Let $G$ be a locally compact group and let $H$ be an open subgroup in $G$. Then the full group $C^*$-algebra of $H$, $C^*(H)$, is a subalgebra of $C^*(G)$ and there is a conditional expectation $$E\colon C^*(G)\to C^*(H),$$ which is induced by restriction $f\in L^1(G) \mapsto f_{|H}\in L^1(H)$ of functions which are integrable w.r.t. the left Haar measure on $G$, see Rieffel, induced representations of $C^*$-algebras, Proposition 1.2.

Note that $E$ isn’t faithful in general, there is an example in the paper right before Definition 1.3.: Consider $G$ a nonamenable discrete group and $H$ an open subgroup consisting of the identity element of $G$. Then there are nonzero elements $c$ in the kernel of the left regular representation of $G$ (since $G$ is not amenable) and they satisfy $E(c^*c)=0$.

My Question: Now, let $G$ be a locally compact amenable group and $H$ be an open compact (amenable) subgroup.

Is then $E$ faithful?

I think yes (I have considered some examples), but I am stuck with a proof.

If I additionally assume $G$ (and $H$) to be discrete I can prove it considering $E$ as a conditional expectation $C_r^*(G)\to C_r^*(H)$ and then it is $\tau_G=\tau_H\circ E$, where $\tau_G$ and $\tau_H$ are the canonical faithful tracial states on $C_r^*(G)$, $C_r^*(H)$ respectively. It follows that $E$ must be faithful.

For the more general case, I thought about trying a similar strategy using the fact that $C_r^*(G_1)$ of a locally compact group $G_1$ which contains a non-trivial amenable open subgroup $H_1$ has a tracial state $\tau^{G_1}$ satisfying $$\tau^{G_1}( \lambda_{G_1}(f))=\int_{H_1}f(s)d\mu(s),$$ see corollary 4.1 in 'embedding theorems in group $C^*$-algebra' by Lee for this fact. If one can check that the tracial state $\tau^{H_1}$ is faithful, then this together with $\tau^{G_1}=\tau^{H_1}\circ E$ implies that $E$ is faithful. But I am stuck with proving faithfulness of the trace. Probably I am on the wrong track..Other strategies regarding my question are welcome.