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Thread: Spectrum of Linear Operator

  1. #1
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    Question Spectrum of Linear Operator

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

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

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

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

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

    Your help is invaluable - I have no idea how to do this.
    Last edited by Nusc; Mar 24th 2009 at 03:06 PM.
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  2. #2
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    The point spectrum $\displaystyle \sigma_p(M_\phi)$ is the set of eigenvalues of $\displaystyle M_\phi$. If $\displaystyle \lambda$ is an eigenvalue, with eigenfunction f, then $\displaystyle M_\phi(f) = \lambda f$. So $\displaystyle \phi(t)f(t) = \lambda f(t)$ for almost all t in X. Thus $\displaystyle (\phi(t)-\lambda)f(t) = 0$ almost everywhere. But f is an eigenfunction, so it cannot be 0 almost everywhere. Therefore there is a set of positive measure on which $\displaystyle \phi(t)=\lambda$. Hence $\displaystyle \sigma_p(M_\phi) = \{\lambda\in\mathbb{C}: \mu(\{t\in X:\phi(t)=\lambda\})>0$.

    The approximate point spectrum $\displaystyle \sigma_{ap}(M_\phi)$ 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 $\displaystyle \lambda$ such that $\displaystyle \mu(\{t\in X:|\phi(t)-\lambda|<\varepsilon\})>0$ for every $\displaystyle \varepsilon>0$. See if you can prove that!
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