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 normed dual $X^*$ is weak* sequentially compact.

There is believed to be a mistake here, possibly a typo. In particular: This is surely wrong if $X$ is a vector space over $\Q$ or some unusual division ring. You can help ProofWiki by reviewing it, and either correcting it or adding some explanatory material as to why you believe it is actually correct after all. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Mistake}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

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:

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

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.