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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.

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try $\psi = 1$. –  Zarrax Jan 30 '13 at 0:22
    
Sorry Zarrax, I forgot to add an hypothesis. I will edit it. –  Tomás Jan 30 '13 at 0:25

1 Answer 1

up vote 2 down vote accepted

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.

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Interesting. What happens if I ask that $\int_{\mathbb{R}^N}\phi=0$. Can we conclude that $\psi\star\phi$ is Schwartz? –  Tomás Jan 30 '13 at 1:07
1  
No I don't think you necessarily can.. you can integrate by parts in the expression for $\psi \ast \phi(x)$ and you end out with a similar expression except with a derivative of $|x - y|^{2-n}$ and an indefinite integral of $\phi$. You can show that for some $\phi$ with $\int \phi = 0$ you have a lower bound $C|x|^{1 - n}$. So you have a better exponent but still not Schwartz. –  Zarrax Jan 30 '13 at 2:18

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