1、Bounded gaps between primes Yitang Zhang Abstract It is proved that lim inf n!1 (p n+1 p n ) 7 10 7 ; where p n is the n-th prime. Our method is a renement of the recent work of Goldston, Pintz and Yildirim on the small gaps between consecutive primes. A major ingredient of the proof is a stronger v
2、ersion of the Bombieri-Vinogradov theorem that is applicable when the moduli are free from large prime divisors only (see Theorem 2 below), but it is adequate for our purpose. Contents 1. Introduction 2 2. Notation and sketch of the proof 3 3. Lemmas 7 4. Upper bound forS 1 16 5. Lower bound forS 2
3、22 6. Combinatorial arguments 24 7. The dispersion method 27 8. Evaluation ofS 3 (r;a) 29 9. Evaluation ofS 2 (r;a) 30 10. A truncation of the sum ofS 1 (r;a) 34 11. Estimation ofR 1 (r;a;k): The Type I case 39 12. Estimation ofR 1 (r;a;k): The Type II case 42 13. The Type III estimate: Initial step
4、s 44 14. The Type III estimate: Completion 48 References 55 11. Introduction Let p n denote the n-th prime. It is conjectured that lim inf n!1 (p n+1 p n ) = 2: While a proof of this conjecture seems to be out of reach by present methods, recently Goldston, Pintz and Yildirim 6 have made signicant p
5、rogress toward the weaker con- jecture lim inf n!1 (p n+1 p n ) 0, then (1.1) will be valid (see 6, Theorem 1). Since the result # = 1=2 is known (the Bombieri-Vinogradov theorem), the gap between their result and (1.1) would appear to be, as said in 6, within a hairs breadth. Until very recently, t
6、he best result on the small gaps between consecutive primes was due to Goldston, Pintz and Yildirim 7 that gives lim inf n!1 p n+1 p n p logp n (log logp n ) 23:5 10 6 : 2This result is, of course, not optimal. The condition k 0 3:5 10 6 is also crude and there are certain ways to relax it. To repla
7、ce the right side of (1.5) by a value as small as possible is an open problem that will not be discussed in this paper. 2. Notation and sketch of the proof Notation p a prime number. a, b, c, h, k, l, m integers. d, n, q, r positive integers. ( q) the von Mangoldt function. j (q) the divisor functio
8、n, 2 (q) = (q). (q) the Euler function. (q) the M obius function. x a large number. L = logx. y;z real variables. e(y) = expf2iy g. e q (y) =e(y=q). jjyjj the distance from y to the nearest integer. m a(q) means m a(modq). c=d means a=d(mod 1) where ac 1(modd). q Q means Q q 2Q. “ any suciently smal
9、l, positive constant, not necessarily the same in each occur- rence. B some positive constant, not necessarily the same in each occurrence. A any suciently large, positive constant, not necessarily the same in each occur- rence. = 1 +L 2A . N the characteristic function of N; N )Z. X l( mod q) a sum
10、mation over reduced residue classes l(modq). C q (a) the Ramanujan sum X l( mod q) e q (la). We adopt the following conventions throughout our presentation. The setH given by (1.3) is assumed to be admissible and xed. We write p for p (H); similar abbreviations will be used in the sequel. Every quan
11、tity depending onH alone is regarded as a constant. For example, the absolutely convergent product S = Y p 1 p p 1 1 p k 0 3is a constant. A statement is valid for any suciently small “ and for any suciently large A whenever they are involved. The meanings of suciently small“ and suciently large“ ma
12、y vary from one line to the next. Constants implied in O or , unless specied, will depend onH, “ and A at most. We rst recall the underlying idea in the proof of 6, Theorem 1 which consists in evaluating and comparing the sums S 1 = X n x (n) 2 (2:1) and S 2 = X n x k 0 X i=1 (n +h i ) (n) 2 ; (2:2)
13、 where (n) is a real function depending onH and x, and (n) = ( logn if n is prime; 0 otherwise: The key point is to prove, with an appropriate choice of , that S 2 (log 3x)S 1 0: (2:3) This implies, for suciently large x, that there is a n x such that the tuple (1.4) contains at least two primes. In
14、 6 the function (n) mainly takes the form (n) = 1 (k 0 +l 0 )! X djP (n) d D (d) log D d k 0 +l 0 ; l 0 0; (2:4) where D is a power of x and P (n) = k 0 Y j=1 (n +h j ): Let ( ;d;c) = X n x n c(d) (n) 1 (d) X n x (n;d)=1 (n) for (d;c) = 1; and C i (d) =fc : 1 c d; (c;d) = 1; P (c h i ) 0(modd)g for
15、1 i k 0 : 4The evaluations of S 1 and S 2 lead to a relation of the form S 2 (log 3x)S 1 = (k 0 T 2 LT 1 )x +O(xL k 0 +2l 0 ) +O(E) for D 0 be a small constant. If D =x 1=4+$ (2:5) and k 0 is suciently large in terms of $, then, with an appropriate choice of l 0 , one can prove that k 0 T 2 LT 1 L k
16、 0 +2l 0 +1 : (2:6) In this situation the errorE can be eciently bounded if the primes have level of distri- bution # 1=2 + 2$, but one is unable to prove it by present methods. On the other hand, for D =x 1=4 “ , the Bombieri-Vinogradov theorem is good enough for boundingE, but the relation (2.6) c
17、an not be valid, even if a more general form of (n) is considered (see Soundararajan 12). Our rst observation is that, in the sums T 1 andT 2 , the contributions from the terms withd having a large prime divisor are relatively small. Thus, if we impose the constraint djP in (2.4), whereP is the prod
18、uct of the primes less than a small power of x, the resulting main term is stillL k 0 +2l 0 +1 with D given by (2.5). Our second observation, which is the most novel part of the proof, is that with D given by (2.5) and with the constraint djP imposed in (2.4), the resulting error X 1 i k 0 X dx 1=2
19、“ , then d can be factored as d =rq (2:8) with the range for r exibly chosen (see Lemma 4 below). Thus, roughly speaking, the characteristic function of the setfd : x 1=2 “ d D 2 ; djPg may be treated as a well factorable function (see Iwaniec 10). The factorization (2.8) is crucial for bounding the
20、 error terms. It suces to prove Theorem 1 with k 0 = 3:5 10 6 5which is henceforth assumed. Let D be as in (2.5) with $ = 1 1168 : Let g(y) be given by g(y) = 1 (k 0 +l 0 )! log D y k 0 +l 0 if yD; and g(y) = 0 if y D; where l 0 = 180: Write D 1 =x $ ; P = Y pD 1 p; (2:9) D 0 = expfL 1=k 0 g; P 0 =
21、Y p D 0 p: (2:10) In the case djP and d is not too small, the factor q in (2.8) may be chosen such that (q;P 0 ) = 1. This will considerably simplify the argument. We choose (n) = X dj(P (n);P) (d)g(d): (2:11) In the proof of Theorem 1, the main terms are not dicult to handle, since we deal with a x
22、ed H. This is quite dierent from 6 and 7, in which various sets H are involved in the argument to derive results like (1.2). By Cauchys inequality, the error (2.7) is eciently bounded via the following Theorem 2. For 1 i k 0 we have X dD 2 djP X c2C i (d) j( ;d;c)j xL A : (2:12) The proof of Theorem
23、 2 is described as follows. First, applying combinatorial argu- ments (see Lemma 6 below), we reduce the proof to estimating the sum ofj( ;d;c)j with certain Dirichlet convolutions . There are three types of the convolutions involved in the argument. Write x 1 =x 3=8+8$ ; x 2 =x 1=2 4$ : (2:13) In t
24、he rst two types the function is of the form = such that the following hold. (A 1 ) = ( (m) is supported on M; j 1 M), j 1 19, (m) j 1 (m)L. 6(A 2 ) = ( (n) is supported on N; j 2 N), j 2 19, (n) j 2 (n)L, x 1x 1 , which may be slightly relaxed, is essential. We say that is of Type III if it is of t
25、he form = N 1 N 2 N 3 such that satises (A 1 ) with j 1 17, and such that the following hold. (A 4 ) N 3 N 2 N 1 ; MN 1 x 1 : (A 5 ) MN 1 N 2 N 3 ; 20 MN 1 N 2 N 3 ) x; 2x): The Type III estimate essentially relies on the Birch-Bombieri result in the appendix to 5 (see Lemma 12 below), which is empl
26、oyed by Friedlander and Iwaniec 5 and by Heath-Brown 9 to study the distribution of 3 (n) in arithmetic progressions. This result in turn relies on Delignes proof of the Riemann Hypothesis for varieties over nite elds (the Weil Conjecture) 4. We estimate each ( ;d;c) directly. However, if one applie
27、s the method in 5 alone, ecient estimates will be valid only for MN 1 x 3=8 5$=2 “ . Our argument is carried out by combining the method in 5 with the factorization (2.8) ( here r is relatively small); the latter will allow us to save a factor r 1=2 . In our presentation, all the (m) and (n) are rea
28、l numbers. 3. Lemmas In this section we introduce a number of prerequisite results, some of which are quoted from the literature directly. Results given here may not be in the strongest forms, but they are adequate for the proofs of Theorem 1 and Theorem 2. Lemma1. Let% 1 (d) and% 2 (d) be the multi
29、plicative functions supported on square-free integers such that % 1 (p) = p ; % 2 (p) = p 1: 7Let T 1 = X d 0 X d 1 X d 2 (d 1 d 2 )% 1 (d 0 d 1 d 2 ) d 0 d 1 d 2 g(d 0 d 1 )g(d 0 d 2 ) and T 2 = X d 0 X d 1 X d 2 (d 1 d 2 )% 2 (d 0 d 1 d 2 ) (d 0 d 1 d 2 ) g(d 0 d 1 )g(d 0 d 2 ): We have T 1 = 1 (k
30、 0 + 2l 0 )! 2l 0 l 0 S(logD) k 0 +2l 0 +o(L k 0 +2l 0 ) (3:1) and T 2 = 1 (k 0 + 2l 0 + 1)! 2l 0 + 2 l 0 + 1 S(logD) k 0 +2l 0 +1 +o(L k 0 +2l 0 +1 ): (3:2) Proof. The sumT 1 is the same as the sumT R (l 1 ;l 2 ;H 1 ;H 2 ) in 6, (7.6) with H 1 =H 2 =H (k 1 =k 2 =k 0 ); l 1 =l 2 =l 0 ; R =D; so (3.1
31、) follows from 6, Lemma 3; the sumT 2 is the same as the sum T R (l 1 ;l 2 ;H 1 ;H 2 ;h 0 ) in 6, (9.12) with H 1 =H 2 =H; l 1 =l 2 =l 0 ; h 0 2H; R =D; so (3.2) also follows from 6, Lemma 3. 2 Remark. A generalization of this lemma can be found in 12. Lemma 2. Let A 1 (d) = X (r;d)=1 (r)% 1 (r) r g
32、(dr) and A 2 (d) = X (r;d)=1 (r)% 2 (r) (r) g(dr): Suppose that dD andj (d)j = 1. Then we have A 1 (d) = # 1 (d) l 0 ! S log D d l 0 +O(L l 0 1+“ ) (3:3) and A 2 (d) = # 2 (d) (l 0 + 1)! S log D d l 0 +1 +O(L l 0 +“ ); (3:4) where # 1 (d) and # 2 (d) are the multiplicative functions supported on squ
33、are-free integers such that # 1 (p) = 1 p p 1 ; # 2 (p) = 1 p 1 p 1 1 : 8Proof. Recall that D 0 is given by (2.10). Since % 1 (r) k 0 (r), we have trivially A 1 (d) 1 + log(D=d) 2k 0 +l 0 ; so we may assumeD=d expf(logD 0 ) 2 g without loss of generality. Writes = +it. For 0 we have X (r;d)=1 (r)% 1
34、 (r) r 1+s =# 1 (d;s)G 1 (s) (1 +s) k 0 where # 1 (d;s) = Y pjd 1 p p 1+s 1 ; G 1 (s) = Y p 1 p p 1+s 1 1 p 1+s k 0 : It follows that A 1 (d) = 1 2i Z (1=L) # 1 (d;s)G 1 (s) (1 +s) k 0 (D=d) s ds s k 0 +l 0 +1 : Note that G 1 (s) is analytic and bounded for 1=3. We split the line of integration into
35、 two parts according tojtj D 0 andjtjD 0 . By a well-known result on the zero-free region for (s), we can move the line segmentf = 1=L;jtj D 0 g to = (logD 0 ) 1 ;jtj D 0 ; where 0 is a certain constant, and apply some standard estimates to deduce that A 1 (d) = 1 2i Z jsj=1=L # 1 (d;s)G 1 (s) (1 +s
36、) k 0 (D=d) s ds s k 0 +l 0 +1 +O(L A ): Note that # 1 (d; 0) =# 1 (d) and # 1 (d;s) # 1 (d) =# 1 (d;s)# 1 (d) X ljd (l)% 1 (l) l (1 l s ): Ifjsj 1=L, then # 1 (d;s) (logL) B , so that, by trivial estimation, # 1 (d;s) # 1 (d)L “ 1 : On the other hand, by Cauchys integral formula, forjsj 1=L we have
37、 G 1 (s) S 1=L: It follows that 1 2i Z jsj=1=L # 1 (d;s)G 1 (s) (1 +s) k 0 (D=d) s ds s k 0 +l 0 +1 1 2i # 1 (d)S Z jsj=1=L (D=d) s ds s l 0 +1 L l 0 1+“ : 9This leads to (3.3). The proof of (3.4) is analogous. We have only to note that A 2 (d) = 1 2i Z (1=L) # 2 (d;s)G 2 (s) (1 +s) k 0 1 (D=d) s ds
38、 s k 0 +l 0 +1 with # 2 (d;s) = Y pjd 1 p 1 (p 1)p s 1 ; G 2 (s) = Y p 1 p 1 (p 1)p s 1 1 p 1+s 1 k 0 ; and G 2 (0) =S. 2 Lemma 3. We have X d 0 we have 1 X d=1 % 1 (d)# 1 (d) d 1+s =B 1 (s) (1 +s) k 0 ; where B 1 (s) = Y p 1 + p (p p )p s 1 1 p 1+s k 0 : Hence, by Perrons formula, X dx 1=4 % 1 (d)#
39、 1 (d) d = 1 2i Z 1=L+iD 0 1=L iD 0 B 1 (s) (1 +s) k 0 x s=4 s ds +O(D 1 0 L B ): Note that B 1 (s) is analytic and bounded for 1=3. Moving the path of integration to 1=3 iD 0 ; 1=3 +iD 0 , we see that the right side above is equal to 1 2i Z jsj=1=L B 1 (s) (1 +s) k 0 x s=4 s ds +O(D 1 0 L B ): Sinc
40、e, by Cauchys integral formula, B 1 (s) B 1 (0) 1=L forjsj = 1=L, and B 1 (0) = Y p p p p 1 1 p k 0 =S 1 ; 10it follows that X d 0, 1 X d=1 % 2 (d)# 2 (d) (d)p s =B 2 (s) (1 +s) k 0 1 with B 2 (s) = Y p 1 + p 1 (p p )p s 1 1 p 1+s k 0 1 ; and B 2 (0) =S 1 . 2 Recall that D 1 andP are given by (2.9),
41、 andP 0 is given by (2.10). Lemma 4. Suppose that dD 2 1 , djP and (d;P 0 )D 1 , we may write d=(d;P 0 ) as d (d;P 0 ) = n Y j=1 p j with D 0 p 1 p 2 :p n D 1 ; n 2: By (3.7), there is a n 0 n such that (d;P 0 ) n 0 Y j=1 p j R and (d;P 0 ) n 0 +1 Y j=1 p j R : The assertion follows by choosing r =
42、(d;P 0 ) n 0 Y j=1 p j ; q = n Y j=n 0 +1 p j ; and noting that r (1=p n 0 +1 )R . 2 Lemma 5. Suppose that 1 i k 0 andj (qr)j = 1. There is a bijection C i (qr)!C i (r)C i (q); c7! (a;b) 11such that c(modqr) is a common solution to c a(modr) and c b(modq). Proof. By the Chinese remainder theorem. 2
43、The next lemma is a special case of the combinatorial identity due to Heath-Brown8. Lemma 6 Suppose that x 1=10 x x “ d and (c;d) = 1. Then for j; 1 we have X N n N 0 n c(d) j (n) N 0 N (d) L j 1 ; the implied constant depending on “, j and at most. Proof. See 11, Theorem 1. 2 The next lemma is (ess
44、entially) contained in the proof of 5, Theorem 4. Lemma 9 Suppose that H;N 2, dH and (c;d) = 1. Then we have X n N (n;d)=1 min H;jjcn=djj 1 (dN) “ (H +N): (3:8) 12Proof. We may assume N H without loss of generality. Writefyg = y y and assume 2 1=H; 1=2. Note thatfcn=dg if and only if bn c(modd) for
45、some b2 (0;d , and 1 f cn=dg if and only if bn c(modd) for some b2 (0;d , Thus, the number of the n satisfying n N, (n;d) = 1 andjjcn=djj is bounded by X q dN q c(d) (q) d “ N 1+“ : Hence, for any interval I of the form I = (0; 1=H; I = 1 1=H; 1); I = ; 0 or I = 1 0 ; 1 with 1=H 10 andj (d 1 )j =j (
46、d 2 )j = 1. Then we have, for any c 1 , c 2 and l, X n N (n;d 1 )=1 (n+l;d 2 )=1 e c 1 n d 1 + c 2 (n +l) d 2 (d 1 d 2 ) 1=2+“ + (c 1 ;d 1 )(c 2 ;d 2 )(d 1 ;d 2 ) 2 N d 1 d 2 : (3:9) Proof. Write d 0 = (d 1 ;d 2 ), t 1 =d 1 =d 0 , t 2 =d 2 =d 0 and d =d 0 t 1 t 2 . Let K(d 1 ;c 1 ;d 2 ;c 2 ;l;m) = X
47、 n d (n;d 1 )=1 (n+l;d 2 )=1 e c 1 n d 1 + c 2 (n +l) d 2 + mn d : We claim that jK(d 1 ;c 1 ;d 2 ;c 2 ;l;m)j d 0 jS(m;b 1 ;t 1 )S(m;b 2 ;t 2 )j (3:10) for some b 1 and b 2 satisfying (b i ;t i ) (c i ;d i ); (3:11) 13where S(m;b;t) denotes the ordinary Kloosterman sum. Note that d 0 , t 1 and t 2 are pairwise coprime. Assume that n t 1 t 2 n 0 +d 0 t 2 n 1 +d 0 t 1 n 2 (modd) and l t 1 t 2 l 0 +d 0 t 1 l 2 (modd 2 ): The conditions (n;d 1 ) = 1 and (n +l;d