Pome99DA2DCCD99FWhen two scalar functionals are uniformly close, their supporting probabilities are close as well. To make this precise, define for two model references $\varphi ,\psi $ in the preceding setting
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11.39.3
The comparison with $B_0$ in proposition 11.39.1 makes this number finite for every such pair. It is a pseudometric: symmetry, vanishing on the diagonal and the triangle inequality follow from the properties of the uniform norm. Distinct singularity types can have distance zero, as example 11.42.3 shows.
Proposition 11.39.4.7DD0741AE5AA For these model references,
where the supremum is over probability measures of finite entropy. The same value is obtained if (11.39.3) uses all bounded measurable $G$, and it does not depend on the smooth positive reference volume.
Proof. By corollary 11.38.5, both duals are finite exactly on the probability measures of finite entropy; rescaling the reference volume to have mass one does not affect finiteness. A uniform bound for $|B_\varphi -B_\psi |$ passes to the defining suprema, so the supremum in the statement is at most $\Delta (\varphi ,\psi )$. Conversely, denote that supremum by $D$. If $\nu \in \mathcal A_\psi (G)$, then $\nu $ has finite entropy, and the dual minimum formula gives
Interchanging the references gives the opposite inequality, so $\Delta (\varphi ,\psi )\leq D$.
For bounded measurable $G$, take continuous approximations with the same uniform bound converging in $L^1(\mathrm{d}V)$; the continuity theorem theorem 11.33.5, applied to each of the two references, extends the scalar bound to $G$. Finally, a change of volume $\mathrm{d}V'=\mathrm{e}^ a\mathrm{d}V$ replaces each $B(G)$ by $B(G+a)$, and translation by $a\in C(X)$ is a bijection of the space of tests, so the supremum is unchanged. $\square$
Proposition 11.39.5.80C599ADD5CB Let $\varphi _ j,\varphi $ be model references in the same background with $\Delta (\varphi _ j,\varphi )\to 0$. Let $G_ j,G$ be uniformly bounded measurable functions with $G_ j\to G$ in $L^1(\mathrm{d}V)$, and choose $\nu _ j\in \mathcal A_{\varphi _ j}(G_ j)$. Then the densities of the $\nu _ j$ are uniformly integrable, every weak cluster point of $(\nu _ j)$ belongs to $\mathcal A_\varphi (G)$, and a weakly convergent subsequence also converges weakly in $L^1$. If $\mathcal A_\varphi (G)$ is a singleton, the whole sequence converges to it in both senses.
Proof. Write $\delta _ j=\Delta (\varphi _ j,\varphi )$. The bounds $|B_{\varphi _ j}-B_\varphi |\leq \delta _ j$ and $|H_{\varphi _ j}-H_\varphi |\leq \delta _ j$, together with the identity $H_{\varphi _ j}(\nu _ j)=B_{\varphi _ j}(G_ j)+\int _ XG_ j\, \mathrm{d}\nu _ j$, give
If $\| G_ j\| _\infty \leq M$, then $H_\varphi (\nu _ j)\leq B_\varphi (0)+2M+2\delta _ j$, so all the measures lie in one finite sublevel set of $H_\varphi $. By theorem 11.35.4, their densities are uniformly integrable, and a weakly convergent subsequence also converges weakly in $L^1$.
Let $\nu $ be the limit of such a subsequence. Uniform integrability, the common bound and convergence in measure give $\int _ X|G_ j-G|\, \mathrm{d}\nu _ j\to 0$; weak $L^1$ convergence applies to the bounded function $G$, and theorem 11.33.5 gives $B_\varphi (G_ j)\to B_\varphi (G)$. Lower semicontinuity of $H_\varphi $ in the last inequality yields
The minimum formula for bounded measurable densities in theorem 11.36.1 gives the reverse inequality, and equality means exactly that $\nu \in \mathcal A_\varphi (G)$. When this set is a singleton, compactness gives convergence of the whole sequence. $\square$
The smaller-background comparison gives an explicit family of references converging to the zero reference.
Corollary 11.39.6.FCFB63397013 Let the model references $\varphi _ j$ satisfy $P_{\varepsilon _ j\theta }(\varphi _ j)\not\equiv -\infty $, where $0<\varepsilon _ j<1$ and $\varepsilon _ j\to 0$. Then $B_{\varphi _ j}\to B_0$ uniformly on $C(X)$. Let $G_ j,G\in C(X)$ with $G_ j\to G$ uniformly, and let $\nu _ j\in \mathcal A_{\varphi _ j}(G_ j)$. Then every weak cluster point of $(\nu _ j)$ belongs to $\mathcal A_0(G)$, the densities of the $\nu _ j$ are uniformly integrable, and a weakly convergent subsequence also converges weakly in $L^1(\mathrm{d}V)$. If $\mathcal A_0(G)$ is a singleton, the whole sequence converges to it in both senses.
Proof. Put $\delta _ j=-n\log (1-\varepsilon _ j)\to 0$. Proposition 11.39.1 gives $\Delta (\varphi _ j,0)\leq \delta _ j$, which is the uniform convergence of $B_{\varphi _ j}$ to $B_0$ on $C(X)$. The assertions about supporting probabilities follow from proposition 11.39.5 with the zero reference as limit. $\square$
For ordinary divisorial types, a comparison of positive residual forms gives a geometric criterion for this convergence of functionals, as the next two sections show. We do not know general convergence criteria for references with degenerate residuals or with arbitrary singularities.
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