Research Article | Open Access

# Powers of Convex-Cyclic Operators

**Academic Editor:**Manuel Maestre

#### Abstract

A bounded operator on a Banach space is convex cyclic if there exists a vector such that the convex hull generated by the orbit is dense in . In this note we study some questions concerned with convex-cyclic operators. We provide an example of a convex-cyclic operator such that the power fails to be convex cyclic. Using this result we solve three questions posed by Rezaei (2013).

#### 1. Introduction and Main Results

Throughout this paper we denote by the algebra of all bounded linear operators on a real or complex infinite dimensional Banach space . An operator is said to be cyclic if there exists a vector (later called cyclic vector for ) such that the linear span of the orbit is dense in . If the orbit is dense itself, without the help of the linear span, then is called hypercyclic and is called hypercyclic for . In the midway stand several notions studied by different authors. For instance, the operator is said to be supercyclic if the projective orbit is dense in . We refer to the books [1, 2] and references therein for further information on hypercyclic operators.

When we sometimes abusively say that a polynomial is a convex polynomial, what we really mean is that , , , and . We will focus our attention on the notion of convex cyclicity introduced by Rezaei in [3]. An operator is said to be convex cyclic if there exists a vector such that the real convex hull of the orbit (denoted by ) is dense in .

In [3] are characterized the convex-cyclic matrices in finite dimension, and the author develops the main properties in the infinite dimensional setting.

A result by Ansari [4] states that if is a hypercyclic operator then is also hypercyclic; this fact is not true for cyclic operators. In this paper we show that Ansari’s result fails also for convex-cyclic operators, solving a question posed in [3].

Another result proved by Bourdon and Feldman on hypercyclic operators says that if the orbit of a vector is somewhere dense, then it is dense (see [5]). From our previous counterexample we can construct a non-convex-cyclic operator such that the has nonempty interior. That is, Bourdon and Feldman’s result is not true in the convex-cyclic setting. Finally we can construct a convex-cyclic operator such that is not weakly hypercyclic; that is, its orbit is not dense in the weak operator topology. The later examples solve Questions 5.5 and 5.6 in [3].

#### 2. Powers of a Convex-Cyclic Operator

The first example of hypercyclic operator on Banach spaces was discovered by Rolewicz (see [6]). Throughout this section or of complex valued sequences. Rolewicz’s operator with is defined on by where denotes the backward shift operator.

Lemma 1. *Set and . For any there exist and a sequence of polynomials such that*(1)* for all ;*(2)* and , and ;*(3)*, .*

*Proof. *Let us denote by the triangle with vertices . Since and , there exists such that for all . Then, there exist barycentric coordinates satisfying and . Then, the polynomials , for all , yield the desired result.

Lemma 2. *Let be a sequence of polynomials satisfying Conditions (1)–(3) of Lemma 1. Then, there exists a dense subset of vectors such that is dense in for all .*

*Proof. *We will use some hypercyclicity criterion version for sequence of operators (see [2, Theorem 3.24]); that is, we will show the existence of two dense subsets and and a sequence of mappings such that (i);(ii);(iii).Let us consider the subsets
which are dense in (see [2, Example 3.2, page 70]).

If with , then
which goes to zero when ; therefore, Condition (i) is fulfilled.

Denoting by , since , , are barycentric coordinates of a triangle, then lies in the degenerate triangle with vertices , in particular .

Let us take with and , and let us define the mapping on as
and we extend linearly on . Clearly as for all and for all . Thus, by the hypercyclicity criterion there exists a dense subset of vectors such that is dense in .

Now, let us prove the main result of this section, which solves Question 5.6 in [3].

Theorem 3. *The operator is convex cyclic on ; however is not.*

* Proof. *If is a polynomial, then . Let us observe that the first coordinate of the powers of are only real numbers. Take . If is the projection on the first coordinate,
which is not dense in . Therefore, is not convex cyclic.

Now, let us prove that is a convex-cyclic operator using a direct application of the Baire category theorem (see, for instance, [2, Theorem 1.57]). Thus is convex cyclic if for any nonempty open subsets , there exists a convex polynomial such that .

Indeed, let and open subsets of , where and , , are nonempty open subsets. Let and with . Set and the sequence of polynomials which guarantees Lemma 1. Hence we have and therefore (this fact will imply that acting on will intersect ). Now we apply Lemma 2 and we obtain a dense subset of hypercyclic vectors for the sequence. Thus there exist and a subsequence such that . Therefore , which yields the desired result.

*Remark 4. *If we take with , using similar arguments as in Theorem 3, we can show that is convex cyclic on but is not (if the operator is convex cyclic but is not).

*Remark 5. *If we consider the Rolewicz operator on real spaces , or , then we can get that the operator () is convex cyclic on or , but clearly is not. Lemma 1 can be adapted clearly to the real case. Now it is well known that if we consider Rolewicz’s operator on real spaces , or , then its complexification can be identified with the same operator on the corresponding spaces of complex sequences. With some slight modification in the proof of Corollary 2.51 in [2] we can obtain that Lemma 2 continues being true on real spaces. The rest of the proof is straightforward.

*Remark 6. *Another difference between hypercyclic operators and convex-cyclic operators is the following: hypercyclic operators are invariant under unimodular multiplications (see [7]); that is, if is hypercyclic, then is also with . However this is not true for convex-cyclic operators; the previous counterexample is convex cyclic; however is not.

Now, let . Let us consider the operator , that is, the same operator provided by Theorem 3 but without the multiplier factor . Then, an easy check shows that the set is contained in the unit disk. Moreover, the set has nonempty interior in ; for instance, contains the triangle with vertices . Using the arguments of Theorem 3 we can find a vector such that the convex orbit is dense in . Therefore the convex orbit has nonempty interior. However the operator is not convex cyclic. This solves Question 5.5 in Rezaei’s paper.

Proposition 7. *Bourdon and Feldman’s result fails for convex-cyclic operators.*

The adjoint of the operator in Theorem 3 has an eigenvalue; therefore, cannot be weakly hypercyclic. This solves Question 5.4 in [3].

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The authors are indebted to the referees for alerting them to some details that they had to take care of in the proof of the first version. Fernando León-Saavedra and María del Pilar Romero de la Rosa were partially supported by Junta de Andalucía FQM-257.

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#### Copyright

Copyright © 2014 Fernando León-Saavedra and María del Pilar Romero-de la Rosa. 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.