Proposition 14.26.143A73F825D91B

Proof. Consider pairs $-1<I<0<J<1$ satisfying

We first show that such pairs exist, with $I$ ranging over an interval. Define

We have $K(0,0)=-2p(0)<0$ and $K(0,1)>0$ by (14.26.13). By continuity, there is $\epsilon >0$ such that $K(I,0)<0<K(I,1)$ for every $I\in [-\epsilon ,0]$. For $0<J<1$,

The intermediate value theorem therefore gives a unique $J=J(I)\in (0,1)$ satisfying (14.26.15) for every $I\in [-\epsilon ,0)$; we may assume $\epsilon <1$.

For any pair satisfying (14.26.15), define the slope on $[-1,1]$ by

and extend it by zero outside that interval. The balance identity makes it continuous at $J$, and continuity at $I$ is immediate. Each piece has bounded derivative, so $h_{I,J}$ is Lipschitz on $\mathbb R$; it vanishes to infinite order at $\pm 1$. Let $G_{I,J}$ be a primitive of $h_{I,J}$. It is $C^{1,1}$ and constant near each end of the real line. Consequently $G_{I,J}(r)$ extends to a $C^{1,1}$ function on $X$ that is smooth near $X\setminus U$.

On the middle interval $[I,J]$ the slope lies between $-p$ and $p$. Indeed,

The first difference vanishes at $I$, and the second vanishes at $J$. Since both are nonincreasing, $h_{I,J}\leq p$ and $h_{I,J}\geq -p$ on $[I,J]$.

Put $\widetilde u_{I,J}=G_{I,J}(r)-\rho $. By (14.26.5), almost everywhere on $U$,

On $I<r<J$, the first two coefficients are nonnegative because $x=p+a$, $y=p+b$ and $-p\leq h_{I,J}\leq p$, and the third is zero. On $-1<r<I$, we have $h_{I,J}=p$ and $b=0$, since $I<0$; so the second coefficient is zero, and the others are nonnegative because $z_0+p'\geq 0$. On $J<r<1$, we have $h_{I,J}=-p$ and $a=0$; so the first coefficient is zero, and the others are nonnegative because $z_0-p'\geq 0$. Outside the band, $h_{I,J}=h_{I,J}'=0$, so the form equals $\omega _0$.

Formula (14.26.16) has no additional term supported on the hypersurfaces $r=I$ and $r=J$: the function $G_{I,J}$ is $C^{1,1}$, so its first derivative is continuous at both transition points, and its distributional Hessian is its bounded almost-everywhere Hessian. The preceding coefficient inequalities therefore prove $\omega +\mathrm{dd}^{\mathrm{c}}\widetilde u_{I,J}\geq 0$ as a current.

By proposition 5.28.5, its Monge–Ampère measure is the pointwise cube of (14.26.16), computed almost everywhere. By the rank-one identities (14.26.3) and (14.26.4), this cube is zero almost everywhere on the band, since one coefficient vanishes there, and it equals $\omega _0^3$ off the band. These are exactly the values of $f\omega _ X^3$ in the preceding construction. Thus $\widetilde u_{I,J}$ solves the equation, and subtracting its supremum gives a normalized solution $u_{I,J}$.

Different values of $I\in [-\epsilon ,0)$ give different normalized solutions. If $I_1<I_2<0$ and $I_1<r<I_2$, then the slope belonging to $I_2$ is $p(r)$, whereas

because $p'>0$ on $(-1,0)$. The two primitives therefore do not differ by a constant. Since the interval $[-\epsilon ,0)$ is uncountable, this proves the first assertion. None of these slopes equals $p$ or $-p$ on all of $(-1,1)$, so the constructed solutions are also distinct from the two smooth solutions of theorem 14.26.1.

Finally, the function

is negative at zero, positive at one, and strictly increasing on $(0,1)$. Its unique zero $t_*$ gives a symmetric pair $I=-t_*$, $J=t_*$ satisfying (14.26.15). Since $z_0$ and $p$ are even, $h_{I,J}$ is odd: on the middle interval it equals $-\int _0^ r z_0(s)\, \mathrm{d}s$. Choose its primitive even. Since $\iota $ changes $r$ into $-r$, if $\rho $ is even, the potential $G_{I,J}(r)-\rho $ and its normalization are fixed by $\iota $. This proves the last assertion. $\square$