# The kernel of all invariant means

Let $G$ be discrete group which is amenable (i.e. it admits an left-invariant mean, i.e. a continuous positive normalised linear functional on $m:\ell^\infty(G) \to \mathbb{R}$ such that $\forall g \in G, m \circ \lambda_g = m$ where $\lambda_g : \ell^\infty(G) \to \ell^\infty(G)$ is the left-regular action of $G$). Let $\mathcal{M}$ be the set of all invariant means on $G$.

Let $c_0(G)$ be the space of bounded functions decreasing to $0$ at $\infty$ (i.e. $f \in c_0(G)$ if $f$ is in the closure of the space of fintily supported functions, i.e. $f \in \ell^\infty(G)$ and $\forall \epsilon >0$ there exists a finite set $F \subset G$ with $\|f\|_{\ell^\infty(G \setminus F)} \leq \epsilon$).

It's not difficult to see that $c_0(G)$ is included in the kernel of all (left- or right- or bi-)invariant mean, provided $G$ is infinite. When $G$ is finite, there is only one such mean, and it is trivial to compute the kernel. However the cardinality of $\mathcal{M}$ is otherwise quite big (if I remember correctly, it is uncountable for $G=\mathbb{Z}$).

$\textbf{Question}$: Assuming $G$ is infinite, is there a description/characterisation of the elements of $\ell^\infty(G)$ which belong to the kernel of all the invariant means, i.e. $\cap_{m \in \mathcal{M}} \ker m$?

-
The trivial answer would be the closed linear span of $\lbrace f-\lambda_g(f) : g \in G,\ f\in\ell_\infty G\rbrace$. –  Narutaka OZAWA Aug 18 '12 at 13:06
Did you compute it for $\mathbb{Z}$? –  Mark Sapir Aug 18 '12 at 13:07
Is this MO question useful to you? mathoverflow.net/questions/65325/… –  Alain Valette Aug 18 '12 at 13:41
@Mark: I'm not sure if this is explicit enough, but take any Folner sequence $F_n$. Then, $f$ has mean zero iff $\lim_{n\to\infty}\sup_{x\in G}|\sum_{y\in F_nx}f(y)|/|F_n|=0$. $\because$ If it's nonzero, then the uniform measure on $F_nx_n$ tends to an invariant mean which is nonzero at $f$. The converse follows from the fact that $f_n(x):=\sum_{y\in F_nx}f(y)/|F_n|$ has the same mean as $f$. –  Narutaka OZAWA Aug 18 '12 at 16:05
Thanks for the comments and references, it makes a good enough answer for me too! –  Antoine Aug 19 '12 at 10:08