%! TEX root = PM.tex % vim: tw=50 % 25/11/2023 10AM \begin{flashcard}[birkhoff-ergodic-thm] \begin{theorem*}[Birkhoff's Ergodic Theorem] \label{birkhoff} \cloze{$(E, \mathcal{E}, \mu)$ $\sigma$-finite and $f \in \Lp[1](\mu)$ and $\theta : E \to E$ is \gls{mp}. $S_0 = 0$ and \[ S_n = S_n(f) = f + f \circ \theta + f \circ \theta^2 + \cdots + f \circ \theta^{n - 1} = f + S_{n - 1} \circ \theta .\] Then there exists a \gls{tinv_fn} $\ol{f}$ with $\mu(|\ol{f}|) \le \mu(|f|)$ such that $\frac{S_n}{n} \to \ol{f}$ \gls{al_ev} as $n \to \infty$.} \end{theorem*} \end{flashcard} \begin{flashcard}[maximal-ergodic-lemma] \begin{lemma*}[Maximal Ergodic lemma] \label{max_erg_lemma} \cloze{For $f \in \Lp[1](\mu)$ and $S^* = \sup_{n \ge 0} S_n(f)$, we must have \[ \int_{\{S^* > 0\}} f \dd \mu \ge 0 .\] } \end{lemma*} \end{flashcard} \begin{proof} \[ \mu(|S_n|) \le \sum_{i = 0}^{n - 1} \mu(|f \circ \theta^i|) \stackrel{\text{$\theta$ \gls{mp}}}{=} \sum_{i = 1}^{n - 1} \mu(|f|) = n |f| \] So $\mu \left( \left| \frac{S_n}{n} \right| \right) \le \mu(|f|)$. So \[ \mu(|\ol{f}|) = \mu \left(\liminf \left| \frac{S_n}{n} \right|\right) \stackrel{\text{\nameref{fatous_lemma}}}{\le} \liminf \mu \left( \left| \frac{S_n}{n} \right| \right) \le \mu(|f|) .\] Note that $\frac{S_n \circ \theta}{n} = \frac{S_{n + 1} - f}{n + 1} \times \frac{n + 1}{n}$, so $\limsup \frac{S_n}{n} \circ \theta = \limsup \frac{S_{n + 1}}{n + 1} = \limsup \frac{S_n}{n}$. Similarly, $\liminf \frac{S_n}{n} \circ \theta = \liminf \frac{S_n}{n}$. For $a < b$, \[ D = D(a, b) = \left\{ \liminf_n \left( \frac{S_n}{n} \right) < a < b < \limsup_n \left( \frac{S_n}{n} \right) \right\} \] is \gls{tinv}. Shall show $\mu(D) = 0$. \[ \Delta = \left\{ \liminf \left( \frac{S_n}{n} \right) < \limsup \left( \frac{S_n}{n} \right) \right\} = \bigcup_{\substack{a, b \in \QQ \\ a < b}} D(a, b) \] Hence if $\mu(D) = 0$ for all $a < b$, then \[ \mu \left( \bigcup_{\substack{a, b \in \QQ \\ a < b}} D(a, b) \right) = 0 \implies \mu(\Delta) = 0 .\] Define \[ \ol{f} = \begin{cases} \liminf \left( \frac{S_n}{n} \right) = \limsup \left( \frac{S_n}{n} \right) & \text{on $\Delta^c$} \\ 0 & \text{on $\Delta$} \end{cases} \] Then $\frac{S_n}{n} \to \ol{f}$ $\mu$ \gls{al_ev} and $\ol{f}$ is \gls{tinv_fn} (as $\liminf \frac{S_n}{n}$ is \gls{tinv_fn} and $\Delta$ is \gls{tinv}). Fix $a < b$. Note that $\theta : D \to D$ by invariance and $\theta$ is $\mu|_D$-\gls{mp}. Also, either $b > 0$ or $a < 0$ (if $a < 0$, change $f$ to $-f$, $b$ to $-a$, $a$ to $-b$, then $b = -a > 0$). So assume $b > 0$ without loss of generality. Shall apply \nameref{max_erg_lemma} on $D$ with $\mu|_D$. For any $B \subseteq D$ measurable and $\mu(B) < \infty$, let $g = f - b\indicator{B}$. Then $g \in \Lp[1]$, and on $D$, \[ S_n(g) = S_n(f) - bS_n(\indicator{B}) \ge S_n(f) - nb > 0 \] for some $n$. Hence, $S^*(g) = \sup_{n \ge 0} S_n(g) > 0$ on $D$. Hence \[ \{S^*(g) . 0\} \cap D = D .\] Thus by \nameref{max_erg_lemma} on $D$, $\int_{\{S^*(g) > 0\} \cap D} g \dd \mu \ge 0$, i.e. \[ 0 \le \int_D (f - b \indicator{B}) \dd \mu = \int_D f \dd \mu - b \mu(B) \] hence $b\mu(B) \le \int_D f \dd \mu$. Since $\mu$ is $\sigma$-finite, there exists $(B_n)$ measurable sets, $B_n \uparrow D$ and $\mu(B_n) < \infty$ for all $n$. Hence \[ b\mu(D) = b\mu(B_n) \le \int_D f \dd \mu < \infty \] (as $f \in \Lp[1]$). Hence $\mu(D) < \infty$. A similar argument applied to $(-f)$ and $(-a)$ will give \[ (-a) \mu(D) \le \int_D(-f) \dd \mu \] (just take $D$ instead of $B$ now, $(-f) - (-a)\indicator{D}$). Hence \[ b\mu(D) \le \int_D f \dd \mu \le a \mu(D) \] But $a < b$. As $\mu(D) < \infty$, hence $\mu(D) = 0$. \end{proof} \begin{flashcard}[von-neumann-erg-thm] \begin{theorem*}[von Neumann's $\Lp$ Ergodic Theorem] \cloze{Let $\mu(E) < \infty$ and $1 \le p < \infty$. Then for $f \in \Lp(\mu)$, \[ \frac{S_n(f)}{n} \to \ol{f} \] in $\Lp$ as $n \to \infty$.} \end{theorem*} \begin{proof} \cloze{ For any $f \in \Lp(\mu)$, $\normp\|f \circ \theta^n\|_p = \normp\|f\|_p$ as $\theta$ is $\mu$ \gls{mp}. So by Minkowski inequality, \[ \left\| \frac{S_n(f)}{n} \right\|_p \le \frac{1}{n} \sum_{i = 0}^{n - 1} \ub{ \normp\|f \circ \theta^1\|_p}_{=\normp\|f\|_p} = \normp\|f\|_p \] Now, let $\eps > 0$ be given, then choose $K < \infty$ such that $\|f - f_K\|_p < \frac{\eps}{3}$ where $f_K = (-K) \vee f \wedge (K)$. \[ \left( \normp\|f - f_K\|_p^p = \int |f - f_K|^p \dd \mu \le \int |f|^p \indicator{|f| > K} \dd \mu \stackrel{\text{\nameref{dct}}}{\longrightarrow} 0 \right) \] But $f_K$ is bounded and $\mu$ a finite measure, so $f_K \in \Lp[1](\mu)$, hence by \nameref{birkhoff}, there exists $\ol{f_K} \in \Lp[1]$ such that $\frac{S_n(f_K)}{n} \to \ol{f_K}$ \gls{al_ev}. Again, $f \in \Lp(\mu) \subset \Lp[1](\mu)$ (as $\mu$ is a finite measure), so by Birkhoff, there exists $\ol{f} \in \Lp[1]$ such that $\frac{S_n(f)}{n} \to \ol{f}$ \gls{al_ev} as $n \to \infty$. Then \begin{align*} \normp\|\ol{f} - \ol{f_K}\|_p^p &= \int |\ol{f} - \ol{f_K}|^p \dd \mu \\ &= \int \liminf_n \left| \frac{S_n(f)}{n} - \frac{S_n(f_K)}{n} \right|^p \dd \mu \\ &= \int \liminf_n \left| \frac{S_n(f - f_K)}{n} \right|^p \dd \mu \\ &\le \liminf \int \left| \frac{S_n(f - f_K)}{n} \right|^p \dd \mu \qquad \text{\nameref{fatous_lemma}} \\ &= \liminf_n \left\| \frac{S_n(f - f_K)}{n} \right\|_p^p \end{align*} Then \begin{align*} \left\| \frac{S_n}{n} - \ol{f} \right\|_p &\stackrel{\text{\nameref{fatous_lemma}}}{\le} \left\| \frac{S_n(f)}{n} - \frac{S_n(f_K)}{n} \right\|_p + \|\ol{f} - \ol{f_K}\|_p + \left\| \frac{S_n(f_K)}{n} - \ol{f_K} \right\|_p \\ &= \ub{\left\| \frac{S_n(f - f_K)}{n} \right\|_p}_{= \normp\|f - f_K\|_p \le \eps/3} + \ub{\|\ol{f} - \ol{f_K}\|_p}_{\le \eps/3} + \ub{\left\| \frac{S_n(f_K)}{n} - \ol{f_K} \right\|_p}_{\to 0} \\ \end{align*} so $\frac{S_n}{n} \to \ol{f}$ in $\Lp$ as $n \to \infty$ as desired.} \end{proof} \end{flashcard}