Dynamics of a random stretch map

Question

Notation: Here $S^1$ denotes the circle, which we view as the unit sphere in $\mathbb C$. We equip the circle with its natural length metric.

Let $\{\epsilon_n\}_{n \geq 1}$ be iid uniformly distributed random variables on $[0, e]$ (with underlying probability space $\Omega$), and define the random maps $T_n: \Omega \times S^1 \to S^1$ by

$$T_n (e^{i \theta}) := e^{i \epsilon_n \theta}.$$

For $x \in S^1$, we write for short $D_n (x) := T_n \dots T_1 (x).$

Question: Is it true that for all $x, y \in S^1$, we have

$$d(D_n (x), D_n (y)) \to 0$$

almost surely?

Answer 1

I think what you're asking is not true.

There is an auxiliary process in the background (implicit in the comment of Martin Hairer and your modification of the question), which is the additive random walk: $$ X_0=0;\ X_{n+1}=X_n+\log \epsilon_n(\omega). $$ This random walk has zero drift, so $\limsup X_n=\infty$.

I'd prefer to describe your maps as taking place on $\Omega\times[0,1)$, where $\Omega$ is the shift space $[0,e]^{\mathbb Z}$ equipped with the product of normalized Lebesgue measure. If a point $\omega\in\Omega$ is written as $(\omega_n)_{n\in\mathbb Z}$, write $\sigma(\omega)$ for the shifted sequence: $\sigma(\omega)_n=\omega_{n+1}$. the map $T$ is given by $$ T(\omega,x)=(\sigma(\omega),\omega_0x\bmod 1). $$ This is a skew product: the first coordinate is just the shift map, and the update rule in the second coordinate depends on the first coordinate. This map $T$ is called a random $\beta$-shift. They have been studied by Buzzi and others. Your $D_n(x)$'s are just the first coordinate of $T^n(x,\omega)$.

Define $n$ to be a reset time if $|D_n(x)-D_n(y)|\ge\frac 1{2e}$. I claim that if $x\ne y$, either there exists $T$ such that $D_T(x)=D_T(y)$ (in which case $D_n(x)=D_n(y)$ for all $n\ge T$), or there are infinitely many reset times. Clearly for any $x$, there are countably many $y$ such that $D_n(x)=D_n(y)$ for some $n$. Assume $x$ and $y$ are such that $D_n(x)\ne D_n(y)$ for all $n$. We claim there are infinitely many reset times, and hence $|D_n(x)-D_n(y)|\not\to 0$.

To see this, let $m\in\mathbb N$. Since $\limsup_{n\to\infty}\omega_m\cdots\omega_{n-1}=\infty$, there exists $n>m$ such that $\omega_m\cdots\omega_{n-1}|x-y|>\frac 1{2e}$.

Either $|D_n(x)-D_n(y)|=\omega_{m}\cdots\omega_{n-1}|x-y|$, in which case $n$ is a reset time; or there exists $m<j<n$ such that $\lfloor\omega_{j-1}D_{j-1}(x)\rfloor \ne \lfloor \omega_{j-1}D_{j-1}(y)\rfloor$ (i.e. $D_{j-1}(x)$ and $D_{j-1}(y)$ lie on opposite sides of a discontinuity of $t\mapsto\epsilon_j t\bmod 1)$. In this second case if $0<|D_{j-1}(x)-D_{j-1}(y)|<\frac1{2e}$ then $|D_j(x)-D_j(y)|>\frac 12$, so there is a reset at time $j$, or if $|D_{j-1}(x)-D_{j-1}(y)|\ge \frac 1{2e}$, then there is a reset at time $j-1$. In all cases, there is a reset time at or before $n$.