$g, f, \hat {f} \in L^{1}(\mathbb R)\cap L^{2}(\mathbb R) \cap C_{0}(\mathbb R) \implies \widehat{(fg)}= \hat{f} \ast \hat{g} ? $ Let $f, g\in L^{1}(\mathbb R)$ and it Fourier transform of $f$, $\hat{f} (y)  = \int _ {\mathbb R} f(x) e^{-2\pi i x \cdot y} dx, \ (y\in \mathbb R)$ and the convolution of $f $ and $g$;  $f\ast g (x)= \int_{\mathbb R} f(y) g(x-y) dy, \ (x\in \mathbb R); $ whenever this integral exists.
It is well-known that,
(1)If $f, g \in L^{1}(\mathbb R)$, then $\widehat{(f\ast g)} = \hat{f}\cdot \hat{g}.$
(2) If $f, g, \hat{f}, \hat{g} \in L^{1}(\mathbb R)$, then, by inversion formula, and using above relation in (1), we get, $\widehat{(fg)}= \hat{f}\ast \hat{g}.$ 
[Indeed, Fourier inversion formula gives us, $\widehat{(\hat{f})}(x) =f(-x)$; and we put, $\check{f}(x)= f(-x)$ and $\Phi(f)=\hat{f}$; now we using  relation in (1), by $\hat{f}$ and $\hat{g}$, in place of $f$ and $g$, we get $\Phi(\hat {f}\ast \hat{g})= \Phi(\hat{f})\cdot\Phi(\hat{g})= \Phi^{2}(f)\cdot\Phi^{2}(g) =\check{f}\check{g}= \check{(fg)}= \Phi^{2}(fg)$; now applying $\Phi^{-1}$ both side, we get, $\widehat{(fg)}= \hat{f}\ast \hat{g}.$]

My question: Let $g\in L^{1}(\mathbb R)\cap L^{2}(\mathbb R) \cap C_{0}(\mathbb R) $ and $f, \hat{f} \in L^{1}(\mathbb R).$ 
  (A) Can we expect $\widehat{(fg)}= \hat{f} \ast \hat{g}$ ? (B) If not, firstly, can we produce counter example, secondly, what optimal(minimal) condition, one can think of on $g$, so this formula will be valid ?

(where $C_{0} (\mathbb R)=\{f:\mathbb R \to \mathbb C: f \text { is continuous on} \ \mathbb R \  \text  {and } \lim_{|x|\to \pm \infty}f(x)=0 \}$) 
(We notice that, by Plancherel theorem, we can say $\hat{g} \in L^{2}(\mathbb R)$; and hence, by Young's inequality, we get, $\hat{f}\ast \hat{g} \in L^{2} (\mathbb R)$)
;Thanks,
 A: Yes, under the given hypotheses, we have
$$\widehat{(fg)} = \hat{f}\ast\hat{g}.\tag{1}$$
Consider the space $E = L^1\cap L^2\cap C_0$ with the norm $\lVert g\rVert = \lVert g\rVert_{L^1} + \lVert g\rVert_{L^2} + \lVert g\rVert_{C_0}$. The subspace $C_c^\infty$ is dense in $E$, and on that, $(1)$ holds for all $f$ such that $f,\hat{f} \in L^1$.
Given $f \in L^1$ with $\hat{f}\in L^1$, consider the linear maps
$$\begin{align}
\alpha &\colon g \mapsto \widehat{(fg)},\\
\beta &\colon g \mapsto \hat{f}\ast \hat{g}
\end{align}$$
from $E$ into $C_0$. Once we saw that they are continuous, it follows that $\alpha = \beta$ since they coincide on a dense subspace.
$\alpha$ factors as
$$E \xrightarrow{g\mapsto g} C_0 \xrightarrow{g\mapsto fg} L^1 \xrightarrow{h \mapsto \hat{h}} C_0,$$
where all parts are continuous: $\lVert g\rVert_{C-0} \leqslant \lVert g\rVert$, $\lVert fg\rVert_{L^1} \leqslant \lVert f\rVert_{L^1}\cdot \lVert g\rVert_{C_0}$, and $\lVert \hat{h}\rVert_{C_0} \leqslant \lVert h\rVert_{L^1}$.
$\beta$ factors as
$$E \xrightarrow{g\mapsto g} L^1 \cap L^2 \xrightarrow{g\mapsto \hat{g}} L^2 \cap C_0 \xrightarrow{h \mapsto \hat{f}\ast h} L^\infty,$$
where the first part is continuous by $\lVert g\rVert_{L^1}+\lVert g\rVert_{L^2} \leqslant \lVert g\rVert$, the second by $\lVert \hat{g}\rVert_{L^2} = \lVert g\rVert_{L^2}$ and $\lVert \hat{g}\rVert_{C_0} \leqslant \lVert g\rVert_{L^1}$, and the last by $\lVert \hat{f}\ast h\rVert_{L^\infty} \leqslant \lVert \hat{f}\rVert_{L^1}\cdot \lVert h\rVert_{C_0}$.
Since $C_0$ is a closed subspace of $L^\infty$, it remains to see that $\mathcal{R}(\beta) \subset C_0$. But we know that $\beta(C_c^{\infty}) \subset C_0$, and by continuity
$$\beta(E) = \beta(\overline{C_c^\infty}) \subset \overline{\beta(C_c^\infty)} \subset \overline{C_0} = C_0.$$
