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I have been seeing various versions of Hahn Banach theorem . There are few confusions that i would like to clarify . 1) In some proof they assume the existence of a sublinear functional and the functional $f$ must satisfy $\le$ condition to the sublinear functional $p$ . Does there always exist a sublinear functional ? In some proofs they don't say anything about the sublinear functional , is it wrong ?

2) I am not able to appreciate the induction , because it assumes the existence of the extension of linear functional and finally says that maximal element should be the Space itself ? I particularly have a problem in applying the first induction step.

I would like to have some knowledge about my confusions and misunderstandings . Thanks

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Question mark key still sticky? – Rudy the Reindeer Nov 24 '12 at 19:52
@MattN. : Comm'on Matt N. You can't make fun of me always :D Its ok . I will try to put only one question mark :) – Theorem Nov 24 '12 at 19:56
Good! : ) ${}{}{}{}{}$ – Rudy the Reindeer Nov 24 '12 at 19:58
Why down votes ? Please can you give me a explanation ? I seriously think this doesn't deserve down voting . – Theorem Nov 24 '12 at 19:59
up vote 2 down vote accepted

For (1), in most applications we are working in a normed space $(X, \|\cdot\|)$, and we usually apply the Hahn-Banach theorem with the sublinear functional $p(x) = c \|x\|$ for some constant $c > 0$ (you can verify this is indeed sublinear). For instance, if we have a bounded linear functional on a subspace $M$ (so that $\|f(x)\| \le c \|x\|$ for all $x \in M$), applying Hahn-Banach with $p(x) = c\|x\|$ produces a bounded extension of $f$ to all of $X$.

For (2), it's hard to be specific without seeing the proof you are looking at, but basically the key step of the proof is that you can extend the functional to a larger subspace (in which the previous subspace has codimension 1). The induction guarantees that you can repeat this extension again and again, and you never have to stop until your subspace is all of $X$, at which point you are done.

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