Lemma 18.3.7 (Almost Subgradient).label 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:
- (1)
$(\phi, \gamma) \le f$.
- (2)
$\dpn{x, \phi}{\lambda}- \gamma = \alpha$.
In particular, $f^{*}\ne \infty$.
Proof. (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 Hahn-Banach Theorem, there exists $\phi \in F$ and $\mu \in \real$ such that
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 $y \in \bracs{f < \infty}$,
so $(\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{E}- \alpha) \le f$ and
$\square$
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