# Separation in dual space

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

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
@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
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 $$\langle x^{*},x \rangle > \sup_{f \in C^{*}} \langle f,x \rangle.$$ 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.

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 $x \in X$.