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# Fixed points and orbits of non-convolution operators

- Fernando León-Saavedra
^{1}Email author and - Pilar Romero-de la Rosa
^{2}

**2014**:221

https://doi.org/10.1186/1687-1812-2014-221

© León-Saavedra and Romero-de la Rosa; licensee Springer. 2014

**Received:**1 July 2014**Accepted:**14 October 2014**Published:**29 October 2014

## Abstract

A continuous linear operator *T* on a Fréchet space *F* is hypercyclic if there exists a vector $f\in F$ (which is called hypercyclic for *T*) such that the orbit $\{{T}^{n}f:n\in \mathbb{N}\}$ is dense in *F*. A subset *M* of a vector space *F* is spaceable if $M\cup \{0\}$ contains an infinite-dimensional closed vector space. In this paper note we study the orbits of the operators ${T}_{\lambda ,b}f={f}^{\prime}(\lambda z+b)$ ($\lambda ,b\in \mathbb{C}$) 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 ${T}_{\lambda ,b}$ is hypercyclic on $H(\mathbb{C})$. We characterize also when the set of hypercyclic vectors for ${T}_{\lambda ,b}$ is spaceable. The fixed point of the map $z\to \lambda z+b$ (in the case $\lambda \ne 1$) plays a central role in the proofs.

## Keywords

- fixed point
- Denjoy-Wolf theorem
- non-convolution operator
- hypercyclic operator
- spaceability

## 1 Introduction

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 $f\in F$ (called hypercyclic vector for *T*) such that the orbit $(\{{T}^{n}f:n\in \mathbb{N}\})$ 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 $M\cup \{0\}$ 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 [1] Chapter 8 or the recent survey [3] and the references therein).

In 1991, Godefroy and Shapiro [4] showed that every continuous linear operator $L:H(\mathbb{C})\to H(\mathbb{C})$ 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 [5]).

In [5], Aron and Markose introduced new examples of hypercyclic operators on $H(\mathbb{C})$ which are not convolution operators. Namely, ${T}_{\lambda ,b}f={f}^{\prime}(\lambda z+b)$, $\lambda ,b\in \mathbb{C}$. In the first section we show that if $\lambda \in \mathbb{D}$ and $b\in \mathbb{C}$ then ${T}_{\lambda ,b}$ is not hypercyclic on $H(\mathbb{C})$. This result together with the results in [5] and [6] shows the following characterization: ${T}_{\lambda ,b}$ is hypercyclic on $H(\mathbb{C})$ if and only if $|\lambda |\ge 1$. Thus, we complete the results of Aron and Markose [5] and Fernández and Hallack [6] characterizing when ${T}_{\lambda ,b}$ ($\lambda ,b\in \mathbb{C}$) is hypercyclic. Let us denote by $HC(T)$ the set of hypercyclic vectors for *T*. In Section 3 we characterize when $HC({T}_{\lambda ,b})$ is spaceable. Namely $HC({T}_{\lambda ,b})$ is spaceable if and only if $|\lambda |=1$. During the proofs, it is essential to take into account the fixed point of the map $z\to \lambda z+b$ ($\lambda \ne 1$).

## 2 Characterizing the hypercyclicity of ${T}_{\lambda ,b}$

The proof of this result follows the ideas of the proof of Proposition 14 in [5].

**Theorem 2.1** *For any* $\lambda \in \mathbb{D}$ *and* $b\in \mathbb{C}$ *and for any* $f\in H(\mathbb{C})$, *the sequence* ${T}_{\lambda ,b}^{n}f\to 0$ *uniformly on compact subsets of* ℂ. *Therefore* ${T}_{\lambda ,b}$ *is not hypercyclic on* $H(\mathbb{C})$.

*Proof*Set $\phi (z)=\lambda z+b$, $\lambda \in \mathbb{D}$ and $b\in \mathbb{C}$. Since $\lambda \ne 1$, $\phi (z)$ has a fixed point ${z}_{0}$. Indeed, ${z}_{0}=\frac{b}{1-\lambda}$. We denote by ${\phi}_{n}(z)$ the sequence of the iterates defined by

*n*th derivative of

*f*. It is well known that if $\lambda \in \mathbb{D}$ then ${z}_{0}$ is an attractive fixed point, that is, ${\phi}_{n}(z)$ converges to the fixed point ${z}_{0}$ uniformly on compact subsets. Indeed, let $R>0$. If $|z|\le R$, then

as $n\to \mathrm{\infty}$. Thus, there exists ${n}_{0}$ such that if $|z|\le R$ then $|{\phi}_{n}(z)-{z}_{0}|<1/2$ for all $n\ge {n}_{0}$.

and since $2n{|\lambda |}^{(n-1)/2}\to 0$ as $n\to \mathrm{\infty}$, we conclude that ${max}_{|z|\le R}|{T}_{\lambda ,b}^{n}f(z)|\to 0$, as $n\to \mathrm{\infty}$, 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 ${n}^{1/2}$ that was missing but that does not break the argument. □

Theorem 13 in [5] and Theorem 2.1 give the following characterization.

**Theorem 2.2** *For any* $\lambda \in \mathbb{C}$ *and* $b\in \mathbb{C}$, *the operator* ${T}_{\lambda ,b}$ *is hypercyclic in* $H(\mathbb{C})$ *if and only if* $|\lambda |\ge 1$.

## 3 Spaceability of the set of hypercyclic vectors for ${T}_{\lambda ,b}$

As stated in [3], there are few non-trivial examples of subsets *M* which are lineable (that is, $M\cup \{0\}$ contains an infinite-dimensional vector space) and are not spaceable. The following result provides the following examples: for $|\lambda |>1$, the set $HC({T}_{\lambda ,b})$ is lineable but it is not spaceable.

Shkarin [7] showed that for the derivative operator *D*, the set of hypercyclic vectors $HC(D)$ is spaceable.

**Theorem 3.1** *For any* $\lambda \in \mathbb{C}$ *and* $b\in \mathbb{C}$, $HC({T}_{\lambda ,b})$ *is spaceable if and only if* $|\lambda |=1$.

*Proof*Firstly, let us suppose that $|\lambda |>1$, and let us prove that $HC({T}_{\lambda ,b})$ does not contain a closed infinite dimensional subspace. Let ${z}_{0}$ be the fixed point of $\phi (z)=\lambda z+b$. Then we consider a sequence of norms defining the topology of $H(\mathbb{C})$. Namely, for $n\in \mathbb{N}$ and $f\in H(\mathbb{C})$, we write

It is easy to see that the above sequence of semi-norms is increasing and defines the original topology on $H(\mathbb{C})$.

- (a)
${p}_{N}(f)>0$, $\mathrm{\forall}f\in HC({T}_{\lambda ,b})$.

- (b)
${p}_{N}({T}_{\lambda ,b}^{n}f)\ge {C}_{n}{p}_{n}(f)$, $\mathrm{\forall}f\in {M}_{n}$.

which are clearly of finite codimension.

and it follows that condition (b) is satisfied with $N=0$ and ${C}_{n}={|\lambda |}^{\frac{{n}^{2}-2n}{4}}\to \mathrm{\infty}$ as $n\to \mathrm{\infty}$, and therefore $HC({T}_{\lambda ,b})$ is not spaceable.

Now, let us suppose that $|\lambda |=1$, and let us prove that $HC({T}_{\lambda ,b})$ is spaceable. Indeed, let us suppose first that $\lambda =1$. If $b=0$ then ${T}_{1,0}=D$, and it was proved by Shkarin [7] that $HC(D)$ is spaceable. If $b\ne 0$ then ${T}_{1,b}=D{e}^{bD}$, so that ${T}_{1,b}=\psi (D)$, where $\psi (z)=z{e}^{bz}$ is an entire function of exponential type that is not a polynomial, and according to Example 10.12 in [[2], p.275], the space $HC({T}_{1,b})$ is spaceable.

*f*of the form

and let us prove that ${T}_{\lambda ,b}^{{n}_{k}}f\to 0$ uniformly on compact subsets as $k\to \mathrm{\infty}$.

In the last step we used inequality (2). This completes the proof of Theorem 3.1. □

## Declarations

### Acknowledgements

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.

## Authors’ Affiliations

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