Hahn-Banach, Order Units, Linear functionals

Dear all,

I've been working on the following problem :

Suppose V is a vector space over R with a positive cone; that is, there is a subset V+ of V, closed under addition and multiplication by non-negative scalars, such that V = V+ - V+ and V+ and -V+ intersect only in {0}. A linear functional f on V is called positive if f is non-negative on V+. We write x <= y if y - x lies in V+. Suppose V has an order unit, that is, there is an element e in V such that for any x in V, there is a real number t > 0 such that tx <= e. Show there is a positive linear functional f on V, with f(e) = 1. Next, suppose x is non-zero in V+. Show there is a positive linear functional f on V, with f(e) = 1 and for which f(x) is strictly positive.

Here's what I've done : I've showed that there is a linear functional f on V such that f(e) = 1, but not necessarily positive. The way I did this was to show that V has a norm given by ||x|| = inf {c : -ce <= x <= ce}, where <= is given by y <= z iff z - y is in V+. (I didn't come up with this norm on my own, I found it somewhere online. Also, I am using a different (and probably equivalent) definition of order unit, which says that e is an order unit if for all x in V there is a c in R such that -ce <= x <= ce.). Then, I showed that -||x||e <= x <= ||x||e. So, ||e|| = 1, and then I know that for any normed vector space W, and any fixed w in W, there is a linear functional g such that g(w) = ||w|| (by Hahn-Banach). Hence, using all of this I show that there's a linear functional f on V such that f(e) = ||e|| = 1. This feels like a lot of work just to show that there's a linear functional f such that f(e) = 1.

Anyway, I never got that f is a positive linear functional. Any thoughts? Is there another way I can go about this problem?

-James
.