# Commuting Separately Quasihomogeneous Small Hankel Operators on Pluriharmonic Bergman Space

Authors: Qi Wu; Chuntao Qin; Yong Chen; Yile Zhao
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We study (semi)commutativity of small Hankel operators with separately quasihomogeneous symbols on the pluriharmonic Bergman space of the unit ball. Some product problems are also concerned.


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## Body

## 1. Introduction

LetBn be the unit ball in Cn and its boundary Sn. Let dv denote the normalized Lebesgue volume measure on the unit ball Bn. L2(Bn,dv) is the Hilbert space of Lebesgue square integrable functions on Bn with inner product: (1)f,g=∫Bnfzgz¯dvz.The Bergman space La2(Bn) is a subspace of L2(Bn,dv) consisting of all holomorphic functions. It is well known that La2(Bn) is a reproducing function space with the reproducing kernel: (2)Kzw=11-w,zn+1for z,w∈Bn. Let P be the orthogonal projection from L2(Bn,dv) onto La2(Bn); then we have (3)Pfz=f,Kz,f∈L2Bn,dv.The pluriharmonic Bergman spaceLh2(Bn) is the closed subspace of L2(Bn,dv) consisting of the pluriharmonic functions on Bn. It is well known that (4)Lh2Bn=La2Bn+La2Bn¯,where La2(Bn)¯={f¯∣f∈La2(Bn)}; then Lh2(Bn) is also a reproducing function space with reproducing kernel:(5)Rzw=Kzw+Kzw¯-1,z,w∈Bn.Let Q denote the orthogonal projection from L2(Bn,dv) onto Lh2(Bn); then (6)Qfz=f,Rz,f∈L2Bn,dv.Using (5), we have(7)Qfz=Pfz+Pf¯z¯-Pf0.LetL∞ be a set of all bounded measurable functions on Bn. Fix a function φ∈L∞. The Toeplitz operator Tφ:Lh2(Bn)→Lh2(Bn) and the small Hankel operator Hφ:Lh2(Bn)→Lh2(Bn) with the symbol φ are defined by (8)Tφf=Qφf,Hφf=QJφf,respectively, where J:L2(Bn,dv)→L2(Bn,dv) is an unitary operator defined by Jf(z)=f(z¯). One can verify that JQ=QJ, so we see the relation between the Toeplitz operator and small Hankel operator is(9)Hφ=JTφ=TJφJ.One can easily check Tφ∗=Tφ¯, so, by the above relation, we have Hφ∗=Hφ∗, and here φ∗(z)=φ(z¯)¯.We need more notations. LetN denote the set of all nonnegative integers. For multi-index α=(α1,α2,…,αn)∈Nn and point z=(z1,z2,…,zn)∈Bn, we write (10)α=α1+α2+⋯+αn,α!=α1!⋯αn!,zα=z1α1⋯znαn.For two multi-indexes α=(α1,α2,…,αn), β=(β1,β2,…,βn)∈Nn, the notations α≥β means that αj≥βj for every j and α⊥β means that (11)α1β1+α2β2+⋯+αnβn=0.Let α-β denote (α1-β1,α2-β2,…,αn-βn). Note that if α-β≥0=(0,0,…,0), then α-β=α-β.For a functionϕ∈L2(Bn,dv), ϕ is said to beradial if ϕ(z)=ϕ(z) andseparately radial if ϕ(z1,z2,…,zn)=ϕ(z1,z2,…,zn).Letr=(r1,r2,…,rn), r=r12+r22+⋯+rn2. For f∈L2(Bn,dv), if there are β,γ∈Nn and separately radial function ϕ such that f(rξ)=ξβξ¯γϕ(r), where ξ∈Sn, then f is called aseparately quasihomogeneous function.The (semi)commutativity of two operators is an important topic in operator theory. In [1], Brown and Halmos completely characterized the commutativity of Toeplitz operators on the classical Hardy space. From then on, many related works on Toeplitz operators or (small) Hankel operators emerged (see, e.g., [2, 3]).For the case on the Bergman space of the unit disk, the commutativity is more subtle than that on the Hardy space. References [4, 5] obtained the Brown-Halmos type theorems for Toeplitz operators with harmonic symbols. Many subsequent works studied these problems for special symbol classes, such as harmonic symbols, radial symbols, or quasihomogeneous symbols; see [6–9], for example.On the harmonic Bergman space, there were some studies focusing on the commuting Toeplitz operators with harmonic symbols [10, 11] or quasihomogeneous symbols [12, 13] and showed that the results obtained are also quite different from the case on the Hardy or Bergman space. Recently, [14] studied the algebraic properties of small Hankel operators on the harmonic Bergman space and got very different commutativity of small Hankel operators compared with the case of Toeplitz operators. So in this paper, we want to continue the work and generalize the commuting small Hankel operators to the higher dimension case.In order to state our main results, we still need the following notations and facts. Denoteτ(Bn) as the set (12)r=r1,…,rn=z1,…,zn:z1,…,zn∈Bn.If ϕ is a bounded separately radial function, then(13)∫Bnϕzdvz=2nn!∫τBnϕrrdr,where rdr=∏i=1nridri. Let R be the set consisting of all ϕ∈L2(B,dv) which is separately radial. Note that for every h∈L2(Bn,dv) we have(14)hrξ=∑β,γ≥0ξβξ¯γϕβ,γr,where every ϕβ,γ∈R. Moreover, if h∈L∞, then also each ξβξ¯γϕβ,γ∈L∞. For details, one may see [15].We will first investigate when two small Hankel operators with a certain symbols commute. For two operatorsT1, T2, their commutator T1T2-T2T1 is denoted by [T1,T2]. So T1 commutes with T2 means [T1,T2]=0.Theorem 1.
Letp,q,s,t∈Nn and ϕ,ψ∈R∩L∞. Then [Hξpξ¯sϕ,Hξqξ¯tψ]=0 if and only if one of the following holds:(i)
ϕ=0 or ψ=0.(ii)
p-s=q-t.Note thatHξpξ¯sϕ∗=Hξpξ¯sϕ¯, so the above theorem implies that Hξpξ¯sϕ is always a normal operator on Lh2(Bn). We also note that it is still open when two Toeplitz operators with separately quasihomogeneous symbols commute.Theorem 2.
Letϕ be bounded radial and h∈L∞. Then [Hϕ,Hh]=0 if and only if one of the following holds: (i)
ϕ=0.(ii)
h is separately radial.(iii)
ϕ is nonzero constant and Jh=h.The above result is also different from the case for two Toeplitz operators with same symbols; see Corollary  5.2 in [15].In order to get the semicommutativity of small Hankel operators (see Corollaries12 and 15 in the next section), we turn to characterize when the product of two small Hankel operators is another small Hankel operator. We obtain the following results.Theorem 3.
Letp,q,s,t∈Nn and ϕ,ψ∈R∩L∞, h∈L∞. Then the following statements are equivalent: (i)
Hξpξ¯sϕHξqξ¯tψ=Hh.(ii)
Hξqξ¯tψHξpξ¯sϕ=Hh.(iii)
ϕ=h=0 or ψ=h=0.Theorem 4.
Letϕ be bounded radial and g,h∈L∞. Then the following statements are equivalent: (i)
HϕHh=Hg.(ii)
HhHϕ=Hg.(iii)
ϕ=g=0 or h=g=0.We would like to point out that the above main results also can apply to answer other related questions. For an example in point, by (9), we have (15)TgTJh-ThTJg=JHgHh-HhHgJ,so one obtains that when [Hg,Hh]=0 can also answer when TgTJh=ThTJg.We will give the proofs of above main results in the next section. Meanwhile, several corollaries also will be deduced.


## 2. Proof of the Main Results

We first recall some known facts.It is well known that a bounded analytic function on the half plane{z∈C:Re⁡z>0} is uniquely determined by its value on an arithmetic sequence of integers. In fact, we have the following classical theorem (see p. 102 of [16]).Lemma 5.
Suppose thatf is a bounded analytic function on {z∈C:Re⁡z>0} which vanishes at the pairwise distinct points z1,z2,…, where inf⁡{zj}>0 and ∑j≥1Re⁡(1/zj)=∞, and then f vanishes identically on {z∈C:Re⁡z>0}.We will need a similar result in higher dimensions which is proved by the above lemma. First, we give the following notations.LetE⊂N2, we say that E satisfiescondition (I) if there exists a sequence αi1i=1∞ such that ∑i=1∞=1/αi1=∞, and, for every fixed αi1, there also exists a sequence αij2j=1∞ such that ∑j=1∞(1/αij2)=∞ and {(αi1,αij2):j=1,2,…}⊂E.One will easily see that for a multi-indexδ∈N2, if E⊂{α∈N2:α≥δ} and let Ec be the complement of E in {α∈N2:α≥δ}, then either E or Ec satisfies the condition (I). Using Lemma 5, one may prove the following (also see Corollary  2.7 in [15]).Lemma 6.
Letp,s∈N2 and g(r) be a bounded function on τ(B2). If the set (16)E=α∈N2:α≥s,∫τB2grr2α+p-srdr=0satisfies condition (I), then g=0.Letα,p,s∈Nn. Set (17)Aϕ;α,p,s≕2nn+α+p-s!∫τBnϕrr2α+2pr-p-srdrα+p-s!,when α+p≥s and (18)Bϕ;s,α,p≕2nn+s-α-p!∫τBnϕrr2sr-p-srdrs-α-p!when α+p≤s. The following lemma is immediately from (9) and Lemma  3.4 of [17] and we will use it frequently.Lemma 7.
Letp,s∈Nn and ϕ∈R; then, for any multi-index α∈Nn, (19)Hξpξ¯sϕzα=Aϕ;α,p,sz¯α+p-sα+p≥s,Bϕ;s,α,pzs-α-pα+p≤s,0others,Hξpξ¯sϕz¯α=Aϕ;α,s,pzα+s-pα+s≥p,Bϕ;p,α,sz¯p-α-sα+s≤p,0others.Note that, for two nonzero multi-indexesp,s∈Nn with p⊥s, then α+p≥s if and only if α≥s and there is no α∈Nn such that α+p≤s. Hence the above lemma gives the following.Lemma 8.
Letp,s∈Nn with p,s≠0 and p⊥s, ϕ∈R; then, for any multi-index α∈Nn, (20)Hξpξ¯sϕzα=Aϕ;α,p,sz¯α+p-sα≥s,0others,Hξpξ¯sϕz¯α=Aϕ;α,s,pzα+s-pα≥p,0others.We are ready to prove the first main result.Proof of Theorem1.
Using Lemma7, direct calculations show that(21)Hξpξ¯sϕHξqξ¯tψzα=Aψ;α,q,tAϕ;α+q-t,s,pzα+q-t+s-pwhen α≥t-q, α≥t-q+p-s and(22)Hξqξ¯tψHξpξ¯sϕzα=Aϕ;α,p,sAψ;α+p-s,t,qzα+p-s+t-qwhen α≥s-p, α≥s-p+q-t. Thus when (23)α≥t-q,t-q+p-s,s-p,q-t+s-p,combining with (21) and (22), it follows from [Hξpξ¯sϕ,Hξqξ¯tψ](zα)=0 that (24)α+q-t+s-p=α+p-s+t-qwhich implies p-s=q-t or(25)Aψ;α,q,tAϕ;α+q-t,s,p=0holds when α≥t-q,α≥t-q+p-s, which induces ϕ=0 or ψ=0. In fact, for the sake of simplicity, we only consider the case n=2. Put (26)E=α∈N2:α≥p+s,Aψ;α,q,t=0,if E satisfies the condition (I); then, by Lemma 6, we can get ψ=0; otherwise Ec will satisfy the condition (I), which by (25) says that the set (27)α∈N2:α≥p+s,Aϕ;α+q-t,s,p=0satisfies the condition (I); thus, by Lemma 6, again we get ϕ=0.
Conversely, ifϕ=0 or ψ=0, it is clear that Hξpξ¯sϕ commutes with Hξqξ¯tψ. Now suppose p-s=q-t. When α≥t-q and α≥0, by (21) and (22), one can easily check(28)Hξpξ¯sϕHξqξ¯tψzα=Hξqξ¯tψHξpξ¯sϕzαsince p-s=q-t. When 0≤α≤t-q, by Lemma 7, we have (29)Hξpξ¯sϕHξqξ¯tψzα=Bψ;t,α,qBϕ;s,t-α-q,pzα,Hξqξ¯tψHξpξ¯sϕzα=Bϕ;s,α,pBψ;t,s-α-p,qzα.It is clear that the above two are equal because p-s=q-t. When α≥0 and t-q≰α≰t-q, both sides of equality (28) are zero, so they are equal. Thus, we conclude Hξpξ¯sϕHξqξ¯tψ=Hξqξ¯tψHξpξ¯sϕ on the analytic part {zα}α∈Nn, which is also true on the coanalytic part {z¯α}α∈Nn with similar arguments when p-s=q-t. The proof is complete.Before we prove Theorem2, we introduce the Mellin transform which is one of the most useful tools in studying our problems. The Mellin transform ϕ^ of a function ϕ∈L1([0,1],rdr) is defined by ϕ^(z)=∫01ϕ(s)sz-1ds. It is known that ϕ^ is a bounded analytic function in the half plane {z:Re⁡z>2}. In addition, for a radial function ϕ, to compute ∫Bnϕ(z)zα2dv using (13), we obtain(30)Aϕ;α,0,0=2n+2αϕ^2n+2α.Letz=(z1,…,zn)=(|z|ξ1,…,zξn)∈Bn, where ξ=(ξ1,…,ξn)∈Sn. Put rj=zj, then ξj=rj/z,j=1,…,n. Let r=(r1,…,rn). For every separately quasihomogeneous function f(z)=f(rξ)=ξβξ¯γϕ(r), there is unique p,s∈Nn with p⊥s such that p-s=β-γ and (31)frξ=ξpξ¯sξ12minβ1,γ1⋯ξn2minβn,γnϕr=ξpξ¯sr1r2minβ1,γ1⋯rnr2minβn,γnϕr≕ξpξ¯sΦr,where Φ is a separately radial function. Thus for each h∈L2(Bn,dv) denoted as the form (14), we may rewrite h=h1+h2+h3, where(32)h1=∑p≥0ξpϕp,0,h2=∑s≥0,s≠0ξ¯sϕ0,s,h3=∑p⊥s,p,s≠0ξpξ¯sϕp,s.Now we can prove the second main result.Proof of Theorem2.
Suppose[Hϕ,Hh]=0. We may write h=h1+h2+h3, where h1, h2, h3 are given by (32). Using Lemmas 7 and 8, direct calculations give that, for every α∈Nn,(33)HϕHhzα=HϕHh1zα+HϕHh2zα+HϕHh3zα=∑p≥0Aϕp,0;α,p,0Aϕ;α+p,0,0zα+p+∑s≥α,s≠αBϕ0,s;s,α,0Aϕ;s-α,0,0z¯s-α+∑0≠s≤αAϕ0,s;α,s,0Aϕ;α-s,0,0zα-s+∑p⊥s,p,s≠0,s≤αAϕp,s;α,p,sAϕ;α+p-s,0,0zα+p-s,(34)HhHϕzα=Aϕ;α,0,0Hhz¯α=∑0≠p≤αAϕp,0;α,0,pAϕ;α,0,0zα-p+∑p≥α,p≠αBϕp,0;p,α,0Aϕ;α,0,0z¯p-α+∑s≥0Aϕ0,s;α,s,0Aϕ;α,0,0zα+s+∑p⊥s,p,s≠0,p≤αAϕp,s;α,s,pAϕ;α,0,0zα+s-p.Comparing the fourth summations in the equalities (33) and (34), it follows from HϕHh(zα)=HhHϕ(zα) that, for every p⊥s,p,s≠0 and s≤α, we have(35)Aϕp,s;α,p,sAϕ;α+p-s;0,0=Aϕs,p;α,p,sAϕ;α,0,0.Similarly, comparing the first summation in the equality (33) with the third one in the equality (34), we then get that, for every p,α≥0,(36)Aϕp,0;α,p,0Aϕ;α+p;0,0=Aϕ0,p;α,p,0Aϕ;α,0,0.
With the same arguments as above, it follows fromHϕHh(z¯α)=HhHϕ(z¯α) that, for every p⊥s,p,s≠0 and s≤α, we have(37)Aϕs,p;α,p,sAϕ;α+p-s;0,0=Aϕp,s;α,p,sAϕ;α,0,0and, for every p,α≥0,(38)Aϕ0,p;α,p,0Aϕ;α+p;0,0=Aϕp,0;α,p,0Aϕ;α,0,0.
Supposeϕ is a nonzero constant; then, by (30), we have (39)Aϕ;α+p-s;0,0=Aϕ;α,0,0≠0,so (35) gives that, for every p⊥s,p,s≠0, when α≥s, (40)Aϕp,s-ϕs,p;α,p,s=0,which induces that ϕp,s=ϕs,p for every p⊥s,p,s≠0 by Lemma 6. Similarly, it follows from (36) we can get ϕp,0=ϕ0,p for every p≠0. Therefore, we obtain Jh=h.
In the following, we supposeϕ is not a constant. It follows from (35) and (37) that, for every p⊥s,p,s≠0, when α≥s,(41)Aϕp,s-ϕs,p;α,p,sAϕ;α+p-s,0,0+Aϕ;α,0,0=0.For the sake of simplicity, we only consider the case n=2. For p⊥s,p,s≠0, put (42)Ep,s=α∈Nn:α≥s,Aϕp,s-ϕs,p;α,p,s=0.If Ep,s satisfies the condition (I), then by Lemma 6 we obtain ϕp,s=ϕs,p; if Ep,s does not satisfy the condition (I), then Ep,sc, the complement of Ep,s in {α∈Nn:α≥s}, must satisfy the condition (I). In particular, we have ∑α∈Ep,sc1/|α|=∞. So by (41) we have, for p⊥s,p,s≠0, (43)Aϕ;α,0,0=-Aϕ;α+p-s,0,0,α∈Ep,sc.That is, when α∈Ep,sc, using (30), we have(44)2n+2αϕ^2n+2α=-2n+2α+2p-2sϕ^2n+2α+2p-2s.Denote(45)Fz=ϕ^z+2n+2p-2s+z+2nz+2n+2p-2sϕ^z+2n,and then F is a bounded analytic function on {z∈C:Re⁡z>2|s|}. Moreover, it follows from (44) that F(2α)=0 when α∈Ep,sc. According to Lemma 5 we thus have F≡0, which implies(46)z+2nϕ^z+2n=-z+2n+2p-2sϕ^z+2n+2p-2sfor every z in {z∈C:Re⁡z>2s}. If p=s, then it follows from the above that 2(z+2n)ϕ^(z+2n)=0, so it is clear that ϕ=0, a contradiction. Without loss of generality, we may assume p>s. Replace z by z+2p-2s in (46) and then compare with (46) to obtain (47)z+2nϕ^z+2n=z+2n+4p-4sϕ^z+2n+4p-4s,and thus, as in the proof of Theorem  6 in [6], the above equation gives that ϕ is a constant, which is also a contradiction.
Therefore, whenϕ is not a constant, then, for every p,s∈Nn with p⊥s and p,s≠0, ϕp,s=ϕs,p. Putting them into (35), we get that, for p⊥s,p,s≠0, (48)Aϕp,s;α,p,sAϕ;α+p-s;0,0-Aϕ;α,0,0=0when α≥s. Since ϕ is not a constant, then similar argument as before we get ϕp,s=0; hence ϕp,s=ϕs,p=0 for p⊥s,p,s≠0. With same arguments, using (36) and (38), we can get ϕp,0=ϕ0,p=0 for p≠0. So h=ϕ0,0, a separately radial function.
Conversely, ifϕ=0, then clearly [Hϕ,Hh]=0; if h is separately radial, then Theorem 1 gives [Hϕ,Hh]=0; if ϕ=c, a nonzero constant, and Jh=h, then using (9) we have (49)HcHh=cJHh=cHJhJ=HJhHc=HhHc,as desired. The proof is complete.Before we prove the last two main results, we should note that, for every small Hankel operatorHh with h∈L∞, it is easy to check that Hh is a complex symmetric operator with complex conjugate C which is defined as Cf(z)=f(z¯)¯ for f∈L2(Bn,dv), that is, Hf∗=CHhC. Hence we have the following analogous result as Proposition  18 in [14].Lemma 9.
Letf,g,h∈L∞. Then HfHg=Hh if and only if HgHf=Hh, and each case implies [Hf,Hg]=0.We also need the following partial result of zero product characterization which has independent interest.Theorem 10.
Letp,s∈Nn, ϕ∈R∩L∞, h∈L∞. Then the following statements are equivalent: (i)
Hξpξ¯sϕHh=0.(ii)
HhHξpξ¯sϕ=0.(iii)
ϕ=0 or h=0.Proof.
By Lemma9 we only need to show (ii) ⇒ (iii). Write h as the form (14). When α≥s-p, by Lemma 7, we have (50)Hξpξ¯sϕzα=Aϕ;α,p,sz¯α+p-s.Using Lemma 7 again we get that the analytic part of HhHξpξ¯sϕ(zα) is given by (51)∑β-γ+p-s≤αAϕ;α,p,sAϕβ,γ;α+p-s,γ,βzα+p-s+γ-β,and hence HhHξpξ¯sϕ(zα)=0 will give that, for each β,γ∈Nn, (52)Aϕ;α,p,sAϕβ,γ;α+p-s,γ,β=0holds for any α≥s-p and α≥β-γ+p-s. Therefore, with the same arguments as done in the proof of Theorem 1 we can get ϕ=0 or ϕβ,γ=0 for each β,γ, as the assertion. We finish the proof.The next rephrasing of above result is a cancellation law for small Hankel operators with separately quasihomogeneous symbols.Corollary 11.
Letp,s∈Nn and 0≠ϕ∈R∩L∞, g,h∈L2. Then the following statements are equivalent: (i)
Hξpξ¯sϕHg=Hξpξ¯sϕHh.(ii)
HgHξpξ¯sϕ=HhHξpξ¯sϕ.(iii)
g=h.Now we turn to prove the third main result.Proof of Theorem3.
By Lemma9 we only need to show (i) ⇒ (iii). Write h as the form (14). Using Lemma 7, direct calculations give that (21) holds when α≥t-q,α ≥t-q+p-s and(53)Hhzα=∑γ-β≤αAϕβ,γ;α,β,γz¯α+β-γ+∑γ-β≥α,γ-β≠αBϕβ,γ;γ,α,βzγ-α-β,and hence when α≥t-q, α≥t-q+p-s, the two equalities (21) and (53) are equal implying that the coanalytic part of Hh(zα) must be zero, that is, A(ϕβ,γ;α,β,γ)=0 for all α≥γ-β, α≥t-q, α≥t-q+p-s.
Now for each fixedβ,γ∈Nn, using Lemma 6, the above tells ϕβ,γ=0, which yields that h=0. Hence we get Hξpξ¯sϕHξqξ¯tψ=Hh=0, which further gives ϕ=0 or ψ=0 by Theorem 10. We complete the proof.Theorem3 answers immediately the following semicommutativity for two small Hankel operators with separately radial symbols.Corollary 12.
Letp,s,q,t∈Nn and ϕ,ψ∈R∩L∞. Then the following statements are equivalent: (i)
Hξpξ¯sϕHξqξ¯tψ=Hξp+qξ¯s+tϕψ.(ii)
Hξqξ¯tψHξpξ¯sϕ=Hξp+qξ¯s+tϕψ.(iii)
ϕ=0 or ψ=0.The next corollary says there are no nonzero idempotent small Hankel operators with separately quasihomogeneous symbols.Corollary 13.
Letp,s∈Nn and ϕ∈R∩L∞. If Hξpξ¯sϕ2=Hξpξ¯sϕ, then ϕ=0.Now it is left to prove the last main result. We need the following lemma which is a generalization of Lemma  19 in [14] to higher dimension case.Lemma 14.
Letg,h∈L∞. Then Th=Hg if and only if g=h=0.Proof.
PutT~h≔PThP and H~g≔PHgP; then T~h and H~g are Toeplitz operator and small Hankel operator on the Bergman space La2(Bn), respectively. It is easy to verify that (54)T~zj¯H~h=H~hT~zj,j=1,2,…,n.Suppose Th=Hg; then T~g=H~h. Applying the above equations we get (55)T~zj¯g=T~zj¯T~g=T~zj¯H~h=H~hT~zj=T~gT~zj=T~zjg,to obtain T~zj¯g-zjg=0, which gives (zj¯-zj)g=0, and so g=0. It follows from Hh=Tg=0 that h=0. The proof is complete.Proof of Theorem4.
SupposingHϕHh=Hg, then by Lemma 9 we get [Hϕ,Hh]=0. So Theorem 2 tells that one of the following cases may occur.
Case (a). (ϕ=0.) In this case, we get ϕ=g=0.
Case (b). (h is a separately radial function.) In this case, by Theorem 3, we have ϕ=g=0 or h=g=0.
Case (c). (ϕ is a nonzero constant and Jh=h.) Without loss of generality, we assume ϕ=1. Combining with (9) we have Hg=H1Hh=JHh=Th. Hence by Lemma 14 we get g=h=0.
The remaining of the proof is clear. We complete the proof.The following semicommutativity is obtained easily by Theorem4.Corollary 15.
Letϕ be bounded radial and g∈L∞. Then the following statements are equivalent: (i)
HϕHg=Hϕg.(ii)
HgHϕ=Hϕg.(iii)
ϕ=0 or g=0.



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