**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](https://books.google.com/books?id=VtSFHDABxMIC&printsec=frontcover&dq=paulsen+completely&hl=en&ei=WhnjS7rPLYXCNaOP5PAC&sa=X&oi=book_result&ct=result#v=onepage&q&f=false).




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When I hear "[Krein-Milman](https://planetmath.org/KreinMilmanTheorem)", 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](https://books.google.com/books?id=4WgnHzOdHt4C&lpg=PP1&dq=paulsen%25252520completely%25252520bounded&pg=PA179#v=onepage&q&f=false) of the Choi-Effros abstract [characterization](https://mathscinet.ams.org/mathscinet-getitem?mr=430809) 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).



What is the "Krein-Milman theorem for cones"?
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 $x\in A$.


