Convolution of $L^1$ and Schwartz function

Question

I had a question regarding convolution. Given $f$ $\in L^1(\mathbb{R})$, is it always possible to find a Schwartz function $g$ such that $f * g$ is Schwartz? I know that such a convolution will always be smooth, but can we make it Schwartz by picking a suitable function $g$? Thanks!

Answer 1

No, that is not always possible. Up to rescaling, you may always assume that the absolute value of an $L^1$ function is a probability density function, and the same clearly holds for Schwartz functions. If the pdf of $X$ is $|f|$ and the pdf of $Y$ is $|g|$, then the pdf of $X+Y$ is $|f|*|g|$. If $g$ is a Schwartz function all the central moments of $Y$ are finite, but if the pdf of $X$ behaves like $\frac{c}{(1+x^2)^{5/2}}$ then only the first three central moments of $X$ are finite, so only the first three moments of $X+Y$ are finite and $|f|*|g|$ cannot be a Schwartz function.