Confusion in definition of peripheral spectrum

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?

• 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. – Carlo Beenakker Jan 31 '18 at 12:21
• @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? – Mark Jan 31 '18 at 12:32

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.