Finite-range lattice log-partition limit

Write $V=L^d$ and $J_0={\sum }_a\mathrm{\parallel }{\mathrm{\Phi }}_a{\mathrm{\parallel }}_{\mathrm{\infty }}$. Each configuration satisfies $\mathrm{\mid }{H}_L(s)\mathrm{\mid }\le J_{0}V$, so $q^V{e}^{-\beta J_{0}V}\le Z_L\le q^V{e}^{\beta J_{0}V}$. Hence $p_L$ is bounded above and below uniformly in $L$.

Fix an integer block length $\mathrm{\ell }>2R$ and write $L=k\mathrm{\ell }+r$ with $0\le r<\mathrm{\ell }$ using [F3]. If $k<0$, then $L=k\mathrm{\ell }+r\le -\mathrm{\ell }+r<0$, contrary to $L\ge 0$, so $k\ge 0$. For the limiting estimate take $L\ge \mathrm{\ell }$, hence $k\ge 1$. Partition ${\mathrm{\Lambda }}_L$ into $k^d$ disjoint $\mathrm{\ell }$-cubes and an outer remainder. Form $H_{L,\mathrm{\ell }}^{\mathrm{c}\mathrm{u}\mathrm{t}}$ by retaining only interactions whose entire support lies in one full cube and assigning zero energy to remainder sites. For a fixed shape, every eligible anchor lies in $[1-R,L+R{]}^d$. If its support crosses an internal block seam, its coordinate normal to that seam lies in at most $2R+1$ integer positions; the other coordinates have at most $(L+2R{)}^{d-1}$ choices. If it is not in such a seam strip but the interaction is omitted, its support meets the outer remainder, so its anchor lies within $\mathrm{\ell }+2R+1$ layers of one of the $d$ outer faces. Counting at most $d(k-1)$ internal seams and $d$ outer faces bounds the omitted anchors per shape by $d(2R+1)(L\mathrm{/}\mathrm{\ell })(L+2R{)}^{d-1}+d(\mathrm{\ell }+2R+1)(L+2R{)}^{d-1}$. Therefore ${\sup }_s\mathrm{\mid }{H}_L(s)-H_{L,\mathrm{\ell }}^{\mathrm{c}\mathrm{u}\mathrm{t}}(s)\mathrm{\mid }\le J_{0}d[(2R+1)(L\mathrm{/}\mathrm{\ell })+\mathrm{\ell }+2R+1](L+2R{)}^{d-1}$.

If two energies on a finite state set differ pointwise by at most $D$, their partition sums satisfy $e^{-\beta D}{Z}^{\mathrm{\prime }}\le Z\le e^{\beta D}{Z}^{\mathrm{\prime }}$, so their log partition sums differ by at most $\beta D$ using [F2]. The cut energy factorizes over the cubes and has zero-energy remainder, hence $Z_{L,\mathrm{\ell }}^{\mathrm{c}\mathrm{u}\mathrm{t}}=Z_{\mathrm{\ell }}^{k^d}{q}^{L^d-k^d{\mathrm{\ell }}^d}$. Divide the comparison by $L^d$ and let $L\to \mathrm{\infty }$ with $\mathrm{\ell }$ fixed; the remainder fraction tends to zero, $k^d\mathrm{/}{L}^d\to {\mathrm{\ell }}^{-d}$, and the energy error has limsup at most $2^{d-1}\beta J_{0}d(2R+1)\mathrm{/}\mathrm{\ell }$.

It follows for every $\mathrm{\ell }>2R$ that ${\mathrm{lim\hspace{0.17em}sup}}_L{p}_L\le p_{\mathrm{\ell }}+2^{d-1}\beta J_{0}d(2R+1)\mathrm{/}\mathrm{\ell }$ and ${\mathrm{lim\hspace{0.17em}inf}}_L{p}_L\ge p_{\mathrm{\ell }}-2^{d-1}\beta J_{0}d(2R+1)\mathrm{/}\mathrm{\ell }$. Thus $0\le {\mathrm{lim\hspace{0.17em}sup}}_L{p}_L-{\mathrm{lim\hspace{0.17em}inf}}_L{p}_L\le 2^d\beta J_{0}d(2R+1)\mathrm{/}\mathrm{\ell }$ for every such $\mathrm{\ell }$. Letting $\mathrm{\ell }\to \mathrm{\infty }$ makes the difference zero; boundedness from step 1.1 then gives a finite limit.

For ${\tilde{H}}_L$, the finite-state comparison of step 2.1 gives $\mathrm{\mid }{L}^{-d}\log {\tilde{Z}}_L-p_L\mathrm{\mid }\le \beta C_b\mathrm{/}L$, which tends to zero. Its limit is therefore the same.