The common terminology we get used to on spectrum of linear operators is
- point spectrum $ \sigma_p $,
- continuous spectrum $ \sigma_c $,
- residual spectrum $ \sigma_r $ .
However, we recently notice an terminology of another type:
- purely point spectrum $ \sigma_{pp} $ ,
- absolutely continuous spectrum $ \sigma_{ac} $ ,
- singularly continuous spectrum $ \sigma_{sc} $ .
The latter type is defined with respect to decomposition measure and orthogonal decomposition of spectral projections. For details, see for instance Chapter 3.3 in
"G. Teschl: Mathematical Methods in Quantum Mechanics - With Applications to Schrödinger Operators (2014)".
Question: What is the relation between above two terminologies, can we give an exact correspondence for them?
Edit: The relation I have found so far is $ \sigma_{pp} = \overline{\sigma_{p}} $.
One possibly useful result: Chapter IV, Proposition 1.16 in
[Engel, Nagel: One-Parameter Semigroups for Linear Evolution Equations (2000)]
However, I can not see clearly how to use this result to explain the connections between the two terminologies.
I managed to read up the definitions in the book you described (link). The two kinds of spectra really encode different things, and I do not believe they should be compared.
The first remark I would make is that the spectra of the "second kind" only make sense for self-adjoint (or normal) operators, that is operators where you have a spectral measure, whereas the spectra of the "first kind" can always be defined. Note however that if $A$ is self-adjoint that then $\sigma_r(A)=\emptyset$ always. That is for spectra of the "first kind" we only have access to $\sigma_p$ and $\sigma_c$ when we want to compare with $\sigma_{pp},\sigma_{ac},\sigma_{sc}$.
The second remark is that by definition $\sigma_p, \sigma_{c},\sigma_r$ are always disjoint, while $\sigma_{pp},\sigma_{ac},\sigma_{sc}$ may all overlap, indeed situations are possible where $\sigma_{pp}(A)=\sigma_{ac}(A)=\sigma_{sc}(A)$ (although not in separable Hilbert spaces, excepting the $0$ Hilbert space).
With these out of the way there are some things that can be said:
If you are interested in these comments then I can describe how you can see them and the examples realising them in more detail. Lets describe a very singular example first: