Lemma 18.4.3.label Let $\dpn{E, F}{\lambda}$ be a duality over $\real$, then:
- (1)
For any non-empty $A \subset E$, let $\ol{\conv}(A)$ be the $\sigma(E, F)$-closed convex hull of $A$, then $H_{A} = H_{\ol{\conv}(A)}$.
- (2)
For any non-empty $A, B \subset E$, $H_{A} \le H_{B}$ if and only if $A$ is contained in the $\sigma(E, F)$-closed convex hull of $B$.
- (3)
For any non-empty $A \subset E$, $I_{A}^{*} = H_{A}$.
- (4)
For any $A \subset E$ non-empty, $\sigma(E, F)$-closed, and convex, $H_{A}^{*} = I_{A}$.
Proof. (1): Since $\ol{\conv}(A) \supset A$, $H_{A} \le H_{\ol{\conv}(A)}$. On the other hand, for each $\phi \in F$, $A \subset \bracs{\phi \le H_A(\phi)}$. Since $\bracs{\phi \le H_A(\phi)}$ is a $\sigma(E, F)$-closed convex set, it contains $\ol{\conv}(A)$. Thus $\ol{\conv}(A) \subset \bracs{\phi \le H_A(\phi)}$ and $H_{\ol{\conv}(A)}(\phi) \le H_{A}(\phi)$.
(2): Using (1), assume without loss of generality that $B$ is $\sigma(E, F)$-closed and convex. Suppose that $H_{A} \le H_{B}$, then $A \subset \bigcap_{\phi \in F}\bracs{\phi \le H_B(\phi)}$. By the Hahn-Banach Theorem, $B = \bigcap_{\phi \in F}\bracs{\phi \le H_B(\phi)}$, so $A \subset B$.
(3): Let $\phi \in F$, then since $I_{A}|_{A^c}= \infty$,
(4): Given that $A$ is $\sigma(E, F)$-closed and convex, $I_{S}$ is convex and lower semicontinuous. By the Fenchel-Moreau Theorem and (3), $I_{A} = I_{A}^{**}= H_{A}^{*}$.$\square$
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