Abstract and Applied Analysis

Volume 2013 (2013), Article ID 473916, 11 pages

http://dx.doi.org/10.1155/2013/473916

Research Article

Product and Commutativity of th-Order Slant Toeplitz Operators

Chaomei Liu¹ and Yufeng Lu²

¹School of Science, Dalian Jiaotong University, Dalian 116028, China

²School of Mathematical Sciences, Dalian University of Technology, Dalian 116024, China

Received 9 November 2012; Revised 12 March 2013; Accepted 14 March 2013

Academic Editor: Miroslaw Lachowicz

Copyright © 2013 Chaomei Liu 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.

The commutativity of th-order slant Toeplitz operators with harmonic polynomial symbols, analytic symbols, and coanalytic symbols is discussed. We show that, on the Lebesgue space and Bergman space, necessary and sufficient conditions for the commutativity of th-order slant Toeplitz operators are that their symbol functions are linearly dependent. Also, we study the product of two th-order slant Toeplitz operators and give some necessary and sufficient conditions.

1. Preliminaries

Throughout this paper, is a fixed positive integer, and . Let be a bounded measurable function on the unit circle , where is the th Fourier coefficient of and is the usual basis of , with being the set of integers. The th-order slant Toeplitz operator with symbol in is defined on as follows:

In the past several decades, slant Toeplitz operators have played outstanding roles in wavelet analysis, curve and surface modelling, and dynamical systems (e.g., see [1–10]). For instance, Villemoes [7] has associated the Besov regularity of solution of the refinement equation with the spectral radius of an associated slant Toeplitz operators and has used the spectral radius of the slant Toeplitz operators to characterize the regularity of refinable functions; Goodman et al. [6] have shown the connection between the spectral radii and conditions for the solutions of certain differential equations that in the Lipschitz classes. However, these mathematicians concentrated mainly on the applications, but these considerations serve as a source of motivation to introduce and study the properties of slant Toeplitz operators.

In 1995, Ho [11–14] began a systematic study of the slant Toeplitz operators on the Hardy space. In [15–17], the authors discussed some properties of th-order slant Toeplitz operator. In [18, 19], the authors defined the slant Toeplitz operator and th-order slant Toeplitz operator on the Bergman spaces, respectively, and studied some properties of these operators.

In this paper, properties of th-order slant Toeplitz operators with harmonic polynomial symbols, analytic symbols, and coanalytic symbols are discussed. We show that, on the Lebesgue space and Bergman space, the necessary and sufficient conditions for the commutativity of th-order slant Toeplitz operators are that their symbol functions are linearly dependent. Meanwhile, we study the product of two th-order slant Toeplitz operators and give some necessary and sufficient conditions.

2. Commutativity of th-Order Slant Toeplitz Operators on

In [17], we investigated the properties of th-order slant Toeplitz operators on and have obtained that for, and commute (essentially commute) if and only if .

Immediately we come up with the following problem.

Could the commutativity of two th-order slant Toeplitz operators be fully characterized by their symbols?

The partial answer to the pervious problem has been obtained in [17]: for and or , and commute (essentially commute) if and only if there exist scalars and , not both zero, such that .

In this section the commutativity of th-order slant Toeplitz operators with analytic symbols and harmonic symbols will be studied. First we discuss the commutativity of two th-order slant Toeplitz operators with analytic symbols.

Proposition 1. Let , then the following statements are equivalent:

(1.1); (1.2)there exist scalars   and , not both zero, such that .

Proof. Begin with the easy direction. First, suppose that (1.2) holds. Without loss of generality, let , so that . Thus, .

To prove the other direction of the proposition, suppose that (1.1) holds; that is, . Let and , then

that is,

where and are both nonnegative integers. Hence, for all nonnegative integers , , and ,

Now we would give the proof in four cases.

Case I. Suppose that . Let and we continue the proof by the induction.

When , from (4) we get that , which means that .

When , from (4) we get that

which means that , since .

Now suppose that for all integers with . Then, observe the connection between and . Let , where and are nonnegative integers with .

When , from (4) we get

From the assumption we have ; that is; .

Hence, from the above discussion we get that for all integers with by the induction, which means that . So, the required result holds.

Case II. Suppose that and . We want to show that for all integers with ; that is, . Suppose that there exists some which is not zero. Without loss of generality, let for all integers with and , where is an integer, then and .

Because , we can get that , which means that

Thus, we have , and, so, since , we have . This leads to a contradiction.

Hence, and the required result holds.

Case III. Suppose that and . Similar to Case II, we can get that . So, the required result holds.

Case IV. Suppose that and . If or , then the required result holds. Otherwise, without loss of generality, let for all integers with and , and let for all integers with and , where and are positive integers. Then, and .

If , then is equal to . The proof is similar to Case I.

If , then is equal to . The proof is similar to Case II.

If , then is equal to . The proof is similar to Case III.

From the preceding Proposition 1, it is evident that Corollary 2 holds.

Corollary 2. Let , then the following statements are equivalent:

(1.1); (1.2) there exist scalars and , not both zero, such that .

From Theorem 2.8 in [17], Proposition 1, and Corollary 2, we can obtain the following Theorem 3.

Theorem 3. Let or , , the following statements are equivalent:

(1.1) and commute; (1.2) and essentially commute; (1.3); (1.4) there exist scalars and , not both zero, such that .

Now we start to study the commutativity of two th-order slant Toeplitz operators with harmonic symbols.

Proposition 4. Let and , where and is a positive integer, then the following statements are equivalent:

(1.1); (1.2)there exist scalars   and , not both zero, such that .

Proof. We begin with the easy direction. First, suppose that (1.2) holds and let without lost of generality, so that . Thus, .

To prove the other direction of the proposition, suppose that (1.1) holds. Since and , then , and

where both and are integers with . Because , we can get that

for any integers with .

Now we start to investigate the connection between and () by induction. Since , without loss of generality, let and let , which also means that .

When , then by (9), we can get that , which means that , since .

Suppose that for any integers with , where . Now we consider the connection between and . Let , where and are both nonnegative integers with .

When , by (9) we get that

From the assumption we get that , which means that .

Hence, by the induction we obtain that ; that is, . So, the required result holds.

Lemma 5. Let , , and , where and are integers and . If , then for any integers with .

Proof. Since , , and , then

Since , we can get that for any integers with ,

for any integers with ,

We want to show that for any integers with . Here are two cases: and . Let , where and are nonnegative integers with .

First Case. If , then . Now we, continue the discussion by induction.

When , by (12), we can get that . So, , since .

When , by (12) we can get that , where and sgn is a sign function. So, , since and .

Suppose that for any integers with . Now, consider the value of .

When , by (12) we get that

where

and is the biggest integer which is not bigger than . Then, by the assumption and (14), we get that , since .

Hence, from the above discussion we obtain that for all integers with by the induction.

Second Case. If , then . Now we continue the discussion by induction.

When , by (12), we can get that . So, , since .

Suppose that for any integers with . Now, consider the value of .

If , by (12) we get that

where

and is the biggest integer which is not bigger than . Then, by the assumption and (16), we get that , since .

If , by (13), we get that

where

and is the biggest integer which is not bigger than . Then, by the assumption and (18), we get that , since .

Hence, from the above discussion we obtain that for all integers with by the induction.

Similarly, we could get that for all integers with .

From Proposition 4 and Lemma 5, it is evident that Proposition 6 holds.

Proposition 6. Let , and , where are integers and , then the following statements are equivalent:

(1.1); (1.2)there exist scalars   and , not both zero, such that .

Theorem 7 is obvious from Theorem 2.8 in [17] and Proposition 6.

Theorem 7. Let , , and , where and are integers and , and the following statements are equivalent:

(1.1) and commute; (1.2) and essentially commute; (1.3); (1.4) there exist scalars and , not both zero, such that .

3. Product of Two th-Order Slant Toeplitz Operators on

In [15, 17], the authors have investigated properties of the product of two th-order slant Toeplitz operators on and have obtained the following result.

Theorem 8 (see [15, 17]). Let , then the following statements are equivalent:

(1.1) is a th-order slant Toeplitz operator; (1.2) is a zero operator; (1.3) is compact; (1.4).

In this section, we will describe properties of the product of two th-order slant Toeplitz operators with analytic symbols and harmonic symbols on by their symbols. First, we start to discuss properties of two th-order slant Toeplitz product with analytic symbols.

Proposition 9. Let , then if and only if or .

Proof. As we know, the “if” direction of the proposition is trivial.

Now suppose that . Since , we have , where and are the Poisson extensions of and , respectively, and they are analytic on the unit disk . Hence, we get that is identically 0 or is identically 0; that is, is identically 0 or is identically 0.

Similarly, we could obtain Corollary 10.

Corollary 10. Let , then if and only if or .

It is obvious that Theorem 11 holds from the preceding analysis.

Theorem 11. Let or , then the following statements are equivalent:

(1.1) is a th-order slant Toeplitz operator; (1.2) is a zero operator; (1.3) is compact; (1.4); (1.5) or .

Now, we start to discuss the properties of two th-order slant Toeplitz product with harmonic symbols.

Proposition 12. Let and , where and are both positive integers, then if and only if or .

Proof. We begin with the easy direction. First, suppose that or ; then it is clear that .

Without loss of generality, let . Otherwise, ; then we can consider the value of and , where and are both integers with and , since and are both polynomial functions. There are four cases: for all integers with ; for all integers with ; and for all integers with ; and for all integers with , where and are both nonnegative integers with and . If the first two cases hold, then the required result holds; if the latter two cases holds, then we have and or and .

Now suppose that and . Since , , then

where and are both integers with and . Because , we get, for any integers with ,

where and are both integers with and . Since , yet we only obtain either or .

First Case. If . Now we want to show that by the induction.

When , by (21), we get that , which means that , since .

When , by (21), we get that , which means that , since .

Now suppose that for any integers with , where is an integer with . Considering the value of , when , by (21), we get that

where and is the biggest integer which is not bigger than . By the assumption and the above equation, we get that ; that is, , since .

From the preceding analysis, by the induction, we can obtain that for any integers with . Hence, .

Second Case. If . Arguing as in the First case, we obtain that .

Proposition 13. Let and , where and are positive integers and , then the following statements are equivalent:

(1.1); (1.2); (1.3)   or .

Proof. We begin with the easy direction. First, suppose that or ; then it is clear that (1.1) and (1.2) hold.

Now suppose that (1.1) holds. Since , , then

where and are both integers with and . Because , we get that for any integers with ,

where and are both integers with and . Since , yet we can obtain that , , or .

First Case. If . Now we start to continue the proof by induction.

When , by (24), we get that , which means that , since .

When , by (24), we get that , which means that , since .

Now suppose that for any integers with , where is an integer with . Considering the value of , when , by (24), we get that

where and is the biggest integer which is not bigger than . By the assumption and the above equation, we get that ; that is, , since .

From the preceding analysis, by the induction we can obtain that for any integers with . Then, by Proposition 12 we get that , since is not 0 identically.

Second Case. If . In the following, we will continue the proof by induction.

When , by (24), we get that , which means that , since .

When , by (24), we get that , where and sgn is a sign function. So, , since and .

Now, suppose that for any integers with , where is an integer with . Considering the value of , when , by (24), we get that

where and is the biggest integer which is not bigger than . By the assumption and the above equation, we get that ; that is, , since .

From the preceding analysis, by the induction we can obtain that for any integers with , which means that .

Third Case. If . A computation analogous to the one done in the second case from which we can get that .

From the above analysis, we have that (1.3) holds.

Arguing as in the pervious discussion, we obtain that if (1.2) holds, then (1.3) is true.

Proposition 14. Let , , where and are both positive integers and , then if and only if or .

Proof. We begin with the easy direction. First, suppose that or ; then it is clear that .

Now suppose that . Since , , then

where and are both integers with and . Because , we get that, for any integers with ,

where and are both integers with and .

Since , without loss of generality, suppose that . We want to continue the proof by induction.

When , by (28), we get that , which means that , since , when , by (28), we get that , which means that , since .

Now suppose that for any integers with , where is an integer with . Considering the value of . When , by (28) we get that

where and is the biggest integer which is not bigger than . By the assumption and the above equation, we get that ; that is, , since .

From the preceding analysis, by the induction we can obtain that for any integers with . Then, by Proposition 13 we get that , since is not 0 identically and is equivalent to .

Form Proposition 3.4 and Theorem 3.1 in [15, 17], we will obtain the following theorem which describes the product of two slant Toeplitz operators with harmonic symbols.

Theorem 15. Let , , where and are both positive integers and , then the following statements are equivalent:

(1.1) is a th-order slant Toeplitz operator; (1.2) is a zero operator; (1.3) is compact; (1.4); (1.5) or .

4. Commutativity of th-Order Slant Toeplitz Operators on Bergman Spaces

Let be the complex plane and let be the unit disk in . Let be the area measure on normalized so that . Let be the space of analytic functions in which consists of Lebesgue measurable functions on with

It is well known that is a closed subspace of the Hilbert space with the inner product and are the orthogonal basis in , where is the set of nonnegative integers. Let be the Banach space of Lebesgue measurable functions on with

Let denote the orthogonal projection from onto . For , the Toeplitz operator on is defined by

and the th-order slant Toeplitz operator on is defined by

where is a bounded linear operator on which is defined as

In this section we will investigate the commutativity of th-order slant Toeplitz operators with coanalytic symbols and harmonic symbols on Bergman space . First, we study the commutativity of th-order slant Toeplitz operators with coanalytic symbols.

Lemma 16. Let be an integer, let , both of which are not 0 identically. If , the following statements are equivalent:

(1.1) for any integers with and ; (1.2) for any integers with and .

Proof. First, suppose that (1.1) holds. Since , we get that and , where , , and . Because and , we get that

so . Since is not 0 identically, without loss of generality, take for any integers with and , then and , where , , and ; so, by (35) we get that

so . Otherwise, if , then by the above equation we get that , which means that , that leads a contradiction; if , then by the above equation we get that , which means that , that leads a contradiction. Hence (1.2) holds.

Similarly, we can obtain the other direction of the Lemma.

Theorem 17. Let , then the following statements are equivalent:

(1.1) and commute; (1.2)there exist scalars and , not both zero, such that .

Proof. First suppose that (1.2) holds; then it is obvious that and commute.

Now suppose that (1.1) holds. So, we get that ; that is, .

Now we continue the discussion in three cases.

First Case. If or . It is obvious that the required result holds.

Second Case. If and . Since , let and , then , , and and , so

Because , we get that

that is, for any integers ,

where and are both nonnegative integers. In the following, we want to continue the proof by the induction.

When , by (39), we get that , so , since . Let , then .

When , by (39), we get that , so .

Suppose that for any integers with , where is a nonnegative integer. Now, consider the connection between and .

When , by (39), we get that

that is,

where and is the biggest integer which is not bigger than . By this assumption, we can obtain that ; that is, , since .

Hence, by the induction, we obtain that for any nonnegative integers from the pervoius discussion; that is, . So, the required result holds.

Third Case. If and are both not 0 identically, and or . Without loss of generality, take for any integers and . By Lemma 16, we get that for any integers and , and

where , and , , so and . Since , yet we can get that . Since and , yet by the second case we get that , where . So, . The required result holds.

Now we are in a position to discuss the commutativity of slant Toeplitz operators with harmonic symbols.

Theorem 18. Let and , where and is an integer, then the following statements are equivalent:

(1.1) and commute; (1.2)there exist scalars and , not both zero, such that .

Proof. First suppose that (1.2) holds. It is obvious that and commute.

Now suppose that (1.1) holds. Let , , , and , then and . Since and commute, we have , that is,

Then, by (43), we get that for any integers with ,

where and are positive integers which are not bigger than . Since , without loss of generality, take . Now we continue the proof by the induction.

When , by (44), we get that , so , since . Let , then .

When , by (44), we get that , so , since .

Suppose that for any integers with . Now, consider the connection between and .

When , by (44), we get that

where and is the biggest integer which is not bigger than . From the assumption we obtain that , so , since .

From the pervous discussion, by the induction we obtain that for any integers with . Hence, .

Since , by (43), we get that

Then,

and we get that ; that is, . So, by (46) we have that and , that is, . Hence, for any integers with , we have

that is,

Since , we have

When , by (50), we get that , so , since . Then we have

Suppose that for any integers with , where . By (50), we get that

Now consider the connection between and .

When , by (52), we get that , so , since .

Form the above discussion, by the induction we can obtain that for any integers with , so .

Since , we have . Hence, the required result holds.

Lemma 19. Let and , where and are integers with and . If and commute, then for any integers with .

Proof. Let , , and , then and . Since and commute, we have ; that is,

Since , let , where is a positive integer. Then, by the pervoius equation, we get that

so , since .

Suppose that for any integers with . Now consider the value of . By the assumption and the above equation we get that

Since , we can get that ; that is,

where and is the biggest integer which is not bigger than . Since , by assumption, we get that .

From the preceding discussion, by the induction we can get that for any integers with .

Theorem 20. Let and , where , and are integers with , then the following statements are equivalent:

(1.1) and commute; (1.2) there exist scalars and , not both zero, such that .

Proof. First suppose that (1.2) holds. It is obvious that and commute.

Now suppose that (1.1) holds. Since , we can get that or .

If , by Theorem 18, we can get the required result.

If , by Lemma 19, we can get that . Let , , , and , then and .

Since and commute, we have ; that is,

By (57) we get that for any integers with ,

where and are positive integers with and . Now we continue the proof by the induction.

When , by (58), we get that , so , since . Let , then .

When , by (58), we get that , so , since .

Suppose that for any integers with . Now consider the connection between and .

When , by (58), we get that

From the assumption we obtain that , so , since .

From the pervious discussion, by the induction we obtain that for any integers with . Hence, .

Since , by (57), we get that

Then,

and we get that ; that is, . So, by (60), we have that and ; that is, . Hence, for any integers with , we have

that is,

Since , we have

When , by (64), we get that , so , since . Then, we have

Suppose that for any integers with , where . By (64), we get that

Now consider the connection between and .

When , by (66), we get that , so , since .

From the pervious discussion, by the induction, we can obtain that for any integers with . Hence,

Let , then , since .

Now we want to show that . Because , yet we get that , which means that

Since , we have ; that is, . Then, we get that for any integers with , ; that is,

By (69), we can get that for any integers with , ; that is, , since .

If , then , which means that , so the required result holds.

If , then . So, by (68), we can get that , that is, . So we get that for any integers with ,

By (70), we can get that for any integers with , , that is, , since .

If , then , which means that , so, the required result holds.

If , then . Then successively by the pervoius method, we can get that for all integers with , which means that . So, the required result holds.

Acknowledgments

The authors thank the referees for several suggestions that improved the paper. This research is supported by NSFC, Items nos. 11271059 and 11226120.

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