15.8 Contact support in a big backgroundACC753805E21

Proof. Subtract a constant from $\rho $ so that $\rho +2\leq h$. Then $\rho +2$ is a competitor, and since the obstacle is continuous, the envelope lies below it everywhere; thus

In particular, $p$ is locally bounded off $Z$. We first prove that $p$ is continuous there; this part does not use the contact assertion.

Apply lemma 5.42.10 to $p-\sup _ Xp$ and add back the constant. The transforms $\Phi _{c,\delta }$ of that lemma are defined on $X$ and, with fixed constants $A,K,C>0$ and $\epsilon (\delta )=C\delta $, satisfy

Off $Z$ the infimum defining $\Phi _{c,\delta }$ is attained at a positive radius, and

For small $\delta $ the first inequality gives $\Phi _{c,\delta }\geq \rho +1$, hence $\rho \leq \Phi _{c,\delta }$ on $X$.

Set

The averaging kernels are positive, have integral one and are supported in balls of radius comparable to $t$, so uniform continuity of $h$ gives $m(\delta )\to 0$. Define

By the choice of $s$, the strict barrier and (15.8.2) show that $v_\delta $ is $\theta $-plurisubharmonic. Since $\rho \leq \Phi _{c,\delta }$,

Thus $v_\delta -m(\delta )$ is a competitor for the envelope, and $v_\delta \leq p+m(\delta )$.

Let $L\Subset X\setminus Z$ be compact. On $L$ the difference $p-\rho $ is bounded above. Since $s\leq Cc$, the identity

therefore gives $\Phi _{c,\delta }-p\leq C_ Lc$ on $L$. At a radius $t$ where the infimum is attained,

Since $c\geq \delta $, this yields $t\geq a_ L\delta $ with a fixed $a_ L>0$. Because $R_ tp+Kt^2$ is increasing in $t$, it follows that

Consequently $R_{a_ L\delta }p\to p$ uniformly on $L$. These averages are smooth for positive radius, even though $p$ is unbounded on $X$. Applying this to compact neighborhoods of the points of $X\setminus Z$ proves that $p$ is continuous on $X\setminus Z$.

We turn to the contact assertion. By the continuity just proved, the set $\{ p<h\} \setminus Z$ is open. Let $x$ be one of its points, and choose a small coordinate ball $B_ r$ centered at $x$ whose closure lies in this set. Let $H$ be the homogeneous replacement of theorem 15.7.2 with boundary values $p|_{\partial B_ r}$. Its domination property gives $H\geq p$ on the ball.

We claim that $H\leq h$ on $B_ r$ if $r$ is small enough. Indeed, in the chosen Euclidean coordinates $z$, with Laplacian $\Delta $, the inequality $\theta +\mathrm{dd}^{\mathrm{c}}H\geq 0$ implies a fixed bound $\Delta H\geq -C$. Comparison with the function $\sup _{\partial B_ r}p+C'(r^2-|z-x|^2)$, with $C'$ fixed large enough, gives

As $r\to 0$, the right-hand side tends to $p(x)$, whereas $\inf _{B_ r}h\to h(x)>p(x)$. This proves the claim for sufficiently small $r$.

Now replace $p$ by $H$ on the ball. The resulting function is $\theta $-plurisubharmonic. To see this, fix $a>0$ and take $\max \{ p,H-a\} $ inside the ball and $p$ outside. Since $H$ and $p$ are continuous with equal boundary values, $H-a<p$ on a collar of $\partial B_ r$, so this is an ordinary local maximum construction. Letting $a\downarrow 0$ gives the function equal to $H$ on $B_ r$ and to $p$ outside, with locally uniform convergence near the boundary of the ball. This function is bounded above by $h$ on all of $X$, so the definition of the envelope forces it to be at most $p$. Thus $p=H$ on $B_ r$, and $\mathrm{MA}_\theta (p)$ vanishes there.

These balls cover $\{ p<h\} \setminus Z$. There the extended measure agrees with the local bounded product, and it does not charge the pluripolar set $Z$. This proves the contact assertion on $X$; no bound on the total mass of $\mathrm{MA}_\theta (p)$ was needed. $\square$

Proof. Using the infimum construction in the proof of proposition 16.28.3, take continuous $f_ j\uparrow f$ with common bounds $m\leq f_ j\leq M$, and shift $\rho $ below $m$. The envelopes $p_ j=P_\theta (f_ j)$ increase and satisfy

Their regularized limit $q=(\lim _ jp_ j)^*$ is therefore $\theta $-plurisubharmonic. Outside a pluripolar set it equals the pointwise limit, which is at most $f$; hence $q\leq p=P_\theta (f)$.

We prove the reverse inequality without using a domination theorem. Let $u$ be any competitor for $P_\theta (f)$, and let $u_ t\leq f$ be the potentials given by lemma 5.40.5, which satisfy this inequality everywhere. For fixed $t,\eta >0$, the sets

are open, increase with $j$, and cover $X$. By compactness, $u_ t\leq f_ j+\eta $ everywhere for all large $j$. Thus $u_ t-\eta $ is a competitor for $p_ j$, and $q\geq u_ t-\eta $. Letting first $\eta \downarrow 0$ and then $t\downarrow 0$, we obtain $q\geq u$ almost everywhere, hence everywhere by lemma 5.24.3. Taking the supremum over all competitors $u$ gives $q\geq p$, so $q=p$.

For fixed $k$, the set $U_ k=\{ p<f_ k\} $ is open. If $j\geq k$, then $p_ j\leq p<f_ k\leq f_ j$ on $U_ k$, so theorem 15.8.1 gives $\mathrm{MA}_\theta (p_ j)=0$ there. On every relatively compact open subset of $X\setminus Z$ the functions $p_ j$ have common bounds. After a common smooth correction and a finite expansion, the local Bedford–Taylor continuity along increasing sequences shows that their measures converge to $\mathrm{MA}_\theta (p)$ there. Since the measure of an open set is at most the lower limit of the measures of the approximating sequence, $\mathrm{MA}_\theta (p)$ vanishes on $U_ k\setminus Z$. The union of these sets is $\{ p<f\} \setminus Z$, and the extended measure does not charge the pluripolar set $Z$. $\square$

Proof. By lemma 5.29.15, there are bounded lower semicontinuous $f_ j$ with common bounds and $f_ j\downarrow f$ quasi-everywhere. The envelopes $p_ j=P_\theta (f_ j)$ decrease to $p=P_\theta (f)$. Indeed, the common lower barrier shows that the limit is not identically $-\infty $; the limit lies below $f$ quasi-everywhere and is thus a competitor for $P_\theta (f)$, while monotonicity gives the other inequality. All these functions have common local bounds off $Z$.

For fixed $k$, put $H_ k=\min \{ 1,(f-p_ k)_+\} $ on $X\setminus Z$, a bounded quasi-continuous function there. If $j\geq k$, then on $\{ H_ k>0\} $ we have $p_ j\leq p_ k<f\leq f_ j$ outside a pluripolar set, so proposition 15.8.4 gives $H_ k\mathrm{MA}_\theta (p_ j)=0$ off $Z$.

Let $g$ be continuous with compact support in $X\setminus Z$. The continuity of bounded products against quasi-continuous weights from section 5.27 applies to the fixed quasi-continuous coefficient $gH_ k$. Consequently $H_ k\mathrm{MA}_\theta (p)=0$ off $Z$. As $k\to \infty $, the sets $\{ H_ k>0\} $ exhaust $\{ p<f\} \setminus Z$, and $Z$ is not charged because it is pluripolar. $\square$

Proof. As in the proof of theorem 16.28.5, the local envelope theorem and a countable Choquet family give $p\leq f$ quasi-everywhere. Set $f_ j=\max (f,-j)$ and $p_ j=P_\theta (f_ j)$. Then $p$ is a competitor for every $f_ j$. Since $f_ j\leq f_1$ and $f_1$ is finite almost everywhere, the finite-sublevel argument in section 5.40 gives a common upper bound for all competitors. Thus the envelopes $p_ j$ are $\theta $-plurisubharmonic and uniformly bounded above. Choose $M$ strictly larger than this bound.

The obstacle $g_ j=\min (f_ j,M)$ is bounded and quasi-continuous. Since $p_ j<M$ and $p_ j\leq f_ j$ quasi-everywhere, $p_ j$ is a competitor for $P_\theta (g_ j)$; the reverse inequality follows from $g_ j\leq f_ j$. Thus $p_ j=P_\theta (g_ j)$, and proposition 15.8.5 gives contact support for $p_ j$ with respect to $g_ j$. Because $p_ j<M$, no contact occurs at the artificial upper cutoff, so

Each $p_ j$ is locally bounded off $Z$, with a lower barrier of the form $\rho -C_ j$. The sequence decreases to $p$: its limit is not identically $-\infty $ because it dominates $p$, and it lies below $f$ quasi-everywhere, so it is itself a competitor for $P_\theta (f)$.

Fix a Hermitian form $\Omega \geq \theta $. For integers $C,k\geq 1$, consider the bounded $\Omega $-plurisubharmonic functions

and set

extended by zero on $Z$. This function is bounded and quasi-continuous. Indeed, off $Z$ the function $p_ k$ is a finite quasi-plurisubharmonic function, truncation by bounded continuous functions preserves quasi-continuity, and the pluripolar set $Z$ can be removed by an open set of arbitrarily small capacity. At a pole of $p$ the second factor vanishes.

For $j\geq k$, on $\{ H_{k,C}>0\} $ we have, outside pluripolar sets,

By plurifine locality, the signed measure $\nu _{j,C}=(\theta +\mathrm{dd}^{\mathrm{c}}v_{j,C})^ n$ therefore agrees there with $\mathrm{MA}_\theta (p_ j)$, and (15.8.7) gives

For fixed $C$, the potentials $v_{j,C}$ have common global bounds and decrease to $v_ C$, so lemma 5.29.17 yields

On the set where $H_{k,C}$ is positive we have $p>-C$, so the restricted product in theorem 13.14.3 identifies $H_{k,C}(\theta +\mathrm{dd}^{\mathrm{c}}v_ C)^ n$ with the positive measure $H_{k,C}\mathrm{MA}_\theta (p)$; on the complement the coefficient is zero. Hence $\mathrm{MA}_\theta (p)$ vanishes on $\{ H_{k,C}>0\} $.

Finally, if $p(x)$ is finite and $p(x)<f(x)$, then $p_ k(x)<f(x)$ for large $k$ and $p(x)>-C$ for large $C$. The countable union of the sets $\{ H_{k,C}>0\} $ therefore covers $\{ p<f\} $ outside $Z\cup \{ p=-\infty \} $. Both excluded sets are pluripolar, and the theorem follows. $\square$