Jérémy Dousselin

Zeros of linear combinations of Dirichlet L-functions on the critical line

Let F be a linear combination of N ≥ 1 Dirichlet L-functions attached to even (or odd) primitive characters. Selberg proved that a positive proportion of non-trivial zeros of F lie on the critical line. Our work here is to provide an explicit lower bound for this proportion. In particular, we show that the lower bound 2.16 × 10 -6 /(N log N ) is admissible for large N .

1 Introduction

Zeros of the Riemann zeta-function on the critical line and the mollification method

In 1838, Dirichlet introduced L-functions as a tool to study primes in arithmetic progression. Soon after, mathematicians understood how central these functions are in number theory, and especially how important it is to understand the distribution of their zeros. For Dirichlet Lfunctions, it is known that their non-trivial zeros lie in a strip, but a crucial hypothesis states that all of these zeros should actually lie on the ℜs = 1/2 line. Many ideas exist to study their zeros on this critical line, and one of them is to study the proportion of an L-function's zeros in the critical strip that are actually on the critical line. Showing that this proportion is non-zero is a thorny question answered by Serlberg. The proof relies on a then recent method: the mollification method.

Let N(T ) be the number of zeros of the Riemann zeta-function ζ(s) that lie in the rectangle {s ∈ C : 0 ≤ ℜ(s) ≤ 1, 0 ≤ ℑ(s) ≤ T }, and let N 0 (T ) be the number of these zeros on the critical line. In 1914, Hardy was the first to prove that infinitely many zeros of ζ(s) are on the critical line. Seven years later, he and Littlewood improved on this [START_REF] Hardy | The zeros of Riemann's Zeta-Function on the critical line[END_REF] and showed that we have N 0 (T ) ≫ T for all large T . However, since N(T ) ≍ T log T (see Theorem 5.24 of [START_REF] Iwaniec | Analytic Number Theory[END_REF]), this lower bound is not strong enough to show that the proportion of non-trivial zeros lying on the critical line (we will call them critical zeros) is positive. It was Selberg [START_REF] Selberg | On the zeros of Riemann's zeta-function[END_REF], in 1942, who refined their argument by studying the zeros of ζ(s)|η(s)| 2 instead of ζ(s), where η(s) is a "mollifier". The role of this mollifying function is to prevent large values of ζ(s) from contributing too much, and hence making it possible to get sharper estimates. We will detail this mollifying process through the next section, when several notations are introduced. With this idea, originally due to Bohr and Landau [START_REF] Bohr | Sur les zéros de la fonction ζ(s) de Riemann[END_REF], Selberg was able to prove that N 0 (T ) ≫ T log T, which proves that κ := lim inf T N 0 (t)/N(T ) > 0. His method actually produces a very small amount of zeros on the critical line, and hence yields a very small lower bound of κ. While we could not find the original proof, it is said (see page 68 of [START_REF] Zhang | A pioneer of analytic number theory in China[END_REF] and §10.9 of [START_REF] Titchmarsh | The theory of the Riemann zeta-function[END_REF]) that Szu-Hoa Min computed the constant given by Selberg's method and proved that κ ≥ 1/60000. This constant have been improved several times, still using the mollification method but not in the

way Selberg did. The first breakthrough in this direction is due to Levison [START_REF] Levinson | At Least One-Third of Zeros of Riemann's Zeta-Function are on σ = 1/2[END_REF] who proved that κ ≥ 1/3, a result then refined by Conrey who showed that κ ≥ 2/5. Currently the record is hold by Pratt, Robles, Zaharescu and Zeindler who showed in 2019 that κ ≥ 5/12 [START_REF] Pratt | More than five-twelfths of the zeros of ζ are on the critical line[END_REF]. In a few words, Levinson's and Selberg's methods are radically different. While Selberg's method detects very little zeros on the line, it is a safe method, for it detects zeros as the sign change of a real function: there is no risk of getting a negative lower bound. However, as explained through Section 3 of [START_REF] Conrey | Critical zeros of lacunary L-functions[END_REF], Levinson's method is more of a gamble. If the mollification is perfect, then one could potentially reach the 100% lower bound. However, too crude estimates would lead to a negative bound for the counting of critical zeros. The counterpart of that risk is the reward, the method producing very high lower bound for the proportion of critical zeros. Note that the original proof of Levison is rather complicated, but a simpler one was given by Young in 2010 [START_REF] Young | A short proof of Levinson's theorem[END_REF].

In the case of a more general Dirichlet L-function both Selberg's method and Levinson's method still work, and the proof is not so different. Conrey, Iwaniec and Soundararajan [START_REF] Conrey | Critical zeros of Dirichlet L-functions[END_REF] went a step further, and used Levison's method to prove that at least 56% of zeros of the family of Dirichlet L-functions are on the critical line. This statement is to be understood as a sort of double average over both the t-aspect and the q-aspect of a Dirichlet L-function, mostly focusing on the latter.

The case of linear combinations of Dirichlet L-functions

The study of zeros of linear combinations of L-functions is motivated by the existence of certain zeta functions (called Epstein zeta functions) which satisfy all properties of L-functions, except the existence of an Euler product representation. It turns out that these zeta functions possess non-trivial zeros off the critical line and hence do not satisfy the Generalized Riemann Hypothesis. However, these zeta functions can be expressed as linear combinations of Hecke L-functions, which is why several number theorists investigated the zeros of general linear combinations of L-functions.

Montgomery conjectured that under natural conditions (notably a certain notion of independence of these L-functions), 100% of the zeros of a linear combination of L-functions lie on the critical line (which implies that the counterexamples to the Generalized Riemann Hypothesis are rare). In their celebrated paper, Bombieri and Hejhal [START_REF] Bombieri | On the distribution of zeros of linear combinations of Euler products[END_REF] proved -under some reasonable, but yet to be proven, hypotheses-that this is true.

Unconditionally, Karatsuba [START_REF] Karatsuba | On the zeros of the Davenport-Heilbronn function lying on the critical line[END_REF] tried to tackle the question and proved that for a specific linear combination of two L-functions, we have N 0 (T ) ≫ T (log T ) 1/2-ε , which narrowly fails to show that the proportion of critical zeros is positive. Surprisingly, a few years later Selberg proved [START_REF] Selberg | Zeros on the critical line of linear combinations of Euler products[END_REF], [START_REF] Selberg | Linear combinations of L-functions and zeros on the critical Line[END_REF] that his method actually applies in the case of "any" linear combination of Dirichlet L-functions. Recently in 2016, Rezvyakova [START_REF] Rezvyakova | On the zeros of linear combinations of L-functions of degree two on the critical line. Selberg's approach[END_REF] proved that Selberg's method also applies in the case of a linear combination of L-functions of degree two attached to automorphic forms.

Our goal here is to provide an explicit lower bound for the proportion of zeros on the critical line for F , a linear combination of Dirichlet L-functions attached to even characters. Apart from trivial zeros implied by the functional equation, the zeros of F lie in a vertical strip. Indeed, F (s) has a series representation for ℜ(s) > 1. Since the first non-zero term dominates the others, F (s) has no zero if ℜ(s) is large enough, and hence we may define σ F := sup{ℜ(s) : F (s) = 0}.

From the functional equation satisfied by F (s), we may deduce the set of zeros on ℜ(s) < 1-σ F , called trivial zeros. The other zeros are confined in the strip 1 -σ F ≤ ℜ(s) ≤ σ F , and we call them non-trivial zeros. We denote by N(T, F ) the number non-trivial zeros with imaginary part in (0, T ), and by N 0 (T, F ) the number of these zeros that are on the critical line. We also write κ F := lim inf T N 0 (2T, F ) -N 0 (T, F ) N(2T, F ) -N(T, F ) .

In order to give an unconditional lower bound for κ F , we will use the mollification method à la Selberg to prove the following:

Theorem 1.1: Let F be a linear combination of N ≥ 1 distinct Dirichlet L-functions attached to even characters. Then for any A > 1/κ, any κ ∈ (0, 1/8) and any θ ∈ (0, 1):

κ F ≥ 2π 1 2A -4N C 1 (A) + C 2 A 3 .
Here, C 1 (A) and C 2 are defined in Theorem 2.1 and implicitly depend on θ and κ. If N = 1, this can be improved by (2.9).

From this, some computations and optimisations lead to the following, more explicit theorem.

Theorem 1.2: Let F be a linear combination of N ≥ 1 distinct Dirichlet L-functions attached to even characters. For small N's, we get the following lower bounds for κ F , as well as the corresponding A and θ. Moreover, when we are interested in very large N ≥ 3, we may state the following: for any given 0 < ε < 1/3, we have

N ≥ 2.9 × 10 -11 ε 3 =⇒ κ F ≥ 2.161 × 10 -6 N log N 1 -3ε - 1.14 log log N log N - 862 log N - 1.2 × 10 7 ε log 2 N .
Note that the case of odd characters is handled similarly, and we only deal with the even ones for the sake of simplicity.

Remark 1.3: Selberg [START_REF] Selberg | Zeros on the critical line of linear combinations of Euler products[END_REF], [START_REF] Selberg | Linear combinations of L-functions and zeros on the critical Line[END_REF] only proved that κ F ≥ c/N 2 , for some c > 0. However, he mentioned that he could strengthen the lower bound to c/(N log N), but he did not provide any idea on how one can achieve such a result. By getting a better error term that him in a crucial lemma (compare our Lemma 4.2 to the Lemma 9 of [START_REF] Selberg | On the zeros of Riemann's zeta-function[END_REF]), we were able to retrieve the result. We will give more technical details later in the proof, but essentially Selberg's estimate corresponding to our lemma was given with an error term of O( √ A), while our error term is O(1). This made him find C 1 (A) ≍ A 3/2 instead of our C 1 (A) ≍ A log A, which explains the improved lower bound.

Acknowledgement: I would like to thank my PhD advisor Youness Lamzouri for suggesting this problem to me, for his time and guidance, and for several improvement suggestions. Un remerciement à Pascal Ciot pour une aide bibliographique bien utile.

Notation and outline of the proof

Now and for the rest of the paper, we write f (T ) g(T ) to say that f (T ) ≤ g(T )(1 + o(1)). Similarly, f (T ) g(T ) means that g(T ) f (T ). The o symbol is to be understood as valid when T → ∞. Several times throughout the paper, the constant implied by the O-symbols may depend on the conductors of the considered L-functions, and we will never explicit this dependence. We will also write f = O(g) if f g. We define log + (x) = max(0, log x).

Let N ≥ 1 be an integer, χ 1 , ..., χ N be even distinct primitive Dirichlet characters to moduli q 1 , ..., q N respectively. We know (see section 5.1 of [START_REF] Iwaniec | Analytic Number Theory[END_REF]) that there is a complex number ε j with

|ε j | = 1 such that φ(s, χ j ) := ε j π -s/2 q s/2 j Γ s 2 L(s, χ) satisfies φ(s, χ j ) = φ(1 -s, χ j ) for all 1 ≤ j ≤ N. Write F (s) := N j=1 c j q 1/4 j ε j q s/2 j L(s, χ j ),
where c j ∈ R * 1 . Standard arguments show (see Theorem 5.8 of [START_REF] Iwaniec | Analytic Number Theory[END_REF] for the case of a single L-function) that

N(2T, F ) -N(T, F ) = T 2π (log T + B F ) + O(log T ), (2.1)
for some B F ∈ R.

The idea is to count zeros of F on the critical line thanks to sign changes of a good function. Let 1 ≤ j ≤ N, and write

ϑ(s) := arg π -s/2 Γ s 2 and X j (s) := ε j q s/2-1/4 j e iϑ(s) L(s, χ j ).
This function X j , thanks to the functional equation given by φ, takes real values on the critical line. Moreover, 1/2 + it is a zero of X j if and only if 1/2 + it is a zero of L. Thus, any sign change of X j (1/2 + i•) implies the existence of a zero of L on the critical line. 1 We chose to normalize each summand by q 1/4 j because q s/2-1/4 j is real on the critical line. This has no impact on the proof since c j are arbitrarily chosen.

Let T be a large real number, and we define (α j (n)) by

L(s, χ j ) -1/2 = n α j (n) n s , for ℜ(s) > 1.
For T ≤ t ≤ 2T , ξ = T κ with κ > 0 to be chosen later 1 , we write

η j (s) := n≤ξ α j (n) n s M(n) =: n β j (n) n s ,
where M is the continuous function

M(x) := M(ξ, x) := 1 log ξ    log ξ if 1 ≤ x ≤ ξ θ , 1 1-θ log(ξ/x) if ξ θ < x ≤ ξ, 0 if x > ξ,
where 0 < θ < 1 is a parameter to be chosen later. Also note that

|β j (n)| ≤ |α j (n)| ≤ 1 since α j (n) = τ -1/2 (n)χ j (n).
Here τ z is the z-th divisor function, defined to be the coefficient of the Dirichlet series of ζ z , satisfying the inequality

|τ z (n)| ≤ τ |z| (n).
Selberg worked with a slightly different function in [START_REF] Selberg | On the zeros of Riemann's zeta-function[END_REF], choosing the continuous function

M Sel (ξ, x) := 1 log ξ log(ξ/x) if 1 ≤ x ≤ ξ, 0 if x > ξ.
This will not heavily impact the rest for the proof, for we have

M(ξ, x) = 1 1 -θ M Sel (ξ, x) -θM Sel (ξ θ , x) . (2.2)
Selberg [START_REF] Selberg | Linear combinations of L-functions and zeros on the critical Line[END_REF] suggested to use this function in the case θ = 1/2 instead of M sel . This is also done in [START_REF] Heath-Brown | Zeros of the riemann zeta-function on the critical line[END_REF], [START_REF] Rezvyakova | On the zeros of linear combinations of L-functions of degree two on the critical line. Selberg's approach[END_REF]. The pros of working with this function are that the upper bound (2.7) below does not depends on N, making it better than the original one found by Selberg in [START_REF] Selberg | On the zeros of Riemann's zeta-function[END_REF], when N is large. Furthermore, the proof of (2.7) is also greatly simplified. The drawback of using M instead of M sel is that the constant term C 2 is rather large, which worsen our lower bound for small N's. See for example the case N = 1, where the lower bound κ ζ ≥ 1/60000 was found admissible, while our lower bound for N = 1 is roughly 10 -7 . However, these results for N = 1 are both vastly improved by Levinson's, and hence the real interest of our Theorem is when N > 1.

One of our improvements is to introduce the parameter θ, which has no impact on the complexity of the proof but has a strong impact on the quality of our lower bound. As the computations of Theorem 1.2 show, it seems that the optimal θ is relatively small, and that it gets smaller as N gets larger. Now, we introduce three integrals that we will use to detect sign changes of X j on the critical line. Let H = A/ log T , where A > 1/κ is a quantity depending only on N, κ and θ to be chosen later. We define 1 One may keep in mind that κ can be taken as close as we want to 1/8, as we will show later.

I j (t, H) := t+H t X j (1/2 + iu)|η j (1/2 + u)| 2 du,

Jérémy Dousselin

NOTATION AND OUTLINE OF THE PROOF

J j (t, H) := t+H t |X j (1/2 + iu)η 2 j (1/2 + u)| du,
and

M j (t, H) := t+H t L(1/2 + iu, χ j )η 2 j (1/2 + iu) du -H.
It is clear that

J j (t, H) ≥ H -|M j (t, H)|, (2.3)
thus if |M j (t, H)| + |I j (t, H)| < H, (2.4)
we get that

J j (t, H) > |I j (t, H)|, (2.5)
which immediately implies that there is a sign change of X j (1/2 + iu) in (t, t + H), and hence a zero of L(1/2 + it, χ j ) in (t, t + H). This detection method works well on short intervals at one condition: the function must not have very large spikes. Indeed, if the curve of X j were composed of one large spike and many smaller spikes, then the area under this curve would be concentrated under the larger spike, making all the remaining area almost undetectable by integral computations, therefore making I j and J j very close even though there could be many sign changes. That is why we mollify the function by a positive factor, so that we get a new function with the same sign changes but much less large oscillations. Without mollification, we would still be able to detect some sign changes, but not accurately enough on such small (of length H ≍ 1/ log T ) intervals. Now to exploit (2.4) and (2.5), we rely on the following estimates, which are the main results of this paper.

Theorem 2.1: Suppose that κ < 1/8 and 0 < θ < 1. Recall that H = A/ log T . Then we have

2T T |I j (t, H)| 2 dt C 1 (A) T (log T ) 2 , (2.6)
Zeros of linear combinations of Dirichlet L-functions on the critical line

and 2T T |M j (t, H)| 2 dt C 2 T (log T ) 2 .
(2.7)

Here,

C 3 = 1 8κ + 3 2 e ̺ θ + e ̺ θ θ (1 -θ) √ ̺ θ π Γ(1/4) Γ(3/4) 4 p 1 + 3p 2 -3p + 1 p 4 -3p 3 + 3p 2 -p , (2.8)
with ̺ θ being the only positive solution of -1 + 2θx + e x(1-θ) (-1 + 2x) = 0, and

C 2 = 6(C 3 + 1 + 2 √ C 3 ) (θκ) 2 .
Furthermore,

C 1 (A) = 8C 2 5 (K 1 A log A + K 2 A + K 3 log A + K 4 )
, where K's are given in (8.20) and under.

Remark 2.2: Rather than focusing on the complex and lengthy explicit expression for C 1 , one should note that C 1 is around A log A (multiplicative constant aside) when A is large. This is the important piece of information conveyed by the expression above. Also note that the estimate (2.7) holds whenever κ < 2/9.

To illustrate how the method works, we will apply it to a single L-function called F . Since we are working with a single L-function, we drop the indices j for simplicity, and we will do it several times later without mentioning this again. Let S be the subset of (T, 2T ) consisting of t's such that |I(t, H)| = J(t, H). Then , where meas is Lebesgue's measure on R. Moreover, by (2.3) and (2.7), we find that

S J(t, H) dt meas(S)H -meas(S)C 2 T (log T ) 2 1/2 .
Thus by combining these we find meas(S)

( C 1 (A) + √ C 2 ) 2 A 2
T.

Now we divide (T, 2T ) into ⌊T /(2H)⌋ pairs of abutting intervals I 1 , I 2 of length H each (except maybe for the last I 2 ). For each pair, there is at least one critical zero of F in I 1 or I 2 unless I 1 ⊂ S. If n 1 is the number of these bad intervals I 1 , then we find

n 1 × H ≤ meas(S) ( C 1 (A) + √ C 2 ) 2 A 2 T,
3 HOW TO HANDLE THE GENERAL CASE and hence

n 1 ( C 1 (A) + √ C 2 ) 2 A 3 T log T.
Therefore there are at least

T 2H -n 1 1 2A - ( C 1 (A) + √ C 2 ) 2 A 3
T log T zeros of F on the critical line in (T, 2T ). By (2.1), this implies that

κ F ≥ 2π 1 2A - ( C 1 (A) + √ C 2 ) 2 A 3
.

(2.9)

Optimising the parameters θ, κ and A would imply the result in the case of a single L-function.

For a general linear combination of L-functions, the idea is to prove that the interval (T, 2T ) can be split into "good" subsets where one of the L-functions dominates the others, and "bad" subsets whose measure we can control. Then, on the good subsets we can work as if we only had a single L-function. We will explain this with further details in the next section.

Remark 2.3: This lower bound is slightly stronger than the general one presented in Theorem 1.1, and the loss comes from a slight technical difficulty arising in the case of a combination of multiple L-functions.

How to handle the general case

We first suppose that Theorem 2.1 is true and we detail how to use it to prove Theorem 1.1. We follow the main lines of approach of Selberg [START_REF] Selberg | Linear combinations of L-functions and zeros on the critical Line[END_REF]. First, we need the following version of Selberg's central limit theorem for a difference of two distinct L-functions.

2T T 1 [a,b] log |L(1/2 + it, χ 1 )| -log |L(1/2 + it, χ 2 )| √ 2π log log T dt = T b a e -πu 2 du+O T log log log 2 T √ log log T .
Proof: Tsang proved a similar result for the Riemann zeta-function in his thesis [START_REF] Tsang | The distribution of the values of the Riemann zeta function[END_REF]. Very few modifications are needed to adapt the proof to the case of a difference of two Dirichlet L-functions.

We will also need the following lemmas.

Lemma 3.2: Fix χ an even Dirichlet character, let k ≥ 1 be fixed, and let x = T 1/(100k) . We have that

2T T log |L(1/2 + it, χ)| -ℜ p<x χ(p) p 1/2+it 2k dt ≪ T B k k 4k ,
Zeros of linear combinations of Dirichlet L-functions on the critical line for some constant B > 0.

Proof: Again, Tsang proved it for the Riemann zeta-function in his thesis (see [START_REF] Tsang | The distribution of the values of the Riemann zeta function[END_REF], Theorem 5.1). The proof remains unchanged for a Dirichlet L-function.

Lemma 3.3 (Lemma 3 of [START_REF] Soundararajan | Moments of the Riemann zeta function[END_REF]): Let T be large and let 2 ≤ x ≤ T . Let k be a natural number such that x k ≤ T / log T . For any complex numbers a(p) we have

2T T p≤x a(p) p 1/2+it 2k dt ≪ T k! p≤x |a(p)| 2 p k . Lemma 3.4: Fix 1 ≤ j ≤ N and k ≥ 1. Define, for t ∈ (T, 2T ) ∆ j (t, H) := 1 H t+H t log |L(1/2 + iu, χ j )| du.
Then, uniformly for 0 ≤ v ≤ H, we have that

2T T (∆ j (t, H) -log |L(1/2 + i(t + v), χ j )|) 2k dt ≪ T B k k 4k ,
for some constant B > 0 that depends on A.

Proof: Put x = T 1/(100k) . Define

Err(t) := log |L(1/2 + it, χ j )| -ℜ p<x χ(p) p 1/2+it .
We have

2T T (∆ j (t, H) -log |L(1/2 + i(t + v), χ j )|) 2k dt ≪ 2T T 1 H t+H t Err(u) du 2k dt + 2T T 1 H t+H t ℜ p<x χ(p) p 1/2+iu du -ℜ p<x χ(p) p 1/2+i(t+v) 2k dt + 2T T Err(t + v) 2k dt. (3.1)
Now remember that H ≤ 1 since T is large. By Hölder's inequality and switching the integrals, the first integral is

≪ 1 H 2T T t+H t Err(u) 2k du dt ≪ 2T +1
T Err(u) 2k du.

Thus, Lemma 3.2 shows that both the first and the third integrals are

≪ T B k k 4k , (3.2)
for some constant B > 0. Now we have to deal with the second integral of (3.1), that we write as

2 := 2T T   ℜ p<x χ(p) 1 H H 0 p -iu du -p -iv p 1/2+it   2k dt.
By Lemma 3.3, we know that we only have to keep the diagonal terms of this:

2 ≪ T k! p<x 1 p p -iH -1 -iH log p -p -iv 2 k . (3.3)
We split the inner sum in two parts, the first over p < e 1/H and the second over the remaining p's. By Mertens' Theorem and Taylor expanding p -iH and p -iv , we find that the contribution of the first part is

≪ p<e 1/H 1 p -iH log p + O(H 2 log 2 p) -iH log p -1 + O(H log p) 2 ≪ p<e 1/H 1 p |H log p| 2 ≪ H p<e 1/H log p p ≪ 1.
Again by Mertens' Theorem, the sum over larger p's is

≪ e 1/H ≤p<x 1 p ≪ log (H log x) ≪ log(H log T ) = log A.
Putting this back in (3.3) and (3.1) together with (3.2), we prove the lemma.

Lemma 3.5: We suppose that κ < 2/9 and 0 < θ < 1. Then we have, for 1 ≤ j ≤ N,

2T T |L(1/2 + it, χ j )η 2 j (1/2 + it)| 2 dt C 3 T where C 3 is defined in (2.8).
The quite technical proof of this lemma is postponed until the end of the paper, for it contains some arguments also found in the proof of (2.6). We can now proceed to the proof of Theorem 1.1.

Proof (of Theorem 1.1): Let δ > 0 be small (δ = 1/10 should be enough), and fix

1 ≤ j = j ′ ≤ N. Let E j,j ′ be the set consisting of t ∈ (T, 2T ) such that | log |L(1/2 + it, χ j )| -log |L(1/2 + it, χ j ′ )|| ≤ (log log T ) δ . If ǫ = (log log T ) δ / √ 2π log log T , then Theorem 3.1 implies that meas(E j,j ′ ) = 2T T 1 [-ǫ,ǫ] log |L(1/2 + it, χ j )| -log |L(1/2 + it, χ j ′ )| √ 2π log log T dt = T ǫ -ǫ e -πu 2 du + O T log log log 2 T √ log log T ≪ T (log log T ) δ-1/2 .

Zeros of linear combinations of Dirichlet L-functions on the critical line

Outside the union of these exceptional sets, (T, 2T ) may be split into subsets on which only one single L-function decisively dominates the others. Of course, for each said subset the dominating L-function may vary. Now we shall see that this dominance is stable over relatively long (compared to 1/ log T ) stretches.

We integrate Lemma 3.4 over 0 ≤ v ≤ H, and we get that

2T T H 0 (∆ j (t, H) -log |L(1/2 + i(t + v), χ j )|) 2k dv dt ≪ HT B k k 4k , (3.4)
for some constant B > 0 and any fixed integer

k ≥ 1. Now write W j (t) the subset of v ∈ [0, H] such that |∆ j (t, H) -log |L(1/2 + i(t + v), χ j )|| > (log log T ) δ/2 . (3.5) Taking k = ⌊6/δ⌋ + 1 so that kδ ≥ 6, (3.4) implies that meas(W j (t)) ≤ H (log log T ) 3 (3.6)
except maybe for a subset of t ∈ (T, 2T ) of measure

≤ 2T T meas(W j (t)) (log log T ) 3 H dt ≤ 2T T H 0 |∆ j (t, H) -log |L(1/2 + i(t + v), χ j )|| 2k (log log T ) 3-kδ H dv dt ≪ T (log log T ) 3 .
Finally by Lemma 3.4 with v = 0, it follows that the set

F j consisting of t ∈ (T, 2T ) such that |∆ j (t, H) -log |L(1/2 + it, χ j )|| > (log log T ) δ/2 (3.7) has measure meas(F j ) ≤ 2T T |∆ j (t, H) -log |L(1/2 + it, χ j )|| (log log T ) δ/2 2k dt ≪ T (log log T ) 6 .
Now if we exclude from (T, 2T ) all the t such that t ∈ E j,j ′ for some 1 ≤ j = j ′ ≤ N, or such that t ∈ F j for some j, or such that meas(W j (t)) > H/(log log T ) 3 for some j, we get that (T, 2T ) except for a subset of size O(T /(log log T ) 1/2-δ ) can be divided into N subsets S j such that for all t ∈ S j and all j ′ = j:

log |L(1/2 + it, χ j )| -log |L(1/2 + it, χ j ′ )| > (log log T ) δ , (3.8)
and such that for any t ∈ S j and any u ∈

H t := (0, H) -N r=1 W r (t): log |L(1/2 + i(t + u), χ j )| -log |L(1/2 + i(t + u), χ j ′ )| = log |L(1/2 + i(t + u), χ j )| -∆ j (t, H) -log |L(1/2 + i(t + u), χ j ′ )| -∆ j ′ (t, H) + ∆ j (t, H) -log |L(1/2 + it, χ j )| + log |L(1/2 + it, χ j ′ )| -∆ j ′ (t, H) + log |L(1/2 + it, χ j )| -log |L(1/2 + it, χ j ′ )| > (log log T ) δ -4(log log T ) δ/2 > 1 2 (log log T ) δ . (3.9)
The first two differences are lower bounded by (3.5), the next two are bounded by (3.7), while the last one is bounded by (3.8). From Lemma 3.5, we see that

t+H t |L(1/2 + iu, χ j )η 2 j (1/2 + iu)| 2 du ≤ H log log T (3.10)
except for a subset of t ∈ (T, 2T ) of measure

≤ 1 H log log T 2T T t+H t |L(1/2 + iu, χ j )η 2 j (1/2 + iu)| 2 du dt ≪ 1 log log T 2T +1 T |L(1/2 + iu, χ j )η 2 j (1/2 + iu)| 2 du ≪ T log log T .
Let us denote by S * j the set S j deprived of those t's. Note that by construction, we have that

N m=1 meas(S * m ) = N m=1 meas(S m ) + O T log log T = T + O T (log log T ) 1/2-δ + T log log T ∼ T. (3.11)
Let F : t → N n=1 c n X n (1/2 + it), and for t ∈ S * j we define

I * j (t, H) := Ht F (t + u)|η 2 j (1/2 + i(t + u))| du and J * j (t, H) := Ht F (t + u)η 2 j (1/2 + i(t + u)) du.
Any sign change of F would imply a zero of e iϑ(1/2+i•) F (1/2 + i•), and hence a critical zero of F . Again, we will use these integrals above to detect sign changes of F . We have, for t ∈ S * j ,

I * j (t, H) = c j Ht X j (1/2 + i(t + u))|η 2 j (1/2 + i(t + u))| du + (N -1)O Ht |L(1/2 + i(t + u), χ j )η 2 j (1/2 + i(t + u))|e -(log log T ) δ /2 du = c j H 0 X j (1/2 + i(t + u))|η 2 j (1/2 + i(t + u))| du + ∪ N r=1 Wr(t) X j (1/2 + i(t + u))|η 2 j (1/2 + i(t + u))| du + O   e -(log log T ) δ /2 t+H t |L(1/2 + iu, χ j )η 2 j (1/2 + iu)| 2 du meas(H t )   . (3.12)
The first equality relies on (3.9), while the second equality comes from an application of the Cauchy-Schwarz inequality. By the Cauchy-Schwarz inequality, (3.6) and (3.10), the second integral of this is

≪ t+H t |L(1/2 + iu, χ j )η 2 j (1/2 + iu)| 2 du N i=1 meas(W i (t)) ≪ H log log T .
The same bound holds for the error term of (3.12). Therefore,

I * j (t, H) = c j I j (t, H) + O H log log T . (3.13) Similarly for t ∈ S * j , J * j (t, H) = |c j |J j (t, H) + O H log log T . (3.14)
Now we need a slight change compared to the case of a single L-function. Indeed, we are now working with sets S * j that are probably not intervals, and this prevents us from using the exact method that we described in Section 2. However, we may easily overcome this difficulty. We fix ε > 0. For any t ∈ (T, 2T ) except for a subset E j of measure

(2 + ε) 2 (C 1 (A) + C 2 )T /(H log T ) 2 = (2 + ε) 2 (C 1 (A) + C 2 )T /A 2 , Theorem 2.1 implies that |I j (t, H)| ≤ H 2 + ε , and |M j (t, H)| ≤ H 2 + ε .
Thus, for any t ∈ S * j \ E j we have by (3.13):

|I * j (t, H)| |c j | = |I j (t, H)| + O H log log T H 2 + ε .
On the other hand by (2.3) and (3.14),

J * j (t, H) |c j | = J j (t, H) + O H log log T ≥ H -|M j (t, H)| + O H log log T H - H 2 + ε .
Combining these two inequalities above, we find that for any

t ∈ S * j \ E j , F (1/2 + i•) has a zero in (t, t + H), since |I * j (t, H)| < J * j (t, H) if
T is large enough. Summing over j and using (3.11), we find that for any t in a subset of (T, 2T ) of measure

j meas(S * j ) -N(2 + ε) 2 (C 1 (A) + C 2 )T A 2 1 -N(2 + ε) 2 C 1 (A) + C 2 A 2 T F (1/2 + i•) has a sign change in (t, t + H).
Again we divide (T, 2T ) into ⌊T /(2H)⌋ pairs of abutting intervals I 1 , I 2 of length H each (except maybe for the last I 2 ). For each pair, there is at least one zero in

I 1 or I 2 unless I 1 is included in a subset E ∞ of measure N(2+ε) 2 C 1 (A)+C 2 A 2
T . If n 1 is the number of these I 1 , then we find

n 1 × H meas(E ∞ ) N(2 + ε) 2 C 1 (A) + C 2 A 2 T,
and hence

n 1 N(2 + ε) 2 C 1 (A) + C 2 A 3 T log T.
Therefore there are at least

T 2H -n 1 1 2A -N(2 + ε) 2 C 1 (A) + C 2 A 3
T log T zeros of F on the critical line in (T, 2T ). ε being as small as we want, we may use (2.1) to conclude.

Preliminary results

We first need a few lemmas to deal with the final proofs. For the remaining proofs, recall that (β(n)) is only supported on integers n ≤ ξ. Thus, we will often not specify the bounds in our summations indexes to lighten notations. Several lemmas among the followings are refinements of lemmas found in [START_REF] Selberg | On the zeros of Riemann's zeta-function[END_REF].

Lemma 4.1: For integers a, b ≥ 1, we put a|b ∞ if p|a =⇒ p|b for all primes p. Let d ≤ T 2κ and q ≥ 1 be positive integers, (q, d) = 1, and let δ|d ∞ . For 0 ≤ v ≤ H a real number, we define

S d (δ, v) := (n,qd)=1 M(nδ)τ -1/2 (nδ) n 1+iv .
Then for any c > 0,

|S d (δ, v)| τ 1/2 (δ) 1 -θ e c + e cθ 2 √ πκc v log T c √ πκ + Γ(1/4) Γ(3/4) dq φ(d)φ(q) log T .
In particular this holds for the optimal c = ρ(v log T, θ), the only positive solution in X of

e (1-θ)X 2X(a + b √ X) -2a -b √ X + 2θX(a + b √ X) -2a -b √ X = 0.
Here,

a := πκv log T , b := Γ(1/4) Γ(3/4
) .

Proof: We know that for any ε > 0,

1 2πi ε+i∞ ε-i∞ x s s 2 ds = 0 if 0 < x ≤ 1, log x if x > 1.
Therefore, by (2.2), our function M can be expressed as the following Mellin representation

M(x) = 1 2πi ℜ(s)=ε x -s 1 (1 -θ) log ξ ξ s -ξ sθ s 2 ds.
Since for any integer n ≥ 1 we have (n, d) = 1 =⇒ (n, δ) = 1, we deduce that

S d (δ, v) = 1 2πi(1 -θ)(log ξ) (n,qd)=1 τ -1/2 (nδ) n 1+iv ℜ(s)=ε (nδ) -s ξ s -ξ sθ s 2 ds = τ -1/2 (δ) 2πi(1 -θ)κ(log T ) ℜ(s)=ε Z(1 + iv + s) ξ s -ξ sθ δ s s 2 ds, (4.1)
where

Z(s) = (n,qd)=1 τ -1/2 (n) n s = p∤qd (1 -p -s ) 1/2 = p|qd (1 -p -s ) -1/2 ζ(s) -1/2 .
We take ε = c/ log T , for some c > 0 to be chosen later. We want to remove the factor ζ(s) by taking advantage of the estimate ζ(1 + s) ≈ 1/s when s is small. Therefore, we will show that the (4.1) integral can be truncated to only keep relatively small imaginary part.

Since

1/ζ(1 + s + iv) ≪ |s| + |v| and on the line ℜ(s) = ε, p|qd (1 -p -(s+1) ) -1/2 ≤ p|qd (1 -p -1 ) -1/2 = dq φ(dq) = dq φ(d)φ(q) , (4.2)
we may bound the contribution of the integral of (4.1) on

|ℑ(s)| > 1/ log log T by O   τ 1/2 (δ) log T d φ(d) ℜ(s)=ε |ℑ(s)|>1/ log log T |s| + |v|| ds| |s| 2   = O τ 1/2 (δ) d φ(d) √ log log T log T . (4.3)
Moreover, we have

ℜ(s)=ε |ℑ(s)|≤1/ log log T (|s| + v) 3/2 |s| 2 | ds| ≪ ℜ(s)=ε |ℑ(s)|≤1/ log log T |s| 3/2 + v 3/2 |s| 2 | ds| ≪ √ ε t≤1/(ε log log T ) 1 (1 + t 2 ) 1/4 dt + H 3/2 ε t≤1/(ε log log T ) 1 1 + t 2 dt ≪ 1 √ log log T + 1 √ log T ≪ 1 √ log log T (4.4)
Using the fact that

1/ζ(1 + z) 1/2 = z + O(z 2 ) = √ z + O(|z| 3/2 ) when z → 0, ℜ(z) > 0,
S d (δ, v) = τ -1/2 (δ) 1 -θ 1 κ(log T )2πi ℜ(s)=ε |ℑ(s)|≤1/ log log T p|qd (1 -p -(1+iv+s) ) -1/2 √ iv + s ξ s -ξ sθ δ s s 2 ds + O τ 1/2 (δ) d φ(d) √ log log T log T . (4.5)
We recall that c = ε log T . Using (4.2), we see that the absolute value of the main term of this is

τ 1/2 (δ) 2πκ(1 -θ) qd φ(d)φ(q) (e cκ + e κcθ ) 1 √ log T ℜ(s)=ε |ℑ(s)|≤1/ log log T √ v + |s| |s| 2 | ds| τ 1/2 (δ) 2πκ(1 -θ) qd φ(d)φ(q) (e cκ + e κcθ ) 1 log T √ v ε R dt 1 + t 2 + 1 √ ε R dt (1 + t 2 ) 3/4 .
Of course, we have R dt/(1 + t 2 ) = π. We let B(x, y) = ∞ 0 s y-1 (1+s) x+y ds be the Beta-function, x, y > 0. By property of this Beta-function (see Theorem II.0.8 of [START_REF] Tenenbaum | Introduction to analytic and probabilistic number theory[END_REF]), we deduce that √ πΓ(1/4)

Γ(3/4) = B(1/4, 1/2) = ∞ 0 s -1/2 (1 + s) 3/4 ds = ∞ 0 2 (1 + t 2 ) 3/4 dt = R 1 (1 + t 2 ) 3/4 dt.
Combining these and going back to (4.5),

|S d (δ, v)| τ 1/2 (δ) 1 -θ e κc + e κcθ 2πκ √ c v log T c π + √ π Γ(1/4) Γ(3/4) dq φ(d)φ(q) log T .
By choosing the optimal c, one concludes.

Lemma 4.2: Fix 0 < v ≤ H. Suppose that ξ/δ ≥ 2,
where δ > 0 is an integer. Put ε = c/ log T , where c > 0. Then for r = 1, 2, 3:

1 2πi ℜ(s)=ε |ℑ(s)|≤1 √ s + iv ξ δ s 1 s r ds = - v 3/2-r 2 √ π ∆ r (log(ξ/δ)v) + O 1 log(ξ/δ) ,
where

∆ 1 (X) := -2X -1/2 +
X 0 e -it -1 t 3/2 dt, and for r = 2, 3:

∆ r (X) := X 0 (X -x) r-2 -2x -1/2 + x 0 e -it -1 t 3/2 dt dx.
Proof: Let L be a curve starting at -∞ -i, joining the ℜ(s) = ε axis horizontally, going up to ε + i as a semi-circle and then returning back to -∞ + i. Zeros of linear combinations of Dirichlet L-functions on the critical line Note that for any s ∈ L, |s| > v. For any integer j and any complex z, we recall the definition

z j := z(z-1)(z-2)...(z-j+1) j! = (-1) j-1 z(1-z)(2-z)...(j-1-z) j!
. Then for any s ∈ L,

√ iv + s = √ s 1 + iv s 1/2 = ∞ j=0 (iv) j s j-1/2 1/2 j .
We chose that branch of √ s + iv which is positive if iv + s is real and positive. Observe that for any integer ℓ ≥ 1:

Γ(ℓ -1/2) = ℓ - 3 2 ℓ - 5 2 × • • • × 3 2 × 1 2 Γ(1/2) = 2Γ(1/2)ℓ!(-1) ℓ-1 1/2 ℓ .
Therefore, for any j ≥ 1, we find that

Γ(j + r -1/2) = j + r - 3 2 j + r - 5 2 × • • • × j - 1 2 Γ(j -1/2) = 2Γ(1/2)j!(-1) j-1 1/2 j j + r - 3 2 j + r - 5 2 × • • • × j - 1 2 .
This holds if j = 0. Let log 2 ≤ u ≤ log ξ. Then by Hankel's formula for 1/Γ together with the above formulas, we find that

1 2πi L √ iv + s e us s r ds = ∞ j=0 (iv) j 1/2 j 1 2πi L e us s j+r-1/2 ds = ∞ j=0 (iv) j 1/2 j u j+r-3/2 Γ(j + r -1/2) = - u r-3/2 2Γ(1/2) ∞ j=0 (-iuv) j j!(j -1/2)(j + 1/2)...(j + r -3/2) . (4.6)
By Taylor expanding the exponential below, we find that

(uv) 1/2 -2(uv) -1/2 + uv 0 e -it -1 t 3/2 dt = (uv) 1/2 -2(uv) -1/2 + ∞ j=1 (-i) j (uv) j-1/2 j!(j -1/2) = ∞ j=0 (-iuv) j j!(j -1/2)
, while for r > 1, Taylor expanding the exponential below together with r -2 successive integrations by parts lead to

(uv) 3/2-r (r -2)! uv 0 (uv -x) r-2 -2x -1/2 + x 0 e -it -1 t 3/2 dt dx = (uv) 3/2-r (r -2)! uv 0 (uv -x) r-2 ∞ j=0 (-i) j x j-1/2 j!(j -1/2) dx = (uv) 3/2-r uv 0 ∞ j=0 (-i) j x j+r-5/2 j!(j -1/2)(j + 1/2)...(j + r -5/2) . = ∞ j=0 (-iuv) j j!(j -1/2)(j + 1/2)...(j + r -3/2)
Using (4.6) and deforming the semi-circle part of L as the segment [ε -i, ε + i], one finds

1 2πi ℜ(s)=ε |ℑ(s)|≤1 √ iv + s e us s r ds = - v 3/2-r 2 √ π ∆ r (uv) + O   ℜ(s)≤ε ℑ(s)=1 √ v + |s| |s| r e uℜ(s) | ds|   . (4.7)
Since we assumed that u ≤ log ξ ≪ 1/ε and since r ≥ 1, the error term is

≪ ε -∞ 1 + (1 + t 2 ) 1/4 (1 + t 2 ) r/2 e ut dt ≪ e uε u ≪ 1 u .
Using (4.7) with u = log(ξ/δ), one concludes.

Lemma 4.3: Let X be a positive real number. We have:

∆ 2 (X) = -4X 1/2 + O 4 3 X 3/2 if X ≤ 1, 2 √ πe -3iπ/4 (X -1) -4 + O 16 3 else. Thus for any X ≥ 0, ∆ 2 (X) = 2 √ πe -3iπ/4 (X -1) -4 + O 16 3 . Furthermore if X ≤ 1, ∆ 3 (X) = - 8 3 X 3/2 + O 8 15 X 5/2 .
Proof: If x ≤ 1 then the inequality |e -it -1| ≤ t, valid for t ≥ 0, implies

x 0 e -it -1 t 3/2 dt = O(2 √ x).
If x > 1 then an integration by parts leads to

-2x -1/2 + x 0 e -it -1 t 3/2 dt = ∞ 0 e -it -1 t 3/2 dt -2x -1/2 - ∞ x e -it -1 t 3/2 dt = ∞ 0 e -it -1 t 3/2 dt + O(2x -3/2 ).
Observe that an integration by parts and a change of variable lead to

∞ 0 e -it -1 t 3/2 dt = -2i ∞ 0 e -it √ t dt = -4i ∞ 0 e -iu 2 du.
The last integral happens to be Fresnel's integral, whose value is known to be e -iπ/4 √ π 2 . Therefore,

∞ 0 e -it -1 t 3/2 dt = 2 √ πe -3iπ/4 .
Zeros of linear combinations of Dirichlet L-functions on the critical line Thus

, if 0 ≤ X ≤ 1, ∆ 2 (X) = X 0 -2x -1/2 + O(2 √ x) dx = -4 √ X + O 4 3 X 3/2 ,
and if X ≥ 1:

∆ 2 (X) = -4+O 4 3 + X 1 2 √ πe -3iπ/4 + O(2x -3/2 ) dx = 2 √ πe -3iπ/4 (X-1)-4+O 4 3 +O(4).
This proves the first part of the lemma.

For the second part of the lemma, we know that for any X ≥ 1,

∆ 2 (X) = 2 √ πe -3iπ/4 (X -1) -4 + O 16 3 .
For X ≤ 1, we have

∆ 2 (X) = -4 √ X + O(4/3) = 2 √ πe -3iπ/4 (X -1) -4 + O -4 √ X -2 √ πe -3iπ/4 (X -1) + 4 + 4/3 .
The part inside of the O symbol is shown to be a decreasing function of X over [0, 1], and hence its maximum value over [0, 1] is equal to |2 √ πe -3iπ/4 + 4| + 4/3 ≤ 16/3, which concludes. We now turn to the last estimate of the lemma. We have

∆ 3 (X) = X 0 (X -x) -2x -1/2 + x 0 e -it -1 t 3/2 dt dx = X 0 ∆ 2 (x) dx.
By the computations above, this concludes.

Lemma 4.4: Let d be an integer, ε = c/ log T with c > 0, and let 0 ≤ v ≤ H. We define

P d (s) = p|d (1 -p -(1+s+iv) ) -1/2 .
Put, for r = 2, 3,

P d (s) = r-1 j=0 s j j! P (j) d (0) + s r R r,d (s)
.

Then for -1 ≤ t ≤ 1, j = 0, 1, 2 we have:

P (j) d (0) = O   p|d (1 + p -3/4 )   , R r,d (ε + it) = O   p|d (1 + p -3/4 )   .
Proof: We consider the rectangle Rec defined by the lines ℜ(s) = ε ± 1/5 and ℑ(s) = ±1 where P d (s) is bounded by p|d (1 -p -1+1/5 ) -1/2 ≪ p|d (1 + p -3/4 ). By Cauchy's integral formula, we find

P (j) d (0) = j! 2πi Rec P d (z) z j+1 dz = O   p|d (1 + p -3/4 )   and R r,d (ε + it) = 1 2πi Rec P d (z) (z -(ε + it))z r dz = O   p|d (1 + p -3/4 )   .
Lemma 4.5: Let d, δ, q ≥ 1 be a positive integers, (d, q) = 1, and 0 < v ≤ 1/ √ log ξ. Then for r = 2, 3, we have

(n,qd)=1 nδ<ξ τ -1/2 (n) log r-1 ξ nδ n 1+iv = - (r -1)! 2 √ π p|qd (1 -p -(1+iv) ) -1/2 v 3/2-r ∆ r (log(ξ/δ)v) + O   p|d (1 + p -3/4 ) (log T ) r-5/2 + (log log T ) r-3/2)   .
Proof: The lemma being true if ξ/δ < 2, we may suppose that ξ/δ ≥ 2. Put ε = c/ log T with c > 0. For x > 0 and r = 2, 3, we have the Mellin transform

1 2πi ε+i∞ ε-i∞ x s s r ds = 0 if 0 < x ≤ 1, (log x) r-1 (r-1)! if x > 1.
Therefore, as in Lemma 4.1,

(n,qd)=1 nδ<ξ τ -1/2 (n) log r-1 ξ nδ n 1+iv = (r -1)! 2πi ε+i∞ ε-i∞ ξ δ s s r (n,dq)=1 τ -1/2 (n) n 1+s+iv ds = (r -1)! 2πi ε+i/ log log T ε-i/ log log T ξ δ s s r Z(1 + s + iv) ds + O d φ(d) (log log T ) r-3/2 = (r -1)! 2πi ε+i ε-i ξ δ s s r P qd (s) √ s + iv ds + O d φ(d) (log T ) r-5/2 + (log log T ) r-3/2) . (4.8)
Note that, by Lemma 9 of [START_REF] Selberg | On the zeros of Riemann's zeta-function[END_REF], we have for, k = 1, 2, 3:

1 2πi ε+i ε-i ξ δ s √ s + iv s k ds = O log(T ) k-3/2 + 1 .
Zeros of linear combinations of Dirichlet L-functions on the critical line

We may now apply this fact, Lemma 4.2 and Lemma 4.4 to get

1 2πi ε+i ε-i ξ δ s s r P qd (s) √ s + iv ds = r-1 j=0 1 2πi P (j) qd (0) j! ε+i ε-i ξ δ s √ s + iv s r-j ds + 1 2πi ε+i ε-i ξ δ s √ s + ivR r,qd (s) ds = -P qd (0) v 3/2-r 2 √ π ∆ r (log(ξ/δ)v) + r-1 j=1 P (j) qd (0)O (log T ) r-j-3/2 + 1 + O   p|d (1 + p -3/4 )   = -P qd (0) v 3/2-r 2 √ π ∆ r (log(ξ/δ)v) + O   p|d (1 + p -3/4 ) log(T ) r-5/2 + 1   . (4.9) Since d/φ(d) ≪ p|d (1 + p -3/4
), one may use (4.8) and (4.9) to conclude.

Corollary 4.6: Let d, q ≥ 1 be co-prime positive integers, 1 ≤ δ ≤ ξ be such that δ|d ∞ , and

0 < v ≤ 1/ log ξ. Then (n,qd)=1 M(nδ)τ -1/2 (nδ) n sin(v log(nδ)) v = O C 4 τ 1/2 (δ) dq φ(d)φ(q) (log ξ) 1/2 + O   τ 1/2 (δ) p|d (1 + p -3/4 )(log T ) 1/3   ,
where

C 4 := 1 3(1 -θ) √ π 9(1 + θ 5/2 ) 5 + 2 max{3 √ 1 -θ, min{max{1, |1 -4θ|} + 3θ √ θ, 1 -θ + 3θ √ 1 -θ}} .
Proof: Multiplying the formula of Lemma 4.5 for r = 2 by log ξ and subtracting the formula for r = 3, we find that for any 0 < u ≤ v:

(n,qd)=1 nδ<ξ τ -1/2 (n) log ξ nδ n 1+iu log(nδ) = -1 2 √ π p|qd (1 -p -(1+iu) ) -1/2 (log ξ)u -1/2 ∆ 2 (log(ξ/δ)u)- 2u -3/2 ∆ 3 (log(ξ/δ)u) + O   p|d (1 + p -3/4 ) log T log log T   .
Thus by Lemma 4.3,

(n,qd)=1 nδ<ξ τ -1/2 (n) log ξ nδ n 1+iu log(nδ) = O   p|d (1 + p -3/4 )(log T ) 4/3   + O 18 15 √ π dq φ(d)φ(q) u(log ξ) 5/2 - 2 3 √ π p|qd (1 -p -(1+iu) ) -1/2 (log ξ -4 log δ) log(ξ/δ).
Using the relation between M sel and M (see (2.2)), we find that

(n,qd)=1 M(nδ)τ -1/2 (nδ) n 1+iu log(nδ) = O 18(1 + θ 5/2 ) (1 -θ)15 √ π τ 1/2 (δ) dq φ(d)φ(q)
u(log ξ) 3/2 (4.10)

+ O 2τ 1/2 (δ) (1 -θ)3 √ π dq φ(d)φ(q) 1 -4 log δ log ξ log ξ δ -θ -4 log δ log ξ log + ξ θ δ + O   τ 1/2 (δ) p|d (1 + p -3/4 )(log T ) 1/3   .
We denote by Q the quantity in absolute value, in the second O. Then, if δ > ξ θ , observe that for any θ ≤ x ≤ 1, we have |1 -4x| ≤ 3, and hence

|Q| ≤ 3 log ξ δ ≤ 3 √ 1 -θ log ξ.
If δ ≤ ξ θ , then observe that for x ∈ [0, θ], we have |1 -4x| ≤ max(1, |1 -4θ|) and |θ -4x| ≤ 3θ, and hence

|Q| ≤ max(1, |1 -4θ|) + 3θ √ θ log ξ.
We may also write the following, since √ a -√ b ≤ √ a -b for any real numbers 0 ≤ a < b:

|Q| ≤ 1 -4 log δ log ξ -θ + 4 log δ log ξ log ξ δ + θ -4 log δ log ξ log ξ θ δ -log ξ δ ≤ (1 -θ) log ξ + 3θ √ 1 -θ log ξ.
Therefore we always have

|Q| ≤ max{3 √ 1 -θ, min{max{1, |1 -4θ|} + 3θ √ θ, 1 -θ + 3θ √ 1 -θ} log ξ} =: C(θ) log ξ.
Multiplying (4.10) by δ -iu , integrating with respect to u from 0 to v, and dividing by v, we find that

(n,qd)=1 M(nδ)τ -1/2 (nδ) nv i (nδ) -iv -1 = O 9(1 + θ 5/2 ) (1 -θ)15 √ π τ 1/2 (δ) dq φ(d)φ(q) v(log ξ) 3/2 + O 2τ 1/2 (δ) (1 -θ)3 √ π dq φ(d)φ(q) C(θ) log ξ + O   τ 1/2 (δ) p|d (1 + p -3/4 )(log T ) 1/3   .
Now by taking the real part of this we get the required result, since v log ξ ≤ 1.

Interlude: integral mean value theorems for Dirichlet L-functions

Before turning to the proofs of (2.6) and Lemma 3.5 -which are quite long and much harder than the previous one-, we need to state some results about integral moments of L-functions.

We will be following Bauer's papers [START_REF] Bauer | Zur Verteilung der Nullstellen der Dirichletschen L-Reihen[END_REF], [START_REF] Bauer | Zeros of Dirichlet L-series on the critical line[END_REF], where he essentially proved our estimates (he studied a slightly different integral, and only in the case α = β in the following theorem). Therefore we will detail how to get the main term in the following theorem -whether α = β or not-, but we will mostly rely on Bauer's papers to estimate the error terms. Note that this section differs from Selberg's original proof, and this allows us to simplify the proof a little and get stronger errors terms in crucial estimates. The consequence of that is that we may take our mollifier to be a bit longer than the one used by Selberg: the length of his mollifier was about T 1/10 , while ours has size of roughly T 1/8 . Theorem 5.1 (see Section 3 of [START_REF] Bauer | Zeros of Dirichlet L-series on the critical line[END_REF]): We put T := qT /(2π). Let q ≥ 1, χ be an even primitive Dirichlet character mod q. Let α = iU/ log T, β = iV / log T be distinct complex numbers, where U and V are real numbers bounded by an absolute constant. Let h, k ≤ T ω , 0 < ω < 1/2, be positive integers and fix 1/2 ≤ c < 1. We have

1 i c+iT c+i L(s + α, χ)L(1 -s -β, χ)h -s k s-1 ds = χ k (h, k) χ h (h, k) T × L(1 -α + β, χ 0 ) T β-α (h, k) 1-α+β (1 -α)h 1-α k 1+β + L(1 + α -β, χ 0 ) (h, k) 1+α-β (1 -β)h 1-β k 1+α + R c (T ; h, k; α, β),
where R c (T ; h, k; α, β) is an error term such that for any small, but fixed, ε > 0 and any bounded complex sequence (θ(n)):

h,k≤T ω θ(h)θ(k)R c (T ; h, k; α, β) ≪ (T 1/2+ω + T 1/3+ω(2-c) )T ε .
(5.1)

If we suppose that α = β, then result holds if the main term is replaced by

χ k (h, k) χ h (h, k) (h, k) h 1-α k 1+α T 1 -α φ(q) q log T (h, k) 2 hk + Q(q, α) ,
for a quantity Q(q, α) defined in (5.13) below.

Proof:

Let M := 1 i c+iT c+i L(s + α, χ)L(1 -s -β, χ)h -s k s-1 ds.
We move the line of integration to the right by the residue theorem, say at c ′ = 1 + ε for some small ε > 0. Using the fact that

L(1/2 + σ + it, χ) ≪ t (1-σ)/3+o(1) if t > 1 and σ ∈ [1/2, 1) (see Theorem 2 of [12]), we get M = 1 i c ′ +iT c ′ +i L(s + α, χ)L(1 -s -β, χ)h -s k s-1 ds + O(T 1/3+ε k ε /h c ). (5.2)
We define

Err 0 := T 1/3+ε k ε /h c , (5.3)
and we write the functional equation in the form

L(s, χ) = Ξ(s, χ)L(1 -s, χ) (5.4)

INTERLUDE: INTEGRAL MEAN VALUE THEOREMS FOR DIRICHLET L-FUNCTIONS

We will use Lemma 1 of [START_REF] Bauer | Zeros of Dirichlet L-series on the critical line[END_REF], which states the following: Let r > 0, and for x ∈ R let e(x) = exp(2πix). We also introduce

E c ′ (r, T ) = T c ′ -1/2 + T c ′ +1/2 |T -r| + T 1/2 , G(χ) = q n=1 χ(n)e n q .
Then this lemma reveals that if r ≤ T , then

c ′ +iT c ′ +i Ξ(1 -s -β, χ)r -s ds = 2πiG(χ)T β q -1 e - r q + O r -1 + r -c ′ E c ′ 2πr q , T ,
and if r > T , then

c ′ +iT c ′ +i Ξ(1 -s -β, χ)r -s ds = O r -c ′ E c ′ 2πr q , T .
Using this lemma together with the functional equation (5.4) in (5.2), we may write that

M = 1 i c ′ +iT c ′ +i L(s + α, χ)L(s + β, χ)Ξ(1 -s -β, χ)h -s k s-1 ds + O(Err 0 ) = 1 ik n,m≥1 χ(n)χ(m) n α m β c ′ +iT c ′ +i k hnm s Ξ(1 -s -β, χ) ds + O(Err 0 ) = 2πG(χ)T β qk nm≤T k/h χ(n)χ(m) n α m β e - nmh qk + O (Err 0 + Err 1 (h, k)) (5.5)
where

Err 1 (h, k) = m,n≥1 1 n ℜ(α) m ℜ(β) k ε (nmh) 1+ε E c ′ (2πnmh/(qk), T ) + m,n:mnh/k≤T 1 mnh . (5.6)
Now we use Perron's formula (see (2.7) of [START_REF] Bauer | Zur Verteilung der Nullstellen der Dirichletschen L-Reihen[END_REF]) to write that for any large x ≥ 1 and for ̺ = max(|α|, |β|):

nm≤T k/h χ(n)χ(m) n α m β e - nmh qk = 1 2πi c ′ +ix c ′ -ix (T k/h) s s n,m χ(n)χ(m) n s+α m s+β e - nmh qk ds (5.7) + O (T k/h) 1+ε x + 2 2̺ (T k/h) 1+2̺ log(2T k/h) x + (2T k/h) 2̺ .
This error term is easily shown to be n -s-α m -s-β = (qK) -2s-α-β D(s).

≪ (T k/h) 1+ε x + 1 =: Err 2 (x, h, k). ( 5
(5.9)

Bauer shows in pages 29-30 of [START_REF] Bauer | Zur Verteilung der Nullstellen der Dirichletschen L-Reihen[END_REF] that

D * (s) = D(s) -E 1 (s) -E 2 (s) + E 3 (s), with E 1 (s) = Kq s+β G(χ)χ(K)χ(-H ′ )L(s + β, χ 0 )ζ(s + α), E 2 (s) = Kq s+α G(χ)χ(K)χ(-H ′ )L(s + α, χ 0 )ζ(s + β), E 3 (s) = Kφ(q)G(χ)χ(K)χ(-H ′ )ζ(s + α)ζ(s + β),
where φ is Euler's totient function and χ 0 is the principal character modulo q. Using the fact that χ is assumed to be even, we find that χ(-H ′ ) = χ(H ′ ). Let R c ′ (x) be the rectangle with vertices at c ′ ± ix and 1/2 ± ix, and define Γ c ′ (x) as the path along the upper, left and lower part of R c ′ (x). Then by using (5.9) the main term of (5.7) becomes

1 2πi R c ′ (x) (T k/h) s s D * (s) + E 1 (s) + E 2 (s) -E 3 (s) (qK) 2s+α+β ds -I(h, k), (5.10)
with

I(h, k) := 1 2πi Γ c ′ (x) (T k/h) s s n,m χ(n)χ(m) n s+α m s+β e - nmh qk ds.
First, we suppose that α = β. Since ζ(s, y) -ζ(s) is regular for 0 < y ≤ 1 -and hence D * is regular-, and since α = β, the only poles of the first integral are simple poles at 1 -α and 1 -β. To compute the residues involved, note that L(s, χ 0 ) = ζ(s) p|q (1 -p -s ) and hence Res s=1 (L(s, χ 0 )) = φ(q)/q. Thus, Res s=1-α (E 2 (s)) = Res s=1-α (E 3 (s)) which implies that

Res s=1-α (E 1 (s)+E 2 (s)-E 3 (s)) = Res s=1-α (E 1 (s)) = Kq 1-α+β G(χ)χ(K)χ(H ′ )L(1-α+β, χ 0 ).
The pole at s = 1 -β is handled similarly, and hence the Residue Theorem yields that the (5.10) integral along the rectangle is equal to

G(χ)χ(K)χ(H ′ ) qK T k h L(1 -α + β, χ 0 ) (T k/h) -α K α-β 1 -α + L(1 + α -β, χ 0 ) (T k/h) -β K -α+β 1 -β .

INTERLUDE: INTEGRAL MEAN VALUE THEOREMS FOR DIRICHLET L-FUNCTIONS

Putting these back in (5.7) and (5.5), and using the fact that G(χ)G(χ) = q, we get

M = χ(K)χ(H ′ ) T hK L(1 -α + β, χ 0 ) T β-α (k/h) -α K α-β 1 -α + L(1 + α -β, χ 0 ) (k/h) -β K -α+β 1 -β (5.11) - 2π qk G(χ)T β I(h, k) + O Err 0 + Err 1 (h, k) + Err 2 (x, h, k) k .
Now, we suppose that α = β. We see that the only difference lies in the fact that the only pole of the integral (5.10) is now of order 2, at s = 1 -α. Let us compute the corresponding residue, that we denote by R. We have

R = d ds E 1 (s) + E 2 (s) -E 3 (s) s(qK) 2s+2α T k h s (s -1 + α) 2 s=1-α = G(χ)χ(K)χ(H ′ ) q 2 K T k h 1-α d dz 2q z p|q (1 -p -z ) -φ(q) (z -α)(qK) 2(z-1) ζ 2 (z) T k h z-1 (z -1) 2 z=1 .
(5.12)

Recall that T = qT /(2/π). The derivative part of this is equal to 2φ(q) log q + 2q p 1 |q

p 2 |q p 1 =p 2 (1 -p -1 2 ) log p 1 p 1 (1 -α) -φ(q) (1 + 2(1 -α) log(qK)) (1 -α) 2 + φ(q) (1 -α) log T k h + Res s=1 (ζ(s) 2 (s -1) 2 ) = φ(q) 1 -α   2 log q + 2 p|q log p p -1 - 1 1 -α -2 log(qK) + log T k h + 2γ   . = φ(q) 1 -α log T k hK 2 + Q(q, α) ,
where γ is the usual Euler-Mascheroni constant and

Q(q, α) := - 1 1 -α + log q 2π + 2γ + 2 p|q log p p -1 . (5.13)
Putting this in (5.12), the Residue Theorem yields that the (5.10) integral along the rectangle is equal to

G(χ)χ(K)χ(H ′ ) T k h -α k qhK T 2π φ(q) 1 -α log T (h, k) 2 hk + Q(q, α) ,
Again, using the equations (5.7) and (5.5) above, we have proved that

M = χ(K)χ(H ′ ) (h, k) h 1-α k 1+α T 1 -α φ(q) q log T (h, k) 2 hk + Q(q, α) - 2π qk G(χ)T α I(h, k) + O Err 0 + Err 1 (h, k) + Err 2 (x, h, k) k .

Zeros of linear combinations of Dirichlet L-functions on the critical line

One should note that we find the same main term found by Matsumoto [START_REF] Matsumoto | The mean square of Dirichlet L-functions[END_REF]. Now we turn the error terms. We have by definition (see (5.3) and (5.8))

h,k≤T ω |Err 0 | ≪ T 1/3+ω(2-c)+2ε , h,k≤T ω |Err 2 (x, h, k)| k ≪ T 2ω + T 1+ω+ε x .
By Lemma 2.2.3 of [START_REF] Bauer | Zur Verteilung der Nullstellen der Dirichletschen L-Reihen[END_REF], the definition (5.6) of Err 1 implies that

h,k≤T ω |Err 1 (h, k)| ≪ T 1/2+ω+2ε .
Now it only remains to estimate the contribution of I(h, k)/k in (5.11). If x is chosen as a large power of T , then Section 5 of [START_REF] Bauer | Zeros of Dirichlet L-series on the critical line[END_REF] shows that for any bounded complex sequence (θ(n)),

h,k≤T ω θ(h)θ(k) I(h, k) k ≪ T 1/2+ω+2ε .
From this, we easily deduce three corollaries. We keep the notation for R c (T ; h, k; α, β). The first corollary is straightforward by taking α = β = 0:

Corollary 5.2: Let h, k ≤ T ω , 0 < ω < 1/2 be integers. Then 1 √ hk 2T T |L(1/2 + it, χ)| 2 h k -it dt = χ k (h, k) χ h (h, k) (h, k) φ(q) q T kh × log T (h, k) 2 hk + Q(q, 0) + 2 log 2 + R 1/2 (T ; h, k; 0, 0).
The second one is proved by partial summation.

Corollary 5.3: Let α = iU/ log T, β = iV / log T be distinct complex numbers, where U and V are real numbers bounded by an absolute constant. Let h, k ≤ T ω , 0 < ω < 1/2, be positive integers. For any smooth function g : (0, ∞) → C, we have:

1 √ hk T 1 L(1/2 + it + α, χ)L(1/2 -it -β, χ) h k -it g(t) dt = χ k (h,k) χ h (h,k) hk × T 1 g(t) L(1 -α + β, χ 0 ) (1 + β -α)(qt/2π) β-α (h, k) 1-α+β (1 -α)h -α k β + L(1 + α -β, χ 0 ) (h, k) 1+α-β (1 -β)h -β k α dt + R(T ; h, k; α, β; g),
where R(T ; h, k; α, β; g) is an error term satisfying

h,k≤T ω θ(h)θ(k)R(T ; h, k; α, β; g) ≪ (T 1/2+ω + T 1/3+3ω/2 )T ε |g(T )| + T 1 |g ′ (t)| dt + 1
for any ε > 0 and any bounded complex sequence (θ(n)). L-FUNCTIONS Proof: Theorem 5.1 proves that for all T > 1 and any distinct complex numbers α, β, we have

F (T ) := 1 √ hk T 1 L(1/2 + it + α, χ)L(1/2 -it -β, χ) h k -it dt = χ k (h,k) χ h (h,k) hk T × L(1 -α + β, χ 0 ) (qT /2π) β-α (h, k) 1-α+β (1 -α)h -α k β + L(1 + α -β, χ 0 ) (h, k) 1+α-β (1 -β)h -β k α + R(T ), (5.14)
where R(T ) is an error term which is R 1/2 (T ; h, k; α, β) when α and β are of the form iU/ log T and iV / log T . Thus if g is a smooth function over (0, ∞), we get by partial integration

1 √ hk T 1 L(1/2 + it + α, χ)L(1/2 -it -β, χ) h k -it g(t) dt = g(T )F (T ) - T 1 g ′ (t)F (t) dt = χ (k/(h, k)) χ (h/(h, k)) hk T 1 g(t) L(1 -α + β, χ 0 )(1 + β -α) (qt/2π) β-α (h, k) 1-α+β (1 -α)h -α k β + L(1 + α -β, χ 0 ) (h, k) 1+α-β (1 -β)h -β k α dt + g(T )R(T ) - T 1 g ′ (t)R(t) dt + O(1).
Now, we take our α and β to be of the wanted form, we let (θ(n)) be a bounded sequence of complex numbers and we fix ε > 0. Then by property of R(T ; h, k; α, β) = R(T ):

h,k≤T ω θ(h)θ(k) g(T )R(T ) - T 1 g ′ (t)R(t) dt + O(1) ≪ (T 1/2+ω + T 1/3+3ω/2 )T ε |g(T )| + T 1 |g ′ (t)|(t 1/2+ω + t 1/3+3ω/2 )t ε dt + T 2ω ≪ (T 1/2+ω + T 1/3+3ω/2 )T ε |g(T )| + T 1 |g ′ (t)| dt + 1 .

This concludes.

In the case where g(t) = (2π/t) iv/2 for some v = V / log T , a choice motivated by Stirling's formula as we shall see later, we find the following.

Corollary 5.4: Let v = V / log T be a real number, V ∈ R * bounded by an absolute constant. Let h, k ≤ T ω , 0 < ω < 1/2 be positive integers. Then 1 √ hk 2T T L(1/2 + it, χ)L(1/2 -it -iv, χ) h k -it 2π t iv/2 dt = 2T χ k (h,k) χ h (h,k) hk (h, k)× L(1 + iv, χ 0 ) q 2 T (h, k) 2 2πk 2 iv/2 2 1+iv/2 -1 2 + iv + L(1 -iv, χ 0 ) T (h, k) 2 2πh 2 -iv/2 2 1-iv/2 -1 2 -iv + R(T ; h, k; v).
Here, R(T ; h, k; v) is an error term satisfying the condition (5.1).

Zeros of linear combinations of Dirichlet L-functions on the critical line 1) . Thus Corollary 5.3 gives

Proof: First, note that |g(T )| + T 1 |g ′ (t)| dt ≪ log T ≪ T o(
1 √ hk T 1 L(1/2 + it + iv, χ)L(1/2 -it, χ) k h -it 2π t -iv/2 dt = χ (h/(h, k)) χ (k/(h, k)) hk × (h, k) T 1 L(1 -iv, χ 0 ) qt(h, k) 2πk -iv + L(1 + iv, χ 0 )(h, k) iv h iv 2π t -iv/2 dt + R(T ; k, h; iv, 0; g) = 2 χ (h/(h, k)) χ (k/(h, k)) hk (h, k) L(1 -iv, χ 0 ) (q(h, k)) 2 2πk 2 -iv/2 T -iv/2+1 -1 2 -iv + L(1 + iv, χ 0 ) (h, k) 2 2πh 2 iv/2 T iv/2+1 -1 2 + iv + R(T ; k, h; iv, 0; g).
Taking the conjugate of this, one finds the expected result.

6 Proof of (2.7)

We start by the easy part of the proof of Theorem 2.1: the estimate on M j (t, H).

Proof (of (2.7)): We fix 1 ≤ j ≤ N, we write χ for χ j , and we denote by L the Dirichlet L-function associated to χ. We also let η = η j be the corresponding mollifying function, and we write M(t, H) instead of M j (t, H).

We first apply Cauchy's residue theorem to the rectangle 1/2 + i(t + H), 3/2 + i(t + H), 3/2 + it, 1/2 + it: since L(s)η(s) -1 has no pole in that rectangle, we have

M(t, H) = t+H t (L(3/2 + iu)η 2 (3/2 + iu) -1) du + 3/2 1/2 (L(σ + it)η 2 (σ + it) -1) dσ - 3/2 1/2 (L(σ + i(t + H))η 2 (σ + i(t + H)) -1) dσ.
Therefore, by the Cauchy-Schwarz inequality, we obtain

|M(t, H)| 2 ≤ 3 t+H t (L(3/2 + iu)η 2 (3/2 + iu) -1) du 2 + 3/2 1/2 (L(σ + it)η 2 (σ + it) -1) dσ 2 (6.1) + 3/2 1/2 (L(σ + i(t + H))η 2 (σ + i(t + H)) -1) dσ 2 .
We start by the first integral. By property of the Dirichlet convolution product, we know that

abc=n χ(a)α(b)α(c) = δ 1 (n), where δ 1 (n) = 1 if n = 1, and δ 1 (n) = 0 otherwise. Therefore, if t ≤ u ≤ t + h, the integrand of the first integral of (6.1) is ∞ n,m,r=1 χ(n)β(m)β(r) (nmr) 3/2+iu -1 = ∞ n=1 abc=n χ(a)β(b)β(c) n 3/2+iu -1 = n≥ξ θ abc=n χ(a)β(b)β(c) n 3/2+iu .
6 PROOF OF (??)

The final equality relies on the fact that

β(k) = α(k) if k < ξ θ . Moreover, abc=n χ(a)β(b)β(c) ≤ abc=n |α(b)α(c)| ≤ abc=n 1 = τ 3 (n).
Note that τ 3 (n) ≪ n o (1) . Consequently, the first squared integral of (6.1) is (1) . (

≪   n≥ξ θ τ 3 (n) n 3/2 log n   2 ≪ ξ -θ/2+o
Next, a well known mean value estimate of Montgomery and Vaughan (see Corollary 3 of [START_REF] Montgomery | Hilbert's Inequality[END_REF]) yields the following:

2T T L(3/2 + it)η 2 (3/2 + it) -1 2 dt ≤ T n≥ξ θ τ 3 (n) 2 n 3 + O(1) T ξ -2θ+o(1) 2 . (6.3)
Now we turn to the second integral. From Lemma 3.5, the Cauchy-Schwarz inequality reveals that

2T T L(1/2 + it)η 2 (1/2 + it) -1 2 dt ≤ (C 3 + 1 + 2 C 3 )T. (6.4)
Thanks to Gabriel's convexity theorem (see Theorem 2 of [START_REF] Gabriel | Some Results Concerning the Integrals of Moduli of Regular Functions Along Certain Curves[END_REF]), we use (6.4) and ( 6.3) to deduce that for all 1/2 ≤ σ ≤ 3/2:

2T T |L(σ + it)η 2 (σ + it) -1| 2 dt ((C 3 + 1 + 2 C 3 )T ) 3/2-σ T ξ -2θ+o(1) 2 σ-1/2 T (C 3 + 1 + 2 √ C 3 ) (2(C 3 + 1 + 2 √ C 3 )) σ-1/2 ξ -(σ-1/2)(2θ+o(1)) .
Therefore, by first applying the Cauchy-Schwarz inequality and then switching the integrals we find

2T T 3/2 1/2 (L(σ + it)η 2 (σ + it) -1) dσ 2 dt ≤ 2T T 3/2 1/2 ξ -(σ 1 -1/2)θ dσ 1 3/2 1/2 ξ (σ 2 -1/2)θ |L(σ + it)η 2 (σ 2 + it) -1| 2 dσ 2 dt T (C 3 + 1 + 2 C 3 ) - ξ -(σ 1 -1/2)θ θ log ξ 3/2 σ 1 =1/2 3/2 1/2 ξ -(σ 2 -1/2)(θ+o(1)) (2(C 3 + 1 + 2 √ C 3 )) σ 2 -1/2 dσ 2 (C 3 + 1 + 2 √ C 3 )T (θ log ξ) 2 (C 3 + 1 + 2 √ C 3 ) (θκ) 2 T (log T ) 2 , since 3/2 1/2 (ξ θ+o(1) /C) -(σ-1/2) dσ = (1 + o(1)
)/(θ log ξ) for any constant C = 0. The same estimate holds for the third integral of (6.1). Therefore, combining these estimates with (6.2), we get that

2T T |M(t, H)| 2 dt 3 T ξ -θ/2+o(1) + 2(C 3 + 1 + 2 √ C 3 ) (θκ) 2 T (log T ) 2 6(C 3 + 1 + 2 √ C 3 ) (θκ) 2 T (log T ) 2 .
Zeros of linear combinations of Dirichlet L-functions on the critical line 7 Proof of Lemma 3.5

We may now prove Lemma 3.5. Again, we fix 1 ≤ j ≤ N, we write χ for χ j , and we denote by L the Dirichlet L-function associated to χ. We also let η = η j be the corresponding mollifying function.

Proof (of Lemma 3.5): We suppose that κ < 2/9 and we fix ε > 0 small enough. We let (γ(n)) be the coefficients such that

η(s) 2 = n γ(n)χ(n) n s . Note that γ(n) = Mτ -1/2 ⋆Mτ -1/2 (n)
, where ⋆ stands for the Dirichlet convolution, and hence is real and is supported on integers n ≤ ξ 2 . By definition, we have that |γ(n

)| ≤ τ 1/2 ⋆ τ 1/2 (n) = 1.
Also note that, if χ 0 is the principal character (mod q), then for any integers h, k:

χ(h)χ(k)χ h (h, k) χ k (h, k) = χ 0 hk (h, k) = χ 0 (hk). (7.1)
Then, by Corollary 5.2 with ω = 2κ, we get

2T T |L(1/2 + it)η 2 (1/2 + it)| 2 dt = h,k γ(h)γ(k)χ(h)χ(k) √ hk 2T T |L(1/2 + it)| 2 h k -it dt = h,k χ 0 (hk)γ(h)γ(k)(h, k) hk φ(q) q T log T (h, k) 2 hk + Q(q, 0) + 2 log 2 + O((T 1/2+2κ + T 1/3+3κ )T ε ) = φ(q) q Σ 1 T (log T + Q(q, 0) + 2 log 2) -Σ 2 T + o(T ), (7.2)
where

Σ 1 := h,k χ 0 (hk)γ(h)γ(k)(h, k) hk , Σ 2 := h,k χ 0 (hk)γ(h)γ(k)(h, k) hk log hk (h, k) 2 .
Now we only have to bound these two sums, as in [10, 24.51]. We have

Σ 1 = d d -1 (h,k)=1 χ 0 (d 2 hk)γ(dh)γ(dk) hk = d d -1 δ µ(δ) h,k χ 0 (d 2 δ 2 hk)γ(dδh)γ(dδk) δ 2 hk = m m -1 δ|m µ(δ) δ h χ 0 (mh)γ(mh) h 2 = d φ(d)A 2 d , (7.3)
where

A d = 1 d h χ 0 (dh)γ(dh) h = h≡0(mod d) χ 0 (h)γ(h) h .
Because of the presence of χ 0 , we may restrict our attention to d co-prime to q. For any positive integers h 1 , h 2 such that h 1 h 2 = h, we let h 1 = δ 1 n, h 2 = δ 2 m, δ 1 δ 2 |d ∞ , and (nm, d) = 1. Note that since δ 1 δ 2 |d ∞ and (q, d) = 1, we have (δ 1 δ 2 , q) = 1. With this decomposition we may write

A d = h≡0(mod d) h 1 h 2 =h M(h 1 )χ 0 (h 1 )τ -1/2 (h 1 )M(h 2 )χ 0 (h 2 )τ -1/2 (h 2 ) h 1 h 2 = δ 1 δ 2 |d ∞ d|δ 1 δ 2 n,m≥1 (nm,d)=1 χ 0 (n)M(δ 1 n)τ -1/2 (δ 1 n)χ 0 (m)M(δ 2 m)τ -1/2 (δ 2 m) δ 1 δ 2 nm = δ 1 δ 2 |d ∞ d|δ 1 δ 2 B d (δ 1 )B d (δ 2 ) δ 1 δ 2 (7.4)
where we have defined

B d (δ) := (n,qd)=1 M(nδ)τ -1/2 (nδ) n
, for simplicity. We may use Lemma 4.1 to get that

|B d (δ)| C 5 τ 1/2 (δ) dq (log T )φ(d)φ(q) , (7.5)
where

C 5 := 1 1 -θ e ̺ θ + e ̺ θ θ 2 √ πκ̺ θ Γ(1/4) Γ(3/4)
and where ̺ θ is the only positive solution of -1

+ 2θx + e x(1-θ) (-1 + 2x) = 0. Since r|d ∞ , d|r τ 1/2 ⋆ τ 1/2 (r) r = r|d ∞ 1 dr = 1 d p|d (1 -p -1 ) -1 = 1 φ(d)
, putting (7.5) in (7.4), we find

|A d | δ 1 δ 2 |d ∞ d|δ 1 δ 2 C 2 5 dq (log T )φ(d)φ(q) τ 1/2 (δ 1 )τ 1/2 (δ 2 ) δ 1 δ 2 ≤ C 2 5 dq (log T )φ(d)φ(q) r|d ∞ , d|r τ 1/2 ⋆ τ 1/2 (r) r = C 2 5 dq (log T )φ(d) 2 φ(q) , (7.6)
Finally, since

A d is zero if (d, q) = 1, inserting (7.6) in (7.3) yields |Σ 1 | C 4 5 q 2 (log T ) 2 φ(q) 2 d≤T 2κ d 2 φ(d) 3 χ 0 (d). (7.7)
Observe that for p a prime and m ≥ 1 an integer, we have

p 2m φ(p m ) 3 = 1 p m p p -1 3 = 1 p m + 1 p m-1 1 p p p -1 3 - 1 p = 1 p m + 1 p m-1 3p 2 -3p + 1 p 4 -3p 3 + 3p 2 -p .
Thus using multiplicativity we find that for d ≥ 1 an integer, we have

d 2 φ(d) 3 = inv ⋆ µ 2 ψ(d),
where ψ is the totally multiplicative function defined on primes by p → (3p 2 -3p + 1)/(p 4 -3p 3 + 3p 2 -p) and inv(d) = 1/d. Thus

d≤T 2κ d 2 φ(d) 3 χ 0 (d) = d≤T 2κ µ 2 (d)ψ(d)χ 0 (d) n≤T 2κ /d χ 0 (n) n ≤ n≤T 2κ χ 0 (n) n ∞ d=1 µ 2 (d)ψ(d) 2κ(log T ) φ(q) q p 1 + 3p 2 -3p + 1 p 4 -3p 3 + 3p 2 -p . (7.8)
Therefore, using (7.7), we get

|Σ 1 | 2κ q φ(q) C 4 5 p 1 + 3p 2 -3p + 1 p 4 -3p 3 + 3p 2 -p (log T ) -1 . (7.9)
Similarly, we will show that Σ 2 ≪ 1 (see [10, 24.50]). First, we have

Σ 2 = d 1 d (h,k)=1
χ 0 (d 2 hk)γ(dh)γ(dk) hk log(hk).

We write log(hk) = log(hkd 2 ) -2 log d, and split Σ 2 in two according to this decomposition, say

Σ 2 = Σ ′ 2 -2Σ ′′ 2 .
Similarly to the case of Σ 1 , we have

Σ ′ 2 = 2 d φ(d)A d Ãd ,
where

Ãd = d|n γ(n)χ 0 (n) log n n = d|n k|n Λ(k) γ(n)χ 0 (n) n = k|n Λ(k) [d,k]|n γ(n)χ 0 (n) n = k Λ(k)A [d,k] .
Here, Λ is von Mangoldt's function, defined by 1 ⋆ Λ = log. Using 8 PROOF OF (??)

Ãd C 2 5 log T q φ(q) k≤T 2κ [d, k]Λ(k) φ 2 ([d, k]) = C 2 5 dq φ 2 (d)φ(q) log T k≤T 2κ kΛ(k)φ 2 ((d, k)) (d, k)φ 2 (k) . ( 7
Note that the last sum is log(T 2κ ) + O(1). Taking P = log T for example, one may inject this in (7.10) to find that:

Ãd 4κ C 2 5 dq φ 2 (d)φ(q) .
Thus, using (7.6) and (7.8), we find .11) As in (7.3), by successively making the variable changes r = d/δ and m = r/k, we find that

Σ ′ 2 4κC 4 5 log T q 2 φ(q) 2 d≤T 2κ d 2 φ(d) 3 χ 0 (d) 8κ 2 C 4 5 q φ(q) p 1 + 3p 2 -3p + 1 p 4 -3p 3 + 3p 2 -p . ( 7
Σ ′′ 2 = d A 2 d δ|d µ(δ) d δ log(d/δ) = d A 2 d r|d k|r Λ(k)rµ(d/r) = d A 2 d k|d Λ(k) m| d k kmµ d km = d A 2 d k|d Λ(k)kφ(d/k).
Using the fact that for integers a, b ≥ 1, φ(a)φ(b) ≤ φ(ab), we find that the inner sum above is Therefore (7.6) and (7.8) yield

≤ φ(d) k|d Λ(k) k φ(k) = φ(d)
Σ ′′ 2 8κ 2 C 4 5 q φ(q) p 1 + 3p 2 -3p + 1 p 4 -3p 3 + 3p 2 -p .
Together with (7.11), (7.9) and (7.2), this concludes the proof.

8 Proof of (2.6)

Finally we turn to the last estimate needed to conclude the proof of Theorem 2.1. We fix 1 ≤ j ≤ N, we write χ for χ j , and we denote by L the Dirichlet L-function associated to χ.

We also let η = η j be the corresponding mollifying function, and we write I(t, H) instead of I j (t, H).

Proof (of (2.6)): Write Y (u) = X(1/2 + iu)|η 2 (1/2 + iu)| for convenience. Using the subconvexity bound L(1/2 + it, χ) ≪ t 1/6+o (1) for any t > 2 and ε > 0 (see Theorem 2 of [START_REF] Kolesnik | On the order of Dirichlet L -functions[END_REF]) and

since |γ(n)| ≤ 1, one gets 2T T t+H t Y (u) du 2 dt = H 0 H 0 2T T Y (u + t)Y (v + t) dt du dv = H 0 H 0 2T +u T +u Y (t)Y (t + v -u) dt du dv = H 0 H 0 2T T Y (t)Y (t -u + v) dt du dv + O   T 1/3+o(1)   n≤ξ 2 |γ(n)| √ n   2   = H 0 H-u -u 2T T Y (t)Y (t + w) dt dw du + O(T 2κ+1/3+o(1) ) = H -H (H -|v|) 2T T Y (t)Y (t + v) dt dv + O(T 2κ+1/3+o(1) ). (8.1)
The last equality is obtained by switching the first two integrals. By Stirling's formula for the Γ-function (see Corollary II.0.13 of [START_REF] Tenenbaum | Introduction to analytic and probabilistic number theory[END_REF]), we may write that for |v| ≤ H:

e iϑ(1/2+it) e -iϑ(1/2+it+iv) = 2π t iv/2 1 + O 1 t .
We let γ(n) := M(n)τ -1/2 (n) for simplicity. For t ∈ (T, 2T ) and |v| ≤ H, we then deduce -again by the subconvexity bound-that

Y (t)Y (t + v) = q -iv/2 L(1/2 + it, χ)L(1/2 -it -iv, χ)|η 2 (1/2 + it)η 2 (1/2 + it + iv)| 2π t iv/2 + O(T 2κ-2/3+o(1) ) = h,k,h ′ ,k ′ L(1/2 + it, χ)L(1/2 -it -iv, χ) β(h)β(k)β(h ′ )β(k ′ ) √ hkh ′ k ′ kk ′ hh ′ it 2πk ′2 tqh ′2 iv/2 + O(T 2κ-2/3+o(1) ) = h,k,h ′ ,k ′ L(1/2 + it, χ)L(1/2 -it -iv, χ) γ(h)γ(k)γ(h ′ )γ(k ′ ) √ hkh ′ k ′ χ(hh ′ )χ(kk ′ ) kk ′ hh ′ it 2πk ′2 tqh ′2 iv/2
+ O(T 2κ-2/3+o (1) ).

Note that over our range of summation, we have hh ′ , kk ′ ≤ T 2κ , and recall (7.1). Integrating this over t, our Corollary 5.4 applied with ω = 2κ yields

2T T Y (t)Y (t + v) dt = 2T h,k,h ′ ,k ′ k ′2 qh ′2 iv/2 χ 0 (kk ′ hh ′ )γ(h)γ(k)γ(h ′ )γ(k ′ ) hkh ′ k ′ (hh ′ , kk ′ )× L(1 + iv, χ 0 ) q 2 T (hh ′ , kk ′ ) 2 2π(kk ′ ) 2 iv/2 2 1+iv/2 -1 2 + iv + L(1 -iv, χ 0 ) T (hh ′ , kk ′ ) 2 2π(hh ′ ) 2 -iv/2 2 1-iv/2 -1 2 -iv + O h,h ′ ,k,k ′ k ′2 h ′2 iv/2
χ(hh ′ )χ(kk ′ ) R(T ; hh ′ , kk ′ ; iv) + O(T 2κ+1/3+o (1) ). (8.2)

8 PROOF OF (??)

By property of R and by letting K ′ = kk ′ and H ′ = hh ′ , the first error term is 1) ,

≪ h,k≤ξ h k iv/2 K ′ ,H ′ ≤ξ 2 K ′2 H ′2 iv/2 χ(H ′ )χ(K ′ )1 k|K ′ 1 h|H ′ R(T ; H ′ , K ′ ; iv) ≪ (T 1/2+2κ + T 1/3+3κ )T o(1) h,k≤ξ 1 ≪ (T 1/2+4κ + T 1/3+5κ )T o(
where 1 n|m = 1 if n|m and 1 n|m = 0 otherwise. Because κ < 1/8, the error term of (8.2) is ≪ T 1-ǫ , for any small ǫ > 0. Using the symmetric roles of the variables, one gets

2T T Y (t)Y (t + v) dt = 4T ℜ h,k,h ′ ,k ′ k ′ h ′ iv χ 0 (hh ′ kk ′ )γ(h)γ(k)γ(h ′ )γ(k ′ ) hkh ′ k ′ (hh ′ , kk ′ ) × L(1 + iv, χ 0 ) qT (hh ′ , kk ′ ) 2 2π(kk ′ ) 2 iv/2 2 1+iv/2 -1 2 + iv + O(T 1-ǫ ) = 4T ℜ K(v)L(1 + iv, χ 0 ) 2 1+iv -1 2 + iv + O(T 1-ǫ ),
where

K(v) := h,k,h ′ ,k ′ χ 0 (hh ′ kk ′ )γ(h)γ(k)γ(h ′ )γ(k ′ ) hkh ′ k ′ (hh ′ , kk ′ ) (hh ′ , kk ′ ) h ′ k qT 2π iv .
Since L(s, χ 0 ) only has a pole of order 1 at s = 1, of residue φ(q)/q, we may expend L as a Laurent series to find that

L(1 + iv, χ 0 ) 2 + iv (2 1+iv/2 -1) = φ(q) 2iqv + O(1),
and hence we may use (8.1) to deduce that

2T T |I(t, H)| 2 dt = 2T φ(q) q H -H (H -|v|)ℜ K(v) iv dv + O T H -H (H -|v|) |K(v)| dv + O(T 1-ǫ ). (8.3)
We fix |v| ≤ H. We will estimate K(v) in a similar manner to the one used to estimate Σ 1 , in the proof of Lemma 3.5. We will also add some arguments of arithmetical nature that may be found in [START_REF] Selberg | On the zeros of Riemann's zeta-function[END_REF]. We will divide the rest of the proof into three parts. First, we will show that K(v) = O(1/ log T ), thus showing that the error term is O(T /(log T ) 3 ). Then, we will prove that ℑ(K(v)/v) is O(1) if |v| ≤ 1/ log ξ. Finally we will bound it for the remaining v's.

We let, for z a complex number, φ z Again we put T = qT /(2π). Using (8.4), we write the following:

K(v) = T iv/2 h,h ′ ,k,k ′ χ 0 (hh ′ kk ′ )γ(h)γ(h ′ )γ(k)γ(k ′ ) hh ′ kk ′ (hh ′ , kk ′ ) 1+iv (h ′ k) iv = T iv/2 d≤T 2κ φ iv (d) h,k,h ′ ,k ′ d|hh ′ d|kk ′ χ 0 (hh ′ kk ′ )γ(h)γ(h ′ )γ(k)γ(k ′ ) h ′1+iv k 1+iv hk ′ = T iv/2 d≤T 2κ φ iv (d)     h,h ′ d|hh ′ χ 0 (hh ′ )γ(h)γ(h ′ ) h ′ h 1+iv     2 . ( 8.6)
We denote the inner sum by A ′ d (v), similarly to the quantity A d introduced in the proof of Lemma 3.5. Again, we may restrict our attention to d that are co-prime to q. We let δ, δ ′ |d ∞ (and hence (δδ ′ , q) = 1) be such that h

= δk, h ′ = δ ′ k ′ for some k, k ′ such that (kk ′ , d) = 1.
With these notations, we have

A ′ d (v) = d|δδ ′ δδ ′ |d ∞ 1 δ 1+iv δ ′   (k,qd)=1 M(kδ)τ -1/2 (kδ) k 1+iv     (k,qd)=1 M(kδ ′ )τ -1/2 (kδ ′ ) k   . (8.7)
The sum inside the first pair of parenthesis is S d (δ, v), and the one in the second pair of parenthesis is simply equal to B d (δ ′ ), these quantities being introduced in Lemma 4.1 and Lemma 3.5 respectively. Then (1) leads to

|S d (δ, v)| C 6 (v log T )τ 1/2 (δ) dq (log T )φ(d)φ(q)
, where

C 6 (v log T ) := 1 1 -θ e ρ(v log T,θ) + e ρ(v log T,θ)θ 2 πκρ(v log T, θ) v log T ρ(v log T, θ) √ πκ + Γ(1/4) Γ(3/4) .
By (7.5), we also have

|B d (δ ′ )| C 5 τ 1/2 (δ ′ ) dq (log T )φ(d)φ(q) .
These inequalities lead to (see the similar computation (7.6)):

|A ′ d (v)| C 5 C 6 (v log T ) d|δδ ′ δδ ′ |d ∞ τ 1/2 (δ)τ 1/2 (δ ′ ) δδ ′ dq (log T )φ(d)φ(q) ≤ C 5 C 6 (v log T ) dq (log T )φ(d) 2 φ(q) . (8.8)
Therefore, by (8.5) and (8.6), we have

K(v) ≪ d≤T 2κ d ρ|d 1 ρ d (log T )φ(d) 2 2 ≪ 1 (log T ) 2 d≤T d 3 ρ|d 1 ρ φ(d) 4 ≪ 1 (log T ) 2 d≤T 1 d p|d (1 -1/p) -5 ≪ 1 (log T ) 2 d≤T 1 d p|d (1 + 1/ √ p) ≪ 1 (log T ) 2 d≤T d -1 ρ|d ρ -1/2 ≤ 1 (log T ) 2 ζ(3/2) d≤T 1 d ≪ 1 log T .
This proves that (8.3) can be written as

2T T |I(t, H)| 2 dt = 2T φ(q) q H -H (H -|v|)ℑ K(v) v dv + O T (log T ) 3 = 4T φ(q) q H 0 (H -v)ℑ K(v) v dv + O T (log T ) 3 .
(8.9)

Now we are interested about this integral over the interval 0 < v < 1/ log ξ, and we fix such a v. Observe that for a, b ∈ C, we have

|ℑ(a 2 b)| = |ℑ(a 2 )ℜ(b) + ℜ(a 2 )ℑ(b)| ≤ |b||ℑ(a 2 )| + |a| 2 |ℑ(b)| ≤ 2|a||b||ℑ(a)| + |a| 2 |ℑ(b)|. Then ℑ K(v) v ≤ d≤T 2κ ℑ T iv/2 φ iv (d) v A ′ d (v) 2 χ 0 (d) ≤ d≤T 2κ ℑ T iv/2 φ iv (d) v |A ′ d (v)| 2 + 2|φ iv (d)||A ′ d (v)| ℑ A ′ d (v) v χ 0 (d). (8.10)
Since sin(x) ≤ x for any x > 0, and by definition of φ iv , we have for d ≤ T 2κ , (d, q) = 1:

ℑ T iv/2 φ iv (d) v = d ρ|d µ(ρ) ρ sin(v log( √ T d/ρ)) v d 1 2 + 2κ log T ρ|d 1 ρ . (8.11) Also, ℑ A ′ d (v) v = h,h ′ d|hh ′ χ 0 (hh ′ )γ(h)γ(h ′ ) vh ′ h sin(-v log h) = d|δδ ′ δδ ′ |d ∞ 1 δδ ′   (k,qd)=1 M(kδ)τ -1/2 (kδ) kv sin(-v log(kδ))     (k,qd)=1 M(kδ ′ )τ -1/2 (kδ ′ ) k   ,
Zeros of linear combinations of Dirichlet L-functions on the critical line and hence by Corollary 4.6 and (7.5)

ℑ A ′ d (v) v d|δδ ′ δδ ′ |d ∞ τ 1/2 (δ) δδ ′   C 4 dq φ(d)φ(q) log ξ + O   p|d (1 + p -3/4 )(log T ) 1/3     × C 5 τ 1/2 (δ ′ ) dq (log T )φ(d)φ(q) ≤ C 4 C 5 √ κ dq φ(d) 2 φ(q) + O   √ d φ(d) 3/2 (log T ) 1/6 p|d (1 + p -3/4 )   .
(8.12)

The last inequality is justified as in (7.6). Collecting (8.11), (8.8), (8.5), (8.12) and injecting them in (8.10), we get (1 + 1/p) -1. Thus

ℑ K(v) v 1 log T 1 2 + 2κ (C 5 C 6 (v log T )) 2 + 2C 4 C 2 5 C 6 (v log T ) √ κ q 2 φ(q) 2 d≤ξ 2 d 3 χ 0 (d) φ(d) 4 ρ|d 1 ρ + O   1 (log T ) 7/6 d≤ξ 2 d 5/2 ρ|d 1 ρ φ(d) 7/2 p|d (1 + p -3/4 )   . ( 8
d≤ξ 2 f (d) d χ 0 (d) = k≤ξ 2 g(k)χ 0 (k) k d≤ξ 2 /k χ 0 (d) d ≤ d≤ξ 2 χ 0 (d) d ∞ k=1 g(k) k 2κ log T φ(q) q p 1 + 5p 5 -6p 4 + 5p 2 -4p + 1 (p -1) 5 p(p + 1
) .

Now we turn to the error term in (8.13). We have

d 5/2 ρ|d 1 ρ φ(d) 7/2 ≪ 1 d p|d (1 -1/p) -9/2 ≪ 1 d p|d (1 + p -3/4 ),
and hence the said error term is

≪ 1 (log T ) 7/6 d≤T 1 d p|d (1 + p -1/2 ) ≪ 1 (log T ) 7/6 d≤T 1 d ρ|d ρ -1/2 ≤ 1 (log T ) 1/6 ζ(3/2).
Therefore for any 0 < v < 1/ log ξ:

ℑ K(v) v 2κC 2 5 q φ(q) p 1 + 5p 5 -6p 4 + 5p 2 -4p + 1 (p -1) 5 p(p + 1) × 1 2 + 2κ C 6 (v log T ) 2 + 2C 4 C 6 (v log T ) √ κ . (8.14)
For commodity we define

C 7 (v log T ) := 1 2 + 2κ C 6 (v log T ) 2 + 2C 4 C 6 (v log T ) √ κ.
Now, we wish to study the quantity ℑ(K(v)/v) when v is larger, namely 1/ log ξ < v < H. We fix d ≤ ξ 2 co-prime to q. Using Lemma 4.5 with r = 2 and using (2.2), one finds that for δ|d ∞ , δ ≤ ξ:

(n,qd)=1 M(nδ)τ -1/2 (nδ) n 1+iv = -τ -1/2 (δ) (1 -θ)2 √ π √ v log ξ p|qd (1 -p -(1+iv) ) -1/2 ∆ 2 v log ξ δ -∆ 2 v log + ξ θ δ + O   τ 1/2 (δ) p|d (1 + p -3/4 )(log T ) -2/3   . (8.15)
For commodity we denote the quantity between brackets by ∆(ξ, δ, θ, v). Putting (8.15) in (8.7), this leads to

A ′ d (v) = d|δδ ′ δδ ′ |d ∞ δ,δ ′ ≤ξ 1 δ 1+iv δ ′   -τ -1/2 (δ) (1 -θ)2 √ π √ v log ξ p|qd (1 -p -(1+iv) ) -1/2 ∆(ξ, δ, θ, v)   B d (δ ′ ) + O     p|d (1 + p -3/4 ) d|δδ ′ δδ ′ |d ∞ τ 1/2 (δ) δδ ′ (log T ) 2/3 |B d (δ ′ )|     .
By (7.5), the error term is

≪ 1 (log T ) 7/6 d φ(d) 1 φ(d) p|d (1 + p -3/4 ) ≪ 1 (log T ) 7/6 1 d p|d (1 + p -3/4 ) 2 .
Using the bound in Lemma 4.3, we deduce that ∆(ξ, δ, θ, v) = O(A), and hence

A ′ d (v) 2 = 1 4(1 -θ) 2 πv(log ξ) 2 p|qd (1 -p -1-iv ) -1        d|δδ ′ δδ ′ |d ∞ δ,δ ′ ≤ξ τ -1/2 (δ)B d (δ ′ )∆(ξ, δ, θ, v) δ 1+iv δ ′        2 (8.16) + O   A (log T ) 13/6 1 d 2 p|d (1 + p -3/4 ) 3 + 1 (log T ) 7/3 1 d 2 p|d (1 + p -3/4 ) 4   .
Zeros of linear combinations of Dirichlet L-functions on the critical line Noticing that φ iv (d) = d 1+iv p|d (1 -p -1-iv ) and using this last equality in (8.6), we find

ℑ K(v) v = O(A/(v(log T ) 7/6 )) + 1 4(1 -θ) 2 π(v log ξ) 2 d≤ξ 2 (d,q)=1 ℑ        p|q (1 -p -1-iv ) -1 T iv/2 d 1+iv        d|δδ ′ δδ ′ |d ∞ δ,δ ′ ≤ξ τ -1/2 (δ)B d (δ ′ )∆(ξ, δ, θ, v) δ 1+iv δ ′        2       
.

Using this together with (8.14) and using the change of variable u = v log T in the following integrals, we find the following

H 0 (H -v)ℑ K(v) v dv 2κC 2
5 q φ(q) p 1 + 5p 5 -6p 4 + 5p 2 -4p + 1 (p -1) 5 p(p + 1)

1/κ 0 (A -u) (log T ) 2 C 7 (u) du + 1 4(1 -θ) 2 π(log ξ) 2 × d≤ξ 2 (d,q)=1 H 1/ log ξ (H -v)ℑ        p|q (1 -p -1-iv ) -1 T iv/2 d 1+iv        d|δδ ′ δδ ′ |d ∞ δ,δ ′ ≤ξ τ -1/2 (δ)B d (δ ′ )∆(ξ, δ, θ, v)/v δ 1+iv δ ′        2        dv (8.17)
Finally we have, for d ≤ ξ 2 , (q, d) = 1:

H 1/ log ξ (H -v)ℑ        p|q (1 -p -1-iv ) -1 T iv/2 d 1+iv        d|δδ ′ δδ ′ |d ∞ δ,δ ′ ≤ξ τ -1/2 (δ)B d (δ ′ )∆(ξ, δ, θ, v)/v δ 1+iv δ ′        2        dv ≤d H 1/ log ξ (H -v)( √ T d) iv p|q (1 -p -1-iv ) -1        d|δδ ′ δδ ′ |d ∞ δ,δ ′ ≤ξ τ -1/2 (δ)B d (δ ′ )∆(ξ, δ, θ, v)/v δ 1+iv δ ′        2 dv ≤d q φ(q) d|δ 1 δ ′ 1 δ 1 δ ′ 1 |d ∞ δ 1 ,δ ′ 1 ≤ξ d|δ 2 δ ′ 2 δ 2 δ ′ 2 |d ∞ δ 2 ,δ ′ 2 ≤ξ τ 1/2 (δ 1 )τ 1/2 (δ 2 ) δ 1 δ ′ 1 δ 2 δ ′ 2 |B d (δ ′ 1 )B d (δ ′ 2 )| × H 1/ log ξ (H -v) √ T d δ 1 δ 2 iv ∆(ξ, δ 1 , θ, v) v ∆(ξ,
if ξ ≥ δ ≥ ξ θ , 2 √ πe -3iπ/4 v(1 -θ) log ξ + O 32 3 if 1 ≤ δ < ξ θ .
Now we split the sums of (8.18) in four, according to whether δ i ≤ ξ θ , i = 1, 2, or not. In the case ξ ≥ δ 1 , δ 2 > ξ θ , the (8.18) The three other cases lead to the same upper bound. Thus, summing (8.18) over d ≤ ξ 2 , (q, d) = 1, we find that the second summand of (8.17) is

(log T ) 2

A -1/κ 1/2 -2κ + 2 (1/2 -2κ) 2 + 256/9 (1 -θ) 2 πκ 2 (Aκ -log(Aκ) -1) This concludes.

+ 32/3 √ π(1 -θ)κ 2 (1 -Aκ + Aκ log(Aκ)) C 2 5 q 2 φ(q) 2 d≤ξ 2 d 2 χ 0 (d) φ(d) log T     d|δδ ′ δδ ′ |d ∞ τ 1/2 (δ)τ 1/2 (δ ′ ) δδ ′     2 1 (log T ) 2 A -1/

Computations

Now it only remains to explain how we did the computations of Theorem 1.2.

Proof (of Theorem 1.2): In the case of small fixed N, we have to compute C 1 . This is easily done by noticing that C 7 (v) is an increasing and positive function of v, for any fixed θ ∈ (0, 1), as standard arguments show. Therefore, it is easy to bound the integrals of K 2 and K 4 by the rectangle method, for example. We chose to bound these integrals by using 100 regular rectangles. Then, the choice of A was done by Sage1 , by finding a positive root of the derivative of

2 2π 1 2A -4N C 1 (A) + C 2 A 3 .
This gives a result for a fixed parameter θ. To find a good θ we split the interval (0, 1) in 10 4 , and then applied the process to each of these θ. The program then returned the one giving the best result.

In the case of large N, we have for some constant C

κ F ≈ 2π 1 2A -4NC A log A A 3 ,
which is optimized when 1/A ≈ NA log A/A 3 , and hence when A is of the form λN log N for some positive number λ. Thus, we may say that

C 1 (A) ≈ 8C 2 5 K 1 A log A ≈ 8C 2 5 K 1 λN log(N) 2 .
Because of the definition of these constants, we should take θ as small as possible, but this will makes C 2 become very large. With this in mind, if N is so large that C 2 becomes negligible, we may write that heuristically

2π 1 2A -4N C 1 (A) + C 2 A 3 ≈ 2π N log N 1 2λ - 32C 2 5 K 1 λ 2 ,
which is optimized by taking λ = 128C 2 5 K 1 = 128C 2 5 (θ)K 1 (θ). Now, we simply inject this in the lower bound for κ F and find how large N should be to compensate the size of C 2 (θ).

We put A = λ(θ)N log N, fix ε > 0 and we take 0 < θ < 1 so that 1/(1 -θ) = 1 + ε.

Observe that (e ̺ θ + e ̺ θ θ )/ √ ̺ θ is an increasing function of θ, and note that for ε < 1/3, we have θ < 1/4. Thus This implies that

C 2 (θ) ≤ 6(C + 3 + 1 + 2 C + 3 ) κ 2 (1 + ε) 6 ε 2 =: (1 + ε) 6 ε 2 C + 2 .
We also take

N ≥ C + 2 (λ -) 3 ε 3 =: N 0 ε 3 (9.

  Figure 1 illustrates this phenomenon by comparing the function ζ(s), the function ζ(s) mollified by η(s), and the function ζ(s) mollified by Selberg's mollifier.

Figure 1 :

 1 Figure 1: Mollification at work on ζ(s). Here we chose the simplest θ = 1/2 and κ = 1/8.

S

  |I(t, H)| dt = S J(t, H) dt, and hence, by (2.6) and the Cauchy-Schwarz inequality S J(t, H) dt meas(S)C 1 (A) T (log T ) 2 1/2

Theorem 3 . 1 :

 31 Let χ 1 and χ 2 be distinct primitive Dirichlet characters. Let a < b be two real numbers and 1 [a,b] be the characteristic function of [a, b]. Then as T → ∞ we have

Figure 2 :

 2 Figure 2: The continuous line is L, and the dotted one is Hankel's contour.

. 8 )

 8 Now we denote the Hurwitz zeta-function by ζ(s, y), 0 < y ≤ 1, and we let H ′ := h/(h, k) and K := k/(h, k). The Hurwitz zeta-function ζ(s, y) is the analytic continuation of the function ∞ n=0 (n + y) -s , defined for ℜ(s) > 1, to the whole complex plan except at s = 1. When s is Zeros of linear combinations of Dirichlet L-functions on the critical line close to 1, we have ζ(s, y) = 1/(s -1) + O y (1), and hence ζ(s, y) -ζ(s) is regular on the whole complex plan. We define s + β) . For s ∈ C, we have n,m χ(n)χ(m) n s+α m s+β e -nmh qk = 1≤ν,µ≤qK χ(ν)χ(µ)e -νµh qk n≡ν(mod qK) m≡µ(mod qK)

( 7 . 6 )

 76 and the fact that for integers a, b ≥ 1 we have φ([a, b])φ((a, b)) = φ(a)φ(b), one gets

. 10 )≤ 1 2P

 101 If ν p is the p-adic valuation, P > 0 is real number, and d ≤ T 2κ , then this last sum is≤ p|d p m ≤T 2κ ,m≥1 (log p)p min(νp(d),m) p m + p∤d p m ≤T 2κ ,m≥1 log d + 4 p≤P p m ≤T 2κ ,m≥1 log p p m + P P -<p≤T 2κ p m ≤T 2κ ,m≥1 log p p m ≤ 2κ log T + 4 log P p≤P 1 p -1 + 1 + O 1 P k≤T 2κ Λ(k) k .

p m |d,m≥1 log p p p - 1 ≤

 1 2φ(d) p m |d,m≥1 log p = 2φ(d) log d ≤ 4κφ(d) log T.

  (n) := m|n µ(m) n m 1+z . By Möbius inversion, it is easy to see that if n ≥ 1 is an integer, then n 1+z = m|n φ z (m). (8.4) Zeros of linear combinations of Dirichlet L-functions on the critical line Note that for any real number |v| ≤ H, |φ iv (d)| ≤ d

. 13 ) 1 ρ 1 4 1 p

 13111 To estimate the double sum in the main term, we let f (d) , which is multiplicative, and we let g = µ ⋆ f , which is also multiplicative. Then for p a prime and m ≥ 2, we have g(p m ) = f (p m ) -f (p m-1 ) = p p-m . For m = 1, we find g(p) = f (p) -

C 5 2 √ πκ̺ 1 5 . 4 ( 1 + ε) 4 p 1 + 3p 2 -

 52154112 (θ) ≤ (1 + ε) e ̺ 1/4 + e ̺ 1/4 /4Moreover, K 1 (θ) = (1 + ε)K 1 (0). Thus we have that(1 + ε) 3 λ -:= 128(C - 5 ) 2 K 1 (0)(1 + ε) 3 ≤ λ(θ) ≤ 128(C + 5 ) 2 K 1 (0)(1 + ε) 3 =: (1 + ε) 3 λ + . (1/4) + e ̺(1/4)/4 )Γ(1/4) π̺(1/4)Γ(3/4) 3p + 1 p 4 -3p 3 + 3p 2 -p =: (1 + ε) 4 C + 3 .

  integral term is equal to

	H 1/ log ξ	u 1/ log ξ	√ δ 1 δ 2 T d	iv	4πiv 2 log	ξ δ 1	log	ξ δ 2	+ O	128 3	√	π(1 -θ)v log ξ +	1024 9	dv v 2 du
	4π(1 -θ) 2 log 2 ξ log( √ T d/(δ 1 δ 2 ))		H 1/ log ξ	 	√ δ 1 δ 2 T d	iu	-	√ δ 1 δ 2 T d	i/ log ξ	  du +	1024 9	H 1/ log ξ	H -v v 2 dv
	+	128 3	√	π(1 -θ) log ξ	H 1/ log ξ	H -v v	dv	(8.19)
	≤ 4π(1 -θ) 2 log 2 ξ	H -1/ log ξ log(T 1/2-2κ )	+	2 log(T 1/2-2κ ) 2 +	1024 9	(H log ξ -log(H log ξ) -1)
	+	128 3	√	π(1 -θ) log ξ	1 log ξ	-H + H log(H log ξ)
	= 4π(1 -θ) 2 κ 2 A -1/κ 1/2 -2κ + 128 3 √ π(1 -	+	2 (1/2 -2κ) 2 +	1024 9	(Aκ -log(Aκ) -1)

θ)(1 -Aκ + Aκ log(Aκ)).

  Here we used(7.5) to bound B d (δ) and (7.8) to estimate the sum over d, which is shown to be Zeros of linear combinations of Dirichlet L-functions on the critical line≤ d≤ξ 2 d 2 χ 0 (d)/φ(d)3 . Using (8.9) and (8.17), we find the expected result :|I(t, H)| 2 dt T (log T ) 2 8C 2 5 (K 1 A log A + K 2 A + K 3 log A + K 4 ) ,

						2T
					T
	where						K 1 :=	p	1 +	3p 2 -3p + 1 p 4 -3p 3 + 3p 2 -p	32/3 √ π(1 -θ)	,	(8.20)
	K 2 := κ + p	p	1 + 1 + p 4 -3p 3 + 3p 2 -p 5p 5 -6p 4 + 5p 2 -4p + 1 (p -1) 5 p(p + 1) 3p 2 -3p + 1 κ 1/2 -2κ	1/κ + 0 (1 -θ) 2 π C 7 (v) dv 256/9	+	32/3 √ π(1 -θ)	(log κ -κ) ,
						K 3 := -	p	1 +	3p 2 -3p + 1 p 4 -3p 3 + 3p 2 -p	256/9 (1 -θ) 2 πκ	,
	and finally
	K 1/κ	vC 7 (v) dv
										0
	+	p	1 +	3p 2 -3p + 1 p 4 -3p 3 + 3p 2 -p	4κ -1/2 (1/2 -2κ) 2 -	256/9 (1 -θ) 2 πκ	(1 + log κ) +	32/3 √ π(1 -θ)κ	.
				κ 1/2 -2κ	+	2 (1/2 -2κ) 2 +	256/9 (1 -θ) 2 πκ 2 (Aκ -log(Aκ) -1)
	+	32/3 √ π(1 -θ)κ 2 (1 -Aκ + Aκ log(Aκ)) C 2 5	q φ(q)	× 2κ	p	1 +	3p 2 -3p + 1 p 4 -3p 3 + 3p 2 -p	.

4 := -κ p 1 + 5p 5 -6p 4 + 5p 2 -4p + 1 (p -1) 5 p(p + 1)

Note that this is precisely here that the quality of our lower bound for the proportion of critical zeros is at stake. Selberg's paper led to an error term ≪ √ v log ξ, and hence the integral of our(8.19) would becomeH 1/ log ξ (H -v)/ √ v dv.This would produce a term ≍ A 3/2 , while our biggest term here is ≍ A log A.

Code available at https://pagepersotan.wordpress.com/articles/.

Except in the case N = 1, where we used (2.9) instead.

Zeros of linear combinations of Dirichlet L-functions on the critical line Finally, note that for 0 ≤ v ≤ 8, θ ∈ (0, 1), we have 1/2 ≤ ρ(v, θ) ≤ 1 and hence

Thus, for N satisfying (9.1), we have

Therefore, for any such N, we find that

Using the fact that 1/(1 + ε) 3 ≥ 1 -3ε, a computation then leads to the expected result.