14
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Out of curiosity, why are you interested in Hölder and not some other kind of uniform modulus of continuity? For example, is the answer known to be negative for Lipschitz continuity? $\endgroup$ – Joonas Ilmavirta Jan 25 '15 at 15:51
  • $\begingroup$ I phrased the question in terms of Holder continuity because this is the weakest condition that would suffice for the application I have in mind. I don't know anything about the Lipschitz case either ... $\endgroup$ – burtonpeterj Jan 25 '15 at 18:26
6
$\begingroup$

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 $h_{\text{top}}(\varphi)\geq\lim_{n \to \infty} \limsup_{\ell \to \infty} \frac{1}{\ell} \log 2^{\ell n}=\infty$.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Can you please indicate how you go from Holder to Lipschitz. $\endgroup$ – Christian Remling Jan 26 '15 at 17:56
  • $\begingroup$ I think this is a wonderful answer, perhaps for the way it uses a dynamical concept to prove a not-obviously-dynamical fact! =o] $\endgroup$ – Ian Morris Jan 26 '15 at 18:24
  • $\begingroup$ @ChristianRemling thanks for your question. In fact that point is more complicated than I originally thought. I think the following construction works: $(X,d)$ is compact, $C>1$ and $\alpha\in (0,1)$ s.t. $d(f(x),f(y))\leq C d(x,y)^\alpha$ and $f$ is a homeo. Given any $\epsilon\in (0,1)$ one defines $$\tilde d(x,y):=\sum_{n\geq 0}\epsilon^n d(f^n(x),f^n(y))$$. Then $d$ and $\tilde d$ are equivalent and $\tilde d(f(x),f(y))\leq \epsilon^{-1}\tilde d(x,y)$. $\endgroup$ – Alejandro Jan 26 '15 at 19:48
  • 4
    $\begingroup$ Thanks, but I'm more confused than before now. You don't seem to be using anything in the last two lines (since $d$ is bounded on $X$, we can always define $\widetilde{d}$). Why doesn't it follow that any $f$ is Lipschitz wrt $\widetilde{d}$? $\endgroup$ – Christian Remling Jan 26 '15 at 21:15

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.