Added basic theory of Lp spaces, alongside some integral stuff.

src/cat/tricks/index.tex

@@ -2,3 +2,4 @@
 \label{chap:tricks}
 
 \input{./src/cat/tricks/dyadic.tex}
+\input{./src/cat/tricks/product.tex}

src/cat/tricks/product.tex

@@ -0,0 +1,25 @@
+\section{Product Inequalities}
+\label{section:product-inequalities}
+
+\begin{lemma}[Young's Inequality]
+\label{lemma:young-inequality}
+    Let $p, q \in (1, \infty)$ such that $1/p + 1/q = 1$, then for any $a, b > 0$,
+    \[
+    ab \le \frac{a^p}{p} + \frac{b^q}{q}
+    \]
+    and for any $\eps > 0$,
+    \[
+    ab \le \eps a^p + \frac{1}{q}(\eps q)^{-q/p}b^q
+    \]
+\end{lemma}
+\begin{proof}
+    Since $x \mapsto \exp(x)$ is convex,
+    \begin{align*}
+        ab &= \exp(\ln(a) + \ln(b)) = \exp\paren{\frac{1}{p}\ln(a^p) + \frac{1}{q}\ln(b^q)} \\
+        &\le \frac{1}{p}\exp(\ln(a^p)) + \frac{1}{q}\exp(\ln(a^q)) = \frac{a^p}{p} + \frac{b^q}{q}
+    \end{align*}
+    For any $\eps > 0$, applying the above inequality to $(\eps p)^{1/p}a$ and $b/(\eps p)^{1/p}$ yields
+    \[
+    ab = (\eps p)^{1/p}a \cdot \frac{b}{(\eps p)^{1/p}} \le \eps a^p + \frac{b^q}{q}(\eps p)^{-(1/p)q} = \eps a^p + \frac{1}{q}(\eps q)^{-q/p}b^q
+    \]
+\end{proof}

src/fa/index.tex

@@ -6,3 +6,4 @@
 \input{./src/fa/lc/index.tex}
 \input{./src/fa/norm/index.tex}
 \input{./src/fa/rs/index.tex}
+\input{./src/fa/lp/index.tex}

src/fa/lp/definition.tex

@@ -0,0 +1,106 @@
+\section{Basic Properties}
+\label{section:lp-basic}
+
+\begin{definition}[$\mathcal{L}^p$ Spaces]
+\label{definition:lp-unequivalence}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, $f: X \to E$ be strongly measurable, and $p \in [1, \infty)$, then $f$ is \textbf{$p$-integrable} if
+    \[
+    \norm{f}_{L^p(X; E)} = \norm{f}_{L^p(\mu; E)} = \norm{f}_{L^p(X, \cm, \mu; E)} = \braks{\int \norm{f}_E^p d\mu}^{1/p} < \infty
+    \]
+    The set $\mathcal{L}^p(X; E) = \mathcal{L}^p(\mu; E) = \mathcal{L}^p(X, \cm, \mu; E)$ is the space of all $p$-integrable functions on $X$.
+\end{definition}
+
+\begin{definition}[Essential Supremum]
+\label{definition:esssup}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $f: X \to E$ be strongly measurable, then $f$ is \textbf{essentially bounded} if
+    \[
+    \norm{f}_{L^\infty(X; E)} = \norm{f}_{L^\infty(\mu; E)} = \norm{f}_{L^\infty(X, \cm, \mu; E)} = \inf\bracs{\alpha \ge 0|\mu(\bracs{f > \alpha}) = 0} < \infty
+    \]
+    In which case, $\norm{f}_{L^\infty(X; E)}$ is the \textbf{essential supremum} of $f$.
+\end{definition}
+
+\begin{definition}[Hölder conjugates]
+\label{definition:holder-conjugates}
+    Let $p, q \in (1, \infty)$, then $p$ and $q$ are \textbf{Hölder conjugates} if
+    \[
+    \frac{1}{p} + \frac{1}{q} = 1
+    \]
+\end{definition}
+
+\begin{lemma}
+\label{lemma:holder-conjugate-gymnastics}
+    Let $p, q \in (1, \infty)$ be Hölder conjugates, then:
+    \begin{enumerate}
+        \item $q = p/(p - 1)$.
+        \item $p = q(p - 1)$.
+    \end{enumerate}
+\end{lemma}
+
+
+\begin{theorem}[Hölder's Inequality, {{\cite[6.2]{Folland}}}]
+\label{theorem:holder}
+    Let $(X, \cm, \mu)$ be a measure space, $E, F$ be a normed vector spaces, $p, q \in [1, \infty]$. If $p, q$ are Hölder conjugates or if $p = 1$ and $q = \infty$, then for any $f \in \mathcal{L}^p(X; E)$ and $g \in \mathcal{L}^q(X; F)$,
+    \[
+    \int \norm{f}_E \norm{g}_F d\mu \le \norm{f}_{L^p(X; E)}\norm{g}_{L^q(X; F)}
+    \]
+\end{theorem}
+\begin{proof}
+    First suppose that $p = 1$ and $q = \infty$. In this case,
+    \[
+    \int \norm{f}_E \norm{g}_F d\mu \le \norm{g}_{L^\infty(X; F)}\int \norm{f}_Ed\mu = \norm{f}_{L^1(X; E)}\norm{g}_{L^\infty(X; F)}
+    \]
+    Now suppose that $p, q \in (1, \infty)$ are Hölder conjugates. Assume without loss of generality that $\norm{f}_{L^p(X; E)} = \norm{g}_{L^q(X; F)} = 1$. By Young's inequality (\ref{lemma:young-inequality}),
+    \[
+    \int \norm{f}_E \norm{g}_F d\mu \le \int \frac{\norm{f}_E^p}{p} + \frac{\norm{g}_F^q}{q} d\mu = \frac{1}{p}\int \norm{f}_E d\mu + \frac{1}{q}\int \norm{g}_F^q d\mu = 1
+    \]
+\end{proof}
+
+\begin{theorem}[Minkowski's Inequality, {{\cite[6.5]{Folland}}}]
+\label{theorem:minkowski}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, $p \in [1, \infty]$, and $f, g \in \mathcal{L}^p(X; E)$, then
+    \[
+    \norm{f + g}_{L^p(X; E)} \le \norm{f}_{L^p(X; E)} + \norm{g}_{L^p(X; E)}
+    \]
+\end{theorem}
+\begin{proof}
+    If $p = 1$, then the theorem holds directly.
+
+    If $p = \infty$, then for any $\alpha > \norm{f}_{L^\infty(X; E)}$ and $\beta > \norm{g}_{L^\infty(X; E)}$, $\alpha + \beta \ge f + g$ $\mu$-almost everywhere, so
+    \[
+    \norm{f + g}_{L^\infty(X; E)} \le \norm{f}_{L^\infty(X; E)} + \norm{g}_{L^\infty(X; E)}
+    \]
+
+    Now suppose that $p \in (1, \infty)$, then $p = q(p - 1)$, and
+    \begin{align*}
+        \norm{f + g}_E^p &\le (\norm{f}_E + \norm{g}_E)\norm{f + g}_E^{p - 1} \\
+        \int\norm{f + g}_E^pd\mu &\le (\norm{f}_{L^p(X; E)} + \norm{g}_{L^p(X; E)})\braks{\int \norm{f + g}_E^{q(p - 1)}d\mu}^{1/q} \\
+        \int\norm{f + g}_E^pd\mu &\le (\norm{f}_{L^p(X; E)} + \norm{g}_{L^p(X; E)})\braks{\int \norm{f + g}_E^{p}d\mu}^{1/q} \\
+        \braks{\int\norm{f + g}_E^pd\mu}^{1/p} &\le \norm{f}_{L^p(X; E)} + \norm{g}_{L^p(X; E)}
+    \end{align*}
+\end{proof}
+
+\begin{definition}[$L^p$ Space]
+\label{definition:lp}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $p \in [1, \infty]$, then $\norm{\cdot}_{L^p(X; E)}$ is a seminorm on $\mathcal{L}^p(X; E)$. The quotient
+    \[
+    L^p(X, \cm, \mu; E) = \mathcal{L}^p(X, \cm, \mu; E)/\bracs{f|f = 0\text{ a.e.}}
+    \]
+    is a normed vector space, known as the $E$-valued \textbf{$L^p$ space} on $(X, \cm, \mu)$.
+\end{definition}
+\begin{proof}
+    By Minkowski's Inequality (\ref{theorem:minkowski}).
+\end{proof}
+
+\begin{proposition}
+\label{proposition:lp-simple-dense}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $p \in [1, \infty)$, then $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$.
+\end{proposition}
+\begin{proof}
+    Let $f \in L^p(X; E)$. By \ref{definition:strongly-measurable}, there exists $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_E \le \norm{f}_E$ and $\norm{f_n - f}_E \to 0$ pointwise as $n \to \infty$.
+
+    For each $n \in \nat$, $\norm{f_n}_E \le \norm{f}_E$, so $\norm{f_n}_{L^p(X; E)} \le \norm{f}_{L^p(X; E)} < \infty$, and $\norm{f_n - f}_E \le 2\norm{f}_E$. By the Dominated Convergence Theorem (\ref{theorem:dct}),
+    \[
+    \limv{n}\int \norm{f_n - f}_E^p d\mu = \int \limv{n}\norm{f_n - f}_E^p d\mu = 0
+    \]
+    Therefore $\norm{f_n - f}_{L^p(X; E)} \to 0$ as $n \to \infty$.
+\end{proof}

src/fa/lp/index.tex

@@ -0,0 +1,4 @@
+\chapter{$L^p$ Spaces}
+\label{chap:lp}
+
+\input{./src/fa/lp/definition.tex}

src/measure/bochner-integral/index.tex

@@ -0,0 +1,4 @@
+\chapter{Bochner Integral}
+\label{chap:bochner-integral}
+
+\input{./src/measure/bochner-integral/strongly.tex}

src/measure/bochner-integral/strongly.tex

@@ -0,0 +1,27 @@
+\section{Strongly Measurable Functions}
+\label{section:strongly-measurable}
+
+\begin{definition}[Strongly Measurable Function]
+\label{definition:strongly-measurable}
+    Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $f: X \to E$, then the following are equivalent:
+    \begin{enumerate}
+        \item For each $\phi \in E^*$, $\phi \circ f$ is $(\cm, \cb_K)$-measurable and $f(X) \subset E$ is separable.
+        \item $f$ is $(\cm, \cb_E)$ measurable and $f(X) \subset E$ is separable.
+        \item There exists a sequence $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that
+        \begin{enumerate}
+            \item[(a)] For each $n \in \natp$, $\norm{f_n}_E \le \norm{f}_E$.
+            \item[(b)] $\norm{f_n(x) - f(x)}_E \to 0$ pointwise as $n \to \infty$.
+        \end{enumerate}
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): TODO
+
+    (2) $\Rightarrow$ (3): By \ref{proposition:measurable-simple-separable-norm}.
+
+    (3) $\Rightarrow$ (1): For each $\phi \in E^*$, $\phi \circ f = \limv{n}\phi \circ f_n$ is measurable by \ref{proposition:limit-measurable}. Since
+    \[
+    f(X) \subset \ol{\bigcup_{n \in \natp}f_n(X)}
+    \]
+    and each $f_n$ is finitely-valued, $f(X)$ is separable.
+\end{proof}

src/measure/index.tex

@@ -5,3 +5,4 @@
 \input{./src/measure/measure/index.tex}
 \input{./src/measure/measurable-maps/index.tex}
 \input{./src/measure/lebesgue-integral/index.tex}
+\input{./src/measure/bochner-integral/index.tex}

src/measure/measure/radon.tex

@@ -131,7 +131,17 @@
     and $\mu = \nu$.
 
     Since $X$ is $\sigma$-compact, $\mu$ is $\sigma$-finite, so $\mu$ is regular by \ref{proposition:radon-regular-sigma-finite}.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:radon-cc-dense}
+    Let $X$ be a LCH space, $\mu: \cb_X \to [0, \infty]$ be a Radon measure, $E$ be a normed space, and $p \in [1, \infty)$, then $C_c(X; E)$ is dense in $L^p(X; E)$.
+\end{proposition}
+\begin{proof}
+    By \ref{proposition:lp-simple-dense}, $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$. Using linearity, it is sufficient to approximate indicator functions of Borel sets with finite measure.
+
+    Let $A \in \cb_X$ and $\eps > 0$. By \ref{proposition:radon-regular-sigma-finite}, there exists $U \in \cn^o(A)$ and $K \subset A$ compact such that $\mu(U \setminus A), \mu(A \setminus K) < \eps/2$. By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $f \in C_c(X; [0, 1])$ such that $f|_K = 1$ and $\supp{f} \subset U$. In which case, for any $y \in E$,
     \[
-    \color{blue}{\int}\llap{\color{green}{\int}}
+    \norm{x \cdot \one_A - x \cdot f}_{L^p(X; E)} \le \norm{x}_E \mu(\bracs{f \ne \one_A})^{1/p} \le \norm{x}_E\mu(U \setminus K)^{1/p} < \eps^{1/p}\norm{x}_E
     \]
 \end{proof}