# Commuting Toeplitz and Hankel Operators on Harmonic Dirichlet Spaces

Authors: Qian Ding; Yong Chen; Yufeng Lu
Journal: Journal of Function Spaces
 (2017)

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## Abstract

On the harmonic Dirichlet space of the unit disk, the commutativity of Toeplitz and Hankel operators is studied. We obtain characterizations of commuting Toeplitz and Hankel operators and essentially commuting (semicommuting) Toeplitz and Hankel operators with general symbols.


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## Body

## 1. Introduction

LetD be the open unit disk in the complex plane C and dA denote the normalized area measure on D. The Sobolev space S is the completion of the space of all smooth functions f on D with norm (1)f=∫DfdA2+∫D∂f∂z2+∂f∂z¯2dA1/2<∞and the inner product of the Sobolev space S is (2)f,g=∫DfdA∫Dg¯dA+∂f∂z,∂g∂z2+∂f∂z¯,∂g∂z¯2,where ·,·2 denotes the inner product in the Hilbert space L2(D,dA).The Dirichlet spaceD is the closed subspace of S consisting of all holomorphic functions on D vanishing at zero and the harmonic Dirichlet space Dh is the closed subspace of S consisting of all harmonic functions on D. There is the relation that Dh=D⊕C⊕D¯, where D¯={f¯:f∈D}. It is well known that each point evaluation in Dh is a bounded linear functional on D, so, for every z∈D, there exists a unique function Rz∈Dh which has the reproducing property (3)fz=f,Rzfor every f∈Dh.SinceDh=D⊕C⊕D¯, there is a relation(4)Rz=Kz+Kz¯+1,where Kz is the reproducing kernel for Dirichlet space D and is given by (5)Kzw=-ln⁡1-z¯w=∑k=1∞z¯kwkk.LetP denote the orthogonal projection of S onto D and Q denote the orthogonal projection of S onto Dh. Since Pg(z)=〈g,Kz〉 for g∈S and z∈D, then by (4) it is easy to see that(6)Qg=Pg+Pg¯¯+g,1for any function g∈S.For a functionφ∈S, the Toeplitz operator Tφ:Dh→Dh with the symbol φ is densely defined by (7)Tφf=Qφffor f∈Dh and φf∈S. The (small) Hankel operator Γψ:Dh→Dh with the symbol ψ is densely defined by (8)Γψg=JQψgfor g∈Dh and ψg∈S, where J is an unitary operator defined by Jf(z)=f(z¯) for f∈S. It is easy to check that QJ=JQ, so Hankel operator has the relation with the Toeplitz operator as follows:(9)Γψ=JTψ=TJψJ.It follows that Γ1=J since T1=I, the identity operator.On the classical Hardy space, Brown and Halmos [1] showed the necessary and sufficient conditions for Toeplitz operator which has the commutativity properties. Also, they obtained the characterization for the product problem of the Toeplitz operators. Their works have been generalized onto the case on the (harmonic) Bergman or Dirichlet space by many authors; see [2–6] and the references therein. Many works related to the product involving Toeplitz or Hankel operators are referred to in [7–12].In recent years, Chen et al. have studied the algebraic properties of Toeplitz operators on the harmonic Dirichlet space ([13]) with general symbols. Later, Feng et al. studied the commutativity of Toeplitz operator and Hankel operator, or two Hankel operators on harmonic Dirichlet space ([14, 15]), and they focused on the operators with harmonic symbols.In the present paper, we continue to study the same characterizing problems for general symbols. In order to handle the general symbols, in the second section, we will give a characterization for when the sum of products of two Toeplitz operators equals a Hankel operator, which is the key to prove our main results (see Proposition5). In the third section, we give the commutativity of Toeplitz and Hankel operators (see Theorem 10) or two Hankel operators (see Theorem 11). We also characterize when the product of two Hankel operators equals another Hankel operator (see Theorem 12), and then, as an consequence, we get the semicommutativity of two Hankel operators (see Corollary 13). In the last section, we study the essential (semi)commutativity of Toeplitz and Hankel operators or two Hankel operators.


## 2. Preliminaries

Forφ∈S, it turns out that φ(reiθ) is absolutely continuous on r∈0,1 for almost every θ∈[0,2π] and absolutely continuous on θ∈[0,2π] for almost every r∈0,1. In particular, the radial limit φ∂D≔limr→1⁡φ(reiθ) exists for almost every θ∈[0,2π]. Moreover, we have φ∂D∈L1(∂D), the space of integrable functions on ∂D, and the Poisson extension of φ∂D belongs to S. See [16, 17] for details and related facts.A nonnegative Borel measureμ on D is called a D-Carleson measure if there exists a constant C>0 such that (10)∫Df2dμ≤Cf2for every f∈D. See [18, 19] for the details. We let M be the space of all U∈S for which u is bounded measurable on D and ∂u/∂z2dA and ∂u/∂z¯2dA are D-Carleson measures, where u is the Poisson extension of U∂D.It is known thatTφ is bounded on the harmonic Dirichlet space Dh if and only if φ∈M (see [13]), so, by (9), Γφ is bounded on Dh if and only if φ∈M.We let(11)Δ0=φ∈S:φ∂D=0.It is easy to see ϕΔ0⊂Δ0 when ϕf∈S for each f∈S, and also ϕ∈Δ0 if and only if ϕ=0 when ϕ∈Dh. Moreover, a decomposition for the Sobolev space S proved in [17, 20] gives the notion that(12)S=Δ0+C⊕D⊕D¯.We start with the following lemma showing that the boundary vanishing property of a symbol gives a simple behavior of the corresponding Toeplitz operator (see [13]).Lemma 1.
Letφ∈Δ0. Then, one has (13)Tφf=∫DφfdAfor every polynomial f∈Dh. In particular, Tφ can be extended to a bounded linear functional on Dh.Note that Lemma1 shows that, for φ∈Δ0, Tφ is at most rank one. It is also the same case for Γφ when φ∈Δ0 by relation (9). In addition, PTφP is the Toeplitz operator on D, denoted by T~φ, and thus, for φ∈M∩Dh, the compactness of T~φ implies φ=0 (see [16, 17]), so, by (9) and Lemma 1, we have(14)Γφis  compact⟺Tφis  compact⟺φ∈Δ0.We also need the following result.Lemma 2.
Letφ,ψ∈M; then the following statements are equivalent:(a)
Γψ-Tφ is compact.(b)
Γψ-Tφ is finite rank.(c)
φ,ψ∈Δ0.Proof.
Letφ=φ0+φ1 and ψ=ψ0+ψ1; here, φ0,ψ0∈Δ0 and φ1,ψ1∈Dh. Note that, by Lemma 1, Γψ0 and Tφ0 are finite rank operators, so we only need to show that Γψ1-Tφ1 is compact or finite rank if and only if φ1=ψ1=0.
IfΓψ1-Tφ1 is compact, then P(Γψ1-Tφ1)P is compact; that is, Γ~ψ1-T~φ1=K; here, Γ~ψ1=PΓψ1P is the (small) Hankel operator on the Dirichlet space D (see [17, 21]), T~φ1=PTφ1P is the Toeplitz operator on D, and K is a compact operator on D.
Claim. T~z¯Γ~ψ1=Γ~ψ1T~z.
In fact, it is easy to check(15)Pz¯mznz=zn-m,n>m,0,n≤m.Let ψ1=ψ1(0)+∑j=1∞(bjzj+b-jz¯j). Then, for n∈Z+, (16)PJψ1zn=∑j>nb-jzj-nby (15). So, for positive integers n and m, we have (17)T~z¯Γ~ψ1zn,zm=z¯PJψ1zn,zm=∑j>nb-jzj-nz¯,zm=∑j>nb-jj-nmzj-n-1z¯,zm-12=b-n-m-1m,Γ~ψ1T~zzn,zm=Jψ1zn+1,zm=∑j=1∞b-jzjz¯n+1,zm=∑j=1∞b-jjmzj-1z¯n+1,zm-12=b-n-m-1m,and hence the claim holds.
SinceΓ~ψ1=T~φ1+K, by the claim, we get (18)T~z¯T~φ1+T~z¯K=T~φ1T~z+KT~z.It is well known that T~z¯T~φ1=T~z¯φ1 and T~φ1T~z=T~zφ1 (see [6, 17, 20]); then, by the above equality, we get (19)T~z¯φ1-zφ1=KT~z-T~z¯K,which gives z¯φ1-zφ1∈Δ0, so φ1=0 because z¯-z≠0 on the boundary of D except at z=±1. Thus, we see that Γψ1 is a compact operator which gives ψ1=0 by (14).
IfΓψ1-Tφ1 is finite rank, then similar arguments give φ1=ψ1=0.
The sufficiency is obvious. The proof is complete.We letP denote the set of all U∈S such that for all integers n≥0(20)∫DUw¯ndA=n+1∫Duw¯ndA,∫DUwndA=n+1∫DuwndA,where u is the Poisson extension of U∂D. Note that, for harmonic function U∈S, we can check that U∈P if and only if U is constant. Also, for U∈M, by (9) and Lemma 1 we see that TU=0 if and only if ΓU=0 and if and only if U∈Δ0∩P. Now, by Lemmas 1 and 2, we can get easily the following result which has independent interest.Corollary 3.
Letφ,ψ∈M. Then, Γψ=Tφ if and only if φ,ψ∈Δ0 and ψ-φ∈P.LetU,V∈M and u,v are the Poisson extensions of U∂D and V∂D, respectively. Then, it is easy to see U-u,V-v∈Δ0. Fix a polynomial f∈Dh. By Lemma 1, we have (21)TVf=Tvf+∫DV-vfdA,QU=TU1=Tu1+∫DU-udA=u+∫DU-udA.It follows from Lemma 1 again that(22)TUTVf=TUTvf+QU∫DV-vfdA=TuTvf+∫DU-uTvfdA+QU∫DV-vfdA=TuTvf+∫DU-uTvfdA+u∫DV-vfdA+∫DU-udA∫DV-vfdA.So,(23)TUTVf-TUTVf0=TuTvf-TuTvf0+u-u0∫DV-vfdA.The above will be used to characterize when the following product of Toeplitz operators equals a Hankel operator:(24)∑l=1NTUlTVl=ΓHfor Ul,Vl,H∈M(1≤l≤N). Here, N is a fixed positive integer. To this end, we also need the following lemma which is easy to verify by (6) (see Lemmas 3.1 and 4.2 in [13] for the details).Lemma 4.
Letu,v∈M be harmonic and write(25)uz=∑k=0∞akzk+∑k=1∞a-kz¯k,vz=∑k=0∞bkzk+∑k=1∞b-kz¯kfor the power series expansions of u,v, respectively. Then, one has (26)TuTvznz-TuTvzn0=Tuvznz-Tuvzn0-nb-nu-u0n+1,TuTvz¯nz-TuTvz¯n0=Tuvz¯nz-Tuvz¯n0-nbnu-u0n+1for every integer n≥0 and z∈D.We now give the following necessary conditions for the sum of products of two Toeplitz operators equal to a Hankel operator which is the key to characterize the related problems.Proposition 5.
LetUl,Vl,H∈M and ul,vl,h be the Poisson extensions of Ul∂D,Vl∂D,H∂D, respectively, 1≤l≤N. Suppose ∑l=1NTUlTVl=ΓH. Then, ∑l=1Nulvl∈Δ0 and h=0. Moreover, if there is l0 such that Vl0∉P, then(27)ul0=∑l≠l0αlul+βfor some constants αl(1≤l≤N,l≠l0) and some constant β.Proof.
Consider power series expansions oful,vl as (28)ulz=∑k=0∞aklzk+∑k=1∞a-klz¯k,vlz=∑k=0∞bklzk+∑k=1∞b-klz¯kfor 1≤l≤N. By (23) and Lemma 4, for every integer n≥0, we have(29)TUlTVlzn-TUlTVlzn0=TulTvlzn-TulTvlzn0+ul-ul0∫DVl-vlzndA=Tulvlzn-Tulvlzn0-nb-nlul-ul0n+1+ul-ul0∫DVl-vlzndA=Tulvlzn-Tulvlzn0+ul-ul0∫DVlzndA-b-nl;here in the last equality we have used the identity (30)∫DvlzndA=b-nln+1for each n and l. On the other hand, for each nonnegative integer n, (31)ΓHzn-ΓHzn0=Γhzn-Γhzn0since H-h∈Δ0. So, from ∑l=1NTUlTVl=ΓH and (29), we get(32)T∑l=1Nulvl-Γhzn-T∑l=1Nulvl-Γhzn0=-∑l=1Nul-ul0∫DVlzndA-b-nlfor integer n≥0.
Similarly, we considerTUlTVlz¯n-TUlTVlz¯n(0) as done in (29) to get(33)TUlTVlz¯n-TUlTVlz¯n0=TulTvlz¯n-TulTvlz¯n0+ul-ul0∫DVl-vlz¯ndA=Tulvlz¯n-Tulvlz¯n0+ul-ul0∫DVlz¯ndA-bnl,so similarly we can get the identity(34)T∑l=1Nulvl-Γhz¯n-T∑l=1Nulvl-Γhz¯n0=-∑l=1Nul-ul0∫DVlz¯ndA-bnlfor integer n≥0.
It follows from (32) and (34) that T∑l=1Nulvl-Γh is a finite rank operator, and thus Lemma 2 gives h=0 and ∑l=1Nulvl∈Δ0, and by Lemma 1 the latter one is (35)T∑l=1Nulvlφ-T∑l=1Nulvlφ0=0for each φ∈Dh. It follows that the left sides of identities (32) and (34) are both zero for each integer n≥0, and so are the right sides of these two identities; that is,(36)∑l=1Nul-ul0∫DVlzndA-b-nl=0,(37)∑l=1Nul-ul0∫DVlz¯ndA-bnl=0for each integer n≥0. If Vl0∉P, then there is integer n0≥0 such that (38)∫DVl0zn0dA-b-n0l0≠0,or∫DVl0z¯n0dA-bn0l0≠0,and hence (36) or (37) gives (27), as desired. The proof is complete.


## 3. Commutativity of Toeplitz and Hankel Operators

As one application of Proposition5, we have the following result.Proposition 6.
Letu1,u2,v1,v2∈M∩Dh. Then, Tu1Tv1=Tu2Tv2 if and only if one of the following statements holds:(a)
u1,u2 or v1,v2 are constants and u1v1=u2v2.(b)
u1 or u2 and v1 or v2 are not constant and there are some constants c1,c2,c3,c4 satisfying c1+c2≠0 such that(39)c1u1+c2u2=c3,c1v2+c2v1=c4,u1v1=u2v2.Proof.
The sufficiency is easy to check and in what follows we prove the necessity:(a)
Ifu1 and u2 both are constants, then u1v1-u2v2 is a harmonic function, so Tu1Tv1=Tu2Tv2 becomes Tu1v1=Tu2v2 which gives u1v1=u2v2. In a similar argument, v1 and v2 are both constants.(b)
Suppose thatv1 is not constant and one of u1 and u2 is not constant. So, v1∉P. It follows from Proposition 5 that(40)u1=αu2+βfor some constants α and β. So,(41)0=Tu1Tv1-Tu2Tv2=Tu2Tαv1-v2+T1Tβv1.Now, if αv1-v2 is not constant, which means αv1-v2∉P, then, by Proposition 5 again, we get that u2 is constant. By (40), we see that u1 is also constant, which is a contradiction. So, αv1-v2 is constant, which combined with (40) and (41) gives (39).
Suppose thatv2 is not constant and one of u1 and u2 is not constant; then, similar arguments will give (39). The proof is complete.By (9) and the above result, we can easily get the following two corollaries which have been proved using different methods in [14] and [15], respectively.Corollary 7.
Letu,v∈M∩Dh. Then, the following statements are equivalent:(a)
TuΓv=ΓvTu.(b)
TuTJv=TJvTJu.(c)
u is constant, or u is not constant, and there are constants α,β such that v=αJu+β and (αu+β)(u-Ju)=0.Corollary 8.
Letu,v∈M∩Dh. Then, the following statements are equivalent:(a)
ΓuΓv=ΓvΓu.(b)
TuTJv=TvTJu.(c)
v is constant and v(u-Ju)=0, or v is not a constant, and there are constants α,β such that u=αv+β and β(v-Jv)=0.As another application of Proposition5, we have the following.Corollary 9.
Letu,v,h∈M∩Dh; then, ΓuΓv=Γh if and only if TJuTv=Γh if and only if u=h=0 or v=h=0.Now, we generalize the above three corollaries to the cases for the general symbols. First, we characterize the commutativity of Toeplitz and Hankel operators.Theorem 10.
LetU,V∈M. Then, TUΓV=ΓVTU if and only if TUTJV=TJVTJU if and only if one of the following statements holds:(a)
IfU∉P, then there are constants α,β such that v=αJu+β, (αu+β)(u-Ju)=0 and, for each φ∈Dh, (42)∫DUQJVφ-JVQJUφdA+u-u0∫DJV-αJU-βφdA=0.(b)
IfV∉P, then there are constants α,β such that u=αJv+β, αJv(v-Jv)=0 and, for each φ∈Dh, (43)∫DUQJVφ-JVQJUφdA+Jv-Jv0∫DαJV-JU+βφdA=0.(c)
IfU,V∈P, then Jv(u-Ju)=0 and, for each φ∈Dh, (44)∫DUQJVφ-JVQJUφdA=0.Proof.
First, we prove the necessity. Note that by Proposition5 we have uJv-JvJu∈Δ0, which means Jv(u-Ju)=0.
(a) IfU∉P, then by Proposition 5 there are constants α,β such that v=αJu+β; this combines with Jv(u-Ju)=0 to get (αu+β)(u-Ju)=0. So, by Corollary 7, we have TuTJv=TJvTJu. Now, by (23), we have (45)0=TUTJV-TJVTJUφ=TUTJV-TJVTJUφ0+u-u0∫DJV-vφdA-Jv-Jv0∫DJU-uφdAfor each φ∈Dh, which combined with v=αJu+β will give (a) because(46)TUTJV-TJVTJUφ0=∫DUQJVφ-JVQJUφdAand α(u-Ju)=0.
(b) IfV∉P, then by Proposition 5 there are constants α,β such that u=αJv+β; this combines with Jv(u-Ju)=0 to get αJv(v-Jv)=0. So, by Corollary 7, we have TuTJv=TJvTJu. The left proof is similar to (a).
(c) Notice that, by Lemma1,  uJv-JvJu∈Δ0 means (47)TuJv-JvJuφ=TuJv-JvJuφ0for each φ∈Dh, so combining with (29), (33), and (46), we can get (c) easily.
The sufficiency is obvious by the above arguments. We complete the proof.With similar and easier arguments, we can get the characterization for commuting of two Hankel operators.Theorem 11.
LetU,V∈M. Then, ΓUΓV=ΓVΓU if and only if TUTJV=TVTJU if and only if one of the following statements holds:(a)
IfU∉P, then there are constants α,β such that v=αu+β, β(u-Ju)=0 and, for each φ∈Dh, (48)∫DUQJVφ-VQJUφdA+u-u0∫DJV-αJU-βφdA=0.(b)
IfV∉P, then there are constants α,β such that u=αv+β, β(v-Jv)=0 and, for each φ∈Dh, (49)∫DUQJVφ-VQJUφdA+v-v0∫DαJV-JU+βφdA=0.(c)
IfU,V∈P, then uJv-vJu∈Δ0 and, for each φ∈Dh, (50)∫DUQJVφ-VQJUφdA=0.Now, we consider when the product of two Hankel operators equals another Hankel operator.Theorem 12.
LetU,V,H∈M. Then, ΓUΓV=ΓH if and only if TJUTV=ΓH if and only if one of the following statements holds:(a)
U,H∈Δ0 and, for each φ∈Dh, (51)∫DJUQVφ-JHφdA=0.(b)
V,H∈Δ0 and, for each φ∈Dh, (52)JU∫DVφdA-∫DJHφdA=0.Proof.
First, assumeTJUTV=ΓH. Then, by Proposition 5, we have Juv∈Δ0 and h=0, and the former one means that Juv=0, so u=0 or v=0.
Ifu=h=0, then U,H∈Δ0. In this case, by Lemma 1, TJUTVφ=ΓHφ for each φ∈Dh which gives (a).
Ifv=h=0, then V,H∈Δ0. In this case, by Lemma 1, TJUTVφ=ΓHφ for each φ∈Dh which gives (b).
The converse is obvious. We complete the proof.SinceUV∈Δ0 with U or V in Δ0, then the following is an easy consequence of the above result which gives the semicommutativity of two Hankel operators.Corollary 13.
LetU,V∈M. Then, ΓUΓV=ΓUV if and only if TJUTV=ΓUV if and only if one of the following statements holds:(a)
U∈Δ0 and, for each φ∈Dh, (53)∫DJUQVφ-JUJVφdA=0.(b)
V∈Δ0 and, for each φ∈Dh, (54)JU∫DVφdA-∫DJUJVφdA=0.


## 4. Essentially Commuting Toeplitz and Hankel Operators

Recall that Lemma1 shows that, for φ∈Δ0, Tφ is at most rank one. It is also the same case for Γφ when φ∈Δ0 by (9). Moreover, if U,V∈M and u,v are the Poisson extensions of U∂D and V∂D, respectively, then it is easy to see that UV=uv+h with h∈Δ0. So, TUV=Tuv+F with F being a finite rank operator, so by (29) and (33) we have the following result which is proved in [13].Lemma 14.
Letφ,ψ∈M and φψ∈M; then, TφTψ-TψTφ and TφTψ-Tφψ are both finite rank operators.Now, we can obtain the conclusions about the compact or finite rank product of Toeplitz and Hankel operators.Theorem 15.
Letφ,ψ∈M; then, the following statements are equivalent:(1)
TφΓψ-ΓψTφ is compact.(2)
TφΓψ-ΓψTφ is finite rank.(3)
TφΓψ-Γφψ is compact.(4)
TφΓψ-Γφψ is finite rank.(5)
ψ∈Δ0 or φ-Jφ∈Δ0.Proof.
First, note that, by (9), we have (55)ΓψTφ-Γφψ=JTψTφ-Tφψ,so ΓψTφ-Γφψ is always finite rank by Lemma 14. It follows from (56)TφΓψ-ΓψTφ=TφΓψ-Γφψ-ΓψTφ-Γφψthat (1)⇔(3) and (2)⇔(4).
In addition,(57)TφΓψ-Γφψ=TφTJψJ-TJφψJ=TφTJψ-TJφJψJ=TφTJψ-TφJψJ+TφJψ-JφJψJ.Again, by Lemma 14, we see that TφΓψ-Γφψ is compact or finite rank if and only if TφJψ-JφJψ is compact or finite rank; the latter one is equivalent to φJψ-JφJψ∈Δ0 by (14). Since φJψ-JφJψ∈Δ0 means φ-Jφ∈Δ0 or Jψ∈Δ0, we get (5). The proof is complete.Theorem 16.
Letφ,ψ,h∈M; then, the following statements are equivalent:(1)
ΓφΓψ-Γh is compact.(2)
ΓφΓψ-Γh is finite rank.(3)
TφΓψ-Th is compact.(4)
TφΓψ-Th is finite rank.(5)
φ,h∈Δ0 or ψ,h∈Δ0.Proof.
By (9), we have (58)TφΓψ-Th=JΓφΓψ-Γh,so (1)⇔(3) and (2)⇔(4). In addition, (59)ΓφΓψ-Γh=ΓφJTψ-Γh=TJφTψ-Γh=TJφTψ-TJφψ+TJφψ-Γh.So, by Lemma 14, ΓφΓψ-Γh is compact or finite rank if and only if TJφψ-Γh is compact or finite rank, and the latter is equivalent to Jφψ,h∈Δ0 by Lemma 2. Since Jφψ∈Δ0 means Jφ∈Δ0 or ψ∈Δ0, we get (5). The proof is complete.The following is the easy conclusion of the above result.Corollary 17.
Letφ,ψ∈M; then, the following statements are equivalent:(1)
ΓφΓψ-Γφψ is compact.(2)
ΓφΓψ-Γφψ is finite rank.(3)
TφΓψ-Tφψ is compact.(4)
TφΓψ-Tφψ is finite rank.(5)
φ∈Δ0 or ψ∈Δ0.



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