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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?2012-07-13
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    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?2012-07-14
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    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 !2012-07-14
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    I am almost certain that Davies's book *does* require conditions on the potential...2012-07-14
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    Moreover, *which* book by E.B. Davies?2012-07-14
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    Dear Yemon- You might be right, but I don't see that he assumes anything on the potential in this particular question... I'm talking about "Spectral theory of differnetial operators" by Davies. This is question 3 in chapter 8 there... Have you got any idea? Hope you'll help me figure it out !!!2012-07-14
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    On page 164 Davies says "We assume that the potentials $V$ satisfy one of the hypotheses of Theorem 8.2.1, Theorem 8.2.3 or Example 8.2.7". It's natural for us to apply the same assumptions to the exercises... By the way the subsequent proof of Lemma 8.3.3 is worded thus: "We observe that $Q(f)\ge 0$ for all $f\in C_c^\infty$ and that such $f$ form a core for $Q$, for reasons which depend upon which assumption we make on $V$. The lemma now follows by an application of the variational formula. QED" A bit on the side of a sketch..2012-07-15

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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).

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    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.2012-07-15
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    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 !2012-07-15
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    @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 http://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 4.2012-07-15
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    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! Thanks2012-07-16