Details on decomposition of spectrum of self-adjoint operators?

Question

Q: What is the relation between the decomposition of spectrum of operators on Banach spaces and the decomposition of spectrum of self-adjoint operators on Hilbert spaces?

Operators on Banach space

In my introduction course to Functional analysis, we defined the decomposition of spectrum for an endomorphism $T$ on a Banach space $X$:

• $\sigma_\mathrm{P}$ – point spectrum; all $\lambda$, where $\ker(T-\lambda) \neq \emptyset$
• $\sigma_\mathrm{comp.}$ – compression spectrum; all $\lambda$, where $\overline{\operatorname{range}(T - \lambda)} \neq X$
• $\sigma_\mathrm{R}$ – residual spectrum; $\sigma_\mathrm{R} = \sigma_\mathrm{comp.} \setminus \sigma_\mathrm{P}$
• $\sigma_\mathrm{approx.}$ – approximate point spectrum; all $\lambda$ such that $\exists\{x_n\}$ $\lVert Tx_n - \lambda x_n \rVert \to 0$
• $\sigma_\mathrm{C}$ – continuous spectrum; $\sigma_\mathrm{C} = \sigma_\mathrm{approx.} \setminus (\sigma_\mathrm{P} \cup \sigma_\mathrm{R})$

Alternatively, $\sigma_\mathrm{C}$ can be characterized by the requirement that:

• $\ker(T-\lambda) = \emptyset$
• $\operatorname{range}(T - \lambda) \neq X$
• $\overline{\operatorname{range}(T - \lambda)} = X$

This decomposition has the nice property, that $\sigma = \sigma_\mathrm{P} \sqcup \sigma_\mathrm{C} \sqcup \sigma_\mathrm{R}$, where “$\sqcup$” means a disjoint union.

Self-adjoint operators on Hilbert space

Reed and Simon define a decomposition of spectrum of a self-adjoint operator $A$ (I believe this generalizes to normal operators too) on a Hilbert space $\mathcal{H}$. They use the Lebegue's decomposition of the operator's spectral measure $\mu$ to define three subspaces of $\mathcal{H}$:

• $\mathcal{H}_\mathrm{p.p.} = \{ \psi \, | \, \mu_\psi \text{ is pure point} \}$
• $\mathcal{H}_\mathrm{a.c.} = \{ \psi \, | \, \mu_\psi \text{ is absolutely continuous} \}$
• $\mathcal{H}_\mathrm{s.c.} = \{ \psi \, | \, \mu_\psi \text{ is singular continuous} \}$

Then they go on to define:

• $\sigma_\mathrm{p.p.} = \{ \text{eigenvalues of }A \}$, I assume that $\overline{\sigma_{p.p.}} = \sigma(A | \mathcal{H}_\mathrm{p.p.})$
• $\sigma_\mathrm{a.c.} = \sigma(A | \mathcal{H}_\mathrm{a.c.})$
• $\sigma_\mathrm{s.c.} = \sigma(A | \mathcal{H}_\mathrm{s.c.})$

In this decomposition, the following equality holds: $\sigma = \overline{\sigma_\mathrm{p.p.}} \cup \sigma_\mathrm{a.c.} \cup \sigma_\mathrm{s.c.}$ (here, the union is not disjoint).

The Question

What is the relation between these two decompositions? The most straightforward observation is that $\sigma_\mathrm{P} = \sigma_\mathrm{p.p.}$, since both of them only include eigenvalues, right? Also, I think I've heard that the residual spectrum of normal operators is empty, but I couldn't find a reliable source for it or prove it myself. If this is true, that would mean that $\sigma_\mathrm{C} = (\overline{\sigma_\mathrm{P}} \cup \sigma_\mathrm{a.c.} \cup \sigma_\mathrm{s.c.}) \setminus \sigma_\mathrm{P}$. Is this it? Or are there some interesting things that can be said about the sets $(\overline{\sigma_\mathrm{P}}\!\setminus\!\sigma_\mathrm{P})$, $(\sigma_\mathrm{a.c.}\!\!\setminus\!\sigma_\mathrm{P})$, $(\sigma_\mathrm{s.c.}\!\!\setminus\!\sigma_\mathrm{P})$?