Theorem 15.33.1.35816840F731 Let $0\leq f\in L^1(\mathrm dV)$ have positive integral. Suppose that $A,\alpha >0$ satisfy
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15.33.2
\begin{equation} \int _ Ef\, \mathrm dV\leq A\operatorname {Cap}_ h(E)^{1+\alpha } \qquad (E\subset X\text{ Borel}). \end{equation}
There is $C>0$, depending only on $(X,h),A,\alpha $, such that every measurable $0\leq g\leq f$ with positive integral satisfies
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15.33.3
\begin{equation} b_ h(g)\geq C\left(\int _ Xg\, \mathrm dV\right)^{-\alpha /(1+\alpha )}, \qquad \rho _ h(g^{1/n})\leq C^{-1/n} \left(\int _ Xg\, \mathrm dV\right)^{\alpha /(n(1+\alpha ))}. \end{equation}
In particular, $f^{1/n}\in \mathcal D_ h^0$. There is $\phi \in \mathrm{PSH}(X,h)\cap C(X)$, with $\sup _ X\phi =0$, such that
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15.33.4
\begin{equation} \mathrm{MA}_ h(\phi )=b_ h(f)f\, \mathrm dV. \end{equation}
The coefficient $b_ h(f)$ and the mass of this measure are finite and strictly positive.
Proof.
Step 1: the coefficient and root tails. Put $g_ j=\min (g,j)$ and let $\phi _ j$ be normalized continuous solutions of
\[ \mathrm{MA}_ h(\phi _ j)=c_ jg_ j\, \mathrm dV, \qquad c_ j=c_ h(g_ j),\qquad \sup _ X\phi _ j=0, \]
given by theorem 14.19.2. The definition (15.28.1) gives $c_ j\downarrow b_ h(g)$. Their measures satisfy a power capacity bound with constant $c_ jA$.
Take competitor zero, $H=1$ and $\varepsilon =1/2$ in lemma 14.19.16, lemma 14.19.14. Choose $D>0$ fixed with $2D<\delta _{1/2}$ and put $U_ D=\{ \phi _ j<\inf _ X\phi _ j+D\} $. These estimates give
\[ D\leq C_\alpha (C_1c_ jA)^{1/n} \operatorname {Cap}_ h(U_ D)^{\alpha /n}, \qquad D^ n\operatorname {Cap}_ h(U_ D) \leq C_1c_ j\int _ Xg_ j\, \mathrm dV. \]
Combining them yields
\[ c_ j^{1+\alpha }\left(\int _ Xg_ j\, \mathrm dV\right)^\alpha \geq c>0. \]
Taking the limit proves the lower bound for $b_ h(g)$. The identity (15.29.2), with its full family of Gauduchon tests, gives the root estimate. Apply it to $g=f\mathbf1_{\{ f>K^ n\} }$ whenever this density is nonzero. Its mass tends to zero; the zero density has root norm zero. Thus the tails of $f^{1/n}$ tend to zero in both $L^ n$ and $\rho _ h$, and proposition 15.32.9 gives $f^{1/n}\in \mathcal D_ h^0$.
Step 2: uniform convergence of a subsequence. Now use $f_ j=\min (f,j)$, $c_ j=c_ h(f_ j)$ and normalized continuous solutions $\phi _ j$. Step 1 gives $b_ h(f)>0$, while $c_ j\leq c_1<\infty $. Hence
\[ \mathrm{MA}_ h(\phi _ j)(E) \leq Ac_1\operatorname {Cap}_ h(E)^{1+\alpha }. \]
The first part of the proof of proposition 14.19.23, using the capacity tail estimate, gives one $M\geq 1$ with $-M\leq \phi _ j\leq 0$ for all $j$.
By theorem 5.25.2, extract a subsequence converging in $L^1(\mathrm dV)$ and almost everywhere. Since $f\, \mathrm dV$ is a fixed finite absolutely continuous measure and the potentials have a common bound, dominated convergence gives
\[ \int _ X|\phi _ j-\phi _ k|\, \mathrm{MA}_ h(\phi _ j) \leq c_1\int _ X|\phi _ j-\phi _ k|f\, \mathrm dV\longrightarrow 0 \]
along this subsequence as both indices tend to infinity. Apply proposition 14.19.23 in both orders, with $H=M$. The subsequence is uniformly Cauchy. Its limit $\phi $ is continuous, $h$-plurisubharmonic by distributional positivity, and has supremum zero.
Step 3: passage of the products. On a coordinate ball choose a fixed smooth $q$ with $\mathrm{dd}^{\mathrm{c}}q\geq h$, and put $S_ j=\mathrm{dd}^{\mathrm{c}}(\phi _ j+q)$ and $S=\mathrm{dd}^{\mathrm{c}}(\phi +q)$. For every $\ell \geq 1$, the calculus of bounded products gives
\[ S_ j^\ell -S^\ell =\mathrm{dd}^{\mathrm{c}}\left[(\phi _ j-\phi ) \sum _{a=0}^{\ell -1}S_ j^ a\wedge S^{\ell -1-a}\right]. \]
The local Chern–Levine–Nirenberg estimate bounds its pairing with a compactly supported smooth test form by $C\| \phi _ j-\phi \| _\infty $. Expand the Hermitian products as in proposition 5.28.2:
\[ \mathrm{MA}_ h(\phi _ j) =\sum _{\ell =0}^ n\binom n\ell S_ j^\ell \wedge (h-\mathrm{dd}^{\mathrm{c}}q)^{n-\ell }. \]
Every term converges, with its smooth correction factor, to the corresponding term for $\phi $. On the other hand, $c_ jf_ j\to b_ h(f)f$ in $L^1(\mathrm dV)$. The two limits prove (15.33.4); its prescribed mass is $b_ h(f)\int _ Xf\, \mathrm dV$.
$\square$
Pome
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