A property of measurable functional calculus

Question

A seemingly simple property of the measurable functional calculus: Let $A$ be a self-adjoint operator on a Hilbert space $H$ and let $P$ be the associated projection-valued measure, such that $A = \int \lambda dP(\lambda)$ and with Borel functional calculus $f \mapsto P(f) = \int f(\lambda) dP(\lambda) =: f(A)$ (which is a $C^\ast$-algebra hommorphism from the bounded Borel functions on $\mathbb R$ to the bounded linear operators on $H$.)

How can we see the property that $Ax = \lambda_0 x$ implies $f(A)x = f(\lambda_0)x$? (for a bounded Borel function $f$). Of course, to show this, we first have to show it for all simple functions. But i do not find a proof, even in the very basic case of $f = 1_\Omega$.

Edit: Ok, i think my question reduced to the following: Why is that $P(1_\Omega)x = P(\Omega)x = \begin{cases} x \text{ if } \lambda_0 \in \Omega\\ 0 \text{ otherwise} \end{cases}$ under the assumption that $Ax = \lambda_0 x$?

Answer 1

You prove it for monomials, then polynomials, then continuous functions. Then you can use Luzin to show that if $f $ is bounded Borel, then $f (A) $ is a wot limit of a net $f_j (A) $ with $f_j $ continuous.

Answer 2

I hope its ok to answer my own question (the reduced one which i labelled with 'Edit'). I would like to know if the following elementary argument works.

The spectral theorem tells us that there is a projection-valued measure $P$ such that $A = \int \lambda dP(\lambda)$. For a fixed vector $x$ this gives a finite Borel measure, defined by $\mu_x(\Omega) = \langle x, P(\Omega) x \rangle$. In terms of this, the quadratic form of $A$ is given by $\langle x, A x \rangle = \int \lambda d\mu_x(\lambda)$.

Since $A$ is self-adjoint and $x$ is an eigenvector of $A$ to eigenvalue $\lambda_0$ we have, using the functional caluclus (mentioned in my question), $0 = \|(A-\lambda_0)x\| = \langle x, (A-\lambda_0)^2 x \rangle = \int_\lambda (\lambda-\lambda_0)^2 d\mu_x$

($d\mu_x$ being the associated spectral measure to the vector $x$) So we see that the measure $d\mu_x$ together with weight $(\lambda-\lambda_0)^2$ is the zero-measure and therefore we can conclude that $d\mu_x = \gamma d\Theta(\lambda-\lambda_0)$ where $\gamma$ is a constant and $d\Theta$ is the dirac delta measure.