resolvent function is analytic

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

Then, the resolvent function (http://planetmath.org/ResolventMatrix) $R_x:ℂ-\sigma (x)⟶𝒜$ defined by $R_x(\lambda )={(x-\lambda e)}^{-1}$ is analytic (http://planetmath.org/BanachSpaceValuedAnalyticFunctions).

Moreover, for each ${\lambda }_0\in ℂ-\sigma (x)$ it has the power series

$R_x(\lambda )={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{R}_x{({\lambda }_0)}^{n+1}{(\lambda -{\lambda }_0)}^n$(1)

where the series converges absolutely for each $\lambda$ in the open disk centered in ${\lambda }_0$ given by

(2)

Proof : Analyticity is defined for functions whose domain is open.

Thus, we start by proving that $ℂ-\sigma (x)$ is an open set in $ℂ$. To do so it is enough to prove that for every ${\lambda }_0\in ℂ-\sigma (x)$ the open disk defined by (2) above is contained in $ℂ-\sigma (x)$.

Let ${\lambda }_0\in ℂ-\sigma (x)$ and $\lambda$ be such that

Then and by the Neumann series (http://planetmath.org/NeumannSeriesInBanachAlgebras) $e-(\lambda -{\lambda }_0)R_x({\lambda }_0)$ is invertible.

Since ${\lambda }_0\notin \sigma (x)$ it follows that $(x-{\lambda }_{0}e)$ is invertible.

Hence, from the equality

$x-\lambda e=x-{\lambda }_{0}e-(\lambda -{\lambda }_0)e=(x-{\lambda }_{0}e)\cdot [e-(\lambda -{\lambda }_0)R_x({\lambda }_0)]$(3)

we conclude that $x-\lambda e$ is also invertible, i.e. $\lambda \in ℂ-\sigma (x)$. Thus $ℂ-\sigma (x)$ is open.

The above proof also pointed out that for every ${\lambda }_0\in ℂ$, $R_x$ is defined in the open disk of radius ${\displaystyle \frac{1}{\parallel R_x({\lambda }_0)\parallel }}$ centered in ${\lambda }_0$.

We now prove the analyticity of the .

Taking inverses on the equality (3) above one obtains

$$R_x(\lambda )={(e-(\lambda -{\lambda }_0)R_x({\lambda }_0))}^{-1}\cdot R_x({\lambda }_0)$$

Again, by the Neumann series (http://planetmath.org/NeumannSeriesInBanachAlgebras), one obtains

$$R_x(\lambda )=\left[\sum _{n=0}^{\mathrm{\infty }}{R}_x{({\lambda }_0)}^n{(\lambda -{\lambda }_0)}^n\right]\cdot R_x({\lambda }_0)=\sum _{n=0}^{\mathrm{\infty }}{R}_x{({\lambda }_0)}^{n+1}{(\lambda -{\lambda }_0)}^n\mathit{\hspace{1em}}\mathrm{\square }$$