\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 \hyperref[Young's inequality]{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 vector 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 vector 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 \hyperref[Minkowski's Inequality]{theorem:minkowski}.
\end{proof}


\begin{proposition}
\label{proposition:dct-lp}
    Let $(X, \cm, \mu)$ be a measure space, $E$ be a Banach space over $K \in \RC$, $p \in [1, \infty)$, $\seq{f_n} \subset L^p(X; E)$, and $f \in L^p(X; E)$. If
    \begin{enumerate}
        \item[(a)] $f_n \to f$ strongly pointwise.
        \item[(b)] There exists $g \in L^p(X) \cap L^+(X)$ such that $\norm{f_n}_E \le g$ for all $n \in \natp$.
    \end{enumerate}

    
    then $f_n \to f$ in $L^p(X; E)$.
\end{proposition}
\begin{proof}
    By assumptions $a$ and $b$, $\norm{f_n - f}_E \le 2g$ for all $n \in \natp$. Since $\seq{f_n} \subset L^p(X; E)$, $f \in L^p(X; E)$, and $g \in L^p(X)$, $\norm{f_n - f}_E^p, g^p \in L^1(X)$ for all $n \in \natp$. By the \hyperref[Dominated Convergence Theorem]{theorem:dct}, 
    \[
    \limv{n}\norm{f_n - f}_{L^p(X; E)}^p = \limv{n}\int \norm{f_n - f}_E d\mu  = 0
    \]
\end{proof}


\begin{proposition}
\label{proposition:lp-simple-dense}
    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector 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 \autoref{definition:strongly-measurable}, there exists $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_E \le \norm{f}_E$ and $\norm{f_n - f}_E \to 0$ strongly pointwise as $n \to \infty$. By the \hyperref[Dominated Convergence Theorem]{proposition:dct-lp}, $f_n \to f$ in $L^p(X; E)$.
\end{proof}

\begin{theorem}[{{\cite[III.6.5]{SchaeferWolff}}}]
\label{theorem:l1-tensor}
    Let $(X, \cm, \mu)$ be a measure space and $E$ be a Banach space over $K \in \RC$, then the map $L^1(X; K) \td{\otimes}_\mu E \to L^1(X; E)$ defined by extending
    \[
    L^1(X; K) \times E \to L^1(X; E) \quad f \otimes x \mapsto x \cdot f
    \]
    
    is an isometric isomorphism. 
\end{theorem}
\begin{proof}
    By (U) of the \hyperref[tensor product]{definition:tensor-product}, the given map admits a unique extension
    \[
    M: L^1(X; K) \otimes E \to L^1(X; E) \quad \sum_{j = 1}^n f_j \otimes x_j \mapsto \sum_{j = 1}^n x_j \cdot f_j
    \]

    Restricting $M$ to the simple functions yields a linear isomorphism
    \[
    M: [L^1(X; K) \cap \Sigma(X; K)] \otimes E \to L^1(X; E) \cap \Sigma(X; E)
    \]

    For any $\phi \in L^1(X; E) \cap \Sigma(X; E)$, write
    \[
    \phi = \sum_{y \in \phi(X) \setminus \bracs{0}}y \cdot \one_{\bracs{\phi = y}} = M\braks{\sum_{y \in \phi(X) \setminus \bracs{0}}\one_{\bracs{\phi = y}} \otimes y}
    \]

    then
    \[
    \normn{M^{-1}\phi}_{L^1(X; K) \otimes E} \le \sum_{y \in \phi(X) \setminus \bracs{0}} \norm{y}_E \cdot \mu\bracs{\phi = y} = \int \norm{\phi}_E d\mu = \norm{\phi}_{L^1(X; E)}
    \]

    On the other hand, for any representation $M^{-1}\phi = \sum_{j = 1}^n a_j \one_{A_j}$, 
    \[
    \normn{\phi}_{L^1(X; E)} \le \sum_{j = 1}^n \norm{a_j}_E \mu(A_j) = \sum_{j = 1}^n \norm{a_j}_E \normn{\one_{A_j}}_{L^1(X; K)}
    \]

    As this holds for all such representations, $\normn{\phi}_{L^1(X; E)} = \normn{M^{-1}\phi}_{L^1(X; K) \otimes E}$. Therefore $M$ restricted to $[L^1(X; K) \cap \Sigma(X; K)] \otimes E$ is an isometry. By \autoref{proposition:lp-simple-dense}, $[L^1(X; K) \cap \Sigma(X; K)] \otimes E$ is dense in $L^1(X; K) \widehat{\otimes}_\pi E$, and $L^1(X; E) \cap \Sigma(X; E)$ is dense in $E$. By the \hyperref[Linear Extension Theorem]{theorem:linear-extension-theorem-normed}, $M$ extends uniquely into the given map on $L^1(X; K) \otimes E$, which then extends into an isometry $L^1(X; K) \otimes E \to L^1(X; E)$. 
\end{proof}



\begin{theorem}[Markov's Inequality]
\label{theorem:markov-inequality}
    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $f: X \to E$ be a Borel measurable function, then
    \begin{enumerate}
        \item For any $\alpha > 0$, 
        \[
        \mu\bracs{|f| \ge \alpha} \le \frac{1}{\alpha}\norm{f}_{L^1(X; E)}
        \]
        \item For $\alpha > 0$ and strictly increasing function $\phi: [0, \infty) \to [0, \infty)$,
        \[
        \mu\bracs{|f| \ge \alpha} \le \frac{1}{\phi(\alpha)}\norm{\phi \circ |f|}_{L^1(X; E)}
        \]
        \item For any $\alpha > 0$,
        \[
        \mu\bracs{|f| \ge \alpha} \le \frac{1}{\alpha^p}\norm{f}_{L^p(X; E)}^p
        \]
    \end{enumerate}
\end{theorem}
\begin{proof}
    (1): For any $\alpha > 0$, 
    \begin{align*}
        \mu\bracs{|f| \ge \alpha} &= \int_{\bracs{f \ge \alpha}} d\mu \le \frac{1}{\alpha}\int |f|d\mu = \frac{\norm{f}_{L^1(X; E)}}{\alpha}
    \end{align*}
    
    (2): By \autoref{lemma:monotone-measurable}, $\phi: [0, \infty) \to [0, \infty)$ is Borel measurable. Since $\phi$ is strictly increasing, $\bracs{|f| \ge \alpha} = \bracs{\phi \circ |f| \ge \phi(\alpha)}$, and the result holds by applying (1) to $\phi \circ |f|$. 

    (3): Since $x \mapsto x^p$ is strictly increasing on $[0, \infty)$, applying (2) to $\phi(x) = x^p$ yields the desired result.
\end{proof}

\begin{proposition}
\label{proposition:lp-in-measure}
    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, $p \in [1, \infty]$, $\seq{f_n} \subset L^p(X; E)$, and $f \in L^p(X; E)$ such that $f_n \to f$ in $L^p$, then $f_n \to f$ in measure.
\end{proposition}
\begin{proof}
    Let $\eps > 0$. If $p < \infty$, then by \hyperref[Markov's Inequality]{theorem:markov-inequality},
    \[
    \limv{n}\mu\bracs{|f_n - f| \ge \eps} \le \limv{n}\frac{1}{\eps^p}\norm{f_n - f}_{L^p(X; E)}^p = 0
    \]
    
    If $p = \infty$, then there exists $N \in \natp$ such that $\norm{f_n - f}_{L^\infty(X; E)} < \eps$ for all $n \ge N$. In which case, $\mu\bracs{|f_n - f| \ge \eps} = 0$ for all $n \ge N$. 
\end{proof}