Consider the interval $[0,1)$ endowed with the metric $$d(x,y):=\min\{|x-y|,1-|x-y|\}.$$ (Topologically, this is the circle!) Let $E_{2}\colon[0,1)\to[0,1)$ be the doubling map defined by $$E_{2}(x):=2x\mod1.$$ Let $f\colon[0,1)\to\mathbb{R}$ be a continuous map (w.r.t. the metric on $[0,1)$ above and the Euclidian metric on $\mathbb{R}$. It is not hard to see that the $n$-periodic points of $E_{2}$ are given by $$\{z(2^{n}-1)^{-1} \ | \ z\in\mathbb{Z}, \ 0\leq z\leq 2^{n}-2\}.$$
Let $x\in[0,1[$ be a point. I want to construct a sequence $(x_{n})$ in $[0,1[$ such that $x_{n}$ is an $n$-periodic point of $E_{2}$ and $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}|f(E_{2}^{k}(x_{n}))-f(E_{2}^{k}(x))|=0.$$
My attempt: Note that the 'circle' $[0,1)$ is compact. So $f$ is uniformly continuous. Also it is easy to see that for each $n\in\mathbb{N}$ we can find a $n$-periodic point $x_{n}$ such that $|x_{n}-x|\leq(2^{n}-1)^{-1}$. So this yields a sequence $(x_{n})$ of $n$-periodic points that converges to $0$. I also proved that $d(E_{2}(p),E_{2}(q))\leq2d(p,q)$ for all $p,q\in[0,1)$. So if we can estimate $d(x_{n},x)<\delta$ for some $\delta>0$, then we can estimate $d(E_{2}^{k}(x_{n}),E_{2}^{k}(x))<2^{k}\delta$ for all $0\leq k\leq n-1$. However, I have no idea how to finish...
Any suggestions are greatly appreciated!