Straightforward computations lead to the following standard property of Fourier transformation: it transforms convolutions into products, i.e. for functions $f$ and $g$ Schwartz class we have $$\widehat{f \star g} = \hat{f} \hat{g} \qquad (1) $$
I am interested in the corresponding property in the more general setting of Fourier transforms on some dual groups. Let $G$ be a reductive group, $\pi$ and automorphic representation of $G$, and define the Fourier transform of functions in the Hecke algebra by $$\widehat{f}(\pi) = \mathrm{tr} \ \pi(f)$$
Property $(1)$ does not always hold in this more general setting. It is easy to check that $\pi(a \star b) = \pi(a) \circ \pi(b)$, hence the property (1) remains true in cases of multiplicty one.
Let $K$ be a (say maximal) compact subgroup and $f$ a function in the Hecke algebra. By definition, $\pi(f) \circ \pi(\mathbf{1}_K)$ is 0 when $\pi$ is ramified, by multiplicity one indeed $\widehat{\mathbf{1}_K}(\pi)=\mathbf{1}_{K-\text{unram}}(\pi)$. In that case, I would like to state property (1), however there is no reason because $$\mathrm{tr}(\pi(f) \circ \pi(\mathbf{1}_K)) = \mathbf{1}_{K-\text{unram}}(\pi) \ \mathrm{tr} \ \pi(f_{|\text{K-unram}}) $$
and this last trace if a priori not $\mathrm{tr} \ \pi(f)$, right? Anyway, my question without those details is:
Given a function $f$, is there a $g$ such that $$\widehat{g \star \mathbf{1}_K} = \hat{f} \cdot \widehat{\mathbf{1}_K}$$
Thanks in advance!
