Hecke Bound for Cusp - Modular Forms 
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*The problem statement, all variables and given/known data


i have a  questions on the piece of lecture notes attached:



*Relevant equations

*The attempt at a solution
I agree 2) of proposition 2.12 holds once we have 1). I thought I understood the general idea of 1), however, my reasoning would hold for $M_k$ it does not depend on $f(t)$ being a cusp and so it must be wrong. This was what I thought was happening:
$q=e^{2\pi i n (u+iv)} \approx e^{-v} $ for large v, and exponential dominates over $v ^ {x}$ ( v>0 as on upper plane )
This would ofc still hold if I included some constant term, I would still get the same quantity is bounded.
can someone please tel me where I have gone wrong with the above reasoning?
 A: I don't think that the reasoning would apply to a form with a constant term.
If $f$ is cusp, we can write $|f(q)|=e^{-2\pi m v}|g(q)|$, where $0<|g(0)|<\infty$ and $m>0$. Then $v^{k/2}|f(q)|= v^{k/2} e^{-2\pi m v}|g(q)|$. Clearly the product $v^{k/2}e^{-2\pi m v}$ has zero limit at infinity, and since $|g(0)|$ is not infinity, we get that the limit of $v^{k/2}|f(q)|$ at infinity is zero. 
If, on the other hand, $f$ is not a cusp form, then $|f(0)|>0$ and so the limit of $v^{k/2}|f(q)|$ at $v=\infty$ will not be dominated by anything going to zero, and so you get an infinite limit, hence $Im(z)^{k/2}f(z)$ would be unbounded on the fundamental domain.
A: It may be handy to have a particular non-cuspidal modular form in mind for examples like this. The prototypical non-cuspidal modular form is the Eisenstein series of weight $2k$
$$ E(\tau) = \sum_{m,n \in \mathbb{Z}^2 - \{(0,0)\}} \frac{1}{(m+n\tau)^{2k}},$$
whose Fourier expansion is
$$ E(\tau) = 2\zeta(2k)\left(1 + c_{2k} \sum_{n \geq 1} \sigma_{2k-1}(n) q^n\right).$$
This expansion looks like a constant term $1$ and a lot of terms with exponential decay in the imaginary part, coming from $q^n$. Your reasoning applies to the exponential decay portion: all terms with exponentials vanish as the imaginary part goes to $\infty$.
But the constant term doesn't, and $\mathrm{Im}(\tau)^k \cdot 2 \zeta(2k)$ clearly goes to $\infty$ as the imaginary part goes to $\infty$.
