A question regarding separation property in $X^*$ where, $X$ is a Banach space

Question

We know that if $X$ is a normed linear space and $A$ is a closed subspace of $X$, then for any $p\in X\setminus A$ there exists a $f\in X^*$ such that, $f(p)\neq 0$ and $f(a)=0\,\forall a\in A$. So this can be applied to the dual $X^*$ also. So for any closed subspace $A$ of $X^*$ and any $f\in X^{*}\setminus A$ there exists a $\Phi\in X^{**}$ such that, $\Phi(f)\neq 0$ and $\Phi(g)=0\,\forall g\in A$. Now if we assume that $X$ is reflexive then such a $\Phi$ is an evaluation map at a point $q\in X$ (say). Then we get, $f(q)\neq 0$ and $g(q)=0\,\forall g\in A$.

Can we say the same even if $X$ is not reflexive?

More precisely does the following holds for any Banach space $X$, which are not necessarily reflexive ?

If $X$ is a Banach space then, for any closed subspace $A$ of $X^*$ and any $f\in X^{*}\setminus A$ does there exist an element $q\in X$ such that, $f(q)\neq 0$ and $g(q)=0\,\forall g\in A$ ?

Answer 1

No. Let $X=\ell^1$, so $X^{*}=\ell^\infty$. Take $A=c_0\subset X^*$. If $q\in X$ and $g(q)=0$ for all $g\in A$, this in particular means that $e_n(q)=q_n$ is zero for all $n$, so $q=0$. So $f(q)$ is impossible.

The assertion becomes true if you require $A$ to be weak$^*$-closed, since then you can use the geometric form of Hahn-Banach to obtain $Q\in X^{**}$ and $c\in\mathbb R$ such that $\def\Re{\operatorname{Re}}$ $$ \Re Q(a)<c<\Re Q(f),\qquad a\in A. $$ Because $A$ is a complex subspace, the only possibility is that $Q(a)=0$ for all $a\in A$. This forces $c>0$. With suitable phase change we can replace $Q$ to get $Q(f)>0$. Finally, we use the result that the dual of $(X^*,w^*)$ is $X$, which means that there exists $q\in X$ such that $Q(g)=g(q)$ for all $g\in X^*$. Then $a(q)=0$ for all $a\in A$ and $f(q)>0$.

The reason the counterexample works in the first paragraph is that $c_0$ is weak$^*$-dense in $\ell^\infty$.

Answer 2

The following argument shows there is no normed space that is not reflexive with this property. Let $X$ be any normed space that is not reflexive. Let $J\colon X\to X^{**}$ be the canonical embedding and let $\varphi\in X^{**}\setminus J(X)$. Then $\ker \varphi$ is a proper closed subspace of $X^{*}$ since $\varphi$ is a non-zero continuous linear functional on $X^{*}$. However, if $x\in X$ is given such that $(Jx)(f) = 0$ for all $f\in \ker \varphi$, then $\ker \varphi \subseteq \ker Jx$ and it follows from this result that there is some $\alpha \in \mathbb{K}$ such that $Jx = \alpha\varphi$. Because $\varphi\not\in J(X)$ and $J$ is injective, we have $\alpha = 0$ and conclude $x = 0$. As a consequence, for any $f_{0}\in X^{*}\setminus \ker \varphi$, there is no $x\in X$ such that $f(x) = 0$ for all $f\in \ker \varphi$ and $f_{0}(x) \neq 0$.

If we look to the weak$^{*}$ topology, it can be seen from here and here that a closed subspace $A\subseteq X^{*}$ fails to have the property you stated precisely when it is a proper weak$^{*}$ dense subspace of $X^{*}$.