16.20 Contact with an arbitrary singular potential99AF407A4E55

Proof. If $u$ is identically $-\infty $, the constants $\phi _ j=-j$ will do, so we assume that $u$ takes a finite value somewhere. Fix a distance $d$ defining the topology of the compact manifold $X$, and put $M=\max _ Xu<\infty $. For each positive integer $j$ define

the maximum is attained because the function in braces is upper semicontinuous in $y$. The $v_ j$ are finite and $j$-Lipschitz: a point where $u$ is finite bounds the maximum from below, and the triangle inequality bounds their differences. They decrease in $j$ and dominate $u$ (take $y=x$). If $u(x)$ is finite, a maximizing point $y_ j$ satisfies $jd(x,y_ j)\leq M-u(x)$, so $y_ j\to x$, and since $v_ j(x)\leq u(y_ j)$, upper semicontinuity gives $v_ j(x)\downarrow u(x)$. If $u(x)=-\infty $, then for every $N$ there is a small ball about $x$ on which $u<-N$, while outside this ball $u(y)-jd(x,y)\leq M-jd(x,y)<-N$ for large $j$. Thus $v_ j(x)\downarrow u(x)$ in this case as well.

The $v_ j$ are only Lipschitz, so we smooth them while keeping monotonicity. Set $\phi _0=M+1$ and $\eta _ j=4^{-j}$, and suppose inductively that $\phi _{j-1}\geq u+\eta _{j-1}$. The continuous function

is at least $u+2\eta _ j$, since $\eta _{j-1}=4\eta _ j$. A continuous function on $X$ can be approximated uniformly by smooth functions: convolution in a chart converges uniformly on compact subsets of the chart (lemma 2.5.1(1)), and a finite smooth partition of unity subordinate to charts (proposition 5.2.6) sums these errors with total weight one. Choose a smooth real $\phi _ j$ with $\| \phi _ j-q_ j\| _\infty <\eta _ j$. Then

These bounds continue the induction, show that the sequence $\phi _ j$ decreases, and, since $v_ j+3\eta _ j\downarrow u$, show that $\phi _ j\downarrow u$, also at the points where $u=-\infty $. $\square$

Proof. We first prove the scalar case, in which all $\theta _ j$, $u_ j$ and $f_ j$ coincide, in three steps of increasing generality (Steps 1–3). The mixed case then follows by polarization (Step 4), and Step 5 proves the assertion about the density.

Step 1: a Kähler background and a plurisubharmonic obstacle. Let $\kappa $ be a Kähler form, let $u$ be $\kappa $-psh, and let $f\geq u$ be a continuous $\kappa $-psh function with bounded metric Laplacian. Write $F=\{ u=f\} $ and $\mu _ a=\langle (\kappa +\mathrm{dd}^{\mathrm{c}}a)^ n\rangle $ for $\kappa $-psh $a$. Since $f$ is bounded, $\mu _ f$ is the Bedford–Taylor product $(\kappa +\mathrm{dd}^{\mathrm{c}}f)^ n$ (proposition 13.4.3, applied with smooth local potentials of $\kappa $). By proposition 16.14.6, applied with $u_ j=f_ j=f$, this measure is given by the weak second derivatives of $f$, so it is the measure $(\kappa +\mathrm{dd}^{\mathrm{c}}f)^ n$ of corollary 16.14.11; we must show that $\mathbf1_ F\mu _ u=\mathbf1_ F\mu _ f$. Take smooth functions $\phi _ j\downarrow u$ as in lemma 16.20.1, and set

Each $\psi _ j$ has bounded metric Laplacian by theorem 16.15.24. Since $u\leq \phi _ j$ and $u\leq f$, the function $u$ is a competitor for $\min \{ \phi _ j,f\} $, and since this obstacle is continuous, proposition 16.11.4(1),(2) give $u\leq \psi _ j\leq f$. The obstacles $\min \{ \phi _ j,f\} $ decrease to $\min \{ u,f\} =u$, so $\psi _ j$ decreases to $P_\kappa (u)$ by proposition 16.11.4(5), and $P_\kappa (u)=u$, because $u$ is itself a competitor and every competitor lies below the upper semicontinuous obstacle $u$ by proposition 16.11.4(2). Moreover $\psi _ j=f$ on $F$, because $u\leq \psi _ j\leq f$. The regular contact identity, proposition 16.14.6, applied on the contact set $\{ \psi _ j=f\} \supset F$, therefore gives

To pass to the limit, choose $C>0$ with $\min _ Xf>-C$, fix $\varepsilon >0$, and put

The bounded $\kappa $-psh functions $\psi _ j^ C$ decrease to $u^ C$, so they converge locally in capacity by proposition 7.5.6. The weights $b_ j$ and $b$ are bounded by one, and they are quasi-continuous, since $\psi _ j^ C$ and $u^ C$ are, by theorem 7.5.2 applied to their sums with smooth local potentials of $\kappa $; moreover $b_ j\to b$ in capacity, because the map $t\mapsto t/(t+\varepsilon )$ on $[0,\infty )$ has Lipschitz constant at most $1/\varepsilon $. The weighted form of theorem 7.5.8, applied with smooth local potentials of $\kappa $ and a finite partition of unity, gives

The measures $\mu _{\psi _ j^ C}$ and $\mu _{\psi _ j}$ agree on the plurifine open set $\{ \psi _ j>-C\} $, by plurifine locality (proposition 13.2.6), and $b_ j$ vanishes on $\{ \psi _ j\leq -C\} $; hence $b_ j\mu _{\psi _ j^ C}=b_ j\mu _{\psi _ j}$. In the same way $b\mu _{u^ C}=b\mu _ u$. On $F$ all the weights take the same continuous value $b_ f=(f+C)/(f+C+\varepsilon )$, which is strictly positive on $X$. Since $\nu $ is carried by $F$, we get $b_ j\mu _{\psi _ j}\geq b_ j\nu =b_ f\nu $ for every $j$, and in the weak limit $b\mu _ u\geq b_ f\nu $. Restricting to $F$, where $b=b_ f$, and dividing by $b_ f$, whose reciprocal is bounded on $X$, we obtain $\mathbf1_ F\mu _ u\geq \nu $. The reverse inequality is the ordered contact inequality of lemma 16.13.1. This proves the scalar identity for a Kähler background and a plurisubharmonic obstacle.

Step 2: obstacles that are psh for a larger background. Let $\theta $ be any closed smooth real $(1,1)$-form, let $u\in \mathrm{PSH}(X,\theta )$, and suppose that $u\leq f$ and that $f$ is $(\theta +A\omega )$-psh for some $A>0$. Increasing $A$, we may assume that $\theta +A\omega $ is Kähler. For every $t>A$, Step 1 applies to the Kähler background $\theta +t\omega $, with the same measuring metric $\omega $. For a continuous real function $g$ on $X$ consider

This is a polynomial in $t$. Indeed, the first integrand expands by the multilinearity of the non-pluripolar product and its compatibility with smooth bounded factors (proposition 13.2.6, proposition 13.4.3), and the second by expanding the wedge products of the weak forms. Every coefficient is a finite measure, by theorem 13.5.4 and corollary 16.14.11. Step 1 gives $Q_ g(t)=0$ for all $t>A$, so every coefficient vanishes, and in particular $Q_ g(0)=0$. Now a finite signed Borel measure $\sigma $ on $X$ with $\int g\, \mathrm d\sigma =0$ for every continuous $g$ vanishes. Indeed, write $\sigma =\sigma ^+-\sigma ^-$ with finite positive Borel measures $\sigma ^\pm $; then $\int g\, \mathrm d\sigma ^+=\int g\, \mathrm d\sigma ^-$ for every continuous $g\geq 0$, so $\sigma ^+\leq \sigma ^-$ and $\sigma ^-\leq \sigma ^+$ on Borel sets by lemma 16.12.1(2), applied with $W=X$. Applied to the difference of the two measures in $Q_ g(0)$, this proves the scalar identity whenever $f$ is $(\theta +A\omega )$-psh for some $A$.

Step 3: general obstacles. Let $f$ be any continuous real function with bounded metric Laplacian, and $u\leq f$. Choose $A>0$ such that $\kappa =\theta +A\omega $ is Kähler, and insert the envelope $v=P_\kappa (f)$:

Here $u\leq v$ because $u$ is $\kappa $-psh and lies below $f$. By theorem 16.15.24, $v$ is a continuous $\kappa $-psh function with bounded metric Laplacian. Step 2, with $v$ as the obstacle, gives

Both $v$ and $f$ belong to $W^{2,p}$ for every finite $p$, by proposition 16.14.9, and the locality of weak Hessians (proposition 7.13.3) shows that their weak second derivatives agree almost everywhere on $\{ v=f\} $. Since $\{ u=f\} =\{ u=v\} \cap \{ v=f\} $, restricting the last identity to $\{ u=f\} $ proves the scalar identity for $f$.

Step 4: the mixed identity. For $\lambda _1,\ldots ,\lambda _ n>0$ put $\theta _\lambda =\sum _ j\lambda _ j\theta _ j$, $u_\lambda =\sum _ j\lambda _ ju_ j$ and $f_\lambda =\sum _ j\lambda _ jf_ j$. Then $u_\lambda $ is $\theta _\lambda $-psh, $f_\lambda $ is continuous with bounded metric Laplacian, $u_\lambda \leq f_\lambda $, and $E\subset \{ u_\lambda =f_\lambda \} $. (Since the $f_ j$ are finite, $E$ contains no pole of any $u_ j$.) We restrict the scalar identity for $u_\lambda $ and $f_\lambda $ to $E$ and expand both sides as measure-valued polynomials in $\lambda $, using, as in Step 2, the multilinearity of the non-pluripolar product (proposition 13.2.6) and of the wedge product of the weak forms. Testing against a continuous function and comparing the coefficients of $\lambda _1\cdots \lambda _ n$, which are $n!$ times the integrals of this function against the two sides of (16.20.3), we see that these integrals agree; by the uniqueness proved in Step 2, we obtain (16.20.3).

Step 5: the density on $E$. For each $j$ choose $A_ j>0$ such that $\theta _ j+A_ j\omega $ is Kähler, and insert the envelope $v_ j=P_{\theta _ j+A_ j\omega }(f_ j)$. Since $u_ j$ is $(\theta _ j+A_ j\omega )$-psh and lies below the continuous function $f_ j$, we have $u_ j\leq v_ j\leq f_ j$ by proposition 16.11.4(1),(2), so $v_ j=f_ j$ on $\{ u_ j=f_ j\} $; as in Step 3, proposition 16.14.9, proposition 7.13.3 show that the weak Hessians of $v_ j$ and $f_ j$ agree almost everywhere there. The positive form $\theta _ j+A_ j\omega +\mathrm{dd}^{\mathrm{c}}v_ j$ has bounded coefficients by theorem 16.15.24, so $\theta _ j+\mathrm{dd}^{\mathrm{c}}f_ j$ has bounded coefficients almost everywhere on $\{ u_ j=f_ j\} $. Hence the density of the mixed product of the weak forms is bounded on $E$. It is nonnegative there, because by (16.20.3) its restriction to $E$ equals the restriction of a positive non-pluripolar product. $\square$

Proof. A constant shift of $V_\theta $ lies below the bounded function $f$, so the competitor family of $P_\theta (f)$ is nonempty, and it is bounded above by $\max _ Xf$. Since $f$ is continuous, $u=P_\theta (f)$ satisfies $u\leq f$ everywhere, by proposition 16.11.4. Theorem 16.20.2, with all $\theta _ j=\theta $, $u_ j=u$ and $f_ j=f$, identifies the restriction of $\langle (\theta +\mathrm{dd}^{\mathrm{c}}u)^ n\rangle $ to $\{ u=f\} $ with that of $(\theta +\mathrm{dd}^{\mathrm{c}}f)^ n$. By theorem 16.13.4, applied to the continuous and hence quasi-continuous obstacle $f$, the measure $\langle (\theta +\mathrm{dd}^{\mathrm{c}}u)^ n\rangle $ is carried by $\{ u=f\} $. This proves the first identity. For $f=0$ the envelope is $V_\theta $ by definition, which gives the second. $\square$