Positive functionals on $\ell^\infty$ 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.
 A: Another interesting class of positive linear functionals on $\ell_\infty$ are $\mathcal F$-limits or limits along an ultrafilter.
Let $(x_n)$ be a real sequence and $\mathcal F$ be a filter on $\mathbb N$. A real number $L$ is $\mathcal F$-limit of this sequence if for each $\varepsilon>0$
$$\{n; |x_n-L|<\varepsilon\}\in\mathcal F.$$
It is known that if the sequence $(x_n)$ is bounded and $\mathcal F$ is an ultrafilter, then $\mathcal F$-limit exists and it is unique. See also this question for the proof of this fact and some references.
If I remember correctly, the $\mathcal F$-limits are precisely the extreme points of the set of all positive normed functionals from $\ell_\infty^*$. (By normed I mean $\lVert f \rVert =1 $.) They are characterized by the property, that they are multiplicative $\varphi(x.y)=\varphi(x).\varphi(y)$. (I.e., if a linear functional $\varphi\in\ell_\infty^*$ is positive, normed and multiplicative, then there is an ultrafilter $\mathcal F$ such that $\varphi$ is $\mathcal F$-limit.)

You may notice that it is not possible to have a functional on $\ell_\infty$ which extends limits and is both shift-invariant and multiplicative.
For the sequence $x=(1,0,1,0,\ldots)$ we have $x+Sx=\overline 1$ (where $S$ denotes the shift operator). If a functional is shift-invariant than $\varphi(x)=\varphi(Sx)=\frac12$.
On the other hand, if a functional $\varphi$ is multiplicative, we get $\varphi(x.Sx)=\varphi(x).\varphi(Sx)=0$, which leads to $(1/2)^2=0$, a contradiction.
A: 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.) 
