# Uniformly converge with equicontinuous family

Let $\{f_n\}$ be a equicontinuous sequence and converge pointwise in a compact set $K$ of $\mathbb{R}^n$. Prove that the sequence converge uniformly in $K$.

My attempt: Since $\{f_n\}$ is equicontinuous, given $\epsilon > 0$, there exist a $\delta >0$ such that $$|f_{i}(x)-f_{i}(y)|<\epsilon$$ if $1\leq i \leq n$ and $|x-y|<\delta.$ In fact, we know that, given $\epsilon>0$ there exist $N \in \mathbb{N}$ such that, as $n \geq N$ then $|f_n - f|<\epsilon$. I must prove that the convergence is uniform.

Since continuous functions are uniformly continuous on compact set, then, given $\epsilon>0$ there exist $\delta (\epsilon)>0$ such that, as $|x-y|<\delta$ we have $|f_{i}(x)-f_{i}(y)|<\epsilon$. So, taking $n \geq N, m\geq N$ there exist a $\delta>0$ such that, as $|x-y|<\delta$ then $$|f_{n}(x) - f_{m}(x)|<|f_{n}(x)-f_{n}(y)| + |f_{n}(y)-f_{m}(y)| + |f_{m}(y)-f_{m}(x)|<3\epsilon,$$

where $|f_{n}(y)-f_{m}(y)|<\epsilon$ by pointwise convergence.

Thus, $\{f_n\}$ is Cauchy, and then $\{f_n\} \rightarrow f$ uniformly.

Is this correct?

No. Observe that you have not used the equicontinuity of the sequence. If your proof were correct, then any sequence of continuous functions that converges pointwise on a compact set would converge uniformly. The error is in the phrase "$|f_n(y)-f_m(y)|<\epsilon$ by pointwise convergence".

Let $f$ be the pointwise limit of $(f_n)$. The way to prove the result is to show, using the Ascoli-Arzèla theorem, that every subsequence of $(f_n)$ has a subsequence that converges uniformly, necessarily to $f$.

No it's not, you're not giving the right justifications. Fix $x$ and $\varepsilon > 0$. $y$ will be as close to $x$ as it has to be.

To bounds $|f_n(x)-f_n(y)|$ independantly of $x$ and $n$ you need uniform equicontinuity, which is a consequence of equicontinuity on a compact set. same thing for $|f_m(y)-f_m(x)|$.

to bounds $|f_n(y)-f_m(y)|$ independantly of y (and thus independantly of x) you'll need uniform convergence of the sequence $(f_n)$, this is not good. An other idea is to chose wisely $y$ : remark that this bound doesn't have to work for every $y$ close to $x$. from first part there exists $\delta > 0$ which doesn't depend on x such that $|x-y|<\delta \Rightarrow |f_n(x)-f_n(y)|<\varepsilon$ and $|f_m(y)-f_m(x)|<\varepsilon$. You juste have to take a finite number of $y_i$ such that $K \subset \bigcup_i B(y_i,\delta)$, this is possible by compacity of $K$. Because the $y_i$ are in finite number and by ponctual convergence, for $n,m$ large enough you'll have $|f_n(y_i)-f_m(y_i)|<\varepsilon$ for all $i$. Note that in this way the $y_i$ will depends on $\varepsilon$. So for $n,m$ large enough and taking $|y_i-x|<\delta$ you get that $|f_n(x)-f_n(x)|<|f_n(x)-f_n(y_i)|+|f_n(y_i)-f_m(y_i)|+|f_m(y_i)-f_m(x)|<3\varepsilon$ uniformly over $K$, so $(f_n)$ is Cauchy.

This is not the nicest way to prove it IMO. Two other ways are the following :

1) use relative compacity of $\{f_n\}$ and a lemma of Cantor : a sequence has a limite $l$ iff from every subsequence you can extract an other subsequence converging to $l$

2) show first that the limit of $\{f_n\}$ must be continous and then that the convergence is uniform.