Journal of Function Spaces and ApplicationsVolume 2012 (2012), Article ID 343194, 21 pagesdoi:10.1155/2012/343194

Research Article

The Multiplication Operator from Spaces to th Weighted-Type Spaces on the Unit Disk

Yongmin Liu¹ and Yanyan Yu²

¹Department of Mathematics, Jiangsu Normal University, Xuzhou 221116, China²School of Mathematics and Physics Science, Xuzhou Institute of Technology, Xuzhou 221111, China

Received 2 February 2012; Revised 11 April 2012; Accepted 30 April 2012

Academic Editor: Amol Sasane

Copyright © 2012 Yongmin Liu and Yanyan Yu. 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.

Let be the space of analytic functions on and . The boundedness and compactness of the multiplication operator from spaces to th weighted-type spaces on the unit disk are investigated in this paper.

1. Introduction

Let denote the space of all analytic functions in the open unit disc of the finite-complex plane , the boundary of , the set of all nonnegative integers and the set of all positive integers. Let be a positive continuous function on (weight) such that and . The th weighted-type spaces on the unit disk , denoted by which were introduced in [1], consist of all such that For , the space becomes the weighted-type space (see, e.g., [2–4]), for the Bloch-type space and for the Zygmund-type space . For , we obtain correspondingly the classical weighted-type space, the Bloch space , and the Zygmund space . Some information on Zygmund-type spaces on the unit disk and some operators on them, for example, in [5–9] and on the unit ball, can be found, for example, in [10, 11]. From now on, we will assume that . Set With this norm, the th weighted-type space becomes a Banach space.

The little th weighted-type space, denoted by , is a closed subspace of consisting of those for which

A positive continuous function on is called a normal if there exist positive numbers , and , such that (see [12]).

For , a function is said to belong to the general function space if where . An is said to belong to if The space was introduced by Zhao in [13]. We can get many function spaces if we take some specific parameters of ; for example (see [13]), and for ; and for ; and ; and . Since for , is the space of constant functions, we assume that .

The multiplication operator is defined by . It is interesting to provide a function theoretic characterization of when induces a bounded or compact composition operator on various spaces (see, e.g., [14–20]). Yu and Liu in [21] studied the boundedness and compactness of the operator from mixed-norm spaces to the Bloch-type space. Stević in [22] studied the boundedness and compactness of the product of the differentiation and composition operator from the space of bounded analytic functions, the Bloch space and the little Bloch space to th weighted-type spaces on the unit disk. Zhang and Xiao in [23] studied the bounded and compact-weighted composition operator from the space to the Bloch-type space in the unit disc. Zhang and Zeng in [24] studied the boundedness and compactness of weighted differentiation composition operators from weighted bergman space to th weighted-type spaces on the unit disk. Ye in [25] studied the boundedness and compactness of the weighted composition operator from into the logarithmic Bloch space on the unit disk. Yang in [26] studied the boundedness and compactness of weighted differentiation composition operators from the space to the Bloch-type space. Yang in [27] studied the boundedness and compactness of the composition operator from the space to th weighted-type spaces on the unit disk. Zhou and Chen in [28] studied the weighted composition operator from the space to the Bloch-type space in the unit ball. Zhu in [29] studied the weighted composition operator from the space to space in the unit ball. Stević in [30, 31] studied the boundedness and compactness of the integral operators between spaces and Bloch-type spaces in the unit ball. This paper focuses on the boundedness and compactness of the operators from to th weighted-type spaces on the unit disk.

From now on, we will always assume that , , , , is normal and . Further, for the sake of simplicity, will always denote an independent constant, which can be different from one display to another.

2. Auxiliary Results

In this section we formulate some auxiliary results which will be used in the proofs of the main results.

Lemma 2.1 (see [13, 27]). Assume that . Then, for each , there is a positive constant , independent of such that and Moreover, if , then .

Lemma 2.2 (see [32, 33]). Let and . Then,

Lemma 2.3 (see [1]). Assume and Then,

By standard arguments (see, e.g., [34] or Lemma  3 in [35]) the following lemma follows.

Lemma 2.4. Assume that . Then, is compact if and only if is bounded and for any bounded sequence in which converges to zero uniformly on compact subsets of as , one has as .

3. The Boundedness and Compactness of from to

In this section, we characterize the boundedness and compactness of .

Theorem 3.1. Assume that and is normal. (a)If , then is bounded if and only if (b)If , then is bounded if and only if (3.1) holds and .(c)If , then is bounded if and only if (3.1) holds and

Proof. Let , ,, . Assume that conditions (3.1) holds. Then for all Since let , then we have , so By the Cauchy integral formula and (3.3), we obtain Note that and are normal, we have hence Similarly, for , we have that Hence, if , by Lemmas 2.1 and 2.2, the Leibnitz formula, (3.1), (3.9), and (3.10), we have that for every and , if , using , we have that for every and , if , by (3.2), we have that for every and . We also have that and for each , Using (3.11), (3.12), (3.13), (3.14), and (3.15) it follows that the operator is bounded.On the other hand, suppose that is bounded, that is, there exists a constant such that for all . Then, we can easily obtain by taking . For a fixed , and constants , set where . A straightforward calculation shows that, for , so It is easy to see that for each and by using the same methods in [23]. By Lemma 2.3, using the same method in [22, 36], we can choose , the corresponding function is denoted by , such that Therefore, From this, we obtain Since is normal and , we get which along with (3.22) implies that (3.1) is necessary for all case. Let , and be bounded. To prove (3.2), we set By a direct calculation, we get thus we have and (see [19, 30]). On the other hand, we have that Moreover, we have that Taking , , if , we have if , we have From (3.29), (3.1), (3.9), and (3.10) we have Using (3.28) and (3.30), it is easy to get that (3.2) holds, finishing the proof of the theorem.

Theorem 3.2. Assume that and is normal. (a)If , then is compact if and only if (b)If , then is compact if and only if (3.31) holds and .(c)If , then is compact if and only if (3.31) holds and

Proof. Assume that conditions (3.31) hold. By Theorem 3.1, is bounded. For any bounded sequence in with uniformly on compact subsets of establish the assertion, it suffices, in view of Lemma 2.4, to show that We assume that . From (3.31) and (3.32), given , there exists a , when , we have By (3.34) and the Cauchy integral formula, we obtain when . Hence, when . Similarly, for , we have that when . Since uniformly on compact subsets of , Cauchy’s estimate gives that converges to 0 uniformly on compact subsets of for each , there exists a such that implies that If , from (3.34), (3.38), (3.39), and Lemmas 2.1 and 2.2, we have when . If , then . By [37, Lemma 3.2], we have By (3.38), (3.39), Lemma 2.1, and , we get when . If , from (3.35), (3.38), (3.39), and Lemmas 2.1 and 2.2, we get when . It follows that the operator is compact. Conversely, assume that is compact. Taking , we get . Let be a sequence in such that as . Taking the test functions , where is defined by (3.17), we will write We obtain that , and for , For , we have that is, converges to 0 uniformly on compact subsets of , using the compactness of , we obtain and consequently (3.31) holds. Assume that and is compact, we only need to prove (3.32) holds. To do this, let be a sequence in such that as . Set Since we have and (see [19, 30]), and uniformly on compact subsets of . Since is compact, we have We also have that Therefore, Taking and , if , we have if , we have From (3.31), (3.35), (3.37), and (3.38), we have that for sufficiently large : Using (3.54) and (3.56), it is easy to get that (3.32) holds. From which we obtain the desired result.

4. The Boundedness of from (or) to

In this section, we characterize the boundedness of .

Theorem 4.1. Assume that and is normal. Then, (1) is bounded if and only if is bounded and (2) is bounded if and only if is bounded and (4.1) holds.

Proof. (1) Assume that is bounded and condition (4.1) holds. Since it follows that Since is normal, by the monotonicity of , for , we have that is, is decreasing on , and for any , there is a such that which implies . Since for and , we have . From (4.3), it follows that For , by applying formula (4.6) to the function , we get when . It follows from (4.1), (4.6), and (4.7) that for . Since, for each polynomial , we have hence, . Since the set of all polynomials is dense in , we have that for every there is a sequence of polynomials such that From this and since the operator is bounded, we have that Since is a closed subspace of , therefore, we have , from which the boundedness of follows. On the other hand, assume that is bounded, then is bounded. By taking the function given by , we obtain as desired.(2) The proof is similar to that of the case (1). We leave the details to the interested reader.

Acknowledgments

The authors acknowledge gratefully the support in part by the National Natural Science Foundation of China (no. 11171285) and the Grant of Natural Science Basic Research of Jiangsu Province of China for Colleges and Universities (nos. 06KJD110175; 07KJB110115). The authors also thank the referees for their thoughtful comments and helpful suggestions which greatly improved the final version of this paper.

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