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Given an amenable group, it is a standard trick to turn a 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 $G$) into a bi-invariant mean (also invariant under pre-composition with the right-regular action of $G$).

From this bi-invariant mean one gets a sequence (or a net, when $G$ is uncountable) of almost invariant probability measures (i.e. $\xi_n \in \ell_1G$ with $\| \lambda_g \xi_n - \xi_n\|_1 \to 0$ and $\| \rho_g \xi_n - \xi_n\|_1 \to 0$).

$\mathbf{Question}$: Does there exists a bi-invariant Folner sequence? (i.e. a sequence of finite set $F_n$ such that $\xi_n = \chi_{F_n} /|F_n|$ is a sequence of almost invariant probability measure)

In other words, is it obvious that the bi-invariant property'' follow through the same proof asleft-'' and ``right-'' do, or, better, does there exist a simpler argument to show such a sequence exists?

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Yes, it is obvious. If you have a sequence of probability measures which is approximately invariant on the left, by convolving these measures on the right with the reflected ones you get an approximately invariant sequence of symmetric measures. By applying to this new sequence the standard "slicing" argument (the same as in the one-sided case) one gets symmetric Folner sets.

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Speaking as one who understands the definitions in the question, but not the answer, I think I can contradict the first sentence. – Anthony Quas Sep 9 '12 at 5:54
I agree with the first sentence of the answer. The rest of the answer is in fact redundant. – Mark Sapir Sep 9 '12 at 9:07
Looks like opinions are divergent :))) – R W Sep 9 '12 at 12:49
Thanks! I'm happy with the answer in any case (though I was hoping for a different argument, in which case the word "simpler" was probably a bad idea). – Antoine Sep 10 '12 at 12:00

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