`Abstract and Applied AnalysisVolume 2012 (2012), Article ID 462482, 18 pageshttp://dx.doi.org/10.1155/2012/462482`
Research Article

## Weighted Composition Operators on the Zygmund Space

Department of Mathematics, Fujian Normal University, Fuzhou 350007, China

Received 25 February 2012; Accepted 19 April 2012

Copyright © 2012 Shanli Ye and Qingxiao Hu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We characterize the boundedness and compactness of the weighted composition operator on the Zygmund space and the little Zygmund space .

#### 1. Introduction

Let be the open unit disk in the complex plane , let be its boundary, and let denote the set of all analytic functions by . For , let

An analytic function is said to belong to the Zygmund space if , and the little Zygmund space consists of all satisfying . From a theorem of Zygmund (see [1, vol. I, page 263] or [2, Theorem 5.3]), we see that if and only if is continuous in the close unit disk and the boundary function such that

for all and all . It can easily proved that is a Banach space under the norm: and that is a closed subspace of . It is easily obtained that For some other information on this space and some operators on it, see, for example, [35].

An analytic self-map induces the composition operator on , defined by for analytic on . It is a well-known consequence of Littlewood’s subordination principle that the composition operator is bounded on the classical Hardy, Bergman, and Bloch spaces (see, e.g., [69]).

Recall that a linear operator is said to be bounded if the image of a bounded set is a bounded set, while a linear operator is compact if it takes bounded sets to sets with compact closure. It is interesting to provide a function theoretic characterization of when induces a bounded or compact composition operator on various spaces. The book [10] contains plenty of information on this topic.

Let be a fixed analytic function on the open unit disk. Define a linear operator on the space of analytic functions on , called a weighted composition operator, by , where is an analytic function on . We can regard this operator as a generalization of a multiplication operator and a composition operator. In recent years, the weighted composition operator has been received much attention and appears in various settings in the literature. For example, it is known that isometries of many analytic function spaces are weighted composition operators (e.g., see [11]). The boundedness and compactness of it has been studied on various Banach spaces of analytic functions, such as Hardy, Bergman, BMOA, Bloch-type spaces, see, for example, [1216]. Also, it has been studied from one Banach space of analytic functions to another, one may see in [1726].

The purpose of this paper is to consider the weighted composition operators on the Zygmund space and the little Zygmund space . Our main goal is to characterize boundedness and compactness of the operators on in terms of function theoretic properties of the symbols and . We also characterize boundedness and compactness of on .

Throughout this paper, constants are denoted by , they are positive and may differ from one occurrence to the other.

#### 2. Auxiliary Results

In order to prove the main results of this paper, we need some auxiliary results.

Lemma 2.1. If , then(i) for every ;(ii) for every .

Proof. Suppose , and , then by (1.4). It follows that hence
One may easily prove (ii) by (1.4). The details are omitted here.

Lemma 2.2. Suppose , then , where .

One may easily obtain it by a calculation.

Lemma 2.3. Suppose is a bounded operator. Then is a bounded operator.

Proof. Suppose is bounded in . It is clear that for any , we have for every . According to Lemma 2.2, we obtain that Then Hence, is a bounded operator.

#### 3. Boundedness of

In this section, we characterize bounded weighted composition operators on the Zygmund space and the little Zygmund space .

Theorem 3.1. Let be an analytic function on the unit disc and an analytic self-map of . Then is bounded on the Zygmund space if and only if and the following are satisfied:

Proof. Suppose is bounded on the Zygmund space . Then we can easily obtain the following results by taking and in , respectively: By (3.3), (3.4), and the boundedness of the function , we get Let in again, in the same way we have Using these facts and the boundedness of the function again, we get
Fix with , we take the test functions: for , where Then we have and by [3], where is not dependent on . Therefore, for all with , we have Let , it follows that Then, by Lemma 2.1 and (3.3), we have Hence
For all with , by (3.7), we haveHence (3.1) holds.
Next, we will show that (3.2) holds. Fix with , we take another test functions: for . It is proved that above, where is not dependent on . Therefore, for all with , we have Let , it follows that Hence By (3.1), Lemma 2.1, and the boundedness of the function , we get
For all with , by (3.5), we haveHence (3.2) holds.
Conversely, suppose that , (3.1) and (3.2) hold. For , by Lemma 2.1, we have the following inequality:This shows that is bounded. This completes the proof of Theorem 3.1.

Theorem 3.2. Let be an analytic function on the unit disc and an analytic self-map of . Then is bounded on the little Zygmund space if and only if , (3.1) and (3.2) hold, and the following are satisfied:

Proof. Suppose that is bounded on the little Zygmund space . Then . Also , thus Since and , we have . Hence (3.24) holds.
Similarly, , thenBy (3.24), and , we get that , that is, (3.23) holds.
On the other hand, by Lemma 2.3 and Theorem 3.1, we obtain that (3.1) and (3.2) hold.
Conversely, letFor all , we have both and as by (1.5). Since , given that , there is a such that , and for all with .
If , it follows that
We know that there exists a constant such that , and for all .
If , it follows thatThus, we conclude that as . Hence for all . On the other hand, is bounded on by Theorem 3.1. Hence is a bounded operator on the little Zygmund space .

The following corollary is just as Theorem  2.2 in [27].

Corollary 3.3. Let be an analytic self-map of . Then is a bounded operator on if and only if

Corollary 3.4. Let be an analytic self-map of . Then is a bounded operator on if and only if , (3.30) and (3.31) hold.

Proof. By Theorem 3.2, is a bounded operator on if and only if , , (3.30) and (3.31) hold. However, by (1.5), implies that . Then, is a bounded operator on if and only if , (3.30) and (3.31) hold.

#### 4. Compactness of

In order to prove the compactness of on the Zygmund space , we require the following lemmas.

Lemma 4.1. Suppose that be a bounded operator on . Then is compact if and only if for any bounded sequence in which converges to 0 uniformly on compact subsets of , we have as .

The proof is similar to that of Proposition 3.11 in [10]. The details are omitted.

Lemma 4.2. Let be a bounded sequence in which converges to 0 uniformly on compact subsets of . Then .

Proof. Let . Given any , there exist such that . If , by Lemma 2.1, it follows that where we use the fact that for all . Then Noting that converges to 0 uniformly on compact subsets of , we get Hence, .

Theorem 4.3. Let be an analytic function on the unit disc and an analytic self-map of . Suppose that be a bounded operator on . Then is compact if and only if the following are satisfied:

Proof. Suppose that is compact on the Zygmund space . Let be a sequence in such that as . Without loss of generality, we may suppose that for all . We take the test functions: where such that . By a direct calculation, we may easily prove that converges to 0 uniformly on compact subsets of . From the proof of Theorem 3.1, we see that . Then is a bounded sequence in which converges to 0 uniformly on compact subsets of . Then by Lemma 4.1. Note that it follows that Then if one of these two limits exits.
On the other hand, letso One may obtain that on compact subsets of by a direct calculation and by the proof of Theorem 3.1. Consequently, is a bounded sequence in which converges to 0 uniformly on compact subsets of . Then by Lemma 4.1. Note that and by Lemma 4.2, it follows that as . Then . The proof of the necessary is completed.
Conversely, Suppose that (i) and (ii) hold. Let be a bounded sequence in which converges to 0 uniformly on compact subsets of . Let . We only prove by Lemma 4.1. This amounts to showing thatBy Lemma 4.2 and bounded on , which implies that , then If , by (3.5), then If , by Lemma 2.1, then Thus, First, letting tend to infinity and subsequently increase to 1, one obtains that as . The third statement is proved similarly.
If , by (3.7), thenIf , then Thus, which also implies that as . This completes the proof of Theorem 4.3.

In order to prove the compactness of on the little Zygmund space , we require the following lemma.

Lemma 4.4. Let . Then is compact if and only if it is closed, bounded, and satisfies

The proof is similar to that of Lemma  1 in [6], we omit it.

Theorem 4.5. Let be an analytic function on the unit disc and an analytic self-map of . Then is compact on the little Zygmund space if and only if and the following are satisfied:

Proof. Assume that (i) and (ii) hold, and . By Theorem 3.2, we know that is bounded on the little Zygmund space . From (ii), we can show that Suppose that with . We obtain that thus, and it follows that hence, is compact on by Lemma 4.1.
Conversely, suppose that is compact on .
First, it is obvious is bounded on , then by Theorem 3.2, we have and that (3.24) holds. On the other hand, by Lemma 4.1 we havefor some .
Next, note that the proof of Theorem 3.1 and the fact that the functions given in (3.8) are in and have norms bounded independently of , we obtain that
Similarly, note that the functions given in (3.16) are in and have norms bounded independently of , we obtain thatfor . So by (4.30) and , it follows that for . However, if , by (3.24), we easily have This completes the proof of Theorem 4.5.

Corollary 4.6. Let be an analytic self-map of . Then is a compact operator on if and only if

In the formulation of corollary, we use the notation on defined by for .

Corollary 4.7. Let be an analytic function on the unit disc . Then the pointwise multiplier is a compact operator if and only if .

#### Acknowledgment

The research was supported by the Natural Science Foundation of Fujian Province, China (Grant no. 2009J01004).

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