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A group $G$ is said to be amenable if for every bounded function $f:G \to \mathbb{R}$, there exists a mean $I(f) \in \mathbb{R}$ such that $I(f)\geq 0$ whenever $f \geq 0$, $$I(\mathbb{1}) =1$$ and is $G$ invariant, that is $$I(f)=I(g \cdot f)$$ where $G$ acts on functionals as follows: $g\cdot f(x)=f(g^{-1}x)$. I am stating the definition mainly to stress the $G$-invariance property here.

Amenable groups have vanishing of bounded cohomology", and the proof uses the mean to construct a coboundary from a cocycle. For instance suppose we have a bounded $2$-cocycle $f:G \times G \to \mathbb{R}$. It satisfies the following relation by definition: for $g_1,g_2,g_3 \in G$, $$f(g_2,g_3)-f(g_1g_2,g_3)+f(g_1,g_2g_3)-f(g_1,g_2)=0$$ Now if we define $h:G \to \mathbb{R}$ as $$h(g)=I(f(g,x))$$ that is, $h(g)$ is the mean of the function $f(g,x)$ (treated as a function in $x$). Then we can check that $d^1h(g_1,g_2)=f(g_1,g_2)$ by properties of the mean. The same argument can be used to show that $H_b^2(G,V)=0$ for reflexive $V$ (that is, $V$ is itself the space of functionals on some space $W$).

My question is, in the above we don't really use any action of $G$ on $V$. We only make $G$ act on functions by translations, but don't use any action on $V$. But in general, cohomology of $G$ wrt $V$ is defined with respect to some action on $G$ on $V$. Let $\rho:G \to U(V)$ be such a unitary action. Then the $2$-cocycle condition would be
$$\rho(g_1)f(g_2,g_3)-f(g_1g_2,g_3)+f(g_1,g_2g_3)-f(g_1,g_2)=0$$

Now too it seems that $h$ defined as the mean of $f$ seems to work as a coboundary. I don't understand how this works out. After all the mean for amenable groups uses only a certain narrow kind of $G$-invariance, so how does it work for every action of $G$ on $V$? I'm clearly missing something here.

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    $\begingroup$ It's hard to guess what you might be missing, because you have not given any source. My best guess is that the theorem being proved is stated using only bounded cohomology with untwisted coefficients, namely coefficients where the group action is trivial: $g \cdot v = v$ for all $g,v$. $\endgroup$
    – Lee Mosher
    Oct 19, 2019 at 16:07

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You're not missing anything: it is true that whenever $G$ is an amenable group, for any dual $G$-module $E$ and any $n \geq 1$ we have $H^n_b(G, E)$. The proof in the discrete case is very easy and you can find it in chapter 3 Frigerio's book "Bounded cohomology of discrete groups": you just use the mean to define a partial homotopy in the cochain complex.

For the continuous case it is more complicated since there are some continuity issues. One would either need to go through uniformly continuous bounded cohomology and define an equivariant projection of the continuous cochain complex to the uniformly continuous one; or use the functorial characterization. You can find all of this in chapter 7 of Monod's book "Continuous bounded cohomology of locally compact groups".

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