Abstract

We characterize the boundedness and compactness of a Toeplitz-type operator on weighted Bergman spaces satisfying the Bekollé-Bonami condition in terms of the Berezin transform.

1. Introduction

Let denote the unit disc in the complex plane, and let denote the normalized Lebesgue measure on . For a given nonnegative integrable function on , we define the weighted Bergman space as the space of all analytic functions on that belong to the weighted space . That is, if is analytic on and satisfies

Weighted Bergman spaces have been studied by several authors in different contexts (see, e.g., [1–5]). Most of the researches about Toeplitz-type operators in this spaces have been done by considering radial weights. In this paper we will consider weights satisfying the the socalled Békollé-Bonami condition.

Definition 1. A function in is said to satisfy the Bekolle-Bonami condition if there exists a constant such that for every interval . Here, denotes the Carleson square:
Condition was introduced by Bekolle and Bonami in [6]. They showed that is necessary and sufficient for the Bergman projection to be bounded on . Condition was used by Luecking in [7, 8] to study Carleson measures in weighted Bergman spaces. Then in [9], Nakazi and Yamada generalized Luecking's results by introducing a more restrictive condition and found, under this condition, a characterization for Carleson measures on spaces. They also gave several examples of weights satisfying this condition. In [10], Constantin generalized Luecking's result and studied Toeplitz-type operators on spaces . The characterization of Carleson measures is as follows.

Theorem 2 (see [7, 10]). Suppose that a.e. in and satisfies the condition; let and let be a positive Borel measure on . Then, the following are equivalent.(1)There exist a constant such that for every polynomial , that is, is a -Carleson measure.(2)There exists and satisfying for all .

Theorem 3 (see [8]). Suppose that satisfies condition ; then, the dual of can be identified with . The pairing is given by

Remark 4. Notice that if satisfies the condition, then it is not hard to show (see, e.g., [8]) that there is a constant such that for any function analytic in and any ; the following holds: Consequently, evaluation functionals are bounded on , and so for every complex number , there exists a function such that for every function ; that is, the functions are the reproducing kernels for . Notice also that the set is dense in .
Moreover, using Theorem 3, we have that there exists a bounded, bijective, linear operator such that for every and . On the other hand, since is a Hilbert space, there exists a bounded, bijective, linear operator such that for every and .
Let denote the evaluation functional at ; then, . On the other hand, if we let , with , denote the Bergman kernel then , and the pairing is given by the product; we have that .
Define by . Then is a bounded operator and . Also, the operator is well defined, and consequently we have that

Definition 5. Let be a positive Borel measure on . Define for each polynomial the Toeplitz operator
We will study the conditions under which Toeplitz operators can be extended to , and we will characterize the boundedness and compactness of Toeplitz operators acting on the spaces in terms of the socalled Berezin transform of the measure .

Definition 6. Let be a positive Borel measure on ; the Berezin transform of is defined as
From now on, we will assume that is a probability measure.
The following result is well known (see [1–5]).

Theorem 7. Suppose that is a finite positive Borel measure on . Then, the following are equivalent here, represents the standard weighted Lebesgue measure , . (a) is bounded on . (b) is a bounded function on . (c) is a Carleson measure for ,
where is the Berezin transform of the Toeplitz operator, here

The proof of this result mainly relies on the behavior of the reproducing kernels when is close to and on the characterization of Carleson measures for the Bergman space. In the case of spaces, such a characterization is known for Carleson measures (see, e.g., [10]), but we do not have an explicit formula for the reproducing kernels; consequently, we will need to use a different technique. This will be an atomic decomposition developed in terms of the reproducing kernels of .

2. Atomic Decomposition for

In this section, we develop a way of expressing functions in as a linear combination of reproducing kernels. Similar problems have been studied in [8, 11]. We will use a similar reasoning as in [8], where the problem is studied in terms of the Bergman kernel. We will assume in the rest of this paper that is a probability measure.

Definition 8. Let . A sequence is called -separated if . Here, denotes the pseudohyperbolic metric: .

Theorem 9 (see [8]). If satisfies condition , then there exists an -separated sequence such that for every function .

Theorem 10. Suppose that satisfies the condition . Then, there exists a sequence which is -separated for some , such that any has the form for some sequence .

Proof. By Theorem 9, we know that there exists an -separated sequence such that . Consider the linear operator defined by Then, , and so is a bounded injective linear operator having closed range. Consequently, the dual mapping is onto. Now, let ; then, for , we have Here, we justify the interchange of integration and summation as follows. Given a sequence and , there exists such that if , then Thus, where
Thus, forms a Cauchy sequence, and consequently it converges in the -norm to Therefore,
Hence, , and since is surjective, then the result holds.

3. Boundedness of Toeplitz Operators

Theorem 11. Suppose that satisfies the condition and that the Toeplitz operator is bounded. Then, the Berezin transform is bounded.

Proof. First, notice that for every finite sum , ; the following holds: So, if is a Cauchy sequence in , it is also a Cauchy sequence in , and so it converges in . Therefore, by Theorem 10, we have that if in the -norm, then in , and consequently for any function , . Moreover, Therefore, Hence, for any , and consequently, .

We will need the following lemma due to Constantin.

Lemma 12 (see [10]). Suppose that satisfies condition ; then,

Proposition 13. If is bounded on , then is a -Carleson measure, where .

Proof. Fix ; if is bounded, then there exists a constant such that for every and consequently and by Theorem 2, we obtain the result.

Proposition 14. Suppose that is a -Carleson measure; then, the Toeplitz operator is bounded.

Proof. For every function, define the linear operator by: . Note that if is a -Carleson measure, then so , , and there exists such that .
We have just proved that for every , there exists such that In particular, taking , we have that for every , Hence, . Moreover, since , then and consequently is bounded.

Corollary 15. For bounded, the following equality holds for every and in :

4. Compactness of Toeplitz Operators

In this section, we will characterize compactness of Toeplitz operators in terms of its Berezin transform. We will use the characterization for vanishing Carleson measures given by Constantin.

Definition 16. A positive Borel measure on is a -vanishing Carleson measure if the inclusion operator is compact.

Theorem 17 (see [10, Theorem 3.3]). Suppose that satisfies the condition. Then is a -vanishing Carleson measure if and only if for every ,

Theorem 18. Suppose that is a positive Borel measure on . Then the following are equivalent:(a) is compact. (b) as . (c) is a -vanishing Carleson.

Proof. (a) (b)
Notice that by Jensen's inequality, we have that and consequently using Lemma 12, we get as . Thus, the family converges to zero weakly in as . We also know from the proof of Theorem 11 that . Consequently, which, since is compact, converges to zero as . (b) (c)
We have that , and by Lemma 12, we obtain that then (c) follows from (b). (c) (a)
By Corollary 15 and Cauchy-Schwarz inequality we have that: and since is a -Carleson measure, then for every , so
Now, if is a sequence in that converges to zero weakly, then by the compactness of we have that converges to zero. Therefore, converges to zero, and so is compact on .

We have used the fact that the family of normalized Bergman kernels converges to zero weakly as . This is also true for the family of normalized reproducing kernels of .

Proposition 19. If satisfies condition , then converges to zero weakly in .

Proof. We use the same notation as in the observation after Theorem 3. The operator maps to . Also, by Lemma 12, we have that which converges to zero uniformly on compact subsets of . Consequently, the family is uniformly bounded on and converges to zero uniformly on compact subsets of . Since is a functional space, then converges to zero weakly on .
Now, the operator is bounded, so converges to zero weakly in .

Acknowledgment

This work was supported by the Pontificia Universidad Javeriana proyect ID PPTA 00004896 and ID PRY 004709.