Alaoglu's Theorem

Theorem
The closed unit ball of the dual of a normed space is compact with respect to the weak* topology.

Proof
Let $X$ be a normed vector space.

Denote by $B$ the closed unit ball in $X$.

Let $X^*$ be the dual of $X$.

Denote by $B^*$ the closed unit ball in $X^*$.

Let $\mathcal F \left({B}\right) = \left[{-1 \,.\,.\, 1}\right]^B$ be the topological space of functions from $B$ to $\left[{-1 \,.\,.\, 1}\right]$.

By Tychonoff's Theorem, $\mathcal F \left({B}\right)$ is compact with respect to the product topology.

We define the restriction map:
 * $R: B^*\to \mathcal F \left({B}\right)$

by $R \left({\psi}\right) = \psi \restriction_B$.

Lemma 1
$R \left({B^*}\right)$ is a closed subset of $\mathcal F \left({B}\right)$.

Lemma 2
$R$ is a homeomorphism from $B^*$ with the weak* topology to its image $R \left({B^*}\right)$ seen as a subset of $\mathcal F \left({B}\right)$ with the product topology.

Proof
Thus by lemma 2, $B^*$ in the weak* topology is homeomorphic with $R \left({B^*}\right)$.

This is a closed subset of $\mathcal F \left({B}\right)$ (by lemma 1) and thus compact.