Theorem - Let $\mathcal{A}$ be a complex Banach algebra with identity element $e$ . Let $x \in \mathcal{A}$ and $\sigma(x)$ denote its spectrum.

Then, the resolvent function $R_x : \mathbb{C}-\sigma(x) \longrightarrow \mathcal{A}$ defined by $R_x(\lambda) = (x-\lambda e)^{-1}$ is analytic.

Moreover, for each $\lambda_0 \in \mathbb{C} - \sigma(x)$ it has the power series representation

(1)

(2)

Thus, we start by proving that $\mathbb{C} - \sigma(x)$ is an open set in $\mathbb{C}$ . To do so it is enough to prove that for every $\lambda_0 \in \mathbb{C} - \sigma(x)$ the open disk defined by (2) above is contained in $\mathbb{C} - \sigma(x)$ .

Let $\lambda_0 \in \mathbb{C} - \sigma(x)$ and $\lambda$ be such that

Then $\|(\lambda - \lambda_0)R_x(\lambda_0)\| < 1$ and by the Neumann series $e - (\lambda - \lambda_0)R_x(\lambda_0)$ is invertible.

Since $\lambda_0 \notin \sigma(x)$ it follows that $(x-\lambda_0 e)$ is invertible.

Hence, from the equality

(3)

The above proof also pointed out that for every $\lambda_0 \in \mathbb{C}$ , $R_x$ is defined in the open disk of radius $\displaystyle \frac{1}{\|R_x(\lambda_0)\|}$ centered in $\lambda_0$ .

We now prove the analyticity of the resolvent function.

Taking inverses on the equality (3) above one obtains

Again, by the Neumann series, one obtains