Closure of sum and multiplication (functional calculus for unbounded linear operators)

Question

Let $A$ be a self-adjoint (unbounded) linear operator in a Hilbert space $H$. Let $E$ be the unique regular resolution of the identity on the Borel subsets of $\mathbb{R}$ such that $$A= \int_\mathbb{R} t dE(t).$$ We then write $$f(A)= \int_\mathbb{R} f dE$$ for a measurable function $f: \mathbb{R}\to \mathbb{C}$ and we have $$f(A)+ g(A)\subseteq (f+g)(A), \quad f(A)g(A)\subseteq (fg)(A).$$ Note that the right hand sides $(f+g)(A)$ and $(fg)(A)$ are closed operators, so this begs the questions:

Is $(f+g)(A)$ the closure of $f(A)+g(A)$? Similarly, is $(fg)(A)$ the closure of $f(A)g(A)$? How to prove this?

I learnt this stuff from chapter 13 in Rudin's functional analysis book, so an answer that uses tools from this chapter is highly appreciated! Of course, any input is welcome.

Answer 1

Yes, this is true. In each case, the argument works by localizing the functions to sets where they are bounded. I will only prove it for sums here.

Let $M_n=\{\lambda\in\mathbb R: \lvert f(\lambda)\rvert+\lvert g(\lambda)\rvert\leq n\}$. Clearly, $E(M_n)H\subset D(f(A))\cap D(f(B))$. For $\xi\in D((f+g)(A))$ let $\xi_n=E(M_n)\xi$. As $M_n\nearrow \mathbb R$, we have $\xi_n\to \xi$ in $H$. Moreover, $$ \lVert (f+g)(A)(\xi-\xi_n)\rVert^2=\lVert (f+g)(A)E(M_n^c)\xi\rVert^2=\int_{\mathbb R}1_{M_n^c}\lvert f(\lambda)+g(\lambda)\rvert^2\,d\langle\xi,E(\lambda)\xi\rangle. $$ This last integral converges to zero by the dominated convergence theorem. Thus $\xi_n\to \xi$ with respect to the graph norm.

Answer 2

In fact, something more general is true. I will treat here the multiplication case, using the same idea as @MaoWao.

Claim: Let $E: \mathcal{F}\to B(H)$ be a resolution of the identity where $(\Omega, \mathcal{F})$ is a measurable space. Let $f,g: \Omega \to \mathbb{C}$ be measurable functions. Then $$\overline{\left(\int fdE\right)\left(\int g dE\right)}= \int (fg)dE.$$

Proof: Let $A = \int f dE$ and $B = \int g dE$. Let $$M_n = \{\omega \in \Omega: |f(\omega)|+|g(\omega)| < n\}.$$ Then $M_n \nearrow \Omega$ so $E(M_n)\nearrow 1_H$ in the strong topology. Note that $$E(M_n)H \subseteq D(AB) = \{\xi \in D(B): B\xi \in D(A)\}.$$ Indeed, let $\xi \in H$. Then $$\int_\Omega |g(\omega)|^2dE_{E(M_n)\xi, E(M_n)\xi}(\omega) = \int_{M_n} |g(\omega)|^2 d E_{\xi, \xi}(\omega) \le n^2 \|\xi\|^2 < \infty$$ so $E(M_n)\xi \in D(B)$. On the other hand, using the fact that $B$ and $E(M_n)$ commute, we get $$\int_\Omega |f(\omega)|^2 dE_{BE(M_n)\xi, BE(M_n)\xi}(\omega)= \int_{M_n} |f(\omega)|^2 dE_{BE(M_n)\xi, BE(M_n)\xi}(\omega) \le n^2 \|BE(M_n)\xi\|^2 < \infty$$ so that $BE(M_n)\xi \in D(A)$. Thus, indeed $E(M_n)H\subseteq D(AB)$.

Next, write $C= \int fg dE$. Let $\xi \in D(C)$. Then $$\|C(\xi- E(M_n)\xi)\|^2 = \|C E(M_n^c)\xi\|^2 = \int_{M_n^c}|(fg)(\omega)|^2 dE_{\xi, \xi}(\omega)\to 0$$ by the dominated convergence theorem (by assumption, $\int_\Omega |f(\omega)g(\omega)|^2 < \infty$, which acts as the dominating function).

It follows that $$(E(M_n)\xi, (AB)E(M_n)\xi) = (E(M_n)\xi, CE(M_n)\xi) \to (\xi, C\xi)$$ in $H \oplus H$. Consequently, elements in the graph of $C$ can be norm-approximated by elements in the graph of $AB$. It follows that the closure of the graph of $AB$ is the graph of $C$, which means exactly that $C$ is the closure of $AB$, as desired.