Theorem 16.22.4FCFA7EB59B34

Proof. We first construct one potential satisfying (16.22.6) as the limit of a decreasing sequence of envelopes, and then show that $E_ a(u,v)$ is the largest solution and has positive mass.

Reduction. Since $u\preceq v$, there are constants $A,B$ with $u-A\leq v-B\leq 0$. If $\eta '$ satisfies (16.22.6) for the pair $u-A$, $v-B$, then $\eta '+aA+(1-a)B$ satisfies it for $u$, $v$; the envelopes of the two pairs differ by the same constant, and masses do not change. We may therefore assume that $u\leq v\leq 0$. Then $m=\int _ X\langle \theta _ v^ n\rangle \geq \int _ X\langle \theta _ u^ n\rangle >0$, and lemma 16.22.1 provides a model potential $\phi \geq v$ with $\int _ X\langle \theta _\phi ^ n\rangle =m$. Since $1-a<0$, we have $au+(1-a)\phi \leq au+(1-a)v$; we first construct a potential below $au+(1-a)\phi $.

A decreasing sequence of envelopes. For $j\geq 1$ put $u_ j=\max \{ u,\phi -j\} $, so that $\phi -j\leq u_ j\leq \phi $; in particular, $u_ j$ and $\phi $ have the same pole set. Put

where $f_ j=-\infty $ on the pole set of $\phi $, as in lemma 16.22.3 for the pair $u_ j\leq \phi $. Outside this pole set, $f_ j=\phi +a(u_ j-\phi )$ lies between $\phi -aj$ and $\phi $. Hence the potential $\phi -aj$ is a competitor, and every competitor is at most $\phi $, so $\eta _ j$ is a potential with

where the masses agree because $\eta _ j\sim \phi \sim u_ j$. Since $u_ j$ decreases with $j$, so do $f_ j$ and $\eta _ j$.

A contact bound. Put $\gamma _ j=a^{-1}\eta _ j+(1-a^{-1})\phi $ and $D_ j=\{ \gamma _ j=u_ j>-\infty \} $. By lemma 16.22.3(2), applied to the pair $u_ j\leq \phi $, we have $\gamma _ j\leq u_ j$ and

A tail estimate. Fix $k>0$ and let $j>k$. On $D_ j$ we have $\eta _ j=\phi +a(u_ j-\phi )$, so there $\eta _ j\leq \phi -ak$ if and only if $u_ j\leq \phi -k$, that is, if and only if $u\leq \phi -k$. The plurifine open set $\{ u>\phi -k\} $ is contained in $\{ u>\phi -j\} $, on which $u_ j=u$; hence $\mu _{u_ j}=\mu _ u$ on $\{ u>\phi -k\} $, and $\mu _{u_ j}(\{ u\leq \phi -k\} )=m-\mu _ u(\{ u>\phi -k\} )$, pole sets having measure zero. Together with (16.22.7) this gives

The envelopes do not collapse. Suppose that $\eta _ j$ decreases to $-\infty $ identically. Then $\sup _ X\eta _ j\to -\infty $ by corollary 10.3.6. Since $\eta _ j\preceq \phi $ and $\phi $ is a model potential, lemma 16.22.2 gives $\eta _ j-\phi \leq \sup _ X\eta _ j$ off the pole set of $\phi $. Hence, for each $k$, $\eta _ j\leq \phi -ak$ everywhere once $j$ is large, and the left side of (16.22.8) equals $\int _ X\langle \theta _{\eta _ j}^ n\rangle =m$. Since $\phi $ is finite wherever $u$ is, $\mu _ u(\{ u>\phi -k\} )\to \int _ X\langle \theta _ u^ n\rangle $ as $k\to \infty $, and (16.22.8) yields $m\leq a^ n\bigl (m-\int _ X\langle \theta _ u^ n\rangle \bigr )$, which contradicts (16.22.5). Consequently $\eta _ j$ decreases to a potential $\eta $. Letting $j\to \infty $ in $\gamma _ j\leq u_ j$ gives $a^{-1}\eta +(1-a^{-1})\phi \leq u$, and since $v\leq \phi $, also (16.22.6).

The largest solution. By lemma 16.22.3(1), applied to the pair $u\leq v$, the potentials satisfying (16.22.6) are exactly the competitors for the obstacle $f=au+(1-a)v$. The potential $\eta $ constructed above is one of them, so part (2) of the lemma shows that $E_ a=E_ a(u,v)=P_\theta (f)$ is the largest potential satisfying (16.22.6), and that $E_ a\leq u$.

Positive mass. Choose $b>a$ that still satisfies (16.22.5). The preceding steps, applied to $b$, give $E_ b=E_ b(u,v)$. The potential

is a competitor for the obstacle $f$ of $E_ a$: quasi-everywhere, $E_ b\leq bu+(1-b)v$, and therefore

Hence $q\leq E_ a$, and the monotonicity of mass together with $\mu _ q\geq (1-a/b)^ n\mu _ u$ yields

If $\int _ X\langle \theta _ u^ n\rangle =\int _ X\langle \theta _ v^ n\rangle $, then (16.22.5) holds for every $a>1$, so $b$ can be taken arbitrarily large, and we obtain $\int _ X\langle \theta _{E_ a}^ n\rangle \geq \int _ X\langle \theta _ u^ n\rangle $. The reverse inequality follows from $E_ a\leq u$. $\square$