16.28.1 Contact sets840B0858DE95

Proof. The solutions exist by theorem 14.7.2. Choose $A\geq 0$ with $\mathrm{dd}^{\mathrm{c}}f\leq A\omega $. At a maximum point of $u_\beta -f$, the second derivative test gives

Evaluating the equation at this point, we obtain

Thus $u_\beta -C_ f/\beta $ is a competitor for $P_\omega (f)$, and $u_\beta \leq p+C_ f/\beta $.

For the lower bound, note that $p$ is continuous by proposition 5.40.3, and choose smooth strictly $\omega $-plurisubharmonic functions $v_ j\downarrow p$ by theorem 5.32.1; by Dini's theorem the convergence is uniform. For any $\delta >0$, choose $j$ so that

Then $v_ j-\delta \leq f$, since $p\leq f$. Since $\omega _{v_ j}>0$ and $X$ is compact, there is $c_ j>0$ such that $\omega _{v_ j}^ n\geq c_ j\omega ^ n$. Put $C_ j=\max (0,-\log c_ j)$. The function

satisfies

Exponential comparison, corollary 14.19.12 with $\lambda =\beta $, gives $w_{j,\beta }\leq u_\beta $. Consequently

Letting first $\beta \to \infty $ with $j$ fixed, so that no lower bound for $c_ j$ uniform in $j$ is needed, and then $\delta \downarrow 0$, we obtain uniform convergence. Bounded Bedford–Taylor continuity then gives weak convergence of the measures.

On a compact subset of the open set $\{ p<f\} $, uniform convergence gives $u_\beta -f\leq -\epsilon $ for some $\epsilon >0$ and all large $\beta $. There the measures $\omega _{u_\beta }^ n$ have densities at most $\mathrm{e}^{-\beta \epsilon }$ with respect to $\omega ^ n$. Testing against continuous functions with compact support in $\{ p<f\} $ and passing to the limit gives $\omega _ p^ n=0$ on this open set. $\square$

Proof. Suppose first that $f$ is continuous. Choose smooth functions $f_ j\to f$ uniformly. The Lipschitz bound for envelopes in proposition 5.40.3 gives $p_ j=P_\omega (f_ j)\to p=P_\omega (f)$ uniformly. Every compact subset of the open set $\{ p<f\} $ has a neighborhood on which $p_ j<f_ j$ for all large $j$. The preceding proposition and convergence of bounded products then give $\omega _ p^ n=0$ on $\{ p<f\} $.

For lower semicontinuous $f$, choose continuous functions $f_ j\uparrow f$ with common upper and lower bounds. For example, for a fixed distance $d$ inducing the topology of $X$, one may take

These functions are Lipschitz and increase to $f$: if convergence failed at $x$, approximate minimizers $y$ would approach $x$, contradicting lower semicontinuity of $f$ at $x$.

The functions $p_ j=P_\omega (f_ j)$ increase and stay bounded, so the regularized limit $p_\infty =(\lim _ jp_ j)^*$ is a bounded $\omega $-plurisubharmonic function. It agrees with $\lim _ jp_ j$ outside a pluripolar set, and since each $p_ j$ lies below $f_ j\leq f$ outside a pluripolar set, we obtain $p_\infty \leq f$ quasi-everywhere. Thus $p_\infty \leq P_\omega (f)$.

For fixed $k$, the set $\{ p_\infty <f_ k\} $ is open, since $p_\infty $ is upper semicontinuous and $f_ k$ is continuous. If $j\geq k$, then $p_ j\leq p_\infty <f_ k\leq f_ j$ there, so the continuous case gives $\omega _{p_ j}^ n=0$ on that open set. Bedford–Taylor continuity along increasing sequences applies after adding a fixed smooth local function whose Hessian dominates $\omega $ and expanding with respect to the resulting smooth form. Together with the lower semicontinuity of the mass of open sets under weak convergence, it yields

Taking the union over $k$ gives zero measure on $\{ p_\infty <f\} $, hence also on $\{ p_\infty <P_\omega (f)\} $ outside a pluripolar set, since $P_\omega (f)\leq f$ quasi-everywhere. Bounded measures of this kind do not charge pluripolar sets, so the domination principle, theorem 14.19.10, gives $p_\infty \geq P_\omega (f)$. The two functions are therefore equal, and the assertion follows. $\square$

Proof. First suppose that $\theta =\omega $ is Hermitian. Choose bounded lower semicontinuous functions $f_ j\downarrow f$ quasi-everywhere, with common bounds, as in lemma 5.29.15. The bounded functions $p_ j=P_\omega (f_ j)$ decrease to $p$. Indeed, their limit is a competitor for $p$, since it lies below $f$ quasi-everywhere, and the reverse inequality follows from monotonicity.

For fixed $k$, put $H_ k=\min \{ 1,(f-p_ k)_+\} $, a bounded quasi-continuous function. Where $H_ k>0$ and $j\geq k$, we have $p_ j\leq p_ k<f\leq f_ j$ outside a pluripolar set. The lower semicontinuous case therefore gives

The convergence of bounded products against quasi-continuous weights in section 5.29 gives $\int _ XH_ k\, \omega _ p^ n=0$. Since the sets $\{ p_ k<f\} $ increase to $\{ p<f\} $, this proves the assertion for Hermitian $\theta $.

For general semipositive $\theta $, set $\theta _ j=\theta +j^{-1}\omega $. The functions $p_ j=P_{\theta _ j}(f)$ have the common bounds $\inf _ Xf\leq p_ j\leq \sup _ Xf$ and decrease. Their limit is $\theta $-plurisubharmonic by distributional convergence and lies below $f$ quasi-everywhere, so it is at most $p=P_\theta (f)$; since $p\leq p_ j$ for all $j$, it equals $p$. For each fixed $k$, the same coefficient $H_ k=\min \{ 1,(f-p_ k)_+\} $ gives

All $p_ j$ are bounded potentials for the fixed Hermitian form $\theta +\omega $. Expanding the products with background $\theta _ j$ with respect to this fixed background, we may use convergence of bounded products and uniform capacity bounds to pass the last identity to the limit, with the quasi-continuous weight $H_ k$; this gives $\int _ XH_ k\, \mathrm{MA}_\theta (p)=0$. Exhausting $\{ p<f\} $ as before completes the proof. $\square$

Proof. The local envelope theorem, applied to a countable family realizing the supremum (Choquet's lemma), gives $p\leq f$ outside a pluripolar set. Choose $M>\sup _ Xp$ and replace $f$ by $\min (f,M)$. This does not change the envelope, since $p$ itself is a competitor for the smaller obstacle. Since $p<M$, the contact set is also unchanged, so it suffices to prove the theorem for the smaller obstacle; that is, we may assume that $f$ is bounded above.

Set $f_ j=\max (f,-j)$ and $p_ j=P_\theta (f_ j)$. The constants $-j$ are competitors, so $p_ j$ is bounded. Theorem 16.28.4 gives $p_ j\leq f_ j$ quasi-everywhere and

The sequence $p_ j$ decreases and stays above $p$. Its limit is therefore $\theta $-plurisubharmonic, it lies below $f$ quasi-everywhere, and so it is a competitor for $p$. Hence $p_ j\downarrow p$.

Fix an integer $k\geq 1$ and $C>0$, and define the bounded nonnegative quasi-continuous function

Its values on pluripolar sets are immaterial. Where this coefficient is positive, $p_ k<f$ and $p>-C$. For $j\geq k$, we then have $p_ j\leq p_ k<f\leq f_ j$ and $p_ j\geq p>-C$ there. Plurifine locality and theorem 16.28.4 therefore imply

The truncated potentials are uniformly bounded and decrease to $\max (p,-C)$, so their measures converge weakly by bounded Bedford–Taylor continuity. The uniform capacity bounds and the weighted convergence in section 5.29 also allow testing against $H_{k,C}$. Hence the last identity holds with $p$ in place of $p_ j$. By plurifine locality, this says that the extended measure of $p$ vanishes on $\{ p_ k<f\} \cap \{ p>-C\} $. Off the pole set of $p$, these sets increase to $\{ p<f\} $ as first $k$ and then $C$ tend to infinity. Continuity of positive measures along increasing sequences proves the assertion, even if the total extended mass is infinite. $\square$

Proof. The envelope lies below both obstacles quasi-everywhere, hence everywhere, since the obstacles are upper semicontinuous. By the contact theorem (theorem 16.28.5 or theorem 15.8.6), its measure is carried by $\{ p=\min (u,v)\} $. Partition this set into $\{ p=u<v\} $ and $\{ p=v\} $, and apply proposition 13.14.6 on each part. All measures vanish on the pluripolar pole sets, which can therefore be discarded throughout. $\square$

Proof. Since $\varphi =P_\theta [\varphi ]\leq 0$, the function $\varphi $ is a competitor for $P_\theta (\varphi +1,0)$. This rooftop is also a competitor for $P_\theta [\varphi ]=\varphi $, being at most $\varphi +1$ and at most zero, so it equals $\varphi $. Now apply proposition 16.28.6 with $u=\varphi +1$ and $v=0$. The contact set $\{ \varphi =\varphi +1<0\} $ is contained in the pole set of $\varphi $, which the extended measure does not charge. Only contact with zero contributes, and this gives the asserted bound. $\square$