Is $f_n \rightrightarrows f $ implies $(f_n(x_n)-f(x_n)) \longrightarrow 0$?

Question

Suppose we have a sequence of functions defined on $S$, $f_n$ that converges uniformly to $f$. Then if $x_n$ is any sequence in the interval can we say that $f_n(x_n)-f(x_n)$ goes to zero?

Efforts:

By definition of uniform convergence, given any positive $\epsilon$ we have, $|f_n-f|<\epsilon$ for every point in $S$ for some $n>N$

Now let $x_n$ be a sequence in interval. In particular take $\epsilon_n=1/n$. There exist $N$ such that $|f_n(x_n)-f(x_n)|<\epsilon_n$ for all $n>N$

But I am not able to proceed from here?

Thanks for help.

Answer 1

Call $E$, the intervel of convergence. Then for any $x_n \in E$ , $$0 \leq \vert f_n(x_n)-f(x_n) \vert \leq \sup_{x \in E} \vert f_n(x)-f(x) \vert \longrightarrow0 $$

Now apply squeeze theorem to conclude the desired result!

Answer 2

Consider the notation $\|f-f_n\|=\sup_{x\in S}|f(x)-f_n(x)|.$ Uniform convergence of $f_n$ to $f$ on $S$ means $\lim_{n\to \infty}\|f-f_n\|=0.$

Now $|f(x_n)-f_n(x_n)|\le \|f-f_n\|$ and $\|f-f_n\|\to 0$ so $|f(x_n)-f_n(x_n)|\to 0.$