Journal of Function Spaces

Volume 2016 (2016), Article ID 1054768, 9 pages

http://dx.doi.org/10.1155/2016/1054768

## Properties of Commutativity of Dual Toeplitz Operators on the Orthogonal Complement of Pluriharmonic Dirichlet Space over the Ball

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

Received 23 October 2015; Accepted 24 December 2015

Academic Editor: Kehe Zhu

Copyright © 2016 Yinyin Hu 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 completely characterize the pluriharmonic symbols for (semi)commuting dual Toeplitz operators on the orthogonal complement of the pluriharmonic Dirichlet space in Sobolev space of the unit ball. We show that, for and pluriharmonic functions, on if and only if and satisfy one of the following conditions: both and are holomorphic; both and are holomorphic; there are constants and , both not being zero, such that is constant.

#### 1. Introduction

For any integer , let denote the open unit ball in . The boundary of is the sphere and the closure of with the Euclidean metric on is denoted by . Let denote the Lebesgue volume measure on the unit ball of , normalized so that the measure of equals 1. The Sobolev space is the completion of the collection of all smooth functions on for whichwhere , is the weak partial derivative. The is a Hilbert space with the inner productwhere denotes the inner product in the Lebesgue space . The Dirichlet space is the closed subspace of consisting of all holomorphic functions, and let denote the orthogonal projection from onto . Then is an integral operator represented bywhere is the reproducing kernel of . By computation, we know thatwhere and is the set of nonnegative integers. The pluriharmonic Dirichlet space is the closed subspace of consisting of all pluriharmonic functions. Let denote the orthogonal projection from onto ; then , where . In fact,LetGiven a function , the multiplication operator , the Toeplitz operator , the Hankel operator , the dual Toeplitz operator , and the dual Hankel operator with symbol are defined, respectively, byThey are all bounded linear operators. Under the decomposition , the multiplication operator is represented asThis shows close relationships among the above four types of operators. Many studies for dual Toeplitz operators offer some insights into the study for Toeplitz operators. So it is reasonable to focus on the dual Toeplitz operators. Although dual Toeplitz operators differ in many ways from Toeplitz operators, they do have some of the same properties. The general problem that we are interested in is the following: what is the relationship between their symbols when two dual Toeplitz operators commute?

For Toeplitz operators, this problem has been studied for a long time. In the case of the classical Hardy space, Brown and Halmos [1] showed that two Toeplitz operators with general bounded symbols commute if and only if either both symbols are analytic, both symbols are conjugate analytic, or a nontrivial linear combination of the symbols is constant.

Initiated by Brown and Halmos’s pioneering work, the problem of characterizing when two Toeplitz operators commute has been one of the topics of constant interest in the study of Toeplitz operators on classical function spaces over various domains. On the Bergman space of the unit disk, Axler and Čučković [2] studied commuting Toeplitz operators with harmonic symbols and obtained a similar result to that of Brown and Halmos. Stroethoff [3] later extended that result to essentially commuting Toeplitz operators. Axler et al. [4] showed that if two Toeplitz operators commute and the symbol of one of them is nonconstant analytic, then the other one must be analytic. Čučković and Rao [5] studied Toeplitz operators that commute with Toeplitz operators with monomial symbols. On the Bergman space of several complex variables, by making use of -harmonic function theory, Zheng [6] characterized commuting Toeplitz operators with pluriharmonic symbols on the Bergman space of the unit ball. Choe and Lee [7–9] studied commuting and essentially commuting Toeplitz operators with pluriharmonic symbols on the unit ball. Lu [10] characterized commuting Toeplitz operators on the bidisk with pluriharmonic symbols. Choe et al. [11] obtained characterizations of (essentially) commuting Toeplitz operators with pluriharmonic symbols on the Bergman space of the polydisk.

The fact that the product of two harmonic functions is no longer harmonic adds some mystery to the study of operators on harmonic Bergman space. Many methods which work for the operators on analytic Bergman space lose their effectiveness on harmonic Bergman space. On the harmonic Bergman space of the unit disk, Ohno [12] first characterized the commutativity of and , where is an analytic function. Choe and Lee [13] studied commuting Toeplitz operator with harmonic symbols and one of the symbols is a polynomial. In [14], Choe and Lee proved that if and supposedly one of them is noncyclic, then if and only if either or is constant. On the pluriharmonic Bergman space of the unit ball, commuting Toeplitz operators were studied in [15, 16].

However, the study on the problem for dual Toeplitz operators started recently. Stroethoff and Zheng [17] characterized the commutativity of dual Toeplitz operators with bounded symbols on the orthogonal complement of the Bergman space of the unit disk and studied algebraic and spectral properties of dual Toeplitz operators. On the Bergman space of the unit ball and the polydisk, commuting dual Toeplitz operators were studied in [18–20]. Yang and Lu [21] gave complete characterization for the (semi)commuting dual Toeplitz operators with harmonic symbols on harmonic Bergman space.

In recent years the Dirichlet space has received a lot of attention from mathematicians in the areas of modern analysis, probability, and statistical analysis. Many mathematicians are interested in function theory and operator theory on the Dirichlet space. Yu and Wu [22, 23] investigated commuting dual Toeplitz operators with harmonic symbols on the Dirichlet space. Yu [24] obtained the commutativity of dual Toeplitz operators with general symbols on Dirichlet space.

In this paper, we want to characterize commuting dual Toeplitz operators with pluriharmonic symbols on the orthogonal complement of the pluriharmonic Dirichlet space in Sobolev space of the unit ball.

We state our main result now. We postpone the proofs of these theorems until Section 3.

Theorem 1. *Suppose that are pluriharmonic functions; then if and only if one of the following statements holds:*(1)*Both and are holomorphic.*(2)*Both and are holomorphic.*(3)*Either or is constant.*

*Theorem 2. Suppose that are pluriharmonic functions, then if and only if one of the following statement holds:(1)Both and are holomorphic.(2)Both and are holomorphic.(3)There are constants and , both not being zero, such that is constant.*

*For , two dual Toeplitz operators with harmonic symbols always commute on the orthogonal complement of harmonic Dirichlet space; that is, holds for all harmonic functions and .*

*A pluriharmonic function in the unit ball is the sum of a holomorphic function and the conjugate of a holomorphic function. It is clear that all pluriharmonic functions on are -harmonic. A good reference for the function theory of the unit ball is Rudin’s book [25].*

*The difficult part of the proof of Theorem 2 is to answer the following question about pluriharmonic functions.*

*Question*. If and are holomorphic functions in , when is pluriharmonic?

*This question is very subtle. If , this question is a special case of Theorem in [6]. In [26], Choe et al. gave a necessary and sufficient condition for this question in Lemma , which is useless to the proof of Theorem 2. In this paper, we give another characterization to the question and induce the proof of Theorem 2.*

*2. Some Lemmas*

*2. Some Lemmas*

*The following Lemma has been known to be true for in [24]. For , the following lemma may be known, but we cannot find its proof; for completeness, we give its proof.*

*Lemma 3. The set of all polynomials in and is dense in .*

*Proof. *We will discuss it in the case of real variables. For and , since and , one can see that the norm of is equivalent to the following norm:where . For any and , by Theorem in [27], there exists a smooth function such that . Choose a constant such that the support set of is contained inIt follows that the support set of is also in . Let be a polynomial such thatfor all in , and letSimilarly, we also can define for . It is obtained thatSimilarly, we have and for any .

Let denote the polynomial . Also we haveSimilar to the above one can see thatfor all in . Thus we have . This completes the proof.

*For two multi-indexes and , the notation means thatThe standard orhonormal basis for consists of the vectors , where is the ordered -tuple that has 1 in the th spot and 0 everywhere else. A direct computation gives that*

*Let and we have the following Lemma.*

*Lemma 4. Set is dense in .*

*Proof. *Since polynomials are dense in by Lemma 3 and is a bounded operator, we get that is dense in .

*The following lemma will be useful for the proof of the main theorem.*

*Lemma 5. Suppose that is holomorphic; then we have , .*

*Proof. *Since is dense in , it suffices to prove for . Since is holomorphic, we have . For , it follows thatin the Dirichlet space. For , a direct computation gives thatwhich is also in . For , it is obtained thatwhere = − !

The last case is similar; we omit the proof. Hence we get that if and is holomorphic, we have . As the same discussion, we can deduce that .

*In the following proposition, we give an answer to the question that when a dual Toeplitz operator equals zero.*

*Proposition 6. Suppose that is a pluriharmonic function. Then if and only if .*

*Proof. *Assume that . LetA direct computation gives that Since , it follows that . The converse part is easy to see.

*If , and are holomorphic functions in , when is -harmonic? In [6], Zheng gives a necessary and sufficient condition for this question. In the following lemma, we give a generalization. For , the inner product of and is defined by .*

*Lemma 7. Suppose that and are holomorphic functions. Then is pluriharmonic if and only if there is unitary matrix:and some such that are constants for , and are constants for .*

*Proof. *To prove the sufficient part, suppose that is the above unitary matrix such that for some , for and for , where are constants. Let and . It follows thatIt is obtained that is pluriharmonic.

Conversely, assume that is pluriharmonic. There exist two holomorphic functions and on such that By complexifying (25) (see Lemma in [28]), for all and in , we get It follows that for and in , we haveThen there is an orthonormal basis of such that for some ,for all . By Gauss elimination, we eliminate and . Then we get the following equations:Since are orthogonal to for , , it follows that and are orthogonal for and . Also the rank of the matrix is . Since the equationshave solutions, it follows that the rank of equals . After orthonormalization, we get orthonormal bases such that for . The case of is similar. Then we get unitary matrix satisfying the lemma.

*3. Proofs of Main Theorems*

*3. Proofs of Main Theorems*

*In this section, we will present the proofs of the main results.*

*Proof of Theorem 1. *If holds, we have the fact that is contained in . It follows that . The desired result follows from the equation . Case is similar. Case is easy to get the desired result.

To prove the necessity, suppose that . Then we have . Since and are pluriharmonic functions, there exist holomorphic functions such that , . Without loss of generality, we assume that . And , . LetBy a direct calculation, we haveSimilarly, we have . Since , it followsHence is obtained. By Theorem in [6], we have implying that one of the following statements holds:(1)Both and are holomorphic.(2)Both and are holomorphic.(3)Either or is constant.(4)There is a nonzero constant such that and are constants.Then it suffices to prove that in condition when both and are not constants. For fixed , letIn the same way, we get the following:Applying Theorem in [6] again, there exist two constants , such thatTherefore we havefor all and .*Case* *1*. If there exist two multi-indexes such that , Then from (37) we haveIf , thenwhich induces for all . That is a contradiction.*Case 2*. If is a monomial for some with . Suppose that such thatLeta direct calculation in the same way above gives It follows that . Then the fact that is a constant implies that which is a contradiction. Hence we get the desired result.

Suppose that are pluriharmonic functions and , where are holomorphic functions. We are ready to prove Theorem 2.

*Proof of Theorem 2. *From the equation , it follows thatThen if and only if .

Assume that . Then for any , we have . It is obtained that By Lemma 5, we have the fact that are holomorphic and are holomorphic. Then we getIt follows thatIf one of is a constant function, without loss of generality, assume that is a constant function; this follows for any ; we getWe have the fact that is pluriharmonic for all . By Theorem in [6], one of the following holds:(1)Both and are constants.(2)Both and are constants.(3)Both and are constants.(4)Both and are constants.(5)There is a nonzero constant such that and are constants.If is a constant function, we have the fact that both and are holomorphic. If is a constant function, then is a constant function. Assume that neither nor is constant. Then for all , is pluriharmonic if and only if one of the following holds:(1)Both and are constants.(2)There is a nonzero constant such that and are constants.Since is holomorphic, . And is not a constant; there exists a multi-index such that . For any multi-index , let . A direct computation givesWe choose a such that . Since , it follows that is not a constant. Then we get that there is a nonzero constant such that is constant. Since , from the fact that is constant, we get , which is a contradiction. Hence if is a constant function, we have either both and are holomorphic or is a constant function.

In the following proof, assume that none of is a constant function. It follows that . By Lemma 7, we get that there is a unitary matrix such that for some , are constants for and are constants for .*Case* *1*. If there exists such that or , it follows that , are constants or , are constants since is a unitary matrix. Hence we get that both and are holomorphic or both and are holomorphic.*Case* *2*. If there exists such that or . We just prove the case of ; the case of is similar. Since are constants for , it follows that there exist a nonzero constant and a constant such thatThen by (46), we getwhich implies thatis pluriharmonic. By Theorem in [6], one of the following holds:(1)Both and are constants.(2)Both and are constants.(3)There is a nonzero constant such that and are constants.If is a constant, it follows easily that . Assume that is not a constant. Then for all , is pluriharmonic if and only if one of the following holds:(1)Both and are constants.(2)There is a nonzero constant such that and are constants.Since is not a constant, similar to the previous proof, we can find such that neither nor is constant, which is a contradiction. Hence we get that .*Case* *3*. For all , we have . For each , there exist constants , and such thatSuppose that and . For multi-index , let ; there exists a holomorphic function such that . Then for all multi-index , we get .

Note that and are not constants. If for every multi-index , . Suppose that and , where . For any multi-index satisfying and , let . Then there exist constants and such thatFrom the above computation, for all nonzero , we have . It follows that . Then for all with and . A direct computation givesIt follows that, for all multi-index , we have . For each , we can find a such that Hence for all nonzero multi-index . Then is a constant and this leads a contradiction.

Then there exists a multi-index such that and For any , let . Similarly, we still have . If , we get . Fix a multi-index , suppose that, for all with , we have . It follows that for all , which implies that is a constant. That is a contradiction. Suppose that there exists a multi-index such that . It follows that and, for all , . If , it follows that . If , we get . We also can find a multi-index with such that . Similarly, for all , if , we have . If , we get . Clearly, . For each nonzero multi-index , we can find a such that , where is a nonzero constant. Then we have , and similar to the proof of Case , we get that .

By Lemma 5, the converse is easy to see. The proof is complete.

*Conflict of Interests*

*Conflict of Interests*

*The authors declare that there is no conflict of interests regarding the publication of this paper.*

*Acknowledgment*

*Acknowledgment*

*This research is supported by NSFC (nos. 11271059, 11271332, 11431011, and 11301047) and NSF of Zhejiang Province (nos. LY14A010013 and LY14A010021).*

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