Let $f$ be any non-zero integral weight (holomorphic) modular form with respect to $SL_2(\mathbb{Z})$ and of weight $k, k\geq 4$. Since it is holomorphic at infinity, for given $\epsilon > 0$, it is bounded above by $a_f(0)+\epsilon$ in some neighbourhood of infinity, say $\{y>y_0\}$. Also, we know that the values of a modular form on the upper half plane are completely determined by its values on the fundamental domain. Now let me I chop off the $\{y > y_0\}$ portion of the fundamental domain, i.e., $\{z \in \mathcal{H} |\;\; |Re(z)| \leq 0.5, |z| \geq 1, Im(z) \leq y_0\}$. Let me call it $\mathcal{F_0}$.Questions:-
1) Isn't this a compact set w.r.t the subspace topology on $\mathcal{H}$ (coming from usual topology of $\mathbb{C}$)? It is closed and bounded right?
2) By continuity of $f$, since f has a maximum and a minimum inside $\mathcal{F_0}$, let us say |f| is bounded above by |f(z_0)| for some $z_0$ in $\mathcal{F_0}$. Then, why isn't $|f|$ bounded by $max\{a_f(0)+\epsilon, |f(z_0)|\}$ everywhere on the Fundamental Domain and hence, everywhere on $\mathcal{H}$. (Of course, I am not asking here why it wouldn't be a constant as of course Liouville cannot be applied here). I am worried because of the assumptions taken in defining the Petersson Inner Product on the space of modular forms, i.e., $|f(z)g(z)| << Im (z)^{-k}$. If the above (being bounded) is true, then clearly, all modular forms satisfy the above assumptions which is absurd. Kindly clarify what is the mistake in (1).