The q-expansion principle at the cusp

Let $p(\tau ):=e^{2\pi i\tau }$. For every $q_0\in D^{\ast }$ there are an open disc $V\subseteq D^{\ast }$ about $q_0$ and a holomorphic $\sigma :V\to \mathfrak{H}$ with $p(\sigma (q))=q$ on $V$. Indeed, choose an angle $\alpha$ with $e^{-i\alpha }{q}_0\mathrm{\notin }(-\mathrm{\infty },0]$ and let $S_{\alpha }=\{q:e^{-i\alpha }q\mathrm{\notin }(-\mathrm{\infty },0]\}$; the function $v(q):=L(e^{-i\alpha }q)+i\alpha$ is holomorphic near $q_0$ by [F2], and $e^{v(q)}=e^{-i\alpha }q\cdot e^{i\alpha }=q$ by the addition law. Taking $V$ a small disc contained in $D^{\ast }\cap S_{\alpha }$ and setting $\sigma (q):=-iv(q)\mathrm{/}(2\pi )$ gives $e^{2\pi i\sigma (q)}=e^{v(q)}=q$, and $\sigma$ is holomorphic by [F3]. Here $\mathrm{\Im }\sigma (q)>0$ holds automatically: $e^{-2\pi \mathrm{\Im }\sigma (q)}=\mathrm{\mid }{e}^{2\pi i\sigma (q)}\mathrm{\mid }=\mathrm{\mid }q\mathrm{\mid }<1$ by [F1], so $\sigma$ maps $V$ into $\mathfrak{H}$.

Define $F(q):=f(\sigma (q))$ for a local section $\sigma$ as in 1.1; this is well defined. If $\sigma ,{\sigma }^{\mathrm{\prime }}$ are two such sections near $q$, then $p(\sigma (q))=q=p({\sigma }^{\mathrm{\prime }}(q))$, so $\sigma (q)-{\sigma }^{\mathrm{\prime }}(q)\in \mathbb{Z}$ by [F1], whence $f(\sigma (q))=f({\sigma }^{\mathrm{\prime }}(q))$ by the periodicity of $f$. Since local sections exist near every $q\in D^{\ast }$, this gives a function $F$ on all of $D^{\ast }$, holomorphic because near each point it is the composition $f\circ \sigma$ of holomorphic functions [F3]. If $\tilde{F}$ is holomorphic on $D^{\ast }$ with $f(\tau )=\tilde{F}(e^{2\pi i\tau })$ for all $\tau$, then for $q\in D^{\ast }$ and a local section $\sigma$ with $p(\sigma (q))=q$ we get $\tilde{F}(q)=\tilde{F}(p(\sigma (q)))=f(\sigma (q))=F(q)$; hence $\tilde{F}=F$ and $F$ is unique.

If $F$ extends holomorphically to $0$, then $F$ is bounded on $\mathrm{\mid }q\mathrm{\mid }<\delta$ for some $\delta >0$; for $\mathrm{\Im }\tau >-\log \delta \mathrm{/}(2\pi )$ one has $\mathrm{\mid }q\mathrm{\mid }<\delta$ by [F1], so $\mathrm{\mid }f(\tau )\mathrm{\mid }=\mathrm{\mid }F(q)\mathrm{\mid }$ is bounded on that half-plane. Conversely, if $\mathrm{\mid }f\mathrm{\mid }\le M$ on $\{\mathrm{\Im }\tau >Y_0\}$, then for $0<\mathrm{\mid }q\mathrm{\mid }<e^{-2\pi Y_0}$ any local section $\sigma (q)$ of 1.1 satisfies $\mathrm{\Im }\sigma (q)=-\log \mathrm{\mid }q\mathrm{\mid }\mathrm{/}(2\pi )>Y_0$ by [F1], so $\mathrm{\mid }F(q)\mathrm{\mid }=\mathrm{\mid }f(\sigma (q))\mathrm{\mid }\le M$; by [F4] the singularity of $F$ at $0$ is removable, $F$ extends holomorphically, and $F(0)={\lim }_{q\to 0}F(q)={\lim }_{\mathrm{\Im }\tau \to \mathrm{\infty }}f(\tau )$ (given $\epsilon >0$ choose $\delta$ with $\mathrm{\mid }F(q)-F(0)\mathrm{\mid }<\epsilon$ for $\mathrm{\mid }q\mathrm{\mid }<\delta$; then $\mathrm{\Im }\tau >-\log \delta \mathrm{/}(2\pi )$ gives $\mathrm{\mid }f(\tau )-F(0)\mathrm{\mid }<\epsilon$). Finally, if the extension exists, either $F-F(0)$ vanishes on a neighbourhood of $0$, in which case take $G:=0$, or it vanishes to finite order $m\ge 1$ and [F5] writes $F(q)-F(0)=qG(q)$ with $G$ holomorphic near $0$; in either case $G$ is bounded near $0$, so $\mathrm{\mid }f(\tau )-F(0)\mathrm{\mid }=\mathrm{\mid }q\mathrm{\mid }\mathrm{\mid }G(q)\mathrm{\mid }\le C{e}^{-2\pi \mathrm{\Im }\tau }$ for all large $\mathrm{\Im }\tau$, which is the asserted $O(e^{-2\pi \delta \mathrm{\Im }\tau })$ with $\delta =1$.

Assume now that $F$ extends holomorphically to $0$. On the annulus $0<\mathrm{\mid }q\mathrm{\mid }<1$ the holomorphic $F$ has Laurent expansion ${\sum }_{n\in \mathbb{Z}}{a}_n{q}^n$, and by [F6] $a_n=\frac{1}{2\pi i}{\oint }_{\mathrm{\mid }q\mathrm{\mid }=\rho }F(q)q^{-n-1}dq$ for every $0<\rho <1$. Since $F$ is holomorphic at $0$, all coefficients $a_n$ with $n<0$ vanish (their principal part is zero, [F4] applied to the Laurent expansion), so $F(q)={\sum }_{n\ge 0}{a}_n{q}^n$ and $f(\tau )=F(e^{2\pi i\tau })={\sum }_{n\ge 0}{a}_n{q}^n$ converges for $\mathrm{\mid }q\mathrm{\mid }<1$.