5
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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 Jul 21 '17 at 12:50
4
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.