PomeLemma 16.28.68.F644817639FD Let $\varphi $ be a model reference. If $u\in \mathcal E_\varphi $ and $u\leq \varphi $, then
\[ w=P_\theta (u-\varphi )\in \mathcal E_0, \qquad w\leq 0, \qquad w+\varphi \leq u. \]
16.28.66 Model references and positive mass97D4C0CD199E
In this section $\theta $ is semipositive and big. Suppose that a potential $u\in \mathrm{PSH}(X,\theta )$ has a competitor in a smaller background $a\theta $, $0<a<1$, that is, $P_{a\theta }(u)\not\equiv -\infty $. Then the big form $(1-a)\theta $ remains available: an envelope in this background has positive mass, and the contact property transfers this mass to $u$. To obtain such competitors for all members of a relative class, we use a reference $\varphi $ satisfying
237D346B8CBA
16.28.67
\begin{equation} P_\theta [\varphi ]=\varphi ,\qquad P_{\varepsilon \theta }(\varphi )\not\equiv -\infty \quad \text{for some }0<\varepsilon <1. \end{equation}
We call such a $\varphi $ a model reference and fix one such $\varepsilon $. Note that $\varphi =P_\theta [\varphi ]\leq 0$. For $u\in \mathcal E_\varphi $, the envelope of $u-\varphi $ belongs to the relative envelope class $\mathcal E_0=\mathcal E_{\mathrm{env}}(X,\theta ,0)$ of the zero potential, and stretching realizes it in an arbitrarily small background. Under a bounded mass hypothesis, section 13.16 uses instead the condition (13.20.5), in which a positive lower volume replaces the condition on the smaller background; the two conditions are compared there.
Proof. Since $\varphi \leq 0$, the potential $u$ lies below the obstacle $u-\varphi $ away from the pole set of $\varphi $, so it is a competitor. Hence the envelope $w$ is not identically $-\infty $. It is at most zero, since $u-\varphi \leq 0$, and it satisfies $w+\varphi \leq u$ off the poles, hence everywhere by lemma 5.24.3. For $b\geq 1$, let $h_ b=Q_ b(u,\varphi )$. Then, off the poles,
\[ h_ b\leq bu-(b-1)\varphi \leq b(u-\varphi ). \]
Since $\theta \geq 0$, the function $b^{-1}h_ b$ is $\theta $-plurisubharmonic, and so it is a competitor for $w$. Hence $h_ b\leq bw$, so $\mathcal A_ b(w,0)$ is nonempty for every $b$. Since also $w\leq 0$, this means exactly that $w\in \mathcal E_0$. $\square$
Lemma 16.28.69.18D2E34A9D42 Let $0<a<1$ and $u\in \mathrm{PSH}(X,\theta )$. If $q=P_{a\theta }(u)$ is not identically $-\infty $, then
\[ \int _ X\mathrm{MA}_\theta (u)\geq (1-a)^ n v_-(\theta ), \]
and the mass on the left is strictly positive, even when $v_-(\theta )=0$.
Proof. The obstacle $u-q$ is nonnegative away from the poles, so by lemma 5.40.2 the envelope $p=P_{(1-a)\theta }(u-q)$ is bounded. By the contact theorem, its measure is carried by $D=\{ p+q=u\} $, and $p+q\leq u$ everywhere. Since $p+q$ is $\theta $-plurisubharmonic, positivity of mixed products and ordered contact imply
\[ \mathrm{MA}_{(1-a)\theta }(p) \leq \mathbf1_ D\mathrm{MA}_\theta (p+q) \leq \mathbf1_ D\mathrm{MA}_\theta (u), \]
where all assertions are first restricted to the complement of the poles. Writing $p=(1-a)\widetilde p$ with $\widetilde p\in \mathrm{PSH}(X,\theta )\cap L^\infty (X)$ and integrating, we obtain the lower bound from the definition of $v_-(\theta )$. Finally, the bounded potential $p$ has strictly positive mass by proposition 16.28.55, since $(1-a)\theta $ is big. $\square$
Theorem 16.28.70.0EA7F76611B5 Let $\varphi $ be a model reference, with $\varepsilon $ as in (16.28.67). For every $u\in \mathcal E_\varphi $ and every $0<\delta <1-\varepsilon $, the envelope $P_{(\varepsilon +\delta )\theta }(u)$ is not identically $-\infty $. Consequently,
\[ \int _ X\mathrm{MA}_\theta (u)\geq (1-\varepsilon )^ n v_-(\theta ), \qquad \int _ X\mathrm{MA}_\theta (u)>0. \]
The mass may be infinite.
Proof. After adding a constant to $u$, we may assume $u\leq \varphi $. Let $w\in \mathcal E_0$ be as in lemma 16.28.68. Since $w\in \mathcal E_0$, the envelope $P_\theta (w/\delta )$ is not identically $-\infty $, and $\delta P_\theta (w/\delta )\leq w$. The potential
\[ P_{\varepsilon \theta }(\varphi )+\delta P_\theta (w/\delta ) \leq \varphi +w\leq u \]
belongs to $\mathrm{PSH}(X,(\varepsilon +\delta )\theta )$, so it is a competitor for $P_{(\varepsilon +\delta )\theta }(u)$. Applying lemma 16.28.69 with $a=\varepsilon +\delta $ and letting $\delta \downarrow 0$ gives the uniform estimate. The positivity follows from the same lemma with any one such $\delta $. $\square$
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