# Strictly Cyclic Functionals, Reflexivity, and Hereditary Reflexivity of Operator Algebras

Authors: Quanyuan Chen; Xiaochun Fang
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

This paper is concerned with strictly cyclic functionals of operator algebras on Banach spaces. It is shown that ifX is a reflexive Banach space and 𝒜 is a norm-closed semisimple abelian subalgebra of B(X) with a strictly cyclic functional f∈X∗, then 𝒜 is reflexive and hereditarily reflexive. Moreover, we construct a semisimple abelian operator algebra having a strictly cyclic functional but having no strictly cyclic vectors. The hereditary reflexivity of an algbra of this type can follow from theorems in this paper, but does not follow directly from the known theorems that, if a strictly cyclic operator algebra on Banach spaces is semisimple and abelian, then it is a hereditarily reflexive algebra.


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## Body

## 1. Introduction

Throughout this paper,X is a complex Banach space, BX stands for the closed unit ball of X and X* is the dual space of X, the space of all continuous linear functionals on X. For a bounded operator A on X, denote by LatA the lattice of all closed invariant subspaces of A and A* the adjoint operator of A. For a subalgebra 𝒜 of B(X), the Banach algebra of all bounded linear operators on X, denote by Lat(𝒜) the lattice of all closed subspaces invariant under every operator in 𝒜. For a set ℒ of subspaces of X, denote by Algℒ the algebra of all operators in B(X) which leave all subspaces in ℒ invariant. An operator algebra 𝒜 is reflexive if(1.1)A=AlgLatA,whereAlgLatA={T∈B(X):LatA⊂LatT}.In [1, 2], Loginov and Sulman introduced the following notion of a reflexive subspace of B(X). If 𝒮 is a subspace (linear manifold) of B(X), we denote by Ref𝒮 the set {A:A∈B(X)andA(x)∈𝒮(x)¯for allx∈X}, where the bar denotes norm closure. The operator subspace 𝒮 is reflexive if Ref𝒮=𝒮. If 𝒮 is a subalgebra of B(X) containing the identity operator I, then Ref𝒮=AlgLat𝒮, and 𝒮 is reflexive as an operator subspace if and only if 𝒮 is reflexive as an operator algebra.In general theory of operator algebras, reflexive algebras seem to play a role somewhat analogous to the role of von Neumann algebras in the study of*-algebras. The problem of finding the conditions under which a (weakly) closed operator algebra is reflexive has an obvious connection with the problem of existence of invariant subspaces for a bounded linear operator or for an operator algebra. A concrete operator algebra of B(X) is reflexive if and only if it is equivalent to Algℒ for some subspace lattice ℒ of X. The reflexivity of an abstract operator algebra has also been studied extensively. Finding conditions for an abstract operator algebra to be reflexive has been the important concern of many scholars and there have been many papers written on the topic. See [1–14] and references therein. In this paper, we are interested in the hereditary reflexivity of operator algebras having strictly cyclic functionals.Let𝒜 be a subalgebra of B(X). If there is a vector x0∈X such that the map α of 𝒜 to 𝒜x0 by α(A)=Ax0 is injective on 𝒜, then the vector x0 is called a separating vector for 𝒜. Let f0∈X*. If the map βf0 of 𝒜 to X* by βf0(A)=A*(f0) is injective on 𝒜, then the functional f0 is called a separating functional for 𝒜. Furthermore, if the subset {A*(f0):A∈𝒜} is norm-closed in X*, then the functional f0 is called a strictly separating functional for 𝒜.𝒜 is called cyclic if 𝒜x0={Ax0:A∈𝒜} is norm-dense in X for some vector x0 in X. 𝒜 is called strictly cyclic if 𝒜x0=X for some vector x0 in X. The vector x0 is called cyclic vector for 𝒜 in the former case and strictly cyclic vector in the latter. Alan Lambert proved [10] that a commutative semisimple strictly cyclic Banach algebra of operators on a Hilbert space is reflexive, and that every strictly cyclic vector for an abelian algebra is a separating vector. Later, Hadwin [7] gave a simple proof that a commutative semisimple strictly cyclic algebra of operators on a Banach space is reflexive. Meanwhile, Hadwin and Nordgren [6] proved that a reflexive operator algebra on a Banach space with a relative strictly separating vector is hereditarily reflexive (a concept initiated and studied by Loginov and Sulman [12]). In [13], Costel Peligrad explicitly stated and proved differently that if a strictly cyclic operator algebra on a Banach space is semisimple and abelian, then it is reflexive and hereditarily reflexive. In the present paper, a similar result is obtained without the hypothesis that the operator algebra 𝒜 is strictly cyclic, provided that 𝒜 has a strictly cyclic functional. Moreover, we construct a semisimple single-generated algebra 𝒜 having a strictly cyclic functional but having no strictly cyclic vectors. An algebra of this type is reflexive by Theorem 3.8 and hereditarily reflexive by Theorem 3.12. However, the (hereditary) reflexivity of an algebra of this type does not follow directly from the results above.The paper is organized as follows. In Section2, we introduce the notion of a (strictly) cyclic functional for operator algebras on a Banach space and develop some properties of strictly cyclic functionals. In Section 3, we clarify the role of the strictly cyclic functional in the reflexivity of operator algebra. The main results of Section 3 are Theorems 3.8 and 3.12. It is shown that a semisimple abelian subalgebra of B(X) with a strictly cyclic functional f0 is reflexive and hereditarily reflexive when X is a reflexive Banach space. The reflexivity of some operator algebras on Banach spaces can follow from Theorem 3.8. In Section 4, we construct a semisimple single-generated algebra 𝒜 having a strictly cyclic functional but having no strictly cyclic vectors. An algebra of this type is reflexive by Theorem 3.8 and hereditarily reflexive by Theorem 3.12. However, the (hereditary) reflexivity of an algebra of this type does not follow directly from the theorems of Costel Peligrad.


## 2. Strict Cyclic Functionals

Definition 2.1.
LetX be a Banach space and let  𝒮 be a subalgebra or a subspace of B(X). Let X* be the dual space of X. If there is a functional f0∈X* such that
(2.1)S*f0={A*(f0):A∈S}={(f0∘A):A∈S}
is norm-dense in X*, then f0 is called a cyclic functional for 𝒮. If there is a functional f0∈X* such that
(2.2)S*f0={A*(f0):A∈S}={(f0∘A):A∈S}=X*,
then f0 is called a strictly cyclic functional for 𝒮.Lemma 2.2 (Open Mapping Lemma).
Suppose thatX is a Banach space and  Y is a normed linear space, and  εf0:X→Y is a bounded operator. Assume that there exist  k>0 and  0<η<1 such that
(2.3)BY⊆k⋅εf0(BX)+ηBY.
Then, εf0 is a surjective open map, and Y is complete.Proposition 2.3.
Let𝒜 be a norm-closed subalgebra of  B(X) and  f0∈X* with  f0≠0. Then, the following are equivalent:
(1)f0 is a strictly cyclic functional for 𝒜;(2) there exists a constantC>0 such that for all g∈BX* there is an operator A∈𝒜 with ∥A∥⩽C such that A*(f0)=g;(3) there areC>0 and 0<η<1 such that for all g∈BX* there is an operator A∈𝒜 with ∥A∥⩽C such that ∥g-A*(f0)∥<η.Proof.
Defineεf0:𝒜→X* by εf0(A)=A*(f0) for A∈𝒜, then εf0 is a bounded linear operator from 𝒜 to X*. Since 𝒜 is norm-closed, it can be viewed as a Banach space. Thus, the implication (1) ⇒ (2) follows immediately from Open Mapping Theorem. The implication (2) ⇒ (3) is trivial. Finally, the implication (3) ⇒ (1) follows from Open Mapping Lemma.The following proposition is similar to Theorem  2 in [15].Proposition 2.4.
Let𝒜 be a norm-closed subalgebra of  B(X) with a strictly cyclic functional  f0∈X*.
(1) Then, every cyclic functionalf for 𝒜 is actually a strictly cyclic functional for 𝒜.(2) Moreover, the set of all (strictly) cyclic functionals for𝒜 is norm-open in X* and coincides with
(2.4)CA={A*(f0):A∈AandA*has a left inverse(mod⁡ker⁡(εf0))inA*},
where εf0 is the map from 𝒜 onto X* by εf0(A)=A*(f0) for every A∈𝒜, and ker(εf0) is the kernel of the map εf0.Proof.
(1) Iff is a cyclic functional for 𝒜, then N=𝒜*f={A*(f):A∈𝒜} is a norm-dense linear manifold of X* and N is 𝒜*-invariant. We will show that f0∈N and thus X*=𝒜*(f0)={A*(f0):A∈𝒜}⊆𝒜*N⊆N.
Letfn be a sequence in N such that {fn} converges uniformly to f0. By Proposition 2.3 (3), we may assume the existence of sequence An such that An*(f0)=f0-fn and limn→∞∥An∥=0. Thus, for n sufficiently large, ∥An*∥<1 and (I-An*)-1=∑k=0∞(An*)k=∑k=0∞(Ank)*. Therefore, (I-An*)-1∈𝒜* and since (I-An*)f0=fn, we have f0=(I-An*)-1fn∈N.
(2) ChooseC as in Proposition 2.3 (2). Let 0<δ<1/C and suppose that f∈X* with ∥f-f0∥<δ. Now if g∈BX*, we choose A∈𝒜 with ∥A∥⩽C and A*(f0)=g. But then, ∥A*(f)-g∥=∥A*(f)-A*(f0)∥⩽Cδ<1. Hence, f is a strictly cyclic functional for the algebra 𝒜 by Proposition 2.3 (3). Thus, the set of all (strictly) cyclic functionals for 𝒜 is uniformly open in X*.
Clearly, ifg=A*(f0)∈𝒞𝒜, then there exists B∈𝒜 such that B*A*=I+L for some L∈ker(εf0). Therefore, f0=B*A*(f0)∈𝒜*(f0). Hence, X*=𝒜*f0⊆𝒜*g, which implies that g is a strictly cyclic functional for 𝒜.
Conversely, ifg=A*(f0) is a cyclic (and, therefore, it is strictly cyclic) functional for 𝒜, then f0∈X*={B*A*(f0):B∈𝒜}. Therefore, there exists B∈𝒜 such that B*A*(f0)=f0, that is, (B*A*-I)f0=0. Hence, B*A*-I=L∈ker(εf0) and B*A*=I-L. Therefore, B* is a left inverse of A* module ker(εf0). Hence, g=A*(f0)∈𝒞𝒜.Remark 2.5.
If𝒜 is a norm-closed commutative subalgebra of B(X) with a strictly cyclic functional f0∈X*, then I∈𝒜. Indeed, let A0∈𝒜 such that A0*(f0)=f0. Then, A*A0*(f0)=A*(f0) for every A∈𝒜. Hence, by commutativity of 𝒜, A0*A*(f0)=A*(f0) for every A∈𝒜. Therefore, A0* is the identity operator on X* and A0=I∈𝒜.


## 3. Hereditary Reflexivity and Strictly Cyclic Functional

Remark 3.1.
LetX be a Banach space and let  𝒮 be a subalgebra or a subspace of B(X). Then, (1) f0∈X* is a strictly cyclic functional for 𝒮 if and only if f0 is a strictly cyclic vector for 𝒮*={T*:T∈𝒮}.
(2)f0∈X* is a cyclic functional for 𝒮 if and only if f0 is a cyclic vector for 𝒮*={T*:T∈𝒮}.Lemma 3.2.
(1) Suppose thatX is a Banach space which is not a Hilbert space. If   T∈B(X), then  σ(T*)=σ(T).
(2) Suppose thatX is a Hilbert space. If T∈B(X), then σ(T*)={λ̅:λ∈σ(T)}.Lemma 3.3.
Suppose thatX is a Banach space and 𝒜 is an abelian subalgebra of B(X). Set 𝒜*={A*:A∈𝒜}. Then, the Banach algebra 𝒜* is semisimple if and only if the Banach algebra 𝒜 is semisimple.Proof.
We may assume that𝒜 is unital for convenience. Let 𝒜* be semisimple and T∈𝒜 (T≠0). Then, T*≠0 and there is a multiplicative linear functional F on 𝒜* such that F(T*)≠0. Since σ(T*)={F(T*):F is a multiplicative linear functional on 𝒜*} by Gelfand transformation, σ(T*) contains a nonzero scalar and σ(T) contains a nonzero scalar by Lemma 3.2. Since σ(T)={G(T):G is a multiplicative linear functional on 𝒜}, it follows that there is a multiplicative linear functional G on 𝒜 such that G(T)≠0. Hence, 𝒜 is also semisimple. Similarly, if 𝒜 is semisimple, then 𝒜* is also semisimple.LetX be a Banach space, and let M be a nonempty subset of X and N be a nonempty subset of X*. The annihilator M⊥ of M and the preannihilator⊥N of N are defined as follows [16]: M⊥={f∈X*:f(x)=0forallx∈M}, ⊥N={x∈X:f(x)=0forallf∈N}. It is obvious that M⊥ is a weak*-closed subspace of X* and ⊥N is a norm-closed subspace of X.⊥(M⊥) is the norm-closure of M in X and (⊥N)⊥ is the weak*-closure of N in X*.Lemma 3.4.
Suppose thatX is a Banach space and  M is a closed subspace of  X and  N is a norm-closed subspace of  X*. Then,
(1)⊥(M⊥)=M;(2) furthermore, ifX is a reflexive Banach space, then (⊥N)⊥=N.Proof.
(1) It is clear.
(2) LetX be a reflexive Banach space and let  J be the natural imbedding map from X into X**. First, we observe that (⊥N)⊥ is the  weak*-closure of N in X* that includes N, that is, N⊆(⊥N)⊥.
Now, we suppose that there existsx0*∈(⊥N)⊥ but x0*∉N. Then, there exists x**∈X** such that x**(x0*)=1 and x**(f)=0 for any f∈N. Since there exists x∈X such that Jx=x** by the reflexivity of the Banach space X, we have that x0*(x)=1 and f(x)=0 for any f∈N. It follows that x0*(x)=1 and x0*∉(⊥N)⊥, which is a contradiction to x0*∈(⊥N)⊥. The proof is complete.Lemma 3.5.
LetA be a bounded linear operator on a reflexive Banach space  X. Then,  N∈LatA* if and only if  ⊥N∈LatA.Proof.
If⊥N∈LatA, then, for any f in N and x in ⊥N, we have that Ax∈(⊥N) and (A*f)(x)=f(Ax)=0. Thus, A*(f)∈(⊥N)⊥=N by Lemma 3.4. Hence, N∈Lat(A*).
Conversely, ifN∈Lat(A*), then for any x in ⊥N and f in N, we have A*f∈N and f(Ax)=(A*f)(x)=0. Thus Ax∈(⊥N). So ⊥N∈LatA.Lemma 3.6.
LetX be a reflexive Banach space and let   𝒜 be a subalgebra of  B(X). Let  𝒜*={A*:A∈𝒜}. If   𝒜* is reflexive, then  𝒜 is a reflexive subalgebra.Proof.
Let𝒜* be reflexive. Suppose that T is a bounded operator leaving invariant all the closed invariant subspaces of 𝒜, that is, Lat𝒜⊂LatT. For each norm-closed invariant subspace N (N⊆X*) of 𝒜*, that is, N∈Lat𝒜*, we have that⊥N∈Lat𝒜. Since Lat𝒜⊂LatT,⊥N∈LatT. It follows that N∈Lat(T*) by Lemma 3.5. Thus, Lat𝒜*⊂LatT*. Since 𝒜* is reflexive, T*∈𝒜*. It follows that T∈𝒜 and 𝒜 is reflexive.Lemma 3.7 ([7, Theorem 8] and [1, 2]).
Suppose thatX is a Banach space and  𝒜 is an abelian subalgebra of  B(X). If  𝒜 is a norm-closed, semisimple, strictly cyclic algebra, then  𝒜 is reflexive.Theorem 3.8.
LetX be a reflexive Banach space and let   𝒜 be an abelian subalgebra of  B(X) with a strictly cyclic functional  f∈X*. If   𝒜 is a norm-closed semisimple algebra, then  𝒜 is reflexive.Proof.
Iff is a strictly cyclic functional for 𝒜, then f is a strictly cyclic vector for 𝒜* by Remark 3.1. Since 𝒜 is semisimple, 𝒜* is also semisimple by Lemma 3.3. It follows that 𝒜* is reflexive by Lemma 3.7. Thus, 𝒜 is reflexive by Lemma 3.6.Definition 3.9 ([1, 2, 12]).
A norm-closed subalgebra𝒜 of B(X) is called hereditarily reflexive if, for every weakly closed subspace ℒ⊆𝒜, every x∈X and T∈B(X), Tx∈ℒx¯ implies T∈ℒ. Here, ℒx¯ denotes the norm closure of {Lx|L∈ℒ} in X.Lemma 3.10.
LetX be a reflexive Banach space and let   𝒜 be a unital abelian subalgebra of  B(X). Let  𝒜*={A*:A∈𝒜}. Then,  𝒜 is hereditarily reflexive if and only if  𝒜* is hereditarily reflexive.Proof.
IfX is a reflexive Banach space, then we only need to show that the hereditary reflexivity of 𝒜 implies the hereditary reflexivity of 𝒜*.
Let𝒜 be hereditarily reflexive. Let ℒ*⊆𝒜* be a norm closed subspace of 𝒜* and T*∈B(X*). We suppose that T*(f)∈ℒ*(f)¯ for every continuous functional f∈X*. Then, for any given x∈X, T*f(x)∈ℒ*(f)¯(x). Here, ℒ*(f)¯ denotes the norm closure of {L*(f)|L∈ℒ} in X*. Let g∈(ℒx¯)⊥. Then, (T*g)(x)∈ℒ*g¯(x) for any x∈X. Since (L*g)(x)=g(Lx)=0 for any L∈ℒ, we have that g(Tx)=(T*g)(x)=0. It follows that Tx∈⊥((ℒx¯)⊥)=ℒx¯ by Lemma 3.4. Thus, T∈ℒ by the hereditary reflexivity of 𝒜. It follows that T*∈ℒ* and 𝒜*  are hereditarily reflexive.Lemma 3.11 ([10, Theorem 10], and [1, 2]).
Suppose thatX is a Banach space and  𝒜 is an abelian subalgebra of  B(X). If  𝒜 is a norm-closed, semisimple, strictly cyclic subalgebra, then  𝒜 is hereditarily reflexive.Theorem 3.12.
LetX be a reflexive Banach space and let   𝒜 be an abelian subalgebra of  B(X) with a strictly cyclic functional  f∈X*. If  𝒜 is a norm-closed semisimple algebra, then  𝒜 is hereditarily reflexive.Proof.
The proof is similar to the proof of Theorem3.8.The following proposition is similar to Theorem  2.3 in [1, 2].Proposition 3.13.
LetX be a Banach space. A subspace  𝒮⊆B(X) is hereditarily reflexive if and only if it is reflexive and every weakly continuous functional on  𝒮 is of the form  A↦f(A(x))(A∈𝒮), where  x∈X and  f∈X*.Proof.
If𝒮 is hereditarily reflexive, then 𝒮 is reflexive by definition. Let F be a weakly continuous functional on 𝒮 and let 𝒦F=kerF, then 𝒦F is a reflexive subspace. Choose B∈𝒮 such that F(B)=1. Since B∉𝒦F=Ref𝒦F, there exists a vector x∈X such that ∥(B-A0)(x)∥>1=F(B) for all A0∈𝒦F. Since F(B)=F(B-A0) for any A0∈𝒦F, by linearity we have the inequality ∥Ax∥⩾F(A) for all A∈𝒮. Define the linear functional ϕ on 𝒮x by ϕ(Ax)=F(A)(A∈𝒮). Then, |ϕ(Ax)|⩽∥Ax∥, that is, ϕ is well defined and is continuous; thus there exists f∈X* such that F(A)=ϕ(Ax)=f(A(x)) for all A∈𝒮.
We now prove sufficiency. LetF be a weakly continuous functional on 𝒮 such that F(A)=f(A(x)) for all A∈𝒮. Clearly, 𝒦F=kerF is a reflexive operator subspace. By Hahn-Banach Theorem, every weakly closed subspace of 𝒮 is the intersection of the kernels of weakly continuous functionals. Since the intersection of any family of reflexive subspaces is reflexive, we have that 𝒮 is hereditarily reflexive.Lemma 3.14.
LetX be a Banach space and let  𝒜 be an abelian subalgebra of  B(X) with a strictly cyclic functional  f0∈X*. Then, the dual space of  𝒜 consists entirely of the maps  A↦F(A*f0)(A∈𝒜), where  F∈X**.Proof.
For anyF∈X**, the map A↦F(A*f0)(A∈𝒜) defines a continuous linear functional on 𝒜.
Conversely, suppose thatg is a continuous linear functional on 𝒜. Define the map C:𝒜→𝒜* by C(A)=A*(A∈𝒜) and define map ρ:𝒜*→X* by ρ(A*)=A*f0(A∈𝒜) (see the graph below). Notice that both map C and map ρ are inverse:(3.1)A⟶CA*⟶ρX*,A⟶CA*⟶ρA*f0.Then (g∘C-1∘ρ-1) is a continuous linear functional on X**. It follows that there exists F∈X** such that g(A)=(g∘C-1∘ρ-1)(f)=F(f), where A∈𝒜 and f=ρ(A*)=A*f0. Thus, g(A)=F(A*f0).Theorem 3.15.
LetX be a Banach space and let  𝒜 be a semisimple abelian subalgebra of  B(X) with a strictly cyclic functional  f0∈X*. Then,  𝒜 is hereditarily reflexive if and only if   X is a reflexive Banach space.Proof.
IfX is a reflexive Banach space, then 𝒜 is hereditarily reflexive by Theorem 3.12. Conversely, if 𝒜 is hereditarily reflexive, then by Proposition 3.13, it is reflexive and every weakly continuous functional on 𝒜 is of the form A↦f(A(x))(A∈𝒜), where x∈X and f∈X*. Notice that 𝒜* is a strictly cyclic commutative subalgebra. For any f∈X*, there exists Af∈𝒜 such that Af*f0=f. Then,
(3.2)f(A(x))=A*f(x)=A*(Af*f0)(x)=(AfA)*(f0)(x)=f0(A(Afx))=f0(Ax1),
wherex1=Afx∈X. It follows that every weakly continuous functional on 𝒜 is of the form A↦f0(A(x1))(A∈𝒜), where x1∈X. 𝒜* and 𝒜 are norm-closed. Then, 𝒜* and X* are isomorphic as Banach spaces. Thus, 𝒜 and X* are isomorphic as Banach spaces. It follows that the set of all weakly continuous linear functionals on 𝒜 is the set of all norm-continuous linear functionals on 𝒜, and by Lemma 3.14, they are of the form A↦F(A*f0)(A∈𝒜), where F∈X**.
By Proposition3.13, 𝒜S is hereditarily reflexive if and only if it is reflexive and every weakly continuous functional on 𝒜 is of the form A↦f(A(x))(A∈𝒜), where x∈X and f∈X*. It follows that X is reflexive.


## 4. An Operator Algebra Having a Strictly Cyclic Functional with No Strictly Cyclic Vectors

Throughout this section, we assume that1<p<∞ and 1/p+1/q=1. Let lp be the Banach space of all absolutely p-summable sequences x={ξn}n=0∞ of complex numbers with ∥x∥=(∑n=0∞|ξn|p)1/p. Let c0 be the space of sequences x=(ξn)n=0∞ such that limn→∞ξn=0 with the norm ∥x∥=supn{|ξn|}. {en}n=0∞ is the natural basis for lp or c0, that is, en=(0,0,…,0,1,0,…) with 1 as the nth component (beginning the indexing at 0). For a sequence {wn}n=1∞ of nonzero complex numbers, let T be the linear transformation on lp defined by(4.1)Te0=0,Ten=wnen-1,forn≥1.ObviouslyT is a unilateral weighted backward shift with weight sequence {wn}n=1∞.Let{fm}m=0∞ be the natural basis for lq, that is, fm=(0,0,…,0,1,0,…) with 1 as the mth component (beginning the indexing at 0). A short computation shows that the adjoint operator T* of T is defined by T*(fm)=wm+1fm+1 for all m=0,1,2,…. Obviously T* is a unilateral weighted forward shift with weight sequence {wn}n=1∞.For an operatorT on a Banach space X, denote by 𝒜T the weakly closed subalgebra generated by T and the identity operator I. An operator T on a Banach space is called strictly cyclic if the weakly closed subalgebra 𝒜T is strictly cyclic.Lemma 4.1.
LetX be a Banach space and let   𝒜 be a subspace of  B(X). If  𝒜 has a strictly cyclic separating vector  x, then there exist  k1>0 and  k2>0 such that
(4.2)k1‖A‖⩽‖Ax‖⩽k2‖A‖for any A∈𝒜.Proof.
Letx be a strictly cyclic separating vector for 𝒜. The evaluation map εx:A↦Ax is bounded below and has dense range (as a mapping from the algebra 𝒜 onto X). It follows that εx and εx-1 are both continuous. The result follows.Remark 4.2.
IfT is strictly cyclic, then the norm topology and the strong topology on 𝒜T coincide. If T has a strictly cyclic functional, then the norm topology and the strong topology on 𝒜T* coincide. Since I∈𝒜T*, 𝒜T* is also the strongly closed subalgebra generated by T* and I.
It is easy to check the following statement.Proposition 4.3.
Any unilateral weighted backward shiftT on  lp(1<p<∞) or  c0 with nonzero weights is not strictly cyclic.Example 4.4.
LetT be a unilateral weighted backward shift on lp(1<p<∞) with weight sequence {wn}n=1∞: wn=(n+1)/n for all n⩾1. Then,
(1)T is not strictly cyclic;(2)T* is strictly cyclic on lq;(3) The weakly closed algebra𝒜T* generated by T* is semisimple. Therefore, 𝒜T is also semisimple by Lemma 3.3;(4)𝒜T* is hereditarily reflexive, and it follows that 𝒜T is hereditarily reflexive from Theorem 3.12.Proof.
(2) Setβ0=1, βn=∏k=1nwk=(2/1)·(3/2)⋯(n+1)/n=n+1(4.3)∑m=0n|βnβmβn-m|p=1+∑m=0n|n+1(m+1)⋅(n-m+1)|p=1+(n+1)p(n+2)p∑m=1n|1m+1+1n-m+1|p≤1+∑m=1n|1m+1+1n-m+1|p≤1+∑m=1n2p(1(m+1)p+1(n-m+1)p)≤1+2p-1∑m=1∞1mp=C.
It follows that supn∑m=0n|βn/βmβn-m|p<∞. Since {|wn|} is monotonically decreasing, T* is strictly cyclic on lq by [17, Theorem  3.2].
(3) It is easy to see that∥Tn∥=∥(T*)n∥=supk∈ℕ|∏i=0n-1wk+i|=supk∈ℕ(1+n/k)=1+n. Then, ∥T∥=∥T*∥=2 and the spectral radius r(T)=r(T*)=limn→∞∥Tn∥1/n=limn→∞(1+n)1/n=1. Since T* is unilateral forward shift on lq, the spectra σ(T)(=σ(T*)) of T contains infinite complex numbers. Notice that the elements of 𝒜T* consist of all P(T*)s and their norm limits [18], where P(·) are all polynomials. Then, the elements of 𝒜T consist of all P(T)s and their norm limits, where P(·) are all polynomials. Then, we have that the Jacobson radical rad(𝒜T)={0}.
In fact, since𝒜T is unital,(4.4)rad(AT)={A∈AT:ABisquasinilpotentforanyB∈AT}.For any A∈𝒜T, there exists a sequence of polynomials Qn(λ) such that ∥Qn(T)-A∥→0. Let A1=P1(T),A2=P2(T)∈𝒜T, where P1(·),P2(·) are polynomials. It is obvious that σ(A1A2)=σ(A1)σ(A2)=P1(σ(T))·P2(σ(T)) by Spectral Theorem. If A1A2 is quasinilpotent, then {0}=σ(A1A2)=σ(A1)σ(A2). It follows that, if A∈rad(𝒜T), then σ(A)·P2(σ(T))={0} for any polynomial P2(λ), and σ(A)={0} since σ(T) contains infinite complex numbers. Thus, Qn(σ(T))→0 and Qn(T)→0, that is, A=0. So 𝒜T is semisimple.Remark 4.5.
LetT be a unilateral weighted backward shift on the reflexive Banach space lp as in Example 4.4. Then, the weakly closed algebra 𝒜T, generated by T and I, is norm-closed, semisimple, and abelian. And 𝒜T has a strictly cyclic functional but has no strictly cyclic vectors. An algebra of this type is reflexive (hereditarily reflexive) by Theorem 3.8 (Theorem 3.12). However, the reflexivity of an algebra of this type does not follow directly from the theorem of Costel Peligrad.Remark 4.6.
If the Banach spaceX is not reflexive, then the algebra 𝒜 described as in Theorem 3.12 is not hereditarily reflexive. We do not know whether the reflexivity of an operator algebra 𝒜⊆B(X) described as in Theorem 3.8 is independent of the reflexivity of the Banach space X. Here, is a relevant example.Example 4.7.
LetS be the unilateral backward shift on c0. That is,
(4.5)S:c0⟶c0,(ξ0,ξ1,ξ2,ξ3,…)↦(ξ1,ξ2,ξ3,…),wherex=(ξ0,ξ1,ξ2,ξ3,…)∈c0.It is obvious that
(4.6)S*:l1⟶l1,(η0,η1,η2,η3,…)↦(0,η0,η1,η2,…),wheref=(η0,η1,η2,η3,…)∈l1,S**:l∞⟶l∞,(ζ0,ζ1,ζ2,ζ3,…)↦(ζ1,ζ2,ζ3,…),wherez=(ζ0,ζ1,ζ2,ζ3,…)∈l∞.Then, (1) Both S and S** are not strictly cyclic. (2) S* is strictly cyclic on l1. (3) The weakly closed algebra 𝒜S* generated by S* is semisimple. Therefore, 𝒜S and 𝒜S** are also semisimple by Lemma 3.3. (4) 𝒜S* is a reflexive algebra, and 𝒜S* is hereditarily reflexive. (5) 𝒜S is a reflexive algebra. (6) 𝒜S is not hereditarily reflexive. (7) 𝒜S** is a reflexive algebra, but 𝒜S** is not hereditarily reflexive.Proof.
(2) Setβ0=1, βn=∏k=1nwk=1. For any positive integers m,n, we have that |βn/βmβn-m|=1. It follows that  supn,m|βn/βmβn-m|<∞. Since {|wn|} is monotonically decreasing, S* is strictly cyclic on l1 by [19] and f0=(1,0,0,0,…) is a strictly cyclic vector for 𝒜S*.
(3) Using the proof similar to the proof of (Example4.4(3)), we have that ∥S∥=∥S*∥=∥S**∥=1 and r(S)=r(S*)=r(S**)=1. Since S* is unilateral forward shift on l1, the spectra σ(S)(=σ(S*)=σ(S**)) of S contains infinite complex numbers. Then, we have that the Jacobson radical rad(𝒜S)={0}. Therefore, 𝒜S, 𝒜S* and 𝒜S** are semisimple.
(4)𝒜S* is reflexive by Lemma 3.7. 𝒜S* is hereditarily reflexive by Lemma 3.11.
(5) Suppose thatT∈B(c0) and Lat(S)⊆Lat(T). Since T*∈B(c0*) and S* has a strictly cyclic vector f0, then there exists T0∈𝒜S such that T*f0=T0*f0. For any fixed A∈𝒜S, let T*A*f0=B*f0.
For eachx∈c0, let Mx=span¯{x,Sx,S2x,…}, where the bar denotes the norm closure. Then, Mx∈Lat(S) and Mx∈Lat(T). We have that Lat(S|Mx)⊆Lat(T|Mx) for every x∈c0. It is obvious that S is not locally algebraic, and there is x∈c0 such that {x,Sx,S2x,…} is linearly independent. Let x0∈c0 such that {x0,Sx0,S2x0,…} is linearly independent. Then, the algebra 𝒜S|Mx0={A|Mx0:A∈𝒜S} has a separating vector x0.
Letφ be an arbitrary homomorphism from 𝒜S to the complex field ℂ and let 𝒦φ={A∈𝒜S:φ(A)=0} be the corresponding maximal ideal of 𝒜S. Define φ*:𝒜S*→ℂ by φ*(A*)=φ(A)(A∈𝒜S). Then, 𝒦φ*={A*∈𝒜S*:φ(A)=0} be the corresponding maximal ideal of 𝒜S*. For any x∈c0, let Nx=𝒦φx={A(x):A∈𝒦φ}. Then, Nx⊆Mx and Nx∈Lat(S)⊆Lat(T). So T(Nx)⊆Nx.
It is obvious thatφ(A)I-A∈𝒦φ and φ(A)x-Ax∈Nx. It follows that φ(A)Tx-TAx∈Nx⊆Mx. In particular, φ(A)Tx0-TAx0∈Mx0. Since x0 is a separating vector for 𝒦φ, we have that φ(A)T-TA∈𝒦φ. Then, φ*(A*)T*-T*A*∈𝒦φ* and φ*(A*)T*(f0)-T*A*(f0)∈𝒦φ*(f0), that is, φ*(A*)T0*(f0)-B*(f0)∈𝒦φ*(f0). So, φ*(A*)T0*-B*∈𝒦φ* and φ(A)T0-B∈𝒦φ. Thus, φ(φ(A)T0-B)=φ(A)φ(T0)-φ(B)=φ(T0A-B)=φ(AT0-B)=0. Since f0 is a separating vector for 𝒦φ*, T0A-B∈𝒦φ and AT0-B∈𝒦φ. Since φ is arbitrary and 𝒜S is semisimple, it follows that B=T0A=AT0, that is, T*A*f0=(AT0)*f0=T0*A*f0 for any A∈𝒜S. Since 𝒜S*f0=X*, T*=T0*, and T=T0∈𝒜S. It follows that 𝒜S is a reflexive algebra.
(6) By Lemma3.14, the dual space of 𝒜S consists entirely of the maps A↦F(A*f0)(A∈𝒜S), where F∈l∞.
Since𝒜S* is strictly cyclic, 𝒜S* and 𝒜S are norm-closed, then 𝒜S* and l1 are isomorphic as Banach spaces. Thus, 𝒜S and l1 are isomorphic as Banach spaces. It follows that the set of all weakly continuous linear functionals on 𝒜S is the set of all norm-continuous linear functionals on 𝒜S.
Similar to Proposition3.13, 𝒜S is hereditarily reflexive if and only if it is reflexive and every weakly continuous functional on 𝒜S is of the form A↦f(A(x))(A∈𝒜S), where x∈c0 and f∈l1. Since the Banach space l1 is not reflexive, there exists F∈l∞ such that F∉J(c0). It follows that 𝒜S is not hereditarily reflexive by the assertion.
(7) Using the proof similar to the proof of (5), we can prove that𝒜S** is a reflexive algebra. Using the proof similar to the proof of (6), we can prove that 𝒜S** is not hereditarily reflexive.



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*Source: 434308-2012-05-09.xml*