Continuum Pressure — Examples

From the coordinate formula in [F1], $\mathrm{\parallel }x{\mathrm{\parallel }}_2=\sqrt{{\sum }_{r=1}^d{x}_r^2}$ is continuous: each coordinate square is continuous by $u^2-v^2=(u-v)(u+v)$, and for $u,v\ge 0$, $\mathrm{\mid }\sqrt{u}-\sqrt{v}\mathrm{\mid }\le \sqrt{\mathrm{\mid }u-v\mathrm{\mid }}$, since, if $u\ge v$, $(\sqrt{u}-\sqrt{v}{)}^2\le (\sqrt{u}-\sqrt{v})(\sqrt{u}+\sqrt{v})=u-v$, with the other case symmetric. Thus the square-root composition is continuous, including at $x=0$. On ${\mathbb{R}}^d\setminus \{0\}$, the function $x\mapsto \mathrm{\parallel }x{\mathrm{\parallel }}_2^{-d-ϵ}$ is therefore continuous by [F2]. The complement of $\{0\}$ is Borel and $\{0\}$ is closed, so the piecewise function is Borel; it is even because $\mathrm{\parallel }-x{\mathrm{\parallel }}_2=\mathrm{\parallel }x{\mathrm{\parallel }}_2$. For $\mathrm{\parallel }x{\mathrm{\parallel }}_2\ge 1$ it satisfies the tail bound with $C=1$ and $R=1$.

Every pair term is nonnegative or $+\mathrm{\infty }$. Thus every finite pair sum satisfies $U_N\ge 0$, so the stability constant is $B=0$. This proves the claimed model-class membership.

From the coordinate formula in [F1], $\mathrm{\parallel }x{\mathrm{\parallel }}_2=\sqrt{{\sum }_{r=1}^d{x}_r^2}$ is continuous: each coordinate square is continuous by $u^2-v^2=(u-v)(u+v)$, and for $u,v\ge 0$, $\mathrm{\mid }\sqrt{u}-\sqrt{v}\mathrm{\mid }\le \sqrt{\mathrm{\mid }u-v\mathrm{\mid }}$, since, if $u\ge v$, $(\sqrt{u}-\sqrt{v}{)}^2\le (\sqrt{u}-\sqrt{v})(\sqrt{u}+\sqrt{v})=u-v$, with the other case symmetric. Thus the square-root composition is continuous, including at $x=0$. The map $r\mapsto r^{-p}$ is continuous on $r>0$ by [F2]. The regions $\{\mathrm{\parallel }x{\mathrm{\parallel }}_2<1\}$ and $\{\mathrm{\parallel }x{\mathrm{\parallel }}_2\ge 1\}$ are Borel, so the piecewise potential is Borel; radiality makes it even. For $\mathrm{\parallel }x{\mathrm{\parallel }}_2\ge 1$, $\mathrm{\mid }{\varphi }_{\mathrm{a}\mathrm{t}\mathrm{t}}(x)\mathrm{\mid }=c\mathrm{\parallel }x{\mathrm{\parallel }}_2^{-d-ϵ}$, so it is power-tempered with $C=c$ and $R=1$.

Put $\delta :=1\mathrm{/}(2\sqrt{d})$. For a fixed particle $q_i$, the half-open cubes $D_k:=q_i+\delta k+[0,\delta {)}^d$, $k\in {\mathbb{Z}}^d$, partition ${\mathbb{R}}^d$. Their Euclidean diameters are at most $\sqrt{d}\delta =1\mathrm{/}2<1$. The example's definition makes every pair at distance less than $1$ forbidden, so centers in a finite-energy configuration are separated by at least $1$. Thus each cube contains at most one center and $D_0$ contains no center other than $q_i$. The cells in max-norm shell $m$ number $(2m+1{)}^d-(2m-1{)}^d$ by [F3]. For $m\ge 1$, factor the difference of powers: $(2m+1{)}^d-(2m-1{)}^d=2{\sum }_{r=0}^{d-1}(2m+1{)}^{d-1-r}(2m-1{)}^r\le 2d{3}^{d-1}{m}^{d-1},$ since $2m+1\le 3m$.

A neighbor of $q_i$ in shell $m=1$ has distance at least $1$ and contributes at most $c$ in absolute value; there are $3^d-1$ such cells. For $m\ge 2$, choose a coordinate $r$ with $\mathrm{\mid }{k}_r\mathrm{\mid }=m$. If $k_r=m$, every point of $D_k$ differs from $q_i$ in coordinate $r$ by at least $\delta m$; if $k_r=-m$, the absolute difference is greater than $\delta (m-1)$. Hence every point of $D_k$ is at Euclidean distance at least $\delta (m-1)$ from $q_i$. By [F2], each such neighbor contributes at most $c{\delta }^{-p}(m-1{)}^{-p}$. The shell count in step 1.2 therefore bounds its total contribution by $2d{3}^{d-1}{m}^{d-1}c{\delta }^{-p}(m-1{)}^{-d-ϵ}\mathrm{.}$

For $m\ge 2$, $m-1\ge m\mathrm{/}2$, so the last bound is at most $2d{3}^{d-1}c{\delta }^{-p}{2}^p{m}^{-1-ϵ}$. Summing over shells and using [F4], every fixed particle satisfies ${\sum }_{j\mathrm{\ne }i}\mathrm{\mid }{\varphi }_{\mathrm{a}\mathrm{t}\mathrm{t}}(q_i-q_j)\mathrm{\mid }\le C_{\ast }:=c(3^d-1)+2d{3}^{d-1}c{\delta }^{-p}{2}^p{\sum }_{m=2}^{\mathrm{\infty }}{m}^{-1-ϵ}<\mathrm{\infty }\mathrm{.}$

For a finite-energy configuration, every pair term is negative and $-U_N={\sum }_{i<j}\mathrm{\mid }{\varphi }_{\mathrm{a}\mathrm{t}\mathrm{t}}(q_i-q_j)\mathrm{\mid }=\frac{1}{2}{\sum }_i{\sum }_{j\mathrm{\ne }i}\mathrm{\mid }{\varphi }_{\mathrm{a}\mathrm{t}\mathrm{t}}(q_i-q_j)\mathrm{\mid }\le \frac{C_{\ast }}{2}N$ by step 3.1. A configuration containing a forbidden pair has $U_N=+\mathrm{\infty }$ and also satisfies the lower bound. Hence the potential is stable with $B=C_{\ast }\mathrm{/}2$. Together with step 1.1, this proves membership in the stated stable power-tempered class.