Norming subsets of the dual ball

Question

Let $X$ be an infinite dimensional Banach space. Suppose $S$ is a subset of $B_{X^*}$, the unit ball of $X^*$, that is norming in the sense that for any $x\in X$, $||x||=\sup_{f\in S}|f(x)|$. Is it true that $B_{X^*}$ is the $w^*$-closure of convex hull of $\pm S$?

Answer 1

Let $C$ be the weak-* closure of the convex hull of $\pm S$. Then $C$ is a closed convex subset of $B_{X^*}$. Let $f \in B_{X^*} \backslash C$. By Hahn-Banach in the locally convex space $X^*$ with its weak-* topology (such that only evaluations at a point of $X$ are linear forms), there is some $x \in X$ and some $a > 0$ such that, for all $g \in S$, $|g(x)| < a < f(x)$.

In particular, $|f(x)| > a \geq \|x\|$, a contradiction.

So $C=B_{X^*}$.

Answer 2

Here is a counter-example in an infinite-dimensional Hilbert space. Let $H$ be an arbitrary infinite-dimensional Hilbert space and consider $H'=\Bbb C\oplus H$. Let $B_H$ be the unit ball in $H$ and denote $[-1,1]\subseteq \Bbb C$ as embedded into $H'$ by $A$ (and denote the vector $1$ in $A$ by $\mathfrak a$). Then define

$$S= \{ \sqrt t\, x + \sqrt{1-t}\, y\,\mid t\in[0,1], x\in A, y\in B_H\}$$

now note that $S=-S$, that $S$ is convex and that $S\subseteq B_{H'}$. We may further charcterise (or alternatively define) $S$ as the norm $≤1$ elements of $H'$ such that the $A$ component is real.

Let $v\in H'$ and split $v=re^{i\varphi}\,\mathfrak{a}+y$ for $y\in H$. Note that $\|v\|^2 = r^2+\|y\|^2$, we define $$v':=\frac{r\,\mathfrak{a}+e^{-i\varphi}y}{\|v\|},$$ which is in $S$. We check that (here the second component of the scalar product is the anti-linear one) $$\langle v, v'\rangle = \frac1{\|v\|}\langle re^{i\varphi}\,\mathfrak{a} + y, r\,\mathfrak{a}+e^{-i\varphi} y\rangle = \frac1{\|v\|}(r^2e^{i\varphi}\,\mathfrak{a}+e^{i\varphi}\|y\|^2)=e^{i\varphi}\|v\|,$$ hence $|\langle v,v'\rangle|=\|v\|$ and $S$ has the norming property.

Now check that $S$ is weak* closed. This is more or less obvious, as if $x_\alpha\in S$ then the $A$ component of $x_n$ is equal to $\langle x_\alpha,\mathfrak a\rangle$, which must be real. If $x_\alpha$ converges weakly to $x$ then $\langle x_\alpha,\mathfrak a\rangle\to \langle x,\mathfrak a\rangle$ must also be real. Since $B_{H'}$ is weak* closed anyway it follows that any limit of an element in $S$ must be in $B_{H'}$ with real $A$ component, hence the limit must lie in $S$.