Is there a residue theorem for holomorphic operator-valued functions? 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!
 A: 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!
A: Under the assumption that this is a finite dimensional problem and $A(\lambda)$ is holomorphic, it can be characterized with the left and right eigenvectors corresponding to the associated nonlinear eigenvalue problem. Let $T(\lambda):=A(\lambda)-\lambda I$ such that we can formula (14) in http://dx.doi.org/10.1016/j.laa.2011.03.030  with $f(\lambda)=2i\pi$. More precisely, if $(\lambda_i,v_i,w_i)$, $i=1,\ldots,p$ are those triples such that $T(\lambda_i)v_i=0$ and $w_i^HT(\lambda_i)=0$ and $\lambda_i\in\operatorname{int}(\Gamma)$ and $w_i^HT'(\lambda_i)v_i=1$. Then,
$$\int_\Gamma (A(\lambda)-\lambda I)^{-1}\,d\lambda=2i\pi\sum_i^pv_iw_i^H.$$
It requires that the eigenvectors can be normalized as $w_i^HT'(\lambda_i)v_i=1$, which is the generic situation and holds in particular if all eigenvalues in $\operatorname{int}(\Gamma)$ are simple. It is also a direct consequence of the earlier result http://dx.doi.org/10.14495/jsiaml.1.52.
