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Thread: spectrum

  1. #1
    Member Mauritzvdworm's Avatar
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    spectrum

    Let $\displaystyle \mathcal{A}$ be the space of all bounded analytic functions in the open unit disk in the complex plane.

    My question is why is die spectrum the closure of the range of $\displaystyle f\in\mathcal{A}$ and not just the normal range?

    $\displaystyle \sigma(f):=\{\lambda\in \mathbb{C}: (\lambda 1 -f) \notin Inv(\mathcal{A}) \}$

    or should I look to the resolvant to find my answer?
    Last edited by Mauritzvdworm; Feb 25th 2010 at 01:18 AM. Reason: typing error
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by Mauritzvdworm View Post
    Let $\displaystyle \mathcal{A}$ be the space of all bounded analytic functions in the open unit disk in the complex plane.

    My question is why is die spectrum the closure of the range of $\displaystyle f\in\mathcal{A}$ and not just the normal range?

    $\displaystyle \sigma(f):=\{\lambda\in \mathbb{C}: (\lambda 1 -f) \notin Inv(\mathcal{A}) \}$

    or should I look to the resolvant to find my answer?
    One way to answer this is to say that the spectrum is a closed (in fact, compact) set. The range of f on the open disk will not be closed.

    Another way to look at it is to notice that if $\displaystyle \lambda$ is in the closure of the range of f, but not in the range itself, then $\displaystyle \lambda 1 -f$ will have an inverse function defined on the disk, but the inverse will not be bounded. So $\displaystyle \lambda 1 -f$ will not have an inverse in $\displaystyle \mathcal{A}$.
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