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 $D$?