Given a real valued potential $V\in L^1(\mathbb{R}^3)$, we say that the Schrodinger operator $-\Delta + V$ has a zero-energy resonance if there exists $\psi\in L^2_{loc}(\mathbb{R}^3)\setminus L^2(\mathbb{R}^3)$ such that $$(-\Delta+V)\psi=0.$$ Is there a simple way to show the existence of a potential $V\in L^{1}(\mathbb{R}^3)\cap L^{3/2}(\mathbb{R}^3)$, $V\neq 0$, such that the corresponding Schrodinger operator actually has a zero-energy resonance?
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$\begingroup$ $V \equiv 0$ and $\psi \equiv 1$? $\endgroup$– Mateusz KwaśnickiCommented Mar 4, 2018 at 18:51
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$\begingroup$ Yes, of course. I meant a non zero potential, sorry. I edited the question $\endgroup$– CapublancaCommented Mar 4, 2018 at 20:56
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1$\begingroup$ Take your favourite $\psi$, set $V(x) = \Delta \psi(x) / \psi(x)$, and see if $V$ is $L^1 \cap L^{3/2}$. If, say, $\psi(x) = 1 + \exp(-|x|^2)$, then $V(x) = (4 |x|^2 - 6) / (1 + \exp(|x|^2))$. $\endgroup$– Mateusz KwaśnickiCommented Mar 4, 2018 at 21:06
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