Let's say I have a function $f:E \to E$ where $E$ is an euclidian space. Moreover there is : $\delta > 0$ such that :
$$\forall (x,y) \in E^2, \mid ||f(x)-f(y)||-||x-y|| \mid \leq \delta$$
Then I want to show that : $\lim_{ n \to \infty} ||\frac{f(nx)}{n} || = ||x||$. Here is what I do : We have $\forall n \in \mathbb{N}$ :
$$\mid ||f(nx)-f(0)||-||nx|| \mid \leq \delta$$
Hence we have (by multiplying the above inequality with $\frac{1}{n}$) : $$\mid ||\frac{f(nx)}{n}-\frac{f(0)}{n}||-||x|| \mid \leq \frac{\delta}{n}$$
Now let's take the limit of the above inequality as $n \to \infty$ : $$\lim_{n \to \infty}\mid ||\frac{f(nx)}{n}-\frac{f(0)}{n}||-||x|| \mid \leq \lim_{n \to \infty}\frac{\delta}{n}$$ Wich is equal to : $$\mid ||\lim_{n \to \infty} \frac{f(nx)}{n}||-||x|| \mid \leq 0$$ So : $$||\lim_{n \to \infty} \frac{f(nx)}{n}|| = ||x||$$
But it appears that this actually false. I don't understand why can someone explain?
Thank you!