Abstract and Applied Analysis

Abstract and Applied Analysis / 2015 / Article

Research Article | Open Access

Volume 2015 |Article ID 209307 | 12 pages | https://doi.org/10.1155/2015/209307

Reducing Subspaces of Some Multiplication Operators on the Bergman Space over Polydisk

Academic Editor: Sung G. Kim
Received27 Nov 2014
Revised14 Feb 2015
Accepted09 Mar 2015
Published01 Apr 2015

Abstract

We consider the reducing subspaces of on , where , , and for . We prove that each reducing subspace of is a direct sum of some minimal reducing subspaces. We also characterize the minimal reducing subspaces in the cases that and , respectively. Finally, we give a complete description of minimal reducing subspaces of on with .

1. Introduction

Denote by the open unit disk in the complex plane and the normalized area measure on . For , denote . For a positive integer , the weighted Bergman space is the space of all holomorphic functions on which are square integrable with respect to the measure . is the Hilbert space with inner product and . In particular, if , then is the weighted Bergman space on . Denote by the set of all the nonnegative integers. For a -dimension multi-index ( means that for any ), write and . Then , where . Obviously, is an orthogonal basis of .

For every bounded analytic function on , the multiplication operator is defined by

Recall that, in a Hilbert space , a (closed) subspace is called reducing subspace of an operator if and . Moreover, is called minimal if does not contain any proper reducing subspaces other than .

Although the definition of multiplication operator seems simple, the invariant subspace lattice is very complicated. Even on the Bergman space , the characterization of invariant subspaces for the Bergman shift remains a very fascinating open problem in operator theory. To get some deeper information about , much effort has been devoted to studying the structure of the reducing subspaces of on (see [1] and its references).

Firstly, it is proved that the multiplication operator , where is the product of two Blaschke factors, has exactly two nontrivial reducing subspaces by Sun and Wang [2] and Zhu [3] independently. On the weighted sequence space , Stessin and Zhu [4] gave a complete description of the reducing subspaces of weighted unilateral shift operators. In particular, they show that has distinct minimal reducing subspaces on . For finite Blaschke product , Hu et al. [5] obtained that has at least a reducing subspace on which the restriction of is unitary equivalent to . Later on, Xu and Yan [6] generalized this result to the weighted Bergman space with . In 2009, Guo et al. [7] proved that if is a Blaschke product of degree 3, then the number of minimal reducing subspaces of is at most . For finite Blaschke product , they also raised a conjecture that the number of nontrivial minimal reducing subspaces of equals the number of connected components of the Riemann surface of over . By different techniques, some partial results are obtained in [810]. Finally, an affirmative answer to the conjecture is given by Douglas et al. [11]. Furthermore, when is an infinite Blaschke product, some relative results are obtained by Guo and Huang in [12, 13].

On , known results about the reducing subspaces of are quite few. If is a monomial, the reducing subspaces of are characterized in [1417]. If , Dan and Huang [18] described the minimal reducing subspaces of and the commutant algebra .

Let be a nonzero reducing subspace of on with . Suppose satisfies . By [16], we know that every must be in . However, on the unweighted Bergman space , it is not true. For example, is a reducing subspace of . But does not belong to .

To know more about how influences the structure of reducing subspaces, we consider the reducing subspaces of over for .

Fix integer and distinct positive integers for . Denote for . In Section 2, we prove that each reducing subspace of is a direct sum of some minimal reducing subspaces. To classify the minimal reducing subspaces, we consider three cases: (i) is irrational; (ii) ; (iii) is rational and . For cases (i) and (ii), we describe the minimal reducing subspaces of . For case (iii), we find that the minimal reducing subspaces of are varied. In Section 3, we give a complete characterization of the reducing subspaces of when the dimension .

2. Reducing Subspaces on

The aim of this section is to give a complete description of the reducing subspaces of on . Denote Define an equivalence on by Write for . For , letClearly, and , where is the partition of by the equivalence . Let be the orthogonal projection from onto .

Theorem 1. Let be a nonzero reducing subspace of on . Then, contains a minimal reducing subspacewhere and with coefficients .

Proof. Let be the orthogonal projection from onto . For abbreviation, we denote .
Firstly, we show that for every . Let . We only need to prove that if , then . If , then ; that is, for . Therefore, If , we find that for . Then,Thus, we get that if , thenSince , we have . Therefore, which implies .
Thus . We also obtain that for any .
Next, we claim that there is a nonzero function in for some positive integer .
Choose a nonzero function in . Let be the minimal integer such that , where is the orthogonal projection from onto . Namely, there exists such that . Then, we can prove that . In fact,(i)if , thenwhere the second equality comes from and the last equality comes from ;(ii)if is out of , then .
Therefore, we get , where is the reducing subspace of induced by . Notice thatHence, we conclude that is a minimal reducing subspace of .

In the following, we will prove that each nonzero reducing subspace of is the orthogonal sum of some minimal reducing subspaces.

Theorem 2. Let be a nonzero reducing subspace of on . Then, where is the reducing subspace of induced by . If , then where is the orthogonal basis of and .

Proof. Denote . Firstly, we know that , since(i);(ii);(iii), for ;(iv), for .
Secondly, we prove that . On the one hand, in the proof of Theorem 1, we get . Then, . On the other hand, if , choose a nonzero function in . Theorem 1 shows that there are and such that . However, for , which is a contradiction.
Finally, we prove that if , then . Choose an orthogonal basis () of the subspace . Theorem 1 shows that . We have that for , sinceLet . Clearly, . Assume that . Take a nonzero function . As in Theorem 1, there is an integer such that and . Since , we have , which is in contradiction with . So we finish the proof.

From this theorem, we know that the reducing subspaces of are determined by the sets . There arises the following question: what are the elements in the set exactly? We begin the research with the case that is irrational.

Lemma 3. If is irrational, then for every .

Proof. Suppose ; that is, , for all . Then, we haveThis is equivalent toWriteClearly, is a polynomial over and for any . Fundamental theorem of algebra shows that , for all . Denote Since is irrational, . Then, implies . So we get and .
Without loss of generality, we may assume . Then there exist nonnegative integers and making If , then , which is in contradiction with the assumption. So .
Equality (21) implies . Then, . Hence, we get and .
Therefore,LetWithout loss of generality, assume . As above, it is easy to get . Applying this process again, we can prove that for .

By Theorems 1 and 2 and Lemma 3, we obtain the following theorem.

Theorem 4. If is irrational, then each reducing subspace of on is a direct sum of some minimal reducing subspaces of the form where .

Proof. Lemma 3 shows that . In light of Theorem 1, we have . For , Theorem 2 implies that , . Thus, .

Next, we consider the case that . Denote by the permutation group of the set . Let for .

Lemma 5. If , then

Proof. Suppose . By the definition of , we haveLet . We have , since is a polynomial on with infinitely many roots. Therefore,For each , there is only one integer such thatthat is, . LetHence .
Conversely, for every , . By definition of , we have Therefore, equality (27) holds, implying . Therefore, .

From this result, we find .

Example 6. LetDenote by the reducing subspace of on induced by . Let(i) for ;(ii) for ;(iii) for .
Then, .

Proof. Let and let . It is easy to check that That is,Similarly,By Lemma 5, we get and . Therefore, . Notice thatwhere , , and . So

If is a nonzero rational number, the structure of minimal reducing subspace turns to be more complicated. In particular, we will study the reducing subspaces of on in the next section.

3. Reducing Subspaces on

Let be rational. We consider the reducing subspaces of on . Recall and ; that is, if and only if for . For every , if , we assume that for . Otherwise, if there exists such that , we can prove that for as in [16]. Since and are decreasing as is increasing, there exist and satisfying and .

This section is organized as follows. Firstly, we consider under the assumption that , , and . Let , , and . We give a description of in the cases that is in , and , respectively. Secondly, we get all the possible cases by symmetry (see Corollaries 11 and 13). Finally, we obtain and Theorem 14.

Lemma 7. Let and let . If satisfies , , and , then one of the following statements holds:(1);(2).

Proof. Let . By definition of , as in Lemma 3, we havefor any . Denotefor . Then,where denotes the disjoint union. Since is not an integer, for .
It is easy to see that Since for and , we haveWithout loss of generality, assumeFirstly, we prove that by contradiction. Otherwise, if , thenSince , we have andTherefore, . Since , it holds that We will find the contradictions under the assumptions (a) and (b) , respectively.
(a) If , then . So andSince , we have By (43) and (48), we get and . It follows thatThus, andBy and equality (48), we conclude thatEqualities (46) and (51) imply . Thus, , which is in contradiction with the assumption.
(b) Suppose . Notice that .
If , then equality (40) implies . Equivalently, and . Hence, , which is impossible.
If , then . It follows that . Since , equality (43) implies . Therefore, and . Then, , which is a contradiction.
Summing up, we must have .
Next, we prove that .
If , then ; that is, In this case, . Otherwise, and , which is in contradiction with .
If , then . It follows that In this case, and . Or else, and , which is in contradiction with . So we get the desired results.

Lemma 8. Fix and . If satisfies , , and , then one of the following statements holds:(1);(2).

Proof. Let for . Then, for . Let for . Then, and for . As in Lemma 7, we assume equality (43) holds. Then, we can prove that or .
Since , we have and . It means for .
If , then statement (1) or statement (2) holds.
If , then and . Equalities (43) and (53) imply thatSince , and are distinct, equality (56) shows thatThen, we have and