Powers of Convex-Cyclic Operators

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.