THE DIFFERENCE BETWEEN CONSECUTIVE PRIMES, II


 Philip Patrick Nicholson
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1 THE DIFFERENCE BETWEEN CONSECUTIVE PRIMES, II R. C. BAKER, G. HARMAN and J. PINTZ [Received May 000; revised 5 November 000]. Introduction Beginning with Hoheisel [8], many authors have found shorter and shorter intervals x x v ; xš that must contain a prime number. The most recent result is v ˆ 0:55: see Baker and Harman [], where the history of the problem is discussed. In the present paper we prove: Theorem. For all x > x 0, the interval x x 0:55 ; xš contains prime numbers. With enough effort, the value of x 0 could be determined effectively. The paper has much in common with []; in particular we use the sieve method of Harman [4, 5]. We no longer use zero density estimates, however, but rather mean value results on Dirichlet polynomials similar to those that give rise to such estimates. Compare, for example, work of Iwaniec and Pintz [9] and Baker, Harman and Pintz []. Much of the improvement over [] arises from the use of Watt's theorem [] on a particular kind of mean value. More accurate estimates for sixdimensional integrals are also used to good effect. There is in addition a device which uses a twodimensional sieve to get an asymptotic formula for a `onedimensionally sieved' set; see Lemmas 6, 7. Unfortunately, these lemmas, which would be of great signi cance for v ˆ 0:5, are not very numerically signi cant when v drops to 0.55; the same applies to the `roãle reversals' discussed below. Let us introduce enough notation to permit an outline of the proof. When E is a nite sequence of positive integers, counted with multiplicity, we write jej for the number of terms of E, and E d ˆfm: dm Eg: Let P z ˆY p; where the symbol p is reserved for a prime variable; and let S E; z ˆjfm E: m; P z ˆ gj: Let v be a positive number, 0:54 < v < 0:55: : p < z Research of the rst author was supported in part by the National Security Agency and the National Science Foundation. 000 Mathematics Subject Classi cation: N05. Proc. London Math. Soc. () 8 (00) 5±56. q London Mathematical Society 00.
2 the difference between consecutive primes 5 Let L ˆ log x; y ˆ x exp L = ; y ˆ x v «, A ˆ x y; x Ç Z and B ˆ x y ; x Ç Z; where «is a suf ciently small positive number. Buchstab's identity is the equation S E; z ˆS E; w w < p < z S E p ; p ; where < w < z; S A; x = counts the primes we are looking for. Our philosophy is to use Buchstab's identity to produce parallel decompositions of S A; x = and S B; x = : S A; x = ˆk j ˆ S B; x = ˆk j ˆ S j l j ˆ k S j l j ˆ k Here S j > 0, Sj > 0 and for j < t < k or j > k we have S j ˆ y Sj o y as x!. Thus S A; x = > y S B; x = k Sj o : y j ˆ t We must thus ensure that not too many sums are discarded, that is, fall into the category t < j < k. Just as in [] we use Buchstab's identity twice to reach the decomposition S A; x = ˆS A; x n 0 S A p ; x n a n 0 < a < = n 0 < a < = n a < a < min a ; a = S j ; S j : S A p p ; p ˆ S S S ; say. : (Here p j ˆ x a j.) We give asymptotic formulae for S and S. The piecewise linear function n... is larger (for given v) than its counterpart in []. From this point on, roãle reversals are employed. To illustrate this, note that S A p p ; p ˆjfp p h A: p j h ) p > p gj: If K is a region in which a > a a, we note that S A p p ; p ˆfhp h A: L log h ; L log p K; a ; a K p j h ) p > h = ; p j h ) p > p g;
3 54 r. c. baker, g. harman and j. pintz leading readily to the formula (in which h ˆ x b ) S A p p ; p ˆ o a ; a K a b ; a K p j h ) p > p x = S A hp ; hp which we term a roãlereversal. The point here is that our asymptotic formulae a m b m S A mn ; x n ˆ y o a y m b n S B mn ; x n : m, M n, N m, M n, N require certain upper bounds on M and N; see Lemmas and. Here m, M means M < m < M; m } M means B M < m < BM; B is a positive absolute constant, which need not have the same value at each occurrence. It will generally be bene cial to attempt as many decompositions as possible. There are two reasons for this. First, if there are several variables, there should often be a combination of variables which satisfy one of our criteria for obtaining an asymptotic formula. Second, if there are many variables, the contribution is already quite small. To see this, note that if represents x n < p n < p n <...< p < x l, then S Bp... p n ; p n Z l Z ˆ y a Z an a... a o... q n L a ˆ n a ˆ n a n ˆ n (compare []). Moreover, a... a q n < and a n Z Z... a a n a n da a... da n a n da...da n a...a n < log l=n n : n!n For v ˆ 0:55 we shall have n > 0:05. Hence the contribution from p < x = 0 (for which one can take n ˆ 8) is at most y log 8 L 8!0:05 o < 0:00000yL : (If `asymptotic formula regions', in the sense of (.4) below, are not discarded, we get a better estimate still.) However, when roãlereversals are used it may not always be bene cial to perform as many decompositions as possible. The reason for this is that with roãlereversals, a sum may be replaced by the difference of two sums, each substantially larger than the original one. If not enough combinations of variables lie in `asymptotic formula regions', we have made matters worse. For example, when decomposing in straightforward fashion we count p...p n m; p j m ) p > p n : When roãlereversals are used we may have p...p n klm; p j k ) p > p r ; p j l ) p > p s ; p j m ) p > p n : The rst expression gives rise to a term a... a q n a n a...a n an ;
4 the difference between consecutive primes 55 while the second leads to a term f f q q a r a s f q a n a...a n a r a s a n for certain expressions f, f and f. The corresponding integral can then be larger than the original term under consideration. The nal decomposition of S, given in 6, arises from Lemmas and, together with formulae of the type S A p... p r ; p r ˆ y o S B y p... p r ; p r :4 a ;... ; a r K discussed in 5. a ;... ; a r K. Application of Watt's theorem Let T ˆ x v «= and T 0 ˆ exp L =. In this section we seek a result of the type Z = it jm s N s K s j jdsj p x = L A : = it 0 where M s and N s are Dirichlet polynomials, M s ˆ a m m s ; N s ˆ b n n s ; m, M and K s is a `zeta factor', that is, K s ˆ k, K k s or n, N log k k s : Note the convention of the same symbol for the polynomial and its `length'. Of course, is a Dirichlet polynomial of length. We shall assume without comment that each Dirichlet polynomial that appears has length at most x and coef cients bounded by a power of the divisor function t: thus, whenever a sequence a m m, M is mentioned, we assume that ja m j < t m B : (This property may be readily veri ed for the particular polynomials employed later.) The bound (.), and any bound in which A appears, is intended to hold for every positive A; the constant implied by the `p' or`o' notation may depend on A, B and «. It is not a long step from (.) to a `fundamental lemma' of the type a m b n S A mn ; w ˆ y o a y m b n S B mn ; w : m, M n, N m, M n, N with This will be demonstrated in. w ˆ exp k, K L log L : :
5 56 r. c. baker, g. harman and j. pintz Lemma. Let N s ˆ p i, P i p...p u s :4 where u < B, P i > w and P...P u < x. Then, for Re s ˆ, where each g i is of the form jn s j < g s... g r s ; with r < L B ; :5 L B Yh i ˆ jn i s j; with h < B; N...N h < x; :6 and among the Dirichlet polynomials N ;...; N h greater than T = are zeta factors. the only polynomials of length Proof. It clearly suf ces to prove (.5) for N s ˆ L n n s n, N where L is von Mangoldt's function. We now obtain the desired result by the identity of HeathBrown [6]. We shall refer to polynomials N s `of type (.4)' to indicate that the hypothesis of Lemma holds for N s. Lemma. If K s is a zeta factor, < U < T, K < 4U and M < T, then Z = iu = iu= jm s j jk s j 4 jdsj p U «M U = : :7 Proof. For K < U = and M < U = this is proved in all essentials by Watt [] in the course of the proof of his main theorem. For K < U = and M > U = we have Z = iu Z jm s j jk s j 4 jdsj p kmk jk s j 4 jdsj = iu= p M «U p M U = «: Now suppose that U = < K < 4U. Using a re ection principle based on [0, Theorem 4.], we may replace K by a zeta factor of length K 0 < U = with error E ˆ O. Thus jkj 4 p jk 0 j 4 jej 4. Since Z = iu = iu= jm s j jej 4 jdsj p Z = iu = iu= p M U U «; the general case of Lemma now follows. jm s j jdsj
6 the difference between consecutive primes 57 Lemma. Let MN N K ˆ x. Suppose that M, N and N are of type (.4) and K s is a zeta factor, K p x = 4. Let M ˆ x a and N j ˆ x b j and suppose that a < v; :8 b b < v a 0 : :9 Here and subsequently a 0 ˆ max a; v. Suppose further that b < 4 v a 0 ; :0 b b < 4 v a 0 : : Then for < U < T, Z U j MN N K it j dt p x = L A : U = : Proof. Suppose rst that 4U < K and write N ˆ MN N. Lemma 5 of [] yields km k p M = L A if M > x «and similar results for N and N.Byan application of Lemmas 4. and 4.8 of [] we obtain kk k p K = U ; kknk p K = U M = N N = L A ˆ x = U L A : Hence the integral in (.) is p U kknk p x = L A : Now suppose that K < 4U. The integral in (.) is Z = Z = 4 Z = 4 jm j jn N j jk N j p x «= 50 M T = N N T = 4 T = 4 N T = = 4 r x g by Lemma and the mean value theorem [, (.)]. Here g ˆ a 0 4 max b b ; v 4 v max 0; b 8 v 5 «: The conditions (.8)±(.) guarantee that g < 5 «. Lemma 4. replaced by: The conclusion of Lemma holds if the hypotheses (.9)±(.) are either b < v or N is a zeta factor; : b < 8 v a 0 : :4 Proof. If either K > 4U,orN > 4U and N is a zeta factor, we may proceed as at the beginning of the proof of Lemma. Thus we may suppose that these cases are excluded. The integral in (.) is at most Z = Z = 4 Z = 4 jm j jn j 4 jkn j 4 p x d ;
7 58 r. c. baker, g. harman and j. pintz where d ˆ a 0 v 0 «4 max 0; 4b v : (If b < v, the mean value theorem yields R jn j 4 p Tx «= 4 ;ifn is a zeta factor and N < 4U, the same bound follows from (.7).) The result now follows, in view of (.4). Lemma 5. Let K s be a zeta factor, K p x = 4. Suppose that M ˆ x a, N ˆ x b, a < v and b < min v 4a0 ; 5 v 4a0 : :5 Suppose further that M s and N s are Dirichlet polynomials of the type (.4). Then Z T j MNK it j dt p x = L A : :6 T 0 Proof. Let U; U Š Ì T 0; T Š. It suf ces to get the above mean value bound over U; U Š. Let a ˆ min v a 0 ; 5 v a 0 : We may suppose that b > v ; since otherwise the result follows from Lemma 4 with b ˆ 0. In view of Lemma, we may suppose that N ˆ N...N t ; where N j ˆ x d j ; d <...< d t and any N j with d j > v is a zeta factor. We now give two cases in which (.6) is valid. Case. There is a subproduct x d of N...N t which is either a zeta factor or has d < v. Moreover, b d < a: If a 0 > v, then a < v a 0 < 8 v a 0 ; while if a 0 < v, then a < 5 v a 0 < 8 v a 0 : Now (.6) follows on applying Lemma 4. Case. There is a subproduct x d of N...N t such that a < d < b a: Let b ˆ min d; b d ; then b a; bš. Let b ˆ b b. Then b b ˆ b b < b a < v a 0 : Moreover, b < v 4a0 ˆ 0 4 v a
8 the difference between consecutive primes 59 and b b < 5 4 b < 4 v 4a0 : Now (.6) follows from Lemma. We may now complete the proof of the lemma. If d t < a, there is evidently a subsum of d... d t in a; aš. Now since if a 0 > v, then a < b a; a < 6v 6a 0 < v < b; while if a 0 < v, then a < 5 9v 6a 0 < v < b: Thus Case holds when d t < a, and of course Case also holds when a < d t < b a. Finally suppose that d t > b a; then we are in Case with d ˆ d t. This completes the proof of Lemma 5.. Sieve asymptotic formulae In this section we establish formulae of the type (.) and use them as a stepping stone to formulae of type (.). In order to link (.) or (.) to the behaviour of Dirichlet polynomials we use the following variant of [, Lemma ]. then Lemma 6. Let F s ˆP k } x c k k s.if Z T jf it j dt p x = L A ; : T 0 k A c k ˆ y y c k O yl A : k B : Lemma 7. Let a and u be positive numbers, w ˆ x = u and D ˆ x a. Suppose that =a < u < log x «: : Then p exp log log w ua ualog ua : :4 d j P w d > D The implied constant is absolute. d Proof. Let r ˆ u log ua =L. We use the simple inequality exp cy < exp c y :5
9 540 r. c. baker, g. harman and j. pintz for c > 0 and 0 < y <. Now d < D r d j P w d > D d j P w d r d ˆ D r Y p r p < w ˆ D r exp < D r exp log p r p < w p < w p r p p < w : :6 p We apply (.5) with y ˆ log p= log w and c ˆ log ua. The last expression in (.6) is at most exp log ua D r exp log p log w p p p < w p < w ˆ exp ualog ua ua O log log w O log w by Mertens' theorems for P p log p and P p. This completes the proof. Lemma 8. Let M s ˆP m, M a m m s, N s ˆP n, N b n n s, M ˆ x a and N ˆ x b, with a < v «and b < min v 4a 0 ; 5 v 4a 0 «: :7 Suppose further that M s and N s are of type (.4). Then (.) holds. Proof. We follow the proof of [, Lemma ]. We must prove (.) with c k ˆ a m b n m d : m; n d j l; d j P w mnlˆ k According to Lemma 5 of HeathBrown [7], m d ˆ m d O d j l; d j P w where g ˆ x «=. Let ck 0 ˆ Then m; n; d j P w ; d < g l 0 mod d ; mnlˆ k m; d, D d j l; d j P w d < g a m b n m d ; c 00 k ˆ c k ˆ k A k A ck 0 O d j l; d j P w g < d < gw! m; n; d j P w g < d < w g; l 0 mod d mnlˆ k k A ck 00 : ja m b n j: Suppose for the moment that MND > x = 4. We now apply Lemma 5 to M s N s K s, where M s ˆ a m m d b md s ; N s ˆ n n s ; K s ˆ k s ; n kmnd} x
10 the difference between consecutive primes 54 and then sum over D ˆ g j ; gš. In each case M has length at most Mx «and Lemma 5 is applicable. We conclude that ck 0 ˆ y ck 0 O yl A : :8 y k A k B We reach the same conclusion with ck 00 in place of ck 0 by modifying M s in an obvious fashion. Finally we bound P k A ck 00 by p y y k B c 00 k p y m; n ja n jjb n j mn d j P w g < d < w g p y ja m j jb n j m n exp u«log u«p yl A ; m; n from Lemma 7. Here w ˆ x = u, so that u ˆ log L. Now (.) follows on assembling this together with (.8). Now suppose that MND < x = 4, so that (.8) and its analogue for ck 00 take the form y a n a y n p yl A : nl A nl B n p x = 4 n p x = 4 This bound is easily established, because the lefthand side is y a n n O y y a y n n n O p ja n j p x = 4 «: n, N n The proof may now be carried through as in the case MND > x = 4. d Lemma 9. Let LMN ˆ x. Let g be a natural number, g < B. Suppose that M ˆ x j ; N ˆ x j ; j j < min 4v ; jj j j < v 8 «; 8g 4 v 4g ; 4g Suppose further that the Dirichlet polynomial L s satis es sup jl it j p L = L A : t T o ; T Š Then F s ˆL s M s N s satis es (.). :9 4gv g 4g : :0 : Proof. This is a variant of Theorem 4 of []. The only modi cation needed to the argument in [] comes in Case (ii), where the expressions I and I must be replaced by I 0 ˆ TM f j = = 4 g TN f j = = 4 g L g gj = g M j = N j = L j = ;
11 54 r. c. baker, g. harman and j. pintz and I 0 ˆ TM f j = = g TN f j = = g TL 4 g 6 gj = 6 g M j = N j = L j = : Here f j ˆmin j; 4 6j ; : a function which has the simple property a f j j < a f 4 4 ˆ 4 a : for any a 6 ; Š. Thus I 0 < T = g MN = 8 g L = ˆ T = g x = 8 g L = = 8 g p x = L A by (.0). Similarly I 0 < T MN = 4 g L = 6 ˆ Tx = 4 g L = 6 = 4 g p x = L A : The desired result follows just as in []. We now require combinatorial lemmas designed to bring Lemma 9 into play after a number of `Buchstab decompositions' of the lefthand side of (.). Lemma 0. Let k > 0, and suppose that Then, writing we have Let 0 < a < «and let h be the least positive integer with a > h v : v a h > a >...> a k > 0; :4 a a... a k a k < v if k > 0: :5 h v a a ˆ max ; h a a... a k < a : h v a h :6 :7 Proof. Suppose rst that k > h. Since the a j are decreasing, (.5) yields a h a k < v; :8 so that a k < v a = h ; a a... a k < v a k < v v a h v a ˆ : h h Now suppose that k < h. We apply (.4) to obtain a a... a k < a h This completes the proof of Lemma 0. v a h h v a ˆ : h :9
12 the difference between consecutive primes Note that the function a of a satis es v < h v h < a < «: 54 :0 Lemma. Make the hypotheses of Lemma 0, and suppose further that a a... a k a k > v: : Then the numbers g ˆ a a... a k and g ˆ a a... a k satisfy jg g j < v : : Proof. We have g g ˆ a a... a k a k > v ˆ v : from (.). If k > h, then we have (.9) from the previous proof. Since a > h v, a k < v a h < v h v h ˆ v : Since the a j are decreasing, g g ˆ a a... a k a k a k a k < v v ˆ v : If k < h, then (.4) yields v a g g < a h ˆ v : h This completes the proof of Lemma. Lemma. Let a 0; Š, 0 < b < min v 4a ; 5 v 4a «: Let M s ˆP m, M a m m s, N s ˆP n, N b n n s ; M ˆ x a and N ˆ x b, where M s and N s are of type (.4). Let I h ˆ h v ; h v ; and write 6v 7 n a ˆmin v a ; a I h 9 h ; h > : Then (.) holds for every n < n a. This result sharpens Lemmas 5 and 6 of []. It is clear that n a > v. The upper bound on b never falls below v «.
13 544 r. c. baker, g. harman and j. pintz Proof of Lemma. omitted. Let The summation conditions m, M and n, N will be w l; z ˆ From Buchstab's identity, w l; z ˆw l; w Here w is given by (.). We must prove that, taking z ˆ x n, c k 0 ˆ if l; P z ˆ ; 0 otherwise. mnlˆ k phˆ l w < p < z a m b n w l; z w h; p : satis es (.). From (.4), c k 0 ˆc 0 k 0 c 00 k 0 c k where c 0 k 0 ˆ a m b n w l; w ; c 00 k 0 ˆ c k ˆ mnlˆ k mnp h ˆ k w < p < z > x v mp = mnp h ˆ k w < p < z mp = < x v a m b n w h ; p ; a m b n w h ; p : We continue the process by applying (.4) to c k. In general, let c k j ˆ a m b n w h j ; p j ; then (.4) gives mp... p j h j ˆ k w < p j <... < p < z mp... p j p = j < x v c k j ˆck 0 j ck 00 j c k j where ck 0 j is obtained from c k j by replacing w h j ; p j by w h j ; w, and c 00 k j ˆ a m b n w h j ; p j : mp... p j p j h j ˆ k w < p j <... < p < z mp... p j p = j < x v < mp... p j p = j :4 For j > L= log w ˆ log L, the sum c k j is empty and decomposition ceases. Each c 0 k j satis es (.). To see this, we write p i ˆ x a i. Then since m < x a, where a I h, we have mp...p j < x a :5 in the sum c 0 k j, by Lemma 0. We now obtain (.) by an appeal to Lemmas 6 and 8, taking M s ˆ a m mp...p j s : mp... p j p = j < x v w < p j <... < p < z
14 the difference between consecutive primes 545 Each c 00 k j satis es (.). For in c 00 k j, we may write h j ˆ p 0...pu 0 where u < log L. It suf ces to prove (.) for the portion c 00 k j; u of c 00 k j arising from a xed value of u. As in the proof of Lemma of [], we may remove the conditions of summation, x v < mp... p j p = j ; p j < p j and p j < p, 0 by the use of Perron's formula, which modi es the coef cients of the Dirichlet polynomials in an acceptable way. (The same process is implicit several times in the rest of the paper.) We apply Lemmas 6 and 9 with g ˆ 5, with M s produced by grouping m; p ;...; p j, L s corresponding to p j, and N s produced by grouping the remaining variables. The restriction to dyadic ranges presents no problem, and we have only to verify (.9)±(.). For (.9) we appeal to Lemma ; the condition (.) derives from one of the summation conditions for c 00 k j. For (.0) we note that j j < n a ; and n a is de ned in such a way that (.0) holds. We complete the proof of (.) for c 00 k j by noting that (.) follows from [, Lemma 5]. We now obtain (.4) for c k 0 on combining the results for the O log L expressions into which we have decomposed it. Lemma. Let M ˆ x a,n ˆ x b and N ˆ x g, where M s, N s and N s are of type (.4) and suppose that a < and either (i) b g < v a «; or (ii) Let g < 4 v a «; b g < v a «; b < v ; g b n ˆ Then (.) holds whenever n < n a. < 8 v 4a «: A n B n : n n ˆ n n, N ; n, N Proof. We follow the proof of Lemma, altering only the discussion of c 0 k j, where N s no longer satis es the requirements of Lemma 8. However, N s ˆN s N s, and at the point in the proof of Lemma 8 where we appeal to Lemma 5, we may substitute an appeal to Lemma in Case (i), or to Lemma 4 in Case (ii). Modi ed in this way, the proof of the necessary variant of Lemma 8 goes through, and we obtain the desired result. 4. The twodimensional sieve For a given positive integer m let us write, suppressing dependence on R, E m ˆfrl: mrl A; r, Rg and de ne F m similarly with A replaced by B.
15 546 r. c. baker, g. harman and j. pintz Lemma 4. Let x = 4 < M < x = and MN R < x «, and suppose M s is of type (.4). Then a m b n S E mn ; w ˆ y a y m b n S F mn ; w O yl A : 4: m, M n, N m, M n, N Proof. As in the proof of Lemma 8, the lefthand side of (4.) is a m b n m d jed mn j O ja m b n j jed mn j : 4: m; n d j P w d < g m; n g < d < g w Let L ˆ x=mn R. For given d j P w, jed mn jˆ ehˆ d ˆ ehˆ d ˆ ef gˆ d r, R e j r r; h ˆ r 0, R = e m f l } L h j l mn rl A! m f f j h l 0 } Lh f j r 0 mn er 0 hl 0 A : 4: r 00, R = ef l 0 } L = fg mn ef gr 00 l 0 A Thus, given any coef cients l d, with jl d j <, a m b n with c k ˆ m; n efg< g w efgj P w l efg m f d < g w d j P w l d jed mn jˆ c k k A m; n mn ef grlˆ k a m b n r, R = ef l } L = fg : 4:4 Let E >, F >, G >, EFG p gw, R ˆ R=EF, L ˆ L=FG, M s ˆ a m m s l efg m f ef g s ; m, M e, E; f, F g, G L s ˆ l s and R s ˆ r s : l } L r } R Lemma and the mean value theorem yield, for T 0 < U < T and max R ; L < 4U, Z U Z = Z = 4 Z = 4 jmr L it j dt < jm j jr j 4 jl j 4 U = p x = 4 v = «p x = «L A :
16 the difference between consecutive primes 547 If R or L is greater than 4U, we get the bound Z U U = jmr L it j dt p x = L A by arguing as at the start of the proof of Lemma. Consequently, F s ˆ c k k s ˆ M s R s L s b n n n s satis es Z T T 0 k } x jf it j dt p x = L A n p x = L A and (.) holds. In particular, the rst summand in (4.) is y a y m b n m; n d j P w d < g and the second is y O ja y m jjb n j m; n Moreover, ja m b n j m; n g < d < g w jb n j n m d jf mn d j O yl A 4:5 g < d < g w jfd mn j p ja m b n j m; n g = < e < g w e j P w jfd mn j O yl A : 4:6 r, R = ef mn ef grl B l } L = fg p y ja m b n j m; n mn ef gr : g < efg< g w r, R = ef Consider, for example, the part of the last sum with e > g = ; this is p y L B e p y L A by Lemma 7. It is now easy to complete the proof on combining this estimate with the formulas (4.5) and (4.6). Lemma 5. Let M } x a with 4 < a < ; let R p x = «. Suppose that M s is of type (.4). Then a m S E m ; x n ˆ y a y m S F m ; x n O yl A m, M m, M for all n < n a. Proof. We must show that c k 0 ˆ a m m, M r, R mrlˆ k w rl; z l
17 548 r. c. baker, g. harman and j. pintz satis es (.), where z ˆ x n. Imitating the proof of Lemma, we nd that this reduces to proving (.) for c 0 k j, c 00 k j, where c 0 k j ˆ rl w ; w ; p... p j c 00 k j ˆ a m mrlˆ k a m mrlˆ k w < p j <... < p < z p... p j j rl mp... p j p = j < x v w < p j <... < p < z p... p j j rl mp... p j p = j < x v < mp... p j p = j rl w ; p p... p j j (with obvious modi cations when j ˆ 0). Let P H denote a sum over all subsets H of f;...; jg; write u H ˆQ i H p i and v H ˆQ i 6 H; i < j p i.forp... p j counted in c 0 k j there is an H such that u H j r and r; v H ˆ. Thus, writing r 0 ˆ r=u H and l 0 ˆ l=v H, we have c 0 k j ˆ w r 0 l 0 ; w : H m a m p ;...; p j r 0, R =u H ; r 0 ; v H ˆ mr 0 p... p j l 0 ˆ k P Inserting the factor K Ç H ˆ 0= j K j P u K j r 0 in place of the condition r 0 ; v H ˆ, we arrive at c 0 k j ˆ H K Ç H ˆ 0= j K j a m w r 00 l 0 ; w m l 0 Here ˆ H K Ç H ˆ 0= Note that, recalling (.5), j K j m 0 A m 0 A m 0 ˆ Q m p i i 6 K m i 6 K p i ˆ m 0 p ;...; p j mr 00 u K p... p j l 0 ˆ k n m 0 n r 00 l 0 ˆ k l 0 r 00, R = u H u K b n w r 00 l 0 ; w : r 00 l 0 a m ; b n ˆ p i i K u K ˆn : 4:7 m 0 n < x = ; nr 00 p R p x = «; 4:8 hence m 0 < x = and m 0 n r 00 p x «in the last expression for c 0 k j. Since there are fewer than j < L possibilities for H and K, Lemma 4 yields (.) for c 0 k j. We may proceed similarly to obtain c 00 k j ˆ j K j H K Ç H ˆ 0= m w < p j <... < p < z r 00, R =u H u K mp... p = j < x v < mp... p = j mr 00 u K p... p j l 0 ˆ k l 0 w r 00 l 0 ; p j :
18 the difference between consecutive primes 549 We approximate this expression by the contribution from K ˆ 0=, c k ˆ w r 00 l 0 ; p j m p ;...; p j H r 00 mr 00 p... p j l 0 ˆ k where the summation conditions for m, p ;...; p j, r 00, l 0 are as in the preceding sum. We have jc 00 k j c k j < ja m 0j b n w r 00 l 0 ; w ˆC k ; H m 0 n l 0 K Ç H ˆ 0= K 6ˆ 0= m 0 n r 00 l 0 ˆ k l 0 r 00, R =u H u K say. Here A m 0 and b n are as in (4.7) and we see that C k ˆ y C y k O yl A ; k A k B while, for some H and K, C k p y L m 0 k B ja m 0j m 0 r 00 < x r 00 l 0 < x l 0 n > w n p y L A : Obviously, then, it suf ces to prove (.) for c k in place of c 00 k j. The argument is now essentially the same as the discussion of c 00 k j at the end of the proof of Lemma and we omit it. This completes the proof of Lemma 5. In the next lemma we use Lemma 5 to obtain a formula of the shape S A p p ; x n ˆ y S B y p p ; x n O yl A p, M p, R p, M p, R 4:9 that would be inaccessible by the method of. Lemma 6. Suppose that M ˆ x a, R < M and M R < x: Then (4.9) holds for n ˆ v. Proof. 5 yields Here In view of Lemma we may suppose that R > x v = «. Lemma p, M S E p ; x n ˆ y y p, M S F p ; x n O yl A : E p ˆfrl: r, R; p rl Ag and F p is de ned similarly with B in place of A.
19 550 r. c. baker, g. harman and j. pintz We have S E p ; x n ˆ jf p 0... p 0 u l: p 0... p 0 u, R; p p 0...p 0 u l A; p, M p, M ˆ p, M ˆ p, M x n < p 0 <... < p 0 u ; p 0... p 0 u, R S A p p 0 ; x n p 0, R p, M x n < p 0 = < R The second sum in the last expression is S ˆ y y p, M x n < p 0 = < R x n < p 0 <...< p 0 u; l; P x n ˆ gj S A p p 0... p 0 u ; x n p 0 < p 0 <... < p 0 u p 0... p 0 u, R = p 0 p 0 < p 0 <... < p 0 u p 0... p 0 u, R = p 0 S A p p 0 p 0... p 0 u ; x n : S B p p 0... p 0 u ; x n O yl A 4:0 by an application of Lemma. For this it suf ces to note that MR = p x =, and R=p p x = p p x = v v = «p x since v > 0:54. (Removal of the condition p 0 < p 0 is covered in [], as pointed out earlier.) Now S A p p 0 ; x n ˆ y S F p ; x n S O yl A p, M p 0, R y p, M ˆ y S B y p p 0 ; x n O yl A p, M p 0, R by an obvious variant of (4.0). This completes the proof of Lemma 6. Lemma 7. Let M < M < M,M M M < x and M > x v, and suppose M s and M s are of type (.4). Then, for 0 < n < v, a m b m S A m p m ; x n m, M p, M m, M ˆ y a y m b m S B m p m ; x n O yl A : m, M Proof. We have m, M p, M M M < M M M = < x = ; M p xm p x v : If M < x v = «, the result follows from Lemma. Suppose now that M > x v = «:
20 the difference between consecutive primes 55 Then M M > M > x = 4. Lemma 5 yields an asymptotic formula for a m b m S E m m ; x n : m, M m, M (Here R ˆ M.) In analogy with (4.0), a m b m S E m m ; x n m ; m ˆ a m b m S A m m p 0... p 0 ; x n u m ; m ˆ m ; m a m b m x n < p 0 <... < p 0 u p 0... p 0 u, M p 0, M m ; m x n < p 0 < M = S A m m p 0 ; x n p 0 < p 0 <... < p 0 u p 0... p 0 u, M = p 0 S A m m p 0... p 0 u ; x n : The second sum in the last expression is y S B m m p 0... p 0 ; x n O yl A : u y m ; m x n < p 0 < M = p 0 < p 0 <... < p 0 u p 0... p 0 u, M = p 0 To see this we divide it into two subsums de ned by (i) m m p 0 < x =, (ii) m m p 0 > x =. If condition (i) holds, then p 0...p 0 u p M x v p x v = v v = «p x since v > 0:54. We now get the desired result from Lemma, with variables regrouped as m ˆ m m p 0 and n ˆ p 0... p 0 u. If condition (ii) holds, then we regroup the variables differently, taking m ˆ m p 0... p 0 u p m M =p 0 p m m M x = p x = ; and n ˆ m p 0 p m M = p xm 5 = p x 5 v = 4 6 «p x v = «: Once again, the desired result follows from Lemma. We may now complete the proof in exactly the same way as the previous lemma. 5. Further asymptotic formulae Let L...L l ˆ x, l >, L j ˆ x a j and a j >«. We shall nd a region of a ;...; a l in which Z T T 0 jl it...l l it j h dt p x h = L A 5: for every A > 0. Here h ˆ or. The case h ˆ is needed for application to
21 55 r. c. baker, g. harman and j. pintz primes in almost all short intervals, which we shall consider in another paper. For h ˆ, (5.) permits us to evaluate... S A p... p l ; p l : p, L p l, L l This is essentially an application of Lemma 6. We have already discussed the removal of the condition b l ; P p l ˆ in counting p... p l b l in the last sum. In order to prove (5.) we need only prove jl it j...l l it j j h p x h = L A 5: j for any set S ˆft ; t ;...g in T 0 ; T Š with jt i t j j > i 6ˆ j. By a simple dyadic decomposition we may assume that L j j = j < jl j it j < Lj j = j where < j j < «and the lefthand inequality is to be deleted when j j ˆ. We recall that the Dirichlet polynomials L j have L j j j < L B.Forj ˆ we need to hypothesize the stronger inequality L j p L A 5: for every A > 0. Now (5.) will follow if we show that S :ˆ jsjl Y B L h j j j p L A 5:4 j (the product Q j runs over j ˆ ;...; l unless otherwise stated). We have at our disposal the bounds ( f as in (.)): jsjl B p max L g j j j j L k j j ; TL g j f j j j L k j f j where k and k are 0 or, and k j is 0 for j >. To see this, apply Lemma of [] tol g j j L k j. We write I j as an abbreviation for jsjl B p L g j j j j L k j j and II j for jsjl B p TL g j f j j j L k j f j : Lemma 8. With the above notation, suppose that h h k < ; v :ˆ k > 0; g g g g Let u ˆ P i > h=g i and suppose that 6 hv < g u h g h g Let b ; c ; a ; c ; b ; c ; a 4 ; c 4 ; a 5 ; b 5 ; c 5 ; a 6 ; b 6 ; c 6 i 6ˆ < hv: h hv > : g i g 5:5 be nonnegative numbers,
22 the difference between consecutive primes 55 a j h=6g ; h=g Š,b j h=6g ; h=g Š, u ˆ h b g c ˆ a h c g ˆ h b g c ˆ a 4 h c g 4 ˆ a r b r c r r > 5 ; 6 g c h hk g 6 g c h hk 6 h hk g 6 h hk g g < k b < g c h hk g < k a < g c h hk < k b g c < h hk g < k a 4 g c 4 < h hk g ; ; 6 g c 5 h < k a 5 k b 5 < g c 5 h ; 6 h < k a 6 k b 6 g c 6 < h: Then (5.) holds whenever a j > g j v j >, a 4 h g b a g c 4 h hk 4g k b > b «v ; 5:6 a 4 h g a a g c 4 h hk 4g k a > a «v ; 5:7 a 4 h g b a g c 4 h hk 4g k b a 4 h g a 4 a g c 4 4 h hk 4g k a 4 > u g ; ; > u h «v ; g 5:8 h «v ; g a 4 h g a 5 a 4 h g b 5 a g c 5 4 h k a 5 k b 5 5:9 > a 5 b 5 «v ; 5:0 and a 4 h g a 6 a 4 h g b 6 a g c 6 4 h k a 6 k b 6 a g u h h g g > u «v 4 hv > u h 5: h «v : 5: g g
23 554 r. c. baker, g. harman and j. pintz Proof. Since a j > gj v for j >, we have I j for j >. There are thus eight cases to consider. Case I, I, I. De ne l by h h v l ˆ g g g g j > j so that 0 < l < h from (5.5). Then S p Y j ˆ L g j g j j j j L k j k j j h = g j L g g j l v = g Y L g Y j g j j j j l = g j L h j j j : j > j Every L j j j in the above product has nonnegative exponent: the exponent is 0 for j < ; h k k lv ˆ h l v g g for j ˆ ; and h l for j >. Since h l v > 0, (5.4) now follows from (5.). Case I, II, I. Since b c ˆ u h=g, we have S p L g j L k j h = g TL g f j L k f j b Y Y L g g j c j > L g j g j j j j h = g j j L h j j j : The monomials in L j j j j 6ˆ, j 6ˆ have product and can be omitted; the corresponding step will be implicit in subsequent cases. Thus S p S j ; j, where S j ; j ˆT b L h j g b f j L g c h hk = g j k b f j < S 4 ; 4 : For the last inequality, we appeal to (.), using 6 h < g b < h 5: and 6 g c h hk < k g b < g c h hk : 5:4 g In subsequent cases, we leave the appeal to (.) implicit but mention the inequalities corresponding to (5.) and (5.4). Finally, (5.6) yields S 4 ; 4 ˆT b L h = 4 g b = L g c = h = 4 hk = 4 g k b = p L A : Case II, I, I. This is similar to the previous case, with L and L interchanged.
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