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Recently in my investigations I faced a problem related to amenable groups. I have no idea if my question is suitable for this site, but after ask some collegues in my department I decided to try to find some help here.

Suppose that $G$ is an amenable group having a mean $ m\in \left(L^{\infty}(G,\mathbb{R}) \right)^{*} $. Is it true that there exist a mean $\tilde{m}\in \left(L^{\infty}(G\times G,\mathbb{R}) \right)^*$ such that for all borelians $B$ we have $$ \tilde{m}(B\times G)=m(B). $$

If the answer is negative could you point me out the counter-example ?

Remark: The answer is positive if $G$ is locally compact group because the mean in $G\times G$ is the product Haar measure of $G$.

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Is $G$ locally compact? Maybe you should write $L^{\infty}(G,\mathbb R)$ then. – Andreas Thom Jun 2 '11 at 6:02
thanks for the suggestion. – user11178 Jun 2 '11 at 21:29
Aren't Fubini-type theorems for finitely additive measures treated in Dunford-Schwartz I? Maybe I'm misremembering and I can't check at the moment, but that's where I'd look first. The treatise on measure theory by Fremlin might also contain some discussion relevant to the present question. – Theo Buehler Jun 2 '11 at 22:57
Your last statement (in the version I am reading) is not right, because if your group is non-compact the Haar measure is infinite and therefore not a "mean" in the sense of amenability of groups. (This is one of the things that makes amenability somewhat subtle/difficult) – Yemon Choi Jun 3 '11 at 5:53
up vote 1 down vote accepted

Can't you argue as follows: given $f\in L^\infty(G\times G)$ for any $a\in L^1(G)$ define $f_a$ by $$ f_a(s) = \int_G f(s,t) a(t) \ dt. $$ Then $f_a\in L^\infty(G,\mathbb R)$ with $\|f_a\| \leq \|f\| \|a\|$. Thus the map $$L^1(G) \rightarrow \mathbb R, \quad a\mapsto m(f_a) $$ is bounded and linear, and so there is some $g=m_f\in L^\infty(G)$ with $$ \int_G g(s) a(s) = m(f_a) \qquad (a\in L^1(G)). $$ Now define $\tilde m$ by $$ \tilde m(f) = m(g). $$

Then, let $x,y\in G$ and define $\hat f(s,t) = f(xs,yt)$, say $\hat f = (x,y)\cdot f$. Then check, for the obvious notation, that $$ \hat f_a = x\cdot f_{y^{-1}\cdot a}, \quad \hat g = y^{-1} \cdot g, \quad \tilde m(\hat f) = m(y^{-1}\cdot g)=m(g) = \tilde m(f), $$ so $\tilde m$ is left invariant. Notice I use that $m$ is left invariant in the calculation of $\tilde m(\hat g)$, and in the calculation of $\hat g$.

Finally, clearly if $f=1$ then $g=1$ and so $\tilde m(f)=1$. Then set $f=\chi_{B\times G}$ so that $g = m(B) 1$ and hence $\tilde m(f) = m(B)$ as you want.

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By the way, what I'm "really" doing is forming $m\otimes m$, but this is a little technically tricky, as it's not 100% obvious how to embed $M^*\otimes M^*$ into $(M\overline\otimes M)^*$, for a von Neumann algebra $M$. But the method I give is utterly standard... – Matthew Daws Jun 2 '11 at 9:07

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