Re: Hahn-Banach, Order Unit question

In article <5220842.1157024553574.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>, james545@xxxxxxxxx says...


Dear all,

I've been working on the following problem (from Hahn-Banach theorem section) :

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



I'm surprised no one has commented on this. I don't know anything about
functional analysis, but I think in this case you want to use the
seminorm version of Hahn-Banach. The function lambda:V -> R given by
lambda(x) = inf{t : x+te is in V+} is a seminorm, and lambda(x)=0 for
all x in V+.

If we let U be the subspace genearated by e, then the functional on U
given by re |-> -r is dominated by lambda, hence extends to a
functional, f, on the whole space that is also dominated by lambda.
It follows that for x in V+, f(x) <= 0, whence -f is the positive
functional you're looking for.


Robert Sheskey



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