From 81e4ef6ffebcdcc4015af1d8b51b5d001f437de0 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Wed, 24 Jun 2026 21:44:58 -0400 Subject: [PATCH] Updated notation and formulation in the Fenchel Moreau theorem. --- src/fa/convex/legendre.tex | 95 ++++++++++++++++++-------------------- 1 file changed, 45 insertions(+), 50 deletions(-) diff --git a/src/fa/convex/legendre.tex b/src/fa/convex/legendre.tex index 536a694..f1ba5cf 100644 --- a/src/fa/convex/legendre.tex +++ b/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, 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) + \dpn{x, \phi}{\lambda} - f^*(\phi) &= \dpn{x, \phi}{\lambda} - \braks{\sup_{z \in E}\dpn{z, \phi}{\lambda} - f(z)} \\ + &= \dpn{x, \phi}{\lambda} + \braks{\inf_{z \in E}f(z) - \dpn{z, \phi}{\lambda}} \\ + &\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^*(y) &= \dpn{x, y}{\lambda} - f(x) + f(x) + f^*(\phi) &= \dpn{x, \phi}{\lambda} \\ + f^*(\phi) &= \dpn{x, \phi}{\lambda} - f(x) \end{align*} if and only if for every $h \in E$, \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} + \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 $y \in \partial f(x)$. @@ -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) \\ - &\le \sup\bracs{\dpn{x, y}{\lambda} - \alpha| y \in F, \alpha \in \real, \dpn{\cdot, y}{\lambda} - \alpha \le f} + 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, \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, \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}{\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) \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 $(x, \alpha) \in E \times \real \setminus A$, $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$.