# Products of Toeplitz Operators on the 2-Analytic Bergman Space

Authors: Bo Zhang; Yixin Yang; Yufeng Lu
Journal: Journal of Function Spaces
 (2021)

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## Abstract

Letf and g be bounded functions, and let Tf and Tg be Toeplitz operators on A22D. We show that if the product TfTg equals zero and one of f and g is a radial function satisfying a Mellin transform condition, then the other function must be zero.


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## Body

## 1. Introduction

LetD be the open unit disk in ℂ equipped with the normalized Lebesgue area measure dAz=1/πdxdy, and let L2=L2D,dA denote the Lebesgue space on D. For n∈ℤ+, let An2 denote the n-analytic Bergman space, that is, the subspaces of L2 consisting of n-differentiable functions such that ∂z¯nf=0, where(1)∂z¯=∂∂z¯=12∂∂x+i∂∂y.As we know,An2 is a Hilbert subspace with the inner product(2)f,g=∫Dfωgω¯dAω,where f, g∈An2.The planar Beurling transform is the singular integral operator given by(3)Sfz=−∫ℂfωω−z2dAω,z∈ℂ.It is well known that the Beurling transform is a unitary operator acting onL2ℂ,dA (see [1], p. 364). For D⊂ℂ, the compression of the Beurling transform to L2 is a bounded linear operator acting on L2 defined by(4)SDfz=−∫Dfωω−z2dAω,fz∈L2.Then-analytic Bergman projection Pn is defined to be the orthogonal projection of L2 onto An2. The singular integral operator SD is related to Pn, and it is known (see [2]) that(5)Pn=I−SDnSD∗n,n∈ℤ+.For a functionu∈L∞, the Toeplitz operator Tu with symbol u on An2 is defined by(6)Tuf=Pnuf,f∈An2.n-analytic functions play an important role in mathematical, and the space An2 has been intensively studied. More details about the structure of these spaces can be found in paper [3–5] and Balk’s book [6].Zero-product problem is a very important question in the operator theory. For Toeplitz operators, we have the general zero-product problem. Namely, iff and g are bounded functions such that TfTg=0, then must one of the functions be zero? Ahern and Cučković (see [7]) obtained an affirmative answer for Toeplitz operators on A12 when one of the functions is radial. Le (see [8, 9]) generalized this result to more than two functions. Cučković and Le (see [10]) gave a positive answer when both functions are harmonic. While the general zero-product problem (even on A12) is still far from being solved, it is known that Toeplitz operators with radial symbols are diagonal with respect to the standard orthonormal basis of A12. However, this is not the case on An2 when n≥2. Then, Cučković and Le (see [10]) raised the following open question:Question 1.
Letf and g be bounded functions, one of which is radial. If TfTg=0 on An2 (or more generally, TfTg has finite rank), must one of these functions be zero?In this paper, we give a partial answer to this question on the 2-analytic Bergman spaceA22. We show that if g is a radial function satisfying a Mellin transform condition, then TfTg=0 if and only if f is a zero function.


## 2. Some Preliminary Results

We adopt the following boundary conditions for the binomial coefficients:(7)n−m=0,wheren=0,±1,±2,⋯andm=1,2,⋯,nn+m=0,wheren=0,1,2,⋯andm=1,2,⋯.An orthogonal basis in the spaceAn2 is given by (see [3, 11])(8)ϕj,k=k+j−11k+j−2!∂k+j−2∂z¯k−1∂zj−1z2−1k+j−2,where k=1,2,⋯andj=1,2,⋯,n. The orthogonal basis can also be written as(9)ϕj,k=k+j−1∑i=0j−1−1ij−1ij+k−i−2j−1zk−i−1z¯j−i−1,where k=1,2,⋯andj=1,2,⋯,n. For n=2, we have the following lemma.Lemma 2.
An orthogonal basis inA22 is given by(10)ϕ1,k=kzk−1,ϕ2,k=k+1kzk−1z¯−k−1zk−2,where k=1,2,⋯.For eachz∈D, since the point evaluation at z is a bounded linear functional on An2, there exists a unique reproducing kernel function Kz,ω∈An2 such that(11)gz=∫DgωKz,ωdAω,z∈D,for everyg∈An2. On 2-analytic Bergman space A22,(12)Kz,ω=∑k=1+∞ϕ1,kzϕ1,k¯ω+∑k=1+∞ϕ2,kzϕ2,k¯ω.The Mellin transformg^ of a function g∈L10,1,rdr is defined by(13)g^z=∫01gssz−1ds.It is easy to see thatg^ is well defined and analytic on the right half-plane z:Rez≥2. Cučković and Rao (see [12]) first used the Mellin transform to study Toeplitz operators on the classical Bergman space.For notational convenience, we defineϕ1,0=ϕ2,0=0 and a0=b0=c0=d0=0. For some Toeplitz operators on 2-analytic Bergman space A22D, we obtain the following lemmas.Lemma 3.
Letg be a bounded radial function. Then, for each p=1,2,⋯, we have(14)∫Dgrωp−1Kz,wdAω=apϕ1,pz+bpϕ2,p+1z,∫Dgrωp−1ω¯Kz,wdAω=cpϕ1,p−1z+dpϕ2,pz,
whereap=2pg^2p, bp=2p+2p+1g^2p+2−pg^2p, cp=2p−1g^2p, and dp=2p+1pg^2p+2−p−1g^2p.Proof.
For eachp=1,2,⋯, since g is a bounded radial function, thus(15)∫Dgrωp−1Kz,wdAω=∫Dgrωp−1∑k=1+∞ϕ1,kzϕ1,k¯ω+∑k=1+∞ϕ2,kzϕ2,k¯ωdAω=∑k=1+∞ϕ1,kz∫Dgrωp−1kω¯k−1dAω+∑k=1+∞ϕ2,kz∫Dgrωp−1ϕ2,k¯ωdAω=2pg^2pϕ1,pz+2p+2p+1g^2p+2−pg^2pϕ2,p+1z,∫Dgrωp−1ω¯Kz,wdAω=∫Dgrωp−1ω¯∑k=1+∞ϕ1,kzϕ1,k¯ω+∑k=1+∞ϕ2,kzϕ2,k¯ωdAω=∑k=1+∞ϕ1,kz∫Dgrωp−1ω¯kω¯k−1dAω+∑k=1+∞ϕ2,kz∫Dgrωp−1ω¯ϕ2,k¯ωdAω=2p−1g^2pϕ1,p−1z+2p+1pg^2p+2−p−1g^2pϕ2,pz.It is well known that radial Toeplitz operators acting onA12 are diagonal, and radial Toeplitz operators acting on An2 can be represented as matrix sequences (see [13]). In the following, we give the exact expression of radial Toeplitz operators on A22.Lemma 4.
Letg be a bounded radial function. Then, for each p∈ℤ+,(16)Tgϕ1,p=papϕ1,pz+bpϕ2,p+1z,Tgϕ2,p=p+1pcp−p−1ap−1ϕ1,p−1z+pdp−p−1bp−1ϕ2,pz.Proof.
Sinceg is a bounded radial function, for each p∈ℤ+, using Lemma 3, we get(17)Tgϕ1,p=Pngpωp−1z=∫Dgrpωp−1Kz,wdAω=papϕ1,pz+bpϕ2,p+1z,Tgϕ2,p=Pngp+1pωp−1ω¯−p−1ωp−2z=p+1∫Dgrpωp−1ω¯−p−1ωp−2Kz,wdAω=p+1p∫Dgrωp−1ω¯Kz,wdAω−p−1∫Dgrωp−2Kz,wdAω=p+1pcpϕ1,p−1z+dpϕ2,pz−p−1ap−1ϕ1,p−1z+bp−1ϕ2,pz=p+1pcp−p−1ap−1ϕ1,p−1z+pdp−p−1bp−1ϕ2,pz.Applying Lemma4, we conclude that radial Toeplitz operators on A22 are not diagonal. The following corollary is an immediate consequence of Lemma 4.Corollary 5.
Letg be a bounded radial function. Then, for each p,q∈ℤ+,(18)Tgϕ1,p,ϕ1,q=pap,ifq=p,0,ifq≠p,Tgϕ1,p,ϕ2,q=pbp,ifq=p+1,0,ifq≠p+1,Tgϕ2,p,ϕ1,q=p+1pcp−p−1ap−1,ifq=p−1,0,ifq≠p−1,Tgϕ2,p,ϕ2,q=p+1pdp−p−1bp−1,ifq=p,0,ifq≠p.


## 3. Products of Two Toeplitz Operators

A bounded functionf is said to be quasihomogeneous of degree k∈ℤ if(19)freiθ=eikθgr,where gr is a radial function (see [14]). For any function f∈L2D,dA, it has the polar decomposition, i.e.,(20)freiθ=∑k∈ℤeikθfkr,where fkr are radial functions in L20,1,rdr (see [12]). A direct calculation gives the following lemma.Lemma 6.
Letf be a bounded function. Then, for each p,q∈ℤ+,(21)fϕ1,p,ϕ1,q=2pqf^q−pp+q,fϕ1,p,ϕ2,q=2pq+1qf^q−p−1p+q+1−q−1f^q−p−1p+q−1,fϕ2,p,ϕ1,q=2p+1qpf^q−p−1p+q+1−p−1f^q−p+1p+q−1,fϕ2,p,ϕ2,q=2p+1q+1pqf^q−pp+q+2+p+q−2pqf^q−pp+q+p−1q−1f^q−pp+q−2.Proof.
For allp,q∈ℤ+, it is easy to verify that(22)fϕ1,p,ϕ1,q=∑k∈ℤeikθfkrpzp−1,qzq−1=∑k∈ℤpqeikθfkrzp−1,zq−1=2pqf^q−pp+q.
Similarly, the rest of the lemma can be proved.When considering the product of two Toeplitz operators, we often use the Mellin convolution. Iff,g∈L10,1,rdr, then their Mellin convolution is given by(23)f∗gr=∫r1frtgtdtt,0≤t<1.The Mellin convolution theorem (see [15]) states that(24)f∗g^s=f^sg^s,and iff and g are bounded, then so is f∗g.It is well known that the Mellin transform is uniquely determined by its value on an arithmetic sequence of integers. The following results (see [15], p. 102, [16]) will be needed later.Theorem 7.
Supposef is a bounded analytic function on z:Rez>0 which vanishes at the pairwise distinct points z1, z2, ⋯, where(25)infzn>0,∑n≥1Re1zn=∞.
Then,f vanishes identically on z:Rez>0.Remark 8.
Using this theorem, we can see that ifg∈L10,1,rdr and if there exists a sequence nkk≥0⊂ℕ such that(26)g^nk=0,∑k≥01nk=∞,then, g^z=0 for all z∈z:Rez>2, by the Müntz-Szasz theorem (see [17], p. 312), g=0.Forp∈ℤ+, we obtain(27)g^p=∫01gssp−1ds.the numbers g^p can also be called the moment Mellin sequence of g. Let(28)Ap=ap−1bp−1pcp−p−1ap−1pdp−p−1bp−1.Ap is closed related to the moment Mellin sequence of g, and we have the following lemma.Lemma 9.
Letp be a fixed positive integer. Then, the following statements hold:(i)
ap=cp=0 if and only if g^2p=0(ii)
bp=0 if and only if p+1g^2p+2−pg^2p=0(iii)
dp=0 if and only if pg^2p+2−p−1g^2p=0(iv)
Ap+1=0 if and only if r2g∗r2g^2p−r4g∗g^2p=0Proof.
From Lemma3, it is easy to check that (i), (ii), and (iii) hold.
To prove (iv), in fact, for a fixedp∈ℤ+,(29)Ap+1=apbpp+1cp+1−papp+1dp+1−pbp=p+1apdp+1−p+1bpcp+1=4p+12pp+2g^2p+4g^2p−g∧2p+22.
It follows thatAp+1=0 if and only if(30)g∧2p+22−g^2p+4g^2p=0.
Usingg^2p+2=r2g^2p, g^2p+4=r4g^2p, and Mellin convolution (24), we get the above equality is equivalent to(31)r2g∗r2g^2p−r4g∗g^2p=0.Lemma 10.
Letg be a bounded radial function. The function g=0 if and only if there exists a sequence pkk≥0⊂ℤ+,(32)∑k≥01pk=∞,suchthatr2g∗r2g^2pk=r4g∗g^2pk.Proof.
If the functiong=0, then r2g∗r2g=r4g∗g=0, for each p∈ℤ+,(33)r2g∗r2g^2p=r4g∗g^2p=0.
This proves the sufficient condition.
Next, we prove the necessary condition. Supposepkk≥0⊂ℤ+,(34)∑k≥01pk=∞,r2g∗r2g^2pk=r4g∗g^2pk.
Using Remark8, we have(35)r2g∗r2g^z=r4g∗g^z,
for allz∈z:Rez>2. Therefore, for each p∈ℤ+,(36)g∧2p+22=g^2p+4g^2p.
That is,g^2pp=1∞ is a geometric sequence. There exists a constant a such that(37)g^2p+2=a·g^2p.
Then,(38)r2g−ag^2p=0.
Since2pp=1∞⊂ℤ+ is a sequence and ∑p=1∞1/2p=∞, by Remark 8, r2−ag=0, which implies g=0.For eachp,q∈ℤ+, let b11p,q=fϕ1,p,ϕ1,q, b12p,q=fϕ1,p,ϕ2,q, b21p,q=fϕ2,p,ϕ1,q, and b22p,q=fϕ2,p,ϕ2,q. Let(39)Bp,q=b11p,qb12p,qb21p+1,qb22p+1,q.The first main result of this paper is the following theorem.Theorem 11.
Letf be a bounded function and g be a bounded radial function. Then, TfTg=0 on A22 if and only if for each p,q∈ℤ+, ApBp−1,q=0.Proof.
Using the fact thatf is a bounded function, we have(40)freiθ=∑k=−∞+∞eikθfkr.
IfTfTg=0, then for each p,q∈ℤ+,(41)TfTgϕ1,p,ϕ1,q=0,TfTgϕ1,p,ϕ2,q=0.
By Lemma4,(42)apfϕ1,p,ϕ1,q+bpfϕ2,p+1,ϕ1,q=0,apfϕ1,p,ϕ2,q+bpfϕ2,p+1,ϕ2,q=0,from which we conclude that(43)ap,bpBp,q=0.
Sincep is arbitrary, it follows that(44)ap−1,bp−1Bp−1,q=0.
Analogously, for eachp,q∈ℤ+, it is easily verified that(45)TfTgϕ2,p,ϕ1,q=0,TfTgϕ2,p,ϕ2,q=0,thus, we get(46)pcp−p−1ap−1,pdp−p−1bp−1Bp−1,q=0.
The above equations are equivalent to(47)ApBp−1,q=0.
This completes the proof of the theorem.Forp=1,2,⋯, firstly if ap=cp=0, then g^2p=0, using Remark 8, we get g=0. Now, if bp=0, then(48)2p+2g^2p+2−2pg^2p=0.Lettingζ=2p, we have(49)ζg^ζ=ζ+2g^ζ+2.It is easy to see that the function ζg^ζ is a periodic function with a period 2. Using the same argument as the one at the end of Section 2 in [12], we conclude that ζg^ζ must be a constant function. Hence,(50)g^ζ=Cζ,where C is a constant and it is clear that g is also a constant. Finally, if dp=0, then(51)2pg^2p+2−2p−2g^2p=0;that is,(52)2pg·r2^2p−2p−2g·r2^2p−2=0.Similarly, we can also conclude thatr2·g is a constant. Thus, if g is a bounded radial function, it must be zero. Finally, we obtain the following lemma.Lemma 12.
Letg be a bounded radial function. Then, the following statements hold:(i)
ap=cp=0 for all p∈ℤ+ if and only if g=0(ii)
bp=0 for all p∈ℤ+ if and only if g is a constant(iii)
dp=0 for all p∈ℤ+ if and only if g=0Remark 13.
In Lemma12, the condition “for all p∈ℤ+” can also be replaced by “a sequence pkk≥0⊂ℤ+ satisfying ∑k≥01/pk=∞.”In Theorem11, if ap=0 or cp=0, or dp=0, then g=0, so it is clear that TfTg=0. If bp=0, then g is a constant; it is also easy to see that if g is not zero and TfTg=0, then f must be zero. If ∣Ap+1∣=0, Ap+1 is not invertible. On the other hand, when Ap+1≠0, then Ap+1 is an invertible matrix. For a bounded radial function g such that Ap+1≠0, if TfTg=0, is it necessary that f=0? The second main theorem of this paper answers this question by giving a sufficient and necessary condition.Theorem 14.
Letg and f be bounded functions and g be a bounded radial function satisfying(53)r2g∗r2g^2p≠r4g∗g^2p,
for eachp∈ℤ+. Then, TfTg=0 on A22 if and only if f=0.Proof.
Iff is a zero function, it is obvious that TfTg=0.
Now, we assumeTfTg=0 and we shall prove f=0. If g is a bounded radial function and for each p∈ℤ+,(54)r2g∗r2g^2p≠r4g∗g^2p,then, by the Mellin convolution theorem (24), it follows that(55)g∧2p+22≠g^2p+4g^2p.
For eachp∈ℤ+,(56)Ap+1=4p+12pp+2g^2p+4g^2p−g∧2p+22.
Applying (55), we get Ap+1≠0, that is, Ap+1 is an invertible matrix. If TfTg=0 and for each q∈ℤ+, we get(57)Ap+1Bp,q=0.
SinceAp+1 is invertible,(58)Bp,q=b11p,qb12p,qb21p+1,qb22p+1,q=0.
Thus,b11p,q=0, by Lemma 6, we have(59)f^q−pp+q=0.
That is,(60)f^kk+2p=0,
wherek=q−p. Since p and q are arbitrary elements in ℤ+, by Remark 8, we obtain fk=0 for all k in ℤ. It follows that f=0. This completes the proof of the theorem.Example 1.
Letg=rm, where m∈ℤ+. Then, for each p∈ℤ+,(61)r2g∗r2g^2p=12p+m+22,r4g∗g^2p=12p+m2p+m+4.
Obviously,r2g∗r2g^2p≠r4g∗g^2p. It is easy to see that TfTrm=0 on A22 if and only if f=0.In the following, we discuss when condition (53) is not satisfied.Case 1.
Ifg is a bounded radial function and for each p∈ℤ+,(62)r2g∗r2g^2p=r4g∗g^2p.
Then, using the Mellin convolution theorem (24), we have(63)g∧2p+22=g^2p+4g^2p.
That is,g^2pp=1∞ is a geometric sequence. Using Lemma 10, we get g must be zero. It is clear that TfTg=0.Case 2.
Ifg is a bounded radial function and for some p∈ℤ+,(64)r2g∗r2g^2p=r4g∗g^2p.(1)
If there exists a sequencepkk≥0⊂ℤ+ satisfying ∑k≥01/pk=∞ such that(65)r2g∗r2g^2pk=r4g∗g^2pk,
then, by using Lemma10, we get that g must be zero function.(2)
If there exists a finite sequencepk⊂ℤ+, or an infinite sequence pk⊂ℤ+ satisfying ∑k≥01/pk<∞, such that(66)r2g∗r2g^2pk=r4g∗g^2pk,
then, the radial functiong may not be zero function. For example, if pk=p1 is finite sequence and p1=1, there exist some nonzero bounded radial functions g such that(67)r2g∗r2g^2=r4g∗g^2.
Letg=ar2+br4, where a,b∈ℝ. Then,(68)g^4=a6+b8;g^6=a8+b10;g^2=a4+b6.
Whena=360, b=−720+1206, a direct calculation shows that condition (67) is satisfied. In this case, we can prove that A2 is not invertible. As(69)a1=2g^2;b1=232g^4−g^2;c1=4g^4;d1=432g^6−g^4,
then(70)A2=a1b12c2−a12d2−b1=2g^2232g^4−g^24g^4−2g^2234g^6−4g^4+g^2.
Sinceg=ar2+br4, it follows from (68) and (70) that(71)A2=a2+b336a+b16a+b23a12+b15.
Whena=360, b=−720+1206, a direct calculation shows that A2=0 and A2 is not invertible.Remark 15.
For a nonzero functiong whose related matrices are Ap, p∈ℤ+, if there exist matrices Bp, p∈ℤ+ such that(i)
Bp are not all zero(ii)
For eachp∈ℤ+, ApBp=0then, we can construct a nonzero function f, such that TfTg=0. The following example solves (i) and (ii) for a fixed p. However, it is still unknown if (i) and (ii) hold for all p∈ℤ+, and we will study this question in the future work.Example 2.
Supposeg=360r2+−720+1206r4. Then(72)A2=−60+406−603+602−60+206−363+482.
AsA2 is not invertible, there exist some nonzero matrix B such that A2B=0. For example,(73)B=6232+232+26.
For eachp=1,2,⋯, analogous to Lemma 3, we define(74)ap′=2pf^2p,bp′=2p+2p+1f^2p+2−pf^2p,cp′=2p−1f^2p,dp′=2p+1pf^2p+2−p−1f^2p,and a0′=b0′=c0′=d0′=0.In Theorem14, if f and g are all bounded radial function and there exists a sequence pkk≥0⊂ℤ+,(75)∑k≥01pk=∞,suchthatr2g∗r2g^2pk≠r4g∗g^2pk,the conclusion is still valid; then, we have the following corollary.Corollary 16.
Letf and g be bounded radial functions. Suppose there exists a sequence pkk≥0⊂ℤ+,(76)∑k≥01pk=∞,suchthatr2g∗r2g^2pk≠r4g∗g^2pk.
IfTfTg=0 on A22, then f=0.Proof.
Forp∈ℤ+, define(77)Bp=pap′pbp′p+2p+1cp+1′−pap′p+2p+1dp+1′−pbp′.
By the hypothesis,f is a bounded radial function, it follows from Lemma 4 and Theorem 11 that TfTg=0 if and only if for each p∈ℤ+,(78)ApBp−1=0.
Letg≠0 and there exists a sequence pkk≥1⊂ℤ+,(79)∑k≥11pk=∞,suchthatr2g∗r2g^2pk≠r4g∗g^2pk.
Then, it follows thatApk+1 is an invertible matrix. Combining this with Apk+1Bpk=0, we get Bpk=0. It follows that(80)apk′=2pkf^2pk=0.
This implies thatf^2pk=0, combing with(81)∑k≥112pk=∞,and using Remark 8, we get f=0.
Forp∈ℤ+, if f and g are bounded radial functions, it follows from Lemma 4 that(82)TfTgϕ1,p=λ11pϕ1,p+λ12pϕ2,p+1,TfTgϕ2,p=λ21pϕ1,p−1+λ22pϕ2,p,where(83)λ11p=papap′+pbp·p+2p+1cp+1′−pap′,λ12p=papbp′+pbp·p+2p+1dp+1′−pbp′,λ21p=p+1pcp−p−1ap−1p−1ap−1′+pdp−p−1bp−1p+1pcp′−p−1ap−1′,λ22p=p+1pcp−p−1ap−1p−1bp−1′+pdp−p−1bp−1p+1pdp′−p−1bp−1′.
IfTfTg has a finite rank, there exists N∈ℤ+, for all p>N, such that(84)TfTgϕ1,p=0,TfTgϕ2,p=0.
As in Corollary16, there exists a sequence pkk≥0 meet the conditions, where pk⊂ℤ+ and pk>N; using properties of Mellin transform, we can obtain that TfTg has a finite rank if and only if f=0.Remark 17.
As in Corollary16, let f and g are bounded radial functions and there exists a sequence pkk≥0⊂ℤ+,(85)∑k≥01pk=∞,suchthatr2g∗r2g^2pk≠r4g∗g^2pk.
Then,TfTg has finite rank if and only if f=0.The following question is the general zero-product problem onAn2 when n≥3.Question 18.
Letf be a bounded function and g be a bounded radial function. Suppose that TfTg=0 on An2 when n≥3, can we obtain any similar conclusions?



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