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I am trying to prove, that a self-adjoint (maybe unbounded) operator has a non-empty spectrum. So far I have argued, that if $\sigma(T)$ would be empty, $T^{-1}$ would be a bounded self-adjoint operator. I now want to show, that $\sigma(T^{-1}) = \{0\}$. Then, because norm and spectralradius are equal for bounded operators it follows $T^{-1}=0$, a contradiction.

For the bold part I have tried the following: For $\lambda \neq 0$ I have to calculate the inverse of $\lambda Id - T^{-1}$ and show that it is bounded. Unfortunately, this appears to be quite difficult. Does someone know how to do this?

Thanks.

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Assume $T$ has empty spectrum. Then $T$ is invertible, $T^{-1}$ is a bounded selfadjoint operator and, for $\lambda \ne 0$, $$ (T^{-1}-\lambda I) =(I-\lambda T)T^{-1}=\lambda(\frac{1}{\lambda}I-T)T^{-1} $$ has bounded inverse $$ \frac{1}{\lambda}T\left(\frac{1}{\lambda}I-T\right)^{-1} $$ So $\sigma(T^{-1})=\{0\}$ because only $\lambda=0$ can be in the spectrum, and it cannot be empty. But that implies $T^{-1}=0$, which is a contradiction.

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    $\begingroup$ Thanks! Shouldn't it be $(I-\lambda T)T^{-1} = \lambda(\frac{1}{\lambda} I - T)T^{-1}$ and therefore the inverse would be $\frac{1}{\lambda} T (\frac{1}{\lambda} I - T)^{-1}$? $\endgroup$ – KennyH Jul 11 '15 at 22:18
  • $\begingroup$ @FlorianW : Yes, sorry about it. I'll correct it. $\endgroup$ – DisintegratingByParts Jul 11 '15 at 22:28
  • $\begingroup$ @TrialAndError: Do you mind adding these two lines to my earlier thread: Empty Spectrum $\endgroup$ – C-Star-W-Star Jul 11 '15 at 22:51
  • $\begingroup$ Hope you don't take it personally: It would be good to mention the argument here too that the spectral radius and norm agree for selfadjoint bounded operators. $\endgroup$ – C-Star-W-Star Jul 11 '15 at 23:02
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    $\begingroup$ The one starting with '(T^-1-lambda I)=...' and the one starting with '1/lambda T(...)^-1'. Thanks. :) $\endgroup$ – C-Star-W-Star Jul 11 '15 at 23:23

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