Fixed points and orbits of non-convolution operators
© León-Saavedra and Romero-de la Rosa; licensee Springer. 2014
Received: 1 July 2014
Accepted: 14 October 2014
Published: 29 October 2014
A continuous linear operator T on a Fréchet space F is hypercyclic if there exists a vector (which is called hypercyclic for T) such that the orbit is dense in F. A subset M of a vector space F is spaceable if contains an infinite-dimensional closed vector space. In this paper note we study the orbits of the operators () defined on the space of entire functions and introduced by Aron and Markose (J. Korean Math. Soc. 41(1):65-76, 2004). We complete the results in Aron and Markose (J. Korean Math. Soc. 41(1):65-76, 2004), characterizing when is hypercyclic on . We characterize also when the set of hypercyclic vectors for is spaceable. The fixed point of the map (in the case ) plays a central role in the proofs.
Let us denote by F a complex infinite dimensional Fréchet space. A continuous linear operator T defined on F is said to be hypercyclic if there exists a vector (called hypercyclic vector for T) such that the orbit is dense in F. We refer to the books [1, 2] and the references therein for further information on hypercyclic operators. From a modern terminology, a subset M of a vector space F is said to be spaceable if contains an infinite-dimensional closed vector space. The study of spaceability of (usually pathological) subsets is a natural question which has been studied extensively (see  Chapter 8 or the recent survey  and the references therein).
In 1991, Godefroy and Shapiro  showed that every continuous linear operator which commutes with translations (these operators are called convolution operators) and which is not a multiple of the identity is hypercyclic. This result unifies two classical results by Birkhoff and MacLane (see the survey ).
In , Aron and Markose introduced new examples of hypercyclic operators on which are not convolution operators. Namely, , . In the first section we show that if and then is not hypercyclic on . This result together with the results in  and  shows the following characterization: is hypercyclic on if and only if . Thus, we complete the results of Aron and Markose  and Fernández and Hallack  characterizing when () is hypercyclic. Let us denote by the set of hypercyclic vectors for T. In Section 3 we characterize when is spaceable. Namely is spaceable if and only if . During the proofs, it is essential to take into account the fixed point of the map ().
2 Characterizing the hypercyclicity of
The proof of this result follows the ideas of the proof of Proposition 14 in .
Theorem 2.1 For any and and for any , the sequence uniformly on compact subsets of ℂ. Therefore is not hypercyclic on .
as . Thus, there exists such that if then for all .
and since as , we conclude that , as , as desired. We point out that this is a refinement of the argument by Aron and Markose. One of the referees chased the constants and recovered the factor that was missing but that does not break the argument. □
Theorem 13 in  and Theorem 2.1 give the following characterization.
Theorem 2.2 For any and , the operator is hypercyclic in if and only if .
3 Spaceability of the set of hypercyclic vectors for
As stated in , there are few non-trivial examples of subsets M which are lineable (that is, contains an infinite-dimensional vector space) and are not spaceable. The following result provides the following examples: for , the set is lineable but it is not spaceable.
Shkarin  showed that for the derivative operator D, the set of hypercyclic vectors is spaceable.
Theorem 3.1 For any and , is spaceable if and only if .
It is easy to see that the above sequence of semi-norms is increasing and defines the original topology on .
which are clearly of finite codimension.
and it follows that condition (b) is satisfied with and as , and therefore is not spaceable.
Now, let us suppose that , and let us prove that is spaceable. Indeed, let us suppose first that . If then , and it was proved by Shkarin  that is spaceable. If then , so that , where is an entire function of exponential type that is not a polynomial, and according to Example 10.12 in [, p.275], the space is spaceable.
and let us prove that uniformly on compact subsets as .
In the last step we used inequality (2). This completes the proof of Theorem 3.1. □
The research was supported by Junta de Andalucía FQM-257. The authors would like to thank the referee for reading our manuscript carefully and for giving such constructive comments, which helped improving the quality of the paper substantially.
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