2
$\begingroup$

I had asked this question on Mathematics Stack Exchange, $2$ days ago but it got no response so I'm asking here.


If $A$ is a closed operator, then the peripheral spectrum of $A$ is defined to be all those points in the spectrum with modulus equal to the spectral radius.

However, I came across a paper which defines it as $$\sigma_\text{per}(A)=\sigma(A)\cap (s(A) + i \mathbb R)$$ where $s(A)$ denotes the spectral bound of $A,$ i.e., $s(A)=\sup\{\text{Re } \lambda:\lambda \in \sigma(A)\}.$

My question is: Are the two definitions equivalent? If not, then what is the reason to define it in this way?

$\endgroup$
  • $\begingroup$ That's not how I read the definition in the cited paper: $\sigma_{\rm per}(A)=\sigma(A)\cap(s(A)+i\mathbb{R})$, so all eigenvalues with real part equal to the spectral bound. $\endgroup$ – Carlo Beenakker Jan 31 '18 at 12:21
  • $\begingroup$ @CarloBeenakker Damn. Sorry for the type. I have corrected it. But even then, I don't see how the two definitions are equivalent. Why would the modulus of the points in peripheral spectrum be equal to the spectral radius? $\endgroup$ – Mark Jan 31 '18 at 12:32
2
$\begingroup$

These are indeed two different definitions, see the discussion in One-parameter Semigroups of Positive Operators. That reference also gives two different names for the two definitions, peripheral spectrum versus boundary spectrum, and shows they are not equivalent. The boundary spectrum contains all eigenvalues with maximal real part, while the peripheral spectrum contains the eigenvalues having maximal absolute value.

In a context where the eigenvalues $\lambda$ give the time dependence $e^{-\lambda t}$ of an amplitude, one is often interested in the decay rate, so one only needs the real part of the eigenvalue and then it makes sense to look at ${\rm Re}\,\lambda$ rather than $|\lambda|$. Here is another reference that uses the spectral bound definition of the peripheral spectrum.

$\endgroup$

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.