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13.23 Rooftops and capacity convergence1D6292952FE6

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A small upper-volume deficit also controls the rooftop envelope of two potentials. For a finite family it suffices that the sum of the deficits stays below the lower volume of a common upper bound. Applying this criterion to stretched potentials gives increasing lower bounds for a suitable subsequence of models. Together with the decreasing upper bounds of the preceding section, they turn convergence for the volume comparison into convergence in capacity.

Lemma 13.23.1.2B568CD72F9D Let $u_1,\ldots ,u_ N,w\in \mathrm{PSH}(X,\theta )$, with $N\geq 1$ and $u_ i\preceq w$. If

\[ \sum _{i=1}^ N\bigl (U(w)-U(u_ i)\bigr ) <\underline{\operatorname {vol}}_\theta (w), \]

then the rooftop envelope $P_\theta (u_1,\ldots ,u_ N)=P_\theta (\min _ iu_ i)$ is not identically $-\infty $.

Proof. Whether a finite rooftop envelope is identically $-\infty $ does not change under bounded changes of its obstacles. By the definition of $U(u_ i)$, we can choose $v_ i\in \mathrm{PSH}(X,\theta )$ with $v_ i\sim u_ i$ whose masses satisfy

\[ \sum _{i=1}^ N\left(U(w)-\int _ X\mu _{v_ i}\right) <\underline{\operatorname {vol}}_\theta (w). \]

Translate so that $v_ i\leq w\leq 0$, and put

\[ v_{i,k}=\max (v_ i,w-k),\qquad p_ k=P_\theta (v_{1,k},\ldots ,v_{N,k}). \]

The common competitor $w-k$ gives $w-k\leq p_ k\leq w$, so $p_ k\sim w$. The $p_ k$ decrease as $k$ increases, and a $\theta $-plurisubharmonic limit would be a competitor for $P_\theta (v_1,\ldots ,v_ N)$. Suppose, to reach a contradiction, that $P_\theta (u_1,\ldots ,u_ N)\equiv -\infty $, so that $P_\theta (v_1,\ldots ,v_ N)\equiv -\infty $ as well. Then the $p_ k$ decrease to $-\infty $, and normalized compactness gives $\sup _ Xp_ k\to -\infty $.

Fix $s>0$ and choose $k>s$ so large that $p_ k\leq -s$. Iterating proposition 16.28.6, which bounds the measure of a rooftop by the measures of its obstacles on the contact sets, and using locality on $\{ v_ i>-s\} $, we obtain

\[ \begin{split} \underline{\operatorname {vol}}_\theta (w) & \leq \int _ X\mu _{p_ k} \leq \sum _{i=1}^ N\int _{\{ v_{i,k}\leq -s\} }\mu _{v_{i,k}}\\ & \leq \sum _{i=1}^ N\left(U(w)-\int _{\{ v_ i>-s\} }\mu _{v_ i}\right). \end{split} \]

Here we used $v_{i,k}\sim w$, and $w-k<-s$, which gives $v_{i,k}=v_ i$ on $\{ v_{i,k}>-s\} =\{ v_ i>-s\} $. Let $s\to \infty $. The measures $\mu _{v_ i}$ are finite and put no mass on pluripolar sets, so the resulting inequality contradicts the choice of the $v_ i$. $\square$

To control the volume of a rooftop envelope, we first stretch each obstacle away from the maximum of the two, and then move both stretched potentials a fixed fraction back toward this maximum. By the lemma, the resulting potentials have a rooftop envelope that is not identically $-\infty $.

Proposition 13.23.2.1542F41C8A51 Let $u_ j,u\in \mathrm{PSH}(X,\theta )$, with $\underline{\operatorname {vol}}_\theta (u)>0$, and suppose that $d_\theta (u_ j,u)\to 0$. For all sufficiently large $j$, the rooftop envelope $p_ j=P_\theta (u_ j,u)$ is not identically $-\infty $, and

\[ d_\theta (p_ j,u)\longrightarrow 0,\qquad \underline{\operatorname {vol}}_\theta (p_ j) \longrightarrow \underline{\operatorname {vol}}_\theta (u). \]

Proof. Put $\delta =\underline{\operatorname {vol}}_\theta (u)$, $m_ j=\max (u_ j,u)$ and $s_ j=d_\theta (u_ j,u)$. After discarding finitely many indices, choose $b_ j\to \infty $ with $b_ j>1$ and $b_ j^ ns_ j<\delta $. The upper-volume deficits of both $u_ j$ and $u$ from $m_ j$ are at most $s_ j$, while $m_ j$ has lower volume at least $\delta $. Thus lemma 13.18.1 shows that the stretching envelopes

\[ G_ j=Q_{b_ j}(u_ j,m_ j),\qquad H_ j=Q_{b_ j}(u,m_ j) \]

are defined. Choose $a\in (0,1)$ with $2nMa<\delta $, and set

\[ g_ j=(1-a)m_ j+aG_ j,\qquad h_ j=(1-a)m_ j+aH_ j. \]

These potentials lie below $m_ j$, because $G_ j\leq u_ j$ and $H_ j\leq u$. Addition and monotonicity give

\[ 2U(m_ j)-U(g_ j)-U(h_ j) \leq 2\bigl (1-(1-a)^ n\bigr )U(m_ j) \leq 2nMa<\delta . \]

Lemma 13.23.1, with $w=m_ j$, therefore shows that $r_ j=P_\theta (g_ j,h_ j)$ is $\theta $-plurisubharmonic. For large $j$, put $t_ j=(ab_ j)^{-1}<1$. Since $r_ j\leq g_ j$ and $r_ j\leq h_ j$, the defining inequalities for $G_ j$ and $H_ j$ show that

\[ (1-t_ j)m_ j+t_ jr_ j\leq u_ j,u. \]

The left-hand side is therefore a competitor for $p_ j$, so $p_ j\not\equiv -\infty $. For either volume, addition and monotonicity give

\[ (1-t_ j)^ n\operatorname {vol}_\theta (u) \leq (1-t_ j)^ n\operatorname {vol}_\theta (m_ j) \leq \operatorname {vol}_\theta (p_ j) \leq \operatorname {vol}_\theta (u). \]

Since $t_ j\to 0$, this proves $\underline{\operatorname {vol}}_\theta (p_ j)\to \underline{\operatorname {vol}}_\theta (u)$; and since $p_ j\leq u$, (13.22.2) gives $d_\theta (p_ j,u)=U(u)-U(p_ j)\to 0$. $\square$

For ordinary models, finite rooftop envelopes are again ordinary models. Their normalization prevents collapse when the number of obstacles tends to infinity.

Theorem 13.23.3.7D5F8FB5CFAB Suppose $u_ j,u\in \mathcal M_\delta (\theta )$ and $d_\theta (u_ j,u)\to 0$. After passing to a subsequence and relabeling it, there are ordinary model potentials $v_ j,w_ j\in \mathcal M_{\delta /2}(\theta )$ such that

\[ w_ j\leq u_ j\leq v_ j,\qquad w_ j\leq u\leq v_ j, \]

$v_ j$ decreases to $u$, $w_ j$ increases with $(\lim _ jw_ j)^*=u$, and

\[ d_\theta (v_ j,u)+d_\theta (w_ j,u)\longrightarrow 0. \]

The original sequence converges to $u$ in capacity.

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

\[ d_\theta (u_ j,u_{j+1})<\delta \, 2^{-(j+1)n},\qquad U(u)-U(r_ j)<\delta \, 4^{-(j+1)n}. \]

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

\[ v_ j=P_\theta \left[\left(\sup _{l\geq j}u_ l\right)^*\right] \searrow u,\qquad d_\theta (v_ j,u)\longrightarrow 0. \]

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

\[ H_ l=Q_{4^ l}(r_ l,u),\qquad K_ l=Q_{2^ l}(r_ l,u) \]

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

\[ (1-2^{-l})u+2^{-l}H_ l\leq K_ l\leq u, \qquad U(u)-U(K_ l)\leq nM2^{-l}. \]

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 $l\geq j$,

\[ Q_{2^ j}(r_ l,u)\geq K_ l. \]

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

\[ (1-2^{-j})u+2^{-j}R_ j^ k \leq w_ j^ k:=P_\theta (r_ j,\ldots ,r_{j+k})\leq u. \]

For either volume,

00C1CF12A199 13.23.4
\begin{equation} (1-2^{-j})^ n\operatorname {vol}_\theta (u) \leq \operatorname {vol}_\theta (w_ j^ k) \leq \operatorname {vol}_\theta (u). \end{equation}

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. $\square$

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