Finite-rank spectral pieces of a self-adjoint compact convolution operator

The spectral theorem [F1] applied to $T=C_{\phi }$ gives that $\mathrm{\Sigma }$ is finite or countably infinite, that every $E_{\lambda }$ is finite dimensional, and that the closed linear span $M$ of ${\bigcup }_{\lambda \in \mathrm{\Sigma }}{E}_{\lambda }$ equals $(\ker C_{\phi }{)}^{\perp }$; distinct eigenspaces are orthogonal by [F2]; [F5] applied to the closed subspace $M$ gives $L^2(K)=M\oplus M^{\perp }$ with $M=(\ker C_{\phi }{)}^{\perp }$, so $M^{\perp }=\ker C_{\phi }$ and hence $L^2(K)=\ker C_{\phi }\oplus M$ as an orthogonal decomposition; since the $E_{\lambda }$ are pairwise orthogonal closed subspaces with closed linear span $M$, the canonical map ${\widehat{⨁}}_{\lambda \in \mathrm{\Sigma }}{E}_{\lambda }\to M$ is a unitary isomorphism by [F4], so $L^2(K)=\ker C_{\phi }\oplus {\widehat{⨁}}_{\lambda \in \mathrm{\Sigma }}{E}_{\lambda }$ is an orthogonal Hilbert-space direct sum; this is (1).

By the choice ${\phi }^{\ast }=\phi$ and [F3], every $E_{\lambda }$ and $\ker C_{\phi }$ is invariant under the right regular representation $\rho$; each $E_{\lambda }$ is closed and finite dimensional by step 1.1, and by [F6] the restriction of $\rho$ to $E_{\lambda }$ is a strongly continuous unitary representation of $K$, hence a finite-dimensional continuous unitary representation; this is (2), and it also exhibits $\rho$ as the Hilbert direct sum of these subrepresentations and the kernel by [F4]. The Axiom of Choice is consumed through the normalized Haar probability, the compact self-adjoint spectral theorem and the cited Hilbert-space suppliers; the argument above is choice-free apart from those inputs.