A Understanding the Difference: Spectra of Unbounded vs. Bounded Operators

[SemM]

Hi, why do unbounded operators and bounded operators differ so much in terms of defining their spectra?

1. The unbounded operator requires a self-adjoint extension to define its spectrum.
2. A bounded one does not require a self-adjoint extension to define the spectral properties.
3. Still the unbounded operator is bounded within its domain and should be defined accordingly.Why can't the spectral properties be defined in the unbounded operators domain by the same method as for bounded operators?

 

[mathwonk]

I am a novice, but it looks as if this answer may help explain the special things that can occur for non self adjoint operators in the unbounded case, i.e. there can be a "residual" spectrum.

https://math.stackexchange.com/questions/662687/spectrum-of-unbounded-operators-spectral-theoremA nice discussion of some general properties of bounded and unbounded operators is in chapter 8 of the book Functional Analysis by Riesz-Nagy. They mention there the necessity of assuming self adjointness in the unbounded case to recover a spectral decomposition, analogous to the bounded case, but refer for the reason to a paper by Von Neumann, written in German, in the 1920's in Math Annalen.

I recommend the expert answer linked above, but to try to summarize, the spectrum of T, as you know, consists of scalars c such that (cI-T) is not invertible. But there are several reasons the inverse is not an admissible operator. Simplest of all, the operator cI-T may not be injective. Or it may be injective but its inverse may not not be bounded. Now unbounded operators are not defined everywhere, so apparently one wants them to be defined at least on a dense set. Thus another problem occurs if cI-T is injective but its range is not dense, so that the inverse is both unbounded and defined only on a non dense set. This is the case that it seems must be ruled out by assuming self adjointness.