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 $\{T^{n}f:n\in \mathbb{N}\}$ 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.

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

$${\phi }_n(z)=\phi \circ \cdots \circ \phi (n\mathrm{times}),$$

an easy computation yields

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

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

$$T_{\lambda ,b}^{n}f(z)={\lambda }^{\frac{n(n-1)}{2}}{f}^{(n)}({\lambda }^{n}z+\frac{(1-{\lambda }^n)b}{1-\lambda })={\lambda }^{\frac{n(n-1)}{2}}{f}^{(n)}({\phi }_n(z)),$$

where $f^{(n)}$ 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, ${\phi }_n(z)$ converges to the fixed point $z_0$ uniformly on compact subsets. Indeed, let $R>0$. If $|z|\le R$, then

$$|{\phi }_n(z)-z_0|=|{\lambda }^{n}z+\frac{(1-{\lambda }^n)b}{1-\lambda }-\frac{b}{1-\lambda }|\le {|\lambda |}^{n}R+\frac{{|\lambda |}^n}{|1-\lambda |}|b|\to 0$$

as $n\to \mathrm{\infty }$. Thus, there exists $n_0$ such that if $|z|\le R$ then $|{\phi }_n(z)-z_0|<1/2$ for all $n\ge n_0$.

If $n\ge n_0$ and $|z|\le R$, we have by the Cauchy inequality

$$\left|f^{(n)}({\phi }_n(z))\right|\le Cn!2^n,\text{where }C=max\{|f(w)|:|w|\le 1\}.$$

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

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

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

$$p_n(f)=\underset{|z-z_0|\le {|\lambda |}^{n/4}}{max}|f(z)|.$$

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

Given the sequence of increasing semi-norms $\{p_n\}$, according to Theorem 10.25 in [2], it is sufficient to find a sequence of subspaces $M_n\subset H(\mathbb{C})$ of finite codimension, positive numbers $C_n\to \mathrm{\infty }$ and $N\ge 1$ satisfying the following:

• (a) $p_N(f)>0$, $\mathrm{\forall }f\in HC(T_{\lambda ,b})$.

• (b) $p_N(T_{\lambda ,b}^{n}f)\ge C_n{p}_n(f)$, $\mathrm{\forall }f\in M_n$.

Indeed, let us consider the subspaces

$$M_n=\{f\in H(\mathbb{C}):f(z_0)=f^{\prime }(z_0)=\cdots =f^{(n-1)}(z_0)=0\},$$

which are clearly of finite codimension.

Notice that ${\phi }_n(z)-z_0={\lambda }^n(z-z_0)$, so that ${\phi }_n(z)$ maps the disk $D(z_0,1)=\{|z-z_0|\le 1\}$ onto $D(z_0,\parallel \lambda {|}^n)$. Hence,

$$\begin{array}{rl}{p}_0\left(T_{\lambda ,b}^{n}f\right)& =\underset{|z-z_0|\le 1}{max}|T_{\lambda ,b}^{n}f(z)|\\ ={|\lambda |}^{\frac{n(n-1)}{2}}\underset{|z-z_0|\le 1}{max}\left|f^{(n)}({\phi }_n(z))\right|\\ ={|\lambda |}^{\frac{n(n-1)}{2}}\underset{|{\phi }_n(z)-z_0|\le {|\lambda |}^{n+1}}{max}\left|f^{(n)}({\phi }_n(z))\right|\\ ={|\lambda |}^{\frac{n(n-1)}{2}}\underset{|w-z_0|\le {|\lambda |}^n}{max}|f^{(n)}(w)|.\end{array}$$

If $f\in M_1$ then $f(z_0)=0$, so that $f(z)={\int }_{[z_0,z]}{f}^{\prime }(\xi )d\xi$. Therefore we have

$$\underset{|z-z_0|\le R}{max}|f(z)|\le R\underset{|z-z_0|\le R}{max}|f^{\prime }(z)|,$$

and it follows easily by induction that if $f\in M_n$ then

$$\underset{|z-z_0|\le R}{max}|f(z)|\le R^n\underset{|z-z_0|\le R}{max}|f^{(n)}(z)|.$$

Thus,

$$\begin{array}{rl}{p}_0\left(T_{\lambda ,b}^{n}f\right)& ={|\lambda |}^{\frac{n(n-1)}{2}}\underset{|w-z_0|\le {|\lambda |}^n}{max}|f^{(n)}(w)|\\ \ge {|\lambda |}^{\frac{n(n-1)}{2}}\underset{|w-z_0|\le {|\lambda |}^{n/4}}{max}|f^{(n)}(w)|\\ \ge {|\lambda |}^{\frac{n(n-1)}{2}}{|\lambda |}^{-n^2/4}\underset{|w-z_0|\le {|\lambda |}^{n/4}}{max}|f(w)|\\ ={|\lambda |}^{\frac{n^2-2n}{4}}{p}_n(f),\end{array}$$

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 $HC(T_{\lambda ,b})$ 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.

Now let us consider the case $\lambda \in \partial \mathbb{D}\setminus \{1\}$. Set $z_0=\frac{b}{1-\lambda }$ the fixed point of $\phi (z)=\lambda z+b$. According to Theorem 10.2 in [2], 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 $H(\mathbb{C})$ on which suitable powers of $T_{\lambda ,b}$ tend to 0. Now the proof mimics some ideas contained in Example 10.13 in [2]. Indeed, for any $n\ge 1$, there is some $C_n>0$ such that

$$x^n\le 2^x\text{for all }x\ge C_n.$$

(1)

Let us consider a strictly increasing sequence of positive integers ${(n_k)}_k$ 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

$$n_j^{n_k}\le 2^{n_j}\text{for }j\ge k+1.$$

(2)

Let us consider $M_0$ the closed subspace of $H(\mathbb{C})$ of all entire functions f of the form

$$f(z)=\sum _{k=1}^{\mathrm{\infty }}{a}_k{(z-z_0)}^{n_k-1},$$

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

We have

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

Notice that $|\lambda |=1$ and the map ${\phi }_{n_k}$ takes the disc $D(z_0,R)$ onto itself, so that

$$\begin{array}{rl}\underset{|z-z_0|\le R}{max}|\left(T^{n_k}f\right)(z)|& =\underset{|z-z_0|\le R}{max}\left|\left(D^{n_k}f\right)({\phi }_{n_k}(z))\right|\\ =\underset{|w-z_0|\le R}{max}|\left(D^{n_k}f\right)(w)|.\end{array}$$

Finally, we have

$$\begin{array}{rl}\underset{|w-z_0|\le R}{max}|\left(D^{n_k}f\right)(w)|& =\underset{|w-z_0|\le R}{max}\left|\sum _{j=k+1}^{\mathrm{\infty }}{a}_j{D}^{n_k}{(w-z_0)}^{n_j-1}\right|\\ \le \sum _{j=k+1}^{\mathrm{\infty }}|a_j|(n_j-1)(n_j-2)\cdots (n_j-n_k)R^{n_j-n_k-1}\\ \le \sum _{j=k+1}^{\mathrm{\infty }}|a_j|n_j^{n_k}{R}^{n_j}\\ \le \sum _{j=k+1}^{\mathrm{\infty }}|a_j|{(2R)}^{n_j}\to 0\text{as }k\to \mathrm{\infty }.\end{array}$$

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