How to show $\left(\hat{f*g}\right)_n=\sum\limits_{n=-\infty}^{\infty}\hat f_n\hat g_n$
By Definition Fourier transform of $f$ is $\displaystyle \hat f_n=\int_0^1f(y)e^{-2\pi i n y}dy$
On the left I have the Fourier transform of the convolution, i.e.
$\displaystyle \left(\hat{f*g}\right)_n=\int_0^1\int_0^1f(y)g(x-y)e^{-2\pi i n x}dydx=\int_0^1f(y)\hat g_ne^{-2\pi in y }dy$
I know that $\int_0^1f(x)\overline{g(x)}dx=\sum\limits_{n=-\infty}^{\infty}\hat f_n \overline{\hat g_n}$
and also using the property $\overline{\hat g_n}=\hat g_{-n}$ I have to show that
$\hat{\left(\hat g_n e^{-2\pi n y }\right)}=g(-y)$
am I correct ? or do I make the problem more confusing ?
First, your formula is wrong. $f*g(x)=\int_0^1 f(y)g(x-y)\,dy,$ so the $n$-th Fourier coefficient of the convolution is $\int_0^1\int_0^1f(y)g(x-y)e^{-2\pi i n x}dydx$. Once you correct that it's just a little trick:
$$\int_0^1\int_0^1f(y)g(x-y)e^{-2\pi i n x}dydx \int_0^1\int_0^1f(y)e^{-2\pi i n y}g(x-y)e^{-2\pi i n (x-y)}dydx\dots$$