Resolution of identity for the multiplication operator

Question

I'm starting my studies in spectral theory and my professor tells me that I should study the resolution of identity. I thought I understood the theory, but when I arrive at the examples ... I can't understand them. So I really appreciate if someone helps me with this:

Consider the self-adjoint operator $\mathcal{M}_ \varphi $ wherw $\varphi : E \rightarrow \mathbb{R}$ acting in $L^2 _\mu (E) $ the function:

$$\Lambda\mapsto P(\Lambda) = \chi_{\varphi^{-1} (\Lambda)} $$ is a resolution of identity. Where $\Lambda$ is a set in the Borel Algebra.

Well, for that I understood, this is the resolution associated with multiplication operator, is this correct? If this is correct, how can I create a resolution identity associate with an operator? Does it exist a hint?

Second thing: The book says

If $f : \mathbb{R} \rightarrow \mathbb{C}$ is a borel function, then $f \mathcal{M}_ \varphi = \mathcal{M}_{f \circ \varphi}.$

How I get it? Using mathematical operations and the theory, how I see this?

Answer 1

The intuition for the spectral measure for a densely-defined, self-adjoint, linear operator $A : \mathcal{D}(A)\subseteq\mathcal{H}\rightarrow\mathcal{H}$ on a Hilbert space $\mathcal{H}$ is that $$ A dE(\lambda)=\lambda dE(\lambda). $$ This is an "infinitesimal" equality that really only has meaning within the context of the spectral integral: $$ \int_{\mathbb{R}}dE(\lambda) = I, \\ A= A\int_{\mathbb{R}}dE(\lambda)=\int_{\mathbb{R}}\lambda dE(\lambda). $$ For $\mu\notin\mathbb{R}$, the resolvent $R(\mu)=(\mu I-A)^{-1}$ is defined, and can be written in terms of the spectral integral as $$ R(\mu)=\frac{1}{\mu I-A}=\int_{\mathbb{R}}\frac{1}{\mu-\lambda}dE(\lambda). $$ This leads to an inversion integral to construct the spectral measure from the resolvent of the self-adjoint $A$: $$ s\mbox{-}\lim_{v\downarrow 0} \frac{1}{2\pi i}\int_{a}^{b}\frac{1}{(u-iv)I-A}-\frac{1}{(u+iv)I-A}du \\ = s\mbox{-}\lim_{v\downarrow 0}\frac{1}{\pi}\int_{a}^{b}\frac{v}{(uI-A)^2+v^2}du \\ = s\mbox{-}\lim_{v\downarrow 0}\frac{1}{\pi}\int_{a}^{b}\int_{\mathbb{R}}\frac{v}{(u-\lambda)^2+v^2}dE(\lambda) du\\ = s\mbox{-}\lim_{v\downarrow 0}\int_{\mathbb{R}}\frac{1}{\pi}\int_a^b\frac{vdu}{(u-\lambda)^2+v^2}dE(\lambda) \\ = \int_{\mathbb{R}}\frac{1}{2}(\chi_{[a,b]}+\chi_{(a,b)})dE(\lambda) = \frac{1}{2}(E[a,b]+E(a,b)) $$ The "s-" is because the integrals converge in the strong operator topology, meaning that the limits exist when you apply the expressions to a vector and then compute the above limits. This is the Herglotz inversion formula for the constructive determination of the spectral measure from the resolvent function of a self-adjoint operator $A$.