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Hahn—Banach theorem says that : Let $E$ a $\mathbb R-$vector space and $p:E\to \mathbb R$ a sub-linear application. Let $G$ a subspace of $E$ and $g:G\to \mathbb R$ a linear application. Then there is $f:E\to \mathbb R$ s.t. $f(x)=g(x)$ for all $x\in G$ and $f(x)\leq p(x)$ for all $x\in E$.

In what this theorem is so important ? Is the fact that $\mathcal N(x):=|p(x)|$ is important or not really ? Since I'm not very used to the linear form and sub-linear application, I don't really see interesting thing behind, so if someone who can tell me more, it would be great.

user380364
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    this has been answered on papa.SE – Andres Mejia Aug 17 '18 at 17:11
  • The essence of Hahn-Banach's theorem is, you can always extend a limited linear functional with the same norm, this is a corollary of the theorem but it shows how strong this theorem is. It is fundamental in functional analysis both in applications and in the very development of theory, it can be shown that it is equivalent to the Axiom of choice. The sub-linear application is not so important, in general it is the very norm of space. – Olecram Aug 17 '18 at 17:19
  • @Olecram I'm no expert on the set theoretic side of things but I'm fairly sure that Hahn-Banach is strictly weaker than choice (in ZF). For example, the ultrafilter lemma implies HB and the ultrafilter lemma is afaik known to be strictly weaker than choice. Edit: the discussion here is relevant. – Rhys Steele Aug 17 '18 at 17:23
  • @Olecram Also, it's just wrong to say that the sublinear functional version isn't important. The sublinear functional version is what allows you to prove all of the geometric forms of HB (usually called Hahn-Banach separation theorems). These separation theorems are actually incredibly useful for many things in functional analysis. – Rhys Steele Aug 17 '18 at 17:31
  • Yes, you can't decuce AC from HB. –  Aug 17 '18 at 17:31

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