Proof of $ran(E(\lambda)) = ker(\lambda-A)^\alpha$

Question

Let $A: H \to H$ be a compact operator on the complex Hilbert space $H$. Let $\lambda \neq 0$ be an Eigenvalue of $A$. Since A is compact, it's Eigenvalues can only accumulate at $0$, so we can find a smooth curve $\Gamma \subset \mathbb{C}$ enclosing no other Eigenvalue than $\lambda$. Then the Risz-Projector $E(\lambda)$ (see also https://en.wikipedia.org/wiki/Riesz_projector) is defined as $$ E(\lambda):= \frac{1}{2\pi i}\int_\Gamma (z-A)^{-1} dz $$ Question: How to prove $ran(E(\lambda)) = ker(\lambda-A)^\alpha$, where $\alpha$ is the ascent multiplicity of $\lambda - A$, i. e. the smallest integer such that $ker(\lambda - A)^\alpha = ker(\lambda - A)^{\alpha + 1}$. $ker(\lambda-A)^\alpha$ is sometimes called the generalized Eigenspace.

Background: This property of Riesz-Projectors is key to the Babuska-Osborn-Theory for approximating non-symmetric Eigenvalue problems. Babuska and Osborn state this property in their paper 'Eigenvalue Problems, 1991' but give no proof (they seem to refer to Dunford and Schwartz but I can't find a proof there either). I found a proof for the special case when $A$ is self-adjoint (in this case $\alpha = 1$) in this book https://link.springer.com/book/10.1007/978-1-4612-0741-2.

In the proof of "$\supset$" (self-adj. case) they show that for $f \in ker(\lambda - A)$ we have $E(\lambda)f = f$. I could generalize this to the non self-adjoint case by looking at Jordan-chains and using that $(\lambda - A)$ and $E(\lambda)$ commute.

However, I don't see how to generalize "$\subset$". Here is how the proof in the self-adjoint case works: They show $$ (\lambda - A) E(\lambda) = \frac{1}{2\pi i}\int_\Gamma (\lambda - A)(z-A)^{-1} dz = \frac{1}{2\pi i}\int_\Gamma (\lambda - z)(z-A)^{-1} dz = 0\\ $$ For the last equality the argument is as follows: We have $||(\lambda - A)^{-1}|| \leq d(\lambda, \sigma(A))^{-1}$ (which is true for self-adjoint operators). Therefore $ |(\lambda - z)| ||(z-A)^{-1}||$ is uniformly bounded on $interior(\Gamma) \backslash \{ \lambda \} $. The equality then follows from Riemann's theorem on removable singularities and Cauchy's theorem.

Any references or ideas are greatly appreciated. Thanks!

Answer 1

Let $\varepsilon >0$ and let $\Gamma_\varepsilon$ be the circle of radius $\varepsilon$ around $\lambda$ that is traversed once in the positive sense.

Then we have for every $n \in \mathbb{N}$: $$(\lambda - A)^n E(\lambda) = \frac{1}{2\pi i} \int_{\Gamma_\varepsilon} (\lambda- z)^n (z-A)^{-1} dz $$

Consider $(\lambda -A)|_{\operatorname{ran} E (\lambda) } : \operatorname{ran} E (\lambda) \to \operatorname{ran} E (\lambda) $. For simplicity let $T= (\lambda -A)|_{\operatorname{ran} E (\lambda) }$.

Using the above we obtain $$ \| T^n \| \leq \sup_{z \in \Gamma_\varepsilon} \|(z-A)^{-1}\| \varepsilon^{n+1}. $$ The $\sup$ is finite for $\varepsilon$ small enough, because $\Gamma_\varepsilon$ is then contained in the resolvent of $A$.

But this implies $$ \lim_{n\to \infty} \| T^n\|^{1/n} =0. $$ Therefore the spectrum of $T$ is $\{0\}$.

As a consequence of $A$ being compact it is true that $\operatorname{ran}E(\lambda) $ is finite dimensional (to be proven below as lemma 1).

Since $T$ is a linear operator between finite dimensional vector spaces whose spectrum is $\{0\}$ it follows that it is nilpotent.

Therefore there is some $k \in \mathbb{N}$ with $$0 = (\lambda-A)^k E(\lambda).$$ If we take the smallest $k$ so that the above equality is true for the first time, then clearly $k \leq \alpha$ and so we conclude that $\operatorname{ran} E(\lambda) \subset \ker (\lambda -A)^\alpha$.

Lemma 1: $\operatorname{ran}E(\lambda) $ is finite dimensional.

Proof: Let $z \neq 0 $ be in the resolvent of $A$. We have $$I = (z-A) (z-A)^{-1} = z (z-A)^{-1} - A (z-A)^{-1}$$ and therefore $$(z-A)^{-1} - z^{-1} =z^{-1} A (z-A)^{-1},$$ but the right hand side is compact since $A$ is compact. Therefore $$ 2 \pi i E(\lambda ) = \int_{\Gamma} (z-A)^{-1}dz =\int_{\Gamma} (z-A)^{-1} - z^{-1} dz = \int_{\Gamma} z^{-1} A (z-A)^{-1} dx $$ is also compact (being the limit in operator norm of compact operators). But $E(\lambda)$ acts as the identity on $\operatorname{ran}E(\lambda)$ since it is a projection, therefore $\operatorname{ran}E(\lambda)$ must be finite dimensional by a standard result.

Answer 2

Consider $A_\lambda$ the restriction of $A$ to $\operatorname{ran} E(\lambda)$. Note that $A_\lambda : \operatorname{ran} E(\lambda) \to \operatorname{ran} E(\lambda)$ thanks to $A$ and $E(\lambda)$ commuting, therefore it makes sense to talk about its eigenvalues.

$A_\lambda$ is an operator between finite-dimensional spaces whose only possible non-zero eigenvalue $\lambda$ since other eigenvalues would be eigenvalues of $A$, yet the eigenspaces of the remaining non-zero eigenvalues lie in the ranges of the other Riesz projections by the $\supset$ part you proved, and these are orthogonal to the range of $E(\lambda)$.
Moreover $0$ is not an eigenvalue of $A_\lambda$ either. Indeed, if we consider $u \in \ker A \cap \operatorname{ran} E(\lambda)$, then we have (I put the anti-linearity on the left variable of the inner product): $$\begin{split}\|u\|^2 &= \left\langle u, \left(\frac{1}{2\pi i}\int_\Gamma (z - A)^{-1}\mathrm{d}z\right)(u)\right\rangle\\ &= \left\langle u, \frac{1}{2\pi i}\int_\Gamma (z - A)^{-1}(u)\mathrm{d}z\right\rangle\\ &= \frac{1}{2\pi i}\int_\Gamma \left\langle u, (z - A)^{-1}(u)\right\rangle\mathrm{d}z\\ &= \frac{1}{2\pi i}\int_\Gamma \left\langle u, \frac{u}{z}\right\rangle\mathrm{d}z\\ &= \frac{1}{2\pi i}\left(\int_\Gamma \frac{1}{z}\mathrm{d}z\right) \|u\|^2 = 0\end{split}$$ where the last equality is due to $0$ not being in the interior of $\Gamma$, and one of the intermediate steps uses the fact that $(z-A)(u) = zu$ hence $u = z(z- A)^{-1}(u)$.

$A_\lambda$ having thus only $\lambda$ as its single eigenvalue, by standard linear algebra there exists an integer $m$ such that $(A_\lambda - \lambda)^m = 0$, and hence such that $((A - \lambda) E(\lambda))^m = (A - \lambda)^m E(\lambda) = 0$. Due to the nature of the algebraic multiplicity $\alpha$, this implies that $(A - \lambda)^\alpha E(\lambda) = 0$, which proves the $\subset$ direction.

(Including this just for transparency: I was partly inspired by OP's answer to the post Why are nonzero eigenvalues of a compact operator poles of its resolvent?)