Added elementary facts about conve functions.

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