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I have been working through Teschl's book "Mathematical Methods in Quantum Mechanics with Applications to Schrodinger Operators" and I am stuck on a problem in Chapter 3. I am trying to prove that if $A$ is a self adjoint operator and $\lambda$ is an eigenvalue for $A$, then using the Borel functional calculus $\chi_{\{\lambda\}}(A)$ is just projection onto the $\lambda$ eigenspace of $A.$

I can prove that $\chi_{\{\lambda\}}(A)\psi = \psi$ when $\psi$ is a normalized eigenvector, so all I need to show is that it takes the orthogonal complement of the eigenspace to $0.$ I have reduced this to proving that for any $\phi,$ the vector $ \chi_{\{\lambda\}}(A)\phi$ is an eigenvector with eigenvalue $\lambda,$ but I'm not sure how to prove this. (Note that $ \chi_{\{\lambda\}}(A)\phi$ may be $0,$ and in fact will be whenever $\phi$ is orthogonal to the $\lambda$ eigenspace even though I can't yet prove that.)

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See the answer to… – user108903 Jan 18 '13 at 21:15
I looked at the link at don't quite understand the answer there. The relevant part seems to be the statement "Since $z\chi_{\{\lambda\}}(z) = \lambda\chi_{\{\lambda\}}(z)$ for all complex numbers $z, A\chi_{\{\lambda\}}(A) = \lambda\chi_{\{\lambda\}}(A)$ which says the range of $\chi_{\{\lambda\}}(A)$ is contained in the eigenspace." I don't even understand what that first equality is saying... – mck Jan 19 '13 at 0:21
$\chi_{\{\lambda\}}$ is a function $\mathbb{C}\to \mathbb{C}$, so the first equation says that $z\chi_{\{\lambda\}}(z)=0$ if $z\ne \lambda$, and $z\chi_{\{\lambda\}}(z)=\lambda$ if $z=\lambda$, which is the same as $\lambda \chi_{\{\lambda\}}(z)$. The second equation follows using the homomorphism property of the functional calculus, and the conclusion about the eigenspace follows. But I think this is the part you had already obtained, that $\chi_{\{\lambda\}}(A)\psi=\psi$ when $\psi$ is in this eigenspace. The trickier remainder of the answer is what you need. – user108903 Jan 19 '13 at 0:28
No it turns out this was the direction I needed. I needed to prove that everything in the range is an eigenvector. Thanks! But one more thing, shouldn't $\chi_{\{\lambda\}}$ be considered as a function from $\mathbf{R} \rightarrow \mathbf{R}?$ – mck Jan 19 '13 at 1:09
Oh yes, so you did. Glad it helped anyway! The $\mathbb{C}$/$\mathbb{R}$ thing is a matter of convention, I think. Maybe if you want to define a functional calculus using only self-adjoint operators then $\mathbb{R}\to\mathbb{R}$ is the way to go; the function really only needs to be defined on the spectrum of the operator. But since you can also make a functional calculus for normal operators and complex Borel functions, I tend to think of them as being functions $\mathbb{C}\to\mathbb{C}$. – user108903 Jan 19 '13 at 1:17

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