Let $G$ be a locally compact group, $C_0(G)$ the $C^*$-algebra of continuous functions on $G$ that vanish at infinity, $C_b(G)$ the $C^*$-algebra of bounded continuous functions on $G$. We know that $C_b(G)$ is the multiplier algebra of $C_0(G)$, and we denote the strict topology on $C_b(G) = \mathcal{M}\bigl( C_0(G) \bigr)$ by $\beta$.

Now for a strongly continuous unitary representation $(\pi, H)$ of $G$, functions of the form $\omega_{\pi,\eta,\xi} : g \in G \to (\pi(g)\eta \mid \xi) \in \mathbb{C}$ are in $C_b(G)$, and we call them matrix coefficients of the representation $\pi$. Since we can form direct sum and tensor product of two strongly continuous unitary representations, as well as the contragredient representation, we see that (the linear span of) matrix coefficients of all strongly continuous unitary representations of $G$ form a $*$-subalgebra $A_0(G)$ of $C_b(G)$.

**Question.** Is $A_0(G)$ strictly dense in $C_b(G)$, i.e. with respect to the $\beta$-topology?

Note that in the compact case, $C_b(G) = C_0(G) = C(G)$ and the $\beta$-topology is the same as the norm topology on the $C^*$-algebra $C(G)$ of continuous functions on $G$. In this case, The answer to the question is affirmative by Peter-Weyl. In the case where $G$ is discrete, then one can check easily that all finitely supported functions on $G$ are already matrix coefficients of the left (or right) regular representation, so the answer to the question is again affirmative. Based on these considerations, here are some sub-questions with some bias on their possible answers.

**Q1.** Does the question have an affirmative answer for unimodular $G$?

**Q2.** Can we construct some counter-example for non-unimodular $G$?

**Q3.** Does the question have an affirmative answer if $G$ is a real Lie group? What if the Lie group $G$ is nilpotent, or solvable, or semisimple/reductive?

Fourier--Stieltjes algebraof $G$, and is usually denoted by $B(G)$. $B(G)$ has been much studied: it turns out to be complete in a natural submultiplicative norm $\fsnorm{\cdot}$, so it is a commutative Banach algebra of functions on $G$. This norm is dominated by the supremum norm. $\endgroup$