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When I read the prove of the Goldstine Theorem(See An Introduction to Banach Space Theory Robert E. Megginson 2.6.26), I find it use the separation theorem for the w* topology without any details.

Suppose $X$ is a normed space, $F$ is a w* closed convex set of $X^*$, $x^*$ is in $X^*$ but not in $F$, then there is $x \in X$, such that $|x^* (x)|> \sup \{\Re y^*(x) \mid y^*\in F\}$.

Can you prove it?

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    Have you studied Theorem 2.2.28, as suggested in the book?2012-06-25
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    Yes,but I do not think the Theorem 2.2.28 is enough, from that, we can just get x** in the X**, but not in the X.2012-06-25
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    I think you are trying to generalize a result that cannot be generalized. In the book, $F$ is not *any* w$^*$-closed convex set. It is the image of the unit ball under the canonical morphism $X \to X^{**}$.2012-06-25
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    I don't have Megginson's book, but all you need for establishing the result you want is the [Hahn-Banach separation theorem](http://en.wikipedia.org/wiki/Hahn–Banach_theorem#Hahn.E2.80.93Banach_separation_theorem) (apply part (ii) of Wikipedia's formulation to the compact convex set $A = \{x^\ast\}$ and the closed convex set $B = F$ in $(X^\ast, w^\ast)$) plus the fact that a $w^\ast$-continuous functional $\lambda$ on $X^\ast$ is of the form $\lambda(y^\ast) = y^\ast(x)$ for a unique $x \in X$ (this is proved e.g. as (*) in my answer [here](http://math.stackexchange.com/a/149998/5363)).2012-06-25
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    @t.b. why don't write this comment as answer? It completely eliminates the gap in Strongart's reasoning.2012-06-25
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    @t.b. How to get the functional λ is w*-continous?2012-06-27
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    It's part of the Hahn-Banach separation theorem (which is proved in any text on topological vector spaces): given two disjoint closed, convex sets, one of which is compact in a locally convex space you can find a continuous linear functional separating them strictly. See e.g. [here](http://books.google.com/books?id=9kXY742pABoC&pg=PA65) (second separation theorem 9.2, page 65 in Schaefer).2012-06-27
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    @t.b That is continuous linear functional, but maybe is not w*-continous.2012-06-27
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    As I said, I apply the theorem to the space $X^\ast$ **equipped with the weak$^\ast$-topology**, so yes, that functional **is** w*-continuous...2012-06-27
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    @t.b The separation functional comes from the dual space,whose topology is strong topology,I do not think you can change it to the weak*-topology.2012-06-28
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    I don't understand the problem. Yes you can switch to the weak$^\ast$-topology (maybe you wanted to ask something else?). Read the statement I linked you to: you can (strictly) separate a compact convex set $A$ from a disjoint closed convex set in *every* locally convex space. The weak$^\ast$-topology is locally convex. So: Equip $X^\ast$ with the weak$^\ast$-topology, then $A = \{x^\ast\}$ is weak$^\ast$-compact, while $B = F$ is given as closed and convex. Then you find a weak$^\ast$-continuous linear functional separating them and which is of the desired form by the answer I linked you to.2012-06-28
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    @t.b Oh, I see!You are right,Thanks.2012-06-29

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