$L^2(\mathbb{R})$ acted on by translation does decompose into a direct integral of one-dimensional representations. So in general you have a direct integral if $G$ is not compact. If $G=SL_2(Q_p)$, you have the same issue. So no direct sum, but a direct integral.

You should look up type 1 groups, for various more complicated issues. The decomposition might not be unique otherwise.

$L^2(\mathbb{R})$ acted on by translation does decompose into a direct integral of one-dimensional representations. So in general you have a direct integral if $G$ is not compact. If $G=SL_2(Q_p)$, you have the same issue. So no direct sum, but a direct integral.

You should look up type 1 groups, for various more complicated issues. The decomposition might not be unique otherwise.

I haven't time to read the full script. I seems that she is using it in the context of supercuspidal representations, which are in fact semisimple in the sense that they decompose into a direct sum (actually a finite sum). Here it is okay.

Murnagahan is certainly wrong if she means unitary = unitarizabile and smooth,e.g. it fails for $C_c^\infty(G)$. Unitary (being a unitary Hilbertspace representation and smooth) implies finite-dimensional, so isn't really interesting.

$L^2(\mathbb{R})$ acted on by translation does decompose into a direct integral of one-dimensional representations. So in general you have a direct integral if $G$ is not compact. You should look up type 1 groups, for various more complicated issues. The decomposition might not be unique otherwise.

$L^2(\mathbb{R})$ acted on by translation does decompose into a direct integral of one-dimensional representations. So in general you have a direct integral if $G$ is not compact. If $G=SL_2(Q_p)$, you have the same issue. So no direct sum, but a direct integral.

You should look up type 1 groups, for various more complicated issues. The decomposition might not be unique otherwise.