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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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You don't need the $\forall f \in C^*$: $f$ is already a dummy variable because of the $\sup$. –  Robert Israel Oct 16 '12 at 6:23
    
@Robert Israel: Thank you for your helping. –  blindman Oct 16 '12 at 6:59
    
See also: math.stackexchange.com/q/149920 –  commenter Oct 16 '12 at 12:59

2 Answers 2

up vote 3 down vote accepted

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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Nice answer! What I can tell from all this is that the weak*-topology forces all continuous linear functionals on $ X^* $ to come from $ X $. My 'inferior' answer below failed to narrow down to this consideration. However, is it true that if we replace 'weak*-closed' by 'closed', then we require $ X $ to be reflexive for the statement to hold? In other words, can we find a counterexample in which the required separating linear functional must come from $ X^{**} \setminus X $? –  Haskell Curry Oct 16 '12 at 6:38
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@HaskellCurry: Let $C$ be the space of sequences in $\ell_1 = (c_0)^\ast$ that sum to zero. This is a norm-closed and weakly closed subspace of $\ell_1$, but it's weak*-dense (it is the kernel of the functional given by a constant sequence in $\ell_\infty \setminus c_0$). If you take any sequence $x$ whose sum isn't zero, you can't separate it by a weak$^\ast$-continuous functional from $C$, so reflexivity is essential for your variant. –  commenter Oct 16 '12 at 12:58
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More generally, if $X$ is a nonreflexive Banach space, let $\phi \in X^{\star\star} \backslash X$, and consider $C^* = \{x^* \in X^*: \phi(x^*) = 0\}$. Then $C^*$ is closed and convex. For any $x \in X$ there is $x^* \in C^*$ such that $x^*(x) \ne 0$ (because otherwise the linear functional $x^* \mapsto x^*(x)$ would be a scalar multiple of $\phi$, which would imply $\phi \in X^{\star\star}$), so $\sup_{x^* \in C^*} x^*(x) = \infty$. –  Robert Israel Oct 16 '12 at 18:26
    
@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^* $). –  Haskell Curry Oct 17 '12 at 2:09

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. –  blindman Oct 16 '12 at 7:00

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