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Hello to everyone, My problem is the following: I have this version of the Hahn-Banach theorem:

Let V be a vector space and let $p:V\rightarrow \mathbb{R}$ be any convex function. Let $W$ be a vector subspace of $V$ and let $L:W\rightarrow \mathbb{R}$ be a linear functional dominated by $p$ on $W$. Then there is a (not generally unique) linear extension $\hat{L}$ of $L$ to $V$ that is dominated by $p$ on $V$. Furthermore $\hat{L}_{|U}=L$.

Does the theorem still hold when $p:V\rightarrow(-\infty,+\infty]$ ? Is someone able to give me a proof or to provide a counter-example that show that the theorem does not hold? And if it does not hold, it is possible to add some conditions that make it still true?

To put it in another way, is the same true if we further relax the hypothesis on $p$ and allow it to be real extended with nontrivial domain, i.e. $\lbrace x\in V: p(x) \in \mathbb{R} \rbrace \neq \emptyset $ ?

Thanks to everyone in advance for helping me.

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"And if it does not hold, it is possible to add some conditions that make it still true?" -- of course it is. But it would help if you thought more, and then told us, what kind of extra conditions you are prepared to impose. –  Yemon Choi Oct 9 '12 at 22:27
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up vote 7 down vote accepted

The Hahn-Banach theorem is wrong for extended real $p$: For $V=\mathbb R^2$ let $p$ be the Minkowski functional of $A= \mathbb R \times (0,\infty)$ (so that $p(x,y)=0$ if $y>0$ and $p(x,y)=\infty$ if $y\le 0$), $W= \mathbb R \times \lbrace 0 \rbrace$, and $L: W\to\mathbb R$ defined by $L(x,0)=x$.

Then $L$ is $p$-dominated but there is no $p$-dominated linear extension.

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Thank you very much for the counter-example. Have you any idea of additional hypothesis that would make the theorem still true with this kind of function? –  alef87 Oct 9 '12 at 21:48
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I think that if p is convex, positively homogeneous and lower semicontinuous, then Hahn-Banach would hold true. –  Luis Silvestre Oct 9 '12 at 23:29
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