#### Abstract

We study some algebraic properties of Toeplitz operators with radial or quasihomogeneous symbols on the pluriharmonic Bergman space. We first give the necessary and sufficient conditions for the product of two Toeplitz operators with radial symbols to be a Toeplitz operator and discuss the zero-product problem for several Toeplitz operators with radial symbols. Next, we study the finite-rank product problem of several Toeplitz operators with quasihomogeneous symbols. Finally, we also investigate finite rank commutators and semicommutators of two Toeplitz operators with quasihomogeneous symbols.

#### 1. Introduction

Let denote the normalized Lebesgue volume measure on the unit ball of . is the Hilbert space of Lebesgue square integrable functions on with the inner product: The Bergman space is the Hilbert space consisting of all holomorphic functions on which are also in . It is well known that is a reproducing function space with reproducing kernel: The pluriharmonic Bergman space is the closed subspace of consisting of the pluriharmonic functions on . It is easy to verify that where . Then is also a reproducing function space with reproducing kernel:

Let denote the orthogonal projection from onto , then Let be the orthogonal projection from onto , then By (4), we have

For a function , the Toeplitz operator with symbol is defined by

In the setting of the classical Hardy space, for and in , Brown and Halmos [1] proved that if and only if one of the following conditions holds: (a) is analytic, (b) is analytic. They also showed that, in both cases . On the Bergman space of the unit disk, conditions (a) and (b) are only sufficient but not necessary. Ahern and Čučković [2] showed that a Brown-Halmos-type result holds for Toeplitz operators with harmonic symbols under certain conditions. Later, Ahern gave the necessary and sufficient conditions for the product of two Toeplitz operators with harmonic symbols to be a Toeplitz operator in [3]. For general symbols, the product problem for Toeplitz operators remains open. Louhichi et al. [4] gave the necessary and sufficient conditions for the product of two quasihomogeneous Toeplitz operators to be a Toeplitz operator.

On the Hardy or Bergman space of several complex variables, the situation is much more complicated. Ding [5] characterized when the product of two Toeplitz operators with bounded symbols on the Hardy space is still a Toeplitz operator. Motivated by the results of Ahern and Čučković [2], Choe et al. solved the product problem for Toeplitz operators with pluriharmonic symbols on the Bergman space of the polydisk [6]. On the Bergman space of the unit ball, Zhou and Dong [7] gave the necessary and sufficient conditions for the product of two Toeplitz operators with radial symbols to be equal to a Toeplitz operator. In [8], they only gave the necessary condition for the product problem of two separately quasihomogeneous Toeplitz operators. Recently, Lu and Zhang characterized when the product of two quasihomogeneous Toeplitz operators is equal to another quasihomogeneous Toeplitz operator in [9].

For the so-called “zero-product” problem, Brown and Halmos [1] easily deduced that if on the Hardy space then either or must be identically zero. A natural and interesting problem on the classical Hardy space of one complex variable is the following zero-product problem: If , whether one of these symbols is identically zero. Guo [10] proved that if , then there exists some such that . In [11], Gu showed that , then there exists some such that . Aleman and Vukotić [12] completely solved the zero-product problem for several Toeplitz operators on the Hardy space. In [2], Ahern and Čučković proved that the result is analogous to that in [1] for two Toeplitz operators with harmonic symbols on the Bergman space of unit disk. Moreover in [13] they proved that if , where is arbitrary bounded and is radial, then either or . Čučković and Louhichi [14] studied finite rank product of several quasihomogeneous Toeplitz operators on the Bergman space of the unit disk. On the Bergman space of the unit ball, Choe and Koo [15] gave a result which is analogous to that in [2] with an assumption about the continuity of the symbols on an open subset of the boundary and solved the zero-product problem for several Toeplitz operators with harmonic symbols that have Lipschitz continuous extensions to the whole boundary. Also, analogous results were proved on polydisk in [16]. Recently, those zero product results have been generalized to finite rank product results in [17]. Le [18] discussed finite rank products of several Toeplitz operators on the weighted Bergman space of the unit ball. Dong and Zhou [8] investigated the zero-product problem of two Toeplitz operators, one of whose symbols is separately quasihomogeneous and the other is arbitrary bounded. In [7] they studied the zero-product problem for several Toeplitz operators with radial symbols. Ding [5] considered the problem of two Toeplitz operators with symbols in on the Hardy space of the polydisk.

For two Toeplitz operators and we define the commutator and semicommutator, respectively, by

On the Hardy space the problem of determining when the commutator or semicommutator has finite rank has been completely solved (see [19, 20]). For the Bergman space the problem seems to be far from solution. Guo et al. [21] completely characterized the finite rank commutator and semicommutator of two Toeplitz operators with bounded harmonic symbols on the Bergman space of the unit disk. Luecking [22] showed that finite rank Toeplitz operators on the Bergman space of the unit disk must be zero. Čučković and Louhichi [14] studied the finite rank semi-commutators and commutators of Toeplitz operators with quasihomogeneous symbols on the Bergman space of the unit disk and obtained different results from the case of harmonic Toeplitz operators. Lu and Zhang [9, 23] characterized finite rank commutators and semicommutators of two quasihomogeneous Toeplitz operators on the Bergman space of the unit ball and the polydisk, respectively.

The fact that the product of two harmonic functions is no longer harmonic adds some mystery in the study of operators on the harmonic Bergman space. The theory of Toeplitz operators on the harmonic Bergman space is quite different from that on . For example, Choe and Lee [24] showed that two analytic Toeplitz operators on commute only when their symbols and the constant function 1 are linearly dependent, but analytic Toeplitz operators always commute on . The following question was raised by Choe and Lee in [24]: If an analytic Toeplitz operator and a coanalytic Toeplitz operator on the harmonic Bergman space commute, then is one of their symbols a constant? To solve this problem, they proved in [25] that if and suppose one of them is noncyclic, then if and only if either or is constant. On the pluriharmonic Bergman space of the unit ball, Lee and Zhu [26] characterized commuting Toeplitz operators with holomorphic symbols and obtained the necessary and sufficient condition for the product of two Toeplitz operators with pluriharmonic symbols to be equal to a Teoplitz operator. Furthermore, they gave a complete description of holomorphic symbols for which the associated Toeplitz operators have zero semicommutator. On the pluriharmonic Bergman space of the polydisk, Choe and Nam [27] obtained results which are parallel to those of [26]. Dong and Zhou [28] studied the problem of when the product of two Toeplitz operators with quasihomogeneous symbols is a Toeplitz operator on the harmonic Bergman space of the unit disk.

Motivated by the recent work of Čučković, Louhichi [14], Zhou and Dong [7, 28], and Lee and Zhu [26] we study Toeplitz operators with radial or quasihomogeneous symbols on the pluriharmonic Bergman space of the unit ball. The present paper is assembled as follows: In Section 2, we introduce some basic properties of the Mellin transform and Mellin convolution. In Section 3, we first give the necessary and sufficient condition for the product of two Toeplitz operators with radial symbols to be a Toeplitz operator. Then we investigate the zero-product problem for several Toeplitz operators with radial symbols. In Section 4, we study the finite rank product problem for several Toeplitz operator with quasihomogeneous symbols. In Sections 5 and 6, we discuss the finite rank commutators and semicommutators of two Toeplitz operators with quasihomogeneous symbols on the pluriharmonic Bergman space.

#### 2. The Mellin Transform and Mellin Convolution

One of the most useful tools in the following calculations is the Mellin transform. The Mellin transform of a function is defined by It is clear that is well defined on the right half-plane and analytic on . It is important and helpful to know that the Mellin transform is uniquely determined by its value on an arithmetic sequence of integers. In fact, we have the following classical theorem [29, page 102].

Theorem 1. *Suppose that is a bounded analytic function on which vanishes at the pairwise distinct points , where*(1)* and*(2)*.**
Then vanishes identically on .*

*Remark 2. *We will often use this theorem to show that if and if there exists a sequence such that
then for all and so .

In the following discussion we need a known fact about the Mellin convolution of two functions. If and are defined on , then their Mellin convolution is defined by The Mellin convolution theorem states that and that, if and are in then so is .

#### 3. Products of Toeplitz Operators with Radial Symbols

For any multi-index , where each is a nonnegative integer, we write , and , .

For two multi-indexes and , the notation means that , and means that . We also define . Moreover, if if .

It is known that is radial if and only if for any unitary transformation of . That is, only depends on . Then for each radial function , we define on by where is a unit vector in . It is obvious that is welldefined. In the following, we will identify an integrable radial function on the unit ball with the corresponding function defined on the interval .

Lemma 3. *Let , is a Toeplitz operator densely defined on , then the symbol map is one to one.*

*Proof. *Assume , we have for all .

Since , then we can imply that , which is equivalent to . Since is dense in , so we get .

A direct calculation gives the following lemma, which will be used constantly in this paper.

Lemma 4. *Let be an integrable radial function on , such that is a bounded operator, then for any multi-index ,
*

*Proof. *For any multi-index , we have
which implies

If , we have . This shows that

Since is a basis for the pluriharmonic Bergman space, we have
By a similar argument, we have .

The following theorem is very simple but essential.

Theorem 5. *Let . Then the following assertions are equivalent.*(1)*For each multi-index , there exists which depends only on such that , .*(2)* is a radial function.*

*Proof. *The implication is obvious fir Lemma 4.

Now, consider the converse implication . By Lemma 3, we just need to prove for any unitary transformation of .

First assume . For any unitary transformation of and , we have . Thus,

By (7), we have
A direct calculation shows that
Similarly, we have
Then we can imply

So . With the same method, we get . Therefore, we can imply on . By Lemma 3 we obtain that and is a radial function.

Corollary 6. *Let and be two square integrable radial functions on such that and are bounded operators. If , then is a radial function.*

*Proof. *According to (14) we have
It follows from Theorem 5 that is a radial function.

Theorem 7. *Let and be two square integrable radial functions on , such that and are bounded operators. Then is equal to the Toeplitz operator if and only if is a solution of the equation
**
where denotes the constant function with value one. *

*Proof. *For each multi-index , it follows from (14) that
are equivalent to
A direct calculation shows , so (27) is equivalent to
By Remark 2, we obtain that (28) is equivalent to (25).

*Example 8. *Let and greater than or equal to −1. Then

In [18] Le discussed finite rank product of Toeplitz operators with general symbols on the weighted Bergman space of the unit ball. Similarly, we will use Mellin transform and Remark 2 to study zero-product problem of Toeplitz operators with radial symbols on pluriharmonic Bergman space.

Theorem 9. *Let be integrable radial functions on , such that be bounded operators, . If , then for some .*

*Proof. *Suppose , then for any multi-index , we have
Using (14), we have

Let . Notice that , hence there exists some such that , then by Remark 2 .

Corollary 10. *Let be an integrable radial function on , such that is a bounded operator. Then if and only if either or .*

* Proof. *If , then . By Theorem 9 we have or . The converse is clear.

#### 4. Product of Toeplitz Operator

In this section, we will show that if the product of several quasihomogeneous Toeplitz operators has finite rank, then at least one of the symbols is equal to zero. This result is analogous to Theorem 3.2 in [18], but we get it in a different way.

*Definition 11. *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 unit sphere and .

Lemma 12. *Let be two multi-indexes and let be an integrable radial function on , such that , and are bounded operators. Then for any multi-index ,
**
Especially, if are two nonzero multi-indexes with , one has
*

* Proof. *Here we only give the proof of the first equation. For multi-indexes and ,
If , we have
Furthermore, hold for all . So we obtain

Note that is still radial and for all , we have
Thus

Especially, if , we know that is equivalent to . Since are nonzero multi-indexes with , there exists such that . This means that there doesn’t exist multi-index such that . So

Theorem 13. *Let be multi-indexes and let be integrable radial functions on such that is bounded . If the product is of finite rank, then for some .*

* Proof. *We denote the product of Toeplitz operators by . By Lemma 12, we have
If , we have
where
and .

Similarly, we have
Then for , we have
where

Since has finite rank, we know that and must have finite rank. Then, there exists a multi-index such that and
which is equivalent to
for all . In fact, we have , in which and , , , , .

Let , we know that . Then (50) implies that
for all . Since is an arithmetic sequence, by Theorem 1, we obtain that there exists some such that .

#### 5. Finite Rank Commutators

In this section, let be two integrable radial functions on . We now pass to investigate the commutator , and .

Theorem 14. *Let be two multi-indexes and let and be two integrable radial functions on such that and are bounded operators. If is nonconstant and has finite rank, then its rank is equal to zero.*

* Proof. *Let denote the commutator . Since has finite rank on , we know that and must have finite rank.

By Lemma 12, we have if ,
If ,
if , , we have .

Since has finite rank, then there must exist , such that and for all .

Similar to the proof of Theorem 4.4 in [7], we can imply or . It follows from (52) and (53) that

Similar to the discussions for , we can imply that for all . In conclusion, the rank of is equal to zero.

Theorem 15. *Let be two nonzero multi-indexes, let and be two integrable radial functions on , such that and are bounded operators. If has finite rank, then the rank of it is less than or equal to the number of such that , and , where . *

* Proof. *Let denote the commutator . By Lemma 12, we have

If has finite rank, there must exist such that
Then, we deduce that
It follows from Remark 2 that
for all . Set , we get that (57) holds for all . Hence for all , we have . Therefore, the rank of is equal to zero.

For , we have if ,
If , , , , we have , if but , , we have , if but , , then .

If or or or ,
where is a constant that depends on .

If has finite rank, use the same method as above, we have
this means , and the rank of is at most equal to the number of such that . In which and . The proof is complete.

*Example 16. *In the case of , we give an example to Theorem 15.

From the proof of Theorem 15 we know that
Below we will construct a radial function , such that
where are multi-indexes, and .

Equation (63) implies that for ,
Now, using Remark 2, we obtain that
for all .

Let be the analytic function defined for by
where denote the gamma function. Then using the well-known identity , (65) implies that
Equation (67) combined with [30, Lemma 6, page 1468] gives us that there exists a constant such that

For a choice of , and , again using the identity , one can see that
Since , , then (68) becomes
Now the proceeding equation and Remark 2 imply that
It is clear that the function is bounded, so Toeplitz operator is bounded.

Finally, by taking the constant to be equal to 1, the radial function satisfies
For , we know that . However, using Lemma 12, it is easy to see that
Therefore the commutator has rank four.

Theorem 17. *Let be two nonzero multi-indexes with , let and be two integrable radial functions on , such that and are bounded operators. If has finite rank, then the rank of it is equal to zero.*

* Proof. *Let denote the commutator . Since , it is easy to see that is equivalent to for all multi-index , and there does not exist multi-index such that . Furthermore, it is impossible that there exists such that and .

If has finite rank, there exists