\section{$L^p$ Inequalities}
\label{section:lp-inequalities}


\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 $\alpha > 0$, 
        \[
        \mu\bracs{\norm{f}_E \ge \alpha} \le \frac{1}{\alpha}\norm{f}_{L^1(X; E)}
        \]
        \item For $\alpha > 0$ and strictly increasing function $\phi: [0, \infty) \to [0, \infty)$,
        \[
        \mu\bracs{\norm{f}_E \ge \alpha} \le \frac{1}{\phi(\alpha)}\norm{\phi \circ |f|}_{L^1(X; E)}
        \]
        \item For any $\alpha > 0$,
        \[
        \mu\bracs{\norm{f}_E \ge \alpha} \le \frac{1}{\alpha^p}\norm{f}_{L^p(X; E)}^p
        \]
    \end{enumerate}
\end{theorem}
\begin{proof}
    (1): For any $\alpha > 0$, 
    \begin{align*}
        \mu\bracs{\norm{f}_E \ge \alpha} &= \int_{\bracs{\norm{f}_E \ge \alpha}} d\mu \le \frac{1}{\alpha}\int |\norm{f}_Ed\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, $\bracs{\norm{f}_E \ge \alpha} = \bracs{\phi \circ \norm{f}_E \ge \phi(\alpha)}$, and the result holds by applying (1) to $\phi \circ \norm{f}_E$. 

    (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{lemma}
\label{lemma:lp-p-diff}
    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, $p \in [1, \infty)$, and $f, g \in L^p(X; E)$, then
    \[
    |\norm{f}_{L^p(X; E)}^p - \norm{g}_{L^p(X; E)}^p| \le p\norm{f - g}_{L^p(X; E)}(\norm{f}_{L^p(X; E)} \vee \norm{g}_{L^p(X; E)})^{p-1}
    \]
\end{lemma}
\begin{proof}
    By \autoref{lemma:power-difference}, 
    \begin{align*}
    |\norm{f}_{L^p(X; E)}^p - \norm{g}_{L^p(X; E)}^p| &\le p|\norm{f}_{L^p(X; E)} - \norm{g}_{L^p(X; E)}|\\
    &\times (\norm{f}_{L^p(X; E)} \vee \norm{g}_{L^p(X; E)})^{p-1} \\
    &\le p\norm{f - g}_{L^p(X; E)}(\norm{f}_{L^p(X; E)} \vee \norm{g}_{L^p(X; E)})^{p-1}
    \end{align*}
    
\end{proof}


\begin{theorem}[Minkowski's Inequality for Integrals]
\label{theorem:minkowski-integral}
    Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be $\sigma$-finite measure spaces, $E$ be a Banach space, and $f: X \times Y \to E$ be $(\cm \otimes \cn)$-strongly measurable. Then:
    \begin{enumerate}
        \item If $E = \real$ and $f \ge 0$, then for any $p \in [1, \infty)$, 
        \[
        \braks{\int_{X}\paren{\int_Y f(x, y)\nu(dy)}^{p}\mu(dx)}^{1/p} \le \int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)} \nu(dy)
        \]
        \item For any $p \in [1, \infty]$, if
        \begin{enumerate}[label=(\alph*)]
            \item For $\nu$-almost every $y \in Y$, $f(\cdot, y) \in L^p(\mu; E)$.
            \item The mapping $y \mapsto \norm{f(\cdot, y)}_{L^p(\nu; E)}$ is in $L^1(\mu; \real)$.
        \end{enumerate}

        then the mapping $x \mapsto \int_Y f(x, y)\nu(dy)$ is in $L^p(\mu; E)$, with
        \[
        \norm{\int_Y f(\cdot, y)\nu(dy)}_{L^p(X, \mu; E)} \le \int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)} \nu(dy)
        \]
    \end{enumerate}
    
\end{theorem}
\begin{proof}[Proof, {{\cite[Theorem 6.19]{Folland}}}. ]
    (1): If $p = 1$, then by \hyperref[Tonelli's Theorem]{theorem:fubini-tonelli}, 
    \[
    \int_X\int_Y f(x, y) \nu(dy)\mu(dx) = \int_Y\int_X f(x, y)\mu(dx)\nu(dy)
    \]

    Now suppose that $p \in (1, \infty)$, assume without loss of generality that $\int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)} \nu(dy) < \infty$. Let $q$ be the Hölder conjugate of $p$ and $\phi \in L^q(\mu; \real) \cap L^+(\mu)$ with $\phi \ge 0$, then by \hyperref[Tonelli's Theorem]{theorem:fubini-tonelli},
    \begin{align*}
        \int_X \int_Y f(x, y)\nu(dy) \phi(x) \mu(dx) &= \int_X\int_Y  f(x, y)\phi(x)\nu(dy) \mu(dx) \\
        &= \int_Y\int_X f(x, y) \phi(x) \mu(dx) \nu(dy)
    \end{align*}
    

    By \hyperref[Hölder's Inequality]{theorem:holder}, 
    \[
    \int_Y\int_X f(x, y) \phi(x) \mu(dx) \nu(dy) \le \norm{\phi}_{L^q(\mu; \real)}\int_Y \norm{f(\cdot, y)}_{L^p(\mu; \real)} \nu(dy)
    \]
    
    As $f \ge 0$ and the above holds for all $\phi \in L^q(\mu; \real) \cap L^+(\mu)$, \autoref{theorem:lp-dual-function} implies that $x \mapsto \int_Y f(x, y)\nu(dy) \in L^p(\mu)$ with 
    \[
    \braks{\int_{X}\paren{\int_Y f(x, y)\nu(dy)}^{p}\mu(dx)}^{1/p} \le \int_{Y}\norm{f(\cdot, y)}_{L^p(X, \mu; E)} \nu(dy)
    \]
\end{proof}