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 $[0, 2\pi]$.

  • 1
    $\begingroup$ Yes $L^2(\Bbb{R/2\pi Z})$ is the same as the $2\pi$-periodic functions $\Bbb{R\to C}$ that are $L^2$ on $[0,2\pi]$. $L^2(S)$ is slightly abstracted as $S$ could also be $\{ z\in \Bbb{C},|z|=1\}$. $L^2(G)$ assumes that you have a Haar measure (which exists if $G$ is locally compact) that is a measure giving an integral satisfying a few axioms: $\int_G f(g)dg$ is finite and non-negative if $f\ge 0$ is bounded & supported on a compact, and $\int_G f(g+x)dg=\int_G f(g)dg$ for all $x\in G$ (group law in additive notation when $G$ is abelian). With $G=\Bbb{Z}$ you'll have $L^2(G)=\ell^2(\Bbb{Z})$ $\endgroup$
    – reuns
    Dec 15, 2021 at 19:06
  • 1
    $\begingroup$ Link to wiki page? $\endgroup$
    – Thomas
    Jan 6, 2022 at 23:55

1 Answer 1


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.


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