A radial continuum phase transition with algebraic decay

Proof. Disjoint balls of radius 1/3 about the points lie in [-1/3,L+1/3]^3, so M_L\le C(L+1)^3. This also bounds the possible cardinalities; compactness of the configuration space for each cardinality shows that the maximum is attained. The integer lattice gives a positive lower asymptotic density.

Fix s>0 and repeat a maximizing configuration in [0,s]^3 by translations in (s+1)\mathbb Z^3. Points in distinct translates are separated by at least one: in a coordinate in which the translates differ, their coordinate separation is at least (s+1)-s=1. Taking all complete translates contained in [0,L]^3 gives

Letting s\to\infty along a sequence realizing the upper asymptotic density makes the right side tend to that upper density. The lower and upper limits therefore agree.□

Proof. If five points formed a clique, rescale their distances by l_t^{-1} and take the four difference vectors from one of them. Their Gram matrix differs entrywise by at most 2\Delta+\Delta^2 from the matrix with diagonal one and off-diagonal entries 1/2. For a diagonal entry this follows from 1\le r^2\le(1+\Delta)^2; for an off-diagonal entry use polarization, \ip{v_i}{v_j}=(\abs{v_i}^2+\abs{v_j}^2-\abs{v_i-v_j}^2)/2. The reference matrix has smallest eigenvalue 1/2. The perturbation has operator norm at most 4(2\Delta+\Delta^2)<1/2, so the Gram matrix would be positive definite and have rank four. Four vectors in \R^3 cannot have this Gram matrix.

For the edge bound, maximize \sum_{\{i,j\}\in E}u_i u_j over u_i\ge0, \sum_i u_i=1, choosing a maximizer with the smallest possible support. If two nonadjacent vertices have positive weights, the objective is affine in their two weights with their sum fixed. Putting their combined weight on one of the vertices cannot decrease the objective, contradicting minimality of the support. The support is therefore a clique of some size r\le4. Its objective is

For m>0, apply this bound to equal weights 1/m. The empty graph case is immediate.□

Proof. The diameter \sqrt3 b of a small cell is less than one. By Lemma 2, among m particles in one cell at least m(m-1)/2-3m^2/8=m^2/8-m/2 pairs have distance outside I_t. Each of these pairs contributes t^2, while all remaining core contributions are nonnegative. Summing over cells proves the first bound, and also proves it when core terms between cells are deleted. For the attractive contribution, including extra diagonal self terms only enlarges the bound, and symmetry gives

This proves (2.6).

There are at most (L/b+2)^3\le216V nonempty cells. Hence \sum_i n_i^2\ge N^2/(216V). Under (2.7), D_0-M/2\ge t^2/16, giving (2.8) with the stated c. The requirements on t are compatible: M grows linearly in t, whereas D_0 grows quadratically. The constants c and c_b are independent of t and of the attraction ranges.□

Proof. The lower bound follows by bounding the integrand from above using (2.8) and using n!\ge(n/\ee)^n for n\ge1. For the other bound, let X_1,\ldots,X_n be independent uniform points in \Lambda_L. Since U^\varphi\le U^{\psi_t} and \int_{\R^3}\psi_t(\abs{x})\,\dd x\le t^2\abs{B(0,2)},

Jensen’s inequality gives \log Z^\varphi_{L,n}\ge n\log L^3-\log n!-\beta Ct^2n^2/L^3. Now use n!\le n^n. At n=0 both bounds are equalities.□

Proof. Enclose \mathcal D in \beta\in[\beta_-,\beta_+] and \abs{h}\le H. Put V=L^3. By (2.10), the logarithm of the summand at n\ge1 is at most

The choice (2.7) implies ct-\zeta/2\ge ct/2. Set \kappa=\beta_-ct^2/2 and B=tH+\beta_+t^2/2+1. For n/V\ge1 the displayed expression is at most -\kappa n^2/V+Bn. Choose D\ge\max(1,2B/\kappa). For n>DV this is at most -\kappa n^2/(2V)\le-(\kappa D/2)n. Summing this geometric majorant proves the assertion with a_0=\kappa D/2.□

Proof. We first prove (2.12), without using a pressure limit or an approximation by finite-range potentials. At d=0, both bounds in (2.10) tend to zero. Fix d>0 and \beta>0. Let H_* be the lower limit of H^\varphi_{L,n}(\beta) as L\to\infty and n/L^3\to d, and choose s_j\to\infty and integers n_j with

The finite volume bounds make H_* finite. Consider any other sequence L\to\infty, n/L^3\to d. For each fixed, sufficiently large j, place k=\lfloor L/(s_j+2)\rfloor^3 cubes of side s_j in \Lambda_L, separated by gaps of width at least two. For any nonnegative integers m_1,\ldots,m_k summing to n, restriction to configurations with these box counts gives

Indeed, interactions between different boxes are nonpositive. There are n!/\prod_i m_i! disjoint assignments of particle labels to the boxes, so the global factor 1/n! leaves exactly the product of their factors 1/m_i!.

We retain the favorable count n_j in most boxes and correct the total particle number in the remaining boxes. Only a vanishing fraction of boxes will change, and (2.10) will bound their canonical costs uniformly. To make this exact for every target count, put

For all sufficiently large j, n_j\ge c_ds_j^3 and q_j-n_j\ge c_ds_j^3. Start with n_j particles in every box. If E=n-kn_j\ge0, fill boxes successively to q_j, with at most one last box given an intermediate integer count. If E<0, empty boxes successively, again with at most one last intermediate count. This changes at most r boxes, where

For fixed large j and then sufficiently large L, the target lies between 0 and kq_j: its mean count divided by s_j^3 tends to d(1+2/s_j)^3<q_j/s_j^3. Thus the adjustment never exhausts the available boxes. The quotient and remainder in division of E by q_j-n_j, or of -E by n_j, give the asserted integer counts.

All counts used have density in [0,4d+4]. By (2.10), their normalized canonical costs have absolute value at most some K<\infty, uniformly for large j. Taking logarithms in (2.15) consequently yields

First let L\to\infty with j fixed. The prefactor tends to (s_j/(s_j+2))^3, and the upper limit of r/k is at most

Then let j\to\infty. This error tends to zero and the preceding upper bound tends to H_*. Every sequence has lower limit at least H_* by its definition; thus every sequence has the same limit. This proves (2.12) with the required order of limits.

For density convexity, fix d_1,d_2\ge0 and \theta\in[0,1]. Choose s_j\to\infty, integers m_{r,j} with m_{r,j}/s_j^3\to d_r for r=1,2, and integers q_j\to\infty. In a cube of side q_j(s_j+2), use q_j^3 separated boxes of side s_j, assigning count m_{1,j} to \lfloor\theta q_j^3\rfloor of them and count m_{2,j} to the rest. The total density tends to \theta d_1+(1-\theta)d_2. The product inequality and the already proved all-sequence limit, applied directly to this diagonal sequence, give

This proof does not presuppose continuity in density. A finite convex function is continuous at positive densities: on any interior compact interval its secant slopes are bounded by secants to two exterior points. Continuity at zero follows from (2.10). Hölder’s inequality makes \log Z^\varphi_{L,n}(\beta) convex in \beta. Passing to the canonical limit proves the stated concavity.

The grand sums are finite by Lemma 4, with \zeta=0. Fix (\beta,h). Choose D there and write V=L^3. Above DV the entire sum tends to zero. Among the remaining \lfloor DV\rfloor+1 terms choose a maximizing count n_L; the maximum is at least one because it includes the vacuum. Therefore

for all sufficiently large L. From any subsequence choose a further subsequence with n_L/V\to d\in[0,D]. The canonical limit bounds its upper limit by hd-H_\varphi(\beta,d). Conversely, for every fixed d\ge0 the single term n=\lfloor dV\rfloor bounds the lower limit by hd-H_\varphi(\beta,d). This proves existence of the pressure and the first duality in (2.14).

The vacuum gives Q_\varphi\ge0. The canonical lower bound and d\log d-d\ge-1 bound the first supremum in (2.14) by a finite quadratic maximum, uniformly over the family on any fixed parameter compact. Hölder’s inequality on the disjoint union of all configuration spaces makes each \log\Xi^\varphi_L jointly convex in (\beta,h); its pointwise limit has the same property.

Finally fix d>0. Convexity of the finite function H_\varphi(\beta,\cdot) gives a supporting slope h there. Explicitly, one may choose h between

The left endpoint is at most the right endpoint by convexity, and both are finite by comparison with secants through fixed points on either side of d. The resulting supporting inequality holds for all v\ge0, so the first duality gives Q_\varphi(\beta,h)=hd-H_\varphi(\beta,d). For all other h the same first duality gives the opposite variational inequality. This proves the second equality in (2.14) without a differentiability or uniqueness assumption.□

Proof. If the region meets no small cell, its grand measure is the vacuum and the assertion is immediate. Otherwise, let the number of cells be positive. Write \alpha=D_0-M/2, u_0=\beta D_0/4, and enclose \mathcal D in \beta\in[\beta_-,\beta_+], \abs{h}\le H. We first check, before taking any maximum, that F_i:=\mathbb E\exp(u_0n_i^2) is finite in each fixed volume. By (2.6), insertion of this exponential into (3.1) bounds its numerator by the product of k one-variable series with exponents

Each spatial integral contributes at most \prod_j(b^3)^{n_j}, with the factors 1/n_j! already in (3.1). The quadratic coefficients are positive in magnitude because

Thus all these series are finite, and the denominator \Xi^v is at least one. In particular, F_{\max}:=\max_iF_i<\infty. To bound this maximum independently of the number of cells, we shall prove

where C_2 depends only on t and \mathcal D. The ratio of attraction strength to core repulsion will therefore make the dependence on neighboring moments sublinear.

Condition on the configuration Y outside cell i. For n particles inside that cell, the part of the energy involving at least one inside particle is bounded below by

This uses the complete core estimate inside the cell, discards nonnegative core terms across cells, and bounds all attractions by their cell suprema. Dividing the conditional weights by the outside-only Boltzmann factor leaves a denominator containing the empty-cell term one. Hence the conditional exponential moment is at most its unnormalized numerator. The inequality

and the factor \exp(u_0n^2) leave the quadratic coefficient D_0/2-J_{ii}/2\ge7D_0/16. Consequently

where the finite constant

depends only on t and \mathcal D and is at least one.

To close this estimate, Cauchy–Schwarz with the weights J_{ij} and the row bound give

Set \theta_j=J_{ij}/M for j\ne i and \theta_\varnothing=1-\sum_{j\ne i}\theta_j\ge0. With the exponent \eta_0 chosen above, \beta G_i^2/D_0\le\eta_0u_0\sum_{j\ne i}\theta_jn_j^2. Apply convexity of the exponential with the additional zero argument of weight \theta_\varnothing, and then the concavity of x\mapsto x^{\eta_0}, to obtain

where the last inequality uses F_{\max}\ge1. Taking expectations in (3.4) and then the maximum over cells yields F_{\max}\le C_2F_{\max}^{\eta_0}, and therefore

This bound is independent of the number of cells. Since \exp(u_0n_i^2)\ge u_0n_i^2 and u_0\ge\beta_-D_0/4, the second moments are uniformly bounded. Cauchy–Schwarz gives the mixed moment bound in (3.2).

Cropping a cell only decreases its volume and the relevant suprema. Every deletion or interpolation allowed in (2.4) retains the full intra-cell core and the same row bound. All steps above therefore apply without change to those systems.□

Proof. Differentiation in \theta is justified directly in finite volume. The empty region is immediate. Put \alpha=D_0-M/2>0. If there are k\ge1 cells and the region has volume V, the common energy bound gives U^{v_\theta}\ge\alpha N^2/k-B_0N, while boundedness of the pair difference gives \abs{\partial_\theta U^{v_\theta}}\le C N^2. Uniformly in \theta and on \mathcal D, the differentiated grand series is therefore dominated by a constant times

The volume and the constant in this domination are fixed only for the differentiation; the resulting estimate below is uniform in volume. Hence

For every configuration,

where the diagonal uses n_i(n_i-1)\le n_i^2. Apply Proposition 3 and integrate over \theta\in[0,1].□

Proof. Interpolate the whole tail at once:

Even for infinitely many terms, the tail converges uniformly and is a bounded continuous pair interaction, of supremum norm at most t\norm{w}_\infty m_k. Thus the derivative exists for each finite configuration and the domination (3.6) justifies differentiation directly, including the infinite endpoint. All intermediate attractions obey the original mass bound.

By the proof of (2.3), the difference matrix can be chosen with

There are at most 216L^3 cells. Dividing (3.5) by tL^3 gives the claimed finite volume upper bound, for instance with C_{\mathcal D,t}=108\beta_+C_1c_b. Its lower bound follows because adding a nonnegative attraction decreases the energy.

Proposition 2 already supplies thermodynamic pressure limits for both endpoints, including the infinite sum. Passing to those limits proves (3.7); convergence of truncation pressures is a consequence, not an input, of this estimate.□

Proof. The pressure duality in Proposition 2 gives, for every u\in\R,

Taking the supremum in u proves (4.2) initially with a supremum. To prove attainment and uniformity, use the lower bound of Lemma 3 in the thermodynamic limit. Since -d\log d+d\le1 for d\ge0, its density objective is at most

This upper bound tends uniformly to -\infty as d\to\infty on every parameter compact, whereas the objective at d=0 is zero. Continuity of H(\beta,\cdot), including at zero, proves attainment and a common bound on all maximizers. Dividing (4.5) by \beta t gives the upper bound

Its coefficients are uniformly bounded for t\ge t_0>1 and \lambda in a fixed compact. Thus the maximizing-density bound is also uniform in sufficiently large t.

If d maximizes (4.2), setting u=d in (4.4) makes u active. Conversely, fix an active u. The density supremum defining Q(\beta,h+\beta au) is attained: its fixed shifted field and the canonical quadratic bound give coercivity. For an attaining density d, (4.4) can equal T_aQ(\beta,h) only if d maximizes (4.2) and (u-d)^2=0. This proves the claimed exact identity of maximizing sets and also attainment in (4.1).

For fixed u, the branch

is convex: its parameter map is invertible and linear, and its last term is linear. The supremum of these branches is convex and is finite by (4.5). If d is a maximizing density, pressure duality and the nonnegativity of the added quadratic give

This proves (4.3), with C_{\mathcal D}=\tfrac12\sup_{\mathcal D}\beta\, U^2. The canonical bound and pressure duality bound Q uniformly on every compact at fixed t; (4.3) does the same for T_aQ. Convexity then gives the asserted local Lipschitz bounds.□

Proof. Compare the active shift u with u+r/(\beta a) in (4.1). Expanding the two quadratic penalties proves (4.6). If (e,d)\in\partial Q(x_u), the supporting inequality gives

Divide separately for r>0 and r<0 and let r\to0 to obtain d=u.□

Proof. For fixed u, put L_u(\beta,h)=(\beta,h+\beta au). The defining support inequalities, applied to L_u and its inverse, give the exact formula

If u is active, Lemma 5 sets d=u, giving the generators in (4.7). Since F_u\le T_aQ everywhere and F_u(\lambda)=T_aQ(\lambda), each such generator supports T_aQ. The convex hull of the displayed union is therefore contained in the left side.

Denote that union by G(\lambda). The active set is bounded by Proposition 4 and closed by continuity of F_u(\lambda) in u, so it is compact. Its shifted points form a compact subset of \Omega. The boundedness and closed graph of \partial Q on this set show that G(\lambda) is compact. Its convex hull is compact as well: in \R^2, each point of a convex hull is a convex combination of at most three points, by eliminating affine dependencies from any larger finite combination.

For the reverse inclusion, fix v_0\in\partial(T_aQ)(\lambda) and a direction z\in\R^2. For small r>0, let \lambda_r=\lambda+rz, choose an active shift u_r there, and choose v_r\in\partial F_{u_r}(\lambda_r). The branch support at the endpoint gives

All u_r are uniformly bounded near \lambda; their shifted points stay in one compact subset of \Omega, and the underlying subgradients of Q are bounded there. Along a sequence r\downarrow0, pass to convergent subsequences of u_r and these subgradients. Continuity of Q and T_aQ makes the limiting shift active at \lambda, and the closed subgradient graph makes the limiting v_r a member of G(\lambda). Thus (4.8) implies

This holds in every direction. Separation from the compact convex set \conv G(\lambda) proves v_0\in\conv G(\lambda).□

Proof. Put S_0=t\sum_j\alpha_jw_{R_j} and S_{\mathrm{new}}=ta w_R. The series for S_0 converges uniformly, is supported in the ball of radius R_0, and has the row bound (2.3), including when there are countably many reference terms. Multiply the reference interaction on pairs belonging to distinct large boxes by a parameter \theta\in[0,1]. Every intermediate interaction retains all intra-small-cell core terms, and its attractive part is bounded by S_0+S_{\mathrm{new}}. This has small-cell row sum at most M. Therefore Proposition 3 and Corollary 1 apply throughout this interpolation, whether or not the intermediate interaction is translation invariant.

The absolute difference is bounded on every pair by

Its row sum over small cells is at most C_b t^2+c_b t(A+1): the core is bounded by t^2 and supported within distance two, and the attraction has the common row bound already proved. This estimate uses an absolute majorant and thus remains valid when \varphi_0 changes sign on a pair of cells.

A nonzero difference row must belong to a small cell within distance R_0+\sqrt3 b of the boundary of its large box. Since R_0\ge2 and s\ge R_0, there are at most C_b s^2R_0 such small cells per large box. The bound follows either by subtracting the volume of the inner cube when the strip is thin, or from s^3\le C s^2R_0 when the strip fills a fixed fraction of the box. Thus the sum of all difference majorants is at most

The finite-volume interpolation bound of Corollary 1, divided by t(ms)^3, proves (5.3).

For a=0 the deleted system consists of independent reference boxes, and its grand partition function is exactly

For completeness, with prescribed box counts (n_z) of total N, the N!/\prod_z n_z! assignments of labels to boxes cancel the global 1/N! and leave \prod_z1/n_z!. Summing the nonnegative product integrals over all (n_z) gives (5.4), including the zero-count terms. Its normalized logarithm equals Q^s_{\varphi_0} for every m. Letting m\to\infty in (5.3) and using only the pressure limit of the full reference proves (5.2).□

Proof. Symmetry follows from w_R(x-y)=w_R(y-x). Let G=\norm{\nabla w}_\infty, so \Lip(w_R)\le GR^{-4}. For a fixed y\in B_{z'}, each displacement x-y' with x\in B_z, y'\in B_{z'} differs from z-y by at most 3s. Consequently

On the infinite grid, at most C(R/s)^3 second boxes can have either term nonzero: all others lie outside the support enlarged by a distance 3s. Integrating over y\in B_{z'} and summing thus gives

The integral is one. This comparison includes boxes meeting internal zero regions of w or crossing its support boundary, because the same global Lipschitz estimate applies there.

Restricting the upper sum to a finite collection only decreases it. For a center with the stated boundary distance, all potentially contributing second boxes are still present in the lower sum. Omitting z'=z loses at most s^3K^-_{zz}\le\norm w_\infty(s/R)^3, which is bounded by \norm w_\infty s/R. Increasing the one constant C_w proves both claims.□

Proof of Theorem 5.1. We first bound the particle counts needed for the upper estimate. For (\beta,h)\in\mathcal D, s\ge1, V=s^3, and n\ge0, put

The reference \varphi_0 belongs to (2.2), and the auxiliary quadratic coefficient satisfies a(1+\eta)\le2(A+1). Apply Lemma 4 on \mathcal D with \zeta=a(1+\eta). It supplies a common cutoff D_{\mathrm{cut}}\ge1 and, after increasing a volume threshold V_0\ge1, the bound

Both D_{\mathrm{cut}} and V_0 depend only on \mathcal D,t. They are fixed before any local occupation constants: the tail lemma uses only the canonical quadratic bound, even with this auxiliary weight.

Let U be the common active-shift bound on \mathcal D from Proposition 4. Choose one compact rectangle \mathcal D'\subset\Omega containing \mathcal D and every point

Only now fix the moment and deletion constants of Proposition 3 and Lemma 6 on \mathcal D'. All these constants are independent of the eventual box side and of the reference.

Put q=R/R_0 and choose

A single lower bound on q, uniform for R_0\ge2, ensures

Partition \Lambda_L, L=ms, into m^3 boxes B_z, and delete reference interactions between different boxes as in Lemma 6. The new attraction remains intact. All limits in m below hold with s,R,R_0, the reference, and the parameters fixed.

Upper bound. Let N_z be the particle count in B_z, and write W_R=\sum_{i<j}w_R(x_i-x_j). Adding nonnegative diagonal self terms, then using symmetry and 2N_zN_{z'}\le N_z^2+N_{z'}^2, gives

Splitting labels among boxes as in (5.4), we obtain

For 0\le n\le D_{\mathrm{cut}}s^3, let d=n/s^3. The canonical term at count n is one of the nonnegative terms in the reference grand sum at the shifted field h+\beta ad. Therefore

All shifted parameters lie in \mathcal D' by (5.8), so the finite-box comparison applies; the bound d\le D_{\mathrm{cut}} controls the extra quadratic. The argument includes n=0, whose term equals one. There are \lfloor D_{\mathrm{cut}}s^3\rfloor+1 such counts; the remaining terms have total at most one by (5.7). Since T_aQ_{\varphi_0}\ge0, taking logarithms in (5.11) yields

Combine this with (5.3) and let m\to\infty in the full physical pressure. This proves the desired upper estimate with the three displayed errors.

Lower bound. Fix an active shift u\in[0,U] for T_aQ_{\varphi_0}(\beta,h). Let \Pi be the product of the m^3 reference grand measures in the boxes, all at (\beta,h+\beta au), and put

The common small-cell moment bound on \mathcal D' gives a uniform bound d_s\le C_{\mathcal D,t}. Indeed each box consists of exactly (s/b)^3 small cells, and Cauchy–Schwarz bounds each mean count by the square root of its uniformly bounded second moment.

The change of field and insertion of the new attraction give the exact identity

where N=\sum_zN_z. The box factorials in \Pi are precisely those obtained by splitting the full grand measure among boxes. All terms are finite by the common quadratic stability bound. Jensen’s inequality therefore gives

Discard the nonnegative contributions to W_R from pairs in one box. For the remaining pairs the kernel infima give

Entire distinct boxes are independent under \Pi, so for z\ne z', \mathbb E_\Pi[N_zN_{z'}]=s^6d_s^2. No independence inside a box is used. Let m_{\mathrm{int}} be the number of centers with distance greater than R+4s from the outer boundary. Nonnegativity of the other rows and Lemma 7 imply

For fixed s,R, m_{\mathrm{int}}/m^3\to1 as m\to\infty. Consequently (5.14) and (5.3), using the known limit of the full physical model, show that

The exact square identity

and (5.2) at the shifted parameter now give

In particular no convergence of d_s to u, or convergence of d_s at all, is required. This proves the same bound at reference coexistence parameters.

Finally, (5.10) gives R_0/s\le2q^{-1/2} and s/R\le q^{-1/2}. For s\ge1,

where the last step uses R_0\ge2 and s\ge\tfrac12R_0\sqrt q. Equations (5.12) and (5.16), with the deletion cost, prove (5.1).

No step imposes a positive lower bound on a, and no error has been divided by a. Only the reference and full physical models require thermodynamic limits; the deleted systems were treated by exact factorization or finite-volume inequalities.□

Proof. Fix a compact parameter set K\subset\Omega, and choose \beta_->0 and H<\infty such that \beta\ge\beta_- and \abs{h}\le H on K. Throughout the proof V=L^3, and I_t=[l_t,(1+\Delta)l_t].

Free configurations and the upper bound. Call a pair bad when its distance is less than one and is not in I_t. A configuration is free if it has no bad pair. In a free configuration, join distinct particles whose distance is less than one. Every edge then has length at most (1+\Delta)l_t. The endpoints of a path of two edges are therefore at distance at most 2(1+\Delta)l_t<1, and hence are joined. Each connected component is consequently a clique with all pair distances in I_t. By Lemma 2, its size is at most four. Choosing one representative from every component gives a set with pairwise distances at least one, so the component count is at most M_L.

Consider the grand integral over free configurations with their energy discarded:

For an ordered list of component sizes m_1,\ldots,m_k, with m_i\in\{1,2,3,4\} and \sum_i m_i=n, the number of ordered partitions of the particle labels into these blocks is n!/\prod_i m_i!. On summing over all ordered size lists, division by k! counts each unordered family of label blocks once. This remains true when sizes repeat: the label blocks themselves are distinct and have exactly k! orderings.

In a block of size m, choose its least label as the root. Its position has volume at most V, and each of the other m-1 particles lies in a ball of volume C l_t^3 about it, where C=4\pi(1+\Delta)^3/3. After cancellation of the global factor 1/n!, the block therefore contributes at most V(C l_t^3)^{m-1}/m!. Discarding all further restrictions within and between components and summing their sizes gives

To pass from free configurations to all configurations, let B(X) be the set of labels incident to a bad pair. There are at least \abs{B(X)}/2 bad pairs, and each has core energy exactly t^2. All other core terms are nonnegative, and X\setminus B(X) is free. Thus the following pointwise bound holds:

Indeed, the summand J=B(X) already bounds the left side. For r=\abs{J} and m=n-r, the label factor is \binom{m+r}{r}/(m+r)!=1/(r!m!). The removed positions integrate to V^r. Tonelli’s theorem and the grand sum now give the exact factorization of the upper bound

Shell endpoints are included in I_t, although their energy may be positive. Pairs at distance exactly one are not edges of the component graph. Both conventions merely discard nonnegative energy in the upper bound, so these arguments hold pointwise, including at the endpoints.

Set a_t(h)=t^{-1}\log\Theta_t(h). The four exponential rates in \Theta_t are

The middle two are convex combinations of the first and last. Their positive prefactors in (6.2) are fixed, so

uniformly for \abs{h}\le H. For k\ge1, the inequality k!\ge(k/\ee)^k, with x=k/V, gives

since x(1-\log x)\le1 for x>0. The same final bound covers k=0. There are only M_L+1=O(1+V) terms in (6.2). First letting L\to\infty, with t fixed, and using Lemma 1 and Proposition 2, we obtain

The final term is at most t^{-1}\ee^{tH-\beta_-t^2/2} on K. Thus (6.5) proves the required uniform upper bound.

Zero-energy configurations and the lower bound. Fix s>0 and 0<\nu\le1. Dilate a maximizing packing in a side-s cube by 1+\nu, and center it in a cube of side

Repeat this pattern in adjacent period cubes. Centers in one period have separation at least 1+\nu; centers in different periods have at least this separation as well. Every center has distance at least 1+\nu from the boundary of its period cube. A cube of side mP contains exactly k=m^3M_s centers.

Choose r\in(0,\nu/16) and let b_0=4\pi r^3/3. At every center z, put the root particle in B(z,r). There are two constructions, with cluster size j=1 or j=4. For j=4, choose vectors a_2,a_3,a_4\in\R^3 such that 0,a_2,a_3,a_4 are the vertices of a regular tetrahedron of side 1+\Delta/2. Fix \delta\in(0,\Delta/24) and require

A root-to-vertex distance differs from (1+\Delta/2)l_t by at most \delta l_t, and another edge differs by at most 2\delta l_t. All six distances consequently lie strictly inside

where the core vanishes. Write b_1=4\pi\delta^3/3>0. The change from positions to the root and the three relative vectors has Jacobian one. For one prescribed label ordering the integration volume at each center is therefore b_0(b_1l_t^3)^3, with positive relative volume b_1^3l_t^9. For j=1, the corresponding volume is b_0.

Every particle lies within r+2l_t of its center. For all sufficiently large t, depending on \nu,r but not on m,\beta,h, these center neighborhoods are disjoint and lie inside their period cubes, and intercluster distances are greater than 1+\nu/2>1+1/t. Hence all configurations just described have exactly zero core energy.

The centers are distinguishable. Assign an unordered set of j particle labels to each of the k centers, and retain just one deterministic ordering of each assigned set, for example the increasing order. There are (jk)!/(j!)^k such assignments. Different assignments send some label to different disjoint center neighborhoods, so their integration events are disjoint. Possible alternative choices of the root within the same set are never summed. Dividing by the canonical factorial (jk)! gives

Using this term in the grand sum and first letting m\to\infty, we find

The vacuum also gives q^{(t)}\ge0. The constants in (6.8) have no dependence on \beta,h, and c_{s,\nu} can be made arbitrarily close to p by taking s large and \nu small. More explicitly, on \abs{h}\le H the two functions jh-3(j-1), j=1,4, have a common finite absolute bound. Given any error tolerance, first choose s,\nu so that replacing c_{s,\nu} by p changes both branches by less than that tolerance, and then take t large enough to make the two logarithmic terms small. Taking the maximum of these branches and zero preserves the uniform error, even where a nonvacuum branch is negative. Together with (6.6), this proves (6.1) on K.□

Proof. Apply the uniform-in-t bound in Proposition 4 to the bare core \psi_t, mass A, and parameter compact K. For all sufficiently large t, every active shift lies in one fixed interval [0,U]. Enlarge U, if necessary, so that 4p\le U.

Write d_1=0,d_2=p,d_3=4p. Completing the square gives the identity

Its supremum over u is P_*(\lambda), and all its maximizing values belong to D_*. The same identity proves (6.10).

The shifted points

form a compact subset of \Omega. Put E_t=\sup_{K'}\abs{q^{(t)}-q_*}, which tends to zero by Proposition 6. Both transforms can be maximized over [0,U], so

If u is active for P_1^t(\lambda), approximation of its objective and of the maximum shows F_*(\lambda,u)\ge P_*(\lambda)-2E_t. By (6.10),

This estimate is uniform on K and requires no strict inequality between the three affine branches.□

Proof. Because g is an interior point of the nondegenerate triangle, the two numbers

are positive. For the first assertion, one may also see positivity directly: if all three inner products were nonpositive, their sum would be zero, so all would vanish, contradicting noncollinearity. Choose

For any unit vector z and any allowed vertices w_i,

The support-function characterization of inclusion of compact convex sets proves (6.14). Every point of the hull of two radius-\tau balls lies within distance \tau of the segment joining their centers. Since \tau<\zeta, this also proves (6.15). Finally, all three planes agree at \lambda_*, and hence for z=\lambda-\lambda_*,

□

Proof. Use the uniform optimizer bound [0,U] from Lemma 8 with K=W_1. All shifted points x=(\beta,h+\beta Au) lie in the fixed compact K' from that proof. Choose one \sigma>0 such that

and set E'_t=\sup_{K''}\abs{q^{(t)}-q_*}\to0. For every (e,d)\in\partial q^{(t)}(x), its two supporting inequalities at x+(\sigma,0) and x-(\sigma,0), together with the independence of q_* from \beta, imply

These are fixed steps on an enlarged compact, so the estimate also covers outer parameters on the boundary of W_1.

At a shifted point corresponding to an active u, Lemma 5 gives d=u for every underlying support. By Proposition 5, the actual active generators of \partial P_1^t(\lambda) are precisely the vectors in the union

For a nearest d_i\in D_*, every such generator w satisfies

Equations (6.12) and (6.20) show that this tends uniformly to zero over all outer parameters, all active optimizers, and all underlying supports. Thus, for all sufficiently large t,

Taking convex hulls proves one inclusion in (6.19); the other follows from convexity of the subdifferential. Finally, (6.18) follows from Lemma 8. All earlier size conditions are lower bounds on t, so they can be imposed at the same time.□

Proof. Fix \lambda=(\beta,h)\in W(s-S_*a), and choose an active shift u for T_aq at \lambda. Let x=(\beta,h+\beta au). The distance from \lambda to the boundary of W(s) is at least S_*a. Since \abs{x-\lambda}\le\beta_{\max}U_*a, every point

lies in W(s). The surplus a leaves these test points in the interior of W(s).

A common density band. For (e,d)\in\partial P(y), the supporting inequality at y, the Lipschitz bound on P, and the bound \norm{P-q}_{W_1}\le\delta give

The last step is the active-point parabola of Lemma 5. Using both signs, (7.1), and the hypothesis on \delta/a, we obtain

The same u works for every y in this neighborhood.

One slope group near the shifted point. The set in the right side of (7.2) is nonempty: the full subdifferential is nonempty, and the assumed hull identity identifies it with the convex hull of that set. If a generator belongs to \clB{v_i}{r}, then

Since the distinct d_i are separated by at least p, at most one index i can satisfy this condition. It is therefore one common index for the entire neighborhood. By convexity of the ball and the assumed hull identity,

Transfer to the physical pressure and its transform. Take v\in\partial q(x) and a unit vector z\in\R^2. At y=x+la z, choose any support of P. Using (7.5) to bound this support and the approximation at x,y, we get

Testing every unit z proves

By Proposition 5, the generating supports of the active branch of T_aq are obtained by adding (au^2/2,0). They all lie in \clB{v_i}{r^+} by the definition of r^+ and \abs{u}\le U_*. The exact convex-hull formula over all active branches gives (7.3).□

Proof. The case n=1 is Proposition 7. Each q_n belongs to the family, and each P_n has the common Lipschitz bound C_*. Apply Lemma 10 inductively with P=P_n, q=q_n, a=a_{n+1}, \delta=\delta_n, r=r_n, and s equal to the radius of W_n. The first condition in (7.12) keeps the parameter radius above 1/8, while the second gives r_n<\tau. The other hypotheses are (7.8) and the first condition in (7.13). The new domain and slope radius are exactly W_{n+1} and r_{n+1}.□

Proof. Minimize P_n(\lambda)-g\cdot\lambda on the compact ball W(1/8). For a boundary point, Lemma 9 and (7.14) give

Thus a minimizer \lambda_n is interior, and g\in\partial P_n(\lambda_n). By Proposition 8, this subdifferential is generated by its intersections with the three balls \clB{v_i}{r_n}. The vector g lies outside the convex hull of any two of the larger balls \clB{v_i}{\tau}. Each of the three intersections must therefore be nonempty. Choose one actual support from each. Their triangle contains \clB{g}{\gamma} by Lemma 9, and their convex hull is contained in \partial P_n(\lambda_n).□

Proof. Since \sum_na_n<A+1 and R_n\ge1, the attraction series converges uniformly. The core and every summand are bounded, continuous, and radial. The common cell estimate of Proposition 1 applies to the sum and gives U^\Phi\ge-B_0N=-(t^2/2)N.

Set \varepsilon=1/32. A nonzero summand at radius r must have R_n\ge r, and then

The core vanishes for r\ge1+1/t<2 and is at most t^2. For every r\ge1, therefore,

Finally,

by the geometric-series bound following (7.9).□

Proof. The field part of either support inequality says that \rho supports Q(\beta_c,\cdot) at h_c. Density duality therefore gives H(\beta_c,\rho)=h_c\rho-Q(\beta_c,h_c). The full support inequality, with s=s_i, reads

Rearrange and take the supremum over h. The reverse duality gives

The function H(\,\cdot\,,\rho) is finite and concave on (0,\infty), so its one-sided derivatives at an interior point exist and are finite. Dividing the last inequality by positive and negative increments yields the stated ordering.□

Proof of Theorem 1.1. Take the potential (7.17) and the constants in Proposition 10. Write g=(g_1,\rho_0), where \rho_0=5p/3>0, and set

For every \rho\in I, the two points

lie in \clB{g}{\gamma}, since their squared distances from g are (\gamma/2)^2+(\rho-\rho_0)^2<\gamma^2/2. By (7.19), both support the same pressure at the same (\beta_c,h_c), independently of \rho. Lemma 11 therefore gives finite one-sided canonical slopes whose difference is at least \gamma for every \rho\in I. The all-sequence canonical limit of Proposition 2 applies in particular to N_L=\lfloor\rho L^3\rfloor for every \beta>0. The specified free energy equals

For each \rho\in I, it is finite for every \beta>0. The finite concave function H_\Phi(\,\cdot\,,\rho) is continuous at \beta_c; hence the common derivative term from 1/\beta cancels in the difference, giving

The potential and the density interval are fixed before the limiting critical parameter is selected, so this one \beta_c works throughout I.□