# Convexity Bound of Rankin-Selberg L-Function

Let $$f,g$$ be primitive modularforms of arbitrary levels $$N_1,N_2$$, trivial nebentypus and same weight $$k$$. Let $$L(f\otimes g,s)=\zeta(2s)\sum_{n\geq1}\frac{\lambda_f(n)\lambda_g(n)}{n^s}$$ be the Rankin-Selberg L-Function where $$\lambda_f$$ and $$\lambda_g$$ are the respective normalized Hecke-Eigenvalues.

I'm looking for a simple upper bound of $$L(f\otimes g,\frac12+it)$$ in the level aspect which holds under general assumptions as given above. In nearly every piece of literature on subconvexity bounds (e.g. Introduction of "Rankin Selberg L-functions in the level aspect" by Kowalski, Michel, Vanderkam) there is the hint that the convexity bound ($$\sqrt N$$ in the level aspect, $$k$$ in the weight aspect) is achieved by applying the Phragmén-Lindelöf principle. As far as I understood, this requires another bound of the function on the edges of the vertical strip $$0<\Re(s)<1$$. So my questions are

• Is there an elementary proof to this problem?
• What bound can I use for $$L(f\otimes g, 1)$$?
• How do I take care of the residue in $$s=1$$ for $$f=\overline{g}$$?
• I'd say : add finitely many Euler factors to $L(f\otimes g, s)$ to obtain a Dirichlet series $H(s)$ with functional equation. Obtain a bound $H(1+it) = O(\log^m(t))$ (in the same way you showed the PNT for $L(f,s)$ ?). Use the functional equation to obtain a bound for $H(it)$. Apply the convexity theorem. I don't know if the implied constants are effective. What the found for $L(f\otimes g,\frac12+it)$ would be useful for ? – reuns Nov 11 '18 at 21:55
• Thanks for your comment. I use the bound as a factor for a more general result which applies not only to newforms but to arbitrary cusp forms. The readers can insert a stronger bound if they want quantitatively better results but I just need an elementarily introduced placeholder instead of a reference to an article of 30 pages about any subconvexity bound. – Nodt Greenish Nov 12 '18 at 10:21
• By the way, $L(s,f \otimes g)$ is not equal to $\sum_{n = 1}^{\infty} \frac{\lambda_f(n) \lambda_g(n)}{n^s}$. Rather, it is equal (up to some Euler factors at bad primes) to this Dirichlet series times $L(2s,\chi_f \chi_g)$, where $\chi_f, \chi_g$ are the nebentypen of $f$ and $g$. – Peter Humphries Nov 12 '18 at 13:06
• Alright, so I need a factor of $\zeta(2s)$. Thanks for that reminder @PeterHumphries, I will watch out for that in my notation. – Nodt Greenish Nov 12 '18 at 13:52
• @reuns I define $H(s)=(2\pi)^{-2s}\Gamma(s+k-1)\Gamma(s)L(s,f\otimes g)$ which allows the functional equation $H(1-s)=H(s)$ for $f\neq g$. So for $L(f\otimes f,s)$ I need another estimation... – Nodt Greenish Nov 12 '18 at 15:17

We use the absolute convergence of $$L(f,s)$$ in $$\sigma>1$$ (or the bound $$|\lambda(n)|\leq \tau(n)$$), the functional equation, Stirling's estimate for the gamma function and the Phragmen-Lindelöf convexity principle.