Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 21H / b / Solution

Assume first that $\overline{\mathcal{F}}$ is compact in the supremum norm. A finite norm-net shows that the norms of its members are bounded, so pointwise boundedness follows. Given $x$ and $\epsilon >0$, choose a finite $\epsilon /3$-net $g_1,\dots ,g_N$ in the closure. Their continuity gives a common neighborhood $U$ of $x$ where $|g_j(y)-g_j(x)|<\epsilon /3$ for every $j$. For any $f\in \mathcal{F}$, choose $g_j$ with $\Vert f-g_j{\Vert }_{\infty }<\epsilon /3$; the triangle inequality then gives $|f(y)-f(x)|<\epsilon$. Thus the family is equicontinuous.

Conversely, suppose pointwise boundedness and equicontinuity. For a prescribed $\epsilon >0$, choose neighborhoods $U_x$ where every $f$ varies from $f(x)$ by less than $\epsilon /3$. Compactness supplies a finite subcover $U_{x_1},\dots ,U_{x_N}$. Pointwise boundedness makes the set of vectors $(f(x_1),\dots ,f(x_N))$ bounded in ${\mathbb{R}}^N$. Partition a bounding box into finitely many boxes of coordinate diameter less than $\epsilon /3$, and choose one function from each occupied box. If $f,g$ belong to the same box and $y\in U_{x_j}$, thenThe chosen functions are therefore a finite $\epsilon$-net in the supremum norm. Thus $\mathcal{F}$ is totally bounded. Its closure is complete and totally bounded, hence compact. To recall the last metric-space fact, successively selecting subsequences in balls of radii $2^{-j}$ produces a Cauchy diagonal subsequence from every sequence; completeness supplies its limit. This proves both directions of the Arzelà-Ascoli theorem.