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.
