For an example, by Alaoglu's theorem, the unit ball of the dual space is weak* compact in weak* topology. Generally speaking, it is not weak* sequential compact, but if we assume it is, my question is, does the limits of the two sense coincide?
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1Can you explain more precisely what you mean by "limits of the two sense"? Which senses? How are you defining them? – Nate Eldredge Apr 22 '13 at 01:18
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For example: ${x^{\asp}{n}}$ is a bounded sequence in the dual space of a banach space $X$, by the Alaoglu's theorem, it will have a limits in the sense of weak* topology. If I assume that it is also sequentially convergent in the sense of for every $x\in X$, we have $x^{\asp}{n}(x)\rightarrow x^{\asp}{0}(x)$, I want to know whether the $x^{\asp}{0}(x)$ is as the same as the limits in weak* topology. – vivian Apr 22 '13 at 01:36
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If $v_n \rightarrow v$ in the weak* topology, $v_n$ need not converge in the usual topology. For example, in the Hilbert space $\mathcal{l}_2$, with orthonormal basis $e_i$, $(e_i,f) \rightarrow 0$ since $\|f\|_2=\Sigma(e_i,f)^2 < \infty$ for all $f \in \mathcal{l}_2$, so $e_i \rightarrow e$ in the weak* topology, but clearly $e_i$ is not convergent in $\mathcal{l}_2$.
Since the weak* topology is weaker than the usual topology, we do have that if $f_i \rightarrow f$ in the usual topology, then $f_i \rightarrow f$ in the weak* topology, and since the weak* topology is Hausdorff, if $f_i \rightarrow f$ strongly and $f_i \rightarrow g$ weakly then the limits coincide.
tmh
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You mean that if there exists the topology for the pointwise convergence of the elements in the dual space, it is stronger than the weak* topology? – vivian Apr 22 '13 at 01:44
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If I understand your question correctly, the answer is this: in any topology, if you have a convergent sequence, then it converges as a net (because given a neighbourhood of the limit, eventually all points are in it).
On the other hand, a convergent net can have infinite subsets (notice that I said "subsets", not "subnets") that have many different accumulation points. To see this, let $\{a_n\}_{n\in\mathbb Z}$ be given by $$ a_n=\begin{cases} q_n,&\ \mbox{ if }n<0\\ 1/(n+1),&\ \mbox{ if }n\geq0 \end{cases} $$ where $\{q_n\}$ is an enumeration of $\mathbb Q\cap[0,1]$. Then $a_n\to0$ as a net with the usual order on $\mathbb Z$, while every point in $[0,1]$ is an accumulation point of the set $\{a_0,a_1,a_{-1},a_2,a_{-2},\ldots\}$
Martin Argerami
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If the normed space is assumed to be separable, the unit ball in the dual space is also weak$^\star$-sequentially compact. In this case norm-bounded subsets of the dual are metrizable, and so weak$^\star$-compactness and weak$^\star$-sequentially compactness on these norm-bounded sets coincide.
A short sketch of the proof for the metrizabilty can be found here.
