Boundary Duhamel estimate for thermodynamic spin dynamics

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.