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Let $X$ be a real Banach space and $X^*$ its dual space. Let $C^*$ be a weak$^*$ closed and convex subset in $X^*$ and $x^*\notin C^*$. Then there exists $x\in X$ such that $ \langle x^*, x\rangle > \sup_{f\in C^*}\langle f, x\rangle. $ I would like to ask whether the statement is true? How can we prove?

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    See also: http://math.stackexchange.com/q/1499202012-10-16

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This is just the separation theorem (see e.g. Rudin, "Functional Analysis", Theorem 3.4(b)) for the locally convex topological vector space $X^*$ with the weak-* topology (whose continuous linear functionals correspond to the points of $X$): slightly more generally you can replace $x^*$ by a weak-* compact convex set disjoint from $C^*$.

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    @RobertIsrael: commenter's observation that $ C^* $ is weak*-dense applies here. Observe that $ C^* $ is a codimension-1 linear subspace of $ X^* $. We now use the basic result that the kernel of a non-zero linear functional on a topological vector space is a closed codimension-1 linear subspace iff the linear functional is continuous. As $ (X^*,\sigma(X^*,X))^* = X $, we see that $ \phi $ is not weak*-continuous. Therefore, $ C^* $ cannot be weak*-closed, which forces it to be weak*-dense (as the codimension of $ C^* $ is $ 1 $, the weak*-closure of $ C^* $ must be all of $ X^* $).2012-10-17
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I don't have the full answer as of now, but the best that I can think of is that if $ X $ is a reflexive Banach space, then the statement is true by the geometric Hahn-Banach Theorem. You can treat $ X^{*} $ as the initial Banach space under consideration and $ C^{*} $ as the closed and convex (and non-empty!) subset of $ X^{*} $. Then if $ x^{*} \notin C^{*} $, one can find an $ x \in X \cong X^{**} $ (by the geometric Hahn-Banach Theorem) such that \begin{equation} \langle x^{*},x \rangle > \sup_{f \in C^{*}} \langle f,x \rangle. \end{equation} Now, there is an error associated with the phrasing of the problem. As you are taking the supremum of the right-hand side of the inequality over $ f \in C^{*} $, there is no need to universally quantify $ f $. On a separate note, if $ C^{*} $ is weak*-closed, then it is automatically closed, so you do not have to add the adjective 'closed' in front of 'weak*-closed'.

If $ X $ is not reflexive, then I think the statement is false. That is because the separating functional that you need to separate $ x^{*} $ and $ C^{*} $ might lie in $ X^{**} \setminus X $. However, I could be wrong, so any opposing opinions are welcome.

Addendum

The above discussion is concerned with the case when $ C^{*} $ is closed with respect to the norm topology on $ X^{*} $. After the latest edit of the problem statement, it is now understood that $ C^{*} $ is to be assumed closed with respect to the weak*-topology on $ X^{*} $. More generally, observe that if we assume $ C^{*} $ to be closed with respect to any locally convex topology $ \mathcal{T} $ that is finer than the weak*-topology (denoted by $ \sigma(X^{*},X) $) but coarser than the Mackey topology (denoted by $ \tau(X^{*},X) $), then the continuous dual of $ (X^{*},\mathcal{T}) $ is still isomorphic to the natural copy of $ X $ in $ X^{**} $. Hence, the separation theorem mentioned by Robert (I call it the geometric Hahn-Banach Theorem) still applies to yield a separating functional $ x \in X $.

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    Dear Sir. Thank you for your construction and comments.2012-10-16