17.5.1 Existence by maximization06665CC7749C

Proof. (1) This is lemma 16.12.1(2) with $W=X$; the second assertion follows by applying the first in both directions.

(2) Since $X$ is compact, weak convergence of measures on $X$ is convergence of the integrals of all continuous functions (see the discussion following definition 5.3.5). Hence $\int _ Xg\, \mathrm d\lambda =\lim _ j\int _ Xg\, \mathrm d\lambda _ j\geq \int _ Xg\, \mathrm d\sigma $ for every continuous function $g\geq 0$ on $X$, and (1) gives $\lambda \geq \sigma $. $\square$

Proof. Let $B=\sup _ X|f|$. The potential $u-|t|B$ is a competitor for the obstacle $u+tf$, and every competitor lies below $u+|t|B$, so $u-|t|B\leq u_ t\leq u+|t|B$. Hence $u_ t\sim u$, and $E(u_ t)\geq E(u)-|t|Bm>-\infty $, so $u_ t\in \mathcal E^1(X,\theta ,\phi )$ by theorem 17.3.14. The obstacle $u+tf$ is upper semicontinuous, so $u_ t\leq u+tf$ everywhere, by proposition 16.11.4(1),(2). In particular $u_0\leq u$; since $u$ is itself a competitor for the obstacle $u$, also $u\leq u_0$, so $u_0=u$.

Next we show that $\mu _{u_{t+s}}\to \mu _{u_ t}$ weakly as $s\to 0$. The obstacles $u+tf$ and $u+(t+s)f$ differ by at most $|s|B$, so proposition 16.11.4(4) gives $u_ t-|s|B\leq u_{t+s}\leq u_ t+|s|B$. In particular all $u_ t$ have the pole set of $u$, and $|u_{t+s}-u_ t|\leq |s|B$ outside it. Since $|\arctan a-\arctan b|\leq |a-b|$ for real $a,b$, and adding the same smooth local potential of $\theta $ to $u_{t+s}$ and $u_ t$ does not change their difference, the distance $d_*$ of definition 13.7.2 between the local potentials of $u_{t+s}$ and $u_ t$ is at most $|s|B$ everywhere. Hence, for every $\delta >0$, the sets where it exceeds $\delta $ are empty once $|s|B<\delta $, and $u_{t+s}\to u_ t$ in extended capacity as $s\to 0$. These potentials have full mass relative to $\phi $, so corollary 13.9.4, applied along every sequence $s_ k\to 0$, gives the asserted weak convergence; in particular $\int _ Xf\, \mu _{u_{t+s}}\to \int _ Xf\, \mu _{u_ t}$.

The obstacle $u+tf$ is bounded above, and it is quasi-continuous, because $u$ is (section 16.11) and $f$ is continuous. Hence theorem 16.13.4 shows that $\mu _{u_ t}$ is carried by the contact set $\{ u_ t=u+tf\} $. On the contact set of $u_{t+s}$ we have $u_{t+s}=u+(t+s)f$ and $u_ t\leq u+tf$, so $u_{t+s}-u_ t\geq sf$ there. On the contact set of $u_ t$ we have $u_ t=u+tf$ and $u_{t+s}\leq u+(t+s)f$, so $u_{t+s}-u_ t\leq sf$ there. The differences are bounded and the pole sets carry no mass, so the tangent bounds of lemma 17.4.10 give, for every real $s$,

Dividing by $s>0$, respectively $s<0$, and letting $s\to 0$, we obtain the right and left derivatives from the weak convergence just proved. Both equal $\int _ Xf\, \mu _{u_ t}$, which depends continuously on $t$ by the same weak convergence. $\square$

Proof. Let $v\in \mathcal D_\nu $. Then $v\leq \phi +C$ for a constant $C$, and $\phi \leq 0$. Thus the negative part of $\phi $ is at most that of $v$ plus $C$, and the positive part of $\phi $ vanishes; hence $\phi \in L^1(\nu )$. If $\mu _ u=\nu $, the tangent bound of lemma 17.4.10 gives

that is, $F_\nu (v)\leq F_\nu (u)$.

Conversely, let $u$ be a maximizer and $f\in C(X,\mathbb R)$. By proposition 17.5.4, the potentials $u_ t=P_\theta (u+tf)$ have bounded difference from $u$ and finite energy, so they belong to $\mathcal D_\nu $; moreover $u_0=u$ and $u_ t\leq u+tf$ everywhere, hence $\nu $-almost everywhere. Therefore

The left side vanishes at $t=0$ and is differentiable there by proposition 17.5.4; since it attains its maximum at $t=0$, its derivative vanishes there. Thus $\int _ Xf\, \mu _ u=\int _ Xf\, \mathrm d\nu $ for every $f\in C(X,\mathbb R)$, and lemma 17.5.3(1) gives $\mu _ u=\nu $. Since $\mu _ u(X)=m$, a maximizer can exist only if $\nu (X)=m$. $\square$

Proof. Put $\nu =g\, \mathrm dV$, a non-pluripolar measure of mass $m$. Every $\theta $-psh potential is integrable for $\mathrm dV$, hence for $\nu $, so $\mathcal D_\nu =\mathcal E^1(X,\theta ,\phi )$; we write $F=F_\nu $. By the shift formula for the energy and $\nu (X)=m$, the functional $F$ is invariant under adding constants, so we may restrict it to potentials $v$ with $\sup _ Xv=0$. For such $v$, lemma 16.22.2 gives $v\leq \phi $, hence $E(v)\leq 0$. Since $\mathrm dV\leq c_ V\, \omega ^ n$ for a constant $c_ V$, proposition 10.3.1 shows that $\int _ X|v|\, \mathrm dV$ is bounded independently of $v$, and $\phi \in L^1(\mathrm dV)$, so

with $C$ independent of $v$. Consequently $\sup F\in [0,C]$, since $F(\phi )=0$ and $F(v)\leq 0+C$.

Let $(v_ j)$ be a normalized maximizing sequence with $F(v_ j)\geq -1$. Then $E(v_ j)=F(v_ j)+\int _ X(v_ j-\phi )g\, \mathrm dV\geq -C-1$. By corollary 17.4.5 a subsequence converges in $L^1(X)$ to some $u\in \mathcal E^1(X,\theta ,\phi )$, and $\sup _ Xu=0$, because the supremum is continuous for $L^1$ convergence of potentials (proposition 10.3.1). The linear term $v\mapsto \int _ X(v-\phi )g\, \mathrm dV$ is continuous for $L^1$ convergence because $g$ is bounded, and the energy is upper semicontinuous by proposition 17.4.4. Hence $F(u)\geq \varlimsup _ jF(v_ j)=\sup F$, so $u$ is a maximizer, and theorem 17.5.5 gives $\mu _ u=\nu $. $\square$

Proof. Let $A=\{ u<v\} $. The comparison principle of proposition 16.25.4 gives

Both potentials are finite $\nu $-almost everywhere, and on $A$ we have $\mathrm e^{\lambda v}>\mathrm e^{\lambda u}$. The difference $\mathrm e^{\lambda v}-\mathrm e^{\lambda u}$ is therefore positive on $A$ and has nonpositive integral over $A$, so $\nu (A)=0$. Hence $\mu _ u(A)=0$, and the domination principle (theorem 16.22.10) gives $u\geq v$ everywhere. If $u$ and $v$ are two solutions, applying this in both directions gives $u=v$. $\square$

Proof. Let $f\in C(X,\mathbb R)$ and $u_ t=P_\theta (u+tf)$ for $t\in \mathbb R$. By proposition 17.5.4, $u_ t\in \mathcal E^1(X,\theta ,\phi )$, $u_0=u$ and $u_ t\leq u+tf$ everywhere. Hence $\mathrm e^{\lambda u_ t}\leq \mathrm e^{\lambda (u+tf)}$, and $F_\lambda (u_ t)\leq F_\lambda (u)$ gives

The left side vanishes at $t=0$. Its energy term is differentiable, with derivative $\int _ Xf\, \mu _{u_ t}$, by proposition 17.5.4. For $|t|\leq 1$ the $t$-derivative $\lambda f\mathrm e^{\lambda (u+tf)}$ of the integrand is bounded by $\lambda \sup _ X|f|\, \mathrm e^{\lambda \sup _ Xu+\lambda \sup _ X|f|}$, which is integrable for the finite measure $\nu $; by the mean value theorem and dominated convergence, the integral is differentiable as well, with derivative $\lambda \int _ Xf\mathrm e^{\lambda (u+tf)}\, \mathrm d\nu $. Since the left side is maximal at $t=0$, its derivative vanishes there: $\int _ Xf\, \mu _ u=\int _ Xf\mathrm e^{\lambda u}\, \mathrm d\nu $. Since $f$ is arbitrary and $\mathrm e^{\lambda u}\nu $ is a finite measure, lemma 17.5.3(1) gives $\mu _ u=\mathrm e^{\lambda u}\nu $. $\square$

Proof. By lemma 17.4.3, the tail envelopes $a_ j=(\sup _{k\geq j}q_ k)^*$ are $\theta $-psh and decrease pointwise to $q$. For each $j$, corollary 5.47.11, applied after adding smooth local potentials of $\theta $, gives $a_ j=\sup _{k\geq j}q_ k$ outside a pluripolar set. The countable union of these sets is pluripolar (theorem 5.47.8), and outside it $q=\lim _ ja_ j=\varlimsup _ jq_ j$. $\square$

Proof. By lemma 17.5.10, $\varlimsup _ jq_ j=q$ outside a pluripolar set, hence $\nu $-almost everywhere. Thus $(q_ j-q)_+\to 0$ $\nu $-almost everywhere, and $\int _ X(q_ j-q)_+\, \mathrm d\nu \to 0$ by bounded convergence, since $0\leq (q_ j-q)_+\leq 2B$ and $\nu $ is finite. Since $|q_ j-q|=2(q_ j-q)_+-(q_ j-q)$, it remains to show that $\int _ Xq_ j\, \mathrm d\nu \to \int _ Xq\, \mathrm d\nu $. We have just seen that $\varlimsup _ j\int _ X(q_ j-q)\, \mathrm d\nu \leq 0$, so it suffices to prove that $L=\varliminf _ j\int _ Xq_ j\, \mathrm d\nu $ is at least $\int _ Xq\, \mathrm d\nu $.

Choose a subsequence $(q_{j_ k})$ with $\int _ Xq_{j_ k}\, \mathrm d\nu \to L$. For bounded Borel functions $y$ on $X$ put $\| y\| =(\int _ Xy^2\, \mathrm d\nu )^{1/2}$. For $l\geq 1$ let $C_ l$ be the set of finite convex combinations of the functions $q_{j_ k}$ with $k\geq l$, and let $d_ l=\inf \{ \| y\| :y\in C_ l\} $. Since $C_{l+1}\subset C_ l$ and $\| y\| \leq B\nu (X)^{1/2}$ for $y\in C_ l$, the numbers $d_ l$ increase to a finite limit. Choose $y_ l\in C_ l$ with $\| y_ l\| ^2\leq d_ l^2+1/l$. For $l\leq l'$ the function $\frac12(y_ l+y_{l'})$ belongs to the convex set $C_ l$, so its norm is at least $d_ l$. The parallelogram identity, obtained by integrating the pointwise identity $(a-b)^2+(a+b)^2=2a^2+2b^2$, therefore gives

and the right side tends to zero as $l\to \infty $, uniformly in $l'\geq l$.

By the Cauchy–Schwarz inequality, $\int _ X|y_ l-y_{l'}|\, \mathrm d\nu \leq \nu (X)^{1/2}\| y_ l-y_{l'}\| $, so we can choose $l_1<l_2<\cdots $ with $\int _ X|y_{l_{i+1}}-y_{l_ i}|\, \mathrm d\nu \leq 2^{-i}$. By monotone convergence the series $\sum _ i|y_{l_{i+1}}-y_{l_ i}|$ has finite integral, hence converges $\nu $-almost everywhere, and therefore $y_{l_ i}$ converges $\nu $-almost everywhere; let $h=\varlimsup _ iy_{l_ i}$, a Borel function with $|h|\leq B$. By bounded convergence, $\int _ Xy_{l_ i}\, \mathrm d\nu \to \int _ Xh\, \mathrm d\nu $. On the other hand, $\int _ Xy_ l\, \mathrm d\nu $ is a convex combination of the numbers $\int _ Xq_{j_ k}\, \mathrm d\nu $ with $k\geq l$, which tend to $L$, so $\int _ Xy_ l\, \mathrm d\nu \to L$. Hence $L=\int _ Xh\, \mathrm d\nu $.

It remains to show that $h=q$ $\nu $-almost everywhere. Each $y_{l_ i}$ is a convex combination of $\theta $-psh functions, hence $\theta $-psh, and $\| y_{l_ i}-q\| _{L^1(X)}\leq \sup _{k\geq l_ i}\| q_{j_ k}-q\| _{L^1(X)}$, which tends to zero. By lemma 17.5.10, applied to the sequence $(y_{l_ i})_ i$, the function $h=\varlimsup _ iy_{l_ i}$ equals $q$ outside a pluripolar set, hence $\nu $-almost everywhere. Thus $L=\int _ Xq\, \mathrm d\nu $, as required. $\square$

Proof. By proposition 10.3.1 there is a constant $C$ with $q_ j,q\leq C$. We show that the functions $\mathrm e^{\lambda q_ j}$ and $\mathrm e^{\lambda q}$, with value zero on the pole sets, are $C_1\omega $-psh for a constant $C_1$ independent of $j$; the preceding lemma then applies to them.

Choose $C_0\geq 0$ with $\theta \leq C_0\omega $, and a finite cover of $X$ by coordinate balls $W'\Subset W$ such that the smaller balls $W'$ still cover $X$, together with smooth functions $h_ W$ near $\overline W$ with $\mathrm{dd}^{\mathrm{c}}h_ W=\theta $. The function $q_ j+h_ W$ is plurisubharmonic on $W$, so for every $\delta >0$ smaller than the distance from $W'$ to the boundary of $W$, its regularization $g_\delta =(q_ j+h_ W)*\rho _\delta $ by the radial mollifier is a smooth plurisubharmonic function on the ball $W'$, and $g_\delta $ decreases to $q_ j+h_ W$ as $\delta \downarrow 0$, by proposition 5.9.2(2). Put $q_{j,\delta }=g_\delta -h_ W$. Then $\mathrm{dd}^{\mathrm{c}}q_{j,\delta }\geq -\theta \geq -C_0\omega $. Moreover $g_\delta (x)$ is an average of values of $q_ j+h_ W\leq C+h_ W$ at points within distance $\delta $ of $x$, so by uniform continuity of $h_ W$ on a compact neighborhood of $\overline{W'}$ in $W$ we have $q_{j,\delta }\leq C+1$ on $W'$ for all small $\delta $, independently of $j$.

For a smooth real function $s$, the chain rule $\partial \bar\partial \, \mathrm e^{\lambda s}=\lambda \mathrm e^{\lambda s}\partial \bar\partial s+\lambda ^2\mathrm e^{\lambda s}\partial s\wedge \bar\partial s$ and the identity $\frac{\mathrm i}{2\pi }\partial s\wedge \bar\partial s=\mathrm ds\wedge \mathrm d^{\mathrm c}s$, which follows from (5.2.10), give

because $\mathrm dq_{j,\delta }\wedge \mathrm d^{\mathrm c}q_{j,\delta }$ is a semipositive $(1,1)$-form. Thus $\mathrm e^{\lambda q_{j,\delta }}$ is $C_1\omega $-psh on $W'$ with $C_1=\lambda C_0\mathrm e^{\lambda (C+1)}$, and so is its decreasing limit $\mathrm e^{\lambda q_ j}$ as $\delta \downarrow 0$, by theorem 5.10.1 applied after adding a smooth local potential of $C_1\omega $; the limit is not identically $-\infty $, since it is nonnegative. Since the balls $W'$ cover $X$, the function $\mathrm e^{\lambda q_ j}$ belongs to $\mathrm{PSH}(X,C_1\omega )$, and the same argument applies to $\mathrm e^{\lambda q}$.

The exponential is Lipschitz with constant $\lambda \mathrm e^{\lambda C}$ on $(-\infty ,C]$, so $\mathrm e^{\lambda q_ j}\to \mathrm e^{\lambda q}$ in $L^1(X)$. These functions take values in $[0,\mathrm e^{\lambda C}]$, and lemma 17.5.11, with the background $C_1\omega $, gives the assertion. $\square$

Proof. By corollary 17.4.5, the set $K_\phi $ of potentials $w\in \mathrm{PSH}(X,\theta )$ with $w\leq \phi $ and $\sup _ Xw=0$ is compact in $L^1(X)$; it contains $\phi $. By corollary 17.5.12, the function $w\mapsto \int _ X\mathrm e^{\lambda w}\, \mathrm d\nu $ is continuous on $K_\phi $. Its values are positive, since the pole set of $w$ is $\nu $-null and $\nu (X)>0$. Hence it has a positive minimum $a$ on $K_\phi $.

Any $v\in \mathcal E^1(X,\theta ,\phi )$ can be written $v=w+c$ with $c=\sup _ Xv$ and $w\in K_\phi $, by lemma 16.22.2. Since $E(w)\leq 0$, the shift formula for the energy gives

The right side tends to $-\infty $ as $c\to \pm \infty $. Since $F_\lambda (\phi )$ is finite, a maximizing sequence $v_ j=w_ j+c_ j$ with $F_\lambda (v_ j)\geq F_\lambda (\phi )-1$ therefore has bounded constants $c_ j$. Moreover $E(v_ j)\geq F_\lambda (v_ j)$, because the exponential term is nonnegative, so $E(w_ j)=E(v_ j)-mc_ j$ is bounded below. By corollary 17.4.5, after passing to a subsequence, $c_ j$ converges and $w_ j$ converges in $L^1(X)$; hence $v_ j\to u$ in $L^1(X)$ for some $u\in \mathcal E^1(X,\theta ,\phi )$. The energy is upper semicontinuous (proposition 17.4.4), and $\int _ X\mathrm e^{\lambda v_ j}\, \mathrm d\nu \to \int _ X\mathrm e^{\lambda u}\, \mathrm d\nu $ by corollary 17.5.12, so $F_\lambda (u)\geq \varlimsup _ jF_\lambda (v_ j)=\sup F_\lambda $, and $u$ is a maximizer. By lemma 17.5.9, $\mu _ u=\mathrm e^{\lambda u}\nu $. Uniqueness in $\mathcal E(X,\theta ,\phi )$ is proposition 17.5.7. $\square$

Proof. Put $g_ j=\inf _{k\geq j}f_ k$, so that $g_ j\uparrow f$ $\nu $-almost everywhere. We first prove that $\mu _{a_ j}\geq g_ j\nu $ for the tail envelopes $a_ j=(\sup _{k\geq j}q_ k)^*$, and then let $j\to \infty $. For $l\geq j$, the finite maximum $\max \{ q_ j,\ldots ,q_ l\} $ has measure at least $g_ j\nu $. Indeed, let $r_1,r_2\in \mathrm{PSH}(X,\theta )$ satisfy $\mu _{r_1},\mu _{r_2}\geq g_ j\nu $, and put $r=\max \{ r_1,r_2\} $. The set $\{ r_1\geq r_2\} $ is the contact set $\{ r_1=r\} $, so lemma 16.13.1, applied to $r_1\leq r$, gives $\mu _ r\geq \mu _{r_1}\geq g_ j\nu $ on $\{ r_1\geq r_2\} $; in the same way $\mu _ r\geq g_ j\nu $ on $\{ r_2>r_1\} $. Hence $\mu _ r\geq g_ j\nu $. Since $\mu _{q_ k}\geq f_ k\nu \geq g_ j\nu $ for $k\geq j$, induction on $l$ proves the claim.

The suprema $\sup _ Xq_ k$ are bounded, by proposition 10.3.1, so the regularized upper envelope $a_ j$ is $\theta $-psh, and it equals $\sup _{k\geq j}q_ k$ almost everywhere (theorem 5.10.2, applied in charts after adding smooth local potentials of $\theta $). The finite maxima $\max \{ q_ j,\ldots ,q_ l\} $ therefore increase almost everywhere to $a_ j$ as $l\to \infty $, and their measures converge weakly to $\mu _{a_ j}$ by corollary 13.8.9. The fixed measure $g_ j\nu \leq \mu _{q_ j}$ is finite, so the inequality passes to the weak limit, by lemma 17.5.3(2):

By lemma 17.4.3, $a_ j$ decreases pointwise to $q$.

To let $j\to \infty $ we truncate. Fix $C>0$ and an index $j_0$. The truncations $a_ j^ C=\max \{ a_ j,V_\theta -C\} $ have uniformly bounded difference from $V_\theta $: they are at least $V_\theta -C$, and since the suprema of the $a_ j$ are bounded, minimality of $V_\theta $ bounds them above by $V_\theta +C'$. They decrease to $q^ C=\max \{ q,V_\theta -C\} $, so they converge to $q^ C$ in extended capacity, by lemma 13.7.3 applied in charts after adding smooth local potentials of $\theta $. Since $q^ C$ has bounded difference from $V_\theta $, it has the mass of $V_\theta $ (corollary 13.8.8), that is, $q^ C\in \mathcal E(X,\theta ,V_\theta )$. Corollary 13.9.4, with $V_\theta $ in place of $\phi $, therefore gives $\mu _{a_ j^ C}\to \mu _{q^ C}$ weakly.

On the set $A_ C=\{ q>V_\theta -C\} $ we have $a_ j\geq q>V_\theta -C$, so $a_ j^ C=a_ j$ on the plurifine open set $\{ a_ j>V_\theta -C\} \supset A_ C$, and locality gives $\mu _{a_ j^ C}\geq \mathbf1_{A_ C}g_{j_0}\nu $ for $j\geq j_0$. Since $\mathbf1_{A_ C}g_{j_0}\nu $ is a fixed finite measure, the inequality passes to the weak limit, by lemma 17.5.3(2), and gives $\mu _{q^ C}\geq \mathbf1_{A_ C}g_{j_0}\nu $. On the plurifine open set $A_ C$ we have $q^ C=q$, so locality gives $\mu _ q\geq \mathbf1_{A_ C}g_{j_0}\nu $. Thus $\mu _ q(E)\geq \int _{E\cap A_ C}g_{j_0}\, \mathrm d\nu $ for every Borel set $E$. Let $j_0\to \infty $ and then $C\to \infty $. The sets $A_ C$ increase to the set where $q$ and $V_\theta $ are finite, whose complement is pluripolar and hence $\nu $-null, so monotone convergence gives $\mu _ q(E)\geq \int _ Ef\, \mathrm d\nu $, that is, $\mu _ q\geq f\nu $. $\square$

Proof. Let $\lambda _ j\downarrow 0$, let $u_ j$ be the solution of $\mu _{u_ j}=\mathrm e^{\lambda _ ju_ j}\nu $ given by theorem 17.5.13, and write $u_ j=w_ j+c_ j$ with $c_ j=\sup _ Xu_ j$. Since $u_ j\preceq \phi $ and $\sup _ Xw_ j=0$, lemma 16.22.2 gives $w_ j\leq \phi $, so $w_ j$ belongs to the set $K_\phi $ of corollary 17.4.5, which is compact in $L^1(X)$. On $K_\phi $ the functions

are continuous, by corollary 17.5.12, and they increase pointwise to the constant $m$, by monotone convergence, since $w\leq 0$ and the pole set of $w$ is $\nu $-null. The convergence is uniform on $K_\phi $ (Dini's theorem): for $\varepsilon >0$ the compact sets $\{ w\in K_\phi :m-H_ j(w)\geq \varepsilon \} $ decrease and have empty intersection, so they are empty for large $j$.

Comparing total masses in the equation gives $m=\mathrm e^{\lambda _ jc_ j}H_ j(w_ j)$, so

Also $\int _ X|1-\mathrm e^{\lambda _ jw_ j}|\, \mathrm d\nu =m-H_ j(w_ j)\to 0$. Hence $r_ j=\mathrm e^{\lambda _ ju_ j}=\mathrm e^{\lambda _ jc_ j}\mathrm e^{\lambda _ jw_ j}\to 1$ in $L^1(\nu )$, and $\mu _{w_ j}=\mu _{u_ j}=r_ j\nu $. Pass to a subsequence along which $r_ j\to 1$ $\nu $-almost everywhere and, by compactness of $K_\phi $, $w_ j\to u$ in $L^1(X)$ for some $u\in K_\phi $. Lemma 17.5.14 gives $\mu _ u\geq \nu $. Since $u\leq \phi $, mass monotonicity gives $\mu _ u(X)\leq m=\nu (X)$. Hence $\mu _ u=\nu $, $u$ has full mass relative to $\phi $, and $\sup _ Xu=0$ because $u\in K_\phi $. $\square$