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013 SMALL GAPS BETWEEN PRIMES

JAMES MAYNARD

Abstract. We introduce a refinement of the GPY sieve method for studying primek-tuplesand small gaps between primes. This refinement avoids previous limitations of the method,and allows us to show that for eachk, the primek-tuples conjecture holds for a positiveproportion of admissiblek-tuples. In particular, lim infn(pn+m − pn) < ∞ for any integerm. We also show that lim inf(pn+1 − pn) ≤ 600, and, if we assume the Elliott-Halberstamconjecture, that lim infn(pn+1 − pn) ≤ 12 and lim infn(pn+2 − pn) ≤ 600.

1. Introduction

We say that a setH = {h1, . . . , hk} of distinct non-negative integers is ‘admissible’ iffor every primep there is an integerap such thatap . h (mod p) for all h ∈ H . We areinterested in the following conjecture.

Conjecture (Primek-tuples conjecture). LetH = {h1, . . . , hk} be admissible. Then thereare infinitely many integers n, such that all of n+ h1, . . . , n+ hk are prime.

Whenk > 1 no case of the primek-tuples conjecture is currently known. Work onapproximations to the primek-tuples conjecture has been very successful in showing theexistence of small gaps between primes, however. In their celebrated paper [5], Goldston,Pintz and Yıldırım introduced a new method for counting tuples of primes, and this allowedthem to show that

(1.1) lim infn

pn+1 − pnlog pn

= 0.

The recent breakthrough of Zhang [9] managed to extend this work to prove

(1.2) lim infn

(pn+1 − pn) ≤ 70 000 000,

thereby establishing for the first time the existence of infinitely many bounded gaps be-tween primes. Moreover, it follows from Zhang’s theorem thethat number of admissible2-tuples contained in [1, x]2 which satisfy the prime 2-tuples conjecture is≫ x2 for largex. Thus in this sense a positive proportion of admissible 2-tuples satisfy the prime 2-tuplesconjecture. The recent polymath project [7] has succeeded in reducing the bound (1.2) to4680, by optimizing Zhang’s arguments and introducing several new refinements.

The above results have used the ‘GPY method’ to study prime tuples and small gapsbetween primes, and this method relies heavily on the distribution of primes in arithmeticprogressions. Givenθ > 0, we say the primes have ‘level of distributionθ’1 if, for anyA > 0, we have

(1.3)∑

q≤xθ

max(a,q)=1

∣

∣

∣

∣π(x; q, a) −

π(x)ϕ(q)

∣

∣

∣

∣≪A

x(log x)A

.

1We note that different authors have given slightly different names or definitions to this concept. For thepurposes of this paper, (1.3) will be our definition of the primes having level of distributionθ.

1

http://arxiv.org/abs/1311.4600v2

2 JAMES MAYNARD

The Bombieri-Vinogradov theorem establishes that the primes have level of distributionθfor anyθ < 1/2, and Elliott and Halberstam [1] conjectured that this could be extended toanyθ < 1. Friedlander and Granville [2] have shown that (1.3) cannot hold withxθ replacedwith x/(log x)B for any fixedB, and so the Elliott-Halberstam conjecture is essentially thestrongest possible result of this type.

The original work of Goldston, Pintz and Yıldırım showed theexistence of boundedgaps between primes if (1.3) holds for someθ > 1/2. Moreover, under the Elliott-Halberstam conjecture one had lim infn(pn+1− pn) ≤ 16. The key breakthrough of Zhang’swork was in establishing a suitably weakened form of (1.3) holds for someθ > 1/2.

If one looks for bounded length intervals containing two or more primes, then the GPYmethod fails to prove such strong results. Unconditionallywe are only able to improveupon the trivial bound from the prime number theorem by a constant factor [4], and evenassuming the Elliott-Halberstam conjecture, the best available result [5] is

(1.4) lim infn

pn+2 − pnlog pn

= 0.

The aim of this paper is to introduce a refinement of the GPY method which removes thebarrier ofθ = 1/2 to establishing bounded gaps between primes, and allows usto showthe existence of arbitrarily many primes in bounded length intervals. This answers thesecond and third questions posed in [5] on extensions of the GPY method (the first havingbeen answered by Zhang’s result). Our new method also has thebenefit that it producesnumerically superior results to previous approaches.

Theorem 1.1. Let m∈ N. We have

lim infn

(pn+m− pn)≪ m3e4m.

Terence Tao (private communication) has independently proven Theorem 1.1 (with aslightly weaker bound) at much the same time. He uses a similar method; the steps aremore-or-less the same but the calculations are done differently. We will indicate some ofthe differences in our proofs as we go along.

We see that the bound in Theorem 1.1 is quite far from the conjectural bound of approx-imatelymlogm predicted by the primem-tuples conjecture.

Our proof naturally generalizes (but with a weaker upper bound) to many subsequencesof the primes which have a level of distributionθ > 0. In particular, we can show corre-sponding results where the primes are contained in short intervals [N,N + N7/12+ǫ ] for anyǫ > 0 or in an arithmetic progression moduloq≪ (logN)A.

Theorem 1.2. Let m ∈ N. Let r ∈ N be sufficiently large depending on m, and letA ={a1, a2, . . . , ar} be a set of r distinct integers. Then we have

#{{h1, . . . , hm} ⊆ A : for infinitely many n all of n+ h1, . . . , n+ hm are prime}#{{h1, . . . , hm} ⊆ A}

≫m 1.

Thus a positive proportion of admissiblem-tuples satisfy the primem-tuples conjecturefor everym, in an appropriate sense.

Theorem 1.3. We havelim inf

n(pn+1 − pn) ≤ 600.

We emphasize that the above result does not incorporate any of the technology used byZhang to establish the existence of bounded gaps between primes. The proof is essentiallyelementary, relying only on the Bombieri-Vinogradov theorem. Naturally, if we assumethat the primes have a higher level of distribution, then we can obtain stronger results.

SMALL GAPS BETWEEN PRIMES 3

Theorem 1.4. Assume that the primes have level of distributionθ for everyθ < 1. Then

lim infn

(pn+1 − pn) ≤ 12,

lim infn

(pn+2 − pn) ≤ 600.

Although the constant 12 of Theorem 1.4 appears to be optimalwith our method in itscurrent form, the constant 600 appearing in Theorem 1.3 and Theorem 1.4 is certainly notoptimal. By performing further numerical calculations ourmethod could produce a betterbound, and also most of the ideas of Zhang’s work (and the refinements produced by thepolymath project) should be able to be combined with this method to reduce the constantfurther.

2. An improved GPY sieve method

We first give an explanation of the key idea behind our new approach. The basic idea ofthe GPY method is, for a fixed admissible setH = {h1, . . . , hk}, to consider the sum

S(N, ρ) =∑

N≤n 0 andwn are non-negativeweights. If we can show thatS(N, ρ) > 0 then at least one term in the sum overn musthave a positive contribution. By the non-negativity ofwn, this means that there must besome integern ∈ [N, 2N] such that at least⌊ρ+1⌋ of then+hi are prime. (Here⌊x⌋ denotesthe largest integer less than or equal tox.) Thus if S(N, ρ) > 0 for all largeN, there areinfinitely many integersn for which at least⌊ρ+1⌋ of then+hi are prime (and so there areinfinitely many bounded length intervals containing⌊ρ + 1⌋ primes).

The weightswn are typically chosen to mimic Selberg sieve weights. Estimating (2.1)can be interpreted as a ‘k-dimensional’ sieve problem. The standard Selbergk-dimensionalweights (which can be shown to be essentially optimal in other contexts) are

(2.2) wn =(

∑

d|∏k

i=1(n+hi )d 1/2.

The new ingredient in our method is to consider a more generalform of the sieveweights

(2.4) wn =(

∑

di |n+hi∀i

λd1,...,dk

)2.

Using such weights withλd1,...,dk is the key feature of our method. It allows us to improveon the previous choice of sieve weights by an arbitrarily large factor, provided thatk is

4 JAMES MAYNARD

sufficiently large. It is the extra flexibility gained by allowingthe weights to depend on thedivisors of each factor individually which gives this improvement.

The idea to use such weights is not entirely new. Selberg [8, Page 245] suggested thepossible use of similar weights in his work on approximations to the twin prime problem,and Goldston and Yıldırım [6] considered similar weights inearlier work on the GPYmethod, but with the support restricted todi < R1/k for all i.

We comment that our choice ofλd1,...,dk will look like

(2.5) λd1,...,dk ≈(

k∏

i=1

µ(di))

f (d1, . . . , dk),

for a suitable smooth functionf . For our precise choice ofλd1,...,dk (given in Proposition4.1) we find it convenient to give a slightly different form ofλd1,...,dk, but weights of theform (2.5) should produce essentially the same results.

3. Notation

We shall viewk as a fixed integer, andH = {h1, . . . , hk} as a fixed admissible set. Inparticular, any constants implied by the asymptotic notation o, O or ≪ may depend onk andH . We will let N denote a large integer, and all asymptotic notation should beinterpreted as referring to the limitN → ∞.

All sums, products and suprema will be assumed to be taken over variables lying in thenatural numbersN = {1, 2, . . . } unless specified otherwise. The exception to this is whensums or products are over a variablep, which instead will be assumed to lie in the primenumbersP = {2, 3, . . . , }.

Throughout the paper,ϕwill denote the Euler totient function,τr (n) the number of waysof writing n as a product ofr natural numbers andµ the Moebius function. We will letǫbe a fixed positive real number, and we may assume without further comment thatǫ issufficiently small at various stages of our argument. We letpn denote thenth prime, and#A denote the number of elements of a finite setA. We use⌊x⌋ to denote the largest integern ≤ x, and⌈x⌉ the smallest integern ≥ x. We let (a, b) be the greatest common divisorof integersa and b. Finally, [a, b] will denote the closed interval on the real line withendpointsa andb, except for in Section 5 where it will denote the least commonmultipleof integersa andb instead.

4. Outline of the proof

We will find it convenient to choose our weightswn to be zero unlessn lies in a fixedresidue classv0 (mod W), whereW =

∏

p≤D0 p. This is a technical modification whichremoves some minor complications in dealing with the effect of small prime factors. Theprecise choice ofD0 is unimportant, but it will suffice to choose

(4.1) D0 = log log logN,

so certainlyW ≪ (log logN)2 by the prime number theorem. By the Chinese remaindertheorem we can choosev0 such thatv0+hi is coprime toW for eachi sinceH is admissible.Whenn ≡ v0 (mod W), we choose our weightswn of the form (2.4). We now wish to

SMALL GAPS BETWEEN PRIMES 5

estimate the sums

S1 =∑

N≤n 0. Letλd1,...,dk be defined in terms of a fixed piecewise differentiablefunction F by

λd1,...,dk =(

k∏

i=1

µ(di)di)

∑

r1,...,rkdi |r i∀i

(r i ,W)=1∀i

µ(∏k

i=1 r i)2

∏ki=1 ϕ(r i)

F

(

log r1logR

, . . . ,log rklogR

)

,

whenever(∏k

i=1 di ,W) = 1, and letλd1,...,dk = 0 otherwise. Moreover, let F be supported onRk = {(x1, . . . , xk) ∈ [0, 1]k :

∑ki=1 xi ≤ 1}. Then we have

S1 =(1+ o(1))ϕ(W)kN(logR)k

Wk+1Ik(F),

S2 =(1+ o(1))ϕ(W)kN(logR)k+1

Wk+1 logN

k∑

j=1

J(m)k (F),

provided Ik(F) , 0 and J(m)k (F) , 0 for each m, where

Ik(F) =∫ 1

0· · ·

∫ 1

0F(t1, . . . , tk)2dt1 . . .dtk,

J(m)k (F) =∫ 1

0· · ·

∫ 1

0

(∫ 1

0F(t1, . . . , tk)dtm

)2

dt1 . . .dtm−1dtm+1 . . .dtk.

We recall that ifS2 is large compared toS1 then using the GPY method we can showthat there are infinitely many integersn such that several of then + hi are prime. Thefollowing proposition makes this precise.

Proposition 4.2. Let the primes have level of distributionθ > 0. Let δ > 0 andH ={h1, . . . , hk} be an admissible set. Let Ik(F) and J

(m)k (F) be given as in Proposition 4.1, and

let Sk denote the set of piecewise differentiable functions F: [0, 1]k → R supported onRk = {(x1, . . . , xk) ∈ [0, 1]k :

∑ki=1 xi ≤ 1} with Ik(F) , 0 and J

(m)k (F) , 0 for each m. Let

Mk = supF∈Sk

∑km=1 J

(m)k (F)

Ik(F), rk =

⌈θMk2

⌉

.

Then there are infinitely many integers n such that at least rk of the n+ hi (1 ≤ i ≤ k) areprime. In particular,lim inf n(pn+rk−1 − pn) ≤ max1≤i, j≤k(hi − h j).

Proof of Proposition 4.2.We letS = S2 − ρS1, and recall that from Section 2 that if wecan showS > 0 for all largeN, then there are infinitely many integersn such that at leasttwo of then+ hi are prime.

6 JAMES MAYNARD

We putR = Nθ/2−ǫ for a smallǫ > 0. By the definition ofMk, we can chooseF0 ∈ Sksuch that

∑km=1 J

(m)k (F0) > (Mk − ǫ)Ik(F0). Using Proposition 4.1, we can then choose

λd1,...,dk such that

S =ϕ(W)kN(logR)k

Wk+1( logRlogN

k∑

j=1

J(m)k (F0) − ρIk(F0) + o(1))

≥ϕ(W)kN(logR)kIk(F0)

Wk+1(( θ

2− ǫ

)(

Mk − ǫ)

− ρ + o(1))

.(4.4)

If ρ = θMk/2− δ then, by choosingǫ sufficiently small, we see thatS > 0 for all largeN.Thus there are infinitely many integersn for which at least⌊ρ + 1⌋ of then+ hi are prime.Since⌊ρ + 1⌋ = ⌈θMk/2⌉ if δ is sufficiently small, we obtain Proposition 4.2. �

Thus, if the primes have a fixed level of distributionθ, to show the existence of many ofthen+hi being prime for infinitely manyn ∈ N we only require a suitable lower bound forMk. The following proposition establishes such a bound for different values ofk.

Proposition 4.3. Let k∈ N, and Mk be as given by Proposition 4.2. Then

(1) We have M5 > 2.(2) We have M105 > 4.(3) If k is sufficiently large, we have Mk > logk− 2 log logk− 2.

We now prove Theorems 1.1, 1.2, 1.3 and 1.4 from Propositions4.2 and 4.3.First we consider Theorem 1.3. We takek = 105. By Proposition 4.3, we haveM105 > 2.

By the Bombieri-Vinogradov theorem, the primes have level of distributionθ = 1/2 − ǫfor anyǫ > 0. Thus, if we takeǫ sufficiently small, we haveθM105/2 > 1. Therefore, byProposition 4.2, we have lim inf(pn+1 − pn) ≤ max1≤i, j≤105(hi − h j) for any admissible setH = {h1, . . . , h105}. By computations performed by Thomas Engelsma (unpublished), wecan choose2H such that 0≤ h1 < . . . < h105 andh105− h1 = 600. This gives Theorem 1.3.

If we assume the Elliott-Halberstam conjecture then the primes have level of distributionθ = 1− ǫ. First we takek = 105, and see thatθM105/2 > 2 for ǫ sufficiently small (sinceM105 > 4). Therefore, by Proposition 4.2, lim infn(pn+2− pn) ≤ max1≤i, j≤105(hi −h j). Thus,choosing the same admissible setH as above, we see lim infn(pn+2 − pn) ≤ 600 under theElliott-Halberstam conjecture.

Next we takek = 5 andH = {0, 2, 6, 8, 12}, with θ = 1− ǫ again. By Proposition 4.3we haveM5 > 2, and soθM5/2 > 1 for ǫ sufficiently small. Thus, by Proposition 4.2,lim inf n(pn+1− pn) ≤ 12 under the Elliott-Halberstam conjecture. This completes the proofof Theorem 1.4.

Finally, we consider the case whenk is large. For the rest of this section, any constantsimplied by asymptotic notation will be independent ofk. By the Bombieri-Vinogradovtheorem, we can takeθ = 1/2− ǫ. Thus, by Proposition 4.3 we have fork sufficiently large

(4.5)θMk

2≥

(14−ǫ

2

)(

logk− 2 log logk− 2)

.

2Explicitly, we can takeH = {0, 10, 12, 24, 28, 30, 34, 42, 48, 52, 54, 64, 70, 72, 78, 82, 90, 94, 100, 112,114, 118, 120, 124, 132, 138, 148, 154, 168, 174, 178, 180, 184, 190, 192, 202, 204, 208, 220, 222, 232, 234,250, 252, 258, 262, 264, 268, 280, 288, 294, 300, 310, 322, 324, 328, 330, 334, 342, 352, 358, 360, 364, 372,378, 384, 390, 394, 400, 402, 408, 412, 418, 420, 430, 432, 442, 444, 450, 454, 462, 468, 472, 478, 484, 490,492, 498, 504, 510, 528, 532, 534, 538, 544, 558, 562, 570, 574, 580, 582, 588, 594, 598, 600}. This set wasobtained from the websitehttp://math.mit.edu/˜primegaps/maintained by Andrew Sutherland.

http://math.mit.edu/~primegaps/

SMALL GAPS BETWEEN PRIMES 7

We chooseǫ = 1/k, and see thatθMk/2 > m if k ≥ Cm2e4m for some absolute constantC(independent ofmandk). Thus, for any admissible setH = {h1, . . . , hk} with k ≥ Cm2e4m,at leastm+ 1 of then + hi must be prime for infinitely many integersn. We can chooseour setH to be {pπ(k)+1, . . . , pπ(k)+k}. This is admissible, since no element is a multipleof a prime less thank (and there arek elements, so it cannot cover all residue classesmodulo any prime greater thank.) This set has diameterpk+π(k) − pπ(k)+1 ≪ k logk. Thuslim inf n(pn+m− pn) ≪ k logk≪ m3e4m if we takek = ⌈Cm2e4m⌉. This gives Theorem 1.1.

We can now establish Theorem 1.2 by a simple counting argument. Givenm, we letk = ⌈Cm2e4m⌉ as above. Therefore if{h1, . . . , hk} is admissible then there exists a subset{h′1, . . . , h

′m} ⊆ {h1, . . . , hk} with the property that there are infinitely many integersn for

which all of then+ h′i are prime (1≤ i ≤ m).We letA2 denote the set formed by starting with the given setA = {a1, . . . , ar} and

for each primep ≤ k in turn removing all elements of the residue class modulop whichcontains the fewest integers. We see that #A2 ≥ r

∏

p≤k(1 − 1/p) ≫m r. Moreover, anysubset ofA2 of sizek must be admissible, since it cannot cover all residue classes modulop for any primep ≤ k. We lets = #A2, and sincer is taken sufficiently large in terms ofm, we may assume thats> k.

We see there are(

sk

)

setsH ⊆ A2 of sizek. Each of these is admissible, and so containsat least one subset{h′1, . . . , h

′m} ⊆ A2 which satisfies the primem-tuples conjecture. Any

admissible setB ⊆ A2 of sizem is contained in(

s−mk−m

)

setsH ⊆ A2 of sizek. Thus there

are at least(

sk

)(

s−mk−m

)−1≫m sm ≫m rm admissible setsB ⊆ A2 of sizem which satisfy the

primem-tuples conjecture. Since there are(

rm

)

≤ rm sets{h1, . . . , hm} ⊆ A, Theorem 1.2holds.

We are left to establish Propositions 4.1 and 4.3.

5. Selberg sieve manipulations

In this section we perform initial manipulations towards establishing Proposition 4.1.These arguments are multidimensional generalizations of the sieve arguments of [3]. Inparticular, our approach is based on the elementary combinatorial ideas of Selberg. Theaim is to introduce a change of variables to rewrite our sumsS1 andS2 in a simpler form.

Throughout the rest of the paper we assume that the primes have a fixed level of distri-butionθ, andR= Nθ/2−ǫ . We restrict the support ofλd1,...,dk to tuples for which the productd =

∏ki=1 di is less thanR and also satisfies (d,W) = 1 andµ(d)

2 = 1. We note that theconditionµ(d)2 = 1 implies that (di , d j) = 1 for all i , j.

Lemma 5.1. Let

yr1,...,rk =(

k∏

i=1

µ(r i)ϕ(r i))

∑

d1,...,dkr i |di∀i

λd1,...,dk∏k

i=1 di.

Let ymax= supr1,...,rk |yr1,...,rk |. Then

S1 =NW

∑

r1,...,rk

y2r1,...,rk∏k

i=1 ϕ(r i)+O

(y2maxN(logR)k

WD0

)

.

Proof. We expand out the square, and swap the order of summation to give

(5.1) S1 =∑

N≤n

8 JAMES MAYNARD

We recall that here, and throughout this section, we are using [a, b] to denote the leastcommon multiple ofa andb.

By the Chinese remainder theorem, the inner sum can be written as a sum over a singleresidue class moduloq = W

∏ki=1[di , ei ] provided that the integersW, [d1, e1], . . . , [dk, ek]

are pairwise coprime. In this case the inner sum isN/q + O(1). If the integers are notpairwise coprime then the inner sum is empty. This gives

S1 =NW

∑′

d1,...,dke1,...,ek

λd1,...,dkλe1,...,ek∏k

i=1[di , ei ]+O

(∑′

d1,...,dke1,...,ek

|λd1,...,dkλe1,...,ek |)

,(5.2)

where∑′ is used to denote the restriction that we requireW, [d1, e1], . . . , [dk, ek] to be

pairwise coprime. To ease notation we will putλmax= supd1,...,dk |λd1,...,dk |. We now see thatsinceλd1,...,dk is non-zero only when

∏ki=1 di < R, the error term contributes

(5.3) ≪ λ2max(∑

d

SMALL GAPS BETWEEN PRIMES 9

This change is invertible. Ford1, . . . , dk with∏k

i=1 di square-free we find that

∑

r1,...,rkdi |r i∀i

yr1,...,rk∏k

i=1 ϕ(r i)=

∑

r1,...,rkdi |r i∀i

(

k∏

i=1

µ(r i))

∑

e1,...,ekr i |ei∀i

λe1,...,ek∏k

i=1 ei

=∑

e1,...,ek

λe1,...,ek∏k

i=1 ei

∑

r1,...,rkdi |r i∀ir i |ei∀i

k∏

i=1

µ(r i) =λd1,...,dk

∏ki=1 µi(di)di

.(5.8)

Thus any choice ofyr1,...,rk supported onr1, . . . , rk with the productr =∏k

i=1 r i square-free and satisfyingr < R and (r,W) = 1 will give a suitable choice ofλd1,...,dk . We letymax = supr1,...,rk |yr1,...,rk |. Now, sinced/ϕ(d) =

∑

e|d 1/ϕ(e) for square-freed, we find bytakingr ′ =

∏ki=1 r i/di that

λmax≤ supd1,...,dk

∏ki=1 di square-free

ymax(

k∏

i=1

di)

∑

r1,...,rkdi |r i∀i

∏ki=1 r i

10 JAMES MAYNARD

We see that there is no contribution fromsi, j with (si, j ,W) , 1 because of the restrictedsupport ofy. Thus we only need to considersi, j = 1 or si, j > D0. The contribution whensi, j > D0 is

≪y2maxN

W

(∑

uD0

µ(si, j)2

ϕ(si, j)2)(∑

s≥1

µ(s)2

ϕ(s)2)k2−k−1

≪y2maxϕ(W)

kN(logR)k

Wk+1D0.

(5.12)

Thus we may restrict our attention to the case whensi, j = 1 ∀i , j. This gives

(5.13) S1 =NW

∑

u1,...,uk

y2u1,...,uk∏k

i=1 ϕ(ui)+O

(

y2maxϕ(W)kN(logR)k

Wk+1D0+ y2maxR

2(logR)4k)

.

We recall thatR2 = Nθ−2δ ≤ N1−2δ andW ≪ Nδ, and so the first error term dominates.This gives the result. �

We now considerS2. We writeS2 =∑k

m=1 S(m)2 , where

(5.14) S(m)2 =∑

N≤n 0 we have

S(m)2 =N

ϕ(W) logN

∑

r1,...,rk

(y(m)r1,...,rk)2

∏ki=1 g(r i)

+O( (y(m)max)2ϕ(W)k−2N(logN)k−2

Wk−1D0

)

+O( y2maxN

(logN)A)

.

Proof. We first expand out the square and swap the order of summation to give

(5.15) S(m)2 =∑

d1,...,dke1,...,ek

λd1,...,dkλe1,...,ek

∑

N≤n

SMALL GAPS BETWEEN PRIMES 11

If either one pair ofW, [d1, e1], . . . , [dk, ek] share a common factor, or if eitherdm or em arenot 1, then the contribution of the inner sum is zero. Thus we obtain

(5.18) S(m)2 =XNϕ(W)

∑′

d1,...,dke1,...,ek

em=dm=1

λd1,...,dkλe1,...,ek∏k

i=1 ϕ([di, ei ])+O

(∑

d1,...,dke1,...,ek

|λd1,...,dkλe1,...,ek |E(N, q))

,

where we have writtenq =W∏k

i=1[di, ei ].We first deal with the contribution from the error terms. Fromthe support ofλd1,...,dk we

see that we only need to consider square-freeq with q < R2W. Given a square-free integerr, there are at mostτ3k(r) choices ofd1, . . . , dk, e1, . . . , ek for whichW

∏ki=1[di, ei ] = r. We

also recall from (5.9) thatλmax≪ ymax(logR)k. Thus the error term contributes

(5.19) ≪ y2max(logR)2k

∑

r 0

(5.20) ≪ y2max(logR)2k(∑

r

12 JAMES MAYNARD

We see the contribution fromsi, j , 1 is of size

≪(y(m)max)2Nϕ(W) logN

(∑

uD0

µ(si1,i2)2

g(si, j)2

≪(y(m)max)2ϕ(W)k−2N(logR)k−1

Wk−1D0 logN.(5.25)

Thus we find that

(5.26) S(m)2 =XNϕ(W)

∑

u1,...,uk

(y(m)u1,...,uk)2

∏ki=1 g(ui)

+O( (y(m)max)2ϕ(W)k−2N(logR)k−2

D0Wk−1)

+O( y2maxN

(logN)A)

.

Finally, by the prime number theorem,XN = N/ logN + O(N/(logN)2). This error termcontributes

(5.27) ≪(y(m)max)2Nϕ(W)(logN)2

(∑

u D0r j . For j , m, the total contribution from

SMALL GAPS BETWEEN PRIMES 13

a j , r j is

≪ ymax(

k∏

i=1

g(r i)r i)(

∑

aj>D0r j

µ(a j)2

ϕ(a j)2)(

∑

am 0. Letγ be a multiplicative function satisfying

0 ≤γ(p)

p≤ 1− A1,

14 JAMES MAYNARD

and

−L ≤∑

w≤p≤z

γ(p) log pp

− κ logz/w ≤ A2

for any 2 ≤ w ≤ z. Let g be the totally multiplicative function defined on primes byg(p) = γ(p)/(p− γ(p)). Finally, let G : [0, 1] → R be a piecewise differentiable functionand let Gmax= supt∈[0,1](|G(t)| + |G

′(t)|). Then

∑

dD0

∑

u1,...,ukD0

1(p− 1)2

(∑

u

SMALL GAPS BETWEEN PRIMES 15

Thus we are left to evaluate the sum

(6.6)∑

u1,...,uk(ui ,W)=1∀i

(

k∏

i=1

µ(ui)2

ϕ(ui)

)

F( logu1

logR, . . . ,

loguklogR

)2.

We can now estimate this sum byk applications of Lemma 6.1, dealing with the sum overeachui in turn. For each application we takeκ = 1 and

γ(p) =

1, p ∤W,

0, otherwise,(6.7)

L≪ 1+∑

p|W

log pp≪ logD0,(6.8)

andA1 andA2 fixed constants of suitable size. This gives

∑

u1,...,uk(ui ,W)=1∀i

(

k∏

i=1

µ(ui)2

ϕ(ui)

)

F( logu1

logR, . . . ,

loguklogR

)2=ϕ(W)k(logR)k

WkIk(F)

+O(F2maxϕ(W)

k(logD0)(logR)k−1

Wk)

.(6.9)

We now combine (6.9) with (6.4) and (6.5) to obtain the result. �

Lemma 6.3. Let yr1,...,rk, F and Fmax be as described in Lemma 6.2. Then we have

S(m)2 =ϕ(W)kN(logR)k+1

Wk+1 logNJ(m)k (F) +O

(F2maxϕ(W)kN(logR)k

Wk+1D0

)

,

where

J(m)k (F) =∫ 1

0· · ·

∫ 1

0

(

∫ 1

0F(t1, . . . , tk)dtm

)2dt1 . . .dtm−1dtm+1 . . .dtk.

Proof. The estimation ofS(m)2 is similar to the estimation ofS1. We first estimatey(m)r1,...,rk.

We recall thaty(m)r1,...,rk = 0 unlessrm = 1 andr =∏k

i=1 r i satisfies (r,W) = 1 andµ(r)2 = 1,

in which casey(m)r1,...,rk is given in terms ofyr1,...,rk by Lemma 5.3. We first concentrate on thiscase wheny(m)r1,...,rk , 0. We substitute our choice (6.3) ofy into our expression from Lemma5.3. This gives

y(m)r1,...,rk =∑

(u,W∏k

i=1 r i)=1

µ(u)2

ϕ(u)F( log r1logR

, . . . ,log rm−1

logR,

logulogR

,log rm+1

logR, . . . ,

log rklogR

)

+O(Fmaxϕ(W) logR

WD0

)

.(6.10)

We can see from this thaty(m)max≪ ϕ(W)Fmax(logR)/W. We now estimate the sum overu in(6.10). We apply Lemma 6.1 withκ = 1 and

γ(p) =

1, p ∤W∏k

i=1 r i ,

0, otherwise,(6.11)

L ≪ 1+∑

p|W∏k

i=1 r i

log pp≪

∑

plogR

log logRlogR

≪ log logN,(6.12)

16 JAMES MAYNARD

and withA1,A2 suitable fixed constants. This gives us

y(m)r1,...,rk = (logR)ϕ(W)

W(

k∏

i=1

ϕ(r i)r i

)

F(m)r1,...,rk +O(Fmaxϕ(W) logR

WD0

)

,(6.13)

where

(6.14) F(m)r1,...,rk =∫ 1

0F( log r1logR

, . . . ,log rm−1

logR, tm,

log rm+1logR

, . . . ,log rklogR

)

dtm.

Thus we have shown that ifrm = 1 andr =∏k

i=1 r i satisfies (r,W) = 1 andµ(r)2 = 1

thenyr1,...,rk is given by (6.13), and otherwisey(m)r1,...,rk = 0. We now substitute this into our

expression(6.15)

S(m)2 =N

ϕ(W) logN

∑

r1,...,rk

(y(m)r1,...,rk)2

∏ki=1 g(r i)

+O( (y(m)max)2ϕ(W)k−2N(logN)k−2

Wk−1D0

)

+O( y2maxN

(logN)A)

.

from Lemma 5.2. We obtain

S(m)2 =ϕ(W)N(logR)2

W2 logN

∑

r1,...,rk(r i ,W)=1∀i

(r i ,r j)=1∀i, jrm=1

(

k∏

i=1

µ(r i)2ϕ(r i)g(r i)r i

)

(F(m)r1,...,rk)2 +O

(F2maxϕ(W)kN(logR)k

Wk+1D0

)

.

(6.16)

We remove the condition that (r i , r j) = 1 in the same way we did when consideringS1.Instead of (6.5), this introduces an error which is of size

≪ϕ(W)N(logR)2F2max

W2 logN

(∑

p>D0

ϕ(p)2

g(p)2p2)(

∑

r

SMALL GAPS BETWEEN PRIMES 17

Remark. If F (t1, . . . , tk) = G(∑k

i=1 ti) for some function G, then Ik(F) and J(m)k (F) simplify

to Ik(F) =∫ 1

0G(t)2tk−1dt/(k− 1)! and J(m)k (F) =

∫ 1

0(∫ 1

tG(v)dv)2tk−2dt/(k− 2)! for each m,

which is equivalent to the results obtained using the original GPY method using weightsgiven by(2.3).

Remark. Tao gives an alternative approach to arrive at his equivalent of Proposition 4.1.His approach is to defineλd1,...,dk in terms of a suitable smooth function f(t1, . . . , tk) as in(2.5). He then estimates the corresponding sums directly using Fourier integrals. This issomewhat similar to the original paper of Goldston, Pintz and Yıldırım[5]. Our functionF corresponds to f(t1, . . . , tk) differentiated with respect to each coordinate.

7. Choice of weight for small k

We letSk denote the set of piecewise differentiable functionsF : [0, 1]k→ R supportedonRk = {(x1, . . . , xk) :

∑ki=1 xi ≤ 1} with Ik(F) , 0 andJ

(m)k (F) , 0 for eachm. We would

like to obtain a lower bound for

(7.1) Mk = supF∈Sk

∑km=1 J

(m)k (F)

Ik(F).

Remark. LetLk denote the integral linear operator defined by

(7.2) LkF(u1, . . . , uk) =k

∑

m=1

∫ 1−∑

i,m ui

0F(u1, . . . , um−1, tm, um+1, . . . , uk)dtm

whenever(u1, . . . , uk) ∈ Rk, and zero otherwise. By standard calculus of variations argu-ments, we find that if F maximizes the ratio

∑km=1 J

(m)k (F)/Ik(F) then F is an eigenfunction

for Lk, and the corresponding eigenvalue is the value of ratio at F.Unfortunately theauthor has not been able to solve the eigenvalue equation forLk when k> 2.

In order to get a suitable lower bound forMk whenk is small, we will consider approx-imations to the optimal functionF of the form

(7.3) F(t1, . . . , tk) =

P(t1, . . . , tk), if ( t1, . . . , tk) ∈ Rk0, otherwise,

for polynomialsP. By the symmetry of∑k

m=1 J(m)k (F) andIk(F), without loss of generality

we can restrict our attention to polynomials which are symmetric functions oft1, . . . , tk.(If F satisfiesLkF = λF thenFσ = F(σ(t1), . . . , σ(tk)) also satisfies this for any permu-tationσ of t1, . . . , tk. Thus the symmetric function which is the average ofFσ over allsuch permutations would satisfy this eigenfunction equation, and so there must be an op-timal function which is symmetric.) Any such polynomial canbe written as a polynomialexpression in the power sum polynomialsP j =

∑ki=1 t

ji .

Lemma 7.1. Let Pj =∑k

i=1 tji denote the j

th symmetric power sum polynomial. Then wehave (

Rk

(1− P1)aPbj dt1 . . .dtk =a!

(k+ jb + a)!Gb, j(k),

where

Gb, j(x) = b!b

∑

r=1

(

xr

)

∑

b1,...,br≥1∑r

i=1 bi=b

r∏

i=1

( jbi)!bi !

is a polynomial of degree b which depends only on b and j.

18 JAMES MAYNARD

Proof. We first show by induction onk that

(7.4)(

Rk

(

1−k

∑

i=1

ti)a

k∏

i=1

taii dt1 . . .dtk =a!

∏ki=1 ai !

(k+ a+∑k

i=1 ai)!.

We consider the integration with respect tot1. The limits of integration are 0 and 1−∑k

i=2 tifor (t2, . . . , tk) ∈ Rk−1. By substitutingv = t1/(1−

∑ki=2 ti) we find

∫ 1−∑k

i=2 ti

0

(

1−k

∑

i=1

ti)a(

k∏

i=1

taii)

dt1 =(

k∏

i=2

taii)(

1−k

∑

i=2

ti)a+a1+1

∫ 1

0(1− v)ava1dv

=a!a1!

(a+ a1 + 1)!

(

k∏

i=2

taii)(

1−k

∑

i=2

ti)a+a1+1

.(7.5)

Here we used the beta function identity∫ 1

0ta(1− t)bdt = a!b!/(a+ b+ 1)! in the last line.

We now see (7.4) follows by induction.By the binomial theorem,

(7.6) Pbj =∑

b1,...,bk∑k

i=1 bi=b

b!∏k

i=1 bi !

k∏

i=1

t jbii .

Thus, applying (7.4), we obtain

(7.7)(

Rk

(1− P1)aPbj dt1 . . .dtk =b!a!

(k+ a+ jb)!

∑

b1,...,bk∑k

i=1 bi=b

k∏

i=1

( jbi)!bi!.

For computationsb will be small, and so we find it convenient to split the summationdepending on how many of thebi are non-zero. Given an integerr, there are

(

kr

)

ways ofchoosingr of b1, . . . , bk to be non-zero. Thus

(7.8)∑

b1,...,bk∑k

i=1 bi=b

k∏

i=1

( jbi)!bi !

=

b∑

r=1

(

kr

)

∑

b1,...,br≥1∑r

i=1 bi=b

r∏

i=1

( jbi)!bi !.

This gives the result. �

It is straightforward to extend Lemma 7.1 to more general combinations of the sym-metric power polynomials. In this paper we will concentrateon the case whenP is apolynomial expression in onlyP1 andP2 for simplicity. We comment the polynomialsGb, jare not problematic to calculate numerically for small values ofb. We now use Lemma 7.1to obtain a manageable expression forIk(F) andJ

(m)k (F) with this choice ofP.

Lemma 7.2. Let F be given in terms of a polynomial P by(7.3). Let P be given in terms ofa polynomial expression in the symmetric power polynomialsP1 =

∑ki=1 ti and P2 =

∑ki=1 t

2i

by P=∑d

i=1 ai(1− P1)bi Pci2 for constants ai ∈ R and non-negative integers bi , ci . Then for

SMALL GAPS BETWEEN PRIMES 19

each1 ≤ m≤ k we have

Ik(F) =∑

1≤i, j≤d

aia j(bi + b j)!Gci+c j ,2(k)

(k+ bi + b j + 2ci + 2c j)!,

J(m)k (F) =∑

1≤i, j≤d

aia j

ci∑

c′1=0

c j∑

c′2=0

(

cic′1

)(

c jc′2

)

γbi ,bj ,ci ,c j ,c′1,c′2Gc′1+c′2,2(k− 1)

(k+ bi + b j + 2ci + 2c j + 1)!,

where

γbi ,bj ,ci ,c j ,c′1,c′2=

bi !b j!(2ci − 2c′1)!(2c j − 2c′2)!(bi + b j + 2ci + 2c j − 2c

′1 − 2c

′2 + 2)!

(bi + 2ci − 2c′1 + 1)!(b j + 2c j − 2c′2 + 1)!,

and where G is the polynomial given by Lemma 7.1.

Proof. We first considerIk(F). We have, using Lemma 7.1,

Ik(F) =(

Rk

P2dt1 . . .dtk =∑

1≤i, j≤d

aia j

(

Rk

(1− P1)bi+bj P

ci+c j2 dt1 . . .dtk

=∑

1≤i, j≤d

aia j(bi + b j)!Gci+c j ,2(k)

(k+ bi + b j + 2ci + 2c j)!.(7.9)

We now considerJ(m)k (F). SinceF is symmetric int1, . . . , tk we see thatJ(m)k (F) is inde-

pendent ofm, and so it suffices to only considerJ(1)k (F). We have

∫ 1−∑k

i=2 ti

0(1− P1)

bPc2dt1 =c

∑

c′=0

(

cc′

)

(

k∑

i=2

t2i)c′

∫ 1−∑k

i=2 ti

0

(

1−k

∑

i=1

ti)b

t2c−2c′

1 dt1

=

c∑

c′=0

(

cc′

)

(P′2)c′(1− P′1)

b+2c−2c′+1∫ 1

0(1− u)bu2c−2c

′

du

=

c∑

c′=0

(

cc′

)

(P′2)c′(1− P′1)

b+2c−2c′+1 b!(2c− 2c′)!

(b+ 2c− 2c′ + 1)!,(7.10)

whereP′1 =∑k

i=2 ti andP′2 =

∑ki=2 t

2i . Thus

(

∫ 1

0Fdt1

)2=

(

d∑

i=1

ai

∫ 1−∑k

j=2 t j

0(1− P1)bi P

ci2

)2

=∑

1≤i, j≤d

aia j

ci∑

c′1=0

c j∑

c′2=0

(

cic′1

)(

c jc′2

)

(P′2)c′1+c

′2(1− P′1)

bi+bj+2ci+2c j−2c′1−2c′2+2

×bi!b j !(2ci − 2c′1)!(2c j − 2c

′2)!

(bi + 2ci − 2c′1 + 1)!(b j + 2c j − 2c′2 + 1)!

.(7.11)

Applying Lemma 7.1 again, we see that

(7.12)(

Rk−1

(1− P′1)b(P′2)

c′dt2 . . .dtk =b!

(k+ b+ c− 1)!Gc,2(k− 1).

Combining (7.11) and (7.12) gives the result. �

20 JAMES MAYNARD

We see from Lemma 7.2 thatIk(F) and∑k

m=1 J(m)k (F) can both be expressed as quadratic

forms in the coefficientsa = (a1, . . . , ad) of P. Moreover, these will be positive definitereal quadratic forms. Thus in particular we find that

(7.13)

∑km=1 J

(m)k (F)

Ik(F)=

aT M2aaT M1a

,

for two rational symmetric positive definite matricesM1,M2, which can be calculated ex-plicitly in terms ofk for any choice of the exponentsbi , ci. Maximizing expressions of thisform has a well-known solution.

Lemma 7.3. Let M1,M2 be real, symmetric positive definite matrices. Then

aT M2aaT M1a

is maximized whena is an eigenvector of M−11 M2 corresponding to the largest eigenvalueof M−11 M2. The value of the ratio at its maximum is this largest eigenvalue.

Proof. We see that multiplyinga by a non-zero scalar doesn’t change the ratio, so we mayassume without loss of generality thataT M1a = 1. By the theory of Lagrangian multipliers,aT M2a is maximized subject toaT M1a = 1 when

(7.14) L(a, λ) = aT M2a − λ(aT M1a − 1)

is stationary. This occurs when (using the symmetricity ofM1,M2)

(7.15) 0=∂L∂ai= ((2M2 − 2λM1)a)i ,

for eachi. This implies that (recalling thatM1 is positive definite so invertible)

(7.16) M−11 M2a = λa.

It then is clear thataT M1a = λ−1aT M2a. �

Proof of parts(i) and(ii ) of Proposition 4.3.To establish Proposition 4.3 we rely on somecomputer calculation to calculate a lower bound forMk. We letF be given in terms of apolynomialP by (7.3). We letP be given by a polynomial expression inP1 =

∑ki=1 ti and

P2 =∑k

i=1 t2i which is a linear combination of all monomials (1− P1)

bPc2 with b+ 2c ≤ 11.There are 42 such monomials, and withk = 105 we can calculate the 42× 42 rationalsymmetric matricesM1 and M2 corresponding to the coefficients of the quadratic formsIk(F) and

∑km=1 Jk(F). We then find

3 that the largest eigenvalue ofM−11 M2 is

(7.17) λ ≈ 4.0020697. . . > 4.

ThusM105 > 4. This verifies part (i) of Proposition 4.3. We comment that by taking arational approximation to the corresponding eigenvector,we can verify this lower boundby calculating the ratio

∑km=1 Jk(F)/Ik(F) using only exact arithmetic.

For part (ii ) of Proposition 4.3, we takek = 5 and

(7.18) P = (1− P1)P2 +710

(1− P1)2 +114

P2 −314

(1− P1).

With this choice we find that

�(7.19) M5 ≥

∑km=1 J

(m)k (F)

I j(F)=

1 417 255708 216

> 2.

3An ancillary MathematicaR© file detailing these computations is available alongside this paper atwww.arxiv.org.

www.arxiv.org

SMALL GAPS BETWEEN PRIMES 21

8. Choice of smooth weight for large k

In this section we establish part (iii ) of Proposition 4.3. Our argument here is closelyrelated to that of Tao, who uses a probability theory proof.

We obtain a lower bound forMk by constructing a functionF which makes the ratio∑k

m=1 J(m)k (F)/Ik(F) large for all largek. We chooseF to be of the form

(8.1) F(t1, . . . , tk) =

∏ki=1 g(kti), if

∑ki=1 ti ≤ 1,

0, otherwise,

for some piecewise differentiable functiong : [0,∞] → R, supported on [0,T]. We seethat with this choiceF is symmetric, and soJ(m)k (F) is independent ofm. Thus we only

need to considerJk = J(1)k (F). Similarly we writeIk = Ik(F).

The key observation is that if the center of mass∫ ∞

0ug(u)2du/

∫ ∞

0g(u)2du of g2 is

strictly less that 1, then for largek we expect that the constraints∑k

i=1 ti ≤ 1 to be able tobe dropped at the cost of only a small error. This is because (by concentration of measure)the main contribution to the unrestricted integralsI ′k =

∫ ∞

0· · ·

∫ ∞

0

∏ki=1 g(kti)

2dt1 . . .dtk and

J′k =∫ ∞

0· · ·

∫ ∞

0(∫ ∞

0

∏ki=1 g(kti)dt1)

2dt2 . . .dtk should come primarily from when∑k

i=1 ti is

close to the center of mass. Therefore we would expect the contribution when∑k

i=1 ti > 1to be small if the center of mass is less than 1, and soIk andJk are well approximated byI ′k andJ

′k in this case.

To ease notation we letγ =∫

t≥0g(t)2dt, and restrict our attention tog such thatγ > 0.

We have

(8.2) Ik =(

Rk

F(t1, . . . , tk)2dt1 . . .dtk ≤

(

∫ ∞

0g(kt)2dt

)k= k−kγk.

We now considerJk. Since squares are non-negative, we obtain a lower bound forJk if werestrict the outer integral to

∑ki=2 ti < 1− T/k. This has the advantage that, by the support

of g, there are no further restrictions on the inner integral. Thus

(8.3) Jk ≥(

t2,...,tk≥0∑k

i=2 ti≤1−T/k

(

∫ T/k

0

(

k∏

i=1

g(kti))

dt1)2

dt2 . . .dtk.

We write the right hand side of (8.3) asJ′k − Ek, where

J′k =(

t2,...,tk≥0

(

∫ T/k

0

(

k∏

i=1

g(kti))

dt1)2

dt2 . . .dtk

=(

∫ ∞

0g(kt1)dt1

)2(∫ ∞

0g(kt)2dt

)k−1= k−k−1γk−1

(

∫ ∞

0g(u)du

)2,(8.4)

Ek =(

t2,...,tk≥0∑k

i=2 ti>1−T/k

(

∫ T/k

0

(

k∏

i=1

g(kti))

dt1)2

dt2 . . .dtk

= k−k−1(

∫ ∞

0g(u)du

)2(

u2,...,uk≥0∑k

i=2 ui>k−T

k∏

i=2

g(ui)2du2 . . .duk.(8.5)

22 JAMES MAYNARD

First we wish to show the error integralEk is small. We do this by comparison with asecond moment. We expect the bound (8.11) forEk to be small if the center of mass ofg2

is strictly less than (k− T)/(k− 1). Therefore we introduce the restriction ong that

(8.6) µ =

∫ ∞

0ug(u)2du

∫ ∞

0g(u)2du

< 1−Tk.

To simplify notation, we putη = (k − T)/(k − 1) − µ > 0. If∑k

i=2 ui > k − T then∑k

i=2 ui > (k− 1)µ and so we have

(8.7) 1≤ η−2( 1k− 1

k∑

i=2

ui − µ)2.

Since the right hand side of (8.7) is non-negative for allui , we obtain an upper bound forEk if we multiply the integrand byη−2(

∑ki=2 ui/(k− 1)− µ)

2, and then drop the requirementthat

∑ki=1 ui > k− T. This gives us

Ek ≤ η−2k−k−1

(

∫ ∞

0g(u)du

)2∫ ∞

0· · ·

∫ ∞

0

(

∑ki=2 ui

k− 1− µ

)2(k

∏

i=2

g(ui)2)

du2 . . .duk.(8.8)

We expand out the inner square. All the terms which are not of the formu2j we can calculateexplicitly in as an expression inµ andγ. We find

∫ ∞

0· · ·

∫ ∞

0

(2∑

2≤i< j≤k uiu j(k− 1)2

−2µ

∑ki=2 ui

k− 1+ µ2

)

du2 . . .duk =−µ2γk−1

k− 1.(8.9)

For theu2j terms we see thatu2j g(u j)

2 ≤ Tujg(u j)2 from the support ofg. Thus

∫ ∞

0· · ·

∫ ∞

0u2j

(

k∏

i=2

g(ui)2)

du2 . . .duk ≤ Tγk−2

∫ ∞

0u jg(u j)2duj = µTγk−1.(8.10)

This gives

Ek ≤ η−2k−k−1

(

∫ ∞

0g(u)du

)2(µTγk−1

k− 1−µ2γk−1

k− 1

)

≤η−2µTk−k−1γk−1

k− 1

(

∫ ∞

0g(u)du

)2.

(8.11)

Since (k−1)η2 ≥ k(1−T/k−µ)2 andµ ≤ 1, we find that putting together (8.2), (8.3), (8.4)and (8.11) we obtain

(8.12)kJkIk≥

(∫ ∞

0g(u)du)2

∫ ∞

0g(u)2du

(

1−T

k(1− T/k− µ)2)

.

To maximize our lower bound (8.12), we wish to maximize∫ T

0g(u)dusubject to the con-

straints that∫ T

0ug(u)2du = µ and

∫ T

0g(u)2du = γ. Thus we wish to maximize the expres-

sion

(8.13)∫ T

0g(u)du− α

(

∫ T

0g(u)2du− γ

)

− β(

∫ T

0ug(u)2du− µ

)

with respect toα, β and the functiong. By the Euler-Lagrange equation, this occurs when∂∂g(g(t) − αg(t)

2 − βtg(t)2) = 0 for all t ∈ [0,T]. Thus we see that

(8.14) g(t) =1

2α + 2βtfor 0 ≤ t ≤ T.

SMALL GAPS BETWEEN PRIMES 23

Since the ratio we wish to maximize is unaffected if we multiplyg by a positive constant,we restrict our attention to functionsg is of the form 1/(1+ At) for t ∈ [0,T] and for someconstantA > 0. With this choice ofg we find that

∫ T

0g(u)du=

log(1+ AT)A

,

∫ T

0g(u)2du=

1A

(

1−1

1+ AT

)

,(8.15)

∫ T

0ug(u)2du=

1A2

(

log(1+ AT) − 1+1

1+ AT

)

.(8.16)

We chooseT such that 1+ AT = eA (which is close to optimal). With this choice we findthatµ = 1/(1− e−A) − A−1 andT ≤ eA/A. Thus 1− T/k− µ ≥ A−1(1− A/(eA − 1)− eA/k).Substituting (8.15) into (8.12) and then using these expressions, we find that

(8.17)kJkIk≥

A1− e−A

(

1−T

k(1− T/k− µ)2)

≥ A(

1−AeA

k(1− A/(eA − 1)− eA/k)2)

,

provided the right hand side is positive. Finally, we chooseA = logk− 2 log logk > 0. Fork sufficiently large we have

(8.18) 1−kT− µ ≥ A−1

(

1−(logk)3

k−

1(logk)2

)

> 0,

and soµ < 1− T/k, as required by our constraint (8.6). This choice ofA gives

(8.19) Mk ≥kJkIk≥ (logk− 2 log logk)

(

1−logk

(logk)2 +O(1)

)

≥ logk− 2 log logk− 2

whenk is sufficiently large. This completes the proof of Proposition 4.3.

9. Acknowledgements

The author would like to thank Andrew Granville, Roger Heath-Brown, Dimitris Kouk-oulopoulos and Terence Tao for many useful conversations and suggestions.

The work leading to this paper was started whilst the author was a D.Phil student at Ox-ford and funded by the EPSRC (Doctoral Training Grant EP/P505216/1), and was finishedwhen the author was a CRM-ISM postdoctoral fellow at the Université de Montréal.

References

[1] P. D. T. A. Elliott and H. Halberstam. A conjecture in prime number theory. InSymposia Mathematica, Vol.IV (INDAM, Rome, 1968/69), pages 59–72. Academic Press, London, 1970.

[2] J. Friedlander and A. Granville. Limitations to the equi-distribution of primes. I.Ann. of Math. (2),129(2):363–382, 1989.

[3] D. A. Goldston, S. W. Graham, J. Pintz, and C. Y. Yıldırım.Small gaps between products of two primes.Proc. Lond. Math. Soc. (3), 98(3):741–774, 2009.

[4] D. A. Goldston, J. Pintz, and C. Y. Yıldırım. Primes in tuples. III. On the differencepn+ν − pn. Funct. Approx.Comment. Math., 35:79–89, 2006.

[5] D. A. Goldston, J. Pintz, and C. Y. Yıldırım. Primes in tuples. I.Ann. of Math. (2), 170(2):819–862, 2009.[6] D. A. Goldston and C. Y. Yıldırım. Higher correlations ofdivisor sums related to primes. III. Small gaps

between primes.Proc. Lond. Math. Soc. (3), 95(3):653–686, 2007.[7] D. H. J. Polymath. A new bound for gaps between primes.Preprint.[8] A. Selberg.Collected papers. Vol. II. Springer-Verlag, Berlin, 1991. With a foreword by K. Chandrasekharan.[9] Y. Zhang. Bounded gaps between primes.Ann. of Math.(2), to appear.

Centre de recherches mathématiques, Université deMontréal, Pavillon André-Aisenstadt, 2920 Chemin dela tour, Room 5357, Montréal (Québec) H3T 1J4

E-mail address: maynardj@dms.umontreal.ca

1. Introduction2. An improved GPY sieve method3. Notation4. Outline of the proof5. Selberg sieve manipulations6. Smooth choice of y7. Choice of weight for small k8. Choice of smooth weight for large k9. AcknowledgementsReferences