Resolvent Mapping is Analytic/Bounded Linear Operator

Theorem
Suppose $B$ is a Banach space, $\mathfrak{L}(B, B)$ is the set of bounded linear operators from $B$ to itself, and $T \in O$. Let $\rho(T)$ be the resolvent set of $T$ in the complex plane. Then the resolvent mapping $f : \rho(T) \to \mathfrak{L}(B,B)$ given by $f(z) = (T - zI)^{-1}$ is analytic and


 * $f'(z) = (T-zI)^{-2}$

Proof
For any $a\in \rho(T)$, define $R_a = (T - aI)^{-1}$. Then we have

Taking limits of both sides and using Norm is Continuous and Resolvent Mapping is Continuous (the latter of which tells us $R_{z+h} \to R_z$ as $h \to 0$), we get


 * $\lim_{h\to 0} \frac{\norm{f(z+h) - f(z) - (T-zI)^{-2}h }_*}{\size h} = \norm{ R_z^2 - R_z^2 }_* = 0$

which is the result.