How to show convolution of an $L^p$ function and a Schwartz function is a Schwartz function

Solution 1:

This is not true. For an example take $f(x) = (1+\|x\|)^{-\alpha}$. If $p\alpha > n$, then $f \in L^p(\mathbb{R}^n)$. Now take a non-negative $\mathcal{C}^\infty$ bump function $g$ with integral $1$, supported in $\|x\|\le 1$. Then $g \in \mathcal{S}$ and it is easy to show that $|(f\star g)(x)| \ge (2+\|x\|)^{-\alpha}$, since the integral is an average of $f$ over the ball of radius $1$ centered at $x$. This implies $f\star g \notin \mathcal S$.

Roughly speaking, convoluting with a Schwarz function can not radically alter the decay at $\infty$, so the best estimates you can hope for are those for smooth $L^p$ functions.