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Claim : A Banach space V is reflexive iff its unit ball B is weakly compact.

So I want to show that 'unit ball is weakly compact => V is reflexive' without using goldstine.

So I'm trying to use the theorem of bipolars.

From the theorem of bipolars we know that $j(B) = j(B)^{oo}$. So if I can show that $B''\subseteq j(B)^{oo}$, then I will have that $j(B)^{oo}= B''$. And hence $j(B) = B''$, and so j is surjective.

Here $j$ is the canonical projection from $V$ to $V''$

So how do I show $B''\subseteq j(B)^{oo}$

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    $\begingroup$ What's wrong with using Goldstine's theorem (which is quite easy)? If you can show that $B'' \subseteq j(B)^{oo}$ then you will have essentially proved Goldstine's theorem unless you used compactness of $j(B)$ again. $\endgroup$ – Martin Mar 20 '13 at 22:08
  • $\begingroup$ There is nothing wrong with Golstine's theorem, I just wanted to show it another way. Do you have a method with doing it with bipolars? $\endgroup$ – user58514 Mar 20 '13 at 22:11
  • $\begingroup$ Well, I was trying point out that the usual proof of Goldstine's theorem proceeds by showing that $B'' \subseteq j(B)^{oo}$, so no, I don't have a proof with bipolars and without Goldstine's theorem. There will at best be some differences in terminology and language. $\endgroup$ – Martin Mar 20 '13 at 22:18
  • $\begingroup$ $\phi \in B'' \Rightarrow \|\phi\| \leq 1 \Rightarrow |\phi(f)| \leq 1 \ \forall f \in B' \\ \Rightarrow Re\phi(f) \leq 1 \ \forall f \in B'$ And if $f \in V'$ such that $Ref(b) \leq 1 \ \forall b \in B$ then it is clear that $|f(b)| \leq 1 \forall b\in B$ Hence $\|f\| \leq 1 $. So $j(B)^{o} \subseteq B'$. Hence $\phi \in j(B)^{oo}\\$ Does this not prove it? And moreover, I am pretty sure I have not proved Goldstine's theorem? $\endgroup$ – user58514 Mar 20 '13 at 22:36
  • $\begingroup$ I was sloppy reading your statement. I initially thought you meant that Golstine's theorem was needed. Now I believe you meant that just a part of it is. How did the proof that you saw show this fact? $\endgroup$ – user58514 Mar 20 '13 at 22:46
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$\phi \in B'' \Rightarrow \|\phi\| \leq 1 \Rightarrow |\phi(f)| \leq 1 \ \forall f \in B' \\ \Rightarrow Re\phi(f) \leq 1 \ \forall f \in B'$

And if $f \in V'$ such that $Ref(b) \leq 1 \ \forall b \in B$ then it is clear that $|f(b)| \leq 1 \forall b\in B$

Hence $\|f\| \leq 1 $. So $j(B)^{o} \subseteq B'$. Hence $\phi \in j(B)^{oo}$

Here is my attempt. Punch holes as you will...

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  • $\begingroup$ +1 that's it. Now combine this with $j(B)\subseteq j(B)^{oo} = \overline{j(B)}^{w\ast}$ and $j(B)\subseteq B''$ and with Alaoglu's theorem and you've proved Goldstine's theorem. $\endgroup$ – Martin Mar 20 '13 at 23:16

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