If $f$ is a modular form and $y^{k/2}f$ stays bounded as $y\rightarrow \infty$, its constant Fourier coefficients at the cusps are zero

Question

Let $f(\tau)$ be a modular form of weight $k$, where $\tau = x + iy \in \mathbb H$. Then $f$ is a cusp form if its constant Fourier coefficients at the cusps are zero. By modularity, it suffices to check at one cusp, say infinity.

Why does the constant Fourier coefficient at $\infty$ vanish if $y^{k/2}f$ stays bounded as $y\rightarrow \infty$?

Answer 1

You mean $f$ is weight $k$-modular for the full modular group. $SL_2(\mathbb{Z}) \setminus\mathbb{H}$ has only one cusp : $SL_2(\mathbb{Z}) i\infty = \mathbb{Q} \cup i\infty$.

That $f$ is analytic on $\mathbb{H}$ and $f(z) = f(z+1)$ implies $$F(q) = f(\frac{\log q}{2i\pi})$$ is analytic for $0 < |q| < 1$ so it has a Laurent expansion valid for $0 < |q|< 1$ $$F(q) = \sum_n a_n q^n, \qquad f(z) = \sum_n^\infty a_n e^{2i \pi nz}$$

Then there are $4$ cases for the behavior at $q=0$

• $F$ has an essential singularity, $F(q) q^k$ is not bounded for any $k$

• $F$ has a pole, $F(q) q^m$ is analytic ($f$ is a meromorphic modular form)

• $F$ is analytic, $a_n=0$ for $n<0$ ($f$ is a modular form)

• $F(0) = 0$ so $F(q) = a_m q^m +O(q^{m+1})$ for some $m \ge 1$ ($f$ is a cusp form)

For any $z$ $$a_n e^{2i \pi nz}= \int_0^1 f(z+x) e^{-2i \pi nx}dx$$ from $f(z) = z^k f(-1/z)$, if $f$ is a cusp form, taking $z= i/n$ we have a bound $a_n= O(n^{k/2})$.

Answer 2

Let $F(q) = \sum\limits_{n=0}^{\infty}a_nq^n$ be the Fourier expansion of $f$ at infinity. So $F(q)$ is holomorphic in the disc $|q| < 1$, and we have $f(z) = F(e^{2\pi i z})$ for all $z \in \mathbb H$. Suppose that for all $x$, $f(x+iy)y^{k/2}$ is bounded as $y$ goes to infinity. You want to show that the constant term $a_0$ is zero.

The only way $f(iy)y^{k/2}$ could be bounded as $y \to \infty$ is if $f(iy) = F(e^{-2\pi y})$ is going to zero.

But $$\lim\limits_{y \to \infty} F(e^{-2\pi y}) = F(0) = a_0$$