TITLE: Let $S$ be the Schwartz class. Show that if $f,g\in S$, then $fg\in S$ and $f*g\in S$, where $*$ denotes convolution.
QUESTION [9 upvotes]: Let $S$ be the Schwartz class. Show that if $f,g\in S$, then $fg\in S$ and $f*g\in S$, where $*$ denotes convolution.

To differentiate $fg$, we may apply Leibniz's rule ( http://en.wikipedia.org/wiki/General_Leibniz_rule ). And then maybe induct on the order of the derivative.
Is there something useful that can be applied to differentiating $f*g$? I guess there's a product rule for convolution. But after using product rule, there's still a convolution sign.

REPLY [2 votes]: You can use the fact that $\widehat{f*g} = \hat f \hat g$. If you know that $f,g \in S$ implies that $\hat f$ and $\hat g$ are in $S$, then use the Fourier inversion formula to get $f * g (x)= (2\pi) ^{-n}~~\widehat{\widehat{f*g}}(-x)$. 
Since $\widehat{f*g} = \hat f \hat g$ and you know that the pointwise product of two functions in $S$ is in $S$, $\widehat{f*g}$ is in $S$, implying that $\widehat{\widehat{f*g}}$ is in the Schwartz space $S$. 
All told, we get that $f * g$ is equal to a constant times a function in $S$.
Notice that everything in $S$ lies in $L^1$, so use of the Inversion formula is valid.