Added section regarding the duality of L^p spaces.

src/fa/lp/definition.tex

@@ -23,11 +23,13 @@
 
 \begin{definition}[Hölder conjugates]
 \label{definition:holder-conjugates}
-    Let $p, q \in (1, \infty)$, then $p$ and $q$ are \textbf{Hölder conjugates} if
+    Let $p, q \in [1, \infty]$, then $p$ and $q$ are \textbf{Hölder conjugates} if
     \[
     \frac{1}{p} + \frac{1}{q} = 1
     \]
 
+    under the identification that $1/\infty = 0$. 
+
 \end{definition}
 
 \begin{lemma}
@@ -40,15 +42,15 @@
 \end{lemma}
 
 
-\begin{theorem}[Hölder's Inequality, {{\cite[6.2]{Folland}}}]
+\begin{theorem}[Hölder's Inequality]
 \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 $g \in \mathcal{L}^q(X; F)$,
+    Let $(X, \cm, \mu)$ be a measure space, $E, F$ be a normed vector spaces, $p, q \in [1, \infty]$ be Hölder conjugates, 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}
+\begin{proof}[Proof, {{\cite[Theorem 6.2]{Folland}}}. ]
     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)}

src/fa/lp/duality.tex

@@ -0,0 +1,230 @@
+\section{Duality of $L^p$ Spaces}
+\label{section:lp-duality}
+
+\begin{lemma}
+\label{lemma:lp-dual-approximation}
+    Let $(X, \cm, \mu)$ be a measure space, $K \in \RC$, $\dpn{E, F}{\lambda}$ be a norming duality of Banach spaces over $K$, and $f: X \to E$ be a strongly measurable function, then there exists $\seq{\phi_n} \subset \Sigma(X, \cm; F)$ such that:
+    \begin{enumerate}
+        \item For each $n \in \natp$, $\norm{\phi_n}_{F} \le 1$.
+        \item For every $n \in \natp$, $|\dpn{f, \phi_n}{\lambda}| \le \norm{f}_{E}$. 
+        \item $|\dpn{f, \phi_n}{\lambda}| \upto \norm{f}_E$ pointwise as $n \to \infty$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    Since $f(X)$ is separable, assume without loss of generality that $E$ is separable. By \autoref{proposition:separable-dual}, there exists $\seq{z_n} \subset \bracsn{z \in F|\ \norm{z}_F \le 1}$ such that for each $y \in F$, $\norm{y}_F = \sup_{n \in \natp}\dpn{y, z_n}{\lambda}$. 
+
+    For each $N \in \natp$ and $x \in X$, let $F_N(x) = 0 \vee \bigvee_{n = 1}^N |\dpn{f(x), z_n}{E}|$, then $0 \le F_N \le \norm{F}_E$ and $F_N \upto \norm{f}_E$ pointwise as $N \to \infty$. 
+    
+    For every $1 \le n \le N$, inductively define
+    \[
+    A_{N, n} = \bracs{x \in X|F_N(x) = |\dpn{f(x), z_n}{E}|} \setminus \bigcup_{k = 1}^{n - 1}A_{N, k}
+    \]
+    
+    Let $\phi_N = \sum_{n = 1}^N z_n \one_{A_{N, n}}$, then $|\dpn{f, \phi_N}{\lambda}| = F_N$, and
+    \begin{enumerate}
+        \item Since $\norm{z_n}_F \le 1$ for all $n \in \natp$, $\norm{\phi_N}_F \le 1$ for each $N \in \natp$. 
+        \item For every $N \in \natp$, $|\dpn{f, \phi_N}{\lambda}| = F_N$, so $|\dpn{f, \phi_N}{\lambda}| \le \norm{f}_E$.
+        \item As $F_N \upto \norm{f}_E$ pointwise as $N \to \infty$, $|\dpn{f, \phi_N}{\lambda}| \upto \norm{f}_E$ pointwise as $N \to \infty$. 
+    \end{enumerate}
+\end{proof}
+
+
+After the duality of $L^p$ and $L^q$ is established for Hölder conjugate exponents for $p, q \in (1, \infty)$, it may be seen that each continuous functional on $L^p$ is "supported" on a $\sigma$-finite set. However, this fact can be established beforehand without the explicit identification. 
+
+\begin{lemma}
+\label{lemma:lp-functional-support}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space over $K \in \RC$, $p \in (1, \infty]$, and $\phi \in L^p(X, \cm, \mu; E)^*$, then there exists a $\sigma$-finite set $A \in \cm$ such that
+    \[
+    \dpn{f, \phi}{L^p(X; E)} = \dpn{\one_A \cdot f, \phi}{L^p(X; E)}
+    \]
+
+    for all $f \in L^p(X, \cm, \mu; E)$. 
+\end{lemma}
+\begin{proof}
+    For each $A \in \cm$, define
+    \[
+    \phi_A: L^p(X; E) \to K \quad \dpn{f, \phi_A}{L^p(X; E)} = \dpn{\one_A \cdot f, \phi}{L^p(X; E)}
+    \]
+
+    Let $f, g \in L^p(X; E)$ and $A, B \in \cm$ such that
+    \begin{enumerate}
+        \item $X = A \sqcup B$. 
+        \item $f|_B = 0$, and $g|_A = 0$. 
+        \item $\norm{f}_{L^p(X; E)} = \norm{g}_{L^p(X; E)} = 1$. 
+    \end{enumerate}
+
+    Suppose that $p \in (1, \infty)$, then for each $t \in \real$, 
+    \begin{align*}
+        \norm{(1 - t)f + tg}_{L^p(X; E)}^p &= \norm{(1 - t)f}_{L^p(X; E)}^p + \norm{tg}_{L^p(X; E)}^p \\
+        &= (1 - t)^p + t^p \\
+        &= (1 - t) \cdot (1 - t)^{p - 1} + t \cdot t^{p - 1}
+    \end{align*}
+
+    Since $p > 1$, for each $t \in (0, 1)$, $(1 - t)^{p - 1}, t^{p - 1} < 1$, so $\norm{(1 - t)f + tg}_{L^p(X; E)}^p < 1$. 
+
+    On the other hand, if $p = \infty$, then
+    \[
+    \norm{(1 - t)f + tg}_{L^\infty(X; E)} = (1 - t) \vee t < 1
+    \]
+
+    Therefore for any $A, B \in \cm$ with $A \cap B = \emptyset$, $\norm{\phi_A}_{L^p(X; E)^*} > 0$, and $\norm{\phi_B}_{L^p(X; E)^*} > 0$,
+    \[
+    \norm{\phi_{A \sqcup B}}_{L^p(X; E)^*} > \norm{\phi_A}_{L^p(X; E)^*} \vee \norm{\phi_B}_{L^p(X; E)^*}
+    \]
+    
+    Now, by \hyperref[density of simple functions]{proposition:lp-simple-dense}, 
+    \[
+    \norm{\phi}_{L^p(X; E)^*} = \sup\bracsn{\norm{\phi_A}_{L^p(X; E)^*}| A \in \cm \ \sigma\text{-finite}}
+    \]
+
+    Let $\seq{A_n} \subset \cm$ such that $\mu(A_n) < \infty$ for all $n \in \natp$, and $\norm{\phi_{A_n}}_{L^p(X; E)^*} \upto \norm{\phi}_{L^p(X; E)^*}$ as $n \to \infty$. Let $A = \bigcup_{n \in \natp}A_n$, then $A$ is $\sigma$-finite and $\norm{\phi_A}_{L^p(X; E)^*} = \norm{\phi}_{L^p(X; E)^*}$. By maximality, there exists no $B \in \cm$ with $B \cap A = \emptyset$ and $\norm{\phi_B}_{L^p(X; E)^*} > 0$, so $\phi = \phi_A$.
+\end{proof}
+
+\begin{theorem}
+\label{theorem:lp-dual-function}
+    Let $(X, \cm, \mu)$ be a measure space, $K \in \RC$, $\dpn{E, F}{\lambda}$ be a norming duality of normed vector spaces over $K$, $p, q \in [1, \infty]$ be Hölder conjugates such that one of the following holds:
+    \begin{enumerate}[label=(\alph*)]
+        \item $p \in (1, \infty]$ and $q \in [1, \infty)$.  
+        \item $p = 1$, $q = \infty$, and $\mu$ is semifinite. 
+    \end{enumerate}
+
+    Let $g: X \to F$ be a strongly measurable function such that for every $f \in \Sigma(X, \cm; E) \cap L^p(X; E)$, $\dpn{f, g}{\lambda} \in L^1(X; E)$, and the mapping
+    \[
+    \phi_g: \Sigma(X, \cm; E) \cap L^p(X; E) \to K \quad f \mapsto \int \dpn{f, g}{\lambda} d\mu
+    \]
+
+    is continuous in the $L^p(X; E)$ norm, then:
+    \begin{enumerate}
+        \item If (a) holds, then $\bracs{g \ne 0}$ is $\sigma$-finite. 
+        \item $g \in L^q(X; F)$ with $\norm{g}_{L^q(X; F)} = \norm{\phi_g}_{L^p(X; E)^*}$. 
+        \item The mapping
+        \[
+        L^q(X; F) \to L^p(X; E)^* \quad g \mapsto \phi_g
+        \]
+
+        is isometric. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Proposition 6.13, Theorem 6.14]{Folland}}}. ]
+    (1): By \autoref{lemma:lp-functional-support}, there exists a $\sigma$-finite set $A \in \cm$ such that for each $f \in L^p(X; E)$, 
+    \[
+    \int \dpn{f, g}{\lambda} d\mu = \int_A \dpn{f, g}{\lambda} d\mu
+    \]
+
+    Therefore $g|_{A^c} = 0$ almost everywhere, and $\bracs{g \ne 0}$ is $\sigma$-finite. 
+
+    (2, truncated): First suppose that $g \in L^q(X; F)$. Assume without loss of generality that $\norm{g}_{L^q(X; F)} = 1$. 
+
+    By \autoref{lemma:lp-dual-approximation}, there exists $\seq{\phi_n} \subset \Sigma(X, \cm; E)$ such that:
+    \begin{enumerate}
+        \item For each $n \in \natp$, $\norm{\phi_n}_{E} \le 1$.
+        \item For every $n \in \natp$, $0 \le |\dpn{g, \phi_n}{\lambda}| \le \norm{g}_{F}$. 
+        \item $|\dpn{g, \phi_n}{\lambda}| \upto \norm{g}_F$ pointwise as $n \to \infty$. 
+    \end{enumerate}
+    
+    (2, a, truncated): Let $\Phi(x) = \norm{g(x)}_F^{q - 1}$\footnote{Under the convention that $0^0 = 1$} for each $x \in X$. If $p < \infty$, then by \autoref{lemma:holder-conjugate-gymnastics}, 
+    \[
+    \norm{\Phi}_{L^p(X; \real)}^p = \int \norm{g}_F^{p(q - 1)}d\mu = \int \norm{g}_F^q d\mu = 1
+    \]
+
+    Otherwise, $\Phi = 1$, and $\norm{\Phi}_{L^\infty(X; \real)} = 1$. 
+
+    Let $\seq{\Phi_n} \subset \Sigma(X, \cm; E) \cap L^p(X; E)$ be non-negative such that $\Phi_n \upto \one_{\bracs{g \ne 0}} \cdot \Phi$. For each $n \in \natp$, $\Phi_n \phi_n \in \Sigma(X, \cm; E) \cap L^p(X; E)$ with $\norm{\Phi_n \phi_n}_{L^p(X; E)} \le 1$. By assumption, 
+    \[
+    |\Phi_n \dpn{\phi_n, g}{H}| = \Phi_n \dpn{\phi_n, g}{H} \cdot \ol{\sgn \dpn{\phi_n, g}{H}} \in L^1(X; \real)
+    \]
+
+    Let $f_n = \Phi_n \phi_n \cdot \ol{\sgn \dpn{\phi_n, g}{H}}$, then by the \hyperref[Monotone Convergence Theorem]{theorem:mct},
+    \begin{align*}
+        \limv{n}\int \dpn{f_n, g}{\lambda}d\mu &= \limv{n}\int \Phi_n \cdot \dpn{\phi_n, g}{\lambda} \cdot \ol{\sgn \dpn{\phi_n, g}{H}}d\mu \\
+        &= \int \Phi \cdot \norm{g}_F d\mu = \int \norm{g}_F^{q - 1}\norm{g}_F d\mu \\
+        &= \norm{g}_{L^q(X; F)}^q = 1
+    \end{align*}
+
+    (2, b, truncated): Let $\alpha \in (0, 1)$, then $\mu\bracs{\norm{g}_F \ge \alpha} > 0$. Since $\mu$ is semifinite, there exists $A \in \cm$ with $A \subset \bracs{\norm{g}_F \ge \alpha}$ and $0 < \mu(A) < \infty$. For each $x \in X$, let $\Phi(x) = \one_A/\mu(A)$, then $\norm{\Phi}_{L^1(X; \real)} = 1$. 
+
+    For every $n \in \natp$, let $f_n = \Phi \phi_n \cdot \ol{\sgn \dpn{\phi_n, g}{H}}$, then $\norm{f_n}_{L^1(X; E)} \le 1$, and by the \hyperref[Monotone Convergence Theorem]{theorem:mct},
+    \begin{align*}
+        \limv{n}\int \dpn{f_n, g}{\lambda}d\mu &= \limv{n}\int \Phi \cdot \dpn{\phi_n, g}{\lambda} \cdot \ol{\sgn \dpn{\phi_n, g}{H}}d\mu \\
+        &= \int \Phi \cdot \norm{g}_F d\mu = \int_A \frac{\norm{g}_F}{\mu(A)} d\mu \ge \alpha
+    \end{align*}
+    
+    As the above holds for all $\alpha \in (0, 1)$, $\norm{\phi_g}_{L^1(X; E)^*} = 1$. 
+
+    (2, general): If (a) holds, then $\bracs{g \ne 0}$ is $\sigma$-finite. In both cases, there exists $\seq{g_n} \subset L^q(X; F)$ such that
+    \begin{enumerate}
+        \item $\norm{g_n}_{F} \upto \norm{g}_F$ as $n \to \infty$.
+        \item For each $n \in \natp$, $\norm{\phi_{g_n}}_{L^p(X; E)^*} \le \norm{\phi_g}_{L^p(X; E)^*}$. 
+    \end{enumerate}
+
+    By the truncated case, for each $n \in \natp$, 
+    \[
+    \norm{g_n}_{L^q(X; F)} = \norm{\phi_{g_n}}_{L^p(X; E)^*} \le \norm{\phi_g}_{L^p(X; E)^*}
+    \]
+
+    If $q < \infty$, then the \hyperref[Monotone Convergence Theorem]{theorem:mct} implies that $\norm{g}_{L^q(X; F)} \le \norm{\phi_g}_{L^p(X; E)^*}$. Otherwise,
+    \[
+    \norm{g}_{L^\infty(X; H)} \le \sup_{n \in \natp}\norm{g_n}_{L^\infty(X; H)} \le \norm{\phi}_{L^1(X; H)^*}
+    \]
+    
+    The above argument shows that the truncation argument was technically not required. By applying the truncated case again, $\norm{g}_{L^q(X; F)} = \norm{\phi_g}_{L^p(X; E)^*}$.
+\end{proof}
+
+
+The typical argument for $L^p$ duality requires using the Radon-Nikodym theorem to extract the function. Since I prefer to not present martingales here, I will only include the Hilbert case. 
+
+\begin{theorem}
+\label{theorem:lp-duality}
+    Let $(X, \cm, \mu)$ be a measure space, $K \in \RC$, $H$ be a Hilbert space over $K$, $p, q \in [1, \infty]$ be Hölder conjugates such that one of the following holds:
+    \begin{enumerate}[label=(\alph*)]
+        \item $p \in (1, \infty)$ and $q \in (1, \infty)$.  
+        \item $p = 1$, $q = \infty$, and $\mu$ is $\sigma$-finite. 
+    \end{enumerate}
+
+    For each $g \in L^q(X, \cm, \mu; H)$, let
+    \[
+    \phi_g: L^p(X, \cm, \mu; H) \to K \quad f \mapsto \int \dpn{f, g}{H} d\mu
+    \]
+
+    then the mapping
+    \[
+    L^q(X, \cm, \mu; H) \to L^p(X, \cm, \mu; H)^* \quad g \mapsto \phi_g
+    \]
+
+    is a conjugate linear isometric isomorphism. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 6.15]{Folland}}}. ]
+    By \autoref{theorem:lp-dual-function}, the given map is isometric. Thus it is sufficient to show that it is surjective. Let $\phi \in L^p(X; H)^*$.
+
+    (Finite): First suppose that $\mu$ is finite, then $\Sigma(X, \cm; H) \subset L^p(X; H)$, and $\phi$ induces an $H$-valued measure on $(X, \cm)$, absolutely continuous with respect to $\mu$. By the \hyperref[Radon-Nikodym Theorem]{theorem:lebesgue-radon-nikodym}, there exists $g \in L^1(X; H)$ such that for each $f \in \Sigma(X, \cm; H)$, 
+    \[
+    \int \dpn{f, g}{H} d\mu = \dpn{f, \phi}{L^p(X; H)}
+    \]
+
+    By \autoref{theorem:lp-dual-function}, $g \in L^q(X; H)$. 
+
+    (Arbitrary): In the case of (a), by \autoref{lemma:lp-functional-support}, there exists a $\sigma$-finite set $A \in \cm$ such that for each $f \in L^p(X; H)$, $\dpn{f, \phi}{L^p(X; H)} = \dpn{\one_A \cdot f, \phi}{L^p(X; H)}$. In the case of (b), $A = X$ is a $\sigma$-finite set satisfying the same restriction condition. 
+
+    Let $\seq{A_n} \subset \cm$ such that $\mu(A_n) < \infty$ for all $n \in \natp$, and $A = \bigsqcup_{n \in \natp}A_n$. By the finite case, there exists $\seq{g_n} \subset L^q(X; H)$ such that for each $n \in \natp$ and $f \in L^p(X; H)$, 
+    \[
+    \int \dpn{f, g_n}{H} d\mu = \dpn{\one_{A_n} \cdot f, \phi}{L^p(X; H)}
+    \] 
+
+    Let $g = \sum_{n = 1}^\infty g_n$. If $q < \infty$, then $g \in L^q(X; H)$ by the \hyperref[Monotone Convergence Theorem]{theorem:mct}. Otherwise,
+    \[
+    \norm{g}_{L^\infty(X; H)} \le \sup_{n \in \natp}\norm{g_n}_{L^\infty(X; H)} \le \norm{\phi}_{L^1(X; H)^*}
+    \]
+    
+    For every $f \in L^p(X; H)$, 
+    \begin{align*}
+    \int \dpn{f, g}{H} d\mu &= \sum_{n = 1}^\infty \int \dpn{f, g_n}{H} d\mu = \sum_{n = 1}^\infty \dpn{\one_{A_n} \cdot f, \phi}{L^p(X; H)} \\
+    &= \dpn{f, \phi}{L^p(X; H)}
+    \end{align*}
+    
+    by the \hyperref[Dominated Convergence Theorem]{theorem:dct}.
+
+    Therefore the mapping is surjective, and hence an isomorphism. 
+\end{proof}
+
+
+

src/fa/lp/index.tex

@@ -2,3 +2,4 @@
 \label{chap:lp}
 
 \input{./definition.tex}
+\input{./duality.tex}

src/fa/norm/hilbert.tex

@@ -352,7 +352,7 @@ A significant property of Hilbert spaces is that every closed subspace is comple
     \phi_x: H \to \complex \quad \phi_x(y) = \dpn{y, x}{E}
     \]
 
-    then the mapping $H \to H^*$ defined by $x \mapsto \phi_x$ is an isometric conjugate linear isomorphism. 
+    then the mapping $H \to H^*$ defined by $x \mapsto \phi_x$ is an conjugate linear isometric isomorphism. 
 \end{theorem}
 \begin{proof}
     By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz} and definition of the norm, $x \mapsto \phi_x$ is an isometric conjugate linear map. 

src/fa/norm/separable.tex

@@ -6,7 +6,7 @@
     Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality of normed vector spaces over $K$ with $E$ being separable, then
     \begin{enumerate}
         \item The closed unit ball $S = \bracsn{y \in F|\ \norm{y}_{F} \le 1}$ is separable with respect to the $\sigma(F, E)$-topology. 
-        \item If the duality is norming, then there exists $\seq{y_n} \subset F$ such that for each $x \in E$, $\norm{x}_E = \sup_{n \in \natp}\dpn{x, y_n}{\lambda}$. 
+        \item If the duality is norming, then there exists $\seq{y_n} \subset F$ such that for each $x \in E$, $\norm{x}_E = \sup_{n \in \natp}|\dpn{x, y_n}{\lambda}|$. 
     \end{enumerate}
 \end{proposition}
 \begin{proof}

src/measure/vector/rn.tex

@@ -15,7 +15,7 @@
     \nu_a(A) = \int_{A} \frac{d\nu_a}{d\mu} d\mu
     \]
 
-    If $\nu \ll \mu$, then $\nu_s = 0$ and $\nu(dx) = \frac{d\nu}{d\mu}\mu(dx)$. In which case, the function $\frac{d\nu}{d\mu}$ is the \textbf{Radon-Nikodym derivative} of $\nu$ with respect to $\mu$.
+    If $\nu \ll \mu$, then $\nu_s = 0$ and $\nu(dx) = \frac{d\nu_a}{d\mu}\mu(dx) = \frac{d\nu_a}{d\mu}\mu(dx)$. In which case, the function $\frac{d\nu}{d\mu}$ is the \textbf{Radon-Nikodym derivative} of $\nu$ with respect to $\mu$.
 \end{theorem}
 \begin{proof}[Proof, {{\cite[Exercise 6.18]{Folland}}}\footnote{I decided to abuse Hilbert spaces for this theorem because it is more fun, and because I will use the Riesz representation theorem twice.}. ]
     (Finite + Positive): First suppose that $\mu$ is finite and $\nu$ is positive. Let $\lambda = \mu + \nu$, then the mapping