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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.

t.b.
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Angelo Lucia
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    The dual space of $\ell^{\infty}$ consists of the (signed) finitely additive measures on $\mathbb{N}$, so you're asking about the construction of (special/notable) positive finitely-additive measures. Non-trivial examples cannot be constructed without using some choice. You may wish to look at my answer here, and also visit this blog post of Tao – t.b. Aug 24 '11 at 11:42
  • Hm. I didn't tell you anything that isn't already on the wiki-page you linked to. A notable class (still a Banach limit) are so-called medial means. – t.b. Aug 24 '11 at 11:45
  • I had looked for questions about "positive operator", but I forgot to look for "non-negative operators"... anyway, I'm perfectly fine with using AC, and I fact I edited my question adding the Banach limits as another class of positive operators. Thank you for the links! – Angelo Lucia Aug 24 '11 at 11:53
  • I didn't mean to imply that you shouldn't use AC, I tried to point out that you can't avoid it. In my skewed perception shift-invariance is the most interesting thing, because it relates to amenability and thus to Banach-Tarski (BT works because $SO(3)$ contains a free group on two generators which is not amenable). You can ask for multiplicativity instead of shift-invariance, but not for both, as is explained on Tao's blog. Maybe Martin Sleziak will chime in and explain further things. – t.b. Aug 24 '11 at 12:09
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    Maybe this trivial observation is a more on-topic remark: Let $1$ be the constant sequence $(1,1,1,\ldots)$. Prove that a linear functional on $\ell^{\infty}$ is positive if and only if $\phi(1) = |\phi|$, so this gives you a means of constructing many examples if you don't care about further properties. – t.b. Aug 24 '11 at 12:29
  • BTW the wikipedia article you linked contains some mistakes - as pointed out on its talk page. It seems that someone changed the article to attain greater generality (to work with complex-valued sequences instead of real-valued) and did not change everything that was need. (To be honest, I knew only about real-valued Banach limits until now.) – Martin Sleziak Aug 24 '11 at 14:07
  • @Theo: I am confused about your trivial observation. If $\phi$ is a positive functional then so is $2\phi$. Did you want to add some additional condition, like $\phi$ extends the usual limit? – Martin Sleziak Aug 24 '11 at 14:11
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    @Martin: I meant what I wrote: $x$ is positive if and only if $| |x| \cdot 1 - x | \leq |x|$. Thus if $\phi(1) = |\phi|$ and $x \geq 0$ we have $||\phi| \cdot |x| - \phi(x)| = |\phi(|x|\cdot 1 - x)| \leq |\phi|\cdot |x|$ but this means that $\phi(x) \geq 0$. The other direction is even easier. Note also that replacing $\phi$ by $2\phi$ multiplies both sides of $\phi(1) = |\phi|$ by $2$. – t.b. Aug 24 '11 at 14:36
  • @Martin: by the way, there is no problem with working with complex sequences and complex-valued functionals (as long as you keep them complex linear). Simply observe that you can decompose everything (sequences and values) into real and imaginary parts (just as with integrals). The underlying theory is the theory of Riesz spaces (Wiki only speaks of the real case, but rest assured...). Recall also the proof of $\mathbb{C}$-linear Hahn-Banach, which essentially works with the real part. – t.b. Aug 24 '11 at 14:46
  • @Theo: Thanks for explaining, I made a very stupid mistake there. I just noticed that you predicted that I will add an answer before I did so. Nice. – Martin Sleziak Aug 24 '11 at 15:58

2 Answers2

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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.

Community
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Martin Sleziak
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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.)

Nate Eldredge
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