Performed more housekeeping for the Legendre transform.

src/fa/convex/legendre.tex

@@ -129,24 +129,26 @@
     Let $E$ be a locally convex space over $\real$ and $f: E \to (-\infty, \infty]$ with $f \ne \infty$, then for each $(x, \alpha) \in \ol{\text{Conv}}(\text{epi}(f))$, $\bracs{x} \times [\alpha, \infty) \subset \ol{\text{Conv}}(\text{epi}(f))$. 
 \end{lemma}
 \begin{proof}
-    First consider $\text{Conv}(\text{epi}(f))$. Let $(x, \alpha), (y, \beta) \in \text{Conv}(\text{epi}(f))$ such that
+    Let
     \[
-    \bracs{x} \times [\alpha, \infty), \bracs{y} \times [\beta, \infty) \subset \text{Conv}(\text{epi}(f))
+    A = \bracsn{(\phi, \alpha) \in \ol{\text{Conv}}(\text{epi}(f))| \bracs{x} \times [\alpha, \infty) \subset \ol{\text{Conv}}(\text{epi}(f))}
     \]
 
-    then for any $t \in [0, 1]$ and $\gamma \ge (1 - t)\alpha + t\beta$, there exists $\alpha' \ge \alpha$ and $\beta' \ge \beta$ such that $\gamma = (1 - t)\alpha' + t\beta'$. In which case, 
+    For each $(x, \alpha), (y, \beta) \in A$, $t \in [0, 1]$, and $\gamma \ge (1 - t)\alpha + t\beta$, there exists $\alpha' \ge \alpha$ and $\beta' \ge \beta$ such that $\gamma = (1 - t)\alpha' + t\beta'$. In which case, 
     \[
-    ((1 - t)x + ty, \gamma) = ((1 - t)x + ty, (1 - t)\alpha' + t\beta') \in \text{Conv}(\text{epi}(f))
+    ((1 - t)x + ty, \gamma) = ((1 - t)x + ty, (1 - t)\alpha' + t\beta') \in \ol{\text{Conv}}(\text{epi}(f))
     \]
 
-    so $\bracs{(1 - t)x + ty} \times [\gamma, \infty] \subset \text{Conv}(\text{epi}(f))$. 
+    so $\bracs{(1 - t)x + ty} \times [\gamma, \infty) \subset \ol{\text{Conv}}(\text{epi}(f))$, $(1 - t)(x + \alpha) + t(y, \beta) \in A$, and $A$ is convex. 
 
-    Since the set of points that satisfy the lemma is convex, and contains $\text{epi}(f)$, the lemma holds for all points in $\text{Conv}(\text{epi}(f))$. 
-
-    Now consider $\ol{\text{Conv}}(\text{epi}(f))$. Let $(x, \alpha) \in \ol{\text{Conv}}(\text{epi}(f))$, $U \in \cn_E(0)$, and $\eps > 0$, then there exists $(y, \beta) \in \text{Conv}(\text{epi}(f))$ such that $x - y \in U$ and $|\alpha - \beta| < \eps$. As such a pair exists for all $U \in \cn_E(0)$ and $\eps > 0$, 
+    Let $(x, \alpha) \in \ol A$, then there exists a net $\langle (x_\gamma, \alpha_\gamma) \rangle_{\gamma \in C} \subset A$ with $(x_\gamma, \alpha_\gamma) \to (x, \alpha)$. In which case, for each $r > 0$, $\langle (x_\gamma, \alpha_\gamma + r) \rangle_{\gamma \in C} \subset A$ and $(x_\gamma, \alpha_\gamma + r) \to (x, \alpha + r)$, so $(x, \alpha + r) \in \ol{\text{Conv}}(\text{epi}(f))$ and
     \[
-    \bracs{x} \times (\alpha, \infty) \subset \ol{\text{Conv}}(\text{epi}(f))
+    \bracs{x} \times [\alpha, \infty) \subset \ol{\text{Conv}}(\text{epi}(f))
     \]
+
+    Thus $(x, \alpha) \in A$ and $A$ is closed. 
+
+    Since $A$ is a closed convex set containing $\text{epi}(f)$, $A = \ol{\text{Conv}}(\text{epi}(f))$. 
 \end{proof}
 
 \begin{lemma}[Almost Subgradient]
@@ -156,6 +158,8 @@
         \item $(\phi, \gamma) \le f$. 
         \item $\dpn{x, \phi}{\lambda} - \gamma = \alpha$. 
     \end{enumerate}
+
+    In particular, $f^{*} \ne \infty$. 
 \end{lemma}
 \begin{proof}
     (1): Since $f$ is convex and $\sigma(E, F)$-lower semicontinuous, $\text{epi}(f)$ is $\sigma(E \times \real, F \times \real)$-closed and convex. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi \in F$ and $\mu \in \real$ such that
@@ -212,30 +216,36 @@
         f^{**}(x) \ge \sup\bracs{\dpn{x, \phi}{\lambda} - \alpha| \phi \in F, \alpha \in \real, \dpn{\cdot, y}{\lambda} - \alpha \le f}
     \]
 
-    (2): By \autoref{lemma:conjugate-function-gymnatics}, $f^{**}$ is lower semicontinuous and convex with $f^{**} \le f$, so $\text{epi}(f^{**}) \supset \text{epi}(f)$ and $\text{epi}(f^{**}) \supset \ol{\text{Conv}}(\text{epi}(f))$. Thus it is sufficient to show that $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$. 
+    (2): By \autoref{lemma:conjugate-function-gymnatics}, $f^{**}$ is lower semicontinuous and convex with $f^{**} \le f$, so $\text{epi}(f^{**}) \supset \text{epi}(f)$ and $\text{epi}(f^{**}) \supset \ol{\text{Conv}}(\text{epi}(f))$. Thus it is sufficient to show that $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$, or equivalently, 
+    \[
+    E \times \real \setminus \ol{\text{Conv}}(\text{epi}(f)) \subset E \times \real \setminus \text{epi}(f)
+    \]
     
-    Let $A = \ol{\text{Conv}}(\text{epi}(f))$ and $(x, \alpha) \in E \times \real \setminus A$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi \in F$ and $\mu \in \real$ such that
+    To this end, let $A = \ol{\text{Conv}}(\text{epi}(f))$ and $(x, \alpha) \in E \times \real \setminus A$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi \in F$ and $\mu \in \real$ such that
     \[
     \sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} - \mu \beta < \dpn{x, \phi}{\lambda} - \mu \alpha
     \]
 
+    Given that the inequality is strict, there exists $\alpha_0 \in (\alpha, \infty)$ such that
+    \[
+    \sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} - \mu \beta \le \dpn{x, \phi}{\lambda} - \mu\alpha_0
+    \]
+
     Since for any $(y, \beta) \in A$, $\beta$ may be arbitrarily large by \autoref{lemma:closed-convex-epigraph}, $\mu \ge 0$. 
 
     In the case that $\mu > 0$, for each $y \in \bracs{f < \infty}$, 
     \begin{align*}
-        \dpn{x, \phi}{\lambda} - \mu\alpha &> \dpn{y, \phi}{\lambda} - \mu f(y) \\
-        -\dpn{y, \phi}{\lambda} + \dpn{x, \phi}{\lambda} - \mu\alpha &> - \mu f(y) \\
-        \dpn{y, \mu^{-1}\phi}{\lambda} - \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha &< f(y)
+        \dpn{x, \phi}{\lambda} - \mu\alpha_0 &\ge \dpn{y, \phi}{\lambda} - \mu f(y) \\
+        -\dpn{y, \phi}{\lambda} + \dpn{x, \phi}{\lambda} - \mu\alpha_0 &\le - \mu f(y) \\
+        \dpn{y, \mu^{-1}\phi}{\lambda} - \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha_0 &\le f(y)
     \end{align*}
 
-    so $(\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{\lambda} - \alpha) \le f$ and
+    so $(\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{\lambda} - \alpha_0) \le f$ and
     \[
-    f^{**}(x) \ge \dpn{x, \mu^{-1}\phi}{\lambda} - \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha \ge \alpha
+    f^{**}(x) \ge \dpn{x, \mu^{-1}\phi}{\lambda} - \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha_0 > \alpha
     \]
 
-    Given that $f^* \ne \infty$, there exists at least one pair $(\phi_0, \gamma_0) \in F \times \real$ such that $(\phi_0, \gamma_0) \le f$. 
-    
-    Now suppose that $\mu = 0$ and let
+    Now suppose that $\mu = 0$. Given that $f^* \ne \infty$, there exists at least one pair $(\phi_0, \gamma_0) \in F \times \real$ such that $(\phi_0, \gamma_0) \le f$. Let
     \[
     \gamma = \sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} < \dpn{x, \phi}{\lambda}
     \]
@@ -250,7 +260,14 @@
         f^{**}(x) \ge \dpn{x, \Phi_t}{\lambda} - \Gamma_t = \dpn{x, \phi_0}{\lambda} - \gamma_0  + t\underbrace{(\dpn{x, \phi}{\lambda} - \gamma)}_{> 0}
     \]
 
-    As the above holds for all $t > 0$, $f^{**}(x) = \infty \ge \alpha$. Since $f^{**}(x) \ge \alpha$ for all $(x, \alpha) \in E \times \real \setminus A$, $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$. 
+    As the above holds for all $t > 0$, $f^{**}(x) = \infty > \alpha$. 
+    
+    Thus $f^{**}(x) > \alpha$ and $(x, \alpha) \not\in \text{epi}(f^{**})$ for all $(x, \alpha) \in E \times \real \setminus A$. Therefore
+    \[
+    E \times \real \setminus A \subset E \times \real \setminus \text{epi}(f^{**})
+    \]
+    
+    and $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$. 
 
      
 \end{proof}