Alaoglu's Theorem

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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 $\mathcal F \left({B}\right)$.


Proof


$\Box$


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


$\Box$


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.

$\blacksquare$


Source of Name

This entry was named for Leonidas Alaoglu.


Sources