Local observables lie in the closed generator domain

If $X=\mathrm{\varnothing }$, [F4] identifies $A=cI$ for some $c\in \mathbb{C}$. By complex linearity and unitality in [F1], ${\alpha }_t(A)=A$, so $A\in D(\delta )$ and $\delta (A)=0$. The displayed commutator sum is empty for $X=\mathrm{\varnothing }$ and is zero for every other localization of this scalar, since $[\mathrm{\Phi }(Z),cI]=0$. For the remaining local-observable argument take $X\mathrm{\ne }\mathrm{\varnothing }$. Every nonzero term $\mathrm{\Phi }(Z)$ meeting $X$ has $Z\subseteq X^{+R}:=\{y\in {\mathbb{Z}}^d:{\mathrm{dist}}_{\mathrm{\infty }}(X,\{y\})\le R\}$, a finite union of finite lattice balls. There are finitely many subsets of $X^{+R}$, so $D_0(A):=\frac{i}{\mathrm{\hslash }}{\sum }_{Z:Z\cap X\mathrm{\ne }\mathrm{\varnothing }}[\mathrm{\Phi }(Z),A]$ is a finite local sum. If $X\subseteq X^{\mathrm{\prime }}$, each term present in the sum for $X^{\mathrm{\prime }}$ but not for $X$ has support disjoint from $X$ and therefore commutes with $A\in {\mathcal{A}}_X$; hence both sums agree. For any two localization regions, compare each with their union to obtain the same conclusion. Thus $D_0(A)$ is independent of the chosen region. For every finite $\mathrm{\Lambda }$ containing $X^{+R}$, finite-volume differentiation gives ${\delta }_{\mathrm{\Lambda }}(A)=\frac{i}{\mathrm{\hslash }}[H_{\mathrm{\Lambda }},A]=D_0(A)$ and ${\alpha }_t^{\mathrm{\Lambda }}(A)-A={\int }_0^t{\alpha }_s^{\mathrm{\Lambda }}(D_0(A))ds\mathrm{.}$ Here the finite-dimensional integral is the coordinatewise Riemann integral; differentiating the finite exponential gives the integrand.

By [F2], ${\alpha }_s^{\mathrm{\Lambda }}(A)$ and ${\alpha }_s^{\mathrm{\Lambda }}(D_0(A))$ converge in norm uniformly on compact time intervals as $\mathrm{\Lambda }$ exhausts the lattice. A continuous Banach-valued function has a Riemann integral because uniform continuity makes its Riemann sums Cauchy in the complete algebra; uniform convergence of the finite-volume integrands lets their integrals pass to the limit. This gives ${\alpha }_t(A)-A={\int }_0^t{\alpha }_s(D_0(A))ds\mathrm{.}$ By [F1], $s\mapsto {\alpha }_s(D_0(A))$ is norm-continuous at $0$; the difference between its average over $[0,t]$ and its value at $0$ is at most ${\sup }_{\mathrm{\mid }s\mathrm{\mid }\le \mathrm{\mid }t\mathrm{\mid }}\mathrm{\parallel }{\alpha }_s(D_0(A))-D_0(A)\mathrm{\parallel }$, which tends to zero. Therefore $A\in D(\delta )$ and $\delta (A)=D_0(A)$.

Let $A_n\in D(\delta )$ with $A_n\to A$ and $\delta (A_n)\to B$ in norm. The integral identities below are immediate at $t=0$. Fix $t\mathrm{\ne }0$ and put $h=t\mathrm{/}N$ for a positive integer $N$; the group law gives the exact telescope ${\alpha }_t(A_n)-A_n={\sum }_{k=0}^{N-1}{\alpha }_{kh}({\alpha }_h(A_n)-A_n)=h{\sum }_{k=0}^{N-1}{\alpha }_{kh}(\delta (A_n))+h{\sum }_{k=0}^{N-1}{\alpha }_{kh}\text{\hspace{0.17em}⁣}(\frac{{\alpha }_h(A_n)-A_n}{h}-\delta (A_n))\mathrm{.}$ As $N\to \mathrm{\infty }$, the second sum has norm at most $\mathrm{\mid }t\mathrm{\mid }\mathrm{\parallel }({\alpha }_h(A_n)-A_n)\mathrm{/}h-\delta (A_n)\mathrm{\parallel }$, which tends to zero by the definition of $\delta (A_n)$. The first sum is a Riemann sum for the continuous function $s\mapsto {\alpha }_s(\delta (A_n))$; its integral exists because the algebra is complete and the function is uniformly continuous on compact intervals. Hence ${\alpha }_t(A_n)-A_n={\int }_0^t{\alpha }_s(\delta (A_n))ds$. Isometry bounds the integral of ${\alpha }_s(\delta (A_n)-B)$ by $\mathrm{\mid }t\mathrm{\mid }\mathrm{\parallel }\delta (A_n)-B\mathrm{\parallel }$, allowing the identity to pass to ${\alpha }_t(A)-A={\int }_0^t{\alpha }_s(B)ds$. Strong continuity bounds the difference between the quotient of this integral and $B$ by ${\sup }_{\mathrm{\mid }s\mathrm{\mid }\le \mathrm{\mid }t\mathrm{\mid }}\mathrm{\parallel }{\alpha }_s(B)-B\mathrm{\parallel }$, which tends to zero. Thus $A\in D(\delta )$ and $\delta (A)=B$, proving that the graph of $\delta$ is closed.