Are there upper bounds on the L function $|L(E,s)|$ for $|s|<C$?

Question

Given some absolute constant $C$ (In my case, $C=4$ would suffice) and an elliptic curve $E/\mathbb{Q}$, are there upper bounds on $|L(E,s)|$ that are uniform for $|s|<C$? Using the functional equation we see that $|L(E,s)|\gg N_{E/\mathbb{Q}}$ for at least some points $s$, and so would it be possible to prove some sort of result of the type

$$\sup_{\substack{N_{E/Q}<B \\ |s|<C}}|L(E,s)|\ll B\tag{1}$$

where the implied constant depends at most on $C$?

The reason I ask this question is that I have a family of elliptic curves $E_n$, and using some simple facts about reductions mod $p$ I have that $E_n(\mathbb{F}_p)=E_{\infty}(\mathbb{F}_p)$ for every $p<n$ and some "final" curve $E_{\infty}$. As a consequence,

$$\lim_{n\to\infty}L(E_n,s)=L(E_{\infty},s)$$

uniformly on $\Re(s)>3/2$. Now, I would like to construct the power series of $L(E_n,s)$ around, say, $s=7/4$. Because of the convergence of $L(E_n,s)$ to $L(E_{\infty},s)$ in $\Re(s)>\frac{3}{2}$ would locally resemble to $L(E_{\infty},s)$. Moreover, if I had some strong result of the type (1) then I could bound the error and conclude that $L(E_n,1)$ was similar (converges to) to $L(E_{\infty},1)$. Since $E_{\infty}$ has rank 0, we can use Gross-Zagier to conclude that $E_n$ has rank 0 as well (at least for large enough $n$).

This sort of argument by power series feels super powerful, and it makes me really wonder whether or not bounds like (1) exist. If they don't I will try to find some on my own.

Answer 1

Assuming the modularity theorem, apply the maximum modulus principle to $$\Lambda(E,s)=N^{s/2}(2\pi)^{-s}\Gamma(s)L(E,s)$$ which is entire and $\Lambda(E,s)=\pm \Lambda(E,2-s)$ (where $N$ is the conductor).

The Hasse bound gives a bound for $\log L(E,s)$ on $\Re(s)=C+2$, this gives a bound for $\Lambda(E,s)$ on $\Re(s)=C+2$ and $\Re(s)=-C$, and since it is entire and rapidly decreasing as $|\Im(s)|\to \infty$ this gives a bound for $\Lambda(E,s)$ on $\Re(s)\in [-C,2+C]$ depending only on $N$ and $C$.

Given a sequence of elliptic curves $E_j$ with conductor $N_j \to \infty$ then $|L(E_j,-1/2)|\to \infty$.

Answer 2

Here is an (I believe optimal) implementation of reuns' answer, for posterity.

Proposition: For any $\epsilon>0$, we have that $|L(E,s)|\ll_{\epsilon,T_0}N^{1/2+\epsilon}$ within the open set $\frac{1}{2}<\Re(s)<\frac{3}{2}$ and $-T_0<\Im(s)<T_0$. The constant implied by the $\ll$ depends only on $\epsilon$ and $T_0$, and not at all on $E$.

proof:

Choosing any $\epsilon>0$, Hasse's bound implies that $|L(E,s)|\ll_{\epsilon}1$ uniformly on $\Re(s)=\frac{3}{2}+\epsilon$. Applying the functional equation, we get that on uniformly $\Re(s)=\frac{1}{2}-\epsilon$

\begin{align*} |\Gamma(s)L(E,s)|&=\left|\frac{N^{(2-s)/2}(2\pi)^{s-2}\Gamma(s)L(E,2-s)}{N^{s/2}(2\pi)^{-s}}\right|\\ &=\left|N^{1-s}\left(2\pi\right)^{2s-2}L(E,2-s)\right|\\ &\ll_{\epsilon} N^{1/2+\epsilon} \end{align*}

Thus, applying the maximum modulus principle to the open set $\frac{1}{2}-\epsilon<\Re(s)<\frac{3}{2}+\epsilon$, $-T\leq\Re(s)\leq T$ as $T\to\infty$ we get that $\left|\Gamma(s)L(E,s)\right|\ll_{\epsilon}N^{1/2+\epsilon}$. Here we used the fact that $\Gamma(s)$ decays extremely quickly as $\Im(s)\to\infty$, and the L function $L(E,s)$ does not grow at a sufficiently fast rate to stop $\Gamma(s)L(E,s)$. from going to 0.

Now, truncating to some height $|\Im(s)|<T_0$ we get that $|L(E,s)|\ll_{\epsilon,T_0}N^{1/2+\epsilon}$ and we are done.