Theorem 15.12.3413B92CD2B72

Proof. The upper bound follows directly from the fixed-point property: since $u\leq 0$ and $u\preceq \varphi $, the function $u$ is a competitor for $P_\theta [\varphi ]=\varphi $.

For the lower bound we construct a subsolution $\psi $ of the type of $\varphi $ whose measure dominates a multiple of $\mathrm{e}^{-\gamma u}\mathrm{MA}_\theta (u)$, and then apply singular domination. Put $\varphi _\varepsilon =P_{\varepsilon \theta }(\varphi )$, so that $\varphi _\varepsilon \leq \varphi \leq 0$. Uniform exponential integrability (lemma 5.26.2) for the normalized potentials $u$ and exponential integrability for the fixed function $\varphi _\varepsilon $ allow us to choose $\gamma >0$ and $1<q<p$ such that

has a uniform $L^ q$ bound. Explicitly, we first choose $q<p$ and apply Hölder's inequality with exponents $p/q$ and $p/(p-q)$; then we choose $\gamma $ so small that the exponential factor is controlled by the Cauchy–Schwarz inequality and the two exponential integrability estimates.

Theorem 15.3.5, applied in the semipositive big background $(1-\varepsilon )\theta $ to the density $1+g$, gives a bounded $v$ with $\sup _ Xv=0$ and a constant $c>0$ such that

Both $\| v\| _\infty $ and $c^{-1}$ are uniformly bounded. Adding the constant one to $g$ makes the root integral of the density at least one, so no lower bound for the integral of $f$ is needed.

Put $h=\varphi _\varepsilon +v\leq \varphi $, which is $\theta $-plurisubharmonic, and define

where the formula for $\psi $ is evaluated off the pole sets. Then $0\leq \chi \leq 1$, $0<\chi '\leq 1$, and $\chi ''>0$. For bounded potentials with $h\leq \varphi $, lemma 5.28.13, applied with $G(x,y)=x+\chi (y-x)$, gives

For unbounded potentials we apply this bounded assertion to $\varphi _ k=\max (\varphi ,-k)$ and $h_ k=\max (h,-k)\leq \varphi _ k$. Since $G$ is nondecreasing in both variables, $G(\varphi _ k,h_ k)$ decreases as $k$ increases. Its limit is $\theta $-plurisubharmonic and lies between $\varphi $ and $\varphi +1$; hence it is not identically $-\infty $, and it equals $\psi $. Locality on common finite levels and positivity of bounded products give

where the second inequality uses $\mathrm{MA}^{\mathrm{ext}}_\theta (h)\geq ((1-\varepsilon )\theta +\mathrm{dd}^{\mathrm{c}}v)^ n\geq cg\, \mathrm{d}V$. These inequalities are first obtained off the pole sets; since pole sets have zero volume, they extend by exhaustion to all Borel sets.

Write $T=-\varphi _\varepsilon \geq 0$. If $C_1\geq 1+\| v\| _\infty $, then $\varphi -h+1\leq T+C_1$. Since

we obtain, with a uniform $d>0$,

Choose $K=2+\gamma ^{-1}|\log d|$. On $\{ u<\psi -K\} $ we have $u<1-K$, because $\psi \leq \varphi +1\leq 1$; hence

The potentials $u$ and $\psi -K$ have the same ordinary singularity type. Moreover, $u\geq \varphi -C_ u$ for some constant $C_ u$, so $P_{\varepsilon \theta }(u)\geq \varphi _\varepsilon -C_ u$ is not identically $-\infty $. Theorem 15.2.2 therefore yields $u\geq \psi -K\geq \varphi -K$. All constants used to choose $K$ are uniform over the family in the theorem; the possibly nonuniform constant $C_ u$ serves only to show that the smaller-background envelope of $u$ is not identically $-\infty $, and it enters no estimate. $\square$