Definition 18.3.2 (Conjugate Function).label Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f: E \to (-\infty, \infty]$ with $f \ne \infty$, then for each $\phi \in F$,
\[\sup_{x \in E}\dpn{x, \phi}{\lambda}- f(x) = \inf\bracsn{\alpha \in \real| (\phi, \alpha) \le f}\]
The mapping
\[f^{*}: F \to (-\infty, \infty] \quad \phi \mapsto \sup_{x \in E}\dpn{x, \phi}{\lambda}- f(x)\]
is the conjugate function of $f$ with respect to the duality $\dpn{E, F}{\lambda}$.
Proof. 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 $x \in E$,
\[\dpn{x, \phi}{\lambda}- f(x) \le \dpn{x, \phi}{\lambda}- \dpn{x, \phi}{\lambda}+ \alpha = \alpha\]
so
\[\sup_{x \in E}\dpn{x, \phi}{\lambda}- f(x) \le \inf\bracsn{\alpha \in \real| (\phi, \alpha) \le f}\]
On the other hand, suppose that $\alpha = \sup_{x \in E}\dpn{x, \phi}{\lambda}- f(x) < \infty$, then
\[\dpn{x, \phi}{\lambda}- \alpha \le f(x)\]
for all $x \in E$. Therefore
\[\sup_{x \in E}\dpn{x, \phi}{\lambda}- f(x) = \inf\bracsn{\alpha \in \real| (\phi, \alpha) \le f}\]
$\square$
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