# Toeplitz Operators with Quasihomogeneous Symbols on the Bergman Space of the Unit Ball

Authors: Bo Zhang; Yufeng Lu
Journal: Journal of Function Spaces and Applications
 (2012)

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## Abstract

We consider when the product of two Toeplitz operators with some quasihomogeneous
symbols on the Bergman space of the unit ball equals a Toeplitz operator with quasihomogeneous symbols. We also characterize finite-rank semicommutators or commutators of two Toeplitz operators with quasihomogeneous symbols.


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## Body

## 1. Introduction

LetdA(z) denote the Lebesgue volume measure on the unit ball 𝔹n of ℂn normalized so that the measure of 𝔹n equals 1. The Bergman space La2(𝔹n) is the Hilbert space consisting of holomorphic functions on 𝔹n that are also in L2(𝔹n,dA). Hence, for each z∈𝔹n, there exists an unique function Kz∈La2(𝔹n) with the following property:
(1.1)f(z)=〈f,Kz〉
for all f∈La2(𝔹n). As is well known that the reproducing kernel Kz is given by
(1.2)Kz(w)=1(1-〈w,z〉)n+1,w∈𝔹n.LetP be the orthogonal projection from L2(𝔹n,dA) onto La2(𝔹n). Given a function φ∈L∞(𝔹n,dA), the Toeplitz operator Tφ: La2(𝔹n)→La2(𝔹n) is defined by the formula
(1.3)Tφ(f)(z)=P(φf)(z)=∫𝔹n‍φ(w)f(w)Kz(w)¯dA(w),
for f∈La2(𝔹n). Since the Bergman projection P has norm 1, it is clear that Toeplitz operators defined in this way are bounded linear operators on La2(𝔹n) and ∥Tφ∥≤∥φ∥∞.We now consider a more general class of Toeplitz operators. ForF∈L1(𝔹n,dA), in analogy to (1.3) we define an operator TF on La2(𝔹n) by
(1.4)TFf(z)=∫𝔹n‍F(w)f(w)Kz(w)¯dA(w),z∈𝔹n.Since the Bergman projectionP can be extended to L1(𝔹n,dA), the operator TF is well defined on H∞, the space of bounded analytic functions on 𝔹n. Hence, TF is always densely defined on 𝔹n. Since P is not bounded on L1(𝔹n,dA), it is well known that TF can be unbounded in general. In [1], Zhou and Dong gave the following definitions, which are based on the definitions on the unit disk in [2].Definition 1.1.
LetF∈L1(𝔹n,dA).(a)
We say thatF is a T-function if (1.4) defines a bounded operator on La2(𝔹n).(b)
IfF is a T-function, we write TF for the continuous extension of the operator (it is defined on the dense subset H∞ of L2(𝔹n)) defined by (1.4). We say that TF is a Toeplitz operator if and only if TF is defined in this way.(c)
If there exists anr∈(0,1), such that F is (essentially) bounded on the annulus {z:r<|z|<1}, then we say F is “nearly bounded.”On the Bergman space of the unit ball, Grudsky et al. [3] gave necessary and sufficient conditions for boundedness of Toeplitz operators with radial symbols. These conditions give a characterization of the radial functions in L1(𝔹n,dA) which correspond to bounded operators and furthermore show that the T-functions form a proper subset of L1(𝔹n,dA) which contains all bounded and  “nearly bounded” functions.We denote the semicommutator and commutator of two Toeplitz operatorsTf and Tg by
(1.5)(Tf,Tg]=Tfg-TfTg,[Tf,Tg]=TfTg-TgTf.In the setting of the classical Hardy space, Brown and Halmos [4] gave a complete characterization for the product of two Toeplitz operators to be a Toeplitz operator. On the Bergman space of the unit disk, Ahern and Čučković [5] and Ahern [6] obtained a similar characterization for Toeplitz operators with bounded harmonic symbols. For general symbols, the situation is much more complicated. Louhichi et al. [2] gave necessary and sufficient conditions for the product of two Toeplitz operators with quasihomogeneous symbols to be a Toeplitz operator and Louhichi and Zakariasy made a further discussion in [7]. But it remains open to determine when the product of two Toeplitz operators is a Toeplitz operator on the Bergman space.The problem of determining when the semicommutator(Tf,Tg] or commutator [Tf,Tg] on the Bergman space has finite-rank seems to be far from solution. The analogous problem on the Hardy space has been completely solved (see [8, 9]). Guo et al. [10] completely characterized the finite-rank semicommutator or commutator of two Toeplitz operators with bounded harmonic symbols on the Bergman space of the unit disk and Luecking [11] showed that finite-rank Toeplitz operators on the Bergman space of the unit disk must be zero. Recently, Čučković and Louhichi [12] studied finite-rank semicommutators and commutators of Toeplitz operators with quasihomogeneous symbols and obtained different results from the case of harmonic Toeplitz operators.Motivated by recent work of Čučković and Louhichi, Zhang et al., and Zhou and Dong (see [1, 12, 13]), we discuss the finite-rank commutator (semicommutator) of Toeplitz operators with more general symbols on the unit ball in this paper. Let p and s be two multi-indexes. A function f∈L1(𝔹n,dA) is called a quasihomogeneous function of quasihomogeneous degree (p,s) if f is of the form ξpξ¯sφ(r) for all ξ in the unit sphere 𝕊n and some function φ(r) defined on the interval [0,1).Letf and g be two nonconstant quasihomogeneous functions (with certain restrictions on their quasihomogeneous degree). In this paper, we investigate the following problems:(1)
Under what conditions doesTfTg=TF hold for some quasihomogeneous function F?(2)
Under what conditions does the semicommutator(Tf,Tg] have finite rank?(3)
Under what conditions does the commutator[Tf,Tg] have finite rank?


## 2. The Mellin Transform and Mellin Convolution

Main tool in this paper will be the Mellin transform. Recall that the Mellin transformφ^ of a function φ∈L1([0,1],rdr) is defined by the equation:
(2.1)φ^(z)=∫01‍φ(s)sz-1ds.
It is easy to check that
(2.2)φ^(z0+p)=zpφ^(z0),
where p≥0 and z0∈ℝ.For convenience, we denoteφ^(z) by φ∧(z) when the form of φ is complicated. It is clear that φ^ is well defined on {z:Re(z)>2} and analytic on {z:Re(z)>2}. It is well known that the Mellin transform φ^ is uniquely determined by its values on {nk}k≥0, where nk∈ℕ and ∑k≥0(1/nk)=∞. The following classical theorem is proved in [14, page 102].Theorem 2.1.
Assume thatf is a bounded analytic function on {z:Re(z)>0} which vanishes at the pairwise distinct points z1, z2,…, where(i)
inf{|zn|}>0  and(ii)
∑n≥1‍Re(1/zn)=∞.Then f vanishes identically on {z:Re(z)>0}.Remark 2.2.
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
(2.3)φ^(nk)=0,∑k≥0‍1nk=∞,
then φ^(z)=0 for all z∈{z:Re(z)>2} and so φ=0.Iff and g are defined on the interval [0,1), then their Mellin convolution is defined by
(2.4)(f*Mg)(r)=∫r1‍f(rt)g(t)dtt.
The Mellin convolution theorem states that
(2.5)(f*Mg^)(r)=f^(r)g^(r),
and that, if f and g are in L1([0,1],rdr), then so is f*Mg.


## 3. Products of Toeplitz Operators with Quasihomogeneous Symbols

For any multi-indexα=(α1,…,αn), where each αi is a nonnegative integer, we will write
(3.1)|α|=α1+⋯+αn,α!=α1!⋯αn!,zα=z1α1⋯znαn
for z=(z1,…,zn)∈𝔹n.Forα=(α1,…,αn) and β=(β1,…,βn), the notation α≽β means that
(3.2)αi≥βi,i=1,…,n
and α⊥β means that
(3.3)α1β1+⋯+αnβn=0.
We also define
(3.4)α-β=(α1-β1,…,αn-βn)
and obtain
(3.5)|α-β|=|α|-|β|,ifα≽β.It is known thatφ is radial if and only if φ(Uz)=φ(z) for any unitary transform U of ℂn. So we have φ(rξ)=φ(rη) for ξ, η∈𝕊n and r∈[0,1). That is, φ(z) depends only on |z|. In this case, we denote φ by φ=φ(r) for convenience. The definition of quasihomogeneous function on the unit disk has been given in many papers (see [2] or [7]), and a similar definition on the unit ball will be given in the following.Definition 3.1.
Letp,s≽0. A function f∈L1(𝔹n,dA) is called a quasihomogeneous function of quasihomogeneous degree (p,s) if f is of the form ξpξ¯sφ where φ is a radial function, that is,
(3.6)f(rξ)=ξpξ¯sφ(r)
for any ξ in the unit sphere 𝕊n and r∈[0,1).The following lemma is from [1] and we will use it often.Lemma 3.2.
Letp, s be two multi-indexes and let φ be a bounded radial function on 𝔹n. Then for any multi-index α,
(3.7)Tξpφ(zα)=2(n+|α|+|p|)φ^(2n+2|α|+|p|)zα+p;Tξ¯sφ(zα)={0ifα≽s,2α!(n+|α|-|s|)!(α-s)!(n-1+|α|)!φ^(2n+2|α|-|s|)zα-sifα≽s;Tξpξ¯sφ(zα)={0ifα+p≽s,2(α+p)!(n+|α|+|p|-|s|)!(α+p-s)!(n-1+|α|+|p|)!φ^(2n+2|α|+|p|-|s|)zα+p-sifα+p≽s.Proposition 3.3.
Letp1,…,pm,s1,…,sm be multi-indexes and let φ1,…,φm be bounded radial functions on 𝔹n. If the product ∏i=0m-1Tξpm-iξ¯sm-iφm-i is of finite rank, then there exists α0 such that
(3.8)∏i=0m-1‍Tξpm-iξ¯sm-iφm-i(zα)=0forα≽α0.Proof.
Denote byS the product of Toeplitz operators ∏i=0m-1Tξpm-iξ¯sm-iφm-i and let rank S=N. For multi-indexes  α≽∑i=1m(pi+si), we have
(3.9)S(zα)=∏i=0m-2‍Tξpm-iξ¯sm-iφm-i(2(α+p1)!(n+|α|+|p1|-|s1|)!(α+p1-s1)!(n-1+|α|+|p1|)!φ1^(2n+2|α|+|p1|-|s1|)zα+p1-s1)=2ma(α)b(|α|)∏i=1m‍φi^(2n+2(|α|+∑j=0i-1‍(|pj|-|sj|))+|pi|-|si|)zα+∑i=1m(pi-si),
where
(3.10)a(α)=∏i=1m((α+∑j=0i-1(pj-sj)+pi)!)∏i=1m((α+∑j=1i(pj-sj))!),b(|α|)=∏i=1m((n+|α|+∑j=1i(|pj|-|sj|))!)∏i=1m((n-1+|α|+∑j=0i-1(|pj|-|sj|)+|pi|)!).
It follows that S(zα)=C(α,p,s)zα+∑i=1m‍(pi-si), where α≽∑i=1m‍(pi+si) and C(α,p,s) is a constant dependent on α, p and s. Thus the set {S(zα):α≽∑i=1m‍(pi+si)}=∨α≽∑i=1m‍(pi+si){zα+∑i=1m‍(pi-si)} contains at most N elements. Let n0=(N,…,N), then there exists α0≽n0+∑i=1m‍(pi+si) such that
(3.11)S(zα)=0foranyα≽α0.Proposition 3.4.
p
1
,
…
,
p
m
,
s
1
,
…
,
s
m, φ1,…,φm are defined as in Proposition 3.3. The product ∏i=0m-1Tξpm-iξ¯sm-iφm-i is of finite rank if and only if φi=0 for some i∈{1,…,m}.Proof.
Using Proposition3.3 and Theorem 2.1, we can easily get the result.This result is analogous to Theorem 3.2 in [15], but we get it in a different way.Similar to the proof of Proposition3.3, we can get a result about finite-rank commutators (semicommutators).Proposition 3.5.
Letp1,p2,s1, and s2 be multi-indexes and let φ1, φ2 be bounded radial functions on 𝔹n. If the commutator [Tξp1ξ¯s1φ1-Tξp2ξ¯s2φ2] (or the semicommutator (Tξp1ξ¯s1φ1-Tξp2ξ¯s2φ2]) is of finite rank, then there exists α0 such that [Tξp1ξ¯s1φ1-Tξp2ξ¯s2φ2](zα)=0 (or (Tξp1ξ¯s1φ1-Tξp2ξ¯s2φ2](zα)=0) for α≽α0.Now we are in a position to characterize when the product of two Toeplitz operators with some quasihomogeneous symbols equals a Toeplitz operator with quasihomogeneous symbols.Whenn=1, if p⊥s, then p=0 or s=0. But when n≥2, there exist nonzero multi-indexes p and s such that p⊥s. In this case, we have the following theorem.Theorem 3.6.
Supposep and s are two nonzero multi-indexes with p⊥s. Let φ1 and φ2 be bounded radial functions on 𝔹n. If there exists a bounded radial function φ such that Tξpφ1Tξ¯sφ2=Tξpξ¯sφ, then φ must be a solution of the equation
(3.12)r2|p|*M⋯*Mr2|p|+2(|s|-1)*M(r|p|+|s|φ)=r2*M⋯*Mr2(|s|-1)*M(r|p|φ1)*M(r|s|φ2).Proof.
Obviously, the equalityTξpφ1Tξ¯sφ2zα=Tξpξ¯sφzα holds for each monomial zα with the multi-index α.
Sincep⊥s, α+p≽s is equivalent to α≽s. By Lemma 3.2, it is easy to check that
(3.13)φ^(2n+2|α|+|p|-|s|)=2c(α)d(|α|)φ^1(2n+2|α|-2|s|+|p|)φ^2(2n+2|α|-|s|),
where
(3.14)c(α)=α!(α+p-s)!(α-s)!(α+p)!,d(|α|)=(n+|α|-|s|)!(n-1+|α|+|p|)!(n-1+|α|)!(n+|α|+|p|-|s|-1)!.
Since p⊥s, we have α!(α+p-s)!=(α+p)!(α-s)!. So (3.13) implies that
(3.15)φ^(2n+2|α|+|p|-|s|)=2d(|α|)φ^1(2n+2|α|-2|s|+|p|)φ^2(2n+2|α|-|s|).
As |s|≥1, we have
(3.16)φ^(2n+2|α|+|p|-|s|)∏i=1|s|(2n+2|α|+2|p|-2i)=φ^1(2n+2|α|-2|s|+|p|)φ^2(2n+2|α|-|s|)∏i=1|s|-1(2n+2|α|-2i).
A direct calculation gives that r2|p|+2i^(2n+2|α|-2|s|)=1/(2n+2|α|-2|s|+2|p|+2i) for 0≤i≤|s|-1. Then we have
(3.17)∏i=0|s|-1‍r2|p|+2i^(2n+2|α|-2|s|)r|p|+|s|φ^(2n+2|α|-2|s|)=∏i=1|s|-1‍r2i^(2n+2|α|-2|s|)r|p|φ^1(2n+2|α|-2|s|)r|s|φ^2(2n+2|α|-2|s|),
or equivalently,
(3.18)(r2|p|*M⋯*Mr2|p|+2(|s|-1)*M(r|p|+|s|φ))∧(2n+2|α|-2|s|)=(r2*M⋯*Mr2(|s|-1)*M(r|p|φ1)*M(r|s|φ2))∧(2n+2|α|-2|s|).
Combining the above equality with Remark 2.2, we get the conclusion.In the following, we give some explicit examples in which Theorem3.6 is applied.Example 3.7.
Supposep=(1,0), s=(0,1), φ1=r, φ2=r, φ is a bounded radial function such that TξprTξ¯sr=Tξpξ¯sφ. Using Theorem 3.6, we can get that φ=1.Example 3.8.
Supposes=0, φ is a bounded radial function such that Tξpφ1Tφ2=Tξpφ, then φ must be a solution of the equation
(3.19)r|p|φ1*Mφ2=χ[0,1]*Mr|p|φ,
where χ[0,1] is the characteristic function of the set [0,1]. For example, suppose p=(1,1), φ1=r, φ2=r4, φ is a bounded radial function such that TξprTr4=Tξpφ, then it follows that |p|=2 and φ=4r2-3r.Louhichi et al. [2] showed that there exist two nontrivial quasihomogeneous Toeplitz operators on the Bergman space of the unit disk such that the product of those Toeplitz operators is also a nontrivial Toeplitz operator, for example TeiθrTe-iθr=T1-log(1/r2). On weighted Bergman space of the unit ball 𝔹n(n>2), Vasilevski [16, 17] showed that there exist parabolic quasihomogeneous (It is clear that the quasihomogeneous function is also a parabolic quasihomogeneous function) symbol Toeplitz operators such that the finite product of those Toeplitz operators is also a Toeplitz operator of this type. However, on the unit ball 𝔹n(n≥2), if p and s are two nonzero multi-indexes which are not orthogonal, we can get that there exist no nontrivial φ1 and φ2 such that Tξpφ1Tξ¯sφ2=Tξpξ¯sφ.Theorem 3.9.
Letp, s be two nonzero multi-indexes which are not orthogonal. Given n≥2, and let φ1 and φ2 be two bounded radial functions on 𝔹n. If there exists a bounded radial function φ such that Tξpφ1Tξ¯sφ2=Tξpξ¯sφ, then φ1=0 or φ2=0.Proof.
IfTξpφ1Tξ¯sφ2=Tξpξ¯sφ, as in Theorem 3.6, we can get
(3.20)φ^(2n+2|α|+|p|-|s|)=2c(α)d(|α|)φ^1(2n+2|α|-2|s|+|p|)φ^2(2n+2|α|-|s|),α≽s,
where
(3.21)c(α)=α!(α+p-s)!(α-s)!(α+p)!,d(|α|)=(n+|α|-|s|)!(n-1+|α|+|p|)!(n-1+|α|)!(n+|α|+|p|-|s|-1)!.
We claim that there exists M0 such that
(3.22)φ^1(2n+2M-2|s|+|p|)φ^2(2n+2M-|s|)=0,forM≥M0.
Sincep is not orthogonal to s, without loss of generality, we can suppose p1s1≠0.
Case1 (n=2). We denote p=(p1,p2), s=(s1,s2). For M>s1+s2+1, let αM=(s1,M-s1) and βM=(s1+1,M-s1-1).
LetF(M)=c(βM)/c(αM)-1. Then
(3.23)F(M)=s1p1M2+((p1+1)(s1+1)(p2-2s2-s1)+(s1+p1+1)(2s1+s2-p2))M+N(M-s1)(M-s1+p2-s2)(s1+p1+1),
where N=(s1+s2)(p2-s2)(s1+1)(p1+1)-(s1+p1+1)s1(s1-p2+s2).
Therefore, the equationF(M)=0 has two solutions at most. It means that there exists M0≥|s|+1 such that F(M)≠0 for any M≥M0. Thus for each M≥M0, we have |αM|=|βM|=M and c(αM)≠c(βM).
Case2  (n≥3). Given a multi-index γ=(γ1,γ2,γ3,…,γn)≽s, let αM=(s1,M-s1,γ3,…,γn) and βM=(s1+1,M-s1-1,γ3,…,γn) for M>s1+s2+1. As in Case 1, we can find an integer M0 with M0≥|s|+1 such that |αM|=|βM|=M and c(αM)≠c(βM) for any M≥M0.
Using (3.20), we have
(3.24)0=(c(αM)-c(βM))d(M)φ^1(2n+2M-2|s|+|p|)φ^2(2n+2M-|s|),
for M>M0. As d(M)≠0 and c(αM)≠c(βM), it is easy to see that (3.22) holds.
LetE1={M∈ℕ:M≥M0 and φ^1(2n+2M-2|s|+|p|)=0} and E2={M∈ℕ:M≥M0 and φ^2(2n+2M-|s|)=0}. Then {M∈ℕ:M≥M0}=E1⋃‍E2. Since
(3.25)∑M≥M0‍1M≤∑M∈E1‍1M+∑M∈E2‍1M,
we know that at least one of the series ∑M∈E1‍(1/M) and ∑M∈E2‍(1/M) diverges. Hence it follows from Remark 2.2 that φ1=0 or φ2=0.


## 4. Finite-Rank Semicommutator

On the unit ball𝔹n(n≥2), we will show that the semicommutator of two Toeplitz operators with some quasihomogeneous symbols is of finite rank if and only if it is zero.Theorem 4.1.
Letp, s be two multi-indexes, and let φ1, φ2 be two integrable radial functions on 𝔹n such that φ1, ξpξ¯sφ2, and ξpξ¯sφ1φ2 are T-functions. If the semicommutator (Tφ1,Tξpξ¯sφ2] has finite rank, then it is equal to zero.Proof.
LetS be the semicommutator (Tφ1,Tξpξ¯sφ2]. If S is finite rank, using Proposition 3.5, there exists α0≽0 such that
(4.1)S(zα)=0∀α≽α0.
Therefore, Lemma 3.2 implies that
(4.2)(2n+2|α|+2|p|-2|s|)φ1^(2n+2|α|+2|p|-2|s|)φ2^(2n+2|α|+|p|-|s|)=φ1φ2^(2n+2|α|+|p|-|s|)
for all α≽α0. Since
(4.3)1(2n+2|α|+2|p|-2|s|)=r^|p|(2n+2|α|+|p|-2|s|),
the above equation is equivalent to
(4.4)r|p|φ1*Mr|s|φ2^(2n+2|α|+|p|-2|s|)=r|p|*Mr|s|φ1φ2^(2n+2|α|+|p|-2|s|).
Note that (r|p|φ1*Mr|s|φ2^) and (r|p|*Mr|s|φ1φ2^) are both analytic on the right half-plane {z:Rez>2} and the sequence {2n+2|α|+|p|-2|s|}α≽α0 is arithmetic. Then Remark 2.2 implies that
(4.5)r|p|φ1*Mr|s|φ2=r|p|*Mr|s|φ1φ2.
Hence, Tφ1Tξpξ¯sφ2=Tξpξ¯sφ1φ2. The proof is complete.Next we will consider when the semicommutator of two quasihomogeneous Toeplitz operators is a finite-rank operator.Remark 4.2.
If the semicommutator(Tξpφ1,Tξsφ2] is of finite rank, following the same process as in Theorem 4.1, we can prove that it must be zero.On the unit disk, Čučković and Louhichi [12] gave an example to show that there exists a nonzero finite rank semicommutator (Teipθf,Te-isθg], where f and g are radial functions. However, the situation on the unit ball 𝔹n(n≥2) is different. Let φ1, φ2 be two integrable radial functions on 𝔹n(n≥2) and p, s be two multi-indexes. Then we will prove that (Tξpφ1,Tξ¯sφ2] is a finite-rank operator if and only if (Tξpφ1,Tξ¯sφ2]=0. Now, we begin with the case that p and s are not orthogonal.Theorem 4.3.
Letp,s be two multi-indexes which are not orthogonal, and let φ1,φ2 be two integrable radial functions on 𝔹n(n≥2) such that ξpφ1, ξ¯sφ2 and ξpξ¯sφ1φ2 are T-functions. If the semicommutator (Tξpφ1,Tξ¯sφ2] has finite rank, then φ1=0 or φ2=0.Proof.
LetS be the semicommutator (Tξpφ1,Tξ¯sφ2]. If S is of finite rank, using Proposition 3.5, we can get that there exists α0 such that
(4.6)S(zα)=0∀α≽α0.
Lemma3.2 gives that
(4.7)α!(n+|α|-|s|)!(α-s)!(n-1+|α|)!2(n+|α|+|p|-|s|)φ1^(2n+2|α|-2|s|+|p|)φ2^(2n+2|α|-|s|)=(α+p)!(n+|α|+|p|-|s|)!(α+p-s)!(n-1+|α|+|p|)!φ1φ2^(2n+2|α|+|p|-|s|)
for all α≽α0.
Sincep is not orthogonal to s, from the proof of Theorem 3.9 and using (4.7), we can get that there exists M0 such that
(4.8)φ^1(2n+2M-2|s|+|p|)φ^2(2n+2M-|s|)=0,∀M≥M0.
Analogous to the proof of Theorem 3.9, it is easy to get φ1=0 or φ2=0.Next, we will show that there exists no nontrivial finite-rank semicommutator(Tξpφ1,Tξ¯sφ2] in the case that p⊥s.Theorem 4.4.
Letp, s be two multi-indexes with p⊥s, and let φ1 and φ2 be two integrable radial functions on 𝔹n such that ξpφ1, ξ¯sφ2 and ξpξ¯sφ1φ2 are T-functions. The semicommutator (Tξpφ1,Tξ¯sφ2] has finite rank if and only if (Tξpφ1,Tξ¯sφ2]=0.Proof.
We only need to prove the necessity. LetS=(Tξpφ1,Tξ¯sφ2] be of finite rank. Since p⊥s, it is easy to see that α+p≽s if and only if α≽s for multi-indexes α. By Lemma 3.2, the following statements hold:(i)
ifα+p≽s, then S(zα)=0;(ii)
ifα≽s, then
(4.9)S(zα)={2(α+p)!(n+|α|+|p|-|s|)!(α+p-s)!(n-1+|α|+|p|)!φ1φ2^(2n+2|α|+|p|-|s|)-2α!(n+|α|-|s|)!(α-s)!(n-1+|α|)!2(n+|α|+|p|-|s|)φ1^(2n+2|α|-2|s|+|p|)×φ2^(2n+2|α|-|s|)2(α+p)!(n+|α|+|p|-|s|)!(α+p-s)!(n-1+|α|+|p|)!}zα+p-s.Combining (i) and (ii) with the assumption that S has finite rank, we get that there exists α0≽s such that
(4.10)S(zα)=0forα≽α0.
To finish the proof, we will prove that S(zα)=0 for α≽s.
Note thatα!(α+p-s)!=(α+p)!(α-s)! for p⊥s and α≽s. Then for each α≽s, (ii) implies that S(zα)=0 if and only if
(4.11)(n+|α|-|s|)!(n-1+|α|)!φ1^(2n+2|α|-2|s|+|p|)φ2^(2n+2|α|-|s|)=(n+|α|+|p|-|s|-1)!2(n-1+|α|+|p|)!φ1φ2^(2n+2|α|+|p|-|s|),
that is,
(4.12)1∏i=1|s|-1(2n+2|α|-2i)φ^1(2n+2|α|-2|s|+|p|)φ^2(2n+2|α|-|s|)=1∏i=1|s|(2n+2|α|+2|p|-2i)φ1φ2^(2n+2|α|+|p|-|s|).
Since χ[0,1]^(n)=1/n, the preceding equality is equivalent to
(4.13)∏i=1|s|-1‍χ[0,1]^(2n+2|α|-2i)φ^1(2n+2|α|-2|s|+|p|)φ^2(2n+2|α|-|s|)=∏i=1|s|‍χ[0,1]^(2n+2|α|+2|p|-2i)φ1φ2^(2n+2|α|+|p|-|s|)
for all α≽s.
By (4.10), we obtain that the equality (4.13) holds for all α≽α0. It is easy to see that {2n+2|α|}α≽α0 is arithmetic. Therefore, by Remark 2.2, we have
(4.14)∏i=1|s|-1‍χ[0,1]^(z-2i)φ^1(z-2|s|+|p|)φ^2(z-|s|)=∏i=1|s|‍χ[0,1]^(z+2|p|-2i)φ1φ2^(z+|p|-|s|)
for all Rez≥2.
In particular ifz=2n+2|α|, with α≽s, we have
(4.15)∏i=1|s|-1‍χ[0,1]^(2n+2|α|-2i)φ^1(2n+2|α|-2|s|+|p|)φ^2(2n+2|α|-|s|)=∏i=1|s|‍χ[0,1]^(2n+2|α|+2|p|-2i)φ1φ2^(2n+2|α|+|p|-|s|).
It follows that the equality (4.13) holds for all α≽s. So S(zα)=0 for all α≽s. Hence, the proof is complete.Example 4.5.
Letp, s be two multi-indexes, φ is a bounded radial function and ap, bs∈ℂ. If (Tφ,Tapzp] is of finite rank, using Theorem 4.1, we obtain that (Tφ,Tapzp]=0. If p is not orthogonal to s and (Tapzp,Tbsz¯s] is of finite rank, so it follows from Theorem 4.3 that ap=0 or bs=0. But if p⊥s, there exist ap0≠0 and bs0≠0 such that (Tap0zp,Tbs0z¯s]=0. In particular, suppose p=(0,0), s=(2,2), ap0=bs0=1, a direct calculation gives that p⊥s and TzpTz¯s=Tzpz¯s, that is, (Tzp,Tz¯s]=0.


## 5. Finite-Rank Commutators

In this section, letφ1,φ2 be two integrable radial functions on 𝔹n. We now pass to investigate the commutator of two quasihomogeneous Toeplitz operators and consider when [Tφ1,Tξpξ¯sφ2], [Tξpφ1,Tξsφ2], or [Tξpφ1,Tξ¯sφ2] have finite rank, respectively.Theorem 5.1.
Letp,s be two multi-indexes with p⊥s, and let φ1,φ2 be two integrable radial functions on 𝔹n such that φ1 and ξpξ¯sφ2 are T-functions. If φ1 is not a constant, then [Tφ1,Tξpξ¯sφ2] is of finite rank if and only if |p|=|s| or φ2=0.Proof.
LetS be the commutator [Tφ1,Tξpξ¯sφ2]. By Lemma 3.2, S(zα)=0 if and only if
(5.1)φ^2(2n+2|α|+|p|-|s|){(n+|α|+|p|-|s|)φ^1(2n+2|α|+2|p|-2|s|)-(n+|α|)φ^1(2n+2|α|)}=0,
for α≽s. If S is of finite rank, using Proposition 3.5, there exists α0≽p+s such that
(5.2)S(zα)=0forα≽α0.
Sincep⊥s, by Theorem 2.1 and following the same process as in Theorem 4.4, we get S=0. Using Theorem 4.4 in [1], we have S=0 if and only if |p|=|s| or φ2=0.
Conversely, if|p|=|s| or φ2=0, then we can easily show that S(zα)=0 for each multi-index α, which implies that Tφ1 and Tξpξ¯sφ2 commute.Remark 5.2.
The same as in Theorem5.1, we can easily prove if the commutator [Tξpφ1,Tξsφ2] is of finite rank, then it must be a zero operator.Next, we give some examples.Example 5.3.
Suppose thatf(z)=apzs, where ap≠0 and s≠0. Let g=ξkφ≠0 be a T-function, where φ is a radial function. Then [Tf,Tg] is of finite rank if and only if Tf and Tg commute. By Theorem 4.9 in [1], we can also get [Tf,Tg] is of finite rank if and only if g is a monomial.On the unit disk, if the commutator[Teipθf,Te-isθg] has finite rank N, then N is at most equal to the quasihomogeneous degree s and a nonzero finite rank commutator has been given in [12]. On the unit ball 𝔹n(n≥2), we will show that the commutator [Tξpφ1,Tξ¯sφ2] has finite rank if and only if Tξpφ1 commutes with Tξ¯sφ2 if and only if φ1=0 or φ2=0.Theorem 5.4.
Letp, s be two nonzero multi-indexes, and let φ1, φ2 be two integrable radial functions on 𝔹n(n≥2) such that ξpφ1 and ξ¯sφ2 are T-functions. If the commutator [Tξpφ1,Tξ¯sφ2] has finite rank, then φ1=0 or φ2=0.Proof.
LetS denote the commutator [Tξpφ1,Tξ¯sφ2]. Applying Lemma 3.2, we get
(5.3)S(zα)=4((α)!(n+|α|-|s|)!(n+|α|+|p|-|s|)(α-s)!(n-1+|α|)!φ^1(2n+2|α|+|p|-2|s|)φ^2(2n+2|α|-|s|)-(α+p)!(n+|α|+|p|-|s|)!(n+|α|+|p|)(α+p-s)!(n-1+|α|+|p|)!×φ^1(2n+2|α|+|p|)φ^2(2n+2|α|+2|p|-|s|)(α)!(n+|α|-|s|)!(n+|α|+|p|-|s|)(α-s)!(n-1+|α|)!)zα+p-s,
for α≽s. If S is finite rank, using Proposition 3.5, there exists α0≽s such that
(5.4)S(zα)=0forα≽α0.
Ifp⊥s, then we have α!(α+p-s)!=(α+p)!(α-s)!. Combining (5.4) and (5.3), we have
(5.5)(n+|α|-|s|)!(n+|α|+|p|-|s|)(n-1+|α|)!φ^1(2n+2|α|+|p|-2|s|)φ^2(2n+2|α|-|s|)=(n+|α|+|p|-|s|)!(n+|α|+|p|)(n-1+|α|+|p|)!φ^1(2n+2|α|+|p|)φ^2(2n+2|α|+2|p|-|s|),
for α≽α0. Analogous to the proof of Theorem 4.8 in [1], it is not difficult to get that φ1=0 or φ2=0.
On the other hand, ifp is not orthogonal to s, then (5.3) and (5.4) imply that
(5.6)φ^1(2n+2|α|+|p|)φ^2(2n+2|α|+2|p|-|s|)=Cαφ^1(2n+2|α|+|p|-2|s|)φ^2(2n+2|α|-|s|),
for α≽α0, where
(5.7)Cα=α!(α+p-s)!(n+|α|-|s|)!(n-1+|α|+|p|)!(α+p)!(α-s)!(n-1+|α|)!(n+|α|+|p|-|s|-1)!(n+|α|+|p|).
Following the same process as in Theorem 3.9, we get φ1=0 or φ2=0, as desired.



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*Source: 414201-2012-10-02.xml*