For a normal operator $T$ on a Hilbert space ${\cal H}$, it is well known that for any continuous complex valued function $f$ on the spectrum of $T$, we have a well-defined operator $f(T) \in B({\cal H})$. 


I have heard that this extends to the unbounded case, but I can't find a precise statement. So if $D$ is a densely-defined unbounded normal operator on ${\cal H}$, does the unbounded version of functional calculus work for any continuous function on the spectrum of $D$?



Unbounded version of continuous functional calculus
You find the precise abstract statment in the Internet Seminar lecture notes of Markus Haase, especially in Chapter 4.
See
<https://www.math.uni-kiel.de/isem21/en/course/phase1>


By the way, the answer is yes.


