The sequence {$f_n$} converges uniformly to a function $f$ on $X$. Prove that $f$ is continuous at $x_0$.

Question

Suppose that for each positive integer $n$, the function $f_n : X \to \mathbb{R}$ is continuous at some $x_0 \in X$, and the sequence {$f_n$} converges uniformly to a function $f$ on $X$. Prove that $f$ is continuous at $x_0$.

$$ |f(x)-f(x_0)| = |f(x)-f_n(x) + f_n(x)-f_n(x_0) + f_n(x_0)-f(x_0)| \ $$ $$ \le |f(x)-f_n(x)| + |f_n(x)-f_n(x_0)| + |f_n(x_0)-f(x_0)| $$

I tried doing it like this but i cannot seem to make the RHS tend to $0$ so that i can show $f(x)$ is continuos at $x_0$

Answer 1

For $\epsilon>0$, pick an $N$ such that $\|f_{N}-f\|_{\infty}<\epsilon/3$. Since $f_{N}$ is continuous at $x_{0}$, there is some $\delta>0$ such that $|f_{N}(x)-f_{N}(x_{0})|<\epsilon/3$ for all $x$ with $0<|x-x_{0}|<\delta$. For such an $x$, we have \begin{align*} |f(x)-f(x_{0})|&\leq|f(x)-f_{N}(x)|+|f_{N}(x)-f_{N}(x_{0})|+|f_{N}(x_{0})-f(x_{0})|\\ &<2\|f_{N}-f\|+\epsilon/3\\ &<\epsilon. \end{align*}

Answer 2

Hint. Given $\epsilon>0$, there is $n\geq0$ such that $$\sup_{x\in X}|f_n(x)-f(x)|<\epsilon/3.$$ This is a useful bound for $|f(x)-f_n(x)|$ and $|f_n(x_0)-f(x_0)|$. Now use the continuity of $f_n$ at $x_0$ for the remaining term $|f_n(x)-f_n(x_0)|$.