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Abstract and Applied Analysis

Volume 2013 (2013), Article ID 151929, 9 pages

http://dx.doi.org/10.1155/2013/151929

## Weighted Differentiation Composition Operators to Bloch-Type Spaces

^{1}Department of Mathematics, Shantou University, Shantou, Guangdong 515063, China^{2}School of Mathematics, Shri Mata Vaishno Devi University, Kakryal, Katra 182320, India

Received 2 February 2013; Accepted 8 April 2013

Academic Editor: Pedro M. Lima

Copyright © 2013 Junming Liu et al. 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 characterized the boundedness and compactness of weighted differentiation composition operators from BMOA and the Bloch space to Bloch-type spaces. Moreover, we obtain new characterizations of boundedness and compactness of weighted differentiation composition operators.

#### 1. Introduction

Let be the open unit disk in the complex plane , the space of all functions holomorphic on , the normalized area measure on , and the space of all bounded holomorphic functions with the norm .

Let . The *-Bloch space * on is the space of all holomorphic functions on such that
The * little **-Bloch space * consists of all such that
Both spaces and are Banach spaces with the norm
and is a closed subspace of . If , they become the classical Bloch space and little Bloch space , respectively. For any , the space consists of functions such that
For information of such spaces, see, for example, [1–4].

For , let be the automorphism of that interchanges and . Let the Green function in with logarithmic singularity at be given by The space consists of all in the Hardy space such that is a Banach space under following norm (see, e.g., [5]):

Let and be holomorphic maps on the open unit disk such that . For a nonnegative integer , we define a linear operator as follows: We call it weighted differentiation composition operators, which was defined in [6, 7]. If and , becomes induced by , defined as , . If and , then is the differentiation operator defined as . If , then we get the weighted composition operator defined as . If and , then reduces to . When , then reduces to differentiation composition operator (also named as product of differentiation and composition operator). If we put , then , the product of multiplication and differentiation operator.

The boundedness and compactness of differentiation composition operator between spaces of holomorphic functions have been studied extensively. For example, Hibschweiler; Portnoy and Ohno studied differentiation composition operator on Hardy and Bergman spaces in [8, 9]; Li; Stević and Ohno studied on Bloch type spaces in [10–12]; Wu and Wulan gave a new compactness criterion of on the Bloch space in [13]. Recently, the weighted differentiation composition operator between different function spaces has also been investigated by several authors (see, for example, [14–21]).

Boundedness, compactness, and essential norm of weighted composition operator between Bloch-type spaces have been studied in [22–24]. Recently, Manhas and Zhao [25] and Hyvärinen and Lindström [26] gave a new characterization of boundedness and compactness of in terms of the norm of (for the compactness of composition operator, see [27, 28]).

Motivated by [13, 25, 26], we study the operator from and Bloch space to Bloch-type spaces.

Throughout this paper, constants are denoted by ; they are positive and not necessarily the same at each occurrence. The notation means that there is a positive constant such that . When and , we write .

#### 2. Some Lemmas

It is well known that . From the definition of the norm, we know Indeed, Girela proved that in Corollary 5.2 of [5]. The following lemma is from Lemma 5 in [29] (see also Lemma 4.12 of [4]).

Lemma 1. *If , then
*

The following lemma may be known, but we fail to find its reference; so we give a proof for the completeness of the paper.

Lemma 2. *Let . Then,
*

* Proof. *Applying Littlewood-Paley identity
and Lemma 1, we have
It follows from the definitions of Bloch space and space that

By Theorem 6.2 of [5] and the proof of Theorem 1 of [30], we have the following lemma.

Lemma 3. *Let be a fixed positive integer and with . If
**
then .*

Lemma 4. *Suppose that is a fixed positive integer. Let , , and
**
If , then there are two positive constants and , depending only on , such that
*

*Proof. *The proof is similar to that of Lemma 2.2 of [13] and is so omitted.

#### 3. Boundedness of

In this section, we characterize the boundedness of from and the Bloch space to Bloch-type spaces.

Theorem 5. *Let , , , and a holomorphic self-map of . Then, the following statements are equivalent: *(a)* is bounded. *(b)* and are bounded. *(c)* is bounded. *(d)* and are bounded. *(e)* is bounded. *(f)* and are bounded. *(g)* and .*(h)

*and .*

*Proof. *It is obvious that , , , and . Thus, we will prove the theorem according to the following steps. (I): , . (II): , . (III): , . (IV): .

(I): , . Suppose that or holds. We choose the test function . By Lemma 2, we get
So
Taking and using the fact that , we have
We now consider the function
It is easy to check that and . Moreover,
Thus, and
We obtain
Thus, for any , we have
Using (21) yields
Combining (26) with (27), we get
We next consider the function
Similarly, we get and
Moreover,
So
and . We have, as above,

Thus, for any ,
Applying (20), we get
Combining (34) with (35) yields
(II): and . Suppose that is bounded or is bounded. Set
If , then for any positive integer , we can find such that
If , then choose the test function . It is clear that . From Lemma 2, we have

So
If , consider the function
where . Let . Then, and
It is easy to see that
So, by Theorems 5.4 and 5.13 of [4], we have and . By Lemma 1 of [31] and Lemma 3, we get . We have
Since is arbitrary, we get . This contradicts the boundedness of and that of .

Now, suppose that is bounded or is bounded. Set
If , then for any positive integer , exists such that
If , then set . The process as above gives
If , consider the function
where . Let . Then, and
Applying Theorems 5.4 and 5.13 of [4] again yields and . We get and
Since is arbitrary, we have . This contradicts the boundedness of .

(III): , . Note that
The desired results follow.

(IV): . Suppose that is true. It follows from Proposition 5.1 of [4] that . So,

Conversely, assume that is true. It is easy to see that
If , then
Hence, is true. From , we obtain that is also true.

From now on, we assume that . For any integer , let
Let with be the smallest positive integer such that . Since is not empty for every integer and . By Lemma 4, for ,
So, is bounded. Similar argument implies
Thus, is bounded. Theorem 5 is proved.

#### 4. Compactness of

The following criterion for the compactness is a useful tool and it follows from standard arguments, for example, Proposition 3.11 of [32] or Lemma 2.10 of [33].

Lemma 6. *Let , , and , or . Suppose that and are in such that . Then, is compact if and only if for any sequence in with , which converges to zero locally uniformly on ; we have .*

We now give the compactness of from and the Bloch space to Bloch-type spaces.

Theorem 7. *Let , , , and a holomorphic self-map of . Then, the following statements are equivalent: *(a)* is compact. *(b)* is compact and is compact. *(c)* is compact. *(d)* is compact and is compact. *(e)* is compact. *(f)* is compact and is compact. *(g)*, ,
*(h)* and .*

* Proof. *The proof is a modification of that of Theorem 5; so we give a sketch of the proof. We will prove the theorem according to the following steps. (I): , . (II): , . (III): , . (IV): .

(I): , . Suppose that or holds. Then by Theorem 5, we have
That is, , .

Let be a sequence in such that as . Now, we consider the function
Simple computation shows that and
It is also easy to check that uniformly on compact subsets of as . Moreover,
We have
By Lemma 6, we get
We next consider the function
Similarly, we get and
It is easy to see that converges to zero uniformly on compact subsets of as and
Thus,
Applying Lemma 6 again, we have
Since is arbitrary, we proved that is true.

(II) , . Suppose that or holds. A similar argument to (I) shows that , . Now, suppose that the equations in are not true. Then, there exists a sequence in and such that as and
Choose a subsequence of if necessary and suppose that . Let
Then, it is easy to check that , , uniformly on compact subsets of and
Thus,
Those contradict the compactness of and .

(III) , . Let be a norm bounded sequence in that converges to zero uniformly on compact subsets of . Let . For , then there exists such that for , we have
Thus, for , we have
where and . Since uniformly on compact subsets of as , we have as . It follows from Lemma 6 that is compact.

Similar as above, we know
From uniformly on compact subsets of , we have and as . So, , are compact.

(IV): . Suppose that is true. Note that and uniformly on compact subsets of as ; by Lemma 6, we have
Conversely, assume that is true. It is easy to see that
If , from , we get that is true. If , as in the proof of Theorem 5, let
And let with be the smallest positive integer such that . For given , there exists a large enough integer with such that
whenever . Let be a norm bounded sequence in that converges to zero uniformly on compact subsets of as . Denote . We get
Then,
where
Since uniformly on compact subsets of , then as . Thus, by Lemma 6, is compact. Similar as above, we can prove that is compact. The proof is complete.

#### Acknowledgments

This work was supported by NNSF of China (Grant no. 11171203) and NSF of Guangdong Province (Grant nos. 10151503101000025 and S2011010004511).

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