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?
Thanks in advance
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? Thanks in advance 


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 RellichKondrachov 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 $\int_{\mathbb{R}^{n} \setminus B_{k}} u_{m}^2 dx \leqslant \frac{1}{k}$. I think that it is quite wellknown, probably except for people who don't have any interest in physics. A good place for delving into such results is fourth volume of Reed's and Simon's Methods of Modern Mathematical Physics  Analysis of Operators. 

