On convolution theorem and Fourier transform 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 $[0, 2\pi]$.
 A: For a locally compact abelian group $G$, the dual group $\hat{G}$ is defined as the set of all homomorphisms to the circle $G\to S^1$. Given any $L^2$ function $f: G\to\mathbb{C}$, the Fourier transform $\hat{f}$ is a function defined on the dual group by the formula $\hat{f}(\chi)=\int_G f(g)\overline{\chi(g)}dg$, where $dg$ denotes Haar measure on $G$. Similarly, the convolution of two functions $f_1,f_2: G\to\mathbb{C}$ is defined by $(f_1*f_2)(g)=\int_G f_1(g-h)f_2(h)dh$
With this setup, the proof of the convolution theorem goes through basically the same way as for a normal Fourier series.
Indeed, given functions $f_1,f_2$, we have
\begin{eqnarray*}
\widehat{f_1*f_2}(\chi)&=&\int_G (f_1*f_2)(g)\overline{\chi(g)}dg\\
& = & \int_{g\in G} \int_{h\in G} f_1(g-h)f_2(h)\overline{\chi(g)}dhdg\\
& = & \int_{h\in G} f_2(h)\overline{\chi(h)}\int_{g\in G}f_1(g-h)\overline{\chi(g-h)}dgdh\\
& = & \int_{h\in G} f_2(h)\overline{\chi(h)}\int_{g\in G}f_1(g)\overline{\chi(g)}dgdh\\
& = & \hat{f_1}(\chi)\hat{f_2}(\chi)
\end{eqnarray*}
In the third equality, we used the fact that $\chi$ is a homomorphism, and in the fourth equality, we used the invariance of the measure $dg$ with respect to translations.
