Abstract

We prove that every bounded linear operator on weighted Bergman space over the polydisk can be approximated by Toeplitz operators under some conditions. The main tool here is the so-called -Berezin transform. In particular, our results generalized the results of K. Nam and D. C. Zheng to the case of operators acting on .

1. Introduction

Let be the unit disk in and be a positive standard weighted probability measure on , where the weighted parameter fulfills and the normalized constant . For a fixed positive integer , the polydisk is the Cartesian product of copies of andis the normalized weighted Lebesgue volume measure on the polydisk . The Bergman space is the set of all analytic functions on in . As is well known forms a closed subspace of and has the structure of reproducing kernel Hilbert space. We denote by the Bergman projection of onto . In case of , is the unweighted Bergman space denoted by . Given an essentially bounded measurable function , we write for the Toeplitz operator with the symbol , which acts on as . That is, the Toeplitz operator is defined as the compression of a multiplication operator on onto the Bergman space. The Toeplitz algebra is the closed subalgebra of generated by , where denotes the algebra of all bounded linear operators on .

Due to their simple structure Toeplitz operators form an important, tractable, and intensively studied subclass in the algebra of all bounded linear operators on . The natural question is whether the Toeplitz algebra is dense in the algebra of all bounded linear operators on the Bergman space. On unweighted Bergman space over the unit disk and even more general domain in , it is proved in [1] that the Toeplitz algebra is dense in the algebra of all bounded linear operators in the sense of strong operator topology (SOT). In general, it is not true if the SOT is replaced by the norm topology.

Nam and Zheng give a criterion for bounded operators approximated by Toeplitz operators on . Since the Berezin transform is a useful tool to study operators on any reproducing kernel Hilbert space, the -Berezin transform for any bounded linear operators acting on was defined in [2]. The operator can be approximated in the norm by Toeplitz operators on the unit ball (see [3]) by using the -Berezin transform. In [4], the ()-Berezin transform for complex-valued regular measures on the weighted -Bergman space over the unit ball was defined and studied. Using it, they show that every can be approximated by certain localized operators and introduce a way to connect the behavior of these localized operators with the Berezin transform. The ()-Berezin transform for general bounded operators acting on the weighted Bergman space was defined in [5] and the authors establish various results on norm approximations via the ()-Berezin transform and describe conditions under which a bounded linear operator can be approximated in norm by Toeplitz operators whose symbols are bounded functions.

In this paper, we will define and study the ()-Berezin transform for general bounded operators acting on the weighted Bergman space in the third section. The ()-Berezin transform of a Toeplitz operator acting on coincides with ()-Berezin transform for considered on the weighted Bergman space . We will show that the ()-Berezin transforms are commuting with each other. In Section 4, we will establish various results on norm approximation by the ()-Berezin transform. More precisely, we describe how to approximate a bounded linear operator on in norm by Toeplitz operators whose symbols are bounded functions which are given as the ()-Berezin transform of the initial operator under some conditions. We would like to point out that these results generalize ideas and theorems in [2] to the case of operators acting on .

2. Preliminaries

Let be the polydisk in equipped with the standard weighted measure (1), where is fixed and for any . For a vector and a positive integer we will employ the notations In addition, if is a positive integer for any and , and are multi-index. Let be the set of nonnegative integers. With , we use the standard notations , and .

As we all know, for all and , where , for , we have and then is the standard orthonormal basis of . The reproducing kernel in is given by for , , and the normalized reproducing kernel . For , let , where , for ; then is an automorphism on that interchanges and . Let ; then Given , introduce the unitary operator on given by , where . It is easy to see that is self-adjoint and so . We have .

For a fixed we define an automorphism on the algebra of all bounded operator on by . In particular, if is a Toeplitz operator, then .

The principle difference between the unit ball and the polydisk is that the later domain is reducible, which involves the tensor product structure of various objects introduced and studied in the paper. In particular, and . Therefore, for the orthonormal basis of and the reproducing kernel in , we have and .

The unitary operator on can be written by . In fact, · . If can be written by , then .

Let denote the class of trace operators on . Given , we write for its trace and recall that the trace norm of is given by . Given , , the rank-one-operator acting on by the formula obviously belongs to . It is easily proved that is in and with norm equal to and . Recall as well that if has rank , then one has the inequality . The pseudo-hyperbolic metric on is defined as .

Throughout the paper and as a convention we will denote by a positive constant depending only on and and appearing in various estimates and whose value may change at each occurrence.

3. The -Berezin Transform

Recall that -Berezin transform for unweighted Bergman space over the unit disk and over the unit polydisk was defined in [6] and [2], respectively. We will follow the recipe in [2] and first introduce some notation. Put so that, for , , we knowFor , , and a positive integer , let .

A generalization of the concept of -Berezin transform to an arbitrary bounded operator on the Bergman space requires a modification of the definition in [2].

Definition 1. We define the -Berezin transform of by It is easy to see that the following pointwise estimate where the constant is independent of ; that is, is a bounded function on with .
In [5], the -Berezin transform of is defined by , for the case of the unit disk . From the point of view of the tensor product structure, given an elementary tensor , its ()-Berezin transform for obviously and naturally has to be defined asUnfortunately, the set of those tensor product operators is not a linear space; that is, for any operator , cannot be written in the form of the tensor product operators. Therefore, we define for any operator with (10), and this coincides with Definition 1. If can be the tensor product form, this definition is the tensor product of ()-Berezin transform for the case of .
As usual we define the ()-Berezin transform of a function byIt is easy to see that . Thus, ()-Berezin transform of a Toeplitz operator acting on coincides with ()-Berezin transform for now considered on the weighted Bergman space .
From the definition of , we have the identity , for , and . The following proposition gives an integral representation of the ()-Berezin transform.

Proposition 2. Let , , and . Then

Proof. For and , we have = and . Using those by (5) and (7), we have

Proposition 3. Let , , and . Then

Proof. We have

For , , put , for . In ([2], P98), the map is a unitary map of and maps to . Let and , for .

Lemma 4. For , , where and , for .

Proof. Since = , then . It is sufficient to show that .
For , and , . Thus for ,

Note that is a complex number of modulus one.

Theorem 5. Let , , and ; then

Proof. By definition, For any , by Proposition 2, Lemma 4, and , we have Then we have .

Lemma 6. Let , and , . If for any , then

Proof. By Theorem 5, we only prove that . Using Proposition 2, Fubini’s theorem, and (11), we have where . Then = , where and are holomorphic functions and for some . Then, it is sufficient to show that , for .

Lemma 7. For any , there is a sequence satisfying for any and , such that converges to pointwise.

Proof. Let denote the algebra of bounded holomorphic functions on . Both the density of in and the density of finite rank operator in the ideal of compact operators on imply that the set is dense in the ideal in the norm topology. Note that is dense in the space of bounded operators on with respect to the strong operator topology. Thus, for each , there is a sequence of finite rank operators converging strongly to . Then Proposition 2 shows that converges to pointwise. To finish the proof we estimate

Proposition 8. Let , and , ; then .

Proof. Let ; then Lemma 7 implies that there is a sequence satisfying ; hence by Lemma 6From (11), we know that and . Furthermore converges to . As a consequence the functions and converge to and , respectively. By the uniqueness of the limit, we have .

Theorem 9. Let ; then there is a constant , such that

Proof. Let ; we have where ; we obtain .

Corollary 10. Let , and ; then

Proof. Let ; choose with whenever , with . If , , by (11), we have Denote by the first integral, and In the first inequality we use that is invariant under the automorphisms and by the Lipschitz continuity of .
Now we estimate the second integral above. It is clear that the right-hand side converges to zero as .

4. Toeplitz Operators to Approximate the Bounded Operators

In this section we will give a criterion for an operator approximated by Toeplitz operators with symbol equal to their ()-Berezin transforms. From Proposition  1.4.10 in [7] there exists Lemma  3.1 in [2].

Lemma 11 (see [2]). Suppose and . Then

Let and let be the conjugate exponent of . Note that the inequalityis equivalent to .

Lemma 11 gives the following lemma.

Lemma 12. Let and and put with , . Then there exists such that for all , andfor all .

Proof. Given , the equality = implies that Thus, let , and apply the ’s inequality According to (34) we have and , for any . Hence inequality (35) follows from Lemma 11.
The second inequality (36) follows from (35) after replacing by , interchanging and , and making use ofwhich holds for all , .

Lemma 13. Let and ; then where is the constant of Lemma 12.

Proof. For , . Thus is the integral operator with kernel function . By classical Schur’s Theorem [8], it is sufficient to prove that there exist positive constants and and a positive measurable function on such that for almost every and for almost every . By Lemma 12, let ; we get the conclusion.

Lemma 14. Let be a bounded sequence in such that as . Then as for any , and uniformly on compact subsets of as .

Proof. To prove the first statement it is sufficient to check that for each multi-index as . Since = , by Proposition 3 and Theorem 5 we have Given a multi-index and , we havePassing to the polar coordinates, the integral part is . Define in the standard way ([9], Formula 8.392), ; then we have In addition, (43) equals Thus we have where is a constant independent of and . In the last estimate we used the boundedness of the sequence and the inequality which easily follows from (3). The first summand above tends to zero as . It is sufficient to estimate the second summand . Estimating the multiple for any in both integrals , for , and , for . By ([9], Formula 8.328.2), and thus there exists such that, for all , . Then The power series in in the last line has radius of convergence equal to 1 and vanishes at 0. Thus the value of becomes small if one takes sufficiently closed to 0.
In order to prove the second statement we use the series representations (3) and (4) again, By the first statement of the lemma the expression uniformly tends to zero as with being already fixed. To estimate we use the Cauchy-Schwarz inequality, In [5], we get ; we finally have . By choosing sufficiently large we can make as small as needed. This ends the proof.

Lemma 15. Let be a bounded sequence in such that as . Assume that, for some , the following inequalities hold: where is independent of . Then as in the -norm.

Proof. By Lemma 13, Then, for , ’s inequality gives Then the first term is less than or equal to which converges to 0 as goes to 1 and the second term tends to 0 as by Lemma 14. Finally, Lemma 13 gives as , proving the statement of the lemma.

Corollary 16. Let such that, for some , the following inequalities holdwhere is independent of . Then as in the -norm.

Proof. Let and by Proposition 8, we have By Corollary 10, the right of equation uniformly tends to 0 as ; that is, . An application of Lemma 15 finishes the proof.

Theorem 17. Let . If there is , such thatwhere is independent of , then as in the -norm.

Proof. Since and , by Corollary 16, we only need to prove that (59) implies (53). Hence, it is sufficient to prove thatBy Lemma 13, we have where is independent of . Let ; then , where is independent of . According to the proof of Corollary 16, we get . Let be an analytic polynomial with ; Lemma 14 implies as . Then, for any and any , there is a sufficiently large such that where is independent of and . Since is arbitrary, we have inequality (60).

Competing Interests

The authors declare that they have no competing interests.

Acknowledgments

This research is supported by NSFC, Item no. 11271059.