Unit ball as an intersection of weakly closed sets.

Question

How can I prove: $B_{E}=\cap_{f\in E', \left\Vert f \right\Vert \leq 1} ${$x\in E : |f(x)| \leq 1$} ?

I thought it was connected to: $ \left\Vert x \right\Vert = sup_{f\in E', \left\Vert f \right\Vert \leq 1} |f(x)|$, but actually I have no idea how to prove it. I need this because I am trying to prove that the unite sphere is never closed in the weak topology and it'd end my proof.

Could anyone help me? Thank you all in advance.

Answer 1

Let $x\in C=\cap_{f\in E'\|f\|\leq 1}\{x\in E:|f(x)|\leq 1\}$, Hahn Banach implies the existence of $f_x$ such that $|f_x(x)|=\|x\|, \|f\|=1$, this implies that $x\in B_E$.

Conversely, let $x\in B_E$, for every $f\in E',\|f\|\leq 1$, we have $|f(x)|\leq \|f\|\|x\|\leq 1$ since $\|f\|\leq 1$ and $\|x\|\leq 1$, we deduce that $x\in C$.

Answer 2

If $x \in B_E$ then for any $f \in E', \|f\| \le 1$ we have

$$|f(x)| \le \|f\|\|x\| \le 1$$

Conversely, if for all $f \in E', \|f\| \le 1$ we have $|f(x)| \le 1$ then

$$\|x\| = \sup_{\substack f\in E' \\ \|f\| \le 1}|f(x)| \le 1$$

so $x \in B_E$.