A weakly modular function $f$ is a modular form if and only if $f$ is bounded in the fundamental domain

Question

We consider modular functions/forms for $SL_2(\Bbb Z)$. I was asked to prove that

A weakly modular function $f$ is a modular form if and only if $f$ is bounded in the fundamental domain $D$.

A function $f: \Bbb H\to\Bbb C$ is said to be a weakly modular function of (even) weight $k$ if $f$ is meromorphic on $\Bbb H$ and satisfies $$f\left(\frac{az+b}{cz+d} \right) = (cz+d)^k f(z) \quad \forall \begin{pmatrix}a & b\\c& d \end{pmatrix} \in SL_2(\Bbb Z)$$ The fundamental domain $D$ is given by $$D = \left\{z\in \Bbb H: \frac{-1}2 \le \operatorname{Re}z\le \frac12, |z| \ge 1 \right\}$$ and a function $f: \Bbb H\to\Bbb C$ is said to be a modular form of weight $k$ if $f$ is a weakly modular function of weight $k$, and it is holomorphic everywhere on $\Bbb H\cup \{\infty\}$. How should I go about proving the claim? Thanks a lot!

Answer 1

1. If $f$ is modular (which, by definition, implies that $f$ is holomorphic on $\mathbb H\cup\{\infty\}$) then $f$ is, in particular, continuous on $D\cup\{\infty\}$, which is compact. Hence $f$ is bounded on $D\cup\{\infty\}$ and so it is bounded on $D$.

2. Conversely, if $f$ is weakly modular and bounded in $D$ then we aim to show that $f$ is holomorphic on $\mathbb H\cup\{\infty\}$. First, as you say, Riemann's theorem on removable singularities tells us that there are no poles in $D\cup\{\infty\} $. Then, for any $z\in\mathbb H$ we can write $z=\frac{aw+b}{cw+d}$ with $w\in D$ and $\begin{pmatrix} a& b\\ c&d\end{pmatrix}\in SL_2(\mathbb Z)$. Then, $$ f(z)=(cw+d)^kf(w). $$ If $w$ is in the interior of $D$ we are done, as this shows that $f(z)$ is bounded in a neighborhood of $z$, and we can rule out the existence of poles by Riemann's theorem again(*). If $w$ is on the boundary, we can anyway cover a full neighborhood of $w$ by finitely many elements of $SL_2(\mathbb Z)$ and so the argument proceeds similarly.

EDIT: (*) If $w$ is in the interior of $D$, let $w\in U\subset D$ with $U$ open and bounded and let $V$ be the image of $U$ under $w\mapsto\frac{aw+b}{cw+d}$. Then $V$ is an open neighborhood of $z$ on which $f$ is bounded: indeed, if $z'\in V$, then there is $w'\in U$ such that $$ f(z') = f(w')(cw'+d)^k $$ which is bounded for all $z'\in V$ because $f$ is bounded on $U$ and because $U$ is bounded. Hence, we have shown that there exists an open neighborhood (namely, $V$) of any point $z\in\mathbb H$ on which $f$ is bounded. (Strictly speaking, as mentioned above, one has to reason slightly differently for points $z$ whose representative in $D$ lies on the boundary of $D$, but that is completely analogous.) Thus any singularity of $f$ on the upper half-plane $\mathbb H$ must be removable.

EDIT(2).

1. The map $w\mapsto \frac{aw+b}{cw+d}$ is an homeomorphism of the Riemann sphere, hence it is an open map (all homeomorphisms are!). Alternative argument: use the open mapping theorem in complex analysis, cf. https://en.wikipedia.org/wiki/Open_mapping_theorem_(complex_analysis)

2. It is convenient to recall the following fact: if $z_0$ is a pole of a meromorphic function $f$, then $\lim_{z\to z_0}|f(z)|=+\infty$ for any path $z\to z_0$. Then, also points at the boundary of $D$ cannot be poles of $f$ (as you can always approach $z\to z_0\in\partial D$ by a path entirely contained in $D$). I apologize, this argument is easier than the hinted one about covering a neighborhood of such points by finitely many $SL_2(\mathbb Z)$-orbits of $D$.

3. Holomorphicity at $\infty$. The fact that $f$ is bounded on $D$, along with the periodicity $f(z+1)=f(z)$, implies that that $f(z)$ is bounded on any half-plane $\Im z\geq k>0$. This ensures holomorphicity at $\infty$ in the usual sense for modular forms.