Added the Vitali convergence theorem.

src/cat/tricks/product.tex

@@ -26,3 +26,34 @@
     \]
 
 \end{proof}
+
+\begin{lemma}
+\label{lemma:power-difference}
+    Let $p \in [1, \infty)$, then for each $a, b \ge 0$, 
+    \[
+    |a^p - b^p| \le p|a - b|(a \vee b)^{p - 1}
+    \]
+\end{lemma}
+\begin{proof}
+    Assume without loss of generality that $0 < a < b$, then
+    \[
+        b^p - a^p = p\int_{a}^b t^{p - 1}dt \le p|a - b|(a \vee b)^{p - 1}
+    \]
+\end{proof}
+
+\begin{lemma}
+\label{lemma:power-sum}
+    Let $p \in [1, \infty)$, then for each $a, b \ge 0$, 
+    \[
+    (a + b)^p \le 2^{p-1}(a^p + b^p)
+    \]
+\end{lemma}
+\begin{proof}
+    Since $t \mapsto t^p$ is convex, 
+    \begin{align*}
+        \braks{\frac{a + b}{2}}^p &\le \frac{a^p}{2} + \frac{b^p}{2} \\
+        \frac{(a+b)^p}{2^p} &\le \frac{a^p}{2} + \frac{b^p}{2} \\
+        (a + b)^p &\le 2^{-1}(a^p + b^p)
+    \end{align*}
+    
+\end{proof}