Abstract and Applied AnalysisVolume 2011 (2011), Article ID 962313, 18 pagesdoi:10.1155/2011/962313

Research Article

Algebraic Properties of Toeplitz Operators on the Polydisk

Bo Zhang,¹ Yanyue Shi,² and Yufeng Lu¹

¹School of Mathematical Sciences, Dalian University of Technology, Dalian 116024, China²School of Mathematical Sciences, Ocean University of China, Qingdao 266100, China

Received 3 August 2011; Revised 29 September 2011; Accepted 2 October 2011

Academic Editor: Alberto d'Onofrio

Copyright © 2011 Bo Zhang et al. 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 discuss some algebraic properties of Toeplitz operators on the Bergman space of the polydisk . Firstly, we introduce Toeplitz operators with quasihomogeneous symbols and property (P). Secondly, we study commutativity of certain quasihomogeneous Toeplitz operators and commutators of diagonal Toeplitz operators. Thirdly, we discuss finite rank semicommutators and commutators of Toeplitz operators with quasihomogeneous symbols. Finally, we solve the finite rank product problem for Toeplitz operators on the polydisk.

1. Introduction

Let be the open unit disk in the complex plane and its boundary the unit circle . For a fixed positive integer , the unit polydisk and the torus are the subsets of which are Cartesian products of copies and , respectively. Let denote the Lebesgue volume measure on the polydisk , normalized so that the measure of equals 1. Let denote the usual Lebesgue space. The Bergman space is the Hilbert space consisting of holomorphic functions on that are also in . Since every point evaluation is a bounded linear functional on , there corresponds to every a unique function which has the following reproducing property: where the notation denotes the inner product in . The function is the well-known Bergman kernel and its explicit formula is given by Here and elsewhere denotes the th component of . The Bergman projection is defined for the Hilbert space orthogonal projection from onto . Given a function , the Toeplitz operator is defined by the formula for all . Since the Bergman projection has norm 1, it is clear that Toeplitz operators defined in this way are bounded linear operators on and .

We now consider a more general class of Toeplitz operators. For , in analogy to (1.3) we define an operator by

Since the Bergman projection can be extended to , the operator is well defined on , where is the space of bounded holomorphic functions on . Hence, is always densely defined on . Since is not bounded on , it is well known that can be unbounded in general. This motivates the following definition, which is based on the definitions on unit ball in [1].

Definition 1.1. Let .(a) is called a -function if (1.4) defines a bounded operator on .(b)If is a -function, one writes for the continuous extension of the operator (it is defined on the dense subset of ) defined by (1.4). is called a Toeplitz operator on .(c)If there exist , , such that is (essentially) bounded on , then one says is “nearly bounded.”

Notice that the -functions form a proper subset of which contains all bounded and “nearly bounded” functions. In this paper, the functions which we considered are all -functions without special introduction. We denote the semicommutator and commutator of two Toeplitz operators and by

The commuting problem and the finite-rank product problem for Toeplitz operators on the Hardy and Bergman spaces over various domains are some of the most interesting problems in operator theory.

For commuting problem, in 1963, Brown and Halmos [2] showed that two bounded Toeplitz operators and on the classical Hardy space commute if and only if (i) both and are analytic, (ii) both and are analytic, or (iii) one is a linear function of the other. On the Bergman space of the unit disk, some similar results were obtained for Toeplitz operators with bounded harmonic symbols or analytic symbols (see [2–4]). The problem of characterizing commuting Toeplitz operators with arbitrary bounded symbols seems quite challenging and is not fully understood until now. In recent years, by Mellin transform, some results with quasihomogeneous symbols (it is of the form , where is a radial function) or monomial symbols were obtained (see [5–7]). On the Hardy and Bergman spaces of several complex variables, the situation is much more complicated. On the unit ball, Toeplitz operators with pluriharmonic or quasihomogeneous symbols were studied in [1, 8–11]. On the polydisk, some results about Toeplitz operators with pluriharmonic symbols were obtained in [10, 12–14].

For finite-rank product problem, Luecking recently proved that a Toeplitz operator with measure symbol on the Bergman space of unit disk has finite rank if and only if its symbols are a linear combination of point masses (see [15]). In [16], Choe extended Luecking’s theorem to higher-dimensional cases. Using those results, Le studied finite-rank products of Toeplitz operators on the Bergman space of the unit disk and unit ball in [17, 18].

Motivated by recent work in [1, 5, 7, 17, 18], we define quasihomogeneous functions on the polydisk and study Toeplitz operators with quasihomogeneous symbols on the Bergman space of the polydisk. The present paper is assembled as follows. In Section 2, we introduce Mellin transform, Toeplitz operators with quasihomogeneous symbols and property (P). In Section 3, we study commutativity of certain quasihomogeneous Toeplitz operators and commutators of diagonal Toeplitz operators. In Sections 4 and 5, we prove that finite rank semicommutators and commutators of Toeplitz operators with quasihomogeneous symbols must be zero operator and we also solve the finite-rank product problem for Toeplitz operators on the Bergman space of the polydisk.

2. Mellin Transform, Toeplitz Operators with Quasihomogeneous Symbols and Property (P)

For any multi-index (here denotes the set of all nonnegative integers), we write and for . The standard orthonormal basis for is , where

For two -tuples of integers and , we define if for all . Similarly, we write if for all and if otherwise. We also define if and .

For any , particularly we write and put , and . Then, , , and .

Recall that a function on is radial if and only if depends only on , that is, for any . For any function , we define the radicalization of by Then, is radial if and only if . For , we have

The main tool in this paper will be the Mellin transform. which is defined by the equation We apply the Mellin transform to functions in ; then, For convenience, we denote by when the form of is complicated. It is clear that is well defined on . Using the Hartogs theorem, for any function , the Mellin transform of is a bounded holomorphic function on .

By calculation, we can get where , , and .

The quasihomogeneous functions have been defined in many spaces (see [5, 7]). In the following, we give a similar definition on the polydisk .

Definition 2.1. Let . A function is called a quasihomogeneous function of degree if is of the form where is a radial function, that is, for any in the torus and .

As in [19], for any -tuple , let is a quasihomogeneous function of degree . It is clear that is a closed subspace of . By Lemma 3.2 in [19], . In particular, for all , if , that is, , then we conclude that , , where .

Lemma 2.2. Let , and let , be radial functions on , such that , , and are all -functions. Then, the following equation holds for every : where and .

Using Lemma 2.2, we can get the following two results: where , , , , , and . It is easy to check that where , , and .

Let be a region in complex plane and holomorphic on . If has a limit point in , such that , then . For functions of several complex variables, the above conclusion does not hold. For example, is analytic on bidisk , point sequence , , has a limit point , and , but is not a zero function on the bidisk. So we need the following definition, which is given in [9, 17].

For any , let be the map defined by the formula for all and . If is a subset of and , we define

As in [9, 17], we say that has property (P) if one of the following statements holds:(1), (2), and , or (3), and, for any , the set has property (P) as a subset of .

Let and be two sets that have property (P). It is not difficult to check that the following statements hold:(1) and have property (P);(2) do not have property (P).

Lemma 2.3. If and does not have property , then is identically zero.

Proof. By the Müntz theorem, we can prove that it is true when (see [7] for more details). Suppose that the conclusion of the lemma holds whenever , where is a positive integer. Consider the case . Since does not have property (P), there must be a , such that does not have property (P). Without loss of generality, taking , then, . For each , . So , for all . For every , let ; then, is an analytic function on and , which does not have property (P). By the induction hypothesis, we have . Thus, on . Therefore, is identically zero.

Theorem 2.4. Let , and let be a -function. Then, the following statements hold. (i)If does not have property (P), then for all .(ii)Let , be the sets that have property (P). If for all , , then for almost all .(iii)If , then is a quasihomogeneous function of degree .

Proof. (i) By direct computation, we have Let Then, . In fact, . Therefore, equality (2.12) shows that for any . That is, does not have property (P). Thus, Lemma 2.3 implies that and for all . (ii) For each , for . Since and have property (P), the subset does not have property (P). By (i), we have for . It is easy to prove that for . So for all , that is, and for almost all .(iii) Since we have If , then . Otherwise, if , then Thus, we get for any . So , this means that there exists such that , that is, is a quasihomogeneous function of degree .

Remark 2.5. Let be as in Theorem 2.4. Then, , where Recall that a densely defined operator on is said to be diagonal if it is diagonal with respect to the standard orthonormal basis. In particular, for , is diagonal if and only if . In this case, , where .

3. Commutativity of Toeplitz Operators

In this section, we study the commutativity of the Toeplitz operators with some special quasihomogeneous symbols and give the characterizations, respectively.

Theorem 3.1. Let be a quasihomogeneous function of degree and . Then, if and only if for any . Moreover, the following statements hold. (i)If , then, for each , . (ii)If , then, for each , .

Proof. Note that, for , The second equality follows that , when .Similarly, Since are the standard orthogonal basis and for , it is easy to check that the following statements are equal:(I);(II), ;(III), ;(IV). Furthermore, Thus, the statements (i) and (ii) hold.

Theorem 3.2. Let , be quasihomogeneous functions of degree and , where . If , then or .

Proof. If are quasihomogeneous functions of degree and , then there exist radial functions and , such that and . If , (2.10) implies that, for all , We claim that there exist with such that It follows that do not have property (P). So we can get or by Lemma 2.3.We only need to prove the claim. Since , there exists , such that . Without losing generality, suppose that . Let ; then, and where , and . Denote and . Note that at least one of the sets and does not have property (P). Since and are analytic on , Lemma 2.3 shows that or .Case 1. If , then where with . Let . Denote by and . Then, at least one of the sets and does not have property (P). By Lemma 2.3 again, we have or . Thus .Case 2. If , then By the same technique, we can get that (3.5) holds when .Similarly, we can find a sequence , where the functions or 0, or 0, and for . Then, (i) , (ii) , and (iii) for every , satisfies (3.5). So we complete the proof.

For , we have the following results.

Theorem 3.3. Let , where and . Let , and . Then, if and only if there exists an analytic function on , such that the function is bounded on and

Proof. As in the proof of Theorem 3.2, it is easy to check that if and only if which is equal to Suppose that there is a function as in this theorem.Note that and , for any . Then, it is easy to check that equality (3.10) holds, that is, .Conversely, if and commute, we will structure an analytic which satisfies the conditions in this theorem.Since for all , the function is analytic on . Note that for , and . Thus, Fix , and let , then, Combining this with Lemma 2.3, we can get that the above equality holds for any . Let ; then, . For each , there exist such that . So Put then, is analytic on and where are all constants. Since and , the set . Thus . That is For each , there exists such that ; then, let . By equality (3.17), we conclude that the function is well defined. Since the function is analytic on , we can prove that is an analytic function on . Let Then, So (3.17) is equal to where and . This completes the proof.

Corollary 3.4. Let be as in Theorem 3.3 and ; then, the following statements hold: (i) if and only if , where ;(ii) if and only if , where .

Proof. (i) By (3.17) we have that is, Then, It follows that there exists such that .On the other hand, if , then Thus, we have .(ii) Can also be proved in the same way.

In [6], Čučkovi and Rao showed that if and is a nonconstant radial function, then implies that is a radial function. However, this is not true if , where . For example, and only for , and it is clear that , but may not be a radial function. Let (if , this set is exactly the set of all non-constant bounded radial functions). In the following, we can give a complete description of .

Theorem 3.5. is a bounded radial function: for each .

Proof. Suppose that and is a radial function. Lemma 2.2 shows that where and It follows that if and only if for any . Let and . Then . The commutativity of and is equivalent to that at least one of and does not have property (P); then, Lemma 2.3 shows that or , where . The rest of the proof is obvious.

Remark 3.6. In Theorem 3.5, particularly if is a radial function such that or , where each is a non-constant radial function, then , so . It follows that is nonempty.

4. Finite Rank Semicommutators and Commutators

Recall that Čučković and Louhichi (see [5]) have found some nonzero finite rank semicommutators of quasihomogeneous symbol Toeplitz operators on the Bergman space of unit disk. In this section, we will show that the finite rank semicommutators and commutators of Toeplitz operators with quasihomogeneous symbols must be zero on with . Our idea is mainly from [17].

Theorem 4.1. Let with , , and let , be radial functions such that , , and are all -functions. If the semicommutator has finite rank, then it must be zero.

Proof. Let denote the semicommutator . For , if is finite rank, by equality (2.9), we have that there exists such that which is equivalent to for . Combining this with Lemma 2.3, we get Hence, In the following, we only need to prove that for all .If , there is a   such that . Without loss of generality, assume that . Then, . For each , let . Since does not have property (P), we have . Therefore, for and . So for .This completes the proof.

We now pass to the commutator of two quasihomogeneous Toeplitz operators. Here the situation is the same as for the semicommutator.

Theorem 4.2. Let with , and let , be radial functions such that and are both -functions. The commutator has finite rank if and only if it is a zero operator.

Proof. Let denote the commutator . For , if has finite rank , by equality (2.10), we have that there exists such that for . As in the proof of Theorem 4.1, the above equation implies that for . Hence, For or , following the same way as above, we can also prove that .This completes the proof.

5. Finite Rank Products of Toeplitz Operators

In [17], the author showed that under certain conditions on the bounded operators and on , if , such that is a finite-rank operator, then must be zero almost everywhere on . On the Bergman space of the polydisk, using the same method as in [17], we can prove Theorem 5.1. Using Theorem 5.1, we get two useful theorems for Toeplitz operators with quasihomogeneous symbols.

Theorem 5.1. Let , be two bounded operators on . Suppose that there is a set which has property (P), such that and . Here, (resp., ) is the linear subspace of spanned by (resp., ). Suppose that such that the operator has finite rank; then is the zero function.

Theorem 5.2. Let and be two positive integers. Let and be quasihomogeneous functions, none of which is the zero function. If such that the operator has finite rank, then is the zero function.

Proof. Let and . Suppose that , , and , , where . By Lemma 2.2, for , we have Define . Since none of the functions is the zero function, the set has property (P).For , we see that . Suppose that such that ; then, So (5.1) implies that for any , . Therefore is contained in the closure of the linear span of in . Now suppose that Then the set has property (P) and, for any , belongs to . Equality (5.1) implies that is a multiple of . So the linear span of is contained in the range of . So there exist subsets and of that have property (P) such that is contained in the closure in of Span() and Span( is a subspace of . Let ; then, Theorem 5.1 implies that is the zero function.

Theorem 5.3. Suppose that the function has the expansion and for all , where , if there is a function , such that has finite rank; then .

Proof. For , By hypothesis, there exists , such that, for any , ; then, . Thus, we have Considering the same argument, we get, for all , Now suppose that has finite rank, and let be the set that spans , where is the space of all holomorphic polynomials in the variable . Then, for any , we see that is a linear combination of , and it follows that is a finite-rank operator. By Theorem 2.4, we conclude that is the zero function.

Acknowledgments

The authors thank the referee for several suggestions that improved the paper. This research is supported by NSFC, Item number 10971020.

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