For a cusp form $f$, the <invariant norm of a modular form>
$$
y^{k/2}|f(x+iy)|
$$
is modular invariant, bounded on $\mathcal F$, and tends to zero at its cusp. Part a therefore makes it bounded throughout $\mathfrak h$: $|f(x+iy)|\leq Cy^{-k/2}$.

The correct PDF expansion is $f(\tau)=\sum_{n\geq1}a_n(f)q^n$. Fourier inversion gives
$$
a_n(f)e^{-2\pi ny}=\int_0^1f(x+iy)e^{-2\pi inx}\,dx,
$$
and hence
$$
|a_n(f)|\leq Ce^{2\pi ny}y^{-k/2}.
$$
Choosing $y=1/n$ proves the <Fourier coefficient bound for a cusp form> $|a_n(f)|\leq C'n^{k/2}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 137 / 2 / b / Solution

Put $\tau=x+iy$. The modular transformation law and
$$
\operatorname{Im}(\gamma\tau)=\frac{y}{|c\tau+d|^2}
$$
show that the <invariant norm of a modular form>
$$
y^{k/2}|f(\tau)|
$$
is invariant under $\Gamma(1)$. On the region from part (a), it is bounded: it is continuous on every truncated region, while the <cusp form> condition makes it tend to zero as $y\to\infty$. Thus
$$
|f(x+iy)|\leq C_0y^{-k/2}
$$
for all $x\in\mathbb R$ and $y>0$.

The <Fourier coefficient> formula on one period gives
$$
a_n=e^{2\pi ny}\int_0^1f(x+iy)e^{-2\pi inx}\,dx,
$$
and hence
$$
|a_n|\leq C_0e^{2\pi ny}y^{-k/2}.
$$
Choosing $y=1/n$ yields
$$
|a_n|\leq C_0e^{2\pi}n^{k/2}.
$$
This proves the <Fourier coefficient bound for a cusp form>.

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 137 / 1 / b / Solution

If $f(\tau)=\sum_{n\geq1}a_nq^n$ is a weight-$k$ level-one cusp form, boundedness of its <invariant norm of a modular form>[invariant norm] gives
$$
|a_n|=O(n^{k/2}).
$$
Integrate $f(x+iy)e^{-2\pi inx}$ over one period and choose $y=1/n$.

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