Added the subdifferential.

src/fa/convex/del.tex

@@ -0,0 +1,59 @@
+\section{Subgradients}
+\label{section:subgradient}
+
+\begin{definition}[Subdifferential]
+\label{definition:subgradient}
+    Let $E$ be a locally convex space over $\real$, $f: E \to (-\infty, \infty]$, $x \in \bracs{f < \infty}$, and $\phi \in E^*$, then $\phi$ is a \textbf{subgradient} of $f$ if for any $h \in E$, 
+    \[
+    f(x + h) \ge f(x) + \dpn{h, \phi}{E}
+    \]
+
+    The set $\partial f(x)$ of all subgradients of $f$ at $x$ is the \textbf{subdifferential of $f$ at $x$}, and the mapping $\partial f$ is the \textbf{subdifferential} of $f$.  
+\end{definition}
+
+\begin{proposition}
+\label{proposition:subgradient-gymnastics}
+    Let $E$ be a locally convex space over $\real$, $f, g: E \to (-\infty, \infty]$, and $x \in \bracs{f, g < \infty}$, then:
+    \begin{enumerate}
+        \item $\partial f(x)$ is a weak*-closed convex set. 
+        \item $\partial(f + g)(x) \supset \partial f(x) + \partial g(x)$. 
+    \end{enumerate}
+\end{proposition}
+% Proof omitted. 
+
+\begin{proposition}
+\label{proposition:subgradient-existence}
+    Let $E$ be a locally convex space over $\real$, $f: E \to (-\infty, \infty]$ be convex, and $x \in \bracs{f < \infty}$, then:
+    \begin{enumerate}
+        \item For any $\phi \in \partial f(x)$ and $h \in E$, 
+        \[
+        \dpn{h, \phi}{E} \le |f(x + h) - f(x)|
+        \]
+        \item If $f$ is continuous at $x$, then $\partial f(x)$ is non-empty, equicontinuous, and weak*-compact. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 4.6]{Clarke}}}. ]
+    (1): By the subgradient inequality, 
+    \[
+    \dpn{h, \phi}{E} \le f(x + h) - f(x) \le |f(x + h) - f(x)|
+    \]
+
+    (2): Since $f$ is continuous at $x$, $\text{epi}(f)^o \ne \emptyset$. Since $(x, f(x)) \not\in \text{epi}(f)^o$, by the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-1}, there exists $\phi \in E^*$ and $\lambda \in \real$ such that for any $(y, \alpha) \in \text{epi}(f)^o$, 
+    \[
+    \dpn{y, \phi}{E} + \lambda \alpha < \dpn{x, \phi}{E} + \lambda f(x)
+    \]
+
+    By continuity of $f$ at $x$, there exists $\alpha_0 \in \real$ such that $\bracs{x} \times [\alpha_0, \infty) \subset \text{epi}(f)^o$. Thus $\lambda < 0$. 
+
+    By \autoref{proposition:convex-interior-closure}, $\text{epi}(f) \subset \ol{\text{epi}(f)^o}$. Therefore for any $(y, \alpha) \in \text{epi}(f)$, 
+    \[
+    \lambda^{-1}\dpn{y, \phi}{E} - \alpha \le \dpn{x, \phi}{E} - f(x)
+    \]
+
+    so $\phi \in \partial f(x) \ne \emptyset$.  
+
+    (3): By \autoref{proposition:subgradient-gymnastics} and the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. 
+\end{proof}
+
+
+

src/fa/convex/index.tex

@@ -2,3 +2,5 @@
 \label{chap:convex-functions}
 
 \input{./def.tex}
+\input{./del.tex}
+\input{./legendre.tex}