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.

Fourier algebraof $G$. This is dense in $C_0(G)$ for the sup norm and hence dense in $L^1(G)$. – Yemon Choi Mar 11 '13 at 21:12containthe Fourier algebra but wouldn't in general be equal to it. However, for the OP's purposes, that still would imply density in $L^1(G)$. – Yemon Choi Mar 11 '13 at 21:19