Fourier Transforms and Schwartz Space

Question

If a function is infinitely differentiable and has a Fourier transform, does it follow that its transform is in Schwartz Space?

Answer 1

If a function is infinitely differentiable and has a Fourier transform, then if this function is not in the Schwartz Space $\mathscr S(R)$, its transform is NOT in $\mathscr S(R)$. For example if a function in $L^1(R)$ is $C^\infty$, but it is not in $\mathscr S(R)$, then it has a FT (all $L^1(R)$ funtions have FT), but it is not in $\mathscr S(R)$. Consider for example $f(x)=1/(1+x^2)$.

The FT maps $\mathscr S(R)$ into $\mathscr S(R)$ in a one to one way, so the FT of functions not in $\mathscr S(R)$ will never be in $\mathscr S(R)$, even if they are $C^\infty$. For a function to be in $\mathscr S(R)$ it not only has to be $C^\infty$, but in addition it and its derivatives have to go to zero at $\pm\infty$ fast enough.