Lemma 10.8.4.ABE17C01E6F9 Let $R$ be a smooth function on a complex manifold $M$ with $0<R<\mathrm e^{-4}$, let $0<\varepsilon <\mathrm e^{-4}$, and put
\[ \begin{gathered} \psi =\log R,\qquad \psi _\varepsilon =\log (\varepsilon +R),\qquad \mu (t)=\varepsilon -t+\log (1-t)\quad (t<1),\\ a_\varepsilon =\mu (\psi _\varepsilon ),\qquad b_\varepsilon =(2-\psi _\varepsilon )^2. \end{gathered} \]
We have $\psi <\psi _\varepsilon <-3$, $a_\varepsilon \geq -\psi _\varepsilon >3$, $1\leq -\mu '(\psi _\varepsilon )\leq 2$ and $a_\varepsilon +b_\varepsilon \leq 5\psi _\varepsilon ^2\leq 5\psi ^2$, and $a_\varepsilon $ and $b_\varepsilon $ are bounded above and below by positive constants depending only on $\varepsilon $.
For every real $(1,1)$-form $\Theta $ on $M$,
\[ a_\varepsilon \Theta -\mathrm i\partial \bar\partial a_\varepsilon -\frac{\mathrm i}{b_\varepsilon }\partial a_\varepsilon \wedge \bar\partial a_\varepsilon =a_\varepsilon \Theta -\mu '(\psi _\varepsilon )\, \mathrm i\partial \bar\partial \psi _\varepsilon . \]
The logarithmic Hessians satisfy
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10.8.5
\begin{equation} \mathrm i\partial \bar\partial \psi _\varepsilon =\frac{R}{\varepsilon +R}\mathrm i\partial \bar\partial \psi +\frac{\varepsilon }{R}\, \mathrm i\partial \psi _\varepsilon \wedge \bar\partial \psi _\varepsilon . \end{equation}
Proof.
(1) Since $\varepsilon +R<2\mathrm e^{-4}<\mathrm e^{-3}$, we have $\log \varepsilon <\psi _\varepsilon <-3$, and $\psi <\psi _\varepsilon $ because $\varepsilon >0$. Next, $a_\varepsilon =\varepsilon -\psi _\varepsilon +\log (1-\psi _\varepsilon )\geq -\psi _\varepsilon $, and $\mu '(t)=-1-(1-t)^{-1}$ lies in $[-2,-1]$ for $t<0$. With $s=-\psi _\varepsilon >3$ we have $a_\varepsilon \leq 1+s+\log (1+s)\leq 1+2s$ and $b_\varepsilon =(2+s)^2\leq 4s^2$, so $a_\varepsilon +b_\varepsilon \leq 1+2s+4s^2\leq 5s^2$, since $s^2-2s-1>0$ for $s\geq 3$; and $s^2\leq \psi ^2$ because $\psi <\psi _\varepsilon <0$. Finally, $\log \varepsilon <\psi _\varepsilon <-3$ gives $3<a_\varepsilon \leq \varepsilon -\log \varepsilon +\log (1-\log \varepsilon )$ and $25<b_\varepsilon \leq (2-\log \varepsilon )^2$.
(2) We have $\mu ''(t)=-(1-t)^{-2}$, hence $-\mu '(t)^2/\mu ''(t)=\bigl ((1-t)+1\bigr )^2=(2-t)^2$, so $b_\varepsilon =-\mu '(\psi _\varepsilon )^2/\mu ''(\psi _\varepsilon )$. By the chain rule, $\mathrm i\partial \bar\partial a_\varepsilon =\mu '(\psi _\varepsilon )\mathrm i\partial \bar\partial \psi _\varepsilon +\mu ''(\psi _\varepsilon )\mathrm i\partial \psi _\varepsilon \wedge \bar\partial \psi _\varepsilon $ and $b_\varepsilon ^{-1}\mathrm i\partial a_\varepsilon \wedge \bar\partial a_\varepsilon =b_\varepsilon ^{-1}\mu '(\psi _\varepsilon )^2\mathrm i\partial \psi _\varepsilon \wedge \bar\partial \psi _\varepsilon =-\mu ''(\psi _\varepsilon )\mathrm i\partial \psi _\varepsilon \wedge \bar\partial \psi _\varepsilon $. The gradient terms cancel, which proves (2).
(3) For $c=\varepsilon $ and for $c=0$ we have $\partial \bar\partial \log (c+R)=(c+R)^{-1}\partial \bar\partial R-(c+R)^{-2}\partial R\wedge \bar\partial R$. Hence $\partial \bar\partial \psi _\varepsilon -\frac{R}{\varepsilon +R}\partial \bar\partial \psi =\bigl (R^{-1}(\varepsilon +R)^{-1}-(\varepsilon +R)^{-2}\bigr )\partial R\wedge \bar\partial R=\frac{\varepsilon }{R}(\varepsilon +R)^{-2}\partial R\wedge \bar\partial R$, and $(\varepsilon +R)^{-2}\partial R\wedge \bar\partial R=\partial \psi _\varepsilon \wedge \bar\partial \psi _\varepsilon $.
$\square$
Pome
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