# Banach-Alaoglu Theorem

(Redirected from Alaoglu's Theorem)

## Theorem

Let $X$ be a separable normed vector space.

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

## Proof 1

The aim of this proof is to show the following:

Given a bounded sequence in $X^*$, there exists a weakly convergent subsequence of that bounded sequence.

Let $\sequence {l_n}_{n \mathop \in \N}$ be a bounded sequence in $X^*$.

Let $\sequence {x_n}_{n \mathop \in \N}$ be a countable dense subset of $X$.

Choose subsequences $\N \supset \Lambda_1 \supset \Lambda_2 \supset \ldots$ such that:

$\forall j \in \N: \map {l_k} {x_j} \to a_j =: \map l {x_j}$

as $k \to \infty$, $k \in \Lambda_j$.

Let $\Lambda$ be the diagonal sequence.

### Lemma 1

$l$ can be extended to an element of $X^*$.

$\Box$

### Lemma 2

$l_k \stackrel {\omega^*} {\to} l$ as $k \to \infty$, $k \in \Lambda$.

$\Box$

## Proof 2

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.