The space C(K,R)
Definition (Space of continuous functions on a compact set)
Let \((X, d)\) be a metric space and \(K \subseteq X\) compact. We define
\[\mathcal{C}(K, \mathbb{R}) = \{ f : K \to \mathbb{R} \;:\; f \text{ continuous}\}\]and, for \(f, g \in \mathcal{C}(K, \mathbb{R})\),
\[d_{\infty}(f, g) = \sup_{x \in K} |f(x) - g(x)|.\]\((\mathcal{C}(K, \mathbb{R}), d_{\infty})\) is a metric space whose convergence coincides with uniform convergence.
Equicontinuity
Definition (Equicontinuous family)
Let \(\{f_{\alpha}\}_{\alpha \in \Omega}\) be a family of functions \(f_{\alpha} : X \to Y\) with \((X, d), (Y, \rho)\) metric spaces. We say that \(\{f_{\alpha}\}_{\alpha \in \Omega}\) is equicontinuous at \(x_{0} \in X\) if for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that
\[d(x_{0}, y) < \delta \implies \rho(f_{\alpha}(y), f_{\alpha}(x_{0})) < \varepsilon \quad \text{for every } \alpha \in \Omega.\]The family is equicontinuous if it is so at every point of \(X\).
Lemma (Compact subsets of \(\mathcal{C}(K, \mathbb{R})\) are equicontinuous)
| Let \(C \subseteq \mathcal{C}(K, \mathbb{R})\) be compact. Then \(C\) is equicontinuous: given \(\varepsilon > 0\) and \(x_{0} \in K\), there exists \(\delta > 0\) such that for every \(f \in C\) and every \(y \in K\) with \(d(x_{0}, y) < \delta\) we have $$ | f(y) - f(x_{0}) | < \varepsilon$$. |
| Proof: By contradiction: if it were to fail, there would exist \(\varepsilon > 0\) and \(x_{0} \in K\) such that for every \(n \in \mathbb{N}\) there exist \(y_{n}, k_{n}\) with \(d(x_{0}, y_{n}) < 1/n\) and $$ | f_{k_{n}}(x_{0}) - f_{k_{n}}(y_{n}) | \geq \varepsilon\(, where\){f_{k_{n}}} \subseteq C\(. By compactness there exists a subsequence\){f_{k_{n_{l}}}}{l \geq 1}\(which converges uniformly to a continuous function\)f : K \to \mathbb{R}\(. To simplify the notation, write\)\tilde{f}{l} := f_{k_{n_{l}}}\(and\)\tilde{y}{l} := y{n_{l}}\(. Then\)\tilde{y}{l} \to x{0}\(and\) | \tilde{f}{l}(x{0}) - \tilde{f}{l}(\tilde{y}{l}) | \geq \varepsilon$$. |
| Since \(\tilde{y}_{l} \to x_{0}\) and \(f\) is continuous, there exists \(l_{0}\) such that $$ | f(\tilde{y}{l}) - f(x{0}) | < \varepsilon/3\(for\)l \geq l_{0}\(. Then, for\)l$$ large, |
| contradicting that $$ | f_{m_{l}}(y_{l}) - f_{m_{l}}(x_{0}) | \geq \varepsilon$$. |
Lemma (Equicontinuity and pointwise convergence to a continuous function give uniform convergence)
Let \(\{f_{n}\}_{n=1}^{\infty}\) be an equicontinuous family of functions \(f_{n} : X \to Y\) and \(K \subseteq X\) compact such that \(\lim_{n \to \infty} f_{n}(x) = f_{0}(x)\) for every \(x \in K\), with \(f_{0}\) continuous on \(K\). Then \(\{f_{n}\}\) converges uniformly to \(f_{0}\) on \(K\).
Proof: Let \(\varepsilon > 0\). By equicontinuity, for each \(x \in K\) there exists \(\delta_{x} > 0\) such that \(y \in B(x, \delta_{x})\) implies \(\rho(f_{n}(x), f_{n}(y)) < \varepsilon/3\) for every \(n \geq 0\). Since \(K\) is compact, there exist \(x_{1}, \dots, x_{m}\) with \(K \subseteq \bigcup_{i=1}^{m} B(x_{i}, \delta_{x_{i}})\). Take \(n_{0}\) such that \(\rho(f_{n}(x_{i}), f_{0}(x_{i})) < \varepsilon/3\) for every \(1 \leq i \leq m\) and every \(n \geq n_{0}\).
For \(y \in K\), there exists \(i\) with \(y \in B(x_{i}, \delta_{x_{i}})\), so
\[\rho(f_{n}(y), f_{0}(y)) \leq \rho(f_{n}(y), f_{n}(x_{i})) + \rho(f_{n}(x_{i}), f_{0}(x_{i})) + \rho(f_{0}(x_{i}), f_{0}(y)) < \tfrac{\varepsilon}{3} + \tfrac{\varepsilon}{3} + \tfrac{\varepsilon}{3} = \varepsilon\]for \(n \geq n_{0}\).
Lemma (Convergence on a dense set for an equicontinuous family yields continuity)
Let \(D \subseteq X\) be dense and \(\{f_{n}\}_{n=1}^{\infty}\) an equicontinuous sequence of functions \(f_{n} : X \to Y\). Suppose that \(\{f_{n}(x)\}_{n=1}^{\infty}\) converges for every \(x \in D\) and that for each \(x \in X\) the set \(\overline{\{f_{n}(x)\}_{n=1}^{\infty}}\) is complete (in particular this holds if \(Y\) is complete, or if \(\overline{\{f_{n}(x)\}_{n=1}^{\infty}}\) is compact). Then \(\{f_{n}(x)\}\) converges for every \(x \in X\) and its limit is continuous.
Proof: Convergence. Let \(x \in X\) and \(\varepsilon > 0\). By equicontinuity there exists \(\delta > 0\) with
\[d(x, y) < \delta \implies \rho(f_{n}(x), f_{n}(y)) < \varepsilon/3 \quad \text{for every } n \geq 1.\]By density there exists \(y \in B(x, \delta) \cap D\) and, since \(\{f_{n}(y)\}\) converges, there exists \(n_{0}\) with \(\rho(f_{n}(y), f_{m}(y)) < \varepsilon/3\) for \(n, m \geq n_{0}\). Then
\[\rho(f_{n}(x), f_{m}(x)) \leq \rho(f_{n}(x), f_{n}(y)) + \rho(f_{n}(y), f_{m}(y)) + \rho(f_{m}(y), f_{m}(x)) < \varepsilon,\]so \(\{f_{n}(x)\}\) is Cauchy. Since \(\{f_{n}(x)\} \subseteq \overline{\{f_{n}(x)\}}\) and the latter is complete by hypothesis, \(\{f_{n}(x)\}\) converges in \(\overline{\{f_{n}(x)\}} \subseteq Y\).
Continuity of the limit. Define \(f(x) = \lim_{n} f_{n}(x)\). For \(d(x, y) < \delta\),
\[\rho(f(x), f(y)) = \lim_{n \to \infty} \rho(f_{n}(x), f_{n}(y)) \leq \tfrac{\varepsilon}{3},\]so \(f\) is continuous at \(x\).
Note (Sufficient hypotheses on the codomain)
The hypothesis “\(\overline{\{f_{n}(x)\}_{n=1}^{\infty}}\) complete” of the previous lemma is satisfied, for example, when \(Y\) is complete (every closed subset of a complete space is complete) or when \(\overline{\{f_{n}(x)\}_{n=1}^{\infty}}\) is compact (since compact \(\Rightarrow\) complete). The second case is the one used in the Arzelà-Ascoli Theorem below.
Lemma (Pointwise convergence on a compact set under equicontinuity implies uniform convergence)
Let \(\{f_{n}\}_{n=1}^{\infty}\) be an equicontinuous family, \(f_{n} : X \to Y\), and \(K \subseteq X\) compact. If \(\lim_{n \to \infty} f_{n}(x) = f_{0}(x)\) exists in \(Y\) for every \(x \in K\), then \(f_{0}\) is continuous and \(f_{n}\) converges uniformly to \(f_{0}\) on \(K\).
Proof: Continuity of \(f_{0}\). Let \(x \in K\) and \(\varepsilon > 0\). By equicontinuity there exists \(\delta > 0\) such that for every \(n \geq 1\), \(d(x, y) < \delta\) implies \(\rho(f_{n}(x), f_{n}(y)) < \varepsilon/3\). Passing to the limit \(n \to \infty\),
\[\rho(f_{0}(x), f_{0}(y)) = \lim_{n \to \infty} \rho(f_{n}(x), f_{n}(y)) \leq \tfrac{\varepsilon}{3} < \varepsilon\]for \(d(x, y) < \delta\), which proves the continuity of \(f_{0}\) without requiring any additional hypothesis on \(Y\).
Uniform convergence. Applying the first equicontinuity lemma (Equicontinuity and pointwise convergence to a continuous function give uniform convergence) with the continuous limit \(f_{0}\) just obtained, \(f_{n} \to f_{0}\) uniformly on \(K\).
The theorem
Theorem (Arzelà-Ascoli)
Let \(X\) be separable and \(\{f_{n}\}_{n=1}^{\infty}\) an equicontinuous family of functions \(f_{n} : X \to Y\). Suppose that \(\overline{\{f_{n}(x)\}_{n=1}^{\infty}}\) is compact for every \(x \in X\). Then there exists a subsequence of \(\{f_{n}\}\) which converges pointwise to a continuous function \(f : X \to Y\). Moreover, the convergence is uniform on every compact set \(K \subseteq X\).
Proof: Let \(D = \{x_{n}\}_{n=1}^{\infty} \subseteq X\) be dense (it exists by separability). Since \(\overline{\{f_{n}(x_{1})\}_{n=1}^{\infty}}\) is compact, there exists a subsequence \(\{f_{n_{k}}^{1}\}_{k}\) such that \(\{f_{n_{k}}^{1}(x_{1})\}\) converges. Likewise \(\overline{\{f_{n_{k}}^{1}(x_{2})\}_{k}}\) is compact (it is a closed set contained in the compact set \(\overline{\{f_{n}(x_{2})\}_{n=1}^{\infty}}\), hence compact), so that there exists a subsequence \(\{f_{n_{k}}^{2}\}\) of \(\{f_{n_{k}}^{1}\}\) such that \(\{f_{n_{k}}^{2}(x_{2})\}\) converges.
Iterating, given \(\{f_{n_{k}}^{m}\}\) with \(\{f_{n_{k}}^{m}(x_{j})\}\) convergent for \(1 \leq j \leq m\), we extract a subsequence \(\{f_{n_{k}}^{m+1}\}\) of \(\{f_{n_{k}}^{m}\}\) such that \(\{f_{n_{k}}^{m+1}(x_{m+1})\}\) converges. By construction, \(\{f_{n_{k}}^{m}(x_{j})\}\) converges for \(1 \leq j \leq m\).
We take the diagonal subsequence \(\{f_{n_{m}}^{m}\}_{m=1}^{\infty}\). For \(j \leq m\), \(\{f_{n_{m}}^{m}(x_{j})\}_{m \geq j}\) is a subsequence of \(\{f_{n_{k}}^{j}\}\) and therefore converges. Thus \(\{f_{n_{m}}^{m}\}\) converges pointwise on \(D\) and is equicontinuous. Moreover, for each \(x \in X\), \(\overline{\{f_{n_{m}}^{m}(x)\}_{m=1}^{\infty}}\) is a closed set contained in the compact set \(\overline{\{f_{n}(x)\}_{n=1}^{\infty}}\), hence compact and in particular complete. By the previous lemma (applied with this completeness condition on the closures of the values), \(\{f_{n_{m}}^{m}\}\) converges on all of \(X\) to a continuous function \(f\), and the convergence is uniform on compact sets.