The question is :
Let be $X$ a separable normed vectorial space and $S$ a subspace de $X$ $ \varphi : S \rightarrow \mathbb{F}$. Show without use Hahn- banach' Theorem (and zorn's lemma) that exists an extension $f$ of $\varphi$ with $\|\varphi\| = \|f\|$.
I have no idea how I start. I only know that $X$ has a subset dense and countable. I think to extend $\varphi$ for the S ' s Closure, but I don't know if "S" is the subset dense in$ X$.
