Characterisation of norm via duality in normed vector space

Question

Let $(F,\| \cdot \|)$ be a normed vector space. Prove that for every $x \in F$ the following characterisation of the norm holds: $$\|x\|=\max\{ f(x) : f \in B_{F'} \} \ \ (*)$$ My attempt: first of all, via Hanh-Banach we prove that for every $x_0 \in F$ with $x_0 \neq 0$ there exists $f_0 \in F'$ with $\| f \|_{F'}=1$ such that $f_0(x_0)=\|x_0 \|$. Now, I have the feeling that this proposition should basically solve the problem, but I'm not sure I really know why (I'm new to functional analysis). If we prove the equality $(*)$ with $\sup$ instead of $\max$, then the previous proposition will ensure that such $\sup$ is in fact a $\max$. We have $\|x\|=f_0(x) \le \sup\{ f(x) : f \in B_{F'} \}$ since $f_0 \in B_{F'}$. I can't figure out how to get the reverse inequality.

Answer 1

For $x_0 \in F$ and $x_0 \ne 0$, let us denote by $M(x_0)$ the set

$M(x_0)=\{ f(x_0) : f \in B_{F'} \} $ and let $m(x_0)= \max M(x_0)$.

By Hahn - Banach we get, as you have shown, a functional $f_0 \in B_{F'}$ such that $f_0(x_0)=||x_0||$. It follows that

(1) $||x_0|| \le m(x_0)$ .

For each $f \in B_{F'}$ we have

$f(x_0) \le |f(x_0)| \le ||f||*||x_0|| \le ||x_0||$.

Therefore

(2) $m(x_0) \le ||x_0||$.

From (1) and (2) the assertion follws.