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In the book Partial Differential Equations by A. Friedman 1969. Part 2 on page 104, in the proof of theorem 2.1 (d).

$A$ is a operator of type $(\psi,M)$ ($-A$ generates a analytic semigroup), and $\Gamma$ is a smooth curve lying in the resolvent set of $-A$, $\lambda\in C$, the integrand $$\frac{1}{2\pi i}\int_{\Gamma}(\lambda I+A)^{-1}Ax\frac{d\lambda}{\lambda}=0,$$ why? How to compute?

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I am not sure what $(\psi,M)$ type is, and you cannot assume that all readers of this site have this 1969 book next to them but usually such things hold when the spectrum is bounded, and the curve $\Gamma$ can be continuously deformed to infinity on the resolvent set. The integrand is analytic in $\lambda$, so by Cauchy's theorem the integral does not depend on $\Gamma$, but on the other hand, a trivial estimate shows that the integral tends to $0$ when $\gamma=${ $z:|z|=R$} and $R\to\infty$. Therefore the integral is $0$.

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  • $\begingroup$ Thank you. I get it. The $\Gamma$ here is not a closed curve, so I can combine it with a part of ball to get a closed curve and then prove the limit of integral on the "ball" is zero. By the wawy, who gave a -1 on my question, so sad. $\endgroup$
    – caoxinru
    Apr 10, 2013 at 9:30
  • $\begingroup$ If $\Gamma$ is not closed, it is unlikely that integral is $0$. The argument I wrote assumes that $\Gamma$ is a closed curve (or several closed curves) which surround the spectrum. I don't know who gave you -1, but I added 1 to cancel it, if it made you so upset:-) $\endgroup$
    – Alexandre Eremenko
    Apr 10, 2013 at 12:01
  • $\begingroup$ I think that voting only in order to offset somebody else's vote is an abuse of the system. You should upvote questions that you think are good, and downvote the ones that you think are bad. The resulting sum should represent the consensus of the community. This particular question is extremely sloppy, with no effort gone into the phrasing, or the grammar, or the spelling. Instead of being sad, the OP should reread his own question at least once. $\endgroup$
    – Alex B.
    Jun 8, 2013 at 10:28
  • $\begingroup$ Alex, You are right. I will not do this anymore. $\endgroup$
    – Alexandre Eremenko
    Jun 9, 2013 at 6:43

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