# Toeplitz Operators on the Dirichlet Space of𝔹n

Authors: HongZhao Lin; YuFeng Lu
Journal: Abstract and Applied Analysis
 (2012)

---

## Abstract

We study the algebraic properties of Toeplitz operators on the Dirichlet
space of the unit ball𝔹n. We characterize pluriharmonic symbol for which the corresponding
Toeplitz operator is normal or isometric. We also obtain descriptions of conjugate holomorphic
symbols of commuting Toeplitz operators. Finally, the commuting problem of Toeplitz operators
whose symbols are of the form zpz¯qϕ(|z|2) is studied.


---

## Body

## 1.    Introduction

For any integern≥1, let 𝔹n={z∈ℂn:|z|<1} denote the open unit ball of ℂn and dm denote the normalized Lebesgue measure on 𝔹n. The Sobolev space w1,2 is defined to be the completion of smooth functions on 𝔹n which satisfy
(1.1)∥f∥2=|∫𝔹n‍fdm|2+∑i=1n‍∫𝔹n‍(|∂f∂zi|2+|∂f∂z-i|2)dm<∞.
The inner product 〈·,·〉 on w1,2 is defined by
(1.2)〈f,g〉=∫𝔹n‍fdm∫𝔹n‍g-dm+∑i=1n‍∫𝔹n‍(∂f∂zi∂g∂zi¯+∂f∂z-i∂g∂z-i¯)dm,∀f,g∈w1,2.
The Dirichlet space 𝒟 of 𝔹n is the closed subspace consisting of all holomorphic functions in w1,2. It is easily verified that each point evaluation is a bounded linear functional on 𝒟. Hence, for each z∈𝔹n, there exists a unique reproducing kernel Kz(w)∈𝒟 such that
(1.3)f(z)=〈f(w),Kz(w)〉,∀f∈𝒟.
Actually, it can be calculated that Kz(w)=1+∑α∈ℤ+n‍(((|α|+n-1)!/|α|n!α!)wαz-α), where α=(α1,…,αn) is a multi-index, αi∈ℤ+, |α|=∑i=1n‍αi and zα=z1α1⋯znαn. For multi-indexes α and β, the notation α≽β means that
(1.4)αi≥βi,i=1,…,n
and α≻β means that α≽β and α≠β.LetP be the orthogonal projection from w1,2 onto 𝒟. By the explicit formula for Kz(w), we have
(1.5)Pψ(z)=〈Pψ,Kz〉=〈ψ,Kz〉=∫𝔹n‍ψdm∫𝔹n‍K¯zdm+∑i=1n‍∫𝔹n‍∂ψ∂wi∂Kz∂wi¯dm(w),∀ψ∈w1,2.
Let Ω={φ∈w1,2:φ,∂φ/∂zi,∂φ/∂z-i∈L∞(𝔹n)}. Given φ∈Ω, the Toeplitz operator Tφ with symbol φ is the linear operator on 𝒟 defined by
(1.6)Tφf=P(φf),∀f∈𝒟.
It is easy to verify that the Toeplitz operator Tφ:𝒟→𝒟 is always bounded, whenever φ∈Ω.The algebric properties of Toeplitz operators on the classical Hardy spaces and Bergman spaces have been well studied, for example, as in [1–5].On the Hardy space of the unit circle, a well-known theorem of Brown and Halmos [1] has shown that two Toeplitz operators with bounded symbols commute if and only if one of the followings holds: (i) both symbols are holomorphic; (ii) both symbols are antiholomorphic; (iii) a nontrivial linear combination of the symbols is constant.On the Bergman space, the commuting problem is more complicated. Axler andC˘uc˘ković [2] proved that Brown-Halmos Theorem also holds for Toeplitze operators with bounded harmonic symbols. However, the corresponding problem for Toeplitz operator with general symbol remains open.In recent years, more and more attention has been paid to the Toeplitz operators on Dirichlet spaces. The algebric properties of the Toeplitz operators on the classical Dirichlet spaces of the unit disc have been investigated intensively in [6–13]. Cao considered Fredholm properties of Toeplitz operators with C1(𝔻) symbols in [6]. Lee showed in [8] that Brown-Halmos's result with harmonic symbols remains vaild on the Dirichlet space of the unit disc. In [12], Duistermaat and Lee gave the following characterizations of the harmonic symbols for which the associated Toeplitz operators are commuting, self-adjoint, or isometric: (1) for a harmonic symbol u∈Ω'={u∈C1(𝔻):u,∂u/∂z,∂u/∂z-∈L∞(𝔻,dA)}, Tu is self-adjoint if and only if u is a real constant function; (2) for a harmonic symbol u∈Ω', Tu is an isometry if and only if u is a constant function of modulus 1; (3) for two harmonic symbols u, v∈Ω', Tu and Tv commute if and only if either u and v are holomorphic or a nontrivial linear combination of u and v is constant on 𝔻. In [13], the corresponding problems have been investigated on the polydisc Dirichlet spaces and similar results have been obtained.Motivated by the work of [12, 13], we study the corresponding problems on the Dirichlet spaces of 𝔹n. In Section 2, we give the characterizations of the pluriharmonic symbol for which the associated Toeplitz operator is self-adjoint or an isometry. In Section 3, we discuss when two Toeplitz operators with conjugate holomorphic symbols commute. At last, we concern with the commuting Toeplitz operators with symbols zpz-qϕ(|z|2).


## 2.   Characterization of Normality and Isometry

In this section, we will give the condition under which Toeplitz operators with pluriharmonic symbols are self-adjoint or isometric on𝒟. Before doing this, we first exhibit some properties of Toeplitz operators on 𝒟.Lemma 2.1.
Letf=∑β∈ℤ+n‍fβzβ∈Ω be holomorphic. Then the following statements hold:(1)
Tf¯1=f(0)¯;(2)
Tf*1=f(0)¯;(3)
Tf¯*1=∫𝔹n‍fKz¯dm=f(0)+∑|β|>0‍((fβ/|β|(n+|β|))zβ).Proof.
By the definition of the Toeplitz operators and the properties of the reproducing kernel, we obtain that(2.1)Tf¯1=〈Tf¯1,Kz〉=〈P(f¯),Kz〉=〈f(0)¯,Kz〉=f(0)¯,Tf*1=〈Tf*1,Kz〉=〈1,fKz〉=∫𝔹n‍fKz¯dm=∫𝔹n‍fKzdm¯=f(0)¯,Tf¯*1=〈Tf¯*1,Kz〉=〈1,f¯Kz〉=∫𝔹n‍fKz¯dm=∫𝔹n‍∑β∈ℤn+‍fβwβ∑α∈ℤ+n‍(|α|+n-1)!|α|n!α!zαw-αdm=f(0)+∑|β|>0‍fβ(|β|+n-1)!|β|n!β!zβ∫𝔹n‍|wβ|2dm=f(0)+∑|β|>0‍fβ(|β|+n-1)!|β|n!β!zβ·n!β!(n+|β|)!=f(0)+∑|β|>0‍fβ|β|(n+|β|)zβ.Lemma 2.2.
Letu=f+g¯∈Ω, where f and g are holomorphic. If Tu is normal, then
(2.2)∑i=1n‍∫𝔹n‍|∂f∂zi|2dm=∑i=1n‍∫𝔹n‍|∂G∂zi|2dm,
where G=Tg¯*(1).Proof.
By assumption, we haveTf+g¯*Tf+g¯=Tf+g¯Tf+g¯*. In particular,
(2.3)〈Tf+g¯*Tf+g¯1,1〉=〈Tf+g¯Tf+g¯*1,1〉.
That is,
(2.4)〈Tf+g¯1,Tf+g¯1〉=〈Tf+g¯*1,Tf+g¯*1〉.
It follows from Lemma 2.1 that
(2.5)〈f+g(0)¯,f+g(0)¯〉=〈f(0)¯+G,f(0)¯+G〉.
Hence,
(2.6)|f(0)+g(0)¯|2+∑i=1n‍∫𝔹n‍|∂f∂zi|2dm=|f(0)¯+G(0)|2+∑i=1n‍∫𝔹n‍|∂G∂zi|2dm.
On the other hand, by the reproducing property and Lemma 2.1, we have
(2.7)G(0)=Tg¯*1(0)=〈Tg¯*1,K0〉=〈Tg¯*1,1〉=g(0).
Then,
(2.8)∑i=1n‍∫𝔹n‍|∂f∂zi|2dm=∑i=1n‍∫𝔹n‍|∂G∂zi|2dm.
This completes the proof.The next lemma shows there is only trivial normal Toeplitz operator with holomorphic (or antiholomorphic) symbols.Lemma 2.3.
Letf=∑β∈ℤ+n‍fβzβ∈Ω be holomorphic. Then the following statements are equivalent:(1)
Tf is normal;(2)
Tf¯ is normal;(3)
Tf+f¯ is normal;(4)
f is a constant function on 𝔹n.Proof.
(
1
)
⇒
(
4
) By Lemma 2.2 (with g=0), we have that
(2.9)∑i=1n‍∫𝔹n‍|∂f∂zi|2dm=∑i=1n‍∫𝔹n‍|∂G∂zi|2dm=0.
This along with the fact that(2.10)∑i=1n‍∫𝔹n‍|∂f∂zi|2dm=∥f-f(0)∥2=∑|β|>0‍|fβ|2∥zβ∥2
proves that fβ=0, for |β|>0. This shows that f is a constant.
(
2
)
⇒
(
4
) Since Tf¯ is normal, it follows from Lemma 2.2 that
(2.11)∑i=1n‍∫𝔹n‍|∂G∂zi|2dm=∑i=1n‍∫𝔹n‍|∂f¯∂zi|2dm=0,
which implies that G is a constant function for G is holomorphic.
On the other hand, Lemma2.1 ensures that
(2.12)G=Tf¯*1=f(0)+∑|β|>0‍fβ|β|(n+|β|)zβ.
It follows that fβ=0, for all |β|>0. Hence f is a constant, as desired.
(
3
)
⇒
(
4
) Suppose Tf+f¯ is normal. Using Lemma 2.2,
(2.13)∑i=1n‍∫𝔹n‍|∂f∂zi|2dm=∑i=1n‍∫𝔹n‍|∂G∂zi|2dm,
where G=Tf¯*(1). We conclude that fβ=0, for |β|>0, since by direct computation and Lemma 2.1(2.14)∑i=1n‍∫𝔹n‍|∂f∂zi|2dm=∑|β|>0‍|fβ|2∥zβ∥2,∑i=1n‍∫𝔹n‍|∂G∂zi|2dm=∑|β|>0‍|fβ|2|β|2(n+|β|)2∥zβ∥2.
Hence f is a constant.
The converse implications are clear. The proof is complete.Since∥Tf¯1∥≥∥Tf¯*1∥ for hyponormal Toeplitz operator Tf¯, using Lemma 2.1, Tf¯ is hyponormal if and only if f is a constant. Consequently, normality of Tf¯ can be replaced by hyponormality in Lemma 2.3.On the Hardy space and the Bergman space, we always haveTu*=Tu-. So it is easy to see that Tu is self-adjoint (i.e., Tu=Tu*) if and only if u is a real-valued function. However, on the Dirichlet space of disc and polydisc, the situations are different because Tu* is not equal to Tu¯ in both cases. In the following, we will study the adjoint of Toeplitz operators with pluriharmonic symbols on the Dirichlet space of 𝔹n.Theorem 2.4.
Letu=f+g¯∈Ω, where f and g are holomorphic. Then Tu*=Tu¯ if and only if u is a constant function.Proof.
First, assume thatu=∑β∈ℤ+n‍aβzβ is holomorphic. Since Tu*=Tu-, for each multi-index α=(α1,α2,…,αn), we have
(2.15)Tu*zα=Tu¯zα,∀|α|>0.
Moreover,
(2.16)〈Tu*zα,1〉=〈Tu¯zα,1〉.
In fact, for |α|>0,
(2.17)〈Tu*zα,1〉=〈zα,Tu1〉=〈zα,u〉=aα¯∥zα∥2=aα¯·|α|n!α!(n+|α|-1)!.
On the other hand,
(2.18)〈Tu¯zα,1〉=〈u¯zα,1〉=∫𝔹n‍u¯zαdm=∫𝔹n‍aα¯|zα|2dm=aα¯·n!α!(n+|α|)!.
Note that
(2.19)|α|n!α!(n+|α|-1)!-n!α!(n+|α|)!>0,
we conclude that, aα=0, for |α|>0.
Second, assume thatu=f+g¯  is the general pluriharmonic symbol and Tu*=Tu¯. In particular, we have
(2.20)(Tf*+Tg¯*)1=Tf¯1+Tg1.
By Lemma 2.1, we get that
(2.21)f(0)¯+∫𝔹n‍gKz¯dm=f(0)¯+g.
Since
(2.22)∫𝔹n‍gKz¯dm=g(0)+∑|β|>0‍gβ|β|(n+|β|)zβ,
where g=∑β∈ℤ+n‍gβzβ, it follows that
(2.23)∑|β|>0‍gβ|β|(n+|β|)zβ=∑|β|>0‍gβzβ.
Equivalently,
(2.24)∑|β|>0‍(1-1|β|(n+|β|))gβzβ=0.
This implies that gβ=0, for |β|>0. So u=f+g(0)¯ is holomorphic. The desired result follows immediately from the previous holomorphic case.
The converse implication is clear. The proof is complete.We now characterize pluriharmonic symbols inducing self-adjoint Toeplitz operators.Theorem 2.5.
Letu=f+g¯∈Ω, where f, g are holomorphic. Then Tu=Tu* if and only if u is a real constant function.Proof.
The “if” part is clear. Supposef=∑β∈ℤ+n‍fβzβ and g=∑β∈ℤ+n‍gβzβ. It follows from Lemma 2.1 that
(2.25)Tu1=Tf+g¯1=f+g(0)¯=f(0)+g(0)¯+∑|β|>0‍fβzβ,Tu*1=Tf+g¯*1=Tf*1+Tg¯*1=f(0)¯+g(0)+∑|β|>0‍gβ|β|(n+|β|)zβ.
Since Tu1=Tu*1, by comparing the coefficients of the above two equations, we have that
(2.26)gβ|β|(n+|β|)=fβ,|β|>0,(2.27)f(0)¯+g(0)=f(0)+g(0)¯.
Letwi=wei, where ei=(0,0,…,0,1︸i,0,…,0). Then we have
(2.28)〈g¯wi,Kz(w)〉=∫𝔹n‍g¯widm+∑j=1n‍∫𝔹n‍∂(g¯wi)∂wj∂Kz(w)∂wj¯dm=∫𝔹n‍∑β∈ℤ+n‍gβwβ¯widm+∫𝔹n‍g¯∂Kz(w)∂wi¯dm=gei¯·∫𝔹n‍|wei|2dm+∫𝔹n‍∑β∈ℤ+n‍gβwβ·∑α≽ei‍(|α|+n-1)!|α|n!α!αiwα-eiz-αdm¯=gei¯n+1+g(0)¯·zei.
This shows that
(2.29)(Tf+g¯wi)(z)=(Tfwi)(z)+(Tg¯wi)(z)=∑β∈ℤ+n‍fβzβ+ei+gei¯n+1+g(0)¯·zei.
On the other hand, if h=∑β∈ℤ+n‍hβwβ is holomorphic, then
(2.30)∫𝔹n‍∂h∂widm=∫𝔹n‍∑β≽ei‍hββiwβ-eidm=hei.
Therefore,
(2.31)∫𝔹n‍∂(fKz(w))∂widm=fei+f(0)·z-ei.
A direct computation shows that
(2.32)∫𝔹n‍g-·∂Kz(w)∂widm=∫𝔹n‍∑α∈ℤ+n‍gαwα¯·∑α≽ei‍(|β|+n-1)!|β|n!β!βiwβ-eiz-βdm=∑α∈ℤ+n‍gα¯(|α+ei|+n-1)!|α+ei|n!(α+ei)!(αi+1)z-α+ei·∫𝔹n‍|wα|2dm=∑α∈ℤ+n‍gα¯z-α+ei|α|+1.
Hence,
(2.33)(Tf+g¯*wi)(z)=〈Tf+g¯*wi,Kz(w)〉=〈wi,(f+g¯)Kz(w)〉=∫𝔹n‍∂((f+g¯)Kz(w))∂wi¯dm=∫𝔹n‍[∂(fKz(w))∂wi+g-·∂Kz(w)∂wi]dm¯=fei¯+f(0)¯·zei+∑α∈ℤ+n‍gαzα+ei|α|+1.
Comparing the expressions of(Tf+g¯wi)(z) and (Tf+g¯*wi)(z), we obtain
(2.34)gβ|β|+1=fβ,∀|β|>0.
It follows from (2.26) and (2.34) that
(2.35)fβ=gβ=0,for|β|>0,n>1,
which, according to (2.27), implies that u=f(0)+g(0)¯ is a real constant function. This complete the proof of the theorem.Note that for Theorem2.5 the assumption “u=f+g¯∈Ω, where f, g are holomorphic” can not be removed. For example, let u=1-|z|2, that is u=1-(z12+z22+⋯+zn2), then by the below Theorem 4.5  Tu=Tu*=0. However, u is not a constant function.Corollary 2.6.
Letu=f+g¯∈Ω, where f and g are holomorphic. Then Tu is a projection operator if and only if u=1 or u=0.Proof.
The “if” part is clear. IfTu is a projection, then Tu=Tu*. Theorem 2.5 implies that u=c where c is a real. Since Tu=Tu2, we see that c2=c. This proves u=1 or u=0.Next, we will characterize pluriharmonic symbols for which the corresponding Toeplitz operator is an isometry.Theorem 2.7.
Letu=f+g¯∈Ω, where f and g are holomorphic. Then the following statements are equivalent:(1)
Tu is unitary;(2)
Tu is isometric;(3)
u is a constant function of modulus 1.Proof.
That(1) implies (2) follows from the fact that unitary operator is isometric.
To prove that(2) implies (3), we denote f=∑β∈ℤ+n‍fβzβ and g=∑β∈ℤ+n‍gβzβ. Recalling the proof of Theorem 2.5, we have that
(2.36)Tf+g¯1=f(0)+g(0)¯+∑|β|>0‍fβzβ,Tf+g¯wi(z)=∑β∈ℤ+n‍fβzβ+ei+gei¯n+1+g(0)¯·zei,Tf+g¯*1=f(0)¯+g(0)+∑|β|>0‍gβ|β|(n+|β|)zβ,(Tf+g¯*wi)(z)=fei¯+f(0)¯·zei+∑β∈ℤ+n‍gβzβ+ei|β|+1.
Calculating the norms of the above items, it follows that
(2.37)∥Tf+g¯1∥2=|f(0)+g(0)¯|2+∑|β|>0‍|fβ|2∥zβ∥2,(2.38)∥Tf+g¯wi(z)∥2=∑|β|>0‍|fβ|2∥zβ+ei∥2+|gei¯n+1|2+|f(0)+g(0)¯|2,(2.39)∥Tf+g¯*1∥2=|f(0)¯+g(0)|2+∑|β|>0‍|gβ|β|(n+|β|)|2∥zβ∥2,(2.40)∥(Tf+g¯*wi)(z)∥2=|fei¯|2+|f(0)¯+g(0)|2+∑|β|>0‍|gβ|β|+1|2∥zβ+ei∥2.
By the assumption, (2.37), (2.38), (2.39), and (2.40) are all equal to 1 since Tu as well as Tu* is an isometry.
Note that∥zβ∥2/[|β|(n+|β|)]2<∥zβ+ei∥2/(1+|β|)2, for |β|>0 and n>1. Comparing (2.39) and (2.40), we obtain that |f(0)+g(0)¯|=1 and gβ=0, for |β|>0 and n>1. Then (2.37) implies that
(2.41)fβ=0,for|β|>0,n>1.
Therefore, u=f(0)+g(0)¯ and |u|=|f(0)+g(0)¯|=1.
Finally, ifu=c is a constant function, by Theorem 2.4, we have Tu*=Tu-. The desired implication (3)⇒(1) follows from the fact that Tu*Tu=Tu¯Tu=M|u|2=1 and TuTu*=TuTu¯=M|u|2=1.


## 3.   Commuting Toeplitz Operators with Conjugate Holomorphic Symbols

In this section, we will study the commuting problems of Toeplitz operators with conjugate holomorphic symbols. By the definition ofTϕ, if ϕ∈Ω is holomorphic, then Tϕ=Mϕ. Therefore, for two holomorphic symbols f, g∈Ω, Tf, and Tg commute. It is natural to ask when Tf¯ and Tg¯ commute, The following theorem shows that Tf¯ commutes with Tg¯ only in the trival case. In this section, we may always assume f=∑β∈ℤ+n‍fβzβ and g=∑β∈ℤ+n‍gβzβ.Theorem 3.1.
Letf∈Ω and g∈Ω be holomorphic. Then Tg¯Tf¯=Tf¯Tg¯ if and only if for α, β, γ≻0,
(3.1)∑γ+β=α‍gγ¯fβ¯(|γ|-|β|)(n+|β|)(n+|γ|)=0.Proof.
Suppose reproducing kernelKz(w)=1+∑|α|>0(‍((|α|+n-1)!/|α|n!α!)wαz-α) = 1+∑|α|>0‍cαwαz-α. Without loss of generality, we may assume f(0)=g(0)=0.
Note that forα≻β≻0,
(3.2)P(z-βzα)=d(α,α-β)zα-β,
where d(α,α-β)=(α!/(n+|α|-1)!)/((α-β)!/(n+|α-β|-1)!). It follows that
(3.3)Tg¯zα=P(g¯wα)=gα¯∥zα∥22+∑γ≺α‍gγ¯d(α,α-γ)zα-γ.
Therefore, we have
(3.4)Tf¯[Tg¯zα]=gα¯∥zα∥22Tf¯(1)+∑γ≺α‍gγ¯d(α,α-γ)(Tf¯zα-γ)=gα¯f(0)¯∥zα∥22+∑γ≺α‍gγ¯fα-γ¯d(α,α-γ)∥zα-γ∥22+∑γ≺α‍∑β≺α-γ‍gγ¯fβ¯d(α,α-γ)d(α-γ,α-γ-β)zα-γ-β=∑γ+β=α‍gγ¯fβ¯n!α!(n+|β|)(n+|α|-1)!+∑γ+β≺α‍gγ¯fβ¯d(α,α-γ-β)zα-γ-β.
Similarly, we have that
(3.5)Tg¯[Tf¯zα]=∑γ+β=α‍fγ¯gβ¯n!α!(n+|β|)(n+|α|-1)!+∑γ+β≺α‍fγ¯gβ¯d(α,α-γ-β)zα-γ-β=∑γ+β=α‍gγ¯fβ¯n!α!(n+|γ|)(n+|α|-1)!+∑γ+β≺α‍gγ¯fβ¯d(α,α-γ-β)zα-γ-β.
Observe that (3.4) and (3.5) have the same coefficients of zα-γ-β, it follows that
(3.6)Tf¯[Tg¯zα]=Tg¯[Tf¯zα]
if and only if
(3.7)∑γ+β=α‍gγ¯fβ¯(|γ|-|β|)(n+|β|)(n+|γ|)·n!α!(n+|α|-1)!=0.
Since n!α!/(n+|α|-1)!>0, the desired result is obtained.Corollary 3.2.
Iffβ=gβ=0 for |β|≠k0, where k0 is a positive integer, then Tf¯Tg¯=Tg¯Tf¯.Proof.
Since each itemgγ¯fβ¯(|γ|-|β|) equals to 0, (3.1) is satisfied. Thus the desired result follows by Theorem 3.1.For example,Tz-(1,1)Tz-(2,0)=Tz-(2,0)Tz-(1,1) since |(1,1)|=|(2,0)|=2. On the Dirichlet space of the unit disc or polydisc, Dusitermaat, Lee, Geng, and Zhou prove that for holomorphic functions f and g, Tg¯Tf¯=Tf¯Tg¯ if and only if f, g, and 1 are dependent, see [12, 13]. However, this is not true on the unit ball Dirichlet space by Corollary 3.2. Indeed, the condition that f, g, and 1 are linearly dependent is sufficient but not necessary for the commuting of Tf¯ and Tg¯.Next, we will discuss when Toeplitz operator with holomorphic symbol and Toeplitz operator with conjugate holomorphic symbol will commute.Theorem 3.3.
Letf∈Ω and g∈Ω be holomorphic. Then Tg¯Tf=TfTg¯ if and only if f or g is a constant function.Proof.
The “if” part implication is obvious. Now supposeTg¯Tf=TfTg¯. For each multi-index α, we have
(3.8)Tf[Tg¯zα]=f·Tg¯zα=∑β≽0‍fβzβ·[gα¯∥zα∥2+∑γ≺α‍gγ¯d(α,α-γ)zα-γ]=gα¯∥zα∥2·f(0)+gα¯∥zα∥2∑β≻0‍fβzβ+∑γ≺α‍∑β≽0‍gγ¯fβd(α,α-γ)zα-γ+β,(3.9)Tg¯[Tfzα]=Tg¯(fzα)=∑ξ=β+α‍g¯ξfβ∥wξ∥2+∑ξ≺α+β‍∑β≽0‍g¯ξfβd(α+β,α+β-ξ)zα+β-ξ.
Assume that f is not a constant function. Hence there exists β0≻0 such that fβ0≠0.
For (3.8), let β=β0 and γ=α≻0, the coefficient of zβ0 is
(3.10)g¯αfβ0n!α!(n+|α|)!.
On the other hand, if we let β=β0 and ξ=α≻0 in (3.9), then the coefficient of zβ0 is
(3.11)g¯αfβ0d(α+β0,β0).
Since
(3.12)n!α!(n+|α|)!≠d(α+β0,β0)
and fβ0≠0, we deduce that
(3.13)g¯α=0,for|α|>0,
which implies that g is a constant function. The proof is complete.


## 4. Commuting Toeplitz Operators with Symbolszpz-qϕ(|z|2)

Zhou and Dong [14] discussed the commuting and zero product problems of Toeplitz operators on the Bergman space of the unit ball in ℂn whose symbols are of the form ξkϕ where ϕ is a radial function. In [15], they generalized the case of the radial symbols to that of the separately quasi-homogeneous symbols. In [16], Grudsky et al. considered weighted Bergman spaces on the unit ball in ℂn. In terms of the Wick symbol of a Toeplitz operator, the complete information about the operator with radial symbols was given. Vasilevski [17] studied the Toeplitz operators with the quasi-radial quasi-homogeneous symbol. For the case of Dirichlet spaces, Chen et al. [18, 19] studied the quasi-radial Toeplitz operaors on the disk. However, little work has been done in the unit ball case. The commuting problem on it is subtle and no general answer is known. Dong and Zhou [15] have shown that any function f in L2(𝔹n,dm) has the decomposition f(z)=∑k∈ℤ‍ξkfk(r), where fk(r) is separately radial. In this section, the commuting and zero product problems of Toeplitz operators Tzpz-qϕ(|z|2),p,q≽0 will be concerned, which may be helpful to the further study of the commuting Toeplitz operators with general symbols. We denote Σ={ϕ:ϕ,ϕ′∈L1([0,1])} and Σ′={ϕ is absolutely continuous on [0,1):ϕ,ϕ′∈L1([0,1))}. In the remaining part of this paper, we will always assume ϕ∈Σ. A direct calculation gives the following lemma.Lemma 4.1.
Letp≽0 and ϕ(|z|2)∈Σ be radial functions. Then
(4.1)Tzpϕ(|z|2)zα={(n+|p+α|-1)ϕ^(n+|p+α|-1)zp+α,p+α≻0,n∫01‍ϕ(r)rn-1dr,p=α=0,
where ϕ^(z)=∫01‍rz-1[ϕ+∫r1‍ϕ′(t)dt]dr.Proof.
To simiplify the statement, we denote reproducing kernelKz(w) by ∑|γ|≥0‍cγwγz-γ. Notice that (∂/∂wi)(wp+αϕ(|w|2))=(pi+αi)ϕwp+α-ei+wp+αϕ′w-i. For p+α≻0, with integration in polar coordinates we have
(4.2)Tzpϕ(|z|2)zα=〈wp+αϕ,Kz(w)〉=∑i=1n‍∫𝔹n‍∂∂wi(wp+αϕ)∂(Kz(w))∂wi¯dm=[∑i=1n‍∫𝔹n‍(pi+αi)2cp+αϕ|wp+α-ei|2dm+∫𝔹n‍(pi+αi)cp+αϕ′|wp+α|2dm]zp+α=[(n+|p+α|-1)∫01‍tn+|p+α|-2ϕ(t)dt+∫01‍tn+|p+α|-1ϕ′dt]zp+α.
Since ∫01‍tn+|p+α|-1ϕ′dt=(n+|p+α|-1)∫01‍tn+|p+α|-2[∫t1‍ϕ′dr]dt with integration by part, the desired result is obvious.
Forp=α=0, it is easy to see that
(4.3)Tzpϕ(|z|2)zα=〈ϕ,Kz(w)〉=∫𝔹n‍ϕdm=n∫01‍ϕ(r)rn-1dr.
The proof is complete.We now characterize the commuting Toeplitz operators whose symbols are of the formzpϕ(|z|2), where p≻0.Theorem 4.2.
Letp, q≻0, ϕ, ψ∈Σ. Tzpϕ(|z|2)Tzqψ(|z|2)=Tzqψ(|z|2)Tzpϕ(|z|2) if and only if (n+|q+α|-1)ψ^(n+|q+α|-1)ϕ^(n+|p+q+α|-1)=(n+|p+α|-1)ϕ^(n+|p+α|-1)ψ^(n + |p+q+α|-1) holds for any multi-index α≽0.Proof.
For any multi-indexα≽0, by Lemma 4.1, it follows that
(4.4)Tzpϕ(|z|2)Tzqψ(|z|2)zα=(n+|q+α|-1)ψ^(n+|q+α|-1)(n+|p+q+α|-1)ϕ^(n+|p+q+α|-1)zp+q+α,Tzqψ(|z|2)Tzpϕ(|z|2)zα=(n+|p+α|-1)ϕ^(n+|p+α|-1)(n+|p+q+α|-1)ψ^(n+|p+q+α|-1)zp+q+α.
Since (n+|p+q+α|-1)>0, the result is followed.A particular case of the above theorem isϕ=ψ. In this case
(4.5)Tzpϕ(|z|2)Tzqϕ(|z|2)=Tzqϕ(|z|2)Tzpϕ(|z|2)
if and only if
(4.6)[(n+|q+α|-1)ϕ^(n+|q+α|-1)-(n+|p+α|-1)ϕ^(n+|p+α|-1)]·ϕ^(n+|p+q+α|-1)=0.
Thus we immediately have the following result.Corollary 4.3.
Letp, q≻0, ϕ∈Σ. If |p|=|q|, then Tzpϕ(|z|2)Tzqϕ(|z|2)=Tzqϕ(|z|2)Tzpϕ(|z|2).Ifϕ is absolutely continuous on [0,1), integrating by parts, one has mϕ^(m)=limr→1-ϕ(r)=ϕ(1-), for any positive integer m. Thus, using Lemma 4.1 one can get the following lemma which will be often used in the sequel.Lemma 4.4.
Letp≽0, ϕ∈Σ′, one has
(4.7)Tzpϕ(|z|2)zα={ϕ(1-)zp+α,p+α≻0;n∫01ϕ(r)rn-1dr,p=α=0.By Lemma4.4, a regular argument shows the results below.Theorem 4.5.
Letpi≻0 and ϕi∈Σ'. Then the followings hold.(1)
Tzp1ϕ1Tzp2ϕ2=Tzp2ϕ2Tzp1ϕ1=Tzp1+p2ϕ1ϕ2.(2)
Tzp1ϕ1×⋯×Tzpkϕk=0 if and only if ϕ1(1-)×⋯×ϕk(1-)=0.(3)
Letpi≠pj for i≠j, Tzp1ϕ1+⋯+Tzpkϕk=0 if and only if each ϕi(1-)=0,1≤i≤k.Before discussing the commutivity of Toeplitz operator with symbolsz-qϕ(|z|2), one needs the following lemma which can be obtained by direct computation.Lemma 4.6.
Let multi-indexq≽0 and ϕ∈Σ. Then
(4.8)Tz-qϕ(|z|2)zα={d(α,α-q)(n+|α|-1)ϕ^(n+|α|-1)zα-q,α≻q;n!q!(n+|q|-1)!∫01rn+|q|-1ϕ(r)dr,α=q;0,α≽q,
where d(α,α-q)=(α!/(n+|α|-1)!)/((α-q)!/(n+|α-q|-1)!) and α≽q means that there exists i0 such that αi0<qi0.Proof.
Forα≻q, We get that
(4.9)Tz-qϕ(|z|2)zα=〈wαw-qϕ,Kz(w)〉=∑i=1n‍∫𝔹n‍∂∂wi(wαw-qϕ)∂(Kz(w))∂wi¯dm=∑i=1n‍∫𝔹n‍(αiwα-eiϕ+wαw-iϕ′)w-q·∑r‍crriwr-eiz-r¯dm=[∑i=1n‍∫𝔹n‍αiϕ|wα-ei|2cα-q(αi-qi)+|wα|2ϕ′cα-q(αi-qi)dm]zα-q=cα-qzα-q|α-q|n!α!(n+|α|-1)![(n+|α|-1)∫01‍tn+|q+α|-2ϕdt+∫01‍tn+|q+α|-1ϕ′dt]=d(α,α-q)(n+|α|-1)ϕ^(n+|α|-1)zα-q.
Forα=q, we have
(4.10)Tz-qϕ(|z|2)zα=〈wαw-qϕ,Kz(w)〉=∫𝔹n‍|zq|2ϕ(|w|2)dm=n!q!(n+|q|-1)!∫01‍tn+|q|-1ϕ(t)dt.
If there exists1≤i≤n such that αi<qi, then
(4.11)Tz-qϕ(|z|2)zα=〈wαw-qϕ,Kz(w)〉=0.
Thus the proof is complete.Note that ifϕ∈Σ', then (n+|α|-1)ϕ^(n+|α|-1)=ϕ(1-). It follows that
(4.12)Tz-qϕ(|z|2)zα=d(α,α-q)ϕ(1-)zα-q,forα≻q.The following theorem gives some properties of the Toeplitz operator with symbolsz-pϕ(|z|2).Theorem 4.7.
Letp, q, pi≻0, pi≠pj for i≠j and ϕ, ψ, ϕi∈Σ'. Then the following assertions hold.(1)
Tz-pϕ(|z|2)Tz-qψ(|z|2)=Tz-qψ(|z|2)Tz-pϕ(|z|2) if and only if ψ(1-)∫01‍rn+|p|-1ϕ(r)dr = ϕ(1-)∫01‍rn+|q|-1ψ(r)dr.(2)
Tz-p1ϕ1×⋯×Tz-pkϕk=0 if and only if one of the following holds:
(i)
ϕ1(1-)=0 and ∫01‍rn+|p1|-1ϕ1(r)dr=0;(ii)
There existsi0 where 2≤i0≤k such that ϕi0(1-)=0.(3)
Tz-p1ϕ1+⋯+Tz-pkϕk=0 if and only if ϕi(1-)=0 and ∫01‍rn+|pi|-1ϕi(r)dr=0 for each i, 1≤i≤k.Proof.
Assertions(2) and (3) are the direct consequence of Lemma 4.6. We only need to prove assertion (1). By Lemma 4.6, for h≻p+q, since ϕ, ψ∈Σ' we have
(4.13)Tz-pϕ(|z|2)Tz-qψ(|z|2)(zh)=d(h,h-q)d(h-q,h-q-p)ϕ(1-)ψ(1-)zh-q-p=d(h,h-q-p)ϕ(1-)ψ(1-)zh-q-p,Tz-qψ(|z|2)Tz-pϕ(|z|2)(zh)=d(h,h-p)d(h-p,h-q-p)ϕ(1-)ψ(1-)zh-q-p=d(h,h-q-p)ϕ(1-)ψ(1-)zh-q-p.
It is obvious that
(4.14)Tz-pϕ(|z|2)Tz-qψ(|z|2)(zh)=Tz-qψ(|z|2)Tz-pϕ(|z|2)(zh)
holds for h≻p+q.
Forh=p+q, we obtain
(4.15)Tz-pϕ(|z|2)Tz-qψ(|z|2)(zh)=d(p+q,p)ψ(1-)n!p!(n+|p|-1)!∫01‍rn+|p|-1ϕ(r)dr=n!(p+q)!(n+|p+q|-1)!ψ(1-)∫01‍rn+|p|-1ϕ(r)dr,Tz-qψ(|z|2)Tz-pϕ(|z|2)(zh)=d(p+q,q)ϕ(1-)n!q!(n+|q|-1)!∫01‍rn+|q|-1ψ(r)dr=n!(p+q)!(n+|p+q|-1)!ϕ(1-)∫01‍rn+|q|-1ψ(r)dr.
Since n!(p+q)!/(n+|p+q|-1)!>0, then the desired result is obvious.In the assertion(1) of Theorem 4.7, if ϕ=ψ=1, then we get
(4.16)ψ(1-)∫01‍rn+|p|-1ϕ(r)dr=1n+|p|,ϕ(1-)∫01‍rn+|q|-1ψ(r)dr=1n+|q|.Therefore, it is easy to get the following corollary.Corollary 4.8.
Letp, q≻0. Then Tz-pTz-q=Tz-qTz-p if and only if |p|=|q|.It is well known thatTϕTψ=Th on the Hardy space if and only if either ϕ- or ψ is holomorphic. However, Lemma 4.6 and Theorem 4.7 implies that a similar result does not hold on the Dirichlet space of the unit ball. Indeed, by the computation in Lemma 4.6 and Theorem 4.7, it is easy to verify that for any p, q≻0, Tz-pTz-qzp+q≠Tz-p+qzp+q.Theorem 4.9.
Letp, q≻0 and ϕ, ψ∈Σ'. Then Tzpϕ(|z|2)Tz-qψ(|z|2)=Tz-qψ(|z|2)Tzpϕ(|z|2) if and only if ϕ(1-)=0 or ψ(1-)=0 and ∫01‍rn+|q|-1ψ(r)dr=0.Proof.
For each multi-indexh≽0, by Lemmas 4.4 and 4.6, we have that
(4.17)Tzpϕ(|z|2)Tz-qψ(|z|2)zh={d(h,h-q)ψ(1-)ϕ(1-)zh+p-q,h≻qn!q!(n+|q|-1)!∫01rn+|q|-1ψ(r)drϕ(1-)zp,h=q0,others,Tz-qψ(|z|2)Tzpϕ(|z|2)zh={d(h+p,h+p-q)ψ(1-)ϕ(1-)zh+p-q,p+h≻qn!q!(n+|q|-1)!∫01rn+|q|-1ϕ(r)drϕ(1-),p+h=q0,others.
Consequently, for h≻q, we conclude
(4.18)d(h,h-q)ψ(1-)ϕ(1-)zh+p-q=d(h+p,h+p-q)ψ(1-)ϕ(1-)zh+p-q.
Since p, q≻0, it is clear that d(h,h-q) can not always equal to d(h+p,h+p-q) for all h≻q. Thus, we get
(4.19)ψ(1-)ϕ(1-)=0.
On the other hand, forh=q, we have
(4.20)n!q!(n+|q|-1)!∫01‍rn+|q|-1ψ(r)drϕ(1-)=d(h+p,h+p-q)ψ(1-)ϕ(1-).
Combining (4.19) and (4.20), we obtain the desired result.Notice the assertion(3) of Theorem 4.5 and the assertion (3) of Theorem 4.7, Theorem 4.9 above shows that Tzpϕ(|z|2)Tz-qψ(|z|2)=Tz-qψ(|z|2)Tzpϕ(|z|2) if and only if Tzpϕ(|z|2)=0 or Tz-qψ(|z|2)=0 holds. That is, Tzpϕ(|z|2) commutes with Tz-qψ(|z|2) only in the trival case.Finally, we will discuss when Toepitze operaorTϕ(|z|2) commute with Tzpz-qψ(|z|2).Theorem 4.10.
Letp, q≻0 and ϕ, ψ∈Σ'. Then the following assertions hold.(1)
Ifp≻q, Tϕ(|z|2)Tzpz-qψ(|z|2)=Tzpz-qψ(|z|2)Tϕ(|z|2) if and only if ψ(1-)=0 or ϕ(1-)=n∫01‍rn-1ϕ(r)dr.(2)
Ifq≻p, Tϕ(|z|2)Tzpz-qψ(|z|2)=Tzpz-qψ(|z|2)Tϕ(|z|2) if and only if ∫01‍rn+|q|-1ψ(r)dr=0 or n∫01‍rn-1ϕ(r)dr=ϕ(1-).(3)
Ifp≽q and q≽p or p=q, Tϕ(|z|2)Tzpz-qψ(|z|2)=Tzpz-qψ(|z|2)Tϕ(|z|2).Proof.
For each multi-indexh≽0, by Lemmas 4.4 and 4.6 we have
(4.21)Tϕ(|z|2)Tzpz-qψ(|z|2)zh={d(p+h,p+h-q)ψ(1-)ϕ(1-)zh+p-q,h+p≻qn!q!(n+|q|-1)!<?displaystyle?>∫01rn+|q|-1ψ(r)dr·n∫01rn-1ϕ(r)dr,h+p=q,Tzpz-qψ(|z|2)Tϕ(|z|2)zh={d(p+h,p+h-q)ψ(1-)ϕ(1-)zh+p-q,h≻0,h+p≻qϕ(1-)n!q!(n+|q|-1)!∫01rn+|q|-1ψ(r)dr,h≻0,h+p=q0,h≻0,h+p≽qn∫01rn-1ϕ(r)drd(p,p-q)ψ(1-)zh+p-q,h=0,p≻qn∫01rn-1ϕ(r)drn!q!(n+|q|-1)!∫01rn+|q|-1ψ(r)dr,h=0,p=q0,h=0,p≽q.Case 1. Suppose p≻q. We have
(4.22)Tϕ(|z|2)Tzpz-qψ(|z|2)zh=d(p+h,p+h-q)ψ(1-)ϕ(1-)zh+p-q,Tzpz-qψ(|z|2)Tϕ(|z|2)zh={d(p+h,p+h-q)ψ(1-)ϕ(1-)zh+p-q,h≻0n∫01rn-1ϕ(r)drd(p,p-q)ψ(1-)zh+p-q,h=0.Tzpz-qψ(|z|2) commutes with Tϕ(|z|2) if and only if
(4.23)d(p,p-q)ψ(1-)ϕ(1-)=d(p,p-q)ψ(1-)n∫01‍rn-1ϕ(r)dr,
which is equivalent to ψ(1-)=0 or ϕ(1-)=n∫01‍rn-1ϕ(r)dr.
Case 2. Suppose q≻p. Note that h+p≻q if and only if h≻q-p. We have
(4.24)Tϕ(|z|2)Tzpz-qψ(|z|2)zh={d(p+h,p+h-q)ψ(1-)ϕ(1-)zh+p-q,h≻q-pn!q!(n+|q|-1)!∫01rn+|q|-1ψ(r)dr·n∫01rn-1ϕ(r)dr,h=q-p0,h≽q-p,Tzpz-qψ(|z|2)Tϕ(|z|2)zh={d(p+h,p+h-q)ψ(1-)ϕ(1-)zh+p-q,h≻q-pϕ(1-)n!q!(n+|q|-1)!∫01rn+|q|-1ψ(r)dr,h=q-p0,h≽q-p.
It follows that Tzpz-qψ(|z|2) commutes with Tϕ(|z|2) if and only if
(4.25)∫01‍rn+|q|-1ψ(r)dr·n∫01‍rn-1ϕ(r)dr=∫01‍rn+|q|-1ψ(r)drϕ(1-),
which is equivalent to ∫01‍rn+|q|-1ψ(r)dr=0 or n∫01‍rn-1ϕ(r)dr=ϕ(1-).
Case 3. Suppose p≽q and q≽p. Let h′={hi}, where hi=max{qi-pi,0} for 1≤i≤n. Then for h≽0, h+p≻q if and only if h≻h′. Thus,
(4.26)Tϕ(|z|2)Tzpz-qψ(|z|2)zh={d(p+h,p+h-q)ψ(1-)ϕ(1-)zh+p-q,h≽h′0,h≽h′,Tzpz-qψ(|z|2)Tϕ(|z|2)zh={d(p+h,p+h-q)ψ(1-)ϕ(1-)zh+p-q,h≽h′0,h≽h′.
It is obvious that Tzpz-qψ(|z|2) commutes with Tϕ(|z|2).
Case 4. Suppose p=q. We have
(4.27)Tϕ(|z|2)Tzpz-qψ(|z|2)zh={d(p+h,p+h-q)ψ(1-)ϕ(1-)zh+p-q,h≻0n!q!(n+|q|-1)!∫01rn+|q|-1ψ(r)dr·n∫01rn-1ϕ(r)dr,h=0,Tzpz-qψ(|z|2)Tϕ(|z|2)zh={d(p+h,p+h-q)ψ(1-)ϕ(1-)zh+p-q,h≻0n∫01rn-1ϕ(r)drn!q!(n+|q|-1)!∫01rn+|q|-1ψ(r)dr,h=0.
It is easy to see that Tzpz-qψ(|z|2) commutes with Tϕ(|z|2). This completes the proof.Corollary 4.11.
Letp⊥q and ϕ, ψ∈Σ′. Then Tϕ(|z|2)Tzpz-qψ(|z|2)=Tzpz-qψ(|z|2)Tϕ(|z|2).Proof.
Note thatp⊥q implies p≽q and q≽p. The desired result is immediately followed by Theorem 4.10.



---

*Source: 958201-2012-11-20.xml*