Added elementary facts about conve functions.

src/fa/convex/def.tex

@@ -0,0 +1,65 @@
+\section{Convexity}
+\label{section:convex-functions}
+
+\begin{definition}[Epigraph]
+\label{definition:epigraph}
+    Let $E$ be a vector space over $\real$, $A \subset E$ be convex, and $f: A \to (-\infty, \infty]$, then the \textbf{epigraph} of $f$ is the set
+    \[
+    \text{epi}(f) = \bracs{(x, y) \in A \times \real| y \ge f(x)}
+    \]
+\end{definition}
+
+
+\begin{definition}[Convex Function]
+\label{definition:convex-function}
+    Let $E$ be a vector space over $\real$, $A \subset E$ be convex, and $f: A \to (-\infty, \infty]$, then the following are equivalent:
+    \begin{enumerate}
+        \item For every $x, y \in E$ and $t \in [0, 1]$, 
+        \[
+        f((1 - t)x + ty) \le (1 - t)f(x) + tf(y)
+        \]
+        \item $\text{epi}(f)$ is convex. 
+    \end{enumerate}
+    
+    If the above holds, then $f$ is \textbf{convex}. 
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $(x, \alpha), (y, \beta) \in \text{epi}(f)$ and $t \in [0, 1]$, then
+    \begin{align*}
+        (1 - t)\alpha + t\beta &\ge (1 - t)f(x) + tf(y) \\
+        &\ge f((1 - t)x + ty)
+    \end{align*}
+
+    so $((1 - t)x + ty, (1 - t)\alpha + t\beta) \in \text{epi}(f)$. 
+
+    (2) $\Rightarrow$ (1): Let $x, y \in A$ and $t \in [0, 1]$, then
+    \[
+    ((1 - t)x + ty, (1 - t)f(x) + tf(y)) \in \text{epi}(f)
+    \]
+
+    so $f((1 - t)x + ty) \le (1 - t)f(x) + tf(y)$. 
+    
+\end{proof}
+
+\begin{lemma}
+\label{lemma:convex-reverse}
+    Let $E$ be a vector space over $\real$, $A \subset E$ be convex, and $f: A \to (-\infty, \infty]$ be convex, then for any $x, y \in E$ and $t \in \real \setminus [0, 1]$ such that $(1 - t)x + ty \in A$, 
+    \[
+    f((1 - t)x + ty) \ge (1 - t)f(x) + tf(y)
+    \]
+\end{lemma}
+\begin{proof}
+    Via an affine transformation, assume without loss of generality that $x = 0$ and $f(x) = 0$. By exchanging $x$ and $y$, assume without loss of generality that $t > 1$. In which case, since $f$ is convex and $y = t^{-1}ty$, $f(y) \le t^{-1}f(ty)$ and $tf(y) \le f(ty)$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:convex-extension}
+    Let $E$ be a vector space over $\real$ and $A \subset E$ be convex, then:
+    \begin{enumerate}
+        \item For any $f, g: A \to (-\infty, \infty]$ convex and $\lambda \ge 0$, $\lambda f + g$ is convex. 
+        \item For any convex functions $\cf \subset (-\infty, \infty]^A$, $\sup_{f \in \cf}f$ is convex. 
+    \end{enumerate}
+\end{proposition}
+
+
+

src/fa/convex/index.tex

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+\chapter{Convex Functions}
+\label{chap:convex-functions}
+
+\input{./def.tex}

src/fa/index.tex

@@ -4,6 +4,7 @@
 
 \input{./tvs/index.tex}
 \input{./lc/index.tex}
+\input{./convex/index.tex}
 \input{./norm/index.tex}
 \input{./rs/index.tex}
 \input{./lp/index.tex}