Problem. Let $H$ be a Hilbert space, and let $A:H\rightarrow H$ be a bounded, linear operator. Suppose $A$ has purely absolutely continuous spectrum and $\sigma_{ac}(A)=[0,1]$. Find the set of $D\subset\mathbb{C}$ such that $\left\|(A-zI)^{-1}\right\|_{L(H)}<1$ for all $z\in D$.
(N.B. As I a do not know how to make sense of absolutely continuous spectrum except when the operator $A$ is self-adjoint, I am assuming the author of the question forgot that detail.)
Solution: If $z\notin[0,1]$, then the function $f(x):=(x-z)^{-1}$ is continuous on $\sigma(A)$. From the continuous functional calculus $\phi:C(\sigma(A))=C([0,1])\rightarrow L(H)$, we see that
$$1=\phi(1)=\phi\left((x-z)(x-z)^{-1})\right)=\phi(x-z)\phi(f)=(A-zI)\phi(f)$$
As the order of the factors was irrelevant in the second equality, we obtain that $\phi(f)=(A-zI)^{-1}$. Since $\phi$ is an isometric homomorphism,
$$\left\|(A-zI)^{-1}\right\|_{L(H)}=\left\|f\right\|_{\infty}=\sup_{x\in[0,1]}\dfrac{1}{\left|x-z\right|}=\dfrac{1}{\text{dist}(z,\sigma(A))}$$
We conclude that the desired set $D$ is the set of all $z\in\mathbb{C}$ with distance $>1$ from the closed unit interval.
I don't see where this solution uses the hypothesis about $A$ only having purely absolutely continuous spectrum. As the continuous functional calculus exists for all self-adjoint operators on $H$, is there an alternative solution which does not use this machinery but instead exploits the spectrum hypothesis?