`International Journal of Mathematics and Mathematical SciencesVolumeΒ 2012, Article IDΒ 970729, 7 pageshttp://dx.doi.org/10.1155/2012/970729`
Research Article

## Fredholm Weighted Composition Operators on Dirichlet Space

School of Mathematics and Computer Science, Shanxi Normal University, Linfen 041004, China

Received 3 June 2012; Accepted 3 August 2012

Copyright Β© 2012 Liankuo Zhao. 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

This paper characterizes bounded Fredholm, bounded invertible, and unitary weighted composition operators on Dirichlet space.

#### 1. Introduction

Let be a Hilbert space of analytic functions on the unit disk . For an analytic function on , we can define the multiplication operator . For an analytic self-mapping of , the composition operator defined on as . These operators are two classes of important operators in the study of operator theory in function spaces [1β3]. Furthermore, for and , we define the weighted composition operator on as

Recently, the boundedness, compactness, norm, and essential norm of weighted composition operators on various spaces of analytic functions have been studied intensively, see [4β9] and so on. In this paper, we characterize bounded Fredholm weighted composition operators on Dirichlet space of the unit disk.

Recall the Dirichlet space that consists of analytic function on with finite Dirichlet integral: where is the normalized Lebesgue area measure on . It is well known that is the only mΓΆbius invariant Hilbert space up to an isomorphism [10]. Endow with norm is a Hilbert space with inner product Furthermore is a reproducing function space with reproducing kernel

Denote . is called the multiplier space of . By the closed graph theorem, the multiplication operator defined by is bounded on . For the characterization of the element in , see [11].

For analytic function on and analytic self-mapping of , the weighted composition operator on is not necessarily bounded. Even the composition operator is not necessarily bounded on , which is different from the cases in Hardy space and Bergman space. See [12] for more information about the properties of composition operators acting on the Dirichlet space.

The main result of the paper reads as the following.

Theorem 1.1. Let and be analytic functions on with . Then is a bounded Fredholm operator on if and only if , bounded away from zero near the unit circle, and is an automorphism of .

If , then the result above gives the characterization of bounded Fredholm composition operator on , which was obtained in [12].

As corollaries, in the end of this paper one gives the characterization of bounded invertible and unitary weighted composition operator on , respectively. Some idea of this paper is derived from [4, 13], which characterize normal and bounded invertible weighted composition operator on the Hardy space, respectively.

#### 2. Proof of the Main Result

In the following, and denote analytic functions on with . It is easy to verify that if is defined on .

Proposition 2.1. Let be a bounded Fredholm operator on . Then has at most finite zeroes in and is an inner function.

Proof. If is a bounded Fredholm operator, then there exist a bounded operator and a compact operator on such that where is the identity operator.
Since we have where is the normalization of reproducing kernel function .
Since is compact and weakly converges to as , as . It follows that there exists constant , , such that for all with . Inequality (2.3) shows that which implies that has no zeroes in , and, hence, has at most finite zeroes in .
Since weakly converges to 0 as , as , that is, It follows from (2.4) that and hence as , that is, is an inner function.

For the proof of the following lemma, we cite Carleson's formula for the Dirichlet integral [14].

Let , be the canonical factorization of as a function in the Hardy space, where , is a Blaschke product, is the singular part of and is the outer part of . Then where is the unit circle, , is the Poisson kernel, and is the singular measure corresponding to .

Lemma 2.2. Let be a bounded operator on , with a finite Blaschke product. Then is bounded.

Proof. Let be the multiplication operator on . Then . Since is a finite Blaschke product, by the Carleson's formula, we have
Since , by the inequality above it is easy to verify that is bounded on if is bounded.

Lemma 2.3. Let be an analytic function on with zero-free. If is a bounded Fredholm operator on , then is univalent.

Proof. If for with , by a similar reasoning as [1, Lemma 3.26], there exist infinite sets and in which is disjoint such that . Hence, which contradicts to that kernel of is finite dimensional.

Corollary 2.4. If is a bounded Fredholm operator on , then is an automorphism of and .

Proof. By Proposition 2.1, has the factorization of with a finite Blaschke product and zero free in . By Lemma 2.2, is a bounded operator on . Since and is a Fredholm operator, is a Fredholm operator also. By Proposition 2.1 and Lemma 2.3, is an univalent inner function, it follows from [1, Corollary 3.28] that is an automorphism of .
Since , is a bounded multiplication operator on , which implies that .

The following lemmas is well-known. It is easy to verify by the fact also.

Lemma 2.5. Let . Then is an invertible operator on if and only if is invertible in .

Lemma 2.6. Let . Then is a Fredholm operator on if and only if is bounded away from the unit circle.

Now we give the proof of Theorem 1.1.

Proof of Theorem 1.1. If is a bounded Fredholm operator on , by Corollary 2.4, and is an automorphism of . Since is invertible, is a Fredholm operator. So is bounded away form the unit circle follows from Lemma 2.6.
On the other hand, if and bounded away from the unit circle, then is a bounded Fredholm operator on . If is an automorphism of , then is invertible. Hence is a bounded Fredholm operator on .

As corollaries, in the following, we characterize bounded invertible and unitary weighted composition operators on .

Corollary 2.7. Let and be analytic functions on with . Then is a bounded invertible operator on if and only if , invertible in , and is an automorphism of .

Proof. Since a bounded invertible operator is a bounded Fredholm operator, the proof is similar to the proof of Theorem 1.1.

Corollary 2.8. Let and be analytic functions on with . is a bounded operator on . Then is a unitary operator if and only if is a constant with and is a rotation of .

Proof. If is a unitary operator, then it must be an invertible operator. By Corollary 2.7, is an automorphism of and is invertible in .
Let be nonnegative integer, . A unitary is also an isometry, so we have
Let such that . Since is an automorphism of , is a finite Blaschke product with zero of order . By Carlesonβs formula for Dirichlet integral, we have Hence, That is, Let , then .
By (2.12), we have and . By (2.9), we obtain is a constant with , which implies that , that is, is a rotation of .
The sufficiency is easy to verify.

Remark 2.9. The key step in the proof of the main result is to analyze zeros of the symbol and univalency of . The following result pointed out by the referee gives a simple characterization of the symbols and for the bounded Fredholm operator on .

Proposition 2.10. Let and be analytic functions on with . is a bounded Fredholm operator on . Then has only finitely many zeros in and is univalent.

Proof. If for , then , which implies that is in the kernel of . Thus if had infinitely many zeros, the kernel of would be infinite dimensional and hence this operator would not be Fredholm.
If for with , by a similar reasoning as [1, Lemma 3.26], there exist infinite sets and in which is disjoint such that . Since has only finitely many zeros in , we can choose infinitely many and such that , . Hence, Since is a Fredholm operator, must be univalent.

#### Acknowledgments

Thanks are for referee for many helpful suggestions which promote the author to think the related issues deeply. This work is supported by YSF of Shanxi (2010021002-2) and NSFC (11201274).

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