Suppose that $ f: \mathbb{R}^n\rightarrow\mathbb{R} $ satisfies $|f(x)| \leqslant C <x>^{-k}$ for $k>2$ (Here $<x>^2 = 1+|x|^2$). And let $u(x) := \int_{\mathbb{R}^n} |x-y|^{2-n}f(y) dy $ be the corresponding Newtonian potential. Show that there exists $C\geqslant 0$ such that : $|u(x)| \leqslant C(<x>^{2-n} + <x>^{2-k})$.
First, I found this type of estimate when $|x|\geqslant\epsilon>0$ but I don't see who I can generalize it for $x\in \mathbb{R}^n$.
Now, I tried :
$|u(x)| \leqslant \sup_{y\in \mathbb{R^n}} |<y>^kF(y)| \int_{\mathbb{R^n}} \frac{<y>^{-k}}{|x-y|^{n-2}} dy \\ \leqslant C v(x)$
Where
$v(x) := \int_{\mathbb{R^n}} \frac{<y>^{-k}}{|x-y|^{n-2}} dy$
Write $x = |x|u$ with $u$ in the unit sphere and change the variable in the integrand by considering $y = |x|\tilde{y}$. We obtain :
$v(x) = |x|^2 \int_{\mathbb{R^n}} \frac{<|x|\tilde{y}>^{-k}}{|u-\tilde{y}|^{n-2}} d\tilde{y} \\ \le <x>^2 \int_{\mathbb{R^n}} \frac{<|x|\tilde{y}>^{-k}}{|u-\tilde{y}|^{n-2}} d\tilde{y} \\ \le <x>^2 \int_{|\tilde{y}|\le2} \frac{<|x|\tilde{y}>^{-k}}{|u-\tilde{y}|^{n-2}} d\tilde{y} + <x>^2 \int_{|\tilde{y}|\ge 2} \frac{<|x|\tilde{y}>^{-k}}{|u-\tilde{y}|^{n-2}} d\tilde{y}$
Estimate of the second term:
For ${|\tilde{y}|\ge 2}$ we have $<|x|\tilde{y}>\ge <x>$ Then $<|x|\tilde{y}>^{-k}\le <x>^{-k}$
Moreover, $|z|>2$ implies $|w-z|>1$ and the integral $\int_{|\tilde{y}|\ge 2} \frac{1}{|u-\tilde{y}|^{n-2}} d\tilde{y}$ is invariant under rotations so it does not depend on $u$. Besides this integral converges because $n>2$. Hence the second term, say $II$ can be estimated by :
$II \le C_2<x>^{2-k}$ where $C_2 = \int_{|\tilde{y}|\ge 2} \frac{1}{|u-\tilde{y}|^{n-2}} d\tilde{y}$.
What remains is the first term. I don't know how to find the term $<x>^{2-n}$
Thank you for your help !