# Spectrum of Linear Operator

• Mar 23rd 2009, 10:25 AM
Nusc
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.
• Mar 24th 2009, 02:05 PM
Opalg
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!