In some proofs of our lecture we used following relation $$\rho(g)u=u*\delta_g$$
The convolution was defined as $(u*v)(h)= \sum_{g\in\Gamma} u(g) v(g^{-1}h)$ for $u,v\in l^2(\Gamma)$ and $h\in\Gamma$ and the right regular representation as $(\rho(g)v)(h)= v(hg)$ for $v\in l^2(\Gamma)$ and $h\in\Gamma$
The problem is that my calculation yields another relation:
$$(u*\delta_g)(h)=\sum_{f\in\Gamma}u(f)\delta_g(f^{-1}h)$$
Every summand is zero, except the one that satisfies $f=hg^{-1}$ (since $g=f^{-1}h \Rightarrow f^{-1}=gh^{-1} \Rightarrow f=hg^{-1}$). Hence it follows
$$(u*\delta_g)(h)= u(hg^{-1})=(\rho(g^{-1})u)(h)$$
Have I made a mistake somewhere? Thanks for your help.
The issue is in your formula for the right regular representation in terms of the convolution product: You state that $$\rho(g)u=u*\delta_g.$$ But the correct formula is $$\rho(g)u=u*\delta_{g^{-1}}.$$ With this, one obtains \begin{align*} (\rho(g)u)(h) &=(u*\delta_{g^{-1}})(h)\\ &=\sum_{f\in\Gamma}u(f)\delta_{g^{-1}}(f^{-1}h)\\ &=u(hg). \end{align*}