15.3 Bounded equations in a semipositive big background9F0629AA9014

Proof. Choose $1<q<p$. Since a fixed quasi-plurisubharmonic function belongs to every finite $L^ r$ space, Hölder's inequality gives a uniform $L^ q$ bound for

By theorem 14.19.2 there are $c>0$ and a continuous $\omega $-plurisubharmonic $v$, normalized by $\sup _ Xv=-1$, such that

The construction in that theorem gives a uniform upper bound for $c$, because in its Gauduchon trace estimate the root integral of $1+g$ is bounded below by the root integral of $1$. The lower-density estimate of theorem 14.19.19 gives $c\| 1+g\| _ q\geq c_ q>0$, so $c$ is also bounded below. The potential estimate of the same theorem then gives $-C\leq v\leq -1$ with a uniform $C$.

Put $h=\rho +av\leq -1$. Then $\theta _ h\geq a\omega _ v$. On $(-\infty ,-1]$ define

and set $s=\chi (h)$, with value $-1$ at the poles of $h$. To justify both the positivity of $\theta +\mathrm{dd}^{\mathrm{c}}s$ and the computation of its Monge–Ampère measure, we first work with the bounded functions $h_ k=\max (h,-k)$, $k\geq 1$. The chain rule for bounded potentials and $\theta \geq 0$ give

These bounded potentials decrease to $s$, so $s\in \mathrm{PSH}(X,\theta )$ and $-1\leq s\leq 0$. On $\{ h>-k\} $, locality and positivity of mixed products give

Letting $k\to \infty $, we obtain this inequality on the set where $h$ is finite; its complement is pluripolar and has zero volume. Since $|h|\leq (-\rho )+aC\leq (1+aC)(-\rho )$, the assertion follows with any common lower bound for $a^ nc(1+aC)^{-2n}$. $\square$

Proof. For $0<\varepsilon \leq 1$, set $\theta _\varepsilon =\theta +\varepsilon \omega $. By the Hermitian existence theorem theorem 14.19.3, there is a continuous solution $v_\varepsilon $ of

Take $s,b$ from lemma 15.3.1, and put $w=s+\lambda ^{-1}\log b$. Positivity of bounded products gives

and bounded exponential comparison (corollary 14.19.12) yields $w\leq v_\varepsilon $. If $0<\varepsilon '<\varepsilon $, then $v_{\varepsilon '}$ is also $\theta _\varepsilon $-plurisubharmonic and

so the same comparison gives $v_{\varepsilon '}\leq v_\varepsilon $. Hence $v_\varepsilon $ decreases to a bounded $\theta $-plurisubharmonic function $v$ with $w\leq v\leq v_1$.

To pass to the limit in the equation, we use the fixed Hermitian background $\Omega =\theta +\omega $ and the expansion

Continuity of bounded products along monotone sequences applies to every term, and the smooth coefficients converge. Dominated convergence applies to the right-hand sides. This proves the equation. Uniqueness follows from bounded domination in the semipositive big background (theorem 16.28.56, proposition 16.28.55) by the usual exponential comparison argument.

For the uniform assertion, note first that the lower barrier $w$ is uniform by lemma 15.3.1. The upper barrier $v_1$ solves an equation in the fixed Hermitian background $\Omega $. The proof of lemma 14.19.26 bounds its supremum using only an upper bound for the density norm and a positive lower bound for the root integral: these two bounds control, respectively, the Hölder term and the denominator in (14.19.27), and they survive the smooth approximation used there. Thus $v_1$, and hence $v$, has a uniform upper bound. $\square$

Proof. Step 1: approximating equations and their constants. For integers $j\geq 1$, put $\omega _ j=\theta +2^{-j}\omega $, and choose continuous normalized solutions

We first bound $c_ j$. Fix a Gauduchon metric $\gamma _0$; when $n=1$ any Hermitian form will do, and the calculation below reduces to Stokes' theorem. The smooth approximation in the proof of theorem 14.19.2, together with the Gauduchon trace, gives a common upper bound: the numerator $\int _ X\omega _ j\wedge \gamma _0^{n-1}$ is bounded above by $\int _ X(\theta +\omega )\wedge \gamma _0^{n-1}$, and the root integral of $f$ is positive and fixed. Here the estimate is applied to the smooth approximating equations before passing to their continuous limits.

For a lower bound, take $s,b$ from lemma 15.3.1. If $c_ j<b$, then

for every constant $A$, and strict domination in the Hermitian background $\omega _ j$ (theorem 14.19.10) would give $\varphi _ j\geq s+A$ for every $A$, which is absurd. Thus $c_ j\geq b$. In fact the constants decrease with $j$: if $k>j$, positivity gives $\mathrm{MA}_{\omega _ j}(\varphi _ k)\geq c_ kf\, \mathrm{d}V$, and the same argument excludes $c_ j<c_ k$. Hence $c_ j\downarrow c>0$.

Step 2: a uniform bound for the potentials. For the Hermitian form $\Omega =\theta +\omega $, all $\varphi _ j$ are normalized $\Omega $-plurisubharmonic functions. Using uniform exponential integrability (lemma 5.26.2) and Hölder's inequality, choose $\gamma >0$ small and $1<q<p$ such that

is bounded in $L^ q$. Since $\varphi _ j\leq 0$, we have $g_ j\geq f$. Theorem 15.3.3, applied with integrability exponent $q$ and exponential exponent $\gamma $, therefore gives uniformly bounded $w_ j\in \mathrm{PSH}(X,\theta )$ with

Choose $C_0\geq \max \{ 1,\sup _ jc_ j\} $ and set $z_ j=w_ j-\gamma ^{-1}\log C_0$. Then

and exponential comparison gives $z_ j\leq \varphi _ j\leq 0$. Thus all $\varphi _ j$ take values in a common bounded interval.

Step 3: the upper envelopes. Extract a subsequence converging in $L^1$ and almost everywhere to the canonical $\theta $-plurisubharmonic representative $\varphi $. We keep the indices $j$; the coefficients of $\omega $ in $\omega _ j$ still decrease to zero. Define

All these functions take values in a common bounded interval. The upper envelopes are $\omega _ j$-plurisubharmonic and decrease to $\varphi $, since the regularized upper envelope agrees almost everywhere with the $L^1$ limit; this also follows from the local Hartogs lemma applied to the bounded tails.

For $\ell \geq j$, positivity gives $\mathrm{MA}_{\omega _ j}(\varphi _\ell )\geq c_\ell f\, \mathrm{d}V\geq cf\, \mathrm{d}V$. By ordered contact (lemma 5.28.3), finite maxima of these functions satisfy the same lower bound. Continuity of bounded products along increasing sequences, followed by the decreasing limit in $j$ and the expansion in the smooth background, gives

Step 4: the lower envelopes. For the reverse inequality we use lower envelopes, which carry a useful exponential weight. For $N\geq j$ put

The obstacle is continuous and bounded, so the measure of $p_{j,N}$ is carried by its contact set. Partition this set into disjoint Borel pieces according to the first index $\ell $ at which $p_{j,N}=\varphi _\ell $. On such a piece, positivity and ordered contact in the background $\omega _\ell $ give

Moreover $p_{j,N}=\varphi _\ell \geq h_ j$ there, and $c_\ell \leq c_ j$. Summing over the pieces therefore yields the global inequality

As $N\to \infty $, the envelopes decrease to $p_ j$. Indeed, their limit lies below each tail obstacle quasi-everywhere and is thus a competitor for $P_\theta (h_ j)$, while the reverse inequality is immediate. Continuity of bounded products along monotone sequences and dominated convergence give

The functions $p_ j$ increase. Let $\varphi '=(\lim _ jp_ j)^*$. Since $h_ j\to \varphi $ almost everywhere, we have $\varphi '\leq \varphi $ for the canonical representatives. Passing to the increasing limit and using (15.3.6), we obtain

Bounded exponential comparison forces $\varphi '\geq \varphi $. Thus $\varphi '=\varphi $, and the last inequality, together with (15.3.6), proves the equation. If necessary, subtracting $\sup _ X\varphi $ gives the required normalization.

Step 5: uniformity and uniqueness of $c$. All bounds used above depend only on the two stated density bounds and the fixed data, which proves the uniform assertion for the constructed solution. Finally, two different positive constants for the same $f$ would contradict strict domination after an arbitrary translation of one of the bounded solutions, exactly as in the lower bound for $c_ j$. This proves the uniqueness of $c$. $\square$