I have the following problem: Let $f\in\mathcal{S}(\mathbb{R}^n)$ and $g\in C^\infty(\mathbb{R}^n)$ such that exists $\alpha>0$ $$ \left|g(x)\right|\leq \frac{1}{1+|x|^\alpha} \quad \forall x\in\mathbb{R}^n $$ I have to prove that either $f\ast g\in\mathcal{S}(\mathbb{R}^n)$ or show why not.
From the hypothesis, my claim is that the convolution doesn't belong to the Schwartz functions. I tried to work with the following proposition. A smooth function $f$ is Schwartz iff for all $N\in\mathbb{N}$ and $\alpha\in\mathbb{N}^n_0$ exists $C>0$ such that $$ |\partial^\alpha f(x)|\leq \frac{C}{(1+|x|)^N} \quad \forall x\in\mathbb{R}^n. $$ Is my negation of the previous proposition correct?
A smooth function $g$ is not a Shchwartz function iff exists $N\in\mathbb{N}$ and exists $\alpha\in\mathbb{N}_0^n$ such that $$ |\partial^\alpha g(x)|> \frac{C}{(1+|x|)^N} \quad \forall x\in\mathbb{R}^n. $$
Also I tried to bound the convolution with a similar idea of the proposition 2.2.7 from Classical Fourier Analysis of Grafakos $$ \int_{\mathbb{R}^n} |f(x-y)||g(y)|\, dy \leq C \int_{\mathbb{R}^n} \frac{1}{(1-|x-y|)^N(1+|y|^\alpha)}\, dy $$ If I define $E = \{y\,:\, |x|/2\leq |y-x|\}$ then $$ \int_E |f(x-y)||g(y)|\, dy \leq \frac{C 2^N}{(1+|x|)^N}\int_E \frac{1}{1+|y|^\alpha}\, dy $$ but I'm stuck with the last integral.
