Let $(X,\mathcal A, \mu)$ be a measure space and let $f \in L^{\infty} (\mu).$ Let $M_f$ denote the multiplication operator on $L^2(\mu).$ Then show that $\lambda \notin \sigma (M_f) \implies \lambda \notin \text {ess ran} f.$
If $(M_f - \lambda)$ is invertible then it is bounded below i.e. there exists $\alpha \gt 0$ such that $\|(M_f - \lambda)h\|_2 \geq \alpha \|h\|_2,$ for all $h \in L^2(\mu).$ Does it anyway imply that $\lambda \notin \text {ess ran} f\ $? Could anybody give me some hint to proceed?
Thanks for your time.
Hint: (This assumes some functional analysis)
Let $B:=\mathscr{B}(L^2(\mu))$ be the set of bounded linear operators on $L^2(\mu)$. This is a unital $C^*$-algebra, and $A:=L^\infty(\mu)$ is a unital $C^*$-subalgebra of $B$ (more precisely, the map $f\mapsto M_f$ is an isometric embedding and $M_1=1_B$). It follows (see 11.29 in Rudin's Functional Analysis, page 296 of the second edition)) that $$ \sigma_A(f)=\sigma_B(M_f) $$ (in fact, $x$ is invertible in $A$ iff it is invertible in $B$).
All this tells you that you can just work with invertibility in $L^\infty(f)$, which is simpler to deal with.