Decaying rate of a convolution between an integrable function and a Schwartz function Suppose $f\in L^1(R^n)$ and $g\in S(R^n)$, where $S(R^n)$ is Schwartz space. Then, Can I have estimation like following?
$$
|[f*g](x)|\leq\frac{1}{(1+|x|)^{s}},
$$
for some $s>n$.
If it is correct, how to prove it? If it is not correct, what additional assumptions does it require?

Of course the answer to the question as stated is "of course not"; the sensible version of the question is whether we have $$
|[f*g](x)|\leq\frac{c}{(1+|x|)^{s}}.
$$
The answer to that question is still no, although it's not so obvious.
Take $n=1$ just to simplify the notation. Choose $g\in\mathcal S(\Bbb R)$ with $g\ge0$ and $g(x)\ge1$ for all $x\in[-1,1]$. Given a sequence $a_n\to\infty$, let $$f=\sum_{n=1}^\infty\frac1{n^2}\chi_{[a_n,a_n+1]}.$$
Then $f*g(a_n)\ge\frac1{n^2}$, so if $a_n$ blows up fast enough, in particular if $(1+|a_n|)^s/n^2\to\infty$, then $(1+|x|)^sf*g(x)$ is not bounded.
Note The same argument shows that $f*g\in C_0$ is the most that can be said about how fast $f*g$ vanishes at infinity: If $\phi\in C_0$ it is not true that $f*g=O(\phi)$ for every $f\in L^1$ and $g\in\mathcal S$.
