Interacting Ensemble Geometry

Since $\psi$ is nonincreasing, $0\le g^d\psi (g)\le \frac{d}{1-2^{-d}}{\int }_{g\mathrm{/}2}^g{r}^{d-1}\psi (r)dr$ for $g>2R_0$: on $[g\mathrm{/}2,g]$, $\psi (r)\ge \psi (g)$, and integration of $r^{d-1}$ gives the displayed coefficient. Both the interval integral and the tail integral in $\eta (g)$ tend to zero as $g\to \mathrm{\infty }$, so $\eta (g)\to 0$.

For $m\ge 2$ and $r\in [(m-1)g,mg]$, monotonicity gives $\psi (r)\ge \psi (mg)$ and $r^{d-1}\ge ((m-1)g{)}^{d-1}$. Hence ${\int }_{(m-1)g}^{mg}{r}^{d-1}\psi (r)dr\ge g^d(m-1{)}^{d-1}\psi (mg)\mathrm{.}$ Since $m^{d-1}\le 2^{d-1}(m-1{)}^{d-1}$, summing over disjoint intervals yields ${\sum }_{m\ge 2}{m}^{d-1}\psi (mg)\le 2^{d-1}{g}^{-d}{\int }_g^{\mathrm{\infty }}{r}^{d-1}\psi (r)dr$. The $m=1$ term equals $g^{-d}{g}^d\psi (g)$, so adding it proves the stated bound because $2^{d-1}\ge 1$.

For integer $m\ge \lceil R_0+3\rceil$, monotonicity gives $\psi (r)\ge \psi (m-1)$ on $[m-2,m-1]$, so ${\int }_{m-2}^{m-1}{r}^{d-1}\psi (r)dr\ge (m-2{)}^{d-1}\psi (m-1)$. Since $m^{d-1}\le c_d(m-2{)}^{d-1}$ for these $m$, summing over disjoint unit intervals proves ${\sum }_{m\ge \lceil R_0+3\rceil }{m}^{d-1}\psi (m-1)<\mathrm{\infty }$.

For each integer $k\ge 1$, let $G_k$ be the least integer greater than $2R_0$ such that $\eta (g)\le k^{-2d}$ for every $g\ge G_k$; it exists by step 1.1 and the well-ordering of the positive integers. Choose a strictly increasing sequence $L_k$ with $L_k\ge k{G}_k$ and $L_k\ge k^2$. For $\mathrm{\ell }\ge L_1$, let $k(\mathrm{\ell })=\max \{k:L_k\le \mathrm{\ell }\}$ and $g(\mathrm{\ell })=\mathrm{\ell }\mathrm{/}k(\mathrm{\ell })$. Then $k(\mathrm{\ell })\to \mathrm{\infty }$, $g(\mathrm{\ell })\ge G_{k(\mathrm{\ell })}$ and $g(\mathrm{\ell })\ge k(\mathrm{\ell })$, while $g(\mathrm{\ell })\mathrm{/}\mathrm{\ell }=1\mathrm{/}k(\mathrm{\ell })$. Therefore ${\left(\frac{\mathrm{\ell }}{g(\mathrm{\ell })}\right)}^d\eta (g(\mathrm{\ell }))\le k(\mathrm{\ell }{)}^{d}k(\mathrm{\ell }{)}^{-2d}=k(\mathrm{\ell }{)}^{-d}\to 0.$ This gives all three gap limits.

Fix distinct points $x,y$ in a finite configuration and a period $L>0$. For $\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}$ larger than $2\mathrm{\mid }x-y{\mathrm{\mid }}_{\mathrm{\infty }}\mathrm{/}L+2R_0\mathrm{/}L$, the triangle inequality gives $\mathrm{\mid }x-y+Lk{\mathrm{\mid }}_{\mathrm{\infty }}\ge L\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}\mathrm{/}2>2R_0$. Grouping lattice indices by $m=\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}$ and applying the sum bound of step 1.2 with gap $g=L\mathrm{/}2$ shows ${\sum }_k\mathrm{\mid }\varphi (x-y+Lk)\mathrm{\mid }<\mathrm{\infty }$ whenever $L\mathrm{/}2>2R_0$. There are finitely many remaining images; any $+\mathrm{\infty }$ term makes the periodic energy $+\mathrm{\infty }$ by definition, while otherwise their finite sum is well-defined. This proves the periodic-image convention used in Superstable regular pair interactions and configurational ensemble notation for all sufficiently large $L$.

Choose an integer $k\ge 1$ with $1\mathrm{/}k\le r_0$. Partition each unit cell $C_i$ into $k^d$ half-open subcubes of side $1\mathrm{/}k$, and denote their occupancies by $n_{i,j}$, $1\le j\le k^d$. Distinct points in one subcube have $\mathrm{\mid }x-y{\mathrm{\mid }}_{\mathrm{\infty }}<1\mathrm{/}k\le r_0$, so their repulsive pair energy is at least $\delta$. Therefore $H_{{\varphi }_r}(\omega )\ge \frac{\delta }{2}{\sum }_{i,j}{n}_{i,j}(n_{i,j}-1)\mathrm{.}$ Writing $\bar{n}=N_{C_i}(\omega )\mathrm{/}{k}^d$, nonnegativity of ${\sum }_j(n_{i,j}-\bar{n}{)}^2$ expands to ${\sum }_j{n}_{i,j}^2\ge N_{C_i}(\omega {)}^2\mathrm{/}{k}^d$, and ${\sum }_{i,j}{n}_{i,j}=N(\omega )$. Thus $H_{{\varphi }_r}(\omega )\ge \frac{\delta }{2k^d}{\sum }_i{N}_{C_i}^2-\frac{\delta }{2}N$.

The stable part obeys $H_{{\varphi }_s}(\omega )\ge -B_{s}N$. Add this inequality to step 1.1 and take $A=\delta \mathrm{/}(2k^d)$ and $B=B_s+\delta \mathrm{/}2$. This proves the cell-quadratic bound. If any term is $+\mathrm{\infty }$, the inequality holds in the extended order; otherwise all sums are finite.

By [F1], $R_{L,\omega }$ is the push-forward of normalized Lebesgue probability on ${\mathrm{\Lambda }}_L$, so it is a probability measure. For a bounded measurable local $f$ and $y\in {\mathbb{R}}^d$, $R_{L,\omega }(f\circ T_y)=L^{-d}{\int }_{{\mathrm{\Lambda }}_L}f(T_{x+y}{\omega }^{(L)})dx\mathrm{.}$ The integrand is $L{\mathbb{Z}}^d$-periodic in $x$. The translated cube ${\mathrm{\Lambda }}_L+y$ is another fundamental region for this lattice; partitioning it and ${\mathrm{\Lambda }}_L$ into their finitely many translated intersections and using translation invariance of Lebesgue measure gives equal integrals. Hence $R_{L,\omega }(f\circ T_y)=R_{L,\omega }(f)$ for every $y$, so $R_{L,\omega }$ is stationary.

For bounded Borel $A$, Tonelli's theorem and the periodic continuation give $R_{L,\omega }(N_A)=L^{-d}{\int }_{{\mathrm{\Lambda }}_L}{\sum }_{q\in \omega }{\sum }_{k\in {\mathbb{Z}}^d}{\mathbf{1}}_A(q+Lk+x)dx\mathrm{.}$ For each fixed $q\in \omega$, the sets ${\mathrm{\Lambda }}_L+Lk$ partition ${\mathbb{R}}^d$ up to null boundaries. Translating variables in each integral therefore makes the inner sum of integrals equal to $\mathrm{\mid }A\mathrm{\mid }$. Summing over the $N$ representatives gives $R_{L,\omega }(N_A)=N\mathrm{\mid }A\mathrm{\mid }\mathrm{/}{L}^d$. Taking $A=[0,1{)}^d$ proves $\rho (R_{L,\omega })=N\mathrm{/}{L}^d$ and finite intensity.

The barycenter ${\bar{C}}_L$ is a probability mixture of the stationary laws $R_{L,\omega }$, so it is stationary. By step 2.1 and Tonelli, $\rho ({\bar{C}}_L)=E_{C_L}[R_{L,\omega }(N_{[0,1{)}^d})]=E_{C_L}N\mathrm{/}{L}^d<\mathrm{\infty }\mathrm{.}$ Thus it belongs to the finite-intensity stationary domain of ${\tau }_L$.

Write $N_i=N_{C_i}(\omega )$. The negative part of each pair potential is bounded both by $2B$, where $B$ is the stability constant of $\varphi$, and, at large separation, by the lower-regular envelope. Set ${\mathrm{\Psi }}_j$ to a finite constant on the finitely many near-cell offsets and to the envelope evaluated at $\mathrm{\mid }j{\mathrm{\mid }}_{\mathrm{\infty }}-1$ on the remaining offsets. The shell count $\mathrm{\mid }\{j:\mathrm{\mid }j{\mathrm{\mid }}_{\mathrm{\infty }}=m\}\mathrm{\mid }\le c_d{m}^{d-1}$ and the cell-envelope conclusion of [F2] give ${\sum }_j{\mathrm{\Psi }}_j<\mathrm{\infty }$. Since $N_i{N}_j\le (N_i^2+N_j^2)\mathrm{/}2$, the total negative pair contribution is at most $C{\sum }_i{N}_i^2$ for a finite $C$.

In case (i), choose an integer $k_0\ge 2$ with $1\mathrm{/}(k_0-1)\le r_1$. Then $\chi (1\mathrm{/}k)$ and $\chi (1\mathrm{/}(k-1))$ are both defined for every $k\ge k_0$. Partition one unit cell into $k^d$ half-open cubes of side $1\mathrm{/}k$, put ${\chi }_k=\chi (1\mathrm{/}k)$, and $\mathrm{\Delta }{\chi }_k={\chi }_k-{\chi }_{k-1}\ge 0$. If the cell contains $N$ points, the identity ${\sum }_j{n}_j^2-N=2{\sum }_j\left(\genfrac{}{}{0px}{}{n_j}{2}\right)$ and nonnegativity of ${\sum }_j(n_j-N\mathrm{/}{k}^d{)}^2$ give at least $\frac{1}{2}(N^2\mathrm{/}{k}^d-N)$ same-subcube pairs at scale $k$. A fixed pair counted at scale $k$ has separation $s<1\mathrm{/}k$, hence $k<1\mathrm{/}s$; summing its nonnegative weights over all scales where it is counted is at most ${\sum }_{k=k_0}^{\lfloor 1\mathrm{/}s\rfloor }\mathrm{\Delta }{\chi }_k\le \chi (s)$. Thus the positive core energy in the cell dominates the sum of these weighted pair counts. With $K_N=\lfloor (N\mathrm{/}2{)}^{1\mathrm{/}d}\rfloor$, each $k\le K_N$ satisfies $N^2\mathrm{/}{k}^d-N\ge N^2\mathrm{/}(2k^d)$, so the core energy is at least $\frac{N^2}{4}{\sum }_{k=k_0}^{K_N}\frac{\mathrm{\Delta }{\chi }_k}{k^d}\mathrm{.}$ Summation by parts writes the coefficient as an initial constant plus ${\sum }_{k=k_0}^{K_N-1}{\chi }_k(k^{-d}-(k+1{)}^{-d})$; and $k^{-d}-(k+1{)}^{-d}\ge d\mathrm{/}(2^{d+1}{k}^{d+1})$. The series ${\sum }_k\chi (1\mathrm{/}k)k^{-d-1}$ diverges: on dyadic intervals $[2^{-j-1},2^{-j}]$, monotonicity bounds the integral above by $C_d{2}^{-jd}\chi (2^{-j-1})$, while the corresponding block $2^{j+1}\le k<2^{j+2}$ contributes at least $c_d{2}^{-jd}\chi (2^{-j-1})$ to that series. The divergent core integral therefore forces the displayed coefficient to tend to infinity.

In case (i), sum the cell core bound from step 1.2 and subtract the negative-pair bound from step 1.1. The resulting coefficient of $N_i^2$ tends to infinity with $N_i$. Above a finite threshold it dominates $2C{N}_i^2$; below that threshold its possible negative remainder is bounded by a constant times $N_i$. Taking $h(N)$ to be half the positive cell lower bound above the threshold and zero below it gives a nondecreasing extended-valued function with $h(N)\mathrm{/}{N}^2\to \mathrm{\infty }$ and $H_{\varphi }\ge {\sum }_{i}h(N_i)-bN$.

In case (ii), choose $k$ with $1\mathrm{/}k<r_h$ and partition every unit cell into $k^d$ subcubes. A finite-energy configuration has at most one point in each subcube, so $N_i\le k^d$ for every $i$. Define $h(N)=0$ for $N\le k^d$ and $h(N)=+\mathrm{\infty }$ for $N>k^d$. If any cell exceeds this occupancy, the configuration has a forbidden pair and $H_{\varphi }=+\mathrm{\infty }$; otherwise stability gives $H_{\varphi }\ge -BN={\sum }_{i}h(N_i)-BN$. This proves the hard-core alternative.

Stability and [F3] give $Z(L,N;\beta )\le e^{\beta BN}{L}^{dN}\mathrm{/}N!$. If $N\mathrm{/}{L}^d\to \rho >0$, the factorial integral bounds [F4] imply $\mathrm{lim\hspace{0.17em}sup}{L}^{-d}\log Z(L,N;\beta )\le \rho (1+\beta B-\log \rho ),$ so $b(\rho )$ and $F(\rho )$ are bounded above on every compact positive-density interval. The vacuum term gives $b(0)=0$: $Z(L,0)=1$ is a lower bound, and the same stability estimate with $N\mathrm{/}{L}^d\to 0$ gives a nonpositive upper limit.

The effective domain of $F$ has nonempty interior. Fix a core cube of side $s>0$ and a gap $g>2R_0$; place one particle in each selected core of a lattice of spacing $s+g$ inside a large box. The absolute interaction from any one occupied core to all others is bounded by $C{\sum }_{m\ge 1}{m}^{d-1}\psi (mg)<\mathrm{\infty }$ by [F1]–[F2], with the gap satisfying the stated hypothesis of the diagonal-gap estimate. Restricting each particle to a subcube of volume $v_0>0$ therefore gives $Z(L,N;\beta )\ge v_0^N{e}^{-C_{0}N}$ for every density strictly below $(s+g{)}^{-d}$, after omitting unused lattice cells. Thus $F$ is finite on a nonempty interval of positive densities; by its definition as a finite-mixture envelope it is concave.

Fix a finite list of core preparations $(s_i,n_i)$ with $0<Z(s_i,n_i;\beta )<\mathrm{\infty }$. For each $i$, choose a gap $g_i=g(s_i)$ from [F2] along a sequence $s_i\to \mathrm{\infty }$ so that $g_i\mathrm{/}{s}_i\to 0$ and the positive interaction between distinct cores in the same lattice packing, divided by core volume, tends to zero. Set tile side $a_i=s_i+g_i$ and effective density $r_i=n_i\mathrm{/}{a}_i^d$. If $M_i$ such tiles carry exactly $n_i$ particles in their central $s_i$-cubes, the same shell count as in the proof of Free-cube grand-pressure limit for decreasing integrable tails gives cross-energy penalty per occupied volume at most $C(n_i\mathrm{/}{s}_i^d{)}^2(s_i\mathrm{/}{g}_i{)}^d\eta (g_i)$, which tends to zero by [F2]. Exact factorization over disjoint core cubes follows because assigning the labeled coordinates to their cores contributes $N!\mathrm{/}{\prod }_i(n_i!{)}^{M_i}$, cancelling the global $N!$.

For any finite convex combination in the definition of $F(\rho )$, choose core preparations with densities tending to its ${\rho }_i$ and rates tending to $b({\rho }_i)$. If their effective densities do not strictly bracket the target $\rho$, add arbitrarily small weights at one attainable density below and one above $\rho$; such densities exist because $\rho$ is interior. Their mixture coefficients can then be perturbed by quantities tending to zero so the effective mean is exactly $\rho$. Place the fixed core types in finitely many macroscopic slabs with these volume fractions, inserting fixed empty layers wider than $R_0$ between slabs. For fixed core sizes and gaps, the distinct-slab cross energy divided by total volume tends to zero: for interaction distances at most $R$, only $O(R{L}^{d-1})$ particles lie in the interface strips, while the remaining contribution is bounded by a constant times $L^d{\sum }_{\mathrm{\mid }j{\mathrm{\mid }}_{\mathrm{\infty }}>R}{\mathrm{\Psi }}_j$; take $L\to \mathrm{\infty }$ and then $R\to \mathrm{\infty }$.

First reserve an outer layer of fixed width $W$. For each large $L$, put ${\rho }_L=N_L\mathrm{/}{L}^d$ and choose the slab fractions in the remaining volume to match the count $N_L$ at density ${\rho }_L{L}^d\mathrm{/}(L^d-O(W{L}^{d-1}))$. This target tends to $\rho$, so the strict bracketing in step 3.1 applies uniformly for large $L$ and absorbs the reserved layer without assuming its one-particle capacity replaces a bulk density. Rounding the finitely many slab and core counts then leaves at most $C_1{L}^{d-1}$ particles unplaced, with $C_1$ depending on the fixed preparations but not on $W$; the error comes only from the finitely many slab faces and incomplete boundary tiles. Choose $W$ large enough to contain at least $C_1{L}^{d-1}$ small cells separated from each other and every core by more than $R_0$. Put each missing particle in one such cell. Its interaction with the fixed preparations is bounded by the same summable tail, so the correction changes log weight by only $O(L^{d-1})$. This constructs exactly $N_L$ particles for every sequence $N_L\mathrm{/}{L}^d\to \rho$, and gives ${\mathrm{lim\hspace{0.17em}inf}}_{L\to \mathrm{\infty }}{L}^{-d}\log Z(L,N_L;\beta )\ge {\sum }_i{\theta }_i\frac{\log Z(s_i,n_i;\beta )-{\mathrm{c}\mathrm{r}\mathrm{o}\mathrm{s}\mathrm{s}}_i}{a_i^d},$ where ${\mathrm{c}\mathrm{r}\mathrm{o}\mathrm{s}\mathrm{s}}_i\mathrm{/}{a}_i^d\to 0$ by step 2.1. Taking preparations approaching the selected convex combination and then the supremum gives $\mathrm{lim\hspace{0.17em}inf}\ge F(\rho )$.

By the definition of $b$, every sequence with density tending $\rho$ has limsup rate at most $b(\rho )\le F(\rho )$. Together with step 4.1 this proves the asserted limit. A finite concave function on an open interval is continuous: fix $a<\rho <b$ in the interval; the chord slopes from $a$ and $b$ bound the difference quotients on a smaller interval around $\rho$, yielding a finite local Lipschitz bound. This establishes finiteness and continuity of $F$ on its effective interior while leaving excluded endpoints undecided.