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Let $f:\mathbb{R}^{3} \rightarrow \mathbb{R}$ be compactly supported and $C^{\infty}$. Let $g:\mathbb{R}^{3} \rightarrow \mathbb{R}$ be defined by $g(x) = \frac{1}{|x|}$ (with $g(0)$ defined arbitarily). Define $h(x)$ by $$ h(x) = (f \ast g)(x) = \int_{\mathbb{R}^{3}} f(y)g(x-y) dy = \int_{\mathbb{R}^{3}} f(x-y)g(y) dy $$

I want to prove that if $\int_{\mathbb{R}^{3}} f = 0$ there is a constant $C > 0$ such that $h(x) \leq C/|x|^2$ for all large $|x|$.

I have been able to prove that $h(x) \leq C/|x|$ for all large $|x|$, but I want to do better. My proof doesn't use that $\int_{\mathbb{R}^{3}} f = 0$ or the full smoothness of $f$. It runs as follows. Since $f$ is compactly supported there is an $R > 0$ such that $f(y)=0$ for all $|y| \geq R$. So $ h(x) = \int_{|y| |y|$, we have $|x|/(|x|-|y|) \leq c$. Hence $$ h(x) \leq \frac{c}{|x|} \int_{|y|

I haven't been able to figure out a way to adjust my method to prove the result I want. Can someone please help me out?

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Suppose that $f(x)=0$ for $|x|\ge R$. Since $\int f(x)\;\mathrm{d}x=0$, for $|x|>2R$, we have $$ \begin{align} \left|\int_{|y|

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    Thanks!. I eventually came up with essentially the same proof.2011-09-24