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!
Best Answer
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 $\operatorname{ran} E(\lambda) \subset \ker (\lambda -A)^\alpha$.
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.