7
$\begingroup$

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.

  • 0
    Maybe in Rudin's book, _Functional Analysis_, more precisely in the chapter about Banach algebras.2012-06-04
  • 0
    "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 http://en.wikipedia.org/wiki/Holomorphic_functional_calculus. There was a related question: http://math.stackexchange.com/questions/15289/why-is-the-following-thing-a-projection-operator2012-06-05
  • 0
    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.2012-06-05
  • 0
    @Jonas Meyer: Thanks for your post. I am aware of the fact that Wikipedia provides a rough overview, but I am seeking for a reference with a detailed treatment of the topic. Unfortunately no such reference is given in the answers of the related question and this question deals with the finite-dimensional case.2012-06-05
  • 0
    @Jonas Meyer: I forgot to state that $A$ is normal or self-adjoint. Sorry for the confusion, I edited the original post.2012-06-05
  • 0
    The fact missing is neither normality nor self-adjointness, but that "the rest of the spectrum" needs to be non-empty, that's what's failing in Jonas's example. Jonas's answer in the linked thread applies to bounded operators on a Banach space, so that's amply sufficient for the needs of your question. The crucial fact entering in the verification is the *[resolvent equation](http://en.wikipedia.org/wiki/Resolvent_formalism)*.2012-06-05
  • 1
    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.2012-06-05
  • 2
    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.2012-06-09
  • 0
    The statement you're looking for is proved (for closed operators) as [Theorem III.6.17](http://books.google.com/books?id=8ji2kN_D3BwC&pg=PA178) on page 178 of Kato's *Perturbation Theory*. In Dunford-Schwartz it follows from Theorem VII.3.20, Corollary VII.3.21 and the last equation in the proof of VII.3.18 (on p.574f of my edition of Volume I). I haven't checked in Taylor. The projection operators you ask about are sometimes called *Riesz projections* and a quick proof of what you ask about is given in [these notes](http://math.arizona.edu/~friedlan/teach/528/proj.pdf) by Lennie Friedlander.2012-06-15
  • 0
    As for an introductory and maybe more modern text than those mentioned above, I'd recommend any functional analysis book treating Banach algebras, like Pedersen's *Analysis Now* or Conway's *Course in functional analysis*, in the latter you want to look at formula 6.9 on page 215.2012-06-15
  • 0
    A last comment: you should ping users using `@username` if you want them to be notified by your comments. I only saw your follow-up by accident, hence the late response.2012-06-15

0 Answers 0