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Let $V$ be a vector subspace of $R^N$, and $l:V \to R$ a linear mapping such that $l(V\bigcap R_{+}^N)\subseteq R_{+}$ (i.e., $l$ is positive).

I have heard that there exists a separating hyperplane sort of argument that allows us to show that $l$ extends to a positive linear functional on $R^N$. I have my own proof, but it is complicated and uses the Bauer-Namioka condition for extension of positive linear functionals on ordered vector spaces of any dimension.

Question: What is this simple separating hyperplane argument that shows that $l$ extends to a positive linear functional on $R^N$? (I believe it should be quite straightforward, but I was not able to construct the right problem to invoke a separation argument...) A reference would be okay as well. Thanks.

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    What you want is also obtainable as a special case of the [Riesz extension theorem](http://en.wikipedia.org/wiki/M._Riesz_extension_theorem).2011-01-23
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    I think that would require additional assumptions on $V$, in particular that the subspace $V$ majorizes and minorizes the whole $R^N$. But think of the simplest case in which $N=2$ and $V$ is the line passing through 0 and (-1,1).2011-01-23
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    @Kevin: you are right. I've made that assumption (which I see you didn't use). But I suspect some modification of it can be made to work also in your case. See the answer I just posted below.2011-01-23
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    In particular, the right problem would be to construct a convex function $\Phi_+$ and a concave function $\Phi_-$ to extend $l$, with $\Phi_+ \geq \Phi_-$ and $\Phi_-$ positive on $\mathbb{R}^N_+$.2011-01-23
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    In the case $V\cap\mathbb{R}^N_+$ is non-empty, you can run Riesz. In the case it is empty, define $\Phi_+$ only on $V - \mathbb{R}^N_+$, and $\Phi_-$ on $V+\mathbb{R}^N_+$. The subsets of $\mathbb{R}^{N+1}$ given by $U_+ = \{ (x,y) | x\in V - \mathbb{R}^N_+, y > \Phi_+(x) \}$ and analogously $U_-$ are convex, disjoint sets. Apply separation hyperplane theorem.2011-01-23
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    @Willie: I believe the question is whether $V\bigcap R^N_{++}$ is empty, in which case I am almost convinced that your argument will work (but to be honest it is not quite simple as thought it would be..., but it is certainly by far the best shot). Thanks.2011-01-23

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(Edit: reading the comments left by Kevin above, Riesz extension requires a bit more than he assumed. But I think that the separating hyperplane argument would boil down to essentially the same trick as outlined below.)

I think you can use the following modified proof of the Riesz extension theorem.

Marcel Riesz Extension Theorem Let $W$ be a vector space (finite dimensional in the following proof; the actual theorem has no such limitations), and $V$ a subspace. Let $F$ be a convex cone in $W$ with the property that for every $w\in W$ there exists $v_+, v_- \in V$ such that $v_+ - w\in F$ and $w - v_- \in F$. Then for any $\phi$ linear on $V$ and positive on $V\cap F$, there exists an extension $\psi$ on $W$ that is positive on $F$.

Proof (sketch) Define $\psi_+(w) = \inf_{v-w\in F, v\in V}\phi(v)$ and $\psi_-(w) = \sup_{w-v\in F, v\in V}\phi(v)$. Check that $\psi_+$ is convex, and $\psi_-(w)$ is concave, and $\psi_+(w) \geq \psi_-(w)$. An application of the separating hyperplane theorem implies that there exists a linear map $\psi$ with $\psi_+ \geq \psi \geq \psi_-$. Since $\psi_+ = \psi_- |_V$, you have that $\psi$ is an extension. By definition $\psi_-|_F > 0$, and so you have the conclusion of the theorem.

(See also this post on Terry Tao's blog for some related concepts.)

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    Your condition on the cone is slightly redundant. It suffices to require that for every $w \in W$ there exists $v_{+} \in V$ such that $v_{+} - w \in F$. Fix such a $v_{+}$ for $w$ and apply the condition for $(-w)$ to get $\tilde{v} \in V$ such that $\tilde{v} + w \in F$. But then $2\tilde{v}+2w \in F$ since $F$ is a convex cone and for the same reason $x = (2\tilde{v} + 2w) + (v_{+} - w) \in F$. Put $v_{-} = - v_{+} - 2\tilde{v} \in V$ then $w - v_{-} = x \in V$.2011-01-23
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    @Theo: thanks. I'll leave the statement as it is because (a) this is how I've seen the theorem stated in the past and (b) I think adding the proof you just gave may be distracting to the main point.2011-01-23
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    Sure, I'll leave my comment, so nothing's lost. I just wanted to point that out. I do agree with (b) and probably (a) is a consequence of the fact that we're in good company concerning this point.2011-01-23
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Perhaps this will help (Exercise 12 seems similar).