The twin prime conjecture, still unsolved, asserts that there are infinitely many primes ${p}$ such that ${p+2}$ is also prime. A more precise form of this conjecture is (a special case) of the Hardy-Littlewood prime tuples conjecture, which asserts that

$\displaystyle \sum_{n \leq x} \Lambda(n) \Lambda(n+2) = (2\Pi_2+o(1)) x \ \ \ \ \ (1)$

as ${x \rightarrow \infty}$, where ${\Lambda}$ is the von Mangoldt function and ${\Pi_2 = 0.6606\dots}$ is the twin prime constant

$\displaystyle \prod_{p>2} (1 - \frac{1}{(p-1)^2}).$

Because ${\Lambda}$ is almost entirely supported on the primes, it is not difficult to see that (1) implies the twin prime conjecture.

One can give a heuristic justification of the asymptotic (1) (and hence the twin prime conjecture) via sieve theoretic methods. Recall that the von Mangoldt function can be decomposed as a Dirichlet convolution

$\displaystyle \Lambda(n) = \sum_{d|n} \mu(d) \log \frac{n}{d}$

where ${\mu}$ is the Möbius function. Because of this, we can rewrite the left-hand side of (1) as

$\displaystyle \sum_{d \leq x} \mu(d) \sum_{n \leq x: d|n} \log\frac{n}{d} \Lambda(n+2). \ \ \ \ \ (2)$

To compute this double sum, it is thus natural to consider sums such as

$\displaystyle \sum_{n \leq x: d|n} \log \frac{n}{d} \Lambda(n+2)$

or (to simplify things by removing the logarithm)

$\displaystyle \sum_{n \leq x: d|n} \Lambda(n+2).$

The prime number theorem in arithmetic progressions suggests that one has an asymptotic of the form

$\displaystyle \sum_{n \leq x: d|n} \Lambda(n+2) \approx \frac{g(d)}{d} x \ \ \ \ \ (3)$

where ${g}$ is the multiplicative function with ${g(d)=0}$ for ${d}$ even and

$\displaystyle g(d) := \frac{d}{\phi(d)} = \prod_{p|d} (1-\frac{1}{p})^{-1}$

for ${d}$ odd. Summing by parts, one then expects

$\displaystyle \sum_{n \leq x: d|n} \Lambda(n+2)\log \frac{n}{d} \approx \frac{g(d)}{d} x \log \frac{x}{d}$

and so we heuristically have

$\displaystyle \sum_{n \leq x} \Lambda(n) \Lambda(n+2) \approx x \sum_{d \leq x} \frac{\mu(d) g(d)}{d} \log \frac{x}{d}.$

The Dirichlet series

$\displaystyle \sum_n \frac{\mu(n) g(n)}{n^s}$

$\displaystyle \sum_n \frac{\mu(n) g(n)}{n^s} = \prod_p (1 - \frac{g(p)}{p^s})$

for ${\hbox{Re} s > 1}$; comparing this with the Euler product factorisation

$\displaystyle \zeta(s) = \prod_p (1 - \frac{1}{p^s})^{-1}$

for the Riemann zeta function, and recalling that ${\zeta}$ has a simple pole of residue ${1}$ at ${s=1}$, we see that

$\displaystyle \sum_n \frac{\mu(n) g(n)}{n^s} = \frac{1}{\zeta(s)} \prod_p \frac{1-g(p)/p^s}{1-p^s}$

has a simple zero at ${s=1}$ with first derivative

$\displaystyle \prod_p \frac{1 - g(p)/p}{1-1/p} = 2 \Pi_2.$

From this and standard multiplicative number theory manipulations, one can calculate the asymptotic

$\displaystyle \sum_{d \leq x} \frac{\mu(d) g(d)}{d} \log \frac{x}{d} = 2 \Pi_2 + o(1)$

which concludes the heuristic justification of (1).

What prevents us from making the above heuristic argument rigorous, and thus proving (1) and the twin prime conjecture? Note that the variable ${d}$ in (2) ranges to be as large as ${x}$. On the other hand, the prime number theorem in arithmetic progressions (3) is not expected to hold for ${d}$ anywhere that large (for instance, the left-hand side of (3) vanishes as soon as ${d}$ exceeds ${x}$). The best unconditional result known of the type (3) is the Siegel-Walfisz theorem, which allows ${d}$ to be as large as ${\log^{O(1)} x}$. Even the powerful generalised Riemann hypothesis (GRH) only lets one prove an estimate of the form (3) for ${d}$ up to about ${x^{1/2-o(1)}}$.

However, because of the averaging effect of the summation in ${d}$ in (2), we don’t need the asymptotic (3) to be true for all ${d}$ in a particular range; having it true for almost all ${d}$ in that range would suffice. Here the situation is much better; the celebrated Bombieri-Vinogradov theorem (sometimes known as “GRH on the average”) implies, roughly speaking, that the approximation (3) is valid for almost all ${d \leq x^{1/2-\varepsilon}}$ for any fixed ${\varepsilon>0}$. While this is not enough to control (2) or (1), the Bombieri-Vinogradov theorem can at least be used to control variants of (1) such as

$\displaystyle \sum_{n \leq x} (\sum_{d|n} \lambda_d) \Lambda(n+2)$

for various sieve weights ${\lambda_d}$ whose associated divisor function ${\sum_{d|n} \lambda_d}$ is supposed to approximate the von Mangoldt function ${\Lambda}$, although that theorem only lets one do this when the weights ${\lambda_d}$ are supported on the range ${d \leq x^{1/2-\varepsilon}}$. This is still enough to obtain some partial results towards (1); for instance, by selecting weights according to the Selberg sieve, one can use the Bombieri-Vinogradov theorem to establish the upper bound

$\displaystyle \sum_{n \leq x} \Lambda(n) \Lambda(n+2) \leq (4+o(1)) 2 \Pi_2 x, \ \ \ \ \ (4)$

which is off from (1) by a factor of about ${4}$. See for instance this blog post for details.

It has been difficult to improve upon the Bombieri-Vinogradov theorem in its full generality, although there are various improvements to certain restricted versions of the Bombieri-Vinogradov theorem, for instance in the famous work of Zhang on bounded gaps between primes. Nevertheless, it is believed that the Elliott-Halberstam conjecture (EH) holds, which roughly speaking would mean that (3) now holds for almost all ${d \leq x^{1-\varepsilon}}$ for any fixed ${\varepsilon>0}$. (Unfortunately, the ${\varepsilon}$ factor cannot be removed, as investigated in a series of papers by Friedlander, Granville, and also Hildebrand and Maier.) This comes tantalisingly close to having enough distribution to control all of (1). Unfortunately, it still falls short. Using this conjecture in place of the Bombieri-Vinogradov theorem leads to various improvements to sieve theoretic bounds; for instance, the factor of ${4+o(1)}$ in (4) can now be improved to ${2+o(1)}$.

In two papers from the 1970s (which can be found online here and here respectively, the latter starting on page 255 of the pdf), Bombieri developed what is now known as the Bombieri asymptotic sieve to clarify the situation more precisely. First, he showed that on the Elliott-Halberstam conjecture, while one still could not establish the asymptotic (1), one could prove the generalised asymptotic

$\displaystyle \sum_{n \leq x} \Lambda_k(n) \Lambda(n+2) = (2\Pi_2+o(1)) k x \log^{k-1} x \ \ \ \ \ (5)$

for all natural numbers ${k \geq 2}$, where the generalised von Mangoldt functions ${\Lambda_k}$ are defined by the formula

$\displaystyle \Lambda_k(n) := \sum_{d|n} \mu(d) \log^k \frac{n}{d}.$

These functions behave like the von Mangoldt function, but are concentrated on ${k}$-almost primes (numbers with at most ${k}$ prime factors) rather than primes. The right-hand side of (5) corresponds to what one would expect if one ran the same heuristics used to justify (1). Sadly, the ${k=1}$ case of (5), which is just (1), is just barely excluded from Bombieri’s analysis.

More generally, on the assumption of EH, the Bombieri asymptotic sieve provides the asymptotic

$\displaystyle \sum_{n \leq x} \Lambda_{(k_1,\dots,k_r)}(n) \Lambda(n+2) \ \ \ \ \ (6)$

$\displaystyle = (2\Pi_2+o(1)) \frac{\prod_{i=1}^r k_i!}{(k_1+\dots+k_r-1)!} x \log^{k_1+\dots+k_r-1} x$

for any fixed ${r \geq 1}$ and any tuple ${(k_1,\dots,k_r)}$ of natural numbers other than ${(1,\dots,1)}$, where

$\displaystyle \Lambda_{(k_1,\dots,k_r)} := \Lambda_{k_1} * \dots * \Lambda_{k_r}$

is a further generalisation of the von Mangoldt function (now concentrated on ${k_1+\dots+k_r}$-almost primes). By combining these asymptotics with some elementary identities involving the ${\Lambda_{(k_1,\dots,k_r)}}$, together with the Weierstrass approximation theorem, Bombieri was able to control a wide family of sums including (1), except for one undetermined scalar ${\delta_x \in [0,2]}$. Namely, he was able to show (again on EH) that for any fixed ${r \geq 1}$ and any continuous function ${g_r}$ on the simplex ${\Delta_r := \{ (t_1,\dots,t_r) \in {\bf R}^r: t_1+\dots+t_r = 1; 0 \leq t_1 \leq \dots \leq t_r\}}$ that had suitable vanishing at the boundary, the sum

$\displaystyle \sum_{n \leq x: n=p_1 \dots p_r} g_r( \frac{\log p_1}{\log n}, \dots, \frac{\log p_r}{\log n} ) \Lambda(n+2)$

was equal to

$\displaystyle (\delta_x+o(1)) \int_{\Delta_r} g_r \frac{x}{\log x} \ \ \ \ \ (7)$

when ${r}$ was odd and

$\displaystyle (2-\delta_x+o(1)) \int_{\Delta_r} g_r \frac{x}{\log x} \ \ \ \ \ (8)$

when ${r}$ was even, where the integral on ${\Delta_r}$ is with respect to the measure ${\frac{dt_1 \dots dt_{r-1}}{t_1 \dots t_r}}$ (this is Dirac measure in the case ${r=1}$). In particular, we have

$\displaystyle \sum_{n \leq x} \Lambda(n) \Lambda(n+2) = (\delta_x + o(1)) 2 \Pi_2 x$

and the twin prime conjecture would be proved if one could show that ${\delta_x}$ is bounded away from zero, while (1) is equivalent to the assertion that ${\delta_x}$ is equal to ${1+o(1)}$. Unfortunately, no additional bound beyond the inequalities ${0 \leq \delta_x \leq 2}$ provided by the Bombieri asymptotic sieve is known, even if one assumes all other major conjectures in number theory than the prime tuples conjecture and its variants (e.g. GRH, GEH, GUE, abc, Chowla, …).

To put it another way, the Bombieri asymptotic sieve is able (on EH) to compute asymptotics for sums

$\displaystyle \sum_{n \leq x} f(n) \Lambda(n+2) \ \ \ \ \ (9)$

without needing to know the unknown scalar ${\delta_x}$, when ${f}$ is a function supported on almost primes of the form

$\displaystyle f(p_1 \dots p_r) = g_r( \frac{\log p_1}{\log n}, \dots, \frac{\log p_r}{\log n} )$

for ${1 \leq r \leq r_*}$ and some fixed ${r_*}$, with ${f}$ vanishing elsewhere and for some continuous (symmetric) functions ${g_r: \Delta_r \rightarrow {\bf C}}$ obeying some vanishing at the boundary, so long as the parity condition

$\displaystyle \sum_{r \hbox{ odd}} \int_{\Delta_r} g_r = \sum_{r \hbox{ even}} \int_{\Delta_r} g_r$

is obeyed (informally: ${f}$ gives the same weight to products of an odd number of primes as to products of an even number of primes, or to put it another way, ${f}$ is asymptotically orthogonal to the Möbius function ${\mu}$). But when ${f}$ violates the parity condition, the asymptotic involves the unknown ${\delta_x}$. This scalar ${\delta_x}$ thus embodies the “parity problem” for the twin prime conjecture (discussed in these previous blog posts).

Because the obstruction to the parity problem is only one-dimensional (on EH), one can replace any parity-violating weight (such as ${\Lambda}$) with any other parity-violating weight and obtain a logically equivalent estimate. For instance, to prove the twin prime conjecture on EH, it would suffice to show that

$\displaystyle \sum_{p_1 p_2 p_3 \leq x: p_1,p_2,p_3 \geq x^\alpha} \Lambda(p_1 p_2 p_3 + 2) \gg \frac{x}{\log x}$

for some fixed ${\alpha>0}$, or equivalently that there are ${\gg \frac{x}{\log^2 x}}$ solutions to the equation ${p - p_1 p_2 p_3 = 2}$ in primes with ${p \leq x}$ and ${p_1,p_2,p_3 \geq x^\alpha}$. (In some cases, this sort of reduction can also be made using other sieves than the Bombieri asymptotic sieve, as was observed by Ng.) As another example, the Bombieri asymptotic sieve can be used to show that the asymptotic (1) is equivalent to the asymptotic

$\displaystyle \sum_{n \leq x} \mu(n) 1_R(n) \Lambda(n+2) = o( \frac{x}{\log x})$

where ${R}$ is the set of numbers that are rough in the sense that they have no prime factors less than ${x^\alpha}$ for some fixed ${\alpha>0}$ (the function ${\mu 1_R}$ clearly correlates with ${\mu}$ and so must violate the parity condition). One can replace ${1_R}$ with similar sieve weights (e.g. a Selberg sieve) that concentrate on almost primes if desired.

As it turns out, if one is willing to strengthen the assumption of the Elliott-Halberstam (EH) conjecture to the assumption of the generalised Elliott-Halberstam (GEH) conjecture (as formulated for instance in Claim 2.6 of the Polymath8b paper), one can also swap the ${\Lambda(n+2)}$ factor in the above asymptotics with other parity-violating weights and obtain a logically equivalent estimate, as the Bombieri asymptotic sieve also applies to weights such as ${\mu 1_R}$ under the assumption of GEH. For instance, on GEH one can use two such applications of the Bombieri asymptotic sieve to show that the twin prime conjecture would follow if one could show that there are ${\gg \frac{x}{\log^2 x}}$ solutions to the equation

$\displaystyle p_1 p_2 - p_3 p_4 = 2$

in primes with ${p_1,p_2,p_3,p_4 \geq x^\alpha}$ and ${p_1 p_2 \leq x}$, for some ${\alpha > 0}$. Similarly, on GEH the asymptotic (1) is equivalent to the asymptotic

$\displaystyle \sum_{n \leq x} \mu(n) 1_R(n) \mu(n+2) 1_R(n+2) = o( \frac{x}{\log^2 x})$

for some fixed ${\alpha>0}$, and similarly with ${1_R}$ replaced by other sieves. This form of the quantitative twin primes conjecture is appealingly similar to the (special case)

$\displaystyle \sum_{n \leq x} \mu(n) \mu(n+2) = o(x)$

of the Chowla conjecture, for which there has been some recent progress (discussed for instance in these recent posts). Informally, the Bombieri asymptotic sieve lets us (on GEH) view the twin prime conjecture as a sort of Chowla conjecture restricted to almost primes. Unfortunately, the recent progress on the Chowla conjecture relies heavily on the multiplicativity of ${\mu}$ at small primes, which is completely destroyed by inserting a weight such as ${1_R}$, so this does not yet yield a viable path towards the twin prime conjecture even assuming GEH. Still, the similarity is striking, and one can hope that further ways to attack the Chowla conjecture may emerge that could impact the twin prime conjecture. (Alternatively, if one assumes a sufficiently optimistic version of the GEH, one could perhaps relax the notion of “almost prime” to the extent that one could start usefully using multiplicativity at smallish primes, though this seems rather wishful at present, particularly since the most optimistic versions of GEH are known to be false.)

The Bombieri asymptotic sieve is already well explained in the original two papers of Bombieri; there is also a slightly different treatment of the sieve by Friedlander and Iwaniec, as well as a simplified version in the book of Friedlander and Iwaniec (in which the distribution hypothesis is strengthened in order to shorten the arguments. I’ve decided though to write up my own notes on the sieve below the fold; this is primarily for my own benefit, but may be useful to some readers also. I largely follow the treatment of Bombieri, with the one idiosyncratic twist of replacing the usual “elementary” Selberg sieve with the “analytic” Selberg sieve used in particular in many of the breakthrough works in small gaps between primes; I prefer working with the latter due to its Fourier-analytic flavour.

— 1. Controlling generalised von Mangoldt sums —

To prove (5), we shall first generalise it, by replacing the sequence ${\Lambda(n+2)}$ by a more general sequence ${a_n}$ obeying the following axioms:

• (i) (Non-negativity) One has ${a_n \geq 0}$ for all ${n}$.
• (ii) (Crude size bound) One has ${a_n \ll \tau(n)^{O(1)} \log^{O(1)} n}$ for all ${n}$, where ${\tau}$ is the divisor function.
• (iii) (Size) We have ${\sum_{n \leq x} a_n = (C+o(1)) x}$ for some constant ${C>0}$.
• (iv) (Elliott-Halberstam type conjecture) For any ${\varepsilon,A>0}$, one has

$\displaystyle \sum_{d \leq x^{1-\varepsilon}} |\sum_{n \leq x: d|n} a_n - C x \frac{g(d)}{d}| \ll_{\varepsilon,A} x \log^{-A} x$

where ${g}$ is a multiplicative function with ${g(p^j) = 1 + O(1/p)}$ for all primes ${p}$ and ${j \geq 1}$.

These axioms are a little bit stronger than what is actually needed to make the Bombieri asymptotic sieve work, but we will not attempt to work with the weakest possible axioms here.

We introduce the function

$\displaystyle G(s) := \prod_p \frac{1-g(p)/p^s}{1-1/p^s}$

which is analytic for ${\hbox{Re}(s) > 0}$; in particular it can be evaluated at ${s=1}$ to yield

$\displaystyle G(1) = \prod_p \frac{1-g(p)/p}{1-1/p}.$

There are two model examples of data ${a_n, C, g}$ to keep in mind. The first, discussed in the introduction, is when ${a_n =\Lambda(n+2)}$, then ${C = 2 \Pi_2}$ and ${g}$ is as in the introduction; one of course needs EH to justify axiom (iv) in this case. The other is when ${a_n=1}$, in which case ${C=1}$ and ${g(n)=1}$ for all ${n}$. We will later take advantage of the second example to avoid doing some (routine, but messy) main term computations.

The main result of this section is then

Theorem 1 Let ${a_n, g, C, G}$ be as above. Let ${\vec k = (k_1,\dots,k_r)}$ be a tuple of natural numbers (independent of ${x}$) that is not equal to ${(1,\dots,1)}$. Then one has the asymptotic

$\displaystyle \sum_{n \leq x} \Lambda_{\vec k}(n) a_n = (G(1)+o(1)) \frac{\prod_{i=1}^r k_i!}{(|\vec k|-1)!} C x \log^{|\vec k|-1} x$

as ${x \rightarrow \infty}$, where ${|\vec k| := k_1 + \dots + k_r}$.

Note that this recovers (5) (on EH) as a special case.

We now begin the proof of this theorem. Henceforth we allow implied constants in the ${O()}$ or ${\ll}$ notation to depend on ${r, \vec k}$ and ${g,G}$.

It will be convenient to replace the range ${n \leq x}$ by a shorter range by the following standard localisation trick. Let ${B}$ be a large quantity depending on ${r, \vec k}$ to be chosen later, and let ${I}$ denote the interval ${\{ n: x - x \log^{-B} x \leq n \leq x \}}$. We will show the estimate

$\displaystyle \sum_{n \in I} \Lambda_{\vec k}(n) a_n = (G(1)+o(1)) \frac{\prod_{i=1}^r k_i!}{(|\vec k|-1)!} C |I| \log^{|\vec k|-1} x \ \ \ \ \ (10)$

from which the original claim follows by a routine summation argument. Observe from axiom (iv) and the triangle inequality that

$\displaystyle \sum_{d \leq x^{1-\varepsilon}: \mu^2(d)=1} |\sum_{n \in I: d|n} a_n - C |I| \frac{g(d)}{d}| \ll_{\varepsilon,A} x \log^{-A} x$

for any ${\varepsilon,A > 0}$.

Write ${L}$ for the logarithm function ${L(n) := \log n}$, thus ${\Lambda_k = \mu * L^k}$ for any ${k}$. Without loss of generality we may assume that ${k_r > 1}$; we then factor ${\Lambda_{\vec k} = \mu_{\vec k} * L^{k_r}}$, where

$\displaystyle \mu_{\vec k} := \Lambda_{k_1} * \dots * \Lambda_{k_{r-1}} * \mu.$

This function is just ${\mu}$ when ${r=1}$. When ${r>1}$ the function is more complicated, but we at least have the following crude bound:

Lemma 2 One has the pointwise bound ${|\mu_{\vec k}| \leq L^{|\vec k|-k_r}}$.

Proof: We induct on ${r}$. The case ${r=1}$ is obvious, so suppose ${r>1}$ and the claim has already been proven for ${r-1}$. Since ${\mu_{\vec k} = \Lambda_{k_1} * \mu_{(k_2,\dots,k_r)}}$, we see from induction hypothesis and the triangle inequality that

$\displaystyle |\mu_{\vec k}| \leq \Lambda_{k_1} * L^{|\vec k| - k_r - k_1} \leq L^{|\vec k| - k_r - k_1} (\Lambda_{k_1} * 1).$

Since ${\Lambda_{k_1}*1 = L^{k_1}}$ by Möbius inversion, the claim follows. $\Box$

We can write

$\displaystyle \Lambda_{\vec k}(n) = \sum_{d|n} \mu_{\vec k}(d) \log^{k_r} \frac{n}{d}.$

In the region ${n \in I}$, we have ${\log^{k_r} \frac{n}{d} = \log^{k_r} \frac{x}{d} + O( \log^{-B+O(1)} x )}$. Thus

$\displaystyle \Lambda_{\vec k}(n) = \sum_{d|n} \mu_{\vec k}(d) \log^{k_r} \frac{x}{d} + O( \tau(x) \log^{-B+O(1)} x )$

for ${n \in I}$. The contribution of the error term to ${O( \tau(x) \log^{-B+O(1)} x )}$ to (10) is easily seen to be negligible if ${B}$ is large enough, so we may freely replace ${\Lambda_{\vec k}(n)}$ with ${\sum_{d|n} \mu_{\vec k}(d) \log^{k_r} \frac{x}{d}}$ with little difficulty.

If we insert this replacement directly into the left-hand side of (10) and rearrange, we get

$\displaystyle \sum_{d \leq x} \mu_{\vec k}(d) \log^{k_r} \frac{x}{d} \sum_{n \in I: d|n} a_d.$

We can’t quite control this using axiom (iv) because the range of ${d}$ is a bit too big, as explained in the introduction. So let us introduce a truncated function

$\displaystyle \Lambda_{\vec k,\varepsilon}(n) := \sum_{d|n} \mu_{\vec k}(d) \log^{k_r} \frac{x}{d} \eta_\varepsilon( \frac{\log d}{\log x} ) \ \ \ \ \ (11)$

where ${\varepsilon>0}$ is a small quantity to be chosen later, and ${\eta_\varepsilon: {\bf R} \rightarrow [0,1]}$ is a smooth function that equals ${1}$ on ${(-\infty,1-4\varepsilon)}$ and equals ${0}$ on ${(1-3\varepsilon,+\infty)}$. Suppose one could establish the following two estimates for any fixed ${\varepsilon>0}$:

$\displaystyle \sum_{n \in I} \Lambda_{\vec k}(n) a_n = \sum_{n \in I} \Lambda_{\vec k,\varepsilon}(n) a_n + O( (\varepsilon+o(1)) C |I| \log^{|\vec k|-1} x ) \ \ \ \ \ (12)$

and

$\displaystyle \sum_{n \in I} \Lambda_{\vec k,\varepsilon}(n) a_n = C Q_{\varepsilon,x} G(1) + o( |I| \log^{|\vec k|-1} x ) \ \ \ \ \ (13)$

where ${Q_{\varepsilon,x}}$ is a quantity that depends on ${\varepsilon, \eta_\varepsilon, \vec k, B, x}$ but not on ${C, g,G}$. Then on combining the two estimates we would have

$\displaystyle \sum_{n \in I} \Lambda_{\vec k}(n) a_n = C Q_{\varepsilon,x} G(1) + (O(\varepsilon) + o(1)) C |I| \log^{|\vec k|-1} x. \ \ \ \ \ (14)$

One could in principle compute ${Q_{\varepsilon,x}}$ explicitly from the proof of (13), but one can avoid doing so by the following comparison trick. In the special case ${a_n=1}$, standard multiplicative number theory (noting that the Dirichlet series ${\sum_n \frac{\Lambda_{\vec k}(n)}{n^s}}$ has a pole of order ${|\vec k|}$ at ${s=1}$, with top Laurent coefficient ${\prod_{j=1}^r k_j!}$) gives the asymptotic

$\displaystyle \sum_{n \in I} \Lambda_{\vec k}(n) a_n = \frac{\prod_{i=1}^r k_i!}{(|\vec k|-1)!} + o(1)) |I| \log^{|\vec k|-1} x$

which when compared with (14) for ${a_n=1}$ (recalling that ${G(1)=C=1}$ in this case) gives the formula

$\displaystyle Q_{\varepsilon,x} = (\prod_{j=1}^r k_j + O(\varepsilon)) |I| \log^{|\vec k|-1} x.$

Inserting this back into (14) and recalling that ${\varepsilon>0}$ can be made arbitrarily small, we obtain (10).

As it turns out, the estimate (13) is easy to establish, but the estimate (12) is not, roughly speaking because the typical number ${n}$ in ${I}$ has too many divisors ${d}$ in the range ${[x^{1-4\varepsilon},1]}$, each of which gives a contribution to the error term. (In the book of Friedlander and Iwaniec, the estimate (13) is established anyway, but only after assuming a stronger version of (iv), roughly speaking in which ${d}$ is allowed to be as large as ${x \exp( -\log^{1/4} x)}$.) To resolve this issue, we will insert a preliminary sieve ${\nu_\varepsilon}$ that will remove most of the potential divisors ${d}$ i the range ${[x^{1-4\varepsilon},1]}$ (leaving only about ${O(1)}$ such divisors on the average for typical ${n}$), making the analogue of (12) easier to prove (at the cost of making the analogue of (13) more difficult). Namely, if one can find a function ${\nu_\varepsilon: {\bf N} \rightarrow {\bf R}}$ for which one has the estimates

$\displaystyle \sum_{n \in I} \Lambda_{\vec k}(n) a_n = \sum_{n \in I} \Lambda_{\vec k}(n) \nu_\varepsilon(n) a_n + O( (\varepsilon+o(1)) C |I| \log^{|\vec k|-1} x ), \ \ \ \ \ (15)$

$\displaystyle \sum_{n \in I} \Lambda_{\vec k}(n) \nu_\varepsilon(n) a_n$

$\displaystyle = \sum_{n \in I} \Lambda_{\vec k,\varepsilon}(n) \nu_\varepsilon(n) a_n + O( (\varepsilon+o(1)) C |I| \log^{|\vec k|-1} x ) \ \ \ \ \ (16)$

and

$\displaystyle \sum_{n \in I} \Lambda_{\vec k,\varepsilon}(n) \nu_\varepsilon(n) a_n = C Q'_{\varepsilon,x} G(1) + o( |I| \log^{|\vec k|-1} x ) \ \ \ \ \ (17)$

for some quantity ${Q'_{\varepsilon,x}}$ that depends on ${\varepsilon, \eta_\varepsilon, \vec k, B, x}$ but not on ${C, g, G,}$, then by repeating the previous arguments we will again be able to establish (10).

The key estimate is (16). As we shall see, when comparing ${\Lambda_{\vec k}(n) \nu_\varepsilon(n)}$ with ${\Lambda_{\vec k,\varepsilon}(n) \nu_\varepsilon(n)}$, the weight ${\nu_\varepsilon}$ will cost us a factor of ${1/\varepsilon}$, but the ${\log^{k_r} \frac{x}{d}}$ term in the definitions of ${\Lambda_{\vec k}}$ and ${\Lambda_{\vec k,\varepsilon}}$ will recover a factor of ${\varepsilon^{k_r}}$, which will give the desired bound since we are assuming ${k_r > 1}$.

One has some flexibility in how to select the weight ${\nu_\varepsilon}$: basically any standard sieve that uses divisors of size at most ${x^{2\varepsilon}}$ to localise (at least approximately) to numbers that are rough in the sense that they have no (or at least very few) factors less than ${x^\varepsilon}$, will do. We will use the analytic Selberg sieve choice

$\displaystyle \nu_\varepsilon(n) := (\sum_{d|n} \mu(d) \psi( \frac{\log d}{\varepsilon \log x} ))^2 \ \ \ \ \ (18)$

where ${\psi: {\bf R} \rightarrow [0,1]}$ is a smooth function supported on ${[-1,1]}$ that equals ${1}$ on ${[-1/2,1/2]}$.

It remains to establish the bounds (15), (16), (17). To warm up and introduce the various methods needed, we begin with the standard bound

$\displaystyle \sum_{n \in I} \nu_\varepsilon(n) a_n = \frac{C|I|}{\varepsilon \log x} (\int_0^1 \psi'(u)^2\ du) G(1) + o(1)), \ \ \ \ \ (19)$

where ${\psi'}$ denotes the derivative of ${\psi}$. Note the loss of ${1/\varepsilon}$ that had previously been pointed out. In the arguments that follows I will be a little brief with the details, as they are standard (see e.g. this previous post).

We now prove (19). The left-hand side can be expanded as

$\displaystyle \sum_{d_1,d_2} \mu(d_1) \mu(d_2) \psi( \frac{\log d_1}{\varepsilon \log x} ) \psi( \frac{\log d_2}{\varepsilon \log x} ) \sum_{n \in I: [d_1,d_2]|n} a_n$

where ${[d_1,d_2]}$ denotes the least common multiple of ${d_1}$ and ${d_2}$. From the support of ${\psi}$ we see that the summand is only non-vanishing when ${[d_1,d_2] \leq x^{2\varepsilon}}$. We now use axiom (iv) and split the left-hand side into a main term

$\displaystyle \sum_{d_1,d_2} \mu(d_1) \mu(d_2) \psi( \frac{\log d_1}{\varepsilon \log x} ) \psi( \frac{\log d_2}{\varepsilon \log x} ) \frac{g(d)}{d} C |I|$

and an error term that is at most

$\displaystyle O_\varepsilon( \sum_{d \leq x^{2\varepsilon}} \tau(d)^{O(1)} | \sum_{n \in I: d|n} a_n - \frac{g(d)}{d} C |I|| ). \ \ \ \ \ (20)$

From axiom (ii) and elementary multiplicative number theory, we have the bound

$\displaystyle \sum_{d \leq x} \tau(d)^{O(1)} | \sum_{n \in I: d|n} a_n - \frac{g(d)}{d} C |I| \ll C |I| \log^{O(1)} x$

so from axiom (iv) and Cauchy-Schwarz we see that the error term (20) is acceptable. Thus it will suffice to establish the bound

$\displaystyle \sum_{d_1,d_2} \mu(d_1) \mu(d_2) \psi( \frac{\log d_1}{\varepsilon \log x} ) \psi( \frac{\log d_2}{\varepsilon \log x} ) \frac{g([d_1,d_2])}{[d_1,d_2]}$

$\displaystyle = \frac{1}{\varepsilon \log x} (\int_0^1 \psi'(u)^2\ du) G(1) + o(\frac{1}{\log x}). \ \ \ \ \ (21)$

The summand here is almost, but not quite, multiplicative in ${d_1,d_2}$. To make it genuinely multiplicative, we perform a (shifted) Fourier expansion

$\displaystyle \psi(u) = \int_{\bf R} e^{-(1+it)u} \Psi(t)\ dt \ \ \ \ \ (22)$

for some rapidly decreasing function ${\Psi}$ (essentially the Fourier transform of ${e^u \psi(u)}$). Thus

$\displaystyle \psi( \frac{\log d}{\varepsilon \log x} ) = \int_{\bf R} \frac{1}{d^{\frac{1+it}{\varepsilon \log x}}} \Psi(t)\ dt,$

and so the left-hand side of (21) can be rearranged using Fubini’s theorem as

$\displaystyle \int_{\bf R} \int_{\bf R} E(\frac{1+it_1}{\varepsilon \log x},\frac{1+it_2}{\varepsilon \log x})\ \Psi(t_1) \Psi(t_2) dt_1 dt_2 \ \ \ \ \ (23)$

where

$\displaystyle E(s_1,s_2) := \sum_{d_1,d_2} \frac{\mu(d_1) \mu(d_2)}{d_1^{s_1}d_2^{s_2}} \frac{g([d_1,d_2])}{[d_1,d_2]}.$

We can factorise ${E(s_1,s_2)}$ as an Euler product:

$\displaystyle E(s_1,s_2) = \prod_p (1 - \frac{g(p)}{p^{1+s_1}} - \frac{g(p)}{p^{1+s_2}} + \frac{g(p)}{p^{1+s_1+s_2}}).$

Taking absolute values and using Mertens’ theorem leads to the crude bound

$\displaystyle E(\frac{1+it_1}{\varepsilon \log x},\frac{1+it_2}{\varepsilon \log x}) \ll_\varepsilon \log^{O(1)} x$

which when combined with the rapid decrease of ${\Psi}$, allows us to restrict the region of integration in (23) to the square ${\{ |t_1|, |t_2| \leq \sqrt{\log x} \}}$ (say) with negligible error. Next, we use the Euler product

$\displaystyle \zeta(s) = \prod_p (1-\frac{1}{p^s})^{-1}$

for ${\hbox{Re} s > 1}$ to factorise

$\displaystyle E(s_1,s_2) = \frac{\zeta(1+s_1+s_2)}{\zeta(1+s_1) \zeta(1+s_2)} \prod_p E_p(s_1,s_2)$

where

$\displaystyle E_p(s_1,s_2) := \frac{(1 - \frac{g(p)}{p^{1+s_1}} - \frac{g(p)}{p^{1+s_2}} + \frac{g(p)}{p^{1+s_1+s_2}})(1 - \frac{1}{p^{1+s_1+s_2}})}{(1-\frac{1}{p^{1+s_1}})(1-\frac{1}{p^{1+s_2}})}.$

For ${s_1,s_2=o(1)}$ with nonnegative real part, one has

$\displaystyle E_p(s_1,s_2) = 1 + O(1/p^2)$

and so by the Weierstrass ${M}$-test, ${\prod_p E_p(s_1,s_2)}$ is continuous at ${s_1=s_2=0}$. Since

$\displaystyle \prod_p E_p(0,0) = G(1)$

we thus have

$\displaystyle \prod_p E_p(s_1,s_2) = G(1) + o(1)$

Also, since ${\zeta}$ has a pole of order ${1}$ at ${s=1}$ with residue ${1}$, we have

$\displaystyle \frac{\zeta(1+s_1+s_2)}{\zeta(1+s_1) \zeta(1+s_2)} = (1+o(1)) \frac{s_1 s_2}{s_1+s_2}$

and thus

$\displaystyle E(s_1,s_2) = (G(1)+o(1)) \frac{s_1s_2}{s_1+s_2}.$

The quantity (23) can thus be written, up to errors of ${o(\frac{1}{\log x})}$, as

$\displaystyle \frac{G(1)}{\varepsilon \log x} \int_{|t_1|, |t_2| \leq \sqrt{\log x}} \frac{(1+it_1)(1+it_2)}{1+it_1+1+it_2} \Psi(t_1) \Psi(t_2)\ dt_1 dt_2.$

Using the rapid decrease of ${\Psi}$, we may remove the restriction on ${t_1,t_2}$, and it will now suffice to prove the identity

$\displaystyle \int_{\bf R} \int_{\bf R} \frac{(1+it_1)(1+it_2)}{1+it_1+1+it_2} \Psi(t_1) \Psi(t_2)\ dt_1 dt_2 = (\int_0^1 \psi'(u)^2\ du)^2.$

But on differentiating and then squaring (22) we have

$\displaystyle \psi'(u)^2 = \int_{\bf R} \int_{\bf R} (1+it_1)(1+it_2) e^{-(1+it_1+1+it_2)u}\Psi(t_1) \Psi(t_2)\ dt_1 dt_2$

and the claim follows by integrating in ${u}$ from zero to infinity (noting that ${\psi'}$ vanishes for ${u>1}$).

We have the following variant of (19):

Lemma 3 For any ${d \leq x^{1-3\varepsilon}}$, one has

$\displaystyle \sum_{n \in I: d|n} \nu_\varepsilon(n) a_n \ll \frac{C|I|}{\varepsilon \log x} \frac{\prod_{p|d} O( \min( \frac{\log p}{\varepsilon \log x}, 1 )^2 )}{d} + R_d \ \ \ \ \ (24)$

where the ${R_d}$ are such that

$\displaystyle \sum_{d \leq x^{1-3\varepsilon}} R_d \ll_A |I| \log^{-A} x \ \ \ \ \ (25)$

for any ${A>0}$. We also have the variant

$\displaystyle \sum_{n \in I: d|n} \nu_\varepsilon(n/d) a_n \ll \frac{C|I|}{\varepsilon \log x} \frac{\prod_{p|d} O(1 ) )}{d} + R_d. \ \ \ \ \ (26)$

If in addition ${d}$ has no prime factors less than ${x^\delta}$ for some fixed ${\delta>0}$, one has

$\displaystyle \sum_{n \in I: d|n} \nu_\varepsilon(n) a_n$

$\displaystyle = \frac{1+o(1)}{d} \frac{C|I|}{\varepsilon \log x} (\int_0^1 \psi'(u)^2\ du) G(1) + O(R_d). \ \ \ \ \ (27)$

Roughly speaking, the above estimates assert that ${\nu_\varepsilon}$ is concentrated on those numbers ${n}$ with no prime factors much less than ${x^\varepsilon}$, but factors ${d}$ without such small prime divisors occur with about the same relative density as they do in the integers.

Proof: The left-hand side of (24) can be expanded as

$\displaystyle \sum_{d_1,d_2} \mu(d_1) \mu(d_2) \psi( \frac{\log d_1}{\varepsilon \log x} ) \psi( \frac{\log d_2}{\varepsilon \log x} ) \sum_{n \in I: [d_1,d_2,d]|n} a_n.$

If we define

$\displaystyle R_d := \sum_{d' \leq x^{1-\varepsilon}: d|d'} \tau(d')^2 |\sum_{n \in I:d'|n} a_n - \frac{g(d')}{d'} C|I||$

then the previous expression can be written as

$\displaystyle \sum_{d_1,d_2} \mu(d_1) \mu(d_2) \psi( \frac{\log d_1}{\varepsilon \log x} ) \psi( \frac{\log d_2}{\varepsilon \log x} ) \frac{g([d_1,d_2,d])}{[d_1,d_2,d]} C|I| + O(R_d),$

while one has

$\displaystyle \sum_{d \leq x^{1-3\varepsilon}} R_d \leq \sum_{d' \leq x^{1-\varepsilon}} \tau(d')^3 |\sum_{n \in I:d'|n} a_n - \frac{g(d')}{d'} C|I||$

which gives (25) from Axiom (iv). To prove (24), it now suffices to show that

$\displaystyle \sum_{d_1,d_2} \mu(d_1) \mu(d_2) \psi( \frac{\log d_1}{\varepsilon \log x} ) \psi( \frac{\log d_2}{\varepsilon \log x} ) \frac{g([d_1,d_2,d])}{[d_1,d_2,d]}$

$\displaystyle \ll \frac{1}{\varepsilon \log x} \frac{\prod_{p|d} O( \min( \frac{\log p}{\varepsilon \log x}, 1 )^2 )}{d}. \ \ \ \ \ (28)$

Arguing as before, the left-hand side is

$\displaystyle \int_{\bf R} \int_{\bf R} E^{(d)}(\frac{1+it_1}{\varepsilon \log x},\frac{1+it_2}{\varepsilon \log x})\ \Psi(t_1) \Psi(t_2) dt_1 dt_2$

where

$\displaystyle E^{(d)}(s_1,s_2) := \sum_{d_1,d_2} \frac{\mu(d_1) \mu(d_2)}{d_1^{s_1}d_2^{s_2}} \frac{g([d_1,d_2,d])}{[d_1,d_2,d]}.$

From Mertens’ theorem we have

$\displaystyle E^{(d)}(s_1,s_2) \ll_\varepsilon \frac{\prod_{p|d} O(1)}{d} \log^{O(1)} x$

when ${\hbox{Re} s_1, \hbox{Re} s_2 = \frac{1}{\varepsilon \log x}}$, so the contribution of the terms where ${|t_1|, |t_2| \geq \sqrt{\log x}}$ can be absorbed into the ${R_d}$ error (after increasing that error slightly). For the remaining contributions, we see that

$\displaystyle E^{(d)}(s_1,s_2) = \frac{\zeta(1+s_1+s_2)}{\zeta(1+s_1) \zeta(1+s_2)} \prod_p E^{(d)}_p(s_1,s_2)$

where ${E^{(d)}_p(s_1,s_2) = E_p(s_1,s_2)}$ if ${p}$ does not divide ${d}$, and

$\displaystyle E^{(d)}_p(s_1,s_2) = \frac{g(p^j)}{p^j} \frac{(1 - \frac{1}{p^{s_1}}) (1 - \frac{1}{p^{s_2}}) (1 - \frac{1}{p^{1+s_1+s_2}})}{(1-\frac{1}{p^{1+s_1}})(1-\frac{1}{p^{1+s_2}})}$

if ${p}$ divides ${d}$ ${j}$ times for some ${j \geq 1}$. In the latter case, Taylor expansion gives the bounds

$\displaystyle |E^{(d)}_p(\frac{1+it_1}{\varepsilon \log x},\frac{1+it_2}{\varepsilon \log x})| \lesssim (1+|t_1|+|t_2|)^{O(1)} \frac{\min( \frac{\log p}{\varepsilon \log x}, 1 )^2}{p}$

and the claim (28) follows. When ${p \geq x^\delta}$ and ${|t_1|, |t_2| \leq \sqrt{\log x}}$ we have

$\displaystyle E^{(d)}_p(\frac{1+it_1}{\varepsilon \log x},\frac{1+it_2}{\varepsilon \log x}) = \frac{1+o(1)}{p^j}$

and (27) follows by repeating the previous calculations. Finally, (26) is proven similarly to (24) (using ${d[d_1,d_2]}$ in place of ${[d_1,d_2,d]}$). $\Box$

Now we can prove (15), (16), (17). We begin with (15). Using the Leibniz rule ${L(f*g) = (Lf)*g + f*(Lg)}$ applied to the identity ${\mu = \mu * 1 * \mu}$ and using ${\Lambda = \mu*L}$ and Möbius inversion (and the associativity and commutativity of Dirichlet convolution) we see that

$\displaystyle L\mu = - \mu * \Lambda. \ \ \ \ \ (29)$

Next, by applying the Leibniz rule to ${\Lambda_k = \mu * L^k}$ for some ${k \geq 1}$ and using (29) we see that

$\displaystyle L \Lambda_k = L \mu * L^k + \mu * L^{k+1}$

$\displaystyle = - \mu * \Lambda * L^k + \Lambda_{k+1}$

and hence we have the recursive identity

$\displaystyle \Lambda_{k+1} = L \Lambda_k + \Lambda *\Lambda_k. \ \ \ \ \ (30)$

In particular, from induction we see that ${\Lambda_k}$ is supported on numbers with at most ${k}$ distinct prime factors, and hence ${\Lambda_{\vec k}}$ is supported on numbers with at most ${|\vec k|}$ distinct prime factors. In particular, from (18) we see that ${\nu_\varepsilon(n) = O(1)}$ on the support of ${\Lambda_{\vec k}}$. Thus it will suffice to show that

$\displaystyle \sum_{n \in I: \nu_\varepsilon(n) \neq 1} \Lambda_{\vec k}(n) a_n \ll (\varepsilon+o(1)) C |I| \log^{|\vec k|-1} x.$

If ${\nu_\varepsilon(n) \neq 1}$ and ${\Lambda_{\vec k}(n) \neq 0}$, then ${n}$ has at most ${|\vec k|}$ distinct prime factors ${p_1 < p_2 < \dots < p_r}$, with ${p_1 \leq x^\varepsilon}$. If we factor ${n = n_1 n_2}$, where ${n_1}$ is the contribution of those ${p_i}$ with ${p_i \leq x^{1/10|\vec k|}}$, and ${n_2}$ is the contribution of those ${p_i}$ with ${p_i > x^{1/10|\vec k|}}$, then at least one of the following two statements hold:

• (a) ${n_1}$ (and hence ${n}$) is divisible by a square number of size at least ${x^{1/10}}$.
• (b) ${n_1 \leq x^{1/5}}$.

The contribution of case (a) is easily seen to be acceptable by axiom (ii). For case (b), we observe from (30) and induction that

$\displaystyle \Lambda_k(n) \ll \log^{|\vec k|} x \prod_{j=1}^k \frac{\log p_j}{\log x}$

and so it will suffice to show that

$\displaystyle \sum_{n_1} (\prod_{p|n_1} \frac{\log p}{\log x}) \sum_{n \in I: n_1 | n} 1_R(n/n_1) a_n \ll (\varepsilon + o(1)) C |I| \log^{-1} x$

where ${n_1}$ ranges over numbers bounded by ${x^{1/5}}$ with at most ${|\vec k|}$ distinct prime factors, the smallest of which is at most ${x^\varepsilon}$, and ${R}$ consists of those numbers with no prime factor less than or equal to ${x^{1/10|\vec k|}}$. Applying (26) (with ${\varepsilon}$ replaced by ${1/10|\vec k|}$) gives the bound

$\displaystyle \sum_{n \in I: d|n} 1_R(n/n_1) a_n \ll \frac{C|I|}{\log x} \frac{1}{n_1} + R_d$

so by (25) it suffices to show that

$\displaystyle \sum_{n_1} (\prod_{p|n_1} \frac{\log p}{\log x}) \frac{1}{n_1} \ll \varepsilon$

subject to the same constraints on ${n_1}$ as before. The contribution of those ${n_1}$ with ${r}$ distinct prime factors can be bounded by

$\displaystyle O(\sum_{p_1 \leq x^\varepsilon} \frac{\log p_1}{p_1 \log x}) \times O(\sum_{p \leq x^{1/5}} \frac{\log p}{p\log x})^{r-1};$

applying Mertens’ theorem and summing over ${1 \leq r \leq |\vec k|}$, one obtains the claim.

Now we show (16). As discussed previously in this section, we can replace ${\Lambda_{\vec k}(n)}$ by ${\sum_{d|n} \mu_{\vec k}(d) \log^{k_r} \frac{x}{d}}$ with negligible error. Comparing this with (16) and (11), we see that it suffices to show that

$\displaystyle \sum_{n \in I} \sum_{d|n} \mu_{\vec k}(d) \log^{k_r} \frac{x}{d} (1 - \eta_\varepsilon(\frac{\log d}{\log x})) \nu_\varepsilon(n) a_n \ll (\varepsilon+o(1)) C |I| \log^{|\vec k|-1} x.$

From the support of ${\eta_\varepsilon}$, the summand on the left-hand side is only non-zero when ${d \geq x^{1-4\varepsilon}}$, which makes ${\log^{k_r} \frac{x}{d} \ll \varepsilon^{k_r} \log^{k_r} x \leq \varepsilon^2 \log^{k_r} x}$, where we use the crucial hypothesis ${k_r > 1}$ to gain enough powers of ${\varepsilon}$ to make the argument here work. Applying Lemma 2, we reduce to showing that

$\displaystyle \sum_{n \in I} \sum_{d|n: d \geq x^{1-4\varepsilon}} \nu_\varepsilon(n) a_n \ll \frac{1+o(1)}{\varepsilon \log x} C |I|.$

We can make the change of variables ${d \mapsto n/d}$ to flip the sum

$\displaystyle \sum_{d|n: d \geq x^{1-4\varepsilon}} 1 \leq \sum_{d|n: d \leq x^{3\varepsilon}} 1$

and then swap the sums to reduce to showing that

$\displaystyle \sum_{d \leq x^{4\varepsilon}} \sum_{n \in I} \nu_\varepsilon(n) a_n \ll \frac{1+o(1)}{\varepsilon \log x} C |I|.$

By Lemma 3, it suffices to show that

$\displaystyle \sum_{d \leq x^{4\varepsilon}} \frac{\prod_{p|d} O( \min( \frac{\log p}{\varepsilon \log x}, 1 )^2 )}{d} \ll 1.$

To prove this, we use the Rankin trick, bounding the implied weight ${1_{d \leq x^{4\varepsilon}}}$ by ${O( \frac{1}{d^{1/\varepsilon \log x}} )}$. We can then bound the left-hand side by the Euler product

$\displaystyle \prod_p (1 + O( \frac{\min( \frac{\log p}{\varepsilon \log x}, 1 )^2}{p^{1+1/\varepsilon \log x}} ))$

which can be bounded by

$\displaystyle \exp( O( \sum_p \frac{\min( \frac{\log p}{\varepsilon \log x}, 1 )^2}{p^{1+1/\varepsilon \log x}} ) )$

and the claim follows from Mertens’ theorem.

Finally, we show (17). By (11), the left-hand side expands as

$\displaystyle \sum_{d \leq x^{1-3\varepsilon}} \mu_{\vec k}(d) \log^{k_r} \frac{x}{d} \eta_\varepsilon(\frac{\log d}{\log x}) \sum_{n \in I: d|n} \nu_\varepsilon(n) a_n.$

We let ${\delta>0}$ be a small constant to be chosen later. We divide the outer sum into two ranges, depending on whether ${d}$ only has prime factors greater than ${x^\delta}$ or not. In the former case, we can apply (27) to write this contribution as

$\displaystyle \sum_{d \leq x^{1-3\varepsilon}} \mu_{\vec k}(d) \log^{k_r} \frac{x}{d} \eta_\varepsilon(\frac{\log d}{\log x}) \frac{1+o(1)}{d} \frac{C|I|}{\varepsilon \log x} (\int_0^1 \psi'(u)^2\ du) G(1)$

plus a negligible error, where the ${d}$ is implicitly restricted to numbers with all prime factors greater than ${x^\delta}$. The main term is messy, but it is of the required form ${C Q'_{\varepsilon,x} G(1)}$ up to an acceptable error, so there is no need to compute it any further. It remains to consider those ${d}$ that have at least one prime factor less than ${x^\delta}$. Here we use (24) instead of (27) as well as Lemma 3 to dominate this contribution by

$\displaystyle \sum_{d \leq x^{1-3\varepsilon}} O( \log^{|\vec k|} x \frac{C|I|}{\varepsilon \log x} \frac{\prod_{p|d} O( \min( \frac{\log p}{\varepsilon \log x}, 1 )^2 )}{d} )$

up to negligible errors, where ${d}$ is now restricted to have at least one prime factor less than ${x^\delta}$. This makes at least one of the factors ${\min( \frac{\log p}{\varepsilon \log x}, 1 )}$ to be at most ${O_\varepsilon(\delta)}$. A routine application of Rankin’s trick shows that

$\displaystyle \sum_{d \leq x^{1-3\varepsilon}} \frac{\prod_{p|d} O( \min( \frac{\log p}{\varepsilon \log x}, 1 ) )}{d} \ll_\varepsilon 1$

and so the total contribution of this case is ${O_\varepsilon((\delta+o(1)) |I| \log^{|\vec k|-1} x)}$. Since ${\delta>0}$ can be made arbitrarily small, (17) follows.

— 2. Weierstrass approximation —

Having proved Theorem 1, we now take linear combinations of this theorem, combined with the Weierstrass approximation theorem, to give the asymptotics (7), (8) described in the introduction.

Let ${a_n}$, ${g}$, ${C}$, ${G}$ be as in that theorem. It will be convenient to normalise the weights ${\Lambda_{\vec k}}$ by ${L^{1-|\vec k|}}$ to make their mean value comparable to ${1}$. From Theorem 1 and summation by parts we have

$\displaystyle \sum_{n \leq x} L^{1-|\vec k|} \Lambda_{\vec k}(n) a_n = (G(1)+o(1)) \frac{\prod_{i=1}^r k_i!}{(|\vec k|-1)!} C x \ \ \ \ \ (31)$

whenever ${\vec k}$ does not consist entirely of ones.

We now take a closer look at what happens when ${\vec k}$ does consist entirely of ones. Let ${1^r}$ denote the ${r}$-tuple ${(1,\dots,1)}$. Convolving the ${k=1}$ case of (30) with ${r-1}$ copies of ${\Lambda}$ for some ${r \geq 1}$ and using the Leibniz rule, we see that

$\displaystyle \Lambda_{(1^{r-1}, 2)} = \frac{1}{r} L \Lambda_{1^r} + \Lambda_{1^{r+1}}$

and hence

$\displaystyle L^{-r} \Lambda_{1^{r+1}} = L^{-r} \Lambda_{(1^{r-1},2)} - \frac{1}{r} L^{1-r} \Lambda_{1^r}.$

Multiplying by ${a_n}$ and summing over ${n \leq x}$, and using (31) to control the ${\Lambda_{(1^{r-1},2)}}$ term, one has

$\displaystyle \sum_{n \leq x} L^{-r} \Lambda_{1^{r+1}}(n) a_n = (G(1)+o(1)) \frac{2}{r!} - \frac{1}{r} \sum_{n \leq x} L^{1-r} \Lambda_{1^{r}}(n) a_n.$

If we define ${\delta_x}$ (up to an error of ${o(1)}$) by the formula

$\displaystyle \sum_{n \leq x} \Lambda(n) a_n = (\delta_x G(1) + o(1)) C x$

then an induction then shows that

$\displaystyle \sum_{n \leq x} L^{1-r} \Lambda_{1^r}(n) a_n = \frac{1}{(r-1)!} (\delta_x G(1) + o(1)) C x$

for odd ${r}$, and

$\displaystyle \sum_{n \leq x} L^{1-r} \Lambda_{1^r}(n) a_n = \frac{1}{(r-1)!} ((2-\delta_x) G(1) + o(1)) C x$

for even ${r}$. In particular, after adjusting ${\delta_x}$ by ${o(1)}$ if necessary, we have ${0 \leq \delta_x \leq 2}$ since the left-hand sides are non-negative.

If we now define the comparison sequence ${b_n := C G(1) (1 + (1-\delta_x) \mu(n))}$, standard multiplicative number theory shows that the above estimates also hold when ${a_n}$ is replaced by ${b_n}$; thus

$\displaystyle \sum_{n \leq x} L^{1-r} \Lambda_{1^r}(n) a_n = \sum_{n \leq x} L^{1-r} \Lambda_{1^r}(n) b_n + o( x )$

for both odd and even ${r}$. The bound (31) also holds for ${b_n}$ when ${\vec k}$ does not consist entirely of ones, and hence

$\displaystyle \sum_{n \leq x} L^{1-|\vec k|} \Lambda_{\vec k}(n) a_n = \sum_{n \leq x} L^{1-|\vec k|} \Lambda_{\vec k}(n) b_n + o( x )$

for any fixed ${\vec k}$ (which may or may not consist entirely of ones).

Next, from induction (on ${j_1+\dots+j_r}$), the Leibniz rule, and (30), we see that for any ${r \geq 1}$ and ${j_1,\dots,j_r \geq 0}$, ${k_1,\dots,k_r}$, the function

$\displaystyle L^{1-j_1-\dots-j_r-|\vec k|} ((L^{j_1} \Lambda_{k_1}) * \dots * (L^{j_r} \Lambda_{k_r})) \ \ \ \ \ (32)$

is a finite linear combination of functions of the form ${L^{1-|\vec k'|} \Lambda_{\vec k'}}$ for tuples ${\vec k'}$ that may possibly consist entirely of ones. We thus have

$\displaystyle \sum_{n \leq x} f(n) a_n = \sum_{n \leq x}f(n) b_n + o( x )$

whenever ${f}$ is one of these functions (32). Specialising to the case ${k_1=\dots=k_r=1}$, we thus have

$\displaystyle \sum_{n_1 \dots n_r \leq x} a_{n} \log^{1-r} n \prod_{i=1}^r (\log n_i/\log n)^{j_i} \Lambda(n_i)$

$\displaystyle = \sum_{n_1 \dots n_r \leq x} b_{n} \log^{1-r} n \prod_{i=1}^r (\log n_i/\log n)^{j_i} \Lambda(n_i) + o(x )$

where ${n := n_1 \dots n_r}$. The contribution of those ${n_i}$ that are powers of primes can be easily seen to be negligible, leading to

$\displaystyle \sum_{p_1 \dots p_r \leq x} a_{n} \log n \prod_{i=1}^r (\log p_i/\log n)^{j_i+1}$

$\displaystyle = \sum_{p_1 \dots p_r \leq x} b_{n} \prod_{i=1}^r (\log p_i/\log n)^{j_i+1} + o(x)$

where now ${n := p_1 \dots p_r}$. The contribution of the case where two of the primes ${p_i}$ agree can also be seen to be negligible, as can the error when replacing ${\log n}$ with ${\log x}$, and then by symmetry

$\displaystyle \sum_{p_1 \dots p_r \leq x: p_1 < \dots < p_r} a_{n} \prod_{i=1}^r (\log p_i/\log n)^{j_i+1}$

$\displaystyle = \sum_{p_1 \dots p_r \leq x: p_1 < \dots < p_r} b_{n} \prod_{i=1}^r (\log p_i/\log n)^{j_i+1} + o(x / \log x).$

By linearity, this implies that

$\displaystyle \sum_{p_1 \dots p_r \leq x: p_1 < \dots < p_r} a_{n} P( \log p_1/\log n, \dots, \log p_r/\log n)$

$\displaystyle = \sum_{p_1 \dots p_r \leq x: p_1 < \dots < p_r} b_{n} P( \log p_1/\log n, \dots, \log p_r/\log n) + o(x / \log x)$

for any polynomial ${P(t_1,\dots,t_r)}$ that vanishes on the coordinate hyperplanes ${t_i=0}$. The right-hand side can also be evaluated by Mertens’ theorem as

$\displaystyle CG(1) \delta_x \int_{\Delta_r} P x + o(x)$

when ${r}$ is odd and

$\displaystyle CG(1) (2-\delta_x) \int_{\Delta_r} P x + o(x)$

when ${r}$ is even. Using the Weierstrass approximation theorem, we then have

$\displaystyle \sum_{p_1 \dots p_r \leq x: p_1 < \dots < p_r} a_{n} g_r( \log p_1/\log n, \dots, \log p_r/\log n)$

$\displaystyle = \sum_{p_1 \dots p_r \leq x: p_1 < \dots < p_r} b_{n} g_r( \log p_1/\log n, \dots, \log p_r/\log n) + o(x / \log x)$

for any continuous function ${g_r}$ that is compactly supported in the interior of ${\Delta_r}$. Computing the right-hand side using Mertens’ theorem as before, we obtain the claimed asymptotics (7), (8).

Remark 4 The Bombieri asymptotic sieve has to use the full power of EH (or GEH); there are constructions due to Ford that show that if one only has a distributional hypothesis up to ${x^{1-c}}$ for some fixed constant ${c>0}$, then the asymptotics of sums such as (5), or more generally (9), are not determined by a single scalar parameter ${\delta_x}$, but can also vary in other ways as well. Thus the Bombieri asymptotic sieve really is asymptotic; in order to get ${o(1)}$ type error terms one needs the level ${1-\varepsilon}$ of distribution to be asymptotically equal to ${1}$ as ${x \rightarrow \infty}$. Related to this, the quantitative decay of the ${o(1)}$ error terms in the Bombieri asymptotic sieve are extremely poor; in particular, they depend on the dependence of implied constant in axiom (iv) on the parameters ${\varepsilon,A}$, for which there is no consensus on what one should conjecturally expect.