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Update: The question is completely answered. I had overlooked a reduction to the self-adjoint case, and the latter can be proved using a Hahn-Banach separation theorem. Thanks to Matthew Daws for first pointing this out to me. Thanks also to Willie Wong for pointing out that I should have asked the author; I later did so, and he confirmed that a Hahn-Banach separation theorem was intended (and generously provided another argument). I want to point out, because this post may highlight a minor error in the book, that it is a great book.


When I hear "Krein-Milman", I think of the result at the link, a standard result in functional analysis. But recently I came across an invocation of "the Krein-Milman theorem (for cones)" which looks more like a Hahn-Banach type result, and I am having trouble tracking down what precisely the author is referring to (and why the result is true).

To make my question precise, I'll have to give the basic setup. Let $V$ be a complex vector space with a conjugate linear involution $*$, a norm that makes $*$ an isometry, and a norm-closed cone $C\subset\{v\in V:v=v^*\}$, meaning $C$ is closed under addition and nonnegative real scalar multiplication, and satisfies $C\cap(-C)=\{0\}$. (Such a $V$ is called an ordered $*$-vector space.)

To paraphrase part of Paulsen's exposition of the Choi-Effros abstract characterization of operator systems1:

If $x$ is in $V\setminus C$, then by the Krein-Milman theorem (for cones) there exists a linear functional $s:V\to\mathbb{C}$ with $s(C)\subseteq[0,\infty)$ and $s(x)<0$.

Because $C$ is a closed convex subset of a normed (and thus locally convex) space, my first inclination was to adapt some version of Hahn-Banach. However, I can't see how to do this while keeping complex linearity and the conclusion of the above claim. (There would be no problem in the real case, e.g. by Theorem 3.4 of Rudin's Functional Analysis, 2nd edition, page 59.) I've looked across the internet and my library's bookshelves to try to find what is meant, with no luck so far.

Question 1: Does anyone have a reference for (or statement and explanation of) the "Krein-Milman theorem for cones"?

Question 2: Am I missing a straightforward argument proving the above claim, for example by an application of Hahn-Banach?

I believe I have faithfully presented enough to cover everything relevant to my question, but the spaces in Paulsen's proof actually have a lot more structure, which you can find at the link above if you think it will help. Hopefully I didn't lose something essential in an effort to not get bogged down--this is clearly a risk because I don't know what theorem is being cited.

1 An operator system is a self-adjoint unital subspace of the algebra of bounded operators on a Hilbert space (or an abstract space that is completely order isomorphic to one of these).

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It just occurred to me: why not just ask him? As far as I can tell, he's alive and well and teaching at Houston. –  Willie Wong May 6 '10 at 23:21
    
Good point. It had occurred to me too, but I suspected I was missing something obvious and wanted a fast response (which I got). Now that the actual mathematical question is behind me, it doesn't seem as pressing. However, if I do ask and learn something, I'll update. –  Jonas Meyer May 6 '10 at 23:32
    
Completely by accident, I've just come across this Wikipedia page: en.wikipedia.org/wiki/… Perhaps Paulsen meant to reference this result? Although it's far from clear to me how it might help (and I've never heard of it before...) –  Matthew Daws May 18 '10 at 20:58
    
Interesting, I didn't know about that either. Thanks. –  Jonas Meyer May 19 '10 at 1:23

2 Answers 2

up vote 3 down vote accepted

Something is wrong with the question, as here's a counter-example. Let $V=c_0$ with the pointwise involution (so this is a commutative C*-algebra). Let C be the obvious cone: the collection of vectors all of whose coordinates are positive. Let $x=(i,0,0,\cdots)$. Then $V^* = \ell^1$, so if $s=(s_n)\in\ell^1$ satisfies $s(C)\subseteq[0,\infty)$, we need that $s_n\geq 0$ for all $n$. But then $s(x)$ is purely imaginary!

So, maybe you also need $x^*=x$. Under this assumption, here's a proof, but it has nothing to do with "Krein-Milman"...

As C is closed, $V\setminus C$ is open, so let A be an open ball about x which doesn't intersect C. Then A and C are disjoint, non-empty, convex, so by Hahn-Banach, as A is open, we can find a bounded linear map $\phi:V\rightarrow\mathbb C$ and $t\in\mathbb R$ with $$ \Re \phi(a) < t \leq \Re \phi(c) $$ for $a\in A$ and $c\in C$. This is e.g. from Rudin's book. As $0\in C$, we see that $t\leq 0$.

Now, we can lift the involution * from V to the dual of V. In particular, define $$ \phi^*(x) = \overline{ \phi(x^*) } \qquad (x\in V)$$

So let $\psi = (\phi+\phi^*)/2$. For $c\in C$, as $c^*=c$, notice that $\psi(c) = \Re \phi(c)$. Hence $0 \leq \psi(c)$ for all $c\in C$. Similarly, as $x^*=x$, we have that $\psi(x) = \Re\phi(x)<t\leq 0$, as $x\in A$.

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Thanks! I still don't know what was meant by Krein-Milman; I had suspected it was a typo, but I just noticed that the same invocation occurs in a previous chapter. The case $x*=x$ isn't quite enough, but all that is really needed is $s(x)\ngeq0$. Thanks again for pointing out that the statement isn't correct. I have to leave the computer now for a while. –  Jonas Meyer May 6 '10 at 13:12
    
I finally got a chance to come back to this. To finish the argument (which as I mentioned in the last comment only requires $s(x)\ngeq0$), I can just break it up into cases depending on whether or not $(x+x^*)/2, (x-x^*)/(2i)$ are in $C$. Perhaps this is what Paulsen had in mind, but he just forgot to mention the reduction (or implicitly left it to the reader). "Krein-Milman" might have just been a slip where "Hahn-Banach" was intended, and it's not too surprising that it would be repeated in 2 very similar arguments (the other being on page 77). Thanks again for the answer. –  Jonas Meyer May 6 '10 at 19:16
    
In retrospect, couldn't we simplify the example by taking $V=\mathbb{C}$? I like your example, but this makes it all the more sad that I didn't realize the problem. –  Jonas Meyer May 6 '10 at 19:53
    
Jonas: Yep, I agree, we could just look at the complex numbers! Also not sure why that didn't occur to me... –  Matthew Daws May 6 '10 at 20:55

I don't know if this is what you are looking for, but this is what comes to mind when you say Krein-Milman for cones. (I happened to be trawling the convex analysis literature for a different result... I am nowhere an expert in this.) The Krein-Milman theorem for cones that I know basically says something like (I maybe missing some details) "a closed convex cone is the convex hull of its extremal rays", and the definition of extremal ray is analogous to the definition of extremal points in convex sets: one indecomposible as a convex combination of two distinct rays.

There are more details in Convex Cones by Fuchssteiner and Lusky. Some papers that you may find useful:

Hope this helps.

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Thank you. I had come across a similar result in my search, but I hadn't seen those particular references, which look interesting. Unfortunately, this doesn't seem to be what Paulsen was getting at, which is part of the reason I was so confused. Nonetheless appreciated. –  Jonas Meyer May 6 '10 at 19:21

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