I am studying Proposition 3.3.3 from "Automorphic Forms and L-functions for the group $GL_n(\mathbb{R})$" by D. Goldfeld, and I need a clarification regarding the behavior of Maass forms as $\text{Im}(z) \to \infty $.
The definition of a Maass form of type $n \in \mathbb{C}$ in this context is as follows:
- It is a function $f:SL_2(\mathbb{Z})\backslash \mathbb{H}\xrightarrow{}\mathbb{C} $ i.e invariant under the action of $ SL_2(\mathbb{Z}) $ on $ \mathbb{H} $,
- $ f $ is a square-integrable (with repspect to $dz=\frac{dx.dy}{y^2}$), non-zero eigenfunction of the hyperbolic Laplacian $\Delta $ on the upper half-plane $\mathbb{H} $ and $\Delta f=n(n-1)f,$
- $ \int_0^1 f(z) \, dx = 0. $
In Proposition 3.3.3, the statement is made that if $ f$ is a Maass form of type 0 or 1, then it must be constant. The justification provided contains the claim that, because $ f $ is a Maass form, it is bounded as $ \text{Im}(z) \to \infty $.
Here is my question:
Why does the fact that $ f $ is a Maass form imply that it is bounded as $ \text{Im}(z) \to \infty $?
I understand that Maass forms are square-integrable, which imposes some constraints on their growth, but I thought that the condition of square-integrability only suggests that $ f$ cannot grow faster than $ y^2 $ as $ y \to \infty $. Why does this lead to $ f $ being bounded, rather than merely preventing rapid growth?
Thank you very much for any help!
