11.39 Changing the prescribed singularities44DF399743E5

Proof. Fix $G$, and let $u_\varphi ,u_\psi $ solve the equations in the ordinary types of $\varphi $ and $\psi $, with constants $c_\varphi ,c_\psi $. Then $u_\varphi \preceq u_\psi $. If $c_\varphi >c_\psi $, the singular strict domination principle applied to $u_\psi $ and $u_\varphi +t$ would give $u_\psi \geq u_\varphi +t$ for every real $t$, which is impossible at a point where both are finite. This proves the monotonicity of $B$, and taking the defining supremum gives the monotonicity of $H$. Since $\varphi \preceq 0$ and the zero potential is itself a model reference, this also proves both upper bounds.

For the lower bound, let $v$ be a bounded solution for $G$ with the zero reference, with constant $c_0$, and put

The two summands are $\varepsilon \theta $- and $(1-\varepsilon )\theta $-plurisubharmonic, respectively, so positivity and homogeneity give

As in the proof of lemma 15.12.2, this inequality is justified first on finite truncation levels and then off the pole set of $P_{\varepsilon \theta }(\varphi )$, which has zero smooth volume. Also $h\preceq u_\varphi $. If $c_\varphi <(1-\varepsilon )^ nc_0$, strict domination would force $u_\varphi \geq h+t$ for every $t$. Thus $c_\varphi \geq (1-\varepsilon )^ nc_0$. Taking logarithms proves the scalar bound, and taking the defining suprema proves the dual bound. $\square$

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. $\square$

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. $\square$

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. $\square$