Choosing a metric in which homeomorphism is Holder continuous

Question

Let $X$ be a compact metrizable space, and let $f:X \to X$ be a homeomorphism. Is it always possible to choose a compatible metric on $X$ in which $f$ is Holder continuous? I've tried some simple tricks like taking any metric $d$ and then defining a new one $d'$ by \begin{equation*} d'(x_1,x_2) = d(f^{-1}(x_1),f^{-1}(x_2)) + d(x_1,x_2) \end{equation*} or \begin{equation*} d'(x_1,x_2) = d(x_1,x_2) + d(f(x_1),f(x_2)) \end{equation*} and I can see no reason for them to be Holder continuous. We can also try to take any metric $d$ and set \begin{equation*} d'(x_1,x_2) = \sup_{n \in \mathbb{Z}} d(f^n(x_1),f^n(x_2)). \end{equation*} but this $d'$ being compatible with the topology is equivalent to the family of iterates $(f^n)_{n \in \mathbb{Z}}$ being equicontinuous.

Answer 1

This is not a complete answer, but topological entropy is not enough to rule out Hölder continuity. The following Hölder continuous automorphism of the Cantor space has infinite topological entropy. Infinite topological entropy is only enough to rule out Lipschitz continuity. Thus the example gives a Hölder continuous function $\varphi: X \to X$ that is not Lipschitz continuous with respect to any equivalent metric on $X$.

Consider $X=\{0,1\}^\mathbb{Z}$ together with the usual metric $d(x,y)=2^{-\min\{|k|\: :\: x_k \neq y_k\}}$.

Define $\varphi:X \to X$ by $\varphi(x)_i=\begin{cases} x_{2i} & i \geq 0\\ x_{-i/2} & i<0, 2\mid i\\ x_{(i-1)/2} & i<0, 2\not\mid i\end{cases}.$

(If one only needs an endomorphism one can take $\psi: 2^{\mathbb{N}_0} \to 2^{\mathbb{N}_0}, \psi(x)_i=x_{2i}$).

This is a homeomorphism. It is Hölder continuous wrt. $d$, since $d(x,y)=2^{-(2n+1)}$ or $d(x,y)=2^{-(2n+2)}$ implies $(x_{-2n},\dots,x_{2n})=(y_{-2n},\dots,y_{2n})$, hence $(\varphi(x)_{-n},\dots,\varphi(x)_{n})=(\varphi(y)_{-n},\dots,\varphi(y)_{n})$ and $d(\varphi(x),\varphi(y))\leq 2^{-(n+1)}$. Therefore $d(\varphi(x),\varphi(y)) \leq d(x,y)^{\frac{1}{2}}$.

To see that $\varphi$ has infinite topological entropy consider the cardinality $S_{n,\ell}$ of $2^{-2n}$ seperated sets with respect to $d_\ell(x,y)=\max_{k=0,\dots,\ell-1}d(\varphi^k(x),\varphi^k(y))$. If $x_i \neq y_i$ for some $i \in I:=\{0,\dots,2n-1\} \cup 2\cdot\{0,\dots,2n-1\} \cup \dots \cup 2^{\ell-1} \cdot\{0,\dots,2n-1\}$, then $d_{\ell}(x,y)>2^{-2n}$. Hence $S_{n,\ell}\geq |I| = 2^{2n+(\ell-1)n}\geq 2^{\ell n}$. Therefore $h_{\text{top}}(\varphi)\geq\lim_{n \to \infty} \limsup_{\ell \to \infty} \frac{1}{\ell} \log 2^{\ell n}=\infty$.

Answer 2

I think the answer is no.

The idea goes as follows: If $f\colon (X,d)\to (X,d)$ is Holder, then there exists a new metric $\tilde d$ equivalent to $d$ and such that $f$ is Lipschitz respect to $\tilde d$. And we know that any Lipschitz map has finite topological entropy (and topological entropy is a topological invariant).

So, if you start with an $f$ such that $h_\mathrm{top}(f)=\infty$, then it seems like it doesn't exists an equivalent metric such that $f$ is Holder.