Lemma 18.3.4.label Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f, g: E \to (-\infty, \infty]$ with $f, g \ne \infty$, then
- (1)
$f^{*}$ is convex and lower semicontinuous.
- (2)
If $f \le g$, then $f^{*} \ge g^{*}$.
- (3)
If $f^{*} \ne \infty$, then $f^{**}\le f$.
Proof. (1): By Proposition 18.1.6 and Proposition 5.22.3.
(3): For each $x \in E$ and $\phi \in F$,
\begin{align*}\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 $\phi \in F$, $f^{**}\le f$.$\square$
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