# 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