Is the Peter-Weyl theorem for $S^{1}$ equivalent to the Fourier Theorem?

Question

recently in one of my Lie Algebra classes the professor mentioned the following:

In the case of $S^{1}$ the Peter-Weyl theorem is equivalent to the Fourier Theorem.

In the beginning I didn't put to much thought to this statement, but when I was reading again my notes I became highly curious, I have been searching for a proof or some argument that helps me understand why this statement is true, but I am not finding anything useful, the only thing that I found was this post Peter-Weyl Theorem on the Sphere but it is for $S^{2}$, and I read the answer, but that didn't help.

Does anybody knows how to prove this statement or where can I find such proof? Thanks in advance.

Answer 1

One form of the Peter-Weyl theorem states that for a compact group $G$, the characters of the (continuous) irreps of $G$ form a basis of the the space of $L^2$ class functions on $G$; that is, the space of $f\in L^2(G)$ with $f(\gamma^{-1} g \gamma) = f(g)$ for all $\gamma\in G$. For $G = S^1$, the situation is simpler. $G$ is abelian, so the class function condition is vacuous; any function on $G$ is a class function. Furthermore, the (continuous) irreps of $S^1$ are just the maps $t \to e^{int}$ for fixed integers $n$. We're thus left with the fact that the maps $e^{int}$ form a basis of $L^2(S^1)$, which is pretty much Fourier analysis. (To clarify, this is a basis in the sense of a Hilbert space: the vectors are orthonormal and span a dense subspace.) There are corollaries and variants of both the Peter-Weyl theorem and Fourier theorems for continuous functions, etc.; but that's the fundamental idea. You might also take a look at Pontryagin duality in the locally compact setting for another variation.