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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 $M_g$

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1 Answers1

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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 $\sigma(M_g) = E_g$.