Hypercyclic and mixing composition operators on ${\mathscr {O}}_M({\mathbb {R}})$

Abstract

In this paper we characterize mixing composition operators acting on the space ${\mathscr {O}}_M({\mathbb {R}})$ of slowly increasing smooth functions. Moreover we relate the mixing property of those operators with the solvability of Abel’s functional equation and we give a sufficient condition for sequential hypercyclicity of composition operators on ${\mathscr {O}}_M({\mathbb {R}})$. This is used to prove that many mixing composition operators are hypercyclic.

1 Introduction

The study of dynamical properties of (continuous linear) operators $T\in L(E)$ on topological vector spaces E has attracted much interest in recent years. While there are few articles dealing with dynamics of operators on non-metrizable spaces, the vast majority of contributions concentrates on the dynamics of operators defined on separable Fréchet spaces. The advantage of completeness and metrizability stems from the applicability of Baire category arguments which are a powerful tool in this context. One prominent example of such a tool is Birkhoff’s transitivity criterion, stating that every topologically transitive operator on a separable Fréchet space is hypercyclic. Recall that T is said to be topologically transitive if for every pair of non-empty, open subsets U, V of E it holds $T^n(U)\cap V\ne \emptyset$ for some $n\in {\mathbb {N}}$, while T is (sequentially) hypercyclic whenever there is $x\in E$ whose orbit $\{T^nx;\,n\in {\mathbb {N}}_0\}$ under T is (sequentially) dense in E. Clearly, on arbitrary Hausdorff topological vector spaces, every hypercyclic operator is topologically transitive. Moreover, T is called (topologically) mixing if for every pair of non-empty, open subsets U, V of E it holds $T^n(U)\cap V\ne \emptyset$ for all sufficiently large $n\in {\mathbb {N}}$, while T is said to be chaotic if it is topologically transitive and if the set of periodic points of T is dense in E. In particular, on Fréchet spaces, mixing operators are sequentially hypercyclic.

The aim of this paper is to study dynamical properties of composition operators acting on the space ${\mathscr {O}}_M({\mathbb {R}})$ of multipliers of the space of rapidly decreasing, smooth functions ${\mathscr {S}}({\mathbb {R}})$ on ${\mathbb {R}}$. More precisely, we are interested in (sequential) hypercyclicity and mixing of composition operators on ${\mathscr {O}}_M({\mathbb {R}})$. Recall that ${\mathscr {O}}_M({\mathbb {R}})$ is given by

$$\begin{aligned} {\mathscr {O}}_M({\mathbb {R}})=\cap _{m=1}^\infty \cup _{n=1}^\infty {\mathscr {O}}_n^m({\mathbb {R}}), \end{aligned}$$

where

$$\begin{aligned} {\mathscr {O}}_n^m({\mathbb {R}}):=\left\{ f\in C^m({\mathbb {R}}):\,|f|_{m,n}:=\sup _{x\in {\mathbb {R}}, 0\le j\le m}(1+|x|^2)^{-n}|f^{(j)}(x)|<\infty \right\} . \end{aligned}$$

The space ${\mathscr {O}}_M({\mathbb {R}})$ is equipped with a natural locally convex topology which makes it a complete, ultrabornological, non-metrizable locally convex space. Hence, ${\mathscr {O}}_M({\mathbb {R}})$ is not a Fréchet space and thus, a mixing operator on ${\mathscr {O}}_M({\mathbb {R}})$ need not be (sequentially) hypercyclic. The study of dynamical properties of composition operators on ${\mathscr {O}}_M({\mathbb {R}})$ was initiated by Albanese, Jordá and Mele. In [1], among other things, they showed that a composition operator $C_\psi :{\mathscr {O}}_M({\mathbb {R}})\rightarrow {\mathscr {O}}_M({\mathbb {R}}), f\mapsto f\circ \psi$ with a smooth symbol $\psi :{\mathbb {R}}\rightarrow {\mathbb {R}}$ is correctly defined (and hence continuous by a standard application of De Wilde’s Closed Graph Theorem) if and only if $\psi \in {\mathscr {O}}_M({\mathbb {R}})$. Moreover, they studied dynamical properties (power boundedness, mean ergodicity) of those operators and showed that the translation operator is mixing on ${\mathscr {O}}_M({\mathbb {R}})$.

Composition operators play an important role in functional analysis. Their dynamical properties on various spaces of functions and sequences were intensively studied over the past decades by many authors, see [4, 5, 11, 12, 17, 22, 24] for (weighted) composition operators on spaces of holomorphic functions, [6] for composition operators on spaces of analytic functions, [21] for weighted composition operators on spaces of smooth functions, [16] for composition operators on spaces of functions defined by local properties, [10] for weighted translation operators acting on the Schwartz space, [15] for weighted composition operators on $L^p$-spaces and weighted spaces of continuous functions, and [6, 8, 19] for hypercyclicity results for non-metrizable locally convex spaces, as well as the references therein.

It is the purpose of this note to complement the results from [1]. In particular, in Theorems 7 and 13 we characterize mixing composition operators $C_\psi$ on ${\mathscr {O}}_M({\mathbb {R}})$ in terms of their symbol $\psi$. Moreover we show in Theorem 15 that this property is closely related to the solvability in ${\mathscr {O}}_M({\mathbb {R}})$ of Abel’s functional equation, i.e. the problem to find for a given symbol $\psi \in {\mathscr {O}}_M({\mathbb {R}})$ a function $H\in {\mathscr {O}}_M({\mathbb {R}})$ which satisfies the equation

$$\begin{aligned} H(\psi (x))=H(x)+1. \end{aligned}$$

Additionally, we give a sufficient condition on the symbol $\psi$ of a composition operator $C_\psi$ on ${\mathscr {O}}_M({\mathbb {R}})$ to be (sequentially) hypercyclic, see Theorem 3. This condition allows to identify the translation operator to be (sequentially) hypercyclic on ${\mathscr {O}}_M({\mathbb {R}})$. Moreover, thanks to Theorem 15, we deduce in Corollary 16 that many mixing composition operators are (sequentially) hypercyclic.

It has been shown by the second author in [21, Theorem 4.2] that a composition operator $C_\psi$ on the Fréchet space of smooth functions $C^\infty ({\mathbb {R}})$ is hypercyclic if and only if it is mixing if and only if $\psi$ has a non-vanishing derivative and no fixed points. Applying standard arguments (see Proposition 2), it is easily seen that for a symbol $\psi \in {\mathscr {O}}_M({\mathbb {R}})$ with topologically transitive $C_\psi$ on ${\mathscr {O}}_M({\mathbb {R}})$, the corresponding composition operator on the space of smooth functions $C^\infty ({\mathbb {R}})$ is topologically transitive as well. We give an example (see Example 10) that the converse implication is not true.

The paper is organized as follows. In Sect. 2, after recalling some topological properties of ${\mathscr {O}}_M({\mathbb {R}})$ which will be relevant for our purpose, we study (sequential) hypercyclicity of composition operators on ${\mathscr {O}}_M({\mathbb {R}})$. In Sect. 3 we characterize mixing composition operators in terms of their symbol while in Sect. 4 we investigate the connection between mixing and the solvability of Abel’s equation in ${\mathscr {O}}_M({\mathbb {R}})$.

Through the paper, by ${\mathbb {N}}=\{1,2,3,\ldots \}$ we denote the set of natural numbers. For a function $\psi :{\mathbb {R}}\rightarrow {\mathbb {R}}$

• we define $\psi _0:{\mathbb {R}}\rightarrow {\mathbb {R}}$ as $\psi _0(x)=x$,

• for every $n\in {\mathbb {N}}$ we define $\psi _n:{\mathbb {R}}\rightarrow {\mathbb {R}}$ inductively via the formula $\psi _n(x)=\psi (\psi _{n-1}(x))$,

• whenever $\psi$ is injective, for every $n\in {\mathbb {Z}}\backslash ({\mathbb {N}}\cup \{0\})$ we define $\psi _n:\psi _{-n}({\mathbb {R}})\rightarrow {\mathbb {R}}$ via the rule: $\psi _n(x)=y$ if and only if $\psi _{-n}(y)=x$.

Finally, for further reference, let us recall Faà di Bruno’s formula which states for smooth functions f, g and $j\in {\mathbb {N}}$

$$\begin{aligned} (f\circ g)^{(j)}(x)=\sum _{\begin{array}{c} i_1,i_2,\ldots , i_j\ge 0 \\ i_1+2i_2+\cdots +ji_j=j \end{array}} \frac{j!}{i_1!i_2!\cdots i_j!}f^{(i_1+i_2+\cdots +i_j)}(g(x)) \cdot \prod _{r=1}^j\left( \frac{g^{(r)}(x)}{r!}\right) ^{i_r}. \end{aligned}$$

For the definition of hypercyclicity, mixing and other unexplained notions from linear dynamics we refer to [13], while we refer to [18] for anything related to functional analysis.

2 Hypercyclicity of composition operators

Obviously, on

$$\begin{aligned} {\mathscr {O}}_n^m({\mathbb {R}})=\left\{ f\in C^m({\mathbb {R}}):\,|f|_{m,n}:=\sup _{x\in {\mathbb {R}}, 0\le j\le m}(1+|x|^2)^{-n}|f^{(j)}(x)|<\infty \right\} \end{aligned}$$

a norm is given by $|\cdot |_{m,n}$ and equipped with this norm, ${\mathscr {O}}_n^m({\mathbb {R}})$ is a Banach space, $m,n\in {\mathbb {N}}$. Additionally, ${\mathscr {O}}^m({\mathbb {R}}):={\text {ind}}_{n\rightarrow \infty }{\mathscr {O}}_n^m({\mathbb {R}})$ is a complete (LB)-space. The space ${\mathscr {O}}_M({\mathbb {R}})$ is endowed with its natural locally convex topology, i.e. ${\mathscr {O}}_M({\mathbb {R}})$ is the projective limit of the (LB)-spaces ${\mathscr {O}}^m({\mathbb {R}})$, $m\in {\mathbb {N}}$, where the linking maps from ${\mathscr {O}}^{m+1}({\mathbb {R}})$ to ${\mathscr {O}}^m({\mathbb {R}})$ are the inclusions.

A fundamental system of continuous seminorms on ${\mathscr {O}}_M({\mathbb {R}})$ is given by

$$\begin{aligned} p_{m,v}(f)=\sup _{x\in {\mathbb {R}}}\max _{0\le j\le m}|v(x) f^{(j)}(x)|,\, f\in {\mathscr {O}}_M({\mathbb {R}}), m\ge 0, v\in {\mathscr {S}}({\mathbb {R}}), \end{aligned}$$

where ${\mathscr {S}}({\mathbb {R}})$ is the space of rapidly decreasing smooth functions (see [14]). In fact it is not difficult to see that ${\mathscr {O}}_M({\mathbb {R}})$ is the space of smooth functions f on ${\mathbb {R}}$ such that $p_{m,v}(f)$ is finite for every $m\ge 0$ and $v\in {\mathscr {S}}({\mathbb {R}})$, as well as the space of multipliers of ${\mathscr {S}}({\mathbb {R}})$. Obviously, ${\mathscr {O}}_M({\mathbb {R}})$ embeds continuously into $C^\infty ({\mathbb {R}})$, and, as is well known, the space of compactly supported smooth functions ${\mathscr {D}}({\mathbb {R}})$ is dense in ${\mathscr {O}}_M({\mathbb {R}})$. Consequently, ${\mathscr {O}}_M({\mathbb {R}})$ is dense in $C^\infty ({\mathbb {R}})$. Below we will use the following property of the topology of ${\mathscr {O}}_M({\mathbb {R}})$ (see [1, Remark 2.2]).

Fact 1

A sequence $(f_n)_{n\in {\mathbb {N}}}$ of functions from ${\mathscr {O}}_M({\mathbb {R}})$ is convergent to f in ${\mathscr {O}}_M({\mathbb {R}})$ if and only if $(f_n)_{n\in {\mathbb {N}}}$ is bounded in ${\mathscr {O}}_M({\mathbb {R}})$ and converges to f in $C^\infty ({\mathbb {R}})$.

Recently, it was shown in [1] that the translation operator on ${\mathscr {O}}_M({\mathbb {R}})$, i.e. $C_\psi :{\mathscr {O}}_M({\mathbb {R}})\rightarrow {\mathscr {O}}_M({\mathbb {R}})$, $f\mapsto f \circ \psi$ with $\psi (x)=x+1$, is mixing. As already mentioned in the introduction, ${\mathscr {O}}_M({\mathbb {R}})$ is not a Fréchet space and thus, Birkhoff’s Transitivity Theorem cannot be applied to conclude hypercyclicity of the translation operator. The main objective of this section is to prove that the translation operator is indeed hypercyclic on ${\mathscr {O}}_M({\mathbb {R}})$.

We start with the following trivial observation.

Proposition 2

For $\psi \in {\mathscr {O}}_M({\mathbb {R}})$ consider the following conditions.

• (i) The composition operator $C_\psi :{\mathscr {O}}_M({\mathbb {R}})\rightarrow {\mathscr {O}}_M({\mathbb {R}})$ is topologically transitive.

• (ii) The composition operator ${\tilde{C}}_\psi :C^\infty ({\mathbb {R}})\rightarrow C^\infty ({\mathbb {R}}), f\mapsto f\circ \psi$ is topologically transitive.

• (iii) $\psi$ has no fixed points and $\psi '(x)>0$ for every $x\in {\mathbb {R}}$.

Then, (i) implies (ii), while (ii) and (iii) are equivalent.

Proof

Since the inclusion $i:{\mathscr {O}}_M({\mathbb {R}})\rightarrow C^\infty ({\mathbb {R}}), f\mapsto f$ is continuous and has dense range, topological transitivity of $C_\psi$ implies the topological transitivity of ${\tilde{C}}_\psi$ (see, [13, Proposition 1.13]). Thus, by [21, Theorem 4.2], $\psi$ has no fixed points and $\psi '(x)\ne 0$ for every $x\in {\mathbb {R}}$. Hence, we either have $\psi '(x)>0$ for every $x\in {\mathbb {R}}$ or $\psi '(x)<0$ for each $x\in {\mathbb {R}}$. Since the latter condition contradicts that $\psi$ has no fixed points, (ii) implies (iii). Another application of [21, Theorem 4.2] shows that (iii) implies (ii). $\square$

In contrast to composition operators on the Fréchet space $C^\infty ({\mathbb {R}})$, the conditions in (i) of the previous Proposition are only necessary but not sufficient for topological transitivity of a composition operator on ${\mathscr {O}}({\mathbb {R}})$, as will be shown in Example 10 below.

Theorem 3

Let $\psi \in {\mathscr {O}}_M({\mathbb {R}})$ be bijective such that $\psi (x)>x$ as well as $\psi '(x)>0$ for every $x\in {\mathbb {R}}$. Additionally, assume that

$$\begin{aligned} \forall \,j\in {\mathbb {N}}\,\exists \,\beta _j\in {\mathbb {R}}, C_j>0, t_j\in {\mathbb {N}}\,\forall \,x\in (\beta _j,\infty ), n\in {\mathbb {N}}:\,|(\psi _{-n})^{(j)}(x)|\le C_j(1+|x|^2)^{t_j}\nonumber \\ \end{aligned}$$

(1)

and

$$\begin{aligned} \forall \,j\in {\mathbb {N}}\,\exists \,\alpha _j\in {\mathbb {R}}, C_j>0, t_j\in {\mathbb {N}}\,\forall \,x\in (-\infty ,\alpha _j), n\in {\mathbb {N}}:\,|(\psi _n)^{(j)}(x)|\le C_j(1+|x|^2)^{t_j}.\nonumber \\ \end{aligned}$$

(2)

Then, $C_{\psi }:{\mathscr {O}}_M({\mathbb {R}})\rightarrow {\mathscr {O}}_M({\mathbb {R}})$ is sequentially hypercyclic.

Before we prove Theorem 3 we make the following comment.

Remark 4

For bijective $\psi \in {\mathscr {O}}_M({\mathbb {R}})$ without fixed points, it either holds $\psi (x)>x$ for every $x\in {\mathbb {R}}$ or $\psi (x)<x$ for each $x\in {\mathbb {R}}$. While Theorem 3 deals with the first case, replacing in hypothesis (1) “$\forall x\in (\beta _j,\infty )$” by “$\forall x\in (-\infty ,\beta _j)$” and in hypothesis (2) “$\forall x\in (-\infty ,\alpha _j)$” by “$\forall x\in (\alpha _j,\infty )$” gives an analogous result for the case $\psi (x)<x, x\in {\mathbb {R}}$. Indeed, let $r(x)=-x$, $x\in {\mathbb {R}}$, be the reflection at the origin. Then, $C_r$ is bijective on ${\mathscr {O}}_M({\mathbb {R}})$ with $C_r^2={\text {id}}_{{\mathscr {O}}_M({\mathbb {R}})}$. Additionally, for $\psi \in {\mathscr {O}}_M({\mathbb {R}})$ we set $\sigma (\psi )=-C_r(\psi )$ so that $\sigma (\sigma (\psi ))=\psi$. Then, we have $C_\psi =C_r\circ C_{\sigma (\psi )}\circ C_r$, so that $C_\psi$ and $C_{\sigma (\psi )}$ are conjugate, in particular, $C_\psi$ is (sequentially) hypercyclic if and only if $C_{\sigma (\psi )}$ is. Obviously, $\psi (x)<x$ for every $x\in {\mathbb {R}}$ precisely when $\sigma (\psi )(x)>x$ for every $x\in {\mathbb {R}}$. Additionally, $\psi$ is bijective if and only if $\sigma (\psi )$ is bijective, and $\left( \sigma (\psi )\right) ^{(j)}(x)=(-1)^{j-1}\psi ^{(j)}(-x)=(-1)^j\sigma (\psi ^{(j)})(x)$.

Proof of Theorem 3

We will explicitly construct a function $g\in {\mathscr {O}}_M({\mathbb {R}})$ whose orbit under $C_\psi$ is sequentially dense in ${\mathscr {O}}_M({\mathbb {R}})$. In order to do so, let $(p_n)_{n\in {\mathbb {N}}}$ be a sequence of compactly supported smooth functions on ${\mathbb {R}}$ such that $\{p_n: n\in {\mathbb {N}}\}$ is dense in ${\mathscr {O}}_M({\mathbb {R}})$ and such that for every $m\in {\mathbb {N}}$ there are infinitely many $n\in {\mathbb {N}}$ with $p_n=p_m$.

Since $\psi$ is bijective, without fixed points, and $\psi (x)>x$ for every x, the sequence $(\psi _n(x))_{n\in {\mathbb {N}}}$ is strictly increasing and tends to infinity while $(\psi _{-n}(x))_{n\in {\mathbb {N}}}$ is strictly decreasing with $\lim _{n\rightarrow \infty }\psi _{-n}(x)=-\infty$. In particular, for every compact subset K of ${\mathbb {R}}$ there is $N\in {\mathbb {N}}$ such that neither $\psi _n(K)$ nor $\psi _{-n}(K)$ intersects K whenever $n\ge N$.

Next, we choose a strictly increasing sequence $(k_n)_{n\in {\mathbb {N}}}$ of nonnegative integers by the following recursive procedure. First we choose $k_1=0$. If $k_1,\ldots , k_n$ have already been chosen, let $k_{n+1}$ be strictly larger than $k_n$ such that the following conditions are satisfied.

• (a) There exists $t\in {\mathbb {R}}$ such that the support of $p_n\circ \psi _{-k_n}$ is contained in $(-\infty , t)$ while the support of $p_{n+1}\circ \psi _{-k_{n+1}}$ is contained in $(t,\infty )$.

• (b) For $1\le l\le n$ and $0\le j\le n+1$ the support of $p_l\circ \psi _{k_{n+1}-k_l}$ is contained in $(-\infty ,\min \{-n-1,\alpha _1,\ldots ,\alpha _{n+1}\})$ and $\left| p_l^{(j)}\left( \psi _{k_{n+1}-k_l}(x)\right) \right| \le |x|$ for every $x\in {\mathbb {R}}$.

• (c) For $1\le l\le n$ and $0\le j\le n+1$ the support of $p_{n+1}\circ \psi _{-(k_{n+1}-k_l)}$ is contained in $(\max \{n+1,\beta _1,\ldots ,\beta _{n+1}\},\infty )$ and $\left| p_{n+1}^{(j)}\left( \psi _{-(k_{n+1}-k_l)}(x)\right) \right| \le |x|$ for every $x\in {\mathbb {R}}$.

It is clear that such a choice of $k_{n+1}$ is possible.

From (a) it follows that the functions $p_n\circ \psi _{-k_n}$, $n\in {\mathbb {N}}$, have pairwise disjoint supports so that by

$$\begin{aligned} \forall \,x\in {\mathbb {R}}:\, g(x)=\sum _{n=1}^\infty p_n\left( \psi _{-k_n}(x)\right) \end{aligned}$$

a smooth function g is defined on ${\mathbb {R}}$. Keeping in mind that for $x\in {\mathbb {R}}$ at most one of the defining summands of g does not vanish at x, an application of condition (c) for $l=1$ combined with hypothesis (1) on $\psi$, and Faà di Bruno’s formula yields for any nonnegative integer m the existence of $C>0$ and $t\in {\mathbb {N}}$ such that for $x\in {\mathbb {R}}$ and $0\le j\le m$

$$\begin{aligned} |g^{(j)}(x)|\le \max \left\{ \left| \left( p_n\circ \psi _{-k_n}\right) ^{(j)}(y)\right| ; 1\le n\le m,\, y\in {\mathbb {R}}\right\} +C(1+|x|^2)^t|x|, \end{aligned}$$

so that $g\in {\mathscr {O}}_M({\mathbb {R}})$.

We claim that $\{C_\psi ^n(g): n\in {\mathbb {N}}\}$ is sequentially dense in ${\mathscr {O}}_M({\mathbb {R}})$. To prove this it is enough to show that the set $\{p_N: N\in {\mathbb {N}}\}$ is contained in the sequential closure of $\{C_\psi ^n(g): n\in {\mathbb {N}}\}$. Thus, we fix $N\in {\mathbb {N}}$. Let $(s_i)_{i\in {\mathbb {N}}}$ be a strictly increasing sequence of nonnegative integers such that $p_N=p_{s_i}$, $i\in {\mathbb {N}}$. Then, for $i\in {\mathbb {N}}$, we have

$$\begin{aligned} C_\psi ^{k_{s_i}}(g)-p_N=\sum _{n=1}^{s_i-1} \left( p_n\circ \psi _{k_{s_i}-k_n}\right) + \sum _{n=s_i+1}^\infty \left( p_n\circ \psi _{-(k_n-k_{s_i})}\right) . \end{aligned}$$

(3)

By condition (b), for $n<s_i$, the support of $p_n\circ \psi _{k_{s_i}-k_n}$ is contained in $(-\infty , -s_i)$. Likewise, for $n>s_i$, condition (c) ensures that the support of $p_n\circ \psi _{-(k_n-k_{s_i})}$ is contained in $(s_i,\infty )$. Hence, both sequences of functions

$$\begin{aligned} \left( \sum _{n=1}^{s_i-1} \left( p_n\circ \psi _{k_{s_i}-k_n}\right) \right) _{i\in {\mathbb {N}}}\quad \text {and}\quad \left( \sum _{n=s_i+1}^\infty \left( p_n\circ \psi _{-(k_n-k_{s_i})}\right) \right) _{i\in {\mathbb {N}}} \end{aligned}$$

(4)

converge to zero in $C^\infty ({\mathbb {R}})$. Thus, Fact 1 combined with (3) will imply $p_N=\lim _{i\rightarrow \infty }C_\psi ^{k_{s_i}}(g)$ in ${\mathscr {O}}_M({\mathbb {R}})$ once we have shown that both sequences in (4) are bounded in ${\mathscr {O}}_M({\mathbb {R}})$, thereby completing the proof.

Since $p_n\circ \psi _{-k_n}$, $n\in {\mathbb {N}}$, have mutually disjoint supports, the summands of the first sequence in (4) have mutually disjoint supports as do the ones of the second sequence. The same arguments which we used to prove that g belongs to ${\mathscr {O}}_M({\mathbb {R}})$ yield that the second sequence in (4) is bounded in ${\mathscr {O}}_M({\mathbb {R}})$. Refering to hypothesis (2) and to condition (b) instead of hypothesis (1) and condition (c), respectively, one shows mutatis mutandis that the first sequence in (4) is bounded in ${\mathscr {O}}_M({\mathbb {R}})$, too. $\square$

Corollary 5

Let $\psi \in {\mathscr {O}}_M({\mathbb {R}})$ be bijective, without fixed points and such that $\psi '(x)>0$ for every $x\in {\mathbb {R}}$ and such that $\{(\psi _n)':\,n\in {\mathbb {Z}}\}$ is bounded in ${\mathscr {O}}_M({\mathbb {R}})$. Then, the composition operator $C_\psi :{\mathscr {O}}_M({\mathbb {R}})\rightarrow {\mathscr {O}}_M({\mathbb {R}})$ is sequentially hypercyclic.

Proof

Since either $\psi (x)>x$ for every $x\in {\mathbb {R}}$ or $\psi (x)<x$ for every $x\in {\mathbb {R}}$ the assertion follows immediately from Theorem 3 and the comment preceding its proof. $\square$

Corollary 6

For $\beta \in {\mathbb {R}}\backslash \{0\}$ and $\psi (x)=x+\beta$ the composition operator $C_\psi$ is sequentially hypercyclic on ${\mathscr {O}}_M({\mathbb {R}})$. Additionally, $C_\psi$ is chaotic.

Proof

The sequential hypercyclicity of $C_\psi$ follows immediately from Corollary 5. Additionally, considering the set $\{\sum _{n\in {\mathbb {Z}}}g(\cdot + nl k_g \beta ); g\in {\mathscr {D}}({\mathbb {R}}), l\in {\mathbb {N}}\}$, where $k_g\in {\mathbb {N}}$ is chosen in such a way that $[\min \text {supp}\,g,\max \text {supp}\,g ]$ and $[k_g+\min \text {supp}\,g,k_g+\max \text {supp}\,g ]$ are disjoint. Then, by standard arguments, this set is dense in ${\mathscr {O}}_M({\mathbb {R}})$ and consists of periodic points for $C_\psi$. Thus, $C_\psi$ is chaotic. $\square$

3 Mixing composition operators

In this section we characterize mixing operators $C_\psi$ acting on ${\mathscr {O}}_M({\mathbb {R}})$ in terms of their symbol $\psi$.

Theorem 7

Let $\psi \in {\mathscr {O}}_M({\mathbb {R}})$ be surjective. Then, the following conditions are equivalent.

• (i) The operator $C_\psi :{\mathscr {O}}_M({\mathbb {R}})\rightarrow {\mathscr {O}}_M({\mathbb {R}})$, $f\mapsto f \circ \psi$ is mixing.

• (ii) $\psi$ is injective with a non-vanishing derivative and without fixed points such that for every $a\in {\mathbb {R}}$ and each $k\in {\mathbb {N}}$, for arbitrary $v\in {\mathscr {S}}({\mathbb {R}})$ it holds

  $$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{x\in \psi _{-n}\left( [\min \{a,\psi (a)\},\max \{a,\psi (a)\}]\right) } \left| v(x)(\psi _n)^{(k)}(x)\right| =0 \end{aligned}$$

  and

  $$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{x\in \psi _{n}\left( [\min \{a,\psi (a)\},\max \{a,\psi (a)\}]\right) } \left| v(x)(\psi _{-n})^{(k)}(x)\right| =0. \end{aligned}$$

• (iii) $\psi$ is injective with a non-vanishing derivative and without fixed points, and there are $a,b\in {\mathbb {R}}$ such that for every $k\in {\mathbb {N}}$ and $v\in {\mathscr {S}}({\mathbb {R}})$ we have

  $$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{x\in \psi _{-n}\left( [\min \{a,\psi (a)\},\max \{a,\psi (a)\}]\right) } \left| v(x)(\psi _n)^{(k)}(x)\right| =0 \end{aligned}$$

  and

  $$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{x\in \psi _{n}\left( [\min \{b,\psi (b)\},\max \{b,\psi (b)\}]\right) } \left| v(x)(\psi _{-n})^{(k)}(x)\right| =0. \end{aligned}$$

Proof

Clearly, (ii) implies (iii). (iii) $\Rightarrow$ (i) Since $\psi$ does not have a fixed point, we have either $\psi (x)>x$ for all $x\in {\mathbb {R}}$, or $\psi (x)<x$ for all $x\in {\mathbb {R}}$. We only consider the case $\psi (x)>x$, the other case is treated, mutatis mutandis, with the same arguments. Thus, $\min \{a,\psi (a)\}=a, \max \{a,\psi (a)\}=\psi (a)$ and $\min \{b,\psi (b)\}=b, \max \{b,\psi (b)\}=\psi (b)$.

Taking into account that $\lim _{n\rightarrow \infty }\psi _n(a)=\infty$ and $\lim _{n\rightarrow \infty }\psi _{-n}(a)=-\infty$ we have ${\mathbb {R}}=\cup _{m\in {\mathbb {Z}}}\left( \psi _m(a),\psi _{m+2}(a)\right)$ and the sequence of open intervals $\left( \psi _m(a),\psi _{m+2}(a)\right) _{m\in {\mathbb {Z}}}$ is a locally finite cover of ${\mathbb {R}}$. Let $(\phi _m)_{m\in {\mathbb {Z}}}$ be a partition of unity on ${\mathbb {R}}$ subordinate to it. Likewise, let $(\eta _m)_{m\in {\mathbb {Z}}}$ be a partition of unity on ${\mathbb {R}}$ subordinate to the locally finite cover of ${\mathbb {R}}$ by the sequence of open intervals $\left( (\psi _m(b),\psi _{m+2}(b))\right) _{m\in {\mathbb {Z}}}$.

Since compactly supported functions are dense in ${\mathscr {O}}_M({\mathbb {R}})$, in view of Kitai’s criterion (see [13, Thm. 12.31]), it is enough to show that for every compactly supported smooth function f the sequences $(f\circ \psi _n)_{n\in {\mathbb {N}}}$ and $(f\circ \psi _{-n})_{n\in {\mathbb {N}}}$ converge to zero in ${\mathscr {O}}_M({\mathbb {R}})$. Note that with f also $f\circ \psi _{-n}$ is a compactly supported smooth function and thus belongs to ${\mathscr {O}}_M({\mathbb {R}})$. When considering $(f\circ \psi _n)_{n\in {\mathbb {N}}}$, respectively $(f\circ \psi _{-n})_{n\in {\mathbb {N}}}$, we may replace f by $\phi _m f$ and $\eta _m f$, respectively, so that without loss of generality $\text {supp}\,f\subset (\psi _m(a),\psi _{m+2}(a))$ and $\text {supp}\,f\subset (\psi _m(b),\psi _{m+2}(b))$, respectively.

We will show that the sequence $(f\circ \psi _n)_{n\in {\mathbb {N}}}$ tends to zero in ${\mathscr {O}}_M({\mathbb {R}})$. To do this, let us fix $v\in {\mathscr {S}}({\mathbb {R}})$ and $k\ge 0$ and we observe

$$\begin{aligned} \sup _{x\in {\mathbb {R}}}\left| v(x)(f\circ \psi _n)^{(k)}(x)\right|= & {} \sup _{x\in [\psi _{m-n}(a),\psi _{m+2-n}(a)]}\left| v(x)(f\circ \psi _n)^{(k)}(x)\right| \nonumber \\\le & {} \sup _{x\in \psi _{-(n-m)}\left( [a,\psi (a)]\right) }\left| v(x)\left( (f\circ \psi _m)\circ \psi _{n-m}\right) ^{(k)}(x)\right| \nonumber \\{} & {} +\sup _{x\in \psi _{-(n-m-1)}\left( [a,\psi (a)]\right) }\left| v(x)\left( (f\circ \psi _{m+1})\circ \psi _{n-m-1}\right) ^{(k)}(x)\right| .\nonumber \\ \end{aligned}$$

(5)

Setting $g=f\circ \psi _m$, in case of $k\ge 1$, for the first summand of the above right hand side, we conclude with Faà di Bruno’s formula

$$\begin{aligned}{} & {} \sup _{x\in \psi _{-(n-m)}\left( [a,\psi (a)]\right) }\left| v(x)\left( (f\circ \psi _m)\circ \psi _{n-m}\right) ^{(k)}(x)\right| \\{} & {} \quad =\sup _{x\in \psi _{-(n-m)}\left( [a,\psi (a)]\right) }\left| \sum _{\begin{array}{c} i_1,i_2,\ldots , i_k\ge 0 \\ i_1+\cdots +ki_k=k \end{array}}\frac{k!}{i_1!\cdots i_k!} g^{(i_1+\cdots +i_k)}\left( \psi _{n-m}(x)\right) v(x)\prod _{r=1}^k\left( \frac{\psi _{n-m}^{(r)}(x)}{r!}\right) ^{i_r}\right| \\{} & {} \quad \le \max _{\begin{array}{c} y\in [a,\psi (a)],\\ 0\le j\le k \end{array}}|g^{(j)}(y)|\sum _{\begin{array}{c} i_1,i_2,\ldots , i_k\ge 0 \\ i_1+\cdots +ki_k=k \end{array}}\frac{k!}{i_1!\cdots i_k!} \sup _{x\in \psi _{-(n-m)}\left( [a,\psi (a)]\right) }\left| v(x)\prod _{r=1}^k\left( \frac{\psi _{n-m}^{(r)}(x)}{r!}\right) ^{i_r}\right| , \end{aligned}$$

which, by the hypotheses on $\psi$ combined with the fact for all $s\in {\mathbb {N}}$ the function $v\in {\mathscr {S}}({\mathbb {R}})$ can be written as a product of s functions from ${\mathscr {S}}({\mathbb {R}})$ (see [23]), tends to zero as n goes to infinity. In case of $k=0$ it follows

$$\begin{aligned} \sup _{x\in \psi _{-(n-m)}\left( [a,\psi (a)]\right) }{} & {} \left| v(x)\left( (f\circ \psi _m)\circ \psi _{n-m}\right) (x)\right| \\{} & {} \quad \le \max _{y\in [a,\psi (a)]}|g(y)|\sup _{x\in \psi _{-(n-m)}\left( [a,\psi (a)]\right) }\left| v(x)\right| , \end{aligned}$$

which clearly converges to zero as n goes to infinity since $v\in {\mathscr {S}}({\mathbb {R}})$.

In the same way one proves that the second summand in the right hand side of (5) converges to zero when n tends to infinity which implies

$$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{x\in {\mathbb {R}}}\left| v(x)(f\circ \psi _n)^{(k)}(x)\right| =0 \end{aligned}$$

for every $k\ge 0$, i.e. $(f\circ \psi _n)_{n\in {\mathbb {N}}}$ converges to zero in ${\mathscr {O}}_M({\mathbb {R}})$. That $(f\circ \psi _{-n})_{n\in {\mathbb {N}}}$ converges to zero in ${\mathscr {O}}_M({\mathbb {R}})$, too, is proved along the same lines.

(i) $\Rightarrow$ (ii) That $\psi$ is injective with a non-vanishing derivative and without fixed points follows from Proposition 2. In order to prove the rest of the properties from (ii), let $a\in {\mathbb {R}}$ be arbitrary. We proceed by induction with respect to k. In what follows, we consider only the case $\psi (x)>x$ for every $x\in {\mathbb {R}}$. In case of $\psi (x)<x$ for every $x\in {\mathbb {R}}$, one only has to replace $[a,\psi (a)]$ by $[\psi (a),a]$ in the arguments below. Let $v\in {\mathscr {S}}({\mathbb {R}})$ and $\varepsilon >0$ be arbitrary. The sets

$$\begin{aligned} U=\{f\in {\mathscr {O}}_M({\mathbb {R}}):|f'(x)|>1 \text { for }x\in [a,\psi (a)]\} \end{aligned}$$

and

$$\begin{aligned} V=\bigg \{f\in {\mathscr {O}}_M({\mathbb {R}}): \sup _{x\in {\mathbb {R}}} \left| v(x)f'(x)\right| <\varepsilon \bigg \} \end{aligned}$$

are non-empty and open in ${\mathscr {O}}_M({\mathbb {R}})$. Since $C_\psi$ is mixing, there exists $N\in {\mathbb {N}}$ with

$$\begin{aligned} C_\psi ^n(U)\cap V\not =\emptyset \quad \text {and}\quad C_\psi ^n(V)\cap U\not =\emptyset \quad \text {for every}\quad n\ge N. \end{aligned}$$

Let $n\ge N$. There are $f,g\in U$ with $f\circ \psi _n\in V$ and $g\circ \psi _{-n}\in V$. We have

$$\begin{aligned} \sup _{x\in \psi _{-n}\left( [a,\psi (a)]\right) } \left| v(x)\psi _n'(x)\right| \le&\sup _{x\in \psi _{-n}\left( [a,\psi (a)]\right) } \left| v(x)f'(\psi _n(x))\psi _n'(x)\right| <\varepsilon \end{aligned}$$

and

$$\begin{aligned} \sup _{x\in \psi _{n}\left( [a,\psi (a)]\right) } \left| v(x)\psi _{-n}'(x)\right| \le&\sup _{x\in \psi _{n}\left( [a,\psi (a)]\right) } \left| v(x)g'(\psi _{-n}(x))\psi _{-n}'(x)\right| <\varepsilon . \end{aligned}$$

This shows that the condition in (ii) holds for $k=1$.

Assume now that the condition in (ii) holds up to $k-1$. To finish the induction, for arbitrary $v\in {\mathscr {S}}({\mathbb {R}})$ we have to show

$$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{x\in \psi _{-n}\left( [a,\psi (a)]\right) } \left| v(x)(\psi _n)^{(k)}(x)\right| =0 \text { and } \lim _{n\rightarrow \infty } \sup _{x\in \psi _{n}\left( [a,\psi (a)]\right) } \left| v(x)(\psi _{-n})^{(k)}(x)\right| =0. \end{aligned}$$

We will show the second assertion, the first is proved in a similar way.

Let $\varepsilon >0$ be arbitrary,

$$\begin{aligned} U=\left\{ f\in {\mathscr {O}}_M({\mathbb {R}}): 1<\left| f^{(l)}(x)\right| <M \text { for } x\in [a,\psi (a)],~ 0\le l\le k\right\} , \end{aligned}$$

where

$$\begin{aligned} M=2\max _{0\le i\le k}\frac{(k+1)!(\psi (a)-a+2)^{k+1-i}}{(k+1-i)!}, \end{aligned}$$

and

$$\begin{aligned}V=\left\{ f\in {\mathscr {O}}_M({\mathbb {R}}):\sup _{x\in {\mathbb {R}}} \left| v(x) f^{(k)}(x)\right| <\frac{\varepsilon }{2}\right\} .\end{aligned}$$

It is clear that U and V are open and non-empty (the polynomial $(x-a+2)^{k+1}$ is in U). Since $C_\psi$ is mixing, there is N such that for all $n\ge N$ we have $C_\psi ^n(V)\cap U\not =\emptyset$. Let $n\ge N$. There is $f\in U$ with $f\circ \psi _{-n}\in V$. Because $f\in U$,

$$\begin{aligned} \sup _{x\in \psi _{n}\left( [a,\psi (a)]\right) } \left| v(x)(\psi _{-n})^{(k)}(x)\right| \le&\sup _{x\in \psi _{n}\left( [a,\psi (a)]\right) } \left| v(x)f'(\psi _{-n}(x))(\psi _{-n})^{(k)}(x)\right| \end{aligned}$$

and, by Faà di Bruno’s formula,

$$\begin{aligned}&\sup _{x\in \psi _{n}\left( [a,\psi (a)]\right) } \left| v(x)\cdot \left( f'(\psi _{-n}(x))(\psi _{-n})^{(k)}(x)- (f\circ \psi _{-n})^{(k)}(x)\right) \right| \\&\quad \le \sup _{x\in \psi _{n}\left( [a,\psi (a)]\right) } |v(x)|\cdot \sum _{\begin{array}{c} i_1,i_2,\ldots , i_{k-1}\ge 0 \\ i_1+2i_2+\cdots +(k-1)i_{k-1}=k \end{array}} \frac{k!\cdot M}{i_1!i_2!\cdots i_{k-1}!}\cdot \prod _{j=1}^{k-1}\left( \frac{\left| \psi _{-n}^{(j)}(x)\right| }{j!}\right) ^{i_j}. \end{aligned}$$

Since for all $s\in {\mathbb {N}}$ every function from ${\mathscr {S}}({\mathbb {R}})$ can be written as a product of s functions from ${\mathscr {S}}({\mathbb {R}})$ (see [23]), the above and the inductive hypothesis imply that for n large enough

$$\begin{aligned} \sup _{x\in \psi _{n}\left( [a,\psi (a)]\right) } \left| v(x)\cdot \left( f'(\psi _{-n}(x))(\psi _{-n})^{(k)}(x)- (f\circ \psi _{-n})^{(k)}(x)\right) \right| <\frac{\varepsilon }{2}. \end{aligned}$$

For n large enough, because $f\circ \psi _{-n}\in V$, we have

$$\begin{aligned} \sup _{x\in \psi _{n}\left( [a,\psi (a)]\right) } \left| v(x)(f\circ \psi _{-n})^{(k)}(x)\right| <\frac{\varepsilon }{2}. \end{aligned}$$

Altogether the above shows for large enough n

$$\begin{aligned} \sup _{x\in \psi _{n}\left( [a,\psi (a)]\right) } \left| v(x)(\psi _{-n})^{(k)}(x)\right| <\varepsilon \end{aligned}$$

which completes the proof. $\square$

Combining Corollary 3 with Theorem 7 we obtain the following result.

Corollary 8

Let $\psi \in {\mathscr {O}}_M({\mathbb {R}})$ be a bijective function with a non-vanishing derivative and without fixed points such that $\{(\psi _n)':\,n\in {\mathbb {Z}}\}$ is bounded in ${\mathscr {O}}_M({\mathbb {R}})$. Then, the composition operator $C_\psi :{\mathscr {O}}_M({\mathbb {R}})\rightarrow {\mathscr {O}}_M({\mathbb {R}})$ is sequentially hypercyclic and mixing.

Proof

By hypothesis, for every $k\in {\mathbb {N}}$ there are $C>0$ and $m\in {\mathbb {N}}_0$ such that $|(\psi _n)^{(k)}(x)|\le C(1+|x|^2)^m$ for every $x\in {\mathbb {R}}$ and $n\in {\mathbb {Z}}$. We consider only the case that $\psi (x)>x$. The case $\psi (x)<x$ is proved along the same lines. Hence, $\lim _{n\rightarrow \infty }\psi _n(a)=\infty$ and $\lim _{n\rightarrow \infty }\psi _{-n}(a)=-\infty$ for each $a\in {\mathbb {R}}$ so that

$$\begin{aligned} \lim _{n\rightarrow \infty }\sup _{x\in \psi _{-n}([a,\psi (a)])}|v(x)(\psi _n)^{(k)}(x)|\le C \lim _{n\rightarrow \infty }\sup _{x\in \psi _{-n}([a,\psi (a)])}|v(x)|(1+|x|^2)^m=0 \end{aligned}$$

as well as

$$\begin{aligned} \lim _{n\rightarrow \infty }\sup _{x\in \psi _{n}([a,\psi (a)])}|v(x)(\psi _{-n})^{(k)}(x)|\le C \lim _{n\rightarrow \infty }\sup _{x\in \psi _{n}([a,\psi (a)])}|v(x)|(1+|x|^2)^m=0 \end{aligned}$$

for every $v\in {\mathscr {S}}({\mathbb {R}})$, $k\in {\mathbb {N}}$. Hence, $C_\psi$ is mixing by Theorem 7. $\square$

Example 9

From Corollary 8 it easily follows that for every $\beta \not =0$ the operator $C_\psi$, where $\psi (x)=x+\beta$, is mixing on ${\mathscr {O}}_M({\mathbb {R}})$. This has already been proved in [1, Proposition 3.6].

Example 10

Let ${\widetilde{\psi }}:[0,1]\rightarrow {\mathbb {R}}$ be a smooth function such that ${\widetilde{\psi }}(x)=3x+1$ for $x\in [0,1/7]$, ${\widetilde{\psi }}(x)=3x-1$ for $x\in [6/7,1]$ and ${\widetilde{\psi }}'(x)>0$ for $x\in [0,1]$ (such a function exists by [20, Lemma 9]). The function $\psi :{\mathbb {R}}\rightarrow {\mathbb {R}}$ defined by the formula

$$\begin{aligned} \psi (x)=\widetilde{\psi (}x-n)+n \quad \text {if}\quad x\in [n,n+1],n\in {\mathbb {Z}}, \end{aligned}$$

belongs to ${\mathscr {O}}_M({\mathbb {R}})$, has no fixed points and a non-vanishing derivative. One can easily calculate that for every $n\in {\mathbb {N}}$

$$\begin{aligned} \psi _{-n}(0)=-n\quad \text {and}\quad (\psi _n)'(\psi _{-n}(0))=\psi _n'(-n)=3^n. \end{aligned}$$

Let now $v\in {\mathscr {S}}({\mathbb {R}})$ be such that $v(x)=e^x$ for $x<0$. Then

$$\begin{aligned} \lim _{n\rightarrow \infty } v(\psi _{-n}(0))(\psi _n)'(\psi _{-n}(0))=\infty . \end{aligned}$$

Thus, by Theorem 7, the operator $C_\psi$ is not mixing on ${\mathscr {O}}_M({\mathbb {R}})$. However it is mixing when considered as an operator acting on $C^\infty ({\mathbb {R}})$ by [21, Theorem 4.2].

In order to give more examples we will need the following technical lemma.

Lemma 11

Let $f\in C^\infty ({\mathbb {R}})$ be such that $\sup _{x\in {\mathbb {R}}}(1+|x|^2)^n|f(x)|<\infty$ for every $n\in {\mathbb {N}}$. Then, there exists $g\in {\mathscr {S}}({\mathbb {R}})$, non-decreasing on $(-\infty ,0]$ and non-increasing on $[0,\infty )$, such that $|f(x)|\le g(x)$ for all $x\in {\mathbb {R}}$.

Proof

We set $s_0=0$ and for every $n\in {\mathbb {N}}$ let

$$\begin{aligned} s_n=\sup _{|x|\ge n-1}|f(x)|. \end{aligned}$$

One easily verifies that $(s_n)_{n\in {\mathbb {N}}}$ is a rapidly decreasing sequence. Let $\varphi :[0,1]\rightarrow {\mathbb {R}}$ be a smooth function which is equal to 1 in a neighborhood of 0, equal to 0 in a neighborhood of 1, and is non-increasing on [0, 1]. We define $g:{\mathbb {R}}\rightarrow {\mathbb {R}}$ by the formula

$$\begin{aligned} g(x)={\left\{ \begin{array}{ll} s_{n+1}+(s_n-s_{n+1})\varphi (x-n), &{} x\in [n,n+1)\text { for some } n\in {\mathbb {N}}\cup \{0\};\\ g(-x), &{}x<0. \end{array}\right. } \end{aligned}$$

It is clear that g has all the requested properties. $\square$

Example 12

Let

$$\begin{aligned} {\widetilde{\psi }}(x)={\left\{ \begin{array}{ll} \sqrt{x^2+1}, &{} x\ge 1,\\ \frac{\sqrt{2}}{2}x+1, &{} x\in [-\sqrt{2},0],\\ -\sqrt{x^2-1}, &{} x\le -\sqrt{3}, \end{array}\right. } \end{aligned}$$

and let $\psi$ be any smooth extension of ${\widetilde{\psi }}$ to ${\mathbb {R}}$ which satisfies $\psi '(x)>0$ for all $x\in {\mathbb {R}}$ (such an extension exists by [20, Lemma 9]).

It is clear that $\psi \in {\mathscr {O}}_M({\mathbb {R}})$. We will show that the composition operator $C_\psi$ is mixing on ${\mathscr {O}}_M({\mathbb {R}})$.

In what follows we will use the following properties of the function $\psi$:

• (1) it is bijective and $\psi (x)>x$ for every $x\in {\mathbb {R}}$;

• (2) for every $x\in {\mathbb {R}}$

  $$\begin{aligned} \lim _{n\rightarrow \infty } \psi _n(x)=\infty \text { and } \lim _{n\rightarrow \infty } \psi _{-n}(x)=-\infty ; \end{aligned}$$

• (3) for every $x\le -\sqrt{2}$ and $n\in {\mathbb {N}}$ we have $\psi _{-n}(x)=-\sqrt{x^2+n}$;

• (4) for every $k\in {\mathbb {N}}$, $a\le b\le -\sqrt{2}$, $v\in {\mathscr {S}}({\mathbb {R}})$

  $$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{x\in [\psi _{-n}(a),\psi _{-n}(b)]} \left| v(x)(\psi _n)^{(k)}(x)\right| =0; \end{aligned}$$

• (5) for every $x\ge 1$ and every $n\in {\mathbb {N}}$ we have $\psi _{n}(x)=\sqrt{x^2+n}$;

• (6) for every $k\in {\mathbb {N}}$, $1\le a\le b$, $v\in {\mathscr {S}}({\mathbb {R}})$

  $$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{x\in [\psi _{n}(a),\psi _{n}(b)]} \left| v(x)(\psi _{-n})^{(k)}(x)\right| =0. \end{aligned}$$

Properties (1), (2), (3) and (5) are easy to verify, we will show now that (4) is satisfied, (6) can be checked in a similar way.

In order to prove that $\psi$ satisfies (4) we can assume (in view of Lemma 11) that v is non-negative and non-decreasing on $(-\infty ,0]$. For every $n\in {\mathbb {N}}$ and $x\in [\psi _{-n}(a),\psi _{-n}(b)]$ we have $\psi _n(x)=-\sqrt{x^2-n}$. By Faà di Bruno’s formula, for every $k\ge 1$, $n\in {\mathbb {N}}$ and $x\in [\psi _{-n}(a),\psi _{-n}(b)]$ we thus have

$$\begin{aligned} (\psi _n)^{(k)}(x)=\sum _{\begin{array}{c} i_1,i_2\ge 0 \\ i_1+2i_2=k \end{array}}C_{k,i_1,i_2} (x^2-n)^{-i_1-i_2+1/2}x^{i_1}, \end{aligned}$$

where the constants $C_{k,i_1,i_2}$ do not depend on n. Therefore

$$\begin{aligned} \sup _{x\in [\psi _{-n}(a),\psi _{-n}(b)]}&\left| v(x)(\psi _n)^{(k)}(x)\right| {\mathop {=}\limits ^{(3)}} \sup _{x\in [-\sqrt{a^2+n},-\sqrt{b^2+n}]} \left| v(x)(\psi _n)^{(k)}(x)\right| \\&\quad \le \sum _{\begin{array}{c} i_1,i_2\ge 0 \\ i_1+2i_2=k \end{array}} |C_{k,i_1,i_2}|v(-\sqrt{b^2+n})b^{-2i_1-2i_2+1}\left( \sqrt{a^2+n}\right) ^{i_1}. \end{aligned}$$

Since $v\in {\mathscr {S}}({\mathbb {R}})$ we get that

$$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{x\in [\psi _{-n}(a),\psi _{-n}(b)]} \left| v(x)(\psi _n)^{(k)}(x)\right| =0. \end{aligned}$$

By (4) and (6), $\psi$ satisfies condition (iii) of Theorem 7 so that $C_\psi$ is mixing on ${\mathscr {O}}_M({\mathbb {R}})$.

We continue with the analogue to Theorem 7 for non-surjective symbol.

Theorem 13

Let $\psi \in {\mathscr {O}}_M({\mathbb {R}})$ be non-surjective. Then, the following conditions are equivalent.

• (i) The operator $C_\psi :{\mathscr {O}}_M({\mathbb {R}})\rightarrow {\mathscr {O}}_M({\mathbb {R}})$, $f\mapsto f \circ \psi$ is mixing.

• (ii) $\psi$ is injective with a non-vanishing derivative and without fixed points such that for every $a\in {\mathbb {R}}$ and each $k\in {\mathbb {N}}$, for arbitrary $v\in {\mathscr {S}}({\mathbb {R}})$ it holds

  $$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{x\in \psi _n\left( [\min \{a,\psi (a)\},\max \{a,\psi (a)\}]\right) } \left| v(x)(\psi _{-n})^{(k)}(x)\right| =0. \end{aligned}$$

• (iii) $\psi$ is injective with a non-vanishing derivative and without fixed points, and there is $a\in {\mathbb {R}}$ such that for each $k\in {\mathbb {N}}$ and for arbitrary $v\in {\mathscr {S}}({\mathbb {R}})$ it holds

  $$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{x\in \psi _n\left( [\min \{a,\psi (a)\},\max \{a,\psi (a)\}]\right) } \left| v(x)(\psi _{-n})^{(k)}(x)\right| =0. \end{aligned}$$

Proof

Obviously, (ii) implies (iii). The implication (i) $\Rightarrow$ (ii) is shown exactly as in the proof of Theorem 7. To prove that (iii) $\Rightarrow$ (i) let us fix non-empty and open sets U and V in ${\mathscr {O}}_M({\mathbb {R}})$ and two compactly supported smooth functions $f\in U$ and $g\in V$. We first consider the case $\psi (x)>x$ for every $x\in {\mathbb {R}}$.

We set $\alpha =-1+\inf \text {supp}\,g$. Moreover, for $n\in {\mathbb {N}}$ we define

$$\begin{aligned}g_n(x)={\left\{ \begin{array}{ll} g(\psi _{-n}(x)), &{} x\in \psi _n({\mathbb {R}}),\\ 0, &{}\text {otherwise}; \end{array}\right. }\end{aligned}$$

so that $g_n$ is a compactly supported smooth function, supported in $\psi _n((\alpha ,\infty ))$. For n large enough we have

$$\begin{aligned} \psi _n((\alpha ,\infty ))\subset (a,\infty )=\cup _{m\in {\mathbb {N}}_0}(\psi _m(a),\psi _{m+2}(a)). \end{aligned}$$

Obviously, the sequence of open intervals $(\psi _m(a),\psi _{m+2}(a))_{m\in {\mathbb {N}}_0}$ is a locally finite cover of $(a,\infty )$. Let $(\phi _m)_{m\in {\mathbb {N}}_0}$ be a partition of unity on $(a,\infty )$ subordinate to this cover. For large enough N, as in the proof of Theorem 7, one shows that $(\phi _m g_n)_{n\in {\mathbb {N}}, n\ge N}$ converges to zero in ${\mathscr {O}}_M({\mathbb {R}})$ for every $m\in {\mathbb {N}}_0$ which implies that the same holds for $(g_n)_{n\in {\mathbb {N}}, n\ge N}$. Therefore $f+g_n\in U$ for n large enough. Furthermore, since $\psi _{n+1}({\mathbb {R}})\subset \psi _n({\mathbb {R}})$ for $n\in {\mathbb {N}}$ and $\displaystyle \cap _{n\in {\mathbb {N}}}\psi _n({\mathbb {R}})=\emptyset$, we have that $C_\psi ^n(f)=0$ for n large enough. Therefore, for n large enough we have $C_\psi ^n(f+g_n)=g\in V$.

In case $\psi (x)<x$, we define ${\tilde{\alpha }}=1+\sup \text {supp}\,g$. Replacing $(\alpha ,\infty )$ by $(-\infty ,{\tilde{\alpha }})$ and $(a,\infty )$ by $(-\infty ,a)$, respectively, the proof is mutatis mutandis the same. $\square$

Example 14

Let

$$\begin{aligned} {\widetilde{\psi }}(x)={\left\{ \begin{array}{ll} e^x, &{} x\le 0;\\ 2x, &{} x\ge 1, \end{array}\right. } \end{aligned}$$

and let $\psi$ be any smooth extension of ${\widetilde{\psi }}$ to ${\mathbb {R}}$ which satisfies $\psi '(x)>0$ for all $x\in {\mathbb {R}}$ (such an extension exists by [20, Lemma 9]). It is clear that $\psi \in {\mathscr {O}}_M({\mathbb {R}})$ and $\psi (x)>x$ for all $x\in {\mathbb {R}}$. Obviously $\psi _n(x)=2^nx$ whenever $x\ge 1$ and for $x\in \psi _n([1,\psi (1)])$ we have $\psi _{-n}(x)=2^{-n}x$. It is straightforward to show that condition (iii) in Theorem 13 is fulfilled for $a=1$. Therefore the composition operator $C_\psi$ is mixing on ${\mathscr {O}}_M({\mathbb {R}})$.

4 A relation to Abel’s equation

In this section we relate the mixing property of composition operators acting on ${\mathscr {O}}_M({\mathbb {R}})$ with the solvability of Abel’s functional equation, i.e. the equation

$$\begin{aligned} H(\psi (x))=H(x)+1 \end{aligned}$$

where $\psi :{\mathbb {R}}\rightarrow {\mathbb {R}}$ is a given function. Solvability of this equation is well-understood in various situations. For example it is known that if $\psi :{\mathbb {R}}\rightarrow {\mathbb {R}}$ is a bijective smooth (or real analytic) function with no fixed points, then this equation has a smooth (real analytic) solution, see [3, 7].

Theorem 15

Let $\psi \in {\mathscr {O}}_M({\mathbb {R}})$ be bijective. The following conditions are equivalent.

• (i) The operator $C_\psi :{\mathscr {O}}_M({\mathbb {R}})\rightarrow {\mathscr {O}}_M({\mathbb {R}})$, $f\mapsto f \circ \psi$ is mixing and for every $v\in {\mathscr {S}}({\mathbb {R}})$

  $$\begin{aligned} \lim _{n\rightarrow \infty }v(\psi _n(0))\cdot n=0 \text { and } \lim _{n\rightarrow \infty }v(\psi _{-n}(0))\cdot n=0. \end{aligned}$$

• (ii) There exists $H\in {\mathscr {O}}_M({\mathbb {R}})$ with a non-vanishing derivative and which satisfies the equation

  $$\begin{aligned} H(\psi (x))=H(x)+1\text { for every }x\in {\mathbb {R}}.\end{aligned}$$

Proof

(i) $\Rightarrow$ (ii) Since $C_\psi$ is mixing, by Proposition 2, the function $\psi$ has no fixed points and has a non-vanishing derivative. In what follows we will assume that $\psi (x)>x$ for every $x\in {\mathbb {R}}$, the other case can be done in a similar way. By [20, Thm. 8] there exists a bijective smooth function H with a non-vanishing derivative and which satisfies the equation

$$\begin{aligned} H(\psi (x))=H(x)+1\text { for every }x\in {\mathbb {R}}. \end{aligned}$$

(6)

We need to show that $H\in {\mathscr {O}}_M({\mathbb {R}})$, i.e. that for every $v\in {\mathscr {S}}({\mathbb {R}})$ and $k\ge 0$

$$\begin{aligned} \sup _{x\in {\mathbb {R}}}\left| v(x)H^{(k)}(x)\right| <\infty . \end{aligned}$$

(7)

In view of Lemma 11 we may assume that v is non-decreasing on $(-\infty ,0)$ and non-increasing on $[0,\infty )$. To prove (7) it is enough to show that

$$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{x\in [\psi _{n}(0),\psi _{n+1}(0)]} \left| v(x)H^{(k)}(x)\right| =0 \end{aligned}$$

(8)

and

$$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{x\in [\psi _{-n-1}(0),\psi _{-n}(0)]} \left| v(x)H^{(k)}(x)\right| =0. \end{aligned}$$

(9)

We will show (8), the proof of (9) is similar.

From (6) it follows that for $n\in {\mathbb {N}}$ and $x\in [\psi _{n}(0),\psi _{n+1}(0)]$ we have

$$\begin{aligned} H(x)=H(\psi _{-n}(x))+n \text { and }H^{(k)}(x)=(H\circ \psi _{-n})^{(k)}(x) \text { for } k\in {\mathbb {N}}.\end{aligned}$$

Thus

$$\begin{aligned} \sup _{x\in [\psi _{n}(0),\psi _{n+1}(0)]} \left| v(x)H(x)\right| =&\sup _{x\in [\psi _{n}(0),\psi _{n+1}(0)]} \left| v(x)(H(\psi _{-n}(x))+n)\right| \\ \le&\left( n+\sup _{x\in [0,\psi (0)]}|H(x)|\right) \cdot v(\psi _n(0)) \end{aligned}$$

and

$$\begin{aligned} \sup _{x\in [\psi _{n}(0),\psi _{n+1}(0)]} \left| v(x)H^{(k)}(x)\right| =&\sup _{x\in [\psi _{n}(0),\psi _{n+1}(0)]} \left| v(x)(H\circ \psi _{-n})^{(k)}(x)\right| . \end{aligned}$$

Therefore (8) follows from the assumptions on $\psi$ and Theorem 7 combined with Faà di Bruno’s formula and the boundedness of $H^{(j)}$ on $[0,\psi (0)]$, $j\in {\mathbb {N}}\cup \{0\}$.

(ii) $\Rightarrow$ (i) Due to $H(\psi (x))=H(x)+1$ for every $x\in {\mathbb {R}}$ it follows that $\psi$ does not have fixed points. Additionally, since $H(\psi _n(x))=H(x)+n$ for every $x\in {\mathbb {R}}$, $n\in {\mathbb {Z}}$, we conclude that $\lim _{n\rightarrow \infty }H(\psi _n(0))=\infty$ and $\lim _{n\rightarrow \infty }H(\psi _{-n}(0))=-\infty$ which implies the surjectivity of H. Since H has non-vanishing derivative we conclude that H is bijective. Moreover, for arbitrary $v\in {\mathscr {S}}({\mathbb {R}})$ we have

$$\begin{aligned} \lim _{n\rightarrow \infty }v\left( \psi _n(0)\right) n=\lim _{n\rightarrow \infty }v\left( \psi _n(0)\right) \left( H\left( \psi _n(0)\right) -H(0)\right) =0 \end{aligned}$$

because $Hv, H(0)v\in {\mathscr {S}}({\mathbb {R}})$ and $\lim _{n\rightarrow \infty }|\psi _n(0)|=\infty$, the latter since $\psi$ has no fixed points. In the same way it follows $\lim _{n\rightarrow \infty }v\left( \psi _{-n}(0)\right) n=0$.

The conditions in (ii) imply that the diagram

commutes and that the operator $C_H$ has dense range (since all compactly supported smooth functions are in its image because H is bijective). Thus $C_\psi$ is quasi-conjugate to the mixing operator $C_{x+1}$ and hence mixing. $\square$

Corollary 16

If $\psi \in {\mathscr {O}}_M({\mathbb {R}})$ is bijective and satisfies the conditions of Theorem 15 (i), then $C_\psi$ is quasi-conjugate to the operator $C_{x+1}$ on ${\mathscr {O}}_M({\mathbb {R}})$. Therefore it is hypercyclic and chaotic.

Remark 17

It is not clear to the authors if for every mixing composition operator $C_\psi$ on ${\mathscr {O}}_M({\mathbb {R}})$, where $\psi$ is bijective, it automatically holds

$$\begin{aligned} \lim _{n\rightarrow \infty }v(\psi _n(0))\cdot n=0 \text { and } \lim _{n\rightarrow \infty }v(\psi _{-n}(0))\cdot n=0 \text { for any }v\in {\mathscr {S}}({\mathbb {R}}). \end{aligned}$$

(10)

If this was the case, then every mixing $C_\psi$ would already be hypercyclic and chaotic by the above corollary.

Condition (10) is satisfied whenever there is $\beta >0$ for which $\psi (x)>x+\beta$ for every $x\in {\mathbb {R}}$. Example 10 shows that the latter is not a sufficient condition for mixing. Example 12 shows that $\lim _{x\rightarrow \infty }(\psi (x)-x)=0$ may happen for a mixing $C_\psi$.

Open problems. Let $\psi \in {\mathscr {O}}_M({\mathbb {R}})$ be such that $C_\psi$ is mixing on ${\mathscr {O}}_M({\mathbb {R}})$.

• 1. Assume additionally that $\psi$ is bijective. Is it true that

  $$\begin{aligned} \lim _{n\rightarrow \infty }v(\psi _n(0))\cdot n=0 \text { and } \lim _{n\rightarrow \infty }v(\psi _{-n}(0))\cdot n=0 \end{aligned}$$

  holds for every $v\in {\mathscr {S}}({\mathbb {R}})$?

• 2. Is $C_\psi$ (sequentially) hypercyclic on ${\mathscr {O}}_M({\mathbb {R}})$?

While we do not know the answer to problem 1, the next theorem shows that the sequence $(\psi _n(0))_{n\in {\mathbb {N}}}$ cannot grow too slowly.

Theorem 18

Let $f\in C^\infty ({\mathbb {R}})$ be real valued such that $\sup _{x\in {\mathbb {R}}}(1+|x|^2)^n|f(x)|<\infty$ for every $n\in {\mathbb {N}}$, $\inf _{x\in {\mathbb {R}}}|1+f'(x)|>0$, and $f'\in {\mathscr {O}}_M({\mathbb {R}})$. Then, for $\psi (x)=x+f(x)$, the operator $C_\psi :{\mathscr {O}}_M({\mathbb {R}})\rightarrow {\mathscr {O}}_M({\mathbb {R}})$ is not topologically transitive.

It should be noted that under the hypotheses of the above theorem $\psi \in {\mathscr {O}}_M({\mathbb {R}})$ so that $C_\psi :{\mathscr {O}}_M({\mathbb {R}})\rightarrow {\mathscr {O}}_M({\mathbb {R}})$ is correctly defined. To prove the above theorem we need the following lemma which is of independent interest.

Lemma 19

Let $f\in C^\infty ({\mathbb {R}})$ be real valued such that $\sup _{x\in {\mathbb {R}}}(1+|x|^2)|f(x)|<\infty$, $\inf _{x\in {\mathbb {R}}}|1+f'(x)|>0$, and $f'\in {\mathscr {O}}_M({\mathbb {R}})$. Let $\psi (x)=x+f(x)$, $x\in {\mathbb {R}}$. Then, $\psi$ is bijective and $g\circ \psi ^{-1}\in {\mathscr {S}}({\mathbb {R}})$ for every $g\in {\mathscr {S}}({\mathbb {R}})$.

Proof

By hypothesis, $\psi '(x)\ne 0$ so that $\psi$ is injective. Moreover, since obviously $\lim _{|x|\rightarrow \infty }|f(x)|=0$, it follows $\lim _{x\rightarrow \pm \infty }\psi (x)=\pm \infty$ so that $\psi$ is bijective. Additionally, for |x| sufficiently large we have

$$\begin{aligned} |\psi (x)|\le |x|+|f(x)|\le |x|+\frac{\sup _{y\in {\mathbb {R}}}(1+|y|^2)|f(y)|}{(1+|x|^2)}\le 2|x| \end{aligned}$$

for x large which implies

$$\begin{aligned} |\psi ^{-1}(x)|\ge \frac{|x|}{2}\ge |x|^{1/k} \end{aligned}$$

(11)

whenever $|x|\ge k$ for some suitable $k\in {\mathbb {N}}$. Obviously, $|\psi ^{-1}(x)|\le (1+|\psi ^{-1}(x)|^2)^{1/2}$, and due to [2], for $n\in {\mathbb {N}}$ there is a polynomial $P_n$ in n variables with integer coefficients such that

$$\begin{aligned} (\psi ^{-1})^{(n)}(x)= & {} \left( \frac{1}{\psi '(\psi ^{-1}(x))}\right) ^{2n-1}P_n\left( \psi '(\psi ^{-1}(x)),\psi ^{(2)}(\psi ^{-1}(x)),\ldots ,\psi ^{(n)}(\psi ^{-1}(x))\right) \\= & {} \left( \frac{1}{1+f'(\psi ^{-1}(x))}\right) ^{2n-1} P_n\left( 1+f'(\psi ^{-1}(x)),f^{(2)}(\psi ^{-1}(x)),\right. \\{} & {} \quad \left. \ldots ,f^{(n)}(\psi ^{-1}(x))\right) . \end{aligned}$$

In particular, since $f'\in {\mathscr {O}}_M({\mathbb {R}})$, for a suitable constant $C>0$ and $k\in {\mathbb {N}}$ it holds for arbitrary $x\in {\mathbb {R}}$

$$\begin{aligned} |(\psi ^{-1})^{(n)}(x)|\le \left( \frac{1}{\inf _{y\in {\mathbb {R}}}|1+f'(y)|}\right) ^{2n-1}C\left( 1+|\psi ^{-1}(x)|^2\right) ^k. \end{aligned}$$

Combining this with (11), an application of [9, Theorem 2.3] proves the claim. $\square$

Proof of Theorem 18

The inclusion ${\mathscr {O}}_M({\mathbb {R}})\hookrightarrow C^\infty ({\mathbb {R}})$ is continuous and has dense range, therefore topological transitivity of $C_\psi$ on ${\mathscr {O}}_M({\mathbb {R}})$ implies that $C_\psi$ is also topologically transitive on $C^\infty ({\mathbb {R}})$. Thus by [21, Theorem 4.2] if $\psi$ has a fixed point or $\psi '(x)=0$ for some $x\in {\mathbb {R}}$, then $C_\psi$ is not topologically transitive.

From now on we will assume that $\psi '(x)\not =0$ for every $x\in {\mathbb {R}}$ and that $\psi$ has no fixed point which implies $f(x)\ne 0$ for every $x\in {\mathbb {R}}$. Moreover, we assume that $f(x)>0$ for all $x\in {\mathbb {R}}$; the proof in case $f(x)<0$ for all $x\in {\mathbb {R}}$ is similar.

Let $g\in {\mathscr {S}}({\mathbb {R}})$ be as in Lemma 11 for f. By Lemma 19 we have $g\circ \psi ^{-1} \in {\mathscr {S}}({\mathbb {R}})$. Therefore, both sets

$$\begin{aligned} U=\{u\in {\mathscr {O}}_M({\mathbb {R}}): 0<u(0)<1 \text { and } 2<u(\psi (0))<3\} \end{aligned}$$

and

$$\begin{aligned} V= \bigg \{v\in {\mathscr {O}}_M({\mathbb {R}}): \sup _{x\in {\mathbb {R}}}|v'(x)g(\psi ^{-1}(x))|<0.5\bigg \} \end{aligned}$$

are open in ${\mathscr {O}}_M({\mathbb {R}})$ and non-empty. If $v\in V$ and $n\ge 1$, then by the Mean Value theorem we get that

$$\begin{aligned} \left| C_\psi ^n(v)(\psi (0))-C_\psi ^n(v)(0)\right| = \left| v(\psi _{n+1}(0))-v(\psi _n(0))\right| =|v'(\xi )f(\psi _n(0))|, \end{aligned}$$

where $\xi \in [\psi _n(0),\psi _{n+1}(0)]$. Using monotonicity of $g\circ \psi ^{-1}$ we get that

$$\begin{aligned} |v'(\xi )f(\psi _n(0))|\le&|v'(\xi )g(\psi _n(0))|\\ =&|v'(\xi )(g\circ \psi ^{-1})(\psi _{n+1}(0))|\\ \le&|v'(\xi )(g\circ \psi ^{-1})(\xi )|<0.5. \end{aligned}$$

Thus $C_\psi ^n(v)\not \in U$ and therefore $C_\psi$ is not topologically transitive. $\square$

Example 20

Obviously, the function $\psi (x)=x+\exp (-x^2/2), x\in {\mathbb {R}},$ belongs to ${\mathscr {O}}_M({\mathbb {R}})$, has no fixed points and satisfies $\psi '(x)\ne 0$, $x\in {\mathbb {R}}$. While the corresponding composition operator $C_\psi$ is hypercyclic/topologically transitive/mixing on $C^\infty ({\mathbb {R}})$ by [21, Theorem 4.2], it is not topologically transitive on ${\mathscr {O}}_M({\mathbb {R}})$ by Theorem 18.