12.10.28 Two attained instances244D24A4396B

Proof. The form is invariant under $z\mapsto 2z$, so it descends to $X$, and it is positive. For $s=|z|^2$ and $F(s)=s^{-1}$, the identity $sF''+2F'=0$ gives $\mathrm{dd}^{\mathrm{c}}\omega =\mathrm{dd}^{\mathrm{c}}(s^{-1})\wedge \mathrm{dd}^{\mathrm{c}}s=0$, while $\omega $ is not closed. Choose a nonconstant smooth real $q$ and small nonzero $\varepsilon $ such that $A=\omega +\mathrm{dd}^{\mathrm{c}}(\varepsilon q)>0$. Put $V=\int _ X\omega ^2$, $\mathrm{d}V=\omega ^2/V$, $f=A^2/\mathrm{d}V$ and $u=\varepsilon q-\sup _ X\varepsilon q$. Invariance of the mass gives $\int _ Xf\, \mathrm{d}V=\int _ XA^2=V$, hence $b_\omega (f)=1$ by theorem 12.10.23.

The form $T=A/V$ satisfies $\mathrm{dd}^{\mathrm{c}}T=0$, $\int _ X\omega \wedge T=1$ and $D_{\sqrt f}(T)=1$. For every smooth strictly positive $Q\in \mathscr G_\omega $, the arithmetic–geometric mean inequality and Stokes' theorem give $D_{\sqrt f}(Q)\leq \int _ XA\wedge Q=1$. Thus $T$ is a smooth strictly positive maximizer, and since $u$ is a threshold solution with supremum zero, corollary 12.10.12 gives $\Phi _ f=u$. This is a bounded nonconstant example. $\square$

Proof. Let

The quadratic form is nonsingular, and $F$ is finite of degree two, with smooth ramification divisor $Z=\{ w=0\} $. Local branch coordinates show that $\mathrm dF$ has rank $n-1$ on $Z$ and $n$ elsewhere. Let $\omega $ be the restriction to $X$ of the normalized Fubini–Study form of $\mathbb P^{n+1}$, and let $A=F^*\omega _{\mathrm{FS}}$, where $\omega _{\mathrm{FS}}$ is the normalized Fubini–Study form of $\mathbb P^ n$. Then

is a smooth function on $X$ with $-\log 2\leq u\leq 0$, $\sup _ Xu=0$ and $A=\omega +\mathrm{dd}^{\mathrm{c}}u$. The change of variables for the branched cover gives $\int _ XA^ n=2$, and Stokes' theorem gives $V=\int _ X\omega ^ n=2$.

Set $\mathrm{d}V=\omega ^ n/V$, $f=A^ n/\mathrm{d}V$, $T=A^{n-1}/V$ and $\gamma =A/V^{1/(n-1)}$. Since $\omega $ is closed, theorem 12.10.23 gives $b_\omega (f)=V/\int _ Xf\, \mathrm{d}V=1$. Stokes' theorem gives $\mathrm{dd}^{\mathrm{c}}T=0$ and $\int _ X\omega \wedge T=1$, and computing the determinant density gives $D_{f^{1/n}}(T)=1$. The arithmetic–geometric mean inequality, which remains valid for semipositive $A$, bounds $D_{f^{1/n}}(Q)$ by $\int _ XA\wedge Q=1$ for every smooth strictly positive $Q\in \mathscr G_\omega $. Thus $T$ is a maximizer, strictly positive off the proper analytic set $Z$, and theorem 12.10.17 gives $\Phi _ f=u$. Its root $\gamma $ has rank $n-1\geq 2$ on $Z$ and is positive elsewhere, as required. $\square$

Proof. By proposition 15.28.7, the constant $c$ for which $\mathrm{MA}_\omega (v)=cf\, \mathrm{d}V$ has a continuous solution is $b_\omega (f)$. The continuous normalized solution is therefore a threshold solution, and it equals $u_0$ by the uniqueness in theorem 12.10.17. For $L^ p$ data the greatest exponential solution coincides with the bounded solution, as explained after theorem 15.35.2. By proposition 15.20.1, applied with the trivial group, the normalized family is uniformly precompact, and every cluster limit solves the threshold equation. Uniqueness identifies every cluster limit with $u_0$, which proves uniform convergence of the whole family. $\square$