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

@@ -0,0 +1,4 @@
+\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}

src/measure/notation.tex

@@ -8,6 +8,7 @@
     $\sigma(\mathcal{E})$ & $\sigma$-algebra generated by $\mathcal{E}$. & \autoref{definition:generated-sigma-algebra} \\
     $\lambda(\mathcal{E})$ & $\lambda$-system generated by $\mathcal{E}$. & \autoref{definition:generated-lambda-system} \\
     $\sigma \otimes \tau$ & Product of ideals. & \autoref{definition:product-ideal} \\
+    $\text{epi}(f)$ & Epigraph of $f$. & \autoref{definition:epigraph} \\ 
     % ---- Measure Theory ----
     $\mathcal{B}_X$ & Borel $\sigma$-algebra on $X$. & \autoref{definition:borel-sigma-algebra} \\
     $\sigma(\{f_i \mid i \in I\})$ & $\sigma$-algebra generated by the maps $\{f_i\}$. & \autoref{definition:generated-sigma-algebra-function} \\

src/topology/main/semicontinuity.tex

@@ -6,6 +6,22 @@
     Let $X$ be a topological space, $f: X \to (-\infty, \infty]$, and $g: X \to [-\infty, \infty)$, then $f$ is \textbf{lower semicontinuous} if for each $a \in \real$, $\bracs{f > \alpha}$ is open, and $g$ is \textbf{upper semicontinuous} if for each $a \in \real$, $\bracs{f < \alpha}$ is open.
 \end{definition}
 
+\begin{proposition}
+\label{proposition:lsc-epigraph}
+    Let $X$ be a topological space, $f: X \to (-\infty, \infty]$, then $f$ is lower semicontinuous if and only if $\text{epi}(f)$ is closed. 
+\end{proposition}
+\begin{proof}
+    Suppose that $f$ is lower semicontinuous. Let $x \in X$ and $y \in (-\infty, \infty)$ with $y < f(x)$, and $\alpha \in (y, f(x))$, then $\bracs{f > \alpha} \in \cn^o_A(x)$, and hence $\bracs{f > \alpha} \times (-\infty, \alpha)$ is a neighbourhood of $(x, y)$ contained in the complement of $\text{epi}(f)$. Therefore $\text{epi}(f)$ is closed. 
+
+    Now suppose that $\text{epi}(f)$ is closed, then for each $\alpha \in (-\infty, \infty)$,
+    \[
+    \bracs{f \le \alpha} = \pi_1[\text{epi}(f) \cap A \times \bracs{\alpha}]
+    \]
+
+    which is closed. Therefore $f$ is lower semicontinuous. 
+\end{proof}
+
+
 \begin{proposition}[{{\cite[Proposition 7.11]{Folland}}}]
 \label{proposition:semicontinuous-properties}
     Let $X$ be a topological space, then