Resolvent Mapping is Continuous/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 $T \in \mathfrak{L}(B, B)$. 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 continuous in the operator norm $\|\cdot\|_*$.

Proof
Pick $z\in\rho(T)$. Since $z\in\rho(T)$, the operator $R_z = (T - zI)^{-1}$ exists and has finite norm $C \geq 0$.

Since Resolvent Set is Open, $z+h\in\rho(T)$ for any $h\in \Bbb C$ smaller than some $\delta > 0$. For such $h$,

Consider $(I - hR_z)^{-1}$. Either $\|R_z\|_* = 0$ so that $\|hR_z\|_* = |h|\|R_z\|_* = 0$ (by Operator Norm is Norm) for all $h$, or else $\|hR_z\|_* < 1$ for $h < 1/\|R_z\|_*$. In either case, $\|hR_z\|_* < 1$ for all sufficiently small $h$.

Therefore, by Invertibility of Identity Minus Operator, for sufficiently small $h$ we have $(I - hR_z)^{-1} = I + hR_z + h^2R_z^2 + \ldots$. Substituting this into $(1)$, we get

as long as $|h| \leq \frac{1}{2\norm{R_z}}$.

The expression on the right-hand side goes to zero as $h$ gets small. Therefore, taking limits, we get

$$ \lim_{h\to0} \norm{ f(z+h) - f(z) }_* = 0. $$

This establishes continuity of $f$ at arbitrary $z\in\rho(T)$.