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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 $$\lim_{\lambda\to\infty} \Vert V(\cdot)(-\Delta+\lambda)^{-1}\Vert =0$$ 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?

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  • $\begingroup$ Are you sure that the Hilbert space is $L^2(\mathbb{R}^3)$ and not $L^2(\Omega)$ with $\Omega$ a bounded subset of $\mathbb{R}^3$? Because in the former case the resolvent $(-\Delta +\lambda)^{-1}, \lambda>0$ is not compact while in the latter case it is. $\endgroup$ – Warlock of Firetop Mountain Jun 22 '18 at 16:37
  • $\begingroup$ @WarlockofFiretopMountain So how about the $L^2(\Omega)$? Because this is an exercise in my textbook, I try to solve it. Sorry, maybe it is wrong, but I am not sure. But the conclusion is compact. $\endgroup$ – Love GQY Jun 23 '18 at 15:28
  • $\begingroup$ What is the textbook? $\endgroup$ – fourierwho Jul 9 '18 at 16:49
  • $\begingroup$ @fourierwho A textbook in China. $\endgroup$ – Love GQY Jul 12 '18 at 5:23
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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 $\lambda\to \infty$

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