A question about a conditional expectation in C*-algebra Let $\Gamma$ be a discrete group. Consider a conditional expectation $\Phi: B(l^{2}(\Gamma))\rightarrow l^{\infty}(\Gamma)$ defined by
$$\Phi(T)=\sum_{g\in \Gamma}e_{g,g}Te_{g,g},$$
where $e_{g,g}$ is the rank-one projection onto $\delta_{g}\in l^{2}(\Gamma)$ (where $\{\delta_{t} : t\in \Gamma\}$ is the canonical orthonormal basis) and the sum is taken in the strong operator topology.
Then, can we verify that $\Phi(\lambda_{s}T\lambda_{s}^{*})=\lambda_{s}\Phi(T)\lambda_{s}^{*}$ ? (Here, the $\lambda_{s}$ denote the left regular representation: $\lambda_{s}(\delta_{t})=\delta_{st}$ for all $s, t\in \Gamma$.) 

You can also do this explictly: for any $h\in \Gamma$,
$$
\Phi(\lambda_s T\lambda_s^*)\delta_h=\sum_g e_{gg}\lambda_s T\lambda_{s^{-1}} e_{gg}\delta_h=e_{hh}\lambda_s T\lambda_{s^{-1}}\delta_h=e_{hh}\lambda_s T\delta_{s^{-1}h}\\=\sum_ge_{hh}\lambda_s\langle T\delta_{s^{-1}h},\delta_g\rangle\delta_g=\sum_g \langle T\delta_{s^{-1}h},\delta_g\rangle e_{hh}\delta_{sg}=\langle T\delta_{s^{-1}h},\delta_{s^{-1}h}\rangle \delta_{h}.
$$
And
$$
\lambda_s\Phi(T)\lambda_{s^{-1}}\delta_h=\sum_g\lambda_se_{gg}Te_{gg}\delta_{s^{-1}h}=\lambda_s e_{s^{-1}h,s^{-1}h}T\delta_{s^{-1}h}=e_{h,s^{-1}h}T\delta_{s^{-1}h}\\
=\sum_g e_{h,s^{-1}h}\langle T\delta_{s^{-1}h},\delta_g\rangle\delta_g
=\langle T\delta_{s^{-1}h},\delta_{s^{-1}h}\rangle\delta_h.
$$
