Theorem 10.1.6 (Alaoglu’s Theorem). Let $E$ be a normed vector space over $K \in \RC$, then $B^{*} = \bracsn{\phi \in E^*| \norm{x}_{E^*} \le 1}$ is compact in the weak*-topology.
Proof. For each $x \in E$, $I_{x} = \bracsn{\dpn{x, \phi}{E}|\phi \in B^*}$ is compact. By Proposition 8.11.10, the closure of $B^{*}$ in $\prod_{x \in E}I_{x}$ is a subset of $\hom(E; K)$. Since $B^{*}$ is bounded, $I_{x} \subset \overline{B_K(0, 1)}$ for all $x \in E$, so the closure of $B^{*}$ is contained in $B^{*}$. By Theorem 4.16.6, $\prod_{x \in E}I_{x}$ is compact. Therefore $B^{*}$ is compact with respect to the weak*-topology.$\square$