A simple question in functional analysis

Question

A classical result, in functional analysis, says that if $T\in B(X)$, the function: $\lambda \rightarrow (\lambda I-T)^{-1}$ is analytic on $\rho(T)$(which is the resolvent set). If I fix an element $x\in X$, then the vector-valued function: $\lambda \rightarrow (\lambda I-T)^{-1}x$ is still analytic?

Answer 1

Yes. Let $R(\lambda)=(\lambda I-T)^{-1}$. Then $$ \lim_{\lambda\rightarrow\lambda_{0}} \left\|\frac{1}{\lambda-\lambda_{0}}\{R(\lambda)-R(\lambda_{0})\}-R'(\lambda_{0})\right\|_{\mathcal{L}(X)}=0 $$ implies $$ \lim_{\lambda\rightarrow\lambda_{0}} \left\|\frac{1}{\lambda-\lambda_{0}}\{R(\lambda)x-R(\lambda_{0})x\}-R'(\lambda_{0})x\right\|_{X}=0. $$ because of the bound $\|Ax\|_{X}\le \|A\|_{\mathcal{L}(X)}\|x\|_{X}$ for any $A\in\mathcal{L}(X)$ and $x\in X$.

In your case, what is more interesting is that the reverse is true: analyticity in the strong operator topology implies analyticity in the uniform operator topology. In fact, analyticity in the weak operator topology implies analyticity in the uniform operator topology, meaning that, if $\lambda \mapsto x^{\star}(A(\lambda)x)$ is holomorphic on an open region $\Omega$ for all $x \in X$ and $x^{\star} \in X^{\star}$, then $\lambda\mapsto A(\lambda) \in \mathcal{L}(X)$ is holomorphic in the topology of $\mathcal{L}(X)$.