# Unbounded version of continuous functional calculus

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 $D$?

• If your $T$ has nonempty resolvent set, then you can use the transformation $z\mapsto \frac{1}{\lambda-z}$ to get a bounded normal operator and a continuous function on its spectrum. – András Bátkai Sep 18 '18 at 18:08
• The general normal case, and even general measurable function, is discussed thoroughly in the book of Schmüdgen ("Unbounded self-adjoint operators on Hilbert space", Springer GTM 265, see Section 5.2 and following.) – Denis Chaperon de Lauzières Sep 18 '18 at 18:30
• @DenisChaperondeLauzières: The work is to get a continuous functional calculus. If you have that (e.g. using Gelfand and Riesz), then you automatically get the measurable one. – András Bátkai Sep 19 '18 at 5:28