# Banach-Alaoglu Theorem

## Theorem

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.

## 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 $\closedint {-1} 1$.

$\map \FF B$

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

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

$\Box$

### Lemma 4

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

$R \sqbrk {B^*}$

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 $R \sqbrk {B^*}$.

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

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

## Proof 3

Let $B_{X^\ast}$ be the closed unit ball in $X^\ast$.

Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.

From the definition of the norm of a bounded linear functional, we have:

$B_{X^\ast} = \set {f : X \to \GF : \cmod {\map f x} \le \norm x \text { and } f \text { is linear} }$

For each $x \in X$, let:

$K_x = \set {\lambda \in \GF : \cmod \lambda \le \norm x}$

Let:

$\ds K = \prod_{x \mathop \in X} K_x$

equipped with the product topology.

Then if $f \in K$ and $x \in X$ we have $\map f x \in K_x$ if and only if $\cmod {\map f x} \le \norm x$.

Then:

$B_{X^\ast} = \set {f \in K : f \text { is linear} }$

Let $\pr_x : K \to \GF$ denote the projection onto the $x$th factor, that is:

$\map {\pr_x} f = \map f x$

for each $f \in K$.

From the definition of the product topology, the product topology on $K$ is the initial topology induced by $\set {\pr_x : x \in X}$.

From Subspace Topology on Initial Topology is Initial Topology on Restrictions, the subspace topology on $B_{X^\ast}$ inherited by $K$ is the initial topology on $B_{X^\ast}$ induced by $\set {\pr_x \restriction_{B_{X^\ast} } : x \in X}$.

Applying Subspace Topology on Initial Topology is Initial Topology on Restrictions to $\struct {X^\ast, w^\ast}$, this is precisely the subspace topology on $B_{X^\ast}$ inherited by the weak-$\ast$ topology on $X^\ast$.

From Tychonoff's Theorem, $K$ is compact.

From Closed Subspace of Compact Space is Compact, it suffices to show that $B_{X^\ast}$ is closed in $\struct {K$.

We may now write:

 $\ds B_{X^\ast}$ $=$ $\ds \set {f \in K : \lambda \map f x + \mu \map f y = \map f {\lambda x + \mu y} \text { for all } x, y \in X, \, \lambda, \mu \in \GF}$ $\ds$ $=$ $\ds \set {f \in K : \map {\paren {\lambda \pr_x + \mu \pr_y - \pr_{\lambda x + \mu y} } } f = 0 \text { for all } x, y \in X, \, \lambda, \mu \in \GF}$ $\ds$ $=$ $\ds \bigcap_{x, y \in X, \, \lambda, \mu \in \GF} \paren {\lambda \pr_x + \mu \pr_y - \pr_{\lambda x + \mu y} }^{-1} \sqbrk {\set 0}$
$\lambda \pr_x + \mu \pr_y - \pr_{\lambda x + \mu y}$ is continuous for each $x, y \in X$ and $\lambda, \mu \in \GF$.

From Continuity Defined from Closed Sets, we have that:

$\paren {\lambda \pr_x + \mu \pr_y - \pr_{\lambda x + \mu y} }^{-1} \sqbrk {\set 0}$ is closed in $K$.

Since the intersection of closed sets is closed, we have that:

$\ds \bigcap_{x, y \in X, \, \lambda, \mu \in \GF} \paren {\lambda \pr_x + \mu \pr_y - \pr_{\lambda x + \mu y} }^{-1} \sqbrk {\set 0}$ is closed in $K$.

So $\struct {B_{X^\ast}, w^\ast}$ is closed in $K$.

From Closed Subspace of Compact Space is Compact, $\struct {B_{X^\ast}, w^\ast}$ is 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.