Q:

Cusp form becomes a bounded function on $\operatorname{SL}_2(\mathbb R)$

I'm reading Gelbart's notes on the decomposition of $L^2(\operatorname{SL}_2(\mathbb Z) \backslash \operatorname{SL}_2(\mathbb R))$, and am stuck on a small detail about reinterpreting modular forms as automorphic forms.  Let $f$ be a cusp form of weight $k$ for $\operatorname{SL}_2(\mathbb Z)$ on the upper half plane $\mathbb H$.  Let $\phi: \operatorname{SL}_2(\mathbb R) \rightarrow \mathbb C$ be the function
$$\phi(g) = f(g.i)(ci+d)^{-k} \tag{$g = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$}$$
I want to say that $\phi$ is a bounded function.  By the Iwasawa decomposition, it suffices to show that the restriction of $\phi$ to the upper triangular matrices is bounded.  That is, I just need to show that $\phi(g)$ is bounded for $g = \begin{pmatrix} y^{1/2} & xy^{-1/2} \\ & y^{-1/2} \end{pmatrix}$.  So it suffices to show that 
$$f(x+iy)y^{k/2}$$
is bounded for all $x \in \mathbb R$ and $y \in (0,\infty)$. The boundedness for $y$ large is not a big deal: we may restrict $x$ to $[0,1]$ and consider the Fourier expansion
$$f(x+iy) = \sum\limits_{n=1}^{\infty} a_ne^{2\pi in (x+iy)}$$ 
We know that as $y$ goes to infinity, $f(x+iy)$ tends to zero uniformly in $x$.  Factoring out a term $e^{-2\pi y}$, we see that $f(x+iy)y^{k/2}$ still tends to zero as $y \to \infty$, since $\lim\limits_{y \to \infty} e^{-2\pi y}y^{k/2} = 0$.
My issue is the boundedness for $y$ small.  How do we know that $f(x+iy)y^{k/2}$ does not blow up as $y$ approaches zero?  I would imagine if $x+iy$ is tending towards a rational number (cusp), we can translate that cusp to $\infty$ using some $\gamma \in \operatorname{SL}_2(\mathbb Z)$ and repeat the same argument.  What if $x$ is irrational though?  

A:

Thanks to reuns for answering.  This is a writeup of his suggestion.  My question is how we know $f(x+iy)y^{k/2}$ is bounded even when $y$ is small.  Letting $$g = \begin{pmatrix} y^{1/2} & xy^{-1/2} \\ & y^{-1/2} \end{pmatrix}$$ we have $\phi(g) = f(x+iy)y^{k/2}$.  There exists $\gamma \in \operatorname{SL}_2(\mathbb Z)$ such that $\gamma.(x+iy) =: x'+iy'$ lies in the usual fundamental domain. Letting
$$g' = \begin{pmatrix} y'^{1/2} & x'y'^{-1/2} \\ & y'^{-1/2} \end{pmatrix}$$
we have $\gamma g.i = g'.i$, so $\gamma g = g'\kappa$ for some $\kappa \in \operatorname{SO}_2(\mathbb R)$.  We have $\phi(g'\kappa) = \xi \phi(g')$ for some $\xi$ on the unit circle, so
$$f(x+iy)y^{k/2} = \phi(g) = \phi(\gamma g) = \phi(g' \kappa) = \xi\phi(g') = \xi f(x'+iy')y'^{k/2}$$
and so $|f(x+iy)y^{k/2}| = |f(x'+iy')y'^{k/2}|$.  So the question of boundedness of $f(x+iy)y^{k/2}$ is determined by $x+iy$ in the fundamental domain, which is in particular away from the real axis.  Then boundedness follows from what I wrote in my question about what happens when $y \to \infty$.  

