Theorem 13.23.37D5F8FB5CFAB

Proof. Step 1: upper bounds. For large $j$, the rooftop envelope $r_ j=P_\theta (u_ j,u)$ is $\theta $-plurisubharmonic by proposition 13.23.2. It is an ordinary model: since $r_ j\leq 0$ and the envelope is monotone, $r_ j\leq P_\theta [r_ j]\leq P_\theta [u_ j]=u_ j$ and likewise $P_\theta [r_ j]\leq u$, so $P_\theta [r_ j]$ is a competitor for the rooftop and equals $r_ j$. The sequence $u_ j$ is $d_\theta $-Cauchy by proposition 13.22.5, and $d_\theta (r_ j,u)=U(u)-U(r_ j)\to 0$ by proposition 13.23.2. Choose a subsequence so that

The tail-maximum construction in the proof of theorem 13.22.7 gives

Indeed, that construction produces the unique $d_\theta $-limit, which here is $u$. The models $v_ j$ satisfy $v_ j\geq u_ j$ and $v_ j\geq u$, and have lower volume at least $\delta $.

Step 2: lower bounds. By lemma 13.18.1 and the choice of the subsequence, the stretching envelopes

are defined. Their defining inequalities show that $(1-2^{-l})u+2^{-l}H_ l$ is a competitor for $K_ l$ and that $K_ l\leq u$. Together with addition and monotonicity, this gives

Choose $j$ so large that $nM2^{1-j}<\delta $. For every $k\geq 0$, the sum of the deficits of $K_ j,\ldots ,K_{j+k}$ from $u$ is less than $\delta $, so lemma 13.23.1 shows that their rooftop envelope is not identically $-\infty $. Since $r_ l\leq u$, the stretching envelope $Q_ b(r_ l,u)$ decreases as $b$ increases; hence, for $l\geq j$,

Thus $R_ j^ k=P_\theta (Q_{2^ j}(r_ j,u),\ldots ,Q_{2^ j}(r_{j+k},u))$ is $\theta $-plurisubharmonic. The convex inequality for each obstacle gives

For either volume,

The argument used for $r_ j$ shows that $w_ j^ k$ is an ordinary model. For fixed $j$ it decreases in $k$ and has supremum zero, so normalized compactness shows that its limit $w_ j$ is $\theta $-plurisubharmonic. The lower-volume conclusion of theorem 13.21.2 passes the lower bound in (13.23.4) to $w_ j$. This bound is positive, so the upper-volume conclusion of the same theorem passes the bound for the upper volume as well.

Step 3: convergence. The $w_ j$ increase, since each is defined by fewer obstacles than the previous one, and $w_ j\leq r_ j\leq u_ j$, $w_ j\leq u$. After discarding finitely many indices, they have lower volume at least $\delta /2$, and (13.23.4) gives $d_\theta (w_ j,u)\leq nM2^{-j}$. By proposition 13.22.4, $w_ j\to u$ in $L^1$, so $(\lim _ jw_ j)^*=u$. The two monotone sequences $v_ j$ and $w_ j$ converge to $u$ in capacity by lemma 5.29.19, and the bounds $w_ j\leq u_ j\leq v_ j$ give $u_ j\to u$ in capacity along the chosen subsequence. Since every subsequence of the original sequence has a further subsequence of this kind, the original sequence converges to $u$ in capacity. $\square$