Proposition 12.10.20.D804C5144A2D Assume the hypotheses of theorem 12.10.17, and let $u_0$ be its unique threshold solution with $\sup _ Xu_0=0$. For real $\lambda >0$, let $u_\lambda $ be the greatest solution of the exponential equation (15.35.1), with $\omega $ in place of $h$, given by corollary 15.35.3, and put
\[ v_\lambda =u_\lambda -\sup _ Xu_\lambda ,\qquad c_\lambda =\mathrm{e}^{\lambda \sup _ Xu_\lambda }. \]
There is a constant $C$ independent of $\lambda >0$ such that
\[ b\leq c_\lambda \leq b\, \mathrm{e}^{C\lambda },\qquad u_0-C\leq v_\lambda \leq 0. \]
Moreover $v_\lambda \to u_0$ in $L^1(X)$ as $\lambda \downarrow 0$.
Proof.
Write $p=f^{1/n}$ and introduce the probability measure
\[ \mathrm{d}\nu =b^{1/n}p k_ T\, \mathrm{d}V. \]
It has mass one because $T$ maximizes. Its density belongs to $L^ n$, since $p\in L^ n$ and the smooth semipositive form $T$ has bounded determinant density $k_ T$. By lemma 5.26.2, $\omega $-plurisubharmonic functions with supremum zero have uniformly bounded moments of every finite order. Hölder's inequality with the exponents $n$ and $n/(n-1)$, which is finite because $n\geq 2$, therefore gives
\[ \int _ X(-v_\lambda )\, \mathrm{d}\nu \leq C. \]
Since $\mathrm{MA}^{\mathrm{ext}}_\omega (v_\lambda )=c_\lambda \mathrm{e}^{\lambda v_\lambda }f\, \mathrm{d}V$, the trace inequality of lemma 13.14.12, applied with the positive regularizations of $T$ and passed to the limit as in the preceding proof, gives $(\omega +\mathrm{dd}^{\mathrm{c}}v_\lambda )\wedge T\geq (c_\lambda /b)^{1/n}\mathrm{e}^{\lambda v_\lambda /n}\, \nu $. Integrating, we obtain
\[ 1\geq (c_\lambda /b)^{1/n} \int _ X\mathrm{e}^{\lambda v_\lambda /n}\, \mathrm{d}\nu \geq (c_\lambda /b)^{1/n}\mathrm{e}^{-C\lambda /n}, \]
where the last step is Jensen's inequality. This proves the upper bound for $c_\lambda $. The lower bound and the comparison
\[ v_\lambda \geq u_0-\lambda ^{-1}\log (c_\lambda /b) \]
follow from corollary 15.35.3 applied with $v=u_0$, and together with the upper bound they give the common lower barrier $u_0-C$.
Take any sequence $\lambda _ j\downarrow 0$. By theorem 5.25.2, a subsequence of $(v_{\lambda _ j})$ converges in $L^1$ to some $v\in \mathrm{PSH}(X,\omega )$ with supremum zero, and proposition 15.28.6 makes $v$ a threshold subsolution. Uniqueness in theorem 12.10.17 therefore gives $v=u_0$. Since every sequence tending to zero has a subsequence with this same $L^1$ limit, the whole family converges.
$\square$
Pome
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