Added the Legendre transform.

refs.bib

@@ -224,3 +224,15 @@
     eprint = {https://math.stackexchange.com/q/392719},
     url = {https://math.stackexchange.com/q/392719}
 }
+
+@book{Clarke,
+  author    = {Clarke, Francis},
+  title     = {Functional Analysis, Calculus of Variations and Optimal Control},
+  series    = {Graduate Texts in Mathematics},
+  volume    = {264},
+  publisher = {Springer},
+  address   = {London},
+  year      = {2013},
+  doi       = {10.1007/978-1-4471-4820-3},
+  isbn      = {978-1-4471-4820-3}
+}

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}
 

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}

src/fa/convex/legendre.tex

@@ -0,0 +1,228 @@
+\section{Conjugate Functions}
+\label{section:legendre}
+
+\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 $y \in F$, 
+    \[
+    \sup_{x \in E}\dpn{x, y}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| \dpn{\cdot, y}{\lambda} - \alpha \le f}
+    \]
+
+    The mapping
+    \[
+    f^*: E^* \to (-\infty, \infty] \quad y \mapsto \sup_{x \in E}\dpn{x, y}{\lambda} - f(x)
+    \]
+
+    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 $x \in E$,  
+    \[
+    \dpn{x, y}{\lambda} - f(x) \le \dpn{x, y}{\lambda} - \dpn{x, y}{\lambda} + \alpha = \alpha
+    \]
+
+    so
+    \[
+    \sup_{x \in E}\dpn{x, y}{\lambda} - f(x) \le \inf\bracsn{\alpha \in \real| \dpn{x, y}{\lambda} - \alpha \le f(x) \forall x \in E}
+    \]
+
+    On the other hand, suppose that $\alpha = \sup_{x \in E} \dpn{x, y}{\lambda} - f(x) < \infty$, then
+    \[
+    \dpn{x, y}{\lambda} - \alpha \le f(x)
+    \]
+
+    for all $x \in E$. Therefore
+    \[
+    \sup_{x \in E}\dpn{x, y}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| \dpn{x, y}{\lambda} - \alpha \le f(x) \forall x \in E}
+    \]
+\end{proof}
+
+\begin{lemma}
+\label{lemma:conjugate-function-gymnatics}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f, g: E \to (-\infty, \infty]$ with $f, g \ne \infty$, then
+    \begin{enumerate}
+        \item $f^*$ is convex and lower semicontinuous. 
+        \item If $f \le g$, then $f^* \ge g^*$. 
+        \item If $f^* \ne \infty$, then $f^{**} \le f$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): By \autoref{proposition:convex-extension} and \autoref{proposition:semicontinuous-properties}. 
+
+    (3): For each $x \in E$ and $y \in F$, 
+    \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, y}{\lambda} + \braks{\inf_{z \in E}f(z) - \dpn{z, y}{\lambda}} \\
+        &\le \dpn{x, y}{\lambda} + f(x) - \dpn{x, y}{\lambda} = f(x)
+    \end{align*}
+
+    As the above holds for all $y \in F$, $f^{**} \le f$.     
+\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 $y \in F$, 
+    \[
+    \dpn{x, y}{\lambda} \le f(x) + f^*(y)
+    \]
+
+    with equality if and only if $y \in \partial f(x)$. 
+\end{theorem}
+\begin{proof}
+    Let $x \in E$ and $y \in F$ with $f(x) < \infty$, then
+    \[
+        f(x) + f^*(y) \ge f(x) + \dpn{x, y}{\lambda} - f(x) = \dpn{x, y}{\lambda}
+    \]
+
+    For the equivalence, 
+    \begin{align*}
+        f(x) + f^*(y) &= \dpn{x, y}{\lambda} \\
+        f^*(y) &= \dpn{x, y}{\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) \\
+        f(x + h) - f(x) &\ge \dpn{h, y}{\lambda} 
+    \end{align*}
+
+    if and only if $y \in \partial f(x)$. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:closed-convex-epigraph}
+    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
+    \[
+    \bracs{x} \times [\alpha, \infty), \bracs{y} \times [\beta, \infty) \subset \text{Conv}(\text{epi}(f))
+    \]
+
+    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, 
+    \[
+    ((1 - t)x + ty, \gamma) = ((1 - t)x + ty, (1 - t)\alpha' + t\beta') \in \text{Conv}(\text{epi}(f))
+    \]
+
+    so $\bracs{(1 - t)x + ty} \times [\gamma, \infty] \subset \text{Conv}(\text{epi}(f))$. 
+
+    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 $\eps > 0$, 
+    \[
+    \bracs{x} \times (\alpha, \infty) \subset \ol{\text{Conv}}(\text{epi}(f))
+    \]
+\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{enumerate}
+        \item There exists $\phi \in E^*$ and $\alpha \in \real$ such that $\dpn{\cdot, \phi}{E} + \alpha \le f$. 
+        \item $f^* \ne \infty$. 
+    \end{enumerate}
+    
+\end{proposition}
+\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
+    \[
+    \sup_{(y, \beta) \in \text{epi}(f)}\dpn{y, \phi}{E} + \mu \beta < \dpn{x, \phi}{E} + \mu \alpha
+    \]
+
+    Since for any $(y, \beta) \in A$, $\beta$ may be arbitrarily large by \autoref{lemma:closed-convex-epigraph}, $\mu < 0$. Thus for each $y \in E$, 
+    \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)
+    \end{align*}
+
+    so $-\dpn{\cdot, \mu^{-1}\phi}{E} + \dpn{x, \mu^{-1}\phi}{E} + \alpha \le f$.
+\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 $(y, \alpha) \in F \times \real$, denote $(y, \alpha) \le f$ if $\dpn{\cdot, y}{\lambda} - \alpha \le f$, then for each $x \in E$,  
+        \[
+        f^{**}(x) = \sup\bracs{\dpn{x, y}{\lambda} - \alpha|(y, \alpha) \in F \times \real, (y, \alpha) \le f}
+        \]
+    \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 $x \in E$, 
+    \[
+    \dpn{x, y}{\lambda} - f^*(y) \le f(x)
+    \]
+
+    so $\dpn{\cdot, y}{\lambda} - f^*(y) \le f$, 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) \\
+    &\le \sup\bracs{\dpn{x, y}{\lambda} - \alpha| y \in F, \alpha \in \real, \dpn{\cdot, y}{\lambda} - \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
+    \[
+    f^{**}(x) \ge \dpn{x, y}{\lambda} - f^*(y) \ge \dpn{x, y}{\lambda} - \alpha
+    \]
+
+    Therefore
+    \[
+        f^{**}(x) \ge \sup\bracs{\dpn{x, y}{\lambda} - \alpha| y \in F, \alpha \in \real, \dpn{\cdot, y}{\lambda} - \alpha \le f}
+    \]
+
+    (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
+    \[
+    \sup_{(y, \beta) \in A}\dpn{y, \phi}{E} + \mu \beta < \dpn{x, \phi}{E} + \mu \alpha
+    \]
+
+    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, \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)
+    \end{align*}
+
+    so $(-\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{E} + \alpha) \le f$ and
+    \[
+    f^{**}(x) \ge -\dpn{x, \mu^{-1}\phi}{E} + \dpn{x, \mu^{-1}\phi}{E} + \alpha \ge \alpha
+    \]
+
+    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
+    \[
+    \gamma = \sup_{(y, \beta) \in A}\dpn{y, \phi}{E} < \dpn{x, \phi}{E}
+    \]
+
+    For each $t > 0$, let $\Phi_t = \phi_0 + t\phi$ and $\Gamma_t = \gamma_0 - t\gamma$, then for each $y \in \bracs{f < \infty}$, 
+    \[
+    \dpn{y, \Phi_t}{E} + \Gamma_t \le f(y) + t\dpn{y, \phi}{E} - t\gamma \le f(y)
+    \]
+
+    so $(\Phi_t, \Gamma_t) \le f$. By (1), 
+    \[
+        f^{**}(x) \ge \dpn{x, \Phi_t}{E} + \Gamma_t = \dpn{x, \phi_0}{E} + \gamma_0  + t\underbrace{\dpn{x, \phi}{E} - \gamma}_{> 0}
+    \]
+
+    As the above holds for all $t > 0$, $f^{**}(x) = \infty \ge \alpha$. Since $f^{**}(x) \ge \alpha$ for all $(x, \alpha) \in E \times \real \setminus A$, $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$. 
+
+    On the other hand, $f^{**} \le f$ by \autoref{lemma:conjugate-function-gymnatics}, so $\text{epi}(f^{**}) \supset \text{epi}(f)$. Since $\text{epi}(f^{**})$ is closed and convex, $\text{epi}(f^{**}) \supset \ol{\text{Conv}}(\text{epi}(f))$. 
+    
+    
+\end{proof}
+
+
+

src/fa/index.tex

@@ -4,11 +4,11 @@
 
 \input{./tvs/index.tex}
 \input{./lc/index.tex}
-\input{./convex/index.tex}
 \input{./norm/index.tex}
 \input{./rs/index.tex}
 \input{./lp/index.tex}
 \input{./order/index.tex}
 \input{./duality/index.tex}
+\input{./convex/index.tex}
 \input{./interpolation/index.tex}
 \input{./notation.tex}

src/fa/lc/convex.tex

@@ -29,7 +29,7 @@
 
 
 
-\begin{lemma}[{{\cite[II.1.1]{SchaeferWolff}}}]
+\begin{lemma}
 \label{lemma:convex-interior}
     Let $E$ be a TVS over $K \in \RC$, $A \subset E$ be convex, $x \in A^o$, and $y \in \ol{A}$, then
     \[
@@ -37,7 +37,7 @@
     \]
 
 \end{lemma}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[II.1.1]{SchaeferWolff}}}. ]
     Fix $t \in (0, 1)$. Using translation, assume without loss of generality that $tx + (1 - t)y = 0$. In which case, $x = \alpha y$ where $\alpha = (1 - t)/t$. By (TVS2), $\alpha A^o \in \cn^o(y)$. Since $y \in \ol{A}$, $\alpha A^o \cap A \ne \emptyset$, so there exists $z \in A^o$ such that $\alpha z \in A$.
 
     Let $\mu = \alpha/(\alpha - 1)$, then since $\alpha < 0$, $\mu \in (0, 1)$ and
@@ -74,11 +74,11 @@
     converges to $tx + (1 - t)y$ by (TVS1) and (TVS2). Hence $tx + (1 - t)y \in \ol{A}$.
 \end{proof}
 
-\begin{proposition}[{{\cite[II.1.3]{SchaeferWolff}}}]
+\begin{proposition}
 \label{proposition:convex-interior-closure}
     Let $E$ be a TVS over $K \in \RC$ and $A \subset E$ be convex. If $A^o \ne \emptyset$, then $\ol{A} = \ol{A^o}$.
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[II.1.3]{SchaeferWolff}}}. ]
     Since $A^o \subset A$, $\ol{A^o} \subset \ol{A}$. Let $x \in A^o$, then for any $y \in \ol{A}$,
     \[
     y \in \ol{\bracs{tx + (1 - t)y|t \in (0, 1)}} \subset \ol{A^o}

src/fa/lc/hahn-banach.tex

@@ -98,12 +98,14 @@
 
 \begin{theorem}[Hahn-Banach, First Geometric Form]
 \label{theorem:hahn-banach-geometric-1}
-    Let $E$ be a TVS over $\real$, $A, B \subset E$ be non-empty convex sets, such that $A \cap B = \emptyset$ and $A$ is open, then there exists $\phi \in E^* \setminus \bracs{0}$ and $\alpha \in \real$ with $A \subset \bracs{\phi \le \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$.
+    Let $E$ be a TVS over $\real$, $A, B \subset E$ be non-empty convex sets, such that $A \cap B = \emptyset$ and $A$ is open, then there exists $\phi \in E^* \setminus \bracs{0}$ and $\alpha \in \real$ with $A \subset \bracs{\phi < \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$.
 \end{theorem}
 \begin{proof}[Proof {{\cite[Theorem 1.6]{Brezis}}}. ]
     Let $C = A - B$, then $C \subset E$ is an open convex set by \autoref{lemma:convex-gymnastics}. Since $A \cap B = \emptyset$, $0 \not\in C$.
 
     By \autoref{lemma:hahn-banach-separation}, there exists $\phi \in E^*$ such that $C \subset \bracs{\phi < 0}$. In which case, for any $a \in A$ and $b \in B$, $\dpb{a - b, \phi}{E} < 0$ implies that $\dpb{a, \phi}{E} < \dpb{b, \phi}{E}$. Let $\alpha \in [\sup_{a \in A}\dpb{a, \phi}{E}, \inf_{b \in B}\dpb{b, \phi}{E}]$, then $A \subset \bracs{\phi \le \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$.
+
+    Since $A$ is open, so is $\phi(A)$. Thus $\phi(A) \subset (-\infty, \alpha]$ implies that $\phi(A) \subset (-\infty, \alpha)$. 
 \end{proof}
 
 \begin{theorem}[Hahn-Banach, Second Geometric Form]

src/fa/norm/separable.tex

@@ -3,14 +3,16 @@
 
 \begin{proposition}
 \label{proposition:separable-dual}
-    Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality of normed vector spaces over $K$ with $E$ being separable, then
+    Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality of normed vector spaces over $K$ with $E$ being separable, and $S = \bracsn{y \in F|\ \norm{y}_F \le 1}$ be the closed unit ball of $F$, then
     \begin{enumerate}
-        \item The closed unit ball $S = \bracsn{y \in F|\ \norm{y}_{F} \le 1}$ is separable with respect to the $\sigma(F, E)$-topology. 
+        \item $S$ is separable with respect to the $\sigma(F, E)$-topology. 
+        \item $S$ is metrisable with respect to the $\sigma(F, E)$-topology. 
+        \item For any $A \subset F$, $A$ is separable with respect to the $\sigma(F, E)$-topology. 
         \item If the duality is norming, then there exists $\seq{y_n} \subset F$ such that for each $x \in E$, $\norm{x}_E = \sup_{n \in \natp}|\dpn{x, y_n}{\lambda}|$. 
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    Let $\seq{x_n} \subset E$ be a dense subset. For each $N \in \natp$, let
+    (1): Let $\seq{x_n} \subset E$ be a dense subset. For each $N \in \natp$, let
     \[
     T_N: S \to \real^N \quad y \mapsto (\dpn{x_1, y}{E}, \cdots, \dpn{x_N, y}{E})
     \]
@@ -23,6 +25,10 @@
     \] 
 
     Thus for each $N \in \natp$, $\dpn{x_n, y_{N, k_N}}{E} \to \dpn{x_n, y}{E}$ as $N \to \infty$. Since $\phi_{N, k_N} \to \phi$ pointwise on a dense subset of $E$, and $\bracsn{y_{N, k_N}|N \in \natp} \subset S$ is uniformly equicontinuous, $\phi_{N, k_N} \to \phi$ in the $\sigma(F, E)$-topology by \autoref{proposition:strong-operator-dense}. 
+
+    (2): Let $\seq{x_n} \subset E$ be a dense subset, then by \autoref{proposition:strong-operator-dense}, the $\sigma(F, E)$-topology on $S$ is induced by $\seq{x_n}$, and hence metrisable by \autoref{theorem:uniform-metrisable}. 
+
+    (3): For any $A \subset E$, $A = \bigcup_{n \in \natp}A \cap nS$. By \autoref{proposition:separable-metric-space}, $A \cap nS$ is separable for each $n \in \natp$. Therefore $A$ is also separable. 
 \end{proof}
 
 \begin{proposition}

src/fa/notation.tex

@@ -49,6 +49,9 @@
     $\mu_G$ & Lebesgue-Stieltjes measure associated with $G$. & \autoref{definition:riemann-lebesgue-stieltjes} \\
     $\int_\gamma f$, $\int_\gamma f(z)dz$ & Path integral of $f$ with respect to $\gamma$. & \autoref{definition:path-integral} \\
     $PI([a, b], \gamma; E)$ & 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} \\
+    $f^*$ & Conjugate function of $f$. & \autoref{definition:conjugate-function} \\ 
     % ---- Interpolation Spaces ---- \\
     $\catc_1$ & Category of compatible couples in $\catc$. & \autoref{definition:compatible-category} \\
     $\cf(E_0, E_1)$ & Calderón space of $(E_0, E_1)$ & \autoref{definition:calderon-space} \\

src/topology/main/semicontinuity.tex

@@ -22,7 +22,7 @@
 \end{proof}
 
 
-\begin{proposition}[{{\cite[Proposition 7.11]{Folland}}}]
+\begin{proposition}
 \label{proposition:semicontinuous-properties}
     Let $X$ be a topological space, then
     \begin{enumerate}
@@ -33,7 +33,7 @@
         \item For any $f: X \to (-\infty, \infty]$ lower semicontinuous, $f$ is Borel measurable. 
     \end{enumerate}
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Proposition 7.11]{Folland}}}. ]
     (1): For any $\alpha \in \real$, 
     \[
     \bracs{f > \alpha} = \begin{cases}