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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 $$ \sigma_{ess}(A+B)=\sigma_{ess}(A).$$

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    $\begingroup$ I have no idea what the notation $V(-\Delta+\lambda)^{-1}$ means. You can prove that the norm of $(-\Delta+\lambda)^{-1}V$ tends to zero using Plancherel. $\endgroup$ – David C. Ullrich Jun 18 '18 at 13:33
  • $\begingroup$ @DavidC.Ullrich Maybe you could understand $\dfrac{V}{-\Delta+\lambda}$? Is it right? But I think this is same as $(-\Delta+\lambda)^{-1}V$. $\endgroup$ – user469065 Jun 18 '18 at 16:03
  • $\begingroup$ @DavidC.Ullrich What is the fourier transform of $(-\Delta+\lambda)^{-1}V$? $\endgroup$ – user469065 Jun 18 '18 at 16:04
  • $\begingroup$ $(|\xi|^2+\lambda)^{-1}\hat f(\xi)$. $\endgroup$ – David C. Ullrich Jun 18 '18 at 16:06
  • $\begingroup$ @DavidC.Ullrich Is it $\mathscr{F}((\Delta+\lambda)^{-1})V)=(\vert \xi \vert ^2 +\lambda)^{-1}\mathscr{F}(V)$? $\endgroup$ – user469065 Jun 18 '18 at 16:14

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