# The separation theorem for the w* topology

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? – Siminore Jun 25 '12 at 12:06
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. – Strongart Jun 25 '12 at 15:15
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 $X \to X^{**}$. – Siminore Jun 25 '12 at 16:53
I don't have Megginson's book, but all you need for establishing the result you want is the Hahn-Banach 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). – t.b. Jun 25 '12 at 18:09
@t.b. why don't write this comment as answer? It completely eliminates the gap in Strongart's reasoning. – Norbert Jun 25 '12 at 23:33