Must a uniformly continuous (under the 2adic metric) surjection $f:\Bbb Z[\frac16]\to\Bbb Z[\frac16]$ which is fixed in $\lvert\cdot\rvert_3$, stabilise in order to approach $0$?
Let us say that some sequence generated by repeated composition of a function $S=(f^n(x):n\in\Bbb N)$, approaches $0$, then its approach stabilises if and only if the sequence with its powers of $2$ factored out $\overline S=(f^n(x)\lvert f^n(x)\rvert_2:n\in\Bbb N)$ is a fixed point, i.e. $\lim_{n\to\infty}\overline S=(\ldots,y,y,y,\ldots)$
Let $X=\Bbb Z[\frac16]$ with the 2-adic topology.
Let some continuous surjection $f:X\to X$ satisfy:
$f:X\setminus0\mapsto X\setminus 0$
$f:0\mapsto0$ (and this is the only fixed point).
- $f$ commutes with both multiplication by $2$ and multiplication by $3$, i.e. $f(2x)=2f(x)$ and $3f(x)=f(3x)$
- $f$ is strictly monotonic in its powers of $2$, i.e. $\lvert f(x)\rvert_2<\lvert x\rvert_2$, and
- $f$ is fixed in its powers of $3$, i.e. $\lvert f(x)\rvert_3=\lvert x\rvert_3$
By the last but one condition, clearly $\lim_{n\to\infty}f^n(x)=0$
It's easy to pick out trivial examples of the above, i.e. $f$ that stabilise as they approach $0$, for example $f(x)=2x$ fulfils all the above conditions and for that the required result is trivial, i.e. that $\overline S=(\ldots x,x,x)$ and therefore $S$ stabilises on approach to $0$.
But is it possible to find a counterexample, or show there is no counterexample to the rule that $\overline S$ always stabilises like this as it approaches the fixed point $0$?
I'm looking hopefully at the following conditions:
- $f$ is continuous with compact support (and therefore, I think, uniformly continuous by Heine Cantor Theorem.)
- $f^n(x)\to0\text{ as }n\to\infty$
- $0$ is a fixed point
- $\overline S$ in some sense measures position orthogonal to $0$
Perhaps therefore $\overline S$ must be a fixed point in some vicinity of $0$
I suspect the proof looks like this: Every $f$ having some approach to $0$ which does not stabilise, necessarily also has some approach in which the powers of $3$ vary.