Banach-Alaoglu Theorem

Theorem
Let $X$ be a separable normed space.

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

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

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

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

Choose subsequences $\N \supset \Lambda_1 \supset \Lambda_2 \supset \ldots$ such that for every $j \in \N$ we have $l_k \left({x_j}\right) \to a_j =: l \left({x_j}\right)$ as $k \to \infty$, $k \in \Lambda_j$.

let $\Lambda$ be the diagonal sequence.

Claim 1
$l$ can be extended to an element of $X^*$.

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

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

We have:
 * $|l(x)| = \lim_{k \to\infty} |l(x_k)| \le \limsup_{k \to \infty}||l_k||_{X^*}||x||_X$

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

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

Claim 2
$l_k \stackrel{\omega^*} {\to} l$ as $k \to \infty$, $k \in \Lambda$.

Proof
Let $X \ni x = \lim_{j \to\infty, \ j \in J} x_j$, where $J$ is some subset of $\N$.

We have then:
 * $|l_k(x) - l(x)|\leq |l_k(x - x_j)| + |l(x - x_j)| + |l_k(x_j) - l(x_j)|\leq$


 * $\leq (\sup_{i \in \Lambda}||l_i||_{X^*} + ||l||_{X^*})||x - x_j||_X + |l_k(x_j) - l(x_j)|\stackrel{k,j \to \infty} {\to} 0$