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Let $X$ be a separable Banach space. If $M$ is a linear manifold in $X^*$ give necessary and sufficient conditions that every functional in $w^*-\mathrm{cl} M$ be the $w^*$- limit of a sequence from M.

My attempt: I suppose $A$ is a w* - closed set and $M=\text{linear span}~ A$. Suppose $\phi \in w^*- \mathrm{cl} M$. Because $w^*- cl M = ||.||-\mathrm{cl} M$, we have $\phi \in ||.||- cl M$ . Thus there is a sequence $\{\phi_n\}$ in $M$ such that $\phi=||.||-\lim \phi_n$, and clearly $\phi=w^*-\lim \phi_n$.

Conversely, suppose $\phi=w^*-\lim \phi_n$ where $\{\phi_n\}$ is a sequence in $M$. Clearly $N(\phi,x_1,...,x_n,\epsilon)\cap M\not=\varnothing$, for every $\epsilon>0$ , $x_i\in X$.

I do not know my proof is correct or not. Also I do not use the condition $M=\text{linear span}~ A$ for the second part. Please help me. Thanks for your help.

Davide Giraudo
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niki
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  • There is a theorem: If A is weak* closed subset of $X^$ and $M=\text{linear span} A$, then M is weak closed iff it is norm closed. To use this theorem I introduced A. – niki Aug 08 '14 at 08:06
  • sorry, There is a mistake in type. A is weak* closed. I correct it – niki Aug 08 '14 at 08:08
  • The question asked me to put conditions on it. so in this case I think it's correct. Is not it? – niki Aug 08 '14 at 08:16
  • Oh, sorry, I misunderstood that you were stating assumptions. I would recommend that you organize your writing to make it clearer what you are attempting to show, but apologies for my misinterpretation. (I'll delete my obsolete comment.) – Jonas Meyer Aug 08 '14 at 08:18
  • It is not true in general, I don't think, that the norm-closure coincides with the weak-closure. If the set is convex, then it coincides with the weak-closure, but this is not good enough for your purpose. Your converse does not prove what you need to prove, either. You need to conclude that $M$ is the linear span of some weak-closed set, which you do not do. Probably, your condition is equivalent to M itself being weak*-closed. – Ben W Aug 08 '14 at 17:01
  • I think $M$ is first countable w.r.t. to weak* topology. Now ue ideas of this answer – Norbert Aug 08 '14 at 18:27
  • hatsoff@ I put the condition "A is weak* closed subset of X∗ and M=linear span A" on the question and want to show every functional in w∗−clM be the w∗- limit of a sequence from M. So I think it's not necessary to show M is the linear span of some weak-closed set. Also I put this condition to use theorem 6.12.9 of conway's functional analysis "If A is weak closed subset of X∗ and M=linear spanA, then M is weak* closed iff it is norm closed." For converse I show if $\phi=w^-\lim \phi_n$ then $\phi\in w^- cl M$. – niki Aug 08 '14 at 19:17

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