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Math Help - Spectrum of Linear Operator

  1. #1
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    Question Spectrum of Linear Operator

    Let
     1 \leq p \leq \infty and let  (X,\Omega, \mu)
    be a \sigma-finite measure space.

    For \phi \in L^\infty(\mu) , define  M_\phi on  L^p(\mu) by  M_\phi f = \phi f \forall f \in L^(X,\Omega,\mu).
    I need to find the following:

     \sigma(M_\phi) , \sigma_ap(M_\phi), and \sigma_p(M_\phi)

    where
     \sigma(M_\alpha) = \{\alpha \in F: M_\alpha-\alpha is not invertible \}
    \sigma_{ap} \equiv \{ \lambda \in C : there is a sequence \{x_n\} in X such that ||x_n|| = 1 for all n and  ||(A-\lambda)x_n||\rightarrow 0 \} and

    and \sigma_p \equiv \{ \lambda in C: ker(A-\lambda) \neq (0)\}

    Your help is invaluable - I have no idea how to do this.
    Last edited by Nusc; March 24th 2009 at 04:06 PM.
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  2. #2
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    The point spectrum \sigma_p(M_\phi) is the set of eigenvalues of M_\phi. If \lambda is an eigenvalue, with eigenfunction f, then M_\phi(f) = \lambda f. So \phi(t)f(t) = \lambda f(t) for almost all t in X. Thus (\phi(t)-\lambda)f(t) = 0 almost everywhere. But f is an eigenfunction, so it cannot be 0 almost everywhere. Therefore there is a set of positive measure on which \phi(t)=\lambda. Hence \sigma_p(M_\phi) = \{\lambda\in\mathbb{C}: \mu(\{t\in X:\phi(t)=\lambda\})>0.

    The approximate point spectrum \sigma_{ap}(M_\phi) is probably harder to describe. I don't offhand know what it is, but I would guess that it is the set of complex numbers \lambda such that \mu(\{t\in X:|\phi(t)-\lambda|<\varepsilon\})>0 for every \varepsilon>0. See if you can prove that!
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