2
$\begingroup$

I have been introduced to the following definition of the essential numerical range of a bounded, linear operator on a separable, infinite-dimensional Hilbert space: $$W_e(T) := \bigcap_{K\in \mathcal K(H)} \overline{W(T+K)}$$ i.e., the essential numerical range $W_e(T)$ of $T$ is the intersection of the closures of numerical ranges of all compact perturbations of $T$.

  • Why is the essential numerical range defined in this way, i.e., what motivates this definition and how is it useful? Even in my wildest dreams, I would have no reason to consider compact perturbations of $T$ in order to define $W_e(T)$.

I have noticed that definitions in operator theory on infinite-dimensional Hilbert spaces are often motivated by finite-dimensional analogs, but I'm not able to draw any connections here. Thanks for any insights!

Note: The numerical range of a (bounded, linear) operator $T\in \mathcal B(\mathcal H)$ is defined as $$W(T) := \{\langle Tx,x\rangle: x\in \mathcal H, \|x\| = 1 \}$$

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ With the caveat that Google searchers are personalised: typing "essential numerical range" into Google shows me the following passage. "The essential numerical range $W_e(T)$ was introduced by Stampfli and Williams for a bounded linear operator in a Hilbert space as the closure of the numerical range of the image of in the Calkin algebra." $\endgroup$
    – Yemon Choi
    Commented Sep 14, 2022 at 12:34
  • $\begingroup$ You might also want to add a link to your MSE version of the same question: math.stackexchange.com/questions/4527855/… $\endgroup$
    – Yemon Choi
    Commented Sep 14, 2022 at 12:35
  • 1
    $\begingroup$ My (wild) guess is that the motivation has to do with what happens with the spectrum: the essential spectrum of a bounded op is invariant under self-adjoint compact perturbations. To give an example: if you let $T$ be the compact operator that sends $x_T$ to itself and everything orthogonal to $x_T$ to 0, its numerical range is $[0,1]$, but its essential range is just $\{0\}$. The connection to "finite dimensional" should be read as "the part of the numerical range that cannot be explained by essentially finite dimensional effects". $\endgroup$
    – Willie Wong
    Commented Sep 14, 2022 at 13:02

0

Reset to default

You must log in to answer this question.