All-positive-temperature one-dimensional Ising DLR uniqueness

Let $r$ be the positive unit eigenvector of $T$ and $w$ an orthogonal unit eigenvector. By [F1] and the eigenvalue formula, $T^n\mathrm{/}{\lambda }_{+}^n=r{r}^{\mathsf{T}}+({\lambda }_{-}\mathrm{/}{\lambda }_{+}{)}^{n}w{w}^{\mathsf{T}}$, with $\mathrm{\mid }{\lambda }_{-}\mathrm{/}{\lambda }_{+}\mathrm{\mid }<1$. Since both entries of $r$ are positive, the denominator entries have a uniform positive lower bound after division by ${\lambda }_{+}^n$.

Fix an interval $[-n,n]$, endpoint spins $a,b$, and a consecutive observed block $s_{\mathrm{\ell }},\dots ,s_r$. Its conditional Gibbs probability is $\frac{(T^{n+\mathrm{\ell }+1}{)}_{a,s_{\mathrm{\ell }}}\left({\prod }_{i=\mathrm{\ell }}^{r-1}{T}_{s_i,s_{i+1}}\right)(T^{n-r+1}{)}_{s_r,b}}{(T^{2n+2}{)}_{a,b}}\mathrm{.}$ Using the uniform spectral expansion in step 1.1, this converges uniformly in $a,b$ to ${\pi }_{s_{\mathrm{\ell }}}{\prod }_{i=\mathrm{\ell }}^{r-1}{Q}_{s_i,s_{i+1}}$. Finite summation over missing intermediate spins gives the same convergence for every cylinder.

The limiting cylinder law is the Markov law $\mu$ in [F1]. The specification in [F3] has the exact finite conditional energy used by the interval laws. By [F2], a subsequential limit of those finite-volume laws is DLR; step 2.1 forces every such limit to equal $\mu$, so $\mu$ is DLR.

If $\nu$ is any DLR state, then for every cylinder $C$, $\nu (C)=\nu {\gamma }_{[-n,n]}(C)$. The conditional probability inside this integral differs uniformly in the exterior endpoints from $\mu (C)$ by a quantity tending to zero by step 2.1. Thus $\nu (C)=\mu (C)$; [F2] gives $\nu =\mu$.