# When one can expect $\widehat{(fg)} = \hat{f} \ast \hat{g}$; $f, g\in L^{1} (G)$?

Let $f, g \in L^{1}(\mathbb T)= L^{1} ([-\pi, \pi))$. We define, the Fourier transform of $f$ as follows: $$\hat{f}(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi} f(t) e^{-int} dt, \ (n\in \mathbb Z).$$ It is well known that Fourier transform takes convolution to point wise product:indeed,by Fubini-Tonelli theorem and change of variable, we can derive, $$\widehat{(f\ast g)} (n) = \hat{f}(n) \cdot \hat{g} (n) , (n\in \mathbb Z).$$

Suppose $fg \in L^{1} (\mathbb T)$ also.

My questions:

(1) To proving $\widehat{(fg)}((n) = \hat{f} (n) \ast \hat{g} (n); (n \in \mathbb Z)$, the hypothesis $fg\in L^{1} (\mathbb T)$ is sufficient; and if yes, how to prove it ? (that is, under what conditions Fourier transform takes point wise multiplication to convolution product ) Or, we need to put some more conditions on $f$ and $g$ ?

(2) Let $f, g \ \ \text{and} \ fg \in L^{1} (\mathbb R)$ and also assume $\hat{f}, \hat{h} \in L^{1} (\mathbb R).$ Can we expect $\widehat{(fg)} = \hat{f} \ast \hat{g}$ ?

(3) Let $G$ be locally compact abelian group and $f, g, fg, \hat{f}, \hat{g} \in L^{1} (G)$. Can we expect similar result in this situation ?

My attempt: Fix $n\in \mathbb Z$. By definition we have, $\widehat{(fg)}((n)= \frac{1}{2\pi}\int_{-\pi}^{\pi} ( f(t)\cdot g(t) ) e^{-int} dt$ and $\hat{f}(n) \ast \hat{g} (n)= \sum_{k\in \mathbb Z} \hat{f} (n-k) \hat{g} (k) = \sum_{k\in \mathbb Z} \{\frac{1}{2\pi}\int_{-\pi}^{\pi} f(t) e^{-i(n-k)t} dt \} \{\frac{1}{2\pi}\int_{-\pi}^{\pi} f(y) e^{-ikt} dy\}$

Thanks,

• On a general locally compact abelian group, the Fourier transform acts $\widehat{\cdot} : L^1(G,\mu) \to \text{C}_\infty(\widehat{G})$ where $\mu$ is a Haar measure on $G$, and Pontryagin duality exists between $G$ and $\widehat{G}$ (so that $G\cong \widehat{\widehat{G}}$). In general we have $\widehat{f*g} = \widehat{f}\cdot \widehat{g}$. If $\widehat{f},\widehat{g} \in L^1(\widehat{G},\nu)$ (where $\nu$ is the dual measure to $\mu$), then the duality allows us to say that $\widehat{\widehat{f}*\widehat{g}} = f\cdot g$. That might get you part of the way to an answer. Commented Feb 11, 2014 at 17:28

Let $f,g$ be tempered distributions on $\mathbb R^n$ such that $g\in \mathscr O_M$, the so-called multipliers space: $g$ is a smooth function such that $$\forall \alpha, \exists N_\alpha\ge 0,\quad \sup_x\vert(\partial^\alpha g)(x)\vert(1+\vert x\vert)^{-N_\alpha}<+\infty.$$ Then the product $fg$ makes sense as a tempered distribution and we may define the convolution of $\hat f\ast\hat g$ as the Fourier transform of $fg$.
A dual point of view: let $F,G$ be tempered distributions on $\mathbb R^n$ such that $$\text{supp } F\times\text{supp } G\ni(x,y)\mapsto x+y\in \mathbb R^n\text{ is proper}.$$ Then the convolution makes sense as a tempered distribution and is defined as $$\langle F\ast G, \phi\rangle=\langle F(x)\otimes G(y), \phi(x+y)\rangle.$$ If $G$ is compactly supported, then $\hat G\in \mathscr O_M$ so that $\hat F\hat G$ makes sense from the above arguments and we have indeed in that case $\widehat{F\ast G}=\hat F\hat G.$ We can use these remarks to answer your question (2): we have for $f,g\in L^1(\mathbb R)$ so that $fg, \hat f, \hat g\in L^1$, the following absolutely converging integrals \begin{multline} \langle\widehat{fg},\phi\rangle_{\mathscr S',\mathscr S}= \langle{fg},\widehat\phi\rangle_{\mathscr S',\mathscr S}=\int f(x) g(x)\hat \phi(x) dx =\iiint \hat f(\xi) \hat g(\eta)e^{2iπ x(\xi+\eta)}\hat \phi(x) dxd\xi d\eta \\=\iint\hat f(\xi) \hat g(\eta)\phi(\xi+\eta) d\xi d\eta \end{multline} and with absolutely converging integrals $\langle\hat f\ast \hat g,\phi\rangle_{\mathscr S',\mathscr S}=\int (\hat f\ast \hat g)(\xi)\phi(\xi) d\xi=\iint\hat f(\xi-\eta)\hat g(\eta)\phi(\xi) d\xi d\eta=\iint\hat f(\xi) \hat g(\eta)\phi(\xi+\eta) d\xi d\eta,$ proving the sought equality. The answer to (3) should be similar.