Upper bound for L-series of modular form from its integral representation

Question

I've read that if $f$ is a cuspidal modular form (that's also an eigenfunction for the Hecke operators) for $SL_{2}(\mathbb{Z})$ of weight $k$, then its L-series $L(w,f)$ satisfies the bound

$$L(w,f) << e^{|w|^{1+\epsilon}}$$

for any $\epsilon > 0$ and $w$ outside of the critical strip. I think I can deduce this from the integral representation $$L(w,f) = \frac{(2\pi)^{w}}{\Gamma(w)}\left[\int_{1}^{\infty}f(iy)y^{w}\frac{dy}{y}+i^{k}\int_{1}^{\infty}f(iy)y^{1-w}\frac{dy}{y}\right]$$ but I'm not exactly sure how. I originally thought about replacing $f(iy)$ with its Fourier series and using Hecke's bound for the fourier coefficients $A_{n}$ (I've let $w = s+(k-1)/2)$ so the Fourier coefficients are constant on average) to control the $f(iy)$ in the integrand and then rewrite both of the integrals as gamma functions but this was ultimately uncessful. There's also the issue of convergence of the infinite series when I interchange the sum and integral. Is there a simple way to get this bound from the integral representation that I'm missing? I can't find much about bounding $1/\Gamma(w)$ on a strip so maybe this is what I would need.

Answer 1

Here is one way to make this rigorous.

We consider bounds in any fixed vertical strip $1 - A \leq \mathrm{Re} s \leq A$ for $A > 0$. It suffices to bound $$ \int_1^\infty f(iy) y^w \frac{dy}{y}, $$ as the other integral is essentially the same. As $f$ is a cuspform, we have that $f(iy) y^A \to 0$ very rapidly as $y \to \infty$ for any fixed $A \in \mathbb{R}$. In particular, for any $\epsilon > 0$ there is some $H = H(A, \epsilon)$ such that $$\lvert f(iy) y^A \rvert \leq \epsilon \qquad (\forall \, y \geq H)$$ and such that $$ \int_H^\infty \lvert f(iy) y^A \rvert \frac{dy}{y} < \epsilon.$$ This type of argument is typically described more when describing that the integral for $\Lambda(s, f)$ converges and has meromorphic continuation.

As $f(iy) y^A$ is continuous and $[1, H]$ is compact, it follows that there is some constant $M = M(A, \epsilon)$ such that $$ \lvert f(iy) y^A \rvert \leq M \qquad (\forall \,y).$$

Thus $$\Big \lvert \int_1^\infty f(iy) y^w \frac{dy}{y} \Big \rvert \leq \int_1^H \lvert f(iy) y^A \rvert \frac{dy}{y} + \epsilon \leq HM + \epsilon.$$

Applied to both integrals, this shows that $\Lambda(w, f)$ is absolutely bounded in vertical strips. We note that this agrees with expectation: we know that this integral behaves roughly like $\Gamma(w) L(w, f)$, which decays very rapidly in vertical strips (due to the gamma function).

Now that we have these bounds in any vertical strip for $\Lambda(w, f)$, we need to recover the statement for $L(w, f)$. This amounts to understanding the reciprocal Gamma function. Fortunately, we know that $1/\Gamma(w)$ has the Weierstrass factorization $$ \frac{1}{\Gamma(z)} = z e^{\gamma z} \prod_{n \geq 1} \big( 1 + \tfrac{z}{n} \big) e^{-z/n}, $$ where $\gamma$ is the Euler-Mascheroni constant. This is in fact a Hadamard factorization and also shows that $\Gamma(w)^{-1}$ is an entire function of order one, and so in particular $$ \frac{1}{\Gamma(w)} \ll e^{\lvert w \rvert^{1 + \epsilon}}. \tag{1}$$

It follows that in any vertical strip, we also have that $L(w, f)$ satisfies the bound $(1)$ (where the implicit constants depend on the strip and $\epsilon$, and essentially depend on what we called $H$ and $M$ above). This isn't quite what you want, but it's enough to get you there.

This bound is sufficient to guarantee the Phragmen-Lindelof convexity principle in vertical strips. In essence, this now allows us to use trivial bounds for $L(w, f)$ for $\mathrm{Re} w > 1$ and $\mathrm{Re} w < 0$ (via the functional equation) to get polynomial bounds of the shape $$ \lvert L(\sigma + it, f) \rvert \ll (1 + t)^{\mu(\sigma)} $$ for an explicit function $\mu$, and where the implicit constants here depend only on easy bounds for $L(w, f)$ (and not on $H$ or $M$ above). In particular, it comes from a trivial bound for $L(1 + \epsilon, f)$ (in the region of absolute convergence) and the constants associated to the growth of the Gamma function (for application to the functional equation).

The function $\mu(\sigma)$ is given for any $\epsilon > 0$ by $$ \mu(\sigma) = \begin{cases} 0 & \sigma > 1, \\ 1 - \sigma + \epsilon & 0 < \sigma < 1, \\ -2 \sigma & \sigma < 0. \end{cases} $$

Collecting these together shows that $L(w, f)$ and $\Lambda(w, f)$ are integral functions of order one, equivalent to your goal bound.

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I'm not sure if there is a more direct way. This method applies to all $L$-functions.

Answer 2

I'd use that $a_n(f)=O(n^k)$

• On $\Re(s) \ge k+2$, $L(s,f)=O(1)$,

then $(2\pi)^{-s} \Gamma(s) L(s,f)$ is invariant under $s\to k-s$ and $\frac{\pi}{\sin(\pi s)} = \Gamma(s)\Gamma(1-s)$ so that

$$L(k-s) = (2\pi)^{k-2s} \Gamma(s)\frac{\sin(\pi (k-s))}{\pi} \Gamma(s+1-k) L(s,f)$$ which is $O(e^{3|s \log s|})=O(e^{|s|^{1+\epsilon}})$.

• On $\Re(s)\in [-2,k+2]$ since $f$ is a cusp form the integral gives that $(2\pi)^{-s} \Gamma(s) L(s,f) = O(1)$ which gives that $L(s,f)=O(e^{\pi |s|/ 2}(1+|s|)^r)=O(e^{|s|^{1+\epsilon}})$.