`Abstract and Applied AnalysisVolume 2011, Article ID 686832, 11 pageshttp://dx.doi.org/10.1155/2011/686832`
Research Article

## Supercyclicity and Hypercyclicity of an Isometry Plus a Nilpotent

1Department of Mathematics, Science and Research Branch, Islamic Azad University, Tehran 1477893855, Iran
2Department of Mathematics, College of Sciences, Shiraz University, Shiraz 71454, Iran
3Department of Mathematics, Payam-e-noor University, Shiraz 71955-1368, Iran

Received 30 December 2010; Accepted 28 March 2011

Copyright © 2011 S. Yarmahmoodi et al. 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

Suppose that is a separable normed space and the operators and are bounded on . In this paper, it is shown that if , is an isometry, and is a nilpotent then the operator is neither supercyclic nor weakly hypercyclic. Moreover, if the underlying space is a Hilbert space and is a co-isometric operator, then we give sufficient conditions under which the operator satisfies the supercyclicity criterion.

#### 1. Introduction

Let be a vector in a separable normed space and an operator on . The orbit of under is defined by

We recall that a vector in is cyclic for an operator on if the closed linear span of is ; it is supercyclic, if the set of all scalar multiples of the elements of is dense in ; also it is said to be (weakly) hypercyclic if is (weakly) dense in . An operator is called cyclic, supercyclic, or (weakly) hypercyclic operator, respectively, if it has a cyclic, supercyclic, or (weakly) hypercyclic vector. Recently, the cyclicity of operators has attracted much attention from operator theorists. For a good source on this topic, see [1]. Hilden and Wallen in [2] proved that isometries on Hilbert spaces with dimension more than one are not supercyclic. Ansari and Bourdon in [3] and Miller in [4] independently proved this fact on Banach spaces. Moreover, recently it is shown in [5] that m-isometric operators on Hilbert spaces, which forms a larger class than isometries, are neither supercyclic nor weakly hypercyclic. In this paper, it is shown that an isometry plus a nilpotent on normed spaces are neither supercyclic nor weakly hypercyclic if they commute. We also discuss this fact when the underlying space is a Hilbert space and the isometry is replaced by a co-isometry. We begin with some elementary properties of such operators. In what follows, as usual, for an operator , , , and are denoted, respectively, the approximate point spectrum, point spectrum, and spectrum of . Also, denotes the open unit disc. Recall that an operator on a normed space is a nilpotent operator of order if and . From now on, we assume that is a nilpotent operator of order unless stated otherwise.

Proposition 1.1. Suppose that is a normed space, and is an isometry such that . If , then (i), (ii), (iii).

Proof. (i) Suppose that . Then it is easily seen that which implies that . Consequently, . Since , a similar argument shows that .
(ii) If , there exits such that . Therefore, Now, if , then ; otherwise, Also, if then ; otherwise, consider and continue this process to conclude that which implies that . Hence, . Moreover, since , using a similar method, we get .
(iii) Let ; then there exists a sequence in such that and Therefore, Suppose that there is so that for all ; then as where which, in turn, implies that .
Now, if (1.7) does not hold, then we can assume, without loss of generality, that satisfies So by (1.5), Now, if there is a constant such that for all , then where which implies that . Otherwise, we can assume, without loss of generality, that as and by (1.5) as . The procedure continues to conclude that . Since , by a similar method .

In the remaining results of this section, the operators and are as in Proposition 1.1.

Corollary 1.2. Suppose that is a normed space. Then is bounded below where .

Proof. Since is an isometry, . In fact, let ; then ; moreover, there exists a sequence in with and so if . Therefore, and so . Now, if , then and so is bounded below.

Corollary 1.3. Suppose that is an infinite dimensional Banach space. Then the operator on is not a compact operator.

Proof. If is a compact operator, then . Thus which contradicts the fact that the spectrum of a compact operator is at most countable.

Proposition 1.4. If the operators and are defined on a normed space , then for every scalar .

Proof. Fix and suppose that for some nonzero vector . By Proposition 1.1, which implies that . Therefore, if , we have Consequently, Since for every , we conclude that . Continue the above process to get , and so .

Corollary 1.5. If is a Hilbert space, then the eigenvectors of corresponding to distinct eigenvalues are orthogonal.

Proof. Let and be eigenvectors of corresponding to distinct eigenvalues and . So, and . By Proposition 1.4, and which implies that . Suppose that denotes the inner product of . Then Replacing by , we obtain ; consequently, But , and so .

Recall that an operator is power bounded if there exists some constant such that for all .

Proposition 1.6. Let be a normed space and . If there is a constant such that for all , then . In particular, if is power bounded, then .

Proof. Since the sequence is bounded, an argument similar to the proof of the Proposition 1.4 shows that .

#### 2. Supercyclicity and Hypercyclicity

We begin this section with a useful lemma.

Lemma 2.1. Let be a normed space. For nonnegative integers , if is a polynomial in with coefficients in of degree , then the sequence is eventually increasing.

Proof. We prove the lemma by induction on , the degree of the polynomial . For , let . It is easily seen that for every Since , there is a positive integer such that This fact coupled with (2.2) implies that for every . Therefore, the sequence is increasing. Suppose that is eventually increasing and let where . Since using the induction hypothesis there is a positive integer such that for every Therefore, for every . Hence, the sequence is increasing.

Theorem 2.2. Suppose that is a normed space, and is an isometry such that . If , then the operator is neither supercyclic nor weakly hypercyclic.

Proof. Let be the completion of and , , and the extensions of , , and on , respectively. Thus, where is an isometry and is a nilpotent operator; moreover, . Also, note that the supercyclicity of the operator implies the supercyclicity of . So we can assume, without loss of generality, that is a Banach space.
As we have seen in the proof of Proposition 1.4, if then and so by Lemma 2.1, the sequence is eventually increasing. Suppose that is a supercyclic vector for . Thus, for any there is a sequence of positive integers and a sequence of scalars such that . Moreover, since the sequence is eventually increasing, we have for large . So letting , we conclude that , for all in . On the other hand, the supercyclicity of implies that it has a dense range and so is invertible. Thus, in light of Proposition 1.1 we see that is invertible. It is easy to see that where Since , by a similar argument the sequence is eventually increasing for every . But is also supercyclic (see [1, Theorem 1.12]); therefore, for every . Thus, is an isometry which implies that it is not a supercyclic operator.
To show that the operator is not weakly hypercyclic, note that for every and every positive integer . If , then there is a nonzero such that because . Now, suppose that is a weakly hypercyclic vector for . Since is weakly dense in and is nonzero, the set is dense in . But for all , which is a contradiction. If , then and so is not a weakly hypercyclic operator.

We remark that there are Banach space isometries which are also weakly supercyclic. Indeed, the unweighted bilateral weighted shift on the space where is weakly supercyclic (see [1, Corollary 10.32]). However, the question that whether an isometry plus a nonzero nilpotent which commute with each other, are weakly supercyclic or not is still an open question.

The following examples show that the commutativity of and is essential in the preceding theorem.

Example 2.3. Let be the standard orthonormal basis for . Define the isometric operator by for all and the weighted shift operator by , where for all integers , for all , and for all . Note that and . Moreover, since , the weighted shift operator is invertible. To see that is supercyclic by Theorem 3.4 of [6], it is sufficient to show that But is finite, because . Furthermore, , because (see [7, pages 299 and 300]). Therefore, (2.15) holds.

Example 2.4. Consider the isometric operator on defined by and the weighted shift operator defined by , where for all , , for , and for . Note that and . Also, since for all , and , we conclude that Furthermore, because Hence, using Corollary 10.27 of [1], we observe that the operator is weakly hypercyclic.

#### 3. A Co-isometry Plus a Nilpotent

From now on, we assume that is a separable Hilbert space with orthonormal basis . Recall that the unilateral shift operator is given by for all and the backward shift operator is defined by and for all . It is known that the operator is supercyclic (see [1, page 9]). It follows that a co-isometry can be supercyclic. In this section, we give sufficient conditions such that the sum of a co-isometry and a nilpotent is supercyclic on .

Theorem 3.1. Suppose that is a co-isometric operator on a Hilbert space . Then is supercyclic if and only if .

Proof. First assume that . Then by the von Neumann-Wold decomposition, for some positive integer (see [8]). Therefore, which is a positive power of a supercyclic operator and so is supercyclic [9]. For the converse, assume that , and let denote the orthogonal projection on . By the von Neumann-Wold decomposition, is a reducing subspace for and is unitary. Since is also unitary, the operator is not supercyclic. Assume that is supercyclic, and is a supercyclic vector for , where and . If , then which is impossible; so . Take , and let be arbitrary. Then there is a nonnegative integer and a scalar such that Hence, is a supercyclic vector for which is impossible.

To prove the next theorem, we need the supercyclicity criterion due to H. N. Salas (see [10], or more generally [11]).

Supercyclicity Criterion
Suppose that is a separable Banach space and is a bounded operator on . If there is an increasing sequence of positive integers , and two dense sets and of such that(1)there exists a function satisfying for all ,(2) for every and ,then is supercyclic.

Theorem 3.2. Suppose that is a co-isometry on a Hilbert space such that . If , then the operator satisfies the supercyclicity criterion.

Proof. By Corollary 1.2, the operator is bounded below and so is left invertible. Consequently, is a right invertible operator. Let. For every , there is a vector in such that . Since , we have which implies that . Hence, and so the operator admits a complete set of eigenvectors that is, for every positive real number , where (see [12], Part (A) of the lemma). Since is bounded below, is invertible. Take . Choose so that , and let Now, if , then as . Finally, for every and every . Thus, the operator satisfies the supercyclicity criterion.

The Hilbert-Schmidt class, , is the class of all bounded operators defined on a Hilbert space , satisfying

where is the norm on induced by its inner product. We recall that is a Hilbert space equipped with the inner product in which denotes the trace of . Furthermore, is an ideal of the algebra of all bounded operators on , see [8]. For any bounded operator on a Hilbert space , the left multiplication operator and the right multiplication operator on are defined by and for all . It is known that an operator B satisfies the supercyclicity criterion if and only if is supercyclic on the space (see [13, page 37]). In the following proposition, we see that an operator may satisfy the supercyclicity criterion although is not a supercyclic operator on .

Proposition 3.3. Suppose that is a Hilbert space and is a co-isometry such that and . Then the operator satisfies the supercyclicity criterion but the operator is not supercyclic on .

Proof. By Theorem 3.2, the operator satisfies the supercyclicity criterion. Moreover, for every we have which implies that is an isometry. Also, if , then . Since , Theorem 2.2 implies that is not supercyclic.

The proof of the following proposition is similar to the proof of the second part of Theorem 2.2, and we omit it.

Proposition 3.4. Suppose that is a normed space and is a co-isometry such that . Then the operator is not weakly hypercyclic.

#### Acknowledgments

The second author is in part supported by a grant from Shiraz University Research Council.

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