Bounded Borel calculus of strongly commuting self-adjoint operators

[Simple functions commute] Let $f={\sum }_i{a}_i{\mathbf{1}}_{{\mathrm{\Lambda }}_i}$ and $g={\sum }_j{b}_j{\mathbf{1}}_{{\mathrm{\Sigma }}_j}$ be complex simple Borel functions over disjoint measurable covers. By [F1], $f(A)={\sum }_i{a}_i{E}_A({\mathrm{\Lambda }}_i)$ and $g(B)={\sum }_j{b}_j{E}_B({\mathrm{\Sigma }}_j)$, so $f(A)g(B)={\sum }_{i,j}{a}_i{b}_j{E}_A({\mathrm{\Lambda }}_i)E_B({\mathrm{\Sigma }}_j)={\sum }_{i,j}{a}_i{b}_j{E}_B({\mathrm{\Sigma }}_j)E_A({\mathrm{\Lambda }}_i)=g(B)f(A),$ because products of scalars commute and each pair of projections commutes by the strong commutation hypothesis.

[Uniform approximation] Let $f,g$ be bounded Borel and fix $\epsilon >0$. By [F4] choose simple $s,t$ with $\mathrm{\parallel }f-s{\mathrm{\parallel }}_{\mathrm{\infty }}\le \epsilon$ and $\mathrm{\parallel }g-t{\mathrm{\parallel }}_{\mathrm{\infty }}\le \epsilon$. By [F1], $\mathrm{\parallel }s(A)-f(A)\mathrm{\parallel }\le \epsilon$, $\mathrm{\parallel }t(B)-g(B)\mathrm{\parallel }\le \epsilon$ and $\mathrm{\parallel }t(B)\mathrm{\parallel }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}+\epsilon$, so the operator norm estimate $\mathrm{\parallel }s(A)t(B)-f(A)g(B)\mathrm{\parallel }\le \mathrm{\parallel }s(A)-f(A)\mathrm{\parallel }\mathrm{\parallel }t(B)\mathrm{\parallel }+\mathrm{\parallel }f(A)\mathrm{\parallel }\mathrm{\parallel }t(B)-g(B)\mathrm{\parallel }\le \epsilon (\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}+\epsilon +\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }})$ gives $s(A)t(B)\to f(A)g(B)$ in operator norm as $\epsilon \downarrow 0$, with the approximants chosen along any sequence ${\epsilon }_n\to 0$.

[Clause (a)] Let $f,g$ be bounded Borel and choose the approximating sequences $s_n\to f$, $t_n\to g$ of step 1.2. By step 1.1, $s_n(A)t_n(B)=t_n(B)s_n(A)$ for every $n$; both sides converge in operator norm to $f(A)g(B)$ and to $g(B)f(A)$ respectively by step 1.2 (applied to the pairs $(s_n,t_n)$ and $(t_n,s_n)$). Limits in operator norm are unique, so $f(A)g(B)=g(B)f(A)$. This is (a).

[Clause (c), first part] Let $f$ be bounded Borel and let $\mathrm{\Lambda }$ be Borel. Applying (a), already proved, to the bounded Borel function ${\mathbf{1}}_{\mathrm{\Lambda }}$ of $A$ and $f$ of $B$ gives $E_A(\mathrm{\Lambda })f(B)=f(B)E_A(\mathrm{\Lambda })$. This is the first assertion of (c).

[Clause (b), membership in the domain] Fix a bounded Borel $f$ and $x\in D(A)$; put $y:=f(B)x$. Put $u_n(\lambda ):=\lambda {\mathbf{1}}_{\{\mathrm{\mid }\lambda \mathrm{\mid }\le n\}}$, so that $A_n=\int u_{n}d{E}_A$ is a bounded Borel function of $A$; clause (a), already proved, applied to the pair of bounded Borel functions $(u_n,f)$ gives $A_{n}f(B)=f(B)A_n$, so by [F1] and [F3] $\int {\lambda }^2{\mathbf{1}}_{\{\mathrm{\mid }\lambda \mathrm{\mid }\le n\}}d{E}_{A,y}=\mathrm{\parallel }{A}_{n}y{\mathrm{\parallel }}^2=\mathrm{\parallel }{A}_{n}f(B)x{\mathrm{\parallel }}^2=\mathrm{\parallel }f(B)A_{n}x{\mathrm{\parallel }}^2\le \mathrm{\parallel }f(B){\mathrm{\parallel }}^2\mathrm{\parallel }{A}_{n}x{\mathrm{\parallel }}^2\le \mathrm{\parallel }f(B){\mathrm{\parallel }}^2\mathrm{\parallel }Ax{\mathrm{\parallel }}^2,$ where the last step uses $\mathrm{\parallel }{A}_{n}x\mathrm{\parallel }\le \mathrm{\parallel }Ax\mathrm{\parallel }$ for $x\in D(A)$. Letting $n\to \mathrm{\infty }$ and using monotone convergence on the left (equivalently the identities of [F3]) gives $\int {\lambda }^{2}d{E}_{A,y}\le \mathrm{\parallel }f(B){\mathrm{\parallel }}^2\mathrm{\parallel }Ax{\mathrm{\parallel }}^2<\mathrm{\infty }$, so $y\in D(A)$.

[Clause (b), action] With $x$ and $y=f(B)x$ as in step 4.1, both $x$ and $y$ lie in $D(A)$. By [F3], $A_{n}x\to Ax$ and $A_{n}y\to Ay$. Using $A_{n}f(B)=f(B)A_n$ from step 4.1, $Ay={\lim }_n{A}_{n}y={\lim }_n{A}_{n}f(B)x={\lim }_{n}f(B)A_{n}x=f(B)Ax,$ the last equality because $f(B)$ is bounded and $A_{n}x\to Ax$. Hence $Af(B)x=f(B)Ax$ for every $x\in D(A)$. Exchanging the roles of $A$ and $B$ gives the other asserted identity, since the strong commutation hypothesis is symmetric.

[Norm limits of functions of $B$] Let $T_n$ be a sequence of bounded Borel functions of $B$ that converges in operator norm to a bounded operator $T$. By steps 4.1-5.1, for every $x\in D(A)$ one has $T_{n}x\in D(A)$ and $A{T}_{n}x=T_{n}Ax$. Thus $T_{n}x\to Tx$ and $A{T}_{n}x=T_{n}Ax\to TAx$. Since the self-adjoint operator $A$ is closed, $Tx\in D(A)$ and $ATx=TAx$. Finite bounded linear combinations are again bounded Borel functions of $B$, so the same domain and action identities hold for them. This completes (b).

[Clause (c), composition] Let $f$ be real-valued and bounded Borel and put $F(\mathrm{\Lambda }):=E_A(f^{-1}(\mathrm{\Lambda }))$ for Borel $\mathrm{\Lambda }\subseteq \mathbb{R}$. Preimages preserve the empty set, the whole space, complements and countable unions, so $F$ is a projection-valued measure. Each $F_x$ is a finite positive Borel measure, hence finite on compact sets, and is regular by [F5]; thus $F$ is a regular PVM. For a complex simple $u={\sum }_i{c}_i{\mathbf{1}}_{D_i}$ the preimages $f^{-1}(D_i)$ are pairwise disjoint, so $\int udF={\sum }_i{c}_{i}F(D_i)={\sum }_i{c}_i{E}_A(f^{-1}(D_i))=\int (u\circ f)d{E}_A$; both sides extend to bounded Borel $u$ by the uniform approximation of [F1] and [F4], giving, for every $x\in H$, the scalar identity $\int ud{F}_x=\int (u\circ f)d{E}_{A,x}$. Taking $u(\lambda )=v_n(\lambda ):={\lambda }^2{\mathbf{1}}_{\{\mathrm{\mid }\lambda \mathrm{\mid }\le n\}}$ and letting $n\to \mathrm{\infty }$ by [F6] gives $\int {\lambda }^{2}d{F}_x=\int f^{2}d{E}_{A,x}=\mathrm{\parallel }f(A)x{\mathrm{\parallel }}^2<\mathrm{\infty }$, where the last equality is [F1] for the bounded $f$; and, choosing $R>\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, $F(\mathbb{R}\setminus [-R,R])=0$, so applying the operator identity to $u(\lambda )=\lambda {\mathbf{1}}_{[-R,R]}(\lambda )$ gives $\int \lambda d{F}_{x,y}=\int fd{E}_{A,x,y}=⟨f(A)x,y⟩$ for all $x,y\in H$. Since $f$ is real-valued, the adjoint-preservation clause of [F2] gives $f(A{)}^{\ast }={\mathrm{\Phi }}_{E_A}(\bar{f})={\mathrm{\Phi }}_{E_A}(f)=f(A)$, so $f(A)$ is self-adjoint, and $f(A)$ is bounded by [F1]. By the uniqueness of the spectral projection-valued measure of a self-adjoint operator in [F3], $F$ is the spectral projection-valued measure of the bounded self-adjoint operator $f(A)$. By the strong commutation hypothesis, $F(\mathrm{\Lambda })E_B(\mathrm{\Sigma })=E_A(f^{-1}(\mathrm{\Lambda }))E_B(\mathrm{\Sigma })=E_B(\mathrm{\Sigma })E_A(f^{-1}(\mathrm{\Lambda }))=E_B(\mathrm{\Sigma })F(\mathrm{\Lambda })$, so $(f(A),B)$ strongly commutes. Applying the same argument once more, now to the pair $(f(A),B)$ and the real-valued bounded Borel function $g$, shows that $(f(A),g(B))$ strongly commutes. This completes (c).