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

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.

• Typo in the title: Should probably be $f_n - f$, not $f-f$. – Tommi Brander Apr 8 at 5:40

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!

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 $$|f(x_n)-f_n(x_n)|\to 0.$$