Continuum Pressure

A nonnegative simple function in the thermodynamics definition is a finite sum of indicators of disjoint measurable sets, hence is measurable and has finite range; conversely, the level sets in [F3] represent every nonnegative finite-range measurable function in that form. Thus the two definitions use the same class of simple minorants.

For any simple minorant $s$, use its canonical level-set representation from [F3]. Both [F1] and [F4] assign the same finite sum ${\sum }_j{c}_j\mu (A_j)$ to $s$, including zero coefficients on sets of infinite measure; [F4] ensures the shared value is representation-independent. Hence the two definitions assign the same value to every minorant.

The two nonnegative integrals are suprema over the same minorants with the same value for each minorant by steps 1.1 and 2.1. Their suprema therefore agree, including when the common value is $+\mathrm{\infty }$.

By [A1] and $\beta >0$, every configuration satisfies $0\le e^{-\beta U_N}\le e^{\beta BN}$, with the value $0$ when $U_N=+\mathrm{\infty }$. Fact [F7] makes the integrand and its restriction to ${\mathrm{\Lambda }}_L^N$ measurable. Let $s={\sum }_{j=1}^r{c}_j{\mathbf{1}}_{A_j}$ be any nonnegative simple minorant of $e^{-\beta U_N}{\mathbf{1}}_{{\mathrm{\Lambda }}_L^N}$, written with disjoint measurable sets and finite coefficients $c_j\ge 0$ as in [L1]. Every $A_j$ with $c_j>0$ is contained in ${\mathrm{\Lambda }}_L^N$, and on it $c_j\le e^{\beta BN}$. For any measurable $E\subseteq {\mathrm{\Lambda }}_L^N$, countable additivity applied to ${\mathrm{\Lambda }}_L^N=E\bigsqcup ({\mathrm{\Lambda }}_L^N\setminus E)$ gives ${\lambda }_d^{\otimes N}(E)\le {\lambda }_d^{\otimes N}({\mathrm{\Lambda }}_L^N)<\mathrm{\infty }$. Applying this first to each $A_j$ and then to their finite disjoint union gives $\int sd{\lambda }_d^{\otimes N}={\sum }_{c_j>0}{c}_j{\lambda }_d^{\otimes N}(A_j)\le e^{\beta BN}{\sum }_{c_j>0}{\lambda }_d^{\otimes N}(A_j)=e^{\beta BN}{\lambda }_d^{\otimes N}\text{\hspace{0.17em}⁣}\left({\bigcup }_{c_j>0}{A}_j\right)\le e^{\beta BN}{\lambda }_d^{\otimes N}({\mathrm{\Lambda }}_L^N)\mathrm{.}$ Taking the supremum over these minorants in [L1] proves the same bound for the restricted integrand's nonnegative integral.

The box and product-measure formulas [F2, F3] give ${\lambda }_d^{\otimes N}({\mathrm{\Lambda }}_L^N)=L^{dN}$. Thus $Z_{L,N}\le (a{e}^{\beta B}{L}^d{)}^N\mathrm{/}N!=(Q{L}^d{)}^N\mathrm{/}N!$ for $N\ge 1$, and this also holds for $N=0$ since $Z_{L,0}=1$.

Summing the nonnegative terms by [F6] and using [F4] gives $1\le {\mathrm{\Xi }}_L\le {\sum }_{N=0}^{\mathrm{\infty }}(Q{L}^d{)}^N\mathrm{/}N!=e^{Q{L}^d}$; the lower bound is the empty-configuration term.

By [F8], $K{L}^d<n_L\le K{L}^d+1$. If $N>n_L$, then $N>K{L}^d$, so $1\le e^{N-K{L}^d}$. Therefore ${\sum }_{N>n_L}{Z}_{L,N}\le e^{-K{L}^d}{\sum }_{N>n_L}\frac{(eQ{L}^d{)}^N}{N!}\le e^{-K{L}^d}{e}^{eQ{L}^d}=e^{(eQ-K)L^d}\le e^{-L^d}\mathrm{.}$ The second inequality extends a nonnegative partial tail to the full exponential series; the equality uses [F5], and the last inequality follows from $K>eQ+1$ and [F5].

For each $M>n_L$, split the $M$th partial sum into the fixed finite prefix ${\mathrm{\Xi }}_L^{\mathrm{c}\mathrm{u}\mathrm{t}}$ and the finite tail from $n_L+1$ through $M$. Taking suprema over $M$ in the nonnegative-series definition [F6] gives ${\mathrm{\Xi }}_L={\mathrm{\Xi }}_L^{\mathrm{c}\mathrm{u}\mathrm{t}}+{\sum }_{N>n_L}{Z}_{L,N}$; subtraction is valid because step 3.1 proves ${\mathrm{\Xi }}_L<\mathrm{\infty }$. Since ${\mathrm{\Xi }}_L\ge 1$, step 3.2 implies ${\mathrm{\Xi }}_L^{\mathrm{c}\mathrm{u}\mathrm{t}}\ge {\mathrm{\Xi }}_L-e^{-L^d}\ge (1-e^{-L^d}){\mathrm{\Xi }}_L$.

Let $m=\mathrm{\parallel }j-j^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}\ge 1$ for two distinct cube indices. In a coordinate where the index difference is $m$, the nearest points of the two intervals are separated by $m(\mathrm{\ell }+g)-\mathrm{\ell }=(m-1)\mathrm{\ell }+mg\ge mg$. Hence every pair of points in $C_j,C_{j^{\mathrm{\prime }}}$ has Euclidean distance at least $mg$.

For $m\ge 1$, [F3] gives the shell count. Use $x^d-y^d=(x-y){\sum }_{r=0}^{d-1}{x}^{d-1-r}{y}^r$ with $x=2m+1$, $y=2m-1$. Since $2m+1\le 3m$, it yields $(2m+1{)}^d-(2m-1{)}^d\le 2d(3m{)}^{d-1}=2d{3}^{d-1}{m}^{d-1}\mathrm{.}$ Therefore $S_{d,ϵ}\le 2d{3}^{d-1}{\sum }_{m\ge 1}{m}^{-1-ϵ}<\mathrm{\infty }$ by [F2], since $1+ϵ>1$.

By [F1] and step 1.1, each such pair contributes at most $C(mg{)}^{-d-ϵ}$ in absolute value, because $mg\ge g\ge R$ and the Euclidean norm dominates the absolute value of any coordinate.

Fix $j\in J$. For each $j^{\mathrm{\prime }}\mathrm{\ne }j$, there are at most $n_j{n}_{j^{\mathrm{\prime }}}\le n^2$ labelled pairs of positions from $C_j$ and $C_{j^{\mathrm{\prime }}}$. Summing the bound from step 2.1 over those pairs and enlarging the finite set of index differences to all nonzero lattice vectors gives at most $C{n}^2{S}_{d,ϵ}{g}^{-d-ϵ}$, since [F3] groups those vectors into the shells whose counts define $S_{d,ϵ}$, and [F2] defines its nonnegative sum. Summing this bound over all $M$ cubes counts each unordered cube pair twice, so dividing by $2$ gives the claimed inequality.

Set $\alpha :=\frac{1}{2}(1+d\mathrm{/}(d+ϵ))$ and $g_{\mathrm{\ell }}:={\mathrm{\ell }}^{\alpha }$. Then $d\mathrm{/}(d+ϵ)<\alpha <1$ and ${\mathrm{\ell }}^d{g}_{\mathrm{\ell }}^{-d-ϵ}={\mathrm{\ell }}^{d-\alpha (d+ϵ)}\to 0$ while $g_{\mathrm{\ell }}\mathrm{/}\mathrm{\ell }\to 0$. Fix a sufficiently large $\mathrm{\ell }\ge 1$ so that $g_{\mathrm{\ell }}\ge R$. For $L\ge \mathrm{\ell }+g_{\mathrm{\ell }}$, set $m_L:=\lfloor L\mathrm{/}(\mathrm{\ell }+g_{\mathrm{\ell }})\rfloor$. Since $L\mathrm{/}(\mathrm{\ell }+g_{\mathrm{\ell }})\ge 1$, [F10] gives $m_L\ge 1$ and $m_L(\mathrm{\ell }+g_{\mathrm{\ell }})\le L<(m_L+1)(\mathrm{\ell }+g_{\mathrm{\ell }})$. Place $M_L=m_L^d$ cubes $C_j=(\mathrm{\ell }+g_{\mathrm{\ell }})j+[0,\mathrm{\ell }{]}^d,j\in \{0,\dots ,m_L-1{\}}^d\mathrm{.}$ They lie inside $[0,L{]}^d$ and have pairwise gaps at least $g_{\mathrm{\ell }}$.

First, if $T$ is a bijection whose map and inverse are measurable and that preserves measure, meaning $\mu (T^{-1}A)=\mu (A)$ for every measurable $A$, then $\mu (TA)=\mu (A)$ as well by applying this identity to $TA$; thus $T^{-1}$ also preserves measure. Such a map preserves nonnegative integrals: for an indicator ${\mathbf{1}}_A$, $\int {\mathbf{1}}_A\circ Td\mu =\mu (T^{-1}A)=\mu (A)=\int {\mathbf{1}}_{A}d\mu$; finite sums give the same equality for simple functions. If $f\ge 0$ is measurable, composition with $T$ sends every simple minorant of $f$ to a simple minorant of $f\circ T$ with the same simple integral. Taking the supremum in [F4] gives $\int fd\mu \le \int f\circ Td\mu$, and applying this inequality to $T^{-1}$ and $f\circ T$ gives equality. The same supremum definition shows monotonicity: $f\le h$ implies $\int fd\mu \le \int hd\mu$. For $N\ge 1$ identical Lebesgue factors, a coordinate permutation preserves the finite product measure: the permutation is a Borel isomorphism, so its push-forward is sigma-finite, and that push-forward agrees with the original measure on rectangles by the rectangle formula in [F6]; finite-product uniqueness, obtained by iterating the uniqueness clause in [F6], makes the measures equal. When $N=0$, there is one empty configuration and one empty assignment, so the equality is immediate. Now fix occupancies $0\le N_j\le n_{\mathrm{\ell }}$ with total $N={\sum }_j{N}_j$. The ordered configurations in the disjoint cubes with exactly these occupancies are partitioned by assignments of the $N$ particle labels to cubes. The number of assignments is $N!\mathrm{/}({\prod }_j{N}_j!)$: order all $N$ labels, then note that each assignment occurs exactly ${\prod }_j{N}_j!$ times from within-cube reorderings. Any two such assignments differ by a permutation of the $N$ labels, and the corresponding coordinate permutation maps one assignment domain to the other. The potential is even, so $U_N$ is invariant under particle-coordinate permutations. The integral invariance just proved therefore makes the assignment-domain integrals equal.

Restrict the grand partition sum in $[0,L{]}^d$ to configurations with no particles outside these cubes and at most $n_{\mathrm{\ell }}$ particles in each cube, where $n_{\mathrm{\ell }}$ is the cutoff from [F2]. If two cubes have index difference $k\mathrm{\ne }0$, the coordinate gap in a coordinate attaining $\mathrm{\parallel }k{\mathrm{\parallel }}_{\mathrm{\infty }}$ is at least $\mathrm{\parallel }k{\mathrm{\parallel }}_{\mathrm{\infty }}{g}_{\mathrm{\ell }}$; [F3] therefore bounds the total absolute inter-cube energy by $D_L\le \frac{C}{2}{M}_L{n}_{\mathrm{\ell }}^2{S}_{d,ϵ}{g}_{\mathrm{\ell }}^{-d-ϵ}\mathrm{.}$ Thus $U_N={\sum }_j{U}_{N_j}^{(j)}+U_{\mathrm{c}\mathrm{r}\mathrm{o}\mathrm{s}\mathrm{s}}$ with $U_{\mathrm{c}\mathrm{r}\mathrm{o}\mathrm{s}\mathrm{s}}\le D_L$, and $e^{-\beta U_N}\ge e^{-\beta D_L}{\prod }_j{e}^{-\beta U_{N_j}^{(j)}}$; this remains true when an internal energy is $+\mathrm{\infty }$, since both sides then vanish.

On each assignment domain from step 2.1, the pointwise lower bound in step 2.2 factors into nonnegative within-cube integrands. Monotonicity from step 2.1 and repeated application of [F5] and the rectangle formula in [F6] factor the resulting lower bound's integral into the product of the translated one-cube integrals. For a cube indexed by $j$ with $N_j\ge 1$, common translation by $(\mathrm{\ell }+g_{\mathrm{\ell }})j$ is a Borel isomorphism mapping $[0,\mathrm{\ell }{]}^{d{N}_j}$ onto $C_j^{N_j}$ and preserves its finite product measure: the push-forward is sigma-finite and, on rectangles, this follows from single-coordinate translation invariance [F7] and the rectangle formula [F6]; uniqueness in [F6] gives equality of the product measures. Pair differences, hence the within-cube integrand, are unchanged by this translation, so the integral invariance proved in step 2.1 identifies the integral over $C_j^{N_j}$ with its origin-centered integral over $[0,\mathrm{\ell }{]}^{d{N}_j}$. When $N_j=0$, the corresponding factor is $Z_{\mathrm{\ell },0}=1$. Multiplying by $a^N\mathrm{/}N!$ cancels the label count from step 2.1, leaving $e^{-\beta D_L}{\prod }_j{Z}_{\mathrm{\ell },N_j}$. Summing over the finitely many occupancy vectors and expanding the finite product of cutoff sums from [F2] gives ${\mathrm{\Xi }}_L\ge e^{-\beta D_L}({\mathrm{\Xi }}_{\mathrm{\ell }}^{\mathrm{c}\mathrm{u}\mathrm{t}}{)}^{M_L}\mathrm{.}$

The integer-part bounds in step 1.1 give $L\mathrm{/}(\mathrm{\ell }+g_{\mathrm{\ell }})-1<m_L\le L\mathrm{/}(\mathrm{\ell }+g_{\mathrm{\ell }})$, so $m_L\mathrm{/}L\to (\mathrm{\ell }+g_{\mathrm{\ell }}{)}^{-1}$ and $M_L\mathrm{/}{L}^d\to (\mathrm{\ell }+g_{\mathrm{\ell }}{)}^{-d}$. By [F2], ${\mathrm{\Xi }}_{\mathrm{\ell }}^{\mathrm{c}\mathrm{u}\mathrm{t}}\ge (1-e^{-{\mathrm{\ell }}^d}){\mathrm{\Xi }}_{\mathrm{\ell }}$, so $\log {\mathrm{\Xi }}_{\mathrm{\ell }}^{\mathrm{c}\mathrm{u}\mathrm{t}}\ge {\mathrm{\ell }}^d{p}_{\mathrm{\ell }}+\log (1-e^{-{\mathrm{\ell }}^d})$. Taking logarithms in step 3.1, dividing by $L^d$, and using step 2.2 gives $p_L\ge \frac{M_L}{L^d}({\mathrm{\ell }}^d{p}_{\mathrm{\ell }}+\log (1-e^{-{\mathrm{\ell }}^d})-\frac{\beta C}{2}{n}_{\mathrm{\ell }}^2{S}_{d,ϵ}{g}_{\mathrm{\ell }}^{-d-ϵ})\mathrm{.}$ Taking the lower limit and using the volume ratio above yields the same bound with $M_L\mathrm{/}{L}^d$ replaced by $(\mathrm{\ell }+g_{\mathrm{\ell }}{)}^{-d}$. The cutoff definition [F2] and $\mathrm{\ell }\ge 1$ give $n_{\mathrm{\ell }}\le K{\mathrm{\ell }}^d+1\le (K+1){\mathrm{\ell }}^d$. Factoring $(\mathrm{\ell }\mathrm{/}(\mathrm{\ell }+g_{\mathrm{\ell }}){)}^d$ and using this estimate in the negative error term therefore gives ${\mathrm{lim\hspace{0.17em}inf}}_{L\to \mathrm{\infty }}{p}_L\ge {\left(\frac{\mathrm{\ell }}{\mathrm{\ell }+g_{\mathrm{\ell }}}\right)}^d[p_{\mathrm{\ell }}+{\mathrm{\ell }}^{-d}\log (1-e^{-{\mathrm{\ell }}^d})-\frac{\beta C}{2}(K+1{)}^2{S}_{d,ϵ}{\mathrm{\ell }}^d{g}_{\mathrm{\ell }}^{-d-ϵ}]\mathrm{.}$

As $\mathrm{\ell }\to \mathrm{\infty }$, the prefactor tends to $1$. Also $e^{-{\mathrm{\ell }}^d}\to 0$, so continuity of $\log$ at $1$ gives $\log (1-e^{-{\mathrm{\ell }}^d})\to 0$; since ${\mathrm{\ell }}^{-d}\to 0$, their product ${\mathrm{\ell }}^{-d}\log (1-e^{-{\mathrm{\ell }}^d})$ tends to zero. The cross-energy error tends to zero because $d-\alpha (d+ϵ)<0$ by step 1.1. The replacement of the prefactor times $p_{\mathrm{\ell }}$ by $p_{\mathrm{\ell }}$ has error at most $Q[1-(\mathrm{\ell }\mathrm{/}(\mathrm{\ell }+g_{\mathrm{\ell }}){)}^d]\to 0$ by [F2]. Thus for every $\eta >0$ there is $T$ such that every $\mathrm{\ell }\ge T$ makes the right side of step 4.1 at least $p_{\mathrm{\ell }}-\eta$.

Let $\bar{p}={\mathrm{lim\hspace{0.17em}sup}}_{\mathrm{\ell }\to \mathrm{\infty }}{p}_{\mathrm{\ell }}$. By [A1], for every $\eta >0$ and every $T\ge 1$ there is an $\mathrm{\ell }\ge T$ with $p_{\mathrm{\ell }}>\bar{p}-\eta$; otherwise the tail supremum from $T$ onward would be at most $\bar{p}-\eta$, contradicting the infimum defining $\bar{p}$. Choose $T$ large enough for step 5.1. Then ${\mathrm{lim\hspace{0.17em}inf}}_{L\to \mathrm{\infty }}{p}_L\ge \bar{p}-2\eta$. Letting $\eta \downarrow 0$ shows $\mathrm{lim\hspace{0.17em}inf}{p}_L\ge \mathrm{lim\hspace{0.17em}sup}{p}_L$.

For any $S,T\ge 1$, choose $L\ge \max (S,T)$. Then ${\inf }_{x\ge S}{p}_x\le p_L\le {\sup }_{x\ge T}{p}_x$. Taking the supremum over $S$ and then the infimum over $T$ proves $\mathrm{lim\hspace{0.17em}inf}{p}_L\le \mathrm{lim\hspace{0.17em}sup}{p}_L$ by [A1]. Together with step 6.1, the two finite tail bounds equal a number $p$. For every $\eta >0$, their definitions give tail thresholds with ${\inf }_{L\ge T_{-}}{p}_L>p-\eta$ and ${\sup }_{L\ge T_{+}}{p}_L<p+\eta$; hence $\mathrm{\mid }{p}_L-p\mathrm{\mid }<\eta$ for all sufficiently large $L$. The limit is finite by [F2]. This proves existence for the specified free-cube reduced grand pressure.