Corollary 18.3.9.label 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 $x \in E$,
\[f(x) = \sup_{n \in \natp}\dpn{x, \phi_n}{\lambda}- \alpha_{n}\]
Proof. For each $(\phi, \alpha) \in F \times \real$, denote $(\phi, \alpha) \le f$ if $\dpn{\cdot, \phi}{\lambda}- \alpha \le f$. By the Fenchel-Moreau Theorem,
\[f(x) = f^{**}(x) = \sup\bracs{\dpn{x, \phi}{\lambda} - \alpha|(\phi, \alpha) \in F \times \real, (\phi, \alpha) \le f}\]
for all $x \in E$. By Proposition 13.4.1,
\[S = \bracs{(\phi, \alpha) \in F \times \real| (\phi, \alpha) \le f}\]
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 $x \in E$,
\[f(x) = \sup_{n \in \natp}\dpn{x, \phi_n}{\lambda}- \alpha_{n}\]
$\square$
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