# Algebraic Properties of Toeplitz Operators on the Polydisk

Authors: Bo Zhang; Yanyue Shi; Yufeng Lu
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We discuss some algebraic properties of Toeplitz operators on the Bergman space of the polydisk𝔻n. Firstly, we introduce Toeplitz operators with quasihomogeneous symbols and property (P). Secondly, we study commutativity of certain quasihomogeneous Toeplitz operators and commutators of diagonal Toeplitz operators. Thirdly, we discuss finite rank semicommutators and commutators of Toeplitz operators with quasihomogeneous symbols. Finally, we solve the finite rank product problem for Toeplitz operators on the polydisk.


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## Body

## 1. Introduction

Let𝔻 be the open unit disk in the complex plane ℂ and its boundary the unit circle 𝕋. For a fixed positive integer n, the unit polydisk 𝔻n and the torus 𝕋n are the subsets of ℂn which are Cartesian products of n copies 𝔻 and 𝕋, respectively. Let dV(z)=dVn(z) denote the Lebesgue volume measure on the polydisk 𝔻n, normalized so that the measure of 𝔻n equals 1. Let Lp=Lp(𝔻n) denote the usual Lebesgue space. The Bergman space A2=A2(𝔻n) is the Hilbert space consisting of holomorphic functions on 𝔻n that are also in L2(𝔻n,dV(z)). Since every point evaluation is a bounded linear functional on A2, there corresponds to every z=(z1,…,zn)∈𝔻n a unique function Kz∈A2 which has the following reproducing property: (1.1)f(z)=〈f,Kz〉,f∈A2,
where the notation 〈·,·〉 denotes the inner product in L2. The function Kz is the well-known Bergman kernel and its explicit formula is given by (1.2)Kz(w)=∏j=1n1(1-wjzj¯)2,w∈Dn.
Here and elsewhere zj denotes the jth component of z. The Bergman projection P is defined for the Hilbert space orthogonal projection from L2 onto A2. Given a function φ∈L∞(𝔻n,dV), the Toeplitz operator Tφ:A2→A2 is defined by the formula(1.3)Tφ(f)(z)=P(φf)(z)=∫Dnf(w)φ(w)Kz(w)¯dV(w)
for all f∈A2. Since the Bergman projection P has norm 1, it is clear that Toeplitz operators defined in this way are bounded linear operators on A2 and ∥Tφ∥≤∥φ∥∞.We now consider a more general class of Toeplitz operators. ForF∈L1(𝔻n,dV), in analogy to (1.3) we define an operator TF by(1.4)TFf(z)=∫DnF(w)f(w)Kz(w)¯dV(w).Since the Bergman projectionP can be extended to L1(𝔻n,dV), the operator TF is well defined on H∞, where H∞ is the space of bounded holomorphic functions on 𝔻n. Hence, TF is always densely defined on A2(𝔻n). Since P is not bounded on L1(𝔻n,dV), it is well known that TF can be unbounded in general. This motivates the following definition, which is based on the definitions on unit ball in [1].Definition 1.1.
LetF∈L1(𝔻n,dV).(a)
F is called a T-function if (1.4) defines a bounded operator on A2.(b)
IfF is a T-function, one writes TF for the continuous extension of the operator (it is defined on the dense subset H∞ of L2(𝔻n)) defined by (1.4). TF is called a Toeplitz operator on A2.(c)
If there existrj∈(0,1), 1≤j≤n, such that F is (essentially) bounded on {z=(z1,z2,…,zn):rj<|zj|<1,1≤j≤n}, then one says F is “nearly bounded.”Notice that theT-functions form a proper subset of L1(𝔻n,dV) which contains all bounded and “nearly bounded” functions. In this paper, the functions which we considered are all T-functions without special introduction. We denote the semicommutator and commutator of two Toeplitz operators Tf and Tg by (1.5)(Tf,Tg]=Tfg-TfTg,[Tf,Tg]=TfTg-TgTf.The commuting problem and the finite-rank product problem for Toeplitz operators on the Hardy and Bergman spaces over various domains are some of the most interesting problems in operator theory.For commuting problem, in 1963, Brown and Halmos [2] showed that two bounded Toeplitz operators Tφ and Tψ on the classical Hardy space commute if and only if (i) both φ and ψ are analytic, (ii) both φ¯ and ψ¯ are analytic, or (iii) one is a linear function of the other. On the Bergman space of the unit disk, some similar results were obtained for Toeplitz operators with bounded harmonic symbols or analytic symbols (see [2–4]). The problem of characterizing commuting Toeplitz operators with arbitrary bounded symbols seems quite challenging and is not fully understood  until now.  In recent years, by Mellin transform, some results with quasihomogeneous symbols (it is of the form eikθϕ, where ϕ is a radial function) or monomial symbols were obtained (see [5–7]). On the Hardy and Bergman spaces of several complex variables, the situation is much more complicated. On the unit ball, Toeplitz operators with pluriharmonic or quasihomogeneous symbols were studied in [1, 8–11]. On the polydisk, some results about Toeplitz operators with pluriharmonic symbols were obtained in [10, 12–14].For finite-rank product problem, Luecking recently proved that a Toeplitz operator with measure symbol on the Bergman space of unit disk has finite rank if and only if its symbols are a linear combination of point masses (see [15]). In [16], Choe extended Luecking’s theorem to higher-dimensional cases. Using those results, Le studied finite-rank products of Toeplitz operators on the Bergman space of the unit disk and unit ball in [17, 18].Motivated by recent work in [1, 5, 7, 17, 18], we define quasihomogeneous functions on the polydisk and study Toeplitz operators with quasihomogeneous symbols on the Bergman space of the polydisk. The present paper is assembled as follows. In Section 2, we introduce Mellin transform, Toeplitz operators with quasihomogeneous symbols and property (P). In Section 3, we study commutativity of certain quasihomogeneous Toeplitz operators and commutators of diagonal Toeplitz operators. In Sections 4 and 5, we prove that finite rank semicommutators and commutators of Toeplitz operators with quasihomogeneous symbols must be zero operator and we also solve the finite-rank product problem for Toeplitz operators on the Bergman space of the polydisk.


## 2. Mellin Transform, Toeplitz Operators with Quasihomogeneous  Symbols and Property (P)

For any multi-indexα=(α1,…,αn)∈ℕn (here ℕ denotes the set of all nonnegative integers), we write aα=α1⋯αn and zα=z1α1⋯znαn for z=(z1,…,zn)∈𝔻n. The standard orthonormal basis for A2 is {eα:α∈ℕn}, where (2.1)eα(z)=(α1+1)⋯(αn+1)zα,α∈Nn,z∈Dn.For twon-tuples of integers α=(α1,…,αn) and β=(β1,…,βn), we define α≻β if αj>βj for all 0≤j≤n. Similarly, we write α≽β if αj≥βj for all 1≤j≤n and α≽β if otherwise. We also define α⊥β if α1β1+⋯+αnβn=0 and α-β=(α1-β1,…,αn-βn).For anyk=(k1,…,kn)∈ℤn, particularly we write k⃗1=(k1,…,k1) and put k*=(|k1|,…,|kn|), k+=(1/2)(k*+k) and k-=(1/2)(k*-k). Then, k+,k-≽0, k=k+-k-, and k+⊥k-.Recall that a functionφ on 𝔻n is radial if and only if φ(z) depends only on (|z1|,|z2|,…,|zn|), that is, φ(eiθ1z1,eiθ2z2,…,eiθnzn)=φ(z1,z2,…,zn) for any θ1,θ2,…,θn∈R. For any function f∈L1(𝔻n,dV), we define the radicalization of f by (2.2)rad(f)(z1,z2,…,zn)=1(2π)n∫02π⋯∫02πf(eit1z1,eit2z2,…,eitnzn)dt1⋯dtn.
Then, f is radial if and only if rad(f)=f. For α∈ℕn, we have (2.3)〈Trad(f)zα,zα〉=〈1(2π)n∫02π⋯∫02πf(eit1z1,eit2z2,…,eitnzn)dt1⋯dtnzα,zα〉=1(2π)n∫02π⋯∫02π∫Dnf(eit1z1,eit2z2,…,eitnzn)zαz¯αdV(z)dt1⋯dtn=1(2π)n∫02π⋯∫02πdt1⋯dtn∫Dnf(w1,w2,…,wn)wαw¯αdV(w)=〈Tfzα,zα〉.The main tool in this paper will be the Mellin transform. which is defined by the equation(2.4)φ̂(z1,…,zn)=∫0∞⋯∫0∞φ(s1,…,sn)s1z1-1⋯snzn-1ds1⋯dsn.
We apply the Mellin transform to functions in L1([0,1]n,r1⋯rndr1⋯drn); then, (2.5)φ̂(z1,…,zn)=∫01⋯∫01φ(s1,…,sn)s1z1-1⋯snzn-1ds1⋯dsn.
For convenience, we denote φ̂(z1,…,zn) by φ∧(z1,…,zn) when the form of φ is complicated. It is clear that φ̂ is well defined on In={z=(z1,z2,…,zn):Rezj>2,j=1,2,…,n}. Using the Hartogs theorem, for any function φ∈L1([0,1]n,r1⋯rndr1⋯drn), the Mellin transform of φ is a bounded holomorphic function on In.By calculation, we can get(2.6)φ̂(z+p)=φ̂(z1+p1,z2+p2,…,zn+pn)=rpφ̂(z1,z2,…,zn)=rpφ̂(z),
where p=(p1,p2,…,pn)≽0, z=(z1,z2,…,zn)∈In, and rp=r1p1r2p2⋯rnpn.The quasihomogeneous functions have been defined in many spaces (see [5, 7]). In the following, we give a similar definition on the polydisk 𝔻n.Definition 2.1.
Letk∈ℤn. A function f∈L1(𝔻n,dV) is called a quasihomogeneous function of degree k if f is of the form ξkφ where φ is a radial function, that is,
(2.7)f(rξ)=ξkφ(r)
for any ξ in the torus 𝕋n and r∈[0,1)n.As in [19], for any n-tuple k∈ℤn, let Hk={f∈L2:fis a quasihomogeneous function of degree k}. It is clear that Hk is a closed subspace of L2. By Lemma 3.2 in [19], L2=⨁s∈ℤnHs. In particular, for all z=(r1ξ1,…,rnξn)∈𝔻, if f∈Hk, that is, f(r1ξ1,…,rnξn)=ξkfk(r1,…,rn), then we conclude that L2(𝔻n,dV)=⨁k∈ℤnξkfk(r1,…,rn), fk∈ℜ, where ℜ={φ:𝔻n→ℂradial|∫[0,1]n|φ(r1,…,rn)|2∏i=1nridri<+∞}.Lemma 2.2.
Letk,l∈ℤn, and let φ, ψ be radial functions on 𝔻n, such that ξkφ, ξlψ, and ξk+lφψ are all T-functions. Then, the following equation holds for every α∈ℕn:
(2.8)Tξkφ(zα)={0ifα⪰k-,2naα+k+1⃗φ̂(2α+k+2⃗)zα+k,ifα⪰k-,
where aα+k+1⃗=(α1+k1+1)⋯(αn+kn+1) and φ̂(2α+k+2⃗)=φ̂(2α1+k1+2,2α2+k2+2,…,2αn+kn+2).Using Lemma2.2, we can get the following two results:(2.9)(Tξkφ,Tξlψ](zα)={0ifα∈E1,2naα+m+1⃗φψ̂(2α+m+2⃗)zα+mifα∈E1c∩E2c,(2naα+m+1⃗φψ̂(2α+m+2⃗)-λ)zα+mifα∈E2,
where m=k+l, λ=4naα+l+1⃗aα+m+1⃗ψ̂(2α+l+2⃗)φ̂(2(α+l)+k+2⃗), E1={α:α≽m-}, E2={α:α≽l-}∩{α:α+l≽k-}, E1c=ℕn∖E1, and E2c=ℕn∖E2. It is easy to check that E1∩E2=ϕ(2.10)[Tξkφ,Tξlψ](zα)={0ifα∈F1c∩F2c,λ1zα+k+lifα∈F1∩F2c,-λ2zα+k+lifα∈F1c∩F2,(λ1-λ2)zα+k+lifα∈F1∩F2,
where λ1=4naα+l+1⃗aα+k+l+1⃗ψ̂(2α+l+2⃗)φ̂(2(α+l)+k+2⃗), λ2=4naα+k+1⃗aα+k+l+1⃗φ̂(2α+k+2⃗)ψ̂(2(α+k)+l+2⃗), F1={α:α≽l-}∩{α:α+l≽k-},F2={α:α≽k-}∩{α:α+k≽l-},F1c=ℕn∖F1 and F2c=ℕn∖F2.LetG be a region in complex plane ℂ and f holomorphic on G. If {zk}k=1∞ has a limit point in G, such that f(zk)=0, then f≡0. For functions of several complex variables, the above conclusion does not hold. For example, f(z1,z2)=z1z2 is analytic on bidisk {(z1,z2):|z1|<1,|z2|<1}, point sequence {(0,1/k)}, k=2,3,…, has a limit point (0,0), and f(0,1/k)=0, but f is not a zero function on the bidisk. So we need the following definition, which is given in [9, 17].For any1≤j≤n, let σj:ℕ×ℕn-1→ℕn be the map defined by the formula σj(s,(α1,…,αn-1))=(α1,…,αj-1,s,αj,…,αn-1) for all s∈ℕ and (α1,…,αn-1)∈ℕn-1. If M is a subset of ℕn and 1≤j≤n, we define (2.11)Mj̃={α̃=(α1,…,αn-1)∈Nn-1:∑s∈N,σj(s,α̃)∈M1s+1=∞}.As in [9, 17], we say that M has property (P) if one of the following statements holds:(1)
M=∅,(2)
M≠∅,n=1, and ∑s∈M1/s<∞, or(3)
M≠∅,n≥2, and, for any 1≤j≤n, the set Mj̃ has property (P) as a subset of ℕn-1.LetM and N be two sets that have property (P). It is not difficult to check that the following statements hold:(1)
M⋂N and M⋃N have property (P);(2)
ℕn∖M do not have property (P).Lemma 2.3.
Ifφ∈L1([0,1]n,r1⋯rndr1⋯drn) and Z(φ̂)={α∈ℕn:φ̂(α)=0} does not have property (P), then φ is identically zero.Proof.
By the Müntz theorem, we can prove that it is true whenn=1 (see [7] for more details). Suppose that the conclusion of the lemma holds whenever n≤N, where N is a positive integer. Consider the case n=N+1. Since Z(φ̂) does not have property (P), there must be a 1≤j≤N+1, such that Z(φ̂)̃j does not have property (P). Without loss of generality, taking j=N+1, then, Z(φ̂)̃N+1≠∅. For each r̃∈Z(φ̂)̃N+1, ∑s∈ℕ,φ̂(r̃,s)=01/(s+1)=∞. So φ̂(r̃,zN+1)=0, for all zN+1∈I1. For every λ∈I1, let φ̂λ(z′)=φ̂(z′,λ); then, φ̂λ is an analytic function on IN and Z(φ̂λ)=Z(φ̂)̃N+1, which does not have property (P). By the induction hypothesis, we have φ̂(z′,λ)=0,z′∈IN. Thus, φ̂(z)=0 on IN+1. Therefore, φ is identically zero.Theorem 2.4.
Letp=(p1,p2,…,pn)∈ℤn, and let f be a T-function. Then, the following statements hold.(i)
IfEp={α;〈Tfzα+p,zα〉=0forallα≽p-} does not have property (P), then 〈Tfzα+p,zα〉=0 for all α≽p-.(ii)
LetE, E′⊆ℕn be the sets that have property (P). If 〈Tfzα,zβ〉=0 for all α∈ℕn∖E, β∈ℕn∖E′, then f(z)≡0 for almost all z∈𝔻n.(iii)
If〈Tfzα,zβ〉=0forallα+p≠β, then f is a quasihomogeneous function of degree p.Proof.
(i) By direct computation, we have(2.12)〈Tfzα+p,zα〉=∫Dnf(z)zα+pz¯αdV(z)=1πn∫[0,1]n∫[0,2π]nf(r1eiθ1,…,rneiθn)∏j=1nrj2αj+|pj|+1∏j=1n(eiθj)pjdθ1⋯dθndr1⋯drn.
Let
(2.13)F(r1,…,rn)=1πn∫[0,2π]nf(r1eiθ1,…,rneiθn)∏j=1n(eiθj)pjdθ1⋯dθn.
Then, F∈L1([0,1]n,r1⋯rndr1⋯drn). In fact, ∥F(r1,…,rn)∥L1([0,1]n,r1⋯rndr1⋯drn)≤∥f∥L1(𝔻n,dV). Therefore, equality (2.12) shows that F̂(2α1+|p1|+2,…,2αn+|pn|+2)=0 for any α=(α1,…,αn)∈ℕn. That is, Z(F̂) does not have property (P). Thus, Lemma 2.3 implies that F≡0 and 〈Tfzα+p,zα〉=0 for all α≽p-.
(ii) For eachl=(l1,…,ln)∈ℕn, 〈Tfzα,zα+l〉=0 for α∈(ℕn∖E)∩(ℕn∖(E′-l)). Since E and E′ have property (P), the subset (ℕn∖E)∩(ℕn∖(E′-l))=(ℕn∖(E∪(E′-l)) does not have property (P). By (i), we have 〈Tfzα,zα+l〉=0 for α∈ℕn. It is easy to prove that 〈Tfzα+l,zα〉=0 for α∈ℕn. So 〈Tfzα,zβ〉=0 for all α,β∈ℕn, that is, Tf=0 and f(z)≡0 for almost all z∈𝔻n.
(iii) Since(2.14)rad(ξ¯pf)(z1,…,zn)=1(2π)n∫[0,2π]n(ξ¯pf)(eit1z1,eit2z2,…,eitnzn)dt1⋯dtn=1(2π)n∫[0,2π]ne-i(t1+θ1)p1⋯e-i(tn+θn)pnf(r1ei(t1+θ1),…,rnei(tn+θn))dt1⋯dtn=1(2π)n∫[0,2π]ne-i(∑j=1ntjpj)f(r1eit1,…,rneitn)dt1⋯dtn,
we have
(2.15)∫Dn[ξprad(ξ¯pf)](z)zαz¯βdV(z)=1(2π)n∫[0,2π]n∫[0,1]n∫[0,2π]ne-i(∑j=1nθj(pj+αj-βj))f(r1eit1,…,rneitn)dθ1⋯dθn×(∏j=1nrjαj+βj+1)e-i(∑j=1ntjpj)dr1⋯drndt1⋯dtn.
If α+p≠β, then ∫𝔻n[ξprad(ξ¯pf)](z)zαz¯βdV(z)=0. Otherwise, if α+p=β, then
(2.16)∫Dn[ξprad(ξ¯pf)](z)zαz¯βdV(z)=1(2π)n∫[0,2π]n∫[0,1]nf(r1eit1,…,rneitn)(∏j=1nrjαj+βj+1)e-i(∑j=1ntjpj)dr1⋯drndt1⋯dtn=∫Dnf(z)zαz¯βdV(z).
Thus, we get ∫𝔻n[ξprad(ξ¯pf)](z)zαz¯βdV(z)=∫𝔻nf(z)zαz¯βdV(z) for any α,β∈ℤn. So ξprad(ξ¯pf)=f, this means that there exists φ(r) such that ξ¯pf(z)=φ(r), that is, f(z)=ξpφ(r) is a quasihomogeneous function of degree p.Remark 2.5.
Letf be as in Theorem 2.4. Then, Tf=∑α∈ℕnw(f,α,α+p)eα+p⊗eα, where
(2.17)w(f,α,α+p)=〈Tfeα,eα+p〉=aα+1⃗aα+p+1⃗∫Dnξpφ(r)zαz¯α+pdV(z)=aα+1⃗aα+p+1⃗φ̂(2α+p+2⃗).
Recall that a densely defined operator on A2(𝔻n) is said to be diagonal if it is diagonal with respect to the standard orthonormal basis. In particular, for f∈L∞(𝔻n), Tf is diagonal if and only if rad(f)=f. In this case, Tf=∑α∈ℕnw(f,α)eα⊗eα, where w(f,α)=〈Tfeα,eα〉=aα+1⃗f̂(2α+2⃗).


## 3. Commutativity of Toeplitz Operators

In this section, we study the commutativity of the Toeplitz operators with some special quasihomogeneous symbols and give the characterizations, respectively.Theorem 3.1.
Letg=ξpφ(r)∈L2(𝔻n) be a quasihomogeneous function of degree p  and f=∑k∈ℤnξkfk(r1,…,rn)∈L2(𝔻n). Then, TfTg=TgTf if and only if TξkfkTg=TgTξkfk for any k∈ℤn.
Moreover, the following statements hold.(i)
IfQ1={α:α+p≽0}∩{α:α+k≽0}∩{α:α+k+p≽0}≠∅, then, for each α∈Q1, φ̂(2α+2k+p+2⃗)f̂k(2α+k+2⃗)=0.(ii)
IfQ2={α:α+k≽0}∩{α:α+p≽0}∩{α:α+k+p≽0}≠∅, then, for each α∈Q2, φ̂(2α+p+2⃗)f̂k(2α+2p+k+2⃗)=0.Proof.
Note that, forα∈ℕn,
(3.1)TgTf(eα)=∑k∈ZnTgTξkfkeα=∑k∈Zn∑β∈Nn〈TgTξkfkeα,eβ〉eβ=∑k∈Zn〈TgTξkfkeα,eα+k+p〉eα+k+p.
The second equality follows that 〈TgTξkfkeα,eβ〉=0, when β≠α+k+p.
Similarly,(3.2)TfTg(eα)=∑k∈ZnTξkfkTgeα=∑k∈Zn∑β∈Nn〈TξkfkTgeα,eβ〉eβ=∑k∈Zn〈TξkfkTgeα,eα+k+p〉eα+k+p.
Since {eα} are the standard orthogonal basis and 〈TgTξkfkeα,eβ〉=〈TξkfkTgeα,eβ〉=0 for β≠α+k+p, it is easy to check that the following statements are equal:(I)
TfTg=TgTf;(II)
〈TgTξkfkeα,eα+k+p〉=〈TξkfkTgeα,eα+k+p〉, α∈ℕn;(III)
TgTξkfkeα=TξkfkTgeα, α∈ℕn;(IV)
TgTξkfk=TξkfkTg.
Furthermore,(3.3)〈TgTξkfkeα,eα+k+p〉=〈Tξkfkeα,eα+k〉〈Tgeα+k,eα+k+p〉={0,α+k⪰0orα+k+p⪰0,aα+1⃗aα+k+p+1⃗aα+k+1⃗φ̂(2α+2k+p+2⃗)f̂k(2α+k+2⃗),α+k⪰0,α+k+p⪰0,〈TξkfkTgeα,eα+k+p〉=〈Tgeα,eα+p〉〈Tξkfkeα+p,eα+k+p〉={0,α+p⪰0orα+k+p⪰0,aα+1⃗aα+k+p+1⃗aα+p+1⃗f̂k(2α+2p+k+2⃗)φ̂(2α+p+2⃗),α+p⪰0,α+k+p⪰0.
Thus, the statements (i) and (ii) hold.Theorem 3.2.
Letf1, f2 be quasihomogeneous functions of degree p and -s, where p≽s≻0. If [Tf1,Tf2]=0, then f1=0 or f2=0.Proof.
Iff1,f2 are quasihomogeneous functions of degree p and -s, then there exist radial functions φ1 and φ2, such that f1=ξpφ1 and f2=ξ¯sφ2. If Tf1Tf2=Tf2Tf1, (2.10) implies that, for all α∈ℕn,
(3.4)aα-s+1⃗φ2̂(2α-s+2⃗)φ1̂(2α-2s+p+2⃗)=aα+p+1⃗φ1̂(2α+p+2⃗)φ2̂(2α+2p-s+2⃗),ifα⪰s,φ1̂(2α+p+2⃗)φ2̂(2α+2p-s+2⃗)=0,ifα⪰s.
We claim that there exist {λk}k=1∞ with ∑1/λk=∞ such that
(3.5)φ1̂(2λk+p1+2,z′)φ2̂(2λk+2p1-s1+2,z′)=0,foranyz′∈In-1.
It follows that Z(φ1̂φ2̂) do not have property (P). So we can get f1=0 or f2=0 by Lemma 2.3.
We only need to prove the claim. Sinces≻0, there exists 1≤j≤n, such that sj≥1. Without losing generality, suppose that j=1. Let λ0=s1-1≥0; then, α0=(λ0,α′)≽s and φ1̂(2λ0+p1+2,α′+p′+2⃗)φ2̂(2λ0+p1+2,α′+2p′-s′+2⃗)=0 where α′∈ℕn-1, p′=(p2,…,pn) and s′=(s2,…,sn). Denote E0={z′∈In-1:φ1̂(2λ0+p1+2,z′)=0} and F0={z′∈In-1:φ1̂(2λ0+p1+2,z′)=0}. Note that at least one of the sets E0 and F0 does not have property (P). Since φ1̂(2λ0+p1+2,z′)=0 and φ2̂(2λ0+p1+2,z′)=0 are analytic on In-1, Lemma 2.3 shows that E0=In-1 or F0=In-1.
Case 1.
IfE0=In-1, then
(3.6)φ1̂(2(α0+s)+p+2⃗)φ2̂(2(α0+s)+2p-s+2⃗)=aα-s+1⃗aα+p+1⃗φ1̂(2φ0+p+2⃗)φ2̂(2α0+s+2⃗)=0,
where α0=(λ0+s,α′) with α′∈Nn-1. Let λ1=λ0+s. Denote by E1={z′∈In-1:φ1̂(2λ1+p1+2,z′)=0} and F1={z′∈In-1:φ1̂(2λ1+p1+2,z′)=0}. Then, at least one of the sets E1 and F1 does not have property (P). By Lemma 2.3 again, we have E1=In-1 or F1=In-1. Thus φ1̂(λ1,z′)φ2̂(λ1,z′)=0,foranyz′∈In-1.Case 2.
IfF0=In-1, then(3.7)φ1̂(2(α0+p)+p+2⃗)φ2̂(2(α0+p)+2p-s+2⃗)=aα-s+1⃗aα+p+1⃗φ1̂(2α0+3p-2s+2⃗)φ2̂(2α0+2p-s+2⃗)=0.
By the same technique, we can get that (3.5) holds when λ1=λ0+p1.
Similarly, we can find a sequenceλk=λ0+u(k)p1+v(k)s1, where the functions u(k)=1 or 0, v(k)=1 or 0, and u(k)+v(k)=1 for k∈ℤ+. Then, (i) λ1≥min{p1,s1}, (ii) ∑k∈ℕ1/λk=+∞, and (iii) for every k≥1, λk satisfies (3.5). So we complete the proof.Forn=2, we have the following results.Theorem 3.3.
Letg(r)=r1m1r2m2, where m1≥0 and m2≥0. Let f(r1,r2)∈L∞([0,1]2), p=(p1,p2)∈ℕ2 and p1·p2≠0. Then, TξpgTξpf=TξpfTξpg if and only if there exists an analytic function on ℂ, such that the function ψ((z2p1-z1p2)/(p12+p22))/((z1+m1)(z2+m2)) is bounded on I2 and
(3.8)f̂(z1,z2)=ψ((z2p1-z1p2)/(p12+p22))(z1+m1)(z2+m2),∀z=(z1,z2)∈I2.Proof.
As in the proof of Theorem3.2, it is easy to check that TξpgTξpf=TξpfTξpg if and only if
(3.9)〈TξpgTξpfeα,eα+2p〉=〈TξpfTξpgeα,eα+2p〉,
which is equal to
(3.10)ĝ(2α+p+2⃗)f̂(2α+3p+2⃗)=f̂(2α+p+2⃗)ĝ(2α+3p+2⃗),∀α∈N2.
Suppose that there is a functionψ as in this theorem.
Note that(3.11)ĝ(z)=∫[0,1]2rm+z-1⃗dr1dr2=1(z1+m1)(z2+m2)≠0,∀z∈I2,
and (2α2+3p2+2)p1-(2α1+3p1+2)p2=(2α2+p2+2)p1-(2α1+p1+2)p2, for any α=(α1,α2)∈ℕ2. Then, it is easy to check that equality (3.10) holds, that is, TξpgTξpf=TξpfTξpg.
Conversely, ifTξpg and Tξpf commute, we will structure an analytic ψ which satisfies the conditions in this theorem.
Sinceĝ(z)≠0 for all z∈I2, the function f̂(z)/ĝ(z) is analytic on I2. Note that |rz-1|<1 for 0<rj<1, j=1,2 and z∈I2. Thus,
(3.12)|f̂(z)|≤∫D2|f|dV(z)=‖f‖L1([0,1]2),|ĝ(z)|≤‖g‖L1([0,1]2).
Fix α0∈ℕ2, and let z0=2α0+p+2⃗, then,
(3.13)f̂(z0)ĝ(z0)=f̂(z0+2p)ĝ(z0+2p)=⋯=f̂(z0+2kp)ĝ(z0+2kp),k=0,1,2,….
Combining this with Lemma 2.3, we can get that the above equality holds for any z0∈I2. Let p⊥=(-p2,p1); then, p⊥p⊥. For each z0∈I2, there exist μ1,μ2∈ℂ such that z0=μ1p+μ2p⊥. So
(3.14)f̂(μ1p+μ2p⊥)ĝ(μ1p+μ2p⊥)=f̂((μ1+2)p+μ2p⊥)ĝ((μ1+2)p+μ2p⊥)=⋯=f̂((μ1+2k)p+μ2p⊥)ĝ((μ1+2k)p+μ2p⊥),k=0,1,2,….
Put
(3.15)F(λ)=f̂(λp+μ2p⊥)ĝ(λp+μ2p⊥)-f̂(μ1p+μ2p⊥)ĝ(μ1p+μ2p⊥),
then, F(λ) is analytic on {z∈ℂ:Re(zp1-μ2p2)>2andRe(zp2+μ2p1)>2} and
(3.16)|F(λ)|≤‖f‖L1(|m1+λp1-μ2p2|⋅|m2+λp2+μ2p1|+C1)≤‖f‖L1(D1|λ|2+D2|λ|+D3),
where C1,D1,D2,D3 are all constants. Since F(μ1+2k)=0 and ∑k=0+∞1/2k=+∞, the set {μ1+2k:k=0,1,2,…}⊆Z(F). Thus F(λ)≡0. That is
(3.17)f̂(λp+μ2p⊥)=f̂(μ1p+μ2p⊥)ĝ(μ1p+μ2p⊥)ĝ(λp+μ2p⊥).
For each μ∈ℂ, there exists λ0∈ℂ such that Re(λ0p1-μp2)>2andRe(λ0p2+μp1)>2; then, let ψ(μ)=f̂(λ0p+μp⊥)/ĝ(λ0p+μp⊥). By equality (3.17), we conclude that the function ψ is well defined. Since the function f̂/ĝ is analytic on I2, we can prove that ψ is an analytic function on ℂ. Let
(3.18)z1=λp1-μ2p2,z2=λp2+μ2p1;
Then,
(3.19)λ=z1p1-z2p2p12+p22,μ2=z2p1-z1p2p12+p22.
So (3.17) is equal to
(3.20)f̂(z1,z2)=ψ((z2p1-z1p2)/(p12+p22))(z1+m1)(z2+m2),
where |ψ((z2p1-z1p2)/(p12+p22))/((z1+m1)(z2+m2))|≤∥f∥L1([0,1]2) and (z1,z2)∈I2. This completes the proof.Corollary 3.4.
Letf,g be as in Theorem 3.3 and p∈ℕ+; then, the following statements hold:(i)
Teipθ1gTeipθ1f=Teipθ1fTeipθ1g if and only if f=r1m1φ(r2), where φ∈L∞([0,1]);(ii)
Teipθ2gTeipθ2f=Teipθ2fTeipθ2g if and only if f=r2m2φ(r1), where φ∈L∞([0,1]).Proof.
(i)  By (3.17) we have
(3.21)∫[0,1]2f(r1,r2)r1z1-1r2z2-1dr1dr2=ψ(z2p)∫[0,1]2r1m1+z1-1r2m2+z2-1dr1dr2,
that is,
(3.22)∫[0,1][∫[0,1]f(r1,r2)r2z2-1dr2-r1m1ψ(z2p)∫[0,1]r2m2+z2-1dr2]r1z1-1dr1=0.
Then,
(3.23)∫[0,1]f(r1,r2)r2z2-1dr2=r1m1ψ(z2p)1m2+z2.
It follows that there exists φ=φ(r2)∈L∞([0,1]) such that f=r1m1φ(r2).
On the other hand, iff=r1m1φ(r2), then
(3.24)f̂(z)ĝ(z+2(p,0))=1m1+z1⋅φ̂(z2)⋅1m1+z1+2p⋅1m2+z2,f̂(z+2(p,0))ĝ(z)=1m1+z1+2p⋅φ̂(z2)⋅1m1+z1⋅1m2+z2.
Thus, we have TξpgTξpf=TξpfTξpg.
(ii) Can also be proved in the same way.In [6], Čučković and Rao showed that if f,g∈L∞(𝔻) and g is a nonconstant radial function, then TfTg=TgTf implies that f is a radial function. However, this is not true if f,g∈L∞(𝔻n), where n≥2. For example, g(z)=g(z1,…,zj-1,|zj|,zj+1,…,zn) and f(z)=f(|z1|,…,|zn|) only for |zj|, and it is clear that TgTf=TfTg, but g(z) may not be a radial function. Let G={g∈L∞(𝔻n):gisradialandforf∈L∞(𝔻n),TfTg=TgTfimpliesthatfisradial}(if n=1, this set is exactly the set of all non-constant bounded radial functions). In the following, we can give a complete description of G.Theorem 3.5.
G={g(z) is a bounded radial function: for each k=(k1,…,kn)≠0,a(2z)ĝ(2z)≠a(2z+2k)ĝ(2z+2k),wherez∈Inandz+k∈In}.Proof.
Suppose thatf(z)=∑k∈ℤnξkfk(r1,…,rn)∈L∞(𝔻n) and g is a radial function. Lemma 2.2 shows that
(3.25)TfTg(zα)=∑k∈Zn2naα+1⃗ĝ(2α+2⃗)Ckf̂k(2α+k+2⃗)zα+k,TgTf(zα)=∑k∈ZnCkf̂k(2α+k+2⃗)2naα+k+1⃗ĝ(2α+2k+2⃗)zα+k,
where α∈ℕn and
(3.26)Ck={0ifα⪰k-,2naα+k+1⃗ifα⪰k-.
It follows that TfTg=TgTf if and only if
(3.27)aα+1⃗ĝ(2α+2⃗)Ckf̂k(2α+k+2⃗)=Ckf̂k(2α+k+2⃗)aα+k+1⃗ĝ(2α+2k+2⃗),
for any k∈ℤn. Let E1={α:(aα+1⃗g-aα+k+1⃗rkg)∧(2α+2⃗)=0} and E2={α:rkf̂k(2α+2⃗)=0}. Then E1∪E2={α;α≽k-}. The commutativity of Tf and Tg is equivalent to that at least one of E1 and E2 does not have property (P); then, Lemma 2.3 shows that fk≡0 or a(2z)ĝ(2z)≠a(2z+2k)ĝ(2z+2k), where z∈Inandz+k∈In. The rest of the proof is obvious.Remark 3.6.
In Theorem3.5, particularly if g is a radial function such that g(z)=∏j=1ngj(zj) or g(z)=∑j=1ngj(zj), where each gj(zj)(1≤j≤n) is a non-constant radial function, then foreachk=(k1,⋯,kn)≠0,a(2z)ĝ(2z)≠a(2z+2k)ĝ(2z+2k),wherez∈Inandz+k∈In, so g∈G. It follows that G is nonempty.


## 4. Finite Rank Semicommutators and Commutators

Recall that Čučković and Louhichi (see [5]) have found some nonzero finite rank semicommutators of quasihomogeneous symbol Toeplitz operators on the Bergman space of unit disk. In this section, we will show that the finite rank semicommutators and commutators of Toeplitz operators with quasihomogeneous symbols must be zero on A2(𝔻n) with n≥2. Our idea is mainly from [17].Theorem 4.1.
Letk,l∈ℤn with n≥2, k+l=m, and let φ, ψ be radial functions such that f1=ξkφ, f2=ξlψ, and ξmφψ are all T-functions. If the semicommutator (Tf1,Tf2] has finite rank, then it must be zero.Proof.
LetS denote the semicommutator (Tf1,Tf2]. For α∈ℕn, if S is finite rank, by equality (2.9), we have that there exists α0≽k-+l- such that
(4.1)S(zα)=2naα+m+1⃗(2naα+l+1⃗ψ̂(2α+l+2⃗)φ̂(2(α+l)+k+2⃗)-φψ̂(2α+m+2⃗))=0forα⪰α0,
which is equivalent to
(4.2)(rk++l+φ)∧(2α+l-k--l-+2⃗)(rk-+l-ψ)∧(2α+l-k--l-+2⃗)=(rl++l-)∧(2α+l-k--l-+2⃗)(rkφψ)∧(2α+l-k--l-+2⃗),
for α≽α0. Combining this with Lemma 2.3, we get
(4.3)(rk++l+φ)∧(z)(rk-+l-ψ)∧(z)=(rl++l-)∧(z)(rkφψ)∧(z),forz∈In.
Hence,
(4.4)S(zα)=0∀α∈E2.
In the following, we only need to prove thatφψ̂(α)=0 for all α∈E1c∩E2c.
IfE1c∩E2c≠∅, there is a j  (0≤j≤n) such that mj-<lj-. Without loss of generality, assume that j=1. Then, {(α1,m2+α2,…,mn+αn):m1-≤α1≤l1-,αj≥0,j=2,…,n}⊆E2. For each m1-≤α1, let Fa1(r2,…,rn)=∫01(φψ)(r1,r1,…,rn)r1a1-1dr1. Since Z(Fa1̂)⊇{(m2+α2,…,mn+αn):(α2,…,αn)∈Nn-1} does not have property (P), we have Fa1̂≡0. Therefore, φψ̂(a1,a2,…,an)=0 for m1-≤a1≤l1- and aj≥mj-,j=2,…,n. So S(zα)=0 for α∈E1c∩E2c.
This completes the proof.We now pass to the commutator of two quasihomogeneous Toeplitz operators. Here the situation is the same as for the semicommutator.Theorem 4.2.
Letk,l∈ℤn with n≥2, and let φ, ψ be radial functions such that f1=ξkφ and f2=ξlψ are both T-functions. The commutator [Tf1,Tf2] has finite rank if and only if it is a zero operator.Proof.
LetS denote the commutator [Tf1,Tf2]. For α∈ℕn, if S has finite rank N, by equality (2.10), we have that there exists α0≽k-+l- such that
(4.5)S(zα)=4naα+k+l+1⃗(aα+l+1⃗ψ̂(2α+l+2⃗)φ̂(2(α+l)+k+2⃗)-aα+k+1⃗φ̂(2α+k+2⃗)ψ̂(2(α+k)+l+2⃗))(zα)=0
for α≽α0. As in the proof of Theorem 4.1, the above equation implies that
(4.6)(χ[0,1]n)∧(2α+2k+2⃗)ψ̂(2α+l+2⃗)φ̂(2(α+l)+k+2⃗)=(χ[0,1]n)∧(2α+2l+2⃗)φ̂(2α+k+2⃗)ψ̂(2(α+k)+l+2⃗)
for α∈F1∩F2. Hence,
(4.7)S(zα)=0∀α∈F4.
For α∈F1∩F2c or α∈F1c∩F2, following the same way as above, we can also prove that S(zα)=0.
This completes the proof.


## 5. Finite Rank Products of Toeplitz Operators

In [17], the author showed that under certain conditions on the bounded operators S1 and S2 on A2(𝔹n), if f∈L2(𝔹n), such that S2TfS1 is a finite-rank operator, then f must be zero almost everywhere on 𝔹n. On the Bergman space of the polydisk, using the same method as in [17], we can prove Theorem 5.1. Using Theorem 5.1, we get two useful theorems for Toeplitz operators with quasihomogeneous symbols.Theorem 5.1.
LetS1, S2 be two bounded operators on A2(𝔻n). Suppose that there is a set 𝕊⊆ℕn which has property (P), such that ker(S2)⊆ℳ¯ and 𝒩⊆ran(S1). Here, ℳ (resp., 𝒩) is the linear subspace of A2(𝔻n) spanned by {zm,m∈𝕊} (resp., {zm,m∈ℕn∖𝕊}). Suppose that f∈L2(𝔻n) such that the operator S2TfS1 has finite rank; then f is the zero function.Theorem 5.2.
LetM and W be two positive integers. Let f1,…,fM and g1,…,gW be quasihomogeneous functions, none of which is the zero function. If f∈L2 such that the operator TfM⋯Tf1TfTgW⋯Tg1 has finite rank, then f is the zero function.Proof.
LetS2=TfM⋯Tf1 and S1=TgW⋯Tg1. Suppose that fj=ξpjφj(r), 1≤j≤M, and gl=ξqlψl(r), 1≤l≤W, where pj,ql∈ℤn. By Lemma 2.2, for α≽∑l=1wql-, we have
(5.1)S1(zα)=(22nW∏j=1Waα+∑l=1jql+1⃗ψĵ(2(α+∑l=1jql)-qj+2⃗))zα+∑j=1Wqj.
Define 𝒥={α∈ℕn:α≽∑l=1Wql-}⋃(⋃j=1W{α∈ℕn:ψĵ(2(α+∑l=1jql)-qj+2⃗)=0}). Since none of the functions ψ1,…,ψW is the zero function, the set 𝒥 has property (P).
Forα∈ℕn∖𝒥, we see that S1(zα)≠0. Suppose that φ∈Aα2 such that S1(φ)=0; then,
(5.2)0=S1(φ)=S1(∑α∈Nn〈φ,zα〉zα)=∑α∈Nn〈φ,zα〉S1zα.
So (5.1) implies that for any α∈ℕn∖𝒥, 〈φ,zα〉=0. Therefore ker(S1) is contained in the closure of the linear span of {zα:α∈𝒥} in Aα2. Now suppose that
(5.3)I={α:α⪰∑l=1Wql-}⋃(Nn⋂(J+∑j=1Wqj)).
Then the set ℐ has property (P) and, for any α∈ℕn∖ℐ, β=α-∑j=1Wqj belongs to ℕn∖𝒥. Equality (5.1) implies that zα=zβ+∑j=1Wqj is a multiple of S1zβ. So the linear span of {zα:α∈ℕn∖ℐ} is contained in the range of S1. So there exist subsets 𝒥 and ℐ of ℕn that have property (P) such that ker(S2) is contained in the closure in Aα2 of Span({zα:α∈𝒥}) and Span({zα:α∈ℕn∖ℐ} is a subspace of S1(Aα2). Let 𝕊=ℐ⋃𝒥; then, Theorem 5.1 implies that f is the zero function.Theorem 5.3.
Suppose that the functionf(z)∈L2(𝔻n) has the expansion
(5.4)f(z)=∑k⪯Mξkfk(r1,…,rn),
and f̂M(l)≠0 for all l≽l0, where l0∈ℕn, if there is a function g(z)∈L2(𝔻n), such that TgTf has finite rank; then g=0.Proof.
Forα∈ℕn,
(5.5)Tf(zα)=∑M⪰k⪰-α2naα+k+1⃗fk̂(2α+k+2⃗)zα+k=2naα+M+1⃗fM̂(2α+M+2⃗)zα+M+∑M⪰k⪰-α,k≠M2naα+k+1⃗fk̂(2α+k+2⃗)zα+k.
By hypothesis, there exists α0∈ℕn, such that, for any α≽α0, f̂M(2α+M+2⃗)≠0; then, f̂M(2α0+M+2⃗)≠0. Thus, we have
(5.6)zα0+M∈Span⁡{Tf(zα0),zα:0⪯α⪯α0+M,α≠α0+M}.Considering the same argument, we get, for all l≽0,
(5.7)zα0+M+l∈Span⁡{Tf(zβ),zα:α0⪯β⪯α0+l,0⪯α⪯α0+M,α≠α0+M}.
Now suppose thatTgTf has finite rank, and let {φ1,…,φN} be the set that spans TgTf(𝒫), where 𝒫 is the space of all holomorphic polynomials in the variable z. Then, for any l∈ℕn, we see that Tgzα0+M+l is a linear combination of {φ1,…,φN}⋃{Tg(zα),where0⪯α⪯α0+Mandα≠α0+M}, and it follows that Tg is a finite-rank operator. By Theorem 2.4, we conclude that g is the zero function.



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*Source: 962313-2011-11-22.xml*