295 lines
14 KiB
TeX
295 lines
14 KiB
TeX
\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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\[
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\sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| (\phi, \alpha) \le f}
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\]
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The mapping
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\[
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f^*: F \to (-\infty, \infty] \quad \phi \mapsto \sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x)
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\]
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is the \textbf{conjugate function} of $f$ with respect to the duality $\dpn{E, F}{\lambda}$.
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\end{definition}
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\begin{proof}
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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 $x \in E$,
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\[
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\dpn{x, \phi}{\lambda} - f(x) \le \dpn{x, \phi}{\lambda} - \dpn{x, \phi}{\lambda} + \alpha = \alpha
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\]
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so
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\[
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\sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) \le \inf\bracsn{\alpha \in \real| (\phi, \alpha) \le f}
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\]
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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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\[
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\dpn{x, \phi}{\lambda} - \alpha \le f(x)
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\]
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for all $x \in E$. Therefore
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\[
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\sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) = \inf\bracsn{\alpha \in \real| (\phi, \alpha) \le f}
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\]
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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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Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f, g: E \to (-\infty, \infty]$ with $f, g \ne \infty$, then
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\begin{enumerate}
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\item $f^*$ is convex and lower semicontinuous.
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\item If $f \le g$, then $f^* \ge g^*$.
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\item If $f^* \ne \infty$, then $f^{**} \le f$.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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(1): By \autoref{proposition:convex-extension} and \autoref{proposition:semicontinuous-properties}.
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(3): For each $x \in E$ and $\phi \in F$,
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\begin{align*}
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\dpn{x, \phi}{\lambda} - f^*(\phi) &= \dpn{x, \phi}{\lambda} - \braks{\sup_{z \in E}\dpn{z, \phi}{\lambda} - f(z)} \\
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&= \dpn{x, \phi}{\lambda} + \braks{\inf_{z \in E}f(z) - \dpn{z, \phi}{\lambda}} \\
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&\le \dpn{x, \phi}{\lambda} + f(x) - \dpn{x, \phi}{\lambda} = f(x)
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\end{align*}
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As the above holds for all $\phi \in F$, $f^{**} \le f$.
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\end{proof}
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\begin{theorem}[Fenchel's Inequality]
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\label{theorem:fenchel-inequality}
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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 any $x \in E$ and $\phi \in F$,
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\[
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\dpn{x, \phi}{\lambda} \le f(x) + f^*(\phi)
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\]
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with equality if and only if $\phi \in \partial f(x)$.
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\end{theorem}
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\begin{proof}
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Let $x \in E$ and $\phi \in F$. Assume without loss of generality that $f(x) < \infty$, then
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\[
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f(x) + f^*(\phi) \ge f(x) + \dpn{x, \phi}{\lambda} - f(x) = \dpn{x, \phi}{\lambda}
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\]
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Now let $x \in E$ and $\phi \in F$ such that $\dpn{x, \phi}{\lambda} = f(x) + f^*(\phi)$, then $f(x), f^*(\phi) < \infty$ and
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\begin{align*}
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f(x) + f^*(\phi) &= \dpn{x, \phi}{\lambda} \\
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f^*(\phi) &= \dpn{x, \phi}{\lambda} - f(x)
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\end{align*}
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By definition,
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\[
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\dpn{x, \phi}{\lambda} - f(x) = f^*(\phi) = \sup_{h \in E}\dpn{x+h, \phi}{\lambda} - f(x+h)
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\]
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so for every $h \in E$,
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\begin{align*}
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\dpn{x, \phi}{\lambda} - f(x) &\ge \dpn{x+h, \phi}{\lambda} - f(x+h) \\
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f(x + h) - f(x) &\ge \dpn{h, \phi}{\lambda}
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\end{align*}
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Thus $\phi$ satisfies the subgradient inequality, and $\phi \in \partial f(x)$.
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On the other hand, if $f(x) < \infty$ and $\phi \in \partial f(x)$, then $f(x + h) - f(x) \ge \dpn{h, \phi}{\lambda}$ for all $h \in E$, and
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\begin{align*}
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f^*(\phi) &= \sup_{h \in E}\dpn{x+h, \phi}{\lambda} - f(x+h) = \dpn{x, \phi}{\lambda} - f(x) \\
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f^*(\phi) + f(x) &= \dpn{x, \phi}{\lambda}
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\end{align*}
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\end{proof}
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\begin{lemma}
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\label{lemma:closed-convex-epigraph}
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Let $E$ be a locally convex space over $\real$ and $f: E \to (-\infty, \infty]$ with $f \ne \infty$, then for each $(x, \alpha) \in \ol{\text{Conv}}(\text{epi}(f))$, $\bracs{x} \times [\alpha, \infty) \subset \ol{\text{Conv}}(\text{epi}(f))$.
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\end{lemma}
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\begin{proof}
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Let
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\[
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A = \bracsn{(\phi, \alpha) \in \ol{\text{Conv}}(\text{epi}(f))| \bracs{x} \times [\alpha, \infty) \subset \ol{\text{Conv}}(\text{epi}(f))}
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\]
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For each $(x, \alpha), (y, \beta) \in A$, $t \in [0, 1]$, 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,
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\[
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((1 - t)x + ty, \gamma) = ((1 - t)x + ty, (1 - t)\alpha' + t\beta') \in \ol{\text{Conv}}(\text{epi}(f))
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\]
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so $\bracs{(1 - t)x + ty} \times [\gamma, \infty) \subset \ol{\text{Conv}}(\text{epi}(f))$, $(1 - t)(x + \alpha) + t(y, \beta) \in A$, and $A$ is convex.
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Let $(x, \alpha) \in \ol A$, then there exists a net $\langle (x_\gamma, \alpha_\gamma) \rangle_{\gamma \in C} \subset A$ with $(x_\gamma, \alpha_\gamma) \to (x, \alpha)$. In which case, for each $r > 0$, $\langle (x_\gamma, \alpha_\gamma + r) \rangle_{\gamma \in C} \subset A$ and $(x_\gamma, \alpha_\gamma + r) \to (x, \alpha + r)$, so $(x, \alpha + r) \in \ol{\text{Conv}}(\text{epi}(f))$ and
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\[
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\bracs{x} \times [\alpha, \infty) \subset \ol{\text{Conv}}(\text{epi}(f))
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\]
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Thus $(x, \alpha) \in A$ and $A$ is closed.
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Since $A$ is a closed convex set containing $\text{epi}(f)$, $A = \ol{\text{Conv}}(\text{epi}(f))$.
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\end{proof}
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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 $(\phi, \gamma) \le f$.
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\item $\dpn{x, \phi}{\lambda} - \gamma = \alpha$.
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\end{enumerate}
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In particular, $f^{*} \ne \infty$.
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\end{lemma}
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\begin{proof}
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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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\[
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\sup_{(y, \beta) \in \text{epi}(f)}\dpn{y, \phi}{E} - \mu \beta < \dpn{x, \phi}{E} - \mu \alpha
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\]
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For any $(y, \beta) \in \text{epi}(f)$, $\beta$ may be arbitrarily large, $\mu \ge 0$. In particular, since $(x, f(x)) \in \text{epi}(f)$, the strict inequality implies that $\mu > 0$. 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{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 $(\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{E} - \alpha) \le f$ 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 $x \in E$,
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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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\item $\text{epi}(f^{**})$ is the $\sigma(E \times \real, F \times \real)$-closed convex hull of $\text{epi}(f)$.
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\item $f^{**}$ is the greatest convex and $\sigma(E, F)$-lower semicontinuous function bounded above by $f$.
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\item $f = f^{**}$ if and only if $f$ is convex and $\sigma(E, F)$-lower semicontinuous.
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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(1): Let $\phi \in F$ such that $f^*(\phi) < \infty$, then for each $x \in E$,
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\[
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\dpn{x, \phi}{\lambda} - f^*(\phi) \le f(x)
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\]
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so $\dpn{\cdot, \phi}{\lambda} - f^*(\phi) \le f$, and
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\begin{align*}
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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) \\
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&\le \sup\bracs{\dpn{x, \phi}{\lambda} - \alpha|(\phi, \alpha) \in F \times \real, (\phi, \alpha) \le f}
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\end{align*}
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On the other hand, let $\phi \in F$ and $\alpha \in \real$ such that $(\phi, \alpha) \le f$, then $f^*(\phi) \le \alpha$, and
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\[
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f^{**}(x) \ge \dpn{x, \phi}{\lambda} - f^*(\phi) \ge \dpn{x, \phi}{\lambda} - \alpha
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\]
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Therefore
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\[
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f^{**}(x) \ge \sup\bracs{\dpn{x, \phi}{\lambda} - \alpha| \phi \in F, \alpha \in \real, \dpn{\cdot, y}{\lambda} - \alpha \le f}
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\]
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(2): 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 $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$, or equivalently,
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\[
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E \times \real \setminus \ol{\text{Conv}}(\text{epi}(f)) \subset E \times \real \setminus \text{epi}(f)
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\]
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To this end, 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 $\phi \in F$, $\mu \in \real$, and $\alpha_0 \in (\alpha, \infty)$ such that
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\[
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\sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} - \mu \beta \le \dpn{x, \phi}{\lambda} - \mu \alpha_0 \le \dpn{x, \phi}{\lambda} - \mu \alpha
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\]
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For any $(y, \beta) \in A$, $\beta$ may be arbitrarily large by \autoref{lemma:closed-convex-epigraph}, so $\mu \ge 0$.
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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_0 &\ge \dpn{y, \phi}{\lambda} - \mu f(y) \\
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-\dpn{y, \phi}{\lambda} + \dpn{x, \phi}{\lambda} - \mu\alpha_0 &\le - \mu f(y) \\
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\dpn{y, \mu^{-1}\phi}{\lambda} - \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha_0 &\le f(y)
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\end{align*}
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so $(\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{\lambda} - \alpha_0) \le f$ and
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\[
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f^{**}(x) \ge \dpn{x, \mu^{-1}\phi}{\lambda} - \dpn{x, \mu^{-1}\phi}{\lambda} + \alpha_0 > \alpha
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\]
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Now suppose that $\mu = 0$. Given that $f^* \ne \infty$, there exists at least one pair $(\phi_0, \gamma_0) \in F \times \real$ such that $(\phi_0, \gamma_0) \le f$. Let
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\[
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\gamma = \sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} < \dpn{x, \phi}{\lambda}
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\]
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For each $t > 0$, let $\Phi_t = \phi_0 + t\phi$ and $\Gamma_t = \gamma_0 + t\gamma$, then for each $y \in \bracs{f < \infty}$,
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\[
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\dpn{y, \Phi_t}{\lambda} - \Gamma_t \le f(y) + t\underbrace{(\dpn{y, \phi}{\lambda} - \gamma)}_{\le 0} \le f(y)
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\]
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so $(\Phi_t, \Gamma_t) \le f$. By (1),
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\[
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f^{**}(x) \ge \dpn{x, \Phi_t}{\lambda} - \Gamma_t = \dpn{x, \phi_0}{\lambda} - \gamma_0 + t\underbrace{(\dpn{x, \phi}{\lambda} - \gamma)}_{> 0}
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\]
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As the above holds for all $t > 0$, $f^{**}(x) = \infty > \alpha$.
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Thus $f^{**}(x) > \alpha$ and $(x, \alpha) \not\in \text{epi}(f^{**})$ for all $(x, \alpha) \in E \times \real \setminus A$. Therefore
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\[
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E \times \real \setminus A \subset E \times \real \setminus \text{epi}(f^{**})
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\]
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and $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$.
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\end{proof}
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\begin{corollary}
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\label{corollary:separable-legendre}
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Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f: E \to (-\infty, \infty]$ with $f \ne \infty$ be convex and lower semicontinuous, then there exists $\seq{(\phi_n, \alpha_n)} \subset F \times \real$ such that for each $x \in E$,
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\[
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f(x) = \sup_{n \in \natp} \dpn{x, \phi_n}{\lambda} - \alpha_n
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\]
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\end{corollary}
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\begin{proof}
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For each $(\phi, \alpha) \in F \times \real$, denote $(\phi, \alpha) \le f$ if $\dpn{\cdot, \phi}{\lambda} - \alpha \le f$. By the \hyperref[Fenchel-Moreau Theorem]{theorem:fenchel-moreau},
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\[
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f(x) = f^{**}(x) = \sup\bracs{\dpn{x, \phi}{\lambda} - \alpha|(\phi, \alpha) \in F \times \real, (\phi, \alpha) \le f}
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\]
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for all $x \in E$. By \autoref{proposition:separable-dual},
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\[
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S = \bracs{(\phi, \alpha) \in F \times \real| (\phi, \alpha) \le f}
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\]
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is separable with respect to $\sigma(F \times \real, E \times \real)$. Therefore there exists $\seq{(\phi_n, \alpha_n)} \subset S$ such that for each $x \in E$,
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\[
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f(x) = \sup_{n \in \natp} \dpn{x, \phi_n}{\lambda} - \alpha_n
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\]
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\end{proof}
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