Why is $\lambda\mapsto(\lambda\textbf{1}-T)^{-1}$ analytic on $\rho(T)$?

Question

Let $T$ be a (closed) linear operator on a Banach space $X$. Denote by $\rho(T)$ the resolvent set of $T$, namely, $$\rho(T)=\{\lambda \in \mathbb{C}:(\lambda\textbf{1}-T)^{-1} \in B(X)\},$$ where $B(X)$ denotes the set of bounded linear operators of $X$ onto $X$.

I am trying to figure out why the mapping $\lambda\mapsto(\lambda\textbf{1}-T)^{-1}$ is analytic on $\rho(T)$.

Can anyone please give me an explanation or point to towards a good reference that shows why this is true?

Answer 1

Let us denote (as usual) the resolvent operator by $R(\lambda,T)$, hence $R(\lambda,T)=(\lambda\textbf{1}-T)^{-1}$. For $\lambda, \mu \in \rho(T)$ with $ \lambda \ne \mu $ one has the Resolvent Equation

$\frac{R(\lambda,T)-R(\mu,T)}{\lambda - \mu}=-R(\lambda,T) \cdot R(\mu,T)$, therefore

$ \lim_{\lambda \to \mu}\frac{R(\lambda,T)-R(\mu,T)}{\lambda - \mu}=-R(\mu,T)^2$.