Theorem 11.47.57444FA94A222

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. $\square$