# Cauchy's integral formula for operators

A supersymmetric transfer matrix and differentiability of the density of states in the one-dimensional Anderson model. Massimo Campanino and Abel Klein. Comm. Math. Phys. 104 (1986), no. 2, 227–241. MR0836001

My problem is p.236.

Let $\Im m \ z \geq 0$ , $T$ and $B(z)$ two operators in $\mathcal{H}_n$ a Hilbert space. There is an operator $R$ where we have:

$$(TB(z))^l1=(I+RB(z)+ \cdots + (RB(z))^l)1.$$

We define $\displaystyle \xi(z)=\lim_{l \to \infty} ((TB(z))^l 1$ and show that $\xi(z) \in \mathcal{H}_n$. We have $TB(z)\xi(z)=\xi(z)$ and $\mathcal{H}_n=\mathbb{C}\xi(z) \oplus \mathcal{H}_n^0$, with $\mathcal{H}_n^0=\{f \in \mathcal{H}_n, f(0)=0\}$.

It diagonalizes $TB(z)$ in the form:

$$TB(z)=\delta_0 \xi(z) \oplus RB(z).$$

Let us fix $E_0 \in \mathbb{R}$ and $\delta>0$ and let $\tau_0^2=\sup\{ \lVert RB(E)^2\rVert_{\mathcal{H}_n^0}; |E-E_0|<\delta\}$. We have $\tau_0^2<1$.

Let $\gamma$ denote the circle $\{z \in \mathbb{C}; |z-1|=\frac{1}{2}(1-\tau_0)\}$ and $\xi_0=\xi(E_0)$.

The author claims that :

For $|E-E_0|<\delta$, $\xi(E)=\frac{1}{2\pi i} \int_\gamma (z-TB(E))^{-1} dz \xi_0$.

It looks like a Cauchy's integral formula for operators. I don't know why this result can be true. Is it linked to a well-known formula ?

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I found what i need here :en.wikipedia.org/wiki/Holomorphic_functional_calculus –  Rémy Thomasse Apr 1 '12 at 21:47