Pome52D96CD9BFA752D96CD9BFA7
The domination principle makes the constant decrease as the density increases, and mixed Monge–Ampère products strengthen this monotonicity to concavity in the logarithmic density. Compactness of the solutions then gives continuity under uniformly bounded $L^1$ perturbations.
For a bounded real measurable function $F$, part (2) of theorem 15.12.1, applied to the bounded density $\mathrm{e}^ F$, gives $u\in \mathrm{PSH}(X,\theta )$ with $u\sim \varphi $ and a unique constant $c_\varphi (F)>0$ such that
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11.33.1
The normalization scalar is
Since the constant is unique, $B_\varphi $ is well defined; it depends on $\varphi $ only through its ordinary singularity type.
Proposition 11.33.2.2770AB03C0D2 Let $F,G\in L^\infty (\mathrm{d}V)$ be real. The functional $B_\varphi $ is decreasing, and for every constant $a\in \mathbb R$,
Consequently
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11.33.3
and $|B_\varphi (G)-B_\varphi (F)|\leq \| G-F\| _\infty $.
Proof. Suppose that $F\leq G$ almost everywhere and, to reach a contradiction, that $c_\varphi (F)<c_\varphi (G)$. Let $u,v\sim \varphi $ solve the two equations. Since $F\leq G$ and the extended measure is unchanged by adding constants, for every constant $a$,
and the ratio is strictly less than one. As $u\sim \varphi $, the envelope $P_{\varepsilon \theta }(u)$ dominates a translate of $P_{\varepsilon \theta }(\varphi )$ and is therefore not identically $-\infty $; moreover $v+a\preceq u$. The singular strict domination principle, theorem 15.2.2, gives $u\geq v+a$ for every $a$, which is impossible at any point where both potentials are finite. Thus $B_\varphi (F)\geq B_\varphi (G)$.
Replacing $F$ by $F+a$ and the constant by $\mathrm{e}^{-a}c_\varphi (F)$ leaves the right-hand side of (11.33.1) unchanged, so uniqueness of the constant proves the translation rule. Finally, $G$ lies between the constant translates $F+\mathop{\rm ess\, inf}_ X(G-F)$ and $F+\mathop{\rm ess\, sup}_ X(G-F)$; monotonicity and the translation rule give (11.33.3). $\square$
The scalar does not depend on the smooth representative of the background: for a smooth real $h$, replacing $\theta $ by $\theta +\mathrm{dd}^{\mathrm{c}}h$ and each potential $u$, including $\varphi $, by $u-h$ preserves the currents and the prescribed ordinary type, and hence leaves the equation and its constant unchanged.
Concavity comes from mixed products. A convex combination of two solutions stays in the prescribed ordinary type, and lemma 13.14.10 bounds its measure below by the corresponding mixed density.
Theorem 11.33.4.8810E18F852C The functional $B_\varphi $ is concave on the real space $L^\infty (\mathrm{d}V)$: for $F,G$ in that space,
Proof. Let $u,v\sim \varphi $ solve the equations for $F$ and $G$, and set $H=tF+(1-t)G$ and $D=c_\varphi (F)^ t c_\varphi (G)^{1-t}$. Lemma 13.14.10 and the arithmetic–geometric mean inequality give
The averaged potential $tu+(1-t)v$ has the ordinary type of $\varphi $. If $c_\varphi (H)<D$, its measure dominates the measure of a solution $w\sim \varphi $ for $H$ by the factor $D/c_\varphi (H)>1$. Applying the singular strict domination principle to $w$ and to every constant translate of the averaged potential, as in the proof of proposition 11.33.2, leads to the same contradiction. Hence $c_\varphi (H)\geq D$, and taking logarithms proves concavity. $\square$
For smooth Hermitian data, this concavity agrees with the second-variation formula of theorem 14.8.1.
Uniformly bounded logarithmic densities give the solutions a common lower barrier, and compactness of the solutions identifies the limiting scalar even when the densities converge only in $L^1$.
Theorem 11.33.5.5E3583383688 If $F_ j,F$ are uniformly bounded real measurable functions and $F_ j\to F$ in $L^1(\mathrm{d}V)$, then
Proof. Choose solutions $u_ j\sim \varphi $ normalized by $\sup _ Xu_ j=0$, and write $c_ j=c_\varphi (F_ j)$. If $|F_ j|\leq M$, monotonicity and the translation rule of proposition 11.33.2 give
In particular, the Monge–Ampère measures of the $u_ j$ are dominated by a fixed multiple of $\mathrm{d}V$, and theorem 15.12.3 gives a constant $C$ such that
From any subsequence, extract a further subsequence with $c_ j\to c>0$ and $u_ j\to u$ in $L^1(X)$. By corollary 15.9.4, the convergence also holds in capacity, $u\sim \varphi $, and the measures converge weakly. On the other hand, the exponential function is Lipschitz on the common range of the $F_ j$, so $c_ j\mathrm{e}^{F_ j}\to c\mathrm{e}^ F$ in $L^1(\mathrm{d}V)$. Therefore
and uniqueness of the constant gives $c=c_\varphi (F)$. Since every subsequence has a further subsequence along which $c_ j\to c_\varphi (F)$, the whole sequence converges, and so do the logarithms. $\square$
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