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If $X$ is a separable normed linear space, then we know that every bounded sequence in $X^*$ has a weak-* convergent subsequence . Can we drop the separability condition , i.e. if we don't assume $X$ is separable, then are there counterexamples ?

Please help. Thanks in advance

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  • $\begingroup$ There are counterexamples; see page 226 here for example. $\endgroup$ – David Mitra Nov 17 '17 at 11:38
  • $\begingroup$ @David Mitra : What is meant by $l_1(\mathbb R)$ on the book ? Isn't it kind of contradictory as I thought $l_1$ is separable ... $\endgroup$ – user495643 Nov 17 '17 at 11:44
  • $\begingroup$ The space of absolutely convergent sums, indexed by $\Bbb R$; so elements are of the form $\sum_{r\in\Bbb R} x_r$. $\endgroup$ – David Mitra Nov 17 '17 at 11:46
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    $\begingroup$ See also, the second answer here. $\endgroup$ – David Mitra Nov 17 '17 at 11:53
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Take $X=l^\infty$, which is not separable. Define the sequence $f_n$ in $(l^\infty)^*$ by $$ f_n (x) = x_n. $$ Then $(f_n)$ is a bounded sequence, in fact, $\|f_n\|_{(l^\infty)^*}=1$.

However, it does not have a weak-star converging subsequence. Let $(f_{n_k})$ denote a subsequence. Then define $x\in l^\infty$ by $$ x_{n_k}=(-1)^k, $$ set all other entries $x_i=0$. Then $$ f_{n_k}(x) = (-1)^k, $$ which is not convergent.

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  • $\begingroup$ Of course, by Alaoglu's theorem. This shows that a subnet of a sequence isn't a subsequence. $\endgroup$ – Jochen Nov 17 '17 at 12:26

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