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Math Help - spectrum

  1. #1
    Member Mauritzvdworm's Avatar
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    spectrum

    Let \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 f\in\mathcal{A} and not just the normal range?

    \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; February 25th 2010 at 02:18 AM. Reason: typing error
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Mauritzvdworm View Post
    Let \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 f\in\mathcal{A} and not just the normal range?

    \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 \lambda is in the closure of the range of f, but not in the range itself, then \lambda 1 -f will have an inverse function defined on the disk, but the inverse will not be bounded. So \lambda 1 -f will not have an inverse in \mathcal{A}.
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