resolvent function is analytic

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 (http://planetmath.org/ResolventMatrix) $R_{x}:\mathbb{C}-\sigma(x)\longrightarrow\mathcal{A}$ defined by $R_{x}(\lambda)=(x-\lambda e)^{-1}$ is analytic (http://planetmath.org/BanachSpaceValuedAnalyticFunctions).

Moreover, for each $\lambda_{0}\in\mathbb{C}-\sigma(x)$ it has the power series

 $\displaystyle R_{x}(\lambda)=\sum_{n=0}^{\infty}R_{x}(\lambda_{0})^{n+1}(% \lambda-\lambda_{0})^{n}$ (1)

where the series converges absolutely for each $\lambda$ in the open disk centered in $\lambda_{0}$ given by

 $\displaystyle|\lambda-\lambda_{0}|<\frac{1}{\|R_{x}(\lambda_{0})\|}$ (2)

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

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

 $|\lambda-\lambda_{0}|<\frac{1}{\|R_{x}(\lambda_{0})\|}$

Then $\|(\lambda-\lambda_{0})R_{x}(\lambda_{0})\|<1$ 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

 $\displaystyle x-\lambda e=x-\lambda_{0}e-(\lambda-\lambda_{0})e=(x-\lambda_{0}% e)\cdot[e-(\lambda-\lambda_{0})R_{x}(\lambda_{0})]$ (3)

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

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 .

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

 $R_{x}(\lambda)=\left[\sum_{n=0}^{\infty}R_{x}(\lambda_{0})^{n}(\lambda-\lambda% _{0})^{n}\right]\cdot R_{x}(\lambda_{0})=\sum_{n=0}^{\infty}R_{x}(\lambda_{0})% ^{n+1}(\lambda-\lambda_{0})^{n}\;\;\;\;\square$
Title resolvent function is analytic ResolventFunctionIsAnalytic 2013-03-22 17:29:36 2013-03-22 17:29:36 asteroid (17536) asteroid (17536) 8 asteroid (17536) Theorem msc 46H05 msc 47A10