PomeLemma 5.40.1.294BB4EEA84F Let $\theta $ be a smooth real $(1,1)$-form on $X$, let $K\subseteq X$ be compact and non-pluripolar, and let $w\in C(K,\mathbb R)$. There is a constant $C=C(\theta ,K,w)$ such that every $a\in \mathrm{PSH}(X,\theta )$ satisfying $a\leq w$ quasi-everywhere on $K$ has $\sup _ X a\leq C$.
5.40 Obstacle envelopes8EA2EE30DE99
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Given an obstacle function, one looks for the largest $\theta $-plurisubharmonic function lying below it. The local upper-envelope theorem constructs this envelope as soon as the competitors have a common upper bound, and normalized compactness provides such a bound for an obstacle that is finite almost everywhere or prescribed on a non-pluripolar compact set. Letting the obstacles $\min (u+C,0)$ increase with $C$ produces the envelope $P_\theta [u]$ of the ordinary singularity type of $u$.
A smooth real $(1,1)$-form $\theta $ is called big if there are $\rho \in \mathrm{PSH}(X,\theta )$ and $\delta >0$ such that
\[ \theta +\mathrm{dd}^{\mathrm{c}}\rho \geq \delta \omega . \]
Here $\omega $ is any Hermitian form; changing it changes $\delta $ but not the condition. The definition does not require $\theta $ to be closed.
Let $\theta $ be a smooth real $(1,1)$-form. For an extended-real function $f$ on $X$, defined at least outside a pluripolar set (the obstacle), write
\[ P_\theta (f)=\left(\sup \{ v\in \mathrm{PSH}(X,\theta ):v\leq f \ \hbox{q.e.}\} \right)^*. \]
As in section 5.27, q.e. (quasi-everywhere) means outside a pluripolar set. The functions $v$ in the supremum are called competitors; the empty supremum is $-\infty $. For obstacles $f_1,\ldots ,f_ N$, we write $P_\theta (f_1,\ldots ,f_ N)=P_\theta (\min _ if_ i)$; when the $f_ i$ belong to $\mathrm{PSH}(X,\theta )$, this is their rooftop envelope. For an upper semicontinuous obstacle, the pointwise constraint $v\leq f$ gives the same competitors: by the local submean inequality and upper semicontinuity of $f$, a quasi-plurisubharmonic function lying below $f$ almost everywhere lies below it everywhere.
If the family of competitors is nonempty and uniformly bounded above, then $P_\theta (f)\in \mathrm{PSH}(X,\theta )$. To apply the local upper-envelope theorem while keeping the exact lower bound $-\theta $, fix a constant positive form $\kappa $ in local coordinates. For each $\epsilon >0$, shrink the coordinate neighborhood and choose a constant form $\theta '$ with $-\epsilon \kappa \leq \theta -\theta '\leq \epsilon \kappa $. A smooth function $r$ with $\mathrm{dd}^{\mathrm{c}}r=\theta '+\epsilon \kappa $ makes every competitor plus $r$ plurisubharmonic, so the local envelope theorem gives $\theta +\mathrm{dd}^{\mathrm{c}}P_\theta (f)\geq -2\epsilon \kappa $ there. Letting $\epsilon $ tend to zero proves the exact bound. The same local theorem shows that the regularization changes the supremum only on a pluripolar set.
The following observation gives the upper bound for an unbounded obstacle. Suppose that $f\leq A$, for a real constant $A$, on a Borel set $E$ with $\operatorname {Vol}(E)=\int _ E\mathrm{d}V>0$, where $\mathrm{d}V$ is the volume form fixed in theorem 5.25.2. If $h$ is a competitor, put $\widetilde h=h-\sup _ Xh$. By theorem 5.25.2,
\[ (\sup _ Xh-A)\operatorname {Vol}(E) \leq \int _ E|\widetilde h|\, \mathrm{d}V\leq C_\theta \quad \hbox{whenever }\sup _ Xh>A. \]
Thus $\sup _ Xh\leq A+C_\theta /\operatorname {Vol}(E)$ for every competitor. In particular, the competitors for a measurable obstacle that is finite almost everywhere have a common upper bound: apply this to a sublevel set $\{ f\leq A\} $ of positive volume.
An obstacle bound on a non-pluripolar compact set also bounds every competitor globally, even when that set has zero smooth volume.
Proof. Otherwise there are such functions $a_ j$ with
\[ M_ j:=\sup _ Xa_ j\geq 2^ j+\max _ Kw. \]
The normalized functions $v_ j=a_ j-M_ j\leq 0$ have uniformly bounded $L^1$ norms by theorem 5.25.2. Choose $A>0$ with $\theta \leq A\omega $. The partial sums
\[ s_ N=\sum _{j=1}^ N2^{-j}v_ j \]
decrease, have uniformly bounded $L^1$ norms, and satisfy $\mathrm{dd}^{\mathrm{c}}s_ N\geq -A\omega $. By corollary 5.25.3, their limit $s$ is a quasi-plurisubharmonic function; in particular, $s$ is not identically $-\infty $.
Let $N_ j\subseteq K$ be a pluripolar set outside which $a_ j\leq w$ on $K$. On $K\setminus \bigcup _ jN_ j$ one has $v_ j\leq -2^ j$ for every $j$, hence $s=-\infty $. The pole set $\{ s=-\infty \} $ of the quasi-plurisubharmonic function $s$ is pluripolar, and so is the countable union of the $N_ j$. Their union would contain all of $K$, contradicting the hypothesis on $K$. $\square$
Suppose that $\mathrm{PSH}(X,\theta )$ is nonempty. Then the envelope of the zero obstacle,
\[ V_\theta =P_\theta (0), \]
belongs to $\mathrm{PSH}(X,\theta )$: the competitors are bounded above by zero, and they include $u-\sup _ Xu$ for every $u\in \mathrm{PSH}(X,\theta )$. In particular,
\[ u-\sup _ Xu\leq V_\theta \qquad (u\in \mathrm{PSH}(X,\theta )). \]
A potential $v\in \mathrm{PSH}(X,\theta )$ has minimal singularities if $u\preceq v$ for every $u\in \mathrm{PSH}(X,\theta )$. By the last inequality, $V_\theta $ has minimal singularities, and every potential $v$ with minimal singularities satisfies $v\sim V_\theta $. If $\theta \geq 0$, then $0$ is a competitor, and hence $V_\theta =0$.
Lemma 5.40.2.AA358481B7F6 Let $\eta $ be a smooth semipositive real $(1,1)$-form on $X$. If an obstacle $f$ is finite almost everywhere and bounded below quasi-everywhere by a constant, then $P_\eta (f)$ is a bounded $\eta $-plurisubharmonic function.
Proof. Since $\eta \geq 0$, constants are $\eta $-plurisubharmonic, so the constant lower bound of $f$ is a competitor. The finite-sublevel argument above gives a common upper bound for all competitors. Hence $P_\eta (f)$ is an $\eta $-plurisubharmonic function lying between these two constants. $\square$
Proposition 5.40.3.630365A8F8B8 If $f\in C^0(X)$, then $P_\omega (f)$ is continuous. Moreover,
\[ \| P_\omega (f)-P_\omega (g)\| _\infty \leq \| f-g\| _\infty \quad (f,g\in C^0(X)). \]
Proof. The constant $\min _ Xf$ is a competitor, so $p=P_\omega (f)$ is bounded. The local envelope theorem gives $p\leq f$ quasi-everywhere, and continuity of $f$ makes this inequality hold everywhere. By theorem 5.32.1, there are smooth $\omega $-plurisubharmonic functions $v_ j\downarrow p$. Fix $\epsilon >0$. The continuous functions $v_ j-f$ decrease to a function that is at most zero, so $v_ j\leq f+\epsilon $ everywhere for all large $j$; otherwise the decreasing compact sets $\{ v_ j-f\geq \epsilon \} $ would have nonempty intersection. Hence $v_ j-\epsilon $ is a competitor, and
\[ p\leq v_ j\leq p+\epsilon . \]
Thus $v_ j\to p$ uniformly, which proves continuity. The last inequality follows from the monotonicity of $P_\omega $ in the obstacle and the identity $P_\omega (g+c)=P_\omega (g)+c$ for constants $c$, applied with $c=\| f-g\| _\infty $. $\square$
Definition 5.40.4.65C9C3635D58 For $u\in \mathrm{PSH}(X,\theta )$, set
\[ P_\theta [u] =\left(\lim _{C\to +\infty }P_\theta (u+C,0)\right)^* =\left(\sup \{ h\in \mathrm{PSH}(X,\theta ):h\preceq u,\ h\leq 0\} \right)^*. \]
We call $u$ an envelope fixed point, or an ordinary model, if $P_\theta [u]=u$; in the Kähler literature such potentials are called model potentials. For $F\in C^\infty (X,\mathbb R)$, replacing the constraint $h\leq 0$ by $h\leq F$ defines $P_{\theta ,F}[u]$.
The two expressions agree, because the family in the second one is the increasing union over $C$ of the families defining $P_\theta (u+C,0)$. They do not change if zero is replaced by $V_\theta $, since every $\theta $-plurisubharmonic function below zero lies below $V_\theta $. The competitors include $u-\sup _ Xu$ and are bounded above by zero, so $P_\theta [u]\in \mathrm{PSH}(X,\theta )$ and $u-\sup _ Xu\leq P_\theta [u]\leq 0$. In particular, $u\preceq P_\theta [u]$; this does not assert that $P_\theta [u]-u$ is bounded.
Directly from the definition, $P_\theta [u]$ depends only on the ordinary singularity type of $u$, and
\[ u\preceq v\quad \Longrightarrow \quad P_\theta [u]\leq P_\theta [v]. \]
Replacing the condition $h\preceq u$ by $h\sim _{\mathcal I}u$ gives the multiplier-ideal envelope $P_\theta [u]_{\mathcal I}$ of definition 18.4.1. This is a different operator; its relation to $P_\theta [u]$ is studied in section 18.2.
For a bounded obstacle in a big background, exceptional sets in the competitor inequality can be removed by a small singular perturbation.
Lemma 5.40.5.44FD5A853A0A Let $\theta $ be big, let $f$ be a bounded function on $X$, and let $u\in \mathrm{PSH}(X,\theta )$ satisfy $u\leq f$ quasi-everywhere. Then there are $u_ t\in \mathrm{PSH}(X,\theta )$, $t>0$, such that $u_ t\leq f$ everywhere and $u_ t\to u$ almost everywhere as $t\downarrow 0$.
Proof. Choose $\rho \in \mathrm{PSH}(X,\theta )$ with $\theta +\mathrm{dd}^{\mathrm{c}}\rho \geq \kappa \omega $ for some $\kappa >0$. The exceptional set $E=\{ u>f\} $ is pluripolar. The global form of Josefson's theorem on compact complex manifolds gives a quasi-plurisubharmonic function $\chi $ with $\chi =-\infty $ on $E$; see [Theorem 1.1.5, SGPT]. After rescaling and adding a constant, we may assume that $\chi \leq 0$ and $\mathrm{dd}^{\mathrm{c}}\chi \geq -\omega $. Subtract a constant from $\rho $ so that $\rho \leq \inf _ X f$. With $s=t/\kappa $, put
\[ u_ t=\frac{u+t\chi +s\rho }{1+s}. \]
All three functions in the numerator are finite almost everywhere, so $u_ t$ is not identically $-\infty $. Moreover,
\[ \theta +\mathrm{dd}^{\mathrm{c}}u_ t =\frac{\theta +\mathrm{dd}^{\mathrm{c}}u+t\mathrm{dd}^{\mathrm{c}}\chi +s(\theta +\mathrm{dd}^{\mathrm{c}}\rho )}{1+s} \geq 0. \]
Outside $E$, both $u$ and $\rho $ are at most $f$, while $\chi \leq 0$, so $u_ t\leq f$. On $E$ the numerator is $-\infty $. Finally, $u_ t\to u$ wherever all three functions are finite, which is a set of full measure. $\square$
For an obstacle prescribed only on a compact set, the same perturbation shows that pluripolar exceptions in the constraint do not change the regularized envelope.
Proposition 5.40.6.F031A16FF76F Let $\theta $ be big, let $K\subseteq X$ be compact and non-pluripolar, and let $\psi $ be a bounded real function on $K$. Set
\[ \begin{aligned} R& =\sup \{ u\in \mathrm{PSH}(X,\theta ):u\leq \psi \text{ on }K\} ,\\ Q& =\sup \{ u\in \mathrm{PSH}(X,\theta ):u\leq \psi \text{ q.e. on }K\} . \end{aligned} \]
Then $R^*=Q^*$ belongs to $\mathrm{PSH}(X,\theta )$ and has minimal singularities; it is bounded if $\theta \geq 0$. Moreover, this regularized envelope does not change if the pointwise constraint is imposed only on $K\setminus E$, for any pluripolar subset $E\subseteq K$.
Proof. Put $m=\inf _ K\psi $ and $M=\sup _ K\psi $. Since $\theta $ is big, $\mathrm{PSH}(X,\theta )$ is nonempty, and $V_\theta \leq 0$; hence $V_\theta +m$ belongs to the family defining $R$. Applying lemma 5.40.1 with the constant obstacle $M$ gives a common upper bound $C$ for the larger family defining $Q$. The upper-envelope argument at the beginning of section 5.40 therefore shows that $R^*$ and $Q^*$ belong to $\mathrm{PSH}(X,\theta )$.
Now fix a competitor $u$ for $Q$ and define a bounded function on $X$ by
\[ f(x)= \begin{cases} \psi (x),& x\in K,\\ \max \{ \sup _ Xu,M\} ,& x\notin K. \end{cases} \]
Then $u\leq f$ quasi-everywhere. By lemma 5.40.5, there are $u_ t\in \mathrm{PSH}(X,\theta )$ satisfying $u_ t\leq f$ everywhere and $u_ t\to u$ almost everywhere. Each $u_ t$ is a competitor for $R$, so $R^*\geq u$ almost everywhere, and lemma 5.24.3 gives this inequality everywhere. Taking the supremum over $u$ and regularizing yields $R^*\geq Q^*$; the inclusion of the competitor families gives the reverse inequality.
Write $F=R^*=Q^*$. Since $F\leq C$, the potential $F-C$ is a competitor for $V_\theta $. Thus
\[ V_\theta +m\leq F\leq V_\theta +C, \]
which proves minimal singularities and, when $\theta \geq 0$, boundedness.
Finally, if $E\subseteq K$ is pluripolar, the family constrained on $K\setminus E$ lies between the families defining $R$ and $Q$, so its regularized supremum also equals $F$. Note that this argument uses the upper bound obtained from the compact set $K$, even when $K\setminus E$ is not compact. $\square$
A potential with minimal singularities stays away from the poles of a strictly positive potential with analytic singularities, in the following sense.
Lemma 5.40.7.4798FCFAC293 Let $\rho \in \mathrm{PSH}(X,\theta )$ have analytic singularities with smooth remainder and satisfy $\theta +\mathrm{dd}^{\mathrm{c}}\rho \geq \varepsilon \omega $ for some $\varepsilon >0$. Set $\Omega =\{ \rho >-\infty \} $. If $\varphi \in \mathrm{PSH}(X,\theta )$ has minimal singularities and is continuous on $\Omega $, then $\varphi -\rho $ is continuous, bounded below, and proper on $\Omega $: all its sublevel sets are compact in $\Omega $.
Proof. Adding constants to $\rho $ and $\varphi $ does not affect the assertions, so we may assume that $\rho \leq 0$ and $\sup _ X\varphi =0$. Since $\varphi $ has minimal singularities, there is a constant $C$ with $\varphi \geq V_\theta -C$. Choose $M\geq 0$ with $\theta \geq -M\omega $. If $a\in (0,1)$ is sufficiently close to one, then
\[ \theta +\mathrm{dd}^{\mathrm{c}}(a\rho ) =a(\theta +\mathrm{dd}^{\mathrm{c}}\rho )+(1-a)\theta \geq \bigl (a\varepsilon -(1-a)M\bigr )\omega >0. \]
Thus $a\rho $ is a competitor for $V_\theta $, and therefore
\[ 0\geq \varphi \geq V_\theta -C\geq a\rho -C, \]
and hence
By hypothesis $\varphi $ is continuous on $\Omega $, and $\rho $ is smooth there because it has analytic singularities with smooth remainder. Hence $\varphi -\rho $ is continuous on $\Omega $, and by (5.40.8) it is bounded below. Since $\rho \to -\infty $ at $X\setminus \Omega $, the sets $\{ \rho \geq -c\} $ are compact subsets of $\Omega $, and (5.40.8) places every sublevel set of $\varphi -\rho $ in one of them. Being closed in $\Omega $, each sublevel set is therefore compact. $\square$
5.40.9 A hyperplane Green envelope on the ball765486939AA3
For a logarithmic pole along a hyperplane section of the unit ball, the envelope can be computed explicitly by restricting to the complex lines normal to the hyperplane. The example also shows how envelopes of bounded obstacles recover the finite part of a logarithmic pole, and that continuity inside the domain does not imply continuity up to the boundary.
In this subsection let $n\geq 2$, let $m\geq 1$ be an integer, and write
\[ B=\{ (\zeta ,w)\in \mathbb C\times \mathbb C^{n-1}:|\zeta |^2+|w|^2<1\} , \qquad D=B\cap \{ \zeta =0\} . \]
Define $G_ m$ as the pointwise supremum of the functions $u\in \mathrm{PSH}(B)$ such that $u\leq 0$ and, near every point of $D$, $u\leq m\log |\zeta |^2+C$ for some local constant $C$. In particular, $u=-\infty $ on $D$. The constants need not stay bounded as the point of $D$ approaches $\partial B$.
Proposition 5.40.10.B986B6AE211B The envelope $G_ m$ is given by
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5.40.11
\begin{equation} G_ m(\zeta ,w)=m\log \frac{|\zeta |^2}{1-|w|^2}. \end{equation}
It is plurisubharmonic and continuous on $B$ with values in $[-\infty ,0)$. The finite part $G_ m-m\log |\zeta |^2$, first defined off $D$, extends smoothly across $D$ as the plurisubharmonic function
\[ H_ m(w)=-m\log (1-|w|^2). \]
Proof. The function $-\log (1-|w|^2)$ is smooth and plurisubharmonic on $|w|<1$: its complex Hessian is
\[ \left(\frac{\delta _{jk}}{1-|w|^2} +\frac{\overline w_ jw_ k}{(1-|w|^2)^2}\right)_{j,k}. \]
Thus the right-hand side of (5.40.11) is plurisubharmonic; it is negative on $B$ and satisfies the pole condition, so it is a competitor. For the reverse inequality, fix a competitor $u$ and a point $w$ with $|w|<1$. The restriction $u(\cdot ,w)$ is either identically $-\infty $ or subharmonic on the disk of radius $R=(1-|w|^2)^{1/2}$. In the latter case, $u(\zeta ,w)-m\log |\zeta |^2$ is subharmonic on the punctured disk and, by the pole condition, bounded above near zero, so it extends as a subharmonic function across zero.
Since $u\leq 0$, this extension is at most $-m\log r^2$ on every circle $|\zeta |=r<R$. The maximum principle and then $r\uparrow R$ give
\[ u(\zeta ,w)\leq m\log \frac{|\zeta |^2}{R^2}. \]
On $D$ both sides are $-\infty $. Taking the supremum over $u$ proves the formula. Near a point of $D$, the function $H_ m$ is bounded and the logarithmic term tends to $-\infty $, which proves continuity there as an extended-real function. $\square$
For a finite continuous obstacle $\Phi $ on $B$, put
\[ P_ B\Phi =\left(\sup \{ v\in \mathrm{PSH}(B):v\leq \Phi \} \right)^*. \]
The local upper-envelope theorem used in section 5.40 applies whenever this family is nonempty and locally uniformly bounded above. Continuity of $\Phi $ ensures that the regularized envelope still satisfies $P_ B\Phi \leq \Phi $.
Proposition 5.40.12.1709375DAB71 For $t>0$, define
\[ \Phi _ t(\zeta ,w)=\min \{ -m\log |\zeta |^2,t\} , \qquad V_ t=P_ B\Phi _ t, \]
with $\Phi _ t=t$ on $D$. Then $0\leq V_ t\leq t$, and
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5.40.13
\begin{equation} -m\log \bigl (1-(1-\mathrm e^{-t/m})|w|^2\bigr ) \leq V_ t(\zeta ,w)\leq H_ m(w). \end{equation}
In particular, $V_ t\uparrow H_ m$ pointwise on all of $B$, including $D$. The convergence is uniform on every compact subset, with
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5.40.14
\begin{equation} 0\leq H_ m-V_ t \leq \frac{m\mathrm e^{-t/m}R^2}{1-R^2} \qquad \text{where }|w|\leq R<1. \end{equation}
Proof. The obstacle is continuous, nonnegative and at most $t$. Zero is a competitor, so the upper-envelope theorem gives $V_ t\in \mathrm{PSH}(B)$ with $0\leq V_ t\leq \Phi _ t\leq t$. Consequently $m\log |\zeta |^2+V_ t$ is a competitor for $G_ m$, and proposition 5.40.10 gives $V_ t\leq H_ m$ off $D$. At a point $(0,w)$, the submean inequality on any sufficiently small circle in the normal line gives the same bound, since $H_ m$ does not depend on $\zeta $.
For the lower bound, set $a_ t=1-\mathrm e^{-t/m}$ and $F_ t(w)=-m\log (1-a_ t|w|^2)$. The Hessian calculation above, with $|w|^2$ replaced by $a_ t|w|^2$, shows that $F_ t$ is plurisubharmonic. Moreover, $0\leq F_ t\leq t$, and $F_ t\leq H_ m\leq -m\log |\zeta |^2$ on $B\setminus D$ because $|\zeta |^2<1-|w|^2$. Thus $F_ t\leq \Phi _ t$ everywhere, and hence $F_ t\leq V_ t$. Finally,
\[ H_ m-F_ t =m\log \left(1+\frac{\mathrm e^{-t/m}|w|^2}{1-|w|^2}\right) \leq \frac{m\mathrm e^{-t/m}|w|^2}{1-|w|^2}. \]
This proves (5.40.14) and the convergence; the monotonicity of $V_ t$ in $t$ follows from that of $\Phi _ t$. $\square$
Example 5.40.15.F5EC436555E6 At every point $p=(0,w_0)\in \partial B$ with $|w_0|=1$, the cluster set of $G_ m$ at $p$, that is, the set of limits of $G_ m$ along sequences in $B$ converging to $p$, is all of $[-\infty ,0]$. Indeed, set $w_ s=(1-s^2)^{1/2}w_0$ for $0<s<1$. For every fixed $0<a<1$, the points $(as,w_ s)$ belong to $B$, tend to $p$ as $s\to 0$, and satisfy $G_ m=m\log a^2$. Choosing $a=a_ s<1$ with $a_ s\to 1$ gives the limit zero, whereas the points $(s^2,w_ s)$ give $G_ m=m\log s^2\to -\infty $. These realize every value in $[-\infty ,0]$, and negativity excludes all others. Thus $G_ m$, although continuous on $B$, has no continuous extension to any point of $\partial B$ in the closure of $D$.
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