Translation-averaged stationary Gibbs state

Choose a DLR state $\mu$. The translated interaction list gives $({\gamma }_{\mathrm{\Lambda }}f)\circ {\theta }_j={\gamma }_{\mathrm{\Lambda }+j}(f\circ {\theta }_j)$ for every bounded measurable $f$, by reindexing each anchor and each local spin in the finite sums. Hence $({\theta }_j\mu ){\gamma }_{\mathrm{\Lambda }}(f)=\mu {\gamma }_{\mathrm{\Lambda }+j}(f\circ {\theta }_j)=\mu (f\circ {\theta }_j)=({\theta }_j\mu )(f)$, so every translate is DLR; by convexity in [F1], the cube average ${\bar{\mu }}_n=\mathrm{\mid }{B}_n{\mathrm{\mid }}^{-1}{\sum }_{j\in B_n}{\theta }_j\mu$ is DLR.

For any fixed shift $k=(k_1,\dots ,k_d)\in {\mathbb{Z}}^d$ and bounded local $f$, cancellation of terms indexed by $B_n\cap (B_n+k)$ gives $\mathrm{\mid }{\bar{\mu }}_n(f)-({\theta }_k{\bar{\mu }}_n)(f)\mathrm{\mid }\le 2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }{B}_n\mathrm{△}(B_n+k)\mathrm{\mid }\mathrm{/}\mathrm{\mid }{B}_n\mathrm{\mid }$. For all sufficiently large $n$, $\mathrm{\mid }{B}_n\mathrm{\mid }=(2n+1{)}^d$ and $\mathrm{\mid }{B}_n\cap (B_n+k)\mathrm{\mid }={\prod }_{q=1}^d(2n+1-\mathrm{\mid }{k}_q\mathrm{\mid })$, so the ratio equals $2(1-{\prod }_{q=1}^d(1-\mathrm{\mid }{k}_q\mathrm{\mid }\mathrm{/}(2n+1)))\to 0$.

By [F2], a subsequence of ${\bar{\mu }}_n$ converges on cylinders to a probability $\bar{\mu }$; since each average is DLR, closedness in [F1] makes its limit DLR. Step 2.1 passes to this limit for every local $f$ and fixed $k$, so $\bar{\mu }$ and ${\theta }_k\bar{\mu }$ agree on all cylinders; uniqueness of measures on cylinders in [F2] gives ${\theta }_k\bar{\mu }=\bar{\mu }$. Thus $\bar{\mu }$ is stationary.