Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 21H (Linear Analysis)

For a compact space $K$, the Arzelà-Ascoli theorem says that a family $\mathcal{F}\subset C(K)$ is relatively compact in the supremum norm if and only if it is pointwise bounded and equicontinuous. In the usual formulation $K$ is compact Hausdorff; the elementary argument below also works for any compact $K$ for real-valued continuous functions.

Pointwise bounded means ${\sup }_{f\in \mathcal{F}}|f(x)|<\infty$ for every $x\in K$. Equicontinuity means that for every $x\in K$ and $\epsilon >0$ there is a neighborhood $U$ of $x$ such that $|f(y)-f(x)|<\epsilon$ for every $y\in U$ and $f\in \mathcal{F}$. Relative compactness means compactness of the closure in $C(K)$, where $\Vert f{\Vert }_{\infty }={\sup }_K|f|$. This is a complete metric vector space, since uniform limits of continuous functions are continuous. Equivalently, every sequence in the family has a uniformly convergent subsequence, with limit in $C(K)$, though not necessarily in the original family.

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.

Set $g_n=f-f_n$. These are nonnegative continuous functions, decrease pointwise, and converge pointwise to zero. For $\epsilon >0$, the open sets $U_n=\{x:g_n(x)<\epsilon \}$ increase and cover $K$. A finite subcover exists by compactness; because the sets are increasing, one $U_N$ already covers $K$. Hence $0\le g_n(x)<\epsilon$ for every $x\in K$ and every $n\ge N$, and $\Vert f-f_n{\Vert }_{\infty }\le \epsilon$. Therefore $\boxed{\Vert f_n-f{\Vert }_{\infty }\to 0}$. This is Dini's theorem; continuity of the limit is essential to the openness used in the proof.