Added basic theory of Lp spaces, alongside some integral stuff.

src/cat/tricks/index.tex

@@ -2,3 +2,4 @@
 \label{chap:tricks}
 
 \input{./src/cat/tricks/dyadic.tex}
+\input{./src/cat/tricks/product.tex}

src/cat/tricks/product.tex

@@ -0,0 +1,25 @@
+\section{Product Inequalities}
+\label{section:product-inequalities}
+
+\begin{lemma}[Young's Inequality]
+\label{lemma:young-inequality}
+    Let $p, q \in (1, \infty)$ such that $1/p + 1/q = 1$, then for any $a, b > 0$,
+    \[
+    ab \le \frac{a^p}{p} + \frac{b^q}{q}
+    \]
+    and for any $\eps > 0$,
+    \[
+    ab \le \eps a^p + \frac{1}{q}(\eps q)^{-q/p}b^q
+    \]
+\end{lemma}
+\begin{proof}
+    Since $x \mapsto \exp(x)$ is convex,
+    \begin{align*}
+        ab &= \exp(\ln(a) + \ln(b)) = \exp\paren{\frac{1}{p}\ln(a^p) + \frac{1}{q}\ln(b^q)} \\
+        &\le \frac{1}{p}\exp(\ln(a^p)) + \frac{1}{q}\exp(\ln(a^q)) = \frac{a^p}{p} + \frac{b^q}{q}
+    \end{align*}
+    For any $\eps > 0$, applying the above inequality to $(\eps p)^{1/p}a$ and $b/(\eps p)^{1/p}$ yields
+    \[
+    ab = (\eps p)^{1/p}a \cdot \frac{b}{(\eps p)^{1/p}} \le \eps a^p + \frac{b^q}{q}(\eps p)^{-(1/p)q} = \eps a^p + \frac{1}{q}(\eps q)^{-q/p}b^q
+    \]
+\end{proof}