# Alaoglu's Theorem

## Theorem

The closed unit ball of the dual of a normed space is compact with respect to the weak* topology.

This has to be rewritten. In particular: Rephrase to match usual presentation |

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

By Tychonoff's Theorem, $\map {\mathcal F} B$ is compact with respect to the product topology.

We define the restriction map:

- $R: B^*\to \map {\mathcal F} B$

by $\map R \psi = \psi \restriction_B$.

### Lemma 1

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

### Proof

$\Box$

### Lemma 2

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

### Proof

$\Box$

Thus by lemma 2, $B^*$ in the weak* topology is homeomorphic with $\map R {B^*}$.

This is a closed subset of $\map {\mathcal F} B$ (by lemma 1) and thus compact.

$\blacksquare$

## Source of Name

This entry was named for Leonidas Alaoglu.

## Sources

- 1989: Ephraim J. Borowski and Jonathan M. Borwein:
*Dictionary of Mathematics*... (previous) ... (next): Entry:**Alaoglu's theorem** - 2010: H.L. Royden and P.M. Fitzpatrick:
*Real Analysis*(4th ed.): $\S 15.1$