Added minkowski's inequality for integrals.

src/fa/lp/ineq.tex

@@ -48,3 +48,50 @@
     \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}
+