aa r X i v : . [ m a t h . N T ] M a y On the Landau-Siegel Zeros Conjecture
Yitang ZhangTable of Content1. Introduction2. The Set Ψ ∗
3. Zeros of L ( s, ψ ) L ( s, ψχ ) in Ω4. The Functions K ± ( s, ψ )5. The Linear Functional Φ( f ; ρ, ψ )6. Solutions to Boundary Value Problems7. Approximation to ϕ
8. The Fundamental Inequality: Preliminary9. Upper Bounds for Θ( k ) and Θ( r )10. Asymptotic Expression for Θ( ϕ )11. The Fundamental Inequality: Completion12. Estimation of E
13. Proof of Theorem 1References
1. Introduction
The main result of this paper is the following
Theorem 1
For any real primitive character χ of modulus D we have L (1 , χ ) > c (log D ) − (log log D ) − (1 . where c > is an effectively computable constant. As a direct consequence of Theorem 1, we have
Theorem 2
Let χ ( mod D ) be as in Theorem 1. Then we have L ( σ, χ ) = 0 (1 . or σ > − c (log D ) − (log log D ) − where c > is an effectively computable constant. With extra effort we can replace the factor (log log D ) − in the above results by one;even the powers of (log D ) − might be slightly improved. However, the present methodfails to achieve the conjectural lower bound L (1 , χ ) ≫ (log D ) − . The major obstacle here is the fact that the trivial bound L ′ ( s, χ ) ≪ (log D ) ( s ∼ , which is applied to derive Theorem 2 from Theorem 1 and to prove Lemma 2.1, can not beessentially sharpened by present methods (see H. Iwaniec and E. Kowalski [IW, Chapter22], for example).Although Theorem 2 does not completely eliminate the Landau-Siegel zeros in theiroriginal definition, our results will be sufficient for various applications in both of theanalytic number theory and algebraic number theory.Throughout, we denote by χ a real primitive character of modulus D with D greaterthan a sufficiently large computable number. Our goal is to derive a contradiction fromthe following fundamental assumption L (1 , χ ) < (log D ) − (log log D ) − . ( A L E ( s ) L E ( s, χ ) and its derivatives at the central point where E is an elliptic curve, L E ( s ) is the L -function associated to E and L E ( s, χ ) is the twisted L -function of E and χ ; the second paper discusses the relationship between the non-vanishing of a family ofautomorphic L -functions at the central point and the Landau-Siegel zeros; the third paperuses evaluation of discrete means over the zeros of the Riemann zeta function to deriveresults on the simple zeros of ζ ( s ).Motivated by their ideas and methods, in this paper we investigate the relationshipbetween the lower bound for L (1 , χ ) and the behaviors of a family of Dirichlet L - functionsin a certain domain Ω. We use Ψ to denote a certain large set of Dirichlet primitivecharacters of which the sizes of the moduli are ∼ Q where Q = exp[(log D ) ] . ψχ to be primitive if ψ ∈ Ψ. On the otherhand, the domain Ω is chosen to be relatively small and it is not closed to the real axis.The methods used in this paper are classical, which mainly include various asymptoticformulae, the large sieve techniques and function approximations.The proof of Theorem 1 is based on three observations to be described below. Some ofthem are closely related to present researches on the Riemann zeta function, in particular,the vertical distribution of the zeros of ζ ( s ). Relations and applications to other problemswill also be discussed. In the following descriptions we henceforth assume that (A1) holds.Our first observation is a ”mollifier interpretation” of the assumption (A1). That is,the function F ( s, ψ ) − L ( s, ψχ )with ψ ∈ Ψ and s ∈ Ω, where F ( s, ψ ) is a (relatively short) partial sum of L ( s, ψ ) L ( s, ψχ ),behaves like a good mollifier of L ( s, ψ ). Here the role played by the factor F ( s, ψ ) − is,roughly speaking, to transform the real character χ into the M¨obius function µ . In fact,our proof is to show, by contradiction, that such a mollifier is too good to exist.In Section 2 and Sectino 3 we prove that the following hold for almost all of thecharacters ψ inΨ:(i). All the zeros of L ( s, ψ ) L ( s, ψχ ) in Ω lie on the critical line.(ii). All the zeros of L ( s, ψ ) L ( s, ψχ ) in Ω are simple.(iii). With a small error, the gap between any pair of consecutive zeros of L ( s, ψ ) L ( s, ψχ )in Ω is asymptotically equal to the average gap π (log Q ) − .The proof of the above result is divided into two steps. Firstly, we formally define asubset Ψ ∗ of Ψ by three inequalities (I1), (I2) and (I3) and show that the assumption(A1) implies that almost all of the characters ψ in Ψ also belong to Ψ ∗ ; secondly, we showthat (i), (ii) and (iii) hold if ψ ∈ Ψ ∗ . It should be remarked that most of the lemmasand propositons of this paper are independent of (A1). However, without assuming (A1)they are not applicable, since, in such a situation, one is unable to determine whether thesubset Ψ ∗ is empty or not.Other properties of L ( s, ψ ) L ( s, ψχ ) with ψ ∈ Ψ ∗ are also obtained, among which themost remarkable one is the relation (see Lemma 3.4) F ( ρ + s, ψ ) − L ( ρ + s, ψ ) L ( ρ + s, ψχ ) ∼ − Q − s , (1 . ρ is a zero of L ( s, ψ ) L ( s, ψχ ) in Ω and | s | ≤ R (log Q ) − where R ∼
130 log log D. ζ ( s ). It is reasonable to believe, under the assumptionof the Riemann Hypothesis, that a significant progress toward Montgomery’s pair corre-lation conjecture [M] or the conjectures on the small gaps between the zeros of ζ ( s ) (seeJ. B. Conrey, A. Ghosh, D. Goldston, S. M. Gonek and D. R. Heath-Brown [CGGGH], J.B. Conrey, A. Ghosh, S. M. Gonek [CGG1] and Y. Zhang [Z1]) which breaks the barrierdue to the limitation of present methods, could be used here to derive a contradictionfrom (iii). As we are unable to do this directly, we consider the functions K ± ( s, ψ ) = F ( s, ψ ) − L ( s ± α, ψ ) L ( s ∓ α, ψχ )with ψ ∈ Ψ ∗ , where α = (log Q ) − . In Section 4 we obtain some properties of the functions K ± ( s, ψ ), including asymptoticformulae whose proofs rely on the gap assertion (iii).Our second observation is a ”linear functional interpretation” of the asymptotic for-mula for the function K + ( ρ + s, ψ ). For f ∈ C [ − ,
1] we defineΦ( f ; s, ψ ) = X n ≤ Q / λ + ( n ) ψ ( n ) n s ϑ (cid:18) Q / n (cid:19) f (1 − α log n ) − X n ≤ Q λ − ( n ) ¯ ψ ( n ) n − s ϑ (cid:18) Q / n (cid:19) f ( α log n − , where ϑ is a certain weight function and where λ ± ( n ) denotes the n th coefficient in theDirichlet series expansion of ζ ( s ± α ) /ζ ( s ∓ α ). Write ϕ s ( x ) = Q xs . In Section 5 it isshown that if ψ ∈ Ψ ∗ , ρ is a zero of L ( s, ψ ) L ( s, ψχ ) satisfying | ρ − s | < | θ | < R (log Q ) − , and K + ( ρ + θ, ψ ) = 0 , (1 . ϕ θ ; ρ, ψ ) ∼ . (1 . f ∈ C [ − ,
1] can be well approximated by a linearcombination of the ϕ θ with θ satisfying (1.4), then the linear functional Φ( f ; ρ, ψ ) willhave a small average order. This naturally leads the the problem of approximating certainfunctions by linear combinations of the ϕ θ with θ satisfying (1.4). The idea of relatingasymptotic formulae for products of L -functions to function approximations is not onlyused in this paper, but also has some other applications. For example, when applied to4he Riemann zeta function and combined with an asymptotic formula for the product of ζ ( s ) and a mollifier, it leads to some interesting results on the zeros of ζ ( s ) that will begiven in our separate paper [Z2].Our last observation, which is the most technical part of this paper, is that if h is alinear combination of the ϕ θ with θ satisfying (1.4), then one may have an asymptoticexpression for h ( x ) with 1 < | x | < h on [ − ,
1] only. In Section 7 we considerthe approximation to ϕ on [ − ,
1] by a linear combination of the exponential functions ϕ θ with θ satisfying (1.4) and a function k which is orthogonal to the solution to a lineardifferential equation of Sturmian type with boundary conditions given in Section 6 (theinner product is introduced in Section 5). Such a property of k is crutial in our proof.Our method actually constructs a linear combination h of the ϕ θ with θ satisfying (1.4)such that, on the interval [ − , ϕ ∼ h + k. Write ˆ h ( x ) = h ( x + ε )with a small positive number ε depending on D . Starting from Section 8, we use twodifferent methods to calculate the quantity Φ(ˆ h ; ρ, ψ ) with ψ ∈ Ψ ∗ and ρ ∈ S ( ψ ) where S ( ψ ) is a set of zeros of L ( s, ψ ) L ( s, ψχ ) in Ω. On one hand, it follows from (1.5) thatΦ(ˆ h ; ρ, ψ ) ∼ . (1 . h ; ρ, ψ ) which involves ϕ . Combining this result with (1.6) we shall, after a lengthyprocess of calculation in Section 8, Section 9 and Section 10 , arrive at a relation of theform R − / ≪ E ( ρ, ψ ) (1 . E ( ρ, ψ ) is complicated. However, it is not difficult to estimate the discrete mean X ψ ∈ Ψ ∗ X ρ ∈ S ( ψ ) E ( ρ, ψ ) , (1 . R − / X ψ ∈ Ψ ∗ X ρ ∈ S ( ψ ) . The Set Φ ∗ Write L = log D, Q = exp[ L ] . (2 . ν ( n ) and υ ( n ) be the arithmetic functions defined by ∞ X ν ( n ) n s = ζ ( s ) L ( s, χ ) , ∞ X υ ( n ) n s = ζ ( s ) − L ( s, χ ) − . (2 . Lemma 2.1
Assume that (A1) holds. Then we have X D ≤ n ≤ Q ν ( n ) n ≪ L − (log L ) − (2 . and X D ≤ n ≤ D ν ( n ) τ ( n ) n ≪ L − (log L ) − . (2 . Proof . We prove (2.4) only, as the proof of (2.3) is analogous. Write D ( s ) = ζ ( s ) − L ( s, χ ) − ∞ X ν ( n ) τ ( n ) n s , so ∞ X ν ( n ) τ ( n ) e − n/x = 12 πi Z (2) D ( s ) ζ ( s ) L ( s, χ ) Γ( s ) x s ds. (2 . χ ( p ) = ± , D p ( s ) in the Eulerproduct representation D ( s ) = Y p D p ( s )satisfies D p ( s ) = ( O ( p − σ ) , if ( p, D ) = 1 , (1 − p − s )(1 + O ( p − σ )) , if p | D = 1 , for σ ≥ σ >
0, the implied constant depending on σ only. Hence, on the half plane σ ≥ / D ( s ) is analytic and it satisfies D ( s ) ≪ Y p | D | − p − s | . (2 . D ≤ x ≤ D . Moving the line of integration to σ = 51 /
100 and applying(2.6) and other well-known estimates, we see that, with an error O ( x / D / ), theright-hand side of (2.5) is asymptotically equal to the residue of the integrand at s = 1,that is equal to 12 πi Z | s − | = η D ( s ) ζ ( s ) L ( s, χ ) Γ( s ) x s ds for any small η > η = L − (log L ) − and assume | s − | = η . By (2.6) we have D ( s ) ≪ L ′ ( w, χ ) ≪ L for | w − | ≤ η we get L ( s, χ ) ≪ η L , so ζ ( s ) L ( s, χ ) ≪ L . (This is a crucial application of the assumption (A1)). Combining all the above estimateswith (2.5) we conclude that X n ≤ x ν ( n ) τ ( n ) ≪ xη L + x / D / ≪ x L − (log L ) − . From this (2.4) follows by partial summation. Let Q denote the set of the integers q ssatisfying Q < q < Q, (6 , q ) = 1 , µ ( q ) = 1 . For any q ∈ Q let Ψ q denote the set of all the primitive characters ψ ( mod q ) such that ψχ is a primitive character ( mod [ q, D ]). LetΨ = [ q ∈Q Ψ q . (2 . X ψ ∈ Ψ ≫ Q . (2 . Lemma 2.2
For any complex numbers a n we have X ψ ∈ Ψ (cid:12)(cid:12)(cid:12)(cid:12) X n ≤ Q a n ψ ( n ) (cid:12)(cid:12)(cid:12)(cid:12) ≪ Q X n ≤ Q | a n | . Let Ω denote the domain |ℜ ( s − s ) | < , |ℑ ( s − s ) | < L s = 12 + iD. (2 . L ( s, ψ ) L ( s, ψχ ) with ψ ∈ Ψ in Ω. Write α = L − (= (log Q ) − ) . (2 . and Ω denote the domainsΩ = (cid:8) s : − α / < ℜ ( s − s ) < , |ℑ ( s − s ) | < L (cid:9) , Ω = (cid:8) s : − (10) − α log L < ℜ ( s − s ) < / , |ℑ ( s − s ) | < L (cid:9) . Lemma 2.3
For any sequence of complex numbers ( a n ) write M (( a n ) , ψ ) = sup s ∈ Ω (cid:26)(cid:12)(cid:12)(cid:12)(cid:12) X n ≤ D a n ψ ( n ) n s (cid:12)(cid:12)(cid:12)(cid:12) (cid:27) , M (( a n ) , ψ ) = sup s ∈ Ω (cid:26)(cid:12)(cid:12)(cid:12)(cid:12) X n ≤ Q a n ψ ( n ) n s (cid:12)(cid:12)(cid:12)(cid:12) (cid:27) . Then we have X ψ ∈ Ψ M (( a n ) , ψ ) ≪ L Q X n ≤ D | a n | n (2 . and X ψ ∈ Ψ M (( a n ) , ψ ) ≪ L / Q X n ≤ Q | a n | n . (2 . Proof
Let R be the rectangle with vertices at s − α / ± i (2 + 20 L ) , s + 2 ± i (2 + 20 L ) . Note that the domain Ω is enclosed in R . By the Cauchy integral formula we get forany s ∈ Ω (cid:12)(cid:12)(cid:12)(cid:12) X n ≤ D a n ψ ( n ) n s (cid:12)(cid:12)(cid:12)(cid:12) = 12 π (cid:12)(cid:12)(cid:12)(cid:12) Z R (cid:18) X n ≤ D a n ψ ( n ) n w (cid:19) dww − s (cid:12)(cid:12)(cid:12)(cid:12) . n − u ≪ n − if n ≤ D and u ≥ / − α / . Hence, by Lemma 2.2 we get for w ∈ R X ψ ∈ Ψ (cid:12)(cid:12)(cid:12)(cid:12) X n ≤ D a n ψ ( n ) n w (cid:12)(cid:12)(cid:12)(cid:12) ≪ Q X n ≤ D | a n | n . On the other hand, note that the total length of R is O ( L ) and | w − s | − < α − / = L if s ∈ Ω and w ∈ R . Gathering these results we obtain (2.11).To Prove (2.12) we choose R as the rectangle with vertices at s − α (cid:0) − log L (cid:1) ± i (2 + 10 L ) , s + 1 ± i (2 + 10 L ) . Note that n − u ≪ L / n − if n ≤ Q and u ≥ / − α (cid:0) − log L (cid:1) , and | w − s | − <α − = L if s ∈ Ω and w ∈ R . The rest of the proof is analogous to that of (2.11). Write F ( s, ψ ) = X n ≤ D ν ( n ) ψ ( n ) n s , G ( s, ψ ) = X n ≤ D υ ( n ) ψ ( n ) n s . By Lemma 2.3, the number of the characters ψ in Ψ such thatsup s ∈ Ω (cid:8)(cid:12)(cid:12) F ( s, ψ ) (cid:12)(cid:12)(cid:9) ≥ L log L , is ≪ L − (log L ) − Q X n ≤ D ν ( n ) n ≪ (log L ) − Q . Clearly the same estimate holds with G ( s, ψ ) in place of F ( s, ψ ). Thus we have Lemma 2.4
The following inequality sup s ∈ Ω (cid:8) | F ( s, ψ ) | + | G ( s, ψ ) | (cid:9) < L log L ( I holds for all but at most O (cid:0) (log L ) − Q (cid:1) characters ψ in Ψ.Now we turn to the function F ( s, ψ ) G ( s, ψ ) − F ( s, ψ ) G ( s, ψ ) − X D Lemma 2.5 Assume that (A1) holds. Then the following inequality sup s ∈ Ω (cid:8) | F ( s, ψ ) G ( s, ψ ) − | (cid:9) < 12 ( I holds for all but at most O ((log L ) − Q ) characters ψ in Ψ.For any primitive character ψ write∆( s, ψ ) = L ( s, ψ ) L (1 − s, ¯ ψ ) . (2 . s, ψ ) = C ( ψ ) (cid:18) kπ (cid:19) / − s Γ (cid:18) − s + a (cid:19) Γ (cid:18) s + a (cid:19) − , where a = (cid:0) − ψ ( − (cid:1) / C ( ψ ) is a complex number depending on ψ andsatisfying | C ( ψ ) | = 1.Suppose ψ ( mod q ) is in Ψ, so ψχ is a primitive character ( mod [ q, D ]). Write∆ ( s, ψ ) = ∆( s, ψ )∆( s, ψχ ) , so L ( s, ψ ) L ( s, ψχ ) = ∆ ( s, ψ ) L (1 − s, ¯ ψ ) L (1 − s, ¯ ψχ ) . (2 . ( s, ψ ) ( ψ ∈ Ψ) satisfies the following relations.∆ ( s, ψ )∆ (1 − s, ¯ ψ ) = 1 , (2 . (cid:12)(cid:12)(cid:12)(cid:12) ∆ (cid:18) 12 + it (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = 1 , (2 . ′ ∆ ( s, ψ ) = − L + O ( L ) , ( | σ | ≤ , | t − D | < L ) , (2 . | ∆ ( s, ψ ) | ≤ Q − σ ( σ ≥ / , t > , (2 . | ∆ ( s, ψ ) | ≤ Q − σ D − σ ( − ≤ σ ≤ / , | t − D | ≤ L ) . (2 . F ( s, ψ ) = F ( s, ψ ) + ∆ ( s, ψ ) F (1 − s, ¯ ψ ) . (2 . L ( s, ψ ) L ( s, ψχ ) by F ( s, ψ ) in the domainΩ . To this end we introduce ς ( x ) = 12 πi Z (1) x s exp( s / dss ( x > . (2 . ǫ → + ς ( ǫ ) = 0 and x s − ǫ s s = Z log x log ǫ e sy dy if 0 < ǫ ≤ x , it follows, by changing the order of integration, that ς ( x ) = lim ǫ → + Z log x log ǫ (cid:26) πi Z (1) exp( sy + s / ds (cid:27) dy (2 . √ π Z log x −∞ exp[ − y ] dy. Hence ς ( x ) = 1 + O (exp[ − (log x ) ]) , if x > , (2 . ς ( x ) = O (exp[ − (log x ) ]) , if x ≤ . Let J ( c ) denote the line segment path from c − i L to c + 10 i L for c ∈ R . Assumethat ψ ∈ Ψ and s ∈ Ω . By the residue theorem we get L ( s, ψ ) L ( s, ψχ ) = I ( s, ψ ) − I ( s, ψ ) + O ( Q − ) , (2 . I ( s, ψ ) = 12 πi Z J (1) L ( s + w, ψ ) L ( s + w, ψχ ) Q w exp( w / dww , ( s, ψ ) = 12 πi Z J ( − − σ ) L ( s + w, ψ ) L ( s + w, ψχ ) Q w exp( w / dww , with Q = Q D − . Here the error term arises from the integrals on the horizontal segments by routine esti-mation. To calculate the integral I ( s, ψ ) we replace the path J (1) by the line u = 1 witha negligible error, and then apply (2.23) getting I ( s, ψ ) = F ( s, ψ ) + E ( s, ψ ) + O ( Q − ) , (2 . E ( s, ψ ) = X D Assume that (A1) holds. Then the following inequality sup s ∈ Ω (cid:8) | L ( s, ψ ) L ( s, ψχ ) − F ( s, ψ ) | (cid:9) < L − / ( I holds for all but at most O ((log L ) − Q ) characters ψ in Ψ.Let Ψ ∗ be the subset of Ψ such that ψ ∈ Ψ ∗ if and only if the inequalities (I1), (I2)and (I3) simultaneously hold. Let Ψ ∗∗ be the complement set of Ψ ∗ in Ψ.From Lemma 2.4, Lemma 2.5 and Lemma 2.6 we dedrive the following Proposition 2.7 Assume that (A1) holds. Then we have X ψ ∈ Ψ ∗∗ ≪ (log L ) − Q . We stress that the set Ψ ∗ is formally defined; by virtue of (2.8), Proposition 2.7 showsthat almost all of the characters ψ in Ψ belong to Ψ ∗ under the assumption (A1) only;otherwise one is unable to determine whether Ψ ∗ is empty or not.We conclude this section by proving several simple results with ψ ∈ Ψ ∗ which will findapplication in the next two sections.A direct consequence of the inequalities (I1) and (I2) is | F ( s, ψ ) | − < L log L ( ψ ∈ Ψ ∗ , s ∈ Ω ) . (2 . Lemma 2.8. Suppose ψ ∈ Ψ ∗ and s ∈ Ω . Then we have F ′ F ( s, ψ ) ≪ L log L . roof. Note that the closed disc | w − s | ≤ α / = L − is contained in Ω , so, by (I1) and(2.34), 12 L − (log L ) − < | F ( w, ψ ) | < L log L , for | w − s | ≤ α / . Thus the lemma follows by Lemma α of E. C. Titchmarsh [T, Section3.9]. Corollary 2.9. Suppose ψ ∈ Ψ ∗ , s, s ′ ∈ Ω and | s − s ′ | ≤ ( L log L ) − . Then we have F ( s, ψ ) F ( s ′ , ψ ) = 1 + O (cid:0) | s − s ′ |L log L (cid:1) . Proof. This follows from the relation F ( s, ψ ) F ( s ′ , ψ ) = exp (cid:26) Z ss ′ F ′ F ( w, ψ ) dw (cid:27) and Lemma 2.8. For ψ ∈ Ψ ∗ define H ( s, ψ ) = ∆ ( s, ψ ) F (1 − s, ¯ ψ ) F ( s, ψ ) . (2 . |H (1 / it, ψ ) | = 1 . (2 . / ≤ σ < / α / and | t − D | < L , then 1 − ¯ s ∈ Ω , so | F (1 − s, ¯ ψ ) | = | F (1 − ¯ s, ψ ) | 6 = 0 , by (2.34), and H ′ H ( s, ψ ) = ∆ ′ ∆ ( s, ψ ) − F ′ F ( s, ψ ) − F ′ F (1 − s, ¯ ψ ) . (2 . Lemma 2.10. Suppose ψ ∈ Ψ ∗ and | t − D | < L . Then we have log |H ( s, ψ ) | = − L (cid:0) O ( L − log L ) (cid:1) for 1 / ≤ σ < / α / , (2 . and H ( s, ψ ) ≪ L Q − σ D σ for σ ≥ / α / . (2 . Proof. Assume 1 / ≤ σ < / α / . By (2.36) we havelog |H ( s, ψ ) | = Z σ / ℜ (cid:26) H ′ H ( u + it, ψ ) (cid:27) du. (cid:12)(cid:12)(cid:12)(cid:12) F ′ F (1 − s, ¯ ψ ) (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) F ′ F (1 − ¯ s, ψ ) (cid:12)(cid:12)(cid:12)(cid:12) . The formula (2.39) follows from (2.18), (2.34) (which trivially holds for σ ≥ / 2) and atrivial bound for F (1 − s, ¯ ψ ). 3. Zeros of L ( s, ψ ) L ( s, ψχ ) in Ω The main result of this section is the following Proposition 3.1 Suppose ψ ∈ Ψ ∗ . Then the following hold.(i). All the zeros of L ( s, ψ ) L ( s, ψχ ) in Ω lie on the critical line.(ii). All the zeros of L ( s, ψ ) L ( s, ψχ ) in Ω are simple.(iii). If { ρ , ρ } is a pair of consecutive zeros of L ( s, ψ ) L ( s, ψχ ) in Ω , then | ρ − ρ | = πα + O (cid:0) L − (log L ) (cid:1) . The proof of Proposition 3.1 is obtained from Lemma 3.2, Lemma 3.3 and Lemma 3.5below.We henceforth assume that ψ ∈ Ψ ∗ in this section. Instead of dealing with the function L ( s, ψ ) L ( s, ψχ ) directly, we consider the function G ( s, ψ ) = L ( s, ψ ) L ( s, ψχ ) F ( s, ψ ) (3 . L ( s, ψ ) L ( s, ψχ ) in Ω by (2.34). The advantage is that thefunction G ( s, ψ ) is directly related to H ( s, ψ ) by (I3).Note that Ω ⊂ Ω and F ( s, ψ ) F ( s, ψ ) = 1 + H ( s, ψ ) . It follows from (I3) and (2.34) that G ( s, ψ ) = 1 + H ( s, ψ ) + O ( L − / log L ) ( s ∈ Ω ) . (3 . / − α < σ < | t − D | < L , then the closed disc | w − s | ≤ α iscontained in Ω . Hence, by (3.2) and the Cauchy integral formula we get G ′ ( s, ψ ) = H ′ ( s, ψ ) + O ( L / log L ) (1 / − α < σ < , | t − D | < L ) . (3 . Lemma 3.2 Suppose ψ ∈ Ψ ∗ , / < σ < and | t − D | < L . Then we have G ( s, ψ ) = 0 . Proof. In the case 1 / α ≤ σ < |H ( s, ψ ) | ≤ e − by Lemma 2.10, so theassertion follows from (3.2). In the case 1 / < σ < / α we have, by (3.3), G (1 − ¯ s, ψ ) − G ( s, ψ ) = H (1 − ¯ s, ψ ) − H ( s, ψ ) + O (cid:0) (2 σ − L / log L (cid:1) . (3 . |H ( s, ψ ) | − = |H (1 − ¯ s, ψ ) | . (3 . |H (1 − ¯ s, ψ ) − H ( s, ψ ) | ≫ (2 σ − L . (3 . G (1 − ¯ s, ψ ) −G ( s, ψ ) = 0, which implies that G ( s, ψ ) = 0by the symmetry of the non-trivial zeros of L ( s, ψ ) L ( s, ψχ ) with respect to the criticalline. Lemma 3.3. Suppose ψ ∈ Ψ ∗ and | t − D | < L . Then we have |G ′ (1 / it, ψ ) | ≫ L . Proof. By (2.37), (2.17) and Lemma 2.8 we get (cid:12)(cid:12)(cid:12)(cid:12) H ′ H (1 / it, ψ ) (cid:12)(cid:12)(cid:12)(cid:12) ≫ L . Combining this with (2.36) and (3.3) we get the assertion. Lemma 3.2 and Lemma 3.3 together imply the assertions (i) and (ii) of Proposition3.1. To prove the gap assertion (iii) of Proposition 3.1 we need a result which will alsoplay an important role in the rest of this paper. Lemma 3.4. Suppose ψ ∈ Ψ ∗ , ρ is a zero of L ( s, ψ ) L ( s, ψχ ) in Ω , and | s | < (10) − α log L .Then we have G ( ρ + s, ψ ) = 1 − Q − s + O (cid:0) L − (log L ) Q − σ (cid:1) . Proof. It follows from (3.2) that H ( ρ, ψ ) = − O ( L − / log L ) . (3 . H ( ρ + s, ψ ) H ( ρ, ψ ) = exp (cid:26) Z s H ′ H ( ρ + w, ψ ) dw (cid:27) (3 . (cid:8) − L s + O ( L − (log L ) ) (cid:9) = Q − s (cid:8) O ( L − (log L ) ) (cid:9) . Note that L − / log L < L − (log L ) Q − σ . From (3.7) and (3.8) it follows that H ( ρ + s, ψ ) = − Q − s + O (cid:8) L − (log L ) Q − σ (cid:9) . (3 . The gap assertion (iii) of Proposition 3.1 is deduced from the following Lemma 3.5. Suppose ψ ∈ Ψ ∗ and ρ is a zero of L ( s, ψ ) L ( s, ψχ ) in Ω . Then the followinghold.(a). G ( ρ + it, ψ ) = 0 for < t ≤ πα − c L − (log L ) , where c > is a sufficiently largeconstant.(b). The function G ( ρ + s, ψ ) has a zero in the disc | s − πiα | < c L − (log L ) . Proof. Assume 0 < t ≤ πα − c L − (log L ) . Then, by (3.3), G ( ρ + it, ψ ) = Z t H ′ ( ρ + iv, ψ ) dv + O ( L / (log L ) t ) . For 0 ≤ v ≤ t we have H ( ρ + iv, ψ ) = − Q − iv + O (cid:0) L − (log L ) (cid:1) by Lemma 3.4 and (3.2), and H ′ H ( ρ + iv, ψ ) = − L + O ( L log L )by (2.37), (2.17) and Lemma 2.8. It follows that G ( ρ + it, ψ ) = 2 L Z t Q − iv dv + O (cid:0) L (log L ) t (cid:1) = i ( Q − it − 1) + O (cid:0) L (log L ) t (cid:1) . (3 . α = L − . Since | Q − it − | = 2 sin( L t ) ≥ (4 /π ) L t, if 0 < t ≤ πα/ , (4 c /π ) L − (log L ) , if πα/ < t ≤ πα − c L − (log L ) , the right-hand side of (3.10) can not be equal to zero as the constant c is sufficientlylarge. This proves (a).To prove (b) we note that 1 − Q − πiα = 0 and | − Q − s | = 2 c L − (log L ) + O (cid:0) L − (log L ) (cid:1) , if | s − πiα | = c L − (log L ) , the implied constant depending on c . It follows, by Lemma3.4, that |G ( ρ + s, ψ ) | > |G ( ρ + πiα, ψ ) | for | s − πiα | = c L − (log L ) . Thus, by the maximum modulus principle we get (b), and complete the proof of Propo-sition 3.1. We conclude this section by proving some consequences of Proposition 3.1 which willfind application in the next section. Lemma 3.6. Suppose ψ ∈ Ψ ∗ , < σ < and | t − D | < L − such that L ( s + w, ψ ) L ( s + w, ψχ ) = 0 , (3 . for any w = u + iv with − (1 / α < u < (21 / α and | v | < (1 / α. Then we have ≪ | L ( s + 2 α, ψ ) || L ( s, ψ ) | ≪ . Proof. We have log | L ( s + 2 α, ψ ) || L ( s, ψ ) | = Z α ℜ (cid:26) L ′ L ( s + u, ψ ) (cid:27) du. If ρ is a zero of L ( s, ψ ) such that |ℑ{ ρ }− t | < 1, then ρ ∈ Ω, so ℜ{ ρ } = 1 / ρ = 1 / iγ . We have ℜ (cid:26) L ′ L ( s + u, ψ ) (cid:27) = X ρ | γ − t | < σ + u − / σ + u − / + ( γ − t ) + O ( L ) (0 ≤ u ≤ α ) , ρ runs through the zeros of L ( s, ψ ). Note that | s + u − ρ | ≫ α for any zero ρ of L ( s, ψ ) by (3.11). Hence, by Proposition 3.1 (iii), the above sum is also O ( L ). Thiscompletes the proof. Lemma 3.7. Suppose ψ ∈ Ψ ∗ , ρ = 1 / iγ is a zero of L ( s, ψ ) in Ω such that | γ − D | < L − . Then we have L ( ρ ± α, ψ ) L ′ ( ρ, ψ ) ≪ α. (3 . Proof. By Proposition 3.1 and the residue theorem, the left-hand side of (3.12) is equalto 12 πi Z | s | =3 α L ( ρ + s ± α, ψ ) L ( ρ + s, ψ ) ds. Moreover, by Proposition 3.1 and Lemma 3.6 we get L ( ρ + s ± α, ψ ) L ( ρ + s, ψ ) ≪ | s | = 3 α . This completes the proof. 4. The Functions K ± ( ρ + s, ψ ) For ψ ∈ Ψ ∗ let S ( ψ ) denote the set of zeros of L ( s, ψ ) L ( s, ψχ ) in the strip | t − D | < ψ ∈ Ψ ∗ and ρ ∈ S ( ψ ) , (4 . R = π (cid:20) log L π (cid:21) − π , ω = αR. (4 . K ± ( ρ + s, ψ ) with | s | ≤ ω ,where K + ( s, ψ ) = L ( s + α, ψ ) L ( s − α, ψχ ) F ( s, ψ ) , K − ( s, ψ ) = L ( s − α, ψ ) L ( s + α, ψχ ) F ( s, ψ ) . (4 . G ( s, ψ ) is defined by (3.1). We have K + ( ρ + s, ψ ) = F ( ρ + s − α, ψ ) F ( ρ + s, ψ ) L ( ρ + s + α, ψ ) L ( ρ + s − α, ψ ) G ( ρ + s − α, ψ ) , (4 . K − ( ρ + s, ψ ) = F ( ρ + s + α, ψ ) F ( ρ + s, ψ ) L ( ρ + s − α, ψ ) L ( ρ + s + α, ψ ) G ( ρ + s + α, ψ ) . (4 . s, ψ ) is defined by (2.13). We have K + ( ρ + s, ψ ) = F (1 − ρ − s, ¯ ψ ) F ( ρ + s, ψ ) ∆( ρ + s + α, ψ )∆( ρ + s − α, ψχ ) K − (1 − ρ − s, ¯ ψ ) , (4 . K − ( ρ + s, ψ ) = F (1 − ρ − s, ¯ ψ ) F ( ρ + s, ψ ) ∆( ρ + s − α, ψ )∆( ρ + s + α, ψχ ) K + (1 − ρ − s, ¯ ψ ) . (4 . Q α = e and Q − σ ≤ e R < L / if | s | ≤ ω . From (4.4), (4.5), Lemma 3.4 andCorollary 2.9 we derive the following Lemma 4.1. Suppose | s | ≤ ω . Then we have K + ( ρ + s, ψ ) K − ( ρ + s, ψ ) = (1 − e Q − s )(1 − e − Q − s ) + O ( L − / R ) . Here we have written R for the factor log L in the error terms in Lemma 3.4 and Corollary2.9.By a familiar property of ∆( s, ψ ) we have∆( ρ + s ± α, ψ )∆( ρ + s ∓ α, ψχ ) ≪ Q − σ ( | s | ≤ ω ) . (4 . Lemma 4.2. Suppose | s | ≤ ω . Then we have K ± ( ρ + s, ψ ) ≪ . if σ ≥ , and Q s K ± ( ρ + s, ψ ) ≪ . if σ ≤ Proof. First we prove (4.9). By the maximum modulus principle, it suffices to show that(4.9) holds for σ ≥ − α, | s | = ω, or σ = − α, | t | ≤ ω. By Lemma 3.6 we see that 1 ≪ (cid:12)(cid:12)(cid:12)(cid:12) L ( ρ + s + α, ψ ) L ( ρ + s − α, ψ ) (cid:12)(cid:12)(cid:12)(cid:12) ≪ σ = − α and | t | ≤ ω or | s | = ω . On the other hand, by Lemma 3.4 we have G ( ρ + s ± α, ψ ) ≪ σ ≥ − α, | s | ≤ ω. σ ≤ | s | ≤ ω . Note that |K ± (1 − ρ − s, ¯ ψ ) | = |K ± ( ρ − ¯ s, ψ ) | . Hence, by (4.6), (4.7), (4.8) and Corollary 2.9 we get K ± ( ρ + s, ψ ) ≪ Q − σ . This proves (4.10). Let C denote the circle | s | = ω . Let T ( ρ, ψ ) = T ( ρ, ψ ) ∪ T ( ρ, ψ ) , where T ( ρ, ψ ) denotes the set of all the zeros of K + ( ρ + s, ψ ) inside C and where T ( ρ, ψ )denotes the set of all the zeros of L ( ρ + s + α, ψ ) in the domain ω < | s | < ω . ByProposition 3.1, each θ ∈ T ( ρ, ψ ) is of the form θ = ± α + πilα + O ( L − R ) , (4 . l ∈ Z and | l | ≤ log L π ;conversely, for any l satisfying the above relations, there exists at most one θ ∈ T ( ρ, ψ )such that either θ = α + πilα + O ( L − R ) or θ = − α + πilα + O ( L − R ) . (4 . − e − Q − θ ≪ L − R if θ ∈ T ( ρ, ψ ) . (4 . K ± ( ρ + s, ψ ) thatessentially rely on the gap assertion (iii) of Proposition 3.1. We introduce ϑ ( x ) = 12 πi Z (1) x s Y ( s ) ds ( x > , (4 . Y ( s ) = 5 α ( Q s/ − Q − s/ ) exp( s / s . Since α ( Q s/ − Q − s/ ) s = Z / − / Q ys dy, 22t follows that ϑ ( x ) = 5 Z / − / ς ( Q y x ) dy with the function ς ( x ) given by (2.21). Hence, by (2.22) and (2.23) we get0 < ϑ ( x ) < , (4 . ϑ ( x ) = 1 + exp[ − (log x − L / ] if x > Q / , (4 . ϑ ( x ) ≪ exp[ − (log x + L / ] if x < Q − / . Let λ + ( n ) and λ − ( n ) be the arithmetic functions defined by ∞ X λ + ( n ) n s = ζ ( s + α ) ζ ( s − α ) , ∞ X λ − ( n ) n s = ζ ( s − α ) ζ ( s + α ) . (4 . Lemma 4.4 Suppose | s | ≤ ω . Then we have K + ( ρ + s, ψ ) = X n ≤ Q / λ + ( n ) ψ ( n ) n ρ + s ϑ (cid:18) Q / n (cid:19) (4 . − Q − s X n ≤ Q λ − ( n ) ¯ ψ ( n ) n − ρ − s ϑ (cid:18) Q / n (cid:19) + O ( L − / ) , and K − ( ρ + s, ψ ) = X n ≤ Q / λ − ( n ) ψ ( n ) n ρ + s ϑ (cid:18) Q / n (cid:19) (4 . − Q − s X n ≤ Q λ + ( n ) ¯ ψ ( n ) n − ρ − s ϑ (cid:18) Q / n (cid:19) + O ( L − / ) . Proof. We prove (4.18) only, as the proof of (4.19) is analogous. By the maximum modulusprinciple, it suffices to prove (4.19) with | s | = 3 ω which is thus assumed, so L ( ρ + s − α, ψ ) = 0 , (4 . L ( ρ + s + α, ψ ) L ( ρ + s − α, ψ ) ≪ . (4 . σ ∗ = max { α, σ + 2 α } . Let Z denote the set of all the zeros of L ( w, ψ ) in the strip | w − ( ρ + s ) | < L − 3. Notethat Z depends on ρ + s and ψ and Z ⊂ Ω. Also, note that the function Y ( w ) has onlya simple pole at w = 0 and the residue is equal to one. Write w = u + iv . Since ℜ{ ρ + s + w − α } < / u = − σ ∗ , by Proposition 3.1, the residue theorem and (4.21), there exists a rectangle R ∗ withvertices at − σ ∗ ± i (cid:0) L + O (1) (cid:1) , ± i (cid:0) L + O (1) (cid:1) such that L ( ρ + s + w + α, ψ ) L ( ρ + s + w − α, ψ ) ≪ . w lies on the horizontal segments of R ∗ (see Lemma 3.6), and such that12 πi Z R ∗ L ( ρ + s + w + α, ψ ) L ( ρ + s + w − α, ψ ) κ ( s, w ) Y ( w ) dw = L ( ρ + s + α, ψ ) L ( ρ + s − α, ψ ) κ ( s, 0) (4 . X ρ ∗ ∈ Z L ( ρ ∗ + 2 α, ψ ) L ′ ( ρ ∗ , ψ ) κ ( s, ρ ∗ − ρ − s + α ) Y ( ρ ∗ − ρ − s + α )where κ ( s, w ) = Q (4 / w − e Q − s Q − (2 / w . Recall that J ( c ) is the line segment path from c − i L to c + 10 i L . We claim that K + ( ρ + s, ψ ) = N + − N − + O ( L − / ) , (4 . N + = 12 πi Z J (1) L ( ρ + s + w + α, ψ ) L ( ρ + s + w − α, ψ ) κ ( s, w ) Y ( w ) dw, N − = 12 πi Z J ( − σ ∗ ) L ( ρ + s + w + α, ψ ) L ( ρ + s + w − α, ψ ) κ ( s, w ) Y ( w ) dw. Since κ ( s, w ) Y ( w ) ≪ Q − if − σ ∗ ≤ u ≤ | v | = 10 L + O (1), it follows from (4.22) that12 πi Z R ∗ L ( ρ + s + w + α, ψ ) L ( ρ + s + w − α, ψ ) κ ( s, w ) Y ( w ) dw = N + − N − + O ( Q − ) , (4 . κ ( s, 0) = G ( ρ + s − α, ψ ) + O (cid:0) L − / R (cid:1) . Combining this result with (4.21), (4.4) and Corollary 2.9 we get L ( ρ + s + α, ψ ) L ( ρ + s − α, ψ ) κ ( s, 0) = K + ( ρ + s, ψ ) + O (cid:0) L − / R (cid:1) . (4 . X ρ ∗ ∈ Z (cid:12)(cid:12) κ ( s, ρ ∗ − ρ − s + α ) Y ( ρ ∗ − ρ − s + α ) (cid:12)(cid:12) ≪ L / . (4 . ρ ∗ ∈ Z . By direct calculation we get κ ( s, ρ ∗ − ρ − s + α ) = Q (4 / ρ ∗ − ρ − s + α ) (cid:0) − Q − ρ ∗ − ρ ) (cid:1) , so κ ( s, ρ ∗ − ρ − s + α ) Y ( ρ ∗ − ρ − s + α ) ≪ α (cid:0) Q − (43 / σ (cid:1)(cid:12)(cid:12) − Q − ρ ∗ − ρ ) (cid:12)(cid:12)(cid:12)(cid:12) ρ ∗ − ρ − s + α (cid:12)(cid:12) . (4 . α ∗ = π L − / . By Proposition 3.1, if ρ ∗ ∈ Z and | ρ ∗ − ρ | < α ∗ , then there exists aninteger l such that | l | ≤ L / and ρ ∗ − ρ = πilα + O ( | l |L − R ) , (4 . − Q − ρ ∗ − ρ ) ≪ | l |L − R ;conversely, if l ∈ Z and | l | ≤ L / , then there exists at most one ρ ∗ ∈ Z satisfying (4.29).Note that | πilα − s + α | ≫ α for any l ∈ Z since | s | = 3 ω . It follows that X ρ ∗ ∈ Z | ρ ∗ − ρ | <α ∗ (cid:12)(cid:12) − Q − ρ ∗ − ρ ) (cid:12)(cid:12)(cid:12)(cid:12) ρ ∗ − ρ − s + α (cid:12)(cid:12) ≪ L − R X | k |≤L / | k || πikα − s + α | ≪ L R . (4 . X ρ ∗ ∈ Z | ρ ∗ − ρ |≥ α ∗ (cid:12)(cid:12) − Q − ρ ∗ − ρ ) (cid:12)(cid:12)(cid:12)(cid:12) ρ ∗ − ρ − s + α (cid:12)(cid:12) ≪ L X k ≥L / k ≪ L / . (4 . Q − (43 / σ < L / < L / . By (4.28), (4.30) and (4.31) we get (4.27), and complete the proof of (4.24).It now remains to calculate the integrals N + and N − . Note that L ( ρ + s + w + α, ψ ) L ( ρ + s + w − α, ψ ) = ∞ X λ + ( n ) ψ ( n ) n ρ + s + w ( u = 1) . To calculate N + we can replace the path J (1) by the line u = 1 and replace the factor κ ( s, w ) by Q (4 / w with negligible errors. It follows that N + = ∞ X λ + ( n ) ψ ( n ) n ρ + s ϑ (cid:18) Q / n (cid:19) + O ( Q − / ) . (4 . N − we first appeal to (2.13), Lemma 3.6 and the relation∆( ρ + s + w + α, ψ )∆( ρ + s + w − α, ψ ) = Q − α + O ( L − ) , (cid:0) w ∈ J ( − σ ∗ ) (cid:1) to get L ( ρ + s + w + α, ψ ) L ( ρ + s + w − α, ψ ) = e − L (1 − ρ − s − w − α, ¯ ψ ) L (1 − ρ − s − w + α, ¯ ψ ) + O ( L − ) , (cid:0) w ∈ J ( − σ ∗ ) (cid:1) , the contribution from the error term to N − being O ( L − / ) by routine estimation. Hence N − = e − πi Z J ( − σ ∗ ) L (1 − ρ − s − w − α, ¯ ψ ) L (1 − ρ − s − w + α, ¯ ψ ) κ ( s, w ) Y ( w ) dw + O ( L − / ) . Next we apply the Cauchy theorem to move the path of integration from J ( − σ ∗ ) to J ( − J ( − 1) by the line u = − e − κ ( s, w ) by − Q − s Q − (2 / w to get N − = − Q − s πi Z ( − L (1 − ρ − s − w − α, ¯ ψ ) L (1 − ρ − s − w + α, ¯ ψ ) Q − (2 / w Y ( w ) dw + O ( L − / ) . (4 . L (1 − ρ − s − w − α, ¯ ψ ) L (1 − ρ − s − w + α, ¯ ψ ) = ∞ X λ − ( n ) ¯ ψ ( n ) n − ρ − s − w ( u = − , πi Z ( c ) x w Y ( w ) dw = − ϑ (1 /x ) for c < , it follows from (4.33) that N − = Q − s ∞ X λ − ( n ) ¯ ψ ( n ) n − ρ − s ϑ (cid:18) Q / n (cid:19) + O ( L − / ) . (4 . n > Q / andthe terms in the sum in (4.34) with n > Q can be neglected. Hence, combining (4.32)and (4.34) with (4.24) we get (4.18). 5. The Linear Functional Φ( f ; ρ, ψ ) For f ∈ C [ − , 1] we define the linear functionalΦ( f ; ρ, ψ ) = X n ≤ Q / λ + ( n ) ψ ( n ) n ρ ϑ (cid:18) Q / n (cid:19) f (1 − α log n ) (5 . − X n ≤ Q λ − ( n ) ¯ ψ ( n ) n − ρ ϑ (cid:18) Q / n (cid:19) f ( α log n − . Let ϕ s ( x ) = Q xs . Note that ϕ s (1 − α log n ) = (cid:18) Qn (cid:19) s , ϕ s ( α log n − 1) = (cid:18) Qn (cid:19) − s , so Φ( ϕ s ; ρ, ψ ) = Q s (cid:26) X n ≤ Q / λ + ( n ) ψ ( n ) n ρ + s ϑ (cid:18) Q / n (cid:19) − Q − s X n ≤ Q λ − ( n ) ψ ( n ) n − ρ − s ϑ (cid:18) Q / n (cid:19)(cid:27) . Hence, by Lemma 4.3 and the inequality | Q s | < L / for | s | ≤ ω we derive the following Lemma 5.1 Suppose | s | ≤ ω . Then we have Φ( ϕ s ; ρ, ψ ) = Q s K + ( ρ + s, ψ ) + O ( L − / ) . n particular, if θ ∈ T ( ρ, ψ ) , then Φ( ϕ θ ; ρ, ψ ) ≪ L − / . For notational simplicity we shall write Φ( f ) for Φ( f ; ρ, ψ ). For any f , f ∈ C [ − , f ∗ f )( x ) = Z x f ( x − y ) f ( y ) dy. (5 . ϕ s ∗ ϕ s ′ = α ϕ s − ϕ s ′ s − s ′ ( s = s ′ ) . (5 . δ = R − / δ = R − / . (5 . d ≥ 425 (in fact, the choice d = 425 is admissible). We define two weightfunctions as follows. ̟ ( x ) = δ − d (1 + δ − x ) d , (5 . ̟ ( x ) = 2 dδ − d (1 + δ + x )(1 + δ − x ) d − . (5 . f , f ∈ C [ − , 1] we define the inner product < f , f > = Z − (cid:2) f ′ ( x ) ¯ f ′ ( x ) ̟ ( x ) + f ( x ) ¯ f ( x ) ̟ ( x ) (cid:3) dx. (5 . k f k = p < f, f > (cid:0) f ∈ C [ − , (cid:1) . (5 . ̟ ( x ) ≫ δ for | x | ≤ ̟ ( x ) ≫ R for | x | ≤ − δ , it follows that Z − f ′ ( x ) | dx ≪ δ − k f k , ( f ∈ C [ − , , (5 . Z − δ − δ f ′ ( x ) | dx ≪ R − k f k , ( f ∈ C [ − , . (5 . f ′ ( x ) replaced by f ( x ). Thus, for any f ∈ C [ − , x ∈ [ − δ , − δ ] such that f ( x ) ≪ R − k f k , so, by the Cauchyinequality and (5.9),max | x |≤ {| f ( x ) |} ≪ R − k f k + Z − f ′ ( y ) | dy ≪ δ − k f k . ( f ∈ C [ − , . (5 . Z − ̟ ( y ) − dy ≪ δ − . Analogously we have max | x |≤ − δ {| f ( x ) |} ≪ R − k f k . ( f ∈ C [ − , . (5 . f ) or its partial sums onaverage when the function f depends on ψ and ρ . In such a situation the typical techniquesof mean value estimates of Dirichlet polynomials with characters fail to apply directly.Our treatment is to separate the terms with n = 1 in Φ( f ) and transform the remainingsums into simple sums by partial summation. To this end we introduceΦ ( f ) = Φ ( f ; ρ, ψ ) = X 1) + Φ ( f ) . (5 . Remark. By virtue of (4.16), in practice, the factors ϑ ( Q / ) and ϑ ( Q / ) in (5.14) canbe replaced by one with negligible errors.For b ≥ a ≥ U + ( a, b ; s, ψ ) = X Q a Assume (A2) holds. Then for any f ∈ C [ − , we have | Φ ( f ) | ≪ k f k log R. Proof. We can assume k f k > η be a positive number tobe specified later. We have (cid:12)(cid:12)(cid:12)(cid:12) X 1) = ̟ (1 − y ). Thus wehave X 6. Solutions to Boundary Value Problems In this section, we introduce a number of functions which are solutions to certainSturmian boundary value problems. All the functions in this section are defined on[ − , g ( x ) = (2 d ) − δ d − (1 + δ − x ) − d . It is direct to verify that g satisfies the differential equation[ g ′ ̟ ] ′ − g ̟ = 0 (6 . g ′ (1) ̟ (1) = − g ′ ( − ̟ ( − 1) = 1 . (6 . A = 4 d δ − d Z (1 + δ − x ) d dx + 2 dδ , and g ( x ) = 2 d ( A δ ) − (1 + δ − x ) − d Z x (1 + δ − y ) d dy. It is direct to verify that the function g satisfies the same differential equation as g andthe boundary condition g ′ (1) ̟ (1) = g ′ ( − ̟ ( − 1) = 1 . (6 . f ∈ C [ − , < f, g > = f (1) + f ( − , (6 . < f, g > = f (1) − f ( − . (6 . k g j k = 2 g j (1) , j = 1 , . 32t follows, by routine calculation, that k g j k = d − δ − + O ( δ d − ) j = 1 , . (6 . ε be areal number satisfying ε = R − / + O ( R − ) , ε ≡ 0( mod 2 π ) . (6 . T ∈ C [ − , 1] such that0 ≤ T ( x ) ≤ , T ′ ( x ) ≪ δ − , T ( x ) = ( , if 1 − δ ≤ | x | ≤ , , if | x | ≤ − δ . Let T ( x ) = sin (cid:0) R (1 + ε − x ) (cid:1) ε − x T ( x ) if x ≥ , T ( x ) = sin (cid:0) R ( − ε − x ) (cid:1) − ε − x T ( x ) if x < . Let T ∈ C [ − , 1] be such that T ( x ) = 0 unless | x − ( − ε ) | < R − , T ( x ) ≪ R , Z − T ( x ) dx = 1 . Let T ( x ) = T ( x ) − T ( x ) . Let g ( x ) be the (unique) solution to the boundary value problem (cid:2) ̟ g ′ (cid:3) ′ − ̟ g = −T ,g ′ (1) = g ′ ( − 1) = 0 . It follows, by partial integration, that < f, g > = Z − f ( y ) T ( y ) dy ( f ∈ C [ − , . (6 . g ( x ) = − ˜ g ( x ) + 12 (cid:2) ˜ g ′ (1) − ˜ g ′ ( − (cid:3) ̟ (1) g ( x ) + 12 (cid:2) ˜ g ′ (1) + ˜ g ′ ( − (cid:3) ̟ (1) g ( x ) , g ( x ) = δ d − (1 + δ − x ) − d Z x (cid:26) Z xy (1 + δ − z ) d dz (cid:27) (1 + δ − y ) − d T ( y ) dy. Note that ( δεR ) − ∼ R / . By partial integration and routine estimation we obtain thefollowing Lemma 6.1 We have k ˜ g k ≪ R / . 7. Approximation to ϕ T ( ρ, ψ ) given in Section 4. Let S ( ρ, ψ ) denote the complexvector space spanned by the exponential functions ϕ θ with θ ∈ T ( ρ, ψ ). For notationalsimplicity we shall write T and S for T ( ρ, ψ ) and S ( ρ, ψ ) respectively.Note that ϕ ( x ) ≡ 1. By the Jensen formula, it can be shown that there exists afunction f ∈ S such that k ϕ − f k ≪ δ. (see P. Borwein and T. Erd´elyi [BE, Chapter 4], for example). This result can not beessentially improved. On the other hand, if one allows f ∈ S + < ϕ s > with s ≪ α and s / ∈ T , then much better approximation to ϕ can be obtained. The main result of thissection is the following Lemma 7.1. Assume (A2) holds. There exist functions h = X θ ∈ T A ( θ ) ϕ θ ∈ S and k ∈ C [ − , such that A ( θ ) ≪ , (7 . h ( x ) ≪ (log R ) , h ′ ( x ) ≪ R log R if | x | ≤ , (7 . B ( s ) := X θ ∈ T A ( θ ) s − θ ≪ δ if s ∈ C , (7 . < k, g > = 0 , k k k ≪ δ, (7 . nd such that k r k = o ( δ ) , (7 . r ( x ) = ϕ ( x ) − h ( x ) − k ( x ) . (7 . Proof. We give a sketch. Write T j for T j ( ρ, ψ ), j = 1 , 2, and let S j be the complex vectorspace spanned by the functions ϕ θ with θ ∈ T j , j = 1 , 2. Let R ∗ be a rectangle withvertices at ± ω + O ( α ) ± iω (log R ) − such that K + ( ρ + s, ψ ) − ≪ R ∗ . Write˜ R ( x ) = 12 πi Z R ∗ ϕ s ( x ) Q s K + ( ρ + s, ψ ) exp (cid:2) s (log R ) /ω (cid:3) s ds, ˜ R ( x ) = 12 πi Z R ∗ ϕ s ( x ) Q s K + ( ρ + s, ψ ) exp (cid:2) s (log R ) /ω (cid:3) s − α ds. Both of ˜ R ( x ) and ˜ R ( x ) are ∼ x ∼ ± 1, while the difference˜ R ( x ) − ˜ R ( x ) = − α πi Z R ∗ ϕ s ( x ) Q s K + ( ρ + s, ψ ) exp (cid:2) s (log R ) /ω (cid:3) s ( s − α ) ds is much closer to zero. Because of the factor exp (cid:2) s (log R ) /ω (cid:3) , the integrals on thehorizontal segments in the expressions for ˜ R j ( x ) can be neglected. On the other hand,by Lemma 4.3 and (A2), we have K + ( ρ + s, ψ ) ∼ s lies on the right side of R ∗ , and K + ( ρ + s, ψ ) ∼ − Q − s if s lies on the left side of R ∗ . It follows that˜ R j ( x ) ∼ πi Z ( ω ) Q ( x − s exp (cid:2) s (log R ) /ω (cid:3) s ds ( s ∼ , (7 . R j ( x ) ∼ πi Z ( − ω ) Q ( x +1) s exp (cid:2) s (log R ) /ω (cid:3) s ds ( s ∼ − . (7 . h ∈ S such that˜ R = K + ( ρ, ψ ) − ϕ + h . (7 . h = X θ ∈ T A ∗ ( θ ) ϕ θ ∈ S A ∗ ( θ ) ≪ , X θ ∈ T A ∗ ( θ ) θ = −K + ( ρ, ψ ) , (7 . k ( x ) := ( h ∗ ˜ R )( x )satisfies < k, g > = 0 and k k k ≪ δ. Let r ( x ) = ( ˜ R ∗ h )( x ) − k ( x ) . It can be shown that k r k = o ( δ ) . On the other hand, by (5.3), (7.9) and (7.10), we see that there exists a function h ∈ S such that ( ˜ R ∗ h )( x ) = ϕ ( x ) − h ( x ) . Combining these results we get the assertion. Remark. The estimate (7.5) will be sufficient for our purpose; in fact a sharper estimatecan be proved. 8. The Fundamental Inequality: Preliminary In this section we assume (A2) holds and start to derive the fundamental inequalitydescribed in Section 1.Recall that ε is given by (6.7). Let C + be the right half of the circle C from − iω to iω , C − the left half of C from iω to − iω . For any f ∈ C [ − , 1] we define three linearfunctionals as follows.Let Φ ∗ ( f ) = Φ ∗ ( f ; ρ, ψ ) = X Q ε 37y Lemma 4.1 and Lemma 4.2 we get1 ≪ |K ± ( ρ + s, ψ ) | ≪ s ∈ C + ) . (8 . | − e ± Q − s | ≫ Q − σ for s ∈ C − , by Lemma 4.1 and Lemma 4.2 we get1 ≪ Q − σ |K ± ( ρ + s, ψ ) | ≪ s ∈ C − ) . (8 . Step 1. We have h (1 + ε ) − e − h ( − ε ) = X θ ∈ T A ( θ )(1 − e − Q − θ ) Q (1+ ε ) θ . By (4.13) and (7.1), the terms with | θ | > ω in the above sum totally contribute O ( L − / ).(This is a crucial application of the gap assertion (iii) of Proposition 3.1). Hence, by theresidue theorem we get h (1 + ε ) − e − h ( − ε ) = 12 πi Z C B ( s )(1 − e − Q − s ) Q (1+ ε ) s ds + O ( L − / ) , (8 . B ( s ) is given by (7.3). Step 2. By Lemma 4.1 we get1 − e − Q − s = K − ( ρ + s, ψ ) K + ( ρ + s, ψ )1 − e Q − s + O ( L − / ) ( s ∈ C ) . We insert this relation into the right-hand side of (8.7) and apply the relation11 − e Q − s = − e − Q s − e − Q s getting h (1 + ε ) − e − h ( − ε ) = J + − J − + O ( L − / ) , (8 . J + = 12 πi Z C + K − ( ρ + s, ψ ) K + ( ρ + s, ψ )1 − e − Q − s B ( s ) Q (1+ ε ) s ds, − = e − πi Z C − K − ( ρ + s, ψ ) K + ( ρ + s, ψ )1 − e Q s B ( s ) Q (3+ ε ) s ds. Step 3. We insert the relation 11 − e Q − s = 1 + e Q − s − e − Q − s into the expression for J + and then split it into two integrals accordingly. By (7.3) androutine estimation, we see that the second integral is O ( δ ). Hence J + = 12 πi Z C + K − ( ρ + s, ψ ) K + ( ρ + s, ψ ) B ( s ) Q (1+ ε ) s ds + O ( δ ) . Analogously, we insert the relation 11 − e Q s = 1 + e Q s − e Q s into the expression for J − and then split it into two integrals accordingly. By (7.3) androutine estimation, we see that the second integral is also O ( δ ). Hence J − = e − πi Z C − K − ( ρ + s, ψ ) K + ( ρ + s, ψ ) B ( s ) Q (3+ ε ) s ds + O ( δ ) . Combining these results with (8.8) we get h (1 + ε ) − e − h ( − ε ) = 12 πi Z C + K − ( ρ + s, ψ ) K + ( ρ + s, ψ ) B ( s ) Q (1+ ε ) s ds (8 . − e − πi Z C − K − ( ρ + s, ψ ) K + ( ρ + s, ψ ) B ( s ) Q (3+ ε ) s ds + O ( δ ) . Step 4. By Lemma 5.1, we can replace the factor K + ( ρ + s, ψ ) Q s in (8.19) by Φ( ϕ s ) withan error much smaller than δ . Hence h (1 + ε ) − e − h ( − ε ) = 12 πi Z C + K − ( ρ + s, ψ )Φ( ϕ s ) B ( s ) Q εs ds (8 . − e − πi Z C − K − ( ρ + s, ψ )Φ( ϕ s ) B ( s ) Q (2+ ε ) s ds + O ( δ ) . tep 5. It follows from (5.3) that h ∗ ϕ s = − α X θ ∈T A ( θ ) s − θ ϕ θ + αB ( s ) ϕ s ( s ∈ C ) , so Φ( h ∗ ϕ s ) = − α X θ ∈T A ( θ ) s − θ Φ( ϕ θ ) + αB ( s )Φ( ϕ s ) ( s ∈ C ) . The first sum on the right-hand side above is O ( L − / ) by (7.1) and Lemma 5.1. Thuswe have B ( s )Φ( ϕ s ) = L Φ( h ∗ ϕ s ) + O ( L / ) ( s ∈ C ) . Inserting this relation into (8.10) and using routine estimation we get (8.4).Now, combining(8.4) with (8.3) we getΘ( h ) ≪ δ . This, together with (7.6), leads to the following Lemma 8.1. Assume (A2) holds. Then we have Θ( ϕ ) − Θ( k ) − Θ( r ) ≪ δ . 9. Upper Bounds for Θ( k ) and Θ( r ) In this section we assume (A2) holds and derive upper bounds for Θ( k ) and Θ( r ) fromthe following Lemma 9.1 Assume (A2) holds. For any f ∈ C [ − , satisfying k f k ≤ δ we have Θ( f ) = (1 − e − ) < f, g > + O ( δ ) + O ( E ( ρ, ψ )) , where E ( ρ, ψ ) = Z − δ (cid:8) |X + (1 + ε − y ; ρ, ψ ) | + |X − (1 + ε − y ; 1 − ρ, ¯ ψ ) | (cid:9) dy (9 . Z − δ − (cid:8) |X + ( − ε − y ; ρ, ψ ) | + |X − ( − ε − y ; 1 − ρ, ¯ ψ ) | (cid:9) dy with the functions X ± to be introduced at the end of this section. Proof. We begin with estimating Φ ∗ ( f ). By virtue of the proof of Lemma 5.3 with aminor modification we have Φ ∗ ( f ) ≪ δ . (9 . T in Section 6,by routine estimation we get f ( − ε ) = Z − f ( x ) T ( x ) dx + O ( R − ) . It follows that Θ( f ) = − (1 − e − ) Z − f ( x ) T ( x ) dx + Ξ( f ) + O ( δ ) . (9 . f ). By the Cauchy theorem,Ξ( f ) = L π Z ω − ω (1 + e − Q it ) K − ( ρ + it, ψ )Φ( f ∗ ϕ it ) Q iεt dt. (9 . f ∗ ϕ it ) = X n ≤ Q / λ + ( n ) ψ ( n ) n ρ ϑ (cid:18) Q / n (cid:19) ( f ∗ ϕ it )(1 − α log n ) − X n ≤ Q λ − ( n ) ¯ ψ ( n ) n − ρ ϑ (cid:18) Q / n (cid:19) ( f ∗ ϕ it )( α log n − , and ( f ∗ ϕ it )( x ) = Z x f ( x − y ) Q iyt dy. (9 . x < f ∗ ϕ it )( x ) = − Z − x f ( x + y ) Q − iyt dy ( x < . (9 . f ∗ ϕ it ) = ( f ∗ ϕ it )(1) − ( f ∗ ϕ it )( − 1) + V ( it ; ρ, ψ ) − V ( it ; ρ, ψ ) + V ( it ; 1 − ρ, ¯ ψ ) , where V ( it ; ρ, ψ ) = Z W ( y ; ρ, ψ ) Q iyt dy, (9 . W ( y ; ρ, ψ ) = X 1) = Z − f ( y ) T ( y ) Q − it (1+ y ) dy (9 . Z f ( y ) Q it (1 − y ) dy + O ( R − / ) . On the other hand, we write the result (4.19) on K − ( ρ + s, ψ ) in Lemma 4.3 as K − ( ρ + it, ψ ) = 1 − Q − it + X 44e get for | z | ≤ δ L π Z ω − ω Q itz (1 + e − Q it ) K − ( ρ + it, ψ ) dt = (1 − e − ) sin( Rz ) πz (9 . X + ( z ; ρ, ψ ) − X − ( z ; 1 − ρ, ¯ ψ ) + O (1) , where X + ( z ; ρ, ψ ) = X Lemma 9.2 Assume (A2) holds. Then we have Θ( ϕ ) = o ( R − / ) + O (cid:0) E ( ρ, ψ ) (cid:1) . 10. Asymptotic Expression for Θ( ϕ ) The aim of this section is to show, apart from some error terms whose average orderswill be estimated in Section 12, thatΘ( ϕ ) ∼ − (1 − e − ) U ( ε ) , where U ( ε ) = 12 πi Z C − Q εs dss . We begin with calculating Ξ( ϕ ). Write U ± ( x ) = 12 πi Z C ± Q xs s ds ( x ∈ R ) . By the residue theorem we get U + ( x ) + U − ( x ) = 1 . (10 . U + ( x ) = 1 + O ((1 + Rx ) − ) , U − ( x ) ≪ (1 + Rx ) − ( x ≥ , (10 . U − ( x ) = 1 + O ((1 + R | x | ) − ) , U + ( x ) ≪ (1 + R | x | ) − ( x ≤ . (10 . ϕ ∗ ϕ s ) = αs (cid:2) − Φ( ϕ ) + Φ( ϕ s ) (cid:3) . Hence Ξ( ϕ ∗ ϕ s ) = − Φ( ϕ ) (cid:2) Ξ − e − Ξ (cid:3) + Ξ − e − Ξ , (10 . = 12 πi Z C + K − ( ρ + s, ψ ) Q εs dss , (10 . = 12 πi Z C − K − ( ρ + s, ψ ) Q (2+ ε ) s dss , (10 . = 12 πi Z C + K − ( ρ + s, ψ ) Φ( ϕ s ) Q εs dss (10 . = 12 πi Z C − K − ( ρ + s, ψ ) Φ( ϕ s ) Q (2+ ε ) s dss . (10 . we insert the formula (4.19) into (10.7), and estimate the contribution fromthe error term to the integral trivially.. Hence, integrating term by term and separatingthe terms with n = 1 in the first sum we getΞ = U + ( ε ) + Y ( ρ, ψ ) − Y (1 − ρ, ¯ ψ ) + O ( L − / ) (10 . Y ( ρ, ψ ) = X 1, and recall the definition of Θ in Section 8. It follows from (10.19)that Θ( ϕ ) = − Φ( ϕ ) + (1 − e − ) U − ( ε )Φ( ϕ ) + Φ ∗ ( ϕ ) (10 . O ( ε ) + O ( E ( ρ, ψ )) . On the other hand, we haveΦ ∗ ( ϕ ) = Φ( ϕ ) − X Assume (A2) holds. Then we have Θ( ϕ ) = (1 − e − ) U − ( ε )Φ( ϕ ) + O ( ε log R ) + O ( E ( ρ, ψ )) . 1. The Fundamental Inequality: Completion It follows from Lemma 5.1, Lemma 4.1 and Lemma 4.2 that | Φ( ϕ ) | ≫ . (11 . U ( ε ) = 12 πi Z − iω − i ∞ Q εs s ds + 12 πi Z − i ∞ iω Q εs s ds. Hence, by a change of variable, U ( ε ) = 1 π Z ∞ R sin( εy ) y dy. Recall that εR ≡ 0( mod 2 π ). Hence, by partial integration, | U ( ε ) | ≫ ( εR ) − ∼ R − / . (11 . Proposition 11.1 (The Fundamental Inequality) Assume that ψ ∈ Ψ ∗ and ρ ∈ S ( ψ ) .Then we have E ( ρ, ψ ) ≫ R − / , (11 . where E ( ρ, ψ ) = 1 R / log R Υ( ρ, ψ ) + E ( ρ, ψ ) + E ( ρ, ψ ) . Proof . If (A2) fails to hold, then we have1 R / log R Υ( ρ, ψ ) ≥ log RR / so (11.3) holds. If (A2) holds, then we have, by (11.1), (11.2), Lemma 9.2 and Lemma10.1, E ( ρ, ψ ) + E ( ρ, ψ ) ≫ R − / so (11.3) holds. Let E = X ψ ∈ Ψ ∗ X ρ ∈ S ( ψ ) E ( ρ, ψ ) , From Propositon 11.1 we derive the following Corollary 11.2. We have E ≫ R − / X ψ ∈ Ψ ∗ X ρ ∈ S ( ψ ) . 2. Estimation of E In this section we estimate E . Our result rests only on the fact that Ψ ∗ is a subset ofΨ. We begin with the following two lemmas. Lemma 12.1 For any complex numbers a n we have X ψ ∈ Ψ ∗ X ρ ∈ S ( ψ ) (cid:12)(cid:12)(cid:12)(cid:12) X n ≤ Q a n ψ ( n ) n ρ (cid:12)(cid:12)(cid:12)(cid:12) ≪ L Q X n ≤ Q | a n | n . Proof. Write P ( s, ψ ) = X n ≤ Q a n ψ ( n ) n s . Assume that ψ ∈ Ψ ∗ . For ρ ∈ S ( ψ ) we have | P ( ρ, ψ ) | ≪ L Z α − α | P ( ρ + it, ψ ) | dt + α Z α − α | P ′ ( ρ + it, ψ ) | dt. (12 . ρ − iα, ρ + iα ] and [ ρ ′ − iα, ρ ′ + iα ] are disjoint if ρ, ρ ′ ∈ S ( ψ ) and ρ = ρ ′ , [ ρ ∈ S ( ψ ) [ ρ − iα, ρ + iα ] ⊂ [ s − i, s + 2 i ] . Hence, by (12.1), X ρ ∈ S ( ψ ) | P ( ρ, ψ ) | ≪ Z D +2 D − (cid:26) L (cid:12)(cid:12)(cid:12)(cid:12) P (cid:18) 12 + it, ψ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) + α (cid:12)(cid:12)(cid:12)(cid:12) P ′ (cid:18) 12 + it, ψ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) (cid:27) dt (12 . P ′ ( s, ψ ) = − X n ≤ Q a n (log n ) ψ ( n ) n s and log n ≤ L for n ≤ Q . Hence, by (12.2) and Lemma 2.2 we complete the proof. Recall that λ ± ( n ) are given by (4.18). From the Euler product representation it followsfor L ≥ | λ + ( p l ) | ≤ λ − ( p l ) = p lα (1 − p − α ) . (12 . Lemma 12.2. For < η ≤ we have X 1. By (12.3) we get λ ± ( p l ) ≪ αp αl log p, so ∞ X λ ± ( p l ) p l ≪ ( α log p ) p . Hence X p ≤ Q η log (cid:26) ∞ X λ ± ( p l ) p l (cid:27) ≪ α X p ≤ Q η (log p ) p ≪ η . This completes the proof. In the estimation of E we shall repeatedly apply the simple bound ϑ ( x ) ≪ ρ, ψ ) are given by (5.15) and note that Z δ y dy̟ (1 − y ) ≪ . By Lemma 12.1 and Lemma 12.2 we get X ψ ∈ Ψ ∗ X ρ ∈ S ( ψ ) Υ( ρ, ψ ) ≪ L Q . (12 . E ( ρ, ψ ) is given by (9.1). To estimate the sum of E ( ρ, ψ ) in E we needthe following consequence of Lemma 12.2 X We have E ≪ L Q R − / (log R ) − . 13. Proof of Theorem 1 Assume q ∈ Q , so q is square-free and every prime divisor of q is greater than three.Let q = p p ....p k be the prime decomposition of q . Then every character ψ mod q can be uniquely writtenas ψ = ψ ψ ....ψ k , where ψ j is a character mod p j . By the definition of the set Ψ q , ψ is in Ψ q if and onlyif ψ j is primitive for ( p j , D ) = 1, and ψ j is non-real for p j | D . Thus we have X ψ ∈ Ψ q Y p | q ( p,D )=1 ( p − Y p | ( q,D ) ( p − . It follows that N := X ψ ∈ Ψ ≫ X q ∈Q q (cid:18) ϕ ( q ) q (cid:19) ≫ Q . (13 . X ρ ∈ S ( ψ ) ≫ L ( ψ ∈ Ψ ∗ ) . (13 . roposition 13.1. Assume (A1) holds. Then we have X ψ ∈ Ψ ∗ X ρ ∈ S ( ψ ) ≫ L Q . Now, under the assumption (A1), a contradiction is immediately derived from Corol-lary 11.2, Proposition 12.3 and Proposition 13.1.The proof of Theorem 1 is complete. Remark. Under the slightly weaker assumption L (1 , χ ) < c (log D ) − with the constant c sufficiently small, it can be shown that X ψ ∈ Ψ ∗ ≫ Q , and thus a contradiction can be derived. Acknowledgements . 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