Consider the Newtonian Potential $$u(x)=\int_{\mathbb{R}^3} \frac{1}{|x-y|} f(y)dy$$, where $\mathbb{R}^3$ is the 3-dimensional Euclidian space. Show that if $|f(y)| \le C{|y|}^{-\alpha}$ for $2<\alpha<3$, then $|u(x)| \le C{|x|}^{2-\alpha}$ for $|x|>1$.
I am puzzled by this estimate. Clearly as long as $x$ is not $0$, the integral near $x$ will diverge to infinity. But the controlling condition for $f(y)$ looks so weak as to make the total estimate for $u(x)$ to hold. Thank you for help!
Simply write $x = \left|x\right| {\omega}$ with ${\omega} \in {S}^{2}$ and use the change of variable $y = \left|x\right| z$. One has
$$\renewcommand{\arraystretch}{2.0} \begin{array}{rcl}\displaystyle \int_{{\mathbb{R}}^{3}}^{}\frac{{\left|y\right|}^{{-{\alpha}}}}{\left|x-y\right|} d y&=&\displaystyle \int_{{\mathbb{R}}^{3}}^{}\frac{{\left|x\right|}^{{-{\alpha}}} {\left|z\right|}^{{-{\alpha}}}}{\left|x-\left|x\right| z\right|} {\left|x\right|}^{3} d z\\ &=&{\left|x\right|}^{2-{\alpha}} \displaystyle \int_{{\mathbb{R}}^{3}}^{}\frac{{\left|z\right|}^{{-{\alpha}}}}{\left|{\omega}-z\right|} d z\\ &=&K_\alpha {\left|x\right|}^{2-{\alpha}} \end{array}$$