Journal of Function Spaces

Volume 2018 (2018), Article ID 6078594, 6 pages

https://doi.org/10.1155/2018/6078594

## On Some Properties of Cowen-Douglas Class of Operators

Department of Mathematics, Payame Noor University, P.O. Box 19395-3697, Tehran, Iran

Correspondence should be addressed to Bahmann Yousefi

Received 9 September 2017; Accepted 29 November 2017; Published 1 February 2018

Academic Editor: Kehe Zhu

Copyright © 2018 Parastoo Heiatian Naeini and Bahmann Yousefi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We will consider multiplication operators on a Hilbert space of analytic functions on a domain . For a bounded analytic function on , we will give necessary and sufficient conditions under which the complement of the essential spectrum of in becomes nonempty and this gives conditions for the adjoint of the multiplication operator belongs to the Cowen-Douglas class of operators. Also, we characterize the structure of the essential spectrum of a multiplication operator and we determine the commutants of certain multiplication operators. Finally, we investigate the reflexivity of a Cowen-Douglas class operator.

#### 1. Introduction

In this section we include some preparatory material which will be needed later.

For a positive integer and a domain , the Cowen-Douglas class consists of bounded linear operators on any fixed separable infinite dimensional Hilbert space with the following properties:(a) is a subset of the spectrum of .(b) for every .(c) for every .(d).

Here Span denotes the closed linear span of a collection of sets in . The classes were introduced by Cowen-Douglas (see [1]), and each element of is called a Cowen-Douglas class operator. By , we mean for some complex domain . For the study of Cowen-Douglas classes , we mention [1–7].

Recall that a bounded linear operator on a Hilbert space is a Fredholm operator if and only if is closed and both and are finite dimensional. We use and to denote, respectively, the spectrum of and the essential spectrum of .

Now let be a separable Hilbert space and let denote the algebra of all bounded linear operators on . Recall that if , then is by definition the lattice of all invariant subspaces of , and is the algebra of all operators in such that . An operator in is said to be reflexive if , where is the smallest subalgebra of that contains and the identity and is closed in the weak operator topology.

Also, if is a Hilbert space of functions analytic on a plane domain , a complex-valued function on for which for every is called a multiplier of and the multiplier on determines a multiplication operator on by , . The set of all multipliers of is denoted by . Clearly , where is the space of all bounded analytic function on . In fact (see [8]).

Let be a Hilbert space of functions analytic on a domain satisfying the following axioms:

*Axiom 1. *For every point , the functional of point evaluation at , is a nonzero bounded linear functional on .

*Axiom 2. *Every function is a multiplier of .

*Axiom 3. *If and , then there is a function such that .

A space satisfying the above conditions is called* Hilbert space of analytic functions on *Ω (see [3, 9]). The Hardy and Bergman spaces are examples for Hilbert spaces of analytic functions on the open unit disk.

Note that, by Axiom 1, there exists a reproducing kernel such that for all . Also, by using Axiom 2 and the closed-graph theorem, the operator of multiplication by , , is a bounded linear operator on . So Axiom 2 says that . If is polynomially bounded on and Ω is the open unit disk, then (see [9, Theorem ]). In the rest of the paper we assume that is a Hilbert space of analytic function on a bounded plane domain .

In this paper, we want to study some properties of operators in . We see that complement of the essential spectrum of a multiplication operator is nonempty if and only if the adjoint of belongs to some . Also, we investigate the intertwining multiplication operators and reflexivity of the multiplication operator on . For some other source on these topics one can see [10–16].

#### 2. Multiplication Operators with Adjoint in and Its Spectra

Recall that if is a Cowen-Douglas class operator, then it should be . For , we would like to give some necessary and sufficient conditions so that becomes a nonempty open set. This implies a sufficient condition for the adjoint of the multiplication operator to be a Cowen-Douglas class operator.

Theorem 1. *Let be a nonconstant function in , , and weakly as . Then there exist a domain and a positive integer such that consists of points (counting multiplicity) for every .*

*Proof. *First note that if , then for some . But by Axiom 1, the functional of evaluation at is a bounded point evaluation; thus the reproducing kernel is defined and we have Thus and clearly . Now let . Then is an invertible element of . But by Axiom 2, we have ; thus is invertible. This implies that ; thus indeed . Now we prove that For this it is sufficient to show that . Let . If , then there exists a sequence such that . By passing to a subsequence if necessary, we may assume that converges to a point in and so by our assumptions weakly. On the other hand we have for all . So we get which contradicts the fact that is Fredholm. Thus we haveNow, let be a connected component of the open set . Since , thus is Fredholm for every in . Also, note that if , then on which is open. Hence and so is injective. Thus for all in . But the index function is continuous from the set of semi-Fredholm operators into with discrete topology; thus, is constant for all in . Put If , then for some and so . Thus . Since a finite subset of points in yields a linearly set independent set of functions in , thus consist of at most points for all in . So for each fixed , there exist in and in such that and for all we have where belongs to and is nonvanishing on Ω. Now by a method used in the proof of [3, Proposition ] we show that the function is also bounded below on . For this choose such that is contained in . Put and thus is a compact subset of and so it has a positive distance to . Now if is not bounded below on , then there exists a sequence in such that as . Since is nonvanishing on implies that , so there exists a positive integer such that for all . Hence for all that is contradiction to . Thus the function is indeed bounded below on . Now since is bounded below and bounded above on it is an invertible element of and so the operator is invertible on because . Thus . Note that since we getBut is injective for all ; thusClearly, is injective; thusfor . Note that, by Axiom 3 on , is one-dimensional (see [17]); thus and therefore consists of exactly points (counting multiplicity) for every and now the proof is complete.

From the proof of Theorem 1, we can conclude the following result.

Corollary 2. *Let be a nonconstant function in and weakly as . Then .*

Note that, by Axiom 3, for every the operator is bounded below on and also the space is one-dimensional (see [3]). So the Hilbert space under consideration, , satisfies the conditions assumed by Zhu in [7].

The following result was stated by Zhu in [7, Proposition ], but its proof is left to readers. For this reason we sketch a proof of this proposition and although our proof might seem more straightforward than the one stated by Zhu, we emphasise that our main idea is given from [7].

Proposition 3. *Suppose and is a domain contained in . If there exists a positive integer such that consists of points (counting multiplicity) for every , then the adjoint of the operator belongs to the Cowen-Douglas class , where .*

*Proof. *Let . Then there exist an invertible function and such thatwhere . Axiom 3 implies that for all is onto (see [17]); thus for all , is onto since is invertible. Also, by Axiom 3, for and so Finally, we note that Now, since is open, and so the proof is complete.

Corollary 4. *Under the conditions of Theorem 1, there exist a positive integer and a domain in the complex plane such that .*

*Proof. *By Theorem 1 and Proposition 3 it is clear.

Now we investigate the converse of Theorem 1.

Corollary 5. *Let be a nonconstant function in . If there exists a domain and a positive integer such that consists of points (counting multiplicity) for every ; then .*

*Proof. *By Proposition 3, the adjoint of the operator belongs to the Cowen-Douglas class , where . Hence for all , is Fredholm and so clearly .

Corollary 6. *Let for some positive integer and a complex domain . If weakly as , then consists of point (counting multiplicity) for every where .*

*Proof. *First note that is Fredholm for all ; thus But by Corollary 2, ; thus, . Now if , then for some and clearly . Since and a finite subset of points in yields a linearly independent set of functions in , thus consist of at most n points for all . Now by the same method used in the proof of Theorem 1, we can see that consists of exactly points (counting multiplicity) for every .

*Example 7. *Consider the Hilbert Bergman space where is the open unit disc in the complex domain. Then holds in Axioms 1, 2, and 3 (see [17, Theorem , page 67]). For the Bergman reproducing kernel function, , clearly we can see that as . So if is a polynomial, then as . But polynomials are dense in ; thus weakly as . Now by Theorem 1 and the proof of Corollary 5, we can see that for some positive integer and a complex domain if and only if .

Proposition 8. *Let be a nonconstant function in and weakly as . Then*

*Proof. *Let ; then is Fredholm. Now we show that is bounded away from zero near By way of contradiction, let be a sequence such that and converges to a point in . Note that by our assumptions weakly and for all . So we get This is a contradiction because is Fredholm. Hence, is bounded away from zero near and so there exists large enough such that This implies that Conversely, if then is bounded away from zero near . Since the zeros of an analytic function are isolated, thus the zeros of are finite. Let be all zeros (counting multiplicity) of in Ω such that Clearly the function is invertible on Ω and so is bounded below. Also, by Axiom 3 on , is Fredholm for all . This implies that is Fredholm and so . So the proof is complete.

#### 3. Intertwining Multiplication Operators

The following characterization of the commutant of is given in Theorem of [2], which is stated for the convenience of the reader. Note that is the reproducing kernel for a coanalytic functional Hilbert space defined in [2].

Theorem 9. *If is in and the operator commutes with , then there exists an analytic function such that (all ) and for every , .*

In the following let be such that if then Also we assume that the composition operator defined by is bounded.

Proposition 10. *Suppose that and there exists a domain such that is a singleton for every . If is odd, and is compact for some natural number ; then for some .*

*Proof. *Note that, by Proposition 3, the adjoint of the operator belongs to the Cowen-Douglas class , where . If , all conditions of Theorem in [5] hold and so there exists such that . For , put Clearly and so by Proposition in [5], there exists such that . But is compact; thus by the Fredholm Alternative Theorem, and so . Hence . Now we show that . PutAnd note that . This implies that , since is analytic and the zeros of are at most countable. Therefore . Now if , then and so by Proposition in [7] the proof is complete. Else, by continuing this manner, we can conclude that which implies that for some.

Proposition 11. *Suppose that and there exists a domain such that is a singleton for every . If is odd, , and is compact for some natural number , then for some.*

*Proof. *If , put Then . Thus for some. But is compact, hence and so . This implies that . Now by Proposition in [5], for some. If , put . Then, clearly from which we can conclude that for some. The compactness of implies that and so .

Thus . Put Hence which implies that . Therefore, . If , then and the proof is complete. If , by continuing this manner, finally we can see that and this completes the proof.

#### 4. Reflexivity in Cowen-Douglas Class of Operators

It is shown in [4] that, under sufficient conditions, an operator in the Cowen-Douglas class can be reflexive, where is a special bounded plane domain. In this section we give some sufficient conditions so that the associated canonical model is reflexive. This answers Question in [9, p. 98]. Indeed, we investigate the reflexivity of , when is an arbitrary bounded domain.

It is well known that every operator in the class is unitarily equivalent to the adjoint of the canonical model associated with a generalized Bergman kernel (g.B.k. for brevity) (see [2, 6]). Actually is the reproducing kernel for a coanalytic functional Hilbert space (briefly ) on which we can define the operator of multiplication by . The operator acting on is called the canonical model associated with . We know that, for every in , is onto and and .

Recall that a compact subset of the plane is a spectral set for a bounded operator if contains and for all rational functions with poles off . Also, an open connected subset of the plane is called a Carathéodory region if its boundary equals the boundary of the unbounded component of .

It is proved in [4] that if is in and is an injective unilateral weighted shift, then is reflexive. Also, it has been shown that if is in , where is a Carathéodory region such that is a spectral set for , then is reflexive (see [4, Theorem ]). This implies that if is a contraction in where is the open unit disk, then is reflexive. Here we want to investigate the reflexivity of on , where is an arbitrary domain in .

Theorem 12. *If is in , where is an arbitrary domain, then there exists a total set such that the weak closure of the set contains .*

*Proof. *Let be a g.B.k. on and let . Then by Theorem 9 and [4, Lemma ], there exists such that for all in . Now let be dense in and choose such that for Put for Define Fix satisfying where for all . Thus . Define Since , . Now clearly is closed subspace of and we have Thus and so . But ,;thus and we get . Therefore and so there exists a sequence of polynomials such that in . Thus in and since for all we get as . But and is invertible; thus for all , as . This implies that for all in the finite linear combinations of that is a total subset of . At this time the proof is complete.

Let , , and be defined as in the proof of Theorem 12. At the end of the proof of Theorem 12, we saw that, for all , as . Now we ask the following question.

*Question 13. *In the proof of Theorem 12, is it true that as ?

If the answer of Question 13 is positive, then for some and we may have the following corollary. Note that the special case of this corollary has been proved as Theorem in [4], only whenever is a Carathéodory region.

Corollary 14. *If is in where is a domain such that is a spectral set for , then is reflexive.*

*Proof. *Let be a g.B.k. on and let . By Theorem 12, there exists a sequence of polynomials such that converges in the finite linear combinations of that is a total subset of , where was a dense set in . Also, as . This implies that for some . Now since is a spectral set for , we conclude that . Since the unit ball of is compact in the weak operator topology, by passing to a subsequence if necessary, we may assume that in the weak operator topology. Therefore, weakly. But Hence, , where, by the proof of Theorem 12, is a function in and satisfies for all in Ω. From this we conclude that , so . Therefore, and is reflexive. This completes the proof.

#### Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

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