Lemma 11.3.3.label Let $E$ be a locally convex space over $\real$, $K \subset E$ be non-empty and compact, and $\phi \in E^{*}$. Let $\alpha = \sup\bracs{\dpn{x, \phi}{E}|x \in K}$, then $A = \bracs{\phi = \alpha}\cap K$ is a non-empty extreme subset of $K$.
Proof. Since $K$ is compact, $\alpha < \infty$ and $A$ is non-empty by Proposition 5.16.2. Let $x \in A$ and $y, z \in K$ such that $x \in (y, z)$. By definition of $\alpha$, $\dpn{y, \phi}{E}= \dpn{z, \phi}{E}= \alpha$. Thus $y, z \in A$ as well.$\square$
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