meanable groups Suppose that the countable group $G$ has a left invariant mean on $l^\infty \left(G\right)$ , $L_l$ , and suppose that $G$ has a right invariant mean on $l^\infty \left(G\right)$ , $L_r$.
Please show that $G$ has a conjugation invariant mean.
I know that exists an isomorphism $\phi:G\rightarrow G$ such that for every $g\in G$ if we denote $l_g:G\rightarrow G$ by $l_g(h)=gh$ and we denote $r_g:G\rightarrow G$ by $r_g(h)=hg$ so we get $\phi\circ l_g = r_g \circ \phi$.
I tried to build $L_c$ (the c is for conjugation) the functions:


*

*$\frac {1}{2} \left(L_l+L_r \right)$

*$\frac {1}{2} \left(L_l\circ \phi+L_r \circ \phi^{-1} \right)$


They both doesn't satisfy (at least I didn't manage to prove) that for every $c_g:G\rightarrow G$ by $c_g(h)=ghg^{-1}$ and for every bounded $f:G\rightarrow \mathbb{R}$  we get $L_c \left(f\circ c_g\right)=L_c(f)$.
Can someone light me up? or maybe I'm in the wrong way at all. any help would be great.

I assume a mean is a linear functional $l^\infty(G)\to\bf R$? I'm not actually familiar with the term.
However two thoughts occur to me:


*

*$L_l$ is $l_g$-invariant, but how does it behave under $r_g$-action?

*The $L$s cannot be applied to each other; what if we reinterpret $L$ as a function of $G$?


These led me to pursue $L_c:f\mapsto L_r(g\mapsto L_l(f\circ r_g))$. Since
$$c_h\circ r_g\,(x)=hxgh^{-1}=r_{gh^{-1}}\circ l_h\,(x)$$
we deduce
$$\begin{array}{ll} L_c\circ c_h\,(f) & =L_r(g\mapsto L_l((f\circ c_h)\circ r_g)) \\ & = L_r(g\mapsto L_l((f\circ r_{gh^{-1}})\circ l_h)) \\ & = L_r(g\mapsto L_l(f\circ r_{gh^{-1}})) \\ & = L_r((g\mapsto L_l(f\circ r_g))\circ r_{h^{-1}}) \\ & = L_r(g\mapsto L_l(f\circ r_g)) \\ & = L_c(f).\end{array}$$
It's fairly likely I've made a mistake, though, given the large number of layers deep we are into composition of functions and operators (I can see diCaprio squinting at me as I type).
