# Schrodinger Operators with diverging Potential

Is it well known that if $H = -\bar{\Delta} + V$ (which is defined over $L^2( \mathbb{R} ^n$ ) and $lim_{|x| \to \infty } = + \infty$, then $H$ has compact resolvent?

Does someone know of any elegant way of proving this?

If we impose some mild conditions on potential then it boils down to compact embeddings of Sobolev spaces. For example, one can assume that $V$ is bounded from below; in that case, for the sake of convenience, I shall consider nonnegative potentials. It is enough to prove compactness of resolvent just for one element of the resolvent set, so I'll take care of $-1$. Let's take $(f_{n})$ to be a sequence of functions satisfying $\|f_n\|_{2} \leqslant 1$. If we denote its image (under the action of resolvent) by $(u_n)$ then we have
$\|\nabla u_{n}\|^2_2 + \|\sqrt{V}u_{n}\|^2_2 = -\langle f_{n}, u_{n} \rangle - \|u_{n}\|^2_2.$
From the above equation, we get $\|u_{n}\|^2 \leqslant \|u_{n}\|$, since LHS is nonnegative. Now we we're in a position to deduce that $\|\nabla u_{n}\|^2_2 \leqslant 1$ and $\|\sqrt{V} u_{n}\|^2_2 \leqslant 1$. Take the ball $B_{k}$ such that $V \geqslant k$ outside. By Rellich-Kondrachov theorem, we can choose a subsequence $u_{n_{1}}$ which converges in $L^{2}(B_{1})$. Then we pick out further subsequences and the diagonal one ends the story, because $\int_{\mathbb{R}^{n} \setminus B_{k}} |u_{m}|^2 dx \leqslant \frac{1}{k}$.
• Dear @Mateusz: You're indeed right...I think I need this conclusion for specific potentials. Mainly ones satisfying one of the following $0 \leq V \in L^1_{loc}$ , $V \in L^p+L^\infty$ . And your solution is indeed suitable for the bounded below case... But does this reslut is also valid for not-necessarily bounded below potentials? (Such that $V \in L^p+L^\infty$ only?) Thanks ! – Jason Mraz Jul 17 '12 at 16:42
• Let me quote one result from the aforementioned book: Let $n \geqslant 3$. Let $V = V_1 + V_2$, where $V_2 \in L^{\frac{n}{2}}+ L^{\infty}$ and $0 \leqslant V_{1} \in L^{1}_{loc}$, $V_1 \to \infty$. Then $H = -\Delta + V_1 + V_2$ has compact resolvent. If we have $V_1 +V_2 = V \in L^{\infty} + L^{p}$ then $V_2 \to \infty$, because $V_1$ is bounded. Let's decompose $V_2 = V_2 \chi_{V_2 < 0} + V_2 \chi_{V_2 \geqslant 0} := V_2' + V_3$; $0 \leqslant V_3 \in L^{1}_{loc}$, $V_3 \to \infty$ and $V_2' \in L^{\frac{n}{2}}$ if $p \geqslant \frac{n}{2}$ (by Hölder). – Mateusz Wasilewski Jul 18 '12 at 8:34