Theorem 15.20.472B88B8B1639

of theorem 15.20.4. We first show that a normalized bounded $K$-invariant solution $u$ coincides with $u_*$, by computing its slope $h$ on the band $|r|\leq 1$ and outside it. Apply lemma 15.20.5. Because $\rho $ and $u$ are invariant under block exchange, $G$ is even, so $h$ is odd and $h(0)=0$.

The slope on the band. On $0<r<1$ we have $x=p>0$ and $y=p+b\geq p>0$. Starting from zero, as long as $-p<h<y$, the equation is $h'=-z_0$. Thus $h(r)=-\int _0^ r z_0(s)\, \mathrm{d}s$. This decreasing function cannot meet the positive upper boundary $y$. Its first meeting with the lower boundary $-p$ occurs at the unique $t_*\in (0,1)$ satisfying

whose existence and uniqueness were proved in proposition 14.26.14.

After that meeting, $h$ must stay on $-p$ until $r=1$. To prove it, suppose a connected component $(a_0,b_0)$ of $\{ h+p>0\} $ lies to its right. Its left endpoint has $a_0\geq t_*$ and $h(a_0)=-p(a_0)<0$. Near that endpoint, $h<0<y$; hence on the component the interior equation applies for as long as $h<0$. It forces $h'=-z_0<0$, so $h$ cannot first reach zero. The equation therefore holds throughout the component. But then

contradicting positivity of $h+p$ immediately to the right of its zero at $a_0$. Thus $h=-p$ on $[t_*,1]$. Oddness determines the entire negative half of the band. The slope on $[-1,1]$ is exactly that of the symmetric explicit solution.

The slope outside the band. Here boundedness of $u$ at the projective ends matters. On $r>1$, $k=xyz>0$. Neither horizontal coefficient can vanish at a finite point of this interval. For the lower one define

Equation (15.20.6) and the smooth lower endpoint show that $\Psi $ is $C^1$. If $h(r_0)=-x(r_0)$, its derivative at that point is $k(r_0)>0$: the $Az$ term vanishes, and the two partial-$r$ terms agree at the common endpoint. This contradicts that the nonnegative $C^1$ function $\Psi $ has an interior minimum zero there. The upper-endpoint integral $\int _{h(r)}^{y(r)}A(r,s)\, \mathrm{d}s$ gives the same contradiction, with derivative $-k(r_0)$, when $h=y$. Thus $-x<h<y$ everywhere on $r>1$, and the smooth ordinary differential equation of the lemma applies.

For $r\geq 2$ the background is Fubini–Study:

Set $a=\sigma +h\in (0,1)$. Its equation is

Hence $a$ is increasing and has a limit in $(0,1]$ at $+\infty $. If that limit were less than one, $h=a-\sigma $ would approach a negative constant and its primitive $G$ would be unbounded below, contrary to boundedness of $u$ and $\rho $. The limit is therefore one. Integrating the equation gives

Taking the limit at infinity gives $C=0$. The scalar function $s^2/2-s^3/3$ is strictly increasing on $(0,1)$, so $a=\sigma $ and $h=0$ for $r\geq 2$. Ordinary uniqueness propagates this zero solution throughout $r>1$, since the coefficients are smooth and the two horizontal coefficients stay positive there. Oddness gives $h=0$ on $r<-1$ as well.

We have proved that $G'$ agrees everywhere with the symmetric explicit slope. Thus $u$ differs from $u_*$ by a constant on $U$. The two quasi-plurisubharmonic functions then agree up to that constant on $X$ by their canonical representatives. Their supremum normalization removes the constant. This proves uniqueness within the bounded $K$-invariant class.

Convergence. Finally proposition 15.20.1 gives uniform precompactness of the normalized positive-exponent family. Every cluster limit solves the zero-exponent equation and is $K$-invariant, because the data are $K$-invariant. The scalar is one for these data. The uniqueness just proved identifies every cluster limit with $u_*$. If convergence of the whole family failed, a sequence staying a fixed positive uniform distance from $u_*$ would have a convergent subsequence, contradicting that identification. This proves uniform convergence and the asserted $C^{1,1}$ regularity of the limit. $\square$