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 $\sequence {l_n}_{n \mathop \in \N}$ 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.

Claim 1

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


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

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

We have:

$\displaystyle \size {\map l x} = \lim_{k \mathop \to \infty} \size {\map l {x_k} } \le \limsup_{k \mathop \to \infty} \norm {l_k}_{X^*} \norm x_X$

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

Since $\set {l_k}_k \mathop \in \N$ 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 $j \in J$:

\(\displaystyle \size {\map {l_k} x - \map l x}\) \(\le\) \(\displaystyle \size {\map {l_k} {x - x_j} } + \size {\map l {x - x_j} } + \size {\map {l_k} {x_j} - \map l {x_j} }\)
\(\displaystyle \) \(\le\) \(\displaystyle \paren {\sup_{i \mathop \in \Lambda} \norm {l_i}_{X^*} + \norm l_{X^*} } \norm {x - x_j}_X + \size {\map {l_k} {x_j} - \map l {x_j} }\)

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 $\map {l_k} {x_j}$ converges to $\map l {x_j}$.

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

$\size {\map {l_k} x - \map l x} < \epsilon$

as requested.


Source of Name

This entry was named for Stefan Banach and Leonidas Alaoglu.