I'm facing the proof, using theorem of Hahn-Banach.
The Theorem is following:
In normed space every linear, continuous functional f on vector subspace $M \subset X$ can be extended to a continuous functional F on X with norm preserved: $\forall_{f \in M'} \exists_{F \in X'}: F|_{M}=f$ and $||F||=||f||$.
Proof:
Let us take $p(x):= ||f|| \cdot ||x||$ for every $x \in X$. Then:
$$ \forall_{x \in M} |f(x)| \leq ||f|| \cdot ||x|| = p(x)$$
Using theorem of Hahn-Banach there exists a functional $F$ extending $f$.
$$ ||F|| \geq ||f|| \text{ (because F extends f),}$$
$$ ||F|| = sup_{||x|| \leq 1} ||Fx|| \leq sup_{||x|| \leq 1} p(x) = sup_{||x|| \leq 1} ||f|| \cdot ||x|| = ||f|| $$
So $||F|| = ||f||$.
And there is my question. Why the $ sup_{||x|| \leq 1} ||Fx|| \leq sup_{||x|| \leq 1} p(x) $ ? And why $p(x)$ isn't in the norm? Should it be?
Regards
Best Answer
Note that $p(x)$ is a seminorm such that $|f(x)|\leq p(x)$, so using the Hahn-Banach theorem there is a linear functional $F$ extending $f$ such that for all $x\in X$ $$|F(x)|\leq p(x)=\|f\|\|x\|.$$ So $$\|F\|=\sup_{\|x\|\leq 1}|F(x)|\leq\sup_{\|x\|\leq 1}p(x)=\|f\|.$$
I'm not sure what you mean with $p(x)$ being in the norm, but it might be useful to note that $p(x)$ is real number, not an element of $x$.