For the following, I say a vector $v>0$ if each of its coordinates is nonnegative and at least one is positive, and $f$ is positive if $v>0$ implies $f(v)\geq 0$. My question is:
Suppose $S$ is a subspace of $\mathbb{R}^n$. Is it true that a nonzero linear functional $f$ on $S$ is positive if and only if its Riesz vector $r_f>0$?
I know this is true on $\mathbb{R}^n$, since if $(e_1,\dots,e_n)$ is the standard basis, then the Riesz vector for $f$ is just $(f(e_1),\dots,f(e_n))$, and the rest follows easily. For me, the difficulty is showing whether it is true or not for an arbitrary subspace $S$ is that I'm not sure if there exists a basis of $S$ consisting of orthonormal, positive vectors.
I thought I could find a subspace that does not have such a nice basis, and then construct a counterexample by tweaking various scalars. So far each attempt has shown that the claim is in fact true, but I can't prove it for sure. Is there a proof or counterexample in the case of an arbitrary subspace $S$ of $\mathbb{R}^n$? Thanks.
Counterexample : $E=\mathbb{R}^3$, $S = \text{Vect}((0,1,1), (1,0,1))$, $f$ is defined by its Riesz vector $r_f = (2,-1,1)$. We can show that $f>0$ :
Let $u=(0,1,1)$ and $v=(1,0,1)$. Every positive element of S can be written as $au+bv$ with $a,b\geq 0$. (This is easy : $au+bv > 0 \Rightarrow (a,a+b,b)>0 \Rightarrow [a>0,b>0]$)
So taking the dot product of these elements with $r_x$ gives a positive number.