# A doubt in proof related to estimating order of magnitude of |c(n)|

I am self studying analytic number theory from Tom M Apostol Modular functions and Dirichlet series in number theory and I couldn't think about an argument in proof related to Modular group.

I am adding image of proof highlighting the argument which I don't understand.

My doubt is in 7 th line of 2nd image ie

$$\phi$$ is bounded in $$R_\Gamma$$ and it has been proved that $$\phi$$ is invariant under $$\Gamma$$ , so how does Apostol deduces $$\phi$$ is bounded in H?

Can someone please give a hint.

Since $$R_\Gamma$$ is a fundamental region for $$\Gamma$$, we have that $$H = \Gamma \cdot R_\Gamma$$. Stated differently, for any point $$z \in H$$, there is an element $$\gamma \in \Gamma$$ such that $$\gamma z = z' \in R_\Gamma$$. This is the defining characteristic of a fundamental domain.
Thus to understand the size of $$\phi(z)$$, you can use that $$\phi(z) = \phi(\gamma z) = \phi(z')$$, where we are using that $$\phi$$ is invariant under $$\Gamma$$. And thus $$\phi$$ being bounded on $$R_\Gamma$$ shows that $$\phi$$ is bounded on $$H$$.
• @davilowryduda according to Apostol z' belongs to closure of fundamental region not fundamental region ( See page 30) . But it is given that $\phi$ is bounded in fundamental region, then you need to prove that $\phi$ is also bounded in closure of fundamental region – Tim Jan 3 '20 at 16:26
• Very well, then prove that $\phi$ is bounded on the closure of the fundamental domain. As $\phi$ is continuous and $\phi \to 0$ as $\mathrm{Im} z \to \infty$, you can consider only a finite region up to some bounded height. Then merely the fact that $\phi$ is continuous and the remaining region is compact shows that $\phi$ is bounded on the whole fundamental domain (and thus on all of $\mathcal{H}$. – davidlowryduda Apr 12 '20 at 17:02