# Eigenprojection as Contour Integral over Resolvent

Let $H$ be a Hilbert space and let $A \in L(H)$ be a bounded linear operator. Assume that $\lambda$ is an eigenvalue of $A$ and assume further that $C_\lambda$ is a simple closed curve in the complex plane that separates $\lambda$ from the rest of the spectrum of $A$. Then $$- \frac{1}{2 \pi i} \int_{C_\lambda}{(A-zI)^{-1}dz}$$ is the projection of $H$ onto the eigenspace of $\lambda$.

Where can I find a proof of this theorem? I am looking for an introductory text that maybe also sheds some light on the theory around this claim, i.e. the definition of the contour integral and the relation of this integral to the measurable functional calculus and the spectral measure of $A$.

EDIT: Assume that $A$ is normal or self-adjoint.

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Maybe in Rudin's book, Functional Analysis, more precisely in the chapter about Banach algebras. –  Davide Giraudo Jun 4 '12 at 12:54
"measurable functional calculus and the spectral measure of $A$": You didn't say that $A$ is normal, so what measurable functional calculus? Another reference (just for a rough overview) is en.wikipedia.org/wiki/Holomorphic_functional_calculus. There was a related question: math.stackexchange.com/questions/15289/… –  Jonas Meyer Jun 5 '12 at 7:15
This is incorrect as stated. E.g., let $A\in B(\ell^2)$ be defined by $A(x_0,x_1,x_2,x_3,x_4\ldots)=(0,0,x_1,\frac{1}{2}x_2,\frac{1}{3}x_3,\frac{1}{4}‌​x_4,\cdots)$. Then $\lambda=0$ is the only element of the spectrum, and it is an eigenvalue with eigenspace $\mathbb C(1,0,0,0,\ldots)$. However, that integral will give you the identity operator. –  Jonas Meyer Jun 5 '12 at 7:27
The books referred to (Dunford-Schwartz, Kato and Taylor) on that Wikipedia page are all very good and contain careful and detailed proofs of what you're asking about. –  t.b. Jun 5 '12 at 8:55
I am aware of the fact that the article on Wikipedia enlists Dunford/Schwarz in the literature list. Since this book has three volumes and ~2500 pages it would be very helpful, if you could be a little more specific. –  Meneldur Jun 9 '12 at 10:53