Adjusted the Legendre sectoin.

src/fa/convex/def.tex

@@ -41,6 +41,31 @@
     
 \end{proof}
 
+\begin{lemma}
+\label{lemma:convex-domain}
+    Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$, then $f$ is convex if and only if $\bracs{f < \infty}$ is convex and $f|_{\bracs{f < \infty}}$ is a convex function. 
+\end{lemma}
+\begin{proof}
+    If $f$ is convex, then for any $x, y \in \bracs{f < \infty}$ and $t \in [0, 1]$, 
+    \[
+    f((1 - t)x + ty) \le (1 - t)f(x) + tf(y) < \infty
+    \]
+
+    so $\bracs{f < \infty}$ is convex, and the restriction $f|_{\bracs{f < \infty}}$ is a convex function. 
+
+    On the other hand, for any $x, y \in E$ and $t \in [0, 1]$, if $f(x) = \infty$ or $f(y) = \infty$, then
+    \[
+    f((1 - t)x + ty) \le = \infty (1 - t)f(x) + tf(y)
+    \]
+
+    Otherwise, $x, y \in \bracs{f < \infty}$, and
+    \[
+    f((1 - t)x + ty) \le (1 - t)f(x) + tf(y)
+    \]
+
+    by convexity of $f|_{\bracs{f < \infty}}$. 
+\end{proof}
+
 \begin{lemma}
 \label{lemma:convex-reverse}
     Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$ be convex, then for any $x, y \in E$ and $t \in \real \setminus [0, 1]$, 

src/fa/convex/legendre.tex

@@ -1,11 +1,20 @@
 \section{Conjugate Functions}
 \label{section:legendre}
 
+\begin{definition}[Affine Minorant]
+\label{definition:affine-minorant}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$, $f: E \to (-\infty, \infty]$, and $(\phi, \alpha) \in F \times \real$, then the pair $(\phi, \alpha)$ is an \textbf{affine minorant} of $f$, denoted $(\phi, \alpha) \le f$, if
+    \[
+    \dpn{x, \phi}{\lambda} - \alpha \le f(x) \quad \forall x \in E
+    \]
+\end{definition}
+
+
 \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 $\phi \in F$, 
     \[
-    \sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| \dpn{\cdot, \phi}{\lambda} - \alpha \le f}
+    \sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| (\phi, \alpha) \le f}
     \]
 
     The mapping
@@ -23,7 +32,7 @@
 
     so
     \[
-    \sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) \le \inf\bracsn{\alpha \in \real| \dpn{x, \phi}{\lambda} - \alpha \le f(x) \forall x \in E}
+    \sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) \le \inf\bracsn{\alpha \in \real| (\phi, \alpha) \le f}
     \]
 
     On the other hand, suppose that $\alpha = \sup_{x \in E} \dpn{x, \phi}{\lambda} - f(x) < \infty$, then
@@ -33,9 +42,23 @@
 
     for all $x \in E$. Therefore
     \[
-    \sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| \dpn{x, \phi}{\lambda} - \alpha \le f(x) \forall x \in E}
+    \sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| (\phi, \alpha) \le f}
     \]
 \end{proof}
+\begin{lemma}
+\label{lemma:conjugate-minorant}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f: E \to (-\infty, \infty]$, then the following are equivalent:
+    \begin{enumerate}
+        \item $f^* \ne \infty$. 
+        \item There exists $(\phi, \alpha) \in F \times \real$ with $(\phi, \alpha) \le f$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $\phi \in \bracsn{f^* < \infty}$, then by \autoref{definition:conjugate-function}, $(\phi, f^*(\phi)) \le f$. 
+
+    (2) $\Rightarrow$ (1): By \autoref{definition:conjugate-function}, $f^*(\phi) \le \alpha$.   
+\end{proof}
+
 
 \begin{lemma}
 \label{lemma:conjugate-function-gymnatics}
@@ -69,24 +92,36 @@
     with equality if and only if $\phi \in \partial f(x)$. 
 \end{theorem}
 \begin{proof}
-    Let $x \in E$ and $\phi \in F$ with $f(x) < \infty$, then
+    Let $x \in E$ and $\phi \in F$. Assume without loss of generality that $f(x) < \infty$, then
     \[
         f(x) + f^*(\phi) \ge f(x) + \dpn{x, \phi}{\lambda} - f(x) = \dpn{x, \phi}{\lambda}
     \]
 
-    For the equivalence, 
+    Now let $x \in E$ and $\phi \in F$ such that $\dpn{x, \phi}{\lambda} = f(x) + f^*(\phi)$, then $f(x), f^*(\phi) < \infty$ and
     \begin{align*}
         f(x) + f^*(\phi) &= \dpn{x, \phi}{\lambda} \\
         f^*(\phi) &= \dpn{x, \phi}{\lambda} - f(x) 
     \end{align*}
 
-    if and only if for every $h \in E$, 
+    By definition, 
+    \[
+    \dpn{x, \phi}{\lambda} - f(x)  = f^*(\phi) = \sup_{h \in E}\dpn{x+h, \phi}{\lambda} - f(x+h)
+    \]
+
+    so for every $h \in E$, 
     \begin{align*}
         \dpn{x, \phi}{\lambda} - f(x) &\ge \dpn{x+h, \phi}{\lambda} - f(x+h) \\
         f(x + h) - f(x) &\ge \dpn{h, \phi}{\lambda} 
     \end{align*}
 
-    if and only if $y \in \partial f(x)$. 
+    Thus $\phi$ satisfies the subgradient inequality, and $\phi \in \partial f(x)$. 
+
+    On the other hand, if $f(x) < \infty$ and $\phi \in \partial f(x)$, then $f(x + h) - f(x) \ge \dpn{h, \phi}{\lambda}$ for all $h \in E$, and
+    \begin{align*}
+    f^*(\phi) &= \sup_{h \in E}\dpn{x+h, \phi}{\lambda} - f(x+h) = \dpn{x, \phi}{\lambda} - f(x) \\
+    f^*(\phi) + f(x) &=  \dpn{x, \phi}{\lambda}
+    \end{align*}
+    
 \end{proof}
 
 \begin{lemma}
@@ -114,38 +149,39 @@
     \]
 \end{proof}
 
-\begin{proposition}
-\label{proposition:lsc-affine-minorant}
-    Let $E$ be a locally convex space over $\real$ and $f: E \to (-\infty, \infty]$ be convex and lower semicontinuous with $f \ne \infty$, then:
+\begin{lemma}[Almost Subgradient]
+\label{lemma:lsc-affine-minorant}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$, $f: E \to (-\infty, \infty]$ be convex and $\sigma(E, F)$-lower semicontinuous, $x \in \bracs{f < \infty}$, and $\alpha < f(x)$, then there exists $(\phi, \gamma) \in E \times \real$ such that:
     \begin{enumerate}
-        \item There exists $\phi \in E^*$ and $\alpha \in \real$ such that $\dpn{\cdot, \phi}{E} + \alpha \le f$. 
-        \item $f^* \ne \infty$. 
+        \item $(\phi, \gamma) \le f$. 
+        \item $\dpn{x, \phi}{\lambda} - \gamma = \alpha$. 
     \end{enumerate}
-    
-\end{proposition}
+\end{lemma}
 \begin{proof}
-    (1): Let $(x, \alpha) \in \bracs{f < \infty} \times \real \setminus \text{epi}(f)$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi \in F$ and $\mu \in \real$ such that
+    (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
     \[
-    \sup_{(y, \beta) \in \text{epi}(f)}\dpn{y, \phi}{E} + \mu \beta < \dpn{x, \phi}{E} + \mu \alpha
+    \sup_{(y, \beta) \in \text{epi}(f)}\dpn{y, \phi}{E} - \mu \beta < \dpn{x, \phi}{E} - \mu \alpha
     \]
 
-    Since for any $(y, \beta) \in A$, $\beta$ may be arbitrarily large by \autoref{lemma:closed-convex-epigraph}, $\mu < 0$. Thus for each $y \in E$, 
+    Since for any $(y, \beta) \in A$, $\beta$ may be arbitrarily small by \autoref{lemma:closed-convex-epigraph}, $\mu > 0$. Thus for each $y \in \bracs{f < \infty}$, 
     \begin{align*}
-        \dpn{x, \phi}{E} + \mu\alpha &> \dpn{y, \phi}{E} + \mu f(y) \\
-        \dpn{x, \mu^{-1}\phi}{E} + \alpha & < \dpn{y, \mu^{-1}\phi}{E} + f(y) \\
-        -\dpn{y, \mu^{-1}\phi}{E} + \dpn{x, \mu^{-1}\phi}{E} + \alpha &< f(y)
+        \dpn{x, \phi}{E} - \mu\alpha &> \dpn{y, \phi}{E} - \mu f(y) \\
+        -\dpn{y, \phi}{E} + \dpn{x, \phi}{E} - \mu\alpha &> -\mu f(y)\\
+        \dpn{y, \mu^{-1}\phi}{E} - \dpn{x, \mu^{-1}\phi}{E} + \alpha &< f(y)
     \end{align*}
 
-    so $-\dpn{\cdot, \mu^{-1}\phi}{E} + \dpn{x, \mu^{-1}\phi}{E} + \alpha \le f$.
+    so $(\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{E} - \alpha) \le f$ and
+    \[
+    \dpn{x, \mu^{-1}\phi}{E} - \dpn{x, \mu^{-1}\phi}{E} + \alpha = \alpha
+    \]
 \end{proof}
 
 
-
 \begin{theorem}[Fenchel-Moreau]
 \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 $(\phi, \alpha) \in F \times \real$, denote $(\phi, \alpha) \le f$ if $\dpn{\cdot, \phi}{\lambda} - \alpha \le f$, then for each $x \in E$,  
+        \item For each $x \in E$,  
         \[
         f^{**}(x) = \sup\bracs{\dpn{x, \phi}{\lambda} - \alpha|(\phi, \alpha) \in F \times \real, (\phi, \alpha) \le f}
         \]
@@ -176,23 +212,25 @@
         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
+    (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 $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$. 
+    
+    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
     \[
-    \sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} + \mu \beta < \dpn{x, \phi}{\lambda} + \mu \alpha
+    \sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} - \mu \beta < \dpn{x, \phi}{\lambda} - \mu \alpha
     \]
 
-    Since for any $(y, \beta) \in A$, $\beta$ may be arbitrarily large by \autoref{lemma:closed-convex-epigraph}, $\mu \le 0$. 
+    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 $y \in E$, 
+    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{x, \mu^{-1}\phi}{\lambda} + \alpha & < \dpn{y, \mu^{-1}\phi}{\lambda} + f(y) \\
-        -\dpn{y, \mu^{-1}\phi}{\lambda} + \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha &< f(y)
+        \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)
     \end{align*}
 
-    so $(-\mu^{-1}\phi, -\dpn{x, \mu^{-1}\phi}{\lambda} - \alpha) \le f$ and
+    so $(\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{\lambda} - \alpha) \le f$ and
     \[
-    f^{**}(x) \ge -\dpn{x, \mu^{-1}\phi}{\lambda} + \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha \ge \alpha
+    f^{**}(x) \ge \dpn{x, \mu^{-1}\phi}{\lambda} - \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha \ge \alpha
     \]
 
     Given that $f^* \ne \infty$, there exists at least one pair $(\phi_0, \gamma_0) \in F \times \real$ such that $(\phi_0, \gamma_0) \le f$. 
@@ -214,7 +252,7 @@
 
     As the above holds for all $t > 0$, $f^{**}(x) = \infty \ge \alpha$. Since $f^{**}(x) \ge \alpha$ for all $(x, \alpha) \in E \times \real \setminus A$, $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$. 
 
-    By \autoref{lemma:conjugate-function-gymnatics}, $f^{**}$ is lower semicontinuous and convex with so $f^{**} \le f$, so $\text{epi}(f^{**}) \supset \text{epi}(f)$ and $\text{epi}(f^{**}) \supset \ol{\text{Conv}}(\text{epi}(f))$.    
+     
 \end{proof}
 
 \begin{corollary}

src/fa/lc/convex.tex

@@ -1,4 +1,4 @@
-\section{Seminorms}
+\section{Convex Sets and Seminorms}
 \label{section:seminorms}
 
 

src/fa/notation.tex

@@ -51,6 +51,7 @@
     $PI([a, b], \gamma; E)$ & Space of path integrable functions with respect to $\gamma$. & \autoref{definition:path-integral} \\ 
     % ---- Convex Functions ---- \\
     $\partial f(x)$ & Subdifferential of $f$ at $x$. & \autoref{definition:subgradient} \\
+    $(\phi, \alpha) \le f$ & $\phi - \alpha \le f$. $(\phi, \alpha)$ is an affine minorant of $f$. & \autoref{definition:affine-minorant} \\ 
     $f^*$ & Conjugate function of $f$. & \autoref{definition:conjugate-function} \\ 
     % ---- Interpolation Spaces ---- \\
     $\catc_1$ & Category of compatible couples in $\catc$. & \autoref{definition:compatible-category} \\