# Why can we assume the existence of extension of linear functional that are dominated by some sub-linear functional?

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.

• 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. – Ittay Weiss Apr 28 '18 at 19:17
• @IttayWeiss I am still confused... why can we just assume that this class is non-empty? – Zheng Xie Apr 28 '18 at 21:08
• 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. – Ittay Weiss Apr 29 '18 at 12:21
• @IttayWeiss thank you very much! I get it – Zheng Xie Apr 29 '18 at 22:30

## 1 Answer

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.

• 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). – Ittay Weiss Apr 29 '18 at 12:20
• Yes, you are right. I have edited the answer. – Martin Argerami Apr 29 '18 at 14:53
• 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? – Zheng Xie Apr 29 '18 at 22:46
• it's clear now, thanks – Zheng Xie Apr 29 '18 at 22:55