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Math Help - Spectrum of linear operators isn't empty

  1. #1
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    Oct 2010
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    Spectrum of linear operators isn't empty

    Hello,

    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
    R_T : \mathbb{C}->L(E) , R_T (z)= (T-z*I)^{-1} is entire.

    Since the Resolvent is bounded, and \|\| T_T (z) \|\| ->0 for  z->\infty
    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 f: \mathbb{C}->\mathbb{C}.
    But in this case our function isn't complex valued.

    Can someone explain it for me?

    Regards
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  2. #2
    Super Member
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    Apr 2009
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    Re: Spectrum of linear operators isn't empty

    It should be clear that if f:\mathbb{C} \rightarrow X is holomorphic then for any l\in X^* we have lf:\mathbb{C} \rightarrow \mathbb{C} is holomorphic, so if f is bounded, and since l is bounded iff it sends bounded sets to bounded sets, then lf is bounded so Liouville's theorem applies and gives lf constant for all l. Assume f is not constant ie. there exist x\neq y such that f(x)\neq f(y) then there is a functional that assigns f(x) to a non-zero complex number (Hahn-Banach), so if f(x),f(y) are linearly dependent we're finished, if not apply Hahn-Banach again assaigning f(y) any other non-zero complex number. Either way we contradict the fact that lf is constant.
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