The forms $A_0,B_0$ already have the required common measure. We obtain a Hermitian background by adding an exact correction $\mathrm{dd}^{\mathrm{c}}\rho $ to their average $\omega _0$ and subtracting $\rho $ from both potentials. The correction must restore positivity at $M_\pm $ while preserving it in the transition regions.
Take $\kappa \in C_ c^\infty ((-2,2))$, equal to one on $[-3/2,3/2]$. For a parameter $\varepsilon >0$ to be chosen, set
\[ \rho (r)=\varepsilon (r^2-1)\kappa (r),\qquad \omega =\omega _0+\mathrm{dd}^{\mathrm{c}}\rho . \]
By (14.26.5), its coefficients on $U$ are $x+\rho '$, $y-\rho '$ and $z+\rho ''$. On $[-3/2,3/2]$ one has $\rho '=2\varepsilon r$ and $\rho ''=2\varepsilon $, so the last coefficient is positive there. For $r\geq 0$, $x+2\varepsilon r$ is positive, including at $r=0$ since $x(0)=p(0)>0$. On $[-3/2,0]$, the function $x$ has a positive minimum $m$, so $x+2\varepsilon r\geq m-3\varepsilon >0$ for small $\varepsilon $. Reflection gives the corresponding estimate for $y-2\varepsilon r$. On the compact region $3/2\leq |r|\leq 2$, all three original coefficients have a positive minimum and the perturbations are $O(\varepsilon )$. For $|r|\geq 2$ and near $X\setminus U$, the form is exactly $\omega _ X$. Hence $\omega $ is strictly positive on $X$ for sufficiently small $\varepsilon >0$.
The form $\omega $ is not closed. Indeed, since $\mathrm{d}R=-\mathrm{d}r\wedge (P-Q)$,
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14.26.11
\begin{equation} \mathrm{d}\omega =\mathrm{d}\omega _0 =\mathrm{d}r\wedge \bigl ((x'-z)P+(y'+z)Q\bigr ). \end{equation}
At $r=0$, $x'=y'=0$ and $z=\chi (0)>0$. Hence $\mathrm{d}\omega =z\, \mathrm{d}r\wedge (Q-P)\ne 0$ there. The nonzero $\mathrm{d}\tau $ component used in (14.26.3) also shows that $\mathrm{d}r\wedge P$ and $\mathrm{d}r\wedge Q$ are linearly independent.
of theorem 14.26.1.
Set $\widetilde u=-w/2-\rho $ and $\widetilde v=w/2-\rho $. Since $\mathrm{dd}^{\mathrm{c}}w=2H$, the construction gives
\[ \omega +\mathrm{dd}^{\mathrm{c}}\widetilde u=A_0\geq 0,\qquad \omega +\mathrm{dd}^{\mathrm{c}}\widetilde v=B_0\geq 0. \]
Subtracting their respective suprema gives $u,v$. Their Hessians are unchanged, and $v-u=w+\text{constant}$ is nonconstant. The function $f=A_0^3/\omega _ X^3=B_0^3/\omega _ X^3$ is smooth, nonnegative and nonzero, and it vanishes on the nonempty open band $|r|<1$. Both potentials are smooth, hence bounded, so they have minimal singularities: for a Hermitian background, the envelope $V_\omega $ of section 5.40 is zero.
$\square$
Every Hermitian background on $\mathbb P^3$ has the bounded mass property of definition 13.14.14: since the Fubini–Study form is closed, Stokes' theorem for bounded potentials shows that all its bounded potentials have the same mass, and proposition 13.14.15 shows that the property does not depend on the metric. This applies to all the backgrounds on $\mathbb P^3$ constructed below.
The linearized equation also explains why the maximum principle argument fails. The form
\[ T=A_0^2+A_0\wedge B_0+B_0^2 \]
satisfies $\mathrm{dd}^{\mathrm{c}}w\wedge T=B_0^3-A_0^3=0$, but $T=0$ on $M_\pm $, where $A_0=B_0=\omega _0$ has rank one. Thus the linear equation for $w$ is not elliptic everywhere, even though both potentials and their common density are smooth.
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