Bernard Damien
email: damien.bernard@math.univ-bpclermont.fr

D Bernard

Small first zeros of L-functions

Keywords: Mathematics Subject Classification : 11M41, 11M50, 11F67 L-function, density theorem, random matrix theory, smallest zero, differential equation with temporal shifts. 1

published or not. The documents may come

Introduction 1.Preview of results

The existence of a deep link between non-trivial zeros of natural families of L-functions and eigenvalues of random matrices has been speculated since Montgomery's work [START_REF] Montgomery | The pair correlation of zeros of the zeta function[END_REF]) in the seventies. So, we are able to assign a classical compact group of matrices to many classical families of L-functions. We can refer to [START_REF] Iwaniec | Low lying zeros of families of L-functions[END_REF], [START_REF] Fouvry | Low-lying zeros of dihedral L-functions[END_REF], [START_REF] Hughes | Linear statistics of low-lying zeros of L-functions[END_REF], [START_REF] Ricotta | Statistics for low-lying zeros of symmetric power L-functions in the level aspect[END_REF] or [START_REF] Miller | One-and two-level densities for rational families of elliptic curves: evidence for the underlying group symmetries[END_REF]. Using the one-level density, Hughes and Rudnick obtained informations about the lowest zero of Dirichlet L-functions ([HR03], section 8). Our aim is to generalise these results.

Extreme low-lying zeros of a natural family of L-functions Let F(Q) be a finite set of L-functions with analytic conductor Q. We build the associate family F = Q≥1 F(Q). We assume Riemann hypothesis for any function in F. We also assume the density theorem for F with test functions Φ satisfying supp Φ ⊂ [-ν; ν] with ν < ν max (F). We write W * [F] for the one-level density for non-trivial low-lying zeros of functions in F. It turns out that the one-level densities that have been identified up to now have always been of the shape W [G] whose Fourier transform is given by

W [G](y) = δ 0 (y) + δ 2 η(y) + ε (1.1)
where δ 0 is the Dirac function, η is defined on R by

η(y) =    1 if |y| < 1 1 2 if |y| = 1 0 if |y| > 1
and δ and ε are given in table 1.

G U O Sp SO + SO - δ 0 0 -1 1 -1 ε 0 1/2 0 0 1
Table 1: Value of (δ, ε).

Finally, for G = Sp, SO + or SO -, let n ≥ 1 be the only integer such that n -1 < ν max (F) ≤ n and consider the equation in λ given by

δ λ cos θ λ - n-1 k=0 U k (λ) sin θ λ -kδ π 2 δα R (k) 2 -1 + εβ R (k) + 2ε λ n-1 k=0 U k (λ) cos θ λ -kδ π 2 = 0. (1.2)
In this equation, let U k be the k-th Chebyshev polynomial of the second kind, (δ, ε) is given in the table 1 and parameters θ λ , α R and β R are defined in lemma 20 (page 35) and proposition 4 (page 38). We prove the following theorem on the smallest non negative imaginary part γf,1 of a non-trivial normalised zero of L(f, .) in F.

Theorem 1 We have

lim sup Q→+∞ min L(f,.)∈F (Q) γf,1 ≤ 1 2ν max (F) ×              1 if W * [F] = W [U ] 4V -1 1 + 2 νmax(F ) if W * [F] = W [O] 4V -1 1 + (δ + 2ε) 2 νmax(F ) if W * [F] = W [Sp], W [SO + ] or W [SO -] and ν max (F) ≤ 1 νmax(F ) π lim R→νmax(F )/2 - λ R if W * [F] = W [Sp], W [SO + ] or W [SO -] and ν max (F) > 1
where λ R is the smallest positive root of equation (1.2) which is not a root of U n U n-1 (with n -1 < ν max (F) ≤ n) and where V is defined by

V : [0, 1 4 [∪] 1 4 , x 1 [ -→ R x -→ tan(2πx) 2πx with x 1 = inf x > 0, tan (2πx) 2πx
= 1 ≈ 0, 71.

Remark 1 We plot the upper bound in theorem 1, denoted

M W * [F ] (ν max (F)), for each W * [F].
From top to bottom, we have

W * [F] = W [Sp], W * [F] = W [U ], W * [F] = W [SO + ], W * [F] = W [O] and W * [F] = W [SO -].
Representative curves of ν max (F) -→ M W * [F ] (ν max (F))

Remark 2 In the orthogonal case, when ν max (F) goes to infinity, we have

2 νmax(F ) V -1 1 + 2 νmax(F ) = √ 6 πνmax(F ) 3/2 1 - 6 5νmax(F ) + o 1 νmax(F ) 3/2 . Representative curve of V -1
Remark 3 In the case 1 < ν max (F) ≤ 2, we can simplify equation (1.2) as follows

(δ + 2ε) 1 -4λ 2 λ (sin λ(1 -R) -2δλ cos λR) -[(δ + 2ε)(1 -R) -1 + 4ε] [cos λ(1 -R) -2δλ sin λR -2λ tan Θ R (sin λ(1 -R) -2δλ cos λR)] = 0 where Θ R = 1 2 R - 1 2 + π 2 1 + δ 2 .
Moreover, this relation always vanishes in λ = 1/2 which is a root of U 2 .

In order to give several instances, let H * k (q) be the set of primitive holomorphic cusp forms of prime level q end even weight k ≥ 2. Let r be a positive integer. We define H r (q) = {L(Sym r f, s), f ∈ H * k (q)} and if r is odd, H ± r (q) = {L(Sym r f, s), f ∈ H * k (q) and ε(Sym r f ) = ±1} .

Then, we have several families of symmetric power L-functions

H r = q prime
H r (q) and H ± r =

q prime H ± r (q).

These families have been studied in [START_REF] Iwaniec | Low lying zeros of families of L-functions[END_REF] (theorem 1.1, for the case r = 1) and [START_REF] Ricotta | Statistics for low-lying zeros of symmetric power L-functions in the level aspect[END_REF] (theorem B). We may sum up some properties of these natural families of L-functions in the following table where ν 1,max (1, k, θ 0 ) = 2 and ν 1,max (r, k, θ 0 ) = 1 -1 2(k -2θ 0 ) 2 r 2 if r ≥ 2 with θ 0 = 7 64 , and for ǫ = ±1, ν ǫ 1,max (1, k, θ 0 ) = 2 and ν ǫ 1,max (r, k, θ 0 ) = inf ν 1,max (r, k, θ 0 ), 3 r(r + 1) if r ≥ 2.

F H r (r even) H r (r odd) H + r (r odd) H - r (r odd) ν max (F ) ν 1,max (r, k, θ 0 ) ν 1,max (r, k, θ 0 ) ν ǫ 1,max (r, k, θ 0 ) ν ǫ 1,max (r, k, θ 0 )

ρ max (F ) 1 r 2 1 r 2 1 2r(r+2) 1 2r(r+2) W [F ] W [Sp] W [O] W [SO + ] W [SO -] W * [F ] W [Sp] W [O] W [SO + ] W [Sp]
N →+∞ min f ∈H * k (N ) γf,1 ≤ V -1 (2) < 0, 19 lim sup N →+∞ min f ∈H + k (N ) γf,1 ≤ 0, 22 lim sup N →+∞ min f ∈H - k (N )
γf,1 ≤ 0, 39 and lim sup

q→+∞ q prime min f ∈H * k (q) γSym r f,1 ≤ 2 ν 1,max (r, k, θ 0 ) V -1 1 + (-1) r+1 2 ν 1,max (r, k, θ 0 ) and if r is odd, ǫ = ±1, we have lim sup q→+∞ q prime min f ∈H * k (q) ε(Sym r f )=ǫ γSym r f,1 ≤ 2 ν ǫ 1,max (r, k, θ 0 ) V -1 1 + ǫ 2 ν ǫ 1,max (r, k, θ 0 )
.

Actually, since H -(N ) = ∅ for large N , we get (see lemma 1 page 9) lim

N →+∞ min f ∈H * k (N )
γf,1 = 0.

Proportion of L-functions which have a small first zero Combining Bienaymé-Chebyshev inequality and statistics for low-lying zeros of symmetric power L-functions, we can obtain a positive proportion of L-functions in our family which have a small first zero. Hughes and Rudnick exposed this phenomenon in the case of Dirichlet L-functions which have a unitary symmetry. Symmetric power L-functions allows us to deal with all currently known symmetry group. H * k (q) denotes the set of primitive holomorphic cusp forms of prime level q and even weight k ≥ 2 and let ω q (f ) be the harmonic weight associated to f in H * k (q). Let also ε(Sym r f ) denotes the sign of the functional equation associated to the L-function L(Sym r f, s).

Theorem 2 Let r be a fixed positive integer. Assuming the Riemann hypothesis for all symmetric power L-functions of order r.

If β ≥ πr 4 4 6π 3 r 4 -24(-1) r πr 2 + 9π -π 3 + 2 √ 6 -π 4 + 6π 4 r 4 + 7π 2 + 12 -24(-1) r π 2 r 2 6π 4 r 4 + 48(-1) r+1 π 2 r 2 + 96 -3π 2 -π 4 then lim inf q→+∞ q prime f ∈H * k (q) γSym r f,1 ≤β ω q (f ) ≥ 1 - π 2 6 16(π 2 + 3)β 4 + 8r 4 (9 -π 2 )β 2 + (3 + π 2 )r 8 [r 6 π 2 -4β 2 r 2 π 2 + 16β 2 (-1) r ] 2 .
Moreover, if r is odd and σ = ±1.

If β r(r+2) √ π ≥ π[24π 2 r 2 (r + 2) 2 + 48σr(r + 2) + 9 -π 2 ] + 2 √ 6 -π 4 + 24π 4 r 2 (r + 2) 2 + 7π 2 + 12 + 48σπ 2 r(r + 2) 24π 4 r 2 (r + 2) 2 + 96σπ 2 r(r + 2) + 96 -3π 2 -π 4 then lim inf q→+∞ q prime 2 f ∈H * k (q) ε(Sym r f )=σ γSym r f,1 ≤β ω q (f ) ≥ 1 - π 2 24 (π 2 + 3)β 4 + 2r 2 (r + 2) 2 (9 -π 2 )β 2 + (π 2 + 3)r 4 (r + 2) 4 [2σβ 2 + π 2 β 2 r(r + 2) -π 2 r 3 (r + 2) 3 ] 2 .
In order to give some examples for small values of r, we plot the graph of the function with parameter β which is associated to the lower bound in the first part of theorem 2. On these representative curves, the critical value of β corresponds to the minimal value of β which appears in theorem 2.

r = 1 r = 2 r = 3 r = 4
Similarly, for a fixed small odd value of r, we plot the lower bounds of theorem 2 in different cases: when there is no restriction on the sign ε(Sym r f ) of the functional equation, when ε(Sym r f ) = +1 and, finally, when ε(Sym r f ) = -1.

First, if r = 1:

ε(Sym r f ) = ±1 ε(Sym r f ) = +1 ε(Sym r f ) = -1
Second, if r = 3:

ε(Sym r f ) = ±1 ε(Sym r f ) = +1 ε(Sym r f ) = -1

Notations

The following notations will be used throughout this paper.

• ⌊x⌋ denotes the floor of the real number x.

• * n≥0 means the sum is running over odd non-negative integers.

• C ∞ c (R) denotes the set of infinitely differentiable functions which are compactly supported.

• For 1 ≤ p < +∞, L p (R) refers to the set of functions f :

R → R such that R |f (x)| p dx < +∞. In this case, we put ||f || p = R |f (x)| p dx 1/p . If f and g are in L 2 (R), let (f, g) L 2 = R f (x)g(x)dx. • If Φ is in L 1 (R), Φ(u) = R Φ(x)e -2iπxu
dx is called the Fourier transform of Φ. When it is allowed, we can apply the inverse transform formula Φ(x) = R Φ(u)e 2iπxu du.

• If f and g are in L 2 (R), f * g(u) = R f (t)g(u -t)dt
is the convolution product of f and g.

• S ν (R) denotes the set of even Schwartz functions whose Fourier transforms are compactly supported in [-R, R] with 0 < R < ν.

Statistics for low-lying zeros

Let F be a natural family of L-functions 1 all of whose satisfies the Riemann hypothesis. Consider a L-function L(f, s) in F with analytic conductor c f . Let

F(Q) = {L(f, s) ∈ F , c f = Q}.
Each zero ρ f of L(f, s) which is on the critical line ℜ(s) = 1/2 can be written ρ f = 1 2 + iγ f , and we denote ρf = 1 2 + iγ f with γf = γ f ln c f 2π the normalized zero.

In order to study the distribution of low-lying zeros of L(f, s), for any test function Φ in S ν (R), we define the low-zeros sum

D[Φ](f ) = γf Φ(γ f )
where the sum is running over the imaginary parts of normalised zeros counted with multiplicity. F(Q) can be seen as a measurable space where measurable sets are all its subsets and which is equipped with the counting probability measure µ F (Q) . D[Φ] is a measurable function on F(Q). However, we are unable to determine the asymptotic behaviour of D[Φ](f ) for a single L-function. As a consequence, we must take into account a family of L-functions in order to obtain a significant result. That is why, we define the one-level density as the expectation of D[Φ]:

E F (Q) (D[Φ]) = 1 |F (Q)| L(f,s)∈F (Q) D[Φ](f )
We also define the variance of the one-level density:

V F (Q) (D[Φ]) = E([D[Φ]-E F (Q) (D[Φ])] 2 ) = 1 |F (Q)| L(f,s)∈F (Q)   D[Φ](f ) - 1 |F (Q)| L(f,s)∈F (Q) D[Φ](f )   2
Our purpose is to find the asymptotic behaviour of

E F (Q) (D[Φ]) and V F (Q) (D[Φ]) when Q goes to infinity. The density conjecture predicts lim Q→+∞ 1 |F(Q)| L(f,s)∈F (Q) D[Φ](f ) = R Φ(t)W [F](t)dt
where W [F] is a density function characterised by F. In order to estimate the one-level density, we convert sums over zeros to sums over primes. Unfortunately, we are able to evaluate these sums over primes only if the support of the test function is small. That's why, density theorems are proved only for test functions in S νmax(F ) (R) with ν max (F) fixed. Currently, the maximal value for ν max (F) is 2 whereas the density conjecture does not predict any restriction on the support of test functions.

Sometimes, we will prefer using the harmonic measure for technical conveniences, rather than the Dirac one. For instance, in the case of symmetric power L-functions. Precisely, if A is a subset of H r (q), let µ h Hr(q) (A) = L(Sym r f,s)∈A w q (f )

1 Our definition of L-function is the one of [IK04] chapter 5.

and

E h Hr(q) (D[Φ]) = f ∈H * k (q) ω q (f )D[Φ](Sym r f ) where ω q (f ) = Γ(k -1) (4π) k-1 f, f q and V h Hr(q) (D[Φ]) = f ∈H * k (q) ω q (f ) D[Φ](Sym r f ) -E h Hr(q) (D[Φ]) 2 .
Then, if r is odd and if A is a subset of H ± r (q), let

µ h H ± r (q) (A) = 2 L(Sym r f,s)∈A w q (f )
and

E h H ± r (q) (D[Φ]) = 2 f ∈H * k (q) ε(Sym r f )=±1 ω q (f )D[Φ](Sym r f ) and V h H ± r (q) (D[Φ]) = 2 f ∈H * k (q) ε(Sym r f )=±1 ω q (f ) D[Φ](Sym r f ) -E h H ± r (q) (D[Φ]) 2 .
Actually, harmonic measures are asymptotic probability measures since we only have lim

q→+∞ q prime f ∈H * k (q)
ω q (f ) = 1 and lim

q→+∞ q prime f ∈H * k (q) ε(Sym r f )=±1 ω q (f ) = 1 2 . (1.3)
The first relation comes from Petersson trace formula ([RR11], proposition 2.2) and the second one is subject to an assumption (Hypothesis N ice(r, f ) of [START_REF] Ricotta | Statistics for low-lying zeros of symmetric power L-functions in the level aspect[END_REF]) we are assuming in order to get density theorems for these families.

1.4 What is the smallest zero of a L-function ?

If L(f, s) is a self-dual L-function, the sign of the functional equation is equal to ±1. Moreover, due to the following observation, we need to define γf,1 which appears in theorems 1 and 2 and to explain the consequences on density theorems.

Lemma 1 ([IK04], proposition 5.1) Let L(f, s) be a self-dual L-function with ε(f ) = -1. Then L f, 1 2 = 0. If F is a natural
family, all of whose L-functions L(f, s) are self-duals and satisfy ε(f ) = -1. Due to the previous lemma, we can denote non-trivial zeros of L(f, s) by

{ρ f,0 } ∪ {ρ f,i , i ∈ Z * }
where ρ f,0 = 1/2 and ρ f,i = 1ρ f,-i if i = 0. Moreover, we have:

... ≤ Im(ρ f,-1 ) ≤ Im(ρ f,0 ) = 0 ≤ Im(ρ f,1 ) ≤ Im(ρ f,2 ) ≤ ...
In the other cases, we use the same notations without ρ f,0 .

We are going to expose its consequences on statistics of low-lying zeros. Let F a natural family of L-functions, we define

D * [Φ](f ) = i∈Z * Φ(γ f,i ). The density function W * [F] is defined to satisfy lim Q→+∞ E F (Q) (D * [Φ]) = R Φ(t)W * [F](t)dt.
(1.4) If all L-functions L(f, s) in F are self-duals and satisfy ε(f ) = -1, we have

E F (Q) (D[Φ]) = Φ(0) + E F (Q) (D * [Φ]) and V F (Q) (D[Φ]) = V F (Q) (D * [Φ]). (1.5)
This phenomenon occurs for the family H -. It has been shown in [START_REF] Iwaniec | Low lying zeros of families of L-functions[END_REF] (equation (1.18)) that the symmetry group associated to H -is SO -. In other words, we have lim

N →+∞ N squarefree E H -(N ) (D[Φ]) = R Φ(t)W [SO -](t)dt ie W [H -] = W [SO -].
As a result, we deduce lim

N →+∞ N squarefree E H -(N ) (D * [Φ]) = R Φ(t)W [Sp](t)dt ie W * [H -] = W [Sp]. (1.6)
Similarly, the family H - r (r odd) has also SO -as symmetry group ([RR11], theorem A). Nevertheless, for the harmonic measure, relation (1.5) becomes

E h H - r (q) (D[Φ]) = 2Φ(0)µ h Hr(q) H - r (q) + E h H - r (q) (D * [Φ])
and

V h H - r (q) (D[Φ]) = V h H - r (q) (D * [Φ]) + 2Φ(0) Φ(0)µ h Hr(q) H - r (q) + E h H - r (q) (D * [Φ]) 1 -2µ h Hr(q) H - r (q) 2 .
Thanks to relation (1.3), we have

lim q→+∞ q prime E h H - r (q) (D[Φ]) = Φ(0) + lim q→+∞ q prime E h H - r (q) (D * [Φ]) and lim q→+∞ q prime V h H - r (q) (D[Φ]) = lim q→+∞ q prime V h H - r (q) (D * [Φ]).
As a result, with theorems A and D of [RR11], we get

lim q→+∞ q prime E h H - r (q) (D * [Φ]) = R Φ(t)W [Sp](t)dt and lim q→+∞ q prime V h H - r (q) (D * [Φ]) = 2 R |u| Φ(u) 2 du. (1.7)
2 Proportion of L-functions with a small smallest zero

In this part, we prove theorem 2. The starting point is the following proposition. We do not give a proof of this result since it is essentially the same than the proof theorem 8.3 from [START_REF] Hughes | Linear statistics of low-lying zeros of L-functions[END_REF].

Proposition 1 Let g be in S R (R) and Φ(x) = (x 2 -β 2 )g 2 (x). Let B(g) = R x 2 g 2 (x)W * [F](x)dx R g 2 (x)W * [F](x)dx .
We assume

lim q→+∞ q prime E h F (q) (D * [Φ]) = R Φ(t)W * [F ](t)dt and lim q→+∞ q prime V h F (q) (D * [Φ]) = V F (Φ).
Then, if β > B(g), we have

lim inf q→+∞ q prime µ h F (q) ({L(f, s) ∈ F; γf,1 < β}) ≥ 1 - V F (Φ) R Φ(t)W * [F](t)dt 2 =: Borne F (β 2 ). (2.1)
Remark 4 The same result holds for all natural family of L-functions with the counting probability measure instead of the harmonic measure.

This proposition gives a result only if the right member term is positive. We denote β min (g) the smallest value of β > B(g) such that this property is satisfied. Thanks to β min (g) > B(g), we may detect a zero. We have:

lim inf q→+∞ q prime min L(f,s)∈F (q)
γf,1 ≤ β min (g) Actually, we prove a better upper bound in theorem 1.

Ricotta and Royer proved in [START_REF] Ricotta | Statistics for low-lying zeros of symmetric power L-functions in the level aspect[END_REF] (theorems A, B and D), that if Φ is in S ρmax(F ) (R), then lim

q→+∞ q prime E h F (q) (D[Φ]) = R Φ(t)W [F](t)dt and lim q→+∞ q prime V h F (q) (D[Φ]) = 2 R |u| Φ(u) 2 du
where ρ max (F) and

W [F] are given in table 2. Since ρ max (F) < 1, we have R Φ(t)W * [F](t)dt = Φ(0) + σ F 2 Φ(0)
with:

F H r (r even) H r (r odd) H + r (r odd) H - r (r odd) σ F (-1) r+1 = -1 (-1) r+1 = 1 1 -1
In order to obtain explicit lower bound in the previous proposition, we specialise relation (2.1) in a fixed test function.

Lemma 2 If 0 < R < 1/2, g 0 (u) = cos πu 2R 1 [-R,R] (u) and Φ(x) = (x 2 -β 2 )g 2 0 (x), then Borne F (β 2 ) = 1 - 2π 2 R 2 3 256(3 + π 2 )R 4 β 4 + 32(9 -π 2 )R 2 β 2 + π 2 + 3 (128σ F R 3 β 2 + 16π 2 R 2 β 2 -π 2 ) 2 and B(g 0 ) = 1 4R 1 + σ F 8R π 2
.

Proof: We have

Φ(u) = 2R-|u| 2 1 16R 2 -β 2 cos πu 2R -1 π 1 16R + β 2 R sin π|u| 2R .1 [-2R,2R] (u) Φ(0) = 1 16R -β 2 R and Φ(0) = -16β 2 R 2 π 2 .
Thanks to several integration by parts, we obtain

R |u| Φ(u) 2 du = 768R 4 β 4 + 3 + 288β 2 R 2 + π 2 -32β 2 R 2 π 2 + 256R 4 β 4 π 2 768π 2 . Lemma 3 Let Borne + F (β 2 ) = max 0; Borne F (β 2 ) . The following table sums up variations of β -→ Borne + F (β 2 )
where

β min (g 0 ) = 1 4R π 2 (3π 2 + 24σ F R -2(π 2 -9)R 2 ) + 4πR 9π 4 + 72σ F Rπ 2 -6(π 4 -7π 2 -12)R 2 3π 4 + 48σ F Rπ 2 + (192 -6π 2 -2π 4 )R 2 . β B(g0) βmin(g0) +∞ Borne + F (β 2 ) 0 -→ 0 ✒ 1 -2π 2 R 2 3 π 2 +3 (π 2 +8Rσ F ) 2
Proof: First, we study the sign of Borne F (X). It is the same as the quadratic polynomial's

Y (X) = 3π 4 -6π 2 R 2 -2π 4 R 2 + X -96R 2 π 4 -768σ F π 2 R 3 -576R 4 π 2 + 64R 4 π 4 +X 2 768R 4 π 4 + 12288σ F R 5 π 2 + 49152R 6 -1536R 6 π 2 -512R 6 π 4 .
Let L(R) be its leading coefficient and ∆(R) its discriminant. We get

∆(R) = -98304(π 4 -7π 2 -12)π 2 <0 R 6 (R -R 1 )(R -R 2 ) with R 1 = min π 2 12σ F + 6(π 2 -3)(π 2 -4) 2(π 4 -7π 2 -12) ; π 2 12σ F -6(π 2 -3)(π 2 -4) 2(π 4 -7π 2 -12) R 2 = max π 2 12σ F + 6(π 2 -3)(π 2 -4) 2(π 4 -7π 2 -12) ; π 2 12σ F -6(π 2 -3)(π 2 -4) 2(π 4 -7π 2 -12) . Numerical values If σ F = -1, then R 1 ≈ -8, 330 and R 2 ≈ 1, 074. If σ F = 1, then R 1 ≈ -1, 074 and R 2 ≈ 8, 330.
In addition,

L(R) = 256(192 -6π 2 -2π 4 ) <0 R 4 (R -R 3 )(R -R 4 ) with R 3 = min π 2 -24σ F + 6π 2 (π 2 + 3) 192 -6π 2 -2π 4 , π 2 -24σ F -6π 2 (π 2 + 3) 192 -6π 2 -2π 4 R 4 = max π 2 -24σ F + 6π 2 (π 2 + 3) 192 -6π 2 -2π 4 , π 2 -24σ F -6π 2 (π 2 + 3) 192 -6π 2 -2π 4 . Numerical values If σ F = -1, then R 3 ≈ -8, 210 and R 4 ≈ 0, 573. If σ F = 1, then R 3 ≈ -0, 573 and R 4 ≈ 8, 210.
Therefore, we always have

R 1 < R 3 < 0 < 1 2 < R 4 < R 2 .
Let W (R) be degree 1 monomial coefficient in Y , then

W (R) = 32π 2 R 2 (-3π 2 -24σ F R + 2(π 2 -9)R 2 ). Since 0 < R < 1/2, we have ∆(R) ≥ 0 and L(R) ≥ 0. The polynomial Y has two real roots, X 1 (R) = -W (R) -∆(R) 2L(R) and X 2 (R) = -W (R) + ∆(R) 2L(R) .
Now, we want compare to X 1 (R), X 2 (R) and B(g

0 ) 2 . Since L(R) is positive , X 1 (R) ≤ X 2 (R). Moreover, [B(g 0 ) 2 -X 1 (R)][B(g 0 ) 2 -X 2 (R)] = π 2 (π 4 + 3π 2 -96)(8π 2 R 2 + 24R 2 + 24σ F π 2 R + 3π 4 ) 8R 2 [π 2 + 8Rσ F ] 2 [-3π 4 -48σ F Rπ 2 + 2(π 4 + 3π 2 -96)R 2 ] 2 >0 (R -R 3 )(R -R 4 ).
As a result, we have

X 1 (R) ≤ B(g 0 ) 2 ≤ X 2 (R) and β min (g 0 ) 2 = X 2 (R). Then, we prove that Borne F (X) is increasing on [X 2 (R); +∞[. We compute dBorne F dX (X) = - 256R 4 B(g 0 ) 6 3π 4 (X -B(g 0 ) 2 ) 3 16R 2 (2Rσ F (π 2 -9) -3π 2 )X -2Rσ F (π 2 + 3) -3π 2 . If 0 < R < 1/2, then 2Rσ F (π 2 -9) -3π 2 < 0. As a consequence: dBorne F dX (X) ≥ 0 ⇐⇒ X ≥ 1 16R 2 2Rσ F (π 2 + 3) + 3π 2 2Rσ F (π 2 -9) -3π 2
Since the right member of this inequality is always negative (when 0 < R < 1/2), then Borne + F is an increasing function.

Proof of theorem 2

We specialise the inequality in proposition 1 in the test function g 0 with 0 < R < ρ max (F). Using lemma 2, if β > B(g 0 ), we have lim inf

q→+∞ q prime µ h F (q) ({L(f, s) ∈ F; γf,1 < β}) ≥ Borne + F (β 2 ).
Thanks to lemma 3, if β > β min (g 0 ), we get lim inf

q→+∞ q prime µ h F (q) ({L(f, s) ∈ F; γf,1 < β}) ≥ Borne F (β 2 ).
To obtain theorem 2 , we evaluate the right member in R = ρ max (F)/2.

The smallest zero

In this part, we prove theorem 1. Our starting point is the following proposition. We do not write the proof that is essentially given in [START_REF] Hughes | Linear statistics of low-lying zeros of L-functions[END_REF] (theorem 8.1).

Proposition 2 The infimum being taken over functions g in S νmax(F )/2 (R), we have

lim sup Q→+∞ min L(f,s)∈F (Q) γf,1 ≤ inf R x 2 g 2 (x)W * [F](x)dx R g 2 (x)W * [F](x)dx . (3.1)
We give a sketch of the proof of theorem 1.

Sketch of the proof

We have to determine an explicit expression of the right member term in (3.1). Let

m := inf g∈Sν (R)\{0} R x 2 g 2 (x)W [G](x)dx R g 2 (x)W [G](x)dx with ν = ν max (F)/2 and W [G] = W * [F].
• In subsection 3.1 (lemma 5), we prove the existence of a function B and a set

H ∞ ν such that m = inf g∈H ∞ ν \{0} B( g). Precisely, we define B by B(h) = 1 4π 2 R h ′ (u) 2 du -δ 2 1 -1 h ′ * h ′ (u)du R h(u) 2 du + δ 2 1 -1 h * h(u)du + ε R h(u)du 2 (3.2)
where (δ, ε) is given in table 1.

• In subsection 3.1 (lemma 6), by a density argument, if

H R = h : R → R even, continuous, C 1 on ] -R, R[, with supp h ⊂ [-R, R] and mR = inf h∈H R \{0} B(h),
we prove m = lim R→ν -mR .

• In subsection 3.1 (lemma 7), by topological arguments, there exists h R in the Sobolev space H 1 0 such that B(h R ) = mR .

• In subsection 3.3 (lemma 10), thanks to Fourier theory, h R satisfies a Volterra equation with temporal shifts

h R (u) = ϕ(u) + δ 2 R u h R (t + 1) -h R (t -1)dt (3.3)
where ϕ is an explicit function which is defined with the unknown parameter mR . Then, h R is in H R .

• Then, we solve the previous Volterra equation in subsection 3.4. As a result, we obtain an explicit expression of h R which also depends on the unknown parameter mR .

• We determine √ mR in subsection 3.5 by solving the equation mR = B(h R ).

• To conclude, we use √ m = lim R→ν - √ mR .

Density and topological arguments

This section is devoted to the proof of the following lemma which sums up the three first steps of the preceding sketch of the proof. Let H 1 0 be the Sobolev space defined by

H 1 0 = u ∈ L 2 (] -R, R[) with u(-R) = u(R) = 0, ∃v ∈ L 2 (R) such that ∀φ ∈ C ∞ c (] -R, R[), R uφ ′ = - R vφ .
If u is in H 1 0 then v is called the weak derivative of u and is denoted by u ′ . The space H 1 0 is equipped with the inner product

u, v H 1 0 = u ′ , v ′ L 2 = R -R u ′ v ′ .
H 1 0 is a reflexive separable Hilbert space ([Bre11], paragraphe VIII.3).

Lemma 4

We have m = lim R→ν -mR . In addition, for each R > 0, there exists h R in H 1 0 such that mR = B(h R ).

Before proving this result, we need to prove some technical lemmas.

A new expression

Lemma 5 Let

H ∞ ν = {h ∈ C ∞ c (R), even and with supp h ⊂ [-R, R] such that 0 < R < ν}. Then, m = inf h∈H ∞ ν \{0} B(h).
Proof: Thanks to Plancherel theorem, inversion formula, Parseval formula and relation (1.1), we have:

R g 2 (x)W [G](x)dx = R g 2 (y) W [G](y)dy = g 2 (0) + δ 2 1 -1 g 2 (y)dy + ε R g 2 (y)dy = R g 2 (x)dx + δ 2 1 -1 g(y) * g(y)dy + εg 2 (0) = R g 2 (y)dy + δ 2 1 -1 g(y) * g(y)dy + ε R g(y)dy 2 Similarly, since xg(x) = -1 2iπ g ′ , we prove R x 2 g 2 (x)W [G](x)dx = 1 4π 2 R g ′ (u) 2 du - δ 2 1 -1 g ′ * g ′ (u)du.
In addition, since the set of Schwartz functions is invariant by Fourier transformation, we have:

g ∈ S ν (R) ⇔ g ∈ H ∞ ν Therefore, m = inf g∈H ∞ ν \{0}
B( g).

A density argument

Lemma 6 We have inf

h∈H ∞ ν \{0} B(h) = lim R→ν -mR .
Proof: We define mollifiers (ρ n ) by

ρ n : R -→ R x -→ n R ρ(t)dt ρ(nx)
where ρ(x) = e

1 |x| 2 -1 if |x| < 1 0 if |x| ≥ 1 .
The function ρ n is non-negative, smooth with supp ρ n ⊂ [-1/n; 1/n] and such that R ρ n (u)du = 1. We recall two properties (see e.g. [START_REF] Brezis | Functional analysis, Sobolev spaces and partial differential equations[END_REF], Theorem 4.22 and Theorem 4.15).

P1 Let 1 ≤ p < ∞, if g ∈ L p (R) then ρ n * g tends to g in L p (R). P2 Let 1 ≤ p ≤ ∞, if f ∈ L 1 (R) and g ∈ L p (R) then f * g ∈ L p (R) and ||f * g|| p ≤ ||f || 1 ||g|| p . Let 0 < R < ν and η > 0. There exists h in H R such that mR ≤ B(h) ≤ mR + η. Let h n = ρ n * h. For large n, h n is in H ∞ ν . We also have h ′ n = ρ n * h ′ . Thanks to property P1, we have lim n→+∞ ||h n || 2 2 = ||h|| 2 2 , lim n→+∞ ||h ′ n || 2 2 = ||h ′ || 2 2 and lim n→+∞ R h n (u)du = R h(u)du.
Thanks to property P2, we may write

1 -1 h n * h n (u) -h * h(u)du ≤ R |h n * h n (u) -h * h(u)| du ≤ ||(h n -h) * h n || 1 + ||h * (h n -h)|| 1 ≤ (||h n || 1 + ||h|| 1 )||h -h n || 1 .
Therefore,

lim n→+∞ 1 -1 h n * h n (u)du = 1 -1 h * h(u)du and similarly lim n→+∞ 1 -1 h ′ n * h ′ n (u)du = 1 -1 h ′ * h ′ (u)du. Then, lim n→+∞ B(h n ) = B(h). There exists h n in H ∞ ν such that | B(h) -B(h n )| ≤ η. As a result, for all η > 0, there exists h n in H ∞ ν such that | mR -B(h n )| ≤ 2η. Then, for all R < ν, we have inf h∈H ∞ ν \{0} B(h) ≤ mR . As a consequence, we have inf h∈H ∞ ν \{0} B(h) ≤ lim R→ν -mR .
On the other side, let h be in

H ∞ ν . There exists R 0 < ν such that supp h ⊂ [-R 0 ; R 0 ]. Since h is in H R 0 and R → mR is decreasing, we get lim R→ν -mR ≤ mR 0 ≤ B(h). Therefore, lim R→ν -mR ≤ inf h∈H ∞ ν \{0} B(h).
Some compact operators Relation (3.2) allows us to extend B to H 1 0 \{0}.

Lemma 7 If R > 0, there exists h R in H 1 0 \{0} such that mR = B(h R ). Proof: Let K be the operator of L 2 (] -R, R[) defined by K[h](u) = δ 2 u+1 u-1 h(t)dt + ε R h(t)dt.
We may write B on the shape

B(h) = h ′ , h ′ L 2 + Kh ′ , h ′ L 2 h, h L 2 + Kh, h L 2 . Since K is a Hilbert-Schmidt operator, it is a compact operator of L 2 (] -R, R[).
Denote by I the identity function of L 2 (] -R, R[), the transformations that have been done in lemma 5 show that, for all

h in L 2 (] -R, R[), (I + K)[h], h L 2 ≥ 0 with equality if and only if h = 0. Since smooth function compactly supported in ] -R, R[ are dense in H 1 0 , the infimum of B over H 1 0 is also equal to mR .
Let (g n ) be a sequence of non zero functions in H R such that lim n→+∞ B(g n ) = mR . We consider h n = g n / g ′ n , g ′ n L 2 . Since for all real number t = 0 and all h in H R , B satisfies B(t.h) = B(h), we have

lim n→+∞ B(h n ) = mR and h ′ n L 2 = h n H 1 0 = 1. Sequences (h n ) and (h ′ n ) are bounded in the Hilbert space L 2 (] -R, R[).
Since the unit ball of L 2 (] -R, R[) is compact for the weak topology, up to consider sub-sequences, there exists

h and k in L 2 (] -R, R[) such that (h n ) (respectively (h ′ n )) tends to h (respectively k) weakly. Moreover, for all function φ in C ∞ c (] -R, R[),
we may write

gφ ′ = lim n→+∞ g n φ ′ = -lim n→+∞ g ′ n φ = -kφ.
Then g belongs to H 1 , g ′ = k and g n tends to g weakly in H 1 . In addition, there exists a compact embedding of H 1 into the set of continuous function on [-R, R] equipped with the norm of uniform convergence ([Bre11], theorem 8.2 et 8.8), we may conclude ([Bre11], 6.1 remark 2) h n converges uniformly to h (precisely to its continuous representative which will be always identified with h). As a result, h is an even functions belonging to

H 1 0 and lim n→+∞ h n L 2 = h L 2 .
Since (h n ) and (h ′ n ) are weakly convergent in L 2 (]-R, R[) and since K is a compact operator, the sequence (Kh n ) and (Kh ′ n ) converge respectively to Kh and Kh ′ strongly in L 2 (] -R, R[). Then, we get ([Bre11], proposition 3.5 (iv))

lim n→+∞ Kh n , h n L 2 = Kh, h L 2 and lim n→+∞ Kh ′ n , h ′ n L 2 = Kh ′ , h ′ L 2 .
To sum up, we have lim n→+∞

(I + K)h ′ n , h ′ n L 2 = 1 + Kh ′ , h ′ L 2 , lim n→+∞ (I + K)h n , h n L 2 = (I + K)h, h L 2 and, since (h ′ n ) tends weakly to h ′ in L 2 , we may deduce h ′ L 2 ≤ lim inf h ′ n L 2 = 1. Furthermore, 1 + Kh ′ , h ′ L 2 is non zero. Indeed, if it was zero, we would have 0 = 1 + Kh ′ , h ′ L 2 ≥ h ′ L 2 + Kh ′ , h ′ L 2 = (I + K)h ′ , h ′ L 2 ≥ 0. So, h ′ L 2 = 1 and (I + K)h ′ , h ′ L 2 = 0, thus h ′ L 2 = 1 and h ′ should be zero ! As a result (I + K)h, h L 2 = 0 because mR is finite and 1 + Kh ′ , h ′ L 2 is non zero. Then h ′ L 2 + Kh ′ , h ′ L 2 (I + K)h, h L 2 = B(h) ≥ mR = lim n→+∞ B(h n ) = 1 + Kh ′ , h ′ L 2 (I + K)h, h L 2 , and h ′ L 2 ≥ 1. Since we already have h ′ L 2 ≤ 1, it comes h ′ L 2 = 1.
To conclude, the function h is non zero and satisfies B(h) = mR .

Remark 5 B is not continuous when B ||•|| H 1 (0, 1) → R is equipped with the weak topology. Indeed, since the unit ball is weakly compact, B should be bounded. However, by considering h n (u) = cos (2n+1)πu 2R 1 [-R,R] (u), we remark that B cannot be bounded.

The proof of lemma 4 comes from lemmas 5, 6 and 7.

Fourier analysis

For technical conveniences, let

B(h) = 16R 2 B(h) and m R = inf h∈H R B(h). Then m R = B(h R ) = 16R 2 mR . Let Ω R =    (c n ) ∈ R N , such that c 2n = 0 and x → n≥0 c n cos πnx 2R is continuous and inH 1 0    .
In this section, we prove some technical lemmas.

Lemma 8 Let

Ψ :

H 1 0 ∩ {continuous, even} -→ Ω R h -→ 1-(-1) n 2R R 0 h(t) cos πnt 2R dt n≥0 .
Then, Ψ is a bijective function.

Proof: For each h in H 1 0 , we associate an even 4R-periodic function h which is defined by: for all x ∈ [0, R], h(x) = h(x) and h(R + x) = -h(Rx).

We also have

h = h • 1 [-R,R] .

representative curve of h representative curve of h

As a result, the mean value of h is equal to zero. Actually, if c n denotes the n-th Fourier coefficient of h, we have

c n = 1 4R 2R -2R h(t)e -iπnt 2R dt = 1 -(-1) n 2R R 0 h(t) cos πnt 2R dt.
Since the Fourier series of a function in H 1 0 is normally convergent, we get

h(u) = h(u) • 1 [-R,R] (u) = 2   n≥0 c n cos πnx 2R   1 [-R,R] (u).
Then Ψ is well defined and the last relation gives us an explicit expression of Ψ -1 .

As a consequence, since B is defined on the set of continuous and even functions in H 1 0 , we may define B on Ω R by B(c) := B(Ψ -1 (c)).

Lemma 9 Let c = (c n ) be in Ω R . If R ≤ 1/2 then B(c) = * n≥0 n 2 c 2 n * n≥0 c 2 n + (δ + 2ε) 8R π 2 * n≥0 (-1) n-1 2 n c n 2 and, if R > 1/2, we have B(c) = * n≥0 n 2 c 2 n -δ 2R m,n≥0 mnµ m,n c m c n * n≥0 c 2 n + δ 2R m,n≥0 λ m,n c m c n + 16Rε π 2 * n≥0 (-1) n-1 2 cn n 2
where, for n and m odd numbers:

λ m,n =    8R 2 mn(-1) m+n 2 π 2 (m 2 -n 2 ) 1 n 2 cos πn 2R -1 m 2 cos πm 2R -8R 2 (-1) m+n 2 mnπ 2 si m = n 2R(2R-1) nπ sin πn 2R -8R 2 π 2 n 2 cos nπ 2R + 8R 2 π 2 n 2 si m = n and µ m,n =    8R 2 (-1) m+n 2 π 2 (m 2 -n 2 ) cos πm 2R -cos πn 2R si m = n -2R(2R-1) nπ sin πn 2R si m = n
Proof: We are giving an explicit expression of B on Ω R . Let h be an even and continuous function in

H 1 0 such that c = (c n ) = Ψ(h).
Thanks to Parseval formula, we have:

1 2R R h(u) 2 du = 1 4R 2R -2R h(u) 2 du = n∈Z |c n | 2 = 2 n≥0 c 2 n 1 2R R h ′ (u) 2 du = 1 4R 2R -2R h′ (u) 2 du = π 2 4R 2 n∈Z n 2 |c n | 2 = π 2 2R 2 n≥0 n 2 c 2 n
In addition, we may write

R h(u)du = R -R h(u)du = 4R π n∈Z (-1) n (2n + 1) c 2n+1 = 8R π n≥0 (-1) n (2n + 1) c 2n+1 .

20

Remember B(h) = 16R 2 B(h) where B(h) is given in (3.2). We obtain an explicit expression of B(c) using

1 -1 h * h(x)dx = R h(x)dx 2 if R ≤ 1 2 4 m≥0 n≥0 c m c n λ m,n if R > 1 2 and 1 -1 h ′ * h ′ (x)dx = 0 if R ≤ 1 2 π 2 R 2 m≥0 n≥0 mnc m c n µ m,n if R > 1 2 .
Hughes and Rudnick proved that m = 1 16ν 2 if G = U . We extend their result.

Corollary 1 If G = O or if G = SO ± and 0 < R ≤ 1/2 then 1 1 + R < m R ≤ 1 1 + 8 π 2 R < 1. If G = SO ± with R > 1/2, then m R < 1. If G = Sp and 0 < R < 1/2, then 1 < m R < 1 1 -8 π 2 R < 2. Proof: Thanks to the Cauchy-Schwarz inequality, if G = O or if G = SO ± with 0 < R ≤ 1/2, we have
B(c) = * n≥0 n 2 c 2 n * n≥0 c 2 n + (δ + 2ε) 8R π 2 * n≥0 (-1) n-1 2 n c n 2 ≥ * n≥0 n 2 c 2 n (1 + R) * n≥0 c 2 n ≥ 1 1 + R
and one of both inequalities is strict. In order to obtain an upper bound for m R , we specialise B in e = (e n ) which is defined by e n = 0 for all n except e 1 = 1. The upper bound comes easily

if G = O, if G = SO ± or if G = Sp and 0 < R ≤ 1/2. If G = SO ± and R > 1/2, we have B(e) = 1 -δ 2R µ 1,1 1 + δ 2R λ 1,1 + 16Rε π 2 = 1 + δ(2R-1) π sin π 2R 1 + δ(2R-1) π sin π 2R + 4Rδ π 2 1 -cos π 2R + 16Rε π 2 < 1.
Finally, if G = Sp and R < 1/2, for all c in Ω R , we have

B(c) = * n≥0 n 2 c 2 n * n≥0 c 2 n -8R π 2 * n≥0 (-1) n-1 2 n c n 2 ≥ * n≥0 n 2 c 2 n * n≥0 c 2 n ≥ 1
and one of both inequalities is strict. Therefore, thanks to lemma 7, we have m R > 1.

Corollary 1 implies, except if G = Sp and R > 1/2, that m R is not the square of an integer. We make the following hypothesis which will be justified page 43.

"Sp hypothesis": If G = Sp and R > 1/2 then m R is not the square of an integer.

Remark 6 In the symplectic case with R > 1/2, by evaluating B in e, we get:

m R ≤ B(e) = 1 -2R-1 π sin π 2R 1 -2R-1 π sin π 2R -4R π 2 1 -cos π 2R = 12R π 2 + O(1)
Therefore, since m R = 16R 2 mR , corollary 1 and proposition 2 show:

lim sup Q→+∞ min L(f,s)∈F (Q) γf,1 ≤ lim R→νmax(F )/2 mR ≪          1 νmax(F ) 1/2 if W * [F] = W [Sp] 1 νmax(F ) if W * [F] = W [SO + ] 1 νmax(F ) 3/2 if W * [F] = W [O] 1 νmax(F ) 2 if W * [F] = W [SO -]
As a result, density conjecture agrees with random matrix prediction.

A Volterra equation with temporal shifts

Since the unitary case has been solved, we assume G is one of the following compact groups O, SO -, SO + or Sp. In this section, we prove the following lemma.

Lemma 10 The optimal test function

h R satisfies, for all 0 ≤ u ≤ R, h R (u) = ϕ(u) + δ 2 R u h R (t + 1) -h R (t -1)dt (3.4)
where (δ, ε) is given in the table 1 and with

ϕ(u) = πk δ,ε 2m R cos √ m R π 2 cos √ m R πu 2R -cos √ m R π 2 • 1 [-R,R] (u)
where

k δ,ε = 2m R π δ 2 R R-1 h R (x)dx + ε R -R h R (x)dx . (3.5)
Remark 7 We identify h R to its continuous representative. As a result, equation (3.4) proves that h R is of class C 1 on ] -R, R[. In other words, the function h R belongs to H R . Before proving this result, we need to prove some technical lemmas. However, we may immediately deduce the following corollary.

Corollary 2 If G = O or if G = SO ± or Sp with R ≤ 1/2, then √ m = 1 ν V -1 1 + (δ + 2ε) 1 ν
where V has been defined in section "Preview of results".

Proof: If G = O, then δ = 0 and lemma 10 gives us immediately

h R = ϕ. Similarly, if G = SO + , SO -or Sp, R ≤ 1/2 and if 0 < t < R then t + 1 / ∈ [-R, R] and t -1 / ∈ [-R, R].
Therefore, in this case, we also have h R = ϕ. We may sum up these remarks by

h R (u) = ϕ(u) = (δ + 2ε) R -R h R (x)dx 2 cos π √ mR 2 cos √ m R πu 2R -cos √ m R π 2 • 1 [-R,R] (u).
As a result, we have

R -R h R (u)du • 1 -(δ + 2ε)R 2 π √ m R tan π √ m R 2 -1 = 0. Since R -R h R (u)du = 0 (otherwise h R = 0), we may conclude 2 π √ m R tan π √ m R 2 = 1 + δ + 2ε R .
The result comes easily from lemmas 5 and 6.

Remark 8 We will use this phenomenon in the next section in order to determine h R in full generality.

Let

S n (t) = sin πnt 2R 1 [-R,R] (t) and C n (t) = cos πnt 2R 1 [-R,R] (t).
Lemma 11 n denotes an odd positive integer. Let h be in H R with c = (c m ) = Ψ(h), then:

m≥0 [mnµ m,n + m R λ m,n ] c m = - 2R nπ (n 2 -m R )(S n * h)(1) + 8R 2 (-1) n-1 2 nπ 2 m R m≥0 (-1) m-1 2 m 1 -cos mπ 2R c m Proof: We have: m≥0 [mnµ m,n + m R λ m,n ] c m = -(n 2 -m R )   2R(2R -1) nπ sin πn 2R c n + 8R 2 nπ 2 m =n m(-1) m+n 2 m 2 -n 2 cos πn 2R c m   - 8R 2 m R n 2 π 2 cos πn 2R c n -m R m≥0 8R 2 (-1) m+n 2 nmπ 2 c m + 8R 2 π 2 m =n n(-1) m+n 2 m(m 2 -n 2 ) (m 2 -m R ) cos πm 2R c m Since n m(m 2 -n 2 ) = 1 n m m 2 -n 2 - 1 m , we may conclude m≥0 [mnµ m,n + m R λ m,n ] c m = -(n 2 -m R )   2R(2R -1) nπ sin πn 2R c n + 8R 2 nπ 2 m =n * (-1) m+n 2 m m 2 -n 2 cos πn 2R -cos πm 2R c m   + 8R 2 m R (-1) n-1 2 nπ 2 m≥0 * (-1) m-1 2 m 1 -cos πm 2R c m .
In addition, we have

(S n * h)(1) = R 1-R h(u) sin πn 2R (1 -u) du = m≥0 * c m R 1-R sin π(m -n)u 2R + πn 2R -sin π(m + n)u 2R - πn 2R du = (2R -1) sin πn 2R c n + 4R π m =n * (-1) m+n 2 m m 2 -n 2 cos πn 2R -cos πm 2R c m . Lemma 12 Let c = (c n ) = Ψ(h R ).
Then, for all odd positive integer n, we have

c n = (-1) n-1 2 k δ,ε n[n 2 -m R ] - δ nπ (S n * h R )(1) = (-1) n-1 2 k δ,ε n[n 2 -m R ] + 2Rδ n 2 π 2 (C n * h ′ R )(1).
Proof: We define a norm on Ω R by

||c|| = +∞ n=0 n|c n | + +∞ n=0 n 2 c 2 n 1/2
and we consider five differentiable functions H, T , S, A and G on (Ω R , || • ||) defined by:

H(c) = n≥0 n 2 c 2 n , T (c) = n≥0 c 2 n , S(c) = n≥0 * (-1) n-1 2 n c n , A(c) = m≥0 n≥0 mnµ m,n c m c n and G(c) = m≥0 n≥0 λ m,n c m c n .
With lemma 9, we may write

B = H -δ 2R A T + δ 2R G + 16Rε π 2 S 2 if R > 1 2 or B = H T + (δ + 2ε) 8R π 2 S 2 if R ≤ 1 2 . Since B(c) = m R , c is a critical point. Therefore, if R > 1/2, we have dB |c = 0 hence: ∀n odd, (n 2 -m R )c n = δ 2R m≥0 [mnµ m,n + m R λ m,n ] c m + ε 16RS(c)m R π 2 (-1) n-1 2 n
Thanks to lemma 11, "Sp hypothesis" and corollary 1, we may conclude

c n = (-1) n-1 2 k δ,ε n[n 2 -m R ] - δ nπ (S n * h R )(1) = (-1) n-1 2 k δ,ε n[n 2 -m R ] + 2Rδ n 2 π 2 (C n * h ′ R )(1) (3.6)
where

k δ,ε = 4Rm R π 2   δ m≥0 * (-1) m-1 2 m 1 -cos mπ 2R c m + 4ε m≥1 * (-1) m-1 2 m c m   = 2m R π δ 2 R R-1 h R (x)dx + ε R -R h R (x)dx . (3.7)
Similarly, c satisfies (3.6) and (3.7) in the case R ≤ 1/2 due to the fact that, if

R ≤ 1/2, we have (S n * h R )(1) = 0 and R R-1 h R (x)dx = R -R h R (x)dx. Proof of lemma 10 Assume 0 ≤ u ≤ R. Since h R (u) = 2 n≥0 c n cos πnu 2R , lemma 12 gives h R (u) = 2k δ,ε n≥0 * (-1) n-1 2 cos πnu 2R n(n 2 -m R ) + 4Rδ π 2 n≥0 * (C n * h ′ R )(1) n 2 cos πnu 2R .
First, using relation 1.444.6 of [START_REF] Gradshteyn | Table of integrals, series, and products. Elsevier[END_REF], we have

n≥0 * (C n * h ′ R )(1) n 2 cos πnu 2R = π 8 R -R π - π|u + t| 2R - π|u -t| 2R h ′ R (1 -t)dt = π 2 8R R u h R (t + 1) -h R (t -1)dt.
Then, using relation 1.445.6 of [START_REF] Gradshteyn | Table of integrals, series, and products. Elsevier[END_REF], we compute

2k δ,ε n≥0 * (-1) n-1 2 cos πnu 2R n(n 2 -m R ) = πk δ,ε 2m R cos √ mRπ 2 cos √ m R πu 2R -cos √ m R π 2 = ϕ(u).

Optimal test Function

In this section, we solve the Volterra equation with temporal shifts which appears in lemma 10. Precisely, we prove that, except for a countable or finite number of value of R, the previous Volterra equation admits one and only one solution in H R . We are giving an explicit expression of h R .

An appropriate partition

For technical convenience, since R will tend to ν (thanks to lemma 6), we may assume that 2R is not an integer. Similarly, since n ≥ 1 is the only integer such that n -1 < 2ν ≤ n, we may assume n-1 2 < R < n 2 , then n = ⌊2R⌋ + 1. Let D be the derivative operator and T , T -1 the shift operators defined by:

D[f ] = f ′ , T [f ](u) = f (u + 1) and T -1 [f ](u) = f (u -1).
We build a partition of [-R, R] which is invariant by T [id], T -1 [id] and by symmetry. We define

a n-2i = R -i if 0 ≤ i ≤ n-1 2 a n-2i-1 = ⌊2R⌋ -R -i if 0 ≤ i ≤ n-2 2 and I 0 =] -a 1 , a 1 [, I k =]a k , a k+1 [, I -k =] -a k+1 , -a k [ for 1 ≤ k ≤ n -1. We have [-R, R] = ⌊2R⌋ k=-⌊2R⌋ I k . Moreover, if |k| < n -2 then T ±1 (I k ) = I k±2 and, if k ∈ {n -2, n -1} then T (I k )∩ supp h R = ∅.

A differential equation with temporal shifts

By derivation and integration, we remark that h R satisfies the previous Volterra equation is equivalent to h R satisfies the following differential equation with temporal shifts: for all u in ] -R, R[,

h ′ R (u) = ϕ ′ (u) - δ 2 [h R (u + 1) -h R (u -1)] . (3.8) Let λ = √ m R π 2R and w = -π 2 k δ,ε 4R √ m R cos π √ m R 2
.

We consider polynomial sequences (T n ) and (U n ) which are defined by induction:

   T 0 = 1 T 1 = X T n+1 = 2XT n -T n-1    U 0 = 1 U 1 = 2X U n+1 = 2XU n -U n-1
Polynomials T n (resp. U n ) are called Chebychev polynomials of first kind (resp. second kind). They satisfy, for all real number θ and all non-negative integer n,

T n (cos θ) = cos nθ and U n (cos θ) = sin(n + 1)θ sin θ .

As a result, we have

T n (X) = n-1 k=0 X -cos π 2n + kπ n , U n (X) = n k=1 [X -θ n (k)] where θ n (k) = cos kπ n + 1 .
In this subsection, we prove the following proposition:

Proposition 3 Even functions h which are compactly supported on [-R, R], C 1 on each I k and which satisfy the differential equation (3.8) on ∪I k are those that satisfy:

• If 0 ≤ k ≤ n -1, then: h| I n-(2k+1) (u) = ⌊ n+1 2 ⌋ j=1 r j (I n-1 )U k (θ n (j)) sin u - n -2k -1 2 θ n (j) - π 2 j + δ n -2k -1 2 + r λ n (k) sin λ u - n -2k -1 2 + θ λ n (k) • If 0 ≤ k ≤ n -2, then: h| I n-2(k+1) (u) = ⌊ n 2 ⌋ j=1 r j (I n-2 )U k (θ n-1 (j)) sin u - n -2(k + 1) 2 θ n-1 (j) - π 2 j + δ n -2(k + 1) 2 + r λ n-1 (k) sin λ u - n -2k -2 2 + θ λ n-1 (k)
where r j (I n-1 ) with 1 ≤ j ≤ n+1 2 and r j (I n-2 ) with 1 ≤ j ≤ n 2 are arbitrary real numbers and where r λ n (k) and θ λ n (k) refer respectively to the modulus and the argument of the complex number

iw λ + δ sin λ U k (λ) U n (λ) e -i(λ+ πδ 2 ) n+1 2 -e i(λ+ πδ 2 ) n-2k-1 2 + U n-k-1 (λ) U n (λ) e i(λ+ πδ 2 ) n+1 2 e -i πδ 2 n-2k-1 2 .
Before proving this result, we need to prove some technical lemmas.

Transformation in a linear differential equation With corollary 2, we may assume δ = 0 (hence δ 2 = 1) and n ≥ 2. Moreover, throughout this section h refers to an even function which is compactly supported on [-R, R], C 1 on each I k and which satisfies the differential equation (3.8) on ∪I k . Let Q k the operator which has the following recursive definition:

     Q 1 = I Q 2 = D + δ 2 (T -1 -T ) Q k+1 = DQ k + 1 4 Q k-1 + δ 2 k T -k + (-1) k T k
Lemma 13 h satisfies the following linear differential equations:

• On I n-1 , if 1 ≤ k ≤ n, h satisfies: i 2 k U k 1 i D [h] = Q k [ϕ ′ ] + δ 2 k T -k [h] • On I n-2 , if 1 ≤ k ≤ n -1, h satisfies: i 2 k U k 1 i D [h] = Q k [ϕ ′ ] + δ 2 k T -k [h] • On I 0 , if 1 ≤ k ≤ ⌊ n+1 2 ⌋, h satisfies: 2 i 2 k T k 1 i D [h] = Q k [ϕ ′ ] + δ 2 k T -k [h] + (-1) k T k [h]
Proof: We prove the first relation by induction for k = 1, ..., n. If k = 1,the differential equation (3.8) gives:

if u ∈ I n-1 , h ′ (u) = ϕ ′ (u) + δ 2 h(u -1).
(3.9)

The result comes easily from the definition of Q 1 and U 1 . If k = 2, we derive relation (3.9) and we apply the differential equation (3.8). We assume the result holds for k and k -1 with 2 ≤ k ≤ n -1. Then, with recursive relation on U k , we get:

i 2 k+1 U k+1 1 i D [h] = 2 i i 2 k+1 DU k 1 i D [h] - i 2 k+1 U k-1 1 i D [h] = D i 2 k U k 1 i D [h] + 1 4 i 2 k-1 U k-1 1 i D [h] = DQ k [ϕ ′ ] + 1 4 Q k-1 [ϕ ′ ] + δ 2 k T -k [h ′ ] + δ 2 k+1 T -(k-1) [h]
since k ≤ n -1 we may apply differential equation (3.8).

= DQ k [ϕ ′ ] + 1 4 Q k-1 [ϕ ′ ] + δ 2 k T -k [ϕ ′ ] + δ 2 k+1 T -(k+1) [h]
Since we apply this relation with u in I n-1 , we have T k+1 [h](u) = 0. Therefore, we have

i 2 k+1 U k+1 1 i D [h] = Q k+1 [ϕ ′ ] + δ 2 k+1 T -(k+1) [h] on I n-1 .
We prove the other relations of this lemma in the same way.

Lemma 14 Let f be a function which is

C 1 on ] -R, R[ and with supp f ⊂ [-R, R]. Then, for all 1 ≤ k ≤ n we have Q k [f | I n-1 ] = i 2 k-1 k-1 j=0 δ i j U k-j-1 1 i D T -j [f | I n-1 ].
Similarly, for all 1 ≤ k ≤ n -1 we have

Q k [f | I n-2 ] = i 2 k-1 k-1 j=0 δ i j U k-j-1 1 i D T -j [f | I n-2 ].
Proof: By induction on k = 1, ..., n. The result comes easily from the definition of Q k if k = 1 or k = 2. We assume the result holds for k and k -

1 with 2 ≤ k ≤ n -1. Since T k [f | I n-1 ] = 0,
we have

Q k+1 = DQ k + 1 4 Q k-1 + δ 2 k T -k = i 2 k   k-1 j=0 δ i j 2D i U k-j-1 1 i D T -j - k-2 j=0 δ i j U k-j-2 1 i D T -j   + δ 2 k T -k = i 2 k k-2 j=0 δ i j 2D i U k-j-1 1 i D -U k-j-2 1 i D T -j + i 2 k δ i k-1 U 1 1 i D T -(k-1) + δ 2 k T -k = i 2 k k j=0 δ i j U k-j 1 i D T -j .
Similarly, we prove the other relation of this lemma.

Using lemmas 13 and 14, we prove that h| I n-1 and h| I n-2 satisfy linear non-homogeneous differential equations with constant coefficients. Precisely, Corollary 3 We have

i n U n 1 i D [h| I n-1 ] = Im iwδ n e iλ(u-n) 1 -(iδe iλ ) n U n (λ) + (iδe iλ ) n+1 U n-1 (λ) λ + δ sin λ and i n-1 U n-1 1 i D [h| I n-2 ] = Im iwδ n-1 e iλ(u-n+1) 1 -(iδe iλ ) n-1 U n-1 (λ) + (iδe iλ ) n U n-2 (λ) λ + δ sin λ . Proof: We have T -n [h| I n-1 ] = 0 and T -(n-1) [h| I n-2 ] = 0.
Thus, with lemma 13, we get

i n U n 1 i D [h| In-1 ] = 2 n Q n [ϕ ′ | In-1 ] and i n-1 U n-1 1 i D [h| In-2 ] = 2 n-1 Q n-1 [ϕ ′ | In-2 ].
Let e λ (u) = we iλu . We have ϕ ′ (u) = Im [e λ (u)]. Since D 2 (e λ ) = (iλ) 2 e λ , thanks to lemma 14, we write

Q n (e λ | In-1 ) = i 2 n-1 n-1 j=0 δ i j U n-j-1 (λ)T -j (e λ | In-1 )
and

Q n-1 (e λ | In-2 ) = i 2 n-2 n-2 j=0 δ i j U n-j-2 (λ)T -j (e λ | In-2 ).
Due to the fact that

n-1 j=0 U j (X)z j = 1 -z n U n (X) + z n+1 U n-1 (X) 1 -2zX + z 2 , (3.10)
we have

Q n (e λ | In-1 ) = δ n-1 ie -iλ 2 n 1 -(iδe iλ ) n U n (λ) + (iδe iλ ) n+1 U n-1 (λ) λδ + sin λ e λ (u -n + 1)
and

Q n (e λ | In-2 ) = δ n-2 ie -iλ 2 n-1 1 -(iδe iλ ) n-1 U n-1 (λ) + (iδe iλ ) n U n-2 (λ) λδ + sin λ e λ (u -n + 2).
We may conclude since

Q n (ϕ ′ ) = Im(Q n (e λ )).
Lemma 15 R → mR is a strictly decreasing function.

Proof: We are assuming that there exist R 1 < R 2 such that mR 1 = mR 2 . Since R → mR is a decreasing function, we may assume there exists an integer n 0 such that

n 0 -1 2 < R 1 < R 2 < n 0 2 . Let λ = λ R 1 = λ R 2 = 2π mR 1 . There exists h R 1 in H R 1 such that B(h R 1 ) = mR 1 = mR 2 .
Thanks to the Volterra equation (3.4), h R 1 | I n 0 -1 (R 2 ) is smooth. Using corollary 3, we have:

i n0 U n0 1 i D [h R1 | In 0 -1(R2 ) ](u) = Im iw R2 δ n0 e iλ(u-n0) 1 -(iδe iλ ) n0 U n0 (λ) + (iδe iλ ) n0+1 U n0-1 (λ) λ + δ sin λ Since supp h R 1 ⊂ [-R 1 , R 1 ], we have h R 1 | I n 0 -1 (R 2 )\I n 0 -1 (R 1 ) = 0. Thus, for all u in I n 0 -1 (R 2 ), we have Im iw R2 δ n e iλ(u-n0) 1 -(iδe iλ ) n0 U n0 (λ) + (iδe iλ ) n0+1 U n0-1 (λ) λ + δ sin λ = 0.
Then, on

I n 0 -1 (R 2 ), h R 1 | I n 0 -1 (R 2 ) satisfies U n 0 1 i D [h R 1 | I n 0 -1 (R 2 ) ](u) = 0 and h R 1 (u) = 0 on I n 0 -1 (R 2 )\I n 0 -1 (R 1 ) = ∅. Using Picard-Lindelöf theorem, we may conclude h R 1 | I n 0 -1 (R 2 ) = 0. Thus, supp h R 1 ⊂ [-a n 0 -1 (R 2 ), a n 0 -1 (R 2 )] and ma n 0 -1 (R 2 ) = mR 2 with a n 0 -1 (R 2 ) = ⌊2R 2 ⌋ -R 2 . As a result, for all R such that a n 0 -1 (R 2 ) < R < R 2 , we have mR = mR 2 . By induction, there exists r k (with r 1 = a n-1 (R 2 )) in ] n 0 -k-1 2 , n 0 -k 2 [ such that for all R in [r k , R 2 [, we have mR = mR 2 . As a result, R → mR is a constant function on ]r n 0 -1 , 1/2[ which contradicts corollary 2. Corollary 4 Except for at most n -1 values of R in ] n-1 2 , n 2 [, we have U n (λ)U n-1 (λ) = 0. Proof: U n (λ)U n-1 (λ) = 0 if λ is one of the n-1 positive roots of U n U n-1 . Since R → λ = 2π √ mR is a strictly decreasing function, U n (λ)U n-1 (λ) = 0 excepted for at most n -1 values of R in ] n-1 2 , n 2 [.
Since R will tends to ν, we may assume U n (λ)U n-1 (λ) = 0.

Lemma 16 There exists 2n-1 complex numbers z 1 (I n-1 ), ..., z n (I n-1 ) and z 1 (I n-2 ), ..., z n-1 (I n-2 )such that

h| I n-1 (u) = Im   n j=1 z j (I n-1 )e iuθn(j) + z λ n (0)e iλ(u-n-1 2 )   and h| I n-2 (u) = Im   n-1 j=1 z j (I n-2 )e iuθ n-1 (j) + z λ n-1 (0)e iλ(u-n-2 2 )   where z λ n (0) = r λ n (0)e iθ λ n (0) = iw (iδ) n U n (λ) e -iλ n+1 2 1 -(iδe iλ ) n U n (λ) + (iδe iλ ) n+1 U n-1 (λ) λ + δ sin λ .
Proof: We solve differential equations of corollary 3. The general solution of the homogeneous equation is any functions which may be written

u -→ Im   n j=1 z j (I n-1 )e iuθn(j)   .
Finally, with corollary 4, it is easy to check that u -→ Im z λ n (0)e iλ(u-n-1 2 ) is a particular solution of our differential equation. Similarly, we success in obtaining an explicit expression of h| I n-2 .

Extension of the optimal test function In lemma 16, we have an explicit expression of h only on I n-2 ∪ I n-1 . Thanks to the differential equation with temporal shifts (3.8), we extend this explicit expression to [-R, R].

Lemma 17 With U -1 = 0, if k = 0, ..., n -1 then:

h| I n-(2k+1) = i δ k T k U k 1 i D [h| I n-1 ] - 2 δ k j=0 i δ j-1 U j-1 1 i D T j [ϕ ′ ]
Similarly, if k = 0, ..., n -2, then:

h| I n-2(k+1) = i δ k T k U k 1 i D [h| I n-2 ] - 2 δ k j=0 i δ j-1 U j-1 1 i D T j [ϕ ′ ]
Proof: We prove the first relation by induction on k = 0, .., n -1. If k = 0, there is nothing to prove. If k = 1, the result comes easily from relation (3.8). Assume the result holds for k -1 and k with 1 ≤ k ≤ n -2. Using relation (3.8), we get

h ′ | I n-(2k+1) (u + 1) = ϕ ′ (u + 1) + δ 2 h| I n-(2k+3) (u) -h| I n-(2k-1) (u + 2) .
Therefore, we have

h| I n-(2k+3) = T 2 [h| I n-(2k-1) ] + 2 δ T D[h| I n-(2k+1) ] -T [ϕ ′ ] .
The result follows easily from the induction hypothesis.

Lemma 18 If 0 ≤ k ≤ n -1, then h| I n-(2k+1) (u) = Im   (iδ) k n j=1 z j (I n-1 )U k (θ n (j))e i(u+k)θn (j) + z λ n (k)e iλ(u-n-2k-1 2 )   and if 0 ≤ k ≤ n -2, then h| I n-2(k+1) (u) = Im   (iδ) k n-1 j=1 z j (I n-2 )U k (θ n-1 (j))e i(u+k)θ n-1 (j) + z λ n-1 (k)e iλ(u-n-2k-2 2 )   where z λ n (k) = r λ n (k)e iθ λ n (k) (3.11) = iw λ + δ sin λ (iδ) k-n e -iλ n+1 2 U k (λ) U n (λ) -e iλ n-2k-1 2 + (iδ) k+1 e iλ n+1 2 U n-k-1 (λ) U n (λ) .
Proof: With lemmas 16 and 17, we may write

h| I n-(2k+1) (u) = Im   n j=1 z j (I n-1 ) i δ k U k (θ n (j))e i(u+k)θn(j) + z λ n (0) i δ k U k (λ)e iλ(u-n-2k-1 2 ) - 2w δ k j=0 i δ j-1 U j-1 1 i D e iλ(u+j)   .
Thanks to relation (3.10), we get

2w δ k j=0 i δ j-1 U j-1 1 i D e iλ(u+j) = 2wδe iλ(u+1) k-1 j=0 i δ j U j (λ)e iλj = iwe iλu 1 -(iδe iλ ) k U k (λ) + (iδe iλ ) k+1 U k-1 (λ) λ + δ sin λ .
Thus we have:

z λ n (0) i δ k U k (λ)e iλ(u-n-2k-1 2 ) - 2w δ k j=0 i δ j-1 U j-1 1 i D e iλ(u+j) = iw λ + δ sin λ e iλ(u-n-2k-1 2 ) (iδ) k-n e -iλ n+1 2 U k (λ) U n (λ) + (iδ) k+1 e iλ n+1 2 U k (λ)U n-1 (λ) U n (λ) -e iλ n-2k-1 2 -(iδ) k+1 e iλ n+1 2 U k-1 (λ)
We may conclude since, if n ≥ 2 and 1 ≤ j ≤ n -1, then

U n-1 U j -U n U j-1 = U n-1-j .
(3.12)

Even conditions h is assumed to be even. We are exploiting this fact in order to obtain some restrictions on complex numbers z j .

Lemma 19 h is given lemma 18. Then h is an even function if and only if we have:

For 1 ≤ j ≤ n+1 2 , z j (I n-1 ) -z n+1-j (I n-1 ) + (iδ) n-1 U n-1 (θ n (j
))e i(n-1)θn (j) z j (I n-1 )z n+1-j (I n-1 ) = 0, and, for 1 ≤ j ≤ n 2 , z j (I n-2 )z n-j (I n-2 ) + (iδ) n-2 U n-2 (θ n-1 (j))e i(n-2)θ n-1 (j) z j (I n-2 )z n-j (I n-2 ) = 0.

Proof: First, using relation (3.8), we prove that h is even if and only if h(u) = h(-u) for all u ∈ I n-2 ∪ I n-1 . Then, due to the fact that z λ n (n -1) = -z λ n (0), if u ∈ I n-1 , we have:

h| In-1 (u) = h I -(n-1) (-u) ⇐⇒ Im   n j=1 z j (I n-1 )e iuθn(j)   = Im   (iδ) n-1 n j=1
z j (I n-1 )U n-1 (θ n (j))e i(n-1)θn(j) e -iuθn(j)

  Since θ n (n + 1 -k) = -θ n (k), we get Im   n j=1 z j (I n-1 )e iuθn(j)   = Im   ⌊ n+1 2 ⌋ j=1 z j (I n-1 ) -z n+1-j (I n-1 e iuθn(j)   and Im   (iδ) n-1 n j=1
z j (I n-1 )U n-1 (θ n (j))e i(n-1)θn(j) e -iuθn(j)

  = -Im   (-iδ) n-1 ⌊ n+1 2 ⌋ j=1
z j (I n-1 )z n+1-j (I n-1 ) U n-1 (θ n (j))e -i(n-1)θn(j) e iuθn(j)   .

Therefore, for all 1 ≤ j ≤ n+1 2 , we get z j (I n-1 )z n+1-j (I n-1 ) + (iδ) n-1 U n-1 (θ n (j))e i(n-1)θn(j) z j (I n-1 )z n+1-j (I n-1 ) = 0.

Corollary 5 For all 1 ≤ j ≤ ⌊ n+1 2 ⌋, there exists a real number r j (I n-1 ) (which may be negative) such that z j (I n-1 )z n+1-j (I n-1 ) = r j (I n-1 )e i[ n-1 2 (θn(j)

+ πδ 2 )+ jπ 2 ] .
For all 1 ≤ j ≤ ⌊ n 2 ⌋, there exists a real number r j (I n-2 ) (which may be negative) such that

z j (I n-2 ) -z n-j (I n-2 ) = r j (I n-2 )e i[ n-2 2 (θn-1(j)+ πδ 2 )+ jπ 2 ] .
Proof: We have

U n-1 (θ n (j)) = U n-1 cos jπ n + 1 = (-1) j+1 .
Therefore, if z j (I n-1 )z n+1-j (I n-1 ) = r j e iθ j , the previous lemma gives r j e iθ j + (iδ) n-1 (-1) j-1 e i(n-1)θ j e -iθ j = 0.

Then r j = 0 or e iθ j = ±e i[ n-1 2 (θn(j)

+ πδ 2 )+ jπ 2 ] .
Proof of proposition 3 From lemma 18, we have for 0 ≤ k ≤ n -1:

h| I n-(2k+1) (u) = Im   (iδ) k ⌊ n+1 2 ⌋ j=1 z j (I n-1 ) -z n+1-j (I n-1 ) U k (θ n (j))e i(u+k)θn(j) + z λ n (k)e iλ(u-n-2k-1 2 )  
Then, corollary 5 gives:

h| I n-(2k+1) (u) = Im   ⌊ n+1 2 ⌋ j=1 r j (I n-1 )U k (θ n (j))e i[(u-n-2k-1 2 )θn(j)-π 2 (j+δ n-2k-1 2 )] +r λ n (k)e i[λ(u-n-2k-1 2 )+θ λ n (k)]
We obtain h| I n-(2k+2) in the same way. To conclude, one can easily check that such a function satisfies the differential equation with temporal shifts (3.8).

Continuity conditions

Since h R satisfies the differential equation (3.8), we have an explicit expression of h R thanks to proposition 3. However, due to the fact that h R is a continuous function, there are some restrictions on coefficients r j which are associated to the explicit expression of h R . In this subsection, we prove that, except for at most a finite number of R in ] n-1 2 , n 2 [, there is one and only one continuous function which satisfies the differential equation (3.8) and we give the exact values of the corresponding coefficients r j .

Lemma 20 h R is a continuous function if and only if coefficients r j (I n-1 ) and r j (I n-2 ) are satisfying the following linear system: for 0 ≤ k ≤ n -1,

⌊ n 2 ⌋ j=1 r j (I n-2 )U k (θ n-1 (j)) sin a n-1 - n -2 2 θ n-1 (j) - π 2 j + δ n -2k -2 2 - ⌊ n+1 2 ⌋ j=1 r j (I n-1 )U k (θ n (j)) sin a n-1 - n -1 2 θ n (j) - π 2 j + δ n -2k -1 2 = r λ n (k) sin λ a n-1 - n -1 2 + θ λ n (k) -r λ n-1 (k) sin λ a n-1 - n -2 2 + θ λ n-1 (k) = r λ U k (λ) sin θ λ -kδ π 2
where r λ and θ λ refer to the modulus and the argument of the complex number:

z λ = -2iwe -iλa n-1 U n (λ)U n-1 (λ) 1 -(iδe iλ ) n U n (λ) + (iδe iλ ) n+1 U n-1 (λ) -2iδe iλ (λ + δ sin λ) = -2iwe -iλa n-1 U n (λ)U n-1 (λ) n-1 k=0 (iδe iλ ) k U k (λ)
Proof: We use the fact that h R is a continuous function on inf I j for j = -(n -1)...n -1. Precisely, h R is continuous on a n-(2k+1) > 0 (ie a n-1 , a n-3 ...)

then for 0 ≤ k ≤ n -2 2 , lim u→a - n-(2k+1) h R | I n-2(k+1) (u) = lim u→a + n-(2k+1) h R | I n-(2k+1) (u). Moreover, h R is continuous on a n-2(k+1) < 0 (ie a -n , a -(n-2) , ...) then for n 2 ≤ k ≤ n -1, lim u→a - n-2(k+1) h R | I n-2(k+1) (u) = lim u→a + n-2(k+1) h R | I n-(2k+1) (u).
Therefore, for 0 ≤ k ≤ n -1, we have

⌊ n 2 ⌋ j=1 r j (I n-2 )U k (θ n-1 (j)) sin a n-1 - n -2 2 θ n-1 (j) - π 2 j + δ n -2k -2 2 - ⌊ n+1 2 ⌋ j=1 r j (I n-1 )U k (θ n (j)) sin a n-1 - n -1 2 θ n (j) - π 2 j + δ n -2k -1 2 = r λ n (k) sin λ a n-1 - n -1 2 + θ λ n (k) -r λ n-1 (k) sin λ a n-1 - n -2 2 + θ λ n-1 (k) .
Reciprocally, if h R is continuous on inf I j , since h R is even and

-inf I j = sup I -j , then h R is a continuous function. if 1 ≤ j ≤ ⌊ n+1 2 ⌋
We consider rows of M R and N R . Precisely, let

M R =    L 0 . . . L n-1    and N R =    L0 . . . Ln-1    .
Thanks to relation (3.13), for 0 ≤ k ≤ ⌊ n-1 2 ⌋, we have

L 2k = k j=0 a 2k,j (-1) k-i L2j , and for 0 ≤ k ≤ ⌊ n -2 2 ⌋, L 2k+1 = k j=0 a 2k+1,j (-1) k-i L2j+1 .
As a result, we have

det M R =   n-1 j=0 2 k   det N R = 2 n(n-1) 2 det N R .
In addition, we may write N R = (n k,j ) with:

n k,j = 1 2i [iδθ n-1 (j)] k e i(Rθ n-1 (j)+Φ j ) -[-iδθ n-1 (j)] k e -i(Rθ n-1 (j)+Φ j ) if 1 ≤ j ≤ ⌊ n 2 ⌋ n k,j+⌊ n 2 ⌋ = 1 2i [iδθ n (j)] k e i(Rθn(j)+ Φj ) -[-iδθ n (j)] k e -i(Rθn(j)+ Φj ) if 1 ≤ j ≤ ⌊ n+1 2 ⌋
Due to the multilinearity of determinant, we have

det N R = (iδ) n(n-1) 2 (2i) n ǫ=(ǫ ′ ,ǫ ′′ ) V [ǫ ′ , ǫ ′′ ](-1) σ(ǫ) e i ⌊ n 2 ⌋ j=1 ǫ ′ (j)(Rθ n-1 (j)+Φ j )+ ⌊ n+1 2 ⌋ j=1 ǫ ′′ (j)(Rθn(j)-Φ)
where the sum is running over functions ǫ : {1, .., n} -→ {-1, +1} which may split in ǫ = (ǫ ′ , ǫ ′′ ) where ǫ

(j) = ǫ ′ (j) if 1 ≤ j ≤ ⌊ n 2 ⌋ and ǫ(j + ⌊ n 2 ⌋) = ǫ ′′ (j) if 1 ≤ j ≤ ⌊ n+1 2 ⌋. In addition, σ(ǫ) = |{1 ≤ j ≤ n; ǫ(j) = -1}|. Finally, V [ǫ ′ , ǫ ′′ ] refers to the determinant of the Vandermonde matrix associated to real numbers (ǫ ′ (1)θ n-1 (1), .., ǫ ′ (⌊ n 2 ⌋)θ n-1 (⌊ n 2 ⌋), ǫ ′′ (1)θ n (1), .., ǫ ′′ (⌊ n+1 2 ⌋)θ n (⌊ n+1 2 ⌋)). Since V [ǫ ′ , ǫ ′′ ] = 0 for all ǫ, we may conclude R -→ det N R vanishes at most a finite number of times on ] n-1 2 ; n 2 [. The same result holds for R -→ det M R since det M R = 2 n(n-1) 2 det N R .
Corollary 6 For all R in ] n-1 2 ; n 2 [ such that det M R = 0, there exists one and only one continuous function satisfying the differential equation with temporal shifts (3.8). Moreover:

• If 1 ≤ j ≤ ⌊ n 2 ⌋, r j (I n-2 ) = r λ n-1 k=0 U k (λ) sin θ λ -kδ π 2 ∆ k+1,j . (3.14) • If 1 ≤ j ≤ ⌊ n+1 2 ⌋, r j (I n-1 ) = -r λ n-1 k=0 U k (λ) sin θ λ -kδ π 2 ∆ k+1,j+⌊ n 2 ⌋ . (3.15)
with ∆ k,j = (-1) k+j det M R M k,j where M k,j denotes the minor of M R obtained by removing from M R its k-th row and j-th column.

Exact value of the minimum

In this section, we finish proving theorem 1. Remember that some cases have been solved in corollary 2. Thanks to the previous section, we have an explicit expression of the optimal test function h R . Nevertheless, this explicit expression depends on the unknown parameter λ which is related to mR by the following relation:

λ 2 = 4π 2 mR
In order to conclude, we solve the equation mR = B(h R ) where mR is the only unknown parameter.

Two equations, one unknown parameter It is technically easier to express λ instead of mR . Thanks to relation (3.2), the relation mR = B(h R ) may be written:

λ 2 = R h ′ R (u) 2 du -δ 2 1 -1 h ′ R * h ′ R (u)du R h R (u) 2 du + δ 2 1 -1 h R * h R (u)du + ε R h R (u)du 2 (3.16)
Furthermore, the relation (3.5) may be written:

w λ cos λR + δ 2 R R-1 h R (x)dx + ε R -R h R (x)dx = 0 (3.17)
As a result, λ is satisfying two equations.

Lemma 22 Relation (3.17) implies relation (3.16).

Proof: The differential equation with temporal shifts give

R h ′ R (u) 2 du - δ 2 1 -1 h ′ R * h ′ R (u)du = R h ′ R (u) h ′ R (u) + δ 2 u+1 u-1 h ′ R (t)dt du = R h ′ R (u)ϕ ′ (u)du.
Thanks to the Volterra equation we have

R h R (u) 2 du + δ 2 1 -1 h R * h R (u)du = R h R (u)ϕ(u)du + δ 2 R h R (u)du R R-1 h R (u)du .
Finally, using an integration by parts, we get

R h ′ R (u)ϕ ′ (u)du -λ 2 R h R (u)ϕ(u)du = -wλ cos λR R h R (u).
Therefore, we may write equation (3.16) as

-λ 2 w λ cos λR + δ 2 R R-1 h R (x)dx + ε R -R h R (x)dx R h R (u) = 0.
As a result, we use relation (3.17) in order to determine λ.

End of the proof of theorem 1 In this subsection, we finish proving theorem 1. Precisely,

Proposition 4 If G = SO + , SO -or Sp with n ≥ 2, then λ R := λ is the smallest positive root of δ λ cos θ λ - n-1 k=0 U k (λ) sin θ λ -kδ π 2 δα R (k) 2 -1 + εβ R (k) + 2ε λ n-1 k=0 U k (λ) cos θ λ -kδ π 2 = 0 (3.18) θ n (j) θ n (j) ∆ k+1,j+⌊ n 2 ⌋ n-1 l=0 U l (θ n (j)) sin π 2 j + δ n -2l -1 2 .
Before proving this result, we need to prove some technical lemmas. We may write h R as a sum of two non-continuous even functions. Let ϕ λ and ψ which are defined on R\{a -n , ..., a n } by:

• If 0 ≤ k ≤ n -1, let ϕ λ | I n-(2k+1) (u) = r λ n (k) sin λ u - n -2k -1 2 + θ λ n (k) .
• If 0 ≤ k ≤ n -2, let ϕ λ | I n-2(k+1) (u) = r λ n-1 (k) sin λ u -n -2k -2 2 + θ λ n-1 (k) .

• supp ϕ λ ⊂ [-R, R].

• If u ∈ R\{a -n , ..., a n }, let ψ(u) = h R (u)ϕ λ (u).

Even though ϕ λ and ψ are not continuous functions, they are smooths on each I k . Furthermore, they satisfy h R (u) = ψ(u) + ϕ λ (u) and ϕ ′′ λ = -λ 2 ϕ λ .

Lemma 23 We have = λRe e iλ(an-1-n-2 2 ) z λ n-1 (k)e iλ(an-1-n-1 2 ) z λ n (k) and, for 1 ≤ k ≤ n -1,

lim u→sup I n-2k-1 ϕ ′ λ | I n-2k-1 (u) - lim u→inf I n-2k
ϕ ′ λ | I n-2k (u) = λRe e iλ(an-n-1 2 ) z λ n (k)e iλ(an-n 2 ) z λ n-1 (k -1) .

The result comes easily from relations (3.11) and (3.12). Similarly, we have: In addition, since z λ n-1 (0) = iδ Un(λ) U n-1 (λ) e iλ/2 z λ n (0) -2wδλRe (e iλR ), we get The result comes easily from relation (3.10).

To conclude, since h R = ψ + ϕ λ , we may easily transform equation (3.17) thanks to both lemmas 23 and 25. Proposition 4 follows immediately due to the fact that r λ = 0 (otherwise h R = 0).

Particular case

In this subsection, we are assuming n = 2 and we give a simpler expression of equation (3.18) than in proposition 4.

Corollary 7 If G = SO + , SO -or Sp and 1 2 < R < 1 then λ R is the smallest positive root of which is not a root of U 1 U 2 and where

Θ R = 1 2 R - 1 2 + π 2 1 + δ 2 .
Proof: The matrix M R which appears in corollary 6 can be written

M R = -1 sin 1 2 1 2 -R -π 2 1 + δ 2 0 sin 1 2 1 2 -R -π 2 1 -δ 2 .
Therefore, det M R = -δ cos Θ R = 0.

We get ∆ 1,1 = -1, ∆ 1,2 = 0, ∆ 2,1 = -δ tan Θ R and ∆ 2,2 = δ cos Θ R . In addition, if k ∈ {1, 2}, we have: In addition, we have

α R (k) = 2(R -1)∆ k+1,1 -2 (cos Θ R + δ sin Θ R ) ∆ k+1
z λ = -2iwe -iλ(1-R) U 1 (λ)U 2 (λ) 1 + iδU 1 (λ)e iλ = -2w U 1 (λ)U 2 (λ) (sin λ(1 -R) -2δλ cos λR + i [cos λ(1 -R) -2δλ sin λR]) .
The result comes easily from these relations.

  Preview of results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Statistics for low-lying zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 What is the smallest zero of a L-function ? . . . . . . . . . . . . . . . . . . . . . 2 Proportion of L-functions with a small smallest zero 3 The smallest zero 3.1 Density and topological arguments . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Fourier analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 A Volterra equation with temporal shifts . . . . . . . . . . . . . . . . . . . . . . . 3.4 Optimal test Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Exact value of the minimum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

  easily from relations (3.14) and (3.15) of corollary 6. Similarly, sinceR -R ψ(u)du = n-1 k=0 In-1 ψ| I n-2k-1 (uk)du + n-2 k=0 In-2 ψ| I n-2k-2 (uk)duwe obtain the second part of this lemma.Lemma 24 For 0 ≤ k ≤ n -2, we havelim u→sup I n-2k-2 ϕ ′ λ | I n-2k-2 (u)lim u→inf I n-2k-1 ϕ ′ λ | I n-2k-1 (u) = λU k (λ)Re (iδ) -k z λ and for 1 ≤ k ≤ n -1, lim u→sup I n-2k-1 ϕ ′ λ | I n-2k-1 (u)lim u→inf I n-2k ϕ ′ λ | I n-2k (u) = λU n-k-1 (λ)Re (iδ) -(n-k-1) z λwhere the complex number z λ is defined in lemma 20.Proof: Since h R satisfies the differential equation with temporal shifts (3.8), we may concludeh ′ R is continuous on ] -R, R[. Therefore, for 0 ≤ k ≤ n -2, we get lim u→sup I n-2k-2 ϕ ′ λ | I n-2k-2 (u)lim u→inf I n-2k-1 ϕ ′ λ | I n-2k-1 (u)

  = (iδ) -(n-1) e -iλ(an-n-1 2 ) U n-1 (λ)z λ , we get lim u→sup In-1 ϕ ′ λ | In-1 (u) = -λRe z λ (iδ) -(n-1) U n-1 (λ)and the previous lemma givesR ϕ λ (u)du = -2 λ n-1 k=0 U k (λ)Re (iδ) -k z λ .

ϕ

  ′ | In-1 (u) + lim u→sup In-2 ϕ ′ | In-2 (u)lim u→inf In-1 ϕ ′ | In-1 (u)lim u→inf In-2 ϕ ′ | In-2 (u)The previous lemma giveslim u→sup In-2 ϕ ′ | In-2 (u)lim u→inf In-1 ϕ ′ | In-1 (u) = λRe (z λ ).

- 2w δλ cos λR - 2 λ r λ cos θ λ + 1 λ

 1 lim u→inf In-2 ϕ ′ λ | In-2 (u) = -2wδλ cos(λR) -λRe U n (λ)z λ (iδ) -n . Re z λ (1 -(iδ) -n U n (λ) + (iδ) -(n+1) U n-1 (λ)) .

  λ(1 -R) -2δλ cos λR) -[(δ + 2ε)(1 -R) -1 + 4ε] [cos λ(1 -R) -2δλ sin λR -2λ tan Θ R (sin λ(1 -R) -2δλ cos λR)] = 0

, 2 β

 2 R (k) = α R (k) + 2 (cos Θ R + δ sin Θ R ) ∆ k+1,2 Therefore:    α R (0) = β R (0) = 2(1 -R) α R (1) = 2δ(1 -R) -2(δ + tan Θ R ) β R (1) = 2δ(1 -R) -4(δ + tan Θ R )As a result, since δε = -ε, we getδ λ cos θ λ -n-1 k=0 U k (λ) sin θ λkδ π 2 δα R (k) 2 -1 + εβ R (k) + 2ε λ n-1 k=0 U k (λ) cos θ λkδ π 2 = δ + 2ε λ cos θ λ -δα R (0) 2 + εβ R (0) -1 + 4ε sin θ λ + 2λδ δα R (1) 2 + εβ R (1) -1 cos θ λ = (δ + 2ε) 1 -4λ 2 λ cos θ λ -[(δ + 2ε)(1 -R) -1 + 4ε] [sin θ λ -2λ tan(Θ R ) cos θ λ ] .

Table 2 :

 2

Several natural families of L-functions

Thus, thanks to theorem 1, we have lim sup

Acknowledgements I would like to thank Frederic Bayart for his availability and all their advices on Sobolev spaces.

Comments on the "Sp hypothesis" In the symplectic case with R > 1/2, our proof of theorem 1 is submitted to the "Sp hypothesis". However, we prove that this theorem is still true even though m 2 R is an odd positive integer. Therefore, throughout this paragraph, we are assuming G = Sp, R > 1/2 and m R = N 2 where N is an odd integer.

Proof: In lemma 12, we prove that for all odd positive integer n,

First, with n = N , we may deduce

Second, for all n = N , we get

The sum in the right member of this equality has been computed in the proof of lemma 10 and we have

Therefore, changing w with

this Volterra equation with temporal shift has been solved in section 3.4. Thus, we get an explicit expression of h R . Now, several cases may occur. First, this explicit expression of h R doesn't satisfy the compatibility equation R R-1 h R (u)du = 0, then the "Sp hypothesis" is true. Second, if h R satisfies the compatibility equation, then on account of the fact that the argument of the complex number z λ is independent of w, λ is a root of equation (3.18). Since m R is the smallest critical value of B, λ is still the smallest root of equation (3.18).