Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
What you want is also obtainable as a special case of the Riesz extension theorem. –  Willie Wong Jan 23 '11 at 15:14
    
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). –  user6141 Jan 23 '11 at 15:30
    
@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. –  Willie Wong Jan 23 '11 at 15:40
    
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 $\mathbb{R}^N_+$. –  Willie Wong Jan 23 '11 at 15:45
    
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. –  Willie Wong Jan 23 '11 at 15:58
show 1 more comment

2 Answers

up vote 2 down vote accepted

(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.)

share|improve this answer
    
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 $w - v_{-} = x \in V$. –  t.b. Jan 23 '11 at 16:17
    
@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. –  Willie Wong Jan 23 '11 at 16:34
    
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. –  t.b. Jan 23 '11 at 16:46
add comment

Perhaps this will help (Exercise 12 seems similar).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.