A subspace of a dual Banach space $X^*$ is norming if and only if its weak$^*$ closure contains a multiple of the unit ball of $X^*$

Question

Definition

A subspace $Z$ of a dual Banach space $X^*$ is said to be norming if there exists $c>0$ such that

$$\sup_{f\in B_Z}|f(x)|\geq c||x||$$

for every $x\in X$ (where $B_Z$ is $B_{X^*}\cap Z$).

Equivalence

I read some time ago (I'm sorry, I can't remember where) that this is equivalent to the fact that there exists $c>0$ such that

$$ cB_{X^*}\subset\overline{B_Z}^{\omega^*}$$

The converse is easy, but the first implication I'm not able to prove. Here's my attempt so far:

Attempt

Suppose there exists $c>0$ such that $\sup_{f\in B_Z}|f(x)|\geq c||x||$. Consider $\varphi\in cB_{X^*}$, and a $\omega^*$-neighbourhood of $\varphi$ given by $\{x_1,\dots,x_n\}\subset X$ and $\varepsilon>0$:

$$U=\{f\in X^*:|(f-\varphi)(x_i)|<\varepsilon , \quad i\in\{1,\dots,n\}\} $$

If we find $f\in U\cap B_Z$ we are done.

For every $x_i$ of our finite list, we have that:

$$|\varphi(x_i)|\leq ||\varphi||~||x_i||\leq c||x_i||\leq\sup_{f\in B_Z}|f(x)| $$

Therefore we can find $f_i\in B_Z$ such that $|\varphi(x_i)|\leq |f(x_i)|$. Then because of this inequality and the fact that $Z$ is a subspace, we can find $g_i\in B_Z$ such that:

$$ \varphi(x_i)=g_i(x_i)\Longrightarrow (\varphi-g_i)(x_i)=0$$

Now I would like to take a linear combination of these $g_i$ to construct a functional in $B_Z\cap U$, but I haven't been able to. If $U$ was a neighbourhood of $0$ then I could do it just multiplying the $g_i$ by $\varepsilon /n$ and dividing by the maximum of $||x_i||$ for example, and then taking the sum of the resulting functionals. But this doesn't work here.

Questions

1. In general, when we have this situation where we have a finite list of functionals in a subspace approximating a non-zero functional $\varphi$ at one point each, can we construct a functional in the same subspace approximating every point at once?

2. How to prove this equivalence? I've searched online but I couldn't find a proof anywhere. I also checked 'Functional Analysis and Infinite-Dimensional Geometry' by Fabian, Habala, Hájek, Montesinos, Pelant, Zizler, but this equivalence is not proven there.

EDIT

I thought of this proof, inspired by the proof that a separating subspace of $X^*$ is $\omega^*$-dense. Suppose there exists $\varphi\in cB_{X^*}$ such that $\varphi\notin\overline{B_Z}^{\omega^*}$. Since $\overline{B_Z}^{\omega^*}$ is a convex closed set, by Hahn-Banach there exists a functional $x$ in $(X^*,\sigma(X^*,X))^*=X$ such that

$$ |\varphi(x)|>\sup_{f\in B_Z}|f(x)|\geq c||x||$$

but $|\varphi(x)|\leq c||x||$, a contradiction.

Even then, the question 1 still stands. My intuition is that there doesn't need to exist a function in the subspace, judging by the argument in the edit, but I don't have a counterexample.

Answer 1

Isn't this just the theorem of bipolars for the space $(X^*,\sigma(X^*,X))$? The definition of norming subspaces is equivalent to $cB^\circ_Z \subseteq B_X$ (where the polar is taken in $X$, the proof is a simple argument using only homogeneity). The latter condition is equivalent to $cB_X^\circ \subseteq B_Z^{\circ\circ}=\overline{B_Z}^{\sigma(X^*,X)}$.