Math. Z. (2008) 258:81–86DOI 10.1007/s00209-007-0158-8 Mathematische Zeitschrift

Derivative-free characterizations of bounded compositionoperators between Lipschitz spaces

Miroslav Pavlovic

Received: 6 August 2006 / Accepted: 15 February 2007 / Published online: 19 April 2007© Springer-Verlag 2007

Abstract We prove that Cφ maps HΛα into HΛβ (0 < α, β ≤ 1) if and only if (1 − |φ|)αbelongs to Λβ(D). In the case β < 1, we give another two equivalent conditions.

Keywords Composition operator · Lipschitz spaces

Mathematics Subject Classification (2000) 47B38 · 30H05 · 30D55 · 46E15 · 47B07

1 Introduction and results

Let D denote the open unit disk of the complex plane, and let φ be a holomorphic map fromD to itself. Denoting the vector space of holomorphic functions on D by H(D), we define thecomposition operator Cφ on H(D) by

(Cφ f )(z) = f (φ(z)).

The Littlewood Subordination Principle shows that the restriction of Cφ to the Hardy or(weighted) Bergman spaces on D is always bounded. (We refer to the books [1] and [10] forthe theory of composition operators.) In this note we consider the Lipschitz space HΛα =H(D) ∩Λα(D), 0 < α ≤ 1, where

Λα(D) ={

f ∈ C(D) : supz,w∈D, z �=w

| f (z)− f (w)||w − z|α < ∞

}.

The space Λα(D) becomes a Banach space when endowed with the norm

‖ f ‖Λα = supD

| f | + supz,w∈D, z �=w

| f (z)− f (w)||w − z|α ,

Supported by MNZŽS Serbia, Project No. ON144010.

M. Pavlovic (B)Faculty of Mathematics, Belgrade, Studentski trg 16, p.p. 550, 11001 Belgrade, Serbiae-mail: pavlovic@matf.bg.ac.yu

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82 M. Pavlovic

and HΛα is a closed subspace of Λα(D). The space Λα(∂D) is defined by replacing D with∂D in the above expressions.

For the theory of the spaces Λα(∂D) and HΛα we refer to Flett’s paper [6] and Duren’sbook [2].

By taking f (z) = z we conclude that

if Cφ maps HΛα into HΛβ, then φ ∈ HΛβ ⊂ A(D) := H(D) ∩ C(D). (1)

However the condition φ ∈ HΛβ is sufficient for Cφ to map HΛα into HΛβ only whenα = 1. This follows from Lemma D below and the following result of Madigan [7] (weinclude a proof for completness, see Sect. 3).

Theorem A Let 0 < α, β ≤ 1, and let φ : D �→ D be a holomorphic function. Then thefollowing conditions are equivalent:

(A) Cφ maps HΛα into HΛβ;(B) There is a constant B such that

|φ′(z)| ≤ B(1 − |φ(z)|)1−α

(1 − |z|)1−β , z ∈ D. (2)

It is our aim to replace (2) by a derivative free condition, namely:

(C) The function (1 − |φ|)α belongs to Λβ(D).

In order to formulate two other conditions we introduce the space Λα,rad(D) : it consistsof those functions ψ ∈ C(D) for which

|ψ(ζ )− ψ(rζ )| ≤ B(1 − r)α, ζ ∈ ∂D, 0 < r < 1,

where B is independent of r and ζ ; see [3]. And recall that the Poisson integral of a functionψ ∈ C(∂D) is the (unique) function P[ψ] ∈ C(D) that is harmonic in D and equal to ψ on∂D. If ψ ∈ C(D), then we write P[ψ] = P[ψ |∂D].Theorem 1 Let 0 < α, β ≤ 1, and let φ : D �→ D be a holomorphic function of classA(D). Then (C) is equivalent to (A). If in addition β < 1, then each of the following threeconditions is equivalent to (A):

(D) The function (1 − |φ|)α belongs to Λβ(∂D) ∩Λβ,rad(D);(D1) The function

(1 − ∣∣φ(

eiθ)∣∣)α is in Λβ(∂D) and there is a constant B such that(

1 − ∣∣φ(reiθ )∣∣)α − (

1 − ∣∣φ(eiθ )∣∣)α ≤ B(1 − r)β, reiθ ∈ D;

(E) The function(1 − ∣∣φ(

eiθ)∣∣)α is in Λβ(∂D) and there is a constant B such that(

1 − |φ(z)|)α − P[(1 − |φ|)α](z) ≤ B(1 − |z|)β, z ∈ D. (3)

Observe that (D1) is formally weaker than (D). Concerning (3) note that the function(1 − |φ|)α is superharmonic and therefore

(1 − |φ|)α − P[(1 − |φ|)α] ≥ 0.

The proof of the theorem can easily be modified to remove the hypothesis φ ∈ A(D), butthis is not important in view of (1).

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Characterizations of bounded composition operators 83

2 Proof of Theorem 1

In order to prove the equivalence (A) ⇐⇒ (C) we use the formula |∇|φ| | = |φ′| to rewrite(2) as ∣∣∇(1 − |φ|)α∣∣(z) ≤ αB(1 − |z|)β−1.

Hence the validity of (A) ⇐⇒ (C) is a consequence of Theorem A and the following.

Lemma 1 Let g(z) = (1 − |φ(z)|)α, 0 < α ≤ 1, where φ : D �→ D is holomorphic. Theng ∈ Λβ(D), 0 < β < 1, if and only if

|∇g(z)| ≤ B(1 − |z|)β−1. (4)

The implication (4) �⇒ g ∈ Λβ(D) is a consequence of the following general fact (see,for example, [9, Lemma 6.4.8]) and the formula

|∇g| = α(1 − |φ|)α−1|φ′|.Lemma A If g is a locally Lipschitz function on D satisfying (4), 0 < β ≤ 1, then g ∈Λβ(D).

For the proof that g ∈ Λβ(D) implies (4) we need the following consequence of the Schwarzlemma (see [8]).

Lemma B If φ is a function holomorphic in D, and z ∈ D, then there is a w0 ∈ D such that|w0 − z| ≤ 1 − |z| and

|φ′(z)| ≤ 2|φ(w0)| − |φ(z)|

1 − |z| .

Proof of Lemma 1 Assuming that g ∈ Λβ(D), we have

(1 − |φ(z)|)α − (1 − |φ(w)|)α ≤ B|z − w|β, z, w ∈ D.

Hence, using the inequality

αxα−1(x − y) ≤ xα − yα, x, y > 0,

we get

α(1 − |φ(z)|)α−1{|φ(w)| − |φ(z)|} ≤ B|z − w|β, z, w ∈ D.

Hence, by Lemma B, for each z ∈ D, we find w0 such that

α(1 − |φ(z)|)α−1|φ′(z)| ≤ 2B|z − w0|β

1 − |z| ,

and |w0 − z| ≤ 1 − |z|. It follows that

α(1 − |φ(z)|)α−1|φ′(z)| ≤ 2B(1 − |z|)β−1,

which concludes the proof because |∇g(z)| = α(1 − |φ(z)|)α−1|φ′(z)|. ��We continue the proof of Theorem 1. The implications (C) �⇒ (D) and (D) �⇒ (D1)

are obvious. For (D1) �⇒ (E) we need another classical fact, due to Hardy and Littlewoodand Privalov.

Lemma C If g ∈ Λβ(∂D), 0 < β < 1, then the Poisson integral of g belongs to Λβ(D).

Now suppose that there holds (D1). Let g = (1 − |φ|)α, and, for z �= 0, let z0 = z/|z|. Leth = P[g]. Since g(z0) = h(z0), we have

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g(z)− h(z) = g(z)− g(z0)+ h(z0)− h(z).

And, since

g(z)− g(z0) ≤ B1|z − z0|β = B1(1 − |z|)β,by hypothesis, and

h(z0)− h(z) ≤ B2|z − z0|β = B2(1 − |z|)β,by Lemma C, we get

g(z)− h(z) ≤ (B1 + B2)(1 − |z|)β .This proves the implication (D1) �⇒ (E).

Finally, it remains to prove “(E) �⇒ (C)” or, what is the same, “(E) �⇒ (B)”. To provethe latter, assume that |w − z| ≤ 1 − |z|. Then

g(z)− g(w) ≤ g(z)− h(w)

= g(z)− h(z)+ h(z)− h(w).

Now we use the hypothesis (E) and Lemma C to get

g(z)− g(w) ≤ B(1 − |z|)β .Hence, as in the proof of Lemma 1,

α(1 − |φ(z)|)α−1{|φ(w)| − |φ(z)|} ≤ B(1 − |z|)β .From this and Lemma B we get

α(1 − |φ(z)|)α−1|φ′(z)| ≤ 2B(1 − |z|)β−1.

This completes the proof of Theorem 1.

3 Proof of Theorem A

This theorem is a consequence of the following well known result of Hardy and Littlewood:

Lemma D A function f holomorphic in D belongs to HΛβ, 0 < β ≤ 1, if and only if

Q( f ) := supz∈D

(1 − |z|)1−β | f ′(z)| < ∞.

Moreover, there is a constant B independent of f such that

Q( f )/B ≤ supz,w∈D

| f (z)− f (w)||w − z|β ≤ B Q( f ).

Assuming (2) we have, for f ∈ HΛα,

|(Cφ f )′(z)| = | f ′(φ(z))||φ′(z)|≤ B1(1 − |φ(z)|)α−1|φ′(z)|≤ B2(1 − |z|)β−1,

and hence Cφ f ∈ Λβ(D).

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Characterizations of bounded composition operators 85

Assume, conversely, that Cφ f ∈ HΛβ for every f ∈ HΛα. Then using Lemma D andapplying the closed graph theorem in the standard way, we find that there is a constant Bindependent of f and z such that

|(Cφ f )′(z)| ≤ B‖ f ‖Λα (1 − |z|)β−1.

Hence, by taking f (z) = (1 − az)α, |a| ≤ 1, we get∣∣(1 − aφ(z))α−1∣∣ |φ′(z)| ≤ B1(1 − |z|)β−1,

with B1 independent of a. Finally we put a = φ(z)/|φ(z)| to get (2). This completes theproof of Theorem A.

4 Remarks on the case α = 1

Let α = 1. Then condition (2) reduces to

|φ′(z)| ≤ B1

(1 − |z|)1−β , z ∈ D,

which is by Lemma D, equivalent to φ ∈ HΛβ. Hence, by Lemma D, the equivalence (B)⇐⇒ (C) ⇐⇒ (D) ⇐⇒ (D1) ⇐⇒ (E) can be stated as follows.

Theorem B Let 0 < β ≤ 1, and let φ : D �→ D be a holomorphic function of class A(D).Then φ ∈ HΛβ if and only if |φ| ∈ Λβ(D). If in addition β < 1, then each of the followingthree conditions is equivalent to φ ∈ HΛβ :

(a) The function |φ| belongs to Λβ(∂D) ∩Λβ,rad(D);(b) The function

∣∣φ(eiθ

)∣∣ is in Λβ(∂D) and there is a constant B such that∣∣φ(eiθ )∣∣ − ∣∣φ(

reiθ )∣∣ ≤ B(1 − r)β, reiθ ∈ D;(c) The function

∣∣φ(eiθ

)∣∣ is in Λβ(∂D) and there is a constant B such that

P[|φ|](z)− |φ(z)| ≤ B(1 − |z|)β, z ∈ D.

Of course, we can remove the condition that φ maps D into D by considering the functionφ1 = φ/M, where M > sup

D|φ|.

Theorem B is known; see [3,8] (β < 1) and [4,5].

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(1972)

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7. Madigan, K.M.: Composition operators on analytic Lipschitz spaces. Proc. Am. Math. Soc. 119, 465–473(1993)

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