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 



Schrodinger Operators with diverging Potential
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 $\int\_{\mathbb{R}^{n} \setminus B\_{k}} |u\_{m}|^2 dx \leqslant \frac{1}{k}$.


I think that it is quite well-known, 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*.


