Theorem 18.3.8 (Fenchel-Moreau).label 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:
- (1)
For each $x \in E$,
\[f^{**}(x) = \sup\bracs{\dpn{x, \phi}{\lambda} - \alpha|(\phi, \alpha) \in F \times \real, (\phi, \alpha) \le f}\] - (2)
$\text{epi}(f^{**})$ is the $\sigma(E \times \real, F \times \real)$-closed convex hull of $\text{epi}(f)$.
- (3)
$f^{**}$ is the greatest convex and $\sigma(E, F)$-lower semicontinuous function bounded above by $f$.
- (4)
$f = f^{**}$ if and only if $f$ is convex and $\sigma(E, F)$-lower semicontinuous.
Proof. (1): Let $\phi \in F$ such that $f^{*}(\phi) < \infty$, then for each $x \in E$,
so $\dpn{\cdot, \phi}{\lambda}- f^{*}(\phi) \le f$, and
On the other hand, let $\phi \in F$ and $\alpha \in \real$ such that $(\phi, \alpha) \le f$, then $f^{*}(\phi) \le \alpha$, and
Therefore
(2): By Lemma 18.3.4, $f^{**}$ 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,
To this end, let $A = \ol{\text{Conv}}(\text{epi}(f))$ and $(x, \alpha) \in E \times \real \setminus A$. By the Hahn-Banach Theorem, there exists $\phi \in F$, $\mu \in \real$, and $\alpha_{0} \in (\alpha, \infty)$ such that
For any $(y, \beta) \in A$, $\beta$ may be arbitrarily large by Lemma 18.3.6, so $\mu \ge 0$.
In the case that $\mu > 0$, for each $y \in \bracs{f < \infty}$,
so $(\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{\lambda}- \alpha_{0}) \le f$ and
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
For each $t > 0$, let $\Phi_{t} = \phi_{0} + t\phi$ and $\Gamma_{t} = \gamma_{0} + t\gamma$, then for each $y \in \bracs{f < \infty}$,
so $(\Phi_{t}, \Gamma_{t}) \le f$. By (1),
As the above holds for all $t > 0$, $f^{**}(x) = \infty > \alpha$.
Thus $f^{**}(x) > \alpha$ and $(x, \alpha) \not\in \text{epi}(f^{**})$ for all $(x, \alpha) \in E \times \real \setminus A$. Therefore
and $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$.$\square$
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