Lemma 23.1.12.14E6CDA9AAC4 Let $0<\eta <1/10$ and define
\[ F_\eta (t,w)=-\eta \int _ t^\infty (k_ r(w)-1)\, \mathrm{d}r. \]
Then $F_\eta $ is smooth and real, and
9CCEEAD6F62C
23.1.13
\begin{align} (F_\eta )_ t=(F_\eta )_{w\bar w} & =\eta (k_ t-1),\\ (\log k_ t)_{w\bar w}& \geq -1/t,\tag{23.1.14}\\ \int _ E|(F_\eta )_ w(t,w)|^2\mathrm{d}A_ E & =2\eta ^2\log (1/t)+O(\eta ^2)\quad (0<t\leq 1). \tag{23.1.15} \end{align}
The error term in (23.1.15) is uniform in $t$ and $\eta $. For fixed nonnegative integers $a,m$ there is also a uniform estimate
464524C306E0
23.1.16
\begin{equation} \| \partial _ t^ aF_\eta (t,\cdot )\| _{C^ m(E)} \leq C_{a,m}\eta \mathrm{e}^{-t/4}\quad (t\geq 1). \end{equation}
Proof.
Integrating the nonconstant Fourier modes gives
\[ F_\eta (t,w)=-\frac{\eta }{\pi ^2} \sum _{(p,q)\ne (0,0)} \frac{\mathrm{e}^{-\pi ^2|\nu _{p,q}|^2t}}{|\nu _{p,q}|^2}\chi _{p,q}(w). \]
This proves smoothness, reality and (23.1.13). The least nonzero value of $\pi ^2|\nu _{p,q}|^2$ is $1/4$, and any fixed number of differentiations introduces coefficients of polynomial growth in $|\nu _{p,q}|$. Factoring out $\mathrm{e}^{-t/4}$ for $t\geq 1$ therefore leaves a convergent Gaussian series, which proves (23.1.16).
To prove (23.1.14), fix a lift of $w$ and give $\Lambda $ the probability weights
\[ P_\lambda = \frac{\mathrm{e}^{-|w-\lambda |^2/t}}{\sum _{\lambda '}\mathrm{e}^{-|w-\lambda '|^2/t}}. \]
Logarithmic differentiation of the Gaussian sum yields
\[ t\partial _{\bar w}\log k_ t=-\sum _\lambda (w-\lambda )P_\lambda , \]
and one more derivative gives
\[ (\log k_ t)_{w\bar w} =-\frac1t+\frac1{t^2} \left(\sum _\lambda |w-\lambda |^2P_\lambda -\left|\sum _\lambda (w-\lambda )P_\lambda \right|^2\right). \]
The expression in parentheses is a variance and is nonnegative.
Finally, Parseval's identity gives, with $\mathcal E_\eta (t)$ denoting the energy in (23.1.15),
\[ \mathcal E_\eta (t)=\frac{\eta ^2}{\pi ^2} \sum _{(p,q)\ne (0,0)} \frac{\mathrm{e}^{-2\pi ^2|\nu _{p,q}|^2t}}{|\nu _{p,q}|^2}, \qquad \mathcal E_\eta '(t)=-2\eta ^2(k_{2t}(0)-1). \]
The Gaussian formula at the origin implies
\[ k_{2t}(0)=\frac1t+ O\left(t^{-1}\mathrm{e}^{-1/(4t)}\right)\quad (0<t\leq 1). \]
To see this, isolate the zero lattice vector and use $\mathrm{e}^{-|\lambda |^2/(2t)}\leq \mathrm{e}^{-1/(4t)}\mathrm{e}^{-|\lambda |^2/4}$ for $\lambda \ne 0$. The error term is integrable near $t=0$, and $\mathcal E_\eta (1)=O(\eta ^2)$. Integrating the derivative between $t$ and $1$ proves (23.1.15).
$\square$
Pome
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