Banach-Alaoglu Theorem/Proof 1

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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.


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 $X^*$.


Lemma 2

$l_k \stackrel {\omega^*} {\to} l$ as $k \to \infty$, $k \in \Lambda$.


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.