Wikipedia says:
On locally compact abelian groups, a version of the convolution theorem holds: the Fourier transform of a convolution is the pointwise product of the Fourier transforms. The circle group $\mathrm S$ with the Lebesgue measure is an immediate example. For a fixed $g$ in $L^1({\mathrm{S}})$, we have the following familiar operator acting on the Hilbert space $L^2(\mathrm{S})$: $$ T{f}(x)=\frac {1}{2\pi }\int _{\mathrm {S} }{f}(y)g(x-y)\,dy $$ The operator $T$ is compact.
Is there a way to prove this?
Caveat. Sadly, I know nothing about group theory, but intuitively I suppose $L^2({\mathrm S})$ means if I have a function, say in $L^2([0, 2\pi])$, the previous convolution makes sense only if I assume the function repeats itself periodically outside $[0, 2\pi]$.