# Functional calculus for unbounded positve operator

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?

## 1 Answer

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 $$G(X)F(X)x=F(X)G(X)x$$.