# 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 $$[0, 2\pi]$$.

• 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 $L^2(G)=\ell^2(\Bbb{Z})$ Commented Dec 15, 2021 at 19:06
• Link to wiki page? Commented Jan 6, 2022 at 23:55

## 1 Answer

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.

• I’m sorry.. does this prove the compactess of the convolution operator? Commented Jan 9, 2022 at 23:07
• @ric.san no. but see math.stackexchange.com/questions/2400645/… Commented Jan 9, 2022 at 23:27