Author:
(1) Yitang Zhang. Table of Links Abstract & Introduction
Notation and outline of the proof
The set Ψ1
Zeros of L(s, ψ)L(s, χψ) in Ω
Some analytic lemmas
Approximate formula for L(s, ψ)
Mean value formula I
Evaluation of Ξ11
Evaluation of Ξ12
Proof of Proposition 2.4
Proof of Proposition 2.6
Evaluation of Ξ15
Approximation to Ξ14
Mean value formula II
Evaluation of Φ1
Evaluation of Φ2
Evaluation of Φ3
Proof of Proposition 2.5 Appendix A. Some Euler products Appendix B. Some arithmetic sums References 3. The set Ψ1 Let ν(n) and υ(n) be given by respectively. It is easy to see that Lemma 3.1. Assume (A) holds. Then Proof. Let which has the Euler product representation For σ ≥ σ0 > 0, by checking the cases χ(p) = ±1 and χ(p) = 0 respectively, it can be seen that and the implied constant depending on σ0. Thus φ(s) is analytic for σ > 1/2 and it satisfies for σ ≥ σ1 > 1/2, the implied constant depending on σ1. The left side of (3.2) is Lemma 3.2. Assume (A) holds. Then we have Proof. As the situation is analogous to Lemma 3.1 we give a sketch only. It can be verified that the function is analytic for σ > 1/2 and it satisfies for σ ≥ σ1 > 1/2, the implies constant depending on σ1. Also, one can verify that This completes the proof. Lemma 3.3. For any s and any complex numbers c(n) we have and Proof. The first assertion follows by the orthogonality relation; the second assertion follows by the large sieve inequality. Let By (3.1) we may write By Cauchy’s inequality and the first assertion of Lemma 3.3 we obtain Thus we conclude Lemma 3.4. The inequality Write Assume that (A) holds. By Cauchy’s inequality, the second assertion of Lemma 3.2 and Lemma 3.1, Thus we conclude Lemma 3.5 Assume that (A) holds. The inequality Let Assume that (A) holds. By Cauchy’s inequality, the first assertion of Lemma 3.2 and Lemma 3.1, Thus we conclude Lemma 3.6. Assume that (A) holds. The inequality We are now in a position to give the definition of Ψ1: Let Ψ1 be the subset of Ψ such that ψ ∈ Ψ1 if and only if the inequalities (3.4), (3.5) and (3.6) simultaneously hold. Proposition 2.1 follows from Lemma 3.4, 3.5 and 3.6 immediately. This paper is available on arxiv under CC 4.0 license. Author: (1) Yitang Zhang. Author: Author: (1) Yitang Zhang. Table of Links Abstract & Introduction Notation and outline of the proof The set Ψ1 Zeros of L(s, ψ)L(s, χψ) in Ω Some analytic lemmas Approximate formula for L(s, ψ) Mean value formula I Evaluation of Ξ11 Evaluation of Ξ12 Proof of Proposition 2.4 Proof of Proposition 2.6 Evaluation of Ξ15 Approximation to Ξ14 Mean value formula II Evaluation of Φ1 Evaluation of Φ2 Evaluation of Φ3 Proof of Proposition 2.5 Abstract & Introduction Abstract & Introduction Notation and outline of the proof Notation and outline of the proof The set Ψ1 The set Ψ1 Zeros of L(s, ψ)L(s, χψ) in Ω Zeros of L(s, ψ)L(s, χψ) in Ω Some analytic lemmas Some analytic lemmas Approximate formula for L(s, ψ) Approximate formula for L(s, ψ) Mean value formula I Mean value formula I Evaluation of Ξ11 Evaluation of Ξ11 Evaluation of Ξ12 Evaluation of Ξ12 Proof of Proposition 2.4 Proof of Proposition 2.4 Proof of Proposition 2.6 Proof of Proposition 2.6 Evaluation of Ξ15 Evaluation of Ξ15 Approximation to Ξ14 Approximation to Ξ14 Mean value formula II Mean value formula II Evaluation of Φ1 Evaluation of Φ1 Evaluation of Φ2 Evaluation of Φ2 Evaluation of Φ3 Evaluation of Φ3 Proof of Proposition 2.5 Proof of Proposition 2.5 Appendix A. Some Euler products Appendix A. Some Euler products Appendix B. Some arithmetic sums Appendix B. Some arithmetic sums References References 3. The set Ψ1 Let ν(n) and υ(n) be given by respectively. It is easy to see that Lemma 3.1. Assume (A) holds. Then Lemma 3.1. Assume (A) holds. Then Proof . Let Proof which has the Euler product representation For σ ≥ σ0 > 0, by checking the cases χ(p) = ±1 and χ(p) = 0 respectively, it can be seen that and the implied constant depending on σ0. Thus φ(s) is analytic for σ > 1/2 and it satisfies for σ ≥ σ1 > 1/2, the implied constant depending on σ1. The left side of (3.2) is Lemma 3.2. Assume (A) holds. Then we have Lemma 3.2. Assume (A) holds. Then we have Proof . As the situation is analogous to Lemma 3.1 we give a sketch only. It can be verified that the function Proof is analytic for σ > 1/2 and it satisfies for σ ≥ σ1 > 1/2, the implies constant depending on σ1. Also, one can verify that This completes the proof. Lemma 3.3. For any s and any complex numbers c(n) we have Lemma 3.3. For any s and any complex numbers c(n) we have and and Proof . The first assertion follows by the orthogonality relation; the second assertion follows by the large sieve inequality. Proof Let By (3.1) we may write By Cauchy’s inequality and the first assertion of Lemma 3.3 we obtain Thus we conclude Lemma 3.4. The inequality Lemma 3.4. The inequality Write Assume that (A) holds. By Cauchy’s inequality, the second assertion of Lemma 3.2 and Lemma 3.1, Thus we conclude Lemma 3.5 Assume that (A) holds. The inequality Lemma 3.5 Assume that (A) holds. The inequality Let Assume that (A) holds. By Cauchy’s inequality, the first assertion of Lemma 3.2 and Lemma 3.1, Thus we conclude Lemma 3.6. Assume that (A) holds. The inequality Lemma 3.6. Assume that (A) holds. The inequality We are now in a position to give the definition of Ψ1: Let Ψ1 be the subset of Ψ such that ψ ∈ Ψ1 if and only if the inequalities (3.4), (3.5) and (3.6) simultaneously hold. Proposition 2.1 follows from Lemma 3.4, 3.5 and 3.6 immediately. This paper is available on arxiv under CC 4.0 license. This paper is available on arxiv under CC 4.0 license. available on arxiv

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