Let $(X,d)$ be a compact metric space, and $\varphi$ be an equicontinuous dynamical system ($\forall\, \varepsilon>0$, there exists $\delta>0$ such that $d(x,y)<\delta\Rightarrow d(\varphi^nx,\varphi^ny)<\varepsilon,\,\forall x,y\in X$ and $n\in\mathbb{Z}$ )
Define a metric $d'(x,y)=\sup_{n\in \mathbb{Z}}d(\varphi^nx,\varphi^ny)$.
I want to prove:
- $d$ and $d'$ are aquivalent.
- $\varphi$ is an isometry with the metric $d'$.
I appreciate any suggestions.
$\def\sup{\mathop{\mathrm{sup}}\limits}\def\Z{\mathbb{Z}}\def\phi{\varphi}$The key is that equicontinuity implies that for every $\epsilon > 0$ there is $\delta > 0$ such that $d'(x, y) \leq \epsilon$ for every $x, y \in X$, since $d(\phi^nx, \phi^ny) < \epsilon$ for every $n\in\Z$ implies $d'(x, y) = \sup_{n\in\Z} d(\phi^nx, \phi^ny) \leq \epsilon$.
For 1. we want to show that $d(x_n, x) \to 0$ iff $d'(x_n, x) \to 0$.
First, suppose that $d(x_n, x) \to 0$. We want to show that $d'(x_n, x) \to 0$. Let $\epsilon > 0$. Take $\delta > 0$ such that $d'(x, y) \leq \frac\epsilon2 < \epsilon$ if $d(x, y)$. Take $n_0$ such that $d(x_n, x) < \delta$ if $n \geq n_0$. Then $d'(x_n, x) < \epsilon$ for $n\geq n_0$. Hence $d'(x_n, x) \to 0$, as desired.
Second, suppose that $d'(x_n, x) \to 0$. We have $d(x_n, x) \leq d'(x_n, x)$, so $d(x_n, x) \to 0$, as desired.
For 2., $$d'(\phi x, \phi y) = \sup_{n\in\Z} d'(\phi^n\phi x, \phi^n\phi y) = \sup_{n\in\Z} d'(\phi^{n+1} x, \phi^{n+1} y) = \sup_{n\in\Z} d'(\phi^{n} x, \phi^{n} y) = d'(x, y),$$ so $\phi$ is an isometry.