# Ascoli-Arzelà theorem

Let $\Omega$ be a bounded subset of $\mathbb{R}^{n}$ and $(f_{k})$ a sequence of functions $f_{k}\colon\Omega\to\mathbb{R}^{m}$. If $\{f_{k}\}$ is equibounded and uniformly equicontinuous then there exists a uniformly convergent subsequence $(f_{k_{j}})$.

A more abstract (and more general) version is the following.

Let $X$ and $Y$ be totally bounded metric spaces and let $F\subset\mathcal{C}(X,Y)$ be an uniformly equicontinuous family of continuous mappings from $X$ to $Y$. Then $F$ is totally bounded (with respect to the uniform convergence metric induced by $\mathcal{C}(X,Y)$).

Notice that the first version is a consequence of the second. Recall, in fact, that a subset of a complete metric space is totally bounded if and only if its closure is compact (or sequentially compact). Hence $\Omega$ is totally bounded and all the functions $f_{k}$ have image in a totally bounded set. Being $F=\{f_{k}\}$ totally bounded means that $\overline{F}$ is sequentially compact and hence $(f_{k})$ has a convergent subsequence.

Title Ascoli-Arzelà theorem AscoliArzelaTheorem 2013-03-22 12:41:00 2013-03-22 12:41:00 paolini (1187) paolini (1187) 13 paolini (1187) Theorem msc 46E15 Arzelà-Ascoli theorem MontelsTheorem