On convolution theorem and Fourier transform

Solution 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.