PomeLemma 23.1.17.7C9989D90F66 For a constant $C_0\geq 1$ depending only on $E$,
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23.1.18
\begin{equation} tk_ t(w)\leq C_0\exp \left( -\frac{|t\partial _{\bar w}\log k_ t(w)|^2}{C_0t}\right) \quad (0<t\leq 1). \end{equation}
Lemma 23.1.177C9989D90F66
Proof. Choose a nearest lattice point $\lambda _0$, and put $d=|w-\lambda _0|$ and $s_\lambda =|w-\lambda |^2-d^2\geq 0$. The distance $d$ is bounded by the diameter of a fundamental rectangle. By the triangle inequality, $s_\lambda \geq |\lambda -\lambda _0|^2/4-C$, and outside a fixed finite set of $\lambda $ we have $s_\lambda \geq |\lambda -\lambda _0|^2/8$. Consequently,
\[ \sum _\lambda (1+|w-\lambda |)\mathrm{e}^{-s_\lambda /t}\leq C, \qquad tk_ t(w)\leq C\mathrm{e}^{-d^2/t}. \]
The probability formula in the previous proof gives, with $G=|t\partial _{\bar w}\log k_ t|$,
\[ G\leq d+\sum _{\lambda \ne \lambda _0}|w-\lambda |\mathrm{e}^{-s_\lambda /t}. \]
If $d\geq 1/4$, this is at most $Cd$. If $d<1/4$, the nonzero lattice distances are at least one and
\[ s_\lambda \geq |\lambda -\lambda _0| (|\lambda -\lambda _0|-2d) \geq \tfrac 12|\lambda -\lambda _0|^2. \]
Summing the resulting Gaussian bound gives $G\leq C(d+\mathrm{e}^{-c/t})$ in all cases. When $G>2C\mathrm{e}^{-c/t}$, we have $d\geq G/(2C)$, and the bound $tk_ t\leq C\mathrm{e}^{-d^2/t}$ gives (23.1.18). In the remaining case $G^2/t$ is bounded, and (23.1.18) follows from $tk_ t\leq C$ after increasing $C_0$. $\square$
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