I am reading the twelfth chapter of Rudin's Functional Analysis and am unable to understand two comments he makes in passing in the section on Resolution of Identity. Let E be a resolution of identity on a measurable space ($\Omega, M$), $H$ be a Hilbert space and $f$ be a complex measurable function on $\Omega$. Then the essential range of $f$ is defined to be the smallest subset of $\mathbb{C}$ that contains $f(p) \, \forall p\in \Omega$ except for those in a subset $\omega$ of $\Omega$ with $E(\omega)=0.$ Here $0$ is the zero operator on $H$. If the essential range of $f$ is bounded, define $\|f\|_{\infty}=sup\{|\lambda|:\lambda \in \text{essential range of }f\}$.
Let $B$ be the set of bounded measurable functions on $\Omega$ and N be the subset of $B$ with $\|f\|_{\infty}=0$. Then $B/N$ is a Banach algebra. The following questions are unclear to me:
- Show that $\|[f]\|=\|f\|_{\infty}$ for $[f]\in B/N$.
- Show that spec$([f])=$essential range of $f$.
I could show that $\|f\|_{\infty}$ is a lower bound for $\|f-g\|, \forall g \in N$. To show that $\|f\|_{\infty}$ is the infimum of the set of $\|f-g\| \, \forall g \in N$, given $\epsilon>0$, we must find $g\in N$ such that $\|f-g\|< \|f\|_{\infty}+\epsilon$. For the given $f$, there exists a subset $\omega$ of $\Omega$ with $E(\omega)=0$. Now let $g(p)=f(p)$ if $p\in \omega$ and $g(p)=0$ otherwise. What can we say about essential range of $g$? My guess is that it is contained in the essential range of $f$. In that case then $g\in N$ and we are done.
For the second question I have been able to show that essential range of $f$ is contained in spec$[f]$ and am still working on the reverse containment.
My confusion is about what exactly essential range is when we do not have a measure but a resolution of identity on the measurable space . I look forward to your help. Thanks.