# $L^2$ norm estimated by $H^{-2}$ norm.

If I have a PDE of the form

$$\left\{\begin{array}{ll} (-\Delta+q)v_1=-(-\Delta+q)v_0, & \text{in}\quad\Omega \\ v_1=0 & \text{on}\quad\partial\Omega \end{array} \right.$$

then is the following estimate correct: $$||v_1||_{L^2(\Omega)}\leq C||-(-\Delta+q)v_0||_{H^{-2}(\Omega)}?$$ Here the function $$v_1$$ solves the above PDE, $$C\geq0$$ and $$v_0$$ is an explicit function. This seems that it should work because of the way that PDE regularity usually works. I think that the form of my PDE is not important. I am also interested to find a reference for this, so I would be very thankful if someone could provide me with one :)

• Just to clarify, $H^{-2}(\Omega)$ is the dual of $H^2_0(\Omega)$, right? – StarBug Aug 23 '19 at 8:54
• @StarBug Yes, you are right. – MathLearner Aug 24 '19 at 6:56
• Just to clarify, I am more interested in the reference than in an actual proof of the fact. I need this for my masters thesis. – MathLearner Aug 24 '19 at 8:51
• I don't think the estimate is true unfortunately. I will try to give a proper answer. – StarBug Aug 24 '19 at 10:04

A simple counter example: Consider on $$\Omega:=B_1$$ some smooth function $$\phi_\epsilon$$ such that $$\phi_\epsilon=1$$ on $$B_{(1-\epsilon)}$$ and $$\phi=0$$ on $$\partial B_1$$. Then (trivially) \begin{align} \Delta \phi_\epsilon = \Delta \phi_\epsilon\quad \text{in } B_1,\qquad \phi_\epsilon=0 \quad \text{on } \partial B_1, \end{align} but \begin{align} &||\Delta \phi_\epsilon||_{H^{-2}} \rightarrow 0\quad\text{as}\quad \epsilon\rightarrow 0\quad\text{and}\\ &||\phi_\epsilon||_2\rightarrow |B_1|^{1/2} \quad\text{as}\quad \epsilon\rightarrow 0. \end{align}