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Let $\Gamma$ be a differentiable Jordan curve in the resolvent set $\rho(A)$ of the self-adjoint operator $A$. How does one show $\chi_\Omega(A) = \int_\Gamma R_A(z) dz$, where $\Omega = Int \Gamma \cap \mathbb R$ and $R_A(z)$ is the resolvent of $A$ at $z$.

Of course, this looks like Cauchy's integral formula (isnt there a $1/2\pi i$-factor missing?). The RHS is a Banach-space valued integral (the operators $R_A(z)$ are bounded operators) and somewhere i've read that function theory easily extends to the setting of general Banach algebras, but i dont feel confident enough to argue why the above formula holds. How could one use the functional calculus from the spectral theorem (for self-adjoint operators) here?

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The spectral theorem gives $$ (\lambda I-A)^{-1} =\int_{\sigma(A)}\frac{1}{\lambda -t}dE(t), \;\;\; \lambda\in\rho(A). $$ If $C$ is a simple closed piecewise smooth positively oriented contour in $\rho(A)$ that encloses part of the spectrum, then $C$ crosses the real axis where the spectral measure is $0$ in a neighborhood of that crossing. So the following integrals make sense \begin{align} \left(\frac{1}{2\pi i}\oint_{C}(\lambda I-A)^{-1} d\lambda\,x,y\right) & = \frac{1}{2\pi i}\oint_{C}((\lambda I-A)^{-1}x,y)d\lambda \\ & = \frac{1}{2\pi i}\oint_{C}\int_{\sigma}\frac{1}{\lambda -t}d(E(t)x,y)d\lambda \\ & = \int_{\sigma}\left(\frac{1}{2\pi i}\oint_{C}\frac{1}{\lambda-t}d\lambda \right)d(E(t)x,y). \end{align} The complex measure $d(E(t)x,y)$ is supported in $\sigma$, which is a positive distance from where $C$ intersects the real axis; and the inner integral is $1$ for $t\in\sigma$ inside $C$ and is $0$ for $t\in\sigma$ and outside $C$. If $\mbox{Int}(C)$ denotes the interior of $C$, then the above becomes $$ \left(\frac{1}{2\pi i}\oint_{C}(\lambda I-A)^{-1} d\lambda\,x,y\right) = (E(\mbox{Int}(C))x,y),\;\;\; x,y \in X, $$ Because this holds for all $x,y \in X$, $$ \frac{1}{2\pi i}\oint_{C}(\lambda I-A)^{-1}d\lambda = E(\mbox{Int}(C)). $$

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