# Every multiplicative linear functional on $\ell^{\infty}$ is the limit along an ultrafilter.

It is well-known that for any ultrafilter $\mathscr{u}$ in $\mathbb{N}$, the map$$a\mapsto \lim_{\mathscr{u}}a$$ is a multiplicative linear functional, where $\lim_{\mathscr{u}}a$ is the limit of the sequence $a$ along $\mathscr{u}$.

I vaguely remember someone once told me that every multiplicative linear functional on $\ell^{\infty}$ is of this form. That is, given a multiplicative linear functional $h$ on $\ell^{\infty}$, there is an ultrafilter $\mathscr{u}$ such that $$h(a)=\lim_{\mathscr{u}}a$$ for all $a\in\ell^{\infty}$.

However, I cannot find a proof to this. I can show that if $h$ is the evaluation at $n$, then $h$ corresponds to the principal ultrafilter centered at $n$, but there are other kinds of multiplicative functionals (all these must vanish on any linear combinations of point masses though).

Can somebody give a hint on how to do this latter case? Thanks!

-

There are several ways of doing this, but I'll go with the most "elementary".

Let $\varphi$ be a nonzero multiplicative functional on $\ell^\infty(\mathbb{N})$. Since $\varphi(1)=\varphi(1^2)=\varphi(1)^2$, we get that $\varphi(1)=1$ (it cannot be zero, because then $\varphi=0$).

Now let $a\in\ell^\infty(\mathbb{N})$ such that $a(n)\in\{0,1\}$ for all $n$. Write $\alpha=\varphi(a)$. As $a(1-a)=0$, we have $$0=\varphi(a(1-a))=\varphi(a)\varphi(1-a)=\alpha(1-\alpha).$$ So either $\alpha=0$ or $\alpha=1$.

Note that we can write $a=1_A$, $A\subset\mathbb{N}$, where $A=\{n: a(n)=1\}$. Now define $$\mathcal U=\{A:\ \varphi(1_A)=1\}.$$ We can see that

• $\mathbb{N}\in\mathcal U$ (since $\varphi(1)=1$)
• $A\in\mathcal U\ \iff\ A^c\not\in\mathcal U$ (because $1_A\,1_{A^c}=0$)
• If $A,B\in\mathcal U$, then $A\cap B\in\mathcal U$ (because $1_{A\cap B}=1_A\,1_B$)
• If $A\in\mathcal U$ and $A\subset B$, then $B\in\mathcal U$ (because $1_A=1_A\,1_B$)

In other words, $\mathcal U$ is an ultrafilter.

Now let $c\in\ell^\infty(\mathbb{N})$ be positive, i.e. $0\leq c\leq 1$. Define sets $$A_j^{(n)}=\{m:\ \frac{j}{2^n}\leq c(m)<\frac{(j+1)}{2^n}\},\ \ j=0,1,\ldots,2^n-1.$$ For fixed $n$, these sets are pairwise disjoint and $$\tag{1}\bigcup_jA_j^{(n)}=\mathbb{N}.$$ As $\mathcal U$ is an ultrafilter, for each $n$ there is exactly one $j(n)$ such that $A_{j(n)}^{(n)}\in\mathcal U$, and none of the others is.

Define $$c_n=\sum_{j=0}^{2^n-1}\,\frac{j}{2^n}\,1_{A_j^{(n)}}.$$ By definition, $\|c-c_n\|\leq 2^{-n}$, so $c_n\to c$ in norm. As $\varphi$ is norm-continuous, we have $\varphi(c)=\lim_n\varphi(c_n)$. And $$\varphi(c_n)=\sum_{j=0}^{2^n-1}\,\frac{j}{2^n}\,\varphi(1_{A_j^{(n)}})=\frac{j(n)}{2^n},$$ so $$\varphi(c)=\lim_n \ c(j(n))=\lim_{\mathcal U}\ c.$$ Last step is to extend by linearity to all of $\ell^\infty(\mathbb{N})$.

-
Argh! I was in the process of typing in an answer along these lines already! Oh well, wasted effort … +1 from me then. – Harald Hanche-Olsen Jul 26 '12 at 19:30
I'm sorry about that. This idea is more or less clear in my mind, because until a few years ago ultrafilters were a complete mistery to me; and then in 2009 I attended a talk by Vladimir Pestov where these ideas were crystal clear. – Martin Argerami Jul 26 '12 at 19:55
This is crystal clear, Martin! – Jon Bannon Jul 26 '12 at 20:00
Heh. No need to apologize. It's all part of the game. – Harald Hanche-Olsen Jul 26 '12 at 20:15
The other immediate way that comes to mind is to notice that $\ell^\infty(\mathbb{N})$ is isomorphic to $C(\beta\mathbb{N})$ (continuous functions on the Stone-Cech compactification of $\mathbb{N}$. Then the multiplicative functionals are the point evaluations; functionals coming from principal filters are the evaluations at points of $\mathbb{N}$, and those coming from free ultrafilters are those evaluations at the corona. But now that I think about it, the computations above are probably still necessary. – Martin Argerami Jul 27 '12 at 1:16