Arzelà–Ascoli theorem

Statement

Let $X$ be a compact metric space and let $C(X)$ be the set of continuous functions $X \to \mathbb{R}$, with the uniform metric $d_\infty(f, g) = \sup_{x \in X} |f(x) - g(x)|$. Let $\mathcal{F} \subseteq C(X)$. Then the closure of $\mathcal{F}$ in $C(X)$ is compact if and only if $\mathcal{F}$ is equicontinuous and uniformly bounded. Here uniformly bounded means that there is a real $M$ with $|f(x)| \le M$ for all $f \in \mathcal{F}$ and $x \in X$.

Remarks

A sequential form: each uniformly bounded and equicontinuous sequence of functions $[a, b] \to \mathbb{R}$ has a subsequence that converges uniformly on $[a, b]$. The proof shows that $\mathcal{F}$ is totally bounded in the complete space $C(X)$, and then uses compact iff complete and totally bounded. For example, the functions $f \colon [0, 1] \to [0, 1]$ that are Lipschitz with constant $1$ form a compact set in $C([0, 1])$. The theorem is a main tool for existence proofs, for example in the Peano existence theorem for ordinary differential equations.

Explanations

L1 human

In $\mathbb{R}^n$, bounded sequences have convergent subsequences. The Arzelà–Ascoli theorem gives the same result for sequences of functions, but you also need equicontinuity. Equicontinuity stops the functions from becoming more and more steep, as $x^n$ does near $1$.