The norm and the spectrum of $B(L^p(X,\mu))$ Let $(X,\mu)$ be a $\sigma$-finite measure space.For $g\in L^{\infty}(X,\mu)$,let $M_g$ be the multiplication operator on $L^p(X,\mu)$ defined by $M_g f=gf$.

I want to know what is the norm and the spectrum in $B(L^p(X,\mu))$ of $M_g$
 A: The norm is equal to the essential supremum $\|g\|_\infty$. To see this use Holder's inequality and then show that the bound is sharp by choosing a sequence approximating an appropriate $\delta$. To answer the question about the spectrum, we need a definition. The essential range of $g$ is defined to be:
$$E_g = \{z \in \mathbb{C}: \mu(g^{-1}(B_\epsilon(z))) > 0 \quad  \forall \epsilon > 0\}$$
We will show that this set is equal to the spectrum $\sigma(M_g)$.


*

*Suppose first that $\lambda \notin E_g$ so there exists $\epsilon > 0$
with $\mu(g^{-1}(B_\epsilon(z))) =0$. Then the function $g'(x) =
   \frac{1}{g(x) - \lambda}$ is bounded by $1/\epsilon$ almost
everywhere and thus $g' \in L^\infty(X,\mu)$. It is simple to check
that $M_{g'}$ is a bounded operator whose inverse is $M_g - \lambda
   I$ so $\lambda \notin \sigma(M_g)$.

*Suppose now that $\lambda \in E_g$ and consider the sequence of sets
$$E_n = g^{-1}(B_{1/n}(\lambda))$$ and their respective
characteristic functions $\chi_n$. It follows that $$\|(M_g - \lambda I)f_n\|_p^p = \int_{E_n} \! | g
   - \lambda|^p \, d\mu$$ $$\le \frac{1}{n^p}\mu(E_n) = \frac{1}{n^p}\|{f_n}\|_p^p $$ By assumption $\mu(E_n) \neq
   0$ and therefore $f_n \neq 0$. Therefore this shows that $M_g - \lambda I$ cannot be
bounded from below, hence cannot be invertible, and so $\lambda \in
   \sigma(M_g)$.
Combining the bullet points shows that $\sigma(M_g) = E_g$.
