Compact-space Laplace principle and exponential tilting

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 ${\nu }_n^F$.