Cusp form for $\text{SL}_2(\mathbb{Z})$ attaining a maximum on $\mathcal{D}=\{z\in\mathbb{H}\colon|\text{Re}(z)|\leq\frac12,\|z\|\geq 1\}$

Question

Let $f:\mathbb{H}\to\mathbb{C}$, $z\mapsto\sum\limits_{i=1}^\infty a_i\exp(2\pi iz)$ be a non-zero weight-$0$ cusp form for $\text{SL}_2(\mathbb{Z})$. Then $|f|$ attains a maximum on $\mathbb{H}$. We can restrict to finding a bound on the fundamental domain $\mathcal{D}=\{z\in\mathbb{H}\colon|\text{Re}(z)|\leq\frac{1}{2},\|z\|\geq 1\}$, since $f$ has weight $0$. Is is true that $f$ being a cusp form implies that $\lim\limits_{y\to\infty}f(x+iy)=0$ uniformly in $x$ for $|x|\leq\frac{1}{2}$? If it holds, I wanted to reason as follows. Pick some $z_0\in\mathcal{D}$ such that $f(z_0)\neq 0$. Take $0<\epsilon<\frac{|f(z)|}{2}$. Then there exists an $M\geq 0$ such that $|f(x+iy)|\leq\epsilon$ for all $x+iy\in\mathcal{D}$ with $y\geq M$. Since the set $\mathcal{D}_M=\{z\in\mathbb{H}\colon|\text{Re}(z)|\leq\frac{1}{2},\text{Im}(z)\leq M,\|z\|\geq 1\}$ is compact, $|f|$ takes a maximum on $\mathcal{D}_M$. Now this maximum cannot be on the line $y=M$, since $|f(x+iM)|\leq\frac{|f(z_0)|}{2}$. Does, it follows that $f$ takes some maximum value on $\mathcal{D}_M^\circ=\{z\in\mathbb{H}\colon|\text{Re}(z)|\leq\frac{1}{2},\text{Im}(z)<M,\|z\|\geq 1\}$. Call this maximum $K$. By constrution of $\mathcal{D}_M$, we have that $|f(z)|<\frac{|f(z_0)|}{2}\leq K$ outside $\mathcal{D}_M$, hereby proven that $|f|$ takes a maximum value on $\mathcal{D}$.

Answer 1

It is indeed true that $\lim_{y \to \infty} f(x + iy) = 0$ uniformly in $x$. There is a holomorphic function $g$ on the punctured unit disk such that $f(z) = g(e^{2\pi iz})$, namely $g(q) = \sum_{n=1}^{\infty}a_nq^n$. Since $f$ is a cusp form we have that the singularity of $g$ at $0$ is removable and $\lim_{q \to 0} g(q) = 0$.

So your proof that $f$ takes a maximum on $\mathcal{D}$ goes through (although you should make sure to choose $M$ such that $M \geq \textrm{Im}(z_0)$ so that you can actually conclude that the maximum on $\mathcal{D}_M$ doesn't occur on the line $y = M$). You can then obtain a contradiction because by the weight $0$ automorphy the maximum of $f$ on $\mathcal{D}$ is also the maximum of $f$ on $\mathbb{H}$. But $f(\mathbb{H})$ is an open set by the open mapping theorem.