From 32ba2483ad8c1035c7e94312cbf75dda1041c30f Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Wed, 24 Jun 2026 18:29:26 -0400 Subject: [PATCH] Added the Legendre transform. --- src/fa/convex/legendre.tex | 228 +++++++++++++++++++++++++++ src/fa/index.tex | 2 +- src/fa/notation.tex | 3 + src/topology/main/semicontinuity.tex | 4 +- 4 files changed, 234 insertions(+), 3 deletions(-) create mode 100644 src/fa/convex/legendre.tex diff --git a/src/fa/convex/legendre.tex b/src/fa/convex/legendre.tex new file mode 100644 index 0000000..02cd050 --- /dev/null +++ b/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} + + + diff --git a/src/fa/index.tex b/src/fa/index.tex index 05a0479..000c810 100644 --- a/src/fa/index.tex +++ b/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} diff --git a/src/fa/notation.tex b/src/fa/notation.tex index aee88af..eaa72bb 100644 --- a/src/fa/notation.tex +++ b/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} \\ diff --git a/src/topology/main/semicontinuity.tex b/src/topology/main/semicontinuity.tex index fa0178a..3c4c5a7 100644 --- a/src/topology/main/semicontinuity.tex +++ b/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}