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what I have done: I can prove that ${\sigma}(T)$ is in the reals by first showing that, for normal $T$, if ther exists positive scalar $t$ with $||Tx||>t||x||$ then $T$ is invertible. Then its quite easy to show that if a we have a non-real number ${\lambda}$ then it is in the resolvent, but I would need it to be in the resolvent at $A$.

I've figured that if I can show that if the boundary of a spectrum is real, then the spectrum is real, then I can prove what I want, but I've no idea if this result is true? Thanks