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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.

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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
up vote 2 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 $\lVert (y/\lVert y\rVert) \rVert=1$).

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

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