#### Abstract

We discuss multiplication operator with a special symbol on the weighted Bergman space of the unit ball. We give the necessary and sufficient conditions for the compactness of multiplication operator on the weighted Bergman space of the unit ball.

#### 1. Introduction

Let denote the unit ball in , and let be the normalized Lebesgue volume measure on . For , we denote by the measure on defined by , where is a normalizing constant such that . For , we write for the norm on and for the inner product on . The Bergman space is the space of holomorphic functions which are square-integrable with respect to measure on . Reproducing kernels and normalized reproducing kernels in are given by respectively, for . For every we have , for all . The orthogonal projection of onto is given byfor and .

Given , the Toeplitz operator , the Hankel operator , and the multiplication operator are given byrespectively. For , we define the Berezin transform of to be the function ; that is,If is bounded, then is a bounded function on . Since the kernels converge weakly to zero as tends , we have that if is compact, then as . The converse (in both cases) is not necessarily true. According to the definition of Berezin transform, the mean oscillation of in the Bergman metric is the function defined on by

For , let be the automorphism of such that and . Thus, we have the change-of-variable formulafor every .

Multiplication operators are one of the most widely studied classes of concrete operators. The study of their behavior on the Hardy and Bergman spaces has generated an extensive list of results in the operator theory and in the theory of function spaces [1â€“6]. One of the useful approaches is the use of the Berezin transform [7â€“11]. This method is motivated by its connections with quantum physics and noncommutative geometry.

In general, Berezin transform plays important role in giving necessary and sufficient conditions for the boundedness and compactness of the Toeplitz operator [12, 13]. However Berezin transform or the mean oscillation is used to obtain the necessary and sufficient conditions for the boundedness and compactness of the Hankel operator or multiplication operator [14, 15]. This work is partially motivated by using Berezin transform to obtain necessary and sufficient conditions for the compactness of multiplication operator on the weighted Bergman space of the unit ball.

Throughout the paper, we will use the letter to denote a generic positive constant that can change its value at each occurrence.

#### 2. Main Results

In this section, we give the necessary and sufficient conditions for the compactness of multiplication operator on the weighted Bergman space of the unit ball. We furthermore obtain the necessary and sufficient conditions for the compactness of Toeplitz operator and Hankel operator.

Theorem 1. Suppose is bounded on . Then is compact operator on if and only if as .

Proof. Suppose as .
Sinceit is clear that . It suffices to prove that the operator is compact by showing that can be approximated by compact operators in the operator norm.
Let . Then , so we have for .
We define for an operator bySince is bounded on , we prove thatThus, the operator is a Hilbert-Schmidt operator. Since is an integral operator with kernel .
By Schurâ€™s test, whenever there exists a positive measurable function on and constants and such thatfor all in , andfor all in , we haveLet . Sinceand HĂ¶lder inequality, it is easy to prove thatwhere , .
Sincethen we obtain where is positive number.
By the above analysis, we get (12) and (13). By Schurâ€™s test we get , where as and does not depend on . So as implies that is compact on .
Suppose is compact on .
Since the kernels converge weakly to zero as tends , then we have converges to zero as tends . So we obtainas .

Let and let ; we say that wheneverNote that does not distinguish constants, while is a norm in . By Theoremâ€‰â€‰5 in [16], we know that is equivalent to (see the definition in [16]).

For any , let denote the subspace of consisting of functions such that

Theorem 2. Suppose and is bounded on . Then the following are equivalent:(a) as ;(b) is compact operator on ;(c) is compact operator on .

Proof. It suffices to prove that and .
. Sincethen we obtain that as if and only if as . By Theorem 1, we obtain that is compact operator on if and only if as .
. It is clear that as if and only ifâ€‰ with symbol is compact operator on in [12]. Since , then it is clear that as if and only if is compact operator on .

Corollary 3. Suppose , is bounded on , and is compact operator on . Then is compact operator on .

Proof. Suppose is compact operator on . So we obtain is compact operator on . Since as , then is compact operator on . Sincewe obtain is compact operator on .

By Lemma 17 and Theoremâ€‰â€‰19 in [16] and Theorem 2, we obtain the following theorem.

Theorem 4. Suppose , , and is bounded on . Then the following are equivalent:(a) is compact operator on ;(b) is compact operator on ;(c) is compact operator on .

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgment

Kan Zhang is partly supported by NSFC (no. 113711821).