 LiYau 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 nonpositive 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 selfcontained question, which does not require first reading an article offline... $\endgroup$– Carlo BeenakkerCommented 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 selfadjoint 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 $q$. $\endgroup$– StudentCommented 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. $\endgroup$– sorrymakerCommented Oct 30, 2021 at 2:50

$\begingroup$ @ nls , you are right, now I think I should have solved this problem by myself, thanks. $\endgroup$– sorrymakerCommented Oct 30, 2021 at 3:01
1 Answer
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. $\endgroup$ Commented Oct 30, 2021 at 2:53