High-temperature susceptibility, correlation summability, and spatial-average concentration

For each $\lambda \in I$, [F1]–[F2] give a proper consistent finite-range Ising specification. Its Dobrushin row constant is at most $c_{\ast }<1$, so [A1] and [F4] give a unique DLR state ${\mu }_{\lambda }$. A lattice translation of ${\mu }_{\lambda }$ is another DLR state for the same translation-invariant kernels; uniqueness therefore makes ${\mu }_{\lambda }$ translation invariant.

Fix finite $\mathrm{\Lambda }$ containing distinct sites $i,j$ and an exterior configuration $\eta$, and write ${\nu }_{\mathrm{\Lambda },\lambda }:={\gamma }_{\mathrm{\Lambda },\lambda }(\cdot \mid \eta )$. Every finite Gibbs weight is positive, so the finite-volume probabilities of ${\sigma }_j=\pm 1$ are positive. Conditional on either value, consistency [F2] identifies the law on $\mathrm{\Lambda }\setminus \{j\}$ with the corresponding finite-volume Gibbs kernel, with the same exterior spins away from $j$ and the spin at $j$ fixed to the selected sign. Apply the finite-volume comparison bound [F7] with the same specification and $b=0$ to these two boundary conditions. For $f={\sigma }_i$, ${\delta }_i(f)=2$ and all other oscillations vanish; the bound is $2{\sum }_{k\in \mathrm{\Lambda }\setminus \{j\}}{D}_{ik}^{\mathrm{\Lambda }\setminus \{j\}}{C}_{kj}(\lambda )$. Each term is a sum of weights of paths from the boundary site $j$ to $i$ whose intermediate vertices avoid $j$, so it is at most $2{\sum }_{n\ge 1}C(\lambda {)}_{ij}^n$.

Let ${\mathrm{\Lambda }}_n=[-n,n{]}^d$ and let ${\nu }_{n,\lambda }:={\gamma }_{{\mathrm{\Lambda }}_n,\lambda }(\cdot \mid {\eta }^{+})$, where ${\eta }_i^{+}=+1$ for all exterior sites. For any fixed cylinder function $g$, choose a finite coordinate set $K$ on which it depends (take any nonempty finite $K$ if $g$ is constant). By the DLR identity, ${\mu }_{\lambda }(g)=\int {\gamma }_{{\mathrm{\Lambda }}_n,\lambda }(g\mid \eta )d{\mu }_{\lambda }(\eta )$ whenever $K\subseteq {\mathrm{\Lambda }}_n$. The finite-volume comparison estimate [F7] with the same specification, $b=0$, and arbitrary boundary conditions bounds the difference between ${\gamma }_{{\mathrm{\Lambda }}_n,\lambda }(g\mid {\eta }^{+})$ and every integrand by ${\sum }_{i\in K}{\delta }_i(g)$ times the total influence-path weight from $i$ to ${\mathrm{\Lambda }}_n^c$. If $R\ge 1$, each such path has at least $q_n:=\lfloor {\mathrm{dist}}_{\mathrm{\infty }}(K,{\mathrm{\Lambda }}_n^c)\mathrm{/}R\rfloor$ edges, and its total weight is at most ${\sum }_{k\ge q_n}{c}_{\ast }^k=c_{\ast }^{q_n}\mathrm{/}(1-c_{\ast })$ by [F9]; $q_n\to \mathrm{\infty }$. If $R=0$, the boundary term is zero. Thus ${\nu }_{n,\lambda }(g)\to {\mu }_{\lambda }(g)$ locally uniformly on $I$.

Put $p={\nu }_{\mathrm{\Lambda },\lambda }({\sigma }_j=+1)$ and let $m_{\pm }={\mathbb{E}}_{{\nu }_{\mathrm{\Lambda },\lambda }}[{\sigma }_i\mid {\sigma }_j=\pm 1]$. Finite Gibbs weights are strictly positive by [F1], so $0<p<1$. Expanding the two-point expectation by the events ${\sigma }_j=\pm 1$ gives ${\mathrm{Cov}}_{{\nu }_{\mathrm{\Lambda },\lambda }}({\sigma }_i,{\sigma }_j)=2p(1-p)(m_{+}-m_{-})$. Since $2p(1-p)\le \frac{1}{2}$, step 1.2 gives $\mathrm{\mid }{\mathrm{Cov}}_{{\nu }_{\mathrm{\Lambda },\lambda }}({\sigma }_i,{\sigma }_j)\mathrm{\mid }\le {\sum }_{n\ge 1}C(\lambda {)}_{ij}^n$. For $i=j$, ${\mathrm{Var}}_{{\nu }_{\mathrm{\Lambda },\lambda }}({\sigma }_i)=1-(\mathbb{E}{\sigma }_i{)}^2\le 1$. By step 2.1, for fixed $i,j$ the finite-volume expectations of ${\sigma }_i$, ${\sigma }_j$, and ${\sigma }_i{\sigma }_j$ converge to those under ${\mu }_{\lambda }$ as the cubes exhaust the lattice. Passing this finite covariance bound to the limit proves it for ${\mu }_{\lambda }$ as well.

For every $n\ge 0$, nonnegativity and the row bound give ${\sum }_{j}C(\lambda {)}_{ij}^{n+1}={\sum }_{k}C(\lambda {)}_{ik}^n{\sum }_j{C}_{kj}(\lambda )\le c_{\ast }{\sum }_{k}C(\lambda {)}_{ik}^n$; induction from $C^0=I$ yields ${\sum }_{j}C(\lambda {)}_{ij}^n\le c_{\ast }^n$. Summing the off-site bound from step 3.1 and using [F9] gives ${\sum }_{j\mathrm{\ne }i}\mathrm{\mid }\mathrm{Cov}({\sigma }_i,{\sigma }_j)\mathrm{\mid }\le {\sum }_{n\ge 1}{\sum }_{j}C(\lambda {)}_{ij}^n\le c_{\ast }\mathrm{/}(1-c_{\ast })$. Adding the diagonal variance, at most $1$, proves the absolute covariance sum bound. If $R\ge 1$ and $\parallel j-i{\parallel }_{\mathrm{\infty }}\ge r$, then $C(\lambda {)}_{ij}^n=0$ for $n<\lceil r\mathrm{/}R\rceil$, because an $n$-step path moves at most $nR$ in the ${\mathrm{\ell }}^{\mathrm{\infty }}$ metric. Summing the remaining row tails with [F9] proves the stated spatial-tail bound. If $R=0$, the zero diagonal and range condition make every off-site coefficient zero, so step 3.1 gives zero off-site covariance.

For each fixed finite set of sites, step 2.1 gives locally uniform convergence of its one-point and two-point expectations under ${\nu }_{n,\lambda }$ to those under ${\mu }_{\lambda }$, hence locally uniform convergence of each covariance. For $i\mathrm{\ne }j$ in ${\mathrm{\Lambda }}_n$, steps 1.2 and 3.1 give $\mathrm{\mid }{\mathrm{Cov}}_{{\nu }_{n,\lambda }}({\sigma }_i,{\sigma }_j)\mathrm{\mid }\le {\sum }_{k\ge 1}C(\lambda {)}_{ij}^k$; the diagonal variance is at most $1$. The same path-tail argument as in step 4.1 therefore bounds, uniformly in $n$ and $\lambda$, the sum of absolute covariances over sites outside any fixed box by a quantity tending to zero as the box grows. Splitting the covariance sum into that finite box and its uniformly small complement proves that the finite sums $S_n(\lambda ):={\sum }_{j\in {\mathrm{\Lambda }}_n}{\mathrm{Cov}}_{{\nu }_{n,\lambda }}({\sigma }_0,{\sigma }_j)$ converge locally uniformly on $I$ to $S(\lambda ):={\sum }_{j\in {\mathbb{Z}}^d}{\mathrm{Cov}}_{{\mu }_{\lambda }}({\sigma }_0,{\sigma }_j)$.

For $m_n(\lambda ):={\mathbb{E}}_{{\nu }_{n,\lambda }}{\sigma }_0$, write $H_{0,n}$ for the interaction part of the stated finite-volume energy and put $M_n={\sum }_{j\in {\mathrm{\Lambda }}_n}{\sigma }_j$; the field energy is $H_{0,n}-(\lambda \mathrm{/}\beta )M_n$. [F6] and the chain rule $d\mathrm{/}d\lambda =(1\mathrm{/}\beta )d\mathrm{/}dt$ give $m_n^{\mathrm{\prime }}(\lambda )={\sum }_{j\in {\mathrm{\Lambda }}_n}{\mathrm{Cov}}_{{\nu }_{n,\lambda }}({\sigma }_0,{\sigma }_j)=S_n(\lambda )$. For the aggregate finite-volume identity, [F5] gives $d(\log Z)\mathrm{/}dt=\beta {\mathbb{E}}_t{M}_n$ and $d^2(\log Z)\mathrm{/}d{t}^2={\beta }^2{\mathrm{Var}}_t(M_n)$. Since $t=\lambda \mathrm{/}\beta$, the chain rule gives $d^2(\log Z)\mathrm{/}d{\lambda }^2={\mathrm{Var}}_{{\nu }_{n,\lambda }}(M_n)$; differentiating $d(\log Z)\mathrm{/}d\lambda ={\mathbb{E}}_{{\nu }_{n,\lambda }}{M}_n$ therefore gives $\frac{d}{d\lambda }{\mathbb{E}}_{{\nu }_{n,\lambda }}{M}_n={\mathrm{Var}}_{{\nu }_{n,\lambda }}(M_n)$. Expanding the square in the variance gives the displayed finite double covariance sum.

Step 2.1 with $g={\sigma }_0$ gives $m_n\to m$ locally uniformly, while step 5.1 gives $m_n^{\mathrm{\prime }}=S_n\to S$ locally uniformly. Each $S_n$ is continuous because it is a finite sum of finite Gibbs expectations, each a finite ratio of exponentials; hence $S$ is continuous as a locally uniform limit. On any closed subinterval of $I$, [F8] applied to $G=m_n$ and derivative $S_n$ gives $m_n(\lambda )-m_n({\lambda }_0)={\int }_{{\lambda }_0}^{\lambda }{S}_n(u)du$. Passing to the locally uniform limits yields $m(\lambda )-m({\lambda }_0)={\int }_{{\lambda }_0}^{\lambda }S(u)du$, so $m^{\mathrm{\prime }}(\lambda )=S(\lambda )$. Since $\lambda =\beta h$, the chain rule gives $dm\mathrm{/}dh=\beta S(\beta h)$.

For finite nonempty $\mathrm{\Lambda }$, expand the variance of ${\bar{\sigma }}_{\mathrm{\Lambda }}$ as $\mathrm{\mid }\mathrm{\Lambda }{\mathrm{\mid }}^{-2}{\sum }_{i,j\in \mathrm{\Lambda }}\mathrm{Cov}({\sigma }_i,{\sigma }_j)$. By the uniform row bound proved in step 4.1, the sum of the absolute values in each row is at most $1\mathrm{/}(1-c_{\ast })$, so the variance is at most $1\mathrm{/}((1-c_{\ast })\mathrm{\mid }\mathrm{\Lambda }\mathrm{\mid })$. Pointwise, ${\mathbf{1}}_{\{\mathrm{\mid }X-\mathbb{E}X\mathrm{\mid }\ge \epsilon \}}\le \mathrm{\mid }X-\mathbb{E}X{\mathrm{\mid }}^2\mathrm{/}{\epsilon }^2$; taking expectations with $X={\bar{\sigma }}_{\mathrm{\Lambda }}$ proves the Chebyshev bound.