Theorem 11.14.7 (Banach-Alaoglu).label Let $E$ be a locally convex space over $K \in \RC$ and $\alg \subset E^{*}$ be equicontinuous, then $\alg$ is relatively compact with respect to $\sigma(E^{*}, E)$.
Proof. For each $x \in E$, $\alg(x) = \bracsn{\dpn{x, \phi}{E}|\phi \in \alg}$ is relatively compact by Proposition 11.14.2. By the Arzelà-Ascoli Theorem,
- (C2)
The $\sigma(E^{*}, E)$-closure of $\alg$ in $\prod_{x \in E}K$ is equicontinuous.
- (C3)
The $\sigma(E^{*}, E)$-closure of $\alg$ in $\prod_{x \in E}K$ is compact.
By Proposition 11.13.8, the $\sigma(E^{*}, E)$-closure of $\alg$ in $\prod_{x \in E}\ol{\alg(x)}$ is a subset of $\hom(E; K)$. Hence the $\sigma(E^{*}, E)$-closure of $\alg$ in $E^{*}$ is compact.$\square$
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