Abstract

A bounded Hilbert space operator is hyponormal if is a positive operator. We consider the hyponormality of Toeplitz operators on a weighted Bergman space. We find a necessary condition for hyponormality in the case of a symbol of the form where and are bounded analytic functions on the unit disk. We then find sufficient conditions when is a monomial.

1. Introduction

Let denote the disk of radius in the complex plane, where is the Lebesgue measure on and . denotes the Hilbert space of complex-valued functions on that are square integrable with respect to . We write . When is analytic on , we have

Denote by the space of analytic functions on such that . It is known that is a Hilbert space [1], and an orthonormal basis is given by . The Toeplitz operator with symbol on is defined by where is bounded and measurable on , is in , and is the orthogonal projection of onto . Hankel operators are defined by , and as before. Recall that a bounded operator on a Hilbert space is hyponormal if is a positive operator. Hyponormality on the Hardy space is studied by Cowen in [2, 3]. Unweighted Bergman spaces on the unit disk and Toeplitz operators on these spaces were considered in [4, 5] and [6]. Hyponormality was first considered by Sadraoui in [7]. An improvement of the necessary condition is due to Ahern and Cuckovic [8]. A new necessary condition in a special case is found by Cuckovic and Curto in [9]. Many results on hyponormality on weighted Bergman spaces treat special cases. We cite for example [1, 10]. In this work for simplicity’s sake, we consider the case and . Under a smoothness assumption, we give a fairly general necessary condition for the hyponormality of Toeplitz operators with a symbol of the form where and are bounded and analytic. We also give sufficient conditions when is a monomial and a polynomial without assuming .

We begin by recalling some general properties of Toeplitz operators.

2. Some General Properties

We assume , are in . Then, we have: (1)(2)(3) if or analytic on

The use of these properties leads to describing hyponormality in more than one form. These are easy to prove by using Douglas lemma [11].

Proposition 1. Let be bounded and analytic on . Then, the following are equivalent: [label=)] (i) is hyponormal(ii)(iii) for any in (iv) for any in (v) where is of norm less than or equal to one

We also need the following lemmas.

Lemma 2. For and integers, we have

Proof. We have

We see that if . For , if and only if . In this case, we get

On the other hand, if , then so . A computation shows so , and the result follows.

Lemma 3. If is bounded analytic on the disk , then the th entry of the matrix of with respect to the orthonormal basis is given by:

Proof. We have:

Thus, by Lemma 2,

Thus, noting that if , we get

We also have by Lemma 2

Then,

The above inner product is nonzero if . We get by Lemma 2

When , we set . We obtain in this case the following corollary.

Corollary 4. If is bounded and analytic on the unit disk , and if . Then, the matrix of with respect to the orthonormal basis

is given by

Recall that the Hardy space is defined by

In the following theorem, the Toeplitz operator on is not necessarily bounded. However, its matrix in the usual orthonormal basis of is defined.

The following theorem is an extension of Lemma 2.4.2 [7] to the weighted Bergman space.

Theorem 5. Assume bounded analytic on the unit disk and . Let be the matrix of the Toeplitz operator on . Then, as .

Proof. The symmetric matrix of with respect to the orthonormal basis is given by (set in Corollary 4 with .)

To simplify, set . Then, for any integer . Set in the first sum, and in the second. With some obvious notations, we get

A computation shows that

It is not difficult to see that and that the following inequality holds: for . We have that since . Writing the sum as an integral, where the measure is the counting measure, and using the dominated convergence theorem, we get that

Clearly, . So we have for , for . Thus,