# Self-adjoint operator has non-empty spectrum.

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.

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.
• @TrialAndError : I just noticed: Why is $\frac{1}{\lambda} T \left( \frac{1}{\lambda} I - T \right)^{-1}$ bounded? It is clear that $\left( \frac{1}{\lambda} I - T \right)^{-1}$ is, but $T$ could be unbounded. – KennyH Aug 6 '15 at 10:10
• @FlorianW : It's defined everywhere and it's closed. That's one way to see it, or write $-T(\frac{1}{\lambda} I-T)^{-1}=\{ (\frac{1}{\lambda}I-T)-\frac{1}{\lambda}I\}(\frac{1}{\lambda}I-T)^{-1}=I-\frac{1}{\lambda}(\frac{1}{\lambda}I-T)^{-1}$. – Disintegrating By Parts Aug 6 '15 at 11:47
• I agree that it is closed. But why is it defined everywhere? The domain of $T$ doesn't necessarily have to be the whole hilbert space! I guess the little calculation is the way to go. – KennyH Aug 6 '15 at 13:22
• @FlorianW : The range of $(\frac{1}{\lambda}I-T)^{-1}$ is the domain of $T$. The domain of $\beta I -T$ is the same as the domain of $T$ and both are closed on the same domain, even selfadjoint if $\beta$ is real. This is something to get used to seeing for unbounded operators. And, yes, the calculation is a good way to go as well. – Disintegrating By Parts Aug 6 '15 at 13:32