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I had an idea on GRH involving a sequence of L-functions having same zeros, then at one step I need a bound on these function and I wonder if this bound is in fact not as hard as GRH itself ?

Let's consider a real primitive Dirichlet character (not principal) $\chi$ of modulo q, and its L-function:

$$L(\chi,s)=\prod_{i=1}^{\infty} (1-\chi(p_i)p_i^{-s})^{-1}$$

For small $\epsilon >0$ we have $|L(\chi,\epsilon)| < M$.

We consider following sequence of L-functions ($p_i$ is the $i^{th}$ prime number):

$$L_N(\chi,s)= L(\chi,s) \prod_{i=1 ; P_i\nmid q}^{N} (1-\chi(p_i)p_i^{-s}) \prod_{j \in (b_1,...,b_k)} (1-\chi(p_j)p_j^{-s})$$

The $(b_1, ...b_k)$ are higher than $N$ and are chosen so that $\prod_{j \in (b_1,...,b_k)} (1-\chi(p_j)p_j^{-\epsilon})$ is sufficiently small (taking enough $p_{b_i}$ such that $\chi(p_{b_i})=1$) to have:

$$|L_N(\chi,\epsilon)|<M$$

Then we have a sequence of L-functions where we have "removed" the first $N$ primes plus some other primes (higher than $p_N$) so that the $L_N(\chi,s)$ functions have same zeros in the critical strip as $L(\chi,s)$ with their sum bounided by the same constant $M$.

Now each partial sum of the L-functions defined are bounded but can we find a general bound for all these partial sums? In other words can we find a constant $K$ such that :

Calling $a_{N,n}$ the coefficient of the Dirichlet $L_N(\chi,s)$ function (these coefficient are in fact coefficient of an induced charcter coming from $\chi$), so that:

$$L_N(\chi,s)= \sum_{n=1}^{\infty} \frac{a_{N,n}}{n^s}$$

We have for all $N$:

$$|\sum_{n<x} \frac{a_{N,n}}{n^{\epsilon}}| < K x^{\epsilon}$$

So is this type of bound finally as hard as GRH itself? Or is this type of bound not true? (note that choice of the $b_i$ provides flexibility to find a good sequence)

Perron's formula can maybe answer the question but up to now the bound I found are not satisfactory. Any idea ?

These partial sums area sums on integer free of small primes but I did not find any good reference. Also no reference on such sequence of L-function having same zeros and their properties. I made previous posts linked to this question:

Evolution of partial sum of a sequence of induced Dirichlet characters and

https://mathoverflow.net/questions/234059/abel-summation-formula-versus-perrons-formula-to-bound-a-partial-sum

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  • $\begingroup$ if I understood well, $L_N(s,\chi) = L(s,\chi_2)$ where $\chi_2$ is a (non-principal and non-primitive) character modulo $lcm(q,p_{b_1},\ldots,p_{b_k}) = q \prod_{j=1}^k p_{b_j}$ ? and it is typically a character sum problem ? $\endgroup$
    – reuns
    Commented Mar 20, 2016 at 21:39
  • $\begingroup$ a simple question : do you have reasons to think that for a fixed $K$, can we find characters modulo $q$ with $q$ arbitrary large, such that $|\sum_{n \le x} \chi(n)| \le K$ ? doesn't the functional equation tell us that $\max_x |\sum_{n \le x} \chi(n)|$ is somewhat related to the modulus of the Gauss sum $|G(\chi)| = |\sum_{n=1}^q \chi(n) e^{2 i \pi n/q}|$, which should be somewhat related to $q$, because the discrete Fourier transform of a $\chi$ is $G(\chi) \overline{\chi}$ ? $\endgroup$
    – reuns
    Commented Mar 20, 2016 at 21:52
  • $\begingroup$ If χ′ is a Dirichlet character modulo N′ such that N and N′ are relatively prime, then $G(\chi\chi^\prime)=\chi(N^\prime)\chi^\prime(N)G(\chi)G(\chi^\prime)$, so in the case of induced characters like mentionned in my question, the Gauss sum of the induced character (induced with principal character of Gauss sum of module 1), I think the Gauss sum does not increase even if the modulus increases. $\endgroup$
    – Bertrand
    Commented Mar 20, 2016 at 22:05
  • $\begingroup$ here $X$ is your character modulo $q$, and $X'$ is the principal character modulo $\prod p_{b_j}$ ? hence $G(\chi \chi') = ?$ $\endgroup$
    – reuns
    Commented Mar 20, 2016 at 22:08
  • $\begingroup$ Make a recurrence: at each step a new character is induced using the principal character modulo $p_i$ which as a Gauss sum of module 1, so, by formula of Gauss sum of characters with modulus relatively prime I think the Gauss sum of the new character does not increase. $\endgroup$
    – Bertrand
    Commented Mar 20, 2016 at 22:23

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