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

Question

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?

Answer 1

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.