Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?


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

share|improve this question
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
show 2 more comments

1 Answer 1

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).

share|improve this answer
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
continuning 3) but how do you see from (2) that we don't have a compact spectrum ? Since I'm studying this material on myself, I'll be glad if you'll be able to help me understand all of these little things... Thanks ! –  Franklin McDeover Jul 15 '12 at 15:40
@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
It was indeed a bit stupid to confuse the resolvent operator and the resolvent set , and to forget about the integration by parts formula. Thanks for explaining it to me! As for (3) - I've re read chapter 4 and went over your answer again, but I still can't understand why the ineqaulity you received implies you don't have discrete spectrum. Will you help me ? (Even if you'll be able to tell me more specifically what should read again, or retry to understand, it'll be very helpful! This entire subject is still very vague to me) Looking forward for your help! Thanks –  Franklin McDeover Jul 16 '12 at 11:36
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.