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?

Table of spectral Decomposition

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*}


The common classification of spectrum is the first one you gave. I'm not sure where $\sigma_{c}'(T)$ would have come from, or why it was defined that way, but it doesn't strike me as a standard or common definition, even though I can see a reason for the alternative definition.

If $T$ is closed, then this alternative definition would probably be better described as "approximate point spectrum," which consists of $\lambda\in\mathbb{C}$ for which there is a sequence of unit vectors $\{ v_{n}\}$, none of which is in $\mathcal{N}(T-\lambda I)$, for which $\lim_{n}\|Tv_{n}-\lambda v_{n}\|=0$. This alternative definition may help explain the authors' motives.

The first classification of spectrum is somewhat lacking when it comes to approximate spectrum because the presence of an eigenvector with eigenvalue $\lambda$ automatically puts $\lambda$ in the point spectrum, even though there may be more to the story for $\lambda$ because it is still possible for there to be approximate eigenvectors with approximate eigenvalue $\lambda$. For example, let $Tf=xf$ on $L^{2}_{\mu}[0,1]$. If $\mu$ has an atom at $\lambda \in [0,1]$, but $\mu([\lambda-\epsilon,\lambda+\epsilon]\setminus\{\lambda\})=0$ for some $\epsilon > 0$, then there is no approximate eigenvector sequence for $T$ with approximate eigenvalue $\lambda$; but if this is not the case, then there is an approximate point spectrum component. Using the first classification, both cases of $T$ are classified with $\lambda$ in their point spectrum ... end of story. But there really is more to the story because $(T-\lambda I)$ may or not may not have a closed range.

  • $\begingroup$ See my edit, the approximate point spectrum is an even bigger part of the spectrum than $\sigma_c'(T)$. $\endgroup$ – whz Sep 16 '14 at 16:53
  • $\begingroup$ @Dominik : Yes, as you have now noted in your new edit, point spectrum is a little too broad if you want to study the fine details of the operator. $\endgroup$ – DisintegratingByParts Sep 16 '14 at 17:11

I think the property of having closed range is simply more useful in certain applications, so you want to single out the 'worst case scenario' by considering those $\lambda$ for which $(T-\lambda)$ fails to have closed range. Note for instance that in the Hilbert space setting, you have the very useful direct sum decomposition $$ \mathcal H= \mathcal R(T-\lambda)\oplus \mathcal N (T^*-\bar \lambda) $$ if and only if $\lambda\notin \sigma '_c(T)$.

On the other hand, the set $\sigma_c(T)$ enjoys the nice property that $(T-\lambda)$ has a densely defined inverse when $\lambda\in \sigma_c$. In a sense, $\sigma_c'(T)$ singles out a poorly-behaved subset of the spectrum, whereas $\sigma_c(T)$ singles out a well-behaved subset of $\sigma_c'(T)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.