Uniform convergence of a sequence of continuous functions

Question

If a sequence of continuous functions $p_n:E \rightarrow \mathbb{R}$ - where $E$ is a complete metric space - converge uniformly to the function $p$, does that mean for any sequence $\{x_n\} \subset E$ such that $x_n \rightarrow x$, $p_n(x_n) \rightarrow p(x)$?

My approach: Fix $n$. Pick $k > n$. $|p_n(x_n)-p(x)| = |p_n(x_n)-p_n(x_k) + p_n(x_k)- p(x)| \leq |p_n(x_n)-p_n(x_k)|+|p_n(x_k)- p(x)| $, which holds for all $k$, so taking limits on $k$, $|p_n(x_n)-p(x)| \leq |p_n(x_n)-p_n(x)|+|p_n(x)- p(x)|$. Here again I'm back to square one as both the argument and the function is changing with $n$ in the first part. Any help is appreciated.

Answer 1

I would notice first that uniform convergence implies $p$ is continuous on $E.$ Thus

$$|p_n(x_n)-p(x)| \le |p_n(x_n)-p(x_n)| + |p(x_n)-p(x)| \le \sup_E |p_n-p| + |p(x_n)-p(x)|.$$

The first term on the right $\to 0$ by uniform convergence, and the second term on the right $\to 0$ by the continuity of $p$ at $x.$

Answer 2

To handle the term $|p_n(x_n)-p_n(x)|$ use equicontinuity at $x$:

Prove that, having fixed $\varepsilon>0$ and $x\in E$, there exists $\delta>0$ such that if $|x-y|<\delta$, then $|f_n(y)-f_n(x)|<\varepsilon$ for all $n\in \mathbb{N}$ (in other words you have continuity, with a $\delta$ that works for every member of your sequence of functions at the same time). Hopefully changing the $x_n$ for $y$ gives a little more clarity as to how to proceed given the uniform convergence.