I know that the spectrum of a compact operator $C$ consists at most of countable set of complex number that accumulates only at $0$. I also know that the resolvent $(\lambda-C)^{-1}$ has a pole at each nonzero element of the spectrum.

Here I meet a serious contradiction. The complex analysis says an analytic function cannot have an infinite (even countable) number of poles in a bounded domain. However, is it possible that a compact operator $C$ has an infinite spectrum so that the poles of the resolvent are infinite in number? I am extremely confused...


We can have an infinite number of poles in a finite region. $\frac 1 {\sin (\frac 1 z)}$ has a pole at each of the points $\frac 1 {n\pi}$.

Note: $0$ is not a pole of the resolvent and there is no disc around it on which the resolvent is analytic except at $0$. So having poles tending to $0$ is not a contradiction to the result you are quoting.

  • $\begingroup$ The Churchill complex variables and application book has an exercise 11 in p.247 that if $R$ is a region inside and on a simple closed contour and $f$ is analytic on $R$ except for poles in the interior of $R$, then $f$ must have at most finite poles. $\endgroup$
    – Keith
    Oct 28 '19 at 9:17
  • $\begingroup$ Then is this exercise wrong? $\endgroup$
    – Keith
    Oct 28 '19 at 9:18
  • $\begingroup$ @Keith In that result a crucial hypothesis is that the singualirities are isolated. Here that condition is not satisfied. $\endgroup$ Oct 28 '19 at 9:19
  • $\begingroup$ But poles are by definition isolated singularities. $\endgroup$
    – Keith
    Oct 28 '19 at 9:21
  • 1
    $\begingroup$ @Keith I don't know about "more severe", but yes, if $0$ is in the spectrum then $0$ is certainly not a pole of the resolvent, since it's not a isolated singularity. $\endgroup$ Oct 28 '19 at 12:30

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