A Primer on the Dirichlet Space

C A M B R I D G E T R AC T S I N M AT H E M AT I C S General Editors ´ S , W. F U LT O N , A . K AT O K , F. K I RWA N , B. BOLLOBA P. S A R NA K , B . S I M O N , B . T OTA RO

203 A Primer on the Dirichlet Space

C A M B R I D G E T R AC T S I N M AT H E M AT I C S GENERAL EDITORS ´ W. FULTON, A. KATOK, F. KIRWAN, P. SARNAK, B. BOLLOBAS, B. SIMON, B. TOTARO A complete list of books in the series can be found at www.cambridge.org/mathematics. Recent titles include the following: 169. Quantum Stochastic Processes and Noncommutative Geometry. By K. B. S INHA and D. GOSWAMI ˜ 170. Polynomials and Vanishing Cycles. By M. TIB AR 171. Orbifolds and Stringy Topology. By A. ADEM , J. LEIDA, and Y. RUAN 172. Rigid Cohomology. By B. LE STUM 173. Enumeration of Finite Groups. By S. R. BLACKBURN, P. M. NEUMANN, and G. VENKATARAMAN 174. Forcing Idealized. By J. ZAPLETAL 175. The Large Sieve and its Applications. By E. KOWALSKI 176. The Monster Group and Majorana Involutions. By A. A. IVANOV 177. A Higher-Dimensional Sieve Method. By H. G. DIAMOND , H. HALBERSTAM, and W. F. GALWAY 178. Analysis in Positive Characteristic. By A. N. KOCHUBEI ´ MATHERON 179. Dynamics of Linear Operators. By F. BAYART and E. 180. Synthetic Geometry of Manifolds. By A. KOCK 181. Totally Positive Matrices. By A. PINKUS 182. Nonlinear Markov Processes and Kinetic Equations. By V. N. KOLOKOLTSOV 183. Period Domains over Finite and p-adic Fields. By J.-F. DAT, S. ORLIK, and M. RAPOPORT ´ 184. Algebraic Theories. By J. A D AMEK , J. ROSICK Y´ , and E. M. VITALE 185. Rigidity in Higher Rank Abelian Group Actions I: Introduction and Cocycle Problem. By A. KATOK and V. NIT¸ IC A˜ 186. Dimensions, Embeddings, and Attractors. By J. C. ROBINSON 187. Convexity: An Analytic Viewpoint. By B. SIMON 188. Modern Approaches to the Invariant Subspace Problem. By I. CHALENDAR and J. R. PARTINGTON 189. Nonlinear Perron–Frobenius Theory. By B. LEMMENS and R. N USSBAUM 190. Jordan Structures in Geometry and Analysis. By C.-H. CHU 191. Malliavin Calculus for L´evy Processes and Infinite-Dimensional Brownian Motion. By H. OSSWALD 192. Normal Approximations with Malliavin Calculus. By I. NOURDIN and G. P ECCATI 193. Distribution Modulo One and Diophantine Approximation. By Y. BUGEAUD 194. Mathematics of Two-Dimensional Turbulence. By S. KUKSIN and A. SHIRIKYAN 195. A Universal Construction for Groups Acting Freely on Real Trees. By I. CHISWELL and ¨ T. MULLER 196. The Theory of Hardy’s Z-Function. By A. IVI C´ 197. Induced Representations of Locally Compact Groups. By E. KANIUTH and K. F. TAYLOR 198. Topics in Critical Point Theory. By K. PERERA and M. SCHECHTER 199. Combinatorics of Minuscule Representations. By R. M. GREEN ´ 200. Singularities of the Minimal Model Program. By J. KOLL AR 201. Coherence in Three-Dimensional Category Theory. By N. G URSKI 202. Canonical Ramsey Theory on Polish Spaces. By V. KANOVEI, M. SABOK, and J. ZAPLETAL 203. A Primer on the Dirichlet Space. By O. E L -FALLAH, K. KELLAY, J. MASHREGHI, and T. R ANSFORD

A Primer on the Dirichlet Space OMAR EL-FALLAH Universit´e Mohammed V-Agdal, Rabat, Morocco KARIM KELLAY Universit´e Bordeaux 1, Bordeaux, France JAVAD MASHREGHI Universit´e Laval, Qu´ebec, Canada THOMAS RANSFORD Universit´e Laval, Qu´ebec, Canada

University Printing House, Cambridge CB2 8BS, United Kingdom Published in the United States of America by Cambridge University Press, New York Cambridge University Press is part of the University of Cambridge. It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning and research at the highest international levels of excellence. www.cambridge.org Information on this title: www.cambridge.org/9781107047525 © Omar El-Fallah, Karim Kellay, Javad Mashreghi and Thomas Ransford 2014 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 2014 Printed in the United Kingdom by CPI Group Ltd. Croydon CR0 4YY A catalogue record for this publication is available from the British Library ISBN 978-1-107-04752-5 Hardback 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.

To: Za¨ınab and Hiba, Anna and Ma¨el, Dorsa, Parisa and Golsa, ´ Julian and Etienne

Contents

Preface

page xi

1

Basic notions 1.1 The Dirichlet space 1.2 Reproducing kernels 1.3 Multiplication 1.4 Composition 1.5 Douglas’ formula 1.6 Weighted Dirichlet spaces Notes on Chapter 1

1 1 4 6 7 8 11 14

2

Capacity 2.1 Potentials, energy and capacity 2.2 Equilibrium measures 2.3 Cantor sets 2.4 Logarithmic capacity Notes on Chapter 2

15 15 19 21 24 27

3

Boundary behavior 3.1 The Cauchy transform 3.2 Beurling’s theorem 3.3 Weak-type and strong-type inequalities 3.4 Sharpness results 3.5 Exponentially tangential approach regions Notes on Chapter 3

29 29 31 35 39 45 48

4

Zero sets 4.1 Zero sets and uniqueness sets 4.2 Moduli of zero sets 4.3 Boundary zeros I: sets of capacity zero 4.4 Boundary zeros II: Carleson sets

50 50 54 61 64

vii

viii

Contents 4.5 Arguments of zero sets Notes on Chapter 4

67 69

5

Multipliers 5.1 Definition and elementary properties 5.2 Carleson measures 5.3 Pick interpolation 5.4 Zeros of multipliers Notes on Chapter 5

71 71 76 84 89 91

6

Conformal invariance 6.1 M¨obius invariance 6.2 Composition operators 6.3 Compactness criteria Notes on Chapter 6

93 93 96 102 106

7

Harmonically weighted Dirichlet spaces 7.1 Dμ -spaces and the local Dirichlet integral 7.2 The local Douglas formula 7.3 Approximation in Dμ 7.4 Outer functions 7.5 Lattice operations in Dμ 7.6 Inner functions Notes on Chapter 7

108 108 110 115 117 122 125 130

8

Invariant subspaces 8.1 The shift operator on Dμ 8.2 Characterization of the shift operator 8.3 Invariant subspaces of Dμ Notes on Chapter 8

132 132 135 140 145

9

Cyclicity 9.1 Cyclicity in Dμ 9.2 Cyclicity in D and boundary zero sets 9.3 The Brown–Shields conjecture 9.4 Measure conditions and distance functions 9.5 Cyclicity via duality 9.6 Bergman–Smirnov exceptional sets Notes on Chapter 9

146 146 151 154 159 166 171 179

Appendix A Hardy spaces A.1 Hardy spaces A.2 Inner and outer functions A.3 The Smirnov class

181 181 183 185

Contents

ix

Appendix B The Hardy–Littlewood maximal function B.1 Weak-type inequality for the maximal function

187 187

Appendix C Positive definite matrices C.1 Basic facts about positive definite matrices C.2 Hadamard products

189 189 190

Appendix D Regularization and the rising-sun lemma D.1 Increasing regularization D.2 Proof of the regularization lemma

193 193 195

References Index of notation Index

197 205 207

Preface

The three classical Hilbert spaces of holomorphic functions in the unit disk are the Hardy, Bergman and Dirichlet spaces. There are several excellent texts covering the Hardy space and the Bergman space. However, to the best of our knowledge, up to now there has been no book devoted to the Dirichlet space. When we began our respective researches into the Dirichlet space, we found ourselves handicapped by the fact that the necessary background information was scattered around the literature, sometimes contained in articles that were difficult to follow. For this reason we began to think about writing an introduction that would be suitable for researchers and graduate students seeking a solid background in the subject. The more we learned about this topic, the more we became convinced that it contains many beautiful ideas that deserve a systematic exposition. The name Dirichlet space derives from its definition in terms of the so-called Dirichlet integral, arising in Dirichlet’s method for solving Laplace’s equation (sometimes called the Dirichlet principle). As far as we can determine, the first appearance of the Dirichlet space under that name dates back to two articles of Beurling and Deny in 1958 and 1959, but in fact the notion existed and had been studied at least since Beurling’s thesis, which was published in 1933 and written even a little earlier. In the years that followed, Beurling and Carleson laid the foundations of the theory and, after their pioneering work, many other distinguished mathematicians made important contributions. Why study the Dirichlet space? Here are a few reasons. (1) The Hardy space corresponds to 2 , the Hilbert space of square-summable sequences. One of the main advantages of thinking of it as a function space is that the shift operator on 2 becomes simply multiplication by z. If one is interested in weighted shifts on 2 , which are very important in operator theory, then one should consider multiplication by z on a weighted function xi

xii

Preface

space. The two most basic non-constant weights lead one immediately to the Dirichlet and Bergman spaces. (2) The Dirichlet integral of a holomorphic function f has a very natural geometric interpretation. It is exactly the area of the image of f , counted according to multiplicity. Seen this way, it is obviously invariant under precomposition with every M¨obius automorphism of the unit disk. It is a remarkable fact that this M¨obius-invariance property characterizes the Dirichlet space among all Hilbert function spaces on the disk. (3) The Dirichlet space is closely related to logarithmic potential theory. In particular, the notions of energy and logarithmic capacity play a prominent role in the theory. This reflects Beurling’s vision of the subject, and yields interesting interactions with physics. (4) The Dirichlet integral is the motivating example of the abstract notion of a Dirichlet form, first introduced by Beurling and Deny in the articles mentioned above. Dirichlet forms have become a fundamental tool in probability and semigroup theory (though this is not an aspect that will be developed in this book). (5) From many points of view, the Dirichlet space is a borderline case. For example, it is very nearly an algebra, but not quite. This borderline nature makes it an interesting and challenging example of a function space. Many important questions remain unsolved, and the Dirichlet space is still an active area of research. What is in the book? To get a quick idea, imagine being presented with a function space on the unit disk. Several standard questions naturally arise. For example: -

What can be said about the boundary behavior of functions in the space? Are there simple characterizations of zero sets and uniqueness sets? What can we say about interpolation? Is the space an algebra? If not, then what are the multipliers? How rich is the operator theory on this function space? For example, can we classify the shift-invariant subspaces? Which functions are cyclic?

In the case of the Hardy space, the answers to all these questions are well known and important. By contrast, in the Dirichlet space, some of the questions have been only partially answered, and even where the complete answers are known, they are more subtle. This is the subject of this book. Perhaps it is also worth mentioning what is not in the book. As it is meant to be a primer, we do not pretend to give an exhaustive treatment of the subject, and certain topics, such as interpolating sequences and the corona problem, have been omitted completely (with much regret). We have decided to

Preface

xiii

restrict ourselves to the classical Dirichlet space, treating other variants such as weighted Dirichlet spaces when they contribute directly to understanding the classical case. The prerequisites are a knowledge of standard complex analysis, measure theory and functional analysis. Also, we have taken for granted a certain familiarity with Hardy spaces, the necessary background being summarized briefly in an appendix. We do however develop the notion of logarithmic capacity ab initio, since it turns up throughout the book. There are exercises at the end of most of the sections, ranging from routine calculations to barely disguised theorems. We have tried our best to attribute results correctly, in notes at the end of each chapter. However, history is sometimes complicated, and we apologize if we have fallen short of our aim. In the course of writing the book, we have benefitted from discussions with many mathematicians. In particular, we thank Alexandru Aleman, Nicola Arcozzi, Sasha Borichev, Håkan Hedenmalm, Stefan Richter, Bill Ross, Kristian Seip and Andrew Wynn. Also we thank J´er´emie Rostand for his help with the illustrations. We are grateful to Roger Astley and his colleagues at Cambridge University Press for their advice and encouragement. Part of this book was written at the CIRM (Luminy), and we express our gratitude to the CIRM for its hospitality. We gratefully acknowledge the financial support of the following granting bodies: CNRST and the Hassan II Academy of Science and Technology (OE), PICS-CNRS (KK), NSERC (JM and TR) and the Canada research chairs program (TR). Of course, we owe a huge debt of gratitude to our spouses, Salma, Nathalie, Shahzad and Line, for supporting us and putting up with us while the book was being written. Last, but not least, we thank our children for constantly reminding us that there are things even more important than mathematics. We dedicate the book to them.

1 Basic notions

In this introductory chapter, we introduce the Dirichlet space. We develop its elementary properties, at the same time pointing the way to the deeper results to be treated in the subsequent chapters.

1.1 The Dirichlet space Let us begin with the fundamental definition. In what follows, D denotes the open unit disk, Hol(D) is the set of all functions holomorphic in D, and dA denotes the area Lebesgue measure on the complex plane C. Definition 1.1.1

Given f ∈ Hol(D), the Dirichlet integral of f is defined by  1 | f  (z)|2 dA(z). D( f ) := π D

The Dirichlet space D is the vector space of f ∈ Hol(D) such that D( f ) < ∞. Clearly D contains all the polynomials and, more generally, all functions holomorphic on D such that f  is bounded on D. We shall see that it also contains many other interesting functions. Our first result is a formula for D( f ) in terms of the Taylor coefficients of f .  Theorem 1.1.2 Let f ∈ Hol(D), say f (z) = k≥0 ak zk . Then  k|ak |2 . (1.1) D( f ) = k≥1

Proof

Writing the area integral in polar coordinates, we have    2 2 1  1 1 2π  k−1  k−1 i(k−1)θ  D( f ) = ka z dA(z) = ka r e     dθ r dr. k k π D k≥1 π 0 0 k≥1 1

2

Basic notions

By Parseval’s formula, for each r ∈ (0, 1),  2π  2   1 k−1 i(k−1)θ  k2 |ak |2 r2k−2 .  kak r e  dθ = 2π 0 k≥1 k≥1 Hence

 D( f ) = 2 0

1



k2 |ak |2 r2k−1 dr =

k≥1



k|ak |2 .



k≥1

From this, we derive the following simple but useful observation. Recall that  the Hardy space H 2 consists of those holomorphic functions f (z) = k≥0 ak zk  such that  f 2H2 := k≥0 |ak |2 < ∞. Corollary 1.1.3

The Dirichlet space D is contained in the Hardy space H 2 .

Proof This follows immediately from the theorem, together with the obvious   fact that k≥0 k|ak |2 < ∞ implies k≥0 |ak |2 < ∞.  We shall make frequent use of this inclusion, exploiting the many known properties of the Hardy space H 2 . A summary of the properties that we shall need can be found in Appendix A. Notice that D is dense in H 2 , because it contains the polynomials. As it is obviously a proper subspace of H 2 , it is not closed in H 2 , so it is not a Hilbert space with respect to  · H2 . We now endow it with a norm, making it a Hilbert space in its own right. For f, g ∈ D, define  1 D( f, g) := f  (z)g (z) dA(z). π D This is a semi-inner product, and clearly D( f, f ) = D( f ). In particular, D( f )1/2 is a semi-norm on D. It is not quite a norm, since D( f ) = 0 whenever f is a constant. To get round this, we define  f, gD :=  f, gH2 + D( f, g)

( f, g ∈ D),

where ·, ·H 2 is the usual inner product on H 2 . This gives a genuine inner product on D, and the corresponding norm  · D is given by  f 2D =  f 2H2 + D( f ) Theorem 1.1.4 norm  · D .

( f ∈ D).

The Dirichlet space D is a Hilbert space with respect to the

  Proof Writing f (z) = k≥0 ak zk , we have  f 2D = k≥0 (k + 1)|ak |2 . Thus the map f → ((k + 1)1/2 ak )k≥0 is an isometry of D onto 2 , the space of square  summable sequences. As 2 is a Hilbert space, so too is D.

1.1 The Dirichlet space

3

This is not the only way to make D a Hilbert space. Another common choice is to take  f 2 := | f (0)|2 + D( f ). This also gives a Hilbert-space norm on D, equivalent to  · D . However, unless stated otherwise, we shall always assume that D carries the norm  · D . Exercises 1.1 1. Let f ∈ D. Prove that  f 2D ≤ | f (0)|2 + 2D( f ).  2. Let f ∈ D, say f (z) = k≥0 ak zk .  (i) Let sn (z) := nk=0 ak zk . Prove that sn D ≤  f D for all n ≥ 0, and that sn − f D → 0 as n → ∞. (ii) Let fr (z) := f (rz). Prove that  fr D ≤  f D for all 0 < r < 1, and that  fr − f D → 0 as r → 1− . 3. Let ( fn )n≥1 ∈ D. Show that, if fn → f locally uniformly on D, then we have D( f ) ≤ lim inf n→∞ D( fn ). Deduce that, if supn D( fn ) < ∞, then f ∈ D.   4. The analytic Wiener algebra is W + := { k≥0 ak zk : k≥0 |ak | < ∞}. Give examples to show that D  W + and that W +  D. 5. The disk algebra A(D) is the set of continuous functions f : D → C that are holomorphic in D. Give examples to show that D  A(D) and that A(D)  D. 6. (This exercise and the next make use of the Hardy spaces H p for general p. For a summary of their properties, see Appendix A.) Let f ∈ Hol(D). (i) Show that, if f  ∈ H 2 , then f ∈ D. (ii) Show that, if f  ∈ H 1 , then f ∈ D. [This is somewhat harder. Use Hardy’s inequality (Theorem A.1.9) to show that, if f  ∈ H 1 , then the  Taylor coefficients ak of f satisfy k |ak | < ∞ and ak = O(1/k).]  (iii) Let f (z) := k≥1 zk /k. Show that f  ∈ H p for all p < 1 but f  D.  7. We know that D ⊂ H 2 . Show that in fact D ⊂ p 0 (see Exercise 1.3.1). Hence, by Theorem 1.1.2 again, D( f 2 ) ≥ C 2

 k≥4

1 = ∞, k log k

and so f 2  D.



We do however have a positive result. In what follows, H ∞ denotes the algebra of bounded holomorphic functions on the unit disk with the norm  f H ∞ := supz∈D | f (z)|. Theorem 1.3.2 norm

The space D ∩ H ∞ is a Banach algebra with respect to the  f D∩H ∞ :=  f H ∞ + D( f )1/2 .

(1.2)

1.4 Composition Proof

7

Let f, g ∈ D ∩ H ∞ . Using Minkowski’s inequality, we have 1  1/2 D( f g)1/2 = | f  g + f g |2 dA π D 1  1/2  1  1/2  2 ≤ | f g| dA + | f g |2 dA π D π D ≤ D( f )1/2 gH ∞ +  f H ∞ D(g)1/2 .

From this, it follows that D∩H ∞ is an algebra, and that (1.2) defines an algebra norm on it. Finally, as both (D,  · D ) and (H ∞ ,  · H∞ ) are complete spaces, it  is easy to check that (D ∩ H ∞ ,  · D∩H∞ ) is complete. In the light of this theorem, it is tempting to believe that D is stable under multiplication by elements of D ∩ H ∞ . We shall see later that this is actually false. The functions that do have this important property are called multipliers, and we shall return to study them in detail in Chapter 5. Exercises 1.3 1. Let α ∈ (0, 1). By comparing the sum with an integral, show that k  j=2

1 (log k)1−α + O(1) as k → ∞. = j(log j)α 1−α

1.4 Composition This short section is based on the following conformal invariance property. Theorem 1.4.1 Let D1 , D2 be domains, let φ : D1 → D2 be a conformal mapping and let f : D2 → C be a holomorphic function. Then   |( f ◦ φ) (z)|2 dA(z) = | f  (w)|2 dA(w). D1

D2

Proof Making the substitution w = φ(z), we have dA(w) = |φ (z)|2 dA(z), whence    | f  (w)|2 dA(w) = | f  (φ(z))|2 |φ (z)|2 dA(z) = |( f ◦ φ) (z)|2 dA(z).  D2

D1

D1

In particular, taking D1 = D and f (z) = z, we obtain the following interpretation of the Dirichlet integral. Corollary 1.4.2 Let φ : D → C be an injective holomorphic map. Then D(φ) equals 1/π times the area of φ(D).

8

Basic notions

Perhaps the most important case of Theorem 1.4.1 is when D1 = D2 = D, in other words, when φ is a conformal automorphism of D. The automorphisms of D are precisely the M¨obius transformations of the form φ(z) = eiθ

a−z 1 − az

(a ∈ D, |eiθ | = 1).

We write Aut(D) for this family of functions. Corollary 1.4.3 If f ∈ Hol(D) and φ ∈ Aut(D), then D( f ◦ φ) = D( f ). Consequently, if f ∈ D, then also f ◦ φ ∈ D. The same argument shows that, if f ∈ D and φ : D → D is any injective holomorphic function, then f ◦ φ ∈ D. To what extent can the hypothesis ‘injective’ be weakened? We shall return to this question in Chapter 6. In the same chapter we shall also see that, quite remarkably, the property of M¨obius invariance described in Corollary 1.4.3 essentially characterizes the Dirichlet space. Exercises 1.4 1. Use Corollary 1.4.2 to give another example of an unbounded function in D.

1.5 Douglas’ formula Let T denote the unit circle. Given f ∈ Hol(D) and ζ ∈ T, we write f ∗ (ζ) := limr→1− f (rζ), whenever this radial limit exists. If f ∈ H 2 , then f ∗ (ζ) exists a.e. on T. Moreover, f ∗ ∈ L2 (T) and f is the Poisson integral of f ∗ . The correspondence f ↔ f ∗ allows us to view the Hardy space as a space of functions on the unit circle, and this turns out to be vital for many applications. There is a formula for D( f ) expressed purely in terms of f ∗ . It is due to Douglas [39]. Theorem 1.5.1 (Douglas’ formula) Let f ∈ H 2 . Then    ∗  f (λ) − f ∗ (ζ) 2 1 D( f ) = 2  |dλ| |dζ|.  λ−ζ 4π T T We shall deduce this formula from a lemma about L2 -Fourier series. Given a function φ ∈ L2 (T), we write φ(k) for its k-th Fourier coefficient, namely  2π 1 φ(eit )e−ikt dt (k ∈ Z). φ(k) := 2π 0

1.5 Douglas’ formula

9

In this notation, Parseval’s formula becomes  2π  1 |φ(eit )|2 dt = | φ(k)|2 . 2π 0 k∈Z Lemma 1.5.2

Let φ ∈ L2 (T). Then      φ(λ) − φ(ζ) 2 1 |dλ| |dζ| = |k| | φ(k)|2 .   λ−ζ 4π2 T T k∈Z

Proof After the change of variables λ = ei(s+t) , ζ = eit , the double integral becomes  2π  2π  i(s+t)  φ(e ) − φ(eit ) 2 1  dt ds.  eis − 1 4π2 0 0 Parseval’s formula, applied to the function ζ → φ(eis ζ) − φ(ζ), gives  2π  1 |φ(ei(s+t) ) − φ(eit )|2 dt = | φ(k)|2 |eiks − 1|2 . 2π 0 k∈Z Hence the double integral equals  2π  1 2π

0

|eiks − 1|2 | φ(k)|2 is ds. |e − 1|2 k∈Z

Finally, we remark that, for each integer k  0,  2π  iks  2π  e − 1 2 1 1 ds = |1 + eis + · · · + ei(|k|−1)s |2 ds = |k|,   2π 0 eis − 1 2π 0 the last equality again from Parseval’s formula. The result follows.



Proof of Theorem 1.5.1 We apply Lemma 1.5.2 with φ = f ∗ . Observe that,  f ∗ (k) = ak if k ≥ 0 and f ∗ (k) = 0 if k < 0. writing f (z) = k≥0 ak zk , we have Hence    ∗   f (λ) − f ∗ (ζ) 2 1 |dλ| |dζ| = k|ak |2 .   λ−ζ 4π2 T T k≥0 By Theorem 1.1.2, this last expression equals D( f ).



As mentioned at the beginning of the section, if f ∈ H 2 , then the radial limit f exists a.e. Indeed, for almost every ζ ∈ T, we have f (z) → f ∗ (ζ) as z → ζ in each non-tangential approach region |z − ζ| < κ(1 − |z|). For functions in D, the same is true even for certain tangential approach regions. We shall prove this as an application of Douglas’ formula. ∗

Theorem 1.5.3 Let f ∈ D. Then, for a.e. ζ ∈ T, we have f (z) → f ∗ (ζ) as z → ζ in each oricyclic approach region |z − ζ| < κ(1 − |z|)1/2 .

10

Basic notions

Figure 1.1 Non-tangential and oricyclic approach regions

If g ∈ H 2 , then lim|z|→1 (1 − |z|2 )|g(z)|2 = 0.  Write g(z) = k≥0 ak zk . By the Cauchy–Schwarz inequality,   |g(z)|2 ≤ |ak |2 |z|2l = g2H2 (1 − |z|2 )−1 (z ∈ D).

Lemma 1.5.4 Proof

k≥0

l≥0

 Hence lim sup|z|→1 (1 − |z|2 )|g(z)|2 ≤ g2H2 . Replacing g by g − nk=0 ak zk , and then letting n → ∞, we deduce that in fact lim|z|→1 (1 − |z|2 )|g(z)|2 = 0.  Proof of Theorem 1.5.3 Let f ∈ D. By Douglas’ formula, we have    ∗  f (λ) − f ∗ (ζ) 2  |dλ| |dζ| < ∞.  λ−ζ T T Hence, for almost every ζ ∈ T, the radial limit f ∗ (ζ) exists and satisfies   ∗  f (λ) − f ∗ (ζ) 2  |dλ| < ∞.  λ−ζ T Fix such a ζ, and define g(z) :=

f (z) − f ∗ (ζ) z−ζ

(z ∈ D).

Then g ∈ Hol(D), the radial limit g∗ exists a.e., and g∗ ∈ L2 (T). Since the denominator of g is an outer function, Smirnov’s maximum principle implies that g ∈ H 2 (see Theorem A.3.8 in Appendix A). Therefore Lemma 1.5.4 applies, and |g(z)|2 = o((1 − |z|2 )−1 ) as |z| → 1. It follows that  |z − ζ|2  | f (z) − f ∗ (ζ)|2 = |z − ζ|2 |g(z)|2 = o 1 − |z|2

(|z| → 1).

Hence f (z) → f ∗ (ζ) as z → ζ in each approach region |z − ζ| < κ(1 − |z|)1/2 . 

1.6 Weighted Dirichlet spaces

11

There is a lot more to be said about boundary values of functions in D. We shall treat this topic in some detail in Chapter 3, where, in particular, we shall see that the tangential approach region in Theorem 1.5.3 can be widened considerably further.

1.6 Weighted Dirichlet spaces In this section we briefly consider the following generalization of the notions of Dirichlet integral and Dirichlet space. Definition 1.6.1 Let w : D → [0, ∞) be a measurable function such that

w(z) dA(z) < ∞. Given f ∈ Hol(D), we define its weighted Dirichlet integral D by  1 | f  (z)|2 w(z) dA(z). Dw ( f ) := π D The corresponding weighted Dirichlet space Dw is the vector space of those f ∈ Hol(D) such that Dw ( f ) < ∞. Obviously, taking w ≡ 1, we recover the standard Dirichlet space D. The following theorem yields another familiar example. Theorem 1.6.2

Let w(z) = log(1/|z|2 ). Then Dw = H 2 and  f 2H 2 = | f (0)|2 + Dw ( f )

Proof Then

( f ∈ H 2 ).

This is a computation like that in Theorem 1.1.2. Let f (z) = 1 Dw ( f ) = π

 

1

 0

1



0

k≥0

ak z k .

2π 

0

=2



 

2  1  kak rk−1 ei(k−1)θ  log 2 dθ r dr r k≥1

k2 |ak |2 r2k−2 log(1/r2 )r dr =

k≥1



|ak |2 .

k≥1



The result follows.

Theorem 1.6.3 Suppose that lim inf |z|→1− w(z)/(1 − |z|) > 0. Then Dw ⊂ H 2 , and Dw is a Hilbert space with respect to the norm  · Dw given by  f 2Dw :=  f 2H2 + Dw ( f ) Proof

( f ∈ Dw ).

(1.3)

By hypothesis, there exist c > 0 and r < 1 such that w(z) ≥ c log(1/|z|2 )

12

Basic notions

for all z with r < |z| < 1. A calculation similar to that in the previous theorem  shows that, if f (z) = k≥0 ak zk , then    1   1  1 | f  (z)|2 c log 2 dA(z) = c 1 − r2k − 2kr2k log |ak |2 . Dw ( f ) ≥ π r 0 such that K(d0 ) > 0. A compact set F can be covered by finitely many compact sets F1 , . . . , Fn of  diameter at most d0 , so cK (F) ≤ cK (F1 ) + · · · + cK (Fn ) ≤ n/K(d0 ) < ∞. The next result is sometimes expressed by saying that capacity is upper semicontinuous. Theorem 2.1.6 Let (Fn )n≥1 be a decreasing sequence of compact subsets of X, and let F := ∩n F n . Then cK (F) = limn cK (Fn ). The proof makes use of the notion of weak*-convergence in P(X). Recall



that μn → μ weak* in P(X) if f dμn → f dμ for each continuous function f on X. Also, every sequence (μn ) in P(X), contains a weak*-convergent subsequence. Lemma 2.1.7 Let (μn ) be a sequence in P(X), and suppose that μn is weak*convergent to μ ∈ P(X). Then lim inf n→∞ IK (μn ) ≥ IK (μ). Proof We first  claim that, if f : X × X → R is a continuous function, then  f dμn dμn → f dμ dμ as n → ∞. Indeed, this is clearly true if f has the form f (x, y) = g(x)h(y) and, using the Stone–Weierstrass theorem, a general continuous f may be uniformly approximated by finite sums of functions of this special form. If K(0) < ∞, then we can apply the claim with f (x, y) := K(d(x, y)), and deduce that IK (μn ) → IK (μ). For the general case, we consider KT (t) := min{K(t), T }. By what we have just proved, IKT (μ) = limn→∞ IKT (μn ). Clearly IKT (μn ) ≤ IK (μn ) for all n, and so IKT (μ) ≤ lim inf n→∞ IK (μn ). Also, by the monotone convergence theorem IKT (μ) → IK (μ) as T → ∞. Hence IK (μ) ≤  lim inf n→∞ IK (μn ), as required. Proof of Theorem 2.1.6 By Theorem 2.1.3 (ii) the sequence cK (Fn ) decreases and cK (Fn ) ≥ cK (F) for all n. We must show that limn→∞ cK (Fn ) ≤ cK (F). If cK (Fn ) = 0 for some n, then we are done. If not, then, for each n, pick μn ∈ P(Fn ) such that IK (μn ) < 1/cK (Fn ) + 1/n. There exists a subsequence of

18

Capacity

the μn which is weak*-convergent to some μ ∈ P(X). Relabeling, if necessary, we may as well suppose that this is the whole sequence. Note that μ is supported in F n for each n, so μ ∈ P(F) and IK (μ) ≥ 1/cK (F). On the other hand, by Lemma 2.1.7, we have IK (μ) ≤ lim inf n→∞ IK (μn ) ≤ limn→∞ 1/cK (Fn ). It  follows that limn→∞ cK (Fn ) ≤ cK (F). Now we extend the definition of capacity to non-compact sets. Definition 2.1.8 Let E be an arbitrary subset of X. • The inner capacity of E is defined by cK (E) := sup{cK (F) : F ⊂ E, F compact}. • The outer capacity of E is defined by c∗K (E) := inf{cK (U) : U ⊃ E, U open in X}. Evidently cK (E) ≤ c∗K (E). A set E is called capacitable if cK (E) = c∗K (E). It is clear that open sets are capacitable, and Theorem 2.1.6 shows that compact sets are capacitable too. In fact, for several of the most important kernels K, it can be shown that every Borel subset of X is capacitable (Choquet’s theorem), but we do not need that result here. The next theorem summarizes some basic properties of c∗K . Theorem 2.1.9 (i) c∗K (∅) = 0. (ii) If E 1 ⊂ E2 , then c∗K (E1 ) ≤ c∗K (E2 ).  (iii) c∗K (∪k≥1 Ek ) ≤ k≥1 c∗K (Ek ). In particular, the outer capacity of a set does not change if we adjoin to it a set of outer capacity zero. Also, a countable union of sets of outer capacity zero is still of outer capacity zero. Proof Parts (i) and (ii) are obvious. For part (iii), we consider first the case when each Ek is an open set Uk . Let F be a compact subset of ∪k Uk . By compactness there exists n ≥ 1 such that F ⊂ U 1 ∪ · · · ∪ Un . We can write F = F1 ∪ · · · ∪ Fn , where each Fk is a compact subset of Uk (see Exercise 2.1.2). Then, using Theorem 2.1.3 (iii), we have cK (F) ≤

n  k=1

cK (Fk ) ≤

n  k=1

cK (Uk ) ≤

 k≥1

cK (Uk ).

 As this holds for each such F, we obtain cK (∪k Uk ) ≤ k cK (Uk ), proving the result in this case. For the general case, we may suppose that c∗K (Ek ) < ∞ for all k, otherwise there is nothing to prove. Let  > 0 and, for each k, let Uk be an open set such

2.2 Equilibrium measures

19

that Uk ⊃ Ek and c(Uk ) < c∗ (Ek ) + /2k . Then ∪k Uk is an open set containing ∪k Ek and, by what we have already proved,   cK (Uk ) ≤ c∗K (Ek ) + . cK (∪k Uk ) ≤ Thus c∗K (∪k E k )

≤



k

k

∗ k cK (E k )+. This holds for all 

> 0, whence the result. 

Corollary 2.1.10 If K(0) = ∞, then c∗K (E) = 0 for every countable set E ⊂ X. Proof The hypothesis that K(0) = ∞ implies that singletons are of capacity zero. The general result follows by applying Theorem 2.1.9 (iii).  Exercises 2.1 1. Let K be a kernel with K  0, and let μ be a finite positive measure on X. Prove that Kμ(x) > 0 for all x ∈ supp μ, and deduce that IK (μ) > 0. 2. Justify the following assertion, used in the proof of Theorem 2.1.9. If F is a compact set and U1 , . . . , Un are open sets such that F ⊂ U1 ∪ · · · ∪ Un , then we can write F = F1 ∪ · · · ∪ Fn , where F1 , . . . , F n are compact sets such that Fk ⊂ Uk for all k.

2.2 Equilibrium measures In this section we consider the important class of measures for which the infimum is attained in (2.1). Definition 2.2.1 Let F be a compact subset of X such that cK (F) > 0. An equilibrium measure for F is a measure ν ∈ P(F) such that IK (μ) ≥ IK (ν) for all μ ∈ P(F). Clearly, if ν is an equilibrium measure for F, then cK (F) = 1/IK (ν). Theorem 2.2.2 Let F be a compact subset of X with cK (F) > 0. Then F has an equilibrium measure. Proof For each n ≥ 1, choose μn ∈ P(F) such that IK (μn ) < 1/cK (F) + 1/n. A subsequence of the μn converges weak* to some ν ∈ P(F). By Lemma 2.1.7, this ν must satisfy IK (ν) ≤ 1/cK (F), so it is an equilibrium measure for F.  For many kernels K, it turns out that the equilibrium measure is unique. We shall not pursue this here. We do however need the following version of a theorem of Frostman, sometimes called the fundamental theorem of potential theory, which details some special properties of the potential of an equilibrium measure. We write supp μ for the (closed) support of a measure μ.

20

Capacity

Theorem 2.2.3 Let F be a compact subset of X such that cK (F) > 0. Let ν be an equilibrium measure for F. Then: (i) Kν(x) ≤ IK (ν) for all x ∈ supp ν, (ii) Kν(x) ≥ IK (ν) for all x ∈ F \ E, where c∗K (E) = 0. Lemma 2.2.4 If μ is a finite positive Borel measure on X, then its potential Kμ is a lower semicontinuous function on X. Proof

Let xn → x0 in X. By Fatou’s lemma, we have   K(d(x0 , y)) dμ(y). lim inf K(d(xn , y)) dμ(y) ≥ n→∞

In other words lim inf n→∞ Kμ(xn ) ≥ Kμ(x0 ), as required.



Proof of Theorem 2.2.3 We first claim that, if μ ∈ P(F) and IK (μ) < ∞, then Kν dμ ≥ IK (ν). In showing this, we may as well suppose that Kν dμ < ∞, otherwise there is nothing to prove. For each t ∈ (0, 1) consider the measure



μt := tμ + (1 − t)ν. A simple calculation using the identity Kμ dν = Kν dμ shows that      Kν dμ − IK (ν) + t2 IK (μ) + IK (ν) − 2 Kν dμ . IK (μt ) = IK (ν) + 2t Also, since μt ∈ P(F), it must satisfy IK (μt ) ≥ IK (ν) for all t ∈ (0, 1). It follows that Kν dμ ≥ IK (ν), as claimed. (i) Suppose, if possible, that Kν(x) > IK (ν) for some x ∈ supp ν. As Kν is lower semicontinuous, there exists a neighborhood N of x such that Kν > IK (ν) on N. As x ∈ supp ν, we must have ν(N) > 0, and hence  Kν dν > IK (ν)ν(N). N

Also, applying the claim at the beginning of the proof to the measure μ(S ) := ν(S \ N)/ν(X \ N), we have  Kν dν ≥ IK (ν)ν(X \ N). X\N

Adding together the two inequalities, we obtain  Kν dν > IK (ν), which is clearly a contradiction. Hence Kν(x) ≤ IK (ν) for all x ∈ supp ν. (ii) For each n ≥ 1, set En := {x ∈ F : Kν(x) ≤ IK (ν) − 1/n}.

2.3 Cantor sets

21

As Kν is lower semicontinuous, En is compact. Suppose, if possible, that cK (En ) > 0 for some n. Then En possesses an equilibrium measure νn . By the claim at the beginning of the proof, we have  Kν dνn ≥ IK (ν). On the other hand, since Kν ≤ IK (ν) − 1/n on En , we have  Kν dνn ≤ IK (ν) − 1/n. This is clearly a contradiction. Thus, in fact, cK (E n ) = 0 for all n. Setting E := ∪n≥1 En , we therefore have c∗ (E) = 0 and Kν(x) ≥ IK (ν) for all x ∈ F \ E. 

2.3 Cantor sets We have seen in Corollary 2.1.10 that, provided K(0) = ∞, all countable sets are of capacity zero. What about uncountable sets? In this section we consider a family of Cantor-type sets that are compact and uncountable. Our aim is to determine which of these sets have capacity zero. To this end, we need a general estimate for capacity, which is of interest in its own right. Definition 2.3.1 Given a compact subset F of X and t > 0, we write NF (t) for the t-covering number of F, namely the smallest number of sets of diameter at most t needed to cover F. Theorem 2.3.2

Let F be a compact subset of X. Then  ∞ |dK(t)| 1 ≥ . cK (F) NF (t) 0

In particular, if the integral diverges, then cK (F) = 0. For the proof we need the following simple lemma. In what follows, we write B(x, t) := {y ∈ X : d(x, y) ≤ t} and K(∞) := limt→∞ K(t). Lemma 2.3.3

Let μ be a probability measure on X. Then  ∞   IK (μ) = μ(B(x, t)) dμ(x) |dK(t)| + K(∞). 0

(2.2)

22

Capacity

Proof

Using Fubini’s theorem, we have  IK (μ) − K(∞) = (K(d(x, y)) − K(∞)) dμ(x) dμ(y)   ∞  |dK(t)| dμ(x) dμ(y) = t=d(x,y)  ∞   dμ(y) dμ(x) |dK(t)| = d(x,y)≤t 0  ∞   = μ(B(x, t)) dμ(x) |dK(t)|.



0

Proof of Theorem 2.3.2 Let μ be a probability measure on F. Let t > 0, let N = N F (t), and let A1 , . . . , AN be a cover of F by closed sets of diameter ≤ t. Set B1 := F ∩ A1 and B j := (F ∩ A j ) \ (A1 ∪ · · · ∪ A j−1 ) ( j ≥ 2). Then B1 , . . . , BN is a Borel partition of F. Also, if x ∈ B j then B j ⊂ B(x, t). Consequently,  N   μ(B(x, t)) dμ(x) μ(B(x, t)) dμ(x) = j=1

≥

N   j=1

Now 1 = μ(F)2 = inequality. Hence

 N j=1

μ(B j )

Bj

2

μ(B j ) dμ(x) =

N 

Bj

≤ N

μ(B j )2 .

j=1

N j=1

 μ(B(x, t)) dμ(x) ≥

μ(B j )2 , by the Cauchy–Schwarz

1 1 = . N NF (t)

Substituting this information into (2.2), we deduce that  ∞ |dK(t)| . IK (μ) ≥ NF (t) 0 As this holds for all probability measures μ on F, the result follows.



We now define the Cantor-type sets that we are going to use. Definition 2.3.4 Let {E nj : 1 ≤ j ≤ 2n , n ≥ 0} be non-empty, compact subsets of X such that, for each n ≥ 1: • the sets Enj ( j = 1, . . . , 2n ) are pairwise disjoint, and j i • each set E n−1 contains precisely two of the sets En . j

n

j

Then the Cantor set corresponding to the data (En ) is E := ∩n≥0 ∪2j=1 En . Clearly E is an uncountable compact set. We do not assume that it is totally disconnected, though this is often the case.

2.3 Cantor sets Theorem 2.3.5

Let E be the Cantor set corresponding to (Enj ). Then  K(dn )  K(en ) 1 ≤ ≤ , n+1 cK (E) n≥0 2n 2 n≥0

23

(2.3)

where dn := max{diam Enj : 1 ≤ j ≤ 2n } j j k k en := min{dist(En+1 , En+1 ) : En+1 , En+1 ⊂ the same Eni , j  k}.

In particular, if the left-hand series in (2.3) diverges, then cK (E) = 0, and if the right-hand series converges, then cK (E) > 0. Proof For convenience, we suppose that K(∞) = 0. This involves no loss of generality since, if (2.3) holds for such a kernel K, then it also holds for K + a for each positive constant a. Clearly, if t ≥ dn then NE (t) ≤ 2n . Hence, by Theorem 2.3.2, we have  ∞  K(d ) − K(d )  K(d ) |dK(t)| 1 n n−1 n = . ≥ ≥ K(d0 ) + n n+1 cK (E) N (t) 2 2 E 0 n≥1 n≥0 This proves the left-hand inequality in (2.3). For the right-hand inequality, we consider a measure μ on E satisfying μ(E nj ) = 1/2n for all j, n (see Exercise 2.3.1). By Lemma 2.3.3,   n  IK (μ) ≤ K(1 ) + μ(B(x, t)) dμ(x) |dK(t)|, n≥1

n+1

where (n )n≥1 is any decreasing sequence which tends to zero. We apply this with n := min{e0 , . . . , en−1 }. If x ∈ E and t < n , then B(x, t) meets Enj for only one j, so μ(B(x, t)) ≤ 1/2n . Consequently  K( ) − K( )  K(e ) n+1 n n ≤ , IK (μ) ≤ K(1 ) + n n 2 2 n≥1 n≥0 the final inequality because K(n+1 ) − K(n ) is either zero or K(en ) − K(n ). By  definition, 1/cK (E) ≤ IK (μ). This gives the right-hand side of (2.3). Exercises 2.3 1. Let E be the Cantor set as constructed above. Prove that there exists a Borel probability measure μ on E such that μ(Enj ) = 1/2n for all j, n. Under what circumstances is μ unique?

24

Capacity

2.4 Logarithmic capacity Though the notion of capacity has been developed in some generality, we shall be interested mostly in the following special case. In what follows, we write log+ (x) := max{log x, 0}. Definition 2.4.1 Let X = T with the chordal metric d(z, w) := |z − w|, and let K(t) := log+ (2/t). The corresponding capacity is called logarithmic capacity. We shall denote it simply by c(·) and the associated outer capacity by c∗ (·). The choice of the constant 2 in log+ (2/t) is largely for convenience, since T has diameter 2. It could be replaced by any other constant A ≥ 2, and the resulting capacity cA would be equivalent in the sense that 1/cA (F) − 1/c(F) = log(A/2) for all F. Another possibility, often seen in the literature, is to take K(t) = log(1/t). This gives a non-positive kernel, but one can nevertheless define    c(F) := exp − inf{IK (μ) : μ ∈ P(F)} . The capacities c and  c are related via the formula 1/c = log(2/ c). In particular,  c(F) = 0 if and only if c(F) = 0. In this section we establish three results specific to logarithmic capacity on the circle. The first is sometimes called the maximum principle for potentials. It applies to any convex kernel, but in particular it holds for K(t) := log+ (2/t). Theorem 2.4.2 Let X = T with the chordal metric d(z, w) := |z − w|, and let K : (0, ∞) → [0, ∞) be a decreasing convex function. If μ is a finite positive Borel measure on T and Kμ ≤ M on supp μ, then Kμ ≤ M on T. Proof Let I be a connected component of T \ supp μ, say I = (eiα , eiβ ), where 0 < β − α ≤ 2π. We shall prove that Kμ ≤ M on I. If α ≤ γ ≤ β, then    θ − γ iγ iθ iγ iθ dμ(θ). K(|e − e |) dμ(e ) = K 2 sin Kμ(e ) = 2 [β,α+2π] Now sin is a concave function on [0, π], and K is convex and decreasing on [0, ∞). Hence, for each θ ∈ [β, α + 2π], the function γ → K(2 sin(θ − γ)/2)) is convex on [α, β] (perhaps infinite at the endpoints). Integrating with respect to μ, we deduce that γ → Kμ(eiγ ) is convex on [α, β]. In particular, we have Kμ(eiγ ) ≤ max{Kμ(eiα ), Kμ(eiβ )} for all γ ∈ (α, β). Now Kμ ≤ M at eiα , eiβ , because both these points belong to supp μ. Therefore Kμ(eiγ ) ≤ M for all γ ∈ (α, β). In other words Kμ ≤ M on I, which is what we wanted to prove.  Combining this result with Theorem 2.2.3, we obtain the following corollary.

2.4 Logarithmic capacity

25

Corollary 2.4.3 Let X = T with the chordal metric d(z, w) := |z − w|, and let K : (0, ∞) → [0, ∞) be a decreasing convex function. Let F be a compact subset of T such that cK (F) > 0. Let ν be an equilibrium measure for F. Then: (i) Kν(x) ≤ IK (ν) for all x ∈ T, (ii) Kν(x) = IK (ν) for all x ∈ F \ E, where c∗K (E) = 0.



The second result that we shall need is a formula for the energy of a measure μ in terms of its Fourier coefficients μ(k), where  μ(k) := e−ikt dμ(eit ) (k ∈ Z). T

Theorem 2.4.4 Let X = T with the chordal metric d(z, w) := |z − w| and let K(t) := log+ (2/t). If μ is a finite positive Borel measure on T, then IK (μ) =

 | μ(k)|2 k≥1

Proof

k

+ μ(T)2 log 2.

We need to prove that   | 1 μ(k)|2 log it dμ(eis ) dμ(eit ) = . is |e − e | k k≥1

(2.4)

Let 0 < r < 1. Then   1 is it − log(1 − rei(s−t) ) dμ(eis ) dμ(eit ) dμ(e ) dμ(e ) = Re log it |e − reis |   k ik(s−t) r e = Re dμ(eis ) dμ(eit ) k k≥1  rk μ(k) μ(k) = Re k k≥1  rk | μ(k)|2 = . k k≥1 The result follows by letting r → 1− on both sides. The passage of the limit inside the integral is justified as follows. If the left-hand side of (2.4) is finite, then we may use the dominated convergence theorem, exploiting the fact that 1/|eit − reis | ≤ 2/|eit − eis | for all r < 1 (see Lemma 1.6.7). If the left-hand side of (2.4) is infinite, then instead we use Fatou’s lemma.  The third result that we need is an estimate for logarithmic capacity in terms of Lebesgue measure on the circle. In what follows, we write |E| for the arclength measure of E.

26

Capacity

Theorem 2.4.5

Let E be a Borel subset of T with |E| > 0. Then c(E) ≥

1 . log(2πe/|E|)

Proof Let K(t) := log+ (2/t). Let F be a compact subset of T with |F| > 0. Let μ be the probability measure on F defined by μ(S ) := |F ∩ S |/|F|. Then  1 2 Kμ(z) = dθ (z ∈ T). log |F| eiθ ∈F |z − eiθ | For a fixed value of z ∈ T, the integral is increased by replacing F with an arc in T of the same length |F|, centered at z. Thus  |F|/2 2 1 dθ log Kμ(z) ≤ |F| −|F|/2 |2 sin(θ/2)|  |F|/2 1 π ≤ log dθ |F| −|F|/2 |θ|  |F|/2π 2π log(1/t) dt = |F| 0 = log(2πe/|F|). It follows that IK (μ) ≤ log(2πe/|F|), and therefore cK (F) ≥ 1/ log(2πe/|F|). This proves the result for compact sets. The result for a general Borel set E now follows easily using the inner regularity of Lebesgue measure, namely |E| = sup{|F| : F ⊂ E, F compact}.  Exercise 2.4.2 below provides an upper bound for the logarithmic capacity of an arc in terms of its measure. In conjunction with Theorem 2.4.5, it shows that c(I)  1/(log(1/|I|)) as the length of the arc I tends to zero. However, no such bound is possible for general closed subsets of T. Indeed, if E is the circular Cantor middle-third set, then |E| = 0, but also c(E) > 0. The latter follows from Theorem 2.3.5, applied with K(t) = log+ (2/t) and en ∼ 3−n−1 (i.e. 3n+1 en → 1 as n → ∞). Exercises 2.4 1. Let X := T and K(t) := log+ (2/t). Let μ1 , μ2 be Borel probability measures on T such that IK (μ j ) < ∞. Show that IK (μ1 ) + IK (μ2 ) − 2IK

 μ 1 + μ2  2

=

μ1 (k) − μ2 (k)|2 1  | . 2 k≥1 k

Use this to show that the equilibrium measure of a compact set of positive logarithmic capacity is unique.

Notes on Chapter 2

27

2. Let X := T and K(t) := log+ (2/t). (i) Show that, if F is a closed subset of T and μ ∈ P(F), then 1 1 ≤ c(F) ≤ . supF Kμ inf F Kμ (ii) Deduce that, if I is closed subarc of T, then c(I) ≤

1 . log(2e/|I|)

(iii) Show that the exact value of c(T) is given by 1  π −1 1 1 c(T) = dθ = . log 2π −π | sin(θ/2)| log 2 3. Let X := T and let K(t) := 1/tα , where 0 < α < 1. The corresponding capacity is called the Riesz capacity of degree α, and is denoted by cα . (i) Show that, if E is a Borel subset of T, then cα (E) ≥ (1 − α)(|E|/π)α . (ii) Let E be the circular middle-third Cantor set. Show that cα (E) = 0 if and only if α ≥ log 2/ log 3. 4. Again, let X := T and let K(t) := 1/tα , where 0 < α < 1. Show that, if μ is a finite positive Borel measure on T, then IK (μ) 

 k≥0

| μ(k)|2 , (1 + k)1−α

where the implied constants depend only on α.

Notes on Chapter 2 §2.1 The material in this section is fairly standard. The abstract approach adopted here is inspired by the treatments in [29] and [70].

§2.2 The notion of equilibrium measure and the fundamental theorem 2.2.3 go back to the thesis of Frostman [48]. In many circumstances, the equilibrium measure is unique; for more on this, see for example [29, §III, Theorem 6] and [61, §III, Proposition 4].

28

Capacity

§2.3 The results in this section are based on the treatments in [29] and [94], though the history of capacity of generalized Cantor sets goes back at least to the paper of Ohtsuka [87]. The formulation of Definition 2.3.4 and Theorem 2.3.5 was inspired by an article of Monterie [81].

§2.4 Logarithmic capacity may be approached in many different ways: via energy [5, 14, 57, 60, 70, 93, 106, 120], potentials [1, 6, 29, 61, 80], r´eduites [12, 38, 59], Green’s functions [51, 65, 85] and transfinite diameter [92]. We have adopted the energy approach. Though these definitions may give rise to different numbers for the logarithmic capacity of a set, they are all equivalent in the sense that any one of them can be made small if any other is sufficiently small. In particular, the sets of logarithmic capacity zero are the same, whichever definition is used.

3 Boundary behavior

Since the Dirichlet space is contained in the Hardy space, it follows that each f ∈ D has non-tangential limits at almost every point of the unit circle, the exceptional set being of Lebesgue measure zero. In fact much more is true: according to a celebrated theorem of Beurling, the exceptional set is even of logarithmic capacity zero. Our primary goal in this chapter is to prove this result, which will play an important role in what follows. Our proof will be based on a representation formula for functions in the Dirichlet space, which is of interest in its own right. Using this formula, we shall deduce not only Beurling’s theorem, but also two other results: the so-called strong-type inequality for capacity, and a theorem on convergence in exponentially tangential approach regions.

3.1 The Cauchy transform We begin with some notation. Definition 3.1.1 Let A := {z ∈ C : 1 < |z| < 2}. We write L2 (A) for the Hilbert space of measurable functions g : A → C such that  1 |g(w)|2 dA(w) < ∞. g2L2 (A) := π A Given g ∈ L2 (A), we define its Cauchy transform Cg : D → C by  g(w) 1 dA(w) (z ∈ D). Cg(z) := π A w−z Using the Cauchy–Schwarz inequality, we see that Cg(z) is well defined for all z ∈ D, and Cg is holomorphic in D. In fact, as the next result shows, it belongs to the Dirichlet space. 29

30

Boundary behavior

Theorem 3.1.2 Proof

If g ∈ L2 (A), then Cg ∈ D and CgD ≤ (3/2)1/2 gL2 (A) .

Let g ∈ L2 (A). For each z ∈ D, we have  1  g(w)   Cg(z) = dA(w) zk = g, φk L2 (A) zk , k+1 π w A k≥0 k≥0

(3.1)

where φk (w) := 1/wk+1 , and ·, ·L2 (A) denotes the inner product on L2 (A). Hence  (k + 1)|g, φk L2 (A) |2 . Cg2D = k≥0

Now (φk )k≥0 is an orthogonal sequence in L2 (A) so, by Bessel’s inequality,  (k + 1)|g, φk L2 (A) |2 ≤ Bg2L2 (A) , k≥0

where B := supk≥0 (k + 1)φk 2L2 (A) . A calculation gives φk 2L2 (A)

1 = π

 A

1 dA(w) = 2 |w|2k+2

 1

2

dr r2k+1

⎧ ⎪ ⎪ ⎨log 4, =⎪ ⎪ ⎩(1 − 4−k )/k,

k = 0, k ≥ 1,

and so 1 ≤ (k + 1)φk 2L2 (A) ≤ 3/2 Therefore B = 3/2 and the result follows.

(k ≥ 0).

(3.2) 

In fact the Cauchy transform maps L2 (A) onto D. This is the next result, which may be viewed as a representation formula for functions in the Dirichlet space. Theorem 3.1.3 Given f ∈ D, there exists g ∈ L2 (A) such that f = Cg and gL2 (A) ≤  f D .  Proof Let f ∈ D, say f (z) = k≥0 ak zk . Let φk (w) := 1/wk+1 , as in the  previous proof, and consider g := k≥0 (ak /φk 2L2 (A) )φk . The sequence (φk ) is an orthogonal sequence in L2 (A), and by (3.2) we have   |ak |2 /φk 2L2 (A) ≤ (k + 1)|ak |2 =  f 2D < ∞, k≥0

k≥0

so the series defining g converges in L2 (A), and gL2 (A) ≤  f D . Furthermore, by (3.1), we have  φk , φ j L2 (A) z j = φk 2L2 (A) zk (z ∈ D, k ≥ 0), Cφk (z) = j≥0

3.2 Beurling’s theorem and therefore Cg(z) =



(ak /φk 2L2 (A) )Cφk (z) =

k≥0



ak zk = f (z)

31

(z ∈ D).

k≥0



This completes the proof. Exercises 3.1

1. Show that, if g(w) = 1/w2 , then Cg(z) = (3/4)z. Deduce that the constant √ 3/2 in Theorem 3.1.2 is sharp. 2. Show that Cg = 0 whenever g(w) = 1/wk (k ≥ 1). 3. Let 0 < α < 1 and let Lα2 (A) to be the set of measurable functions g : A → C such that  1 g2L2 (A) := |g(w)|2 (|w|2 − 1)−α dA(w) < ∞. α π A Show that the Cauchy transform is a bounded linear map C : Lα2 (A) → Dα , and prove that it is surjective.

3.2 Beurling’s theorem Recall that c(E) denotes the logarithmic capacity of E ⊂ T, and that c∗ (E) is the corresponding outer capacity. Our main goal in this section is to prove the following theorem. Theorem 3.2.1 (Beurling’s theorem) Let f ∈ D. Then there exists E ⊂ T with c∗ (E) = 0 such that, if ζ ∈ T \ E, then f ∗ (ζ) := limr→1− f (rζ) exists, and f (z) → f ∗ (ζ) as z → ζ inside each region |z − ζ| < κ(1 − |z|). A property is said to hold quasi-everywhere (q.e.) on T, if it holds everywhere on T \ E where c∗ (E) = 0. Thus Beurling’s theorem can be summarized by saying that each f ∈ D has non-tangential limits quasi-everywhere on T. For the proof of Beurling’s theorem, it is helpful to extend the Cauchy transform to the unit circle as follows. Definition 3.2.2 Given g ∈ L2 (A), we define its maximal Cauchy transform  : T → [0, ∞] by Cg  |g(w)| 1  dA(w) (ζ ∈ T). Cg(ζ) := π A |w − ζ|

32

Boundary behavior

 If ζ ∈ T and Cg(ζ) < ∞, then we set 1 Cg(ζ) := π Theorem 3.2.3 Proof

 A

g(w) dA(w). w−ζ

 is lower semicontinuous on T. Let g ∈ L (A). Then Cg 2

Let ζn → ζ0 in T. By Fatou’s lemma,   1 |g(w)| |g(w)| 1 lim inf dA(w) ≥ dA(w). n→∞ π A |w − ζn | π A |w − ζ0 |

 0 ), as required.  n ) ≥ Cg(ζ In other words, lim inf n→∞ Cg(ζ



 plays the role of a sort of maximal function. The next theorem shows that Cg Theorem 3.2.4

 < ∞. Then Let g ∈ L2 (A), let ζ ∈ T, and suppose that Cg(ζ)  |z − ζ|   |Cg(z)| ≤ 1 + Cg(ζ) (z ∈ D), (3.3) 1 − |z|

and Cg(z) → Cg(ζ) as z → ζ in each region |z − ζ| < κ(1 − |z|). Proof

For z ∈ D, we have   |g(w)| |w − ζ|   1 |Cg(z)| ≤ dA(w) ≤ sup Cg(ζ). π A |w − z| w∈A |w − z|

Now, if z ∈ D and w ∈ A, then |z − ζ| |z − ζ| |w − ζ| |w − z| + |z − ζ| ≤ ≤1+ ≤1+ . |w − z| |w − z| |w − z| 1 − |z| This gives (3.3). Turning now to the second part of the theorem, for δ > 0 let us define gδ (w) := g(w)1{1 0. As Cg(ζ) < ∞, the dominated conver δ (ζ) → 0 as δ → 0. Hence, for each κ > 0, gence theorem implies that Cg lim sup |Cg(z) − Cg(ζ)| = 0.

z→ζ |z−ζ|≤κ(1−|z|)



This gives the result.

 Theorem 3.2.4 begs the question: on how large a set can Cg(ζ) be infinite? This is where logarithmic capacity enters the picture. Theorem 3.2.5

Let g ∈ L2 (A). Then  > t) ≤ Ag2 2 /t2 c(Cg L (A)

(t > 0),

where A is an absolute constant. The key to the proof of this theorem is the following elementary lemma. Lemma 3.2.6

Proof

There exists B ≥ 2 such that, for all ζ1 , ζ2 ∈ T,  dA(w) 1 B ≤ 2 log . π A |w − ζ1 ||w − ζ2 | |ζ1 − ζ2 |

Making the change of variable z := (w − ζ1 )/(ζ2 − ζ1 ), we obtain   dA(w) dA(z) = 1 0. Thus c∗ (Cg L (A)  = ∞) = 0.  Letting t → ∞, we get c∗ (Cg Finally, we are in a position to prove Beurling’s theorem. Proof of Theorem 3.2.1 Let f ∈ D. From Theorem 3.1.3 we have f = Cg for some g ∈ L2 (A). By Theorem 3.2.4, Cg has non-tangential limits at every   ζ ∈ T for which Cg(ζ) < ∞. And by Corollary 3.2.7, we have Cg(ζ) < ∞ quasi-everywhere. The proof is complete.  Exercises 3.2 1. The aim of this exercise is to establish an analogue of Beurling’s theorem for the weighted Dirichlet spaces Dα . Fix α with 0 < α < 1. We write cα for the Riesz capacity of degree α, introduced in Exercise 2.4.3. (i) Show that there exists a constant Bα > 0 such that, for all ζ1 , ζ2 ∈ T,  Bα (|w|2 − 1)−α 1 dA(w) ≤ . π A |w − ζ1 ||w − ζ2 | |ζ1 − ζ2 |α [Hint: By symmetry, it is enough to estimate the integral on the set A := A ∩ {w : |w − ζ1 | ≤ |w − ζ2 |}. Writing w = ζ1 + reiθ , show that the integral on A is majorized by   2π 1 3 (2r| cos θ|)−α r dr dθ, π r=0 θ=0 r max{r, |ζ1 − ζ2 |/2} and deduce the required estimate.] (ii) Let Lα2 (A) be the space introduced in Exercise 3.1.3. Show that there exists a constant Aα > 0 such that, for all g ∈ Lα2 (A),  > t) ≤ Aα g2 2 /t2 cα (Cg L (A) α

(t > 0).

(iii) Deduce that, if f ∈ Dα , then it has a non-tangential limit at each point of T outside a set of outer cα -capacity zero.

3.3 Weak-type and strong-type inequalities The proof of Beurling’s theorem above yields a lot more information about the boundary values of functions in D. In particular, the following important result is more or less an immediate consequence.

36

Boundary behavior

Theorem 3.3.1 (Weak-type inequality for capacity) Let f ∈ D. Then c∗ (| f ∗ | > t) ≤ A f 2D /t2

(t > 0),

(3.5)

where A is an absolute constant. Proof Using Theorem 3.1.3, we can find a function g ∈ L2 (A) such that  > t} is open f = Cg and gL2 (A) ≤  f D . By Theorem 3.2.3 the set {Cg 2  in T, and by Theorem 3.2.5 we have c(Cg > t) ≤ AgL2 (A) /t2 , where A is an  q.e. on T. Assembling the pieces, absolute constant. Finally, | f ∗ | = |Cg| ≤ Cg 

we obtain (3.5).

Recall that | · | denotes Lebesgue measure on T, and that it is related to logarithmic capacity via Theorem 2.4.5. Corollary 3.3.2

Let f ∈ D. Then   ∗ 2 2 {| f | > t} ≤ Ae−Bt / f D

(t > 0),

where A, B > 0 are absolute constants. Proof



Combine Theorems 3.3.1 and 2.4.5.

We saw in Exercise 1.1.7 that D ⊂ H p for all p < ∞. This amounts to saying that, if f ∈ D, then | f ∗ | p ∈ L1 (T) for all p < ∞. The next result strengthens this considerably. Corollary 3.3.3 Proof Then

Let f ∈ D. Then exp(| f ∗ |2 ) ∈ L1 (T).

Assume first that  f 2D < B, where B is the constant in Corollary 3.3.2. 

| f ∗ (eiθ )|2

(e T

  − 1) dθ = =

0 T∞ 

 = ≤

| f ∗ (eiθ )|

t=0 ∞

t=0 ∞

2

2tet dt dθ 2

{| f ∗ |>t}

2tet dθ dt

2tet |{| f ∗ | > t}| dt 2

2Atet e−Bt 2

2

/ f 2D

dt

t=0

< ∞. For the general case, write f = p + g, where p is a polynomial and g2D < B/2. Then | f ∗ |2 ≤ 2|p|2 + 2|g∗ |2 q.e. on T, whence   ∗ iθ 2 2 ∗ iθ 2 e| f (e )| dθ ≤ e2p∞ e2|g (e )| dθ < ∞.  T

T

3.3 Weak-type and strong-type inequalities

37

This result is close to being optimal. If p > 2, then there exists f ∈ D such that exp(| f ∗ | p )  L1 (T) (see Exercise 3.3.3). The weak-type inequality, Theorem 3.3.1, can be strengthened a little further. This extra strength turns out to be vital for certain applications, for example, the characterization of Carleson measures to be derived in §5.2. Theorem 3.3.4 (Strong-type inequality for capacity) Let f ∈ D. Then  ∞ c∗ (| f ∗ | > t) dt2 ≤ A f 2D , (3.6) 0

where A is an absolute constant. The strategy of the proof is the same as for the weak-type inequality. We first establish the following inequality for the Cauchy transform. Theorem 3.3.5

Let g ∈ L2 (A). Then  ∞  > t) dt2 ≤ Ag2 2 , c(Cg L (A)

(3.7)

0

where A is an absolute constant. Proof Once again, it will be convenient to work with the capacity cK , where K(t) := log+ (B/t) and B is the constant in Lemma 3.2.6. We shall prove that, if g ∈ L2 (A), then  ∞  > t) dt2 ≤ 8g2 2 . cK (Cg (3.8) L (A) 0

If so, then, since c is majorized by a multiple of cK , the same inequality holds with cK by c, and 8 replaced by another absolute constant.  > 2k }. Let n ≥ 1 and, for k = −n, . . . , n, let Fk be a compact subset of {Cg n k Set μ := k=−n 2 cK (Fk )νk , where νk is an equilibrium measure for Fk . Clearly,  T

 Cg(ζ) dμ(ζ) ≥

n 

22k cK (Fk ).

k=−n

On the other hand, by (3.4) we have   Cg(ζ) dμ(ζ) ≤ gL2 (A) (2IK (μ))1/2 . T

Now, by Corollary 2.4.3, we have Kνk ≤ IK (νk ) on T. Hence, for all j, k,   Kνk dν j ≤ min{IK (ν j ), IK (νk )}, Kν j dνk =

38

Boundary behavior

and therefore IK (μ) = ≤ =

n  n  k=−n j=−n n  n  k=−n j=−n n n  

 j

k

2 cK (F j )2 cK (Fk )

T

Kν j dνk

2 j cK (F j )2k cK (Fk ) min{IK (ν j ), IK (νk )} 2 j 2k min{cK (F j ), cK (Fk )}

k=−n j=−n

≤2 ≤4

k n  

2 j 2k cK (Fk )

k=−n j=−n n  2k

2 cK (Fk ).

k=−n

If we put all this together, then we obtain n 

n 1/2   22k cK (Fk ) ≤ gL2 (A) 8 22k cK (Fk ) ,

k=−n

k=−n

whence n 

22k cK (Fk ) ≤ 8g2L2 (A) .

k=−n

As this holds for all such compact sets Fk , we get n 

 > 2k ) ≤ 8g2 2 . 22k cK (Cg L (A)

k=−n

Finally, letting n → ∞, we obtain ∞ 

 > 2k ) ≤ 8g2 2 , 22k cK (Cg L (A)

k=−∞



and this implies (3.8).

Proof of Theorem 3.3.4 By Theorem 3.1.3, there exists a function g ∈ L2 (A)  q.e. on T, we have such that f = Cg and gL2 (A) ≤  f D . As | f ∗ | = |Cg| ≤ Cg  ∞  ∞ ∗ ∗ 2  > t) dt2 ≤ Ag2 2 ≤ A f 2 , c (| f | > t) dt ≤ c(Cg D L (A) 0

which establishes (3.6).

0



3.4 Sharpness results

39

Exercises 3.3 1. Show that the strong-type inequality (3.6) implies the weak-type inequality (3.5) (with the same constant A). 2. Show that the strong-type inequality (3.6) holds with A := 8 + (40 + log(3/2))/ log 2 < 70. 3. Let p > 2, and define  f (z) := log

1 1/p 1−z

(z ∈ D).

Show that D( f ) < ∞, but that exp(| f ∗ | p )  L1 (T). 4. The aim of this exercise is to establish versions of the strong- and weak-type inequalities for the weighted Dirichlet spaces Dα . Fix α with 0 < α < 1. We use the same notation as in Exercise 3.2.1. In particular, cα denotes the Riesz capacity of degree α. (i) Show that, if g ∈ Lα2 (A), then  ∞  > t) dt2 ≤ Aα g2 2 , cα (Cg L (A) 0

α

where Aα is a positive constant depending only on α. (ii) Deduce that, if f ∈ Dα , then  ∞ α  f 2D , c∗α (| f ∗ | > t) dt2 ≤ A α 0

α is a positive constant depending only on α. where A (iii) Deduce that, if f ∈ Dα , then α  f 2D /t2 c∗α (| f ∗ | > t) ≤ A α

(t > 0).

3.4 Sharpness results To what extent are the theorems of the preceding sections sharp? For example, in Beurling’s theorem, is capacity zero the right condition for the exceptional set? Is the strong-type inequality really the best result of its kind? In this section we address these questions by proving partial converses to Beurling’s theorem and the strong-type inequality. We begin with a converse to Beurling’s theorem.

40

Boundary behavior

Theorem 3.4.1 Let E be a closed subset of T of logarithmic capacity zero. Then there exists f ∈ D such that limz→ζ Re f (z) = ∞ for all ζ ∈ E. The function f may be chosen to be continuous on D \ E. For the proof, we need a lemma describing the functions that will be used as the basic building blocks in the proof of the theorem. Lemma 3.4.2 Let K(t) := log+ (2/t) and let μ be a probability measure on T with IK (μ) < ∞. Define    2 fμ (z) := dμ(eit ) log (z ∈ D). 1 − ze−it Then: (i) (ii) (iii) (iv)

fμ ∈ D and D( fμ ) = IK (μ) − log 2, | Im fμ (z)| ≤ π/2 on D, 0 ≤ Re fμ (z) ≤ log(2/ dist(z, supp μ)) on D, Re fμ∗ (ζ) ≥ Kμ(ζ) q.e. on T.

Proof

(i) Clearly fμ is holomorphic in D, with Taylor expansion given by fμ (z) = log 2 +

 μ(k) k≥1

k

zk .



μ(k)|2 /k = IK (μ) − log 2, where the last equality comes Hence D( fμ ) = k≥1 | from Theorem 2.4.4. In particular D( fμ ) < ∞, so fμ ∈ D. (ii) If |w| < 1, then the argument of 1 − w lies between −π/2 and π/2. Consequently, if z ∈ D and eit ∈ T, then the imaginary part of log(2/(1 − ze−it )) lies between −π/2 and π/2. The result follows easily from this. (iii) Taking real parts, we have  2 Re f (z) = log it dμ(eit ). |e − z| The double inequality follows easily from this. (iv) Let ζ ∈ T be a point where f ∗ (ζ) exists. Letting z → ζ in the previous equation, and using Fatou’s lemma, we get  2 ∗ dμ(eit ) = Kμ(ζ). log it  Re f (ζ) ≥ |e − ζ| Proof of Theorem 3.4.1 Throughout the proof, we work with the logarithmic kernel K(t) = log+ (2/t) and the corresponding capacity c(·).

3.4 Sharpness results

41

By hypothesis c(E) = 0. Therefore, using Theorem 2.1.6, we can choose a decreasing sequence of closed neighborhoods En of E in T such that  c(En )1/2 < ∞. n

For each n ≥ 1, let νn be an equilibrium measure for En , define fνn as in Lemma 3.4.2, and set  f (z) := c(En ) fνn (z) (z ∈ D). n≥1

From Lemma 3.4.2 (ii) and (iii), it is clear that this series converges locally uniformly in D. We shall show that f satisfies the conclusions of the theorem. First, let us prove that f ∈ D. By Lemma 3.4.2 (i), we have D( fνn ) ≤ IK (νn ) = 1/c(En ), the last equality by definition of equilibrium measure. As fνn (0) does not change with n, it follows that  fνn D ≤ A/c(En )1/2 , for some absolute constant A. Therefore   c(En ) fνn D ≤ A c(En )1/2 < ∞, (3.9) n

n

so the series defining f converges in D. In particular, f ∈ D, as claimed. Next we examine boundary values. Using Lemma 3.4.2 (iv), we see that Re fν∗n ≥ Kνn q.e. on T. Also, by Corollary 2.4.3, we have Kνn = IK (νn ) = 1/c(En ) q.e. on En . Combining these observations, we deduce that Re fν∗n ≥ 1/c(E n )

q.e. on En .

Since Re fνn is a positive harmonic function on D, it is the Poisson integral of a finite positive measure on T, the absolutely continuous part of which is (Re fν∗n (ζ))|dζ|/2π. We have just seen that Re fν∗n ≥ 1/c(En ) q.e. (and hence a.e.) on En , which is a neighborhood of E. It follows that the unrestricted limits of Re fνn satisfy lim inf Re fνn (z) ≥ 1/c(En ) z→ζ

(ζ ∈ E).

(3.10)

Therefore, for each N ≥ 1, lim inf Re f (z) ≥ z→ζ

N 

c(E n ) lim inf Re fνn (z) ≥

n=1

z→ζ

N 

1=N

(ζ ∈ E).

n=1

As N is arbitrary, we get lim Re f (z) = ∞ z→ζ

(ζ ∈ E).

Finally, we show how to modify the construction so that, in addition, f is

42

Boundary behavior

continuous on D \ E. Let E n , νn and fνn be exactly as before. In addition, we choose a sequence rn increasing to 1 such that Re fνn (rn ζ) ≥ 1/2c(En )

(ζ ∈ E).

 Such a choice is possible by (3.10). Set fn (z) := fνn (rn z) and f := n c(En ) fn . The Dirichlet norm of fn is no larger than that of fνn so, just as before, f ∈ D. Also, for every N, we have lim inf Re f (z) ≥ z→ζ

N 

c(En ) Re fn (ζ) ≥

n=1

N 

1/2 = N/2

(ζ ∈ E),

n=1

so limz→ζ Re f (z) = ∞ for all ζ ∈ E. Finally, the terms fn in the sum defining f are all continuous on D, and the sum itself converges locally uniformly on D \ E, because, by parts (ii) and (iii) of Lemma 3.4.2, | fn (z)| = | fνn (rn z)| ≤ π/2 + log(2/ dist(rn z, En ))

(z ∈ D).

Therefore f is continuous on D \ E.



Now we turn to the strong-type inequality. The equivalence between (i) and (iii) in the next theorem demonstrates that, in certain circumstances at least, the strong-type inequality is sharp. The equivalence with (ii) will prove useful later, in Chapter 9, when we come to study cyclicity. We write d to denote arclength distance on T, and Et := {ζ ∈ T : d(ζ, E) ≤ t}. Theorem 3.4.3 Let E be a closed subset of T, and let η : (0, π] → (0, ∞) be a continuous, decreasing function such that η(0) := limt→0+ η(t) = ∞. The following are equivalent. (i) There exists f ∈ D such that lim inf | f (z)| ≥ η(d(ζ, E)) z→ζ

(ζ ∈ T).

(ii) For each  > 0, there exists f ∈ D such that | Im f | <  on D and lim inf Re f (z) ≥ η(d(ζ, E)) z→ζ

(ζ ∈ T).

(iii) The function η satisfies 

π

c(E t ) |dη2 (t)| < ∞.

0

Proof

The implication (ii)⇒(i) is obvious.

(3.11)

3.4 Sharpness results

43

For the implication (i)⇒(iii), we observe that, if f satisfies (i), then necessarily | f ∗ | ≥ η(t) q.e. on Et , so  π  ∞  π   c(Et ) |dη2 (t)| ≤ c∗ | f ∗ | ≥ η(t) |dη2 (t)| = c∗ (| f ∗ | ≥ s) ds2 . 0

η(π)

0

The last integral is finite by the strong-type inequality, Theorem 3.3.4. Finally, we turn to the implication (iii)⇒(ii). The construction of the function f follows the same general lines as in the proof of Theorem 3.4.1, but instead of using the relatively crude estimate (3.9), we exploit the Hilbert-space structure of D with the aid of the following lemma. Lemma 3.4.4 Let (hn )n≥1 be a sequence of vectors in a Hilbert space (H, ·)  such that (hn − hm ) ⊥ hm whenever n ≥ m. Then n≥1 hn /hn 2 converges in H  if and only if n≥1 n/hn 2 < ∞. We have hn , hm  = hm 2 whenever n ≥ m, and hence n n n n  hk 2   hmin{k,l} 2  2k − 2m + 1  = = .  hk 2 hk 2 hl 2 k=m hk 2 k=m k=m l=m   Thus, if k k/hk 2 converges, then the partial sums of k hk /hk 2 form a  Cauchy sequence, and therefore converge in H. Conversely, if k hk /hk 2 converges in H, then its partial sums are bounded in norm, so the calculation  above (with m = 1) shows that the partial sums of k k/hk 2 are bounded,  whence k k/hk 2 < ∞.  Proof

We now return to the proof of the implication (iii)⇒(ii) in Theorem 3.4.3. Let n0 be a positive integer with n0 ≥ η(π). For n ≥ n0 , set δn := η−1 (n) and cn := c(Eδn ). We then have  π   δn−1  c(Et ) |dη2 (t)| ≥ c(E δn ) |dη2 (t)| = cn (n2 − (n − 1)2 ), 0

n>n0

so condition (iii) implies that

δn

n>n0



ncn < ∞.

(3.12)

n≥n0

Increasing n0 , if necessary, we can further suppose that  cn < 2/π.

(3.13)

n≥n0

For n ≥ n0 , let νn be an equilibrium measure for Eδn , and let fνn be defined as in Lemma 3.4.2. Finally, define  cn fνn (z) (z ∈ D). f (z) := n0 + n≥n0

44

Boundary behavior

Clearly this series converges locally uniformly in D. We shall show that f satisfies all the requirements of (ii). We begin by proving that f ∈ D. It will be convenient to work with the norm  ·  defined by h2 := |h(0)|2 + D(h), which is equivalent to the usual norm  · D on D. Let hn := fνn − fνn (0) = fνn − log 2. Using Lemma 3.4.2, we have  1 2 dνn (λ) dνn (ζ). hn  = D( fνn ) = log |λ − ζ| By polarization, the associated inner product ·, · satisfies  1 hn − hm , hm  = d(νn − νm )(λ) dνm (ζ) log |λ − ζ|   Kνm dνm , = Kνm dνn − where, as usual, K(t) := log+ (2/t). By Corollary 2.4.3, we have

Kνm = 1/cm q.e. on Eδm . Hence, if n ≥ m, then Kνm dνn = 1/cm = Kνm dνm , and consequently hn − hm , hm  = 0. Lemma 3.4.4 thus applies. Now hn 2 = IK (νn ) − log 2 = 1/cn − log 2.  Thus (3.12) implies that n n/hn 2 < ∞, and consequently, by Lemma 3.4.4,   2 n hn /h converges in D. This in turn implies that n cn hn converges in D,  whence also n cn fνn . Thus f ∈ D, as claimed. Next, we show that | Im f | <  on D. Indeed, for all z ∈ D, we have   | Im f (z)| ≤ cn | Im fνn (z)| ≤ cn π/2 < , n≥n0

n≥n0

where we used successively the definition of f , Lemma 3.4.2 (ii) and (3.13). Finally we prove that lim inf z→ζ Re f (z) ≥ η(d(ζ, E)) for all ζ ∈ T. This is obvious if d(ζ, E)) ≥ δn0 , since Re f ≥ n0 everywhere in D. So assume that d(ζ, E) < δn0 , and let N be the integer such that δN+1 ≤ d(ζ, E) < δN . By Lemma 3.4.2 (iv), for all n we have Re fν∗n ≥ Kνn = 1/cn q.e. on Eδn . Hence, Re f ∗ ≥ n0 +

N  n=n0

cn Re fν∗n ≥ n0 +

N 

1 = N + 1 q.e. on EδN .

n=n0

Just as in the proof of Theorem 3.4.1, as Re f is a positive harmonic function, it is at least as large as the Poisson integral of its radial boundary values, and we have just seen that these boundary values are at least N + 1 q.e. (hence a.e.) on EδN . As ζ is an interior point of EδN , it follows that lim inf Re f (z) ≥ N + 1 = η(δN+1 ) ≥ η(d(ζ, E)). z→ζ

The proof is complete.



3.5 Exponentially tangential approach regions

45

Exercises 3.4 1. Use Theorem 3.4.3 to show that there exists f ∈ D such that lim Re f (z) ≥ log log z→ζ

1 |1 − ζ|

(ζ ∈ T).

2. Show how to modify the construction of f in Theorem 3.4.3 so as to ensure that, in addition, f is continuous on D \ E. 3. Let E be a closed subset of T, and let η : (0, 2] → (0, ∞) be a decreasing convex function such that  2 |Et |η (t)2 dt < ∞. (3.14) 0

(i) Let g(w) := −3wη (dist(w, E)/3) (w ∈ A). Show that g ∈ L2 (A). (ii) Let f := Cg + η(1/3). Show that f ∈ D and that Re f (z) ≥ η(dist(z, E))

(z ∈ D).

(3.15)

[This exercise is closely related to Theorem 3.4.3. It can be shown that the hypothesis (3.14) is stronger than (3.11), but the corresponding conclusion (3.15) is valid on all of D, not just the boundary. For more details, we refer to [46, §7].]

3.5 Exponentially tangential approach regions In this section we shall consider limits along certain tangential approach regions. We have already seen one result of this kind. In Theorem 1.5.3, we showed that, if f ∈ D, then for almost every ζ ∈ T we have f (z) → f ∗ (ζ) as z → ζ in the oricyclic region |z − ζ| < κ(1 − |z|)1/2 . Using the techniques developed in this chapter, we can now prove the following much stronger version of this result, in which the approach region is exponentially tangential. Theorem 3.5.1 Let f ∈ D. Then, for a.e. ζ ∈ T, we have f (z) → f ∗ (ζ) as z → ζ in each region  |z − ζ| < κ log

1 −1 1 − |z|

(κ > 0).

The proof is based on the representation formula Theorem 3.1.3, and on a slightly more sophisticated version of the estimate (3.3). In order to state the estimate, we need to recall the notion of maximal function.

46

Boundary behavior

Given h ∈ L1 (T), its Hardy–Littlewood maximal function Mh : T → [0, ∞] is defined by  1 (Mh)(ζ) := sup |h(eiθ )| dθ (ζ ∈ T). δ>0 2δ |θ−arg ζ| t}| ≤ 6πhL1 (T) /t

(t > 0).

(3.16)

For more details, we refer to Appendix B. Let g ∈ L2 (A) and let ζ ∈ T. Then   3 1/2  |Cg(z)| ≤ 2Cg(ζ) + 2 (Mhg )(ζ)|z − ζ| log 1 − |z|

Lemma 3.5.2

where



2

hg (ζ) :=

|g(rζ)|2 r dr

(z ∈ D),

(3.17)

(ζ ∈ T).

1

Proof

Fix z ∈ D. Clearly |Cg(z)| ≤

1 π

 A

|g(w)| dA(w). |w − z|

We shall estimate the right-hand side by splitting the domain of integration into two regions. Let δ := |z − ζ| and set Aδ := A ∩ {w : |w − ζ| ≤ 2δ}. For w ∈ A \ Aδ , we have  w − ζ   ≤ |w − ζ| ≤ 2,  w−z |w − ζ| − δ and consequently   1 |g(w)| 2|g(w)| 1  dA(w) ≤ dA(w) = 2Cg(ζ). π A\Aδ |w − z| π A |w − ζ| In the region Aδ , we apply the Cauchy–Schwarz inequality:  1/2  1  1/2 1  |g(w)| 1 1 2 |g(w)| dA(w) dA(w) . dA(w) ≤ π Aδ |w − z| π Aδ π Aδ |w − z|2 Now Aδ is contained in the sector {1 < r < 2, |θ − arg ζ| < πδ} (see Figure 3.1). Hence   1 1 |g(w)|2 dA(w) ≤ hg (eiθ ) dθ ≤ 2δ(Mhg )(ζ). π Aδ π |θ−arg ζ| 0 small enough to ensure that both  δ > }| <  and |{Mhgδ > }| < . It follows that |{Cg lim sup | f (z) − f ∗ (ζ)| ≤ 3 + 2(κ)1/2 z→ζ z∈Ωκ (ζ)

for all ζ ∈ T outside a set of measure 2. Letting  → 0, we deduce that lim f (z) = f ∗ (ζ)

z→ζ z∈Ωκ (ζ)

for all ζ ∈ T outside a set of measure zero.



We conclude by remarking that Theorem 3.5.1 is optimal, in the sense that the tangential approach region cannot be widened any further. This statement will be made precise in Exercise 4.2.1.

Notes on Chapter 3 §3.1 The representation formula 3.1.3 is due to Dyn kin [41]. The simple proof given here is taken from Borichev [20]. We remark that it is also possible to represent a function f ∈ D as a Cauchy-type transform of a function in L2 (D). Indeed, if we write  1 g(w) dA(w) (z ∈ D), (Cg)(z) := π D 1 − wz then f (z) = f (0) + zC( f  ). This formula was implicitly used in proving Theorem 1.6.6.

§3.2 Beurling’s theorem (for radial limits) appeared in [16]. Our approach follows Borichev [20].

§3.3 The weak-type inequality for capacity is due to Beurling [16]. He had earlier obtained Corollary 3.3.2 in his thesis [15], showing that, if f (0) = 0 and D( f ) ≤ 1, then for each 2 t > 0 we have |{| f ∗ | > t}| ≤ 2πe1−t . This was later strengthened by Chang and Marshall [34] (see also [47] and [75]), who proved that   exp(| f ∗ |2 ) dθ : f (0) = 0, D( f ) ≤ 1 < ∞. sup T

The strong-type inequality for capacity is due to Hansson [54], extending earlier results of Maz ya and of Adams. Our proof is adapted from that given in the book of Adams and Hedberg [1]. This book also contains a detailed account of the history of the strongtype inequality for capacity.

Notes on Chapter 3

49

§3.4 The basic construction in the proof of Theorem 3.4.1 appears in Carleson’s paper [26, p. 332]. The modification required to produce a function continuous on D \ E is due to Brown and Cohn [23]. Theorem 3.4.3 is from [43]. Exercise 3.4.3 is based on a result in [46, §7].

§3.5 Theorem 3.5.1 is due to Nagel, Rudin and Shapiro [83]. They also showed that the result is optimal, in the sense that the tangential approach region cannot be widened any further (see Exercise 4.2.1). Our approach is an adaptation of their ideas and those in [20]. Actually this theorem is the simplest of a variety of results of this kind, with a playoff between the width of the approach region, the size of the exceptional set, and the space of functions in question. For more on this we refer to [20, 21, 82, 84, 121, 122].

4 Zero sets

If f is a function holomorphic on a domain D and f  0, then its zeros are isolated in D and of finite multiplicity. We can thus write them as a finite or infinite sequence (zn ), each zero repeated according to its multiplicity. If f belongs to the Hardy space H 2 and f  0, then its zeros (zn ) form a Blaschke sequence; in other words, they satisfy the Blaschke condition  (1 − |zn |) < ∞. (4.1) n

Conversely, given a sequence (zn ) in D satisfying (4.1), there exists f ∈ H 2 whose zero set is precisely (zn ), including multiplicities. Indeed, f may even be chosen to be a Blaschke product, so that f ∈ H ∞ . The situation for the Dirichlet space D is rather more complicated. Since D ⊂ H 2 , the Blaschke condition (4.1) is clearly necessary for (zn ) to be the zero set of some f ∈ D, but it turns out not to be sufficient. Indeed, no satisfactory necessary and sufficient condition is known. The results that are known suggest that the problem is complex. In this chapter we shall present a number of the most important of these theorems. We shall also examine the distribution of zeros on the boundary.

4.1 Zero sets and uniqueness sets Definition 4.1.1 A sequence (zn )n≥1 in D is a zero set for D if there exists f ∈ D such that f (zn ) = 0 for all n and f (z)  0 for all other z ∈ D. Repetitions are permitted. In that case, the number of times an element zn repeats in the sequence represents the order of the zero of f at that point. A sequence (zn )n≥1 in D is a uniqueness set for D if the only function f ∈ D satisfying f (zn ) = 0 for all n is f ≡ 0. 50

4.1 Zero sets and uniqueness sets

51

We begin with a simple class of examples of zero sets. Theorem 4.1.2 Let (xn ) be a sequence in [0, 1) such that Then (xn ) is a zero set for D.



n (1

− xn ) < ∞.

Proof Since the (xn ) satisfy the Blaschke condition, there exists a Blaschke product B with zeros at precisely the points (xn ). By logarithmic differentiation, 1 − xn2 B (z)  = B(z) (z − xn )(1 − xn z) n

(z ∈ D).

Multiplying up by B and using Lemma 1.6.7, we have |B (z)| ≤

 1 − x2  8(1 − xn ) n ≤ 2 |1 − xn z| |1 − z|2 n n

(z ∈ D).

It follows that, if we define f (z) := (1 − z)2 B(z), then f  is bounded, and in  particular f ∈ D. Clearly f has zeros at precisely the points (xn ). One might think that it would have been even simpler to take f := B. However, infinite Blaschke products never belong to the Dirichlet space (see Exercise 4.1.1), so multiplication by some kind of factor is necessary. The following theorem extends this idea to the case where the zeros of the Blaschke product no longer necessarily lie on a radius. Theorem 4.1.3 Let f ∈ H 2 and let B be a Blaschke product. Let (zn ) be the zero set of B. Then  1  1 − |zn |2 | f ∗ (ζ)|2 |dζ|. (4.2) D(B f ) = D( f ) + 2 2π |ζ − z | n T n Proof Suppose first that B(z) = z. It is clear from Theorem 1.1.2 that D(z f ) = D( f ) +  f 2H2 , which gives (4.2) in this case. Next, suppose that B(z) = c(z1 − z)/(1 − z1 z), where 0 < |z1 | < 1 and |c| = 1. Using the M¨obius invariance of the Dirichlet integral, Corollary 1.4.3, we have D(B f ) = D(z( f ◦ B−1 )) = D( f ◦ B−1 ) +  f ◦ B−1 2H2  1 = D( f ) + | f ∗ (ζ)|2 |B (ζ)| |dζ| 2π T  1 1 − |z1 |2 | f ∗ (ζ)|2 |dζ|. = D( f ) + 2π T |ζ − z1 |2 This proves (4.2) in this case. The case where B is a finite Blaschke product follows by induction.

52

Zero sets

Finally, for the general case, let us write bn for the product of the first n terms in B, and Bn for the product of the remaining terms. By what we have already proved for finite Blaschke products, for each n we have  n  1 1 − |zk |2 ∗ 2 | f (ζ)| |dζ|. D(bn f ) = D( f ) + 2π T |ζ − zk |2 k=1 Since bn f → B f locally uniformly on D, we get D(B f ) ≤ lim inf n→∞ D(bn f ). Hence  ∞  1 1 − |zk |2 ∗ 2 D(B f ) ≤ D( f ) + | f (ζ)| |dζ|. 2π T |ζ − zk |2 k=1 For the reverse inequality, note that D(B f ) = D(bn Bn f ) = D(Bn f ) +

 n  1 1 − |zk |2 ∗ 2 | f (ζ)| |dζ|. 2π T |ζ − zk |2 k=1

Here we have used the fact that |B∗n | = 1 a.e. on T. Also, since Bn f → f locally uniformly on D, we have lim inf n→∞ D(Bn f ) ≥ D( f ). Hence  ∞  1 1 − |zk |2 ∗ 2 | f (ζ)| |dζ|. D(B f ) ≥ D( f ) + 2π T |ζ − zk |2 k=1 This completes the proof.



We shall see later that this is actually a special case of a much more general result (Theorem 7.6.9). As a first application, we prove that, in the Dirichlet space, the notions of zero set and uniqueness set are complementary. Theorem 4.1.4 A sequence (zn ) in D is a zero set for D if and only if it is not a uniqueness set. Proof The ‘only if’ is obvious. For the ‘if’, suppose that (zn ) is not a uniqueness set. Then there exists g ∈ D such that g(zn ) = 0 for all n and g  0. The problem is that, in addition to the (zn ), there may be some additional zeros (wn ), either a finite or infinite sequence. In the case of an infinite sequence, they satisfy the Blaschke condition, because they are zeros of a function in H 2 . Let B be the Blaschke product with zeros at the (wn ), and let f := g/B. Then f ∈ H 2 and has zeros exactly at the (zn ). Finally, by Theorem 4.1.3, we have D( f ) ≤ D(B f ) = D(g) < ∞, and so f ∈ D.  The formula (4.2) has another interesting consequence. Recall that we write log+ (x) := max{log(x), 0}.

4.1 Zero sets and uniqueness sets Theorem 4.1.5

If (zn ) is a zero set for D, then   1 − |zn |2  log+ |dζ| < ∞. |ζ − zn |2 T n

53

(4.3)

Proof By assumption, there is a function in D with zero set (zn ). We can factorize this function as B f , where B is the Blaschke product whose zero set is (zn ), and f ∈ H 2 \ {0}. By Theorem 4.1.3, we have   1 − |zn |2 ∗ 2 1 | f (ζ)| |dζ|. D(B f ) = D( f ) + 2π T n |ζ − zn |2 In particular, the integral on the right-hand side is finite. Since log+ x ≤ x for all x ≥ 0, it follows that   1 − |zn |2  log+ | f ∗ (ζ)|2 |dζ| < ∞. 2 |ζ − zn | T n Also, because f ∈ H 2 \ {0}, we necessarily have  1 log+ ∗ 2 |dζ| < ∞. | f (ζ)| T As log+ (ab) ≤ log+ a + log+ b for all a, b ≥ 0, we deduce that (4.3) holds.



Although the condition (4.3) is difficult to interpret in general, it does allow us to exhibit sequences satisfying the Blaschke condition (4.1) that are uniqueness sets for D. These sequences may even be chosen to be convergent.  Theorem 4.1.6 Let (δn ) be a sequence in (0, 1) such that n δn < ∞ and   n δn log(1/δn ) = ∞. For n ≥ 1, set rn := 1 − δn and θn := k≥n δk . Then (rn eiθn ) is a uniqueness set for D. Proof

If θ ∈ (θn+1 , θn ), then |eiθ − rn eiθn | ≤ |θ − θn | + (1 − rn ) ≤ 2δn ,

and so |eiθ

1 − rn2 δn 1 ≥ = . − rn eiθn |2 4δ2n 4δn

Consequently, for each n, we have  θn  θn   1 − rn2  1 − rk2  + + log log dθ ≥ dθ |eiθ − rk eiθk |2 |eiθ − rn eiθn |2 θn+1 θn+1 k  1  . ≥ δn log+ 4δn

54

Zero sets

Hence, finally,  2π 0

log+

 k

  1  1 − rk2  δn log+ dθ ≥ = ∞. iθ iθ 2 k 4δn |e − rk e | n 

By Theorem 4.1.5, it follows that (rn eiθn ) is a uniqueness set.

Comparing Theorems 4.1.2 and 4.1.6, we see that there exists a sequence (zn ) converging to 1 such that (|zn |) is a zero set but (zn ) itself is not. Thus the arguments of (zn ) play a role in determining whether it is a zero set for D. We shall return to this topic in §4.5. Exercises 4.1 1. Use formula (4.2) to show that no infinite Blaschke product belongs to D.  2. Show that, in Theorem 4.1.6, we can equally well take θn := k≥n ηk δk , where (ηk ) is any positive sequence such that ηk  1 as k → ∞. 3. For n ≥ 3, let   1 ei/ log n . zn := 1 − n(log n)2 Show that (zn )n≥3 is a uniqueness set for D. [Use the previous exercise.]

4.2 Moduli of zero sets Is there a condition on the moduli |zn | guaranteeing that the sequence (zn ) is a zero set for D? As we have just seen, though the Blaschke condition (4.1) is necessary, it is not sufficient. A condition that is sufficient is given by the following theorem. Theorem 4.2.1

Let (zn ) be a sequence in D \ {0} such that   1 −1 log < ∞. 1 − |zn | n

(4.4)

Then (zn ) is a zero set for D. For example, if |zn | = 1 − e−n , then (zn ) satisfies (4.4). On the other hand, if |zn | = 1 − e−n , then (zn ) does not satisfy (4.4), even though it does satisfy the Blaschke condition. Thus, in order to satisfy (4.4), the sequence (zn ) must approach the boundary ‘very fast’. The condition (4.4), though stringent, is optimal in the following sense. 2

4.2 Moduli of zero sets Theorem 4.2.2

55

Let (rn ) be a sequence in (0, 1) such that   log n

1 −1 = ∞. 1 − rn

(4.5)

Then there exists a sequence (θn ) such that (rn eiθn ) is a uniqueness set for D. We now turn to the proofs of these two theorems, beginning with the second. Proof of Theorem 4.2.2 Let (In )n≥1 be contiguous arcs in T such that |In | = 1/ log(1/(1 − rn )) for all n. Let eiθn be the midpoint of In for each n, and set zn := rn eiθn . We shall show that (zn ) is a uniqueness set for D. If ζ ∈ In , then  |zn − ζ| ≤ |zn − eiθn | + |eiθn − ζ| < 2 log

1 −1 . 1 − rn

In other words, if we set   Ω(ζ) := z ∈ D : |z − ζ| < 2 log

1 −1  , 1 − |z|

 then zn ∈ Ω(ζ) for all ζ ∈ In . Now, condition (4.5) implies that n |In | = ∞, so almost every ζ ∈ T lies in infinitely many In . Thus, for almost every ζ ∈ T, the set Ω(ζ) contains infinitely many (zn ). Also, by Theorem 3.5.1, if f ∈ D, then for almost every ζ ∈ T, we have f (z) → f ∗ (ζ) as z → ζ in Ω(ζ). Putting these two facts together, we see that, if f ∈ D and f (zn ) = 0 for all n, then f ∗ = 0 a.e. on T, and consequently f ≡ 0. Thus (zn ) is indeed a uniqueness set.  The rest of this section is devoted to the proof of Theorem 4.2.1. The strategy will be to express the problem in terms of reproducing kernels, thereby reducing it to a geometric question about Hilbert spaces. Indeed, suppose that f ∈ D and f (zn ) = 0 for all n but f  0. Replacing f by f /zm , we may assume that f (0)  0. Thus, in terms of reproducing kernels, we have  f, kzn D = 0 for all n and  f, 1D  0. Obviously, for this to be possible, it is necessary that 1 lie outside the closed linear span M of {kzn : n ≥ 1}. Conversely, if 1 does indeed lie outside M, then writing f := 1 − P1, where P : D → M denotes the orthogonal projection, we have f  0 and f ∈ M ⊥ , so f (zn ) =  f, kzn D = 0 for all n. For the time being, we leave aside the Dirichlet space, and concentrate on the problem of estimating the distance between an element of a Hilbert space and a subspace. We shall make extensive use of the notion of positive definite and semi-definite matrices. For the definition and relevant background on this subject, see Appendix C.

56

Zero sets

Lemma 4.2.3 Let H be a complex Hilbert space, and let x1 , . . . , xn ∈ H. Let ⎛ ⎞ ⎜⎜⎜x1 , x1  x1 , x2  . . . x1 , xn ⎟⎟⎟ ⎜⎜ .. .. ⎟⎟⎟⎟ .. G := ⎜⎜⎜⎜⎜ ... . . . ⎟⎟⎟⎟ . ⎜⎝ ⎠ xn , x1  xn , x2  . . . xn , xn  Then G is positive semi-definite. Further, it is positive definite if and only if the vectors x1 , . . . , xn are linearly independent. Proof

Let λ1 , . . . , λn ∈ C. Then   2    xi , x j λi λ j = λi xi , λ j x j =  λi xi  ≥ 0. i, j

i

j

i

Thus G is positive semi-definite. It fails to be positive definite if and only if  there is a choice of (λi ) not all zero such that  i λi xi  = 0, in other words, if and only if x1 , . . . , xn are linearly dependent vectors.  The matrix G is called the Gram matrix corresponding to x1 , . . . , xn . We shall denote its determinant by G(x1 , . . . , xn ). The next result gives a distance formula in terms of Gram determinants. Lemma 4.2.4 Let H be a Hilbert space, let x1 , . . . , xn ∈ H be linearly independent vectors, and let M := span{x1 , . . . , xn }. Then, for all x ∈ H, dist(x, M)2 = Proof

G(x, x1 , . . . , xn ) . G(x1 , . . . , xn )

Fix x ∈ H, and consider the vector y ∈ H given by the formula   x1 ... xn   x x , x x , x  . . . x , x  1 1 1 1 n  y :=  . . ..  . . .. ..  .. .    xn , x xn , x1  . . . xn , xn 

Clearly we have x, y = G(x, x1 , . . . , xn ). Also, each inner product xi , y is a determinant with two identical rows, and so xi , y = 0 (i = 1, . . . , n). Thus y ∈ M ⊥ . Expanding the determinant defining y by the first row, we see that y = G(x1 , . . . , xn )x + w, where w is a linear combination of x1 , . . . , xn . Therefore, if we set z := y/G(x1 , . . . , xn ) and t := −w/G(x1 , . . . , xn ), then x = t + z is a decomposition of x as a sum of a vector in M and a vector in M ⊥ . Consequently, dist(x, M)2 = z2 = x, z =

G(x, x1 , . . . , xn ) x, y = . G(x1 , . . . , xn ) G(x1 , . . . , xn )



4.2 Moduli of zero sets

57

Though this lemma gives an exact formula for the distance, the Gram determinants may be rather difficult to compute in practice, and so the following simpler estimate is useful. Lemma 4.2.5 Let H be a Hilbert space, let x1 , . . . , xn ∈ H be linearly independent vectors such that xi , x j   0 for all i, j, and let M := span{x1 , . . . , xn }. Let x ∈ H \ {0}, and suppose that the n × n matrix H = (hi j ), defined by hi j := 1 −

xi , xx, x j  x, xxi , x j 

(i, j ∈ {1, . . . , n}),

is positive semi-definite. Then dist(x, M)2 ≥ x2

n 

h j j.

j=1

Proof Let us calculate the determinant G(x, x1 , . . . , xn ) using row operations. Label the rows from 0 to n, and for each i ∈ {1, . . . , n}, subtract xi , x/x, x times row 0 from row i. The resulting matrix, which of course has the same determinant, has x, x as its first entry, and all the other entries in the leftmost column are zero. Hence G(x, x1 , . . . , xn ) = x, x det(A), where A is the n × n matrix given by  xi , xx, x j   xi , x Ai j = xi , x j  − x, x j  = xi , x j  1 − . x, x x, xxi , x j  Thus A is equal to the Hadamard product G ◦ H, where G is the matrix defined in Lemma 4.2.3. By that lemma, G is positive definite. Also H is positive semi-definite, by assumption. It follows by Oppenheim’s inequality (see  Theorem C.2.3 in Appendix C) that det(A) ≥ det(G) nj=1 h j j . Putting this all together, we obtain G(x, x1 , . . . , xn ) = x, x det(A) ≥ x2 G(x1 , . . . , xn )

n 

h j j.

j=1

Combining this with the result of the Lemma 4.2.4, we obtain the desired conclusion.  Now we return to the Dirichlet space. As indicated at the beginning of the proof, the idea is to apply the preceding theory, in particular Lemma 4.2.5, to H = D with x = k0 = 1 and x j = kz j ( j = 1, . . . , n), where kz j is the reproducing kernel corresponding to z j . To be able to apply Lemma 4.2.5, we need to know that the hypotheses are satisfied. That is the subject of the next lemma. Lemma 4.2.6

Let z1 , . . . , zn be distinct points in D \ {0}. Then

58

Zero sets

(i) kz1 , . . . , kzn are linearly independent, (ii) kzi , kz j D  0 for all i, j, (iii) the matrix H = (hi j ) defined by hi j := 1 −

kzi , 11, kz j 

(i, j ∈ {1, . . . , n}),

1, 1kzi , kz j 

is positive semi-definite. For the proof, we require one more lemma.  Lemma 4.2.7 Let f ∈ Hol(D), say f (z) = n≥0 an zn . Suppose that a0 = 1 and that an > 0 and a2n ≤ an−1 an+1 for all n ≥ 1. Then  1 = bn zn (z ∈ D), (4.6) 1− f (z) n≥0 where bn ≥ 0 for all n.

 Proof Since f (0)  0, the function 1 − 1/ f (z) has a Taylor expansion n bn zn   in a neighborhood of 0. Multiplying up, we get ( n an zn )(1 − n bn zn ) = 1 in a neighborhood of 0, and equating coefficients then gives n 

an− j b j = an

(n ≥ 1).

(4.7)

j=0

It follows that n n−1 1  1  an− j b j = an− j−1 b j an j=0 an−1 j=0

(n ≥ 2).

Rearranging this last equation, and recalling that a0 = 1, we get bn =

n−1 

an− j−1 b j

j=0

 an an− j  − an−1 an− j−1

(n ≥ 2).

By assumption a j /a j−1 increases with j, so the term in parentheses is nonnegative. Also, we have b0 = 0 and b1 = a1 > 0. By induction it follows that  bn ≥ 0 for all n. Formula (4.7) now shows that bn ≤ an for all n, so n bn zn is holomorphic on D, and the relation (4.6) holds on the whole of D.   Proof of Lemma 4.2.6 (i) Suppose that nj=1 λ j kz j = 0. If f ∈ D, then   λ j kz j 0 = f, j

D

=

n 

λ j f (z j ).

j=1

In particular, taking f to be a polynomial with f (z1 ) = 1 and f (z j ) = 0 ( j ≥ 2),

4.2 Moduli of zero sets

59

we deduce that λ1 = 0. Likewise λ2 , . . . , λn = 0. Hence kz1 , . . . , kzn are linearly independent. (ii) For all i, j ∈ {1, . . . , n}, we have  1  1 log kzi , kz j D = kzi (z j ) =  0. zi z j 1 − zi z j (iii) For i, j ∈ {1, . . . , n}, we have hi j = 1 −

kzi , 1D 1, kz j D 1, 1D kzi , kz j D

=1−

log

zi z j  1 . 1−zi z j

By Lemma 4.2.7, applied to f (z) := (1/z) log(1/(1 − z)) = have  z 1− bn zn (z ∈ D), = 1 log( 1−z ) n≥0

 n≥0

zn /(n + 1), we

where bn ≥ 0 for all n. Therefore, given λ1 , . . . , λn ∈ C, we have n  n  n n n  2     hi j λi λ j = bn zni znj λi λ j = bn  λi zni  ≥ 0. i=1 j=1

i=1 j=1 n≥0

n≥0

i=1



This shows that H is positive semi-definite. Finally, we are ready to assemble all the pieces.

Proof of Theorem 4.2.1 Assume first that all the (zn ) are distinct. By Lemmas 4.2.5 and 4.2.6, for each n ≥ 1 we have n ! 2   |z j |2 " dist 1, span{kz1 , . . . , kzn } ≥ 1−  1  . log 1−|z j |2 j=1 Letting n → ∞, we deduce that dist(1, M)2 ≥

∞ ! 

1−

j=1

log

|z j |2 "  1  , 1−|z j |2

where M denotes the closed linear span of {kzn : n ≥ 1}. The condition (4.4) ensures that the infinite product converges to a positive limit, so dist(1, M) > 0. Thus, if we set f := 1 − P1, where P denotes the orthogonal projection of D onto M, then f  0, and f ∈ M ⊥ , so f (zn ) =  f, kzn D = 0 for all n. This proves the theorem in this case. We note, for future reference, that f (0) =  f, 1D = 1 − P1, 1D = 1 − P12D =  f 2D , and that  f 2D = dist(1, M)2 ≥

n !  1− j=1

|z j |2 "  1  . log 1−|z j |2

60

Zero sets

There remains the case where some of the (zn ) are repeated. For this we use a small trick. Let  > 0, and let (wn ) be another sequence in D, chosen so that • |wn | = |zn | for all n, • |wn − zn | <  for all n, • the (wn ) are distinct. By the argument in the preceding paragraph, there exists f ∈ D \ {0} such that f (wn ) = 0 for all n. The argument shows, moreover, that f can be chosen so that ∞ ! n !  |w j |2 "  |z j |2 " f (0) =  f 2D ≥ 1− 1−  1  =  1  . log 1−|w j |2 log 1−|z j |2 j=1 j=1 Thus, if we set g := f / f (0), then sup>0 g D < ∞, and of course g (wn ) = 0 for all n and g (0) = 1. Using Theorem 1.2.1, we see that the family (g )>0 is locally uniformly bounded, and therefore a normal family. Hence there is a sequence of  tending to zero such that the corresponding functions g converge locally uniformly on D. The limit function g is holomorphic, and by Hurwitz’s theorem g(zn ) = 0 for all n, even including multiplicities. Further g(0) = 1, so g  0. Finally, by Fatou’s lemma, D(g) ≤ sup D(g ) < ∞, so g ∈ D. This completes the proof.  Another proof of this theorem will be outlined in Exercise 5.4.3. Exercises 4.2 1. [This exercise shows that the approach region in Theorem 3.5.1 is optimal.] Let ψ : (0, 1) → (0, ∞) be a function such that ψ(t) log(1/t) → ∞ as t → 0+ . (i) Construct a sequence (rn ) in (0, 1) such that   1 −1 log < ∞ but ψ(1 − rn ) = ∞. 1 − rn n n (ii) Let (In )n≥1 be contiguous arcs in T with |In | = ψ(1 − rn ) for all n, let eiθn be the midpoint of In and set zn := rn eiθn . Show that, for almost all ζ ∈ T, the set Ω(ζ) := {z : |z − ζ| < ψ(1 − |z|)} contains infinitely many (zn ). (iii) Show that there exists f ∈ D with zero set (zn ). (iv) Deduce that the limit lim f (z) fails to exist for almost all ζ ∈ T. z→ζ z∈Ω(ζ)

2. Prove that the coefficients (bn ) in Lemma 4.2.7 satisfy

 n≥0

bn ≤ 1.

4.3 Boundary zeros I: sets of capacity zero

61

3. Let (zn ) be a Blaschke sequence in D. Define  1 − |zn |2   E := ζ ∈ T : 0, |y| < 1/(n4 (1 + x)2 )}. Let F n : D → Ωn be the Riemann mapping, normalized so that Fn (0) = 1/n2 and limr→1 Re F n (rζn ) = ∞. Note that D(Fn ) = area(Ωn )/π = 2/(πn4 )

(n ≥ 1),

and so (n ≥ 1). Fn 2D ≤ |Fn (0)|2 + 2D(Fn ) ≤ (1 + 4/π)/n4   Consequently n Fn D < ∞, and the series n≥1 Fn converges in D to a

62

Zero sets

function F. Since Re Fn ≥ 0 for all n, we clearly have Re F ≥ 0. Set f := e−F . Then f is bounded and nowhere zero on D. Further, since | f  | = |F  e−F | ≤ |F  |, we have f ∈ D. Finally, for each ζn ∈ E, lim | f (rζn )| = lim e− Re F(rζn ) ≤ lim e− Re Fn (rζn ) = 0. r→1

r→1

Thus f has all the required properties.

r→1



The preceding result applies to arbitrary countable subsets of T. If we restrict ourselves just to closed sets, then we can also treat uncountable sets of capacity zero. The argument is basically similar, the role of the Riemann mappings now being played by certain functions already constructed in the previous chapter. As usual, c(·) denotes logarithmic capacity, as defined in Chapter 2. Also, we write A(D) for the disk algebra, namely the algebra of continuous functions f : D → C that are holomorphic in D. Theorem 4.3.2 Let E be a closed subset of T with c(E) = 0. Then there exists f ∈ D ∩ A(D) such that E = {z ∈ D : f (z) = 0}. Proof According to Theorem 3.4.1, there exists a function F continuous on D \ E and holomorphic in D such that F ∈ D and limz→ζ Re F(z) = ∞ for all ζ ∈ E. Set f := e−F . Then f ∈ A(D) and f −1 ({0}) = E. Also, since f  = − f F  , we have D( f ) ≤  f 2∞ D(F) < ∞, so f ∈ D.  Thus ‘capacity zero’ is a sufficient condition for a closed set to be a boundary zero set. We shall see in the next section that it is not necessary. However, we do have the following partial converse to Theorem 4.3.2. Theorem 4.3.3 Let E be a closed subset of T with c(E) > 0. Then there exists a positive sequence (λn ) converging to 0 such that ∪n≥1 (eiλn E) is a boundary uniqueness set in the following sense: if f ∈ D ∩ H ∞ and f ∗ = 0 q.e. on ∪n≥1 (eiλn E), then f ≡ 0. Proof Since c(E) > 0, there exists a probability measure μ on E whose energy  μ(k)|2 /k < ∞. Let (ρk ) be a IK (μ) is finite. By Theorem 2.4.4, we have k≥1 |  2 μ(k)|2 /k < ∞. For t > 0, positive sequence such that ρk → ∞ but still k ρk | define min{2, kt} ω(t) := sup . ρk k≥1 Since ρk → ∞, it follows that ω(t) → 0 as t → 0+ . We may thus choose a   sequence (ln )n≥1 satisfying the conditions n ln = 1 and n ln log ω(ln ) = −∞  (see Exercise 4.3.1). Finally, we set λn := k≥n lk . We shall prove that, with this choice of (λn ), the set ∪n≥1 (eiλn E) is a uniqueness set for D ∩ H ∞ .

4.3 Boundary zeros I: sets of capacity zero

63

Let f ∈ D ∩ H ∞ with f ∗ = 0 q.e. on ∪n≥1 (eiλn E). Define g : D → C by  g(z) := f (zζ) dμ(ζ) (z ∈ D). E

Clearly g ∈ H ∞ , and by the dominated convergence theorem,  g∗ (eiλn ) = f ∗ (eiλn ζ) dμ(ζ) = 0 (n ≥ 1), E ∗

the last equality because f vanishes q.e. (and hence μ-a.e.) on eiλn E. Now the Taylor coefficients of g are related to those of f by g(k) = f (k) μ(−k) = f (k) μ(k)

(k ≥ 0).

By the Cauchy–Schwarz inequality, it follows that   1/2  | μ(k)|2 2 1/2 ρk | g(k)|ρk ≤ k| f (k)|2 | < ∞. k k≥1 k≥1 k≥1 Hence g is continuous on D, and satisfies  |2 sin k(θ1 − θ2 )/2| |g(eiθ1 ) − g(eiθ2 )| ≤ | g(k)|ρk ≤ Cω(|θ1 − θ2 |), ρk k≥1  where C := k≥1 | g(k)|ρk . In particular, since g(eiλn ) = 0, we have |g(eiθ )| ≤ Cω(ln ) and hence  1  iθ log |g(e )| dθ = 0

n≥1

λn λn+1

(λn+1 < θ < λn ),

log |g(eiθ )| dθ ≤ log C +



ln log ω(ln ) = −∞.

n≥1

As g is bounded, this implies that g ≡ 0, and in particular f (0) = g(0) = 0. If we repeat the argument successively with f (z)/zk , k = 0, 1, 2, . . . , we find that f (k) (0) = 0 for all k. Hence f ≡ 0.  Corollary 4.3.4 There exists a closed subset F of T of measure zero such that, if f ∈ D ∩ H ∞ and f ∗ = 0 q.e. on F, then f ≡ 0. Proof Let E be the circular Cantor middle-third set. Then c(E) > 0, so there exists a sequence (λn ) tending to zero such that ∪n (eiλn E) is a uniqueness set in the sense of the theorem. Set F := ∪n (eiλn E) ∪ E. Then F is a closed set of  measure zero and it too is a uniqueness set for D ∩ H ∞ . In fact, though the boundedness of f was needed in the proof of Theorem 4.3.3, both the theorem and its corollary remain true without it. One way to see this is to show that every function in D can be expressed as the quotient of two bounded functions in D (see Exercise 7.5.1).

64

Zero sets Exercises 4.3

1. Prove the following fact, needed in the proof of Theorem 4.3.3. Given a function ω : (0, ∞) → (0, ∞) such that limt→0+ ω(t) = 0, there exists a   positive sequence (ln ) such that n ln = 1 and n ln log ω(ln ) = −∞.

4.4 Boundary zeros II: Carleson sets Given a closed subset E of T of logarithmic capacity zero, Theorem 4.3.2 yields a function f ∈ D ∩ A(D) having E as its zero set. Can the degree of smoothness of f be further improved? For example, can f be chosen so that f  is continuous up to the boundary? It turns out that a new phenomenon intervenes. For n ≥ 1, let us write n A (D) := { f ∈ A(D) : f (n) ∈ A(D)}. Note that An+1 (D) ⊂ An (D) for all n ≥ 1, and that A1 (D) ⊂ D ∩ A(D). It is well known that the boundary zero sets of functions in A(D) are precisely the closed sets of Lebesgue measure zero. For n ≥ 1, however, the zero sets of An (D) are more restricted. Beurling [16] discovered the following necessary condition, and Carleson [26] proved that it is also sufficient. Definition 4.4.1 A Carleson set is a closed subset E of T such that    1 |dζ| < ∞. log dist(ζ, E) T

(4.8)

The following theorem establishes the necessity of this condition. Theorem 4.4.2 Let f ∈ A1 (D) with f  0, and let E := {ζ ∈ T : f (ζ) = 0}. Then E is a Carleson set. Proof Let M := supD | f  |. Then | f (z) − f (w)| ≤ M|z − w| for all z, w ∈ D. In particular, | f (z)| ≤ M dist(z, E) for all z ∈ D. It follows that   log | f (ζ)| |dζ| ≤ 2π log M + log(dist(ζ, E)) |dζ|. T

T

On the other hand, since f ∈ H 2 \ {0}, we certainly have  log | f (ζ)| |dζ| > −∞. T

Therefore (4.8) holds. Now we turn to sufficiency.



4.4 Boundary zeros II: Carleson sets

65

Theorem 4.4.3 Let E be a Carleson set and let n ≥ 1. Then there exists an outer function f ∈ An (D) such that E = {ζ ∈ T : f (ζ) = 0}. Furthermore, f may be chosen so that f (k) ≡ 0 on E for k = 0, . . . , n and so that f can be continued holomorphically across every point of T \ E. In particular, Carleson sets are zero sets of functions in D, and these functions can be chosen to be C n -smooth up to the boundary. Note that the conditions of being a Carleson set and of having capacity zero are independent (see Exercise 4.4.4), so neither one of the Theorems 4.3.2 and 4.4.3 implies the other. Proof Let (I j ) j≥1 be the connected components of T \ E and, for each j, let α j , β j be the endpoints of I j . Define φ : T → R+ by ⎧ ⎪ ⎪ ⎨|(ζ − α j )(ζ − β j )|, ζ ∈ I j , φ(ζ) := ⎪ ⎪ ⎩0, ζ ∈ E. Note that dist(ζ, E)2 ≤ φ(ζ) ≤ 2 dist(ζ, E), so by (4.8) log φ ∈ L1 (T). Let N ≥ 1 and let f be the outer function such that | f ∗ | = φN a.e. on T. Explicitly,  N  ζ+z log φ(ζ) |dζ| (z ∈ D). f (z) := exp 2π T ζ − z We shall prove that this f works provided that N > 2n. We first show that f extends holomorphically across every point of T \ E. For each j, the function (1 − z/α j )(1 − z/β j ) is outer, so we can write it as  1  ζ+z  (1 − z/α j )(1 − z/β j ) = exp log |(ζ − α j )(ζ − β j )| |dζ| (z ∈ D). 2π T ζ − z Dividing this into the formula for f , we obtain the decomposition f (z) = (1 − z/α j )N (1 − z/β j )N eNh j (z) where 1 h j (z) := 2π

 T\I j

(z ∈ D),

(4.9)

  φ(ζ) ζ+z log |dζ|. ζ−z |(ζ − α j )(ζ − β j )|

Notice that the integral defining h j can be taken over T \ I j , because the integrand vanishes on I j . This shows that h j extends holomorphically across the arc I j , and therefore, by (4.9), so too does f . Repeating this for every I j , we deduce that f extends holomorphically across each point of T \ E, as claimed. Next, we establish the following estimate for f : | f (z)| ≤ 2N dist(z, E)N

(z ∈ D).

(4.10)

66

Zero sets

Let w ∈ E. For each ζ ∈ T we have φ(ζ) ≤ 2|ζ − w|, and so  N  ζ + z log(2|ζ − w|) |dζ| = 2N |z − w|N | f (z)| ≤ exp Re 2π T ζ−z

(z ∈ D).

As this holds for all w ∈ E, we obtain (4.10). In particular, f extends continuously to D and f = 0 on E. Lastly, we seek analogous estimates for f (k) . For this, we re-use the decomposition (4.9). Differentiating k times the formula for h j gives    (2ζ)k! φ(ζ) 1 h(k) |dζ|, (z) = log j k+1 2π T\I j (ζ − z) |(ζ − α j )(ζ − β j )| and consequently |h(k) j (z)| ≤

Ck! dist(z, T \ I j )k+1

(z ∈ D),

where C is a constant independent of j, k. Applying Leibniz’ formula to (4.9), we deduce that | f (k) (z)| ≤

Ck | f (z)| dist(z, T \ I j )2k

(z ∈ D),

where the Ck are constants depending on k, but not on j. Repeating for each I j , we obtain Ck | f (z)| | f (k) (z)| ≤ (z ∈ D). dist(z, E)2k In combination with the estimate (4.10), this yields | f (k) (z)| ≤ 2N Ck dist(z, E)N−2k

(z ∈ D).

Thus, provided that N > 2n, the functions f, f  , . . . , f (n) all extend continuously to D and are equal to zero on E.  Exercises 4.4 1.

(i) Show that a closed subset of a Carleson set is again a Carleson set. (ii) Show that the union of two Carleson sets is again a Carleson set. 2. Let E be a closed subset of T, and let (Ik ) be the components of T \ E. Show  that E is a Carleson set if and only if |E| = 0 and k |Ik | log(1/|Ik |) < ∞. 3. Which of the following subsets of T are Carleson sets? (i) {ei/n : n ≥ 1} ∪ {1}. (ii) {ei/ log n : n ≥ 2} ∪ {1}. (iii) {eiθ : θ ∈ C}, where C is the Cantor middle-third set.

4.5 Arguments of zero sets

67

4. Give examples of closed subsets E of T to show that neither one of the conditions ‘E is a Carleson set’ and ‘E is of logarithmic capacity zero’ implies the other. [Hint: Use the preceding exercise.]

4.5 Arguments of zero sets We now return to the problem of describing zero sets of D in the interior of the unit disk. Let us recall what we have seen so far. Given a sequence (rn ) in (0, 1), there are three possibilities.  • If n (1 − rn ) = ∞, then every sequence (zn ) with |zn | = rn is a uniqueness set for D (because it is true even for H 2 ).  • If n 1/ log(1/(1 − rn )) < ∞, then every sequence (zn ) with |zn | = rn is a zero set for D (Theorem 4.2.1).   • If n (1 − rn ) < ∞ and n 1/ log(1/(1 − rn )) = ∞, then there exist sequences (zn ) and (zn ) with |zn | = |zn | = rn for all n, such that (zn ) is a zero set and (zn ) is a uniqueness set (see Theorems 4.1.2 and 4.2.2.) Thus, in the third case, the arguments of the (zn ) play a critical role in determining whether (zn ) is a zero set or a uniqueness set. This is the subject of this section. The following theorem is a substantial generalization of Theorem 4.1.2, and is in some sense dual to Theorem 4.2.1. Theorem 4.5.1 Let E be a Carleson set. If (zn ) is a Blaschke sequence such that zn /|zn | ∈ E for all n, then (zn ) is a zero set for D. The theorem is sharp in the following sense. Theorem 4.5.2 Let (θn )n≥1 be a sequence such that {eiθn : n ≥ 1} is not a Carleson set. Then there exists a positive Blaschke sequence (rn ) such that (rn eiθn ) is a uniqueness set for D. We now turn to the proofs of the two theorems. Proof of Theorem 4.5.1 By Theorem 4.4.3, since E is a Carleson set, there exists f ∈ A1 (D) such that f = 0 on E. In particular, we have f ∈ D and | f (z)| ≤ C dist(z, E) for some constant C. Given a Blaschke sequence (zn ), let B be the Blaschke product with zero set (zn ). Clearly (B f )(zn ) = 0 for all n, and we shall now show that B f ∈ D.

68

Zero sets By Theorem 4.1.3, we have  1  1 − |zn |2 D(B f ) = D( f ) + | f (ζ)|2 |dζ|. 2π T |ζ − zn |2 n

Since f ∈ D we certainly have D( f ) < ∞. Also, since | f (z)| ≤ C dist(z, E) and zn /|zn | ∈ E, we have | f (ζ)| ≤ C|ζ − zn /|zn || ≤ 2C|ζ − zn | for all n, where the last inequality comes from Lemma 1.6.7. Consequently,  1  1 − |zn |2  | f (ζ)|2 |dζ| ≤ 4C 2 (1 − |zn |2 ) < ∞. 2 2π |ζ − z | n T n n Thus D(B f ) < ∞, and the proof is complete.



For the proof of Theorem 4.5.2, we need a lemma. Lemma 4.5.3 Let E be a subset of T whose closure is not a Carleson set. Then there exists a sequence in E which converges and whose closure is not a Carleson set. Proof

By assumption, we have   log T

 1 |dζ| = ∞. dist(ζ, E)

By the compactness of T, there exists ζ0 ∈ T such that, for every neighborhood I of ζ0 ,    1 log |dζ| = ∞. dist(ζ, E) I Let (In ) be a decreasing sequence of neighborhoods of ζ0 whose intersection is equal to {ζ0 }. For each n, we may pick a finite subset Fn of E ∩ In such that    1 |dζ| ≥ n. log dist(ζ, Fn ) In Let S := ∪n Fn . Then S ⊂ E and S is a sequence converging to ζ0 . Finally, S contains Fn for each n, so       1 1 log |dζ| ≥ sup log |dζ| = ∞, dist(ζ, S ) dist(ζ, Fn ) n T T and thus the closure of S is not a Carleson set.



Proof of Theorem 4.5.2 As {eiθn : n ≥ 1} is not a Carleson set, Lemma 4.5.3 guarantees that there is a convergent subsequence of the (eiθn ) whose closure is still not a Carleson set. Relabeling, we may as well suppose that this is the whole sequence. Furthermore, since the union of two Carleson sets is again a

Notes on Chapter 4

69

Carleson set, we can suppose that (eiθn ) converges to its limit from one side. Relabeling again, we may thus assume that (θn ) is either increasing or decreasing. We shall suppose that it is decreasing, the proof for the increasing case being practically the same, with obvious modifications. Finally, there is no loss of generality in supposing that 0 < θn+1 < θn < 1 for all n and that θn → 0.  For n ≥ 1, set δn := θn − θn+1 . Clearly n δn < ∞. Also, since the closure of  {eiθn : n ≥ 1} is not a Carleson set, we have n δn log(1/δn ) = ∞. We are now in the situation of Theorem 4.1.6. By that result, if we set rn := 1−δn , then (rn eiθn )  is a uniqueness set for D, and obviously (rn ) is a Blaschke sequence. Exercises 4.5 1. Show that there exist zero sets for D of the form ((1 − n−2 )eiθn ) for which θn → 0 arbitrarily slowly.

Notes on Chapter 4 §4.1 Theorems 4.1.2 and 4.1.3 are due to Carleson [25]. Theorems 4.1.5 and 4.1.6 are due to Caughran [31].

§4.2 Theorem 4.2.1 is due to H. Shapiro and Shields [110], and Theorem 4.2.2 is due to Nagel, Rudin and J. Shapiro [83]. Slightly weaker versions of both results had earlier been obtained by Carleson in [25]. A different proof of Theorem 4.2.2, due to Richter, Ross and Sundberg, can be found in [101]. Their construction yields a uniqueness set that accumulates at just one point of the unit circle. Lemma 4.2.7 is a result of Kaluza (see [55, p. 68 and p. 91]). Exercise 4.2.3 is taken from [63], where it is also shown that there exist sequences (zn ) that are zero sets for D, but for which E is of positive logarithmic capacity.

§4.3 Theorem 4.3.1 is due to Rudin [105]. Theorem 4.3.2 is a result of Brown and Cohn [23], improving an earlier theorem of Carleson [26, Theorem 4]. In fact, rather more is true: given a closed set E of capacity zero and given a continuous function h on E, there exists f ∈ D ∩ A(D) such that f |E = h. This interpolation theorem is due to Peller and Hruˇscˇ ev [90]; another proof, due to Koosis, can be found in [123, IIIE.6]. Theorem 4.3.3 is due to Carleson [25].

70

Zero sets

§4.4 Carleson sets are also referred to in the literature as Beurling–Carleson sets and even as Beurling–Carleson–Hayman sets. An account of their history can be found in Shapiro’s article [111], which also contains a lot of further information about them. Theorem 4.4.2 is due to Beurling [16] and Theorem 4.4.3 is due to Carleson [26]. Carleson’s theorem can be extended to yield a function f ∈ A∞ (D) := ∩n≥1 An (D) with zero set E: see [69, 86, 118]. In [26], Carleson also proved the following result in the other direction: if E is a closed subset of T which is not a Carleson set, and if, in addition, for some α ∈ (0, 1), the Riesz capacity cα satisfies cα (E ∩ I) ≥ const.|I| for all arcs I with centers belonging to E, then E is a uniqueness set for D. Further extensions of this result were obtained by Maz ya and Havin [79]; see also [62]. We also mention the article of Malliavin [74], in which the author characterizes boundary uniqueness sets for D in terms of a modified capacity, defined using a complex-valued kernel.

§4.5 Theorems 4.5.1 and 4.5.2 are due to Caughran [32], though the proofs here are a bit different. The article [63] contains generalizations of Theorem 4.5.1 where, instead of stipulating that the arguments of the zn belong to E, one assumes merely that they converge to E suitably fast. Another result along these lines can be found in the article of Bogdan [19]. The proof of Lemma 4.5.3 is taken from [68]; a further proof can be found in [78].

5 Multipliers

The Dirichlet space D is not an algebra. We saw this in §1.3, where we also proved that D ∩ H ∞ is an algebra, even a Banach algebra with respect to a suitable norm. This is reminiscent of the case of Hardy spaces: H 2 is not an algebra, but H ∞ is one. However, there is also a significant difference between the Dirichlet and Hardy cases. The algebra H ∞ derives much of its importance from the fact that its elements are multipliers of H 2 , in other words, that h ∈ H∞, f ∈ H2 ⇒ h f ∈ H2. On the other hand, it turns out that not all elements of D ∩ H ∞ are multipliers of D. The functions that are multipliers of D form an algebra in their own right which, for many purposes, is the correct analogue of H ∞ in the Dirichlet-space setting. In this chapter we shall study the multipliers of D, attempting to characterize them, and establishing Dirichlet-space analogues of a number of basic results about H ∞ concerning zeros and interpolation.

5.1 Definition and elementary properties Definition 5.1.1 A multiplier of D is a function h : D → C such that h f ∈ D whenever f ∈ D. The set of multipliers is an algebra, called the multiplier algebra of D. We denote it by M(D). Clearly every multiplier of D is itself a member of D. Indeed, if h ∈ M(D), then h = h1 ∈ D. In the other direction, every polynomial is a multiplier of D. More generally, if h ∈ Hol(D) and h is bounded on D, then h is a multiplier. To see this, note that, given f ∈ D, we have (h f ) = h f +h f  = h (z f ) +(h−zh ) f  , 71

72

Multipliers

and so D(h f ) ≤ 2h 2H∞ D(z f ) + 2h − zh 2H∞ D( f ) < ∞. Further examples of multipliers are given in Exercise 5.1.3. Given h ∈ M(D), we write Mh : D → D for the multiplication operator Mh ( f ) := h f Theorem 5.1.2 itself.

( f ∈ D).

If h ∈ M(D), then Mh is a continuous linear map from D to

Proof We use the closed graph theorem. Suppose that fn → 0 in D and that Mh ( fn ) → g in D. We need to show that g = 0. For each z ∈ D, evaluation at z is a continuous linear functional on D, so g(z) = limn Mh ( fn )(z) =  h(z) limn fn (z) = 0. Hence g = 0, and we are done. This theorem allows us to define the multiplier norm  · M(D) on M(D) by hM(D) := Mh  = sup{h f D :  f D ≤ 1}

(h ∈ M(D)).

Taking f = 1, we see that hM(D) ≥ hD for all h ∈ M(D). Theorem 5.1.3

(M(D),  · M(D) ) is a Banach algebra.

Proof It is obviously a normed algebra; all that is left to prove is completeness. Suppose that (hn ) is a Cauchy sequence in M(D). Then (Mhn ) is a Cauchy sequence in the Banach algebra of B(D) of bounded linear operators on D, so there exists T ∈ B(D) such that Mhn − T B(D) → 0. Thus hn f − T ( f )D → 0 for each f ∈ D. In particular hn − T (1)D → 0. So, if we set h := T (1), then h ∈ D and hn → h pointwise on D. It follows that T ( f ) = h f for all f ∈ D.  Hence h ∈ M(D), and hn − hM(D) = Mhn − T B(D) → 0 as n → ∞. The next result establishes an important connection between multipliers and reproducing kernels. We recall from §1.2 that, if  1  1 kw (z) := log (z ∈ D), zw 1 − zw then kw ∈ D and f (w) =  f, kw D

(w ∈ D).

Also, given an operator T on a Hilbert space, we write T ∗ for its adjoint. Theorem 5.1.4

If h ∈ M(D), then Mh∗ (kw ) = h(w)kw

(w ∈ D).

Conversely, given a bounded operator T : D → D such that the kw are eigenvectors of T ∗ , there exists h ∈ M(D) such that T = Mh .

5.1 Definition and elementary properties Proof

73

Let w ∈ D. If f ∈ D, then

 f, Mh∗ (kw )D = Mh ( f ), kw D = h f, kw D = h(w) f (w) =  f, h(w)kw D . As this holds for all f ∈ D, we deduce that Mh∗ (kw ) = h(w)kw . For the converse, let h(w) be the complex conjugate of the eigenvalue of T ∗ corresponding to the eigenvector kw . Then, for each f ∈ D, we have (T f )(w) = T f, kw D =  f, T ∗ kw D = h(w) f, kw D = h(w) f (w). Hence h is a multiplier and T = Mh .



This result, though simple, has many useful consequences. One is given in Exercise 5.1.4, and another will appear when we come to study interpolation in §5.3. Here, we shall use it to establish an important property of multipliers. Theorem 5.1.5

If h ∈ M(D), then h is bounded and hH∞ ≤ hM(D) .

Proof Let h ∈ M(D). Then Mh∗ is a bounded operator on D, so its eigenvalues are bounded, and of modulus no more than Mh . By Theorem 5.1.4, it follows  that the values of h are bounded, and of modulus no more than hM(D) . We have now seen that M(D) ⊂ D∩H ∞ . Also, we showed in Theorem 1.3.2 that D ∩ H ∞ is a Banach algebra. So it is very tempting to guess that this inclusion is in fact an equality, all the more so since the analogous result for the Hardy space (namely M(H 2 ) = H ∞ ) is actually true. However, for the Dirichlet space, this guess turns out to be false, as we now show. Theorem 5.1.6

There exists h ∈ D ∩ H ∞ such that h  M(D). k

Proof For k ≥ 0, let nk := 22 . For k ≥ 1 and nk−1 ≤ n < nk , let an := 1/(nk k2 ).  We shall show that h(z) := n≥2 an zn satisfies the conclusions of the theorem. First, we note that     1   1 an = an = (nk − nk−1 ) 2 ≤ < ∞, nk k k2 n≥2 k≥1 n ≤nt}  ∞ = μ{|Cg| > t} dt2 . 0

We therefore need to estimate μ{|Cg| > t}. For this, we use Theorem 3.2.4. In the notation of that theorem, we have  |z − ζ|   Cg(ζ) (z ∈ D, ζ ∈ T). |Cg(z)| ≤ 1 + 1 − |z|  Hence, if |Cg(z)| > t, then Cg(ζ) > t/3 for all ζ in the arc with center z/|z| and  > t/3}). By (5.12), it follows that length 2(1 − |z|). Thus {|Cg| > t} ⊂ S ({Cg  > t/3})) ≤ Ac(Cg  > t/3). μ{|Cg| > t} ≤ μ(S ({Cg Feeding this information back into our previous estimate, we obtain   ∞  ∞  > t/3) dt2 = 9A  > s) ds2 . | f |2 dμ ≤ Ac(Cg c(Cg D

0

0

By the strong-type inequality for capacity, Theorem 3.3.5,  ∞  > s) ds2 ≤ Bg2 2 , c(Cg L (A) 0

where B is an absolute constant. Hence, finally,  | f |2 dμ ≤ 9ABg2L2 (A) ≤ 9AB f 2D . D

Thus μ is a Carleson measure for D, satisfying (5.11) with C = 9AB. Exercises 5.2 1. Prove the following inequality, used in the proof of Theorem 5.2.3: |k(z, w)| ≤ 2 log

2 +π |1 − wz|

(z, w ∈ D).

2. Show that the conditions (5.10) and (5.12) are equivalent to one another.



5.2 Carleson measures

83

3. Let μ be a finite positive measure on [0, 1). Using Theorem 5.2.6, show that μ is a Carleson measure for D if and only if    1  μ (1 − t, 1) = O (t → 0). log(1/t) Deduce that the necessary condition (5.7) in Theorem 5.2.5 (i) is sharp. 4. Give an example of a Carleson measure μ for D that does not satisfy (5.4). [Hint: Use Exercise 3.] 5. Let μ be a finite positive measure on D.   k k  (i) Let

f ∈ D, say

f (z) := k ak z , and set f (z) := k |ak |z . Show that 2 2 | f | dμ ≤ D |  f | d μ, where  μ is the measure on [0, 1) given by D  μ(S ) := μ({z : |z| ∈ S }) (ii) Deduce that, if    μ {z : |z| > 1 − t} = O

(S ⊂ [0, 1)).

1  log(1/t)

(t → 0),

then μ is a Carleson measure for D. [Hint: Apply Exercise 3 to  μ.] 6. Let φ : (0, 2π] → R+ be a continuous strictly increasing function such

2π that φ(x)/x is strictly decreasing and 0 (φ(x)/x) dx = ∞. The purpose of this question is to construct a finite positive measure μ on D that satisfies (5.8) but is not a Carleson measure for D, thereby showing that Theorem 5.2.5 (ii) is sharp. (i) For each n ≥ 0, set ln := φ−1 (2−n ). Show that ln+1 < ln /2 for all n and  that n 2−n log(1/ln ) = ∞. (ii) Let E be the Cantor subset of T associated to the sequence (ln ) (thus E = ∩n E n , where En is a union of 2n disjoint arcs each of length ln ). Show that E is of logarithmic capacity zero. [Use Theorem 2.3.5.] (iii) Let σ be the Cantor–Lebesgue measure associated to E, namely the unique probability measure supported on E giving weight 2−n to each of the arcs making up En . Show that σ(I) ≤ 4φ(|I|) for each arc I. (iv) Let (δn ) be any decreasing sequence in (0, 1). Let μn be the measure on D defined by μn ((1 − δn )B) := σ(B) (so μn is just σ, scaled to live  on the slightly smaller circle |z| = 1 − δn ). Finally, let μ := n≥0 2−n μn . Show that, for each arc I, we have μ(S (I)) ≤ 2σ(I), and deduce that μ satisfies (5.8). (v) Show that the set Eδn := {ζ ∈ T : d(ζ, E) ≤ δn } is a union of closed arcs I1 , . . . , Ik satisfying μ(∪k1 S (I j )) ≥ 2−n . Deduce that, if (5.10) holds, then necessarily 2−n ≤ Ac(E δn ). Conclude that, if (δn ) is chosen to

84

Multipliers decrease fast enough, then (5.10) fails and μ is not a Carleson measure for D.

5.3 Pick interpolation The classic Pick interpolation problem is to determine whether, given distinct points z1 , . . . , zn ∈ D and points w1 , . . . , wn ∈ D, there exists a holomorphic map h : D → D such that h(z j ) = w j for all j. As is well known, a necessary and sufficient condition for h to exist is that the n × n matrix (ai j ) defined by ai j :=

1 − wi w j 1 − zi z j

(i, j ∈ {1, . . . , n})

(5.13)

be positive semi-definite. In this section, we shall prove an analogue of this result for the Dirichlet space. To formulate this analogue, we first re-write the condition that h maps D into D simply as hH∞ ≤ 1. Now, as was mentioned in the introduction to the chapter, the natural counterpart of H ∞ in the Dirichlet-space setting is the multiplier algebra M(D). If we adopt this point of view, then the Dirichletspace analogue of the Pick problem can be stated as follows. Given distinct points z1 , . . . , zn ∈ D and points w1 , . . . , wn ∈ D, determine whether there exists h ∈ M(D) with hM(D) ≤ 1 such that h(z j ) = w j for all j. This is the problem that we are going to solve. Theorem 5.3.1 Let z1 , . . . , zn ∈ D be distinct points and let w1 , . . . , wn ∈ D. Then there exists h ∈ M(D) with hM(D) ≤ 1 such that h(z j ) = w j for all j if and only if the n × n matrix (ai j ) defined by ai j := (1 − wi w j )kzi , kz j D

(i, j ∈ {1, . . . , n})

(5.14)

is positive semi-definite. The condition (5.14) is the natural analogue of (5.13). Indeed, if we replace kzi , kz j D by the inner product in H 2 of the H 2 -reproducing kernels for zi , z j , then we obtain exactly (5.13). Proof of Theorem 5.3.1 The ‘only if’ is very simple. Assume that h exists. Let Mh : D → D be the multiplication operator Mh ( f ) := h f , and let Mh∗ be its Hilbert-space adjoint. Then we have Mh∗  = Mh  = hM(D) ≤ 1. Also, by Theorem 5.1.4, each reproducing kernel kz j is an eigenvector of Mh∗ with

5.3 Pick interpolation

85

eigenvalue h(z j ) = w j . Hence, for all λ1 , . . . , λn ∈ C, we have n n  2 2   0 ≤  λ j kz j D −  Mh∗ ( λ j kz j )D j=1

j=1

n n  2  2 =  λ j kz j D −  λ j w j kz j D j=1

=

n  n 

j=1

λi λ j (1 − wi w j )kzi , kz j D .

i=1 j=1

This shows that the matrix with entries ai j := (1 − wi w j )kzi , kz j D is positive semi-definite. The proof of the ‘if’ is a bit more involved. We shall need the following auxiliary result from operator theory. Lemma 5.3.2 (Parrott’s lemma) Let H be a finite-dimensional Hilbert space. Let R be a rank-one operator on H and let A be an arbitrary operator on H. Then min A + λR = max{AP, QA}, λ∈C

(5.15)

where P and Q are the orthogonal projections of H onto ker R and (im R)⊥ respectively. Proof A simple compactness argument shows that the minimum is attained. Suppose that it is attained at λ = λ0 say, and set B := A + λ0 R. Clearly min A + λR = min B + λR = B. λ∈C

λ∈C

Also, since RP = QR = 0, we have AP = BP and QA = QB. Thus, with the assumption that B + λR ≥ B for all λ ∈ C, we need to prove that B = max{BP, QB}. As P = Q = 1, we certainly have B ≥ max{BP, QB}. Suppose, if possible, that this inequality is strict. We shall show that this leads to a contradiction. Let y0 ∈ H be a unit vector for which By0  = B. We claim that B attains its norm precisely on the one-dimensional subspace spanned by y0 . Indeed, the set where B attains its norm is the subspace ker(B∗ B − B2 I), and if this subspace had dimension two or more, then it would have a non-zero intersection with ker R, on which B never attains its norm since BP < B. The claim is established.

86

Multipliers We next show that By0 , Ry0  = 0.

(5.16)

Fix θ ∈ R. For each  > 0, let y ∈ H be a unit vector on which B + eiθ R attains its norm. Since B + eiθ R ≥ B, it follows that (B + eiθ R)y 2 ≥ B2

( > 0).

Expanding this out and dividing through by , we obtain   ( → 0+ ). 2 Re e−iθ By , Ry  + O() ≥ 0 By compactness, there exists a sequence n → 0 such that the corresponding vectors yn converge. The limit vector is a unit vector in H on which B attains its norm so, by the claim above, it is necessarily a unimodular multiple of y0 . Letting  → 0 through this sequence (n ), we therefore obtain   2 Re e−iθ By0 , Ry0  ≥ 0. As θ is arbitrary, it follows that (5.16) holds. There are now two possibilities: either Ry0 = 0 or Ry0  0. In the first case, we have y0 ∈ ker R, which implies By0  = BPy0  ≤ BP < B. In the second case, thanks to (5.16) we have By0 ∈ (im R)⊥ , which implies By0  = QBy0  ≤ QB < B. Either way, B does not attain its norm on y0 , contradicting our choice of y0 . The lemma is proved.  We now return to the proof of the ‘if’ part of Theorem 5.3.1. Recall that we are given z1 , . . . , zn ∈ D and w1 , . . . , wn ∈ D such that the matrix (ai j ) defined in (5.14) is positive semi-definite, and our objective is to construct a multiplier h with hM(D) ≤ 1 such that h(z j ) = w j for all j. To this end, we define an operator T on the subspace Hn of D spanned by {kz1 , . . . , kzn } by setting T (kz j ) := w j kz j

( j = 1, . . . , n).

The calculation in the first part of the proof shows that the assumption that (ai j ) is positive semi-definite is exactly equivalent to the fact that T  ≤ 1. Now let zn+1 be any point of D distinct from z1 , . . . , zn . Given w ∈ D, we define an operator T w on Hn+1 := span{kz1 , . . . , kzn+1 } by setting ⎧ ⎪ ⎪ ⎨w j kz j , j = 1, . . . , n, T w (kz j ) := ⎪ ⎪ ⎩wkz , j = n + 1. n+1

5.3 Pick interpolation

87

Evidently T w is an extension of T . Our aim is to show that, with an appropriate choice of w, we still have T w  ≤ 1. For this, we are going to use Lemma 5.3.2. Fix a vector u ∈ Hn+1  Hn such that kzn+1 , uD = 1, and define a rank-one operator R : Hn+1 → Hn+1 by ( f ∈ Hn+1 ).

R( f ) :=  f, uD kzn+1

Notice that T w = T 0 + wR for all w ∈ C. Therefore, by Lemma 5.3.2, min T w  = max{T 0 P, QT 0 }, w∈C

where P, Q are the orthogonal projections of Hn+1 onto ker R and (im R)⊥ respectively. We shall estimate each of T 0 P and QT 0  in turn. First we consider T 0 P. This is the same thing as the restriction of T 0 to Hn , which is exactly T . As remarked earlier, T  ≤ 1, so we have T 0 P ≤ 1. Now we turn to QT 0 . This is a little more complicated. Let us note first that QT 0 (kzn+1 ) = 0. A basis for Hn+1  Ckzn+1 is given by  kz j := Q(kz j ) = kz j −

kz j , kzn+1 D kzn+1 , kzn+1 D

kzn+1

( j = 1, . . . , n).

A simple calculation shows that kz j ) = w j QT 0 ( kz j

( j = 1, . . . , n).

Thus QT 0  ≤ 1 if and only if the n × n matrix (ci j ) is positive semi-definite, where kzi ,  kz j D (i, j ∈ {1, . . . , n}). ci j := (1 − wi w j ) Now a further calculation shows that  kzi ,  kz j D = Q(kzi ), Q(kz j )D = kzi , Q(kz j )D = kzi , kz j D −

kz j , kzn+1 D kzi , kzn+1 D

kzn+1 , kzn+1 D  kzi , kzn+1 D kzn+1 , kz j D  = kzi , kz j D 1 − . kzi , kz j D kzn+1 , kzn+1 D Consequently, ci j = ai j bi j , where ai j := (1 − wi w j )kzi , kz j D and bi j := 1 −

kzi , kzn+1 D kzn+1 , kz j D . kzi , kz j D kzn+1 , kzn+1 D

The matrix (ai j ) is positive semi-definite, by assumption, and the matrix (bi j )

88

Multipliers

was shown to be positive semi-definite in Lemma 4.2.6. Their Hadamard product (ci j ) is therefore positive semi-definite by the Schur product theorem (see Theorem C.2.2 in Appendix C), and we conclude that QT 0  ≤ 1, as we wanted. In summary, we have shown that there exists wn+1 ∈ C such that the extended operator T wn+1 is still a contraction. Moreover, as wn+1 is an eigenvalue of a contraction, necessarily wn+1 ∈ D. We now iterate this argument. Let (z j ) be a dense sequence of distinct points in D, beginning with the given n-tuple z1 , . . . , zn . By applying the above procedure repeatedly, we can construct successively wn+1 , wn+2 , . . . ∈ D such that, for each m ≥ n, the operator T defined on the span of {kz j : 1 ≤ j ≤ m} by T (kz j ) := w j k j

(1 ≤ j ≤ m)

is always a contraction. As the set {kz j : j ≥ 1} spans a dense subspace of D, the operator T extends by continuity to the whole of D, still remaining a contraction. By construction, all the reproducing kernels kz j are eigenvectors of T . As (z j ) is dense in D and the map z → kz : D → D is continuous (see Exercise 5.3.1), it follows that kz is an eigenvector of T for all z ∈ D. By Theorem 5.1.4, there exists a multiplier h of D such that T = Mh∗ . Clearly this h satisfies hM(D) ≤ 1 and  h(z j ) = w j ( j = 1, . . . , n). The ‘if’ part of the theorem is sometimes expressed by saying that the Dirichlet space has the Pick property. Strictly speaking, this is a property not just of the space, but of the particular norm  · D , or equivalently, of the reproducing kernel kw . Indeed, if we were to replace  · D by the equivalent Hilbert-space norm on D given by  f 2 := | f (0)|2 + D( f )

( f ∈ D),

(5.17)

then (D,  · ) would no longer have the Pick property (see Exercise 5.3.3). Exercises 5.3 1. As usual, let kw be the reproducing kernel for D at w. Prove that the map w → kw : D → D is continuous. 2. Show that, if h ∈ M(D) with hM(D) ≤ 1, then  h(z ) − h(z ) 2 |kz1 , kz2 D |2 1 2    ≤1− kz1 2D kz2 2D 1 − h(z1 )h(z2 )

(z1 , z2 ∈ D).

3. Let  ·  be the Hilbert-space norm on D given by (5.17).

5.4 Zeros of multipliers

89

(i) Show that the associated reproducing kernel kw is given by  kw (z) = 1 + log

1  1 − wz

(w ∈ D, z ∈ D).

(ii) Show that the pairs (z1 , z2 ) := (0, z0 ) and (w1 , w2 ) := (0, w0 ) satisfy the condition (5.14) with respect to this kernel if and only if   (1 − |w0 |2 ) 1 + log

1  ≥ 1. 1 − |z0 |2

(5.18)

 (iii) Let h(z) := k≥0 ak zk be a multiplier of D whose multiplier norm with  respect to  ·  is at most 1. Show that k≥0 (k + 1)|ak |2 ≤ 1. Deduce that, if h(0) = 0 and h(z0 ) = w0 , then 1 |z0 | |w0 | ≤ √ 2 1 − |z0 |

(z ∈ D).

(5.19)

(iv) Show that there exists a pair (z0 , w0 ) satisfying (5.18) but not (5.19). Conclude that (D,  · ) does not have the Pick property.

5.4 Zeros of multipliers If f ∈ H 2 and f  0, then its zeros satisfy the Blaschke condition, and the corresponding Blaschke product is a function in H ∞ with the same zeros as f . The following theorem is the analogous result for the Dirichlet space. Theorem 5.4.1

Given f ∈ D, there exists h ∈ M(D) with the same zero set.

We note straightaway the following consequence. Corollary 5.4.2

The union of two zero sets for D is again a zero set.

Proof By the theorem, the two sets in question are the zero sets of multipliers h1 , h2 . Their union is the zero set of the product h1 h2 , which belongs to D.  We now turn to the proof of Theorem 5.4.1. In contrast to the case of Hardy spaces, we cannot simply build h using a Blaschke product: indeed the Dirichlet space contains no infinite Blaschke products (see Exercise 4.1.1). Instead, we use the Pick interpolation theorem from the previous section. Proof of Theorem 5.4.1 We can suppose that  f D = 1. Let (z j ) j≥1 be the zeros of f , counted according to multiplicity. Fix z0 ∈ D such that f (z0 )  0.

90

Multipliers

Set w0 := f (z0 )/kz0 2D and w j := 0 ( j ≥ 1). Then, for each n ≥ 1 and each (n + 1)-tuple λ0 , λ1 , . . . , λn ∈ C, we have n  n 

n 2  λi λ j (1 − wi w j )kzi , kz j D =  λ j kz j  − |λ0 |2 | f (z0 )|2

i=0 j=0

j=0 n n 2   2  =  λ j kz j  −  f, λ j kz j D  ≥ 0. j=0

j=0

By Theorem 5.3.1, there exists hn ∈ M(D) with hn M(D) ≤ 1 such that hn (z0 ) = f (z0 )/kz0 D and hn (z j ) = 0 ( j = 1, . . . , n). Then, for all n, we have hn H ∞ ≤ hn M(D) ≤ 1, so (hn )n≥1 is a normal family, and some subsequence converges locally uniformly on D. The limit function h is also a multiplier with hM(D) ≤ 1 (see Exercise 5.1.2), and clearly h(z0 ) = f (z0 )/kz0 D  0 and h(z j ) = 0 ( j ≥ 1). This is not quite the end of the story, since it is possible that h has more zeros than f . However, we can simply divide them out to obtain the desired function (see Exercise 5.4.1).  We shall give another proof of this theorem, in a slightly more general form, at the end of Chapter 8 (see Corollary 8.3.10). Also in Chapter 9, we shall establish an analogue for boundary zero sets (see Theorem 9.2.6). Exercises 5.4 1. Let h ∈ Hol(D) and B be a Blaschke product. Show that, if hB ∈ M(D), then also h ∈ M(D) and hM(D) ≤ hBM(D) . [Use Theorem 4.1.3.] 2. Let Z be a zero set for D and let z0 ∈ D \ Z. Define cD := inf{ f D : f ∈ D, f |Z = 0, f (z0 ) = 1}, cM(D) := inf{hM(D) : h ∈ M(D), h|Z = 0, h(z0 ) = 1}. Prove that cM(D) /cD = kz0 D . [The proof of Theorem 5.4.1 gives the inequality ≤. For the reverse inequality, consider f := hkz0 /kz0 2D .] 3. [Another proof of Theorem 4.2.1.] Let (zn ) be a sequence in D \ {0} such that   1 −1 < ∞. log 1 − |zn | n

Notes on Chapter 5

91

(i) For each n, show that there exists hn ∈ M(D) with hn M(D) ≤ 1 such that hn (zn ) = 0 and |hn (0)| ≥ 1 − (ii) Let h := h(0)  0.

 n

log

|zn |2  1 . 1−|zn |2

hn . Show that h ∈ M(D) and h(zn ) = 0 for all n, but

Notes on Chapter 5 §5.1 Theorem 5.1.6 is based on an example of Taylor [119]. Another proof that not every bounded function in D is a multiplier can be found in [117]. Exercise 5.1.3 is a result of Brown and Shields [24].

§5.2 The notion of Carleson measure (for H p -spaces) was introduced by Carleson in his solution of the corona problem [28]. Theorem 5.2.2 is based on a result of Arcozzi, Rochberg and Sawyer in [10]. Theorem 5.2.5 (ii) was obtained by Wynn in [124] under the additional assumption that φ is a concave function, and the sharpness result of Exercise 5.2.6 was also established in [124]. The proofs given here are from [42]. Theorem 5.2.6 is due to Stegenga [117]. As outlined in Exercise 5.2.3, Stegenga’s theorem leads to a simple characterization of those Carleson measures for D supported on [0, 1), but this result can also be obtained directly, see [114]. Exercise 5.2.5 is taken from [102]. There are a number of other characterizations of Carleson measures for D, though none of them are particularly simple. For example, Arcozzi, Rochberg and Sawyer have shown that a finite positive measure μ on D is a Carleson measure for D if and only if there exists a constant C such that, for all closed arcs I in T,   Re k(z, w) dμ(w) dμ(z) ≤ Cμ(S (I)). S (I)

S (I)

For a discussion of this and other characterizations, we refer to the articles [9, 10, 11, 64].

§5.3 Theorem 5.3.1 is due to Agler [2], and Lemma 5.3.2 is a special case of a theorem of Parrott [89]. In fact the Dirichlet space enjoys a stronger property, the so-called complete Pick property, which corresponds to a version of Theorem 5.3.1 for matrix-valued multipliers. Shimorin [116] has shown that the weighted Dirichlet space Dw has the complete Pick property whenever w is a positive superharmonic function (even in the absence

92

Multipliers

of an explicit formula for the reproducing kernel). The book of Agler and McCarthy [4] contains a detailed account of the complete Pick property for general reproducing kernels.

§5.4 The proof of Theorem 5.4.1 is taken from the preprint of Marshall and Sundberg [76], which is also the source for Exercises 5.4.2 and 5.4.3. The main purpose of [76] was to establish a characterization of interpolating sequences for D. This result, which was also proved independently by Bishop [17] and later simplified by Bøe [18], can be stated as follows. Given a sequence of distinct points (zn )n≥1 in D, we have  $ # h(z1 ), h(z2 ), h(z3 ), . . . : h ∈ M(D) = ∞  if and only if n kzn −2 D δzn is a Carleson measure for D and sup m,n mn

|kzn , kzm D | < 1. kzn D kzm D

For a detailed treatment, we refer to the book of Seip [109].

6 Conformal invariance

Back in §1.4, we saw that if φ is a M¨obius automorphism of the unit disk, then D( f ◦ φ) = D( f ) for all f ∈ D. Despite its simplicity, this observation has already proved useful, notably in the proof of Theorem 4.1.3. In this chapter, we start by showing that this M¨obius-invariance property essentially characterizes the Dirichlet space. We then go on to study the problem of determining which other maps φ : D → D leave D invariant in the sense that f ∈ D ⇒ f ◦ φ ∈ D. This leads to the notion of composition operators on D, to which we provide a brief introduction.

6.1 M¨obius invariance Let Aut(D) denote the group of holomorphic automorphisms of the unit disk, namely the M¨obius maps of the form φ(z) := eiθ

a−z 1 − az

(a ∈ D, eiθ ∈ T).

There are many interesting spaces X of holomorphic functions on D that are M¨obius invariant in the sense that f ∈ X, φ ∈ Aut(D) ⇒ f ◦ φ ∈ X. Some of them are even Hilbert spaces. For example, both H 2 and D have this property (for the case of H 2 , see Exercise 6.1.2). However, rather remarkably, the stronger property that D( f ◦ φ) = D( f )

( f ∈ D, φ ∈ Aut(D))

essentially characterizes D among Hilbert space of holomorphic functions. Here is a precise statement of the theorem. 93

94

Conformal invariance

Theorem 6.1.1 Let H be a vector subspace of Hol(D) and let ·, · be a semiinner product on H. Assume the following. (i) (ii) (iii) (iv)

If f ∈ H and φ ∈ Aut(D), then f ◦ φ ∈ H and  f ◦ φ, f ◦ φ =  f, f .  f 2 := | f (0)|2 +  f, f  defines a Hilbert-space norm  ·  on H. Convergence in this norm implies pointwise convergence in D. The space H contains at least one non-constant function.

Then H = D, and there is a constant c such that  f, f  = cD( f ) for all f ∈ H.  Proof We claim that, if f = k≥0 ak zk ∈ H, then ak zk ∈ H for each k, with ak zk  ≤  f . To see this, fix k ≥ 0 and consider the sequence of functions 1 f (e2πi j/n z)e−2πi jk/n n j=1 n

gn (z) :=

(n ≥ 1).

(6.1)

By assumption (i), each gn ∈ H, and satisfies gn  ≤  f . By assumption (ii), (H,  · ) is a Hilbert space, so there exists a subsequence (gn j ) that converges weakly to some g ∈ H with g ≤  f . For each fixed z ∈ D, we have  2π 1 f (eiθ z)e−ikθ dθ = ak zk . (6.2) lim gn (z) = n→∞ 2π 0 By assumption (iii), evaluation at z is a continuous linear functional with respect to the norm  · , so it follows that g(z) = ak zk for all z ∈ D. This justifies the claim. By assumption (iv), the space H contains a non-constant function f0 . Then  f0 = k≥0 ak zk with ak  0 for at least one k > 0. By the claim, it follows that zk ∈ H for this k. Using assumption (i), we deduce that φkt ∈ H, where φt (z) := (t − z)/(1 − tz) (0 < t < 1). The Taylor series of φkt is φt (z)k = φt (0)k + kφt (0)k−1 φt (0)z + · · · = tk + ktk−1 (t2 − 1)z + · · · . In particular, the Taylor coefficient of z is non-zero so, re-applying the claim, we deduce that z ∈ H. By assumption (i) again, we have φt ∈ H. The Taylor series of φt is  φt (z) = t + (t − 1/t) t jz j. j≥1

In particular, all the Taylor coefficients are non-zero so, using the claim yet again, we have zk ∈ H for all k ≥ 0. The monomials zk are mutually orthogonal with respect to ·, ·. This is a simple consequence of assumption (i). Indeed, given j, k ≥ 0 with j  k, choose θ with ei( j−k)θ  1. By assumption (i), we then have z j , zk  = (eiθ z) j , (eiθ z)k  = ei( j−k)θ z j , zk ,

6.1 M¨obius invariance

95

whence z j , zk  = 0. Next, we compute zk , zk . For this, we use the part of the claim which says  that, if f = k ak zk ∈ H, then ak zk  ≤  f  for all k. Applying this with f = φt , we obtain (t − 1/t)tk zk  ≤ φt  = (|φt (0)|2 + φt , φt )1/2 = (t2 + z, z)1/2 . In particular, taking t = 1 − 1/k, it follows that k2 − k (1 − 1/k)−k (1 + z, z)1/2 = O(k) (k → ∞). 2k − 1  Thus, if 0 < t < 1, then the series k≥0 tk zk converges absolutely in H. This permits us to expand the following inner product in powers of z: zk  ≤

1 − tφt , 1 − tφt  =

  1 − t2 1 − t2 , = (1 − t2 )2 t2k zk , zk  (0 < t < 1). 1 − tz 1 − tz k≥0

On the other hand, by assumption (i), we also have 1 − tφt , 1 − tφt  = 1 − tz, 1 − tz = 1, 1 + t2 z, z

(0 < t < 1).

If we compare the right-hand sides of the last two expressions and equate powers of t, then we obtain z, z − 21, 1 = z, z, z , z  − 2z k

k

k−1

,z

k−1

 + zk−2 , zk−2  = 0

(k ≥ 2).

Solving this recurrence gives zk , zk  = kz, z for all k ≥ 0. To summarize, we have now proved that H contains all polynomials p, and that p, p = cD(p), where c := z, z. We next show that D is a closed subspace of H and that  f, f  = cD( f ) for all f ∈ D. Given f ∈ D, let fn denote the n-th partial sum of the Taylor series of f . Since the differences fn − fm are polynomials vanishing at 0, we have  fn − fm 2 = cD( fn − fm ) for all m, n. Thus ( fn ) is a Cauchy sequence with respect to  · . By assumption (ii), there exists  f ∈ H such that  fn −  f  → 0.  f = f. By assumption (iii), this implies that fn → f pointwise in D, so in fact  We conclude that f ∈ H and that  f, f  = limn→∞  fn , fn  = limn→∞ cD( fn ) = cD( f ). Since D is complete, it is necessarily closed in H. The last thing to prove is that D is the whole of H. For this, we consider H  D, the orthogonal complement of D in H. Using assumption (i), it is easy to check that, if f ∈ H D, then f ◦φ ∈ H D for all φ of the form φ(z) = eiθ z. Thus, the same argument as that given at the beginning of the proof shows that,  if f = k ak zk ∈ H  D, then ak zk ∈ H  D for all k. But since ak zk ∈ D, this

96

Conformal invariance

implies that ak = 0 for all k, and so f = 0. Thus H  D = {0}, and D is the whole of H, as desired.  Exercises 6.1 1. Find the Taylor series expansion of the function gn defined in (6.1). Use this series to give another proof of (6.2). 2. Let φ ∈ Aut(D), say φ(z) = eiθ (a − z)/(1 − az), where |a| < 1. (i) Show that, if f is holomorphic in a neighborhood of D, then   1 − |a|2 | f (λ)| |dλ| = |( f ◦ φ)(ζ)| |dζ|. |1 − aζ|2 T T (ii) Deduce that  f ◦ φ2H2 ≤

1 + |a| 2  f H 2 . 1 − |a|

(iii) Deduce that the same inequality holds for all f ∈ H 2 .

6.2 Composition operators Given a holomorphic map φ : D → D, we write Cφ : Hol(D) → Hol(D) for the composition operator defined by C φ ( f ) := f ◦ φ

( f ∈ Hol(D)).

In this section, we address the problem of determining those functions φ for which Cφ maps D back into D. Let us begin with some elementary observations. If Cφ (D) ⊂ D, then the linear map Cφ : D → D is automatically continuous. This is a simple consequence of the closed graph theorem. We shall write C φ  for its operator norm. Also, for this to happen, φ itself must belong to D, since φ = Cφ (z). In the other direction, we saw in §1.4 that, if φ : D → D is univalent, then Cφ (D) ⊂ D. To study non-univalent φ, we introduce the counting function nφ . Given a holomorphic map φ : D → D and w ∈ D, we write nφ (w) for the number (possibly infinite) of solutions of the equation φ(z) = w, counted according to multiplicity. If φ is non-constant, then nφ is a lower semicontinuous function on D. In particular it is measurable. Theorem 6.2.1 Let φ : D → D be a holomorphic function. Then Cφ (D) ⊂ D if and only if there exists a constant B such that   ( f ∈ D). (6.3) | f  |2 nφ dA ≤ B | f  |2 dA

6.2 Composition operators

97

For the proof, we need a generalization of the change-of-variable formula for non-univalent functions. Lemma 6.2.2 Let φ : D → D be a holomorphic map and let u : D → [0, ∞] be a positive measurable function. Then   2 (u ◦ φ)|φ | dA = unφ dA. D

D

Proof We may assume that φ is non-constant, otherwise the result is obvious. Then Z := {z ∈ D : φ (z) = 0} is a countable set, and D \ Z may be written as a countable union of disjoint rectangles (R j ) on each of which φ is a diffeomorphism onto its range. By the usual change-of-variable formula, for each j we have   (u ◦ φ)|φ |2 dA = u dA. φ(R j )

Rj

Summing over j, we deduce that    2 (u ◦ φ)|φ | dA = u 1φ(R j ) dA. D\Z

D

j

Now Z, being countable, is of measure zero. Therefore the left-hand side equals

 (u ◦ φ)|φ |2 dA. Also j 1φ(R j ) = nφ on D \ φ(Z), and φ(Z), being countable, D is of measure zero. Thus the right-hand side equals D unφ dA. Putting together these facts, we obtain the result.  By the lemma, applied with u := | f  |2 , we have  1 | f  |2 nφ dA ( f ∈ D). D( f ◦ φ) = π D

Proof of Theorem 6.2.1

Therefore, if (6.3) holds, then D( f ◦ φ) < ∞ for all f ∈ D, and so Cφ (D) ⊂ D. Conversely, as already remarked, if C φ (D) ⊂ D, then Cφ : D → D is a bounded linear map, and consequently, Cφ ( f )D ≤ Cφ   f D

( f ∈ D).

This implies that D( f ◦ φ) ≤ 2Cφ 2 D( f )

( f ∈ D).

By Lemma 6.2.2 again, it follows that   1 1 | f  |2 nφ dA ≤ 2Cφ 2 | f  |2 dA π D π D which shows that (6.3) holds with B := 2Cφ 2 .



98

Conformal invariance

This theorem plays much the same sort of role for composition operators that Theorem 5.1.7 did for multipliers. Just like Theorem 5.1.7, it is unsatisfactory inasmuch as the criterion (6.3) is hard to check in practice. By analogy with the multiplier case, we are therefore led to study measures μ on D satisfying   B | f  |2 dμ ≤ | f  |2 dA ( f ∈ D), π D D or, equivalently,



B |h| dμ ≤ π D



2

D

|h|2 dA

(h ∈ B),

where B is the Bergman space on D, defined by    1 B := h ∈ Hol(D) : h2B := |h|2 dA < ∞ . π D In other words, using the language of §5.2, we seek to characterize Carleson measures for B. Fortunately, this turns out to be somewhat easier than in the case of Carleson measures for D. We recall from §5.2 that, given an arc I ⊂ T, the corresponding Carleson box S (I) is defined by S (I) := {reiθ : eiθ ∈ I, 1 − |I| < r < 1}. Theorem 6.2.3 Let μ be a finite positive measure on D. The following statements are equivalent. (i) There exists a constant B such that  |h|2 dμ ≤ Bh2B

(h ∈ B).

D

(ii) For w ∈ D,

 D

  dμ(z) = O (1 − |w|2 )−2 |1 − wz|4

(|w| → 1).

(iii) For arcs I ⊂ T, μ(S (I)) = O(|I|2 )

(|I| → 0).

For the proof, we need an elementary lemma. If h ∈ B, then  (1 − |w|2 )2 3 |h(z)|2 ≤ |h(w)|2 dA(w) π D |1 − wz|4

Lemma 6.2.4

(z ∈ D).

(6.4)

6.2 Composition operators Proof

99

For z ∈ D and n ≥ 0, we have  (1 − |w|2 )2 wn dA(w) (1 − wz)4 D   wn (1 − |w|2 )2 (k + 3)(k + 2)(k + 1)(wz)k dA(w) = D

k≥0

  = (k + 3)(k + 2)(k + 1)zk k≥0 n

0

1



2π

rn+k (1 − r2 )2 ei(n−k)θ r dθ dr

0

= πz /3. By linearity and approximation, it follows that, for all f ∈ Hol(D) ∩ L1 (D, dA),  3 (1 − |w|2 )2 f (z) = f (w) dA(w) (z ∈ D). π D (1 − wz)4 In particular, this holds for f = h2 , where h ∈ B. Inequality (6.4) follows.



Proof of Theorem 6.2.3 We first show that (i) implies (ii). Let h := ψ , where ψ(z) := (z − w)/(1 − wz). A computation gives |h(z)|2 = (1 − |w|2 )2 /|1 − wz|4 and h2B = D(ψ) = D(z) = 1. Therefore, if (i) holds, then  (1 − |w|2 )2 dμ(z) ≤ B (w ∈ D). 4 D |1 − wz| Thus (ii) holds. Next we show that (ii) implies (i). For this, we use Lemma 6.2.4. By that lemma, if h ∈ B, then      (1 − |w|2 )2 3 |h(z)|2 dμ(z) ≤ |h(w)|2 dA(w) dμ(z) 4 D D π D |1 − wz|   dμ(z)  3 2 2 2 dA(w). = |h(w)| (1 − |w| ) 2 π D D |1 − wz| Therefore, if (ii) holds, then there exists a constant B such that   1 2 |h| dμ ≤ B|h|2 dA = Bh2B (h ∈ B), π D D and so (i) holds. Now we show that (ii) implies (iii). Given an arc I ⊂ T with |I| < 1/4, choose w ∈ D so that |w| = 1 − |I| and w/|w| is the midpoint of I. If z ∈ S (I), then elementary estimates give |1 − wz| ≤ 3|I|, and so   μ(S (I)) dμ(z) dμ(z) ≥ ≥ . 4 4 (3|I|)4 D |1 − wz| S (I) |1 − wz|

100

Conformal invariance

Consequently, if (ii) holds, then there exists a constant B such that B B μ(S (I)) ≤ ≤ . (3|I|)4 (1 − |w|2 )2 |I|2 It follows that μ(S (I)) ≤ 34 B|I|2 , and therefore (iii) holds. Finally, we prove that (iii) implies (ii). Let w ∈ D with |w| > 1/2. Using Fubini’s theorem, we have    ∞ dμ(z) = 4t−5 dt dμ(z) 4 D |1 − wz| D t=|1−wz|  ∞ = 4t−5 dμ(z) dt t=0 {z:|1−wz|≤t}  ∞ = μ{z ∈ D : |1 − wz| ≤ t}4t−5 dt. t=0

Now {z ∈ D : |1 − wz| ≤ t} = {z ∈ D : |z − 1/w| ≤ t/|w|}. This set is contained in S (I) for some arc I with |I| = 8t, and it is even empty if t < 1 − |w|. Thus, if (iii) holds, then ⎧ ⎪ ⎪ t < 1 − |w|, ⎨0, μ{z ∈ D : |1 − wz| ≤ t} ≤ ⎪ ⎪ ⎩Ct2 , t ≥ 1 − |w|, where C is a constant independent of w. Feeding this into the previous estimate, we obtain  ∞  2C 8C dμ(z) ≤ Ct2 4t−5 dt = ≤ . 4 2 (1 − |w|) (1 − |w|2 )2 D |1 − wz| t=1−|w| 

This proves that (ii) holds.

We can now read off the characterization of those φ for which Cφ (D) ⊂ D. Theorem 6.2.5 Let φ : D → D be a holomorphic map such that φ ∈ D. The following statements are equivalent. (i) C φ (D) ⊂ D. (ii) For w ∈ D,  nφ (z)   1 dA(z) = O (1 − |w|2 )−2 π D |1 − wz|4 (iii) For arcs I ⊂ T, 1 π

(|w| → 1).

 nφ dA = O(|I|2 ) S (I)

(|I| → 0).

(6.5)

6.2 Composition operators

101

Proof By Lemma 6.2.2, the fact that φ ∈ D implies that dμ := nφ dA is a finite measure on D. The result is therefore an immediate consequence of Theorems 6.2.1 and 6.2.3 combined.  As an application of this theorem, we derive a simple sufficient condition for Cφ (D) ⊂ D. It is part (ii) of the following result. Theorem 6.2.6 all k ≥ 1.

Let φ : D → D be a holomorphic map such that φk ∈ D for

(i) If Cφ (D) ⊂ D, then D(φk ) = O(k) as k → ∞. (ii) If D(φk ) = O(1) as k → ∞, then C φ (D) ⊂ D. Proof

(i) If C φ (D) ⊂ D, then

D(φk ) ≤ φk 2D = C φ (zk )2D ≤ Cφ 2 zk 2D = C φ 2 (k + 1)

(k ≥ 0).

(ii) Assume now that D(φk ) ≤ B for all k ≥ 1. We shall verify that (6.5) holds. For each w ∈ D, we have   nφ (z) nφ (z) 24 1 dA(z) ≤ dA(z) π D |1 − wz|4 π D |1 − w2 z2 |4   24 (k + 2)(k + 1)k|w2 z2 |k−1 nφ (z) dA(z) = π D k≥1   4 2 k−1 1 ≤ 6.2 k|w | k2 |z|2k−2 nφ (z) dA(z) π D k≥1  k|w2 |k−1 D(φk ). = 6.24 ≤

k≥1 4

6.2 B . (1 − |w|2 )2 

Hence (6.5) holds, as required.

In Theorem 5.1.4 we saw that the multiplication operators could be characterized as those operators on D whose adjoints have the reproducing kernels as eigenvectors. The following result is the analogue for composition operators. Theorem 6.2.7

If φ : D → D is holomorphic and Cφ (D) ⊂ D, then C φ∗ (kw ) = kφ(w)

(w ∈ D).

Conversely, if T : D → D is a bounded operator such that T ∗ maps each reproducing kernel to a reproducing kernel, then there exists a holomorphic map φ : D → D such that T = Cφ .

102

Conformal invariance

Proof

Assume that Cφ (D) ⊂ D. Let w ∈ D. For each f ∈ D, we have  f, C φ∗ (kw )D = Cφ ( f ), kw D = f (φ(w)) =  f, kφ(w) D .

It follows that Cφ∗ (kw ) = kφ(w) . For the converse, suppose that, for each w ∈ D, we have T ∗ (kw ) = kφ(w) , where φ : D → D is some unknown function. For each f ∈ D, we then have (T f )(w) = T ( f ), kw D =  f, T ∗ (kw )D =  f, kφ(w) D = f (φ(w))

(w ∈ D).

In particular, taking f (z) = z, we see that φ = T (z) ∈ Hol(D). The calculation  above then shows that T ( f ) = Cφ ( f ) for all f ∈ D, whence the result. If φ : D → D is holomorphic and Cφ (D) ⊂ D, then    1   1 = O log (|z| → 1). log 1 − |φ(z)|2 1 − |z|2

Corollary 6.2.8

Proof From the theorem, kφ(z) 2D = Cφ∗ (kz )2D ≤ Cφ 2 kz 2D (z ∈ D). Recalling that kw 2D = kw (w) = |w|−2 log(1/(1 − |w|2 )), we obtain the result.  Exercises 6.2 1. Let φ : D → D be a non-constant holomorphic map. Show that its counting function nφ is lower semicontinuous on D. 2. Let μ be a finite positive measure on D. By examining the proof of Theorem 6.2.3, prove that the following ‘little o’ statements are equivalent. (i) For w ∈ D,

 D

  dμ(z) = o (1 − |w|2 )−2 |1 − wz|4

(|w| → 1).

(ii) For arcs I ⊂ T, μ(S (I)) = o(|I|2 )

(|I| → 0).

6.3 Compactness criteria Now that we have characterized those φ for which C φ : D → D is a bounded operator, it seems only natural to ask when Cφ is compact. An answer is provided by the following analogue of Theorem 6.2.5, in which the ‘big O’ conditions are replaced by corresponding ‘little o’ ones. Theorem 6.3.1 Let φ : D → D be a holomorphic map such that φ ∈ D. The following statements are equivalent.

6.3 Compactness criteria (i) Cφ : D → D is compact. (ii) For w ∈ D,  nφ (z)   1 dA(z) = o (1 − |w|2 )−2 4 π D |1 − wz| (iii) For arcs I ⊂ T, 1 π

103

(|w| → 1).

 nφ dA = o(|I|2 )

(|I| → 0).

S (I)

Proof The equivalence between (ii) and (iii) follows from Exercise 6.2.2. We shall prove the equivalence between (i) and (ii). Suppose first that (i) holds. For w ∈ D \ {0}, set ψw (z) :=

|w| w − z w 1 − wz

(z ∈ D).

Then ψw 2D = ψw 2H2 +D(ψw ) = 2 for all w ∈ D\{0}, and ψw → 1 pointwise as |w| → 1. Hence ψw → 1 weakly in D as |w| → 1. As Cφ is compact, it follows that Cφ (ψw ) → Cφ (1) in norm as |w| → 1, in other words, that ψw ◦φ−1D → 0 as |w| → 1. This implies that D(ψw ◦φ) → 0 as |w| → 1. Also, by Lemma 6.2.2,  1 (1 − |w|2 )2 D(ψw ◦ φ) = nφ (z) dA(z) (w ∈ D \ {0}). π D |1 − wz|4 Putting these facts together, we deduce that (ii) holds. Now suppose that (ii) holds. Let ( fk ) be a sequence such that fk → 0 weakly in D. We need to show that Cφ ( fk ) → 0 in norm. Since Cφ ( fk )2D ≤ | fk (φ(0))|2 + 2D( fk ◦ φ) and fk → 0 pointwise on D, it suffices to show that D( fk ◦ φ) → 0 as k → ∞. We now estimate this quantity. Using Lemmas 6.2.2 and 6.2.4, we have  1 | f  (z)|2 nφ (z) dA(z) D( fk ◦ φ) = π D k    1 3 (1 − |w|2 )2  ≤ | fk (w)|2 dA(w) nφ (z) dA(z) 4 π D π D |1 − wz|    3 1 (1 − |w|2 )2 nφ (z) dA(z) | fk (w)|2 dA(w). = 4 π D π D |1 − wz| Since (ii) holds, given  > 0, there exists r ∈ (0, 1) such that  (1 − |w|2 )2 1 nφ (z) dA(z) <  (r < |w| < 1). π D |1 − wz|4

104

Conformal invariance

Also, it is clear that, if |w| ≤ r, then   nφ (z) 1 D(φ) 1 (1 − |w|2 )2 nφ (z) dA(z) ≤ dA(z) = . π D |1 − wz|4 π D (1 − r)4 (1 − r)4 Substituting these estimates into the inequality for D( fk ◦ φ), we obtain   D(φ)  3 3 D( fk ◦ φ) ≤ | fk (w)|2 dA(w) + | f (w)|2 dA(w). π r t) log dt < ∞, 1−t 1−t 0 and this is, in a certain sense, best possible. The article [71] contains further results along these lines.

7 Harmonically weighted Dirichlet spaces

In §1.6, we introduced the notion of a weighted Dirichlet integral. Given a measurable function w : D → (0, ∞), we defined  1 | f  (z)|2 w(z) dA(z) ( f ∈ Hol(D)), Dw ( f ) := π D Dw := { f ∈ Hol(D) : Dw ( f ) < ∞}. In this chapter we are going to examine in some detail a special class of weights, namely those w that are harmonic functions on D. There are (at least) two reasons why these particular weights are important. Firstly, a general harmonic weight w can be synthesized as an average of extreme harmonic weights, which turn out to be particularly simple to treat. In particular, this is true of the constant weight w ≡ 1, which corresponds to the standard Dirichlet integral. We shall see in the course of the chapter that this idea is a very useful tool. Secondly, the harmonically weighted spaces Dw play a central role in the theory of shift-invariant subspaces of the Dirichlet space. This will be the subject of the next chapter.

7.1 Dμ -spaces and the local Dirichlet integral It is well known that every positive harmonic function w on D can be represented as the Poisson integral Pμ of a finite (positive, Borel) measure μ on T, namely  1 − |z|2 w(z) = Pμ(z) := dμ(ζ) (z ∈ D). 2 T |z − ζ| This leads us to make the following definition. 108

7.1 Dμ -spaces and the local Dirichlet integral

109

Definition 7.1.1 Let μ be a finite measure on T with μ(T) > 0. The associated harmonically weighted Dirichlet integral and harmonically weighted Dirichlet space are defined respectively by  1 | f  (z)|2 Pμ(z) dA(z) ( f ∈ Hol(D)), Dμ ( f ) := π D Dμ := { f ∈ Hol(D) : Dμ ( f ) < ∞}. Clearly Dμ is a subspace of Hol(D) containing the polynomials. The following result tells us a bit more. Theorem 7.1.2 Let μ be a finite measure on T with μ(T) > 0. Then Dμ ⊂ H 2 , and Dμ is a Hilbert space with respect to the norm  · Dμ given by  f 2Dμ :=  f 2H2 + Dμ ( f )

( f ∈ Dμ ).

Proof This is actually a special case of Theorem 1.6.3. All we have to show is that lim inf |z|→1 Pμ(z)/(1 − |z|) > 0. And this is indeed true, because  1 + |z| Pμ(z) μ(T) = > 0 (z ∈ D).  dμ(ζ) ≥ 2 1 − |z| 4 T |z − ζ| There are three special cases particularly worthy of note. The first is when μ = m, normalized Lebesgue measure on T. In this case we have Pm ≡ 1, so Dm ( f ) = D( f ), the standard Dirichlet integral. The second case is when μ = 0. This is not actually covered by the definition, but, in view of the last result, it is logical to extend Definition 7.1.1 to this case by defining D0 ( f ) := 0 and D0 := H 2 . Finally, we consider the case when μ = δζ , the Dirac measure at ζ ∈ T. In this case, we write Dζ ( f ) in place of Dδζ ( f ). We thus have  1 − |z|2 1 Dζ ( f ) = | f  (z)|2 dA(z). π D |z − ζ|2 This is called the local Dirichlet integral of f at ζ, and the corresponding space Dζ is called the local Dirichlet space at ζ. Theorem 7.1.3

Let μ be a finite measure on T. Then  Dμ ( f ) = Dζ ( f ) dμ(ζ) ( f ∈ Hol(D)). T

In particular, D( f ) =

1 2π

(7.1)

 T

Dζ ( f ) |dζ|

( f ∈ Hol(D)).

(7.2)

110

Harmonically weighted Dirichlet spaces

Proof

By Fubini’s theorem:    1 − |z|2 1  2 | f (z)| dμ(ζ) dA(z) Dμ ( f ) = 2 π D T |z − ζ|     1 − |z|2 1 = | f  (z)|2 dA(z) dμ(ζ) π D |z − ζ|2 T Dζ ( f ) dμ(ζ). = T

This gives (7.1), and (7.2) follows by taking μ to be Lebesgue measure.



Thanks to this useful result, it frequently suffices to treat Dζ ( f ) rather than Dμ ( f ) for general μ. This can help simplify the notation. Exercises 7.1 1. Verify directly that Dμ (zn ) = nμ(T). 2. Let f ∈ Hol(D) and suppose that f  is bounded on D. Show that f ∈ Dμ for every finite measure μ on T, and that Dμ ( f ) ≤  f  2∞ μ(T). 3. Let ( fn ), f ∈ Hol(D), and suppose that fn → f locally uniformly on D. Show that, for each finite measure μ on T, Dμ ( f ) ≤ lim inf Dμ ( fn ). n→∞

4. Let μ be a probability measure on T. (i) Given f ∈ Hol(D) and λ ∈ T, define fλ (z) := f (λ−1 z). Show that  1 Dμ ( fλ ) |dλ| = D( f ). 2π T Deduce that, if f ∈ D, then fλ ∈ Dμ for at least one λ ∈ T. (ii) Deduce that Dμ  H ∞ . (iii) Deduce that Dμ is not an algebra.

7.2 The local Douglas formula Recall that, given f ∈ Hol(D) and ζ ∈ T, we write f ∗ (ζ) for the radial limit limr→1− f (rζ), whenever it exists. In §1.5, we established a formula of Douglas expressing D( f ) in terms of f ∗ . We now seek to extend this formula to the local Dirichlet integral Dζ ( f ). As a by-product, we shall also derive a formula

7.2 The local Douglas formula

111

for Dζ ( f ) in terms of the Taylor coefficients of f , analogous to Theorem 1.1.2. Both are consequences of the following basic theorem, which establishes a criterion for membership in the local Dirichlet space Dζ . Theorem 7.2.1 Let f ∈ Hol(D) and let ζ ∈ T. Then Dζ ( f ) < ∞ if and only if f (z) = a + (z − ζ)g(z), where g ∈ H 2 and a ∈ C. In this case Dζ ( f ) = g2H2 , and f ∗ (ζ) = a. Moreover, f (z) → f ∗ (ζ) as z → ζ in each oricyclic approach region |z − ζ| < κ(1 − |z|2 )1/2 (κ > 0). For the proof, we need two lemmas. n−1 k z . Then (pn )n≥1 is an Lemma 7.2.2 For n ≥ 1, define pn (z) := nzn − k=0 orthogonal basis of H 2 , and pn 2H2 = n(n + 1) for all n. m−1 m Proof Let 1 ≤ m < n. Writing pm (z) = k=0 (z − zk ), we have pn , pm H2 =

m−1 m−1    (−1 + 1) = 0. pn , zm H 2 − pn , zk H2 = k=0

k=0

Thus (pn )n≥1 is an orthogonal sequence in H 2 . If f ∈ H 2 and f ⊥ (pn )n≥1 , then (writing p0 := 0) we have pn − pn−1 = n(zn − zn−1 ), and so  f, zn H 2 −  f, zn−1 H2 =  f, pn − pn−1 H 2 /n = 0

(n ≥ 1).

Therefore all the Taylor coefficients of f are equal, which, together with the fact that f ∈ H 2 , implies that f = 0. Thus (pn )n≥1 is an orthogonal basis of H 2 . n−1 2 1 = n(n + 1).  Finally, for each n we have pn 2H2 = n2 + k=0 To state the second lemma, we introduce the space    1 2 |h(z)|2 (1 − |z|2 ) dA(z) < ∞ . A := h ∈ Hol(D) : hA := π D

(7.3)

Note that A is a Hilbert space. The monomials (zn )n≥0 form an orthogonal basis of A, and zn 2A = 1/(n + 1)(n + 2) for all n. Also, convergence in A implies local uniform convergence in D (see Exercise 7.2.1). Lemma 7.2.3

Define T : Hol(D) → Hol(D) by T g(z) :=

((z − 1)g(z)) z−1

(g ∈ Hol(D)).

Then the restriction of T to H 2 maps H 2 isometrically onto A. Proof Let (pn )n≥1 be the sequence defined in Lemma 7.2.2. A simple calculation shows that (z−1)pn (z) = nzn+1 −(n+1)zn +1, and thus T pn (z) = n(n+1)zn−1 .

112

Harmonically weighted Dirichlet spaces

Therefore (T pn )n≥1 is an orthogonal sequence in A. Furthermore, for each n ≥ 1, T pn 2A = n2 (n + 1)2 zn−1 2A = n(n + 1) = pn 2H2 . It follows that T gA = gH 2 whenever g is a finite linear combination of (pn ). For a general g ∈ H 2 , we argue as follows. Since (pn )n≥1 spans a dense subspace of H 2 , we can find a sequence (gk ) in H 2 such that each gk is a finite linear combination of pn and gk − gH2 → 0. By what we have already proved, T gk − T gl A = gk − gl H 2 for all k, l, so (T gk ) is a Cauchy sequence in A. As A is complete, there exists h ∈ A such that T gk − hA → 0. Then both gk → g and T gk → h locally uniformly in D, so the continuity properties of T imply that h = T g. Thus T gA = lim T gk A = lim gk H2 = gH 2 . k→∞

k→∞

We conclude that T is an isometry from H 2 into A. Finally, since the image of T is closed in A and contains the sequence  (zn−1 )n≥1 , it follows that T maps H 2 onto A. Proof of Theorem 7.2.1 We shall prove the result for ζ = 1. The general case follows by a simple change of variable z → ζz. By definition, D1 ( f ) < ∞ means that  1 1 − |z|2 | f  (z)|2 dA(z) < ∞. π D |z − 1|2 This is equivalent to saying that f  (z)/(z−1) ∈ A, where A is the space defined in (7.3). By Lemma 7.2.3, this happens if and only if there exists g ∈ H 2 such that f  (z)/(z − 1) = T g(z). This in turn means that f  (z) = ((z − 1)g(z)) , or equivalently, that f (z) = a + (z − 1)g(z) for some a ∈ C. This establishes the basic equivalence. Now assume that f (z) = a + (z − 1)g(z), where g ∈ H 2 . Using the fact that T is an isometry from H 2 to A, we have  f  (z) 2  = D1 ( f ). g2H 2 = T g2A =  z−1 A Also, by Lemma 1.5.4 we have lim|z|→1 (1 − |z|2 )|g(z)| = 0, and so  |z − 1|2  | f (z) − a|2 = |z − 1|2 |g(z)|2 = o 1 − |z|2

(|z| → 1).

Therefore f (z) → a as z → 1 in any approach region |z − 1| < κ(1 − |z|2 )1/2 .  Corollary 7.2.4

If f ∈ Dμ , then f ∗ (ζ) exists μ-a.e. on T.

7.2 The local Douglas formula

113

Proof Let f ∈ Dμ . By Theorem 7.1.3, Dζ ( f ) < ∞ for μ-almost every ζ ∈ T, and by Theorem 7.2.1, f ∗ (ζ) exists at each such ζ.  It is now a simple matter to derive a version of Douglas’ formula for Dζ . Theorem 7.2.5 (Local Douglas formula) Let f ∈ H 2 and let ζ ∈ T. If f ∗ (ζ) exists, then  1 | f ∗ (λ) − f ∗ (ζ)|2 |dλ|. (7.4) Dζ ( f ) = 2π T |λ − ζ|2 Otherwise Dζ ( f ) = ∞. Of course, the original Douglas formula Theorem 1.5.1 follows upon integrating both sides of (7.4) with respect to Lebesgue measure and using (7.2). Proof of Theorem 7.2.5 Suppose first that Dζ ( f ) < ∞. Then f ∗ (ζ) exists by Theorem 7.2.1, and Dζ ( f ) = g2H2 , where g(z) := ( f (z) − f ∗ (ζ))/(z − ζ). So (7.4) certainly holds in this case. It remains to prove that, if f ∗ (ζ) exists and Dζ ( f ) = ∞, then the right-hand side of (7.4) is infinite. Consider g(z) := ( f (z) − f ∗ (ζ))/(z − ζ). The numerator of g is in H 2 , and the denominator is an outer function. If we had g∗ ∈ L2 (T), then by the Smirnov maximum principle (see Theorem A.3.8 in Appendix A) it would follow that g ∈ H 2 , and Theorem 7.2.1 would imply that Dζ ( f ) < ∞, contrary to hypothesis. Therefore g∗ L2 (T) = ∞, which amounts to saying that the right-hand side of (7.4) is infinite, as required.  Using Theorem 7.2.1, or rather the ideas in its proof, we can also derive a formula for Dζ ( f ) in terms of the Taylor coefficients of f .  Theorem 7.2.6 Let f ∈ Hol(D), say f (z) = k≥0 ak zk , and let ζ ∈ T. Then n  2  1  kak ζ k  . (7.5) Dζ ( f ) = n(n + 1) k=1 n≥1 Proof Making the change of variable z → ζz, we can reduce to the case ζ = 1. So we just need to prove that n 2  1  (7.6) D1 ( f ) =  kak  . n(n + 1) k=1 n≥1 Suppose first that D1 ( f ) < ∞. By Theorem 7.2.1, f (z) = a+(z−1)g(z), where g ∈ H 2 and D1 ( f ) = g2H2 . We can compute g2H2 via Parseval’s formula, using the orthogonal basis (pn )n≥1 of H 2 furnished by Lemma 7.2.2:  |g, pn  2 |2  |g, pn  2 |2 H H g2H2 = . = 2 n(n + 1) p  n H2 n≥1 n≥1

114

Harmonically weighted Dirichlet spaces n−1  bk . Also, as Now, writing g(z) = k≥0 bk zk , we have g, pn H 2 = nbn − k=0 f (z) = a + (z − 1)g(z), we have a0 = a − b0 and ak = bk−1 − bk for all k ≥ 1. It  follows that g, pn H 2 = − nk=1 kak . Combining these facts, we obtain (7.6). Now suppose that D1 ( f ) = ∞. We need to show that the right-hand side of (7.6) is infinite too. Suppose, if possible, that it is finite. Then the series g := −

n 

kak

n≥1 k=1



pn n(n + 1)

converges in H 2 . Applying the operator T defined in Lemma 7.2.3 to both sides, we obtain n   kak zn−1 , T g(z) = − n≥1 k=1

the series converging in A, and hence locally uniformly in D. This implies that ((z − 1)g(z)) = −(z − 1)

n 

∞   kak zn−1 = kak zk−1 = f  (z).

n≥1 k=1

k=1

It follows that f (z) = a + (z − 1)g(z) for some constant a, and so D1 ( f ) < ∞, contrary to hypothesis. This completes the proof.  Exercises 7.2 1. Let A be defined by (7.3). Show that, if h ∈ A, then |h(z)|2 ≤

2h2A (1 − |z|2 )3

(z ∈ D).

Deduce that convergence in A implies local uniform convergence in D. 2. (i) Show that, if f ∈ Hol(D), then, for every ζ ∈ T,  f − f (0)2H2 ≤ 4Dζ ( f ). (ii) Let f (z) := (z − 1)(z − z2 + z3 + · · · + (−1)n+1 zn ). Verify that D1 ( f ) = n and that  f 2H2 = 4n − 2. Deduce that the constant 4 in part (i) is sharp. 3. Show that integrating formula (7.5) with respect to Lebesgue measure leads to (1.1).  4. Let f ∈ Hol(D), say f (z) = k≥0 ak zk , and suppose that Dζ ( f ) < ∞. By Theorem 7.2.1, f (z) = f ∗ (ζ) + (z − ζ)g(z), where g ∈ H 2 .   (i) Writing g(z) = k≥0 bk zk , show that nk=0 ak ζ k = f ∗ (ζ) − bn ζ n+1 .  k ∗ (ii) Deduce that ∞ k=0 ak ζ converges to f (ζ).

7.3 Approximation in Dμ

115

(iii) Show that Dζ ( f ) =

∞ 2    ak ζ k  .

(7.7)

n≥0 k=n+1

(iv) We now have two apparently different formulas for Dζ ( f ) in terms of the Taylor coefficients of f , namely (7.5) and (7.7). Show directly that the right-hand sides of these formulas are in fact equal.  5. (Converse to previous exercise.) Let f ∈ Hol(D), say f (z) = k≥0 ak zk , and  let ζ ∈ T. Prove that, if k≥0 ak ζ k converges and ∞ 2    ak ζ k  < ∞, n≥0 k=n+1

then Dζ ( f ) < ∞.

7.3 Approximation in Dμ Given f ∈ Hol(D) and 0 < r < 1, let us write fr for the r-dilation of f , namely fr (z) := f (rz). Each of the functions fr is then holomorphic in a neighborhood of D. It is obvious from the formula (1.1) that, if f ∈ D, then  fr − f D → 0 as r → 1− . The corresponding result for Dμ is also true, but trickier to prove. This is our aim in this section. Theorem 7.3.1 Let μ be a positive finite measure on T, and let f ∈ Dμ . Then  fr − f Dμ → 0 as r → 1− . The key to the proof is the following estimate. Lemma 7.3.2 Let μ be a positive finite measure on T, and let f ∈ Dμ . Then Dμ ( fr ) ≤ Dμ ( f ) for 0 < r < 1. This in turn depends on the following lemma. Lemma 7.3.3 Let f ∈ H 2 and let w ∈ D. Then   | f ∗ (λ) − f (w)|2 | f ∗ (λ)|2 − | f (w)|2 |dλ| = |dλ|. 2 |λ − w| |λ − w|2 T T Proof

The difference between the two sides is  2| f (w)|2 − 2 Re( f (w) f ∗ (λ)) |dλ|. |λ − w|2 T

(7.8)

(7.9)

116

Harmonically weighted Dirichlet spaces

Now u(z) := | f (w)|2 − Re( f (w) f (z)) is a harmonic function on D, which is equal to the Poisson integral of its radial boundary values u∗ . In particular,  1 − |w|2 ∗ 1 u (λ) |dλ| = u(w) = 0. 2π T |λ − w|2 Since u∗ (λ) = | f (w)|2 − Re( f (w) f ∗ (λ)) a.e. on T, it follows that the integral (7.9) vanishes, as desired.  Proof of Lemma 7.3.2 It suffices to show that Dζ ( fr ) ≤ Dζ ( f ) for all ζ ∈ T. The result then follows by integrating both sides with respect to dμ(ζ). Fix ζ ∈ T. We may suppose that Dζ ( f ) < ∞. Then, by Lemma 7.2.3, we have f (z) = a + (z − ζ)g(z), where g ∈ H 2 and g2H 2 = Dζ ( f ). Let r < 1. A simple computation shows that fr (z) = b + (z − ζ)hr (z), where b := f (rζ) and h(z) := r

(z − ζ)g(z) − (rζ − ζ)g(rζ) . z − rζ

By Lemma 7.2.3 again, it follows that Dζ ( fr ) = hr 2H2 ≤ h2H2 . Now    (λ − ζ)g∗ (λ) − (rζ − ζ)g(rζ) 2 1 r2  h2H2 =  |dλ| 2π T λ − rζ    (λ − ζ)g∗ (λ) 2 1 ≤ r2   |dλ| 2π T λ − rζ ≤ g2H2 , where the first inequality comes from Lemma 7.3.3, and the second one holds because, using Lemma 1.6.7, we have |r(λ − ζ)| 2r ≤ ≤1 |λ − rζ| 1+r

(λ ∈ T).

Putting all these estimates together, we obtain Dζ ( fr ) ≤ Dζ ( f ), as desired. Proof of Theorem 7.3.1



Using the parallelogram identity, we have

Dμ ( fr − f ) + Dμ ( fr + f ) = 2Dμ ( f ) + 2Dμ ( fr )

(0 < r < 1).

By Lemma 7.3.2, Dμ ( fr ) ≤ Dμ ( f ) for all r ∈ (0, 1). Also, by Fatou’s lemma, lim inf r→1 Dμ ( fr + f ) ≥ Dμ (2 f ) = 4Dμ ( f ). Hence lim supr→1 Dμ ( fr − f ) ≤ 0. Therefore, finally,    lim−  fr − f 2Dμ = lim−  fr − f 2H 2 + Dμ ( fr − f ) = 0 + 0 = 0. r→1

r→1

We conclude with a simple but important application of this result. Corollary 7.3.4

The polynomials are dense in Dμ .

7.4 Outer functions

117

Proof Let f ∈ Dμ and let  > 0. By Theorem 7.3.1, there exists r ∈ (0, 1) such that  fr − f Dμ < /2. Fix this r, and let sn fr be the n-th partial sum of the Taylor expansion of fr around zero. Since fr is holomorphic on a neighborhood of D, the sequences (sn fr )n≥0 and (sn fr )n≥0 converge uniformly on D to fr and fr respectively, and it follows that sn fr − fr Dμ < /2 if n is large enough. Then sn fr is a polynomial and sn fr − f Dμ < .  It is not in general true, however, that the Taylor series of f ∈ Dμ converges to f in the norm of Dμ . See Exercise 7.3.2 below. Exercises 7.3 1. Let f ∈ Hol(D) and let ζ ∈ T. Show that Dζ ( fr ) is an increasing function of r ∈ (0, 1), and that Dζ ( fr ) → Dζ ( f ) as r → 1− . 2. Given f ∈ Hol(D), denote by sn f the n-th partial sum of the Taylor expansion of f about zero. (i) Show that, if f (z) := zn − zn+1 , then  f 2D1 = 3 and sn f 2D1 = n + 1. (ii) Deduce that there exists f ∈ D1 such that supn≥0 sn f D1 = ∞. In particular, sn f → f in D1 . [Hint: Banach–Steinhaus theorem.]

7.4 Outer functions Recall that f ∈ Hol(D) is an outer function if it is of the form   1  ζ+z log φ(ζ) |dζ| (z ∈ D), f (z) = exp 2π T ζ − z where φ : T → R+ is a function such that log φ ∈ L1 (T). In this case the radial limit f ∗ (ζ) := limr→1− f (rζ) exists for a.e. ζ ∈ T, and | f ∗ | = φ a.e. Also f ∈ H 2 if and only if φ ∈ L2 (T). For more information about outer functions, see Appendix A.2. An outer function f is uniquely determined by | f ∗ |, and this is frequently how it is specified in practice. It is therefore desirable to be able to express D( f ), or more generally Dμ ( f ), purely in terms of | f ∗ | (as opposed to f ∗ ). Notice that Douglas’ formula does not achieve this, because the difference | f ∗ (λ) − f ∗ (ζ)| depends not only on the moduli of f ∗ (λ) and f ∗ (ζ), but also on their arguments, arguments which may be difficult to determine in practice. A formula for D( f ) of the type that we seek was discovered by Carleson, and subsequently extended to the case of Dζ ( f ) by Richter and Sundberg. Here are their results.

118

Harmonically weighted Dirichlet spaces

Theorem 7.4.1 (Carleson’s formula) Let f ∈ H 2 be an outer function. Then   1 (| f ∗ (λ)|2 − | f ∗ (ζ)|2 )(log | f ∗ (λ)| − log | f ∗ (ζ)|) D( f ) = 2 |dλ| |dζ|. 4π T T |λ − ζ|2 Theorem 7.4.2 (Richter–Sundberg formula) Let f ∈ H 2 be an outer function, and let ζ ∈ T be a point such that f ∗ (ζ) exists. Then  | f ∗ (λ)|2 − | f ∗ (ζ)|2 − 2| f ∗ (ζ)|2 log | f ∗ (λ)/ f ∗ (ζ)| 1 Dζ ( f ) = |dλ|. (7.10) 2π T |λ − ζ|2 Remarks (i) For certain classes of functions, Carleson’s formula can be used to derive estimates for the Dirichlet integral which, though not exact formulas, are of a much simpler form. A useful example of this kind, for so-called distance functions, will be given in Theorem 9.4.3. (ii) Formula (7.10) contains an ambiguity if f ∗ (ζ) = 0. In this case, the product | f ∗ (ζ)|2 log | f ∗ (λ)/ f ∗ (ζ)| should be interpreted as being zero. (iii) The integrands in both formulas are non-negative, so the integrals are well defined, possibly infinite. (iv) If f  H 2 or if the radial limit f ∗ (ζ) does not exist, then Dζ ( f ) = ∞, as we already saw in Theorem 7.2.1. (v) Of course, given a finite measure μ on T, we can get a formula for Dμ ( f ) as usual, by integrating both sides of (7.10) with respect to dμ(ζ). The case when μ is Lebesgue measure leads to the following proof of Theorem 7.4.1. Proof of Theorem 7.4.1 Since f ∈ H 2 , it follows that f ∗ (ζ) exists and is nonzero for a.e. ζ ∈ T. Thus (7.10) holds for a.e. ζ ∈ T. Integrating both sides with respect to normalized Lebesgue measure, we find that   | f ∗ (λ)|2 − | f ∗ (ζ)|2 − 2| f ∗ (ζ)|2 log | f ∗ (λ)/ f ∗ (ζ)| 1 |dλ| |dζ|. D( f ) = 2 4π T T |λ − ζ|2 If we exchange the roles of λ and ζ, then we also have   1 | f ∗ (ζ)|2 − | f ∗ (λ)|2 − 2| f ∗ (λ)|2 log | f ∗ (ζ)/ f ∗ (λ)| |dλ| |dζ|. D( f ) = 2 4π T T |λ − ζ|2 Taking the average of these two expressions for D( f ), we obtain the formula in the statement of the theorem.  It remains to prove Theorem 7.4.2. This will occupy us for much of the rest of the section. The main idea is to re-express the local Douglas formula (7.4) purely in terms of | f ∗ |. Our starting point in this endeavor is Lemma 7.3.3. Recall that it says that, if f ∈ H 2 and w ∈ D, then   | f ∗ (λ) − f (w)|2 | f ∗ (λ)|2 − | f (w)|2 |dλ| = |dλ|. (7.11) 2 |λ − w| |λ − w|2 T T

7.4 Outer functions

119

The plan is to take w = rζ in both sides of (7.11) and let r → 1. We begin with the left-hand side. Let f ∈ H 2 and let ζ ∈ T. Then  | f ∗ (λ) − f (rζ)|2 |dλ| = 2πDζ ( f ). lim r→1 T |λ − rζ|2

Lemma 7.4.3

Proof

We first show that  | f ∗ (λ) − f (rζ)|2 lim inf |dλ| ≥ 2πDζ ( f ). r→1 |λ − rζ|2 T

(7.12)

We may assume that the left-hand side is finite. Pick a sequence rn → 1 such that the family of functions gn (z) := ( f (z) − f (rn ζ))/(z − rn ζ) satisfies  1 | f ∗ (λ) − f (rζ)|2 |dλ|. lim gn 2H 2 = lim inf n→∞ r→1 2π T |λ − rζ|2 A subsequence (gnk ) then converges weakly in H 2 to a limit g. In particular, gnk (0) → g(0), so the corresponding subsequence of scalars f (rnk ζ) converges to a limit a, and we have f (z) = a + (z − ζ)g(z). Using Theorem 7.2.1, we deduce that Dζ ( f ) = g2H2 ≤ lim gnk 2H2 , k→∞

which gives (7.12). Now we show that lim sup r→1

 T

| f ∗ (λ) − f (rζ)|2 |dλ| ≤ 2πDζ ( f ). |λ − rζ|2

(7.13)

We may assume that the right-hand side is finite. Then, in particular, f ∗ (ζ) exists. Lemma 7.3.3, applied with f replaced by f − f ∗ (ζ), gives   | f ∗ (λ) − f (w)|2 | f ∗ (λ) − f ∗ (ζ)|2 |dλ| ≤ |dλ| (w ∈ D). |λ − w|2 |λ − w|2 T T It follows that lim sup r→1

 T

 | f ∗ (λ) − f (rζ)|2 | f ∗ (λ) − f ∗ (ζ)|2 |dλ| ≤ lim sup |dλ| 2 |λ − rζ| |λ − rζ|2 r→1 T  | f ∗ (λ) − f ∗ (ζ)|2 |dλ|, ≤ |λ − ζ|2 T

where the second inequality comes from the dominated convergence theorem (justified using the inequality of Lemma 1.6.7). This establishes (7.13).  Now we turn our attention to the right-hand side of (7.11). It is here that we need to assume that f is outer.

120

Harmonically weighted Dirichlet spaces

Lemma 7.4.4 Let f ∈ H 2 be an outer function, and let ζ ∈ T be a point such that limr→1 | f (rζ)| = 1. Then   | f ∗ (λ)|2 − | f (rζ)|2 | f ∗ (λ)|2 − 1 − 2 log | f ∗ (λ)| |dλ| = |dλ|. lim 2 r→1 T |λ − rζ| |λ − ζ|2 T Proof Since f is an outer function, log | f | is the Poisson integral of log | f ∗ |. It follows that  log | f ∗ (λ)/ f (rζ)| |dλ| = 0 (0 < r < 1). (7.14) |λ − rζ|2 T Hence,  | f ∗ (λ)|2 − | f (rζ)|2 |dλ| |λ − rζ|2 T  | f ∗ (λ)|2 − | f (rζ)|2 − 2| f (rζ)|2 log | f ∗ (λ)/ f (rζ)| = |dλ| |λ − rζ|2 T

(0 < r < 1).

Now | f (rζ)| → 1 as r → 1. Also, the integrands on the right-hand side are all positive, because x − 1 − log x ≥ 0 for all x > 0. Therefore we may apply Fatou’s lemma, to deduce that   | f ∗ (λ)|2 − | f (rζ)|2 | f ∗ (λ)|2 − 1 − 2 log | f ∗ (λ)| lim inf |dλ| ≥ |dλ|. r→1 |λ − rζ|2 |λ − ζ|2 T T In the other direction, we begin by observing that, for all a, b > 0, a2 − 1 − 2 log a = (a2 − b2 ) + (b2 − 1 − 2 log b) − 2 log(a/b) ≥ a2 − b2 − 2 log(a/b), and thus | f ∗ (λ)|2 − 1 − 2 log | f ∗ (λ)| ≥ | f ∗ (λ)|2 − | f (rζ)|2 − 2 log | f ∗ (λ)/ f (rζ)|). The last term on the right-hand side satisfies (7.14). Therefore we have   | f ∗ (λ)|2 − 1 − 2 log | f ∗ (λ)| | f ∗ (λ)|2 − | f (rζ)|2 |dλ| ≥ |dλ| (0 < r < 1). |λ − rζ|2 |λ − rζ|2 T T It follows that   | f ∗ (λ)|2 − | f (rζ)|2 | f ∗ (λ)|2 − 1 − 2 log | f ∗ (λ)| |dλ| ≤ lim sup |dλ| lim sup 2 |λ − rζ| |λ − rζ|2 r→1 r→1 T T  | f ∗ (λ)|2 − 1 − 2 log | f ∗ (λ)| |dλ|. ≤ |λ − ζ|2 T Here the last inequality follows from the dominated convergence theorem if the final integral is finite (justified again using Lemma 1.6.7), and is obvious anyway if the integral is infinite. 

7.4 Outer functions

121

Now we can prove the Richter–Sundberg formula. Proof of Theorem 7.4.2 Suppose first that f ∗ (ζ)  0. Replacing f by f / f ∗ (ζ), we can suppose that f ∗ (ζ) = 1. Set w = rζ in (7.11), and let r → 1 using Lemmas 7.4.3 and 7.4.4. This yields  | f ∗ (λ)|2 − 1 − 2 log | f ∗ (λ)| |dλ|, 2πDζ ( f ) = |λ − ζ|2 T giving (7.10) in this case. If f ∗ (ζ) = 0, then the right-hand side of (7.10) is the same as the right-hand side of (7.4), so the result follows directly from Theorem 7.2.5 in this case.  The final result of this section complements the Richter–Sundberg formula. Theorem 7.4.5 Let f be an outer function (not necessarily in H 2 ). Let ζ ∈ T, let a ≥ 0, and suppose that  | f ∗ (λ)|2 − a2 − 2a2 log | f ∗ (λ)/a| |dλ| < ∞. |λ − ζ|2 T Then Dζ ( f ) < ∞, and f ∗ (ζ) exists and satisfies | f ∗ (ζ)| = a. Remark a = 0.

The term 2a2 log | f ∗ (λ)/a| should be interpreted as being zero if

Proof The case when a > 0 can be reduced to a = 1 by considering f /a. There are thus two cases to consider, namely a = 0 and a = 1. If a = 0, then the hypothesis becomes  | f ∗ (λ)|2 |dλ| < ∞. 2 T |λ − ζ| Thus f (z) = (z − ζ)g(z) where g ∈ H 2 , and so by Theorem 7.2.1 we have Dζ ( f ) < ∞ and f ∗ (ζ) = 0, proving the result in this case. For the rest of the proof, we suppose that a = 1. The hypothesis becomes  | f ∗ (λ)|2 − 1 − 2 log | f ∗ (λ)| |dλ| < ∞. (7.15) |λ − ζ|2 T The first step is to show that (7.15) implies that f ∈ H 2 . For this, we use the inequality x2 − 1 − 2 log x ≥ (x − 1)2 , which, together with (7.15), gives  (| f ∗ (λ)| − 1)2 |dλ| < ∞. |λ − ζ|2 T This easily implies that f ∗ ∈ L2 (T). Hence, by Smirnov’s maximum principle (see Theorem A.3.8 in Appendix A), we do indeed have f ∈ H 2 .

122

Harmonically weighted Dirichlet spaces

Next, we show that limr→1 | f (rζ)| = 1. For this, let us recall that, since f is outer, log | f | is the Poisson integral of log | f ∗ |. Hence, if φ : R → R+ is any positive convex function, then, by Jensen’s inequality, we have    1 − r2   1 φ log | f ∗ (λ)| |dλ| φ log | f (rζ)| ≤ 2π T |λ − rζ|2  1 − r2 4 ≤ φ(log | f ∗ (λ)|) |dλ|. 2π |λ − ζ|2 T Let us apply this inequality with φ(x) := e2x − 1 − 2x, which is indeed both positive and convex. The last integral is then finite by (7.15). It follows that limr→1 φ(log | f (rζ)|) = 0, which in turn forces limr→1 | f (rζ)| = 1, as claimed. To conclude, we use successively Lemmas 7.4.3, 7.3.3 and 7.4.4 to obtain  | f ∗ (λ) − f (rζ)|2 2πDζ ( f ) = lim |dλ| r→1 T |λ − rζ|2  | f ∗ (λ)|2 − | f (rζ)|2 = lim |dλ| r→1 T |λ − rζ|2  | f ∗ (λ)|2 − 1 − 2 log | f ∗ (λ)| |dλ|. = |λ − ζ|2 T In particular Dζ ( f ) < ∞. By Theorem 7.2.1 f ∗ (ζ) exists and, from what we have already proved, | f ∗ (ζ)| = limr→1 | f (rζ)| = 1. The proof is complete.  Exercises 7.4 1. Let f ∈ Hol(D) be an outer function. Show that, if f  H2 , then Carleson’s formula (Theorem 7.4.1) continues to hold in the sense that   (| f ∗ (λ)|2 − | f ∗ (ζ)|2 )(log | f ∗ (λ)| − log | f ∗ (ζ)|) |dλ| |dζ| = ∞. |λ − ζ|2 T T [Hint: Use Theorem 7.4.5.]

7.5 Lattice operations in Dμ Let ∨, ∧ denote the usual lattice operations in R+ , namely x∨y := max{x, y} and x ∧ y := min{x, y}. We extend these operations to outer functions as follows. Definition 7.5.1 Given outer functions f, g, we define f ∨ g and f ∧ g to be the unique outer functions satisfying the following relations a.e. on T: |( f ∨ g)∗ (ζ)| = | f ∗ (ζ)| ∨ |g∗ (ζ)|, |( f ∧ g)∗ (ζ)| = | f ∗ (ζ)| ∧ |g∗ (ζ)|.

7.5 Lattice operations in Dμ

123

We shall show that the Dμ -spaces are stable under these operations. In particular, this is true of D. Theorem 7.5.2 Let μ be a finite measure on T and let f, g be outer functions in Dμ . Then f ∨ g and f ∧ g belong to Dμ , and Dμ ( f ∨ g) ≤ Dμ ( f ) + Dμ (g), Dμ ( f ∧ g) ≤ Dμ ( f ) + Dμ (g). The proof depends on the results of the previous section and the following elementary lemma. Lemma 7.5.3

Define F : (0, ∞) × [0, ∞) → R+ by ⎧ ⎪ ⎪ ⎨ x − y − y log(x/y), x > 0, y > 0, F(x, y) := ⎪ ⎪ ⎩ x, x > 0, y = 0.

Then, for all x1 , x2 ∈ (0, ∞) and y1 , y2 ∈ [0, ∞), we have F(x1 ∨ x2 , y1 ∨ y2 ) ≤ F(x1 , y1 ) + F(x2 , y2 ), F(x1 ∧ x2 , y1 ∧ y2 ) ≤ F(x1 , y1 ) + F(x2 , y2 ).

Proof We first establish a monotonicity property of F. A simple computation gives ∂F/∂x = 1 − y/x and ∂F/∂y = log(y/x). It follows that the line y = x is the valley of the graph of F in the following sense: • x → F(x, y) is decreasing for x ≤ y and increasing for x ≥ y, • y→ F(x, y) is decreasing for y ≤ x and increasing for y ≥ x. For the main part of the proof, we can suppose without loss of generality that x1 ≤ x2 . If y1 ≤ y2 , then the result is obvious. Suppose that y1 ≥ y2 . Using the monotonicity property of F above, we then have ⎧ ⎪ ⎪ ⎨F(x1 , y1 ), if x2 ≤ y1 , F(x1 ∨ x2 , y1 ∨ y2 ) = F(x2 , y1 ) ≤ ⎪ ⎪ ⎩F(x2 , y2 ), if x2 ≥ y1 , and

⎧ ⎪ ⎪ ⎨F(x1 , y1 ), F(x1 ∧ x2 , y1 ∧ y2 ) = F(x1 , y2 ) ≤ ⎪ ⎪ ⎩F(x2 , y2 ),

if x1 ≤ y2 , if x1 ≥ y2 .

As F is non-negative, the terms on the right-hand side are all bounded above  by F(x1 , y1 ) + F(x2 , y2 ).

124

Harmonically weighted Dirichlet spaces

Proof of Theorem 7.5.2 Since f, g ∈ Dμ , we have Dζ ( f ) + Dζ (g) < ∞ for μ-almost every ζ ∈ T. Fix such a ζ. In particular f ∗ (ζ) and g∗ (ζ) both exist. Define a := | f ∗ (ζ)| ∨ |g∗ (ζ)|. Then, using Lemma 7.5.3 and Theorem 7.4.2, we have  |( f ∨ g)∗ (λ)|2 − a2 − 2a2 log |( f ∨ g)∗ (λ)/a| |dλ| |λ − ζ|2 T  F(| f ∗ (λ)|2 ∨ |g∗ (λ)|2 , | f ∗ (ζ)|2 ∨ |g∗ (ζ)|2 ) = |dλ| |λ − ζ|2 T   F(| f ∗ (λ)|2 , | f ∗ (ζ)|2 ) F(|g∗ (λ)|2 , |g∗ (ζ)|2 ) ≤ |dλ| + |dλ| |λ − ζ|2 |λ − ζ|2 T T = 2πDζ ( f ) + 2πDζ (g) < ∞. By Theorem 7.4.5, it follows that ( f ∨g)∗ (ζ) exists and satisfies |( f ∨g)∗ (ζ)| = a. Substituting this information into (7.10) and repeating the calculation above, we obtain that Dζ ( f ∨ g) ≤ Dζ ( f ) + Dζ (g). Integrating with respect to dμ(ζ) gives Dμ ( f ∨ g) ≤ Dμ ( f ) + Dμ (g). An exactly analogous argument works for f ∧ g.  In general f ∈ Dμ does not imply f 2 ∈ Dμ (see Exercise 7.1.4). However, we do have the following restricted version. Theorem 7.5.4 Let μ be a finite measure on T, and let f ∈ Dμ be outer. Then f ∧ f 2 ∈ Dμ and Dμ ( f ∧ f 2 ) ≤ 4Dμ ( f ). For the proof, we need another elementary lemma, similar to Lemma 7.5.3. Lemma 7.5.5 Define F : (0, ∞) × [0, ∞) → R+ as in Lemma 7.5.3. Then, for all x ∈ (0, ∞) and y ∈ [0, ∞), we have F(x ∧ x2 , y ∧ y2 ) ≤ 4F(x, y). Proof Set G(x, y) := 4F(x, y) − F(x ∧ x2 , y ∧ y2 ). We show that G(x, y) ≥ 0. There are four cases to consider. Case 1: x ≥ 1 and y ≥ 1. In this case, we have G(x, y) = 4F(x, y) − F(x, y) = 3F(x, y) ≥ 0. Case 2: x ≥ 1 and y < 1. Writing Gy for ∂G/∂y, for all t < 1 we have Gy (x, t) = −4(1 − t) log(x/t) − 2t log x ≤ 0. It follows that G(x, y) ≥ G(x, 1) ≥ 0, where the last inequality is from Case 1.

7.6 Inner functions

125

Case 3: x < 1 and y ≤ 1. For all t < 1 we have Gy (x, t) = 4(1 − t) log(t/x), which has the same sign as t − x. It follows that G(x, y) ≥ G(x, x) = 0, where the last equality holds since F(t, t) = 0 for all t. Case 4: x < 1 and y > 1. For t > 1, we have G y (x, t) = 3 log t − 2 log x ≥ 0. It follows that G(x, y) ≥ G(x, 1) ≥ 0, 

where the last inequality is from Case 3.

Proof of Theorem 7.5.4 This is just like the proof of Theorem 7.5.2, using Lemma 7.5.5 instead of Lemma 7.5.3.  Exercises 7.5 1. Let f ∈ Dμ be an outer function. Show that f = f1 / f2 , where f1 , f2 are bounded functions in Dμ . [Hint: Take f1 := f ∧ 1 and f2 := 1/( f ∨ 1).]

7.6 Inner functions Theorems 7.4.1 and 7.4.2 are valid only for outer functions: just consider what happens when f (z) = zn , for example. In this section we shall derive extensions of these formulas, also due to Carleson and to Richter and Sundberg, which are valid for all functions in H 2 . For these, we need to take into account inner factors. Recall that a function θ is inner if it is a bounded holomorphic function on D such that |θ∗ | = 1 a.e. on T. Every f ∈ H 2 \ {0} has a unique factorization f = fi fo , where fi , fo ∈ H 2 , fi is inner and fo is outer. For more about inner functions, see Appendix A.2. Our starting point is a result which permits us to factor out inner functions in local Dirichlet integrals. Theorem 7.6.1

Let f ∈ H 2 , let θ be an inner function, and let ζ ∈ T. Then Dζ ( f θ) = Dζ ( f ) + | f ∗ (ζ)|2 Dζ (θ).

(7.16)

Remarks (i) If the radial limit f ∗ (ζ) does not exist, then Dζ ( f ) = ∞, and the right-hand side of (7.16) should be interpreted as being ∞. (ii) If the radial limit f ∗ (ζ) = 0, then the product | f ∗ (ζ)|2 Dζ (θ) should be interpreted as being 0, even if Dζ (θ) = ∞. Similar remarks will apply to all the subsequent applications of this result.

126

Harmonically weighted Dirichlet spaces

Proof of Theorem 7.6.1 Using Lemma 7.3.3, for all w ∈ D we have  | f ∗ (λ)θ∗ (λ) − f (w)θ(w)|2 |dλ| |λ − w|2 T  | f ∗ (λ)θ∗ (λ)|2 − | f (w)θ(w)|2 |dλ| = |λ − w|2 T   | f ∗ (λ)θ∗ (λ)|2 − | f (w)|2 | f (w)|2 − | f (w)θ(w)|2 |dλ| + |dλ| = |λ − w|2 |λ − w|2 T T   ∗ 2 | f ∗ (λ)|2 − | f (w)|2 |θ (λ)| − |θ(w)|2 2 |dλ| + | f (w)| |dλ| = |λ − w|2 |λ − w|2 T T   ∗ | f ∗ (λ) − f (w)|2 |θ (λ) − θ(w)|2 2 = |dλ| + | f (w)| |dλ|. |λ − w|2 |λ − w|2 T T Put w = rζ, let r → 1, and apply Lemma 7.4.3. If f ∗ (ζ) exists and is non-zero, then we straightaway obtain the result. If f ∗ (ζ) = 0, then Dζ ( f θ) = Dζ ( f ) directly from Theorem 7.2.5. Finally, if f ∗ (ζ) does not exist, then Dζ ( f ) = ∞, and since the chain of equalities above implies that Dζ ( f θ) ≥ Dζ ( f ), we have  Dζ ( f θ) = ∞ as well. Corollary 7.6.2 Let f ∈ H 2 and let θ be an inner function. If θ f ∈ Dμ , then f ∈ Dμ , and Dμ ( f ) ≤ Dμ (θ f ). Proof By the theorem, Dζ ( f ) ≤ Dζ ( f θ) for all ζ ∈ T. Integrate both sides with respect to μ to get the result.  The following special case is particularly worthy of note. Corollary 7.6.3 Let μ be a finite measure on T. If f ∈ Dμ and if f = fi fo is its inner-outer factorization, then fo ∈ Dμ and Dμ ( fo ) ≤ Dμ ( f ). Proof

Apply the Corollary 7.6.2 with f, θ replaced by fo , fi respectively.



Here is one further consequence of Theorem 7.6.1, which will prove useful later on. We recall that the notation f ∧ g was defined in the previous section. Corollary 7.6.4 Let μ be a finite measure on T, let f ∈ H2 be an outer function and let θ be an inner function. If θ f ∈ Dμ , then θ( f ∧ 1) ∈ Dμ and Dμ (θ( f ∧ 1)) ≤ Dμ (θ f ). Proof

By Theorems 7.6.1 and 7.5.2, for μ-almost every ζ ∈ T, we have Dζ (θ( f ∧ 1)) = Dζ (θ)(| f ∗ (ζ)| ∧ 1)2 + Dζ ( f ∧ 1) ≤ Dζ (θ)| f ∗ (ζ)|2 + Dζ ( f ) = Dζ (θ f ).

Integrating with respect to μ gives the result.



7.6 Inner functions

127

Theorem 7.6.1 effectively reduces the problem of computing local Dirichlet integrals to one of treating inner and outer functions separately. We already know how to handle outer functions, so now we turn our attention to inner functions. We shall derive three formulas for the local Dirichlet integral of an inner function. Theorem 7.6.5

Let θ be an inner function, and let ζ ∈ T. Then 1 − |θ(rζ)|2 . r→1 1 − r2

Dζ (θ) = lim Proof

By Lemma 7.4.3, 1 Dζ (θ) = lim r→1 2π

 T

|θ∗ (λ) − θ(rζ)|2 |dλ|. |λ − rζ|2

Now, by Lemma 7.3.3, for each w ∈ D we have  ∗  ∗ 2 1 1 |θ (λ) − θ(w)|2 |θ (λ)| − |θ(w)|2 |dλ| = |dλ| 2π T 2π T |λ − w|2 |λ − w|2  1 − |θ(w)|2 1 1 − |θ(w)|2 |dλ| = . = 2π T |λ − w|2 1 − |w|2 Putting these facts together, we obtain the result.



Our second formula for Dζ (θ) is even simpler, but supposes that ζ is a regular point of θ, in other words, that θ extends to be holomorphic in a neighborhood of ζ. Theorem 7.6.6 of θ. Then

Let θ be an inner function, and let ζ ∈ T be a regular point Dζ (θ) = |θ (ζ)|.

Proof Replacing θ(z) by αθ(βz), where α, β are suitable unimodular constants, we may suppose that ζ = 1 and that θ(1) = 1. Then, as r → 1, 1 − |θ(r)|2 1 − |1 + θ (1)(r − 1) + o(r − 1)|2 = = Re θ (1) + o(1 − r). 1 − r2 1 − r2 Hence, by Theorem 7.6.5, D1 (θ) = Re θ (1). Finally, since |θ| = 1 on an arc of T around 1, we must have θ (1) ≥ 0, so Re θ (1) = |θ (1)|.  By Theorem A.2.4 in Appendix A, an inner function θ can be written in a unique way as Blaschke product times a singular inner function. In other words,  an an − z     ζ+z dσ(ζ) (z ∈ D), (7.17) exp − θ(z) = c |an | 1 − an z T ζ −z n

128

Harmonically weighted Dirichlet spaces

where c is a unimodular constant, (an ) is a sequence (finite or infinite) in D  such that n (1 − |an |) < ∞, and σ is a finite positive Borel measure on T which is singular with respect to Lebesgue measure. (It is possible that a finite number of an are zero. Our convention is that an /|an | = 1 for these terms.) Our final formula for Dζ (θ) is given in terms of the data (an ) and σ. Theorem 7.6.7 ζ ∈ T,

Let θ be the inner function given by (7.17). Then, for each

Dζ (θ) =

 1 − |an |2 n

|ζ − an

|2

 +

T

2 dσ(λ). |λ − ζ|2

(7.18)

Proof We may take c = 1. If σ({ζ}) > 0, then θ∗ (ζ) = 0, and Theorem 7.6.5 implies that Dζ (θ) = ∞. Also clearly dσ(λ)/|λ − ζ| = ∞. Thus (7.18) certainly holds in this case. Now suppose that σ({ζ}) = 0. For k ≥ 1, let θk be the inner function defined by   |a | a − z     λ+z n n θk (z) := dσ(λ) (z ∈ D). exp − an 1 − an z |λ−ζ|≥1/k λ − z |a −ζ|≥1/k n

Then ζ is a regular point of θk , so by the preceding theorem Dζ (θk ) = |θk (ζ)|. A computation using logarithmic derivatives (see Exercise 7.6.2) gives  1 − |an |2  2 |θk (ζ)| = + dσ(λ). (7.19) 2 2 |ζ − an | |λ−ζ|≥1/k |λ − ζ| |a −ζ|≥1/k n

As k → ∞, the right-hand side tends to the right-hand side of (7.18). It remains to show that Dζ (θk ) → Dζ (θ) as k → ∞. To see this, observe first that θk → θ locally uniformly on D, and so by Fatou’s lemma lim inf Dζ (θk ) ≥ Dζ (θ). k→∞

On the other hand, by Theorem 7.6.1, applied with θk playing the role of f and θ/θk playing the role of θ, we have Dζ (θ) ≥ Dζ (θk ) for all k, and hence lim sup Dζ (θk ) ≤ Dζ (θ). k→∞

Putting these facts together gives Dζ (θk ) → Dζ (θ) and completes the proof.  We can now derive the formulas of Carleson and Richter–Sundberg mentioned at the beginning of the section.

7.6 Inner functions

129

Theorem 7.6.8 (Richter–Sundberg formula) Let f ∈ H 2 \ {0} with innerouter factorization f = fi fo . Suppose that fi (z) is given by (7.17). Let ζ ∈ T be a point such that fo∗ (ζ) exists. Then Dζ ( f ) = Io + Ib + I s ,

(7.20)

where  | fo∗ (λ)|2 − | fo∗ (ζ)|2 − 2| fo∗ (ζ)|2 log | fo∗ (λ)/ fo∗ (ζ)| 1 |dλ|, Io := 2π T |λ − ζ|2  1 − |a |2 n Ib := | f ∗ (ζ)|2 , 2 o |ζ − a n| n  2 dσ(λ)| fo∗ (ζ)|2 . I s := 2 T |λ − ζ| Proof

By Theorem 7.6.1, we have Dζ ( fi fo ) = Dζ ( fo ) + | fo∗ (ζ)|2 Dζ ( fi ).

Theorem 7.4.2 gives a formula for Dζ ( fo ) and Theorem 7.6.7 gives one for Dζ ( fi ). The result follows.  Theorem 7.6.9 (Carleson’s formula) Let f ∈ H 2 \ {0} with inner-outer factorization f = fi fo . Suppose that fi is given by (7.17). Then D( f ) = Jo + Jb + J s , where   1 (| f ∗ (λ)|2 − | f ∗ (ζ)|2 )(log | f ∗ (λ)| − log | f ∗ (ζ)|) |dλ| |dζ|, 2 4π T T |λ − ζ|2   1 1 − |an |2 ∗ 2 Jb := | f (ζ)| |dζ|, 2π T n |ζ − an |2   2 1 Js := | f ∗ (ζ)|2 dσ(λ) |dζ|. 2π T T |λ − ζ|2 Jo :=

Proof It suffices to integrate both sides of (7.20) with respect to normalized Lebesgue measure on T, symmetrize as in the proof of Theorem 7.4.1, and  remark that | fo∗ | = | f ∗ | a.e. We conclude with a simple application that generalizes Exercise 4.1.1. Corollary 7.6.10 The only inner functions in D are finite Blaschke products.

130 Proof

Now

Harmonically weighted Dirichlet spaces Applying Theorem 7.6.9 with f = fi and fo = 1 gives     1 − |an |2 2 1 1 D( fi ) = |dζ| + dσ(λ) |dζ|. 2π T n |ζ − an |2 2π T T |λ − ζ|2  T

2 |dζ| = ∞ |λ − ζ|2

so D( fi ) can only be finite if σ = 0. Also  1 − |w|2 1 |dζ| = 1 2π T |ζ − w|2

(λ ∈ T),

(w ∈ D),

so D( fi ) can only be finite if the set of zeros (an ) is finite. In this case, fi is a finite Blaschke product, and D( fi ) is equal to the degree of fi .  Exercises 7.6 1. Let f ∈ H 2 , let θ be an inner function, and suppose that θ f ∈ Dμ . Show that, for each n ≥ 1, we have θn f ∈ Dμ and that Dμ (θn f ) ≤ nDμ (θ f ). 2. Let θ1 , θ2 be inner functions. Suppose that ζ ∈ T is a regular point for both θ1 , θ2 . Prove that |(θ1 θ2 ) (ζ)| = |θ1 (ζ)| + |θ2 (ζ)|. Use this to prove formula (7.19). 3. Let θ be the inner function defined by z − 1 θ(z) := exp z+1

(z ∈ D),

and set f := 1/θ. The formula (7.16) clearly fails in this case. Where does the proof break down? 4. Using Theorem 7.4.5, show that Theorems 7.6.8 and 7.6.9 hold for all functions of the form f = fi fo , where fi is inner and fo is outer, irrespective of whether f ∈ H 2 . [This set of functions is the so-called Smirnov class; see Appendix A.3.]

Notes on Chapter 7 §7.1 The Dμ spaces were introduced by Richter in [96], as part of his analysis of 2-isometries (more on this in the next chapter). Aleman [7] introduced a more general class of Dirichlet spaces with positive superharmonic weights, which includes both the Dμ spaces and the Dα spaces (0 < α < 1) mentioned in §1.6.

Notes on Chapter 7

131

§7.2 The local Dirichlet integral was studied systematically by Richter and Sundberg in their article [98], on which a large part of this chapter is based. In particular, they established the local Douglas formula (7.4). In fact, they took this formula as the definition of local Dirichlet integral, showing subsequently that it is indeed equivalent to the one given given earlier.

§7.3 The main approximation theorem 7.3.1 was obtained by Richter in [96]. He established a weaker version of Lemma 7.3.2, in which Dμ ( fr ) ≤ CDμ ( f ) for some unknown constant C. Richter and Sundberg [98] showed that this inequality holds with C = 4, and Aleman [7] improved this to C = 5/2. Finally Sarason [108] gave a proof with C = 1 by identifying the local Dirichlet spaces Dζ as a special case of so-called de Branges– Rovnyak spaces. The proof given here is an adaptation of his idea.

§7.4 Theorem 7.4.1 was originally obtained (with a different proof) by Carleson [27]. The local version, Theorem 7.4.2, is due to Richter and Sundberg [98].

§7.5 The results of this section are taken from the paper [99] of Richter and Sundberg.

§7.6 Theorem 7.6.9 was originally obtained (with a different proof) by Carleson [27]. The local version, Theorem 7.6.8, together with the other theorems of this section, are due to Richter and Sundberg [98]. Sarason had already obtained a precursor of Theorem 7.6.5 in [107], and had remarked on the close connection with angular derivatives and the Julia–Carath´eodory theorem. Finally, we remark that several of the themes already studied for D have analogues for Dμ that are the subjects of recent research. These include boundary behavior [53], zero sets [52], Carleson measures [33, 35] and Pick interpolation [116].

8 Invariant subspaces

‘Invariant subspaces’ here refers to the closed subspaces invariant under the shift operator f → z f . The invariant subspaces of H 2 were completely classified by Beurling, who showed that they are of the form θH 2 for some inner function θ (together with the zero subspace). It is no exaggeration to say that Beurling’s theorem is one of the cornerstones of the whole theory of Hardy spaces. Our aim in this chapter is to obtain an analogous result in the Dirichlet space D. One of the principal differences between H 2 and D is that, whereas the shift operator acts as an isometry on H 2 , this is no longer true on D. Instead, it is a so-called 2-isometry. One of our main tasks will be to study 2-isometries in general, and to prove a representation theorem for them. This theorem brings into play the Dμ -spaces introduced in the previous chapter. We shall then exploit the representation theorem to study the structure of invariant subspaces of D and (because it involves virtually no extra work) those of Dμ .

8.1 The shift operator on Dμ Let μ be a finite measure on T, and let Dμ be the corresponding harmonically weighted Dirichlet space defined in the previous chapter. Definition 8.1.1 The shift operator on Dμ is the operator Mz defined by (Mz f )(z) := z f (z)

( f ∈ Dμ ).

Our first task is to show that Mz is indeed a bounded operator on Dμ . Theorem 8.1.2

Let f ∈ Dμ . Then the following statements hold.

(i) The radial limit f ∗ (ζ) exists μ-almost everywhere on T. 132

8.1 The shift operator on Dμ

133

(ii) f ∗ ∈ L2 (μ) and  f ∗ L2 (μ) ≤ (1 + μ(T)1/2 ) f Dμ . (iii) z f ∈ Dμ and z f 2Dμ =  f 2Dμ +  f ∗ 2L2 (μ) .

(8.1)

Proof (i) This was already proved in Corollary 7.2.4. (ii) Define g(z) := ( f (z) − f (0))/z. Since f ∈ H 2 , we also have g ∈ H 2 . Thus we can apply Theorem 7.6.1 to g, with θ(z) = z, to obtain Dζ (zg) = Dζ (g) + |g∗ (ζ)|2 Dζ (z). As Dζ (z) = 1 and f = zg + f (0), this is equivalent to Dζ ( f ) = Dζ (g) + | f ∗ (ζ) − f (0)|2 . Integrating with both sides respect to μ, we deduce that  | f ∗ (ζ) − f (0)|2 dμ(ζ) ≤ Dμ ( f ). T

∗

Hence f ∈ L (μ) and 2

 f ∗ L2 (μ) ≤ Dμ ( f )1/2 + | f (0)|μ(T)1/2 ≤ (1 + μ(T)1/2 ) f Dμ . (iii) Re-apply Theorem 7.6.1 directly to f , still with θ(z) = z, to obtain Dζ (z f ) = Dζ ( f ) + | f ∗ (ζ)|2 . Integrating both sides of this equation with respect to μ gives Dμ (z f ) = Dμ ( f ) +  f ∗ 2L2 (μ) , which is equivalent to (8.1). In particular, Dμ (z f ) < ∞. Corollary 8.1.3 Proof

Mz is a bounded operator on Dμ with Mz  ≤

√

 2 + μ(T)1/2 .

By parts (ii) and (iii) of the theorem,

√ z f 2Dμ ≤ (1 + (1 + μ(T)1/2 )2 ) f 2Dμ ≤ ( 2 + μ(T)1/2 )2  f 2Dμ .

This is equivalent to the statement of the corollary.



In the light of this result, it is natural to ask when Mφ : f → φ f is a bounded operator on Dμ . This is equivalent to asking whether φ is a multiplier on Dμ . We shall not pursue this question in the same depth as we did for D in Chapter 5, contenting ourselves with the following sufficient condition. Theorem 8.1.4 Let f ∈ Dμ and let φ ∈ Hol(D) with φ ∈ H ∞ . Then φ f ∈ Dμ , and Dμ (φ f ) ≤ 2φ2H∞ Dμ ( f ) + 2φ 2H∞  f ∗ 2L2 (μ) .

(8.2)

134

Invariant subspaces

The proof depends on a technical estimate. Lemma 8.1.5 Let f ∈ H 2 and let φ ∈ H ∞ . Let ζ ∈ T be a point such that f ∗ (ζ) exists. Then Dζ (φ f ) ≤ 2φ2H∞ Dζ ( f ) + 2Dζ (φ)| f ∗ (ζ)|2 , and Dζ (φ)| f ∗ (ζ)|2 ≤ 2φ2H∞ Dζ ( f ) + 2Dζ (φ f ). Proof If f ∗ (ζ) = 0, then clearly Dζ (φ f ) ≤ φ2H∞ Dζ ( f ). So we may as well suppose that f ∗ (ζ)  0 and that Dζ (φ) < ∞, which implies that φ∗ (ζ) exists. We then have  1 |φ∗ (λ) f ∗ (λ) − φ∗ (ζ) f ∗ (ζ)|2 |dλ| Dζ (φ f ) = 2π T |λ − ζ|2   1 | f ∗ (λ) − f ∗ (ζ)|2 |φ∗ (λ) − φ∗ (ζ)|2  ≤ + 2| f ∗ (ζ)|2 2|φ∗ (λ)|2 | dλ| 2 2π T |λ − ζ| |λ − ζ|2 ≤ 2φ2H∞ Dζ ( f ) + 2| f ∗ (ζ)|2 Dζ (φ). This gives the first inequality. The second inequality is proved the same way.  Proof of Theorem 8.1.4 Since φ ∈ H ∞ , we have Dζ (φ) ≤ φ 2H ∞ for all ζ ∈ T (see Exercise 7.1.2). Clearly also φ ∈ H ∞ . By Lemma 8.1.5, it follows that, for μ-almost every ζ ∈ T, Dζ (φ f ) ≤ 2φ2H∞ Dζ ( f ) + 2φ 2H∞ | f ∗ (ζ)|2 . Integrating both sides with respect to dμ(ζ), we get (8.2).



This theorem does not extend to multiplication by H ∞ -functions or even by inner functions. Indeed, we have already seen that the only inner functions in D are finite Blaschke products (Corollary 7.6.10). However, it is always possible to divide by inner functions, in the sense of Corollary 7.6.2. This will prove useful at several points later in the chapter. Exercises 8.1 1. Let S ∗ : H 2 → H 2 be the adjoint of the shift operator on H 2 , namely S ∗ f (z) := ( f (z) − f (0))/z. Let ζ ∈ T and let f ∈ Dζ . (i) Show that S ∗ f ∈ Dζ and that S ∗ f Dζ ≤  f Dζ . (ii) Show that S ∗n f Dζ → 0 as n → ∞. [Hint: Use Corollary 7.3.4.] (iii) Show that Dζ ( f ) = Dζ (S ∗ f ) + |(S ∗ f )∗ (ζ)|2 .

8.2 Characterization of the shift operator (iv) Deduce that Dζ ( f ) =



135

|(S ∗k f )∗ (ζ)|2 .

k≥1

[Compare this formula with the one obtained in Exercise 7.2.4.]

8.2 Characterization of the shift operator We begin this section by establishing some further properties of (Mz , Dμ ). Theorem 8.2.1

(Mz , Dμ ) satisfies the following.

(i) Mz2 f 2Dμ − 2Mz f 2Dμ +  f 2Dμ = 0 for all f ∈ Dμ . (ii) ∩n≥0 Mzn (Dμ ) = {0}. (iii) dim(Dμ  Mz (Dμ )) = 1. Proof

(i) Using (8.1), we have z2 f 2Dμ − z f 2Dμ = z f ∗ 2L2 (μ) =  f ∗ 2L2 (μ) = z f 2Dμ −  f 2Dμ .

(ii) If f ∈ ∩n≥0 Mzn (Dμ ), then f is holomorphic on D with a zero of infinite order at the origin, so f = 0. (iii) If f ∈ Dμ and f (0) = 0, then f = zg, where g ∈ Dμ by Corollary 7.6.2. Thus Mz (Dμ ) is precisely the set of f ∈ Dμ such that f (0) = 0, which is a closed subspace of codimension one.  The three properties of Theorem 8.2.1 were singled out because they completely characterize (Mz , Dμ ). This is the conclusion of the following theorem. Theorem 8.2.2 Let T be a bounded operator on a Hilbert space H such that: (i) T 2 x2 − 2T x2 + x2 = 0 for all x ∈ H; (ii) ∩n≥0 T n (H) = {0}; (iii) dim(H  T (H)) = 1. Then there exists a unique finite measure μ on T such that (T, H) is unitarily equivalent to (Mz , Dμ ). An operator T that satisfies property (i) is called a 2-isometry, and one that satisfies property (ii) is called analytic. We shall see in the course of the proof that, under the conditions (i) and (ii), property (iii) is equivalent to T being cyclic (i.e., there exists a vector x such that {T n x : n ≥ 0} spans a dense subspace of H). Thus Theorem 8.2.2 is sometimes stated in the form that every cyclic, analytic, 2-isometry is unitarily equivalent to (Mz , Dμ ) for some μ.

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The proof of Theorem 8.2.2 will take up the rest of the section. The first step is to establish an elementary but important property of 2-isometries. Lemma 8.2.3 If (T, H) is a 2-isometry, then T x ≥ x for all x ∈ H. Consequently, the image of each closed subspace of H under T is again a closed subspace of H. Proof

As T is a 2-isometry, T 2 x2 − T x2 = T x2 − x2

(x ∈ H).

Replacing x by T j x for j = 0, . . . , k − 2, we find that T k x2 − T k−1 x2 = T x2 − x2

(k ≥ 1).

Therefore, if we sum from k = 1 to n, we obtain T n x2 − x2 = n(T x2 − x2 )

(n ≥ 1).

(8.3)

In particular, T x2 − x2 ≥ −x2 /n. Letting n → ∞ gives the result.



Theorem 8.2.4 (Wandering subspace theorem) Let (T, H) be an analytic 2-isometry. Then {T n (H  T (H)) : n ≥ 0} spans a dense subspace of H. Proof By Lemma 8.2.3, the operator T ∗ T − I is positive. Therefore T ∗ T is invertible, and L := (T ∗ T )−1 T ∗ is a left inverse of T . The operator Q := T L is the orthogonal projection onto T (H), and P := I − Q is the orthogonal projection onto H  T (H). Note also that L is a contraction: indeed, using Lemma 8.2.3 again, we have Lx ≤ T Lx = Qx ≤ x

(x ∈ H).

Let x ∈ H. Then, for each n ≥ 1, x − T n Ln x =

n−1  k=0

(T k Lk x − T k+1 Lk+1 x) =

n−1 

T k PLk x ∈

n−1 

k=0

T k P(H).

k=0

The right-hand side lies in the span of {T k (H  T (H)) : k ≥ 0}. Thus it suffices to show that some subsequence of (T n Ln x)n≥0 converges weakly to zero. Replacing x by Ln x in (8.3), we have T n Ln x2 − Ln x2 = n(T Ln x2 − Ln x2 ) = n(QLn−1 x2 − Ln x2 ) ≤ n(Ln−1 x2 − Ln x2 ).

8.2 Characterization of the shift operator Divide by n and sum:  T n Ln x2 − Ln x2 n≥1

n

137

≤ x2 < ∞.

We infer that lim inf n→∞ (T n Ln x2 − Ln x2 ) = 0. Also, as L is a contraction, the sequence (Ln x) is bounded. Therefore some subsequence of (T n Ln x) is bounded, and so some further subsequence of (T n Ln x) converges weakly to an element y ∈ H. As the sequence (T n H)n≥0 is decreasing, it follows that y ∈ ∩n≥0 T n H. By assumption, T is analytic, so ∩n≥0 T n H = {0}. Therefore y = 0, and the proof is complete.  Corollary 8.2.5 Let (T, H) be an analytic 2-isometry. Then T is cyclic if and only if dim(H  T (H)) = 1. In this case, each non-zero vector of H  T (H) is cyclic for T . Proof If dim(H  T (H)) = 1, then by the theorem it is clear that each nonzero vector of H  T (H) is cyclic for T . Conversely, it is obvious that, if T has a cyclic vector x, then H is spanned by T (H) ∪ {x}, so dim(H  T (H)) ≤ 1. Moreover, T (H) is a proper, closed subspace of H, so in fact dim(H  T (H)) = 1.  Applying Corollary 8.2.5 to (Mz , Dμ ), we obtain that 1 is a cyclic vector, thereby recovering the fact that polynomials are dense in Dμ (Corollary 7.3.4). However, this is not really a new proof, because the density of polynomials in Dμ is used implicitly in showing that (Mz , Dμ ) is a 2-isometry. The next step is to construct the measure μ that appears in Theorem 8.2.2. Theorem 8.2.6 Let (T, H) be a 2-isometry and let a ∈ H  T (H) be an element of norm 1. Then there exists a unique positive finite Borel measure μ on T such that, for all polynomials p, p(T )a = pDμ . For the proof, we need a lemma. Lemma 8.2.7 Let (T, H) be a 2-isometry and let a ∈ H. Then there exists a positive finite Borel measure μ on T such that, for every polynomial p,  |p|2 dμ. T p(T )a2 − p(T )a2 = T

Proof Denote by ·, · the inner product on H, and let us define a second sesquilinear form on H by the formula [x, y] := T x, T y − x, y

(x, y ∈ H).

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Note that [x, x] = T x2 − x2 ≥ 0 for all x, by Lemma 8.2.3, so [·, ·] is a semi-inner product on H. Further, the fact that T is a 2-isometry translates to [T x, T x] = [x, x], in other words, T is an isometry with respect to [·, ·]. By polarization, it follows that [T x, T y] = [x, y] for all x, y ∈ H. Define a linear functional Λ on the trigonometric polynomials by stipulating that Λ(einθ ) = Λ(e−inθ ) := [T n a, a]

(n ≥ 0).

From the property that [T x, T y] = [x, y], it then follows that Λ(ei( j−k)θ ) = [T j a, T k a]

( j, k ≥ 0).

We claim that Λ is a positive linear functional. Indeed, by the Fej´er–Riesz theorem, every non-negative trigonometric polynomial is of the form |p|2 for  some analytic polynomial p, and if p(z) = nj=0 c j z j , then    Λ(|p|2 ) = Λ c j ck ei( j−k)θ = c j ck [T j a, T k a] = [p(T )a, p(T )a] ≥ 0. j,k

j,k

The claim is justified. As the trigonometric polynomials are uniformly dense in C(T), there is a (unique) extension of Λ to a positive linear functional on C(T). By the Riesz representation theorem for continuous functions, there is a (unique) positive finite Borel measure μ on T such that Λ(φ) = φ dμ for all φ ∈ C(T). In particular, if p is an analytic polynomial, then  |p|2 dμ = Λ(|p|2 ) = [p(T )a, p(T )a] = T p(T )a2 − p(T )a2 . Thus μ has the required property, and the proof of the lemma is complete.



Proof of Theorem 8.2.6 First we prove uniqueness. Let μ and ν be two finite Borel measures on T satisfying the conclusion of the theorem. Then we have pDμ = pDν for every polynomial p. Using (8.1), we deduce that, for every polynomial p,   2 |p| dμ = |p|2 dν. T

T

By polarization, it follows that, for every pair of polynomials p, q,   pq dμ = pq dν. T

T

In particular, taking p(z) = z and q(z) = zm , we find that μ and ν have exactly the same Fourier coefficients, and consequently μ = ν. n

8.2 Characterization of the shift operator

139

Now we turn to the proof of existence. By Lemma 8.2.7, there exists a finite positive measure μ on T such that, for every polynomial p,  zp2Dμ − p2Dμ = |p|2 dμ = T p(T )a2 − p(T )a2 . (8.4) T

Define a sesquilinear form on polynomials by (p, q) := p(T )a, q(T )a − p, qDμ . Equation (8.4) translates to (zp, zp) = (p, p), which, by polarization, implies that (zp, zq) = (p, q)

for all polynomials p, q.

(8.5)

Also, since a ∈ H  T (H), we have a, T n a = 0 for all n ≥ 1, and for n = 0 we have a, a = a2 = 1. Together, these imply (1, zn ) = 0

(n ≥ 0).

(8.6)

Combining (8.5) and (8.6), we see that (zn , zm ) = 0 for all m, n ≥ 0, and hence that (p, q) = 0 for all polynomials p, q. In particular, (p, p) = 0, which amounts  to saying that p(T )a2 = p2Dμ , as desired. Finally, we have all the ingredients necessary to prove Theorem 8.2.2. Proof of Theorem 8.2.2 We begin with the uniqueness of μ. The conclusion of the theorem is that there is a unitary operator U : Dμ → H such that U −1 T U = Mz . Then p(T )U = U p(Mz ) for every polynomial p, and in particular p(T )U(1) = U p(Mz )(1) = U(p). Thus, writing a := U(1), we have a ∈ H  T (H) and p(T )a = U(p). In particular, p(T )a = pDμ . Therefore the uniqueness of μ follows from the uniqueness part of Theorem 8.2.6. Now we turn the existence of μ. By assumption dim(H  T (H)) = 1. Pick a ∈ H  T (H) with a = 1. By Theorem 8.2.6, there exists a unique positive finite measure μ on T such that p(T )a = pDμ for all polynomials p. Thus, if we define U(p) := p(T )a, then U is an isometry from the space of polynomials, thought of as a subspace of Dμ , into the Hilbert space H. As the polynomials are dense in Dμ , the operator U extends to an isometry of the whole of Dμ onto a closed subspace of H. Moreover, by Corollary 8.2.5, the vector a is a cyclic vector for T , which implies that the range of U is dense in H. Thus in fact U is a surjective isometry, and therefore a unitary operator. Finally, since U −1 T U(p) = zp, it follows that U −1 T U = Mz . Hence (T, H) is unitarily  equivalent to (Mz , Dμ ).

140

Invariant subspaces Exercises 8.2

√ 1. Let (T, H) be a 2-isometry that is not an isometry. Show that T n   n as n → ∞. [Hint: Use formula (8.3).] 2. Let (T, H) be a 2-isometry. Set H1 := ∩n≥0 T n (H) and H2 := H  H1 . (i) (ii) (iii) (iv) (v)

Show that T (H1 ) = H1 , and deduce that T |H1 is invertible. Show that (T |H1 )−1 is a 2-isometry Deduce that T |H1 is unitary. Show that T (H2 ) ⊂ H2 . Conclude that T = U ⊕ S , where U is unitary on H1 and S is an analytic 2-isometry on H2 .

8.3 Invariant subspaces of Dμ Let μ be a finite measure on T. Our goal in this section is to describe the closed shift-invariant subspaces of Dμ . It is convenient to introduce a little notation. Definition 8.3.1 Given a bounded operator T on a Hilbert space H, we write Lat(T, H) for the family of closed subspaces M of H such that T (M) ⊂ M. The elements of Lat(T, H) form a lattice under the operations M ∩ N and M + N (whence the notation). We seek to describe Lat(Mz , Dμ ). From the very definition, it is clear that if M ∈ Lat(Mz , Dμ ), then pM ⊂ M for every polynomial p. We extend this observation to more general functions. Theorem 8.3.2 Let M ∈ Lat(Mz , Dμ ) and let φ ∈ H ∞ . Then φM ∩ Dμ ⊂ M. Proof Let f ∈ M and suppose that φ f ∈ Dμ . We need to show that φ f ∈ M. Suppose first that φ is holomorphic in a neighborhood of D. Let φn be the n-th partial sum of the Taylor series of φ. Clearly φn f ∈ M for every n. Also φn → φ and φn → φ uniformly on D, so by Theorem 8.1.4, φn f − φ f Dμ → 0. Hence φ f ∈ M in this case. Now suppose merely that φ ∈ H ∞ . For 0 < r < 1, we consider φr (z) := φ(rz). Then φr is holomorphic in a neighborhood of D, so, by what we just proved, φr f ∈ M. We claim that suprt If we combine the inequalities (8.8) and (8.9), then we obtain   1 (|φ∗ |2 − t2 ) dm ≤ 2 (|φ∗ |2 − t2 ) dm. ∗ ∗ t |φ |>t |φ |>t This holds for each t > 1, which is only possible if |φ∗ | ≤ 1 m-a.e. on T. Thus φ is bounded and φH ∞ ≤ 1. (ii) As already mentioned in (i), we have M = φDμφ , where dμφ = |φ∗ |2 dμ. As φ is bounded, Dμφ ⊃ Dμ . Therefore φDμ ⊂ φDμφ = M ⊂ Dμ . It follows that φ is a multiplier of Dμ . Assume again that φDμ = 1. By (the proof of) Theorem 8.3.7, if f ∈ Dμ , then φ f Dμ =  f Dμφ . Also, since φH ∞ ≤ φDμ ≤ 1, we have   Dζ ( f ) dμφ (ζ) ≤ Dζ ( f ) dμ(ζ) = Dμ ( f ). Dμφ ( f ) = T

T

Therefore  f Dμφ ≤  f Dμ , and so, finally, φ f Dμ ≤  f Dμ .



As a corollary, we obtain a generalization to Dμ of Theorem 5.4.1. Corollary 8.3.10 Given f ∈ Dμ , there exists a multiplier φ of Dμ such that f and φ have the same zero set in D. Proof By Theorem 8.3.9, there is a multiplier φ of Dμ such that [ f ]Dμ = [φ]Dμ . Since f and φ generate the same closed invariant subspace of Dμ , they must have the same zero set in D.  Exercises 8.3 1. Let M, N ∈ Lat(Mz , Dμ ) with M, N  {0}. Show that M ∩ N  {0}. [Hint: Show first that M ∩ H ∞  {0} and N ∩ H ∞  {0}.]

Notes on Chapter 8

145

Notes on Chapter 8 §8.1 The basic properties of the shift operator on Dμ were established by Richter in [96]. Lemma 8.1.5 is due to Richter and Sundberg [98]. The formula in Exercise 8.1.1 is taken from [115].

§8.2 The results in this section are due to Richter [95, 96]. The terminology 2-isometry was introduced by Agler [3]. The problem of characterizing those operators unitarily equivalent to Mz on spaces of analytic functions has been studied intensively. We mention in particular a remarkable result of Cowen and Douglas [37] (see also the paper of Carlsson [30]) that characterizes those operators unitarily equivalent to the shift operator Mz on some Hilbert space of analytic functions on a domain in C.

§8.3 The crucial Theorem 8.3.6 was first obtained in the case of D by Richter and Shields [97], and subsequently extended to Dμ by Richter and Sundberg [99]. The remaining results in this section are due to Richter [96] and to Richter and Sundberg [98, 99]. The proof of Theorem 8.3.9 is taken from the thesis of Aleman [7]. Olin and Thomson [88] defined an operator T to be cellular-indecomposable if we have M ∩ N  {0} whenever M, N ∈ Lat(T, H) \ {0}. Thus Exercise 8.3.1 shows that (Mz , Dμ ) is cellular-indecomposable. This can be used to prove that dim(M  zM) ≤ 1 for M ∈ Lat(Mz , Dμ ). The paper of Bourdon [22] contains some more general results along these lines.

9 Cyclicity

In the previous chapter we saw that every (closed) invariant subspace of Dμ is cyclic, in other words, that it is generated by a single function in Dμ . In this chapter, we shall take the opposite point of view: starting with f ∈ Dμ , can we identify the invariant subspace that it generates? It is easy to see that g belongs to this invariant subspace if and only if there is a sequence of polynomials (pn ) such that pn f − gDμ → 0, but in practice it is often difficult to determine which g have this property. Therefore we seek other, more explicit descriptions of the invariant subspace generated by f . In particular, we pose the question: which functions f generate the whole of Dμ ? A complete answer to this is not known, even for the special case of the Dirichlet space D. In fact the characterization of those functions cyclic for D is the subject of a conjecture involving the logarithmic capacity of boundary zero sets. We shall present some partial solutions to this conjecture.

9.1 Cyclicity in Dμ We begin by formalizing the notions above. Let μ be a finite positive measure on T, and let Dμ be the associated harmonically weighted Dirichlet space. As usual, if μ is normalized Lebesgue measure, then Dμ is just the classical Dirichlet space D, so all the results of this section apply in particular to D. Definition 9.1.1 Given S ⊂ Dμ , the invariant subspace generated by S is the smallest (closed) invariant subspace of Dμ that contains S , namely the intersection of all M ∈ Lat(Mz , Dμ ) such that S ⊂ M. We denote it by [S ]Dμ . In the case when S is a singleton { f }, we write simply [ f ]Dμ . Finally, we say f is cyclic for Dμ if [ f ]Dμ = Dμ . We begin with a useful reformulation of Theorem 8.3.2. 146

9.1 Cyclicity in Dμ Theorem 9.1.2

147

Let f, g ∈ Dμ with |g| ≤ | f | on D. Then [g]Dμ ⊂ [ f ]Dμ .

Proof Let M := [ f ]Dμ and φ := g/ f . Then f ∈ M, φ ∈ H ∞ and φ f = g ∈ Dμ , so by Theorem 8.3.2, φ f ∈ M. In other words g ∈ [ f ]Dμ . It follows that  [g]Dμ ⊂ [ f ]Dμ . The next result describes [ f ]Dμ in terms of the inner-outer factorization of f . Recall that, if f ∈ Dμ and if f = fi fo is its inner-outer factorization, then also fo ∈ Dμ and  fo Dμ ≤  f Dμ (see Corollary 7.6.3). Theorem 9.1.3 Then

Let f ∈ Dμ and let f = fi fo be its inner-outer factorization. [ f ]Dμ = fi [ fo ]Dμ ∩ Dμ = [ fo ]Dμ ∩ fi H 2 .

For the proof, we need two lemmas. The first of these formalizes an idea that we have already used a few times. Lemma 9.1.4 Let M be a closed subspace of Dμ . Suppose that there exists a sequence ( fn ) in M such that supn Dμ ( fn ) < ∞ and fn → f pointwise in D. Then f ∈ M. Proof Using the result of Exercise 7.2.2, we have  fn − fn (0)2H2 ≤ 4Dμ ( fn ), and consequently  fn 2Dμ = | fn (0)|2 +  fn − fn (0)2H2 + Dμ ( fn ) ≤ | fn (0)|2 + 5Dμ ( fn ). It follows that supn  fn Dμ < ∞. Therefore there exists a subsequence ( fn j ) weakly convergent in Dμ . As weak convergence implies pointwise convergence in D, the weak limit must be f . In particular f ∈ Dμ . Finally, as M  is weakly closed in Dμ , we have f ∈ M. Lemma 9.1.5 Let θ be an inner function and let h be an outer function. Suppose that both θh, θh2 ∈ Dμ . Then θh ∈ [θh2 ]Dμ . Proof By Corollary 7.6.2 we have h, h2 ∈ Dμ . Also, by Theorem 7.5.4 we have Dμ (h ∧ h2 ) ≤ 4Dμ (h). Let n ≥ 1, replace h by nh throughout, and then divide by n2 to obtain Dμ (h ∧ nh2 ) ≤ 4Dμ (h). A calculation similar to that in the proof of Corollary 7.6.4 then shows that Dμ (θ(h ∧ nh2 )) ≤ 4Dμ (θh). On the other hand, for all n we have |θ(h ∧ nh2 )| ≤ n|θh2 |, so by Theorem 9.1.2 we get θ(h ∧ nh2 ) ∈ [θh2 ]Dμ . Now apply Lemma 9.1.4 to deduce that θh ∈ [θh2 ]Dμ .  Proof of Theorem 9.1.3

We shall prove the three inclusions

[ f ]Dμ ⊂ ([ fo ]Dμ ∩ fi H 2 ) ⊂ ( fi [ fo ]Dμ ∩ Dμ ) ⊂ [ f ]Dμ .

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The first inclusion is easy. Indeed, [ f ]Dμ ⊂ [ fo ]Dμ by Theorem 9.1.2, and obviously [ f ]Dμ ⊂ [ f ]H2 = fi H 2 . Hence [ f ]Dμ ⊂ [ fo ]Dμ ∩ fi H 2 . For the second inclusion, we begin by remarking that, by Theorem 8.3.7, [ fo ]Dμ = φDμφ for some φ ∈ Dμ . Further, since fo is outer, φ must be outer as well. Thus, if g ∈ [ fo ]Dμ ∩ fi H 2 , then g = φh, where h ∈ Dμφ and h ∈ fi H 2 . By Corollary 7.6.2, it follows that h = fi k, where k ∈ Dμφ . Therefore g = fi φk ∈ fi φDμφ = fi [ fo ]Dμ . Clearly also g ∈ Dμ . Thus [ fo ]Dμ ∩ fi H 2 ⊂ fi [ fo ]Dμ ∩ Dμ . For the third inclusion, we note that fi [ fo ]Dμ ∩ Dμ is a closed invariant subspace of Dμ , so it has the form [ψ]Dμ for some ψ ∈ Dμ . By Theorem 8.3.9, we may further suppose that ψ is a bounded multiplier of Dμ . Let ψ = ψi ψo be the inner-outer factorization of ψ. Since both f ∈ [ψ]Dμ ⊂ [ψ]H2 = ψi H 2 and ψ ∈ fi [ fo ]Dμ ⊂ fi H 2 , we must have ψi = fi . It follows that ψo ∈ [ fo ]Dμ . Thus there exist polynomials (pn ) such that pn fo → ψo in Dμ . As ψ is a multiplier, pn fo ψ → ψo ψ in Dμ . Also, again using the fact that ψi = fi , we have pn fo ψ = pn ψo f ∈ [ f ]Dμ . Consequently ψo ψ ∈ [ f ]Dμ . By Lemma 9.1.5, ψ ∈ [ψψo ]Dμ . Therefore ψ ∈ [ f ]Dμ . As ψ generates fi [ fo ]Dμ ∩ Dμ , we conclude that fi [ fo ]Dμ ∩ Dμ ⊂ [ f ]Dμ .  Theorem 9.1.3 effectively reduces the general problem of identifying [ f ]Dμ to the special case where f is an outer function. In the Hardy space, every outer function f satisfies [ f ]H 2 = H 2 . This is no longer true in Dμ (a simple example is given in Exercise 9.1.1) or even in D (more on this in the next section). In the rest of the section, we gather together a few general results about invariant subspaces of Dμ generated by outer functions. The first of these allows us in many cases to restrict our attention to bounded functions. Theorem 9.1.6

Let f be an outer function in Dμ . Then [ f ∧ 1]Dμ = [ f ]Dμ .

Proof The inclusion [ f ∧ 1]Dμ ⊂ [ f ]Dμ is immediate from Theorem 9.1.2. Also, since | f ∧ n| ≤ n| f ∧ 1|, the same theorem shows that f ∧ n ∈ [ f ∧ 1]Dμ for each n, and, using Theorem 7.5.2, we have Dμ ( f ∧ n) ≤ Dμ ( f ) for all n. By Lemma 9.1.4, f ∈ [ f ∧ 1]Dμ .  Theorem 9.1.7 Let f be an outer function, let α > 0, and suppose f, f α ∈ Dμ . Then [ f α ]Dμ = [ f ]Dμ . Proof We first prove the result under the additional assumption that f is bounded. Then f n ∈ Dμ for all n, because Dμ ∩ H ∞ is an algebra. If α ≥ 1 and n ≥ α, then Theorem 9.1.2 shows that [ f n ]Dμ ⊂ [ f α ]Dμ ⊂ [ f ]Dμ . Also, rek peated application of Lemma 9.1.5 shows that [ f ]Dμ ⊂ [ f 2 ]Dμ ⊂ · · · ⊂ [ f 2 ]Dμ for all k. Combining these inclusions we obtain [ f ]Dμ = [ f α ]Dμ for all α ≥ 1. The case α < 1 follows by switching the roles of f and f α .

9.1 Cyclicity in Dμ

149

Finally, if f is unbounded, then we use Theorem 9.1.6 twice over, as follows: [ f ]Dμ = [ f ∧ 1]Dμ = [( f ∧ 1)α ]Dμ = [ f α ∧ 1]Dμ = [ f α ]Dμ . Theorem 9.1.8



Let f, g be outer functions in Dμ . Then [ f ∨ g]Dμ = [ f, g]Dμ .

Proof By Theorem 9.1.2, we have f, g ∈ [ f ∨g]Dμ and so [ f, g]Dμ ⊂ [ f ∨g]Dμ . For the reverse inclusion, we use Theorem 8.3.7 to write [ f, g]Dμ = φDμφ , where φ ∈ Dμ . Then f = φ f1 and g = φg1 , where f1 , g1 ∈ Dμφ . As f, g are outer, so are φ, f1 , g1 . By Theorem 7.5.2, we have f1 ∨ g1 ∈ Dμφ . Hence f ∨ g = φ( f1 ∨ g1 ) ∈ φDμφ = [ f, g]Dμ , and so finally [ f ∨ g]Dμ ⊂ [ f, g]Dμ .  Theorem 9.1.9 Let f, g be outer functions, and suppose that f, g, f g ∈ Dμ . Then [ f g]Dμ = [ f ]Dμ ∩ [g]Dμ . Proof We begin by showing that [ f ]Dμ ∩ [g]Dμ ⊂ [ f g]Dμ . As [ f ]Dμ ∩[g]Dμ is an invariant subspace, it must be of the form [φ]Dμ , where φ is a (bounded) multiplier of Dμ . We may suppose that φ∞ < 1. Also, since [ f ]Dμ ∩ [g]Dμ contains at least one outer function, for example ( f ∧ 1)(g ∧ 1), it follows that φ must be outer. Consequently [φ2 ]Dμ = [φ]Dμ , by Theorem 9.1.7. Thus, if we can show that φ2 ∈ [ f g]Dμ , then we will have proved the inclusion [ f ]Dμ ∩ [g]Dμ ⊂ [ f g]Dμ . Since φ ∈ [ f ]Dμ ∩ [g]Dμ , there are sequences of polynomials (pn ) and (qn ) such that pn f → φ and qn g → φ in Dμ . Writing pn f = (pn f )i (pn f )o for the inner-outer factorization, we set fn := (pn f )i ((pn f )o ∧ 1). By Corollary 7.6.4 we have Dμ ( fn ) ≤ Dμ (pn f ) for all n. As pn f → φ in Dμ , it follows that supn Dμ ( fn ) < ∞. Likewise, defining gn := (gn g)i ((qn g)o ∧ 1), we obtain supn Dμ (gn ) < ∞. Since  fn ∞ ≤ 1 and gn ∞ ≤ 1 for all n, we deduce that supn Dμ ( fn gn ) < ∞. For each n, there exists a constant Cn such that | fn gn | ≤ Cn | f g|, so fn gn ∈ [ f g]Dμ . Clearly fn gn converges pointwise in D to φ2 . Applying Lemma 9.1.4, we get φ2 ∈ [ f g]Dμ , as desired. Now we turn to the proof of the reverse inclusion [ f g]Dμ ⊂ [ f ]Dμ ∩ [g]Dμ . This follows directly from Theorem 9.1.2 if f, g are both bounded, so the proof boils down to relaxing the boundedness of f, g. We do this in two stages. Suppose first that f is unbounded, but g is still bounded. Then, for each n we have ( f ∧ n)g ∈ [ f ]Dμ ∩ [g]Dμ . We claim that supn Dμ (( f ∧ n)g) < ∞. If so, then Lemma 9.1.4 shows that f g ∈ [ f ]Dμ ∩ [g]Dμ . To justify the claim, we use Lemma 8.1.5. According to that lemma, for μ-almost every ζ ∈ T, we have Dζ (( f ∧ n)g) ≤ 2g2H ∞ Dζ ( f ∧ n) + 2Dζ (g)|( f ∧ n)∗ (ζ)|2 ≤ 2g2H ∞ Dζ ( f ) + 2Dζ (g)| f ∗ (ζ)|2 ≤ 6g2H ∞ Dζ ( f ) + 4Dζ ( f g).

150

Cyclicity

Integrating with respect to dμ(ζ) gives Dμ (( f ∧ n)g) ≤ 6g2H ∞ Dμ ( f ) + 4Dμ ( f g). In particular, supn Dμ (( f ∧ n)g) < ∞, justifying the claim. Now suppose both f and g are unbounded. Note that f (g∧1) = f g∧ f ∈ Dμ . Hence, by what we have just proved, [ f (g ∧ 1)]Dμ ⊂ [g]Dμ . Clearly we have | f g ∧ 1| ≤ | f (g ∧ 1) ∨ g|. Hence, from Theorems 9.1.6, 9.1.2 and 9.1.8, we get [ f g]Dμ = [ f g ∧ 1]Dμ ⊂ [ f (g ∧ 1) ∨ g]Dμ = [ f (g ∧ 1), g]Dμ ⊂ [g]Dμ . By symmetry [ f g]Dμ ∈ [ f ]Dμ as well, whence the result. Corollary 9.1.10



If f ∈ Dμ is invertible in Dμ , then it is cyclic for Dμ .

Proof To say that f is invertible means that 1/ f ∈ Dμ . This forces both f and 1/ f to be outer functions. By Theorem 9.1.9, we then have [ f ]Dμ ∩ [1/ f ]Dμ = [1]Dμ = Dμ . In particular [ f ]Dμ = Dμ .  The final result of the section is the companion to Theorem 9.1.8 and also generalizes Theorem 9.1.6. Theorem 9.1.11 [ f ]Dμ ∩ [g]Dμ .

Let f and g be outer functions in Dμ . Then [ f ∧ g]Dμ =

Proof By Theorem 9.1.2 we have [ f ∧ g]Dμ ⊂ [ f ]Dμ ∩ [g]Dμ . The reverse inclusion follows from [ f ]Dμ ∩ [g]Dμ = [ f ∧ 1]Dμ ∩ [g ∧ 1]Dμ = [( f ∧ 1)(g ∧ 1)]Dμ ⊂ [ f ∧ g]Dμ , where we used successively Theorems 9.1.6, 9.1.9 and 9.1.2.



Exercises 9.1 1. Let μ = δ1 , the Dirac mass at {1}. Let M := { f ∈ Dμ : f ∗ (1) = 0}. Show that M is a closed invariant subspace of (Mz , Dμ ). Deduce that [z − 1]Dμ  Dμ , even though z − 1 is an outer function. 2. Let f ∈ Dμ , and suppose also that log f ∈ Dμ . (i) Show that f is an outer function. (ii) Show that, for all α ∈ (0, 1), we have   Dμ ( f α ) ≤ α2 Dμ ( f ) + Dμ (log f ) . (iii) Deduce that f α → 1 in Dμ as α → 0+ . (iv) Deduce that f is cyclic for Dμ . 3. Give an example to show that a function can be cyclic for Dμ without being invertible in Dμ .

9.2 Cyclicity in D and boundary zero sets

151

4. Let f, g ∈ Dμ , and suppose also that f g ∈ Dμ . (i) Show that, if at least one of f, g is outer, then [ f g]Dμ = [ f ]Dμ ∩ [g]Dμ . (ii) Give an example to show that, if neither f nor g is outer, then we may have [ f g]Dμ  [ f ]Dμ ∩ [g]Dμ .

9.2 Cyclicity in D and boundary zero sets The results of the previous section relate cyclic subspaces of Dμ to others but, with the exception of Corollary 9.1.10, they do not explicitly identify [ f ]Dμ . This is because the answer depends on the measure μ. To proceed further, we need to know what μ is. In this section, we consider the important case where μ is Lebesgue measure, and so Dμ = D, the classical Dirichlet space. We begin with a basic example. Theorem 9.2.1

Let p be a polynomial. Then [p]D = { f ∈ D : f (w j ) = 0, j = 1, . . . , m},

where w1 , . . . , wm are the zeros of p inside D, counted according to multiplicity. In particular, if p has no zeros in D, then it is cyclic for D. Proof Let p = pi po be the inner-outer factorization of p. Here the inner factor pi is the finite Blaschke product with zeros at w1 , . . . , wm , and the outer factor po is a polynomial all of whose zeros lie in C \ D. By Theorem 9.1.3, [p]D = [po ]D ∩ pi H 2 = [po ]D ∩ { f ∈ H 2 : f (w j ) = 0, j = 1, . . . , m}. It therefore suffices to show that [po ]D = D. Let us write po (z) = c(z − a1 ) . . . (z − an ), where a1 , . . . , an ∈ C \ D, and c ∈ C with c  0. By Theorem 9.1.9, we have [po ]D = [c]D ∩ [z − a1 ]D ∩ · · · ∩ [z − an ]D . Clearly [c]D = D. So it suffices to show that [z − a]D = D for all a ∈ C \ D.  Fix a ∈ C \ D. Let g ∈ D  [z − a]D , say g(z) = n≥0 bn zn . Then we have g, zn−1 (z − a)D = 0 for all n ≥ 1. This translates to (n + 1)bn − nabn−1 = 0 for all n ≥ 1, and consequently (n + 1)bn = an b0 for all n ≥ 0. Thus   (n + 1)|bn |2 = |a|2n |b0 |2 /(n + 1), g2D = n≥0

n≥0

and this can only be finite if b0 = 0, whence g = 0. Therefore [z−a]D = D.



152

Cyclicity

One way to interpret this result is that a polynomial is cyclic for D if and only if it is an outer function. One might guess that the same is true for general functions in D (as is the case in H 2 ) but, as we shall see, this turns out to be false. The reason is that zero sets on the boundary also have a role to play. To discuss this further, it is convenient to introduce some notation. Recall that q.e. means quasi-everywhere, namely outside a set of outer logarithmic capacity zero. Definition 9.2.2 Given an arbitrary subset E of T, we write DE := { f ∈ D : f ∗ = 0 q.e. on E}. Clearly DE = D if and only if c∗ (E) = 0. Also DE = {0} whenever |E| > 0. We shall return to study the intermediate cases later. The following theorem provides the connection between boundary zero sets and invariant subspaces. Theorem 9.2.3

For every subset E ⊂ T, we have DE ∈ Lat(Mz , D).

Proof It is clear that DE is a subspace of D and that it is Mz -invariant. The point at issue is whether it is closed in D. Let ( fn ) be a sequence in DE , and suppose that fn → f in D. By the weaktype inequality for capacity Theorem 3.3.1, for each t > 0 we have c∗ (E ∩ {| f ∗ | > t}) ≤ c∗ (| f ∗ − fn∗ | > t) ≤ A f − fn 2D /t2 , where A is an absolute constant. Let n → ∞ to get c∗ (E ∩ {| f ∗ | > t}) = 0. As this holds for all t > 0, it follows that f ∗ = 0 q.e. on E, in other words, f ∈ DE .  Thus DE is closed in D, as claimed. Corollary 9.2.4 Proof

If f ∈ D, then [ f ]D ⊂ DE , where E := {ζ ∈ T : f ∗ (ζ) = 0}.

Clearly f ∈ DE , and by the theorem DE ∈ Lat(Mz , D).

Corollary 9.2.5



If f is cyclic for D, then f is outer and c∗ ( f ∗ = 0) = 0.

Proof The first conclusion follows from (the easy part of) Theorem 9.1.3, and the second one follows from Corollary 9.2.4.  As a simple application of these ideas, we can prove an analogue of Theorem 5.4.1 for boundary zero sets. Theorem 9.2.6 Given f ∈ D, there exists a multiplier φ of D such that the boundary zero sets { f ∗ = 0} and {φ∗ = 0} differ by a set of outer logarithmic capacity zero.

9.2 Cyclicity in D and boundary zero sets

153

Proof By Theorem 8.3.9 there exists φ ∈ M(D) such that [ f ]D = [φ]D . Corollary 9.2.4 then tells us that f ∗ = 0 q.e. on {φ∗ = 0} and φ∗ = 0 q.e. on { f ∗ = 0}. In other words, the symmetric difference of { f ∗ = 0} and {φ∗ = 0} is of outer capacity zero.  In §4.3 and §4.4 we gave some constructions of boundary zero sets for D. What do these constructions tell us about cyclicity? We shall derive two consequences. The first shows that the conclusion in Corollary 9.2.5 cannot be improved (at least for f ∈ D ∩ A(D)), and the second shows that no part of the conclusion is redundant. Theorem 9.2.7 Let E be a closed subset of T such that c(E) = 0. Then there exists f ∈ D ∩ A(D) such that f = 0 on E and f is cyclic for D. Proof The existence of f , without the cyclicity statement, was established in Theorem 4.3.2. using the construction in Theorem 3.4.1. We shall now prove that f , as constructed there, is cyclic.  The function f was built as f = exp(− j≥1 F j ), where F j ∈ D ∩ A(D)   with Re F j ≥ 0 and j F j D < ∞. Let fn := exp(− j>n F j ). Then we have   D( fn ) ≤ D( j>n F j ) ≤ ( j>n F j D )2 → 0, and fn → 1 boundedly on D, so   fn − 1D → 0. Also, for each n, we have fn / f = exp( nj=1 F j ), which is a bounded function, so fn ∈ [ f ]D . Letting n → ∞, we deduce that 1 ∈ [ f ]D .  Theorem 9.2.8 cyclic for D.

There exists an outer function f ∈ D ∩ A(D) which is not

Proof Let E be the circular Cantor middle-third set. Then E is a Carleson set, so by Theorem 4.4.3, there exists an outer function f ∈ D ∩ A(D) whose zero set is precisely E. Also, c(E) > 0, so by Corollary 9.2.5 it follows that f is not cyclic for D.  Exercises 9.2 1. For n ≥ 1, set Ln :=

n k=1

1/k and

pn (z) :=

n   Lk  k−1 1− z . Ln k=1

Show by direct computation that 1 − (1 − z)pn (z)2D = 1/Ln . Deduce that (1 − z) is cyclic for D.

154

Cyclicity

2. Let E be a countable subset of T (not necessarily closed). Show that there exists a cyclic function f ∈ D such that limr→1 f (rζ) = 0 for all ζ ∈ E. [Hint: Use Theorem 4.3.1.] 3. Let f ∈ D be a univalent function such that f (z)  0 for all z ∈ D. (i) Prove that f is an outer function. [Hint: By Theorem A.1.10 in Appendix A, we have 1/ f ∈ H p for all p < 1/2.] (ii) Use the change of variable w := f (z) to show that, for 0 < α < 1, D( f α ) ≤ α + α2 D( f ). (iii) Deduce that f α → 1 in D as α → 0+ . (iv) Deduce that f is cyclic for D. (v) Deduce that c∗ ( f ∗ = 0) = 0.

9.3 The Brown–Shields conjecture In the previous section, we saw that there are two obstacles to cyclicity in D: inner factors and zero sets on the boundary. Are there any others? We formulate this question more precisely, as follows. Problem 9.3.1

Let f ∈ D be outer and let E := { f ∗ = 0}. Is [ f ]D = DE ?

The inclusion [ f ]D ⊂ DE always holds, by Corollary 9.2.4. The issue is whether [ f ]D ⊃ DE . Problem 9.3.1 seems to have been raised first by Brown and Shields in [24]. They also conjectured that the following special case should always hold. Conjecture 9.3.2 (Brown–Shields) then f is cyclic for D.

If f ∈ D is outer and c∗ ( f ∗ = 0) = 0,

In other words, the two necessary conditions for cyclicity in Corollary 9.2.5 are between them also sufficient. This conjecture (and a fortiori Problem 9.3.1) are still open. In this section and the next, we present some partial solutions, starting with the following result. Theorem 9.3.3

Let f ∈ D be an outer function, and define E := {ζ ∈ T : lim inf | f (z)| = 0}. z→ζ

If g ∈ D and |g(z)| ≤ C dist(z, E) on D, then g ∈ [ f ]D .

(9.1)

9.3 The Brown–Shields conjecture

155

Theorem 9.3.3 falls short of being a true solution to Problem 9.3.1 for two reasons. Firstly, the set E is defined differently. In the conjecture it is the zero set of f ∗ , and in Theorem 9.3.3 it is the zero set of lim inf | f |, which may in principle be a larger set. Of course, the two sets agree if f (or even just | f |) is continuous up to the boundary. However, there are examples of outer functions in D where the two sets disagree, and the set in (9.1) may even be the whole unit circle (see Exercise 9.3.2). Secondly, in the theorem g is assumed to satisfy the Lipschitz-type estimate |g(z)| ≤ C dist(z, E), rather than the much weaker condition g∗ = 0 q.e. on E defining membership of DE . The Lipschitz condition forces E to be a Carleson set (see Exercise 9.3.3), so it is quite restrictive. To prove the theorem, we require the following ‘fusion lemma’. We write d for arclength distance on T, and | · | for Lebesgue measure on T. Lemma 9.3.4 Let f1 , f2 ∈ D be outer functions such that | f j∗ (ζ)| ≤ Cd(ζ, E), where E is a closed subset of T. Let f be an outer function such that, on each component of T \ E, either | f ∗ | = | f1∗ | a.e. or | f ∗ | = | f2∗ | a.e. Then f ∈ D, and D( f ) ≤ D( f1 ) + D( f2 ) +

 C 2 π2  C 2 π2 log . 2 | f1 (0) f2 (0)|

The precise value of the bound is unimportant. What matters is that it is independent of E. Proof By scaling, we can suppose that C = 1/π. This implies that | f j | ≤ 1. Also, |E| = 0 and f ∈ H 2 . Thus, by Carleson’s formula (Theorem 7.4.1),  1 (| f ∗ (λ)|2 − | f ∗ (ζ)|2 )(log | f ∗ (λ)| − log | f ∗ (ζ)|) D( f ) = |dλ| |dζ| 4π2 |λ − ζ|2   T×T + +2 , = U1 ×U1

U2 ×U 2

U1 ×U 2

where U j is the union of the components of T \ E on which | f ∗ | = | f j∗ | a.e. By Carleson’s formula again, the first two terms on the right-hand side are bounded above by D( f1 ) and D( f2 ) respectively. It remains to estimate the third term, namely   (| f1∗ (λ)|2 − | f2∗ (ζ)|2 )(log | f1∗ (λ)| − log | f2∗ (ζ)|) 1 |dλ| |dζ|. (9.2) 2π2 U1 U2 |λ − ζ|2 If λ ∈ U1 and ζ ∈ U 2 , then there must be a point of E between them, so d(λ, ζ) ≥ d(λ, E) + d(ζ, E), and consequently   | f ∗ (λ)|2 − | f ∗ (ζ)|2  ≤ π−2 d(λ, E)2 + π−2 d(ζ, E)2 ≤ π−2 d(λ, ζ)2 ≤ 1 |λ − ζ|2 . 1 2 4

156

Cyclicity

Therefore the expression in (9.2) is bounded above by    1 log | f ∗ (λ)| − log | f ∗ (ζ)| |dλ| |dζ| 1 2 2 8π U1 ×U 2   1 1  1 log ∗ + log ∗ |dλ| |dζ| ≤ 2 | f1 (λ)| | f2 (ζ)| 8π T×T 1 1 1 1 + log , = log 2 | f1 (0)| 2 | f2 (0)| where we have used the facts that | f j∗ | ≤ 1 and f j is outer ( j = 1, 2). This gives the required estimate.  Proof of Theorem 9.3.3 We can assume that f is bounded. Otherwise replace f by f ∧ 1 and use Theorem 9.1.6. This does not change the set E (see Exercise 9.3.1). Note that, as g is also bounded, it follows that f g ∈ D. Let (In )n≥1 be the connected components of T \ E. For each n, let gn be the outer function such that ⎧ ∗ ⎪ ⎪ a.e. on ∪ j≤n I j , ⎨|g |, ∗ |gn | = ⎪ ⎪ ⎩| f ∗ g∗ |, a.e. on ∪ j>n I j . Clearly gn → go pointwise on D, where go is the outer factor of g. Also, by Lemma 9.3.4, we have supn D(gn ) < ∞. We claim that gn ∈ [ f ]D for all n. If so, then by Lemma 9.1.4 we obtain go ∈ [ f ]D , and hence also g ∈ [ f ]D , thereby proving the theorem. It remains to establish the claim that gn ∈ [ f ]D . Fix n. Let (Jk )k≥1 be an increasing sequence of compact sets, each a union of n arcs, such that ∪k≥1 Jk = ∪nj=1 I j . For each k, let pk be the monic polynomial of degree 2n with zeros at the endpoints of Jk , and let hk be the outer function such that ⎧ ⎪ ⎪ a.e. on Jk , ⎨|pk g∗ |, ∗ |hk | = ⎪ ⎪ ⎩|pk f ∗ g∗ |, a.e. on T \ Jk . By Lemma 9.3.4 we have supk D(hk ) < ∞. Also h∗k / f ∗ is bounded a.e. on Jk (because d(Jk , E) > 0) and a.e. on T \ Jk (because |h∗k / f ∗ | = |pk g∗ | there). Therefore hk ∈ [ f ]D for all k. Now hk converges pointwise to pgn , where p is a monic polynomial of degree 2n with zeros in T. By Lemma 9.1.4, pgn ∈ [ f ]D . Using Theorems 9.1.9 and 9.2.1, we have [pgn ]D = [p]D ∩ [gn ]D = [gn ]D . Hence, finally, gn ∈ [ f ]D , as claimed.  The next result is a partial solution to Conjecture 9.3.2. As usual, we write Et := {ζ ∈ T : d(ζ, E) ≤ t}, where d is arclength distance on T.

9.3 The Brown–Shields conjecture Theorem 9.3.5

157

Let f ∈ D be an outer function, and define E as in (9.1). If  1/e log log(1/t) c(Et ) dt < ∞, (9.3) t log(1/t) 0

then f is cyclic for D. Once again, this result falls short of being a full solution to Conjecture 9.3.2 for two reasons. The first, just as before, is that E is defined differently in the theorem and in the conjecture. The second reason is that condition (9.3) is stronger than c(E) = 0. Indeed, c(E) = 0 is equivalent to limt→0+ c(Et ) = 0 (this follows from Theorem 2.1.6), whereas (9.3) amounts to saying that c(Et ) → 0 at least at a certain rate as t → 0+ . Theorem 9.3.5 suffers from a further defect. The condition (9.3) is rather difficult to verify in practice. It does hold whenever E is finite (see Exercise 9.3.4), and with some work one can also show that it holds for certain very thin Cantor-type sets, but in general it is rather unwieldy. We shall return to this problem in the next section, replacing the capacity condition by a measure condition that is easier to handle. For the time being, let us concentrate on the proof of the theorem. The idea is to use (9.3) to construct a ‘nice’ function g lying in [ f ]D , and then to deform g into the constant function 1 while staying inside [ f ]D . The deformation is carried out using the following lemma. Lemma 9.3.6 Let h ∈ D be a function such that | Im h| < π/6 on D. Let Ω := {λ ∈ C \ {0} : | arg(λ)| < π/6} and, for λ ∈ Ω, define gλ (z) := exp(−λeh(z) )

(z ∈ D).

Then gλ ∈ D for all λ ∈ Ω, the map λ → gλ : Ω → D is holomorphic, and gλ − 1D → 0 as λ → 0 from within Ω. Proof For λ ∈ Ω and z ∈ D, we have | arg(λeh(z) )| < π/6 + π/6 = π/3. Hence Re(λeh ) ≥ |λeh |/2 on D, and consequently, |gλ | ≤ exp(−|λeh |/2) ≤ 1 and |gλ | ≤ |h ||λeh | exp(−|λeh |/2) ≤ 2|h |. It immediately follows that gλ ∈ D. If λn → λ0 in Ω, then gλn → gλ0 and gλn → gλ0 pointwise on D, and the estimates above allow us to apply the dominated convergence theorem to show that gλn − gλ0 D → 0. Thus λ → gλ is continuous as a map : Ω → D. The

158

Cyclicity

usual argument involving Cauchy’s theorem and Morera’s theorem then shows that this map is in fact holomorphic. Finally, let (λn ) be a sequence in Ω tending to 0. It is clear that gλn → 1 and gλn → 0 pointwise on D. Using the dominated convergence theorem once  again, we deduce that gλn − 1D → 0, as required. Proof of Theorem 9.3.5 We apply Theorem 3.4.3. Define η(t) := log log(1/t) for t ∈ (0, 1/e), and set it equal to zero for t ≥ 1/e. Then  π  1/e 2 log log(1/t) dt, c(Et ) |dη2 (t)| = c(Et ) t log(1/t) 0 0 which is finite by (9.3). Therefore, by Theorem 3.4.3, there exists a function h ∈ D such that | Im h| < π/6 and lim inf Re h(z) ≥ η(d(ζ, E)) z→ζ

(ζ ∈ T).

Define Ω and gλ as Lemma 9.3.6. As in the proof of the lemma, we have |gλ | ≤ exp(−|λeh |/2) on D. Hence, for each λ ∈ Ω and ζ ∈ T, lim sup |gλ (z)| ≤ exp(−|λ|eη(d(ζ,E)) /2) ≤ d(ζ, E)|λ|/2 . z→ζ

By the maximum principle, it follows that |gλ (z)| ≤ Cλ dist(z, E)|λ|/2 , where C λ is a constant depending on λ (see Exercise 9.3.5). In particular, if λ ∈ Ω and |λ| ≥ 2, then |gλ (z)| ≤ Cλ dist(z, E), so by Theorem 9.3.3 we have gλ ∈ [ f ]D . Since λ → gλ : Ω → D is holomorphic, it follows that gλ ∈ [ f ]D for all λ ∈ Ω. Finally, since gλ − 1D → 0 as λ → 0, we obtain 1 ∈ [ f ]D . Thus f is cyclic.  Exercises 9.3 1. Let f ∈ D be an outer function. Show that   lim inf |( f ∧ 1)(z)| = min lim inf | f (z)|, 1 z→ζ

z→ζ

(ζ ∈ T).

2. The aim of this question is to construct an outer function f ∈ D for which the set E defined in (9.1) is the whole unit circle. (i) Construct h ∈ D \ {0} whose zeros (wn ) in D accumulate at every point of T. [Hint: Use Theorem 4.2.1.]  (ii) Let f be the outer factor of h. Show that n | f (wn )|2 < ∞. [Hint: Use Theorem 4.1.3 and the fact that |h|2 is a subharmonic function.] (iii) Deduce that lim inf z→ζ | f (z)| = 0 for all ζ ∈ T. 3. Let E be a closed subset of T. Suppose that there exists g ∈ Hol(D), g  0 such that |g(z)| ≤ C dist(z, E). Show that E is a Carleson set.

9.4 Measure conditions and distance functions

159

4. Let E be a finite subset of T. Show c(Et )  1/ log(1/t) as t → 0+ , and deduce that that (9.3) holds for this E. 5. Let g ∈ Hol(D), and suppose that lim sup |g(z)| ≤ Cd(ζ, E)α

(ζ ∈ T),

z→ζ

where E is a closed subset E of T, and C, α are positive constants. By applying the maximum principle to g(z)/(z − a)α , where a ∈ E, show that |g(z)| ≤ C(π/2)α dist(z, E)α

(z ∈ D).

9.4 Measure conditions and distance functions The criterion (9.3) is expressed in terms of c(Et ), a quantity that is difficult to estimate for all but the simplest sets E. In this section we establish a sufficient condition for cyclicity expressed in terms of |E t |, the Lebesgue measure of Et , which is somewhat easier to handle. Theorem 9.4.1 Let f ∈ D be an outer function, and set E := {ζ ∈ T : lim inf z→ζ | f (z)| = 0}. Suppose that, for some σ > 0, |Et | = O(tσ ) (t → 0), and that

 0

π

dt = ∞. |Et |

(9.4)

(9.5)

Then f is cyclic for D. The condition (9.4) implies that E is a Carleson set (see Exercise 9.4.1). Condition (9.5) implies that E is of logarithmic capacity zero and, for certain Cantor-type sets, it is actually equivalent to E being of capacity zero (see Exercises 9.4.2 and 9.4.3). Neither of the conditions implies the other. The basic idea of the proof of Theorem 9.4.1 is the same as that for Theorem 9.3.5, namely to use Theorem 9.3.3 to construct a ‘nice’ function belonging to [ f ]D , and then to deform this function into 1 while staying inside [ f ]D . This time, however, the deformation process is somewhat more complicated. It takes place within a certain family of functions called distance functions, which we now proceed to define. In what follows, d denotes the arclength distance on T.

160

Cyclicity

Definition 9.4.2 Let E be a closed subset of T of Lebesgue measure zero, and let w : (0, π] → R+ be a continuous function such that   log w(d(ζ, E)) |dζ| < ∞. (9.6) T

The distance function corresponding to w, E is the outer function fw,E satisfy∗ (ζ)| = w(d(ζ, E)) a.e. on T, namely: ing | fw,E fw,E (z) := exp

 1  ζ+z  log w(d(ζ, E)) |dζ| 2π T ζ − z

(z ∈ D).

(9.7)

One of the virtues of distance functions is that, under a certain concavity condition, their Dirichlet integrals are particularly easy to estimate. Theorem 9.4.3 Let w, E be as in Definition 9.4.2, and let fw,E be the corresponding distance function. Suppose further that w is increasing, and that there exists γ > 2 such that t → w(tγ ) is a concave function. Then  π w (t)2 |Et | dt, (9.8) D( fw,E ) ≤ Cγ 0

where Cγ is a constant depending only on γ. For the proof, we need an inequality that can be used to check conditions like (9.6). Lemma 9.4.4 Let E be a closed subset of T of Lebesgue measure zero, and let φ : T → [0, ∞) be a positive, measurable function. Then   π |Et | dt. φ(d(ζ, E)) |dζ| ≤ φ(t) t T 0 Proof

Let (I j ) be the connected components of T \ E. Then  T

where ψ := 2

φ(d(ζ, E)) |dζ| =

  2 j

|I j |/2 0

 φ(t) dt =

π

φ(t)ψ(t) dt,

(9.9)

0



1[0,|I j |/2] , a positive, decreasing function. In particular, taking

δ φ = 1[0,δ] , we get |Eδ | = 0 ψ(t) dt ≥ δψ(δ), so ψ(t) ≤ |E t |/t for all t. Substituting this inequality back into (9.9) gives the result.  j

Proof of Theorem 9.4.3 We use Carleson’s formula for the Dirichlet integral of an outer function, Theorem 7.4.1. Let ζ1 , ζ2 denote points of T, and write

9.4 Measure conditions and distance functions

161

δ j := d(ζ j , E). In this notation, Carleson’s formula becomes  1 (w2 (δ1 ) − w2 (δ2 ))(log w(δ1 ) − log w(δ2 )) |dζ1 | |dζ2 | 2 4π |ζ1 − ζ2 |2 T2  1 (w2 (δ1 ) − w2 (δ2 ))(log w(δ1 ) − log w(δ2 )) |dζ1 | |dζ2 | ≤ 16 d(ζ1 , ζ2 )2 T2  1 (w2 (δ1 ) − w2 (δ2 ))(log w(δ1 ) − log w(δ2 )) ≤ |dζ1 | |dζ2 |. 8 d(ζ1 , ζ2 )2 δ1 ≥δ2

D( fw,E ) =

For convenience, let us extend w to the whole of R+ by defining w(t) := w(π) for t > π. We then have w(δ1 ) ≤ w(δ2 + d(ζ1 , ζ2 )), and consequently   π 2 (w (δ2 + s) − w2 (δ2 ))(log w(δ2 + s) − log w(δ2 )) 1 ds |dζ2 | 4 T 0 s2   |Et | 1 π π (w2 (t + s) − w2 (t))(log w(t + s) − log w(t)) ds dt, ≤ 2 4 0 0 t s

D( fw,E ) ≤

where, for the last step, we used Lemma 9.4.4. To estimate this, we now exploit the concavity assumption on w. This assumption amounts to saying that t → w (t)t1−1/γ is decreasing. Thus 

t+s

w (t + s) − w (t) = 2

2

2w(u)w (u) du

t



t+s

≤

2w(t + s)w (t)(t/u)1−1/γ du

t

  = 2γw(t + s)w (t)t (1 + s/t)1/γ − 1 . The concavity assumption also implies that w(t)/t1/γ is decreasing. Therefore  log w(t + s) − log w(t) =

t+s

t



t+s

w (u) du w(u)

u1/γ du w(u) u t  t+s (t + s)1/γ du t1−1/γ w (t) ≤ w(t + s) u t 1/γ (1 + s/t) log(1 + s/t). = tw (t) w(t + s) =

u1−1/γ w (u)

162

Cyclicity

Combining these estimates, we obtain  π 2 (w (t + s) − w2 (t))(log w(t + s) − log w(t)) ds s2 0  π   ds 2γw (t)2 t2 (1 + s/t)1/γ − 1 (1 + s/t)1/γ log(1 + s/t) 2 ≤ s 0  ∞   dx = w (t)2 t 2γ (1 + x)1/γ − 1 (1 + x)1/γ log(1 + x) 2 . x 0 Since γ > 2, the integral converges, say to Cγ . Plugging this into the estimate for D( fw,E ) yields  Cγ π  2 D( fw,E ) ≤ w (t) |Et | dt, 4 0 

giving the desired result.

The final ingredient in the proof of Theorem 9.4.1 is a regularization lemma. Lemma 9.4.5 Let a > 0, let β ∈ (0, 1] and let φ : (0, a] → (0, ∞) be a function such that • φ(t)/t is decreasing, • 0 < φ(t) ≤ tβ for all t ∈ (0, a],

a • 0 dt/φ(t) = ∞. Then, given α ∈ (0, β), there exists a function ψ : (0, a] → (0, ∞) such that • ψ(t)/tα is increasing, • φ(t) ≤ ψ(t) ≤ tβ for all t ∈ (0, a],

a • 0 dt/ψ(t) = ∞. The proof of this result is a real-variable argument involving the so-called rising-sun lemma. It is fairly long so, in order not to interrupt the flow, we postpone it to Appendix D. Proof of Theorem 9.4.1 Let f be the function in the statement of the theorem. Our aim is to prove that 1 ∈ [ f ]D . Let σ be as in the statement of the theorem. We can suppose that σ < 1/2. Fix α with 1/2 < α < (1 + σ)/2, and set w(t) := t1−α . By Lemma 9.4.4,    π log w(d(ζ, E))| |dζ| ≤ C | log t|tσ−1 dt < ∞, T

0

so we may construct the distance function g := fw (for brevity, we have

9.4 Measure conditions and distance functions

163

dropped E from the notation). We claim that g ∈ [ f ]D . Indeed, by Theorem 9.4.3, applied with γ = 1/(1 − α), we have  π D(g) ≤ C t−2α tσ dt < ∞, 0

so g ∈ D. As g is bounded, it follows that gn ∈ D for all n ≥ 1. Also, if n is larger than 1/(1 − α), then gn satisfies |gn (z)| ≤ C dist(z, E). Consequently, by Theorem 9.3.3, we have gn ∈ [ f ]D . Also, from Theorem 9.1.7, we have [gn ]D = [g]D , and thus g ∈ [ f ]D , as claimed. The rest of the proof consists of showing that 1 ∈ [g]D . We shall achieve this by constructing a family of functions wδ : (0, π] → R+ for 0 < δ < 1, such that the corresponding distance functions fwδ belong to [g]D , converge pointwise to 1 as δ → 0, and satisfy lim inf δ→0 D( fwδ ) < ∞. If such a family exists, then by Lemma 9.1.4 we have 1 ∈ [g]D , as desired. Here is the construction. Fix β with α < β < (1 + σ)/2, and define a function φ : (0, π] → R+ by φ(t) := min{|Et |, tβ }. This function satisfies the hypotheses of the regularization lemma, Lemma 9.4.5, so there exists a function α ψ : (0, π] → R+ satisfying the conclusions of that

π lemma, namely: ψ(t)/t is β increasing, φ(t) ≤ ψ(t) ≤ t for all t ∈ (0, π], and 0 dt/ψ(t) = ∞. Note that, for 0 < t < 1, we have ψ(t) ≥ φ(t) ≥ t. For 0 < δ < 1, define wδ : (0, π] → R+ by ⎧ α 1−α ⎪ 0 < t ≤ δ, δ t /ψ(δ), ⎪ ⎪ ⎪

π ⎪ ⎨ wδ (t) := ⎪ Aδ − log t ds/ψ(s), δ < t ≤ ηδ , ⎪ ⎪ ⎪ ⎪ ⎩1, ηδ < t ≤ π. Here Aδ , ηδ are constants, chosen to make wδ a continuous function satisfying 0 ≤ wδ ≤ 1 (see Figure 9.1).

Figure 9.1 The function wδ

164

Cyclicity

Just as before, wδ satisfies (9.6), and so we may define the corresponding distance function fwδ . We now check that fwδ has the required properties. We claim that there exists γ > 2 such that, for all sufficiently small δ > 0, the function t → wδ (tγ ) is concave on (0, π]. Assume this for the moment. Then Theorem 9.4.3 applies and, for all small δ, we have fwδ ∈ D. Moreover, since wδ (t)/t1−α is bounded, so too is fwδ /g. Therefore, by Theorem 9.1.2, fwδ ∈ [g]D . The fact that fwδ → 1 pointwise as δ → 0 is an easy consequence of the assertion that limδ→0 ηδ = 0, which we now prove. Given  > 0, if ηδ > , then wδ () < 1, in other words  π  π ds ds δ log − − log < 1. ψ(δ) δ ψ(s)  ψ(s) As δ → 0, the left-hand side tends to infinity. Thus ηδ ≤  for all sufficiently small δ. The last property to be checked is that lim inf δ→0 D( fwδ ) = 0. This requires a bit more work. By Theorem 9.4.3 again, for all δ small enough,  π D( fwδ ) ≤ C wδ (t)2 |Et | dt, 0

where C is a constant independent of δ. We examine this integral separately on (0, δ) and (δ, ηδ ). Let us begin with (δ, ηδ ). Here we have  ηδ  π  ηδ ds −2 |Et | wδ (t)2 |Et | dt = dt. ψ(t)2 δ δ t ψ(s) Note that if |E t | ≤ tβ then ψ(t) ≥ |Et |, whereas if |Et | > tβ then ψ(t) = tβ . The last integral is therefore majorized by  ηδ  π  ηδ  π ds −2 1 ds −2 Ctσ dt dt + ψ(t) t2β δ δ t ψ(s) t ψ(s)    π ds −1  π ds −2 σ+1−2β ≤ +C ηδ , ηδ ψ(s) ηδ ψ(s) and this tends to zero as δ → 0. Now we consider what happens on (0, δ). Here we have  δ  δ δ2α wδ (t)2 |Et | dt = t−2α |Et | dt. ψ(δ)2 0 0 If |Et | ≤ tβ for all t ∈ (0, δ), then |Et |/tα ≤ ψ(t)/tα ≤ ψ(δ)/δα , and so,  δ  δ δ2α 1 δα 1 δ −2α ≤ . t |E | dt ≤ t−α dt = t ψ(δ) 0 1 − α ψ(δ) 1 − α ψ(δ)2 0

9.4 Measure conditions and distance functions

165

On the other hand, if |Et | > tβ for a sequence t = δn tending to zero, then ψ(δn ) = δβn for all n, and consequently  δn  δn δ2α δ2α C n n −2α δ1+σ−2β t |Et | dt ≤ 2β t−2αCtσ dt = , n 1 + σ − 2α ψ(δn )2 0 δn 0 which tends to zero as n → ∞. Putting everything together, we get lim inf δ→0 D( fwδ ) = 0, as we wanted. All that remains is to establish the claim about concavity. Fix γ > 2 with 1 − 1/γ < α. Our aim is to prove that t1−1/γ wδ (t) is decreasing. On (0, δ) we have t1−1/γ wδ (t) = Ct−ν , where ν := α + 1/γ − 1 > 0. This is certainly decreasing. On (δ, ηδ ) we have   t1−1/γ  π ds −1 tα  ν π ds −1 1−1/γ  t t wδ (t) = = . ψ(t) t ψ(s) ψ(t) t ψ(s)

π Now ψ(t)/tα is increasing. Also, the derivative of t → tν t ds/ψ(s) has the same sign as  π ds t − , ν ψ(s) ψ(t) t which is positive if t is small enough. Thus t1−1/γ wδ (t) is decreasing on (δ, ηδ ) provided that δ is small enough. Lastly, at t = δ, we need that the left derivative of wδ exceeds the right derivative, which boils down to the inequality  π ds 1 ≥ , 1−α δ ψ(s) and this certainly holds for small δ, since the left-hand side tends to infinity as δ → 0. In summary, we have shown that t1−1/γ wδ (t) is decreasing on (0, π] if δ is small enough. The claim about concavity is proved, and with it, the theorem.  Exercises 9.4 1. Let E be a closed subset of T of measure zero. (i) Let φ : (0, π] → R+ be a positive, decreasing, differentiable function. Show that   π φ(d(ζ, E)) |dζ| = |φ (t)||Et | dt + 2πφ(π). T

0

(ii) Deduce that E is a Carleson set if and only if 2. Let E be a closed subset of T.

π 0

(|Et |/t) dt < ∞.

166

Cyclicity (i) Let NE (t) be the t-covering number of E, namely the number of sets of diameter t needed to cover E, where diameter is measured using the arclength metric. Show that (t/2)NE (t) ≤ |E t | ≤ 2tNE (t)

(0 < t ≤ π).

[Hint for the left-hand inequality: Consider a maximal disjoint collection of closed arcs of diameter t/2 all of which meet E.] (ii) Deduce that the logarithmic capacity c(E) satisfies  π 1 1 ≥ dt. c(E) 2|E t| 0 [Hint: Use Theorem 2.3.2, taking into account the fact that logarithmic capacity is defined using the chordal metric on T rather than the arclength metric.] 3. Let (ln )n≥0 be a sequence in (0, 2π) such that λ := supn≥0 (ln+1 /ln ) < 1/2, and let E be the associated circular Cantor set. (Thus, we begin with a closed arc of length l0 , remove an open arc from the middle to leave two closed arcs of length l1 , remove open arcs from their middles to leave four arcs of length l2 , etc.; then E is the intersection of the resulting nested sequence of sets.) (i) Show that |E t | = O(tσ ) as t → 0, where σ := 1 − log 2/ log(1/λ). (ii) Show that the following statements are equivalent: • c(E) = 0,  • n≥0 2−n log(1/ln ) = ∞, π • 0 dt/|Et | = ∞. [Hint: Use the preceding exercise and Theorem 2.3.5.] (iii) Deduce that, if f ∈ D ∩ A(D) vanishes precisely on E, then f is cyclic for D if and only if f is outer and c(E) = 0.

9.5 Cyclicity via duality By the Hahn–Banach theorem, if the only element of the dual space of D that vanishes on [ f ]D is zero, then [ f ]D = D and f is cyclic for D. This suggests analyzing further those elements of the dual space of D that vanish on [ f ]D , which is what we are going to do in this section. There are various ways of representing the dual of D. One, of course, is as D itself, via the usual Hilbert-space duality, defined using the (sesquilinear) inner product ·, ·D . However, we shall instead identify the dual of D with a certain

9.5 Cyclicity via duality

167

Bergman-type space Be , using a complex bilinear pairing ·, ·. We begin by describing this pairing in detail. In what follows, C∞ denotes the Riemann sphere. Definition 9.5.1 Let De := {z ∈ C∞ : |z| > 1}. We write Be for the space of all φ ∈ Hol(De ) such that  1 dA(z) |φ(z)|2 < ∞. φ2Be := π De |z|2 We can express φBe in terms of the Laurent coefficients of φ as follows. Proposition 9.5.2 Let φ ∈ Hol(De ). If φ(∞)  0, then φBe = ∞. If φ(∞) = 0,  say φ(z) = k≥0 bk /zk+1 , then φ2Be =

 |b |2 k . k + 1 k≥0

(9.10)

Proof The first statement follows from the fact that dA(z)/|z|2 is an infinite measure on De . For the second, writing the area integral in polar coordinates, we have    1 bk 2 dA(z) φ2Be =   π De k≥0 zk+1 |z|2  ∞  2π  2 dr  1 −k−1 −i(k+1)θ  = e  bk r  dθ . π 1 r 0 k≥0 By Parseval’s formula, for each r ∈ (0, 1),  2π  2   1 −k−1 −i(k+1)θ  e |bk |2 r−2k−2 .  bk r  dθ = 2π 0 k≥0 k≥0 Hence

 φ2Be

=2 1

∞



|bk |2 r−2k−3 dr =

k≥0

 |bk |2 . k+1 k≥0



This result immediately shows that (Be ,  · Be ) is a Hilbert space. Also, the appearance of the k + 1 in the denominator in (9.10) strongly suggests defining the following pairing.   Definition 9.5.3 If f (z) = k≥0 ak zk ∈ D and φ(z) = k≥0 bk /zk+1 ∈ Be , then we write  ak bk .  f, φ := k≥0

168

Cyclicity

Proposition 9.5.4 defined, and

For all f ∈ D and φ ∈ Be , the pairing  f, φ is well | f, φ| ≤  f D φBe .

Proof

By the Cauchy–Schwarz inequality,   1/2 1/2  |ak bk | ≤ (k + 1)|ak |2 |bk |2 /(k + 1) . k≥0

k≥0



k≥0

In fact it is easy to see that the pairing ·, · establishes an isometric isomorphism between Be and the dual of D. This prompts the following definition. Definition 9.5.5 Given S ⊂ D, we write S ⊥ := {φ ∈ Be :  f, φ = 0 for all f ∈ S }. By the Hahn–Banach theorem, f is cyclic for D if and only if [ f ]⊥D = {0}. We are therefore interested in calculating [ f ]⊥D . The following theorem is the central result of this section. Recall that A(D) denotes the disk algebra and H p denotes the Hardy spaces on D. Theorem 9.5.6 Let f ∈ D ∩ A(D), and define E := {z ∈ D : f (z) = 0}. Let φ ∈ Be , and suppose that φ ∈ [ f ]⊥D . Then (i) φ extends to be holomorphic in C∞ \ E, (ii) f φ extends to be holomorphic on D, and f φ ∈ ∩0 1, we have    uk /λk+1 , φ = uk , φ/λk+1 = bk /λk+1 = φ(λ), (λ − u)−1 , φ = k≥0

k≥0

k≥0

the passage to the limit in the second equality justified by Proposition 9.5.4. The next step is to extend φ from De to a larger domain. For this, we use some abstract functional analysis. Observe that D ∩ A(D) is a Banach algebra with respect to the norm hD∩A(D) := D(h)1/2 + hA(D) . Let I be the closed ideal in D ∩ A(D) generated by f . Every element of I is a limit in D ∩ A(D) (and hence in D) of a sequence (pn f ), where pn are polynomials. It follows that I ⊂ [ f ]D . Since φ ∈ [ f ]⊥D , it vanishes on I, and therefore it induces a

9.5 Cyclicity via duality

169

continuous linear functional  φ on the quotient algebra (D ∩ A(D))/I via the formula  h,  φ := h, φ

(h ∈ D ∩ A(D)).

(Here  h is shorthand for the coset h + I. Also, we are abusing notation slightly in continuing to use ·, · for this quotient pairing.) In conjunction with (9.11), this gives φ φ(λ) = (λ −  u)−1 , 

(|λ| > 1).

(9.12)

Notice that the right-hand side of this last formula defines a holomorphic conu), where σ( u) denotes the spectrum of tinuation of φ to the whole of C∞ \ σ(  u in the (Banach) quotient algebra (D ∩ A(D))/I. The last step is to show that σ( u) ⊂ E. Let λ ∈ σ( u). By the Gelfand theory, there exists a character (multiplicative linear functional) χ on (D ∩ A(D))/I such that χ( u) = λ. Then, for any polynomial p, we have χ( p) = χ(p( u)) = p(χ( u)) = p(λ). Since polynomials are dense in D ∩ A(D), it follows that χ(  f ) = f (λ). But  f = 0, since f ∈ I. Consequently f (λ) = 0, and so λ ∈ E, as claimed. This completes the proof of part (i). (ii) The basis for this part of the proof is another formula for φ, or more precisely for f φ, but this time inside D. The formula is f (λ)φ(λ) =  f, (λ − u)−1 φ

(λ ∈ D \ E).

(9.13)

Notice that Be is stable under multiplication by bounded holomorphic functions on De , so (λ − u)−1 φ ∈ Be if |λ| < 1, and the formula at least makes sense. Also, once (9.13) is established, the right-hand side clearly provides a holomorphic continuation of f φ to the whole of D. To prove (9.13), we start from (9.12). Fix λ ∈ D \ E. We shall construct an explicit inverse for ( u − λ) in (D ∩ A(D))/I. Define g(z) :=

f (z) − f (λ) z−λ

(z ∈ D \ {λ}).

Clearly λ is a removable singularity and, after its removal, g ∈ D ∩ A(D). Also (u − λ)g = f − f (λ), so in the quotient algebra ( u − λ) g = − f (λ), and hence −1 f (λ)( u − λ) = − g. Plugging this into (9.12), we obtain f (λ)φ(λ) = − g,  φ = −g, φ. We now derive an expansion for g in powers of λ. Define T : D → D by

170

Cyclicity

(T h)(z) := (h(z) − h(0))/z. Clearly T acts as a contraction on D. Now   λk T k+1 f = f + λk (uT k+1 f − T k f ) (u − λ) k≥0

k≥0

= f−



λk (T k f )(0)

k≥0

= f − f (λ), the last equation arising from the fact that (T k f )(0) is the k-th Taylor coefficient  of f . A comparison with the definition of g reveals that g = k≥0 λk T k+1 f . Substituting this expression for g into the previous formula for f (λ)φ(λ), we get   f (λ)φ(λ) = − λk T k+1 f, φ = − λk  f, φ/uk+1  =  f, (λ − u)−1 φ. k≥0

k≥0

This establishes (9.13) and, as already remarked, shows that f φ ∈ Hol(D). It remains to prove that f φ ∈ ∩0 4, then |gλ (z)| ≤ C dist(z, E)2 , and Theorem 9.6.6 then shows that pgλ , φ = 0 for all polynomials p. By the identity principle, it follows that pgλ , φ = 0 for all λ ∈ Ω and all p. Letting λ → 0, it follows that p, φ = 0 for all p. As polynomials are dense in D, we deduce that φ = 0. Thus E is Bergman–Smirnov exceptional.  Let E be a closed subset of T such that, for some σ > 0,  π dt = ∞. and |Et | = O(tσ ) (t → 0) |E t| 0

Theorem 9.6.9

Then E is a Bergman–Smirnov exceptional set. Proof Just as in the proof of Theorem 9.4.1, the assumptions on E allow us to construct a cyclic function g for D such that |g(z)| ≤ C dist(z, E) on D, where  > 0. Replacing g by a high enough power gn , we may further suppose that  = 2. By Theorem 9.6.6, we then have HolBe ,N + (C∞ \ E) ⊂ [g]⊥D = {0}. Therefore E is Bergman–Smirnov exceptional.  The next result extends our library of Bergman–Smirnov exceptional sets still further, using a quite different technique. Theorem 9.6.10 Let E be a closed subset of T. Suppose that E contains a Bergman–Smirnov exceptional set E  such that E \ E  is countable. Then E is itself Bergman–Smirnov exceptional. The following special case is particularly worthy of note. Corollary 9.6.11 Every countable closed subset of T is a Bergman–Smirnov exceptional set.  The proof of Theorem 9.6.10 uses the following decomposition lemma.

9.6 Bergman–Smirnov exceptional sets

177

Lemma 9.6.12 Let E1 , E2 be disjoint closed subsets of T, and E := E1 ∪ E2 . Then every φ ∈ HolBe ,N + (C∞ \ E) can be decomposed as φ = φ1 + φ2 , where φ j ∈ HolBe ,N + (C∞ \ E j ) for j = 1, 2. Proof Using Cauchy’s integral formula in the standard way, we can decompose φ as φ = φ1 + φ2 , where φ j ∈ Hol(C∞ \ E j ) ( j = 1, 2). Adding and subtracting a constant, we may also ensure that φ1 (∞) = φ2 (∞) = 0. The point is to show that φ j |De ∈ Be and that φ j |D ∈ N + . We begin with φ j |De . Let U1 , U2 be bounded disjoint open neighborhoods of E1 , E2 respectively. Then   1 dA(z) 1 dA(z) |φ1 (z)|2 + |φ1 (z)|2 . φ1 2Be = 2 π De ∩U1 π De \U1 |z| |z|2 The second integral is finite, because φ1 is bounded on De \ U1 and φ1 (z) = O(1/z) as |z| → ∞. The first integral is finite because φ1 = φ − φ2 , where φ|De ∈ Be and φ2 is bounded on U1 . It follows that φ1 |De ∈ Be . Likewise for φ2 . Now we consider φ j |D . To show that φ j |D ∈ N + , we use the characterization of N + as the set of h ∈ Hol(D) such that {ζ → log+ |h(rζ)| : 0 < r < 1} is uniformly integrable on T (see Theorem A.3.7 in Appendix A). Once again, let U1 , U2 be disjoint open neighborhoods of E1 , E 2 . On T \ U 1 , the family {log+ |φ1 (rζ)| : 0 < r < 1} is uniformly bounded, so it is certainly uniformly integrable. On T ∩ U1 , log+ |φ1 (rζ)| = log+ |φ(rζ) − φ2 (rζ)| ≤ log+ |φ(rζ)| + log+ |φ2 (rζ)| + log 2, where the right-hand side is the sum of a uniformly integrable family (because φ|D ∈ N + ) and a uniformly bounded family. Putting this all together we deduce that {log+ |φ1 (rζ)| : 0 < r < 1} is uniformly integrable on T, and hence that  φ1 |D ∈ N + . The case of φ2 is similar. Proof of Theorem 9.6.10 Let φ ∈ HolBe ,N + (C∞ \E). Let F be the closed subset of E consisting of those points across which φ has no holomorphic continuation. Clearly we then have φ ∈ HolBe ,N + (C∞ \ F). We claim that F ⊂ E  . Indeed, let us suppose the contrary. Then F \ E  is non-empty, countable and closed in T \ E  . By the Baire category theorem, F contains an isolated point ζ. Set F1 := {ζ} and F 2 := F \ {ζ}. By Lemma 9.6.12, we can write φ = φ1 +φ2 , where φ j ∈ HolBe ,N + (C∞ \ F j ). As F1 is a singleton, it is Bergman–Smirnov exceptional, and so φ1 = 0. Consequently φ = φ2 , which is holomorphic at ζ, contradicting the definition of F. The claim is justified. Since F ⊂ E  and E  is Bergman–Smirnov exceptional, so too is F, and therefore, finally, φ = 0, as required. 

178

Cyclicity

All these examples tend to suggest that the converse of Theorem 9.6.3 is true. Hedenmalm and Shields [58] posed this formally as a problem. Problem 9.6.13 (Hedenmalm–Shields) Let E be a closed subset of T such that c(E) = 0. Must E be a Bergman–Smirnov exceptional set? This problem is still open. Note that, thanks to Theorem 9.6.2, an affirmative answer would imply the Brown–Shields conjecture, at least for those f continuous up to the boundary. We conclude by remarking that, if we replace N + by B, the usual Bergman space on D, then HolBe ,B (C∞ \ E) = {0} precisely when E is of logarithmic capacity zero. A proof is sketched in Exercise 9.6.3 below. This rules out the possibility of answering Problem 9.6.13 negatively by using the Cauchy transform of a measure on T, as in the proof of Theorem 9.6.3. Exercises 9.6 1. Let f ∈ D ∩ A(D) be an outer function and let E := {ζ ∈ T : f (ζ) = 0}. Let g ∈ D and suppose that |g(z)| ≤ C dist(z, E)2 on D. Show that [ f ]⊥D ⊂ HolBe ,N + (C∞ \ E) ⊂ [g]⊥D , and deduce that g ∈ [ f ]D . (Compare Theorem 9.3.3.) 2. Show that the union of two Bergman–Smirnov exceptional sets is again a Bergman–Smirnov exceptional set. 3. Let E be a closed subset of T of logarithmic capacity zero, and let f ∈ HolBe ,B (C \ E). (i) Let s > 0 and set E s := {ζ ∈ T : d(ζ, E) ≤ s}. Let K(t) = log+ (2/t) and let ν s be the equilibrium measure for E s . Prove that    1  1 f (rζ) − f (ζ/r) |dζ| = 0. lim− Kν s (ζ) − r→1 2π T c(E s ) (ii) Deduce that lim−

r→1

1 2π



  f (0) . Kν s (ζ) f (rζ) − f (ζ/r) |dζ| = c(E s) T

(iii) Using the result of Theorem 2.4.4, prove that, for 0 < r < 1:    1    Kν s (ζ) f (ζ/r) |dζ| ≤  f |De B c(E s )−1/2 , e 2π T  1      Kν s (ζ) f (rζ) |dζ| ≤ 2 f |D B c(E s )−1/2 + | f (0)|/c(T). 2π T (iv) By combining (ii) and (iii) and letting s → 0, deduce that f (0) = 0. (v) Repeating with f replaced by zn f (n ∈ Z), conclude that f ≡ 0.

Notes on Chapter 9

179

Notes on Chapter 9 §9.1 The study of cyclicity in D was instigated by Brown and Shields in [24]. Their results were subsequently improved and extended to Dμ by Richter and Sundberg, whose article [99] is the source for the results in this section. Exercise 9.1.2 is based on a result in [13].

§9.2 Theorem 9.2.3 goes back to Carleson [26]. The proof given here is from [24], where it is attributed to J. Shapiro. Theorem 9.2.7 is due to Brown and Cohn [23]. Exercise 9.2.1 is based on a result in [13]. The first four parts of Exercise 9.2.3 are taken from [99]. Part (v) is a result of Beurling [16].

§9.3 As mentioned in the text, Conjecture 9.3.2 is due to Brown and Shields [24]. The proof of Theorem 9.3.3 is from [44], adapted from a technique of Korenblum [67]. Theorem 9.3.5 is from [43]. Exercise 9.3.2 is based on a suggestion of Richter (private communication).

§9.4

 Theorem 9.4.3 can be made more precise as follows. Let us write nE (t) := 2 j 1{|I j |>2t} where (I j ) are the components of T \ E. Then, under the same hypotheses as in the theorem, we have  π D( fw,E )  w (t)2 tnE (t) dt, 0

where the implied constants depend only on γ. There are also versions of this result for the case γ ≤ 2, as well as for when w is a decreasing function. For more on distance functions, we refer to the article [44], on which this section is based.

§9.5 The principal results in this section are due to Hedenmalm and Shields [58]. Using a somewhat different technique, Richter and Sundberg [100] showed that these results can be generalized to functions not necessarily belonging to the disk algebra, provided that one defines E := {ζ : lim inf z→ζ | f (z)| = 0}.

§9.6 The notion of Bergman–Smirnov exceptional set was introduced by Hedenmalm and Shields in [58]. They proved that such sets are of logarithmic capacity zero, and posed the question as to whether the converse is true. They also showed that every countable

180

Cyclicity

compact subset of T is Bergman–Smirnov exceptional. The first uncountable examples were given in [43]. Theorem 9.6.8 as well as the results used in its proof are taken from this source. Lemma 9.6.5 is a result of Taylor and Williams [118, Lemma 5.8], formulated here in different terms, and with a slightly different proof. Exercise 9.6.3 is a result of Kahane and Salem [61].

Appendix A Hardy spaces

This appendix is intended as a brief summary of the basic elements of the theory of Hardy spaces. For the most part, we state results without proof. Unless explicit references are given, proofs can be found in [104, Chapter 17], which provides a succinct introduction to the subject. The reader seeking more details is referred to the specialized texts on Hardy spaces [40, 50, 66, 77].

A.1 Hardy spaces For 0 < p < ∞, the Hardy space H p is defined by  2π   p H := f ∈ Hol(D) : sup | f (reiθ )| p dθ < ∞ .

Definition A.1.1

0 0 for all k. Conversely, suppose that det(Ak ) > 0 for 1 ≤ k ≤ n. By induction, the matrix An−1 is positive definite. Let M be the sum of the eigenspaces corresponding to all the negative eigenvalues of A. As An−1 is positive definite, we must have M ∩ (Cn−1 × {0}) = {0}, and consequently dim M ≤ 1. Therefore A has at most one negative eigenvalue. If it has exactly one negative eigenvalue, then det(A) < 0, contrary to hypothesis. Therefore all the eigenvalues are positive, and A is positive definite.  Note that having det(Ak ) ≥ 0 for all k ∈ {1, . . . , n} does not imply that A is positive semi-definite. Consider, for example, the matrix ( ) 0 0 A := . 0 −1

C.2 Hadamard products Definition C.2.1 Let A = (a jk ) and B = (b jk ) be complex n × n matrices. The Hadamard product of A and B is the n × n matrix A ◦ B := (c jk ), where c jk = a jk b jk ( j, k ∈ {1, . . . , n}). Thus the Hadamard product is obtained simply by coordinatewise multiplication. Theorem C.2.2 (Schur product theorem) If A and B are positive semi-definite matrices, then A ◦ B is also positive semi-definite. If both A and B are positive definite, then so is A ◦ B. Proof Consider the representation (C.2) and write U = (ui j ). Then, for each i and j, n  σk uik u jk . ai j = k=1

Therefore, for any choice of λ1 , . . . , λn ∈ C, we have n  n 

ai j bi j λi λ j =

i=1 j=1

n  n  n 

σk uik u jk bi j λi λ j

i=1 j=1 k=1

=

n  k=1

σk

n n   i=1 j=1

 bi j (λi uik )(λ j u jk ) .

Positive definite matrices

191

Since B is positive semi-definite, the double sum in brackets is non-negative. Also, since A is positive semi-definite, we have σk ≥ 0 for each k. Hence n 

ai j bi j λi λ j ≥ 0.

i, j=1

Thus A ◦ B is positive semi-definite. If both A and B are positive definite, then there exists t > 0 such that A − tI and B − tI are both positive semi-definite. Writing     A ◦ B = (A − tI) ◦ B + tI ◦ (B − tI) + t2 I, exhibits A ◦ B as a sum of three positive semi-definite matrices, the last of which is positive definite. It follows that A ◦ B is itself positive definite.  Theorem C.2.3 (Oppenheim’s inequality) definite n × n matrices. Then

Let A and B be positive semi-

det(A ◦ B) ≥ det(A) det(I ◦ B). Proof By Theorem C.2.2 the matrix A ◦ B is positive semi-definite and so det(A ◦ B) ≥ 0. This proves the result in the case when det(A) = 0, so we may as well assume that det(A) > 0. This means that A is positive definite. Let J be the n × n matrix all of whose entries are zero except for the very last one, which equals 1. Then, for each t ∈ R, we have (A − tJ)k = Ak

(1 ≤ k ≤ n − 1),

and expansion with respect to the last row shows that det(A − tJ) = det(A) − t det(An−1 )

(t ∈ R).

Thus, by Sylvester’s criterion Theorem C.1.2, the matrix A − tJ is positive definite provided that t < det(A)/ det(An−1 ). For all such t, Theorem C.2.2 tells us that (A−tJ)◦B is positive semi-definite, and hence det((A−tJ)◦B) ≥ 0. After expansion with respect to the bottom row, this last inequality is equivalent to det(A ◦ B) ≥ det(An−1 ◦ Bn−1 )tbnn . As this holds for t < det(A)/ det(An−1 ), it also holds for t = det(A)/ det(An−1 ), whence det(A ◦ B) det(An−1 ◦ Bn−1 ) ≥ bnn . det(A) det(An−1 ) Iterating this gives det(A ◦ B)  ≥ b j j, det(A) j=1 n

192

Appendix C 

which is the desired inequality.

Taking B = I in this theorem, we immediately obtain the following corollary. Corollary C.2.4 (Hadamard’s inequality) If A is an n × n positive semidefinite matrix, then n  det(A) ≤ a j j. j=1

Remark Oppenheim’s inequality can be strengthened as follows. If A and B are positive semi-definite n × n matrices, then det(A ◦ B) + det(A) det(B) ≥ det(A) det(I ◦ B) + det(A ◦ I) det(B). This inequality is known as Schur’s inequality. We do not stop to give a proof since the result is not needed in the book.

Appendix D Regularization and the rising-sun lemma

In this appendix we prove a regularization lemma, Lemma 9.4.5, used in establishing Theorem 9.4.1. The principal tool is the notion of the increasing regularization of a function, and we begin by examining this separately.

D.1 Increasing regularization Definition D.1.1 Given a function u : [0, ∞) → [0, ∞), we define its increasing regularization  u : [0, ∞) → [0, ∞) by  u(x) := inf{u(y) : y ≥ x}

(x ≥ 0).

Clearly  u is increasing and  u ≤ u. Also,  u is maximal with these two properties, in the sense that, if v is any increasing function with v ≤ u, then also v ≤ u. We shall need two results about increasing regularizations. The first is a version of the so-called rising-sun lemma of F. Riesz. Lemma D.1.2 Let u : [0, ∞) → [0, ∞) be a function that is lower semicontinuous and right-continuous. Let  u be the increasing regularization of u and define U := {x ∈ [0, ∞) :  u(x) < u(x)}. Then U is open in [0, ∞). Further, if a, b are the endpoints of any component of U, then u(a) ≥ u(b). Proof Let x ∈ U. Then there exists y > x such that u(y) < u(x). By lower semicontinuity u(y) < u(x ) for all x in a neighborhood of x. All such x also belong to U. Thus U is open in [0, ∞). Now let a, b be the endpoints of a component of U. Since U is open in [0, ∞), we have b  U, and hence u(y) ≥ u(b) for all y ≥ b. Let x ∈ (a, b). As u is lower semicontinuous on the compact set [x, b], its minimum on this set is attained, at x0 say. We then have u(y) ≥ u(x0 ) for all y ≥ x0 , which implies that 193

194

Appendix D

 u(x0 ) = u(x0 ) and so x0  U. The only possibility is that x0 = b. Thus u ≥ u(b) on [x, b], and in particular u(x) ≥ u(b). Finally, letting x → a and using the right-continuity of u, we obtain u(a) ≥ u(b).  In the rising-sun terminology, the set U corresponds to the shade. The second result that we need, taken from [44], is an estimate for the proportion of [0, ∞) that stays in the sun. Recall that the lower density of a Borel set B ⊂ [0, ∞) is defined by    B ∩ [0, x] . ρ− (B) := lim inf x→∞ x Lemma D.1.3 Let u : [0, ∞) → [0, ∞) be a positive function such that u(x)−x is decreasing. Set S := {x ∈ [0, ∞) :  u(x) = u(x)}. Then S is a Borel set and u(x) . x Proof As u(x) − x is decreasing, it follows that u1 (x) := limy↓x u(y) exists for all x. The function u1 is both lower semicontinuous and right-continuous, and u1 (x) − x is decreasing. Further, we have both u1 = u and  u1 =  u except on countable sets. Thus, we may as well suppose from the outset that u is lower semicontinuous and right-continuous, so that Lemma D.1.2 applies. We may also suppose that u(x) → ∞ as x → ∞ for, if not, then necessarily lim inf x→∞ u(x)/x = 0, and there is nothing to prove. As a consequence of this supposition, S is necessarily unbounded. Let y ∈ S . Let I1 , . . . , In be a finite set of components of U := [0, ∞) \ S lying in [0, y]. Let a j , b j be the endpoints of I j . Renumbering, if necessary, we may suppose that 0 ≤ a1 < b1 < · · · < an < bn ≤ y. Then ρ− (S ) ≥ lim inf x→∞

|I1 ∪ · · · ∪ In | =

n  (b j − a j ) j=1

≤

n  

b j − u(b j ) − a j + u(a j )



j=1

≤ y − u(y) + u(0), where, for the first inequality we used Lemma D.1.2, and for the second the fact that u(x) − x is decreasing. As this holds for any such set of components, it  follows that U ∩ [0, y] ≤ y − u(y) + u(0). Recalling that U is the complement of S , we deduce that   S ∩ [0, y] ≥ u(y) − u(0) (y ∈ S ). Now, given x ∈ [0, ∞), let y be the smallest element of S such that y ≥ x.

Regularization and the rising-sun lemma Then

  S ∩ [0, x] x

=

  S ∩ [0, y] x

≥

  S ∩ [0, y] y

≥

195

u(y) − u(0) . y

It follows that lim inf x→∞ |S ∩ [0, x]|/x ≥ lim inf y→∞ u(y)/y, which is what we had to prove. 

D.2 Proof of the regularization lemma For convenience, we re-state the regularization lemma, Lemma 9.4.5. Lemma D.2.1 Let a > 0, let β ∈ (0, 1] and let φ : (0, a] → (0, ∞) be a function such that • φ(t)/t is decreasing, • 0 < φ(t) ≤ tβ for all t ∈ (0, a], a • 0 dt/φ(t) = ∞. Then, given α ∈ (0, β), there exists a function ψ : (0, a] → (0, ∞) such that • ψ(t)/tα is increasing, • φ(t) ≤ ψ(t) ≤ tβ for all t ∈ (0, a], a • 0 dt/ψ(t) = ∞. The proof is based on increasing regularization, as developed in the previous section, and on the following simple result about sets of positive lower density. Lemma

→ [0, ∞) be a positive decreasing function such

∞ D.2.2 Let v : [0, ∞) that 0 v(x) dx = ∞. Then B v(x) dx = ∞ for every Borel set B ⊂ [0, ∞) with ρ− (B) > 0. Proof Suppose that ρ− (B) > 0. Then there exists λ > 0 such that, for all sufficiently large x,    B ∩ [0, x] ≥ λx. Fix a > 1/λ. Then, for all sufficiently large k,    v(x) dx ≥ v(ak ) B ∩ [ak−1 , ak ] k−1 k B∩[a ,a ]    ≥ v(ak )  B ∩ [0, ak ] − ak−1   ≥ v(ak ) λak − ak−1 .

196

Appendix D

Also, for all k,

 [ak ,ak+1 ]

v(x) dx ≤ v(ak )(ak+1 − ak ).

Hence, for all sufficiently large k,   λ − 1/a v(x) dx ≥ v(x) dx. a − 1 [ak ,ak+1 ] B∩[ak−1 ,ak ]

Summing over these k, we deduce that B v(x) dx = ∞.



Proof of Lemma D.2.1 By a simple change of scale, we can reduce to the case where a = 1. This will simplify the notation in what follows. Define u : [0, ∞) → [0, ∞) by the formula α 1 log φ(e−x ) − x 1−α 1−α The properties of φ are reflected in u as follows:

(x ≥ 0).

u(x) := −

φ(t)/t decreasing

⇐⇒

φ(t) ≤ tβ

⇐⇒

dt =∞ φ(t)

⇐⇒

 0

1

u(x) − x decreasing, β−α u(x) ≥ x, 1−α  ∞ e(1−α)(u(x)−x) dx = ∞. 0

Let  u : [0, ∞) → [0, ∞) be the increasing regularization of u, and let us define ψ : (0, 1] → [0, ∞) to be the function satisfying the formula 1 α log ψ(e−x ) − x (x ≥ 0). 1−α 1−α The desired properties of ψ correspond to properties of  u as follows:  u(x) = −

ψ(t)/tα increasing

⇐⇒

φ(t) ≤ ψ(t) ≤ tβ  1 dt =∞ ψ(t) 0

⇐⇒ ⇐⇒

 u(x) increasing, β−α x, u(x) ≥  u(x) ≥ 1−α  ∞ e(1−α)(u(x)−x) dx = ∞. 0

It thus suffices to prove these three properties of  u. The first two are obvious. For the third, we remark that, writing S := {x ∈ [0, ∞) :  u(x) = u(x)}, we have  ∞   e(1−α)(u(x)−x) dx ≥ e(1−α)(u(x)−x) dx = e(1−α)(u(x)−x) dx. 0 (1−α)( u(x)−x)

S

S

is a decreasing function and, by Lemma D.1.3, we have Also e ρ− (S ) ≥ (β − α)/(α − 1) > 0. Therefore, by Lemma D.2.2, the last integral diverges, and the proof is complete. 

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Index of notation

dA A(D) A(De ) An (D) A A◦B Aut(D) B Be B(x, r) c(·) c∗ (·) cα (·) cK (·) c∗K (·) Cg  Cg Cμ C C∞ Cφ D De diam dist D D( f ) ·, ·D  f D [ f ]D [ f ]⊥D DE Dw

area Lebesgue measure, 1 disk algebra, 3 disk algebra on the exterior of the unit disk, 170 functions such that f (n) belongs to the disk algebra, 64 annulus {z : 1 < |z| < 2}, 29 Hadamard product of A and B, 190 group of automorphisms of D, 8 Bergman space, 98 Bergman space on the exterior of the unit disk, 167 closed ball of center x and radius r, 21 logarithmic capacity, 24 outer logarithmic capacity, 24 Riesz capacity of degree α, 27 capacity with respect to kernel K, 16 outer capacity with respect to kernel K, 18 Cauchy transform of g, 29 maximal Cauchy transform of g, 31 Cauchy transform of measure μ, 183 complex plane, 1 Riemann sphere, 167 operator of composition with φ, 96 open unit disk, 1 exterior of the closed unit disk, 167 diameter, 16 distance, 56 Dirichlet space, 1 Dirichlet integral of f , 1 Dirichlet inner product of f, g, 2 Dirichlet norm of f , 2 closed invariant subspace of D generated by f , 146 see Definition 9.5.5, 168 the set of f ∈ D such that f ∗ = 0 q.e. on E., 152 weighted Dirichlet space, 11

205

206 Dα Dμ Dμ ( f ) [ f ]Dμ Dζ Dζ ( f ) |E| Et f ∗ (ζ) fi fo fr fw,E f ∨g f ∧g G(x1 , . . . , xn ) H K Hol(D) HolBe ,N + (C∞ \ E) Hp H2 H∞ IK (μ) Kμ kw 2 L2 (A) Lat(T, H) log+ (x) M(D) Mh Mh Mz μ(k) N N+ NF nφ P(F) Pμ q.e. ρ− (B) S (I) supp μ T  u W+

Index of notation weighted Dirichlet space with power weight, 13 harmonically weighted Dirichlet space, 109 harmonically weighted Dirichlet integral of f , 109 closed invariant subspace of Dμ generated by f , 146 local Dirichlet space at ζ, 109 local Dirichlet integral of f at ζ, 109 arclength measure of E, 25 the set of points at a distance at most t from E, 42 radial limit of f at ζ, 182 inner-outer factorization of f , 184 r-dilation of f , 115 distance function corresponding to w, E, 160 outer function determined by max{| f ∗ |, |g∗ |}, 122 outer function determined by min{| f ∗ |, |g∗ |}, 122 determinant of the Gram matrix of x1 , . . . , xn , 56 orthogonal complement of K in H, 87 holomorphic functions on D, 1 see Definition 9.6.1, 171 Hardy space, 181 the Hardy space, 182 bounded holomorphic functions on the unit disk, 181 energy of μ with respect to kernel K, 16 potential of μ with respect to kernel K, 15 reproducing kernel for D at w, 5 square summable sequences, 2 square-integrable functions on the annulus A, 29 T -invariant closed subspaces of H, 140 maximum of log x and zero, 24 multiplier algebra of D, 71 Hardy–Littlewood maximal function of h, 187 operator of multiplication by h, 72 shift operator, 132 Fourier coefficients of μ, 25 Nevanlinna class, 181 Smirnov class, 185 covering number of F, 21 counting function of φ, 96 Borel probability measures on F, 16 Poisson intgeral of μ, 108 quasi-everywhere, 31 lower density of B, 194 Carleson box corresponding to I, 78 closed support of μ, 19 unit circle, 8 increasing regularization of u, 193 analytic Wiener algebra, 3

Index

Blaschke product, 50–54, 67, 89, 90, 127, 129, 130, 134, 151, 184, 185 Blaschke sequence, 50, 61, 67, 69 Bøe, B., 92 Bogdan, K., 70 Borichev, A., 48 Bourdon, P. S., 145 Brown, L., 49, 69, 91, 154, 179 Brown–Shields conjecture, 154–159, 178 partial solution to, 154, 156

Adams, D. R., 48 Agler, J., 14, 91, 92, 145 Aleman, A., 130, 131, 145 algebra, 6, 7, 12–14, 71, 110, 142, 148, 181 Banach algebra, 6, 12, 71–73, 168, 170 disk algebra, 3, 62, 168, 179 on exterior of disk, 170 multiplier algebra, 71, 84 quotient algebra, 169 subalgebra, 185 Wiener algebra, 3 analytic operator, 135, 136, 137, 140, 141 approach region, 49 non-tangential, 9, 182 oricyclic, 9, 45, 111 tangential, 9, 11, 45 exponentially, 29, 45–48 widest possible, 48, 49, 60 Arazy, J., 106 Arcozzi, N., 14, 91 automorphism of unit disk, 8, 93

Cantor set, 21–23, 63, 83, 153, 157, 159, 166 Cantor–Lebesgue measure, 83 capacitable set, 18 capacity, 15, 16, 34, 157 capacity of Cantor set, 23, 26–28 capacity of countable set, 19 estimate in terms of diameter, 16 estimate in terms of Lebesgue measure, 25 inner capacity, 18 is upper semicontinuous, 17 logarithmic capacity, see logarithmic capacity outer capacity, 18 Riesz capacity, see Riesz capacity strong-type inequality for, see strong-type inequality for capacity weak-type inequality for, see weak-type inequality for capacity Carleson box, 78, 98 Carleson measure, 37, 76, 78, 91, 131 for Bergman space, 98, 107 for Dirichlet space, 76–84, 92 characterization of, 80, 91 dual formulation of, 76

Baire category theorem, 177 Banach space, 181 Banach–Steinhaus theorem, 5, 117 Bergman space, 98, 107, 167, 178 Bergman–Smirnov exceptional set, 171–179 Bessel’s inequality, 30 Beurling’s theorem on boundary limits, 29, 31–35, 48 converse to, 39 on invariant subspaces, 132, 143, 184 Beurling, A., 14, 48, 70, 132, 179 Bishop, C. J., 92 Blaschke condition, 50, 50, 51–54, 89

207

208 necessary condition for, 79 sufficient condition for, 77, 79 Carleson set, 61, 64, 67, 68, 70, 153, 155, 158, 159, 165 Carleson’s formula, 118, 129, 155, 160 Carleson, L., 49, 69, 70, 76, 91, 117, 125, 131, 179 Carlsson, M., 145 Cauchy transform, 29–31, 37, 48 extension to unit circle, 31 maximal, 31 of a measure, 170, 172, 178, 183 Cauchy’s integral formula, 177 Cauchy’s theorem, 158 Cauchy–Schwarz inequality, 4, 10, 13, 22, 29, 34, 46, 63, 77, 143, 168, 174 Caughran, J. G., 69, 70 cellular-indecomposable operator, 145 Chang, S.-Y. A., 48 change-of-variable formula, 97 character, 169 Choquet’s theorem, 18 closed graph theorem, 14, 72, 76, 96 Cohn, W., 49, 69, 179 composition operator, 93, 96–107 boundedness of, 100 compactness of, 102, 107 Hilbert–Schmidt, 104, 107 in Schatten class, 107 concave function, 24, 91, 160, 164 contraction, 88, 136, 137, 170 convergence in norm, 5, 43, 94, 103, 117, 141 locally uniform, 41, 42, 44, 60, 90, 111, 114 pointwise, 141, 147, 149, 156, 163 uniform, 117 weak, 94, 119, 136, 137, 140, 141, 147 weak*, 17–19 convex function, 24, 45, 122 corona problem, 76, 91 countable set as Bergman–Smirnov exceptional set, 176, 179 as boundary zero set, 61 as set of capacity zero, 19 counting function, 96, 107 covering number, 21, 166 Cowen, C. C., 107 Cowen, M. J., 145 cyclic function, 146, 146–171, 176, 185

Index operator, 135, 137 subspace, 143, 146, 151 vector, 137, 139 de Branges–Rovnyak space, 131 dilation, 115 Dirichlet integral, 1, 108, 109 Carleson’s formula for, 118, 129, 155, 160 conformal invariance of, 7 Douglas’ formula for, 8–11, 113, 117 formula in terms of Taylor coefficients, 1 interpretation as area, 7 local, see local Dirichlet integral of distance function, 160 of outer function, 118 weighted, 11, 108 harmonically weighted, 109 Dirichlet space, 1–3 characterization via M¨obius invariance, 93 inner product, 2 local, 109, 131 norm, 2 representation formula, 29, 30, 45, 48 reproducing kernel, see reproducing kernel weighted, 11–14, 35, 39, 91 harmonically weighted, see harmonically weighted Dirichlet space disk algebra, 3, 62, 168, 179 on exterior of disk, 170 distance function, 118, 159–166, 179 Dirichlet integral of, 160 dominated convergence theorem, 25, 33, 63, 119, 120, 157, 158 Douglas’ formula, 8–11, 113, 117 local Douglas formula, 110, 113, 118, 131 Douglas, J., 14 Douglas, R. G., 145 dual, 67, 76, 166, 168 Dyn kin, E. M., 48 energy, 16, 28, 62, 172 formula in terms of Fourier coefficients, 25 equilibrium measure, 19–21, 43, 81 potential of, 19 uniqueness of, 26, 27 factorization canonical, 184, 185 inner-outer, 125, 126, 129, 141, 143, 147–149, 151, 184

Index Fatou’s lemma, 12, 20, 25, 32, 40, 60, 116, 120, 128 Fatou’s theorem, 182 Fej´er–Riesz theorem, 138 Fields medal, 14 Fisher, S. D., 106 Frostman’s theorem, 19 Frostman, O., 27 Fubini’s theorem, 22, 34, 100, 110 fusion lemma, 155 Gallardo-Guti´errez, E. A., 107 Gelfand theory, 169 Gonz´alez, M. J., 107 Gram matrix, 56 Green’s function, 28 Hadamard product, 57, 88, 190 Hadamard’s inequality, 192 Hahn–Banach theorem, 166, 168 Hansson, K., 48 Hardy space, 2, 3, 8, 13, 29, 50, 71, 73, 76, 78, 89, 132, 148, 168, 181–186 Hardy’s inequality, 3, 182 Hardy–Littlewood maximal function, 46, 187–188 harmonic function, 13, 14, 41, 44, 108, 116, 183 subharmonic function, 158 superharmonic function, 91 harmonically weighted Dirichlet integral, 109 harmonically weighted Dirichlet space, 108–110, 132, 146 polynomials dense in, 116, 137 Hastings, W. W., 107 Hausdorff–Young inequality, 3 Havin, V. P., 70 Hedberg, L. I., 48 Hedenmalm, H., 178, 179 Hilbert space, 2–4, 11, 14, 29, 43, 55–57, 72, 85, 88, 93, 94, 104, 109, 111, 135, 139, 140, 145, 166, 167, 182 Hilbert–Schmidt operator, 104 H¨older’s inequality, 3 Hruˇscˇ ev, S. V., 69 ideal, 168 increasing regularization, 193–195, 196 inner factor, 125, 151, 154 inner function, 125–130, 132, 134, 143, 147, 170, 183–185, 185

209

local Dirichlet integral of, 127–128 singular inner function, 127, 184, 185 inner product, 2, 5, 30, 56, 166 semi-, 2, 138 inner-outer factorization, 125, 126, 129, 141, 143, 147–149, 151, 184 interpolating sequence, 92 invariant subspace, 108, 132, 140–144, 146, 148–150, 152, 184, 185 generated by a set, 146 invertible, 150 isometry, 2, 112, 132, 138–140 2-isometry, 130, 132, 135, 136–138, 140, 141, 145 Jensen’s inequality, 122 Julia–Carath´eodory theorem, 131 Kahane, J.-P., 180 Kaluza, T., 69 kernel, 15 convex, 24 Koebe function, 183 Koosis, P., 69 Korenblum, B. I., 179 lattice operations, 122 applied to outer functions, 122 Lebesgue measure, 1, 25, 26, 29, 36, 61, 64, 81, 109, 110, 113, 114, 118, 128, 129, 144, 146, 151, 155, 159, 160, 182, 184 Cantor–Lebesgue measure, 83 Lef`evre, P., 107 Li, D., 107 linear functional, 138, 168 continuous, 4, 14, 72, 94, 170 multiplicative, 169 positive, 138 local Dirichlet integral, 109, 131 of inner function, 127–128 Douglas formula for, 110, 113, 118, 131 formula in terms of Taylor coefficients, 113, 115 of outer function, 118 Richter–Sundberg formula for, 118, 121, 129 local Dirichlet space, 109, 111, 131 local Douglas formula, 110, 113, 118, 131 logarithmic capacity, 24–27, 28, 29, 31, 33, 34, 40, 61, 62, 69, 80, 83, 105, 106, 146, 152, 159, 166, 171, 178, 179

210 lower density, 194, 195 Luecking, D. H., 107 MacCluer, B. D., 107 Malliavin, P., 70 Marshall, D. E., 48, 92 maximum principle, 12, 158, 159, 173, 174 for potentials, 24 Smirnov’s, 10, 113, 121, 175, 186 two-sided, 172 Maz ya, V. G., 48, 70 McCarthy, J. E., 14, 92 Minkowski’s inequality, 7 M¨obius invariance, 8, 51, 93–96 characterization of Dirichlet space via, 93 M¨obius transformation, 8, 12, 174 monotone convergence theorem, 17 Monterie, M. A., 28 Morera’s theorem, 158 multiplication operator, 72, 84, 101 multiplier, 7, 71–75, 133, 143, 152 boundedness of, 73 characterization of, 74, 76 zero set of, 89–91 multiplier algebra, 71, 84 Nagel, A., 49, 69 Nevanlinna class, 181, 185 non-tangential limit, 9, 29, 31, 35, 182 norm, 2, 3, 6, 11, 12, 14, 42, 71, 88, 94, 109, 137, 168, 181 attainment of, 85, 86 algebra norm, 7 multiplier norm, 72, 89 operator norm, 96 semi-norm, 2 sup-norm, 170 normal family, 60, 90 Ohtsuka, M., 28 Olin, R. F., 145 Oppenheim’s inequality, 57, 191, 192 orthogonal basis, 111, 113 orthogonal projection, 55, 59, 85, 87, 136 orthonormal basis, 104, 105 outer factor, 151, 156, 158, 186 outer function, 10, 65, 113, 117–127, 143, 144, 147–150, 152–160, 166, 170, 171, 178, 183–185, 185 Dirichlet integral of, 118 lattice operations applied to, 122

Index local Dirichlet integral of, 118 parallelogram identity, 116 Parrott’s lemma, 85 Parrott, S., 91 Parseval’s formula, 2, 9, 113, 167 Peller, V. V., 69 Pick interpolation, 5, 84–89, 89, 131 Pick property, 88, 89 complete Pick property, 91 Plateau problem, 14 Poisson integral, 8, 13, 41, 44, 108, 116, 120, 122, 182, 186 positive definite matrix, 55–57, 189–192 positive semi-definite, 55–59, 84, 85, 87, 88, 189–192 potential, 15, 20, 28 maximum principle for, 24 of equilibrium measure, 19 quasi-everywhere, 31, 34, 105, 152 Queff´elec, H., 107 radial limit, 8–10, 48, 105, 110, 117, 118, 125, 132, 174, 182, 185 regular point, 127, 128, 130 regularization lemma, 162, 163, 195 representation formula, 29, 30, 45, 48 reproducing kernel, 4–5, 14, 55, 57, 72, 76, 79, 84, 88, 89, 92, 101 Richter, S., 69, 117, 125, 130, 131, 145, 179 Richter–Sundberg formula, 118, 121, 129 complement to, 121 Riemann mapping, 61, 62 Riemann sphere, 167 Riesz capacity, 15, 27, 35, 39, 70, 106 Riesz representation theorem for continuous functions, 138 for Hilbert space, 4 rising-sun lemma, 162, 193 Rochberg, R., 14, 91 Rodr´ıguez-Piazza,L., 107 Ross, W. T., 14, 69 Rudin, W., 49, 69 Salem, R., 180 Sarason, D., 131 Sawyer, E. T., 14, 91 Schur product theorem, 88, 190 Schur’s inequality, 192

Index Schwarz inequality, see Cauchy–Schwarz inequality Seip, K., 92 sesquilinear form, 137, 139 Shapiro, H. S., 69 Shapiro, J. H., 49, 69, 70, 107, 179 Shields, A. L., 69, 91, 145, 154, 178, 179 shift operator, 132–135, 145, 184 characterization of, 135–140, 145 Shimorin, S. M., 91 singular inner function, 127, 184, 185 Smirnov class, 130, 170, 185 characterization of, 185 Smirnov’s maximum principle, 10, 113, 121, 175, 186 spectral theorem, 189 Stegenga, D. A., 91 Stone–Weierstrass theorem, 17 strong-type inequality for capacity, 29, 35–39, 43, 48, 82 in weighted Dirichlet space, 39 sharpness of, 42 subharmonic function, 158 Sundberg, C., 69, 92, 117, 125, 131, 145, 179 superharmonic function, 91 Sylvester’s criterion, 189, 191 Taylor, B. A., 180 Taylor, G. D., 91 Thomson, J. E., 145

211

transfinite diameter, 28 trigonometric polynomial, 138 uniqueness set, 50–54, 67, 69 on boundary, 62, 70 unitary matrix, 189 unitary operator, 139, 140, 142, 143 univalent function, 154, 182 wandering subspace theorem, 136 weak-type inequality for capacity, 35–39, 48, 105, 152 in weighted Dirichlet space, 39, 106 for Hardy–Littlewood maximal function, 46, 187 weight harmonic weight, 13, 108 power weight, 13 superharmonic weight, 130 Wick, B. D., 14 Wiener algebra, 3 Williams, D. L., 180 Wynn, A., 91 zero set, 5, 50, 50–54, 89, 90, 131, 144, 153, 182 arguments of, 67–70 moduli of, 54–61 of multiplier, 89–91 on boundary, 50, 61–67, 70, 146, 152–155