Commuting Toeplitz and Hankel Operators on Harmonic Dirichlet Spaces

where ⟨⋅, ⋅⟩2 denotes the inner product in the Hilbert space L2(D, dA). The Dirichlet space D 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 thatDh = D ⊕ C ⊕D, whereD = {f : f ∈ D}. It is well known that each point evaluation inDh is a bounded linear functional on D, so, for every z ∈ D, there exists a unique function Rz ∈ Dh which has the reproducing property f (z) = ⟨f, Rz⟩ (3) for every f ∈ Dh. SinceDh = D ⊕ C ⊕D, there is a relation Rz = Kz + Kz + 1, (4) whereKz is the reproducing kernel for Dirichlet spaceD and is given by


Introduction
Let D be the open unit disk in the complex plane C and  denote the normalized area measure on D. The Sobolev space S is the completion of the space of all smooth functions  on D with norm and the inner product of the Sobolev space S is where ⟨⋅, ⋅⟩ 2 denotes the inner product in the Hilbert space  2 (D, ).
The Dirichlet space D is the closed subspace of S consisting of all holomorphic functions on D vanishing at zero and the harmonic Dirichlet space D ℎ is the closed subspace of S consisting of all harmonic functions on D. There is the relation that D ℎ = D ⊕ C ⊕ D, where D = { :  ∈ D}.It is well known that each point evaluation in D ℎ is a bounded linear functional on D, so, for every  ∈ D, there exists a unique function   ∈ D ℎ which has the reproducing property for every  ∈ D ℎ .Since D ℎ = D ⊕ C ⊕ D, there is a relation where   is the reproducing kernel for Dirichlet space D and is given by Let  denote the orthogonal projection of S onto D and  denote the orthogonal projection of S onto D ℎ .Since () = ⟨,   ⟩ for  ∈ S and  ∈ D, then by (4) it is easy to see that  () =  () +  () + ⟨, 1⟩ for any function  ∈ S. For a function  ∈ S, the Toeplitz operator   : D ℎ → D ℎ with the symbol  is densely defined by for  ∈ D ℎ and  ∈ S. The (small) Hankel operator Γ  : D ℎ → D ℎ with the symbol  is densely defined by for  ∈ D ℎ and  ∈ S, where  is an unitary operator defined by () = () for  ∈ S. It is easy to check that  = , so Hankel operator has the relation with the Toeplitz operator as follows: It follows that Γ 1 =  since  1 = , 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][3][4][5][6] and the references therein.Many works related to the product involving Toeplitz or Hankel operators are referred to in [7][8][9][10][11][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 Proposition 5).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.
We let It is easy to see Δ 0 ⊂ Δ 0 when  ∈ S for each  ∈ S, and also  ∈ Δ 0 if and only if  = 0 when  ∈ D ℎ .Moreover, a decomposition for the Sobolev space S proved in [17,20] gives the notion that 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.Let  ∈ Δ 0 .Then, one has for every polynomial  ∈ D ℎ .In particular,   can be extended to a bounded linear functional on D ℎ .
Note that Lemma 1 shows that, for  ∈ Δ 0 ,   is at most rank one.It is also the same case for Γ  when  ∈ Δ 0 by relation (9).In addition,    is the Toeplitz operator on D, denoted by T , and thus, for  ∈ M ∩ D ℎ , the compactness of T implies  = 0 (see [16,17]), so, by (9) and Lemma 1, we have We also need the following result.
Lemma 2. Let ,  ∈ M; then the following statements are equivalent: Proof.Let  =  0 +  1 and  =  0 +  1 ; here,  0 ,  0 ∈ Δ 0 and  1 ,  1 ∈ D ℎ .Note that, by Lemma 1, Γ  0 and   0 are finite rank operators, so we only need to show that is the (small) Hankel operator on the Dirichlet space D (see [17,21]), T 1 =   1  is the Toeplitz operator on D, and  is a compact operator on D.

Claim. T𝑧 Γ𝜓
In fact, it is easy to check Let by (15).So, for positive integers  and , we have and hence the claim holds.
Since Γ 1 = T 1 + , by the claim, we get It is well known that T T 1 = T 1 and T 1 T = T 1 (see [6,17,20]); then, by the above equality, we get which gives  1 −  1 ∈ Δ 0 , so  1 = 0 because  −  ̸ = 0 on the boundary of D except at  = ±1.Thus, we see that Γ  1 is a compact operator which gives  1 = 0 by (14). If The sufficiency is obvious.The proof is complete.
We let P denote the set of all  ∈ S such that for all integers  ≥ 0 where  is the Poisson extension of | D .Note that, for harmonic function  ∈ S, we can check that  ∈ P if and only if  is constant.Also, for  ∈ M, by (9) and Lemma 1 we see that   = 0 if and only if Γ  = 0 and if and only if  ∈ Δ 0 ∩ P. Now, by Lemmas 1 and 2, we can get easily the following result which has independent interest.Corollary 3. Let ,  ∈ M.Then, Γ  =   if and only if ,  ∈ Δ 0 and  −  ∈ P.
Let ,  ∈ M and , V are the Poisson extensions of | D and | D , respectively.Then, it is easy to see −, −V ∈ Δ 0 .Fix a polynomial  ∈ D ℎ .By Lemma 1, we have It follows from Lemma 1 again that The above will be used to characterize when the following product of Toeplitz operators equals a Hankel operator: for   ,   ,  ∈ M (1 ≤  ≤ ).Here,  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. Let , V ∈ M be harmonic and write for the power series expansions of , V, respectively.Then, one has for every integer  ≥ 0 and  ∈ 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. Let   ,   ,  ∈ M and   , V  , ℎ be the Poisson extensions of   | D ,   | D , | D , respectively, 1 ≤  ≤ .Suppose ∑  =1       = Γ  .Then, ∑  =1   V  ∈ Δ 0 and ℎ = 0.Moreover, if there is  0 such that   0 ∉ P, then for some constants   (1 ≤  ≤ ,  ̸ =  0 ) and some constant .
Proof.Consider power series expansions of   , V  as for 1 ≤  ≤ .By (23) and Lemma 4, for every integer  ≥ 0, we have here in the last equality we have used the identity for each  and .On the other hand, for each nonnegative integer , since  − ℎ ∈ Δ 0 .So, from ∑  =1       = Γ  and (29), we get for integer  ≥ 0.
Similarly, we consider         −         (0) as done in (29) to get so similarly we can get the identity for integer  ≥ 0. It follows from ( 32) and (34) that  ∑  =1   V  − Γ ℎ is a finite rank operator, and thus Lemma 2 gives ℎ = 0 and ∑  =1   V  ∈ Δ 0 , and by Lemma 1 the latter one is for each  ∈ D ℎ .It follows that the left sides of identities ( 32) and (34) are both zero for each integer  ≥ 0, and so are the right sides of these two identities; that is, for each integer  ≥ 0. If   0 ∉ P, then there is integer  0 ≥ 0 such that and hence (36) or (37) gives ( 27), as desired.The proof is complete.

Commutativity of Toeplitz and Hankel Operators
As one application of Proposition 5, we have the following result.
and only if one of the following statements holds: (b)  1 or  2 and V 1 or V 2 are not constant and there are some constants  1 ,  2 ,  3 ,  4 satisfying Proof.The sufficiency is easy to check and in what follows we prove the necessity: In a similar argument, V 1 and V 2 are both constants.
(b) Suppose that V 1 is not constant and one of  1 and  2 is not constant.So, V 1 ∉ P. It follows from Proposition 5 that for some constants  and .So, Now, if V 1 − V 2 is not constant, which means V 1 − V 2 ∉ P, then, by Proposition 5 again, we get that  2 is constant.
By (40), we see that  1 is also constant, which is a contradiction.So, V 1 − V 2 is constant, which combined with (40) and (41) gives (39).Suppose that V 2 is not constant and one of  1 and  2 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. Let , V ∈ M ∩ D ℎ .Then, the following statements are equivalent: (c)  is constant, or  is not constant, and there are constants ,  such that V =  +  and ( + )( − ) = 0.
Corollary 8. Let , V ∈ M ∩ D ℎ .Then, the following statements are equivalent: (c) V is constant and V( − ) = 0, or V is not a constant, and there are constants ,  such that  = V +  and (V − V) = 0.
As another application of Proposition 5, we have the following.
for each  ∈ D ℎ , 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.Proof.First, assume     = Γ  .Then, by Proposition 5, we have V ∈ Δ 0 and ℎ = 0, and the former one means that V = 0, so  = 0 or V = 0.
The converse is obvious.We complete the proof.
Since  ∈ Δ 0 with  or  in Δ 0 , then the following is an easy consequence of the above result which gives the semicommutativity of two Hankel operators.

Corollary 13 .
Let ,  ∈ M.Then, Γ  Γ  = Γ  if and only if     = Γ  if and only if one of the following statements holds: (a)  ∈ Δ 0 and, for each  ∈ D ℎ , ∫ D [ () − ]  = 0. (53) (b)  ∈ Δ 0 and, for each  ∈ D ℎ ,  ∫ D   − ∫ D   = 0. (54) [18,19]19]for the details.We let M be the space of all  ∈ S for which  is bounded measurable on D and |/| 2  and |/| 2  are D-Carleson measures, where  is the Poisson extension of | D .It is known that   is bounded on the harmonic Dirichlet space D ℎ if and only if  ∈ M (see