# Proposed proofs for weak convergence question

I have the following question and two proposed proofs. Please advise if these proofs are adequate and which of the two is better. Thanks.

Question: Let $V$ be a reflexive, separable Banach space. Take $\{u_{k}\}_{k}$ be a bounded sequence in $V$. We want to show that there is a subsequence and $u \in V$ such that $u_{k} \rightharpoonup u$. There are two ways to prove this:

Proofs:

1. Let $\{u_{k}\}_{k}$ be a bounded set in $V$. $V$ is reflexive so we can identify $V$ and $V^{**}$. Also since $V$ is reflexive and separable it follows that $V^{*}$ is reflexive and separable. Using Banach Selection Principle(which states: in a Banach space with a separable predual, any bounded sequence contains a weakly* convergent subsequence.) it follows that since $V^{**}$ has a separable predual that any bounded sequence contains a weakly* convergent subsequence. So then there exists a subsequence such that ${u_{k}}_{l} \rightharpoonup u$ in $V$.

2. The following proof only uses reflexivity of Banach space $V$. Let $\{u_{k}\}_{k}$ be a bounded sequence in $V$. We use Kakutani Theorem(Let $V$ be a Banach space. Then $V$ is reflexive iff $B_{V} = {x \in V: \Vert x \Vert \leq 1}$ is compact in the weak topology of $V$). Since $\{u_{k}\}_{k}$ is a bounded set in $V$ it follows that $\Vert u_{k} \Vert \leq M$ for some $M > 0$ and for all $k \in \mathbb{N}$, therefore $\Vert \frac{u_{k}}{M} \Vert \leq 1$ for all $n$. By the Kakutani Theorem it follows then that $(\frac{u_{k}}{M})_{k}$ has a weakly convergent subsequence in $V$. Therefore $(u_{k})_{k}$ has a weakly convergent subsequence.

Or, from the separability of $V^\ast$ - which follows from the separability of $V$ and reflexivity - deduce that the weak topology on the closed unit ball of $V$ is metrizable, hence the closed unit ball is weakly sequentially compact.
• They are, at the bottom of things, the same proof. The Banach selection principle just asserts the sequential compactness of the closed unit ball of $V^\ast$ under the conditions, which is a direct consequence of the metrizability of the subspace topology of the weak* topology on the closed unit ball of $V^\ast$ (compact by Banach-Alaoglu) when $V$ is separable. My taste prefers $$V^{\ast\ast}\text{ separable}\Rightarrow V^\ast \text{ separable} \Rightarrow B_V \text{ metrizable},$$ where $B_V$ is the closed unit ball of $V$. But that's taste. – Daniel Fischer Feb 23 '14 at 13:34