#### Abstract

We discuss the boundedness and compactness of the weighted composition operator from mixed norm space to Bloch-type space on the unit ball of .

#### 1. Introduction

Let be the class of all holomorphic functions on and the collection of all the holomorphic self-mappings of , where is the unit ball in the -dimensional complex space . Let denote the Lebesegue measure on normalized so that and the normalized rotation invariant measure on the boundary of . For , let be the radial derivative of .

A positive continuous function on is called normal (see, e.g., [1]) if there exist three constants , and , such that for In the rest of this paper we always assume that is normal on , and from now on if we say that a function is normal we will also suppose that it is radial on , that is, for .

Let , , and be normal on . is said to belong to the mixed norm space if is a measurable function on and , where If , then is just the space is measurable function on .

Let . If , then is just the weighted Bergman space . In particular, is Bergman space if and . Otherwise, if and , then is the Dirichlet-type space.

For , , let ; it is easy to see that the mixed norm space , written by , consists of all such that

Now is said to belong to Bloch-type space if where is the complex gradient of .

It is clear that is a Banach space with norm . For , we denote where

It was proved that , , and are equivalent for in [2, 3].

Let , ; the composition operator induced by is defined by and the weighted composition operator is defined by for . We can regard this operator as a generalization of a multiplication operator and a composition operator . That is, when , we obtain and when we obtain .

It is interesting to provide a function theoretic characterization when and induce a bounded or compact weighted composition operator between some spaces of holomorphic functions on . Recently, this operator is well studied by many papers; see, for example, [3–17] and their references therein. In particular, Stević [18] gave some conditions of weighted composition operators between mixed-norm spaces and spaces on the unit ball. Zhou and Chen [19] discussed weighted composition operators from to Bloch-type spaces on the unit ball. More recently, the weighted composition operator from Bers-type space to Bloch-type space on the unit ball was studied in [6]. Now in this paper, we will continue this line of research and characterize the boundedness and compactness of the weighted composition operator acting from mixed-norm spaces to Bloch-type space on the unit ball of . The paper is organized as follows. In Section 2, we give some lemmas. The main results are given in Section 3.

Throughout the remainder of this paper, will denote a positive constant; the exact value of which will vary from one appearance to the next. The notation means that there is a positive constant such that .

#### 2. Some Lemmas

Lemma 1. *Assume that , , and . Then there is a positive constant which is independent of such that
*

*Proof. *We first prove (10). By the monotonicity of the integral means and [20, Theorem 1.12] we have that
from which the desired result (10) follows.

Next we prove (11). By the monotonicity of the integral means, using the well-known asymptotic formula (e.g., [21, Theorem 2]), we obtain that
By [20, Theorem 1.12], it follows that
Then the desired result (11) follows. This completes the proof.

From the above lemma, when , then For , , denote the Bergman metric of by

Lemma 2. *Let and . Then
**
for all , where denotes the Jacobian matrix of and
*

*Proof. *Let . If , the desired result is obvious. If , from the definition of ,
Thus
The desired result follows from (20). The proof is completed.

The proof of the next lemma is standard; see, for example, [4, Proposition 3.11]. Hence, it is omitted.

Lemma 3. *Assume that , , is a normal function, and , . Then is compact if and only if for any bounded sequence in which converges to zero uniformly on compact subsets of as ; then , as .*

Lemma 4. *For and , one has
*

*Proof. *
This completes the proof.

#### 3. The Boundedness and Compactness of

Theorem 5. *Assume that , , is a normal function, and , . Then is bounded if and only if
*

*Proof *
*Sufficiency.* Assume that (23) and (24) hold. Then for any , if for , by Lemma 1 and Lemma 2, it follows that
When for . From (23) we can easily obtain
Combining (25) and (26), the boundedness of follows.*Necessity.* Suppose that is bounded. Firstly, we assume that and , where and .

If , where , choose the function

By [20, Theorem 1.12] and Lemma 4 we have that
Then and . Moreover, and
Thus
By the definition of and (30) it follows that
This shows that when , (24) follows.

On the other hand, if . For , let and , when ; otherwise when . Take
By [20, Theorem 1.12] and Lemma 4 we obtain that
Hence and . Moreover and
Similar to the proof of (30), we obtain that
It follows from (35) that
That is, when , (24) follows. Combining the above two cases, the desired result (24) holds.

For the general situation, we can use some unitary transform to make and we can prove (11) by taking the function . By the linearity of the unitary transform , , and the normalized rotation invariant measure on the boundary , we get that

Next we prove (23). Set the function
for fixed and . Then,
By [20, Theorem 1.12], it follows that
Applying Lemma 4 we have that
Therefore , and . Besides,
Therefore,
It follows from (43) and (24) that
Combining (44) and (45), the desired result (23) holds. This completes the proof.

Theorem 6. *Assume that , , is a normal function, and , . Then is compact if and only if the followings are all satisfied:*(a)* and for ;*(b)*(c)*

*Proof *
*Sufficiency.* Suppose that , , and hold. Then for any , there is , such that
when .

Let be any sequence which converges to uniformly on compact subsets of satisfying . Then and converge to uniformly on . Hence

If and , by Lemma 1 and Lemma 2, we have