Lemma 15.14.1.869D3138741D Fix $p>1$ and $L>0$. There is $m>0$ such that, whenever
there exists $v\in \mathrm{PSH}(X,\theta )$ with
The constant $m$ depends only on the fixed geometry and on $\rho ,p,L$.
869D3138741DLemma 15.14.1.869D3138741D Fix $p>1$ and $L>0$. There is $m>0$ such that, whenever
there exists $v\in \mathrm{PSH}(X,\theta )$ with
The constant $m$ depends only on the fixed geometry and on $\rho ,p,L$.
Proof. Choose $1<q<p$. By Hölder's inequality and the finiteness of all polynomial moments of the fixed quasi-plurisubharmonic function $\rho $, the functions $h=g(2-\rho )^{2n}$ have uniformly bounded $L^ q$ norms. Lemma 15.3.1, applied in the Hermitian background $\omega $, gives $s\in \mathrm{PSH}(X,\omega )$ with $-1\leq s\leq 0$ and
for a fixed $m>0$. Put $u_1=\rho +s$. Then $\theta _{u_1}\geq \omega _ s$ and $u_1\leq \rho \leq \eta $. On $\Omega $ define
so that $1\leq t\leq 2-\rho $. Lemma 5.28.13, applied on relatively compact subsets of $\Omega $ with $a=\eta $, $h=u_1$ and $\chi (x)=(1-x)^{-1}$, gives
Since $\eta \leq v\leq \eta +1$, lemma 5.24.7 extends $v$ across $Z$, and the bounds extend by lemma 5.24.3. On $\Omega $, positivity of the local products gives
Neither side charges $Z$, so the inequality holds on $X$. $\square$
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