\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}



