Let $f$ and $g$ be two cuspforms on $\Gamma \backslash \mathbb{H}$. They could be Maass cuspforms, or holomorphic modular forms. Let us say that they are holomorphic and also that $\Gamma = \operatorname{SL}\_2(\mathbb{Z})$ for simplicity. The product $f \overline g$ is not necessarily a cuspidal function on $\Gamma \backslash \mathbb{H}$ although it decays even faster than $f$ or $g$ as it approaches a cusp.
One can see that the function is not cuspidal by taking $f = g$ and noting that,
$$
\int\_0^1 |f|^2(x+iy) dx \neq 0.
$$


I guess this is not a question but rather a surprised statement about cuspidality not being synonymous with vanishing at cusps.


It seems like the two statements are equivalent only when the function in question is also an eigenfunction of the Laplacian.


Could you elaborate on this issue, for example could you find a cuspidal function, which does not vanish at the cusps? Of course the function has to be not an eigenfunction.



Product of two cuspforms is not a cuspform
I don't think it's possible to find a "nice" (say, smooth) function $f \in L\_2(\Gamma \backslash \mathbb{H})$ such that $(1) \int\_0^{1} f(x+iy) dx = 0$ for all $y > 0$ and $\lim\_{y\rightarrow \infty} f(x+iy) \neq 0$. This may be total overkill, but consider the spectral decomposition of such an $f$, namely
$$(2) \qquad f(z) = \sum\_{j} \langle f, u\_j \rangle u\_j(z) + \frac{1}{4\pi } \int\_{\mathbb{R}} \langle E(\cdot, 1/2 + it), f\rangle E(z, 1/2 + it) dt.$$
By unfolding, the inner product of $f$ with the Eisenstein series $E(z,s)$ is zero by the assumption (1); initially this is easy for the real part of $s$ large but then follows by analytic continuation.
By inserting (2) into (1) we see that $\langle f, u\_0 \rangle = 0$, that is $f$ is orthogonal to the constant eigenfunction. Now in (2) take $z= x+iy$ with $y$ large. Each term in the sum is very small since all the Maass forms vanish at the cusp, and the projections of $f$ onto the constant eigenfunction and the Eisenstein series are zero.


