Proposition 17.6.1DC1056C1A4A1

Proof. The envelopes $\phi _\varepsilon $ decrease to a $\theta $-psh limit $q\geq P_\theta [u]$; the point is the reverse inequality, which we obtain by showing that $\{ q>P_\theta [u]+\delta \} $ has arbitrarily small volume for every $\delta >0$.

Step 1: the limit exists. Put $\phi =P_\theta [u]$. Since $u\leq 0$, we have $u\leq V$, so $w_\varepsilon =u+\varepsilon (V-u)$ decreases to $u$ as $\varepsilon \downarrow 0$ (at the poles of $u$ all these functions equal $-\infty $). In particular $u\leq w_\varepsilon \leq 0$. By the monotonicity of $P_\theta [\cdot ]$ in proposition 16.21.2(1) and the bound $P_\theta [\cdot ]\leq V$ of definition 16.21.1, we get $\phi \leq \phi _\varepsilon \leq V$, and $\phi _\varepsilon $ decreases as $\varepsilon $ decreases. Hence $q=\lim _{\varepsilon \downarrow 0}\phi _\varepsilon $ exists pointwise; as a decreasing limit of $\theta $-psh potentials bounded below by $\phi $, it is $\theta $-psh (corollary 10.3.6), and $\phi \leq q\leq V$. Since $\phi $ and $V$ are integrable, dominated convergence gives $\phi _\varepsilon \to q$ in $L^1(X)$. It remains to prove that $q\leq \phi $.

Step 2: masses and a normalized tail bound. For $0\leq k\leq n$ consider the finite non-pluripolar measures

Since $\theta _{w_\varepsilon }=(1-\varepsilon )\theta _ u+\varepsilon \theta _ V$, multilinearity (proposition 13.2.6) gives

Here $a_0=\int _ X\langle \theta _ V^ n\rangle >0$ by proposition 13.9.2, so $m_\varepsilon \geq \varepsilon ^ na_0>0$. By the mass preservation in proposition 16.21.2(3) and by theorem 16.22.11, $\phi _\varepsilon $ is a model potential of mass $m_\varepsilon $.

In the rest of the proof, sets defined by inequalities between potentials, such as $\{ u<V-A\} $, are understood pointwise, with values in $[-\infty ,+\infty )$. The pole sets of all potentials involved are pluripolar, hence null for all the non-pluripolar measures below and for $\omega ^ n$. For $A>0$ put

The maximum is over a finite set which contains $k=0$. As $A\to \infty $, the sets $\{ u<V-A\} $ decrease to a subset of the pole set of $u$, which is $\nu _ k$-null; since the measures $\nu _ k$ are finite, continuity from above gives $r(A)\downarrow 0$. Moreover $w_\varepsilon -V=(1-\varepsilon )(u-V)$, so $\{ w_\varepsilon <V-A\} \subset \{ u<V-A\} $. Using the expansion of $\mu _{w_\varepsilon }$ and the bound $\nu _ k(\{ u<V-A\} )\leq r(A)a_ k$, valid for every $k$ (when $a_ k=0$ both sides vanish), we obtain the estimate

which is uniform in $\varepsilon $.

Step 3: rooftops. For fixed $A>0$ put

The potential $w_\varepsilon $ is a common minorant of $w_\varepsilon +A$ and $V$, so $w_\varepsilon \leq p_{\varepsilon ,A}$. On the other hand $p_{\varepsilon ,A}\leq V\leq 0$ and $p_{\varepsilon ,A}\leq w_\varepsilon +A$, so $p_{\varepsilon ,A}$ is a competitor for $P_\theta [w_\varepsilon ]$. Hence $w_\varepsilon \leq p_{\varepsilon ,A}\leq \phi _\varepsilon $, and since $w_\varepsilon $ and $\phi _\varepsilon $ both have mass $m_\varepsilon $, the monotonicity of mass (theorem 13.8.7) shows that $p_{\varepsilon ,A}$ has mass $m_\varepsilon $ as well; thus $p_{\varepsilon ,A}\in \mathcal E(X,\theta ,\phi _\varepsilon )$. As $w_\varepsilon +A$ decreases to $u+A$, proposition 16.11.6 gives $p_{\varepsilon ,A}\downarrow p_ A$ as $\varepsilon \downarrow 0$. Likewise $u\leq p_ A\leq V\leq 0$ and $p_ A\leq u+A$, so $p_ A$ is a competitor for $P_\theta [u]$, and $p_ A\leq \phi $.

Step 4: auxiliary solutions. Apply theorem 17.5.6 with the model potential $\phi _\varepsilon $ of mass $m_\varepsilon $ in place of $\phi $, the volume form $\omega ^ n$ and the constant density $m_\varepsilon /\Omega $. It gives $\chi _\varepsilon \in \mathcal E^1(X,\theta ,\phi _\varepsilon )$ with

By lemma 16.22.2, $\chi _\varepsilon \leq \phi _\varepsilon $. By proposition 10.3.1, normalized $\theta $-psh potentials are uniformly bounded in $L^1$, so there is a constant $C_1$, independent of $\varepsilon $, with $\int _ X(-\chi _\varepsilon )\, \omega ^ n\leq C_1$.

Step 5: a volume bound uniform in $\varepsilon $. Fix $\delta >0$ and $B>1$, put $\tau =\min \{ 1/2,\delta /(4B)\} $, and define

Since $\phi _\varepsilon \leq 0$, the complement of $G_{\varepsilon ,B}$ lies in $\{ \chi _\varepsilon \leq -B\} $, whose volume is at most $C_1/B$ by Markov's inequality. Moreover,

and since $\chi _\varepsilon $ and $\phi _\varepsilon $ have the same mass, the monotonicity of mass places $\psi _\varepsilon $ in $\mathcal E(X,\theta ,\phi _\varepsilon )$. On $G_{\varepsilon ,B}$ we have $\psi _\varepsilon -\phi _\varepsilon =\tau (\chi _\varepsilon -\phi _\varepsilon )-\delta /2>-\tau B-\delta /2\geq -3\delta /4$, whereas $p_{\varepsilon ,A}<\phi _\varepsilon -\delta $ on $E_{\varepsilon ,A,\delta }$. Hence

By multilinearity, $\mu _{\psi _\varepsilon }\geq \tau ^ n\mu _{\chi _\varepsilon }=\tau ^ n(m_\varepsilon /\Omega )\omega ^ n$. As $p_{\varepsilon ,A}$ and $\psi _\varepsilon $ both belong to $\mathcal E(X,\theta ,\phi _\varepsilon )$, the comparison principle of proposition 16.25.4 applies to them, and we obtain

To estimate the right side, we use the contact inequality for rooftops, corollary 16.13.5:

The second contact set does not meet $\{ p_{\varepsilon ,A}<\psi _\varepsilon \} $: at a common point we would have $V<\psi _\varepsilon \leq \phi _\varepsilon -\delta /2\leq V-\delta /2$. At a point of $\{ p_{\varepsilon ,A}<\psi _\varepsilon \} $ in the first contact set, $w_\varepsilon +A<\psi _\varepsilon \leq V-\delta /2$, hence $w_\varepsilon <V-A$. Therefore, by (17.6.2),

Dividing by the positive number $\tau ^ nm_\varepsilon /\Omega $ and adding the volume of the complement of $G_{\varepsilon ,B}$, we arrive at

Step 6: passage to the limit. Take a sequence $\varepsilon _ j\downarrow 0$. If $q(x)>p_ A(x)+\delta $, then $\phi _{\varepsilon _ j}(x)\geq q(x)$ and $p_{\varepsilon _ j,A}(x)\downarrow p_ A(x)$, so $x\in E_{\varepsilon _ j,A,\delta }$ for all large $j$. Thus

and Fatou's lemma together with (17.6.3) shows that the volume of $\{ q>p_ A+\delta \} $ is at most $C_1/B+\Omega \tau ^{-n}r(A)$. Since $p_ A\leq \phi $, the same bound holds for the set $\{ q>\phi +\delta \} $, which depends neither on $A$ nor on $B$, whereas $\tau $ depends on $B$ and $\delta $ but not on $A$. Letting first $A\to \infty $, so that $r(A)\to 0$, and then $B\to \infty $, we see that $\{ q>\phi +\delta \} $ has zero volume. As $\delta >0$ was arbitrary, $q\leq \phi $ almost everywhere, and hence everywhere, because both are $\theta $-psh (see the beginning of section 16.11). Together with $\phi \leq q$, this gives $q=\phi $. $\square$