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Commuting Separately Quasihomogeneous Small Hankel Operators on Pluriharmonic Bergman Space
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.
Let be the unit ball in and its boundary . Let denote the normalized Lebesgue volume measure on the unit ball . is the Hilbert space of Lebesgue square integrable functions on with inner product: The Bergman space is a subspace of consisting of all holomorphic functions. It is well known that is a reproducing function space with the reproducing kernel: for . Let be the orthogonal projection from onto ; then we have
The pluriharmonic Bergman space is the closed subspace of consisting of the pluriharmonic functions on . It is well known that where ; then is also a reproducing function space with reproducing kernel:Let denote the orthogonal projection from onto ; then Using (5), we have
Let be a set of all bounded measurable functions on . Fix a function . The Toeplitz operator and the small Hankel operator with the symbol are defined by respectively, where is an unitary operator defined by . One can verify that , so we see the relation between the Toeplitz operator and small Hankel operator isOne can easily check , so, by the above relation, we have , and here .
We need more notations. Let denote the set of all nonnegative integers. For multi-index and point , we write For two multi-indexes , , the notations means that for every and means that Let denote . Note that if , then .
For a function , is said to be radial if and separately radial if .
Let , . For , if there are and separately radial function such that , where , then is called a separately quasihomogeneous function.
The (semi)commutativity of two operators is an important topic in operator theory. In , 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,  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. Denote as the set If is a bounded separately radial function, thenwhere . Let be the set consisting of all which is separately radial. Note that for every we havewhere every . Moreover, if , then also each . For details, one may see .
We will first investigate when two small Hankel operators with a certain symbols commute. For two operators , , their commutator is denoted by . So commutes with means .
Theorem 1. Let and . Then if and only if one of the following holds:(i) or .(ii).
Note that , so the above theorem implies that is always a normal operator on . We also note that it is still open when two Toeplitz operators with separately quasihomogeneous symbols commute.
Theorem 2. Let be bounded radial and . Then if and only if one of the following holds: (i).(ii) is separately radial.(iii) is nonzero constant and .
The above result is also different from the case for two Toeplitz operators with same symbols; see Corollary 5.2 in .
In order to get the semicommutativity of small Hankel operators (see Corollaries 12 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. Let and , . Then the following statements are equivalent: (i).(ii).(iii) or .
Theorem 4. Let be bounded radial and . Then the following statements are equivalent: (i).(ii).(iii) or .
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 so one obtains that when can also answer when .
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 is uniquely determined by its value on an arithmetic sequence of integers. In fact, we have the following classical theorem (see p. 102 of ).
Lemma 5. Suppose that is a bounded analytic function on which vanishes at the pairwise distinct points , where and , and then vanishes identically on .
We will need a similar result in higher dimensions which is proved by the above lemma. First, we give the following notations.
Let , we say that satisfies condition (I) if there exists a sequence such that , and, for every fixed , there also exists a sequence such that and .
One will easily see that for a multi-index , if and let be the complement of in , then either or satisfies the condition (I). Using Lemma 5, one may prove the following (also see Corollary 2.7 in ).
Lemma 6. Let and be a bounded function on . If the set satisfies condition (I), then .
Lemma 7. Let and ; then, for any multi-index ,
Note that, for two nonzero multi-indexes with , then if and only if and there is no such that . Hence the above lemma gives the following.
Lemma 8. Let with and , ; then, for any multi-index ,
We are ready to prove the first main result.
Proof of Theorem 1. Using Lemma 7, direct calculations show thatwhen , andwhen , . Thus when combining with (21) and (22), it follows from that which implies orholds when , which induces or . In fact, for the sake of simplicity, we only consider the case . Put if satisfies the condition (I); then, by Lemma 6, we can get ; otherwise will satisfy the condition (I), which by (25) says that the set satisfies the condition (I); thus, by Lemma 6, again we get .
Conversely, if or , it is clear that commutes with . Now suppose . When and , by (21) and (22), one can easily checksince . When , by Lemma 7, we have It is clear that the above two are equal because . When and , both sides of equality (28) are zero, so they are equal. Thus, we conclude on the analytic part , which is also true on the coanalytic part with similar arguments when . The proof is complete.
Before we prove Theorem 2, we introduce the Mellin transform which is one of the most useful tools in studying our problems. The Mellin transform of a function is defined by . It is known that is a bounded analytic function in the half plane . In addition, for a radial function , to compute using (13), we obtain
Let , where . Put , then . Let . For every separately quasihomogeneous function , there is unique with such that and where is a separately radial function. Thus for each denoted as the form (14), we may rewrite , where
Now we can prove the second main result.
Proof of Theorem 2. Suppose . We may write , where , , are given by (32). Using Lemmas 7 and 8, direct calculations give that, for every ,Comparing the fourth summations in the equalities (33) and (34), it follows from that, for every and , we haveSimilarly, comparing the first summation in the equality (33) with the third one in the equality (34), we then get that, for every ,With the same arguments as above, it follows from that, for every and , we haveand, for every ,Suppose is a nonzero constant; then, by (30), we have so (35) gives that, for every , when , which induces that for every by Lemma 6. Similarly, it follows from (36) we can get for every . Therefore, we obtain .
In the following, we suppose is not a constant. It follows from (35) and (37) that, for every , when ,For the sake of simplicity, we only consider the case . For , put If satisfies the condition (I), then by Lemma 6 we obtain ; if does not satisfy the condition (I), then , the complement of in , must satisfy the condition (I). In particular, we have . So by (41) we have, for , That is, when , using (30), we haveDenoteand then is a bounded analytic function on . Moreover, it follows from (44) that when . According to Lemma 5 we thus have , which impliesfor every in . If , then it follows from the above that , so it is clear that , a contradiction. Without loss of generality, we may assume . Replace by in (46) and then compare with (46) to obtain and thus, as in the proof of Theorem 6 in , the above equation gives that is a constant, which is also a contradiction.
Therefore, when is not a constant, then, for every with and , . Putting them into (35), we get that, for , when . Since is not a constant, then similar argument as before we get ; hence for . With same arguments, using (36) and (38), we can get for . So , a separately radial function.
Conversely, if , then clearly ; if is separately radial, then Theorem 1 gives ; if , a nonzero constant, and , then using (9) we have as desired. The proof is complete.
Before we prove the last two main results, we should note that, for every small Hankel operator with , it is easy to check that is a complex symmetric operator with complex conjugate which is defined as for , that is, . Hence we have the following analogous result as Proposition 18 in .
Lemma 9. Let . Then if and only if , and each case implies .
We also need the following partial result of zero product characterization which has independent interest.
Theorem 10. Let , , . Then the following statements are equivalent: (i).(ii).(iii) or .
Proof. By Lemma 9 we only need to show (ii) (iii). Write as the form (14). When , by Lemma 7, we have Using Lemma 7 again we get that the analytic part of is given by and hence will give that, for each , holds for any and . Therefore, with the same arguments as done in the proof of Theorem 1 we can get or for each , 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. Let and , . Then the following statements are equivalent: (i).(ii).(iii).
Now we turn to prove the third main result.
Proof of Theorem 3. By Lemma 9 we only need to show (i) (iii). Write as the form (14). Using Lemma 7, direct calculations give that (21) holds when ,α andand hence when , , the two equalities (21) and (53) are equal implying that the coanalytic part of must be zero, that is, for all , , .
Now for each fixed , using Lemma 6, the above tells , which yields that . Hence we get , which further gives or by Theorem 10. We complete the proof.
Theorem 3 answers immediately the following semicommutativity for two small Hankel operators with separately radial symbols.
Corollary 12. Let and . Then the following statements are equivalent: (i).(ii).(iii) or .
The next corollary says there are no nonzero idempotent small Hankel operators with separately quasihomogeneous symbols.
Corollary 13. Let and . If , then .
Now it is left to prove the last main result. We need the following lemma which is a generalization of Lemma 19 in  to higher dimension case.
Lemma 14. Let . Then if and only if .
Proof. Put and ; then and are Toeplitz operator and small Hankel operator on the Bergman space , respectively. It is easy to verify that Suppose ; then . Applying the above equations we get to obtain , which gives , and so . It follows from that . The proof is complete.
Proof of Theorem 4. Supposing , then by Lemma 9 we get . So Theorem 2 tells that one of the following cases may occur.
Case (a). (.) In this case, we get .
Case (b). ( is a separately radial function.) In this case, by Theorem 3, we have or .
Case (c). ( is a nonzero constant and .) Without loss of generality, we assume . Combining with (9) we have . Hence by Lemma 14 we get .
The remaining of the proof is clear. We complete the proof.
The following semicommutativity is obtained easily by Theorem 4.
Corollary 15. Let be bounded radial and . Then the following statements are equivalent: (i).(ii).(iii) or .
Conflicts of Interest
The authors declare that they have no conflicts of interest.
The first three authors were supported by NSFC (nos. 11771401 and 11471113) and ZJNSFC (no. LY14A010013). The last author was supported by ZJNSFC (no. LY13A010021).
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