Abstract and Applied AnalysisVolume 2012 (2012), Article ID 434308, 12 pagesdoi:10.1155/2012/434308

Research Article

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

Quanyuan Chen1,2 and Xiaochun Fang¹

¹Department of Mathematics, Tongji University, Shanghai 200092, China²College of Information, Jingdezhen Ceramic Institute, Jingdezhen, Jiangxi 333403, China

Received 25 September 2011; Revised 4 March 2012; Accepted 14 March 2012

Academic Editor: Gerd Teschke

Copyright © 2012 Quanyuan Chen and Xiaochun Fang. 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.

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 is a norm-closed semisimple abelian subalgebra of B(X) with a strictly cyclic functional , then 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, is a complex Banach space, stands for the closed unit ball of and is the dual space of , the space of all continuous linear functionals on . For a bounded operator on , denote by the lattice of all closed invariant subspaces of and the adjoint operator of . For a subalgebra of , the Banach algebra of all bounded linear operators on , denote by ) the lattice of all closed subspaces invariant under every operator in . For a set of subspaces of , denote by Alg the algebra of all operators in which leave all subspaces in invariant. An operator algebra is reflexive if

In [1, 2], Loginov and Sulman introduced the following notion of a reflexive subspace of . If is a subspace (linear manifold) of , we denote by the set , where the bar denotes norm closure. The operator subspace is reflexive if . If is a subalgebra of containing the identity operator , then , and is reflexive as an operator subspace if and only if 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 is reflexive if and only if it is equivalent to for some subspace lattice of . 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 be a subalgebra of . If there is a vector such that the map of to by is injective on , then the vector is called a separating vector for . Let . If the map of to by is injective on , then the functional is called a separating functional for . Furthermore, if the subset is norm-closed in , then the functional is called a strictly separating functional for .

is called cyclic if is norm-dense in for some vector in . is called strictly cyclic if for some vector in . The vector is called cyclic vector for 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 is strictly cyclic, provided that has a strictly cyclic functional. Moreover, we construct a semisimple single-generated algebra 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 with a strictly cyclic functional is reflexive and hereditarily reflexive when 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 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 be a Banach space and let be a subalgebra or a subspace of . Let be the dual space of . If there is a functional such that is norm-dense in , then is called a cyclic functional for . If there is a functional such that then is called a strictly cyclic functional for .

Lemma 2.2 (Open Mapping Lemma). Suppose that is a Banach space and is a normed linear space, and is a bounded operator. Assume that there exist and such that Then, is a surjective open map, and is complete.

Proposition 2.3. Let be a norm-closed subalgebra of and with . Then, the following are equivalent: (1) is a strictly cyclic functional for ;(2) there exists a constant such that for all there is an operator with such that ;(3) there are and such that for all there is an operator with such that .

Proof. Define by for , then is a bounded linear operator from to . Since 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 be a norm-closed subalgebra of with a strictly cyclic functional . (1) Then, every cyclic functional for is actually a strictly cyclic functional for .(2) Moreover, the set of all (strictly) cyclic functionals for is norm-open in and coincides with where is the map from onto by for every , and is the kernel of the map .

Proof. (1) If is a cyclic functional for , then is a norm-dense linear manifold of and is -invariant. We will show that and thus .Let be a sequence in such that converges uniformly to . By Proposition 2.3 (3), we may assume the existence of sequence such that and . Thus, for sufficiently large, and . Therefore, and since , we have .(2) Choose as in Proposition 2.3 (2). Let and suppose that with . Now if , we choose with and . But then, . Hence, is a strictly cyclic functional for the algebra by Proposition 2.3 (3). Thus, the set of all (strictly) cyclic functionals for is uniformly open in .Clearly, if , then there exists such that for some . Therefore, . Hence, , which implies that is a strictly cyclic functional for .Conversely, if is a cyclic (and, therefore, it is strictly cyclic) functional for , then . Therefore, there exists such that , that is, . Hence, and . Therefore, is a left inverse of module . Hence, .

Remark 2.5. If is a norm-closed commutative subalgebra of with a strictly cyclic functional , then . Indeed, let such that . Then, for every . Hence, by commutativity of , for every . Therefore, is the identity operator on and .

3. Hereditary Reflexivity and Strictly Cyclic Functional

Remark 3.1. Let be a Banach space and let be a subalgebra or a subspace of . Then, (1) is a strictly cyclic functional for if and only if is a strictly cyclic vector for .(2) is a cyclic functional for if and only if is a cyclic vector for .

Lemma 3.2. (1) Suppose that is a Banach space which is not a Hilbert space. If , then . (2) Suppose that is a Hilbert space. If , then .

Lemma 3.3. Suppose that is a Banach space and is an abelian subalgebra of . Set . Then, the Banach algebra is semisimple if and only if the Banach algebra is semisimple.

Proof. We may assume that is unital for convenience. Let be semisimple and (). Then, and there is a multiplicative linear functional on such that . Since is a multiplicative linear functional on by Gelfand transformation, contains a nonzero scalar and contains a nonzero scalar by Lemma 3.2. Since is a multiplicative linear functional on , it follows that there is a multiplicative linear functional on such that . Hence, is also semisimple. Similarly, if is semisimple, then is also semisimple.

Let be a Banach space, and let be a nonempty subset of and be a nonempty subset of . The annihilator of and the preannihilator of are defined as follows [16]: , . It is obvious that is a -closed subspace of and is a norm-closed subspace of . is the norm-closure of in and is the -closure of in .

Lemma 3.4. Suppose that is a Banach space and is a closed subspace of and is a norm-closed subspace of . Then, (1) =; (2) furthermore, if is a reflexive Banach space, then .

Proof. (1) It is clear.(2) Let be a reflexive Banach space and let be the natural imbedding map from into . First, we observe that is the -closure of in that includes , that is, .Now, we suppose that there exists but . Then, there exists such that and for any . Since there exists such that by the reflexivity of the Banach space , we have that and for any . It follows that and , which is a contradiction to . The proof is complete.

Lemma 3.5. Let be a bounded linear operator on a reflexive Banach space . Then, if and only if .

Proof. If , then, for any in and in , we have that and . Thus, by Lemma 3.4. Hence, .Conversely, if , then for any in and in , we have and . Thus . So .

Lemma 3.6. Let be a reflexive Banach space and let be a subalgebra of . Let . If is reflexive, then is a reflexive subalgebra.

Proof. Let be reflexive. Suppose that is a bounded operator leaving invariant all the closed invariant subspaces of , that is, . For each norm-closed invariant subspace () of , that is, , we have that. Since ,. It follows that by Lemma 3.5. Thus, . Since is reflexive, . It follows that and is reflexive.

Lemma 3.7 ([7, Theorem 8] and [1, 2]). Suppose that is a Banach space and is an abelian subalgebra of . If is a norm-closed, semisimple, strictly cyclic algebra, then is reflexive.

Theorem 3.8. Let be a reflexive Banach space and let be an abelian subalgebra of with a strictly cyclic functional . If is a norm-closed semisimple algebra, then is reflexive.

Proof. If is a strictly cyclic functional for , then is a strictly cyclic vector for by Remark 3.1. Since is semisimple, is also semisimple by Lemma 3.3. It follows that is reflexive by Lemma 3.7. Thus, is reflexive by Lemma 3.6.

Definition 3.9 ([1, 2, 12]). A norm-closed subalgebra of is called hereditarily reflexive if, for every weakly closed subspace , every and , implies . Here, denotes the norm closure of in .

Lemma 3.10. Let be a reflexive Banach space and let be a unital abelian subalgebra of . Let . Then, is hereditarily reflexive if and only if is hereditarily reflexive.

Proof. If is a reflexive Banach space, then we only need to show that the hereditary reflexivity of implies the hereditary reflexivity of .Let be hereditarily reflexive. Let be a norm closed subspace of and . We suppose that for every continuous functional . Then, for any given , . Here, denotes the norm closure of in . Let . Then, for any . Since for any , we have that . It follows that by Lemma 3.4. Thus, by the hereditary reflexivity of . It follows that and are hereditarily reflexive.

Lemma 3.11 ([10, Theorem 10], and [1, 2]). Suppose that is a Banach space and is an abelian subalgebra of . If is a norm-closed, semisimple, strictly cyclic subalgebra, then is hereditarily reflexive.

Theorem 3.12. Let be a reflexive Banach space and let be an abelian subalgebra of with a strictly cyclic functional . If is a norm-closed semisimple algebra, then 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 be a Banach space. A subspace is hereditarily reflexive if and only if it is reflexive and every weakly continuous functional on is of the form , where and .

Proof. If is hereditarily reflexive, then is reflexive by definition. Let be a weakly continuous functional on and let , then is a reflexive subspace. Choose such that . Since , there exists a vector such that for all . Since for any , by linearity we have the inequality for all . Define the linear functional on by (). Then, , that is, is well defined and is continuous; thus there exists such that for all .We now prove sufficiency. Let be a weakly continuous functional on such that for all . Clearly, is a reflexive operator subspace. By Hahn-Banach Theorem, every weakly closed subspace of is the intersection of the kernels of weakly continuous functionals. Since the intersection of any family of reflexive subspaces is reflexive, we have that is hereditarily reflexive.

Lemma 3.14. Let be a Banach space and let be an abelian subalgebra of with a strictly cyclic functional . Then, the dual space of consists entirely of the maps , where .

Proof. For any , the map () defines a continuous linear functional on .Conversely, suppose that is a continuous linear functional on . Define the map by () and define map by () (see the graph below). Notice that both map and map are inverse:Then is a continuous linear functional on . It follows that there exists such that , where and . Thus, .

Theorem 3.15. Let be a Banach space and let be a semisimple abelian subalgebra of with a strictly cyclic functional . Then, is hereditarily reflexive if and only if is a reflexive Banach space.

Proof. If is a reflexive Banach space, then is hereditarily reflexive by Theorem 3.12. Conversely, if is hereditarily reflexive, then by Proposition 3.13, it is reflexive and every weakly continuous functional on is of the form , where and . Notice that is a strictly cyclic commutative subalgebra. For any , there exists such that . Then, where . It follows that every weakly continuous functional on is of the form , where . and are norm-closed. Then, and are isomorphic as Banach spaces. Thus, and are isomorphic as Banach spaces. It follows that the set of all weakly continuous linear functionals on is the set of all norm-continuous linear functionals on , and by Lemma 3.14, they are of the form , where .By Proposition 3.13, is hereditarily reflexive if and only if it is reflexive and every weakly continuous functional on is of the form , where and . It follows that is reflexive.

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

Throughout this section, we assume that and . Let be the Banach space of all absolutely -summable sequences of complex numbers with . Let be the space of sequences such that with the norm . is the natural basis for or , that is, with 1 as the th component (beginning the indexing at 0). For a sequence of nonzero complex numbers, let be the linear transformation on defined by

Obviously is a unilateral weighted backward shift with weight sequence .

Let be the natural basis for , that is, with 1 as the th component (beginning the indexing at 0). A short computation shows that the adjoint operator of is defined by for all . Obviously is a unilateral weighted forward shift with weight sequence .

For an operator on a Banach space , denote by the weakly closed subalgebra generated by and the identity operator . An operator on a Banach space is called strictly cyclic if the weakly closed subalgebra is strictly cyclic.

Lemma 4.1. Let be a Banach space and let be a subspace of . If has a strictly cyclic separating vector , then there exist and such that for any .

Proof. Let be a strictly cyclic separating vector for . The evaluation map is bounded below and has dense range (as a mapping from the algebra onto ). It follows that and are both continuous. The result follows.

Remark 4.2. If is strictly cyclic, then the norm topology and the strong topology on coincide. If has a strictly cyclic functional, then the norm topology and the strong topology on coincide. Since , is also the strongly closed subalgebra generated by and .It is easy to check the following statement.

Proposition 4.3. Any unilateral weighted backward shift on or with nonzero weights is not strictly cyclic.

Example 4.4. Let be a unilateral weighted backward shift on with weight sequence : for all . Then,(1) is not strictly cyclic;(2) is strictly cyclic on ;(3) The weakly closed algebra generated by is semisimple. Therefore, is also semisimple by Lemma 3.3;(4) is hereditarily reflexive, and it follows that is hereditarily reflexive from Theorem 3.12.

Proof. (2) Set , It follows that . Since is monotonically decreasing, is strictly cyclic on by [17, Theorem  3.2].(3) It is easy to see that . Then, and the spectral radius . Since is unilateral forward shift on , the spectra of contains infinite complex numbers. Notice that the elements of consist of all s and their norm limits [18], where are all polynomials. Then, the elements of consist of all s and their norm limits, where are all polynomials. Then, we have that the Jacobson radical .In fact, since is unital,For any , there exists a sequence of polynomials such that . Let , where are polynomials. It is obvious that by Spectral Theorem. If is quasinilpotent, then . It follows that, if , then for any polynomial , and since contains infinite complex numbers. Thus, and , that is, . So is semisimple.

Remark 4.5. Let be a unilateral weighted backward shift on the reflexive Banach space as in Example 4.4. Then, the weakly closed algebra , generated by and , is norm-closed, semisimple, and abelian. And 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 is not reflexive, then the algebra described as in Theorem 3.12 is not hereditarily reflexive. We do not know whether the reflexivity of an operator algebra described as in Theorem 3.8 is independent of the reflexivity of the Banach space . Here, is a relevant example.

Example 4.7. Let be the unilateral backward shift on . That is, It is obvious that Then, (1) Both and are not strictly cyclic. (2) is strictly cyclic on . (3) The weakly closed algebra generated by is semisimple. Therefore, and are also semisimple by Lemma 3.3. (4) is a reflexive algebra, and is hereditarily reflexive. (5) is a reflexive algebra. (6) is not hereditarily reflexive. (7) is a reflexive algebra, but is not hereditarily reflexive.

Proof. (2) Set , . For any positive integers , we have that . It follows that . Since is monotonically decreasing, is strictly cyclic on by [19] and is a strictly cyclic vector for .(3) Using the proof similar to the proof of (Example 4.4(3)), we have that and . Since is unilateral forward shift on , the spectra of contains infinite complex numbers. Then, we have that the Jacobson radical . Therefore, , and are semisimple.(4) is reflexive by Lemma 3.7. is hereditarily reflexive by Lemma 3.11. (5) Suppose that and . Since and has a strictly cyclic vector , then there exists such that . For any fixed , let .For each , let , where the bar denotes the norm closure. Then, and . We have that for every . It is obvious that is not locally algebraic, and there is such that is linearly independent. Let such that is linearly independent. Then, the algebra has a separating vector .Let be an arbitrary homomorphism from to the complex field and let be the corresponding maximal ideal of . Define by . Then, be the corresponding maximal ideal of . For any , let . Then, and . So .It is obvious that and . It follows that . In particular, . Since is a separating vector for , we have that . Then, and , that is, . So, and . Thus, . Since is a separating vector for , and . Since is arbitrary and is semisimple, it follows that , that is, for any . Since , , and . It follows that is a reflexive algebra.(6) By Lemma 3.14, the dual space of consists entirely of the maps , where .Since is strictly cyclic, and are norm-closed, then and are isomorphic as Banach spaces. Thus, and are isomorphic as Banach spaces. It follows that the set of all weakly continuous linear functionals on is the set of all norm-continuous linear functionals on .Similar to Proposition 3.13, is hereditarily reflexive if and only if it is reflexive and every weakly continuous functional on is of the form , where and . Since the Banach space is not reflexive, there exists such that . It follows that is not hereditarily reflexive by the assertion.(7) Using the proof similar to the proof of (5), we can prove that is a reflexive algebra. Using the proof similar to the proof of (6), we can prove that 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).

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