Lemma 16.18.193E8FFBE7FEB

Proof. For $\varepsilon >0$ we minimize instead the penalized function

For fixed $z$ it tends to $+\infty $ at both ends of the interval: as $s\to -\infty $ the linear term $-cs$ dominates the logarithm, because $H$ is bounded below. Its second derivative in $s$ is $H_{ss}+\varepsilon /(b-s)^2>0$, so it has a unique minimizer $s_\varepsilon (z)$, a nondegenerate critical point. By the implicit function theorem, $s_\varepsilon $ and the minimum value $v_\varepsilon (z)=H_\varepsilon (z,s_\varepsilon (z))$ are smooth. The critical equation is

so at the minimizer $0\leq H_ s\leq c$ and $0\leq \sigma :=\varepsilon /(b-s)\leq c$. Since $\operatorname{Re}\zeta $ is pluriharmonic, the joint Levi form of the barrier $-\varepsilon \log (b(z)-\operatorname{Re}\zeta )$ is $\varepsilon \, \mathrm i\partial \phi \wedge \bar\partial \phi /\phi ^2-\sigma \, \mathrm i\partial \bar\partial b$, where $\phi =b(z)-\operatorname{Re}\zeta $; the first term is nonnegative, and the second is at least $-\sigma C_ b\widehat\omega \geq -cC_ b\widehat\omega $. Hence, at the minimizer, the joint Levi form of $H_\varepsilon $ is bounded below by $-\pi _ D^*(\Lambda +cC_ b\widehat\omega )$.

We now use the Schur complement argument of Kiselman's minimum principle (theorem 5.12.1); it needs the Levi bound only at the minimizer, not elsewhere in the fiber. At the minimizer write the joint complex Hessian of $H_\varepsilon $ in block form $\bigl (\begin{smallmatrix}P& y/2\\ y^*/2& a/4\end{smallmatrix}\bigr )$, where $P=(\partial ^2H_\varepsilon /\partial z_ j\partial \bar z_ k)$, $y=(\partial ^2H_\varepsilon /\partial z_ j\partial s)$ and $a=\partial ^2H_\varepsilon /\partial s^2>0$; here we used that $\partial _\zeta =\partial _{\bar\zeta }=\frac12\partial _ s$ on functions of $z$ and $s=\operatorname{Re}\zeta $. Since $\partial _ sH_\varepsilon (z,s_\varepsilon (z))=0$, the chain rule gives $\partial _{z_ j}v_\varepsilon (z)=(\partial _{z_ j}H_\varepsilon )(z,s_\varepsilon (z))$, and differentiating the critical equation in $\bar z_ k$ gives $\bar y_ k+a\, \partial _{\bar z_ k}s_\varepsilon =0$, because $H_\varepsilon $ is real. Hence $\partial _{z_ j}\partial _{\bar z_ k}v_\varepsilon =P_{jk}+y_ j\partial _{\bar z_ k}s_\varepsilon $, that is, the complex Hessian of $v_\varepsilon $ is $P-a^{-1}yy^*$.

To bound it from below, let $M$ be the coefficient matrix of $\Lambda +cC_ b\widehat\omega =\mathrm i\sum _{j,k}M_{jk}\, \mathrm dz_ j\wedge \mathrm d\bar z_ k$. By the Levi bound at the minimizer, the block matrix $\bigl (\begin{smallmatrix}P+M& y/2\\ y^*/2& a/4\end{smallmatrix}\bigr )$ is positive semidefinite, and evaluating its Hermitian form at the vector $(\xi ,-2a^{-1}y^*\xi )$, $\xi \in \mathbb C^ n$, gives $\xi ^*(P+M-a^{-1}yy^*)\xi \geq 0$. Thus

It remains to pass to the limit $\varepsilon \to 0$. On a compact neighborhood in the base, $b$ is bounded, and evaluating at $s=b-1$ gives a uniform upper bound for $v_\varepsilon $ when $0<\varepsilon \leq 1$. Since $H$ is bounded below and $-cs-\log (b-s)$ grows linearly as $s\to -\infty $, the minimizers $s_\varepsilon $ are bounded below uniformly; the minimizers of the unpenalized problem have the same lower bound. Both problems may therefore be restricted to a common compact interval below their varying upper end point. The unpenalized minimum $v$ is finite, and it is continuous: after the substitution $s=b(z)-t$ the upper end point becomes fixed, the range of $t$ is a fixed compact interval, and a minimum of a continuous function over a fixed compact interval depends continuously on the parameter.

On this range the penalty is bounded below by a quantity that tends uniformly to zero, so $v_\varepsilon \geq v-o(1)$. For an upper bound, evaluate $H_\varepsilon $ at an unpenalized minimizer if it lies at least $\sqrt\varepsilon $ below $b$, and otherwise at $s=b-\sqrt\varepsilon $. In the first case the penalty is at most $\varepsilon |\log \sqrt\varepsilon |$; in the second, the unpenalized function changes by $o(1)$, uniformly, by its uniform continuity on the compact range. Consequently $v_\varepsilon \to v$ locally uniformly, and the inequalities for $\mathrm i\partial \bar\partial v_\varepsilon $ pass to the distributional limit. $\square$