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 \} $.


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?

  • 2
    $\begingroup$ Why don't you prove that $$\sup \{|l(x)| : l \in E^∗, \|l\| ≤ 1 \} = \sup \{|l(x)| : l \in E^∗, \|l\| = 1 \}$$ then the answer to your question would be: we can use either definition, and in practice use whichever one is more convenient at the time. $\endgroup$
    – GEdgar
    Commented Jul 21, 2017 at 12:50

1 Answer 1


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!


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