Let $T$ be a bounded operator acting on a Banach space $X$. The point spectrum $\sigma_p(T)$ is of $T$ is defined to be $$\sigma_p(T):=\{\lambda\in\mathbb C~|~T-\lambda\text{ has nonempty kernel}\}$$ and the residual spectrum $\sigma_r(T)$ is defined to be $$\sigma_r(T):=\{\lambda\in\mathbb C~|~T-\lambda\text{ has empty kernel but Ran}(T-\lambda)\text{ is not even dense in }X\}.$$ For the continuous spectrum there seem to exist two non-equivalent definitions. For example, Dunford-Schwartz defines $$\sigma_c(T):=\{\lambda\in\mathbb C~|~T-\lambda\text{ has empty kernel but Ran}(T-\lambda)\text{ is a dense proper subset of } X\}$$ whereas for example Kato or Schmüdgen define \begin{align*}\sigma_c'(T)&:=\{\lambda\in\mathbb C~|~\text{Ran}(T-\lambda)\text{ is not a closed subset of } X\}\\ \sigma_r'(T)&:=\{\lambda\in\mathbb C~|~T-\lambda\text{ has a bounded inverse defined only on a proper subset of } X\}. \end{align*}
Does anyone have insights on the difference of these definitions? What are the advantages? The former one gives a partition of the spectrum, so that's kind of nice, but why is the second definition (more?) useful?
Edit: I just created a table that categorizes parts of the spectrum in terms of the questions:
- Is $K_\lambda:=$Ker$(T-\lambda)$ zero or nonzero?
- Is $R_\lambda:=$Ran$(T-\lambda)$ closed or not?
- Is $R_\lambda:=$Ran$(T-\lambda)$ dense or not?
According to this table the point spectrum can be subdivided into four subtypes and the residual spectrum can be subdivided into two subtypes. The alternative definitions would correspond to \begin{align*}\sigma_c'(T)&=\sigma_{c}(T)\cup\sigma_{r,1}(T)\cup\sigma_{p,2}(T)\cup\sigma_{p,3}(T)\\ \sigma_r'(T)&=\sigma_{r,1}(T) \end{align*}
The approximate point spectrum and compression spectrum on the other hand are given by \begin{align*} \sigma_{ap}(T)&=\sigma_{c}(T)\cup\sigma_{r,1}(T)\cup\sigma_{p,1}(T)\cup\sigma_{p,2}(T)\cup\sigma_{p,3}(T)\cup\sigma_{p,4}(T)\\ \sigma_{com}(T)&=\sigma_{r,1}(T)\cup\sigma_{r,2}(T)\cup\sigma_{p,3}(T)\cup\sigma_{p,4}(T). \end{align*}