Large-deviation variational mechanics and ensemble inequivalence

For each integer $m\ge 1$, put $U_m:=\{x\in K:\mathrm{\mid }F(x)\mathrm{\mid }<m\}$. These sets are open by continuity of $F$: if $x\in U_m$, continuity gives a neighborhood on which $\mathrm{\mid }F(y)-F(x)\mathrm{\mid }<m-\mathrm{\mid }F(x)\mathrm{\mid }$, hence $\mathrm{\mid }F(y)\mathrm{\mid }<m$. They cover $K$ because every real $\mathrm{\mid }F(x)\mathrm{\mid }$ is less than some positive integer. By the finite-open-cover meaning of compactness in [A2], finitely many $U_{m_1},\dots ,U_{m_k}$ cover $K$, with $k\ge 1$ since $K$ is nonempty. Let $M:=\max \{m_1,\dots ,m_k\}$; then $-M<F(x)<M$ for every $x\in K$. Thus $F$ is bounded above and below. This uses the explicitly defined cover and a maximum of finitely many integers, without any simultaneous choice of pointwise neighborhoods. Since $I$ is not identically $+\mathrm{\infty }$, choose $x_0\in K$ with $I(x_0)<+\mathrm{\infty }$. Therefore $F(x_0)-I(x_0)\le c\le {\sup }_{K}F<+\mathrm{\infty }$, so $c$ is finite. The integrand $e^{nF}$ is bounded above and bounded below by the positive constant $e^{n{\inf }_{K}F}$, and ${\mu }_n$ is a probability measure, hence $0<Z_n<+\mathrm{\infty }$ and ${\nu }_n^F$ is a probability measure.

Let $G\subseteq K$ be open, let $x\in G$ with $I(x)<+\mathrm{\infty }$, and let $\epsilon >0$. Continuity of $F$ gives an open neighborhood $U$ with $x\in U\subseteq G$ and $F(y)>F(x)-\epsilon$ for every $y\in U$. Hence ${\int }_G{e}^{nF}d{\mu }_n\ge e^{n(F(x)-\epsilon )}{\mu }_n(U)$. By [A1], the logarithm of this integral divided by $n$ has liminf at least $F(x)-\epsilon -{\inf }_{U}I\ge F(x)-\epsilon -I(x)$. Taking the supremum over finite-rate $x\in G$ and then $\epsilon \downarrow 0$ proves the open-set integral lower bound ${\mathrm{lim\hspace{0.17em}inf}}_n{n}^{-1}\ln {\int }_G{e}^{nF}d{\mu }_n\ge {\sup }_{x\in G}(F(x)-I(x))$.

Let $C\subseteq K$ be closed and put $c_C:={\sup }_{x\in C}(F(x)-I(x))$, with $\sup \mathrm{\varnothing }=-\mathrm{\infty }$. If $C=\mathrm{\varnothing }$, its integral is zero and the required bound is immediate. Since $F$ is bounded and $I\ge 0$, either $c_C$ is finite or $c_C=-\mathrm{\infty }$. Suppose first that $c_C$ is finite, and fix $\epsilon >0$. For each $x\in C$ with $I(x)<+\mathrm{\infty }$, lower semicontinuity of $I$ and continuity of $F$ give an open neighborhood $V_x$ on which $F\le F(x)+\epsilon$ and $I\ge I(x)-\epsilon$. Choose a ball $U_x$ centered at $x$ with ${\bar{U}}_x\subseteq V_x$. For each $x\in C$ with $I(x)=+\mathrm{\infty }$, choose $M>{\sup }_{K}F-c_C+1$; lower semicontinuity gives an open neighborhood $V_x$ on which $I>M$, and choose such a ball $U_x$. Take the family of all centered balls satisfying these conditions; it covers $C$. A closed subset $C$ of compact $K$ is compact: adjoining $K\setminus C$ to any relative open cover of $C$ gives an open cover of $K$. Thus finitely many admissible balls suffice, and only finitely many centers and neighborhood witnesses are selected. Put $C_j:=C\cap {\bar{U}}_{x_j}$ for the resulting centers. Each $C_j$ is closed. At a finite-rate center, ${\sup }_{C_j}F-{\inf }_{C_j}I\le F(x_j)-I(x_j)+2\epsilon \le c_C+2\epsilon ;$ at an infinite-rate center, ${\sup }_{C_j}F-{\inf }_{C_j}I\le {\sup }_{K}F-M<c_C-1.$ The integral over $C$ is bounded by the finite sum ${\sum }_j{e}^{n{\sup }_{C_j}F}{\mu }_n(C_j)$. Applying [A1] to each closed $C_j$ and using that a finite sum has logarithmic limsup no greater than its largest summand gives ${\mathrm{lim\hspace{0.17em}sup}}_n\frac{1}{n}\ln {\int }_C{e}^{nF}d{\mu }_n\le c_C+2\epsilon \mathrm{.}$ Letting $\epsilon \downarrow 0$ proves the localized bound in this case. If $c_C=-\mathrm{\infty }$, every point of $C$ has infinite $I$. For any real $M$, lower semicontinuity gives, at each $x\in C$, an open neighborhood $V_x$ with $I>M$; take all centered balls with closures in such neighborhoods and then a finite subcover of $C$. The corresponding closed pieces $C_j$ then satisfy ${\sup }_{C_j}F-{\inf }_{C_j}I\le {\sup }_{K}F-M$. The same finite-sum and LDP argument gives a logarithmic limsup at most ${\sup }_{K}F-M$. Since $M$ is arbitrary, this limsup is $-\mathrm{\infty }=c_C$. Thus in all cases ${\mathrm{lim\hspace{0.17em}sup}}_n\frac{1}{n}\ln {\int }_C{e}^{nF}d{\mu }_n\le {\sup }_{x\in C}(F(x)-I(x))\mathrm{.}$

Taking $G=K$ in step 1.2 and $C=K$ in step 1.3 gives ${\mathrm{lim\hspace{0.17em}inf}}_n{n}^{-1}\ln Z_n\ge c$ and ${\mathrm{lim\hspace{0.17em}sup}}_n{n}^{-1}\ln Z_n\le c$, proving the log-Laplace limit. The function $I_F=I-F+c$ is lower semicontinuous, is nonnegative by the definition of $c$, and has compact sublevel sets because those sets are closed in compact $K$. To prove attainment, set $f:=F-I$, which is upper semicontinuous and bounded above by ${\sup }_{K}F$. For every integer $k\ge 1$, the set $A_k:=\{x\in K:f(x)\ge c-1\mathrm{/}k\}$ is closed and nonempty by the definition of $c$; these sets are nested. Compactness gives ${\bigcap }_{k\ge 1}{A}_k\mathrm{\ne }\mathrm{\varnothing }$: otherwise their open complements would cover $K$, and a finite subcover would contradict that a nested finite intersection is nonempty. Every point of this intersection has $f(x)\ge c-1\mathrm{/}k$ for all $k$, hence $f(x)\ge c$ and therefore $f(x)=c$. Equality $I_F(x)=0$ holds exactly when $F(x)-I(x)=c$, so its nonempty zero set is precisely the stated maximizing set.

For open $G\subseteq K$, apply the open-set integral lower bound from step 1.2 and subtract the log-normalizer limit in step 2.1. Since ${\sup }_G(F-I)-c=-{\inf }_G{I}_F$, this is the LDP lower bound for ${\nu }_n^F$.

For closed $C\subseteq K$, apply the closed-set integral upper bound from step 1.3 and subtract the log-normalizer limit in step 2.1. Since ${\sup }_C(F-I)-c=-{\inf }_C{I}_F$, this is the LDP upper bound for ${\nu }_n^F$.

Enumerate $\mathcal{A}=\{a_1,\dots ,a_d\}$, with $d\ge 1$, and identify ${\mathbb{R}}^{\mathcal{A}}$ with Euclidean ${\mathbb{R}}^d$. Write $\mathrm{\parallel }x{\mathrm{\parallel }}_1:={\sum }_{i=1}^d\mathrm{\mid }{x}_i\mathrm{\mid }$ and $\mathrm{\parallel }x{\mathrm{\parallel }}_2:=({\sum }_{i=1}^d{x}_i^2{)}^{1\mathrm{/}2}$; the declared ${\mathrm{\ell }}^1$ topology on the simplex is the topology from $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_1$. The simplex is closed because it is the intersection of the closed coordinate constraints $0\le q(a_i)\le 1$ and the closed affine constraint ${\sum }_{i}q(a_i)=1$; these are closed since coordinate maps and their finite sum are continuous. It is bounded because $\mathrm{\parallel }q{\mathrm{\parallel }}_2^2={\sum }_{i}q(a_i{)}^2\le {\sum }_{i}q(a_i)=1$. By [F6] it is compact in the Euclidean topology. For every $x\in {\mathbb{R}}^d$, $\mathrm{\parallel }x{\mathrm{\parallel }}_2\le \mathrm{\parallel }x{\mathrm{\parallel }}_1$ because $({\sum }_i\mathrm{\mid }{x}_i\mathrm{\mid }{)}^2\ge {\sum }_i{x}_i^2$, and $\mathrm{\parallel }x{\mathrm{\parallel }}_1\le d\mathrm{\parallel }x{\mathrm{\parallel }}_2$ because ${\sum }_i\mathrm{\mid }{x}_i\mathrm{\mid }\le d{\max }_i\mathrm{\mid }{x}_i\mathrm{\mid }\le d\mathrm{\parallel }x{\mathrm{\parallel }}_2$. These inequalities show that the Euclidean and ${\mathrm{\ell }}^1$ metrics have the same open sets, so the simplex is compact in its declared topology. By continuity of $h$ and $D$, each fiber $h^{-1}(\{u\})$ and each sublevel set $\{q:D(q\mathrm{\parallel }p)\le r\}$ is closed in the simplex and hence compact by [F7]. The continuous-image and extreme-value conclusions in [F8] therefore apply to these sets. In particular, [F4] makes $q\mapsto D(q\mathrm{\parallel }p)+\beta h(q)$ continuous on the compact simplex, so it attains a finite minimum $m_{\beta }$. The rate $D$ is nonnegative by [F5]. The empirical macrostate law under $p^{\otimes n}$ has rate $D(q\mathrm{\parallel }p)$ by [F2]. For every Borel $B\subseteq \mathrm{\Delta }(\mathcal{A})$, the canonical macrostate probability is exactly $p_{n,\beta }(L_n\in B)=\frac{{\int }_B{e}^{-n\beta h(q)}d{\mu }_n(q)}{{\int }_{\mathrm{\Delta }(\mathcal{A})}{e}^{-n\beta h(q)}d{\mu }_n(q)},$ where ${\mu }_n$ is the base law of $L_n$; this follows by summing the word weights in [F1] over the fiber $L_n^{-1}(B)$. Thus it is the tilt in [F3] with $F=-\beta h$, and its rate is $D+\beta h-m_{\beta }$. Its zero set is ${\mathcal{M}}_{\beta }$. Let $U$ be any open neighborhood of ${\mathcal{M}}_{\beta }$ and put $C:=\mathrm{\Delta }(\mathcal{A})\setminus U$. If $C=\mathrm{\varnothing }$, the probability of the complement is zero. Otherwise $C$ is a nonempty closed subset of the compact simplex, hence compact by [F7]. The continuous rate is positive at every point of $C$ because its zero set is ${\mathcal{M}}_{\beta }\subset U$; by [F8] it attains a strictly positive minimum on $C$, and the LDP upper bound makes the canonical macrostate probability of that complement exponentially small.

Let $q\in {\mathcal{M}}_{\beta }$ and put $u=h(q)$. If $r\in H_u$, then $D(r)+\beta u\ge D(q)+\beta u$, so $q\in {\mathcal{M}}_u$ and $D(q)=d(u)$. For any $v$ in the energy range, minimizing over $H_v$ gives $d(v)+\beta v\ge d(u)+\beta u$, equivalently $s(v)-\beta v\le s(u)-\beta u$. Thus $u$ maximizes the stated entropy-minus-energy expression.

The fiber $H_u:=\{q\in \mathrm{\Delta }(\mathcal{A}):h(q)=u\}$ is nonempty and closed, hence compact by step 1.1 and [F7], because $u$ is in the range of continuous $h$. By [F4] and [F8], $D$ attains its finite minimum $d(u)$ on $H_u$, so ${\mathcal{M}}_u$ is nonempty. For fixed $\delta >0$, put $B_{\delta }:=\{q:\mathrm{\mid }h(q)-u\mathrm{\mid }<\delta \}$; it is open and contains $H_u$. By [F2], $-{\inf }_{B_{\delta }}D\le {\mathrm{lim\hspace{0.17em}inf}}_n\frac{1}{n}\ln p^{\otimes n}(L_n\in B_{\delta })\le {\mathrm{lim\hspace{0.17em}sup}}_n\frac{1}{n}\ln p^{\otimes n}(L_n\in B_{\delta })\le -{\inf }_{{\bar{B}}_{\delta }}D\mathrm{.}$ For each matching radius ${\delta }_j$ in the Statement the two infima agree, so the inner limit exists and equals their negative common value. The lower bound is finite because $B_{{\delta }_j}$ contains a finite-rate point of $H_u$; hence the conditioning event has positive probability for all sufficiently large $n$.

Write $d_{\delta }:={\inf }_{{\bar{B}}_{\delta }}D$ for any $\delta >0$; since $H_u\subseteq {\bar{B}}_{\delta }$, $d_{\delta }\le d(u)$. Fix $\epsilon >0$ and let $K_{\epsilon }:=\{q\in \mathrm{\Delta }(\mathcal{A}):D(q\mathrm{\parallel }p)\le d(u)-\epsilon \}$. By [F4], this set is closed and hence compact by [F7]; it is disjoint from $H_u$ by the definition of $d(u)$. Thus its compact image $h(K_{\epsilon })$ does not contain $u$ by [F8]. If it is nonempty, the continuous function $v\mapsto \mathrm{\mid }v-u\mathrm{\mid }$ on that compact image attains a positive minimum by [F8]; continuity follows from $\mid \mathrm{\mid }v-u\mathrm{\mid }-\mathrm{\mid }w-u\mathrm{\mid }\mid \le \mathrm{\mid }v-w\mathrm{\mid }$. Therefore the image is separated from $u$; if it is empty the conclusion is immediate. For every sufficiently small $\delta$, the closed shell ${\bar{B}}_{\delta }$ misses $K_{\epsilon }$, so $d_{\delta }\ge d(u)-\epsilon$. Together with $d_{\delta }\le d(u)$, this proves $d_{\delta }\to d(u)$ as $\delta \downarrow 0$, in particular along the matching sequence ${\delta }_j$. Step 2.1 now gives the iterated shell limit $-d(u)=s(u)$.

Let $U$ be an open neighborhood of ${\mathcal{M}}_u$ and put $C:=\mathrm{\Delta }(\mathcal{A})\setminus U$, which is closed and hence compact by step 1.1 and [F7]. If $C\cap H_u=\mathrm{\varnothing }$, then $h(C)$ is compact by [F8] and does not contain $u$; if $C$ is nonempty, the continuous function $v\mapsto \mathrm{\mid }v-u\mathrm{\mid }$ attains a positive minimum there by [F8], using $\mid \mathrm{\mid }v-u\mathrm{\mid }-\mathrm{\mid }w-u\mathrm{\mid }\mid \le \mathrm{\mid }v-w\mathrm{\mid }$. Thus sufficiently small shells miss $C$; this is also immediate when $C=\mathrm{\varnothing }$. Otherwise, $C\cap H_u$ is a nonempty closed subset of the compact simplex and is compact by [F7]. Continuity of $D$ and [F8] give $d_C:={\min }_{C\cap H_u}D>d(u)$, since $C$ contains no constrained minimizer. Choose $0<\eta <d_C-d(u)$ and set $K:=\{q\in C:D(q\mathrm{\parallel }p)\le d(u)+\eta \}$; it is closed and hence compact by [F4] and [F7], and it is disjoint from $H_u$. Therefore $h(K)$ is compact by [F8] and does not contain $u$. If empty it is already separated from $u$; otherwise, the continuous distance function $v\mapsto \mathrm{\mid }v-u\mathrm{\mid }$ attains a positive minimum there by [F8], so $h(K)$ is separated from $u$. Hence for all sufficiently small $\delta$, ${\inf }_{C\cap {\bar{B}}_{\delta }}D\ge d(u)+\eta$. For every sufficiently small matching radius ${\delta }_j$, we also have $d_{{\delta }_j}<d(u)+\eta \mathrm{/}2$. The LDP upper bound [F2] on the closed set $\overline{C\cap B_{{\delta }_j}}\subseteq C\cap {\bar{B}}_{{\delta }_j}$ bounds the numerator's logarithmic limsup exponent by $-d(u)-\eta$. By step 2.1 the denominator's logarithmic limit exponent is $-d_{{\delta }_j}$, so subtracting gives a conditional logarithmic limsup at most $d_{{\delta }_j}-d(u)-\eta <-\eta \mathrm{/}2$. Thus for every sufficiently large $j$ the conditional probability tends to zero exponentially as $n\to \mathrm{\infty }$, which gives the asserted iterated concentration limit.

Conversely, let $u$ maximize $s(v)-\beta v$ and let $q\in {\mathcal{M}}_u$. For any $r\in \mathrm{\Delta }(\mathcal{A})$, put $v=h(r)$. Then $D(r)+\beta v\ge d(v)+\beta v\ge d(u)+\beta u=D(q)+\beta u$, so $q\in {\mathcal{M}}_{\beta }$. This proves both the equality of energy sets and the macrostate criterion. Rearranging $s(v)-\beta v\le s(u)-\beta u$ gives exactly $s(v)\le s(u)+\beta (v-u)$; hence the criterion is the stated upper-supporting-line condition, with no formal inverse-transform assumption.

Write $q(+1)=(1+m)\mathrm{/}2$ and $q(-1)=(1-m)\mathrm{/}2$. Substituting into [F1] with $p(+1)=p(-1)=1\mathrm{/}2$ gives $D(q\mathrm{\parallel }p)=\frac{1+m}{2}\ln (1+m)+\frac{1-m}{2}\ln (1-m),$ with the zero-mass endpoint term omitted. Since $q$ depends continuously on $m\in [-1,1]$ and [F8] makes $D(q\mathrm{\parallel }p)$ continuous on the whole simplex, adding $\beta h(q)=-\beta J{m}^2\mathrm{/}2$ shows that ${\mathrm{\Phi }}_{\beta }$ is continuous on $[-1,1]$; [F8] supplies the boundary limit $t\ln t\to 0$ behind the convention $0\ln 0=0$. The formula is even in $m$.

For $\mathrm{\mid }m\mathrm{\mid }<1$ define $a(m):=\mathrm{atanh}(m):=\frac{1}{2}\ln \frac{1+m}{1-m}$. By [F3]–[F5] and direct differentiation, $a^{\mathrm{\prime }}(m)=\frac{1}{1-m^2},{\mathrm{\Phi }}_{\beta }^{\mathrm{\prime }}(m)=a(m)-\beta Jm\mathrm{.}$ For $0<m<1$ put $g(m):=a(m)\mathrm{/}m$ and define $r:[0,1)\to \mathbb{R}$ by $r(m):=m\mathrm{/}(1-m^2)-a(m)$. Then $g^{\mathrm{\prime }}(m)=r(m)\mathrm{/}{m}^2$, $r(0)=0$, and $r^{\mathrm{\prime }}(m)=2m^2\mathrm{/}(1-m^2{)}^2>0$ for $0<m<1$. The mean value theorem [F6] applies to $r$ on $[0,m]$ because $r$ is continuous there and differentiable on $(0,m)$; it gives $r(m)>0$, hence $g^{\mathrm{\prime }}(m)>0$.

The same theorem [F6] applied to $a$ on $[0,m]$ gives $g(m)=a^{\mathrm{\prime }}(c_m)$ for some $c_m\in (0,m)$; as $m\downarrow 0$, $c_m\downarrow 0$ and $a^{\mathrm{\prime }}(c_m)\to 1$, so ${\lim }_{m\downarrow 0}g(m)=1$. As $m\uparrow 1$, $(1+m)\mathrm{/}(1-m)\to +\mathrm{\infty }$, and the increasing onto logarithm [F3] gives $a(m)\to +\mathrm{\infty }$, hence $g(m)\to +\mathrm{\infty }$. Thus $g$ is continuous, strictly increasing from limiting value $1$ to $+\mathrm{\infty }$ on $(0,1)$.