0
$\begingroup$

I am confused about something related to Hahn-Banach. According to my book, one corollary of H-B is that for $X$ a real or complex normed space, there exists $f \in X'$ such that $\|f\| = 1$ and $f(x) = \|x\|$.

But, is $f$ then linear? Because for the norm we have $f(\alpha x) = \|\alpha x\| = |\alpha|\|x\|$, and $f(x+y) = \|x+y\| \leq \|x\| + \|y\|$, how can this give linearity? Since $X' = B(X,\mathbb{F}) \subseteq L(X,\mathbb{F})$, $f$ should be linear right?

$\endgroup$
2
$\begingroup$

It is not the case that $f(x)=\|x\|$ for every $x\in X$. The corollary goes like this: Given a particular $x_0\in X$, there exists $f\in X'$ such that $\|f\|=1 $ and $f(x_0)=\|x_0\|$.

$\endgroup$
  • $\begingroup$ Indeed! So this is no problem for linearity right? $\endgroup$ – Sigurd Apr 15 at 12:26
  • $\begingroup$ Excatly. No problem at all. $\endgroup$ – uniquesolution Apr 15 at 12:26
  • $\begingroup$ Thanks for the clarification, I already thought it was something small I overlooked but that's cleared up now $\endgroup$ – Sigurd Apr 15 at 12:27
2
$\begingroup$

The equation $f(x)=\|x\|$ holds for a particular $x$, not for all $x$. $f$ itself depends on the given vector $f$.

$\endgroup$

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.