I have been reading Kato's perturbation theory book and on spectral theory in general, and I have a question regarding the separation of spectrum (c.f. Kato, Perturbation theory for linear operators, Theorem III.6.18). I'll explain the general setup below.

$X$ is a complex Banach space and $\mathcal{L}:X\to X$ is a bounded operator. For $z\in\sigma(\mathcal{L})$ which is isolated, he defines the spectral projection as $\pi_z=\frac{1}{2i\pi}\int_C (wI-\mathcal{L})^{-1}dw\in B(X)$, where $C$ is a small circle around $z$, and $B(X)$ is the space of bounded operators on $X$.

When $\mathcal{L}$ is finite rank, he mentions that $\pi_z$ is the projection onto the generalized eigenspace associated to $z$.

Does this naturally extend to operators which are not finite rank; for example if $z$ is an isolated eigenvalue such that $\pi_z$ has finite dimensional range, then can we say $\pi_z$ projects onto the generalised eigenspace of $z$?

  • 1
    $\begingroup$ Yes, exactly. You can find that in the book of Dunford and Schwarz ("Linear Operators, Part I"). The assumption on the finite-dimensional range of the projection is crucial here and cannot be dropped. $\endgroup$
    – amsmath
    Commented Aug 30, 2019 at 20:45
  • $\begingroup$ Thanks for the reference! :) $\endgroup$
    – Artur
    Commented Aug 30, 2019 at 20:53
  • $\begingroup$ You're welcome. $\endgroup$
    – amsmath
    Commented Aug 30, 2019 at 21:06

1 Answer 1


If $z$ is an isolated point of the spectrum, then $\pi_z$ is well-defined by the integral around $z$ that encloses no other points of the spectrum. And, $$ (\mathcal{L}-zI)\pi_z = \frac{1}{2\pi i}\oint_{C}(\mathcal{L}-zI)(wI-\mathcal{L})^{-1}dw \\ = \frac{1}{2\pi i}\oint_{C}(w-z)(wI-\mathcal{L})^{-1}dw $$ The same is true for all positive powers of $\mathcal{L}-zI$, i.e., $(\mathcal{L}-zI)^n$ is turned into $(w-z)^{n}$ inside the integral for $n=1,2,3,\cdots$: $$ (\mathcal{L}-zI)^n\pi_z =\frac{1}{2\pi i}\oint_{C}(w-z)^{n}(wI-\mathcal{L})^{-1}dw. $$ These give the negative coefficients in the Laurent series expansion of the resolvent around $z$. And $(\mathcal{L}-zI)$ implements a ladder from one coefficient to the next. $(\mathcal{L}-zI)^n\pi_z$ is eventually found to be $0$ by making assumptions about rank.


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