I am trying to prove this statement but I have problems when dealing with the dual of $l^\infty$. I have found a characterization of the weak star convergence in terms of the boundedness of the norm of the sequence, but I don't know how to use it... Thanks in advance
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Are you considering filters/nets, or just sequences? – Daniel Fischer May 09 '17 at 14:36
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Just sequences. – Axel May 09 '17 at 14:37
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Have you heard of Banach limits? – Daniel Fischer May 09 '17 at 14:38
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Yes Daniel Fischer, I know them, as an extension of the functional limit (when i restrict myself to convergent functions c).. But I don't see yet the connection here. – Axel May 09 '17 at 14:47
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Such a thing is an element of $(\ell^{\infty})^{\ast} \setminus \ell^1$. So if you can find a bounded sequence $(x_n)$ in $\ell^{\infty}$ such that $\langle x_n, y_n\rangle \to 0$ for all $y\in \ell^1$, but $\Lambda x_n \not\to 0$ for a Banach limit $\Lambda$ … – Daniel Fischer May 09 '17 at 14:51
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Yes, I see. Thank you Daniel! – Axel May 09 '17 at 14:53
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In general, let $E$ be a Banach space and let $B_E$ (resp. $B_{E^*}$) be the unit ball of $E$ (resp. $E^*$) in te norm topology, there exists three very useful results:
1) (Theorem of Banach-Alaoglu) $B_{E^*}$ is compact in the weak* topology.
2) (Characterization of reflexivity) $E$ is reflexive if and only if $B_E$ is compact in the weak topology.
3) $E$ is reflexive if and only if $E^*$ is reflexive.
Now, let $E$ be a non-reflexive Banach space (take $l^1$ for example), then $E^*$ ($l^\infty$ in this example) is non-reflexive and $B_{E^*}$ is not compact in the weak topology, thus, there exists a sequence $(f_n)\subset B_{E^*}$ such that doesn't have weak convergent subsequences, however, by the theorem of Banach-Alaoglu, there exist a weak* convergent subsequence $(f_{n_k})\subset(f_n)$.
Ludwik
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1I think it's a little more subtle, since failure of compactness means there's a net without a convergent subnet, but this doesn't necessarily mean there is a sequence without a convergent subsequence. – Jose27 Dec 28 '19 at 04:23
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Oh, you're right! Since $l^\infty$ is non-separable, I can't guarantee any relation between sequential compactness and compactness in the weak topology. – Ludwik Dec 28 '19 at 05:31