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Jul 18, 2012 at 8:34 comment added Mateusz Wasilewski 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).
Jul 17, 2012 at 16:42 comment added Jason Mraz 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 !
Jul 13, 2012 at 22:07 vote accept Jason Mraz
Jul 13, 2012 at 18:38 history answered Mateusz Wasilewski CC BY-SA 3.0