Banach-Alaoglu Theorem/Proof 2

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Theorem

Let $X$ be a separable normed vector space.

Then the closed unit ball in its dual $X^*$ is weak* sequentially compact.


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 $\map \FF B = \closedint {-1} 1^B$ be the topological space of functions from $B$ to $\closedint {-1} 1$.

By Tychonoff's Theorem, $\map \FF B$ is compact with respect to the product topology.

We define the restriction map:

$R: B^* \to \map \FF B$

by:

$\map R \psi = \psi \restriction_B$


Lemma 3

$\map R {B^*}$ is a closed subset of $\map \FF B$.

$\Box$


Lemma 4

$R$ is a homeomorphism from $B^*$ with the weak* topology to its image $\map R {B^*}$ seen as a subset of $\map \FF B$ with the product topology.

$\Box$


Thus by Lemma 4, $B^*$ in the weak* topology is homeomorphic with $\map R {B^*}$.

This is a closed subset of $\map \FF B$ (by Lemma 3) and thus compact.

$\blacksquare$


Also known as

The Banach-Alaoglu Theorem is also known just as Alaoglu's Theorem.


Source of Name

This entry was named for Stefan Banach and Leonidas Alaoglu.