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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|<R} f(y)g(x-y) dy. $ Since $g(x-y)=1/|x-y| \geq 0$ and since $f(y) \leq f^{+}(y) := \max\{f(y),0\}$ everywhere, we have $$ h(x) \leq \int_{|y|<R} \frac{f^{+}(y)}{|x-y|} dy \leq \sup_{|y|<R} \frac{1}{|x-y|} \int_{|y|<R} f^{+}(y) dy \leq \frac{1}{|x|} \sup_{|y|<R} \frac{|x|}{|x|-|y|} \int_{|y|<R} f^{+}(y) dy $$ Since $$ \frac{|x|}{|x|-|y|} \leq c \quad \text{ if and only if } \quad |x| \geq \frac{c}{c-1}|y|, $$ if $|x| \geq \frac{c}{c-1}R$ and $R > |y|$, we have $|x|/(|x|-|y|) \leq c$. Hence $$ h(x) \leq \frac{c}{|x|} \int_{|y|<R} f^{+}(y) dy = \frac{C}{|x|} $$ for $|x| \geq \frac{c}{c-1}R$.

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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up vote 2 down vote accepted

Suppose that $f(x)=0$ for $|x|\ge R$. Since $\int f(x)\;\mathrm{d}x=0$, for $|x|>2R$, we have $$ \begin{align} \left|\int_{|y|<R} \frac{f(y)}{|x-y|}\mathrm{d}y\right| &=\left|\int_{|y|<R} f(y)\left(\frac{1}{|x-y|}-\frac{1}{|x|}\right)\mathrm{d}y\right|\\ &\le\|f\|_{L^1}\left\|\frac{1}{|x-y|}-\frac{1}{|x|}\right\|_{L^\infty(|y|<R)}\\ &=\|f\|_{L^1}\left(\frac{1}{|x|-R}-\frac{1}{|x|}\right)\\ &=\|f\|_{L^1}\frac{R}{(|x|-R)|x|}\\ &\le\|f\|_{L^1}\;2R/|x|^2 \end{align} $$ Note that no smoothness of $f$ is used. The support of $f$ and its $L^1$-norm are the only things used about $f$ (other than $\int f(x)\;\mathrm{d}x=0$). The decay of the derivative of $1/|x|$ is what is important.

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Thanks!. I eventually came up with essentially the same proof. – maxpower Sep 24 '11 at 1:00

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