Let $f\in L^2(\mathbb{R}^3)$ be a function with compact support and define $v:\mathbb{R}^3\to\mathbb{R}$ by $$v(x)=\int_{\mathbb{R}^3}\frac{f(y)}{|x-y|}dy$$
Is true that $v\in W^{2,2}(\mathbb{R}^3)$ and $$\|v\|_{2,2}\leq C\|f\|_2$$
for some positive constant $C$?
Thank you
Let $\mathrm{supp}(f)\subset B(0,R) \subset\subset \Omega$ for some open and bounded $\Omega$ such that $\mathrm{dist}(B,\partial\Omega) \geq h>0$, then $$v(x)=\int_{\mathbb{R}^3}\Gamma(x-y)f(y)dy = \int_{\Omega}\Gamma(x-y)f(y)dy.$$ By $L^p$-estimate of Poisson equation (Calderon-Zygmund, GT theorem 9.9, $1<p<\infty$): $$ \|D^2 v\|_{L^p(\Omega)} \leq C\|f\|_{L^p(\Omega)}. $$ Moreover, the trace of $v$ is zero on $\partial \Omega$ (you might need to prove this using a mollifier sequence argument for $f$), similar argument applies for $\nabla v$ for $$\nabla v(x) = \int_{\Omega}\nabla_x\Gamma(x-y)f(y)dy.$$
Now by Friedrichs' inequality: $$ \|v\|_{L^p(\Omega)}\leq C\|\nabla v\|_{L^p(\Omega)}\leq C\|D^2 v\|_{L^p(\Omega)}, $$ which implies $v\in W^{2,p}(\Omega)$, then extending $v$ to $ W^{2,p}(\mathbb{R}^3)$ will finish the job.