# Step in proof of Goldstine Theorem

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 $$\exists \ x\in X, ||x|| such that $$\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.
• You don't need a Banach space for this to work. Any normed space will do fine. Apr 1, 2020 at 17:09
• 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 ...
– Meep
Apr 1, 2020 at 17:21
• 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. Apr 1, 2020 at 17:30
• @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.
– Meep
Apr 1, 2020 at 18:20
• Is this Metric Embeddings with Zsak? Apr 13, 2020 at 13:52

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 $$x’\in B_X\cap U$$.
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 $$\overline{B_{X}}^*$$.