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  • Li-Yau 1983_Article
  • The second part of above paper used the discrete eigenvalues of $\frac{-\Delta}{q}$ where $q>0$ to proof the the number of non-positive eigenvalues of Schrödinger operator $-\Delta+V$ can be bounded by the $L_{\frac{n}{2}}$-norm of $V^-$.
  • My question is: under what condition of $q$ can we proof the spectrum of $\frac{-\Delta}{q}$ is discrete.
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    $\begingroup$ you'll probably get a better response if you ask a self-contained question, which does not require first reading an article off-line... $\endgroup$ Oct 29, 2021 at 15:34
  • $\begingroup$ In order to show that the spectrum is discrete, you consider the inverse of the operator $-\Delta/q$ and show that it is compact and self-adjoint over some Hilbert Space. There is a standard example involving the Laplacian in Brezis' book on Functional Analysis. You can just modify that proof and deduce what conditions are required for $q$. $\endgroup$
    – Student
    Oct 29, 2021 at 15:39
  • $\begingroup$ @Carlo Beenakker,sorry, I didn't explain my question clearly, actually my question has nothing to do with this paper. $\endgroup$
    – sorrymaker
    Oct 30, 2021 at 2:50
  • $\begingroup$ @ nls , you are right, now I think I should have solved this problem by myself, thanks. $\endgroup$
    – sorrymaker
    Oct 30, 2021 at 3:01

1 Answer 1

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Assuming $q>0$ the Schroedinger operator $-\Delta/q$ is associated to the form $a(u,v)=\int_{\mathbb R^n} \nabla u \cdot \nabla v$ in $L^2(\mathbb R^n, q\, dx)$. The form domain consists of all $u\in L^2(\mathbb R^n, q\, dx)$ such that $u \in \dot H^1:=\{u \in L^{2^*}(\mathbb R^n), \nabla u \in L^2(\mathbb R^n)\} $ and the discreteness of the spectrum is equivalent to the compactness of the embedding of the form domain into $L^2(\mathbb R^n, q\, dx)$.

This follows when the map $$T:\dot H^1 \to L^2(\mathbb R^n), \quad Tu=q^{1/2}u$$ is compact, which is true whenever $q \in L^{n/2}$.

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  • $\begingroup$ @Giorgio Metafune thanks, I get it. $\endgroup$
    – sorrymaker
    Oct 30, 2021 at 2:53

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