TITLE: Positive functionals on $\ell^\infty$
QUESTION [5 upvotes]: A continuous linear functional $\varphi: \ell^\infty \to \mathbb{R}$ is said to be positive if
$$ x \ge 0 \rightarrow \varphi(x) \ge 0 \quad \forall x \in \ell^\infty.$$
If $\varphi$ is in $\ell^1 \subset {(\ell^\infty)}^{*}$, then exists $(\alpha_n) \in \ell^1$ such that  $\varphi(x) = \sum \alpha_n x_n$. If $(\alpha_n) \ge 0$, then $\varphi$ is positive.
If $\varphi$ is a Banach limit, then it's positive.
Are there other good/simple/notable classes of positive operators in ${(\ell^\infty)}^*$?
[edit] As stated in the comments below, feel free to give examples and constructions based on the Axiom of Choice.

REPLY [4 votes]: It may be worth observing that the cone of positive functionals is weak-* closed.  Combining this with the Banach-Alaoglu theorem gives another technique for producing positive functionals: any norm-bounded set of positive functionals has a weak-* limit point, which is again a positive functional.
For example, if $\phi_n$ is the evaluation functional $\phi_n(x) = x_n$, then any weak-* limit point $\phi$ of the sequence $\{\phi_n\}$ is a positive functional.  These have the interesting property that $\phi(x)$ is always a limit point (in particular, a subsequential limit) of the bounded sequence $\{x_n\}$.  These are rather different from Banach limits.  (Consider $x = (1,0,1,0,\dots)$; a Banach limit $\psi$ must have $\psi(x) = 1/2$, but $\phi(x)$ is either 0 or 1.)