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Let $G$ be a countable amenable group and $f\in\ell^\infty(G)$. Denote by $L,R,I$ respetively the sets of left-, right- and bi-invariant means on $G$. Denote by $M_L(f)$ (resp. $M_R(f),M_I(f)$) be the sets of values attained by the integral $\int f(x)d\mu(x)$, when $\mu$ goes over $L$ (resp. $R,I$).

Question: Is it always true that $M_L(f)=M_R(f)=M_I(f)$? Counterexample?

Thanks in advance,


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Write $G^{op}$ for $G$ with the multiplication reversed. Then $M_L(G^{op}) = M_R(G)$. But $G^{op} \cong G$ by $x \mapsto x^{-1}$, so $M_L(G) = M_R(G)$. Or am I overlooking something? – Tom Leinster Jan 23 '12 at 19:23
@Tom: You are overlooking the fact that these sets depend on a given function $f$. – Alain Valette Jan 23 '12 at 19:32
Oh, silly me. Thanks. – Tom Leinster Jan 23 '12 at 19:40
I'll edit the post to clarify that those sets depend on $f$. – Valerio Capraro Jan 23 '12 at 19:56
That's very nice of you, but I think it was perfectly clear before; I was just making a stupid mistake. – Tom Leinster Jan 23 '12 at 20:06
up vote 6 down vote accepted

For a given amenable group $G$, these sets will coincide for all $f \in \ell^\infty(G)$ if and only if the sets $L$, $R$, and $I$ coincide. Just notice that if $f \in \ell^\infty(G)$, and $g \in G$ then we have $M_L( f - \lambda_g(f) ) = \{ 0 \}$.

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As an addendum, for countable groups $L=R=I$ if and only if every conjugacy class is finite. See Paterson, AMENABLE GROUPS FOR WHICH EVERY TOPOLOGICAL LEFT INVARIANT MEAN IS INVARIANT, Pacific J. Math (1979). So for instance, the group of permutations on a countable set that fix all but finitely many elements is a counter-example (for which $f$?) – Valerio Capraro Jan 23 '12 at 20:59

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