THE DIFFERENCE BETWEEN CONSECUTIVE PRIMES, II

[Pages:31]THE DIFFERENCE BETWEEN CONSECUTIVE PRIMES, II

R. C. BAKER, G. HARMAN and J. PINTZ

[Received 3 May 2000; revised 15 November 2000]

1. Introduction

Beginning with Hoheisel [8], many authors have found shorter and shorter intervals x ? xv; x that must contain a prime number. The most recent result is v 0:535: see Baker and Harman [1], where the history of the problem is discussed. In the present paper we prove:

Theorem 1. For all x > x 0, the interval x ? x0:525; x contains prime numbers.

With enough effort, the value of x0 could be determined effectively. The paper has much in common with [1]; 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 [2]. Much of the improvement over [1] arises from the use of Watt's theorem [11] on a particular kind of mean value. More accurate estimates

for six-dimensional integrals are also used to good effect. There is in addition a device which uses a two-dimensional sieve to get an asymptotic formula for a `one-dimensionally sieved' set; see Lemmas 16, 17. Unfortunately, these lemmas,

which would be of great signi?cance for v 0:53, are not very numerically signi?cant when v drops to 0.525; the same applies to the `ro?le 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

Ed fm: dm P Eg:

Let

Pz p;

p< z

where the symbol p is reserved for a prime variable; and let

SE; z jfm P E: m; Pz 1gj:

Let v be a positive number, 0:524 < v < 0:535:

1:1

Research of the ?rst author was supported in part by the National Security Agency and the National Science Foundation. 2000 Mathematics Subject Classi?cation: 11N05.

Proc. London Math. Soc. (3) 83 (2001) 532?562. q London Mathematical Society 2001.

the difference between consecutive primes

533

Let L log x; y1 x exp?3L 1 = 3 ; y x v ?,

A x ? y; x ? Z and B x ? y1; x ? Z;

where ? is a suf?ciently small positive number.

Buchstab's identity is the equation

SE; z SE; w ?

SEp; p;

w< p p2gj:

If K is a region in which a1 > 1 ? a1 ? a2, we note that

SAp1 p 2 ; p 2 fhp 2 h1 P A: L ?1 log h1; L ?1 log p 2 P K;

a 1 ; a 2 P K

p j h1 A p > h11 = 2; p j h A p > p 2g;

534

r . c. baker , g. harman and j. pintz

leading readily to the formula (in which h x b3)

a1; a 2 P K

SAp1 p 2 ;

p2

1

o1

1 ? a 2 ? b 3 ; a 2 P K

S Ah p 2 ;

x hp2

1 = 2

pjhAp >p2

which we term a ro?le-reversal. The point here is that our asymptotic formulae

m,M

n,N

am

bm

SAmn;

xn

y y1

1

o1

m,M

am bn SBmn ; x n

n,N

1:3

require certain upper bounds on M and N; see Lemmas 12 and 13. Here m , M means M < m < 2M; m } M means B ?1M < 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 < pn < pn?1 < . . . < p1 < xl, then

? SBp1... pn ; pn

y1 L

1

o1

l

a1 n

a1

a2 n

.

.

.

an?1

an n

q

1

?

a1

?. an

.

.

?

an

da1 a1

. .

. .

. dan . an2

(compare [1]). Moreover,

q

1 ? a1 ? . . . ? an an

0:05. Hence the contribution from p1 < x1=10 (for which one can take n 8) is at most

y L

log 28 8! 0:05

1

o1

<

0:000002yL ?1:

(If `asymptotic formula regions', in the sense of (1.4) below, are not discarded, we get a better estimate still.)

However, when ro?le-reversals are used it may not always be bene?cial to perform as many decompositions as possible. The reason for this is that with ro?le-reversals, 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

p1 . . . pn m; p j m A p > pn: When ro?le-reversals are used we may have

p1 . . . pn klm; p j k A p > pr ; p j l A p > ps; p j m A p > pn:

The ?rst expression gives rise to a term

q

1 ? a1 ? . . . ? an an

a

1

.

.

.

1 an

?

1

a

2 n

;

the difference between consecutive primes

535

while the second leads to a term

q

f1 ar

q

f2 as

q

f3 an

1 a1 . . . an ar as an

for certain expressions f1, f2 and f3. The corresponding integral can then be larger than the original term under consideration.

The ?nal decomposition of S2 , given in ? 6, arises from Lemmas 12 and 13, together with formulae of the type

a1; ... ; ar P K

SAp1... pr ;

pr

y y1

1

o1

a1; ... ; ar P K

SBp1... pr ;

pr

1:4

discussed in ? 5.

2. Application of Watt's theorem

Let T x 1 ? v ? ? = 2 and T0 expL 1 = 3. In this section we seek a result of

the type

1=2iT jMsN sK sj jdsj p x 1 = 2L ? A

2:1

1 = 2 i T0

where Ms and Ns are Dirichlet polynomials,

Ms

am m ? s; Ns

bn n?s;

m,M

n,N

and Ks is a `zeta factor', that is,

Ks k ? s or

log kk ? s:

k,K

k,K

Note the convention of the same symbol for the polynomial and its `length'. Of course, 1 is a Dirichlet polynomial of length 1. 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 amm , M is mentioned, we assume that

j amj < tmB:

(This property may be readily veri?ed for the particular polynomials employed

later.) The bound (2.1), 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 (2.1) to a `fundamental lemma' of the type

m,M

am

n,N

bn

SAmn ;

w

y y1

1

o1

m,M

am

n,N

bn

SBmn ;

w

2:2

with

w exp

L log L

:

2:3

This will be demonstrated in ? 3.

536

r . c. baker , g. harman and j. pintz

Lemma 1. Let

Ns

p1 . . . pu? s

pi , Pi

2:4

where

u < B,

Pi

>

w

and

P1 . . . Pu

< x.

Then,

for

Re s

1 2

,

jNsj < g1s . . . grs; with r < L B;

2:5

where each gi is of the form

h L B jNi sj;

i1

with h < B; N1 . . . Nh < x;

2:6

and among the Dirichlet polynomials greater than T 1=2 are zeta factors.

N1; . . . ; Nh

the

only

polynomials

of

length

Proof.

It clearly suf?ces to prove (2.5) for

Ns Lnn? s

n,N

where L is von Mangoldt's function. We now obtain the desired result by the identity of Heath-Brown [6].

We shall refer to polynomials Ns `of type (2.4)' to indicate that the hypothesis of Lemma 1 holds for Ns.

Lemma 2. If Ks is a zeta factor, 1 < U < T, K < 4U and M < T, then

1=2iU j Msj 2j Ksj 4j dsj p U 1 ?1 M 2U ? 1 = 2 :

1=2iU =2

2:7

Proof. For K < U 1=2 and M < U 1=2 this is proved in all essentials by Watt [11] in the course of the proof of his main theorem. For K < U 1 =2 and M > U 1=2

we have

1=2iU

jMsj2jKsj4j dsj p kMk12 jKsj4jdsj

1=2iU =2

p M 1?U p M 2U 1=2?:

Now suppose that U 1=2 < K < 4U. Using a re?ection principle based on [10, Theorem 4.13], we may replace K by a zeta factor of length K H < U 1=2 with error E O1. Thus j K j 4 p j K H j4 jE j 4. Since

1=2iU

1=2iU

jMsj2j Ej4jdsj p

j Msj 2jdsj

1=2iU =2

1=2iU =2

p M U U ?;

the general case of Lemma 2 now follows.

the difference between consecutive primes

537

Lemma 3. Let MN1 N 2 K x. Suppose that M, N1 and N2 are of type (2.4) and Ks is a zeta factor, K p x3 =4. Let M xa and Nj x bj and suppose that

a < v;

2:8

b1

1 2

b

2

<

1 2

1

v

?

a

H:

Here and subsequently aH maxa; 1 ? v. Suppose further that

2:9

b2

<

1 4

1

3v

?

a

H;

2:10

b1

3 2

b

2

<

1 4

3

v

?

a

H:

Then for 1 < U < T,

U

U

=

2

jM

N1N

2

K

1 2

i

t

j

d

t

p

x

1

=

2L

?

A:

2:11 2:12

Proof. Suppose ?rst that 4U yields k M k1 p M 1 = 2L ? A if M

< >

K x?

and and

write N MN1 N 2. Lemma 5 similar results for N1 and N2.

of By

[2] an

application of Lemmas 4.2 and 4.8 of [11] we obtain

k

K

k1

p

K 1= U

2

;

k K N k1

p

K 1=2 U

M 1 = 2N1 N 21 = 2L ? A

x1=2 U

L ? A:

Hence the integral in (2.12) is

p U k K N k1 p x 1 = 2L ? A:

Now suppose that K < 4U. The integral in (2.12) is

1 = 2

1 = 4

1 = 4

jMj2

j N12N 2j 2

jK 2N2j2

p

x ? = 50M

T

1 = 2N12N 2

T

1 = 4

T

1 = 41

N

2 2

T

? 1 = 2 1 = 4

r xg

by Lemma 2 and the mean value theorem [2, (3.3)]. Here

g

1 2

a

H

1 4

max2

b1

b2

;

1

?

v

1 4

1

?

v

max0;

1 2

b

2

?

1 8

1

?

v

?

1 25

?

:

The

conditions

(2.8)?(2.11)

guarantee

that

g<

1 2

?

1 25

?

.

Lemma 4. replaced by:

The conclusion of Lemma 3 holds if the hypotheses (2.9)?(2.11) are

either

b1

<

1 2

1

?

v

or

N1

is

a

zeta

factor;

b

2

<

1 8

1

3v

?

1 2

a

H:

2:13 2:14

Proof. If either K > 4U , or N1 > 4U and N1 is a zeta factor, we may proceed as at the beginning of the proof of Lemma 3. Thus we may suppose that these

cases are excluded. The integral in (2.12) is at most

1 = 2

1 = 4

1 = 4

jMj2

j N1j 4

jK N2j4

p xd;

538

r . c. baker , g. harman and j. pintz

where

(If

b1

<

1 2

1

d

1 2

a

H

1 2

1

?

v

? v, the mean value

?

1 10

?

1 4

max0;

theorem yields

4b

2

?

1 2

1

?

v:

jN1j4 p T x?=4;

if

N1

is

a

zeta

factor and N1 < 4U, the same bound follows from (2.7).) The result now follows,

in view of (2.14).

Lemma 5. Let Ks be a zeta factor, K p x3=4. Suppose that M x a, N x b, a < v and

b

<

min

1 2

3v

1?

4aH ;

1 5

3

v?

4 a H :

2:15

Suppose further that Ms and Ns are Dirichlet polynomials of the type (2.4). Then

T

T0

jM

N

K

1 2

it

j

d

t

p

x

1

=

2L

?

A

:

2:16

Proof.

Let

1 2

U;

U

?

T0

;

T

.

It

suf?ces

to

get

the

above

mean

value

bound

over

1 2

U;

U .

Let

a

min2v ?

2aH;

1 5

1

?

3v 2aH :

We may suppose that

b

>

1 2

1

?

v;

since otherwise the result follows from Lemma 4 with b2 0. In view of Lemma 1, we may suppose that

N N1 . . . Nt ;

where

Nj

xdj; d1 < . . . < dt

and

any

Nj

with

dj

>

1 2

1

?

v

is

a

zeta

factor.

We now give two cases in which (2.16) is valid.

Case 1. There is a subproduct xd of N1 . . . Nt which is either a zeta factor or

has

d

<

1 2

1

?

v.

Moreover ,

b ? d < a:

If

aH

>

1 12

13

v

?

1,

then

a

<

2v

?

2a

H

<

1 8

1

3v

?

1 2

a

H;

while

if

aH

<

1 12

13v

?

1,

then

a

<

1 5

1

?

3v

2

a

H

<

1 8

1

3v

?

1 2

a

H

:

Now (2.16) follows on applying Lemma 4. Case 2. There is a subproduct xd of N1 . . . Nt such that

a < d < b ? a:

Let

b2

mind;

b?

d;

then

b2

P a;

1 2

b.

Let

b1

b ? b2.

Then

b1

1 2

b2

b

?

1 2

b2

<

b

?

1 2

a

<

1 2

v

1

?

aH:

Moreover ,

b

2

<

1 2

1 2

3v

1

?

4

aH

1 4

3v

1

?

a

H

the difference between consecutive primes

539

and

b1

3 2

b2

<

5 4

b

<

1 4

3

v

?

4aH

:

Now (2.16) follows from Lemma 3.

We may now complete the proof of the lemma. If dt < a, there is evidently a subsum of d1 . . . dt in a; 2a. Now

2a < b ? a;

since

if

aH

>

1 12

13

v

? 1,

then

3a

<

6v

?

6a

H

<

1 2

1

?

v

<

b;

while

if

aH

<

1 12

13v

?

1,

then

3a

<

1 5

3

?

9v

6aH

<

1 2

1

?

v

<

b:

Thus Case 2 holds when dt < a, and of course Case 2 also holds when a < dt < b ? a.

Finally suppose that dt > b ? a; then we are in Case 1 with d dt . This completes the proof of Lemma 5.

3. Sieve asymptotic formulae

In this section we establish formulae of the type (2.2) and use them as a stepping stone to formulae of type (1.3). In order to link (2.2) or (1.3) to the behaviour of Dirichlet polynomials we use the following variant of [2, Lemma 11].

Lemma 6. then

Let Fs k } x ck k ? s. If

T

T0

j

F

1 2

i

t

j

d

t

p

x

1

=

2L

?

A;

ck

kPA

y y1

ck O yL ? A:

kPB

3:1 3:2

Lemma 7. Let a and u be positive numbers, w x1= u and D xa. Suppose that

1= a < u < log x1 ? ?:

3:3

Then

d j Pw

1 d

p

explog

log

w

2

ua

?

u

a

log

u

a:

d >D

3:4

The implied constant is absolute.

Proof. Let r u log ua=L. We use the simple inequality expcy ? 1 < expc ? 1y

3:5

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