Vtyjkyuk

Let $Y$ be a finite volume quotient of the upper half plane. I can show that a weight $0$ nonharmonic Maass form is cusp iff it is square integrable.

One direction uses only local information near a cusp: if $f$ vanishes at $mathfrak a$, then $f$ is $L^2$ in a neighborhood of $mathfrak a$. (Using the Fourier-Whittaker expansion.)

The other implication ($L^2$ at all cusps implies vanishing at all cusps) uses global information. The point is that unless we know something about the eigenvalue $lambda$ (with $Delta f + lambda f = 0$), if for example $lambda = s(1-s)$ with $sigma = Re(s) in (frac12, 1)$ there could be a term $y^1-s$ in the Fourier expansion which is $L^2$ but does not vanish at $infty$. I can only conclude that $lambda in mathbb R$ if I know that $f$ is globally $L^2$: the Laplacian is positive on a complete Riemannian manifold.

Is it nonetheless true that if a nonharmonic Maass form is $L^2$ near a cusp, then it vanishes at that cusp?

For harmonic Maass forms this of course not true, there are constant functions, for example. I'm unable to construct a counterexample; I can get a Maass form whose constant term involves (say) $y^1/4$ by analytically continuating the real analytic Eisenstein series, but then there is a term with $y^3/4$ as well.