A relationship between Fourier transform and convolution. If $f , g , \hat{f} \hat{g} \in L^1(\mathbb{R})$, I have to prove that
$$
\int_{\mathbb{R}} \hat{f}(y) \hat{g}(y) e^{2 \pi i x y} \, dy = (f*g)(x)
$$
for almost all $x \in \mathbb{R}$. I use the next definition for Fourier transform and for the convolution:
$$
\hat{f}(x) = \int_{\mathbb{R}} f(y) e^{- 2 \pi i x y} \, dy \quad \mbox{ and } \quad (f*g)(x) = \int_{\mathbb{R}} f(x - y) g(y) \, dy\mbox{.}
$$
I tried to use Fubini and change of variable theorems and I didn't obtain any result. Maybe I didn't change correctly the variables when I tried to show the equality using change of variable theorem. Thank you very much.

You can apply definition : by Fourier transform of
$\displaystyle(f*g)(x) = \int_{\mathbb{R}} f(x - t) g(t) dt$
we have
$$\hat{(f*g)}(y) = \int_{\mathbb{R}}\,\left(\int_{\mathbb{R}} f(x - t) g(t) dt\right)e^{- 2 \pi i x y} \, dx=\int_{\mathbb{R}}\,\int_{\mathbb{R}} f(x - t) g(t) e^{- 2 \pi i x y} \,dt dx$$
take substitution $x-t=u$ 
$$\int_{\mathbb{R}}\,\int_{\mathbb{R}} f(u) g(t) e^{- 2 \pi i (t+u) y} \,du dy=\int_{\mathbb{R}} f(u)  e^{- 2 \pi i u y} \,du \int_{\mathbb{R}} g(t)  e^{- 2 \pi i t y} \,du =\hat{f}(y) \hat{g}(y)$$
