I came across a result in a paper(Proposition 2.2 on the second page in https://www.ams.org/journals/tran/1988-306-02/S0002-9947-1988-0933321-3/S0002-9947-1988-0933321-3.pdf) which states:

If $X$ is reflexive and $\mathcal{T}$ is bounded, then $R\sigma(A) \cap i \mathbb{R} = P\sigma(A) \cap i \mathbb{R}$.

Here $\mathcal{T}$ is a semigroup in $X$, generated by $A$.

There is a proof but I am still confused. How is it even possible that an $\lambda \in i\mathbb{R}$ can belong both to point spectrum and to residual spectrum? Wouldn't that mean that for an $r \in \mathbb{R}$ the operator $r i - A$ is both injective and not injective at the same time? Since the point spectrum are those complex numbers $\lambda$ where $\lambda - A$ is not injective and the residual spectrum are those where $\lambda - A$ is injective but doesnt have dense range. It is definitely not meant as an equality of empty sets since there are bounded semigroups which have eigenvalues on the imaginary axis.

Can anyone point out the flaw in my thinking?


As we can read in the paragraph after Proposition 2.1, $R\sigma(A)$ is by definition the set of all $\lambda \in\mathbb C$ such that $\operatorname{range}(\lambda -A)$ is not dense in $X$.

Therefore, injectivity is not part of the definition.

The same definition is used, for example, in Engel's book.

  • $\begingroup$ Thank you so much, I don't know how I could overread this information so many times! $\endgroup$
    – lsir
    Mar 13 '19 at 5:05
  • $\begingroup$ @lessir Believe it or not, I overread the same sentence in the same paper (many times) some time ago. $\endgroup$
    – Pedro
    Mar 13 '19 at 13:32

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