Example 12.10.29E97964925666

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. $\square$