Let $\mathscr{H}=L^2({\mathbb{R}^3})$ be Hilbert space, suppose $V\in L^2({\mathbb{R}^3}), \lambda>0$, show that $$\lim_{\lambda\to\infty} \Vert V(-\Delta+\lambda)^{-1}\Vert =0$$
I try to use this method How to get $[(-\Delta+\lambda)^{-1}u](x)=\int_{\mathbb{R}^3}\frac{\exp(-\sqrt{\lambda}\vert x-y \vert )}{4\pi \vert x-y \vert}u(y)dy$ But I do not understand the reasons. It would imply $\sigma_{ess}(-\Delta+V)=[0,\infty)$, which using the Wely's theorem:
Suppose $A$ is a self-adjoint operator in Hilbert space, $B$ is symmetric if $B$ is $A-$compact then $$ \sigma_{ess}(A+B)=\sigma_{ess}(A).$$