陈景润定理数学证明存在错误吗?

1973年，陈景润定理证明论文审稿人王元先生在稿件审查意见栏目里面写了“证明无误”字样，据此，陈景润的1+2论文得以正式发表。。

三十年之后，反动文人王晓明想“整死”陈景润。要是在陈景润定理证明中发现了错误，那么。反动文人王晓明就要“翻天”
了。

2007年，美国知名数学家Jie Wu发表学术论文，重新证明了陈景润定理，发现“证明无误”.

本文附件是：Jie Wu论文的全文，白纸黑字，可供读者参阅。

袁萌陈启清 2月7日

附件：陈景润定理最新第三方证明如下（英文原文）
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem
By Jie Wu（1925- ）
To cite this version: Jie Wu. Chen’s double sieve, Goldbach’s conjecture and the twin prime problem. Acta Arithmetica, Instytut Matematyczny PAN, 2004, 114 (no. 3), pp.215–273. <hal-00145781>
hal-00145781, version 1 - 11 May 2007
Acta Arithmetica 114 (2004), 215–273
Chen’s double sieve,
Goldbach’s conjecture and the twin prime problem
By Jie Wu
Abstrac（摘要）. For every even integer N, denote by D(N) and D1,2(N) the number of representations of N as a sum of two primes and as a sum of a prime and an integer having at most two prime factors, respectively. In this paper, we give a new upper bound for D(N) and a new lower bound for D1,2(N), which improve the corresponding results of Chen. We also obtain similar results for the twin prime problem
Contents
§ 1. Introduction ........ 1
§ 2. Preliminary lemmas ........ 5
§ 3. Chen’s double sieve ..... 9
§ 4. Weighted inequalities for sieve function ............. 15
§ 5. Functional inequalities between H and h ............. 22
§ 6. Proofs of Propositions ...... 28
§ 7. Proof of Theorem 1......... 30
§ 8. Proof of Theorem 3 ..............32 § 9. Chen’s system of weights . 33
§ 10. Proofs of Theorems 2 and 5 ............ 38
§ 11. Proof of Theorem 4 ............ 42 References .......... 47
§ 1. Introduction
Let (n) be the number of all prime factors of the integer n with the convention (1) = 0. For an even integer N > 4, we de ne D(N) as the number of representations of N as a sum of two primes: D(N) := |{p 6 N : (N p) = 1}|, where and in what follows, the letter p, with or without subscript, denotes a prime number. The well known Goldbach conjecture can be stated as D(N) > 1 for every even integer N > 4. A more precise version of this conjecture was proposed by Hardy & Littlewood [15]:
(1.1) D(N) ～2Θ(N) (N →∞),
2000 Mathematics Subject Classi cation: 11P32, 11N35, 11N05.
2 J. Wu
where
Θ(N) :=
CNN (logN)2
and CN := Y p|N,p>2
p 1 p 2 Y p>2
1 1 (p 1)2 . Certainly, the asymptotic formula (1.1) is extremely di cult. Although the lower bound problem remains open, the upper bound problem has a rich history. In 1949 Selberg [25] proved
(1.2) D(N) 6 {16 + o(1)}Θ(N) with the help of his well known λ2-upper bound sieve. By applying Linnik’s large sieve method, C.D. Pan [20] in 1964 improved 16 to 12. In 1966, Bombieri & Davenport [1] obtained 8 instead. Their proof is based on the linear sieve formulas and the mean value theorem of Bombieri– Vinogradov. It seems very di cult to prove (1.2) with a constant strictly less than 8 by the method in [1]. Firstly the linear sieve formulas (see Lemma 2.2 below) (1.3) XV (z)f logQ logz + error 6 S(A;P,z) 6 XV (z)F logQ logz + error
are the best possible in the sense that taking
A = Bν :={n : 1 6 n 6 x, (n) ≡ ν (mod2)} (ν = 1,2),
the upper and lower bounds in (1.3) are respectively attained by ν = 1 and ν = 2 (see [14], page 239). Secondly it is hopeless to try to improve the level of distribution 1 2 in Bombieri– Vinogradov’s theorem. In 1978, Chen [10] introduced a new idea in Selberg’s sieve and proved
(1.4) D(N) 6 7.8342Θ(N) (N > N0).
His sieve machine involves two variables and is quite complicated. Roughly speaking, for the sequence A = {N p : p 6 N} he introduced two new functions h(s) and H(s) such that (1.3) holds with f(s) + h(s) and F(s) H(s) in place of f(s) and F(s), respectively. The key innovation is to prove h(s) > 0 and H(s) > 0 via three weighted inequalities (see [10], (23), (47), (64), (90), and (91)). It is worth pointing out that he did not give complete proofs for these three inequalities. Among the three inequalities, the third one is the most complicated (with 43 terms) and it seems quite di cult to reconstruct a proof. Indeed, combining any one of these three inequalities with the Chen–Iwaniec switching principle (see [7] and [16]) leads to a constant less than 8. In order to derive a better result, Chen further introduced a very complicated iterative method. In 1980, C.B. Pan [19] applied essentially the rst weighted inequality of Chen to get 7.988. According to [22], Chen’s proof is very long and somewhat di cult to follow, but his idea is clear. In this paper, inspired by the ideas in [26] we shall rst try to give a more comprehensive treatment on Chen’s double sieve and prove an upper bound sharper than (1.4).
Theorem 1. For su ciently large N, we have
D(N) 6 7.8209Θ(N).
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 3
The improvement comes from a new weighted inequality (see Lemma 4.2 below), which is still quite complicated with 21 terms, but much simpler than Chen’s third and more powerful than his second and third inequality. Recently Cai & Lu [6] give another weighted inequality (with 31 terms), which is simpler but slightly weaker than Chen’s third.
One way of approaching the lower bound problem in (1.1) is to give a non trivial lower bound for the quantity D1,2(N) := |{p 6 N : (N p) 6 2}|. In this direction, Chen [7] proved, by his system of weights and switching principle, the following famous theorem: Every su ciently large even integer can be written as sum of a prime and an integer having at most two prime factors. More precisely he established
(1.5) D1,2(N) > 0.67Θ(N) (N > N0).
Then Halberstam & Richert [14] obtained a better constant 0.689 in place of 0.67 by a careful numerical calculation. As they indicated in [14], it would be interesting to know whether a more elaborate weighting procedure could be adapted to the purpose of (1.5). This might lead to numerical improvements and could be important. In 1978 Chen improved the constant 0.689 of Halberstam & Richert to 0.7544 and to 0.81 by two more elaborate systems of weights ([8], [9]). Very recently by improving Chen’s weighting device Cai and Lu [5] obtained 0.8285, which they described as being near to the limit of what could be obtained by the method employed. The second aim of this paper is to propose a larger constant.
Theorem 2. For su ciently large N, we have
D1,2(N) > 0.836Θ(N).
The proof of Theorem 2 is based on a modi ed version of Chen’s weights (see Lemma 9.2 below), the linear sieve and the mean value theorems of Pan & Ding [21] and of Fouvry [11].
A conjecture of the same nature is the twin prime problem, which can be stated as π2(x) :=|{p 6 x : (p + 2) = 1}|→∞ (x →∞). Similar to (1.1), Hardy & Littlewood [15] conjectured (1.6) π2(x) ～ Π(x) (x →∞), where Π(x) := Cx (logx)2 and C := 2Y p>2 1 1 (p 1)2 . The methods of Selberg, Pan, Bombieri & Davenport and Chen work in a similar way and give upper bounds of this type (1.7) π2(x) 6 {a + o(1)}Π(x), where the constant a is half of the corresponding constant in the Goldbach problem. Due to the sieve of Rosser–Iwaniec and mean value theorems of Bombieri, Fouvry, Friedlander and Iwaniec, the history of (1.7) is much richer than that of (1.2). We refer the reader to [26] and [6] for a detailed historical description of this problem. In particular Wu [26] obtained 3.418 in place of a + o(1) by placing these new mean value theorems in Chen’s method. The main di culty for applying these mean value theorems in [26] is to not destroy the fact that the error terms are a ected by well factorisable coe cients. Recently Cai & Lu [6] improved the constant 3.418 to 3.406. Our argument in proving Theorem 1 allows us to give a better result.
4 J. Wu
Theorem 3. For su ciently large x, we have
π2(x) 6 3.3996Π(x).
As an analogue of Theorem 2, Chen [7] proved that
(1.8) π1,2(x) > 0.335Π(x) (x > x0),
where
π1,2(x) :=|{p 6 x : (p + 2) 6 2}|. The constant 0.335 was improved by many mathematicians. Like (1.7), the history of (1.8) is much richer than that of (1.5). A detailed historical description on this problem can be found in the recent paper of Cai [3]. In particular he obtained 1.0974 in place of 0.335, which is an improvement of Wu’s constant 1.05 [26]. Here we can propose a slightly better result.
Theorem 4. For su ciently large x, we have
π1,2(x) > 1.104Π(x).
Remark 1. (i) Theorems 1 and 3 show that the principal terms in the linear sieve formulas can be improved in the special cases A = {N p : p 6 N} or A = {p + 2 : p 6 x} (see the end of Section 3). This seems to be interesting and important. Our argument is quite general, which works for all sequences satisfying the Chen–Iwaniec switching principle. (ii) Certainly we could obtain a better constant than 3.3996 in Theorem 3 if we used mean value theorems of ([11], Corollary 2), ([12], Lemma 6) and ([18], Proposition) as in the proof of Theorem 4. But the numerical computation involved would be quite complicated. The Chen theorem in short intervals was rst studied by Ross [23]. Let α ∈ (0,1) be a xed constant and de ne, for θ ∈ (0,1), x > 2 and even integer N > 4, D1,2(N,θ) := |{αN 6 p 6 αN + Nθ : (N p) 6 2}|, π1,2(x,θ) := |{x 6 p 6 x + xθ : (p + 2) 6 2}|. He proved (see [28]) that for θ > 0.98, N > N0(θ) and x > x0(θ),
D1,2(N,θ) Ξ(N,θ), π1,2(x,θ) Π(x,θ),
where
Ξ(N,θ) :=
Nθ (logN)2 Y p|N,p>2
p 1 p 2 Y p>2
1 1 (p 1)2
and
Π(x,θ) :=
2xθ (logx)2 Y p>2
1 1 (p 1)2 . The constant 0.98 was further improved to 0.973 by Wu [28], to 0.9729 by Salerno & Vitolo [24] and to 0.972 by Cai & Lu [4]. Our method allows us to take a smaller exponent.
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 5
Theorem 5. For every θ > 0.971, N > N0(θ) and x > x0(θ), we have
D1,2(N,θ) > 0.012Ξ(N,θ), π1,2(x,θ) > 0.006Π(x,θ).
Acknowledgement. The author would like to thank E. Fouvry for his generous help in writing this article, and the referee for his very careful reading of the manuscript.
§ 2. Preliminary lemmas This section is devoted to present the formula of the Rosser–Iwaniec linear sieve and some mean value theorem on the distribution of primes in arithmetic progressions, which will be needed later. Before stating these results, it is necessary to recall some de nitions. Let k be a positive integer and τk(n) the number of ways of writing n as the product of k positive integers. An arithmetical function λ(q) is of level Q and of order k if
λ(q) = 0 for q > Q and |λ(q)| 6 τk(q) for q > 1.
We say that λ is well factorable if for every decomposition Q = Q1Q2 (Q1,Q2 > 1) there exist two arithmetical functions λ1 and λ2 of level Q1,Q2 and of order k such that λ = λ1 λ2. Lemma 2.1. If λ′ is an arithmetical function of level Q′ (6 Q) and of order k′, then λ λ′ is well factorable of level QQ′ and of order k + k′. Let A be a nite sequence of integers and P a set of prime numbers. For z > 2, we put P(z) :=Qp<z,p∈P p and de ne the sieve function S(A;P,z) :=|{a ∈A : (a,P(z)) = 1}|. If d is a square-free integer with all its prime factors belonging to P, we denote by Ad the set of elements of A divisible by d and we write the following approximate formula
(2.1) |Ad| =
w(d) d
X + r(A,d),
where X > 1 is independent of d, and w(d) is a multiplicative function satisfying
(2.2) 0 6 w(p) < p for p ∈P.
We also de ne
V (z) := Y p<z
1 w(p) p
and suppose that there exists an absolute constant K > 1 such that
(2.3)
V (z1) V (z2)
6
logz2 logz1 1 + K logz1 (z2 > z1 > 2).
The formula of the Rosser–Iwaniec linear sieve [17] is stated as follows.
6 J. Wu
Lemma 2.2. Let 0 < ε < 1 8 and 2 6 z 6 Q1/2. Under the assumptions (2.1), (2.2) and (2.3), we have (2.4) S(A;P,z) 6 XV (z) F logQ logz + E +X l<L X q|P(z) λ+ l (q)r(A,q) and (2.5) S(A;P,z) > XV (z) f logQ logz + E X l<L X q|P(z) λ l (q)r(A,q). In these formulas, L depends only on ε, the λ± l are well factorable functions of order 1 and of level Q, and E ε + ε 8eK/(logQ)1/3. The functions F,f are de ned by
(2.6)
F(u) = 2eγ/u, f(u) = 0 (0 < u 6 2), (uF(u))′ = f(u 1), (uf(u))′ = F(u 1) (u > 2), where γ is Euler’s constant.
As usual, we denote by μ(q) Mo¨bius’ function, (q) Euler’s function and ν(q) the number of distinct prime factors of q. De ne π(y;q,a,m) := X mp6y mp≡a(modq) 1, li(y) :=Z y 2 dt logt and E0(y;q,a,m) := π(y;q,a,m) li(y/m) (q) . The next lemma is due to Pan & Ding [21], which implies Bombieri–Vinogradov’s theorem. Here we state it in the form of ([22], Corollary 8.12). Lemma 2.3. Let f(m) 1 and α ∈ (0,1]. Let r1(y) be a positive function depending on x and satisfying r1(y) xα, y 6 x. Let r2(m) be a positive function depending on x and y, and satisfying mr2(m) x, m 6 xα, y 6 x. Then for every A > 0, there exists a constant B = B(A) > 0 such that Xq 6√x/(log x)B μ(q)23ν(q) max y6x max (a,q)=1

X m6x1 α (m,q)=1 f(m)E0(y;q,a,m)

x (logx)A , Xq 6√x/(log x)B μ(q)23ν(q) max y6x max (a,q)=1

X m6x1 α (m,q)=1 f(m)E0(mr1(y);q,a,m)

x (logx)A , Xq 6√x/(log x)B μ(q)23ν(q) max y6x max (a,q)=1

X m6x1 α (m,q)=1 f(m)E0(mr2(m);q,a,m)

x (logx)A .
In order to prove Theorem 5, it is necessary to generalize the mean value theorem of Pan & Ding in short intervals. Such a result was established by Wu ([27], theorem 2).
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 7
Lemma 2.4. Let f(m) 1, ε be an arbitrarily small positive number and de ne H(y,h,q,a,m) := π(y + h;q,a,m) π(y;q,a,m) li((y + h)/m) li(y/m) (q)
.
Then for any A > 0, there exists a constant B = B(A) > 0 such that X q6Q μ(q)23ν(q) max (a,q)=1 max h6xθ max x/2<y6x

X m6M,(m,q)=1 f(m)H(y,h,q,a,m)

xθ (logx)A
for x > 10, 3 5 + ε 6 θ 6 1, Q = xθ 1/2/(logx)B and M = x(5θ 3)/2 ε. In the proofs of Theorem 3 and 4, we shall need some mean value theorems with well factorable or almost well factorable coe cients. Let M > 1, N > 1 and X := MN. Let {αm} and {βn} be two sequences of order k supported in [M,2M] and [N,2N] respectively. We also suppose the conditions below: (i) For all B, the equality Xn ≡n0(modk),(n,d)=1 βn = 1 (k) X (n,dk)=1 βn + OB,kNτk(d)/(log2N)B holds for d > 1, k > 1 and (k,n0) = 1. (ii) If n has a prime factor p with p < exp{(loglogn)2}, then βn = 0. The following result is an immediate consequence of Corollary 2 of [11], Lemma 6 of [12] and the proposition of [18].
Lemma 2.5. Under the conditions (i) and (ii) above, for any A and for any ε > 0 we have X( q,a)=1 λ(q) X mn≡a(modq) αmβn 1 (q) X (mn,q)=1 αmβn ε,A X (logX)A uniformly for |a| 6 (logX)A and ν := logN/logX (ε 6 ν 6 1 ε). Here λ(q) is a well factorisable function of order 1 and of level Q := Xθ(ν) ε, where θ(ν) is given by
θ(ν) =

6 5ν 10 for ε 6 ν 6 1 15, 1 2 + ν for 1 15 6 ν 6 1 10, 5 2ν 8 for 1 10 6 ν 6 3 14, 3+2ν 6 for 3 14 6 ν 6 1 4, 2 ν 3 for 1 4 6 ν 6 2 7, 2+ν 4 for 2 7 6 ν 6 2 5, 1 ν for 2 5 6 ν 6 1 2, 1 2 for 1 2 6 ν 6 1 ε. Proof. The value (6 5ν)/10 in [ε,1/15] comes from Corollary 2 (ii) of [11]. The intervals [1/15,1/10] and [1/10,3/14] follow from the proposition of [18] by decomposing λ = λ1 λ2 with Q1 = Q = x1/2 ε, Q2 = R = Nx ε
and
Q1 = Q = x5/8 εN 5/4, Q2 = R = Nx ε,
respectively. The remaining case is Lemma 6 of [12].
8 J. Wu
The next lemma is Corollary 2 (i) of [11]. This is the rst result, which is valid uniformly for |a| 6 X and has the level of distribution > 1 2. Lemma 2.6. Under the conditions (i) and (ii) above, for any A and for any ε > 0 we have X( q,a)=1 λ(q) X mn≡a(modq) αmβn 1 (q) X (mn,q)=1 αmβn ε,A X (logX)A uniformly for |a| 6 X and ε 6 ν := logN/logX 6 1 10. Here λ(q) is a well factorisable function of order 1 and of level Q := X5(1 ν)9 ε.
As usual de ne
π(y;q,a) := X p6y,p≡a(modq)
1.
The following result is due to Bombieri, Friedlander & Iwaniec ([2], theorem 10).
Lemma 2.7. Let λ be a well factorable function of order k and of level Q = x4/7 ε. For any ε > 0 and any A, we have uniformly for x > 3 and |a| 6 (logx)A, X( q,a)=1 λ(q) π(x;q,a) li(x) (q) ε,k,A x (logx)A .
When we use the weighted inequality, some coe cients are merely “almost well factorable”. So we need the following results, due to Fouvry & Grupp ([13], theorem 2 and the corollary).
Lemma 2.8. Let λ be a well factorable function of level Q1 and of order k, ξ an arithmetical function satisfying the conditions |ξ(q2)| 6 logx and ξ(q2) = 0 (q2 > Q2) and let Λ be the von Mangoldt function. Then we have for any integer a, any ε > 0 and any A > 0, X( q1q2,a)=1 λ(q1)ξ(q2) π(x;q1q2,a) li(x) (q1q2) a,ε,k,A x (logx)A ,
so long as one of the following three conditions is true:
Q2 6 Q1, Q1Q2 6 x4/7 ε,(C .1) Q2 > Q1, Q1Q6 2 6 x2 ε,(C .2) ξ(q) = Λ(q), Q1Q2 6 x11/20 ε, Q2 6 x1/3 ε.(C .3)
The next two lemmas also are useful when we apply the switching principle.
Lemma 2.9 ([26], Lemma 7). Let λ be a well factorable function of level Q := x4/7 ε and of order k. Let η > 0 and {εi}16i6r be real numbers such that εi > η, ε1 + ε2 + + εr = 1.
Then for any integer a, any ε > 0 and any A > 0, we have X( q,a)=1 λ(q) X p1 pr≡a(modq) xεi<pi62xεi (16i6r) 1 1 (q) X (p1 pr,q)=1 (xεi<pi62xεi (16i6r)
1 a,ε,k,A x (logx)A
.
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 9
Lemma 2.10 ([26], Lemma 12). Let x > 2 and y = x1/u. Then X n6x p|n p>y 1 = x logy ω(u) + O x (logy)2 , where ω(u) is Buchstab’s function de ned by ω(u) = 1/u (1 6 u 6 2) and uω(u) ′ = ω(u 1) (u > 2).Moreover we have ω(u) 6 0.561522 (u > 3.5) and ω(u) 6 0.567144 (u > 2).
§ 3. Chen’s double sieve We shall sieve the sequence
A := {N p : p 6 N}. Let δ > 0 be a su ciently small number and k ∈ Z. Put Q := N1/2 δ, d := Q/d, L := logN, Wk := Nδ1+k. Let be a real number with 1 +L 4 6 < 1 + 2L 4. We put P(N) := {p : (p,N) = 1} and denote by π[Y,Z) the characteristic function of the set P(N)∩[Y,Z). For k ∈ Z+ and N > 2, let Uk(N) be the set of all arithmetical functions σ which can be written as the form σ = π[V1/ ,V1) π[Vi/ ,Vi), where i is an integer with 0 6 i 6 k, and V1,...,Vi are real numbers satisfying
(3.1)

V 2 1 6 Q, V1V 2 2 6 Q, V1 Vi 1V 2 i 6 Q, V1 > V2 > > Vi > Wk. By convention, σ is the characteristic function of the set {1} if i = 0. From this de nition and Lemma 2.1, we see immediately the following result. Lemma 3.1. (i) We have Uk(N) Uk+1(N) for k ∈ Z+. (ii) Let σ = π[V1/ ,V1) π[Vi/ ,Vi) ∈ Uk(N). Then σ is well factorable of level V := V1 Vi and of order i. If λ is well factorable of level Q/V and of order 1, then σ λ is well factorable of level Q and of order k + 1.
Let F and f be de ned as in (2.6) and let (3.2) A(s) := sF(s)/2eγ and a(s) := sf(s)/2eγ,
We introduce the notation Φ(N,σ,s) :=X d σ(d)S(Ad;P(dN),d1/s), Θ(N,σ) := 4li(N)X d
σ(d)CdN (d)logd
.
For k ∈ Z+, N0 > 2 and s ∈ [1,10], we de ne Hk,N0(s) and hk,N0(s) as the supremum of h > ∞ such that for all N > N0 and σ ∈ Uk(N) one has the following inequalities Φ(N,σ,s) 6 {A(s) h}Θ(N,σ) and Φ(N,σ,s) > {a(s) + h}Θ(N,σ) respectively. From this de nition, we deduce immediately the following result.
10 J. Wu Lemma 3.2. For k ∈ Z+,N > N0,s∈ [1,10] and σ ∈ Uk(N), we have (3.3) Φ(N,σ,s) 6 {A(s) Hk,N0(s)}Θ(N,σ) and
(3.4) Φ(N,σ,s) > {a(s) + hk,N0(s)}Θ(N,σ). Obviously Hk,N0(s),hk,N0(s) are decreasing on N0, and they are also decreasing on k by Lemma 3.1. Hence we can write
Hk(s) := lim N0→∞
Hk,N0(s),
H(s) := lim k→∞
Hk(s),
hk(s) := lim N0→∞
hk,N0(s),
h(s) := lim k→∞
hk(s).
Lemma 3.3. For N > 2 and σ = π[V1/ ,V1) π[Vi/ ,Vi) ∈ Uk(N), we have L 5k δ,k X d σ(d)/d δ,k 1,(3 .5) X d σ(d) δ,k V1 Vi,(3 .6) Θ(N,σ) δ,k N/L5k+2.(3 .7) Proof. Let σ = π[V1/ ,V1) π[Vi/ ,Vi) ∈ Uk(N). We have (3.8) X d σ(d) d = Y 16j6i X pj∈P(N)∩[Vj/ ,Vj) 1 pj . The prime number theorem of the formPp6x 1 = li(x) + O(xe 2(log x)1/2) implies Xp j∈P(N)∩[Vj/ ,Vj) 1 pj = X Vj/ 6pj<Vj 1 pj X Vj/ 6pj<Vj,pj|N 1 pj = log logVj log(Vj/ ) + O e log1/2(Vj/ ) + L Vj logL . Therefore our choice of and (3.1) give us (3.9) X pj∈P(N)∩[Vj/ ,Vj) 1/pj δ,k L 5. Now (3.5) follows from (3.8) and (3.9). Since σ(d) 6= 0 implies d 6 V1 Vi, the second inequality in (3.5) implies (3.6). Noticing Θ(N,σ) NL 2Pd σ(d)/d, we obtain (3.7) by the rst inequality in (3.5). Proposition 1. For k ∈ Z+ and s ∈ [1,10], we have Hk(s) > 0 and hk(s) > 0.
Proof. We shall prove only the rst inequality. The second one can be treated similarly. Let σ = π[V1/ ,V1) π[Vi/ ,Vi) ∈ Uk(N). We use Lemma 2.2 with
X =
li(N) (d)
, w(p) = p/(p 1) if p∈P(N), 0 otherwise
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 11 to estimate σ(d)S(Ad;P(dN),d1/s). By Merten’s formula and (3.1), we can infer that for any ε > 0 (3.10) V (d1/s) = {1 + Oδ,k(ε)}2sCdN eγ logd .
By using Lemma 2.2 and (3.10), we deduce σ(d)S(Ad;P(dN),d1/s) 6 4li(N) σ(d)CdN (d)logd A log(Q/V ) logd1/s + Oδ,k(ε) (3 .11) +X l<L σ(d) X q|P(d1/s) λ+ l (q)r(Ad,q), where λ+ l (q) is well factorable of level Q/V with V := V1 Vi and of order 1. If σ(d) 6= 0, we have d ∈ [V/ i,V ], which implies 0 6 logV logd 6 ilog 6 2kL 4. From this we deduce that Alog(Q/V )/logd1/s = A(s)+Oδ,k(ε). Inserting (3.11) and summing over d, we obtain
(3.12) Φ(N,σ,s) 6 {A(s) + Oδ,k(ε)}Θ(N,σ) + R, where R :=X l<LX d σ(d) X q|P(d1/s) λ+ l (q)r(Ad,q). Let q | P(d1/s). It is clear that μ(q)2 = 1 and (Nd,q) = 1. Thus we have r(Ad,q) = |Adq| li(N)/ (dq) = π(N;dq,N) li(N)/ (dq). Hence we can see, by using Lemmas 3.1(ii) and 2.3, that R ε X q6Q τk+1(q)|π(N;dq,N) li(N)/ (dq)|(3 .13) δ,k,ε N/L5k+3.
From (3.7), (3.12) and (3.13), we deduce
Φ(N,σ,s) 6 {A(s) + Oδ,k(ε)}Θ(N,σ), which implies, by the de nition of Hk,N0(s), for any ε > 0 and su ciently large N0
Hk,N0(s) > Oδ,k(ε). First making N0 →∞ and then ε → 0, we obtain Hk(s) > 0. Proposition 2. For 2 6 s 6 s′ 6 10, we have h(s) > h(s′) +Z s′ 1 s 1 H(t) t dt and H(s) > H(s′) +Z s′ 1 s 1 h(t) t dt.
12 J. Wu
Proof. We shall only prove the rst inequality as the second one can be established in the same way. Let k > 0 and σ = π[V1/ ,V1) π[Vi/ ,Vi) ∈ Uk(N). By Buchstab’s identity, we write (3.14) Φ(N,σ,s) = Φ(N,σ,s′) X d σ(d) X d1/s′ 6p<d1/s S(Adp;P(dN),p).
Next we shall give an upper bound for the last double sums S. The idea is to prove that the characteristic function of dp belongs to Uk+1(N). Thus S can be estimated by a function Hk+1,N0. We put V := V1 Vi, V := Q/V and αj := V 1/s′ j. Let r be the integer satisfying αr 6 V 1/s < αr+1. Noticing that σ(d) 6= 0 V 1/s′ 6 d1/s′ and V 1/s 6 d1/s, we deduce S 6X d σ(d) X α06p<αr S(Adp;P(dN),p) + R1(3 .15) = X 16j6rX d,p σ(d)π[αj 1,αj)(p)S(Adp;P(dpN),(dp)1/s ) + R1,
where s := logd/logp 1 and R1 :=X d
σ(d) X αr6p<d1/s S(Adp;P(dN),p).
We would prove that σ π[αj 1,αj) ∈ Uk+1(N). It su ces to verify that V1,V2,...,Vi,αj satisfy (3.1) for j 6 r. If Vi > αj, then V1V2 Viα2 j 6 V V 2/s = Q2/sV 1 2/s 6 Q and αj > V 1/s′ > V 1/s′ i > W1/s′ k > Wk+1. If V1 > > Vl > αj > Vl+1 > > Vi, we have V1 VlαjVl+1 V 2 n 6 V α2 j 6 V V 2/s 6 Q for l < n 6 i. Thus σ π[αj 1,αj) ∈ Uk+1(N). Since s depends on d and p, we replace it by a suitable quantity independent of d and p such that we can use (3.3) with Hk+1,N0. For this we introduce s1 := log(V /αj)/logαj, s2 := log(V /αj i 1)/logαj 1. Noticing that σ(d)π[αj 1,αj)(p) 6= 0 s1 6 s 6 s2, we deduce from (3.15) that S 6 X 16j6rX d,p σ(d)π[αj 1,αj)(p)S(Adp;P(dpN),(dp)1/s1) + R1 + R2
where R2 := X 16j6rX d,p
σ(d)π[αj 1,αj)(p) S(Adp;P(dpN),(dp)1/s ) S(Adp;P(dpN),(dp)1/s1) .
Now we can use (3.3) in Lemma 3.2 to write S 6 X 16j6r {A(s1) Hk+1,N0(s1)}Θ(N,σ π[αj 1,αj)) + R1 + R2 6 4li(N)X d σ(d)CdN (d)logd X α06p<αr A(s ) Hk+1,N0(s ) (p 2)(1 logp/logd) + R1 + R2 6 4li(N)X d σ(d)CdN (d)logd X d1/s′ 6p<d1/s A(s ) Hk+1,N0(s ) (p 2)(1 logp/logd) + R1 + R2 + R3,
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 13
where we have used the fact that A(s) Hk+1,N0(s) is increasing on s, and the notation R3 := 4li(N)X d σ(d)CdN (d)logd X V 1/s′ 6p<d1/s′ A(s ) Hk+1,N0(s ) (p)(1 logp/logd) .
Applying the prime number theorem, an integration by parts shows that
Xd 1/s′ 6p<d1/s
A(s ) Hk+1,N0(s ) (p 2)(1 logp/logd)
=Z s′ 1 s 1
A(t) Hk+1,N0(t) t
dt + Oδ,k(ε).
Hence (3.16) S 6 Z s′ 1 s 1
A(t) Hk+1,N0(t) t
dt + Oδ,k(ε) Θ(N,σ) + R1 + R2 + R3. It remains to estimate R1,R2,R3. Observing that σ(d) 6= 0 V/ i 6 d < V , we have d1/s 6 V 1/s i/s. Thus log(logd1/s/logαr) 6 log(1 + log 1+i/s/log(V 1/s/ )) δ,k L 5. By using the prime number theorem and the previous estimate, we have R1 X d σ(d) X αr6p<d1/s N/dp(3 .17) NL 5X d σ(d)/ (d) δ,k Θ(N,σ)/L3.
Similarly we can show that
(3.18) R3 δ,k Θ(N,σ)/L3.
By the de nition of R2, we easily see that R2 X d σ(d) X α06p<αr X (dp)1/s 6p′<(dp)1/s1
N/dpp′.
Using a similar preceding argument, we can show that
R2 δ,k
N L4 X d
σ(d) (d)logd X V 1/s′ 6p<V 1/s
1 p
(3.19)
δ,k
Θ(N,σ) L3
.
Combining (3.16)–(3.19), we obtain the desired upper bound, for N > N0, (3.20) S 6 Z s′ 1 s 1 A(t) Hk+1,N0(t) t dt + Oδ,k(ε) Θ(N,σ). Inserting it in (3.14), estimating the rst sum on the right-hand side of (3.14) by (3.9) and noticing the relation a(s′) a(s) =Z s′ s A(t 1)dt,
14 J. Wu
we nd that, for N > N0(ε,δ,k), Φ(N,σ,s) >na(s) + hk,N0(s′) +Z s′ 1 s 1
Hk+1,N0(t) t
dt + Oδ,k(ε)oΘ(N,σ). From the de nition of hk,N0(s), we deduce that, for any ε > 0 and for su ciently large N0, hk,N0(s) > hk,N0(s′) +Z s′ 1 s 1 Hk+1,N0(t) t dt + Oδ,k(ε). Taking N0 →∞ and then ε → 0, we obtain hk(s) > hk(s′) +Z s′ 1 s 1 Hk+1(t) t dt which implies the required inequality. This completes the proof.
Corollary 1. The function H(s) is decreasing on [1,10]. The function h(s) is increasing on [1,2] and is decreasing on [2,10].
Proof. According to the de nition, we easily see that Hk,N0(s) is decreasing on [1,3] since A(s) = 1 for 1 6 s 6 3. Thus H(s) is also decreasing on [1,3]. When 3 6 s 6 10, the required result follows immediately from Propositions 2 and 1. Similarly the de nition of hk,N0(s) and the fact that a(s) = 0 for 1 6 s 6 2 show that h(s) is increasing on [1,2]. Propositions 2 and 1 imply that h(s) is decreasing on [2,10]. This concludes the proof. The central results in this section are Propositions 3 and 4 below. Before stating it, it is necessary to introduce some notation. Let 1 6 s 6 3 6 s′ 6 5 and s 6 κ3 6 κ2 6 κ1 6 s′. De ne α1 := κ1 2, α4 := s′ s′/κ2 1, α7 := s′ s′/κ1 s′/κ3, α2 := s′ 2, α5 := s′ s′/κ3 1, α8 := s′ s′/κ1 s′/κ2, α3 := s′ s′/s 1, α6 := s′ 2s′/κ2, α9 := κ1 κ1/κ2 1. Let 1[a,b](t) be the characteristic function of the interval [a,b]. We put σ(a,b,c) :=Z b a log c t 1 dt t , σ0(t) := σ(3,t + 2,t + 1) 1 σ(3,5,4) . We can prove that H(s) satis es some functional inequalities. Proposition 3. For 5 > s′ > 3 > s > 2 and s′ s′/s > 2, we have (3.21) H(s) > Ψ1(s) +Z 3 1 H(t)Ξ1(t;s)dt, where Ψ1(s) is de ned as in Lemma 5.1 below and Ξ1(t;s) = Ξ1(t;s,s′) is given by
Ξ1(t;s) :=
σ0(t) 2t
log 16 (s 1)(s′ 1)
+
1[α2,3](t) 2t
log (t + 1)2 (s 1)(s′ 1)
+
1[α3,α2](t) 2t
log t + 1 (s 1)(s′ 1 t) .
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 15
Proposition 4. Let 5 > s′ > 3 > s > 2 and s 6 κ3 < κ2 < κ1 6 s′ satisfy
s′ s′/s > 2, 1 6 αi 6 3 (1 6 i 6 9), α1 < α4, α5 < α8.
Then we have (3.22) H(s) > Ψ2(s) +Z 3 1
H(t)Ξ2(t;s)dt,
where Ψ2(s) is de ned as in Lemma 5.2 below, and Ξ2(t;s) = Ξ2(t;s,s′,κ1,κ2,κ3) is given by
Ξ2(t;s) :=
σ0(t) 5t
log 1024 (s 1)(s′ 1)(κ1 1)(κ2 1)(κ3 1)
+
1[α2,3](t) 5t
log (t + 1)5 (s 1)(s′ 1)(κ1 1)(κ2 1)(κ3 1)
+
1[α9,α1](t) 5t
log t + 1 (κ2 1)(κ1 1 t)
+
1[α5,α2](t) 5t
log t + 1 (κ3 1)(s′ 1 t)
+
1[α3,α2](t) 5t
log t + 1 (s 1)(s′ 1 t)
+
1[α1,α2](t) 5t
log (t + 1)2 (κ1 1)(κ2 1)
+
1[α7,α5](t) 5t(1 t/s′)
log s′2 (κ1s′ s′ κ1t)(κ3s′ s′ κ3t)
+
1[α5,α8](t) 5t(1 t/s′)
log s′(s′ 1 t) κ1s′ s′ κ1t
+
1[α6,α8](t) 5t(1 t/s′)
log s′ κ2s′ s′ κ2t
+
1[α8,α2](t) 5t(1 t/s′)
log(s′ 1 t).
We shall prove these two propositions in Section 6. It is easy to see that Ξi(t;s) is positive and that for s ∈ [1,3) there exist parameters s′,κi such that Ψi(s) > 0. Therefore H(s) > 0 for s ∈ [1,3) and then Proposition 2 implies that h(s) > 0 for s ∈ [1,3). In Sections 7 and 8, we shall give numeric solution of (3.20) and (3.21), and prove Theorems 1 and 3.
§ 4. Weighted inequalities for sieve function The aim of this section is to present two weighted inequalities for sieve function. The rst is essentially due to Chen ([10], (23)). The second is new, which is not only much simpler than the third weighted inequality of Chen ([10], (64), (90) and (91)) but also more powerful. Lemma 4.1. Let 1 6 s < s′ 6 10. For N > 2, k > 0 and σ ∈ Uk(N), we have 2Φ(N,σ,s) 6X d σ(d)( 1 2 + 3) + Oδ,k(N1 η),
16 J. Wu
where η = η(δ,k) > 0 and i = i(d) is given by 1 := 2S(Ad;P(dN),d1/s′), 2 := X d1/s′ 6p<d1/s (p,N)=1 S(Adp;P(dN),d1/s′), 3 := XXX d1/s′ 6p1<p2<p3<d1/s (p1p2p3,N)=1 S(Adp1p2p3;P(dp1N),p2).
Proof. By the Buchstab identity, we have 2S(Ad;P(dN),d1/s) = 1 2 X d1/s′ 6p<d1/s (p,N)=1 S(Adp;P(dN),p),(4 .1) Xd 1/s′ 6p1<d1/s (p1,N)=1 S(Adp1;P(dN),p1) = 0 + XX d1/s′ 6p16p3<d1/s (p1p3,N)=1 S(Adp1p3;P(dp1N),p3),(4 .2) Xd 1/s′ 6p3<d1/s (p3,N)=1 S(Adp3;P(dN),p3) = 2 XX d1/s′ 6p1<p3<d1/s (p1p3,N)=1 S(Adp1p3;P(dN),p1),(4 .3)
where
0 := X d1/s′ 6p1<d1/s (p1,N)=1
S(Adp1;P(dN),d1/s).
Inserting (4.2)–(4.3) into (4.1), dropping the term 0 (which is non-negative) and replacing p1 6 p3 by p1 < p3, we nd that 2S(Ad;P(dN),d1/s) 6 1 2 + 1,
where
1 := XX d1/s′ 6p1<p3<d1/s (p1p3,N)=1 S(Adp1p3;P(dN),p1) S(Adp1p3;P(dp1N),p3) (4 .4) = XXX d1/s′ 6p16p2<p3<d1/s (p1p2p3,N)=1 S(Adp1p2p3;P(dp1N),p2).
By the inequality S(Adp2 1p3;P(dN),p1) N/dp2 1p3 and the fact that d > Wk, we easily see XXd 1/s′ 6p1<p3<d1/s S(Adp2 1p3;P(dN),p1) XX d1/s′ 6p1<p3<d1/s N/dp2 1p3 δ,k N1 η/d
for some η = η(δ,k) > 0. Inserting it in (4.4), we obtain that
(4.5) 1 = 3 + Oδ,k(N1 η/d).
Finally we complete the proof with (3.5).
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 17 Lemma 4.2. Let 1 6 s 6 κ3 < κ2 < κ1 6 s′ 6 10. For N > 2, k > 0 and σ ∈ Uk(N), we have 5Φ(N,σ,s) 6X d σ(d)(Γ1 Γ2 Γ3 Γ4 + Γ5 + + Γ21) + Oδ,k(N1 η),
where η = η(δ,k) > 0 and Γi = Γi(d) is given by
Γ1 := 4S(Ad;P(dN),d1/s′) + S(Ad;P(dN),d1/κ1), Γ2 := X d1/s′ 6p<d1/s (p,N)=1 S(Adp;P(dN),d1/s′), Γ3 := X d1/s′ 6p<d1/κ2 (p,N)=1 S(Adp;P(dN),d1/s′), Γ4 := X d1/s′ 6p<d1/κ3 (p,N)=1 S(Adp;P(dN),d1/s′), Γ5 := XX d1/s′ 6p1<p2<d1/κ2 (p1p2,N)=1 S(Adp1p2;P(dN),d1/s′), Γ6 := X X d1/s′ 6p1<d1/κ1,d1/κ2 6p2<d1/κ3 (p1p2,N)=1 S(Adp1p2;P(dN),d1/s′), Γ7 := XX d1/s′ 6p1<p2<d1/κ1 (p1p2,N)=1 S(Adp1p2;P(dN),p1), Γ8 := X X d1/s′ 6p1<d1/κ1 6p2<d1/κ2 (p1p2,N)=1 S(Adp1p2;P(dN),p1), Γ9 := X X X d1/κ16p1<p2<p3<d1/κ3 (p1p2p3,N)=1 S(Adp1p2p3;P(dN),p2), Γ10 := X X X d1/κ16p1<p2<d1/κ2 6p3<d1/s (p1p2p3,N)=1 S(Adp1p2p3;P(dN),p2), Γ11 := X X X d1/κ16p1<d1/κ2 6p2<p3<d1/κ3 (p1p2p3,N)=1 S(Adp1p2p3;P(dN),p2), Γ12 := X X X d1/s′ 6p1<p2<d1/κ1,d1/κ3 6p3<d1/s (p1p2p3,N)=1 S(Adp1p2p3;P(dN),p2), Γ13 := X X X d1/s′ 6p1<d1/κ1 6p2<d1/κ26p3<d1/s (p1p2p3,N)=1 S(Adp1p2p3;P(dN),p2), Γ14 := X X X d1/s′ 6p1<d1/κ1,d1/κ2 6p2<p3<d1/s (p1p2p3,N)=1 S(Adp1p2p3;P(dN),p2),
18 J. Wu Γ15 := X X X d1/κ1 6p1<d1/κ2 6p2<d1/κ36p3<d1/s (p1p2p3,N)=1
S(Adp1p2p3;P(dN),p2),
Γ16 := X X X X d1/κ2 6p1<p2<p3<p4<d1/κ3 (p1p2p3p4,N)=1
S(Adp1p2p3p4;P(dN),p3),
Γ17 := X X X X d1/κ2 6p1<p2<p3<d1/κ3 6p4<d1/s (p1p2p3p4,N)=1
S(Adp1p2p3p4;P(dN),p3),
Γ18 := X X X X d1/κ2 6p1<p2<d1/κ36p3<p4<d1/s (p1p2p3p4,N)=1
S(Adp1p2p3p4;P(dN),p3),
Γ19 := X X X X d1/κ1 6p1<d1/κ2,d1/κ3 6p2<p3<p4<d1/s (p1p2p3p4,N)=1
S(Adp1p2p3p4;P(dN),p3),
Γ20 := X X X X X d1/κ2 6p1<d1/κ3 6p2<p3<p4<p5<d1/s (p1p2p3p4p5,N)=1
S(Adp1p2p3p4p5;P(dN),p4),
Γ21 := X X X X X X d1/κ3 6p1<p2<p3<p4<p5<p6<d1/s (p1p2p3p4p5p6,N)=1
S(Adp1p2p3p4p5p6;P(dN),p5).
Proof. Let S := S(Ad;P(dN),d1/s). By using the Buchstab identity, we have 2S = 2S(Ad;P(dN),d1/s′) X d1/s′ 6p<d1/s (p,N)=1 S(Adp;P(dN),p)(4 .6) Γ3 + XX d1/s′ 6p1<p2<d1/κ2 (p1p2,N)=1 S(Adp1p2;P(dN),p1) X d1/κ2 6p<d1/s (p,N)=1 S(Adp;P(dN),p) =: 2S(Ad;P(dN),d1/s′) E1 Γ3 + D′ 1 E2.
We can also write, always by Buchstab’s identity, S = S(Ad;P(dN),d1/s′) X d1/s′ 6p<d1/κ3 (p,N)=1 S(Adp;P(dN),p)(4 .7) X d1/κ36p<d1/s (p,N)=1 S(Adp;P(dN),p).
But we have Xd 1/s′ 6p<d1/κ3 (p,N)=1
S(Adp;P(dN),p) = Γ4 Γ7 Γ8 X X d1/κ16p1<p2<d1/κ3 (p1p2,N)=1
S(Adp1p2;P(dN),p1)
Γ6 + X X X d1/s′ 6p1<p2<d1/κ1,d1/κ2 6p3<d1/κ3 (p1p2p3,N)=1
S(Adp1p2p3;P(dN),p1).
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 19
Inserting these relations into (4.7), it yields that S = S(Ad;P(dN),d1/s′) Γ4 + Γ6 + Γ7 + Γ8(4 .8) X d1/κ3 6p<d1/s (p,N)=1 S(Adp;P(dN),p) + X X d1/κ1 6p1<p2<d1/κ3 (p1p2,N)=1
S(Adp1p2;P(dN),p1)
X X X d1/s′ 6p1<p2<d1/κ1,d1/κ2 6p3<d1/κ3 (p1p2p3,N)=1
S(Adp1p2p3;P(dN),p1)
=: S(Ad;P(dN),d1/s′) Γ4 + Γ6 + Γ7 + Γ8 E3 + D2 E4. Similar to (4.8), we can prove that S = S(Ad;P(dN),d1/s′) Γ2 + Γ5 X X X d1/s′ 6p1<p2<p3<d1/κ2 (p1p2p3,N)=1 S(Adp1p2p3;P(dN),p1)(4 .9) +n X X d1/s′ 6p1<d1/κ2 6p2<d1/s (p1p2,N)=1 + X X d1/κ26p1<p2<d1/s (p1p2,N)=1 oS(Adp1p2;P(dN),p1) =: S(Ad;P(dN),d1/s′) Γ2 + Γ5 E5 + D′′ 1 + D′′′ 1 . Finally we write (4.10) S = S(Ad;P(dN),d1/κ1) n X d1/κ1 6p<d1/κ3 (p,N)=1 + X d1/κ36p<d1/s (p,N)=1 oS(Adp;P(dN),p).
For p1 < d1/κ3 < d1/s, we have S(Adp;P(dN),p) > X p6p1<d1/κ3 (p1,N)=1
S(Adpp1;P(dN),p1) + X d1/κ3 6p1<d1/s (p1,N)=1
S(Adpp1;P(dN),p1).
This implies Xd 1/κ1 6p<d1/κ3 (p,N)=1
S(Adp;P(dN),p) > X X d1/κ1 6p1<p2<d1/κ3 (p1p2,N)=1
S(Adp1p2;P(dN),p2)
+ X X d1/κ1 6p1<d1/κ3 6p2<d1/s (p1p2,N)=1
S(Adp1p2;P(dN),p2).
Inserting it into (4.10), we obtain S 6 S(Ad;P(dN),d1/κ1) X X d1/κ1 6p1<p2<d1/κ3 (p1p2,N)=1 S(Adp1p2;P(dN),p2)(4 .11) X X d1/κ1 6p1<d1/κ3 6p2<d1/s (p1p2,N)=1 S(Adp1p2;P(dN),p2) X d1/κ3 6p<d1/s (p,N)=1 S(Adp;P(dN),p) =: S(Ad;P(dN),d1/κ1) E6 E7 E3.
20 J. Wu
Now by adding up the inequalities (4.6), (4.8), (4.9) and (4.11) and by noticing the estimate D2 E6 6 Γ9 + Oδ,k(N1 η/d), we get (4.12) 5S 6 Γ1 Γ2 Γ3 Γ4 + Γ5 + + Γ9 + 2 + Oδ,k(N1 η/d)
where
2 := D1 E1 E2 2E3 E4 E5 E7
and
D1 := D′ 1 + D′′ 1 + D′′′ 1 = X X d1/s′ 6p1<p2<d1/s (p1p2,N)=1
S(Adp1p2;P(dN),p1).
Clearly we have
E1 > XX d1/s′ 6p1<p2<d1/s (p1p2,N)=1
S(Adp1p2;P(dN),p2).
Thus an application of Bechstab’s identity gives us D1 E1 6 X X X d1/s′ 6p1<p2<p3<d1/s (p1p2p3,N)=1 S(Adp1p2p3;P(dN),p2) + Oδ,k(N1 η/d).
From this, we can deduce D1 E1 E5 6 X X X d1/s′ 6p1<p2<p3<d1/s (p1p2p3,N)=1
S(Adp1p2p3;P(dN),p2)
X X X d1/s′ 6p1<p2<p3<d1/κ2 (p1p2p3,N)=1
S(Adp1p2p3;P(dN),p2) + Oδ,k(N1 η/d)
= X X X d1/s′ 6p1<p2<d1/κ2 6p3<d1/s (p1p2p3,N)=1
S(Adp1p2p3;P(dN),p2)
+ X X X d1/s′ 6p1<d1/κ2 6p2<p3<d1/s (p1p2p3,N)=1
S(Adp1p2p3;P(dN),p2)
+ X X X d1/κ2 6p1<p2<p3<d1/s (p1p2p3,N)=1
S(Adp1p2p3;P(dN),p2) + Oδ,k(N1 η/d)
=: D3 + D4 + D5 + Oδ,k(N1 η/d).
We have
D3 E4 6 Γ10 + Γ12 + Γ13. By splitting D4 into 4 subsums, we have D4 = Γ11 + Γ14 + Γ15 + X X X d1/κ1 6p1<d1/κ2,d1/κ3 6p2<p3<d1/s (p1p2p3,N)=1
S(Adp1p2p3;P(dN),p2).
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 21
Similarly by splitting E7 into 2 subsums, we have E7 = E8 + X X d1/κ16p1<d1/κ2,d1/κ36p2<d1/s (p1p2,N)=1
S(Adp1p2;P(dN),p2)
> E8 + X X X d1/κ16p1<d1/κ2,d1/κ36p2<p3<d1/s (p1p2p3,N)=1
S(Adp1p2p3;P(dN),p3),
where
E8 := X X d1/κ2 6p1<d1/κ3 6p2<d1/s (p1p2,N)=1
S(Adp1p2;P(dN),p2).
By noticing that X X Xd 1/κ1 6p1<d1/κ2,d1/κ3 6p2<p3<d1/s (p1p2p3,N)=1
S(Adp1p2p3;P(dN),p2) S(Adp1p2p3;P(dN),p3)
= Γ19 + Oδ,k(N1 η/d),
we can deduce
D4 E7 6 Γ11 + Γ14 + Γ15 + Γ19 E8 + Oδ,k(N1 η/d).
Since
E2 > X X X d1/κ2 6p1<p2<p3<d1/s (p1p2p3,N)=1
S(Adp1p2p3;P(dN),p3),
we have
D5 E2 6 X X X X d1/κ26p1<p2<p3<p4<d1/s (p1p2p3p4,N)=1
S(Adp1p2p3p4;P(dN),p3) + Oδ,k(N1 η/d)
=: D6 + Oδ,k(N1 η/d).
Similarly E3 > X X X X d1/κ3 6p16p2<p3<p4<d1/s (p1p2p3p4,N)=1
S(Adp1p2p3p4;P(dN),p4) =: E′ 3,
E3 > X X X X X d1/κ3 6p16p2<p3<p4<p5<d1/s (p1p2p3p4p5,N)=1
S(Adp1p2p3p4p5;P(dN),p5) =: E′′ 3,
E8 > X X X X d1/κ2 6p1<d1/κ3 6p26p3<p4<d1/s (p1p2p3p4,N)=1
S(Adp1p2p3p4;P(dN),p4) =: E′ 8,
D6 = Γ16 + Γ17 + Γ18 + X X X X d1/κ2 6p1<d1/κ3 6p2<p3<p4<d1/s (p1p2p3p4,N)=1
S(Adp1p2p3p4;P(dN),p3)
+ X X X X d1/κ3 6p1<p2<p3<p4<d1/s (p1p2p3p4,N)=1
S(Adp1p2p3p4;P(dN),p3)
=: Γ16 + Γ17 + Γ18 + D′ 6 + D′′ 6.
22 J. Wu
Since
D′ 6 E′ 8 = Γ20 + Oδ,k(N1 η/d)
and
D′′ 6 E′ 3 E′′ 3 = X X X X X d1/κ3 6p1<p2<p3<p4<p5<d1/s (p1p2p3p4p5,N)=1
S(Adp1p2p3p4p5;P(dN),p4)
X X X X X d1/κ36p1<p2<p3<p4<p5<d1/s (p1p2p3p4p5,N)=1
S(Adp1p2p3p4p5;P(dN),p5) + Oδ,k(N1 η/d)
= Γ21 + Oδ,k(N1 η/d),
we have
D6 2E3 E8 6 Γ16 + Γ17 + Γ18 + D′ 6 + D′′ 6 E′ 3 E′′ 3 E′ 8 = Γ16 + Γ17 + Γ18 + Γ20 + Γ21 + Oδ,k(N1 η/d).
Combining these estimations leads to the following inequalities
(4.13)
2 6 D3 + D4 + D5 E2 2E3 E4 E7 6 Γ10 + + Γ15 + Γ19 + D6 2E3 E8 + Oδ,k(N1 η/d) 6 Γ10 + + Γ21 + Oδ,k(N1 η/d). Now the desired result follows from (4.12) and (4.13).
§ 5. Functional inequalities between H(s) and h(s) In this section, we start from two weighted inequalities for the sieve function to deduce two functional inequalities between H(s) and h(s). They will be used to prove Propositions 3 and 4 in the next section. Lemma 5.1. For 5 > s′ > 3 > s > 2 and s′ s′/s > 2, we have
H(s) > Ψ1(s) +
1 2Z 1 1/s′ 1 1/s
h(s′t) t(1 t)
dt + H(s′),
where Ψ1(s) = Ψ1(s,s′) is given by Ψ1(s) := Z s′ 1 2 log(t 1) t
dt +
1 2Z 1 1/s′ 1 1/s
log(s′t 1) t(1 t)
dt I1(s)
and I1(s) = I1(s,s′) is given by
I1(s) := max φ>2 Z Z Z 1/s′6t6u6v61/s
ω φ t u v u dtdudv tu2v
.
Proof. Our starting point is the inequality in Lemma 4.1. We need to estimate all terms in the right-hand side of this inequality. Firstly, (3.3) of Lemma 3.2 gives us (5.1) X d σ(d) 1 6 2 A(s′) Hk,N0(s′) Θ(N,σ).
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 23
Secondly, by an argument similar to the proof of (3.20), we can prove, for any ε > 0 and N > N0(ε,δ,k), (5.2) X d σ(d) 2 > Z 1 1/s′ 1 1/s a(s′t) + hk+1,N0(s′t) t(1 t) dt ε Θ(N,σ). Finally we apply the switching principle to estimatePd σ(d) 3. For this, we introduce E := {e : e = dnp1p2, σ(d) 6= 0, (n,p1,p2) satis es (5.3) below}, where
(5.3) d1/s′ 6 p1 < p2 < d1/s, (p1p2,dN) = 1, n 6 N/dp1p2 2, (n, NP(p2)/p1) = 1;
and
B := {b : b = N ep3, e ∈E, p2 < p3 6 κ(d,e)}, where κ(d,e) := min{N/e, d1/s}. The set B is a multiset and an element b may occur more than once. ClearlyPd σ(d) 3 does not exceed the number of primes in the set B. Thus X d σ(d) 3 6 S(B;P(N),Q1/2) + O(Q1/2). In the setE, d is not determined uniquely by e. This causes technical di culty. In order to avoid it, we de ne E′ and B′, similar to E and B, with the condition (n, NP(p2)/p1) = 1 replaced by (n, dNP(p2)/p1) = 1 and E by E′ respectively. Obviously the di erence S(B;P(N),Q1/2) S(B′;P(N),Q1/2) is 6X d σ(d) X X X d1/s′ 6p1<p2<p3<d1/s X n6N/dp1p2p3 (n,d)=1 1 δ,k N1 η
where η = η(δ,k) > 0. Hence (5.4) X d σ(d) 3 6 S(B′;P(N),Q1/2) + Oδ,k(N1 η). In order to estimate S(B′;P(N),Q1/2), we use Theorem 5.2 in [14] with X := X e∈E′ X p2<p36κ(d,e) 1, w(q) = q/ (q) if μ(q)2 = (q,N) = 1 0 otherwise to write (5.5) S(B;P(N),Q1/2) 6 8(1 + ε)CNX logN + O(R1 + R2),
where
R1 := X q<Q q|P(Q1/2)
3ν1(q)

X e∈E′( e,q)=1
X p2<p36κ(d,e) ep3≡N(modq)
1
1 (q) X p2<p36κ(d,e)
1

,
R2 := X q<Q q|P(Q1/2)
3ν1(q)/ (q) X e∈E′( e,q)>1
Xp 2<p36κ(d,e)
1.
24 J. Wu We rst estimate R2. Noticing that for e ∈ E′ we have e 6 N1 η and the smallest prime factor of e is > min{p1,Vk/ } > W1/s′ k , we can deduce
(5.6)
R2 6 N X q6Q
3ν1(q) (q) X e6N1 η (e,q)>W1/s′ k
1 e
NL X m>W1/s′ k
1 m X q6Q q≡0(modm)
3ν1(q) (q)
NW1/3s′ k X m>W1/s′ k
1 m X q6Q q≡0(modm)
1 q
NW1/2s′ k X m>W1/s′ k
1 m2
δ,k
Θ(N,σ) L3
. Next we estimate R1. Let g(a) := Pe∈E′,e=a 1. Obviously for each e = dnp1p2 ∈ E′, theintegers d,n,p1,p2 are pairwisely coprime. Therefore they are uniquely determined by e. Thus g(a) 6 1 and there are some injections r0(e) = d and r(e) = p2. Then we have R1 X q6Q (q,N)=1 μ(q)23ν1(q)

X a∈I(a) (a,q)=1 g(a) X p2<p36κ(d,a) ap3≡N(modq) 1 1 (q) X p2<p36κ(d,a) 1

, where I(a) :=r0(a)r0(a)2/s′, N/r0(a)1/s′ . Since d1/s′ < r(e) < d1/s and er(e) 6 N, we can write R1 R(1) 1 + R(2) 1 + R(3) 1 ,
where
R(1) 1 := X q6Q (q,N)=1 μ(q)23ν1(q)

X a∈I1(a) (a,q)=1
g(a) X p36r0(a)1/s ap3≡N(modq)
1
1 (q) X p36r0(a)1/s
1

,
R(2) 1 := X q6Q (q,N)=1 μ(q)23ν1(q)

X a∈I2(a) (a,q)=1
g(a) X p36N/a ap3≡N(modq)
1
1 (q) X p36N/a
1

,
R(3) 1 := X q6Q (q,N)=1 μ(q)23ν1(q)

X a∈I(a) (a,q)=1
g(a) X p36r(a) ap3≡N(modq)
1
1 (q) X p36r(a)
1

,
and I1(a) :=r0(a)r0(a)2/s′, N/r0(a)1/s , I2(a) :=N/r0(a)1/s, N/r0(a)1/s′ . Applying Lemma 2.3 yields R(j) 1 δ,k N/L5k+5 for j = 1,2,3. Hence (5.7) R1 Θ(N,σ)/L3.
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 25
Replacing (n, NP(p2)/p1) = 1 by (n, NP(p2)) = 1 in the de nition of X, the di erence is δ,k NL2/dd1/s′ δ,k N/L2d. Thus we can obtain, by Lemma 2.10, that X =X d σ(d)n X X X d1/s′ 6p1<p2<p3<d1/s (p1p2,dN)=1 Xn 6N/dp1p2p3 (n,NP(p2))=1 1 + Oδ,k(N/L2d)o 6 (1 + ε)X d σ(d) X X X d1/s′ 6p1<p2<p3<d1/s Nω(log(N/dp1p2p3)/logp2) dp1p2p3 logp2 + Oδ,k N L2d . By applying the prime number theorem, we can deduce X 6 (1 + ε)NX d σ(d) dlogd Z Z Z 1/s′6t6u6v61/s ω φd,N t u v u dtdudv tu2v , where φd,N := log(N/d)/logd. Obviously σ(d) 6= 0 implies φd,N > 2. Thus (5.8) X 6 (1 + ε)I1(s)NX d σ(d) dlogd 6 (1 + ε)I1(s)NX d σ(d) (d)logd . Combining (5.4)–(5.8) and noticing CN 6 CdN, we obtain, for any ε > 0 and N > N0(ε,δ,k), (5.9) X d σ(d) 3 6 {2I1(s) + ε}Θ(N,σ). Inserting (5.1), (5.2) and (5.9) into the inequality of Lemma 4.1 and noticing that (5.10) A(s′) = A(s) +Z s′ 1 2 log(t 1) t dt, a(s′t) = log(s′t 1), we nd that Φ(N,σ,s) 6 A(s) Ψ1(s) 1 2Z 1 1/s′ 1 1/s hk+1,N0(s′t) t(1 t) dt Hk,N0(s′) + ε Θ(N,σ). By the de nition of Hk,N0(s), we must have
Hk,N0(s) > Ψ1(s) +
1 2Z 1 1/s′ 1 1/s
hk+1,N0(s′t) t(1 t)
dt + Hk,N0(s′) ε.
Making N0 →∞ and then ε → 0 yields
Hk(s) > Ψ1(s) +
1 2Z 1 1/s′ 1 1/s
hk+1(s′t) t(1 t)
dt + Hk(s′).
Now it remains to take k →∞ to get the desired result. Lemma 5.2. For 5 > s′ > 3 > s > 2, s′ s′/s > 2 and s 6 κ3 < κ2 < κ1 6 s′, we have
H(s) > Ψ2(s) +
4 5
H(s′) + 1 5
H(κ1) +
1 5Z 1 1/s′ 1 1/s
h(s′t) t(1 t)
dt
+
1 5Z 1 1/s′ 1 1/κ2
h(s′t) t(1 t)
dt +
1 5Z 1 1/s′ 1 1/κ3
h(s′t) t(1 t)
dt
+
1 5Z 1/κ2 1/s′
dt t Z 1/κ2 t
H(s′ s′t s′u) u(1 t u)
du
+
1 5Z 1/κ1 1/s′
dt t Z 1/κ3 1/κ2
H(s′ s′t s′u) u(1 t u)
du
+
1 5Z 1/κ1 1/s′
dt t Z 1/κ2 t
H((1 t u)/t) u(1 t u)
du,
26 J. Wu
where Ψ2(s) = Ψ2(s,s′,κ1,κ2,κ3) is given by
Ψ2(s) :=
2 5Z s′ 1 2
log(t 1) t
dt
2 5Z κ1 1 2
log(t 1) t
dt
1 5Z κ2 1 2
log(t 1) t
dt
+
1 5Z 1 1/s′ 1 1/s
log(s′t 1) t(1 t)
dt +
1 5Z 1 1/κ1 1 1/κ3
log(κ1t 1) t(1 t)
dt
2 5
21 X i=9
I2,i(s)
and I2,i(s) = I2,i(s,s′,κ1,κ2,κ3) is given by
I2,i(s) := max φ>2 ZD2,i ω φ t u v u dtdudv tu2v
(i = 9,...,15),
I2,i(s) := max φ>2 ZD2,i ω φ t u v w v dtdudvdw tuv2w
(i = 16,...,19),
I2,20(s) := max φ>2 ZD2,20 ω φ t u v w x w dtdudv dwdx tuvw2x
,
I2,21(s) := max φ>2 ZD2,21 ω φ t u v w x y x dtdudv dwdxdy tuvwx2y
.
The sets D2,i (9 6 i 6 21) are de ned as follows: D2,9 := {(t,u,v) : 1/κ1 6 t 6 u 6 v 6 1/κ3}, D2,10 := {(t,u,v) : 1/κ1 6 t 6 u 6 1/κ2 6 v 6 1/s}, D2,11 := {(t,u,v) : 1/κ1 6 t 6 1/κ2 6 u 6 v 6 1/κ3}, D2,12 := {(t,u,v) : 1/s′ 6 t 6 u 6 1/κ1, 1/κ3 6 v 6 1/s}, D2,13 := {(t,u,v) : 1/s′ 6 t 6 1/κ1 6 u 6 1/κ2 6 v 6 1/s}, D2,14 := {(t,u,v) : 1/s′ 6 t 6 1/κ1, 1/κ2 6 u 6 v 6 1/s}, D2,15 := {(t,u,v) : 1/κ1 6 t 6 1/κ2 6 u 6 1/κ3 6 v 6 1/s}, D2,16 := {(t,u,v,w) : 1/κ2 6 t 6 u 6 v 6 w 6 1/κ3}, D2,17 := {(t,u,v,w) : 1/κ2 6 t 6 u 6 v 6 1/κ3 6 w 6 1/s}, D2,18 := {(t,u,v,w) : 1/κ2 6 t 6 u 6 1/κ3 6 v 6 w 6 1/s}, D2,19 := {(t,u,v,w) : 1/κ1 6 t 6 1/κ2,1/κ3 6 u 6 v 6 w 6 1/s}, D2,20 := {(t,u,v,w,x) : 1/κ2 6 t 6 1/κ3 6 u 6 v 6 w 6 x 6 1/s}, D2,21 := {(t,u,v,w,x,y) : 1/κ3 6 t 6 u 6 v 6 w 6 x 6 y 6 1/s}. Proof. By (3.3) of Lemma 3.2, we have (5.11) X d σ(d)Γ1 6 4A(s′) + A(κ1) 4Hk,N0(s′) Hk,N0(κ1) Θ(N,σ). Similar to (3.20), we can prove X d σ(d)Γ2 > Z 1 1/s′ 1 1/s a(s′t) + hk+1,N0(s′t) t(1 t) dt ε Θ(N,σ),(5 .12) X d σ(d)Γ3 > Z 1 1/s′ 1 1/κ2 a(s′t) + hk+1,N0(s′t) t(1 t) dt ε Θ(N,σ),(5 .13) X d σ(d)Γ4 > Z 1 1/s′ 1 1/κ3 a(s′t) + hk+1,N0(s′t) t(1 t) dt ε Θ(N,σ).(5 .14)
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 27 Similar to (3.20) and in view of A(s′ s′t s′u) = A((1 t u)/t) = 1, we have X d σ(d)Γ5 6 Z 1/κ2 1/s′ dt t Z 1/κ2 t 1 Hk+2,N0(s′ s′t s′u) u(1 t u) du + ε Θ(N,σ),(5 .15) X d σ(d)Γ6 6 Z 1/κ1 1/s′ dt t Z 1/κ3 1/κ2 1 Hk+2,N0(s′ s′t s′u) u(1 t u) du + ε Θ(N,σ),(5 .16) X d σ(d)Γ7 6 Z 1/κ1 1/s′ dt t Z 1/κ1 t 1 Hk+2,N0((1 t u)/t) u(1 t u) du + ε Θ(N,σ),(5 .17) X d σ(d)Γ8 6 Z 1/κ1 1/s′ dt t Z 1/κ2 1/κ1 1 Hk+2,N0((1 t u)/t) u(1 t u) du + ε Θ(N,σ).(5 .18) We have also, for i = 9,...,21, (5.19) X d σ(d)Γi 6 {2I2,i(s) + ε}Θ(N,σ).
As before, inserting (5.11)–(5.19) into the inequality of Lemma 4.2 and using the de nition of Hk,N0(s), we can deduce (5.20) 5Hk,N0(s) > A(s,s′) + B(s,s′) ε, where A(s,s′) := 5A(s) 4A(s′) A(κ1) + 4Hk,N0(s′) + Hk,N0(κ1) +Z 1 1/s′ 1 1/s a(s′t) t(1 t) dt +Z 1 1/s′ 1 1/κ2 a(s′t) t(1 t) dt +Z 1 1/s′ 1 1/κ3 a(s′t) t(1 t) dt Z 1/κ2 1/s′ Z 1/κ2 t dtdu tu(1 t u) Z 1/κ1 1/s′ Z 1/κ3 t dtdu tu(1 t u) 2 21 X i=9 I2,i(s) and B(s,s′) := Z 1 1/s′ 1 1/s +Z 1 1/s′ 1 1/κ2 +Z 1 1/s′ 1 1/κ3 hk+1,N0(s′t) t(1 t) dt +Z 1/κ2 1/s′ dt t Z 1/κ2 t Hk+2,N0(s′ s′t s′u) u(1 t u) du +Z 1/κ1 1/s′ dt t Z 1/κ3 1/κ2 Hk+2,N0(s′ s′t s′u) u(1 t u) du +Z 1/κ1 1/s′ dt t Z 1/κ2 t Hk+2,N0((1 t u)/t) u(1 t u) du. For a > b > 2, we have Z 1/b 1/a dt t Z 1/b t du u(1 t u) =Z 1/b 1/a du u Z u 1/a dt t(1 t u) =Z 1/b 1/a log(a 1 au) log(1/u 2) u(1 u) du =Z 1 1/a 1 1/b log(at 1) t(1 t) dt Z a 1 b 1 log(t 1) t dt,
28 J. Wu
where we have used the change of variables t = 1 u and t = 1/u 1 respectively. Similarly for a > b > c > d > 2, we have Z 1/b 1/a dt t Z 1/d 1/c du u(1 t u) =Z 1 1/c 1 1/d log(at 1) t(1 t) dt Z 1 1/c 1 1/d log(bt 1) t(1 t)
dt.
By using these two relations and (5.10), a simple calculation shows A(s,s′) = 5Ψ2(s) + 4Hk,N0(s′) + Hk,N0(κ1). Inserting this into (5.20) and making N → ∞, ε → 0 and k → ∞, we obtain the desired inequality. This completes the proof.
§ 6. Proofs of Propositions 3 and 4 We rst prove a preliminary lemma. Let 1[a,b](t) be the characteristic function of the interval [a,b]. We put σ(a,b,c) :=Z b a log c t 1 dt t , σ0(t) := σ(3,t + 2,t + 1) 1 σ(3,5,4) . Lemma 6.1. Let 3 6 s′ 6 5, 0 < a < b < 1 and 2 6 ac < bc 6 4. Then we have h(4) >Z 3 1 H(t) σ0(t) t dt.(6 .1) H(s′) >Z 3 1 H(t) σ0(t) t log 4 s′ 1 + 1[s′ 2,3](t) t log t + 1 s′ 1 dt,(6 .2) Z b a h(ct) t(1 t) dt > log b ab a ab Z 3 1 H(t) σ0(t) + 1[bc 1,3](t) t dt(6 .3) +Z 3 1 H(t) 1[ac 1,bc 1](t) t log (1 a)(t + 1) a(c 1 t) dt. Proof. By Proposition 2, we have H(s′) >Z 4 s′ 1 h(u) u du >Z 4 s′ 1 h(4) +Z 3 u 1 H(t) dt t du u (6.4) = h(4)log 4 s′ 1 +Z 3 1 H(t) 1[s′ 2,3](t) t log t + 1 s′ 1 dt. From Proposition 2 and (6.4), we deduce h(4) >Z 5 3 H(v) v dv > h(4)σ(3,5,4) +Z 3 1 H(t) σ(3,t + 2,t + 1) t dt, which implies the inequality (6.1). The inequality (6.2) follows immediately from (6.4) and (6.1). By using Proposition 2, we have Z b a h(ct) t(1 t) dt = cZ bc ac h(u) u(c u) du > cZ bc ac du u(c u) h(4) +Z 3 u 1 H(t) dt t = h(4) +Z 3 1 H(t) 1[bc 1,3](t) t dt log b ab a ab +Z 3 1 H(t) 1[ac 1,bc 1](t) t log (1 a)(t + 1) a(c 1 t) dt, which combines (6.1) to give (6.3). This completes the proof.
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 29
Proof of Proposition 3. By using Lemma 6.1, a simple calculation shows
1 2Z 1 1/s′ 1 1/s
h(s′t) t(1 t)
dt + H(s′) >Z 3 1
H(t)Ξ1(t;s)dt,
which, together with Lemma 4.1, implies the desired result. Proof of Proposition 4. From (6.1)–(6.3), we can deduce Z 1 1/s′ 1 1/s h(s′t) t(1 t) dt +Z 1 1/s′ 1 1/κ2 h(s′t) t(1 t) dt +Z 1 1/s′ 1 1/κ3 h(s′t) t(1 t) dt + 4H(s′) + H(κ1)(6 .5) >Z 3 1 H(t) 1[α2,3](t) t log (t + 1)5 (s 1)(s′ 1)(κ1 1)(κ2 1)(κ3 1) dt +Z 3 1 H(t) σ0(t) t log 1024 (s 1)(s′ 1)(κ1 1)(κ2 1)(κ3 1) dt +Z 3 1 H(t) 1[α5,α2](t) t log t + 1 (κ3 1)(s′ 1 t) dt +Z 3 1 H(t) 1[α4,α2](t) t log t + 1 (κ2 1)(s′ 1 t) dt +Z 3 1 H(t) 1[α3,α2](t) t log t + 1 (s 1)(s′ 1 t) dt +Z 3 1 H(t) 1[α1,α2](t) t log t + 1 κ1 1 dt. By the change of variable v = s′(1 t u), we have Z 1/κ2 1/s′ dt t Z 1/κ2 t H(s′ s′t s′u) u(1 t u) du =Z 1/κ2 1/s′ dt t Z s′(1 2t) s′(1 1/κ2 t) s′H(v) v(s′ s′t v) dv. Interchanging the order of integration and a simple calculation show that Z 1/κ2 1/s′ dt t Z 1/κ2 t H(s′ s′t s′u) u(1 t u) du =Z 3 1 H(t) 1[α6,α4](t) t(1 t/s′) log s′ κ2s′ s′ κ2t (6 .6) + 1[α4,α2](t) t(1 t/s′) log(s′ 1 t) dt. Similarly we can prove (6.7) Z 1/κ1 1/s′ dt t Z 1/κ3 1/κ2 H(s′ s′t s′u) u(1 t u) du =Z 3 1 H(t)L1(t)dt where L1(t) := 1[α7,α5](t) t(1 t/s′) log s′2 (κ1s′ s′ κ1t)(κ3s′ s′ κ3t) + 1[α5,α8](t) t(1 t/s′) log s′(s′ 1 t) κ1s′ s′ κ1t + 1[α8,α4](t) t(1 t/s′) log (s′ 1 t)(κ2s′ s′ κ2t) s′ , and (6.8) Z 1/κ1 1/s′ dt t Z 1/κ2 t H((1 t u)/t) u(1 t u) du =Z 3 1 H(t)L2(t)dt
30 J. Wu
with
L2(t) :=
1[α9,α1](t) t
log t + 1 (κ2 1)(κ1 1 t)
+
1[α1,α4](t) t
log t + 1 κ2 1 + 1[α4,α2](t) t
log(s′ 1 t).
Now by inserting (6.5)–(6.8) into Lemma 5.2, we easily deduce the required result.
§ 7. Proof of Theorem 1 We need to resolve the functional inequalities (3.21) and (3.22). It seems very di cult to give the exact solutions, because we only know that H(s) is decreasing. Next we shall give a numeric lower bound for solution by using discretion, which is su cient to prove Theorem 1. Put s0 := 1 and si := 2 + 0.1×(i + 1) for i = 1,...,9. Since H(s) is decreasing on [1,10], Proposition 4 allows us to deduce
(7.1) H(si) > Ψ2(si) +
9 X j=1
ai,jH(sj),
where
ai,j :=Z sj sj 1
Ξ2(t,si)dt (i = 1,...,4; j = 1,...,9).
Similarly Proposition 3 implies
(7.2) H(si) > Ψ1(si) +
9 X j=1
ai,jH(sj),
where
ai,j :=Z sj sj 1
Ξ1(t,si)dt (i = 5,...,9; j = 1,...,9)
Table 1. Choice of parameters
i si s′ i κ1,i κ2,i κ3,i Ψ1(si) Ψ2(si)
1 2.2 4.54 3.53 2.90 2.44 0.015826357
2 2.3 4.50 3.54 2.88 2.43 0.015247971
3 2.4 4.46 3.57 2.87 2.40 0.013898757
4 2.5 4.12 3.56 2.91 2.50 0.011776059
5 2.6 3.58 0.009405211
6 2.7 3.47 0.006558950
7 2.8 3.34 0.003536751
8 2.9 3.19 0.001056651
9 3.0 3.00 0.000000000
The parameters s′ i, κ1,i, κ2,i and κ3,i are chosen such that Ψ1(si) or Ψ2(si) is maximal.
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 31
We put
A :=
a1,1 a1,9 . . . . . . a9,1 a9,9

, H :=
H(s1) . . . H(s9)

, B :=

Ψ2(s1) . . . Ψ2(s4) Ψ1(s5) . . . Ψ1(s9)

.
Then (7.1) and (7.2) can be written as
(7.3) (I A)H > B,
where I is the unit matrix. In order to resolve (7.3), we rst solve the system of linear equations
(7.4) (I A)X = B,
by using Maple and obtain
X =

0.0223939 0.0217196 0.0202876 0.0181433 0.0158644 0.0129923 0.0100686 0.0078162 0.0072943

.
From (7.3) and (7.4), we deduce that
(I A)(H X) > 0. Since all elements of (I A) 1 are positive, it follows that
H > X.
In particular we have
H(2.2) > 0.0223939. Now taking σ = {1} and s = 2.2 in (3.3) of Lemma 3.2, we nd, for δ su ciently small, N0 su ciently large and N > N0, D(N) 6 SA;P(N),N(1/2 δ)/2.2 = Φ(N,{1},2.2) 6 A(2.2) Hk,N0(2.2) 4CNli(N) log(N1/2 δ) 6 8(1 0.0223938)Θ(N) 6 7.82085Θ(N).
This completes the proof of Theorem 1.
32 J. Wu
Remark 2. (i) The constant s1 = 2.2 comes from the fact that Ψ2(s) attains the maximal value at s = s1 (approximately). Since H(s) is decreasing on [1,10], we have H(2.1) > 0.0223939. In order to obtain a better lower (which leads to a smaller constant than 7.82085), we must look for a new weighted inequality (as in Lemma 4.1 and 4.2) such that the corresponding main term Ψ(2.1) has a lager lower bound than 0.015826357. (ii) If we divide the interval [2,3] into more subintervals than 9, it is certain that we can obtain a better result. But the improvement is very minuscule.
§ 8. Proof of Theorem 3 In the case of the twin primes problem, we need to sieve the following sequence B := {p + 2 : p 6 x}. Thinking to Lemmas 2.7 and 2.9, we have 4 7 for the level of distribution in place 1 2 in the Bombieri–Vinogradov theorem. Thus we can take Q := x4/7 δ and d := Q/d in the de nitions described in Section 3. As before, we can prove the corresponding Propositions 3 and 4 with the following modi cation: In the de nition of Ψ1(s) we add a factor 7 8 before I1(s), and in the de nition of Ψ2(s) we replace the factor 2 5 before the sum by 7 20. When we use the switching principle to treat the terms 3 and Γi for 5 6 i 6 21, the related error terms can be estimated by using Lemma 2.9 which has 4 7 for the level of distribution (see [26], page 380).
Table 2. Choice of parameters
i si s′ i κ1,i κ2,i κ3,i Ψ1(si) Ψ2(si)
1 2.1 4.93 3.62 2.86 2.34 0.020914508
2 2.2 4.91 3.62 2.85 2.33 0.020399717
3 2.3 5.00 3.63 2.82 2.30 0.019005124
4 2.4 4.52 3.64 2.87 2.40 0.016618139
5 2.5 3.72 0.013597508
6 2.6 3.62 0.010644985
7 2.7 3.49 0.007155027
8 2.8 3.35 0.003741586
9 2.9 3.19 0.001087780
10 3.0 3.00 0.000000000
As before we can prove
H(2.1) > 0.0287118.
Thus for δ su ciently small, x0 su ciently large and x > x0, we have π2(x) 6 SB;P(2),x(1/2 δ)/2.1 6 3.5(1 0.0287117)Π(x) 6 3.39951Π(x).
This completes the proof of Theorem 3.
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 33
§ 9. Chen’s system of weights Let A :={N p : p 6 N}, P(q) := {p : (p,q) = 1}. The inequality (9.1) below appeared in [9] (page 479, (11)) with (κ,σ) = ( 1 12, 1 3.047),( 1 9.2, 1 3.41) without proof. Cai [3] gave a proof with an extra assumption 3σ + κ > 1. Here we present a proof without Cai’s assumption. This removal is important in our argument.
Lemma 9.1. Let 0 < κ < σ < 1 3. Then we have (9.1) D1,2(N) > S(A;P(N),Nκ) 1 2S1 S2 1 2S3 + 1 2S4 + O(N1 κ),
where Si = Si(κ,σ) (1 6 i 6 4) are de ned by S1 := X Nκ6p<Nσ (p,N)=1 S(Ap;P(N),Nκ), S2 := X X Nσ6p1<p2<(N/p1)1/2 (p1p2,N)=1 S(Ap1p2;P(Np1),p2), S3 := X X Nκ6p1<Nσ6p2<(N/p1)1/2 (p1p2,N)=1 S(Ap1p2;P(Np1),p2), S4 := X X X Nκ6p1<p2<p3<Nσ (p1p2p3,N)=1 S(Ap1p2p3;P(Np1),p2).
Proof. Clearly the desired inequality (9.1) is equivalent to (9.2) D1,2(N) > X a∈A,(a,P(Nκ))=11 1 2s1(a) s2(a) 1 2s3(a) + 1 2s4(a) + O(N1 κ),
where
s1(a) := X Nκ6p<Nσ p|a,(p,N)=1
1,
s2(a) := X X Nσ6p1<p2<(N/p1)1/2 p1p2|a,(p1p2,N)=1 p|(a/p1p2) p>p2
1,
s3(a) := X X Nκ6p1<Nσ6p2<(N/p1)1/2 p1p2|a,(p1p2,N)=1 p|(a/p1p2) p>p2
1,
s4(a) := X X X Nκ6p1<p2<p3<Nσ p1p2p3|a,(p1p2p3,N)=1 p|(a/p1p2p3) p>p2
1.
Let
δ (a) := 1 if (a) 6 2, 0 otherwise.
34 J. Wu
Then it is easy to see D1,2(N) > X a∈A,(a,P(Nκ))=1
δ (a) = X a∈A,(a,P(Nκ))=1
μ(a)2δ (a) + O(N1/2),
where we have used the fact that X a∈A (a,P(Nκ))=1 {1 μ(a)2}δ (a) 6 X Nκ6p6N1/2
1 N1/2.
Similarly if we write
δ(a) := 1 1 2s1(a) s2(a) 1 2s3(a) + 1 2s4(a),
we can show that Xa ∈A,(a,P(Nκ))=1
δ(a) = X a∈A, (a,P(Nκ))=1
μ(a)2δ(a) + O(N1 κ).
Thus in order to prove (9.2) it su ces to verify that
(9.3) δ (a) > δ(a)
for a ∈A, μ(a)2 = 1 and (a,P(Nκ)) = (a,N) = 1. We rst observe that (9.3) is trivial if (a) 6 2, since δ (a) = 1 and s4(a) = 0 in this case. It remains to show that δ(a) 6 0 in all other cases, which can be veri ed as follows: If (a) = 3 and s1(a) = 0, then s2(a) = 1 and s3(a) = s4(a) = 0. Thus δ(a) = 0. If (a) = 3 and s1(a) = 1, then s3(a) = 1 and s2(a) = s4(a) = 0. Thus δ(a) = 0. If (a) = 3 and s1(a) = 2, then s2(a) = s3(a) = s4(a) = 0. Thus δ(a) = 0. If (a) = 3 and s1(a) = 3, then s2(a) = s3(a) = 0 and s4(a) = 1. Thus δ(a) = 0. If (a) > 4 and s1(a) = 1, then s3(a) = 1 and s2(a) = s4(a) = 0. Thus δ(a) = 0. If (a) > 4 and s1(a) = 2, then s2(a) = s3(a) = s4(a) = 0. Thus δ(a) = 0. If (a) > 4 and s1(a) > 3, then s2(a) = s3(a) = 0 and s4(a) = s1(a) 2. Thus δ(a) = 0. This completes the proof.
The main di erence between (9.1) and Chen’s other weighted inequalities (see (34) of [7] and page 425 of [8]) is the additional positive term S4. However a direct application of sieve to S4 leads to zero contribution. In order to take advantage of S4, Chen used (9.1) with two di erent couples of parameters (κ,σ). Then an agreeable application of the Buchstab identity and switching principle leads to some compensation. This idea was also used by Cai & Lu [4] and Cai [3]. Here we make some modi cations of their argument such that this process is more powerful.
Lemma 9.2. Let 0 < κ1 < κ2 < ρ < σ2 < σ1 < 1 3 such that 3κ1 + ρ > 1 2. Then we have 4D1,2(N) > 4S(A;P(N),Nκ1) Υ1 Υ2 Υ3 + Υ4 + Υ5 + Υ6 2Υ7(9 .4) 2Υ8 Υ9 Υ10 + Υ11 + Υ12 Υ13 Υ14 + Υ15 + O(N1 κ1),
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 35
where
Υ1 := X Nκ1 6p<Nκ2 (p,N)=1
S(Ap;P(N),p),
Υ2 := X Nκ1 6p<Nσ1 (p,N)=1
S(Ap;P(N),Nκ1),
Υ3 := X Nκ1 6p<Nσ2 (p,N)=1
S(Ap;P(N),Nκ1),
Υ4 := X X Nκ1 6p1<p2<Nκ2 (p1p2,N)=1
S(Ap1p2;P(N),Nκ1),
Υ5 := X X Nκ1 6p1<Nκ26p2<N1/2 2κ1/p1 (p1p2,N)=1
S(Ap1p2;P(N),Nκ1),
Υ6 := X X Nκ1 6p1<Nκ2,Nρ6p2<Nσ2 (p1p2,N)=1
S(Ap1p2;P(N),p1),
Υ7 := X X Nσ1 6p1<p2<(N/p1)1/2 (p1p2,N)=1
S(Ap1p2;P(Np1),p2),
Υ8 := X X Nσ2 6p1<p2<(N/p1)1/2 (p1p2,N)=1
S(Ap1p2;P(Np1),p2),
Υ9 := X X Nκ1 6p1<Nσ16p2<(N/p1)1/2 (p1p2,N)=1
S(Ap1p2;P(Np1),p2),
Υ10 := X X Nκ26p1<Nσ26p2<(N/p1)1/2 (p1p2,N)=1
S(Ap1p2;P(Np1),Nσ1),
Υ11 := X X X Nκ26p1<p2<p3<Nσ2 (p1p2p3,N)=1
S(Ap1p2p3;P(N),p2),
Υ12 := X X X Nκ26p1<Nσ2,Nσ16p2<p3<(N/p1)1/2 (p1p2p3,N)=1
S(Ap1p2p3;P(Np1),p2),
Υ13 := X X X X Nκ16p1<p2<p3<p4<Nκ2 (p1p2p3p4,N)=1
S(Ap1p2p3p4;P(N),p2),
Υ14 := X X X X Nκ16p1<p2<p3<Nκ26p4<N1/2 2κ1/p3 (p1p2p3p4,N)=1
S(Ap1p2p3p4;P(N),p2),
Υ15 := X X X X Nκ26p1<Nσ26p2<p3<p4<Nσ1 (p1p2p3p4,N)=1
S(Ap1p2p3p4;P(N),p3).
Proof. The inequality (9.1) with (κ,σ) = (κ2,σ2) implies
(9.5) 2D1,2(N) > 2S(A;P(N),Nκ2) S1(κ2,σ2) 2Υ8 S3(κ2,σ2) + Υ11 + O(N1 κ2).
36 J. Wu
Buchstab’s identity, when applied three times, gives the equality 2S(A;P(N),Nκ2) = 2S(A;P(N),Nκ1) Υ1 X Nκ16p<Nκ2 (p,N)=1 S(Ap;P(N),Nκ1) + Υ4 X X X Nκ1 6p1<p2<p3<Nκ2 (p1p2p3,N)=1 S(Ap1p2p3;P(N),p1). Similarly a simple application of Buchstab’s identity yields S1(κ2,σ2) = X Nκ2 6p<Nσ2 (p,N)=1 S(Ap;P(N),Nκ1) X X Nκ1 6p1<Nκ2 6p2<Nρ (p1p2,N)=1 S(Ap1p2;P(N),p1) Υ6. Clearly p1 < Nσ2 and σ2 < σ1 < 1 3 imply that Nσ1 < (N/p1)1/2. Thus by Buchstab’s identity, we can write S3(κ2,σ2) = X X Nκ2 6p1<Nσ2 6p2<Nσ1 (p1p2,N)=1 S(Ap1p2;P(Np1),Nσ1) + X X X Nκ2 6p1<Nσ26p2<p3<Nσ1 (p1p2p3,N)=1 S(Ap1p2p3;P(Np1),p3) + O(N1 σ2) + X X Nκ2 6p1<Nσ2,Nσ1 6p2<(N/p1)1/2 (p1p2,N)=1 S(Ap1p2;P(Np1),Nσ1) X X X Nκ2 6p1<Nσ2,Nσ1 6p2<p3<(N/p1)1/2 (p1p2p3,N)=1 S(Ap1p2p3;P(Np1),p2) = Υ10 Υ12 + X X X Nκ2 6p1<Nσ26p2<p3<Nσ1 (p1p2p3,N)=1 S(Ap1p2p3;P(Np1),p3) + O(N1 σ2).
Inserting these into (9.5), we nd that (9.6) 2D1,2(N) > 2S(A;P(N),Nκ1) Υ1 Υ3+Υ4+Υ6 2Υ8 Υ10+Υ11+Υ12+ +O(N1 κ2), where := X X X Nκ1 6p1<p2<p3<Nκ2 (p1p2p3,N)=1 S(Ap1p2p3;P(Np1),p1) X X X Nκ2 6p1<Nσ26p2<p3<Nσ1 (p1p2p3,N)=1 S(Ap1p2p3;P(Np1),p3) + X X Nκ1 6p1<Nκ2 6p2<Nρ (p1p2,N)=1 S(Ap1p2;P(N),p1). Next we shall further decompose . In view of 3κ1 + ρ > 1 2, we have Nρ > N1/2 2κ1/p1 provided p1 > Nκ1. Thus Buchstab’s identity allows us to write X XN κ1 6p1<Nκ2 6p2<Nρ (p1p2,N)=1 S(Ap1p2;P(N),p1) > Υ5 X X X Nκ1 6p1<p2<Nκ26p3<N1/2 2κ1/p2 (p1p2p3,N)=1 S(Ap1p2p3;P(N),p1).
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 37
Thus we have
> Υ5 + 1,
where
1 := X X X Nκ1 6p1<p2<p3<Nκ2 (p1p2p3,N)=1
S(Ap1p2p3;P(N),p1)
X X X Nκ1 6p1<p2<Nκ26p3<N1/2 2κ1/p2 (p1p2p3,N)=1
S(Ap1p2p3;P(N),p1)
X X X Nκ2 6p1<Nσ26p2<p3<Nσ1 (p1p2p3,N)=1
S(Ap1p2p3;P(Np1),p3).
Now the inequality (9.6) becomes
2D1,2(N) > 2S(A;P(N),Nκ1) Υ1 Υ3 + Υ4 + Υ5 + Υ6(9 .7) 2Υ8 Υ10 + Υ11 + Υ12 + 1 + O(N1 κ2).
The inequality (9.1) with (κ,σ) = (κ1,σ1) gives us
(9.8) 2D1,2(N) > 2S(A;P(N),Nκ1) Υ2 2Υ7 Υ9 + S4(κ1,σ1) + O(N1 κ1).
Adding (9.7) to (9.8) yields
4D1,2(N) > 4S(A;P(N),Nκ1) Υ1 Υ2 Υ3 + Υ4 + Υ5 + Υ6(9 .9) 2Υ7 2Υ8 Υ9 Υ10 + Υ11 + Υ12 + 2 + O(N1 κ1),
where
2 := 1 + X X X Nκ1 6p1<p2<p3<Nσ1 (p1p2p3,N)=1
S(Ap1p2p3;P(N),p2).
Clearly all domains of summation in the three terms on the right-hand side of 1 are distinct and are contained in the domain of summation of the last triple sums on the right-hand side of 2 (since 3κ1 + σ1 > 3κ1 + ρ > 1 2). Therefore we have 2 > X X X Nκ1 6p1<p2<p3<Nκ2 (p1p2p3,N)=1 S(Ap1p2p3;P(N),p1) S(Ap1p2p3;P(N),p2) X X X Nκ16p1<p2<Nκ2 6p3<N1/2 2κ1/p2 (p1p2p3,N)=1 S(Ap1p2p3;P(N),p1) S(Ap1p2p3;P(N),p2) + X X X Nκ26p1<Nσ26p2<p3<Nσ1 (p1p2p3,N)=1 S(Ap1p2p3;P(N),p2) S(Ap1p2p3;P(N),p3) = Υ13 Υ14 + Υ15 + O(N1 κ1).
Combining this with (9.9), we obtain the required result. This completes the proof.
38 J. Wu
Remark 3. Lemmas 9.1 and 9.2 are also valid for A′ :={p + 2 : p 6 x}, A′′ := {p + 2 : x < p 6 x + xθ}, A′′′ := {N p : αN < p 6 αN + Nθ}, if we make some suitable modi cations. For example, we have
4π1,2(x) > 4S(A′;P(2),xκ1) Υ′ 1 Υ′ 2 Υ′ 3 + Υ′ 4 + Υ′ 5 + Υ′ 6 2Υ′ 7(9 .10) 2Υ′ 8 Υ′ 9 Υ′10 + Υ′11 + Υ′12 Υ′13 Υ′14 + Υ′15 + O(N1 κ1), where Υ′ j is similarly de ned as Υj with the di erence that A is replaced by A′, P(N) by P(2), P(Np1) byP(2p1), (N/p1)1/2 by (x/p1)1/2, N1/2 2κ1/p2 by x4/7 2κ1/p2 (in Υ5 and Υ14), Nρ by xρ, Nκi by xκi, Nσi by xσi and that the conditions (p,N) = 1, (p1p2,N) = 1, (p1p2p3,N) = 1 and (p1p2p3p4,N) = 1 are eliminated. The assumption on the parameters is
0 < κ1 < κ2 < ρ < σ2 < σ1 < 1 3, 3κ1 + ρ > 4 7.
The last condition is necessary in the proof of > Υ′ 5 + 1.
§ 10. Proofs of Theorems 2 and 5 For simplicity, we write L := logN and use B to denote a suitable positive constant determined by Lemma 2.3. We shall estimate all terms Υi in the inequality (9.4). For this we suppose that
(10.1) 1 12 = κ1 < κ2 6 1 8, 1 4 = ρ < σ2 < σ1 < 1 3, 3σ1 + κ1 > 1, 2σ1 + σ2 + κ2 > 1.
1 Lower bound of S(A;P(N),Nκ1) We apply (2.5) of Lemma 2.2 with X = li(N), w(p) = p/ (p) if p ∈P(N), 0 otherwise,
z = Nκ1, Q =
√N LB
.
Since |λ± l (q)| 6 1, Lemma 2.3 with the choice f(1) = 1 and f(m) = 0 if m > 2 implies that

X l<L X q|P(z) λ± l (q)r(A,q)

ε X q6√N/LB μ(q)2 max y6N max (a,q)=1

π(y;q,a) li(y) (q)

ε N L3 . In view of V (z) ～ 2e γCN/logz (γ is the Euler constant) and CN 1, we can deduce (10.2) S(A;P(N),Nκ1) > {F0 + O(ε)}Θ(N) with F0 := 2f(1/2κ1)/κ1eγ. 2 Upper bounds of Υ1, Υ2 and Υ3 We apply (2.4) of Lemma 2.2 with
X =
li(N) (p)
, w(p) = p/ (p) if p ∈P(N), 0 otherwise,
z = p, Q =
√N pLB
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 39
to S(Ap;P(N),p). The contribution of the error term in (2.4) is ε X Nκ16p<Nκ2,(p,N)=1 X q6√N/pLB,q|P(p) |r(A,pq)| ε X d6√N/LB μ(d)2 max y6N max (a,d)=1

π(y;d,a) li(y) (d)

ε
N L3
by Lemma 2.3 with the same choice of f as above. Thus
Υ1 6 {1 + O(ε)}N L X Nκ1 6p<Nκ2
V (p) (p)
F log(√N/p) logp + O N L3 .
The standard procedure for replacing sums over primes by integrals yields
(10.3) Υ1 6 {F1 + O(ε)}Θ(N),
where
F1 :=
2 eγ Z κ2 κ1
F(1/2t 1) t2
dt =
4 eγ Z 1/2κ1 1 1/2κ2 1
F(t)dt.
Similarly we can prove
(10.4) Υi 6 {Fi + O(ε)}Θ(N) (i = 2,3),
where
F2 :=
4 eγ Z (1/2 κ1)/κ1 (1/2 σ1)/κ1
F(t) 1 2κ1t
dt, F3 :=
4 eγ Z (1/2 κ1)/κ1 (1/2 σ2)/κ1
F(t) 1 2κ1t
dt.
3 Lower bounds of Υ4 and Υ5 As before we can deduce, from Lemmas 2.2 and 2.3, that
(10.5) Υi > {Fi + O(ε)}Θ(N) (i = 4,5),
where
F4 :=
2 κ1eγ Z κ2 κ1
dt t Z κ2 t
f 1/2 t u κ1 du u
,
F5 :=
2 κ1eγ Z κ2 κ1
dt t Z 1/2 2κ1 t κ2
f 1/2 t u κ1 du u
.
We have used the following fact to remove the condition (p1p2,N) = 1: X p1|Np 1>Nκ XN κ6p2<Nσ N p1p2 + X Nκ6p1<Nσ X p2|Np 2>Nκ N p1p2 N1 κL.
4 Upper bounds of Υi for i = 7,8,9,10,13,14 We shall only majorize Υ7 and the others can be treated similarly. Since σ1 > 1 4, the quantity Υ7 is equal to the number of primes p 6 N such that N p = p1p2p3 with Nσ1 6 p1 < p2 < (N/p1)1/2, p3 > p2 and (p1p2p3,N) = 1. De ne M := {m : m = p1p2, Nσ1 6 p1 < p2 < (N/p1)1/2, (p1p2,N) = 1}, B := {b : b = N mp 6 N, m ∈M, p 6 N/m}.
40 J. Wu
It is clear that
Υ7 6 S(B;P(N),N1/2) + O(N1/2). By applying (2.4) of Lemma 2.2 with
X = X m∈M
li N m , w(p) = p/ (p) if p ∈P(N), 0 otherwise,
Q =
√N LB
,
we obtain
(10.6) Υ7 6
8CNX logN {1 + O(ε)}+ Oε√N + R3 + R4 ,
where
R3 := X q6√N/LB (q,N)=1
μ(q)2

Xm ∈M (q,m)=1
X mp6N mp≡N(modq)
1
li(N/m) (q)

,
R4 := X q6√N/LB,(q,N)=1
μ(q)2 (q) X m∈M,(q,m)>1
li N m . Let f(m) be the characteristic function of M. Since m 6 N3/4 for m ∈M, Lemma 2.3 implies (10.7) R3 = X q6√N/LB (q,N)=1 μ(q)2

Xm 6N5/6 (q,m)=1 f(m) X mp6N mp≡N(modq) 1 li(N/m) (q)

N L3 . Noticing that (d,m) > 1 implies (d,m) > Nσ1 for m ∈M, we have
R4
N L X q6√N
μ(q)2 (q) X m6N3/4,(q,m)>Nσ1
1 m

N L X q6N
μ(q)2 (q) X d|q,d>Nσ1
1 d X n6N3/4/d
1 n
N X q6√N
μ(q)2 (q) X d|q,d>Nσ1
1 d
N X Nσ1<d6N
1 d X l6√N/d
μ(dl)2 (dl)
N X Nσ1<d6N
μ(d)2 d (d) X l6√N/d
μ(l)2 (l)
.
Since the function μ(n)2/ (n) is multiplicative and μ(pν)2/ (pν) = 1/(p 1) for ν = 1 and = 0 for ν > 2, it is plain to see that X l6t μ(n)2 (n) logt.
Thus
(10.8) R4 N1 σ1L2.
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 41
By the prime number theorem, we obtain
(10.9)
X ={1 + o(1)} X X Nσ16p1<p2<(N/p1)1/2
N p1p2 log(N/p1p2)
={1 + o(1)}
N LZ 1/3 σ1
log(1/t 2) t(1 t)
dt
={1 + o(1)}
N LZ 1/σ1 1 2
log(t 1) t
dt.
Inserting (10.7)–(10.9) into (10.6) yields
(10.10) Υ7 6 {F7 + O(ε)}Θ(N), where F7 := 8Z 1/σ1 1 2 log(t 1) t dt. Similarly we can prove that
(10.11) Υi 6 {Fi + O(ε)}Θ(N) (i = 8,10), where F8 := 8Z 1/σ2 1 2 log(t 1) t dt, F10 := 8Z σ2 κ2 log(1/σ2 1 t/σ2) t(1 t) dt. [We need to use the assumption 2σ1 + σ2 + κ2 > 1 in Υ10.] For the terms Υ9,Υ13 and Υ14 with p1 6 N1/10, we can apply Lemma 2.6 instead of Lemma 2.3. A similar argument allows us to show that
(10.12) Υi 6 {Fi + O(ε)}Θ(N) (i = 9,13,14), where
F9 :=
36 5 Z 1/10 κ1
log(1/σ1 1 t/σ1) t(1 t)2
dt + 8Z σ1 1/10
log(1/σ1 1 t/σ1) t(1 t)
dt,
F13 :=
36 5 Z 1/10 κ1
dt1 t1(1 t1)Z κ2 t1
dt2 t2 2 Z κ2 t2
dt3 t3 Z κ2 t3
ω 1 t1 t2 t3 t4 t2 dt4 t4
+ 8Z κ2 1/10
dt1 t1 Z κ2 t1
dt2 t2 2 Z κ2 t2
dt3 t3 Z κ2 t3
ω 1 t1 t2 t3 t4 t2 dt4 t4
,
F14 :=
36 5 Z 1/10 κ1
dt1 t1(1 t1)Z κ2 t1
dt2 t2 2 Z κ2 t2
dt3 t3 Z 1/2 2κ1 t3 κ2
ω 1 t1 t2 t3 t4 t2 dt4 t4
+ 8Z κ2 1/10
dt1 t1 Z κ2 t1
dt2 t2 2 Z κ2 t2
dt3 t3 Z 1/2 2κ1 t3 κ2
ω 1 t1 t2 t3 t4 t2 dt4 t4
.
[We need to use the assumption 3σ1 + κ1 > 1 in Υ9 and Lemma 2.10 in Υ13 and Υ14.] By inserting (10.2)–(10.5), (10.10)–(10.12) and by using the trivial lower bounds Υi > 0 (i = 6,11,12,15) into (9.4), we get the following inequality
D1,2(N) > {F(κ1,κ2,ρ,σ2,σ1) + O(ε)}Θ(N),
42 J. Wu
where
F(κ1,κ2,ρ,σ2,σ1) := 1 4(4F0 F1 F2 F3 + F4 + F5 2F7 2F8 F9 F10 F13 F14).
Taking κ1 = 1 12, κ2 = 29 250, ρ = 1 4, σ2 = 141 500 and σ1 = 41 125, a numerical computation gives us
F(κ1,κ2,ρ,σ2,σ1) > 1 4(4×13.473613 3.891854 20.432098 17.327241 + 0.697375+ 2.118119 2×0.004609 2×0.434368 5.161945 5.468377 0.023310 0.182860) > 0.83607.
[For the integrals F13 and F14, we make use of ω(u) 6 0.561522 for u > 3.5.] This completes the proof of Theorem 2.
Theorem 5 can be proved in the same way. The only di erence is to replace Lemmas 2.3 and 2.6 by Lemma 2.4. Here, the choice of parameters is
(θ,κ1,κ2,ρ,σ2,σ1) = (0.971, (2θ 1)/12, 0.111, (2θ 1)/4, 0.271,0.313).
§ 11. Proof of Theorem 4 The proof of Theorem 4 is very similar to that of Theorem 2. But we must use Lemmas 2.5, 2.7 and 2.8 in place of Lemmas 2.3 and 2.6. In order to take the advantage of these lemmas, we must carry out a more careful and delicate analysis. Thus the proof will be slightly complicated. Suppose that the parameters satisfy the following conditions:
2 21 = κ1 < κ2 6 1 7, 2 7 = ρ < σ2 < 29 100 < σ1 < 1 3, 3σ1 + κ1 > 1, 2σ1 + σ2 + κ2 > 1.
1 Lower bounds of S(A′;P(2),xκ1) By (2.5) of Lemma 2.2 and Lemma 2.7, we can easily prove
(11.1) S(A′;P(2),xκ1) > {G0 + O(ε)}Π(x) with G0 := f(4/7κ1)/κ1eγ.
2 Upper bounds of Υ′ 1, Υ′ 2 and Υ′ 3 We divide the interval [xκ1,xκ2] into O(L) subintervals of the form [P,2P) and apply (2.4) of Lemma 2.2 to S(A′ p;P(2),p) for p ∈ [P,2P). We have S(A′ p;P(2),p) 6 {1 + O(ε)}x L V (p) (p) F log(Q/P) logp +X l<L X q|P(p) λ+ l (q)r(A′,pq) where Q = x4/7 ε and λ+ l (q) is well factorable of level Q/P and of order 1. Denote by πP the characteristic function of the primes in the interval [P,2P). Noticing that P 6 xκ2 P 6 Q/P, Lemma 2.1 shows that πP λ+ l is well factorable of level Q and of order 2. Thus Lemma 2.7 allows us to deduce that XP 6p<2PX l<L X q|P(p) λ+ l (q)r(A′,pq) ε x/(logx)4
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 43
and
Υ′ 1 6 {1 + O(ε)}x logx X xκ1 6p<xκ2
V (p) (p)
F log(Q/p) logp + O x (logx)4 . Since V (p) ～e γC/logp, the prime number theorem implies that
(11.2) Υ′ 1 6 {G1 + O(ε)}Π(x) with G1 :=
7 4eγ Z 4/7κ1 1 4/7κ2 1
F(t)dt.
We divide the interval of summation [xκ1,xσ1] of Υ′ 2 into three parts:
[xκ1,x2/7 ε], [x2/7 ε,x29/100], [x29/100,xσ1],
and use (2.4) of Lemma 2.2 to handle each sum. As before we apply Lemma 2.7, the condition (C.2) and (C.3) of Lemma 2.8, respectively, to control the corresponding error terms. We nd
(11.3) Υ′ 2 6 {G2 + O(ε)}Π(x),
where
G2 :=
1 κ1eγ Z 4/7κ1 1 2/7κ1
F(t) 4/7κ1 t
dt +Z 2/7κ1 13/50κ1
F(t) 2/κ1 t
dt +Z 13/50κ1 (11/20 σ1)/κ1
F(t) 11/20κ1 t
dt .
Similarly
Υ′ 3 6 {G3 + O(ε)}Π(x),
where
(11.4) G3 :=
1 κ1eγ Z (4/7 κ1)/κ1 2/7κ1
F(t) 4/7κ1 t
dt +Z 2/7κ1 (2 6σ2)/κ1
F(t) 2/κ1 t
dt .
3 Lower bounds of Υ′ 4 and Υ′ 5 In view of 3κ2 6 3 7 < 4 7, a similar argument proving (11.2) implies that
(11.5) Υ′ 4 > {G4 + O(ε)}Π(x),
where
G4 :=
1 κ1eγ Z κ2 κ1
dt t Z κ2 t
f 4/7 t u κ1 du u
.
Our assumptions on κ1 and ρ imply that p2 1p2 6 x4/7 ε and p2 1 6 x4/7 ε. As before we can apply (2.5) of Lemma 2.2 and Lemma 2.7 to get
(11.6) Υ′ 5 > {G5 + O(ε)}Π(x),
where
G5 :=
1 κ1eγ Z κ2 κ1
dt t Z 4/7 2κ1 t κ2
f 4/7 t u κ1 du u
.
4 Upper bounds of Υ′ i for i = 7,8,9,10,13,14 We shall apply the technique of [12]. Since σ1 > 2 7, the quantity Υ′ 7 is equal to the number of primes p 6 x such that p + 2 = p1p2p3 with xσ1 6 p1 < p2 < (x/p1)1/2 and p3 > p2.
44 J. Wu
Introduce the set
B := {b 2 : b = p1p2p3 6 x, xσ1 6 p1 < p2 < p3}.
Then we have
Υ′ 7 = S(B;P(2),x1/2) + O(x1/2). Let := 1 +L 4. We cover the set B by cuboids B(t1,t2,t3) := b 2 : b = p1p2p3 6 x, pi ∈ [ ti, ti+1) for 1 6 i 6 3 where ti are integers satisfying xσ1 6 t1 6 t2 6 t3 and t1+t2+t3+3 6 x. In view of x2/7 6 p2 6 x(1 σ2)/2 6 x2/5, Lemma 2.5 with the choice αm = 1 if m = p1p3 0 otherwise , βn = 1 if n = p2 0 otherwise
implies the inequality X( q,2)=1
λ+ l (q) |B(t1,t2,t3)q| |B(t1,t2,t3)| (q) x (logx)18
,
where λ+ l (q) is well factorable of order 1 and of level Q = xθ(t2) with θ(t2) = (2 + t2)/4. Thus we nd by (2.4) of Lemma 2.2, S(B(t1,t2,t3);P(2),x1/2) 6 2C{1 + O(ε)} θ(t2)L |B(t1,t2,t3)|+ O x (logx)18 . Since the number of cuboids B(t1,t2,t3) is O(logx)15 , we have X( t1,t2,t3) |B(t1,t2,t3)| θ(t2) = X X xσ16p1<p26(x/p1)1/2 4x p1p2 log(x/p1p2)(2 + logp2/logx) + O x (logx)2
=
4x{1 + O(ε)} logx Z Z σ16t6u6(1 t)/2
dtdu tu(1 t u)(2 + u)
.
Combining these estimates, we obtain
(11.7) Υ′ 7 6 {G7 + O(ε)}Π(x),
where
G7 := 8 Z Z σ16t6u6(1 t)/2
dtdu tu(1 t u)(2 + u)
.
Analogously we have
(11.8) Υ′ 8 6 {G8 + O(ε)}Π(x),
where
G8 := 8 Z Z σ26t6u6(1 t)/2
dtdu tu(1 t u)(2 + u)
.
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 45
For Υ′ 9, the assumption 3σ1 + κ1 > 1 allows us to write Υ′ 9 = S(B′;P(2),x1/2) + O(x1/2)
with (11.9) B′ := {b 2 : b = p1p2p3 6 x, xκ1 6 p1 < xσ1 6 p2 < p3}. We decompose B′ = B′ 1 ∪ ∪B′ 6, where B′ 1, ..., B′ 6 are de ned as in (11.9) but we add respectively the extra conditions p1 6 x1/10 in B′ 1; p1 > x1/10 and p1p2 6 x1/2 in B′ 2; p1 > x1/10 and p 2 1 p8 2 > x3 in B′ 3; p2 > x2/5 and p 2 1 p8 2 6 x3 in B′ 4; x(1 σ2)/2 < p2 6 x2/5 and p1p2 > x1/2 in B′ 5; p2 6 x(1 σ2)/2 and p1p2 > x1/2 in B′ 6. Again we can use Lemma 2.5 with θ(logp1/logx) = (1 + 2logp1/logx)/2 for B′ 1; θ(logp1/logx) = (5 2logp1/logx)/8 for B′ 2 and B′ 3; θ(logp2/logx) = 1 logp2/logx for B′ 4; θ(logp2/logx) = (2 + logp2/logx)/4 for B′ 5 and B′ 6. Then we have
(11.10) Υ′ 9 6 {G9 + O(ε)}Π(x), where G9 = 4 Z Z κ16t6σ16u6(1 t)/2 t61/10 dtdu tu(1 t u)(1 + 2t) + 16 Z Z κ16t6σ16u6(1 t)/2 t>1/10, t+u61/2 dtdu tu(1 t u)(5 2t) + 16 Z Z κ16t6σ16u6(1 t)/2 t>1/10,8u>2t+3 dtdu tu(1 t u)(5 2t) + 2 Z Z κ16t6σ16u6(1 t)/2 u>2/5,8u62t+3 dtdu tu(1 t u)(1 u) + 8 Z Z κ16t6σ16u6(1 t)/2(1 σ1)/26u62/5, t+u>1/2 dtdu tu(1 t u)(2 + u) + 8 Z Z κ16t6σ16u6(1 t)/2 u6(1 σ1)/2,t+u>1/2 dtdu tu(1 t u)(2 + u) .
46 J. Wu
Similarly in view of the assumption 2σ1 + σ2 + κ2 > 1, we can prove (11.11) Υ′10 6 {G10 + O(ε)}Π(x), where G10 = 16 Z Z κ26t6σ26u6(1 t)/2 t+u61/2 dtdu tu(1 t u)(5 2t) + 16 Z Z κ26t6σ26u6(1 t)/2 8u>2t+3 dtdu tu(1 t u)(5 2t) + 2 Z Z κ26t6σ26u6(1 t)/2 2/56u6(2κ2+3)/8 dtdu tu(1 t u)(1 u) + 2 Z Z κ26t6σ26u6(1 t)/2u >(2κ2+3)/8,8u62t+3 dtdu tu(1 t u)(1 u) + 8 Z Z κ26t6σ26u6(1 t)/21 /2 κ26u62/5,t+u>1/2 dtdu tu(1 t u)(2 + u) + 8 Z Z κ26t6σ26u6(1 t)/2(1 σ2)/26u61/2 κ2,t+u>1/2 dtdu tu(1 t u)(2 + u) + 8 Z Z κ26t6σ26u6(1 t)/2 u6(1 σ2)/2,t+u>1/2 dtdu tu(1 t u)(2 + u) . More easily we can prove that (11.12) Υi 6 {Gi + O(ε)}Θ(N) (i = 13,14), where G13 := 4Z 1/10 κ1 dt1 t1(1 + 2t1)Z κ2 t1 dt2 t2 2 Z κ2 t2 dt3 t3 Z κ2 t3 ω 1 t1 t2 t3 t4 t2 dt4 t4 + 16Z κ2 1/10 dt1 t1(5 2t1)Z κ2 t1 dt2 t2 2 Z κ2 t2 dt3 t3 Z κ2 t3 ω 1 t1 t2 t3 t4 t2 dt4 t4
,
G14 := 4Z 1/10 κ1
dt1 t1(1 + 2t1)Z κ2 t1
dt2 t2 2 Z κ2 t2
dt3 t3 Z 4/7 2κ1 t3 κ2
ω 1 t1 t2 t3 t4 t2 dt4 t4
+ 16Z κ2 1/10
dt1 t1(5 2t1)Z κ2 t1
dt2 t2 2 Z κ2 t2
dt3 t3 Z 4/7 2κ1 t3 κ2
ω 1 t1 t2 t3 t4 t2 dt4 t4
.
Inserting these estimations and the trivial lower bounds Υ′ i > 0 (i = 6,11,12,15) into (9.10), we obtain π1,2(x) > {G(κ1,κ2,ρ,σ2,σ1) + O(ε)}Π(x), where G(κ1,κ2,ρ,σ2,σ1) := 1 4(4G0 G1 G2 G3 + G4 + G5 2G7 2G8 G9 G10 G13 G14).
Chen’s double sieve, Goldbach’s conjecture and the twin prime problem 47
Taking κ1 = 2 21, κ2 = 13 100, ρ = 2 7, σ2 = 36 125 and σ1 = 332 1000, a numerical computation gives us G(κ1,κ2,ρ,σ2,σ1) > 1 4(4×5.894705 1.611441 7.921437 6.736885 + 0.270916+ 0.913995 2×0.000124 2×0.145114 1.790090 1.930545 0.006814 0.059690) > 1.10409.
[For the integrals F13 and F14, we make use of ω(u) 6 0.561522 for u > 3.5 and ω(u) 6 0.567144 for u > 2.] This completes the proof of Theorem 4.
References
[1] E. Bombieri & H. Davenport, Small di erences between prime numbers, Proc. Roy. Soc. Ser. A 239 (1966), 1–18.
[2] E. Bombieri, J.B. Friedlander & H. Iwaniec, Primes in arithmetic progressions to large moduli, Acta Math. 156 (1986), 203–251.
[3] Y.C. Cai, A remark on Chen’s theorem, Acta Arith. 102 (2002), 339–352.

[4] Y.C. Cai & M.G. Lu, Chen’s theorem in short intervals, Acta Arith. 91 (1999), 311–323.
[5] Y.C. Cai & M.G. Lu, On Chen’s theorem, in: Analytic number theory (Beijing/Kyoto, 1999), 99–119, Dev. Math. 6, Kluwer Acad. Publ., Dordrecht, 2002.
[6] Y.C. Cai & M.G. Lu, On the upper bound for π2(x), Acta Arith., to appear.
[7] J.R. Chen, On the representation of a large even integer as the sum of a prime and the product of at most two primes, Sci. Sinica 16 (1973), 157–176.
[8] J.R. Chen, On the representation of a large even integer as the sum of a prime and the product of at most two primes (II), Sci. Sinica 21 (1978), 421–430.
[9] J.R. Chen, Further improvement on the constant in the proposition ‘1+2’: On the representation of a large even integer as the sum of a prime and the product of at most two primes (II) (in Chinese), Sci. Sinica 21 (1978), 477–494.
[10] J.R. Chen, On the Goldbach’s problem and the sieve methods, Sci. Sinica 21 (1978), 701–739.
[11] E. Fouvry, Autour du th′eor`eme de Bombieri–Vinogradov, Ann. scient. Ec. Norm. Sup. 20 (1987), 617–640.
[12] E. Fouvry & F. Grupp, On the switching principle in sieve theory, J. reine angew. Math. 370 (1986), 101–125.
[13] E. Fouvry & F. Grupp, Weighted sieves and twin prime type equations, Duke J. Math. 58 (1989), 731–748.
[14] H. Halberstam & H.-E. Richert, Sieve Methods, Academic Press, London, 1974.
[15] G.H. Hardy & J.E. Littlewood, Some problems of ‘partitio numerorum’ III : On the expression of a number as a sum of primes, Acta Math. 44 (1923), 1–70.
[16] H. Iwaniec, Primes of the type φ(x, y) + A where φ is a quadratic form, Acta Arith. 21 (1972), 203–234.
[17] H. Iwaniec, A new form of the error term in the linear sieve, Acta Arith. 37 (1980), 307–320.
[18] H. Iwaniec & J. Pomykala, Sums and di erence of quartic norms, Mathematik 40 (1993), 233–245.
48 J. Wu
[19] C.B. Pan, On the upper bound of the number of ways to represent an even integer as a sum of two primes, Sci. Sinica 23 (1980), 1367–1377.
[20] C.D. Pan, A new application of the Yu.V. Linnik large sieve method, Chinese Math. Acta 5 (1964), 642–652.
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[22] C.D. Pan & C.B. Pan, Goldbach Conjecture, Science Press, Beijing, China, 1992.
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[27] J. Wu, Th′eor`emes g′en′eralis′es de Bombieri–Vinogradov dans les petits intervalles, Quart. J. Math. Oxford (2), 44 (1993), 109–128. [28] J. Wu, Sur l’′equation p +2 = P2 dans les petits intervalles, J. London Math. Soc. (2), 49 (1994), 61–72.
Institut Elie Cartan UMR 7502 UHP-CNRS-INRIA Universit′e Henri Poincar′e (Nancy 1) 54506 Vand uvre–l`es–Nancy FRANCE e–mail: wu

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