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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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