Let $G$ be a locally compact subgroup, $\mu$ a Haar-measure. For $f \in L^1(G)$, and for $\pi$ a unitary, topology irreducible, representation of $G$ on an Hilbert space $H_\pi$, it is customary to define a continuous operator $\pi(f)$ on $H_\pi$ by $$\pi(f) v = \int_G f(g) \pi(g)(v) d\mu(g).$$
Let us say here that $f \in L^1(G)$ is of trace class if for every unitary, absolutely irreducible representation $\pi$ of $G$, the operator $\pi(f)$ is of trace class. (side question: is there a name outside for this notion?).
I have many question about this notions, but let me give one:
Is it true that the space of trace class functions on $G$ is dense in $L^1(G)$? If not, is it true for a separable type I locally compact group ?
I know of certain larges classes of group for which it is true. Compact group and abelian locally compact groups are trivial examples, since all irreducible $\pi$'s are finite dimensional. Real and $p$-adic Lie group are other examples since smooth compact support functions are trace class (IIRC) with the usual meaning of smooth ($C^\infty$ in the real case, locally constant in the $p$-adic case), and are dense. Yet this is proved in a rather ad hoc way, for example in the real case (Duflo-Labesse) by showing that such a function $f$ is the convolution of two such functions $f_1$ and $f_2$, so that $\pi(f)=\pi(f_1)\pi(f_2)$ and $\pi(f_1), \pi(f_2)$ are Hilbert-Schmiddt. Also the adelic case (necessary for the global trace formula) follows from the real and $p$-adic case.
Yet I don't know if this is true in general, or at least for a large abstract class of locally compact group (defined with a property such as type I, not as a list of example such as "real Lie groups"), with if possible a uniform proof.
If the question is too difficult, or the answer is no, does that help if we ask the same question for functions that are reduced trace class, i.e. such that $\pi(f)$ is of trace class for all irreducible $\pi$'s in the reduced spectrum (i.e. in the support of the regular representation).
PS: Clearly, the condition of being a trace class does not change if we replace $L^1(G)$ by $C^\ast(G)$ with its natural norm. Hence one of the tag.