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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.

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Proof 2 is incomplete. Without the separability assumption, you are facing the fact that the closed unit ball is in general not metrizable in the weak topology, and thus you cannot deduce sequential compactness (the existence of [weakly] convergent subsequences) from the compactness.

You could cite the Eberlein-Shmulian theorem to obtain the fact that the closed unit ball of a reflexive Banach space is weakly sequentially compact.

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.

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  • $\begingroup$ Okay thanks for your response. Using one of the corrections to proof 2 you suggested, which proof would you suggest I use for this result? $\endgroup$ – user100431 Feb 23 '14 at 13:14
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    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Feb 23 '14 at 13:34

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