# Algebraic Properties of Toeplitz Operators on the Pluriharmonic Bergman Space

Authors: Jingyu Yang; Liu Liu; Yufeng Lu
Journal: Journal of Function Spaces and Applications
 (2013)

---

## 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.


---

## Body

## 1. Introduction

Letdν denote the normalized Lebesgue volume measure on the unit ball Bn of Cn. L2(Bn,dν) is the Hilbert space of Lebesgue square integrable functions on Bn with the inner product:
(1)〈f,g〉=∫Bn‍f(z)g(z)¯dv(z).
The Bergman space La2(Bn) is the Hilbert space consisting of all holomorphic functions on Bn which are also in L2(Bn,dv). It is well known that La2(Bn) is a reproducing function space with reproducing kernel:
(2)Kz(w)=1(1-〈w,z〉)n+1,z,w∈Bn.
The pluriharmonic Bergman space Lh2(Bn) is the closed subspace of L2(Bn,dv) consisting of the pluriharmonic functions on Bn. It is easy to verify that
(3)Lh2(Bn)=La2(Bn)+La2(Bn)¯,
where La2(Bn)¯={f¯∣f∈La2(Bn),f(0)=0}. Then Lh2(Bn) is also a reproducing function space with reproducing kernel:
(4)Rz(w)=Kz(w)+Kz(w)¯-1,z,w∈Bn.LetQ denote the orthogonal projection from L2(Bn,dv) onto Lh2(Bn), then
(5)Qφ(z)=〈φ,Rz〉,φ∈L2(Bn,dv).
Let P be the orthogonal projection from L2(Bn,dv) onto La2(Bn), then
(6)Pφ(z)=〈φ,Kz〉,φ∈L2(Bn,dv).
By (4), we have
(7)Qφ(z)=Pφ(z)+P(φ¯)(z)¯-Pφ(0).For a functionφ∈L∞(Bn,dv), the Toeplitz operator Tφ:Lh2(Bn)→Lh2(Bn) with symbol φ is defined by
(8)Tφ(f)=Q(φf)(z)=∫Bn‍f(w)φ(w)Rz(w)¯dv(w).In the setting of the classical Hardy space, forf and g in L∞(T), Brown and Halmos [1] proved that TfTg=Th 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(Dn) 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 TfTg=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 Tφ1Tφ2⋯Tφn=0, whether one of these symbols is identically zero. Guo [10] proved that if Tφ1Tφ2⋯Tφ5=0, then there exists some i such that φi=0. In [11], Gu showed that Tφ1Tφ2⋯Tφ6=0, then there exists some i such that φi=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 TfTg=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(Tn) on the Hardy space of the polydisk.For two Toeplitz operatorsTφ and Tψ we define the commutator and semicommutator, respectively, by
(9)[Tφ,Tψ]=TφTψ-TψTφ,(Tφ,Tψ]=Tφψ-TφTψ.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 onLa2. For example, Choe and Lee [24] showed that two analytic Toeplitz operators on Lh2 commute only when their symbols and the constant function 1 are linearly dependent, but analytic Toeplitz operators always commute on La2. 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 TfTg¯=Tg¯Tf 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 transformφ^ of a function φ∈L1([0,1],rdr) is defined by
(10)φ^(z)=∫01‍φ(s)sz-1ds.
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 thatf is a bounded analytic function on {z:Rez>0} which vanishes at the pairwise distinct points z1,z2,…, where(1)
inf{|zn|}>0 and(2)
∑n≥1‍Re(1/zn)=∞.
Thenf vanishes identically on {z:Rez>0}.Remark 2.
We will often use this theorem to show that ifφ∈L1([0,1],rdr) and if there exists a sequence (nk)k≥0⊂ℕ such that
(11)φ^(nk)=0,∑k≥0‍1nk=∞,
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. Iff and g are defined on [0,1), then their Mellin convolution is defined by
(12)(f*Mg)(r)=∫r1‍f(rt)g(t)dtt,0≤r<1.
The Mellin convolution theorem states that
(13)f*Mg^(s)=f^(s)g^(s),
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-indexα=(α1,…,αn), where each αi is a nonnegative integer, we write |α|=α1+⋯+αn,  α!=α1!⋯αn! and zα=z1α1⋯znαn, z=(z1,…,zn)∈Bn.For two multi-indexesα=(α1,…,αn) and β=(β1,…,βn), the notation α≽β means that αi≥βi,i=1,2,…n, and α⊥β means that α1β1+⋯+αnβn=0. We also define α-β=(α1-β1,…,αn-βn). Moreover, if |α-β|=|α|-|β| if α≽β.It is known thatφ is radial if and only if φ(Uz)=φ(z) for any unitary transformation U of Cn. 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 Cn. 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.
Letφ∈L2(Bn,dv), Tφ is a Toeplitz operator densely defined on Lh2(Bn), then the symbol map φ→Tφ is one to one.Proof.
AssumeTφ=0, we have Q[φf]=0 for all f∈Lh2(Bn).
Sincezβ∈Lh2(Bn), then we can imply that 〈φzα,zβ〉=0, which is equivalent to ∫Bn‍φzαz¯βdv(z)=0. Since {zαz¯β:α,βareanymulti-indexes} is dense in L2(Bn,dv), so we get φ=0.A direct calculation gives the following lemma, which will be used constantly in this paper.Lemma 4.
Letψ be an integrable radial function on Bn, such that Tψ is a bounded operator, then for any multi-index α,
(14)Tψ(zα)=2(n+|α|)ψ^(2n+2|α|)zα,Tψ(z¯α)=2(n+|α|)ψ^(2n+2|α|)z¯α.Proof.
For any multi-indexβ, we have
(15)〈Tψzα,zβ〉=〈ψzα,zβ〉={0α≠β;2n!α!(n-1+|α|)!ψ^(2n+2|α|)α=β,〈zα,zβ〉={0α≠β;n!α!(n+|α|)!α=β,
which implies
(16)〈Tψzα,zβ〉=2(n+|α|)ψ^(2n+2|α|)〈zα,zβ〉.
Ifβ≻0, we have 〈Tψzα,z¯β〉=0=〈zα,z¯β〉. This shows that
(17)〈Tψzα,z¯β〉=〈zα,z¯β〉∀β≻0.
Since{zα}α≽0⋃{z¯α}α≻0 is a basis for the pluriharmonic Bergman space, we have
(18)Tψ(zα)=2(n+|α|)ψ^(2n+2|α|)zα.
By a similar argument, we have Tψ(z¯α)=2(n+|α|)ψ^(2n+2|α|)z¯α.The following theorem is very simple but essential.Theorem 5.
Letφ∈L2(Bn,dv). Then the following assertions are equivalent.(1)
For each multi-indexα, there exists λ|α|∈C which depends only on |α| such that Tφ(zα)=λ|α|zα, Tφ(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 Tφ∘U=Tφ for any unitary transformation U of Cn.
First assumeTφ(zα)=λ|α|zα. For any unitary transformation U of Cn and z∈Bn, we have U-1z∈Bn. Thus,
(19)Tφ(U-1z)α=λ|α|(U-1z)α.
By (7), we have
(20)Tφ∘U(zα)=P[φ(Uz)zα]+P[φ(Uz)¯z¯α]¯-P[φ(Uz)zα](0).
A direct calculation shows that
(21)P[φ(Uz)zα](w)=∫Bn‍φ(Uz)zα(1-〈w,z〉)n+1dv(z)=∫Bn‍φ(z)(U-1z)α(1-〈w,U-1z〉)n+1dv(z)=∫Bn‍φ(z)(U-1z)α(1-〈Uw,z〉)n+1dv(z)=P[φ(z)(U-1z)α](Uw).
Similarly, we have
(22)P[φ(Uz)¯zα¯](w)=P[φ(z)¯(U-1z)α¯](Uw),P[φ(Uz)zα](0)=P[φ(z)(U-1z)α](0).
Then we can imply
(23)Tφ∘U(zα)(w)=P[φ(z)(U-1z)α](Uw)+P[φ(z)(U-1z)α¯](Uw)-P[φ(z)(U-1z)α](0)=Tφ(U-1z)α(Uw)=[λ|α|(U-1z)α](Uw)=λ|α|wα=Tφ(zα)(w).
SoTφ∘U(zα)=Tφ(zα). With the same method, we get Tφ∘U(z¯α)=Tφ(z¯α). Therefore, we can imply Tφ∘U=Tφ on Lh2(Bn). By Lemma 3 we obtain that φ∘U=φ and φ is a radial function.Corollary 6.
Letφ1 and φ2 be two square integrable radial functions on Bn such that Tφ1 and Tφ2 are bounded operators. If Tφ1Tφ2=Tψ, then ψ is a radial function.Proof.
According to (14) we have
(24)Tφ1Tφ2(zα)=(2n+2|α|)2φ^1(2n+2|α|)φ^2(2n+2|α|)zα,Tφ1Tφ2(z¯α)=(2n+2|α|)2φ^1(2n+2|α|)φ^2(2n+2|α|)z¯α.
It follows from Theorem 5 that ψ is a radial function.Theorem 7.
Letφ1 and φ2 be two square integrable radial functions on Bn, such that Tφ1 and Tφ2 are bounded operators. Then Tφ1Tφ2 is equal to the Toeplitz operator Tψ if and only if ψ is a solution of the equation
(25)1*Mψ=φ1*Mφ2,
where 1 denotes the constant function with value one.Proof.
For each multi-indexα, it follows from (14) that
(26)Tφ1Tφ2(zα)=Tψ(zα),Tφ1Tφ2(z¯α)=Tψ(z¯α)
are equivalent to
(27)12n+2|α|ψ^(2n+2|α|)=φ^1(2n+2|α|)φ^2(2n+2|α|).
A direct calculation shows 1^(2n+2|α|)=1/(2n+2|α|), so (27) is equivalent to
(28)1*Mψ^(2n+2|α|)=φ1*Mφ2^(2n+2|α|).
By Remark 2, we obtain that (28) is equivalent to (25).Example 8.
Letl and m greater than or equal to −1. Then
(29)T|z|lT|z|m={ll-mT|z|l-ml-mT|z|ml≠m,T|z|l(1+lln|z|)l=m.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.
Letφi be integrable radial functions on Bn, such that Tφi be bounded operators, i=1,…,N. If Tφ1⋯TφN=0, then φi=0 for some i.Proof.
SupposeTφ1⋯TφN=0, then for any multi-index α, we have
(30)Tφ1⋯TφN(zα)=0Tφ1⋯TφN(z¯α)=0.
Using (14), we have
(31)φ^1(2n+2|α|)⋯φ^N(2n+2|α|)=0.
LetEi={|α|∈ℕ:φ^i(2n+2|α|)=0}. Notice that E1∪⋯∪EN=ℕ, hence there exists some i such that ∑|α|∈Ei‍(1/|α|)=∞, then by Remark 2  φi=0.Corollary 10.
Letφ be an integrable radial function on Bn, such that Tφ is a bounded operator. Then Tφ2=Tφ if and only if either φ=0 or φ=1.Proof.
IfTφ2=Tφ, then TφTφ-1=0. By Theorem 9 we have φ=0 or φ=1. The converse is clear.


## 4. Product ofn 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.
Letp,s≽0. A function f∈L1(Bn,dv) is called a quasihomogeneous function of degree (p,s) if f is of the form ξpξ¯sφ, where φ is a radial function, that is,
(32)f(rξ)=ξpξ¯sφ(r),
for any ξ in the unit sphere Sn and r∈[0,1).Lemma 12.
Letp,s be two multi-indexes and let φ be an integrable radial function on Bn, such that Tξpφ,Tξ¯pφ, and Tξpξ¯sφ are bounded operators. Then for any multi-index α,
(33)Tξpξ¯sφ(zα)={2(p+α)!(n+|p|+|α|-|s|)!(p+α-s)!(n-1+|α|+|p|)!×φ^(2n+2|α|+|p|-|s|)zp+α-sp+α≽s,2s!(n+|s|-|α|-|p|)!(s-α-p)!(n-1+|s|)!×φ^(2n+|s|-|p|)z¯s-α-pp+α⪯s,0p+α≱s,p+α≰s,Tξpξ¯sφ(z¯α)={2(s+α)!(n+|s|+|α|-|p|)!(s+α-p)!(n-1+|α|+|s|)!×φ^(2n+2|α|+|s|-|p|)z¯s+α-ps+α≽p,2p!(n+|p|-|α|-|s|)!(p-α-s)!(n-1+|p|)!×φ^(2n+|p|-|s|)zp-α-ss+α⪯p,0s+α≱p,s+α≰p.
Especially, if p,s are two nonzero multi-indexes with p⊥s, one has
(34)Tξpξ¯sφ(zα)={2(p+α)!(n+|p|+|α|-|s|)!(p+α-s)!(n-1+|α|+|p|)!×φ^(2n+2|α|+|p|-|s|)zp+α-sα≽s,0α≱s.(35)Tξpξ¯sφ(z¯α)={2(s+α)!(n+|s|+|α|-|p|)!(s+α-p)!(n-1+|α|+|s|)!×φ^(2n+2|α|+|s|-|p|)z¯s+α-pα≽p,0α≱p.Proof.
Here we only give the proof of the first equation. For multi-indexesα and β,
(36)〈P[ξpξ¯sφzα],zβ〉=∫Bn‍ξpξ¯sφ(z)zαz¯βdv(z)=∫[0,1)‍2nr2n+|α|+|β|-1φ(r)dr∫Sn‍ξpξ¯sξαξ¯βdσ(ξ)={2n!(p+α)!(n-1+|α|+|p|)!×φ^(2n+2|α|+|p|-|s|)p+α≽s,β=p+α-s,0p+α≽s,β≠p+α-s,0p+α≱s.
If p+α≽s, we have
(37)〈zp+α-s,zβ〉=∫Bn‍zp+α-sz¯βdv(z)=∫[0,1)‍2nr2n+|p|+|α|-|s|+|β|-1dr×∫Sn‍ξp+α-sξ¯βdσ(ξ)={n!(p+α-s)!(n+|α|+|p|-|s|)!β=p+α-s,0β≠p+α-s.
Furthermore, 〈P[ξpξ¯sφzα],z¯β〉=0=〈zp+α-s,z¯β〉 hold for all β≻0. So we obtain
(38)P[ξpξ¯sφzα]={2(p+α)!(n+|p|+|α|-|s|)!(p+α-s)!(n-1+|α|+|p|)!×φ^(2n+2|α|+|p|-|s|)zp+α-sp+α≽s,0p+α≱s.
Note thatφ¯ is still radial and φ^(n)¯=φ¯^(n) for all n∈N, we have
(39)P[ξ¯pξsφ¯z¯α]={2s!(n+|s|-|α|-|p|)!(s-α-p)!(n-1+|s|)!×φ^¯(2n+|s|-|p|)zs-α-pp+α⪯s,0p+α≰s.
Thus
(40)Tξpξ¯sφ(zα)=P[ξpξ¯sφzα]+P[ξ¯pξsφ¯z¯α]¯-P[ξpξ¯sφzα](0)={2(p+α)!(n+|p|+|α|-|s|)!(p+α-s)!(n-1+|α|+|p|)!×φ^(2n+2|α|+|p|-|s|)zp+α-sp+α≽s,2s!(n+|s|-|α|-|p|)!(s-α-p)!(n-1+|s|)!×φ^(2n+|s|-|p|)z¯s-α-pp+α⪯s,0p+α≰s,p+α≱s.
Especially, ifp⊥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 pi≠0,si=0. This means that there doesn’t exist multi-index α such that p+α⪯s. So
(41)Tξpξ¯sφ(zα)={2(p+α)!(n+|p|+|α|-|s|)!(p+α-s)!(n-1+|α|+|p|)!×φ^(2n+2|α|+|p|-|s|)zp+α-sα≽s,0α≱s.Theorem 13.
Letpi,si be multi-indexes and let φi be integrable radial functions on Bn such that Tξpiξ¯siφi is bounded i=1,2,…,m. If the product Tξpmξ¯smφm⋯Tξp1ξ¯s1φ1 is of finite rank, then φi=0 for some i∈{1,2,…,m}.Proof.
We denote the product of Toeplitz operatorsTξpmξ¯smφm⋯Tξp1ξ¯s1φ1 by S. By Lemma 12, we have
(42)S(zα)=∏i=m2‍Tξpiξ¯siφi×{2(p1+α)!(n+|p1|+|α|-|s1|)!(p1+α-s1)!(n-1+|α|-|p1|)!×φ1^(2n+2|α|+|p1|-|s1|)zp1+α-s1p1+α≽s1,2s1!(n+|s1|-|p1|-|α|)!(s1-p1-α)!(n-1+|s1|)!×φ1^(2n+|s1|-|p1|)z¯s1-α-p1p1+α⪯s1,0p1+α≱s1,p1+α≱s1.
If p1+α≽s1,p2+p1+α≽s1+s2,…,pm+⋯+p1+α≽s1+⋯+sm, we have
(43)S(zα)=2ma(α)b(|α|)×∏i=1m‍φi^[2n+2(|α|+∑j=0i-1‍(|pj|-|sj|))+|pi|-|si|]×zα+∑i=1mpi-si,
where
(44)a(α)=∏i=1m[(α+∑j=0i-1(pj-sj)+pi)!]∏i=1m[(α+∑j=1i(pj-sj))!],(45)b(|α|)=∏i=1m[(n+|α|+∑j=1i(|pj|-|sj|))!]∏i=1m[(n-1+|α|+∑j=0i-1(|pj|-|sj|)+|pi|)!],p0=0 and s0=0.
Similarly, we have(46)S(z¯α)=∏i=m2‍Tξpiξ¯siφi×{2(s1+α)!(n+|s1|+|α|-|p1|)!(s1+α-p1)!(n-1+|α|-|s1|)!×φ1^(2n+2|α|+|s1|-|p1|)zs1+α-p1s1+α≽p1,2p1!(n+|p1|-|s1|-|α|)!(p1-s1-α)!(n-1+|p1|)!×φ1^(2n+|p1|-|s1|)z¯p1-α-s1s1+α⪯p1,0s1+α≱p1,s1+α≱p1.
Then for s1+α≽p1,s2+s1+α≽p1+p2,…sm+⋯+s1+α≽p1+⋯+pm, we have
(47)S(z¯α)=2ma′(α)b′(|α|)×∏i=1m‍φi^[2n+2(|α|+∑j=0i-1‍(|sj|-|pj|))+|si|-|pi|]×zα+∑i=1msi-pi,
where
(48)a′(α)=∏i=1m[(α+∑j=0i-1(sj-pj)+si)!]∏i=1m[(α+∑j=1i(sj-pj))!],b′(|α|)=∏i=1m[(n+|α|+∑j=1i∑‍(|sj|-|pj|))!]∏i=1m[(n-1+|α|+∑j=0i-1(|sj|-|pj|)+|si|)!].
SinceS has finite rank, we know that {S(zα)}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,…,pm+⋯+p1+α0≽s1+⋯+sm and
(49)S(zα)=0∀α≽α0,
which is equivalent to
(50)φm^[2n+2|α|+2∑j=0m-1‍(|pj|-|sj|)+|pm|-|sm|]⋯φ1^[2n+2|α|+|p1|-|s1|]=0,
for all α≽α0. In fact, we have α0=(t1,…,tn), in which tj=max{k1(j),k2(j),…,km(j)} and kij=max{(s1(j)-p1(j)+s2(j)-p2(j)+⋯+si(j)-pi(j)),0}, si=(si(1),…,si(n)), pi=(pi(1),…,pi(n)), i∈{1,2,…,m}, j∈{1,2,…,n}.
Letl=min{|p1|-|s1|,2(|p1|-|s1|)+|p2|-|s2|,…,2∑j=1m-1‍(|pj|-|sj|)+|pm|-|sm|}, we know that 2|α|+l≥0,forallα≽α0. Then (50) implies that
(51)(r2(∑j-1m-1(|pj|-|sj|))+|pm|-|sm|-lφm)⋀(2n+2|α|+l)⋯(r|p1|-|s1|-lφ1)⋀(2n+2|α|+l)=0,
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 φi=0.


## 5. Finite Rank Commutators

In this section, letφ,ψ be two integrable radial functions on Bn. We now pass to investigate the commutator [Tψ,Tξpξ¯sφ],[Tξpφ,Tξsψ] and [Tξpφ,Tξ¯sψ].Theorem 14.
Letp,s be two multi-indexes and let φ and ψ be two integrable radial functions on Bn such that Tψ and Tξpξ¯sφ are bounded operators. If ψ is nonconstant and [Tψ,Tξpξ¯sφ] has finite rank, then its rank is equal to zero.Proof.
LetS denote the commutator [Tψ,Tξpξ¯sφ]. Since S has finite rank on Lh2(Bn), we know that {S(zα)}α≽0 and {S(z¯α)}α≻0 must have finite rank.
By Lemma12, we have if p+α≽s,
(52)S(zα)=2(p+α)!(n+|p|+|α|-|s|)!(p+α-s)!(n-1+|p|+|α|)!×φ^(2n+2|α|+|p|-|s|)×[2(n+|p|+|α|-|s|)ψ^(2n+2|p|+2|α|-2|s|)-2(n+|α|)ψ^(2n+2|α|)]zp+α-s.
If p+α⪯s,
(53)S(zα)=2s!(n+|s|-|α|-|p|)!(s-α-p)!(n-1+|s|)!φ^(2n+|s|-|p|)×[2(n+|s|-|α|-|p|)ψ^(2n+2|s|-2|α|-2|p|)-2(n+|α|)ψ^(2n+2|α|)]z¯s-α-p,
if p+α≱s, p+α≰s, we have S(zα)=0.
Since{S(zα)}α≽0 has finite rank, then there must exist α0, such that p+α0≽s and S(zα)=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
(54)S(zα)=0∀α≽0.
Similar to the discussions forS(zα), we can imply that S(z¯α)=0 for all α≽0. In conclusion, the rank of S is equal to zero.Theorem 15.
Letp,s be two nonzero multi-indexes, let φ and ψ be two integrable radial functions on Bn, such that Tξpφ and Tξsψ are bounded operators. If [Tξpφ,Tξsψ] has finite rank, then the rank of it is less than or equal to the number of α such that α⪯p+s, and {[Tξpφ,Tξsψ](zα)}α≽0⋃{[Tξpφ,Tξsψ](z¯α)}α≻0⊂⋁α⪯p+s{zp+s-α}, where ⋁α⪯s+p{zs+p-α}=span{zs+p-α:α⪯s+p}.Proof.
LetS denote the commutator [Tξpφ,Tξsψ]. By Lemma 12, we have
(55)S(zα)=2(n+|α|+|p|+|s|)×[2(n+|α|+|s|)ψ^(2n+2|α|+|s|)×φ^(2n+2|α|+|p|+2|s|)-2(n+|α|+|p|)×ψ^(2n+2|α|+2|p|+|s|)×φ^(2n+2|α|+|p|)]zp+α+s.
If{S(zα)}α≽0 has finite rank, there must exist k≥0 such that
(56)S(zα)=0,∀|α|≥k.
Then, we deduce that
(57)1^(2n+2|p|+2|α|)φ^(2n+2|s|+2|α|+|p|)×ψ^(2n+|s|+2|α|)=1^(2n+2|s|+2|α|)φ^(2n+2|α|+|p|)×ψ^(2n+|s|+2|α|+2|p|),|α|≥k.
It follows from Remark 2 that
(58)r2|p|1^(z)r2|s|+|p|φ^(z)r|s|ψ^(z)=r2|s|1^(z)r|p|φ^(z)r2|p|+|s|ψ^(z),
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(zα)=0. Therefore, the rank of {S(zα)}α≽0 is equal to zero.
ForS(z¯α), we have if α≽p+s,
(59)S(z¯α)=4α!(n+|α|-|p|-|s|)!(α-p-s)!(n-1+|α|)!×[|p|(n+|α|-|s|)!|p|(n-1+|α|-|s|)!ψ^(2n+2|α|-|s|)×φ^(2n+2|α|-2|s|-|p|)-(n+|α|-|p|)!(n-1+|α|-|p|)!×ψ^(2n+2|α|-2|p|-|s|)×φ^(2n+2|α|-|p|)|p|(n+|α|-|s|)!|p|(n-1+|α|-|s|)!]z¯α-p-s.
If α≱p,  α≰p,  α≱s,  α≰s, we have S(z¯α)=0, if α≽p but α≱p+s,  α≰p+s, we have S(z¯α)=0, if α≽s but α≱p+s,  α≰p+s, then S(z¯α)=0.
Ifs⪯α⪯p+s or p⪯α⪯p+s or α⪯p or α⪯s,
(60)S(z¯α)=C(α)zp+s-α,
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
(61)S(z¯α)=0,∀α≽p+s,
this means {S(z¯α)}α≽0⊂⋁α⪯p+s{zp+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 ofn=2, we give an example to Theorem 15.
From the proof of Theorem15 we know that
(62)TξprmTξsf(zα)=TξsfTξprm(zα),∀α≽0.
Below we will construct a radial function f, such that
(63)TξprmTξsf(z¯α)=TξsfTξprm(z¯α),∀α≽p+s,
where p,s are multi-indexes, ξ∈S2 and m≥0.
Equation (63) implies that for α≽p+s,
(64)r-|s|-2|p|f^(2|α|+4+2|p|)r-|s|-2|p|f^(2|α|+4)=(2|α|+4-2|p|)(2|α|+4-|p|+m-2|s|)(2|α|+4-2|s|)(2|α|+4-|p|+m).
Now, using Remark 2, we obtain that
(65)r-|s|-2|p|f^(z+2|p|)r-|s|-2|p|f^(z)=(z-2|p|)(z-|p|+m-2|s|)(z-2|s|)(z-|p|+m),
for all Rez≥2|p|+2|s|+4.
LetF be the analytic function defined for Rez≥2|p|+2|s| by
(66)F(z)=Γ((z-2|p|)/2|p|)Γ((z-|p|-2|s|+m)/2|p|)Γ((z-2|s|)/2|p|)Γ((z-|p|+m)/2|p|),
where Γ denote the gamma function. Then using the well-known identity Γ(z+1)=zΓ(z), (65) implies that
(67)r-|s|-2|p|f^(z+2|p|)r-|s|-2|p|f^(z)=F(z+2|p|)F(z),forRez>2|p|+2|s|.
Equation (67) combined with [30, Lemma 6, page 1468] gives us that there exists a constant c such that
(68)r-|s|-2|p|f^(z)=cF(z)forRez>2|p|+2|s|.
For a choice ofp=(2,0),s=(0,1), and m=6, again using the identity Γ(z+1)=zΓ(z), one can see that
(69)F(z)=4(z-2)z(z-4)=2[1z+1z-4].
Since 1^(z)=1/z, r-4^(z)=1/(z-4), then (68) becomes
(70)r-5f^(z)=c[1^(z)+r-4^(z)],forRez>6.
Now the proceeding equation and Remark 2 imply that
(71)f(r)=c(r5+r),wherecisaconstant.
It is clear that the function f is bounded, so Toeplitz operator Tξ(0,1)f is bounded.
Finally, by taking the constantc to be equal to 1, the radial function f(r)=r5+r satisfies
(72)Tξ(2,0)r6Tξ(0,1)(r5+r)(zα)=Tξ(0,1)(r5+r)Tξ(2,0)r6(zα),∀α≽0,Tξ(2,0)r6Tξ(0,1)(r5+r)(z¯α)=Tξ(0,1)(r5+r)Tξ(2,0)r6(z¯α),∀α≽(2,1).
For α=(2,1),α1=(1,1),α2=(0,1),α3=(2,0),α4=(1,0), we know that αi⪯α,(i=1,2,3,4). However, using Lemma 12, it is easy to see that
(73)Tξ(2,0)r6Tξ(0,1)(r5+r)(z¯(1,1))=536z(1,0),Tξ(0,1)(r5+r)Tξ(2,0)r6(z¯(1,1))=0,Tξ(2,0)r6Tξ(0,1)(r5+r)(z¯(0,1))=1645z(2,0),Tξ(0,1)(r5+r)Tξ(2,0)r6(z¯(0,1))=0,Tξ(2,0)r6Tξ(0,1)(r5+r)(z¯(2,0))=0,Tξ(0,1)(r5+r)Tξ(2,0)r6(z¯(2,0))=845z(0,1),Tξ(2,0)r6Tξ(0,1)(r5+r)(z¯(1,0))=0,(74)Tξ(0,1)(r5+r)Tξ(2,0)r6(z¯(1,0))=59z(1,1).
Therefore the commutator [Tξ(2,0)r6,Tξ(0,1)(r5+r)] has rank four.Theorem 17.
Letp,s be two nonzero multi-indexes with p⊥s, let φ and ψ be two integrable radial functions on Bn, such that Tξpφ and Tξ¯sψ are bounded operators. If [Tξpφ,Tξ¯sψ] has finite rank, then the rank of it is equal to zero.Proof.
LetS denote the commutator [Tξpφ,Tξ¯sψ]. 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(zα)}α≽0 has finite rank, there exists α0≽0 such that
(75)S(zα)=0,∀α≽s+α0.
Since p⊥s, we have α!(p+α-s)!=(α+p)!(α-s)!. So
(76)S(zα)=α!(α-s)![(n+|α|-|s|)!(n-1+|α|)!2(n+|α|+|p|-|s|)×ψ^(2n+2|α|-|s|)×φ^(2n+2|α|-2|s|+|p|)-(n+|α|+|p|-|s|)!(n-1+|p|+|α|)!2(n+|α|+|p|)×ψ^(2n+2|α|+2|p|-|s|)×φ^(2n+2|α|+|p|)(n+|α|-|s|)!(n-1+|α|)!]zp+α-s.
As in the proof of Theorem 15, we have S(zα)=0 for all α≽s, this means that the rank of {S(zα)}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.
Letp,s be two multi-indexes and let φ and ψ be two integrable radial functions on Bn, such that Tψ,Tξpξ¯sφ are bounded operators.(a)
Ifs≽p, (Tψ,Tξpξ¯sφ] has finite rank, then its rank is at most equal to the number of α such that α⪯s-p and {(Tψ,Tξpξ¯sφ](zα)}α≽0⋃{(Tψ,Tξpξ¯sφ](z¯α)}α≻0⊆⋁α⪯s-p{z¯s-p-α}.(b)
Ifs⪯p, (Tψ,Tξpξ¯sφ] has finite rank, then its rank is at most equal to the number of α such that α⪯p-s and {(Tψ,Tξpξ¯sφ](zα)}α≽0⋃{(Tψ,Tξpξ¯sφ](z¯α)}α≻0⊆⋁α⪯p-s{zp-s-α}.(c)
Ifs≱p,  s≰p, and (Tψ,Tξpξ¯sφ] has finite rank, then its rank is equal to zero.Proof.
LetS denote the semicommutator (Tψ,Tξpξ¯sφ].
Ifs+α≽p, we compute that
(77)S(z¯α)=2(s+α)!(n+|s|+|α|-|p|)!(s+α-p)!(n-1+|s|+|α|)!×[φψ^(2n+2|α|+|s|-|p|)-2(n+|s|+|α|-|p|)×ψ^(2n+2|s|+2|α|-2|p|)×φ^(2n+2|α|+|s|-|p|)φψ^]z¯s+α-p.
Ifs+α⪯p,
(78)S(z¯α)=2p!(n+|p|-|α|-|s|)!(p-α-s)!(n-1+|p|)!×[φψ^(2n+|p|-|s|)-2(n+|p|-|α|-|s|)×ψ^(2n+2|p|-2|α|-2|s|)×φ^(2n+|p|-|s|)φψ^]zp-α-s,
if s+α≱p 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
(79)S(z¯α)=0∀α≽p-s+α0.
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{zp-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≱p,s≰p). For s=(s1,…,sn), p=(p1,…,pn), without loss of generality, we assume s1<p1 but si≥pi(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,…,sn-pn), it is clear that p+α′≽s. So for any s≱p and s≰p, there exists α0=(max{(s1-p1),0},…,max{(sn-pn),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≱α0S(z¯α)=0.
If{S(z¯α)}α≻0 has finite rank, there exists α′≽0 such that
(80)S(z¯α)=0,∀α≽α0+α′.
As in the proof of Case 1, we can imply that
(81)S(z¯α)=0,∀α≽α0.
This means that the rank of {S(z¯α)}α≻0 is equal to zero.
Similar to the discussions forS(z¯α), we obtain that(1)
ifs≽p, the rank of S(zα) is at most equal to the number of α such that α⪯s-p, and {S(zα)}α≽0⊆⋁α⪯s-p{z¯s-p-α},(2)
ifs⪯p, the rank of S(zα) is equal to zero,(3)
ifs≱p and s≰p, then the rank of S(zα) is equal to zero. The proof is completed.Remark 19.
(1) Ifp,s are two nonzero multi-indexes with p⊥s, and if (Tψ,Tξpξ¯sφ] has finite rank, then the rank of it is equal to zero.
(2) Ifp=0, (Tψ,Tξ¯sφ] has finite rank, then the rank of it is at most equal to the number of α such that α⪯s and {(Tψ,Tξ¯sφ](zα)}α≽0⋃{(Tψ,Tξ¯sφ](z¯α)}α≻0⊆⋁α⪯s{z¯s-α}.
(3) If(Tψ,Tξpφ] has finite rank, then the rank of it is at most equal to the number of α such that α⪯p and {(Tψ,Tξpφ](zα)}α≽0⋃{(Tψ,Tξpφ](z¯α)}α≻0⊆⋁α⪯p{zp-α}.Example 20.
In the case ofn=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
(82)Tr6Tξ(1,1)r-2(zα)=Tξ(1,1)r4(zα),∀α≽0,Tr6Tξ(1,1)r-2(z¯α)=Tξ(1,1)r4(z¯α),∀α≽(1,1).
Using Lemma 12, we have
(83)Tr6Tξ(1,1)r-2(z¯(1,0))=14z(0,1),Tξ(1,1)r4(z¯(1,0))=15z(0,1),Tr6Tξ(1,1)r-2(z¯(0,1))=14z(1,0),Tξ(1,1)r4(z¯(0,1))=15z(1,0)
so the rank of (Tr6,Tξ(1,1)r-2] is equal to two.Theorem 21.
Letp,s be two nonzero multi-indexes, let φ and ψ be two integrable radial functions, such that Tξpφ,Tξsψ are bounded operators. If (Tξpφ,Tξsψ] has finite rank, then its rank is at most equal to the number of α such that α⪯p+s, and {(Tξpφ,Tξsψ](zα)}α≽0⋃{(Tξpφ,Tξsψ](z¯α)}α≻0⊆⋁α⪯p+s{zp+s-α}.Proof.
The proof of this Theorem is similar to that of Theorem15. Here, we omit it.Example 22.
In case ofn=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
(84)Tξ(1,0)r2Tξ(1,1)r6(zα)=Tξ(2,1)r8(zα),∀α≽0,Tξ(1,0)r2Tξ(1,1)r6(z¯α)=Tξ(2,1)r8(z¯α),∀α≽(2,1).
However,
(85)Tξ(1,0)r2Tξ(1,1)r6(z¯(2,0))=0,Tξ(2,1)r8(z¯(2,0))=115z(0,1),Tξ(1,0)r2Tξ(1,1)r6(z¯(1,1))=121z(1,0),Tξ(2,1)r8(z¯(1,1))=115z(1,0),Tξ(1,0)r2Tξ(1,1)r6(z¯(1,0))=427z(1,1),Tξ(2,1)r8(z¯(1,0))=415z(1,1),Tξ(1,0)r2Tξ(1,1)r6(z¯(0,1))=427z(2,0),Tξ(2,1)r8(z¯(0,1))=215z(2,0),
Therefore the rank of (Tξ(1,0)r2,Tξ(1,1)r6] is equal to four.Theorem 23.
Letp,s be two nonzero multi-indexes with p⊥s, let φ and ψ be two integrable radial functions on Bn, such that Tξpφ and Tξ¯sψ are bounded operators. If (Tξpφ,Tξ¯sψ] has finite rank, then the rank of it is equal to zero.Proof.
The proof of this Theorem is similar to that of Theorem17.



---

*Source: 578436-2013-03-04.xml*