A problem about Convolution

Question

If $\phi\in C_0^{\infty}(\mathbb{R}^N)$ and $\psi\in L_{loc}^1(\mathbb{R}^N)$ is defined by $\psi(x)=|x|^{2-N}$, $N\geq 3$ , does $\phi\star\psi$ is in the Schwartz space?

Note: $\star$ stands for convolution.

Answer 1

Suppose that $\phi(x)$ is nonnegative and is supported in $|x| < R$. Look at the expression $$\psi \ast \phi (x) = \int_{{\mathbf R}^n}|x - y|^{2 - n}\phi(y)\,dy$$ If $|x| > 2R$, in order for the integrand to be nonzero you need $|y| < R$ which implies $|x - y| \leq |x| + R < 2|x|$. Thus since $\phi(x)$ is nonnegative you have $$\psi \ast \phi (x) \geq \int_{{\mathbf R}^n}(2|x|)^{2 - n}\phi(y)\,dy$$ $$= (2|x|)^{2-n}||\phi||_{L^1}$$ This does not decay fast enough to be Schwartz.