Lemma 23.1.51.21E7B1647FA9 Uniformly in $\eta $ and $w$, $F_\eta (t,w)\leq C\eta $ for $t>0$, and $F_\eta (t,w)\geq -2\eta \log (1/t)-C\eta $ for $0<t\leq 1$. Uniformly on $\sqrt t\leq |w|\leq 2\sqrt t$ near the origin of $E$, as $t\to 0$,
Lemma 23.1.5121E7B1647FA9
\begin{align*} F_\eta (t,w)& =2\eta \log t+O(\eta ),\\ t k_ t(w)& =2\mathrm e^{-|w|^2/t}+O(\mathrm e^{-1/(Ct)}),\\ t\partial _{\bar w}\log k_ t(w)& =-w+O(\mathrm e^{-1/(Ct)}). \end{align*}
Proof. For the upper bound use $k_ r\geq 0$ on $(0,1]$ and (23.1.16). The nonnegative Fourier coefficients imply $k_ r(w)\leq k_ r(0)$, and Gaussian summation gives $k_ r(0)\leq 2/r+C r^{-1}\mathrm e^{-1/(2r)}$ for $r\leq 1$. Its integrable error proves the lower bound. On the indicated annulus the nonzero-lattice terms and their derivatives are exponentially small relative to the zero-lattice term, proving the last two identities. For the first, split the defining integral at a fixed small $r_0$. On $[t,r_0]$ the nonzero-lattice contributions have bounded integral, and the difference between $(2/r)\mathrm e^{-|w|^2/r}$ and $2/r$ has integral at most $2|w|^2/t\leq 8$. The remaining integral is uniformly bounded. $\square$
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