Added the Legendre transform.

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/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}