PomeD8DD45551DB2D8DD45551DB2
To pass to the limit in the supporting inequality, we need to control the mass that supporting probabilities can put on small sets. The convex dual of $B_\varphi $ provides this control: the measures in each of its sublevel sets are uniformly absolutely continuous. This allows the logarithmic density and the supporting measure to vary simultaneously.
Let $\mathcal P(X)$ denote the set of Borel probability measures on $X$, with the weak topology. Define
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11.35.1
The value $+\infty $ is allowed, while taking $G=0$ shows that $H_\varphi \geq B_\varphi (0)>-\infty $.
Proposition 11.35.2.0C21983AC42D The functional $H_\varphi $ is convex and weakly lower semicontinuous. For every $F\in C(X)$,
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11.35.3
The minimizers are exactly $\mathcal A_\varphi (F)$.
Proof. Each function of $\nu $ in the supremum defining $H_\varphi $ is affine and weakly continuous, so their supremum is convex and lower semicontinuous. By definition, $H_\varphi (\nu )-\int _ XF\, \mathrm{d}\nu \geq B_\varphi (F)$, with equality precisely when
This is the defining inequality of $\mathcal A_\varphi (F)$, and the minimum is attained because $\mathcal A_\varphi (F)$ is nonempty by theorem 11.34.2. $\square$
Theorem 11.35.4.6830F8DE9750 For every finite real $L$, the set
is weakly compact. Its measures are uniformly absolutely continuous with respect to $\mathrm{d}V$: for every $\epsilon >0$ there is a $\delta >0$ such that
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11.35.5
for every Borel set $E$. In particular, writing $\mathrm{d}\nu =h_\nu \mathrm{d}V$, the densities in this set are uniformly integrable in $L^1(\mathrm{d}V)$. On $\mathcal K_ L$, weak convergence of measures also implies weak convergence of their densities in $L^1(\mathrm{d}V)$.
Proof. Weak compactness follows from lower semicontinuity and compactness of $\mathcal P(X)$. We may assume that the sublevel set is nonempty, so that $L\geq B_\varphi (0)$. Fix $t>0$. By theorem 11.33.5, there is $\delta _ t>0$ such that
Indeed, otherwise there would be a sequence of such functions $\chi $ tending to zero in $L^1$ with $B_\varphi (t\chi )\leq B_\varphi (0)-1$, contrary to that theorem. For $\nu \in \mathcal K_ L$, the defining supremum gives
Choose $t$ so large that $(L-B_\varphi (0)+1)/t<\epsilon /2$. If $\mathrm{d}V(E)<\delta _ t/2$, outer regularity gives an open neighborhood $O$ of $E$ with $\mathrm{d}V(O)<\delta _ t$. For any compact $K\subset E$, choose a continuous cutoff $0\leq \chi \leq 1$ equal to one on $K$ and supported in $O$. The last inequality gives $\nu (K)<\epsilon /2$, uniformly in $\nu \in \mathcal K_ L$, and inner regularity yields $\nu (E)\leq \epsilon /2<\epsilon $. This proves (11.35.5).
In particular, every measure in the sublevel set is absolutely continuous, with a density of integral one. Together with the uniform small-set estimate, this is exactly uniform integrability on the finite measure space $(X,\mathrm{d}V)$.
Finally, suppose that $\nu _ j\to \nu $ weakly within the sublevel set. For a bounded measurable $G$, choose continuous $G_ k$ with $|G_ k|\leq \| G\| _\infty $ and $G_ k\to G$ in $\mathrm{d}V$-measure; such approximations exist by Lusin's theorem. Uniform absolute continuity gives
To see this, split $X$ into $\{ |G_ k-G|\leq a\} $ and its complement; the integral over the first set is at most $a$, and the integral over the second is at most $2\| G\| _\infty $ times a uniformly small measure. For fixed $k$, weak convergence applies to $G_ k$. Letting first $j\to \infty $ and then $k\to \infty $ proves convergence of the integrals against $G$, which is exactly weak $L^1(\mathrm{d}V)$ convergence of the densities. In particular, every sequence in the sublevel set has a subsequence whose densities converge weakly in $L^1$. $\square$
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