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ISRN Mathematical Analysis
Volume 2013 (2013), Article ID 492052, 6 pages
http://dx.doi.org/10.1155/2013/492052
Research Article

Norm of a Volterra Integral Operator on Some Analytic Function Spaces

1College of Mathematics and Information Science, Henan Normal University, Xinxiang 453007, China
2Department of Mathematics, Jiaying University, Meizhou, Guangdong 514015, China

Received 27 June 2013; Accepted 30 July 2013

Academic Editors: O. Miyagaki and M. Winter

Copyright © 2013 Hao Li and Songxiao Li. 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 an analytic function in the unit disc . The Volterra integral operator is defined as follows: In this paper, we compute the norm of on some analytic function spaces.

1. Introduction

Let be the unit disk of complex plane and the class of functions analytic in . Denote by the normalized Lebesgue area measure in and the Green function with logarithmic singularity at ; that is, , where is the Möbius transformation of .

Let . The is the space of all functions such that From [1, 2], we see that = BMOA, the space of all analytic functions of bounded mean oscillation. When , the space is the same and equal to the Bloch space , which consists of all for which See [3, 4] for the theory of Bloch functions.

For , the -Bloch space, denoted by , is the space of all such that It is clear that for .

Let and let . The mean Lipschitz space consists of those functions for which It is obvious that is just the Bloch space , which is contained in for every . Note that increases with . We refer to [5] for more information of mean Lipschitz spaces.

For , we say that an belongs to the growth space if It is easy to see that .

For , an is said to belong to the space if

For , the Besov space is defined to be the space of all analytic functions in such that

Let . The Volterra integral operators and are defined as follows: It is easy to see that where denotes the multiplication operator; that is, . If is a constant, then all results about , , or are trivial. In this paper, we assume that is a nonconstant. Both operators have been studied by many authors. See [623] and the references therein.

Norms of some special operators, such as composition operator, weighted composition operator, and some integral operators, have been studied by many authors. The interested readers can refer [13, 2432], for example. Recently, Liu and Xiong studied the norm of integral operators and on the Bloch space, Dirichlet space, BMOA space, and so on in [13]. In this paper, we study the norm of integral operator on some function spaces in the unit disk.

2. Main Results

In this section, we state and prove our main results. In order to formulate our main results, we need some auxiliary results which are incorporated in the following lemmas.

Lemma 1 (see [5, page 144]). If   , then , , and the inequality is sharp for each fixed .

Lemma 2. Let and . For any , the following one has: where is any point in .

Proof. For any , taking and the subharmonicity of , we get and so For any , let . Replacing by and applying the change of variable formula give the following: The proof is complete.

Theorem 3. Let . The integral operator is bounded on if and only if . Moreover, one has

Proof. If , by (4), we have Thus .
On the other hand, denote . Given any , there exists such that . Let Then we have . In fact, taking and and using Poisson integral, we get Taking , we obtain So and . Thus Theorem 2.6 in [5] yields By Theorem 2.12 in [5], we have so the arbitrariness of gives and the proof is complete.

Lemma 3 in [13] gives the norm of on Dirichlet space. Here, we consider the norm of on -Dirichlet space .

Theorem 4. Let and . Then is bounded on if and only if . Moreover, one has

Proof. First, we assume that . Let . Then (6) gives and so we have .
Now we need only to show the reverse inequality. Denote . Given any , there exists such that . Let where is any path in from to . By Theorem 13.11 in [33, page 274], we know is an analytic function in and . Also it is easy to check that . Indeed, by using the method of the proof of Lemma 4.2.2 in [4], we have Let , and so . Thus by Lemma 2 we have Since is arbitrary, we get which implies the desired result.

Theorem 5. Let and let . The integral operator is bounded on if and only if . Moreover, one has

Proof. If , then by (7), we have and so .
Now we need only to show the reverse inequality. Denote . Given any , there exists such that . Let We see that . Indeed, Let . Then . Thus by Lemma 2, we have Since is arbitrary, we get . The proof is complete.

Theorem 6. Let and let . The integral operator is bounded from to if and only if . Moreover, one has

Proof. If , then by (3), we have Hence .
For the converse, denote . Given any , there exists such that . Set where is any path in from to . By Theorem 13.11 in [33, page 274], we know that is an analytic function in and , and it is easy to check that . Thus Since is arbitrary, we obtain the desired result. The proof is complete.

Theorem 7. Let . The integral operator is bounded from to if and only if . Moreover, one has

Proof. If , then by Lemma 1, we have and hence .
For the converse, denote . Given any , there exists such that . Let Then by the proof of Theorem 3, we see that . In the meantime, we know that , which gives Since is arbitrary, we get the desired result. The proof is complete.

Finally, we consider the norm of from to some Banach spaces.

Theorem 8. If , then the following assertions hold.(1)Let . The integral operator is bounded from space to space if and only if satisfies Moreover, one has (2)Let and let . The integral operator is bounded from space to space if and only if satisfies Moreover, one has (3)Let . The integral operator is bounded from space to space if and only if satisfies . Moreover, one has (4)Let . The integral operator is bounded from space to space if and only if satisfies . Moreover, one has

Proof. The assertion (1) will be proved only here, and the conclusions of (2), (3), and (4) follow by using the similar arguments to that used in proving (1), and so the proofs are omitted.
If , then by (1), we have and so
For the converse, let . It is easy to see that . Thus The desired result follows by (47) and (48). The proof is complete.

Acknowledgments

Hao Li is supported by the National Natural Science Foundation of China (no. 11126284). Songxiao Li is supported by the project of Department of Education of Guangdong Province (no. 2012KJCX0096).

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