Fixed up convention in convex functions.

src/fa/convex/def.tex

@@ -3,16 +3,16 @@
 
 \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
+    Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$, then the \textbf{epigraph} of $f$ is the set
     \[
-    \text{epi}(f) = \bracs{(x, y) \in A \times \real| y \ge f(x)}
+    \text{epi}(f) = \bracs{(x, y) \in E \times \real| y \ge f(x)}
     \]
 \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:
+    Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$, then the following are equivalent:
     \begin{enumerate}
         \item For every $x, y \in E$ and $t \in [0, 1]$, 
         \[
@@ -32,7 +32,7 @@
 
     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
+    (2) $\Rightarrow$ (1): Let $x, y \in E$ and $t \in [0, 1]$, then
     \[
     ((1 - t)x + ty, (1 - t)f(x) + tf(y)) \in \text{epi}(f)
     \]
@@ -43,7 +43,7 @@
 
 \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 $(1 - t)x + ty \in A$, 
+    Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$ be convex, then for any $x, y \in E$ and $t \in \real \setminus [0, 1]$, 
     \[
     f((1 - t)x + ty) \ge (1 - t)f(x) + tf(y)
     \]
@@ -52,12 +52,31 @@
     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-differential}
+    Let $E$ be a vector space over $\real$, $f: E \to (-\infty, \infty]$ be convex, and $x \in \bracs{f < \infty}$, then for any $h \in E$,
+    \[
+    \lim_{t \downto 0} \frac{f(x + th) - f(x)}{t} = \inf_{t > 0} \frac{f(x + th) - f(x)}{t}
+    \]
+
+    exists in $[-\infty, \infty]$. 
+\end{proposition}
+\begin{proof}
+    Let $0 < s \le t$, then since $f$ is convex, 
+    \begin{align*}
+        f(x + sh) &\le \paren{1 - \frac{s}{t}}f(x) + \frac{s}{t}f(x + th) \\
+        f(x + sh) - f(x) &\le \frac{s}{t}[f(x + th) - f(x)] \\
+        \frac{f(x + sh) - f(x)}{s} &\le \frac{f(x + th) - f(x)}{t}
+    \end{align*}
+\end{proof}
+
+
 \begin{proposition}
 \label{proposition:convex-extension}
-    Let $E$ be a vector space over $\real$ and $A \subset E$ be convex, then:
+    Let $E$ be a vector space over $\real$, 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 $f, g: E \to (-\infty, \infty]$ 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. 
+        \item For any convex functions $\cf \subset (-\infty, \infty]^E$, $\sup_{f \in \cf}f$ is convex. 
     \end{enumerate}
 \end{proposition}