Let $P_\Omega$ be a projection valued measure and let $R_A(z)=(A-z)^{-1}$ be the resolvent map. It can be shown that $$R_A(z)=-\sum_{j=0}{\frac{A^j}{z^{j+1}}}$$

whenever this series is defined.

My question is, given a self adjoint operator $A$, and the fact that we know that $A$ = $\int \lambda P(d\lambda)$, how to proof that $$ R_A(z) = \int\frac{1}{\lambda-z}dP(d\lambda)$$

This is assumed in Frederic Schuller lecture 11


This follows from a general property of integrals with respect to projection-valued measures associated to self-adjoint operators. Namely, if $P$ is the unique projection-valued measure such that $$A=\int\lambda \,P(d\lambda) $$ then for all $f:\mathbb{R}\to \mathbb{C}$ Borel measurable and bounded on $\sigma(A)$, in symbols $f\in B(\sigma(A))$, we have $$f(A)=\int f(\lambda)\,P(d\lambda) \qquad (\star)$$ Where $f(A)$ is defined in the sense of the (Borel) functional calculus (see e.g. these notes, pag. 178). Even if you do not know the functional calculus, it suffices for you to know that if $z\notin \sigma(A)$ (otherwise $R_A(z)$ is not defined) and $f(\lambda)=f_z(\lambda)=(\lambda-z)^{-1}$, then we have $f_z(A)=(A-z)^{-1}$, as one would obtain by formally replacing $\lambda$ with $A$ in the expression of $f_z$.

Indeed, let us check that $f_z\in B(\sigma(A))$. Since $z\notin \sigma(A)$, $f_z$ is continuous on $\sigma(A)$ and since $\sigma(A)$ is compact, $f_z$ is also bounded. Therefore, substituting in $(\star)$ we obtain $$(A-z)^{-1}=\int \frac{1}{\lambda-z}\,P(d\lambda) $$ as desired.


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