15.28 Subsolutions for an individual density72CDE5ADD348

Proof. Subtracting a common constant, which does not change the extended measures, we may assume that all potentials are nonpositive. Pass to a subsequence along which $s_ j\to s$ and $g_ j\to g$ almost everywhere, and set

Then $0\leq a_ j\uparrow g$ almost everywhere, and $w_ j\in \mathrm{PSH}(X,h)$ decreases to $s$.

Step 1: $\mathrm{MA}^{\mathrm{ext}}_ h(w_ j)\geq a_ j\, \mathrm{d}V$. Fix $j$. For $N\geq j$, the finite maximum $m_ N=\max _{j\leq \ell \leq N}s_\ell $ satisfies $\mathrm{MA}^{\mathrm{ext}}_ h(m_ N)\geq a_ j\, \mathrm{d}V$: on the set $\{ m_ N=s_\ell \} $, proposition 13.14.6 gives $\mathrm{MA}^{\mathrm{ext}}_ h(m_ N)\geq \mathrm{MA}^{\mathrm{ext}}_ h(s_\ell )\geq g_\ell \, \mathrm{d}V\geq a_ j\, \mathrm{d}V$, and these sets cover $X$. As $N\to \infty $, the functions $m_ N$ increase, are bounded above by zero, and have regularized limit $w_ j$. By lemma 5.29.19, $m_ N\to w_ j$ in capacity, so the damped measures $\mathrm{e}^{m_ N}\mathrm{MA}^{\mathrm{ext}}_ h(m_ N)$ converge weakly to $\mathrm{e}^{w_ j}\mathrm{MA}^{\mathrm{ext}}_ h(w_ j)$. Since $m_ N\to w_ j$ almost everywhere and $a_ j\leq g$, dominated convergence gives $\mathrm{e}^{m_ N}a_ j\to \mathrm{e}^{w_ j}a_ j$ in $L^1(\mathrm{d}V)$. Passing to the limit in $\mathrm{e}^{m_ N}\mathrm{MA}^{\mathrm{ext}}_ h(m_ N)\geq \mathrm{e}^{m_ N}a_ j\, \mathrm{d}V$ gives

The pole set of $w_ j$ has zero volume and is not charged by $\mathrm{MA}^{\mathrm{ext}}_ h(w_ j)$. Dividing by the positive weight off this set gives $\mathrm{MA}^{\mathrm{ext}}_ h(w_ j)\geq a_ j\, \mathrm{d}V$.

Step 2: passage to $s$. Since $w_ j$ decreases to $s\in \mathrm{PSH}(X,h)$, lemma 5.29.19 gives $w_ j\to s$ in capacity, and the damped measures of $w_ j$ converge weakly to $\mathrm{e}^ s\mathrm{MA}^{\mathrm{ext}}_ h(s)$. Moreover $0\leq \mathrm{e}^{w_ j}a_ j\leq g$ and $\mathrm{e}^{w_ j}a_ j\to \mathrm{e}^ sg$ almost everywhere, so dominated convergence and Step 1 give

Dividing by $\mathrm{e}^ s$ on $\{ s>-N\} $ and letting $N\to \infty $ gives $\mathrm{MA}^{\mathrm{ext}}_ h(s)\geq g\, \mathrm{d}V$ off the pole set of $s$. Both measures give zero mass to the pole set, which proves the lemma. $\square$

Proof. If $\mathrm{MA}^{\mathrm{ext}}_ h(u)\geq af\, \mathrm{d}V$ with $a>0$, then also $\mathrm{MA}^{\mathrm{ext}}_ h(u)\geq af_ k\, \mathrm{d}V$ for every $k$, and lemma 15.27.1 gives $a\leq c_ h(f_ k)$. Letting $k\to \infty $ gives $a\leq b_ h(f)$. If $b_ h(f)=0$, the coefficient zero is attained by the zero potential.

Suppose now that $b=b_ h(f)>0$, and let $u_ k$ be continuous solutions of

By theorem 5.25.2, a subsequence $u_{k_ j}$ converges in $L^1$ to some $u\in \mathrm{PSH}(X,h)$ with $\sup _ Xu=0$. The densities $c_ h(f_{k_ j})f_{k_ j}$ are dominated by $c_ h(f_1)f$ and converge to $bf$ almost everywhere, hence in $L^1$. Lemma 15.28.3 now gives $\mathrm{MA}^{\mathrm{ext}}_ h(u)\geq bf\, \mathrm{d}V$, so the maximum is attained at $a=b$. $\square$

Proof. The family is nonempty by theorem 15.28.4. By theorem 5.25.2, an $L^1$ limit of members of $\mathcal S_ f$ is $h$-plurisubharmonic with supremum zero, and lemma 15.28.3, with $g_ j=g=bf$, shows that it satisfies the measure inequality. Thus $\mathcal S_ f$ is a closed subset of the compact set of normalized $h$-plurisubharmonic functions, hence compact. It is also stable under finite maxima: if $s_1,s_2\in \mathcal S_ f$, then $\max (s_1,s_2)$ has supremum zero, and proposition 13.14.6, applied on the sets where the maximum equals $s_1$ or $s_2$, gives $\mathrm{MA}^{\mathrm{ext}}_ h(\max (s_1,s_2))\geq bf\, \mathrm{d}V$.

By Choquet's lemma there is a sequence $(s_ i)_{i\geq 1}$ in $\mathcal S_ f$ with $(\sup _ is_ i)^*=\Phi _ f$. Fix $s_0\in \mathcal S_ f$. The finite maxima $t_ N=\max (s_0,s_1,\ldots ,s_ N)$ belong to $\mathcal S_ f$, increase with $N$, and satisfy $s_0\leq t_ N\leq 0$; their regularized limit is $\Phi _ f$. Hence $t_ N\to \Phi _ f$ almost everywhere and, by dominated convergence, in $L^1$. Lemma 15.28.3 gives $\mathrm{MA}^{\mathrm{ext}}_ h(\Phi _ f)\geq bf\, \mathrm{d}V$, and $s_0\leq \Phi _ f\leq 0$ forces $\sup _ X\Phi _ f=0$. Thus $\Phi _ f\in \mathcal S_ f$, and by construction it dominates every member of $\mathcal S_ f$.

Finally, suppose $v_{\lambda _ j}\to v$ in $L^1$ with $\lambda _ j\downarrow 0$. The measures $\mathrm{MA}^{\mathrm{ext}}_ h(v_{\lambda _ j})=\mathrm{e}^{\lambda _ ju_{\lambda _ j}}f\, \mathrm{d}V$ have densities converging in $L^1$ to $bf$, by corollary 15.35.3. Applying lemma 15.28.3 to these densities and potentials shows that $v\in \mathcal S_ f$. $\square$

Proof. Suppose that $f\in L^ p$ with $p>1$, and let $v$ be a continuous solution of $\mathrm{MA}_ h(v)=c_ h(f)f\, \mathrm{d}V$. It is a subsolution with coefficient $c_ h(f)$, so theorem 15.28.4 gives $c_ h(f)\leq b_ h(f)$. Conversely, lemma 15.27.1, applied to a threshold subsolution, gives $b_ h(f)\leq c_ h(f)$.

For general $f$, the continuous solution for $f_ k$ has total mass $c_ h(f_ k)\int _ Xf_ k\, \mathrm{d}V$, which is at least $v_-(h)$ by the definition of the lower mass. Letting $k\to \infty $ and using monotone convergence gives the lower bound for $b_ h(f)$. If $h$ is Kähler, Stokes' theorem for bounded potentials gives instead the identity $c_ h(f_ k)\int _ Xf_ k\, \mathrm{d}V=\int _ Xh^ n$, and the same limit proves the last assertion. $\square$

Proof. We construct $f=(1+Q)/M$, where $Q\geq 0$ concentrates on closed nowhere dense sets in such a way that it has essential infimum zero on every nonempty open set.

Step 1: the pieces. Let $(U_ i)_{i\geq 1}$ be a countable basis of nonempty open subsets of $X$. We construct inductively numbers $\alpha _ j>0$, nonzero bounded densities $q_ j$, and closed nowhere dense sets $K_ j\subset X$ such that

Start with $\alpha _1=1/4$. Once $\alpha _ j$ has been chosen, theorem 12.15.4, theorem 12.2.3 and $v_-(h)=0$ give a smooth strictly positive probability density $g_ j$ with $c_ h(g_ j)<\alpha _ j2^{-j}$. Removing from $X$ an open dense set of small volume, we obtain a closed nowhere dense set $K_ j$ for which $q_ j=g_ j\mathbf1_{K_ j}$ is nonzero and, since $g_ j$ is bounded, arbitrarily close to $g_ j$ in $L^2$. By lemma 15.27.2 with $p=2$, we can therefore choose $K_ j$ so that $c_ h(q_ j)\leq \alpha _ j2^{-j}$.

After $K_1,\ldots ,K_ N$ have been chosen, put

each set inside the minimum is open and nonempty, since a finite union of closed nowhere dense sets is closed and nowhere dense. Choose

and continue. In particular,

Step 2: the concentrating part. The function $Q=\sum _ j\alpha _ jq_ j$ is nonnegative and integrable. We claim that its essential infimum on every nonempty open set is zero. Since every nonempty open set contains some $U_ i$, it suffices to consider $U_ i$. Take $N\geq i$ and remove $K_1,\ldots ,K_ N$ from $U_ i$. On the remaining set, whose volume is at least $m_ N$, the first $N$ summands of $Q$ vanish. By Markov's inequality, the set where the tail $\sum _{j>N}\alpha _ jq_ j$ exceeds $2^{-N}$ has volume at most

Hence $Q\leq 2^{-N}$ on a subset of $U_ i$ of positive volume. Letting $N\to \infty $ proves the claim.

Step 3: $b_ h(f)=0$. Set $M=\int _ X(1+Q)\, \mathrm{d}V$ and $f=(1+Q)/M$, a probability density with $f\geq 1/M$. If $\mathrm{MA}^{\mathrm{ext}}_ h(u)\geq af\, \mathrm{d}V$ with $a>0$, then $\mathrm{MA}^{\mathrm{ext}}_ h(u)\geq (a\alpha _ j/M)q_ j\, \mathrm{d}V$, and lemma 15.27.1 gives

This is impossible. Hence $b_ h(f)=0$ by theorem 15.28.4.

Step 4: $C_ h^{\mathrm{sm}}(f)>0$. Let $g$ be a smooth strictly positive probability density and put $R=\| g/f\| _{L^\infty }$. Then $g\leq R(1+Q)/M$ almost everywhere, and this forces $g\leq R/M$ everywhere. Otherwise, by continuity, $g>R(1+\varepsilon )/M$ on a nonempty open set for some $\varepsilon >0$, while $Q<\varepsilon $ on a subset of positive volume of this set by Step 2, a contradiction. The order and homogeneity of the scalar in lemma 15.27.1 now give $c_ h(g)\geq c_ h(R/M)=(M/R)c_ h(1)$, that is,

The constant density $g=1$ attains this value, since $\operatorname *{ess\, inf}_ XQ=0$ gives $\| 1/f\| _{L^\infty }=M$. Thus $C_ h^{\mathrm{sm}}(f)=M c_ h(1)>0$, as required. $\square$