How to show that $V(\cdot)(-\Delta+\lambda)^{-1}$ is compact? 
Let $\mathscr{H}=L^2({\mathbb{R}^3})$ be Hilbert space, suppose $V\in L^2({\mathbb{R}^3}), \lambda>0$, show that $$\lim_{\lambda\to\infty} \Vert V(\cdot)(-\Delta+\lambda)^{-1}\Vert =0$$ Suppose $V\in L^2({\mathbb{R}^3})$, how to show that $\lim_{\lambda\to\infty} \Vert V(-\Delta+\lambda)^{-1}\Vert =0$

How to show that $V(\cdot)(-\Delta+\lambda)^{-1}$ is compact?
Why is it a Hilbert-Schmidt operator?
 A: I'm pehaps not able to answer the question, but maybe some of my work is useful for you,
The greens, G(x,y) function to $A=-\Delta + \lambda^2$ is 
$$
(2\pi)^{3/2}\left ( \frac{\lambda}{|x-y|} \right )^{1/2} K_{1/2}(\lambda |x-y|)
$$ see https://en.wikipedia.org/wiki/Green%27s_function, now
$$
K_{1/2}(z) = \sqrt{\frac{\pi}{2 z}}e^{-z}
$$ (https://en.wikipedia.org/wiki/Bessel_function#Modified_Bessel_functions:I.CE.B1.2C_K.CE.B1), Hence the greens function is 
$$
\frac{C}{|x-y|} e^{-\lambda |x-y|}
$$ Hence
$$
\langle V A^{-1}, v \rangle = \int\int V(x) \frac{C}{|x-y|} e^{-\lambda |x-y|}v(y)\,dx \,dy
$$ which is
$$
\langle V A^{-1}, v \rangle = \int V(x) \int_0^{\infty}  \frac{C_2}{r} e^{-\lambda r}\left ( \int_{S_1} v(x+r*\hat e)\sin(\theta)d\phi d\theta \right )r dr dx
$$Now r cancels and we can rearenge the integral and note by Cauchy Schwarz
$$
\left | \int V(x)v(x + r \hat e)\sin(\theta() dx \right | \leq ||V|| ||v||
$$ Hence apply the triangle inequality and what's left after noting that the integral of $\phi,\theta$ is bounded that
$$
|\langle V A^{-1}, v \rangle| \leq C_3|v||V|\int_0^\infty e^{- \lambda r}  \, dr =  C_3||v||||V|| / \lambda
$$
So the dual norm yields,
$$
\sup_{|v|\leq 1} |\langle V A^{-1}, v \rangle| \leq C_3 ||V|| / \lambda
$$ and this goes to zero as $\lambda\to \infty$
