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Let $(G,\cdot)$ be a group and $\phi:G\rightarrow\mathbb R$ bounded. Let me say that the pair $(G,\phi)$ is amenable if there is a finitely additive probability measure $\mu$ on $G$ such that for all $y\in G$

$$ \int \phi(x)d\mu(x)=\int \phi(x\cdot y)d\mu(x)=\int\phi(y\cdot x)d\mu(x) $$

Question: Does there exist a non-amenable group such that the pair $(G,\phi)$ is amenable for all $\phi\in\ell^\infty(G)$?

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  • $\begingroup$ How about the free group with 2 generators? $\endgroup$
    – user6976
    Commented Sep 4, 2011 at 15:05
  • $\begingroup$ I have no idea. Is it trivial that for any $\phi$ there is such a measure? $\endgroup$
    – Valerio Capraro
    Commented Sep 4, 2011 at 15:28
  • $\begingroup$ Did you try any $\phi$ at all? $\endgroup$
    – user6976
    Commented Sep 4, 2011 at 15:57
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    $\begingroup$ There are ergodic theorems for free group actions (Bufetov, Alexander I. Convergence of spherical averages for actions of free groups. Ann. of Math. (2) 155 (2002), no. 3, 929–944.) Perhaps that can help. But I would suggest that you ask the simplest question for which you do not know the answer first. How about the indicator function of a generator? $\endgroup$
    – user6976
    Commented Sep 4, 2011 at 17:49
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    $\begingroup$ I think you should try to consider the indicator case yourself. It should be an easy exercise. Then you will get an idea what to do next. $\endgroup$
    – user6976
    Commented Sep 4, 2011 at 23:32

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The answer is no, no such non-amenable group can exist. It follows from Justin Moore's answer to his own question that a single characteristic function can witness the non-amenability of a group.

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  • $\begingroup$ many thanks for the link. I will try to get into the details tomorrow. $\endgroup$
    – Valerio Capraro
    Commented Sep 11, 2011 at 1:42

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