#### 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.

#### 1. Introduction

Let be the open unit disk in the complex plane and denote the normalized area measure on . The Sobolev space is the completion of the space of all smooth functions on with norm and the inner product of the Sobolev space is where denotes the inner product in the Hilbert space .

The Dirichlet space is the closed subspace of consisting of all holomorphic functions on vanishing at zero and the harmonic Dirichlet space is the closed subspace of consisting of all harmonic functions on . There is the relation that , where . It is well known that each point evaluation in is a bounded linear functional on , so, for every , there exists a unique function which has the reproducing property for every .

Since , there is a relationwhere is the reproducing kernel for Dirichlet space and is given by

Let denote the orthogonal projection of onto and denote the orthogonal projection of onto . Since for and , then by (4) it is easy to see thatfor any function .

For a function , the Toeplitz operator with the symbol is densely defined by for and . The (small) Hankel operator with the symbol is densely defined by for and , where is an unitary operator defined by for . It is easy to check that , so Hankel operator has the relation with the Toeplitz operator as follows:It follows that since , 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 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.

#### 2. Preliminaries

For , it turns out that is absolutely continuous on for almost every and absolutely continuous on for almost every . In particular, the radial limit exists for almost every . Moreover, we have , the space of integrable functions on , and the Poisson extension of belongs to . See [16, 17] for details and related facts.

A nonnegative Borel measure on is called a -Carleson measure if there exists a constant such that for every . See [18, 19] for the details. We let be the space of all for which is bounded measurable on and and are -Carleson measures, where is the Poisson extension of .

It is known that is bounded on the harmonic Dirichlet space if and only if (see [13]), so, by (9), is bounded on if and only if .

We let It is easy to see when for each , and also if and only if when . Moreover, a decomposition for the Sobolev space 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 . Then, one has for every polynomial . In particular, can be extended to a bounded linear functional on *

Note that Lemma 1 shows that, for , is at most rank one. It is also the same case for when by relation (9). In addition, is the Toeplitz operator on , denoted by , and thus, for , the compactness of implies (see [16, 17]), so, by (9) and Lemma 1, we have

We also need the following result.

Lemma 2. *Let ; then the following statements are equivalent:*(a)* is compact.*(b)* is finite rank.*(c)*.*

*Proof. *Let and ; here, and . Note that, by Lemma 1, and are finite rank operators, so we only need to show that is compact or finite rank if and only if .

If is compact, then is compact; that is, ; here, is the (small) Hankel operator on the Dirichlet space (see [17, 21]), is the Toeplitz operator on , and is a compact operator on .*Claim*. .

In fact, it is easy to checkLet . Then, for , by (15). So, for positive integers and , we have and hence the claim holds.

Since , by the claim, we get It is well known that and (see [6, 17, 20]); then, by the above equality, we get which gives , so because on the boundary of except at . Thus, we see that is a compact operator which gives by (14).

If is finite rank, then similar arguments give .

The sufficiency is obvious. The proof is complete.

We let denote the set of all such that for all integers where is the Poisson extension of . Note that, for harmonic function , we can check that if and only if is constant. Also, for , by (9) and Lemma 1 we see that if and only if and if and only if . Now, by Lemmas 1 and 2, we can get easily the following result which has independent interest.

Corollary 3. *Let . Then, if and only if and .*

Let and are the Poisson extensions of and , respectively. Then, it is easy to see . Fix a polynomial . By Lemma 1, we have It follows from Lemma 1 again thatSo,

The above will be used to characterize when the following product of Toeplitz operators equals a Hankel operator: for . 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 in [13] for the details).

Lemma 4. *Let be harmonic and writefor the power series expansions of , respectively. Then, one has for every integer and .*

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 and be the Poisson extensions of , respectively, . Suppose . Then, and . Moreover, if there is such that , thenfor some constants and some constant .*

*Proof. *Consider power series expansions of as for . By (23) and Lemma 4, for every integer , we havehere in the last equality we have used the identity for each and . On the other hand, for each nonnegative integer , since . So, from and (29), we getfor integer .

Similarly, we consider as done in (29) to getso similarly we can get the identityfor integer .

It follows from (32) and (34) that is a finite rank operator, and thus Lemma 2 gives and , and by Lemma 1 the latter one is for each . It follows that the left sides of identities (32) and (34) are both zero for each integer , and so are the right sides of these two identities; that is,for each integer . If , then there is integer such that and hence (36) or (37) gives (27), as desired. The proof is complete.

#### 3. Commutativity of Toeplitz and Hankel Operators

As one application of Proposition 5, we have the following result.

Proposition 6. *Let . Then, if and only if one of the following statements holds:*(a)* or are constants and .*(b)* or and or are not constant and there are some constants satisfying such that*

*Proof. *The sufficiency is easy to check and in what follows we prove the necessity:(a)If and both are constants, then is a harmonic function, so becomes which gives . In a similar argument, and are both constants.(b)Suppose that is not constant and one of and is not constant. So, . It follows from Proposition 5 thatfor some constants and . So,Now, if is not constant, which means , then, by Proposition 5 again, we get that is constant. By (40), we see that is also constant, which is a contradiction. So, is constant, which combined with (40) and (41) gives (39).

Suppose that is not constant and one of and 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 . Then, the following statements are equivalent:*(a)*.*(b)*.*(c)* is constant, or is not constant, and there are constants such that and .*

Corollary 8. *Let . Then, the following statements are equivalent:*(a)*.*(b)*.*(c)* is constant and , or is not a constant, and there are constants such that and .*

As another application of Proposition 5, we have the following.

Corollary 9. *Let ; then, if and only if if and only if or .*

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. *Let . Then, if and only if if and only if one of the following statements holds:*(a)*If , then there are constants such that , and, for each , *(b)*If , then there are constants such that , and, for each , *(c)*If , then and, for each , *

*Proof. *First, we prove the necessity. Note that by Proposition 5 we have , which means .

(a) If , then by Proposition 5 there are constants such that ; this combines with to get . So, by Corollary 7, we have . Now, by (23), we have for each , which combined with will give (a) becauseand .

(b) If , then by Proposition 5 there are constants such that ; this combines with to get . So, by Corollary 7, we have . The left proof is similar to (a).

(c) Notice that, by Lemma 1, means for each , 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. *Let . Then, if and only if if and only if one of the following statements holds:*(a)*If , then there are constants such that , and, for each , *(b)*If , then there are constants such that , and, for each , *(c)*If , then and, for each , *

Now, we consider when the product of two Hankel operators equals another Hankel operator.

Theorem 12. *Let . Then, if and only if if and only if one of the following statements holds:*(a)* and, for each , *(b)* and, for each , *

*Proof. *First, assume . Then, by Proposition 5, we have and , and the former one means that , so or .

If , then . In this case, by Lemma 1, for each which gives (a).

If , then . In this case, by Lemma 1, for each which gives (b).

The converse is obvious. We complete the proof.

Since with or in , then the following is an easy consequence of the above result which gives the semicommutativity of two Hankel operators.

Corollary 13. *Let . Then, if and only if if and only if one of the following statements holds:*(a)* and, for each , *(b)* and, for each , *

#### 4. Essentially Commuting Toeplitz and Hankel Operators

Recall that Lemma 1 shows that, for , is at most rank one. It is also the same case for when by (9). Moreover, if and are the Poisson extensions of and , respectively, then it is easy to see that with . So, with being a finite rank operator, so by (29) and (33) we have the following result which is proved in [13].

Lemma 14. *Let and ; then, and 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. *Let ; then, the following statements are equivalent:*(1)* is compact.*(2)* is finite rank.*(3)* is compact.*(4)* is finite rank.*(5)* or .*

*Proof. *First, note that, by (9), we have so is always finite rank by Lemma 14. It follows from that and .

In addition, Again, by Lemma 14, we see that is compact or finite rank if and only if is compact or finite rank; the latter one is equivalent to by (14). Since means or , we get (5). The proof is complete.

Theorem 16. *Let ; then, the following statements are equivalent:*(1)* is compact.*(2)* is finite rank.*(3)* is compact.*(4)* is finite rank.*(5)* or .*

*Proof. *By (9), we have so and . In addition, So, by Lemma 14, is compact or finite rank if and only if is compact or finite rank, and the latter is equivalent to by Lemma 2. Since means or , we get (5). The proof is complete.

The following is the easy conclusion of the above result.

Corollary 17. *Let ; then, the following statements are equivalent:*(1)* is compact.*(2)* is finite rank.*(3)* is compact.*(4)* is finite rank.*(5)* or .*

#### Conflicts of Interest

The authors declare that they have no conflicts of interest.

#### Acknowledgments

The first and the third authors were supported by NSFC (11671065). The second author was supported by NSFC (11471113) and ZJNSFC (LY14A010013, LY13A010021). This work was started during the first and the second authors’ visit to the Department of Mathematics and Statistics, State University of New York at Albany, in the year 2014. They wish to express their great gratitude to the institution for hospitality and Professors Kehe Zhu and Rongwei Yang for helpful discussions.