Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\varepsilon>0$ be small and \begin{align*} \widetilde{F}_{\varepsilon}(\xi):=\frac{4}{\varepsilon}\sum_{0<\gamma\leq \varepsilon^{-2}}\frac{\sin(\gamma \xi)\sin \frac{\gamma \varepsilon}{2}}{\gamma^2}, \end{align*} where the sum is taken over the imaginary parts $\gamma$ of the zeros of $L(s,\chi_1)$, and $\chi_1$ is the real Dirichlet character $\pmod q$ for a fixed $q$ (we may assume the generalized Riemann hypothesis if that is of any help in what follows). I am reading Rubinstein and Sarnak's article Chebychev's Bias (conserning the unevenness of distribution of primes into residue classes), where they need lower bounds for this quantity. First, the estimate $\widetilde{F}_{\varepsilon}(\frac{\varepsilon}{2})\gg \log \varepsilon^{-1}$ is easy: \begin{align*} \widetilde{F}_{\varepsilon}(\frac{\varepsilon}{2})=4\varepsilon \sum_{0<\gamma\leq \varepsilon^{-2}}\left(\frac{\sin\frac{\gamma \varepsilon}{2}}{\gamma \varepsilon}\right)^2\geq 4\varepsilon \sum_{0<\gamma\leq \varepsilon^{-1}}\frac{1}{3^2}\gg \log \varepsilon^{-1}, \end{align*} where we just discarded most of the terms, which are non-negative, and used the asymptotic formula for the number of zeros of the $L$-functions. But then the paper claims that the estimate $\widetilde{F}_{\varepsilon}(\varepsilon)\gg \log \varepsilon^{-1}$ follows trivially. Nevertheless, I cannot see at all how to deduce that claim.

A natural attempt to bound $\widetilde{F}_{\varepsilon}(\varepsilon)$ would be to use $\sin 2x=2\sin x \cos x$ to get \begin{align*} \widetilde{F}_{\varepsilon}(\varepsilon)=\frac{8}{\varepsilon}\sum_{0<\gamma\leq \varepsilon^{-2}}\frac{\sin^2 \frac{\gamma \varepsilon}{2}\cos \frac{\gamma \varepsilon}{2}}{\gamma^2}. \end{align*} However, as this sum goes all the way to $\varepsilon^{-2}$, we cannot just discard terms to use our trivial inequality for $\widetilde{F}_{\varepsilon}(\frac{\varepsilon}{2})$. Indeed, It seems that no simple estimate could work because of the oscillatory nature of the summands and the fact that if the sines and cosines are estimated to $1$ in the range $[\varepsilon^{-1},\varepsilon^{-2}]$, the bound we get for the sum in this range is already of the same order as the sum over the range $[0,\varepsilon^{-1})$.

After various attempts, the only approach I found that might work would be to use an asymptotic formula for $\gamma_n$, i.e. the $n$th zero, with error of order $O(\frac{1}{\log n})$, and then we would end up with some very complicated trigonometric sum, though there would no longer be a summation over the zeros. Then we could use partial summation to the new sum, and hopefully get a bound $o(\log \varepsilon^{-1})$ for the sum over the range $[\varepsilon^{-1},\varepsilon^{-2}]$. Is there some (much) simpler way to get the bound, as Rubinstein and Sarnak assert? If not, how could one prove $\widetilde{F}_{\varepsilon}(\varepsilon)\gg \log \varepsilon^{-1}$?

EDIT: It is of course okay to propose a different way of fixing the issue in the paper, though I would still be curious to know how to prove such an estimate.

share|improve this question
1  
I would think that they overlooked something, which is on the other hand not serious. That is, instead of (2.15) I would proceed with $\widetilde{F}_{2\varepsilon}(\varepsilon)\gg \log \varepsilon^{-1}$ and see if the argument goes through with little alterations. –  GH from MO Mar 24 at 20:48
1  
Yes, it seems that it works! Lemma 2.4 might change a little bit but I can't see any substantial difference. –  Joni Teräväinen Mar 25 at 7:14
2  
@GHfromMO I consulted one of the authors, and he says that there is a typo in the line in question, and your way is the right way to go, by choosing the $\xi$ variable, and then using almost periodicity (the OP seems to have confirmed that this does, indeed, work). –  Igor Rivin Mar 25 at 11:39
    
I am glad it works that way! –  GH from MO Mar 25 at 12:20

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.