i found this theorem in a book, but i can't understand the proof given there:
The thm. says, every linear Operator on a Banach-space: T:E->E has nonempty spectrum.
The proof is very short:
If T would have an empty spectrum, then the resolvent
Since the Resolvent is bounded, and for
by the thm. of Liouville, the resolvent is constant.
My Question is, why the Thm. of Liouville works in this case?
I know that this thm. is correct for functions .
But in this case our function isn't complex valued.
Can someone explain it for me?