# Hahn Banach theorem with no dominating sublinear functional

Let $V$ be a vector space and $M$ be subspace of it. If $f$ is a linear functional on $M$, is it possible to extend it to the whole space $V$? If we have a sublinear functional $p$ on $V$ dominating $f$ on $M$, then by Hahn Banach we know that there is an extension. I would like to know if the same statement is valid with out the hypothesis of domination sublinear functional. I would like to understand the role of sublinear functional in the proof of Hahn Banach. Thanks

• Sure it's possible, extend a basis of $M$ to a basis of $V$, and define the extension arbitrarily on the basis vectors not in $M$. The domination by a sublinear functional is something that makes the existence of an extension more difficult to prove. The important result is that often, you have a continuous extension. – Daniel Fischer Aug 18 '13 at 17:54
• what if $V$ is infinite dimensional and there is no basis? – chandu1729 Aug 18 '13 at 17:56
• @chandu1729 The axiom of choice (or Zorn's lemma if you prefer) allows you to always find bases of vector spaces. Again to echo Daniel, the crucial thing about Hahn-Banach is the relationship with continuity. – Sharkos Aug 18 '13 at 17:59
• The nice thing about dominating the sublinear functional is that many times we can get information about the norm of the extension, about continuity, etc. An extension always exists, but doesn't always give all the information or characteristics we want. – rfauffar Aug 18 '13 at 19:05
• Look for "Hamel basis", a term often used for infinite-dimensional spaces. – GEdgar Aug 18 '13 at 19:15

## 1 Answer

I would like to understand the role of sublinear functional in the proof of Hahn Banach.

The primary purpose is to ensure that when a space has a norm, we can extend a bounded functional to another bounded linear functional. Much of functional analysis revolves around bounded linear functionals.

As pointed out in comments, if we do not care about boundedness/majorization, the result is also true, and easier to prove. Also, not as useful.