# Proving Compactness of Resolvent Of an Operator

I have encountered the following functional analysis question, which I can't figure out how to prove: Prove that if $H:= - \bar{ \Delta } +V$ on $L^2 (\mathbb{R}^n)$ , and $lim_{|x| \to \infty} V(x) = + \infty$ , then $H$ has compact resolvent.

Can someone help me figure out how to prove this exercise?

Thanks

( $H:= - \bar{ \Delta }$ stands for the closure of the laplacian)

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What is $V$? And which limit are you taking? – Davide Giraudo Jul 13 '12 at 22:25
I presume the missing function in the limit is V. It might help to know the regularity of V - is it just measurable or something better? – user31373 Jul 14 '12 at 3:29
You're right. Inside the limit is V! By Davies' book, no further assumptions need to be made on this potential... Can you help me ? Thanks ! – Franklin McDeover Jul 14 '12 at 7:23
I am almost certain that Davies's book does require conditions on the potential... – user16299 Jul 14 '12 at 9:03
Moreover, which book by E.B. Davies? – user16299 Jul 14 '12 at 9:04

You want to show that the essential spectrum of $H$ is empty. Suppose it contains some number $\lambda\in\mathbb R$. Lemma 8.4.1 provides us, for every $\epsilon>0$, with an infinite-dimensional space $L_\epsilon$ in which $\|Hf-\lambda f\|\le \epsilon \|f\|$. Taking the inner product of $Hf-\lambda f$ with $f$, we obtain $$(1)\qquad \qquad \int (|\nabla f|^2+(V-\lambda)|f|^2)\le \epsilon\int |f|^2,\qquad f\in L_\epsilon$$ Now split $V-\lambda-\epsilon$ into the positive and negative parts, $V-\lambda-\epsilon=V_+-V_-$. The important thing is that $V_-$ is compactly supported. Rewriting (1) as $$(2)\qquad\qquad \int (|\nabla f|^2+V_+|f|^2)\le \int V_-|f|^2,\qquad f\in L_\epsilon$$ we run into trouble because $-\Delta$ has discrete spectrum on bounded domains (Theorem 6.2.3).
Dear Leonid, 1) The Resolvent of an operator is the complement of the spectrum of the operator. If the essential spectrum is empty, our spectrum must be discrete. How does this imply the compactness of our resolvent set? 2) When you say you're taking inner product, which inner product are you taking? I can't see how you get the inequality (1) from the regular $L^2$ inner-product... 3) When taking the inner product, you're taking it on the whole $\mathbb{R}^n$ . But in (2), youre saying that we get a contradiction. I guess it's because $V_-$ is compacttly supported. – Franklin McDeover Jul 15 '12 at 15:38
@FranklinMcDeover 1) You are confusing the resolvent set with resolvent operator. The resolvent set is open by definition, so it can't be compact unless it's empty (which does not happen for self-adjoint operators). See en.wikipedia.org/wiki/Resolvent_formalism 2) I took the $L^2$ inner product, followed by integration by parts: $\int (-\Delta f)f=\int |\nabla f|^2$. Sorry, but I'm not sure you ready to deal with 3) yet. I suggest re-reading earlier chapters, in particular 4. – user31373 Jul 15 '12 at 16:50