Banach-Alaoglu Theorem/Lemma 2

Lemma for Banach-Alaoglu Theorem
Let $X$ be a separable normed vector space.

Let $X^*$ be the dual space of $X$.

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.

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

Proof
Let:


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

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

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

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.