Algebraic Properties of Toeplitz Operators on the Pluriharmonic Bergman Space

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 𝑑𝜈 denote the normalized Lebesgue volume measure on the unit ball 𝐵_𝑛 of 𝐶^𝑛. 𝐿²(𝐵_𝑛,𝑑𝜈) is the Hilbert space of Lebesgue square integrable functions on 𝐵_𝑛 with the inner product: 𝑓,𝑔=_𝐵𝑛𝑓(𝑧)𝑔(𝑧)𝑑𝑣(𝑧). (1) The Bergman space 𝐿²_𝑎(𝐵_𝑛) is the Hilbert space consisting of all holomorphic functions on 𝐵_𝑛 which are also in 𝐿²(𝐵_𝑛,𝑑𝑣). It is well known that 𝐿²_𝑎(𝐵_𝑛) is a reproducing function space with reproducing kernel: 𝐾_𝑧(𝑤)=1(1−⟨𝑤,𝑧⟩)^𝑛+1,𝑧,𝑤∈𝐵_𝑛. (2) The pluriharmonic Bergman space 𝐿²_ℎ(𝐵_𝑛) is the closed subspace of 𝐿²(𝐵_𝑛,𝑑𝑣) consisting of the pluriharmonic functions on 𝐵_𝑛. It is easy to verify that 𝐿²_ℎ𝐵_𝑛=𝐿²_𝑎𝐵_𝑛+𝐿²_𝑎𝐵_𝑛, (3) where 𝐿²_𝑎(𝐵_𝑛)={𝑓∣𝑓∈𝐿²_𝑎(𝐵_𝑛),𝑓(0)=0}. Then 𝐿²_ℎ(𝐵_𝑛) is also a reproducing function space with reproducing kernel: 𝑅_𝑧(𝑤)=𝐾_𝑧(𝑤)+𝐾_𝑧(𝑤)−1,𝑧,𝑤∈𝐵_𝑛. (4)

Let 𝑄 denote the orthogonal projection from 𝐿²(𝐵_𝑛,𝑑𝑣) onto 𝐿²_ℎ(𝐵_𝑛), then 𝑄𝜑(𝑧)=𝜑,𝑅_𝑧,𝜑∈𝐿²𝐵_𝑛,𝑑𝑣. (5) Let 𝑃 be the orthogonal projection from 𝐿²(𝐵_𝑛,𝑑𝑣) onto 𝐿²_𝑎(𝐵_𝑛), then 𝑃𝜑(𝑧)=𝜑,𝐾_𝑧,𝜑∈𝐿²𝐵_𝑛,𝑑𝑣. (6) By (4), we have 𝑄𝜑(𝑧)=𝑃𝜑(𝑧)+𝑃𝜑(𝑧)−𝑃𝜑(0). (7)

For a function 𝜑∈𝐿^∞(𝐵_𝑛,𝑑𝑣), the Toeplitz operator 𝑇_𝜑∶𝐿²_ℎ(𝐵_𝑛)→𝐿²_ℎ(𝐵_𝑛) with symbol 𝜑 is defined by 𝑇_𝜑𝑓=𝑄𝜑𝑓(𝑧)=_𝐵𝑛𝑓(𝑤)𝜑(𝑤)𝑅_𝑧(𝑤)𝑑𝑣(𝑤). (8)

In the setting of the classical Hardy space, for 𝑓 and 𝑔 in 𝐿^∞(𝑇), Brown and Halmos [1] proved that 𝑇_𝑓𝑇_𝑔=𝑇_ℎ if and only if one of the following conditions holds: (a) 𝑔 is analytic, (b) 𝑓 is analytic. They also showed that, in both cases ℎ=𝑓𝑔. 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 𝐻²(𝐷^𝑛) 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 𝑇_𝑓𝑇_𝑔=0 on the Hardy space 𝐻²(𝑇) then either 𝑓 or 𝑔 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 𝑇_𝜑1𝑇_𝜑2⋯𝑇_𝜑𝑛=0, whether one of these symbols is identically zero. Guo [10] proved that if 𝑇_𝜑1𝑇_𝜑2⋯𝑇_𝜑5=0, then there exists some 𝑖 such that 𝜑_𝑖=0. In [11], Gu showed that 𝑇_𝜑1𝑇_𝜑2⋯𝑇_𝜑6=0, then there exists some 𝑖 such that 𝜑_𝑖=0. 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 𝑇_𝑓𝑇_𝑔=0, where 𝑓 is arbitrary bounded and 𝑔 is radial, then either 𝑓≡0 or 𝑔≡0. 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 𝐿²(𝑇^𝑛) on the Hardy space of the polydisk.

For two Toeplitz operators 𝑇_𝜑 and 𝑇_𝜓 we define the commutator and semicommutator, respectively, by 𝑇_𝜑,𝑇_𝜓=𝑇_𝜑𝑇_𝜓−𝑇_𝜓𝑇_𝜑,𝑇_𝜑,𝑇_𝜓=𝑇_𝜑𝜓−𝑇_𝜑𝑇_𝜓. (9)

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 𝐿²_𝑎. For example, Choe and Lee [24] showed that two analytic Toeplitz operators on 𝐿²_ℎ commute only when their symbols and the constant function 1 are linearly dependent, but analytic Toeplitz operators always commute on 𝐿²_𝑎. 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 𝑓,𝑔∈𝐻^∞ and suppose one of them is noncyclic, then 𝑇_𝑓𝑇_𝑔=𝑇_𝑔𝑇_𝑓 if and only if either 𝑓 or 𝑔 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 𝜑 of a function 𝜑∈𝐿¹([0,1],𝑟𝑑𝑟) is defined by 𝜑(𝑧)=¹₀𝜑(𝑠)𝑠^𝑧−1𝑑𝑠. (10) It is clear that 𝜑 is well defined on the right half-plane {𝑧∶Re𝑧≥2} and analytic on {𝑧∶Re𝑧>2}. It is important and helpful to know that the Mellin transform 𝜑 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 𝑓 is a bounded analytic function on {𝑧∶Re𝑧>0} which vanishes at the pairwise distinct points 𝑧₁,𝑧₂,…, where(1)inf{|𝑧_𝑛|}>0 and(2)∑_𝑛≥1Re(1/𝑧_𝑛)=∞.
Then 𝑓 vanishes identically on {𝑧∶Re𝑧>0}.

Remark 2. We will often use this theorem to show that if 𝜑∈𝐿¹([0,1],𝑟𝑑𝑟) and if there exists a sequence (𝑛_𝑘)_𝑘≥0⊂ℕ such that 𝜑𝑛_𝑘=0,_𝑘≥01𝑛_𝑘=∞, (11) then 𝜑(𝑧)=0 for all 𝑧∈{𝑧∶Re(𝑧)>2} and so 𝜑=0.

In the following discussion we need a known fact about the Mellin convolution of two functions. If 𝑓 and 𝑔 are defined on [0,1), then their Mellin convolution is defined by 𝑓∗_𝑀𝑔(𝑟)=¹_𝑟𝑓𝑟𝑡𝑔(𝑡)𝑑𝑡𝑡,0≤𝑟<1. (12) The Mellin convolution theorem states that 𝑓∗_𝑀𝑔(𝑠)=𝑓(𝑠)𝑔(𝑠), (13) and that, if 𝑓 and 𝑔 are in 𝐿¹([0,1],𝑟𝑑𝑟) then so is 𝑓∗_𝑀𝑔.

3. Products of Toeplitz Operators with Radial Symbols

For any multi-index 𝛼=(𝛼₁,…,𝛼_𝑛), where each 𝛼_𝑖 is a nonnegative integer, we write |𝛼|=𝛼₁+⋯+𝛼_𝑛,  𝛼!=𝛼₁!⋯𝛼_𝑛! and 𝑧^𝛼=𝑧^𝛼1₁⋯𝑧^𝛼𝑛_𝑛, 𝑧=(𝑧₁,…,𝑧_𝑛)∈𝐵_𝑛.

For two multi-indexes 𝛼=(𝛼₁,…,𝛼_𝑛) and 𝛽=(𝛽₁,…,𝛽_𝑛), the notation 𝛼≽𝛽 means that 𝛼_𝑖≥𝛽_𝑖,𝑖=1,2,…𝑛, and 𝛼⟂𝛽 means that 𝛼₁𝛽₁+⋯+𝛼_𝑛𝛽_𝑛=0. We also define 𝛼−𝛽=(𝛼₁−𝛽₁,…,𝛼_𝑛−𝛽_𝑛). Moreover, if |𝛼−𝛽|=|𝛼|−|𝛽| if 𝛼≽𝛽.

It is known that 𝜑 is radial if and only if 𝜑(𝑈𝑧)=𝜑(𝑧) for any unitary transformation 𝑈 of 𝐶^𝑛. That is, 𝜑(𝑧) only depends on |𝑧|. Then for each radial function 𝜑, we define 𝜑 on [0,1) by 𝜑(𝑟)=𝜑(𝑟𝜉) where 𝜉 is a unit vector in 𝐶^𝑛. It is obvious that 𝜑 is welldefined. In the following, we will identify an integrable radial function 𝜑 on the unit ball with the corresponding function 𝜑 defined on the interval [0,1).

Lemma 3. Let 𝜑∈𝐿²(𝐵_𝑛,𝑑𝑣), 𝑇_𝜑 is a Toeplitz operator densely defined on 𝐿²_ℎ(𝐵_𝑛), then the symbol map 𝜑→𝑇_𝜑 is one to one.

Proof. Assume 𝑇_𝜑=0, we have 𝑄[𝜑𝑓]=0 for all 𝑓∈𝐿²_ℎ(𝐵_𝑛).
Since 𝑧^𝛽∈𝐿²_ℎ(𝐵_𝑛), then we can imply that ⟨𝜑𝑧^𝛼,𝑧^𝛽⟩=0, which is equivalent to ∫_𝐵𝑛𝜑𝑧^𝛼𝑧^𝛽𝑑𝑣(𝑧)=0. Since {𝑧^𝛼𝑧^𝛽:𝛼,𝛽areanymulti-indexes} is dense in 𝐿₂(𝐵_𝑛,𝑑𝑣), so we get 𝜑=0.

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

Lemma 4. Let 𝜓 be an integrable radial function on 𝐵_𝑛, such that 𝑇_𝜓 is a bounded operator, then for any multi-index 𝛼, 𝑇_𝜓𝑧^𝛼=2(𝑛+|𝛼|)𝜓(2𝑛+2|𝛼|)𝑧^𝛼,𝑇_𝜓𝑧^𝛼=2(𝑛+|𝛼|)𝜓(2𝑛+2|𝛼|)𝑧^𝛼. (14)

Proof. For any multi-index 𝛽, we have 𝑇_𝜓𝑧^𝛼,𝑧^𝛽=𝜓𝑧^𝛼,𝑧^𝛽=0𝛼≠𝛽;2𝑛!𝛼!(|𝛼!𝜓(|𝛼𝑧𝑛−1+|)2𝑛+2|)𝛼=𝛽,_𝛼,𝑧^𝛽=0𝛼≠𝛽;𝑛!𝛼!(𝑛+|𝛼|)!𝛼=𝛽, (15) which implies 𝑇_𝜓𝑧^𝛼,𝑧^𝛽=2(𝑛+|𝛼|)𝜓(2𝑛+2|𝛼|)𝑧^𝛼,𝑧^𝛽. (16)
If 𝛽≻0, we have ⟨𝑇_𝜓𝑧^𝛼,𝑧^𝛽⟩=0=⟨𝑧^𝛼,𝑧^𝛽⟩. This shows that 𝑇_𝜓𝑧^𝛼,𝑧^𝛽=𝑧^𝛼,𝑧^𝛽∀𝛽≻0. (17)
Since {𝑧^𝛼}_𝛼≽0⋃{𝑧^𝛼}_𝛼≻0 is a basis for the pluriharmonic Bergman space, we have 𝑇_𝜓𝑧^𝛼=2(𝑛+|𝛼|)𝜓(2𝑛+2|𝛼|)𝑧^𝛼. (18) By a similar argument, we have 𝑇_𝜓(𝑧^𝛼)=2(𝑛+|𝛼|)𝜓(2𝑛+2|𝛼|)𝑧^𝛼.

The following theorem is very simple but essential.

Theorem 5. Let 𝜑∈𝐿²(𝐵_𝑛,𝑑𝑣). Then the following assertions are equivalent.(1)For each multi-index 𝛼, there exists 𝜆_|𝛼|∈𝐶 which depends only on |𝛼| such that 𝑇_𝜑(𝑧^𝛼)=𝜆_|𝛼|𝑧^𝛼, 𝑇_𝜑(𝑧^𝛼)=𝜆_|𝛼|𝑧^𝛼.(2)𝜑 is a radial function.

Proof. The implication (2)⇒(1) is obvious fir Lemma 4.
Now, consider the converse implication (1)⇒(2). By Lemma 3, we just need to prove 𝑇_𝜑∘𝑈=𝑇_𝜑 for any unitary transformation 𝑈 of 𝐶^𝑛.
First assume 𝑇_𝜑(𝑧^𝛼)=𝜆_|𝛼|𝑧^𝛼. For any unitary transformation 𝑈 of 𝐶^𝑛 and 𝑧∈𝐵_𝑛, we have 𝑈⁻¹𝑧∈𝐵_𝑛. Thus, 𝑇_𝜑𝑈⁻¹𝑧^𝛼=𝜆_|𝛼|𝑈⁻¹𝑧^𝛼. (19)
By (7), we have 𝑇_𝜑∘𝑈𝑧^𝛼=𝑃𝜑(𝑈𝑧)𝑧^𝛼+𝑃𝜑(𝑈𝑧)𝑧^𝛼−𝑃𝜑(𝑈𝑧)𝑧^𝛼(0). (20) A direct calculation shows that 𝑃𝜑(𝑈𝑧)𝑧^𝛼(𝑤)=_𝐵𝑛𝜑(𝑈𝑧)𝑧^𝛼(1−⟨𝑤,𝑧⟩)^𝑛+1(𝑧)=𝑑𝑣_𝐵𝑛𝜑(𝑧)𝑈⁻¹𝑧^𝛼1−𝑤,𝑈⁻¹𝑧^𝑛+1(𝑧)=𝑑𝑣_𝐵𝑛𝜑(𝑧)𝑈⁻¹𝑧^𝛼(1−⟨𝑈𝑤,𝑧⟩)^𝑛+1(𝑧)𝜑(𝑧𝑈𝑑𝑣=𝑃)⁻¹𝑧^𝛼(𝑈𝑤). (21) Similarly, we have 𝑃𝜑(𝑈𝑧)𝑧^𝛼(𝑤)=𝑃𝜑(𝑧)𝑈⁻¹𝑧^𝛼),𝑃𝜑()𝑧(𝑈𝑤𝑈𝑧^𝛼(0)=𝑃𝜑(𝑧)𝑈⁻¹𝑧^𝛼(0). (22) Then we can imply 𝑇_𝜑∘𝑈𝑧^𝛼(𝑤)=𝑃𝜑(𝑧)𝑈⁻¹𝑧^𝛼(𝑈𝑤)+𝑃𝜑(𝑧)𝑈⁻¹𝑧^𝛼)𝜑(𝑧)𝑈(𝑈𝑤−𝑃⁻¹𝑧^𝛼(0)=𝑇_𝜑𝑈⁻¹𝑧^𝛼()=𝜆𝑈𝑤_|𝛼|𝑈⁻¹𝑧^𝛼(𝑈𝑤)=𝜆_|𝛼|𝑤^𝛼=𝑇_𝜑𝑧^𝛼(𝑤). (23)
So 𝑇_𝜑∘𝑈(𝑧^𝛼)=𝑇_𝜑(𝑧^𝛼). With the same method, we get 𝑇_𝜑∘𝑈(𝑧^𝛼)=𝑇_𝜑(𝑧^𝛼). Therefore, we can imply 𝑇_𝜑∘𝑈=𝑇_𝜑 on 𝐿²_ℎ(𝐵_𝑛). By Lemma 3 we obtain that 𝜑∘𝑈=𝜑 and 𝜑 is a radial function.

Corollary 6. Let 𝜑₁ and 𝜑₂ be two square integrable radial functions on 𝐵_𝑛 such that 𝑇_𝜑1 and 𝑇_𝜑2 are bounded operators. If 𝑇_𝜑1𝑇_𝜑2=𝑇_𝜓, then 𝜓 is a radial function.

Proof. According to (14) we have 𝑇_𝜑1𝑇_𝜑2𝑧^𝛼=(2𝑛+2|𝛼|)²𝜑₁(2𝑛+2|𝛼|)𝜑₂(2𝑛+2|𝛼|)𝑧^𝛼,𝑇_𝜑1𝑇_𝜑2𝑧^𝛼=(2𝑛+2|𝛼|)²𝜑₁(2𝑛+2|𝛼|)𝜑₂(2𝑛+2|𝛼|)𝑧^𝛼. (24) It follows from Theorem 5 that 𝜓 is a radial function.

Theorem 7. Let 𝜑₁ and 𝜑₂ be two square integrable radial functions on 𝐵_𝑛, such that 𝑇_𝜑1 and 𝑇_𝜑2 are bounded operators. Then 𝑇_𝜑1𝑇_𝜑2 is equal to the Toeplitz operator 𝑇_𝜓 if and only if 𝜓 is a solution of the equation 1∗_𝑀𝜓=𝜑₁∗_𝑀𝜑₂, (25) where 1 denotes the constant function with value one.

Proof. For each multi-index 𝛼, it follows from (14) that 𝑇_𝜑1𝑇_𝜑2𝑧^𝛼=𝑇_𝜓𝑧^𝛼,𝑇_𝜑1𝑇_𝜑2𝑧^𝛼=𝑇_𝜓𝑧^𝛼 (26) are equivalent to 1|𝛼|𝜓(|𝛼=𝜑2𝑛+22𝑛+2|)₁(2𝑛+2|𝛼|)𝜑₂(2𝑛+2|𝛼|). (27) A direct calculation shows 1(2𝑛+2|𝛼|)=1/(2𝑛+2|𝛼|), so (27) is equivalent to 1∗_𝑀𝜓(2𝑛+2|𝛼|)=𝜑₁∗_𝑀𝜑₂(2𝑛+2|𝛼|). (28) By Remark 2, we obtain that (28) is equivalent to (25).

Example 8. Let 𝑙 and 𝑚 greater than or equal to −1. Then 𝑇_|𝑧|^𝑙𝑇_|𝑧|^𝑚=𝑙𝑇𝑙−𝑚^|𝑧|𝑙−𝑚𝑇𝑙−𝑚^|𝑧|𝑚𝑇𝑙≠𝑚,_{|𝑧|𝑙(1+𝑙ln|𝑧|)}𝑙=𝑚. (29)

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 𝜑_𝑖 be integrable radial functions on 𝐵_𝑛, such that 𝑇_𝜑𝑖 be bounded operators, 𝑖=1,…,𝑁. If 𝑇_𝜑1⋯𝑇_𝜑𝑁=0, then 𝜑_𝑖=0 for some 𝑖.

Proof. Suppose 𝑇_𝜑1⋯𝑇_𝜑𝑁=0, then for any multi-index 𝛼, we have 𝑇_𝜑1⋯𝑇_𝜑𝑁𝑧^𝛼=0𝑇_𝜑1⋯𝑇_𝜑𝑁𝑧^𝛼=0. (30) Using (14), we have 𝜑₁(2𝑛+2|𝛼|)⋯𝜑_𝑁(2𝑛+2|𝛼|)=0. (31)
Let 𝐸_𝑖={|𝛼|∈ℕ∶𝜑_𝑖(2𝑛+2|𝛼|)=0}. Notice that 𝐸₁∪⋯∪𝐸_𝑁=ℕ, hence there exists some 𝑖 such that ∑_{|𝛼|∈𝐸𝑖}(1/|𝛼|)=∞, then by Remark 2  𝜑_𝑖=0.

Corollary 10. Let 𝜑 be an integrable radial function on 𝐵_𝑛, such that 𝑇_𝜑 is a bounded operator. Then 𝑇²_𝜑=𝑇_𝜑 if and only if either 𝜑=0 or 𝜑=1.

Proof. If 𝑇²_𝜑=𝑇_𝜑, then 𝑇_𝜑𝑇_𝜑−1=0. By Theorem 9 we have 𝜑=0 or 𝜑=1. The converse is clear.

4. Product of 𝑛 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 𝑝,𝑠≽0. A function 𝑓∈𝐿¹(𝐵_𝑛,𝑑𝑣) is called a quasihomogeneous function of degree (𝑝,𝑠) if 𝑓 is of the form 𝜉^𝑝𝜉^𝑠𝜑, where 𝜑 is a radial function, that is, 𝑓(𝑟𝜉)=𝜉^𝑝𝜉^𝑠𝜑(𝑟), (32) for any 𝜉 in the unit sphere 𝑆_𝑛 and 𝑟∈[0,1).

Lemma 12. Let 𝑝,𝑠 be two multi-indexes and let 𝜑 be an integrable radial function on 𝐵_𝑛, such that 𝑇_𝜉^𝑝_𝜑,𝑇_𝜉^𝑝_𝜑, and 𝑇_𝜉𝑝𝜉^𝑠_𝜑 are bounded operators. Then for any multi-index 𝛼, 𝑇_𝜉𝑝𝜉^𝑠_𝜑𝑧^𝛼=2𝑝+𝛼!𝑛+𝑝+|𝛼|−|𝑠|!!|𝛼|+𝑝!×𝜑|𝛼|+𝑝−|𝑠|𝑧𝑝+𝛼−𝑠𝑛−1+2𝑛+2^{𝑝+𝛼−𝑠}|𝑠|−|𝛼|−𝑝!𝑝+𝛼≽𝑠,2𝑠!𝑛+!(|𝑠!×𝜑|𝑠|−𝑝𝑠−𝛼−𝑝𝑛−1+|)2𝑛+𝑧^{𝑠−𝛼−𝑝}0̸𝑇𝑝+𝛼⪯𝑠,𝑝+𝛼≥𝑠,𝑝+𝛼≰𝑠,_{𝜉𝑝𝜉𝑠𝜑}𝑧^𝛼=2(𝑠+𝛼)!𝑛+|𝑠|+|𝛼|−𝑝!!(|𝛼|+|𝑠!×𝜑|𝛼|+|𝑠|−𝑝𝑠+𝛼−𝑝𝑛−1+|)2𝑛+2𝑧^{𝑠+𝛼−𝑝}𝑝−|𝛼|−|𝑠|!𝑠+𝛼≽𝑝,2𝑝!𝑛+!𝑝!×𝜑𝑝−|𝑠𝑧𝑝−𝛼−𝑠𝑛−1+2𝑛+|^{𝑝−𝛼−𝑠}0̸𝑠+𝛼⪯𝑝,𝑠+𝛼≥𝑝,𝑠+𝛼≰𝑝. (33) Especially, if 𝑝,𝑠 are two nonzero multi-indexes with 𝑝⟂𝑠, one has 𝑇_𝜉𝑝𝜉^𝑠_𝜑𝑧^𝛼=2𝑝+𝛼!𝑛+𝑝+|𝛼|−|𝑠|!!|𝛼|+𝑝!×𝜑|𝛼|+𝑝−|𝑠|𝑧𝑝+𝛼−𝑠𝑛−1+2𝑛+2^{𝑝+𝛼−𝑠}̸𝑇𝛼≽𝑠,0𝛼≥𝑠._{𝜉𝑝𝜉𝑠𝜑}𝑧^𝛼=2(𝑠+𝛼)!𝑛+|𝑠|+|𝛼|−𝑝!!(|𝛼|+|𝑠!×𝜑|𝛼|+|𝑠|−𝑝𝑠+𝛼−𝑝𝑛−1+|)2𝑛+2𝑧^{𝑠+𝛼−𝑝}̸𝛼≽𝑝,0𝛼≥𝑝. (34)(35)

Proof. Here we only give the proof of the first equation. For multi-indexes 𝛼 and 𝛽, 𝑃𝜉^𝑝𝜉^𝑠𝜑𝑧^𝛼,𝑧^𝛽=_𝐵𝑛𝜉^𝑝𝜉^𝑠𝜑(𝑧)𝑧^𝛼𝑧^𝛽(𝑧)=𝑑𝑣_[0,1)2𝑛𝑟^{2𝑛+|𝛼|+|𝛽|−1}𝜑(𝑟)𝑑𝑟_𝑆𝑛𝜉^𝑝𝜉^𝑠𝜉^𝛼𝜉^𝛽(𝜉)=!𝑑𝜎2𝑛!𝑝+𝛼|𝛼|+𝑝!×𝜑|𝛼|+𝑝−|𝑠|0̸𝑛−1+2𝑛+2𝑝+𝛼≽𝑠,𝛽=𝑝+𝛼−𝑠,𝑝+𝛼≽𝑠,𝛽≠𝑝+𝛼−𝑠,0𝑝+𝛼≥𝑠. (36) If 𝑝+𝛼≽𝑠, we have 𝑧^{𝑝+𝛼−𝑠},𝑧^𝛽=_𝐵𝑛𝑧^{𝑝+𝛼−𝑠}𝑧^𝛽(𝑧)=𝑑𝑣_[0,1)2𝑛𝑟^{2𝑛+|𝑝|+|𝛼|−|𝑠|+|𝛽|−1}×𝑑𝑟_𝑆𝑛𝜉^{𝑝+𝛼−𝑠}𝜉^𝛽(𝜉)=!𝑑𝜎𝑛!𝑝+𝛼−𝑠𝑛+|𝛼|+𝑝−|𝑠|!𝛽=𝑝+𝛼−𝑠,0𝛽≠𝑝+𝛼−𝑠. (37) Furthermore, ⟨𝑃[𝜉^𝑝𝜉^𝑠𝜑𝑧^𝛼],𝑧^𝛽⟩=0=⟨𝑧^{𝑝+𝛼−𝑠},𝑧^𝛽⟩ hold for all 𝛽≻0. So we obtain 𝑃𝜉^𝑝𝜉^𝑠𝜑𝑧^𝛼=2𝑝+𝛼!𝑛+𝑝+|𝛼|−|𝑠|!!|𝛼|+𝑝!×𝜑|𝛼|+𝑝−|𝑠|𝑧𝑝+𝛼−𝑠𝑛−1+2𝑛+2^{𝑝+𝛼−𝑠}̸𝑝+𝛼≽𝑠,0𝑝+𝛼≥𝑠. (38)
Note that 𝜑 is still radial and 𝜑(𝑛)=𝜑(𝑛) for all 𝑛∈𝑁, we have 𝑃𝜉^𝑝𝜉^𝑠𝜑𝑧^𝛼=2𝑠!𝑛+|𝑠|−|𝛼|−𝑝!!(|𝑠!×𝑠−𝛼−𝑝𝑛−1+|)𝜑2𝑛+|𝑠|−𝑝𝑧^{𝑠−𝛼−𝑝}𝑝+𝛼⪯𝑠,0𝑝+𝛼≰𝑠. (39) Thus 𝑇_𝜉𝑝𝜉^𝑠_𝜑𝑧^𝛼=𝑃𝜉^𝑝𝜉^𝑠𝜑𝑧^𝛼+𝑃𝜉^𝑝𝜉^𝑠𝜑𝑧^𝛼−𝑃𝜉^𝑝𝜉^𝑠𝜑𝑧^𝛼0)=2!𝑝+|𝛼|−|𝑠|!(𝑝+𝛼𝑛+!|𝛼|+𝑝!×𝜑|𝛼|+𝑝−|𝑠𝑧𝑝+𝛼−𝑠𝑛−1+2𝑛+2|^{𝑝+𝛼−𝑠}|𝑠|−|𝛼|−𝑝!𝑝+𝛼≽𝑠,2𝑠!𝑛+!(|𝑠!×𝜑|𝑠|−𝑝𝑠−𝛼−𝑝𝑛−1+|)2𝑛+𝑧^{𝑠−𝛼−𝑝}0̸𝑝+𝛼⪯𝑠,𝑝+𝛼≰𝑠,𝑝+𝛼≥𝑠. (40)
Especially, if 𝑝⟂𝑠, we know that 𝑝+𝛼≽𝑠 is equivalent to 𝛼≽𝑠. Since 𝑝,𝑠 are nonzero multi-indexes with 𝑝⟂𝑠, there exists 𝑖∈{1,2,…,𝑛} such that 𝑝_𝑖≠0,𝑠_𝑖=0. This means that there doesn’t exist multi-index 𝛼 such that 𝑝+𝛼⪯𝑠. So 𝑇_𝜉𝑝𝜉^𝑠_𝜑𝑧^𝛼=2𝑝+𝛼!𝑛+𝑝+|𝛼|−|𝑠|!!|𝛼|+𝑝!×𝜑|𝛼|+𝑝−|𝑠|𝑧𝑝+𝛼−𝑠𝑛−1+2𝑛+2^{𝑝+𝛼−𝑠}̸𝛼≽𝑠,0𝛼≥𝑠. (41)

Theorem 13. Let 𝑝_𝑖,𝑠_𝑖 be multi-indexes and let 𝜑_𝑖 be integrable radial functions on 𝐵_𝑛 such that 𝑇_{𝜉𝑝𝑖𝜉}^𝑠_𝑖𝜑𝑖 is bounded 𝑖=1,2,…,𝑚. If the product 𝑇_{𝜉𝑝𝑚𝜉}^𝑠𝑚_𝜑𝑚⋯𝑇_{𝜉𝑝1𝜉}^𝑠_1𝜑1 is of finite rank, then 𝜑_𝑖=0 for some 𝑖∈{1,2,…,𝑚}.

Proof. We denote the product of Toeplitz operators 𝑇_{𝜉𝑝𝑚𝜉}^𝑠𝑚_𝜑𝑚⋯𝑇_{𝜉𝑝1𝜉}^𝑠_1𝜑1 by 𝑆. By Lemma 12, we have 𝑆𝑧^𝛼=²_𝑖=𝑚𝑇_{𝜉𝑝𝑖𝜉}^𝑠_𝑖𝜑𝑖×2𝑝₁+𝛼!𝑛+𝑝₁+|𝛼|−𝑠₁!𝑝₁+𝛼−𝑠₁!𝑛−1+|𝛼|−𝑝₁!×𝜑₁2𝑛+2|𝛼|+𝑝₁−𝑠₁𝑧^{𝑝1+𝛼−𝑠1}𝑝₁+𝛼≽𝑠₁,2𝑠₁!𝑛+𝑠₁−𝑝₁−|𝛼|!𝑠₁−𝑝₁−𝛼!𝑛−1+𝑠₁!×𝜑₁2𝑛+𝑠₁−𝑝₁𝑧^{𝑠1−𝛼−𝑝1}𝑝₁+𝛼⪯𝑠₁,0𝑝₁+𝛼≥𝑠̸₁,𝑝₁+𝛼≥s̸₁. (42) If 𝑝₁+𝛼≽𝑠₁,𝑝₂+𝑝₁+𝛼≽𝑠₁+𝑠₂,…,𝑝_𝑚+⋯+𝑝₁+𝛼≽𝑠₁+⋯+𝑠_𝑚, we have 𝑆𝑧^𝛼=2^𝑚𝑎(𝛼)𝑏𝛼×(||)_𝑚𝑖=1𝜑_𝑖2𝑛+2|𝛼|+^𝑖−1_𝑗=0𝑝_𝑗−𝑠_𝑗+𝑝_𝑖−𝑠_𝑖×𝑧^{𝛼+∑𝑚𝑖=1𝑝𝑖−𝑠𝑖}, (43) where 𝑎(𝛼)=∏^𝑚_𝑖=1𝛼+∑^𝑖−1_𝑗=0𝑝_𝑗−𝑠_𝑗+𝑝_𝑖!∏^𝑚_𝑖=1𝛼+∑^𝑖_𝑗=1𝑝_𝑗−𝑠_𝑗!𝑏𝛼=∏,(||)^𝑚𝑖=1𝑛+|𝛼|+∑^𝑖_𝑗=1𝑝_𝑗−𝑠_𝑗!∏^𝑚_𝑖=1𝑛−1+|𝛼|+∑^𝑖−1_𝑗=0𝑝_𝑗−𝑠_𝑗+𝑝_𝑖!, (44)(45)𝑝₀=0 and 𝑠₀=0.
Similarly, we have 𝑆z^𝛼=²_𝑖=𝑚𝑇_{𝜉𝑝𝑖𝜉}^𝑠_𝑖𝜑𝑖×2𝑠₁+𝛼!𝑛+𝑠₁+|𝛼|−𝑝₁!𝑠₁+𝛼−𝑝₁!𝑛−1+|𝛼|−𝑠₁!×𝜑₁2𝑛+2|𝛼|+𝑠₁−𝑝₁𝑧^{𝑠1+𝛼−𝑝1}𝑠₁+𝛼≽𝑝₁,2𝑝₁!𝑛+𝑝₁−𝑠₁−|𝛼|!𝑝₁−𝑠₁−𝛼!𝑛−1+𝑝₁!×𝜑₁2𝑛+𝑝₁−𝑠₁𝑧^{𝑝1−𝛼−𝑠1}𝑠₁+𝛼⪯𝑝₁,0s₁+𝛼≥𝑝̸₁,𝑠₁+𝛼≥𝑝̸₁. (46) Then for 𝑠₁+𝛼≽𝑝₁,𝑠₂+𝑠₁+𝛼≽𝑝₁+𝑝₂,…𝑠_𝑚+⋯+𝑠₁+𝛼≽𝑝₁+⋯+𝑝_𝑚, we have 𝑆𝑧^𝛼=2^𝑚𝑎^′(𝛼)𝑏^′𝛼×(||)_𝑚𝑖=1𝜑_𝑖2𝑛+2|𝛼|+^𝑖−1_𝑗=0𝑠_𝑗−𝑝_𝑗+𝑠_𝑖−𝑝_𝑖×𝑧^{𝛼+∑𝑚𝑖=1𝑠𝑖−𝑝𝑖}, (47) where 𝑎^′(𝛼)=∏^𝑚_𝑖=1𝛼+∑^𝑖−1_𝑗=0𝑠_𝑗−𝑝_𝑗+𝑠_𝑖!∏^𝑚_𝑖=1𝛼+∑^𝑖_𝑗=1𝑠_𝑗−𝑝_𝑗!,𝑏^′𝛼=∏(||)^𝑚𝑖=1𝑛+|𝛼|+∑^𝑖_𝑗=1∑𝑠_𝑗−𝑝_𝑗!∏^𝑚_𝑖=1𝑛−1+|𝛼|+∑^𝑖−1_𝑗=0𝑠_𝑗−𝑝_𝑗+𝑠_𝑖!. (48)
Since 𝑆 has finite rank, we know that {𝑆(𝑧^𝛼)}_𝛼≽0 and {𝑆(𝑧^𝛼)}_𝛼≻0 must have finite rank. Then, there exists a multi-index 𝛼₀ such that 𝑝₁+𝛼₀≽𝑠₁,𝑝₂+𝑝₁+𝛼₀≽𝑠₁+𝑠₂,…,𝑝_𝑚+⋯+𝑝₁+𝛼₀≽𝑠₁+⋯+𝑠_𝑚 and 𝑆𝑧^𝛼=0∀𝛼≽𝛼₀, (49) which is equivalent to 𝜑_𝑚2𝑛+2|𝛼|+2^𝑚−1_𝑗=0𝑝_𝑗−𝑠_𝑗+𝑝_𝑚−𝑠_𝑚⋯𝜑₁2𝑛+2|𝛼|+𝑝₁−𝑠₁=0, (50) for all 𝛼≽𝛼₀. In fact, we have 𝛼₀=(𝑡₁,…,𝑡_𝑛), in which 𝑡_𝑗=max{𝑘₁^(𝑗),𝑘₂^(𝑗),…,𝑘_𝑚^(𝑗)} and 𝑘^𝑗_𝑖=max{(𝑠₁^(𝑗)−𝑝₁^(𝑗)+𝑠₂^(𝑗)−𝑝₂^(𝑗)+⋯+𝑠_𝑖^(𝑗)−𝑝_𝑖^(𝑗)),0}, 𝑠_𝑖=(𝑠_𝑖⁽¹⁾,…,𝑠_𝑖^(𝑛)), 𝑝_𝑖=(𝑝_𝑖⁽¹⁾,…,𝑝_𝑖^(𝑛)), 𝑖∈{1,2,…,𝑚}, 𝑗∈{1,2,…,𝑛}.
Let 𝑙=min{|𝑝₁|−|𝑠₁|,2(|𝑝₁|−|𝑠₁|)+|𝑝₂|−|𝑠₂|,…,2∑^𝑚−1_𝑗=1(|𝑝_𝑗|−|𝑠_𝑗|)+|𝑝_𝑚|−|𝑠_𝑚|}, we know that 2|𝛼|+𝑙≥0,forall𝛼≽𝛼₀. Then (50) implies that 𝑟^{2(∑𝑚−1𝑗−1(|𝑝𝑗|−|𝑠𝑗|))+|𝑝𝑚|−|𝑠𝑚|−𝑙}𝜑_𝑚^⋀(|𝛼|)⋯𝑟2𝑛+2+𝑙_{|𝑝1|−|𝑠1|−𝑙}𝜑₁^⋀(2𝑛+2|𝛼|+𝑙)=0, (51) for all 𝛼≽𝛼₀. Since {2𝑛+2|𝛼|+𝑙}_𝛼≽𝛼0 is an arithmetic sequence, by Theorem 1, we obtain that there exists some 𝑖 such that 𝜑_𝑖=0.

5. Finite Rank Commutators

In this section, let 𝜑,𝜓 be two integrable radial functions on 𝐵_𝑛. We now pass to investigate the commutator [𝑇_𝜓,𝑇_𝜉𝑝𝜉^𝑠_𝜑],[𝑇_𝜉^𝑝_𝜑,𝑇_𝜉^𝑠_𝜓] and [𝑇_𝜉^𝑝_𝜑,𝑇_𝜉^𝑠_𝜓].

Theorem 14. Let 𝑝,𝑠 be two multi-indexes and let 𝜑 and 𝜓 be two integrable radial functions on 𝐵_𝑛 such that 𝑇_𝜓 and 𝑇_𝜉𝑝𝜉^𝑠_𝜑 are bounded operators. If 𝜓 is nonconstant and [𝑇_𝜓,𝑇_𝜉𝑝𝜉^𝑠_𝜑] has finite rank, then its rank is equal to zero.

Proof. Let 𝑆 denote the commutator [𝑇_𝜓,𝑇_𝜉𝑝𝜉^𝑠_𝜑]. Since 𝑆 has finite rank on 𝐿²_ℎ(𝐵_𝑛), we know that {𝑆(𝑧^𝛼)}_𝛼≽0 and {𝑆(𝑧^𝛼)}_𝛼≻0 must have finite rank.
By Lemma 12, we have if 𝑝+𝛼≽𝑠, 𝑆𝑧^𝛼=2𝑝+𝛼!𝑛+𝑝+|𝛼|−|𝑠|!!𝑝+|𝛼|!×𝜑|𝛼|+𝑝−|𝑠|×2𝑝+|𝛼|−|𝑠|𝜓𝑝|𝛼||𝑠|(|𝛼𝜓(|𝛼𝑧𝑝+𝛼−𝑠𝑛−1+2𝑛+2𝑛+2𝑛+2+2−2−2𝑛+|)2𝑛+2|)^{𝑝+𝛼−𝑠}. (52) If 𝑝+𝛼⪯𝑠, 𝑆𝑧^𝛼=2𝑠!𝑛+|𝑠|−|𝛼|−𝑝!!(|𝑠!𝜑|𝑠|−𝑝×2|𝑠|−|𝛼|−𝑝𝜓|𝑠||𝛼|𝑝(|𝛼𝜓(|𝛼𝑠−𝛼−𝑝𝑛−1+|)2𝑛+𝑛+2𝑛+2−2−2−2𝑛+|)2𝑛+2|)𝑧^{𝑠−𝛼−𝑝}, (53) if 𝑝+𝛼≥𝑠̸, 𝑝+𝛼≰𝑠, we have 𝑆(𝑧^𝛼)=0.
Since {𝑆(𝑧^𝛼)}_𝛼≽0 has finite rank, then there must exist 𝛼₀, such that 𝑝+𝛼₀≽𝑠 and 𝑆(𝑧^𝛼)=0 for all 𝛼≽𝛼₀.
Similar to the proof of Theorem  4.4 in [7], we can imply 𝜑=0 or |𝑝|=|𝑠|. It follows from (52) and (53) that 𝑆𝑧^𝛼=0∀𝛼≽0. (54)
Similar to the discussions for 𝑆(𝑧^𝛼), we can imply that 𝑆(𝑧^𝛼)=0 for all 𝛼≽0. In conclusion, the rank of 𝑆 is equal to zero.

Theorem 15. Let 𝑝,𝑠 be two nonzero multi-indexes, let 𝜑 and 𝜓 be two integrable radial functions on 𝐵_𝑛, such that 𝑇_𝜉^𝑝_𝜑 and 𝑇_𝜉^𝑠_𝜓 are bounded operators. If [𝑇_𝜉^𝑝_𝜑,𝑇_𝜉^𝑠_𝜓] has finite rank, then the rank of it is less than or equal to the number of 𝛼 such that 𝛼⪯𝑝+𝑠, and {[𝑇_𝜉^𝑝_𝜑,𝑇_𝜉^𝑠_𝜓](𝑧^𝛼)}_𝛼≽0⋃{[𝑇_𝜉^𝑝_𝜑,𝑇_𝜉^𝑠_𝜓](𝑧^𝛼)}_𝛼≻0⊂⋁_{𝛼⪯𝑝+𝑠}{𝑧^{𝑝+𝑠−𝛼}}, where ⋁_{𝛼⪯𝑠+𝑝}{𝑧^{𝑠+𝑝−𝛼}}=span{𝑧^{𝑠+𝑝−𝛼}∶𝛼⪯𝑠+𝑝}.

Proof. Let 𝑆 denote the commutator [𝑇_𝜉^𝑝_𝜑,𝑇_𝜉^𝑠_𝜓]. By Lemma 12, we have 𝑆𝑧^𝛼|𝛼|+𝑝+|𝑠×2(|𝛼|+|𝑠𝜓(|𝛼|+|𝑠×𝜑|𝛼|+𝑝|𝑠||𝛼|+𝑝×𝜓|𝛼|𝑝+|𝑠|×𝜑|𝛼|+𝑝𝑧=2𝑛+|𝑛+|)2𝑛+2|)2𝑛+2+2−2𝑛+2𝑛+2+22𝑛+2^{𝑝+𝛼+𝑠}. (55)
If {𝑆(𝑧^𝛼)}_𝛼≽0 has finite rank, there must exist 𝑘≥0 such that 𝑆𝑧^𝛼=0,∀|𝛼|≥𝑘. (56) Then, we deduce that 1𝑝|𝛼|𝜑|𝑠||𝛼|+𝑝×𝜓(|𝑠||𝛼=1(|𝑠||𝛼𝜑|𝛼|+𝑝×𝜓|𝑠||𝛼|𝑝,|𝛼|2𝑛+2+22𝑛+2+22𝑛++2|)2𝑛+2+2|)2𝑛+22𝑛++2+2≥𝑘. (57) It follows from Remark 2 that 𝑟^2|𝑝|1(𝑧)𝑟^{2|𝑠|+|𝑝|}𝜑(𝑧)𝑟^|𝑠|𝜓(𝑧)=𝑟^2|𝑠|1(𝑧)𝑟^|𝑝|𝜑(𝑧)𝑟^{2|𝑝|+|𝑠|}𝜓(𝑧), (58) for all Re𝑧>2𝑘. Set 𝑧=2𝑛+2|𝛼|, we get that (57) holds for all |𝛼|≥0. Hence for all 𝛼≽0, we have 𝑆(𝑧^𝛼)=0. Therefore, the rank of {𝑆(𝑧^𝛼)}_𝛼≽0 is equal to zero.
For 𝑆(𝑧^𝛼), we have if 𝛼≽𝑝+𝑠, 𝑆𝑧^𝛼=4𝛼!𝑛+|𝛼|−𝑝−|𝑠|!!(|𝛼!×(|𝛼|−|𝑠!𝛼−𝑝−𝑠𝑛−1+|)𝑛+|)(|𝛼|−|𝑠!𝜓(|𝛼|−|𝑠×𝜑|𝛼||𝑠|−𝑝−|𝛼|−𝑝!𝑛−1+|)2𝑛+2|)2𝑛+2−2𝑛+|𝛼|−𝑝!×𝜓|𝛼|𝑝−|𝑠|×𝜑|𝛼|−𝑝𝑛−1+2𝑛+2−22𝑛+2𝑧^{𝛼−𝑝−𝑠}. (59) If 𝛼≥𝑝̸,  𝛼≰𝑝,  𝛼≥𝑠̸,  𝛼≰𝑠, we have 𝑆(𝑧^𝛼)=0, if 𝛼≽𝑝 but 𝛼≥𝑝̸+𝑠,  𝛼≰𝑝+𝑠, we have 𝑆(𝑧^𝛼)=0, if 𝛼≽𝑠 but 𝛼≥𝑝̸+𝑠,  𝛼≰𝑝+𝑠, then 𝑆(𝑧^𝛼)=0.
If 𝑠⪯𝛼⪯𝑝+𝑠 or 𝑝⪯𝛼⪯𝑝+𝑠 or 𝛼⪯𝑝 or 𝛼⪯𝑠, 𝑆𝑧^𝛼=𝐶(𝛼)𝑧^{𝑝+𝑠−𝛼}, (60) where 𝐶(𝛼) is a constant that depends on 𝛼.
If {𝑆(𝑧^𝛼)}_𝛼≻0 has finite rank, use the same method as above, we have 𝑆𝑧^𝛼=0,∀𝛼≽𝑝+𝑠, (61) this means {𝑆(𝑧^𝛼)}_𝛼≽0⊂⋁_{𝛼⪯𝑝+𝑠}{𝑧^{𝑝+𝑠−𝛼}}, and the rank of 𝑆(𝑧^𝛼) is at most equal to the number of 𝛼 such that 𝛼∈𝐺. In which 𝐺={𝛼∶𝑠⪯𝛼⪯𝑝+𝑠}⋃{𝛼∶𝑝⪯𝛼⪯𝑝+𝑠}⋃{𝛼∶𝛼⪯𝑝}⋃{𝛼∶𝛼⪯𝑠} and 𝐺⊆{𝛼⪯𝑝+𝑠}. The proof is complete.

Example 16. In the case of 𝑛=2, we give an example to Theorem 15.
From the proof of Theorem 15 we know that 𝑇_𝜉^𝑝_𝑟𝑚𝑇_𝜉^𝑠_𝑓𝑧^𝛼=𝑇_𝜉^𝑠_𝑓𝑇_𝜉^𝑝_𝑟𝑚𝑧^𝛼,∀𝛼≽0. (62) Below we will construct a radial function 𝑓, such that 𝑇_𝜉^𝑝_𝑟𝑚𝑇_𝜉^𝑠_𝑓𝑧^𝛼=𝑇_𝜉^𝑠_𝑓𝑇_𝜉^𝑝_𝑟𝑚𝑧^𝛼,∀𝛼≽𝑝+𝑠, (63) where 𝑝,𝑠 are multi-indexes, 𝜉∈𝑆₂ and 𝑚≥0.
Equation (63) implies that for 𝛼≽𝑝+𝑠, 𝑟^{−|𝑠|−2|𝑝|}𝑓2|𝛼|+4+2𝑝𝑟^{−|𝑠|−2|𝑝|}𝑓(2|𝛼|)=2|𝛼|𝑝2|𝛼|𝑝|𝑠|+4+4−2+4−+𝑚−2(2|𝛼||𝑠2|𝛼|𝑝.+4−2|)+4−+𝑚 (64) Now, using Remark 2, we obtain that 𝑟^{−|𝑠|−2|𝑝|}𝑓𝑧+2𝑝𝑟^{−|𝑠|−2|𝑝|}𝑓(𝑧)=𝑧−2𝑝𝑧−𝑝+𝑚−2|𝑠|(|𝑠𝑝,𝑧−2|)𝑧−+𝑚 (65) for all Re𝑧≥2|𝑝|+2|𝑠|+4.
Let 𝐹 be the analytic function defined for Re𝑧≥2|𝑝|+2|𝑠| by 𝐹(𝑧)=Γ𝑧−2𝑝/2𝑝Γ𝑧−𝑝−2|𝑠|+𝑚/2𝑝Γ(𝑧−2|𝑠|)/2𝑝Γ𝑧−𝑝+𝑚/2𝑝, (66) where Γ denote the gamma function. Then using the well-known identity Γ(𝑧+1)=𝑧Γ(𝑧), (65) implies that 𝑟^{−|𝑠|−2|𝑝|}𝑓𝑧+2𝑝𝑟^{−|𝑠|−2|𝑝|}𝑓(𝑧)=𝐹𝑧+2𝑝𝐹(𝑧),forRe𝑧>2𝑝+2|𝑠|. (67) Equation (67) combined with [30, Lemma 6, page 1468] gives us that there exists a constant 𝑐 such that 𝑟^{−|𝑠|−2|𝑝|}𝑓(𝑧)=𝑐𝐹(𝑧)forRe𝑧>2𝑝+2|𝑠|. (68)
For a choice of 𝑝=(2,0),𝑠=(0,1), and 𝑚=6, again using the identity Γ(𝑧+1)=𝑧Γ(𝑧), one can see that 𝐹(𝑧)=4(𝑧−2)𝑧()1𝑧−4=2𝑧+1.𝑧−4 (69) Since 1(𝑧)=1/𝑧, 𝑟⁻⁴(𝑧)=1/(𝑧−4), then (68) becomes 𝑟⁻⁵𝑓(𝑧)=𝑐1(𝑧)+𝑟⁻⁴(𝑧),forRe𝑧>6. (70) Now the proceeding equation and Remark 2 imply that 𝑓(𝑟)=𝑐𝑟⁵+𝑟,where𝑐isaconstant. (71) It is clear that the function 𝑓 is bounded, so Toeplitz operator 𝑇_𝜉^(0,1)_𝑓 is bounded.
Finally, by taking the constant 𝑐 to be equal to 1, the radial function 𝑓(𝑟)=𝑟⁵+𝑟 satisfies 𝑇_𝜉^(2,0)_𝑟6𝑇_𝜉^(0,1)_(𝑟5+𝑟)𝑧^𝛼=𝑇_𝜉^(0,1)_(𝑟5+𝑟)𝑇_𝜉^(2,0)_𝑟6𝑧^𝛼𝑇,∀𝛼≽0,_{𝜉(2,0)𝑟6}𝑇_𝜉^(0,1)_(𝑟5+𝑟)𝑧^𝛼=𝑇_𝜉^(0,1)_(𝑟5+𝑟)𝑇_𝜉^(2,0)_𝑟6𝑧^𝛼,∀𝛼≽(2,1). (72) For 𝛼=(2,1),𝛼₁=(1,1),𝛼₂=(0,1),𝛼₃=(2,0),𝛼₄=(1,0), we know that 𝛼_𝑖⪯𝛼,(𝑖=1,2,3,4). However, using Lemma 12, it is easy to see that 𝑇_𝜉^(2,0)_𝑟6𝑇_𝜉^(0,1)_(𝑟5+𝑟)𝑧^(1,1)=5𝑧36^(1,0),𝑇_𝜉^(0,1)_(𝑟5+𝑟)𝑇_𝜉^(2,0)_𝑟6𝑧^(1,1)𝑇=0,_{𝜉(2,0)𝑟6}𝑇_𝜉^(0,1)_(𝑟5+𝑟)𝑧^(0,1)=16𝑧45^(2,0),𝑇_𝜉^(0,1)_(𝑟5+𝑟)𝑇_𝜉^(2,0)_𝑟6𝑧^(0,1)𝑇=0,_{𝜉(2,0)𝑟6}𝑇_𝜉^(0,1)_(𝑟5+𝑟)𝑧^(2,0)𝑇=0,_{𝜉(0,1)(𝑟5+𝑟)}𝑇_𝜉^(2,0)_𝑟6𝑧^(2,0)=8𝑧45^(0,1),𝑇_𝜉^(2,0)_𝑟6𝑇_𝜉^(0,1)_(𝑟5+𝑟)𝑧^(1,0)𝑇=0,_{𝜉(0,1)(𝑟5+𝑟)}𝑇_𝜉^(2,0)_𝑟6𝑧^(1,0)=59𝑧^(1,1). (73)(74) Therefore the commutator [𝑇_𝜉^(2,0)_𝑟6,𝑇_𝜉^(0,1)_(𝑟5+𝑟)] has rank four.

Theorem 17. Let 𝑝,𝑠 be two nonzero multi-indexes with 𝑝⟂𝑠, let 𝜑 and 𝜓 be two integrable radial functions on 𝐵_𝑛, such that 𝑇_𝜉^𝑝_𝜑 and 𝑇_𝜉^𝑠_𝜓 are bounded operators. If [𝑇_𝜉^𝑝_𝜑,𝑇_𝜉^𝑠_𝜓] has finite rank, then the rank of it is equal to zero.

Proof. Let 𝑆 denote the commutator [𝑇_𝜉^𝑝_𝜑,𝑇_𝜉^𝑠_𝜓]. Since 𝑝⟂𝑠, it is easy to see that 𝑝+𝛼≽𝑠 is equivalent to 𝛼≽𝑠 for all multi-index 𝛼, and there does not exist multi-index 𝛼^′ such that 𝑝+𝛼^′⪯𝑠. Furthermore, it is impossible that there exists 𝛼₀ such that 𝛼₀⪯𝑠 and 𝑠⪯𝛼₀+𝑝.
If {𝑆(𝑧^𝛼)}_𝛼≽0 has finite rank, there exists 𝛼₀≽0 such that 𝑆𝑧^𝛼=0,∀𝛼≽𝑠+𝛼₀. (75) Since 𝑝⟂𝑠, we have 𝛼!(𝑝+𝛼−𝑠)!=(𝛼+𝑝)!(𝛼−𝑠)!. So 𝑆𝑧^𝛼=𝛼!()!(|𝛼|−|𝑠!𝛼−𝑠𝑛+|)(|𝛼!2|𝛼|+𝑝−|𝑠|×𝜓(|𝛼|−|𝑠×𝜑|𝛼||𝑠|+𝑝−|𝛼|+𝑝−|𝑠|!𝑛−1+|)𝑛+2𝑛+2|)2𝑛+2−2𝑛+𝑝+|𝛼|!2|𝛼|+𝑝×𝜓|𝛼|𝑝−|𝑠|×𝜑|𝛼|+𝑝𝑧𝑛−1+𝑛+2𝑛+2+22𝑛+2^{𝑝+𝛼−𝑠}. (76) As in the proof of Theorem 15, we have 𝑆(𝑧^𝛼)=0 for all 𝛼≽𝑠, this means that the rank of {𝑆(𝑧^𝛼)}_𝛼≽0 is equal to zero.
Similarly, we can prove that the rank of {𝑆(𝑧^𝛼)}_𝛼≻0 is equal to zero.

6. Finite Rank Semicommutators

We will discuss the semicommutators of two Toeplitz operators with quasihomogeneous symbols, the results in this section are different from those in [9, 14].

Theorem 18. Let 𝑝,𝑠 be two multi-indexes and let 𝜑 and 𝜓 be two integrable radial functions on 𝐵_𝑛, such that 𝑇_𝜓,𝑇_𝜉𝑝𝜉^𝑠_𝜑 are bounded operators.(a)If 𝑠≽𝑝, (𝑇_𝜓,𝑇_𝜉𝑝𝜉^𝑠_𝜑] has finite rank, then its rank is at most equal to the number of 𝛼 such that 𝛼⪯𝑠−𝑝 and {(𝑇_𝜓,𝑇_𝜉𝑝𝜉^𝑠_𝜑](𝑧^𝛼)}_𝛼≽0⋃{(𝑇_𝜓,𝑇_𝜉𝑝𝜉^𝑠_𝜑](𝑧^𝛼)}_𝛼≻0⊆⋁_{𝛼⪯𝑠−𝑝}{𝑧^{𝑠−𝑝−𝛼}}.(b)If 𝑠⪯𝑝, (𝑇_𝜓,𝑇_𝜉𝑝𝜉^𝑠_𝜑] has finite rank, then its rank is at most equal to the number of 𝛼 such that 𝛼⪯𝑝−𝑠 and {(𝑇_𝜓,𝑇_𝜉𝑝𝜉^𝑠_𝜑](𝑧^𝛼)}_𝛼≽0⋃{(𝑇_𝜓,𝑇_𝜉𝑝𝜉^𝑠_𝜑](𝑧^𝛼)}_𝛼≻0⊆⋁_{𝛼⪯𝑝−𝑠}{𝑧^{𝑝−𝑠−𝛼}}.(c)If 𝑠≥𝑝̸,  𝑠≰𝑝, and (𝑇_𝜓,𝑇_𝜉𝑝𝜉^𝑠_𝜑] has finite rank, then its rank is equal to zero.

Proof. Let 𝑆 denote the semicommutator (𝑇_𝜓,𝑇_𝜉𝑝𝜉^𝑠_𝜑].
If 𝑠+𝛼≽𝑝, we compute that 𝑆𝑧^𝛼=2(𝑠+𝛼)!𝑛+|𝑠|+|𝛼|−𝑝!!(|𝑠|+|𝛼!×|𝛼|+|𝑠|−𝑝|𝑠|+|𝛼|−𝑝×𝜓|𝑠||𝛼|𝑝×𝜑|𝛼|+|𝑠|−𝑝𝑠+𝛼−𝑝𝑛−1+|)𝜑𝜓2𝑛+2−2𝑛+2𝑛+2+2−22𝑛+2𝑧^{𝑠+𝛼−𝑝}. (77)
If 𝑠+𝛼⪯𝑝, 𝑆𝑧^𝛼=2𝑝!𝑛+𝑝−|𝛼|−|𝑠|!!𝑝!×𝑝−|𝑠|𝑝−|𝛼|−|𝑠|×𝜓𝑝|𝛼||𝑠|×𝜑𝑝−|𝑠|𝑧𝑝−𝛼−𝑠𝑛−1+𝜑𝜓2𝑛+−2𝑛+2𝑛+2−2−22𝑛+^{𝑝−𝛼−𝑠}, (78) if 𝑠+𝛼≥𝑝̸ and 𝑠+𝛼≰𝑝, then 𝑆(𝑧^𝛼)=0. Next we consider three cases.
Case  1 (𝑝≽𝑠). If {𝑆(𝑧^𝛼)}_𝛼≻0 has finite rank, there exists 𝛼₀≽0 such that 𝑆𝑧^𝛼=0∀𝛼≽𝑝−𝑠+𝛼₀. (79) Similar to the proof of Theorem 15, we deduce that 𝑆(𝑧^𝛼)=0 for all 𝛼≽𝑝−𝑠. Therefore, the rank of {𝑆(𝑧^𝛼)]_𝛼≻0 is at most equal to the number of 𝛼 such that 𝛼⪯𝑝−𝑠, and {𝑆(𝑧^𝛼)}_𝛼≻0⊆⋁_{𝛼⪯𝑝−𝑠}{𝑧^{𝑝−𝑠−𝛼}}.
Case  2 (𝑠≽𝑝). Since 𝑠≽𝑝, we know that 𝑠+𝛼≽𝑝 for any multi-index 𝛼. This means that (77) for all 𝛼≽0. If {𝑆(𝑧^𝛼)}_𝛼≻0 has finite rank, again using the idea of proof of theorem, we conclude that the rank of {𝑆(𝑧^𝛼)}_𝛼≻0 is equal to zero.
Case  3 (𝑠≥𝑝̸,𝑠≰𝑝). For 𝑠=(𝑠₁,…,𝑠_𝑛), 𝑝=(𝑝₁,…,𝑝_𝑛), without loss of generality, we assume 𝑠₁<𝑝₁ but 𝑠_𝑖≥𝑝_𝑖(𝑖≥2). Since 𝑠₁<𝑝₁, hence 𝑝₁+𝛿>𝑠₁ for any nonnegative integer 𝛿. So there does not exist multi-index 𝛼, such that 𝑝+𝛼⪯𝑠. But if 𝛼′=(0,𝑠₂−𝑝₂,…,𝑠_𝑛−𝑝_𝑛), it is clear that 𝑝+𝛼^′≽𝑠. So for any 𝑠≥𝑝̸ and 𝑠≰𝑝, there exists 𝛼₀=(max{(𝑠₁−𝑝₁),0},…,max{(𝑠_𝑛−𝑝_𝑛),0}), such that 𝑝+𝛼₀≽𝑠, but there does not exist 𝛼^′ such that 𝑝+𝛼^′⪯𝑠. Hence, we have (77) holding for 𝛼≽𝛼₀, and for 𝛼≥𝛼̸₀𝑆(𝑧^𝛼)=0.
If {𝑆(𝑧^𝛼)}_𝛼≻0 has finite rank, there exists 𝛼^′≽0 such that 𝑆𝑧^𝛼=0,∀𝛼≽𝛼₀+𝛼^′. (80) As in the proof of Case 1, we can imply that 𝑆𝑧^𝛼=0,∀𝛼≽𝛼₀. (81) This means that the rank of {𝑆(𝑧^𝛼)}_𝛼≻0 is equal to zero.
Similar to the discussions for 𝑆(𝑧^𝛼), we obtain that(1)if 𝑠≽𝑝, the rank of 𝑆(𝑧^𝛼) is at most equal to the number of 𝛼 such that 𝛼⪯𝑠−𝑝, and {𝑆(𝑧^𝛼)}_𝛼≽0⊆⋁_{𝛼⪯𝑠−𝑝}{𝑧^{𝑠−𝑝−𝛼}},(2)if 𝑠⪯𝑝, the rank of 𝑆(𝑧^𝛼) is equal to zero,(3)if 𝑠≥𝑝̸ and 𝑠≰𝑝, then the rank of 𝑆(𝑧^𝛼) is equal to zero. The proof is completed.

Remark 19. (1) If 𝑝,𝑠 are two nonzero multi-indexes with 𝑝⟂𝑠, and if (𝑇_𝜓,𝑇_𝜉𝑝𝜉^𝑠_𝜑] has finite rank, then the rank of it is equal to zero.
(2) If 𝑝=0, (𝑇_𝜓,𝑇_𝜉^𝑠_𝜑] has finite rank, then the rank of it is at most equal to the number of 𝛼 such that 𝛼⪯𝑠 and {(𝑇_𝜓,𝑇_𝜉^𝑠_𝜑](𝑧^𝛼)}_𝛼≽0⋃{(𝑇_𝜓,𝑇_𝜉^𝑠_𝜑](𝑧^𝛼)}_𝛼≻0⊆⋁_𝛼⪯𝑠{𝑧^𝑠−𝛼}.
(3) If (𝑇_𝜓,𝑇_𝜉^𝑝_𝜑] has finite rank, then the rank of it is at most equal to the number of 𝛼 such that 𝛼⪯𝑝 and {(𝑇_𝜓,𝑇_𝜉^𝑝_𝜑](𝑧^𝛼)}_𝛼≽0⋃{(𝑇_𝜓,𝑇_𝜉^𝑝_𝜑](𝑧^𝛼)}_𝛼≻0⊆⋁_𝛼⪯𝑝{𝑧^𝑝−𝛼}.

Example 20. In the case of 𝑛=2, we give an example to Theorem 18 (Remark 19). Similar to Example 16, there exist 𝜓=𝑟⁶,  𝜑=𝑟⁻² and 𝑝=(1,1) such that 𝑇_𝑟6𝑇_𝜉^(1,1)_𝑟−2𝑧^𝛼=𝑇_𝜉^(1,1)_𝑟4𝑧^𝛼𝑇,∀𝛼≽0,_𝑟6𝑇_𝜉^(1,1)_𝑟−2𝑧^𝛼=𝑇_𝜉^(1,1)_𝑟4𝑧^𝛼,∀𝛼≽(1,1). (82) Using Lemma 12, we have 𝑇_𝑟6𝑇_𝜉^(1,1)_𝑟−2𝑧^(1,0)=14𝑧^(0,1),𝑇_𝜉^(1,1)_𝑟4𝑧^(1,0)=15𝑧^(0,1),𝑇_𝑟6𝑇_𝜉^(1,1)_𝑟−2𝑧^(0,1)=14𝑧^(1,0),𝑇_𝜉^(1,1)_𝑟4𝑧^(0,1)=15𝑧^(1,0) (83) so the rank of (𝑇_𝑟6,𝑇_𝜉^(1,1)_𝑟−2] is equal to two.

Theorem 21. Let 𝑝,𝑠 be two nonzero multi-indexes, let 𝜑 and 𝜓 be two integrable radial functions, such that 𝑇_𝜉^𝑝_𝜑,𝑇_𝜉^𝑠_𝜓 are bounded operators. If (𝑇_𝜉^𝑝_𝜑,𝑇_𝜉^𝑠_𝜓] has finite rank, then its rank is at most equal to the number of 𝛼 such that 𝛼⪯𝑝+𝑠, and {(𝑇_𝜉^𝑝_𝜑,𝑇_𝜉^𝑠_𝜓](𝑧^𝛼)}_𝛼≽0⋃{(𝑇_𝜉^𝑝_𝜑,𝑇_𝜉^𝑠_𝜓](𝑧^𝛼)}_𝛼≻0⊆⋁_{𝛼⪯𝑝+𝑠}{𝑧^{𝑝+𝑠−𝛼}}.

Proof. The proof of this Theorem is similar to that of Theorem 15. Here, we omit it.

Example 22. In case of 𝑛=2, we give an example to Theorem 21. As we construct Example 16, let 𝜑=𝑟²,  𝜓=𝑟⁶, and 𝑝=(1,0), 𝑠=(1,1), we have 𝑇_𝜉^(1,0)_𝑟2𝑇_𝜉^(1,1)_𝑟6𝑧^𝛼=𝑇_𝜉^(2,1)_𝑟8𝑧^𝛼𝑇,∀𝛼≽0,_{𝜉(1,0)𝑟2}𝑇_𝜉^(1,1)_𝑟6𝑧^𝛼=𝑇_𝜉^(2,1)_𝑟8𝑧^𝛼,∀𝛼≽(2,1). (84) However, 𝑇_𝜉^(1,0)_𝑟2𝑇_𝜉^(1,1)_𝑟6𝑧^(2,0)𝑇=0,_{𝜉(2,1)𝑟8}𝑧^(2,0)=1𝑧15^(0,1),𝑇_𝜉^(1,0)_𝑟2𝑇_𝜉^(1,1)_𝑟6𝑧^(1,1)=1𝑧21^(1,0),𝑇_𝜉^(2,1)_𝑟8𝑧^(1,1)=1𝑧15^(1,0),𝑇_𝜉^(1,0)_𝑟2𝑇_𝜉^(1,1)_𝑟6𝑧^(1,0)=4𝑧27^(1,1),𝑇_𝜉^(2,1)_𝑟8𝑧^(1,0)=4𝑧15^(1,1),𝑇_𝜉^(1,0)_𝑟2𝑇_𝜉^(1,1)_𝑟6𝑧^(0,1)=4𝑧27^(2,0),𝑇_𝜉^(2,1)_𝑟8𝑧^(0,1)=2𝑧15^(2,0), (85) Therefore the rank of (𝑇_𝜉^(1,0)_𝑟2,𝑇_𝜉^(1,1)_r6] is equal to four.

Theorem 23. Let 𝑝,𝑠 be two nonzero multi-indexes with 𝑝⟂𝑠, let 𝜑 and 𝜓 be two integrable radial functions on 𝐵_𝑛, such that 𝑇_𝜉^𝑝_𝜑 and 𝑇_𝜉^𝑠_𝜓 are bounded operators. If (𝑇_𝜉^𝑝_𝜑,𝑇_𝜉^𝑠_𝜓] has finite rank, then the rank of it is equal to zero.

Proof. The proof of this Theorem is similar to that of Theorem 17.

Acknowledgments

The authors would like to thank the referees for their excellent suggestions. This research is supported by NSFC (Grant nos. 10971020, and 11271059) and Research Fund for the Doctoral Program of Higher Education of China.