# Fixed points and orbits of non-convolution operators

## 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 $\left\{{T}^{n}f:n\in \mathbb{N}\right\}$ is dense in F. A subset M of a vector space F is spaceable if $M\cup \left\{0\right\}$ contains an infinite-dimensional closed vector space. In this paper note we study the orbits of the operators ${T}_{\lambda ,b}f={f}^{\prime }\left(\lambda z+b\right)$ ($\lambda ,b\in \mathbb{C}$) 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 $H\left(\mathbb{C}\right)$. 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.

## 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 $\left(\left\{{T}^{n}f:n\in \mathbb{N}\right\}\right)$ 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 $M\cup \left\{0\right\}$ 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 $L:H\left(\mathbb{C}\right)\to H\left(\mathbb{C}\right)$ 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 $H\left(\mathbb{C}\right)$ which are not convolution operators. Namely, ${T}_{\lambda ,b}f={f}^{\prime }\left(\lambda z+b\right)$, $\lambda ,b\in \mathbb{C}$. 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 $H\left(\mathbb{C}\right)$. This result together with the results in  and  shows the following characterization: ${T}_{\lambda ,b}$ is hypercyclic on $H\left(\mathbb{C}\right)$ if and only if $|\lambda |\ge 1$. Thus, we complete the results of Aron and Markose  and Fernández and Hallack  characterizing when ${T}_{\lambda ,b}$ ($\lambda ,b\in \mathbb{C}$) is hypercyclic. Let us denote by $HC\left(T\right)$ the set of hypercyclic vectors for T. In Section 3 we characterize when $HC\left({T}_{\lambda ,b}\right)$ is spaceable. Namely $HC\left({T}_{\lambda ,b}\right)$ 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 .

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

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

${\phi }_{n}\left(z\right)=\phi \circ \cdots \circ \phi \phantom{\rule{1em}{0ex}}\left(n\phantom{\rule{0.25em}{0ex}}\mathrm{times}\right),$

an easy computation yields

${\phi }_{n}\left(z\right)={\lambda }^{n}z+\frac{1-{\lambda }^{n}}{1-\lambda }b.$

Let us observe that the iterates of the operator ${T}_{\lambda ,b}$ have the form

${T}_{\lambda ,b}^{n}f\left(z\right)={\lambda }^{\frac{n\left(n-1\right)}{2}}{f}^{\left(n\right)}\left({\lambda }^{n}z+\frac{\left(1-{\lambda }^{n}\right)b}{1-\lambda }\right)={\lambda }^{\frac{n\left(n-1\right)}{2}}{f}^{\left(n\right)}\left({\phi }_{n}\left(z\right)\right),$

where ${f}^{\left(n\right)}$ denotes the 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, ${\phi }_{n}\left(z\right)$ converges to the fixed point ${z}_{0}$ uniformly on compact subsets. Indeed, let $R>0$. If $|z|\le R$, then

$|{\phi }_{n}\left(z\right)-{z}_{0}|=|{\lambda }^{n}z+\frac{\left(1-{\lambda }^{n}\right)b}{1-\lambda }-\frac{b}{1-\lambda }|\le {|\lambda |}^{n}R+\frac{{|\lambda |}^{n}}{|1-\lambda |}|b|\to 0$

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

If $n\ge {n}_{0}$ and $|z|\le R$, we have by the Cauchy inequality

Now, it follows from Stirling’s formula that $n!\le e{n}^{n+1/2}{e}^{-n}$. Hence, if $|z|\le R$ and $n\ge {n}_{0}$, then

$|{T}_{\lambda ,b}^{n}f\left(z\right)|\le Cn!{2}^{n}{|\lambda |}^{\frac{n\left(n-1\right)}{2}}\le Ce{n}^{1/2}{\left(\frac{2n{|\lambda |}^{\left(n-1\right)/2}}{e}\right)}^{n},$

and since $2n{|\lambda |}^{\left(n-1\right)/2}\to 0$ as $n\to \mathrm{\infty }$, we conclude that ${max}_{|z|\le R}|{T}_{\lambda ,b}^{n}f\left(z\right)|\to 0$, 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  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 $H\left(\mathbb{C}\right)$ if and only if $|\lambda |\ge 1$.

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

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

Shkarin  showed that for the derivative operator D, the set of hypercyclic vectors $HC\left(D\right)$ is spaceable.

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

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

${p}_{n}\left(f\right)=\underset{|z-{z}_{0}|\le {|\lambda |}^{n/4}}{max}|f\left(z\right)|.$

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

Given the sequence of increasing semi-norms $\left\{{p}_{n}\right\}$, according to Theorem 10.25 in , it is sufficient to find a sequence of subspaces ${M}_{n}\subset H\left(\mathbb{C}\right)$ of finite codimension, positive numbers ${C}_{n}\to \mathrm{\infty }$ and $N\ge 1$ satisfying the following:

1. (a)

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

2. (b)

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

Indeed, let us consider the subspaces

${M}_{n}=\left\{f\in H\left(\mathbb{C}\right):f\left({z}_{0}\right)={f}^{\prime }\left({z}_{0}\right)=\cdots ={f}^{\left(n-1\right)}\left({z}_{0}\right)=0\right\},$

which are clearly of finite codimension.

Notice that ${\phi }_{n}\left(z\right)-{z}_{0}={\lambda }^{n}\left(z-{z}_{0}\right)$, so that ${\phi }_{n}\left(z\right)$ maps the disk $D\left({z}_{0},1\right)=\left\{|z-{z}_{0}|\le 1\right\}$ onto $D\left({z}_{0},\parallel \lambda {|}^{n}\right)$. Hence,

$\begin{array}{rl}{p}_{0}\left({T}_{\lambda ,b}^{n}f\right)& =\underset{|z-{z}_{0}|\le 1}{max}|{T}_{\lambda ,b}^{n}f\left(z\right)|\\ ={|\lambda |}^{\frac{n\left(n-1\right)}{2}}\underset{|z-{z}_{0}|\le 1}{max}|{f}^{\left(n\right)}\left({\phi }_{n}\left(z\right)\right)|\\ ={|\lambda |}^{\frac{n\left(n-1\right)}{2}}\underset{|{\phi }_{n}\left(z\right)-{z}_{0}|\le {|\lambda |}^{n+1}}{max}|{f}^{\left(n\right)}\left({\phi }_{n}\left(z\right)\right)|\\ ={|\lambda |}^{\frac{n\left(n-1\right)}{2}}\underset{|w-{z}_{0}|\le {|\lambda |}^{n}}{max}|{f}^{\left(n\right)}\left(w\right)|.\end{array}$

If $f\in {M}_{1}$ then $f\left({z}_{0}\right)=0$, so that $f\left(z\right)={\int }_{\left[{z}_{0},z\right]}{f}^{\prime }\left(\xi \right)d\xi$. Therefore we have

$\underset{|z-{z}_{0}|\le R}{max}|f\left(z\right)|\le R\underset{|z-{z}_{0}|\le R}{max}|{f}^{\prime }\left(z\right)|,$

and it follows easily by induction that if $f\in {M}_{n}$ then

$\underset{|z-{z}_{0}|\le R}{max}|f\left(z\right)|\le {R}^{n}\underset{|z-{z}_{0}|\le R}{max}|{f}^{\left(n\right)}\left(z\right)|.$

Thus,

$\begin{array}{rl}{p}_{0}\left({T}_{\lambda ,b}^{n}f\right)& ={|\lambda |}^{\frac{n\left(n-1\right)}{2}}\underset{|w-{z}_{0}|\le {|\lambda |}^{n}}{max}|{f}^{\left(n\right)}\left(w\right)|\\ \ge {|\lambda |}^{\frac{n\left(n-1\right)}{2}}\underset{|w-{z}_{0}|\le {|\lambda |}^{n/4}}{max}|{f}^{\left(n\right)}\left(w\right)|\\ \ge {|\lambda |}^{\frac{n\left(n-1\right)}{2}}{|\lambda |}^{-{n}^{2}/4}\underset{|w-{z}_{0}|\le {|\lambda |}^{n/4}}{max}|f\left(w\right)|\\ ={|\lambda |}^{\frac{{n}^{2}-2n}{4}}{p}_{n}\left(f\right),\end{array}$

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 $HC\left({T}_{\lambda ,b}\right)$ is not spaceable.

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

Now let us consider the case $\lambda \in \partial \mathbb{D}\setminus \left\{1\right\}$. Set ${z}_{0}=\frac{b}{1-\lambda }$ the fixed point of $\phi \left(z\right)=\lambda z+b$. According to Theorem 10.2 in , since ${T}_{\lambda ,b}$ satisfies the hypercyclicity criterion for the full sequence of natural numbers, it suffices to exhibit an infinite dimensional closed subspace ${M}_{0}$ of $H\left(\mathbb{C}\right)$ on which suitable powers of ${T}_{\lambda ,b}$ tend to 0. Now the proof mimics some ideas contained in Example 10.13 in . Indeed, for any $n\ge 1$, there is some ${C}_{n}>0$ such that

(1)

Let us consider a strictly increasing sequence of positive integers ${\left({n}_{k}\right)}_{k}$ satisfying ${n}_{k+1}\ge {C}_{{n}_{k}}$. If $j\ge k+1$, then ${n}_{j}\ge {n}_{k+1}\ge {C}_{{n}_{k}}$, therefore by (1) we have

(2)

Let us consider ${M}_{0}$ the closed subspace of $H\left(\mathbb{C}\right)$ of all entire functions f of the form

$f\left(z\right)=\sum _{k=1}^{\mathrm{\infty }}{a}_{k}{\left(z-{z}_{0}\right)}^{{n}_{k}-1},$

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

We have

$\left({T}^{{n}_{k}}f\right)\left(z\right)={\lambda }^{\frac{{n}_{k}\left({n}_{k}-1\right)}{2}}\left({D}^{{n}_{k}}f\right)\left({\phi }_{{n}_{k}}\left(z\right)\right).$

Notice that $|\lambda |=1$ and the map ${\phi }_{{n}_{k}}$ takes the disc $D\left({z}_{0},R\right)$ onto itself, so that

$\begin{array}{rl}\underset{|z-{z}_{0}|\le R}{max}|\left({T}^{{n}_{k}}f\right)\left(z\right)|& =\underset{|z-{z}_{0}|\le R}{max}|\left({D}^{{n}_{k}}f\right)\left({\phi }_{{n}_{k}}\left(z\right)\right)|\\ =\underset{|w-{z}_{0}|\le R}{max}|\left({D}^{{n}_{k}}f\right)\left(w\right)|.\end{array}$

Finally, we have

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

## References

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## 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.

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Correspondence to Fernando León-Saavedra.

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