We have $-\Delta+\lambda$ is invertible on $C_{c}^{\infty}$ and using Fourier transform, then $(-\Delta+\lambda)^{-1}u$ is in Schwartz space for $u\in C_{c}^{\infty}(\mathbb{R}^3)$. Using Green's function, $$[(-\Delta+\lambda)^{-1}u](x)=\int_{\mathbb{R}^3}\frac{\exp(-\sqrt{\lambda}\vert x-y \vert )}{4\pi \vert x-y \vert}u(y)dy. \ (*)$$
How to get the (*)? Moreover, we have $$[V(-\Delta+\lambda)^{-1}u](x)=\int_{\mathbb{R}^3}\vert V(x) \vert \frac{\exp(-\sqrt{\lambda}\vert x-y \vert )}{4\pi \vert x-y \vert}u(y)dy, \ (**)$$
where the kernel is Hilbert-Schmidt kernel $$K_{\lambda}(x,y)=\vert V(x) \vert \frac{\exp(-\sqrt{\lambda}\vert x-y \vert )}{4\pi \vert x-y \vert}\in L^2(\mathbb{R}^6).$$
Since $K_{\lambda}(x,y)\in L^2(\mathbb{R}^6)$, $V(-\Delta+\lambda)^{-1}$ is compact.(From $V$ is $(-\Delta)$-compact.
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