Abstract and Applied Analysis

Volume 2013 (2013), Article ID 186326, 5 pages

http://dx.doi.org/10.1155/2013/186326

## Hyponormal Toeplitz Operators on the Dirichlet Spaces

School of Mathematical Sciences, Dalian University of Technology, Dalian 116024, China

Received 28 June 2013; Revised 2 October 2013; Accepted 24 October 2013

Academic Editor: Natig M. Atakishiyev

Copyright © 2013 Puyu Cui and Yufeng Lu. 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 completely characterize the hyponormality of bounded Toeplitz operators with Sobolev symbols on the Dirichlet space and the harmonic Dirichlet space.

#### 1. Introduction

Let be the open unit disk in the complex plane and be the normalized Lebesgue area measure on . and denote the essential bounded measurable function space and the space of square integral functions on with respect to , respectively. The Bergman space consists of all analytic functions in . The Sobolev space is the space of functions with the following norm: is a Hilbert space with the inner product The Dirichlet space consists of all analytic functions in with . The Sobolev space is defined by with the norm Let be the orthogonal projection of onto . is an integral operator represented by where is the reproducing kernel of . For , the Toeplitz operator with symbol is defined by is a bounded operator for on .

Yu gave a decomposition of the Sobolev space in [1]. Let be the set of all the following polynomials: where and run over a finite subset of and . Let denote the closure of in , and let denote . Since the set of all polynomials in and is dense in , there is the following decomposition:

Since and by the above decomposition, it follows that, if , then , where , , (the space of the analytic functions on ) with .

For the space , there is the following proposition.

Proposition 1 (see [1]). *Let . Then .*

A bounded linear operator on a Hilbert space is called hyponormal if is a positive operator. There is an extensive literature on hyponormal Toeplitz operators on (the Hardy space on ) [2–4]. The corresponding problems for the Toeplitz operators on the Bergman space have been characterized in [5–9]. In the case of the Dirichlet space and the harmonic Dirichlet space, Lu and Yu proved that there are no nonconstant hyponormal Toeplitz operators with certain symbols [10]. In this paper, we completely characterize the Toeplitz operators with on Dirichlet space and harmonic Dirichlet space .

#### 2. Case on the Dirichlet Space

In this section, the hyponormality of with on will be discussed.

Theorem 2. *Let with , , and . Then is hyponormal on if and only if . *

*Proof. *By Proposition 1, we only need to prove the necessity with .

Let . Simple calculations imply that

Furthermore,

Therefore

Similarly, we have
Denote for . Since is hyponormal, we have
For , implies that
Hence
Letting , since and are convergent and is disconvergent, we get . Similarly, by choosing , we get for . Note that implies that . Thus for and the proof is finished.

The following corollary generalizes Theorems 1 and 2 in [10]. Denote where is the space of the bounded analytic functions on .

Corollary 3. *Let . Then is hyponormal on if and only if is a constant function. *

#### 3. Case on the Harmonic Dirichlet Space

In this section, we will characterize the hyponormality of with on .

The harmonic Dirichlet space consists of all harmonic functions in . It is a closed subspace of , and hence it is a Hilbert space with the following reproducing kernel:

Let be the orthogonal projection of onto . is an integral operator represented by For , the Toeplitz operator with symbol is defined by is a bounded operator for on (see [11]).

Theorem 4. *Let with , , and . Then is hyponormal on if and only if . *

*Proof. *By Proposition 1, we only need to prove the necessity with .

Let . Since is hyponormal on , we have . Note that

Thus

Similarly, we have
For , let and for . It follows that
Therefore,
For every , we have
where . Letting , Since and are convergent and ( is fixed) is disconvergent, we get for . The proof is finished.

The following corollary generalizes Theorem 3 in [10].

Corollary 5. *Suppose that with . Then is hyponormal on if and only if is a constant function.*

#### Acknowledgments

This research is supported by NSFC (no. 11271059) and Research Fund for the Doctoral Program of Higher Education of China.

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