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In Brezis's Functional Analysis, there is a corollary

Corollary 3.30. Let $E$ be a separable Banach space and let $\left(f_{n}\right)$ be a bounded sequence in $E^*$. Then there exists a subsequence $\left(f_{n_{k}}\right)$ that converges in the weak topology $\sigma\left(E^*, E\right)$.

I would like to make sure if there is not any result proving the converse or a counter-example?

Francesca
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    What kind of converse are you interested in ? A weakly converging sequence is bounded. – daw Feb 28 '24 at 20:39
  • Hi @daw :D. Well, by Eberlein-Smulian Theorem we have the characterization: If E is Banach Space. Then E is reflexive $\iff; \forall (x_n) \in E$ bounded and there exists $(x_{n_k})\subset (x_n)$ such that $x_{n_k}$ converges weakly in $\sigma(E, E')$. Hence, I was wondering if there is a similar result for separable spaces instead of reflexive (in this case considering a weak star topology). – Francesca Feb 28 '24 at 21:06
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    If I understand correctly, the statement you are asking about is whether the following is true: If every bounded sequence in $E^{}$ has a subsequence which converges in the weak topology, then $E$ is separable. Is this indeed what you are asking? – Dean Miller Feb 28 '24 at 22:14
  • @DeanMiller Exactly. Is exactly this, thank you. – Francesca Feb 28 '24 at 22:36

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Let $X$ be a reflexive Banach space. As $X$ is reflexive, $X^{*}$ is also reflexive. As $X^{*}$ is reflexive, the closed unit ball of $X^{*}$ is compact in the weak topology on $X^{*}$, and consequently any scalar multiple of the closed unit ball of $X^{*}$ is compact in the weak topology on $X^{*}$.

Now take a bounded sequence in $X^{*}$. As this sequence is bounded, it is contained in a scalar multiple of the closed unit ball of $X^{*}$, which is compact in the weak topology on $X^{*}$. By the Eberlein-Šmulian theorem, the sequence has a subsequence which converges in the weak topology on $X^{*}$. As $X^{*}$ is reflexive, the weak and weak* topologies on $X^{*}$ coincide. So it follows that this sequence has a subsequence which converges in the weak* topology on $X^{*}$.

This holds for any reflexive Banach space $X$. As there exist reflexive Banach spaces which are not separable (for example, $\ell_{2}(\Gamma )$ where $\Gamma$ is an uncountable set), it follows that there are non-separable Banach spaces where every bounded sequence has a subsequence which converges in the weak* topology.

As an extra note, it is worth mentioning that if $X$ is a Banach space, you have that any bounded sequence in $X^{*}$ has a subsequence which converges in the weak* topology on $X^{*}$ if $X$ is reflexive or separable. You also have that the closed unit ball of $X^{*}$ is always compact in the weak* topology on $X^{*}$. However, it is not always the case that the closed unit ball of $X^{*}$ is sequentially compact with respect to the weak* topology on $X^{*}$. A counterexample can be obtained in the case where $X = \ell_{\infty}$. For details, see this post. So although compact and sequentially compact subsets coincide in the weak topology on a Banach space by the Eberlein-Šmulian theorem, compact and sequentially compact subsets do not coincide in the weak* topology on the dual of a Banach space in general. This also shows that there is no analogue of the Eberlein-Šmulian theorem with respect to the weak* topology.

Dean Miller
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    Thank you very much Dean. Looks like that these spaces $\ell_{2}$ and $\ell_{\infty}$ are good to think about counter examples... – Francesca Feb 29 '24 at 02:52
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    You're welcome! When you want to prove that reflexivity is required for some result, typically the spaces $c_{0}$, $\ell_{1}$, $\ell_{\infty}$ and $C([0,1])$ are good places to start looking in order to find counterexamples when your Banach space is not reflexive. That is quite useful for when you are looking at the weak and weak* topologies because of how those concepts are connected to reflexivity. – Dean Miller Feb 29 '24 at 06:15