1
$\begingroup$

Let $f$ be a holomorphic or Maass cusp form for $SL(2,Z)$. Define $L(s,f)=\sum_{n\ge 1}\frac{a_f(n)}{n^s}$, for $\Im s$ sufficiently large. Then $$L(-1+it,f)\ll_f \log^c q(f)t$$ holds, for some conatant $c$ ? Where $q(f)$ is the analytic conductor, $|t|>2$.

We ever see the bound like $L(1/2,f)$, i want to know the corresponding bound for $L(s+it)$ with $\Re s<0$. I search the H. Iwaniec and E. Kowalski's book, it seems there are no records. Please, if some references talk about it, show me their names.

$\endgroup$
1
  • $\begingroup$ Use the functional equation and Stirling's formula. $\endgroup$
    – Lucia
    Oct 14, 2013 at 2:41

1 Answer 1

5
$\begingroup$

This is not true. By standard bounds $|L(2+it,f)|\asymp_f 1$, hence by the functional equation and standard bounds for the gamma function $|L(-1+it,f)|\asymp_f |t|^{3/2}$ for $|t|>1$. Here $A\asymp_f B$ means that $c_1 B<A<c_2 B$ with positive constants $c_1$ and $c_2$ depending on $f$. Note that I normalize $L$-functions to have center $1/2$, as is customary in analytic number theory and in the theory of automorphic forms.

In general, for a principal automorphic $L$-function $L(s,\pi)$ we have the convexity bound $$ L(\sigma+it,\pi) \ll_{\sigma,\epsilon} C(1/2+it,\pi)^{\max(1/2-\sigma,1/2-\sigma/2,0)+\epsilon},\qquad \sigma\in\mathbb{R}, $$ where $C(1/2+it,\pi)$ is the analytic conductor of $L(1/2+it,\pi)$, and this is hard to improve. In particular, this bound is not known in general with the factor $C(1/2+it,\pi)^\epsilon$ replaced by $\log^AC(1/2+it,\pi)$, although it is certainly known in the special case you are considering.

It is known that the Grand Riemann Hypothesis implies the Grand Lindelöf Hypothesis, which would replace $\max(1/2-\sigma,1/2-\sigma/2,0)$ above by $\max(1/2-\sigma,0)$, and the factor $C(1/2+it,\pi)^\epsilon$ by $\log^AC(1/2+it,\pi)$.

$\endgroup$
4
  • 1
    $\begingroup$ Actually the convexity bound on the $1/2$ line does hold without the $C^{\epsilon}$ factor. This is due to Heath-Brown arxiv.org/abs/0809.1752 , which appeared in Acta Arithmetica. $\endgroup$
    – Lucia
    Oct 14, 2013 at 18:25
  • $\begingroup$ @Lucia: Thanks for this comment. Let us mention that Heath-Brown requires some control on the Langlands parameters at the various primes, but these are known by the work of Luo-Rudnick-Sarnak. The bound I displayed is more general, it relies on the work of Molteni (which in turn needs the Luo-Rudnick-Sarnak bounds if I recall correctly). $\endgroup$
    – GH from MO
    Oct 14, 2013 at 19:02
  • $\begingroup$ Brilliant! Thanks for your reply. In IK'book, $(5.20)$ states $L(\sigma+it,f)\ll_{\epsilon,f}C(\frac{1}{2}+it,f)^{\frac{1-\sigma}{2}+\epsilon}$, when $0\le\sigma\le 1$ and $t$ is large. $C^\epsilon$ factor may be not omitted in the convexity bound above. $\endgroup$
    – H.Flip
    Oct 17, 2013 at 8:04
  • $\begingroup$ @Houfei: The bound I stated is stronger and more general than (5.20) that you quote. First, the implied constant only depends on $\sigma$ (not on $\pi$). In fact this dependence is continuous in $\sigma$. Second, I stated it for all $\sigma$ (not just $0\leq\sigma\leq 1$). $\endgroup$
    – GH from MO
    Oct 20, 2013 at 1:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.