ResolventFunctionIsAnalytic 2007-08-22 13:35:34 2007-12-07 20:16:41 Theorem resolvent matrix % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here {\bf 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 \PMlinkname{resolvent function}{ResolventMatrix} $R_x : \mathbb{C}-\sigma(x) \longrightarrow \mathcal{A}$ defined by $R_x(\lambda) = (x-\lambda e)^{-1}$ is \PMlinkname{analytic}{BanachSpaceValuedAnalyticFunctions}. Moreover, for each $\lambda_0 \in \mathbb{C} - \sigma(x)$ it has the power series \PMlinkescapetext{representation} \begin{align} R_x(\lambda) = \sum_{n=0}^{\infty} R_x(\lambda_0)^{n+1}(\lambda -\lambda_0)^n \end{align} where the series converges absolutely for each $\lambda$ in the open disk centered in $\lambda_0$ given by \begin{align} |\lambda - \lambda_0| < \frac{1}{\|R_x(\lambda_0)\|} \end{align} {\bf 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 \begin{displaymath} |\lambda - \lambda_0| < \frac{1}{\|R_x(\lambda_0)\|} \end{displaymath} Then $\|(\lambda - \lambda_0)R_x(\lambda_0)\| < 1$ and by the \PMlinkname{Neumann series}{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 \begin{align} x-\lambda e = x- \lambda_0 e - (\lambda - \lambda_0)e = (x-\lambda_0 e)\cdot [e - (\lambda - \lambda_0)R_x(\lambda_0)] \end{align} 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 \PMlinkescapetext{resolvent function}. Taking inverses on the equality (3) above one obtains \begin{displaymath} R_x(\lambda) = (e-(\lambda-\lambda_0)R_x(\lambda_0))^{-1} \cdot R_x(\lambda_0) \end{displaymath} Again, by the \PMlinkname{Neumann series}{NeumannSeriesInBanachAlgebras}, one obtains \begin{displaymath} 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 \end{displaymath}