Resolvent Mapping is Continuous
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Theorem
Bounded Linear Operator
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 $\|\cdot\|_*$.
Banach Algebra
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.