TITLE: Every multiplicative linear functional on $\ell^{\infty}$ is the limit along an ultrafilter.
QUESTION [15 upvotes]: 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!

REPLY [3 votes]: 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 $|\varphi(c)|\leq \|c\|$.)