Banach-Alaoglu Theorem/Lemma 4

Lemma for Banach-Alaoglu Theorem
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:
 * $\map R \psi = \psi \restriction_B$

$R$ 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.

Proof
Firstly, $R$ is an injection.

Indeed, let $\psi_1, \psi_2 \in B^\ast$ such that $\map R {\psi_1} = \map R {\psi_2}$.

Then, for all $x \in X \setminus \set 0$ we have:

In addition, $\map {\psi_1} 0 = \map {\psi_2} 0 = 0 $.

Thus, $\psi_1 = \psi_2$.

Second, we show that $R^{-1}$ is continuous.

Recall and

Let $\psi_0 \in B^\ast$, $x \in X$, and $r \in \R_{>0}$ be arbitrary.

Let
 * $\map N {\psi_0 ; x ; r } := \set { \psi \in B^\ast : \size {\map \psi x - \map {\psi_0} x } < r } $

Now, it suffices to check that $R \sqbrk { \map N {\psi_0 ; x ; r } }$ is open in $\map \FF B$.

This can be seen as:

where:
 * $\map {I_r} {\map {\psi_0} x} := \openint {\map {\psi_0} x - r} {\map {\psi_0} x + r} \cap \closedint {-1} 1$

Third, $B^\ast$ is a Hausdorff space in view of Weak-* Topology is Hausdorff.

Finally, $R \sqbrk {B^\ast}$ is compact by Closed Subspace of Compact Space is Compact.

Thus we can apply Continuous Bijection from Compact to Hausdorff is Homeomorphism to:
 * $R^{-1} : R \sqbrk {B^\ast} \to B^\ast$.