# Banach-Alaoglu Theorem/Proof 3

## 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

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:

\(\ds B_{X^\ast}\) | \(=\) | \(\ds \set {f \in K : \lambda \map f x + \mu \map f y = \map f {\lambda x + \mu y} \text { for all } x, y \in X, \, \lambda, \mu \in \GF}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \set {f \in K : \map {\paren {\lambda \pr_x + \mu \pr_y - \pr_{\lambda x + \mu y} } } f = 0 \text { for all } x, y \in X, \, \lambda, \mu \in \GF}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \bigcap_{x, y \in X, \, \lambda, \mu \in \GF} \paren {\lambda \pr_x + \mu \pr_y - \pr_{\lambda x + \mu y} }^{-1} \sqbrk {\set 0}\) |

From Linear Combination of Continuous Functions valued in Topological Vector Space is Continuous:

- $\lambda \pr_x + \mu \pr_y - \pr_{\lambda x + \mu y}$ 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.

$\blacksquare$