11.47 Envelopes for singular radial obstaclesAEE10281DB3D

Proof. Put $p=\mathrm e^ x/(1+\mathrm e^ x)$, $T=1-\log p$ and $a=\alpha T^{\alpha -1}$. Differentiating (11.46.3) gives

Since $T\geq 1$ and $0<a\leq \alpha <1$, these formulas give the asserted positivity and bounds. On the left, $A(x)=1-(1-x)^\alpha +o(1)$ and $A-xA'$ has leading term $-(1-\alpha )(-x)^\alpha $. On the right, $A=x+(1-\alpha )\mathrm e^{-x}+O(\mathrm e^{-2x})$ and $A'=1-(1-\alpha )\mathrm e^{-x}+O(\mathrm e^{-2x})$. The tangent at $x$ is $y\mapsto A'(x)y+A(x)-xA'(x)$, so both uniform limits follow. Finally $A'-1<0$ gives the decreasing right-end limit. $\square$

Proof. We first identify the full envelope, then examine contact and the two ends. Every supporting slope of $C_ F$ lies in $[0,1]$, by the left growth $g\to -\infty $ and the right growth $g=x+o(1)$. The sandwich forces its asymptotic slopes to be $0,1$.

The competitor family is invariant under the unitary group $U(k)$ acting on the affine coordinates. Its regularized supremum is therefore radial. Adding smooth local potentials of $\kappa $ gives a locally upper-bounded family of plurisubharmonic functions, so the envelope is $\kappa $-plurisubharmonic. The competitor $u_\alpha -K_ F$ prevents collapse, and upper semicontinuity of the obstacle preserves the everywhere bound after regularization, including at $o$. A radial plurisubharmonic total potential $B(\log r)$ has convex profile $B$: restriction to a complex line gives this from radial subharmonicity. Positivity in a direction orthogonal to $z$ gives $B'\geq 0$. Conversely a convex nondecreasing profile has nonnegative tangential and radial Hessian eigenvalues $B'/r$ and $B''/r$, first for smooth profiles and then by convex smoothing on annuli. The envelope's total profile is consequently a convex minorant of $g$, and is at most $C_ F$.

The same criterion makes $C_ F(\log r)$ plurisubharmonic on the punctured chart. Its upper bound near $o$ and (11.47.4) give a removable extension with value $-\infty $. Since $C_ F'\leq 1$, the function $C_ F-x$ is nonincreasing and, by the sandwich, has a finite right-end limit. The proposed $E_ F$ is thus locally bounded above near infinity. Add a local smooth potential of $\kappa $ and apply lemma 5.24.7 across the hyperplane. This gives a global competitor, hence the formula. The sandwich proves ordinary equivalence without subtracting the infinite values at $o$.

On a compact interval in the free set the gap $g-C_ F$ is positive. Replacing $C_ F$ on a sufficiently short subinterval by its chord preserves convexity at both endpoints and stays below $g$. Unless $C_ F$ is locally affine this would give a larger minorant. It follows that it is affine on each free component. At a contact point $y$, domination by $g$ gives $C'_{F,-}(y)\geq g'(y)\geq C'_{F,+}(y)$, and convexity reverses the outer inequality. Thus $C_ F$ is differentiable at contacts, with derivative $g'$, and it is differentiable on free intervals as well. Its derivative is continuous. For $a<b$ outside one common free interval, move $a$ to its right contact endpoint and $b$ to its left contact endpoint whenever necessary. These points $a',b'\in [a,b]$ have the same respective derivatives. An unbounded free interval lacking the needed endpoint contains both original points, in which case the difference is zero. Otherwise

which proves (11.47.6).

For all sufficiently negative $x$, the tangent to $A$ at $x$ lies below $g$ on $\operatorname {supp}F$, by lemma 11.47.2, and below $A=g$ elsewhere. Hence $C_ F=A$ on a left tail. The residual is zero near $o$.

For the other end, put $L_ F=\inf _{\mathbb R}(g(x)-x)\leq 0$. If this infimum is attained at $y$, the affine minorant $x+L_ F$ touches $g$ there. The bound $C_ F'\leq 1$ forces $C_ F=x+L_ F$ for $x\geq y$. A negative infimum is always attained, since $g-x$ tends to $+\infty $ on the left and to zero on the right. With $R=1/r$, the envelope on this tail is

If $L_ F=0$ is not attained and $F\ne 0$, the continuous function $g(y)-y$ has a positive minimum on $\operatorname {supp}F$. The right tangents of $A$ approach the line $y$ uniformly there, so every sufficiently late tangent lies below $g$ and gives $C_ F=A$ at its contact point. Thus $C_ F=A$ on a right tail; for $F=0$ this holds everywhere. Both alternatives give a smooth residual near infinity, because $R$ and $u_\alpha $ are smooth there.

On the intervening finite annulus,

Composing with the smooth coordinate function $x=\log r$ gives $C^{1,1}$ regularity. The exact end formulas extend it over the whole projective space. $\square$

Proof. On an annulus, the radial flux formula (11.46.4), followed by convex smoothing and bounded Bedford–Taylor continuity, gives the displayed measure. Continuity of $C_ F'$ excludes boundary atoms. The non-pluripolar truncations discard $o$, and smoothness at infinity excludes hyperplane mass. The asymptotic slopes $0,1$ therefore give mass one. An affine profile has zero radial eigenvalue and hence zero top-degree measure. Finally, if $\delta =\| F-G\| _\infty $, subtracting $\delta $ from a competitor for $F$ gives a competitor for $G$, and conversely. The resulting value bound for the envelopes gives (11.47.9) off $o$, and continuity of the residuals extends it across both ends. $\square$

Proof. The left tangents of $A$ lie below $A-M$ on $[a,b]$ once their contact point is less than a common $a_0<a$. They are minorants of every $A+F$, so all residuals vanish on the same neighborhood of $o$. The sandwich and slope bound give uniform real Hessian bounds on any fixed finite annulus, since $L_2\leq \| A''\| _\infty +M$. It remains to control infinity, where the right contact point need not stay bounded.

Put $B=\max (b,0)+1$. For $x>B$, the obstacle is $A$. At contact $C_ F'=A'$. On a free interval with finite right contact endpoint $d$, its constant slope is $A'(d)\geq A'(x)$; an unbounded right free interval has slope one. Thus

The locally Lipschitz functions $C_ F',A'$ agree on the contact set, so their derivatives agree almost everywhere there. Indeed, a differentiable Lipschitz function has derivative zero at almost every density point of its zero set. On the free set $C_ F''=0$. The reference formula consequently gives

In a chart $Z_ j\ne 0$, $j\geq 1$, use $\zeta =Z_0/Z_ j$ and $y=(Z_ i/Z_ j)_{i\geq 1,i\ne j}$. Then

On fixed relatively compact chart pieces, $|\nabla R|\leq c\sqrt R$ and $|\nabla ^2R|\leq c$. Write the envelope as $\mathcal E_ F(R)=C_ F(-\log R)+\log R-\log (1+R)$ for $R>0$. Its derivatives are

The tail bounds give $|(\mathcal E_ F)_ R|\leq 2$ and $|(\mathcal E_ F)_{RR}|\leq c_\alpha /R+1$. The real chain rule therefore bounds its Hessian uniformly, and its gradient tends to zero as $\zeta \to 0$. Also $L_ F=\lim (C_ F-x)\in [-M,0]$, and integrating $d_ F\leq \mathrm e^{-x}$ gives $|C_ F(x)-x-L_ F|\leq \mathrm e^{-x}$. Hence $\mathcal E_ F-L_ F=O(R)$, so the extended envelope is differentiable on the hyperplane with zero gradient. The Hessian bound makes the gradient Lipschitz off the hyperplane. Apply it along straight coordinate segments and use the continuous zero gradient at a crossing; on segments contained in the hyperplane the gradient is zero. This proves a uniform $C^{1,1}$ extension. Subtract the fixed smooth $u_\alpha $ near infinity and join this estimate to the common pole neighborhood and a fixed annulus.

For the stability estimate put $Z=R_ F-R_ G$. By (11.47.9), $\| Z\| _\infty \leq \delta $, and its gradient has a uniform Lipschitz bound. On a smaller fixed chart, a real directional increment of length $h$ gives $|\partial _ e Z|\leq 2\delta /h+Kh$. Taking $h$ of order $\sqrt\delta $, or enlarging the constant when limited by the chart radius, yields $\| \nabla Z\| _\infty \leq c\sqrt\delta $. For points a distance $t$ apart,

Dividing by $t^\gamma $ and splitting at $t=\sqrt\delta $ gives the estimate in a finite atlas. When $\delta =0$, the value bound gives equality. This argument does not assert convergence in $C^{1,1}$. $\square$

Proof. Every competitor restricts to a projective slice as a $\kappa $-plurisubharmonic function or as $-\infty $ identically. To check the exact lower bound when $H$ is nonclosed, choose a smooth quadratic correction $q$ with $\mathrm{dd}^{\mathrm{c}}q\geq H$ on a coordinate polydisk and let $v_\epsilon =(v+q)*\eta _\epsilon -q$ on a smaller polydisk, where $\eta _\epsilon $ is the rescaling of a fixed smooth nonnegative radial probability kernel. Radial plurisubharmonic convolutions decrease to the canonical $v+q$, and

Restrict this smooth inequality to a fixed slice. The corrected restrictions decrease either to $-\infty $ or to a plurisubharmonic function in local $L^1$; in the latter case the background error tends to zero distributionally. Subtracting $q$ gives the exact $\kappa $ bound, and these canonical restrictions patch on the connected slice. The everywhere obstacle constraint now gives $v\leq \operatorname {pr}_1^*E_ F$ on every slice. Upper regularization preserves this bound. Conversely every base competitor lifts, since $h_0>0$, proving equality of the full envelopes.

The bounded truncations $\max (E_ F^ H,-j)$ are pullbacks. By proposition 5.28.2, lemma 5.28.4, dimensional multilinearity leaves only the term with $k$ projective factors and $m$ factor entries. Restrict to strict superlevel sets and let $j\to \infty $ to obtain the stated non-pluripolar measure and its finite mass. The residual regularity and uniform estimates pull back in a fixed smooth product atlas. Finally $E_ F^ H\geq u_\alpha -K_ F$. At $t=1$ take the stretching witness $v_1=E_ F^ H$; for $t>1$ take $v_ t=u_\alpha -tK_ F$, so that

Together with ordinary domination this is definition 16.28.61, including the canonical values at the common pole. $\square$

Proof. Take $Y=(\mathbb C^2\setminus \{ 0\} )/\langle \zeta \mapsto \mathrm e\zeta \rangle $. With $s=\log |\zeta |^2$, the form $h_0=\mathrm{dd}^{\mathrm{c}}s+\mathrm ds\wedge \mathrm d^{\mathrm c}s$ is invariant under the quotient. Its two summands are positive on the tangential and radial complex directions, respectively, so it is Hermitian. Moreover $\mathrm dh_0=-\mathrm ds\wedge \mathrm{dd}^{\mathrm{c}}s\ne 0$. The Hopf surface is diffeomorphic to $S^3\times S^1$ and has odd first Betti number. Thus $X=\mathbb P^2\times Y$ is a non-Kähler fourfold with $\mathrm dH\ne 0$, and proposition 11.47.14 applies. $\square$