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Question: Let $X$ be a normed space and $X^*$ be its continuous dual of $X.$ Assume that $(x_\alpha^*)_\alpha$ is a bounded net in $X^*.$ Is it true that there exists a cluster point $x^*$ in $X^*$ such that $(x_\alpha^*)_\alpha$ converges to $x^*?$

My attempt: Recall that Banach Alaoglu states that $B_{X^*}$ is weak$^*$-compact. If $\|x_n^*\|\leq 1$ for all $n\geq 1,$ then there exists a subnet $(x^*_{\alpha_\beta})$ of $(x_\alpha^*)_\alpha$ such that the subnet converges to some $x^*\in B_{X^*}.$ However, I am not able to do this for the whole net.

Idonknow
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  • Take a fixed $x^* \in X^$ and set $x^_n := n , x^*$. I doubt that this has a cluster point. – gerw Feb 19 '19 at 06:39
  • @gerw pardon my edit. I forgot to include bounded in my question. So we need a bkunded net. – Idonknow Feb 19 '19 at 08:17
  • Which topology are you using on $X^$? Is it the operator-norm topology or the weak- topology? And, most of all, why do you expect that any bounded net would converge? This is even not true when $X$ is 1-dimensional... – Sangchul Lee Feb 19 '19 at 09:56
  • @Sangchul Lee I am using weak star topology on $X^*$. I do not expect that bounded net would converge.i just want to get a cluster point from a bounded net. Clearly it holds in finite dimensional spaces, right? – Idonknow Feb 19 '19 at 10:12
  • I originally assumed that you know the existence of convergent subnet, which rendered me puzzled as you seemed asking either about what you arleady know or about the convergence of the whole net. But, given that you used the word 'subset' (which I originally misread as subnet), this seems not true at all. In such case, I would like to refer to a more general fact that a topological space is compact if and only if any net has a convergent subnet. (For instance, see this posting.) – Sangchul Lee Feb 19 '19 at 11:01
  • I apologise for my typo. It should be subnet instead of subset. Yes, I would like to know the convergence of the whole net. I know that the net has convergence subnet due to weak star compactness. – Idonknow Feb 19 '19 at 11:05
  • Then I am puzzled again... What if we consider the sequence $x_n^* = (-1)^n$ in $X = \mathbb{C} \simeq X^$ (or more precisely, $x_n^ = (x \mapsto (-1)^n x)$? This is certainly bounded but not convergent in weak-* topology on $X^*$ (which coincide with the usual topology since $X$ is finite-dimensional). – Sangchul Lee Feb 19 '19 at 11:12
  • Take this: $X = X^* = \mathbb R$ and the sequence $x_n = (-1)^n$. This sequence is also a net and does not converge. – gerw Feb 19 '19 at 11:12
  • Gerw and Sangchul Lee, both examples show that there is a cluster point of some subsequence but the whole sequence does not have ant cluster point. – Idonknow Feb 19 '19 at 13:20

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