Uniformly bounded finite-range quantum spin interaction

Fix $B\in {\mathcal{A}}_Y$ and define $C_B^{\mathrm{\Lambda }}(S,s):={\sup }_{D\in {\mathcal{A}}_S,D\mathrm{\ne }0}\frac{\mathrm{\parallel }[{\alpha }_s^{\mathrm{\Lambda }}(D),B]\mathrm{\parallel }}{\mathrm{\parallel }D\mathrm{\parallel }}$ for every nonempty $S\subseteq \mathrm{\Lambda }$, with all operators viewed in ${\mathcal{A}}_{\mathrm{\Lambda }\cup Y}$. For $A\in {\mathcal{A}}_X$, put $H_X^{\mathrm{\Lambda }}={\sum }_{\mathrm{\varnothing }\mathrm{\ne }Z\subseteq X}\mathrm{\Phi }(Z)$ and $A_X(s)=e^{-is{H}_X^{\mathrm{\Lambda }}\mathrm{/}\mathrm{\hslash }}A{e}^{is{H}_X^{\mathrm{\Lambda }}\mathrm{/}\mathrm{\hslash }}$. The finite matrix exponential is unitary because its adjoint is $e^{is{H}_X^{\mathrm{\Lambda }}\mathrm{/}\mathrm{\hslash }}$ and the two commuting exponential series multiply to $I$; hence $A_X(s)$ ranges over all of ${\mathcal{A}}_X$ and preserves norm. Differentiating finite matrix exponentials and using the Jacobi identity, the terms contained in $X$ cancel and the terms disjoint from $X$ commute with $A_X(s)$. With $K_X(s)={\sum }_{Z\subseteq \mathrm{\Lambda }:Z\cap X\mathrm{\ne }\mathrm{\varnothing },Z\subseteq ̸X}{\alpha }_s^{\mathrm{\Lambda }}(\mathrm{\Phi }(Z))$, the remaining commutator $f(s)=[{\alpha }_s^{\mathrm{\Lambda }}(A_X(s)),B]$ satisfies $f^{\mathrm{\prime }}(s)=\frac{i}{\mathrm{\hslash }}[K_X(s),f(s)]-\frac{i}{\mathrm{\hslash }}{\sum }_{\begin{array}{c}Z\subseteq \mathrm{\Lambda }\\ Z\cap X\mathrm{\ne }\mathrm{\varnothing },Z\subseteq ̸X\end{array}}[{\alpha }_s^{\mathrm{\Lambda }}(A_X(s)),[{\alpha }_s^{\mathrm{\Lambda }}(\mathrm{\Phi }(Z)),B]]\mathrm{.}$ For the homogeneous equation, the finite matrix ODE $U^{\mathrm{\prime }}(s)=(i\mathrm{/}\mathrm{\hslash })K_X(s)U(s)$ has a solution on every compact interval: its Picard series converges uniformly because the $n$th iterated integral is bounded by $(T{\sup }_s\mathrm{\parallel }{K}_X(s)\mathrm{\parallel }\mathrm{/}\mathrm{\hslash }{)}^n\mathrm{/}n!$. Each $K_X(s)$ is self-adjoint, so differentiating $U(s{)}^{\ast }U(s)$ gives zero; since $U(0)=I$, $U(s{)}^{\ast }U(s)=I$. In finite dimension this makes $U(s)$ unitary, so conjugation by its propagator preserves the matrix norm. This argument also applies after reversing the time parameter, because the reversed homogeneous generator is still skew-adjoint. Set $\sigma =1$ if $t\ge 0$ and $\sigma =-1$ if $t<0$, and write $s=\sigma u$ for $0\le u\le \mathrm{\mid }t\mathrm{\mid }$. The equation for $f(\sigma u)$ has homogeneous commutator generated by $\sigma i{K}_X(\sigma u)\mathrm{/}\mathrm{\hslash }$ and a source multiplied by $\sigma$; the source sign does not change its norm. Variation of constants therefore gives the same norm estimate for either sign of $t$, with intermediate dynamics evaluated at signed times: $C_B^{\mathrm{\Lambda }}(X,t)\le C_B^{\mathrm{\Lambda }}(X,0)+\frac{2}{\mathrm{\hslash }}{\sum }_{\begin{array}{c}Z\subseteq \mathrm{\Lambda }\\ Z\cap X\mathrm{\ne }\mathrm{\varnothing }\end{array}}\mathrm{\parallel }\mathrm{\Phi }(Z)\mathrm{\parallel }{\int }_0^{\mathrm{\mid }t\mathrm{\mid }}{C}_B^{\mathrm{\Lambda }}(Z,\sigma u)du\mathrm{.}$ We enlarged the sum from crossing terms to all terms meeting $X$, which only increases its nonnegative right-hand side.

Iterate this integral inequality. All intermediate times retain sign $\sigma$; in absolute-time parameters the iterated region is $0\le u_n\le \cdots \le u_1\le \mathrm{\mid }t\mathrm{\mid }$, whose volume is $\mathrm{\mid }t{\mathrm{\mid }}^n\mathrm{/}n!$. Thus the same factorial weights apply for either sign of time. The zero-length term is at most $2\mathrm{\parallel }B\mathrm{\parallel }{\mathbf{1}}_{X\cap Y\mathrm{\ne }\mathrm{\varnothing }}$. A length-$n$ contribution, $n\ge 1$, is a chain $Z_1,\dots ,Z_n\subseteq \mathrm{\Lambda }$ with $Z_1\cap X\mathrm{\ne }\mathrm{\varnothing }$, $Z_{j+1}\cap Z_j\mathrm{\ne }\mathrm{\varnothing }$, and $Z_n\cap Y\mathrm{\ne }\mathrm{\varnothing }$, weighted by ${\prod }_{j=1}^n\mathrm{\parallel }\mathrm{\Phi }(Z_j)\mathrm{\parallel }$ and $(2\mathrm{\mid }t\mathrm{\mid }\mathrm{/}\mathrm{\hslash }{)}^n\mathrm{/}n!$. The first interaction sum is at most $\kappa \mathrm{\mid }X\mathrm{\mid }$; each later sum is at most $\kappa \mathrm{\mid }{Z}_j\mathrm{\mid }\le \kappa M$, so the total length-$n$ weight is at most $\kappa \mathrm{\mid }X\mathrm{\mid }(\kappa M{)}^{n-1}$. After $N$ iterations, the unresolved remainder is bounded by $2\mathrm{\parallel }B\mathrm{\parallel }(2\mathrm{\mid }t\mathrm{\mid }\mathrm{/}\mathrm{\hslash }{)}^{N+1}\kappa \mathrm{\mid }X\mathrm{\mid }(\kappa M{)}^N\mathrm{/}(N+1)!$; it tends to zero. Hence no infinite path expansion is assumed without convergence.

If $X\cap Y=\mathrm{\varnothing }$, any contributing length-$n$ chain satisfies ${\mathrm{dist}}_{\mathrm{\infty }}(X,Y)\le nR$: choose a site in each successive nonempty intersection and apply the triangle inequality, using ${\mathrm{diam}}_{\mathrm{\infty }}{Z}_j\le R$ at each of the $n$ steps. Thus its indicator is at most $e^{\mu (nR-{\mathrm{dist}}_{\mathrm{\infty }}(X,Y))}$. Since $M\ge 1$, $\kappa \mathrm{\mid }X\mathrm{\mid }(\kappa M{)}^{n-1}\le \mathrm{\mid }X\mathrm{\mid }(\kappa M{)}^n$. Summing the resulting exponential series gives $C_B^{\mathrm{\Lambda }}(X,t)\le 2\mathrm{\parallel }B\mathrm{\parallel }\mathrm{\mid }X\mathrm{\mid }{e}^{-\mu {\mathrm{dist}}_{\mathrm{\infty }}(X,Y)}(e^{2\kappa M{e}^{\mu R}\mathrm{\mid }t\mathrm{\mid }\mathrm{/}\mathrm{\hslash }}-1)\mathrm{.}$ If $X\cap Y\mathrm{\ne }\mathrm{\varnothing }$, then the distance is zero and the zero-length term together with the same series is bounded by $2\mathrm{\parallel }B\mathrm{\parallel }\mathrm{\mid }X\mathrm{\mid }{e}^{2\kappa M{e}^{\mu R}\mathrm{\mid }t\mathrm{\mid }\mathrm{/}\mathrm{\hslash }}$. In either case, multiplying by $\mathrm{\parallel }A\mathrm{\parallel }$ proves the stated estimate.

For $t=0$, both sides of the stated estimate vanish. For $t\mathrm{\ne }0$, put $\sigma =\mathrm{sign}(t)\in \{-1,1\}$, and for $s$ between $0$ and $t$, set $G(s)={\alpha }_s^{{\mathrm{\Lambda }}^{\mathrm{\prime }}}({\alpha }_{t-s}^{\mathrm{\Lambda }}(A))$. Differentiating the finite matrix exponentials gives $G^{\mathrm{\prime }}(s)=\frac{i}{\mathrm{\hslash }}{\alpha }_s^{{\mathrm{\Lambda }}^{\mathrm{\prime }}}\text{\hspace{0.17em}⁣}([H_{{\mathrm{\Lambda }}^{\mathrm{\prime }}}-H_{\mathrm{\Lambda }},{\alpha }_{t-s}^{\mathrm{\Lambda }}(A)])\mathrm{.}$ Indeed, the derivative of ${\alpha }_s^{{\mathrm{\Lambda }}^{\mathrm{\prime }}}$ contributes $(i\mathrm{/}\mathrm{\hslash })[H_{{\mathrm{\Lambda }}^{\mathrm{\prime }}},\cdot ]$, while differentiating ${\alpha }_{t-s}^{\mathrm{\Lambda }}$ contributes $-(i\mathrm{/}\mathrm{\hslash })[H_{\mathrm{\Lambda }},\cdot ]$; the outer map ${\alpha }_s^{{\mathrm{\Lambda }}^{\mathrm{\prime }}}$ is unitary conjugation and hence isometric. Integrating from $0$ to $t$, reversing the orientation for $t<0$, and using $u=\mathrm{\mid }t-s\mathrm{\mid }$ gives intermediate evolution time $t-s=\sigma u$. Expanding $H_{{\mathrm{\Lambda }}^{\mathrm{\prime }}}-H_{\mathrm{\Lambda }}$ therefore yields $\mathrm{\parallel }{\alpha }_t^{{\mathrm{\Lambda }}^{\mathrm{\prime }}}(A)-{\alpha }_t^{\mathrm{\Lambda }}(A)\mathrm{\parallel }\le \frac{1}{\mathrm{\hslash }}{\int }_0^{\mathrm{\mid }t\mathrm{\mid }}{\sum }_{\begin{array}{c}Z\subseteq {\mathrm{\Lambda }}^{\mathrm{\prime }}\\ Z\subseteq ̸\mathrm{\Lambda }\end{array}}\mathrm{\parallel }[\mathrm{\Phi }(Z),{\alpha }_{\sigma u}^{\mathrm{\Lambda }}(A)]\mathrm{\parallel }du\mathrm{.}$ Terms with $Z\cap \mathrm{\Lambda }=\mathrm{\varnothing }$ commute with ${\alpha }_{\sigma u}^{\mathrm{\Lambda }}(A)\in {\mathcal{A}}_{\mathrm{\Lambda }}$ and contribute zero; only interaction sets crossing the boundary of $\mathrm{\Lambda }$ remain.

For each crossing $Z$, apply [F2] to $A$ and $\mathrm{\Phi }(Z)$ at the signed time $\sigma u$. Its bound depends on $\mathrm{\mid }\sigma u\mathrm{\mid }=u$, so the same positive integrand estimate holds for either sign of $t$. For any $y\in Z\setminus \mathrm{\Lambda }$, choose $z\in Z$ minimizing the distance from $X$ to $Z$. Since ${\mathrm{diam}}_{\mathrm{\infty }}Z\le R$, the triangle inequality gives ${\mathrm{dist}}_{\mathrm{\infty }}(X,\{y\})\le {\mathrm{dist}}_{\mathrm{\infty }}(X,Z)+R$, so $e^{-\mu {\mathrm{dist}}_{\mathrm{\infty }}(X,Z)}\le e^{\mu R}{e}^{-\mu {\mathrm{dist}}_{\mathrm{\infty }}(X,\{y\})}\mathrm{.}$ Summing over every exterior $y$ overcounts each crossing interaction and therefore gives an upper bound. By the definition of $\kappa$, ${\sum }_{\begin{array}{c}Z\subseteq {\mathrm{\Lambda }}^{\mathrm{\prime }}\\ Z\subseteq ̸\mathrm{\Lambda }\\ Z\cap \mathrm{\Lambda }\mathrm{\ne }\mathrm{\varnothing }\end{array}}\mathrm{\parallel }\mathrm{\Phi }(Z)\mathrm{\parallel }{e}^{-\mu {\mathrm{dist}}_{\mathrm{\infty }}(X,Z)}\le \kappa e^{\mu R}{\sum }_{y\mathrm{\notin }\mathrm{\Lambda }}{e}^{-\mu {\mathrm{dist}}_{\mathrm{\infty }}(X,\{y\})}\mathrm{.}$ The displayed inequality is independent of ${\mathrm{\Lambda }}^{\mathrm{\prime }}$; integrating the time factor in [F2] over an interval of length $\mathrm{\mid }t\mathrm{\mid }$ proves the claimed estimate.

For each $r\ge 1$, the max-metric shell around one lattice site has $(2r+1{)}^d-(2r-1{)}^d$ sites. A finite union bound over $x\in X$ shows that ${\sum }_{y\in {\mathbb{Z}}^d}{e}^{-\mu {\mathrm{dist}}_{\mathrm{\infty }}(X,\{y\})}<\mathrm{\infty }$, since a polynomial shell count is summable against $e^{-\mu r}$. Given $\epsilon >0$, choose a finite set of sites whose omitted tail is below $\epsilon$; every sufficiently large exhausting $\mathrm{\Lambda }$ contains that finite set. Thus the exterior tail tends to zero, uniformly for $\mathrm{\mid }t\mathrm{\mid }\le T$ after replacing $\mathrm{\mid }t\mathrm{\mid }$ by $T$ in the prefactor.

Fix local $A\in {\mathcal{A}}_X$; if its displayed localization is empty, embed the scalar $A\in \mathbb{C}I$ into any one-site algebra, so take $X\mathrm{\ne }\mathrm{\varnothing }$. Let $C_n=\{x\in {\mathbb{Z}}^d:d_{\mathrm{\infty }}(x,0)\le n\}$. For all sufficiently large $n$, $C_n\supseteq X$, and for $m\ge n$ the boundary estimate [F3] bounds $\mathrm{\parallel }{\alpha }_t^{C_m}(A)-{\alpha }_t^{C_n}(A)\mathrm{\parallel }$ by its right side with $\mathrm{\Lambda }=C_n$. That exterior exponential tail tends to zero as $n\to \mathrm{\infty }$, uniformly for $\mathrm{\mid }t\mathrm{\mid }\le T$ because its time factor is bounded there. Hence ${\alpha }_t^{C_n}(A)$ is Cauchy uniformly on compact time intervals; completeness in [A1], [A2], [F1], and [F2] gives a limit ${\alpha }_t(A)\in \mathcal{A}$. Given any other increasing finite exhaustion ${\mathrm{\Lambda }}_j$, choose $n$ so large that both the canonical limit error and the [F3] bound for $C_n$ are below $\epsilon \mathrm{/}2$. Since $C_n$ is finite and ${\mathrm{\Lambda }}_j\uparrow {\mathbb{Z}}^d$, eventually $C_n\subseteq {\mathrm{\Lambda }}_j$; applying [F3] to this nested pair proves ${\alpha }_t^{{\mathrm{\Lambda }}_j}(A)\to {\alpha }_t(A)$ uniformly on $[-T,T]$. Thus the limit is independent of the exhaustion. The finite-matrix premise [A1] is supplied by Finite matrix C-star algebra and density-state duality; this proof uses that earlier result rather than reproving it.

Every finite-volume map is an isometric unital $\ast$-automorphism by conjugation with the unitary matrix exponential in [F4]. Norm limits of these complex-linear maps preserve addition and scalar multiplication, so the limit on local observables is complex-linear. The limits also preserve the unit, adjoints, products on local observables, and norm. For products, if local sequences $A_n\to A$ and $B_n\to B$ are used, submultiplicativity gives $\mathrm{\parallel }{A}_n{B}_n-AB\mathrm{\parallel }\le \mathrm{\parallel }{A}_n\mathrm{\parallel }\mathrm{\parallel }{B}_n-B\mathrm{\parallel }+\mathrm{\parallel }{A}_n-A\mathrm{\parallel }\mathrm{\parallel }B\mathrm{\parallel }$, so multiplication passes continuously to the completion; the adjoint is norm-continuous as well. By density of ${\mathcal{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ and the isometry bound, each ${\alpha }_t$ extends uniquely to an isometric complex-linear unital $\ast$-homomorphism of $\mathcal{A}$.

For local $A$, ${\alpha }_0(A)={\lim }_{\mathrm{\Lambda }}{\alpha }_0^{\mathrm{\Lambda }}(A)=A$, and density extends ${\alpha }_0={\mathrm{id}}_{\mathcal{A}}$ to all $A\in \mathcal{A}$. For local $A$ and fixed $s,t$, choose local $C$ with $\mathrm{\parallel }C-{\alpha }_t(A)\mathrm{\parallel }<\epsilon$. For large $\mathrm{\Lambda }$, both ${\alpha }_t^{\mathrm{\Lambda }}(A)$ is within $\epsilon$ of ${\alpha }_t(A)$ and ${\alpha }_s^{\mathrm{\Lambda }}(C)$ is within $\epsilon$ of ${\alpha }_s(C)$. By isometry, replacing ${\alpha }_t^{\mathrm{\Lambda }}(A)$ with $C$ in ${\alpha }_s^{\mathrm{\Lambda }}{\alpha }_t^{\mathrm{\Lambda }}(A)$ costs at most $2\epsilon$, and replacing ${\alpha }_s^{\mathrm{\Lambda }}(C)$ by ${\alpha }_s(C)$ costs at most $\epsilon$. The finite-volume group law ${\alpha }_s^{\mathrm{\Lambda }}{\alpha }_t^{\mathrm{\Lambda }}(A)={\alpha }_{s+t}^{\mathrm{\Lambda }}(A)$ therefore gives $\mathrm{\parallel }{\alpha }_s{\alpha }_t(A)-{\alpha }_{s+t}(A)\mathrm{\parallel }\le 4\epsilon$ after taking limits. Letting $\epsilon \downarrow 0$ proves the group law on local $A$; density extends it to every $A\in \mathcal{A}$. In particular, ${\alpha }_{-t}$ is the inverse of ${\alpha }_t$, so each is bijective and hence an automorphism.

If $A\in {\mathcal{A}}_X$ is local, finite-volume differentiation gives $\mathrm{\parallel }{\alpha }_t^{\mathrm{\Lambda }}(A)-A\mathrm{\parallel }\le \frac{\mathrm{\mid }t\mathrm{\mid }}{\mathrm{\hslash }}\mathrm{\parallel }[H_{\mathrm{\Lambda }},A]\mathrm{\parallel }\le \frac{2\kappa \mathrm{\mid }X\mathrm{\mid }\mathrm{\parallel }A\mathrm{\parallel }}{\mathrm{\hslash }}\mathrm{\mid }t\mathrm{\mid },$ because only terms meeting $X$ contribute and ${\sum }_{Z\cap X\mathrm{\ne }\mathrm{\varnothing }}\mathrm{\parallel }\mathrm{\Phi }(Z)\mathrm{\parallel }\le \kappa \mathrm{\mid }X\mathrm{\mid }$. Passing to the limit gives continuity at $t=0$ for local $A$. For general $A\in \mathcal{A}$, choose local $C$ close to $A$ and use $\mathrm{\parallel }{\alpha }_t(A)-A\mathrm{\parallel }\le 2\mathrm{\parallel }A-C\mathrm{\parallel }+\mathrm{\parallel }{\alpha }_t(C)-C\mathrm{\parallel }$. The group law gives continuity at every time.

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.