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.