# 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?

• 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. – Warlock of Firetop Mountain Jun 22 '18 at 16:37
• @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. – Love GQY Jun 23 '18 at 15:28
• What is the textbook? – fourierwho Jul 9 '18 at 16:49
• @fourierwho A textbook in China. – Love GQY Jul 12 '18 at 5:23

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 $\lambda\to \infty$