Resolvent Set of Element of Banach Algebra is Open

Theorem

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

Let $x \in A$.

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

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Then $\map {\rho_A} x$ is open.

Proof

Without loss of generality suppose that $A$ is unital, swapping $A$ for its unitization if necessary.

Let $\map G A$ be the group of units of $A$.

Define $S : \C \to A$ by:

$\map S \lambda = \lambda {\mathbf 1}_A - x$

From Resolvent Mapping is Continuous: Continuous, $S$ is continuous.

From Group of Units in Unital Banach Algebra is Open, $\map G A$ is open.

From the definition of the resolvent set, we have:

$\map {\rho_A} x = \set {\lambda \in \C : \lambda {\mathbf 1}_A - x \in \map G A}$

That is:

$\map {\rho_A} x = S^{-1} \sqbrk {\map G A}$

Since $S$ is continuous, and $\map G A$ is open, $S^{-1} \sqbrk {\map G A}$ is open.

We conclude that $\map {\rho_A} x$ is open.

$\blacksquare$