Banach-Alaoglu Theorem

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Let $X$ be a separable normed space.

Then the closed unit sphere in its dual $X^*$ is weak* sequentially compact.


We have to show the following: given a bounded sequence in $X^*$, there is a weakly convergent subsequence.

Let $\left\langle{l_n}\right\rangle_{n \mathop \in \N}$ be a bounded sequence in $X^*$.

Let $\left\langle{x_n}\right\rangle_{n \mathop \in \N}$ be a countable dense subset of $X$.

Choose subsequences $\N \supset \Lambda_1 \supset \Lambda_2 \supset \ldots$ such that:

$\forall j \in \N: l_k \left({x_j}\right) \to a_j =: l \left({x_j}\right)$

as $k \to \infty$, $k \in \Lambda_j$.

Let $\Lambda$ be the diagonal sequence.

Claim 1

$l$ can be extended to an element of $X^*$.


$l$ can be extended in the obvious way to a linear function on $M = \operatorname{span} \left\{{x_j}\right\}_{j \mathop \in \N}$.

We extend it to a functional in $X^*$ by pointwise limit (notice that $M$ is dense in $X$).

We have:

$\displaystyle \left\vert{l \left({x}\right)}\right\vert = \lim_{k \mathop \to \infty} \left\vert{l \left({x_k}\right)}\right\vert \le \limsup_{k \mathop \to \infty} \left\Vert{l_k}\right\Vert_{X^*} \left\Vert{x}\right\Vert_X$

where $x_k \to x$ as $k \to \infty$.

Since $\left\{{l_k}\right\}_k \mathop \in \N$ was bounded, $l$ is bounded and thus continuous.


Claim 2

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



$\displaystyle X \ni x = \lim_{\substack {j \mathop \to \infty \\ j \mathop \in J} } x_j$

where $J$ is some subset of $\N$.

We have then, for every $j \in J$:

\(\displaystyle \left\vert{l_k \left({x}\right) - l \left({x}\right)}\right\vert\) \(\le\) \(\displaystyle \left\vert{l_k \left({x - x_j}\right)}\right\vert + \left\vert{l \left({x - x_j}\right)}\right\vert + \left\vert{l_k \left({x_j}\right) - l \left({x_j}\right)}\right\vert\)
\(\displaystyle \) \(\le\) \(\displaystyle \left({\sup_{i \mathop \in \Lambda} \left\Vert{l_i}\right\Vert_{X^*} + \left\Vert{l}\right\Vert_{X^*} }\right) \left\Vert{x - x_j}\right\Vert_X + \left\vert{l_k \left({x_j}\right) - l \left({x_j}\right)}\right\vert\)

Now given $\epsilon >0$, we find a $j \in J$ such that the first term is less than $\epsilon / 2$.

For fixed $j$, we have by construction of $l$ that $l_k \left({x_j}\right)$ converges to $l \left({x_j}\right)$.

Therefore, we can find a $k_\epsilon$ such that for $k \ge k_\epsilon$ we have:

$\left\vert{l_k \left({x}\right) - l \left({x}\right)}\right\vert < \epsilon$

as requested.


Source of Name

This entry was named for Stefan Banach and Leonidas Alaoglu.