# On Some Properties of Cowen-Douglas Class of Operators

Authors: Parastoo Heiatian Naeini; Bahmann Yousefi
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We will consider multiplication operators on a Hilbert space of analytic functions on a domainΩ⊂C. For a bounded analytic function φ on Ω, we will give necessary and sufficient conditions under which the complement of the essential spectrum of Mφ in φΩ becomes nonempty and this gives conditions for the adjoint of the multiplication operator Mφ 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.


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## Body

## 1. Introduction

In this section we include some preparatory material which will be needed later.For a positive integern and a domain U⊂C, the Cowen-Douglas class Bn(U) consists of bounded linear operators T on any fixed separable infinite dimensional Hilbert space X with the following properties:(a)
U is a subset of the spectrum of T.(b)
Ranλ-T=X for every λ∈U.(c)
dim⁡ker⁡λ-T=n for every λ∈U.(d)
Spanker⁡λ-T:λ∈U=X.Here Span denotes the closed linear span of a collection of sets inX. The classes BnU were introduced by Cowen-Douglas (see [1]), and each element of BnU is called a Cowen-Douglas class operator. By Bn, we mean Bn(U) for some complex domain U. For the study of Cowen-Douglas classes Bn, we mention [1–7].Recall that a bounded linear operatorA on a Hilbert space is a Fredholm operator if and only if ranA is closed and both ker⁡A and ker⁡A∗ are finite dimensional. We use σ(A) and σe(A) to denote, respectively, the spectrum of A and the essential spectrum of A.Now letH be a separable Hilbert space and let B(H) denote the algebra of all bounded linear operators on H. Recall that if A∈B(H), then Lat(A) is by definition the lattice of all invariant subspaces of A, and AlgLat(A) is the algebra of all operators B in B(H) such that LatA⊂LatB. An operator A in B(H) is said to be reflexive if AlgLatA=WA, where W(A) is the smallest subalgebra of B(H) that contains A and the identity I and is closed in the weak operator topology.Also, ifH is a Hilbert space of functions analytic on a plane domain Ω, a complex-valued function φ on Ω for which φf∈H for every f∈H is called a multiplier of H and the multiplier φ on H determines a multiplication operator Mφ on H by Mφf=φf, f∈H. The set of all multipliers of H is denoted by MH. Clearly MH⊂H∞(Ω), where H∞(Ω) is the space of all bounded analytic function on Ω. In fact φ∞≤Mφ (see [8]).LetH be a Hilbert space of functions analytic on a domain Ω⊂C satisfying the following axioms:Axiom 1.
For every pointω∈Ω, the functional of point evaluation at ω, is a nonzero bounded linear functional on H.Axiom 2.
Every functionφ∈H∞(Ω) is a multiplier of H.Axiom 3.
Iff∈H and fλ=0, then there is a function g∈H such that z-λg=f.A spaceH satisfying the above conditions is calledHilbert space of analytic functions onΩ (see [3, 9]). The Hardy and Bergman spaces are examples for Hilbert spaces of analytic functions on the open unit disk.Note that, by Axiom1, there exists a reproducing kernel kw∈H such that fω=f,kω for all f∈H. Also, by using Axiom 2 and the closed-graph theorem, the operator of multiplication by φ, Mφ, is a bounded linear operator on H. So Axiom 2 says that MH=H∞Ω. If Mz is polynomially bounded on H and Ω is the open unit disk, then MH=H∞Ω (see [9, Theorem 1]). In the rest of the paper we assume that H is a Hilbert space of analytic function on a bounded plane domain Ω.In this paper, we want to study some properties of operators inBn. We see that complement of the essential spectrum of a multiplication operator Mφ is nonempty if and only if the adjoint of Mφ belongs to some Bn. Also, we investigate the intertwining multiplication operators and reflexivity of the multiplication operator on Bn. For some other source on these topics one can see [10–16].


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

Recall that ifT is a Cowen-Douglas class operator, then it should be σT∖σeT≠∅. For φ∈H∞Ω, we would like to give some necessary and sufficient conditions so that σMφ∖σeMφ becomes a nonempty open set. This implies a sufficient condition for the adjoint of the multiplication operator Mφ to be a Cowen-Douglas class operator.Theorem 1.
Letφ be a nonconstant function in H∞Ω, σMφ∖σeMφ≠∅, and kz/kz→0 weakly as dist(z,∂Ω)→0. Then there exist a domain V⊂φ(Ω) and a positive integer n such that Ω∩φ-1(λ) consists of n points (counting multiplicity) for every λ∈V.Proof.
First note that ifλ∈φΩ, then λ=φω for some ω∈Ω. But by Axiom 1, the functional of evaluation at ω is a bounded point evaluation; thus the reproducing kernel kω is defined and we have (1)Mφ∗kω=φω-kω.Thus λ∈σ(Mφ) and clearly φ(Ω)-⊂σ(Mφ). Now let λ∉φΩ-. Then φ-λ is an invertible element of H∞Ω. But by Axiom 2, we have MH=H∞Ω; thus Mφ-λ is invertible. This implies that σ(Mφ)⊂φ(Ω)-; thus indeed σMφ=φ(Ω)-. Now we prove that (2)σMφ∖σeMφ=φΩ∖σeMφ.For this it is sufficient to show that ∂φΩ⊂σeMφ. Let λ∉σeMφ. If λ∈∂φΩ, then there exists a sequence znn⊂Ω such that φzn→λ. By passing to a subsequence if necessary, we may assume that znn converges to a point in ∂Ω and so by our assumptions kzn/kzn→0 weakly. On the other hand we have (3)Mφ-λ∗kznkzn=φzn--λ-kznkznfor all n∈N. So we get (4)Mφ-λ∗kznkzn⟶0which contradicts the fact that Mφ-λ∗ is Fredholm. Thus we have(5)σMφ∖σeMφ=φΩ∖σeMφ.Now, let V be a connected component of the open set φΩ∖σeMφ. Since V∩σeMφ=∅, thus Mφ-λ is Fredholm for every λ in V. Also, note that if Mφ-λf=0, then f=0 on Ω∖(φ-λ)-10 which is open. Hence f≡0 and so Mφ-λ is injective. Thus (6)indexMφ-λ∗=dim⁡ker⁡Mφ-λ∗for all λ in V. But the index function is continuous from the set of semi-Fredholm operators into Z∪±∞ with discrete topology; thus, index(Mφ-λ)∗ is constant for all λ in V. Put (7)dim⁡ker⁡Mφ-λ∗=n.If z∈V, then λ=φ(λ0) for some λ0∈Ω and so Mφ∗kλ0=λ-kλ0. Thus kλ0∈ker⁡(Mφ-λ)∗. Since a finite subset of points ω in Ω yields a linearly set independent set of functions kω in H, thus Ω∩φ-1(λ) consist of at most n points for all λ in V. So for each fixed λ∈V, there exist λ1,λ2,…,λm in Ω and n1,n2,…,nm in N such that m≤n and for all z∈Ω we have (8)φz-λ=ψzz-λ1n1z-λ2n2⋯z-λmnm,where ψ belongs to H∞Ω and is nonvanishing on Ω. Now by a method used in the proof of [3, Proposition 3.1] we show that the function ψ is also bounded below on Ω. For this choose r>0 such that B(λ,r)- is contained in V. Put K=φ-1(B(λ,r)),- and thus K is a compact subset of Ω and so it has a positive distance δ to ∂Ω. Now if ψ is not bounded below on Ω, then there exists a sequence zi in Ω-λ1,λ2,…,λn such that ψ(zi)→0 as i→∞. Since ψ is nonvanishing on Ω implies that φzi→λ, so there exists a positive integer N such that φzi∈B(λ,r)- for all i>N. Hence zi∈K for all i>N that is contradiction to zi→∂Ω. Thus the function ψ is indeed bounded below on Ω. Now since ψ is bounded below and bounded above on Ω it is an invertible element of H∞Ω and so the operator Mψ is invertible on H because MH=H∞Ω. Thus indexMψ=0. Note that since (9)Mφ-λ=MψMz-λ1n1Mz-λ2n2⋯Mz-λmnm,we get(10)indexMφ-λ=∑j=1mnjindexMz-λj.But Mφ-λ is injective for all λ∈V; thus(11)indexMφ-λ=-dim⁡kerMφ-λ∗=-n.Clearly, Mz-λj is injective; thus(12)indexMz-λj=-dim⁡kerMz-λj∗for j=1,…,m. Note that, by Axiom 3 on H, kerMz-λj∗ is one-dimensional (see [17]); thus ∑j=1mnj=n and therefore Ω∩φ-1(λ) consists of exactly n points (counting multiplicity) for every λ∈V and now the proof is complete.From the proof of Theorem1, we can conclude the following result.Corollary 2.
Letφ be a nonconstant function in H∞Ω and kz/kz→0 weakly as dist(z,∂Ω)→0. Then ∂φ(Ω)⊂σeMφ.Note that, by Axiom3, for every λ∈Ωthe operator Mz-λ is bounded below on H and also the space H⊖z-λH is one-dimensional (see [3]). So the Hilbert space under consideration, H, satisfies the conditions assumed by Zhu in [7].The following result was stated by Zhu in [7, Proposition 5.2], 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.
Supposeφ∈H∞Ω and V is a domain contained in φΩ. If there exists a positive integer n such that Ω∩φ-1(λ) consists of n points (counting multiplicity) for every λ∈V, then the adjoint of the operator Mφ:H→H belongs to the Cowen-Douglas class Bn(U), where U=z-:z∈V.Proof.
Letλ=φω∈V. Then there exist an invertible function ψ∈H∞Ω and z1,z2,…,zm∈Ω∩φ-1(λ) such that(13)Mφ-λ∗=Mψ∗Mz-z1∗k1Mz-z2∗k2⋯Mz-zm∗km,where ∑i=1mki=n. Axiom 3 implies that for all i=1,…,m,Mz-zi∗ is onto (see [17]); thus for all λ∈V, Mφ-λ∗ is onto since Mψ is invertible. Also, by Axiom 3, dim[⁡kerMz-zi∗]=1 for i=1,…,m and so (14)dim⁡ker⁡Mφ-λ∗=∑i=1mkidim⁡ker⁡Mz-zi∗=∑i=1mki=n. Finally, we note that (15)Spankω:ω∈φ-1V⊂Spanker⁡Mφ-λ∗:λ∈V.Now, since φ-1V is open, Spankω:ω∈φ-1(V)=H and so the proof is complete.Corollary 4.
Under the conditions of Theorem1, there exist a positive integer n and a domain U in the complex plane such that Mφ∗∈Bn(U).Proof.
By Theorem1 and Proposition 3 it is clear.Now we investigate the converse of Theorem1.Corollary 5.
Letφ be a nonconstant function in H∞Ω. If there exists a domain V⊂φ(Ω) and a positive integer n such that Ω∩φ-1(λ) consists of n points (counting multiplicity) for every λ∈V; then V⊂σ(Mφ)∖σeMφ.Proof.
By Proposition3, the adjoint of the operator Mφ:H→H belongs to the Cowen-Douglas class Bn(U), where U=z-:z∈V. Hence for all λ∈U, Mφ∗-λ is Fredholm and so clearly V⊂σ(Mφ)∖σeMφ.Corollary 6.
LetMφ∗∈Bn(U) for some positive integer n and a complex domain U. If kz/kz→0 weakly as dist(z,∂Ω)→0, then Ω∩φ-1(λ) consists of n point (counting multiplicity) for every λ∈V where V=z-:z∈U.Proof.
First note thatMφ∗-λ is Fredholm for all λ∈U; thus (16)V⊂σMφ∖σeMφ=φΩ-∖σeMφ.But by Corollary 2, ∂φΩ⊂σeMφ; thus, V⊂φ(Ω). Now if λ∈V, then λ=φ(ω) for some ω∈Ω and clearly kω∈kerMφ-λ∗. Since dim⁡ker⁡Mφ-λ∗=n and a finite subset of points ω in Ω yields a linearly independent set of functions kω in H, thusΩ∩φ-1(λ) consist of at most n points for all λ∈V. Now by the same method used in the proof of Theorem 1, we can see that Ω∩φ-1(λ) consists of exactly n points (counting multiplicity) for every λ∈V.Example 7.
Consider the Hilbert Bergman spaceLa2(D) where D is the open unit disc in the complex domain. Then La2(D) holds in Axioms 1, 2, and 3 (see [17, Theorem 8.5, page 67]). For the Bergman reproducing kernel function, kz, clearly we can see that kz→∞ as dist(z,∂D)→0. So if p is a polynomial, then (17)p,kzkz=pzkz⟶0 as dist(z,∂D)→0. But polynomials are dense in La2D; thus kz/kz→0 weakly as dist(z,∂D)→0. Now by Theorem 1 and the proof of Corollary 5, we can see that Mφ∗∈Bn(U) for some positive integer n and a complex domain U if and only if σMφ∖σeMφ≠∅.Proposition 8.
Letφ be a nonconstant function in H∞Ω and kz/kz→0 weakly as dist(z,∂Ω)→0. Then(18)σeMφ=∩nφz∈Ω:distz,∂Ω<1n.Proof.
Letλ∉σeMφ; then Mφ-λ is Fredholm. Now we show that φ-λ is bounded away from zero near ∂Ω. By way of contradiction, let znn⊂Ω be a sequence such that φzn→λ and znn converges to a point in ∂Ω. Note that by our assumptions kzn/kzn→0 weakly and (19)Mφ-λ∗kznkzn=φzn--λ-kznkznfor all n∈N. So we get (20)Mφ-λ∗kznkzn⟶0.This is a contradiction because Mφ-λ∗ is Fredholm. Hence, φ-λ is bounded away from zero near ∂Ω and so there exists m∈N large enough such that (21)inf⁡φz-λ:distz,∂Ω<1m>0.This implies that (22)λ∉∩nφz∈Ω:distz,∂Ω<1n.Conversely, if(23)λ∉∩nφz∈Ω:distz,∂Ω<1n, then φ-λ is bounded away from zero near ∂Ω. Since the zeros of an analytic function are isolated, thus the zeros of φ-λ are finite. Let λ1,λ2,…,λj be all zeros (counting multiplicity) of φ-λ in Ω such that (24)φz-λ=ψzz-λ1z-λ2⋯z-λj.Clearly the function ψ is invertible on Ω and so Mψ is bounded below. Also, by Axiom 3 on H, Mz-λj is Fredholm for all i=1,…,j. This implies that Mφ-λ is Fredholm and so λ∉σeMφ. So the proof is complete.


## 3. Intertwining Multiplication Operators

The following characterization of the commutantT′ of T is given in Theorem 3.7 of [2], which is stated for the convenience of the reader. Note that K is the reproducing kernel for a coanalytic functional Hilbert space K defined in [2].Theorem 9.
IfS is in Bn(Ω) and the operator X commutes with S, then there exists an analytic function Φ:Ω→Bn(Cn) such that XKλ,·=Kλ,·Φ(λ) (all λ∈Ω) and for every f∈K, X∗f·=(Φ·)∗f(·).In the following letΩ be such that if λ∈Ω then -λ∈Ω. Also we assume that the composition operator C-z:H→H defined by C-zf=f(-z) is bounded.Proposition 10.
Suppose thatφ∈H∞Ω and there exists a domain V⊂φ(Ω) such that Ω∩φ-1(ω) is a singleton for every ω∈V. If φ is odd, SMφ2=Mφ2S and SMφ2n-1-Mφ2n-1S is compact for some natural number n; then S=Mh for some h∈H∞Ω.Proof.
Note that, by Proposition3, the adjoint of the operator Mφ:H→H belongs to the Cowen-Douglas class B1(U), where U=z-:z∈V. If n=1, all conditions of Theorem 4 in [5] hold and so there exists h∈H∞Ω such that S=Mh. For n>1, put (25)T1=SMφ2n-1-Mφ2n-1S.Clearly T1Mφ=-MφT1 and so by Proposition 3 in [5], there exists h∈H∞Ω such that T1=MhC-z. But Mh=MhC-z∘C-z is compact; thus by the Fredholm Alternative Theorem, h=0 and so T1=0. Hence SMφ2n-1=Mφ2n-1S. Now we show that SMφ2n-3=Mφ2n-3S. Put(26)T2=SMφ2n-3-Mφ2n-3S.And note that Mφ2T2=0. This implies that T2=0, since φ is analytic and the zeros of φ are at most countable. Therefore SMφ2n-3=Mφ2n-3S. Now if n=2, then SMφ=MφS and so by Proposition 4.1 in [7] the proof is complete. Else, by continuing this manner, we can conclude that SMφ=MφS which implies that S=Mφ for someh∈H∞Ω.Proposition 11.
Suppose thatφ∈H∞Ω and there exists a domain V⊂φ(Ω) such that Ω∩φ-1(ω) is a singleton for every ω∈V. If φ is odd, SMφ2=Mφ2S, and SMφ2n-1+Mφ2n-1S is compact for some natural number n, then S=MhC-z for someh∈H∞Ω.Proof.
Ifn=1, put (27)T1=SMφ+MφS. Then T1Mφ=MφT1. Thus T1=Mh1 for someh1∈H∞Ω. But Mh1 is compact, hence h1=0 and so T1=0. This implies that SMφ=-MφS. Now by Proposition 3 in [5], S=MhC-z for someh∈H∞Ω. If n>1, put T2=SMφ2n-1+Mφ2n-1S. Then, clearly MφT2=T2Mφ from which we can conclude that T2=Mh1 for someh1∈H∞Ω. The compactness of Mh1 implies that h1=0 and so T2=0.
ThusSMφ2n-1=-Mφ2n-1S. Put (28)T3=SMφ2n-3+Mφ2n-3S.Hence Mφ2T3=0 which implies that T3=0. Therefore, SMφ2n-3=-Mφ2n-3S. If n=2, then SMφ=-MφS and the proof is complete. If n>2, by continuing this manner, finally we can see that SMφ=-MφS and this completes the proof.


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

It is shown in [4] that, under sufficient conditions, an operator T in the Cowen-Douglas class Bn(Ω) can be reflexive, where Ω 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 5.6 in [9, p. 98]. Indeed, we investigate the reflexivity of Bn(Ω), when Ω is an arbitrary bounded domain.It is well known that every operator in the classBn(Ω) is unitarily equivalent to the adjoint of the canonical model associated with a generalized Bergman kernel (g.B.k. for brevity) K (see [2, 6]). Actually K is the reproducing kernel for a coanalytic functional Hilbert space KK (briefly K) on which we can define the operator Tz- of multiplication by z-. The operator T=Tz-∗ acting on K is called the canonical model associated with K. We know that, for every λ in Ω, T-λ is onto and (29)ker⁡T-λ=ran⁡Kλ,·=Kλ,·ξ:ξ∈Cn,and dim⁡ker⁡T-λ=n.Recall that a compact subsetF of the plane is a spectral set for a bounded operator A if F contains σA and f(A)≤supz∈F⁡f(z) for all rational functions f with poles off F. Also, an open connected subset G of the plane is called a Carathéodory region if its boundary equals the boundary of the unbounded component of C-G-.It is proved in [4] that if T is in B1(Ω) and T∗ is an injective unilateral weighted shift, then T is reflexive. Also, it has been shown that if T is in Bn(Ω), where Ω is a Carathéodory region such that σT=Ω- is a spectral set for T, then T is reflexive (see [4, Theorem 2]). This implies that if T is a contraction in Bn(D) where D is the open unit disk, then T is reflexive. Here we want to investigate the reflexivity of T on Bn(Ω), where Ω is an arbitrary domain in C.Theorem 12.
IfT is in Bn(Ω), where Ω⊂C is an arbitrary domain, then there exists a total set Y such that the weak closure of the set pTy:pisapolynomial,y∈Y contains AlgLat(T).Proof.
LetK be a g.B.k. on Ω and let X∈AlgLat(T). Then by Theorem 9 and [4, Lemma 1], there exists ψ∈H∞Ω such that XKλ,·=ψ(λ)Kλ,· for all λ in Ω. Now let F=λnn=1∞ be dense in Ω and choose ξi∈Cn such that Kλi,·ξi≠0 for i=1,2,…. Put Ki=∨Kλi,·ξi for i=1,2,…. Define (30)K∞′=K1⊕K2⊕K3⊕⋯=⨁i=1∞Ki,T∞=TK1⊕TK2⊕TK3⊕⋯=⨁i=1∞Ti,X∞=X⊕X⊕X⊕⋯=⨁i=1∞Xi. Fix f=⨁i=1∞ciK(λi,·)ξi satisfying (31)∑i=1∞ci2Kλi,·ξi2<∞, where ci≠0 for all i. Thus f∈K∞′. Define (32)M=cl⨁i=1∞pTciKλi,·ξi:pisapolynomial.Since f∈M, M≠∅. Now clearly M is closed subspace of K∞′ and we have (33)T∞f=⨁i=1∞ciTKλi,·ξi=⨁i=1∞ciλiKλi,·ξi.
ThusT∞f∈M and so M∈LatT∞. But LatT⊆LatX,;thus LatT∞⊆LatX∞ and we get M∈LatX∞. Therefore X∞f∈M and so there exists a sequence pnn of polynomials such that(34)⊕icipnTciKλi,·ξi⟶X∞f=⊕iciψTciKλi,·ξi in K∞′. Thus ⊕icipnλi-ψλiKλi,·ξi→0 in K∞′ and since for all i(35)⊕icipnλi-ψλiKλi,·ξi≥cipnλi-ψλiKλi,·ξi,we get supicipnλi-ψλiKλi,·ξi→0 as n→∞. But (36)cipnλi-ψλiKλi,·ξi2=cipnλi-ψλi2Kλi,λiξi,ξi=ci2pnλi-ψλi2Kλi,λi1/2ξi2 and Kλi,λi is invertible; thus for all i, (pn-ψ)λi→0 as n→∞. This implies that (pn(T)-X)g→0 for all g in the finite linear combinations of (37)Y=Kλi,·ξ:i∈N,ξ∈Cn that is a total subset of K. At this time the proof is complete.Letψ, pnn, and F=λii be defined as in the proof of Theorem 12. At the end of the proof of Theorem 12, we saw that, for all i, (pn-ψ)λi→0 as n→∞. Now we ask the following question.Question 13.
In the proof of Theorem12, is it true that supi(pn-ψ)λi→0 as n→∞?If the answer of Question13 is positive, then pnF≤M for some M>0 and we may have the following corollary. Note that the special case of this corollary has been proved as Theorem 2 in [4], only whenever Ω is a Carathéodory region.Corollary 14.
IfT is in Bn(Ω) where Ω⊂C is a domain such that σT=Ω- is a spectral set for T, then T is reflexive.Proof.
LetK be a g.B.k. on Ω and let X∈AlgLatT. By Theorem 12, there exists a sequence of polynomials pnn such that pn(T) converges in the finite linear combinations of (38)Kλ,·ξ:λ∈F,ξ∈Cn that is a total subset of K, where F=λnn=1∞ was a dense set in Ω. Also, supi(pn-ψ)(λi)→0 as n→∞. This implies that pnΩ=pnF≤M for some M>0. Now since σT=Ω- is a spectral set for T, we conclude that pn(T)≤M. Since the unit ball of B(K) is compact in the weak operator topology, by passing to a subsequence if necessary, we may assume that pn(T)→A in the weak operator topology. Therefore, pnTKλ,·ξ→AKλ,·ξ weakly. But (39)pnTKλ,·ξ=pnλKλ,·ξ⟶ψλKλ,·ξ.Hence, AKλ,·ξ=ψλKλ,·ξ, where, by the proof of Theorem 12, ψ is a function in H∞Ω and satisfies XKλ,·=ψ(λ)Kλ,· for all λ in Ω. From this we conclude that A=X, so X∈WT. Therefore, AlgLat(T)⊂WT and T is reflexive. This completes the proof.



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