analysis - Inequality for $|f(x_n)-f(x_n+(y_n-x_n)|$.

If $f: \mathbb{R} \to \mathbb{R}$ is a function and $a_n$ a sequence of real numbers.We have $f_{a_n(x)}=f(x+a_n)$.Show that if for any zero sequence $a_n$ we have $f_{a_n} \to f$ uniformly,then $f$ is uniformly continuous.

To show that $f$ is uniformly continuous,it suffives to show that if $x_n,y_n$ are two sequences of real numbers and $x_n-y_n \to 0$,then $f(x_n)-f(y_n) \to 0$.

$|f(x_n)-f(y_n)|=|f(x_n)-f(x_n+(y_n-x_n)| \leq$ what?

We know that $f_{a_n} \to f$ uniformly, that means that $\exists n_0$ such that $\forall n \geq n_0: |f_{a_n}-f|_{\mathbb{R}}= \sup_{x \in \mathbb{R}} |f_{a_n}-f| \leq \epsilon$

How can I find an inequality for $|f(x_n)-f(x_n+(y_n-x_n)|$ ?

Answer

You've basically got it. Just let $a_n=y_n-x_n$. This is your zero sequence (which I assume means a sequence that converges to $0$). Then

$$|f(x_n)-f(y_n)|=|f(x_n)-f(x_n+a_n)=|f(x_n)-f_{a_n}(x_n)|\le\sup_{x\in\mathbb{R}}|f(x)-f_{a_n}(x)|\le\epsilon$$