I want to prove the following proposition: owning to the $ H^2_{loc}(\mathbb{R}^n)$ regularity when solving $\Delta^{(0)}_Vu=f\in L^2$, the compactness of the resolvent of $\Delta^{(0)}_V$ is a consequence of $\forall$ $u\in C^\infty _0(\mathbb{R}^n)$
$\|R(x)u\|^2=<u,R(x)^2>\le c[<u,\Delta^{(0)}_V u> +\|u\|^2]$ with $\lim_{x\mapsto \infty} R(x)=+\infty$
This is what I tried to do :
we suppose that $\Delta^{(0)}_V$ has not a compact resolent hence since it is a self adjoint operator thre existe an orthonormal sequence $u_k$ such that $<u_k,\Delta^{(0)}_V u_k>$ is bounded hence the RHS is bounded but since $\lim_{x\mapsto \infty} R(x)=+\infty$ we have $\lim_{x\mapsto \infty}\|R(x)u_k\|^2=+\infty$ thus the LHS goes to $+\infty$ contradiction I want to know if my answer is true. Thanks in advance.