Finite-range Dobrushin uniqueness and boundary influence

Fix a finite $\mathrm{\Lambda }$ and an exterior condition $\xi$. On $A^{\mathrm{\Lambda }}$, define the random-scan heat-bath chain that chooses one site uniformly and resamples it from ${\gamma }_{\{i\}}(\cdot \mid x_{\mathrm{\Lambda }\setminus \{i\}}\xi )$. Properness and consistency imply that the finite-volume law ${\gamma }_{\mathrm{\Lambda }}(\cdot \mid \xi )$ is stationary for this chain: applying a one-site kernel to it leaves it unchanged by ${\gamma }_{\mathrm{\Lambda }}{\gamma }_{\{i\}}={\gamma }_{\mathrm{\Lambda }}$.

Any two stationary laws for the same-boundary chain have a stationary coupling: start with any coupling of them, evolve the pair using a common random site and the maximal coupling [F1], and take a convergent subsequence of Cesàro averages using [F4]. The marginals remain the given stationary laws. If $q_i$ is the stationary disagreement probability, then stationarity makes it equal to the mismatch probability on updating $i$. Telescoping the one-site conditional laws over the finitely many differing internal spins gives $q_i\le {\sum }_{j\in \mathrm{\Lambda }}{c}_{ij}{q}_j$. Therefore ${\max }_i{q}_i\le c{\max }_i{q}_i$, so all $q_i=0$. Thus the finite-volume Gibbs law is the unique stationary law; this proof needs no positivity assumption on the one-site conditionals.

For two boundary conditions $\xi ,\zeta$, start the pair chain at the product of the two finite-volume Gibbs laws and use common-site maximal updates. Its marginal laws remain stationary by step 1.1; Cesàro averaging and [F4] therefore give a stationary coupling whose marginals are those Gibbs laws. For a current pair $X,Y$, telescoping the differing internal and exterior spins one at a time bounds the update mismatch at site $i$ by ${\sum }_{j\in \mathrm{\Lambda }}{c}_{ij}{\mathbf{1}}_{\{X_j\mathrm{\ne }{Y}_j\}}+b_i$, where $b_i={\sum }_{j\mathrm{\notin }\mathrm{\Lambda }}{c}_{ij}{\mathbf{1}}_{\{{\xi }_j\mathrm{\ne }{\zeta }_j\}}$. Under the stationary coupling, if $q_i=\mathrm{Pr}(X_i\mathrm{\ne }{Y}_i)$, random-scan stationarity gives $q_i$ equal to the mismatch probability on updating $i$, hence $q_i\le {\sum }_{j\in \mathrm{\Lambda }}{c}_{ij}{q}_j+b_i$. For $\mathrm{\Lambda }=\mathrm{\varnothing }$ there are no coordinates to bound.

Let $C_{\mathrm{\Lambda }}=(c_{ij}{)}_{i,j\in \mathrm{\Lambda }}$. Iterating the inequality in step 2.1 gives $q\le C_{\mathrm{\Lambda }}^n\mathbf{1}+{\sum }_{k=0}^{n-1}{C}_{\mathrm{\Lambda }}^{k}b$. Every row sum of $C_{\mathrm{\Lambda }}$ is at most $c<1$, so $\mathrm{\parallel }{C}_{\mathrm{\Lambda }}^n\mathbf{1}{\mathrm{\parallel }}_{\mathrm{\infty }}\le c^n\to 0$. A nonzero term of $(C_{\mathrm{\Lambda }}^{k}b{)}_i$ is an influence path of $k$ internal edges followed by one edge to the exterior; each edge changes ${\mathrm{\ell }}^{\mathrm{\infty }}$-distance by at most $R$ by [F3]. For $R\ge 1$ and ${\mathrm{dist}}_{\mathrm{\infty }}(i,{\mathrm{\Lambda }}^c)\ge r(R+1)$, such a path has at least $r+1$ edges, hence $k\ge r$. Since $0\le b_i\le c<1$, its total weight is bounded by ${\sum }_{k\ge r}{c}^k=c^r\mathrm{/}(1-c)$ by [F5]. If $R=0$, all off-diagonal influences vanish and the boundary term is zero, so the same bound holds.

For a bounded local $f$, let ${\delta }_i(f)$ be its maximum change when only spin $i$ changes. Coordinate-by-coordinate coupling gives $\mathrm{\mid }{\gamma }_{\mathrm{\Lambda }}(f\mid \xi )-{\gamma }_{\mathrm{\Lambda }}(f\mid \zeta )\mathrm{\mid }\le {\sum }_{i\in \mathrm{supp}f}{\delta }_i(f)q_i$. Step 3.1 makes this bound tend to zero uniformly over the boundary conditions as $\mathrm{\Lambda }$ exhausts the lattice.

Let $\mu ,\nu$ be DLR states, which exist by [F2]. Integrate the uniform boundary difference from step 4.1 against $\mu (d\xi )\nu (d\zeta )$; using $\mu {\gamma }_{\mathrm{\Lambda }}=\mu$ and $\nu {\gamma }_{\mathrm{\Lambda }}=\nu$ gives $\mu (f)=\nu (f)$ for every bounded local $f$. Cylinder indicators are local, so the measures agree on a generating family and hence are equal.