Let $G$ be a second-countable locally compact group. We consider its left-regular representation $\lambda^G: G\to\mathcal{U}(L^2(G))$ given via $\lambda^G_g(\xi)(h)=\xi(g^{-1}h)$.
Recall: Given any continuous unitary representation $\pi: G\to\mathcal U(\mathcal H_\pi)$ on a Hilbert space, a so-called diagonal matrix coefficient of $\pi$ is a function of the form $g\mapsto \langle \pi(g)\xi\mid\xi\rangle$ for some $\xi\in\mathcal H_\pi$. We say that another representation $\rho$ is weakly contained in $\pi$, if every diagonal matrix coefficient of $\rho$ can be approximated by sums of diagonal matrix coefficients of $\pi$, uniformly in compact subsets of $G$. Two unitary representations that weakly contain each other are called weakly equivalent. (See here for more info and related references.)
It is well-known that $\rho$ and $\pi$ are weakly equivalent if and only if the two induced C*-algebra representations $C^*(G)\to\cal B(H_\pi)$ and $C^*(G)\to\cal B(H_\rho)$ have the same kernel.
My question: If $H\subseteq G$ is a closed subgroup, is the restriction $\lambda^G|_H$ weakly equivalent to $\lambda^H$?
In more C*-algebraic terms: Does the obvious unitary representation $H\to\mathcal{M}(C_r^*(G))$ induce a well-defined inclusion $C_r^*(H)\subseteq\mathcal{M}(C_r^*(G))$?
I have the impression that the answer to my question is probably painfully obvious to the experts, but I am too much of a novice in these things to see right away how the proof would look like. Of course this is quite obvious for discrete groups (and various other special cases), but that's not what I'm after. Is there perhaps a literature reference that would directly answer this question in general? My own quick search did not amount to much, but that doesn't have to mean anything.