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Suppose we have a Hilbert space $\mathscr{H}$ and a bounded linear map $T\in\mathscr{B(H)}$ NOT necessarily self-adjoint. There seems to be loads of definitions of the essential spectrum of $T$. My question is, whether in the Hilbert space setting the following are equivalent:

  1. $\lambda$ is such that $T-\lambda{I}$ is not Fredholm

  2. $\lambda$ is in $\sigma(T)\backslash\sigma_{d}(T)$ where $\sigma_{d}(T)$ denotes the set of isolated points of the spectrum (discrete spectrum).

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  • $\begingroup$ The essential spectrum is typically defined within the Fredholm theory, that is your (1) above. Cf. "Linear operators and their spectra" by Brian Davies, page 118 (freely available online, by the way). See also "Spectral theory of operators on Hilbert spaces" by C. Kubrusly, page 145. For non-normal operators definitions (1) and (2) are no longer equivalent, but I don't know examples.. $\endgroup$
    – mforets
    Feb 14, 2016 at 18:52
  • $\begingroup$ You need to define $\sigma_d$ more carefully otherwise 1 and 2 are not equivalent for self-adjoint operators either! Set $T=I$ for instance. $\endgroup$
    – Simon
    May 31, 2016 at 2:13
  • $\begingroup$ As mentioned by @mforets, Davies' book used to be online at nms.kcl.ac.uk/brian.davies/web_page/LOTS.html. This seems to be offline now though. I found a copy on the Internet Archive. Here you can also find the web supplement which has 100 pages of additional material. $\endgroup$ Jun 26, 2022 at 3:27

1 Answer 1

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  1. Semi Fredholm class is $\mathcal{SF}$, which is the union of left semi Fredholm class and right semi Fredholm class. Fredholm class is $\mathcal{F}$, which is the intersection between left semi Fredholm class and right semi Fredholm class.
  2. Fredholm index: $\mbox{ind}(T)$ of $T\in\mathcal{SF}$ (Semi Fredholm) is defined by $\mbox{ind}(T)=\dim \ker(T)-\dim \ker(T^*).$
  3. $\sigma_0(T)=\{\lambda \in \sigma(T): \lambda I-T\in \mathcal{F} \mbox{ and ind}(\lambda I-T)=0\}$.
  4. Let $T\in \mathcal{B}(\mathcal{H})$ and $\sigma_w(T)$ be the Weyl spectrum of $T$. Then $\sigma(T)=\sigma_w(T)\cup \sigma_0(T)$.
  5. Let $T$ to be self adjoint operator. Then $\sigma_d(T)=\sigma_0(T)$ and $\sigma_w(T)=\sigma_e(T)$. Therefore $\sigma(T)=\sigma_e(T)\cup \sigma_d(T)$. Here $\sigma_e(T)$ is the essential spectrum of $T$.
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