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In my lectures, we gave a proof of Goldstine's theorem

$\overline{B_X}^{w*}=B_{X**}$

where $B_X$ is the norm-closed unit ball of the Banach space X, and $w*$ is the weak-star topology.

Now to prove this, we used a lemma regarding 'local reflexivity':

Let $\phi \in B_{X**}$ and $||\phi|| < M$ and $E\subset X^*, \ dimE <\infty$. Then $\exists \ x\in X, ||x||<M$ such that $\hat{x}|_E=\phi|_E$ where $\hat{x}$ denotes the canonical embedding of $X$ into $X^{**}$.

Now the part I am having trouble with is showing that $B_{X**} \subset \overline{B_X}^{w*}$. In particular, we said:

  • Take $\phi \in B_{X**}$ and a weak* open neighbourhood of $\phi$, i.e. pick some $f_1, …, f_n \in X^*$ and $\epsilon >0$, and take the set $U=\{\psi \in X^{**} | \ | (\psi - \phi )f_i| < \epsilon \forall i \in [n]\}$
  • Now by the local reflexivity lemma, we have an $x\in X$ such that $\hat{x}(f_i) = \phi (f_i)$ for all $i$, hence $\hat{x}\in U$.
  • Now if $||x||\leq 1$ then we are done since $\hat{x}\in B_X \cap U$. THIS IS THE PART I DO NOT UNDERSTAND! (The rest of the proof goes on that if this is not the case, we can normalise our $x$ to have something which works.

So I simply do not get how $\hat{x}\in B_X \cap U \implies \phi \in \overline{B_X}^{w*}$.

I have tried:

  • Thinking about interiors instead. So suppose $\phi \notin \overline{B_X}^{w*}$, then $\phi \in int^{w*}(U - B_X)$ So I am hoping that all $\eta \in int^{w*}(U - B_X)$ have norm greater than 1, so I get a contradiction? I know that $B_{X**}$ is w* closed, so maybe this makes the point obvious. But for some reason I am not seeing how.
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  • $\begingroup$ You don't need a Banach space for this to work. Any normed space will do fine. $\endgroup$
    – J. De Ro
    Apr 1, 2020 at 17:09
  • $\begingroup$ Hi @ε-δ Sorry if I am missing your point, but I don't see how that helps? I haven't tried to use the completeness of X with respect to the norm ... $\endgroup$
    – Meep
    Apr 1, 2020 at 17:21
  • $\begingroup$ I'm not claiming my comment is helpful. Just that the assumption that $X$ is Banach is unnecessary. I am not used to your notation, but if you know the Banach separation theorem (for separating compact/closed sets) I can provide you an easy alternative proof. $\endgroup$
    – J. De Ro
    Apr 1, 2020 at 17:30
  • $\begingroup$ @ε-δ Oh OK. We were just proving this for the case of Banach spaces (even if it applies more widely) because it was a step towards the metric characterisation of superreflexivity of Banach spaces (Ribe Program). I have seen a post here using the Hahn Banach separation, which I am familiar with- thank you for suggesting it! But I am really trying to understand the reasoning here. That I don't get it means there is some gap in my understanding that I need to fill. $\endgroup$
    – Meep
    Apr 1, 2020 at 18:20
  • $\begingroup$ Is this Metric Embeddings with Zsak? $\endgroup$ Apr 13, 2020 at 13:52

2 Answers 2

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The proof idea is clear. In order to show that $B_{X^{**}} \subset \overline{B_X}^{w^*}$, for any $\phi \in B_{X^{**}}$ and a weak* open neighborhood $U$ of $\phi$, we have to find $x\in B_X\cap U$. We have that $x\in U$, so if $\|x\|\leq 1$ then we are done since $\hat{x}\in B_X \cap U$. If $\|x\|>1$ then we have to construct an other point $x’\in B_X\cap U $.

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I think I now get this proof. The point is that the open neighbourhood we constructed about $\phi$ is completely arbitrary. In particular, given our $\phi \in B_{X**}$, then

  • Claim for contradiction $\phi \not\in \overline{B_{X}}^*$

  • So by definition of clusre, we have a weak star open neighbourhood about $\phi$, say $U$, for which $U \cap \overline{B_{X}}^* = \emptyset$

  • But we have shown using local reflexivity, that given any weak star open neighbourhood of $\phi$ it contains an $\hat{x}\in B_X \cap U$ (the part of the proof I didn't add is that if $||x||\not \leq 1 $ when found by lemma 2, we can normalise it to find something that is).

I.e. no weak star open neighbourhood of $\phi$ is disjoint from $B_X$, and so $\phi$ is in the closure $\overline{B_{X}}^*$.

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