• 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.
  • 2
    $\begingroup$ 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


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 $q \in L^{n/2}$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.