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Discussion of reductive groups: For reductive groups over local fields, you can replace a.e. w.r.t. to Plancherel measure by every smooth, admissible representation, but I guess you are well aware of that. Some reps are not tempered and do not contribute to the Plancherel formula, so this far more. The notion of smooth in the sense I indicate above coincides with the usual one in the non-archimedean world/totally disconnected groups. Be careful, general Lie groups are not type I. For the adelic points of a reductive group over a global field, you need an argument along the lines of Is a reductive adelic group a Type I group?, but the situation is similar to that of reductive local groups.

Discussion of reductive groups: For reductive groups over local fields, you can replace a.e. w.r.t. to Plancherel measure by every smooth, admissible representation, but I guess you are well aware of that. Some reps are not tempered and do not contribute to the Plancherel formula, so this far more. The notion of smooth in the sense I indicate above coincides with the usual one in the non-archimedean world/totally disconnected groups. Be careful, general Lie groups are not type I. For the adelic points of a reductive group over a global field, you need an argument along the lines of Is a reductive adelic group a Type I group?, but the situation is similar to that of reductive local groups.

Discussion of Plancherel: As Yemon Choi points out, a necessity of the Plancherel theorem (check e.g. http://www.encyclopediaofmath.org/index.php/Unitary_representation) is that the $tr\; \pi (f\ast f^*)$ is well-defined for every $f \in L^1(G) \cap L^2(G)$ and a.e. $\pi$ with respect to the Plancherel measure $\mu$. The Plancherel theorem states then for $G$ be a unimodular separable locally compact group of type I $$ \int\limits_G |f(g)|^2 d g = \int\limits_{unitary \; dual} tr\; \pi( f \ast f^*) d \mu(\pi).$$ There exist extensions, which drop unimodular, type I or seperable as stated in the above link, but don't know how the extensions look like, so that's why I worked with rather special $G$ here.

Discussion of the Plancherel theorem: A necessity of the Plancherel theorem (check e.g. http://www.encyclopediaofmath.org/index.php/Unitary_representation) is that the $tr\; \pi (f\ast f^*)$ is well-defined for every $f \in L^1(G) \cap L^2(G)$ and a.e. $\pi$ with respect to the Plancherel measure $\mu$. The Plancherel theorem states then for $G$ be a unimodular separable locally compact group of type I $$ \int\limits_G |f(g)|^2 d g = \int\limits_{unitary \; dual} tr\; \pi( f \ast f^*) d \mu(\pi).$$ There exist extensions, which drop unimodular, type I or seperable as stated in the above link, but don't know how the extensions look like, so that's why I worked with rather special $G$ here.

$$4 \cdot S \ast T^\ast = (S+T) \ast (S+T)^\ast - (S-T) \ast (S-T)^\ast$$ $$ - i(S-iT) \ast (S-iT)^* + (S+iT) \ast (S+iT)^*.$$ Applying that $ \pi ( (S+T) \ast (S+T)^\ast ), \dots$ are trace class by Plancherel for a.e. $\pi$ yields the theorem. Q.e.d.

$$4 \cdot S \ast T^\ast = (S+T) \ast (S+T)^\ast - (S-T) \ast (S-T)^\ast$$ $$ - i(S-iT) \ast (S-iT)^* + i(S+iT) \ast (S+iT)^*.$$ Applying that $ \pi ( (S+T) \ast (S+T)^\ast ), \dots$ are trace class by Plancherel for a.e. $\pi$ yields the theorem. Q.e.d.