15.10 A bounded subsolution theorem898B7B54B3D0

Proof. Write $\nu =\mathrm{MA}_\theta (\psi )$, a finite non-pluripolar measure.

Step 1: smooth strictly positive density. First suppose $f$ is smooth and strictly positive. Choose a Hermitian form $\Omega \geq \theta $ and, by theorem 5.32.1, smooth $\Omega $-plurisubharmonic functions $\psi _ j\downarrow \psi $. The envelopes

lie between $\psi $ and $\psi _ j$, so they are uniformly bounded and decrease to $\psi $. Contact support and ordered contact in the background $\Omega $ give

In particular, each $\mathrm{MA}_\theta (q_ j)$ has a bounded density with respect to smooth volume. Moreover $\mathrm{MA}_\theta (q_ j)(X)>0$ by proposition 16.28.55. Theorem 15.3.3 therefore gives bounded $u_ j\in \mathrm{PSH}(X,\theta )$ with

Since $f$ is bounded above and away from zero, we can choose $C$ such that $\mathrm{e}^{\lambda (q_ j-C)}f\leq 1\leq \mathrm{e}^{\lambda (q_ j+C)}f$ for all $j$. Then $q_ j-C$ and $q_ j+C$ are a subsolution and a supersolution, and bounded exponential comparison gives $q_ j-C\leq u_ j\leq q_ j+C$.

Extract an $L^1$ limit $u$, which is again bounded. The moving domination

and corollary 15.9.4, applied with the reference zero, give convergence in capacity. Continuity of bounded products under convergence in capacity, with the quasi-continuous weights $\mathrm{e}^{\lambda u_ j}f$, then gives

Step 2: bounded measurable density. Choose smooth positive functions $f_ j$ with a common upper bound $L>0$ such that $f_ j\to f$ in $L^1(\nu )$. Such an approximation exists: by regularity of the finite measure $\nu $, $f$ can be approximated by continuous functions on large compact sets, and continuous functions by smooth ones uniformly; truncating to a fixed bounded interval and adding a positive constant tending to zero preserves the bounds and the convergence. Step 1 gives bounded $u_ j$ with

For a fixed $C$ we have $\mathrm{e}^{\lambda (\psi -C)}f_ j\leq 1$ for all $j$, so $\psi -C$ is a subsolution and $u_ j\geq \psi -C$. The upper bound requires a separate argument, because $\inf _ Xf_ j$ may tend to zero.

The construction in Step 1 also gives the mixed inequality

To prove it without a mixed inequality for arbitrary singular measures, we use the approximants $u_{j,k}$ and $q_ k$ of Step 1, for which

and both Monge–Ampère measures have densities with respect to smooth volume. Lemma 5.28.7 applies to the pair $(u_{j,k},q_ k)$. For fixed $j$, both sequences are uniformly bounded and converge in capacity to $(u_ j,\psi )$, so continuity of bounded products against quasi-continuous weights passes the mixed inequality to the limit and yields (15.10.2).

Put $M_ j=\sup _ Xu_ j$, $v_ j=u_ j-M_ j$ and $c_ j=\mathrm{e}^{\lambda M_ j}$. Suppose the $M_ j$ were unbounded above. After extraction, $M_ j\to +\infty $ and $v_ j\to v$ in $L^1(X)$, where $v\in \mathrm{PSH}(X,\theta )$ is normalized. By lemma 5.44.2, $\mathrm{e}^{\lambda v_ j/n}\to \mathrm{e}^{\lambda v/n}$ in $L^1(\nu )$. Together with the inequalities

this gives

the last integral is positive because $\mu (X)>0$ and $\nu $ does not charge the pole set of $v$. Integrating (15.10.2) and applying lemma 5.29.20, we find some $a>0$ such that, along the extraction,

contradicting $M_ j\to +\infty $. Thus the $u_ j$ are uniformly bounded above as well as below.

Extract an $L^1$ limit $u$. Now $\mathrm{MA}_\theta (u_ j)\leq C''\nu $, so corollary 15.9.4 gives convergence in capacity and weak convergence of the Monge–Ampère measures. The right-hand sides converge in total variation, by the convergence of exponentials against the fixed measure $\nu $ and the estimate

Consequently $\mathrm{MA}_\theta (u)=\mathrm{e}^{\lambda u}\mu $. Uniqueness follows from bounded exponential comparison. $\square$