Functional calculus for unbounded positve operator

Question

In Conway's book, the author has said the following (Theorem 4.10). Let $X$ be a self-adjoint positive unbounded operator. Denote $F(X)$ by the functional calculus for any Borel measurable function $F:\sigma(X)\to\mathbb C.$ It is easy to see that $F(X)$ is indeed a bounded operator if $F$ is bounded. But in Theorem 4.0, he has mentioned (without proof) that $(FG)(X)=F(X)G(X)=G(X)F(X)$ if $F$ is bounded and $G$ is any Borel function defined on $\sigma(X).$ My doubt is how can one obtain $F(X)G(X)=G(X)F(X)$? Since the domain of $F(X)G(X)$ is just domain of $G(X)$ and the domain of $G(X)F(X)$ is the full Hilbert space and these two may not match! can someone shed any light on this?

Answer 1

Your observation is correct: You must show that $F(X)$ preserves the domain of $G(X)$ so that $F(X)G(X)x=G(X)F(X)x$ makes sense for all $x\in\mathcal{D}(G(X))$. Note that $x\in\mathcal{D}(G(X))$ iff $$ \int |g(\lambda)|^2 d\|E(\lambda)x\|^2 < \infty, $$ which is equivalent to assuming $\lim_{r\uparrow\infty} \|G_r(X)x\| < \infty$, where $G_r(\lambda)=\chi_{|G|\le r}(\lambda)G(\lambda)$. So, if $x\in\mathcal{D}(G(X))$ and $F$ is bounded, it follows that $F(X)x\in\mathcal{D}(G(X))$ and $$ G_r(X)F(X)x=F(X)G_r(X)x $$ leads to the conclusion that $F(X)x\in\mathcal{D}(G(X))$ and $G(X)F(X)x=F(X)G(X)x$.