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The following problem keeps bothering me:

Let $H:=-\Delta+V$ be a Schrödinger-Operator in $\mathbb{R}^n$, where $V$ is a Kato-Potential of type $K_n$, which especially yields that $H$ is e.s.a. on $C_0^\infty$ and the resolvent operator is an Integral operator (see e.g. Simon- Schrödinger Semigroups). Furthermore, let $V$ be $C^\infty$ everywhere but on a finite set of points, say $N$, where $V$ is discontinuous.

Now let $\varphi$ be a test function, i.e. smooth and vanishing together with all derivatives on $N$. (In my case, $\varphi$ is furthermore a Schwartz function, but I don't think this plays a role in this question.) Furthermore, let $\lambda$ be in the resolvent set of $H$ and $\psi=R_\lambda(H)\varphi$, which is equivalent to $\psi$ being a strong $\mathcal{L}^2$-solution to the PDE $$ (-\Delta+V-\lambda)\psi=\varphi $$ By elliptic regularity, I instantly get that $\psi$ is $C^\infty$ on $\mathbb{R}^n-N$.

I want to know, how $\psi$ behaves on $N$. My Intuition tells me, that $\psi$ has to vanish together with all derivatives on $N$, just as $\varphi$ does. However, I am not able to provide a rigorous proof for this. I am quite sure that I am overlooking something trivial, this has to be solvable by standard methods. Does anyone have an idea?

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  • $\begingroup$ I'm not sure if the following example is relevant: If $V = -\frac{2}{x^2}$ then $x^2$ solves $(-\Delta + V)x^2=0$. Not all derivatives of $x^2$ vanish at $0$. $\endgroup$ Dec 10, 2014 at 17:05
  • $\begingroup$ @Otis: Thank you for your reply. I am really not sure, I also stumbled over this combination, but I think (hope) it is not relevant as $0$ is in the spectrum of this Schrödinger operator and furthermore $x^2$ is not an $L^2$-function. $\endgroup$
    – Daniel
    Dec 11, 2014 at 14:16

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