h −1 2 (1 − r )

=0

almost everywhere, so that by (b) of Theorem 2.1, ω ⊥ h 2 . If (2.3) holds, then for small t h(t) ≤ h 2 (t) and thus (2.4) holds for c < c3 .

Appendix K gives more examples where (2.4) holds. The extremal constant C1 = inf{C : (2.2) holds} is not known. However the above proof of (2.2) shows C1 < 6C2 where C2 = inf{C : (1.1) holds} and the proof of Theorem 1.1 yields C2 ≤ 1, but the value of C2 is not known. See Bañuelos and Moore [1999]. Bañuelos [1986] and Lyons [1990] have other proofs that C2 ≤ 1. The extremal sup{c : (2.4) holds for some domain}

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275

is also not known. Theorem 2.1 itself is a consequence of the next lemma. Lemma 2.3. Let ϕ be the conformal mapping from D to a simply connected domain and let E ⊂ ∂D be a Borel set. (a) If lim inf

(1 − r )|ϕ (r ζ )| ≤A h −1 (1 − r )

almost everywhere on E, then there exists E 1 ⊂ E such that |E \ E 1 | = 0 and h (ϕ(E 1 )) ≤ 4 A|E|. (b) There are constants c1 > 0 and q > 0, not depending on E, such that if lim inf

(1 − r )|ϕ (r ζ )| >B>0 h −1 (1 − r )

(2.6)

almost everywhere on E, then h (ϕ(E)) ≥ c1

B 1

(1 + B) 2

|E|q > 0.

Each of the six implications of Theorem 2.1 follows immediately from either part (a) or part (b) of Lemma 2.3, and the remaining details of the proof of Theorem 2.1 are left to the reader. Proof of Lemma 2.3(a). The proof of (a) is almost the same as the proof of Theorem VI.5.2. We may assume A > 0. Fix α = max{2, 21A } and deﬁne I (z) = {ζ ∈ ∂D : |z − ζ | < α(1 − |z|)}. Let {z n } be a sequence in D such that (1 − |z n |)|ϕ (z n )| ≤ Ah −1 (1 − |z n |), and such that {z n } is nontangentially dense on E. Fix δ > 0. By the Vitali covering lemma there is a subsequence {z k } of {z n } so that the intervals I (z k ) are pairwise disjoint, (1 − |z k |)|ϕ (z k )| < and |E \

&

δ , 2α

I (z k )| = 0.

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Take wk = ϕ(z k ), rk = αdist(wk , ∂), Bk = B(wk , rk ), and

& Vδ = ∂ ∩ Bk . Then rk ≤ 2α(1 − |z k |)|ϕ (z k )| < δ and h(rk ) ≤ h 2α Ah −1 (1 − |z k |) . If A ≤ 41 , then h(rk ) ≤ (1 − |z k |), and if A > 14 , then since

h(t) t

is decreasing by (2.1),

h(rk ) ≤ 4 A(1 − |z k |). In either case we obtain

h(rk ) ≤ 4 A

|I (z k )| ≤ 4 A|E|.

Therefore V = V1/m satisﬁes h (V ) ≤ 4 A|E|. On the other hand, because α ≥ 2, Lemma VI.5.3 implies that |E \ ϕ −1 (V )| = 0, and thus (a) holds for E 1 = E ∩ ϕ −1 (V ). For the proof of (b) we need three additional lemmas. Lemma 2.4. Let ϕ be the conformal mapping from D onto a simply connected domain , and let γ be a crosscut of with endpoints w1 , w2 ∈ ∂. Set ζ j = ϕ −1 (w j ), j = 1, 2. If σ is the geodesic in D connecting ζ1 to ζ2 , let z σ ∈ σ satisfy |z σ | = inf σ |z|. Then diam(γ ) ≥ c(1 − |z σ |)|ϕ (z σ )|,

(2.7)

for some absolute constant c > 0. Proof. Applying a Möbius transformation to the disc and linear map to , we may suppose that σ = (−1, 1), z σ = 0, ϕ(0) = 0 and ϕ (0) = 1. We need to prove diam(γ ) ≥ c > 0. Let Br = {|z| ≤ r }. By the Koebe one-quarter theo1 . On the other rem, B 1 ⊂ . If B(w1 , diam(γ )) ∩ B 1 = ∅, then diam(γ ) ≥ 20 4

5

hand, if B(w1 , diam(γ )) ∩ B 1 = ∅, then ϕ −1 (γ ) separates B 1 ⊂ ϕ −1 (B 1 ) 24

5

from a semicircle, say T + = ∂D ∩ {Imz > 0}. Thus dD (B 1 , T + ) ≥ dC (B 1 , B(ζ1 , diam(γ )) ≥ log 24

and (2.7) holds.

6

6

C , diam(γ )

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277

The second lemma is from Carleson’s fundamental paper [1973]. It is the key to many of the deeper results in this chapter. Lemma 2.5. Let be a simply connected domain. Fix w0 ∈ and let ζ0 ∈ ∂. For |w0 − ζ0 | 0 0 take 1 M2 . k0 = k0 (M) = 1 + π contains Then there exist r0 = r0 (M) > 0 such that if r < r0 , then ∩ ∂ D N≤

1 2π k0 log log 2 r

crosscuts γ1 , . . . , γ N of , such that each γ j separates w0 from a continuum β j ⊂ ∂ and ω(D ∩ ∂ \

N &

βj ) < r M ,

(2.8)

j=1

where ω(E) = ω(w0 , E, ).

D D α1,2

ζ0

γ2 α1,1

U1 β 1

σ w0

γ1

Figure VIII.1 Proof of Lemma 2.5.

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connecting w0 to ∂. Let {γ j } be the set Proof. Fix one curve σ ⊂ \ D of component arcs of ∩ ∂ D having the property that there exists a curve through P ⊂ connecting w0 to w j ∈ ∩ D such that P ﬁrst crosses ∩ ∂ D γ j . By Exercise I.13, \ γ j consists of two simply connected components; / U j . Because is simply connected take U j to be that component with w0 ∈ = ∅, U j ∩ Uk = ∅ whenever γ j = γk . Also by Exercise I.13, and ∂ \ D β j = (∂U j ) \ γ j ⊂ ∂ is a continuum and γ j separates β j from w0 . Now U j ∂ D is a union of arcs τ j,k , and by Exercise I.13, each τ j,k separates w0 from a continuum α j,k ⊂ β j . Let α j = k α j,k and let j be the family of paths in joining σ to α j . Then by Theorem IV 5.3 and Exercise VI.7, ω j = ω(α j ) ≤

8 −π λ( j ) , e π

(2.9)

even if is not a Jordan domain or α j is not connected. Let j be the set of to ∪k τ j,k . Then every path in j contains a path from paths in U j joining ∂ D j , so that by the extension rule, λ( j ) ≤ λ( j ). By the parallel rule, j

2π 1 1

= ≤ . λ( j ) log 2 ∂ D, ∂ D d D\D

Hence for k = 1, 2, . . . , # # j:

1 $ 1 2π 1 ≥ k log ≤ λ( j ) log 2 r k log r1

and

$ # 8 1 2π k log . # j : ω j ≥ r kπ ≤ π log 2 r 2 1 M Then since k0 = 1 + π , ω j ≤ π8 r π k0

∞ 1 8 2π ωj ≤ log (k + 1)r πk ≤ r M log 2 r π k=k0

if r < r0 and if r0 is small. Let

# $ 8 N = # j : ω j ≥ r πk0 . π

(2.10)

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279

2π 1 Then N ≤ k0 log 2 log r by (2.10). Select γ1 , . . . , γ N to have the largest ω j . Then

ω(D ∩ ∂ \

N &

βj ) ≤

ωj ≤ r M .

ω j ≤ π8 r πk0

1

Lemma 2.6. Assume the L-function h(t) satisﬁes (2.1) and

2/3 Ct ≤ h(t) ≤ ct exp log(1/t) .

(2.11)

Then there is C2 > 0 such that if 0 < t < C2 then B

h(t/B) ≤ h(t),

(2.12)

t ≤ C3 h(t). log(1/t)

(2.13)

1

(1 + B) 2 and there is C3 < ∞ such that

log(1/t)h

Proof. By (2.1), h(t)/t is decreasing

and (2.12) holds if B ≤ 1. Assume B > 1. Set x = 1/t and g(x) = log h(t) t . Then by (2.11) C ≤ g(x) ≤ (log x)2/3 for large x. Then since g(x) and (log x)2/3 are both L-functions, Theorem 19 of Hardy [1954] gives 2 (log x)−1/3 3x for x large and integration then yields (2.12) and (2.13). g (x) ≤ c

Proof of Lemma 2.3(b). To prove (b) we may assume that h(t) satisﬁes (2.11). Indeed, since (2.11) holds for every measure function of the form h 1 (t) = te

! C log

1 t

log log log

1 t

we will have established Theorem 2.2 if we prove (b) under the additional assumption (2.11). On the other hand, if (2.11) fails, then by Hardy’s theorem on L-functions, h 1 (t) = o(h(t)). But then we can use Theorem 2.2 (which would have been proved for the measure function h 1 ) to obtain h 1 (ϕ(E)) > 0 and h (ϕ(E)) = ∞ whenever E ⊂ ∂D has |E| > 0. Therefore part (b) of Lemma 2.3 will also hold when (2.11) fails.

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Now assume (2.6) holds for E with |E| > 0. Fix δ > 2. By (VII.1.6) and Theorem VII.2.1 there is a constant c > 0 so that for ζ ∈ E (1 − |z|)|ϕ (z)| > cB > 0. h −1 (1 − |z|)

lim inf

z∈δ (ζ )

1

Let δn (ζ ) = δ (ζ ) ∩ {z : |z| > 1 − n1 }. For sufﬁciently large n, the closed set E n = {ζ ∈ E :

inf 1

z∈δn (ζ )

(1 − |z|)|ϕ (z)| 1 ≥ (cB + )} ⊂ E h −1 (1 − |z|) n

satisﬁes |E| < 2|E n |, and we may replace E by E n and B by B = cB + n1 . Cover ϕ(E) by discs Dν of radius rν < r0 such that h(rν ) ≤ 2h (ϕ(E)). (2.14) ν ) < 1 , By choosing r0 sufﬁciently small we can assume that 2π ω(∂ D n dist(ϕ(0), ϕ(E)) > 4r0 , and h(r0 )/r0 > B, since h(t)/t is decreasing. Then (ν) ν for by the M = 1 case of Lemma 2.5 there are subarcs γ j ⊂ ∂ D 1 ≤ j ≤ N (ν) ≤ (ν)

such that γ j and

1 2π log( ), log 2 rν (ν)

is a crosscut of separating w0 from a continuum β j

ω(Dν ∩ ∂ \

N& (ν)

(ν)

β j ) ≤ rν ≤

1

h(rν ) . B

⊂ ∂

(2.15)

On the other hand, if ϕ(ζ1 ) and ϕ(ζ2 ) are the endpoints of an arc β j(ν) , then ν ) ≤ 1 . By discarding some β (ν) if necessary, we will |ζ1 − ζ2 | < 2π ω(∂ D (ν)

j

n

suppose that β j ∩ E = ∅. Thus z σ ∈ E, since δ > 2, where z σ is the point closest to the origin on the geodesic connecting ζ1 and ζ2 . By (2.6) and Lemma 2.4 c2 (ν) (ν) h −1 (ω(β j )) ≤ h −1 (1 − |z σ |) ≤ diamγ j , B so that N (ν) j=1

(ν)

h −1 (ω(β j )) ≤

2c2 rν . B

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3. The Number of Bad Discs (ν)

Fix ν, set t j = h −1 (ω(β j )), t = N (ν)

(ν)

ω(β j ) =

j=1

Because N (ν) ≤

2π log 2

h(t j ) =

h(t) t

h(t j ) +

t j ≤s

h(t j ).

t j >s

log(1/t), (2.13) gives

t j ≤s

t j ≤ 2c2 rν /B, and s = t/ log 1/t. Then

j=1

h(t j ) ≤

Moreover, since

N (ν)

2c r

2π t 2 ν log(1/t)h ≤ C4 h . log 2 log(1/t) B

is decreasing, (2.13) also gives

h(t j ) =

t j ≥s

2c r h(t j ) h(s) 2 ν tj ≤ tj ≤ C3 h . t s B j t ≥s t ≥s j

j

Therefore by (2.12) N (ν)

1

(ν)

ω(β j ) ≤ C4

j=1

(1 + B) 2 h(rν ). B

(2.16)

Now (2.14), (2.15), and (2.16) yield 1

|E| = ω(ϕ(E)) ≤ C1

(1 + B) 2 h (ϕ(E)), B

which is assertion (b).

3. The Number of Bad Discs Let be a simply connected domain, normalized so that ∞ ∈ and diam(∂) = 2, and write ω(E) = ω(∞, E, ). It follows from Beurling’s projection theorem that ω(B(ζ, ρ)) ≤ ρ 1/2 ,

(3.1)

for all ζ ∈ ∂ and all ρ. The power of ρ in inequality (3.1) is sharp if ∂ is a line segment with one end at ζ , because then ω(B(ζ, ρ)) ≥ cρ 1/2 for ρ ≤ 1. In their [1994] paper, Carleson and Makarov estimated the number of times (3.1) can almost fail. For ε > 0 deﬁne N (, ε, ρ) to be the maximum cardinality of sets of the form 1 ζk ∈ ∂ : B(ζk , ρ) are pairwise disjoint and ω(B(ζk , ρ)) ≥ ρ 2 +ε .

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Theorem 3.1. There are constants A and K such that for all N (, ε, ρ) ≤ Aρ −K ε .

(3.2)

The constant K ∗ = inf{K : (3.2) holds for some constant A} will play a leading role in the later sections of this chapter and determining K ∗ is an important open problem. Proof. Fix δ small and consider the annuli A j (ζ ) = {z : δ j < |z − ζ | < δ j−1 }, having moduli m = and write

1 2π log 1/δ. Let

mj =

inf

E j denote any subarc of ∩ {|z − ζ | = δ j }

{E j ,E j−1 }

d∩A j (E j , E j−1 )

and X j (ζ ) = m j − m. Then X j ≥ 0 and X j = 0 if and only if A j ∩ ∂ lies on a line segment with endpoint ζ. Lemma 3.2. Given ε > 0, there is n(ε) > 0 such that if n > n(ε), ρ = δ n and

1 ω B(ζ, ρ) ≥ ρ 2 +ε , then n−1

X j (ζ ) ≤

j=1

3 1 εn log . π δ

Proof. We use Lemma 2.5 with w0 = ∞ and M = ρ −ε ≥ 1 +

(3.3)

1 2

+ 2ε. Take n so large that

2π 1 log . log 2 ρ

Then by Lemma 2.5 there is an arc E ⊂ {|z − ζ | = 2ρ} ∩ that separates ∞ from a continuum β ⊂ ∂ for which 1

ω(β) = ω(β ∩ B(ζ, 2ρ)) ≥

1

ρ 2 +ε − ρ 2 +2ε 2π log 2

log

1 ρ

1

≥ ρ 2 +2ε .

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3. The Number of Bad Discs Take J = {|z − ζ | = 1}. Then J meets ∂ so that by Theorem IV.5.3, 1 1 1 dist (E, J ) ≤ 1 + + 2ε log . π 2 ρ

Set E n = E. Because is simply connected we can by reverse induction (i.e., for j = n − 1, n − 2, . . . , 2, 1) ﬁnd arcs E j ⊂ ∩ {|z − ζ | = δ j } such that E j separates E j+1 from J . Then by the serial rule, n−1

dist∩A j (E j , E j−1 ) ≤ dist (E, J ).

j=1

Hence n−1 (n − 1) 1 1 1 1 Xj ≤ 1 + log + + 2ε log , 2π δ π 2 ρ j=1

and (3.3) holds when n ≥

1 2ε

+ 1.

Lemma 3.3. There are absolute constants σ > 0 and η > 0 such that if ζ ∈ ∂, k ≥ 2, and X 1 (ζ ) ≤ δ kσ , then for all ζ ∈ ∂ ∩ A1 (ζ ) and all 2 ≤ j ≤ k, 1 X j (ζ ) ≥ η log . δ

(3.4)

θ Aj

(ζ ) ∂

ζ

ζ A1 (ζ )

Figure VIII.2 Proof Lemma 3.3. Proof. Since diam(∂) = 2 the continuum ∂ meets both boundary components of A1 (ζ ). Let θ be the smallest angle of a sector of A1 (ζ ) that contains

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A1 (ζ ) ∩ ∂. Then by Corollary V.4.3 or Exercise V.4, X 1 (ζ ) ≥ c

θ3 (log 1δ )2

,

where c is an absolute constant. Consequently θ<

δk 2

(3.5)

if σ > 3 and δ is sufﬁciently small. But now if ζ ∈ A1 (ζ ) ∩ ∂ and 2 ≤ j ≤ k, then the annulus A j (ζ ) ∩ ∂ contains two crosscuts of A j (ζ ) separated by angle π3 . Note this is even true for |ζ − ζ | > 1 − δ because diam(∂) = 2. 1 . That gives (3.4) with η = 10π To complete the proof of Theorem 3.1, we let D be any closed disc of radius 1 and let N = N (, n, λ, δ) be the maximum number of points ζ p ∈ D ∩ ∂ such that |ζ p − ζq | ≥ δ n , p = q

(3.6)

and n−1

X j (ζ p ) ≤ λ.

(3.7)

j=1

Also let N (n, λ, δ) = sup N (, n, λ, δ). {,D}

Fix δ small. We claim there exist constants C1 and C2 such that N (n, λ) ≡ N (n, λ, δ) ≤ C1 eC2 λ .

(3.8)

By Lemma 3.2, (3.8) implies (3.2) with K = π3 C2 . We prove (3.8) by induction on n. Because diam(∂) = 2, we clearly have N ≤ Cδ −2n and we obtain (3.8) for n ≤ 2 by taking C1 sufﬁciently large. Now assume n ≥ 3 and assume (3.8) holds for all k < n. Take ζ1 , . . . ζn ∈ ∂ satisfying (3.6) and (3.7). Let σ and η be the constants in Lemma 3.3. There are three cases. Case 1: min D∩∂ X 1 (ζ ) > δ 2σ .

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3. The Number of Bad Discs

In this case n−1

X j (ζ p ) ≤ λ − δ 2σ .

j=2

1 2

Cover D ∩ ∂ by c δ discs of radius δ. Consider one such disc and suppose it contains points ζ1 , · · · , ζm that satisfy (3.6) and (3.7). Rescaling to a disc of radius 21 , we see by induction that m ≤ N (n − 1, λ − δ 2σ ) ≤ C1 eC2 (λ−δ

2σ )

.

Hence N (, n, λ) ≤ c

1 2 δ

e−C2 δ C1 eC2 λ ≤ C1 eC2 λ , 2σ

if C2 ≥ C2 (δ). Case 2: min D∩∂ X 1 (ζ ) ≤ δ nσ . Fix ζ ∈ D ∩ with X 1 (ζ ) ≤ δ nσ . Then by Lemma 3.3, X j (z) ≥ η log 1δ for 2 ≤ j ≤ n and all z ∈ A1 (ζ ) ∩ ∂. Thus if λ < (n − 1)η log 1δ , then all ζ p ∈ D ∩ ∂ with (3.6) and (3.7) satisfy ζ p ∈ B(ζ, δ), and by a rescaling N (, n, λ) ≤ N (n − 1, λ) ≤ C1 ec2 λ . On the other hand, if λ ≥ (n − 1)η log 1δ , then by (3.6) N (, n, λ) ≤ c

1 2n δ

2nλ

≤ ce (n−1)η ≤ C1 eC2 λ ,

if C1 and C2 are sufﬁciently large. Case 3: δ nσ < min D∩∂ X 1 (ζ ) ≤ δ 2σ . Let the minimum be 2 attained at ζ ∈ D ∩ ∂, and choose k ≥ 2 so that X 1 (ζ ) ∈ δ σ (k+1) , δ σ k . Then by Lemma 3.3, X 2 , . . . , X k ≥ η log

1 δ

k on A1 (ζ ) ∩ ∂. By (3.5) we can cover D ∩ ∂ \ B(ζ, δ) with c 1δ discs of radius δ k , each centered in A1 (ζ ). Then by rescaling and induction, N (, n, λ) ≤ N (n − 1, λ − δ kσ ) + c ≤ C1 e

C2 (λ−δ kσ )

≤ C 1 eC2 λ provided C2 ≥ C2 (δ).

+c

1 k δ

1 k δ

1 N (n − k, λ − (k − 1)η log ) δ 1

C1 eC2 (λ−(k−1)η log δ )

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4. Brennan’s Conjecture and Integral Means Spectra Let ϕ be a conformal mapping from the disc D onto a simply connected domain . Brennan’s conjecture is that 2− p |ϕ | d xd y = |(ϕ −1 ) | p d xd y < ∞ (4.1)

D

( 43 , 4),

/ the integral (4.1) diverges for the whenever < p < 4. When p ∈ z . When p = 2 the integral has value π. When Koebe function ϕ(z) = (1−z) 2 4 3

< p < 3, (4.1) is a consequence of the distortion estimate (I.4.17). See Metzger [1973]. When 43 < p < 2, (4.1) follows from the theorem of Prawitz, Exercise I.23(d), and a simple integration. Using the methods in Carleson [1973], Brennan [1978] established (4.1) in the range 3 0. In Appendix L we give Pommerenke’s [1985a] elementary proof of (4.1) for p < 3.399. Brennan’s conjecture is true for many special types of domains; see Exercise 8 and the paper [1998] of Barañski, Volberg, and Zdunik. Brennan’s conjecture is a point on a spectrum of questions about integral means of derivatives of univalent functions. Let ϕ be univalent on D, let t ∈ R, and deﬁne #

1 β $ . (4.2) βϕ (t) = inf β : |ϕ (r eiθ )|t dθ = O 1−r 5 3

The function βϕ (t) is the integral means spectrum of ϕ. Clearly βϕ (t) ≥ 0. By Hölder’s inequality, βϕ (t) is a convex and continuous function. By the distortion theorem (I.4.17),

βϕ (t) − s, if s < 0, (4.3) βϕ (t + s) ≤ βϕ (t) + 3s, if s > 0. The Koebe function k(z) =

z (1−z)2

has integral means spectrum ⎧ ⎨ 3t − 1, if t ≥ 13 , βk (t) = 0, if −1 ≤ t < 13 , ⎩ |t| − 1, if t < −1.

(4.4)

Deﬁne the (unbounded) universal integral means spectrum to be B(t) = sup βϕ (t) : ϕ is univalent . Thus B(t) is the smallest number such that for all ε > 0 and all univalent ϕ there is constant C(ε, ϕ) such that

1 B(t)+ε |ϕ (r eiθ )|t dθ ≤ C(ε, ϕ) . 1−r

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Theorem I.4.5 yields the upper bound

3t, if t ≥ 0, B(t) ≤ |t|, if t < 0, and (4.4) shows that

287

(4.5)

⎧ ⎨ 3t − 1, if t ≥ 31 , B(t) ≥ 0, if −1 ≤ t < 13 , ⎩ |t| − 1, if t < −1.

(4.6)

Feng and MacGregor [1976] proved B(t) = 3t − 1 for t ≥ 2/5. See Exercise 7. Theorem 4.2 below shows that B(t) = |t| − 1 for t < −K /2, and Appendix L includes Pommerenke’s [1985a] proof that 1 1/2 1 ≡ γ (t) ≤ 3t 2 + 7|t|3 (4.7) − t + 4t 2 B(t) ≤ − + t+ 2 4 for all t. 3t − 1 |t| |t| − 1 γ

3t

− K2 −1 0 13 25 Figure VIII.3 Known bounds for B(t). Lemma 4.1. Brennan’s conjecture holds if and only if B(−2) = 1. Proof. Assume B(−2) = 1 and let 2 < p < 4. Then B(2 − p) < α < 1 by convexity, so that for all ϕ 1 α |ϕ (r eiθ )|2− p dθ ≤ C . 1−r Then Fubini’s theorem gives (4.1). Conversely, if (4.1) holds for p < 4 then because the means |ϕ (r eiθ )|2− p dθ

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are increasing in r , |ϕ (r eiθ )|2− p dθ ≤

1 1−r

|ϕ (z)|2− p d xd y

{r <|z|<1}

and B(2 − p) ≤ 1. Therefore by continuity B(−2) ≤ 1, while (4.4) implies B(−2) ≥ 1. The proof of Lemma 4.1 also shows sup{ p : (4.1) holds} = inf{ p : B(2 − p) ≤ 1}. Theorem 4.2. Let K be a constant for which the conclusion (3.2) of Theorem 3.1 holds. Then for t < −K /2, B(t) = |t| − 1. Deﬁne K ∗ = inf{K : (3.2) holds for some constant A}.

(4.8)

By Theorem 4.2, Brennan’s conjecture will be true if K ∗ ≤ 4. We will see conversely in Section 5 that Brennan’s conjecture is false if K ∗ > 4. Section 5 also contains an example that shows B(t) > |t| − 1 for t ∈ (−2, −1). Therefore K ∗ ≥ 4 and Brennan’s conjecture is equivalent to the conjecture that K ∗ = 4. Proof. Fix t < −K /2 and assume ∞ ∈ and diam(∂) = 2. Let ϕ : D → be univalent and satisfy ϕ(0) = ∞. For 1 − r small we partition the circle C 1−r arcs I j of equal length δ, with 1−r {|z| = r } into 1−r 16 ≤ δ < 8 and write z j for the midpoint of I j . Because || log(ϕ )||B ≤ 6, we have |ϕ |t dθ ∼ δ|ϕ (z j )|t . (4.9) Ij

Let n and k be positive integers with n large and set A(k) = { j : δ

(k+1) n

k

< |ϕ (z j )| ≤ δ n }

and N (k) = card A(k). By the distortion theorem, |ϕ (z j )| ≥ 1−r 8 > δ, so N (k) = 0 for k ≥ n. By Koebe’s estimate if j ∈ A(k) then δ δ ϕ B(z j , ) ⊃ B ϕ(z j ), |ϕ (z j )| 2 8 k+1 1 δ 1+ n δ 1+ n ⊃ B ϕ(z j ), ⊃ ϕ B(z j , ) . 8 32

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289

1

k

1 (1+ n ) k Set ρ = 10 δ and ε = 21 − n+k . Since δ n ∼ 1 for large n, the discs {B(ϕ(z j ), ρ)} j∈A(k) are pairwise disjoint and 1

δ 1+ n ), {|z| < 1 − r } ω ∞, B(ϕ(z j ), ρ), ≥ ω 0, B(z j , 40

1

≥ Cδ = Cρ 2 +ε . By Theorem 3.1

N ( j) ≤ Aδ

−K 2

(1− nk )

.

j≥k

Therefore by (4.9) and a summation by parts, n−1 k |ϕ (r eiθ )|t dθ ≤ 2π + C δ n t+1 N (k) k=0 n−1 K

k t K k ≤ 2π + C Aδ 1− 2 1 + δ n t (1 − δ − n )δ 2 n

≤C

t t + K /2

k=0 |t|−1

1 1−r

,

since t < −K /2.

5. β Numbers and Polygonal Trees Let ϕ be the conformal map from D onto a simply connected domain and let limits ζ1 = ζ2 be points of ∂D at which ϕ has nontangential

ϕ(ζ1 ) and ϕ(ζ2 ). For ε > 0 let Bε (ζ j ) be the component of ∩ B ϕ(ζ j ), ε that contains some tail segment {ϕ(r ζ j ) : r0 < r < 1} of the image arc {ϕ(r ζ j ) : 0 ≤ r ≤ 1}. (If ϕ(ζ j ) = ∞, we let Bε (ζ j ) be the corresponding component of ∩ {|z| > 1/ε}.) Let ε = ε (, ζ1 , ζ2 ) be the family of paths in joining Bε (ζ1 ) to Bε (ζ2 ), and let ε∗ be the set of paths in C joining B(ϕ(ζ1 ), ε) to B(ϕ(ζ2 ), ε). If ε < δ let j = j (ε, δ) = {γ ∩ Bδ (ζ j ) : γ ∈ ε }, j = 1, 2. Then by the basic rules for extremal lengths we have λ(ε ) ≥ λ(δ ) + λ( 1 ) + λ( 2 ),

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and λ( j ) ≥

1 log δ/ε , j = 1, 2. 2π

Then 0 ≤ λ(δ ) − λ(δ∗ )

1 log δ/ε + λ( 1 ) − 2π

1 log δ/ε + λ( 2 ) − 2π ≤ λ(ε ) − λ(ε∗ ) + O(δ 2 + ε2 ),

(5.1)

where we get O(δ 2 + ε2 ) by comparing C∗ \ {B(ϕ(ζ1 ), ε) ∪ B(ϕ(ζ2 ), ε)} to an annulus bounded by concentric circles. Therefore the limit

lim λ(ε ) − λ(ε∗ ) ε→0

exists in [0, ∞]. Deﬁne ∗

β = β(, ζ1 , ζ2 ) = lim e−2π(λ(ε )−λ(ε )) . ε→0

(5.2)

Then 0 ≤ β(, ζ1 , ζ2 ) ≤ 1. δ If β > 0 then for ε < δ small the two middle terms in (5.1) are small,

and if ε is ﬁxed, then by Corollary V.4.3, E j ∩ B(ϕ(ζ j ), δ) \ B(ϕ(ζ j ), ε) is conﬁned to a narrow sector with vertex ϕ(ζ j ), where E j is the connected component of ∂ ∩ B(ϕ(ζ j ), δ) which contains ϕ(ζ j ). Consequently β(, ζ1 , ζ2 ) > 0 can hold for at most one point ζ j ∈ ϕ −1 (ϕ(ζ j )) and we can unambiguously deﬁne

β(, ϕ(ζ1 ), ϕ(ζ2 )) = β(, ζ1 , ζ2 ).

ϕ(ζ j ) ε

ϕ(r ζ j ) δ

Figure VIII.4

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291

It is clear that β(, ζ1 , ζ2 ) is invariant under Möbius transformations, and we will always take ζ2 = ∞. Lemma 5.1. Suppose ψ : H → is a conformal mapping such that ψ(z) → 1, z2

(z → ∞).

(5.3)

Let x ∈ R and assume that ψ(x) = a, ψ (x) = 0, and ψ (x) exists. Then β(, a, ∞) =

2 |ψ (x)|

.

(5.4)

Proof. If ε is small, ψ −1 (Bε (x)) is approximately a half-disc of radius while ψ −1 (|z| = 1ε ) is approximately the semicircle of radius ε * , 1 1 |ψ (x)| 21 λ(ε ) ≈ log π ε 2

− 12

!

2ε |ψ (x)|

. Then

while λ(ε∗ ) ≈

1 1 log 2 , 2π ε

and therefore (5.4) holds.

For example, suppose ∂ is a ﬁnite connected union of line segments with extreme points a1 , . . . , an and ∞. In this case we say ∂ is a polygonal tree. ∂ a3

a2 a1

a5

a4

a7

a6 Figure VIII.5 Polygonal tree.

Let ψ : H → satisfy (5.3) and let a j = ψ(x j ). Then β(, a j , ∞) =

2 . |ψ (x j )|

On the other hand, if a ∈ / {a j }, β(, a, ∞) = 0.

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When ∂ is a polygonal tree we write β j = β(, a j , ∞). Example 5.2. Let 0 < t < 1 and set % ψ(z) = (z 2 − 1)(z 2 − t 2 ). Then ψ is the conformal map from H to a domain bounded by the half line [−t, ∞] and the segment on the imaginary axis [−ib, ib] where b = (1 − t 2 )/2, and ψ satisﬁes (5.3). ib −t −ib Figure VIII.6 This example has three non-zero β numbers, and by (5.4) they are β1 = β(, −t, ∞) =

2t 1 + t2

and β2 = β3 = β(, ±ib, ∞) =

1 1 − t2 . 2 1 + t2

Note that

β j2 =

1 2t 2 <1 + 2 (1 + t 2 )2

since 0 < t < 1, but that for p < 2,

p

βj = =

2 p t p + 21− p (1 − t 2 ) p (1 + t 2 ) p 1 + 21− p (1 − t) p

+ o((1 − t) p )

(5.5) >1

if 1 − t is small. Theorem 5.3. Let K ∗ be deﬁned by (4.8) and let p ≥ 1. Then the following are equivalent: p (a) β j ≤ 1 for every polygonal tree. (b) K ∗ ≤ 2 p. (c) B(− p) = p − 1.

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From (5.5) and the theorem it follows that K ∗ ≥ 4. Consequently we have the following corollary. Corollary 5.4. The following are equivalent: 2 (a ) β j ≤ 1 for every polygonal tree. (b ) K ∗ = 4. (c ) Brennan’s conjecture is true. Proof of Theorem 5.3. The implication (b) ⇒ (c) is Theorem 4.2, and we will prove (c) ⇒ (a) in Section 6. (a) ⇒ (b): Assume (b) is false and let p < q < K ∗ /2. By the deﬁnition of ∗ K there exists, for any A > 0, C > 0, ε > 0, and ρ > 0, a simply connected domain such that ∞ ∈ and diam(∂) = 2 and there exist

1 2q+1 (5.6) N ≥ Aρ −2qε log ρ points ζ j ∈ ∂ such that |ζ j − ζk | ≥ 2ρ and 1

ω(∞, B(ζ j , ρ) ∩ ∂, ) ≥ ρ 2 +ε . By Lemma 2.5 there is an arc γ j ⊂ ∩ ∂ B(ζ j , 2ρ) having endpoints in ∂ such that γ j separates ∞ from a continuum α j ⊂ ∂ and 1

ω(∞, α j , ) ≥ c

ρ 2 +ε log ρ1

.

Here c denotes a constant having possibly different values at different occurrences. Let ϕ : D → be conformal with ϕ(0) = ∞. Then ϕ −1 (γ j ) is an arc in D with endpoints on ∂D. Let σ j be the hyperbolic geodesic in D having the same endpoints as ϕ −1 (γ j ) and let z j be the midpoint of σ j . Then 1

1 − |z j | ∼ 1 (σ j ) ≥ ω(∞, α j , ) ≥ c

ρ 2 +ε log ρ1

≡ η.

(5.7)

Set r j = (1 − |z j |)/2 and B j = B(z j , r j ). By Koebe’s estimate and the Gehring–Hayman Theorem, Exercise III.15, diamϕ(B j ) ∼ dist(ϕ(z j ), ∂) ≤ c diam(ϕ(σ j )) ≤ c diam(γ j ).

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Therefore ϕ(B j ) ⊂ B(ζ j , cρ),

(5.8)

and hence no point lies in more than C1 balls B j . Also, by (I.4.17) 1 − |z j | ≤ c|ϕ (z j )| ≤

cρ , 1 − |z j |

(5.9)

and by (5.7) and (5.9) we have B j ⊂ {z : 1 − cρ 1/2 ≤ |z| ≤ 1 −

η }. 2

(5.10)

Moreover, (5.9) gives

2+q

Bj

|ϕ |−q d xd y ≥ cr j2 |ϕ (z j )|−q ≥ c

rj

ρq

,

so that by (5.10), (5.8), and (5.6) N η2+q 1 2q+1 η2+q −q −2qε |ϕ (z)| d xd y ≥ ≥ c Aρ . log 1 C1 ρ q ρ ρq 1−cρ 2 ≤|z|≤1− η2 Consequently there exists R, 1

1 − cρ 2 < R < 1 −

η 2

(5.11)

such that 1 q−1 1 1 2q+1 η2+q |ϕ (Reiθ )|−q dθ ≥ cρ − 2 Aρ −2qε log ≥ cA q ρ ρ 1− R (5.12) if ε is small. From (5.12) it follows easily that (c) ⇒ (b); i.e., that B(− p) > p − 1 if 2 p < K ∗ ; but let us continue and prove (a) ⇒ (b). The left-hand side of (5.12) increases with R and 1 − 1+R 2 is comparable to 1

1 − R. Thus we may suppose that R = e− n ∼ 1 − n1 and that ϕ is analytic in a neighborhood of D with ϕ = 0 in D. Replacing ϕ with −ϕ(−z) if necessary, we may suppose that q−1 3π 2 1 dθ 1 |ϕ (e− n eiθ )|−q ≥ c A. (5.13) π n 2π 2 Let ϕ1 be the conformal map of D onto D \ [ 1e , 1] such that ϕ1 (0) = 0 and

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295

ϕ(1) = 1e . Then

1

ϕn (z) = ϕ1 (z n ) n ≡ z

ϕ1 (z n ) zn

1 n

is a conformal map of D onto D \ ∪σk where {σk } are n equally spaced radial slits. Set ζk = e2πik/n , k = 1, . . . , n. Then ϕn = 0 at {ζk } and ϕn (ζk ) = Rζk . Moreover 1

|ϕn (ζk )| = n|ϕ1 (1)||ϕ1 (1)|1− n ∼ cn. Let ϕ0 be the conformal map of H onto D \ [0, 1) given by ϕ0 (z) =

(z +

√

1 z 2 − 1)2

,

and set ψ = C0 ϕ ◦ ϕn ◦ ϕ0 , where 1

C0 =

1

4ϕ1 (0) n 4e− n ∼ . lim z→∞ zϕ(z) diam(∂)

ϕ R

1

∂

ϕn ◦ ϕ0 (H) Figure VIII.7 Then lim z→∞ ψ(z)/z 2 = 1 and βk =

2 |ψ (x

k )|

=

2

|ϕ (Rζ

, k )||ϕn (ζk )||ϕ0 (x k )|

where xk = ϕ0−1 (ζk ). Since |ϕ0 (ϕ −1 (eiθ ))| ≤ c1 < ∞ on { π2 ≤ θ ≤

3π 2 },

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βk ≥

c n|ϕ (Rζ

k )|

provided π2 ≤ arg ζk ≤ 3π 2 . By (5.13) and Koebe’s estimate q−1 q 1 1 1 βk ≥ c ≥ c A, n |ϕ (Rζk )|q 2n where the sums are taken over those k for which π2 ≤ arg ζk ≤ 3π 2 . Approximate ψ(R) by a polygonal tree. Then (a) is violated because A is large, c is a constant and p < q.

6. The Dandelion Construction and (c) ⇒ (a) Let T be a polygonal tree with extreme points ∞ and a1 , a2 , . . . , a N . Let a0 ∈ T be such that the inﬁnite segment (a0 , ∞] is a relatively open subset of T in the direction −eiθ0 , and set T0 = T \ (a0 , ∞]. The β numbers are dilation invariant, and we may suppose that T0 ⊂ B(a0 , 1). Choose ε0 > 0 so that for each j ≥ 1 T ∩ B(a j , ε0 ) = [a j , a j − ε0 eiθ j ) is a line segment and B(a j , ε0 ) ∩ B(ak , ε0 ) = ∅ if j = k. Set T1 = T and for δ < 21 ε0 set &

a j + δei(θ j +θ0 ) (T0 − a0 ) . T2 = T1 ∪ B(a1,2 , δε0 ) a1

a1

T a0

a0 a2,2,2 a2

a2,1 a2 B(a2 , ε0 )

Figure VIII.8 Dandelion construction.

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297

Then T2 is a new polygonal tree with extreme points ∞ and the N 2 ﬁnite points a j1 , j2 = a j1 + δei(θ j1 +θ0 ) (a j2 − a0 ). For each pair j1 , j2 , T2 ∩ B(a j1 , j2 , δε0 ) = [a j1 , j2 , a j1 , j2 − δε0 eiθ j1 , j2 ) is again a line segment, and the balls {B(a j1 , j2 , δε0 )} are pairwise disjoint. We continue, attaching a translation and rotation of δ 2 T0 at the N 2 ﬁnite extreme points of T2 to obtain another polygonal tree T3 ⊃ T2 with N 3 ﬁnite extreme points, which we label a j1 , j2 , j3 . This construction can be iterated indeﬁnitely, yielding at stage n a tree Tn ⊃ Tn−1 having N n ﬁnite extreme points a j1 , j2 ,..., jn , and Tn+1 is constructed from Tn by attaching rotations and translations of T0 dilated by the factor δ n . Note that B(a j1 ,..., jn , δ n−1 ε0 ) ⊂ B(a j1 ,..., jn−1 , δ n−2 ε0 ) since δ n−1 ε0 + δ n−1 < δ n−2 ε0 and so for each n the balls {B(a j1 ,..., jn , δ n−1 ε0 )} are pairwise disjoint. Set n = C∗ \ Tn , T∞ =

&

Tn ,

n

and =

&

n

◦

= C∗ \ T∞ .

We call T∞ the dandelion generated by T . Note, however, that T∞ also depends on the choice of the dilation factor δ. Write J = ( j1 , j2 , . . . jn ) for a multi-index of length |J | = n in {1, 2, . . . N }n , and deﬁne a J = a j1 , j2 ,..., jn ,

β J = β(n , a J , ∞),

nJ = n ∪

& |I |=n I = J

B(a I , 2δ n ),

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and J = ( j1 , j2 , . . . , jn−1 ). By the extension rule 3J ≡ β(nJ , a J , ∞) ≤ β J , β and by the proof of (5.1) β J ≤ β J β jn (1 + O(δ)).

(6.1)

The next lemma shows that the reverse inequality to (6.1) almost holds if the dilation factors are small enough. Lemma 6.1. Given α > 0 there exists δ = δ(α) such that for every multi-index J of length n, α 3 3J ≥ (1 − )β3 (6.2) β jn Jβ 2 and 3J ≥ (1 − α)n β j1 β j2 . . . β jn . βJ ≥ β

(6.3)

Proof. We may choose δ small enough that α )β j , 2 for j = 1, . . . , N . By the extension rule and induction, (6.2) implies (6.3) and we only prove (6.2). Because β numbers are translation invariant, we may assume a0 = 0. Choose δ so small that 4 1 1 δ ≤ ε0 , log 2π c 1 − α2 j βj ≡ β(1 , a j , ∞) ≥ (1 −

where c is the constant given in Exercise IV.10. By construction, ∂nJ ∩ {z : δ n−1 < |z − a J | < δ n−2 ε0 } is a radial line segment and

J \ B(a J , 2δ n−1 ). nJ \ B(a J , 2δ n ) ⊃ n−1 3

Set = {z : |z − a J | = δ n− 2

√

ε0 }. If ε is small then by Exercise IV.10 and

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299

the extension rule, 1 λ(ε ) ≡ dnJ (∂ B(a J , ε), ∂ B(0, )) ε

1 1 δ 4 ≤ dnJ (∂ B(a J , ε), ) + dnJ (, ∂ B(0, )) + c ε ε0 1 1 1 ≤ dnJ (∂ B(a J , ε), ) + d J (, ∂ B(0, )) + log n−1 ε 2π 1−

α 2

.

Removing a radial slit from an annulus does not affect the extremal distance between its bounding circles, so that as in the proof of (5.1) 1 1 d J , ∂ B(0, ) −dC , ∂ B(0, ) n−1 ε ε 1 1 ≤ d J ∂ B(a J , ε), ∂ B(0, ) −dC ∂ B(a J , ε), ∂ B(0, ) + O(ε), n−1 ε ε and

dnJ ∂ B(a J , ε), −dC ∂ B(a J , ε), . .

ε0 ε0 ε ε ) −dC ∂ B(a jn , n−1 ), ∂ B(a0 , ) = d jn ∂ B(a jn , n−1 ), ∂ B(a0 , 1 δ δ δ δ

ε δ n−1 ≤ d jn ∂ B(a jn , n−1 ), ∂ B(a0 , ) 1 δ ε

ε δ n−1 − dC ∂ B(a jn , n−1 ), ∂ B(a0 , ) + O(ε) δ ε and

1 1 dC ∂ B(a J , ε), +dC , ∂ B(0, ) = dC ∂ B(a J , ε), ∂ B(0, ) + O(ε). ε ε Thus α 3 3J = lim e−2π(λ(ε )−λ(ε∗ )) ≥ β3 β jn β J (1 − ), ε→0 2 for δ sufﬁciently small, which implies (6.2).

Proof of (c) ⇒ (a): Fix p > 0 and assume (a) is false. Then there is a polygonal tree T with ﬁnite extreme points a1 , a2 , . . . , a N such that N j=1

p

β j = A > 1.

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We make the dandelion construction with dilation parameter δ so small, for n ≥ 2, that Lemma 6.1 holds with α satisfying (1 − α) p A > 1.

(6.4)

Let ψ : H → be a conformal map satisfying (5.3) and set 1 + z : D → . ϕ(z) = ψ i 1−z For a multi-index J = ( j1 , . . . , jn ), let ϕ J : D → nJ , with ϕ J (0) = ϕ(0) and ϕ J (1) = ∞. If f = ϕ −1 , f J = ϕ −1 J , and ζ J = f J (a J ) then 1 8 C |ϕ J (ζ J )| = ≤ . (6.5) 4 3 3 |1 − ζ J | β J βJ By construction B(a J , δ n−1 ε0 ) ∩ Tn is a line segment terminating at a J , so that by the Schwarz reﬂection principle, n−1 1 2δ ε0 2 z + ... (6.6) f J (a J + δ n−1 ε0 z 2 ) = ζ J + ϕ J (ζ J ) is univalent on D. By the growth theorem (I.4.16) n−1 1 2δ ε0 2

n−1 2 |z| 1 + O(|z|) , | f J (a J + δ ε0 z ) − ζ J | = ϕ J (ζ J )

(6.7)

uniformly in J . Set U J = nJ \ B(a J , δ n ) ⊂ , V J = f J (U J ) ⊂ D. Tn ζJ δn aJ AJ

UJ

fJ f J (A J )

Figure VIII.9 The map f J near a J .

D \ VJ

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301

If τ J : V J → D is conformal with τ J (0) = 0 and τ J (1) = 1, then by Schwarz’s lemma 1 − | f (z)| ≥ 1 − |τ J ◦ f J (z)|

(6.8)

for z ∈ U J . Set A J = {z : 2δ n < |z − a J | < 3δ n and dist(z, T ) ≥ δ n }. Then for z ∈ A J , dist(z, ∂U J ) ≥ c dist(z, T ), so that by (6.8) and Koebe’s estimate, Theorem I.4.3, | f (z)| ≥ c|(τ J ◦ f J ) (z)|.

(6.9)

By (6.6) and the distortion theorem (I.4.17), n−1 1 2 d f J (a J + δ n−1 ε0 z 2 ) ≥ 1 2δ ε0 dz 7 |ϕ (ζ J )| for |z| ≤ 1/2, and hence by (6.5) | f J (z)| ≥ C

3J β δn

1 2

,

(6.10)

for z ∈ A J . It follows from (6.7) that ∂ V J ∩ D is approximately a semi-circle centered at ζ J , so that by the distortion theorem there is a constant c1 independent of J so that |τ J (w)| ≥ c1 for w ∈ f J (A J ), and so by (6.9), (6.10), and (6.11) 1 3J 2 β , | f (z)| ≥ C n δ for z ∈ A J . Set W = f (∪|J |=n A J ) ⊂ D.

(6.11)

(6.12)

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Then by the distortion theorem, (6.12), (6.3), and the binomial theorem, (1 − |w|) p−2+2ε |ϕ (w)|− p d xd y W = (1 − | f (z)|) p−2+2ε | f (z)| p+2 d xd y |J |=n

AJ

≥C

|J |=n

≥ Cδ nε

(δ n ) p−2+2ε | f (z)|2 p+2ε d xd y AJ

3J p+ε β

|J |=n

1

ε

≥ C δ (1 − α)

p+ε

N

p+ε 2n

βj

.

j=1

By (6.4) ε

δ (1 − α)

p+ε

N

p+ε

βj

>1

j=1

for ε sufﬁciently small, so that (1 − |w|) p−2+2ε |ϕ (w)|− p d xd y = +∞. D

(6.13)

1 Because 0 (1 − r )−1+ε dr < ∞ for ε > 0, (6.13) implies B(− p) > p − 1 and (c) ⇒ (a).

7. Baernstein’s Example on the Hayman–Wu Theorem Exercise VII.5 described the improvement, by Fernández, Heinonen and Martio [1989], of the Hayman–Wu theorem: There is p > 1 and C > 0 such that whenever is a simply connected domain, f : → D is conformal, and L is a line, | f | p d x ≤ C. (7.1) L∩

The example = D \ [0, 1) shows that (7.1) fails when p = 2. A natural parallel exists between (7.1) and the Brennan problem; and in Havin, Hruššëv, and Nikol’skii [1984], Baernstein conjectured that (7.1) should hold for all p < 2. However, in [1989] Baernstein employed a dandelion construction to show that (7.1) fails for some p < 2.

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303

Consider the tree 1

1

T = {(1 + 2− 2 )eiθ : 0 ≤ θ ≤ π } ∪ {i y : y ≥ 1 + 2− 2 } and its truncation 1

T0 = {(1 + 2− 2 )eiθ : 0 ≤ θ ≤ π } 1

at the fork point a0 = i(1 + 2− 2 ). Let δ > 0 be a constant to be determined later. Let T1 = T and by induction assume Tk has 2k ﬁnite extreme points a J , all of which fall on the line # k $ 1 δ − δ . Imz = −(1 + 2− 2 ) 1−δ Deﬁne &

Tk+1 = Tk ∪ a J + δ k (T0 − a0 ) , and note that the induction hypothesis then holds for Tk+1 . Set T∞ = Tk , 1 δ = C∗ \ T∞ , k = C∗ \ Tk ⊂ , and L = {Imz = −(1 + 2− 2 ) 1−δ }. Take conformal maps f : → D and f k : k → D such that f (−i) = f k (−i) = 0. This construction differs from the dandelion construction in Section 6 because T is not polygonal, but it is not hard to see that all of the estimates from Section 6 still hold. Alternatively, approximate T0 by a polygonal tree such that the ﬁnal 1 line segments ending at z = ±(1 + 2− 2 ) are vertical and such that inequality (7.3) below holds.

a0

R L Figure VIII.10

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Theorem 7.1. If δ is sufﬁciently small there exists p, 0 < p < 2 such that | f | p d x = ∞. (7.2) L 1

Proof. The domain 1 = C \ T has two β numbers, β1 = β(1 , 1 + 2− 2 , ∞) 1 and β2 = β(1 , −(1 + 2− 2 ), ∞). Set √ % 1 2z − 1 1+z , ψ3 (z) = z 2 − 1, ψ4 (z) = i(1 + 2− 2 ) , ψ1 = z 2, ψ2 = z+1 1−z and ψ = ψ4 ◦ ψ3 ◦ ψ2 ◦ ψ1 . Then ψ is a conformal map of H onto C \ T and lim

z→∞ 1

ψ(z) =1 z2

1

and ψ(±2− 4 ) = ±(1 + 2− 2 ). By explicit calculation, √ 1 1 |ψ (2− 4 )| = |ψ (−2− 4 )| = 8( 2 − 1), so that

√ 1+ 2 > 1. (7.3) A ≡ β1 + β2 = 2 We will use the notation from Section 6. Choose α > 0 so that (1 − α)A > 1. By Lemma 6.1 with p = 1, and the binomial theorem, 3J ≥ [(1 − α)A]n , (7.4) β |J |=n

for δ sufﬁciently small. Take such δ, ﬁx n, and for |J | = n set IJ = L ∩ AJ . Then |I J | ≥ Cδ n and the intervals {I J : |J | = n} are pairwise disjoint. By (6.12) p 3J 2 n β p | f | dx ≥ c n δ , δ IJ and by (7.4) if p ≤ 2 1 2n p p | f | dx = | f | p d x ≥ C δ 1− 2 (1 − α)A . ∪I J

|J |=n

IJ

But there exists p0 = p0 (δ) < 2 such that if p > p0 p

δ 1− 2 (1 − α)A > 1

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7. Baernstein’s Example on the Hayman–Wu Theorem / L p (L). and therefore f ∈

305

Notes See Zygmund [1959] and M. Weiss [1959] for more about lacunary series similar to (1.2). Makarov’s original [1985] proof of Theorem 1.1 used Fourier series instead of Hardy’s inequality. For Theorem 2.1 see Makarov [1987] and [1990]. Theorem 2.2 is from Makarov [1985], but we have followed the proof in Makarov [1987]. A different proof by Rohde [1988] can also be found in Pommerenke [1991]. In [1987] Makarov extended his compression theorem VI.5.1 to other Hausdorff dimensions. If ϕ is univalent on D and if E ⊂ ∂D, then for all p > 0, dimϕ(E) ≥

p dimE . βϕ (− p) + p + 1 − dimE

See Exercise 5 for the proof. Pommerenke’s proof of (4.1) for p < 3.399 given Appendix L below has been improved by Bertilsson [1998] and [1999] to yield p < 3.421. Shimorin’s recent paper [2003] establishes Brennan’s conjecture up to p < 3.7858; Hedenmalm and Shimorin [2004] have improved Shimorin’s estimate further. Deﬁne the bounded universal integral means spectrum to be Bb (t) = sup βϕ (t) : ϕ is univalent and ϕ(D) is bounded . Thus Bb (t) ≤ B(t). Makarov [1998] proves B(t) = max Bb (t), 3t − 1 . See Exercise 9. Consequently there is a transition point t ∗ such that

3t − 1, if t ≥ t ∗ , B(t) = Bb (t), if t < t ∗ , and by Exercise 7, 13 < t ∗ < 25 . By Exercise 9(a), Bb (t) = t − 1 for t ≥ 2. Much deeper is the Jones and Makarov [1995] result that Bb (t) = t − 1 + O(t − 2)2 , as t → 2, which implies Bb (2) = 1 and Bb (2) = 1. Near t = 0 the lower bound Bb (t) = B(t) ≥ 0.117t 2 , from Makarov [1986a] and Rohde [1989], complements the upper bound (4.7), and very recently the upper bound near t = 0 has been improved by Hedenmalm

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and Shimorin [2004] to Bb (t) ≤ 0.437t 2 . Carleson and Jones [1992] proved that # $ log |nan | :ϕ= an z n is bounded and univalent , Bb (1) = sup lim sup log n n→∞ and conjectured that Bb (1) = 41 . Using the computer program “Zipper”, see Marshall [1993], they obtained the estimate Bb (1) ≥ 0.21. Computer experiments by Kraetzer [1996] for polynomial Julia sets also support the Carleson– Jones conjecture, but the best results known to date without computer assistance are 0.17 ≤ Bb (1) ≤ 0.4884 from Pommerenke [1975], Duren [1983] and Grinshpan and Pommerenke [1997], though Hedenmalm and Shimorin [2004] have improved the upper bound to 0.46 with the assistance of computer calculations. Brennan’s conjecture and the Carleson–Jones conjecture would both follow from the beautiful and general conjecture in Kraetzer [1996]: t2 , for |t| ≤ 2. (BCJK) 4 Pommerenke [1999] calls (BCJK) the Brennan–Carleson–Jones–Kraetzer conjecture. Note that g(t) = t 2 /4 is the only quadratic polynomial that satisﬁes both g(0) = g (0) = 0 and g(2) = g (2) = 1. It is not yet known if Bb (t) = Bb (−t). Binder [1997], [1998a] studied the counterpart of (BCJK) for complex parameters t: If ϕ is bounded and univalent, is 1 |t|2 +ε 4 |ϕ (r eiθ )t |dθ ≤ C(ϕ, ε) 1−r Bb (t) =

for all ε > 0 and all t ∈ C with |t| < 2? Makarov [1998] characterized those functions β(t) that have the form βϕ (t) for some bounded univalent function ϕ. See Bertilsson [1999], Hedenmalm and Shimorin [2004], Makarov [1998], Pommerenke [1999], and the recent Jones [2005] for more complete discussions of Brennan’s conjecture, integral means spectra, and related topics.

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Exercises and Further Results

Exercises and Further Results 1. Let g(z) =

∞

k

k=0

z 2 , z ∈ D and Sn (ζ ) = lim sup √

n

k=0 ζ

2k ,

ζ ∈ ∂D. Prove

|Sn (ζ )| =1 n log log n

(1.3)

holds if and only if |g(r ζ )| lim sup ! = 1. 1 1 r →1 log 1−r log log log 1−r 2. Prove Hardy’s identity (1.6). 3. Pommerenke [1986b] shows that (1.1) fails with constant C = 0.685 by studying g(z) = C1

∞

n

z 15

n=1

with C1 chosen so that ||g||B = 1. 4. Let ∞ ∈ be a simply connected domain, let ζ0 ∈ ∂, let K ⊂ be a continuum, and let α > 0 and ε > 0. There is δ0 > 0 so that if 0 < δ < δ0 and ω(∞, B(ζ0 , δ), ) ≥ δ α , then there is a crosscut β ⊂ ∂ B(ζ0 , 2δ) of such that α+ε 1 dist (K , β) ≤ log . π δ Hint: Use the proof of Lemma 2.5. 5. Let ϕ be a conformal map from D to a simply connected domain . The following results are in Makarov [1987]. See also Pommerenke [1991]. (a) Let 0 < t ≤ 1. Show that Lemma 2.5 is still true when the conclusion (2.8) is replaced with Mr t (ϕ −1 (D ∩ ∂ \

N &

β j )) < Ct r M .

j=1

Hint: Follow the proof of Lemma 2.5 but use Pﬂuger’s theorem V.3.4. and Corollary D.3 from Appendix D to replace (2.9) with Mr t (α j ) ≤ Ct e−π λ( j ) . (b) For all Borel E ⊂ ∂D and all p > 0, dimϕ(E) ≥

p dimE . βϕ (− p) + p + 1 − dimE

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We may assume is a Jordan domain. Take t < 1, set s=

pt >t βϕ (− p) + p + 1 − t

and τ=

ps <s<1 p+s

and assume Mr s (ϕ(E)) = 0. Cover ϕ(E) by discs Dν = B(ζν , rν ) so that s 2π 1 rν < ε. For M = s let γ jν and β jν , 1 ≤ j ≤ N (ν) ≤ log 2 log( rν ), be the crosscuts and continua given by the version of Lemma 2.5 proved in (a). Then by (a)

Mr t (ϕ −1 (Dν ∩ ∂ \

ν

Nν &

β jν )) < Ct ε.

j=1

I jν

= and z νj = (1 − |I jν |)cνj where cνj is On the other hand, let the center of I jν . Then by Hölder’s inequality and Exercise IV.8, Nν

ϕ −1 (β jν )

τ

|I jν ||ϕ (z νj )|

j=1

Nν τ

1 1−τ ≤ C log( ) diamγ jν rν j=1

1 1−τ τ ≤ C log( ) rν rν because the γ jν are disjoint subarcs of ∂ B(ζν , 2rν ). But taking r = 1 − |I jν | we also have ν −p ν −p ν − p−1 |ϕ(r eiθ )|− p dθ ≤ C|I jν |(−β(− p)− p−1−η) |I j | |ϕ (z j )| ≤ C|I j | for any η > 0. Together, these inequalities yield |I jν |t ≤ Cε, ν

j

which proves the assertion. (c). Use (b) and (4.7) to prove dimϕ(E) ≥ 1 whenever dimE = 1 and show this result implies Theorem VI.5.1. 6. Prove (4.5) and (4.6). 7. Let ϕ be a univalent function on D and let t > 2/5. Then 1 |ϕ (r eiθ )|t dθ ≤ C(t, ϕ) . (1 − r )3t−1 In other words, B(t) = 3t − 1. For t > 1/2, use Exercise I.23(d). For 2/5 < t ≤ 1/2, set p = 2/(2 − t), q = 2/t, and ﬁx an integer k ≥ 2 such

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309

1/k that 2 pt (1 − 1/k) > 1. Then write g(z) = ϕ(z k) and use Hölder’s inequality, but estimate |g(r eiθ )| pt (k−1) dθ and |g (r eiθ )|2 dθ with care. See Feng and MacGregor [1976] for the details. 8. (a) Let ϕ be a conformal mapping from D onto a simply connected domain such that ϕ(0) = 0. The domain and the map ϕ are said to be starlike if the segment [0, w] ⊂ whenever w ∈ . This holds if and only if zϕ (z) Re >0 ϕ(z) on D. More generally, ϕ and are called close-to-convex if there exists starlike ψ such that zϕ (z) Re >0 (E.1) ψ(z) on D, and ϕ is close-to-convex if and only if C \ is a union of closed halflines meeting only at their endpoints. See Duren [1983] or Pommerenke [1991] for the proofs of these two geometric results. (b) Let ϕ be starlike. Then for 0 < q < 2, |ϕ (z)|−q d xd y < ∞. D

Thus Brennan’s conjecture is true for starlike univalent functions. For the proof, use (E.1) and the facts that if F(z) is analytic on D and ReF(z) > 0, then F ∈ H p for all p < 1 by Exercise A.4 and |F(z)| ≤ C/(1 − |z|) by the Schwarz lemma. (c) If ϕ is close-to-convex and 43 < q < 4, then |ϕ (z)|−q d xd y < ∞. D

Hint: Extend the proof of (b). Brennan [1978] attributes this result to B. Dahlberg and J. Lewis. (d) If ϕ is close-to-convex and E ⊂ ∂D is a Borel set, then dimE . 2 − dimE (e) If 0 < α < 1, there is a starlike domain bounded by a rectiﬁable Jordan curve and E ⊂ ∂D such that α (E) > 0 and α dimϕ(D) ≤ . 2−α See Makarov [1987] for (d) and (e). dimϕ(E) ≥

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9. The bounded universal integral means spectrum is # $ Bb (t) = sup βψ (t) : ψ is univalent and ψ(D) is bounded . Thus Bb (t) ≤ B(t). (a) Suppose = ψ(D) is bounded. Then

1 4 dist ψ(z), ∂ |ψ (z)| ≤ = o 1 − |z|2 1 − |z| and

1 1 = O . |ψ (z)| 1 − |z| Consequently when ψ is bounded, βψ (t ± h) ≤ βψ (t) + s, s > 0. On the other hand, for r ≥ 21 , iθ 2 (1 − r ) |ψ (r e )| dθ ≤ 2

1 2π

r

|ψ (teiθ )|2 dθ tdt ≤ 2π Area(),

0

so that βψ (2) ≤ 1. It follows that Bb (t) ≤ t − 1 for t ≥ 2. Pommerenke [1999]. (b) Let ⊂ C∗ be a simply connected domain such that ∂ is bounded and let ϕ : D → be conformal. Then βϕ (t) ≤ Bb (t), for all real t. Hint: Consider the restrictions of ϕ to two subarcs of the circle. (c) Let ψ be bounded and univalent and let δ = 1 − r be small. Partition δ and let z k be the center of Ik . Then {|z| = r } into arcs Ik with |Ik | = 2πr 1 a t−1 j |ψ (r eiθ )|dθ ∼ N (a j ), (E.2) δ j

where N (a) = #{Ik : |ψ (z k )| ∼ δ −a }, and aj =

j log 1δ

.

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Exercises and Further Results

Because the sum (E.2) has O log 1δ terms, it compares to its largest term and we obtain

log N (a j ) . (E.3) βψ (t) = lim sup sup a j t − 1 + log 1δ δ→0 j Moreover, let = ψ(D), let α ≥ 21 , and deﬁne f (α) = lim lim sup η↓0

log N (ρ, α, η)

ρ↓0

log ρ1

,

where N (ρ, α, η) = #{disjoint B = B(ζ, ρ), ζ ∈ ∂, ρ α+η ≤ ω(B) ≤ ρ α−η }. Then for α =

1 1−a ,

f (α) log N (a) . ∼ α log 1δ

(E.4)

Hence (E.3) gives βψ (t) = sup α

and

f (α) − t −1+t α

F(α) − t Bb (t) = sup −1+t α α

(E.5)

where F(α) =

sup

bounded

f (α).

It is clear that f (α) and F(α) are nondecreasing. See Makarov [1998] for (E.5) and several related results. (d) Makarov [1998] also showed (E.6) B(t) = max Bb (t), 3t − 1 . Therefore there is a transition point t ∗ such that

3t − 1, if t ≥ t ∗ , B(t) = Bb (t), if t < t ∗ , and by Exercise 7, 13 < t ∗ < 25 . To prove (E.6), we can assume by (b) that ϕ = ψ1 where ψ is a bounded univalent function such that ψ(0) = 1. We consider the arcs Ik and their

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centers z k from (c). For 0 < b ≤ 2 and −1 < a < 1 be of the form deﬁne

j log

1 δ

and

N (a, b) = #{(a, b) : |ψ(z k )| ∼ δ b , |ψ (z k )| ∼ δ −a }. Then

|ϕ (r eiθ )|t dθ ∼

1 (a+2b)t−1 (a,b)

δ

N (a, b).

In this sum the number of terms is O (log 1δ )2 , and therefore we have

log N (a, b) βϕ (t) = lim sup sup (a + 2b)t − 1 + . log 1δ δ→0 (a,b)

(E.7)

To estimate N (a, b), ﬁx a, b, and δ. By Lemma 2.5 we can assume {z : |z| = 1 − δ, |ψ(z)| ∼ δ b , |ψ (z)| ∼ δ −a } is an arc I of length |I | = N (a, b)δ = δ τ and center c I . Let ρ = δ 1−a−b , let z I = (1 − |I )c I , and deﬁne T (z) =

z + zI 1 + zI z

and (z) = ψ

ψ(T (z)) . δb

| ∼ 1 and |ψ | ∼ δ τ −a−b . Partition T −1 (I ) into N (a, b) Then on T −1 (I ), |ψ 1−τ these arcs of diameter (and harmonic measure) ∼ δ 1−τ = ρ 1−a−b . Under ψ 1−a−b = ρ. It follows that arcs have images of diameter ∼ δ log N (a, b) log

1 δ

≤ (1 − a − b)F

1−τ . 1−a−b

(E.8)

By the Beurling projection theorem the worst case of (E.8) is given by τ = b2 , and then (E.7) and (E.5) yield (E.6). 10. (a) Let T be a polygonal tree with ﬁnite extreme points a1 , a2 , . . . , a N . Suppose that f : D → = C \ T is a conformal map such that f (ζk ) = ak . Prove that 2π N 1 dθ 1 . = lim (1 − r 2 ) iθ 2 r →1 | f (r e )| 2π | f (ζk )|2 0 k=1

Prove an analogous statement with 2 replaced by p.

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Exercises and Further Results

(b) Set βk = β(, ak , ∞), where = C \ T and let ψ : H → be a conformal map such that ψ(z)/z 2 → 1 as z → ∞. For ﬁxed t > 0, deﬁne 1+w f (w) = f t (w) = ψ it · : D → . 1−w Prove lim lim (1 − r )

t→∞ r →1

2 N f (0) dθ = 64 βk2 . f (r eiθ ) 2π

2π

2

0

Prove

k=1

lim lim (1 − r )

r →1 t→∞

2 f (0) dθ = 64. f (r eiθ ) 2π

2π

2

0

f (0))/ f (0)

tends to the Koebe function uniformly on Note that ( f (w) − compact subsets of the disc as t → ∞. Consequently, if the Koebe function is a local maximum for the mean of 1/| f |2 on each circle of radius r < 1, then Brennan’s conjecture is true. See Bertilsson [1999], p. 85, for the best result in this direction. 11. Let ψ : H → be a conformal mapping. (a) For every x ∈ R and every α > 0, the nontangential limit z − x Jψ (x) = lim α (x)z→x ψ (z) exists and is ﬁnite. Write α (∞) = {x + i y ∈ H : |x| ≤ αy}. Then also ψ (z) Jψ (∞) = lim α (∞)z→∞ z exists and is ﬁnite. (b) If Jψ (x) > 0 and Jψ (∞) > 0, then limα (x)z→x ψ(z) = a exists and is ﬁnite, and limα (∞)z→∞ ψ(z) = ∞, and β(, a, ∞) = Jψ (x)Jψ (∞). (c) Conversely, if a ∈ ∂ and β(, a, ∞) > 0, then there exists ψ : H → such that limα (0)z→0 ψ(z) = a, Jψ (0) > 0, and Jψ (∞) > 0. See Bertilsson [1999] for (a), (b), and (c). 12. For p > 0, B(− p) = p − 1 if and only if there is a constant C such that for all normalized univalent functions ϕ(z) = z + a2 z 2 + . . . and all r < 1, 1 p−1 |ϕ (r eiθ )|− p dθ ≤ C . 1−r

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This observation is due to Bertilsson [1999]. Its proof is included in the argument following (5.12) together with (c) implies (a) of Theorem 5.3. 13. (a) It is a deep and difﬁcult theorem of Astala [1994] that if F : D → D is p K -quasiconformal, then for p = K2K −1 , the Jacobian J F is weak L : C . λp Eremenko and Hamilton [1995] have an alternate proof, also quite deep. (b) Let be a simply connected domain. Assume there exists G : → D such that G is K -quasiconformal and locally Lipschitz, Area{z ∈ D : J F (z) > λ} ≤

lim sup w→z

|G(z) − G(w)| ≤ M < ∞, |z − w|

for all z ∈ . Then whenever ψ : → D is conformal map and p < |ψ (z)| p d xd y < ∞.

2K K −1 ,

G −1 , change variables and use (a). In particular, Bren-

Hint: Write F = ψ ◦ nan’s conjecture is true if, for every ε > 0 and every simply connected domain , there is a locally Lipschitz 2 + ε quasiconformal map of onto D. See Bishop [2002a] and [2002b]. (c) Suppose ϕ : D → is a conformal map, and suppose there exists a K -quasiconformal h : D → D such that h is bilipschitz with respect to the hyperbolic metric on D and such that for some C < ∞, 1 − |h(z)| ≤ C|ϕ (z)|(1 − |z|) for all z ∈ D. Then G = h ◦ ϕ −1 satisﬁes the hypotheses in part (b). See Bishop [2002a] and Exercise 2, Appendix J.

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IX Inﬁnitely Connected Domains

Let be a domain such that ∞ ∈ and Cap(∂) > 0. In this chapter we study the relation between Hausdorff measures and harmonic measure on ∂. When is ﬁnitely connected, the problem is well understood by Theorem VII.2.2 and by simple comparisons, and thus we assume that ∂ has inﬁnitely many components. In Section 1 we give the proof by Batakis of the Makarov–Volberg theorem that if = C \ K and K is a certain type of Cantor set then there exists F ⊂ K such that ω(F, ) = 1 but dimHaus (F) < dimHaus (K ). In Section 2 we prove the result of Carleson [1985] and Jones and Wolff [1986] that if K = C∗ \ is totally disconnected and satisﬁes some reasonable geometric hypotheses then there is F ⊂ K such that ω(F, ) = 1 but dimHaus (F) < 1. Then in Section 3 we discuss the [1988] Jones–Wolff theorem that for every there exists F ⊂ ∂ with ω(F) = 1 but dimHaus (F) ≤ 1.

1. Cantor Sets Fix numbers 0 < a < b < 1/2. For every sequence {an } with a ≤ an ≤ b

(1.1)

form the Cantor set K = K ({an }) as follows: + K = Kn , where K 0 = [0, 1] × [0, 1], K 1 = 4j=1 Q 1j ⊂ K 0 , and each Q 1j is a square of side σ1 = a1 containing a corner of Q 0 = K 0 . At stage n & Q nJ , Kn =

315

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where Q nJ is a square of side σn = a1 a2 · · · an having its sides parallel to the axes, and J = ( j1 , j2 , . . . , jn ) is a multi-index of length |J | = n with n+1 jk ∈ {1, 2, 3, 4}. Then Q nJ ⊃ {Q n+1 J, j : j = 1, . . . , 4}, and the Q J, j are components of K n+1 containing the four corners of Q nJ . See Figure IX.1. n V(J,3) n W(J,3)

Q n(J,1)

Q n−1 J

Q n(J,4)

Q n+1 (J,2,3)

Q n(J,2) Figure IX.1

It is clear from (1.1) and Chapter III that Cap(K ) > 0. It is also clear that K ({an }) has Hausdorff dimension α({an }) = sup β > 0 : lim inf 4n σnβ = ∞ n→∞ (1.2) = inf β > 0 : lim inf 4n σnβ = 0 n→∞

because the Hausdorff measure β (K ) can be estimated using coverings by the squares Q nJ . Set = C∗ \ K and write ω(F) = ω(∞, F, ). Theorem 1.1. There exists F ⊂ K such that ω(F) = 1

(1.3)

dimHaus (F) < dimHaus (K ).

(1.4)

but

The proof of Theorem 1.1 depends on four lemmas. Write K Jn = K ∩ Q nJ and take the squares W Jn = (1 + c1 )Q nJ V Jn = (1 + c2 )Q nJ

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concentric with Q nJ , where 0 < c1 < c2 are constants that depend only on a and b and c2 is so small that for some constant c3 > 0 dist(V Jn1 , V Jn2 ) ≥ c3 σn > 0 whenever J1 = J2 . Lemma 1.2. There exist constants δ1 > 0 and δ2 > 0, depending only on a and b, such that for all z ∈ ∂ W Jn ω(z, K Jn , ) ≥ ω(z, K Jn , V Jn \ K Jn ) ≥ δ1 ,

(1.5)

and for all compact E ⊂ K Jn ω(z, E, ) ≥ ω(z, E, V Jn \ K Jn ) ≥ δ2 ω(z, E, ).

(1.6)

Proof. Let g(z, w) be Green’s function for V Jn . Then by a change of scale there are constants C1 and C2 such that g(z, ζ ) > C1

(1.7)

whenever z ∈ ∂ W Jn and ζ ∈ K Jn , and g(z, ζ ) ≤ log

1 + log σn + C2 |z − ζ |

(1.8)

whenever z, ζ ∈ K Jn . Let μ be the equilibrium distribution for K Jn and consider the Green potential U (z) = g(z, ζ )dμ(ζ ). Then U (z) is a positive harmonic function on V Jn \ K Jn . Since Cap(K Jn ) ≥ C3 σn by Chapter III, (1.8) gives 0 < U (z) ≤ C4 on K Jn . On the other hand, U (z) = 0 on ∂ V Jn . Therefore on V Jn \ K Jn we have ω(z, K Jn , ) ≥ ω(z, K Jn , V Jn \ K Jn ) ≥ U (z)/C4 ≥

C1 C4

C1 by (1.7), and (1.5) holds with δ1 = C . A very similar argument was used in 4 Chapter III. In (1.6) the leftmost inequality is obvious. To prove the rightmost inequality take z 0 ∈ ∂ W Jn so that ω(z 0 , E, ) = sup∂ W n ω(z, E, ). Then by the maxiJ mum principle and (1.5) n n ω(z 0 , E, V J \ K J ) = ω(z 0 , E, ) − ω(ζ, E, )dω(z, ζ, V Jn \ K Jn ) ∂ V Jn

≥ ω(z 0 , E, ) − (1 − δ1 )ω(z 0 , E, ) = δ1 ω(z 0 , E, )

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and (1.6) follows by Harnack’s inequality.

Choose the index jn+1 so that for every J = ( j1 , j2 , . . . , jn ), Q n+1 (J,1) is the

upper right corner of Q nJ and Q n+1 (J,4) is the lower left corner. Lemma 1.3. There is η > 0 such that for all n ≥ 2,

n n ω(∞, K 1,1,...,1,1 , ) ≥ (1 + η)ω(∞, K 1,1,...,1,4 , ).

Proof. For j = 1, 4 write n u j (z) = ω(z, K 1,1,...,1, j , )

and n−1 ∗ n U j (z) = ω(z, K 1,1,...,1, j , C \ K 1,1,...,1 ). n−1 n n that separates K 1,1,...,1,1 from K 1,1,...,1,4 Let L be the diagonal of K 1,1,...,1 n and let H be the half-plane bounded by L and containing K 1,1,...,1,4 . Then on n−1 W = H \ K 1,1,...,1 n U4 (z) − U1 (z) = ω(z, K 1,1,...,1,4 , W ).

L H n K (1,...,1)

n K (1,...,4)

n ∂ W(1,...,4)

W

Figure IX.2 The proof of Lemma 1.3. Therefore U4 − U1 > 0

(1.9)

U4 − U1 > c4 > 0

(1.10)

n−1 on W , while on K 1,1,...,1,4

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with constant c4 independent of n, by Lemma 1.2 and Harnack’s inequality. Now u 1 (z) − u 4 (z) = U1 (z) − U4 (z) − (U1 (ζ ) − U4 (ζ ))dω(z, ζ, ). n−1 K \K 1,1,...,1

Hence by (1.9) and (1.10) n−1 u 1 (∞) − u 4 (∞) ≥ c4 ω(∞, K 1,1,...,1,4 , ).

On the other hand, by Lemma 1.2 and Harnack’s inequality n−1 , ) ≥ c5 ω(z, K 1,1,...,1,4 n on ∂ W1,1,...,1,4 . Therefore u 1 (∞) ≥ (1 + c4 c5 )u 4 (∞) by the maximum prin ciple and the lemma holds with η = c4 c5 .

The next lemma is from Carleson [1985] and Makarov and Volberg [1986]. Lemma 1.4. Let be a domain such that ∞ ∈ , and let A1 ⊂ B1 ⊂ A2 ⊂ B2 ⊂ . . . An ⊂ Bn be simply connected Jordan domains such that for each j j = Bj \ A j ⊂ and mod( j ) ≥ α > 0.

(1.11)

Let u and v be positive harmonic functions on such that u = v = 0,

on

∂ \ A1 .

Then there exists C = C(α) > 0 and q = q(α) < 1 such that on \ Bn , u(z)4v(z) 4 (1.12) − 1 ≤ Cq n . u(∞) v(∞)

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∂ B3

1

3 ∞

2

∂ A3

Figure IX.3 Proof. By (1.11) there is a conformal mapping ϕ j : j → {1 < |z| < 1 + β j } with ϕ(∂ A j ) = {|z| = 1} and β j ≥ β = e2π α − 1. For k = 1, 2 set

kβ j γ j(k) = ϕ j−1 {|z| = 1 + } 3 and

4 max{u(z) v(z) : z ∈ γ j(k) } = . 4 min{u(z) v(z) : z ∈ γ j(k) }

a j(k)

By Harnack’s inequality there is C = C(α) such that (k)

1 ≤ aj

≤ C.

We claim there is δ3 > 0 such that (2)

aj

(1)

≤ aj

(1)

1 − δ3 (a j

− 1) .

(1.13)

Let us accept (1.13) for the moment and complete the proof of (1.12). By the maximum principle, (1)

(2)

(1)

a1 ≥ a1 ≥ a2 ≥ . . . ≥ an(2) (2)

because u = v = 0 on ∂ \ A1 . Write a j

= 1 + b j . Then by (1.13)

1 + b j+1 ≤ (1 + b j )(1 − δb j ) ≤ 1 + (1 − δ)b j

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and bn ≤ C(1 − δ)n−1 . Since on \ Bn 5 u(z) u(∞) −1 (1 + bn ) ≤ ≤ (1 + bn ), v(z) v(∞) we therefore obtain (1.12). To prove (1.13) we can assume minγ (1) j

u(z) v(z)

(1)

= 1. Then on γ j ,

(1)

v(z) ≤ u(z) ≤ a j v(z) = (1 + b)v(z) where b > 0 by Harnack’s inequality. Set b ∩ {u ≤ (1 + )v(z)} 2 b (1) E 2 = γ j ∩ {u > (1 + )v(z)} 2 (1)

E1 = γj

(1)

and let λz denote harmonic measure at z for the component of \ γ j taining

γ j(2) .

con-

γ j(2) ,

Since u = v = 0 on ∂ \ A1 we have, for z ∈ b v(ζ )dλz (ζ ) + (1 + b) v(ζ )dλz (ζ ) u(z) ≤ (1 + ) 2 E1 E2

and

v(z) =

E 1 ∪E 2

v(ζ )dλz (ζ ).

Hence b u(z) ≤ (1 + b)v(z) − λz (E 1 ) min v(ζ ) E1 2 and since by Harnack’s inequality v(z) ≤ C1 min E 1 v(ζ ), with C1 = C1 (α), we obtain C1 bλz (E 1 ) (1) . u(z) ≤ v(z)a j 1 − 2(1 + b) Similarly, b u(z) ≥ v(z) + λz (E 2 ) min v(ζ ) E2 2 ≥ v(z) 1 + C2 (α)λz (E 2 ) , and we conclude that (2)

aj

(1)

≤ aj

(1)

1 − C3 (α)bλz (γ j ) ≤ a j(1) 1 − C4 (α)b .

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Lemma 1.5. There is N = N (a, b) such that for every n and J there exists J = J (J ) = (J, jn+1 , . . . , jn+N ) such that 1 ω(K Jn ) . 3 4N Proof. Fix J and N and for j = 1, 4 write )≤ ω(K Jn+N

(1.14)

n+N v j (z) = ω(z, K J,1,...,1, j , )

and n+N n n Vj (z) = ω(z, K J,1,...,1, j , V J \ K J ).

Let z ∈ ∂ W Jn . By Lemma 1.4, V (z)4V (z) V (z)4V (z) 4 4 1 1 4 4 − 1 ≤ − 1 + Cq N . v1 (z) v4 (z) v1 (∞) v4 (∞) But V (z)4V (z) v4 (z) V1 (z) V4 (z) 4 1 4 − 1 = − V4 (z) v1 (z) v1 (z) v4 (z) v4 (z) v4 (z) v4 (ζ ) v1 (ζ ) − = dω(z, ζ, V Jn \ K Jn ) V4 (z) ∂ V Jn v4 (z) v1 (z) 4 v4 (z) v4 (ζ ) v1 (ζ ) v1 (z) 4 = − 1 dω(z, ζ,V Jn \ K Jn ) V4 (z) ∂ V n v4 (z) v4 (ζ ) v4 (z) J

1 ≤ (1 + Cq N )Cq N , δ2 by Lemmas 1.2 and 1.4. If we take J = (1, 1, . . . , 1) we get v1 (∞) ≥ 1 + η, v4 (∞) by Lemma 1.3 and if N is large, η V1 (z) ≥1+ V4 (z) 2 by the two preceding estimates. However, VV41 (z) (z) does not depend on J . Therefore we have v1 (∞) η >1+ v4 (∞) 4

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1. Cantor Sets

for all J if N is large. Consequently inf J

ω(K Jn+N ) 4 N +1 −N ≤ N +1 4 ω(K Jn ) 4 +η

and (1.14) holds if N is replaced by N p , where p = p(η, N ).

Conclusion of Proof of Theorem 1.1. On any K Jn take two probabilities P1 and P2 such that 1 −N −N and P2 (K Jn+N . 4 (J ) ) = 4 3 Write J = Jn (z) when z ∈ K Jn . Let n 1 be large, take n j+1 = n j + j and set P1 (K Jn+N (J ) ) =

(n +m)N +N

#

Ej = z ∈ K :

j #{m ≤ j : z ∈ K J (J (z))

}

j

≥

2 −N $ 4 . 3

By the elementary theory of large deviations, Durrett [1996], pp. 70-76, there are constants C1 and C2 > 0 (depending on N ) such that if n 1 is large P1 (E j ) ≤ e−C1 j

(1.15)

P2 (E j ) ≥ 1 − e−C2 j .

(1.16)

and

Set F =

∞

k=1

Ak , where Ak =

∞ +

K \ Ej .

(1.17)

j=k

Then K\F=

∞ ∞ & +

Ej ,

k=1 j=k

and by (1.14) and (1.15) ω(K \ F) ≤ lim

k→∞

∞

e−C1 j = 0.

j=k

On the other hand, since the E j are independent it follows from (1.16) and (1.17) that for j > k, Ak can be covered by eC2 (n k −n j ) 4n j N squares of side σn j N . Let α = dimHaus (K ), let a be as in (1.1), and let 0 < γ < α satisfy e−C2

1 N (β−γ ) a

<1

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for some β > α. Then by (1.1) and (1.2), γ

γ (Ak ) ≤ lim eC2 (n k −n j ) 4n j N σn j N = 0. j→∞

Therefore γ (F) = 0.

2. For Certain , dim ω < 1 Let K ⊂ C be compact and let = C∗ \ K . We assume there are constants α, C1 , and C2 such that for all z 0 ∈ K and all 0 < r < diam(K ) Cap(B(z 0 , r ) ∩ K ) ≥ C1r,

(2.1)

and assume there is a ring domain Ar = A(z 0 , r ) ⊂ ∩ {r < |z − z 0 | < αr } bounded by rectiﬁable Jordan curves 1 and 2 such that ( j ) ≤ C2 r

(2.2)

dist(1 , 2 ) ≥ αr.

(2.3)

and

Ar

r/α

r

Figure IX.4 Theorem 2.1. Assume K satisﬁes (2.1), (2.2), and (2.3). Then there exists δ = δ(α, C1 , C2 ) > 0 and there exists F ⊂ K such that ω(∞, F, ) = 1 but dimHaus (F) ≤ 1 − δ.

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325

The proof of Theorem 2.1 consists of three steps. We ﬁrst construct a dyadic approximation of K by sets Q n, j called cubes. Next we employ a stopping time to select certain of the cubes Q n, j . Then a lower bound for the harmonic measure of the stopped Q n, j is obtained from an analysis of the critical points of Green’s function. Assume diam(K ) = 1 and set ω(E) = ω(∞, E ∩ K , ). Throughout the proof c and C denote constants which may change with each occurrence but which depend only on α, C1 , and C2 . Lemma 2.2. For n = 1, 2, . . . there exist simply connected Jordan domains Q n, j such that & K ⊂ Q n, j , (2.4) j

Q n, j ⊂ {z : dist(z, K ) < 4−n },

(2.5)

dist(∂ Q n, j , K ) ≥ c4−n ,

(2.6)

diam(Q n, j ) ≤ c

4−n , α

(2.7)

dist(Q n, j , Q n,k ) ≥ c4−n ,

(2.8)

dist(∂ Q n, j , ∂ Q n+1,k ) ≥ c4−n .

(2.9)

and

Moreover, ∂ Q n, j is a rectiﬁable Jordan curve of class C 3 with arc length parameterization z = z(s) satisfying d3z d2z (2.10) 2 ≤ C4n and 3 ≤ C42n . ds ds

Q n+1,k

Figure IX.5

Q n, j

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Proof. Let {Sn, j } denote the components of {z : dist(z, K ) < 4−n } and let z 0 ∈ K ∩ Sn, j . Then by (2.3) Sn, j is contained in the bounded component of −n C∗ \ A(z 0 , 4α ), and for k = j, Sn,k ∩ A(z 0 ,

4−n ) = ∅. α

By (2.2), (2.3), and Exercise IV.5(d), mod(A(z 0 ,

log Rn, j 4−n )) = ≥ Cα. α 2π −n

Let f n, j be a conformal mapping from {1 < |z| < Rn, j } onto A(z 0 , 4α ) and let % −n γn, j be the core curve f n, j ({z : |z| = Rn, j }) of A(z 0 , 4α ). Then by Koebe’s theorem there is c = c(α) such that dist(γn, j , K ) ≥ c4−n

(2.11)

and γn, j satisﬁes the smoothness condition (2.10). Let {Un, j } be the bounded components of the complements of the components of γn, j . Then each Un, j is simply connected. Let {Vn, j } be the set of Un, j such that K ∩ Un, j = ∅ and Un, j is maximal with respect to set inclusion. Then by (2.11) and the smoothness of γn, j there exists Q n, j ⊂ Vn, j satisfying (2.4)–(2.10) such that K ∩ Vn, j ⊂ Q n, j . Like dyadic squares, the cubes Q n, j have the property Q m,k ⊂ Q n, j or Q m,k ∩ Q n, j = ∅ if m > n. Furthermore K = ∂ =

+& n

Q n,k .

k

Now ﬁx integers M and m and set B M = Q n,k : n ≤ m, ω(Q n,k ) ≥ M4−n and Q n,k is maximal , G M = Q m, j : Q m, j ⊂ Q n,k for all Q n,k ∈ B M and = C∗ \

& & (Q n,k ) ∪ (Q m, j ) ⊂ . BM

GM

= g ). ω(E) = ω(∞, E, Write G(z) (z, ∞) and

(2.12)

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327

Lemma 2.3. There is a constant c = c(α) such that if Q n, j ∈ B M ∪ G M , then ω(Q n, j ) ≤ cω(Q n, j ),

(2.13)

, and on ∂ ∂G ≤ cM. (2.14) ∂n n, j ⊂ Q n, j and n, j ⊃ Q n, j with K ∩ Q Proof. By Lemma 2.2 there is open Q −n n, j by n, j , Q n, j ) ≥ c4 . Then ω(z, K ∩ Q n, j , ) ≥ c at all z ∈ ∂ Q dist(∂ Q (2.1) and the proof of Lemma 1.2. Then (2.13) follows by the maximum prinn, j . ) on ∂ \Q ciple, since ω(z, Q n, j , ) ≥ 0 = ω(z, Q n, j , −n+1 for all Q n,k ∈ B M ∪ G M and by (2.10) By construction ω(Q n,k ) ≤ M4 sup

∂ Q n,k

∂G ∂G ≤ c inf . ∂ Q n,k ∂n ∂n

Hence by (2.13) and Lemma 2.2, sup

∂ Q n,k

∂G 2π c ≤ ω(∂ Q n,k ) ≤ cM, ∂n (∂ Q n,k )

so that (2.14) holds. The heart of the proof of Theorem 2.1 is the identity ∂G ∂G 1 j ), log ds = γ + G(z 2π ∂ ∂n ∂n

(2.15)

j

and γ = − log Cap(∂ ) is Robin’s in which {z j } is the set of critical points of G constant for ∂ . See Exercise III.2 for the proof. We assume Cap(K ) < 1 and take M so large that γ > 0 in (2.15). Then (2.14) and (2.15) yield j ) ≤ log(cM). (2.16) G(z j

Now let N = N (α) be a positive integer to be determined later and deﬁne U N = Q n, j : Q p,k ⊂ Q n, j and Q p,k ∈ B M ∪ G M ⇒ p > n + N . If N (α) is sufﬁciently large, it follows from (2.1), (2.6), and (2.7) that whenever Q n, j ∈ U N there exist two disjoint cubes Q n+N ,k1 , Q n+N ,k2 ∈ / B M ∪ G M such that Q n+N ,k1 ∪ Q n+N ,k2 ⊂ Q n, j . Lemma 2.4. There exist constants c and N such that

(2.17)

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(a) If Q n, j ∈ U N then ≥ c inf G(z) ω(Q n, j ).

(2.18)

∂ Q n, j

(b) If Q n+N ,k ⊂ Q n, j , then ≤ c inf G ∂ Qn, j

≤ 1 inf G. sup G 2 ∂ Q n, j ∂ Q n+N ,k

(2.19)

n, j \ Q n, j ⊂ n, j ⊃ Q n, j such that Q and Proof. (a) There is Q n, j , ∂ Q n, j ) ≤ c dist(∂ Q n, j , ∂ ). dist(∂ Q n, j and by Harnack’s in) ≥ c on ∂ Q By (2.1) and Lemma 2.2, ω(z, Q n, j ∩ ∂ n, j . Hence ζ ) > C for all z ∈ Q n, j and ζ ∈ ∂ Q equality and a comparison G(z, for z ∈ Q n, j , G(z, ζ )dω ω(Q n, j ). G(z) ≥ n, j (∞, ζ ) ≥ c \ Q n, j ∂Q

(b) Write Q n, j = Q 0 ⊃ Q 1 ⊃ Q 2 ⊃ . . . ⊃ Q N = Q n+N ,k where Q p = Q n+ p,k for some k. By (2.1) and Lemma 2.2 there is η > 0 such that for all p and all z ∈ ∂ Q p , ) ≤ 1 − η. η < ω(z, ∂ Q p−1 , Q p−1 ∩ Hence by induction ) ≤ (1 − η) N . η N < ω(z, ∂ Q 0 , Q 0 ∩ Also, by Lemma 2.2 and Harnack’s inequality, ≤ C inf G(z), sup G(z) ∂ Q0

∂ Q0

= 0 on Q 0 ∩ ∂ , so that since G ≤ sup G ≤ C(1 − η) N inf G, cη N inf G ∂ Q0

∂ Q0

∂ QN

and if C(1 − η) N < then (2.19) holds.

1 , 2

(2.20)

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329

Fix N so that (2.17) and (2.20) both hold, and note that the lower bound in (2.19) holds with c = c(α, N ). such that Lemma 2.5. If Q n, j ∈ U3N , there is a critical point z n, j of G & z n, j ∈ Q n, j \ Q n+3N ,k and n, j ) ≥ c ω(Q n, j ). G(z Proof. There are 4 cubes Q n, j ⊃ Q n+N , j1 ⊃ Q n+2N , j2 ⊃ Q n+3N , j3 , which / B M ∪ G M and by Lemma 2.4 we relabel as Q 1 ⊃ Q 2 ⊃ Q 3 ⊃ Q 4 , so that Q k ∈ < 1 inf G. sup G 2 ∂ Qk ∂ Q k+1 < b} ⊂ . b = inf ∂ Q 4 G(z), and U = Q 1 ∩ {a < G Take a = sup∂ Q 1 G(z), 2 3 = a} and By (2.17) and (2.19) Q \ Q contains a component of ∂U ∩ {G in U and by (2.18) hence by Exercise II.17 there is a critical point z n, j of G and (2.19) n, j ) ≥ c inf G ≥ c G(z ω(Q n, j ). ∂ Q n, j

Proof of Theorem 2.1. By (2.8) and (2.12) each Q p, j ∈ B M ∪ G M falls in p − N cubes Q n,k ∈ U M . Therefore by (2.16) and (2.18), c n ω(Q n,k ) + c m ω(Q m,k ) ≤ log(cM), BM

GM

and we can take m large and get

ω(Q m,k ) ≤

GM

Set

so that

# $ ω(Q n,k ) B ∗M = Q n,k ∈ B M : ω(Q n,k ) ≤ 4

ω(Q n,k ) ≤

B M \B ∗M

Then

1 . 4

B ∗M

ω(Q n,k ) ≥

1 . 4

1 2

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IX. Inﬁnitely Connected Domains

and on B ∗M 1 ω(Q n,k ) ≤ ≤c 4 ω(Q n,k ) by (2.13). Therefore

ω(Q n,k ) ≥

B ∗M

1 . 2c

Each Q n, j ∈ B ∗M can be covered by a ball of radius c4−n ≤ rn, j ≤ C4−n by Lemma 2.2 and M4−n ≤ ω(Q n, j ) ≤ M4−n+1 by the deﬁnition of B M . Therefore if M is large, 1 C log M ≤ rn, j log rn, j M and

rn, j ≥

ω(Q n, j ) M

≥

1 . 2M

Choose β so that

rn, j ≥

rn, j ≥β

1 . 4M

Then β ≥ M −c

because by (2.21) log β1 2M

≤

rn, j log

rn, j <β

1 rn, j

≤

C log M . M

Consequently

(rn, j )1−δ ≤ β −δ

rn, j ≥β

while

rn, j ≥β

c ω(Q n, j ) ≤ C M 1−c δ , M

ω(Q n, j ) ≥ c.

(2.21)

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331

Taking δ > 0 small, we see there exists, for n = 1, 2, . . . a sequence Bn, j of balls of radii rn, j such that & ω( Bn, j ) ≥ c j

and

rn, j ≤ 2−n .

j

Then E=

∞ & ∞ & +

Bn, j

m=1 n=m j

satisﬁes ω(E) ≥ c and dimHaus (E) ≤ 1 − δ. By Harnack’s inequality there is c such that ω(z, E, ) ≥ c at all z ∈ ∪ j ∂ Q 1, j . Now for every Q n, j given by Lemma 2.2 this construction yields a set E n, j ⊂ K ∩ Q n, j such that inf ω(z, E n, j , ) ≥ c

∂ Q n, j

and dimHaus (E n, j ) ≤ 1 − δ. Now set F=

&

E n, j .

n, j

Then dimHaus (E) ≤ 1 − δ, and by the maximum principle ω(z, F, ) ≥ c for all z ∈ , which implies that ω(F) = 1.

3. For All , dim ω ≤ 1 Let K ⊂ C be a compact set of positive capacity and set = C∗ \ K .

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Theorem 3.1. There exists F ⊂ ∂ such that ω(F) = 1 and dimHaus (F) ≤ 1. The proof consists of two lemmas and a domain modiﬁcation construction. be closed squares with the same center ζ0 , with sides Lemma 3.2. Let Q ⊂ Q = R(Q). Assume parallel to the axes, and with sidelengths (Q) and ( Q) \ Q ⊂ Q and E = Q\ has Cap(E) > 0. Fix ε > 0 and let B be a closed disc with center ζ0 such that

1+ε = ((Q))−ε e−(1+ε)γ E . (3.1) e−γ B = Cap(B) = ((Q))−ε Cap(E) Set = ( ∪ Q) \ B and ). ω = ω(∞, · , Then if R = R(ε) is sufﬁciently large, ω(B) ≥ C(ε)ω(E),

(3.2)

ω(A) ≥ ω(A).

(3.3)

and for all A ⊂ ∂ \ Q,

A Q

A Q B

E Q

Q

Figure IX.6

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333

\ Q ⊂ . We may assume are open because Q Proof. The sets ∪ Q and Cap(∂ \ Q) > 0, and after a change of scale we may assume (Q) = 1. Let g(z, ζ ) = log

1 + h(z, ζ ) |z − ζ |

be Green’s function for ∪ E. Then

R − 1 log |ζ − ζ0 |dω\ Q(z, ζ ) ≥ log h(z, ζ0 ) = , 2 ∂\ Q

(3.4)

and |∇ζ h(z, ζ )| = O

1

R when ζ ∈ B ∪ E. Let μ E and μ B be the equilibrium distributions for E and B and ﬁx z 0 ∈ ∂ Q. Then on B,

1 u(z) = g(z, ζ )dμ B (ζ ) = γ B + h(z 0 , ζ0 ) + O , R B and on E,

1

g(z, ζ )dμ E (ζ ) = γ E + h(z 0 , ζ0 ) + O

v(z) = E

and u(z) ≥ v(z) + O Furthermore u = v = 0 on ∂ \ Q cause γ E > 0). Hence on \ Q, ) ≥ ω(z, B, ≥ ≥

u(z)

1

R

R

.

(beon \ Q

1

γ B + h(z 0 , ζ0 ) + O

v(z) + O R1

R

1

γ B + h(z 0 , ζ0 ) + O

R

1

γ E + h(z 0 , ζ0 ) + O

1 ω(z, E, ),

R

(1 + ε)γ E + h(z 0 , ζ0 ) + O

R

by (3.4), and with (3.4) that implies (3.2). To prove (3.3) we assume ζ0 = 0. Write U = {|z| < 1 − z ζ g(z, ζ ) = log z R ζR R − R

R 2 },

let

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be Green’s function for U , let g B be Green’s function for U \ B, and let g E be Green’s function for U \ E. Then on U \ B g(w, ζ )dωU \B (z, w) g B (z, ζ ) = g(z, ζ ) −

∂B

≥ g(z, ζ ) − log 2 + and on U \ E

C2 ω(z, ∂ B, U \ B), R

g E (z, ζ ) = g(z, ζ ) −

∂E

g(w, ζ )dωU \E (z, w)

C1 ω(z, E, U \ E). ≤ g(z, ζ ) − log 2 + R If |z| = R/2 and if R is large, then by hypothesis (3.1) ω(z, B, U \ B) ≤ ω(z, E, U \ E), and hence ∂g B ∂g E ≥ ∂nζ ∂nζ

(3.5)

at ζ ∈ bU. Now suppose ω(z, A) ω(z 0 , A) = sup = λ > 1. ω(z 0 , A) |z|=R/2 ω(z, A) Then by the maximum principle λ ω(z, A) − ω(z, A) > 0 on ∂U, because λ ω > ω on A. However, using (3.5) at z 0 then yields 0 = λ ω(z 0 , A) − ω(z 0 , A) 1 ∂g B (z 0 , ζ ) ∂g E (z 0 , ζ ) 1 = λ ω(ζ, A)ds(ζ ) − ω(ζ, A)ds(ζ ) 2π ∂U ∂n 2π ∂U ∂n 1 ∂g B (z 0 , ζ )

> λ ω(ζ, A) − ω(ζ, A) ds(ζ ) 2π ∂n ≥ 0, which is a contradiction, and therefore (3.3) follows from the maximum prin ciple. Lemma 3.3. Write g(z) = g (z, ∞). Suppose ⊂ {|z| < 1} is a ﬁnite union of Jordan curves that separate K = C \ from ∞ and suppose there are con-

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stants c1 , . . . c N , possibly not distinct, such that =

335 N

j=1 j

where

j ⊂ {g(z) = c j } is a Jordan curve. Then I () =

1 2π

∂g log |∇g|ds ≥ − log 2. ∂n

(3.6)

2

1 K

3

Figure IX.7 Proof. Replacing K by {g ≤ ε} for small ε, we can assume is a ﬁnitely connected domain with smooth boundary. Let ζk be the critical points of g that lie outside . Then by Green’s theorem, almost exactly as in Exercise III.2, 1 ∂ g(ζk ) + g log |∇g|ds + γ I () = 2π j ∂n k j cj ∂ g(ζk ) + log |∇g|ds + γ , = 2π j ∂n k

j

where γ = γ () ≤ γ (K ) is Robin’s constant for . Let α j be the (ﬁnite) number of critical points of g inside j and let β j be the (ﬁnite) number of components of K inside j . Then α j = β j − 1, by Exercise II.17 and ∂ 1 log |∇g|dz = α j − β j = −1 2π j ∂n by Green’s theorem. Therefore I () =

k

g(ζk ) −

j

cj + γ .

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Now write n(c) for the number of components of {g = c} outside . Then ∞ g(ζk ) = (n(c) − 1)dc 0

and j

∞

cj = 0

χ {c≤c j } (c)dc ≤

j

c0

n(c)dc, 0

where c0 = sup{c : n(c) > 1}. Hence I () ≥ γ − c0 . But since ⊂ {|z| < 1}, all critical points fall into {|z| < 1}, and if |z| < 1 then log |z − ζ |dω(ζ ) ≤ γ + log 2.

g(z) = γ + K

Hence c0 ≤ γ + log 2 and (3.6) follows.

Proof of Theorem 3.1. We now modify the domain = C∗ \ K . We assume K ⊂ {|z| < 1/2}. Fix ε > 0 and ﬁx an integer R so that R > 2 + R(ε) where R(ε) is given by Lemma 3.2. Choose a large constant M and a small constant ρ > 0 so that M ≤ log ρ1 . Consider the grid G of dyadic squares having side ρ and lower left corners (m + ni)ρ. Partition G into R 2 subfamilies G p,q , where 1 ≤ p, q ≤ R, by the rule G Q ∈ G p,q ⇐⇒ (m, n) ≡ ( p, q) mod (R × R). Write K p,q =

&

K ∩ Q,

G p,q

p,q = C∗ \ K p,q , and ω p,q (E) = ω(∞, E, p,q ). Then ω(E) ≤

ωi, j (E ∩ K p,q ),

and to prove Theorem 3.1 we may assume at the start that = p,q for some p, q. To modify = p,q we alternate two constructions:

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disc construction: Let Q be a square of side . Replace E = K ∩ Q by a closed ball B with the same center as Q and with capacity

1+ε Cap(B) = −ε CapE . with a new harmonic measure Replacing E by B yields a new domain ω and (3.2) and (3.3) hold for ω if ∂ ∩ (R Q \ Q) = ∅. annulus construction: Given a square Q of side (Q) with sides parallel to = ∪ (R Q \ Q)◦ and K by K \ (R Q \ Q)◦ . This the axes, replace by ) which satisﬁes yields a new harmonic measure ω = ω(∞, · , ω(A) ≥ ω(A) whenever A ∩ (R Q \ Q) = ∅. To begin, take = p,q and perform the disc construction on every square Q j ∈ G p,q . This yields a new domain 1 , whose complement is a union of discs B j , and a new harmonic measure ω1 . Now choose a maximal dyadic square Q 1 such that (Q 1 ) ≥ ρ and ω1 (Q 1 ) ≥ M(Q 1 ). If no such Q 1 exists, the domain modiﬁcation stops. If Q 1 exists, perform the annulus construction on Q 1 , obtaining a new 1 , and then perform the disc construction on Q 1 to 1 , replacing K ∩ Q 1 by a disc B 1 . Denote by 2 and ω2 the new domain and harmonic measure thusly obtained. Next choose a maximal dyadic square Q 2 such that (Q 2 ) ≥ ρ, Q 2 ⊂ Q 1 , and ω2 (Q 2 ) ≥ M(Q 2 ). If no Q 2 exists, stop. If Q 2 exists, perform the annulus construction on Q 2 . However, if B 1 ∩ ∂(R Q 2 \ Q 2 ) = ∅, do not remove B 1 ∩ R Q 2 \ Q 2 from 2 . Then perform the disc construction on Q 2 to 2 , replacing K ∩ Q 2 by a disc B 2 , and getting a new domain 3 with harmonic measure ω3 . By induction assume we have replaced squares Q 1 , . . . , Q n−1 by discs 1 B , . . . , B n−1 and formed new domains 1 , . . . , n . Choose a maximal dyadic square Q n such that (Q n ) ≥ ρ, Q n ⊂ Q j , for all j ≤ n − 1, and ωn (Q n ) ≥ M(Q n ).

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If no Q n exists, stop. If Q n exists, perform the annulus construction on Q n to n . However, if j ≤ n − 1 and B j ∩ ∂(R Q n \ Q n ) = ∅, do not remove B j ∩ R Q n \ Q n from n . Then perform the disc construction on Q n to n , replacing K ∩ Q n by a disc B n , and getting a new domain n+1 with harmonic measure ωn+1 . Because there are only ﬁnitely many candidate squares and no square Q j is repeated, the modiﬁcation stops after a ﬁnite number of steps at a ﬁnal domain ∗ = C \ K ∗ with harmonic measure ω∗ . The complement K ∗ is the union of ﬁnitely many disjoint nonremoved discs {B k : k ∈ S} ∪ {B j : j ∈ T }. If k > j and B j ∩ ∂(R Q k \ Q k ) = ∅, then B j ⊂ 2Q j and (Q k ) ≥ (Q j ). Hence B j ∩ R(ε)Q k = ∅, and (3.2) and (3.3) hold for each stage of the construction. Consequently ω∗ (B k ) ≥ C(ε)ω(Q k ) for all k ∈ S, and ω∗ (Q j ) ≥ C(ε)ω(Q j ) for all j ∈ T. By the construction ω∗ (Q k ) ≥ cM(Q k )

(3.7)

for all k ∈ S, and ω∗ ({|z − z 0 | < r }) ≤ C Mr if z 0 ∈ Q = Q j or Q j and r ≥ (Q). Lemma 3.4. K ⊂

& S

2R Q k ∪

&

Qj.

(3.8)

T

Proof. Let Q ∈ G p,q and set E = K ∩ Q. If Q = Q j , j ∈ T then (3.8) is obvious. If not there is a ﬁrst index j1 such that Q 0 ⊂ R Q j1 \ Q j1 and Q 0 gets removed. If j1 ∈ S, then (3.8) is true, and if j1 ∈ / S there is a ﬁrst index j j j 1 2 2 j2 > j1 such that Q ⊂ R Q \ Q . However, in that case (Q j2 ) ≥ 2(Q j1 ), because if (Q k ) = (Q j ) then Q j ⊂ R Q k ⇐⇒ Q k ⊂ R Q j . Since there exists an increasing chain j1 < j2 < . . . < jn with jn ∈ S and Q jk ⊂ R Q jk+1 \ Q jk+1 , we conclude that Q ⊂ 2R Q jn .

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Now let Q = Q k , k ∈ S or Q = Q j , j ∈ T, let B be the corresponding disc and assume z ∈ 2Q \ B. Then Green’s function g(z) = g∗ (z, ∞) has the form

log |z − ζ |dω∗ (ζ ) +

g(z) =

K ∗ \B

B

log |z − ζ |dω∗ + γ (K ∗ )

= u(z) + v(z) + γ (K ∗ ). By (3.7) and a partial integration dω∗ (ζ ) 1 |∇v(z)| ≤ ≤ C M log . ∗ (Q) K \B |z − ζ | Let r denote the distance to the center z 0 of Q, and let r (B) be the radius of B. Then simple calculations give dω∗ (ζ ) |∇u(z)| ≤ B |z − ζ | and −

∂u = ∂r ≥

Re B

(z − ζ )z dω∗ (ζ ) |z − ζ |2 |z|

ω∗ (B) r (B)ω∗ (B) . − |z − z 0 | |z − z 0 |2

Set α = α(B)Max

ω∗ (B) M 2 log

1 (Q)

, 2r (B) .

If α > 2r (B), then | ∂u ∂r | > |∇v| on the circle σ = {|z − z 0 | = α} and the level set g = inf σ g(z) is a closed curve separating B from ∞. On this level curve |∇g(z)| ≤

1 ω∗ (B) ≤ C M 2 log . α (Q)

(3.9)

If α = 2r (B) we take σ = ∂ B ⊂ {g = 0} for the level curve. Then because R(ε)Q \ B ⊂ ∗ , sup∂ B |∇g| ≤ C inf ∂ B |∇g| by reﬂection, and (3.9) also holds for σ . Now = S∪T σ separates K ∗ from ∞ and 1 ∂g 1 ω∗ (B) log+ C M 2 log log+ |∇g|ds ≤ C 2π ∂n (Q) S∪T

≤ C log log

1 ρ

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because M ≤ log ρ1 . Therefore by Lemma 3.3, ∂g 1 1 log |∇g| ds ≤ C log log . 2π ∂n ρ Consider the measure function h(r ) = r 1+ε and the Hausdorff content Mh deﬁned in Appendix D. By (3.7) & Mh (K ∩ 2R Q k ) ≤ (2R)1+ε ((Q k ))1+ε S

S

C 1+ε ∗ k C R 1+ε ≤ ω (Q ) ≤ . M M S

Let r j be the radius of B j , j ∈ T, and set T1 = { j ∈ T : ω∗ (B j ) ≥ ρ ε/2 r j }. By (3.1) and (D.11) from Appendix D Cap(K ∩ Q ) j Mh (K ∩ Q j ) ≤ r j1+ε log rj for any j ∈ T. Therefore

& (K ∩ Q j ) ≤ C(1 + ε)ρ ε/2 ω∗ (B j ) ≤ Cρ ε/2 . Mh T1

T1

Finally, we have ω

& (K ∩ Q j ) ≤ T2

1 ∗ ω (B j ), C(ε) T2

ε

and by (3.9)|∇g| ≤ ρ 2 on ∂ B j . Consequently 1 ∂g ω∗ (B j ) = ds 2π ∂ B j ∂n T2 T2 ∂g C log |∇g| ds ≤ 1 ε log ρ ∂ B j ∂n ≤

T2 C log log ρ1 ε log ρ1

.

Therefore, with ε ﬁxed and η > 0 given, we can choose M and ρ so that there exists A ⊂ K such that Mh (A) < η while ω(K \ A) < η. That means ω ⊥ h , and the Theorem follows.

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Notes

Notes Section 1 follows the elementary arguments in Batakis [1996]. Earlier Makarov and Volberg [1986] had given an explicit formula for dim ω = inf{α : ω ⊥ α }

(N.1)

in terms of the critical points of Green’s function. See also Volberg [1992], [1993] and Lyubich and Volberg [1995]. These papers rely on a beautiful connection between information theory and harmonic measure for self-similar sets originally due to Carleson [1985] and to Manning [1984] in the case of Julia sets that we regretfully must omit from this book. See Makarov [1990a] and [1998] for complete details and several further related results. Theorem 2.1 is an unpublished result of Jones and Wolff [1986]. Carleson [1985] had obtained a similar theorem for certain Cantor sets using information theory and the identity (2.15) on Green’s function critical points. Jones and Wolff later made a direct proof of the more general Theorem 2.1, still using (2.15). The Green’s function identity (2.15) itself ﬁrst appeared in Ahlfors’ [1947] paper on analytic capacity. See Widom [1969], [1971a], and [1971b] and Jones and Marshall [1985] for other applications. In the substantial paper [1993], Wolff extended Theorem 3.1 to show that for all there is F ⊂ ∂ having σ − ﬁnite linear measure and ω(F) = 1. Makarov’s paper [1998] and the recent work [2003] by Binder, Makarov, and Smirnov explore the connection between other Hausdorff measures and harmonic measure in general domains. Let ω be harmonic measure for some domain and deﬁne the lower pointwise dimension of ω at z ∈ ∂ as αω (z) = lim inf δ→0

log ω(B(z, δ)) . log δ

Set f ω (α) = dim{z ∈ ∂ : αω (z) ≤ α}

(N.2)

and deﬁne the universal dimension spectra to be (α) = sup f ω (α)

and SC (α) =

sup

simply connected

f ω (α).

Makarov [1998] conjectures that for all α ≥ 1, (α) = SC (α).

(N.3)

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Conjecture (N.3) is connected to the Brennan conjecture because Makarov [1998] proves that SC (α) = inf α(1 + Bb (t)) + (1 − α)t , (α ≥ 1), t≥0

where Bb (t) is the bounded universal integral means spectrum deﬁned in Exercise VIII.9. To prove (N.3) it is enough to work with certain Cantor sets. Let D be a simply connected domain, let {D j } be a ﬁnite collection of pairwise disjoint Jordan domains such that D j ⊂ D, and suppose & F: Dj → D is an analytic function such that each restriction F D j is univalent. Write F (n) = F ◦ F ◦ . . . ◦ F for the n-th iterate of F. The set K =

∞ +

Fn

−1 &

Dj

n=1

is called a conformal Cantor set. Carleson and Jones [1992] and Makarov [1998] show that (N.3) holds if and only if f ω (α) ≤ SC (α)

(N.4)

whenever C∗ \ is a conformal Cantor set. In [2002] Binder, Makarov, and Smirnov established (N.4) for those conformal Cantor sets for which F is a polynomial.

Exercises and Further Results 1. (a) Let K be a Cantor set satisfying the hypotheses of Theorem 1.1. Then there exists σ < dimHaus (K ) such that there is F ⊂ K with dimHaus (F) = σ and ω(F) = 1, but such that if E ⊂ K has dimHaus (E) < σ then ω(E) = 0. (b) However, if the deﬁnition of the Cantor set K at the beginning of Section 1 is generalized so that each square Q nJ is replaced by four corner squares Q nJ, j , j = 1, 2, 3, 4, having sidelengths (Q nJ, j ) = a nJ (Q nJ ) that depend on J but satisfy 0 < a ≤ a nJ ≤ b < 1/2,

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then there is an example such that dimHaus (ω) = dimHaus (K ). This is due to Bishop and independently to Ancona and Batakis. (c) If the Cantor set K is constructed with an = δ, 0 < δ ≤ 1/4, harmonic measure on K had dimension greater than 1 − Cδ for some constant C. This result is due to Ancona. See Batakis [1996] for the proofs of all these results. 2. Let be a domain such that for some η > 0 ω(w, B(w, 2dist(w, ∂)), ) ≥ η

(E.1)

for all w ∈ . Then harmonic measure ω for satisﬁes ω ⊥ h

(E.2)

for every measure function h(t) such that lim

t→0

h(t) = 0. t

The proof outlined below is a simple variation of the original Makarov argument given in Section VI.5. (a) Let ϕ : D → be a universal covering map (see Ahlfors [1973], Fisher [1983], or Gamelin [2001].) Let {z n } ⊂ D be nontangentially dense on ∂D and let wn = ϕ(z n ). Then

& ω B(wn , 2dist(wn , ∂)) = 1. Hint: Consider u ◦ ϕ when 1 − u is the harmonic measure above and use Exercises III.12 and VI.11. (b) Let ϕ : D → be a universal cover. If w0 = ϕ(z 0 ) then dist(w0 , ∂) ≤ |ϕ (z 0 )|(1 − |z 0 |2 ). Hint: Since ϕ has no zeros the monodromy theorem yields an inverse map ψ : B(w, dist(w, ∂)) → D such that ψ ◦ ϕ(w) = w and ψ(w0 )) = z 0 . Now apply the Schwarz lemma. Also see Lemma X.5.5. (c) Use parts (a) and (b) and the proof of Theorem VI.5.2 to show that (E.1) implies (E.2). We do not know who ﬁrst discovered this argument, which is mentioned in the introduction of Jones and Wolff [1988]. 3. Condition (E.1) has many equivalents and a compact set K ⊂ C is called uniformly perfect if = C ∗ \ K satisﬁes any one of them. Exercise III.11 gave a capacity condition equivalent to (E.1). Thirteen more equivalent conditions are listed below.

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IX. Inﬁnitely Connected Domains (i) There is a constant c > 0 such that if ϕ : D → is a universal cover and if w0 = ϕ(z 0 ) then dist(w0 , ∂) ≥ c|ϕ (z 0 )|(1 − |z 0 |2 ). In other words, the inequality in Exercise 2(b) can be reversed. See Pommerenke [1979]. (ii) Let ϕ : D → be a universal cover map. The expression dρ(w) =

|ϕ (z)||dz| , ϕ(z) = w 1 − |z|2

which is independent of the choices of z or ϕ, is the inﬁnitesimal form of the hyperbolic metric on . In other word, the hyperbolic distance between z 1 and z 2 is ρ(z 1 , z 2 ) = inf dρ(w) γ

γ

where the inﬁmum is taken over all arcs γ in that connect z 1 to z 2 . Now K is uniformly perfect if and only if there is a constant c > 0 such that |dw| ρ(z 1 , z 2 ) ≥ c inf . γ dist(w, ∂) γ (iii) There is a constant C > 0 such that every closed curve in not homotopic to a point in has hyperbolic length a least C. See Pommerenke [1979]. (iv) There is a constant C such that if R ⊂ is a ring and each component of C∗ \ R meets K , then mod(R) ≤ C. See Beardon and Pommerenke [1979]. (v) There is a constant C such that if A ⊂ is an annulus and each component of C∗ \ A meets K , then mod(A) ≤ C. Use (iii) and Exercise IV.5. (vi) If ϕ : D → is a universal cover, then g = log(ϕ ) is a Bloch function. Pommerenke [1979]. (vii) There is a constant c > 0 such that if ϕ : D → is a universal cover and if ϕ(z 1 ) = ϕ(z 2 ) then ρ(z 1 , z 2 ) ≥ c. Pommerenke [1979]. (viii) There is a constant δ > 0 such that if ϕ : D → is a universal cover and if ϕ −1 (w) = {z j }, then - z j − z k inf ≥ δ. k 1 − zk z j j; j =k

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Pommerenke [1984]. This the famous Carleson [1958] condition that {z j } be an interpolating sequence for H ∞ . It also has many equivalents. See Garnett [1981], for example. (ix) The domain has a strong barrier; i.e., there is a nonconstant positive superharmonic function U (z) on and a constant c > 0 such that in the distribution sense cU (z) U (z) + ≤ 0. dist(z, ∂)2 See Ancona [1986]. (x) There is a constant C such that Hardy’s inequality holds: If u(z) ∈ C 1 has compact support contained in , then |u(z)| 2 ∇u(z) 2 d xd y. d xd y ≤ C dist(z, ∂)

Ancona [1986]. (xi) Let ϕ : D → be a universal cover map. Then K is uniformly perfect if and only if there is a constant C such that whenever D ⊂ is a disc, every component of ϕ −1 (D) is a C-quasidisc. See Heinonen and Rohde [1995]. (xii) Let g(z, ζ ) be Green’s function for . There is ε > 0 and C < ∞ such that for all ζ ∈ g(z, ζ )1+ε < C. ∇z g(z,ζ )=0

See González [1992]. (xiii) Again let ϕ : D → be a universal covering map. A fundamental domain for ϕ is a set F ⊂ D such that ϕ(F) = and ϕ is one-toone on F. Then K is uniformly perfect if and only if there exists a fundamental domain F for ϕ such that ∂F is a quasicircle. González [1992]. The equivalence of (i), (ii), (v), (vi), and (viii) is independently due to Behrens [1976]. 4. For 1 ≤ j, k ≤ N let B j,k be a closed disc with center ( Nj , Nk ) and radius 1 , and set K = B j,k and = C∗ \ K . Assume all discs B j,k have r j,k < 2N the same harmonic measure at ∞, ω(∞, ∂ B j,k , ) = Prove there is C independent of N so that r j,k ≤ C.

1 . N2

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Hint: Use Chapter III. 5. (a) Let ω be harmonic measure for some domain and deﬁne dimω by (N.1). Prove that if α = dimω, then f ω (α) = α, where f ω is deﬁned by (N.2). (b) Using the estimate Bb (t) ≤ Ct 2 from Chapter VIII, prove that Conjecture (N.3) implies (α) ≤ α − c(α − 1)2 for 1 ≤ α ≤ 2.

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X Rectiﬁability and Quadratic Expressions

First we introduce three topics, two old and one new. (i) The ﬁrst topic is the classical Lusin area function. The Lusin area function gives another description of H p functions and another almost everywhere necessary and sufﬁcient condition for the existence of nontangential limits. The area function is discussed in Section 1. In Appendix M we prove the Jerison–Kenig theorem that the area function determines the H p class of an analytic function on a chord-arc domain. (ii) The second topic is the characterizations of subsets of rectiﬁable curves in terms of certain square sums. These theorems, from Jones [1990], are proved in Sections 2 and 3. (iii) The third topic is the Schwarzian derivative, which measures how much an analytic function deviates from a Möbius transformation. Section 4 is a brief introduction to the Schwarzian derivative. Then we turn to the chapter’s main goal, an exposition of the two papers [1990] and [1994] by Bishop and Jones. In Section 5 the Schwarzian derivative is estimated by the Jones square sums and by a second related quantity. In Section 6 rectiﬁable quasicircles are characterized by a quadratic integral akin to the Lusin function but featuring the Schwarzian derivative. In Sections 7, 8, and 9 the same quadratic integral gives new criteria for the existence of angular derivatives and further characterizations of BMO domains. Section 10 brings together most of the ideas from the chapter to prove a local version of the F. and M. Riesz theorem, and in Section 11 this F. and M. Riesz theorem leads us to the most general form of the Hayman–Wu theorem.

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1. The Lusin Area Function Let F(z) be an analytic function on D. The AF(ζ ) = Aα F(ζ ) =

area function of F is 1 2 F (z) d xd y 2 ,

α (ζ )

where α > 1 and

α (ζ ) = z ∈ D : |z − ζ | < α(1 − |z|) .

2 AF(ζ ) is called the area function because AF(ζ ) is the area of F(α (ζ )) when values are counted with their multiplicities. We will need a famous theorem of Calderón [1965]. Theorem 1.1. Let F(z) be analytic in D and let 0 < p < ∞. Then F ∈ H p if and only if AF ∈ L p (∂D), and there is a constant c = c( p, α) such that c||AF|| p ≤ ||F − F(0)|| H p ≤ c−1 ||AF|| p . p

p

p

(1.1)

A complete proof of Calderón’s theorem will be given in Appendix M. See also Calderón [1965] or Stein [1970] and [1993]. In this section we prove Calderón’s theorem for p = 2 on certain cone domains and we prove the local theorem that modulo sets of measure zero, AF(ζ ) < ∞ if and only if F has a nontangential limit at ζ . On D the p = 2 case of Calderón’s theorem has a simple Fourier series proof. By Fubini’s theorem

2 AF(ζ ) ds = |F (z)|2 {ζ : z ∈ α (ζ )} d xd y, ∂D

D

and by the deﬁnition of α 1 (1 − |z|2 ) ≤ {ζ : z ∈ α (ζ )} ≤ c(α)(1 − |z|2 ). c(α) Hence

∂D

2 AF(ζ ) ds ≤ c(α)

|F (z)|2 (1 − |z|2 )d xd y

D

≤ c2 (α)

(1.2)

∂D

2 AF(ζ ) ds.

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1. The Lusin Area Function Now write F(z) =

an z n . Then ||F − F(0)||2H 2 = 2π

∞

|an |2

n=1

and

∞ 2 F (z) (1 − |z|2 )d xd y = π

D

n=1

n |an |2 . n+1

Therefore (1.1) holds at p = 2 and 2 1 2 F (z) (1 − |z|2 )d xd y ≤ 1 ||F − F(0)||2 2 . (1.3) ||F − F(0)|| H 2 ≤ H 4 2 D

The case p = 2 also has a simple geometric proof using (1.2) and Green’s theorem. See Exercise 8 of Chapter II. Let U be a simply connected domain such that ∂U is a rectiﬁable curve, let F(z) be analytic on U , and let ϕ be a conformal map from D onto U . By deﬁnition we say F(z) is in H p (U ), 0 < p < ∞, if 1

F ϕ(z) (ϕ (z)) p ∈ H p (D). When F ∈ H p (U ) we take p

||F|| H p (U ) =

∂D

iθ p iθ F ϕ(e ) |ϕ (e )|dθ.

By the F. and M. Riesz theorem, ϕ has non-zero nontangential limit almost everywhere on ∂D, and it follows that every F(z) ∈ H p (U ) has nontangential limit F(ζ ) arc length almost everywhere on ∂U and iθ p iθ p F(ζ ) p ds. F ϕ(e ) |ϕ (e )|dθ = ||F|| H p (U ) = ∂D

∂U

p ||F|| H p (U )

does not depend on the choice of ϕ. Consequently the norm We deﬁne a special cone domain to be a cone domain having either the form & U= β (ζ ), (1.4) E

where E is a compact subset of ∂D, or the form & U= βh (ζ ), E

(1.5)

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where E is a compact subset of an arc I ⊂ ∂D such that |I | ≤ h < 1 and where βh (ζ ) = β (ζ ) ∩ {z : 1 − |z| < h} with β > 2. See Figure X.1. When U is a special cone domain, we deﬁne the central point of U in case (1.4) by cU = 0 and in case (1.5) by 3h )c I 4 where c I is the center of the arc I. Every special cone domain U is connected and U cU if U = ∅ because β > 2 and h ≥ |I |. cU = (1 −

E h

cU

Figure X.1 The special cone domain U =

E

βh (ζ ).

When U is a domain and z ∈ U, we write d(z) = dist(z, ∂U ) and we deﬁne the cone (relative to U ) at ζ ∈ ∂U to be α (ζ, U ) = {z ∈ U : |z − ζ | < αd(z)}. ζ

ζ

α (ζ, U )

U

ζ Figure X.2 α (ζ, U ) and α (ζ , U ).

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If F(z) is analytic on U we also deﬁne the area function 1 F (z) 2 d xd y 2 . AF(ζ ) = Aα F(ζ, U ) = α (ζ,U )

Theorem 1.2. Let U be a special cone domain and let F(z) be analytic on U . Then F ∈ H 2 (U ) if and only if AF ∈ L 2 (∂U , ds) and if and only if

2 F (z) d(z)d xd y < ∞.

(1.6)

U

There exist constants C1 = C1 (β) > 0 and C2 > 0 such that 2 2 F (z) d(z)d xd y ≤ C2 ||F||2 2 . (1.7) C1 ||F||2H 2 (U ) ≤ F(cU ) + H (U ) U

Furthermore, F ∈

H 2 (U )

if and only if 2 F (z) d(z)3 d xd y < ∞ U

and there is C3 = C3 (β) > 0 such that 2 2 F (z) d(z)d xd y ≤ |F (cU )|2 + F (z) d(z)3 d xd y C3−1 U

≤ C3

U

2 F (z) d(z)d xd y.

(1.8)

U

Of course when U = D (1.6) is nothing but the case p = 2 of Calderón’s theorem 1.1 above. Theorem 1.2 is a special case of a result from Kenig’s thesis [1980], which generalized Calderón’s theorem for 0 < p < ∞ to the more general setting of Lipschitz domains. Later Jerison and Kenig [1982a] extended Calderón’s theorem to domains bounded by chord-arc curves. See Appendix M. The proof of Theorem 1.2 hinges on three geometric properties of special cone domains: (i) For all α > 1 there are constants c1 = c1 (α, β, h) and c2 = c2 (α, β, h) such that c1 d(z) ≤ 1 ({ζ ∈ ∂U : z ∈ α (ζ, U )}) ≤ c2 d(z).

(1.9)

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U

αd(z)

ζ z

Figure X.3 = α (ζ, U ) is shaded. (ii) There exists δ = δ(β, h) such that B(cU , δ) ⊂ U.

(1.10)

(iii) Let ϕ : D → U be a conformal mapping satisfying ϕ(0) = cU . Then ϕ (eiθ )eiθ

iθ π dϕ − − arg ϕ(e ) − cU = arg arg ϕ(eiθ ) − cU dθ 2 (1.11) π ≤ τ (β, h) < . 2 Inequality (1.11) gives control almost everywhere on ∂U on the angle between the normal vector and the position vector emanating from cU . τ ζ cU

Figure X.4 The angle τ in (1.11).

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Proof of Theorem 1.2. Again Fubini’s theorem gives

(AF(ζ ))2 ds = |F (z)|2 1 {ζ ∈ ∂U : z ∈ α (ζ, U )} d xd y, ∂U

U

and then (1.9) yields

2 AF(ζ ) ds ≤ c(α, U ) |F (z)|2 d(z)d xd y ∂U

U

(1.12)

≤ c (α, U )

∂U

2 AF(ζ ) ds.

L 2 (∂U ) if and only if (1.6) holds. Notice (1.12) is the analogue

Therefore AF ∈ of (1.2) for the special cone domain U. To prove (1.7) deﬁne |F(ζ )|2 ds = A=

∂D

∂U

F ◦ ϕ 2 |ϕ |dθ,

F ◦ ϕ (z) 2 |ϕ (z)|(1 − |z|2 )d xd y,

B= D

and D= D

2 2 ϕ (z) F ◦ ϕ √ (1 − |z|2 )d xd y. ϕ (z)

By the Koebe theorem,

B≤8

2 F (z) d(z)d xd y ≤ 8B.

U

We may assume F(cU ) = 0. By (1.3) we have % 2 d

A≥2 F(ϕ(z)) ϕ (z) (1 − |z|2 )d xd y dz D 2 1 ≥ (F ◦ ϕ) (z) |ϕ (z)|(1 − |z|2 )d xd y − 2 D ϕ (z) 2 F ◦ ϕ 2 √ × (1 − |z|2 )d xd y ϕ (z) D

=B−

D . 2

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Write g = log(ϕ ). By (1.11), |Img| = | arg(ϕ )| < 2π, so that by Exercise 3 of Appendix F, g ∈ BMO and ϕ 2 (1 − |z|)d xd y = |g (z)|2 (1 − |z|)d xd y ϕ is a Carleson measure with norm bounded by a ﬁxed constant C. Therefore (F ◦ ϕ)(z) 2 |ϕ (z)||g (z)|2 (1 − |z|2 )d xd y ≤ C A D= D

and B ≤ (1 + 2C)A with C independent of β and h. The proof that A ≤ C(β)B will exploit the inequality τ (β, h) < Because τ (β, h) < π2 and |ϕ(eiθ )| ≤ 1, we have 2 ϕ (eiθ )eiθ F ◦ ϕ A≤ dθ ϕ(eiθ ) − cU ∂D iθ iθ F ◦ ϕ 2 e ϕ (e ) dθ . ≤ sec τ ϕ(eiθ ) − c ∂D

π 2

in (1.11).

U

Hence by Green’s theorem (see Exercise 8 of Chapter II), 1 2 zϕ (z) |F ◦ ϕ| log d xd y . A ≤ sec τ ϕ(z) − cU |z| D

Write h(z) =

and because

zϕ (z)

. Then 1 |(F ◦ ϕ) |2 |h| log d xd y A ≤ 4 sec τ |z| D 1 + 4 sec τ (F ◦ ϕ)(F ◦ ϕ) h log d xd y, |z| ϕ(z)−cU

D

z ϕ(z)−cU

≤ c(δ) by the maximum principle,

1 A ≤ 4c(δ)(sec τ )B + 4 sec τ (F ◦ ϕ)(F ◦ ϕ) h log d xd y, |z| D

again by Exercise II.8. Now by (1.11), h(z) = e V (z) where |ImV | ≤ τ <

π 2,

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and h = V h. Therefore by Cauchy–Schwarz 1 (F ◦ ϕ)(F ◦ ϕ) h log d xd y |z| D 1 = (F ◦ ϕ) (F ◦ ϕ)V |h| log d xd y |z| D

"1 2 2 1 ≤ (F ◦ ϕ) |h| log d xd y |z| D

"1 2 2 1 2 × F ◦ ϕ |V | |h| log d xd y |z| D

1 ≤ c(δ)B 2

"1

2 √ 2 2 1 ◦ ϕ) h |V | log . d xd y (F |z| D

But since |ImV | ≤ τ |V |2 log

1 d xd y |z|

is a Carleson measure with constant bounded by c(β), and we conclude that 1

1

A ≤ C(β)(B + B 2 A 2 ), so that (1.7) holds. To prove (1.8) note that because |F (z)|2 is subharmonic we have 2 F (z) d(z)d xd y, F (cU ) 2 ≤ c1 (β, h) U

by property (ii) and 2 F (z) ≤ 2 π η4 for z ∈ U and η < d(z). Hence 2 F (z) d(z)3 ≤

Now if |w − z| <

d(z) 2

F (w) 2 dudv

B(z,η)

32 π d(z)

F (w) 2 dudv.

B(z, d(z) 2 )

then d(z) 3 ≤ d(w) ≤ d(z) 2 2

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so that by Fubini’s theorem, 2 3 F (z) d(z)3 d xd y ≤ 32 F (w) 2 d xd ydudv π 2d(w) U U B(w,d(w)) 2 F (w) d(w)dudv, ≤ 48 U

which establishes the right-hand inequality in (1.8). To prove the other inequality in (1.8) we again use property (ii) to get 2 2 F (z) d(z)3 d xd y |F (z) − F (cU )| d(z) ≤ C(δ) U

B(cU , 2δ ).

for z ∈

Therefore

|F (z) − F (cU )|2 d(z)d xd y ≤ c (δ)

B(cU , 2δ )

|F (z)|2 d(z)3 d xd y.

U

z−cU . Then When z ∈ U \ B(cU , 2δ ) write z 0 = cU + 2δ |z−c U| z 2 F (z) − F (z 0 ) 2 ≤ F (w) ds z z0 z 3 F (w) 2 d(w) 23 ds ≤ d(w)− 2 ds z0 z0 z − 21 F (w) 2 d(w) 23 ds ≤ C (β)d(z) z0

because by trigonometry d(w) ≥ d(z) + c2 (β)|z − w| if w ∈ [z 0 , z].

U

z w z0 cU

Figure X.5

w∗

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w Write w ∗ for the projection of w ∈ U to ∂U along the radius [0, |w| ]. Then again by trigonometry

d(z) ≤ c3 (β)|z − w ∗ | when z ∈ [w, w∗ ]. Using polar coordinates we then obtain |F (z) − F (cU )|2 d(z)d xd y U

2 ≤ C(δ) F (w) d(w)3 d xd y U

+ C (β)

F (w) 2 d(w) 23

U ∩{|z−cU |> 2δ }

≤ C (β)

w∗

w

1

|z − w∗ | 2 dr dudv

.

F (w) 2 d(w)3 dudv,

U

which proves the other half of (1.8).

We say two statements P and Q about points on ∂D are almost everywhere equivalent if the symmetric difference {P true}{Q true} has measure zero. For example, if u(z) is a harmonic function on D and if α > 1, then by Privalov’s theorem, “u ∗α (ζ ) < ∞” is almost everywhere equivalent to “u has nontangential limit at ζ ”. The next theorem is the parallel result for area functions. Theorem 1.3. Let F(z) be an analytic function on D and let α > 1. Then the statements AF(ζ ) < ∞ and F has nontangential limit at ζ are almost everywhere equivalent. A simple corollary of Theorem 1.3 and Plessner’s theorem is that Theorem 1.3 is also true when F(z) is a harmonic function. The martingale analogue of Theorem 1.3 is the Lévy theorem J.3 in Appendix J. One half of Theorem 1.3 is due to Marcinkiewicz and Zygmund [1938] and the other half to Spencer [1943].

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The proof of Theorem 1.3 depends on a fourth geometric property of the special cone domains having the form (1.4) and of their cones α (ζ, U ) : If ζ ∈ E = ∂U ∩ ∂D and if z ∈ U , then d(z) ≤ 1 − |z| and hence α (ζ, U ) ⊂ α (ζ ). Conversely, suppose z ∈ α (ζ ) ⊂ U and z ∗ ∈ ∂β (ζ ) satisfy |z − z ∗ | ≤ d(z), with α < β. Then 1 − |z| ≤ d(z) + 1 − |z ∗ | and |z ∗ − ζ | ≤ |z − ζ | + d(z). Hence |z − ζ | |z ∗ − ζ | |z − ζ | + d(z) ≤ d(z) + ≤ d(z) + , α β β so that α (ζ ) ⊂ γ (ζ, U ) ⊂ U

(1.13)

when γ =

1+ 1 α

−

1 β 1 β

U

.

(1.14)

ζ

α (ζ ) cU = 0 γ (ζ, U )

Figure X.6 Cones satisfying (1.13). Proof of Theorem 1.3. First we prove the Marcinkiewicz–Zygmund half. Let K ⊂ ∂D be compact and assume F has nontangential limit at every ζ ∈ K . Fix

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α > 1. Take β > α and deﬁne γ by (1.14). Taking K smaller, we can assume F is bounded on the cone domain & U= β (ζ ). ζ ∈K

Thus F ∈

H 2 (U )

and by Theorem 1.2

Aγ F(ζ, U ) < ∞

almost everywhere on K . Therefore by (1.13) Aα F(ζ ) < ∞ almost everywhere on K . To prove Spencer’s half of the theorem, we can assume that for all eiθ ∈ E, |F (z)|2 d xd y ≤ N α (eiθ )

for some constant N . Let K ⊂ E be a compact set such that for ﬁxed δ, |E ∩ (θ − h, θ + h)| ≥ 3/4 (1.15) 2h whenever 0 < h < δ. By the theorem on points of density, it is enough to prove F has nontangential limit almost everywhere on K . Set & α (eiθ ). U= eiθ ∈K

Then by (1.15) there is c(δ) > 0 such that 2π N ≥ |F (z)|2 d xd ydθ K α (eiθ )

= U

|F (z)|2 K ∩ {θ : z ∈ α (θ )} d xd y

≥ c(δ)

|F (z)|2 (1 − |z|2 )d xd y.

U

≥ c (δ)

|F (z)|2 dist(z, ∂U )d xd y.

U

Therefore F ∈ H 2 (U ) by Theorem 1.2, and F has nontangential limits almost everywhere on K by the F. and M. Riesz theorem.

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Theorem 1.4. Let F(z) be analytic in D and let α > 1. Then 2 F (z) (1 − |z|2 )2 d xd y < ∞

(1.16)

α (ζ )

is almost everywhere equivalent to F has nontangential limit at ζ. Proof. We outline two proofs. The ﬁrst proof uses Theorem 1.2. Suppose 2 F (z) (1 − |z|2 )2 d xd y ≤ M α (ζ )

for all ζ ∈ E. Given ε > 0 and c < 1, there exist a compact set E 0 ⊂ E and a δ0 > 0 such that |E \ E 0 | < ε and |B(ζ, δ) ∩ E| ≥ cδ for all ζ ∈ E 0 and all δ < δ0 . Set U=

&

(1.17)

α (ζ ).

E0

By (1.17),

E ∩ {|ζ − z| < α(1 − |z|)} ≥ c (1 − |z|2 )

when z ∈ U and 1 − |z| ≤ αδ . Therefore 2 2 F (z) dist(z, ∂U )3 d xd y ≤ F (z) (1 − |z|2 )3 d xd y U

U

≤C

2 F (z) (1 − |z|2 )2 d xd y

E α (ζ )

≤ C M. Then by Theorem 1.2, F ∈ H 2 (U ) and F has nontangential limits almost everywhere on E 0 . Conversely, assume F has nontangential limit at every ζ ∈ E. Fix β > α. Given ε > 0 there is a compact set E 0 ⊂ E such that |E \ E 0 | < ε and F is bounded on & U= β (ζ ). E0

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Thus F ∈ H 2 (U ) and by Theorem 1.2 |F (z)|2 d(z)3 d xd y < ∞. U

Let U0 =

&

α (ζ ).

E0

If z ∈ U0 , then 1 − |z|2 ≤ c(α, β)d(z). Consequently by Fubini’s theorem |F (z)|2 (1 − |z|2 )2 d xd y E 0 α (ζ )

=

|F (z)|2 (1 − |z|2 )2 {ζ ∈ E 0 : z ∈ α (ζ )} d xd y

U0

≤C

|F (z)|2 d(z)3 d xd y,

U

and (1.16) holds almost everywhere on E 0 .

The second proof combines Theorem 1.3 with the pointwise result below. Lemma 1.5. Let α > 1 and let F(z) be analytic on α (ζ ). (a) If 1 < β < α there is C(α, β) such that for all ζ ∈ ∂D, 2 2 F (z) (1 − |z|2 )2 d xd y ≤ C(α, β) F (z) d xd y. β (ζ )

α (ζ )

(b) There is Cα such that for all ζ ∈ ∂D, 2 2 F (z) d xd y ≤ Cα F (0) 2 + Cα F (z) (1 − |z|2 )2 d xd y. α (ζ )

α (ζ )

The proof of Lemma 1.5, which is very like the proof of (1.8), is an exercise for the reader.

2. Squares Sums and Rectiﬁability Let E be a bounded set. We seek geometric conditions that are necessary and sufﬁcient for E to lie on a rectiﬁable curve. Clearly one necessary condition is 1 (E) < ∞, and if E is connected then 1 (E) < ∞ if and only if E is a

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subset of a rectiﬁable curve. In fact, a connected set E is the trace of a (selfintersecting) rectiﬁable closed curve of length if and only if 1 (E) ≤ /4; see Falconer [1985] or David and Semmes [1991]. However, there exist sets E such that 1 (E) = 0 but E is not contained in any rectiﬁable curve; see Exercise 3. For z ∈ E and t > 0, deﬁne "

1 sup dist(w, L) : L is a line through z . β E (z, t) = inf t E∩B(z,t) Thus β E (z, t) is a scale invariant measure of the least deviation E ∩ B(z, t) has from lines through the point z. Because z ∈ E, the width of the narrowest strip containing E ∩ B(z, t) is comparable to tβ E (z, t). t 2tβ E (z, t)

z

E

Figure X.7 Theorem 2.1. There is a constant C such that if E is a bounded set and diam(E) dt β E2 (z, t) ≤ K (2.1) t 0 for all z ∈ E, then there is a rectiﬁable curve such that E ⊂ and () ≤ CeC K diam(E). Theorem 2.5 will provide a converse of Theorem 2.1: If E ⊂ and is a rectiﬁable curve, then (2.1) holds at 1 almost every point of E. See Exercise 7 for a reﬁnement of Theorem 2.1. To simplify the proof of Theorem 2.1 we introduce a dyadic variant of β E (z, t). When Q is a dyadic square of side (Q), write a Q for the concentric square with side (a Q) = a(Q) and deﬁne "

1 β E (a Q) = (2.2) inf sup dist(z, L) : L is any line . (a Q) E∩a Q

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Then 2β E (a Q)(a Q) is the width of the narrowest strip containing E ∩ a Q. Lemma 2.2. There are constants c1 , c2 , and c3 such that if z ∈ E, then for all d > 0, d dt c3 β E2 (z, t) ≤ β E2 (3Q) : (Q) ≤ d, z ∈ Q t 0 (2.3) d dt 2 ≤ c1 + c2 β E (z, t) . t 0 Proof. By a change of scale we may assume d = 1. Note that if Q is a dyadic square containing z, then √ 3Q ⊂ B(z, 2 2(Q)) and β E (3Q) ≤

√ √ 2 2 β E (z, 2 2(Q)). 3

(2.4)

Furthermore, if t ≤ s then s β E (z, t) ≤ β E (z, s). t Write Q n = Q n (z) for the unique dyadic square # $ Q n = j2−n ≤ x < ( j + 1)2−n , k2−n ≤ y < (k + 1)2−n

(2.5)

such that z ∈ Q n . Then by (2.4) and (2.5) ∞ n=0

β E2 (3Q n ) ≤

∞ 8 n=0

9

3

β E2 (z, 2−n+ 2 )

5 ∞ −n+ 32 2 2 2 dt ≤ β E (z, t) −n+ 23 9 log 2 t n=0 2 1 dt 5 32 2 β (z, t) + log 2 . ≤ 9 log 2 0 E t 2

To prove the lower estimate, note that when we reﬂect a strip S containing E ∩ 3Q about a line through z ∈ E ∩ Q parallel to the sides of S we obtain a strip having centerline through z and not more than twice as wide as S. Since B(z, (Q)) ⊂ 3Q, we therefore have 1 β E (z, (Q)), 6 and it follows from (2.5) and (2.6) that β E (3Q) ≥

(2.6)

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∞

∞

β E2 (3Q n )

n=0

1 2 ≥ β E (z, 2−n ) 36 n=0 ∞ 2−n dt 1 β E2 (z, t) ≥ −n−1 144 log 2 t n=0 2 1 1 dt ≥ β 2 (z, t) . 144 log 2 0 E t

Proof of Theorem 2.1. In the proof we assume diam(E) ≤ 1. Then by (2.1) and Lemma 2.2 β E2 (3Q) ≤ c1 + c2 K (2.7) (Q)≤1,z∈Q

for all z ∈ E. We ﬁrst describe the basic construction underlying the proof. Let S be a closed rectangle with side lengths L and β L, where β ≤ 1, and assume E meets each side of S. Partition each L side of S into three segments of length L/3, and deﬁne the middle third M(S) as that subrectangle with each L side replaced by its middle third. There are two cases. Case 1: There exists a point p ∈ E ∩ M(S). See Figure X.8. Cut S into two rectangles A0 and A1 by the line through p orthogonal to the L sides of S, and let α j L be the length of the projection of A j onto an L side of S. Then 1/3 ≤ α j ≤ 2/3. Cover A j ∩ E by a rectangle Sj such that A j ∩ E meets each side of Sj and such that the shortest side of Sj is as small as possible. Then S0 ∩ S1 ∩ E = ∅ since p ∈ S0 ∩ S1 . Let L j be the length of the longest side of Sj . By the Pythagorean theorem, L j ≤ (α 2j + β 2 )1/2 L ≤ (1 + 5β 2 )α j L . A1

M(S)

E βL

S S1

A0

(2.8)

p S0 L Figure X.8 Case 1.

E

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Case 2: E ∩ M(S) = ∅. Let A0 and A1 be the two components of S \ M(S). See Figure X.9. Cover A j ∩ E by a rectangle Sj such that each side of Sj meets A j ∩ E and Sj has shortest side as small as possible. Join S0 to S1 by the shortest segment T having endpoints in E ∩ Sj . Write α0 = α1 = 1/2, write L j for the length of the longest side of Sj , and write |T | for the length of T . Then (2.8) still holds and |T | ≤ (1 + β 2 )L .

(2.9)

A1 M(S) S1

A0 T

L1 βL

E

S0

L

Figure X.9 Case 2. In both cases α0 + α1 = 1,

(2.10)

and there is a constant C such that Sj has shortest side of length β j L j and β j ≤ Cβ E (3Q)

(2.11)

for any dyadic Q such that Q ∩ Sj = ∅ and diam(Sj ) ≤ (Q) < 2 diam(Sj ), because in that case Sj ⊂ 3Q. To start, assume E ⊂ Q 0 = [0, 1] × [0, 1], but assume no smaller dyadic Q contains E. Let S0 be a rectangle such that E ⊂ S0 , E meets all four sides of S0 , and S0 has smallest shortest side. Write its side lengths as L 0 and β L 0 , where β ≤ 1 and β ≤ Cβ(3Q 0 ). Now perform the basic construction on S0 , getting two rectangles S0,0 and S0,1 , and (in Case 2) a line segment T0 . Continue. After n steps we have rectangles S I , I = (i 1 , i 2 , . . . , i n ), i j = 0, 1, with longsides L I , and we have weights α I ∈ [1/3, 2/3] and factors β I ≤ 1. For these rectangles, conditions (2.8), (2.10), and (2.11) hold. If Case 2 is applied to S I we also get a segment TI of length |TI |, for which (2.9) holds. After N = 25 steps the diameter of a rectangle S I drops by at least 1/2. Indeed, if we begin with a rectangle that has side lengths 1 and β ≤ 1 and

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diameter d0 = (1 + β 2 )1/2 , then in the worst case the construction yields a rectangle having sides α and (α 2 + β 2 )1/2 and diameter d1 = (2α 2 + β 2 )1/2 . Since α ≤ 2/3, we have d1 ≤ (17/18)1/2 d0 , and since (17/18)25/2 ≤ 1/2, the claim holds. Consequently each dyadic square Q can occur at most N times during iterated applications of the basic construction. For a multi-index I = (i 1 , i 2 , . . . , i n ) of size |I | = n and for m < n, write Im = (i 1 , i 2 , ...., i m ). Let {I (1) , . . . , I ( p) } be a set of multi-indices having size |I ( j) | ≤ n such that I ( j) ⊂ I (k) for j = k. Then by (2.8) and (2.10), L I ( j) ≤ L 0 (1 + 5β I2m ) α Im . |I |=n 1≤m≤n

1≤m≤n

By (2.11) and the hypothesis (2.7), (1 + 5β I2m ) ≤ Ce5c2 N K , 1≤m≤n

and by (2.10)

-

α Im = 1.

|I |=n 1≤m≤n

Therefore

L I ( j) ≤ CeC K

(2.12)

for some new constant C, and in particular L I ≤ CeC K .

|I |=n

Next we estimate |TI |. Write Mk = {|TI | : Case 2 has been used k times on I1 , I2 , . . . , In = I }. To estimate Mk note that there is a constant β0 < 1 such that if we use Case 2 and L 0 + L 1 ≥ 0.9L , then β ≥ β0 and L 0 + L 1 ≤ C L . Furthermore, by (2.7) the alternative L 0 + L 1 ≥ 0.9L can only hold for at most N subindices Im . Therefore by (2.9) and (2.12),

Mk ≤ (1 + β02 )C N (0.9)k−N CeC K and

Mk ≤ C e C K .

(2.13)

k

At the nth stage of the construction we have a connected union n of rectangles and segments such that E ⊂ n . By (2.12) and (2.13) there is a contin-

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uous map σn : [0, 1] → n such that σn ([0, 1]) has bounded length, such that σn ([0, 1]) meets all four sides of each rectangle S I , |I | = n, and such that σn+1 is built by modifying σn on each σn−1 (S I ), with |I | = n. Then lim σn deﬁnes a rectiﬁable curve containing E. An earlier characterization of subsets of rectiﬁable curves was also given by Jones in [1990]. For any set E deﬁne β 2 (E) = diam(E) + {β E2 (3Q)(Q) : Q dyadic, (Q) ≤ diam(E)}. Theorem 2.3. Suppose E is a bounded set. Then E is contained in a rectiﬁable curve if and only if β 2 (E) is ﬁnite. Moreover, there are constants c1 , c2 so that c1 β 2 (E) ≤ inf 1 () ≤ c2 β 2 (E). ⊃E

(2.14)

Theorem 2.3 is sometimes called the “traveling salesman theorem” because it bounds the total distance the salesman must travel to visit a ﬁnite set E of cities when each pair of cities in E is connected by a straight road. Exercise 6 describes the minimal spanning tree problem and the simple “greedy algorithm” which gives a shortest spanning tree. However the greedy algorithm gives no a priori bound like (2.14) for arbitrarily large sets. See Garey and Johnson [1979] and Lawler, Lenstra, Rinnooy Kan and Shmoys [1990] for more information about spanning trees and the classical traveling salesman problem. We ﬁrst prove the right-hand inequality of (2.14), that there exists a curve with 1 () ≤ c2 β 2 (E). The argument is like the proof of Theorem 2.1. Let Rn be the sum of the diameters of the rectangles at stage n. From (2.8) we have L 0 + L 1 ≤ L + Cβ E2 (3Q)(Q)

(2.15)

for any Q such that Q ∩ E ∩ S = ∅ and diam(S) ≤ (Q) ≤ 2 diam(S). As noted before, each such Q can occur at most N = 25 times during iterations of the construction, so that Rn ≤ Cβ 2 (E). To estimate the lengths of the segments created by applications of Case 2, note again that if Case 2 is used and if L 0 + L 1 ≥ 0.9L, then β ≥ β0 and |T | ≤ Cβ E2 (3Q)(Q). Thus the sum of the lengths of the middle segments created from applications of Case 2 with β ≥ β0 is at most Cβ 2 (E). Now write Rn = I n + I I n where In is the sum of the lengths of the rectangles at stage n to which Case 1 will be applied or for which β ≥ β0 , and where I In is the sum of the lengths of

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the rectangles to which Case 2 will be applied and for which β < β0 . Let Tn+1 denote the sum of the lengths of the middle segments created in Case 2 with β < β0 . Then β 2 (3Q)(Q) + 0.9I In , Rn+1 ≤ In + C where the sum is taken over all Q which intersect a rectangle R, created at stage n and such that diam(R) ≤ (Q) < 2 diam(R). Moreover by (2.9), β 2 (3Q)(Q), (0.1)Tn+1 ≤ (0.1)I In + C and so Rn+1 + (0.1)Tn+1 ≤ Rn + C

Thus (0.1)Tn+1 ≤ (Rn − Rn+1 ) + C

β 2 (3Q)(Q).

β 2 (3Q)(Q),

and because Rn is bounded, Tn ≤ C diam(E) + β 2 (3Q)(Q). n

Q dyadic

The left-hand inequality in (2.14) will be proved in the next section. Here we only treat the special case of a Lipschitz graph. Lemma 2.4. Let be the graph of a Lipschitz function. Then (2.14) holds for all E ⊂ . Proof. We use an elegant argument from Jones [1990]. Assume that, after a translation and a dilation, = 0 ≤ x ≤ 1, y = f (x) , where f (0) = f (1) and f (x2 ) − f (x1 ) ≤ M|x2 − x1 |. It is enough to show the left side of (2.14) holds for E = . Set 6 j ( j + 1) 7 , I jn = n , 2 2n and nj = f (I jn ), and let Jjn be the segment 6 j ( j + 1) 7 Jjn = f ( n ), f ( ) . 2 2n

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J2n+1 j+1

J2n+1 j Jjn I jn

Figure X.10 n+1 n Then J2n+1 j , J2 j+1 , and Jj are the three sides of a triangle and # $ n+1 ∪ J δn, j = 2n sup dist(z, Jjn ) : z ∈ J2n+1 j 2 j+1

satisﬁes n+1 n −n 2 (J2n+1 j ) + (J2 j+1 ) − (Jj ) ≥ c2 δn, j ,

with c = c(M) > 0, again by the Pythagorean theorem. Therefore 2 c2−m δm,k ≤ 2().

(2.16)

m,k

Now set ∞ βn, j = 2n sup dist(z, Jjn ) : z ∈ nj ≤ 2n−m sup δm,k : Ikm ⊂ I jn . m=n

Write p = m − n and

sup δm,k : Ikm ⊂ I jn = δn+ p,k (n, j).

Then (2.16) and Minkowski’s inequality give #

2 2−n βn, j

$1 2

≤

n, j

≤ ≤

#

2−n

∞

2 $ 21

2− p δn+ p,k (n, j)

p=0 n, j ∞ # 2 2−n 2−2 p δn+ p,k (n, p=0 n, j ∞ # $1 p 2 2 2− 2 2−m δm,k p=0 m,k

$1 j)

2

(2.17)

1 ≤ 2c () 2 . Extend f to [−1, 2] by deﬁning f (x ± 1) = f (x), and let (t) be the translate {y = f (x + t)}. We get numbers βn, j (t) that also satisfy (2.17). But then

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for ﬁxed n,

β (3Q)2 ≤ C

(Q)=2−n−2

1

−1

j

βn, j (t)2 dt,

because with probability 1/2 there exists j such that {Rez − t : z ∈ Q} ⊂ In, j . That gives (2.14) for . We make two observations about the construction. First, if E is a ﬁnite set, then the curve constructed consists of line segments joining the points of E, and these segments are the TI from Case 2. Second, if E is connected, then by (2.14) C1 β 2 (E) ≤ 1 (E) ≤ C2 β 2 (E). If E is connected it is not necessary to construct the middle segments TI in Case 2. Without the TI , every stage of the construction yields a union of rectangles, each meeting E, and this union is connected because E is connected. Because the rectangles are shrinking, the constructed set is exactly E, although the parameterized curve may trace portions of E several times. The next result, from Bishop and Jones [1994], is a converse to Theorem 2.1. Theorem 2.5. Let be a Jordan curve. Then, except for sets of zero 1 measure, z is a tangent point of if and only if 1 dt β2 (z, t) < ∞. (2.18) t 0 Proof. Write Tn() for the set of tangent points of . In Chapter VI we proved that & Tn() = Kn , n

where, after an afﬁne transformation, K n = 0 ≤ x ≤ 1; y = f (x) = g(x) , where f (x) and g(x) are Lipschitz functions, and f (x) ≤ g(x), 0 < x < 1. Moreover, there is a function h(x) such that f (x) ≤ h(x) ≤ g(x), on 0 < x < 1, and there is a neighborhood Vn of K n such that Vn ∩ = y = h(x) : 0 < x < 1 .

(2.19)

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By Lemma 2.4, β 2f (3Q)(Q) < ∞ and βg2 (3Q)(Q) < ∞.

(2.20)

But whenever K n ∩ Q = ∅,

β2 (3Q) ≤ C β 2f (3Q) + βg2 (3Q) .

(2.21)

Indeed, we may assume that βg (3Q) is small and that β f (3Q) ≤ βg (3Q) by symmetry. Let S be a strip of width βg (3Q)(Q) containing 3Q ∩ {y = g(x)} and let T be a parallel strip having width Mβg (3Q)(Q) and having the same top edge as S. When M is large, 3Q ∩ {y = f (x)} ⊂ T (for otherwise by (2.19) β f (3Q) > βg (3Q)) and consequently 3Q ∩ {y = h(x)} ⊂ T and (2.21) holds with C = C(M). Now by (2.20) and (2.21) 2 β∩V (3Q)(Q) < ∞. n (Q)<1

Then (2.3) and integrating the left side of (2.18) against arc length on K n yields inequality (2.18) at almost every point of K n . Conversely, let E = {z ∈ : (2.18) holds}. Then by Theorem 2.1 and the construction in Chapter VI, & E = E0 ∪ En , n

where 1 (E 0 ) = 0 and where, after an afﬁne transformation, E n ⊂ {(x, f n (x)) : 0 < x < 1}, and f n is a Lipschitz function. We may suppose that z = (0, f n (0)), that f n is differentiable at x = 0, that f n (0) = 0, and that (0, f n (0)) is a point of density of E n ∩ . See Figure X.11. ir/2

−r

y = εx

r

Figure X.11

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For ε ﬁxed and for r small, {x ∈ (−r, r ) : | f n (x) | < ε|x|, (x, f n (x)) ∈ } 2r and

0

But then

r

β2 (0, t)

≥ (1 − ε)

dt < ε. t

ir r

−ir r , ∪B , =∅ ∩ B 2 4 2 4

because otherwise β (z, t) > 14 for all t ∈ ( r2 , r ). Therefore z is a cone point for both components of C \ , and almost all such points lie in Tn(). Theorem 2.5 is an improvement of Corollary VI.6.4. Let 1 and 2 be the components of C \ , let tθ j (t) be the length of an arc of j ∩ {z : |z − w| = t} that separates w from a point z j ∈ j and let ε(w, t) = max{|π − θ j (t)| : j = 1, 2}. Then clearly ε(w, t) ≤ Cβ(w, t). Therefore Theorem 2.5 implies that 1 dt ε2 (z, t) < ∞ t 0

(2.22)

at 1 almost every tangent point of , which is the assertion of Corollary VI.6.4. To our knowledge, it has not been determined if has a tangent at almost every point where (2.22) holds.

3. A Decomposition Theorem To complete the proof of the left-hand inequality in (2.14) we use the following decomposition theorem, also from Jones [1990]. We call a simply connected domain an M-Lipschitz domain if, after a translation and a dilation, 0 ∈ and ∂ = {r (θ )eiθ : 0 ≤ θ ≤ 2π },

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3. A Decomposition Theorem

where |r (θ1 ) − r (θ2 )| ≤ M|θ1 − θ2 | and 1 ≤ r (θ ) ≤ 1. M +1 Theorem 3.1. There is a constant M such that whenever is a simply connected domain with 1 (∂) < ∞ there exists a rectiﬁable curve such that \ =

∞ &

j ,

(3.1)

j=0

such that each j is an M-Lipschitz domain, and such that 1 (∂ j ) ≤ M1 (∂).

(3.2)

j

Proof. The proof is in two steps. The ﬁrst step is to use a corona construction as in Garnett [1981] or David and Semmes [1991] to obtain (3.1) and (3.2) with the Lipschitz domains j replaced by domains U j bounded by uniformly chord-arc curves. The second step is to further decompose each chord-arc domain U j into M-Lipschitz domains. √ Let ϕ be a conformal map from D onto . Write F = ϕ and g = log(ϕ ). The proof will use the familiar inequality ||g||B ≤ 6

(3.3)

many times. By the F. and M. Riesz theorem, F ∈ H 2 (see Exercise VI.1) and by (1.3) or Exercise II.8, 1 1 2 |ϕ (z)||g (z)| log d xd y = 4 |F (z)|2 log d xd y |z| |z| D D (3.4) ≤ 2||F ||2H 2 ≤ 21 (∂). Set D0 = {|z| < 21 } and U0 = ϕ(D0 ). Then by Theorem I.4.5, there is a constant M independent of such that ∂U0 is an M-Lipschitz domain. Since ϕ ∈ H 1 , we also have 1 (∂U0 ) ≤ 1 (∂).

(3.5)

Next form the dyadic Carleson boxes Q = {r eiθ : 1 − 2−n ≤ r < 1, π j2−(n+1) ≤ θ < π( j + 1)2−(n+1) },

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0 ≤ j < 2n+1 , of sidelength (Q) = 2−n and their top halves T (Q) = Q ∩ {1 − 2−n ≤ r < 1 − 2−(n+1) } & = Q \ {Q : Q ⊂ Q, Q = Q}. Write z Q for the center of T (Q).

zQ

Q

Figure X.12 Fix ε > 0 to be determined later and consider a Carleson box Q. If sup |g(z) − g(z Q )| ≥ ε,

T (Q)

(3.6)

we say Q is a type 0 box and deﬁne D(Q) = T (Q). If Q is of type 0, then again by (3.3) ∂U Q = ∂ϕ(D Q ) is a chord-arc curve with chord-arc constant bounded by a ﬁxed constant C0 . If (3.6) fails for Q, deﬁne G(Q) to be the set of maximal Carleson boxes Q ⊂ Q for which sup |g(z) − g(z Q )| ≥ ε,

T (Q )

and deﬁne D(Q) = Q \

&

(3.7)

Q.

G(Q)

Then ∂D(Q) is a chord-arc curve with chord-arc constant at most 4 and sup |g(z) − g(z Q )| ≤ ε.

D(Q)

(3.8)

It follows that the domain U Q = ϕ(D(Q)) is bounded by a chord-arc curve with constant at most 5 if ε is small.

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3. A Decomposition Theorem Q Q D(Q)

Q

Figure X.13 Write {D j , j ≥ 1} for the set of maximal (with respect to set inclusion) regions D(Q). Then the family {D0 } ∪ {D j , j ≥ 1} is pairwise disjoint. Write U j = ϕ(D j ), j = 0, 1, . . . . To prove (3.2) we ﬁrst show that 1 (∂U j ) ≤ M1 (∂). (3.9) j≥0

If D j = D(Q) for Q of type 0, then by (3.3) |ϕ (z)|ds 1 (∂U j ) = ∂ T (Q)

≤ c(Q)|ϕ (z Q )| 1 |ϕ (z)||g (z)|2 log d xd y. ≤c |z| T (Q)

Since 1 almost every point lies in at most 2 sets ∂U j , (3.4) then gives 1 (∂U j ) ≤ 21 (∂). (3.10) type 0

If Q is not of type 0, we say Q has type 1 if 1 (∂D ∩ ∂D Q ) ≥

(Q) , 2

and we say Q has type 2 if (3.11) fails. When D j = D(Q) for Q of type 1, (3.8), (3.3), and (3.11) give |ϕ (z)|ds ≤ C |ϕ (z)|ds. 1 (∂U j ) = ∂D j

bD∩∂D(Q)

(3.11)

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Therefore

1 (∂U j ) ≤ C

type 1

∂D

|ϕ (z)|ds ≤ 21 (∂).

(3.12)

Now assume D j = D(Q) where Q is a type 2 box. Then D j is a chord-arc domain and ∂D j is an Ahlfors regular curve, with constant independent of j. Let {Jk } denote the top edges of the boxes in G(Q). Then 1 (∂D(Q)) 1 (Jk ) ≥ 12 because Q is of type 2. By (3.3), (3.7), and equicontinuity there are δ > 0 and E k ⊂ Jk such that |g(z) − g(z Q )| ≥ δ on E k and 1 (E k ) ≥ δ1 (Jk ). Consequently |F(z) − F(z Q )|2 ds ≥ C(δ)(Q), |F(z Q )|2 ∂D(Q) and

1 (∂U j ) =

∂D(Q)

|F(z)| ds ≤ C (δ) 2

∂D(Q)

|F(z) − F(z Q )|2 ds.

We would like to estimate the latter integral via (1.7), but unfortunately D(Q) is not a Lipschitz domain. However D(Q) is a chord-arc domain, and hence Theorem M.1 of Appendix M gives1 |F(z) − F(z Q )|2 ds ∂D(Q) (3.13)

≤C |F (z)|2 1 B z, 2dist(z, ∂D(Q)) d xd y D(Q)

Now because ∂D(Q) is Ahlfors regular, (3.13) gives 2 |F(z) − F(z Q )| ds ≤ C |F (z)|2 dist(z, ∂D(Q))d xd y ∂D(Q)

D(Q)

|F (z)|2 log

≤C D(Q)

Therefore type2

1 (∂U j ) ≤ C D

|F (z)|2 log

1 d xd y. |z|

1 d xd y ≤ C1 (b). |z|

1 For a proof of (3.13) using Theorem 1.2. only, see Exercise 9.

(3.14)

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Together (3.5), (3.10), (3.12), and (3.14) yield (3.9). We have already showed that U0 = ϕ(D0 ) is an M-Lipschitz domain. If U j = ϕ(T (Q)) for a type 0 box Q, partition T (Q) into eight polar rectangles Sk by dividing the top edge of T (Q) into four equal intervals and halving its radial side. Then by (3.3) ϕ(Sk ) is an M-Lipschitz domain and 1 (∂ϕ(Sj )) ≤ C1 (∂U j ). j

Finally, suppose U j = ϕ(D(Q)) for a box Q not of type 0. The construction is easier to explain in the upper half-plane H and we assume Q is a square in H with base and interval I ⊂ R. See Figure X.14. Whenever J = Jk is the top edge of a box Q k ∈ G(Q) we construct a tree T = T (Q k ). Let S 1 be a segment of slope 1 joining the left endpoint of J to R and let S 2 be a segment of slope −1 joining the right endpoint of J to R, and deﬁne T 1 = J ∪ S1 ∪ S2. Assume by induction that T n ⊃ T n−1 has been deﬁned and that T n ∩ {Imz = 4−n (J )} 1 ⊂ T of slope ±1. Let S 2 consists of 2n points z n, j situated on segments Sn, n j n, j √ −n 1 be the segment of length 2 4 (J ) orthogonal to Sn, j that joins z n, j to R 2 n and set T n+1 = T n ∪ j Sn, n T satisﬁes j . Then T = T (Q k ) = ∞ √

2n 4−n (Q k ). 1 (T ) = 1 + 2 2 + n=1

Q Jk Qk I Figure X.14

(3.15)

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Write Dj \

&

T (Q k ) =

&

U j,k .

G(Q)

Then by (3.8) each ϕ(U j,k ) is an M-Lipschitz domain, and by (3.15) and (3.8), 1 (∂ϕ(U j,k )) ≤ C|ϕ (z Q )|(Q), k

which establishes (3.1) and (3.2).

Corollary 3.2. There exits a constant M < ∞ such that if is a connected ⊃ such plane set with 1 () < ∞, then there exists a connected plane set ) ≤ M1 (), the bounded components D j of C \ are M-Lipschitz that 1 ( , and the boundary of the unbounded component of domains with ⊂ ∂D j √ C\ is a circle at least 3 21 () units from . For the proof apply the theorem to each √ bounded component of the union of , a line segment and a circle at least 3 21 () units from . Now let be connected with 1 () < ∞, let {D j } be the Lipschitz domains given by Corollary 3.2 and write j = ∂D j and δ j = diamD j . Let Q be any dyadic square and deﬁne F(Q) = { j : j ∩ 3Q = ∅, δ j ≥ (Q)}, and G(Q) = { j : j ∩ 3Q = ∅, δ j < (Q)}. Then we have the following lemma. Lemma 3.3. There is a constant C1 such that if (Q) ≤ diam(), 1 β2 (3Q) ≤ C1 β2 j (3Q) + C1 2 Area(D j ). (Q) F (Q)

(3.16)

G(Q)

Proof. We may assume (Q) = 1 and β (3Q) > 0, so that 3Q ∩ j = ∅ for 2 some j and 3Q ⊂ D j . If F(Q) = ∅ then G(Q) Area(D j ) ≥ 9 (Q). Thus we can assume there exists 1 ∈ F(Q). Let L be a line such that d = sup dist(z, L) ≤ β1 (3Q)(3Q), 1 ∩3Q

and let z 0 ∈ ∩ 3Q have maximal distance d0 = dist(z 0 , 1 ). Let z 1 ∈ 1 have 1 minimal distance to z 0 and let z 2 = z 0 +z 2 . There are three cases, as shown in Figure X.15.

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L

1

z1

j

2 B

d1

z0

L

1

1

z3

Dj Case 1

Case 2

Case 3

Figure X.15 Case 1: F(Q) = {1 }. Then B = B(z 2 ,

& d0 Dj )⊂ 2 G(Q)

and hence β2 (Q) ≤ (d + d0 )2 ≤ 8β2 1 (2Q) + 2d02 ≤ 8β2 1 (2Q) +

8 Area(D j ). π G(Q)

Case 2: F(Q) = {1 , 2 } for disjoint D1 and D2 . In this case we may assume β2 j (3Q) < ε0 , j = 1, 2, since otherwise (3.16)

holds with C1 ∼ ε0−2 . Let d1 = sup2 ∩3Q dist(z, 1 ). Then if ε0 is sufﬁciently small β (3Q) ≤ β1 (3Q) + β2 (3Q) + d1 because for j = 1, 2, D j is an M-Lipschitz domain with diameter δ j ≥ (Q). Also because D2 is an M-Lipschitz domain there exists z 3 ∈ (3Q) \ D1 ∪ D2 such that dist(z 3 , j ) ≥ cd1 , j = 1, 2. Consequently there is constant C2 such that

β2 (3Q) ≤ C2 β2 1 (3Q) + β2 2 (3Q) + C2 d12

≤ C2 β2 1 (3Q) + β2 2 (3Q) + C1 Area(D j ). G(Q)

Case 3: F(Q) contains at least three distinct j . Then because each D j is an M-Lipschitz domain there exists at least one j ∈ F(Q) such that β j (3Q) ≥ C1 , and (3.16) is clear in this case.

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To ﬁnish the proof of Theorem 2.3, let be a rectiﬁable curve and let { j } be as in Corollary 3.2. By Lemma 2.4 β2 j (3Q)(Q) ≤ c2 ( j ). Q

2−n

If δ j < there are at most four dyadic squares Q such that (Q) = 2−n and D j ∈ G(Q). Hence 1 1 Area(D j ) = Area(D j ) (Q) (Q) Q

G(Q)

D j ∈G(Q)

j

≤4

Area(D j )

≤8

(2m δ j )−1

m=0

j

∞

Area(D j )(δ j )−1

j

≤C

( j ) ≤ C M().

j

Therefore by Lemma 3.3 β 2 () ≤ C 1 ().

4. Schwarzian Derivatives The Schwarzian derivative of a locally univalent analytic function is ϕ 1 ϕ (z) 2 Sϕ(z) = (z) − ϕ 2 ϕ (z) ϕ (z) 3 ϕ (z) 2 = . − ϕ (z) 2 ϕ (z)

(4.1)

A calculation with the ﬁrst expression in (4.1) shows that if T (z) =

az + b , cz + d

where ad − bc = 0, is a Möbius transformation, then ST (z) = 0.

(4.2)

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381

Conversely, if Sϕ = 0 on some disc, then for constants a and A = 0, ϕ (z) 2 = , ϕ (z) a−z

ϕ (z) =

A , (a − z)2

and ϕ(z) is a Möbius transformation. A calculation with the second expression in (4.1) gives

2 S(ϕ ◦ ψ)(z) = Sϕ(ψ(z)) ψ (z) + Sψ(z), (4.3) and therefore S(T ◦ ϕ)(z) = Sϕ(z)

(4.4)

whenever T is a Möbius transformation. If ϕ is a locally univalent meromorphic function we deﬁne

Sϕ = S 1/ϕ in a neighborhood of a pole of ϕ. By (4.4), this deﬁnition is consistent with the original deﬁnition (4.1) of Sϕ. If ϕ is deﬁned on the disc D and if T ∈ M is a self map of D, then by (4.2) and (4.3)

2 2 2

1 − |z|2 S(ϕ ◦ T )(z) = 1 − |z|2 Sϕ(T (z)) T (z) 2

= 1 − |T (z)|2 Sϕ(T (z)) . Consequently the expression

2 1 − |z|2 Sϕ(z)

(4.5)

is a conformal invariant on D. Theorem 4.1. If ϕ is univalent on D, then

2 1 − |z|2 Sϕ(z) ≤ 6.

(4.6)

Conversely, if ϕ is analytic on D and if

2 sup 1 − |z|2 Sϕ(z) < 2,

(4.7)

D

then ϕ is univalent and ∂ϕ(D) is a quasicircle. Proof. Suppose ϕ is univalent on D. By (4.4) we may assume ϕ(z) = z + a2 z 2 + a3 z 3 + . . . .

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By the conformal invariance of (4.5) we only need to prove (4.6) at z = 0, where it becomes 3 ϕ (0) − (ϕ (0))2 = |6a3 − 6a22 | ≤ 6. 2 1 n = 1z + ∞ But since the function F(z) = ϕ(z) n=0 bn z is univalent, the area theorem, Lemma I.4.2, gives |a3 − a22 | = |b1 | ≤ 1, which proves (4.6). We omit the proof of the converse (4.7). Nehari ﬁrst proved ϕ is univalent in [1949], and it is the famous theorem of Ahlfors and Weill [1962] that ∂ϕ(D) is a quasicircle. See also Ahlfors [1963] and Lehto [1987]. The constants 6 and 2 2 2 in (4.6) and (4.7) are both sharp. However, if supD (1 − |z| ) Sϕ(z) ≤ 2, then ϕ is univalent (Nehari [1949]), and if (1 − |z|2 )2 |Sϕ(z)| < 2 for all z ∈ D, then ∂ϕ(D) is a Jordan curve (Gehring and Pommerenke [1984]). We will also need the local version of Theorem 4.1. Theorem 4.2 (Ahlfors). Let U be a K -quasidisc and let ψ : U → D be conformal. Then there is ε = ε(K ) > 0 such that if ϕ is meromorphic on U and if 1 − |ψ(z)|2 2 Sϕ(z) < ε, sup |ψ (z)| U then ϕ extends to a quasiconformal mapping of C ∗ . In particular, ϕ is one-to-one on U and ∂ϕ(U ) is a quasicircle. We will not prove Theorem 4.2 either. See Ahlfors [1963] or Lehto [1987]. Lehto [1987] has a much fuller discussion of Theorems 4.1 and 4.2, which lead to the Bers identiﬁcation of universal Teichmüller space with an open connected subset of the space of Schwarzians of univalent functions under the norm

2 ||ϕ||S = sup 1 − |z|2 Sϕ(z) . D

Gehring [1977] proved a converse of Theorem 4.2: If U is a simply connected domain for which the conclusion of Theorem 4.2 holds for some ε > 0, then U is a quasidisc. We always write

(4.8) g(z) = log ϕ (z)

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when ϕ is univalent on D. Theorem 4.1 is reminiscent of Theorem VII.2.1, which said that for ϕ univalent ϕ (z) (1 − |z|2 )|g (z)| = (1 − |z|2 ) ≤ 6, (4.9) ϕ (z) and (4.7) resembles Becker’s [1972] converse: If ||g||B = sup(1 − |z|2 )|g (z)| < 1, D

then ϕ is univalent. In fact, the interplay between the two conformal invariants (1 − |z|2 )|g (z)| and

2 1 − |z|2 Sϕ(z)

2 will be crucial throughout this chapter. 2 Note that if (1 − |z| )|g (z)| is small on 2 Sϕ(z) is also small on compact subsets a hyperbolic ball B, then 1 − |z| z−z 1 of B. However, if ϕ(z) = z−z , then Sϕ = 0, but 2

g (z) =

ϕ (z) 2 = ϕ (z) z2 − z

(4.10)

is not small. Lemma 4.3. Let ε > 0, a > 0, b > 0, and c > max(a, b) be given. Then there is δ = δ(ε, a, b, c) such that if ϕ is univalent on the hyperbolic ball {z : ρ(z, z 0 ) < c}, and if

2 1 − |z|2 Sϕ(z) < δ (4.11) on the hyperbolic ball {z : ρ(z, z 0 ) < a}, then for some Möbius transformation T , z − z z − z 0 0 (4.12) + ϕ(z) − T −1 <ε T (ϕ(z)) − 1 − z0 z 1 − z0 z on the hyperbolic ball {z : ρ(z, z 0 ) < b}. Proof. We may assume z 0 = 0. There is a unique Möbius transformation T so that T ◦ ϕ(0) = 0, (T ◦ ϕ) (0) = 1 and (T ◦ ϕ) (0) = 0,

(4.13)

and we replace ϕ by T ◦ ϕ. By (4.8) 1 Sϕ(z) = g (z) − (g (z))2 , 2

(4.14)

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and under the substitution g = −2

h , h

(4.14) becomes Sϕ h = 0. 2 This differential equation has a unique solution satisfying h +

h(0) = 1 and h (0) = 0, and the solution is obtained by power series: If Sϕ(z) =

∞

an z n ,

n=0

and h(z) = 1 +

∞

bn z n ,

n=2

then for n ≥ 2, an−2 bk an−2−k n(n − 1)bn = − − . 2 2 n−2

(4.15)

k=2

If δ is sufﬁciently small, then by (4.11) all an are small for small n. Thus for any τ > 0 there is δ > 0 so that (4.11) and (4.15) give sup |h(z) − 1| < τ.

ρ(z,0)
(4.16)

Unwinding ϕ from h now gives (4.12) for ε, if (4.16) holds for τ = τ (ε).

5. Geometric Estimates of Schwarzian Derivatives Let ϕ be the conformal map of D onto a bounded simply connected domain . In this section we give two different geometric estimates for the invariant

2 Schwarzian 1 − |w|2 |Sϕ(w)|. The two estimates are not equivalent, but both estimates measure how much deviates from a half-plane or disc when is observed from the vantage point ϕ(w). The second estimate uses the numbers β E from Section 1 while the ﬁrst estimate uses a more function theoretic quantity η (z θ , t) deﬁned immediately below.

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Fix eiθ ∈ ∂D such that the nontangential limit z θ = ϕ(eiθ ) exists. For t > 0, let L be the set of lines L such that (a) L ∩ B(z θ , 2t ) = ∅, and such that one component Wt (θ ) of B(z θ , 2t) \ L satisﬁes both (b) Wt (θ ) ⊂ , and (c) when ϕ{(r eiθ : 0 < r < 1} is oriented by increasing r , Wt (θ ) contains the ﬁrst arc of ϕ({r eiθ : 0 < r < 1}) ∩ B(z θ , 2t) \ L , If L = ∅, deﬁne η (z θ , t) =

$ # 1 inf sup dist(z, ∂) : z ∈ L ∩ B(z θ , 2t) , 2t L∈L

and if L = ∅, put η (z θ , t) = 1. ϕ(r eiθ )

ϕ(r eiθ )

∂

∂ Wt (θ )

Figure X.16 On the left η ∼ β; on the right η = 1 while β is small. See also Exercise 12 for some remarks about the relation between η (z, t) and β (z, t). Theorem 5.1. Let z 0 = ϕ(w0 ) ∈ and let z θ = ϕ(eiθ ) ∈ ∂ satisfy d = |z θ − z 0 | ≤ 2dist(z 0 , ∂). Then for s, 0 < s < 1, there is C = C(s), depending only on s, such that ∞

2 1 − |w0 |2 Sϕ(w0 ) ≤ C η (z θ , 2k d)2−ks .

(5.1)

k=0

Proof. We may assume z 0 = w0 = 0 and d = 1. Write ηk = η (z θ , 2k d). Fix δ = δ(s), 0 < δ < δ0 where δ0 will be determined later. We can suppose η0 ≤ δ, because otherwise (4.6) gives (5.1) with constant C = 6δ . Since is bounded, limk→∞ ηk = 1 and there exists a ﬁrst integer N such that η N +1 > δ. Let L 0 be a line with L 0 ∩ B(z θ , 21 ) = ∅ such that L 0 gives the minimum η0 . The line L 0 exists because η0 ≤ δ. Let H0 be the component of C \ L 0 such

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that H0 ∩ B(z θ , 2) ⊂ . Then z 0 ∈ H0 by the ﬁrst arc condition (c), because z 0 = ϕ(0) ∈ B(z θ , 2). For k ≤ N , let L k be a line with L k ∩ B(z θ , 2k−1 ) = ∅ such that L k gives the minimum ηk . Again L k exists because ηk ≤ δ. Let Hk be that component of C \ L k with Hk ∩ B(z θ , 2k+1 ) ⊂ . Then by condition (c) Hk ∩ Hk−1 ∩ B(z θ , 2k−1 ) = ∅. Write Bk = B(z θ , 2k ) and Ak = Bk+1 \ Bk . Set V0 = H0 and for 1 ≤ k ≤ N ,

&

(Hj ∩ A j ) ∪ (Hk \ Bk ) ∩ . Vk = int H0 ∩ B(z θ , 2) ∪ 1≤ j≤k−1

H2 \ B2 H1 ∩ A1 L2

z0

∂

L1 z 0∗

∂

Figure X.17 The region V2 . / Vk if z 0∗ Then Vk is a simply connected domain, Vk ⊂ , and z 0 ∈ Vk but z 0∗ ∈ is the reﬂection of z 0 through L 0 = ∂ V0 . By the distortion theorem, |ϕ (0)| ≥ dist(z 0 , ∂) ≥ 1/2, and if δ0 is small then + σ = ϕ({|w| = 1/4}) ⊂ Vk . Set G(z) = g (z, z 0 ) and G k (z) = gVk (z, z 0 ). Then the theorem is an immediate consequence of the next three lemmas: Lemma 5.2. There is a constant C, independent of ϕ, such that Sϕ(0) ≤ C sup G(z) − G 0 (z) . σ

Lemma 5.3. If z ∈ σ, then |G N (z) − G 0 (z)| ≤ C(s)

∞ k=0

ηk 2−ks .

(5.2)

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Lemma 5.4. If z ∈ σ, then |G(z) − G N (z)| ≤ C(s)

N −1

ηk 2−ks + C(s)2−N s .

(5.3)

k=0

Proof of Lemma 5.2. By (4.6) we can assume ε = sup G(z) − G 0 (z) σ

is small, because otherwise the lemma holds with C = 6ε . Let T (w) = the conformal map of D onto V0 . Then sup − log |ϕ −1 (z)| + log |T −1 (z)| = ε,

z 0∗ w w−1

be

σ

so that on

ϕ −1 (σ )

= {w : |w| = 1/4}, T −1 (ϕ(w)) 1−ε ≤ ≤ 1 + ε. w

Because T −1 (0) = 0, it follows via Taylor series that sup T −1 (ϕ(w)) − λw ≤ Cε |w|≤1/8

for some constant λ with |λ| = 1 and therefore that Sϕ(0) = S(T −1 ◦ ϕ)(0) ≤ Cε,

which proves Lemma 5.2. Proof of Lemma 5.3. First notice that by the deﬁnition of η(z, t),

dist L k ∩ ∂ B k , L k−1 ∩ ∂ B k ≤ C2k (ηk + ηk−1 ).

(5.4)

Let r (r ) be the length of VN ∩ {|z| = r }. For k ≥ 1 and 2k ≤ r ≤ 2k+1 we have {|z| = r } ⊂ Ak−1 ∪ Ak ∪ Ak+1 , so that by (5.4)

(r ) − π ≤ C(ηk−1 + ηk + ηk+1 ) ≤ Cδ,

and (r ) ≤ π + Cδ. Now choose δ so that

π ≥ s. π + Cδ

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Then for z ∈ σ and k ≥ 2, Theorem IV.6.1 gives us * , 2k−1 dr ω(z, Ak ∩ ∂ VN , VN ) ≤ C exp −π r (r ) 1 * , 2k−1 −π dr ≤ C exp ≤ C2−ks . π + Cδ 1 r

(5.5)

In the same way, we also get ω(z, (∂ VN ) \ B N , VN ) ≤ C2−N s . The function

z − z∗ 0 G 0 (z) = log z − z0

is harmonic on C \ {z 0 , z 0∗ } and when k ≥ 1, sup Ak |∇G 0 (z)| ≤ C2−k since |z 0 − z θ | ≤ 2. If z ∈ Ak ∩ ∂ VN , then min(k,N )

dist(z, ∂ V0 ) ≤ C2k

ηn ,

n=0

and therefore G 0 (z) ≤ C

k

ηn .

n=0

But then since G N − G 0 is harmonic on VN , (5.5) gives G 0 (ζ )dω(z, ζ ) G N (z) − G 0 (z) = ∂ VN

≤C

N min(k,N ) n=0

k=0

≤C

ηk

k

≤ C(s)

ηn 2−ks

2−ns

n≥k ∞

ηk 2−ks ,

k=0

and that proves (5.2). Proof of Lemma 5.4. Let be a curve in VN such that for z ∈ ∩ Ak c1 dist(z, L k ) ≤ 2k (ηk−1 + ηk + ηk+1 ) ≤ c2 dist(z, L k ),

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provided ηk−1 + ηk + ηk+1 > 0, and such that if ζk (s) is the arc length parameterization of ∩ Ak , then |ζk | ≤ c, and |ζk | ≤ c2−k . This curve exists because of (5.4). If ηk−1 + ηk + ηk+1 = 0, then ∂ ∩ Ak+1 falls in a line segment and in this case let ∩ Ak be a parallel segment such that dist( ∩ Ak , ∂) ≥ δ2k .

D

B2 B1 z0

L2 ∂

L1

z 0∗

∂

Figure X.18 Let D ⊂ VN be the simply connected domain such that z 0 ∈ D and such that ∂D consists of two arcs 1 = B N +1 ∩ and 2 ⊂ ∂ B N +1 . Partition ∂D into arcs I j such that dist(I j , ∂) ∼ diam(I j ). Then if I j ⊂ 1 ∩ Ak , diam(I j ) ≤ c2k (ηk−1 + ηk + ηk+1 ). Let z ∈ D ∩ Ak satisfy dist(z, ) ∼ 2k . Then for I j ⊂ Ak ∩ 1 , a simple comparison using the smoothness of ∩ Ak yields ω(z, I j , D) ≤ C diam(I j )2−k . By the proof of (5.5), ω(z 0 , Ak ∩ 1 , D) ≤ C2−ks . Therefore, ω(z 0 , I j , D) ≤ C diam(I j )2−k 2−ks ≤ C2−ks (ηk−1 + ηk + ηk+1 ). On the other hand, if I j ⊂ 2 , then

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also by the proof of (5.5). Altogether these inequalities give us ω(z 0 , I j , D) ≤ 2−N s +

N

ηk 2−ks .

(5.6)

k=0

Let Jj = ϕ −1 (I j ). Then sup G(z) = sup sup log ∂D

j

Jj

1 ∼ sup sup(1 − |w|). |w| j Jj

(5.7)

Let w ∗ ∈ ϕ −1 (∂D) satisfy 1 − |w ∗ | = supϕ(∂D) (1 − |w|) and say w∗ ∈ Jn . Let Jn±1 denote the two arcs adjacent to Jn and set J = Jn−1 ∪ Jn ∪ Jn+1 . Then if w ∈ ϕ −1 (∂D) \ J, we have |w| ≥ |w ∗ | and ρ(w, w ∗ ) ≥ C. Set

U = ϕ −1 (D) ∩ {|w| < |w∗ |} ∪ {ρ(w, w ∗ ) < C} .

ϕ −1 (D) U

J Figure X.19 Then by (5.6) and (5.7), sup G(z) ≤ C1 (1 − |w ∗ |) ≤ C2 ω(0, J, U) ∂D

≤ C2 ω(z 0 , In−1 ∪ In ∪ In+1 , D) N −1

≤ C2 2−N s + ηk 2−ks . k=0

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Therefore, since ⊃ VN ⊃ D ⊃ σ , sup |G − G N | = sup(G − G N ) σ

σ

≤ sup(G − G N ) ≤ sup G(z) ∂D

≤ C2 2−N s +

∂D

N −1

ηk 2−ks .

k=0

That proves (5.3).

The second estimate of 1 − |w|2 )2 Sϕ(w) uses the β numbers from Section 2. It will also be valid in the more general situation in which ϕ is replaced by a covering map. Let E be a compact set having at least three points and let ψ : D → C∗ \ E be a universal covering map, which by deﬁnition means that ψ is analytic, maps D onto C∗ \ E, and is locally univalent, i.e., ψ (z) = 0. (It is a famous theorem of Koebe that such ψ exist when E has at least three points, see Ahlfors [1973], Fisher [1983], or Gamelin [2001].) Universal covering maps satisfy a weakened form of Theorem I.4.3.

Lemma 5.5. If ψ(w0 ) = z 0 , and if ψ is univalent on B(w0 , η(1 − |w0 |2 )), then dist(z 0 , E) ≤ |ψ (w0 )|(1 − |w0 |2 ) ≤

4 dist(z 0 , E). η

Proof. We may take z 0 = 0 and w0 = 0. Then the right-hand inequality is the ψ(ηw) −1 Koebe estimate (I.2.11) applied to ηψ (0) . By the monodromy theorem, ψ has a single valued univalent branch on B(0, dist(0, E)) satisfying ψ −1 (0) = 0, and the left-hand inequality is the Schwarz lemma for this branch of ψ −1 . When z 0 ∈ / E, and let Q = Q 0 be the smallest dyadic square such that z 0 ∈ Q and dist(z 0 , E) ≤ (Q). Let Q n be that dyadic square of side 2n (Q) such that Q ⊂ Q n and set βn = β E (10Q n ) as deﬁned in (2.2). Let Sn be a closed strip of width 10 · 2n+1 βn (Q) such that E ∩ 10Q n ⊂ Sn , let L n be the axis of Sn , and let E n∗ be the orthogonal projection of E ∩ 10Q n onto L n . Let m n be the length of the largest interval in (L n ∩ 10Q n ) \ E n∗ and set mn , δn = βn + γn . γn = γ E (Q n ) = (Q n )

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Clearly γn measures the largest gap in E n∗ . Theorem 5.6. Fix 0 < η < 1 and 0 < s < 1. Let ψ : D → C∗ \ E be a univerψ(w0 ) = z0 . Then there is C = C(η, s) sal covering map and w0 ∈ D satisfy

such that if ψ is univalent on B w0 , η(1 − |w0 |) , then ∞ 2 δn 2−ns . 1 − |w0 |2 Sψ(w0 ) ≤ C

(5.8)

n=0

Proof. The argument resembles the proof of Theorem 5.1. Fix δ = δ(s).We can suppose δ ≤ δ0 where δ0 is to be determined later, because otherwise (4.6) gives (5.8) with C = 6/δ. If N denotes the ﬁrst integer such that δ N +1 > δ then N > 1 and N is ﬁnite, because eventually γn is large. We assume that w0 = 0 and that (Q 0 ) = 1. Let S0 be a closed strip of width 2β0 such that E ∩ 10Q 0 ⊂ S0 . Since β0 < δ, z 0 ∈ / S0 . Let H0 be the component of C \ S0 with z 0 ∈ H0 . For n < N , let Sn be a closed strip of width 2n+1 βn such that Sn ∩ Q n = ∅ and E ∩ 10Q n ⊂ Sn , and let Hn be that component of C \ Sn such that Hn ∩ Hn−1 ∩ Q (n−1) = ∅. The component Hn exists because 2βn + βn−1 < 1, and Hn is unique because diam(E) ≥ 2n . This time write An = 5Q n \ 5Q n−1 . Set V0 = H0 and for 1 ≤ n ≤ N , Vn = int (H0 ∩ 5Q 0 ) ∪ (H1 ∩ A1 ) ∪ . . . ∪ (Hn ∩ An ) . Then Vn is a simply connected domain, Vn ⊂ (C∗ \ E), and z 0 ∈ Vn but if z 0∗ is the reﬂection of z 0 through ∂ V0 , then z 0∗ ∈ / Vn . The analogue of (5.4) holds because βn + βn−1 is small. We continue to assume w0 = z 0 = 0. By the univalence hypothesis and the distortion theorem, |ψ (0)| ≥ dist(z 0 , E) and + ψ({|w| ≤ η/8}) ⊂ Vn . Set σ = ψ({|w| = η/8}).

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0 is the Möbius transformation from Deﬁne G n = gVn (z, z 0 ). If T (w) = w−1 D to the half-plane V0 with T (0) = 0, then 1 G 0 (T (w)) = log . w

Since ψ (w) = 0, there exists, by the monodromy theorem, a unique branch of ψ −1 deﬁned on VN and satisfying ψ −1 (0) = 0. Using ψ −1 , we deﬁne the harmonic function g(z), z ∈ VN \ {0} by 1

(5.9) G ψ(w) = log w on ψ −1 (VN ). At this point the proofs of Lemmas 5.2, 5.3, and 5.4 can be repeated with ϕ replaced by ψ and with ηk replaced by δk to produce a proof of Theorem 5.6. Note that (5.7) holds for the new G because of the deﬁnition (5.9). We leave the details to the reader.

6. Schwarzian Derivatives and Rectiﬁable Quasicircles Let ϕ be a conformal mapping from D onto a bounded Jordan domain . We

1 write F = ϕ 2 and, as always, g = log(ϕ ). Then 1 F = Fg , 2 F = and

1 1 2 1

F g + (g ) = F Sϕ + (g )2 , 2 2 2

(6.1)

2 2 |ϕ | Sϕ ≤ 2|ϕ | Sϕ + (g )2 + 2|ϕ ||g |4 ≤ 8|F |2 + 8|F |2 |g |2 .

Because ||g||B ≤ 6, we therefore have 2 2 3 |ϕ (z)| Sϕ(z) (1 − |z| ) d xd y ≤ 8 |F |2 (1 − |z|2 )3 d xd y D

+ 8 · 36

|F |2 (1 − |z|2 )d xd y.

By Theorem 1.2 both terms on the right are bounded by C||F||2H 2 = ||ϕ || H 1 and by the F. and M. Riesz theorem, ||ϕ || H 1 = (). Therefore we have the

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following lemma. Lemma 6.1. There is a constant C such that if ∂ is rectiﬁable, then |ϕ (z)||Sϕ(z)|2 (1 − |z|2 )3 d xd y ≤ C(∂). D

When ∂ is a quasicircle, Bishop and Jones [1994] proved the converse of Lemma 6.1. Theorem 6.2. If is a bounded quasidisc, then ∂ is rectiﬁable if and only if |ϕ (z)||Sϕ(z)|2 (1 − |z|2 )3 d xd y < ∞, D

and there is a constant C1 such that −1 C1 (∂) ≤ diam() + |ϕ (z)||Sϕ(z)|2 (1 − |z|2 )3 d xd y D

(6.2)

≤ C1 (∂). The constant C1 depends only on the quasiconformality constant K of the quasicircle ∂. See Theorem VII.3.3. Proof. By Lemma 6.1 and the trivial estimate diam() ≤ (∂), the right-hand inequality of (6.2) holds, even when ∂ is not a quasicircle. To prove the left-hand inequality we must come to grips with the set where |g (z)| is large but |Sϕ(z)| is small. Let {Sj } denote the Whitney squares for and ﬁx ε > 0 and δ > 0. We write Sj ∈ Bε,δ and say Sj is a bad square if sup |g (z)|(1 − |z|2 ) ≥ ε,

(6.3)

sup Sϕ(z) (1 − |z|2 )2 ≤ δ.

(6.4)

ϕ −1 (Sj )

but

ϕ −1 (Sj )

Lemma 6.3. There is ε0 > 0 and C > 0, such that if 0 < ε < ε0 and δ > 0, then |ϕ |ds ≤ C dist(ϕ(0), ∂) + C (Sj ) ∂D

+ C(1 + δ −2 )

D

Bε,δ

2 |ϕ (z)| Sϕ(z) (1 − |z|2 )3 d xd y

(6.5)

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{ϕ −1 (Sj ) : Sj ∈ Bε,δ }. Using Koebe’s theorem and using

1 Theorem 1.2 twice with F = ϕ 2 , we get |ϕ (z)|ds ∼ dist(ϕ(0), ∂) + |ϕ (z)||g (z)|2 (1 − |z|2 )d xd y, (6.6)

Proof. Write B =

∂D

D

and by (6.1) |ϕ (z)|ds ≤ ∂D

C0 dist(ϕ(0), ∂) + C0 |F (0)|2 (6.7) (z))2 2 (g + C0 |ϕ (z)| Sϕ(z) + (1 − |z|2 )3 d xd y, 2 D

for some constant C0 . Since ||g||B ≤ 6 we have |F (0)|2 =

|ϕ (0)||g (0)|2 ≤ 36 dist(ϕ(0), ∂), 4

so that (6.7) gives |ϕ (z)|ds ≤ C1 dist(ϕ(0), ∂) ∂D 2 |ϕ (z)| Sϕ(z) (1 − |z|2 )3 d xd y + C1 D

+ C1

|ϕ (z)||g (z)|4 (1 − |z|2 )3 d xd y

B

+ C1

|ϕ (z)||g (z)|4 (1 − |z|2 )3 d xd y,

D\B

for some constant C1 . To estimate the integral over B we use ||g||B ≤ 6 and the Koebe theorem to get |ϕ (z)| 4 2 3 4 |ϕ (z)||g (z)| (1 − |z| ) d xd y ≤ 6 d xd y 1 − |z|2 B

Bε,δ

≤ 64 C1

ϕ −1 (Sj )

(Sj ).

Bε,δ

For the remaining integral we note that on D \ B either (6.3) fails, or (6.4) fails but still |g (z)|(1 − |z|2 ) ≤ 6. Furthermore, if (6.4) fails for Sj , then by

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Theorem 4.1 and the Schwarz lemma, there are constants c1 and c2 such that |Sϕ(z)|(1 − |z|2 )2 ≥ c1 δ on a disc E j ⊂ ϕ −1 (Sj ) with |E j | ≥ c2 |ϕ −1 (Sj )|. Because sup |ϕ (z)| ≤ c3 inf |ϕ (z)| ϕ −1 (Sj )

ϕ −1 (Sj )

we consequently have |ϕ (z)||g (z)|4 (1 − |z|2 )3 d xd y D\B

≤ε

2

|ϕ (z)||g (z)|2 (1 − |z|2 )d xd y

D

64 C + 2 δ

2 |ϕ (z)| Sϕ(z) (1 − |z|2 )3 d xd y.

D

Therefore |ϕ (z)|ds ≤ C1 dist(ϕ(0), ∂) ∂D

64 + C1 1 + 2 δ + 6 C1 4

2 |ϕ (z)| Sϕ(z) (1 − |z|2 )3 d xd y

D

(Sj ) + C1 ε

j

2

|ϕ (z)||g (z)|2 (1 − |z|2 )d xd y.

D

Then by (6.6) and (6.8), 2 |ϕ (z)|ds ≤ C1 dist(ϕ(0), ∂) + C2 ε ∂D

(6.8)

∂D

|ϕ (z)|ds

2 64 |ϕ (z)| Sϕ(z) (1 − |z|2 )3 d xd y + C1 1 + 2 δ D 4 (Sj ), + 6 C1

Bε,δ

and whenever C2 ε2 < 1, (6.5) holds with C =

64 C 1 . 1−C2 ε2

Now assume is a quasidisc with constant K . By a theorem of Ahlfors [1963], there is A = A (K ) such that whenever ⊂ D is a circle, ϕ() satisﬁes the Ahlfors three-point condition (VII.2.5) with constant A .

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397

When T ∈ M, replacing ϕ by ϕ ◦ T does not change the integral in (6.2) and so we can assume dist(ϕ(0), ∂) ∼ diam().

(6.9)

In that case we claim that if δ is sufﬁciently small, then every bad square is a bounded hyperbolic distance from ϕ(0). By (6.9) and Lemma 6.3, that claim will prove the theorem. Suppose Sj is a bad square such that ϕ(z 0 ) ∈ Sj where 1 − |z 0 |2 is small and where (6.3) holds at z 0 . If δ is small, then by Lemma 4.3 there is a Möbius transformation z − z1 Tz = A z − z2 such that

T − ϕ < ε

(6.10)

on the hyperbolic ball B = {z : ρ(z, z 0 ) ≤ b} and T (z ) ε 0 (1 − |z 0 |2 ) ≥ . T (z 0 ) 2 Hence by (4.10), the pole z 2 of T satisﬁes dist(z 2 , ∂ B) ≤ |z 2 − z 0 | ≤

4 (1 − |z 0 |2 ), ε

while for b = b(ε) ﬁxed so that sinh(2b) = ε −2 , ∂ B has euclidian radius

r ≥ ε −2 (1 − |z 0 |) + O (1 − |z 0 |)3 . Consequently, if 1 − |z 0 | is small, there are adjacent arcs I ⊂ ∂ B and J ⊂ ∂ B with (I ) = (J ) ≥ √2ε (1 − |z 0 |) such that C (T (I )) ≥ . (T (J )) ε But by (6.10), that contradicts (VII.2.5) for ϕ(∂ B).

7. The Bishop–Jones H 2 −η Theorem 1

When is not a quasidisc the condition 2 |ϕ (z)| Sϕ(z) (1 − |z|2 )3 d xd y < ∞ D

(7.1)

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of Theorem 6.2 no longer implies that ∂ is rectiﬁable. For example, if is a half-plane, then Sϕ = 0 but ∂ is not rectiﬁable. However in [1994] Bishop and Jones obtained a sharp substitute theorem, and the proof of this theorem is the key to the deeper results in this chapter. Theorem 7.1. Let ϕ be the conformal mapping from D onto . If 2 B = |ϕ (0)| + |ϕ (z)| Sϕ(z) (1 − |z|2 )3 d xd y < ∞, D 1

then for any η, 0 < η < 21 , ϕ ∈ H 2 −η , and ||ϕ ||

1

H 2 −η

≤ C(η)B.

(7.2)

In particular, if the Bishop–Jones integral (7.1) is ﬁnite, then ϕ has non-zero angular derivative almost everywhere on ∂D, the cone points of ∂ have full harmonic measure relative to , and ω 1 by Theorem VI.4.2. Theorem 7.1 is sharp; again the counterexample is the map ϕ from D to a half-plane. For the applications to come, Theorem 7.1 will be less important than its local versions, Theorem 7.2 and Corollary 7.3. Recall we always write g = log(ϕ ). Theorem 7.2. Let E ⊂ ∂D be compact. Let U = ζ ∈E β (ζ ) be a cone domain, let 1 < α < β, and let ε > 0. Let ϕ be the conformal mapping from U onto a simply connected domain and assume that at every ζ ∈ J ⊂ E, 2 |ϕ (z)| Sϕ(z) (1 − |z|2 )2 d xd y ≤ N < ∞. (7.3) α (ζ )

Then there is C(N , ε) < ∞ and there exists J0 ⊂ J such that |J0 | ≥ (1 − ε)|J | and |ϕ (z)||g (z)|2 d xd y < C(N , ε) (7.4) α (ζ )

at every ζ ∈ J0 .

Corollary 7.3. Let E ⊂ ∂D be compact. Let U = ζ ∈E β (ζ ) be a cone domain, and let 1 < α < β. Let ϕ be the conformal mapping from U onto a simply connected domain . Then at almost every ζ ∈ E for which 2 |ϕ (z)| Sϕ(z) (1 − |z|2 )2 d xd y < ∞ (7.5) α (ζ )

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|ϕ (z)||g (z)|2 d xd y < ∞.

399

(7.6)

α (ζ )

The corollary follows by sending ε → 0 and N → ∞ in Theorem 7.2. Since

1 F(z) = ϕ (z) 2 satisﬁes |F (z)|2 = |ϕ (z)||g (z)|2 , Theorem 1.3 and Corollary 7.3 imply ϕ has a non-zero angular derivative at almost every point where (7.5) holds. The converse assertion, that (7.5) holds almost everywhere (7.6) holds, is

1 very easy. Since g = log(ϕ ) and F = ϕ 2 we have |ϕ ||g |2 = 4|F |2 , so that (7.6) implies 2 F (z) d xd y < ∞, α (ζ )

and by Lemma 1.5,

2 F (z) (1 − |z|)2 d xd y < ∞

δ (ζ )

for any δ, 1 < δ < α. Also recall from Section 6, |ϕ ||Sϕ|2 ≤ 8|F |2 + 8|g |2 |F |2 . Since ||g||B ≤ 6, we conclude that 2 |ϕ (z)| Sϕ(z) (1 − |z|2 )2 d xd y < ∞ δ (ζ )

if (7.6) holds at ζ, and a point of density argument then implies (7.5) holds almost everywhere that (7.6) holds. The proof of Theorem 7.2 resembles the main argument in the proof of Theorem 7.1 and we will prove Theorem 7.1 ﬁrst and then indicate the changes needed to get Theorem 7.2. Proof of Theorem 7.1.. We take B = 1. Set (ζ ) = z : |z − ζ | < 1 − |z| . Given η, 0 < η < 21 and given λ ≥ λ0 = λ0 (η) > 1, we construct a region R ⊂ D and a compact set E ⊂ ∂D such that & (ζ ) ⊂ R, (7.7) ζ ∈E

∂D \ E ≤ Cλ−1+2η ,

(7.8)

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and

|ϕ ||g |2 (1 − |z|2 )d xd y ≤ Cλ,

(7.9)

R

where the constant C in (7.8) and (7.9) depends on η but not on λ. Assume that we have built sets R and E satisfying (7.7), (7.8), and (7.9).

1 Write p = 21 − η and as usual take F = ϕ 2 . Then ϕ ∈ H p if and only if F ∈ H 2 p and ||ϕ || H p = ||F||2H 2 p . Recall the area function AF(ζ ) =

1 F (z) 2 d xd y 2 (ζ )

=

1 4

|ϕ (z)||g (z)|2 d xd y

1 2

.

(ζ )

By Theorem 1.1 F ∈ H 2 p if and only if AF ∈ L 2 p and ||F|| H 2 p ≤ C p ||AF||2 p . Since {ζ : z ∈ (ζ )} ≤ c(1 − |z|2 ), we have by (7.7) (AF)2 dθ ≤ C |ϕ (z)||g (z)|2 (1 − |z|2 )d xd y, E

R

and thus by (7.8) and (7.9), {θ : AF(θ ) > t} ≤ ∂D \ E + 1 (AF)2 dθ t2 E λ ≤ Cλ−1+2η + C 2 , t 2

(7.10)

for all λ ≥ λ0 . Take λ = t 2 p+1 so that the two terms on the extreme right of (7.10) are equal. Then by (7.10) there is t0 = t0 (λ0 ) such that ∞ 2p ||AF||2 p = 2 p t 2 p−1 {θ : AF(θ ) > t} dt 0 ∞ 4 p2 −4 p−1 t0 2π t 2 p−1 dt + C t 2 p+1 dt ≤ 2p 0

≤ C(η) if 0 < η < 1.

t0

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That proves (7.2), and the proof of Theorem 7.1 is reduced to constructing sets R and E that satisfy (7.7), (7.8), and (7.9). We ﬁrst construct R. To start put {|z| < 21 } ⊂ R and notice that we have |ϕ (z)||g (z)|2 (1 − |z|2 )d xd y ≤ C|ϕ (0)| (7.11) |z|< 21

because ||g||B ≤ 6. As in Section 3 take the dyadic Carleson boxes Q = {r eiθ : 1 − 2−n ≤ r < 1, π j2−(n+1) ≤ θ < π( j + 1)2−(n+1) }, 0 ≤ j < 2n+1 , of sidelength (Q) = 2−n and their top halves T (Q) = Q ∩ {1 − 2−n ≤ r < 1 − 2−(n+1) } & = Q \ {Q : Q ⊂ Q, Q = Q}, and write z Q for the center of T (Q). Fix δ and ε to be determined later. Say Q ∈ L, for large, if (7.12) sup (1 − |z|2 )2 Sϕ(z) > δ. T (Q)

When Q is large, deﬁne D(Q) = T (Q). Say Q ∈ G, for good, if Q ∈ / L and sup (1 − |z|2 )|g (z)| < ε.

T (Q)

Say Q ∈ B, for bad, if Q ∈ / L and Q is not good, i.e., sup (1 − |z|2 )|g (z)| ≥ ε.

T (Q)

⊃ Q, ( Q) = 2(Q) satisﬁes If Q ∈ / L, we call Q maximal if the next bigger Q 1 / L. When Q ∈ M Q ∈ L or if (Q) = 2 . Write M for the set of maximal Q ∈ deﬁne & D(Q) = Q \ {Q ⊂ Q, Q ∈ L}. Then

z:

& 1 D(Q), ≤ |z| < 1 = 2 L∪M

and the sets under this union are disjoint.

(7.13)

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X. Rectiﬁability and Quadratic Expressions Q1 Q2 D(Q 2 )

Q3 Figure X.20 Q 1 ∈ L; Q 2 , Q 3 , ∈ M. To complete the construction of R we consider four cases and we estimate the contribution to (7.9) in each case. Case I: Q ∈ L. Put D(Q) = T (Q) ⊂ R and pass to the next level of boxes Q ⊂ Q with (Q ) = 21 (Q). Since ||g||B ≤ 6, Theorem 4.1 and the Schwarz lemma yield |ϕ (z)||g (z)|2 (1 − |z|2 )d xd y T (Q)

≤C

6 2 δ

2 |ϕ (z)| Sϕ(z) (1 − |z|2 )3 d xd y

(7.14)

T (Q)

for every Case I box. Case II: Q ∈ G ∩ M.

1 Put D(Q) ⊂ R and pass to the maximal Q ⊂ Q. Recall that F = ϕ 2 and 4|F |2 = |ϕ ||g |2 . To estimate the contribution to (7.9) in Case II we need the inequality |ϕ (z)|g (z)|2 (1 − |z|2 )d xd y D(Q)

=4

|F (z)|2 (1 − |z|2 )d xd y ≤ C((Q))3 |F (z Q )|2

(7.15)

D(Q)

+C

|F (z)|2 (1 − |z|2 )3 d xd y.

D(Q)

Because inequality (7.15) very much resembles (1.7) of Theorem 1.2 its proof is left as an exercise. Remembering that 2|F | = |F||Sϕ + (g )2 | and that

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||g||B ≤ 6, we obtain from (7.15) |ϕ (z)|g (z)|2 (1 − |z|2 )d xd y D(Q)

≤ C(Q)|ϕ (z Q )| + C

2 |ϕ (z)| Sϕ(z) (1 − |z|2 )3 d xd y

(7.16)

D(Q)

|ϕ (z)||g (z)|4 (1 − |z|2 )3 d xd y

+C D(Q)

in Case II. We need the following lemma. Lemma 7.4. Assume δ < Q ∈ G.

ε2 2 . If

Q ∈ G, if Q ⊂ Q, and if Q ∩ D(Q) = ∅, then

Accept Lemma 7.4 for a moment. Then we can bound the last term in (7.16) using 4 2 3 2 |ϕ (z)||g (z)| (1 − |z| ) d xd y ≤ ε |ϕ (z)||g (z)|2 (1 − |z|2 )d xd y. D(Q)

D(Q)

Therefore if < 1 we have |ϕ (z)||g (z)|2 (1 − |z|2 )d xd y Cε2

D(Q)

≤ C(Q)|ϕ (z Q )| + C

2 |ϕ (z)| Sϕ(z) (1 − |z|2 )3 d xd y.

(7.17)

D(Q)

Consider the ﬁrst term C(Q)|ϕ (z Q )| on the right side of (7.17). Because ( Q) = 2(Q) and (7.12) holds Q is maximal, either (Q) = 1/2 or Q ⊂ Q, In the ﬁrst case for Q. (Q)|ϕ (z Q )| ≤ c|ϕ (0)|, and this ﬁrst case can occur for at most four squares Q. In the second case Theorem 4.1 and the Schwarz lemma give Sϕ(z) 2 (1 − |z|2 )3 d xd y ≥ cδ 2 (Q). T (Q)

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Because ||g||B ≤ 6 it follows that the ﬁrst term on the right-hand side of (7.17) is majorized by a constant multiple of the second term, and hence that |ϕ (z)||g (z)|2 (1 − |z|2 )d xd y D(Q)

≤

C δ2

2 |ϕ (z)| Sϕ(z) (1 − |z|2 )3 d xd y.

(7.18)

D(Q)

Thus for Case II boxes (7.18) holds with at most four exceptions, when we have the additional term c|ϕ (0)|, and for (7.9) this additional term is harmless if λ0 ≥ 1. Proof of Lemma 7.4. Let r1 eiθ ∈ T (Q ). There is r0 ≥ 21 with r0 eiθ ∈ T (Q) and [r0 eiθ , r1 eiθ ] ⊂ D(Q). Set r = sup s ≤ r1 : (1 − t 2 )|g (teiθ )| ≤ ε on [r0 , s] . Then since (1 − t 2 )2 |Sϕ(teiθ )| ≤ δ (4.14) gives δ + ε2 d ε < 2 2 (1 − t ) dt 1 − t 2 2

|g (teiθ )| ≤ on [r0 , r ]. Therefore

(1 − r 2 )|g (r eiθ )| < ε and r = r1 .

The remaining two cases concern the bad boxes Q ∈ B. We begin with a lemma that shows the bad boxes are sparsely distributed. Lemma 7.5. Given ε > 0 and an integer n > 0, there exist C = C(ε) and δ = δ(ε, n) such that if Q is a bad box for which sup (1 − |z|2 )2 |Sϕ(z)| ≤ δ,

T (Q)

and if B(Q) = Q ∩ {1 − |z| ≥ 2−n (Q)} ∩ {(1 − |z|2 )|g (z)| ≥ ε}, then there exists a hyperbolic geodesic σ such that sup ρ(z, σ ) < C, B(Q)

(7.19)

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405

where ρ denotes hyperbolic distance in D. Moreover, given η > 0 there is δ > 0 such that if (7.19) holds for δ, then 1 − |z | 2−η ϕ (z ) 1 − |z | 2+η 0 1 0 ≤ (7.20) ≤ 1 − |z 1 | ϕ (z 0 ) 1 − |z 1 | if z 0 ∈ σ ∩ T (Q) and z 1 ∈ σ ∩ Q ∩ {2n/2 (Q) ≤ 1 − |z 1 | ≤ 2−n (Q)}. Proof. If δ is small, then by Lemma 4.3 there is a Möbius transformation T such that |ϕ − T | is small on Q ∩ {1 − |z| ≥ 2−n (Q)}, so small in fact that T (z) ε (1 − |z|2 )2 − g (z) ≤ T (z) 2 on Q ∩ {1 − |z| ≥ 2−n (Q)}. Let z 0 ∈ T (Q) ∩ {|g (z)|(1 − |z|2 ) ≥ ε}, and let z ∗ be the pole of T . Then by (4.10) the conclusions of Lemma 7.5 all hold when ∗ σ is the circular arc or thogonal to ∂D that passes through z 0 and |zz ∗ | , because they all hold for T .

Now ﬁx n ∼ 10 and β = β(ε, δ) ∼ δ 2 . Case III: Q ∈ B ∩ M and 2 |ϕ (z)| Sϕ(z) (1 − |z|2 )3 d xd y ≥ β(Q)|ϕ (z Q )|. J= D(Q)

Let σ = σ Q be the geodesic of Lemma 7.5 and choose points w j ∈ σ ∩ B(Q), j = 0, 1, . . . , j0 such that w0 = z 0 and 1 − |w j | = 2− j (1 − |z 0 |). If possible, we let j ∗ be the least j ≤ j0 such that

(1 − |w j |)|ϕ (w j )| ∼ λJ ≥ λβ(Q)|ϕ (z Q ))|.

(7.21)

j≤ j ∗

If j ∗ exists we take Q ∗ = Q ∗ (Q) ⊂ Q, Q ∗ ∈ B, such that w j ∗ ∈ T (Q ∗ ) and deﬁne D(Q) = D(Q) \ Q ∗ . If j ∗ exists, then by (7.21) and the upper bound in (7.20), (Q ∗ (Q)) ≤ Cλ−1+η (Q).

(7.22)

If no such j ∗ exists and if w j0 ∈ Q ⊂ Q with Q ∈ B, we deﬁne Q 1 = Q and D(Q) = D(Q) \ Q 1 . If no such j ∗ exists and if no such Q exists, we take D(Q) = D(Q). In each instance we put D(Q) ⊂ R.

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D(Q)

Q ∗ (Q) σ

zQ

Figure X.21 By Lemmas 7.5 and 7.4 there is a cone with vertex ζ ∈ σ ∩ ∂D such that = ∅, then there exists maximal dyadic if Q ∈ B, Q ⊂ Q, and Q ∩ D(Q) Q j ∈ B, Q j ⊂ Q, Q j ∩ D(Q) = ∅ such that T (Q j ) ∩ = ∅. Write |ϕ (z)||g (z)|2 (1 − |z|2 )d xd y = + . D(Q)

D(Q)\

D(Q)∩

\ By extending the radial edges of each maximal Q j to we partition D(Q) \ with uniformly bounded chordinto chord-arc domains D j ⊃ Q j ∩ D(Q) arc constants. Choose z j ∈ ∩ ∂D j . Then by the proof for Case II and Harnack’s inequality, 2 2 |ϕ (z)||g (z)| (1−|z| )d xd y = |ϕ (z)||g (z)|2 (1 − |z|2 )d xd y D(Q)

Dj

|ϕ (z)||g (z)|2 (1−|z|2 )d xd y

+ D(Q)∩

≤

|ϕ (z)||Sϕ(z)|2 (1−|z|2 )3 d xd y

Dj

+

(1 − |z j |2 )|ϕ (z j )|

+

|ϕ (z)||g (z)|2 (1−|z|2 )d xd y.

D(Q)∩

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The values 1 − |z j | decrease geometrically and each z j is a bounded hyperbolic distance from the geodesic σ. Therefore by Harnack’s inequality and (7.21), (1 − |z j |2 )|ϕ (z j )| ≤ C (1 − |w j |2 )|ϕ (w j )| ≤ CλJ. j≤ j ∗

Harnack’s inequality also gives |ϕ (z)||g (z)|(1 − |z|2 )d xd y ≤ C (1 − |w j |2 )|ϕ (w j )| ≤ CλJ. D(Q)∩

Since λ > 1, we therefore have |ϕ (z)||g (z)|2 (1 − |z|2 )d xd y D(Q)

≤ Cλ

2 |ϕ (z)| Sϕ(z) (1 − |z|2 )3 d xd y

(7.23)

D(Q)

if Q is a Case III box and if j ∗ exists. If no such j ∗ exists and if there is no Q ⊂ Q with Q ∈ B and w j0 ∈ Q , we stop the sum at j0 and we still obtain (7.23). Finally, if no j ∗ exists but if w j0 ∈ Q ∈ B, Q ⊂ Q, we repeat the construction with Q replaced by Q 1 = Q and with a possibly new geodesic σ1 containing w j0 ∈ Q 1 , possibly constructing a new Q ∗1 or a new Q 1 = Q 2 , and we obtain (7.23) for 1 ). We repeat the construction until we reach a case where a square D(Q Q ∗m is deﬁned or a case where neither Q ∗m nor Q m is deﬁned. If we reach a square Q ∗m we deﬁne Q ∗ (Q) = Q ∗m . Then (7.22) holds for Q ∗ (Q). We put j ) into R. Note that (7.23) holds for each D(Q) = D(Q) \ Q ∗ (Q) = D(Q set D(Q j ) and that these sets are disjoint. Case IV: Q ∈ B ∩ M and 2 |ϕ (z)| Sϕ(z) (1 − |z|2 )3 d xd y < β|ϕ (z Q )|(Q). J= D(Q)

Since β ≤ this case can only occur if (Q) = 1/2, and thus for at most four boxes Q. Deﬁne Q ∗ by Q ∗ ∩ σ = ∅ and cδ 2 ,

(Q ∗ )|ϕ (z Q ∗ )| ∼ λ(Q)|ϕ (z Q )|

(7.24)

and take D(Q) = D(Q) \ Q ∗ . If no such Q ∗ exists take D(Q) = D(Q). Put D(Q) ⊂ R and do not consider any smaller Q ⊂ Q. The argument used in

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|ϕ (z)||g (z)|2 (1 − |z|2 )d xd y ≤ Cλ|ϕ (0)|,

D(Q)

which is good enough because there are at most four such Q. Proof of (7.7) and (7.9). By (7.13) we have & R = D \ {Q ∗ (Q) : Q ∈ B ∩ M}. Since |ϕ (0)| ≤ 6, (7.11), (7.14), (7.18), (7.24), and the many cases of (7.23) give (7.9) for R provided λ ≥ B. Let I ∗ (Q) ⊂ ∂D be the base of Q ∗ (Q) and set & (7.25) E = ∂D \ {3I ∗ (Q) : Q ∈ B ∩ M}. Then (7.7) holds on E. Proof of (7.8). To prove (7.8) we need an additional lemma. Lemma 7.6. Given η > 0 there is C = C(η) such that if Q ∈ B ∩ M and if Q ∗ (Q) exists, then (Q) ≤ Cλη (∂D(Q) ∩ ∂D).

(7.26)

Proof of Lemma 7.6. By hypothesis and by (7.20) there exists Q ∗∗ , Q ∗ (Q) ⊂ Q ∗∗ ⊂ Q, such that |ϕ (z Q ∗∗ )|(Q ∗∗ ) ∼ λη J,

(7.27)

and by the lower bound in (7.20) (Q ∗∗ ) ≥ Cλ−η (Q).

(7.28)

Q ∗∗

∩ D(Q). By (7.27) and Corollary Consider the chord-arc domain = 1 η I.4.4, diam(ϕ()) ≥ Cλ J. By Theorem M.1 applied to F = (ϕ ) 2 and by the proof of (7.22), (ϕ(∂)) ≤ C|ϕ (z Q ∗∗ )|(Q ∗∗ ) + CλJ ≤ C |ϕ (z Q ∗∗ )|(Q ∗∗ ). Let E = {Q ∈ L : ∂ T (Q ) ∩ ∂ = ∅} and A = ∂ ∩ E ∂ T (Q ). Then by (7.12) and the Schwarz lemma, |ϕ (z)|ds (ϕ(A)) = A 2 (7.29) ≤C |ϕ (z)| Sϕ(z) (1 − |z|2 )3 d xd y ≤ J. E T (Q )

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Since ∂ϕ() has length and diameter comparable to λη J it follows from (7.29) and the Lavrentiev estimate (5.1) of Chapter VI that if λ ≥ λ0 (η) then ω(z Q ∗∗ , A, ) = ω(ϕ(z Q ∗∗ , ϕ(A), ϕ()) is small. Then since is a chord-arc (A) domain with bounded constants, we conclude that (Q ∗∗ ) is small and hence that (∂ ∩ ∂D) ≥ c(Q ∗∗ ),

and with (7.28) this implies (7.26).

Finally, note that because the sets {∂D(Q) ∩ ∂D : Q ∈ B ∩ M} are pairwise disjoint, (7.26) and (7.22) give the inequality (7.8). Proof of Theorem 7.2. A point of density argument shows that there exists J1 ⊂ J , |J1 | ≥ (1 − 3ε )|J | such that if W = J1 β (ζ ), then |ϕ (z)||Sϕ(z)|2 (1 − |z|2 )3 d xd y W

≤C

|ϕ (z)||Sϕ(z)|2 (1 − |z|2 )2 d xd yds(ζ ) ≤ C N .

J1 α (ζ )

Set D=

&

β (ζ )

ζ ∈J1

and V=

&

{T (Q) : T (Q) ⊂ D}.

Then V ⊂ D. Deﬁne L, M, G, and B as in the proof of Theorem 7.1, but include only those T (Q) such that T (Q) ⊂ D. For such Q deﬁne & D(Q) = V ∩ Q \ {Q ⊂ Q ∩ D : Q ∈ L}. Then the proof of Theorem 7.1 yields a set E ⊂ J1 , deﬁned by (7.25) and a region R ⊂ V so that (7.7), (7.8), and (7.9) hold, and so that |E| ≥ (1 − 3ε )|J | and another point of density argument gives J0 ⊂ E for which |J0 | ≥ (1 − ε)|J | and for which (7.4) holds.

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8. Schwarzians and BMO Domains Recall that a simply connected domain is called a BMO domain if the mapping function ϕ : D → satisﬁes g = log(ϕ ) ∈ BMO. The results in this chapter yield two characterizations of BMO domains that complement Theorem VII.5.3. Theorem 8.1. The following are equivalent. (a) is a BMO domain. (b) There exist δ > 0 and C > 0 such that for all z 0 ∈ there is a subdomain U ⊂ such that (i) z 0 ∈ U, (ii) ∂U is rectiﬁable and (∂U) ≤ Cdist(z 0 , ∂), and (iii) ω(z 0 , ∂ ∩ ∂U, U) ≥ δ. (c) There exist δ > 0 and C > 0 such that for all z 0 ∈ there is a subdomain U ⊂ such that (i) z 0 ∈ U and dist(z 0 , ∂) ≤ C dist(z 0 , ∂U), (ii) ∂U is chord-arc with constant at most C and (∂U) ≤ C dist(z 0 , ∂), and (iii) (∂ ∩ ∂U) ≥ δ dist(z 0 , ∂). (d) |Sϕ(z)|2 (1 − |z|2 )3 d xdy is a Carleson measure on D. (e) There exist δ > 0 and C > 0 such that for every z 0 ∈ D there exists a Lipschitz domain V ⊂ D such that (i) z 0 ∈ V, (ii) ω(z 0 , ∂V ∩ ∂D, V) ≥ δ, and |ϕ (z)||Sϕ(z)|2 (1 − |z|2 )3 d xd y ≤ C|ϕ (z 0 )|(1 − |z 0 |2 ).

(iii) V

Proof. Theorem VII.5.3 had the implications (a) ⇒ (b) ⇒ (c) ⇒ (a). Here we use the arguments of the previous section to treat (a) ⇐⇒ (d), (a) ⇒ (e), and (e) ⇒ (b). (a) ⇒ (d): This was ﬁrst observed by Zinsmeister [1984]. By (4.14) we have |S(ϕ)| ≤

|g |2 + |g |. 2

It follows from (a) that |g (z)|2 (1 − |z|2 )d xd y is a Carleson measure and hence |g (z)|4 (1 − |z|2 )3 d xd y is also a Carleson measure, because g ∈ B. For any

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analytic function we have |g (z)|2 (1 − |z|2 )3 d xd y ≤ C |g (z)|2 (1 − |z|2 )d xd y, (Q) T

T (Q)

(Q) = {z : dist(z, T (Q)) ≤ (Q)/4}. Thus |g (z)|2 (1 − |z|2 )3 d xd y where T is also a Carleson measure and (d) holds. (d) ⇒ (a): This is due to Astala and Zinsmeister [1991]. Because (a) and (d) are invariant under conformal self maps of D, to prove (a) it is enough to show (8.1) |g(eiθ ) − g(0)|2 dθ ≤ C. See Garnett [1981]. Let

A=

|g (z)|2 (1 − |z|2 )d xd y

D

and

B=

|g (z)|2 (1 − |z|2 )3 d xd y.

D

Then since

|g (0)|

≤ 6 we have by Fourier series, B ≤ 12 A ≤ 3B + 63 π

(8.2) C .

and by Theorem 1.2 it will be enough to show B ≤ By (4.14) we have 1 |Sϕ(z)|2 (1 − |z|2 )3 d xd y + |g (z)|4 (1 − |z|2 )3 d xd y. B≤2 2 D

D

Set U = {z ∈ D : |g (z)|(1 − |z|2 ) ≤ 1/2}. Then by (8.2) A B 9 |g (z)|4 (1 − |z|2 )3 d xd y ≤ ≤ + π. 4 12 2

(8.3) (8.4)

U

Set V = {T (Q) : supT (Q) |Sϕ(z)|(1 − |z|2 )2 ≥ δ} where δ = δ(1/4, 10) is given by Lemma 7.5. Then because ||g||B ≤ 6, C 4 2 3 |g (z)| (1 − |z| ) d xd y ≤ 2 |Sϕ(z)|2 (1 − |z|2 )3 d xd y. (8.5) δ V

V

If T (Q) \ (U ∪ V ) = ∅ then by Lemma 7.5, T (Q ) ⊂ (U ∪ V ) for more than half of the Carleson boxes Q ⊂ Q with (Q ) = 2−10 (Q). Consequently

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D \ (U ∪ V ) ⊂ T (Q j ), where {Q j } is a sequence of Carleson boxes with (Q j ) ≤ C , with C independent of ϕ. Hence 4 2 3 |g (z)| (1 − |z| ) d xd y ≤ C (1 − |z|2 )−1 d xd y ≤ C . j

D\(U ∪V )

T (Q j )

(8.6)

Together (8.3), (8.4), (8.5), and (8.6) give us

9 C 23 |Sϕ(z)|2 (1 − |z|2 )3 d xd y + π + C , B ≤ 2+ 2 24 δ 4 D

which establishes (8.1). (d) and (a) ⇒ (e): Because (e) is invariant under Möbius self maps of D, we can suppose z 0 = 0. Then let V be the Lipschitz region constructed in the proof of (a) ⇒ (b) from Theorem VII.5.3 and note that |ϕ | is bounded above and below on V. Then use (d). (e) ⇒ (b): We may assume z 0 = ϕ(0). We repeat the proof of Theorem 7.1, with D replaced by the Lipschitz domain V given by (e), just as we did in the of Theorem 7.2. We obtain a Lipschitz domain R ⊂ V such that proof |ds < ∞, with 0 ∈ R and with ω(0, ∂D ∩ ∂R, R) ≥ δ . Then (b) holds |ϕ ∂R 2 for U = ϕ(R).

9. Angular Derivatives Let ϕ be a conformal mapping from D onto a simply connected domain and let G = {ζ ∈ ∂D : ϕ has an angular derivative at ζ and |ϕ (ζ )| = 0}. In this section we give several almost everywhere characterizations of G. By Theorem VI.6.1 we know that ζ ∈ G is almost everywhere equivalent to ϕ(ζ ) is a cone point of .

(9.1)

Furthermore, for any α > 1, ζ ∈ G is almost everywhere equivalent to 2 ϕ (z) d xd y < ∞, (9.2) α (ζ )

by the theorem of Marcinkiewicz, Zygmund, and Spencer, Theorem 1.3 above. Bishop and Jones [1994] give several other almost everywhere characterizations of G.

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Theorem 9.1. Let ϕ be a conformal map from D onto a simply connected domain and let g = log(ϕ ). Then the following are almost everywhere equivalent on ∂D: (a) ϕ has non-zero angular derivative at ζ. 2 |ϕ (z)| Sϕ(z) (1 − |z|2 )2 d xd y < ∞. (b) α (ζ )

|ϕ (z)||g (z)|2 d xd y < ∞.

(c)

α (ζ ) 1

(d)

0

(e)

η2 (ϕ(ζ ), t)

dt < ∞. t

Sϕ(z) 2 (1 − |z|2 )2 d xd y < ∞.

α (ζ )

Proof. The road map is: (a) ⇒ (b) ⇒ (c) ⇒ (a); then (d) ⇒ (a) ⇒ (d); and ﬁnally (b) ⇒ (e) ⇒ (a). (a) ⇒ (b): Suppose K ⊂ ∂D is a compact set such that sup |ϕ (z)| ≤ M

α (ζ )

for ﬁxed M < ∞ for all

eiθ

∈ K . Form the cone domain & U= α (eiθ ). K

Then

{ζ : z ∈ α (ζ )} ≤ cα (1 − |z|),

and Fubini’s theorem again gives 2 |ϕ (z)| Sϕ(z) (1 − |z|2 )2 d xd ydθ K α (eiθ )

≤

2 |ϕ (z)| Sϕ(z) (1 − |z|2 )3 d xd y.

(9.3)

U

But the integral on the right in (9.3) is ﬁnite by Lemma 6.1 because ∂ϕ(U ) is rectiﬁable. Therefore (b) holds almost everywhere on K . (b) ⇒ (c): This is Corollary 7.3.

1 (c) ⇒ (a): Because |ϕ ||g |2 = 4|F |2 , where F = ϕ 2 , Theorem 1.3 and Theorem VI.2.3 show ϕ has non-zero nontangential limit almost everywhere that (c) holds.

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(d) ⇒ (a): Fix α > 1 large. If (d) holds on a set E of positive measure, then for any ε > 0 there exists δ > 0 and compact K ⊂ E such that δ dt η2 (ϕ(ζ ), t) < ε t 0 and |E \ K | < ε. Then by Theorem 4.1 and the Cauchy–Schwarz inequality, Sϕ(z) (1 − |z|2 )2 ≤ Cε 21 , (9.4) on V =

&

1

1

α (ζ ) ∩ {dist(ϕ(z), ∂) ≤ δ 2 ε 4 }.

K 1 2

1

Let U = V ∩ B(ζ, δ ε 4 ) where ζ is a point of density of K . If α is sufﬁciently large, then U is a quasidisc and by Theorem 4.2, ϕ(U ) is a quasidisc if ε is sufﬁciently small. But then by Exercise 12, β(ϕ(ζ ), t) ≤ Cη(ϕ(ζ ), t) on K . Therefore by (d), by Theorem 2.5, and by the F. and M. Riesz theorem, ϕ(∂U ) has a tangent at ϕ(ζ ) for almost every ζ ∈ K . Hence (9.1) and (a) hold almost everywhere on K . (a) ⇒ (d): Now suppose the angular derivative ϕ (ζ ) exists and is continuous on a compact set K ⊂ ∂D. It is enough to show (d) holds almost everywhere on K . Let δ > 0 be small and cover K by arcs I j = {|ζ − ζ j | ≤ δ} and form the cone domains & U j = {|z| ≥ 1 − δ} ∩ 2 (ζ ). ζ ∈K ∩I j

ϕ

If δ is small then U j is connected and is continuous and almost constant on U j , so that γ j = ϕ(∂U j ) is a chord-arc curve. By Theorem 2.5 1 dt βγ2j (z θ , t) < ∞ (9.5) t 0 almost everywhere on K ∩ I j . Let (9.5) hold at eiθ ∈ K ∩ I j and let t be so small that βγ j (z θ , 2t) < 1/16. Let S be a strip of width 4tβγ j (z θ , 2t) containing γ j ∩ B(z θ , 2t). Then one side L of S satisﬁes (a), (b), and (c) in the deﬁnitions of ηϕ(U j ) (z θ , t) and of η (z θ , t) because ϕ(U j ) ⊂ . Let z ∈ L ∩ B(z θ , 2t) and let z ∗ be the orthogonal (to L) projection of z onto γ j . Then dist(z, ∂) ≤ |z − z ∗ | + dist(z ∗ , γ j ∩ ϕ(K )), and by the deﬁnition of U j and the continuity of ϕ on U j , dist(z ∗ , γ j ∩ ϕ(K )) ≤ Ctβγ j (z θ , t).

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Therefore η (z θ , t) ≤ Cβγ j (z θ , t) and by (9.5), (d) holds at eiθ . (b) ⇒ (e): This is clear because (b) ⇒ (a). (e) ⇒ (a): We follow the proof of Theorem 8.1, (d) ⇒ (a). Assume (e) holds on a set E. Then there exists a compact set K ⊂ E such that when U = K α (ζ ), Sϕ(z) 2 (1 − |z|2 )3 d xd y < ∞. (9.6) U

Recall that g = log(ϕ ). We will show (9.6) implies 2 g (z) d(z)3 d xd y < ∞ B=

(9.7)

U

where d(z) = dist(z, ∂U ). Since |g (0)| ≤ 6, it will then follow from Theorem 1.2 that g, and also ϕ , have a ﬁnite non-zero nontangential limit almost everywhere on K . By (4.14) 2 4 1 3 g (z) d(z)3 d xd y, Sϕ(z) d(z) d xd y + B≤2 2 U

U

and by Theorem 1.2 2 g (z) d(z)d xd y ≤ C(α)(B + |g (0)|2 ). A= U

Set U1 = U ∩ {z : |g (z)|d(z) ≤ C(α)−1/2 }. Then 2 4 1 g (z) d(z)3 d xd y ≤ B + |g (0)| . 2 2 2 U1

For ﬁxed δ > 0, set U2 = U ∩ {z : Sϕ(z) (1 − |z|2 )2 ≥ δ}. Then because ||g||B ≤ 6, 4 4 g (z) d(z)3 d xd y ≤ g (z) (1 − |z|2 )3 d xd y U2

U2

C ≤ 2 δ

U2

Sϕ(z) 2 (1 − |z|2 )3 d xd y.

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Finally, set U3 = U \ (U1 ∪ U2 ). By Lemma 7.5, there is a sequence {Q j } of Carleson boxes such that (Q j ) < ∞ and U3 ⊂ T (Q j ). But then g (z)|4 (1 − |z|2 )3 d xd y ≤ 64 (Q j ) < ∞, U3

and we obtain 2 B

C Sϕ (1 − |z|2 )3 d xd y + 64 ≤ 2+ 2 (Q j ). 2 δ

Therefore (9.7) holds.

10. A Local F. and M. Riesz Theorem Theorem 10.1. Let be a simply connected domain, let z 0 ∈ satisfy dist(z 0 , ∂) ≥ 1, and let be a connected set with 1 () ≤ M < ∞. Given δ > 0 there exists ε = ε(M, δ) > 0 such that if E ⊂ ∩ ∂ and ω(z 0 , E, ) > δ, then 1 (E) > ε. The F. and M. Riesz theorem is a simple corollary of Theorem 10.1. In [1980] Øksendal conjectured Theorem 10.1 and proved it where is a line. See also Exercise III.14 and Øksendal [1981]. Shortly thereafter Kaufman and Wu [1982] proved the theorem when is a chord-arc curve. But the full result was not proved until Bishop and Jones [1990]. The difﬁculty in the general case is to make effective use of the hypothesis 1 () ≤ M, and for this we will use Theorem 2.3. Proof. Let ψ : D → C∗ \ E be the universal covering map, ψ(0) = z 0 . By the monodromy theorem there is a simply connected G ⊂ D such that 0 ∈ G, ψ is univalent on G, and ψ(G) = . By hypothesis ω(0, ∂G ∩ ∂D, G) > δ.

(10.1)

Then by Beurling’s projection theorem dist(0, ∂G) ≥ c(δ),

(10.2)

|ψ (0)| ≤ c (δ).

(10.3)

and hence by Lemma 5.5,

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10. A Local F. and M. Riesz Theorem

Let G denote the Fuchsian group {γ ∈ M : ψ ◦ γ = ψ} and let I be the identity of G, I (z) = z. The normal fundamental domain of G is deﬁned by F = {z ∈ D : |γ (z)| < 1 for all γ ∈ G \ {I }}, and ψ is one-to-one on F. Write G \ {I } = {γ j } and γ j (z) = |λ j | = 1 and γ j (0) = a j . Then

λ j z+a j 1+a j λ j z ,

where

1 − |a j |2 |1 + a j λ j z|2 and the fundamental domain F has the form D \ B j where each B j is a disc orthogonal to ∂D. |γ j (z)| =

Lemma 10.2. The normal fundamental domain F satisﬁes c(δ) 2

(10.4)

|∂D ∩ ∂F| ≥ 2π δ.

(10.5)

dist(0, ∂F) ≥ and

There exist β(δ ) > 1 and a compact set K ⊂ ∂D ∩ ∂F such that |K | ≥ π δ and

&

β (ζ ) ⊂ F.

(10.6)

(10.7)

K

Proof. By (10.2), |a j | ≥ c(δ) for all j, and since |a j | 1 1 inf{|z| : |γ j (z)| = 1} = −1≥ − , 2 |a j | |a j | 2 that gives (10.4). By Lemma I.2 from Appendix I and Plessner’s theorem, ωG almost every point of ∂D ∩ ∂G is a cone point for G. Write K G for the set of cone points of ∂G. Then by (10.1) and a comparison, |∂D ∩ K G | ≥ 2π δ. Because ∂D ∩ K G ∩ γ j (K G ) = ∅ for all γ j , we have ∂D ∩ ∂F ∩ γ j −1 (K G \ ∂F) ∂D ∩ ∂F ≥ ∂D ∩ ∂F ∩ K G + ≥ ∂D ∩ K G , because |(γ j −1 ) | ≥ 1 on γ j (∂F) ∩ (∂G \ ∂F). That gives (10.5).

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Write I j = B j ∩ ∂D. By (10.4) we have |I j | ≤ π − δ < π, and if ε > 0 is small there is β > 1 such that if Jj is the arc concentric with I j having |Jj | = (1 + ε)|I j |, then β (ζ ) ∩ B j = ∅ for all ζ ∈ ∂D \ Jj . Choose δ and set ε = 2(1−δ) & K = ∂D \ Jj . Then (10.7) holds and (10.6) follows from (10.5). That proves Lemma 10.2. Fix 1 < α < β and form the cone domain & U= α (ζ ). K

If η = η(α, β) is sufﬁciently small, then B(w, η(1 − |w|)) ⊂ F for all w ∈ U. We claim that 2 |ψ (z)| Sψ(z) (1 − |z|2 )3 d xd y ≤ C M. (10.8) U

If (10.8) holds then by Fubini’s theorem 2 |ψ (z)| Sψ(z) (1 − |z|2 )2 d xd yds(ζ ) < C M, K α (ζ )

and by Theorem 7.2 there exist N < ∞ and K 0 ⊂ K such that |K 0 | > |K |/2 √ and F = ψ has area function A(F) ≤ N at all ζ ∈ K 0 . Take & α (ζ ). U0 = K0

Then by Theorem 1.2, ψ(U0 ) is a Jordan domain having 1 (∂ψ(U0 )) ≤ C(|ψ (0)| + N ) ≤ C (M, δ), ω(z 0 , E ∩ ψ(U0 )) ≥ δ (α, δ) by (10.6), and dist(z 0 , ∂ψ(U0 )) ≥ c . Therefore by (10.3) and Lavrentiev’s theorem, Exercise VI.2, 1 (E) ≥ c(δ, 1 ()), and Theorem 10.1 is proved, provided (10.8) holds. Proof of (10.8). Let {Q j } denote the Whitney squares in C \ E. We may assume each Q j is a dyadic square. Let Q nj denote the dyadic square Q nj ⊃ Q j such

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10. A Local F. and M. Riesz Theorem that (Q nj ) = 2n (Q j ), and let N (Q j ) be the ﬁrst n such that δn (Q j ) = β E (Q nj ) + γ E (Q nj ) ≥ 10−3 . If z ∈ U ∩ ψ −1 (Q j ) then by Theorem 5.6, N (Q j ) δn (Q j )2−ns ≡ δ(Q j ), (1 − |z| ) Sψ(z) ≤ Cs

2 2

n=0

for any s < 1. Therefore 2 |ψ (z)| Sψ(z) (1 − |z|2 )3 d xd y ≤ δ 2 (Q j ) Qj

U

But by Lemma 5.5

U ∩ψ −1 (Q j )

|ψ (z)| d xd y. (1 − |z|2 )

U ∩ψ −1 (Q j )

|ψ (z)| d xd y ≤ C(Q j ), (1 − |z|2 )

and we obtain 2 |ψ (z)| Sψ(z) (1 − |z|2 )3 d xd y U

≤C

δ 2 (Q j )(Q j )

Qj

= 2C

(Q j ) N

2

(Q j )

(10.9)

n=0

j

+ 2C

β E (Q nj )2−ns

(Q j ) N j

γ E (Q nj )2−ns

2

(Q j ).

n=0

The two sums on the right of (10.9) will be estimated using the next two lemmas. Write Q for the set of all dyadic squares. Lemma 10.3. For n ≥ 0, β E2 (Q nj )(Q j ) : N (Q j ) ≥ n ≤ 2n β E2 (Q)(Q) ≤ C2n (). Q

Proof. For each Q ∈ Q there are at most 4n dyadic squares Q j such that Q nj = Q and for each of them (Q j ) = 2−n (Q). Therefore

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β E2 (Q nj )(Q j ) : N (Q j ) ≥ n =

β E2 (Q)

(Q j ) : Q nj = Q, N (Q j ) ≥ n

Q

≤

β E2 (Q)

{(Q j ) : Q nj = Q}

Q

≤ 2n

β E 2 (Q)(Q) ≤ C2n (),

Q

by Theorem 2.3.

We take s > 1. Then by Lemma 10.3 and the Cauchy–Schwarz inequality, (Q j ) N j

β E (Q nj )2−ns

2

(Q j )

n=0

≤

≤

1 −s 1−2 n

β E2 (Q nj )2−ns (Q j )

{ j:N (Q j )≥n}

1 (1−s)n 2 β E2 (Q)(Q) 1 − 2−s n

(10.10)

Q

≤ C(). That bounds half of the right side of (10.9). Lemma 10.4. For n ≥ 0, {γ 2 (Q nj )(Q j ) : N (Q j ) ≥ n} ≤ C2n (). Proof. We can assume γ (Q nj ) ≥ 105 β(Q nj )

(10.11)

because Lemma 10.3 bounds the sum over the other Q nj . Using the Whitney squares of C \ E, we partition \ E into disjoint subsets Ik such that c1 dist(Ik , E) ≤ diam(Ik ) ≤ c2 dist(Ik , E) for absolute constants c1 and c2 . For each Q = Q nj satisfying (10.11) there is k = k(Q) such that Ik ⊂ 10Q and γ (Q) ≤ c

diam(Ik ) . (Q)

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421

Then for each k and there is a bounded number of Q ∈ Q such that k(Q) = k and (Q) = . Because (Q) ≥ 10 diam(Ik ), that gives k(Q)=k

∞

1 C C 2−ν ≤ ≤ . (Q) diam(Ik ) diam(Ik )

(10.12)

ν=0

Again because there are at most 4n squares Q j with Q nj = Q ∈ Q, we have by (10.11) {γ 2 (Q nj )(Q j ) : N (Q j ) ≥ n and (10.11) holds} = γ 2 (Q) {(Q j ) : Q nj = Q} (10.11) holds

≤ 2n

γ 2 (Q)(Q)

(10.11) holds

≤ C2n

(10.11) holds

= C2n

≤ C2

2

diam(Ik )

{Q:k(Q)=k}

k

n

2 diam(Ik (Q)) (Q)

1 (Q)

diam(Ik ) ≤ C2n (),

k

because the Ik are disjoint.

To estimate the second sum in (10.9), we keep s > 1 and repeat the proof of (10.10) to get (Q j ) N j

γ E (Q nj )2−ns

2

(Q j ) ≤ C().

n=0

That concludes the proof of (10.9) and therefore (10.8).

11. Ahlfors Regular Sets and the Hayman–Wu Theorem Recall that a compact set is Ahlfors regular if there exists a constant A > 0 such that for every disc B(z, r ), 1 ( ∩ B(z, r )) ≤ Ar. Theorem 11.1. Let be a compact connected set. Then is Ahlfors regular if

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and only if there is a constant C() such that 1 (ϕ −1 ( ∩ )) ≤ C()

(11.1)

for every simply connected domain and every conformal mapping ϕ from D onto . It is easy to see that (11.1) implies is Ahlfors regular; take ϕ to be a linear map from D to B(z, r ). The proof of the converse uses Theorem 10.1 and Lemma 11.2 below. Lemma 11.2. Let be an Ahlfors regular compact connected set and let be a simply connected domain. Let S = {z j } be a sequence in ∩ such that sup inf ρ(z, z j ) ≤ M.

(11.2)

j

Set

# dist(z j , ∂) $ , K j = z : |z − z j | ≤ 2 and assume S has a ﬁnite partition S = S1 ∪ . . . ∪ S N , so that if z j ∈ Sn and & Kk , j = \ {z k ∈Sn ,k = j}

then inf ω(z j , ∂ ∩ ∂ j , j ) ≥ δ > 0.

(11.3)

j

Then for any conformal ϕ : D → , 1 (ϕ −1 ( ∩ )) ≤ C(A, N , M, δ). ∂

ϕ −1 ()

ϕ Figure X.22

∂

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Proof. Write w j = ϕ −1 (z j ) and D j = {w ∈ D : ρ(w, w j ) ≤ M}. Then by (11.2), 1 (ϕ −1 ( ∩ )) ≤ 1 (D j ∩ ϕ −1 ( ∩ )), while by the Koebe estimates in Theorem I.4.3 and the Ahlfors regularity of , 1 (D j ∩ ϕ −1 ( ∩ )) ≤ C(A, M)(1 − |w j |). Thus the lemma will be proved if we show 1 − |w j | ≤ C.

(11.4)

Sn

We can assume ρ(w j , wk ) = ρ(z j , z k ) ≥ 1 whenever z j = z k by taking M ≥ 2 in (11.2) and by removing any unnecessary z j . Then the discs ∗j = {w : ρ(w, w j ) ≤ 1/3} are disjoint. Take 0 < η <

1 8

to be determined and set

j = {w : ρ(w, w j ) ≤ η}. Since η < 18 we have j ⊂ ϕ −1 (K j ) by the estimates in Section I.4. When proving (11.4) we may assume that { j : z j ∈ Sn } is ﬁnite and that 0∈ / ∪ j . Consider the harmonic function & u(z) = u n (z) = ω(z, ∂D, D \ { j : z j ∈ Sn }). By Green’s theorem, u(0) =

1 2π

∂D

u(eiθ )dθ −

1 ∂u 1 log ds, 2π ∂ j ∂n |z| Sn

where the normal derivatives are exterior to j . Hence ,* , * ∂u 1 1 1 u(eiθ )dθ − u(0) ≤ 1. ds ≤ inf log j |z| 2π ∂ j ∂n 2π ∂ D Sn

(11.5)

But inf log j

1 ≥ c4 (1 − |w j |) |z|

and in a moment we will prove the inequality 1 ∂u ds ≥ c5 . 2π ∂ j ∂n Therefore (11.4) follows from (11.5) and (11.6).

(11.6)

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j

D Figure X.23 The proof of (11.6). To prove (11.6) we use the functions & U j (z) = ω(z, ∂D, D \ {k , z k ∈ Sn \ {z j }}) and Vj (z) = ω(z, ∂ j , D \ j ). By the maximum principle u(z) ≥ U j (z) − Vj (z) if z ∈ D \ ∪{k : z k ∈ Sn }. All the annuli D \ j have the same modulus, so that when z ∈ ∂∗j , Vj (z) = ε(η), where limη→0 ε(η) = 0. By our hypothesis (11.3) and by Harnack’s inequality U j (z) ≥ c6 = c6 (δ) on ∂∗j . Thus we can choose η = η(δ) so small that u(z) ≥ c7 = c6 /2 on

&

∗j .

Sn

Let A j be the annulus (∗j )◦ \ j . Then u(z) ≥ c7 ω(z, ∂∗j , A j ) on A j , so that ∗

∂ω(z, j , A j ) ∂u ≥ c7 on ∂ j . ∂n ∂n

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But since all annuli A j have the same modulus, we have ∂ω(z, ∗j , A j ) 1 ds = C(η) > 0. 2π ∂ j ∂n Therefore (11.6) holds with c5 = C(η)c7 .

The moment he saw this proof of Lemma 11.2, Burgess Davis came up with a simple Brownian motion proof. See Exercise 15. Proof of Theorem 11.1. Let {Q j } be the set of Whitney squares for such that Q j ∩ = ∅ and ﬁx z j ∈ Q j ∩ . Given T large, we partition {z j } into N = N (T ) subsets S1 , . . . , S N such that ρ(z j , z k ) ≥ T whenever z j , z k ∈ Sn and j = k. To prove Theorem 11.1 we verify condition (11.3) of Lemma 11.2 for S1 ∪ S2 ∪ . . . ∪ S N .

3j

z j γk

∂

Figure X.24 The proof of Theorem 11.1. 3j be the simply connected component of j ∩ B(z j , 10 diam Q j ) that Let contains z j . By the Beurling projection theorem \ B(z j , 10 diam Q j ), 3j ) ≤ 1/4. ω(z j , ∂ Therefore (11.3) will hold if 3j ) ≤ 1/4, ω(z j , (∂ j \ ∂) ∩ B(z j , 10 diam Q j ),

(11.7)

because (11.7) implies 3j ) ≥ 1/2. ω(z j , ∂ ∩ ∂ j , j ) ≥ ω(z j , ∂ ∩ B(z j , 10 diam Q j ), To establish (11.7) we let γk be a line segment joining Q k ∩ to ∂ such that (γk ) ≤ 4 diam(Q k ). Now since is Ahlfors regular we can, given ε > 0,

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take T so large that 1 (B(z j , 10 diam Q j ) ∩ (∂ j ∪ On the other hand, 3j ) ∪ j = ( ∩

&

γk )) ≤ ε diamQ j .

&

3j } {γk : z k ∈ ∂

is a connected set and ( j ) ≤ C diamQ j . Therefore Theorem 10.1 implies inequality (11.7).

Notes The books of Zygmund [1959] and Stein [1970], [1993] give more complete discussions of the Lusin area function. Our proof of Theorem 1.2 is from Coifman, Jones, and Semmes [1989]. We have derived Theorem 1.3 from Theorem 1.2, but Chapter XIV of Zygmund [1959] has a proof more like the arguments in our Chapter VI. Sections 2 and 3 are based on Jones [1990]. For further results and applications see Jones [1989], [1991]; Bishop and Jones [1997a], [1997b], [1997c]; David [1991], [1998]; David and Semmes [1991], [1993], [1997]; Fang [1994]; Garnett, Jones and Marshall [1992]; Mattila, Melnikov and Verdera [1996]; Okikiolu [1992]; and Pajot [2002]. We have barely touched the Schwarzian derivative. Lehto [1987] has an excellent introduction. See Thurston [1986] for a completely geometrical explanation and see Chuaqui and Osgood [1998] and Osgood [1997] for more recent results. The recent work of Choi [2004] includes a proof of Theorem 10.1 that uses cone domain constructions instead of β numbers. Walden [1994] has the L p version of Theorem 1.1: For some p > 1, |(ϕ −1 ) (w)| p |dw| < ∞. ∩

Walden’s theorem generalizes the Fernández, Heinonen and Martio theorem from Exercise VII.7. Lemma 11.2 is from Garnett, Gehring and Jones [1983]. See also Garnett [1986] and Exercise 15.

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Exercises and Further Results 1. (a) If f ∈ H 2 (D) and g(z) ∈ H ∞ (D) then by Green’s theorem and the product rule, 1 2 2 | f (z)| |g (z)| (1 − |z|)d xd y ≤ | f (z)|2 |g (z)|2 log d xd y |z| D

D

≤ C||g||2∞ || f ||2H 2 . (b) Use (a) to prove (1.7) without using Carleson measures. See Gamelin [1980] and Coifman, Jones and Semmes [1989]. 2. (a) Prove Lemma 1.5. (b) More generally, prove that for n ≥ 2 and α > 1, n−1 2 F (z) d xd y ≤ C(n, α) |F ( j) (0)|2 1

α (ζ )

+ C(n, α)

(n) 2 F (z) (1 − |z|)2n−2 d xd y.

α (ζ )

(c) If n ≥ 2 and 1 < β < α, there is C(n, α, β) such that (n) 2 2 F (1 − |z|)2n−2 d xd y ≤ C(n, α, β) F (z) d xd y. β (ζ )

α (ζ )

3. Let K be the Cantor set obtained from [0, 1] by repeatedly removing middle halves, and let E = K × K . Then 1 (E) < ∞ but E is not contained in a rectiﬁable curve. Also construct an example with 1 (E) = 0. Hint: Alter −n the construction so that at stage n there are 4n squares of side 4n . 4. (a) Show that the integral in (2.1) is inﬁnite at z = 0 when E is the union of two radii of the unit circle that meet at an angle less than π . (b) Prove that if 1 dt β E2 (z, t) < ∞, t 0 then limt→0 β E (z, t) = 0. (c) Suppose L n are lines through z with βn = β E (z, 2−n ) =

1 2−n

sup

E∩B(z,2−n )

dist(w, L n ).

If L n converges to a line L and βn → 0, then E has a tangent at z.

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1 (d) Prove that if βn < ∞ (equivalently 0 β E (z, t)/tdt < ∞) then E has a tangent at z. 5. (a) Construct a Jordan curve such that 0 ∈ and ε(0, t) = 0, t < 1 but β (0, t) ≥ β0 > 0 t < 1. (b) Construct a Jordan curve such that 0 ∈ , 1 dt β 2 (0, t) < ∞ t 0 but such that fails to have a tangent at z = 0. 6. Let E be a ﬁnite set of points in C. Deﬁne a path from z 1 ∈ E to z n ∈ E to be a collection {[z i , z i+1 ] : 1 ≤ i ≤ n − 1} of line segments with endpoints {z i }n1 ⊂ E, and deﬁne a circuit to be a closed path (i.e. z 1 = z n ) such that z i = z j , for 1 ≤ i < j ≤ n − 1 and with n ≥ 4. A spanning tree G for E is a collection of line segments with endpoints in E such that G contains a path connecting every pair of points z 1 , z 2 ∈ E, and such that G contains no circuit in E. A tour of E is a closed path containing every point of E. The greedy algorithm is the following construction of a spanning tree for E : Choose any z 1 ∈ E. Let z 2 ∈ E satisfy |z 1 − z 2 | = minz∈E\{z 1 } |z − z 1 |, and put the line segment e1 = [z 1 , z 2 ] into G. Having chosen distinct points z 1 , . . . , z m−1 in E, choose z m ∈ E \ {z 1 , . . . , z m−1 } so that for some k < m, |z m − z k | = min{|z − z j | : z ∈ E \ {z 1 , . . . , z m−1 } and 1 ≤ j ≤ m − 1}, and put the line segment em = [z k , z m ] into G. Continue this process until G = E. (a) Prove that the greedy algorithm constructs a spanning tree of minimal possible length. Hint: Let M be a minimal spanning tree and let G be the spanning tree constructed by the greedy algorithm. Let k be the largest integer so that e1 , . . . , ek ∈ G ∩ M. If we add ek+1 to M, then M contains a circuit all of whose segments have length no more that |ek+1 |. But by the construction of G, this circuit must contain and edge e with exactly one vertex in the collection of endpoints of e1 , . . . , ek and length at least |ek+1 |. Replacing e with ek+1 , we obtain another minimal spanning tree M with e1 , . . . , ek , ek+1 ∈ G ∩ M . (b) Construct a tour T of E with (T ) ≤ 21 (G). Hint: Do induction on the number of edges. (c) Prove that every tour of minimal length contains a spanning tree. (d) If K is a connected set containing E with 1 (K ) minimal, then K contains a closed rectiﬁable curve with () ≤ 21 (K ).

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(e) Use the above to show that 2 min{1 (K ) : K connected ⊃ E} ≥ min{() : closed curve ⊃ E} = min{(T ) : T tour of E}

(E.1)

> 1 (G) ≥ min{1 (K ) : K connected ⊃ E}, where G is the spanning tree constructed by the greedy algorithm. A connected set of minimal length containing E does not necessarily consist of line segments with vertices in E, but by (E.1) the greedy algorithm constructs a connected set G ⊃ E of length no more that twice the minimal possible length. The greedy algorithm together with the construction in (b) constructs a tour of length no more than twice the length of a shortest tour. (f) Show that the method of Section 2 constructs a spanning tree for E. Each segment comes from applications of Case 2. (g) If E is a compact set in C contained in a rectiﬁable curve , then replacing each arc in \ E with a straight line segment with endpoints in E, we obtain a rectiﬁable curve which is not longer than . Given ε > 0, choose z 1 , . . . , z m ∈ E so that if z ∈ E then |z − z j | < ε for some j. Applying the greedy algorithm to {z 1 , . . . , z m } we then obtain a tree G ε such that 1 (G ε ) ≤ (), and dist(z, G ε ) < ε for every z ∈ E. Construct a tour Tε of G ε with (Tε ) ≤ 1 (G ε ). Show that there is a limiting tour T = limε j →0 Tε j with (T ) ≤ 2(). Constructing a tour of minimal length containing a ﬁnite set E is called the traveling salesman problem. The greedy algorithm together with the construction in (b) gives a tour of length no more than twice the minimal possible length. 7. Recall from Chapter VII that a curve is Ahlfors regular if there is a constant A > 0 such that for every disc B(z, r ), 1 ( ∩ B(z, r )) ≤ Ar. Let E ⊂ C be compact. Prove E is contained in an Ahlfors regular curve if and only if there is a constant A such that for any z ∈ E and any 0 < R < diamE, R dt β E2 (w, t) d1 (w) ≤ A R. t E∩B(z,R) 0 See Jones [1990]. Related results can be found in David and Semmes [1991] and Pajot [2002]. 8. This exercise outlines the proof of the following result from Bishop and Jones [1997a].

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Theorem. Let E be a compact connected set such that β E (3Q) ≥ β0 > 0 for all dyadic squares Q such that Q ∩ E = ∅ and (Q) ≤

(E.2) diam(E) . 3

dim(E) ≥ 1 + cβ0 2 ,

Then (E.3)

where c is a constant independent of β0 . (a) When E is a compact connected set deﬁne βk2 (E) = diam(E) + β E2 (3Q)(Q) : Q is dyadic and (Q) ≥ 2−k , and

1 k (E) = inf () : is rectiﬁable and inf dist(z, ) ≤ 2−k− 2 . E

Then C1 βk2 (E) ≤ k (E) ≤ C2 βk2 (E), where the constants C1 and C2 do not depend on β0 . Hint: Apply Theorem 2.3 to &# $ Q : Q dyadic, (Q) = 2−k , and Q ∩ E = ∅ . k (E) = ∂ (b) Let E be a compact connected set, let Q N be a dyadic square of side such that Q ∩ E = ∅. Then for n > N , 3Q contains at least 2−N ≤ diam(E) 3 n−N dyadic squares Q n of side 2−n such that Q n ∩ E = ∅. 2 (c) Now let E be a compact connected set such that (E.2) holds for every . Fix a dyadic square dyadic Q such that Q ∩ E = ∅ and (Q) ≥ diam(E) 3 diam(E) −N Q N of side 2 ≤ such that Q N ∩ E = ∅. Then by (b), 3 β N2 +k (Q N ∩ E) ≥ kβ0 2 2−N . Apply (a) to E Q = (E ∩ 3Q N ) ∪ ∂ Q to show 3Q N contains at least Nk = ckβ0 2 2k dyadic squares {Q j } such that (Q j ) = 2−N −k ,

3Q j ∩ 3Q j = ∅, j = j , and Q j ∩ E = ∅,

(E.4)

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where c is an absolute constant. Hint: Consider N +k (E Q ). (d) We continue to assume the compact connected set E satisﬁes (E.2). Fix k = [ β12 ] and note that by (E.4) 0

Nk ≥ 2(1+cβ0

2 )k

≡ δ,

for some constant c. Assume diam(E) ≥ 3. By (c) and induction, there are generations Gn = {Q nj }, n = 0, 1, . . . of dyadic squares of side 2−nk such that ∈ Gn−1 , (i) For n ≥ 1 each Q nj ∈ Gn is a subset of some Q n−1 j (ii) Each Q n−1 ∈ Gn−1 contains exactly Nk distinct squares Q nj ∈ Gn , j and (iii) Q nj ∩ E = ∅ for all Q nj ∈ Gn . (e) If the compact set E satisﬁes (i)–(iii), then dim(E) ≥ 1 + cβ0 2 .

(E.5)

Hint: Deﬁne the measure μ by μ(Q nj ) = Nk −n ≤ 2−(1+cβ0

2 )nk

.

Then μ has support E and μ(E) = 1. Moreover, there is a constant C = C(k) such that Frostman’s Theorem D.1 can be applied to Cμ to yield 1+cβ0 2 (E) > 0. (f) Except for the value of c in (E.5) the above theorem is sharp. If β > 0, this can be seen by a variation on the von Koch snowﬂake construction. 9. Let Q be a dyadic Carleson box in D, let z Q be the center of its top half T (Q), let G(Q) be a family of pairwise disjoint Carleson boxes Q ⊂ Q, and let D(Q) = Q \ G(Q) Q . Let F ∈ H 2 have no zeros and || log F||B ≤ 6. This exercise outlines a proof without using Appendix M of the inequality |F(z) − F(z Q )|2 ds ≤ C |F (z)|2 dist(z, ∂D(Q))d xd y. ∂D(Q)

D(Q)

The construction is simpler in the upper half-plane and for this reason we assume the box Q and its subboxes Q are dyadic squares in H that have one side in R. Let {Jk } be the set of maximal vertical line segments in ∂D(Q) ∩ G(Q) ∂ Q . For each, Jk ⊂ ∂ Q (k) for a unique Q (k) ∈ G(Q). Let Wk ⊂ Q (k) be the triangle bounded by Jk , a segment of slope ±2, and a segment Ik of length (J2k ) on the top edge of Q (k). Deﬁne & U = D(Q) ∪ (Wk ∪ Jk ∪ Ik ). k

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Then U is a Lipschitz domain and the proof of Theorem 1.2 can be adapted to yield (1.7) for U. Now since log F is a Bloch function it is not hard to show |F(z) − F(z Q )|2 ds ≤ C |F(z) − F(z Q )|2 ds. ∂D(Q)

∂U

On the other hand, 2 |F (z)| dist(z, ∂U)d xd y ≤ 2 |F (z)|2 dist(z, ∂D(Q))d xd y. U

D(Q)

Hint: The reﬂections of the Wk through the Jk are pairwise disjoint. 10. Let ϕ be univalent 2 on D and let g(z) = log(ϕ (z)). If ||g||B is small, then 2 supD 1 − |z| |Sϕ| is small, but not conversely. On the other hand if

2 supD 1 − |z|2 |Sϕ| is small, then there is a Möbius transformation T such that (T ◦ ϕ)

1 − |z|2 (T ◦ ϕ) is small. 11. Suppose D ⊂ D is a special cone domain and let ϕ be univalent on D. Then Lemma 4.3 holds for ϕ: Given ε > 0, a > 0, and b > 0, there exists δ = δ(ε, a, b) such that if

2 dist(z, ∂D) Sϕ(z) < δ on the hyperbolic ball {z : ρD (z, z 0 ) < a}, then for some Möbius transformation T , z − z z − z 0 0 + ϕ(z) − T −1 <ε T (ϕ(z)) − 1 − z0 z 1 − z0 z on the hyperbolic ball {z : ρD (z, z 0 ) < b}. 12. Let be a simply connected Jordan domain and let z ∈ ∂. (a) If is a quasidisc, prove 1 η (z, t) ≤ β∂ (z, t) ≤ Cη (z, t) C for small t, where the constant C depends only on the quasiconformality of ∂. (b) For any constant M make an example for which β∂ (z, t) ≥ Mη (z, t).

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(c) For any constant M make an example for which η (z, t) ≥ Mβ∂ (z, t). 13. (a) Let D ⊂ D be a special cone domain and let ϕ be a conformal map from D onto a quasidisc . Write d(z) = dist(z, ∂D). Then 2 1 (∂) ≤ diam() + |ϕ (z)| Sϕ(z) d(z)3 d xd y. C1 D

(b) Let D be a chord-arc domain with 0 ∈ D and dist(0, ∂D) ≥ c diam(D). Write d(z) = dist(z, ∂D). Let F(z) be analytic on D. Verify that F ∈ H 2 (D) if and only if |F (z)|2 d(z)d xd y < ∞, D

and that

C||F||2H 2 (D) ≤ |F(0)|2 +

|F (z)|2 d(z)d xd y ≤

D

and C

||F||2H 2 (D)

≤ |F(0)| + |F (0)| + 2

2

1 ||F||2H 2 (D) C

|F (z)|2 d(z)3 d xd y

D

1 ≤ ||F||2H 2 (D) , C where C and C depend only on the chord-arc constant of D and the constant c above. Hint: Show (1.9) holds for D, and then use Theorem M.1 from Appendix M. (c) Let D be as in part (b) and let ϕ be a conformal mapping from D to a domain such that ∂ is rectiﬁable. Then 2 |ϕ (z)| Sϕ(z) d(z)3 d xd y ≤ C(∂). D

14. Prove inequality (7.15) without using Theorem M.1. 15. (B. Davis) Let {w j } ⊂ D satisfy ρ(w j , wk ) ≥ 1, let j = {ρ(w, w j ) ≤ 18 }, and assume &

inf ω w j , ∂D, D \ (E.6) k ≥ δ > 0, j

k = j

just as in Lemma 11.2. Let Pz denote probability on the space of Brownian paths Bt starting at z and let E denote integration with respect to P0 . By

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Kakutani’s theorem the set E j = {Bt hits j before hitting ∂D} has P0 (E j ) ≥

c(1 − |w j |) . 2π

(E.7)

Set N = χ E j . The random variable N counts the number of j visited by Bt before Bt reaches ∂D. By (E.7), (1 − |w j |) ≤ 2π E(N ). In terms of Pz our hypothesis (E.6) says that &

& inf Pz Bt hits j before hitting ∂D = inf Pz Ek z∈ j

z∈ j

k = j

k = j

(E.8)

≤ 1 − δ < 1. Let τ1 = inf{t : Bt ∈ j } be the ﬁrst time Bt hits j . Assuming that τn has been deﬁned and that Bτn ∈ j (n) , let & j } τn+1 = inf{t > τn : Bt ∈

j = j (n)

be the time of the ﬁrst visit to a different disc j . Then the functions τn and τD = inf{t : Bt ∈ ∂D} are measurable with respect to P0 . Now let M = max{n : τn ≤ τD }. Then M − 1 is the number of trips from one j to another taken by Bt before it hits ∂D. Therefore M ≥ N and E(N ) ≤ E(M) =

∞

P0 (M ≥ n).

n=0

Now by (E.8) and the independence of Brownian increments, P0 (M ≥ n + 1) = P0 (M ≥ n + 1; Bτn ∈ j ) j

≤ (1 − δ)

P0 (M ≥ n; Bτn ∈ j )

j

= (1 − δ)P0 (M ≥ n), and by induction ∞ 2π (1 − |w j |) ≤ 2π (1 − δ)n−1 = . δ n=1

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A. Hardy Spaces If 0 < p < ∞, then by deﬁnition an analytic function f (z) on the unit disc D is in the Hardy space H p if p | f (r eiθ )| p dθ = || f || H p < ∞. sup (A.1) 0
For p = ∞ the Hardy space H ∞ is deﬁned to be the space of bounded analytic functions on D with norm || f ||∞ = supD | f (z)|. Write fr (z) = f (r z) and suppose 0 < r < s < 1. Then || fr ||∞ ≤ || f s ||∞ . If p < ∞, then 2π p | fr (z)| ≤ P r z | f s (eiθ )| p dθ 0

s

because the function | f (z)| p is subharmonic. It then follows from Fubini’s theorem that || fr || p is an increasing function of r whenever 0 < p ≤ ∞. Theorem A.1. Assume 1 ≤ p < ∞ and let f ∈ H p . Then (a) The function f (z) has a nontangential limit f (eiθ ) almost everywhere dθ and 2π f (r eiθ ) − f (eiθ ) p dθ = 0. (A.2) lim r →1 0

(b) If f ≡ 0, then

log f (eiθ ) dθ > −∞.

(A.3)

The case p > 1 of part (a) follows from Exercise I.11 of Chapter I, and the signiﬁcance of Theorem A.1 lies in the case p = 1 and the inequality (A.3).

435

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From part (a) it follows that

f (z) = lim f (r z) = r →1

Pz (θ ) f (eiθ )dθ,

and since || fr || p is increasing in r , we obtain p

p

|| f || H p = || f || L p (∂ D) for all f ∈ H p . Theorem A.1 is also true for f ∈ H p for all p < 1. See Exercise A.1. Part (a) of Theorem A.1 is due to F. and M. Riesz [1916], and part (b) is from F. Riesz [1923]. Part (b) has the following important corollary. Corollary A.2. If f ∈ H 1 and f ≡ 0, then | f (eiθ )| > 0 almost everywhere. The proof of Theorem A.1 will use Theorem A.3 below, which is called Hardy–Littlewood maximal theorem for H p functions: Theorem A.3. Let α > 1 be ﬁxed, let 0 < p ≤ ∞, and let f ∈ H p . Then the nontangential maximal function f α ∗ (ζ ) = sup | f (z)| α (ζ )

satisﬁes || f α ∗ || p ≤ Aα || f || H p ,

(A.4)

where the constant Aα depends only on α. We ﬁrst derive Theorem A.1 from Theorem A.3 and then return to the proof of Theorem A.3. Proof of Theorem A.1(a). First suppose p = 1. By (A.1) and the Banach– Alaoglu theorem, there are rn → 1 so that f (rn eiθ )dθ converges weak-star to a ﬁnite complex measure μ on ∂D. Then iθ Pz (θ ) f (rn e )dθ = Pz (θ )dμ(θ). f (z) = lim f (rn z) = lim n→∞

n→∞

Write dμ = f dθ + dν, where f (eiθ ) ∈ L 1 (dθ ) and ν is singular to Lebesgue measure. Then iθ f (z) = Pz (θ ) f (e )dθ + Pz (θ )dν(θ ),

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and by Fatou’s theorem Pz (θ ) f (eiθ )dθ has nontangential limit f (eiθ ) almost everywhere. In a moment we will prove that Pz (θ )dν = 0 (A.5) lim α (ζ )z→ζ

almost everywhere. That will mean f (z) has nontangential limit f (eiθ ) almost everywhere, and (A.2) will then follow from (A.4) and dominated convergence. To prove (A.5) we can assume ν > 0. Fix α > 1 and ε > 0. Take a compact set K ⊂ ∂D so that |K | <ε and ν(∂D \ K ) < ε, and write ν1 = χ K ν and ν2 = ν − ν1 . Then lim z→ζ Pz (θ )dν1 (θ ) = 0 for all ζ ∈ ∂D \ K . The proofs of Lemmas I.2.2 and I.2.4 show that # $ 3 + 6α (3 + 6α)ε ζ ∈ ∂D : sup Pz (θ )dν2 (θ ) > λ ≤ dν2 ≤ . λ λ α (ζ ) Taking λ = ε1/2 then gives # $ 1/2 1/2 ζ ∈ ∂D : lim sup Pz (θ )dν(θ ) ≥ ε ≤ |K | + (3 + 6α)ε α (ζ )z→ζ

≤ ε + (3 + 6α)ε1/2 , so that (A.5) holds almost everywhere. If 1 < p < ∞, part (a) is easier. There are rn → 1 so that frn (θ ) = f (rn eiθ ) converges weakly in L p to some function f ∈ L p . Then f (z) = lim f (rn z) = n→∞

Pz (θ ) f (eiθ )dθ

p because Pz ∈ L q , q = p−1 , and then Fatou’s theorem, (A.4), and dominated convergence yield (A.2).

Proof of Theorem A.1(b). Part (b) follows from part (a). We may suppose f (0) = 0, by dividing f by z n in case f (0) = 0. Then for r < 1 the familiar Jensen formula gives dθ log | f (0)| ≤ log | f (r eiθ )| 2π dθ dθ = log+ | f (r eiθ )| − log− | f (r eiθ )| , 2π 2π where log− x = (log(1/x))+ = max(− log x, 0). By dominated convergence

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and part (a),

lim

r →1

log+ | f (r eiθ )|dθ =

log+ | f (eiθ )|dθ,

while by Fatou’s lemma, − iθ log | f (e )|dθ ≤ lim inf log− | f (r eiθ )|dθ. r →1

Therefore we have the important inequality dθ log | f (0)| ≤ log | f (eiθ )| , 2π

(A.6)

which implies (A.3).

Proof of Theorem A.3.The proof will be divided into two cases, p ≥ 2 and 0 < p < 2. In the ﬁrst case, p ≥ 2, f is the Poisson integral of an L p function f (eiθ ) so that by Lemma I.2.2, f α ∗ (eiθ ) ≤ (1 + 2α)M f (eiθ ),

(A.7)

where M f is the Hardy–Littlewood maximal function of f (eiθ ). That means (A.4) is a consequence of the basic inequality ||M f || p ≤ C p || f || p ,

1 < p ≤ ∞,

(A.8)

which we will √ establish in a moment. In (A.8) the constants C p depend on p, but C p ≤ 2 6 for p ≥ 2. Except for its range of p, (A.8) is just (A.4) with f α∗ replaced by M f , and for this reason (A.8) is also known as the Hardy– Littlewood maximal theorem. The second case, 0 0 set m(λ) = {M f > λ} . The function m(λ) is called the distribution function of M f . By Fubini’s theorem, 2π # M f (eiθ ) ∞ $ p p−1 pλ dλ dθ = pλ p−1 m(λ)dλ. (A.9) (M f ) dθ = 0

0

0

We also have m(λ) ≤

3|| f ||1 , λ

(A.10)

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from Lemma I.2.4. Now we use an important trick, due to Marcinkiewicz, to improve inequality (A.10). Set f 1 = f χ {| f |> λ } , 2

and f 2 = f χ {| f |≤ λ } . 2

Then M f ≤ M( f 1 ) + M( f 2 ), while M( f 2 ) ≤ || f 2 ||∞ ≤ λ2 , so that by (A.10), λ 6 m(λ) ≤ M( f 1 ) > | f |dθ. ≤ 2 λ {| f |> λ2 } Then from (A.9) and Fubini’s theorem we get ∞ p ||M f || p ≤ 6 pλ p−2

{| f |> λ2 }

0

iθ = 6 f (e )

| f |dθ dλ

2| f (eiθ )|

pλ p−2 dλdθ

0

3 p2 p = | f | p dθ. p−1

3p 1 p is decreasing we also have the bound That proves (A.8), and because p−1 √ C p ≤ 2 √6 for p ≥ 2. Together (A.8) and (A.7) prove (A.4) when p ≥ 2 with Aα ≤ 2 6(1 + 2α). The Marcinkiewicz argument is also valid when 1 < p < 2, but it unfortu 3p 1 p which are unbounded as p → 1. nately gives constants 2 p−1 Case II. 0 < p ≤ 2. We can assume f ≡ 0. Take 0 < r < 1 so that f has no zeros on {|z| = r } and set 1 Pz (θ ) log | f (r eiθ )|dθ. Ur (z) = 2π Then log | f (r z)| ≤ Ur (z) on D. Since Ur is real and harmonic on D, there is r is analytic and r such that Ur + i U a unique harmonic conjugate function U r (0) = 0. Write U p

gr (z) = e 2 (Ur +i Ur ) .

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Then | f (r z)| p ≤ |gr (z)|2 and p |gr |2 dθ = | f (r eiθ )| p dθ ≤ || f || p .

(A.11)

Let r → 1, and let g be a weak limit of gr in L 2 . Then g ∈ H 2 because Poisson 2/ p kernels are in L 2 and ||g|| H 2 ≤ || f || H p by (A.11). Moreover, | f (z)| = lim | f (r z)| ≤ |g(z)|2/ p . r →1

Therefore

( f α∗ ) p

≤

(gα∗ )2

∈

L 1,

and

|| f α∗ || p ≤ ||gα∗ ||2 . 2/ p

Thus (A.4) holds for 0 < p < 2 with the same bound on Aα as for the case p = 2. Exercise A.1. (a) Suppose 0 < p < 1, and suppose f ∈ H p . Prove f has nontangential limit f (eiθ ) almost everywhere and if f (0) = 0, 2π dθ log | f (eiθ )| . log | f (0)| ≤ 2π 0 (b) On the other hand, if E ⊂ ∂D is compact, if |E| = 0, and if p < ∞, then there is f (z) ∈ H p such that limr →1 | f (r eiθ )| = ∞ everywhere on E. Exercise A.2. (a) For p > 1 and α > 1, there exists C( p, α) > 0 such that if f ∈ L p (∂D) and u = u f then ||u ∗α || p ≤ C( p, α)|| f || p . / L 1 (∂D). (b) However there is f ∈ L 1 (∂D) such that sup0 1 and if u is the Poisson integral of f ∈ L p . It also follows from Exercise I.19(a) that p | f (z)| dμ ≤ C | f (eiθ )| p dθ if p > 0 and if f ∈ H p . These results are from Carleson [1958] and [1962]. Exercise A.4. If F(z) is analytic on D and if ReF(z) > 0 for all z ∈ D, then F ∈ H p for all p < 1. Hint: Use Zygmund’s theorem on conjugate functions

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441

from Chapter II. Somewhat deeper is the weak-type estimate |{θ : |Fα∗ | > λ}| ≤

C . λ

See for example Garnett [1981].

B. Mixed Boundary Value Problems This appendix shows how to solve a mixed boundary value problem on a domain by reducing it to the Dirichlet problem on a Riemann surface double of . Let be a bounded ﬁnitely connected Jordan domain with ∂ = m j=1 j , 1+α where each j is of class C and j ∩ k = ∅ if j = k. Let v represent the steady-state temperature of a thin homogeneous plate in the shape of . Then v is harmonic in and the loss of heat through an arc J on ∂ is given by ∂v ds, J ∂n where n is the unit outer normal to ∂. Given a function f on ∂, the Neumann problem is to ﬁnd v ∈ C 1 () such that v is harmonic on and ∂v/∂n = f on ∂. In other words, ﬁnd a steady-state temperature on the plate that produces a prescribed heat ﬂow f across the boundary. By Green’s theorem f must satisfy the necessary condition ∂v f ds = ds = 0. (B.1) ∂ ∂ ∂n Except for issues of smoothness, (B.1) is the only condition necessary for the Neumann problem. Theorem B.1. Suppose is a ﬁnitely connected Jordan domain such that 1+α for some α > 0. Let β > 0, and suppose f ∈ C β (∂) satisﬁes ∂ ∈C 1 ∂ f ds = 0. Then there is a harmonic function v ∈ C () such that ∂v = f ∂n on ∂, and v is unique up to an additive constant.

(B.2)

If f satisﬁes (B.1) and is continuous but not necessarily Hölder continuous, there is an elementary argument that yields a weaker version of the conclusion ∂v/∂n = f. See Exercise B.1. Proof of Theorem B.1.The uniqueness is easy: we may assume f = 0 and, via a conformal mapping, we may assume ∂ consists of analytic curves. Each

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ζ ∈ ∂ has a neighborhood V and conformal map ϕ : V → D so that ϕ(V ∩ ∂) = R ∩ D and ϕ(V ∩ ) = D ∩ {Imz > 0}. Then v ◦ ϕ −1 = Im H for some function H analytic on D ∩ {Imz > 0}. We can take ReH = 0 on R ∩ D, since (ReH )x = (v ◦ ϕ −1 ) y = 0 on R ∩ D, by Corollary II.4.5 and the Cauchy–Riemann equations. By the Schwarz reﬂection principle, H extends to be analytic on D, with Im H (z) = Im H (z).

(B.3)

Hence v extends to be harmonic on a neighborhood W of . But then by (B.3), supW v(z) is attained on and v is constant. To prove there exists harmonic v ∈ C 1 () satisfying (B.2), ﬁrst suppose that is simply connected and let γ (s) denote arc length parameterization of ∂ with positive orientation. Deﬁne s F(γ (s)) = f (γ (t))dt, 0

and let u be the solution to the Dirichlet problem with boundary data F. Then ∂u ∂s = f on ∂, by Corollary II.4.6, and u = ReH for some analytic function H , so that by Corollary II.4.5, v = −Im H ∈ C 1+ε (), for some ε > 0. Then by continuity and the Cauchy–Riemann equations, ∂u ∂F ∂v = = = f ∂n ∂s ∂s on ∂, and (B.2) holds. When is not simply connected, this deﬁnition may not give a single valued function F, and not every harmonic function has a single valued conjugate function. To get around those difﬁculties, we need two lemmas. Lemma B.2. Let be a bounded, ﬁnitely connected Jordan domain. Write ∂ = m j=1 j , where j ∩ k = ∅ for j = k and each j is a Jordan curve. Let ω j (z) = ω(z, j , ). If u is a harmonic function on , then there exist unique a1 , . . . , am−1 such that u(z) +

m−1

a j ω j (z)

j=1

has a single valued conjugate function on .

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Proof. By Lemma II.2.2 we may assume each j is an analytic curve. If γ u /∂s is a single valued function on is a closed C 1 Jordan curve in , then ∂ γ , where u is a locally deﬁned harmonic conjugate of u, and there is a single valued conjugate function u on the full domain if and only if ∂ u ds = 0 γ ∂s for all such Jordan curves. If γ1 and γ2 are C 1 Jordan curves homologous in , then by Cauchy’s theorem, ∂ u ∇u · n ds = Im (u x − iu y )dz = 0, ds = γ1 −γ2 ∂s γ1 −γ2 γ1 −γ2 since u x − iu y is analytic on . Let γ j ⊂ be homologous to j and deﬁne the period of u over j by ∂u u) = ds. Pj ( γ j ∂n u ) does not depend on the choice of γ j . By Green’s We have just seen that the Pj ( theorem, m

Pj ( u ) = 0,

j=1

so that u has a single valued conjugate if and only if u ) = 0 for j = 1, . . . , m − 1. Pj ( By Schwarz reﬂection each ω j extends to be harmonic on a neighborhood of . Set ∂ω j ds. (B.4) αk, j = k ∂n Then Lemma B.2 is proved if we can ﬁnd a1 , . . . , am−1 such that m−1

αk, j a j = −Pk ( u ),

1 ≤ k ≤ m − 1,

j=1

and the next lemma ensures that such a1 , . . . , am−1 exist. Lemma B.3. The matrix {αk, j }m−1 j,k=1 is nonsingular.

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Proof. If αk, j b j = 0 then, as we have seen, there is an analytic function F on such that ReF =

m−1

bj ωj .

j=1

But then ∂ImF/∂n = 0 on ∂, and by the (already proven) uniqueness part of Theorem B.1, ImF is constant on . Hence F is constant, and since ReF = 0 on m , we have b j = ReF| j = 0, and thus {αk, j } is nonsingular. We return to the proof of the existence of v in Theorem B.1. By Lemma B.3, there exist b1 , . . . , bm−1 so that m−1

f1 = f −

j=1

satisﬁes

bj

∂ω j ∂n

k

f 1 ds = 0,

(B.5)

for k = 1, . . . , m − 1. Since ∂ f ds = 0 and ∂ ∂ω j /∂n = 0 , the equality (B.5) also holds when k = m. On each j deﬁne s f 1 (ζ j (t))dt, F(ζ j (s)) = 0

where ζ j (s) is the arc length parameterization of j with positive orientation. Then by (B.5), F is well deﬁned on ∂ and F ∈ C 1+ε (∂) for some ε > 0. By Corollary II.4.6 there exists u 1 ∈ C 1+ε () with ∂u 1 /∂s = f 1 . Find a1 , . . . , am−1 , by Lemma B.2, so that u2 = u1 +

m−1

a j ω j = ReH

j=1

for some function H analytic in . Let v2 = −Im H . If ζ ∈ ∂, there is a simply connected domain N ⊂ bounded by a C 1+α curve, so that ∂ N ∩ ∂ is a neighborhood of ζ in ∂. By Corollary II.4.5, v2 ∈ C 1+ε () for some ε > 0. Again continuity and the Cauchy–Riemann equations yield ∂u 2 ∂u 1 ∂F ∂v2 = = = = f1 ∂n ∂s ∂s ∂s Now let v = v2 +

m−1 j=1

bj ωj .

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Then ∂ω j ∂v ∂v2 ∂ω j bj bj = + = f1 + = f, ∂n ∂n ∂n ∂n m−1

m−1

j=1

j=1

and Theorem B.1 is proved.

The solutions to the Neumann and Dirichlet problems combine to solve mixed boundary value problems. Theorem B.4. Let be a ﬁnitely connected Jordan domain with ∂ ∈ C 1+α β for some α > 0. Let f 1 ∈ C(∂), let β > 0, and let f 2 ∈ C (∂) be such that ∂ f 2 ds = 0. Let J be a ﬁnite union of open arcs in ∂ with pairwise disjoint closures. Then there is a harmonic function u ∈ C 1 ( ∪ J ) ∩ C() such that u = f 1 on ∂ \ J,

(B.6)

and ∂u = f 2 on J. ∂n If ∂ \ J = ∅, the function u is unique.

(B.7)

Proof. J J J

J J R

τ (J )

τ (J )

d

τ () τ (J ) Figure B.1. The doubled Riemann surface d . Form the doubled Riemann surface d by attaching two copies of along the set J . In other words, assume ⊂ {Imz > 0}, let τ (z) = z map to the reﬂected domain τ () and set d = ∪ J ∪ τ ( ∪ J ) / ∼, where the union denotes a disjoint union and where z ∼ w if and only if w = τ (z) = z and z ∈ J. The sets and τ () are declared open sets in d . To deﬁne a basic neighborhood of p ∈ J , let W be a simply connected neighborhood of

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p in such that W ∩ ∂ ⊂ J and let ϕ : W ∩ → D+ = D ∩ {Imz > 0} be a conformal map so that the continuous extension of ϕ to W ∩ ∂ satisﬁes ϕ(W ∩ ∂) = (−1, 1). The companion neighborhood τ (W ∩ ) in τ () is (−1, 1) so that τ (z) corresponds identiﬁed with the reﬂection D− of D+ through to ϕ(z). The full disc V = W ∪ τ (W ) / ∼ then deﬁnes a neighborhood of p in d , with local coordinate ( ϕ(z), z ∈ W, (z) = ϕ(z), z ∈ τ (W ). Thus, f is analytic on V if and only if f ◦ −1 is analytic on D. For example, if is the unit disc D, then d is conformally equivalent to C∗ \ (∂D \ J ). Notice that if v is harmonic on V and if v(τ (z)) = v(z), then ∂v ◦ ϕ −1 /∂ y = 0 on (−1, 1), and since ∂ ∈ C 1+α , it then follows from Theorem II.4.3 that ∂v/∂n = 0 on J ∩ ∂ W, where the derivative is now taken with respect to the original coordinates on W . We may assume ∂ consists of analytic curves. Choose a domain 1 ⊂ with ∂ \ ∂1 = J, such that d \ τ (1 ) is (conformally equivalent to) a domain 2 ⊃ with 2 ∩ ∂ = J . On d we can apply the Schwarz alternating method from the proof of Theorem II.1.1, with σ and σ augmented by ∂1 ∩ and ∂2 \ ∂, respectively, and therefore the Dirichlet problem can be solved on d . By Theorem B.1, there is a u 1 ∈ C 1 () with ∂u 1 /∂n = f 2 on ∂. Let u 2 solve the Dirichlet problem on d for the boundary values f 1 − u 1 on ∂d . Then, by the remarks above, ∂u 2 /∂n = 0 on J , and therefore u 1 + u 2 satisﬁes conditions (B.6) and (B.7). To prove uniqueness when ∂ \ J = ∅, we can suppose f 1 = f 2 = 0. Then if u satisﬁes (B.6) and (B.7), u extends to be harmonic on d with u = 0 on ∂d . Hence u = 0 by the maximum principle. We do not know the origin of the idea of reducing mixed boundary value problems on to Dirichlet problems on Riemann surface doubles of . Ohtsuka [1970] credited conversations between Ohtsuka and Strebel. See also Tsuji [1959], p.31. Exercise B.1. (a) Let f be real valued and continuous on ∂D, and assume f dθ = 0. Then there is v(z) ∈ C 1 (D), harmonic on D, with ∂v/∂r = f on ∂D if and only if f is continuous on ∂D. 2π (b) If f ∈ C(∂D) is real valued and if 0 f (eit )dt = 0, set 2π iθ e +z θ dθ f (eit )dt . H (z) = iθ e −z 0 2π 0

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Then

2π

H (z) = −

0

and i ∂H (z) = − ∂r r

2 log eiθ − z − θ i

2π

0

f (eiθ )

eiθ + z −1 eiθ − z

f (eiθ )

dθ , 2π

dθ . 2π

Thus v = −Im H ∈ C(D) and ∂v/∂r = f on ∂D. This gives an alternate proof of Theorem B.1 when = D without the assumption that f ∈ C β . However, it does not prove v ∈ C 1 and in fact v may not be C 1 in this situation. Exercise B.2. Use Green’s function to ﬁnd the kernel N (z, ζ ), z ∈ , ζ ∈ ∂, such that under the conditions of Theorem B.1, v(z) = N (z, ζ ) f (ζ )ds. ∂

m−1 Exercise B.3. Prove that the matrix αk, j j,k=1 deﬁned by (B.4) is symmetric, αk, j = α j,k .

C. The Dirichlet Principle A real valued function on ∂D is piecewise continuous if f is continuous except on a ﬁnite set E but f has ﬁnite left and right limits at each point of E. A real valued function u(z) is piecewise smooth on a domain if u(z) is continuous on and if u(z) is continuously differentiable on \ J, where J = J (u) is a ﬁnite union of (closed or open) analytic arcs in . The Dirichlet integral of u or the squared Dirichlet norm of u is 2 D(u) = D (u) = |∇u| d xd y = |∇u|2 d xd y.

\J

If u = ReF and F is analytic on , then D(u) = |F |2 d xd y = Area F(),

where each value F(z) is counted according to its multiplicity. If D(u) < ∞ and D(v) < ∞, the Dirichlet inner product is ∇u · ∇v d xd y = (u x vx + u y v y )d xd y. D(u, v) =

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It satisﬁes the Cauchy–Schwarz inequality, D(u, v) ≤

%

% D(u) D(v),

and D(u ± v) = D(u) + D(v) ± 2D(u, v).

(C.1)

Given f ∈ C(∂), set D f = v : v is piecewise smooth in and lim v(z) = f (ζ ), for all ζ ∈ ∂ . z→ζ

The Dirichlet principle is the assertion that the extremal problem D(u) = inf{D(v) : v ∈ D f },

(C.2)

has unique solution u ∈ D f , and that u = u f is the solution of the Dirichlet problem on for the boundary value f . Unfortunately, this formulation of the Dirichlet principle is not true. Here is a counterexample: If f ∈ C(∂D) has Fourier series +∞

f (e ) ∼ iθ

an einθ ,

(C.3)

−∞

then its Poisson integral is u(r eiθ ) =

+∞

an r |n| einθ ,

0 ≤ r < 1,

−∞

and D(u) = 2π

+∞

|n||an |2 .

(C.4)

−∞

Hence there exist f ∈ C(∂D) such that D(u) = ∞. For example, f (eiθ ) =

∞ 1 cos(2k θ ) k2 k=1

is in C(∂D), but D(u) = π

∞ k 2 k=1

k4

= ∞.

However this example is the only thing wrong with Dirichlet principle. We shall see that (C.2) holds whenever D f ∩ v : D(v) < ∞ = ∅.

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Lemma C.1. Let f be piecewise continuous on ∂D and let u be the Poisson integral of f . Let E ⊂ ∂D be the ﬁnite set of discontinuities of f , and set D f = v : v bounded and piecewise smooth on D, lim v(z) z→ζ = f (ζ ), for all ζ ∈ ∂D \ E . If D f ∩ v : D(v) < ∞ = ∅, then D(u) < ∞, and u is the unique element of D f of minimal Dirichlet norm. That is, D(u) = inf{D(v) : v ∈ D f }, and if v ∈ D f satisﬁes D(v) = D(u), then v = u. Proof. Let f have Fourier series (C.3) and set u N (r e ) = iθ

N

an r |n| einθ .

−N

Then D(u N ) < ∞. If v ∈ D f and D(v) < ∞, then D(v − u N ) < ∞, and by Green’s theorem and dominated convergence, 2π ∂u N (v − u N )(r eiθ ) (r eiθ )dθ D(u N , v − u N ) = lim r →1 0 ∂r 2π ∂u N iθ = (v − u N )(eiθ ) (e )dθ. ∂r 0 But on ∂D, v − u N = f − u N has Fourier series |n|≥N +1 an einθ , so that by the Parseval identity, D(u N , v − u N ) = 0. Consequently D(v) = D(u N ) + D(v − u N )

(C.5)

by (C.1), and D(u N ) ≤ D(v). Since |∇u N | → |∇u| on D, Fatou’s lemma gives D(u) ≤ lim inf D(u N ) ≤ D(v) and we conclude both that D(u) < ∞ and that u has minimal Dirichlet norm in D f . To prove that u is the unique element of D f of minimal Dirichlet norm, note that when D(u) < ∞, (C.4) implies that D(u − u N ) → 0, and consequently

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that D(v − u N ) → D(v − u). Hence by (C.5) we have D(v) = D(u) + D(v − u)

(C.6)

whenever v ∈ D f and D(v) < ∞. Therefore if v ∈ D f and D(v) = D(u), then D(v − u) = 0 and v − u is constant on . But then since v(z) − u(z) → 0 when z → ζ ∈ ∂D \ E, that means v = u. Because the Dirichlet integral is a conformal invariant, Lemma C.1 also holds with any simply connected Jordan domain in place of D. Lemma C.2. Let I be an open arc on ∂D, and let f be a bounded continuous function on I . Set C f = v : v is piecewise smooth on D and lim v(z) = f (ζ ), for all ζ ∈ I . z→ζ

If

C f ∩ v : D(v) < ∞ = ∅,

then C f contains a unique element u(z) of minimal Dirichlet norm and u(z) solves the mixed boundary value problem: ⎧ u = 0, on , ⎪ ⎨ u = f, on I, (C.7) ⎪ ⎩ ∂u = 0, on ∂D \ I . ∂r The set ∂D \ I is called a free boundary because nothing is required of a function in C f on ∂D \ I . Proof. Conformally map D to D+ = D ∩ {Imz > 0} so that ∂D \ I is mapped to [−1, 1] ⊂ ∂D+ . Because Dirichlet integrals are conformally invariant and because the map is conformal across ∂D \ I , this does not change the extremal problem inf{D(v) : v ∈ C f } nor the condition (C.7) expected of its solution. Continue to write f (ζ ) for the boundary data on ∂D+ \ [−1, 1]. Let U (z) be the Poisson integral on D of

f (ζ ), Imζ ≥ 0, g(ζ ) = f (ζ ), Imζ < 0. Then U satisﬁes (C.7) on D+ and D(U ) = inf{D(V ) : V ∈ Dg } by Lemma C.1. If v ∈ C f were continuous up to (−1, 1), its reﬂection

v(z), Imz ≥ 0, V (z) = v(z), Imz < 0,

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would be in Dg and the lemma would follow from Lemma C.1. However, because v may not be continuous up to (−1, 1) we must ﬁrst make some minor adjustments.

Wε

Figure C.1 ! For ε > 0, set γε = D+ ∩ {z : |z + εi | = 1 + ε12 }, and . . # i 1 $ + # 1$ i Wε = z : z + < 1 + 2 z : z − < 1 + 2 , ε ε ε ε and for z ∈ Wε , let z ∗ ∈ γε satisfy Rez ∗ = Rez. For small δ > 0, δ 2 |∇v| dsdε ≤ C |∇v|2 d xd y, 0

γε

W δ ∩D +

and there is ε, 0 < ε < δ, so that ε γε |∇v|2 ds ≤ δ. Set ⎧ ⎨ v(z), z ∈ D+ \ Wε , Vδ (z) = v(z), z ∈ D+ \ Wε ; ⎩ v(z ∗ ), z ∈ Wε . Then Vδ ∈ Dg , so that D(U ) ≤ D(Vδ ). But DD+ (v) =

1 lim DD (Vδ ), 2 δ→0

and hence DD+ (U ) ≤ DD+ (v). The proof of uniqueness is the same as for Lemma C.1. If D(v) = DD+ (U ), then v − U is constant on D+ , and because lim z→ζ (v(z) − U (z)) = 0 for all ζ ∈ ∂D ∩ {Imz > 0}, then v = U .

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Theorem C.3. Let be a ﬁnitely connected Jordan domain, let E be a nonempty ﬁnite union of open arcs on ∂ and let f ∈ C(E). Set C f = v : v is piecewise smooth on , lim v(z) = f (ζ ), for all ζ ∈ E , z→ζ

and assume

C f ∩ v : D(v) < ∞ = ∅.

Then the extremal problem D(u) = inf{D(v) : v ∈ C f }

(C.8)

has unique solution u, and u satisﬁes u = 0 on , and u = f on E. Moreover, if for some α > 0, ∂ \ E consists of C 1+α arcs, then ∂u (C.9) = 0 on ∂ \ E. ∂n Proof. After a conformal mapping, we may assume ∂ consists of analytic Jordan curves. By Theorem II.4.3 this assumption does not change (C.9). Let {u n } ⊂ C f be a minimizing sequence for the extremal problem (C.8), so that D(u n ) → inf{D(v) : v ∈ C f }. Then by (C.1),

un + um

un − um D(u n ) + D(u m ) +D = , 2 2 2 and D(u n − u m ) → 0 because C f is a convex set. Hence there is a vector (α, β) ∈ L 2 () such that # $ (u n )x − α 2 + (u n ) y − β 2 d xd y → 0. (C.10) D

Moreover, (α, β) is independent of the minimizing sequence {u n }. Fix an open disc D such that D ⊂ , and let Un be the solution of the Dirichlet problem on D with boundary value u n , and replace u n ∈ C f by Un χ D + u n χ \D ∈ C f .

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Then the new sequence {u n } is still a minimizing sequence for (C.8). The familiar estimate for harmonic functions 1 1 2 4 2 2 |U (z)| ≤ |U (z)| d xd y , sup 2 πr {|z−z 0 |
which follows from the Cauchy–Schwarz inequality and the mean value property, now shows that both sequences {(u n )x } and {(u n ) y } converge to harmonic functions, uniformly on compact subset of D. Since D is any disc in , we conclude that α and β are (after alterations on a set of measure zero) harmonic functions on and ∂α ∂ 2un ∂β = lim = . n ∂y ∂ x∂ y ∂x We claim there exists a harmonic function u such that ∂u ∂u

, = α, β . ∇u = ∂x ∂y Now (C.11) holds if

(C.11)

γ

αd x + βdy = 0

for every smooth simple closed curve γ ⊂ . Fix such a curve γ and take a simply connected domain W such that ∂ W is piecewise analytic, such that W ⊂ , and such that γ \ W is a single point p. See Figure C.2.

γ p W

Figure C.2 As before, we can suppose the minimizing functions u n are harmonic on a neighborhood of p. In W let Un be the solution of the Dirichlet problem for the boundary value u n . Then by Corollary II.4.6, ∇Un is bounded on γ ∩ W. Set vn = Un χ W + u n χ \W . Then vn is also a minimizing sequence, and by

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dominated convergence

∂vn ∂vn αd x + βdy = lim dx + dy = lim u n ( p) − u n ( p) = 0. n n ∂y γ γ ∂x Therefore there exists a harmonic function u(z) satisfying (C.11). Because E = ∅ and (C.10) holds, an integration shows lim u n (z) exists for all z ∈ . Then since ∇u n → ∇u on , we may assume lim u n (z) = u(z) for all z ∈ . We now claim that u ∈ C f and that u satisﬁes (C.9). Let ζ ∈ E. There is δ > 0 such that W = ∩ B(ζ, δ) is a Jordan domain whose boundary consists of one arc γ1 = B(ζ, δ) ∩ ∂ ⊂ E and a second arc γ2 = ∩ ∂ B(ζ, δ). On W let vn be the solution of the Dirichlet problem with boundary value u n ∈ C(∂ W ), and let v be the solution with boundary value f χ γ1 + u χ γ2 . Then vn → v and ∇vn → ∇v on W . By Lemma C.1, DW (vn ) ≤ DW (u n ), and in fact by (C.6) DW (u n ) − DW (vn ) = DW (u n − vn ). But DW (vn ) + D (u n ) − DW (u n ) ≥ β = lim D (u n ), n

because vn χ W + u n χ \W ∈ C f , and therefore DW (u n − vn ) → 0. With (C.10) that gives DW (u − vn ) → 0, so that again by Fatou’s lemma, DW (u − v) ≤ lim inf DW (u − vn ) = 0. n

Hence u − v is constant on W , and because v = u on γ2 , u = v in W . Consequently lim u(z) = f (ζ ),

z→ζ

and u ∈ C f . To verify (C.9) let γ1 be a component of ∂ \ E and let γ2 ⊂ be a Jordan arc connecting the endpoints of γ1 so that γ1 ∪ γ2 bounds a Jordan domain W ⊂ . Because u ∈ C f has minimal Dirichlet norm and functions in C f are unrestricted on γ1 , DW (u) = inf DW (v) : v piecewise smooth on W, v = u on γ2 and Lemma (C.2) now implies ∂u/∂n = 0 on γ1 .

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Corollary C.4. Let be a ﬁnitely connected Jordan domain, and let E and F be ﬁnite unions of open arcs on ∂ such that E ∩ F = ∅. Then the extremal problem # inf D (v) : v piecewise smooth on , $ (C.12) lim sup v(z) ≤ 0, lim inf v(z) ≥ 1 z→ζ ∈E

z→ζ ∈F

has a unique solution, which satisﬁes u(z) = 0 on ,

(C.13)

u = 0 on E and u = 1 on F.

(C.14)

and

If ∂ consists of C 1+α curves, then ∂u (C.15) = 0 on ∂ \ (E ∪ F). ∂n Proof. This is a corollary of Theorem B.4 and of Theorem C.3. We will give both proofs; a third proof will be given in Appendix H. By conformal invariance, ∂ can be assumed to consist of analytic Jordan curves. To derive the corollary from Theorem C.3, let ε > 0 and take h ∈ C 1 (R) so that h(x) = x on [0, 1], and so that 0 ≤ h ≤ 1 and −ε < h < 1 + ε on R. When v is in the class (C.12), so also is h ◦ v and D(h ◦ v) ≤ D(v). Therefore the extremum (C.12) is the same for the subclass in which lim z→ζ ∈E v(z) = 0 and lim z→ζ ∈F v(z) = 1, and by Theorem C.3 the unique extremal for the subclass is the function u satisfying (C.13) and (C.14). To obtain the corollary from Theorem B.4 we follow Courant’s book [1950]. Let u(z) be the solution given by Theorem B.4 of the mixed boundary value problem (C.13), (C.14), and (C.15). Then u(z) is a harmonic measure on a Riemann surface double of , so that ∂u/∂n is real analytic on the open set E ∪ F ⊂ ∂. At an endpoint ζ of E ∪ F, |∇u| = O(|z − ζ |−1/2 ). Indeed, assume that ζ ∈ ∂ E and that there is a neighborhood W of ζ and a conformal map ψ of W onto D with ψ(ζ ) = 0, ψ(E ∩ W ) = [0, 1), and ψ(W ∩ ) = D+ or D− . The function U (w) = u ◦ ψ −1 (w 2 ) can be reﬂected from a quarter disc ﬁrst across the imaginary axis and then the real axis, to yield a function harmonic in D \ {0}. Because u is bounded, the singularity at w = 0 is removable, and U (w) = ImF(w) with F analytic on D and F(0) = 0. Because u > 0 on , F has only a simple zero and F(w) = aw + O(|w|2 ), with a > 0. Thus % % u(z) = ImF( ψ(z)) = a(Im ψ(z)) + O(|z − ζ |),

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and |∇u(z)| = O(|ψ(z)|−1/2 ). Hence for any v in the class (C.12) and any sequence n ↑ of domains with smooth boundaries such that n ⊂ , we have ∂u D (v − u, u) = lim Dn (v − u, u) = lim (v − u) ds n→∞ n→∞ ∂ ∂n n ∂u = (v − u) ds. ∂n E∪F Because u is a harmonic measure we have 0 ≤ u(z) ≤ 1 by Exercise H.5(a), and hence ∂u/∂n ≤ 0 on E and ∂u/∂n ≥ 0 on F. Therefore D (v − u, u) ≥ 0, and D (v) = D (u) + D (v − u) + 2D (v − u, u) ≥ D (u), with equality holding if and only if v = u.

D. Hausdorff Measure The two-dimensional Hausdorff measure of a set E ⊂ C is deﬁned to be 2 (E) = π1 Area(E), where 2 Area(E) = inf πr j : E ⊂ B(z j , r j ) (D.1) is the area of E. To deﬁne other Hausdorff measures we replace πr 2 in (D.1) by other functions of r . A measure function is a continuous and increasing function h(r ) on [0, ∞). With respect to the measure function h, the Hausdorff content of a set E is Mh (E) = inf h(r j ) : E ⊂ B(z j , r j ) . (D.2) The Hausdorff content Mh assigns ﬁnite mass to every bounded set, but the content Mh is usually not ﬁnitely additive. For example if h(r ) = 2r, then Mh ([0, 1]) = Mh ([0, 1] + iε) = 1, while

% Mh [0, 1] ∪ ([0, 1] + iε) = 1 + ε2 . To obtain a measure, we change the deﬁnition (M.2) by requiring that the cover of E be comprised of small balls only. Deﬁne h(r j ) : E ⊂ B(z j , r j ) and r j ≤ δ . δh (E) = inf If δ2 < δ1 , then δh2 ≥ δh1 because δh2 entails fewer covers. Consequently the

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limit h (E) = lim δh (E) δ→0

exists in [0, ∞], and h (E) is the deﬁnition of the h-Hausdorff measure of E. When h(r ) = r α , we call h (E) the α-dimensional Hausdorff measure of E and write h (E) = α (E). For example, 2 (E) = π1 Area(E) as was noted above, 0 (E) is the number of points in E, and if E lies on a rectiﬁable curve, then 1 (E) is half the arc length of E. For any measure function h, h is a metric outer measure in the sense of Carathéodory: h (A ∪ B) = h (A) + h (B), if dist(A, B) > 0, and therefore h is a countably additive measure for which all Borel sets are measurable. See Wheeden and Zygmund [1977] or Folland [1984]. Note that if α < β, then δ β−α δα (E) ≥ δβ (E), so that α (E) < ∞ ⇒ β (E) = 0.

(D.3)

The Hausdorff dimension of E is dim(E) = inf{α : α (E) = 0}. By (D.3) we have

α (E) =

0, if α > dim(E), ∞, if α < dim(E).

Exercise D.1 shows that for E ⊂ C, dim(E) can be any number in [0, 2]. Hausdorff content can be used to ﬁnd dim(E). Clearly Mh (E) ≤ h (E).On the other hand, if Mh (E) = 0, then for any ε > 0 there is a cover B(z j , r j ) of E such that h(r j ) < ε, and r j ≤ inf{t : h(t) ≥ ε} → 0 as ε → 0 because h is increasing. It follows that h (E) = 0. Thus for any measure function h, Mh (E) = 0 if and only if h (E) = 0,

(D.4)

dim(E) = inf{α : Mr α (E) = 0}.

(D.5)

and

There is a dyadic version of Mh that is easier to work with. For integers j, k,

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and n, form the dyadic square j j +1 k k+1 Q = n, n × n, n . 2 2 2 2 It has side length (Q) = 2−n and lower left corner ( j2−n , k2−n ). For each n, the dyadic squares Q with (Q) = 2−n are a disjoint paving of the plane. Every with ( Q) = 2(Q). dyadic square Q is contained in a unique dyadic square Q Therefore the dyadic squares have this important property: If Q 1 and Q 2 are two dyadic squares, then ⎧ ⎨ Q1 ⊂ Q2, Q 2 ⊂ Q 1 , or ⎩ Q 1 ∩ Q 2 = ∅. Now deﬁne m h (E) = inf

h((Q p )) : E ⊂

Q p , and each Q p is a dyadic square .

Every dyadic square of side r is contained in a ball of radius r while each ball of radius r is contained in 25 dyadic squares Q with r/2 < (Q) ≤ r . Therefore Mh (E) ≤ m h (E) ≤ 25Mh (E),

(D.6)

and (D.4) and (D.5) also hold with m h in place of Mh . Using m h yields an easy proof that every E ⊂ C has Hausdorff dimension at most two. Partition the unit square Q into 4n smaller squares Q j such that (Q j ) = 2−n . Then if α > 2, lim (2−n )α = lim 2n(2−α) = 0, n

j

n

and m r α (Q) = 0. Thus α (Q) = 0 and since α is countably additive, α (C) = 0. This shows that dim(E) ≤ 2 for all E ⊂ C. The following theorem is useful for computing dimensions and estimating capacities. Theorem D.1 (Frostman). Let h be a measure function. If μ is a positive measure such that μ(B(z, r )) ≤ h(r ) for all z and r , then μ(E) ≤ h (E)

(D.7)

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for every Borel set E. Conversely, if E is a compact set, and if Mh (E) > 0, then there exists a positive measure μ, supported on E, such that (D.7) holds for all z and r and such that Mh (E) . (D.8) 225 Proof. Suppose μ(B(z, r )) ≤ h(r ) for all z and r . If E ⊂ B(z j , r j ) = B j then & μ(B j ) ≤ h(r j ), μ(E) ≤ μ( B j ) ≤ μ(E) ≥

and that proves the ﬁrst assertion of the theorem. Now suppose E is compact, and without loss of generality suppose E is contained in the unit square. For each n ≥ 0, deﬁne the measure μn so that for every dyadic Q n with (Q n ) = 2−n ,

h(2−n ), if Q n ∩ E = ∅, μn (Q n ) = 0, if Q n ∩ E = ∅, and such that on Q n the measure μn is absolutely continuous to area measure and has constant density. Each Q n−1 with (Q n−1 ) = 2−n+1 is the union of 4 squares Q n, j with (Q n, j ) = 2−n . If μn (Q n−1 ) = μn (Q n, j ) > h(2−n+1 ), j

then reduce the density on each Q n, j so that μn (Q n−1 ) = μn (Q n, j ) = h(2−n+1 ), (n−1)

. Now repeat this process for squares Q k with and call this new measure μn (k) (Q k ) = 2−k and obtain measures {μn }, for k = n − 2, . . . , 0. We arrive at (0) (0) a measure μn satisfying μn (Q) ≤ h((Q)) for all dyadic squares Q with (Q) ≥ 2−n . (0) The sequence {μ(0) n } has a subsequence {μn j } which converges weak-star

−n to a measure μ. Then because μ(0) n has support in {z : dist(z, E) ≤ 2 · 2 }, supp μ ⊂ E. Because weak-star convergence implies μ(U ) ≤ lim inf μn (U ) for all open U, we have μ(Q) ≤ 9h((Q)) for every dyadic square Q. Therefore since each B(z, r ) is contained in 25 dyadic squares Q with r/2 < (Q) ≤ r , we obtain (D.6) for the new measure μ/225. (0) By construction, the measure μn is supported on a union of dyadic squares Q with (Q) = 2−n such that either

μ(0) n (Q) = h((Q)),

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with or Q is contained in a larger dyadic square Q μ(0) n ( Q) = h(( Q)). Thus μ(0) n (E) ≥ inf

h(Q j ),

(D.9)

where the inﬁmum is taken over all disjoint collections of dyadic squares {Q j } satisfying (Q j ) ≥ 2−n and Q j ⊃ E. Then by (D.9) and (D.6) μ(E) = lim μ(0) n (E) ≥ m h (E) ≥ Mh (E), and dividing μ by 225 yields (D.8). Corollary D.2. Let h be a measure function such that 1 h(r ) dr < ∞. r 0 If E is a compact set such that if h (E) > 0, then , * diam(E) h(r ) 225 dr . Cap(E) ≥ diam(E) exp − Mh (E) 0 r

(D.10)

(D.11)

Proof. If (D.10) holds and if h (E) > 0 then by (D.4), Mh (E) > 0 and by Theorem D.1, there is a measure μ, supported on E, for which (D.7) and (D.8) hold. Fix z ∈ E and write m(r ) = μ(B(z, r )). Then 1 log dμ(ζ ) Uμ (z) = |ζ − z| E diam(E) 1 log dm(r ), = lim ε→0 ε r and by a partial integration,

diam(E) m(r ) 1 + lim dr Uμ (z) ≤ μ(E) log diam(E) ε→0 ε r diam(E) h(r ) 1 + dr. ≤ μ(E) log diam(E) r 0

Hence by Theorem III.4.1, diam(E) μ h(r ) 1 1 ≤ dr + log , γ (E) ≤ I μ(E) μ(E) 0 r diam(E) which yields (D.11).

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Corollary D.2 shows that Cap(E) is positive if dim(E) > 0. The next corollary is a quantitative version of that result. Corollary D.3. If 0 < α ≤ 2, there is a constant C = Cα > 0 such that if E is compact then 1

(D.12) Cap(E) ≥ C Mr α (E) α . Proof. By a change of scale z → λz, λ > 0, we may assume Mr α (E) = 1. Let μ be a measure on E satisfying μ(B(z, r )) ≤ r α for all balls B(z, r ) and 1 . Then by a partial integration μ(E) ≥ 225 1 Uμ (z) ≤ log dμ(ζ ) |ζ − z| {|ζ −z|≤1} 1 μ((B(z, r )) dr = r 0 1 1 r α−1 dr = , ≤ α 0 and γ (E) ≤ 225α by Theorem III.4.1. That gives (D.12) with Cα ≥ e−225α . Theorem D.4. Let

1 −1 h(r ) = log . r If Cap(E) > 0, then h (E) = ∞.

(D.13)

Theorem D.4 is due to Erdös and Gillis [1937]. It is almost a converse to Corollary D.2. For example, when p > 1 and 0 < r < 1/3, the measure function h(r ) = (log 1/r )−1 (log log 1/r )− p , satisﬁes the hypothesis (D.10) of Corollary D.2. Proof. If Cap(E) > 0, then lim

r →0 E∩B(z,r )

log

1 dμ E (ζ ) = 0 |ζ − z|

almost everywhere μ E . Then by Egoroff’s theorem there exists compact J ⊂ E, such that μ E (J ) > 0 and there exists a positive decreasing function K (r ) such that log 1/r lim = 0, r →0 K (r )

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and

K (|ζ − z|)dμ E (ζ ) ≤ M ≤ ∞.

sup z∈J

E

Suppose r j ≤ δ and J⊂

N &

B(z j , r j ).

1

Then μ E (J ) ≤

μ E (B(z j , r j )) ≤ M

log 1/r 1 ≤ M sup h(r j ), K (r j ) K (r ) r ≤δ

and sending δ → 0 shows h (J ) = ∞.

Exercise D.1. (a) The Cantor set, deﬁned in Example III.4.7, has Hausdorff dimension log 2 α= . log 3 Recall that K = K n where K 0 = [0, 1] and K n is obtained from K n−1 by removing the middle one-third of each interval in K n−1 . If δ = 3−n /2, then we can cover K n by 2n balls of radius δ and δα (K ) ≤ δα (K n ) ≤ 2n

3−n α 2

= 2−α .

Letting n → ∞, we obtain α (K ) ≤ 2−α and dim(K ) ≤ α. For the opposite inequality, let μn be a constant multiple of one-dimensional Lebesgue measure restricted to K n . Choose the constant so that each of the 2n intervals I in K n has mass μn (I ) = 2−n . Let μ be a weak-star limit of μn . Then μ(K ) = lim μ(K m ) = lim lim μn (K m ) = 1. m

m

n

Let B = B(z, r ) and choose m so that 3−m ≤ 2r < 3−m+1 . The distance between two intervals in K m−1 is at least 3−m+1 , so B intersects at most one interval in K m−1 . Then for n > m, μn (B) ≤ 2−m+1 ≤ 2α+1r α .

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D. Hausdorff Measure Letting n → ∞, we obtain μ(B) ≤ 2α+1r α , and by Theorem D.1, 1 = μ(K ) ≤ α (K )2α+1 .

Thus 2−α−1 ≤ α (K ) ≤ 2−α and dim(K ) ≥ α. A more careful estimate shows that α (K ) = 2−α . An alternate proof can be found in Falconer [1985]. (b) If, when we construct K n from K n−1 , we remove the middle intervals of relative length 1 − 2δ, 0 < δ < 1/2, instead of the middle one-third, then we obtain dim(E) = log 2/ log 1/δ. (c) We describe a general construction of planar Cantor sets. Let K 0 be a square in R2 with edge length l0 = 1 and let l j+1 < l j /2, j = 0, 1, . . . . The (closed) set K j consists of 4 j squares of side length l j , each sitting in a corner of a square in K j−1 . The sets K 1 , K 2 , and K 3 are shown in Figure D.1.

K1

Let K =

K2 Figure D.1

K3

K j and let h be a measure function with h(l j ) = 4− j ,

(D.14)

1 ≤ h (K ) ≤ 1. 36

(D.15)

for j = 0, 1, . . . Then

To prove (D.15), cover K j by 4 j discs of radius l j . Then for δ = l j , δh (K ) ≤ δh (K j ) ≤ 4 j h(l j ) = 1. Letting j → ∞, we obtain h (K ) ≤ 1. To prove the lower bound on h (K ), let μ j be a constant multiple of area measure restricted to K j . Choose the constant so that each of the 4 j squares S in K j has mass μ j (S) = 4− j .

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Let μ be a weak-star limit of μ j . Then μ(K ) = lim μ(K m ) = lim lim μ j (K m ) = 1. m

m

j

Let B = B(z, r ) and choose m so that lm ≤ r < lm−1 . Then B intersects at most nine disjoint squares of side lm−1 and hence at most nine squares in K m−1 . Thus for n > m μn (B) ≤ 9 · 4−m+1 ≤ 36h(r ). Letting n → ∞, we obtain μ(B) ≤ 36h(r ) and by Theorem D.1 1 = μ(K ) ≤ 36h (K ), and (D.15) holds. (d) If l j = a j , 0 < a < 1/2, then the Hausdorff dimension of K = K a is α = dim(K a ) =

log 4 . log 1/a

This shows that the Hausdorff dimension can be any number in [0, 2]. (e) More generally, if α = sup{t : lim inf 4 j l jt = ∞} = inf{t : lim inf 4 j l jt = 0}, then dim(K ) = α. (f) Let = C∗ \ K , with K as in part (d). Prove g (z, ∞) is Hölder continuous, and ﬁnd its best Hölder exponent. Exercise D.2. (a) Assume the compact set E can be covered by A(r ) balls of radius at most r and 1 1 dr = ∞. (D.16) 0 r A(r ) Then Cap(E) = 0. Hint: Without loss of generality, we may assume that A(r ) is a decreasing function of r , and that E ⊂ B(0, 1/2). Then log 1/|z − ζ | ≥ 0 for all z, ζ ∈ E. Assume Cap(E) > 0, and let μ be the equilibrium distribution for E. Then 1 log I (μ) = dμ(ζ )dμ(z) |z − ζ| E E 1 1 log dm(z, r )dμ(z), = r E 0

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D. Hausdorff Measure where m(z, r ) = μ(B(z, r )). Show ε 1 1 log m(z, ε)dμ(z) ≤ log dm(z, r )dμ(z) → 0, ε r E E 0 as ε → 0, and

m(z, r ) dr dμ(z) r E 0 ∞ 2−n dr ≥ m(z, 2−n−1 )dμ(z) . −n−1 r 2 E

I (μ) =

1

(D.17)

n=0

Now cover E by An+2 = A(2−n−2 ) balls B(z k , 2−n−2 ). We may suppose that no point is in more than 16 balls, by modifying the collection of balls. Then by (D.17) An+2 ∞ 2−n 1 dr m(z, 2−n−1 )dμ(z) I (μ) ≥ r 2−n−1 16 B(z k ,2−n−2 ) n=0

≥

k=1

∞

2−n

−n−1 n=0 2

An+2 1 dr m(z k , 2−n−2 )2 16 r k=1

∞ An+2 log 2 = m(z k , 2−n−2 )2 . 16 n=0 k=1

But n+2

A

1 = μ(E)2 ≤

2

μ(B(z k , 2−n−2 )

k=1

Hence ∞ > I (μ) ≥

An+2

≤ An+2

m(z k , 2−n−2 )2 .

k=1

1 ∞ 1 log 2 1 ≥C dr, 16 An+2 r A(r ) 0 n=0

which contradicts (D.16). (b) Let K be the set constructed in Exercise D.1(c). Then Cap(K ) > 0 if and only if ∞ m=1

4−m log

1 < ∞. lm

(D.18)

Hint: Suppose (D.17) holds. As in (D.14) let h be a measure function with

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h(lm ) = 4−m . Then 1 ∞ h(r ) lm−1 1 h(lm−1 ) log =3 4−m log < ∞. dr ≤ r lm lm 0 m m=1

Since h (K ) > 0, Corollary D.2 implies Cap(K ) > 0. But if (D.9) fails we can cover K by 4m balls of radius lm , so let A be decreasing with A(lm ) = 4m . Then 1 ∞ 1 lm−1 4−m log = ∞, dr ≥ r A(r ) lm 0 m=1

and by part (a), Cap(K ) = 0. (c) Cantor sets can also be used to show that (D.10) is necessary in Corollary D.2. Suppose h is a measure function such that h −1 (t/4) < h −1 (t)/2 and 1 h(r ) dr = ∞, r 0 then there exists a compact set K with h (K ) > 0 and Cap(K ) = 0. Hint: Take lm = h −1 (4−m ), construct the Cantor set as in D.1(c) and apply part (b). Further examples show that no complete description of capacity can be given in terms of Hausdorff measures. See Carleson [1967a] for (a), (b), and (c).

E. Transﬁnite Diameter and Evans Functions Transﬁnite diameter is a generalization of diameter that gives a geometric interpretation of capacity. Let E be a compact set. For z 1 , . . . , z n ∈ E set * , 2 n(n−1) |z i − z j | . (E.1) Dn (z 1 , . . . , z n ) = 1≤i< j≤n

The exponent pn = n(n − 1)/2 is chosen because the product Dn has pn factors, so that Dn is homogeneous of degree 1: Dn (λz 1 , . . . , λz n ) = λDn (z 1 , . . . , z n ),

λ > 0.

The diameter of order n is deﬁned by dn = dn (E) =

max

z 1 ,...,z n ∈E

Dn (z 1 , . . . , z n ).

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E. Transﬁnite Diameter and Evans Functions The usual diameter of E is d2 (E). A second quantity related to dn is 1 1 min log . m n (E) = max z 1 ,...,z n ∈E z∈E n |z − z j |

(E.2)

j

Theorem E.1. If E is compact then Cap(E) = lim dn (E) = lim e−m n (E) . n

n

The limit d∞ (E) = limn dn (E) is called the transﬁnite diameter of E. Proof. We ﬁrst compare (E.1) and (E.2). Choose z 1 , . . . , z n ∈ E so that Dn (z 1 , . . . , z n ) = dn (E) and let 1 1 Tj (z) = log . n−1 |z − z i | i;i = j

Write log

1 2 1 log = . dn (E) n(n − 1) |z j − z i |

(E.3)

i, j:i< j

If we view {z i : i = j} as ﬁxed, then the terms in (E.3) which entail z j are 1 2 log . n(n − 1) |z j − z i | i:i = j

Thus z j is chosen so as to minimize Tj on E, and hence Tj (z j ) ≤ m n−1 (E). This implies n 1 1 1 1 log = log dn (E) n n−1 |z j − z i | j=1

i:i = j

n 1 Tj (z j ) ≤ m n−1 (E). = n j=1

Now let μn =

δz n . Then min log I N (μn ) ≡ 1 n

≤

1 n2

≤ log

1 , N dμn (z)dμn (ζ ) |z − ζ | 1 N log + |z j − z i | n

i, j:i = j

N 1 + . dn (E) n

(E.4)

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Take a subsequence {μn j } converging weak-star to a measure ν ∈ P(E). Then 1 1 min log , N dν(z)dν(ζ ) ≤ lim inf log . I N (ν) ≡ n→∞ |z − ζ | dn (E) Letting N → ∞ we obtain by Theorem III.4.1 γ (E) ≤ I (ν) ≤ lim inf log n→∞

1 . dn (E)

(E.5)

Suppose now that Cap(E) > 0, and let μ be the equilibrium distribution for E. Then for any z 1 , . . . , z n ∈ E n n 1 1 1 1 log log ≤ dμ(z) inf z∈E n |z − z j | n |z − z j | E j=1

=

1 n

j=1 n

Uμ (z j ) ≤ γ (E),

j=1

and hence m n (E) ≤ γ (E).

(E.6)

By (E.4), (E.5) and (E.6), γ (E) = lim log n

1 = lim m n (E), n dn (E)

which proves the theorem when Cap(E) > 0. If Cap(E) = 0, take E j decreasing to E such that Cap(E j ) > 0. Then by (E.4) e−m n−1 (E) ≤ dn (E), and by (E.5) lim sup dn (E) ≤ lim sup dn (E j ) ≤ e−γ (E j ) , n→∞

n→∞

and the Theorem follows.

The extremal (E.2) can be used to construct, for any compact set E of zero capacity, a potential that is inﬁnite exactly on E. Theorem E.2 (Evans). Suppose E is compact. If Cap(E) = 0 there exists a probability measure μ supported on E such that the potential U = Uμ of μ satisﬁes lim U (z) = +∞,

z→ζ0

(E.7)

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for all ζ0 ∈ E. Conversely, if there exist a function U harmonic on C \ E and a constant K > 0 such that (E.7) holds and U (z) ≥ K log

1 |z|

(E.8)

for |z| sufﬁciently large, then Cap(E) = 0. See Evans [1933]. A potential Uμ satisfying (E.7) is called an Evans function for E. Proof. Suppose Cap(E) = 0. Choose z 1 , . . . , z n ∈ E so that n 1 1 min log = m n (E), z∈E n |z − z j | j=1

and let μn =

1 n

n

j=1 δz j .

Then Uμn (z) = log E

1 dμ(ζ ) ≥ m n (E) → ∞. |z − ζ |

Choose a subsequence {Un j } of {Un } so that Un j (z) ≥ 2 j for all z ∈ E, and let μ=

∞

2− j μn j .

j=1

Then Uμ (z) =

∞

2− j Un j (z)

j=1

satisﬁes (E.7). Conversely, suppose there is U harmonic in C \ E satisfying (E.7) and (E.8) and suppose that Cap(E) > 0. Let μ be the equilibrium distribution for E and let Uμ be its potential. Then by the maximum principle U (z) − 2K Uμ (z) ≡ +∞, for z ∈ C \ E, which is impossible. Thus Cap(E) = 0.

Exercise E.1. If f is analytic in a region , then the cluster set of f at ζ ∈ ∂ is Cl( f, ζ ) = w : there exists {z n } ⊂ , with lim z n = ζ and lim f (z n ) = w . If K ⊂ ∂ then the cluster set of f at K is & Cl( f, ζ ). Cl( f, K ) = ζ ∈K

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Suppose f is bounded and analytic on D \ K where K is compact and Cap(Cl( f, K )) = 0. Prove f extends to be analytic and bounded on D. Radó [1924] proved this result in the case Cl( f, K ) = {0}. Hint: First show we may suppose f is bounded and analytic on C \ K , by considering f (ζ ) dζ f 1 (z) = . ∂ B(0,1−2ε) ζ − z 2πi By Cauchy’s integral formula f − f 1 extends to be bounded and analytic on C \ K . When f is bounded and analytic in a neighborhood of ∞, f extends to be analytic at ∞, and we may suppose that f (∞) ∈ Cl( f, K ) by replacing f with f ◦ ϕ for some linear fractional transformation ϕ if needed. Now let U be an Evans function for E = Cl( f, K ) and consider V = U ◦ f . Then V is constant by Theorem E.2 and Corollary III.8.4, and hence f is constant.

F. Martingales, Brownian Motion, and Kakutani’s Theorem Let (X, M, P) be a probability space, that is, a measure space with P(X ) = 1, and let N ⊂ M be a sub σ -algebra. For f ∈ L 1 (X, M, P) there is a unique N -measurable function F such that χ S f d P = χ S Fd P (F.1) for all S ∈ N . The function F is called the conditional expectation of f given N , and it is written as F = E( f |N ). To prove F exists (and is unique almost everywhere) we use the Radon– Nikodym theorem: The signed measure N S → ν(S) = χ S f d P is absolutely continuous to P on the σ -algebra N . Hence there is an N -measurable F so that ν(S) = χ S Fd P, and F is unique almost everywhere. We will have F = f , unless f happens to be N -measurable. If f ∈ L 2 (X, M, P), then F is the orthogonal projection of f onto the closed subspace L 2 (X, N , P) of N -measurable functions. Example F.1. Let (X, M, P) = ([0, 2π ), B, dθ/2π ), where B is the algebra

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of Borel sets. Let 0 = θ0 < θ1 < . . . < θn = 2π , set I j = [θ j−1 , θ j ), and let N be the σ -algebra generated by {I1 , I2 , . . . , I N }. The N -measurable functions are the step functions constant on the I j , and E( f |N ) =

N

αj χ Ij ,

j=1

where αj =

1 θ j − θ j−1

f dθ Ij

is the average of f over I j . Now let M1 ⊂ M2 ⊂ . . . ⊂ M be an increasing sequence of σ -algebras such that ∞ &

Mn

generates M.

(F.2)

n=1

Suppose f n is Mn -measurable, f n ∈ L 1 (X, Mn , P). We say the sequence { f n } is a martingale (relative to {Mn }) if for all n f n = E( f n+1 |Mn ).

(F.3)

From the uniqueness of conditional expectations it follows that f n = E( f n+k |Mn ),

k > 0.

(F.4)

Martingales are ubiquitous in probability theory, see Chung [1974] or Doob [1974]. Given the sequence Mn , we can always make a martingale by starting with f ∈ L 1 (X, M, P) and setting f n = E( f |Mn ).

(F.5)

It is a theorem that a martingale { f n } has the form (F.5) if and only if the sequence { f n } is uniformly integrable. We need only a weak form of that theorem. Lemma F.2. Let 1 < p < ∞ and let { f n } be a martingale relative to {Mn } such that sup f n p < ∞. n

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Then there is f ∈ L p (X, M, P) such that f n = E( f |Mn ) and f n converges to f weakly in L p . Proof. From { f n } take a subsequence converging weakly in L p to some limit f . For S ∈ Mn , χ S ∈ L q , q = p/( p − 1) and hence (F.4) yields χ S f n d P = lim χ S f n j d P = χ S f d P. j

Hence f n = E( f |Mn ). By (F.2) the limit point f does not depend on the subsequence { f n } and that means { f n } converges weakly to f . Doob’s martingale theorem says that an L 1 bounded martingale converges almost everywhere. We will prove the theorem in the special case of L p bounded martingales, and for L 1 bounded martingales we only prove lim sup | f n | < ∞ a.e. Both results hinge on a crucial inequality, called the weak-L1 estimate for the maximal function, which is the martingale variant of the estimate for the Hardy–Littlewood maximal function, Lemma I.2.4. Lemma F.3. Let { f n } be a martingale such that sup f n 1 = M < ∞. n

Set f ∗ (x) = sup | f n (x)|. n

Then for all λ > 0, P( f ∗ > λ) ≤

M . λ

(F.6)

Proof. Fix λ and deﬁne N (x) = inf{n : | f n (x)| > λ}. Then Sn ≡ {x : N (x) = n} = {x : | f n (x)| > λ} ∩

n−1 + k=1

{x : | f k (x)| ≤ λ} ∈ Mn .

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Therefore ∗

λP( f > λ) = lim m

≤ lim m

≤ lim m

m

λP(Sn )

n=1 m n=1 m

χ Sn | f n |d P χ Sn | f m |d P

n=1

≤ lim f m 1 ≤ M, m

The second inequality above follows from

| f n | = |E( f m |Mn )| ≤ E(| f m | Mn ).

(F.7)

So (F.6) is proved.

Incidentally, the function N (x) is called a stopping time, which by deﬁnition means that {N (x) ≤ n} ∈ Mn . Corollary F.4. If { f n } is a martingale such that supn f n 1 < ∞, then lim sup | f n | < ∞ almost everywhere. Proof. This is trivial because P(lim sup | f n | > 1/ε) ≤ P( f ∗ > 1/ε) ≤ Mε for all ε > 0.

Theorem F.5. Let p > 1 and let { f n } be a martingale satisfying sup f n p < ∞. n

Let f be the weak limit provided by Lemma F.3. Then f n (x) → f (x) for almost all x. Proof. By Hölder’s inequality we may assume p < ∞. This case of the martingale theorem is easier because we already have f n = E( f |Mn ) for some f ∈ L p . We may suppose f is real. The argument is like the proof of Fatou’s theorem, I.2.1. By (F.6) f (x) = lim sup f n (x) − lim inf f n (x)

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is ﬁnite almost everywhere and in fact P( f (x) > ε) ≤

2 ε

| f |d P.

(F.8)

Let D = {g ∈ L p : g is Mn measurable for some n}. Then D is dense in L p by (F.2) and g = 0 when g ∈ D. Now take g ∈ D such that f − g p < ε2 . Then f = f −g and so (F.8) yields, via Hölder’s inequality, 2 2 f − g1 ≤ ε2 < 2ε. ε ε Consequently f = 0 almost everywhere and F(x) = lim f n (x) exists almost everywhere. Because { f n } is L p bounded for p > 1 it follows that f n − F p → 0, by Egoroff’s theorem and the weak convergence of f n to f . Hence F = f a.e. and the theorem is proved. P( f > ε) ≤

Now let be a bounded domain in C and for each z ∈ let (z) be the circle 1 (z) = {w : |w − z| = dist(z, ∂)}. 2 For ﬁxed z 0 ∈ we deﬁne a martingale z 0 , z 1 , . . . , inductively as follows. Take z n+1 ∈ (z n ), n ≥ 0 such that z n+1 − z n is uniformly distributed, which means that for A ⊂ (z n ), P(z n+1 ∈ A) =

length(A) , n ≥ 0. length((z n ))

One often meets a martingale before seeing its σ -algebras Mn ; to deﬁne them for {z n } let X = {(eiθ1 , eiθ2 , . . .)} be an inﬁnite product of unit circles, let M be the Borel σ -algebra on X and 8 let P be the inﬁnite product measure dθ j /2π . Set M0 = {φ, X } and let Mn be the smallest σ -algebra for which θ1 , θ2 , . . . , θn are measurable. Then 1 dist(z n , ∂)eiθn+1 , 2 and {z n } is a martingale relative to {Mn } because 2π 1 eiθ dθ = 0. 2π 0 z n+1 = z n +

Also note that (F.9) implies z n+1 ∈ and z n+1 = z n .

(F.9)

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z0

475

z1 z2

zn

Figure F.1 Because is bounded, the martingale theorem ensures that z ∞ = lim z n n

exists almost everywhere on X . Clearly z ∞ ∈ , and because dist(z n , ∂) = 2|z n+1 − z n | → 0, we have z ∞ ∈ ∂ whenever z ∞ exists. Kakutani’s famous theorem identiﬁes the hitting distribution of z ∞ with the harmonic measure of z 0 on ∂. Theorem F.6. Let z 0 ∈ and let E ⊂ ∂ be a Borel set. Then ω(z 0 , E, ) = P(z ∞ ∈ E)

(F.10)

where z ∞ is the limit of the martingale (F.9) started at z 0 . For example, let be the (unbounded) domain {z : Rez > 0, |Imz| < 1}. Let z 0 = 0 and let E = ∂ ∩ {Rez ≥ n}. From Theorem F.6 one can see w(z 0 , E, ) ≤ C1 e−c2 n because if k ≤ Rez j ≤ k + 1, then P(Rez ∞ < k + 2) has a positive lower bound independent of k. Though Theorem F.6 has been stated only for bounded domains, we can use it on ∩ {|z| < N } and let N → ∞. The precise estimate πn 4 π sinh( )e− 2 + O(e−πn ) π 2 follows from conformally mapping to the upper half-plane.

w(z 0 , E, ) =

Proof of Theorem F.6. The key observation is that if u is harmonic in , then by the mean value property u n = u(z n ) is a martingale. First we use the

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observation to show P(z ∞ is a regular point of ∂) = 1.

(F.11)

Suppose (F.11) is false. Then there is a compact set K , consisting only of irregular points, such that P(z ∞ ∈ K ) > 0. By Kellogg’s theorem Cap(K ) = 0 and by Evans’ theorem there is u(z) harmonic on C\K such that u(z) → ∞ if z → ζ ∈ K . Adding a constant, we may suppose u > 0 on . Then the positive martingale u n = u(z n ) is L 1 bounded (since |u n |d P = u n d P = u(z 0 )) but lim u n = ∞ on {z ∞ ∈ K }, contradicting Lemma F.3. Now let f ∈ C(∂) and let u be the solution of the Dirichlet problem for the boundary data f . Then u n = u(z n ) is a bounded martingale and by (F.11) u n → f (z ∞ ) almost everywhere with respect to P. Hence by Lemma F.2 and the uniqueness of martingale limits, u n = E( f (z ∞ )|Mn ) and in particular f (ζ )dw(z 0 , ζ ) = u(z 0 ) = u 0 = f (z ∞ )d P. ∂

Approximating characteristic functions by continuous functions now yields (F.10). The martingale {z n } is a discrete variant of Brownian motion. Let us give, without proof, a brief description of Brownian motion. Let W be the set of continuous paths B(t), t ≥ 0, in the plane having common initial point B(0) = z 0 . By a theorem of Wiener there is a probability measure P on W with respect to which B(t) will almost surely meet any circle r (z 0 ) = {z : |z − z 0 | = r } at some ﬁnite time and the location of the ﬁrst hitting of r (z 0 ) is uniformly distributed: length(A) P(ﬁrst hit of r (z) ∈ A) = . 2πr All of the increments B(t + s) − B(s) have the same distribution as B(t) − z 0 and if 0 ≤ t0 < t1 < · · · < tn then the increments B(t1 ) − B(t0 ), B(t2 ) − B(t1 ), . . . , B(tn ) − B(tn−1 ) are stochastically independent:

+ P(B(t j+1 ) − B(t j ) ∈ Sj ). P {B(t j+1 ) − B(t j ) ∈ Sj }) =

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With an appropriate σ -algebra M, the measure space (W, M, P) is called Brownian motion, and a random (with respect to P) function B(t) is called a Brownian path. Again let be a bounded domain and let z 0 ∈ . Then τ1 = τ1 (B) = inf{t : B(t) ∈ (z 0 )} is M-measurable and so is z 1 = B(τ1 ). Similarly, τ2 = τ2 (B) = inf{t > τ1 : B(t) ∈ (z 1 ) = (B(τ1 ))} and z 2 = B(τ2 ) are also measurable. In fact there is an inﬁnite sequence 0 < τ1 < · · · of random times such that z n = B(τn ). Thus our martingale z n coincides with B(t) observed at the discrete sequence of times τn . Now let τ∞ = inf{t : B(t) ∈ ∂}. Because almost every path B(t) is continuous, τ∞ = lim τn and B(τ∞ ) = lim z n = z ∞ . Therefore Theorem F.6 can be restated as w(z 0 , E, ) = P(B(τ∞ ) ∈ E), or in words, the harmonic measure at z 0 of the Borel set E is the probability that Brownian motion, started at z 0 , ﬁrst exits in E. We do not know how to show directly that f (z ∞ )d P u(z 0 ) = is a harmonic function of z 0 without using a prior solution to the Dirichlet problem. Thus we cannot use (F.10) as an alternate deﬁnition of the solution of the Dirichlet problem. Had we worked with Brownian motion B(t) instead of the discrete version z n , there would have been no such difﬁculty because then u(z) would have the mean value property for all discs contained in . Exercise F.1. Let p > 1 and let f n = E( f, Mn ) be an L p bounded martingale. (a) Show that limn || f n || p = || f || p . (b) Use (a) to show || f n − f || p −→ 0. (c) Prove the maximal theorem || f ∗ || p ≤ A p || f || p ,

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and use it to give another proof of (b). Exercise F.2. In the probability space dθ [0, 2π ), M, , 2π where M is the Borel sets, let Mn be the σ -algebra generated by the dyadic intervals 6 j2π ( j + 1)2π In, j = , , 0 ≤ j < 2n . 2n 2n Thus Mn consists of unions of the dyadic intervals In, j , and Mn ⊂ Mn+1 . For f ∈ L 1 ([0, 2π )), set f n = E( f Mn ) and for I = In, j , 1 f dθ, fI = |I | I so that fn =

f In, j χ In, j .

j

We say f ∈ BMO(M), for dyadic bounded mean oscillation, if sup E(| f − f n | Mn ) = || f ||BMO(M) < ∞. n

Equivalently, || f ||BMO(M)

1 = sup |I | I =In, j

| f − f I |dθ. I

(a) Prove f ∈ BMO(M) if and only 1 | f (eiθ ) − α|dθ < ∞. sup inf I =In, j α∈C |I | I (b) Assume || f ||BMO(M) = 1, and let λ > 2. Prove the John–Nirenberg theorem: There are constants C and c, independent of f , such that for all n, P({θ : | f (eiθ ) − f n (eiθ )| > λ Mn }) ≡ E(χ {θ:| f − f n |<λ} Mn ) (F.12) ≤ Ce−cλ . Equivalently, sup

I ∈Mn

|{θ ∈ I : | f (eiθ ) − f I | > λ}| ≤ Ce−cλ . |I |

Hint: Fix n, set N0 = n, and form the stopping time N1 (θ ) = inf{m > n : | f m (θ ) − f n (θ )| > 2}.

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Then by Chebychev’s inequality, 1 P(N1 < ∞ Mn ) ≡ E(χ {N1 <∞} Mn ) ≤ . 2 Now deﬁne N2 (θ ) = inf{m > N1 (θ ) : | f m (θ ) − f N1 (θ )| > 2}, and continue. The BMO assumption yields P(Nk < ∞ Mn ) ≤ 2−k , and when 2k < λ ≤ 2(k + 1) that gives (F.12). (c) Without {Mn }, we say f ∈ L 1 ((0, 2π ]) has bounded mean oscillation, f ∈ BMO, if for all intervals I ⊂ ∂D, 1 | f − f I |dθ ≤ || f ||BMO < ∞. |I | I See Exercise 21 of Chapter I. Suppose f ∈ BMO. Prove that if I ⊂ J are intervals such that |J | = 2n |I |, then | f J − f I | ≤ cn|| f ||BMO . (d) Prove that if || f ||BMO ≤ 1, then for λ > 2 and for any interval I , |{θ ∈ I : | f (eiθ ) − f I | > λ}| ≤ Ce−cλ . |I | Repeat the argument of (b), using subintervals of I having lengths 2−n |I |. (e) Let p > 1. There is C p such that if f ∈ BMO, then 1 p sup inf | f − α| p dθ ≤ C p || f ||BMO . |I | α∈ C I I This is immediate from (d). See John and Nirenberg [1961]. Exercise F.3. Let u(z) be the Poisson integral of f ∈ L 1 (∂D). (a) Prove f ∈ BMO if and only if f (eiθ ) − u(a) 2 Pa (θ )dθ < ∞. sup a∈D ∂ D

χ Hint: Compare the Poisson kernel to the box kernel |I I| as in Figure I.4 from Chapter I and use the John–Nirenberg theorem. (b) Using (a) and Green’s theorem, show f ∈ BMO if and only if 1 − az d xd y < ∞. |∇u|2 log sup z−a a∈D D

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See Exercise II.8. (c) Using (b) and the identity 2 2 z − a 2 = (1 − |a| )(1 − |z| ) , 1− 1 − az |1 − az|2

show f ∈ BMO if and only if |∇u|2 (1 − |z|2 )d xd y is a Carleson measure. It follows that f ∈ BMO if and only if its conjugate function f ∈ BMO. Fefferman and Stein [1972]. (d) It follows from (c) that BMO is conformally invariant. If f (z) is analytic on D and if f (eiθ ) ∈ BMO, then for every Möbius transformation z−a , a ∈ D, T z = 1−az || f ◦ T ||BMO ≤ C|| f ||BMO . For detailed proofs see, for example, Garnett [1981].

G. Carleman’s Method Let be a rectangle with sides parallel to the axes, let z 0 ∈ and let E be the right edge of . If ﬁnitely many horizontal slits are removed from , then the right side of inequality (IV.6.3) does not change, but the harmonic measure ω(z 0 , E) becomes smaller. This defect in Theorem IV.6.1 is circumvented in Carleman’s [1933] version of the theorem. Theorem G.1 (Carleman). Let ⊂ C be a domain, let x = ∩ {Rez = x} and let E b = ∂ ∩ {Rez ≥ b}. Suppose |x | ≤ M < ∞, and let (x) denote the length of the longest interval in x . Assume z 0 = x0 + i y0 ∈ B(z 0 , r0 ) ⊂ . Then for b > x0 ω(z 0 , E b , ) ≤

2πr0 9M 2

b

t

exp 2π x0

x0

" − 21 dx . dt (x)

(G.1)

The function (x) is measurable because it is lower semicontinuous. The main improvement of (G.1) over (IV.6.3) is the replacement of θ (x) with the smaller (x). Further comparisons will be given after the proof of Theorem G.1. The proof uses Wirtinger’s inequality: If g and g are real valued continuous functions on the interval (a, b) and if g(a) = g(b) = 0, then 2 b b π 2 (g ) d x ≥ g 2 d x. (G.2) b−a a a

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G. Carleman’s Method

Wirtinger’s inequality (G.2) is proved by examining the Fourier series of g. Lemma III.3.2 is very similar to Wirtinger’s inequality. Carleman’s idea is to ﬁnd a differential inequality for the Dirichlet integral of the harmonic measure ω(z, E b , ). To prove Theorem G.1, we may suppose that is bounded, that ∂ consists of ﬁnitely many analytic Jordan curves, and that inf{Rez : z ∈ } = 0. Write ω(z) = ω(z, E b , ) and deﬁne t A(t) = |∇ω|2 d yd x. x

0

By our smoothness assumptions on ∂, the function A(t) is continuously dif ferentiable. Write x = ix where {ix } are the connected components of x . Lemma G.2. (Carleman’s differential inequality). For x ∈ (0, b), A (x) ≥

2π A(x). (x)

Proof. By the assumptions on ∂, ω2x dy + A (x) = x

x

(G.3)

ω2y dy

∂ix ,

for x ∈ (0, b). Because ω = 0 on Wirtinger’s inequality gives π 2 2 ω y dy ≥ ω2 dy, |ix | ix ix and hence

x

ω2y dy

≥

By Green’s theorem

π (x)

2 x

ω2 dy.

A(x) =

x

ωωx dy,

and by the Cauchy–Schwarz inequality 9 2 2 ωx ≥ A (x) x

Thus A2 A ≥ + 2 x ω dy

and that proves the lemma.

π (x)

x

ω2 dy.

2 x

(G.4)

ω2 dy ≥

2π A, (x)

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The Dirichlet integral A(t) is connected to the harmonic measure ω(z 0 ) via the function ϕ(x) = ω2 dy. x

By Harnack’s inequality ω ≥ ω(z 0 )/3 on B(z 0 , r0 /2), so that 9ϕ(x0 ) . r0

ω2 (z 0 ) ≤

(G.5)

On the other hand, because ω = 0 on ∂, ϕ (x) = 2 ωωx dy. x

By (G.4),

ϕ

= 2 A and so Carleman’s differential inequality (G.3) reads ϕ (x) 2π ≥ . ϕ (x) (x)

Now set μ(x) = 2π/(x) and ψ(x) =

x

t

exp 0

dμ dt,

0

so that 2π ψ = . ψ (x) Then (G.3) can be rewritten as ϕ ψ ϕ log = − ≥ 0. ψ ϕ ψ Therefore ϕ /ψ is non-decreasing. Because ψ > 0, we obtain ϕ (x)ψ (t) ≤ ϕ (t)ψ (x) whenever 0 < x < t. Because ϕ(0) = ψ(0) = 0, integrating the above inequality from 0 to x gives ϕ(x)ψ (t) ≤ ϕ (t)ψ(x), and integrating again from x to t then gives ϕ(x)ψ(t) ≤ ϕ(t)ψ(x),

(G.6)

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whenever 0 < x < t. Increasing in {Rez < x0 } increases ω but does not change the right side of (G.1), so we may assume that μ(x) = 2π/(x) = 2π/M on x < x0 . Then M 2π x0 /M − 1) (e 2π

ψ(x0 ) = and ψ(b) = ψ(x0 ) +

b

x0

exp x0

2π ≥ ψ(x0 ) 1 + M

" μ(s)ds exp

0

b

t

exp x0

"

"

t

μ(s)ds dt

x0

(G.7)

μ(s)ds dt .

x0

Now ϕ(b) ≤ |b | ≤ M, so that by (G.5), (G.6), and (G.7) 9ϕ(z 0 ) ≤ r0 9M ≤ 1+ r0

ω(z 0 )2 ≤

9M ψ(x0 ) · r0 ψ(b)

t " −1 b 2π exp μ(s)ds dt M x0 x0

(G.8)

and taking the square root in (G.8) gives us (G.1). To compare (G.1) and (IV.6.3), note that

b

t

exp 2π x0

x0

t " " b dx dx exp 2π dt ≥ dt (x) b−ε x0 (x) "

b−ε dx , ≥ ε exp 2π (x) x0

so that ω(z 0 , E b ) ≤

9M 2 2πr0 ε

21

exp −π

b−ε x0

" dx . (x)

(G.9)

If (x) = θ (x) inequality (G.9) is weaker than (IV.6.3) because the upper limit is b − ε rather than b. If (x) ≥ 2π δ > 0 for all x > (1 − δ)b > 0, then we can make a closer comparison with (IV.6.3), for in that case

" b b dx exp −2π dt ≥ δ 1 − e−b , x0 t (x)

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and thus ω(z 0 , E b ) ≤

9M 2 2πr0 δ(1 − e−b )

21

exp −π

b x0

dx . (x)

(G.10)

Of course, either (G.9) or (G.10) is enough to give Theorem V.1.1.

Notes Tsuji [1959] proved (G.9) in polar coordinates. Our presentation follows Carleman [1933] and Haliste [1965].

H. Extremal Distance in Finitely Connected Jordan Domains Several of the results in Chapter IV for simply connected domains have reformulations valid for ﬁnitely connected Jordan domains. We begin with the description of the extremal distance d (E, F) in Theorem IV.4.1. Assume is a ﬁnitely connected Jordan domain, let E and F be ﬁnite unions of closed arcs on ∂, and assume E ∩ F = ∅. In that case we call (, E, F) a triple. The extension of Theorem IV.4.1 is easy when there is one component of ∂ such that E ∪ F ⊂ 1 , because in that case there is a comformal mapping of onto a rectangle with horizontal line segments removed so that E and F correspond to the vertical ends of the rectangle if and only if there is a subarc σ ⊂ 1 such that E ⊂ σ and F ∩ σ = ∅. For the proof we can assume is bounded by analytic curves. Take two copies of and form the Riemann surface double d by attaching the copies along ∂ \ (E ∪ F). Let ω(z) = ω(z, Fd , d ) be the harmonic measure of the double Fd of F in ω be the (locally deﬁned) harmonic conjugate of ω in . Then just d , and let as in the proof of Theorem IV.4.1, ∂ ω/∂s = 0 on each component j of ∂ for which j ∩ (E ∪ F) = ∅. Therefore ∂ ω ds = 0 ∂s j and hence ϕ = ω + i ω is single valued and analytic on . The remainder of the argument is exactly like the proof of Theorem IV.4.1. There is a similar ﬁnitely connected version

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of Theorem IV.4.2; see Exercise H.1. There is also a generalization of Theorem IV.4.1 that holds for all triples (, E, F), but now the slit rectangle must be replaced by a union of rectangles pasted together along their edges to form a Riemann surface. We formulate the result in terms of the real part of the conformal map to the Riemann surface. For motivation, let us look back at the proof of Theorem IV.4.1 when ∂ is an analytic Jordan curve. The extremal metric is ρ0 (z) = |ϕ (z)| = |∇ω(z)|, where ω = Reϕ and ϕ maps onto a rectangle. By reﬂection, the function ω is continuous on and harmonic on a neighborhood of \ P, where P is the set of endpoints of the arcs in E ∪ F. Moreover, ω solves the mixed Dirichlet–Neumann problem: ω(z) = 0,

ω(z) =

z ∈ ,

(H.1)

z ∈ E, z ∈ F,

0, 1,

(H.2)

and ∂ω(z) = 0, z ∈ ∂ \ (E ∪ F). (H.3) ∂n Theorem H.1. Suppose is a ﬁnitely connected Jordan domain and suppose E and F are disjoint closed subsets of ∂, each consisting of a ﬁnite many subarcs of ∂. Then the extremal distance between E and F is given by −1 |∇ω|2 d xd y, d (E, F) = D(ω) ≡

where ω is a solution to (H.1) and (H.2). Extremal metrics exist and are given by constant multiples of ρ = |∇ω| a.e. dxdy. If ∂ consists of C 1+ε curves then ω is the unique solution to (H.1), (H.2), and (H.3). The conjugate extremal distance satisﬁes d ∗ (E, F) = 1/d (E, F). Theorem H.1 also implies Corollary C.4 of Appendix C. Corollary H.2. Suppose is a ﬁnitely connected Jordan domain and suppose E and F are disjoint closed subsets of ∂, each consisting of a ﬁnite many subarcs of ∂. Then ω is the unique function in C 1 () satisfying # 2 |∇ω| d xd y = inf |∇v|2 d xd y : v ∈ C 1 (), $ (H.4) lim sup v(z) ≤ 0, lim inf v(z) ≥ 1 . z→ζ ∈E

z→ζ ∈F

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Indeed, for v in the right side of (H.4), |∇v| is an admissible metric for the problem d (E, F) and L(, |∇v|) ≥ 1. Since L(, |∇ω|) = 1, and since |∇ω| is the extremal metric, we must have A(, |∇v|) ≥ A(, |∇ω|), which is (H.4). Theorem IV.2.1 implies ω is the unique minimizing function for (H.4). This proof only needs v ∈ C() to be piecewise C 1 . Proof of Theorem H.1. Before starting, we return to the conformal mapping ϕ constructed in the proof of Theorem IV.4.1. As z traces ∂, each excised horizontal line segment on the boundary of ϕ() is traced twice by ϕ(z) and each vertical line segment is traced once. The points in R which are endpoints of the horizontal slits correspond to the points in ∂ where ∇ω = 0; these are ω at the critical points. the critical points of ω. Let {vi } denote the values of If we remove the curves {z : ω(z) = vi }, we partition into ﬁnitely many ω is an smaller regions j . The image of each j under the map ϕ = ω + i uncut rectangle of length 1. The shortest curves for the extremal distance on the slit rectangle are the horizontal line segments not meeting the excised slits. In other words, the shortest curves in are those level sets { ω(z) = c} whose images do not meet the excised slits. In the general case the deﬁnition of ω is the same. We can assume ∂ consists of analytic Jordan curves. By Corollary II.4.6, this assumption will not change (H.3) when ∂ consist of C 1+ε curves. If ∂ \ (E ∪ F) = ∅, take two copies of joined along ∂ \ (E ∪ F), forming the Riemann surface double d , and let Fd consist of those boundary curves of ∂d corresponding to F. If ∂ \ (E ∪ F) = ∅, we let d = and Fd = F. Let ω(z) be the harmonic measure of Fd in d . As before, we work with the restriction of ω(z) to . From the proof of Theorem IV.4.1 it is clear that the (locally deﬁned) function ϕ = ω + i ω is continuous up to ∂ and that ω = Reϕ satisﬁes (H.1), (H.2), and (H.3). In fact, by the Schwarz reﬂection principle, ω is the only solution to (H.1)–(H.3). Moreover, except on the set P = endpoints of E ∪ F , ϕ is analytic on ∂, and (IV.4.2), (IV.4.3), and (IV.4.4) hold. In our more general case, ω is not single valued. For each ζ ∈ \ P there is a conformal map f ζ deﬁned in a neighborhood Uζ of ζ with f ζ (ζ ) = 0 and a constant aζ so that ω + i ω = aζ + f ζk

(H.5)

for some k ≥ 1. If k ≥ 2, we call ζ a critical point of order k–1. Near ζ , the level set {z : ω(z) = ω(ζ )} consists of 2k analytic arcs which meet at ζ separated by equal angles of π/k, and these level lines do not depend on the choice of ω that was made. Each ζ ∈ for which |∇ω| = 0 then lies on a maximal analytic curve along which ω + i ω can be continued analytically with

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ω constant and parameterized so that ω is increasing. Call these curves level curves of ω. We claim that each such level curve will either terminate at a critical point of ω or at a point of E ∪ F. To see this, ﬁrst note that if α is a level curve then α cannot be closed since ω is continuous and increasing on α. If Uζ ∩ α has inﬁnitely many components, then we can ﬁnd an arc β in Uζ with both endpoints on α and on which ω is constant. A subarc of α together with β then forms a closed curve, along which either ω is increasing or constant. This contradicts the continuity of ω. By compactness, the level curve α must either terminate at a critical point or terminate at a point of ∂. If α terminates at a point ζ ∈ ∂ \ {E ∪ F}, then since ω is constant on ∂ ∩ Uζ , the local description (H.5) implies ζ is a critical point of ω. This proves the claim. We can also show that α cannot terminate at a point ζ ∈ P. To see this note that % (H.6) ϕ(z) = ϕ(ζ ) + a1 (z − ζ ) + . . . . Indeed, since ∂ is analytic, there is a neighborhood W of ζ and a conformal map ψ of the unit disc D onto W so that ψ(0) = ζ and ψ({z ∈ D : Imz > 0}) = W ∩ . On the quarter disc Q = {z ∈ D : Imz > 0, Rez > 0}, f (z) ≡ ϕ(ψ(z 2 )) is analytic and on each segment in ∂ Q ∩ D, either Re f or Im f is constant. Thus f extends by two reﬂections to be analytic in D with Re f ≡ 0 or Re f ≡ 1 on a diameter. By the argument used to establish (IV.4.3), f (0) = 0. Thus ϕ has an analytic extension to W \ (E ∪ F), on which (H.6) holds. Moreover, near ζ the level set {z : ω(z) = ω(ζ )} consists of one analytic arc terminating at ζ , namely the subarc of ∂ \ (E ∪ F) contained in W . In particular, a level curve of ω in cannot terminate at ζ . We will call the level lines { ω = vi } passing through the critical points of ω the critical level lines. The critical level lines partition into smaller regions j . See Figure H.1. The boundary of j consists of ﬁnitely many analytic arcs γ j,k , on the relative interior of which either ∂ω ∂ω = 0 and =0 ∂s ∂n

ω = vi } if γ j,k ⊂ {

(H.7)

or ∂ω ∂ω (H.8) = 0, and = 0, if γ j,k ⊂ E ∪ F. ∂s ∂n Every endpoint of γ j,k is either a critical point or a point of E ∪ F. We will show that ϕ = ω + i ω is single valued on j and that ϕ( j ) is either a rectangle or

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an annulus. This will follow from the argument principle applied to the (single valued) analytic function ϕ = ωx − iω y . E

1

2

R1

3

E

F 4

ϕ

R2 ϕ(E)

5

ϕ(F)

R3 R4 R5

F 0

1

Figure H.1 ϕ( j ) = R j . We shall need an extended version of the argument principle, and we digress brieﬂy to prove it. Suppose W is a bounded, ﬁnitely connected Jordan domain and suppose each boundary curve is piecewise continuously differentiable. Orient ∂ W so that it has index 1 about each point of W . We say ∂ W has an interior angle of θ0 ∈ [0, 2π ] at ζ0 if the argument of the tangent vector, m arg dz ds , increases by π − θ0 at ζ0 . Let F = {ζk }k=1 be a ﬁnite subset of ∂ W and let Wε = W \ ∪k {z : |z − ζk | ≤ ε}. If g is continuous on ∂ W \ F then we deﬁne the principal value integral as g(z)dz = lim g(z)dz, PV ∂W

ε→0 ∂ Wε ∩∂ W

provided the limit exists. For example, the total increase in the argument of the tangent vector around ∂ W is given by d dz P V Im log dz + (π − θk ) = 2π(1 − (N − 1)), (H.9) ds ∂ W dz where N is the number of components of ∂ W and the sum on the left side is taken over the ﬁnite set of points where the interior angle θk is not equal to π . If f is meromorphic on W , then we say f has a singularity at ζ0 ∈ ∂ W of order α0 ∈ R if α0 f (z) − lim (H.10) W z→ζ0 f (z) z − ζ0 exists. Thus if f (z) ∼ (z − ζ0 )α0 then f has a singularity of order α0 at ζ0 . Lemma H.3 (extended argument principle). Suppose W is a bounded, ﬁnitely connected Jordan domain and suppose each boundary curve in ∂ is piecewise continuously differentiable. Let P = {ζk }m k=1 ⊂ ∂ W and suppose ∂ W has

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interior angle θk at ζk , k = 1, . . . , m, and interior angle π at all ζ ∈ ∂ W \ P. If f is meromorphic in a neighborhood of W \ P, having no zeros or poles on ∂ W \ P, and if f has a singularity of order αk at ζk , k = 1, . . . , m, then m f (z) dz αk θk P V Im = (Nz − N p ) + , (H.11) f (z) 2π 2π ∂W k=1

where N z and N p respectively are the number of zeros and poles of f in W , counting multiplicity. For example, if ∂ W is smooth and if f is meromorphic on W then f (z) dz P V Im = (Nz − N p ) + (Nzb − N pb )/2, ∂ W f (z) 2π where Nzb and N pb are the zeros and poles of f on ∂ W , counting multiplicity. Proof. By the usual argument principle, if ε > 0 is small, f (z) dz = 2πi(Nz − N p ). ∂ Wε f (z)

(H.12)

Set σkε = {z : |z − ζk | = ε} ∩ W . Then by the deﬁnition of θk dz lim Im = −θk . ε→0 σkε z − ζk By (H.10)

lim Im

ε→0

σkε

f (z) dz αk θk =− . f (z) 2π 2π

(H.13)

Summing (H.13) over k and subtracting the result from (H.12) proves the lemma. To apply this lemma on the region W = j , recall that ϕ is analytic and non-zero on j . Moreover on the relative interior of each analytic arc γ j,k in dϕ d ∂ j , arg dϕ ds is constant by (H.7) and (H.8). Thus ds arg ds = 0 and d dϕ log ds 0 = P V Im ds ∂ j ds d

= P V Im log ϕ (z) dz (H.14) ∂ j dz d dz + P V Im log dz. ds ∂ j dz

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Let ζ1 , . . . , ζm be the endpoints of the arcs γ j,k in ∂ j , and let θi and αi be the corresponding interior angles of j and singularities of ϕ . Then by (H.6), (H.9), (H.11), and (H.14) 0=

m

αi θi + 2π(2 − N ) +

i=1

m

(θi − π ),

i=1

and hence 2π(N − 2) =

m

αi θi + θi − π.

(H.15)

i=1

There are three kinds of endpoints ζi of γ j,k : (i) If ζi is an endpoint of E ∪ F then by (H.6) ϕ has a singularity of order ω that contains ζi is an arc of αi = −1/2. By (H.6) the only level curve of ∂ \ (E ∪ F) and so ∂ j has an interior angle θi = π at ζi . In this case, then αi θi + θi − π = −π/2. (ii) If ζi ∈ (E ∪ F)◦ , then ϕ is analytic and non-zero in a neighborhood of ζi . ω = c2 } to the other and Thus ∂ j turns from one of the sets {w = c1 } or { ∂ j has interior angle θi = π/2 at ζi . In this case αi θi + θi − π = −π/2 again. (iii) Finally if ζi is a critical point of ω, then by (H.5) ∂ j has an interior angle θi = π/k, k ≥ 2, and ϕ has a singularity αi = k − 1. In this case αi θi + θi − π = 0. If K j denotes the number of endpoints of γ j,k in E ∪ F ∩ ∂ j , and if N j denotes the number of boundary components of j , then (H.15) yields π 2π(N j − 2) + K j = 0, 2 with K j ≥ 0 and N j ≥ 1 integers. Hence either N j = 2 and K j = 0 or N j = 1 and K j = 4. The ﬁrst case occurs only when is conformally equivalent to an annulus with boundary curves E and F, as in Theorem IV.4.2. Thus we may suppose the second case holds, so that j is simply connected and E ∪ F ∩ ∂ j consists of two intervals. Then ω must be the restriction to j of the harmonic measure of F ∩ ∂ j in the Riemann surface double of j . The proof of Theorem IV.4.1 shows that ϕ = ω + i ω is a conformal map of j onto a rectangle. As in Figure H.1, we can view ϕ as a conformal map of j to a stack of rectangles of length 1. Now apply the length–area proof of Example IV.1.1 with = 1 and h equal to the sum of the heights of the rectangles. Equality holds when the metric ρ is constant on the union of rectangles because ω is continuous on and changes by 1 along any curve connecting E to F. The

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extremal distance between E and F is then the reciprocal of the sum of the heights of the rectangles. By conformal invariance, the extremal metric on is |∇ω| and the level lines of ω which do not meet the critical points are the shortest curves for the extremal distance from E to F. The Dirichlet integral D(ω) is the sum of the areas of the rectangles, all of which have length 1, and thus d (E, F) = D(ω)−1 . The conjugate extremal distance is also the sum of the heights of the rectangles, since any curve separating E from F must separate the ends of each of the rectangles. Thus ∗ d (E, F) = 1/d (E, F).

We remark that the function ϕ = ω + i ω in the preceding proof can be regarded as a conformal map of onto a Riemann surface formed by identifying the appropriate edges of the rectangles obtained. As an aside, we note that the preceding proof divided the region into subregions so that equality holds in the parallel rule. The reasoning with the extended argument principle in the proof of Theorem H.1 also gives a count of the critical points of ω in , which will be useful in the examples considered below. Suppose ∂ consists of disjoint analytic Jordan curves and that E and F are ﬁnite unions of closed arcs on ∂. Let Nc = the number of components of ∂, Ne = the number of endpoints of arcs in E ∪ F N z = the number of critical points of ω in , and Nzb = the number of critical points of ω on ∂, where a critical point of order k is counted k times. Then we have the following proposition. Proposition H.4. Nzb Ne = + (Nc − 2). 2 4 Proof. The argument used to establish (H.14) on j also applies to . Because ∂ is analytic, all interior angles are equal to π . At a critical point of order k on ∂, ϕ has a singularity αi = k and at an endpoint of E ∪ F, ϕ has a singularity αi = −1/2 by (i) above. Proposition H.4 now follows from (H.9), (H.14), and Lemma H.3. Nz +

Example H.5. Let be the unit disc D and let E and F consist of two arcs each, interleaved on ∂D. In this case ω(z) = ω(z, F, C∗ \ (E ∪ F)). By Proposition H.4, there are three possible cases: Either ω has one critical point in D of multiplicity one, or ω has two critical points on ∂D, each with multiplicity

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one, or ω has a single critical point of order two on ∂D. In the ﬁrst case the critical level lines of ω that pass through this critical point will terminate on E and F, dividing D into four quadrilaterals. See Figure H.2. The function ω + i ω will be a conformal map from each quadrilateral to a rectangle of unit length. The edges of the rectangles will be identiﬁed so as to form a surface equivalent to the Riemann surface for z 2 over D. The Dirichlet integral D(ω) is the sum of the areas of the four rectangles. In other words, the extremal distance is the reciprocal of the sum of the four heights of the rectangles. The curves in that correspond to one vertical line segment in each rectangle are the extremal curves for the conjugate extremal distance. The conjugate extremal distance is the sum of the heights of the rectangles. For the second case, note that by (H.5), ω changes direction along ∂ at a critical point of order k = 1, while on each interval in ∂D \ (E ∪ F) ω varies from 0 to 1 (or from 1 to 0). Hence in the second case both critical points are in the same interval in ∂D \ (E ∪ F). From each critical point a critical level line enters D. Since these level lines must divide D into quadrilaterals, each containing a portion of E and F, one critical level line terminates in E and the other terminates in F, dividing D into three quadrilaterals. In the third case ω has a double critical point ζ ∈ ∂D \ (E ∪ F). Two critical level lines enter D from ζ ; one terminates in E, and the other in F. In this case, D is also divided into three quadrilaterals.

E

F ϕ = ω + i ω zc

ϕ(E)

ϕ(F)

E F 0

Reϕ(z c )

1

Figure H.2 Example H.6. Let be the unit disc D with two smaller discs removed. Let F = ∂D and let E = ∂ \ F. By Proposition H.4, there will be one critical point in , with two critical level lines passing through it. One critical level line of ω will connect the two inner discs. The other level line will have both endpoints on ∂D. They divide into two quadrilaterals. The function ω + i ω

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is a conformal map of each quadrilateral onto a rectangle of unit length. Again, the extremal distance is the reciprocal of the sum of the heights of the two rectangles. The conjugate extremal distance, that is, the extremal length of the curves that separate E from F, is the sum of the heights of the rectangles. See Figure H.3.

F

ϕ(z c )

F E

E

ϕ

ϕ(E)

ϕ(F)

ϕ(E)

ϕ(F)

zc ϕ(z c ) 0

1

Figure H.3 If is a ﬁnitely connected Jordan domain and if E is a ﬁnite union of arcs on one component of ∂, the upper estimate in Theorem IV.5.3 still holds. Theorem H.7. Let be a ﬁnitely connected Jordan domain, and let E be a ﬁnite union of arcs contained in one component 1 of ∂. Suppose σ is a Jordan arc in C connecting z 0 to 1 \ E and let λ ≡ d\σ (σ, E). Then 8 −π λ . e π Notice that σ ∩ need not be connected. The proof of Theorem H.7 is almost the same as the proof of Theorem IV.5.3. With no loss of generality, assume that 0 = z 0 ∈ ⊂ D and that the component of ∂ containing E is the unit circle. Let 1 = {z ∈ D : z 2 ∈ }. Now follow the proof of Theorem IV.5.3, using a conformal map of 1 onto a rectangle with horizontal line segments removed, as in the discussion in the ﬁrst paragraph of this appendix. Theorem H.7 also yields a more general version of Theorem IV.6.1. ω(z 0 , E, ) ≤

Theorem H.8. Suppose is a ﬁnitely connected Jordan domain and suppose E ⊂ {z ∈ : Rez ≥ b}. Let z 0 ∈ with x0 = Rez 0 < b and assume that for x0 < x < b, I x ⊂ {z ∈ : Rez = x} separates z 0 from E. If θ (x) ≡ (I x ) is measurable then b dx 8 ω(z 0 , E) ≤ exp −π . π x0 θ (x)

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z0

I x2

E

I x2

E

I x1 E

I x1 x0

x1

x2

b

R

Figure H.4 be the component of Proof. Let \ {z : Rez = b} = ∂ ∩ {z : Rez = b}. As in the proof of Theothat contains z 0 and let E is a ﬁnite union of arcs on ∂ . Let rem IV.6.2, we may suppose that E σ ⊂ {z : Rez = Rez 0 } be a curve (not necessarily contained in ) connecting By the maximum principle containing E. z 0 to the component of ∂ ). ω(z 0 , E, ) ≤ ω(z 0 , E, By Theorem H.7 and (IV.6.2) ω(z 0 , E) ≤

b dx 8 exp −π . π x0 θ (x)

The next result uses Theorem H.1 to compute capacity using extremal distance. Theorem H.9. Suppose is a ﬁnitely connected Jordan domain and suppose ∞ ∈ . Let C R = {z : |z| = R}. Then γ = lim 2π d (C R , ∂) − log R R→∞

where γ is Robin’s constant for ∂ deﬁned by Cap(∂) = e−γ . As we saw in Section V.3, 2πd (C R , ∂) − log R increases with R, and hence by Theorem H.9, if ∂ ⊂ {z : |z| < R}, Cap(∂) ≤ Re−2πd (C R ,∂) . Any choice of a metric ρ will give a lower bound for the extremal distance and hence an upper bound for capacity. By the symmetry rule, the Ahlfors–Beurling

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Theorem V.3.2 is a special case of Theorem H.9. Proof. Let g(z) = g(z, ∞, ) denote Green’s function for with pole at ∞, and recall that g(z) = log |z| + γ + o(1),

(H.16)

Let FR = {z ∈ : g(z) = log R + γ } and let ε > 0. Then for sufﬁciently large R, by (H.16), C R = {|z| = R} lies between the curves FR(1−ε) and FR(1+ε) and thus d (FR(1−ε) , ∂) ≤ d (C R , ∂) ≤ d (FR(1+ε) , ∂). To prove the theorem, it therefore sufﬁces to show that 2πd (FR , ∂) = log R + γ .

(H.17)

Let R = {z ∈ : g(z) < log R + γ }. Then ω(z, FR , R ) =

g(z) . log R + γ

As in the proof of Theorem H.1, remove the level curves of ω that pass through the critical points of g, and obtain conformal maps of simply connected subdomains onto a stack of rectangles. The extremal distance, d (FR , ∂), is the reciprocal of the sum of the heights of the rectangles. Then because ∂ g /∂s = −∂g/∂n < 0 on FR , the sum of the heights of the rectangles is ∂ g ds − ∂s FR . log R + γ Since g − C is Green’s function for {z : g(z) − C > 0}, by (2.12) of Chapter II applied to the constant function u ≡ 1 we have, ∂ g ∂g ds = − = −2π, FR ∂s FR ∂n Thus d (FR , ∂) = proving (H.17) and the theorem follows.

log R + γ , 2π

The reduced extremal distance on ﬁnitely connected Jordan domains is related to Green’s function on a doubled Riemann surface in the same way that extremal distance is related to harmonic measure on a doubled Riemann surface. To see the connection between Theorem V.3.2 and Theorem H.9, suppose is

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a ﬁnitely connected Jordan domain, suppose E is contained in one component of ∂, and let z 0 ∈ . Let d be the Riemann surface double of obtained by attaching two copies of along ∂ \ E. As in the proof of Theorem V.3, there is a conformal map f of onto a region of the form A = C∗ \ (D ∪

n &

L j ),

j=1

where L 1 , . . . , L n are radial line segments, f (E) = ∂D, and f (z 0 ) = ∞. The map f is given by − log | f | = gd (z, z 0 ) + gd (z, z 0 ), where z 0 and z 0 denote the two points in d corresponding to z 0 ∈ . As in the rectangle example, the radial slits do not affect dD (C R , ∂D), so by Theorem H.9 and the deﬁnition of reduced extremal distance γA (H.18) δ (z 0 , E) = δ A (∞, ∂D) = − . 2π When is a Jordan domain, we can explicitly ﬁnd the right side of (H.18) and that is how we proved Theorem V.3.2. The analogue of Theorem H.1 for reduced extremal distance is in Exercise V.2.

Notes The proof of Theorem H.1 in Ahlfors [1973] is similar to the proof given here, except in that proof is cut with level lines of ω, instead of ω, giving rectangles and annuli of various sizes. In the case of slit rectangles as in Section IV.4, this corresponds to making vertical, instead of horizontal, cuts through the ends of the slits. The proof of Theorem H.1 given here derives (H.1)–(H.3) from the existence of harmonic measure on d , and these are the only properties of harmonic measure that were used. The proof of Corollary H.2 shows that the problem (H.4) of minimizing Dirichlet norm and the more general problem of ﬁnding d (E, F) have the same solution. Exercise H.1. (a) Formulate and prove a version of Theorem IV.4.1 for ﬁnitely connected Jordan domains such that E ∪ F is contained in one component of ∂. (b) Formulate and prove the analogue of Theorem IV.4.2 for ﬁnitely connected domains such that E and F are contained in different components of ∂. Exercise H.2. In Theorem IV.6.1 and Theorem H.8, if I x contains a ﬁnite union of intervals I j of length θ j (x), each of which separates z 0 from F, then deﬁning

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ρ = 1/θ j (x) on I j improves the estimate in (IV.6.3) by replacing 1/θ(x) with 1/θ j (x). Give an example where this gives an improvement to the estimate (G.1) of Carleman. See Haliste [1965]. Exercise H.3. Let be the unit disc with two discs B(r, ε) and B(−r, ε) removed. Derive Example H.6 directly in this special case by applying the map z 2 , then using Example IV.1.2 and the symmetry rule. Exercise H.4. Show that it is possible to have two distinct critical points on ∂D in Example H.5, and show that it is possible to have one critical point of order 2 on ∂D. Exercise H.5. Extend Theorem H.9 to any domain whose boundary has positive capacity.

I. McMillan’s Twist Point Theorem Theorem I.1. Let ϕ be the conformal mapping from D onto a simply connected domain . At almost every ζ ∈ ∂D either (a) ϕ has angular derivative, or

a ﬁnite non-zero (b) arg ϕ(z) − ϕ(ζ ) is unbounded above and below on every arc σ ⊂ D having endpoint ζ.

Note that since ϕ(z) − ϕ(ζ ) = 0, continuous branches of arg ϕ(z) − ϕ(ζ ) are deﬁned in D. If ϕ(ζ1 ) = ϕ(ζ2 ) and if (a) holds at ζ1 , then ϕ(ζ1 ) is a cone point for and (b) cannot hold at ζ2 . Thus it is consistent to call ϕ(ζ ) a twist point if (b) holds at ζ. By Theorem VI.6.1, (a) holds almost everywhere on {ζ : ϕ(ζ ) is a cone point for }, while

ω 1 ω

on ϕ {ζ : (a) holds} and ω ⊥ 1 on the set of twist points of ω. However, we are not saying 1 (twist points) = 0, because that is not true. By Theorem VI.4.1, the twist point theorem only says that the twisting described in (b) occurs at almost every Plessner point for ϕ . Theorem I.1 is from McMillian [1969]. The proof of Theorem I.1 consists of two lemmas. The ﬁrst lemma is from Pommerenke [1975], page 555; see also Pommerenke [1991], page 142.

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Lemma I.2. Fix α > 2, and let E be the set of ζ ∈ ∂D such that the nontangential limit ϕ(ζ ) exists and

sup arg ϕ(z) − ϕ(ζ ) < ∞. (I.1) α (ζ )

Then almost everywhere on E

sup arg ϕ (z) < ∞.

α (ζ )

(I.2)

Neither the hypotheses depends on which con (I.1) nor the conclusion

(I.2) tinuous branches of arg ϕ(z) − ϕ(ζ ) and arg ϕ (z) that we have chosen. But throughout the proof let us ﬁx a continuous function h(z, w) on D × D so that ( arg ϕ(w)−ϕ(z) , if z = w, w−z h(z, w) = if z = w. arg(ϕ (z)), For ζ ∈ ∂D set h(z, ζ ) = arg

ϕ(ζ ) − ϕ(z) ζ −z

=

ϕ(w) − ϕ(z) arg . α (ζ )w→ζ w−z

The estimate ϕ(w) − ϕ(z) ≤ (1 − |z|2 )ϕ (z)

lim

w−z 1−zw , w−z 2 1 − 1−zw

is only the conformally invariant form of inequality (4.14) of Chapter I. It shows that π (I.3) |h(z, w) − h(z, z)| < 2 whenever |w − z| < (1 − |z|2 )/10. Proof. If the lemma is false, then there is M < ∞ and measurable E 1 ⊂ E with |E 1 | > 0 such that sup h(z, ζ ) ≤ M, α (ζ )

but sup h(z, z) = ∞

α (ζ )

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I. McMillan’s Twist Point Theorem

for all ζ ∈ E 1 . We may assume ζ = 1 is a point of density of E 1 . Then there exist α (1) z n → 1 such that lim arg ϕ (z n ) = ∞.

n→∞

Set J = ∂ D ∩ {Rez ≥ 0}, and Tn (z) =

z n z + |z n | . |z n | 1 + |z n |z

Then Jn = Tn (J ) is the arc on ∂D, with center circle through z n .

zn |z n | ,

cut off by an orthogonal

Tn

J

Figure I.1 Because 1 is a point of density, |Jn ∩ E 1 | → 1, |Jn | and since on J 1 − |z n |2 ≤ |Tn (z)| ≤ 1 − |z n |2 , 4 we have |J ∩ Tn−1 (E 1 )| → 1. |J | Set ψn (z) =

z n ϕ(Tn (z)) − ϕ(z n ) . |z n | (1 − |z n |2 )ϕ (z n )

zn

Jn

E1

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Then ψn is a normalized univalent function and by Corollary V.3.6, there exists ζn ∈ J ∩ Tn−1 (E 1 ) so that 1 |ψn (r ζn )|dr ≤ λ, 0

1 on with constant λ independent of n. By the distortion theorem, |ψn (z)| > 13 1 |z| ≥ 10 , and hence 1 ψn arg ψn (ζn ) − arg ψn ( ζn ) ≤ (I.4) (r ζn ) dr ≤ 13λ. 1 ψn 10 10 ζn ). Then by (I.4), Set wn = Tn ( 10

h(z n , Tn (ζn )) − h(z n , z n ) − h(z n , wn ) − h(z n , z n ) w −z (I.5) n n , ≤ 13λ + arg Tn (ζn ) − z n π n which is the invariant form of (I.4). Then because arg Tnw(ζnn−z )−z n ≤ 2 and

|wn − z n | ≤ (1 − |z n |2 )/10, (I.3) and (I.5) yield π h(z n , Tn (ζn )) − h(z n , z n ) ≤ 13λ + . 2 Because z n ∈ α (Tn (ζn )) and Tn (ζn ) ∈ E 1 , that means |h(z n , Tn (ζn ))| ≤ M as (I.2) asserts. Write Arg(ϕ(z) − ϕ(ζ )) for that continuous branch of arg(ϕ(z) − ϕ(ζ )) in D satisfying

0 ≤ Arg ϕ(0) − ϕ(ζ ) ≤ 2π. The second lemma is from McMillan’s paper [1969]. Lemma I.3. If ϕ(z) has nontangential limit ϕ(ζ ) at ζ and if σ ⊂ D is an arc with endpoint ζ such that (I.6) sup Arg ϕ(z) − ϕ(ζ ) ≤ M < ∞, z∈σ

then

lim sup Arg ϕ(z) − ϕ(ζ ) < ∞. Dz→ζ

(I.7)

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Proof. Let z 0 ∈ D be the initial point of σ. We can assume that ζ = 1 and that |ϕ(z 0 ) − ϕ(1)| > 1. But we do not assume that ϕ(z) → ϕ(1) when σ z → 1. Set Cn = {w : |w − ϕ(1)| = 2−n }. Each component of ϕ −1 (Cn ) is a Jordan arc in D at each end of which ϕ(z) has a limit in Cn ∩ ∂. Hence by Lindelöf’s theorem, each component of ϕ −1 (Cn ) has two distinct endpoints on ∂D \ {1}. Let Vn be that component of D \ ϕ −1 (Cn ) such that for some rn < 1, [rn , 1) ⊂ Vn ,

(I.8)

and set γn = D ∩ ∂ Vn . Now σ ∩ Vn = ∅ ϕ −1 (C

because otherwise the component of would have endpoint ζ = 1. Therefore

n)

(I.9) which separates σ from [rn , 1)

σ ∩ γn = ∅ / Vn . By (I.8), since z 0 ∈ D ∩ V n ⊂ V0 , and hence γn ⊂ V0 . Set n = V0 ∩ ϕ −1 (Cn ) ⊃ γn . Then by (I.9), σ ∩ n = ∅, and there exists an arc σn ⊂ σ ∩ V0 joining a point on γ0 to a point on n . Therefore ϕ(σn ) ⊂ Un = {2−n < |w − ϕ(1)| < 1}, and ϕ(σn ) joins wn ∈ ϕ(γ0 ) to wn ∈ Cn . On U n \ ϕ(σn ) the function An (w) = arg(w − ϕ(1)) + − has a continuous branch. Let A− n (w) and An (w) = An (w) + 2π be the two boundary values of An at w ∈ σn . Then by (I.6) and the maximum principle, An (w) − A− n (wn ) ≤ 2π + sup arg(w − ϕ(1)) − arg(wn − ϕ(1)) ϕ(σn ) (I.10) ≤ 2M.

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Connect w , w ∈ ϕ(γ0 ) by an open Jordan arc J0 ⊂ ϕ(V0 ), which is a connected open set, and let W be the bounded component of C \ (C0 ∪ J ) not containing ϕ(1). On the closed Jordan domain W , there is a continuous branch of arg(w − ϕ(1)) which matches Arg(w − ϕ(1)) on J. But on C0 ∩ ∂ W , which is an arc of the circle C0 , arg(w − ϕ(1)) cannot increase by 2π. Consequently Arg(w − ϕ(1)) − Arg(w − ϕ(1)) ≤ 2π and because σ ∩ γ0 = ∅,

sup Arg w − ϕ(1) ≤ M + 2π, γ0

(I.11)

by the hypothesis (I.6). Finally, let w ∈ ϕ(V0 ) \ ϕ(σ ), and let α ⊂ ϕ(V0 ) be a Jordan arc joining w to some w ∈ γ0 . For n large, α ∪ {w} ⊂ Un . If Wn is that component of ϕ(V0 ) ∩ Un with w ∈ Wn , then ϕ(γ0 ) ∩ ∂ Wn contains an arc of C0 . Hence Wn contains a second arc β which joins w to w ∈ ϕ(γ0 ) so that β ∩ ϕ(σ ) = ∅. w

C0

β ϕ(z 0 )

w w ϕ(σ )

J0 Cn

Figure I.2 Then by (I.10) and (I.11), Arg(w − ϕ(1)) − Arg(w − ϕ(1)) = An (w) − An (w ) ≤ 2M + 2π, and by the deﬁnition of V0 , we obtain (I.7).

Proof of Theorem I.1. We already know (a) holds whenever ϕ has a non-zero ﬁnite nontangential limit. Thus it is enough to show (b) holds almost everywhere on Plessner points of ϕ ∩ ζ : ϕ(ζ ) exists .

But for almost every such ζ , |Arg ϕ(z) − ϕ(ζ ) | is not bounded, by Lemma I.2, and thus almost every such ζ is a twist point by Lemma I.3.

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J. Bloch Martingales and the Law of the Iterated Logarithm Let g be a Bloch function and set

z

G(z) =

g(w)dw. 0

Let I = (a, b) be a subarc of ∂D. Then by Theorem VII.1.3 and a partial integration, the limit 1 −i dG(r eiθ ) iθ g I = lim g(r e )dθ = lim e−iθ dθ r →1 |I | I r →1 r |I | I dθ (J.1) −ib ib −ia ia 1 −ie G(e ) + ie G(e ) e−iθ G(eiθ ) dθ + = |I | |I | I exists. If I and J are adjacent intervals of equal length, then by (J.1) g I ∪J =

1

gI + gJ . 2

(J.2)

In the probability space

dθ , 2π where M is the Borel sets, let Mn be the σ -algebra generated by the dyadic intervals 6 j2π ( j + 1)2π In, j = , , 0 ≤ j < 2n , 2n 2n with n ﬁxed. Thus Mn is the set of unions of arcs In, j , again with n ﬁxed, and Mn ⊂ Mn+1 . Then [0, 2π ), M,

gn (ζ ) =

n −1 2

g In, j χ In, j (ζ )

0

is measurable with respect to Mn , and gn = E(gn+1 |Mn ) by (J.2). Therefore gn , Mn is a martingale. Martingales with respect to the dyadic σ -algebras Mn are called dyadic martingales.

Lemma J.1. There is a constant C such that if g ∈ B, then

g (1 − 2−n )ζ − gn (ζ ) ≤ CgB ,

(J.3)

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almost everywhere on ∂D and gn+1 − gn ≤ CgB .

(J.4)

Proof. The argument is like the proof of Theorem VII.1.3. Deﬁne # $ Tn, j = r ζ : 1 − 2−n ≤ r ≤ 1 − 2−(n+1) , ζ ∈ In, j . Then Tn, j has bounded hyperbolic diameter, so that by Theorem VII.1.2, sup g(z) − g(w) ≤ C1 gB . (J.5) z,w∈Tn, j

Tn, j

Figure J.1 On the other hand, by (VII.1.9) and the proof of (J.1),

1 g (1 − 2−n )eiθ dθ ≤ C1 gB g In, j − |In, j | In, j and together (J.5) and (J.6) imply both (J.3) and (J.4).

(J.6)

A real valued dyadic martingale { f n } is called a Bloch martingale if there exists g ∈ B such that f n (ζ ) = Regn (ζ ) for all n. Bloch martingales have a very simple characterization: Lemma J.2. Let { f n } be a real dyadic martingale. Then { f n } is a Bloch martingale if and only if { f n } has bounded jumps f n, j − f n, j+1 ≤ C, (J.7)

where f n, j = f n I . n, j

The jump condition (J.7) implies that { f n } has bounded differences | f n − f n+1 | ≤ C.

(J.8)

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However, (J.8) is weaker than (J.7) because nearly half the pairs In, j , In, j+1 do not have a common mother In−1,k . Proof. When { f n } is a Bloch martingale, (J.7) follows from (J.5). Now suppose { f n } satisﬁes (J.7), and note that (J.7) implies an inequality better than itself, namely ( f n ) I − ( f n ) I (J.9) m, j m, j+ p ≤ C p. Indeed if m ≥ n, (J.9) is the same as (J.7), while if m < n, (J.9) follows from (J.7) and the martingale relation for { f n }. Assume f 0 = 0 and set θ h n (θ ) = f n (t)dt. 0

Then

h n (θ + 2−k ) + h n (θ − 2−k ) − 2h n (θ ) = 2−k ( f n )[θ,θ+2−k ] − ( f n )[θ−2−k ,θ]

and expressing [θ, θ + 2−k ] and [θ − 2−k , θ ] as unions of dyadic intervals, we conclude from (J.9) that h n (θ + 2−k ) + h n (θ − 2−k ) − 2h n (θ ) ≤ C2−k . It then follows via the proof of Theorem II.3.3 that h n ∗ ≤ C1 , with constant C1 independent of n. By Theorem II.3.3, ∗ is an equicontinuous family, and therefore h(eiθ ) = lim h n (eiθ ) exists and h∗ ≤ C. But then u h ) satisﬁes Regn = f n , so that { f n } is a Bloch martingale. g = (u h + i By Lemma J.2 and Theorem VII.2.1, questions about the boundary behavior of univalent functions become questions about discrete dyadic martingales that satisfy the jump condition (J.7). We will follow Makarov [1990a] and derive the Bloch function law of the iterated logarithm, Theorem VIII.1.1, from the martingale law of the iterated logarithm. In the martingale law of the iterated logarithm, energy is measured by square functions. When { f n , Fn } is a martingale with E(| f n |2 ) < ∞, write n f = n = f n − f n−1 ,

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for the martingale difference and deﬁne the square function as Sn = Sn ( f ) = E

n ( j )2 F j−1 j=1

2 = Sn−1 + E ( f n2 − 2 f n f n−1 + f n−1 ) Fn−1

2 . = Sn−1 + E f n2 Fn−1 − f n−1

(J.10)

By its deﬁnition the square function Sn is predictable, i.e., measurable with respect to the recent past Fn−1 , and by (J.10) f n2 − Sn is a martingale. By the Cauchy–Schwarz inequality, Sn is a non-decreasing function of n. In fact, these three properties determine Sn uniquely. See Exercise 1. When {Fn } is the dyadic ﬁltration {Mn }, as it is with Bloch martingales, (n )2 = ( f n − f n−1 )2 is predictable and (J.10) simpliﬁes into Sn = nj=1 ( j )2 . Theorem J.3 (Lévy). Let { f n } be a real martingale. Then { f n } has a ﬁnite limit almost everywhere on {S = limn Sn < ∞}. This is the martingale analogue of Theorem X.1.3. See Lévy [1937]. Proof. Fix M > 0 and let τ = τ (x) = inf{n − 1 : Sn ≥ M}. Because Sn is predictable, τ is a stopping time, which means that {x : τ (x) ≤ n} ∈ Mn . Set

f n , if n < τ , Fn = f τ , if n ≥ τ. Then Fn is a martingale, called the stopped martingale, and Sn (F) ≤ M. The general identity,

E ( f n − f 0 )2 = E(Sn ), which holds because the differences j are orthogonal, shows that sup Fn 2 < ∞. n

Therefore Fn has a ﬁnite limit almost everywhere by the martingale theorem. But since M is arbitrary and f n = Fn on {S < M}, that means f n has a ﬁnite limit almost everywhere on {S < ∞}. Theorem J.4. Let { f n } be a real dyadic martingale. Then fn lim sup √ ≤ 1, 2Sn log log Sn n→∞

(J.11)

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almost everywhere on {S = limn Sn = ∞}. Conversely, there is an absolute constant c > 0 such that if n ( f ) is bounded and if S = ∞ almost everywhere, then fn lim sup √ ≥c (J.12) 2Sn log log Sn n→∞ almost everywhere. Actually (J.12), with c = 1, holds almost everywhere on {S = ∞}. See Stout [1974]. Before giving the proof of Theorem J.4, we use it and Theorem J.3 to give a proof of Theorem VIII.1.1. Proof of Theorem VIII.1.1. Suppose g ∈ B and gB = 1. Set f n = Regn . Then by (J.8), Sn ≤ Cn. By Theorem J.4 when S < ∞ and by Theorem J.5 when S = ∞ we have % | f n | ≤ C n log log n for all large n. If 2−n+1 ≤ 1 − r < 2−n , then (J.3) gives |Reg(r eiθ ) − f n (θ )| ≤ C1 , and therefore

(J.13)

|Reg(r eiθ )| ≤ C log

1 1 log log log (1 − r ) (1 − r )

almost everywhere.

Proof of Theorem VIII.1.2. Suppose ||g||B ≤ 1 and g satisﬁes (VIII.1.4). Write gr (z) = g(r z). Then by Exercise II.8 and (VIII.1.4), ||Re(gr − g(0))||22 =

1 ||gr − g(0)||22 2

≥2

|gr (z)|2 (1 − |z|2 )d xd y

≥ c(β, M) log

1 , 1−r

and for any given A > 0 there exists R < 1 so that sup |Re(g(z) − g(0))| ≥ A.

|z|≤R

But since (VIII.1.4) is conformally invariant that means for every z ∈ D there is w ∈ D such that ρ(z, w) ≤ ρ(0, R) and |Re(g(w) − g(z))| ≥ A. Now take A ≥ 5C1 where C1 is the constant in (J.13) and let N ≈ ρ(0, R). Then by

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(J.13) we see that for every n and a.a. θ there are n ≤ n 1 ≤ n 2 ≤ n + 2N so that | f n 1 (θ ) − f n 2 (θ )| ≥ C1 . Consequently sup

{k:|n−k|≤2N }

|k f | ≥ C =

C1 . 2N

Therefore Sn ≥ Cn for some constant C, and (VIII.1.5) follows from (J.3) and (J.12). The proof of Theorem J.4 will use exponential transforms. The exponential transform of a real dyadic martingale { f n } is exp f n . j=1 cosh j ( f )

Z n = 8n

We f0 = 0 and take Z 0 = 1. When { f n } is a dyadic martingale,

assume 8 E en Mn−1 = cosh(n ) and the denominator nj=1 cosh j is measurable with respect to Mn−1 . Therefore

E(en |Mn−1 ) E Z n |Mn−1 = Z n−1 = Z n−1 , cosh(n ) and Z n is a martingale. In particular, E(Z n ) =

Zn

dθ = 1. 2π

For α > 0 deﬁne N = N (θ ) = Nα (θ ) = inf{n : f n ≥ α}. Then N is a stopping time, i.e., {θ : N (θ ) = n} ∈ Mn . If gn is a sequence of Mn -measurable functions, the function g N = g N (θ) (θ ) = gk χ {N =k} k

is deﬁned on {N < ∞} and g N is M-measurable. If {gn } is a martingale, then G N ∧n =

n−1

gk χ {N =k} + gn χ {N ≥n}

k=0

is also a martingale. Lemma J.5. Let f n be a real dyadic martingale with f 0 = 0, let 0 < α < β, and set N = Nα . Then E α,β = {N < ∞ and S N ≤ β}

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satisﬁes

2 − α2β E α,β ≤ e .

(J.14)

Moreover, there are constants p, 0 1 such that if |k f | ≤ 1 and α > α0 , then {S N ≤ α 2 } ≥ p. (J.15) Proof. Set t = α/β and let Z n be the exponential transform of the martingale t f n . Then Z n ≥ et fn e−t since cosh t ≤ et

2 S /2 n

,

2 /2

. Therefore α2 dθ t 2 dθ exp tα − β ZN ≤ . e 2β E α,β = 2 2π 2π E α,β E α,β

On the other hand, since Z N = lim Z N ∧n on E α,β and Z n ≥ 0, we have dθ dθ ZN ≤ lim inf Z N ∧n ≤ 1, 2π 2π E α,β and that proves (J.14). 2 To prove (J.15) note that cosh t ≥ ec0 t , if 0 ≤ t ≤ 1, where c0 = log cosh 1, 2 and hence Z n ≤ et fn e−c0 t Sn . Therefore dθ dθ ≤ et (α+1) exp (−c0 t 2 S N ∧n ) , 1= Z N ∧n 2π 2π if |k f | ≤ 1. But S N ∧n → S N even when N = ∞, and hence dθ t (α+1) exp (−c0 t 2 S N ) , 1≤e 2π or dθ 2 ≤ 1 − e−t (α+1) . 1 − e−c0 t SN 2π Hence and by Chebychev’s inequality −t (α+1) {S N > β} ≤ 1 − e 2 1 − e−c0 t β

when β > α. When β = α 2 and α > 2 we get (J.15) by taking t =

2 c0 α .

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Proof of Theorem J.4. We prove the upper bound (J.10) ﬁrst. Fix ε > 0 and set "

% E = S = ∞, f n > (1 + ε) 2Sn log log Sn inﬁnitely often . We show |E| = 0. Deﬁne the stopping time Tk = inf{n : Sn ≥ (1 + ε)k }. Then lim Tk = ∞ on {S = ∞}, so that if # $ % E k = S = ∞, there exists n ∈ [Tk , Tk+1 ), with f n > (1 + ε) 2Sn log log Sn , then E⊂

+&

Ek .

n k>n

But by (J.14), (1+ε) (1 + ε)2 2(1 + ε)k log log((1 + ε)k ) 1 . = |E k | ≤ exp − 2(1 + ε)k+1 k log(1 + ε) Hence |E| ≤

1+ε 1 1 lim = 0. n→∞ log(1 + ε) k (1+ε) k>n

In proving the lower bound (J.11) we can assume |n f | ≤ 1. Set τk = inf{n : Sn > k}. Then k ≤ Sτk ≤ k + 1 and limk τk = ∞ almost everywhere. Set Fk = f τk . Now we claim that by (J.15) there exists p > 0 such that for all m and all sufﬁciently large k, # √ $ p (J.16) Fm+k − Fm > k ≥ . 2 Indeed, set An = {τm = n}. Then An is a union of dyadic intervals An, j ∈ Mn , and we can regard f n+s χ An, j as a dyadic martingale, indexed by s, with respect √ to the probability measure |A1n, j | χ An, j d x. We apply (J.15), with α = k, β = k, and N = N√k to each such martingale f n+s χ An, j , and recall that for a dyadic martingale, f s − f 0 has the same probability distribution as − f s + f 0 . Then we get k # # $ √ $ √ $ +# N =s Fm+k − Fm > k = Fm+k − Fm > k s=1

≥

k $ 1 # $ p 1 # N =s = N ≤k ≥ , 2 2 2 s=1

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511

which is (J.16). Fix N ≥ 2 and partition (N j−1 , N j ] into 6 log j 7 K = log 2p integer intervals, each of length approximately k = (N j − N j−1 )

log

2 p

log j

.

Applying (J.16) in each interval and using the independence of martingale differences gives # log j $ 1 1 Nj (J.17) ≥ pK ≥ , FN j − FN j−1 ≥ 2 2 j log p because the event (J.17) is contained in an intersection of K independent events of the form (J.16). For large N we now repeat the argument leading to (J.16), but using (J.11) instead of (J.15). We conclude that for sufﬁciently large j, ! log j 2 1 j N log j < Nj , (J.18) FN j−1 ≤ √ 4 log 2p N except on a set of small measure, because Sn ≤ N j−1 + 1 when τ N j−1 = n. Because (J.17) and (J.18) describe events that are independent in j, the Borel– Cantelli lemma gives (J.12) with constant c = ! 1 2 . 4 log

p

Exercise J.1. Let { f n , Fn } be a martingale and set n

2 E j F j−1 . Sn = j=1

Prove that (i) Sn is Fn−1 measurable, (ii) Sn is nondecreasing in n, and (iii) f n2 − Sn is a martingale with respect to {Fn }. Also prove (i), (ii), and (iii) determine Sn uniquely. Exercise J.2. (a) Let f n be a Bloch martingale. There exists a quasisymmetric homeomorphism h : ∂D → ∂D and a constant C < ∞ such that for every dyadic interval In, j , |h(In, j )| ≤ C|In, j | exp( f In, j ).

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The constant of quasisymmetry (deﬁned in (VII.3.5)) of h depends only on the Bloch norm of f n . See Bishop [2002a]. (b) There is a constant K < ∞ such that whenever ϕ : D → is a conformal map there exists a K -quasiconformal h : D → D such that h is bilipschitz with respect to the hyperbolic metric on D and such that for some C < ∞, 1 = |h(z)| ≤ C|ϕ (z)|(1 − |z|) for all z ∈ D. Use (a). If K can be taken arbitrarily close to 2, then Brennan’s conjecture is true. See Bishop [2002a] and Exercise VIII.13.

K. A Dichotomy Theorem Let p(z) = z d + ad−1 z d−1 + . . . + a0 be a monic polynomial of degree d ≥ 2 and let p (n) denote the iterate p (n) (z) = p ◦ p . . . ◦ p(z). For p(z) the basin of attraction of ∞ is = z : lim p (n) (z) = ∞ . n→∞

The basin is an open connected subset of C ∗ and is simply connected if and only if p (z) = 0, z ∈ .

(K.1)

If p (z) = 0 for some z ∈ , then ∂ has inﬁnitely many components. When p(z) = z 2 + b, (K.1) holds if and only if b belongs to the famous Mandelbrot set. For these facts see Carleson and Gamelin [1993]. Exercise III.8 gave some general results about harmonic measure on . Here we will prove a very precise theorem when (K.1) holds for z 2 + b. Theorem K.1. Assume (K.1) holds and let ω = ω(∞, ., ). Then either (a) ω 1 ω, or !

(b) There exists c(ω) such that if h c (t) = t exp c log 1t log log log 1t , then ω h c ,

f or c > c(ω),

(K.2)

ω ⊥ h c ,

f or c ≤ c(ω).

(K.3)

and

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513

Theorem K.1 is due to Przytycki, Urba´nski, and Zdunik [1989] and [1991], and their results are stronger in several ways: A simply connected domain is called an RB-domain, for repelling boundary, if is hyperbolic and if there is an analytic map g deﬁned on a neighborhood U of ∂ such that g(∂) ⊂ ∂, g(U ∩ ) ⊂ , and

+

g −n (U ∩ ) = ∂.

n≥1

When P(z) satisﬁes (K.1), the basin of attraction of ∞ is an RB-domain. Przytycki, Urba´nski, and Zdunik proved Theorem K1 for RB-domains. When z 2 + b satisﬁes (K.1), they also showed c(ω)2 is a real analytic and subharmonic function of the parameter b. In case (a), even with an RB-domain, Zdunik [1991] obtained a much better conclusion: ∂ is either an analytic arc or a real analytic Jordan curve. Przytycki, Urba´nski, and Zdunik use Gibbs measures and other ideas from dynamical systems, as well as Theorem K.3 below. A second slightly less general proof is in Chapter III of Makarov [1990a]. We present a third proof, due also to Makarov. Proof. Assume (K.1) holds. Then there is a conformal mapping ϕ : D → with 1 ϕ(z) = + . . . z that conjugates p(z) to the function z d : p(ϕ(z)) = ϕ(z d ), z ∈ D.

(K.4)

See page 65 of Carleson and Gamelin [1993]. Set −1 g(z) = log 2 , z ϕ (z) and z d−1 p (ϕ(z)) f (z) = log . d Then by (K.4), g(z) − g(z d ) = f (z)

(K.5)

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or g(z) =

∞

k

f (z d ).

(K.6)

k=0

Let Dn be the σ -algebra generated by the d-adic subintervals of [−π, π ) : 6 j2π ( j + 1)2π In, j = , , 0 ≤ j ≤ dn, dn dn and write [h]n = E(h Dn ) whenever h ∈ L 1 . Lemma K.2. Assume (K.1) and let f be deﬁned by (K.5). Then there are C < ∞ and β < 1 such that for all n, || f − [ f ]n ||2 < Cβ n .

(K.7)

Proof. Write c(I ) for the center of I ∈ Dn and set a I = (1 − |I |)c(I ). Because [ f ]n is an orthogonal projection we have 2 dθ 2 . || f − [ f ]n ||2 ≤ f − f (a I ) 2π I Dn

Write

p (z) = d

d−1 -

(z − c j ).

j=1

Then almost everywhere on I ,

1 f (ζ ) − f (a I ) = (d − 1) log 1 − |I | so that 2 f (ζ )− f (a I ) 2 ≤ d(d − 1) 2

For α > 1/2, set

(

1 log 1 − |I |

+

d−1 j=1

ϕ(ζ ) − c j log ϕ(a ) − c I

j

,

) 2 d−1 ϕ(ζ ) − c j 2 + log . ϕ(a I ) − c j j=1

" |ϕ(ζ ) − ϕ(a I )| α J = J (I ) = ζ ∈ I : ≤ |I | . |ϕ (a I )|

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Then since | log z|2 ≤ |z − 1|2 + |1/z − 1|2 for all z ∈ C, we have 2 | f (eiθ ) − f (a I ) dθ J / 2 d−1 ϕ(eiθ − ϕ(a I ) 2 1 d(d − 1)2 + |I | log( ≤ ϕ(a ) − c 2 1 − |I | I j j=1 J 0 ϕ(eiθ − ϕ(a I ) 2 dθ + ϕ(eiθ ) − c j 2π / 0 ϕ (a I ) 2 ϕ (a I ) 2 dθ −3n 2α ≤ C(d) d + |I | ϕ(a ) + ϕ(eiθ ) 2π I J j / 0 ϕ (a I ) 2 −3n 2α ≤ C(d) d + |I | |I | . ϕ(a I ) j

Thus by Exercise I.22, f (eiθ ) − f (a I ) 2 dθ 2π I ∈Dn J /

0 ϕ ((1 − |I |)eiθ ) 2 dθ + C|I | ≤ C(d) d iθ ∂ D ϕ((1 − |I |)e ) 2π 1 ≤ C (d)|I |2α−1 log . 1 − |I | −2n

2α

(K.8)

On the other hand, the Beurling projection theorem shows that , * 1−α |I \ J | α ≤ ω a I , {|ϕ(ζ ) − ϕ(a I )| ≥ |I | |ϕ (a I )|}, D ≤ C|I | 2 , |I | while by the John–Nirenberg theorem and Exercise F.2, 1 f (eiθ ) − f (a I ) 4 dθ ≤ L 4 . |I | I 2π Therefore f (eiθ ) − f (a I ) 2 dθ ≤ |I \ J |1/2 (C4 |I |)1/2 ≤ (CC4 )1/2 |I | 5−α 4 , 2π I \J and

Dn

I \J

| f − f (a I )|2

1−α

dθ ≤ C d −n 4 . 2π

Take α = 59 . Then (K.8) and (K.9) imply (K.7) with β = d

(K.9) − 15

.

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Conclusion of the proof of Theorem K.1. Because k E( f (eid θ ) Dk ) = 0,

(K.10)

(K.7) shows that ρk =

f (eid θ ) f (eiθ ) k

dθ 2π

satisﬁes |ρk | ≤ Cβ k || f ||2 ,

(K.11)

and the series σ2 =

∞

| f |2

dθ +2 2π

f (eid θ ) f (eiθ ) k

k=1

dθ 2π

(K.12)

converges absolutely. Set Sn (θ ) =

n

f (eid θ ). k

k=0

Then n + 1 − k ||Sn ||22 = || f ||22 + 2 ρk , n+1 n+1 n

(K.13)

k=1

so that ||Sn ||22 = σ2 n→∞ n + 1 lim

and 0 ≤ σ 2 < ∞. The dichotomy in Theorem K.1 is between the two cases σ 2 = 0 and σ 2 > 0. Case (a): Assume σ 2 = 0. Then by (K.13), ||Sn ||22

= (n + 1)σ − 2(n + 1) 2

∞

ρk − 2

k=n+1

= 0 − An − Bn , and by (K.8), An → 0, while |Bn | is bounded. Therefore sup ||Sn ||2 < ∞,

n k=1

kρk

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517

and by (K.6), g(z) is the real part of an H 2 function. Hence ϕ has a non-zero nontangential limit almost everywhere, and by Theorem V.4.2. ω 1 ω. Again, Zdunik [1991] has a much stronger conclusion. Case (b): Assume σ 2 > 0. We use the law of the iterated logarithm, in the following form: Theorem K.3. Suppose the real function f ∈ L 2 satisﬁes condition (K.11) and let σ 2 be deﬁned by (K.12). If σ 2 > 0, then almost everywhere, √ Sn = 2 σ. lim sup √ (K.14) n log log n We omit the proof of Theorem K.3, which relies on a result of Ibragimov [1962] on the central limit theorem for weakly dependent random variables. See Resnik [1968], Theorem 2.2, and pages 79–82 of Philipp and Stout [1975]. Returning to the proof of Theorem K.1, we let ζ ∈ I ∈ Dn . Then n 1 j dj Sn (ζ ) − [Sn ]n (ζ ) = f (eid θ )dθ f (ζ ) − |I | I j=0

=

n

j j f (ζ d ) − [ f ]n− j (ζ d ) ,

j=0

so that by (K.7), sup ||Sn − [Sn ]n ||2 ≤ ∞ n

and

Sn − [Sn ]n =0 lim sup √ n log log n

almost everywhere. Let |z| = r satisfy 1 − d −n < r ≤ 1 − d −n−1 . Then √ n log log n 1 ! −→ √ log d 1 1 log 1−r log log log 1−r g(z) − [g]n (z/|z|) ≤ C, while by (K.10), [g]n = [Sn ]n .

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Therefore (K.14) gives

g(r eiθ )

lim sup ! log

1 1−r

log log log

1 1−r

=

2 σ log d

(K.15)

almost everywhere and (K.2) and (K.3) both follow from (K.15) and Theorem VIII.2.1, exactly as in the proof of Theorem VIII.2.2.

L. Two Estimates on Integral Means Theorem L.1. If ϕ is univalent on D and if p ≤ 1.399, then 1 d xd y < ∞. p D |ϕ (z)| Proof. The Schwarzian derivative of ϕ is deﬁned to be ϕ 1 ϕ 2 Sϕ = − . ϕ 2 ϕ The Schwarzian derivative satisﬁes (1 − |z|2 )2 |Sϕ (z)| ≤ 6, by Theorem X.4.1.

−1/2 Moreover, if we set h = ϕ = an z n then 1 3 |h (z)| = Sϕ (z)h(z) ≤ |h(z)|. 2 (1 − |z|2 )2 Thus

|an |2 n 2 (n − 1)2 r 2n−4 =

2π

|h (r eiθ )|2

0

Set s = r 2 and y(s) =

dθ 2π

2π dθ 9 |h(r eiθ )|2 ≤ (1 − r 2 )4 0 2π 9 |an |2 r 2n . = 2 4 (1 − r ) |an |2 s n . Then for s0 sufﬁciently close to 1, y (4) (s) ≤

9.01 y(s), (1 − s)4

provided s0 ≤ s < 1. The function Y (s) = C(1 − s)−λ satisﬁes Y (4) (s) =

9.01 Y (s), (1 − s)4

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where λ is the positive root of λ(λ + 1)(λ + 2)(λ + 3) = 9.01. Choosing C sufﬁciently large, we may suppose that Y ( j) (s0 ) > y ( j) (s0 ), for j = 0, 1, 2, 3. Thus s t4 t 3 t 2 9.01 Y (s) − y(s) ≥ (Y (t1 ) − y(t1 ))dt1 dt2 dt3 dt4 . (1 − t1 )4 s0 s0 s0 s0 Note that if Y (t1 ) − y(t1 ) > 0 for s0 ≤ t1 < s, then Y (s) − y(s) > 0 and hence Y (s) − y(s) > 0 for s0 ≤ s < 1. Thus 2π 1 C dθ . (L.1) = y(r 2 ) ≤ iθ |ϕ (r e )| 2π (1 − r 2 )λ 0 In other words, we have β(−1) ≤ λ. By the distortion theorem |ϕ (r eiθ )| ≥ C(1 − r ), and hence by (L.1), 1 1 C r dθdr ≤ dr < ∞, |ϕ (r eiθ )| p (1 − r ) p−1+λ 0 D

provided p < 2 − λ. A calculation shows that 0.600 < λ < 0.601, and the theorem follows. Theorem L.2. If ϕ is univalent on D then 1 1/2 1 βϕ (t) ≤ − + t + . − t + 4t 2 2 4 Proof. By Hardy’s identity (VIII.1.6) for t = 2 p, 2π 1 |ϕ (r eiθ )|t dθ I (r ) = I (r, ϕ ) = 2π 0 satisﬁes r

t2 d r I (r ) = dr 2π

2π 0

(L.2)

ϕ (r eiθ ) |ϕ (r eiθ )|t r iθ dθ. ϕ (r e )

I (r )

≥ 0, this yields 2π ϕ (z) t2 2r 2 2r 2 2 2 |ϕ (r eiθ )|t z − + r I (r ) ≤ dθ. 2π 0 ϕ (z) 1 − r2 1 − r2

Since

By (I.4.20) zϕ (z) 4r 2r 2 , − ≤ ϕ (z) 1 − r2 1 − r2

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and so we obtain 4r 2 t 2 (4 + r 2 ) r 2 I (r ) ≤ I (r ) (1 − r 2 )2 2π 2r 2 zϕ (z) 4r 2 t2 |ϕ (r eiθ )|t Re dθ. − + 2 2π 0 ϕ (z) 1 − r 1 − r2 But by deﬁnition r I (r ) =

t 2π

Therefore (L.3) yields r 2 I (r ) ≤

2π

|ϕ (r eiθ )|t Re

0

(L.3)

zϕ (z) ϕ (z)

dθ.

4r 2 t 2 (4 + r 2 ) 4r 3 t I (r ) + I (r ), (1 − r 2 )2 1 − r2

and given ε > 0 there is r0 = r0 (ε) so that if r0 < r < 1, I (r ) ≤

3t 2 + ε 2t + ε I (r ) + I (r ). (1 − r )2 1−r

(L.4)

Let g(r ) = c(1 − r )−a where a is a positive solution of a(a + 1) = (2t + ε)a + 3t 2 + ε and where c > 0 is so large that g(r0 ) > I (r0 ) and g (r0 ) > I (r0 ). Then since g (r ) =

3t 2 + ε 2t + ε g(r ) + g (r ), (1 − r )2 1−r

(L.4) and a little calculus shows I (r ) < g(r ) on r0 < r < 1. Sending ε → 0 then gives (L.2). Exercise L.1. Use Theorem L.1 to prove B(−2) ≤ λ + 1 < 1.601.

Notes These results are from Pommerenke [1985a]. See Bertilsson [1998], [1999], Shimorin [2003] and Hedenmalm and Shimorin [2004] for improvements on Theorem L.1.

M. Calderón’s Theorem and Chord-Arc Domains In this appendix we prove Theorem X.1.1 for domains bounded by chord-arc curves. Let f (z) be analytic on a simply connected domain , let ζ ∈ ∂, and

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521

let α > 1. We recall the deﬁnitions of the cone α (ζ ) = α (ζ, ) = z ∈ : |z − ζ | < α dist(z, ∂) , the maximal function f α∗ (ζ ) = sup | f (z)|, α (ζ )

and the area function Aα f (ζ ) =

| f (z)|2 d xd y

1 2

.

α (ζ )

The following theorem is from Jerison and Kenig [1982a]. Theorem M.1. Let be a domain bounded by a chord-arc curve, let α > 1, let 0 < p < ∞, and let f (z) be analytic on . Then f ∈ H p () if and only if Aα f ∈ L p (∂, ds), and for z 0 ∈ there is a constant c = c(z 0 , , p, α) such that c||Aα f || L p (∂) ≤ || f − f (z 0 )|| H p () ≤ c−1 ||Aα f || L p (∂) . p

p

p

Let ϕ : D → be a conformal mapping with ϕ(0) = z 0 . The proof depends on three crucial properties of the chord-arc domain : (i) There exist β = β(α, , z 0 ) > 1 and γ = γ (α, , z 0 ) > 1 such that for all ζ ∈ ∂D, ϕ(β (ζ )) ⊂ α (ϕ(ζ )) ⊂ ϕ(γ (ζ )).

(M.1)

This holds because ∂ is a quasicircle, and β and γ depend only on the dilatation of the quasiconformal extension of ϕ. See Exercise VII.8. (ii) On ∂D the measures dμ(θ) = |ϕ (eiθ )|dθ and dθ are A∞ -equivalent. This is proved in Section VII.4. (iii) There is q > 0 such that 1 ∈ H q (D). ϕ (z)

(M.2)

This is also proved in Section VII.4. A careful reading of the proof of Theorem M.1 below will show that the constant c(α, p, z 0 , ) depends only on α, p, the ratio dist(z 0 , ∂)/diam, and the constants implicit in the properties (i), (ii), and (iii) of cited above. Lemma M.2. Let be a chord-arc domain, let α > 1, and let 0 < p < ∞. Suppose f (z) is analytic on . Then f ∈ H p () if and only if f α∗ ∈ L p (∂, ds),

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and there exists a constant c = c(, α, p) such that c|| f ∗ || p ≤ || f || H p () ≤ c−1 || f ∗ || p . p

p

p

(M.3) 1

Proof. Assume f ∈ H p (); that is, assume F(z)(ϕ (z)) p ∈ H p (D), where F(z) = f ◦ ϕ(z). Then F has nontangential limit a.e. and p |F(eiθ )| p |ϕ (eiθ )|dθ = || f || H p () < ∞. Let us ﬁrst suppose p ≥ p◦ = q+1 q , where q is given in (M.2). Then by Hölder’s inequality, with exponents p◦ and q + 1, we have q 1 1 p

1 1 dθ q+1 |F(r eiθ )|dθ ≤ |F(r eiθ )||ϕ (r eiθ )| p◦ ◦ dθ p◦ iθ ϕ (r e )

1

1− 1 |ϕ (r eiθ )|dθ p p◦ ≤ |F(r eiθ )| p |ϕ (r eiθ )|dθ p q 1

1 dθ q+1 , × ϕ (r eiθ ) where the last inequality holds because p ≥ p◦ and ϕ ∈ H 1 . Consequently F ∈ H 1 (D) and the Poisson formula holds for F(z). Therefore by (i), p || f α∗ || L p (∂) ≤ |Fγ∗ | p |ϕ |dθ ≤ Cγ |M F| p |ϕ |dθ. ∂D

∂D

But now by Section VII.4, |ϕ (eiθ )|dθ (see Garnett [1981], page 255) |M F| p |ϕ |dθ ≤ C ∂D

∈ A p◦ so that by Muckenhoupt’s theorem

∂D

|F| p |ϕ |dθ = C|| f || H p () , p

and the left-hand inequality in (M.3) holds when p ≥ p◦ . 1

p

Now suppose p < p◦ . Because F(z)(ϕ (z)) p ∈ H 1 we can write F = Bg p◦ where B(z) is a Blaschke product, where g(z) has no zeros in D, and where p◦ |F| = |g| p on ∂D. Then (Fγ∗ ) p ≤ (gγ∗ ) p◦ and as before we have p || f α∗ || L p (∂) ≤ |Fγ∗ | p |ϕ |dθ ≤ |gγ∗ | p◦ |ϕ |dθ ∂D ∂D p p0 |g| |ϕ |dθ = Cγ |F| p |ϕ |dθ = Cγ || f || H p () . ≤ Cγ ∂D

∂D

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Conversely, assume f α∗ ∈ L p (∂, ds) and ﬁx r < 1. Partition the circle {|z| = r } into arcs I j such that |I j | ≤ (β − 1)(1 − r ), where β is deﬁned by (M.1). Then I j ⊂ β (ζ ) z : z ∈ I j }, and for all ζ ∈ I j∗ = { |z| |ϕ |ds ≤ C(, z 0 ) dist(ϕ(I j ), ∂) ≤ C(, z 0 ) |ϕ |dθ, (ϕ(I j )) = I j∗

Ij

where C(, z 0 ) depends only on the dilatation of the quasiconformal extension of ϕ. Then iθ p iθ ∗ p |F(r e )| |ϕ (r e )|dθ ≤ C |Fβ | |ϕ |dθ ≤ C | f α∗ | p ds, j

I j∗

∂

and that yields the right-hand inequality in (M.3).

By Lemma M.2 and properties (i) and (iii), Theorem M.1 is an immediate consequence of the following result, which is the A∞ -weights version of Calderón’s theorem. It is due to Gundy and Wheeden [1974]. Theorem M.3. Let w(eiθ ) be an A∞ weight, let α > 1, and let 0 < p < ∞. Then there is C = C(α, w) such that whenever F(z) is analytic on D and F(0) = 0, C||Aα F|| L p (wdθ) ≤ ||Fα∗ || L p (wdθ) ≤ C −1 ||Aα F|| L p (wdθ) . p

p

p

(M.4)

Proof. Write μ = w(eiθ )dθ. By Exercise I.11 from Chapter I, ||Fα∗ || L p (μ) ≤ ||Fβ∗ || L p (μ) ≤ Cα,β ||Fα∗ || L p (μ) p

p

p

when β > α. That means we can prove Theorem M.3 by comparing ||Aα F|| to ∗ || and ||F ∗ || to ||A F||. ||F2α 2α α We now claim the following lemma. Lemma M.4. Given ε > 0 and κ > 1, there exists δ < 1 such that for all λ > 0 and all analytic F(z) satisfying F(0) = 0 ∗

≤ δλ ≤ εμ A∗α F > λ . (M.5) μ A∗α F > κλ ∩ F2α Before we prove Lemma M.4, let us use it to derive the left-hand inequality in (M.4). Replace F(z) by F(r z), r < 1, so that the next three integrals below

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converge. By (M.5) we have p

||Aα F|| L p (μ) = pκ p

∞

0 ∞

λ p−1 μ Aα F > κλ )dλ

∗ λ p−1 μ F2α > δλ dλ + εpκ p ≤ pκ p ∞0

× λ p−1 μ Aα F > λ dλ κ 0 p p p ∗ = ||F2α F|| L p (μ) + εκ p ||Aα F|| L p (μ) . δ Then taking εκ p ≤

and sending r → 1 yield κ p κ p p p ∗ p ||Aα F|| L p (μ) ≤ 2 ||F2α || L p (μ) ≤ 2Cα,2α ||Fα∗ || L p (μ) , δ δ which is the left-hand side of (M.4). Many authors call (M.5) and similar inequalities “good-λ inequalities”, despite the fact that (M.5) is true for all λ > 0. Exercise M.2 includes a short history of the term “good-λ inequality". 1 2

Proof of Lemma M.4. Write the open set {Aα F > λ} as the union of disjoint j open arcs {Jj } and partition each Jj into closed arcs {Ik } with disjoint interiors j j such that |Ik | = dist(Ik , ∂D \ Jj ). Ik

Ij Figure M.1 The family {Ik } =

&∞ j=1

j

{Ik } has the three properties { Aα F > λ} =

&

Ik ,

Iko ∩ I jo = ∅ when j = k, and dist(Ik , { Aα F ≤ λ}) = |Ik |. For this reason {Ik } is called the Whitney decomposition of { Aα F > λ}. By the third property there exists ζk with dist(ζk , Ik ) = |Ik | and Aα F(ζk ) ≤ λ.

(M.6)

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525

The main step in proving (M.5) is to show that whenever we are given κ > 1 and η > 0, there exists δ > 0 such that for all Ik Ik ∩ Aα F > κλ ∩ F ∗ ≤ δλ ≤ η Ik ∩ Aα F > λ . (M.7) 2α Indeed, if η = η(ε) is small, (M.7) and the A∞ condition (VII.4.2) then give

∗

μ Ik ∩ Aα F > κλ ∩ F2α ≤ δλ ≤ εμ Ik ∩ Aα F > λ , and a summation over {Ik } gives inequality (M.5). Now let us prove (M.7). Fix I = Ik and suppose |I | < 41 . Set ∗ E = I ∩ Aα F > κλ ∩ F2α ≤ δλ , & U= α (ζ ), E

U0 = U ∩ {1 − |z| ≤ 2|I |}, U1 = U ∩ {1 − |z| > 2|I |}, and W =

&

2α (ζ ).

E

Then

κ 2 λ2 |E| ≤ E

=

AF(ζ )2 ds |F (z)|2 ζ ∈ E : z ∈ α (ζ ) d xd y

U

≤

2 α

U0

|F (z)|2 (1 − |z|)d xd y + |E|

(M.8) |F (z)|2 d xd y,

U1

because

ζ ∈ E : z ∈ α (ζ ) ≤ min |E|, 2 (1 − |z|) . α To bound the integral over U0 , we form the larger domain W0 = W ∩ 1 − |z| ≤ 4|I | .

(M.9)

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E U0 W0

Figure M.2 Note that W0 is a special cone domain and that 1 − |z| ≤ c(α) dist(z, ∂ W0 ). if z ∈ U0 ⊂ W0 , by trigonometry. Therefore by Theorem X.1.2, |F (z)|2 (1 − |z|)d xd y ≤ c(α) |F (z)|2 dist(z, ∂ W0 )d xd y U0

W0

≤ C(α)

∂ W0

|F(z) − F(z W0 )|2 ds

(M.10)

≤ C (α)(2δλ)2 |I |.

To bound the integral over U1 , set U1 = U1 ∩ α (ζk ) ∪ U1 \ α (ζk ) . Then we have |E| |F (z)|2 d xd y ≤ |E| |F (z)|2 d xd y ≤ λ2 |E|. (M.11) U1 ∩α (ζk )

α (ζk )

On the other hand, if z ∈ U, then by a calculation B(z, 1−|z| 4 ) ⊂ W. Conse) and hence quently |F| ≤ δλ on B(z, 1−|z| 4 |F (z)| ≤ Therefore

4δλ . 1 − |z|

|F (z)| d xd y ≤ (4δλ) |I |

|E|

2

U1 \α (ζk )

2

U1 \α (ζk )

d xd y . (1 − |z|)2

But since dist(ζk , I ) = |I |, writing the last integral in polar coordinates gives d xd y ≤ C, (1 − |z|)2 U1 \α (ζk )

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M. Calderón’s Theorem and Chord-Arc Domains

and thus

|E|

|F (z)|2 d xd y ≤ 4C(δλ)2 |I |,

(M.12)

U1 \α (ζk )

Together, (M.8), (M.10), (M.11), and (M.12) give us δ 2 δ 2

1 2 |I | + C (α) |I |, |E| ≤ C (α) 1− κ κ κ and we obtain (M.7).

To prove the right-hand inequality in (M.4), we ﬁx λ > 0 and now let {Ik } be the Whitney decomposition of {Fα∗ > λ}. Write 2Ik for the arc concentric with Ik having length |2Ik | = 2|Ik |, and note that no point falls in more than three arcs 2Ik . Lemma M.5. Given η > 0 and κ > 2, there exist δ > 0 such that for each Ik , Ik ∩ F ∗ > κλ ∩ M χ k ≤ 1 ≤ η Ik , (M.13) α 2 where M is the Hardy–Littlewood maximal operator and χ k is the characteristic function χ k = χ 2Ik ∩{ζ :A2α F(ζ )>δλ} . Before proving Lemma M.5, let us show how it gives the right-hand inequality in (M.4). Again replace F(z) by F(r z), r < 1 so that the integrals converge. Then given ε > 0 there exists η = η(ε) > 0 so that by (M.13) and the A∞ -condition,

1 ≤ εμ Ik . μ Ik ∩ Fα∗ > κλ ∩ M χ k ≤ 2 By Exercise VII.14, w ∈ A p◦ for some p◦ > 1, so that by Chebychev’s inequality,

1 p ≤ C(μ)2 p◦ ||χ k || L◦p◦ (μ) μ Ik ∩ M χ k > 2

. = C(μ)2 p◦ μ 2Ik ∩ ζ : A2α F(ζ ) > δλ Consequently,

μ Fα∗ > κλ ≤ εμ Fα∗ > λ + 3C(μ)2 p◦ μ ζ : A2α F(ζ ) > δλ ,

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and hence ∗ p || L p (μ) ||F2α

≤

p Cα,2α ||Fα∗ || L p (μ)

∞

= pκ

∞

p 0

λ p−1 μ Fα∗ > λ dλ + 3C(μ)2 p◦ pκ p ≤ εpκ p ∞0

× λ p−1 μ A2α > δλ dλ 0 κ p p p p = εκ ||Fα∗ || L p (μ) + 3C(μ)2 p◦ ||A2α F|| L p (μ) . δ Again taking εκ p <

λ p−1 μ Fα∗ > κλ dλ

and sending r → 1 give κ p p ∗ p ||F2α || L p (μ) ≤ C(μ)2 p◦ +1 ||A2α F|| L p (μ) , δ and this inequality is the right-hand side of (M.4). 1 2

Proof of Lemma M.5. Fix I = Ik and assume |I | < 41 . Take 1 E 0 = I ∩ Fα∗ > κλ ∩ M χ k ≤ , 2 1 E = I ∩ Mχk ≤ , 2 E 1 = 2I ∩ A2α ≤ δλ , & U = {1 − |z| < 2|I |} ∩ 2α (ζ ), E

and U1 =

&

2α (ζ ).

E1

Then

|F (z)|2 ζ ∈ E 1 : |z − ζ | ≤ 2α(1 − |z|) d xd y

U1

≤

|F (z)|2 d xd yds ≤ 2δ 2 λ2 |I |.

E 1 2α (ζ )

Now if z ∈ U there is ζ ∗ ∈ E with |ζ ∗ − z| ≤ 2α(1 − |z|). Therefore ζ : A2α F(ζ ) > δλ ∩ |ζ − ζ ∗ | < 1 − |z| ≤ 1 − |z| , 4 8 and 1 − |z| . (M.14) | ζ ∈ E 1 : |z − ζ | < 2α(1 − |z|) ≥ 16

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529

Inequality (M.14), which is the reverse of (M.9), is the reason the function χ M k was introduced. Because of (M.14), we have |F (z)|2 dist(z, ∂U )d xd y ≤ |F (z)|2 (1 − |z|)d xd y U

U

|F (z)|2 ζ ∈ E 1 : |z − ζ | (M.15)

≤ 16 U1

≤ 2α(1 − |z|) d xd y ≤ 32δ 2 λ2 |I |. The domain U is a special cone domain, and so (M.15) implies |F − F(cU )|2 ds ≤ C(α)δ 2 λ2 |I |. ∂U

(M.16)

Now let ζ ∈ E 0 ⊂ E. By trigonometry there exists β = β(α) > 1 such that α (ζ ) ∩ {1 − |z| < |I |} ⊂ β (ζ, U ). See Figure X.6. Take z ∈ α (ζ ) ∩ {1 − |z| ≥ |I |}. By trigonometry there is z = r z, r (α) ≤ r ≤ 1 such that z ∈ α (ζk ), where dist(ζk , I ) = |I | and Fα∗ (ζk ) ≤ λ. In particular, we have |F(z )| ≤ λ. When w ∈ [z , z], a calculation shows B(w, (1−|w| 4 ) ⊂ 2α (ζ ) so that 1 16 4δλ 2 2 |F (w)| ≤ |F | d xd y ≤ 1 . 2 π(1 − |w|) π 2 |I | 1−|w| B(w,

4

)

Then an integration gives |F(z)| ≤ (1 + cδ)λ.

(M.17)

Because (M.17) holds at z = cU , we conclude that

sup |F(z) − F(cU )| ≥ κ − 2(1 + cδ) λ β (ζ,U )

for every ζ ∈ E 0 . But then by (M.16) and Lemma M.2 |E 0 | ≤

C(α)δ 2 |I | , (κ − 2(1 + cδ))2

and (M.13) is proved.

Exercise M.1. Prove Lemma M.5 and inequality (M.7) in the case |I | > 41 . Exercise M.2. (a) Suppose (X, μ) is a measure space and suppose f is a nonnegative μ-measurable function. Let p > 0, κ > 1 and 0 < ε < 1, and suppose

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εκ p < 1. Call λ > 0 good if εμ({ f > λ}) < μ({ f > κλ}). If G = {good λ}, then p

|| f || L p (μ) ≤

(1 − ε)κ p 1 − εκ p

pλ p−1 μ({ f > λ})dλ. G

(b) Now consider the inequality ∗

≤ δλ ≤ εμ A∗α F > λ μ A∗α F > κλ ∩ F2α

(M.5)

and set f = A∗α F. Note that (M.5) is trivial if λ ∈ / G. On the other hand, by part (a) any inequality of the form ∗ > K λ}), λ ∈ G μ({ f > λ}) ≤ Cμ({F2α p ||Aα F|| L p (μ)

(M.18)

∗ || p C||F2α L p (μ)

yields ≤ which was the objective of (M.5). Of course, (M.5) also implies (M.18). In their original papers [1970] and [1972] Burkholder and Gundy worked with inequalities like (M.18) for λ ∈ G (but without calling them “good"). See Burkholder [1973] and Gundy and Wheeden [1974] for related inequalities, and Bañuelos and Moore [1999] for a more recent discussion. (c) In general, for analytic functions F(z) on the unit disc with F(0) = 0, no inequality of the form Aα F > λ ≤ C F ∗ > N λ Kα is valid for all λ > 0. Exercise M.3. Let be a locally rectiﬁable Jordan curve such that ∞ ∈ and let 1 and 2 be the components of C∗ \ . (a) Prove that j is a BMO domain if and only if there is a constant C |F (z)|2 dist(z, )d xd y ≤ |F(ζ )|2 ds

j

for all F bounded and analytic on j satisfying F(z) = O( 1z ), z → ∞. (b) Prove that is a chord-arc curve if and only for both j = 1 and j = 2, |F(z)|2 ds ≤ C |F (z)|2 dist(z, )d xd y

j

for all F bounded and analytic on j satisfying F(z) = O( 1z ), z → ∞. These results are due to Bruna and González [1999].

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554

CY535/Garnett

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12:30

Author Index

Adams, D. R., 66, 112, 531 Adams, R. A., 66, 531 Ahlfors, L. V., xiii, xiv, 3, 27, 73, 113, 149, 150, 153, 157, 163, 193, 194, 230, 234, 241, 243–246, 254, 256, 259–261, 267, 341, 343, 376, 382, 391, 396, 421–423, 425, 429, 494, 496, 531, 532, 534, 543, 546, 552 Alexander, H., 222, 531 Ancona, A., 343, 345, 531 Anderson, J. M., 257, 532 Astala, K., 257, 314, 411, 532 Baernstein II, A., 25, 34, 36, 124, 150, 194, 269, 302, 532 Bagemihl, F., 205, 224, 532 Balogh, Z., 157, 165, 166, 193, 532 Bañuelos, R., 274, 530, 532, 533 Bara´nski, K., 286, 533 Barth, K. F., 532, 533 Batakis, A., 315, 341, 343, 533 Beardon, A. F., 344, 533 Becker, J., 70, 235, 236, 383, 533 Behrens, M., 345, 533 Benedicks, M., 533 Bennett, C., 66, 533 Bertilsson, D., xiii, 193, 199, 305, 306, 313, 314, 520, 534 Betsakos, D., 123, 534 Beurling, A., xiv, 100, 105, 107, 108, 118, 122, 123, 125, 127, 128, 145, 149, 150, 152, 155, 157, 159, 160, 163, 187, 193–195, 213, 214, 218, 243, 244, 259, 281, 312, 416, 425, 494, 515, 531, 534, 538, 549 Bieberbach, L., 26, 534, 538, 551 Binder, I., 306, 341, 342, 534 Bishop, C. J., xiii, 203, 211, 220, 221,

225–227, 257, 261, 262, 314, 343, 347, 370, 394, 397, 398, 412, 416, 426, 429, 512, 534, 535 Bonk, M., 123, 157, 165, 166, 193, 532, 535 Bourgain, J., 535 Brannan, D. A., 532, 533, 535, 548, 553 Brelot, M., 112, 535 Brennan, J., xiii, xiv, 269, 286–288, 293, 302, 305, 306, 309, 314, 342, 512, 533–535, 537, 548 Brolin, H., 114, 535 Browder, A., 219, 535 Brown, L., 227, 535 Bruna, J., 530, 535 Burdzy, K., 194, 198, 536, 537, 540, 550 Burkholder, D., 530, 536 Calderón, A. P., 348, 351, 520, 523, 536, 537, 551 Campbell, D. M., 536 Carathéodory, C., 1, 13, 29, 37, 39, 43, 51, 66, 178, 200, 201, 238, 457, 536 Caratheodory, C., 13 Carleman, T., 150, 193, 480–482, 484, 497, 536 Carleson, L., xiii, xiv, 31–33, 66, 107, 112, 114, 115, 119, 121, 213, 220, 221, 269, 270, 277, 281, 286, 306, 315, 319, 341, 342, 345, 354, 355, 373, 374, 401, 410–412, 416, 427, 431, 440, 466, 480, 512, 513, 534–537, 553 Carroll, T. F., 194, 537 Chang, S.-Y. A., 537 Choi, S., 227, 426, 537 Choquet, G., 535 Chuaqui, M., 426, 537 Chung, K. L., 269, 471, 537

555

CY535/Garnett

0 521 47018 8

December 6, 2007

556

12:30

Author Index

Cima, J. A., 34, 537 Clunie, J. G., 257, 532, 533, 536, 548, 553 Coifman, R. R., 247, 257, 426, 427, 537 Courant, R., 455, 537 David, G., 362, 373, 426, 429, 538, 546, 552 de Branges, L., 538, 539 Denjoy, A., 157, 193, 533, 535, 538, 543 Denker, M., 538 Doob, J. L., 112, 471, 472, 538 Du Plessis, N., 538 Duren, P. L., 25, 26, 34, 36, 125, 238–240, 306, 309, 532, 538, 548, 549 Durrett, R., 323, 538 Dynkin, E. B., 538 Erdös, P., 461, 538 Eremenko, A., 314, 538 Essèn, M., 123, 538 Evans, G. C., 466, 468–470, 476, 539 Falconer, K. J., 362, 463, 539 Fang, X., 426, 539 Fefferman, C., 34, 247, 480, 534, 537, 539 Fejér, L., 127, 539 Feller, W., 241, 269, 272, 539 Feng, J., 287, 309, 539 Fernández, J. L., 124, 260, 302, 426, 532, 539 Fisher, S. D., 343, 391, 539 FitzGerald, C. H., 539 Flinn, B. B., 124, 539 Folland, G. B., xiii, 457, 539 Frank, N. H., 551 Freedman, D., 540 Frostman, O., 431, 458, 540 Fuchs, W., 150, 540 Gamelin, T. W., xiii, 114, 115, 219, 343, 391, 427, 512, 513, 537, 540 Gardiner, F., 540 Gardiner, S., 194, 540 Garey, M. R., 367, 540 Garnett, J. B., xv, 32, 34, 119, 219, 221, 247, 268, 345, 373, 411, 426, 441, 480, 522, 535, 540 Garsia, A., 25, 540 Gehring, F. W., 122, 166, 242, 257, 259, 260, 293, 382, 426, 532, 540–542, 548, 549 Gillis, J., 461, 538 ´ Goldstein, V. M., 541 Goluzin, G. M., 25, 541

González, M. J., 345, 530, 536, 541 Granados, A., 539 Grinshpan, B., 306, 541 Grötzsch, H., 150, 151, 541 Gundy, R., 523, 530, 536, 541 Gunning, R. C., 541 Hag, K., 540 Haliste, K., 123, 484, 497, 538, 541 Hall, T., 108, 124, 125, 541, 542 Hamilton, D. H., 314, 538, 539 Hardy, G. H., 9, 35, 36, 53, 200, 247, 270–272, 279, 305, 307, 345, 435, 436, 438, 472, 519, 527, 532, 541, 543, 547 Hausdorff, F., 542 Havin, V. P., 302, 542 Hayman, W. K., 1, 23, 25, 34–36, 122, 124, 125, 166, 222, 269, 293, 302, 347, 421, 532, 533, 535, 536, 540, 542, 549, 550, 553 Hedberg, L. I., 66, 112, 531 Hedenmalm, H., 305, 306, 520, 542 Heinonen, J., 112, 123, 124, 242, 260, 302, 345, 426, 532, 539, 542, 548, 549 Heins, M., 542 Helms, L. L., 542 Hersch, J., 150, 542 Hobson, E. W., 225, 543 Hoffman, K., 227, 543 Hruššëv, S. V., 302, 542 Hunt, R. A., 247, 543 Ibragimov, I. A., 517, 543 Itô, K., 543 Jenkins, J. A., 180, 193, 196, 543 Jerison, D. S., 119, 121, 241, 244, 245, 249, 257, 261, 347, 351, 521, 535, 543 John, F., 253, 254, 262, 478, 479, 515, 543 Johnson, D. S., 367, 540 Jones, P. W., xiii, 34, 121, 214, 220, 221, 238, 257, 268, 270, 272, 305, 306, 315, 341–343, 347, 367, 368, 370, 372, 394, 397, 398, 412, 416, 426, 427, 429, 535, 537, 540, 543, 544 Kahane, J.-P., 240, 244, 544 Kakutani, S., 73, 434, 470, 475, 544 Katznelson, Y., 232, 544 Kaufman, R., 213, 416, 544 Keldysh, M. V., 257, 544 Kellogg, O. D., 50, 62, 66, 97, 476, 544

CY535/Garnett

0 521 47018 8

December 6, 2007

12:30

Author Index

Kenig, C. E., 119, 121, 241, 244, 245, 249, 257, 261, 347, 351, 521, 535, 543, 544 Kilpeläinen, T., 112, 542 Klemes, I., 532 Koebe, P., 18, 19, 23, 25, 26, 30, 36, 88, 249, 250, 252, 264, 276, 286, 288, 293, 296, 301, 313, 326, 353, 391, 395, 423, 545 Koskela, P., 123, 165, 242, 535, 542 Kraetzer, P., 306, 545 Kraus, W., 545 Lévy, P., 545 Lamperti, J., 545 Landau, E., 545 Landkof, N. S., 112, 545 Latfullin, T. G., 541 Lavrentiev, M. A., 202, 213, 222, 246, 249, 250, 257, 418, 537, 544, 545 Lawler, E. L., 367, 545 Lehto, O., 152, 241, 242, 382, 426, 540, 545 Lelong-Ferrand, J., 545 Lenstra, J. K., 367, 545 Lindeberg, J. W., 545 Lindelöf, E., 2, 3, 27, 38, 40, 67, 102, 158, 176, 206, 501, 543, 545 Linnik, Ju. V., 543 Littlewood, J. E., 9, 53, 247, 271, 436, 438, 472, 527, 541, 545 Lohwater, A. J., 224, 545 Lyons, T., 274, 545 Lyubich, M. Yu., 341, 545 MacGregor, T. H., 287, 309, 539 MacManus, P., 257, 546 Maitland, B. J., 546 Makarov, N. G., xiii, xiv, 121, 200, 213, 214, 216, 221, 257, 269, 270, 272, 273, 281, 305–307, 309, 311, 315, 319, 341–343, 505, 513, 534, 537, 544–546, 550 Manning, A., 341, 546 Marcinkiewicz, J., 357, 358, 412, 439, 546 Marshall, D. E., 66, 125, 127, 150, 154, 155, 186, 194, 198, 306, 341, 426, 536, 537, 540, 544, 546, 547 Martio, O., 112, 124, 260, 262, 302, 426, 539–542, 547 Matsumoto, K., 547 Mattila, P., 25, 426, 547 Maxwell, J., 193, 547 McKean, H. P. Jr., 543, 547 McMillan, J. E., xiv, 200, 210, 211, 213, 219, 228, 497, 500, 547, 548

557

Melnikov, M., 426, 547 Metzger, T., 286, 547 Meyer, Y., 257, 537 Milloux, H., 119, 547 Moore, C. N., 274, 530, 532, 533 Mountford, T. S., 547 Muckenhoupt, B., 247, 522, 543, 547 Müller, P. F. X., 544 Näkki, R., 123, 262, 542, 547 Nehari, Z., 382, 541, 547 Nevanlinna, R., 103, 534, 547 Nikol’skii, N. K., 302, 542 Nirenberg, L., 253, 254, 478, 479, 515, 543 O’Neill, M. D., 228, 548 Ohtsuka, M., 150, 153, 156, 446, 547 Oikawa, K., 180, 543 Okikiolu, K., 426, 548 Osgood, B. G., 426, 532, 537, 541, 548, 549 Ostrowski, A., 177, 178, 180, 181, 187, 193, 196, 207, 239, 548 Pajot, H., 426, 429, 548 Petersen, K. E., 537, 548 Pfeffer, W. F., 533 Pﬂuger, A., 164, 307, 548 Phillip, W., 548 Piranian, G., 213, 219, 240, 547, 548 Pitt, L. D., 257, 532 Plessner, A. I., xiv, 200, 203, 205, 207, 357, 417, 497, 502, 548 Pommerenke, Ch., xiii–xv, 34, 70, 168, 200, 216, 221, 236, 238, 253, 257, 258, 263, 267, 268, 270, 286, 287, 305–307, 309, 310, 344, 345, 382, 497, 520, 532, 533, 539, 541, 548, 549 Port, C. S., 547 Prawitz, H., 35, 267, 286, 549 Privalov, I. I., xiv, 52, 53, 200, 203–205, 224, 357, 549 Pruss, A. R., 123, 549 Przytycki, F., 513, 549 Radó, T., 470, 549 Ransford, T., 112, 549 Rayleigh, L. (Strutt, J. W.), 150, 193, 550 Resnik, M. K., 517, 550 Revelt, H., 550 Riesz, F., xiv, 113, 200, 204, 205, 209, 210, 213, 223, 347, 349, 359, 373, 393, 414, 416, 436, 550

CY535/Garnett

558

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Author Index

Riesz, M., xiv, 200, 204, 205, 209, 210, 213, 223, 347, 349, 359, 373, 393, 414, 416, 436, 550 Rinnoy Kan, A. H. G., 545 Rodin, B., 180, 184, 194, 550 Rohde, S., 23, 36, 123, 165, 305, 345, 532, 535, 542, 549, 550 Rossi, H., 227, 543 Salem, R., 269, 544, 550 Sarason, D., 268, 550 Sarvas, J., 262, 547 Sastry, S., 187, 188, 194, 551 Schiffer, M. M., 537, 538 Schober, G., 34, 537, 551 Seidel, W., 205, 224, 532, 545 Semmes, S. W., 257, 362, 373, 426, 427, 429, 537, 538, 551 Shapiro, H. S., 238, 239, 538 Sharpley, R., 66, 533 Shields, A., 227, 238, 239, 535, 538 Shimorin, S., 305, 306, 520, 542, 551 Slater, J. C., 551 Smirnov, S., 268, 341, 342, 534, 544, 546 Smirnov, V., 239, 240, 251, 252, 257, 267, 268, 538, 544 Smith, W., 546 Solynin, A. Y., 123, 534 Spencer, D. C., 357, 359, 412, 551 Stein, E. M., 8, 25, 32, 34, 66, 348, 426, 480, 539, 551 Stout, W., 507, 517, 548, 551 Strutt, J. W., See Rayleigh, L. Sundberg, C., 125, 150, 154, 155, 547 Teichmüller, O., xiv, 152, 157, 167, 173, 183, 382, 532, 540, 545, 551 Thurman, R. E., 228, 548 Thurston, W. P., 426, 551 Torchinsky, A., 247, 551 Triebel, H., 66, 551 Tsuji, M., xiv, 73, 112, 238, 446, 484, 551 Tukia, P., 551

Urba´nski, M., 513, 538, 549 Väisälä, J., 260, 262, 541, 547 de la Vallée Poussin, C., 112, 552 Verdera, J., 426, 547 Virtanen, K. I., 152, 241, 545 Vodopýanov, S. K., 541 Volberg, A. L., 286, 315, 319, 341, 533, 545, 546, 552 Walden, B. L., 426, 552 Warschawski, S. E., 71, 180, 184, 193, 194, 549, 550, 552 Weill, G., 382, 531 Weiss, M., 305, 552 Weitsman, A., 542 Wermer, J., 112, 219, 534, 535, 552 Wheeden, R. L., xiii, 25, 247, 457, 523, 530, 541, 543, 552 Widom, H., 341, 552 Wiener, N., xiv, 73, 89, 90, 93, 97, 112, 113, 117, 476, 552 Wilson, J. M., 537 Wolff, J., 193, 553 Wolff, T. H., xiii, 214, 315, 341, 343, 537, 540, 544, 553 Wu, J.-M. G., 1, 23, 25, 28, 36, 124, 213, 222, 269, 302, 347, 416, 421, 532, 542, 544, 550, 553 Yang, S., 260, 540, 553 Yushkevich, A. A., 538 Zdunik, A., 286, 513, 517, 533, 549, 553 Zeller, K., 227, 535 Zinsmeister, M., 254, 257, 267, 410, 411, 532, 539, 550, 553 Zygmund, A., xiii, 25, 51, 56, 57, 66, 225, 231, 238, 248, 261, 267, 269, 305, 357, 358, 412, 426, 440, 457, 532, 546, 550–553 Øksendal, 119, 120, 125, 213, 416, 553 Øyma, 23, 222, 553

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Symbol Index

Symbol A A∞ , A p A(, ρ) AF(ζ ) αω (z) β(, ζ1 , ζ2 ) β E (z, t) β E (a Q) β 2 (E) βϕ (t) B

Name and Page set of accessible points, 206 weights, 247 ρ-area, 130 area function of F, 348 lower pointwise dimension, 341 β numbers, 290 362 363 Jones square sum, 367 integral means spectrum, 286 the set where ϕ has nontangential limit zero, 207 B(t) universal integral means spectrum, 286 B(t) Brownian path, 477 Bb (t) bounded universal integral means spectrum, 305 B(ζ, δ),Bδ (ζ) disc, 14, 48 B Bloch space, 229 B0 little Bloch space, 258 ||g||B Bloch norm, 229 BMO bounded mean oscillation, 34 BMO(M) bounded mean oscillation, 478 C complex plane C α , C k+α Lipschitz class, 52,55,62 C∗ extended plane, cU central point, 350 Cap(E) logarithmic capacity of a compact set, 74–75 of a Borel set, 85 Cl( f, ζ ) cluster set, 469 ∂ boundary of , δD (z 1 , z 2 ) pseudohyperbolic metric, 23 D unit disc, 5 d (E, F) extremal distance, 130

Symbol ∗ (E, F) d d(z)

Name and Page conjugate extremal distance, 131 distance from z to the boundary, 350 dn (E) diameter of order n, 466 d∞ (E) transﬁnite diameter, 467 reduced extremal distance, 162 δ(z 0 , E) diam diameter dim(E) Hausdorff dimension, 457 dim ω dimension of harmonic measure, 341 dist(A, B) euclidean distance, 17 dist (z 1 , z 2 ) euclidean distance in , 165 ds arc length, 129 |dz| hyperbolic metric, 16 2 1−|z|

D(u) D(u, v) E∗ E( f |N ) F G γ (E) γG (E) α (ζ ) αh (ζ ) βε (ζ, θ ) α (ζ, U ) g (z, w)

G

559

Dirichlet norm, 447 Dirichlet inner product, 447 Laplacian circular projection, 86 conditional expectation, 470 fundamental domain, 345 normal, 417 Fuchsian group, 417 Robin’s constant, 74–75 Green capacity, 111 cone, 6 truncated cone in D, 203 truncated cone in C, 173 cone relative to U , 350 Green’s function with pole at w on a ﬁnitely connected domain, 41 on an arbitrary domain, 75 the set where a conformal map has non-zero angular derivative, 207

CY535/Garnett

0 521 47018 8

560

Symbol G μ (z) h(r ) H H∞ Hp H p (U ) I (ν) Im K = K () K∗ (Q) λ () λ ()−1 λ(z 0 , E) 2 (E) h (E) α (E) |·| f C α M f (ζ ) M M m(λ) mod() M(G) m h (E) Mh (E) μE ∇ ∇2 η (z θ , t) || · || n N (, ε, ρ) ω(z, E, H) ω(z, E, D) ω(z, E, )

ω f (δ)

December 6, 2007

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Symbol Index

Name and Page Green potential on H, 108 on D, 111 measure function, 456 upper half-plane, 1 bounded analytic functions, 435 Hardy space, 435 Hardy space on domain U , 349 energy integral, 79 imaginary part set of cone points, 208 growth rate of number of bad discs, 288 length of the base of Q, 31 extremal length, 130 module, 133 146 two-dimensional Hausdorff measure, 456 h-Hausdorff measure, 457 α-dimensional Hausdorff measure, 457 length or absolute value Lipschitz norm, 52 Hardy–Littlewood maximal function, 9 σ -algebra, 470 group of self maps of D, 16 distribution function, 438 module of a ring domain, 151 reduced modulus, 168 dyadic Hausdorff content, 458 Hausdorff content, 456 equilibrium distribution, on C 1+α Jordan curves, 78 on a compact set, 83 gradient 2nd order gradient, 56 385 norm unit outer normal, 43 number of bad discs, 281 harmonic measure in half-plane, 2–4 harmonic measure in disc, 5 harmonic measure, in a Jordan domain, 13 in a ﬁnitely connected Jordan domain, 39 in an arbitrary domain, 90 modulus of continuity, 70

Symbol d (, E, F) Pf

Name and Page doubled Riemann surface, 445 triple, 484 Perron solution to the Dirichlet problem, 73 Pz (t) Poisson kernel for H, 4 Pz (θ ) Poisson kernel for D, 5 Pz (ζ ) Poisson kernel for z ∈ at ζ ∈ ∂, 45 Pz∗ (θ ) smallest even decreasing majorant of Pz (θ ), 9 Pj ( u) period of u , 443 (α) universal dimension spectra, 341 PV principal value integral, 488 Q box, 31 Q (w1 , w2 ) quasihyperbolic distance, 20 R real numbers, 1 Re real part ρ(z 1 , z 2 ) hyperbolic distance in D, 16 ρ (w1 , w2 ) hyperbolic distance in , 20 ρ metric, 130 ρ-length ρ-length, 130 S(z, w) spherical distance, 258 Sn ( f ) square function, 506 Sϕ(z) Schwarzian derivative, 380 end of the proof S standard strip, 176 suppμ closed support of μ, 96 T (Q) top half of Q, 107 θ (x) 148 Tn() tangent points of , 211 u conjugate function or harmonic conjugate of u, 50 u ∗α (ζ ) nontangential maximal function, 7 uf solution to the Dirichlet problem on H, 3 (Poisson integral, 4) on D, 7 (Poisson integral, 5) on a Jordan domain, 13 on a ﬁnitely connected Jordan domain, 38 Uμ (z) logarithmic potential, 77 VMO vanishing mean oscillation, 268 VMOA analytic and vanishing mean oscillation, 268 (X, M, P) probability space, 470 (W, M, P) Brownian motion, 477 Z∗ Zygmund class, 56 f Z ∗ Zygmund norm, 56 Zn exponential transform, 508 ∅ empty set

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Subject Index

Deﬁnitions are found on boldface page numbers absolutely continuous on lines, 241 accessible point, 206, 218, 559 and nontangential limit, 206 Ahlfors condition, 234, 244, 245 Ahlfors regular, 246, 376, 421–423, 425 and BMO domains, 254, 267 and Hayman–Wu theorem, 421 and Jones square sums, 429 John domain, See Smirnov domain quasicircle, See chord-arc curve Ahlfors’ theorem on quasidiscs, 382 Ahlfors–Beurling theorem, 163, 193, 495 A∞ -condition, See A∞ -equivalent A∞ -equivalent, 246, 247, 265, 525, 527 and reverse Hölder condition, 247 harmonic measure and arc length, 249–253 to arc length, 246, 254 A∞ -weight, 247, 255, 260, 266, 267, 523 almost everywhere equivalent, 208, 357, 360, 412 alternating method, Schwarz, See Schwarz alternating method analytic arc, 42, 44, 447, 486, 487, 489, 513 analytic Jordan curve, 42, 166 and conformal maps, 70 and ﬁnitely connected domain, 42, 43, 68 angular derivative, xiii, xiv, 157, 175, 550, 552 almost everywhere, 256, 347, 398, 399, 414, 497 and cones, 200 and conformality, 175, 207, 225 and extremal distance, 180 and geometric conditions, 185, 197, 198 dx and θ(x) , 184

and nontangential limit of derivative, 175 and rectiﬁable curves, 202, 203 and snowﬂake, 212 and vertical limit, 176 non-zero, 173, 180, 207, 219, 220, 238, 239 nonexistence, 239 positive, 180, 181, 183, 184 problem, 194, 536, 537, 540, 551 angular derivative at +∞, 185, 186, 187, 192, 193 and geometric condition, 186 annulus construction, 337, 338 annulus example, 132, 179, 488, 490 annulus, slit, See slit annulus A p -weight, xiv, 247, 252–254, 266 and reverse Hölder condition, 247 area, 130, 191, 559 area function, xiv, 347, 348, 351, 418, 426, 559 and H p , 400 and Brownian motion, 532 and Hardy space, 348, 521 and nontangential limits, 357, 521, 533, 541 and second derivative, 361 and the nontangential maximal function, 523 area theorem, 19, 169, 382, 548 argument principle, extended, 488, 491 asymptotic value, 157, 158 and growth, 158 asymptotically conformal, 258 asymptotically smooth, 268 B, See Bloch space B0 , See little Bloch space bad box, 401, 404

561

CY535/Garnett

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562

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Subject Index

bad discs, number of, 281, 282, 284, 285 bad square, 394, 397 Baernstein’s example, 25, 269, 302 Balogh–Bonk theorem, 157, 166, 193 barrier, 94 strong, See strong barrier base, 31, 32, 34, 377, 408, 560 basin of attraction, 114, 512, 513, 534, 549 β numbers, 290, 290–306, 313 and Brennan’s conjecture, 293 and number of bad discs, 292, 293 and second derivative, 291 and universal integral means spectrum, 292 Beurling metric, 160, 187 Beurling shove theorem, 123, 534, 538, 549 Beurling’s criterion for extremal metrics, 152 Beurling’s projection theorem, 105, 108, 281, 416, 534 and compression, 213 and harmonic measure of a ball, 107, 118 Beurling’s theorem on regular points, 100 Beurling–Ahlfors condition, 243, 244 bilipschitz, 255, 256, 259, 261, 314 1

Bishop–Jones H 2 −η theorem, 347, 397 Bishop–Jones, theorems of, xiii, 385, 392, 394 Bloch function, 229, 257, 344, 432, 544, 548 and BMO, 253 and boundary behavior, 230, 550 and gap series, 230, 232, 269 and nontangential limits, 231 and normality, 230, 532 and quasicircles, 235 and the law of the iterated logarithm, 269, 270, 503, 505, 549 and univalence, xiv, 232–241, 532, 548 Lipschitz in hyperbolic metric, 230 Bloch martingale, 504, 505, 506, 511 and bounded jumps, 504 Bloch norm, 229, 512, 559 Bloch space, 229, 559 BMO, See bounded mean oscillation BMO domain, 254, 253–257, 267, 347, 410 and Carleson measures, 410 and chord-arc domain, 255, 410 and H ∞ (), 530 boundary Harnack inequality, 28, 553 boundary smoothness, 49, 59–65 boundary value problems, mixed, See mixed Dirichlet–Neumann problem bounded mean oscillation, 34, 479, 480, 532, 537, 542–544, 559

and A∞ , 254, 267 and Carleson measures, 354 dyadic, 478, 479 bounded universal integral means spectrum, 305, 310, 342, 559 box, Carleson, See Carleson box Brennan’s conjecture, xiii, xiv, 269, 286, 313, 533–535, 537, 548 and Carleson–Jones’ conjecture, 306 and integral means spectrum, 286, 306, 342 and known p, 305, 518, 520 and Kraetzer’s conjecture, 306 and number of bad discs, 288, 293 and polygonal trees, 293 and quasiconformal maps, 314, 512, 535 and the Hayman–Wu theorem, 302 and universal integral means spectrum, 287 for close-to-convex functions, 309 for starlike functions, 309 Brownian motion, 425, 477, 532, 534–536, 544, 548, 553 and angular derivatives, 194 and harmonic measure, 475, 477 and martingales, 476 and the Dirichlet problem, 73, 477 and the Hayman–Wu theorem, 433, 434 Brownian path, 433, 476, 477, 543, 559 Calderón’s theorem, 348, 351, 523, 551 C α , 0 < α < 1, 52, 559 and Poisson integral, 53 C k+α curve, 62, 65 C k+α function, 55, 62, 65, 559 and distance to the boundary, 70 and scale of spaces, 57, 66 Cantor set, 87, 88, 212, 315, 427, 462, 463 and dim ω, 316 and capacity, 88, 100 and harmonic measure, xiii, 533, 537, 545, 552 and Hausdorff dimension, 316, 462, 464 and regular points, 100 and regularity, 100 conformal, 342 square, 315, 341–343, 465, 466 capacity and circular projection, 86 and contractions, 86 and estimate of harmonic measure, 104 and extremal distance, 494, 497 and harmonic measure, 73, 78

CY535/Garnett

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Subject Index

and Hausdorff content, 461 and Hausdorff measure, 460, 461, 466 and length, 76, 166 and logarithmic potential, 78 and reduced extremal distance, 163 and regular points, 97, 98 and Robin’s constant, 74 and subadditivity, 101, 116 and transﬁnite diameter, 466, 467 and uniformly perfect sets, 119, 343 and Wiener series, 97 Green, See Green capacity logarithmic, 559 of a Borel set, 85 of a compact set, 75, 163 of a disc, 74, 337 of a union of Jordan curves, 74 of an interval, 76 of set of irregular points, 97 zero, See polar set Carathéodory’s theorem, 1, 13, 29, 39, 43, 178, 200, 201 and the Dirichlet problem, 13, 51, 66 Carleman’s differential inequality, 481, 482, 536 Carleman’s estimate, 150, 480, 497 Carleman’s method, 193, 480–484 Carleson box, 560 dyadic, See dyadic box on D, 31, 32, 33, 249, 412, 416 on H, 32, 377 Carleson measure and BMO, 480 and BMO domain, 410, 411 on D, 31, 32, 33, 354, 355, 440 on H, 32, 107 Carleson, conjecture of, 220 Carleson, lemma of, 277 Carleson, theorems of, xiii, 315, 324, 341 Carleson–Makarov, theorems of, xiii, xiv, 269, 282, 292, 293 central point, 350, 559 chord-arc curve, 246, 374, 414, 537 and L 2 estimates, 536 and A∞ , 246–253, 260 and Calderón’s theorem, 351, 520, 521 and F. and M. Riesz theorem, 416 and harmonic measure, 246, 251, 253 and H ∞ (), 530 and Lipschitz curve, 246 and Lipschitz decomposition, 373

563

and quasicircle, 251, 253 and quasiconformal mapping, 551 and quasidisc, 261 and Smirnov domain, 251, 252 and the Cauchy integral, 551 ﬁrst appearance, 257 chord-arc domain, 254, 520 and BMO domain, 255, 257, 410 circuit, 428 and electrical resistance, 138 circular projection, 86, 559 and capacity, 89 and harmonic measure, 105, 108 and Hayman–Wu theorem, 24 close-to-convex, 309 cluster set, 469, 559 completely invariant, 114 compress, 187, 212, 213, 214, 305 conditional expectation, 470, 471, 559 cone and angular derivative, 200 and chord-arc domain, 406, 521 and Hayman–Wu theorem, 36 and Lindelöf’s theorem, 67 and tangent, 175 and tents, 215 in D, 6, 7, 8, 50, 559 in , 47, 48, 71 truncated, See truncated cone cone domain, 204 and Calderón’s theorem, 348 special, See special cone domain cone domain construction, 200, 204, 205, 359 and angular derivative, 413, 414 and Bishop–Jones theorem, 398 and BMO domain, 256 and local F. and M. Riesz theorem, 418, 426 cone point, 207, 412, 497 and harmonic measure, 227 and tangent, 208 preimage, 497 set of, 208, 217, 220, 262, 398, 417 two-sided, 372 cone relative to U , 350, 559 conformal at a boundary point, 174, 175, 176, 207, 536 and geometric condition, 177 and inner tangents, 177 of D, 203, 204, 207, 211, 225, 239 of H, 177, 178, 180 conformal Cantor set, 342

CY535/Garnett

564

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Subject Index

conjugate extremal distance, 131, 139, 559 a metric for, 159 and equipotential lines, 138 and harmonic measure, 159 and QED domain, 260 and reciprocal of extremal distance, 140, 150, 485 and sum of heights of rectangles, 491–493 and the parallel rule, 137 for a rectangle, 132, 139 for an annulus, 133, 142 lower bound for, 187 conjugate function, 37, 50, 50–59, 439, 541, 543, 560, see also harmonic conjugate and BMO, 480 and conformal map, 60 and exponential integrability, 51 and Herglotz integral, 64 and no angular derivative, 207 and smoothness, 53 and Zygmund class, 57 and Zygmund’s theorem, 441 single valued, 442, 443 continuous tangent, 60, 61, 62, 71 covering lemma, 11 covering lemma, Vitali, See Vitali covering lemma critical level lines, 487, 492 critical point of Green’s function, 113, 325, 327, 329, 335, 336, 341, 544 of harmonic measure, 71, 486, 486–493, 495, 497 crosscut, 29, 161, 223, 262 and parallel rule, 137 and serial rule, 136 vertical, 179, 183 cubes, 325, 326, 327, 329 curve, 129, 131, 135, 139 d-adic subintervals, 514 dandelion, 296, 297, 300, 302, 303 decomposition into Lipschitz domains, 373, 378 Denjoy problem Ahlfors’ solution, 157, 193 Beurling’s solution, 193, 194 Carleman’s solution, 193 diameter of order n, 466, 559 dichotomy theorem, 512 dilatation, 241, 249, 251, 521, 523

dimension, See Hausdorff dimension Dini continuous, 70, 71 Dirichlet inner product, 447, 559 Dirichlet integral, 447, 450, 537, 546 of harmonic measure, 481, 482, 491, 492 Dirichlet norm, 447, 449, 450, 454, 496, 559 Dirichlet principle, 66, 448, 449, 450, 452, 455, 537 and extremal distance, 485 Dirichlet problem, 3, 37, 73, 441, 446 and Brownian motion, 73, 477 and Carathéodory’s theorem, 13 and mixed boundary problems, 445 and Perron solution, xiv, 73, 560 numerical solution, 66 on a doubled Riemann surface, 446 on a Jordan domain, 13 solution, 448, 453, 454, 533, 560 solution and Green’s function, 92 solution on D, 5, 7, 452 solution on H, 3 solution on a doubled Riemann surface, 142, 446 solution on a ﬁnitely connected Jordan domain, 37, 40, 41, 66 solution on a Jordan domain, 38, 59, 64, 442 Wiener solution, xiv, 89, 89–97, 117, 476, 552 disc construction, 336, 337, 338 distortion theorem, 22, 386, 392, 500 and bad discs, 288, 301, 302 and Brennan’s conjecture, 286 and growth of univalent functions, 35 and integral means spectrum, 286, 519 distribution function, 438, 560 doubled Riemann surface, 150, 445, 495, 560 and extremal distance, 142, 484, 486 and harmonic measure, 455, 490 and mixed boundary value problems, 441, 446 and reduced extremal distance, 496 doubling measure, 244, 245, 261, 265 and A∞ -equivalence, 246, 247, 265 dyadic BMO, See bounded mean oscillation dyadic box, 32, 107–111, 125, 373, 401–409, 458, see also dyadic square in H, 431 dyadic Hausdorff content, 458, 560 dyadic interval, 265, 266, 478, 503, 505, 510, 511 dyadic martingale, 503, 504–506, 508, 510

CY535/Garnett

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Subject Index

dyadic square, 336, 337, 458 and Frostman’s theorem, 459, 460 and Hausdorff content, 458 and Jones square sums, See Jones square sums energy integral, 73, 79, 82, 560 and polar sets, 85 and positive deﬁnite, 79 energy, ﬁnite, 79 equilibrium distribution, 91, 95, 96, 98, 99, 101, 103, 105, 109, 115–117, 195, 317, 333, 464, 468, 469, 560 and Cantor set, 88 and Green’s function, 85 and harmonic measure, 73, 78 and uniqueness, 83 on a ﬁnite union of smooth curves, 78 on a set of positive capacity, 83 equilibrium potential, 83, see also potential Erdös-Gillis theorem, 461 euclidean distance, 17, 559 euclidean distance in , 165 euclidean length, 129, 161 Evans function, 469, 470 and polar set, 112, 468 Evans’ theorem, 468, 476 expand, 187, 212, 213, 214 exponential transform, 508, 509, 560 extended argument principle, 488, 491 extension rule, 134, 135, 136, 151, 154, 162, 165, 171, 179, 182, 278, 298, 299 and electrical resistence, 138 extremal distance, 129, 130, 139, 140, 142, 150, 162, 164, 559 and angular derivative, 157, 180 and capacity, 494 and distance from a point to an arc, 145 and electrical resistance, 138 and harmonic measure, 129, 143, 146, 159 and quasiconformal maps, 243 and shortest curves, 486, 491 and slit annulus, 142, 299 and slit rectangle, 129, 140 examples of, 491, 492 for a rectangle, 131, 132 for an annulus, 132, 142 in a ﬁnitely connected domain, 484, 485, 491–493, 495 lower bounds for, 146, 189–191, 494 to a set on ∂D, 164

565

upper bounds for, 187 extremal distance domain, M-, 260, 540, 541, 553 extremal length, xiv, 130, 149, 560, see also extremal distance and conjugate extremal distance and angular derivative, 180 and annulus example, 133 and β numbers, 289 and conformal invariance, 130, 132, 139, 150 and length–area method, 16 and moduli, 130 and path family, 135 and rectangle example, 131 and Teichmüller’s Modulsatz, 193 precursor, 128, 193 extremal metric, 130, 153, 543 and Beurling’s criterion, 152, 153 and critical points, 491 and Dirichlet principle, 485 and ﬁeld lines, 138 and gradient, 147, 195, 485, 486, 491 and magnitude of electric ﬁeld, 138 and modulus of derivative, 130, 132, 140, 142, 150 and nonexistence, 153 and slit annulus, 142 and slit rectangle, 139 and uniqueness, 133, 134, 153 for a rectangle, 132 for a ring domain, 133 for an annulus, 133 Fatou set, 114 Fatou’s theorem, 1, 6, 7, 8, 12, 13, 48, 204, 437, 473 and angular derivative, 202 ﬁnitely connected Jordan domain, 37, 37–72 free boundary, 450 Frostman’s theorem, 431, 458 fundamental domain, 345, 541, 559 normal, 417, 559 Gehring–Hayman theorem, 122, 166, 293 geodesic, hyperbolic, See hyperbolic geodesic good λ, 524, 530 good box, 401 greedy algorithm, 367, 428, 429 Green capacity, 111, 123 Green energy, 111 Green potential, 105, 106, 123, 124, 317, 560

CY535/Garnett

566

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Subject Index

and Hall’s lemma, 125 and Wiener series, 112, 117 on D, 111 on H, 108 Green’s function, 41, 46, 66, 75, 114, 117, 163, 317, 334, 339, 495, 548, 559 and capacity, 74 and conformal map to D, 41, 68 and equilibrium potential, 85 and harmonic measure, 37, 45, 65 and periods of harmonic conjugate, 69 and reduced extremal distance, 164 and regular points, 95 and Robin’s constant, 163 and symmetry, 43, 78 and the Neumann kernel, 447 and uniformly perfect sets, 345 and uniqueness, 78 on D, 105, 123 on H, 108 on a ﬁnitely connected domain, 41 on doubled Riemann surface, 495 on non-Jordan domains, 68 growth theorem, 22, 300 H ∞ , H p , H p (U ), See Hardy space Hall’s lemma, 108, 124, 125, 542 radial version, 125 Hardy identity, 270 Hardy space H p , xiii, 35, 200, 427, 435, 436, 440, 537–539, 560 H p (U ), 240, 349, 521, 522, 544, 560 H ∞ , 427, 435, 560 and area function, 347, 348, 521 and boundary values non-zero a.e., 436 and Carleson measures, 440 and derivative of conformal map, 71 and integrability of log | f |, 435 and nontangential limit, 435 and the nontangential maximal function, 436, 521 and univalent functions, 35, 309, 400 on a chord-arc domain, 521 Hardy–Littlewood maximal function, 8, 9, 25, 49, 438, 472, 537, 547 and nontangential maximal function, 9 weak-type L 1 estimate, 11, 12, 49 Hardy–Littlewood maximal theorem, 271, 436, 438, 477 harmonic conjugate, 50, 56, 143, 163, 443, 484

and periods, 443 harmonic measure, 39, 90 absolute, 103 and A p -weight, 248 and arc length, 47, 65, 249, 250, 252, 253 and Cantor sets, See Cantor set and capacity, 78, 104 and chord-arc curves, 246 and circular projection, 105 and cone points, 227 and extremal distance, 129, 143, 146, 159 and Hausdorff dimension, 214, 315, 316, 324, 332, 341–343, 346, 535, 544, 549 and Hausdorff measure, 538, 546, 549, 550, 553 and 1 , xiv, 201, 202, 208, 212, 217, 218, 221, 416, 512 and 2 , 119 and α , xiii, 213, 214, 305, 341 and h , 213, 272, 273, 275, 343, 512 and logarithmic potential, 78 and radial projection, 125 critical points, 71 for a Jordan domain, 13 lower bounds for, 104, 105, 108, 118, 119, 123–125, 127, 143, 145, 151, 154, 155, 159–161, 164 of a small set, 91 of irregular points, 97 of set of accessible points, 206 on D, 5 on H, 2, 4 upper bounds for, 104, 106, 108, 123, 143, 145, 146, 148, 149, 151, 154, 155, 161, 164, 222, 224, 281, 480, 483, 484, 493 Harnack’s inequality, 27, 40, 90, 318–321, 328, 331, 406, 407, 424, 482 boundary, 28, 553 on H, 5 Harnack’s principle, 27, 39, 40, 75 Hausdorff content, 340, 456, 457, 560 and capacity, 461 Hausdorff dimension, 206, 457, 458, 534, 535, 542, 559 Hausdorff measure, α-dimensional, 315, 316, 457, 560 Hausdorff measure, h-, xiii, xiv, 269, 272, 457, 560 Hausdorff measure, two-dimensional, 456, 560 Hayman–Wu theorem, 1, 23, 124, 532 and L p version, 269, 302

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Subject Index

and Ahlfors regular sets, 347, 421 and best constant, 36, 222, 553 Herglotz integral, 52, 56, 57, 64, 154 and smoothness, 55 hyperbolic distance, 20, 21, 257, 397, 560 in D, 16, 230, 405, 407 in a plane domain, 344 hyperbolic geodesic, 17, 264, 280 in D, 17, 230, 276, 293, 404, 405, 407 hyperbolic metric, 1, 559 and Lipschitz functions, 30 and the universal cover, 344 on D, 16, 230, 314, 512 inner normal, 173, 174 vertical, 174, 176–178, 180, 184, 185, 196 inner tangent, 173, 174–178, 180, 184, 185, 196, 203, 208, 211, 212, 225 and conformality, 175, 177 and tangent, 203 for snowﬂake, 239 integral means spectrum, 286, 306, 534, 542, 549, 559 bounded universal, 305, 310, 342, 559 for Koebe function, 286 universal, 286, 545, 559 bound for, 519 upper dx θ(x) , 129, 150, 160 and angular derivative, 184 and extremal distance, 148, 160, 161, 184, 185, 197 and harmonic measure, 148, 160, 161, 494 interior angle, 488, 489–491 irregular point, 93, 97, 476 iterated logarithm, law of, See law of the iterated logarithm Jenkins–Oikawa–Rodin–Warschawski theorem, 180 Jerison–Kenig theorem, 347, 351, 521 John domain, 245, 262, 263–265, 540 Ahlfors regular, See Smirnov domain and quasicircle, 263 John–Nirenberg theorem, 253, 254, 478, 479, 515 Jones square sums, xiv, 378–380, 419, 420, 427, 428, 430, 559 and Ahlfors regular, 429 and rectiﬁability, xiii, 347, 361–372 and rectiﬁable sets, 430 and Schwarzian derivative, 347, 384, 391, 392, 419

567

Jones–Wolff, theorems of, xiii, 315, 324, 332 Jordan domain, 13, 1–36 ﬁnitely connected, 37 Julia set, 114, 306, 341, 534, 546 Kakutani’s theorem, 73, 434, 475 Kellogg, theorems of, 50, 62, 66, 97, 476, 552 Koebe function, 18, 19, 23, 30, 36, 286, 313 Koebe one-quarter theorem, 18, 88, 264, 276, 326, 395 Koebe’s estimate, 19, 25, 252, 264, 288, 293, 296, 301 invariant form, 20 Kraetzer’s conjecture, 306 L-function, 272, 279 large box, 401 Lavrentiev curve, See chord-arc curve Lavrentiev’s estimate, 213, 222, 250 Lavrentiev’s theorem, 249, 418 law of the iterated logarithm converse of Makarov’s, 270 for harmonic functions, 533 for Kleinian groups, 535 for lacunary series, 552 for stationary processes, 550 Makarov’s, 214, 269, 505, 545, 546, 549 martingale, 505, 506, 507, 517 length–area principle, 16, 149, 156, 193, 490 level curve of ω, 487, 490, 495 Lévy’s theorem, 357, 506 Lindelöf’s maximum principle, 2, 27, 38, 102 Lipschitz class, 52, 559 and scale of spaces, 57, 66 Lipschitz constant, 246 and A p -weight, 248 Lipschitz curve, 234, 246, 248 and chord-arc curve, 246 Lipschitz domain, 72, 351, 410, 412, 432, 544, 553 and Green’s theorem, 72 Lipschitz domain, M-, 372, 373, 377–379 Lipschitz function, 56, 198, 208, 368, 370, 371 and conjugate function, 71 Lipschitz graph, 72, 185, 186, 209, 368, 537 Lipschitz norm, 52, 560 little Bloch space, 258, 550, 559 locally connected, 29, 263 logarithmic capacity, See capacity logarithmic potential, See potential logarithmico-exponential function, 272 lower pointwise dimension, 341, 559

CY535/Garnett

568

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Subject Index

Makarov, theorems of, xiii, xiv, 200, 213, 214, 216, 269, 270, 272, 273, 305, 307, 309, 311, 342, 343, 505, 513 Makarov–Volberg theorem, 315, 316 martingale, xiii, 471, 473, 474, 476, 477, 503, 505, 506, 508–511, 536, 540 and a.e. convergence, 357, 473, 506 and Brownian motion, 476, 477 and law of the iterated logarithm, 272 and maximal function, 472 and uniform integrability, 471 and weak convergence, 471 martingale theorem, 473, 475, 506 Doob’s, 472 maximal box, 374, 401 maximum principle, Lindelöf’s, See Lindelöf’s maximum principle maximum principle, strong form, 102, 103 McMillan, theorems of, xiv, 200, 208, 210, 211, 228, 497, 500 measure function, 213, 272, 279, 340, 456, 457, 458, 460, 461, 463, 465, 466, 560 measure preserving transformation, 115 metric, 130 mixed Dirichlet–Neumann problem, 150, 485 and Dirichlet principle, 450, 452, 455 and Riemann surface double, 441, 445, 446 Möbius transformation, 90, 261, 291, 347, 380, 381, 383, 393, 397, 405, 432 and Schwarzian derivative, 383 of D, 16, 123, 124, 276, 412, 480 module of a curve family, 130, 133, 560 of a ring domain, 133, 151, 243, 245, 261, 424, 425 modulus, See module modulus of continuity, 70, 560 modulus, reduced, See reduced modulus molliﬁcation argument, 80 Neumann problem, 441 and mixed boundary problems, 445 solution to, 441 nontangential limit, 7, 174 and accessible point, 206 and angular deriviative, 175 and area function, 357, 533, 541 and boundary differentiability, 28, 232 and conformal at a boundary point, 174 and harmonic measure, 12 and non-Plessner points, 204, 205

and nontangentially bounded, 204 and Poisson integral, 12, 28, 29 and second derivative, 360 of a conjugate function, 50 of a univalent function, 127, 206 of an H p function, 347, 435, 440, 522 of bounded harmonic function, 47 of the Poisson integral of a measure, 7 nontangential maximal function, 7, 8, 25, 256, 436, 521, 533, 537, 541, 560 and Hardy–Littlewood maximal function, 9 and the area function, 523 weak-type L 1 estimate, 7, 441 nontangentially accessible domains, 245 nontangentially bounded, 203, 204, 224 nontangentially dense a.e., 214, 216, 218, 227, 275, 343 normal function, 230, 532 normal fundamental domain, 417, 559 normalized univalent function, 18, 35, 76, 88, 165, 233, 313, 500 and length of radius image, 166 number of bad discs and β numbers, 292 and universal integral means spectrum, 288, 292 Ostrowski’s theorem, 177, 178, 180, 181, 187, 193, 196, 207, 239 p.p., 114 parallel rule, 136, 137, 172, 193, 278, 491 and electrical resistance, 138 equality in, 151 path, 278, 428 path family, 129, 133–137, 150, 152, 242, 278, 289 period, 69, 443, 560 Pﬂuger’s theorem, 164, 307 Pick’s theorem, 25, 30 piecewise continuous, 37, 40, 447, 449, 488 piecewise smooth, 447, 448–450, 452, 454, 455, 486 Plessner point, 205, 497, 502 Plessner’s theorem, xiv, 200, 205, 207, 357, 417 Poisson integral, 47 growth of gradient, 53 on D, 5, 7, 9, 12, 26–29, 31, 34 on H, 4, 26 Poisson kernel, 37, 45

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Subject Index

and box kernel, 479 on D, 5, 28 on H, 4 polar set, 85, 104, 111, 114, 123, 127, 206 and Dirichlet problem, 91 and Evans function, 468 and harmonic functions, 102 and harmonic measure, 103 and subharmonic functions, 104 and the energy integral, 85 polygonal tree, 291, 292, 296, 297, 299, 303, 312 and Brennan’s conjecture, 293 and number of bad discs, 293 Pommerenke’s theorem, xiv, 217 potential, 77, 96, 165, 195, 560 and capacity, 78 and harmonic measure, 78 and Robin’s constant, 85, 95 Prawitz’s theorem, 35, 267, 286 predictable square function, 506 principal value integral, 488, 560 Privalov’s theorem, xiv, 200, 204, 205, 224, 357 probability space, 470, 478, 503, 560 Przytycki–Urba´nski-Zdunik theorem, 513 pseudohyperbolic metric, 23, 559 QED domain, See extremal distance domain quasicircle, xiv, 234, 235, 241, 245, 252, 257, 259, 260, 261, 265, 394, 521, 533, 539, 541 Ahlfors regular, See chord-arc curve and A∞ -weights, 267 and bilipschitz map, 261 and Bloch functions, 235, 236 and BMO, 255, 256 and chord-arc curve, 251, 253 and fundamental domain, 345 and harmonic measure, 241, 244 and John domain, 245, 263 and quasiconformal mappings, 243 and Schwarzian derivative, 347, 381, 382, 394 and welding map, 550 and Zygmund function, 261, 262 seventeen characterizations, 257, 532 quasiconformal extension, 244, 245, 249, 251, 252, 259, 261, 521, 523, 531, 541 and Schwarzian derivative, 382 quasiconformal mapping, 241, 242, 244, 245, 256, 259–261, 314, 512, 532, 534, 535,

569

538, 542, 551 and Ahlfors condition, 244 and Beurling–Ahlfors condition, 243 and extremal distance, 242, 243 and Hausdorff dimension, 541 and images of large circles, 243 and images of small circles, 242 and inverse map, 242 and module of a ring domain, 243 and quasicircles, 243 quasiconformal reﬂection, 256, 553 quasidisc, 235, 239, 245, 260, 261, 267, 345, 397, 414, 432, 433 and Ahlfors condition, 396 and QED domain, 260 and quasiconformal extensions, 244 and Schwarzian derivative, 382, 394 quasihyperbolic distance, 20, 560 quasisymmetric function, 243, 511, 552 RB-domain, 513 rectangle example, 131, 133, 139, 143, 145, 147, 150, 182, 183, 196, 198, 486, 488, 490–493, 495, 496 and conjugate extremal distance, 132 rectangle, slit, See slit rectangle reduced extremal distance, 157, 162, 163, 164, 496, 559 and conformal invariance, 163 and euclidean distance in , 165 and Green’s function, 495 to a set on ∂D, 163 reduced modulus, 168, 560 regular point, 93, 94–100, 115–118, 476 and barriers, 93 and boundary in a neighborhood, 95 and capacity, 73 and circular projection, 100 and Green’s function, 95 and sufﬁcient conditions, 100, 101 repelling boundary domain, See RB-domain reverse Hölder condition, 247, 266 and A∞ -equivalent, 247 and A p -weight, 247 reverse Jensen’s inequality, 253 ρ-length, 130, 138, 560 Riemann surface, from slit rectangles, 485 Riesz, F. and M., theorem, xiv, 200, 202, 204, 205, 209, 210, 213, 223, 347, 349, 359, 373, 393, 414 Riesz, F. and M., theorem, local, 416

CY535/Garnett

570

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Subject Index

ring domain, 133, 142, 151, 152, 167, 243, 324 Robin’s constant, 91, 113, 327, 335, 559 and capacity, 74, 75 and extremal distance, 494 and reduced extremal distance, 163 and regular points, 95 and the energy integral, 83 and the equilibrium potential, 78, 85 for a compact set, 75, 163 for a union of Jordan curves, 74 Sastry’s lemma, 187, 188 Schwarz alternating method, 37, 40, 446 Schwarzian derivative, xiv, 347, 380, 426, 518, 535, 540, 547, 548, 551, 560 and H p space of derivative, 398 and angular derivatives, 413 and Bloch functions, 383 and BMO domain, 410 and Carleson measure, 410 and geometric estimate, 385, 392 and Möbius transformations, 381, 383 and norm, 382 and quasicircles, 381 and rectiﬁable curves, 394 and rectiﬁable quasicircles, 394 and univalence, 381, 382 invariant, 384 separating curve, 30, 131, 151, 156, 161, 167, 183, 219, 228, 258, 276–278, 282, 283, 293, 318, 334, 339, 372, 491, 493, 496, 501 and conjugate extremal distance, 131, 139 and extremal distance, 146, 148, 149 separating ring domain, 151, 152, 245 serial rule, 135, 136, 137, 145, 147, 150, 162, 167, 283 and electrical resistence, 138 approximate equality in, 153, 167, 168, 170, 183, 187, 191 equality in, 150 singularity, 455, 488, 489–491 slit annulus, 142, 299 slit rectangle, 129, 139, 140, 146, 155, 480, 484, 485, 493, 496 Smirnov domain, 240, 251, 267, 268, 544 and chord-arc curve, 251, 252 non-, 239, 240, 257, 267, 268, 538 spanning tree, 367, 428, 429 special cone domain, 349, 350, 353, 433, 526, 529 and Hardy space, 351

and Möbius transformations, 432 geometric properties, 351, 358 spherical distance, 258, 560 square function, 505, 506, 560 and exponential integrability, 509 standard strip, 176, 181, 560 starlike, 309 stochastically independent, 476 stopped martingale, 506 stopping time, 247, 325, 473, 478, 506, 508, 510 and A∞ -equivalence, 247 strip domain, 146, 147, 179, 551 strong barrier, 345, 532, 539 strongly mixing, 115 symmetry rule, 137, 151, 171, 183, 196, 495, 497 tangent, 60, 61, 427 and existence a.e., 220 Teichmüller’s Modulsatz, xiv, 152, 157, 167, 183 three circles theorem, 67 tour, 428, 429 transﬁnite diameter, 112, 466, 467, 538, 559 and capacity, 467 traveling salesman problem, 367, 429, 544 traveling salesman theorem, 367 triple, 484, 485, 560 truncated cone in C, 173, 174, 176, 212, 239, 268, 559 in D, 203, 559 twist point, 211, 218, 497, 502 twist point theorem, 211, 218, 225, 497 type 0 box, 374, 375, 377 type 1 box, 375, 376 type 2 box, 375, 376 uniformly perfect, 119, 343, 344, 345, 549 thirteen equivalent conditions, 343–345 univalent function, 18 and Bloch space, 232 and no non-zero angular derivative, 238 universal dimension spectra, 341, 560 universal integral means spectrum, 286, 545, 559 β numbers, 292 and number of bad discs, 292 universal integral means spectrum, bounded, 305, 310, 342, 559 vanishing mean oscillation, 268, 550, 560

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Subject Index

vertical inner normal, See inner normal vertical projection, 86 Vitali cover, 28 Vitali covering lemma, 25, 28, 216, 217, 275 VMO, See vanishing mean oscillation VMOA, 268, 548, 560 von Koch snowﬂake, 211, 212, 227, 235, 239, 431 and Hausdorff dimension, 212 weak-L 1 estimate for maximal function, 472 weak-type 1-1, 7, 11, 12, 49, 441 welding map, 245, 259, 261, 267, 534, 551 and quasicircles, 550 Whitney decomposition, 524, 527 Whitney square, 21, 22, 25, 31, 394, 418, 420, 425 Wiener series, 73, 97, 115 Wiener’s solution to the Dirichlet problem, xiv, 89, 89–97, 117, 552 Wirtinger’s inequality, 480, 481

zipper algorithm, 66, 306, 546 Z ∗ , See Zygmund class Zygmund class, 56, 57, 231, 560 and growth of gradient, 57 and scale of spaces, 57, 66 Zygmund function, 56, 241 and A∞ -weight, 267 and Bloch space, 231 and conjugate function, 57 and nowhere differentiability, 240 and quasicircles, 261, 262 and tangents, 225 and univalent functions, 238 example of, 225 Zygmund norm, 56, 57, 58, 231, 240 Zygmund’s theorem, 51, 248, 441 Øksendal conjecture on ω and 1 , 416 on ω and α , α > 1, 213

571