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

Question

It is well-known that for any ultrafilter $\mathscr{u}$ in $\mathbb{N}$, the map\begin{equation}a\mapsto \lim_{\mathscr{u}}a\end{equation} 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 \begin{equation} h(a)=\lim_{\mathscr{u}}a \end{equation} 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!

Answer 1

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 $$ A_j^{(n)}=\{m:\ \frac{j}{2^n}\leq c(m)<\frac{(j+1)}{2^n}\},\ \ j=0,1,\ldots,2^n. $$ 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 (if $A\cup B=\mathbb N$, then either $A\in\mathcal U$ or $B=A^c\in\mathcal U$; by induction, this applies to arbitrary partitions of $\mathbb N$).

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 $\ell^\infty(\mathbb{N})$.

Answer 2

Let me give a slightly different approach to the second half of Martin Argerami's answer. Suppose $\varphi:\ell^\infty\to\mathbb{C}$ is a multiplicative linear functional, and as in Martin Argerami's answer define $\mathcal{U}=\{A:\varphi(1_A)=1\}$ and prove that $\mathcal{U}$ is an ultrafilter. Now we just need to prove that $$\varphi(c)=\lim\limits_\mathcal{U} c$$ for any $c\in\ell^\infty$.

To prove this, let $L=\lim\limits_\mathcal{U} c$ and fix $\epsilon>0$. Since $L$ is the limit of $c$ along $\mathcal{U}$, the set $$A=\{n:|c(n)-L|<\epsilon\}$$ is in $\mathcal{U}$. Let $d=c\cdot 1_A$. Since $\varphi$ is multiplicative, $\varphi(d)=\varphi(c)\varphi(1_A)=\varphi(c)\cdot 1=\varphi(c)$. But by definition of $A$, $\|d-L1_A\|\leq\epsilon$. Thus $$|\varphi(c)-L|=|\varphi(c)-\varphi(L1_A)|=|\varphi(d)-\varphi(L1_A)|=|\varphi(d-L1_A)|\leq \|\varphi\|\epsilon.$$ Since $\epsilon$ was arbitrary, we conclude that $\varphi(c)=L$.

(Note that this argument, like Martin Argerami's, assumes $\varphi$ is bounded. You don't actually need to assume this; you can prove it. Indeed, $\|c\|$ can be described as the least $r\in [0,\infty)$ such that for all $\lambda\in\mathbb{C}$ such that $|\lambda|> r$, $\lambda-c$ has a multiplicative inverse. Since $\varphi$ preserves multiplicative inverses, $\lambda-\varphi(c)\neq 0$ whenever $|\lambda|> \|c\|$, so $|\varphi(c)|\leq \|c\|$.)