From 83072f4ba4628128bf9b65351fbe27826acc51cd Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Thu, 25 Jun 2026 13:03:51 -0400 Subject: [PATCH] Adjusted the Legendre sectoin. --- src/fa/convex/legendre.tex | 84 +++++++++++++++++++++++++------------- src/fa/notation.tex | 1 + 2 files changed, 56 insertions(+), 29 deletions(-) diff --git a/src/fa/convex/legendre.tex b/src/fa/convex/legendre.tex index 6cd6a79..4709822 100644 --- a/src/fa/convex/legendre.tex +++ b/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} @@ -126,38 +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 $x \in E$, \[ f^{**}(x) = \sup\bracs{\dpn{x, \phi}{\lambda} - \alpha|(\phi, \alpha) \in F \times \real, (\phi, \alpha) \le f} \] @@ -188,23 +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 $y \in \bracs{f < \infty}$, + 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 $(\phi_0, \gamma_0) \le f$. @@ -226,7 +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} diff --git a/src/fa/notation.tex b/src/fa/notation.tex index 2719d97..cea2534 100644 --- a/src/fa/notation.tex +++ b/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} \\