1
$\begingroup$

In the proof of Hahn-Banach Theorem, we first form the collection of all linear functionals on vector subspaces that extend a given functional f and that are dominated by a sub-linear functional p.

I was wondering why can we assume the existence of such extensions?

Can anybody help? I am a beginner in functional analysis. Thanks.

$\endgroup$
  • $\begingroup$ The given functional $f$ is assumed to be bounded by $p$, so that $f$ belongs to the class you describe, hence it is not empty. $\endgroup$ – Ittay Weiss Apr 28 '18 at 19:17
  • $\begingroup$ @IttayWeiss I am still confused... why can we just assume that this class is non-empty? $\endgroup$ – Zheng Xie Apr 28 '18 at 21:08
  • $\begingroup$ You’re not just assuming the set is not empty. The set is defined as the bounded extensions of f. Since f itself is a bounded extension of itself, the set contains at least f. $\endgroup$ – Ittay Weiss Apr 29 '18 at 12:21
  • $\begingroup$ @IttayWeiss thank you very much! I get it $\endgroup$ – Zheng Xie Apr 29 '18 at 22:30
0
$\begingroup$

In HB you start with a functional $f$ defined on a subspace $W$ of $V$, and $|f|\leq p$ on $W$. Then, as you say, you consider the collection of pairs $(W',f')$ such that $W\subset W'$ and $f'$ extends $f$. Such collection is always nonempty because it contains the pair $(W,f)$.

As it turns out, the usual proofs of HB include the specific proof that if $v_0\not\in W$, there exists an extension $\tilde f$ of $f$ to $\operatorname{span}\{V,v_0\}$ such that $|\tilde f|\leq p$. So one can extend "one-by-one" dimension wise. Of course we need Zorn to deal with the infinite-dimensional case.

$\endgroup$
  • $\begingroup$ I disagree that this key element is needed to show a nonempty chain exists. The steps of the proof are: 1) Define the relevant set (which is not empty since f extends itself); 2) Showing that every chain has an upper bound (pretty much the usual argument); 3) Showing the guaranteed maximal element is defined on the entire space (which is where the key step is used by assuming the negation and obtaining a larger extension). $\endgroup$ – Ittay Weiss Apr 29 '18 at 12:20
  • $\begingroup$ Yes, you are right. I have edited the answer. $\endgroup$ – Martin Argerami Apr 29 '18 at 14:53
  • $\begingroup$ Thank you so much! It seems that we can extend it "one-by-one" dimension wise as you said by $f'(w+cv_0)=f(w)+cf(v_0)$, is it correct? But for infinite-dimensional case how can we show that this chain has an upper bound? $\endgroup$ – Zheng Xie Apr 29 '18 at 22:46
  • $\begingroup$ it's clear now, thanks $\endgroup$ – Zheng Xie Apr 29 '18 at 22:55

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.