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Thread: Spectrum of linear operators isn't empty

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

    Since the Resolvent is bounded, and $\displaystyle \|\| T_T (z) \|\| ->0$ for$\displaystyle z->\infty$
    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 $\displaystyle f: \mathbb{C}->\mathbb{C}$.
    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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    Re: Spectrum of linear operators isn't empty

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