On kernel of an unbounded operator

Question

If spectral measure of $A$ is nonatomic, for which class of measurable (with respect to spectral measure) functions we have ker$f(A)=0.$

Answer 1

Regarding my comment: Take $\varphi: [0,1] \to [0,1]$ to be the Cantor-Lebesgue function and $\psi$ be its pseudoinverse. The spectral measure of the multiplication operator $A = M_\psi$ on $L^2([0,1])$ is suppported on the Cantor set. Therefore composing $A$ with a function that is zero on the Cantor set but nonzero everywhere else will give the zero operator. It is even possible to take $f$ continuous.

As a positive result, if $f$ is continuous and ever zero and $A$ bounded, since the spectrum of $A$ is compact, there is a minimal $\delta < |f(z)|$ and therefore $\sigma(f(A)) \subset \mathbb{C} \setminus B_{\delta}(0)$ and therefore it is invertible.

The subtle problem is what happens when $f$ is continuous and nonzero, $A$ unbounded with closed noncompact $\sigma(A)$ and over the spectrum we have that the infimum of $|f(z)|$ is zero. In that case you have to see that the spectral measure $E$ satisfy that the intersection of the spaces $$ H_n = \mathrm{range} \, E\Big\{ z : |f(z)| \leq \frac1{n} \Big\} $$ is $\{0\}$, which holds if $A$ is densely defined.