Reference-relative entropy and Gibbs variational inequality

For $x>0$, set $h(x)=x-1-\log x$. By [F1], $h^{\mathrm{\prime }}(x)=1-1\mathrm{/}x$, which is negative for $0<x<1$ and positive for $x>1$; hence $h(x)\ge h(1)=0$, with equality only at $x=1$. On $\{p>0\}$ put $x=q\mathrm{/}p$ to obtain $p\log (p\mathrm{/}q)\ge p-q$. The negative part of the left integrand is integrable: where $q>p$, writing $x=q\mathrm{/}p>1$ gives $p\log (q\mathrm{/}p)=q(\log x)\mathrm{/}x\le q$, since $h(x)\ge 0$ implies $\log x\le x-1$ and $(x-1)\mathrm{/}x\le 1$. Therefore its integral is well-defined as an extended real number, and $D(p\mathrm{\parallel }q)\ge {\int }_{\{p>0\}}(p-q)dm=1-{\int }_{\{p>0\}}qdm\ge 0.$

If $D=0$, both nonnegative gaps in the last inequalities vanish: $q$ has no mass on $\{p=0\}$, and $h(q\mathrm{/}p)=0$ $p$-almost everywhere. The strict equality case in step 1.1 gives $q=p$ on $\{p>0\}$, hence $p=q$ almost everywhere. Conversely, if $p=q$ almost everywhere then the defining integrand is zero and $D=0$.

In the Gibbs case, [F2] and [A2] give $\log q=-\beta H-\log Z$ $p$-almost everywhere, so all terms below are integrable and $D(p\mathrm{\parallel }q)=\int p\log pdm+\beta ⟨H{⟩}_p+\log Z=\beta (⟨H{⟩}_p+{\beta }^{-1}\int p\log pdm-F)\mathrm{.}$ Since $S[p]=-k\int p\log pdm$ and ${\beta }^{-1}=kT$, the middle term is $-TS[p]$. Nonnegativity and the equality condition follow from steps 1.1 and 2.1.

If a density gives positive mass to $\{q=0\}$, its extended $D$ is $+\mathrm{\infty }$ by definition; otherwise step 1.1 applies after removal of null sets. Thus ${\mathcal{F}}_{\mathrm{e}\mathrm{x}\mathrm{t}}\ge F$ for every probability density. The density $q$ is allowed here without any energy or entropy integrability assumption: directly $D(q\mathrm{\parallel }q)=0$, so ${\mathcal{F}}_{\mathrm{e}\mathrm{x}\mathrm{t}}[q]=F$. Step 2.1 gives uniqueness. Step 3.1 proves agreement with the original finite formula on $\mathcal{A}$, without assigning any meaning to an undefined difference $+\mathrm{\infty }-\mathrm{\infty }$.

Let $Q(B):={\int }_{B}qdm$. Positive finite $Z$ implies $H$ is finite $Q$-almost everywhere: $H=+\mathrm{\infty }$ has zero Gibbs weight, while $H=-\mathrm{\infty }$ can occur only on an $m$-null set, since its exponential weight is infinite. Set $A_n:=\{\mathrm{\mid }H\mathrm{\mid }\le n\}$ and $c_n:=Q(A_n)$. Then $c_n\uparrow 1$. Indeed the disjoint sets $A_1,A_2\setminus A_1,\dots$ cover the finite-energy set, so countable additivity of the probability measure $Q$ makes their partial sums $c_n$ converge to $1$. For every sufficiently large $n$, $c_n>0$ and $p_n:=q{\mathbf{1}}_{A_n}\mathrm{/}{c}_n$ is a probability density. On its support $\mathrm{\mid }H\mathrm{\mid }\le n$ and $\log p_n=-\beta H-\log Z-\log c_n$ is bounded, so $p_n\in \mathcal{A}$. Moreover $p_n\mathrm{/}q=1\mathrm{/}{c}_n$ on that support, hence $D(p_n\mathrm{\parallel }q)=-\log c_n⟶0.$ The limit follows from the continuity of the logarithm at $1$ supplied by [F1]. Step 3.1 now gives finite free energies $F-{\beta }^{-1}\log c_n\to F$, proving the stated infimum. Equality for any $p\in \mathcal{A}$ forces $p=q$ almost everywhere by step 2.1; consequently it is attained exactly when $q\in \mathcal{A}$, which is precisely the two stated integrability conditions.