Introduction to function spaces on the disk 8680593370

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Miroslav Pavlovi´c

Introduction to Function Spaces on the Disk

Matematiˇcki institut SANU Beograd 2004

Miroslav Pavlovi´c Faculty of Mathematics Belgrade University 11001 Belgrade, p.p. 550 Serbia

Typeset by the author in LATEX. c M. Pavlovi´c and Matematiˇcki institut SANU.

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without prior permission of the copyright owner.

To Mirjana and Pavle

Contents Preface 1 Quasi-Banach spaces 1.1 Quasinorm and p-norm . . . . 1.2 Linear operators . . . . . . . 1.3 Open mapping, closed graph 1.4 F -spaces . . . . . . . . . . . . 1.5 The spaces ℓp . . . . . . . . .

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2 Interpolation and maximal functions 2.1 The Riesz/Thorin theorem . . . . . . . . . . . 2.2 Weak Lp -spaces and Marcinkiewicz’s theorem 2.3 Maximal function and Lebesgue points . . . . 2.4 The Rademacher functions . . . . . . . . . . . 2.5 Nikishin’s theorem . . . . . . . . . . . . . . . 2.6 Nikishin and Stein’s theorem . . . . . . . . . 2.7 Banach’s principle . . . . . . . . . . . . . . .

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3 Poisson integral 3.1 Harmonic functions . . . . . . . . . . 3.2 Borel measures and the space h1 . . 3.3 Radial limits of the Poisson integral 3.4 The spaces hp and Lp (T) . . . . . . 3.5 The Littlewood/Paley theorem . . . 3.6 Harmonic Schwarz lemma . . . . . .

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38 38 42 45 49 51 53

4 Subharmonic functions 4.1 Basic properties . . . . . . . . . . . . 4.2 Properties of the mean values . . . . 4.3 Integral means of univalent functions 4.4 The subordination principle . . . . . 4.5 The Riesz measure . . . . . . . . . . 4.6 A Littlewood/Paley theorem . . . .

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55 55 60 62 64 67 70

5 Classical Hardy spaces 5.1 Basic properties . . . . . . . . . . 5.2 The space H 1 . . . . . . . . . . . 5.3 Blaschke product . . . . . . . . . 5.4 Inner and outer functions . . . . 5.5 Composition with inner functions

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6 Conjugate functions 6.1 Harmonic conjugates . . . . . . . . . . . . . 6.2 Riesz projection theorem . . . . . . . . . . . 6.3 Applications of the projection theorem . . . 6.4 Aleksandrov’s theorem . . . . . . . . . . . . 6.5 Strong convergence in H 1 . . . . . . . . . . 6.6 Quasiconformal harmonic homeomorphisms

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92 92 96 99 100 101 104

7 Maximal functions, interpolation, coefficients 7.1 Maximal theorems . . . . . . . . . . . . . . . . 7.2 Maximal characterization of H p . . . . . . . . . 7.3 “Smooth” Ces` aro means . . . . . . . . . . . . . 7.4 Interpolation of operators on Hardy spaces . . 7.5 On the Hardy/Littlewood inequality . . . . . . 7.6 On the dual of H 1 . . . . . . . . . . . . . . . .

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110 110 114 116 119 123 127

8 Bergman spaces: Atomic decomposition 8.1 Bergman spaces . . . . . . . . . . . . . . 8.2 Reproductive kernels . . . . . . . . . . . 8.3 The Coifman/Rochberg theorem . . . . 8.4 Coefficients of vector-valued functions .

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9 Subharmonic behavior 143 9.1 Subharmonic behavior and Bergman spaces . . . . . . . . . . . . . . . . . 143 9.2 The space hp , p < 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 9.3 Subharmonic behavior of smooth functions . . . . . . . . . . . . . . . . . 148 10 Lipschitz spaces 10.1 Lipschitz spaces of first order . . . . 10.2 Lipschitz condition for the modulus . 10.3 Lipschitz spaces of higher order . . . 10.4 Growth of derivatives . . . . . . . .

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154 154 158 160 162

11 Lacunary series 11.1 Lacunary series in H p . . . . . . . . . . . . . . 11.2 Karamata’s theorem and Littlewood’s theorem 11.3 Lacunary series in C[0, 1] . . . . . . . . . . . . 11.4 Lp -integrability of lacunary series on (0, 1) . . .

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170 170 172 176 178

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Bibliography

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Index

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Preface This text contains some facts, ideas, and techniques that can help or motivate the reader to read books and papers on various classes of functions on the disk and the circle. The reader will find several well known, fundamental theorems as well as a number of the author’s results, and new proofs or extensions of known results. Most of assertions are proved, although sometimes in a rather concise way. A number of assertions are named by Exercise, while certain assertions are collected in Miscellaneous or Remarks; most of them can be treated by the reader as exercises. The reader is assumed to have good foundation in Lebesgue integration, complex analysis, functional analysis, and Fourier series, which means in particular that he/she had a good training through these areas. It is of some importance that the reader can accept the following: Throughout this text, constants are often given without computing their exact values. In the course of a proof, the value of a constant C may change from one occurrence to the next. Thus, the inequality 2C 6 C is true even if C > 0. Acknowledgment I want to express my appreciation to those who pointed out to me several typos as well as suggestions for improvement. In particular, I want to mention the detailed comments from Professor Miroljub Jevti´c and Professor Miloˇs Arsenovi´c. I also want to express my deep gratitude to Mathematical Institute of Serbian Academy of Arts and Sciences and to the Faculty of Economics, Finance and Administration, Belgrade, for financial support.

Cvetke and Belgrade, 20 March – 22 April 2003.

6

1

Quasi-Banach spaces

In this text we mention only two examples of locally convex spaces: h(Ω), the space of all complex-valued functions harmonic in Ω ⊂ C, and its subspace H(Ω) consisting of analytic functions. For our purposes, the class of locally bounded spaces is more important. By Kolmogorov’s theorem, the intersection of this class with the class of locally convex spaces consists precisely of normable spaces. The topology of a locally bounded space can be described by a “quasinorm”; conversely, a “quasinormable” space is locally bounded. In the class of quasi-Banach spaces, there hold the “basic principles of functional analysis.” A concise discussion of these principles is contained in Section 1.3 and, in the context of F -spaces, in 1.4. Some properties of ℓp are stated, without proofs, in Section 1.5.

1.1

Quasinorm and p-norm

Let X be a (complex) vector space. A functional k · k : X 7→ [0, ∞) is called a quasinorm if the following conditions hold: kf + gk 6 K(kf k + kgk),

(1.1)

where K (> 1) is a constant independent of f, g ∈ X; and kf k > 0 (f 6= 0),

kλf k = |λ| kf k (λ ∈ C).

(1.2)

The couple (X, k · k) is then called a quasinormed space. The standard example are Lebesgue spaces: if µ is a positive measure defined on a sigma-algebra of subsets of a set S, then the space Lp (µ) = Lp (S, µ) = Lp (S) (0 < p 6 ∞) consists of all measurable complex-valued functions f on S for which Z 1/p kf k = kf kp = |f |p dµ < ∞, S

with the usual interpretation in the case p = ∞. When p < 1, this functional is not a norm but satisfies (1.1) with K = 21/p−1 and, moreover, kf + gkp 6 kf kp + kgkp .

(1.3)

A functional satisfying (1.3) and (1.2) is called a p-norm. From (1.3) it follows that kf1 + f2 + · · · + fn kp 6 kf1 kp + kf2 kp + · · · + kfn kp . A similar inequality holds in the general case although a quasinorm need not be a p-norm for any p. 7

8

1 Quasi-Banach spaces

1.1.1 Lemma If k · k is a quasinorm on X, then there exist constants p ∈ (0, 1) and C 6 4 such that
kf1 + f2 + · · · + fn kp 6 C kf1 kp + kf2 kp + · · · + kfn kp (1.4) for every finite sequence f1 , . . . , fn ∈ X.

From this one can deduce that X is “p-normable” for some p > 0. 1.1.2 Theorem (Aoki/Rolewicz) If k · k is a quasi-norm on X, then there is p > 0 and a p-norm ||| · ||| on X such that kf k/C 6 |||f ||| 6 kf k, f ∈ X, where C is independent of f . The p-norm is defined by |||f ||| = inf

n n X j=1

kfj kp

1/p

:f=

n X j=1

o fj ,

where the infimum is taken over all finite sequences {fj } ⊂ X. Proof of Lemma. Take p so that (2K)p = 2, where K is the constant from (1.1), and define the functional H on X in the following way: H(0) = 0 and H(f )p = 2k if 2k−1 6 kf kp < 2k for some integer k. Since kf kp 6 H(f )p 6 2kf kp ,

(1.5)

inequality (1.4) is a consequence of the inequality kf1 + · · · + fn kp 6 2

H(f1 )p + · · · + H(fn )p




.

The latter holds for n = 1. If n > 2, we consider two cases. (i) Let the summands H(fj ) be mutually distinct and arranged in decreasing order. Then we have H(fj )p 6 21−j H(f1 )p

(1 6 j 6 n).

(1.6)

From (1.1) it follows that kf + gk 6 2K max{kf k, kgk}, whence, by (1.5), kf1 + · · · + fn k 6 max{(2K)j H(fj ) : 1 6 j 6 n}. Because of (1.6) and the choice of p, it turns out that kf1 + · · · + fn kp 6 2H(f1 )p , which implies the required inequality. (ii) Assume that the sequence H(fj ) contains at least two equal elements; for example, let H(f1 ) = H(f2 ) = 2m . Then 2m−1 6 kf kp1 , kf kp2 < 2m . Since kf1 + f2 kp 6 (2K)p max{kf1 kp , kf2 kp } = 2 max{kf1 kp , kf2 kp } 6 2m+1 ,

1.1 Quasinorm and p-norm

9

we have H(f1 +f2 )p 6 2m+1 = H(f1 )p +H(f2 )p . This and the induction hypothesis imply k(f1 + f2 ) + · · · + fn kp 6 2 62

62

H(f1 + f2 )p + · · · + H(fn )p




H(f1 )p + H(f2 )p + · · · + H(fn )p
kf1 kp + kf2 kp + · · · + kfn kp .




✷

The space X is endowed with the structure of a topological vector space by declaring “a neighborhood of zero” to mean “a set containing {f : kf k < 1/n} for some n = 1, 2, . . . .”(∗) This topology is metrizable, according to the Aoki/Rolewicz theorem; namely, if a p-norm ||| · ||| is equivalent to the original quasinorm, then the formula d(f, g) = |||f − g|||p defines a metric that induces the same topology. The space X need not be locally convex(†) ; it is locally bounded because the neighborhoods {f : kf k < 1/n} are bounded in the sense of theory of topological vector spaces. On the other hand, it is known that a locally bounded vector topology can be described by a quasinorm (cf. [87]). 1.1.3 Exercise On the space Lp (0, 1) (0 < p < 1), there is not an equivalent q-norm for 1 > q > p. The same holds for the sequence space ℓp . Quasi-Banach and p-Banach spaces A quasinormed space X is called a quasiBanach space if it is complete, which means that a sequence {fn } ⊂ X is convergent if (and only if) kfm − fn k → 0 as m, n → ∞. If X is p-normed and complete, then X is said to be p-Banach. 1.1.4 P Proposition Let X be p-normed. of the P Then X is complete iff convergence P series kfn kp implies convergence of f . If X is complete and f converges, n n

p P∞

P∞ then there holds the inequality n=1 fn 6 n=1 kfn kp .

1.1.5 Proposition Let {fjk } (j, k > 1) be a double sequence in a p-Banach   ∞ ∞ P ∞ ∞ P P P P kfjk kp < ∞, then the iterated series space X. If fjk fjk and j,k

j=1 k=1

k=1 j=1

converge and have the same sum.

1.1.6 Exercise (Peck [81]) Let (X, k · k) be a complex p-normed space of dimension n < ∞. By a theorem of Carathe´odory, every point from the convex hull of the unit ball can be represented as a convex combination of 2n points from the ball (because the real dimension is 2n). This can be used to show that there exists a norm k · kn on X such that kf kn 6 kf k 6 (2n)1/p−1 kf kn . Note that (2n)1/p−1 is not the best constant, at least for n = 1. (∗) The “ball” {f : kf k < 1} need not be an open set. Therefore a quasinorm, in contrast to a p-norm, need not be continuous. (†) For example, the space Lp (0, 1), 0 < p < 1, is not locally convex.

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1 Quasi-Banach spaces

1.2

Linear operators

In the class of quasinormed spaces, continuity and boundedness of linear operators are equivalent. In the space L(X, Y ), of continuous linear operators from X to Y , the quasinorm is defined by kT k := supkf k61 kT f k. The space L(X, Y ) is complete iff so is Y . An operator T ∈ L(X, Y ) is said to be invertible if it is bijective and its inverse is continuous. 1.2.1 Proposition Let X be a quasi-Banach space and T ∈ L(X, X) such an operator that kI − T k < 1, where I is the identity operator. Then T is invertible
−1 and there holds the inequality kT −1 kp 6 C 1 − kI − T kp , where C and p are the constants from Lemma 1.1.1. P∞ Proof. Consider the series k=0 (I − T )k . From inequality (1.4), applied to the space L(X, X), we get

X

p n X

n

k

(I − T ) kI − T kpk . 6 C

k=m

k=m

Therefore the series P∞ converges; denote its sum by S. Then we have ST = T S = I and kSkp 6 C k=0 kI − T kpk , which was to be proved. ✷ The following statement is important although its proof is very simple.

1.2.2 Theorem Let X and Y be quasi-Banach spaces and E a dense subset of X. Let Tn ∈ L(X, Y ) be a sequence such that supn kTn k < ∞. If the limit limn→∞ Tn f exists for all f ∈ E, then it exists for all f ∈ X and the operator T f := limn→∞ Tn f is linear and continuous. 1.2.3 Exercise Let T be a continuous linear operator from a quasi-normed space X to quasi-normed space Y , and let E be a subset of X such that the linear hull of E is dense in X. If Y0 is a closed subspace of Y such that T (E) ⊂ Y0 , then T (X) ⊂ Y0 . q-Banach envelope In the general case, a quasi-Banach space is embedded into many q-Banach spaces; the “smallest” of them is called the q-Banach envelope of X. To be more precise, define the functional Nq (0 < q 6 1) on X in the following way:  1/q X  X fj = f , (1.7) : kfj kq Nq (f ) = inf j

j

where the infimum is taken over the set of finite sequences {fj } ⊂ X. This functional is a “ q-seminorm”, i.e., satisfies the conditions {Nq (f + g)}q 6 {Nq (f )}q + {Nq (g)}q ,

Nq (λf ) = |λ| Nq (f ).

1.2 Linear operators

11

The set {f ∈ X : Nq (f ) = 0} =: Ker Nq is a closed subspace of X. If Ker Nq = {0}, i.e., if Nq is a q-norm, then the “completion” of the space (X, Nq ) is a qBanach space and is called the q-Banach envelope of X; denote it by [X]q . According to the Aoki/Rolewicz theorem, always there exists a q such that X = [X]q , with equivalent quasinorms. A simple but illustrative example is X = ℓp ; then [X]q = ℓq (p < q 6 1) and the corresponding quasinorms are equal (see 1.2.8 and 1.2.6). It is much more difficult to identify the envelops of the Hardy space H p (see Theorem 8.3.5). The importance of the space [X]q lies in the fact that every operator from X to an arbitrary q-Banach space extends to an operator on [X]q ; more precisely: 1.2.4 Proposition Let X possess the q-Banach envelope ( i.e., let Nq be a qnorm) and let Y be an arbitrary q-Banach space. If T ∈ L(X, Y ), then there exists a unique operator S ∈ L([X]q , Y ) such that Sf = T f for all f ∈ X. The following fact is useful in identifying the envelope: 1.2.5 Proposition Let X be continuously embedded intoPa q-Banach space Y in ∞ such P∞ a wayq that everyq f ∈ Y can be represented as f = n=1 fn , fn ∈ X, with n=1 kfn kX 6 Ckf kY , where C does not depend of f . Then Y = [X]q (with equivalent quasinorms). Proof. The space X is a dense subset of Y . Since X is dense in [X]q , we see that it suffices to prove that the q-norms k · kY and Nq are equivalent on X. P Let f = fj , where {fj } is a finite sequence in X. Then kf kqY 6

X

kfj kqY 6 C q

X

kfj kqX .

Taking the infimum over {fj } ⊂ X, we get kf kY 6 CNq (f ). (Incidentally this shows that Nq is a q-norm.) P∞ the reverse inequality, let f ∈ X. Then f = n=1 fn , where P∞ P∞To prove q q . Since N (f ) 6 kf k , we get 6 Ckf k N (f )q 6 Ckf kqY . k q n n X q n n Y X n=1 n=1 kf P P Hence fn converges to f in [X]q to f , and Nq (f )q 6 Nq (fn )q 6 Ckf kqY , which completes the proof. ✷ Miscellaneous 1.2.6 The functional Nq is a q-norm on X iff there is a q-Banach space Y such that L(X, Y ) separates points in X. The latter means that for every f 6= 0 there is T ∈ L(X, Y ) with T f 6= 0. 1.2.7 The dual of a quasi-Banach space X is X ∗ = L(X, C). If X ∗ separates points in X, then the Banach envelope of X is equal to the completion of the normed space (X, N ), where N (f ) = sup{|Λf | : Λ ∈ X ∗ , kΛk 6 1}.

12

1 Quasi-Banach spaces

1.2.8 If X = Lp (0, 1), 0 < p < 1 and 1 > q > p, then Nq (f ) = 0 for all f ∈ X. This is connected with the relation L(X, Y ) = {0}, where Y is an arbitrary qBanach space.

1.3

Open mapping, closed graph

Let X, Y be a pair of complete spaces such that X is a dense subset of Y , which means that each member of Y can be approximated by members of X. This does not imply that members of a ball K1 ⊂ Y can be approximated by members of any fixed ball K2 ⊂ X, i.e., that K 2 ⊃ K1 . (K 2 = the closure of K2 in the topology of Y .) Namely, as the following theorem states, if K 2 ⊃ K1 , then X = Y . 1.3.1 Theorem Let X and Y be quasi-Banach spaces. Let T ∈ L(X, Y ) be such that the closure of T (B), where B = {f ∈ X : kf k < 1}, contains a neighborhood of zero in Y . Then the mapping T : X 7→ Y is open and the operator Tb : X/ Ker T 7→ Y is invertible.

A mapping is open if it maps open sets onto open sets. If T ∈ L(X, Y ), then the operator Tb ∈ L(X/ Ker T, Y ) is defined by Tb(f + Ker T ) = T f . The quasinorm in X/Z is defined by kf + Zk = inf{kf − gk : g ∈ Z}. Proof. Because of the Aoki/Rolewicz theorem, we can suppose that X and Y are p-normed for some p < 1. Let δ > 0 and U = {f ∈ X : kf kp < δ}. From the hypotheses of the theorem it follows that there are balls Un = {f ∈ X : kf kp < δn } and Vn = {g ∈ Y : kgkp < εn }, n > 1, limn εn = 0, such that Vn ⊂ T (Un ), ∞ X δn < δ.

(1.8) (1.9)

n=1

We will prove that V1 ⊂ T (U ); then it will be easy to complete the proof. Let g ∈ V1 . It follows from (1.8) that there exists f1 ∈ U1 such that g−T f1 ∈ V2 . Similarly, there is f2 ∈ U2 such that (gP− T f1 ) − T f2 ∈ V3 . Continuing in this way, n we get g − k=1 T fk ∈ Vn+1 , fk ∈ Uk . It followsPthat P∞the sequence of relations the series k fk g = n=1 T fn . And since kfk kp < δk , inequality (1.9) implies thatP ∞ converges; denote its sum by f . Thus we have g = T f and kf kp 6 n=1 kfn kp < δ, which was to be proved. ✷ The open mapping theorem 1.3.2 Theorem Let X and Y be complete spaces and T ∈ L(X, Y ). If T is onto, then T is open. In particular, T is invertible if it is onto and one-to-one. Proof. Let U be the unit ball in X. Choose a zero-neighborhood W so that W − W ⊂ U . By the hypothesis, the space Y is the union of the sets T (nW )

1.3 Open mapping, closed graph

13

(n > 1). By Baire’s category theorem, the closure at least of one of them has nonempty interior, which implies that T (W ) contains an open set V 6= ∅. Then V − V is a neighborhood of zero and there hold the inclusions V − V ⊂ T (W ) − T (W ) ⊂ T (W ) − T (W ) ⊂ T (U ). Now the desired result follows from Theorem 1.3.1.

✷

1.3.3 Exercise A subspace E of a quasi-Banach space X is said to have the Hahn/Banach extension property (HBEP) if each λ ∈ E ∗ has an extension Λ ∈ X ∗ . If E has HBEP, then Λ can be chosen so that kΛkX ∗ 6 CkλkE ∗ , where C is independent of λ. As a special case of the open mapping theorem we have: 1.3.4 Theorem (on equivalent norms) Let k · k1 and k · k2 be quasinorms on a a vector space X, and let kf k1 6 kf k2 for every f ∈ X. If X is complete with respect to both quasinorms, then there exists a constant C < ∞ such that kf k2 6 Ckf k1 for all f ∈ X. The uniform boundedness principle 1.3.5 Theorem (Banach/Steinhauss) Let X and Y be quasi-Banach spaces, and let {As } ⊂ L(X, Y ) be a family of operators. If sups kAs f k < ∞, for all f ∈ X, then sups kAs k < ∞. In particular the limit of an everywhere convergent sequence of bounded operators is a bounded operator. Proof. Let kf k2 = kf kX + sups kAs f kY (f ∈ X). From the hypotheses it follows that the functional k · k2 is a quasinorm on X. It is not hard to prove that the space (X, k · k2 ) is complete and therefore the conclusion follows from ✷ Theorem 1.3.4. 1.3.6 Corollary Let B : X ×Y 7→ Z be a separately continuous bilinear operator, where X, Y, Z are quasi-Banach spaces. Then there is a constant C < ∞ such that kB(f, g)kZ 6 Ckf kX kgkY for all f ∈ X, g ∈ Y . “Separately continuous” means that every operator of the form f 7→ B(f, g) (f ∈ X) or g 7→ B(f, g) (g ∈ Y ) is continuous. Shauder basis A sequence {en : n > 1} in a quasi-Banach space X is called a Shauder basis of X if P to each f ∈ X there corresponds a unique scalar sequence {λn (f )} such that ∞ f = n=1 λn (f )en , the series converging in the topology of X.

1.3.7 Proposition If {en : n > 1} is a Shauder basis of X, then the functionals Pn λn are continuous and the linear operators Sn : X 7→ X defined by Sn f = k=1 λk (f )ek are uniformly bounded.

14

1 Quasi-Banach spaces

Proof. Let |||f ||| = supn kSn f k. Since kf − Sn f k → 0, we have kf k 6 K|||f |||, where K is the constant from (1.1), and therefore, by Theorem 1.3.4, it is enough to prove that X is complete with respect to the quasinorm ||| · |||. Let {fj }∞ j=1 be a Cauchy sequence in ||| · |||. This implies, because of the completeness of k · k, that there is a sequence gn such that sup kSn fj − gn k → 0

n>1

as j → ∞,

(1.10)

and that for every k the sequence {λk (fj )}∞ j=1 converges; let γk = limj λk (fj ). Since the functional λk is linear and the space Sn (X) is finite-dimensional, it follows that λk (gn ) = limP j λk (Sn fj ) = limj λk (fj ) = γk for k 6 n, and λk (gn ) = 0 for k > n. n Hence gn = k=1 γk ek . OnPthe other hand, (1.10) implies that {gn } converges in ∞ k · k to some g. Thus g = n=1 γn en , whence gn = Sn g. Returning to (1.10) we see that |||fj − g||| → 0 as j → ∞, which was to be proved. ✷ 1.3.8 Exercise A sequence {en : n > 1} of nonzero vectors in a quasi-Banach space X is a Shauder basis of X if and only if the following conditions are satisfied: (a) The closed linear Pm Pnspan of {en } is X; (b) There is a constant K such that k j=0 aj ej k 6 Kk j=0 aj ej k for all scalar sequences {aj } and m < n.

The closed graph theorem

1.3.9 Theorem Let T : X 7→ Y be a linear operator, where X and Y are complete spaces. Then T is continuous if the following condition is satisfied: For every sequence {fn } ⊂ X such that fn tends to 0 ∈ X and T fn tends to some g ∈ Y we have g = 0. Proof. It follows from the hypotheses that X is complete with respect to the ✷ quasinorm kf k2 = kf kX + kT f kY so we can apply Theorem 1.3.4.

1.4

F -spaces

The closed graph theorem remains valid in a wider class of spaces, the so called F -spaces. By the term “F -norm” on a vector space X we mean a functional N : X 7→ [0, ∞) satisfying: (a) N (f ) = 0 =⇒ f = 0; (b) N (f + g) 6 N (f ) + N (g); (c) N (λf ) 6 N (f ) for |λ| 6 1, and lim N (λf ) = 0.

λ→0

(1.11)

The formula d(f, g) = N (f − g) defines an invariant metric on X and the topology induced by this metric is vectorial, which means in particular that multiplication by scalars is continuous on C × X. In the case where the metric d is complete, the space X is called an F -space. A p-Banach space can be treated as an F -space by introducing the F -norm N (f ) = kf kpX .

1.4 F -spaces

15

Besides, if X is a locally convex space whose topology is given by a sequence of seminorms pn (n = 1, 2, . . . ), then the formula N (f ) =

∞ X 2−n pn (f ) 1 + pn (f ) n=1

defines an F -norm on X that induces the same topology. As an example one can take the space h(D) consisting of all harmonic functions on the unit disk D ⊂ C as well as its analytic analogue H(D). These spaces are endowed with the topology of uniform convergence on compact subsets of D. This topology can be given by the sequence of norms pn (f ) = max|z|6rn |f (z)|, where rn ∈ (0, 1) is an arbitrary sequence tending to 1. Concerning the requirement (1.11), which guarantees the continuity of scalar multiplication, it is useful to consider the case of the Nevanlinna class N (D). This class consists of the functions f ∈ H(D) for which Z π
1 log 1 + |f (reiθ )| dθ < ∞. (1.12) N (f ) := sup 0 1, cf. [52, Theorem I.2.7]) and Stiles [96] (p < 1, cf. [35, Proposition 2.9]): 1.5.6 Theorem Every bounded linear operator from ℓq to ℓp (0 < p < q < ∞) is compact; the same is true for linear operators from c0 to ℓp . (Moreover, if p < q 6 1, then every operator from a q-Banach space to ℓp (p < q) is compact.) Consequently, no space of the class ℓp , 0 < p < ∞, and c0 is isomorphic to a subspace of another member of this class. On the other hand, if 0 < q < p 6 2, then Lp (0, 1) is isometrically isomorphic to a subspace of Lq (0, 1) (see [52, Theorem II.3.4]).

17

2

Lebesgue spaces: Interpolation and maximal functions

Thorin’s proof of two variants of the Riesz/Thorin theorem is in Section 2.1; as an example, we prove the Hausdorff/Young theorem. The simplest version of Marcinkiewicz’s interpolation theorem is proved in Section 2.2; as an example, we prove Paley’s theorem on Fourier coefficients (Theorem 2.2.5). Section 2.3 contains the Hardy/Littlewood maximal theorem with application to Lebesgue points. In Section 2.4 we prove Khintchine’s inequality, which says that the subspace of Lp (0, 1), 0 < p < ∞, spanned by the Rademacher functions is isomorphic with ℓ2 . The rest of this chapter is devoted to the proof of Nikishin’s theorem. This theorem states, in particular, that if a bounded linear operator T maps Lp (T) into Lq (T), where 0 < q < p 6 2, then actually T maps Lp (T) into the weak Lebesgue space Lp,∞ (T r A), where T r A is of arbitrarily small measure (see Theorem 2.5.2). If in addition T “commutes with rotations”, then we can take A = ∅. Also, we prove the so called Banach’s principle and the theorem on a.e. convergence (Theorems 2.7.1 and 2.7.2).

2.1

The Riesz/Thorin theorem

The proof of various variants of the Riesz/Thorin (convexity) theorem can be found, e.g., in the books [7, Ch. IV §2] and [8]. The case of bilinear forms Let γ = (γ1 , . . . , γm ) and δ = (δ1 , . . . , δn ) be sequences of positive real numbers. For a ∈ Cm , b ∈ Cn and p, q > 0 let kakγ,p =

X m j=1

p

|aj | γj

1/p

,

kbkδ,q =

X n

k=1

p

|bk | δk

2.1.1 Theorem Let 0 < p0 , q0 , p1 , q1 6 ∞. Let B(a, b) =

m,n X

Bjk aj bk ,

j,k=1

18

a ∈ Cm , b ∈ Cn ,

1/q

.

2.1 The Riesz/Thorin theorem

19

where Bjk ∈ C, and suppose that |B(a, b)| 6 M0 kakγ,p0 kbkδ,q0 ,

|B(a, b)| 6 M1 kakγ,p1 kbkδ,q1

for all a ∈ Cm , b ∈ Cn . If 1 η 1−η + , = p p0 p1

1 η 1−η + = q q0 q1

and 0 < η < 1,

then |B(a, b)| 6 M01−η M1η kakγ,p kbkδ,q . In other words, if M (α, β) = sup{ |B(α, β)| : kakγ,1/α 6 1, kbkδ,1/β 6 1}, then the function log M (α, β) is convex in the quadrant α > 0, β > 0. Proof. (Throughout the proof we omit the indices γ, δ.) Let p, q < ∞, and kakp = 1 and kbkq = 1.

(2.1)

Define a(z) ∈ Cm and b(z) ∈ Cn by a(z)j = |aj |p/p(z) ei arg aj where

1−z z 1 = + p(z) p0 p1

and b(z)k = |bk |q/q(z) ei arg bk , and

1 1−z z = + . q(z) q0 q1

The function F (z) := M0z−1 M1−z B(a(z), b(z)) is entire, as a sum of exponential functions. We consider the restriction of F to the strip Π = {z : 0 6 Re z 6 1}. We have, for t ∈ R, |a(it)j | = |aj |p/p0 ,

|a(1 + it)j | = |aj |p/p1 ,

|b(it)k | = |bk |q/q0 ,

|b(1 + it)k | = |bk |q/q1 ,

whence, in view of (2.1), ka(it)kp0 = 1, ka(1 + it)kp1 = 1,

kb(it)kq0 = 1, kb(1 + it)kq1 = 1.

It follows that |F (it)| = M0−1 |B(a(it), b(it))| 6 1 and similarly |F (1 + it)| 6 1 for every t ∈ R. Thus the function F is analytic and bounded on Π and |F | 6 1 on the boundary of Π. It follows that |F | 6 1 on Π and in particular |B(a, b)| 6 M01−η M1η , which completes the proof in the case p, q < ∞. The remaining case is similar. ✷

20

2 Interpolation and maximal functions

The case of linear operators 2.1.2 Theorem Let (R, µ) and (S, ν) be sigma-finite measure spaces and let 1 6 p0 , q0 , p1 , q1 6 ∞. Let T be a (complex-)linear operator defined on µ-simple functions on R and taking values in the set of all complex ν-measurable functions, and let kT f kq0 6 M0 kf kp0 and kT f kq1 6 M1 kf kp1 for all µ-simple functions f on R. If η 1 η 1−η 1−η 1 + , + and 0 < η < 1, = = p p0 p1 q q0 q1 then kT f kq 6 M01−η M1η kf kp for all µ-simple functions f . Remark. Concerning the validity of this theorem in the entire first quadrant, that is, for 0 < pk , qk 6 ∞, see [7, page 281]. R Proof of Theorem. We consider the bilinear form B(f, g) = S (T f )g dν, where f, g are simple functions on R, S, respectively, and use the formula kT f kq = sup{ |B(f, g)| : kgkq′ = 1, g simple}, Pm Pn where 1/q ′ + 1/q = 1. Let f = j=1 aj Kj , g = k=1 bk Hk , where Kj and Hk are sequences of “pairwise disjoint” characteristic functions. Then straightforward calculation shows that we can apply Theorem 2.1.1 with the indices p0 , q0′ , p1 , q1′ , Z Z Z Hk dν. Kj dµ, δk = Bj,k = (T Kj )Hk dν, γj = S

R

S

The details are left to the reader. ✷ The following form of the preceding theorem is perhaps more convenient in application. 2.1.3 Theorem Let (R, µ) and (S, ν) be sigma-finite measure spaces and let 1 6 p0 , q0 , p1 , q1 6 ∞. Let T be a linear operator defined on the complex space Lp0 (R, µ)+Lp1 (R, µ) and taking values in the set of all complex ν-measurable functions, and let kT f kq0 6 M0 kf kp0 , kT f kq1 6 M1 kf kp1 for all f ∈ Lp0 , f ∈ Lp1 , respectively. If η 1−η 1 + , = p p0 p1

1 η 1−η + = q q0 q1

and 0 < η < 1,

then T is a bounded operator from Lp (R, µ) into Lq (S, ν) and kT f kq 6 M01−η M1η kf kp . Proof. We shall consider the case where the measure µ is finite and p0 < p1 6 ′ ∞. Then Lp (µ) ⊂ Lp0 (µ). With the above notation, let g ∈ Lq (ν) be a simple p function and let f ∈ L (µ) be arbitrary. Choose a sequence fn of simple functions on R such that kfn − f kp → 0. Then kfn − f kp0 → 0 and therefore Z T (fn − f )g dν → 0, S

2.1 The Riesz/Thorin theorem

21

′

because T : Lp0 7→ Lq0 is continuous on Lp0 (µ) and g ∈ Lq0 . Hence Z 1−η η T (f )g dν 6 M0 M1 kf kp kgkq′ . S

The result follows.

✷

The case of real-linear operators For the validity of the conclusion kT f kq 6 M01−η M1η kf kp in the Riesz/Thorin theorem, it is essential that the operator T be complex-linear, i.e., that T (λf ) = λT f for all complex scalars λ (see [7, Ch. 4, Example 1.3]). For real spaces, the conclusion of the theorem remains valid if pk 6 qk (k = 0, 1). Otherwise, we have kT f kq 6 2M01−η M1η kf kp . However, if T is a positive linear operator, then Theorem 2.1.3 remains valid for any pk , qk > 1. The Hausdorff/Young theorem Let T denote the unit circle of the complex plane. For a function f ∈ L1 (T), let fb(n) be the Fourier coefficients of f , 1 fb(n) = 2π

Z

π

f (eiθ )e−inθ dθ.

−π

2.1.4 Theorem If f ∈ Lp (T), 1 6 p 6 2,  X ∞

n=−∞

′ |fb(n)|p

1/p′

6 kf kp ,

(2.2)

where 1/p + 1/p′ = 1. In other words: If f ∈ Lp (T) and {bn } is a two-sided ℓp P b sequence, 1 6 p 6 2, then the series bn f (n) is absolutely convergent and there holds the inequality ∞ X b (2.3) bn f (n) 6 kf kp k{bn }kp . n=−∞

Proof. The theorem is true for p = 1 because |fb(n)| 6 kf k1 , and is true for p = 2 because of Parseval’s formula. Then the result is obtained by the Riesz/Thorin theorem. ✷ ′

p 2.1.5 Exercise If p > 2 and {bn }∞ −∞ ∈ ℓ , then there exists a unique function
′ P∞ p′ 1/p g ∈ Lp (T) such that gb(n) = bn for all n and kgkp 6 . n=−∞ |bn |

22

2 Interpolation and maximal functions

2.2

Weak Lp -spaces and Marcinkiewicz’s theorem

The space Lp,∞ Let Ω be a measure space with a (positive) sigma-finite measure µ. The weak Lp space Lp,∞ (µ), 0 < p < ∞, consists of those measurable functions f on Ω for which
1/p kf kp,∞ := sup λ · µ(f, λ) < ∞, 0 λ} . Chebyshev’s inequality, Z 1 µ(g, λ) 6 |g| dµ, λ Ω

shows that Lp ⊂ Lp,∞ , while the formula Z ∞ Z µ(g, λ) d(λq ) |g|q dµ = Ω

(2.4)

0

(proved by means of Fubini’s theorem) implies Lp,∞ ⊂ Lq for q < p, if µ is finite. The quantity k · kp,∞ is a norm for no p, but we have kf + gkp,∞ 6 Cp (kf kp,∞ + kgkp,∞ )

(Cp = 2max(1/p,1) ),

and hence k · kp,∞ is a (complete) quasinorm. It is interesting, however, that if p = 1, then the space need not be locally convex (if, for example, Ω = [0, 1] with Lebesgue measure), although it can be q-renormed for every q < 1. For p > 1 the space is locally convex, and for p < 1 it is p-convex, i.e., there is an equivalent p-norm on it.(∗) 2.2.1 Exercise There hold the inequalities µ(f1 + f2 , λ1 + λ2 ) 6 µ(f1 , λ1 ) + µ(f2 , λ2 ), µ(f1 f2 , λ1 λ2 ) 6 µ(f1 , λ1 ) + µ(f2 , λ2 ). Marcinkiewicz’s theorem Quasilinear operators Let T be an operator acting from a vector space X to the set of all nonnegative measurable functions defined on a measure space (Ω, µ). Then T is called a quasilinear operator if there exists a constant K such that T (f + g) 6 K(T f + T g)

(f, g ∈ X).

If K = 1, then T is said to be subadditive. If an operator S with values in the set of finite measurable functions on Ω is linear, then the operator T f = |Sf | is subadditive. (∗) For

further information see Kalton [36]. See also [7] for the general theory of weak Lp spaces.

2.2 Weak Lp -spaces and Marcinkiewicz’s theorem

23

2.2.2 Theorem Let µ and σ be sigma-finite measures on Ω and S, respectively, let 0 < p < q 6 ∞ and let T be a quasilinear operator from Lp (σ) + Lq (σ) to the set of all nonnegative µ-measurable functions. Assume there exist constants C1 and C2 , independent of f , such that kT f kp,∞ 6 C1 kf kp , kT f kq,∞ 6 C2 kf kq .

(2.5) (2.6)

Then for every s ∈ (p, q) there exists a constant C independent of f such that kT f ks 6 Ckf ks .

(2.7)

In the case q = ∞ inequality (2.6) should be interpreted as kT f k∞ 6 C2 kf k∞ . Weak type and strong type If T satisfies (2.5), i.e., if T maps Lp into Lp,∞ and is continuous at zero, then we say that T is of weak type (p, p); if (2.7) holds, then T is of strong type (s, s). Proof of Theorem 2.2.2 We consider the case where K = C1 = C2 = 1 and q < ∞, leaving the remaining cases to the reader. We have to deduce the inequality Z Z s |T f | dµ 6 C |f |s dσ Ω

S

from two “weak” inequalities: 1 λp

Z

|f |p dσ,

(2.8)

1 µ(T f, λ) 6 q λ

Z

|f |q dσ.

(2.9)

µ(T f, λ) 6

S

S

To show this we represent the function f in the form f = gλ + hλ , where ( f (ζ), if |f (ζ)| > λ, gλ (ζ) = 0, if |f (ζ)| < λ. Since T f 6 T (gλ ) + T (hλ ), we have µ(T f, λ) 6 G(λ) + H(λ), where G(λ) = µ(T gλ , λ/2)

and H(λ) = µ(T hλ , λ/2).

It follows from (2.8) and (2.9) that Z Z G(λ) 6 (2/λ)p |gλ |p dσ = (2/λ)p S

|f |>λ

|f |p dσ

(2.10)

24

2 Interpolation and maximal functions

and H(λ) 6 (2/λ)q

Z

S

Now we use the formula Z Z s |T f | dµ = s

|hλ |q dσ = (2/λ)q

∞

µ(T f, λ)λ

s−1

dλ 6 s

∞

0

0

Ω

Z

s−1

Z

|f |6λ

|f |q dσ.


G(λ) + H(λ) λs−1 dλ.

Multiplying inequality (2.10) by sλ and then integrating over λ ∈ (0, ∞) we get Z ∞ Z ∞ Z  s−1 p s−p−1 G(λ)λ dλ 6 s2 λ s |f |p dσ dλ 0

0

= s2p

Z Z S

s2p = s−p

Z

S

|f |>λ

|f |

0

 λs−p−1 dλ |f |p dσ

|f |s dσ.

The analogous inequality for H(λ) is proved in a similar way.

✷

Marcinkiewicz’s theorem for L log+ L 2.2.3 Theorem Let µ and σ be finite measures on Ω and S, respectively, let 1 < q 6 ∞ and let T be a quasilinear operator from L1 (σ) to the set of all nonnegative µ-measurable functions. If T satisfies (2.5)(p = 1) and (2.6), then Z Z T f dµ 6 K1 + K2 |f | log+ |f | dσ, (2.11) Ω

S

where K1 and K2 are independent of f . The class of those σ-measurable functions f on S for which the integral on the right hand side of (2.11) is finite is denoted by L log+ L(S). The proof is similar to that of Theorem 2.2.2. Other variants of Marcinkiewicz’s theorem can be found in Zygmund [100, Ch. XII§4]; the proof of some of them is much more difficult. Paley’s theorem The implication f ∈ Lp (T) =⇒

∞ X

n=−∞

|fb(n)|p < ∞ ′

(1 < p < 2),

which is a weak form of the Hausdorff/Young theorem, was improved by Hardy and Littlewood; namely: P∞ 2.2.4 Theorem If f ∈ Lp (T), 1 < p < 2, then n=0 (n + 1)p−2 (c∗n )p < ∞, where {c∗n } is the decreasing rearrangement of the sequence {fb(n)}.

2.3 Maximal function and Lebesgue points

25

An application of Marcinkiewicz’s theorem yields a more general result, due to Paley: 2.2.5 Theorem Let (Ω, µ) be a finite measure space and let {ϕn }∞ 1 be an orthonormal sequence in L2 (Ω, µ) such that supn kϕn k∞ < ∞. Then ∞ X

n=1

np−2 |an |p 6 Ckf kpp ,

where 1 < p < 2 and an =

R

Ω

f ∈ Lp (Ω, µ),

f ϕn dµ.

Proof. Let µ(Ω) = 1 and supn kϕn k∞ = K. Define the measure σ on N, the set of positive integers, by σ({n}) = n−2 . Define the operator T : L1 (Ω, µ) 7→ L0 (N, σ) by (T f )(n) = nan . Bessel’s inequality implies that T is of strong type (2, 2). To prove that T is of weak type (1, 1) and therefore to conclude the proof (by Marcinkiewicz’s theorem), observe that |an | 6 Kkf k1 . Hence, if kf k1 = 1, we have σ{n : |T f (n)| > λ} 6 σ{n : Kn > λ} 6 which concludes the proof.

X

n−2 6 CK min(1, 1/λ),

n>λ/K

✷

2.2.6 Exercise Let (Ω, µ) be a sigma-finite measure space and let {ϕn }∞ 1 be an orthonormal sequence in L2 (Ω, µ) such that kϕn k 6 M nγ , where M and γ are positive constants. Then for 1 < p < 2 there holds the inequality ∞ X

n=1

2.3

n(γ+1)(p−2) |an |p 6 Ckf kpp .

Maximal function and Lebesgue points

The maximal function of a 2π-periodic function φ ∈ L1 (−π, π) is the (2π-periodic) function Mφ defined as 1 (Mφ)(θ) = sup 2h 0 1, then Mφ ∈ Lp (−π, π) and kMφkp 6 Cp kφkp , where Cp depends only of p. By | · · · | we denote the Lebesgue measure on the line. Proof. Assertion (b) is obtained from (a) by Marcinkiewicz’s theorem. To prove (a), let φ ∈ L1 (−π, π), let E = {θ ∈ (−π, π) : Mφ(θ) > 1} and let K be a compact subset of E. It suffices to find an absolute constant C such that |K| 6 Ckφk1 . By the definition of Mφ and the compactness of E,R there are S intervals Ii (i = 1, . . . , n) such that Ii ⊂ (−2π, 2π), K ⊂ Ii and |Ii | 6 Ii |φ(t)| dt. Assume that the sequence |Ii | is decreasing. Let J1 = I1 . Let J2 = Ik , where k is the smallest i for which Ii ∩ J1 = ∅. Then let J3 = Im , where m is the smallest i > k such that Ii ∩ (J1 ∪ J2 ) = ∅. Continuing in this S waySwe find a sequence Jj ⊂ (−2π, 2π) of pairwise disjoint intervals such that Ii ⊂ Jj∗ , where, for each j, Jj∗ is the interval “concentric” with Jj and |Jj∗ | = 3|Jj |. It follows that (1/3)|K| 6

X j

|Jj | 6

XZ j

which gives the desired inequality with C = 6.

Jj

|φ(t)| dt, ✷

Lebesgue points The maximal theorem has many important applications. It is useful, for example, in proving almost everywhere convergence. Usually, we can easily prove a.e. convergence for a dense set of functions, and then use the maximal theorem to interchange the limits. Here we consider the existence of Lebesgue points. The Lebesgue point of a function φ is a point x ∈ R such that 1 h→0 2h lim

Z

h

−h

φ(t + x) − φ(x) dt = 0.

The set of all Lebesgue points of f is called the Lebesgue set of f . 2.3.2 Theorem If a 2π-periodic function φ is integrable on (−π, π), then almost every point in R is a Lebesgue point for φ. 2.3.3 Corollary The inequality |φ(θ)| 6 (Mφ)(θ) holds almost everywhere. Proof of Theorem. The operator T φ(x) = lim sup h→0

1 2h

Z

h

−h

φ(t + x) − φ(x) dt

2.3 Maximal function and Lebesgue points

27

satisfies: (a) T (φ1 + φ2 ) 6 T φ1 + T φ2 ; (b) T φ 6 |φ| + M φ ; (c) T g = 0 if g is continuous. Let φ ∈ L1 (−π, π), λ > 0 and ε > 0. Choose a continuous function g so that kφ − gk1 < ε. From (a) we get T φ 6 T g + T (φ − g) = T (φ − g) and then, from (b), by Theorem 2.3.1 and Chebyshev’s inequality, we get {θ : T (φ − g)(θ) > λ} 6 {θ : |φ − g|(θ) > λ/2} + {θ : M(φ − g)(θ) > λ/2} 4π 2C 2(2π + C)ε kφ − gk1 + kφ − gk1 6 . λ λ λ Thus {θ : T (φ − g)(θ) > λ} = 0, for every λ > 0, because ε is arbitrary. 6

✷

Non-periodic case

Let φ be a locally integrable function defined on Rn , n > 1. The maximal function M φ is defined on Rn by Z 1 φ(ξ) dVn (ξ), (M φ)(z) = sup n (2.14) r>0 r |ξ−z| 1 is a constant. We shall prove that kφn kp 6 κp on the length of {ak }n0 . In the case n = 1 we have kφn kp =

(2.17)

kφn k2 by induction

 |a + a |p + |a − a |p 1/p 0 1 0 1 6 κ1/2 p kφn k2 , 2

because of (2.17). Let n > 2. Then, as is easily verified, Z

0

1

X p Z n
1 1 a r (t) dt = |a0 + ψ(t)|p + |a0 − ψ(t)|p dt, j j 2 0 j=0

(2.18)

30

2 Interpolation and maximal functions

where ψ(t) =

n P

ak rk−1 (t) =

n−1 P

ak+1 rk (t). From this and (2.17) it follows that

k=0

k=1

Z

0

1

X p Z n aj rj (t) dt 6

0

j=0

1

|ψ(t)|2 + κp |a0 |2


p/2

dt.


p/2 , which can be seen by using the binomial The last integral is 6 kψk2p +κp |a0 |2 formula, or by using Jensen’s inequality for the concave function x 7→ (x2/p + 1)p/2 , 1/2 x > 0. On the other hand, by induction hypothesis we have kψkp 6 κp kψk2 , which implies p Z 1 X n
p/2 aj rj (t) dt 6 κp kψk22 + κp |a0 |2 . 0

j=0

This completes the proof because κp kψk22


2 1/2

+ κp |a0 |

=

κ1/2 p

X n j=0

2

|aj |

1/2

.

✷

Miscellaneous 2.4.2 What is essential in the above proof is that rn is defined by rn (t) = r(2n−1 t), n > 1, where r ∈ L∞ (R) is a 1-periodic function such that r(t + 1/2) ≡ −r(t). As an example we can take r(t) = ei 2πt to get Paley’s inequality; this inequality will be discussed in a different context (see Theorems 11.1.1 and 11.1.3, page 170). 2.4.3 The Lp -closure of the linear span of the Rademacher functions is complemented in Lp , for 1 < p < ∞. P 2.4.4PLet {fj }j>1 be a finite sequence in a vector space X, let Ft = j>1 rj (t)fj , Ft = j>1 εj rj (t)fj , where εj = ±1. If X0 is a subset of X, then {t ∈ [0, 1] : Gt ∈ X0 } = {t ∈ [0, 1] : Ft ∈ X0 } .

2.5

Nikishin’s theorem

While Marcinkiewicz’s theorem enables us to deduce a strong inequality from two weak inequalities, by using Nikishin’s theorem we can obtain a weak inequality from a “very weak” one. Before stating this theorem we introduce some terminology. The space L0 and sublinear operators Let (Ω, µ) be a finite measure space. The vector space L0 (µ) = L0 (Ω, µ) of all finite measurable functions on Ω becomes an F -space when endowed with the F -norm Z |f | N0 (f ) = dµ. 1 + |f | Ω

2.5 Nikishin’s theorem

31

Convergence in L0 (µ) is equivalent to convergence in measure, i.e., fn → 0 iff limn µ(fn , λ) = 0 for every λ > 0. This can be deduced from the formula N0 (g) =

Z

0

∞

µ(g, λ) dλ, (1 + λ)2

(2.19)

which can easily be deduced from (2.4)(q = 1) by taking g = |f |/(1 + |f |). An operator T that maps a quasi-normed space X to the set of nonnegative measurable functions is said to be sublinear if (almost everywhere): (a) T (f + g) 6 T f + T g for f, g ∈ X; (b) T (λf ) = |λ| T f for f ∈ X and λ ∈ C. If T f is finite a.e. for all f , then we can treat it as an operator from X to L0 . Since (a) and (b) imply |T f − T g| 6 T (f − g), we see that continuity of T at the origin implies continuity of T on all of X. Spaces of type p A quasi-Banach space X is of type p (0 < p 6 2) if there exists a constant K such that

p Z 1 X X

rj (t)fj kfj kp (2.20)

dt 6 K 0

j>0

j

for every finite sequence {fj } ⊂ X, where rj are the Rademacher functions; recall that rj (t) = sign sin(2j tπ). Every p-Banach space (0 < p 6 1) satisfies (2.20) with K = 1. And by application of Khintchine’s inequality (2.16), one proves the following: 2.5.1 Theorem Lp has type min(p, 2) (p > 0). From Khintchine’s inequality also follows that the notion of type p has no sense for p > 2 because if X = C, then the integral in (2.20) is “proportional” to the ℓ2 -norm of the sequence {fj }. Nikishin’s theorem In order to state Nikishin’s theorem let TB f = χB T f , where χB is the characteristic function of B ⊂ Ω. 2.5.2 Theorem (Nikishin [66]) Let X be a quasi-Banach space of type p, 0 < p 6 2, and let T : X 7→ L0 (Ω, µ) be sublinear and continuous. Then for every ε > 0 there exists a measurable set B ⊂ Ω such that µ(Ω r B) < ε and TB maps X to Lp,∞ and is continuous. Lemmas For the proof we need two lemmas. The first lemma provides a characterization of continuity of T by means of “weak” inequalities.

32

2 Interpolation and maximal functions

2.5.3 Lemma Let T : (X, k · k) 7→ L0 (µ) be a sublinear operator on a quasiBanach space X. Then T is continuous iff there exists a decreasing function c(λ), λ > 0, such that limλ→∞ c(λ) = 0 and µ(T f, λkf k) 6 c(λ)

(λ > 0, f ∈ X).

(2.21)

Therefore, Nikishin’s theorem says that under some conditions inequality (2.21) can be improved so that c(λ) = C/λp outside a set of arbitrarily small measure. Proof. Let T be continuous. Then there exists a decreasing function ε(λ) defined for λ > 0 such that ε(λ) → 0 (λ → ∞), and that there holds the implication kf k = 1/λ =⇒ N0 (T f ) < ε(λ). On the other hand, (2.19) implies µ(T f, 1) 6 2N0 (T f ) because the function µ(T f, λ) is decreasing. Thus if kf k = 1/λ, then µ(T f, λkf k) 6 2ε(λ) =: c(λ), and hence µ(T f, λkf k) = µ(T f /λkf k, 1) 6 c(λ) for every λ > 0. Conversely, if (2.21) is satisfied, let Z ∞ c(λ/δ) ǫ(δ) = dλ. (1 + λ)2 0 From (2.19) it follows that there holds the implication kf k < δ =⇒ N0 (T f ) < ǫ(δ), which implies that T is continuous, because ǫ(δ) → 0 (δ → 0).

✷

2.5.4 Lemma Let the hypotheses of Theorem 2.5.2 be satisfied. Then its conclusions hold if (and only if) there exists a decreasing function c(λ), λ > 0, such that limλ→∞ c(λ) = 0 and µ(sup T fj (ω), λ) 6 c(λ) (2.22) j

for all sequences {fj } ⊂ X such that

P

j

kfj kp 6 1.

This lemma can also be interpreted in the following way: The conclusions of Theorem 2.5.2 hold iff the operator Sf = supj T fj maps the space ℓp (X) into L0 (µ) and is continuous. The space ℓp (X) consists of those sequences {fn }∞ 1 ⊂ X for which 1/p X ∞ p < ∞. k{fn }k = kfn k n=1

Proof. We will consider “if” part.(†) Let ε > 0 and choose R > 0 so that c(R) < ε, where c(λ) is the function from (2.22). We say that a measurable set F (†) Concerning the opposite direction, which is not needed in our text, one could see Wojtaszczyk [98, Ch. III.H]; the proof of “if” part is essentially the same as in that book.

2.5 Nikishin’s theorem

33

satisfies condition (W ) if there exists f ∈ X, kf k 6 1, such that µ(F ) T f (ω)p > Rp

(ω ∈ F ).

(W )

If the collection, say F, of such sets is empty, then for fixed λ > 0 and f ∈ X, kf k 6 1, we take F = {ω ∈ Ω : T f (ω) > λ} to get µ(F ) T f (ω)p 6 Rp (ω ∈ F ), whence µ(F )λp 6 Rp . i.e., µ{ω ∈ Ω : T f (ω) > λ} 6 (R/λ)p , and this means that T maps X into Lp,∞ and is continuous. Therefore we can suppose that F is nonempty. Let (Fj ) be a maximal collection of pairwise disjoint measurable sets satisfyingP (W ) and let fj ∈ X be the corresponding vectors. SPutting cj = µ(Fj )1/p p Fj ). From (2.22) it we have j kcj fj k 6 1 and supj T (cj fj )(ω) > R (ω ∈ follows that µ(∪Fj ) 6Sc(R) < ε. Since the collection is maximal, there holds the inequality µ ω ∈ Ω r Fj : T f (ω) > λ 6 (R/λ)p , for every λ > 0. This means that the operator TB , where B = Ω r ∪Fj , maps X into Lp,∞ , which concludes the proof. ✷ Proof of Theorem 2.5.2. First observe that Lemma 2.5.4 remains true if we assume that the sequence fj is finite. Let fj ∈ X be such a sequence, and let X kfj kp 6 1. (2.23) j>1

According to Lemma 2.5.4, it is enough to find a decreasing function c1 (λ) defined for λ > 0 such that limλ→∞ c1 (λ) = 0 and µ(Eλ ) 6 c1 (λ)

(λ > 0), (2.24) P where Eλ = {ω ∈ Ω : maxj T fj > λ}. Let gt = j>1 rj (t)fj (0 6 t 6 1), where rj are the Rademacher functions. Then 2rk (t)fk = gt + gtk , where gtk = −r1 (t)f1 − · · · − rk−1 (t)fk−1 + rk (t)fk − . . . .

Therefore there holds the inequality 2T (fk ) 6 T (gt ) + T (gtk ) almost everywhere on [0, 1]. Since { t ∈ [0, 1] : T gt (ω) > η} = { t ∈ [0, 1] : T gtk (ω) > η} { t ∈ [0, 1] : T gt (ω) > T fk (ω) } > 1/2 for for every η > 0 (see 2.4.4), we see that all ω ∈ Ω and k > 1. Thus we have {t ∈ [0, 1] : T gt (ω) > λ} > 1/2, ω ∈ Eλ , so it follows that Z {t ∈ [0, 1] : T gt (ω) > λ} dµ(ω). µ(Eλ )/2 6 Eλ

This inequality is sufficient the proof in the case when X is p-Banach, R to conclude R p < 1. Namely, replacing Eλ by Ω and applying the formula Z Z Z {t ∈ [0, 1] : T gt (ω) > λ} dµ(ω) = dt dµ(ω) Ω

=

Z

0

1

dt

Z

T gt (ω)>λ

dµ(ω) =

Z

0

Ω

T gt (ω)>λ

1

µ{ω ∈ Ω : T gt (ω) > λ} dt,

34

2 Interpolation and maximal functions

we get µ(Eλ )/2 6

Z

0

1

µ{ω ∈ Ω : T gt (ω) > λ} dt.

Now Lemma 2.5.3 and the inequality kgt k 6 1, which holds for every t, give µ(Eλ ) 6 2c(λ/2), where c(λ) is the function from Lemma 2.5.3. So we have proved (2.24) in this special case. Let X be a space of type p. Then the inequality kgt k 6 1 can be false for some t, but from (2.23) and (2.20) it follows that {t : kgt k > λ} 6 K/λp . (2.25) In order to exploit this fact, we start from the inequality √ √ {t : T gt (ω) > λ} 6 {t : kgt k > λ} + {t : T gt > kgt k λ} .

Hence, by integration over ω ∈ Ω and using (2.25) and Fubini’s theorem as above, we get Z {t : T gt (ω) > λ} dµ(ω) µ(Eλ )/2 6 Ω

6 Kλ−p/2 µ(Ω) +

Z

0

1

√  µ ω : T gt (ω) > kgt k λ dt.

√ The last integral is 6 c( λ/2), because of Lemma 2.5.3, so we have (2.24) again. ✷

2.6

Nikishin and Stein’s theorem

This theorem tells us that under additional, algebraic, conditions we can take B = Ω in Theorem 2.5.2. We restrict ourselves to the case of the multiplicative group T. 2.6.1 Theorem (Nikishin/Stein) Let X be a space of type p ∈ (0, 2], and let ζ 7→ fζ be a mapping of the unit circle T to the set of all isometric endomorphisms of X. If T : X 7→ L0 (T) is sublinear, continuous and “commutes with rotations”, i.e., for every ζ ∈ T satisfies the condition (T fζ )(ω) = (T f )(ωζ), ω ∈ T a.e., then T (X) ⊂ Lp,∞ (T) and T is continuous as an operator from X to Lp,∞ (T). In the case where X is Lp (T) or Lp (D), we put fζ (z) = f (ζz). Proof. Let dµ(ζ) = |dζ|/2π. Theorem 2.5.2 guarantees that there is a set B so
p that µ(B) > 0 and µ Aλ ∩ B 6 C (kf k/λ) , where Aλ = {ω ∈ T : |(T f )(ω)| > λ}, and C is independent of f and λ. If we put fζ (ζ
∈ T) instead of f and apply the p hypotheses of the theorem, we get µ (ζ −1 Aλ ) ∩ B 6 C (kf k/λ) . Integrating this with respect to dµ(ζ) and using the formula Z
µ (ζ −1 Aλ ) ∩ B dµ(ζ) = µ(Aλ )µ(B), (2.26) T

2.6 Nikishin and Stein’s theorem

35

 p C kf k we get µ(Aλ ) 6 , which was to be proved. µ(B) λ To verify formula (2.26), we write the left-hand side as Z Z Z
χζ −1 A (ω) dµ(ω) dµ(ζ) µ ζ −1 A ∩ B dµ(ζ) = T

T

B

(χ is the characteristic function), and then apply Fubini’s theorem together with the relation χζ −1 A (ω) = χω−1 A (ζ). ✷

A theorem on multipliers As a nice application of the Nikishin/Stein theorem and the Marcinkiewicz theorem we have the following. 2.6.2 Theorem Let T : L1 (T) 7→ L0 (T) be a continuous linear operator that commutes with rotations. Then T is of weak type (1, 1) and of strong type (p, p) for every p ∈ (1, ∞), and there exists a bounded sequence mn (−∞ < n < ∞) such that ∞ X T f (eiθ ) = mn fb(n)einθ (2.27) n=−∞

for every trigonometric polynomial f . Further, if f ∈ Lp , p > 1, then

for every integer n.

Tcf (n) = mn fb(n)

(2.28)

Proof. Since Lp is of type p for 0 < p 6 2, Nikishin/Stein theorem tells us that the operator T with the above properties is of weak type (p, p) for p = 1 and p = 2. Hence, by Marcinkiewicz’s theorem, T is of strong type (p, p) for 1 < p < 2. To prove the rest, let g(w) = wn , w ∈ T, for a fixed integer n. By the hypothesis, for every ζ ∈ T we have ζ n (T g)(w) = (T g)(ζw) for a.e., w ∈ T. The function T g belongs to L1 because g ∈ L2 and T is of strong type (p, p) for p ∈ (1, 2). It follows that if φ ∈ L∞ , then Z Z (T g)(w)φ(w) |dw| = ζ −n (T g)(ζw)φ(w) |dw| T

T

for every ζ ∈ T. Integrating this with respect to ζ and using Fubini’s theorem, we get Z Z Z 1 (T g)(w)φ(w) |dw| = φ(w) |dw| ζ −n (T g)(ζw) |dζ| 2π T T T Z Z 1 n = φ(w)w |dw| ζ −n (T g)(ζ) |dζ|. 2π T T

36

2 Interpolation and maximal functions

Hence (T g)(w) = wn

Z

T

ζ −n (T g)(ζ) |dζ| =: mn wn

for a.e. w ∈ T;

this proves formula (2.27). The validity of (2.28) can then be deduced from the Weierstrass theorem that the trigonometric polynomials are dense in Lp . It remains to prove that T is of strong type (q, q) for q > 2. By Marcinkiewicz’s theorem (or by the Riesz/Thorin theorem), we can assume that q > 2. Let f, g be trigonometric polynomials. Then, in view of (2.27), we have Z π Z π 1 1 (T f )(eiθ )g(e−iθ ) dθ = f (eiθ )(T g)(e−iθ ) dθ. 2π −π 2π −π Using this and the fact that T is of strong type (p, p) for p = q/(q − 1), we conclude that kT f kq 6 Ckf kq , (2.29)

where C is independent of f . Now let f ∈ Lq be arbitrary, and let fn be a sequence of trigonometric polynomials such that kfn − f kq → 0. The validity of (2.29) for trigonometric polynomials implies kT fn kq 6 Cq kf kq .

(2.30)

Since kfn − f k1 6 kfn − f kq and T is continuous from L1 to L0 , we see that T fn → T f in measure; after extracting a subsequence we can assume that T fn → T f almost everywhere. Now Fatou’s lemma and (2.30) give kT f kq 6 Cq kf kq , which was to be proved. ✷

2.7

Banach’s principle

The following fact, known as Banach’s principle, plays an important role in applications of Theorem 2.6.1 to maximal operators. 2.7.1 Theorem Let X be a quasi-Banach space, let Tn (n > 1) be a sequence of continuous linear operators from X to L0 (Ω, µ), and let Tmax f (ω) := sup |Tn f (ω)| < ∞ n>1

for almost all ω ∈ Ω. Then the operator Tmax : X 7→ L0 (Ω, µ) is continuous. Proof.(‡) Let L0 (ℓ∞ ) denote the set of all functions F = (f1 , f2 , . . . ) : Ω 7→ ℓ∞ with measurable coordinates. The following two facts, the proof of which is left to the reader, imply the validity of the theorem. (a) With the F -norm Z kF (ω)k∞ dµ(ω), 1 + kF (ω)k∞ Ω (‡) See

[7, theorem IV.5.7], where a proof is given based on Baire’s theorem.

2.7 Banach’s principle

37

the set L0 (ℓ∞ ) is an F -space. (b) The operator T g = (T1 g, T2 g, . . . ) maps X into L0 (ℓ∞ ) and has closed graph. ✷ Theorem on a.e. convergence 2.7.2 Theorem Suppose that the conditions of Theorem 2.7.1 are satisfied. If the limit lim Tn f (ω) := T f (ω) a.e. n→∞

exists and is finite for every f from a dense subset of X, then it exists for every f ∈ X, and T is continuous as an operator from X to L0 . Proof. Let X0 denote the dense subset. Consider the following sublinear operator on X : Sf (ω) = lim sup |Tm f (ω) − Tn f (ω)| m,n→∞

(ω ∈ Ω).

By Banach’s principle, this operator is continuous because Sf 6 2Tmax f . By Lemma 2.5.3, we have µ(Sf, ε) 6 c(ε/kf k)

(ε > 0, f ∈ X),

(2.31)

where c(λ) → 0 as λ → ∞. On the other hand, since Sg = 0 for g ∈ X0 , we have S(f ) = S(f − g) for all f ∈ X, g ∈ X0 . From this and (2.31) it follows that µ(Sf, ε) 6 c(ε/kf − gk k)

(ε > 0),

where, for a fixed f ∈ X, we have chosen a sequence gk ∈ X0 so that kf − gk k → 0 (k → ∞). Thus µ(Sf, ε) = 0 for every ε > 0. The result follows. ✷

3

Poisson integral

P Harmonic functions occur in Fourier analysis naturally. If P cn einθ is the ∞ 1 iθ Fourier series of a function φ ∈ L (T), then the function f (re ) = −∞ cn r|n| einθ is harmonic in D. The function f = P [φ] is called the Poisson integral of φ. In this chapter we are mainly concerned with some classical results on radial limits of P [φ], or, what is the same, on the Abel summability of the Fourier series of φ, where φ belongs to Lp (T), 1 6 p 6 ∞, or C(T). However, there are much stronger results in this direction, and the reader should consult Zygmund’s book. For example, in many cases we can replace Abel’s method of summability by the method (C, α), α > 0. But, in contrast to the case of (C, α), Abel’s method leads us to considering not only the radial but also the non-tangential limits of f (z). At the end we prove the Littlewood/Paley inequality and a variant of Schwarz lemma for harmonic functions (Theorems 3.5.1 and 3.6.1)

3.1

Harmonic functions

A complex-valued function f , defined on a domain Ω ⊂ C, is said to be harmonic if it is of class C 2 and ∆f ≡ 0 in Ω, where ∆ denotes the Laplacian, ∆f = here 1 ∂f = ∂f (z) = ∂z 2

∂2f ∂2f ∂2f + = 4 ∂x2 ∂y 2 ∂z∂ z¯



∂f ∂f −i ∂x ∂y



(z = x + iy);

∂f ¯ (z) = 1 = ∂f ∂ z¯ 2

,



∂f ∂f +i ∂x ∂y



.

Thus f is harmonic iff the function ∂f /∂z is analytic. This implies the following. 3.1.1 Theorem A harmonic function f defined on a simply connected domain Ω can be represented in the form f (z) = h(z)+g(z), z ∈ Ω, where h and g are analytic and uniquely determined up to an additive constant; conversely, if f = h + g¯, where h and g are analytic, then f is harmonic. Using this theorem one can deduce various properties of harmonic functions from the corresponding properties of analytic functions. For instance, the composition of a harmonic function with an analytic function is harmonic. Uniqueness theorem As a further consequence of Theorem 3.1.1 we have: If f is harmonic in a simply connected domain Ω and f = 0 in an open subset of Ω, then f = 0 in Ω. 38

3.1 Harmonic functions

39

Series expansion If f is harmonic in D = {z : |z| < R}, then there are unique P∞ complex numbers fb(k), −∞ < k < ∞, such that f (reiθ ) = k=−∞ fb(k)r|k| eikθ , with the series converging uniformly and absolutely on every compact subset of D. Conversely, if such a series converges in D, then it converges uniformly on compact subsets of D and its sum is harmonic in D. Heine/Borel property The set of all functions harmonic in a domain Ω and endowed with the topology of uniform convergence on compact subsets of Ω is denoted by h(Ω). By H(Ω) we denote the subspace of h(Ω) consisting of analytic functions. Both h(Ω) and H(Ω) are complete and have the Heine/Borel property. The latter means: If a sequence fn ∈ h(Ω), resp. fn ∈ H(Ω), is uniformly bounded on compact subsets, then there is a subsequence tending uniformly on compact subsets to a harmonic, resp. analytic, function. Mean value property A characteristic property of harmonic functions is the mean value property on circles: Z π 1 f (a + Reiθ ) dθ, {z : |z − a| 6 R} ⊂ Ω. (3.1) f (a) = 2π −π Green’s formula If f is harmonic, then (3.1) can be deduced from, say, Green’s formula, a special case of which can be stated as follows: If F is a C 2 -function in D = {z : |z| < R}, then Z π Z 1 d 1 F (reiθ ) dθ = (∆F )(z) dm(z) (3.2) dr 2π −π 2πr |z| 0

and

(3.4)

n=1

1 2π

Z

π

P (r, t) dt = 1.

(3.5)

−π

The Poisson kernel has the reproducing property: 3.1.5 Theorem If a function f continuous on D is harmonic in D, then Z π 1 f (reiθ ) = P (r, θ − t)f (eit ) dt (reiθ ∈ D). 2π −π

(3.6)

3.1 Harmonic functions

41

Proof. The equality 1 f (0) = 2π

Z

T

f (ζ) |dζ|

(T = ∂D)

follows from the mean value property on the circles |z| = ρ < 1 and the continuity of f . Applying this equality to the harmonic function f ◦ ϕa , where ϕa (z) = we get

a−z , 1−a ¯z

|a| < 1, |z| 6 1,

Z

1 f (a) = 2π

T

f (ϕa (ζ)) |dζ|.

Now introduce the substitution ζ = ϕa (w), i.e., w = ϕa (ζ). Since 1 − |a|2 |dw|, |1 − a ¯w|2

|dζ| = we get f (a) = which is another form of (3.6).

1 2π

Z

T

P (¯ aw)f (w) |dw|,

✷

The Poisson integral of a function The Poisson integral of a function φ ∈ L1 (T) is the harmonic function P [φ] defined by Z π 1 iθ P [φ](re ) = P (r, θ − t)φ(eit ) dt (reiθ ∈ D). 2π −π The proof of Theorem 3.1.5 shows that Z 1 P [φ](z) = φ(ϕz (w)) |dw| 2π T

(z ∈ D).

(3.7)

The Poisson integral can be used to solve the Dirichlet problem for the disk: 3.1.6 Theorem If φ is a continuous function defined on T, then φ has a unique continuous extension to D that is harmonic in D; this extension equals P [φ]. An immediate consequence is the well known Weierstrass theorem on approximation by trigonometric polynomials: The set of all trigonometric polynomials is dense in each of the spaces C(T), Lp (T) (0 < p < ∞). Proof of Theorem 3.1.6. The uniqueness follows from Theorem 3.1.5. Let f = P [φ]. From (3.7) and the continuity of the function φ, by using, for instance, the dominated convergence theorem, we get limD∋ z→1 f (z) = φ(1), which proves that the extension is continuous at the point z = 1; and so on. ✷

42

3 Poisson integral

The above proof, although very short, has a disadvantage in that it is based on very special properties of the Poisson kernel. The standard proof depends on (3.5) and the following: lim sup P (r, t) = 0 for all δ ∈ (0, π) (see [86, 100, 22, 46]). r→1 δδ

(|t| > δ),

|P ′ (r, t)| · |γ(t)| dt = 0.

Now we apply the inequality |t P ′ (r, t)| 6 2P (r, t), |t| 6 π, to obtain 1 L 6 lim sup π r→1−

Z

γ(t) P (r, t) dt. t −δ δ

Finally, (3.14) follows from this and the hypothesis γ(t)/t → 0, t → 0.

✷

3.3.2 (symmetric derivative) In Proposition 3.3.1, the ordinary derivative γ(θ + t) − γ(θ − t) γ ′ (θ) can be replaced by the symmetric derivative lim . t→0 2t Fatou’s theorem Proposition 3.3.1 is of elementary character and only deeper results on differentiability of functions give deeper results on the existence of radial limits. Namely, by the Lebesgue theorem, the derivative γ ′ (θ) exists for almost all θ (see [86]), which along with Proposition 3.3.1 shows that the Poisson/Stieltjes integral has finite radial limits almost everywhere. Returning to the Poisson integral of a measure, we can state a more precise form of this result as follows.(∗) 3.3.3 Theorem (Fatou) If µ ∈ M (T), then P [µ] has radial limits almost everywhere. Besides lim− P [µ](reiθ ) = φ(eiθ ) r→1

for almost all θ, where φ(eiθ ) dθ is the absolutely continuous part of µ. (∗) Rudin considers the case of “M -harmonic” functions on the complex ball, [84, Ch. V]. Much more information on the classical case can be found in Garnett [22, Ch. I] and, of course, Zygmund [100, Ch. III].

3.3 Radial limits of the Poisson integral

47

Recall that the measure µ is uniquely represented as dµ(eiθ ) = φ(eiθ ) dθ + dµs (eiθ ), where φ is an integrable function and µs is a singular measure (the theorem of Lebesgue and Radon/Nikodym). As a special case of Theorem 3.3.3 we have: 3.3.4 Corollary Every bounded harmonic function on D has radial limits almost everywhere. On the other hand, this special case is sufficient to prove the qualitative part of Fatou’s theorem, i.e., to prove the existence of the limits for h1 -functions (see the proof of Corollary 3.3.9). Another special case: 3.3.5 Corollary A measure µ ∈ M (T) is singular iff lim− P [µ](reiθ ) = 0 almost r→1 everywhere. 3.3.6 Exercise If u is the Poisson integral of a singular measure, then Z 2π lim |u(reiθ )|p dθ = 0 for 0 < p < 1. r→1

0

Theorems of Kolmogorov and Carleson P∞ |n| inθ b Since P [φ](reiθ ) = n=−∞ φ(n)r e , Theorem 3.3.3 shows that the Fourier series of an integrable function φ is almost everywhere summable by the Abel/Poisson method with the sum equal to φ. In connection with this, let us mention that Kolmogorov proved the existence of an everywhere divergent Fourier series (cf. [44, 100]). Carleson [12] proved that the Fourier series of an L2 -function converges almost everywhere, which is generalized to the case p > 1 by Hunt [31]. Nontangential limits Slightly modifying the proof of Proposition 3.3.1 one can prove the “nontangential” variant of Fatou’s theorem: 3.3.7 Theorem If f ∈ h1 , then for almost all ζ ∈ T there exists the limit ∢ lim f (z) := z→ζ

lim

Uζ ∋z→ζ

f (z),

where Uζ is the convex envelope of the union {z : |z| < ρ} ∪ {ζ}, and ρ < 1 is fixed. Proof. The key property of the set Uζ is: reiθ ∈ Uζ =⇒ |θ − arg ζ| 6 const(1 − r).

(3.15) ′

In order to prove that ∢ limz→1 P S[γ](z) = 0. under the hypothesis γ (0) = 0, we start from (3.11); we get Z δ 1 L := ∢ lim sup |P S[γ](z)| 6 ∢ lim sup |P ′ (r, θ − t)| · |γ(t)| dt 2π iθ z→1 re →1 −δ

48

3 Poisson integral

(0 < δ < 1). For a given ε > 0, take δ so that |γ(t)/t| < ε for |t| < δ, whence Z

δ

−δ

|P ′ (r, θ − t)| · |γ(t)| dt 6 ε

Z

δ

−δ

|t P ′ (r, θ − t)| dt.

Finally, by using the property (3.15) we find |t P ′ (r, θ − t)| 6 const P (r, θ − t) (reiθ ∈ U1 ); thus 1 L 6 Cε 2π

Z

π

−π

P (r, θ − t) dt = Cε,

etc.

✷

Radial =⇒ nontangential The above theorem can also be deduced from the following. 3.3.8 Theorem Suppose f is bounded and analytic in D, and ζ ∈ T. Then the existence of the radial limit of f at ζ implies the existence of the nontangential limit at ζ. 3.3.9 Corollary If f ∈ H(D) and u = Re f ∈ h1 , then f has nontangential limits at almost every point ζ ∈ T. Proof of Corollary. We can assume that u is positive. Then the function 1/(1 + f ) is analytic and bounded in D. The result follows. ✷ Proof of Theorem. Let ζ = 1 and limr→1 f (r) = 0. If f fails to have the limit 0 within Uζ , then there exist ε > 0 and sequences tn → 0, 0 < tn < 1, and wn , |wn | < ρ0 < 1, such that |f (tn wn + 1 − tn )| > ε

for all n.

(3.16)

Consider the functions fn (w) = f (tn w + 1 − tn ), w ∈ D. The sequence fn is uniformly bounded in D and therefore there exists a subsequence, denote it by fn , that converges to a function g ∈ H(D) uniformly on |w| < ρ0 ; this means that fn (wn ) − g(wn ) → 0. By the hypothesis, we have f (tn r + 1 − tn ) → 0 for 0 < r < 1, which implies that g(r) = 0 for 0 < r < 1, whence g(w) = 0 for all w ∈ D. Thus ✷ fn (wn ) → 0, which contradicts (3.16). Lindel¨ of’s theorem Theorem 3.3.8 is a special case of Lindel¨of’s theorem: Let f ∈ H ∞ (D) and let γ : [0, 1) 7→ D be a continuous curve such that γ(t) → 1 as t → 1. If there exists limt→1 f (γ(t)) = L, then f has nontangential limit L at the point 1. For the proof see [84, Theorem 8.4.1].

3.4 The spaces hp and Lp (T)

3.4

49

The spaces hp and Lp (T)

The harmonic Hardy space hp (0 < p 6 ∞) is defined as hp = { f ∈ h(D) : kf kp = sup Mp (r, f ) < ∞ }, r 1, and f ∈ h(D), then Mp (r, f ) increases with r. Moreover, if p > 1, then Mp (r, f ) is strictly increasing unless f = const.

It should be noted, however, that if p < 1 and f is positive, then Mp (r, f ) is decreasing. Proof. The “increasing” property can be deduced from the subharmonicity of the function |f |p (see Theorem 4.2.1), or by application of Minkowski’s inequality in continuous form (“norm of integral 6 integral of norm”) to the formula(†) Z π 1 iθ f (λre ) = P (λ, t)fr (θ + t) dt (0 < λ < 1), 2π −π where fr (t) = f (reit ). To prove the second assertion, let p > 1 and suppose that Mp (r1 , f ) = Mp (r2 , f ) for some 0 6 r1 < r2 < 1. Let Z π 1 |f (zeit )|p dt = Mpp (|z|, f ), |z| < r2 . u(z) = 2π −π The function u is subharmonic in the disk |z| < r2 and attains its maximum at z = r1 . It follows that u=const, by the maximum principle (see Theorem 4.1.7), whence Z π 1 |f (reit )|p dt, 0 < r < ρ. |f (0)|p = 2π −π (†) The case p = ∞ is trivial, while in the case 1 6 p < ∞ we can apply Jensen’s inequality as well.

50

3 Poisson integral

Since

Z π 1 Z π p 1 it |f (0)| = f (re ) dt 6 |f (reit )|p dt 2π −π 2π −π p

(by Jensen’s inequality or by H¨ older’s inequality), we see that Z π 1 Z π p 1 f (reit ) dt = |f (reit )|p dt, r < r2 . 2π −π 2π −π

This implies that f (reit ) depends only on r, when r < r2 , and therefore f (z) = ✷ const for |z| < 1; see 3.1.2.

Radial limits Since kφk2 =

∞  X

n=−∞

2 b |φ(n)|

1/2

,

b where φ ∈ L2 , as well as fb(n) = φ(n) provided f = P [φ], the Poisson integral acts as an isometric isomorphism from L2 onto h2 . It is very important that this fact extends to the case 1 < p 6 ∞. On the other hand, as we have seen, the operator P : L1 7→ h1 is not onto. 3.4.2 Theorem The function f belongs to hp (1 < p 6 ∞) iff it is equal to the Poisson integral of some function φ ∈ Lp (T). And if f = P [φ], φ ∈ Lp (T), then kf kp = kφkp 1 lim− r→1 2π

Z

(1 6 p 6 ∞),

π

−π

|f (reiθ ) − φ(eiθ )|p dθ = 0

(1 6 p < ∞)

(3.17)

and lim f (reiθ ) = φ(eiθ )

r→1−

almost everywhere.

The proof is very similar to the proof of Theorem3.2.3 and will be omitted here. The Poisson kernel shows that an h1 -function need not be equal to the Poisson integral of the boundary function. However, (3.17) implies the following. 3.4.3 Corollary If f ∈ hp , p > 1, then f = P [f∗ ], where f∗ (eiθ ) = lim− f (reiθ ). r→1

The Poisson kernel also shows that boundedness of f∗ does not imply boundedness of f . However, if p > 1, we have the following. 3.4.4 Corollary Let a function f ∈ h(D) have radial limits f∗ (eiθ ) almost everywhere and f∗ ∈ L∞ (T). If f ∈ hp for some p > 1, then f ∈ h∞ and kf k∞ = kf∗ k∞ .

3.5 The Littlewood/Paley theorem

51

Exercises 3.4.5 [72] The Poisson kernel satisfies: Mpp (r, P ) = Mqq (r, P ) (q = 1 − p) and Mp (r, P ) ≍ (1 − r),  ≍ (1 − r) log

for 0 < p < 1/2;  2 e , for p = 1/2; 1−r ≍ (1 − r)1/p−1 , for p > 1/2.

3.4.6 The inclusion hp ⊂ h(D) (1 6 p 6 ∞) is compact, i.e., every closed ball of the space hp are compact in the topology of uniform convergence on compact subsets of D. 3.4.7 Let z ∈ D. The norm of the linear functional z 7→ f (z) on the space hp (1 6 p < ∞) is equal to Kp (|z|)(1 − |z|2 )−1/p , where 1/q  Z π 1 it 2q−2 (1/p + 1/q = 1). |1 − re | dt Kp (r) = 2π −π Observe that K2 (r) = (1 + r2 )1/2 .

3.5

The Littlewood/Paley theorem

Here we consider the simplest variant of the Littlewood/Paley theorem. 3.5.1 Theorem (a) If u ∈ hp , 2 6 p < ∞, then Z |∇u(z)|p (1 − |z|)p−1 dA(z) < ∞ K :=

(3.18)

D

and there holds the inequality Z |∇u(z)|p (1 − |z|)p−1 dA(z) 6 Cp kukpp . D

(b) If u is harmonic in D and satisfies condition (3.18) for some 1 < p < 2, then u ∈ hp and we have kukpp 6 Cp (K + |u(0)|p ). The case p > 2 There are many proofs of (a). For example, it is possible to apply the Riesz/Thorin interpolation theorem; namely, the operators (S1 u)(z) = (1 − |z|)

∂u ∂z

and

(S2 u)(z) = (1 − |z|)

∂u ∂ z¯

map hp into Lp (D, dA/(1 − |z|)) for p = ∞ and p = 2. An elementary but rather long proof, based on local estimates deduced from the Hardy/Stein identity, was found by Luecking [55]. We shall present a proof that is both elementary and short. We only need three simple lemmas on positive harmonic functions.

52

3 Poisson integral

3.5.2 Lemma If u is a positive harmonic function in D, then |∇u(z)| 6

2u(z) . 1 − |z|

Proof. Applying the inequality |∇u(0)| 6 2u(0) (Theorem 3.2.6) to the function w 7→ u(z + (1 − |z|)w), we get the result. ✷ 3.5.3 Lemma If u ∈ hp is a real-valued function and 1 < p < ∞, then there are nonnegative functions h1 and h2 from hp such that u = h1 − h2 ,

kukpp = kh1 kpp + kh2 kpp .

Proof. Let g1 (ζ) = max{u(ζ), 0} and g2 (ζ) = max{−u(ζ), 0} for ζ ∈ T. Then, because of Theorem 3.4.2, the required conditions are satisfied by the functions h1 = P [g1 ] and h2 = P [g2 ]. ✷ 3.5.4 Lemma Let u > 0 belong to hp , 1 < p < ∞. Then Z 1 p(p − 1) p p dA. up−2 |∇u(z)|2 log kukp = |u(0)| + 2 |z| D Proof. This is easily deduced from Green’s formula and the formula ∆(up ) = p(p − 1)up−2 |∇u|2 .

✷

Proof of Theorem 3.5.1(a). It is enough to consider real-valued functions. Because of Lemma 3.5.3, we can suppose that u is positive. Then Lemmas 3.5.4 and 3.5.2 give Z p2 − p kukpp > |∇u|2 22−p |∇u|p−2 (1 − |z|)p−1 dA, 2 D which implies the desired conclusion. The case 1 < p < 2 We shall apply the method of “dualization.” Let Xp be the (real) subspace of hp consisting of real-valued functions. Let Yp be the space of real-valued harmonic functions that satisfy (3.18) or, equivalently, Z  1 p−1 dA(z) < ∞; |∇u(z)|p log |z| D

the norm is introduced in the obvious way. From the proof of (a) it follows that Xq ⊂ Yq (1/q+1/p = 1), the inclusion being continuous. From this we can conclude that Yp ⊂ Xp provided that we have the inclusions Yp ⊂ (Yq )∗ and (Xq )∗ ⊂ Xp with respect to the same bilinear form. The results of Section 3.4 can be formulated in terms of real Lp and hp spaces. As a consequence we we get (Xq )∗ = Xp with respect to the bilinear form Z π 1 u(eiθ ) v(eiθ ) dθ. (u, v)0 = 2π −π

3.6 Harmonic Schwarz lemma

53

On the other hand, by H¨ older’s inequality, we have Yp ⊂ (Yq )∗ with respect to a different form, namely Z 1 (u, v)1 = u(0)v(0) + ∇u(z) · ∇v(z) log dA(z). |z| D Fortunately, from Green’s formula and the equality ∆(uv) = 2∇u · ∇v it follows that (u, v)0 = (u, v)1 provided that u and v are harmonic in a neighborhood of the closed disk. Then, by H¨ older’s inequality, we get |(u, v)0 | = |(u, v)1 | 6 kukYp kvkYq . Now we use assertion (a) to get |(u, v)0 | 6 CkukYp kvkXq , where C depends only on q. And since sup{|(u, v)0 | : kvkXq 6 1} = kukXp , we see that kukp 6 CkukYp , provided the function u is harmonic in a neighborhood of the closed disk. In the general case, we apply this inequality to the functions u(ρz), ρ → 1, and this completes the proof for 1 < p < 2.

3.6

Harmonic Schwarz lemma

If f ∈ h(D), |f | 6 1, and f (0) = 0, then there holds the sharp inequality |f (z)| 6

4 arctan |z| . π

See [6]; see also 4.4.6. 3.6.1 Theorem Let f be a complex-valued function harmonic in D such that |f (z)| 6 1 for z ∈ D. Then there holds the inequality 1 − |z|2 4 f (0) 6 arctan |z|. (3.19) f (z) − 2 1 + |z| π

If equality occurs for some z ∈ D r {0}, then there are real constants α and β such that f (z) = eiα κ(eiβ z) for all z ∈ D, where κ(z) =

2 1−z 2 2r sin θ arg = arctan π 1+z π 1 − r2

(z = reiθ ).

Proof. We start from the formula Z π 1 − r2 1 1 − r2  1 − r2 f (0) = − f∗ (eiθ ) dθ f (r) − 2 2 1+r 2π −π 1 + r − 2r cos θ 1 + r2 Z π (1 − r2 )2r 1 cos θ = f∗ (eiθ ) dθ. 1 + r2 2π −π 1 + r2 − 2r cos θ From this and the hypothesis |f | 6 1 we get (1 − r2 )2r 1 Z π 1 − r2 | cos θ| (r) − f (0) dθ. f 6 1 + r2 1 + r2 2π −π 1 + r2 − 2r cos θ

54

3 Poisson integral

So we have to compute the integral Z π 1 | cos θ| J= dθ. 2π −π 1 + r2 − 2r cos θ We have 1 J= 2π

Z

1 = 2π

Z

= Since

we see that

Z

1 π

π/2

−π/2 π/2

−π/2

Z

π/2

0



 cos θ cos θ + dθ 1 + r2 − 2r cos θ 1 + r2 + 2r cos θ

2(1 + r2 ) cos θ dθ (1 + r2 )2 − 4r2 cos2 θ

2(1 + r2 ) cos θ dθ (1 − r2 )2 + 4r2 sin2 θ

1 + r2 2(1 + r2 ) cos θ 2r sin θ dθ = arctan , 2 r(1 − r2 ) 1 − r2 (1 − r2 )2 + 4r2 sin θ J=

1 + r2 2r 1 + r2 arctan = 2 arctan r. 2 2 r(1 − r ) 1−r r(1 − r2 )

Combining all these results we get (3.19) for z = r. If equality holds for a fixed z = r > 0, then, as the above inequalities show, we have |f∗ | = 1 a.e., and f∗ (eiθ ) cos θ = eiα g(θ), where α is a constant and g > 0. It follows that ( 1, θ ∈ (−π/2, π/2), −iα iθ e f∗ (e ) = −1, θ ∈ (π/2, 3π/2). Hence f (reiθ ) =

1 2π

Z

= eiα

P (r, θ − t)f∗ (eiθ ) dt

−π

1 =e 2π iα

π

Z

π/2

−π/2

(1 − r2 )4r cos(t − θ) dθ (1 − r2 )2 + 4r2 sin2 θ

2 2r cos θ arctan = eiα κ(ireiθ ). π 1 − r2

✷

¯ (0)| 6 4/π with 3.6.2 With the hypotheses of Theorem 3.6.1 we have |∂f (0)| + |∂f iα iβ equality iff f (z) ≡ e κ(e z) for some α, β ∈ R. Therefore, if f is in addition real-valued, then |∇f (0)| 6 4/π.

4

Subharmonic functions

A real-valued C 2 -function u is subharmonic iff ∆u > 0. If u is not of class C 2 , then u is subharmonic iff it is the limit of a decreasing sequence of subharmonic functions of class C 2 . The importance of subharmonic functions for spaces of analytic and harmonic functions lies in the fact that if f is analytic (resp. harmonic), then |f |p is subharmonic for every p > 0 (resp. p > 1). This chapter contains concise proofs of the basic properties such as the maximum principle, local integrability, approximation by smooth functions, the subordination principle. The discussion of the integral means (Sections 4.2, 4.3) follows H¨ormander’s book [30](∗) and includes Prawitz’ theorem 4.3.1 and, as consequences, Koebe’s one-quarter theorem, and Bieberbach’s theorem. In Section 4.5 we prove a weak version of Riesz’ representation theorem. Section 4.6 is devoted to the proof of a Littlewood/Paley theorem for subharmonic functions.

4.1

Basic properties

A function u : D 7→ [−∞, ∞), where D is a subdomain of the complex plane, is said to be subharmonic if it is upper semicontinuous, i.e., u(a) > lim sup u(z) z→a

for all a ∈ D,

(4.1)

and for every a ∈ D there exists R > 0 such that {z : |z − a| < R} ⊂ D and Z π 1 u(a + ρeiθ ) dθ for every 0 < ρ < R. (4.2) u(a) 6 2π −π Upper semicontinuity implies boundedness from above, which guarantees the existence of the integral in (4.2). From (4.2) it follows that for every a ∈ D there exists a sequence zn → a such that u(a) 6 u(zn ), which implies lim supz→a u(z) = u(a). In particular, u is continuous at a if u(a) = −∞. There are discontinuous subharmonic functions; e.g., the function u(z) =

∞ X log |z − 2−n | 2n n=1

is subharmonic in the entire plane and is discontinuous at zero. Because of the mean value property, every real-valued harmonic function is both subharmonic and superharmonic. (∗) The

only difference is in the proof of Prawitz’ theorem.

55

56

4 Subharmonic functions

A function u is said to be superharmonic if −u is subharmonic. In the case of C 2 -functions there is a simple criterion of subharmonicity deduced from Green’s formula: A function u ∈ C 2 (D) is subharmonic iff ∆u > 0 in D.

From this and the formula ∆(u ◦ ϕ)(z) = (∆u)(ϕ(z)) |ϕ′ (z)|2 , where ϕ is an analytic function, we get: The composition u ◦ ϕ is subharmonic if u is subharmonic and ϕ is analytic. In the general case this assertion can be reduced to the “smooth” case by approximating an arbitrary subharmonic function by smooth ones (Theorem 4.1.15, later on). An important example of a subharmonic function taking the value −∞ is the function log |z − a|. More generally: 4.1.1 Theorem If f is analytic in D, then the function log |f | is subharmonic in D, and |f |p is subharmonic for every p > 0. New examples can be produced by using the following assertions: 4.1.2 Theorem The sum and the maximum of a finite sequence of subharmonic functions are subharmonic functions. The same holds for the limit of a decreasing sequence of subharmonic functions. 4.1.3 Theorem Let φ be an increasing convex function that is defined and continuous on an interval I ⊂ [−∞, +∞). If v is subharmonic and takes its values in I, then the function u = φ(v) is subharmonic. In particular, u is subharmonic in the following cases: (a) u = |h|p , where p > 1 and h is harmonic; (b) u = v p , p > 1, where v is subharmonic and nonnegative. Exercises 4.1.4 [59] If p > 0 and f (z) =

Pn

k=m

ak z k , then for 0 < p < ∞ we have

rn Mp (1, f ) 6 Mp (r, f ) 6 rm Mp (1, f )

0 < r < 1.

4.1.5 Let h 6≡ 0 be a real-valued harmonic function and 0 < p < 1. The function |h|p is subharmonic iff h is constant. The function |h|p is superharmonic iff h has no zeros. 4.1.6 Let u1 , . . . , un be a sequence of nonnegative functions subharmonic in D. If p > 1, then the function u := (up1 + · · · + upn )1/p is subharmonic in D. If uk = |fk |, where fk are analytic, then u is subharmonic for every p > 0.

4.1 Basic properties

57

The maximum principle The simplest variant of the maximum principle says: 4.1.7 Theorem A nonconstant subharmonic function cannot attain its maximum inside the domain. In particular, a nonconstant harmonic function attains neither the maximum nor the minimum inside the domain. Proof. Let u be subharmonic in a domain D and let M denote the set of points in D where u attains its maximum. Because of the semicontinuity, M is closed. Let us prove that M is open as well, which will imply M = D or M = ∅, so the proof will be finished. Let a ∈ M . Then (4.2) implies that, for R small enough and for all ρ < R, we have u(a) = u(a + ρζ) almost everywhere on the circle |ζ| = 1. From this and (4.1) it follows that u(a) 6 u(a + ρζ) everywhere; thus, u(a) = u(a + ρζ) everywhere, i.e., {z : |z − a| < R} ⊂ M . ✷ 4.1.8 Corollary If u is upper semicontinuous on D and subharmonic in D, then max{u(z) : z ∈ D} = max{u(ζ) : ζ ∈ ∂D}. 4.1.9 Corollary Let D be a bounded domain, let u be a function subharmonic in D and upper semicontinuous on D, and let h be a real-valued function harmonic in D and continuous on D. If u 6 h on ∂D, then u 6 h in D; besides, u < h if u is not harmonic. 4.1.10 Corollary If a function is both subharmonic and superharmonic, then it is harmonic. Proof. If u and −u are subharmonic in D, then u is continuous. Let K be a disk, relatively compact in D, and let h be harmonic and h = u on ∂K. Then u 6 h and −u 6 −h and K. ✷ Corollary 4.1.9 gives a partial explanation of the term “subharmonic.” Moreover, we have: 4.1.11 Theorem Let u be an upper semicontinuous function in a domain D, with values in [−∞, ∞). Then u is subharmonic iff for each disk K with K ⊂ D, and each function h continuous on K and harmonic in K, the condition u 6 h on ∂K implies u 6 h in K. Proof. Necessity follows from Corollary 4.1.8 applied to the function u − h. Let D ⊂ D and let u be an upper semicontinuous function satisfying the above condition. Let φ be an arbitrary continuous function on ∂D such that φ > u on ∂D. Then the function Z π 1 h(reiθ ) = P (r, θ − t)φ(eit ) dt 2π −π

58

4 Subharmonic functions

equals the continuous extension of φ to D (Theorem 3.1.6). Hence u 6 h in D and in particular u(0) 6 h(0), i.e., 1 u(0) 6 2π

Z

π

φ(eit ) dt.

−π

Being upper semicontinuous, the function u(eit ) is the infimum of a family of continuous functions; therefore, Z π 1 u(eit ) dt. u(0) 6 2π −π Applying this inequality to the functions z 7→ u(a + ρz), we find that there holds (4.2), and this was to be proved. ✷ Local integrability If u is subharmonic in D, then u(a) 6

1 πρ2

Z

|z−a| −∞. (Otherwise, b is in the interior of the complement of E.) The disk G0 = {|z − a| < ε/2} contains b and u is integrable on G0 because of (4.3). Hence, b ∈ E, which means that E is closed as well. ✷ Approximation by smooth functions 4.1.15 Theorem Let u 6≡ −∞ be subharmonic in a domain D. Then there exists an increasing sequence of open sets Dn , whose union is D, and a decreasing sequence of subharmonic functions un ∈ C ∞ (Dn ) tending to u.

4.1 Basic properties

59

Proof. Let ω(z) = ω0 (|z|) be a nonnegative “radial” function of class C ∞ (C) with compact support in D such that Z ω(w) dm(w) = 1. D

For ε > 0 small enough consider the sets Dε = {z : dist(z, C r D) > ε} and the functions Z Z
ε−2 ω (w − z)/ε u(w) dm(w), ω(w)u(z + εw) dm(w) = uε (z) = C

C

where u ≡ 0 outside of D. Then uε is finite (because of local integrability of u), subharmonic and of class C ∞ in Dε . From the formula Z π Z 1 u(z + rεeit ) dt rω0 (r) dr uε (z) = 2 −π

0

and the inequality Z π

u(z + rεeit ) dt 6

Z

π

u(z + rδeit ) dt,

δ < ε,

−π

−π

(see Theorem 4.2.1 and its proof), it follows that u(z) 6 uδ (z) 6 uε (z), δ < ε, z ∈ Dε . And since u is bounded from above on compact subsets, we can apply the “limsup” variant of Fatou’s lemma; we get Z π Z 1 lim sup u(z + rεeit ) dt. rω0 (r) dr lim sup uε (z) 6 2 ε→0

0

−π

ε→0

Hence lim sup uε (z) 6 u(z), which completes the proof. ε→0

✷

Miscellaneous 4.1.16 Let u 6≡ −∞ be upper semicontinuous on D and subharmonic in D. Then Z π 1 P (r, θ − t)u(eit ) dt (0 < r < 1). u(reiθ ) 6 2π −π 4.1.17 If a real-valued function h is semicontinuous and satisfies the condition Z π 1 h(a + reiθ ) dθ, h(a) = 2π −π locally, then h is harmonic. 4.1.18 If an upper semicontinuous function u satisfies the condition Z 1 u(z) dm(z), u(a) 6 2 πr |z−a| 0, is convex of log r if the function x 7→ ϕ(ex ) is convex. In other words, ϕ(r) is convex of log r if there holds the inequality ϕ(r11−λ r2λ ) 6 (1 − λ)ϕ(r1 ) + λϕ(r2 ),

0 < λ < 1.

If ϕ is of class C 2 , then it is convex of log r iff ϕ′′ (r) + ϕ′ (r)/r > 0. 4.2.1 Theorem Let u be subharmonic in the disk |z| < R. Then the function Z π 1 I(r, u) = u(reiθ ) dθ (0 < r < R) 2π −π is finite, increasing and convex of log r. The same holds for the function I∞ (r, u) = max u(reit ). 06t62π

Proof. If u is subharmonic, then so is the function Z π 1 u(zeiθ ) dθ = I(|z|, u) (|z| < R). I(z) := 2π −π For fixed 0 < r1 < r2 define the harmonic function h(z) = a log |z| + b by h(rj ) = I(rj ). Since I(z) = h(z) on the boundary of the annulus r1 6 |z| 6 r2 , we have I(z) 6 h(z) for r1 < |z| < r2 . From this it follows that I(r, u) is convex of log r. That I(r, u) increases with r follows from the fact that the function ϕ(x) = I(r, ex ) is convex and bounded for −∞ < x < log R, and this completes the proof in case of I(r, u). In case of I∞ the proof is similar; we define h by h(z) = I∞ (rj ) for |z| = rj . ✷ The function I(r, u) need not be increasing if u is defined and subharmonic in an annulus; a simple example is the function u(z) = − log |z| which is subharmonic (and harmonic) in the annulus C r {0}. 4.2.2 Theorem Let u be subharmonic in the annulus ρ < |z| < R. Then the function Z π 1 u(reiθ ) dθ (ρ < r < R) I(r, u) = 2π −π

is finite and convex of log r.

Remark. The same holds for the function I∞ (r, u). Proof. Since u is locally integrable and Z Z r1 Z π u(z) dm(z) = r dr u(reiθ ) dθ, r0 0. Then the function I(r, eu ) is logarithmically convex of log r in the interval ρ < r < R. Proof. The function v(z) = ec log |z|+u(z) is subharmonic in the annulus ρ < |z| < R for every c > 0. Hence, the function I(ex , v) = ecx I(ex , eu ) is convex for every c > 0. The result follows. ✷ 4.2.4 Corollary If f is a function analytic in the annulus ρ < |z| < R, then the function 1/p  Z π 1 |f (reiθ )|p dθ Mp (r, f ) = 2π −π is logarithmically convex of log r in the interval ρ < r < R, for every 0 < p 6 ∞. In the case p = ∞ this is Hadamard’s three circles theorem. The case p < ∞ was discussed by Hardy [25] in the first paper from “theory of Hardy spaces.” Miscellaneous 4.2.5 [30, Theorem 3.2.17] If u is subharmonic in the plane and satisfies the condition Z π 1 u(reiθ ) dθ = o(log r) (r → ∞), 2π −π then u harmonic. 4.2.6 (Liouville’s theorem [30, Theorem 3.2.24]) If u is subharmonic in all of C and u(z) 6 o(log |z|) (z → ∞), then u is a constant. 4.2.7 If u > 0 is subharmonic in the annulus ρ < |z| < R and p > 1, then the function Mp (r, u) = {I(r, up )}1/p is convex of log r for ρ < r < R.

61

62

4 Subharmonic functions

4.3

Integral means of univalent functions

A function f defined on D is said to be univalent if it is analytic and one-to-one. The leading example is the Koebe function f (z) = z/1 − z)2 mapping D to C slit from −1/4 to −∞ along the real axis. Prawitz’ theorem 4.3.1 Theorem Let f : D 7→ C be a univalent function and f (0) = 0. Then for every p > 0 the function Z π 1 Jp (r) = Jp (r, f ) = |f (reiθ )|−p dθ, 0 < r < 1, 2π −π is decreasing. What is interesting here is that the function u = |f |−p is subharmonic in the annulus D r {0} but not in D, because u(0) = +∞. Also, the function −u is not subharmonic in D and therefore we cannot apply Theorem 4.2.1. Proof. We have 2πJ

′ p (r)

Z

2π

|f (reiθ )|−p−2 Re{f (reiθ )f ′ (reiθ )eiθ } dθ Z |f (ζ)|−p−2 f (ζ)f ′ (ζ) dζ = −(p/r) Im = −p

0

|ζ|=r

= −(p/r) Im = −(p/r)

Z

Z

Γr

Γr

|w|−p−2 w ¯ dw

(w = u + iv)

|w|−p−2 (u dv − v du),

where Γr is the image under f of the circle |ζ| = r; the curve Γr is oriented positively. Now we apply Green’s formula to the domain Ωr,R bounded by Γr and the circle |w| = R, where R > max|z|=r |f (z)|. Since ∂(|w|−p−2 u) ∂(−|w|−p−2 v) − = −p|w|−p−2 , ∂u ∂v we have Z Z |w|−p−2 (u dv − v du) − |w|=R

Γr

|w|−p−2 (u dv − v du) = −p

The first integral is equal to 2πR−p , and therefore ZZ J ′p (r) = −(p/r)R−p − (p2 /2πr)

Ωr,R

This concludes the proof.

✷

ZZ

Ωr,R

|w|−p−2 du dv.

|w|−p−2 du dv

4.3 Integral means of univalent functions

63

Distortion theorems 4.3.2 Theorem (Bieberbach) If f is a univalent function in D, then |f ′′ (0)| 6 4|f ′ (0)|. Proof. [30] We can assume that f (0) = 0 and f ′ (0) = 1. Then
f (z)−1/2 = z −1/2 1 − f ′′ (0)z/4 + z 2 h(z) ,

where h is analytic in D. By Theorem 4.3.1, case p = 1, the function
J1 (r) = r−1 1 + |f ′′ (0)/4|2 r2 + r4 M22 (r, h)

is decreasing. Hence the function r−1 + |f ′′ (0)/4|2 r is decreasing, and hence |f ′′ (0)/4| 6 1. ✷ Remark. The famous theorem of de Branges [17] states that if f is univalent in D, then f (n) (0) n > 1. 6 n|f ′ (0)|, n! This was conjectured by Bieberbach. 4.3.3 Theorem (Koebe) If f is univalent in D, then f (D) contains the disk of radius |f ′ (0)|/4 centered at f (0). Proof. [30] Let f (0) = 0. If w is not in the range of f , then the function g(z) = 1/(f (z) − w) is univalent in D and hence |g ′′ (0)| 6 4|g ′ (0)|. It follows that |f ′′ (0) + 2f ′ (0)2 /w| 6 4|f ′ (0)|, whence 2|f ′ (0)2 /w| 6 4|f ′ (0)| + |f ′′ (0)| 6 8|f ′ (0)|. Thus |w| > |f ′ (0)/4|, and this concludes the proof. ✷ 4.3.4 Corollary If f is a conformal mapping of D ⊂ C onto G, then dist(f (z), ∂G) |f ′ (z)| 6 6 4|f ′ (z)|, 4 dist(z, ∂D)

z ∈ D.

4.3.5 Theorem If f is univalent in D, then 1−r |f ′ (z)| 1+r 6 6 , (1 + r)3 |f ′ (0)| (1 − r)3

|z| = r.

 z−w  , Proof. If we apply Bieberbach’s theorem to the function g(w) = f 1 − z¯w we get f ′′ (z) r 2r2 − . 6 z ′ f (z) 1 − r2 1 − r2  f ′′  ∂ Since r log |f ′ | = Re z ′ we see that ∂r f 2r − 4 ∂ 2r + 4 6 . log |f ′ | 6 2 1−r ∂r 1 − r2

The desired result is obtained by integration.

✷

64

4 Subharmonic functions f (z) − f (0) |f ′ (0)| 4.3.6 Corollary If f is univalent in D, then . 6 z (1 − |z|)2 f (z) − f (0) |f ′ (0)| . 4.3.7 Exercise If f is univalent in D, then > z (1 + |z|)2

4.3.8 Exercise If f is univalent in D, then where M (r) = max |f (z)|, m(r) = min |f (z)|. |z|=r

p p M (r)−p 6 −Jp′ (r) 6 m(r)−p , r r

|z|=r

Mean growth If p > 0, then the function Ip (r) = Ip (r, f ) = J−p (r, f ) is increasing, and this fact does not depends on the hypothesis that f is univalent (Theorem 4.2.1). However, the proof of Theorem 4.3.1 gives additional information on Ip (r), namely: If f is univalent in D, f (0) = 0 and f ′ (0) = 1, then Ip′ (r) =

p2 p p R − r 2πr

ZZ

Ωr,R

|w|p−2 du dv.

This implies that Ip′ (r) 6 (p/r)M (r)p . Combining this with Corollary 4.3.6 we get: 4.3.9 Theorem If f is univalent in D and 0 < p < 1/2, then Ip (r, f ) is bounded, and  1  I1/2 (r, f ) = O log , r → 1. 1−r

4.4

The subordination principle

Let F be a univalent function defined in D. A function f analytic in D is said to be subordinate to F if f (D) ⊂ F (D) and f (0) = F (0). In other words, f is subordinate to F if f (z) = F (ω(z)), where |ω(z)| 6 |z|, z ∈ D, and ω is analytic. In this form the notion of subordination is defined for arbitrary functions. This notion is important because of the following theorem, known as Littlewood’s subordination principle. 4.4.1 Theorem If a function u is subordinate to a subharmonic function U , then 1 2π

Z

π

−π

u(reiθ ) dθ 6

1 2π

Z

π

U (reiθ ) dθ

(0 < r < 1).

(4.4)

−π

In the simplest case ω(z) = ρz, 0 < ρ < 1, this theorem reduces to Theorem 4.2.1. Proof. We can assume that U is continuous. Let h be a function harmonic in Dr = {|z| < r}, continuous on the closure and equal to U on the boundary. Then

4.4 The subordination principle

U 6 h on Dr and, hence, u(z) = U (ω(z)) 6 h(ω(z)) for |z| = r. It follows that Z π Z π Z π

1 1 1 u(reiθ ) dθ 6 h ω(reiθ ) dθ = h ω(0) = h(0) = h(reiθ ) dθ. 2π −π 2π −π 2π −π

This concludes the proof because h(reiθ ) = U (reiθ ). ✷ For various applications of the subordination principle in the theory of univalent functions we refer the reader to Duren [19, Ch. 6]. We will consider two examples which cannot be found in [19].

4.4.2 Theorem (Kolmogorov/Smirnov) If f ∈ H(D) and Re f ∈ h1 , then f ∈ hp , i.e., Z π 1 |f (reiθ )|p dθ < ∞ for every p ∈ (0, 1). sup r 0. Then f is subordinate to the univalent function F (z) = c(1 + z)/(1 − z), c = Re f (0). Applying the subordination principle we get the following: Z π Z π Z π 1 + reiθ p 1 + eiθ p 1 iθ p p 1 p 1 dθ |f (re )| dθ 6 c dθ 6 c 2π −π 2π −π 1 − reiθ 2π −π 1 − eiθ Z π/2 u(0)p p2 =c (0 < p < 1). (cot θ)p dθ = π 0 cos(pπ/2) If f is arbitrary, then we can use Theorem 3.2.5(b) to reduce the proof to the preceding case. ✷ 4.4.3 Remark The Kolmogorov/Smirnov theorem can be proved in the following way. Let Re f > 0. Then Re(f p ) = |f |p cos(arg f ) > |f |p cos(πp/2), and hence Z π Z π  1 1 1 1 f (reiθ ) p dθ 6 Re{f (0)p }. Re f (reiθ )p dθ = 2π −π cos(πp/2) 2π −π cos(πp/2) Our next example is the case p 6 1 of the following theorem of Ahern [1].

4.4.4 Theorem If f ∈ H(D) and 0 < |f (z)| < 1 for all z ∈ D, then for every p > 0 we have Z π 1 (1 − |f (reiθ )|)p dθ > cp (1 − r)1/2 , 2π −π where cp is a positive constant. Ahern’s proof is based on a highly nontrivial analysis of singular measures, which enabled him to treat the case p > 1. Here we use the subordination principle to discuss the case p = 1/2. It turns out that then Ahern’s theorem can be improved. On the other hand, it seems that application of the subordination principle is limited to the case p 6 1.

65

66

4 Subharmonic functions

4.4.5 Theorem With the hypotheses of the previous theorem, we have Z π
1/2 1 2 1 − |f (reiθ )| dθ > c(1 − r)1/2 log , 2π −π 1−r where c is a positive constant.

Proof. The analytic function a(z) = − log f (z) maps D into the right half-plane and therefore is subordinate to the function λ(1 + z)/(1 − z), where λ = a(0) > 0. It follows that f (z) is subordinate to  1 + z . Sλ (z) = exp − λ 1−z

The function −(1 − |z|)1/2 is subharmonic for |z| < 1, and therefore, by the subordination principle, Z π Z π
1/2
1/2 1 1 iθ 1 − |f (re )| dθ > 1 − |Sλ (reiθ )| dθ. 2π −π 2π −π

In order to estimate this integral we use the inequality x 2x 6 1 − e−x 6 , 1+x 1+x

x > 0.

It follows that Z π Z π Z π 1 (1 − r)1/2 λP (r, θ) 1/2 iθ 1/2 dθ ≍ (1 − |Sλ (re )|) dθ ≍ dθ 2 1/2 2π −π 1 + λP (r, θ) 0 (1 − r + θ ) 0 √ ✷ Introducing the change θ = t 1 − r we conclude the proof. Miscellaneous 4.4.6 (Harmonic Schwarz lemma) If u ∈ h(D) is real valued, |u| 6 1, and u(0) = 0, then u is subordinate to the function U (z) =

1+z 2 arg . π 1−z

4 Hence |u(z)| 6 arctan |z|, and, by the subordination principle, M2 (r, u) 6 r, π 0 < r < 1. 4.4.7 (Rogosinski’s theorem [19, §6.2]) Let f (z) = F (ω(z)), where F is anaPn Pn lytic in D, and ω is as above. Let Fn (z) = k=0 Fb(k)z k and fn (z) = k=0 fb(k)z k . Then Fn (ω(z)) = fn (z) + O(z n+1 ). Therefore, by the subordination principle and Parseval’s formula, If f is subordinate to F ∈ H(D), then n X

k=0

|fb(k)|2 r2k 6

n X

k=0

|Fb(k)|2 r2k ,

0 < r < 1.

4.5 The Riesz measure

67

4.4.8 With the hypotheses of Theorem 4.4.4, we have Z π 1 (1 − |f (reiθ )|)p dθ > cp (1 − r)p for 0 < p < 1/2. 2π −π 4.4.9 [19] If U = |f |p , 0 < p < ∞, then strict equality holds for 0 < r < 1 in (4.4) unless f is constant or ω(z) = αz, |α| = 1.

4.5

The Riesz measure

Let C02 (D) be the set of all C 2 -functions with compact support in D. If u is subharmonic in D, then the functional Z u∆ϕ dm, ϕ ∈ C02 (D) ϕ 7→ D

is positive and therefore there exists a unique positive Borel measure µ on D such that Z Z u∆ϕ dm, ϕ ∈ C02 (D). ϕ dµ = D

D

This measure is called the Riesz measure of u. If u is of class C 2 , then dµ = (∆u) dm. The Riesz measure of the function log |z − a| is equal to 2πδa , where δa denotes the Dirac measure at the point a. The existence and uniqueness of the Riesz measure can be proved by using Riesz’ theorem on representation of positive linear functionals on C0 (D).(†) Green’s formula Here we will consider in some detail a generalization of Green’s formula. 4.5.1 Theorem Let u be subharmonic in DR = {z : |z| < R} and let u(0) > −∞. Then Z π Z 1 1 r dµ(z), (4.5) u(reiθ ) dθ − u(0) = log 2π −π 2π |z| −∞. (†) For more information on the Riesz measure we refer to H¨ ormander [30] and Hayman/Kennedy [27]. For a discussion on Riesz’ representation of positive linear functionals on C0 (D), see Rudin [86, Ch. 2].

68

4 Subharmonic functions

Jensen’s formula The well known Jensen’s formula is one of special cases of (4.5). Namely, if a function f is analytic in DP R and f (0) 6= 0, then the Riesz measure of the function log |f (z)| is equal to 2π k δak , where ak are the zeros of f . This and (4.5) give Jensen’s formula, 1 2π

Z

π

−π

log |f (reiθ )| dθ = log |f (0)| +

X

log

|ak | r. Let g be a C 2 -function in DR such that g = 0 in a neighborhood of the circle |z| = r and g 6 Gρ in DR . Then 1 I(r, un ) − I(ρ, un ) > 2π

Z

1 g(z)∆un (z) dm(z) = 2π DR

Z

un (z)∆g(z) dm(z).

DR

Here we apply the dominated convergence theorem, which is possible because |un | 6 |u| + |u1 | and the functions u and u1 are locally integrable (Theorem 4.1.12), to get Z Z 1 1 I(r, u) − I(ρ, u) > u(z)∆g(z) dm(z) = g(z) dµ(z). 2π DR 2π DR Now we take an increasing sequence of the functions g which tends to Gρ and get I(r, u) − I(ρ, u) >

1 2π

Z

Gρ (z) dµ(z).

DR

The reverse inequality is proved in a similar way. Finally, let ρ tend to 0.

✷

Riesz’ representation formula A slight modification of the above proof yields the following variant of Theorem 4.5.1.

4.5 The Riesz measure

69

4.5.2 Theorem If u is subharmonic in a neighborhood of the closed unit disk, then Z 1 G(z, w) dµ(w), (4.6) u(z) = h(z) + 2π D for every z ∈ D, where Z π 1 P (r, θ − t)u(eit ) dt, h(reiθ ) = 2π −π w−z µ is the Riesz measure of u, and G(z, w) = log . 1 − wz ¯ Miscellaneous

4.5.3 Let u be subharmonic in D. Then for every ε ∈ (0, 1/4) there exists a, |a| < 1/4, such that   Z π Z π 1 1 u(a + εeiθ ) dθ − u(a) 6 Cε2 u(eiθ ) dθ − u(0) , 2π −π 2π −π where C is an absolute constant. 4.5.4 Let ϕ : D 7→ C be a conformal mapping and let u be subharmonic in D. Then Z Z f (ϕ−1 (w)) dµu (w), f (z) dµu◦ϕ (z) = ϕ(D)

D

where f is an arbitrary positive (Borel) function on D, and µ is the Riesz measure of the corresponding function. For instance, if v(z) = u(a + rz) (a ∈ C, r > 0), then Z Z
f (z) dµv (z) = f (w − a)/r dµu (w). D

a+rD

4.5.5 If u subharmonic in the unit disk, then there holds the formula Z Z
f (z) dµu◦ϕa (z) = f ϕa (w) dµu (w) (|a| < 1), D

D

a−z and f is positive and Borel. where ϕa (z) = 1−a ¯z

4.5.6 If f is analytic and p > 0, then the Riesz measure of |f |p is absolutely continuous and equal to p2 p−2 ′ 2 |f | |f | dm. 2 If p > 2, then the function |f |p−2 |f ′ |2 is subharmonic; from this and Green’s formula we can deduce that the function Z π √ 1 |f ( reiθ )|p dθ r 7→ 2π −π is convex.

70

4 Subharmonic functions

4.6

A Littlewood/Paley theorem

For a function u defined in D we write I(u) = sup I(r, u), where, as above, 0 0 be subharmonic in D and let µ be the Riesz measure of u. If q > 1 and I(uq ) < ∞, then there holds the inequality Z  q
(1 − |z|)−1 µ(Eε (z)) dm(z) 6 Cq I(uq ) − u(0)q , (4.7) D

where ε = 1/6 and Eε (z) = {w : |w − z| < ε(1 − |z|)}.

If in addition ∆u (u ∈ C 2 ) is a subharmonic function, then Z ∆u dm > πε2 (1 − |z|)2 ∆u(z), µ(Eε (z)) = Eε (z)

which leads to the following: 4.6.2 Theorem Let u > 0 be a subharmonic function of class C 2 (D) such that its Laplacian is a subharmonic function. If q > 1 and I(uq ) < ∞, then Z
q
(1 − |z|)2q−1 ∆u(z) dm(z) 6 Cq I(uq ) − u(0)q . (4.8) D

The classical inequality of Littlewood and Paley (Theorem 3.5.1) is a special case of (4.8). Namely, if p > 2 and I(|h|p ) < ∞, where f is a real-valued harmonic function in D, then we take u = h2 and q = p/2, and get Z
(1 − |z|)p−1 |∇h|p dm 6 Cp I(|h|p − |h(0)|p ) . D

In the case q < 1 we have the following theorem, the proof of which is omitted here (cf. [79]).

4.6.3 Theorem Let 0 < q < 1 and let u > 0 be a C 2 -function such that both uq and ∆u are subharmonic. If Z (1 − |z|)2q−1 (∆u)q dm < ∞, D

q

then I(u ) < ∞ and there holds the inequality Z I(uq ) − u(0)q 6 Cq (1 − |z|)2q−1 (∆u)q dm(z). D

4.6 A Littlewood/Paley theorem

71

Jevti´c [32] extended the above theorems to the case of M-harmonic functions on the unit ball in Cn . Local estimates for the Riesz measure In what follows we suppose that u is a nonnegative subharmonic function defined in D and denote by µ the Riesz measure of u. As we have seen, there holds the formula Z 1 r I(r, u) − u(0) = log dµ(z) (0 < r < 1) (4.9) 2π rD |z|

(see Theorem 4.5.1).

4.6.4 Lemma There holds the equality I(u) − u(0) =

1 2π

Z

D

log

1 dµ(z). |z|

Proof. Write (4.9) as 1 I(r, u) − u(0) = 2π

Z

Kr (z) log

D

r dµ(z), |z|

where Kr (z) is the characteristic function of the disk rD. Since Kr (z) log(r/|z|) increases with r, we have Z 1 r lim Kr (z) log lim I(r, u) − u(0) = dµ(z). 2π D r→1− |z| r→1− Since I(r, u) increases, we see that I(u) = limr→1− I(r, u).

✷

4.6.5 Lemma Let q > 1 and let µq be the Riesz measure of uq . Then {µ(E)}q 6 Cq µq (5E)

(4.10)

for every disk E such that 6E ⊂ D. The constant Cq depends only of q. If E is a disk of radius R, then rE denotes the concentric disk of radius Rr. Proof. By translation, the proof reduces to the case where E is centered at zero. Then, since µ(E) = ν((1/r)E), where ν is the Riesz measure of the function u(rz), we can assume that the radius of E is fixed, e.g., E = εD, ε = 1/6. Using the simple inequalities
q
q I(r, u) − u(0) 6 I(r, u) − u(0)q
q and I(r, u) 6 I(r, uq ), which hold because q > 1, we see from (4.9) (applied to u and uq ) that q  Z Z r r 1 1 dµ(z) 6 dµq (z). log log 2π rD |z| 2π rD |z|

72

4 Subharmonic functions

Letting r = 4ε, we get

Z

{µ(2εD)}q 6 C

4εD

|z|−1 dµq (z),

(4.11)

where we have applied the estimate log(4ε/|z|) > log 2 for |z| < 2ε and log(4ε/|z|) 6 1/|z|. Therefore, in order to prove (4.10) we have to remove |z|−1 . To do this, we translate the “center” of (4.11) to get Z |z − a|−1 dµq (z) {µ(2εDa )}q 6 C 4εDa

for a ∈ εD, where Da = {z : |z − a| < 1}. Since εD ⊂ 2εDa and 4εDa ⊂ 5εD, we see that Z {µ(εD)}q 6 C |z − a|−1 dµq (z). 4εDa

Now we integrate this inequality over the disk εD, with respect to dm(a), and apply Fubini’s theorem, which finishes the proof because Z sup |z − a|−1 dm(a) < ∞. ✷ z∈D

εD

Proof of Theorem 4.6.1 From (4.10) it follows that Z Z  q (1 − |z|)−1 µ(Eε (z)) dm(z) 6 C (1 − |z|)−1 µq (E5ε (z)) dm(z).

(4.12)

D

D

Further, from

µq (E5ε (z)) =

Z

dµq (w)

E5ε (z)

and Fubini’s theorem it follows that the right-hand side of (4.12) equals Z Z (1 − |z|)−1 dm(z), dµq (w) D

G(w)

where G(w) = {z : |z − w| < 5ε(1 − |z|)}. Since z ∈ G(w) implies |z| − |w| < 5ε(1 − |z|), whence 1 − |w| < (1 + 5ε)(1 − |z|), we see that Z (1 − |z|)−1 dm(z) 6 (1 + 5ε) m(G(w)) (1 − |w|)−1 . G(w)

And since (1 − 5ε)(1 − |z|) < 1 − |w| for z ∈ G(w), we have m(G(w)) 6 C ′ (1 − |w|)2 , where C ′ = π(5ε/(1 − 5ε))2 . Combining all these results we see that Z Z  q (1 − |z|)−1 µ(Eε (z)) dm 6 Cq (1 − |w|) dµq (w). D

D

This completes the proof of (4.7) because of Lemma 4.6.4 and the inequality 1− |w| 6 log(1/|w|).

5

Classical Hardy spaces

There are various equivalent definitions of H p -spaces. If p > 1, the shortest way to introduce H p is by identifying it with a subspace of Lp (T), b H p (T) = {φ ∈ Lp (T) : φ(−n) = 0 for n > 1}.

Thus H p (T) coincides with the closure in Lp (T) of the set of all analytic polynomials; this can be used to define H p (T) for 0 < p < 1. In this text, we define H p as a subclass of H(D), see (5.1). In view of Riesz’ projection theorem 6.2.1, H p is isomorphic with hp for 1 < p < ∞. Because of the theorem of Burkholder, Gundy and Silverstein, see Theorem 7.2.1, one can define H p as a space of harmonic functions (6= hp for p 6 1) for every p > 0, which is used to extend H p -theory to several real variables (cf. [93]). This chapter contains the standard facts on radial limits and factorization; an exception is Section 5.5, where we consider the composition of an H p -function with an inner function. Our approach slightly differs from that in other texts [18, 22, 46, 83, 86, 99] in that we first prove the Hardy/Littlewood decomposition lemma (Lemma 5.1.7), and then deduce the radial limits theorem and F. and M. Riesz’ theorems, without appealing to the Blaschke products.

5.1

Basic properties

The Hardy space H p (0 < p 6 ∞) is defined as the subspace of hp consisting of analytic functions, H p = {f ∈ H(D) : kf kp = supr 0. If p > 1, then H p is a Banach space, and if 0 < p < 1, it is a p-Banach space. The completeness is proved in the standard way. The first step is the continuity of the inclusion H p ⊂ H(D), which follows from the following lemma. 5.1.1 Lemma If f ∈ H p , 0 < p 6 ∞, then |f (z)| 6 (1 − |z|2 )−1/p kf kp .

(5.2)

The space H p is not normable for p < 1. On the other hand, it follows from the lemma that the dual of H p separates points in H p . 5.1.2 Corollary If f ∈ H p , then Mq (r, f ) 6 (1 − r2 )1/q−1/p kf kp 73

(q > p).

(5.3)

74

5 Classical Hardy spaces

Proof. This follows from (5.2) and the following: Z π  q−p 1 Mqq (r, f ) = Mpp (r, f ). |f (reiθ )|q−p |f (reiθ )|p dθ 6 sup |f (reiθ )| 2π −π θ

✷

Taking q = 1 > p and r = 1 − (1/n) in (5.3), we get, via the inequality M1 (r, f ) > |fb(n)|rn , one of many results of Hardy and Littlewood. 5.1.3 Corollary If f ∈ H p (0 < p < 1), then |fb(n)| 6 Cp kf kp (n + 1)1/p−1 , where Cp depends only on p.

Proof of Lemma 5.1.1. Assume that f is analytic in a neighborhood of the closed disk. Then, for p < ∞, Z 1 kf kpp = |f (ζ)|p |dζ|. 2π T z−w . By the substitution ζ = ϕ(ξ) we get 1 − z¯w Z Z 1 1 |f (ϕ(ξ))|p |ϕ′ (ξ)| |dξ| = |f (ϕ(ξ))ϕ′ (ξ)1/p |p |dξ|. kf kpp = 2π T 2π T

For fixed z ∈ D let ϕ(w) =

The function inside the last integral is subharmonic and therefore kf kpp > |f (ϕ(0))ϕ′ (0)1/p |p = |f (z)|p (1 − |z|2 ), which was to be proved.

✷

5.1.4 Theorem For every p > 0 the space H p is complete. Proof. Let {fn } be a Cauchy sequence in H p . This means that for every ε > 0 there exists N such that Mp (r, fn −fm ) < ε for every r and m, n > N . From (5.1.1) it follows {fn } is a Cauchy sequence in H(D) and hence {fn } converges uniformly on compact subsets to some function f ∈ H(D). Letting m tend to ∞, we get Mp (r, fn − f ) 6 ε for n > N and all r, which implies kfn − f kp 6 ε for n > N . ✷ 5.1.5 Exercise If 0 < p < ∞ and f is analytic in D, then Mp (r, f ) is strictly increasing unless f =const; see Proposition 3.4.1. In particular, if Mp (r, f ) = |f (0)| for some r > 0, then f =const. 5.1.6 Exercise For a fixed z ∈ D, equality occurs in (5.2) iff f (w) = c where c is a constant.



1 − |z|2 (1 − z¯w)2

1/p

,

5.1 Basic properties

75

A decomposition lemma Various properties of a zero-free H p -function can be deduced from the corresponding properties of the function f 2/p ∈ H 2 . The following lemma of Hardy and Littlewood is often used to reduce the general case to the zero-free case. 5.1.7 Lemma If f ∈ H p , p > 0, then there exist functions g and h without zeros in D such that f = g + h, kgkp 6 kf kp and khkp 6 kf kp . For example, in proving an inequality of the form kT f kX 6 Ckf kp , f ∈ H p , where X is a quasinormed space and T : H p 7→ X a linear operator, we can suppose that f has no zeros in D. Proof. Assume, at first, that f 6≡ 0 is analytic in a neighborhood of the closed disk and that f has at least one zero in D. Then the number of zeros of f is finite; denote the zeros by a1 , . . . , am (counting multiplicity). Let m Y z − ak A(z) = 1−a ¯k z k=0

and define g and h as follows: g = (A − 1)f /2A, h = (A + 1)f /2A; we have f = g + h. Neither g nor h have zeros in D because |A| < 1 in D and the function f /A has no zeros in D. Both h and g are analytic in a neighborhood of the closed disk because so are A − 1, A + 1 and f /A. And since |A| = 1 on T, we have |g| 6 |f | and |h| 6 |f | on T, which proves the lemma in that special case. If f is arbitrary, let fn (z) = f (rn z), where rn = 1 − 1/n (or any sequence tending to 1). By the preceding, we have a decomposition fn = gn + hn with the the desired properties. Since kgn kp 6 kfn kp 6 kf kp

(5.4)

(and similarly for h), then, according to Lemma 5.1.1, the sequences {gn } and {hn } are bounded on compact subsets of D. Therefore, passing to subsequences, we can assume that gn and hn tend uniformly on compact subsets to analytic functions g and h, respectively. By Hurwitz’ theorem, the function g is either without zeros or g ≡ 0 because gn have no zeros. The same holds for h. But it is not true that g ≡ 0 because this and f = g + h imply f ≡ h, which is impossible because f 6≡ 0 and f has zeros. Finally, from (5.4) it follows that Mp (r, gn ) 6 kf kp for every r < 1, and hence Mp (r, g) 6 kf kp . ✷ Radial limits Since H 1 ⊂ h1 , we see from the Riesz/Herglotz theorem and Fatou’s theorem that every function f ∈ H 1 has radial limits almost everywhere. However, the hypothesis that f is analytic improves the properties of the boundary function substantially (see, e.g., Theorems 5.1.8 and 5.2.1).

76

5 Classical Hardy spaces

5.1.8 Theorem Let f ∈ H p , 0 < p 6 ∞. Then almost everywhere on T there exist radial limits f∗ (eiθ ) = lim− f (reiθ ) , and there hold the relations r→1

lim

r→1−

1 2π

Z

π

−π

kf kp = kf∗ kp

(p 6 ∞)

|f (reiθ ) − f∗ (eiθ )|p dθ = 0

(5.5) (p < ∞).

(5.6)

Proof. When p > 1, we may appeal to Theorem 3.4.2. Let f ∈ H p and 1/2 < p 6 1. Observe that (5.5) is implied by (5.6). Next, because on the lemma on decomposition, we may assume that f has no zeros. Then the function g = f 1/2 ∈ H 2p is well defined. Since 2p > 1, the function g, so the function f = g 2 , has radial limits. Let f∗ denote the boundary function and fr (eiθ ) = f (reiθ ). Then kfr − f∗ kp = k(gr − g∗ )(gr + g∗ )kp 6 kgr − g∗ k2p kgr + g∗ k2p , where we have applied the Cauchy/Schwarz inequality. Since kgr − g∗ k2p tends to 0 and kgr + g∗ k2p is bounded, we can conclude that there holds (5.6) for p > 1/2. In the same way we reduce the case p > 1/4 to the case p > 1/2, etc. ✷ The set of all harmonic polynomials is not dense in hp for p 6 1. However, we have: 5.1.9 Theorem If 0 < p < ∞, then the set of all (analytic) polynomials is dense in H p . Proof. This follows from (5.6) and the fact that kfr − sn fr kp → 0 (n → ∞), for every fixed r ∈ (0, 1), where sn g denotes the partial sum of the Taylor series of g. ✷ The Poisson integral of log |f∗ | If f is analytic in a neighborhood of the closed disk, then we have log |f | 6 P [log |f∗ |] because of the subharmonicity of log |f |. 5.1.10 Theorem Let f 6≡ 0 belong to H p (p > 0). Then log |f∗ | ∈ L1 (T) and there holds Z π 1 P (r, θ − t) log |f∗ (eit )| dt, 0 6 r < 1. (5.7) log |f (reiθ )| 6 2π −π Before proving this theorem we note two consequences. Smirnov’s maximum principle There are unbounded functions f ∈ H(D) for 1 + z  . which the boundary function belongs to L∞ (T); one of them is f (z) = exp 1−z However: 5.1.11 Theorem (Smirnov) If f ∈ H p , p > 0, and if f∗ ∈ L∞ , then f ∈ H ∞ and kf k∞ = kf∗ k∞ .

5.1 Basic properties

77

In the case p > 1 this theorem is contained in Corollary 3.4.4, while in the general case it is a consequence of (5.7), or of the weaker inequality Z π 1 |f (reiθ )|p 6 P (r, θ − t)|f∗ (eit )|p dt (0 6 r < 1) (5.8) 2π −π and (5.5). It is worthwhile to note that (5.8) can be deduced immediately from (5.6) and the inequality |f (ρz)|p 6 P [ |fρ |p ](z). In fact, inequality (5.7), together with Jensen’s inequality for the function x 7→ ex , implies a more general fact, namely: 5.1.12 Theorem (Smirnov) If f ∈ H p and f∗ ∈ Lq for some q > p, then f ∈ H q . Uniqueness theorem An immediate consequence of Theorem 5.1.10: 5.1.13 Theorem If f ∈ H p and f∗ (eiθ ) = 0 on a set of positive measure, then f (z) = 0 for every z ∈ D. Proof of Theorem 5.1.10 Let f ∈ H p and, say, f (0) 6= 0. Since log+ x 6 xp /p, x > 0, it follows that Z π 1 log+ |f (reiθ )| dθ 6 C (0 < r < 1), 2π −π where C is a constant. We also have Z π Z π Z π 1 1 log |f (reiθ )| dθ = 2 1 log+ |f (reiθ )| dθ − log |f (reiθ )| dθ. 2π −π 2π −π 2π −π

The last summand is 6 − log |f (0)| because of the subharmonicity of log |f |. Hence, Z π 1 log |f (reiθ )| dθ 6 2C − log |f (0)|, (5.9) 2π −π

and Fatou’s lemma concludes the proof that log |f∗ | ∈ L1 (T). To prove (5.7) we start from the inequality log |f (ρz)| 6 P [log |fρ |](z),

where fρ (eiθ ) = f (ρeiθ ),

(5.10)

ρ < 1, z ∈ D, which holds for an arbitrary f ∈ H(D) because log |f | is subharmonic. Since log x = log+ x − log− x, we have, from (5.10), log |f (z)| 6 lim sup P [log+ |fρ |](z) − lim inf P [log− |fρ |](z). ρ→1

ρ→1

And since | log+ x − log+ y| 6 |x − y|p /p, x, y > 0, it follows from (5.6) that Z π + 1 log |f (ρeiθ )| − log+ |f∗ (eiθ )| dθ = 0; lim− ρ→1 2π −π

(5.11)

78

5 Classical Hardy spaces

therefore lim sup P [log+ |fρ |](z) = P [log+ |f∗ |](z). Now we deduce (5.7) from (5.11) ρ→1

by means of Fatou’s lemma.

✷

Remark. For another proof, see 7.1.5.

5.2

The space H 1

The results of this section, due to F. and M. Riesz, Privalov, and Smirnov(∗) , show how much H 1 differs from h1 . The Poisson integral of the boundary function A function belonging to h1 need not be equal to the Poisson integral of the boundary function. However: 5.2.1 Theorem If f ∈ H 1 , then f∗ ∈ L1 and f = P [f∗ ]. This is easily deduced from the relation f (rz) = P [fr ](z) (r < 1), by means of (5.6). 5.2.2 Exercise (Cauchy’s integral formula) If f ∈ H 1 , then 1 f (z) = 2πi

Z

T

f∗ (ζ) dζ ζ −z

(z ∈ D).

Now we are in position to prove the famous theorem of F. and M. Riesz: 5.2.3 Theorem Let µ be a complex Borel measure on T such that for every n = 1, 2, . . . . Then µ is absolutely continuous.

R

T

ζ n dµ(ζ) = 0

Proof. Let f = P [µ]. Then f ∈ h1 and fb(k) = µ b(k) for every k ∈ Z (Proposition 3.2.1). Hence, the condition of the theorem implies that f is analytic. Hence f ∈ H 1 so, according to Theorem 5.2.1, we have f = P [f∗ ]. In view of the injectivity of the Poisson integral (Theorem 3.2.2), it follows that dµ(eiθ ) = f∗ (eiθ ) dt. ✷ Bounded variation =⇒ absolute continuity 5.2.4 Theorem If f ∈ H 1 and if the boundary function is almost everywhere equal to a function of bounded variation, then f has absolutely continuous extension to D. (†) (∗) Further information, as well as references and historical comments, can be found in Zygmund [100, Ch. VII§§8–10] and Duren [18, Ch. III] (†) i.e., a continuous extension that is absolutely continuous on T.

5.2 The space H 1

79

Proof. Let f∗ = γ a.e., γ ∈ BV [−π, π]. Then f = P [γ], by Theorem 5.2.1. We have ∂f = P S[γ](reiθ ) − k · P (r, θ + π), g(reiθ ) := ∂θ where P S[γ] is the Poisson/Stiltjes integral of γ (see (3.10) and (3.13) ). Then, by the Riesz/Herglotz theorem, g ∈ H 1 and, by Theorem 5.2.1, g = P [g∗ ]; thus g = P S[G], where Z θ g∗ (eit ) (θ ∈ R). G(θ) = 0

Applying (3.13) again, with the obvious change of notation, and taking into account that the function G is 2π-periodic because g(0) = 0, we get Z π ∂ 1 ∂f = P S[G](reiθ ) = P (r, θ − t)G(t) dt. ∂θ ∂θ 2π −π It follows that f (reiθ ) = const + which concludes the proof.

1 2π

Z

π

−π

P (r, θ − t)G(t) dt,

✷

In a similar way one proves the following: 5.2.5 Theorem The derivative of a function f ∈ H(D) belongs to H 1 iff f has absolutely continuous extension to D. The boundary function of the function (∂/∂θ)f (reiθ ) = ireiθ f ′ (reiθ ), if f ∈ H 1 , is equal to (d/dθ)f∗ (eiθ ). Conformal mappings 5.2.6 Theorem Let f be a conformal mapping of D onto a domain G whose boundary, ∂G, is a Jordan curve. Then f ′ ∈ H 1 iff ∂G is rectifiable. If ∂G is rectifiable, then Z π

|∂G| =

and |∂G| > 2

P∞

n=0

−π

|f ′ (eiθ )| dθ

(5.12)

|fb(n)|, where |∂G| is the length of ∂G.

The last inequality follows from (5.12) and the inequality ∞ X |fb(n)| 6 πkf k1 , n+1 n=0

(5.13)

due to Hardy (see 5.3.7). Proof. Let f ′ ∈ H 1 . The function f can be extended as a continuous function to D, and the extended function is a homeomorphism between D and G (theorem of Carathe´ odory, [100, Ch. VII, §10]). By Theorem 5.2.5, this extension is absolutely

80

5 Classical Hardy spaces

continuous on T and therefore ∂G is parameterized by the absolutely continuous function Γ(θ) = f∗ (eiθ ) (|θ| 6 π). Hence the length of ∂G is equal to Z

π

−π

|Γ′ (θ)| dθ.

Now formula (5.12) follows from theorem 5.2.5. Conversely, let ∂G be rectifiable, |∂G| = 2π, and let γ(θ) (0 6 θ 6 2π) be the arclength parameterization of ∂G. The function f has the radial limits ϕ(θ) = limr→1− f (reiθ ) on a set S ⊂ [0, 2π], |S| = 2π. Since the extended mapping is homeomorphic, the function ϕ “increases”, i.e., there exists an increasing function t : S 7→ [0, 2π] such that ϕ(θ) = γ(t(θ)), θ ∈ S. We extend the function t(θ) as an increasing function on [0, 2π], and the corresponding extension of ϕ is of bounded variation on [0, 2π]. Now Theorem 5.2.4 shows that f has absolutely continuous extension to D. Finally, f ′ ∈ H 1 , by Theorem 5.2.5.(‡) ✷

5.3

Blaschke product

If a function f ∈ H(D) has an infinite number of zeros, a1 , a2 , . . . , then 1−|an | → 0. If f ∈ H p , there holds more: 5.3.1 Theorem If {an } (n > 1) is the sequence of zeros of a function f ∈ H p P∞ (p > 0), then the Blaschke condition is satisfied: n=1 (1 − |an |) < ∞. Conversely, if a sequence {an } ⊂ D satisfies the Blaschke condition, then the product ∞ Y an − z |an | B(z) = 1−a ¯n z an n=1 converges in D and the function B is analytic and has the properties: (a) The sequence of zeros of B, including repetitions for multiplicities, coincides with {an }; (b) |B(z)| 6 1 for |z| < 1; (c) |B(eiθ )| = 1 almost everywhere. The function B(z) is called a Blaschke product; Q∞in the case an = 0 the ratio |an |/an is interpreted as −1. Note that B(0) = n=1 |an |, and this product converges iff the Blaschke condition is satisfied. By the term a Blaschke product we also mean a function of the form B(z) =

k Y an − z |an | 1 −a ¯n z an n=1

with k > 1 as well as the function B(z) ≡ 1. (‡) so we have proved the implication r“. . . =⇒ f ′ ∈ H 1 ” without appealing to the theorem of Carathe´ odory.

5.3 Blaschke product

81

Proof. Let f ∈ H p (p > 0) and, say, f (0) 6= 0. From the inequality   Z π Z π 1 1 log |f (reiθ )|p dθ 6 log |f (reiθ )|p dθ , 2π −π 2π −π which is obtained by an application of Jensen’s inequality for the concave function log x, it follows that Z π 1 log |f (reiθ )| dθ 6 log kf kp . 2π −π On the other hand, there holds (Jensen’s) formula Z π X 1 r log |f (reiθ )| dθ = log |f (0)| + . log 2π −π |ak | |ak | 1. 2π −π Finally, since |B(eiθ )| 6 1 almost everywhere, we have (c).

✷

Riesz’ factorization theorem 5.3.2 Theorem Let {an } be the sequence of zeros of f ∈ H p (p > 0) and B(z) the corresponding Blaschke product. Then f /B belongs to H p and kf /Bkp = kf kp . Consequently, every H p -function can be represented as f = Bg, where B is a Blaschke product (finite or infinite), the function g has no zeros in D and kf kp = kgkp .

82

5 Classical Hardy spaces

Proof. With the above hypotheses, let Bk (z) =

k Y an − z |an | . 1 −a ¯n z an n=1

Then the function f /Bk belongs to H p because it is analytic in D and |Bk (z)| > 1/2 near T. Since |f /Bk | = |f | on T, we have kf /Bk kp = kf kp . Hence 1 2π

Z

π

−π

|f (reiθ )| dθ 6 kf kp |Bk (reiθ )|

for every r < 1. By Fatou’s lemma we get Z π |f (reiθ )| 1 dθ 6 kf kp , 2π −π |B(reiθ )| which implies kf /Bkp 6 kf kp . The reverse holds because |f /B| > |f |.

✷

5.3.3 Exercise If f ∈ H p , then there exist functions g, h ∈ H 2p such that f = gh.

Some inequalities Isoperimetric inequality Let G be a domain with rectifiable boundary. Then there holds the (isoperimetric) inequality |G| 6 |∂G|2 /4π, and equality occurs iff G is a disk. This inequality can be rewritten as 2  Z Z π 1 ′ 2 ′ iθ |f (e )| dθ , |f | dA 6 2π −π D where f is a conformal mapping of D onto G. Here dA denotes the Lebesgue measure on D normalized so that the measure of D is 1. Thus the isoperimetric inequality is a consequence of the following theorem of Carleman. 5.3.4 Theorem If f ∈ H p , p > 0, then Z |f |2p dA 6 kf k2p p .

(5.14)

D

Equality occurs iff f (z) = c(1 − az)−2/p , where c and a are complex constants, |a| < 1. Proof. Let f ∈ H p . Then we can write f = Bg 2/p , where B is a Blaschke product and g is in H 2 and has no zeros in D. Then Z Z 2 2p |g 2 |2 dA and kf k2p |f | dA 6 p = kgk2 . D

D

5.3 Blaschke product

83

We have Z Since

∞ n 2 X X 2 2 An g dA = , where An = gb(k)b g (n − k) . n + 1 D n=0 k=0

An 6 (n + 1)

n X

k=0

we have

Z

D

|b g (k)|2 |b g (n − k)|2 ,

n ∞ X X 2 2 g dA 6 |b g (k)|2 |b g (n − k)|2 = kgk22 . n=0 k=0

This proves the inequality. If in (5.14) equality holds, then we see from the preceding that B = 1 and that there exists a sequence λn such that gb(k)b g (n − k) = λn for 0 6 k 6 n. This implies that g(z) = c(1 − az)−1 ; etc. ✷ 5.3.5 Exercise [62] If f ∈ H p , p > 0, and n = 2, 3, . . . , then

1/np  Z 6 kf kp . (n − 1) |f (z)|np (1 − |z|2 )n−2 dA(z) D

It is not known whether this holds for other values of n. p 5.3.6 Exercise Let q > p > 0 and r = p/q. If q/p is an integer and f ∈ H p , then  1/q Z π 1 iθ q |f (re )| dθ 6 kf kp . 2π −π Inequalities of Riesz/Fej´ er and Hilbert If g is a function analytic in D, then a special case of the Riesz/Zygmund theorem (see (6.5)) states that Z 1 |g ′ (r)| dr 6 πkg ′ k1 . −1

′

Replacing here g by f and using Riesz’ factorization we get the Riesz/Fejer inequality: Z 1 |f (r)|p dr 6 πkf kpp (f ∈ H p , p > 0). (5.15) −1

In particular, if p = 2 and f (z) = X

m,n>0

P∞

n=0

an z 2n (an > 0), then (5.15) yields

∞ X am an |an |2 . 6π m + n + (1/2) n=0

(5.16)

84

5 Classical Hardy spaces

This inequality, known as Hilbert’s inequality, can be deduced immediately from the equality Z π Z 1 f (eiθ )2 eiθ dθ, f (r)2 dr = i 0

−1

a consequence of Cauchy’s integral theorem. From (5.16) it follows that 1/2 1/2  X X ∞ ∞ X am bn 2 2 . |bn | |an | 6π m + n + (1/2) n=0 n=0

(5.17)

m,n>0

Hardy’s inequality From Hilbert’s inequality we can obtain a slightly improved version of Hardy’s inequality (5.13). Namely: 5.3.7 Theorem If f ∈ H 1 , then ∞ X

|fb(n)| 6 πkf k1 . n + (1/2) n=0

(5.18)

Proof. Let f ∈ H 1 and let f = Bg be the Riesz’ factorization of f . Then the functions F = Bg 1/2 and G = g 1/2 belong to H 2 and kf k1 = kF k22 = kGk22 . Let b ak = |Fb(k)| and bk = |G(k)|. Then we have n ∞ X X X 1 am bn |fb(n)| 6 . ak bn−k = n + (1/2) n + (1/2) m + n + (1/2) n=0 n=0 ∞ X

k=0

m,n>0

Now we use (5.17) to get ∞ X

|fb(n)| 6 πkF k2 kGk2 = πkf k1 . n + (1/2) n=0

✷

5.3.8 Remark In the case p = 2 the isoperimetric inequality (5.3.4) can be written as ∞ X |fb(n)|2 6 kf k21 . (5.19) n + 1 n=0

It is interesting to compare this inequality with (5.18). In general, convergence Pb |f (n)|/(n + 1), with f ∈ H(D), does not imply convergence of of the series P b |f (n)|2 /(n + 1). However, if f ∈ H 1 , then |fb(n)| 6 kf k1 , and therefore (5.18) implies a weak form of (5.19), namely ∞ X |fb(n)|2 6 πkf k21 . n + 1 n=0

On the other hand, (5.19) implies kf k21 − |f (0)|2 > (1/2)|f ′ (0)|2 , which cannot be deduced from (5.18).

5.4 Inner and outer functions

5.4

85

Inner and outer functions

Inner-outer factorization Riesz’ factorization theorem can be refined by introducing inner and outer functions. Suppose that a function f ∈ H p has no zeros in D. Then log |f | is harmonic and belongs to h1 , which follows from (5.9). Nevertheless inequality (5.7) may be  1 + z strict; this is the case if, e.g., f is the so called atomic function exp − . The 1−z atomic function satisfies the following: (a) |S(z)| 6 1 for z ∈ D;

(b) |S∗ (eiθ )| = 1 almost everywhere.

An analytic function satisfying (a) and (b) is called an inner function.

Singular inner functions If an inner function has no zeros, then it is called a singular inner function. The following theorem describes a connection between singular inner functions and singular measures. 5.4.1 Theorem A function S ∈ H(D) is a singular inner function iff there exists a nonnegative singular measure σ on T such that   Z ζ +z 1 ic dσ(ζ) , (5.20) S(z) = e exp − 2π T ζ − z where c is a real constant. Proof. If S is inner, then the function u = log |S| is negative and, by the Riesz/Herglotz theorem, there exists a nontrivial positive measure σ ∈ M (T) such that u = −P [σ]. The measure is singular because u(reiθ ) → 0, r → 1 (Corollary 3.3.5). Then, by passing to “analytic completion”, we get Z 1 ζ +z log S(z) = − dσ(ζ) + ic, 2π T ζ − z where c is a real constant, and this implies (5.20). The rest of the proof is simpler and we omit it. ✷ Outer functions

A function F is called an outer function if   Z 1 ζ +z ic log ψ(ζ) |dζ| , F (z) = e exp 2π T ζ − z

(5.21)

where ψ > 0 is a measurable function such that log ψ ∈ L1 (T). By the theorems of Fatou and Privalov/Plessner, F has radial limits and we have |F∗ (ζ)| = ψ(ζ), ζ ∈ T a.e. If f ∈ H p , then, in view of (5.7), we can define F by (5.21) with ψ = |f∗ |. The function ω = f /F is then inner and we have the factorization f = ωF . This leads to Smirnov’s factorization theorem.

86

5 Classical Hardy spaces

5.4.2 Theorem Every function f ∈ H p , p > 0, admits a representation f = BSF , where B is a Blaschke product, S is a singular inner function and F is an outer function. We have kf kp = kF kp , |f (z)| 6 |F (z)|, and |f∗ | = |F∗ |. We call F and I = BS the outer factor and the inner factor of f , respectively. The factorization f = IF is called the inner-outer factorization of f . This factorization is unique if we require, for instance, that I(0) is a positive real number. Exercises 5.4.3 A positive, nonconstant harmonic function u is equal to the Poisson integral of a singular measure iff there exists an inner function ω such that u(z) = Re

1 + ω(z) . 1 − ω(z)

5.4.4 If f is a nonconstant inner function, then there holds strict inequality in (5.7). On the other hand, if a function f ∈ H p is outer, then Z π 1 P (r, θ − t) log |f∗ (eit )| dt (0 6 r < 1). log |f (reiθ )| = 2π −π Conversely, if this equality holds for a fixed reiθ ∈ D, then f is outer. In particular, a function f ∈ H p is outer iff Z π 1 log |f∗ (eit )| dt. log |f (0)| = 2π −π 5.4.5 Every outer function can be represented as the ratio of two bounded outer functions. Consequently, every H p -function is the ratio of two bounded analytic functions. 5.4.6 If f ∈ H p and 1/f ∈ H p for some p > 0, then f is outer. By Theorem 4.4.2, f is in H p if Re f > 0. Since Re(1/f ) = f¯ /|f |2 , we see that f is outer if Re f > 0. Addendum: Riesz’ representation theorem The factorization theorems of Riesz and Smirnov are closely related to the Riesz’ representation theorem for subharmonic functions. We formulate this theorem following H¨ormander [30, §3.3]. 5.4.7 Theorem (a) A function u 6≡ −∞, subharmonic in D, can be represented in the form Z w−z 1 dµ(w), log u(z) = h(z) + (5.22) 2π D 1 − wz ¯

where h is a function harmonic in D and µ is a positive measure in D, iff the function I(r, u) (0 < r < 1) is bounded from above, which is equivalent to the requirement

5.4 Inner and outer functions

87

that u possesses a harmonic majorant in D. The function h and the measure µ are uniquely determined by u; h is the smallest harmonic majorant of u and µ is equal to the Riesz measure of u. (b) Let µ be a positive measure on D. If the integral in (5.22) converges for some z ∈ D, then Z (1 − |z|) dµ(z) < ∞. (5.23) D

Conversely, this condition implies that the integral in (5.22) defines a subharmonic function 6≡ −∞, the smallest harmonic majorant of which is ≡ 0. When specialized to the case u = log |f |, f ∈ H(D), this theorem yields the following extension of Theorem 5.3.2. 5.4.8 Theorem Let f ∈ H(D), f 6≡ 0, and Z π 1 log |f (reiθ )| dt 6 C 2π −π

(0 < r < 1).

(5.24)

If {an } is the sequence of zeros of f , then the Blaschke condition is satisfied and f = Bg, where B is the corresponding Blaschke product and g has no zeros in D. The smallest harmonic majorant of log |B| is ≡ 0. Further, if f admits a factorization f = f1 f2 with |f1 (z)| 6 1 and f2 (z) 6= 0 for z ∈ D, then (5.24) holds and |f1 (z)| 6 |B(z)|, |f2 (z)| > |g(z)|. Here we only note that if u = log |f |, f ∈ H(D), then condition (5.23) reduces P to the Blaschke condition because the Riesz measure of log |f | is equal to 2π δan . Beurling’s approximation theorem

This theorem can be viewed as a generalization of the fact that the set, Q, of all polynomials is dense in H p for 0 < p < ∞. 5.4.9 Theorem Let f ∈ H p , 0 < p < ∞. Then the closed linear span in H p of the set Qf = {qf : q ∈ Q} is equal to BSH p , where BS is the inner factor of f . In particular, if f is outer, then the set Qf is dense in H p . Proof. [30] Since |BS| 6 1, we see that it suffices to prove that QF is dense in H p . Proving this reduces to proving that Q ⊂ QF because Q is dense in H p . Now we see that the proof of the theorem reduces to the proof that 1 ∈ QF . Let   Z ζ +z 1 ic F (z) = e exp λ(ζ) |dζ| , 2π T ζ − z   Z 1 ζ +z ic Fs (z) = e exp λ(ζ) |dζ| . 2π λ(ζ) 1. 5.5.1 Proposition Let ω be an inner function. If φ and g are Borel measurable functions on T such that φ = g a.e., then φ ◦ ω∗ = g ◦ ω∗ a.e. Consequently, if φn are Borel functions on T such that φn → f a.e., then φn ◦ ω∗ → f ◦ ω∗ a.e. Proof. If ω(0) = a, then ω = ϕ ◦ (ϕ ◦ ω), where ϕ(z) = (a − z)/(1 − a ¯z) and ϕ ◦ ω(0) = 0. Therefore we can assume that ω(0) = 0. Let E = {eiθ ∈ T : φ(eiθ ) 6= g(eiθ )}. This is a set of measure zero, and we have to prove that the set F = {eiθ ∈ T : ω∗ (eiθ ) ∈ E}

S is of measure zero. To prove P this let ε > 0, let Eε = iθn In , whereiθIn ⊂ T are closed arcs such that E ⊂ Eε , n |In | < ε, and let Fε = {e ∈ T : ω∗ (e ) ∈ Eε } We shall prove that Z 2π Z 2π iθ Kn (eiθ ) dθ = |In |, (5.25) Kn (ω∗ (e )) dθ = 0

0

5.5 Composition with inner functions

89

where Kn is the characteristic function of In . This implies that |Fε | 6

XZ n

2π

Kn (ω∗ (eiθ )) dθ < ε,

0

and this implies that F is of measure zero, because F ⊂ Fε for all ε > 0. To prove (5.25), when n is fixed, we choose a sequence φj ∈ C(T) such that φj (eit ) tends to Kn (eit ) for every t and |φj (eit )| 6 1 for every t. Then φj (ω∗ (eit )) → Kn (ω∗ (eit )), as j → ∞, so we can apply the dominated convergence theorem to reduce the proof to the formula Z

2π

φ(ω∗ (eiθ )) dθ =

Z

2π

φ(eiθ ) dθ,

0

0

φ ∈ C(T).

(5.26)

Finally, this is reduced to the case where φ is a trigonometric polynomial. The details are left to the reader. ✷ The formula (5.26) extends to arbitrary φ ∈ L1 (T). 5.5.2 Theorem [85, 94] If φ is a Borel function of class L1 (T) and ω is an inner function with ω(0) = 0, then φ ◦ ω∗ ∈ L1 (T) and Z

0

2π

φ(ω∗ (eiθ )) dθ =

Z

2π

φ(eiθ ) dθ.

0

Proof. In order to reduce the proof to the case φ ∈ C(T), we can suppose that φ is a positive real function. The sequence min{φ(ζ), n} increases to φ(ζ) everywhere, so the proof reduces to the case where φ is bounded. If φ is bounded, then we choose a bounded sequence φn ∈ C(T) such that φn → φ a.e.; by Proposition 5.5.1, we have φn ◦ ω∗ → φ ◦ ω∗ a.e. The result follows. ✷ 5.5.3 Theorem If φ is a Borel function of class L1 (T) and ω is an inner function, then P [φ ◦ ω∗ ] = P [φ] ◦ ω. Proof. It suffices to consider the case where φ ∈ C(T). Then the functions P [φ ◦ ω∗ ] and P [φ] ◦ ω are harmonic and bounded so it suffices to prove that their boundary functions coincide almost everywhere. Since P [φ] is continuous on the closed disk, we have limr→1 P [φ](ω(reiθ )) = φ(ω∗ (eiθ )) a.e. On the other hand, limr→1 P [φ ◦ ω](reiθ ) = (φ ◦ ω∗ )(eiθ ) a.e., and this completes the proof. ✷ 5.5.4 Corollary If ω and I are inner functions, then so is the composition I ◦ ω, and (I ◦ ω)∗ = I∗ ◦ ω∗ a.e. If in addition I is singular, then so is I ◦ ω. Proof. This follows from the relations: P [I∗ ◦ ω∗ ] = P [I∗ ] ◦ ω = I ◦ ω and P [(I ◦ ω)∗ ] = I ◦ ω. ✷

90

5 Classical Hardy spaces

Stephenson’s theorems Combining the above results one easily proves the following. 5.5.5 Theorem If f ∈ H p , p > 0, and ω is inner with ω(0) = 0, then f ◦ ω ∈ H p , (f ◦ ω)∗ = f∗ ◦ ω∗ , and kf ◦ ωkp = kf kp . If f = IF is the inner-outer factorization of f , then f ◦ ω = (I ◦ ω)(F ◦ ω) is the inner-outer factorization of f ◦ ω. We conclude this section by proving the implication f ◦ ω ∈ H p =⇒ f ∈ H p . 5.5.6 Theorem If f ∈ H(D) and ω is an inner function, then f ◦ ω ∈ H p implies f ∈ H p. Proof. Assume that ω(0) = 0. Let f ◦ ω ∈ H p , p > 0, let u = |f |p and v = |f ◦ ω|p , and let h = P [v∗ ]. We know that v 6 h; see (5.8). Let D = rD, where r, 0 < r < 1, is fixed. Then u 6 M on D for some constant M < ∞. Put Ω = ω −1 (D). For 0 < ρ < 1, let Eρ = {ζ ∈ T : ω(ρζ) ∈ D}. Since 1 − |ω(ρζ)| → 0 a.e. as ρ → 1, we see, using Egorov’s theorem, that limρ→1 |Eρ | = 0. Hence we can choose ρ so that the following is true: If ϕ is the bounded harmonic function in the disk ρD whose values are M on ρEρ and 0 on the rest of ρEρ , then ϕ(0) < 1.

(5.27)

Since u is subharmonic and continuous in D, there exists a function u1 ∈ C(D) harmonic in D such that u 6 u1 6 M and u1 = u at every point of ∂D. By the mean value property, Z π Z π 1 1 iθ u1 (0) = u(re ) dθ = |f (reiθ )|p dθ. (5.28) 2π −π 2π −π Since u ◦ ω 6 h in D, we have that u1 ◦ ω 6 h on ∂Ω ∩ ρD. Consider the function u1 ◦ ω − h on the closure of the set Ωρ = Ω ∩ ρD. At boundary points of Ωρ that lie in ρD we have u1 ◦ω −h 6 0. The other boundary points of Ωρ lie in ρEρ , and there u1 ◦ ω − h 6 u1 ◦ ω 6 M. Thus u1 ◦ ω − h 6 ϕ on ∂Ωρ . Since these functions are harmonic in Ωρ and Ωρ ∋ 0, it follows from (5.27) that u1 (0) = u1 (ω(0)) 6 h(0)+1. ✷ Now the desired result follows from (5.28). Approximation by inner functions Let ω be an inner function with ω(0) = 0. If f ∈ H 2 , then, by Rogosinski’s theorem (see 4.4.7) and Stephenson’s theorem, we have ∞ X

k=n

|fb(k)|2 >

∞ X

k=n

|Fb(k)|2 ,

where f = F ◦ ω. This fact can be expressed in terms of best approximation.

5.5 Composition with inner functions

91

Let Qn , n > 0, denote the set of all holomorphic polynomials of degree at most n. For a function f ∈ H p , let En (f )p = inf g∈Qn kf −gkp . Then the above inequality can be stated as En (f ◦ ω)2 > En (f )2 , f ∈ H 2 . It is interesting that this extends to the case 1 6 p 6 ∞. 5.5.7 Theorem [64] Let 1 6 p 6 ∞, f ∈ H p , and let ω be an inner function with ω(0) = 0. Then En (f ◦ ω)p > En (f )p . Proof. Let (f, h) =

1 2π

Z

π

f (eiθ )h(eiθ ) dθ.

−π

We identify H p with H p (T). From the general theory of best approximation in Banach spaces we know that En (f )p = sup{ |(f, h)| : h ∈ Lq (T), (Qn , h) = 0, khkq 6 1} (1/p + 1/q = 1), where (Qn , h) = 0 means that (g, h) = 0 for every g ∈ Qn . Now we apply this formula to f ◦ ω and use the following facts: (a) If h ∈ Lq (T), then h ◦ ω ∈ Lq (T) and kh ◦ ωkq = 1, and (b) if (h, Qn ) = 0, then h ◦ ω = 0. Fact (a) follows from Theorem 5.5.2, while the proof of (b) is straightforward—it is enough to observe that (h, Qn ) = 0 iff b h(j) = 0 for 0 6 j 6 n. We get En (f ◦ ω)p > sup{ |(f ◦ ω, h ◦ ω)| : h ∈ Lq (T), (Qn , h) = 0, khkq 6 1}.

This concludes the proof because |(f ◦ω, h◦ω)| = |(f, h)|, by Theorem 5.5.2.

✷

5.5.8 Remark If we change the notation and denote by En (f )p the best Lp approximation of f ∈ Lp (T) by trigonometric polynomials of degree 6 n, then Theorem 5.5.7 remains valid. However, the above method heavily depends on the Hahn/Banach theorem and cannot be applied to the case p < 1. It would be interesting to study this case.

6

Conjugate functions

The main theorems of this chapter are the Privalov/Plessner theorem on the existence of radial limits of conjugate functions and the existence of the Hilbert operator (Theorem 6.1.1), and the Riesz theorem that if u is in hp , 1 < p < ∞, then so is its harmonic conjugate (Section 6.2). The rest of the chapter is devoted to some related results. The equality Lp (T) = H p (T) + H p (T), 0 < p < 1, due to Aleksandrov, is in Section 6.4. Section 6.5 contains a theorem on strong convergence in H 1 . In Section 6.6 we characterize harmonic quasiconformal homeomorphisms of the unit disk via the Hilbert transformation of the derivative of the boundary function.

6.1

Harmonic conjugates

To each f ∈ h(D) there corresponds the harmonic conjugate fe ∈ h(D), fe(reiθ ) = −i

∞ X

(sign n) fb(n)r|n| einθ .

n=−∞

If f is real-valued, then fe is uniquely determined by the conditions: (a) fe is realvalued, (b) f + ife is analytic, and (c) fe(0) = 0. For an arbitrary f ∈ h(D) there holds e fe = −f + f (0).

(6.1)

gf = Im(f − f (0)). And if f is analytic, then fe = −i(f − f (0)), and Re 1 The function conjugate to P [φ], φ ∈ L (T), equals Z π Z π   1 1 it iθ e e P (r, θ − t)φ(e ) dt = Pe(r, t) φ(θ − t) − φ(θ + t) dt, P [φ](re ) = 2π −π 2π 0

where we write φ(x) instead of φ(eix ). Here Pe denotes the conjugate Poisson kernel, 1+z 2z = Im , Pe(z) = Im 1−z 1−z

i.e., Pe(r, θ) = Pe(reiθ ) =

This kernel does not belong to h1 . The kernels P and Pe are connected by the formula Pe(r, θ) −

1 − r P (r, θ) 1 =− . tan(θ/2) tan(θ/2) 1 + r 92

2r sin θ . 1 + r2 − 2r cos θ

(6.2)

6.1 Harmonic conjugates

93

Note that 1/tan(θ/2) = Pe(1, θ). It is known that Pe[φ] need not be in h1 (see6.1.4) but the Kolmogorov/Smirnov theorem 4.4.2 says that there holds the implication f ∈ h1 =⇒ fe ∈ hp , 0 < p < 1.(∗) The Privalov/Plessner theorem

6.1.1 Theorem If φ ∈ L1 (T), then there exist and are equal the following limits (a.e.): Z 1 π φ(θ − t) − φ(θ + t) dt. lim− Pe[φ](reiθ ) and lim+ 2 tan(t/2) r→1 ε→0 π ε The existence of limr→1− Pe[φ](reiθ ) is contained in Corollary 3.3.9. This theorem guarantees the existence of the improper integral Z 1 π φ(θ − t) − φ(θ + t) iθ e dt. φ(e ) = π 0+ 2 tan(t/2)

There exists such a function φ ∈ C(T) that this integral converges absolutely for no θ. It is even more interesting that there exists a function φ ∈ C(T) such that the improper integral Z π φ(θ + t) − φ(θ) dt 2 tan(t/2) + 0 diverges for every θ (see [100, p. 133–4]).

The Hilbert operator The function φe is said to be conjugate with φ and the operator H taking φ to φe is called the Hilbert operator.(†) The Hilbert operator maps L1 into Lp , for every p < 1, but not into L1 , so in the general case the Poisson integral of φe has no sense. However, as we will prove later on (see Theorem 6.1.3), e = Pe[φ]. if φe ∈ L1 , then P [φ] Proof of Theorem 6.1.1. It suffices to prove the relation   Z π φ(θ − t) − φ(θ + t) 1 iθ e dt = 0, lim P [φ](re ) − 2π 1−r tan(t/2) r→1−

(6.3)

under the hypothesis that θ is a Lebesgue point of φ. We write the difference under limr→1− in (6.3) as I1 (r) + I2 (r), where Z 1−r   1 I1 (r) = Pe(r, t) φ(θ − t) − φ(θ + t) dt. 2π 0 Since |Pe(r, t)| 6 2/(1 − r), |t| 6 1 − r, we have Z 1−r 1 φ(θ − t) − φ(θ + t) dt → 0 |I1 (r)| 6 π(1 − r) 0 (∗) See

also Kolmogorov’s theorem 7.1.10. 2 tan(t/2) is replaced by t.

(†) Usually

(r → 1),

94

6 Conjugate functions

because θ is a Lebesgue point. In the case of the integral  Z π    1 1 e φ(θ − t) − φ(θ + t) dt I2 (r) = P (r, t) − 2π 1−r tan(t/2)

we use the formula (6.2); it follows that 1 Pe(r, t) − 6 const. P (r, t) tan(t/2) Thus

|I2 (r)| 6 C

1 2π

Z

π

−π

(1 − r < |t| < π).

P (r, t) φ(θ − t) − φ(θ + t) dt.

Now the hypothesis that θ is a Lebesgue point and the following lemma imply that I2 (r) → 0 (r → 1), and this completes the proof. ✷ 6.1.2 Lemma If θ is a Lebesgue point of a function ψ ∈ L1 , then Z π 1 P (r, t)|ψ(θ + t) − ψ(θ)| dt = 0. lim sup r→1− 2π −π Proof. This can be deduced from Proposition 3.3.1 by taking Z t |ψ(θ + x) − ψ(θ)| dx. ✷ γ(t) = 0

The Poisson integral of the conjugate function e = Pe[φ]. 6.1.3 Theorem If φ ∈ L1 and φe ∈ L1 , then P [φ]

Proof. Assume that φ is real-valued. By the Kolmogorov/Smirnov theorem, the function f = P [φ] + iPe[φ] belongs to H p for p < 1. Now Smirnov’s theorem 5.1.12 tells us that f ∈ H 1 , and hence f = P [f∗ ], by Theorem 5.2.1. Finally, e by the theorems of Fatou and Privalov/Plessner, we see that since f∗ = φ + iφ, and the result follows.

e P [φ] + iPe[φ] = P [f∗ ] = P [φ] + iP [φ], ✷

Miscellaneous 6.1.4 P∞ If {ak } is a convex sequence tending to 0, then the sum of the series 1 n=1 an cos nθ is positive for every θ ∈ (−π, π), and its sum belongs to L (T) (see [42, Theorem 4.1]). In particular, the function φ(eiθ ) =

∞ X

(log n)−1 cos nθ

n=2

is in L1 , while the function conjugate to φ is equal to not in L1 .

P∞

−1 n=2 (log n)

sin nθ and is

6.1 Harmonic conjugates

95

6.1.5 Using (6.1) one can prove the following: If φe ∈ L1 , then Z π 1 ee φ = −φ + φ(eiθ ) dθ. 2π −π

The Privalov/Plessner theorem can be stated in the following form:

6.1.6 Theorem Let φ ∈ L1 (T) and let Φ(θ) be the indefinite integral of the function θ 7→ φ(eiθ ). Then the improper integral Z 1 π Φ(θ + t) + Φ(θ − t) − 2Φ(θ) − dt π 0+ 4 sin2 (t/2) e exists for all θ and is equal to φ(θ) almost everywhere. A more general variant states:

Let γ ∈ BV [−π, π] and let γ(t + 2π) − γ(t) = const. Then the integral Z 1 π γ(θ + t) + γ(θ − t) − 2γ(θ) dt − π 0+ 4 sin2 (t/2) and the limit

g lim P S[γ](reiθ )

r→1−

exist and are equal almost everywhere. See Zygmund [100, Ch. III §§7-8, IV §3 and VII §1]. The Riesz/Zygmund inequality As we have seen, if g ∈ h1 , then the conjugate function ge need not belong to h1 . A result of Riesz and Zygmund [100, Ch. IV, (6.28)] states that if g ∈ h1 , then Z

ge(reit ) dr 6 πkgk1 . r −1

In other words:

1

(6.4)

6.1.7 Theorem If g ∈ h(D) and (∂g/∂θ) ∈ h1 , then Z



∂g ∂g

(reit ) dr 6 π . ∂r ∂θ 1 −1 1

(6.5)

Proof. We can assume that g is harmonic in a neighborhood of D. The function v = r∂g/∂r is conjugate to the function u = ∂g/∂θ and therefore Z π   1 Im F (reit ) u(ei(θ−t) ) dt, (6.6) v(reiθ ) = 2π −π

96

6 Conjugate functions

where F (z) = 2z/(1 − z). Now (6.5) can be deduced from (6.6) and the equation Z

1

−1

|r−1 Im F (reit )| dr = π

(0 < |t| < π),

(6.7)

by using Fubini’s theorem. In order to prove (6.7) let 0 < t < π, and apply Cauchy’s integral theorem to the function F (z)/z on the semidisk {reiθ : 0 6 r 6 1, t 6 θ 6 t + π} (which does not contain the point z = 1). We get Z

F (reit ) dr = −i r

Z

Im F (reit ) dr = − r

Z

1

−1

Hence

Z

1

−1

t+π

F (eiθ ) dθ.

t t+π

Re F (eiθ ) dθ.

t

Now (6.7) follows from Re F (eiθ ) = −1 and Im F (reit )/r > 0.

✷

Geometric interpretation Let γ be a function continuous and of bounded variation on [−π, π], let γ(π) = γ(−π), and f (reiθ ) = P [γ]. The function f can be treated as a continuous mapping of the closed disk to the complex plane. 6.1.8 Corollary If f is a homeomorphism, then length of f ([−1, 1]) 6 (1/2) × length of ∂f (D). The hypothesis that the curve z = γ(t), π 6 t 6 π, is a Jordan curve is not sufficient for f to be a homeomorphism. A sufficient condition is described by Choquet’s theorem [13]: 6.1.9 Theorem A sufficient condition for f to be a homeomorphism is that the curve z = γ(t) (|t| 6 π) is a convex Jordan curve.

6.2

Riesz projection theorem

The operator R+ acting from h(D) to H(D) according to the rule (R+ u)(z) = P ∞ b(n)z n is called the Riesz projector. n=0 u The projection theorem

It is a direct consequence of Parseval’s theorem that R+ acts as an orthogonal projection from h2 onto H 2 . This fact was generalized by M. Riesz in the following way. 6.2.1 Theorem If 1 < p < ∞, then R+ acts as a bounded projection from hp onto H p .

6.2 Riesz projection theorem

By the Kolmogorov/Smirnov theorem, we can treat R+ as an operator from L1 (T) to H p (T) = {f∗ : f ∈ H p } ⊂ L0 (T), for p < 1. From the projection theorem and Theorem 3.4.2 it follows that for every φ ∈ Lp (T) (1 < p < ∞) there exists b b b a unique function ψ ∈ Lp (T) such that ψ(n) = φ(n) for n > 0 and ψ(n) = 0 p for n < 0. This enables us to treat R+ as an operator from L (T) to H p (T), 1 < p < ∞. However, the Riesz projection does not map L1 into H 1 (T) (see 6.1.4); furthermore, H 1 is not complemented in L1 , i.e., there is no bounded projection from L1 to H 1 (see [87]).(‡) The conjugate functions theorem Since the Riesz projector is connected with conjugate function in a simple way, namely R+ u = u(0) + (u + ie u)/2, the Riesz theorem can be stated as follows: 6.2.2 Theorem If u ∈ hp , p > 1, then u e ∈ hp and there exists a constant Cp such that ke ukp 6 Cp kukp . In view of the connection between conjugate functions and the Hilbert operator (Privalov/Plessner theorem, 6.1.1), we have:

6.2.3 Theorem The Hilbert operator maps Lp (T) to Lp (T) for 1 < p < ∞. In the case p = 2, Theorems 6.2.2 and 6.2.1 follow from Parseval’s formula; we have ke uk22 = kuk22 − |u(0)|2 . If a proof is known either for 1 < p < 2 or for p > 2, then the general case can be treated by duality. Let us mention four “short proofs.” (i) The operator R+ : L1 (T) 7→ L0 (T) is of strong type (2, 2) and, by Kolmogorov’s theorem (Theorem 7.1.10), of weak type (1, 1). Therefore we can apply Marcinkiewicz’s theorem. (ii) If 1 < p < 2, we can use Theorem 2.6.2 and the existence of the radial limits of u e.

(iii) If 1 < p < 2, then, as noted after Theorem 7.2.1, we can use Theorems 7.2.1 and 7.1.2.

(iv) If p = 2n , for some positive integer n, then we can easily deduce the validity of Theorem 6.2.2 from the case p = 2; see the proof of Theorem 6.2.6. Then, for arbitrary p > 2, we can apply the Riesz/Thorin theorem. Here we present an elementary proof, due to P. Stein, based on the Hardy/Stein identities. (‡) If every subspace of a Banach space X is complemented in X, then X is isomorphic to a Hilbert space [51].

97

98

6 Conjugate functions

Hardy/Stein identities There is a way to express the H p -norm as a double integral. For example, from Green’s formula (3.3) and the formula ∆(|f |2 ) = 4|f ′ |2 it follows that Z 1 2 2 dA(z). (6.8) kf k2 = |f (0)| + 2 |f ′ (z)|2 log |z| D Concerning other values of p, we consider only the cases that are sufficient to prove Riesz’ projection theorem. 6.2.4 Lemma Let 0 < p < ∞. A function f ∈ H(D) belongs to H p if and only if Z 1 p2 dA < ∞. HSp (f ) := |f (0)|p + |f |p−2 |f ′ |2 log 2 D |z| Moreover we have kf kpp = HSp (f ). Proof. If f has no zeroes in D, then it suffices to apply (6.8) to the function f p/2 . If p > 2, then the function |f |p is of class C 2 so we can apply Green’s formula. In the general case one applies Green’s formula to the functions (|f |2 + ε)p/2 , ε > 0, and then let ε tend to 0 (see also 4.5.6). ✷ The formula kf kpp = HSp (f ) is known as the Hardy/Stein identity. There holds an analogous formula for real-valued harmonic functions. We only need the case of positive functions. 6.2.5 Lemma Let f ∈ H(D) and let u = Re f belong to hp , 1 < p < ∞. Then Z p(p − 1) 1 kukpp = |u(0)|p + dA. (6.9) up−2 |f ′ |2 log 2 |z| D Proof. In the case where u > 0 this reduces to Lemma 3.5.4. In the general case one considers the functions (u2 + ε)p/2 . ✷ Proof of Riesz’ theorems We shall prove Theorem 6.2.2. We may suppose that u is real-valued, and then the theorem can be stated as follows. 6.2.6 Theorem Let f be analytic in D, and let 1 < p < ∞. If Re f ∈ hp , then f ∈ H p and there holds the inequality kf kpp 6 Cp (k Re f kpp + |f (0)|p ). If f is a conformal mapping of the disk onto the domain G = {z : 0 < Re z < 1}, then Re f ∈ h∞ but f is not in H ∞ ; therefore, the theorem does not hold for p = ∞. For the case p = 1 see 6.1.4. Proof of Theorem 6.2.6. Consider first the case 1 < p 6 2. Let u = Re f ∈ hp . In view of Lemma 3.5.3, we may suppose that u > 0 and, as in the proof of

6.3 Applications of the projection theorem

99

Lemma 5.1.7, that f is analytic in a neighborhood of D. Then from Lemmas 6.2.4 and 6.2.5, together with the inequality up−2 > |f |p−2 , it follows that kf kpp − |f (0)|p 6

p (kukpp − |u(0)|p ), p−1

which gives the desired result for 1 < p 6 2. Let 2 < p 6 4 and let f be analytic in a neighborhood D. Then the function g = −if 2 belongs to H q , q = p/2, and we have Re g = 2uv, where v = Im f , and therefore, by the preceding case, kf kpp = kgkqq 6 Cp k2uvkqq . On the other hand, Cauchy/Schwarz inequality gives kuvkqq 6 (kukp kvkp )p/2 . Hence kf kpp 6 p/2

p/2

p/2

Cp 2q kukp kf kp . Since kf kp is finite, we can divide both sides by kf kp , and this yields the result for 2 6 p 6 4; etc. ✷ 6.2.7 Remark If 0 < p < 1, then (6.9) does not hold, but we still have Z p(1 − p) 1 dA = u(0)p − Mpp (1, u) 6 u(0)p , up−2 |f ′ |2 log 2 |z| D

provided u is positive and continuous on the closed disk. When combined with Lemma 6.2.4, this yields another proof of Theorem 4.4.2.

6.3

Applications of the projection theorem

The projection theorem has many important applications. For example, the trigonometric system is a Shauder basis in Lp (T) for 1 < p < ∞; in other words, the system of the functions r|n| einθ is a Shauder basis in hp .(§) n P ikθ b φ(k)e , where 6.3.1 Theorem Let φ ∈ Lp (T), 1 < p < ∞, and φm,n (eiθ ) = k=m m and n are integers, m < n. Then

kφm,n kp 6 Cp kφkp kφ − φm,n kp → 0 as n → ∞, m → −∞.

(6.10) (6.11)

Proof. Let ek (eiθ ) = eikθ . Then φm,n = em R+ (e−m φ) − en R+ (e−n φ). From this and Theorem 6.2.1 we obtain (6.10), and from (6.10) and the Weierstrass ✷ approximation theorem we obtain (6.11). p Now we can determine the dual of H for 1 < p < ∞. 6.3.2 Theorem If 1 < p < ∞, then the dual of H p is isomorphic to H q (1/p + 1/q = 1) with respect to the bilinear form Z π ∞ X 1 fb(n)b g (n)r2n . f (re−iθ )g(reiθ ) dθ = lim (f, g) = lim r→1− r→1− 2π −π n=0

(§) However, there are spaces, e.g., Bergman, in which this system is not a basis although there holds the analogue of the projection theorem.

100

6 Conjugate functions

Proof. Let Λ be a bounded linear functional on H p and Λ1 be the Hahn/Banach extension to hp . By the projection theorem and Theorem 3.4.2, there exists a function g ∈ hq such that kΛ1 k = kg1 kq and Λ1 f = (f, g1 ) for f ∈ hp . Hence, Λf = (f, g1 ) for f ∈ H p . The function g = R+ g1 belongs to hq and we have (f, g) = (f, g1 ) for f ∈ H p (because f is analytic), and this proves the inclusion (H p )∗ ⊂ H q . The reverse inclusion is proved by using H¨older’s inequality. ✷ 6.3.3 Exercise (isomorphism Lp to H p ) If 1 < p < ∞, then the formula (T u)(z) =

∞ X

n=0 p

u b(n)z 2n + p

defines an isomorphism of h onto H .

∞ X

n=1

u b(−n)z 2n−1

6.3.4 Exercise (Parseval’s formula) If f ∈ Lp (T) and g ∈ Lq (T), where 1/p + 1/q = 1 and 1 < p < ∞, then the series ∞ X

n=−∞

g (n) r|n| einθ fb(n)b

converges uniformly in D, and there holds Parseval’s formula: Z π ∞ X 1 g (n). f (eiθ )g(eiθ ) dθ = fb(n)b 2π −π n=−∞

6.4

Aleksandrov’s theorem

Relation (5.5) shows that Lp (T) contains an isometric copy of H p (p > 0); denote this subspace by H p (T). Thus H p (T) = {f∗ : f ∈ H p }. If p > 1, then H p (T) can be described in the following way: b H p (T) = {φ ∈ Lp (T) : φ(−n) = 0 for n > 1}.

In the case p < 1, this fact does not hold, simply because the Fourier coefficients are not defined; then H p (T) is equal to the Lp -closure of b T+ = {φ ∈ T : φ(−n) = 0 for n > 1},

where T is the set of all trigonometric polynomials. Let H p (T) = {φ : φ ∈ H p (T)}. One of consequences of the projection theorem is that Lp (T) = H p (T)+H p (T), 1 < p < ∞. This fact was extended by Aleksandrov [4, 3] to the case p < 1. However, in that case, the decomposition is not unique (up to an additive constant) because the intersection H p ∩ H p is equal to the linear span of the set of the functions ga (ζ) = 1/(1 − aζ) (a ∈ T, ζ ∈ T) (see [4, 3]). 6.4.1 Theorem (Aleksandrov) If f ∈ Lp (T), p < 1, then there are functions f1 ∈ H p (T), f2 ∈ H p (T), such that f = f1 + f2 and kf1 kp + kf2 kp 6 Cp kf kp .

6.5 Strong convergence in H 1

101

Proof. Let X denote the direct sum of the spaces H p and H p . Consider the operator T : X 7→ Lp , T (f1 , f2 ) = f1 + f2 . For every trigonometric polynomial f , kf kp 6 1, we will find (f1 , f2 ) so that f = T (f1 , f2 ), where k(f1 , f2 )k 6 Cp (and Cp depends only P of p) and then the result will follow from Theorem 1.3.1. Let f = |k|6n ak eikθ and γ(ζ) = ζ n . Then γf and γ f¯ belong to H p . Put

ϕ = γ/(γ − 1). Then ϕ ∈ H p ∩ H p and ϕ + ϕ¯ = 1. Now let ϕt (ζ) = ϕ(ζeit ), ¯ t . Routine t ∈ R, and gt = f ϕt , ht = f¯ϕt . Then gt and ht are H p and f = gt + h calculation shows that Z π 1 kgt kpp dt = kϕkpp kf kpp , 2π −π

which means that there exists t such that kgt kp 6 kϕkp . For this value of t we ¯ t , and this have kht kpp 6 kf kpp + kgt kpp 6 1 + kϕkpp . Finally, we take f1 = gt , f2 = h completes the proof. ✷ Kalton’s theorem Let B p denote the space of functions f ∈ H(D) such that kf kBp :=



|f (0)| +

Z

′

D

p

p−1

|f (z)| (1 − |z|)

1/p < ∞. dA(z)

By the Littlewood/Paley theorem, we have B p ⊂ H p for 1 < p < 2. This inclusion remains valid for p 6 1, which follows from the inequality Mpp (rn+1 , f ) − Mpp (rn , f ) 6 Cp 2−np Mpp (rn+1 , f ′ ),

rn = 1 − 2−n ,

see Proposition 7.1.6. Therefore the following result of Kalton [37] improves Theorem 6.4.1. 6.4.2 Theorem If f ∈ Lp (T), p < 1, then there are functions g ∈ B p , h ∈ B p , such that f = g∗ + h∗ and kgkBp + khkBp 6 Cp kf kp .

6.5

Strong convergence in H 1

For a function f analytic in D let Pn f =

n 1 X 1 sj f, An j=0 j + 1

where

An =

n X j=0

1 j+1

(n = 0, 1, 2, . . .)

and sj f are the partial sums of the Taylor series of f . It is well known that ksn f k 6 C An kf k and that An is “best possible”. A direct consequence is that n

1 X 1 ksj f k 6 Ckf k log n log n j=0 j + 1

(n > 2).

(6.12)

102

6 Conjugate functions

where C is an absolute constant. It turns out, however, that there holds the stronger inequality n 1 X 1 ksj f k 6 Ckf k (f ∈ H 1 , n > 2). (6.13) log n j=0 j + 1 Moreover, we have the following characterization of the space H 1 . 6.5.1 Theorem [92, 76] For a function f analytic in D the following assertions are equivalent: f ∈ H1 ; n 1 X 1 sup ksj f k < ∞ ; j+1 n An j=0 sup kPn f k < ∞.

(6.14) (6.15)

n

Remark. It follows from the proof that the quantities occurring in (6.14) and (6.15) are “proportional” to the original norm in H 1 ; in particular there holds (6.13). Since the polynomials are dense in H 1 , we have the following consequence: 6.5.2 Theorem If f ∈ H 1 , then lim n

n 1 X 1 kf − sj f k = 0 An j=0 j + 1

and, consequently, n 1 X 1 ksj f k = kf k. lim n An j+1 j=0

6.5.3 Corollary [76] If f ∈ H 1 , then lim inf kf − sn f k = 0. n→∞

1 There are functions such that limn kφ − sn φk = ∞; such an example is P∞ φ ∈ L −1/2 iθ given by φ(e ) = j=2 (log j) cos jθ. Since the sequence (log j)−1/2 is convex, the function belongs to L1 (see 6.1.4). Furthermore, one can show that kf −sn f k > c(log n)1/2 , c = const. > 0. We omit the details. By means of Fatou’s lemma, from Corollary 6.5.3 we obtain:

6.5.4 Corollary [76] If φ ∈ H 1 (T), then lim inf |φ(eiθ ) − sn φ(eiθ )| = 0 a.e. n→∞

On the other hand, there exists a function φ ∈ H 1 (T) whose Fourier series diverges almost everywhere (see [100, Ch. VIII, theorem (3.5)]).

6.5 Strong convergence in H 1

103

Konyagin’s theorem The above corollary does not extend to L1 ; Konyagin [45] proved the following improvement of Kolmogorov’s theorem: If {ψ(m)} is a sequence of positive numbers such that √ √ ψ(m) = o( ln m/ ln ln m) as m → ∞, then there exists a function φ ∈ L1 (T) such that lim sup sm φ(eiθ )/ψ(m) = ∞ for all θ ∈ T. m→∞

Proof of Theorem 6.5.1 It is obvious that (6.14) implies (6.15). To prove that the condition f ∈ H 1 implies (6.14) let f ∈ H 1 and for fixed n > 2 and w ∈ D define the function g ∈ H 1 by g(z) = (1 − rz)−1 f (rwz) (|z| 6 1), P∞ j j where r = 1 − 1/n. We have g(z) = j=0 sj f (w)r z . Applying Hardy’s inequality (Theorem 5.3.7) we get ∞ X j=0

∞

X 1 1 |sj f (w)|rj = |b g (j)| 6 πkgk. j+1 j+1 j=0

Since rj = (1 − 1/n)j > c for 0 6 j 6 n, where c > 0 is an absolute constant, we have Z 2π n X 1 |sj f (w)| 6 (π/c)kgk = (1/2c) |1 − reit |−1 |f (rweit )|dt. j + 1 0 j=0 Integrating this inequality over the circle |w| = 1 we find Z 2π n X 1 |1 − reit |−1 dt, ksj f k 6 (1/2c)kf k j + 1 0 j=0 where we have used Fubini’s theorem. Finally, using the estimate Z 2π 1 |1 − reit |−1 dt 6 C log = C log n, 1−r 0 we see that (6.13) holds and therefore that (6.14) is implied by f ∈ H 1 . Let f be analytic in D. From the uniform convergence of sn f on compact sets it follows that Pn f 7→ f uniformly on compact subsets of D. Assuming that kPn f k 6 1 for each n, we have M1 (r, Pn f ) 6 1 for all n and r < 1. This implies, via the uniform convergence of Pn f on the circles |z| = r, that M1 (r, f ) 6 1 for every r < 1, which means that kf k 6 1. Thus we have proved that (6.15) implies f ∈ H 1 , and this completes the proof. ✷

104

6 Conjugate functions

Remarks 6.5.5 Inequality (6.12) is optimal in L1 in the sense that log n cannot be replaced by any ψ(n) (independent of f ) such that ψ(n) = o(log n). To see this one takes f to be the Poisson kernel, then let r tend to 1 and use the norm estimate for the Dirichlet kernel. 6.5.6 Using Fejer’s theorem one shows, by summation by parts, that if f ∈ h1 , then supn kPn f k < ∞, where Pn is extended to harmonic functions in the obvious way. Conversely, if f is harmonic in D and supn kPn f k < ∞, then f ∈ h1 .

6.6

Quasiconformal harmonic homeomorphisms(§)

Throughout this section we denote by ϕ a continuous increasing function on R such that ϕ(t + 2π) − ϕ(t) ≡ 2π, so that the function γ(t) = eiϕ(t) is 2π-periodic and continuous, and of bounded variation on [0, 2π]. We consider the harmonic mapping f defined on D = {z : |z| < 1} by f (z) =

1 2π

Z

π

−π

P (r, θ − t)γ(t) dt

(z = reiθ ).

(6.16)

By Choquet’s theorem (Theorem 6.1.9), f is a homeomorphism of D onto D. Conversely, every orientation-preserving homeomorhism f : D 7→ D, harmonic in D, can be represented in the form (6.16).(¶) A consequence of Choquet’s theorem and a result of Lewy [48] is that the Jacobian of f is strictly positive in D, i.e., ¯ (z)|2 > 0 Jf (z) = |∂f (z)|2 − |∂f

(z ∈ D).

(6.17)

Being harmonic, the mapping f can be represented as f (z) = h(z) + g(z), g(0) = 0, where h and g are analytic in D and uniquely determined by f . We can rewrite (6.17) as g ′ (z) (6.18) ′ < 1 (z ∈ D). h (z)

Characterization theorem

We characterize those ϕ for which f is quasiconformal, i.e., for which (6.18) can be improved to g ′ (z) k = sup ′ < 1. (6.19) z∈D h (z) (§) This

section is almost identical to the paper [80]. properties of harmonic homeomorphisms are described in [14].

(¶) Various

6.6 Quasiconformal harmonic homeomorphisms

105

6.6.1 Theorem The mapping f is quasiconformal iff the function ϕ is bi-Lipschitz and the Hilbert transformation of ϕ′ is essentially bounded on R. In other words, f is quasiconformal iff ϕ is absolutely continuous and satisfies the conditions:

Z ess sup θ∈R

π

+0

ess inf ϕ′ > 0, ess sup ϕ′ < ∞,

ϕ′ (θ + t) − ϕ′ (θ − t) dt < ∞. t

(6.20) (6.21) (6.22)

The proof that the three conditions are sufficient is short; we simply compute the radial limits of the modulus of the bounded analytic function g ′ /h′ and apply the maximum modulus principle. The necessity proof is more complicated and depends on Mori’s theorem in theory of quasiconformal mappings (cf. Ahlfors [2]), which states that if Φ is a quasiconformal homeomorphism of D, then |Φ(z1 ) − Φ(z2 )| 6 C|z1 − z2 |α where

1−k , α= 1+k

(z1 , z2 ∈ D),

(6.23)

¯ ∂Φ(z) k = sup z∈D ∂Φ(z)

and C depends only on f (0)(k) . The mapping |z|α (z/|z|) shows that the exponent α cannot be improved in the class of arbitrary k-quasiconformal homeomorphisms. However, it follows from our proof (see (6.32) ) that if Φ is harmonic, then it satisfies the ordinary Lipschitz condition (with Lipschitz constant depending on k). Combining this with Heinz’ inequality [29] |h′ (z)|2 + |g ′ (z)|2 > 1/π 2 , z ∈ D, which holds if f (0) = 0, we get the following. 6.6.2 Theorem If the mapping f is quasiconformal, then it is bi-Lipschitz, i.e., there is a constant L < ∞ such that f (z1 ) − f (z2 ) 1 6 L (z1 , z2 ∈ D). 6 L z1 − z2 Note that an arbitrary bi-Lipschitz homeomorphism is quasiconformal.

Boundary values of the derivatives In calculating the boundary values of the analytic functions h′ and g ′ it is useful to use the formulas  fθ (z)  1 (6.24) h′ (z) = ∂f (z) = e−iθ fr (z) − i 2 r   ¯ (z) = 1 eiθ fr (z) + i fθ (z) , g ′ (z) = ∂f (6.25) 2 r (k) C

= 16 if Φ(0) = 0.

106

6 Conjugate functions

where fθ = ∂f /∂θ, fr = ∂f /∂r. The derivatives fr and fθ are connected by the simple but fundamental fact that the function rfr is equal to the harmonic conjugate of fθ . It follows from (6.16) that fθ equals the Poisson–Stieltjes integral of γ = eiϕ : Z π 1 iθ P (r, θ − t) dγ(t). fθ (re ) = 2π −π Hence, by Fatou’s theorem, the radial limits of fθ exist almost everywhere and limr→1− fθ (reiθ ) = γ0′ (θ) a.e., where γ0 is the absolutely continuous part of γ. It turns out that if γ is absolutely continuous, then limr→1− fr (reiθ ) = H(γ ′ )(θ), a.e. Absolute continuity The function γ, of course, need not be absolutely continuous. However: If Z π 1 fr (ρeiθ ) dθ < ∞, sup ρ Since

1 A(θ). 2K

(6.30)

Z π
1 1 − cos(ϕ(θ + t) − ϕ(θ)) dt 4π −π
 1
1 1 − Re e−iϕ(θ) f (0) > 1 − |f (0)| , = 2 2

A(θ) >

we get ess inf ϕ′ (θ) > 0. Thus condition (6.20) is satisfied. In order to verify (6.21) we use the inequality 2 Z π  ϕ(θ + t) − ϕ(θ) dt ϕ′ (θ) 6 C t −π

(6.31)

(C is an absolute constant) which is obtained from (6.29). Assume first that ϕ is of class C 2 and choose θ so that ϕ′ (θ) = max ϕ′ =: M . Let 0 < β < 1. It follows from (6.31) that 2−β Z π  ϕ(θ + t) − ϕ(θ) M 6C M β dt, t −π

108

6 Conjugate functions

whence M 1−β 6 C

Z

π

−π



ϕ(θ + t) − ϕ(θ) t

2−β

dt.

Now we apply Mori’s inequality (6.23) to deduce that Z π
2−β 1−k . tα−1 dt, α = M 1−β 6 C1 1+k 0

Choose β so that (α − 1)(2 − β) > −1, which is possible because (α − 1)(2 − β) → α − 1 > −1 as β → 1− , to get max ϕ′ 6 C2 , where C2 depends only on K. From this and (6.30) we get A(θ) 6 2KC2 and hence, by (6.29) and (6.26), |h′ (eiθ )| 6 C3 . The function h′ (z) is continuous on the closed disk because the function γ = eiϕ is C 2 , so we have |h′ (z)| 6 C3 (z ∈ D), (6.32) and the constant C3 depends only on K. In the general case consider the mappings fn , of D onto D, defined by fn (z) = f (wn (z))/rn = hn (z) + gn (z)

(rn = 1 − 1/n, n > 2),

where wn is the conformal mapping of D onto Gn = f −1 (rn D), wn (0) = 0, wn′ (0) > 0. Since the boundary of Gn is an analytic Jordan curve, the mapping wn can be continued analytically across ∂D, which implies that fn has a harmonic extension across ∂D. Since also ′ ′ gn (g ◦ wn )wn′ = h′ (h′ ◦ wn )w′ 6 k, n n

we can appeal to the preceding special case to conclude that |h′ (wn (z))| |wn′ (z)|/rn 6 C3 , where C3 is independent of n and z. And since Gn ⊂ Gn+1 and ∪Gn = D, we can apply Carath´eodory’s convergence theorem (Theorem 6.6.3 below): wn (z) tends to z, uniformly on compacts, whence wn′ (z) → 1 (n → ∞). Thus inequality (6.32) holds in the general case. Using this and (6.26) we get ϕ′ (θ)+|B(θ)| 6 C4 . Finally, it remains to apply (6.28). The Carath´ eodory convergence theorem Here we prove the simplest variant of Carath´eodory’s theorem; for the general form see [23, Ch. II§5] as well as [19]. 6.6.3 Theorem Let fn : D 7→ D be a sequence of univalentSfunctions such that fn (0) = 0, fn′ (0) > 0, fn (D) ⊂ fn+1 (D) for every n, and fn (D) = D. Then fn (z) → z uniformly on compact subsets. Proof. The set {fn } is relatively compact in H(D) and hence it is enough to prove that every H(D)-convergent subsequence of {fn } converges to the function ϕ(z) = z. Therefore we can assume that fn tends, uniformly on compact subsets, to some function f ∈ H(D). Clearly f (D) ⊂ D and fn′ (0) → f ′ (0). Let Dρ =

6.6 Quasiconformal harmonic homeomorphisms

{z : |z| 6 ρ}, ρ < 1. Since ∪fn (D) = D and Dρ is compact, we see that fn (D) contains Dρ for n > n0 , where n0 is large enough. The function g(w) = fn−1 (ρw) maps D into D and g(0) = 0 and hence g ′ (0) = ρ/fn′ (0) 6 1, for n > n0 . It follows that fn′ (0) → 1 and hence f ′ (0) = 1. Hence f (z) = z, by Schwarz’ lemma, and the proof is complete. ✷ A problem Let QCH = {f∗ : f is a q.c. harmonic homeomorhism of D}. Is QCH a group with respect to composition? The set of all quasiconformal harmonic homeomorphisms of D is not a group because the composition of two harmonic functions need not be harmonic. On the other hand, the set of all quasiconformal transformations of D is a group (cf. [2]). Miscellaneous 6.6.4 [56] The mapping f = P [eiϕ ] is quasiconformal if ϕ ∈ C 1 (R), min ϕ′ > 0 and Z π ω(t) dt < ∞, (6.33) t 0

where ω(t) = sup{ |ϕ′ (x) − ϕ′ (y)| : |x − y| < t} is the modulus of continuity of ϕ′ . Condition (6.33), known as Dini’s condition (applied to ϕ′ ), is sufficient but not necessary for the Hilbert transformation of ϕ′ to belong to L∞ . 6.6.5 Let G = ϕ(D), where ϕ : D 7→ C is a univalent function such that ϕ′ (0) > 0. Let fn : D 7→ G be a sequence of univalent functions such that fn (0) = ϕ(0), fn′ (0) > 0, fn (D) ⊂ fn+1 (D) for every n, and ∪fn (D) = G. Then fn (z) → ϕ(z) uniformly on compact subsets. 6.6.6 Martio [56] proved that condition (6.21) implies that Jf (z) is bounded above on D. The converse is true because Jf (z) = Im( fr (z)fθ (z) ), whence Jf (eiθ ) = ϕ′ (θ)A(θ). Therefore f satisfies a Lipschitz condition on the boundary iff Jf (z) 6 C (z ∈ D), where C is a constant, or, equivalently, |f (E)| 6 C|E| for any measurable set E ⊂ D. Clearly, this does not imply a Lipschitz condition on D. The mapping f satisfies a Lipschitz condition on D iff both ϕ′ and Hϕ′ are bounded.

109

7

Maximal functions, interpolation, coefficients This chapter concerns two fundamental facts:

(1) (maximal theorem) The operator Mrad defined on h(D) by (Mrad u)(eiθ ) = sup |u(reiθ )| 0 0. (2) (maximal characterization) Let p ∈ (0, ∞). An analytic function f belongs to H p iff Mrad (Re f ) belongs to Lp (T). In Section 7.3 we state a theorem of Hardy and Littlewood on (C, α)-means in H p and prove an analogous result concerning “smooth” Ces´ aro means. A Marcinkiewicz theorem for H p , 0 < p < ∞, with applications to Taylor coefficients, is in Section 7.4. In Section 7.5 we consider some extensions of Hardy and Littlewood’s theorem on Taylor coefficients. Section 7.6 contains some remarks on the dual of H 1 .

7.1

Maximal theorems

Radial maximal function Let u be a complex-valued function defined on D. The radial maximal function of u is the function Mrad u defined on T by (Mrad u)(ζ) = sup |u(rζ)| 0 1) into L1,∞ (T) (resp. Lp (T)), and is continuous. It is worthwhile to note that this theorem can be proved without appealing to the main maximal theorem. Namely, we can use Nikishin’s theorem, Banach’s principle, and Fatou’s theorem, but Proposition 7.1.1 is of independent interest. Using the properties of subharmonic functions, we can easily deduce the “subharmonic” version of Theorem 7.1.2. 7.1.3 Theorem (subharmonic maximal) Let u ba a nonnegative, continuous, subharmonic function on D, and let Z π sup u(reiθ )p dθ =: Ap < ∞, where p > 1. 0 0, we have Z π
p 1 Mrad f (eiθ ) dθ 6 Cp kf kpp . 2π −π

7.1.5 Remark We can use the complex maximal theorem to deduce  p (5.7) from (5.10). Namely, from the inequality log |f (ρeiθ )| 6 Mrad f (eiθ ) /p, and  p the complex maximal theorem it follows that Mrad f is an integrable majorant of the family log |fρ | and therefore we may apply Fatou’s lemma to inequality (5.10).

112

7 Maximal functions, interpolation, coefficients

A further example: 7.1.6 Proposition If f ∈ H(D) and 0 < p < 1, then Mpp (ρ, f ) − Mpp (r, f ) 6 Cp (ρ − r)p Mpp (ρ, f ′ )

(0 < r < ρ < 1).

Proof. From the inequality |f (ρeiθ )| − |f (reiθ )| 6 (ρ − r) sup{|f ′ (seiθ )| : r < s < ρ}, it follows that |f (ρeiθ )|p − |f (reiθ )|p 6 (ρ − r)p sup{|f ′ (seiθ )|p : r < s < ρ}. Then the desired inequality is obtained by integration in θ and using the maximal theorem. ✷ Nontangential maximal function Theorems 7.1.2, 7.1.3, and 7.1.4 remain true if we replace the radial maximal function by the nontangential maximal function M∗ u, M∗ u(ζ) = sup |u(z)| z∈Uζ

(ζ ∈ T),

where r0 < 1 is fixed and Uζ = the convex hull of {|z| 6 r0 } ∪ {ζ}. These variants are proved by means of the following. 7.1.7 Proposition If φ ∈ L1 (T) and f = P [φ], then M∗ f (ζ) 6 C Mφ(ζ), ζ ∈ T, where C depends only on r0 . The proof is very similar to that of Proposition 7.1.1 and we omit it; see the proof of Theorem 3.3.7. Although the pointwise estimate M∗ u(eiθ ) 6 const Mrad u(eiθ ) is not valid, we still have: 7.1.8 Theorem (Fefferman/Stein [21]) If 0 < p < ∞ and u > 0 is subharmonic in D, then Z 2π Z 2π iθ p {Mrad u(eiθ )}p dθ. {M∗ u(e )} dθ 6 Cp 0

0

Proof. The proof is based on the inequality Z Cq q u(z) 6 2 u(w)q dm(w), δ |w−z| λ}, and m(λ) = |Gλ |. Assume we have proved that Z Z 2 u2 dθ + 2λ2 m(λ), λ > 0. (7.3) v dθ 6 Eλ

Eλ

Multiplying this by qλ−q−1 , q = 2 − p > 0, and then integrating from λ = 0 to ∞, and using Fubini’s theorem, we get Z 2π Z 2π Z 2π 2q (u∗ )2−q dθ v 2 (u∗ )−q dθ 6 u2 (u∗ )−q dθ + 2 − q 0 0 0 Hence

Z

2π 2

∗ p−2

v (u )

dθ 6 Cp

Z

2π

(u∗ )p dθ,

p < 2,

0

0

where Cp = 1 + 2(2 − p)/p. To obtain (7.2) assume that by Jensen’s inequality, Z

0

2π p

∗ −p

|v| (u )

∗ p

(u ) dθ

2/p

6

Z

=

0

0

(u∗ )p dθ = 1. Then,

2π

{|v|p (u∗ )−p }2/p (u∗ )p dθ

0

Z

R 2π

(7.4)

2π

|v|2 (u∗ )p−2 dθ;

now (7.2) follows from (7.4). Proof of (7.3). We can suppose that 0 < |Eλ | < 2π. Let F (z) = sup |u(ζz)|, z ∈ C. ζ∈A

The set Fλ = {z ∈ C : F (z) 6 λ} ∩ D is nonempty and simply connected because the sup{F (z), |z|/λ} is subharmonic in C. Also, Fλ contains the set Hλ = S function {eiθ A : θ ∈ Eλ }. Let Γλ = (∂Hλ ) r Eλ . By Cauchy’s integral formula, we have Z 1 f (z)2 dz = f (0)2 = 0, 2πi ∂Hλ z whence −

Z

f (eiθ )2 dθ =

f (eiθ )2 (dθ + idr/r).

Γλ

Eλ

Taking the real parts we get Z Z (v 2 − u2 ) dθ = Eλ

Z

Γλ

(u2 − v 2 ) dθ −

Z

2uv dr/r.

Γλ

Since |dr/rdθ| 6 1 on Γλ and 2|uv| 6 u2 + v 2 , we conclude that Z Z Z 2u2 dθ u2 dθ + v 2 dθ 6 Eλ

Eλ

Γλ

116

7 Maximal functions, interpolation, coefficients

Now the desired result follows from the inequality Z Z Z u2 dθ 6 λ2 dθ = λ2 dθ = λ2 m(λ). Γλ

Γλ

Gλ

This completes the proof of the theorem. ✷ Remark. The above proof is a slight modification of Koosis’ proof (see Garnett [22]). The only difference is in that Koosis uses (7.3) to show that Z λ 2 |{θ : |v(θ)| > λ}| 6 m(λ) + 2 s m(s) ds. λ 0

7.3

“Smooth” Ces` aro means

In contrast to the case 1 < p < ∞, the sequence {z n } is not a Shauder basis in H p for p ∈ (0, 1]. Hardy and Littlewood proved that this sequence is a (C, α) basis in H p for α > 1/p − 1 (p 6 1) (for a proof, see [67]). Instead, we shall construct the “smooth” Ces` aro basis, which has an advantage in that it is “universal”, i.e., independent of p (Theorem 7.3.4). Before, we state the theorem of Hardy and Littlewood. The Ces` aro means of order α > −1 of an analytic function f are defined by n

σnα f (z) =

Γ(n + 1) X Γ(α + n + 1 − k) b f (k)z k , Γ(α + n + 1) Γ(n + 1 − k) k=0

where Γ is the Euler gamma function. In particular n  X k b f (k)z k . σn1 f (z) = 1− n+1 k=0

Define the maximal operator σmax by

α (σmax f )(ζ) = sup |σnα f (ζ)| n

(ζ ∈ T).

It should be noted that the nontangential maximal function M∗ f is dominated α by a constant multiple of σmax f ; in the case α = 1, this follows from the inequality |f (ζz)| 6 and this follows from the formula

|1 − z|2 (σ 1 f )(ζ), (1 − |z|)2 max

f (ζz) = (1 − z)2

∞ X

(n + 1)(σn1 f )(ζ) z n .

n=0

α f kp 6 Cp kf kp , 7.3.1 Theorem If 0 < p 6 1, α > 1/p−1, and f ∈ H p , then kσmax α α iθ iθ lim kf − σn f kp = 0, and lim σn f (e ) = f (e ) a.e. n→∞

n→∞

In the case 1 < p < ∞, there is a much deeper result, due to Carleson[12] and Hunt[31], which states that the above relations hold for the partial sums of f ∈ hp (see p. 47).

7.3 “Smooth” Ces` aro means

117

The polynomials Wn Let ψ be a complex-valued C ∞ -function with compact support in R. Define the trigonometric polynomials Wn , n > 1, by Wn (eit ) = Wnψ (eit ) =

X

|k| 0, then there are functions h ∈ H ∞ and g ∈ H p such that  |f | λ  ∗ and , |h∗ | 6 Cλ min λ |f∗ | Z |f∗ (ζ)|p |dζ|, kgkpp 6 C ζ∈T, |f∗ (ζ)|>λ

where C depends only on p.

7.4 Interpolation of operators on Hardy spaces

121

Proof. Let λ > 0 and define the functions α on T, and F on D by    |f∗ | p/2 1 α = max 1, and F = . λ P [α] + iPe[α]

Since P [α] > 1 in D, we have 0 < |F | 6 1 in D. Therefore the function G = 1 − (1 − F 4/p )2/p

is well defined, analytic and bounded in D. We claim that the functions h = Gf and g = (1 − G)f satisfy the desired conditions. Since |F∗ | 6 1/α and |G| 6 C|F |4/p (by Schwarz’ lemma), we have    |f∗ | −2 , |h∗ | 6 C|f∗ | min 1, λ

and this gives the desired estimate for h. On the other hand, |g∗ | 6 C|1 − F∗ |2/p |f∗ |  2/p α − 1 |H(α − 1)| 6C |f∗ | + α α

6 C(1 − 1/α)2/p |f∗ | + Cλ |H(α − 1)|2/p ,

where H is the Hilbert operator. Since α = 1 on the set {|f∗ | < λ}, and H is bounded on L2 , we see that Z Z kgkpp 6 C (1 − 1/α)2 |f∗ |p + Cλp |H(α − 1)|2 T

T

6C 6C This completes the proof.

Z

|f∗ |>λ

|f∗ |p + Cλp

|f∗ |>λ

|f∗ |p .

Z

Z

T

(α − 1)2

✷

Application to Taylor coefficients As an example, consider the operator (T f )(n) = (n + 1) sup |fb(k)|, 06k6n

By Corollary 5.1.3, for every p ∈ (0, 1) we have

(T f )(n) 6 Cp (n + 1)1/p kf kp ,

f ∈ H p , n > 0.

f ∈ H p.

Define the measure µ on Ω = {0, 1, 2, . . .} by µ({n}) = (n + 1)−2 . Then, arguing as in the proof of Theorem 2.2.5 we find that T satisfies (7.9) for every p ∈ (0, 1). Hence, by Theorem 7.4.1, T maps H p into Lp (Ω, µ), for every p ∈ (0, 1). Thus we have proved:

122

7 Maximal functions, interpolation, coefficients

7.4.3 Theorem [57] If f ∈ H p , 0 < p < 1, then

∞ P

n=0

(n + 1)p−2 sup |fb(k)|p < ∞. k6n

As a corollary we have part (a) of the following theorem of Hardy and Littlewood. 7.4.4 Theorem (a) If f ∈ H p , 0 < p < 1, then |fb(n)| = o(n1/p−1 ). (b) Assertion (a) is optimal in the following sense: If |fb(n)| = O(ψn ) for every f ∈ H p , where ψn > 0, then there is a constant c > 0 such that ψn > cn1/p−1 .

Proof. To prove (b) assume that |fb(n)| = O(ψn ). Consider the space ℓ∞ ψ of all scalar sequences {an } for which k{an }kψ := supn ψn−1 |an | < ∞. By the hypothesis we have H p ⊂ ℓ∞ ψ . To prove that the inclusion operator is continuous assume that (1) fj → f in H p , and (2) {fbj (n)} → {an } in ℓ∞ , as j → ∞. It follows from ψ

assertion (a) and (1) that fbj (n) → fb(n), j → ∞, for every n. On the other hand, (2) implies that fbj (n) → an for every n. Hence an = fb(n) for every n, and therefore the closed graph theorem tells us that the inclusion is continuous. In particular, we have supn ψn−1 |fbr (n)| 6 Cp kfr kp , 0 < r < 1, where fr (z) = (1 − rz)−2/p and Cp depends only on p. Since fbr (n) =



 −2/p (−1)n rn > cn2/p−1 n

and kfr kp = (1 − r2 )−1/p ,

we conclude that ψn−1 n2/p−1 rn 6 C(1 − r)−1/p , where C is independent of r and n. Now the desired result is obtained by taking r = 1 − 1/n. ✷ Lq -integrability of Mp (r, f ) We have proved that if f ∈ H p , p < ∞, then   Mq (r, f ) = O (1 − r)1/q−1/p

(r → 1)

for q > p,

(7.10)

see (5.3). Then, using the fact that the polynomials are dense in H p , we can prove that (7.10) remains valid if we replace “O” by “o”. This is further improved by inequality (7.11) below because Mp (r, f ) increases with r. 7.4.5 Theorem (Hardy/Littlewood) If f ∈ H p , 0 < p < ∞ and ∞ > q > p, then Z

0

1

(1 − r)−p/q Mqp (r, f ) dr 6 Ckf kpp ,

where C depends only on p, q. From (5.3) and (7.11) we get the following:

(7.11)

7.5 On the Hardy/Littlewood inequality

123

7.4.6 Corollary If f ∈ H p , 0 < p < ∞, ∞ > q > p and s > p, then Z

0

1

(1 − r)sα−1 Mqs (r, f ) dr 6 Ckf ksp ,

where α = 1/p − 1/q (> 0) and C is independent of f . In the case s = q, this can be written in the form: 7.4.7 Corollary If f ∈ H p , 0 < p < ∞ and q > p, then Z |f (z)|q (1 − |z|2 )q/p−2 dA(z) 6 Cp,q kf kqp . D

In connection with this corollary see 5.3.5. Proof of Theorem 7.4.5. We can suppose that q < ∞ because √ M∞ (r, f ) 6 (1 − r)−1/q Mq ( r, f ).

Define the (quasilinear) operator T : H p 7→ C(0, 1) by (T f )(r) = (1 − r)−1/q Mq (r, f ),

0 < r < 1.

We will prove that T maps H s into Ls,∞ (0, 1) for s = p and s = q. By Theorem 7.4.1, this will imply that T maps H s into Ls (0, 1) for p < s < q; this will conclude the proof. In the case s = p we use inequality (5.3); we get T f (r) 6 A(1 − r)−1/p kf kp ,

(7.12)

where A is a positive constant. If kf kp = 1, then it follows from (7.12) that

{r ∈ (0, 1) : T f (r) > λ} 6 {r ∈ (0, 1) : A(1 − r)−1/p > λ} = min(1, (A/λ)p ),

which proves that T is of weak type (p, p). In the case s = q the desired conclusion follows from the inequality Mq (r, f ) 6 kf kq . The proof is complete. ✷ Remark. For a more elementary proof, see [18].

7.4.8 Exercise If f is harmonic in D, then (7.11) holds for q > p > 1.

7.5

On the Hardy/Littlewood inequality

As a consequence of Theorems 7.4.3 and 2.2.4 we have another result of Hardy and Littlewood.

124

7 Maximal functions, interpolation, coefficients

7.5.1 Theorem If f ∈ H p , 0 < p < 2, then K :=

∞ X

n=0

(n + 1)p−2 |fb(n)|p 6 Cp kf kpp .

(7.13)

If a function f ∈ H(D) satisfies the condition K < ∞, for some p > 2, then f ∈ H p and kf kpp 6 Cp K. In the case 1 < p < 2, this theorem is, of course, weaker than Theorem 2.2.4. The following proof contains, however, an improvement of (7.13) in other direction. Proof. In the case q = 2 > p, inequality (7.11) can be written as p/2 X Z 1 ∞ (1 − r)−p/2 |fb(n)|2 r2n dr 6 Cp kf kpp . (7.14) 0

n=0

Let f ∈ H p , 0 < p < 2. We use (7.14) and Lemma 7.5.4 below to get p/2  ∞ X X |fb(k)|2 6 Cp kf kpp . 2−n(1−p) 2−n n=0

(7.15)

k∈In

Since card(In ) = 2n for n > 1, and the function t 7→ tp/2 , t > 0, is concave, we have  p/2 X X 2−n > 2−n |fb(k)|2 |fb(k)|p , k∈In

k∈In

and this together with the previous relation implies (7.13). To discuss the case p > 2 we consider the space Xp ⊂ H(D) defined by condition (7.13). It is easily seen that (Xp )∗ = Xq , 1/p + 1/q = 1, with respect to the bilinear form ∞ X fb(n)b g (n) (f, g) = n=0

p ∗

q

Since (H ) = H , 1 < p < ∞, with respect to the same form, we deduce from H p ⊂ Xp , p < 2, that Xq ⊂ H q , q > 2. The proof is complete. ✷ Inequality (7.15) can be improved by combining Theorem 7.4.5 with the Hausdorff/Young theorem. For example, in the case p = 1 we get: 7.5.2 Theorem [57] If f ∈ H 1 , then 1/q ∞  X X q −n b < ∞, |f (k)| 2 n=0

k∈In

0 < q < ∞.

(7.16)

Note that the quantity in (7.16) increases with q. In the case q = 1, (7.16) reduces to ∞ X |fb(n)| < ∞. n+1 n=0

7.5 On the Hardy/Littlewood inequality

125

Addendum: Littlewood’s conjecture In connection with the so called Littlewood’s conjecture, Z 2π X n iλ θ k e dθ > c log n, 0

k=1

the following extension of Theorem 7.5.1 (p = 1) was proved in [65]. 7.5.3 Theorem If {λn }∞ of positive integers, 1 is a strictly increasing sequence P∞ and f ∈ H 1 is such that supp fb ⊂ {λn : n > 1}, then n=1 n−1 |fb(λn )| 6 Ckf k1 , where C is independent of {λn } and f . Lp -integrability of power series with positive coefficients 7.5.4 Lemma Let α > −1, 0 < q < ∞, and In = {k : 2n 6 k < 2n+1 } for n > 1, I0 = {0, If {an } is a sequence of nonnegative numbers such that the P1}. ∞ series G(r) = n=0 an rn converges for every r ∈ (0, 1), then the following three conditions are equivalent and the corresponding quantities are “proportional”: Z 1 (1 − r)α G(r)q dr < ∞, 0

∞ X

n=0 ∞ X

2−(α+1)n

−α−2

(n + 1)

n=0

X

ak

k∈In X n

k=0

q

ak

< ∞,

q

< ∞.

In the case of the function G(r) = supn>0 an rn the expressions should be replaced by supk∈In ak , sup06k6n ak , respectively.

P

k∈In

ak ,

Pn

k=0

ak

This lemma is an immediate consequence of its special case: 7.5.5 Lemma Let {ak } be a sequence of nonnegative real numbers such that P∞ 2k the series F (r) = converges for every r ∈ (0, 1). Let α > −1 and k=0 ak r 0 < q < ∞. Then the conditions Z 1 ∞ X 2−(α+1) aqk < ∞ (1 − r)α F (r)q dr < ∞ and B := A := 0

k=0

are equivalent. There holds the inequality B/C 6 A 6 CB, where C depends only k on α, q. The same holds for the function F (r) = supk ak r2 . Proof. The proof of the inequality A > B/C is very simple (see, e.g., the proof of Theorem 11.1.1, p. 171)(∗) . To prove the reverse inequality, observe first that (∗) It is, however, difficult to prove that this inequality remains hold if we drop the hypothesis an > 0; see Theorems 11.4.1 and 11.4.2.

126

7 Maximal functions, interpolation, coefficients

the case q 6 1 is trivial; we integrate the inequality (1 − r)α F (r)q 6 (1 − r)α

∞ X

k

aqk rq2 .

k=0

In the case q > 1 we can use Jensen’s inequality (as in [58]), or proceed as follows. Write the inequality A 6 CB as kT ({an })(r)kLq (µ) 6 Cq k{an }kℓq , where T ({an })(r) = (1 − r)β

∞ X

n

2βn an r2 ,

β > 0,

n=0

and dµ(r) = dr/(1 − r). It is easily verified that T maps ℓq to Lq (µ) for q = 1, ∞, so we can apply the Riesz/Thorin theorem. ✷ Miscellaneous 7.5.6 If α > 0 and let {an } be a nonnegative sequence, then the following conditions are equivalent: ∞ X

n=0

  an rn = O (1 − r)−α

(r → 1− ),

and

X

ak = O(2αn )

k∈In

(n → ∞).

The equivalence remains true if we replace “O” by “o.” 7.5.7 If f ∈ H p , 1 < p < 2, and {λn }∞ 1 is a strictly increasing sequence of positive P∞ p−2 b p integers, then n=1 n |f (λn )| < ∞. 7.5.8 [59] Lemma 7.5.4 can be generalized in the following way: P b For an analytic function f let ∆n f (z) = f (k)z k . Let 1 < p < ∞, 0 < q < ∞, k∈In α > 0. Then Z 1 ∞ X
q (1 − r)qα−1 Mpq (r, f ) dr < ∞ if and only if 2−αn k∆n f kp < ∞. 0

n=0

In the case q = ∞ there holds the equivalence Mp (r, f ) = O(1 − r)−α ⇐⇒ k∆n f kp = O(2αn ). This can be deduced from Lemma 7.5.5 in the following way. We have Mp (r, f ) 6

∞ X

n=0

Mp (r, ∆n f ),

1 6 p 6 ∞,

and, by Riesz’ projection theorem, Mp (r, f ) > cp supn>0 Mp (r, ∆n f ), 1 < p < ∞. It remains to apply 4.1.4.

7.6 On the dual of H 1

127

7.5.9 Let 1 < p < ∞, f ∈ H(D), and let the sequence fb(n) be decreasing. Then, f ∈ H p if and only if ∞ X (n + 1)p−2 |fb(n)|p < ∞. n=0

This can be deduced from 7.5.8 and the Littlewood/Paley theorem.

7.5.10 Assertion 7.5.8 does not hold for p ∈ (0, 1) ∪ {∞}. Then we can use Theorem 7.3.4 to replace ∆n f by Γn ∗ f , where Γn is a sequence of polynomials Γn , n > 0, satisfying, for all p > 0, b n ) ⊂ [2n−1 , 2n+2 ], n > 1, supp(Γ ∞ X Γn ∗ f, f ∈ H(D), f= n=0

kΓn ∗ f kp 6 Cp kf kp ,

f ∈ H p,

kΓn kp ≍ 2n(1−1/p) .

A more complicated polynomials can be constructed by means of Theorem 7.3.1 (cf. [69]). Such polynomials play a central role in calculating multipliers for various spaces of analytic functions (cf. [33, 70]).

7.6

On the dual of H 1

A function φ ∈ L1 (T) is called a function of bounded mean oscillation if Z Z it 1 φ(e ) − φI dt = kφk∗ < ∞, where φI = 1 φ(eit ) dt, sup |I| I I⊂T |I| I

and the supremum is taken over all subarcs of T. The class BMO = {φ : kφk∗ < ∞} is normed by kφkBMO = kφkL1 + kφk∗ . The intersection of BMO with H 1 (T) is denoted by BMOA. The famous theorem of Fefferman states that BMOA is isomorphic to the dual of H 1 (T). This means the following: A function φ ∈ H 1 (T) belongs to BMOA if and only if 1 Z π (7.17) sup φ(eit )Q(e−it ) dt = kφk′∗ < ∞, 2π kQk1 61 −π

where the supremum is taken over the set of all polynomials in H 1 . Moreover, the norm k · k′∗ is equivalent to the original norm in BMOA. The proof of Fefferman’s theorem as well as of various other properties of BMOA can be found in Garnett [22, Ch. VI]. Here we can accept (7.17) as a definition of BMOA. Also, we can treat BMOA as a space of analytic functions on D, via the Poisson integral. Then Theorem 7.5.2 leads to the following.

128

7 Maximal functions, interpolation, coefficients

7.6.1 Theorem [57] Let {an }∞ 0 be a sequence of complex numbers satisfying  1/q X n −n q sup 2 2 < ∞, for some q > 1. (7.18) |ak | n>0

k∈In

Then the function f (z) =

P∞ 0

an z n belongs to BMOA.

Note that (7.18) is satisfied if an = O(1/n). 7.6.2 Remark It is interesting that condition (7.18) is sufficient for the validity of the implication lim−

r→1

∞ X

an rn exists =⇒

∞ X

an converges;

n=0

n=0

see Theorem 11.2.2 and Corollary 11.2.3. Miscellaneous 7.6.3 The space BMOA is closely related to two important spaces of functions— the space H ∞ and the Bloch space B = {f ∈ H(D) : supz∈D (1 − |z|2 )|f ′ (z)| < ∞}. k′∗ 6 kf k∞ so H ∞ ⊂ BMOA. This inclusion is proper It is clear from (7.17) that kfP ∞ because the function f (z) = n=1 z n /(n + 1) is in BMOA but not in H ∞ . The inclusion BMOA ⊂ B can be deduced from the easily proved relations Z Z π 1 1 it −it z ) log dA(z), φ(e )Q(e ) dt = f (0)Q(0) + 2 f ′ (z)Q′ (¯ 2π −π |z| D and kQk1 6 |Q(0)| + C

Z

D

|Q′ (z)| dA(z),

where CPis an absolute constant. The inclusion is proper. Namely, the function n f (z) = z 2 is in B. On the other hand, since H 1 ⊂ H p for 1 < p < ∞, and ′ (H p )∗ = H p (Theorem 6.3.2), we see that BMOA ⊂ H p for every p < ∞ and therefore f is not in BMOA because f is not in H 2 . 7.6.4 Fefferman’s duality theorem can be expressed in terms of multipliers, namely: BMOA = (H 1 , H ∞ ) := {g ∈ H(D) : g ∗ f ∈ H ∞ for all f ∈ H 1 }. It was proved in [61] that (H 1 , BMOA) = B. 7.6.5 If f ∈ H 1 and g ∈ BMOA, then f (eit )g(e−it ) need not be integrable on [−π, π]. Example: 1−z 1 − z 1 − z −2 , g(z) = log log f (z) = z 2 . 1+z 1+z 1+z

8

Bergman spaces: Atomic decomposition

In this chapter we consider the Coifman/Rochberg theorem on the atomic decomposition of Bergman spaces (Theorems 8.3.1, 8.3.4) as well as some applications, due to Kalton [37], to the theory of vector-valued analytic functions (Theorems 8.4.2, 8.4.3). Some technical simplifications in the proof of the Coifman/Rochberg theorem are made. Kalton’s results are formulated in a more precise form (Theorems 8.4.3, 8.4.5). Further topics in the theory of Bergman spaces can be found in the book of Hedenmalm, Korenblum and Zhu [28].

8.1

Bergman spaces

Let 0 < p < ∞. The Bergman space Ap is the subspace of Lp (D, dA) consisting of analytic functions. The quasinorm can be written as kf kp = kf kpAp = 2

Z

1

Ip (r, f ) rdr,

where

Ip (r, f ) =

0

1 2π

Z

π

−π

|f (reiθ )|p dθ.

8.1.1 Proposition For p ∈ (0, ∞) there hold the following: (a) For every z ∈ D the functional f 7→ f (z) is continuous on Ap ; (b) The space Ap is complete; (c) If f ∈ Ap , and f̺ (z) = f (̺z), then f̺ ∈ Ap and kf − f̺ k → 0 (̺ → 1); (d) The set of all polynomials is a dense subset Ap . Proof. The function |f |p is subharmonic, which means that Z 1 p |f (w)|p dA(w) |f (z)| 6 2 R |w−z| 1 is discussed in a similar way. (d) Let f ∈ Ap , ε > 0 and 0 < ̺ < 1. The Taylor series of f̺ converges uniformly on D, which implies kf̺ − sn k 6 ε for n large enough, where sn are the partial sums of the Taylor series of fρ . Now (d) can easily be deduced from (c). ✷ Exercises

8.1.2 The assertion (c) of Proposition 8.1.1 holds and for an arbitrary function f ∈ Lp (D). 8.1.3 The inclusion Aq ⊂ Ap is compact for q > p. Remark. It is known that Ap is isomorphic to ℓp (see Theorem 1.5.1). And since every operator from ℓq to ℓp , q > p, is compact (see Theorem 1.5.6), every operator from Aq u Ap , q > p, is compact. 8.1.4 There holds the inequality |f (z)| 6 (1 − |z|2 )−2/p ||f ||Ap . See Lemma 5.1.1. Pn 8.1.5 For every p > 0 and every polynomial f (z) = k=m ak z k there holds the inequality 1 1 kf kH p 6 kf kAp 6 kf kH p . n+1 m+1

8.2

Reproductive kernels

Let Kp (w, z) =

1 (1 − wz) ¯ 2/p+1

(z, w ∈ D).

Let dµp denote the measure defined over D by dµp (w) = cp (1 − |w|2 )2/p−1 dA(w), where Z p 1 = (1 − |w|2 )2/p−1 dA(w) = . cp 2 D The following theorem shows that da Kp has the reproducing property. 8.2.1 Theorem Let 0 < p 6 1 and f ∈ Ap . Then for every z ∈ D the function w 7→ f (w)Kp (w, z) belongs to L1 (D, dµp ) and there holds the formula Z Kp (w, z)f (w) dµp (w). (8.2) f (z) = D

8.2 Reproductive kernels

131

Proof. If kf kAp = 1 and z ∈ D, then, according to (8.1),

|f (w)| = |f (w)|p |f (w)|1−p 6 |f (w)|p (1 − |w|)−2/p+2 ,

which implies that the function w 7→ f (w)Kp (w, z) is in L1 (D, dµp ). It follows that the functional Z Λz f = Kp (w, z)f (w) dµp (w) D p

is well defined and bounded on A . Since the functional f 7→ f (z) is bounded (Proposition 8.1.1(a)) and the polynomials are dense in Ap , we see that the proof of (8.2) reduces to the proof that Λz f = f (z) for f (z) = z n , n > 0. However this follows from the formulas ∞ X 1 Γ(α + k + 1) = zk , (1 − z)α+1 Γ(α + 1)Γ(k + 1) k=0 Z Z Γ(α + 1)Γ(k + 1) (1 − |w|2 )α−1 dA(w), (1 − |w|2 )α−1 |w|2k dA(w) = Γ(α + k + 1) D D R which are true for α > 0, and the formula D wm w ¯ n dA(w) = 0 (m 6= n). ✷ Remark. A more interesting proof of (8.2) can be found in [63]. 8.2.2 Lemma Let f ∈ Ap , 0 < p 6 1 and let Z |Kp (w, z)| |f (w)| dµp (w) g(z) = D

(z ∈ D).

Then g ∈ Lp (D, dA) and there holds the inequality kgkLp 6 Cp kf kAp . Proof. Let E be a partition of the unit disk into disjoint sets E with the following properties: diam(E) 1 6 6C C 1 − |w|

(w ∈ E),

1 area(E) 6C (w ∈ E), 6 C (1 − |w|)2 1 − |ζ| 1 6 6C (ζ, w ∈ E), C 1 − |w| where C is an absolute constant. Such a family consists of the sets n 1 2πj 2π(j + 1) o 1 , z : k+1 < 1 − |z| 6 k , k+2 6 arg z < 2 2 2 2k+2

where k = 0, 1, 2, . . . , 0 6 j < 2k+2 . Let E = {En : n > 1}. Then Z ∞ X 2/p−1 (1 − |wn |) |f (w)Kp (w, z)| dA(w) g(z) 6 C 6C

n=1 ∞ X

n=1

En

(1 − |wn |)2/p+1 sup |f (w)Kp (w, z)|, w∈En

132

8 Bergman spaces: Atomic decomposition

where {wn } is an arbitrary sequence such that wn ∈ En and C is a constant depending only on p. It follows that p

g(z) 6 C

∞ X

n=1

(1 − |wn |)2+p sup |f (w)Kp (w, z)|p . w∈En

For a fixed z the function p f (w) F (w) = |f (w)Kp (w, z)| = 2/p+1 (1 − w¯ z) p

is subharmonic in D because the function f (w)/(1 − w¯ z )2/p+1 is analytic with respect to w. Therefore Z 1 F (w) 6 F (ζ) dA(ζ), A(Dw ) Dw where Dw ⊂ D is an arbitrary disk centered at w. From this and the properties of the family E we get Z |f (w)|p 1 dA(w), sup F 6 C 2 (1 − |wn |) Bn |1 − wz| ¯ 2+p En where Bn is the union of those E ∈ E for which the set E ∩ E n is nonempty. Combining these with the inequality Z C 1 dA(z) 6 2+p ¯ (1 − |w|)p D |1 − wz| (see Lemma 8.2.3 below), we get Z

p

g(z) dA(z) 6 C

D

∞ Z X

n=1

Bn

|f (w)|p dA(w).

Now our result follows from the easily checked fact that each Bn contains at most N members of the family E, with N being independent of n. ✷ 8.2.3 Lemma Let Is (a) =

1 2π

Z

π

−π

|1 − a ¯eiθ |−s dθ,

Js (a) =

Z

D

|1 − a ¯w|−s dA(w).

(∗)

:  −s+1  (1 − |a|) 2 Is (a) ≍ log 1−|a|   1

There hold the relations

(∗) “A

for s > 1 for s = 1 for s < 1

(a ∈ D),

≍ B ” means that the ratio A/B lies between two positive constants.

8.3 The Coifman/Rochberg theorem

133

 −s+2  (1 − |a|) 2 Js (a) ≍ log 1−|a|   1

Proof. Since Js (a) = 2 ρ = |a| < 1. We have 1 Is (a) = π ≍

8.3

1 π

R1 0

Z

0

(a ∈ D).

for s < 2

Is (ra) dr, it suffices to discuss the case of Is . Let

π

 −s/2 (1 − r)2 + 4r sin2 (θ/2) dθ

0

Z

for s > 2 for s = 2

π

(1 − r + θ)−s dθ.

✷

The Coifman/Rochberg theorem

It is the idea of Coifman and Rochberg [15] to represent a member of Ap as a sum of “atoms” by replacing the integral in (8.2) by a Riemannian sum over a sufficiently fine partition of the disk. They proved the “decomposition theorem” for every p > 0 and for a class of domains in Cn , in particular on balls. Here we consider the case p < 1 because this case, as was shown by Kalton [37], is of fundamental importance in the theory of vector-valued analytic functions. 8.3.1 Theorem (on atomic decomposition) Let 0 < p 6 1. Then (a) There exists a sequence {wn } in D and a constant C such that for every f ∈ Ap there exists a sequence {an } ⊂ ℓp with the properties ∞ X

1 − |wn |2 (1 − w ¯n z)2/p+1 n=1 Z ∞ X |an |p 6 C |f |p dA.

f (z) =

n=1

an

(8.3)

D

(b) The function f of the form (8.3), where {an } ∈ ℓp , belongs to Ap and there holds kf kAp 6 Ck{an }kp . For the proof we need a lemma. 8.3.2 Lemma There exists a constant C such that |Kp (w, z) − Kp (ζ, z)| 6 C for all E ∈ E, w, ζ ∈ E and z ∈ D.

|w − ζ| |Kp (w, z)| diam E

134

8 Bergman spaces: Atomic decomposition

Proof. By the mean value theorem, we have |Kp (w, z) − Kp (ζ, z)| 6 (2/p + 1)|w − ζ| sup |1 − a ¯z|−2/p−2 a∈E

6 (2/p + 1) 6 Cp

|w − ζ| sup |Kp (a, z)| 1 − |a| a∈E

|w − ζ| sup |Kp (a, z)|. diamE a∈E

On the other hand, ¯ (w − a)z |w − a| 1 − 1 − wz 6 C. = 6 1−a ¯z 1 − a ¯z 1 − |a|

Hence |1 − wz| ¯ 6 C|1 − a ¯z| and hence |Kp (a, z)| 6 C|Kp (w, z)| for a, w ∈ E. The result follows. ✷ Proof of Theorem 8.3.1. Assertion (b) follows from the boundedness of the sequence kψk kAp , where 1 − |wk |2 ψk (z) = . (1 − w ¯k z)1+2/p Let us prove (a). Let ε > 0. Dividing each E ∈ E into N subsets, where N is a sufficiently large integer independent of E, we can represent D as a disjoint union D1 ∪ D2 ∪ . . . , where D1 , D2 , . . . are subsets of D with the properties: diam(Dn ) ε 6 6 C1 ε C1 1 − |w|

and

area(Dn ) ε2 6 6 C1 ε2 C1 (1 − |w|)2

(w ∈ Dn ).

Let {wn } be a sequence such that wn ∈ Dn . Define the operator T by T f (z) =

∞ X

n=1

where

bn (1 − |wn |2 )Kp (wn , z)

1 bn = 1 − |wn |2

Z

(z ∈ D),

f (w) dµp (w).

Dn

Proceeding as in the proof of Lemma 8.2.2, we can prove that T maps Ap into A . In order to conclude the proof it suffices to prove that T is an isomorphism for ε small enough and that Z ∞ X |f |p dA. |bn |p 6 C p

D

n=1

The proof of the latter is similar to that of Lemma 8.2.2. To prove the rest we start from the relation ∞ Z X f (z) − T f (z) = (Kp (w, z) − Kp (wn , z))f (w) dµp (w). n=1

D

8.3 The Coifman/Rochberg theorem

135

From this, by Lemma 8.3.2, we get |T F (z) − f (z)| 6 Cε = Cε

∞ Z X

n=1

Z

D

Dn

|f (w)| |Kp (z, w)| dµp (w)

|f (w)| |Kp (z, w)| dµp (w).

Now Lemma 8.2.2 shows that ||T f − f || 6 Cp ε||f ||. Finally we take ε = 1/2Cp and apply Proposition 1.2.1. ✷ It is of importance that the sequence wn in (8.3) can be chosen from the annulus {w : r < |w| < 1}, where r < 1 is fixed. 8.3.3 Theorem (Kalton [38]) Let 0 < r < 1 and ψ ∈ Ap . Then ψ can be represented as ψ(z) =

∞ X

k=0

αk

1 − |ζk |2 2

(1 − ζk z) p +1

,

where

r 6 |ζk | < 1,

X ∞

k=0

p

|αk |

1/p

6 C||ψ||Ap ,

and C depends only on r and p. In contrast to Theorem 8.3.1, here we do not assert {ζk } is independent of ψ (though this is probably true). Proof. If Γ is an arbitrary set, then the space ℓp (Γ) consists of those functions ω 7→ aω for which 1/p X < ∞. |aω |p kak := ω∈Γ

Let Γ = {w : r 6 |w| < 1}. Define the operator S : ℓp (Γ) 7→ Ap by S({aw })(z) =

X

w∈Γ

aw

1 − |w|2 . (1 − wz) ¯ 2/p+1

Let E = S(ℓp ) and Λ an arbitrary (not necessarily continuous) linear functional on Ap such that Λ(E) = 0. We want to prove Λ = 0 on Ap , which implies that S is onto; this proves the desired result, according to the open mapping theorem. Let F be the linear hull of the vectors ϕk (z) = 1/(1 − w ¯k z)2/p+1 , |wk | < r, where {wk } is the sequence from Theorem 8.3.1. If f ∈ Ap , then f = g + h, where g ∈ E, h ∈ F and ||g|| + ||h|| 6 C||f ||. Since dim(F ) < ∞, we see that Λ is bounded on F , i.e., there exists a constant C1 such that |Λh| 6 C1 ||h||, which implies that |Λf | = |Λh| 6 C2 ||f ||, and this means that Λ is bounded on Ap . Take φ(w) = Λ(ψw ), where ψw (z) = 1/(1 − wz) ¯ 2/p+1 . The function φ is antianalytic and φ(w) = 0 for r 6 |w| < 1 because Λ(E) = 0. It follows that φ(w) = 0 for every w ∈ D and in particular Λϕk = 0 for every k. By Theorem 8.3.1, this implies Λ = 0 on Ap , which was to be proved. ✷

136

8 Bergman spaces: Atomic decomposition

The case p > 1 If 1 < p < ∞, then the formulation of the Coifman/Rochberg theorem is similar to that of Theorem 8.3.1; we only have to replace (8.3) by f (z) =

∞ X

n=1

an

(1 − |wn |2 )2/q , (1 − w ¯n z)2

where 1/p + 1/q = 1 (see [99]). The case of weighted spaces The weighted Bergman space Apβ , 0 < p < ∞, β > −1, consists of those f ∈ H(D) for which kf kp,β :=

Z

D

1/p |f (z)|p (1 − |z|2 )β dA(z) < ∞.

8.3.4 Theorem Let 0 < p 6 1, β > −1 and γ > 0. (a) There exists a sequence {wn } in D and a constant C such that every f ∈ Apβ can be represented as f (z) =

∞ X

n=1

an

(1 − |wn |2 )γ (1 − w ¯n z)γ+(β+2)/p

(8.4)

with k{an }kℓp 6 Ckf kp,β . (b) Every function f of the form (8.4) with {an } ∈ ℓp belongs to Apβ and kf kp,β 6 Ck{an }kℓp . This theorem is proved in the same way as its particular case, Theorem 8.3.1 (β = 0, γ = 1); the key is in the formula f (z) = cs

Z

D

f (w)(1 − |w|2 )s−2 dA(w) (1 − wz) ¯ s

(s = γ + (β + 2)/p),

which holds for f ∈ Apβ . Envelops of Hardy spaces We have defined the q-Banach envelope of a quasi-Banach space (see p. 10). Now we can give a nontrivial example. 8.3.5 Theorem If 0 < p < q 6 1, then the q-Banach envelope of the Hardy space H p is equal to Aqβ , β = q/p − 2.

8.4 Coefficients of vector-valued functions

137

Proof. The space X = H p is embedded into the q-Banach space Y = Aqβ P∞ (see 7.4.7). On the other hand, every f ∈ Y can be represented as f = n=1 fn , where (1 − |wn |2 )γ fn = an (1 − w ¯n z)γ+1/p P and ( |an |q )1/q 6 Ckf kY . Lemma 8.2.3 shows that Z

π

−π

1 C dθ 6 , iθ γp+1 |1 − w ¯n e | (1 − |wn |)γp

which implies ∞ X

n=1

kfn kqX 6 C

∞ X

n=1

|an |q 6 Ckf kY .

Now the result follows from Proposition 1.2.5.

8.4

✷

Coefficients of vector-valued analytic functions

A function F : Ω 7→ X, where Ω is a domain in C, is said to be analytic if every point in Ω admits a neighborhood in which f can be expanded into a power series with X-valued coefficients. In the case where Ω is the unit disk, it turns out that the P∞ analyticity implies the existence of vectors Fb(n) such that F (z) = n=0 Fb(n) z n , |z| < 1, with uniform convergence on compact subsets. The vectors Fb(n), uniquely determined by F , are called the Taylor coefficients and, as one expects, satisfy the condition lim supn→∞ kFb(n)k1/n 6 1. On the other hand, if {fn } is a sequence of vectors in X, then the condition lim sup kfn k1/n 6 1

(8.5)

n→∞

P∞ is necessary and sufficient for the series n=0 fn z n to converge for every z ∈ D. In the case of convergence, the sum of that series is analytic in D. Therefore the set of the functions F : D 7→ X analytic in D can be identified with the set of the formal power series satisfying (8.5), i.e., with the set of the power series converging in D. We will denote this set by H(D, X). Derivatives The derivative of a function f ∈ H(D, X) is defined by means of P∞ power series, F ′ (z) = n=1 nFb(n) z n−1 , or by the formula F ′ (z) = lim

w→z

F (w) − F (z) . w−z

(8.6)

It should be noted, however, that the existence of the limit (8.6) for an arbitrary function F does not guarantee that F is analytic.

138

8 Bergman spaces: Atomic decomposition

Hadamard product The Hadamard product of a (scalar) function ψ ∈ H(D) and a function F ∈ H(D, X) is defined by (ψ ∗ F )(z) =

∞ X

n=0

b Fb(n) z n . ψ(n)

The proof that ψ ∗ F belongs to H(D, X) is straightforward. There holds z F ′ (z) = (ψ ∗ F )(z),

where

ψ(z) =

z . (1 − z)2

8.4.1 Proposition Let F ∈ H(D, X), where X is a p-Banach space, and let P∞ the series k=1 ψk (z) = ψ(z), where ψj ∈ H(D), converge uniformly on compact subsets. Then ∞ X (ψ ∗ F )(z) = (ψk ∗ F )(z) (|z| < 1). k=1

Proof. Let |z| = r < 1, FN (z) = Then

PN

j=1 (ψj

∗ F )(z), and RN (z) =

kFN (z) − ψ ∗ F (z)kp = kRN ∗ F (z)kp 6

∞ X

AN (j),

P∞

j=N +1

ψj (z). (8.7)

j=0

bN (j)|p kFb(j)kp |z|jp . The sequence RN (z) is uniformly bounded where AN (j) = |R on compact subsets and therefore for every ρ < 1 there exists a constant C = C(ρ) bN (j)| 6 C/ρj for all N and j. From this and the inequality kFb(j)k 6 such that |R K(ρ)/ρj , we√get AN (j) 6 M (ρ)(r/ρ2 )jp where M (ρ) = C(ρ)p K(ρ)p . Thus by taking ρ2 = r, we see that the sequence AN has a summable majorant. On the other hand, from the hypothesis RN (z) → 0, uniformly on compact subsets, it follows AN (j) → 0 (N → ∞), for every j. Therefore we can apply the dominated convergence theorem: lim

N →∞

∞ X

AN (j) =

j=0

∞ X j=0

lim AN (j) = 0.

N →∞

Now the desired assertion follows from (8.7).

✷

Inequalities for the coefficients Let H ∞ (X) denote the set of all bounded functions F ∈ H(D, X); the quasinorm is given by kF k∞,X = sup|z| 1).

(8.9)

Before proving the theorem consider a simple example. Let f ∈ H p , 0 < p < 1. Define the function F : D 7→ H p by F (w)(z) = f (zw) (z, w ∈ D). The function P∞ b n n F belongs to H(D, H p ) because F (w) = n=0 f (n)en w , en (z) = z , and the series converges for |w| < 1. The function F is bounded; moreover kF k∞ = kf kH p . An application of Theorem 8.4.3 gives |fb(n)| 6 Cp (n + 1)1/p−1 kf kH p because ken kH p = 1. Thus, in this special case, Theorem 8.4.3 reduces to the well known result of Hardy and Littlewood (Corollary 5.1.3). Proof of Theorem. In the case n = 1 the assertion is contained in Theorem 8.4.2 because F ′ (0) = Fb(1). Let n > 2. We use the atomic decomposition again. In this situation we choose G in (8.8) so that D2/p G = F ′ . Then we take ψ(w) = wn in (8.8) and get X An Fb(n + 1)wn = αk (1 − |zk |2 )F ′ (zk w), where

An =

(n + 1)! . (2/p + 1)...(2/p + n)

Since An behaves as n1−2/p , it follows that X np−2 kFb(n + 1)kp 6 C |αk |p (1 − |zk |2 )p kF ′ (zk )kp . From this and Lemma 8.4.2 it follows

np−2 kFb(n + 1)kp 6

Finally since

we get (8.9).

✷

X

p

|αk | 6 C

Z

D

X

|αk |p kF kp∞,X .

|z|np dA(z) =

2c np + 2

Vector-valued Hardy spaces Theorem 8.4.3 can be used for its own improvement. For a quasi-Banach space X let H p (X) (0 < p < ∞) denote the space of analytic functions F : D 7→ X such that 1/p  Z π 1 iθ p kF (re )k dθ < ∞. kF kp,X = sup 0 0 and A(z) = P∞ ak z k . Then Q ∗ F (w) = Proof. Let ak = Q(n k=0 Pn b(k)wk . Thus Q ∗ F (w) is the n-th coefficient in the Taylor expansion a F n−k k=0 of the function A(z)F (zw), which implies, by Theorem 8.4.4, kQ ∗ F (w)kpX 6 Cn1−p sup

0 1, then (9.1) holds with Cp = 1; if u = |f |, f analytic, then Cp = 1 for every p > 0. In the case where u = |g|, g harmonic, Theorem 9.1.1 is a formal consequence of the following theorem of Hardy and Littlewood [26]: 9.1.2 Theorem (Hardy/Littlewood) If g = Re f , where f is analytic in D, then Z Z   (9.2) |f |p dA 6 Cp | Im f (0)|p + |g|p dA . D

D

143

144

9 Subharmonic behavior

Indeed, since the function |f |p is subharmonic for every p > 0, and since |g(0)|p = |f (0)|p , we see that (9.2) implies (9.1). However, it seems more natural first to prove Theorem 9.1.1 and then deduce Theorem 9.1.2. Proof of Theorem 9.1.1 First proof.(∗) If we apply (9.1) to the disk of radius 1 − |z| centered at z ∈ D, we get Z p −2 up dA. (9.3) u(z) 6 Cp (1 − |z|) D

Hence, it is natural to consider the maximum of the function F (z) = (1−|z|)2 u(z)p , which exists under the condition that u is bounded on D. We shall assume that u is bounded(†) and that Z up dA = 1. (9.4) D

Then the function F is above semicontinuous on the closed disk and equal to zero on the boundary. Consequently, F attains its maximum at a point a ∈ D. Now we apply the sub-mean-value property of u on the disk Da of radius (1 − |a|)/2 centered at a. It follows that Z −2 u(a) 6 4(1 − |a|) u dA. Da

Writing u = up up−1 , we see that, in view of (9.4), (1−|a|)2 u(a) 6 4(Ma )1−p , where Ma is the supremum of u on Da . Assume we have found a constant Kp such that Ma 6 Kp u(a); then (1 − |a|)2 u(a) 6 4(Kp )1−p u(a)1−p , i.e., after multiplying by u(a)p−1 , u(0)p = F (0) 6 F (a) 6 4(Kp )1−p , which implies (9.1). In order to prove Ma 6 Kp u(a), let z ∈ Da . Then |z| − |a| 6 (1 − |a|)/2, whence 1 − |a| 6 2(1 − |z|). By the definition of a, there holds the inequality F (z) 6 F (a), i.e., 2  1 − |a| p u(a)p . u(z) 6 1 − |z| From these two inequalities it follows that Ma 6 41/p u(a), which completes the proof. ✷ Remark. Let u : D 7→ R be a Borel function that satisfies the condition Z C u(a)p 6 2 up dA r Dr (a) (∗) In the case of the modulus of a harmonic function, the proof is given by Kuran [47] and Fefferman and Stein [21]. Fefferman and Stein’s proof can be found in Koosis [46] and Garnett [22]. Here we present two proofs from [75]. (†) If u is not bounded, then we apply (9.1) to the functions u(ρz).

9.1 Subharmonic behavior and Bergman spaces

145

for some p > 0, provided that Dr (a) = {z : |z − a| < r} ⊂ D. Then this condition is satisfied for every p > 0. Second proof. This proof is applicable only to the case where u = |g|, g harmonic. This time we start from the inequality |∇g(a)| 6

K sup{|g(z)| : |z − a| < r}, r

(9.5)

which is valid whenever the disk Dr (a) = {z : |z − a| < r} is contained in D. (The constant K is independent of r and a.) From Lagrange’s theorem and inequality (9.5) it follows that u(a) 6 u(z) + K(t/r) sup u (s = t + r) Ds (a)

provided z ∈ Dt (a) and Ds (a) ⊂ D. The point a is chosen in the same way as in the first proof (with other hypothesis for u). Then, we choose t and r so that s = (1 − |a|)/2. Then u(w)p 6 4u(a)p for w ∈ Ds (a) and therefore u(a) 6 u(z) + Kp (t/r)u(a),

z ∈ Dt (a),

where Kp = 41/p . Next we choose t and r so that Kp (t/r) = 1/2, and get the inequality u(a)p 6 2p u(z)p , z ∈ Dt (a). Integrating this inequality in z ∈ Dt (a), we get Z t2 u(a)p 6 2p

up dA 6 2p .

Dt (a)

Since t = cp (1 − |a|), we see that u(0)p 6 (1 − |a|)2 u(a)p 6 2p /c2p , which concludes the proof. ✷ Proof of Theorem 9.1.2. Let p < 1. We start from the inequality Z ′ p |f (0)| 6 C |g(w)|p dA(w)

(ε = 1/2),

(9.6)

εD

which follows from (9.5) and (9.1); write this as Z ′ p |f (0)| 6 C |g(w)|p dτ (w)

(ε = 1/2),

εD

where dτ (w) = (1 − |w|2 )−2 dA(w). The measure dτ invariant with respect to conformal automorphisms, which means that Z Z G dτ G ◦ ϕa dτ = D

D

146

9 Subharmonic behavior

for every positive Borel function G, where ϕa (z) =

a−z 1−a ¯z

(|a| < 1).


The automorphism ϕa has the property ϕa ϕa (z) ≡ z. Hence, applying (9.6) to the function f ◦ ϕa , we get Z Z p ′ p 2 p |f (z)|p dτ (z), (9.7) |g(ϕa (w)| dτ (w) = C |f (a)| (1 − |a| ) 6 C Hε (a)

εD

where Hε (a) = ϕa (εD) = {z : |ϕa (z)| < ε}. Integrating inequality (9.7) and changing the order of integration, we obtain Z Z Z dA(a). (9.8) |f (z)|p dτ (z) |f ′ (a)|p (1 − |a|2 )p dA(a) 6 C Hε (z)

D

D

The set Hε (z) is a hyperbolic disk centered at z. The euclidean area of Hε (z) is “proportional” to (1 − |z|2 )2 , so (9.8) implies Z Z (9.9) |f ′ (a)|p (1 − |a|2 )p dA(a) 6 Cp |g(a)|p dA(a). D

D

Now it is enough to prove the inequality Z Z   |f |p dA 6 Cp |f (0)|p + |f ′ (a)|p (1 − |a|2 )p dA(a) ; D

(9.10)

D

indeed, (9.2) is implied by (9.9), (9.10) and (9.1). In order to prove (9.10), we start from Z 1 Z ∞ X 2−n Mpp (rn , f ) |f |p dA = Mpp (r, f )r dr 6 2 0

D

n=0

= 2|f (0)|p +

∞ X

n=0


2−n Mpp (rn+1 , f ) − Mpp (rn , f ) ,

where rn = 1 − 2−n . Hence, by Proposition 7.1.6, Z

D

|f |p dA 6 2|f (0)|p + Cp

∞ X

2−n 2−np Mpp (rn+1 , f ′ ),

n=0

from which (9.10) is obtained immediately. In the case p > 1 we can proceed in a similar way; we only have to modify the proof of (9.10) (see [61]). Also, we can proceed as in the proof of Riesz’ theorem 6.2.6, i.e., if 1 6 p 6 2 apply the preceding result to the function f 2 ; etc. ✷

9.2 The space hp , p < 1

147

9.1.3 Remark As Theorem 9.1.2 shows, the space hAp is “self-conjugate” for every p > 0, in contrast to the case of hp , which is self-conjugate only for p > 1. Using Riesz’ theorem on conjugate functions, we deduced that the sequence en (reiθ ) = r|n| einθ is a Shauder basis of hp (Theorem 6.3.1), which implies that {en } is a basis of hAp for p > 1. However, this sequence is not a basis of hAp for p 6 1. More precisely, for every p 6 1 there exists a function f ∈ Ap the Taylor series of which does not converge in Ap . 9.1.4 Remark The space hAp is equal to the direct sum of its subspaces Ap and A¯p0 = {f¯ : f ∈ Ap , f (0) = 0}, for every p > 0. This can be used to show that hAp and Ap are isomorphic for every p > 0 (see 6.3.3).

9.2

The space hp , p < 1

Recall that the harmonic Hardy space hp , 0 < p < ∞, is defined by the requirement kukp := sup Mp (r, u) < ∞. r c log (0 < r < 1) ? (9.14) 1−r Hardy and Littlewood [26] proved that the answer is positive provided that p = 1/N , N = 2, 3, . . . ; their example is u(z) = DN P (z) =

∂N P (reiθ ), ∂θN

where P is the Poisson kernel. It is clear that u satisfies condition (9.13) for p = 1/N . That condition (9.14) is satisfied follows from the estimate (e.g., [89]) |DN P (z)| 6 C

1 − |z|2 . |1 − z|N +1

It should be noted that solving Problem A leads to solution of B. Namely, condition (9.13) implies condition (9.14). Indeed, if (9.13) is satisfied, then |fb(n)| > c(n + 1)1/p−1 , so the conclusion follows from the inequality Mpp (r, f ) > c

∞ X

n=0

(see Theorem 7.5.1).

(n + 1)p−2 |fb(n)|p rnp

9.2.3 Exercise (Hardy/Littlewood) Let u = Re f , f ∈ H(D), α > 0, and 0 < p < ∞. Then the following three conditions are equivalent: Mp (r, f ) = O(1 − r)−α ;

9.3

Mp (r, u) = O(1 − r)−α ;

Mp (r, f ′ ) = O(1 − r)−α−1 .

Subharmonic behavior of smooth functions

In this section we present some results that are closely related to Theorem 9.1.1. Unless otherwise stated, we consider real-valued functions defined on a proper subdomain G of C. For the proofs of slightly more general results we refer to [75, 77]. For generalizations to the case of the so called M-harmonic functions, see [73, 34].

9.3 Subharmonic behavior of smooth functions

149

The class SH(G). For a constant K > 1, let SH K (G) denote the class of nonnegative, continuous functions u on G such that Z u(z) 6 Kr−2 u dm Dr (z)

whenever Dr (z) := {w : |w − z| < r} ⊂ G. We put SH(G) =

S

K>1

SHK (G).(‡)

As remarked on page 144, the first proof of Theorem 9.1.1 gives the following result: 9.3.1 Proposition Let p > 0. If u ∈ SH K (G), then up ∈ SH C (G), where C is a constant depending only on K and p. Classes of C 1 functions The class HC 1 (G) property

The second proof of Theorem 9.1.1 (p. 145) was based on the |∇f (z)| 6 Kr−1 sup |f |,

Dr (z) ⊂ G,

Dr (z)

(9.15)

1 where K > 1 is a constant independent of Dr (z) ⊂ G. We denote by HCK (G) the class of all locally Lipschitz functions f on G satisfying (9.15). Note that (9.15) is implied by |∇f (z)| 6 K|f (z)|/δG (z), δG (z) = dist(z, ∂G), (9.16)

which is a restriction on the growth of f and is therefore stronger than (9.15). 9.3.2 Example (a) It is a simple but important fact that condition (9.16) is satisfied if f is a positive function harmonic in G. (b) If f = |g ′ |, where g is a univalent function in G = D, then the proof of Theorem 4.3.5 shows that |∇f (z)| = |g ′′ (z)| 6 1 The class OCK (G)

3f (z) . 1 − |z|2

1 is the subclass of HC2K (G) consisting of those f such that

|∇f (z)| 6 Kr−1 Of (z, r),

Dr (z) ⊂ G,

where Of (z, r) is the oscillation of f on Dr (z), Of (z, r) = sup{ |f (w) − f (z)| : w ∈ Dr (z) }. 1 9.3.3 Example If a function f : G 7→ R is convex or harmonic, then f ∈ OCK (G) x 1 for some K independent of f . In particular the function f (x+iy) = e is in HC (G), where G is the right half-plane, but f does not satisfy (9.16). (‡) In

defining other classes we proceed in a similar way.

150

9 Subharmonic behavior

9.3.4 Example As a further example, we have that |f | ∈ OC 1 (G), if f is analytic in G. (See Lemma 10.2.4.) 1 9.3.5 Theorem (a) If f ∈ HCK (G), then |f | ∈ SH C (G), where the constant C depends only on K. (b) If f ∈ OC 1 (G), then both |f | and |∇f | belong to SH(G).

We note two consequences of assertion (a). 9.3.6 Corollary Let p > 0. A function f ∈ C 1 (G) belongs to HC 1 (G) if and only if there is a constant K such that Z p −2−p |∇f (z)| 6 Kr |f |p dm, 0 < r < δG (z). Dr (z)

Let Op f (z, r) = p

n

1 |Dr (z)|

Z

Dr (z)

the L -oscillation over Dr (z).

o1/p , |f (w) − f (z)|p dm(w)

9.3.7 Corollary Let p > 0. A function f belongs to OC 1 (G) if and only if |∇f (z)| 6 Kr−1 Op f (z, r), 0 < r < δG (z), for some constant K. This is deduced from the preceding corollary by considering the functions f − const. Proof of Theorem 9.3.5. The proof of (a) is the same as the second proof of Theorem 9.1.1, p. 145. It remains to prove that f ∈ OC 1 (G) implies |∇f | ∈ SH(G). 1 Let f ∈ OCK (G). By Theorem 9.3.1, it suffices to prove that, for some q, the function |∇f |q belongs to SH(G). This can be reduced to proving that Z |∇f |q dm |∇f (0)|q 6 C D

provided that D ⊂ G. By Corollary 9.3.7, we have Z |f (z) − f (0)| dm(z). |∇f (0)| 6 K D

On the other hand, |f (z) − f (0)| 6 |z| whence

Z

D

|f (z) − f (0)| dm(z) 6

Z

0

Z

1

0

1

dr

|∇f (rz)| dr, Z

D

|∇f (rz)| |z| dm(z).

Hence, by the change z = w/r and Fubini’s theorem, Z Z 1 Z |∇f (0)| 6 K |∇f (w)| dm(w) r−3 |w| dr 6 K |∇f (w)||w|−1 dm(w). D

|w|

D

9.3 Subharmonic behavior of smooth functions

151

Now the required inequality is proved by H¨older’s inequality with the indices q = 3 and q ′ = 3/2, using the fact that the function w 7→ |w|−1 belongs to the space L3/2 (D, dm). ✷ Classes of C 2 -functions The class HC 2 (G)

consists of those f ∈ C 2 (G) for which |∆f (z)| 6 Kr−1 sup |∇f | + K0 r−2 sup |f | Dr (z)

(9.17)

Dr (z)

for some constants K, K0 . The condition |∆f (z)| 6 K|∇f (z)|δG (z)−1 + K0 |f (z)|δG (z)−2 implies (9.17) with the same values of K and K0 . The class OC 2 (G)

consists of those f ∈ C 2 (G) such that |∆f (z)| 6 Kr−1 sup |∇f |

(9.18)

Dr (z)

for some constant K. In particular, OC 2 (G) contains every function f for which |∆f (z)| 6 K|∇f (z)|/δG (z),

z ∈ G.

By Lagrange’s theorem, f belongs to OC 2 (G) iff |∆f (z)| 6 Kr−1 sup |∇f | + K0 r−2 O(f, r)

for some K and K0 .

Dr (z)

9.3.8 Example Condition (9.18) is satisfied if f is a harmonic function on an arbitrary domain G. Let f be an eigenfunction of ∆, i.e., ∆f ≡ λf for some constant λ. Assuming that the closure of rD = Dr (0) is contained in G, we have Z π Z 1 d 2 ∆f dm. f (reiθ ) dθ = r−1 2π −π dr rD Hence r∆f (0) = 2

1 2π

Z

π

−π

d f (reiθ ) dθ − r−1 dr

Z

rD

(f − f (0)) dm,

and hence |∆f (0)| 6 2r−1 sup |∇f | + r|λ| sup |∇f |. rD

rD

Applying this to the function w 7→ f (w +z) we conclude that f ∈ OC 1 (G) provided G is bounded. On the other hand, if f (x+iy) = sin x, and G is the right half-plane, then ∆f = −f but f is not in OC 2 (G).

152

9 Subharmonic behavior

9.3.9 Theorem The following inclusions hold: (a) HC 2 (G) ⊂ HC 1 (G); and (b) OC 2 (G) ⊂ OC 1 (G). 9.3.10 Corollary A function f ∈ C 2 (G) belongs to HC 2 (G) if and only if there is a constant K such that |∆f (z)| 6 Kr−2 sup |f |. Dr (z)

The proof of Theorem 9.3.9 is based on the following lemmas. 9.3.11 Lemma If f : Dr (z) → R is a C 2 -function, then |∇f (z)| 6 2r−1 sup |f | + (2/3) r sup |∆f |. Dr (z)

(9.19)

Dr (z)

This inequality is a consequence of the formula Z π Z 1 1 ∂f iθ −iθ (0) = f (e )e dθ − ∆f (w)(1 − |w|2 )w−1 dm(w), ∂z 2π −π 4π D where f is a C 2 -function defined in a neighborhood of the closed unit disk. The verification of this formula can be reduced to the case of f (z) = z m z¯n , where m, n are nonnegative integers. 9.3.12 Lemma Let F1 , F2 and F3 be nonnegative, continuous functions on G such that, for some constant K, F1 (z)/K 6 r−1 sup F2 + r sup F3 Dr (z)

and

(9.20)

Dr (z)

F3 (z)/K 6 r−1 sup F1 + r−2 sup F2 Dr (z)

(9.21)

Dr (z)

whenever Dr (z) ⊂ G. Then there is a constant C = C(K) such that F1 (z) 6 Cr−1 sup F2 .

(9.22)

Dr (z)

Proof. By translations the proof of (9.22) reduces to the case z = 0. Let the closure of Dε (0) be contained in G and F2 6 1 on Dε (0). (In the general case we consider the functions Fi /A, where A is chosen so that F2 (z) 6 A for all z ∈ Dε (0).) Choose z ∈ Dε (0) so that F1 (w)(ε − |w|) 6 F1 (z)(ε − |z|) for all w ∈ Dε (0). This implies that F1 (w) 6 2F1 (z) for w ∈ Dδ (z), where δ = (ε − |z|)/2. Now we use the hypotheses to find w in Dδ (z) so that F1 (z)/K 6 r−1 +(Kr/t)F1 (w)+Krt−2 for all r, t > 0 such that r + t = δ, which implies F1 (z)/K 6 r−1 + (2Kr/t)F1 (z) + Krt−2 . Now choose r, t so that r + t = δ and 2Kr/t = 1/2K, which implies that r = c1 (ε − |z|), t = c2 (ε − |z|) for some ci = ci (K), to obtain F1 (z)/K 6 F1 (z)/2K + −2 K1 (ε − |z|)−1 , where K1 = c−1 1 + c1 c2 . Hence F1 (0)ε 6 F1 (z)(ε − |z|) 6 2KK1 , and this concludes the proof. ✷ Proof of Theorem 9.3.9. Let f satisfy (9.17). We may assume that K > 2 and K0 > 2. Define the functions F1 (z) = |∇f (z)|, F2 (z) = |f (z)|, F3 (z) =

9.3 Subharmonic behavior of smooth functions

153

|∆f (z)|. Then (9.20) is satisfied because of (9.19), and (9.21) is satisfied because of (9.17). Hence f ∈ HC 1 (G), by Lemma 9.3.12. This proves assertion (a). To prove (b) let f ∈ OC 2 (G). Applying (a), together with its proof, to the functions f − c we find a constant K1 independent of z, r, c so that |∇f (z)| 6 K1 r−1 supDr (z) |f − c|. Finally we take c = f (z) to finish the proof. ✷ Vector-valued functions The preceding notions and results can be extended to vector-valued functions. For example, if f = (f1 , f2 ) is a function from G to R2 , then we replace |∇f (z)| by kDf (z)k, where Df (z) : R2 7→ R2 is the derivative of f at z. The Laplacian of f is defined as ∆f = (∆f1 , ∆f2 ). Applying Lemma 9.3.11 to the functions af1 + bf2 , a2 + b2 = 1, we find that kDf (z)k 6 2r−1 sup |f | + (2/3) r sup |∆f |. Dr (z)

Dr (z)

It turns out that Theorem 9.3.9 remains valid. As an application we note a sufficient condition for a C 3 -function to be in OC 2 (G). 9.3.13 Theorem A real valued C 3 -function f belongs to OC 2 (G) if there are constants K1 and K2 such that |∇(∆f )(z)| 6 K1 r−1 sup ∇2 (f ) + K2 r−2 sup |∇f |. Dr (z)

Here ∇2 (f ) =

(9.23)

Dr (z)

 2 1/2 ∂ f 2 ∂ 2 f 2 ∂ 2 f 2 . 2 + 2 + ∂x ∂y ∂x∂y

Proof. Suppose that a real-valued function f satisfies (9.23). Since ∇(∆f ) = ∆(∇f ) and kD(∆f )k ≍ ∇2 (f ), we see that condition (9.23) means that ∇f ∈ HC 2 (G). Therefore (9.23) implies ∇f ∈ HC 1 (G), by Theorem 9.3.9, which means kD(∇f )(z)k 6 Kr−1 supDr (z) |∇f | for some constant K. Since obviously |∆f | 6 const · kD(∇f )k, it follows that f ∈ OC 2 (G). ✷ Remark. Condition (9.23) implies the formally stronger condition |∇(∆f )(z)| 6 Kr−2 sup |∇f |. Dr (z)

9.3.14 Corollary A C 4 -function f : G → R belongs to OC 2 (G) if so does ∆f . Consequently a C ∞ -function f belongs to OC 2 (G) if so does ∆k f for some integer k. In particular every polyharmonic function of finite order belongs to OC 2 . A function f is polyharmonic of order k, where k is a positive integer, if ∆k f ≡ 0. For the theory of polyharmonic functions we refer to [5]. Proof. Let ∆f ∈ OC 2 . Then ∆f ∈ HC 1 , by Theorem 9.3.9, which means that |∇(∆f )(z)| 6 Kr−1 supDr (z) |∆f |. Now the desired conclusion follows from Theorem 9.3.13. ✷

10

Lipschitz spaces

In this chapter we are concerned with some results which relate the modulus of smoothness of n-th order of a function g ∈ C(T) with the n-th tangential derivative of the Poisson integral of g. In Section 10.1 we consider the case n = 1 (Theorem 10.1.1); as an application we prove Privalov’s theorem on conjugate functions (Theorem 10.1.3 and 10.1.4). In Section 10.2 we use Section 10.1 to prove some results on the Lipschitz condition for the modulus of an analytic function. The case n > 2 is considered in the next two sections.

10.1

Lipschitz spaces of first order

The space Λα (K) If K is a bounded subset of C, then, by definition, Λα (K) (0 < α 6 1) is the set of all complex-valued functions g on K such that |g(z) − g(w)| 6 C |z − w|α

(z, w ∈ K),

where C is a constant independent of z, w. The space Lip(ω, K) More generally, let ω be a majorant, i.e., a continuous increasing function on [0, t0 ], where t0 is large enough, such that ω(0) = 0 and that ω(t)/t decreases (t > 0). Then the space Lip(ω, K) is defined by the requirement |g(z) − g(w)| 6 C ω(|z − w|).

(10.1)

The norm is given by Cg + |g(a)|, where Cg = C (> 0) is the smallest constant satisfying (10.1) and a is any fixed point from K; with this norm the space Lip(ω, K) is Banach. Since maxK |g| 6 |g(a)| + Cg ω(diam K), we see that the inclusion Lip(ω, K) ⊂ L∞ (K) is continuous. Lipschitz condition and the tangential derivative In the case K = T condition (10.1) is equivalent to |g(eit ) − g(eiθ )| 6 C ω(|t − θ|), t, θ ∈ R (the value of C need not be the same). In particular, this means that the “classical” Lipschitz space Λ1 (T) consists of absolutely continuous functions g ∈ C(T) with the bounded derivative (d/dθ)g(eiθ ), and that therefore is isomorphic to L∞ (T). Since Z π Z π d ∂ it P (r, θ − t) g(e ) dt = P (r, θ − t)g(eit ) dt, dt ∂θ −π −π we can conclude that g belongs to Λ1 (T) iff D(P [g]) belongs to h∞ , where D is the operator of “tangential” differentiation, (Du)(reiθ ) := ∂u(reiθ )/∂θ. This can be generalized in the following way. 154

10.1 Lipschitz spaces of first order

155

10.1.1 Theorem Let ω be a majorant such that ω(t) tα

is almost increasing for some α > 0.

(10.2)

Then a function u, harmonic in D, is equal to the Poisson integral of some function g ∈ Lip(ω, T) iff there exists a constant C such that |Du(z)| 6 C

ω(1 − |z|) 1 − |z|

for all z ∈ D.

(10.3)

A real function ϕ, defined on some interval, is called almost increasing if there exists a constant C such that x < y implies ϕ(x) < Cϕ(y). (An almost decreasing function is defined similarly.) Let A(D) denote the disk-algebra, i.e., the set of functions that are analytic in D and have a continuous extension to the boundary. Since Df (reiθ ) = ireiθ f ′ (reiθ ), f ∈ A(D), we have the following consequence. 10.1.2 Corollary Let ω be as in the theorem and let f ∈ A(D). The boundary function f∗ belongs to Lip(ω, T) iff |f ′ (z)| 6 C

ω(1 − |z|) 1 − |z|

(z ∈ D).

This last condition implies f ∈ A(D). Theorem 10.1.1 is an immediate consequence of the following assertions which will be proved later on in a more general situation (see Propositions 10.4.4 and 10.4.5); the direct proof is rather simple. If u = P [g], g ∈ C(T), then M∞ (r, Du) 6 C(1 − r)−1 ω(g, 1 − r), 0 < r < 1, where ω(g, t) is the modulus of continuity of g. R1 On the other hand, if u ∈ h(D) and 0 M∞ (r, Du) dr < ∞, then u = P [g] for some g ∈ C(T) satisfying Z 1 ω(g, t) 6 C M∞ (r, Du) dr, 0 < t < 1. 1−t

Privalov’s theorem on conjugate functions The condition g ∈ Λ1 (T) does not imply that the radial derivative ∂P [g]/∂r is bounded (in contrast to the tangential derivative) because the functions ∂u/∂θ and r∂u/∂r represent an arbitrary pair of harmonic conjugates. However, under an additional hypothesis on the majorant we have the following theorem of Privalov: 10.1.3 Theorem Let (10.2) be satisfied and suppose that ω(t)/tβ

is almost decreasing for some β < 1.

(10.4)

If u = P [g] and g ∈ Lip(ω, T), then the conjugate function u e has a continuous extension to D and its boundary function belongs to Lip(ω, T).

156

10 Lipschitz spaces

In other words, If conditions (10.2) and (10.4) are satisfied, then the Hilbert operator maps Lip(ω, T) to Lip(ω, T). Proof of Theorem 10.1.3. Let u be real-valued. The function v = Du is harmonic and therefore |∇v(z)| 6 2R−1 sup{ |v(w)| : |w − z| < R}, where R = (1 − |z|)/2. Using the hypotheses and Theorem 10.1.1, we get |∇v(z)| 6 C

ω(1 − |z|) . (1 − |z|)2

Since |∇v| > |∂v/∂θ|/r = |∂ 2 u/∂θ2 |/r and ∂ 2 u/∂θ2 = −r2 ∂ 2 u/∂r2 , it follows that ∂2u ω(1 − r) , 0 < r < 1, 2 (reiθ ) 6 C ∂r (1 − r)2 where C is independent of reiθ . Integrating this from r = 0 and using (10.4) we get ∂u ω(1 − r) , (reiθ ) 6 C ∂r 1−r which together with (10.3) gives |∇e u(z)| = |∇u(z)| 6 C

ω(1 − |z|) . 1 − |z|

(10.5)

e ∈ Lip(ω, D) (see Lemma 10.1.6 below). ✷ Finally, (10.2) and (10.5) imply u According to the above proof, we have the following theorem on extension of a Lipschitz condition from T to D. 10.1.4 Theorem (a) Let ω satisfy (10.2) and (10.4). Then the function g belongs to Lip(ω, T) iff the Poisson integral of g belongs to Lip(ω, D). (b) If f ∈ A(D), then (10.2) is sufficient for the validity of the implication f∗ ∈ Lip(ω, T) ⇒ f ∈ Lip(ω, D). 10.1.5 Remark There are cases where a Lipschitz condition on the circle extends to the disk with saving the corresponding Lipschitz constant. For example, if f ∈ A(D), 0 < α 6 1 and |f (ζ)−f (η)| 6 |ζ −η|α , ζ, η ∈ T, then |f (z)−f (w)| 6 |z −w|α , z, w ∈ D. The latter is equivalent to |f (z) − f (w)| 6

|z − w| |1 − wz| ¯ 1−α

(z, w ∈ D),

and this implies |f ′ (z)| 6 (1 − |z|2 )α−1 , z ∈ D.

10.1.6 Lemma If a C 1 -function u : D 7→ C satisfies (10.5) and ω satisfies the (Dini) condition Z x ω(t) ω1 (x) = dt < ∞, x > 0, (10.6) t 0 then u ∈ Lip(ω1 , D).

10.1 Lipschitz spaces of first order

157

Proof (cf. [84, Lemma 6.4.8]). Let ω(1 − |z|) , 1 − |z|

|∇u(z)| 6

z ∈ D.

Let |a| 6 |b| 6 1. By Lagrange’s theorem, |u(a) − u(b)| 6

ω(1 − |c|) |a − b|, 1 − |c|

where c = (1 − λ)a + λb for some λ ∈ (0, 1). Since |c| 6 |b| and ω(t)/t decreases, we see that ω(1 − |b|) ω(1 − |c|) 6 , 1 − |c| 1 − |b| hence |u(a) − u(b)| 6 ω(|a − b|) 6 ω1 (|a − b|), under the condition |a − b| 6 1 − |b|. If 1 − |b| 6 |a − b| 6 1 − |a|, then |u(a) − u(b)| 6 |u(a) − u(b′ )| + |u(b′ ) − u(b)|, where b′ = (1 − δ)b/|b|, δ = |a − b|. Using Lagrange’s theorem as above we get |u(a) − u(b′ )| 6

ω(δ) ω(1 − |b′ |) |a − b′ | = |a − b′ | 6 ω(δ) 6 ω1 (δ). ′ 1 − |b | δ

In the case of |u(b′ ) − u(b)|, we have ′

|u(b ) − u(b)| 6

Z

|b|

|b′ |

ω(1 − t) dt 6 1−t

Z

1

1−δ

ω(1 − t) dt = ω1 (δ). 1−t

Finally, if δ > 1 − |a|, we use the inequality |u(a) − u(b)| 6 |u(a) − u(a′ )| + |u(a′ ) − u(b′ )| + |u(b′ ) − u(b)|, where a′ = (1 − δ)a/|a|, and then proceed in a similar way as above. ✷ 10.1.7 Remark If ω satisfies (10.6), then ω1 is a concave majorant. Let ω2 (x) = x

Z

1

x

ω(t) dt, t2

0 < x < 1.

This function is not increasing but has the properties: ω2 is concave, ω2 (t)/t is decreasing and ω2 (0+) = 0. It follows that ω2 is a majorant near 0. We have Z x ω2 (t) dt = ω1 (x) + ω2 (x), 0 < x < 1. t 0 Then Lemma 10.1.6 and the proof of Privalov’s theorem yield: Suppose that ω satisfies the Dini condition. (a) If u = P [g] and g ∈ Lip(ω, T), then the conjugate function u e has a continuous extension to D, and, moreover, u and u e belong to Lip(ω3 , D), where ω3 (x) = ω1 (x) + ω2 (x). (b) If f ∈ A(D) and f∗ ∈ Lip(ω, T), then f ∈ Lip(ω1 , D).

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10 Lipschitz spaces

10.1.8 Remark A majorant ω is said to be regular if there exists a constant C such that Z x Z 1 ω(t) ω(t) dt + x dt 6 Cω(x) (0 < x < 1). t t2 0 x It is easily verified that condition (10.2)&(10.4) implies that ω is regular. The converse is true as well. Moreover, ω satisfies (10.2) iff ω1 (x) 6 Cω(x) for some constant C; and ω satisfies (10.4) iff ω2 (x) 6 Cω(x) for some constant C.

10.2

Lipschitz condition for the modulus

A function f ∈ H(D) satisfies the condition |f (z) − f (w)| 6 |z − w| in D iff |f ′ | 6 1 in D. the corresponding Lipschitz condition for |f | is satisfied On the other hand, iff ∇|f | 6 1. Since ∇|f | = |f ′ |, we conclude that there holds the relation f ∈ Λ1 (D) ⇐⇒ |f | ∈ Λ1 (D),

f ∈ H(D).

This is the simplest case of the following theorem. 10.2.1 Theorem Let ω satisfy the Dini condition (10.6). If f ∈ H(D) and |f | ∈ Lip(ω, D), then f ∈ Lip(ω1 , D). The theorem states, in particular, that if |f | ∈ Lip(ω, D), and ω satisfies (10.6), then f ∈ A(D). On the other hand, there exists a function f ∈ H(D) r A(D) such that the function |f | has a continuous extension to the closed unit disk. To show this, we use the known fact that there exists a bounded analytic function u + iv such that u is continuous on D, while v has no continuous extension to D. Then there are a point ζ ∈ T, two sequences {zn } ⊂ D and {wn } ⊂ D tending to ζ, and two points a, b ∈ C (a 6= b) such that v(zn ) → a and v(wn ) → b. We can assume that eia 6= eib since otherwise we can consider the function (u + iv)/λ for a suitable λ > 0. Then the desired function is f = exp(u + iv). As a consequence of the above theorem we have the following result of Dyakonov. 10.2.2 Theorem [20] Let ω satisfy (10.2). A function f ∈ H(D) belongs to the space Lip(ω, D) iff |f | belongs to the same space. Before stating another theorem of Dyakonov recall that the function |f | is subharmonic and consequently P [|f∗ |](z) − |f (z)| > 0, for all z ∈ D. We also know that P [|f∗ |](z) − |f (z)| → 0 as |z| → 1, (†) so it is a natural question how fast the convergence in (†) can be. It turns out that this is closely related to Lipschitz condition for f , i.e., to growth of the first derivative. 10.2.3 Theorem (Dyakonov [20]) Let ω satisfy (10.2) and (10.4), f ∈ A(D). Then the following conditions are equivalent:

10.2 Lipschitz condition for the modulus

159

(i) |f∗ | ∈ Lip(ω, T) and |f (eiθ )| − |f (reiθ )| 6 Cω(1 − r) for some constant C; (ii) |f∗ | ∈ Lip(ω, T) and P [|f∗ |](z) − |f (z)| 6 Cω(1 − |z|) for some constant C; (iii) f ∈ Lip(ω, D). Dyakonov’s proofs are based on theorems on pseudoanalytic continuation and theorems on division by inner functions. Here we present the proof from [78]. The key is in connection between the modulus of the first derivative and the oscillation of the modulus of the function. 10.2.4 Lemma Let Dz = {w : |w − z| 6 1 − |z|}, f ∈ A(D). Then
1 (1 − |z|)|f ′ (z)| 6 sup |f (w)| − |f (z)| 2 w∈Dz

(z ∈ D).

Proof. Let Mz = sup{|f (w)| : w ∈ Dz }. If z = 0 and M0 = 1, then Schwarz’ lemma gives |f ′ (0)| 6 1 − |f (0)|2 6 2(1 − |f (0)|), which is the required inequality in a special case. In the general case we apply this special case to the function F (ζ) = f (z + ζ(1 − |z|))/Mz , ζ ∈ D. ✷ Proof of Theorem 10.2.1 Assuming |f | ∈ Lip(ω, D), we have |f (w)| − |f (z)| 6 Cω(|w − z|) 6 Cω(1 − |z|) for every z ∈ D and w ∈ Dz . Taking the supremum over w ∈ Dz and using Lemma 10.2.4, we get |f ′ (z)|(1 − |z|) 6 2Cω(1 − |z|). Now the result follows from Lemma 10.1.6. ✷ Proof of Theorem 10.2.3. [78] The implication (iii) ⇒ (i) is trivial. Consider the implication (i) ⇒ (ii). Assuming (i), let h(z) = P [|f∗ |](z), f ∈ Lip(ω, D) =: X and |f∗ | ∈ Lip(ω, T). Then h ∈ X, by Theorem 10.1.4, and |f | ∈ X because f ∈ X. Hence h(z) − |f (z)| = h(z) − |f (z/|z|)| + |f (z/|z|)| − |f (z)| = h(z) − h(z/|z|) + |f (z/|z|)| − |f (z)| 6 Cω(1 − |z|) for z ∈ D r {0}, which implies (ii). In order to prove that (ii) implies (iii), we start from the inequality |f (w)| − |f (z)| 6 h(w) − |f (z)| = h(w) − h(z) + h(z) − |f (z)|, valid for z ∈ D and w ∈ Dz . From the hypothesis |f∗ | ∈ Lip(ω, T) and Theorem 10.1.4 it follows that h(w) − h(z) 6 Cω(|w − z|) 6 Cω(1 − |z|),

w ∈ Dz .

By (ii), we have h(z) − |f (z)| 6 Cω(1 − |z|), so we get |f (w)| − |f (z)| 6 Cω(1 − |z|),

w ∈ Dz .

The proof is now concluded as in the case of Theorem 10.2.2.

✷

160

10 Lipschitz spaces

10.3

Lipschitz spaces of higher order

Moduli of smoothness If h is a complex-valued function defined on R, then ∆nt h (n is a positive integer, t ∈ R) denotes the n-th symmetric difference with step t : ∆1t h(θ) = h(θ + t) − h(θ)

(θ ∈ R),

and ∆nt h = ∆1t ∆nt h

(n > 2).

In particular, ∆2t h(θ) = h(θ + 2t) − 2h(θ + t) + h(θ). In the general case there holds the formula n   X n (−1)n−k h(θ + kt). ∆nt h(θ) = k k=0

If g is a function on the unit circle, then ∆nt g is defined by ∆nt g(eiθ ) = ∆nt h(θ), where h(θ) = g(eiθ ). For fixed n and t, ∆nt is a linear operator which preserves C(T); we have k∆nt gk 6 2n kgk, where k · k = k · k∞ denotes the max-norm in C(T). The modulus of smoothness of order n is defined by ωn (g, t) = sup{k∆ns gk : |s| < t}, t > 0, g ∈ C(T). Lipschitz spaces Let φ be a positive function on (0, π] and let n be an integer. The Lipschitz space Lipn (φ) consists of those functions g ∈ C(T) for which ωn (g, t) = O(φ(t)),

t → 0;

kgkφ,n = |b g (0)| + sup

ωn (g, t) . φ(t)

the norm is defined by

0 0, belongs to Λ∗ for α = 1, and to Λα (T) for α < 1. It is known that for α 6 1 the function g is nowhere differentiable (Weierstrass, Hardy).

10.3 Lipschitz spaces of higher order

161

The spaces h∞,n (ψ) Let h∞ (ψ) denote the class of the functions u ∈ h(D) for which u(z) = O (ψ (1/(1 − r))), r = |z| → 1− , where ψ(x), x > 1, is a positive nonincreasing function that grows slower than some power of x. The latter means β that, for some β > 0, the function ψ(x)/x is almost decreasing. Such functions
γ are, for example ψ(x) = xα log(2x) , where α > 0 and β ∈ R, or α = 0 and β > 0. In the simplest(∗) case, when ψ(x) ≡ 1, we have h∞ (ψ) = C(T), and then, as we have seen, the Poisson integral acts as an isometrical isomorphism from C(T) onto hC(D) = C(D) ∩ h(D) (see Theorem 3.1.7). Hence every subclass of C(T) can be regarded as a subclass of hC(D), and conversely. Following the paper [71], we shall show that any h∞ (ψ) is isomorphic to some Lipschitz space via the tangential derivative of sufficiently large order. Under the above hypothesis on ψ, we define kukψ = |u(0)| + sup

0 1). xα

(Uα )

Concerning the function φ, we shall assume that it positive and almost increasing on (0, 1] and that there exists β > 0 such that φ(t) is almost decreasing (0 < t < 1). tβ (∗) with

respect to ψ

(Uβ0 )

162

10 Lipschitz spaces

10.4

Growth of derivatives

The following theorem is classical and is due Hardy and Littlewood, Zygmund, and Privalov (see [18, Ch. 2, §§1,2], [100], and [67]). 10.4.1 Theorem Let 0 < a 6 n (n = 1, 2, . . . ) and u ∈ h(D). Then the following two conditions are equivalent: u ∈ hC(D)

& ωn (u∗ , t) = O(ta ) ;

n

a−n

D u(z) = O((1 − |z|)

).

(10.7) (10.8)

If in addition a < n and u = Re f , f ∈ H(D), then the condition f (n) (z) = O((1 − |z|)a−n ) is equivalent to each of the preceding. Recall that hC(D) = C(D) ∩ h(D). In this section we extend Theorem 10.4.1 in the following way. 10.4.2 Theorem If φ(t)/tn almost decreases and φ(t)/tb almost increases, for some b > 0, then the Poisson integral acts as an isomorphism of Lipn (φ) onto h∞,n (ψ), where ψ(x) = xn φ(1/x), x > 1. Remark. The analogous assertion for the pair lipn (φ), ho,n (ψ) is also valid. This theorem is contained in the following. 10.4.3 Theorem Let n be a positive integer and let ψ and φ satisfy conditions (U ) and (Un0 ). Then the following assertions are equivalent: (a) Lipn (φ) = h∞,n (ψ); (b) There are constants α < n and C such that ψ satisfies (Uα ) and φ(t)/C 6 tn ψ(1/t) 6 Cφ(t),

0 < t < 1.

Condition (Un0 ) is natural because the modulus of smoothness has the property that ωn (g, t)/tn almost decreases (Lemma 10.4.7). However, the theorem cannot be applied to the case φ(t) =

1 (1 + log(1/t))b

(b > 0),

and we do not know whether the space Lipn (φ) is isomorphic to some of the spaces h∞ (ψ). Proof of Theorem 10.4.3 The implication (b)⇒(a) is a consequence of the following two propositions, which will be proved later on.

10.4 Growth of derivatives

163

10.4.4 Proposition If u ∈ hC(D), then

M (r, Dn u) 6 C(1 − r)−n ωn (u∗ , 1 − r), 0 < r < 1,

(10.9)

where C < ∞ depends only on n (n = 1, 2, . . . ). For the proof see page 166. 10.4.5 Proposition If u ∈ h(D) and Z 1 (1 − r)n−1 M (r, Dn u) dr < ∞,

(10.10)

0

then u ∈ hC(D) and

ωn (u∗ , t) 6 C

Z

1

1−t

(1 − r)n−1 M (r, Dn u) dr,

0 < t < 1,

(10.11)

where C depends only on n. For the proof see page 168. Let n be a fixed integer. Condition (Uα ) from the preceding section can be written as ψ(y) 6 C(y/x)α ψ(x), y > x > 1. (Uα ) Hence (Uα ), α < n, implies Z ∞ ψ(y)y −n−1 dy 6 Cx−n ψ(x),

x > 1,

(An )

x

which is part of the following lemma.(†)

10.4.6 Lemma The function ψ satisfies (An ) iff there exists α < n such that ψ satisfies (Uα ). Proof. We need to prove that (An ) implies (Uα ) for some α < n. Let ψ satisfy (An ), and let Z ∞ ψ(y)y −n−1 dy, x > 1. F (x) = x

It is easy to see that cF (x) 6 x−n ψ(x) 6 CF (x), x > 1, and hence it suffices to find b > 0 such that xb F (x) is nonincreasing for x > 1. Choose b so that F (x) 6 (1/b)ψ(x)x−n , x > 1, which can be written as F (x) 6 −(1/b)xF ′ (x). This implies that the derivative of xb F (x) is 6 0, which concludes the proof. ✷ Now the implication (b)⇒(a) of Theorem 10.4.3 is obtained immediately from Propositions 10.4.4, 10.4.5, Lemma 10.4.6 and the formula Z 1 Z ∞ (1 − r)n−1 ψ(1/(1 − r)) dr, 0 < t < 1. ψ(y)y −n−1 dy = 1/t

1−r

For the proof of the implication (a)⇒(b) we need some more lemmas. (†) Such

assertions are often encountered in the theory of regularly varying functions (see [88]).

164

10 Lipschitz spaces

10.4.7 Lemma If g ∈ C(T), then the function ωn (t)/tn is almost decreasing for t > 0. Proof. The lemma is easily deduced from the inequality ωn (g, 2t) 6 2n ωn (g, t),

t > 0,

(10.12)

while (10.12) can be proved by means of the formula X ∆nt g(eiθ ) = gˆ(j)(eijt − 1)n eijθ

(10.13)

|j| kgk∞ . Proof. From (10.13) it follows that (−1)n Hence kg − gˆ(0)k∞ 6



 1 g(eiθ ) − gˆ(0) = 2π

1 2π

Z

Z

π

−π

(∆nt g)(eiθ ) dt.

π

−π

k∆nt k∞ dt 6 ωn (g, π) = ωn (gk , π/k).

✷

10.4.9 Lemma Let X = Lipn (φ) or h∞,n (ψ). Then there hold the assertions: (i) If a sequence {um } ⊂ X converges in norm to u, then um (z) → u(z) uniformly on compact subsets of D. (ii) X is complete. Proof. Let X = Lipn (φ). According to Lemmas 10.4.8 (k = 1) and 10.4.7, kukX > |u(0)| + ωn (u∗ , π)/φ(π) > cku∗ k∞ = ckuk∞ ,

u ∈ X.

This shows that X is continuously embedded into C(T) = hC(D), which implies (i). The same fact can be used to prove the completeness of X. In the case when X = h∞,n (ψ) the proof is even simpler. ✷ 10.4.10 Lemma Let φ and ψ satisfy the conditions (U ) and (Un0 ). Let uk (z) = z k , where |z| 6 1 and k = 1, 2, . . . Then kn ψ(k) 1 ≍ φ(1/k)

kDn uk kψ ≍

(k > 1),

(10.14)

kukφ,n

(k > 1).

(10.15)

10.4 Growth of derivatives

165

Proof. The proof of (10.14) is straightforward. In order to prove (10.15) we use the equality (∆nt uk∗ (w)) = wk (eikt − 1)n , w ∈ T. Hence ωn (uk∗ , t) = 2n sup{| sin(ks/2)|n : 0 < s 6 t},

t > 0,

and therefore kuk kφ,n = 2n sup{| sin(ks/2)|n /φ(t) : s 6 t 6 1, 0 < s 6 1}, Since φ(t) > φ(s)/C for 0 < s 6 t 6 1, we have kuk kφ,n 6 C sup{| sin(ks/2)|n /φ(s) : 0 < s 6 1}, where C is independent of k. If 1/k 6 s 6 1, then | sin(ks/2)|n /φ(s) 6 C/φ(1/k) because the function 1/φ is almost decreasing. If 0 < s 6 1/k, then | sin(ks/2)|n /φ(s) 6 2−n k n sn /φ(s) 6 C k n (1/k)n /φ(1/k) because sn /φ(s) is almost increasing. Thus kukφ,n 6 C/φ(1/k). The proof of the reverse inequality is simpler. ✷ Now we ready to prove the the implication (a)⇒(b) in Theorem 10.4.3. Let Lipn (φ) = h∞,n (ψ). It follows from Lemma 10.4.9 and the closed graph theorem that kDn ukψ ≍ kuk kφ,n , k > 1, where uk is as in Lemma 10.4.10. Hence, by Lemma 10.4.10, φ(1/k) ≍ (1/k)n ψ(k), k > 1, which yields φ(t) ≍ tn ψ(1/t),

0 < t 6 1,

(10.16)

and this is part of (b). In order to prove that (a) implies (Uα ) for for some α < n, we consider the functions Uk (z) = k −n ψ(k)z k +

∞ X j=2


(jk)−n ψ(jk) − ψ((j − 1)k) z jk ,

z ∈ D.

Assume, as we may, that ψ is nondecreasing. Then n

k

M (r, D Uk ) 6 ψ(k)r +

∞ X j=2

6 ψ(k)rk +

∞ X

ψ(jk) − ψ((j − 1)k) jk−1 X

j=2 p=(j−1)k

= ψ(k)rk +

∞ X j=k

= (1 − r)

∞ X

p=k




rjk

ψ(p + 1) − ψ(p)

ψ(p + 1) − ψ(p)




rp+1


ψ(p) rp 6 C ψ 1/(1 − r) .




rp+1

166

10 Lipschitz spaces

It follows that {Uk } is a norm bounded sequence in h∞,n (ψ). Now we use the inclusion h∞,n (ψ) ⊂ Lipn (φ) to conclude that the functions Uk are continuous on the closed disk and ωn (Uk∗ , t) 6 Cφ(t),

0 < t < 1,

(10.17)

where C is independent of t, k. On the other hand, by Lemmas 10.4.7 and 10.4.8, Cωn (Uk∗ , 1/k) > ωn (Uk∗ , π/k) > kUk∗ k∞ ∞ X
j −n ψ(jk) − ψ((j − 1)k) = k −n ψ(k) + k −n j=2

= k −n

∞ X j=1


j −n − (j + 1)−n ψ(jk).

Hence, by (10.17), (10.16) and (U ), k −n

∞ X j=1


j −n−1 ψ (j + 1)k 6 Cφ(1/k) 6 Ck −n ψ(k)

and therefore Z Z ∞ y −n−1 ψ(y) dy = k −n

∞

y −n−1 ψ(yk) dy 6 Ck −n ψ(k),

k = 1, 2, . . . .

1

k

It is easily verified that this implies (An ). Thus ψ satisfies (Uα ) for some α < n (Lemma 10.4.6), and this concludes the proof of Theorem 10.4.3. Proof of Proposition 10.4.4 In proving Proposition 10.4.4 and Proposition 10.4.5 we shall use the inequalities M (r, Dn+1 f ) 6 C(1 − r)−1 M ((1 + r)/2, Dn u)

M (r, f (n+1) ) 6 C(1 − r)−1 M ((1 + r)/2, Dn u),

(10.18)

where u = Re f , f ∈ H(D), and n > 0. Equivalently: if Dn u is bounded in D, then |Dn+1 f (z)| 6 C(1 − |z|)−1 kDn uk∞ ,

|f (n+1) (z)| 6 C(1 − |z|)−1 kDn uk∞ ,

where C is independent of f and z. The proof is left to the reader as an exercise. In proving Proposition 10.4.4 we may assume that u is real-valued and harmonic in a neighborhood of the closed disk. For fixed r < 1 let h(θ) = ur (θ) = u(reiθ ). Then Z n h(n) (θ + x1 + · · · + xn ) dx1 . . . dxn , (10.19) (∆t h)(θ) = tE

10.4 Growth of derivatives

167

where tE is the n-dimensional cube [0, t]n . Hence h(n) (θ) = (Dn u)(reiθ )tn Z
n h(n) (θ + x1 + · · · + xn ) − h(n) (θ) dx1 . . . dxn . = (∆t )(θ) − tE

Since |h

(n)

(θ + x) − h

(n)

we get M (r, D

n+1

Z (θ)| =

x

h

(n+1)

0

u) 6

k∆nt ur k∞

Z

+

(θ + y) dy 6 M (r, Dn+1 )x,

tE

x = x1 + . . . xn ,

M (r, Dn+1 u)(x1 + · · · + xn ) dx1 . . . dxn

= k∆nt ur k∞ + (n/2)M (r, Dn+1 )tn+1 ,

0 < r < 1, t > 0.

The function ∆nt u defined by (∆nt u)(reiθ ) = (∆nt ur )(θ) is harmonic on the closed disk and therefore k∆nt ur k∞ 6 k∆nt u∗ k 6 ωn (u∗ , t), t > 0. These inequalities together with (10.18) yield M (r, Dn u) 6 t−n ωn (u∗ , t) + Kt(1 − r)−1 M ((1 + r)/2, Dn u)

(10.20) −n

(t > 0, 0 < r < 1,) where K depends only on n. Let A(r) = (1 − r) 0 < r < 1. It follows from (10.20) that

M (r, Dn u),

A(r) 6 t−n (1 − r)n ω(t) + 2n Kt(1 − r)−1 A((1 + r)/2), where ω(t) = ω(u∗ , t). Choose an integer m so that 2n K 6 (1/4)2m and take t = a(1 − r), a = 2−m . Then we have A(r) 6 a−m ω(1 − r) + (1/4)A((1 + r)/2), 0 < r < 1. Integrating this inequality from ρ (< 1) to 1, and introducing appropriate substitutions, we get Z 1 Z 1 Z 1−ρ A(r) dr A(r) dr 6 a−m ω(t) dt + (1/2) ρ

(1+ρ)/2

0

6 a−m

Z

1−ρ

ω(t) dt + (1/2)

0

Hence, since the integral

R1

Z

1

A(r) dr.

ρ

A(r) dr is finite, Z Z 1 −m A(r) dr 6 a (1/2) ρ

ρ

1−ρ

ω(t) dt.

0

Now (10.9) follows from the inequalities Z 1−ρ Z ω(t) dt 6 (1 − ρ) ω(1 − ρ), M (r, Dn+1 u)(1 − ρ)n+1 6 (n + 1) 0

1

A(r) dr,

ρ

which are valid because the functions ω and M are increasing. Thus the proof of Proposition 10.4.4 is finished.

168

10 Lipschitz spaces

Proof of Proposition 10.4.5 Let u = Re f , where f is analytic u D. Then Z n X 1 1 f (k) (rz) k k (1 − s)n z n+1 f (n+1) (sz) ds f (z) = z (1 − r) + k! n! r

(10.21)

k=0

(z ∈ D, 0 < r < 1). Denoting the sum by fr,n we have Z 1 1 |f (z) − fr,n (z)| 6 (1 − s)n M (s, f (n+1) ) ds. n! r From this and (10.18) it follows that (10.10) implies kf − fr,n k∞ → 0 (r → 1− ). Since the functions fr,n (r < 1) are continuous on the closed disk, we see that (10.10) implies the continuity of f , and consequently of u, on the closed disk. In order to prove (10.11) let ur (θ) = u(reiθ ), 0 < r 6 1. Then (10.11) is equivalent to Z 1 n (1 − s)n−1 M (s, Dn u) ds, 0 < t < 1. (10.22) k∆t u1 k∞ 6 C 1−t

Let r = 1 − 2t, 0 < t < 1/4. Then k∆nt u1 k 6 k∆nt (u1 − ur )k + k∆nt ur k. It follows from (10.19) and the “increasing” property of M (r, Dn u) that Z 1 (1 − s)n−1 M (s, Dn u) ds, k∆nt ur k 6 tn M (r, Dn u) 6 n 1−t

and side

therefore we have to prove that k∆nt (u1 − ur )k is dominated by of (10.22). Since k∆nt (u1 − ur )k 6 k∆nt (f1 − fr )k, it is enough Z

k∆nt (f1 − fr )k 6 C

the right-hand to prove that

1

1−t

(1 − s)n−1 M (s, Dn u) ds.

To prove this write (10.21) in the form f1 (θ) − fr (θ) = H(θ) + 1 H(θ) = n! We have

Z

r

n X

k=1

hk (θ)(1 − r)k /k!,

where

1

(1 − s)n ei(n+1)θ f (n+1) (seiθ ) ds,

and hk (θ) = f (k) (reiθ )eikθ .

Z 2n 1 (1 − s)n M (s, f (n+1) ) ds n! r Z 1 (1 − s)n−1 M ((1 + s)/2, Dn u) ds 6C

k∆nt Hk 6 2n kHk 6

r

n

=2 C

Z

1

1−t

(1 − s)n−1 M (s, Dn u) ds,

10.4 Growth of derivatives

169

where we have applied (10.18). In order to estimate k∆nt hk k, let m = n − k + 1 (1 6 k 6 n) and observe that (10.19) implies (m)

k−1 k−1 m k∆nt hk k = k∆tk−1 ∆m k∆m t khk k. t hk k 6 2 t hk k 6 2 (m)

From this and the inequality khk k 6 C(1 − r)−1 M ((1 + r)/2, Dn u) (see (10.18) ) it follows that Z 1 (1 − s)n−1 M (s, Dn u) ds, k∆nt hk k 6 Ctn−k M (1 − t, Dn u) 6 Ct−k 1−t

where C is independent of t. Combining all the above results yields (10.11) for 0 < t < 1/4. If t > 1/4, we can apply Lemma 10.4.7 to reduce (10.11) to the case 0 < t < 1/4, and this completes the proof. ✷ Misclellaneous 10.4.11 (conjugate functions) If ψ(x), x > 1, satisfies (U ) (p. 161) and there exists a constant β > 0 such that ψ(x)/xβ

almost increases

(10.23)

then the space h∞ (ψ) is “self-conjugate”, i.e., there holds u ∈Ph∞ (ψ) ⇒ u e ∈ ∞ h∞ (ψ), or, what is the same, the Riesz projector (R+ u)(z) = n=0 u b(n)z n acts from h∞ (ψ) to h∞ (ψ). From this and Theorem 10.4.3 it follows that the same holds for Lipn (φ) provided there exist constants β < n and α > 0 such that φ(t)/tα is almost increasing and φ(t)/tβ is almost decreasing. It was proved in [91] that if (U ) holds, then (10.23) is necessary for h∞ (ψ) to be self-conjugate. 10.4.12 If a < n, then the derivative Dn u in Theorem 10.4.1 may be replaced by each of the derivatives ∂ n u/∂ j r∂ n−j θ (j = 0, 1, . . . , n), see 10.4.11. If 0 < a < n, then Theorem 10.4.1 remains true when “O” is replaced by “o”. 10.4.13 A real function g ∈ C(T) belongs to Λ∗ iff the Cauchy integral 1 f (z) := 2π satisfies |f ′′ (z)| 6

Z

π

−π

C 1 − |z|

u(eiθ ) dθ, 1 − ze−it (|z| < 1).

This condition implies that f ∈ A(D) and that the boundary function f∗ belongs to Λ∗ .

11

Lacunary series

In the first section we consider Paley’s theorems on lacunary series in H p (Theorems 11.1.1 and 11.1.3). The rest of the chapter is devoted to the proof of a generalized version of the Gurarij/Matsaev theorem on Lp (0, 1)-integrability of lacunary power series (see Theorems 11.4.2 and 11.3.1). Since the proof is based on the ideas from Karamata’s proof of Littlewood’s tauberian theorem, we included a short discussion of Karamata’s ideas (Section 11.2).

11.1

Lacunary series in H p

A sequence {nk }k>1 of positive real numbers is called lacunary if it satisfies the condition nk+1 > 1. inf k>1 nk P The corresponding series ak xnk is then called a lacunary series.

11.1.1 Theorem (Paley) Let nk be a lacunary sequence of positive integers. If f ∈ H 1 , then X 1/2 ∞ 2 b , kf k1 > c |f (nk )| k=1

where c > 0 is a constant independent of f .

This theorem does not extend to the case of L1 (T), and this can be seen from 6.1.4. Theorem 11.1.1 will be deduced from the following result. 11.1.2 Theorem (Hardy/Littlewood) Let f be analytic in D. Then there hold the assertions: (a) If f ∈ H p , 0 < p 6 2, then K :=

Z

0

1

Mp2 (r, f ′ )(1 − r) dr < ∞

(11.1)

and there exists a constant Cp such that K 6 Cp kf k2p . (b) If 2 6 p < ∞, then (11.1) imply f ∈ H p and kf k2p 6 Cp (K + |f (0)|2 ). Proof. (a) Let 0 < p 6 2. In view of Lemma 5.1.7, it is enough to discuss the case where f has no zeros in D. Further, we can assume that kf kp = 1. Let 170

11.1 Lacunary series in H p

171

g = f p/2 . Then kf kpp = kgk22 and Mpp (r, f ′ ) = (2/p)p

1 2π

Z

π

−π

|g(reiθ )|2−p |g ′ (reiθ )|p dθ.

Hence, by H¨older’s inequality with the indices 2/(2 − p), 2/p, we get Mpp (r, f ′ ) 6 (2/p)p M22−p (r, g)M2p (r, g ′ ). Since M2 (r, g) 6 kgk2 = 1, we see that Mp2 (r, f ′ ) 6 (2/p)2 M22 (r, g ′ ), and this reduces the proof to the easily proved inequality Z

1

0

M22 (r, g ′ )(1 − r)dr 6 Ckgk22 .

(b) As remarked in the proof of Lemma 6.2.4, the function |f |p is of class C 2 so we can apply Green’s formula to prove that the Hardy/Stein identity, kf kpp

p2 = |f (0)| + 2 p

Z

D

|f |p−2 |f ′ |2 log

1 dA, |z|

holds for every f ∈ H p . Let g(z) = f (ρz), 0 < ρ < 1. By H¨older’s inequality we get Z π 1 |g(reiθ )p−2 |g ′ (reiθ )|2 dθ 6 Mpp−2 (r, g)Mp2 (r, g ′ ) 6 kgkp−2 Mp2 (r, g ′ ). p 2π −π Hence kgkpp

2

6 |g(0)|

kgkp−2 p

Multiplying this inequality by desired result. ✷

+p

2

kgk2−p p

kgkpp−2

Z

1

0

Mp2 (r, g ′ )r log

1 dr. r

and then letting ρ tend to 1, we get the

Proof of Theorem 11.1.1. Let λ = inf k>1 nnk+1 . From Theorem 11.1.2 it k follows that ∞ Z rm+1 X kf k1 > c1 M12 (r, f ′ )(1 − r) dr, m=1

rm

where rm = 1 − λ−m , c1 =const> 0. For each m, the block Im = [λm , λm+1 ) contains at most one member of the sequence {nk }. Therefore we can suppose that nm ∈ Im for all m. Since M1 (r, f ′ ) > n|fb(n)|rn−1 , we have Z

rm+1

rm

M12 (r, f ′ )(1 − r) dr > n2m |fb(nm )|2

Now the proof is easily completed.

✷

Z

rm+1

rm

r2nm −2 (1 − r) dr.

172

11 Lacunary series P∞ 11.1.3 Theorem (Paley) Let the series f (z) = k=1 ak z nk , P where {nk } is a ∞ lacunary sequence, converge in D. Then f ∈ H p (0 < p < ∞) iff k=1 |ak |2 < ∞. There exists a constant C = Cp > 0 such that C −1 k{ak }k2 6 kf kp 6 Ck{ak }k2 .

(11.2)

Proof. In the case 1 6 p < 2, the result is an immediate consequence of Paley’s theorem. Let 0 < p < 1. Let f be analytic in a neighborhood of the closed disk. Then, by means of the Cauchy/Schwarz inequality, we get Z (2−p)/2 (2−p)/2 kf k1 = |f |p/2 |f |1−p/2 6 kf kp/2 6 kf kp/2 . p kf k2−p p kf k2 p/2

p/2

Since kf k1 > ckf k2 , we see that c kf k2 6 kf kp . If f is arbitrary, then we apply this inequality to the functions fρ (ρ → 1) and this completes the proof in the case 0 < p < 2. Let 2 < p < ∞ and q = p/(p − 1). It follows from Paley’s theorem that Pb the operator P , (P f )(z) = f (nk )z nk , is bounded from H q to H 2 . The dual ∗ operator P is formally equal to P , and since (H q )∗ = H p (Theorem 6.3.2), we have kP f kp 6 Cp kf k2 for f ∈ H 2 . Hence kf kp 6 Cp kf k2 if P f = f . In the case p > 2, the theorem can also be deduced from Theorem 11.1.2(b) by using 7.5.5. ✷ Miscellaneous 11.1.4 Inequality (11.2), known as Paley’s inequality, holds for every p ∈ (0, ∞) under the more general assumption that f is an arbitrary trigonometric polynomial with lacunary coefficients. On the other hand, in the case p = ∞, there holds Sidon’s theorem [100, Theorem VI.(6.1)]: 11.1.5 Theorem A trigonometric series with lacunary coefficients is the Fourier series of a bounded function iff the sequence of coefficients is absolutely summable. For further properties of lacunary series see Zygmund [100, Ch. V, §§6,7].

11.2

Karamata’s theorem and Littlewood’s theorem

Recall Abel’s theorem: Let ∞ ∞ X X f (z) := sn z n Ak z k = (1 − z) k=0

n=0

(|z| < 1, z ∈ C),

(11.3)

11.2 Karamata’s theorem and Littlewood’s theorem

173

where Pn {Ak } (k > 0) is a sequence of vectors in a Banach space (X, | · |), and sn = k=0 Ak . If there exists the limit limn→∞ sn =: s ∈ X, then limr→1− f (r) = s. In other words, convergence of a series implies its summability by Abel’s method; the converse is not true. Tauber proved that if the “tauberian“ condition limn→∞ n An = 0 is satisfied, then the existence of limr→1 f (r) =: l implies limn→∞ sn = l. Littlewood [54] improved Tauber’s theorem by replacing Tauber’s condition by (∗) sup (k + 1)|Ak | < ∞. (11.4) k>0

The original proof of Littlewood is very complicated. Karamata [39] found a simple approach to Littlewood’s theorem, which enabled him to improve and generalize some more tauberian theorems, mainly due to Hardy and Littlewood [39, 41, 40]. Karamata’s theorem Condition (11.4) and boundedness of the function f (r) = (1−r) imply boundedness of the sequence sn (Lemma 11.2.4).

P∞

k=0 sk r

k

on (0, 1)

11.2.1 Theorem (Karamata) If the sequence sn is bounded and the condition ∞ P lim− (1 − r) sn rn = S is satisfied, then r→1

n=0

lim− (1 − r)

r→1

∞ X

sk ψ(rk )rk = S

Z

1

ψ(t) dt,

(11.5)

0

k=0

where ψ is an arbitrary Riemann(†) integrable function on [0, 1]. Proof of Littlewood’s theorem. In order to deduce Littlewood’s theorem from Karamata’s theorem, i.e., to prove that the conditions (11.4) and lim−

r→1

∞ X

Ak r k = S

(11.6)

k=0

P imply convergence of the series Ak , we choose ψ so that rψ(r) = 1/(λ − 1) for e−λ 6 r < e−1 , ψ(r) = 0 otherwise, where λ > 1. Then from (11.5) we get, by taking r = e−1/n , [λn] X 1 sk = S. (11.7) lim n→∞ (λ − 1)n k=n+1

On the other hand, 

   [λn] [λn] X X [λn] − n 1 1 sk − sn = (sk − sn ). − 1 sn + (λ − 1)n (λ − 1)n (λ − 1)n

(∗) See

k=n+1

Theorem 11.2.2. cannot be replaced by “Lebesgue”.

(†) “Riemann”

k=n+1

174

11 Lacunary series

From this and condition (11.4) it follows that  lim sup n→∞

 [λn] X 1 sk − sn 6 lim sup (λ − 1)n n→∞ k=n+1

max

n 1 is arbitrary.

✷

Proof of Karamata’s theorem. Observe that the expression (1 − r)

∞ X

ψ(rk ) rk =

k=0

∞ X

k=0

ψ(rk )(rk − rk+1 )

is a Riemannian sum of the function ψ(r). The diameter of the underlying partition is equal to 1 − r and therefore Z 1 ∞ X ψ(t) dt. ψ(rk ) rk = lim− (1 − r) r→1

0

k=0

It follows that condition (11.6) can be written as lim (1 − r)

r→1−

Replacing r by r

1+α

∞ X

k=0

(sk − S)rk = 0.

, α > 0, we find that lim− (1 − r)

r→1

and this implies lim− (1 − r)

r→1

∞ X

k=0

∞ X

k=0

(sk − S)rαk rk = 0,

(sk − S)P (rk ) rk = 0,

where P is an arbitrary polynomial. Hence ∞ X (sk − S)ψ(rk ) rk lim sup (1 − r) r→1−

k=0

∞ X
(sk − S) ψ(rk ) − P (rk ) rk 6 lim sup (1 − r) r→1−

k=0

∞ X ψ(rk ) − P (rk ) rk 6 lim sup (1 − r) r→1−

=M

Z

0

1

k=0

ψ(t) − P (t) dt,

where M = supn>0 |sn |. This concludes the proof because the polynomials are dense in L1 (0, 1). ✷

11.2 Karamata’s theorem and Littlewood’s theorem

175

On Littlewood’s theorem The proof of Littlewood’s theorem shows that condition An = O(1/n) can be weakened. Here we state a special case of a very general result of Karamata [40]. P∞ 11.2.2 Theorem Let the function (11.3) have property (11.6). The series 0 An converges if there exists a function δ(λ), λ > 1, such that limλ→1+ δ(λ) = 0, and k X Aj 6 δ(λ) for all λ > 1. lim sup max n→∞ n 1,

(11.9)

k=n+1

where C is independent of n. It should be observed that condition (11.8) is necessary for convergence of an arbitrary series and is weaker than the condition lim sup n→∞

[λn] X

j=n+1

|Aj | 6 δ(λ),

while the latter is weaker than (11.9); namely, if (11.9) holds, then by H¨older’s inequality, [λn] X

j=n+1

|Aj | 6

[λn]  X

j=n+1

|Aj |q

1/q

([λn] − n)1−1/q 6 C(λ − 1)1−1/q ,

so we can take δ(λ) = C(λ − 1)1−1/q . Theorem 11.2.2 is an immediate consequence of the above proof of Littlewood’s theorem and the following lemma. 11.2.4 Lemma If the function f is bounded and j X Ak < ∞, sup max

n>0 n6j62n

(11.10)

k=n

then the sequence sn bounded.

We leave the proof of this lemma to the interested reader. We only note that one can start from the inequality n +1 ∞ 2X X |f (rn ) − s2n | 6 Aj (1 − rnj ) + Aj rnj ,

j=0

j=2n

(11.11)

176

11 Lacunary series P2n where rn = 1 − 2−n . If (11.10) is strengthened to supn>0 j=n |Aj | < ∞, then the proof becomes shorter; namely, then it easy to show that the right-hand side of (11.11) is bounded. Miscellaneous 11.2.5 The “uniform” version of Dirichlet/Jordan test says: The Fourier series of a 2π-periodic continuous function f (θ) of bounded variation converges uniformly on R. Let Ak (θ) = ak cos kθ + bk sin kθ (θ ∈ R, k > 1), where ak , bk are the Fourier coefficients of f (θ). From Theorem 3.1.6 it follows that

∞

a0 X k

Ak r − f lim− +

=0 2 r→1 ∞ k=1

If in addition f is of bounded variation on [0, 2π], then kAk k∞ = O(1/k), so the conclusion of the test can be obtained by Littlewood’s theorem. 11.2.6 Equality (11.5) holds if the sequence sn is real and bounded from above. In his proof Karamata appeals to the following: If ψ is a Riemann integrable function, then for every ε > 0 there are polynomials P (r) and Q(r) such that
R1 P (r) < ψ(r) < Q(r) and 0 Q(r) − P (r) dr < ε. Pn 11.2.7 Let the sequence sn = k=0 Ak be bounded and let condition (11.6) be satisfied. If ϕ is an absolutely continuous function on the segment [0, 1] and ϕ(0) = 0, then ∞ X Ak ϕ(rk ) = ϕ(1) lim− f (r). lim− r→1

11.3

r→1

k=0

Lacunary series in C[0, 1]

We continue to denote by {Ak } a sequence of vectors in a Banach space X. We consider the series ∞ X Ak rλk , (11.12) L(r) = k=0

where λk is a lacunary sequence, i.e., a sequence satisfying λk+1 = q > 1. k>0 λk inf

(11.13)

The following theorem is taken from Gurarij/Matsaev [24]. 11.3.1 Theorem If there exists the limit S := limr→1− L(r), and is finite, then P∞ the series k=0 Ak converges.

11.3 Lacunary series in C[0, 1]

177

Proof. First we prove that the hypotheses imply sup |Ak | 6 C M,

(11.14)

k>0

where M = sup0 0) a nonnegative real function that satisfies the condition λa µc F (x, y) 6 F (λx, µy) 6 λb µd F (x, y)

(0 < λ, µ < 1),

(11.15)

where a, b, c, d are positive constants (a > b, c > d). We write F ∈ ∆(a, b; c, d) or F ∈ ∆ according to whether the values of a, b, c, d are or are not important. The simplest example is F (x, y) = xα y p (α > 0, p > 0). Further examples can be found by considering the functions F (x, y) = xα Φ(xβ y), where Φ : [0, ∞) 7→ [0, ∞) is a function for which there are positive constants γ and δ such that Φ(t)/tγ decreases and Φ(t)/tδ increases. 11.4.1 Theorem If F ∈ ∆, and L is given by (11.12), then the following conditions are equivalent: Z

0

1

(1 − r)−1 F (1 − r, |L(r)|) dr < ∞;

(11.16)

∞  1  X F , |An | < ∞. λn n=0

(11.17)

In the case where F (x, y) = xy p , this reduces to the following theorem of Gurarij and Matsaev [24]:

11.4 Lp -integrability of lacunary series on (0, 1)

179

11.4.2 Theorem For every p ∈ (0, ∞), the following conditions are equivalent: Z

0

∞ X

1

|L(r)|p dr < ∞;

n=0

(1/λn )|An |p < ∞.

We shall deduce Theorem 11.4.1 from a weaker result, namely: 11.4.3 Proposition There holds the inequality Z

0

1

(1 − r)−1 F (1 − r, |L(r)|) dr > c0 sup F (1/λn , |An |),

(11.18)

n>0

where c0 is a positive constant. Proof. Let F ∈ ∆(a, b; c, d). The integral in (11.18) is > Z

1

(1 − r)−1 F (1 − r, |L(r)|r1/d ) dr/r,

0

and since the series L(r)r1/d is lacunary, it follows that in (11.18) we can write r−1 dr instead of dr. Next, assuming that λk+1 /λk > q > 1 for every k, let P (r) =

[r(1 − r)]N , [ρ(1 − ρ)]N

where N is an integer and ρ satisfies the condition 2−q < ρ < 1/2 < 1 − ρ < 2−1/q . Choose ε so small that 2−q(1−ε) < ρ < 2−(1+ε) ,

2−(1−ε) < 1 − ρ < 2−(1+ε)/q .

For every δ > 0 we can choose N so that for 2−(1+ε) < r < 2−(1−ε) ,

P (r) > 1,

for 0 < r < 2−q(1−ε) ,

0 < P (r) 6 δr,

0 < P (r) 6 δ(1 − r), for 2−(1+ε)/q < r < 1. It is easily checked that Int : =

Z

1 −1

0

6C

(1 − r)

Z

0

1

 X ∞ λk dr Ak P (r ) F 1 − r, r 

0


dr (1 − r)−1 F 1 − r, |L(r)| , r

where C depends only of P and F . Therefore to prove (11.18) it is enough to find a polynomial P so that Int > c0 supn>0 F (1/λn , |An |). Suppose that An → 0

180

11 Lacunary series

and take ν so that F (1/λν , |Aν |) > F (1/λn , |An |) for all n. We shall prove the (formally) stronger inequality Intν (F ) > c0 sup F (1/λn , |An |),

(11.19)

n>0

where  X ∞ λk dr Intν (F ) = , Ak P (r ) (1 − r) F 1 − r, r Jν 0
Jν = 2−(1+ε)/λν , 2−(1−ε)/λν . Z

−1



There hold the implication

k > ν, r ∈ Jν =⇒ rλk 6 2−q(1−ε) ,

k < ν, r ∈ Jν =⇒ rλk > 2−(1+ε)/q .

(11.20)

Now we pass to the proof of (11.19). Assume first that F ∈ ∆(a, b; 1, d). This means that the function y 7→ F (x, y)/y (y > 0) decreases, which implies that F (x, y1 + y2 ) 6 F (x, y1 ) + F (x, y2 ), whence Z
dr (1 − r)−1 F 1 − r, P (rλν )|Aν | Intν (F ) > r Jν XZ
dr − (1 − r)−1 F 1 − r, P (rλk )|Ak | . r Jν k6=ν

By the relations (11.20) and the properties of P , for δ small enough, one can prove that the subtrahend is small with respect to F (1/λν , |Aν |), while the minuend is > K F (1/λν , |Aν |), where K is independent of δ, and this concludes the proof in the special case. (Details are omitted.) On the other hand, applying Minkowski’s inequality to the preceding inequality, we get, for p > 1,  Z  1/p
dr 1/p Intν (F p ) > (1 − r)−1 F p 1 − r, P (rλν )|Aν | r Jν  Z X
dr 1/p , − (1 − r)−1 F p 1 − r, P (rλk )|Ak | r Jν k6=ν

which is enough to complete the proof because an arbitrary function of class ∆ can be represented as F p , where F ∈ ∆(a, b; 1, d). ✷ In order to deduce Theorem 11.4.1 from Proposition 11.4.3 we need two technical lemmas. 11.4.4 Lemma If α > 0, β > 0, then Z 1 (1 − ρ)α−1 (ρ − r)β−1 dρ 6 Cα,β (1 − r)α+β−1 r

(0 < r < 1).

(11.21)

11.4 Lp -integrability of lacunary series on (0, 1)

181

Proof. Let α < √1 and β < 1.(‡) We split the interval√ (0, 1) by the point √ ρ = r. If r < ρ < r, then (1 − ρ)α−1 (ρ − r)β−1 6 (1 − r)α−1 (ρ − r)β−1 . If √ √ r < ρ < 1, then (1 − ρ)α−1 (ρ − r)β−1 6 (1 − ρ)α−1 ( r − r)β−1 . The result follows by integration of these inequalities over the corresponding intervals. ✷ 11.4.5 Lemma If G is a positive measurable function on (0, 1) and γ > 0, then Z ρ Z 1 Z 1 (ρ − r)γ/2−1 G(r) dr, (1 − ρ)γ/2−1 dρ (1 − r)γ−1 G(r) dr > c0 0

0

0

where c0 is a positive constant depending only on γ.

Proof. We can apply Fubini’s theorem and appeal to inequality (11.21). ✷ Proof of Theorem 11.4.1. Let F satisfy condition (11.15) and assume there holds (11.16). Put G(r) = (1 − r)−b F (1 − r, |L(r)|). According to Lemma 11.4.5 we have Z ρ Z 1 Z 1 (ρ − r)b/2−1 G(r) dr. (1 − ρ)b/2−1 dρ (1 − r)−1 F (1 − r, |L(r)|) dr > c0 0

0

0

On the other hand, the inner integral equals Z 1 (1 − r)b/2−1 G(ρr) dr ρb/2 0

= ρb/2

Z

1

(1 − r)b/2−1 (1 − ρr)−b F (1 − ρr, |L(ρr)|) dr,

0

> ρb/2

Z

0

which is

1

(1 − r)−b/2−1 F (1 − r, |L(ρr)|) dr.

(Here we used the fact that x−b F (x, y) increases with x, which follows from (11.15).) Further, the function x−b/2 F (x, y) belongs to ∆(a − b/2, b/2; c, d) so we can apply Proposition 11.4.3; we get Z 1 λn (1 − r)−b/2−1 F (1 − r, |L(ρr)|) dr > c λb/2 n F (1/λn , |An |ρ ). 0

From these inequalities and the estimates (1 − ρ)−γ > c0

∞ X

λγn ρλn ,

we get Z Z 1 −1 (1 − r) F (1 − r, |L(r)|) dr > c0 0

where

n=0

0

1

dρ

∞ X

n=0

γ = 1 − b/2 > 0,

λn λn1−b/2 ρλn +b/2 λb/2 n F (1/λn , |An |ρ ),

whence, by condition (11.15), we see that (11.17) holds. The proof of the implication (11.17) =⇒ (11.16) is much simpler and we omit it. The reader could see the ✷ papers [60] and [58]. (‡) Only

this case will be used in the proof of the theorem.

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Index (An ), 163 Ap , Bergman space, 16 Apβ , weighted Bergman space, 136 absolute continuity of f∗ , 78 approximation by inner functions, 91 approximation by smooth functions, 58 atomic decomposition, 133, 135, 136

Formula Cauchy integral, 78 Green, 39, 67 Jensen, 68, 81 Parseval, 100 Riesz representation, 69 Function almost increasing, 155 atomic, 85 conjugate, 93, 97 harmonic conjugate, 92 inner, 85, 89 inner, singular, 85 logarithmically convex, 61 main maximal, 25 nontangential maximal, 112 of bonded mean oscillation, 127 outer, 85, 86 polyharmonic, 153 Rademacher, 28 radial maximal, 110 semicontinuous, 55 subharmonic, 55 superharmonic, 56 Weierstrass, 160

BV [a, b], 44 B, Bloch space, 128 BMO, 127 Banach envelope, 10 Blaschke condition, 80 Blaschke product, 80 Bourgain’s lemma, 120 C02 (D), 67 c0 , 16 complemented subspaces of ℓp , 17 conformal mappings, 79 convexity of mean values, 60, 61 Dr (z), 149 Du = ∂u/∂θ, 154 ∆(a, b; c, d), 178 ∆n t , symmetric difference, 160 dµp , 130 dA = dm/π, 82 decomposition of H p functions, 75 density of polynomials in H p , 76 density of rational functions in Lp , 27 Dini condition, 156 Dirichlet problem, 41 Dirichlet/Jordan test, 176 disk-algebra, 155 dm, Lebesgue measure in C, 39

H, Hilbert operator, 93 H(D, X), all X-valued analytic functions, 137 H(Ω), all analytic functions, 39 HC 1 , 149 HC 2 (G), 151 H ∞ (X), 138 H p , Hardy space, 73 H p (T), 100 H p (T), 100 h(Ω), all harmonic functions, 39 hAp , harmonic Bergman space, 143 hC(D) = C(D) ∩ h(D), 42 h1 , 43 hp , harmonic Hardy space, 49 h∞ (ψ), 161 h∞,n (ψ), 161 Hadamard product, 138, 141 Hardy/Stein identity, 98

Eε (z), 70 E, 131 fe, conjugate function, 92, 93 f∗ , boundary function, 50 F -norm, 14

187

M, maximal function, 25 majorant, 154 maximum principle, 57 Smirnov, 76 mean value property, 39 moduli of smoothness, 160 multiplier, 15 multipliers, 35

harmonic Schwarz lemma, 53, 66 Heine/Borel property, 39 I(r, u), mean values, 60 I(u), 70 Ip (r, f ), 129 Inequality Chebyshev, 22 Hardy, 79, 84 Hardy/Littlewood, 124 Harnack, 44 Heinz, 105 Hilbert, 83, 84 isoperimetric, 82, 83 Khintchine, 29 Littlewood/Paley, 51, 70 Paley, 30, 172 Riesz/Fejer, 83 Riesz/Zygmund, 95 integral means, 49 of univalent functions, 62 interpolation for Hardy spaces, 119 isometry Lp to hp , 50 isomorphism Ap to ℓp , 16

Nq (f ), 10 N (D), Nevanlinna class, 15 N + (D), Smirnov class, 15 N0 (f ), quasinorm in L0 , 30 Nevanlinna class, 15 OC 1 , 149 OC 2 (G), 151 Of (z, r), 149 Op f (z, r), 150 ωn (g, t), modulus of smoothness, 160 Operator bilinear, 13 from ℓp to ℓq , 17 Hilbert, 93, 97, 156 invertible, 10, 12 of strong type, 23 of weak type, 23 quasilinear, 22 subadditive, 22 sublinear, 31

Jf (z), 104 Jp (r, f ), 62 Kp (w, z), 130 Lp,∞ , weak Lebesgue space, 22 Λ∗ , Zygmund class, 160 Λα (K), Lipschitz space, 154 Lip(ω, K), Lipschitz space, 154 Lipn (φ), Lipschitz space, 160 λ∗ , little Zygmund class, 160 L(r), 176 L0 , all measurable functions, 30 lacunary series, 125, 172, 176 Lp -integrability, 179 Lebesgue point, 26, 27 Lebesgue set, 26, 27 Littlewood’s conjecture, 125

P (z) = P (r, θ), Poisson kernel, 40 P S, Poisson/Stieltjes integral, 44 P [·], Poisson integral, 41 Pn f , 101 Pe , conjugate Poisson kernel, 92 p-norm, 7 Poisson integral, 41 of f∗ , 50, 78 of a Borel measure, 42 of log|f∗ |, 76 of the conjugate function, 94 Poisson kernel, 40, 51 conjugate, 92 Poisson/Stieltjes integral, 44 Principle Banach, 36 Littlewood subordination, 64 maximum, 57, 76

M (T), Borel measures, 42 M∗ u, nontangential maximal function, 112 Mp (r, f ), integral mean, 49 Mrad u, radial maximal function, 110 µ(f, λ), distribution function, 22

188

maximum modulus, 40 uniform boundedness, 13

on coefficients in BMO, 128 on radial and nontangential limits, 48 Abel, 172 Ahern, 65 Aleksandrov, 100 Aoki/Rolewicz, 8 Banach/Steinhauss, 13 Beurling approximation, 87 Bieberbach, 63 Burkholder, Gundy and Silverstein, 114 Carath´eodory convergence, 108 Choquet, 96 closed graph, 14 Coifman/Rochberg, 133 complex maximal, 111 Dyakonov, 158 Fatou on nontangential limits, 47 Fatou on radial limits, 46 Fefferman, 127 Fefferman/Stein, 112 Gurarij/Matsaev, 176, 179 Hadamard’s three circles, 61 Hardy, 61 Hardy/Littlewood, 122, 170 on hp , p < 1, 147, 148 on Cesaro means, 116 on coefficients, 24, 122 on conjugate functions, 148 on conjugates, 143 Harnack, 44 Hausdorff/Young, 21 Hurwitz, 44 inner-outer factorization, 86 Kalton, 101 Karamata, 173, 176 Koebe, 63 Kolmogorov, 113 Kolmogorov/Smirnov, 65 Konyagin, 103 Lebesgue points, 26 Lindel¨ of, 48 Liouville, 61 Littlewood tauberian, 173, 175 Littlewood/Paley, 51, 70 local integrability, 58 Marcinkiewicz, 23, 24, 119 maximal, 26, 116, 119

q-Banach envelope, 10 of the space H p , 136 quasinorm, 7 R+ , Riesz projector, 96 rj (t), Rademacher functions, 28 Radial limits of hp functions, 50 of H p -functions, 76 of conjugate function, 93 of the Poisson integral, 45, 46 reproductive kernels, 130 Riesz measure, 67 Riesz projector, 96 SH, 149 α σmax f , 116 σnα f , 116 Shauder basis, 13, 99 Smirnov class, 15 Space M (T), 42 L0 , 30 hC(D), 42 h1 , 43 h∞,n (ψ), 161 ho,n (ψ), 161 F -space, 14 h∞,n (ψ), 161 p-Banach, 9 Bergman, 16, 129 Bergman harmonic, 143 Bergman, weighted, 136 BMO, 127 Hardy, 73 Hardy harmonic, 49 Lipschitz, 154 Lipschitz, of higher order, 160 of type p, 31 quasi-Banach, 9 weak Lebesgue, 22 Zygmund, 160 subordination, 64 symmetric difference, ∆n t , 160 Theorem

189

McGehee, Pigno and Smith, 125 Mori, 105 Nikishin, 31 Nikishin/Stein, 34 on a.e. convergence, 37 on coefficients in F ∈ H ∞ (X), 140 on coefficients in H 1 , 124 on quasiconformal harmonic homeomorphisms, 105 on subharmonic behavior, 143 on the dual of H p , 99 open mapping, 12 Paley on Fourier coefficients, 25 on lacunary series, 170 Prawitz, 62 Privalov on conjugates, 155, 157 Privalov/Plessner, 93, 95 radial maximal, 111 Riesz factorization, 81 Riesz on conjugate functions, 97 Riesz projection, 96 Riesz representation, 86 Riesz the brothers, 78 Riesz/Herglotz, 43, 44 Riesz/Thorin, 18, 20 Rogosinski, 66 Sidon, 172 Stephenson on composition, 90 subharmonic maximal, 111 Tauber, 173 uniqueness, 38, 77 Weierstrass approximation, 41 (U ), 161 (Uα ), 161 (Uβ0 ), 161 Wn (polynomials), 117 Wmax , 117 weak type, 23 [X]q , 11

190