Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Hope this won't turn out to be a stupid question. If I want to apply the Hahn-Banach lemma to prove the Hahn-Banach theorem, then I want to define a function $p$ that provides an upper bound on my functional defined on a linear subspace $Y \subset X$. Say, I want to extend $f \in Y^\ast$. Then the notes define $p(x) = \|f\| \|x\|$. I don't quite believe this works since if $\|y\| \lt 1$ we might not have $f(y) \leq p(y) = \|f\| \|y\| = \sup_{\|x\|=1} |f(x)| \|y\|$. What am I missing? Thanks for your help.

share|improve this question
    
Yes. :) But now that you've fixed it, could you provide an example where we do not have the inequality? –  tomasz Aug 14 '12 at 15:11
    
@tomasz Well, given that the notes should be correct it should be impossible. But to me the thing looks like $f(y) \leq K$ where $K$ is a constant. Multiplying $K$ by something small makes it look as if I could break the inequality. –  Rudy the Reindeer Aug 14 '12 at 15:13

1 Answer 1

up vote 1 down vote accepted

The thing you're probably missing is the fact that the left hand side depends on the norm of $y$. Maybe you will be more convinced if you write it that way (for $y\neq 0$, if $y$ is zero both sides are zero, so the inequality holds): $$f(y)=f(y/\lVert y\rVert)\cdot\lVert y\rVert\leq \lVert f\rVert\lVert y\rVert$$ (notice that $\lVert (y/\lVert y\rVert) \rVert=1$).

share|improve this answer
    
Oh dear. I hate it when I ask obviously stupid questions. Thank you! –  Rudy the Reindeer Aug 14 '12 at 15:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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