Assume that $f\in\mathscr{E}'(\mathbb{R}^3)$ (distribution with compact support), $u=E*f$ is the solution of $$-\triangle u=f.$$ Prove that:$$\left|u(x)+\frac{\left<f,1\right>}{4\pi|x|}\right|\leq\frac{C}{|x|^2},$$as $|x|\to \infty.$ Here $C$ is a postive constant.
In fact, I think it is a physical inequality(If you take $f\in \mathscr{L}^1(\mathbb{R}^3)$, $\left<f,1\right>=\int_{\mathbb{R}^3}f(x){\operatorname{d}}x$ will be the total elements). And it may be the concretization of some inequalities in Sobolev space.