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 $D$?

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    $\begingroup$ 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. $\endgroup$ – András Bátkai Sep 18 '18 at 18:08
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    $\begingroup$ 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.) $\endgroup$ – Denis Chaperon de Lauzières Sep 18 '18 at 18:30
  • $\begingroup$ @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. $\endgroup$ – András Bátkai Sep 19 '18 at 5:28

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.

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  • $\begingroup$ Thanks for the answer and the reference. I must admit I am surprised. I expected that to generalise to the unbounded case one would need to "pay a price" . . . however, it seems that this is not the case. $\endgroup$ – Dave Shulman Sep 18 '18 at 20:40

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