# On the Schrödinger equation and the eigenvalue problem

• 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.
• you'll probably get a better response if you ask a self-contained question, which does not require first reading an article off-line... Oct 29, 2021 at 15:34
• 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 $q$. Oct 29, 2021 at 15:39
• @Carlo Beenakker，sorry, I didn't explain my question clearly, actually my question has nothing to do with this paper. Oct 30, 2021 at 2:50
• @ nls , you are right, now I think I should have solved this problem by myself, thanks. Oct 30, 2021 at 3:01

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