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Let $G$ be a discrete group which is amenable (i.e. it admits an left-invariant mean, i.e. a continuous positive normalised linear functional $m:\ell^\infty(G) \to \mathbb{R}$ 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 finitely supported functions, i.e. $f \in \ell^\infty(G)$ and $\forall \epsilon >0$ there exists a finite set $F \subset G$ with $\lVert f\rVert_{\ell^\infty(G \setminus F)} \leq \epsilon$).

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 $G=\mathbb{Z}$).

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. $\bigcap_{m \in \mathcal{M}} \ker m$?

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    $\begingroup$ The trivial answer would be the closed linear span of $\lbrace f-\lambda_g(f) : g \in G,\ f\in\ell_\infty G\rbrace$. $\endgroup$
    – Narutaka OZAWA
    Commented Aug 18, 2012 at 13:06
  • $\begingroup$ Did you compute it for $\mathbb{Z}$? $\endgroup$
    – user6976
    Commented Aug 18, 2012 at 13:07
  • $\begingroup$ Is this MO question useful to you? mathoverflow.net/questions/65325/… $\endgroup$
    – Alain Valette
    Commented Aug 18, 2012 at 13:41
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    $\begingroup$ @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 $f$. $\endgroup$
    – Narutaka OZAWA
    Commented Aug 18, 2012 at 16:05
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    $\begingroup$ Thanks for the comments and references, it makes a good enough answer for me too! $\endgroup$
    – ARG
    Commented Aug 19, 2012 at 10:08

2 Answers 2

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(Essentially from Narutaka Ozawa's comments)

Let $K = \bigcap_{m \in \mathcal{M}} \ker m$. The trivial answer would be that $K$ is the closed linear span of $$S =\{f-\lambda_g(f):\;g\in G,\;f\in\ell^\infty(G) \}.$$

For a non-amenable group this is all of $\ell^\infty(G)$. For an amenable group with Følner net $(F_i)$, then for any $f \in K$
$$\lim_{i\to\infty}\sup_{x\in G}\frac{\left|\sum_{y\in F_ix}f(y)\right|}{|F_i|}=0.$$ Indeed, if it's nonzero, we can find $(x_i)$ such that $$\limsup\frac{\left|\sum_{y\in F_ix_i}f(y)\right|}{|F_i|}>0;$$ then the uniform measure on $F_ix_i$ accumulates to an invariant mean that is non-zero at $f$.

Hence, if $f \in K$, then $$f_i(x)=\sum_{y\in F_ix}\frac{f(y)}{|F_i|}$$ has the same mean as $f$ for every invariant mean. A consequence of the above is that $\|f_i\|_{\ell^\infty} \to 0$. Thus $f - f_i \in S$ and $f-f_i \to f$ which shows that $K \subset \overline{S}$.

The inclusion $\overline{S} \subset K$ is a direct consequence of the invariance of the mean.

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  • $\begingroup$ I made quite a big edit, please feel free to roll back if I messed it up. Reading this proof after a few year, the crucial point that elements of $K$ are limits of elements of $S$ was hard to understand. $\endgroup$
    – ARG
    Commented Mar 7, 2022 at 13:01
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    $\begingroup$ @ARG it's fine; just prefer $\bigcap$ (bigcap) for multiple intersections to $\cap$ (cap) which is only for intersections as binary operator. $\endgroup$
    – YCor
    Commented Mar 7, 2022 at 13:06
  • $\begingroup$ ah, sorry... I usually use \displaystyle (which normally turns the $\cap$ into $\bigcap$ automatically) but forgot to do it here $\endgroup$
    – ARG
    Commented Mar 7, 2022 at 13:15
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    $\begingroup$ @ARG, \displaystyle doesn't convert $\cap$ \cap to $\bigcap$ \bigcap; see $\cap\bigcap\displaystyle\cap\bigcap$ \cap\bigcap\displaystyle\cap\bigcap. You always need to use it manually when appropriate. $\endgroup$
    – LSpice
    Commented Mar 7, 2022 at 14:22
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    $\begingroup$ @Lspice oh! cool, thanks for the latex-info! As for the math: it is not trivial to me! (that's most of the content of the answer). (1) any element of the kernel satisfies (*) (the equation with $\displaystyle \lim_{i \to \infty} \sup_{x \in G} ... =0$) (2) so do the $f_i$ (because they also have 0 mean. (3) $f-f_i \in S$ (it's a convex combination of some $f - \lambda_g f$) (4) because of (2), the $f-f_i$ tend to $f$ (for this you have to interpret (2) as saying that the $f_i \to 0$ in norm). (5) this is how you show $K$ is contained in the closure of $S$. $\endgroup$
    – ARG
    Commented Mar 7, 2022 at 14:28
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The answer is in this paper: Yuji Takahashi, Functions with a unique mean value and amenability. Proc. Amer. Math. Soc. 121 (1994), 775-777 (Freely accessible here on AMS' site)

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    $\begingroup$ Actually, Yuji Takahashi only cites the Theorem 1.1 (and proves something nice in the context) in Rosenblatt and Yang which essentially answers the question. (There it is formulated as a result on functions which always have the same means, so you have to consider the constant function.) The Paper of Rosenblatt and Yang is also freely availiable. $\endgroup$
    – ARG
    Commented Mar 7, 2022 at 13:06
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    $\begingroup$ @ARG, to protect against re-organisations of journal sites (Springer is a particularly bad offender, for example), it is often good to link to the DOI when possible: Rosenblatt and Yang - Functions with a unique mean value. $\endgroup$
    – LSpice
    Commented Mar 7, 2022 at 14:25

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