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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.

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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
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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
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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

2 Answers 2

up vote 5 down vote accepted

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.

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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.)

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