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?


  • $\begingroup$ Are we aware that $\mathcal F$-$\lim(x_n)$ always exists if $(x_n)$ is bounded and $\mathcal F$ is ultrafilter? $\endgroup$ – Berci Mar 29 '13 at 0:17
  • $\begingroup$ (Meanwhile I found the answer on it: assume $A<x_n<B$ for all $n$, then split the interval $[A,B]$ on $N$ equal pieces $I_1,..,I_N$, and define $U(i):=\{n\mid x_n\in I_i\}$, then $U(i)\in\mathcal F$ for exactly one $i$, so we will get a decreasing sequence of intervals..) $\endgroup$ – Berci Mar 29 '13 at 0:48
  • $\begingroup$ in general, if $(x_n)$ lies inside a compact set then $\mathcal F-\lim(x_n)$ exists, for every $\mathcal F$ ultrafilter. $\endgroup$ – Rafael Mar 29 '13 at 4:04
  • $\begingroup$ Maybe you can find something useful among the references about $\mathcal F$-limits listed here. There was also this question, see also this discussion in chat. $\endgroup$ – Martin Sleziak Oct 9 '15 at 4:43

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.

  • $\begingroup$ Banach-Alaoglu is actually equivalent to the ultrafilter lemma, so in that sense this argument doesn't avoid the use of ultrafilters! $\endgroup$ – Qiaochu Yuan Mar 29 '13 at 0:53
  • $\begingroup$ thanks guys ! i appreciate it $\endgroup$ – Rafael Mar 29 '13 at 4:17

$\lim_{\mathcal F}$ is of course continuous, and has norm $1$, as $$|\lim_{\mathcal F}(x_n)|\le \|(x_n)\|_\infty\,,$$ and on constant sequences it holds with equality.

  • $\begingroup$ yeah, i found out that $\mathcal F - \lim x_n$ is in the closure of $\{x_n\}$, so it follows that $\lvert \mathcal F - \lim x_n \rvert \le \sup \lvert x_n \rvert $ $\endgroup$ – Rafael Mar 29 '13 at 4:09

I have done similar work before, which is included in the paper: Chao You, A note on $\tau$-convergence, $\tau$-convergent algebra and applications, Topology Appl. 159, No. 5, 1433-1438 (2012). It may happen that you cannot define a multiplicative linear functional on $l^{\infty}(\mathbb{N})$, but on a subalgebra of it. I used the concept of multiplier there. Hopefully, this will be useful to you. Good luck!

  • 1
    $\begingroup$ thanks :) i found out that an ultrafilter is free iff is contains all cofinite sets. this made linearity and multiplicativity pretty much like the analogous result for sequences, and proving that it actually extends the usual limit also became straightforward $\endgroup$ – Rafael Mar 29 '13 at 4:15

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