# Algebraic Properties of Quasihomogeneous and  Separately Quasihomogeneous Toeplitz Operators on the Pluriharmonic Bergman Space

Authors: Hongyan Guan; Liu Liu; Yufeng Lu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study some algebraic properties of Toeplitz operator with quasihomogeneous or separately quasihomogeneous symbol on the pluriharmonic Bergman space of the unit ball inℂn. We determine when the product of two Toeplitz operators with certain separately quasi-homogeneous symbols is a Toeplitz operator. Next, we discuss the zero-product problem for several Toeplitz operators, one of whose symbols is separately quasihomogeneous and the others are quasi-homogeneous functions, and show that the zero-product problem for two Toeplitz operators has only a trivial solution if one of the symbols is separately quasihomogeneous and the other is arbitrary. Finally, we also characterize the commutativity of certain quasihomogeneous or separately quasihomogeneous Toeplitz operators.


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## Body

## 1. Introduction

Forn≥1, let ℂn be the cartesian product of n copies of ℂ. For any points z=(z1,z2,…,zn) and w=(w1,w2,…,wn) in ℂn, we use the notions 〈z,w〉=z1w1¯+z2w2¯+⋯+znwn¯ and |z|=|z1|2+|z2|2+⋯+|zn|2 for the inner product and the associated Euclidean norm. Let 𝔹n denote the open unit ball which consists of points z∈ℂn with |z|<1 and let 𝕊n denote the unit sphere. Let dV be the normalized Lebesgue volume measure on 𝔹n and let dσ be the normalized surface area measure on 𝕊n. L2(𝔹n,dV) is the Hilbert space consisting of all Lebesgue square integrable functions on 𝔹n with the inner product
(1)〈f,g〉=∫𝔹n‍f(z)g(z)¯dV(z).The Bergman spaceLa2(𝔹n) is the closed subspace consisting of the analytic functions in L2(𝔹n,dV). Let P be the orthogonal projection from L2(𝔹n,dV) onto La2(𝔹n), then P can be expressed by
(2)(Pf)(z)=〈f,Kz〉=∫𝔹n‍f(w)1(1-〈z,w〉)n+1dV(w),
where Kz(w)=1/(1-〈w,z〉)n+1 is the Bergman reproducing kernel.A functionf is said to be pluriharmonic if and only if f satisfies that DjD¯kf=0(j,k=1,2,…,n), where Dj=∂/∂zj and D¯j=∂/∂z¯j (see page 9 of [1]). The pluriharmonic Bergman space, denoted by Lh2(𝔹n), is the closed subspace of L2(𝔹n,dV) consisting of all the pluriharmonic functions on 𝔹n. It is well known that Lh2(𝔹n) is also a Hilbert space. We will write Q for the orthogonal projection from L2(𝔹n,dV) onto Lh2(𝔹n). It is easy to verify that each point evaluation is a bounded linear functional on Lh2(𝔹n). It follows that Lh2(𝔹n) is also a reproducing function space with reproducing kernel:
(3)Rz(w)=Kz(w)+Kz(w)¯-1,z,w∈𝔹n,Qf(z)=Pf(z)+Pf¯(z)¯-Pf(0),f∈L2(𝔹n,dV).For a functionϕ∈L∞(𝔹n,dV), we define the Toeplitz operator Tϕ:Lh2(𝔹n)→Lh2(𝔹n) with symbol ϕ by
(4)Tϕ(f)=Q(ϕf),f∈Lh2(𝔹n).For product problem, on the Hardy space, Brown and Halmos [2] showed that if f and g are bounded functions on the unit circle, then TfTg is another Toeplitz operator if and only if either f¯ or g is analytic. In the setting of the Bergman space, the condition that either f¯ or g is analytic is still sufficient, but it is no longer necessary. Ahern and Čučković [3] showed that a Brown-Halmos type result holds for Toeplitz operators with harmonic symbols on La2(𝔹n). In [4], Ahern characterized when the product of two Toeplitz operators with harmonic symbols is a Toeplitz operator. Later in [5], Louhichi et al. gave the necessary and sufficient conditions for the product of two quasihomogeneous Toeplitz operators to be a Toeplitz operator. Recently, Dong and Zhou [6] characterized when the product of quasihomogeneous Toeplitz operators is a Toeplitz operator on the harmonic Bergman space of the unit disk.The situation is more complicated on the Hardy, Bergman, and harmonic Bergman spaces of several complex variables. In 2003, Ding [7] discussed the product problem for two Toeplitz operators with bounded symbols on the Hardy space H2(𝔻n). After that, Choe et al. [8] solved the product problem for pluriharmonic Toeplitz operators on the Bergman space of the polydisk. On the Bergman space of the unit ball, Zhou and Dong [9] determined when the product of two radial Toeplitz operators is a Toeplitz operator. Later in [10], they discussed the product problem for two separately quasihomogeneous Toeplitz operators. In Zhang and Lu’s paper [11], they characterized the product problem for two Toeplitz operators with quasihomogeneous symbols. On the pluriharmonic Bergman space, Yang et al. [12] gave the necessary and sufficient conditions for the product of two radial Toeplitz operators to be a Toeplitz operator.For zero-product problem, on the Hardy space, Brown and Halmos [2] proved that if f,g∈L∞(𝕋) such that TfTg=0, then one of the symbols must be the zero function. Motivated by this result, Guo [13] showed that Tϕ1Tϕ2⋯Tϕn=0 implies that ϕi=0 for some i, when n=5. After that, Gu [14] proved that for n=6, the result in [13] is also true. Recently, Aleman and Vukotić [15] completely solved the zero-product problem for several Toeplitz operators on the Hardy space. On the Bergman space of the unit disk, Ahern and Čučković [3] obtained that the result is analogous to that in [2] for the zero-product problem of two harmonic Toeplitz operators. Furthermore, they got that TfTg=0 implies f=0 or g=0, where f is arbitrary bounded and g is radial in [16]. In 2003, Čučković [17] proved that if f∈L∞(𝔻) such that TfTzj-z¯l=0, where j,l are both positive integers, then f=0. Later in [18], Louhichi et al. considered the zero-product problem for f,g∈L∞(𝔻,dA) with g=∑k=-∞N‍eikθgk, where gk is a bounded radial function and N is a positive integer. On the Bergman space of the unit ball, Dong and Zhou [10] investigated the zero-product problem of two Toeplitz operators, one of whose symbols is separately quasihomogeneous and the other is arbitrary bounded. Bauer and Vasilevski [19] considered the zero-product problem and a more general problem of zero finite sum of finite products of Toeplitz operators. Recently, Yang et al. [12] discussed the zero-product problem for several radial Toeplitz operators on the pluriharmonic Bergman space of the unit ball.For commuting problem, Brown and Halmos [2] firstly considered the commutativity of two Toeplitz operators on the Hardy space. They showed that two bounded Toeplitz operators Tϕ and Tψ commute if and only if (1) both ϕ and ψ are analytic, (2) both ϕ and ψ are coanalytic, or (3) one is a linear function of the other. On the Bergman space of the unit disk, Axler and Čučković [20] obtained that the same result is also true for Toeplitz operators with bounded harmonic symbols. In [21], Čučković and Rao used the Mellin transform to study the commutativity of two Toeplitz operators on La2(𝔻) and described those operators which commute with Teipθrm for (p,m)∈ℕ×ℕ. Later in [22], Louhichi and Zakariasy characterized commuting Toeplitz operators on La2(𝔻) with quasihomogeneous symbols. On the Bergman space of the unit ball, Zheng [23] studied commuting Toeplitz operators with pluriharmonic symbols. Recently, extending Vasilevski's results in [24, 25], Quiroga-Barranco and Vasilevski gave the description of many (geometrically defined) classes of commuting Toeplitz operators of the unit ball in [26, 27]. Zhou and Dong [9] studied the commuting problem for quasihomogeneous Toeplitz operators. In 2012, Dong and Zhou [28] and Louhichi and Zakariasy [29] characterized the commuting Toeplitz operators with radial or quasihomogeneous symbols on the harmonic Bergman space of the unit disk. In papers [19, 30–33], the authors studied the wide classes of (nongeometrically defined) commutative Banach algebras generated by Toeplitz operators of the Bergman spaces on the unit ball.Motivated by recent results of the unit ball in [9, 10, 12], in this paper, on the pluriharmonic space of the unit ball, we first characterize the product of two Toeplitz operators with certain separately quasihomogeneous symbols to be a Toeplitz operator. Next, we solve the zero-product problem for several Toeplitz operators when one of the symbols is separately quasihomogeneous and the others are quasihomogeneous and show that zero-product problem for two Toeplitz operators with certain symbols has only a trivial solution. At last, the commutativity of certain (separately) quasihomogeneous Toeplitz operators is also discussed.


## 2. Preliminaries

Letℕ denote the set of all nonnegative integers. For any α=(α1,α2,…,αn)∈ℕn, for any point z=(z1,z2,…,zn)∈𝔹n, we write
(5)|α|=α1+α2+⋯+αn,α!=α1!α2!,…,αn!,zα=z1α1z2α2,…,znαn.For two multi-indexesα=(α1,α2,…,αn), β=(β1,β2,…,βn)∈ℕn, the notations α≽β and α⊥β mean that αi≥βi for every i and α1β1+α2β2+⋯+αnβn=0. Let α-β denote (α1-β1,α2-β2,…,αn-βn). Moreover, if α≽β, |α-β|=|α|-|β|.For a functionϕ∈L1(𝔹n,dV), ϕ is said to be radial if and only if ϕ(Uz)=ϕ(z) for any unitary transformation U of ℂn; ϕ is said to be separately radial if and only if ϕ(Uz)=ϕ(z) for any unitary transformation U of ℂn with a diagonal matrix. This implies that a radial function has a form ϕ(z)=ϕ(|z|) and a separately radial function has a form ϕ(z)=ϕ(|z1|,…,|zn|).Using radial function and separately radial function, we give the following definition.Definition 1.
Letp,s∈ℕn with p⊥s and f∈L1(𝔹n,dV).
(I)f is called a quasihomogeneous function of degree (p,s) if f has the following decomposition:
(6)f(|r|ξ)=ξpξ¯sϕ(|r|)
for any ξ∈𝕊n, r=(r1,r2,…,rn) and |r|=|r1|2+|r2|2+⋯+|rn|2, where ri=|zi| for 1≤i≤n and ϕ is a radial function. In this case, Tf is called quasihomogeneous Toeplitz operator of degree (p,s).
(II)f is called a separately quasihomogeneous function of degree (p,s) if f has the following decomposition:
(7)f(|r|ξ)=ξpξ¯sϕ(r)
for any ξ∈𝕊n, where ϕ is a separately radial function. In this case, Tf is called separately quasihomogeneous Toeplitz operator of degree (p,s).
We now recall some useful results from [34]. Denote by τ(𝔹n) the base of 𝔹n, that is,
(8)τ(𝔹n)={r=(r1,r2,…,rn)=(|z1|,|z2|,…,|zn|):z=(z1,z2,…,zn)∈𝔹n}.
If ϕ is a bounded separately radial function, we get that
(9)∫𝔹n‍ϕ(z)dV(z)=2nn!∫τ(𝔹n)‍ϕ(r)rdr,
where rdr=∏i=1nridri.
Letℛ={ϕ:𝔹n→ℂisseparatelyradial:∫τ(𝔹n)‍|ϕ(r)|2rdr<∞}. Dong and Zhou [10] showed that for f∈L2(𝔹n,dV), f has the decomposition
(10)f(|r|ξ)=∑p⊥s,p,s∈ℕn‍ξpξ¯sfp,s(r),fp,s∈ℛ.
They also proved the following result.Lemma 2.
Letf(z)=∑p⊥s,p,s∈ℕnξpξ¯sfp,s(r)∈L∞(𝔹n,dV); then ξpξ¯sfp,s(r) is bounded on 𝔹n for multi-indexes p,s∈ℕn with p⊥s.In order to get our main results, we need to introduce the Mellin transform, which is defined for any functionϕ∈L1([0,1],rdr) by the formula
(11)ϕ^(z)=∫01‍ϕ(r)rz-1dr.
It is well known that ϕ^ is well defined on the right half-plane {z:Rez≥2} and is analytic on {z:Rez>2}. It is helpful that the Mellin transform is uniquely determined by its value on an arithmetic sequence of integers. In fact, we have the following result (see [35, page 102]).Lemma 3.
Suppose thatϕ is a bounded analytic function on {z:Rez>0} which vanishes at the pairwise distinct points z1,z2,…, where(I)
inf{|zn|}>0, and(II)
∑n≥1Re(1/zn)=∞.Then ϕ vanishes identically on {z:Rez>0}.Remark 4.
We will often use Lemma3 to show that if ϕ∈L1([0,1],rdr) and if there exists a sequence {nk}⊂ℕ such that
(12)ϕ^(nk)=0,∑k≥0‍1nk=∞,
then ϕ^(z)=0 for all z∈{z:Rez>2} and so ϕ=0.In this paper, we will need a similar result in higher dimensions. Now we give the following definition.Definition 5.
LetE be a subset of ℤ+2; one says that E satisfies condition (I) if the following statement holds:
(I) there exists a sequence{αi(1)}i=1∞ such that ∑i=1∞‍=1/αi(1)=∞, and for every fixed αi(1), there also exists a sequence {αj(i)(2)}j(i)=1∞ such that ∑j(i)=1∞‍1/αj(i)(2)=∞ and {(αi(1),αj(i)(2)):j(i)=1,2,…}⊂E.Remark 6.
It follows from Definition5 that for a multi-index δ∈ℕ2, if E is a subset of {α∈ℤ+2:α≽δ} and if Ec is the complement of E in {α∈ℤ+2:α≽δ}, then either E or Ec satisfies condition (I).In this paper, we will often use Lemmas 4 and 12 in [12] and Lemma 2.1 and Corollary 2.7 in [10] which can be stated as follows.Lemma 7.
Letϕ be an integrable radial function on 𝔹n such that Tϕ is a bounded operator then for any multi-index α,
(13)Tϕ(zα)=2(n+|α|)!ϕ^(2n+2|α|)zα,Tϕ(z¯α)=2(n+|α|)!ϕ^(2n+2|α|)z¯α.Lemma 8.
Letp,s be two multi-indexes and let ϕ be an integrable radial function on 𝔹n such that Tξpϕ, Tξ¯sϕ and Tξpξ¯sϕ are bounded operators. Then for any multi-index α,
(14)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-ps≽p+α,s≠p+α,0s⋡p+α,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-sp≽s+α,p≠s+α,0p⋡s+α,s+α⋡p.
In particular, if p,s are two nonzero multi-indexes with p⊥s, one has
(15)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;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.Lemma 9.
Letα,β∈ℕn with α≠β and let t∈[0,1). If ϕ is a bounded separately radial function on 𝔹n, then
(16)∫𝕊n‍ϕ(tξ)ξαξ¯βdσ(ξ)=0.Lemma 10.
Letp,s∈ℕ2 and let g(r) be a bounded function on τ(𝔹2). If the set
(17)E={α∈ℤ+2:α≽s,∫τ(𝔹2)‍g(r)r2α+p-srdr=0}
satisfies condition (I), then g=0.


## 3. The Product of Toeplitz Operators with Separately Quasihomogeneous Symbols

We start this section with the following result.Lemma 11.
Letϕ be a bounded separately radial function on 𝔹n; then for any multi-index α∈ℕn,
(18)Tϕ(zα)=2n(n+|α|)!∫τ(𝔹n)‍ϕ(r)r2αrdrα!zα,Tϕ(z¯α)=2n(n+|α|)!∫τ(𝔹n)‍ϕ(r)r2αrdrα!z¯α.Proof.
For multi-indexesα,β∈ℕn, it is well known that
(19)〈zα,zβ〉={0α≠β,n!α!(n+|α|)!α=β.
According to (9) and Lemma 9, we get that
(20)〈Tϕzα,zβ〉=〈ϕzα,zβ〉=∫𝔹n‍ϕ(z)zαz¯βdV(z)={0α≠β,2nn!∫τ(𝔹n)‍ϕ(r)r2αrdrα=β.
Similarly, if β≽0, β≠0, we obtain 〈Tϕzα,z¯β〉=0=〈zα,z¯β〉. Since {zβ}β≽0∪{z¯β}β≽0,β≠0 is a basis of the pluriharmonic Bergman space, we have
(21)Tϕ(zα)=2n(n+|α|)!∫τ(𝔹n)‍ϕ(r)r2αrdrα!zα.
By a similar argument, one can deduce that Tϕ(z¯α)=((2n(n+|α|)!∫τ(𝔹n)‍ϕ(r)r2αrdr)/α!)z¯α. This completes the proof.The following theorem is crucial for us to get our main results.Theorem 12.
Letf be a bounded function on 𝔻. Then the following conditions are equivalent:(a)
for anyα∈ℕn, there exists λα∈ℂ such that Tf(zα)=λαzα;(b)
for anyα∈ℕn, there exists λα∈ℂ such that Tf(z¯α)=λαz¯α.Proof.
Assume (a) holds; that is,Tf(zα)=λαzα. For any multi-index β∈ℕn, it follows from (19) that
(22)〈Tfz¯α,z¯β〉=〈fz¯α,z¯β〉=〈fzβ,zα〉=〈Tfzβ,zα〉=λβ〈zβ,zα〉={0α≠β,λα〈zα,zα〉α=β=λα〈z¯α,z¯β〉,〈Tfz¯α,zβ〉=〈fz¯α,zβ〉=〈fz0,zα+β〉=〈Tfz0,zα+β〉=λ0〈z0,zα+β〉={0α≠0orβ≠0,λ0α=β=0=λα〈z¯α,zβ〉.{zβ}β≽0∪{z¯β}β≽0 which is a basis of the pluriharmonic Bergman space implies that Tf(z¯α)=λαz¯α.
By a similar argument, one can show that (b) implies (a), which completes the proof.Using Theorem12, we give the necessary and sufficient condition when a bounded function is a separately radial function.Theorem 13.
Letϕ∈L∞(𝔹n,dV). Then the following statements are equivalent:(a)
for anyα∈ℕn, there exists λα∈ℂ such that Tϕ(zα)=λαzα;(b)
ϕ is a separately radial function.Proof.
It is easy to show that (b) implies (a) by Lemma11.
Now suppose (a) holds. That is,Tϕ(zα)=λαzα. Then for any unitary transformation U of ℂn with diagonal matrix and z∈𝔹n, we get U-1z∈𝔹n. Hence, one can obtain Tϕ(U-1z)α=λα(U-1z)α. A direct calculation shows that
(23)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 get that
(24)P[ϕ(Uz)zα¯](w)=P[ϕ(z)(U-1z)α¯](Uw),P[ϕ(Uz)zα](0)=P[ϕ(z)(U-1z)α](0).
It follows that
(25)Tϕ∘U(zα)(w)=P[ϕ(Uz)zα](w)+P[ϕ(Uz)zα¯]¯(w)-P[ϕ(Uz)zα](0)=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)=λα(U-1z)α(Uw)=λαwα=Tϕ(zα)(w).
By Theorem 12, we have Tϕ∘U(z¯α)=λαz¯α=Tϕ(z¯α). Consequently, one can get that Tϕ∘U=Tϕ and so ϕ∘U=ϕ. Then ϕ is a separately radial function. This completes the proof.A direct calculation gives the following lemma, which we will use often.Lemma 14.
Letp,s be two multi-indexes and let ϕ∈ℛ and ξpξ¯sϕ∈L∞(𝔹n,dV). Then for any multi-index α∈ℕn,
(26)Tξpξ¯sϕ(zα)={(((p+α-s)!∫τ(𝔹n)‍)-1(∫τ(𝔹n)‍2n(n+|α|+|p|-|s|)!×∫τ(𝔹n)‍ϕ(r)r2α+2p×|r|-(|p|+|s|)rdr∫τ(𝔹n)‍)×((p+α-s)!∫τ(𝔹n)‍)-1)zp+α-sp+α≽s,(((s-α-p)!∫τ(𝔹n)‍)-1(∫τ(𝔹n)‍2n(n+|s|-|α|-|p|)!×∫τ(𝔹n)‍ϕ(r)r2s×|r|-(|p|+|s|)rdr∫τ(𝔹n)‍)×((s-α-p)!∫τ(𝔹n)‍)-1)z¯s-α-ps≽p+α,s≠p+α,0s⋡p+α,p+α⋡s;Tξpξ¯sϕ(z¯α)={(((s+α-p)!∫τ(𝔹n)‍)-1(∫τ(𝔹n)‍2n(n+|α|+|s|-|p|)!×∫τ(𝔹n)‍ϕ(r)r2α+2s×|r|-(|p|+|s|)rdr∫τ(𝔹n)‍)×((s+α-p)!∫τ(𝔹n)‍)-1)z¯s+α-ps+α≽p,(((p-α-s)!∫τ(𝔹n)‍)-1(∫τ(𝔹n)‍2n(n-|α|+|p|-|s|)!×∫τ(𝔹n)‍ϕ(r)r2p×|r|-(|p|+|s|)rdr∫τ(𝔹n)‍)×((p-α-s)!∫τ(𝔹n)‍)-1)zp-α-sp≽s+α,p≠s+α,0p⋡s+α,s+α⋡p.
Furthermore, if p,s are two nonzero multi-indexes with p⊥s, one has
(27)Tξpξ¯sϕ(zα)={(((p+α-s)!∫τ(𝔹n)‍)-1(∫τ(𝔹n)‍2n(n+|α|+|p|-|s|)!×∫τ(𝔹n)‍ϕ(r)r2α+2p×|r|-(|p|+|s|)rdr∫τ(𝔹n)‍)×((p+α-s)!∫τ(𝔹n)‍)-1((p+α-s)!∫τ(𝔹n)‍)-1)zp+α-sp+α≽s,0p+α⋡s;Tξpξ¯sϕ(z¯α)={(((s+α-p)!∫τ(𝔹n)‍)-1(∫τ(𝔹n)‍2n(n+|α|+|s|-|p|)!×∫τ(𝔹n)‍ϕ(r)r2α+2s×|r|-(|p|+|s|)rdr∫τ(𝔹n)‍)×((s+α-p)!∫τ(𝔹n)‍)-1)z¯s+α-ps+α≽p,0s+α⋡p.Proof.
Here, we only prove that (26) and (27) hold. For any multi-index β∈ℕn, if p+α⋡s, then there exists i such that αi+pi<si. Hence, p+α≠β+s. It follows from Lemma 9 that
(28)〈P[ξpξ¯sϕzα],zβ〉=∫𝔹n‍ξpξ¯sϕ(z)zαz¯βdV(z)=∫01‍2n|r|2n-1|r||α|+|β|d|r|×∫𝕊n‍ϕ(|r|ξ)ξp+αξ¯s+βdσ(ξ)=0.
For α+p≽s, using Lemma 9, (9), and (19), we get
(29)〈P[ξpξ¯sϕzα],zβ〉=∫𝔹n‍ξpξ¯sϕ(z)zαz¯βdV(z)={0β≠p+α-s,2nn!∫τ(𝔹n)‍ϕ(r)r2α+2p×|r|-(|p|+|s|)rdrβ=p+α-s,〈zp+α-s,zβ〉={0β≠p+α-s,n!(p+α-s)!(n+|α|+|p|-|s|)!β=p+α-s.
Moreover, for β≽0,β≠0, we have 〈P[ξpξ¯sϕzα],z¯β〉=0. It follows that for p+α≽s, 〈P[ξpξ¯sϕzα],z¯β〉=〈zp+α-s,z¯β〉=0. Thus, one can obtain that
(30)P[ξpξ¯sϕzα]={(×((p+α-s)!∫τ(𝔹n)‍)-1(∫τ(𝔹n)‍2n(n+|α|+|p|-|s|)!×∫τ(𝔹n)‍ϕ(r)r2α+2p×|r|-(|p|+|s|)rdr∫τ(𝔹n)‍)×((p+α-s)!∫τ(𝔹n)‍)-1×((p+α-s)!∫τ(𝔹n)‍)-1)zp+α-sp+α≽s,0p+α⋡s.
Note that ϕ¯ is still a separately radial function; by a similar argument, we get that
(31)P[ξ¯pξsϕ¯z¯α]={(((s-α-p)!∫τ(𝔹n)‍)-1(∫τ(𝔹n)‍2n(n+|s|-|α|-|p|)!×∫τ(𝔹n)‍ϕ¯(r)r2s×|r|-(|p|+|s|)rdr∫τ(𝔹n)‍)×((s-α-p)!∫τ(𝔹n)‍)-1((s-α-p)!∫τ(𝔹n)‍)-1)zs-α-ps≽p+α,0s⋡p+α.
It follows from above two equations that
(32)Tξpξ¯sϕ(zα)={(((p+α-s)!∫τ(𝔹n)‍)-1(∫τ(𝔹n)‍2n(n+|α|+|p|-|s|)!×∫τ(𝔹n)‍ϕ(r)r2α+2p×|r|-(|p|+|s|)rdr∫τ(𝔹n)‍)×((p+α-s)!∫τ(𝔹n)‍)-1((p+α-s)!∫τ(𝔹n)‍)-1)zp+α-sp+α≽s,(((s-α-p)!∫τ(𝔹n)‍)-1(∫τ(𝔹n)‍2n(n+|s|-|α|-|p|)!×∫τ(𝔹n)‍ϕ(r)r2s×|r|-(|p|+|s|)rdr∫τ(𝔹n)‍)×((s-α-p)!∫τ(𝔹n)‍)-1((s-α-p)!∫τ(𝔹n)‍)-1)z¯s-α-ps≽p+α,s≠p+α,0s⋡p+α,p+α⋡s.
Furthermore, if p⊥s and p,s are nonzero multi-indexes, there exists i∈{1,2,…,n} such that pi>0 and si=0. This implies that there does not exist α∈ℕn such that s≽p+α. It follows that (27) holds. This completes the proof.Theorem 15.
Letp,s∈ℕn be two nonzero multi-indexes with p⊥s and let ϕ be a bounded function on 𝔹n. Then the following conditions are equivalent:(a)
for anyα∈ℕn, there exists λα∈ℂ such that
(33)Tϕ(zα)={0p+α⋡s,λαzp+α-sp+α≽s;(b)
ϕ is a separately quasihomogeneous function of degree (p,s).Proof.
It is obvious that (b) implies (a). Now assume (a) holds. For any multi-indexβ∈ℕn, we have
(34)〈Tz¯pzsϕ(zα),zβ〉=〈ϕzα+s,zp+β〉=〈Tϕzα+s,zp+β〉=〈λα+szα+p,zp+β〉={0α≠β,λα+sn!(α+p)!(n+|α|+|p|)!α=β.
Similarly, for β≽0 and β≠0, we get that 〈Tz¯pzsϕ(zα),z¯β〉=0=〈zα,z¯β〉. In light of (19), one can deduce that
(35)Tz¯pzsϕ(zα)=λα+s(α+p)!(n+|α|)!α!(n+|α|+|p|)!zα.
It follows from Theorem 13 that z¯pzsϕ is a separately radial function. Let z¯pzsϕ(z)=ψ(z), where ψ is a separately radial function on 𝔹n. It follows that
(36)ϕ(z)=ψ(r)r-(2p+2s)|r|(|p|+|s|)ξpξ¯s,
which implies that ϕ is a separately quasihomogeneous function of degree (p,s).Now we discuss when the product of two Toeplitz operators with certain symbols is a Toeplitz operator.Theorem 16.
Letϕ1,ϕ2 be two bounded separately radial functions on 𝔹n. If there exists a bounded function h such that Tϕ1Tϕ2=Th, then h is also a separately radial function on 𝔹n.Proof.
It follows from Lemma11 that for α∈ℕn(37)Th(zα)=Tϕ1Tϕ2(zα)=2n(n+|α|)!∫τ(𝔹n)‍ϕ2(r)r2αrdrα!·2n(n+|α|)!∫τ(𝔹n)‍ϕ1(r)r2αrdrα!zα.
In virtue of Theorem 13, we get that h is a separately radial function. This completes the proof.Theorem 17.
Letp,s∈ℕn be two nonzero multi-indexes with p⊥s and let ϕ1,ϕ2∈ℛ such that ξ¯sϕ1,ξpϕ2∈L∞(𝔹n,dV). If there exists a bounded function h such that Tξ¯sϕ1Tξpϕ2=Th, then h is a separately quasihomogeneous function of degree (p,s).Proof.
For any multi-indexα∈ℕn, using Lemma 14, one can obtain that
(38)Th(zα)=Tξ¯sϕ1Tξpϕ2(zα)={λαzα+p-sα+p≽s,0α+p⋡s,
where λα=(2n(n+|α|+|p|-|s|)!∫τ(𝔹n)‍ϕ1(r)r2α+2p|r|-|s|rdr/(p+α-s)!)·(2n(n+|α|+|p|)!∫τ(𝔹n)‍ϕ2(r)r2α+2p|r|-|p|rdr/(p+α)!). It follows from Theorem 15 that h is a separately quasihomogeneous function of degree (p,s).Using the same technique, we give the following result and omit the proof.Theorem 18.
Letp,s∈ℕn be two nonzero multi-indexes with p⊥s and let ϕ1,ϕ2∈ℛ such that ϕ1,ξpξ¯sϕ2∈L∞(𝔹n,dV). If there exists a bounded function h such that Tϕ1Tξpξ¯sϕ2=Th, then h is a separately quasihomogeneous function of degree (p,s).


## 4. The Zero-Product Problem of Toeplitz Operators with Quasihomogeneous and Separately Quasihomogeneous Symbols

In this section, we will study the zero-product problem for several Toeplitz operators when one of the symbols is separately quasihomogeneous and the others are homogeneous, and show that the zero-product problem for two Toeplitz operators with certain symbols has only a trivial solution on the pluriharmonic Bergman space of the unit ball.Theorem 19.
Letpi,si(1≤i≤m)∈ℕn be nonzero multi-indexes with pi⊥si and let ϕ1,ϕ2,…,ϕm-1 be square integrable radial functions and ϕm∈ℛ such that ξpiξ¯siϕi∈L∞(𝔹n,dV). Then
(39)Tξp1ξ¯s1ϕ1Tξp2ξ¯s2ϕ2…Tξpmξ¯smϕm=0
if and only if ϕi=0 for some i.Proof.
SupposeTξp1ξ¯s1ϕ1Tξp2ξ¯s2ϕ2…Tξpmξ¯smϕm=0. Then for multi-index α∈ℕn,
(40)Tξp1ξ¯s1ϕ1Tξp2ξ¯s2ϕ2…Tξpmξ¯smϕm(zα)=0,Tξp1ξ¯s1ϕ1Tξp2ξ¯s2ϕ2…Tξpmξ¯smϕm(z¯α)=0.
It follows from Lemmas 8 and 14 that
(41)Tξp1ξ¯s1ϕ1Tξp2ξ¯s2ϕ2…Tξpmξ¯smϕm(zα)={Cαzα+∑j=0m-1pm-j-∑j=0m-1sm-jα+pm≽sm,…,α+∑j=0m-1‍pm-j≽∑j=0m-1‍sm-j,0others,
where
(42)Cα=(∫τ(𝔹n)‍2n(n+|α|+|pm|-|sm|)!×∫τ(𝔹n)‍ϕm(r)r2α+2pm|r|-(|pm|+|sm|)rdr∫τ(𝔹n)‍)×((pm+α-sm)!)-1··∏i=1m-1‍{(2(α+∑j=0i‍pm-j-∑j=0i-1‍sm-j)!×(n+|α|+∑j=0i‍|pm-j|-∑j=0i‍|sm-j|)!)×((α+∑j=0i‍pm-j-∑j=0i‍sm-j)!×(n-1+|α|+∑j=0i‍|pm-j|-∑j=0i-1‍|sm-j|)!)-1×ϕm-i^(2n+2|α|+2∑j=0i-1‍|pm-j|-2∑j=0i-1‍|sm-j|+|pm-i|-|sm-i|)}.
If Tξp1ξ¯s1ϕ1Tξp2ξ¯s2ϕ2…Tξpmξ¯smϕm(zα)=0, then, for any multi-index α∈ℕn such that α+pm≽sm,…,α+∑j=0m-1‍pm-j≽∑j=0m-1‍sm-j, we have
(43)∫τ(𝔹n)ϕm(r)r2α+2pm|r|-(|pm|+|sm|)rdr‍·∏i=1m-1‍ϕm-i^(2n+2|α|+2∑j=0i-1‍|pm-j|-2∑j=0i-1‍|sm-j|+|pm-i|-|sm-i|∑j=0i-1‍)=0.
Now we are ready to show that ϕi=0 for some i. For the sake of simplicity, we will consider the case of n=2. Let E={α≽∑j=0m-1‍sm-j:∫τ(𝔹n)‍ϕm(r)r2α+2pm|r|-(|pm|+|sm|)rdr=0}. If E satisfies condition (I), then ϕm=0 by Lemma 10. Otherwise, let Ec denote the complement of E in {α∈ℤ+2:α≽∑j=0m-1‍sm-j}, and it follows from Remark 6 that Ec satisfies condition (I). Now let
(44)M={α+pm≽sm,…,α+∑j=0m-1‍pm-j≽∑j=0m-1‍sm-j:∫τ(𝔹n)‍ϕm(r)r2α+2pm×|r|-(|pm|+|sm|)rdr≠0∑j=0m-1‍}.
It is obvious that M⊃Ec, which implies that M satisfies condition (I). Furthermore, we get that
(45)∑α∈M‍1|α|=∞.
Set
(46)Mi={α+pm≽sm,…,α+∑j=0m-1‍pm-j≽∑j=0m-1‍sm-j:ϕi^(2n+2|α|+2∑j=0m-i-1‍×|pm-j|-2∑j=0m-i-1‍|sm-j|+|pi|-|si|∑j=0m-i-1‍)=0}
for 1≤i≤m-1. According to (43), we obtain that M⊂∪i=1m-1Mi. Hence there exists some i such that
(47)∑α∈Mi‍1|α|=∞.
By Remark 4, one can deduce that ϕi=0 for some i(1≤i≤m-1). Moreover, if ϕi=0 for some i(1≤i≤m), then Tξp1ξ¯s1ϕ1Tξp2ξ¯s2ϕ2…Tξpmξ¯smϕm(z¯α)=0.
Conversely, it is obvious that ifϕi=0 for some i, then
(48)Tξp1ξ¯s1ϕ1Tξp2ξ¯s2ϕ2…Tξpmξ¯smϕm=0.
This completes the proof.The following result is a partial answer to the zero-product problem for two Toeplitz operators onLh2(𝔹n).Theorem 20.
Letf(z)=∑p,s∈ℕn,p≠0,s≠0,p⊥s‍ξpξ¯sfp,s(r)∈L∞(𝔹n,dV), where fp,s(r)∈ℛ. Let g=ξp*ξ¯s*ϕ(r)∈L∞(𝔹n,dV), where p*,s* are two nonzero indexes with p*⊥s* and ϕ∈ℛ. Then TfTg=0 if and only if either f=0 or g=0.Proof.
By Lemma2, we get that ξpξ¯sfp,s(r) is bounded for any multi-indexes p,s∈ℕn,p≠0,s≠0,p⊥s.
IfTfTg=0, then for any multi-index α∈ℕn,  TfTg(zα)=0 and TfTg(z¯α)=0. It follows from (27) that
(49)Tξpξ¯sfp,s(r)(zα)=c(p,s,α)∫τ(𝔹n)‍fp,s(r)r2α+2p×|r|-(|p|+|s|)rdrzα+p-s,
where
(50)c(p,s,α)={0p+α⋡s,2n(n+|α|+|p|-|s|)!(p+α-s)!p+α≽s
and consequently,(51)Tf(zα)=∑p,s∈ℕn,p≠0,s≠0,p⊥s‍c(p,s,α)×∫τ(𝔹n)‍fp,s(r)r2α+2p|r|-(|p|+|s|)rdrzα+p-s.TfTg(zα)=0 together with (27) implies that for any multi-indexes p,s∈ℕn, p≠0, s≠0, and p⊥s,
(52)∫τ(𝔹n)‍fp,s(r)r2α+2p*-2s*+2p|r|-(|p|+|s|)rdr·∫τ(𝔹n)‍ϕ(r)r2α+2p*|r|-(|p*|+|s*|)rdr=0
for all α≽2s*+s. Next we will prove that either ϕ=0 or fp,s=0 for p,s∈ℕn, p≠0, s≠0, and p⊥s. Similarly as the proof of Theorem 19, we will prove it in the case of n=2. Let
(53)Ep,s={α∈ℤ+2:α≽2s*+s,∫τ(𝔹n)fp,s(r)r2α+2p*-2s*+2p‍×|r|-(|p|+|s|)rdr=0∫τ(𝔹n)‍}.
If Ep,s satisfies condition (I), noting that
(54)|fp,s(r)rp+s|r|-(|p|+|s|)|=|ξpξ¯sfp,s(r)|<∞,
then fp,s=0 by Lemma 10. Otherwise, let Ep,sc denote the complement of Ep,s in {α∈ℤ+2:α≽2s*+s}; then Ep,sc satisfies condition (I) by Remark 6. It follows from (52) that
(55)∫τ(𝔹n)‍ϕ(r)r2α+2p*|r|-(|p*|+|s*|)rdr=0,∀α∈Ep,sc.
Using Lemma 10 again, we obtain that ϕ=0. Moreover, if either f=0 or ϕ=0, then TfTg(z¯α)=0. Hence, TfTg=0 implies that either f=0 or ϕ=0.
The converse implication is obvious. This completes the proof.Corollary 21.
Letf(z)=ξpξ¯sfp,s(r)∈L∞(𝔹n,dV), where p,s∈ℕn, p≠0, s≠0, p⊥s, and fp,s(r)∈ℛ. Then Tf2=Tf implies that either f=0 or f=1.


## 5. The Commutativity of Toeplitz Operators with Quasihomogeneous and Separately Quasihomogeneous Symbols

In this section, we will consider the commuting problem for two Toeplitz operators with certain symbols.Theorem 22.
Letp,s∈ℕn be two nonzero multi-indexes with p⊥s let ψ∈ℛ such that ξpξ¯sψ∈L∞(𝔹n,dV), and let ϕ be a bounded radial function on 𝔹n. If ϕ is nonconstant, then TϕTξpξ¯sψ=Tξpξ¯sψTϕ if and only if either |p|=|s| or ψ=0.Proof.
It follows from Lemmas7 and 14 that TϕTξpξ¯sψ=Tξpξ¯sψTϕ if and only if
(56)∫τ(𝔹n)‍ψ(r)r2α+2p|r|-(|p|+|s|)rdr×(2n+2|α|+2|p|-2|s|)ϕ^×(2n+2|α|+2|p|-2|s|)=∫τ(𝔹n)‍ψ(r)r2α+2p|r|-(|p|+|s|)rdr×(2n+2|α|)ϕ^(2n+|α|)
for p+α≽s and
(57)∫τ(𝔹n)‍ψ(r)r2α+2s|r|-(|p|+|s|)rdr×(2n+2|α|+2|s|-2|p|)ϕ^×(2n+2|α|+2|s|-2|p|)=∫τ(𝔹n)‍ψ(r)r2α+2s|r|-(|p|+|s|)rdr×(2n+2|α|)ϕ^(2n+|α|)
for s+α≽p. As in the proof of Theorem 5.1 in [10], for p+α≽s, (56) and the property that ϕ is nonconstant imply that either |p|=|s| or ψ=0. Furthermore, if either |p|=|s| or ψ=0, it is easy to show that (57) holds for s+α≽p. Hence, Tϕ commutes with Tξpξ¯sψ implies either |p|=|s| or ψ=0.
Conversely, if either|p|=|s| or ψ=0, one can get that (56) and (57) hold. That is, TϕTξpξ¯sψ=Tξpξ¯sψTϕ. This completes the proof.As forn=1, a separately radial function is a radial and it follows that Toeplitz operators with separately radial symbols commute, there is no contradiction with an extension of a result in [29] to the case of n≥2. It is given in [29] that a Toeplitz operator on Lh2(𝔻) with radial symbol commutes with another Toeplitz operator if that operator also has a radial symbol. The following theorem will show that this result is not true on Lh2(𝔹n)(n≥2).Theorem 23.
Forn≥2, let ψ(z) be a bounded separately radial function and let ϕ(z) be a bounded radial function on 𝔹n. Then Tψ  and Tϕ commute.Proof.
The proof is similar to that of Theorem22, so we omit it.Next, we will give a description of commuting quasihomogeneous Toeplitz operators the with same degree.Theorem 24.
Letp,s∈ℕn be two nonzero multi-indexes with p⊥s,|p|≠|s| and let ϕ,ψ be two square integrable radial function on 𝔹n such that ξpξ¯sϕ,ξpξ¯sψ∈L∞(𝔹n,dV). Then Tξpξ¯sϕTξpξ¯sψ=Tξpξ¯sψTξpξ¯sϕ if and only if ϕ=Cψ for some constant C.Proof.
It follows from Lemma8 that
(58)Tξpξ¯sϕTξpξ¯sψ=Tξpξ¯sψTξpξ¯sϕ
if and only if
(59)ϕ^(2n+2|α|+3|p|-3|s|)ψ^(2n+2|α|+|p|-|s|)=ψ^(2n+2|α|+3|p|-3|s|)ϕ^×(2n+2|α|+|p|-|s|)
for all α+2p≽2s and
(60)ϕ^(2n+2|α|+3|s|-3|p|)ψ^(2n+2|α|+|s|-|p|)=ψ^(2n+2|α|+3|s|-3|p|)ϕ^×(2n+2|α|+|s|-|p|)
for all α+2s≽2p. If |p|>|s|, according to Lemma 3, then (59) holds which implies that
(61)ϕ^(z+2|p|-2|s|)ψ^(z)=ψ^(z+2|p|-2|s|)ϕ^(z),z∈{z:Rez>0}.
It follows from Lemma 6 in [36] that ϕ=Cψ. Furthermore, if ϕ=Cψ, then (60) holds. Otherwise, if |p|<|s|, by a similar argument, (60) implies that ϕ=Cψ and so (59) holds.
The converse implication is obvious. This completes the proof.Theorem 25.
Letf(z)=∑s∈ℕn‍ξ¯sfs(r)∈L∞(𝔹n,dV), where fs(r)∈ℛ. Let ϕ be a bounded separately radial function on 𝔹n. Then TfTϕ=TϕTf if and only if Tξ¯sfs(r)Tϕ=TϕTξ¯sfs(r) for every multi-index s∈ℕn.Proof.
Using Lemmas11 and 14, one can get that for any multi-index α∈ℕn(62)TfTϕ(zα)=μα∑s∈ℕn‍Tξ¯sfs(r)(zα)=μα[∑s∈ℕn,α≽s‍Tξ¯sfs(r)(zα)+∑s∈ℕn,s≽α,s≠α‍Tξ¯sfs(r)(zα)+∑s∈ℕn,α⋡s,s⋡α‍Tξ¯sfs(r)(zα)]=μα[∑s∈ℕn,α≽s‍((∫τ(𝔹n)‍2n(n+|α|-|s|)!×∫τ(𝔹n)‍fs(r)r2α|r|-|s|rdr)×((α-s)!)-1((∫τ(𝔹n)‍2n(n+|α|-|s|)!)zα-s+∑s∈ℕn,s≽α,s≠α‍((∫τ(𝔹n)‍2n(n+|s|-|α|)!×∫τ(𝔹n)‍fs(r)r2s×|r|-|s|rdr∫τ(𝔹n)‍)×((s-α)!)-1((∫τ(𝔹n)‍2n(n+|s|-|α|)!)z¯s-α∑s∈ℕn,α≽s‍],
where μα=2n(n+|α|)!∫τ(𝔹n)‍ϕ(r)r2αrdr/α!. It follows that
(63)〈TfTϕ(zα),zβ〉={0α⋡β,2nμα(n+|β|)!∫τ(𝔹n)‍fα-β(r)r2α|r||β|-|α|rdrβ!×〈zβ,zβ〉α≽β
for β≽0 and
(64)〈TfTϕ(zα),z¯β〉=2nμα(n+|β|)!∫τ(𝔹n)‍fα+β(r)r2α+2β|r|-(|β|+|α|)rdrβ!×〈zβ,zβ〉
for β≽0 and β≠0.
By a similar argument, one can deduce that(65)〈TϕTf(zα),zβ〉={0α⋡β,2nνβ(n+|β|)!∫τ(𝔹n)‍fα-β(r)r2α|r||β|-|α|rdrβ!×〈zβ,zβ〉α≽β
for β≽0 and
(66)〈TϕTf(zα),z¯β〉=2nνβ(n+|β|)!∫τ(𝔹n)‍fα+β(r)r2α+2β|r|-(|β|+|α|)rdrβ!×〈zβ,zβ〉
for β≽0 and β≠0, where νβ=2n(n+|β|)!∫τ(𝔹n)‍ϕ(r)r2βrdr/β!.
Now suppose thatTfTϕ=TϕTf. It follows that TfTϕ(zα)=TϕTf(zα) and TfTϕ(z¯α)=TϕTf(z¯α) for any multi-index α∈ℕn. Using Lemmas 11 and 14 again, we have
(67)〈Tξ¯sfsTϕ(zα),zβ〉={((∫τ(𝔹n)‍2nμα(n+|β|)!×∫τ(𝔹n)‍fα-β(r)r2α×|r||β|-|α|rdr∫τ(𝔹n)‍)×(β!)-1((∫τ(𝔹n)‍2nμα(n+|β|)!)〈zβ,zβ〉β=α-s,0others={〈TfTϕ(zα),zβ〉β=α-s,0others={〈TϕTf(zα),zβ〉β=α-s,0others={((∫τ(𝔹n)‍2nνβ(n+|β|)!×∫τ(𝔹n)‍fα-β(r)r2α×|r||β|-|α|rdr∫τ(𝔹n)‍)×(β!)-1((∫τ(𝔹n)‍2nμα(n+|β|)!)〈zβ,zβ〉β=α-s,0others=〈TϕTξ¯sfs(zα),zβ〉
for any multi-indexes β∈ℕn and s∈ℕn and
(68)〈Tξ¯sfsTϕ(zα),z¯β〉={((∫τ(𝔹n)‍2nμα(n+|β|)!×∫τ(𝔹n)‍fα+β(r)r2α+2β×|r|-(|β|+|α|)rdr∫τ(𝔹n)‍)×(β!)-1((∫τ(𝔹n)‍2nμα(n+|β|)!)〈zβ,zβ〉β=s-α,0others={〈TfTϕ(zα),z¯β〉β=s-α,0others={〈TϕTf(zα),z¯β〉β=s-α,0others={((∫τ(𝔹n)‍2nνβ(n+|β|)!×∫τ(𝔹n)‍fα+β(r)r2α+2β×|r|-(|β|+|α|)rdr∫τ(𝔹n)‍)×(β!)-1((∫τ(𝔹n)‍2nμα(n+|β|)!)〈zβ,zβ〉β=s-α,0others=〈TϕTξ¯sfs(zα),z¯β〉
for any multi-indexes β∈ℕn, β≠0, and s∈ℕn. By the same technique, we get that
(69)〈Tξ¯sfsTϕ(z¯α),zβ〉=〈Tξ¯sfsTϕ(z¯α),zβ〉
for any multi-indexes β∈ℕn and s∈ℕn, and
(70)〈Tξ¯sfsTϕ(z¯α),z¯β〉=〈Tξ¯sfsTϕ(z¯α),z¯β〉
for any multi-indexes β∈ℕn, β≠0, and s∈ℕn. Hence Tξ¯sfsTϕ=TϕTξ¯sfs.
Conversely, ifTξ¯sfsTϕ=TϕTξ¯sfs, it is easy to show that TfTϕ=TϕTf. This completes the proof.It is well known thatTϕ*=Tϕ¯, so one can easily get the following result.Theorem 26.
Letf(z)=∑p∈ℕn‍ξpfp(r)∈L∞(𝔹n,dV), where fp(r)∈ℛ. Let ϕ be a bounded separately radial function on 𝔹n. Then TfTϕ=TϕTf  if and only if Tξpfp(r)Tϕ=TϕTξpfp(r) for every multi-index p∈ℕn.



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*Source: 252037-2013-11-06.xml*