Arzela-Ascoli Theorem

Def'n. Equicontinuous

A family of function \(\mathcal F \subset C(K,\mathbb R^m)\) is equicontinuous at \(a\in K\)

\[\forall f \in \mathcal F. \forall \epsilon > 0. \exists \delta > 0. \|x-a\|<\delta \Rightarrow \|f(x)-f(a)\|<\epsilon\]

Then \(\mathcal F\) is equicontinuous on \(K\) is \(\mathcal F\) is equicontinuous \(\forall a\in K\).

Def'n. Uniformly equicontinuous

\(\mathcal F\) is uniformly equicontinuous if

\[\forall \epsilon > 0. \exists \delta > 0. \forall f\in\mathcal F. \forall x,y\in K. \|x-y\|<\delta \Rightarrow \|f(x)-f(y)\|<\epsilon\]

Theorem 1

\(\mathcal G := \{g_n\}_{n\geq 1}\cup \{g\}\) where \(g_n \in C(K,\mathbb R^m), g_n\rightarrow^{u.c.}g\), then \(\mathcal G\) is equicontinuous.

Then, for each \(g_k \in \{g_1,g_2,...,g_N\}\), take \(\delta_k\) by continuity of each \(g_k\), take \(\delta = \min\{\delta_1,..,\delta_k, \delta'\}\)

\[\forall f \in \mathcal F. \forall \epsilon > 0. \exists \delta > 0. \|x-a\|<\delta \Rightarrow \|f(x)-f(a)\|<\epsilon\]

Example 1

\(f_n(x)=x^n, x\in [0,1], \mathcal F = \{f_n\}_{n\geq 1}\) is not equicontinuous at 1.

proof. Take \(\epsilon =1/2\), let \(\delta > 0\), wlog, \(\delta < 1\).
Take \(y = 1-\delta/2<1\), then \(y^n \rightarrow 0\), hence we can take \(N\) s.t. \(\forall n\geq N, 1-y^n > 1/2\)
Therefore, \(|1-y|=\delta/2 < \delta\) but \(|f_n(1)-f_n(y)|= 1-y^n > 1/2\)

Lemma 1. Compact Implies Equicontinuous

If \(\mathcal F\) compact, then \(\mathcal F\) equicontinuous on \(K\).

proof. Suppose \(\mathcal F\) is not equicontinuous,
take \(a\in K, \epsilon > 0\) s.t. \(\forall n\geq 1. \exists f_n\in\mathcal F. \exists x_n\in K\) s.t. \(\|x_n-a\|<1/n\) but \(\|f_n(x_n) - f_n(a)\| \geq \epsilon\), hence we construct sequences \(\{x_n\}, \{f_n\}\)
Then, any subset of \(\{f_n\}\) cannot be equicontinuous \((i)\)
However, since \(\mathcal F\) is compact, take \(\{f_{n_k}\}\) converges uniformly to some \(f\in\mathcal F\), and \(\{f_{n_k}\}\cup \{f\}\) is equicontinuous, this contradicts with \((i)\)

Lemma 2. Equicontinuous implies uniformly equicontinuous

proof. Suppose \(\mathcal F\) not u.e.c.
Take \(\epsilon > 0\) s.t. \(\forall n\geq 1, \exists x_n,y_n \in K. \exists f_n\in \mathcal F. \|x_n-y_n\| < 1/n \land \|f_n(x_n)-f_n(y_n)\|\geq \epsilon\) hence we construct sequence \(\{x_n\}, \{y_n\}, \{f_n\}\)
Since \(K\) is compact, take \(x_{n_k}\rightarrow a \in K\), then \(y_{n_k} = x_{n_k}-(x_{n_k}-y_{n_k})\rightarrow a\)

\(\mathcal F\) is equicontinous at \(a\in K\Rightarrow \exists \delta > 0, \|f_n(x)-f_n(a)\|\leq \epsilon/2\) for all \(\|x-a\|<\delta, f\in\mathcal F\).
Since \(x_{n_k}\rightarrow a, y_{n_k}\rightarrow a, \exists M\in \mathbb N, \forall m \geq M. \|x_{n_m}-a\|<\delta\land \|y_{n_m}-a\|<\delta\)
Therefore, \(\forall m\geq M\)

\[\begin{align*} \|f_{n_m}(x_{n_m})- f_{n_m}(y_{n_m})\|&\leq \|f_{n_m}(x_{n_m})- f_{n_m}(a)\| + \|f_{n_m}(a)- f_{n_m}(y_{n_m})\|\\&< 2(\epsilon/2)\\&=\epsilon \end{align*}\]

contradicts with assumption

Def'n. Totally bouded

\(S\subseteq K\) is an \(\epsilon\)-net of \(K\) if \(K\subseteq \cup_{a\in S}B_\epsilon(a)\)
\(K\) is totally bounded if it has a finite \(\epsilon\)-net \(\forall \epsilon > 0\)

Lemma 3. Bounded Implies Totally Bounded

If \(K\subseteq \mathbb R^m\) bounded, then totally bounded.

proof. Let \(\epsilon > 0\), choose \(N, N \leq \min\{\epsilon, \frac{1}{\sqrt m}\}\)
\(K\) bounded, hence \(\exists L > 0, \forall x = (x_1,...,x_m)\in K, |x_i|\leq L, \forall i\)
Let \(F = \{\frac{k_i}{2N^2}\}_{k_i\in \mathbb Z}\subseteq [-L,L]\), then \(F\) is a finite \(\frac{1}{2N^2}\)-net for \([-L,L]\).
Let \(A = \{x_1,...,x_m\}\subseteq \mathbb R^m\) s.t. \(x_i \in F, \forall i\)
Let \(\tilde A = \{a\in A: B_{\epsilon/2}(x)\cap K \neq \emptyset\}\). Then, for each \(x\in \tilde A\), choose \(x_a \in B_{\epsilon/2}(a)\cap K\).
Take \(x = (x_1,...,x_m)\in K\) for each \(i = 1,...,m, \exists a_i \in F\) s.t. \(|x_i-a_i|<\frac{1}{2N^2}\)
Then, \(\|x-a\| = \sqrt{\sum |x_i-a_i|^2} < \sqrt{\frac{m}{4N^4}} = \sqrt M/2N^2 \leq N/2N^2 \leq (2N)^{-1} < \epsilon/2\)
Also, \(B_{\epsilon/2}(a)\cap K\neq \emptyset, a\in \tilde A\), so that \(x_a\) is defined.
Then, \(\|x-x_a\|\leq \|x-a\|+\|x-x_a\|< \epsilon/2+\epsilon/2 = \epsilon\)

Lemma 4

If \(K\) bounded, then \(K\) contains a sequence \(\{x_i\}_{i\geq 1}\) dense in \(K\). Moreover, \(\forall \epsilon > 0. \exists N\in\mathbb N\) s.t. \(\{x_i\}_{i\leq N}\) is an \(\epsilon\)-net for \(K\).

proof. For each \(k\geq i\), let \(B_k\) be a finite \(k^{-1}\)-net.
Take \(\{x_i\}\) be the sequence which lists the \(B_k\) consecutively, i.e. \(x_0,...,x_{N_0}\in B_0, x_{N_0+1}, ...,x_{N_1} \in B_1,..\)
Let \(x\in K\), then \(\forall k\geq 1. \exists n_k\) s.t. \(x_{n_k}\in B_k\) and \(\|x-x_{n_k}\|< k^{-1}\), hence dense.
Also, given \(\epsilon > 0\), choose \(k > \epsilon^{-1}, \{x_i\}_{i\leq N_k}\) is a \(\epsilon\)-net.

Thrm. Arzela-Ascoli Theorem

\(\mathcal F \subseteq C(K,\mathbb R^m)\) is compact IFF closed, bounded, euicontinuous

\(\Rightarrow\) proof. Suppose not closed, then take \(\{f_n\}\subseteq \mathcal F\) s.t. \(f_n\rightarrow f\not\in \mathcal F\) contradicts with compactness.
Suppose not bounded, take \(\{f_n\}\subseteq \mathcal F\) that \(\|f_n\|_\infty\rightarrow\infty\) contradicts
and Lemma 1

\(\Leftarrow\) proof. Fix \(\{f_n\} \subseteq \mathcal F\), by the Lemma 4, \(\exists \{x_i\}\subseteq K\) s.t. \(\forall \epsilon > 0. \exists \{x_i\}_{i\leq N}\) is a \(\epsilon\)-net.
WTF \(\{f_{n_k}\} \subseteq \{f_n\}\) s.t. \(f_{n_k}(x_i)\rightarrow^{k}L_i, \forall 1\leq i\leq N\).
Let \(A_0=\mathbb N\), since \(\{f_n(x_1)\}_{n\in A_0}\) bounded, by Bolzano-Weierstrass Theorem, take the convergent subsequence, i.e. \(A_1\subseteq A_0, \lim_{n\in A_1}f_n(x_1) =L_1\).
Inductively take \(\mathcal A = A_0\supseteq A_1\supseteq A_2\supseteq ...\) be a decreasing sequence, s.t. \(\lim_{n\in A_i}f_n(x_i)=L_i\)
Then, for each \(k\geq 1\), let \(n_k\) be the \(k\)th element of \(A_k\), i.e.

\[\begin{matrix} &A_1 &n_1 & & \\ &A_2 & & n_2 &\\ &A_3 & & &n_3 \\ &... \end{matrix}\]

Since \(A_n\) is decreasing, for each \(i\geq 1\), there are at most \(i-1\) elements are not in \(A_i\).
In particular, this implies \(\lim_k f_{n_k}(x_i) = \lim_{n\in A_i} f_n(x_i) = L_i\)
Let \(g_k = f_{n_k}\), let \(\epsilon > 0\), since \(\mathcal F\) is equicontinuous, i.e. uniform equicontinuous. Take \(\delta > 0. \|x-y\|<\delta\Rightarrow\|f(x)-f(y)\|<\epsilon/3, \forall f \in \mathcal F\)
By definition of \(\{x_i\}\), take \(N\in\mathcal N, \{x_1,...,x_n\}\) is a \(\delta\)-net.
Since \(\lim_{k\rightarrow\infty} g_k(x_i)\) exists for all \(i\geq 1\).
\(\exists M\in \mathbb N\) s.t. \(\forall k,l\geq M. \forall 1\leq i\leq N\Rightarrow \|g_n(x_i)-g_l(x_i)\|<\epsilon/3\)
Since \(\{x_1,...,x_N\}\) is a \(\delta\)-net, \(\exists i, \|x_i-x\|<\delta\)

\[\|g_k(x)-g_l(x)\|\leq \|g_k(x)-g_k(x_i)\| + \|g_k(x_i)-g_l(x_i)\| + \|g_l(x_i)-g_l(x)\|<\epsilon\]

Since \(\{g_k\}\) is uniform Cauchy, and \(C(K,\mathbb R^m)\) is complete, \(g_k\rightarrow g\in C(K,\mathbb R^m)\)
Since \(\mathcal F\) closed, \(g\in\mathcal F\), therefore, compact.

Note that we only used closed at the very end. Therefore, If \(\{f_n\}\subseteq C([a,b])\) and is bounded and equicontinuous, then it has a convergent subsequence.