Lemma 6.5.3.EE99921A3F33 Let $0<\varepsilon <1/4$. For $\zeta \in \mathbb C$ with $|\zeta |<1/4$ put
\[ s=-\log (|\zeta |^2+\varepsilon ^2),\qquad \eta =s+\log s,\qquad \psi _\varepsilon =-\log \eta . \]
Then $\psi _\varepsilon $ is a smooth, bounded, strictly subharmonic function on the disc $\Delta _{1/4}$. Let $h=|\partial _{\bar\zeta }\psi _\varepsilon |^2/(\psi _\varepsilon )_{\zeta \bar\zeta }$. Then
\[ \begin{split} (1-h)\mathrm e^{\psi _\varepsilon }& \ge (6s^2)^{-1},\\ h& \le \frac{2}{-\log (2\varepsilon ^2)}\quad \text{for }|\zeta |\le \varepsilon ,\\ \frac{\mathrm e^{\psi _\varepsilon }}{|\zeta |^2(\psi _\varepsilon )_{\zeta \bar\zeta }} & \le 25/4\quad \text{for }\varepsilon /2\le |\zeta |\le \varepsilon . \end{split} \]
In particular $h<1$ everywhere, and the bound for $h$ on the disc $|\zeta |\le \varepsilon $ tends to zero as $\varepsilon \to 0$.
Proof.
Since $|\zeta |^2+\varepsilon ^2<1/8$, we have $s>\log 8>2$, so $\eta >s>2$ and $\psi _\varepsilon <0$ is well defined. For fixed $\varepsilon $, $s$ ranges in a bounded interval, so $\psi _\varepsilon $ is bounded; and $\psi _\varepsilon =G(|\zeta |^2)$ for a function $G$ that is smooth near $[0,1/16)$, so $\psi _\varepsilon $ is smooth.
For $\zeta \ne 0$ put $x=|\zeta |^2$, $t=\log x$ and $b=x/(x+\varepsilon ^2)\in (0,1)$, and denote derivatives with respect to $t$ by primes. Then $s'=-b$ and $b'=b(1-b)$, so $s''=-b(1-b)$. With $D=1+1/s$ we have $\eta '=s'+s'/s=-bD$ and $D'=-s'/s^2=b/s^2$. Hence $\psi _\varepsilon '=-\eta '/\eta =bD/\eta $ and $\psi _\varepsilon ''=(b'D+bD')/\eta -bD\eta '/\eta ^2$, that is,
\[ \psi _\varepsilon '=\frac{bD}{\eta },\qquad \psi _\varepsilon ''=A+B,\quad A=\frac{b^2D^2}{\eta ^2},\quad B=\frac{b(1-b)D}{\eta }+\frac{b^2}{\eta s^2}>0. \]
For a function $g$ of $t=\log |\zeta |^2$ we have $\partial _{\bar\zeta }g=g'/\bar\zeta $ and $\partial _\zeta \partial _{\bar\zeta }g=g''/|\zeta |^2$. Hence $(\psi _\varepsilon )_{\zeta \bar\zeta }=\psi _\varepsilon ''/x>0$ and $h=(\psi _\varepsilon ')^2/\psi _\varepsilon ''=A/(A+B)<1$ for $\zeta \ne 0$. In general $\partial _{\bar\zeta }\psi _\varepsilon =G'(|\zeta |^2)\zeta $ and $(\psi _\varepsilon )_{\zeta \bar\zeta }=G''(|\zeta |^2)|\zeta |^2+G'(|\zeta |^2)$, so at $\zeta =0$ we have $\partial _{\bar\zeta }\psi _\varepsilon =0$ and $(\psi _\varepsilon )_{\zeta \bar\zeta }=G'(0)$, where $G'(x)=\psi _\varepsilon '/x=D/(\eta (x+\varepsilon ^2))>0$; so $(\psi _\varepsilon )_{\zeta \bar\zeta }>0$ and $h=0$ there. Thus $\psi _\varepsilon $ is strictly subharmonic, and $h<1$ everywhere.
For the first estimate let $\zeta \ne 0$. Since $B\ge b^2/(\eta s^2)$, we have $A/B\le D^2s^2/\eta =(s+1)^2/\eta $, and therefore
\[ (1-h)\mathrm e^{\psi _\varepsilon }=\frac{B}{\eta (A+B)} \ge \frac1{\eta +(s+1)^2}\ge \frac1{6s^2}, \]
because $\eta \le 2s\le 2s^2$ and $(s+1)^2\le 4s^2$ for $s\ge 1$. At $\zeta =0$ we have $(1-h)\mathrm e^{\psi _\varepsilon }=1/\eta \ge 1/(6s^2)$ as well. For the second estimate let $0<x\le \varepsilon ^2$. Then $b\le 1/2$, and $B\ge b(1-b)D/\eta $ gives
\[ h\le \frac AB\le \frac{bD}{\eta (1-b)}\le \frac D\eta =\frac{s+1}{s\eta }\le \frac2s, \]
since $s+1\le 2s\le 2\eta $; moreover $s\ge -\log (2\varepsilon ^2)$, because $x+\varepsilon ^2\le 2\varepsilon ^2$. Together with $h(0)=0$, this proves the second estimate. For the third, let $\varepsilon /2\le |\zeta |\le \varepsilon $. Then $1/5\le b\le 1/2$, so $\eta \psi _\varepsilon ''\ge \eta B\ge b(1-b)D\ge b(1-b)\ge 4/25$, and $\mathrm e^{\psi _\varepsilon }/(|\zeta |^2(\psi _\varepsilon )_{\zeta \bar\zeta })=1/(\eta \psi _\varepsilon '')\le 25/4$.
$\square$
Pome
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