Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm wondering whether there is such a thing as a "residue theorem for holomorphic operator-valued functions". More precisely, I want to evaluate an integral of the form

$P:=\int_{\Gamma} (A(\lambda) - \lambda)^{-1} d \lambda$

where each $A(\lambda)$ is a closed operator and $\Gamma$ encloses an eigenvalue $\lambda_0$ of the holomorphic operator pencil $A(\lambda) - \lambda$, i.e., for some eigenfunction $u_0$ we have $A(\lambda_0)u_0 - \lambda_0u_0=0$.

In the case of a $\lambda$-independent $A$ the operator $P$ is (up to a constant) the well-known Riesz Projection corresponding to $\lambda_0$. If (and how) this can be generalized to a $\lambda$-nonlinear eigenvalue problem is precisely what my question is concerned with.

Thanks for any help in advance!

share|improve this question
    
+1 cool question –  draks ... Mar 14 '12 at 18:47

1 Answer 1

up vote 4 down vote accepted

Beyond the fairly standard discussion of resolvents, there is a reasonable "Cauchy theory" for vector-valued holomorphic and meromorphic functions, with values in a quasi-complete locally convex topological vector space. Rudin's Functional Analysis discusses the Frechet-space-valued case fairly thoroughly, with some abstractions. My functional analysis notes at http://www.math.umn.edu/~garrett/m/fun/ include discussion of quasi-completeness, weak-and-strong holomorphy, etc. Bourbaki's "Integration" talks about vector-valued integrals in this generality, too.

The potentially delicate point is what "meromorphy" means for a vector-valued function $F$: basically, there must be a scalar-valued holomorphic function $f$ so that $f\cdot F$ is holomorphic. That is, one must avoid bad singularities. Granting that, the residue calculus works as well as one could reasonably hope!

share|improve this answer
    
Thanks very much, that is of great help to me! –  user26895 Mar 15 '12 at 8:49
    
@user26895 Very good. :) –  paul garrett Mar 15 '12 at 12:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.