Adjusted the Legendre sectoin.
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Bokuan Li
@@ -1,11 +1,20 @@
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\section{Conjugate Functions}
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\label{section:legendre}
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\begin{definition}[Affine Minorant]
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\label{definition:affine-minorant}
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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
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\[
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\dpn{x, \phi}{\lambda} - \alpha \le f(x) \quad \forall x \in E
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\]
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\end{definition}
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\begin{definition}[Conjugate Function]
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\label{definition:conjugate-function}
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Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f: E \to (-\infty, \infty]$ with $f \ne \infty$, then for each $\phi \in F$,
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\[
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\sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| \dpn{\cdot, \phi}{\lambda} - \alpha \le f}
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\sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| (\phi, \alpha) \le f}
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\]
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The mapping
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@@ -23,7 +32,7 @@
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so
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\[
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\sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) \le \inf\bracsn{\alpha \in \real| \dpn{x, \phi}{\lambda} - \alpha \le f(x) \forall x \in E}
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\sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) \le \inf\bracsn{\alpha \in \real| (\phi, \alpha) \le f}
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\]
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On the other hand, suppose that $\alpha = \sup_{x \in E} \dpn{x, \phi}{\lambda} - f(x) < \infty$, then
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@@ -33,9 +42,23 @@
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for all $x \in E$. Therefore
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\[
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\sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| \dpn{x, \phi}{\lambda} - \alpha \le f(x) \forall x \in E}
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\sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| (\phi, \alpha) \le f}
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\]
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\end{proof}
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\begin{lemma}
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\label{lemma:conjugate-minorant}
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Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f: E \to (-\infty, \infty]$, then the following are equivalent:
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\begin{enumerate}
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\item $f^* \ne \infty$.
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\item There exists $(\phi, \alpha) \in F \times \real$ with $(\phi, \alpha) \le f$.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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(1) $\Rightarrow$ (2): Let $\phi \in \bracsn{f^* < \infty}$, then by \autoref{definition:conjugate-function}, $(\phi, f^*(\phi)) \le f$.
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(2) $\Rightarrow$ (1): By \autoref{definition:conjugate-function}, $f^*(\phi) \le \alpha$.
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\end{proof}
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\begin{lemma}
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\label{lemma:conjugate-function-gymnatics}
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@@ -126,38 +149,39 @@
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\]
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\end{proof}
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\begin{proposition}
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\label{proposition:lsc-affine-minorant}
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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:
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\begin{lemma}[Almost Subgradient]
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\label{lemma:lsc-affine-minorant}
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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:
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\begin{enumerate}
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\item There exists $\phi \in E^*$ and $\alpha \in \real$ such that $\dpn{\cdot, \phi}{E} + \alpha \le f$.
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\item $f^* \ne \infty$.
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\item $(\phi, \gamma) \le f$.
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\item $\dpn{x, \phi}{\lambda} - \gamma = \alpha$.
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\end{enumerate}
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\end{proposition}
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\end{lemma}
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\begin{proof}
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(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
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(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
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\[
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\sup_{(y, \beta) \in \text{epi}(f)}\dpn{y, \phi}{E} + \mu \beta < \dpn{x, \phi}{E} + \mu \alpha
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\sup_{(y, \beta) \in \text{epi}(f)}\dpn{y, \phi}{E} - \mu \beta < \dpn{x, \phi}{E} - \mu \alpha
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\]
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Since for any $(y, \beta) \in A$, $\beta$ may be arbitrarily large by \autoref{lemma:closed-convex-epigraph}, $\mu < 0$. Thus for each $y \in E$,
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Since for any $(y, \beta) \in A$, $\beta$ may be arbitrarily small by \autoref{lemma:closed-convex-epigraph}, $\mu > 0$. Thus for each $y \in \bracs{f < \infty}$,
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\begin{align*}
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\dpn{x, \phi}{E} + \mu\alpha &> \dpn{y, \phi}{E} + \mu f(y) \\
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\dpn{x, \mu^{-1}\phi}{E} + \alpha & < \dpn{y, \mu^{-1}\phi}{E} + f(y) \\
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-\dpn{y, \mu^{-1}\phi}{E} + \dpn{x, \mu^{-1}\phi}{E} + \alpha &< f(y)
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\dpn{x, \phi}{E} - \mu\alpha &> \dpn{y, \phi}{E} - \mu f(y) \\
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-\dpn{y, \phi}{E} + \dpn{x, \phi}{E} - \mu\alpha &> -\mu f(y)\\
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\dpn{y, \mu^{-1}\phi}{E} - \dpn{x, \mu^{-1}\phi}{E} + \alpha &< f(y)
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\end{align*}
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so $-\dpn{\cdot, \mu^{-1}\phi}{E} + \dpn{x, \mu^{-1}\phi}{E} + \alpha \le f$.
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so $(\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{E} - \alpha) \le f$ and
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\[
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\dpn{x, \mu^{-1}\phi}{E} - \dpn{x, \mu^{-1}\phi}{E} + \alpha = \alpha
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\]
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\end{proof}
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\begin{theorem}[Fenchel-Moreau]
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\label{theorem:fenchel-moreau}
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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:
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\begin{enumerate}
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\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 $x \in E$,
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\item For each $x \in E$,
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\[
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f^{**}(x) = \sup\bracs{\dpn{x, \phi}{\lambda} - \alpha|(\phi, \alpha) \in F \times \real, (\phi, \alpha) \le f}
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\]
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@@ -188,23 +212,25 @@
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f^{**}(x) \ge \sup\bracs{\dpn{x, \phi}{\lambda} - \alpha| \phi \in F, \alpha \in \real, \dpn{\cdot, y}{\lambda} - \alpha \le f}
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\]
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(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
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(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 $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$.
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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
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\[
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\sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} + \mu \beta < \dpn{x, \phi}{\lambda} + \mu \alpha
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\sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} - \mu \beta < \dpn{x, \phi}{\lambda} - \mu \alpha
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\]
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Since for any $(y, \beta) \in A$, $\beta$ may be arbitrarily large by \autoref{lemma:closed-convex-epigraph}, $\mu \le 0$.
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Since for any $(y, \beta) \in A$, $\beta$ may be arbitrarily large by \autoref{lemma:closed-convex-epigraph}, $\mu \ge 0$.
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In the case that $\mu < 0$, for each $y \in \bracs{f < \infty}$,
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In the case that $\mu > 0$, for each $y \in \bracs{f < \infty}$,
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\begin{align*}
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\dpn{x, \phi}{\lambda} + \mu\alpha &> \dpn{y, \phi}{\lambda} + \mu f(y) \\
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\dpn{x, \mu^{-1}\phi}{\lambda} + \alpha & < \dpn{y, \mu^{-1}\phi}{\lambda} + f(y) \\
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-\dpn{y, \mu^{-1}\phi}{\lambda} + \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha &< f(y)
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\dpn{x, \phi}{\lambda} - \mu\alpha &> \dpn{y, \phi}{\lambda} - \mu f(y) \\
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-\dpn{y, \phi}{\lambda} + \dpn{x, \phi}{\lambda} - \mu\alpha &> - \mu f(y) \\
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\dpn{y, \mu^{-1}\phi}{\lambda} - \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha &< f(y)
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\end{align*}
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so $(-\mu^{-1}\phi, -\dpn{x, \mu^{-1}\phi}{\lambda} - \alpha) \le f$ and
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so $(\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{\lambda} - \alpha) \le f$ and
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\[
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f^{**}(x) \ge -\dpn{x, \mu^{-1}\phi}{\lambda} + \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha \ge \alpha
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f^{**}(x) \ge \dpn{x, \mu^{-1}\phi}{\lambda} - \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha \ge \alpha
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\]
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Given that $f^* \ne \infty$, there exists at least one pair $(\phi_0, \gamma_0) \in F \times \real$ such that $(\phi_0, \gamma_0) \le f$.
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@@ -226,7 +252,7 @@
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As the above holds for all $t > 0$, $f^{**}(x) = \infty \ge \alpha$. Since $f^{**}(x) \ge \alpha$ for all $(x, \alpha) \in E \times \real \setminus A$, $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$.
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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 $\text{epi}(f^{**}) \supset \ol{\text{Conv}}(\text{epi}(f))$.
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\end{proof}
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\begin{corollary}
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