# R.H.S. of Poisson equation localized $\Rightarrow$ Solution localized

Let $\Omega\subset\mathbb{R}^d$ a connected open set (which is not necessarily bounded). Assume that $f\in C_0^\infty(\Omega)$ with $\operatorname{supp}(f)\subset K,$ with $K$ compactand let $u$ be a solution to the equation $$-\Delta u+u=f\quad\text{in }\Omega,\\ \quad\quad u\in H^1_0(\Omega).$$ I'm looking for a statement of the following type: For every $\epsilon>0$ there exists a $R>0$ (depending only on $K$ but not on $f$) such that $$\|u\|_{L^2(\Omega\setminus B_R(0))}\leq \epsilon\|f\|_{L^2}.$$

This is easily obtained for $d=3$, $\Omega=\mathbb{R}^3$ using the fundamental solution, but the general case is not clear to me...

• Are you sure about your three-dimensional result? If I take $t \, f$ on the right-hand side, the solution is $t \, u$. But then $t \nearrow \infty$ contradicts your estimate. – gerw Mar 3 '17 at 12:46
• Oh, I forgot to say that $f$ is suppoosed to stay bounded in $L^2$. I corrected it in the text. – Frank Mar 3 '17 at 14:01
• If your result is clear in d=3 on the whole space from a fundamental solution calculation, why is it not clear in d>3 on the whole space? (d<3 are obviously rather different.) – Ian Mar 3 '17 at 15:02
• Maybe I'm not very well informed, but I don't know the fundamental solution explicitly for $d>3$. – Frank Mar 3 '17 at 15:05
• I forgot that you have that +u in there changing matters, but the technique is still the same, it's just solving it on the Fourier side and computing an integral. Anyway, that still doesn't tell you how to handle other $\Omega$s, which I don't know how to do. – Ian Mar 3 '17 at 15:52