hi here $\sigma$ is spectrum,if $(X,\mu)$ is topological measure owed with borel measure and $$ r_{ess}(f) = \{w \in \mathbb{R}_+ : \mu(f^{-1}(B(w, \epsilon))) > 0\}. $$ i have proved that if f is continious and X is compact then $r_{ess}(f)=range(f)$
so if we assume $H=L^{2}([0,1])$ and $f=C^{0}([0,1])$, X is compact how can we prove that $\sigma(M_{f})=range(f)$ where $M_{f}$ is multiplication operator ,i showed that if $(\lambda-f)\phi$ is not invertible where $\phi \in L^{2}([0,1])$ then $\lambda$ must be in range(f) but i dont see how to do the converse i think the idea is to use that in this case i i use that $r_{ess}(f)=range(f)$ and last question what i dont assume that neither $f=C^{0}([0,1])$ nor X is compact $\sigma(f) $ would not be the \range(f) but why? thanks