I'm trying to understand the Fourier transform on compact groups but run into technical problems regarding the equivalence classes of unitary representations. Here is my train of thought:
$G$ is a compact group. For every $f\in L^1(G)$ we may define the Fourier transform $$\widehat{f}(\pi) := \int_G\ f(x) \pi(x)^*\ dx$$
where $\pi$ is ... an individual irreducible, unitary representation or a representative of unitary equivalence class of such representations? In the latter case I guess it would be more formal to write $$\widehat{f}([\pi]) := \int_G\ f(x) [\pi](x)^*\ dx$$
where $[\pi]$ stands for the whole unitary equivalence class. But then... what is the codomain of $[\pi]$? We know that individual representatives of $[\pi]$ are unitary representations $\pi_1 : G \longrightarrow U(H_{\pi_1}),$ $\pi_2 : G \longrightarrow U(H_{\pi_2})$... and for each pair there exists a unitary operator $T:H_{\pi_1}\longrightarrow H_{\pi_2}$ such that $T\pi_1(x)T^{-1} = \pi_2(x)$ for every $x\in G$.But what about $[\pi]$? Is this a function from $G$ to $U(H)$ where $H$ is some Hilbert space constructed from all those $H_{\pi_1}, H_{\pi_2}, ...$?
Assuming my guess is correct (i.e., the domain of the Fourier transform is the set of equivalence classes rather than individual representations) I'm having trouble proving that $\widehat{f}$ is well-defined. It is my understanding that the Fourier transform should be the same on every representative of the equivalence class, but if I take $\pi_2 \in [\pi_1]$ then it is not clear to me how $$\widehat{f}(\pi_1) = \int_G\ f(x) \pi_1(x)^*\ dx$$
and $$\widehat{f}(\pi_2) = \int_G\ f(x) T\pi_1(x)^*T^{-1}\ dx$$
are ''the same''. Thx in advance for any comments and suggestions!