14.29.12 A positive equation background38D1973448C7

As in the first example, an exact correction makes the average form strictly positive without changing the two solution forms. Here the coefficient of $R$ is already positive, and the correction restores the missing coefficient of $P$ or $Q$ at each endpoint. Let $\varepsilon >0$ be a parameter, restricted below.

Let $h(r)=(r^2-1)\eta (r)$, extend it by zero on the tails, and set $\rho =\varepsilon h$. The coefficients of $\omega =\omega _0+\mathrm{dd}^{\mathrm{c}}\rho $ are

On $|r|\leq 3/2$, we have $h'=2r$ and $h''=2$. For $0\leq r\leq 3/2$, $x+2\varepsilon r>0$, since the only zero of $x$ there is at $r=1$, where the correction is positive. On $[-3/2,0]$, $x$ has a positive minimum $m_0$, so positivity holds if $3\varepsilon <m_0$. Reflection proves positivity of the second coefficient, and the third is $z+2\varepsilon >0$. On the compact transition region $3/2\leq |r|\leq 2$, all original coefficients have a common positive lower bound $m_1$. If $M_1$ bounds $|h'|+|h''|$, taking also $\varepsilon <m_1/(M_1+1)$ preserves their positivity there. For $|r|\geq 2$, the background equals $f_*(t)^{1/3}\omega _ X>0$. Hence $\omega $ is strictly positive on $X$.

The correction is exact, so $\mathrm{d}\omega =\mathrm{d}\omega _0$. At $r=0$, where $x=x_ b$ and $y=y_ b$, evenness of $S_*$ gives $x'(0)=-1/2$ and $y'(0)=1/2$. Hence (14.26.11) gives

Indeed, if $\mathrm{d}r\wedge (Q-P)=0$, wedging with $P$ and $\mathrm d^{\mathrm c}r$ would contradict $P\wedge Q\wedge R>0$. Thus the Hermitian background $\omega $ is not closed.

Now put $\widetilde u=-F/2-\rho $ and $\widetilde v=F/2-\rho $. Then $\omega +\mathrm{dd}^{\mathrm{c}}\widetilde u=A_0$, $\omega +\mathrm{dd}^{\mathrm{c}}\widetilde v=B_0$. Subtracting their respective suprema gives $u,v$ with $\sup _ Xu=\sup _ Xv=0$, and (14.29.2) holds. Their difference is nonconstant because $F$ is nonconstant.