Banach-Alaoglu Theorem/Proof 2

From ProofWiki
Jump to navigation Jump to search


Let $X$ be a separable normed vector space.

Then the closed unit ball in its normed dual $X^*$ is sequentially compact with respect to the weak-$\ast$ topology.


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 $X^*$.


$\map \FF B = \closedint {-1} 1^B$

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:

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


$\forall \psi \in B^*: \map R \psi = \psi \restriction_B$

Lemma 3

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


Lemma 4

$R$ is a homeomorphism from $B^*$ with the weak* topology to its image:

$R \sqbrk {B^*}$

seen as a subset of $\map \FF B$ with the product topology.


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

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

By the Eberlein-Šmulian Theorem, this is sequentially compact.

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.