Differential operator applied to convolution
Solution 1:
You have the right idea for proving $D^\alpha (g*f) = (D^\alpha g)*f $ "formally" using the Fourier transform. You're also right that you need to show that both sides of the equation are well-defined. The right side is clearly well-defined since $g\in \mathcal{S}$. To show that $g*f\in C^\infty$, use the Fourier-side decay condition: if $\widehat{g*f}$ decays faster than any power of $k$, then $g*f\in C^\infty$. This should be easy to show using $\widehat{g*f} = \hat{g}\hat{f}$ and the fact that $g\in \mathcal{S}$.