Let $G$ be a compact abelian group. Then for any $\phi\in C(G)$ one can define an "action" on $L^2(G)$ via an integral operator $T_\phi$ as follows: $$T_\phi(f)(h)=\int_G \phi(g) f(hg)\, dg$$
Let $R$ denote the $\mathbb{C}$-algebra of these operators. (It is not hard to see that $R$ is adjoint-closed, commutative, and consists of compact operators.)
Proposition. $L^2(G)=\widehat{\bigoplus}_{0\ne \mu\colon R\to\mathbb{C}} L^2(G)_{\mu}$ of simultaneous eigenspaces $$L^2(G)_\mu=\{f\in L^2(G)\mid T_\phi f=\mu(T_\phi) f \text{ for all } T_\phi\in R\}$$
This is corollary 3.6.1 in Garrett's notes (See page 7). (I believe there are some typos in the notes; I hope they are fixed above.)
Now, while I can follow the derivation of this proposition, I'm not sure how it can be applied to the simplest case namely $G=S^1$. In other words, what are $\mu$ and $L^2(S^1)_\mu$ explicitly? (By Corollary 4.0.1 in the same notes, I can say that each $L^2(S^1)_\mu$ can be decomposed into a direct sum of one dimensional spaces given by characters of $S^1$ or simply the exponential functions.)