We continue the discussion of sieve theory from Notes 4, but now specialise to the case of the *linear sieve* in which the sieve dimension is equal to , which is one of the best understood sieving situations, and one of the rare cases in which the precise limits of the sieve method are known. A bit more specifically, let be quantities with for some fixed , and let be a multiplicative function with

for all primes and some fixed (we allow all constants below to depend on ). Let , and for each prime , let be a set of integers, with for . We consider finitely supported sequences of non-negative reals for which we have bounds of the form

for all square-free and some , and some remainder terms . One is then interested in upper and lower bounds on the quantity

The fundamental lemma of sieve theory (Corollary 19 of Notes 4) gives us the bound

This bound is strong when is large, but is not as useful for smaller values of . We now give a sharp bound in this regime. We introduce the functions by

where we adopt the convention . Note that for each one has only finitely many non-zero summands in (6), (7). These functions are closely related to the Buchstab function from Exercise 28 of Supplement 4; indeed from comparing the definitions one has

for all .

Exercise 1 (Alternate definition of )Show that is continuously differentiable except at , and is continuously differentiable except at where it is continuous, obeying the delay-differential equationsfor , with the initial conditions

for and

for . Show that these properties of determine completely.

For future reference, we record the following explicit values of :

We will show

Theorem 2 (Linear sieve)Let the notation and hypotheses be as above, with . Then, for any , one has the upper boundif is sufficiently large depending on . Furthermore, this claim is sharp in the sense that the quantity cannot be replaced by any smaller quantity, and similarly cannot be replaced by any larger quantity.

Comparing the linear sieve with the fundamental lemma (and also testing using the sequence for some extremely large ), we conclude that we necessarily have the asymptotics

for all ; this can also be proven directly from the definitions of , or from Exercise 1, but is somewhat challenging to do so; see e.g. Chapter 11 of Friedlander-Iwaniec for details.

Exercise 3Establish the integral identitiesand

for . Argue heuristically that these identities are consistent with the bounds in Theorem 2 and the Buchstab identity (Equation (16) from Notes 4).

Exercise 4Use the Selberg sieve (Theorem 30 from Notes 4) to obtain a slightly weaker version of (12) in the range in which the error term is worsened to , but the main term is unchanged.

We will prove Theorem 2 below the fold. The optimality of is closely related to the parity problem obstruction discussed in Section 5 of Notes 4; a naive application of the parity arguments there only give the weak bounds and for , but this can be sharpened by a more careful counting of various sums involving the Liouville function .

As an application of the linear sieve (specialised to the ranges in (10), (11)), we will establish a famous theorem of Chen, giving (in some sense) the closest approach to the twin prime conjecture that one can hope to achieve by sieve-theoretic methods:

Theorem 5 (Chen’s theorem)There are infinitely many primes such that is the product of at most two primes.

The same argument gives the version of Chen’s theorem for the even Goldbach conjecture, namely that for all sufficiently large even , there exists a prime between and such that is the product of at most two primes.

The discussion in these notes loosely follows that of Friedlander-Iwaniec (who study sieving problems in more general dimension than ).

** — 1. Optimality — **

We first establish that the quantities appearing in Theorem 2 cannot be improved. We use the parity argument of Selberg, based on weight sequences related to the Liouville function.

We argue for the optimality of ; the argument for is similar and is left as an exercise. Suppose that there is for which the claim in Theorem 2 is not optimal, thus there exists such that

for as in that theorem, with sufficiently large.

We will contradict this claim by specialising to a special case. Let be a large parameter going to infinity, and set . We set , then by Mertens’ theorem we have . We set to be the residue class , thus (3) becomes

and (14) becomes

Now let be a small fixed quantity to be chosen later, set , and let be the sequence

This is clearly finitely supported and non-negative. For any , we have

from the multiplicativity of . If , then , and then by the prime number theorem for the Liouville function (Exercise 41 from Notes 2, combined with Exercise 18 from Supplement 4) we have

(say), annd hence the remainder term in (15) is of size

As such, the error term in (16) may be absorbed into the term, and so

Now we count the left-hand side. Observe that is supported on those numbers that are the product of an odd number of primes (possibly with repetition), in which case . To be coprime to , all these primes must be at least ; since we are restricting , we thus must have . The left-hand side of (18) may thus be written as

This expression may be computed using the prime number theorem:

Exercise 6Show that the expression (19) is equal to .

Since is continuous for , we obtain a contradiction if is sufficiently small.

Exercise 7Verify the optimality of in Theorem 2. (Hint:replace by in the above arguments.)

** — 2. The linear sieve — **

We now prove the forward direction of Theorem 2. Again, we focus on the upper bound (12), as the lower bound case is similar.

Fix . Morally speaking, the most natural sieve to use here is the (upper bound) *beta sieve* from Notes 4, with the optimal value of , which for the linear sieve turns out to be . Recall that this sieve is defined as the sum

where is the set of divisors of with , such that

for all odd . From Proposition 14 of Notes 4 this is indeed an upper bound sieve; indeed we have

where is the set of divisors of with , such that

for all odd . Now for the key heuristic point: if lies in the support of , then the sum in (20) mostly vanishes. Indeed, if is such that and for some and odd , then one has for some that is not divisible by any prime less than . On the other hand, from (21), (22) one has

and

which (morally) implies from the sieve of Erathosthenes that is prime, thus and so is not in the support of . As such, we expect the upper bound sieve to be extremely efficient on the support of , which when combined with the analysis of the previous section suggests that this sieve should produce the desired upper bound (12).

One can indeed use this sieve to establish the required bound (12); see Chapter 11 of Friedlander-Iwaniec for details. However, for various technical reasons it will be convenient to modify this sieve slightly, by increasing the parameter to be slightly greater than , and also by using the fundamental lemma to perform a preliminary sifting on the small primes.

We turn to the details. To prove (12) we may of course assume that is suitably small. It will be convenient to worsen the error in (12) a little to , since one can of course remove the logarithm by reducing appropriately.

Set and . By the fundamental lemma of sieve theory (Lemma 17 of Notes 4), one can find combinatorial upper and lower bound sieve coefficients at sifting level supported on divisors of , such that

We will use the upper bound sieve as a preliminary sieve to remove the for ; the lower bound sieve plays only a minor supporting role, mainly to control the difference between the upper bound sieve .

Next, we set such that

for all odd ; in particular, for all . By Proposition 14 of Notes 4, this is indeed an upper bound sieve for the primes dividing :

Multiplying this with the second inequality in (24) (this is the method of *composition of sieves*), we obtain an upper bound sieve for the primes up to :

Multiplying this by and summing in , we conclude that

Note that each product appears at most once in the above sum, and all such products are squarefree and at most . Applying (3), we thus have

Thus to prove (12), it suffices to show that

for sufficiently large depending on . Factoring out the summation using (23), it thus suffices to show that

Now we eliminate the role of . From (1), (5) and Mertens’ theorem we have

for large enough. Also, if , then and all prime factors of are at least . Thus has at most prime factors, each of which are at least . From (1) we then have

The contribution of the error term is then easily seen to be for large enough, and so we reduce to showing that

One can proceed here by evaluating the left-hand side directly. However, we will proceed instead by using the weight function from the previous section. More precisely, we will evaluate the expression

in two different ways, where is as before (but with the role of now replaced by the function ). Firstly, since , we see from the argument used to establish (17) that

(say). Since each appears at least once, we can thus write (26) as

which upon factoring the sum using (23) and Mertens’ theorem

Thus to verify (25), it will suffice to show that (26) is of the form

for sufficiently large.

To do this, we abbreviate (26) as

By Proposition 14 of Notes 4, we can expand as

where for any , is the collection of all divisors of with such that

for all with the same parity as . For technical reasons, we will also impose the additional inequality

for all with the opposite parity as ; this follows from (30) when , but is an additional constraint when and is even, but in the above identity is odd, so this additional constraint is harmless. For similar reasons we impose the inequality

which follows from (30) or (31) except when , but then this inequality is automatic from our hypothesis , which implies if is chosen small enough.

Inserting the identity (28), we can write (26) as

where

and

We first estimate . By(24), we can write as the sum of

(where we bound by ). The error may be rearranged as

which by (23) is of size for small enough. As for the main term (33), we see from Exercise 6 (and the arguments preceding that exercise) that this term is equal to for sufficiently large. Thus, to obtain the desired approximation for (26), it will suffice to show that

Next, we establish an exponential decay estimate on the :

Lemma 8For sufficiently large depending on , we havefor all and some absolute constant .

*Proof:* (Sketch) Note that if is in , then and all prime factors are at least , thus we may assume without loss of generality that .

We bound

Note that if lies in , then

thanks to (29), (32). From this and the fundamental lemma of sieve theory we see (Exercise!) that

and so it will suffice to show that

By the prime number theorem, the left-hand side is bounded (Exercise!) by as , where

and is the set of points with ,

and such that

for all with the same parity as , and

for all . It thus suffices to prove the bound

for all and some absolute constant .

We use an argument from the book of Harman. Observe that vanishes for , which makes the claim (35) routine for (Exercise!) for sufficiently large. We will now inductively prove (35) for all odd . From the change of variables , we obtain the identity

where when is odd and when is even (Exercise!). In particular, if is odd and (35) was already proven for , then

One can check (Exercise!) that the quantity is maximised at , where its value is less than (in fact it is ) if is small enough. As such, we obtain (35) if is sufficiently close to .

Finally, (35) for even follows from the odd case (with a slightly larger choice of ) by one final application of (36).

Exercise 9Fill in the steps marked (Exercise!) in the above proof.

In view of this lemma, the total contribution of with for some sufficiently large is acceptable. Thus it suffices to show that

whenever is odd.

By (24), we can write as

plus an error of size

Arguing as in the treatment of the term, we see from (23) that the error term is bounded by

as desired for large enough, since and . Thus it suffices to show that

If and appear in the above sum, then we have where has no prime factor less than , has an even number of prime factors, and obeys the bounds

thanks to (29). Note that (29) also gives , and thus (since and ) we see that if is small enough and is large enough. This forces to either equal , or be the product of two primes between and . The contribution of the case is bounded by , which is acceptable. As for the contribution of those that are the product of two primes, the prime number theorem shows that there are at most

values of that can contribute to the sum, and so this contribution to is at most

but by (34) the sum here is for large enough, and the claim follows. This completes the proof of (12).

Exercise 10Establish the lower bound (13) in Theorem 2. (Note that one can assume without loss of generality that , which will now be needed to ensure (31) when .)

** — 3. Chen’s theorem — **

We now prove Chen’s theorem for twin primes, loosely following the treatment in Chapter 25 of Friedlander-Iwaniec. We will in fact show the slightly stronger statement that

for sufficiently large , where is the set of all numbers that are products of at most two primes, and . Indeed, after removing the (negligible) contribution of those that are powers of primes, this estimate would imply that there are infinitely many primes such that is the product of at most two primes, each of which is at least .

Chen’s argument begins with the following simple lower bound sieve for :

Lemma 11If , then

*Proof:* If has no prime factors less than or equal to , then and the claim follows. If has two or more factors less than or equal to , then and the claim follows. Finally, if has exactly one factor less than or equal to , then (as ) it must be of the form for some , or it is divisible by for some , and the claim again follows.

In view of this sieve (trivially managing the contribution of or of the case where , and using the restriction of to be coprime to ), it suffices to show that

for sufficiently large , where

and

We thus seek sufficiently good lower bounds on and sufficiently good upper bounds on and . As it turns out, the linear sieve, combined with the Bombieri-Vinogradov theorem, will give bounds on with numerical constants that are sufficient for this purpose.

We begin with . We use the lower bound linear sieve, with equal to the residue class for all , so that is the residue class . We approximate

where is the multiplicative function with and for . From the Bombieri-Vinogradov theorem (Theorem 17 of Notes 3) we have

(say) if for some small fixed . Applying the lower bound linear sieve (13), we conclude that

where

We can compute an asymptotic for :

as , where is the twin prime constant.

From (11) we have . Sending slowly to zero, we conclude that

Now we turn to . Here we use the upper bound linear sieve. Let be as before. For any dividing and , we have

where and are as previously. We apply the upper bound linear sieve (12) with level of distribution , to conclude that

We sum over . Since is at most , and each number less than or equal to has at most prime factors, we have

The error term is thanks to (39). Since , we thus have

for sufficiently large , thanks to Exercise 12. We can compute the sum using Exercise 37 of Notes 1, to obtain

which by (10) and sending slowly gives

A routine computation shows that

Finally, we consider , which is estimated by “switching” the sieve to sift out small divisors of , rather than small divisors of . Removing those with , as well as those that are powers of primes, and then shifting by , we have

where is the finitely supported non-negative sequence

Here we are sifting out the residue classes , so that .

The sequence has good distribution up to level :

Proposition 13One haswhere is as before, and

(say), with as before.

*Proof:* Observe that the quantity in (42) is bounded above by if the summand is to be non-zero. We now use a finer-than-dyadic decomposition trick similar to that used in the proof of the Bombieri-Vinogradov theorem in Notes 3 to approximate as a combination of Dirichlet convolutions. Namely, we set , and partition (plus possibly a little portion to the right of ) into consecutive intervals each of the form for some . We similarly split (plus possibly a tiny portion of ) into intervals each of the form for some . We can thus split as

Observe that for each there are only choices of for which the summand can be non-zero. As such, the contribution of the diagonal case can be easily seen to be absorbed into the error, as can those cases where the product set is not contained completely in . If we let be the set of triplets obeying these properties, we can thus approximate by , where is the Dirichlet convolution

From the general Bombieri-Vinogradov theorem (Theorem 16 of Notes 3) and the Siegel-Walfisz theorem (Exercise 64 of Notes 2) we see that

where

(say) and

This gives the claim with replaced by the quantity ; but by undoing the previous decomposition we see that this quantity is equal to up to an error of (say), and the claim follows.

Applying the upper bound sieve (12) (with sifting level ), we thus have

and hence by (10) and Exercise 12

for sufficiently large.

Note that

From the prime number theorem and Exercise 37 of Notes 1, we thus have

(In fact one also has a matching lower bound, but we will not need it here.) We thus conclude that

where

The left-hand side of (38) is then at least

One can calculate that , and the claim follows.

Exercise 14Establish Chen’s theorem for the even Goldbach conjecture.

Remark 15If one is willing to use stronger distributional claims on the primes than is provided by the Bombieri-Vinogradov theorem, then one can use a simpler sieve than Chen’s sieve to establish Chen’s theorem, but then the required distributional theorem will then either be conjectural or more difficult to establish than the Bombieri-Vinogradov theorem. See Chapter 25 of Friedlander-Iwaniec for further discussion.

## 15 comments

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29 January, 2015 at 4:16 pm

Eytan PaldiIt seems that in the RHS of (6)-(7), “” (in the integrands) should be ““.

[Corrected, thanks – T.]29 January, 2015 at 5:32 pm

Eytan PaldiAlso in the line below (7).

[Corrected, thanks – T.]29 January, 2015 at 7:41 pm

arch1In the last two integrals preceding Exercise 9, the upper integration limit is misplaced.

[Corrected, thanks -T.]29 January, 2015 at 7:50 pm

arch13 lines up from (37): we see from (23) *that the error*(?) is bounded by

[Corrected, thanks -T.]1 February, 2015 at 8:46 am

Lior SilbermanIn the proof of Chen’s Theorem, in the paragraph “We begin with “, is said to be the class of -2 mod rather than mod .

[Corrected, thanks – T.]31 March, 2015 at 7:58 am

KaraskasJust after the fragment “and the Siegel-Walfisz theorem (Exercise 64 of Notes 2) we see that [Formula] where” it should probably be log^{-10}x instead of log^{-100x}.

[Corrected, thanks – T.]31 March, 2015 at 1:20 pm

KaraskasRight now it’s log^{-10x} instead of log^{-10}x.

[Corrected, thanks – T.]10 April, 2015 at 3:07 pm

KaraskasChen’s theorem section: shouldn’t it be under the big sigma in the definition of ? The same thing reappears a few lines later.

Just after “Now we turn to “, what happens if p divides d?

[I was not able to locate this issue – T.]11 April, 2015 at 12:43 am

KaraskasIn the same fragment it probably should be instead of , shouldn’t it? That would also explain my last question.

[Corrected, thanks – T.]11 April, 2015 at 9:53 am

KaraskasThe issue wasn’t there anymore while I was writting about it. I think you had repaired it a few days before and I forgot to refresh the website. Sorry for the confusion.

By the way, thank you for writing posts about sieve theory. Some details and ideas are much easier for me to follow right now.

14 April, 2015 at 2:49 pm

KaraskasThe formula under “Applying the upper bound sieve” fragment: it should be instead of in one place.

[Corrected, thanks – T.]15 December, 2015 at 5:04 pm

KaraskasI have a little confusion: before the equation (40) we send {\varepsilon} to 0 slowly but in Theorem 2 we have “if {D} is sufficiently large depending on {\varepsilon, s, c}.” How should I understand this?

15 December, 2015 at 6:23 pm

KaraskasIt seems to me like we are using two different epsilons there. By the way, is it possible to relax conditions in Theorem 2 and let depend on z?

Theorem 2 also reminds me the Jurkat-Richert theorem, although we do not have any restriction on the sieve dimension. How do these two theorems are related to each other?

Sorry for doubling the post and using LaTeX wrongly in the previous one.

15 December, 2015 at 7:13 pm

Terence TaoFor any fixed choice of , the largeness condition on required for Theorem 2 will apply if is large enough, thus

whenever is sufficiently large depending on (this allows for the o(1) errors to be absorbed into the error). In particular, for any , one has

whenever is sufficiently large depending on (by choosing appropriately depending on ). This implies that

as .

Another way to think about it as follows: if one has a bound of the form

as for any fixed choice of , then one automatically has the improvement

as , by setting equal to a function of that goes to zero sufficiently slowly (slow enough so that the error in (1) still goes to zero even though is now varying).

The Jurkat-Richert theorem has a slightly better error term replacing , and a slightly worse error term for the errors, but can easily be used as a substitute for Theorem 2 for the purpose of proving Chen’s theorem.

30 December, 2015 at 6:32 pm

KaraskasI have one problem with Lemma 11. How do we rule out the numbers of the form , where and and ?

I saw an another proof of Chen’s theorem in which the author (Nathanson, I guess) used the sum instead of . By this trick we can easily avoid thinking about numbers which are not square-free.

Are these two approaches equivalent in some easy way which I cannot see right now?

I would also like to thank you for your last answer.

[Oops, this case does need to be inserted, but it is easily dealt with (with , it contributes from trivial estimates). -T.]