# uniformly convergent sequence of functions on a compact space

There's an exercise on Kaplansky's textbook that says:

Let $\{f_i\}$ and $f$ be continuous real functions on a compact metric space $M$. Prove that $f_i$ converges uniformly to $f$ if and only if the following is true: When a sequence $\{x_n\}_{n \in \mathbb N}$ converges to $x_0$, then $f_1(x_1), f_2(x_2),...$ converges to $f(x_0)$.

I have to prove two implications, I've started by:

Suppose $f_n \rightrightarrows f$. Let $\{x_n\}_{n \in \mathbb N}$ : $x_n \to x_0$ and let $\epsilon>0$, I want to show there is some $N_\epsilon$: $\forall n\geq N_\epsilon$ $|f_n(x_n)-f(x_0)|<\epsilon$. By hypothesis, there exists $n_0: \forall n\geq n_0$, $\forall x \in M$ $\space$$|f_n(x)-f(x_0)|<\frac{\epsilon}{2}$. I got stuck here, I think I must use that $f_n$ and $f$ are continuous.

With the suggestion I could do this: $|f_n(x_n)-f(x_0)|=|f_n(x_n)-f(x_n)+f(x_n)-f(x_0)|\leq |f_n(x_n)-f(x_n)|+|f(x_n)-f(x_0)|$.

$f$ is continuous, then, there is $\delta>0 :d(y,x_0)<\delta \implies |f(y)-f(x_0)|<\frac{\epsilon}{2}$. By hypothesis, $x_n \to x_0$, so there is $n_0 : \forall n\geq n_0, d(x_n,x_0)<\delta \implies |f(x_n)-f(x_0)|<\frac{\epsilon}{2}$. Now, $f_n \rightrightarrows f$, this means that there exists $n_1$ such that $\forall n\geq n_1$ $\forall x \in X$ $|f_n(x)-f(x_0)|<\frac{\epsilon}{2}$, in particular, the inequality holds for $x=x_n$ for every $n$. If I take $N_{\epsilon}=max\{n_0,n_1\} \implies |f_n(x_n)-f(x_0)|<\epsilon$.

I welcome any suggestions or hints to prove the other implication, I haven't used the fact that $X$ is compact, I will have to use that.

Here is a start. ($$\implies$$)
$$|f_n(x_n)-f(x)| = |f_n(x_n)-f(x_n)-f(x)+f(x_n) |$$
$$\leq |f_n(x_n)-f(x_n)|+ |f(x)-f(x_n)| < \dots\,.$$
• Thanks for the help. I have problems with $|f_n(x_n)-f_n(x)|$: for each $f_n$, I know how to pick appropiately $N: \forall n\geq N$, $|f_n(x_n)-f_n(x)|<\frac{\epsilon}{2}$ (I use that $f_n$ is continuous and that $x_n \to x$), but I can't or I don't know how to control all the functions $f_n$ at the same time. Nov 22 '13 at 2:48
• @user100106: Note that, $f_n$ are continuous functions on a compact set. Nov 22 '13 at 3:06
• I could prove it with this (similar to yours): $|f_n(x_n)-f(x)| = |f_n(x_n)-f(x_n)+f(x_n)+f(x)|\leq |f_n(x_n)-f(x_n)|+|f(x_n)+f(x)|$, I think I have to use compactness for the other implication. Nov 22 '13 at 15:58
• @user100106: Yes. Good job! You have uniform continuity of $f_n$. Nov 23 '13 at 21:06