Lower bounds for Petersson inner products of cuspforms with integral Fourier coefficients Let $N\geq 1$ be an integer and let $S_2(\Gamma_0(N))$ be the cusp forms of weight 2 for the usual congruence subgroup $\Gamma_0(N)\subset SL_2(\mathbb Z)$. 
Let $a_n(f)$ denote the n-th Fourier coefficient of $f\in S_2(\Gamma_0(N))$, $n\geq 1$. 
Let $X_0(N)$ be a smooth projective model over $\mathbb Q$ of the modular curve associated to $\Gamma_0(N)$ and let $$(f,g)=\int_{X_0(N)}f(z)\overline{g(z)}dxdy,$$ $z=x+iy$, be the Petersson inner product of $f,g\in S_2(\Gamma_0(N))$.
Question (a) Does there exist a constant $c>0$, depending at most on $\Gamma_0(N)$ (or $X_0(N)$), with the following property: Suppose $f\in S_2(\Gamma_0(N))$ is an eigenform with $a_1(f)=1$  which lies in the new part, $g\in S_2(\Gamma_0(N))$ and  $a_n(f),a_n(g)\in\mathbb Z, n\geq 1$. If $(f,g)>0$, then $$(f,g)\geq c.$$
(b) Can one find such a constant $c>0$ with an explicit dependence on $N$.
(c) Can one find such a constant $c>0$ which is absolute.
My main interest is when $f$ in the above question is in addition a newform for $\Gamma_0(N)$ with $a_1(f)=1$, but I don't know if this assumption simplifies things. Further,  David Loeffler mentions below that $(f,g)$ is related to a residue of a certain $L$-series at $s=2$. 
 A: In Schulze-Pillot, Yenrice "Petersson Products of bases of spaces of cusp forms" Theorem 11, they mention a bound which might answer b). 
For a newform $f$ of level $N$ it states
$$\langle f,f\rangle\geq \bigg(4\pi \mathrm{e}^{4\pi} N\cdot \prod_{p \mid N}\left(1+\frac{1}{p}\right)\bigg)^{-1}\,.$$
This is achieved by the estimation
$$\int_{\mathcal F} \left|f(z)\right|^2 y^{k-2}\,\mathrm{d}x\,\mathrm{d}y\geq \int_1^\infty \exp \left(-4\pi y\right)\,\mathrm{d}y\,,$$
where $\mathcal F$ is the usual fundamental domain. Consider that $\left[\Gamma_1:\Gamma_0(N)\right]=N\cdot \prod_{p \mid N}\left(1+\frac{1}{p}\right)$ is the normalizing factor for the Petersson inner product.
A: I typed a comment but the formatting wouldn't come out right, so here it is as an answer!
I cannot work out why you expect the "Plancherel or Parseval type" formula to work. Does it not bother you a little that $a_n(f)$ and $a_n(g)$ are perfectly capable of being integers for all $n$, so your series is obviously divergent?
Much better is to consider the series
$$ L(f, g, s) = \sum_{n \ge 1} a_n(f) \overline{a_n(g)} n^{-s},$$
which converges for $Re(s) > 2$ (this is not so easy to see, but it is easy to show that it converges for $Re(s) \gg 0$). This has meromorphic continuation to all $s \in \mathbb{C}$ with a pole at $s = 2$ at which the residue is (maybe up to a normalizing constant depending on $N$ that I have forgotten) the Petersson product $\langle f, g \rangle$.
But this does not help you to get lower bounds on $\langle f, g \rangle$ as far as I can see. Some quite grotty things can happen, e.g. if $f$ is an eigenform and there is another newform $f'$ with $f = f'$ modulo some integer $N$, then one can take $g = (f - f')/N$, and this will be integral but its Petersson product with $f$ will be $\langle f, f \rangle / N$. So the issue of bounding Petersson products below is quite closely related to the issue of congruences between eigenforms.
A: It's not hard to see that the answer to (a) is yes. There is a basis of $S_2^{\textrm{new}}(\Gamma_0(N))$ consisting of newforms. These newforms come into Galois orbits $\{f^\sigma\}_{\sigma}$. Here $\sigma$ runs through the embeddings of $K_f$ into $\mathbf{C}$, where $K_f$ is the field generated by the Fourier coefficients of $f$. A basic fact is that the $\mathbf{C}$-span of a given Galois orbit is generated by cusp forms with integral coefficients. This follows from considering the forms $\sum_{\sigma} \sigma(a) f^\sigma$ where $a$ runs through a $\mathbf{Z}$-basis of the ring of integers of $K_f$.
It follows that if the newform $f$ has integral coefficients, then its orthogonal complement $f^\perp$ is generated by cusp forms with integral coefficients. Now consider the linear map $\lambda_f : S_2(\Gamma_0(N),\mathbf{Z}) \to \mathbf{R}$ given by $\lambda_f(g)=(f,g)$. By the previous remark, the kernel of $\lambda_f$ has rank one less than the rank of $S_2(\Gamma_0(N),\mathbf{Z})$, which implies that the image of $\lambda_f$ is of the form $c_f \mathbf{Z}$ for some $c_f >0$. Thus we can take for $c$ the minimum of all the $c_f$. In fact $c_f = (f,f)/m_f$ for some integer $m_f$ measuring the congruences of $f$ with other cusp forms.
