Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\rho$ denote the resolvent of a closed operator and if $\lambda \in \rho(A)$, define $R(\lambda,A) := (\lambda I -A)^{-1}$. If $\mu$ is sufficiently small ($|\lambda-\mu|<\frac{1}{\|R(\lambda,A)\|}$), then the assertions is, that $R(\mu,A)$ can be represented via a power series in the following way: $$R(\mu,A) = \sum_{n=0}^{\infty} (\lambda-\mu)^nR(\lambda,A)^{n+1}$$, which converges absolutely in $\mathcal{B}(X,[D(A)])$ ($[D(A)]$ means $D(A)$ equipped with the graph norm) and uniformly in $B(\lambda,\frac{\delta}{\|R(\lambda,A)\|})$ for each $\delta \in (0,1)$. proving the uniform convergence I would use Weierstrass' convergence criterion, but actually I don't know how to prove the absolute convergence with respect to the operator topology. Greetings

share|cite|improve this question
Yes $X$ is a Banach space. – David Dec 19 '12 at 14:25

Your Answer


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

Browse other questions tagged or ask your own question.