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

Abstract

This paper is concerned with strictly cyclic functionals of operator algebras on Banach spaces. It is shown that if X 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.

1. Introduction

Throughout this paper, 𝑋 is a complex Banach space, 𝐵_𝑋 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 if𝒜=AlgLat𝒜,whereAlgLat𝒜={𝑇∈𝐵(𝑋)∶Lat𝒜⊂Lat𝑇}. (1.1)

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)forall𝑥∈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.

Let 𝒜 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 𝛽_𝑓0 of 𝒜 to 𝑋^∗ by 𝛽_𝑓0(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 Section 2, 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. Let 𝑋 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 𝒮^∗𝑓₀=𝐴^∗𝑓₀∶𝐴∈𝒮=𝑓₀∘𝐴∶𝐴∈𝒮 (2.1) is norm-dense in 𝑋^∗, then 𝑓₀ is called a cyclic functional for 𝒮. If there is a functional 𝑓₀∈𝑋^∗ such that 𝒮^∗𝑓₀=𝐴^∗𝑓₀∶𝐴∈𝒮=𝑓₀∘𝐴∶𝐴∈𝒮=𝑋^∗, (2.2) then 𝑓₀ is called a strictly cyclic functional for 𝒮.

Lemma 2.2 (Open Mapping Lemma). Suppose that 𝑋 is a Banach space and 𝑌 is a normed linear space, and 𝜀_𝑓0∶𝑋→𝑌 is a bounded operator. Assume that there exist 𝑘>0 and 0<𝜂<1 such that 𝐵_𝑌⊆𝑘⋅𝜀_𝑓0𝐵_𝑋+𝜂𝐵_𝑌. (2.3) Then, 𝜀_𝑓0 is a surjective open map, and 𝑌 is complete.

Proposition 2.3. Let 𝒜 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 constant 𝐶>0 such that for all 𝑔∈𝐵_𝑋∗ there is an operator 𝐴∈𝒜 with ‖𝐴‖⩽𝐶 such that 𝐴^∗(𝑓₀)=𝑔;(3) there are 𝐶>0 and 0<𝜂<1 such that for all 𝑔∈𝐵_𝑋∗ there is an operator 𝐴∈𝒜 with ‖𝐴‖⩽𝐶 such that ‖𝑔−𝐴^∗(𝑓₀)‖<𝜂.

Proof. Define 𝜀_𝑓0∶𝒜→𝑋^∗ by 𝜀_𝑓0(𝐴)=𝐴^∗(𝑓₀) for 𝐴∈𝒜, then 𝜀_𝑓0 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. Let 𝒜 be a norm-closed subalgebra of 𝐵(𝑋) with a strictly cyclic functional 𝑓₀∈𝑋^∗.
(1) Then, every cyclic functional 𝑓 for 𝒜 is actually a strictly cyclic functional for 𝒜.(2) Moreover, the set of all (strictly) cyclic functionals for 𝒜 is norm-open in 𝑋^∗ and coincides with 𝒞_𝒜=𝐴^∗𝑓₀∶𝐴∈𝒜and𝐴^∗hasaleftinversemodker𝜀_𝑓0𝑖𝑛𝒜^∗, (2.4) where 𝜀_𝑓0 is the map from 𝒜 onto 𝑋^∗ by 𝜀_𝑓0(𝐴)=𝐴^∗(𝑓₀) for every 𝐴∈𝒜, and ker(𝜀_𝑓0) is the kernel of the map 𝜀_𝑓0.

Proof. (1) If 𝑓 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.
Let 𝑓_𝑛 be a sequence in 𝑁 such that {𝑓_𝑛} converges uniformly to 𝑓₀. By Proposition 2.3 (3), we may assume the existence of sequence 𝐴_𝑛 such that 𝐴_𝑛^∗(𝑓₀)=𝑓₀−𝑓_𝑛 and lim_𝑛→∞‖𝐴_𝑛‖=0. Thus, for 𝑛 sufficiently large, ‖𝐴_𝑛^∗‖<1 and (𝐼−𝐴_𝑛^∗)⁻¹=∑^∞_𝑘=0(𝐴_𝑛^∗)^𝑘=∑^∞_𝑘=0(𝐴_𝑛^𝑘)^∗. Therefore, (𝐼−𝐴_𝑛^∗)⁻¹∈𝒜^∗ and since (𝐼−𝐴_𝑛^∗)𝑓₀=𝑓_𝑛, we have 𝑓₀=(𝐼−𝐴_𝑛^∗)⁻¹𝑓_𝑛∈𝑁.
(2) Choose 𝐶 as in Proposition 2.3 (2). Let 0<𝛿<1/𝐶 and suppose that 𝑓∈𝑋^∗ with ‖𝑓−𝑓₀‖<𝛿. Now if 𝑔∈𝐵_𝑋∗, 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, if 𝑔=𝐴^∗(𝑓₀)∈𝒞_𝒜, then there exists 𝐵∈𝒜 such that 𝐵^∗𝐴^∗=𝐼+𝐿 for some 𝐿∈ker(𝜀_𝑓0). 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, if 𝑔=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(𝜀_𝑓0) and 𝐵^∗𝐴^∗=𝐼−𝐿. Therefore, 𝐵^∗ is a left inverse of 𝐴^∗ module ker(𝜀_𝑓0). Hence, 𝑔=𝐴^∗(𝑓₀)∈𝒞_𝒜.

Remark 2.5. If 𝒜 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 𝐴^∗₀(𝑓₀)=𝑓₀. Then, 𝐴^∗𝐴^∗₀(𝑓₀)=𝐴^∗(𝑓₀) for every 𝐴∈𝒜. Hence, by commutativity of 𝒜, 𝐴^∗₀𝐴^∗(𝑓₀)=𝐴^∗(𝑓₀) for every 𝐴∈𝒜. Therefore, 𝐴^∗₀ is the identity operator on 𝑋^∗ and 𝐴₀=𝐼∈𝒜.

3. Hereditary Reflexivity and Strictly Cyclic Functional

Remark 3.1. Let 𝑋 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 that 𝑋 is a Banach space which is not a Hilbert space. If 𝑇∈𝐵(𝑋), then 𝜎(𝑇^∗)=𝜎(𝑇).
(2) Suppose that 𝑋 is a Hilbert space. If 𝑇∈𝐵(𝑋), then 𝜎(𝑇^∗)={𝜆∶𝜆∈𝜎(𝑇)}.

Lemma 3.3. Suppose that 𝑋 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 that 𝒜 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.

Let 𝑋 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 that 𝑋 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, if 𝑋 is a reflexive Banach space, then (^⟂𝑁)^⟂=𝑁.

Proof. (1) It is clear.
(2) Let 𝑋 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 exists 𝑥^∗₀∈(^⟂𝑁)^⟂ but 𝑥^∗₀∉𝑁. Then, there exists 𝑥^∗∗∈𝑋^∗∗ such that 𝑥^∗∗(𝑥^∗₀)=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 𝑥^∗₀(𝑥)=1 and 𝑓(𝑥)=0 for any 𝑓∈𝑁. It follows that 𝑥^∗₀(𝑥)=1 and 𝑥^∗₀∉(^⟂𝑁)^⟂, which is a contradiction to 𝑥^∗₀∈(^⟂𝑁)^⟂. The proof is complete.

Lemma 3.5. Let 𝐴 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, if 𝑁∈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. Let 𝑋 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. Let 𝒜^∗ 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 that 𝑋 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. Let 𝑋 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. If 𝑓 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 subalgebra 𝒜 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. Let 𝑋 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. If 𝑋 is a reflexive Banach space, then we only need to show that the hereditary reflexivity of 𝒜 implies the hereditary reflexivity of 𝒜^∗.
Let 𝒜 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 that 𝑋 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. Let 𝑋 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 Theorem 3.8.

The following proposition is similar to Theorem  2.3 in [1, 2].

Proposition 3.13. Let 𝑋 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. If 𝒮 is hereditarily reflexive, then 𝒮 is reflexive by definition. Let 𝐹 be a weakly continuous functional on 𝒮 and let 𝒦_𝐹=ker𝐹, then 𝒦_𝐹 is a reflexive subspace. Choose 𝐵∈𝒮 such that 𝐹(𝐵)=1. Since 𝐵∉𝒦_𝐹=Ref𝒦_𝐹, there exists a vector 𝑥∈𝑋 such that ‖(𝐵−𝐴₀)(𝑥)‖>1=𝐹(𝐵) for all 𝐴₀∈𝒦_𝐹. Since 𝐹(𝐵)=𝐹(𝐵−𝐴₀) for any 𝐴₀∈𝒦_𝐹, 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. Let 𝐹 be a weakly continuous functional on 𝒮 such that 𝐹(𝐴)=𝑓(𝐴(𝑥)) for all 𝐴∈𝒮. Clearly, 𝒦_𝐹=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. Let 𝑋 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 any 𝐹∈𝑋^∗∗, the map 𝐴↦𝐹(𝐴^∗𝑓₀)(𝐴∈𝒜) defines a continuous linear functional on 𝒜.
Conversely, suppose that 𝑔 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:𝒜⟶^𝐶𝒜⟶^∗𝜌𝑋^∗,𝐴⟶^𝐶𝐴⟶^∗𝜌𝐴^∗𝑓₀. (3.1)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. Let 𝑋 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. If 𝑋 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 𝐴_𝑓∈𝒜 such that 𝐴^∗_𝑓𝑓₀=𝑓. Then, 𝑓(𝐴(𝑥))=𝐴^∗𝑓(𝑥)=𝐴^∗𝐴^∗_𝑓𝑓₀(𝑥)=𝐴_𝑓𝐴^∗𝑓₀(𝑥)=𝑓₀𝐴𝐴_𝑓𝑥=𝑓₀𝐴𝑥₁, (3.2)
where 𝑥₁=𝐴_𝑓𝑥∈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 Proposition 3.13, 𝒜_𝑆 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 that 1<𝑝<∞ and 1/𝑝+1/𝑞=1. Let 𝑙^𝑝 be the Banach space of all absolutely 𝑝-summable sequences 𝑥={𝜉_𝑛}^∞_𝑛=0 of complex numbers with ‖𝑥‖=(∑^∞_𝑛=0|𝜉_𝑛|^𝑝)^1/𝑝. Let 𝑐₀ be the space of sequences 𝑥=(𝜉_𝑛)^∞_𝑛=0 such that lim_𝑛→∞𝜉_𝑛=0 with the norm ‖𝑥‖=sup_𝑛{|𝜉_𝑛|}. {𝑒_𝑛}^∞_𝑛=0 is the natural basis for 𝑙^𝑝 or 𝑐₀, that is, 𝑒_𝑛=(0,0,…,0,1,0,…) with 1 as the 𝑛th component (beginning the indexing at 0). For a sequence {𝑤_𝑛}^∞_𝑛=1 of nonzero complex numbers, let 𝑇 be the linear transformation on 𝑙^𝑝 defined by𝑇𝑒₀=0,𝑇𝑒_𝑛=𝑤_𝑛𝑒_𝑛−1,for𝑛≥1. (4.1)

Obviously 𝑇 is a unilateral weighted backward shift with weight sequence {𝑤_𝑛}^∞_𝑛=1.

Let {𝑓_𝑚}^∞_𝑚=0 be the natural basis for 𝑙^𝑞, that is, 𝑓_𝑚=(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^∗(𝑓_𝑚)=𝑤_𝑚+1𝑓_𝑚+1 for all 𝑚=0,1,2,…. Obviously 𝑇^∗ is a unilateral weighted forward shift with weight sequence {𝑤_𝑛}^∞_𝑛=1.

For an operator 𝑇 on a Banach space 𝑋, denote by 𝒜_𝑇 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 𝒜_𝑇 is strictly cyclic.

Lemma 4.1. Let 𝑋 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 𝑘₁‖𝐴‖⩽‖𝐴𝑥‖⩽𝑘₂‖𝐴‖ (4.2)for any 𝐴∈𝒜.

Proof. Let 𝑥 be a strictly cyclic separating vector for 𝒜. The evaluation map 𝜀_𝑥∶𝐴↦𝐴𝑥 is bounded below and has dense range (as a mapping from the algebra 𝒜 onto 𝑋). It follows that 𝜀_𝑥 and 𝜀_𝑥⁻¹ are both continuous. The result follows.

Remark 4.2. If 𝑇 is strictly cyclic, then the norm topology and the strong topology on 𝒜_𝑇 coincide. If 𝑇 has a strictly cyclic functional, then the norm topology and the strong topology on 𝒜^∗_𝑇 coincide. Since 𝐼∈𝒜^∗_𝑇, 𝒜^∗_𝑇 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 shift 𝑇 on 𝑙^𝑝(1<𝑝<∞) or 𝑐₀ with nonzero weights is not strictly cyclic.

Example 4.4. Let 𝑇 be a unilateral weighted backward shift on 𝑙^𝑝(1<𝑝<∞) with weight sequence {𝑤_𝑛}^∞_𝑛=1: 𝑤_𝑛=(𝑛+1)/𝑛 for all 𝑛⩾1. Then,
(1) 𝑇 is not strictly cyclic;(2) 𝑇^∗ is strictly cyclic on 𝑙^𝑞;(3) The weakly closed algebra 𝒜_𝑇∗ generated by 𝑇^∗ is semisimple. Therefore, 𝒜_𝑇 is also semisimple by Lemma 3.3;(4) 𝒜_𝑇∗ is hereditarily reflexive, and it follows that 𝒜_𝑇 is hereditarily reflexive from Theorem 3.12.

Proof. (2) Set 𝛽₀=1, 𝛽_𝑛=∏^𝑛_𝑘=1𝑤_𝑘=(2/1)⋅(3/2)⋯(𝑛+1)/𝑛=𝑛+1^𝑛_𝑚=0𝛽_𝑛𝛽_𝑚𝛽_𝑛−𝑚^𝑝=1+^𝑛_𝑚=0𝑛+1(𝑚+1)⋅(𝑛−𝑚+1)^𝑝()=1+𝑛+1^𝑝(𝑛+2)^𝑝𝑛_𝑚=11+1𝑚+1𝑛−𝑚+1^𝑝≤1+^𝑛_𝑚=11+1𝑚+1𝑛−𝑚+1^𝑝≤1+^𝑛_𝑚=12^𝑝1(𝑚+1)^𝑝+1(𝑛−𝑚+1)^𝑝≤1+2^∞𝑝−1_𝑚=11𝑚^𝑝=𝐶. (4.3) It follows that sup_𝑛∑^𝑛_𝑚=0|𝛽_𝑛/𝛽_𝑚𝛽_𝑛−𝑚|^𝑝<∞. Since {|𝑤_𝑛|} is monotonically decreasing, 𝑇^∗ is strictly cyclic on 𝑙^𝑞 by [17, Theorem  3.2].
(3) It is easy to see that ‖𝑇^𝑛‖=‖(𝑇^∗)^𝑛‖=sup_𝑘∈ℕ|∏^𝑛−1_𝑖=0𝑤_𝑘+𝑖|=sup_𝑘∈ℕ(1+𝑛/𝑘)=1+𝑛. Then, ‖𝑇‖=‖𝑇^∗‖=2 and the spectral radius 𝑟(𝑇)=𝑟(𝑇^∗)=lim_𝑛→∞‖𝑇^𝑛‖^1/𝑛=lim_𝑛→∞(1+𝑛)^1/𝑛=1. Since 𝑇^∗ is unilateral forward shift on 𝑙^𝑞, the spectra 𝜎(𝑇)(=𝜎(𝑇^∗)) of 𝑇 contains infinite complex numbers. Notice that the elements of 𝒜^∗_𝑇 consist of all 𝑃(𝑇^∗)s and their norm limits [18], where 𝑃(⋅) are all polynomials. Then, the elements of 𝒜_𝑇 consist of all 𝑃(𝑇)s and their norm limits, where 𝑃(⋅) are all polynomials. Then, we have that the Jacobson radical rad(𝒜_𝑇)={0}.
In fact, since 𝒜_𝑇 is unital,rad𝒜_𝑇=𝐴∈𝒜_𝑇∶𝐴𝐵isquasinilpotentforany𝐵∈𝒜_𝑇. (4.4)For any 𝐴∈𝒜_𝑇, there exists a sequence of polynomials 𝑄_𝑛(𝜆) such that ‖𝑄_𝑛(𝑇)−𝐴‖→0. Let 𝐴₁=𝑃₁(𝑇),𝐴₂=𝑃₂(𝑇)∈𝒜_𝑇, 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(𝒜_𝑇), then 𝜎(𝐴)⋅𝑃₂(𝜎(𝑇))={0} for any polynomial 𝑃₂(𝜆), and 𝜎(𝐴)={0} since 𝜎(𝑇) contains infinite complex numbers. Thus, 𝑄_𝑛(𝜎(𝑇))→0 and 𝑄_𝑛(𝑇)→0, that is, 𝐴=0. So 𝒜_𝑇 is semisimple.

Remark 4.5. Let 𝑇 be a unilateral weighted backward shift on the reflexive Banach space 𝑙^𝑝 as in Example 4.4. Then, the weakly closed algebra 𝒜_𝑇, generated by 𝑇 and 𝐼, is norm-closed, semisimple, and abelian. And 𝒜_𝑇 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 space 𝑋 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. Let 𝑆 be the unilateral backward shift on 𝑐₀. That is, 𝑆∶𝑐₀⟶𝑐₀,𝜉₀,𝜉₁,𝜉₂,𝜉₃,…↦𝜉₁,𝜉₂,𝜉₃,…,where𝑥=𝜉₀,𝜉₁,𝜉₂,𝜉₃,…∈𝑐₀. (4.5)It is obvious that 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^∞. (4.6)Then, (1) Both 𝑆 and 𝑆^∗∗ are not strictly cyclic. (2) 𝑆^∗ is strictly cyclic on 𝑙¹. (3) The weakly closed algebra 𝒜_𝑆∗ generated by 𝑆^∗ is semisimple. Therefore, 𝒜_𝑆 and 𝒜_𝑆∗∗ are also semisimple by Lemma 3.3. (4) 𝒜_𝑆∗ is a reflexive algebra, and 𝒜_𝑆∗ is hereditarily reflexive. (5) 𝒜_𝑆 is a reflexive algebra. (6) 𝒜_𝑆 is not hereditarily reflexive. (7) 𝒜_𝑆∗∗ is a reflexive algebra, but 𝒜_𝑆∗∗ is not hereditarily reflexive.

Proof. (2) Set 𝛽₀=1, 𝛽_𝑛=∏^𝑛_𝑘=1𝑤_𝑘=1. For any positive integers 𝑚,𝑛, we have that |𝛽_𝑛/𝛽_𝑚𝛽_𝑛−𝑚|=1. It follows that sup_𝑛,𝑚|𝛽_𝑛/𝛽_𝑚𝛽_𝑛−𝑚|<∞. Since {|𝑤_𝑛|} is monotonically decreasing, 𝑆^∗ is strictly cyclic on 𝑙¹ by [19] and 𝑓₀=(1,0,0,0,…) is a strictly cyclic vector for 𝒜_𝑆∗.
(3) Using the proof similar to the proof of (Example 4.4(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(𝒜_𝑆)={0}. Therefore, 𝒜_𝑆, 𝒜_𝑆∗ and 𝒜_𝑆∗∗ are semisimple.
(4) 𝒜_𝑆∗ is reflexive by Lemma 3.7. 𝒜_𝑆∗ is hereditarily reflexive by Lemma 3.11.
(5) Suppose that 𝑇∈𝐵(𝑐₀) and Lat(𝑆)⊆Lat(𝑇). Since 𝑇^∗∈𝐵(𝑐^∗₀) and 𝑆^∗ has a strictly cyclic vector 𝑓₀, then there exists 𝑇₀∈𝒜_𝑆 such that 𝑇^∗𝑓₀=𝑇^∗₀𝑓₀. For any fixed 𝐴∈𝒜_𝑆, let 𝑇^∗𝐴^∗𝑓₀=𝐵^∗𝑓₀.
For each 𝑥∈𝑐₀, let 𝑀_𝑥=span{𝑥,𝑆𝑥,𝑆²𝑥,…}, where the bar denotes the norm closure. Then, 𝑀_𝑥∈Lat(𝑆) and 𝑀_𝑥∈Lat(𝑇). We have that Lat(𝑆|𝑀_𝑥)⊆Lat(𝑇|𝑀_𝑥) 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 𝒜_𝑆|𝑀_𝑥0={𝐴|𝑀_𝑥0∶𝐴∈𝒜_𝑆} has a separating vector 𝑥₀.
Let 𝜑 be an arbitrary homomorphism from 𝒜_𝑆 to the complex field ℂ and let 𝒦_𝜑={𝐴∈𝒜_𝑆∶𝜑(𝐴)=0} be the corresponding maximal ideal of 𝒜_𝑆. Define 𝜑^∗∶𝒜^∗_𝑆→ℂ by 𝜑^∗(𝐴^∗)=𝜑(𝐴)(𝐴∈𝒜_𝑆). Then, 𝒦^∗_𝜑={𝐴^∗∈𝒜^∗_𝑆∶𝜑(𝐴)=0} be the corresponding maximal ideal of 𝒜^∗_𝑆. For any 𝑥∈𝑐₀, let 𝑁_𝑥=𝒦_𝜑𝑥={𝐴(𝑥)∶𝐴∈𝒦_𝜑}. Then, 𝑁_𝑥⊆𝑀_𝑥 and 𝑁_𝑥∈Lat(𝑆)⊆Lat(𝑇). So 𝑇(𝑁_𝑥)⊆𝑁_𝑥.
It is obvious that 𝜑(𝐴)𝐼−𝐴∈𝒦_𝜑 and 𝜑(𝐴)𝑥−𝐴𝑥∈𝑁_𝑥. It follows that 𝜑(𝐴)𝑇𝑥−𝑇𝐴𝑥∈𝑁_𝑥⊆𝑀_𝑥. In particular, 𝜑(𝐴)𝑇𝑥₀−𝑇𝐴𝑥₀∈𝑀_𝑥0. Since 𝑥₀ is a separating vector for 𝒦_𝜑, we have that 𝜑(𝐴)𝑇−𝑇𝐴∈𝒦_𝜑. Then, 𝜑^∗(𝐴^∗)𝑇^∗−𝑇^∗𝐴^∗∈𝒦^∗_𝜑 and 𝜑^∗(𝐴^∗)𝑇^∗(𝑓₀)−𝑇^∗𝐴^∗(𝑓₀)∈𝒦^∗_𝜑(𝑓₀), that is, 𝜑^∗(𝐴^∗)𝑇^∗₀(𝑓₀)−𝐵^∗(𝑓₀)∈𝒦^∗_𝜑(𝑓₀). So, 𝜑^∗(𝐴^∗)𝑇^∗₀−𝐵^∗∈𝒦^∗_𝜑 and 𝜑(𝐴)𝑇₀−𝐵∈𝒦_𝜑. 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 𝒦^∗_𝜑, 𝑇₀𝐴−𝐵∈𝒦_𝜑 and 𝐴𝑇₀−𝐵∈𝒦_𝜑. Since 𝜑 is arbitrary and 𝒜_𝑆 is semisimple, it follows that 𝐵=𝑇₀𝐴=𝐴𝑇₀, that is, 𝑇^∗𝐴^∗𝑓₀=(𝐴𝑇₀)^∗𝑓₀=𝑇^∗₀𝐴^∗𝑓₀ for any 𝐴∈𝒜_𝑆. Since 𝒜^∗_𝑆𝑓₀=𝑋^∗, 𝑇^∗=𝑇^∗₀, and 𝑇=𝑇₀∈𝒜_𝑆. It follows that 𝒜_𝑆 is a reflexive algebra.
(6) By Lemma 3.14, the dual space of 𝒜_𝑆 consists entirely of the maps 𝐴↦𝐹(𝐴^∗𝑓₀)(𝐴∈𝒜_𝑆), where 𝐹∈𝑙^∞.
Since 𝒜_𝑆∗ is strictly cyclic, 𝒜_𝑆∗ and 𝒜_𝑆 are norm-closed, then 𝒜_𝑆∗ and 𝑙¹ are isomorphic as Banach spaces. Thus, 𝒜_𝑆 and 𝑙¹ are isomorphic as Banach spaces. It follows that the set of all weakly continuous linear functionals on 𝒜_𝑆 is the set of all norm-continuous linear functionals on 𝒜_𝑆.
Similar to Proposition 3.13, 𝒜_𝑆 is hereditarily reflexive if and only if it is reflexive and every weakly continuous functional on 𝒜_𝑆 is of the form 𝐴↦𝑓(𝐴(𝑥))(𝐴∈𝒜_𝑆), 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 𝒜_𝑆 is not hereditarily reflexive by the assertion.
(7) Using the proof similar to the proof of (5), we can prove that 𝒜_𝑆∗∗ is a reflexive algebra. Using the proof similar to the proof of (6), we can prove that 𝒜_𝑆∗∗ is not hereditarily reflexive.

Acknowledgments

The authors thank deeply the reviewers that led us to the articles [6, 7]. The authors would also like to express their hearty thanks to the referees for their very helpful comments and suggestions which improved the article greatly. Q. Chen would like to express deep gratitude to Professor Don Hadwin and Changli Tao for their hearty help during the work. This paper was supported by the National Nature Science Foundation of China (Grant 11071188).