Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 3 / 21H (Linear Analysis)

A family $F$ of continuous functions on a compact space $K$ is equicontinuous if, for every $x$ and every $\epsilon >0$, there is a neighbourhood $U$ of $x$ such that $|f(y)-f(x)|<\epsilon$ for every $y$ in $U$ and every $f$ in $F$. The Arzela-Ascoli theorem says that $F$ has compact closure in $C(K)$ with the uniform norm if and only if it is equicontinuous and pointwise bounded.

Apply Arzela-Ascoli theorem on $[-m,m]$. Pointwise boundedness and equicontinuity give a subsequence converging uniformly there. Successively extract subsequences for $m=1,2,\dots$, and take the diagonal subsequence. It converges uniformly on every closed bounded interval. Its limits on nested intervals agree, so they define $f$ on the real numbers; the locally uniform limit of continuous functions is continuous.

Fix $x$ and $\epsilon >0$. Since $\mathcal{K}$ is compact in the uniform norm, choose a finite $\epsilon /3$-net $f_1,\dots ,f_r$. By continuity, there is a neighbourhood $U$ of $x$ on which $|f_i(y)-f_i(x)|<\epsilon /3$ for every $i$. For $f$ in $\mathcal{K}$, choose $f_i$ with $\Vert f-f_i{\Vert }_{\infty }<\epsilon /3$. The triangle inequality gives $|f(y)-f(x)|<\epsilon$ on $U$, uniformly in $f$. Thus $\mathcal{K}$ is equicontinuous.

A locally compact Hausdorff space is regular. Cover $K$ by finitely many open sets whose compact closures lie in $U$; their union $V$ satisfies $K\subset V\subset \overline{V}\subset U$, with $\overline{V}$ compact. The compact Hausdorff space $\overline{V}$ is normal, so the Urysohn lemma gives a continuous function $g:\overline{V}\to [0,1]$ equal to $1$ on $K$ and $0$ on $\overline{V}\setminus V$. Extend $g$ by zero outside $\overline{V}$. The boundary values make the extension continuous, and its support is a compact subset of $U$.