Resolvent Mapping is Continuous

Theorem

Let $B$ be a Banach space.

Let $\mathfrak{L}(B, B)$ be the set of bounded linear operators from $B$ to itself.

Let $T \in \mathfrak{L}(B, B)$.

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 $\|\cdot\|_*$.

Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra over $\C$.

Let ${\mathbf 1}_A$ be the identity element of $A$.

Let $x \in A$.

Let $\map {\rho_A} x$ be the resolvent set of $x$ in $A$.

Define $R : \map {\rho_A} x \to A$ by:

$\map R \lambda = \paren {\lambda {\mathbf 1}_A - x}^{-1}$

Then $R$ is continuous.