# Banach-Alaoglu Theorem/Lemma 4

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## 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 $\closedint {-1} 1$

- $\map \FF B$

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:

\(\ds \map {\psi_1} x\) | \(=\) | \(\ds \norm x \map {\psi_1} {\frac x {\norm x} }\) | Definition of Linear Functional | |||||||||||

\(\ds \) | \(=\) | \(\ds \norm x \map {\map R {\psi_1} } {\frac x {\norm x} }\) | as $\frac x {\norm x} \in B$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \norm x \map {\map R {\psi_2} } {\frac x {\norm x} }\) | by hypothesis | |||||||||||

\(\ds \) | \(=\) | \(\ds \norm x \map {\psi_1} {\frac x {\norm x} }\) | as $\frac x {\norm x} \in B$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \map {\psi_2} x\) | Definition of Linear Functional |

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

Thus, $\psi_1 = \psi_2$.

$\Box$

In view of Inverse of Homeomorphism is Homeomorphism, it suffices to show that:

- $R^{-1} : R \sqbrk {B^\ast} \to B^\ast$

is a homeomorphism.

Observe that $X^\ast$ is Hausdorff space by Weak-* Topology is Hausdorff.

Thus $B^\ast$ is also Hausdorff space by Subspace of Hausdorff Space is Hausdorff.

Recall that $\map \FF B$ is compact.

In addition, by Lemma 3, $R \sqbrk {B^\ast}$ is closed.

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

Thus, in view of Continuous Bijection from Compact to Hausdorff is Homeomorphism, it suffices to show that:

- $R^{-1} : R \sqbrk {B^\ast} \to B^\ast$

is continuous.

Recall Definition of Weak-* Topology and Definition of Topology Generated by Synthetic Sub-Basis

Let $\psi_0 \in X^\ast$, $x \in X$, 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 } $

Observe, if $x \ne 0$, we have:

- $\map N {\psi_0 ; x ; r } = \map N {\psi_0 ; \frac x {\norm x} ; \frac r {\norm x} }$

Without loss of generality, therefore, we may assume $x \in B$.

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

This can be seen as:

\(\ds R \sqbrk { \map N {\psi_0 ; x ; r } }\) | \(=\) | \(\ds \set { \map R \psi : \psi \in B^\ast,\; \map \psi x \in \openint {\map {\psi_0} x - r} {\map {\psi_0} x + r} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \set { \map R \psi : \psi \in B^\ast,\; \map \psi x \in \map {I_r} {\map {\psi_0} x} }\) | since $\size {\map \psi x} \le 1$, as $\psi \in B^\ast$ and $x \in B$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \set { \map R \psi : \psi \in B^\ast,\; \map {\map R {\psi} } x \in \map {I_r} {\map {\psi_0} x} }\) | Definition of $R$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \set { \map R \psi : \psi \in B^\ast,\; \map {\pr_x} {\map R \psi} \in \map {I_r} {\map {\psi_0} x} }\) | Definition of $\pr_x$ | |||||||||||

\(\ds \) | \(=\) | \(\ds R \sqbrk {B^\ast} \cap \pr_x^{-1} \sqbrk {\map {I_r} {\map {\psi_0} x} }\) |

where:

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

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$\blacksquare$