# Math Help - Spectrum of Linear Operator

1. ## 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.

2. 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!