How do I Show that $R:Res(T)\rightarrow B(v),R(\lambda)=(T-\lambda I)^{-1}$ is a continuous function. I know that $R(\lambda)-R(\mu)=(\lambda-\mu)R(\lambda)R(\mu)$ and that the resolvent set is open in C. But I can't seem to find a bound on $R(\lambda)$.
Any ideas?
We want to show that as $\lambda_n \to \lambda$, $R(\lambda_n) \to R(\lambda)$. Since $$R(\lambda) - R(\lambda_n) = (\lambda_n - \lambda) R(\lambda)R(\lambda_n)$$ it will clearly suffice to show that $\|R(\mu)\|$ is bounded in some neighbourhood of $\lambda$.
So fix $\lambda \in \rho(T)$ (the resolvent set of $T$). By the usual Von Neumann series argument, if $|\mu - \lambda| < \|R(\lambda)\|$ then $\mu \in \rho(T)$ with $$R(\mu) = \sum_{j=0}^\infty R(\lambda)^{j+1} (\lambda - \mu)^j.$$ If we additionally impose, say, $|\mu - \lambda| < \frac{1}{2} \|R(\lambda)\|$ then $$\|R(\mu)\| \leq \sum_{j=0}^\infty \|R(\lambda)\|^{j+1} |\lambda - \mu|^j \leq 2 \|R(\lambda)\|$$ and so $\|R(\mu)\|$ is uniformly bounded in a neighbourhood of $\lambda$, as desired.