Modulus of Continuity
Definition 1
- $f \in C[0, \infty)$,
Proposition 2
- (1) If $\delta_{1} \le \delta_{2}$, \(m^{T}(f, \delta_{1}) \le m^{T}(f, \delta_{2})\),
- (2) If \(m^{T}(f, \delta_{1} + \delta_{2}) \le m^{T}(f, \delta_{1}) + m^{T}(f, \delta_{2})\),
- (3) $f$ is uniformly continuous if and only if $\lim_{\delta \searrow 0} m^{T}(f, r) = 0$,
- (4) If $\lambda > 0$, $m^{T}(f, \lambda \delta) < (1 + \lambda) m^{T}(f, \delta)$,
- (5)
(6) $m^{T}(f, \delta)$ is continuous in $f \in C[0, T]$ for fixed $\delta$.
proof
proof of (1)
Since $B(\delta_{1}) \subseteq B(\delta_{2})$,
\[\begin{eqnarray} \sup \{ \abs{f(x) - f(y)} \mid (x, y) \in B(\delta_{1}) \} \le \sup \{ \abs{f(x) - f(y)} \mid (x, y) \in B(\delta_{2}) \} \nonumber \end{eqnarray}\]proof of (3)
(only if)
Let $\epsilon > 0$ be fixed. Since $f$ is uniformly continuous, there exists $\delta > 0$ such that
\[\abs{ x - y } < \delta \Rightarrow \abs{ f(x) - f(y) } < \frac{\epsilon}{2} .\]Thus,
\[\begin{eqnarray} \sup \{ \abs{ f(x) - f(y) } \mid \abs{ x - y } < \delta \} & \le & \frac{\epsilon}{2} \nonumber \\ & < & \epsilon \nonumber \end{eqnarray}\]Since $\epsilon$ was chosen arbitrary, the proof completed.
(if)
Let $\epsilon > 0$ be fixed. By assumption,
proof of (4)
proof of (5)
By triangle inequality,
\[\begin{eqnarray} \forall x, y \in [0, T], \ \abs{x - y} < \delta, \ \abs{f(x) - f(y)} & \le & \abs{f(x) - g(y)} + \abs{g(y) - g(x)} + \abs{g(x) - f(y)} \nonumber \end{eqnarray}\]Hence
\[\begin{eqnarray} \forall x, y \in [0, T], \ \abs{x - y} < \delta, \ \abs{f(x) - f(y)} & \le & \abs{f(x) - g(y)} + m^{T}(g, \delta) + \abs{g(x) - f(y)} \nonumber \\ & \le & \abs{f(x) - g(y)} + m^{T}(g, \delta) + \norm{ f - g }_{\infty, [0, T]} \nonumber \\ & \le & \norm{ f - g }_{\infty, [0, T]} + m^{T}(g, \delta) + \norm{ f - g }_{\infty, [0, T]} \nonumber \\ & \le & 2 \norm{ f - g }_{\infty, [0, T]} + m^{T}(g, \delta) . \nonumber \end{eqnarray}\]Thus,
\[\begin{eqnarray} & & m^{T}(f, \delta) \le 2 \norm{ f - g }_{\infty, [0, T]} + m^{T}(g, \delta) \nonumber \\ & \Leftrightarrow & m^{T}(f, \delta) - m^{T}(g, \delta) \le 2 \norm{ f - g }_{\infty, [0, T]} \end{eqnarray} .\]Similary,
\[\begin{eqnarray} & & m^{T}(g, \delta) \le 2 \norm{ f - g }_{\infty, [0, T]} + m^{T}(f, \delta) \nonumber \\ & \Leftrightarrow & m^{T}(g, \delta) - m^{T}(f, \delta) \le 2 \norm{ f - g }_{\infty, [0, T]} . \nonumber \end{eqnarray}\]Therefore, the proof of (5) completes.
(6)
Let $(f_{n}) \in C[0, \infty]$ be a sequence such that
\[\norm{ f_{n} - f }_{\infty, [0, T]} \rightarrow 0 .\]By (5),
\[\abs{ m^{T}(f_{n}, \delta) - m^{T}(f, \delta) } \le 2 \norm{ f_{n} - f }_{\infty, [0, T]} \rightarrow 0 .\]Therefore, the proof (6) completes.
Theorem Arzela-Ascoli
- $A \subseteq C[0, \infty)$,
- $\bar{A}$,
- closure of $A$,
The following conditions are euqivalent;
(1) $\bar{A}$ is compact
(2)
\[\begin{eqnarray} \sup_{\omega \in A} \abs{\omega(0)} & < & \infty \label{arzela_ascoli_theorem_condition_01} \\ \forall T > 0, \ \lim_{\delta \rightarrow 0} \sup_{\omega \in A} m^{T}(\omega, \delta) = 0 \label{arzela_ascoli_theorem_condition_02} \end{eqnarray}\]proof
(1) $\Rightarrow$ (2)
\[\begin{eqnarray} G_{n} & := & \{ \omega \in C[0, \infty) \mid \abs{\omega(0)} < n \} \nonumber \\ \bar{A} & \subseteq & \bigcup_{n \in \mathbb{N}} G_{n} . \nonumber \end{eqnarray}\]Since $\bar{A}$ is compact, there exists $n \in \mathbb{N}$ such that $\bar{A} \subseteq G_{n}$. Let $\epsilon > 0$ be fixed. Let
\[\delta > 0, \ K_{\delta} := \{ \omega \in \bar{A} \mid m^{T}(\omega, \delta) \ge \epsilon \} .\]Since $\omega^{T}(\omega, \delta)$ is continuous in $\omega \in \Omega$, $K_{\delta}$ is closed and \(K_{\delta} \subseteq \bar{A}\). A closed subset of compact set is also compact. It follows that $K_{\delta}$ is compact. By proposition, \(\lim_{\delta \rightarrow 0} m^{T}(\omega, \delta) = 0\). It follows that
\[\bigcap_{\delta > 0} K_{\delta} = \emptyset .\]Thus, there exists $\delta_{\epsilon} > 0$ such that $K_{\delta_{\epsilon}} = \emptyset$. This proves $(\Rightarrow)$ part.
(1) $\Leftarrow$ (2)
Since $C[0, \infty)$ is a metric space, we will show that every sequence \(\{\omega_{n}\}_{n=1}^{\infty} \subseteq A\) has a convergent subsequence. Let \(\{\omega_{n}\}_{n \in \mathbb{N}} \subseteq A\) be fixed. By \(\eqref{arzela_ascoli_theorem_condition_01}\), there exists $\delta_{1} > 0$ such that
\[\sup_{\omega \in A} m^{T}(\omega, \delta_{1}) < 1 .\]It follows that for $m \in \mathbb{N}$ and $t (0, T]$ with $(m - 1) \delta_{1} < t \le ((m \delta_{1}) \wedge T)$,
\[\begin{eqnarray} \abs{\omega(t)} & \le & \abs{\omega(0)} + \sum_{k=1}^{m-1} \abs{ \omega(k\delta_{1}) - \omega((k - 1)\delta_{1}) } + \abs{ \omega(t) - \omega((m - 1) \delta_{1}) } \nonumber \\ & \le & \abs{\omega(0)} + \sum_{k=1}^{m-1} m^{T}(\omega, \delta_{1}) + m^{T}(\omega, \delta_{1}) \nonumber \\ & \le & \abs{\omega(0)} + m \nonumber . \end{eqnarray}\] \[\begin{eqnarray} \forall r \in \mathbb{Q}_{\ge 0}, \ \omega_{n}(r) & \le & \abs{\omega_{n}(0)} + m & \le & \sup_{\omega \in A} \abs{\omega(0)} + m \nonumber \end{eqnarray}\]Hence \(\{\omega_{n}(r)\}\) is bounded for each \(r \in \mathbb{Q}_{\ge 0}\). Let \(\{r_{0}, r_{1}, \ldots\}\) be an enumeration of \(\mathbb{Q}_{\ge 0}\). Since \(\{\omega_{n}(r_{0})\}\) is bounded, there exists a subsequence \(\{\omega_{n}^{(0)}\}_{n \in \mathbb{N}}\) converging to a limit denoted \(\omega(r_{0})\). From \(\{\omega_{n}^{(k-1)}\}\), we recursively define a subsequence \(\{\omega_{n}^{(k)}\}\) converging to a limit denoted \(\omega(r_{k})\). As a consequence of this process, we can define diagonal sequence
\[\{ \bar{\omega}_{n} \}_{n \in \mathbb{N}} := \{ \omega_{n}^{(n)} \}_{n \in \mathbb{N}}\]such that
\[\forall r \in \mathbb{Q}_{\ge 0}, \ \omega_{n}(r) \rightarrow \omega(r) .\]