13.18 Volume deficits and relative full massE8D1B2CE6F8F

Proof. Adding constants to $u$ and $v$ only translates the competitors, so we may assume $v\leq 0$. Write $V=\overline{\operatorname {vol}}_\theta (v)$. Since $D<b^{-n}\delta $, the definition of the upper volume gives $w\in \mathrm{PSH}(X,\theta )$ with $w\sim u$ and

After a translation of $w$ we may also assume $w\leq v$. For positive integers $j$, put

The function $v-bj$ is a competitor for $h_ j$, and the defining convex inequality together with $w_ j\leq v$ gives $h_ j\leq v$. Thus $v-bj\leq h_ j\leq v$, so $w_ j\sim v$ and $h_ j\sim v$, and $h_ j$ decreases as $j$ increases. We claim that the decreasing limit $h$ of the $h_ j$ is not identically $-\infty $. If it were, theorem 5.25.2 would give $\sup _ Xh_ j\to -\infty $.

By theorem 15.8.6, the measure of $h_ j$ is carried by the contact set $C_ j=\{ b^{-1}h_ j+(1-b^{-1})v=w_ j\} $. Positivity of the mixed products and ordered contact for $b^{-1}h_ j+(1-b^{-1})v\leq w_ j$ give

Fix a positive integer $k$. If $h\equiv -\infty $, then $h_ j\leq -bk$ everywhere for all large $j>k$. Since $v\leq 0$, every point of $C_ j$ then satisfies $w_ j\leq -k$, hence $w\leq -k$. Moreover $w_ j=w$ on $\{ w>-k\} \subset \{ w>v-j\} $, so locality gives

Here the first inequality uses $h_ j\sim v$, and the last uses $w_ j\sim v$. Letting $k\to \infty $ contradicts (13.18.2), since the finite non-pluripolar measure $\mathrm{MA}_\theta (w)$ gives no mass to the pole set of $w$. This proves the claim. Consequently $h\in \mathrm{PSH}(X,\theta )$, and $b^{-1}h+(1-b^{-1})v\leq w$. If $w\leq u+C$, then $h-bC\in \mathcal A_ b(u,v)$, which proves the lemma. $\square$

Proof. Let $C_\pi $ be the uniform bound for Lelong numbers of $\pi ^*\theta $-plurisubharmonic functions on $Z$ given by lemma 5.38.8. Since $u\preceq v$, the difference of Lelong numbers is nonnegative, and it is at most $\nu (u\circ \pi ,z)\leq C_\pi $. This proves the assertion when $D\geq \delta $.

If $D<\delta $, then by lemma 13.18.1 every $b>1$ with $b^ nD<\delta $ admits some $h\in \mathcal A_ b(u,v)$. Pulling back the defining convex inequality and taking Lelong numbers gives

For $D>0$, let $b\uparrow (\delta /D)^{1/n}$; for $D=0$, let $b\to \infty $. Taking generic values along a prime divisor proves the divisorial assertion. $\square$

Proof. First let $u\in \mathcal E_{\mathrm{env}}(X,\theta ,\varphi )$. For each $a\in (0,1)$, a member of $\mathcal A_ b(u,\varphi )$ with $b=(1-a)^{-1}$ gives $h\in \mathrm{PSH}(X,\theta )$ such that $a\varphi +(1-a)h\leq u$. By theorem 13.17.1, proposition 13.17.2 and homogeneity, for either volume,

Letting $a\to 1$ proves that both volumes of $u$ equal those of $\varphi $; this direction does not use positive lower volume.

Conversely, let $u\in \mathcal E_{\mathrm{vol}}(X,\theta ,\varphi )$. Then $D=0$ in lemma 13.18.1 with $v=\varphi $, so $\mathcal A_ b(u,\varphi )\neq \varnothing $ for every $b>1$. For $b=1$ the family contains $u$. This is the defining condition of $\mathcal E_{\mathrm{env}}(X,\theta ,\varphi )$. $\square$