Updated notation and formulation in the Fenchel Moreau theorem.

src/fa/convex/legendre.tex

@@ -3,37 +3,37 @@
 
 \begin{definition}[Conjugate Function]
 \label{definition:conjugate-function}
-    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f: E \to (-\infty, \infty]$ with $f \ne \infty$, then for each $y \in F$, 
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f: E \to (-\infty, \infty]$ with $f \ne \infty$, then for each $\phi \in F$, 
     \[
-    \sup_{x \in E}\dpn{x, y}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| \dpn{\cdot, y}{\lambda} - \alpha \le f}
+    \sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| \dpn{\cdot, \phi}{\lambda} - \alpha \le f}
     \]
 
     The mapping
     \[
-    f^*: E^* \to (-\infty, \infty] \quad y \mapsto \sup_{x \in E}\dpn{x, y}{\lambda} - f(x)
+    f^*: F \to (-\infty, \infty] \quad \phi \mapsto \sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x)
     \]
 
     is the \textbf{conjugate function} of $f$ with respect to the duality $\dpn{E, F}{\lambda}$. 
 \end{definition}
 \begin{proof}
-    Fix $y \in F$. Let $\alpha \in  \real$ such that $\dpn{x, y}{\lambda} - \alpha \le f(x)$ for all $x \in E$, then for any $x \in E$,  
+    Fix $\phi \in F$. Let $\alpha \in  \real$ such that $\dpn{x, \phi}{\lambda} - \alpha \le f(x)$ for all $x \in E$, then for any $x \in E$,  
     \[
-    \dpn{x, y}{\lambda} - f(x) \le \dpn{x, y}{\lambda} - \dpn{x, y}{\lambda} + \alpha = \alpha
+    \dpn{x, \phi}{\lambda} - f(x) \le \dpn{x, \phi}{\lambda} - \dpn{x, \phi}{\lambda} + \alpha = \alpha
     \]
 
     so
     \[
-    \sup_{x \in E}\dpn{x, y}{\lambda} - f(x) \le \inf\bracsn{\alpha \in \real| \dpn{x, y}{\lambda} - \alpha \le f(x) \forall x \in E}
+    \sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) \le \inf\bracsn{\alpha \in \real| \dpn{x, \phi}{\lambda} - \alpha \le f(x) \forall x \in E}
     \]
 
-    On the other hand, suppose that $\alpha = \sup_{x \in E} \dpn{x, y}{\lambda} - f(x) < \infty$, then
+    On the other hand, suppose that $\alpha = \sup_{x \in E} \dpn{x, \phi}{\lambda} - f(x) < \infty$, then
     \[
-    \dpn{x, y}{\lambda} - \alpha \le f(x)
+    \dpn{x, \phi}{\lambda} - \alpha \le f(x)
     \]
 
     for all $x \in E$. Therefore
     \[
-    \sup_{x \in E}\dpn{x, y}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| \dpn{x, y}{\lambda} - \alpha \le f(x) \forall x \in E}
+    \sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| \dpn{x, \phi}{\lambda} - \alpha \le f(x) \forall x \in E}
     \]
 \end{proof}
 
@@ -49,41 +49,41 @@
 \begin{proof}
     (1): By \autoref{proposition:convex-extension} and \autoref{proposition:semicontinuous-properties}. 
 
-    (3): For each $x \in E$ and $y \in F$, 
+    (3): For each $x \in E$ and $\phi \in F$, 
     \begin{align*}
-        \dpn{x, y}{\lambda} - f^*(y) &= \dpn{x, y}{\lambda} - \braks{\sup_{z \in E}\dpn{z, y}{\lambda} - f(z)} \\
+        \dpn{x, \phi}{\lambda} - f^*(\phi) &= \dpn{x, \phi}{\lambda} - \braks{\sup_{z \in E}\dpn{z, \phi}{\lambda} - f(z)} \\
-        &= \dpn{x, y}{\lambda} + \braks{\inf_{z \in E}f(z) - \dpn{z, y}{\lambda}} \\
+        &= \dpn{x, \phi}{\lambda} + \braks{\inf_{z \in E}f(z) - \dpn{z, \phi}{\lambda}} \\
-        &\le \dpn{x, y}{\lambda} + f(x) - \dpn{x, y}{\lambda} = f(x)
+        &\le \dpn{x, \phi}{\lambda} + f(x) - \dpn{x, \phi}{\lambda} = f(x)
     \end{align*}
 
-    As the above holds for all $y \in F$, $f^{**} \le f$.     
+    As the above holds for all $\phi \in F$, $f^{**} \le f$.     
 \end{proof}
 
 \begin{theorem}[Fenchel's Inequality]
 \label{theorem:fenchel-inequality}
-    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f: E \to (-\infty, \infty]$ with $f \ne \infty$, then for any $x \in E$ and $y \in F$, 
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f: E \to (-\infty, \infty]$ with $f \ne \infty$, then for any $x \in E$ and $\phi \in F$, 
     \[
-    \dpn{x, y}{\lambda} \le f(x) + f^*(y)
+    \dpn{x, \phi}{\lambda} \le f(x) + f^*(\phi)
     \]
 
-    with equality if and only if $y \in \partial f(x)$. 
+    with equality if and only if $\phi \in \partial f(x)$. 
 \end{theorem}
 \begin{proof}
-    Let $x \in E$ and $y \in F$ with $f(x) < \infty$, then
+    Let $x \in E$ and $\phi \in F$ with $f(x) < \infty$, then
     \[
-        f(x) + f^*(y) \ge f(x) + \dpn{x, y}{\lambda} - f(x) = \dpn{x, y}{\lambda}
+        f(x) + f^*(\phi) \ge f(x) + \dpn{x, \phi}{\lambda} - f(x) = \dpn{x, \phi}{\lambda}
     \]
 
     For the equivalence, 
     \begin{align*}
-        f(x) + f^*(y) &= \dpn{x, y}{\lambda} \\
+        f(x) + f^*(\phi) &= \dpn{x, \phi}{\lambda} \\
-        f^*(y) &= \dpn{x, y}{\lambda} - f(x) 
+        f^*(\phi) &= \dpn{x, \phi}{\lambda} - f(x) 
     \end{align*}
 
     if and only if for every $h \in E$, 
     \begin{align*}
-        \dpn{x, y}{\lambda} - f(x) &\ge \dpn{x+h, y}{\lambda} - f(x+h) \\
+        \dpn{x, \phi}{\lambda} - f(x) &\ge \dpn{x+h, \phi}{\lambda} - f(x+h) \\
-        f(x + h) - f(x) &\ge \dpn{h, y}{\lambda} 
+        f(x + h) - f(x) &\ge \dpn{h, \phi}{\lambda} 
     \end{align*}
 
     if and only if $y \in \partial f(x)$. 
@@ -145,76 +145,71 @@
 \label{theorem:fenchel-moreau}
     Let $\dpn{E, F}{\lambda}$ be a duality over $\real$, and $f: E \to (-\infty, \infty]$ with $f \ne \infty$ and $f^{*} \ne \infty$, then:
     \begin{enumerate}
-        \item For each $(y, \alpha) \in F \times \real$, denote $(y, \alpha) \le f$ if $\dpn{\cdot, y}{\lambda} - \alpha \le f$, then for each $x \in E$,  
+        \item For each $(\phi, \alpha) \in F \times \real$, denote $(\phi, \alpha) \le f$ if $\dpn{\cdot, \phi}{\lambda} - \alpha \le f$, then for each $x \in E$,  
         \[
-        f^{**}(x) = \sup\bracs{\dpn{x, y}{\lambda} - \alpha|(y, \alpha) \in F \times \real, (y, \alpha) \le f}
+        f^{**}(x) = \sup\bracs{\dpn{x, \phi}{\lambda} - \alpha|(\phi, \alpha) \in F \times \real, (\phi, \alpha) \le f}
         \]
+        \item $\text{epi}(f^{**})$ is the $\sigma(E \times \real, F \times \real)$-closed convex hull of $\text{epi}(f)$. 
+        \item The biconjugate $f^{**}$ is the greatest convex and $\sigma(E, F)$-lower semicontinuous function bounded above by $f$. 
+        \item $f = f^{**}$ if and only if $f$ is convex and $\sigma(E, F)$-lower semicontinuous.  
     \end{enumerate}
-
-    If $F = E^*$, then
-    \begin{enumerate}[start=1]
-        \item The epigraph $\text{epi}(f^{**})$ is the closed convex hull of $\text{epi}(f)$. 
-        \item The biconjugate $f^{**}$ is the greatest convex and lower semicontinuous function bounded above by $f$. 
-        \item $f = f^{**}$ if and only if $f$ is convex and lower semicontinuous.  
-    \end{enumerate}
-    
 \end{theorem}
 \begin{proof}
-    (1): Let $y \in F$ such that $f^*(y) < \infty$, then for each $x \in E$, 
+    (1): Let $\phi \in F$ such that $f^*(\phi) < \infty$, then for each $x \in E$, 
     \[
-    \dpn{x, y}{\lambda} - f^*(y) \le f(x)
+    \dpn{x, \phi}{\lambda} - f^*(\phi) \le f(x)
     \]
 
-    so $\dpn{\cdot, y}{\lambda} - f^*(y) \le f$, and
+    so $\dpn{\cdot, \phi}{\lambda} - f^*(\phi) \le f$, and
     \begin{align*}
-    f^{**}(x) &= \sup_{y \in F} \dpn{x, y}{\lambda} - f^*(y) = \sup_{\substack{y \in F \\ f^*(y) < \infty}} \dpn{x, y}{\lambda} - f^*(y) \\
+    f^{**}(x) &= \sup_{\phi \in F} \dpn{x, \phi}{\lambda} - f^*(\phi) = \sup_{\substack{\phi \in F \\ f^*(\phi) < \infty}} \dpn{x, y}{\lambda} - f^*(\phi) \\
-    &\le \sup\bracs{\dpn{x, y}{\lambda} - \alpha| y \in F, \alpha \in \real, \dpn{\cdot, y}{\lambda} - \alpha \le f}
+    &\le \sup\bracs{\dpn{x, \phi}{\lambda} - \alpha|(\phi, \alpha) \in F \times \real, (\phi, \alpha) \le f}
     \end{align*}
     
-    On the other hand, let $y \in F$ and $\alpha \in \real$ such that $\dpn{\cdot, y}{\lambda} - \alpha \le f$, then $f^*(y) \le \alpha$, and
+    On the other hand, let $\phi \in F$ and $\alpha \in \real$ such that $(\phi, \alpha) \le f$, then $f^*(\phi) \le \alpha$, and
     \[
-    f^{**}(x) \ge \dpn{x, y}{\lambda} - f^*(y) \ge \dpn{x, y}{\lambda} - \alpha
+    f^{**}(x) \ge \dpn{x, \phi}{\lambda} - f^*(\phi) \ge \dpn{x, \phi}{\lambda} - \alpha
     \]
 
     Therefore
     \[
-        f^{**}(x) \ge \sup\bracs{\dpn{x, y}{\lambda} - \alpha| y \in F, \alpha \in \real, \dpn{\cdot, y}{\lambda} - \alpha \le f}
+        f^{**}(x) \ge \sup\bracs{\dpn{x, \phi}{\lambda} - \alpha| \phi \in F, \alpha \in \real, \dpn{\cdot, y}{\lambda} - \alpha \le f}
     \]
 
     (2): 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
     \[
-    \sup_{(y, \beta) \in A}\dpn{y, \phi}{E} + \mu \beta < \dpn{x, \phi}{E} + \mu \alpha
+    \sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} + \mu \beta < \dpn{x, \phi}{\lambda} + \mu \alpha
     \]
 
     Since for any $(y, \beta) \in A$, $\beta$ may be arbitrarily large by \autoref{lemma:closed-convex-epigraph}, $\mu \le 0$. 
 
     In the case that $\mu < 0$, for each $y \in E$, 
     \begin{align*}
-        \dpn{x, \phi}{E} + \mu\alpha &> \dpn{y, \phi}{E} + \mu f(y) \\
+        \dpn{x, \phi}{\lambda} + \mu\alpha &> \dpn{y, \phi}{\lambda} + \mu f(y) \\
-        \dpn{x, \mu^{-1}\phi}{E} + \alpha & < \dpn{y, \mu^{-1}\phi}{E} + f(y) \\
+        \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha & < \dpn{y, \mu^{-1}\phi}{\lambda} + f(y) \\
-        -\dpn{y, \mu^{-1}\phi}{E} + \dpn{x, \mu^{-1}\phi}{E} + \alpha &< f(y)
+        -\dpn{y, \mu^{-1}\phi}{\lambda} + \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha &< f(y)
     \end{align*}
 
-    so $(-\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{E} + \alpha) \le f$ and
+    so $(-\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha) \le f$ and
     \[
-    f^{**}(x) \ge -\dpn{x, \mu^{-1}\phi}{E} + \dpn{x, \mu^{-1}\phi}{E} + \alpha \ge \alpha
+    f^{**}(x) \ge -\dpn{x, \mu^{-1}\phi}{\lambda} + \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha \ge \alpha
     \]
 
     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
     \[
-    \gamma = \sup_{(y, \beta) \in A}\dpn{y, \phi}{E} < \dpn{x, \phi}{E}
+    \gamma = \sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} < \dpn{x, \phi}{\lambda}
     \]
 
     For each $t > 0$, let $\Phi_t = \phi_0 + t\phi$ and $\Gamma_t = \gamma_0 - t\gamma$, then for each $y \in \bracs{f < \infty}$, 
     \[
-    \dpn{y, \Phi_t}{E} + \Gamma_t \le f(y) + t\dpn{y, \phi}{E} - t\gamma \le f(y)
+    \dpn{y, \Phi_t}{\lambda} + \Gamma_t \le f(y) + t\dpn{y, \phi}{\lambda} - t\gamma \le f(y)
     \]
 
     so $(\Phi_t, \Gamma_t) \le f$. By (1), 
     \[
-        f^{**}(x) \ge \dpn{x, \Phi_t}{E} + \Gamma_t = \dpn{x, \phi_0}{E} + \gamma_0  + t\underbrace{(\dpn{x, \phi}{E} - \gamma)}_{> 0}
+        f^{**}(x) \ge \dpn{x, \Phi_t}{\lambda} + \Gamma_t = \dpn{x, \phi_0}{\lambda} + \gamma_0  + t\underbrace{(\dpn{x, \phi}{\lambda} - \gamma)}_{> 0}
     \]
 
     As the above holds for all $t > 0$, $f^{**}(x) = \infty \ge \alpha$. Since $f^{**}(x) \ge \alpha$ for all $(x, \alpha) \in E \times \real \setminus A$, $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$.