Uniform exterior-pressure limit for tempered configurations

Put $S_n=\{i\in {\mathbb{Z}}^d:\mathrm{\mid }i{\mathrm{\mid }}_{\mathrm{\infty }}\le n\}$, $v_n=\mathrm{\mid }{S}_n\mathrm{\mid }=V_n$, $N_i(\omega )=\mathrm{\mid }\omega \cap C_i\mathrm{\mid }$, $T_n(\omega )={\sum }_{i\in S_n}{N}_i(\omega {)}^2$, and $T_n^h(\omega )={\sum }_{i\in S_n}h(N_i(\omega ))$. Choose $m_0\ge \lceil R_0+1\rceil$ and $M_0\ge \max \{2B,\psi (m_0-1)\}$. Set ${\gamma }_m=M_0$ for integers $m<m_0$ and ${\gamma }_m=\psi (m-1)$ for $m\ge m_0$. Then $\gamma$ is nonincreasing, dominates the negative part at every cell offset, dominates $\mathrm{\mid }\varphi \mathrm{\mid }$ at offsets $m\ge m_0$, and ${\sum }_{j\in {\mathbb{Z}}^d}{\gamma }_{\mathrm{\mid }j{\mathrm{\mid }}_{\mathrm{\infty }}}<\mathrm{\infty }$ by [F1]–[F2]. Let ${\mathrm{\Delta }}_m={\gamma }_m-{\gamma }_{m+1}\ge 0$ and $v_m=(2m+1{)}^d$. Since ${\gamma }_r={\sum }_{m\ge r}{\mathrm{\Delta }}_m$, Tonelli gives ${\sum }_m{\mathrm{\Delta }}_m{v}_m={\sum }_j{\gamma }_{\mathrm{\mid }j{\mathrm{\mid }}_{\mathrm{\infty }}}<\mathrm{\infty }$. Thus ${\delta }_k:=\frac{1}{2}{\sum }_{m\ge k}{\mathrm{\Delta }}_m{v}_m\to 0$.

For an interior configuration $\omega \subset {\mathrm{\Lambda }}_n$ and any $\zeta \in \mathrm{\Omega }(t)$, the long-offset cell sum satisfies ${\sum }_{i\in S_n,j\in {\mathbb{Z}}^d:\mathrm{\mid }i-j{\mathrm{\mid }}_{\mathrm{\infty }}\ge k}{\gamma }_{\mathrm{\mid }i-j{\mathrm{\mid }}_{\mathrm{\infty }}}{N}_i(\omega )N_j(\zeta )\le {\delta }_k(T_n(\omega )+2^{d}t{V}_n)\mathrm{.}$ Indeed, expand ${\gamma }_r={\sum }_{m\ge r}{\mathrm{\Delta }}_m$ and enlarge the offset set in the $m$th term to $j\in S_m+i$. With $S_{n,m}(\zeta )={\sum }_{j\in {\mathbb{Z}}^d}{N}_j(\zeta {)}^2\mathrm{\#}\{i\in S_n:j\in S_m+i\},$ the inequality $ab\le (a^2+b^2)\mathrm{/}2$ bounds that term by $\frac{1}{2}{\mathrm{\Delta }}_m(v_m{T}_n+S_{n,m})$. The multiplicity is at most $\min (v_n,v_m)$ and vanishes unless $j\in S_{n+m}$, so $S_{n,m}\le \min (v_n,v_m)t{v}_{n+m}\le 2^{d}t{v}_n{v}_m$. Summing proves the estimate.

Set $g(r)=r\sqrt{h(r)}$ for $r\in {\mathbb{N}}_0$ and $g^{\ast }(s)={\sup }_{r\in {\mathbb{N}}_0}\{rs-g(r)\}$. Then $g(r)\mathrm{/}{r}^2\to \mathrm{\infty }$ and $g(r)\mathrm{/}h(r)\to 0$, so $g^{\ast }(s)\mathrm{/}{s}^2\to 0$: for each $a>0$, eventually $g(r)\ge r^2\mathrm{/}(4a)$, and completing the square gives $g^{\ast }(s)\le a{s}^2+C_a$. Also $rs\le g(r)+g^{\ast }(s)$. For a fixed $k\ge m_0$, every near cross pair has its inside cell in $S_n\setminus S_{n-k}$ and its exterior cell in $S_{n+k}\setminus S_n$; its negative energy is at most $M_0$. Counting the finitely many offsets of size below $k$, for every $a>0$ we obtain $W_{\mathrm{n}\mathrm{e}\mathrm{a}\mathrm{r}}^{-}\le C_k(a{T}_n^h+C_a\mathrm{\mid }{S}_n\setminus S_{n-k}\mathrm{\mid }+at{v}_{n+k}+C_a\mathrm{\mid }{S}_{n+k}\setminus S_n\mathrm{\mid })\mathrm{.}$ Here $g(r)\le ah(r)+C_a$ and $g^{\ast }(s)\le a{s}^2+C_a$ were used, with ${\sum }_{j\in S_{n+k}\setminus S_n}{N}_j(\zeta {)}^2\le t{v}_{n+k}$. First choose $k$ large in step 2.1 so that its far negative-energy bound is at most $ϵ(T_n^h+V_n)\mathrm{/}2$. For this fixed $k$, both boundary-cell counts are $O_k(n^{d-1})$ and $v_{n+k}\mathrm{/}{v}_n\to 1$, so choosing $a$ small and then $n$ large makes the near bound at most $ϵ(T_n^h+V_n)\mathrm{/}2$. The far estimate uses $T_n\le T_n^h+q^2{V}_n$ for a fixed $q$ with $h(r)\ge r^2$ when $r\ge q$. Consequently, for each $0<ϵ<1$ and all sufficiently large $n$, $W_n(\omega ,\zeta )\ge -ϵT_n^h(\omega )-ϵV_n$ uniformly over $\zeta \in \mathrm{\Omega }(t)$.

For the lower bound, restrict interior particles to ${\mathrm{\Lambda }}_{n-k}$, leaving the corridor empty. Choose $k\ge m_0$, so every interior-exterior cell offset is in the far tail. The two-sided estimate of step 2.1 gives $\mathrm{\mid }W(\omega ,\zeta \setminus {\mathrm{\Lambda }}_n)\mathrm{\mid }\le {\delta }_k(T_{n-k}(\omega )+2^{d}t{V}_{n-k})\mathrm{.}$ With $T_{n-k}\le T_{n-k}^h+q^2{V}_{n-k}$, choose $k$ large enough that the right side is at most $ϵ(T_{n-k}^h+V_n)$. Since $T_{n-k}^h\le H_{n-k,\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}}+bN$, integration over the restricted positions yields ${\mathrm{\Xi }}_{n,\zeta }(z,\beta )\ge e^{-\beta ϵV_n}\mathrm{\Xi }(2(n-k)+1;z{e}^{-\beta ϵb},\beta (1+ϵ))$ uniformly over $\zeta$. For fixed $ϵ$, the selected $k$ is fixed as $n\to \mathrm{\infty }$, and the removed corridor volume fraction vanishes.

By [F3], $T_n^h\le H_{n,\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}}+bN$. For each $0<ϵ<1$ and all sufficiently large $n$, step 3.1 therefore gives ${\mathrm{\Xi }}_{n,\zeta }(z,\beta )\le e^{\beta ϵV_n}{\sum }_{N\ge 0}\frac{z^N}{N!}{\int }_{{\mathcal{S}}_N({\mathrm{\Lambda }}_n)}{e}^{-\beta H_{n,\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}}+\beta ϵT_n^h}d{q}^N$ and ${\sum }_{N\ge 0}\frac{z^N}{N!}{\int }_{{\mathcal{S}}_N({\mathrm{\Lambda }}_n)}{e}^{-\beta H_{n,\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}}+\beta ϵT_n^h}d{q}^N\le \mathrm{\Xi }(2n+1;z{e}^{\beta ϵb},\beta (1-ϵ))\mathrm{.}$ The inequality is uniform over $\zeta \in \mathrm{\Omega }(t)$.

Divide the bounds in steps 4.1–3.2 by $V_n$ and let $n\to \mathrm{\infty }$; [F4] supplies the perturbed free-cube pressure limits. These limits are finite convex functions of $(\beta ,\log z)$ by Hölder's inequality [F5], since finite-volume log partition functions are convex and pointwise limits preserve convexity. A finite convex function is continuous on the open parameter domain $\beta >0$, $z>0$. Let $ϵ\downarrow 0$ in both uniform bounds. Their upper and lower limits coincide with $p(z,\beta )$ uniformly over $\mathrm{\Omega }(t)$. For $t=0$, every exterior configuration in $\mathrm{\Omega }(0)$ is empty, so the equality with the free-wall pressure is immediate.