Let $X$ be a normed space. I want to show that - using Hahn Banach - for every $x$ we have
$$||x|| = \sup\{|f(x)| : f \in X^*, ||f|| \leq 1\}$$
One direction is easy:
$$|f(x)| \leq ||f|| \cdot ||x|| \leq ||x||$$
Therefore
$$||x|| \geq \sup\{|f(x)| : f \in X^*, ||f|| \leq 1\}$$
For the other direction one uses Hahn-Banach.
Defining $g: \mathbb{R}x \to \mathbb{R}$ with $g(tx) = t||x||$$
We can conclude by Hahn-Banach that there exists $g': X \to \mathbb{R}$ such that $||g'|| = ||g||$. We know that $||g|| = 1$ therefore $||g'|| = 1$.
If I now choose $f = g'$, how can I derive that $|f(x)| = ||x||$, to show that $||x|| \leq \sup\{|f(x)| : f \in X^*, ||f|| \leq 1\}$