Journal of Function Spaces and Applications

Volume 2012, Article ID 343194, 21 pages

http://dx.doi.org/10.1155/2012/343194

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

^{1}Department of Mathematics, Jiangsu Normal University, Xuzhou 221116, China^{2}School 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