I'm trying to prove that there exists a multiplicative linear functional in $\ell_\infty^*$ that extends the limit funcional that is defined in $c$ (i.e., im looking for a linear functional $f \colon \ell_\infty \to \mathbb K$ such that $f( (x_n * y_n) ) = f (x_n) f(y_n)$, for every $(x_n), (y_n) \in \ell_\infty$, and such that $f((x_n)) = \lim x_n$ , if $(x_n)$ converges).
I found a lot of references saying that it exists but I can't find a detailed proof. The usual Hahn-Banach approach doesn't work because I get a shift invariant functional, and thats inconsistent with multiplicavity. The references I found suggest that using ultrafilters to define the limit should work. I found this interesting proof of the reciprocal: every multiplicative linear functional in $\ell_\infty^*$ is a limit along an ultrafilter:
Every multiplicative linear functional on $\ell^{\infty}$ is the limit along an ultrafilter. it assumes $\mathbb K = \mathbb R$, but I could adapt so I think (assuming I made no mistakes) it works for $\mathbb C$ as well.
In sum, I'm looking for a proof of the result here
http://planetmath.org/BasicPropertiesOfALimitAlongAFilter
but it should work for $\mathbb C$ and the $(x_n) \mapsto \mathcal F -\lim (x_n)$ functional should be continuous. Is there a good reference for this? or is it just trivial?
thanks
Best Answer
This argument avoids the use of ultrafilters.
Let $e_n \in \ell_\infty^*$ be the evaluation map $e_n(x) = x_n$. The set $\{e_n\}$ is contained in the unit ball of $\ell_\infty^*$, which by Alaoglu's theorem is weak-* compact, hence $\{e_n\}$ has a weak-* cluster point; call it $f$. I claim this $f$ has the properties you desire.
It is easy to check that the set of multiplicative linear functionals is weak-* closed in $\ell_\infty^*$; each $e_n$ is multiplicative and hence so is $f$.
For each $x \in \ell^\infty$, let $\pi_x : \ell_\infty^* \to \mathbb{C}$ be the evaluation functional $\pi_x(g) = g(x)$. By definition of the weak-* topology, $\pi_x$ is weak-* continuous. Suppose $x \in c \subset \ell_\infty$ is convergent, with $x_n \to a$. By continuity of $\pi_x$, $f(x) = \pi_x(f)$ must be a cluster point of $\{\pi_x(e_n)\} = \{x_n\}$. But $x$ is a convergent sequence so the only cluster point of $\{x_n\}$ is $a$. Thus $f(x) = a$.
$f$ has another interesting property: since $\mathbb{C}$ is metric, all cluster points in $\mathbb{C}$ are subsequential limits. Thus for any $x \in \ell^\infty$, $f(x)$ is a subsequential limit of $\{x_n\}$. For instance, if $x_n = (-1)^n$, $f(x)$ must be either -1 or 1, whereas a Banach limit must assign $x$ the limit value of 0. A corollary of this is that for any real sequence $\{x_n\}$, we must have $\liminf x_n \le f(x) \le \limsup x_n$.
It is also interesting to note that $f$ is an element of $\ell_\infty^*$ that cannot correspond to an element of $\ell^1$. So this gives an alternate proof that $\ell^1$ is not reflexive.
If you like, you can instead produce $f$ as a limit of a subnet of $\{e_n\}$, or a limit of $\{e_n\}$ along an ultrafilter. In any case it is a cluster point.