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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. enter image description here

enter image description here

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.

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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 $H$.

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  • $\begingroup$ @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 $\endgroup$ – Tim Jan 3 '20 at 16:26
  • $\begingroup$ 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 $\mathcal{H}$. $\endgroup$ – davidlowryduda Apr 12 '20 at 17:02
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    $\begingroup$ thank you very much!! $\endgroup$ – Tim Apr 12 '20 at 17:38

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