Table of Contents
Journal of Operators
Volume 2015, Article ID 718257, 10 pages
http://dx.doi.org/10.1155/2015/718257
Research Article

Composition Operators from -Bloch Space to -Bloch Space on the Fourth Cartan-Hartogs Domains

School of Mathematics and Statistics, Jiangsu Normal University, Xuzhou, Jiangsu 221116, China

Received 15 June 2015; Revised 22 August 2015; Accepted 17 September 2015

Academic Editor: Ruhan Zhao

Copyright © 2015 Jianbing Su and Chao Zhang. 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 obtain new generalized Hua’s inequality corresponding to , where denotes the fourth Cartan-Hartogs domain in . Furthermore, we introduce the weighted Bloch spaces on and apply our inequality to study the boundedness and compactness of composition operator from to for and .

1. Introduction

The study of composition operators on various Banach spaces of analytic functions has been a long active field in complex and functional analysis. The composition operators as well as related operators known as the weighted composition operators between the Bloch space and Lipschitz space were investigated in [1, 2] in the case of the unit disk. The study of the composition operators on the Bloch space was given in [3] for the polydisc, in [4, 5] for the unit ball, and in [69] for the bounded symmetric domains.

In 1930s, all irreducible bounded symmetric domains were divided into six types by E. Cartan. The first four types of irreducible domains are called the classical bounded symmetric domains. The other two types, called exceptional domains, consist of one domain each (a - and -dimensional domain). In 2000, Yin constructed four kinds of domains corresponding to the classical bounded symmetric domains, called the Cartan-Hartogs domains [10]. It is known that the Cartan-Hartogs domains are nonhomogeneous domains except the unit ball. So it is different from the bounded symmetric domains. The fourth Cartan-Hartogs domains, denoted by , can be expressed as where is the fourth classical bounded symmetric domains [11] and is the transpose of .

For simplicity, we will write for if no ambiguity can arise.

Let be a holomorphic self-map of . The class of all holomorphic functions on is denoted by . The composition operator on is defined by for all and .

In 1955, Hua in [12] proved an inequality: if , are complex matrices and , are both Hermitian positive definite matrices, then Equality holds if and only if .

In 2015, Su et al. obtained generalized Hua’s inequality corresponding to the first Cartan-Hartogs domain (see Theorem   in [13]). From Theorem   in [13], it is easy to get more precise inequality (see Lemma 4). Furthermore, we obtain new generalized Hua’s inequality corresponding to (see Lemma 5).

In this paper, we define the -Bloch space as the space that consists of all such that where

It is clear that is a set of constant functions when , so we assume that .

In this paper, we will obtain some results about the composition operators for the case of the weighted Bloch space on the fourth Cartan-Hartogs domain. In Section 2, we state several auxiliary results most of which will be used in the proofs of the main results. In Sections 3 and 4, we establish the main results of the paper. We give the sufficient conditions and necessary conditions for the boundedness (in Section 3) and the compactness (in Section 4) of composition operator from to , where .

2. Some Lemmas

In order to obtain our main results, we need the following lemmas.

Lemma 1. If then for all and . Where ,

Proof. From (5), there exists a constant such that whenever , where is a positive number.
Set . It is easy to know that is a compact subset of . Thus, there exists a constant such that . SoTherefore, there exists a constant such that for all and . The proof is completed.

Lemma 2. Let be a compact subset of . Then, there exists a constant such that for all .

Proof. When , we have Since and , .
When , denote . For any compact subset , there exists a constant such that .
For any , , we have Furthermore, Thus, So we have Letting , we can get for all . The proof is completed.

Lemma 3. Let and be a compact subset of . Then, there exists a constant such that

Proof. By Lemma 2, there exists a constant such that The proof is completed.

Lemma 4. Let , , and . If , , , and , then

Proof. Set , . Obviously, . So we can get (21) by the process of the proof on Theorem   in [13].

The following conclusion of Lemma 5 is new generalized Hua’s inequality corresponding to .

Lemma 5. Let , , and . If and , then

Proof. When , there exists an orthogonal matrix (see [11]) such that where .
Set Then, From Lemma 4, we can get The proof is completed.

Lemma 6. The composition operator from to is compact if and only if as for every bounded sequence in such that uniformly on every compact subset of .

Proof. Assume that from to is compact. Let be a bounded sequence in such that uniformly on every compact subset of . Suppose as . Then, there exists a subsequence of such that . Since is compact, there exists a subsequence of the bound subsequence , still denoted as , such that , . For any compact set , there exists a constant depending only on such that Thus, uniformly on compact subset . For , there exists a constant such thatwhenever and .
Note that uniformly on compact subset ; then, for the above , there exists a constant such thatwhenever and . Let ; from (28) and (29), we get whenever and . So ; furthermore, on . This is a contradiction and we have .
Conversely, let be a bounded sequence in with . Then, there exists a subsequence of and as . Thus, Therefore, is compact. The proof is completed.

3. The Boundedness of Composition Operators

Theorem 7. Ifthen is bounded.
Conversely, if is bounded, thenwhere , , , and .

and see Lemma 1.

Proof. Let . Then, By Lemma 1 and condition (32), there exists a constant such that for all . So By Lemma 3, we have . Thus, So we get that is bounded.
For the conversion, assume that is a bounded operator with for all .
If , we use a family of test functions in which is defined by Then,Thus,On one hand, we have It is easy to know that Obviously, . Sowhere .
On the other hand, we can get where Thus, Set ; by (39), we have Furthermore, we have If , set Then, For the same reason, it can be proved that (33) holds. The details are omitted here. The proof is completed.

4. The Compactness of Composition Operators

Theorem 8. Ifthen is compact.
Conversely, if is compact, thenwhere , , , , , and ; see Theorem 7.

Proof. Let be a bounded sequence in with and uniformly on compact subsets of . Furthermore, by Weierstrass Theorem, it is easy to show that uniformly on compact subsets of . Thus, for any , there exists such that whenever . Then, Write , and then is a compact subset of .
Let ; then, there exists a constant such that Therefore, there exists a constant such that SoSince as , from (56) and (59) we obtain By Lemma 6, we know that is compact.
For the conversion, assume that (54) fails; then, there exists a sequence in with as and such that for all .
We will construct a family of functions satisfying the following three conditions:(I) is a bounded sequence in ;(II) tends to zero uniformly on compact subsets of ;(III).This contradicts with the compactness of . Hence, we prove that (54) is necessary for that being compact.
If , setWe have It is easy to know that Combining with Lemma 5, we can get , where . Thus, is a bounded sequence in . This means that satisfies condition (I).
By Lemma 5,