TITLE: Corollary of Hahn-Banach theorem regarding the norm
QUESTION [6 upvotes]: Let $E$ be a normed space. Then, for each $x \in E$ we  can define its norm as 
$\|x\| = \sup \{|l(x)| : l \in E^∗, \|l\| ≤ 1 \} $.
Proof: 
Let $x \in E$ and set $S = \sup\{|l(x)| : l \in E^∗, \|l\| ≤ 1 \} \leq \|x\|.$ 
On the one-dimensional subspace $F = \mathrm{span}(x)$ we define a functional $f$ by $f(\lambda x)= \lambda \|x\| $ for all $\lambda  \in \mathbb{R}$ 
it follows that $f \in F^*$ with $\|f\| = 1$.  Hence, by the Hahn-Banach Theorem,
there exists some $l \in E^*$ with $l|_F\equiv f$ and $\|l\| = 1$. In particular this implies  $|l(x)| = |f(x)| = \|x\|$. This implies that $S = \|X\|$ and that the supremum is attained.
Now, why is it, that there a $\leq$ sign in the set of wich we take the  supremum, since the proof uses a functional that  clearly has norm equal to 1?

REPLY [5 votes]: It doesn't matter that there is a $\leq$ inside the set which we are taking the supremum over since the supremum is always achieved, i.e. $$\|x\|=\sup\{|l(x)|~:~l\in E^*,~\|l\|\leq 1\}=\max\{|l(x)|~:~l\in E^*,~\|l\|\leq 1\}.$$ 
Since you know exactly what the linear functional looks like and that it has $\|l\|=1$ you can replace the $\leq$ with an $=$ if you wanted!