Zero free regions for Dirichlet series Christophe Delaunay, Emmanuel Fricain, Elie Mosaki, Olivier Robert
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Christophe Delaunay, Emmanuel Fricain, Elie Mosaki, Olivier Robert. Zero free regions for Dirichlet series. Transactions of the American Mathematical Society, American Mathematical Society, 2013, 365 (6), pp.3227-3253. hal-00552945
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HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. ZERO FREE REGIONS FOR DIRICHLET SERIES
CHRISTOPHE DELAUNAY, EMMANUEL FRICAIN, ELIE MOSAKI, AND OLIVIER ROBERT
Abstract. In this paper, we are interested in explicit zero-free discs for some Dirichlet series and we also study a general Beurling-Nyman criterion for L- functions. Our results generalize and improve previous results obtained by N. Nikol- ski and by A. de Roton. As a concrete application, we get, for example, a Beurling- Nyman type criterion for the Siegel zero problem.
1. Introduction In this article, we are interested in zero-free regions for functions that are obtained s as meromorphic continuation of Dirichlet series s n 1 ann− . Such study arises naturally in various fields of mathematics such as7→ functional≥ analysis and number theory. P The particular case of the Riemann zeta function has been most studied, and is related to the Riemann Hypothesis, asserting that the zeta function does not vanish on the half-plane (s) > 1/2. Several attempts have been made in the direction of solving or reformulatingℜ this conjecture. In his thesis B. Nyman [Nym50] gave a reformulation of the Riemann Hypothesis by means of functional analysis. More precisely let denotes the fractional part and let be the set of functions {·} N n θ f(x)= c j , j x j=1 X where 0 < θ 1, c C and j ≤ j ∈ n
(1.1) cjθj =0. j=1 X Then Nyman proved that the Riemann Hypothesis holds if and only if the character- istic function χ of the interval (0, 1) belongs to the closure of in L2(0, 1). Later (0,1) N 2010 Mathematics Subject Classification. 11M26, 30H10. Key words and phrases. Dirichlet series, Beurling–Nyman criterion, Hardy spaces, zeros of L- functions. This work was supported by the ANR project no. 07-BLAN-0248 ”ALGOL”, the ANR project no. 09-BLAN-005801 ”FRAB” and the ANR project no. 08-BLAN-0257 ”PEPR”. 1 2 CHRISTOPHE DELAUNAY, EMMANUEL FRICAIN, ELIE MOSAKI, AND OLIVIER ROBERT on A. Beurling [Beu55] extended this result by proving that if 1 ˜ λ ˜ 2 λ where dr(λ) = dist(x , Kr) is the distance in L ((0, 1), dx/x) between x and the subspace K˜r. In the case λ = r =1/2, if d˜r(λ) = 0 the region (1.2) corresponds to the half-plane (µ) > 1/2 and Nikolski recovers Nyman’s result (note that in this ˜ ℜ 2 case, d1/2(1/2) = 0 if and only if χ(0,1) belongs to the closure of in L (0, 1); see Proposition 7.3). N More recently, A. de Roton [dR07a] generalized Nyman’s work to the Selberg class of Dirichlet functions and reformulated the Generalized Riemann Hypothesis. Finally let us mention that in [BDBLS00] and [Bur02] interesting lower bounds are ♯ obtained. First the authors proved in [BDBLS00] that if 0 < λ 1 and λ is the closed span in L2(0, + ) of functions ≤ N ∞ n θ f(x)= c j , x> 0, j x j=1 X where λ < θj 1, cj C, then the Riemann Hypothesis holds if and only if ≤ ∈ ♯ limλ 0 d(λ) = 0, where d(λ) = dist(χ(0,1), λ). In other words, we can drop the condition→ (1.1) in the Nyman’s theorem. FurthermoreN they proved that 1 lim inf d(λ) log > 0. λ 0 λ → s | | Then this lower bound was improved by J.F. Burnol in [Bur02] and generalized by de Roton in [dR06] and [dR09] for the Selberg class. ZERO FREE REGIONS FOR DIRICHLET SERIES 3 The aim of our article is to make a further investigation of Nikolski’s work in a more general situation including, in particular, the Selberg class. Associated to some auxiliary function ϕ: [0, ) C, we introduce a class of spaces K (in particular, ∞ → r in the case of the Riemann zeta function, Kr contains K˜r for a special choice of ϕ). Then, for a large class of Dirichlet series, we obtain explicit zero-free regions that are larger than Nikolski’s regions in the case of zeta. For some well chosen λ parameters, if dist(x ,Kr) = 0 our zero-free regions correspond to (µ) >r and we recover (and in fact improve a little bit) the results of de Roton concℜ erning the re- formulation of the Generalized Riemann Hypothesis. Finally, we investigate several explicit applications. In particular, we give zero-free regions for the zeta function and Dirichlet L-functions ; it should be pointed out that the domains obtained have the property of being explicit, but they do not have the ambition of competing with the classical non-zero regions that are involved, for instance, in the prime number theorem. An other application we obtain is somehow a Beurling-Nyman’s criterion for the Siegel’s zero problem of Dirichlet L-functions. As far as we know, it seems to be a new criterion concerning this question. 2. Notations and statements of the main results For a generic s C, we denote by σ (respectively t) its real (respectively imagi- ∈ nary) part so that we have s = σ + it. For s C, we denote by Πs the half-plane defined by ∈ Π = Π = z C : (z) > σ . s σ { ∈ ℜ } For the sequel, we fix a Dirichlet series a L(s)= n ns n 1 X≥ satisfying the following conditions: For every ε> 0, we have a nε. • n ≪ε There exists σ0 < 1 such that the function s L(s) admits a meromorphic • continuation to (s) > σ with a unique pole7→ of order m at s = 1. ℜ 0 L The function s (s 1)mL L(s) is analytic with finite order in Π . • 7→ − σ0 The growth condition on the coefficients (an)n implies that L(s) is an absolutely convergent Dirichlet series for (s) > 1. The second condition tells us that the ℜ function (s 1)mL L(s) can be analytically continued in some half-plane Π which − σ0 contains Π1. Note that we do not require neither an Euler product nor a functional equation for L(s). We also consider an auxiliary function ϕ : [0, + [ C satisfying the following conditions: ∞ −→ 4 CHRISTOPHE DELAUNAY, EMMANUEL FRICAIN, ELIE MOSAKI, AND OLIVIER ROBERT ϕ is supported on [0, 1]. • ϕ is locally bounded on (0, 1). • σ0 ϕ(x)= O(x− ) when x 0. • σ1 → ϕ(x)= O((1 x)− ) when x 1−, for some σ < 1/2. • − → 1 The fact that σ1 < 1 is sufficient for the integral 1 dt ϕ(t)ts t Z0 to be absolutely convergent for (s) > σ . Hence, the Mellin transform ϕ of ϕ ℜ 0 is analytic on Πσ0 . The condition σ1 < 1/2 will become clearer in Lemma 3.3. Recall that the (unnormalized) Mellin transform of a Lebesgue-measurable functionb ϕ : [0, + [ C is the function ϕ defined by ∞ → + ∞ dt ϕ(s)= ϕ(t)ts (s C), b t ∈ Z0 2 whenever the integral isb absolutely convergent. Let H (Πσ) be the Hardy space of analytic functions f : Π C such that f < where σ → k k2 ∞ 1 + 2 ∞ 2 f 2 = sup f(x + it) dt . k k x>σ | | Z−∞ Then the (normalized) Mellin transform 2 dt 2 : L (0, 1), t1−2σ H (Πσ) M ∗ ϕ −→ 1 ϕ 7−→ √2π is a unitary operator (use the Paley-Wiener’s theorem and the change of vari- ables going from the Fourier transform to the Mellin transform,b for instance see 2 dt [Nik02, p. 166]). Here, we write L (0, 1), t1−2σ for the subspace of functions in 2 dt ∗ L (0, + ), 1 2 that vanish almost everywhere on (1, + ). We define ∞ t − σ ∞ n ψ(u) = res(L(s)ˆϕ(s)us,s = 1) a ϕ (u R ), − n u ∈ + n σ0. The following gives a criterion for this property.∞ Theorem 2.1. Let r > σ0. If mL 1, we assume furthermore that r = 1. The following are equivalent: ≥ 6 2 du (i) The function ψ belongs to L ((1, + ), 1+2 ). ∞ u r (ii) The function t L(r + it)ˆϕ(r + it) belongs to L2(R). 7−→ ZERO FREE REGIONS FOR DIRICHLET SERIES 5 We will see in Corollary 3.4 that the conditions (i) and (ii) above are satisfied whenever r > 1. Furthermore, Theorem 2.1 is a generalization of [dR07a, Proposi- tion 3.3] when ϕ = χ(0,1) and r =1/2. We see that the second condition of the theorem above depends on some growth estimates of L in vertical strip which is, in its turn, linked with questions related to the convexity bound and to the Lindel¨of hypothesis. In particular, if L is a function in the Selberg class, then one can prove that L satisfies the generalized Lindel¨of hypothesis if and only if for every k N, we have ∈ Lk( 1 + it) 2 2 R (2.1) t 1 L ( ). 7−→ 2 + it ∈ Note that with the special choice of ϕ = χ(0,1), thenϕ ˆ(s)=1/s and the con- dition (ii) above for r = 1/2 means exactly that (2.1) is satisfied for k = 1. Moreover, in [dR07a] using the functional equation, it is shown that the condition 2 du ψ L ((1, + ), u2 ) (in the case when ϕ = χ(0,1)) is necessary for the generalized Riemann∈ Hypothesis∞ for L-functions in the Selberg class. In [dR07b], it is also shown that the condition on ψ (still with ϕ = χ(0,1)) is satisfied for L-functions in the Selberg class of degree less than 4. n Fix an integer m 0 and let W = n 1(0, 1] . We say that α W is of length n if ≥ ≥ ∈ α belongs to (0, 1]n. For each α in W , its length is denoted by ℓ(α). Now let α W and c Cℓ(α), we say that A =(α,cS) is an m-admissible sequence if ∈ ∈ ℓ(α) (2.2) c α (log α )k = 0 for all 0 k m 1 . j j j ≤ ≤ − j=1 X Furthermore, A is said to be non-trivial if c Cℓ(α) (0,..., 0) . It is easy to prove (see Lemma 4.2 and Lemma∈ 4.1)\{ that for any} fixed ℓ > m, there are infinitely many non-trivial m-admissible sequences of length ℓ. We also easily see that every A =(α,c) is 0-admissible. Note that the notion of admissible sequences had been introduced in [dR07a] and it is a generalization of condition (1.1) which appears in Nyman’s theorem and which corresponds to m = 1. 2 du From now on, we assume that r is chosen such that ψ L ((1, + ), 1+2 ). ∈ ∞ u r Then we associate to each mL-admissible sequence A = (α,c) the function fA,r defined by ℓ(α) r σ0 αj f (t)= t − c ψ , t> 0. A,r j t j=1 2 Xdt We will show that fA,r L ((0, 1), t1−2σ0 ) and we let ∈ ∗ K = span f : A a m -admissible sequence r { A,r L } 6 CHRISTOPHE DELAUNAY, EMMANUEL FRICAIN, ELIE MOSAKI, AND OLIVIER ROBERT and λ¯ 2σ0 d (λ) = dist t − χ ,K (λ Π ), r (0,1) r ∈ σ0 where the (closed) span and the distance are taken with respect to the space 2 dt 2 L ((0, 1), t1−2σ0 ). Let us remark that we trivially have dr(λ) 1/(2 (λ) 2σ0). ∗One of our main theorems is the following which gives≤ zero-freeℜ discs− for the function L. Theorem 2.2. Let λ Π . Then L does not vanish on r σ + D (λ), where ∈ σ0 − 0 r µ λ D (λ) := µ C : − < 1 2( (λ) σ )d2(λ) . r ∈ µ + λ¯ 2σ − ℜ − 0 r 0 − p Note that the zero-free regions which appear in Theorem 2.2 are discs (empty 2 if dr(λ)=1/(2 (λ) 2σ0)) or half-planes. More precisely, for λ = a + ib Πσ0 (a > σ , b R)ℜ and R− [0, 1], then the set ∈ 0 ∈ ∈ µ λ µ C : − < R ∈ µ + λ¯ 2σ − 0 2 a+R (a 2σ0) − is the open (euclidean) disc whose center is Ω = 1 R2 , b and radius is − 2R(a σ0) − 1 R2 if R [0, 1[; if R = 1 this set is the half-plane Πσ0 . In both cases, we easily − ∈ see that this set is contained in the half-plane Πσ0 . As a corollary of the proof of Theorem 2.2, we obtain an other explicit version. Corollary 2.3. Let λ Π . Then L does not vanish on the disc ∈ σ0 µ λ f (λ) r σ + µ C : − < 2( (λ) σ )| A,r | , − 0 ∈ µ + λ¯ 2σ ℜ − 0 f ( 0 A,r 2 ) − p kd k for any m -admissible sequence A. L Note that taking L(s) = ζ(s), ℓ(α) = 2 and ϕ = χ(0,1) (so that σ0 = σ1 = 0) we recover exactly the results of Nikolski ([Nik95]). Now, taking ϕ(t) = (1 σ1 − t)− χ(0,1)(t), we obtain larger zero-free discs whenever (λ) is large enough. We refer the reader to Section 7 for further discussions aboutℑ the Rieman zeta function and other examples. We will see in Theorem 4.3 that n λ+r σ0 f (λ)= L(λ + r σ )ˆϕ(λ + r σ ) c α − , λ Π . A,r − 0 − 0 j j ∈ σ0 j=1 ! X Hence Corollaryd 2.3 can be understood as follows: let λ Π such that L(λ + r ∈ σ0 − σ0) = 0; then there is a small neighborhood of λ + r σ0 free of zeros for L. Of course,6 this is an obvious consequence of the continuity− of the function L but the interest of Corollary 2.3 is that it gives an explicit neighborhood where the function ZERO FREE REGIONS FOR DIRICHLET SERIES 7 L does not vanish; moreover, this explicit neighborhood is expressed in terms of the values of L and in particular it does not use any estimates of the derivatives. Finally, we obtain a general Beurling-Nyman type theorem. Theorem 2.4. Suppose that the function ϕˆ does not vanish on the half-plane Πr, log ϕˆ(x+r σ0) | − | that lim supx + x = 0 and that a1 = 0. Then the following assertions are equivalent:→ ∞ 6 (1) The function L does not vanish on the half-plane Πr. (2) There exists λ Π such that d (λ)=0. ∈ σ0 r (3) For all λ Πσ0 , we have dr(λ)=0. ∈ 2 1 2σ0 (4) We have Kr = L ((0, 1), dt/t − ). ∗ Taking L(s) = ζ(s), ℓ(α) = 2, r = λ = 1/2 and ϕ = χ(0,1) (so that we can take σ1 σ0 = σ1 = 0), we obtain Nyman’s theorem. Now, taking ϕ(t) = (1 t)− χ(0,1)(t) we extend the results of [dR07a] (see also Section 7 for more details).− An other consequence of Theorem 2.2 is, in a way, a Beurling-Nyman criterion for Dirichlet L-functions. More precisely, let χ be a Dirichlet character with conductor q and L(χ, s) its L-function. Then, for 1/2 r< 1, we define d by ≤ r 1 2 2 1 ℓ 1 r r dt dr = min t − t cj χ(n) ℓ,c,α 0 − t Z j=1 n<αj /t X X ℓ ℓ where the minimum is taken over all ℓ 0, c = (cj) C and α = (αj) (0, 1] . One can show (see Proposition 7.6) that≥d2 < 1/(2 2∈r). We have ∈ r − Theorem 2.5. If 1 C2 d2 r ≤ 2 2r − 2(log q)2(1 r)3 − − for some (absolute) constant C and some 1/2 r 1, then L(χ, σ) does not vanish in the real-interval σ> 1 C/ log q. ≤ ≤ − In order to obtain the criterion for the Siegel zero problem, then we consider all Dirichlet characters χ and an absolute constant C independant of χ. The next section is devoted to the proof of Theorem 2.1 and to the study of the function ψ. Section 4 will focus on the admissible sequences and the functions fA,r. Theorem 2.2 and Corollary 2.3 will be proven in Section 5 and Theorem 2.4 will be proven in Section 6. Some explicit examples will be studied in Section 7 in which we will prove Theorem 2.5. 8 CHRISTOPHE DELAUNAY, EMMANUEL FRICAIN, ELIE MOSAKI, AND OLIVIER ROBERT 3. The function ψ and proof of theorem 2.1 We define the functions ψ1 and ψ2 by s ψ1(u) = res(L(s)ˆϕ(s)u ,s =1) (u R+), n ∈ ψ (u) = a ϕ (u R ), 2 n u ∈ + n mL p k (3.1) L(s)ˆϕ(s)= − H(s), (s 1)k − Xk=1 − with H analytic in Πσ0 . Lemma 3.1. We have 1 1 s 1 (3.2) H(s)= ψ t − dt = φˆ(t), (s) > 1, t ℜ Z0 where φ(t)= ψ(1/t)χ(0,1)(t). Proof of Lemma 3.1. On the one hand, we have + ∞ s 1 (3.3) L(s)ˆϕ(s)= ψ (u)u− − du, (s) > 1. 2 ℜ Z1 Indeed, this equality comes from the following computation and Fubini’s theorem: 1 1 s an s 1 t dt L(s)ˆϕ(s) = ϕ(t)t − dt = a ϕ(t) ns n n t n 1 ! 0 n 1 0 ≥ Z ≥ Z X + X ∞ n s 1 = a ϕ u− − du n u n 1 n ≥ Z X+ ∞ n s 1 = anϕ u− − du 1 u Z n 1 and then ℜ 1 an σ 1 | | ϕ(t) t − dt < + . nσ | | ∞ n 1 0 X≥ Z ZERO FREE REGIONS FOR DIRICHLET SERIES 9 On the other hand, for each 1 k m , we have ≤ ≤ L us u(log u)k 1 (3.4) res ,s =1 = − , (s 1)k (k 1)! − − and an easy induction argument gives + 1 ∞ k 1 s 1 (log u) − u− du = . (k 1)! (s 1)k − Z1 − Hence, with the notation of (3.1), the equality + mL ∞ s 1 p k − − − (3.5) ψ1(u)u du = k . 1 (s 1) Z Xk=1 − follows by linearity. Equations (3.3) and (3.5) imply mL + 1 p k ∞ s 1 1 s 1 − − − − H(s)= k L(s)ˆϕ(s)= ψ(u)u du = ψ t dt. (s 1) − 1 0 t Xk=1 − Z Z Remark 3.2. From (3.1) and (3.4), we easily see that we can write ψ1(t)= tP (log t) where P is a polynomial of degree < m (P 0 if m = 0). More precisely we have L ≡ L mL p k k 1 ψ (t)= t − (log t) − . 1 (k 1)! Xk=1 − Lemma 3.3. The function s H(s) is of finite order on Πr. Moreover, for all σ > 1, the function t H(σ +7→it) belongs to L2(R) and 7→ 1 H(σ + it) 2dt = O for σ + . 1 2σ1 R | | σ → ∞ Z − Proof of Lemma 3.3. The function s H(s) is bounded on some neighborhood V1 of s = 1. The functions s p−j7→and (by assumption) s L(s) are of finite 7→ j (s 1)j 7→ order on Π r V . Moreover, s ϕ−ˆ(s) is bounded on the closure of Π since σ0 1 P7→ r 1 dx ϕˆ(s) ϕ(x) xr , (s) r, | | ≤ | | x ℜ ≥ Z0 σ0 σ1 and the last integral is convergent because r > σ and ϕ(x) = O(x− (1 x)− ). 0 − Hence we can conclude that H is of finite order on Πr, which proves the first part of the lemma. 10 CHRISTOPHE DELAUNAY, EMMANUEL FRICAIN, ELIE MOSAKI, AND OLIVIER ROBERT Let σ > 1 be a fixed real number. We easily check that, for every 1 j mL, the j 2 ≤ ≤ function t (σ 1+ it)− belongs to L (R) and we have 7−→ − 1 2 1 (3.6) j dt = O 2j 1 , σ + . R (σ + it 1) σ → ∞ Z − − On the other hand, using Plancherel’s formula and the fact that σ1 < 1/2 and σ> 1 > σ0, we have 1 dx (3.7) ϕˆ(σ + it) 2dt =2π ϕ(x) 2x2σ < + . R | | | | x ∞ Z Z0 More precisely 1 dx 1 (3.8) ϕ(x) 2x2σ = O , σ + . | | x σ1 2σ1 → ∞ Z0 − σ0 σ1 Indeed, we have ϕ(x)= O(x− (1 x)− ) and − 1 σ0 σ1 2 2σ dx x− (1 x)− x = β(2(σ σ ), 1 2σ ) | − | x − 0 − 1 Z0 where β(s, z) is the beta function. Using Stirling’s formula, we have Γ(1 2σ ) β(2(σ σ ), 1 2σ ) − 1 , σ + . 0 1 1 2σ1 − − ∼ (2σ) − → ∞ So we get the estimate (3.8). Therefore t ϕˆ(σ + it) belongs to L2(R) and we have 7−→ 1 (3.9) ϕˆ(σ + it) 2dt = O , σ + . 1 2σ1 R | | σ → ∞ Z − It remains to notice that s L(s) is bounded in Πσ (since by hypothesis σ > 1) and then the second part of7→ the lemma follows immediately from (3.6), (3.9) and (3.1). Now we can prove Theorem 2.1. 1 Assume that (i) is satisfied. Then the function φ(t) = ψ t χ(0,1)(t) belongs to 2 dt ˆ 2 L ((0, 1), 1 2 ) and thus the function G := φ belongs to H (Π ). The analytic t − r r continuation∗ principle implies that the equality (3.2) is satisfied for every s Π , ∈ r that is H(s)= φˆ(s)= G(s), s Π . Since G H2(Π ), we know that the function ∈ r ∈ r G∗, defined by G∗(t) := lim G(σ + it), σ r →> exists almost everywhere on R and belongs to L2(R). But G∗(t) = lim H(σ + it)= H(r + it), σ r →> ZERO FREE REGIONS FOR DIRICHLET SERIES 11 because H is continuous on the closed half-plane (s) r. Therefore t H(r+it) 2 ℜ ≥ 7−→ belongs to L (R). It remains to notice that for every 1 k mL, the function k 2 ≤ ≤ t (r 1+ it)− belongs to L (R) (this is where we have to assume that r = 1 7−→ − 6 whenever mL 1). Thus according to (3.1), we get that t L(r + it)ˆϕ(r + it) belongs to L2(R≥). 7→ Conversely assume that (ii) is satisfied. Then the function t H(r + it) belongs to L2(R). 7−→ Let σ2 > max(1,r). The function H is analytic on Ω := s : r< (s) < σ2 and continuous on the closure of Ω. Using the Hardy, Littlewood{ and Ingℜham’s theorem} [HIP27, Theorem 7], we deduce from Lemma 3.3 that sup H(σ + it) 2dt < + . r σ σ2 R | | ∞ ≤ ≤ Z 2 Hence, the second part of Lemma 3.3 gives that H belongs to H (Πr). Since we 2 2 dt 2 dt have H (Πr) = L (0, 1), t1−2r , there exists φ1 L (0, 1), t1−2r such that M ∗ ∈ ∗ H(s)= φˆ (s), for every s Π . Thus for (s) > max(1,r), we have φˆ (s)= H(s)= 1 ∈ r ℜ 1 φˆ(s). By injectivity of the Mellin transform, we get that 1 ψ χ (t)= φ(t)= φ (t). t (0,1) 1 1 2 dt Thus t ψ t χ(0,1)(t) belongs to L (0, 1), t1−2r , which implies that ψ belongs 2 7−→ du ∗ to L (1, + ), 1+2r and that concludes the proof of Theorem 2.1. ∞ u 2 du Corollary 3.4. Let r> 1. Then ψ L (1, + ), 1+2 and we have ∈ ∞ u r σ1 1/2 ψ = O r − , as r + . k k2 → ∞ Proof. Let r > 1 be a fixed real number. By Lemma 3.3, the function t H(r+it) 2 7−→ j belongs to L (R) and for every 1 j mL, the function t (r 1+it)− belongs also to L2(R). Therefore it follows≤ from≤ (3.1) that t L7−→(r + it−)ˆϕ(r + it) belongs 2 2 7−→ du to L (R) and Theorem 2.1 implies that ψ L ((1, + ), 1+2 ). Moreover, if we let ∈ ∞ u r 12 CHRISTOPHE DELAUNAY, EMMANUEL FRICAIN, ELIE MOSAKI, AND OLIVIER ROBERT as before φ(t)= ψ(1/t)χ(0,1)(t), we have by Plancherel’s formula + 1 2 ∞ du 1 dt ψ 2 = ψ(u) 2 = ψ k k2 | | u1+2r t t1 2r Z1 Z0 − 2 = φ L2((0,1), dt ) k k ∗ t1−2r 1 = φˆ(r + it) 2 dt 2π R | | Z 1 = H(r + it) 2 dt, 2π R | | Z and Lemma 3.3 gives the result. Remark 3.5. In fact, if L(s) has some nice arithmetical properties (e.g. satisfying a Wiener-Ikehara type theorem), one can expect that the main contributions of ψ1 will be compensated by the main contributions of ψ2 and so that the function ψ 2 du will belong to the space L (1, + ), 1+2 for smaller value of r. For example, if ∞ u r L(s) = ζ(s) and if ϕ(t) = χ0,1(t) for all t (0, 1), then ψ(u) = u u +1 (so ∈ 2 du − ⌈ ⌉ ψ(u)= u for almost all u (1, )) and ψ L (1, + ), 1+2 for all r > 0. In { } ∈ ∞ ∈ ∞ u r this caseϕ ˆ(s)=1/s and t ζ(r + it) / r + it L2(R) for all r > 0 as expected by Theorem 2.1. We will discuss7→ | about| other| examples| ∈ in Section 7. Remark 3.6. Assume that ϕ is bounded at t = 1 (which corresponds to σ1 = 0). Thus, for every α> 1, we have ψ(t)= ψ (t) ψ (t)= O(tα), t + . Indeed, on 1 − 2 → ∞ the one hand, as we have seen ψ1(t)= tP (log t), where P is a polynomial of degree < mL. On the other hand, let α > 1 be a fixed real number. Then there exists a constant C > 0 such that α ϕ(t) Ct− , t (0, 1). | | ≤ ∈ Hence we get n t α ψ (t) a ϕ C a | 2 | ≤ | n| t ≤ | n| n n 4. Admissible sequences and the functions fA,r The following results will be useful in the sequel. ZERO FREE REGIONS FOR DIRICHLET SERIES 13 Lemma 4.1. Let (α,c) be an m-admissible sequence. (1) For any polynomial P of degree d < m, we have ℓ(α) cjαjP (log αj)=0. j=1 X (2) For any positive real number λ1 and for any real number λ2 maxj(αj), λ1 1 λ1 ≥ both ((αj )j, (cjαj− )j) and ((αj/λ2)j,c) are m-admissible sequences. The proof is easy and follows immediately from the definition and so we omit it. For an m-admissible sequence A =(α,c) we define the entire function ℓ(α) g (s)= c αs (s C). A j j ∈ j=1 X Using (2.2), we notice that these functions satisfy (4.1) g(k)(1)=0 (0 k m 1). A ≤ ≤ − Lemma 4.2. Let m be a nonnegative integer. Then we have the followings (1) For every integer ℓ m + 1 and every 0 < α1 < α2 < < αℓ 1, there exists c Cℓ such≥ that A = (α,c) is an m-admissible··· sequence≤ and ∈ g(m)(1) =0. Furthermore we can choose c =0. A 6 ℓ 6 (2) For every s1 C r 1 , there exists an m-admissible sequence A such that g (s ) =0. ∈ { } A 1 6 m+1 Proof. (1) Let ℓ m + 1, then there is a unit vector (b1, b2,...,bm, bℓ) C such that ≥ ∈ 1 ...... 1 1 b1 0 . . (log α1) ...... (log αm) (log αℓ) . . (4.2) . . . = , . . . bm 0 (log α )m ...... (log α )m (log α )m b 1 1 m ℓ ℓ since the corresponding van der Monde matrix is invertible. Setting bj , 1 j m αj ≤ ≤ cj = 0, m +1 j ℓ 1 ≤ ≤ − bℓ , j = ℓ. αℓ 14 CHRISTOPHE DELAUNAY, EMMANUEL FRICAIN, ELIE MOSAKI, AND OLIVIER ROBERT (m) Then A = (α,c) is an m-admissible sequence and gA (1) = 1. Furthermore we easily see that b = 0. Indeed if b = 0 then we have ℓ 6 ℓ 1 ...... 1 b1 0 (log α ) ...... (log α ) . . 1 m . . . . . = . , . . . . (log α )m 1 ...... (log α )m 1 1 − m − bm 0 and using once more that the corresponding van der Monde matrix isinvertible, we get b1 = b2 = = bm = bℓ = 0, which contradicts (4.2). Therefore bℓ = 0 and then c = 0. ··· 6 ℓ 6 (2) Let A =(α,c) be a non trivial m-admissible sequence. For 0 <λ< 1, consider g (s)= g (1 λ + λs). Note that Aλ A − ℓ 1 λ λ s gAλ (s)= cjαj− (αj ) , j=1 X λ 1 λ and, by Lemma 4.1, Aλ = ((αj )j, (cjαj− )j) is an m-admissible sequence. Now let s1 = 1; then there necessarily exists 0 <λ< 1 such that gAλ (s1) = 0, since gA is analytic6 and non identically zero. 6 2 du Theorem 4.3. Let r > σ such that ψ L ((1, + ), 1+2 ) and let A = (α,c) be 0 ∈ ∞ u r an mL-admissible sequence. We define ℓ(α) r σ0 αj f (t)= t − c ψ (t> 0). A,r j t j=1 X Then we have: (1) fA,r(t)=0, if t> maxj αj. 2 dt (2) fA,r(t) L ((0, 1), t1−2σ0 ). (3) For (s∈) >∗ σ , we have ℜ 0 (4.3) f (s)= L(s + r σ )ˆϕ(s + r σ )g (s + r σ ). A,r − − 0 − 0 A − 0 Proof. (1) As before, we write ψ = ψ ψ . From the remark 3.2, we have ψ (u)= d 1 − 2 1 uP (log u) where P is a polynomial of degree < mL. Then ℓ(α) ℓ(α) ℓ(α) α α α c ψ j = c ψ j c ψ j j t j 1 t − j 2 t j=1 j=1 j=1 X X X ℓ(α) ℓ(α) α α nt = c j P log( j ) c a ϕ . j t t − j n α j=1 j=1 j X X n<αXj /t ZERO FREE REGIONS FOR DIRICHLET SERIES 15 If t> maxj(αj) the second sum is 0 because αj/t < 1. Furthermore, by Lemma 4.1, (α/t,c) is an mL-admissible sequence and the first sum is also zero. r σ0 αj 2 dt (2) Each of the terms t − ψ t belongs to the space L ((0, 1), t1−2σ0 ). Indeed, if α := max α , we have k k j j α 2 α 2 k k 2r 2σ0 αj dt k k 2r 1 αj t − ψ = t − ψ dt 1 2σ0 0 t t − 0 t Z Z + 2r ∞ 2 du = αj ψ (u) αj | | u1+2r Z kαk 1 + du ∞ du = α2r ψ(u) 2 + α2r ψ(u) 2 j α 1+2r j 1+2r j | | u 1 | | u Z kαk Z 1 + du ∞ du = α2r ψ (u) 2 + α2r ψ(u) 2 , j α 1 1+2r j 1+2r j | | u 1 | | u Z kαk Z since ψ(u)= ψ1(u) for u< 1. Now, by hypothesis, the second integral is finite and αj the first integral is also finite because ψ1(u)= uP (log u) and α > 0. Furthermore, we see that k k (4.4) 1/2 ℓ(α) 1 r 2 du fA,r 2 dt cjα ψ1(u) + ψ 2 du . L∗((0,1), 1 2 ) j min α 1+2r L ((1,+ ), 1+2 ) k k t − σ0 ≤ | | j j | | u k k ∞ u r j=1 Z maxj αj X (3) For (s) > 1, we have by (3.3) ℜ + ∞ s 1 L(s)ˆϕ(s)= ψ2(u)u− − du. Z1 Hence ℓ(α) ℓ(α) + s ∞ u − du L(s)ˆϕ(s) c αs = c ψ (u) j j j α 2 u j=1 j=1 1 j X X Z ℓ(α) αj α dt = c ψ j j 2 t t1 s j=1 0 − X Z ℓ(α) 1 α dt = c ψ j , j 2 t t1 s j=1 0 − X Z the last equality follows from ψ (α /t) = 0 if α < t 1. 2 j j ≤ 16 CHRISTOPHE DELAUNAY, EMMANUEL FRICAIN, ELIE MOSAKI, AND OLIVIER ROBERT Using once more Lemma 4.1, we have ℓ(α) ℓ(α) α α c ψ j = c ψ j , j t − j 2 t j=1 j=1 X X whence ℓ(α) ℓ(α) 1 α dt L(s)ˆϕ(s) c αs = c ψ j j j − j t t1 s j=1 0 j=1 − X Z X 1 σ0 r dt − = t fA,r(t) 1 s − 0 t − Z 1 s+σ0 r 1 = f (t)t − − dt − A,r Z0 = f (s + σ r), (s) > 1. − A,r 0 − ℜ So, for (s) > 1+σ0 r, we have fA,r(sb)= L(s+r σ0)ˆϕ(s+r σ0)gA(s+r σ0). By theℜ analytic continuation− principle, the− equality holds− for (s−) > σ . (Note− that ℜ 0 from (4.1) the pole of L(s + r dσ0) is killed by the zero of gA(s + r σ0) at s =1 r + σ .) − − − 0 5. Zeros free regions for Dirichlet series Before proving Theorem 2.2 we need some well known tools concerning the Hardy 2 space H (Πσ0 ). For these we refer to [Hof62, Chapter 8]. Actually, in [Hof62], 2 the following facts are stated for the space H (Π0) but it is easy to obtain the 2 corresponding results for H (Πσ0 ), for instance using the unitary map h(z) 2 2 7−→ h(z σ0) from H (Πσ0 ) onto H (Π0). Recall− that if h H2(Π ), then ∈ σ0 h∗(σ0 + it) := lim h(σ + it) σ σ0 →> exists for almost every t R (with respect to the Lebesgue measure). Moreover 2 ∈ we have h∗ L (σ0 + iR) and h 2 = h∗ 2. We can therefore identify (unitarely) 2 ∈ k k 2 k k R H (Πσ0 ) with a (closed) subspace of L (σ0 +i ). In the following we use the symbol 2 h not only for the function in H (Πσ0 ) but also for its ”radial ” limit (in other words 2 we forget the star). This identification enables us to consider H (Πσ0 ) as an Hilbert space, with scalar product given by + ∞ (5.1) h, g = h(σ + it)g(σ + it) dt, h, g H2(Π ). h i2 0 0 ∈ σ0 Z−∞ ZERO FREE REGIONS FOR DIRICHLET SERIES 17 Now for λ Π , we have the following integral representation ∈ σ0 + 1 ∞ h(σ + it) h(λ)= 0 dt, 2π λ σ0 it Z−∞ − − so that we can write, using (5.1), (5.2) h(λ)= h, k , h λi2 2 where kλ is the function in H (Πσ0 ) defined by 1 1 (5.3) kλ(z) := , z Πσ0 . 2π z 2σ0 + λ¯ ∈ − 2 The function kλ is called the reproducing kernel of H (Πσ0 ) and we have 1 (5.4) k 2 = k ,k = k (λ)= . k λk2 h λ λi2 λ 4π( (λ) σ ) ℜ − 0 Recall now a useful property of factorization for functions in the Hardy space. 2 Lemma 5.1. Let h H (Πσ0 ) and µ Πσ0 such that h(µ)=0. Then there exists g H2(Π ) such that∈ h = g and∈ ∈ σ0 k k2 k k2 z µ h(z)= − g(z), z Π . z +¯µ 2σ ∈ σ0 − 0 Proof. See the results in [Hof62, Chapter 8, pp. 132] and apply the transform h(z) h(z σ0). 7−→ − Proof of Theorem 2.2 Denote by Er = Kr. Since the Mellin transform is a 2 dt 2 M unitary map from L ((0, 1), t1−2σ0 ) onto H (Πσ0 ), we have ∗ E = span 2 (h : A a m -admissible sequence) , r H (Πσ0 ) A,r L where h (s)= f (s), for (s) > σ . It follows from Theorem 4.3 that A,r M A,r ℜ 0 1 hA,r(s)= L(s + r σ0)ˆϕ(s + r σ0)gA(s + r σ0), (s) > σ0. −√2π − − − ℜ Now assume that there is µ Π such that L(µ + r σ ) = 0. Since µ Π , we ∈ σ0 − 0 ∈ σ0 get that hA,r(µ) = 0, for every mL-admissible sequence A. Thus, for all h Er with 2 ∈ h 2 = 1, we have h(µ) = 0. Since h H (Πσ0 ), we know from Lemma 5.1 that ktherek is g H2(Π ) such that g = ∈h = 1 and ∈ σ0 k k2 k k2 z µ h(z)= − g(z) (z Π ). z +¯µ 2σ ∈ σ0 − 0 Hence with Cauchy-Schwarz inequality and (5.2), we deduce that λ µ λ µ λ µ h(λ) = − g(λ) − g k = − k , | | λ +¯µ 2σ | | ≤ λ +¯µ 2σ k k2k λk2 λ +¯µ 2σ k λk2 0 0 0 − − − 18 CHRISTOPHE DELAUNAY, EMMANUEL FRICAIN, ELIE MOSAKI, AND OLIVIER ROBERT so µ λ sup h(λ) ¯− kλ 2. h Er | | ≤ µ + λ 2σ0 k k ∈ h 2=1 − k k By contraposition, we have proved that L does not vanish on sup f Er h(λ) µ λ h∈2=1 | | r σ + µ C : − < k k . − 0 ∈ µ + λ¯ 2σ k 0 λ 2 − k k It remains to prove that 2 sup h Er h(λ) h∈2=1 | | k k =1 2( (λ) σ )d2(λ). k 2 − ℜ − 0 r k λk2 To show this equality, first remark that sup h(λ) = sup h, kλ 2 = PEr kλ 2, h Er | | h Er |h i | k k h∈2=1 h∈2=1 k k k k 2 where PE denotes the orthogonal projection onto the closed subspace E of H (Πσ0 ). Then Pythagoras’ Theorem implies that 2 2 2 2 ⊥ sup h(λ) = PEr kλ 2 = kλ 2 PEr kλ 2. h Er | | k k k k −k k h∈2=1 k k Hence 2 sup h Er h(λ) 2 h∈2=1 | | dist (kλ, Er) k k =1 . k 2 − k 2 k λk2 k λk2 But for (s) > σ , we have ℜ 0 1 ¯ 1 ¯ 1 1 λ 2σ0 λ 2σ0 s 1 √ t − χ(0,1)(t) (s)= t − t − dt = ¯ = 2πkλ(s), M √2π 0 √2π λ 2σ0 + s Z − 2 dt according to (5.3). Since the Mellin transform is an isometry from L ((0, 1), t1−2σ0 ) 2 ∗ onto H (Πσ0 ), we obtain 1 λ¯ 2σ0 dr(λ) dist(kλ, Er)= dist(t − χ(0,1)(t),Kr)= . √2π √2π We conclude the proof of Theorem 2.2 using k 2 = 1 . λ 2 4π( (λ) σ0) k k ℜ − Proof of Corollary 2.3 It follows from the proof of Theorem 2.2 and (5.4) that L ZERO FREE REGIONS FOR DIRICHLET SERIES 19 does not vanish on C µ λ r σ0 + µ : ¯− < 4π( (λ) σ0) sup h(λ) . − ∈ µ + λ 2σ0 ℜ − h Er | | ∈ − p h 2=1 k k So in particular (with h = hA,r), we get that L does not vanish on µ λ h (λ) r σ + µ C : − < 4π( (λ) σ )| A,r | , − 0 ∈ µ + λ¯ 2σ ℜ − 0 h 0 A,r 2 − p k k which proves corollary 2.3 because h = f = 1 f . A,r M A,r √2π A,r The following could be interesting in applications. d Corollary 5.2. Let M be a subspace of Kr and λ Πσ0 . Then L does not vanish on the disc ∈ 2 µ λ 2 λ¯ 2σ0 r σ + µ C : − < 1 2( (λ) σ )dist (t − χ , M) . − 0 ∈ µ + λ¯ 2σ − ℜ − 0 (0,1) ( 0 ) − λ¯ 2σ0 λ¯ 2σ0 Proof. It is sufficient to note that dist(t − χ(0,1), M) dist(t − χ(0,1),Kr) = ≥ dr(λ) and then apply Theorem 2.2. Adapting the proof of Theorem 2.2, we could obtain immediately the following generalization. N Theorem 5.3. Let k and λ Πσ0 . The function L does not have any zero of order greater or equal∈ to k on ∈ µ λ 1 r σ + µ C : − < 1 2( (λ) σ )d2(λ) 2k . − 0 ∈ µ + λ¯ 2σ − ℜ − 0 r 0 − 6. A Beurling–Nyman type theorem for Dirichlet series One of the main steps for proving Theorem 2.4 is to prove that the space Er is a 2 closed subspace of H (Πσ0 ) that is invariant under multiplication operator τv, v 0, and then to apply Lax–Beurling’s theorem. Before recalling this theorem, we≥ give some notations and results. We refer to [Nik02, Part A, Chap. 2 & 6] for more details. For v R, let τ be the operator of multiplication on L2(σ + iR) defined by ∈ v 0 ivt 2 (τ f)(σ + it)= e− f(σ + it) (f L (σ + iR)). v 0 0 ∈ 0 We also denote by H∞(Πσ0 ) the Hardy space of bounded analytic functions on Πσ0 ; 2 as in H (Πσ0 ), functions in H∞(Πσ0 ) admit ”radial” limits at almost every points of the boundary (s)= σ and if f H∞(Π ) and f ∗ is its boundary limits, then ℜ 0 ∈ σ0 20 CHRISTOPHE DELAUNAY, EMMANUEL FRICAIN, ELIE MOSAKI, AND OLIVIER ROBERT R f ∗ L∞(σ0 +i ) and f ∗ = f (= supz Πσ f(z) ). A function Θ in H∞(Πσ0 ) ∈ k k∞ k k∞ ∈ 0 | | is said to be inner if Θ(σ0 + it) = 1 for almost every point t R. | | 2 ∈ It is easy to see that if Θ is inner and E =ΘH (Πσ0 ), then E is a closed subspace of H2(Π ) invariant by τ , v 0. Lax–Beurling’s theorem gives the converse. σ0 v ≥ 2 Theorem 6.1 (Lax–Beurling). Let E be a closed subspace of H (Πσ0 ) such that τvE E, v 0. Then there is an inner function Θ H∞(Πσ0 ) unique (up to a ⊂ ∀ ≥ 2 ∈ constant of modulus one) such that E =ΘH (Πσ0 ). Remark 6.2. When E is spanned by a family of functions, we can precise a little bit 2 more the conclusion of Theorem 6.1. Indeed, let E be a closed subspace of H (Πσ0 ) 2 spanned by a family of functions (fi)i I , fi H (Πσ0 ) and let fi = higi be the ∈ ∈ factorization of fi in inner factor hi and outer factor gi. If τvE E, v 0, then E = ΘH2(Π ), where Θ = gcd(h : i I) is the greatest common⊂ inner∀ ≥ divisor of σ0 i ∈ the family (hi)i I . ∈ Proof of Theorem 2.4 First of all, note that 1 (6.1) h− ( 0 )= s Π : L(s + r σ )=0 , { } { ∈ σ0 − 0 } h Er \∈ where we recall that Er = Kr = span 2 (hA,r : A a mL-admissible sequence) , M H (Πσ0 ) and 1 hA,r(s)= L(s + r σ0)ˆϕ(s + r σ0)gA(s + r σ0), (s) > σ0. −√2π − − − ℜ Indeed, the first inclusion 1 s Π : L(s + r σ )=0 h− ( 0 ) { ∈ σ0 − 0 } ⊂ { } h Er \∈ is trivial. For the converse inclusion, it is sufficient to notice that according to Lemma 4.2, the only common zeros to functions g (s + r σ ) is s =1 r + σ but A − 0 − 0 this zero is compensated by the singularities of L(s + r σ0) at this point. The proofs of (4) = (3) and (3) = (2) are trivial. − ⇒ ⇒ (2) = (1): let λ Πσ0 such that dr(λ) = 0. Then according to Theorem 2.2, the function⇒ L does not∈ vanish on µ λ r σ + µ C : − < 1 , − 0 ∈ µ + λ¯ 2σ 0 − and it follows from the paragraph after Theorem 2.2 that this region is precisely the half-plane Πr. ZERO FREE REGIONS FOR DIRICHLET SERIES 21 2 dt (1) = (4): denote by Sβ the ”shift” on L ((0, 1), t1−2σ0 ) defined by ⇒ ∗ σ0 t (S f)(t) := β− f , 0 < β 1, 0 t 1. β β ≤ ≤ ≤ 2 dt It is clear that Sβ is a unitary operator on L ((0, 1), t1−2σ0 ). We will show that ∗ (6.2) S K K (0 < β 1). β r ⊂ r ≤ If A =(α,c) is a m -admissible sequence and 0 t 1, we have L ≤ ≤ σ0 t (S f )(t)=β− f β A,r A,r β ℓ(α) r σ0 σ0 t − αjβ =β− cjψk βr σ0 t − j=1 r X =β− fA′,r(t), where A′ = (α′,c) is the mL-admissible sequence with α′ = (αj′ )j = (βαj)j and so f ′ K . Hence for every admissible sequence A, we have proved that S f K . A ,r ∈ r β A,r ∈ r Since Sβ is a bounded operator and the functions fA,r span the subspace Kr, we 2 deduce (6.2). Therefore we obtain that Er = Kr is a closed subspace of H (Πσ0 ) M 1 which is invariant under the semi-group of operators Sβ − , 0 < β 1. Now let us show that M M ≤ 1 (6.3) S − = τ , M βM v 2 2 dt 1 with v = log β. If h H (Πσ0 ) and if f L ((0, 1), t1−2σ0 ) is such that − h = f, we have for− (s) > σ ∈, ∈ ∗ M ℜ 0 1 1 ∞ s 1 ( Sβ − h)(s)= (Sβf)(t)t − dt M M √2π 0 Z + 1 ∞ σ0 t s 1 = β− f t − dt √2π β 0 Z + s σ0 1 ∞ s 1 =β − f(u)u − du √2π Z0 s σ0 s σ0 =β − ( f)(s)= β − h(s). M Hence we have 1 it it log β ( S − h)(σ + it)= β h(σ + it)= e h(σ + it)=(τ h)(σ + it), M βM 0 0 0 v 0 with v = log β, which proves (6.3). We therefore obtain that τvEr Er, for all v 0 and− then Lax–Beurling’s Theorem (see Theorem 6.1) implies that⊂ there is an ≥ 2 inner function Θ in the half-plane Πσ0 such that Er =ΘH (Πσ0 ). Moreover we know that Θ = BS, where B is a Blaschke product and S is a singular inner function 22 CHRISTOPHE DELAUNAY, EMMANUEL FRICAIN, ELIE MOSAKI, AND OLIVIER ROBERT ([Nik02]). But according to Remark 6.2, the zeros of B coincide with the set of common zeros of functions h Er. Hence it follows from (6.1) and the hypothesis that B has no zeros. In other∈ words, B 1. We will now show that S 1. First note that since h is analytic on (s) >≡ 2σ r and since S is a common≡ inner A,r ℜ 0 − divisor of all functions in Er, it follows that S can be continued analytically through a(s σ0) the axis σ0 + iR. In particular, this forces S to be of the form S(s) = e− − , for some a 0 (see for instance [Nik02, Part A, Chap. 4 & 6]). Now let hA,r be a function in ≥E and write h = Sh, with h H2(Π ). r A,r ∈ σ0 Lemma 6.3. Let h H2(Π ). Then ∈ σ0 log h(σ) lim sup | | 0. σ + σ ≤ → ∞ Proof of lemma 6.3. Using (5.2) and (5.4), we get h h(σ) h k = k k2 , | |≤k k2k σk2 2√π(σ σ )1/2 − 0 for every σ > σ0. Hence log h(σ) O(1) 1 log(σ σ ) | | − 0 , σ ≤ σ − 2 σ which gives the result letting σ + . → ∞ On the one hand, according to the previous lemma, we have log h (x) log S(x) (6.4) lim sup | A,r | lim sup | | = a 0. x + x ≤ x + x − ≤ → ∞ → ∞ And on the other hand, writing ℓ(α) x+r σ0 h (x)= L(x + r σ )ˆϕ(x + r σ ) c α − , A,r − 0 − 0 j j j=1 X we can assume, by Lemma 4.2, that 0 < α1 < < αℓ(α) = 1 and that cℓ(α) = 0. Then we have ··· 6 h (x) a ϕˆ(x + r σ )c , x + . | A,r |∼| 1|| − 0 ℓ(α)| → ∞ Hence we can find x0 > 1 such that a h (x) | 1| ϕˆ(x + r σ )c , (x > x ), | A,r | ≥ 2 | − 0 ℓ(α)| 0 which gives log h (x) log 2 + log a + log c + log ϕˆ(x + r σ ) , (x > x ). | A,r | ≥ − | 1| | ℓ(α)| | − 0 | 0 ZERO FREE REGIONS FOR DIRICHLET SERIES 23 Therefore we obtain that log h (x) log ϕˆ(x + r σ ) lim sup | A,r | lim sup | − 0 | =0. x + x ≥ x + x → ∞ → ∞ 2 Using (6.4), we conclude that a = 0. Hence S 1, which gives Er = H (Πσ0 ), that 2 dt ≡ is Kr = L ((0, 1), t1−2σ0 ). ∗ Remark 6.4. In fact, with the hypothesis on ϕ, one can easily show that we always log ϕˆ(x+r σ0) | − | have lim supx + x 0. Note that there exists some ϕ for which the lim sup is negative.→ ∞ ≤ 7. Some examples and applications 7.1. The Riemann zeta function. Let 1 ζ(s)= ( (s) > 1), ns ℜ n 1 X≥ be the Riemann zeta function. Then it is well known that ζ can be meromorphically continued in the whole plane C, with a unique pole of order 1 at the point s = 1 ([Tit86]). Thus, the function ζ satisfies our hypothesis with mL = 1 (and for instance, σ = 0). Now let us consider the function ϕ defined on [0, + [ by 0 ∞ σ1 (1 t)− if 0 t< 1 ϕ(t)= − ≤ , 0 if t 1 ( ≥ where σ1 < 1/2 is fixed. Γ(s)Γ(1 σ1) Then an elementary computation shows thatϕ ˆ(s)= − , (s) > 0. Hence, Γ(s+1 σ1) − ℜ Γ(1 σ ) (7.1) ϕˆ(s) − 1 as t = (s) . | | ∼ t 1 σ1 | | |ℑ |→∞ | | − We now illustrate1 the use of Theorem 2.1 in order to determine the values of r such that du ψ L2 (1, ), . ∈ ∞ u1+2r An obvious computation shows that u 1 ψ(u)= , u> 0. 1 σ − n σ1 1 n Let µ = µ(1/2) be a convexity bound for ζ(1/2+ it) ([Ten08]), so that we have for all ε> 0, 1 (1 2µ)r+ε O t 2 − − if 0 r 1/2, ζ(r + it) = ε | | ≤ ≤ 2µ(1 r)+ε | | (Oε t − if 1/2 r < 1. | | ≤ We know that µ< 1/4. We get from (7.1) O t 1 (2 4µ)r+2σ1+ε if 0 r 1/2, ζ(r + it)ˆϕ(r + it) 2 = ε − − − | | 2+2σ1+4µ(1 r)+ε ≤ ≤ | | O t − − if 1/2 r < 1. ( ε | | ≤ So a direct application of Theorem 2.1 gives: 2 du Proposition 7.1. With the notation above, then ψ L (1, ), u1+2r if one the following holds: ∈ ∞ max(0, σ /(1 2µ)) 2 du Theorem 7.2. Let r =1. Then ψ L ((1, ), u1+2r ) if and only if r > max(0, σ1). Moreover, if the condition6 holds we∈ have ∞ 2 ψ L2((1, ), du ) C(σ1)ζ(1+2(r σ1)), k k ∞ u1+2r ≤ − 23−2σ1 22−σ1 1 where C(σ )= 2 + 2 + . 1 (3 2σ1)(1 σ1) (1 σ1) 1 2σ1 − − − − Proof. In the case of the zeta function and for the previous choice of ϕ, we have u n σ1 ψ(u)= 1 − (u> 0). 1 σ − − u 1 n Hence 2 σ1 2 ∞ u n − du ψ 2 = 1 2r+1 k k 1 1 σ1 − − u u Z − n k 1 1 σ1 1 σ1 − 1 σ1 1 (u k) − u − σ1 (u k + 1) − (7.2) + − (u n)− − . uσ1 1 σ ≤ 1 σ − − ≤ 1 σ 1 1 n=1 1 − − X − In particular, one has 1 σ1 k u − σ1 1 (u n)− = + Oσ1 (1) u (k,k + 1], k 1 1 σ − − −(u k)σ1 ∈ ≥ − 1 n=1 − X so for each k, the integral 2 k+1 1 σ1 k u − σ1 (u n)− du k 1 σ1 − − Z − n=1 X converges if and only if 2σ1 < 1. Moreover, if 2σ1 < 1 these integrals are uniformly bounded with respect to k because using (7.2), we have 2 k+1 1 σ1 k k+1 1 σ1 2 u − σ1 (u k + 1) − 1 (u n)− du − + du 1 σ − − ≤ 1 σ (u k)σ1 Zk 1 n=1 Zk 1 − X − − 1 1 σ1 2 (u + 1) − 1 + du = C(σ1 ) ≤ 1 σ uσ1 Z0 1 − e 26 CHRISTOPHE DELAUNAY, EMMANUEL FRICAIN, ELIE MOSAKI, AND OLIVIER ROBERT From this, we deduce that for r > σ1 k 2 ∞ k+1 1 σ1 2 u − σ1 du ψ = (u n)− k k2 1 σ − − u1+2r 2σ1 k 1 n=1 − Xk=1 Z − X k 2 ∞ k+1 1 σ1 1 u − σ1 ( u n)− du 1+2r 2σ1 ≤ k − k 1 σ1 − − k=1 Z − n=1 X X ∞ 1 C(σ ) < + . 1 1+2r 2σ1 ≤ k − ∞ Xk=1 e As far as C(σ1) is concerned, we have the following bound 1 2 2σ1 1 1 σ1 1 (u + 1) − 2 (1 + u) − du Ce(σ1)= du + du + (1 σ )2 1 σ uσ1 u2σ1 Z0 − 1 − 1 Z0 Z0 1 (u + 1)2 2σ1 22 σ1 1 du 1 du e − du + − + ≤ (1 σ )2 1 σ uσ1 u2σ1 Z0 − 1 − 1 Z0 Z0 23 2σ1 22 σ1 1 − + − + = C(σ ), ≤ (3 2σ )(1 σ )2 (1 σ )2 1 2σ 1 − 1 − 1 − 1 − 1 which proves the first part of the theorem. On the other hand, suppose that σ > 0. (If σ 0 there is nothing to prove 1 1 ≤ since by hypothesis r > σ0 and σ0 = 0 in our case). It sufficies to find U(σ1) > 0 such that 2 k+1 1 σ1 k u − σ1 (u n)− du U(σ1) k 1 σ1 − − ≥ Z − n=1 X uniformely in k. In that case, we shall have k 2 ∞ k+1 1 σ1 2 u − σ1 du ψ = (u n)− k k2 1 σ − − u1+2r 2σ1 k 1 n=1 − Xk=1 Z − X k 2 ∞ k+1 1 σ1 1 u − σ1 (u n)− du 1+2r 2σ1 ≥ (k + 1) − k 1 σ1 − − k=1 Z − n=1 X X ∞ 1 U(σ ) 1 1+2r 2σ1 ≥ (k + 1) − Xk=1 hence r > σ1. ZERO FREE REGIONS FOR DIRICHLET SERIES 27 It now remains to prove 2 k+1 1 σ1 k u − σ1 (u n)− du U(σ1) k 1 σ1 − − ≥ Z − n=1 X uniformely in k. From (7.2), we see that we have in fact 1 σ1 k u − σ1 1 κ(u) (u n)− = + u (k,k+1], k 1, σ1 < 1/2 , 1 σ − − −(u k)σ1 1 σ ∈ ≥ 1 n=1 1 − X − − 1 σ1 where 0 κ(u) (u k + 1) − for all u (k,k + 1]. We deduce≤ that≤ for− any 0 <δ< 1 ∈ 2 2 k+1 1 σ1 k k+δ 1 σ1 k u − σ1 u − σ1 (u n)− du (u n)− du 1 σ − − ≥ 1 σ − − k 1 n=1 k 1 n=1 Z − X Z − X k+δ 1 σ1 1 (u k + 1) − 2 − du ≥ (u k)2σ1 − (1 σ )(u k)σ1 Zk − − 1 − k+δ 1 (1 + δ)1 σ1 2 − du ≥ (u k)2σ1 − (1 σ )(u k)σ1 Zk − − 1 − 1 2σ1 1 σ1 1 σ1 δ − 2(1 + δ) − δ − ≥ 1 2σ − (1 σ )2 − 1 − 1 1 σ1 σ1 1 2σ1 1 2(1 + δ) − δ δ − . ≥ 1 2σ − (1 σ )2 − 1 − 1 We have the expected result by choosing 1 σ1 σ1 1 2σ1 1 2(1 + δ) − δ U(σ ) := δ − 1 1 2σ − (1 σ )2 − 1 − 1 with δ > 0 small enough so that 1 σ1 σ1 2(1 + δ) − δ 1 2 < . (1 σ1) 1 2σ1 − − 7.1.1. Zero-free discs for ζ. Let r > max(0, σ1), r = 1. Then according to The- 2 du 6 orem 7.2, the function ψ L (1, ), u1+2r . Let A = (α,c) be a 1-admissible sequence, which means that∈ ∞ ℓ(α) cjαj =0, j=1 X 28 CHRISTOPHE DELAUNAY, EMMANUEL FRICAIN, ELIE MOSAKI, AND OLIVIER ROBERT with 0 <αj 1, cj C. Take λ C, (λ) > 0. It follows from Theorem 2.2 that the Riemann≤ zeta function∈ s ζ∈(s) doesℜ not vanish in the disc 7−→ µ λ r + µ C : − < 1 2 (λ)d2(λ) , ∈ µ + λ¯ − ℜ r ¯ p where d (λ) = dist(tλχ ,K ), with K = span(f : A a 1-admissible sequence). r (0,1) r r A,r If we take only admissible sequences of length 2, we can easily prove the following. Proposition 7.3. We have α α K = span tr(c ψ 1 + c ψ 2 ), c α + c α =0, 0 <α 1 = r 1 t 2 t 1 1 2 2 j ≤ α 1 span tr(ψ αψ ), 0 <α 1 . t − t ≤ In particular, taking ϕ = χ(0,1) (that is σ1 = 0), we recover the subspace K˜r (see the Introduction) and the result of Nikolski [Nik95, Theorem 0.1]. We will now try to give more explicit zero free discs for the Riemann zeta function. Applying Corollary 2.3 gives that s ζ(s) does not vanish in the disc 7→ ℓ(α) λ+r µ λ ζ(λ + r)ˆϕ(λ + r) j=1 cjαj r + µ C: − < 2 (λ) . ∈ µ + λ¯ ℜ f 2 A,r L∗((0,P1),dt/t) p k k But the estimate (4.4) gives ℓ(α) r 1 cjα + ψ 2 du if 0 Proposition 7.4. Let max(0, σ1) ZERO FREE REGIONS FOR DIRICHLET SERIES 29 1 σ1 Since ϕˆ(λ + r) = O(1/ (λ) − ) we obtain larger disc than in ([Nik95]) whenever | | |ℑ | (λ) is large enough and σ1 > 0. We emphasize on the fact that our zero-free discs ℑare explicit. For example, if we take α =1/4 in (7.3), we obtain the following zero-free region for zeta which can be improved but has the merit to be quite explicit: Corollary 7.5. Let max(0, σ1) F (λ, r, σ1)= 1 λ+r 1 2 (λ) 4 4 Γ(λ + r)Γ(1 σ1) ℜ − − ζ(λ + r) . 1 r 1 1 | | + Cp(σ1)ζ(1+2( r σ1)) + Γ(λ + r +1 σ1) 4 4 (1 σ1)√2 2r − − − − p Recall that the discs in Corollary 7.5 are euclidean discs of center (x, y ) and radius R where: 1+ F (λ, r, σ )2 x = r + (λ) 1 , y = (λ) ℜ 1 F (λ, r, σ )2 ℑ − 1 2 (λ)F (λ, r, σ ) R = ℜ 1 . 1 F (λ, r, σ )2 − 1 Taking, for example, λ =0.01+50i, r =0.49 and σ1 =0.4 then a simple evaluation 1 of F (λ, r, σ1) implies that ζ has no zero in the disc of center 2 + 50i and radius 6 1 1 3.75 10− . Note that ζ has a zero at s = +49.773 ...i and at s = +52.970 ...i. × 2 2 7.2. Dirichlet L-functions. Let χ be a non trivial Dirichlet character of conductor s cond(χ) = q and L(χ, s) = n 1 χ(n)n− be the (degree 1) associated Dirichlet function. This function has an analytic≥ continuation to the whole complex plane. P For simplicity, we take σ1 = σ0 = 0 and 1 if 0 t< 1, ϕ(t)= ≤ 0 if t 1. ( ≥ 2 du So we have ψ(u)= n 0. The admissibility− condition for the sequence A is empty and∞ we may take P ℓ r fA,r(t)= t cj χ(n) j=1 X n<αXj /t 30 CHRISTOPHE DELAUNAY, EMMANUEL FRICAIN, ELIE MOSAKI, AND OLIVIER ROBERT for all A =(α,c) where ℓ 0, α (0, 1]ℓ and c Cℓ. Let r such that 1/2 r < 1 and λ =1 r. For simplicity,≥ we∈ write d = d (λ∈) so that ≤ − r r 2 1 ℓ 2 1 r r dt dr = min t − t cj χ(n) cj ,αj 0 − t Z j=1 n<αj /t X X Furthermore, we will write dr,χ if we need to specify the dependance on χ. Remark 2 that we have trivially dr 1/(2 2r) (just take all the cj to be 0). In fact, we can be a little bit more precise.≤ − Proposition 7.6. We have d2 < 1/(2 2r). r − Proof. Let c C and 0 <α 1. A short calculation gives ∈ ≤ 2 1 2 1 r r dt d t − ct χ(n) = r ≤ − t Z0 n<α/t X 2 1 ∞ du 2 2r+1 2α (cL(χ, 1)) + c α χ(n) 1+2r . 2 2r − ℜ | | 1 u − Z n 2 2 1 2(1 r)dr (1 2(1 r)dr) σ > 1 − − −2 − − . − p dr (We can easily check that the disc (7.4) contains σ = 1 and it is known that L(χ, σ) = 0 for σ 1.) Hence, L(χ, σ) does not vanish on 6 ≥ (7.5) σ> 1 (1 r) 1 2(1 r)d2. − − − − r We expect (by the Riemann hypothesisp for Dirichlet functions) that dr = 0 which would imply that L(χ, s) does not vanish on Πr. If we had “only” 1 C2 d2 r ≤ 2 2r − 2(log q)2(1 r)3 − − ZERO FREE REGIONS FOR DIRICHLET SERIES 31 for some (absolute) constant C, it would imply (using (7.5)) that L(χ, σ) does not vanish in the real-interval σ > 1 C/ log q. That proves Theorem 2.5 and we also get immediately the following. − Theorem 7.7. If there exists an absolute constant C such that for all (real) char- acter χ there exists r [1/2, 1) with ∈ 1 C2 d2 < , r,χ 2 2r − 2(log cond(χ))2(1 r)3 − − then there is no Siegel’s zero for Dirichlet L-functions. The previous theorem can be seen as a Beurling-Nyman criterion for Siegel’s conjecture. 7.2.1. Explicit zero-free discs for L(χ, s). We now apply Corollary 2.3. Note that for the special choice of ϕ we have done (ϕ = χ(0,1)), we haveϕ ˆ(s)=1/s. Moreover, since the function ψ is bounded, say ψ B, we have k k ≤ ψ L2((1, ),du/u1+2r ) B/√2r. k k ∞ ≤ Then Corollary 2.3 asserts that s L(χ, s) does not vanish in the disc 7→ µ λ L(χ, λ + r) √2r r + µ C: − < 2 (λ)| | . ∈ µ + λ¯ ℜ λ + r B ( ) p | | Let us remark that the P´olya-Vinogradov’s theorem implies that B 2√q log q where q = cond(χ) is the conductor of χ. Note also that there are some≤ improve- ments of the P´olya-Vinogradov’s inequality for some characters (see for example [GS07]). s 7.3. The Selberg class. Let L(s) = n 1 ann− be a L-function in the Selberg ≥ class ([dR07a]). We denote by d its degree and by mL the order of its pole at s = 1.S P As for the Riemann zeta function, we take σ1 (1 t)− if 0 t< 1 ϕ(t)= − ≤ 0 if t 1 ( ≥ where σ1 < 1/2. By the Phragmen-Lindel¨of principle, we have that L(r + it) = d (1 r)+ε O (t 2 − ) for 0 In particular, for r =1/2, this gives σ1 < 1/2 d/4 which is better than σ1 < d/4 obtained in [dR07a]. Nevertheless, we should− mention that for d< 4, we can in− fact take σ1 = 0 (see [dR07b]). References [BD03] L. B´aez-Duarte. A strengthening of the Nyman-Beurling criterion for the Riemann hypothesis. Atti Accad. Naz. Lincei Cl. Sci. Fis. Mat. Natur. Rend. Lincei (9) Mat. Appl., 14(1):5–11, 2003. [BDBLS00] L. B´aez-Duarte, M. Balazard, B. Landreau, and E. Saias. Notes sur la fonction ζ de Riemann. III. Adv. Math., 149(1):130–144, 2000. [Beu55] A. Beurling. A closure problem related to the riemann zeta-function. Proc. Nat. Acad. USA, 41(5):312–314, 1955. 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Christophe Delaunay,Universite´ de Lyon; Universite´ Lyon 1; Institut Camille Jordan CNRS UMR 5208; 43, boulevard du 11 Novembre 1918, F-69622 Villeurbanne E-mail address: [email protected] Emmanuel Fricain, Universite´ de Lyon; Universite´ Lyon 1; Institut Camille Jor- dan CNRS UMR 5208; 43, boulevard du 11 Novembre 1918, F-69622 Villeurbanne E-mail address: [email protected] Elie Mosaki,Universite´ de Lyon; Universite´ Lyon 1; Institut Camille Jordan CNRS UMR 5208; 43, boulevard du 11 Novembre 1918, F-69622 Villeurbanne E-mail address: [email protected] Olivier Robert Universite´ de Lyon, F-42023, Saint-Etienne, France; Universite´ de Saint-Etienne, F-42000, Saint-Etienne, France; Laboratoire de mathematiques´ (LAMUSE, EA 3989) 23, rue du Dr P. Michelon F-42000, Saint-Etienne France; E-mail address: [email protected]