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,