# Coefficients of cusp forms satisfy $a_n=O(n^k)$

Suppose f is a modular cusp form of weight 2k. I want to show that $$a_n=O(n^k)$$ where $$f(s)=\sum_{n=1}^\infty a_ne^{2\pi ins}$$

I was reading through the proof of the statement in the following link(Theorem 1.3): https://www2.math.ethz.ch/education/bachelor/seminars/ws0607/modular-forms/Dirichlet_Series.pdf

However, I came across the following statement in the proof:

Here $$\phi(x)$$ is a function defined on the upper half plane with $$f$$ being the modular function and $$Im(s)=y$$. The condition that $$\phi(x)\rightarrow 0$$ as $$y\rightarrow 0$$ only implies that $$\phi(x)$$ is bounded for some $$y>y_0$$. How does that translate to $$\phi$$ being bounded on the entire half plane?

Edit: Hecke Bound for Cusp - Modular Forms The link provides a lemma which states that $$\phi(s)$$ is in variant under the action of $$\gamma \in SL_2(\mathbb Z)$$.

I think this means if we can find a $$\gamma$$ that acts by moving up every element in the half plane we are done. So does anyone have such a $$\gamma$$ in mind?

For $$\gamma\in SL_2(\Bbb{R})$$,$$\Im(\gamma(z))=|cz+d|^{-2}\Im(z)$$
Thus if $$f\in M_k(\Gamma)$$ then $$\phi(z)=|f(z)|\Im(z)^{k/2}$$ satisfies $$\forall \gamma\in \Gamma,\qquad \phi(\gamma(z))= \phi(z)$$ Whence it suffices that $$\phi$$ is bounded at the finitely many cusps (which happens when $$f$$ is a cusp form) to get that it is bounded everywhere.
• Thanks for the answer. i know that $\phi (z)$ is bounded for $I(z)>\tau$ ($\tau$ large enough). So if there is a $\gamma$ that increases the imaginary part of z i.e $Im(\gamma z)>Im(z)$ that would suffice to show that $\phi$ is bounded on the upper half plane right? Commented Jan 16, 2020 at 13:45
• Do you know of any such $\gamma$ Commented Jan 16, 2020 at 13:46
• The upper half plane is covered by the $SL_2(Z)$ translates of the fundamental domain $\Re(z)\in [-1/2,1/2],|z|\ge 1$ en.wikipedia.org/wiki/… Commented Jan 16, 2020 at 13:50
• Thanks for the answer but I would like to clarify a few things since I'm new to modular forms. By the fundamental domain - ($R(z)\in [-1/2,1/2],|z|\geq 1$) - do you mean the set of representatives of the orbits generated by the group action of $SL_2(\mathbb Z)$ on the half plane? If so this means that every element in the half plane can be reached by the action on the fundamental domain. Since the fundamental domain can be bounded the entire half plane is bounded? Commented Jan 16, 2020 at 14:20
• Since $\phi$ is $SL_2(Z)$ invariant and bounded on the fundamental domain then it is bounded everywhere Commented Jan 16, 2020 at 14:23