The theorem below is the Borel functional calculus for bounded normal operators defined on complex Hilbert space. 
I'm now studying the case for unbounded normal operators which are densely defined on a complex Hilbert space. It seems that the main definition can be generalised. However, for the properties of that definition, few issues related to the domain of the operator appears if we want generalise the same property. For, example as for the property (iv) above, the commutative property $ST =TS$ should include the assertion that the domain of $ST$ equals that of $TS.$
My question is: If we assume that $T,S$ commute on the common domain and that $f(T) (D(S)) \subset D(S),$ can we prove that $$Sf(T)x = f(T)Sx, \forall x\in D(S)$$
Thank you in advanced