11.34.15 Divisorial smoothingA54217F2E13D

Proof. The function $W$ is convex: both one-sided derivatives at zero equal one, and the derivative of $2\chi $ increases from one to two on $[0,\infty )$. Since $\omega _1=\mathrm{dd}^{\mathrm{c}}\chi (x)$ on the affine chart, $W-2\chi $ is $2\omega _1$-plurisubharmonic there. Since $W$ has slope one for $x\leq 0$ and agrees with $2\chi $ for $x\geq 0$, the function $W-2\chi $ extends to a global $2\omega _1$-plurisubharmonic function, with a logarithmic pole along $D$ and equal to zero where $|z|\geq 1$. The asserted bounds follow from $\psi =x-\chi $: the difference $\varphi -\psi $ equals $2\log 2-\chi $ for $x\leq 0$ and $\chi -x$ for $x\geq 0$.

For $C>0$ write $v_ C=P_\theta (\min \{ \psi +C,0\} )$. Rotations of the first factor and the transitive unitary action on the second preserve the whole competitor class, so the envelope is radial and independent of the second factor. The function $V_ C=2\chi +v_ C$ is then convex in $x$ and satisfies $V_ C\leq \chi +x+C$. This upper bound as $x\to -\infty $ forces the slope of $V_ C$ at $-\infty $ to be at least one, and by convexity all its slopes are at least one. For $x<0$,

For $x\geq 0$, the zero obstacle gives $V_ C\leq 2\chi =W$. Consequently $v_ C\leq \varphi $. If $C\geq 2\log 2$, the bounds already proved make $\varphi $ a competitor, so $v_ C=\varphi $. Letting $C\to \infty $ proves $P_\theta [\psi ]=\varphi $. Since $\varphi \sim \psi $ and the ordinary envelope depends only on the singularity type, also $P_\theta [\varphi ]=\varphi $.

Finally

Since $\psi \leq \varphi $, the function $\psi $ is a competitor for $P_{(3/4)\theta }(\varphi )$, which is therefore not identically $-\infty $. $\square$

Proof. On $X\setminus D$, the function $w=v-\psi $ is bounded, because $v\sim \varphi \sim \psi $, and satisfies $\beta +\mathrm{dd}^{\mathrm{c}}w\geq 0$. After adding local potentials of the closed form $\beta $, lemma 5.24.7 extends $w$ across $D$; the extensions agree on overlaps and give a bounded $\beta $-plurisubharmonic function on $X$. Conversely, for every bounded $w\in \mathrm{PSH}(X,\beta )$, the function $\psi +w$ is $\theta $-plurisubharmonic with current $\beta +\mathrm{dd}^{\mathrm{c}}w+[D]$, and it has the ordinary type of $\varphi $. The decomposition is unique, since two such functions $w$ agree off $D$ and hence everywhere.

Put $T=\beta +\mathrm{dd}^{\mathrm{c}}w$. The bounded-product construction gives

and Stokes' theorem against the constant function yields $\int _ XT^2=4$. The product gives no mass to $D$ by proposition 5.28.2; we also give a direct cutoff estimate, which will be used again below. Shift $w$ so that $0\leq w\leq M$, and set $f_ t=\max \{ \psi /t,-1\} $ for $t>0$. Choose smooth convex functions $\sigma _\epsilon $ with derivative in $[0,1]$, equal to $-1$ below $-1-\epsilon $ and to the identity above $-1+\epsilon $, converging uniformly to $\max \{ \cdot ,-1\} $. Then $f_{t,\epsilon }=\sigma _\epsilon (\psi /t)$ is globally smooth, constant near $D$, and

Testing the distributional identity against $f_{t,\epsilon }$, using positivity of $T+\beta $, and letting $\epsilon \to 0$, we obtain

Here the last integral equals four by Stokes' theorem. The right-hand side tends to zero as $t\to \infty $, since $f_ t\to 0$ off $D$ and $\beta ^2(D)=0$; as $-f_ t=1$ on $D$, this gives $T^2(D)=0$. Off $D$ the currents $\theta +\mathrm{dd}^{\mathrm{c}}v$ and $T$ agree, so locality and the fact that the extended measure puts no mass on the pole set prove (11.34.19).

For the Hermitian background $\beta $, theorem 14.19.2 with density $\mathrm e^ F$ gives a bounded solution $w$ and a constant $c>0$, and the correspondence above turns it into a solution $\psi +w$ in the ordinary type of $\varphi $. Every potential of that type has mass four, so integration forces $c=4/\int _ X\mathrm e^ F\mathrm dV$, which is the formula for $B_\varphi $. Jensen's inequality for the probability $\nu _ F=\mathrm e^ F\mathrm dV/\int _ X\mathrm e^ F\mathrm dV$ gives

Thus $\nu _ F$ is a supporting probability. For any supporting probability, testing its inequality at $G=F\pm th$ and differentiating at $t=0$ shows that its integral against every $h\in C(X)$ equals that of $\nu _ F$.

For a smooth real $h$ and sufficiently small $|t|$, we have $\beta +t\mathrm{dd}^{\mathrm{c}}h>0$, so $\psi +th$ stays in the same ordinary type and its residual measure is $(\beta +t\mathrm{dd}^{\mathrm{c}}h)^2$, with derivative $2\mathrm{dd}^{\mathrm{c}}h\wedge \beta $ at $t=0$. Since $\beta $ is closed, the constant weight $\rho =1/4$ annihilates this derivative after integration, and $Q=\rho \beta $ gives the asserted probability. This normalization uses the residual mass: since $\int _ X\theta \wedge Q=3/2$, the cofactor $Q$ is not a normalized Gauduchon current for $\theta $ in the sense of section 11.7. $\square$

Proof. Smoothness and monotone convergence follow directly from the defining formula, also on $D$, where $\psi =-\infty $. Off $D$, set

The chain rule gives $\Omega _{\delta ,g}=A_\delta +R_\delta $, where

Thus $A_\delta \geq \omega _1+\omega _2$ and $R_\delta \geq 0$, and the inequality $\Omega _{\delta ,g}\geq \omega $ extends across $D$ by continuity. Closedness of $\theta $ and Stokes' theorem give the mass eight. Since $\Omega _{\delta ,g}$ is closed, the Gauduchon condition $\mathrm{dd}^{\mathrm{c}}(\rho \Omega _{\delta ,g})=0$ reduces to the elliptic equation $\mathrm{dd}^{\mathrm{c}}\rho \wedge \Omega _{\delta ,g}=0$ on the connected surface, whose smooth solutions are constant by the maximum principle. Normalization against $\Omega _{\delta ,g}^2$ gives $\rho =1/8$ and the stated adjoint cofactors and probabilities.

The rank-one form $R_\delta $ has square zero, hence $\Omega _{\delta ,g}^2=A_\delta ^2+2A_\delta \wedge R_\delta $. The bounded coefficients of $A_\delta $ converge off $D$ to those of $\beta $, which gives $A_\delta ^2\to 4\mathrm dV$. Write $N=\mathrm d\psi \wedge \mathrm d^{\mathrm c}\psi $. In $A_\delta \wedge R_\delta $, the part in which $R_\delta $ contributes only derivatives along the first factor (the normal part) is

Terms mixing a derivative along the first factor with one along the second vanish for degree reasons. The remaining tangential term is $\lambda (1-\lambda )(2-\lambda )\omega _1\wedge \mathrm dg\wedge \mathrm d^{\mathrm c}g$, which tends to zero by dominated convergence.

The normal coefficients can be computed in polar coordinates. Put $s=\delta \mathrm e^{g}$ and $t=|z|^2$. Then

After angular integration the radial measure of $\lambda (1-\lambda )N$ is

It is bounded by $\mathrm dy/(1+y)^2$ and tends to that measure, whose integral is one. With an additional factor $1-\lambda $, the rescaled measure becomes $\mathrm dy/(1+(1+s)y)^3$, which is bounded by $\mathrm dy/(1+y)^3$ and has limiting integral $1/2$. The mass of both measures outside every fixed tube around $D$ tends to zero, uniformly in the point of the second factor because $g$ is bounded. Testing against continuous functions therefore gives

Combining the normal and the remaining terms proves

Dividing by eight gives the second line of (11.34.22). The limit is positive since $4\omega _2+\gamma \geq 3\omega _2$, and its mass on $D$ is $4/8=1/2$ by Stokes' theorem on the second factor. Taking $g=0$ and a small nonconstant smooth $g$ gives different limits, since $\mathrm{dd}^{\mathrm{c}}g$ vanishes identically only for constant $g$.

The difference of the potentials satisfies

off $D$. The right-hand side tends to zero and is dominated by an integrable logarithmic pole, which proves $L^1$ convergence. Continuity of distributional differentiation then gives the limit of the cofactors. A locally integrable function times the smooth form $\beta $ has absolutely continuous coefficients, so it cannot produce the singular coefficients of $[D]$.

For convergence in capacity, use the capacity $\operatorname {Cap}_\omega $ of section 5.29; after subtracting one, its test potentials range over $v\in \mathrm{PSH}(X,\omega )$ with $-1\leq v\leq 0$. For such $v$, apply the cutoff argument of (11.34.20) with $\beta $ replaced by $\omega $ and $w=v+1$. Stokes' theorem gives $\int _ X\omega _1\wedge (2\omega +\mathrm{dd}^{\mathrm{c}}v)=2$, so

The same bound shows that $D$ has capacity zero. For $\eta >0$, the set of points off $D$ where $u_{\delta ,g}-\psi >\eta $ is contained in

This is a sublevel set of $\psi $ whose level tends to $-\infty $ as $\delta \downarrow 0$. The capacity bound for such sublevel sets, together with $\operatorname {Cap}_\omega (D)=0$, proves convergence in capacity.

Finally, $\Omega _{\delta ,g}\geq \omega $ gives $F_{\delta ,g}\geq -\log 2$. In the affine chart $|\mathrm d\psi |_\omega ^2\leq C/r$, while $g$ and its first two derivatives are bounded. The decomposition above yields $\Omega _{\delta ,g}\leq C_ g(1+r^{-1})\omega $, hence

Every finite power of this logarithmic bound is integrable. Smooth convergence off $D$ gives $F_{\delta ,g}\to 0$ there, and dominated convergence proves convergence in $L^ p$. $\square$

Proof. The capacity is the one of section 5.29, and $\beta \leq 2\omega $ gives $\operatorname {Cap}_\beta \leq 4\operatorname {Cap}_\omega $. Set $a_ j=(w_ j+M)/(2M)$ and $S_ j=\beta +\mathrm{dd}^{\mathrm{c}}a_ j$. Then $0\leq a_ j\leq 1$ and

Positivity of bounded mixed products justifies these inequalities. Thus $\int _ E\beta \wedge T_ j\leq 2M\operatorname {Cap}_\beta (E)$ for every Borel set $E$, and the same bound holds for $T$. By closedness and Stokes' theorem, $\int _ X\beta \wedge (T_ j+T)=8$. Now test the identity

against $f$. On $X\setminus E_{j,\eta }$ the potential difference has absolute value at most $\eta $, and on $E_{j,\eta }$ at most $2M$. Consequently

Dividing by four proves (11.34.25). Letting first $j\to \infty $ with $\eta $ fixed and then $\eta \downarrow 0$ shows that $\int _ Xf\, \mathrm d\nu _ j\to \int _ Xf\, \mathrm d\nu $. All $\nu _ j$ and $\nu $ have mass one by proposition 11.34.18, and smooth functions are uniformly dense in $C(X)$, so $\nu _ j\rightharpoonup \nu $.

The constant test potential in the definition of capacity gives $4\mathrm dV(E)\leq \operatorname {Cap}_\beta (E)$. Splitting the potential difference over the same sets yields

Taking the limits in the same order proves $L^1$ convergence, and hence convergence of the currents $Q_ j$ by continuity of distributional differentiation. The limiting cofactor satisfies $\mathrm{dd}^{\mathrm{c}}Q=0$, $\int _ X\beta \wedge Q=1$, and $T\wedge Q=\nu $ as a bounded product. For smooth real $h$, the signed measure $2\mathrm{dd}^{\mathrm{c}}h\wedge T$ is the formal linearization of the residual operator at $w$ in the direction $h$, and its integral against the constant $1/4$ vanishes by Stokes' theorem. It is the derivative of the residual measure along $w+th$ whenever these functions remain $\beta $-plurisubharmonic.

If $T_ j$ is smooth and strictly positive, the maximum principle on the connected surface makes its Gauduchon multiplier constant, and the mass four normalizes it to $1/4$. Finally, if $\psi +w$ solves the equation at $F$, then $\nu =T^2/4=\mathrm e^ F\mathrm dV/\int _ X\mathrm e^ F\mathrm dV$, which is the unique supporting probability by proposition 11.34.18. $\square$