The Hardy-Littlewood Method

Second Edition

R. C. Vaughan

(I think we should start at chapter 3. About typo errors, please contact blue_239k@outlook.com)

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3

Goldbach′s problemsGoldbach's\text{ }problems

3.1 The ternary Goldbach problem

Vinogradov's attack on Goldbach's ternary problem follows the pattern of the previous chpater, but this time with

f(α)=∑p≤n(log⁡p)e(αp)f(\alpha)=\sum_{p\le n}(\log p)e(\alpha p)

(3.1)

The poor current state of knowledge concerning the distribution of primes in arithmatic progressions demands that the major arcs be rather sparse. The principal difficulty then lies on the minor arcs and the establishment of a suitable anologue of Weyl's inequality.

Let BB denote a positive constant, and for sufficiently large write

P=(log⁡n)BP = (\log n)^{B}

(3.2)

When 1≤a≤q≤P1\le a\le q\le P and (a,p)=1(a,p)=1, let

$$\mathfrak{M}(q,a) = \{\alpha:|\alpha-a/q|\le Pn^{-1}\}$$

(3.3)

denote a typical major arc and write M\mathfrak{M} for thier union. Since nn is large, the major arcs are disjoint and lie in

U=(Pn−1,1+Pn−1]\mathscr{U} = (Pn^{-1},1+Pn^{-1}]

Let m=U\M\mathfrak{m} = \mathscr{U}\backslash\mathfrak{M}. Then, by (3.1),

$$\begin{align*} R(n)&=\int_{\mathscr{U}}f(\alpha)^3e(-n\alpha)\text{d}\alpha\\ &=\int_{\mathfrak{M}}f(\alpha)^3e(-n\alpha)\text{d}\alpha+\int_{\mathfrak{m}}f(\alpha)^3e(-n\alpha)\text{d}\alpha \end{align*}$$

(3.4)

where

$$R(n)=\sum_{\substack{p_1,p_2,p_3\\\\p_1+p_2+p_3=n}}(\log p_1)(\log p_2)(\log p_3)$$

(3.5)

The treatment of the minor arcs rests principally on the following theorem

Theorem 3.1\textbf{Theorem 3.1}\quadSuppose that (a,q)=1,q≤n(a,q)=1, q\le n and ∣α−a/q∣≤q−2|\alpha-a/q|\le q^{-2}. Then

$$f(\alpha) \ll (\log n)^4(nq^{-1/2}+n^{4/5}+n^{1/2}q^{1/2})$$

ProofProof\quad Let

τx=∑d∣xd≤Xμ(d)\tau_x=\sum_{\substack{d|x\\\\d\le X}}\mu(d)

where μ\mu is Möbius function. Then taking X=n2/5X=n^{2/5} and λ(x,y)=Λ(y)e(αxy)\lambda(x,y)=\Lambda(y)e(\alpha xy) in the identity

$$\begin{align*} &\sum_{X<y\le n}\lambda(1,y)+\sum_{X<x\le n}\sum_{X<y\le n/x}\tau_x\lambda(x,y)\\ &=\sum_{d\le X}\sum_{X<y\le n/d}\sum_{z\le n/(yd)}\mu(d)\lambda(dz,y) \end{align*}$$

gives

f(α)=S1−S2−S3+O(n1/2)f(\alpha)=S_1-S_2-S_3+O(n^{1/2})

where

$$S_1=\sum_{x\le X}\sum_{\substack{y\le n/x\\\text{ }}}\mu(x)(\log y)e(\alpha xy)\\ S_2=S_1=\sum_{x\le X^{2}}\sum_{y\le n/x}c_xe(\alpha xy)\quad\text{with }c_x=\sum_{\substack{d\le X,y\le X\\dy=x}}\mu(d)\Lambda(y)\\ S_3=\sum_{\substack{x>X,y>X\\\\xy\le n}}\tau_x\Lambda(y)e(\alpha xy)$$

Here Λ\Lambda is von Mangoldt's function, and the identity follows by observing that τx=0(1<x≤X)\tau_x=0\quad(1<x\le X) and inverting the order of summation.

The inner sum in S1S_1 is

$$\mu(x)\int_1^{n/x}\sum_{\gamma<y\le n/x}e(\alpha xy)\frac{\text{d}\gamma}{\gamma}$$

and cx≪log⁡xc_x\ll \log x. Hence

$$S_1,S_2\ll(\log n)\sum_{x\le X^2}\min(n/x, ||\alpha x||^{-1})$$

Therefore, by Lemma 2.2,

S1,S2≪(log⁡n)2(nq−1+n4/5+q)S_1,S_2\ll (\log n)^2(nq^{-1}+n^{4/5}+q)

Thus it remains to estimate S3S_3.

Let $\mathscr{A}=\{X,2X,4X,\dots,2^kX:2^kX^2<n\le 2^{k+1}X^2\}$

Then

S3=∑Y∈AS(Y)S_3=\sum_{Y\in \mathscr{A}}S(Y)

where

$$S(Y)=\sum_{Y<x\le 2Y}\sum_{X<y\le n/x}\tau_x\Lambda(y)e(\alpha xy)$$

By cauchy's inequality,

$$|S(Y)|^2\ll (\sum_{x\le 2Y}d(x)^2)\sum_{Y<x\le 2Y}|\sum_{X<y\le n/x}\Lambda(y)e(\alpha xy)|^2$$

It is easily shown that ∑x≤Zd(x)2≪Z(log⁡2Z)3\underset{x\le Z}{\sum}d(x)^2\ll Z(\log 2Z)^3. Hence

$$|S(Y)|^2\ll Y(\log n)^5\sum_{y\le n/Y}\sum_{z\le n/Y}\min(Y, ||\alpha (y-z)||^{-1})$$

Thus, by Lemma 2.2,

∣S(Y)∣2≪n(log⁡n)6(nq−1+Y+n/Y+q)|S(Y)|^2\ll n(\log n)^6(nq^{-1}+Y+n/Y+q)

which gives

$$\begin{align*} S_3&\ll\sum_{Y\in\mathscr{A}}(\log n)^3(nq^{-1/2}+n^{1/2}Y^{1/2}+nY^{-1/2}+n^{1/2}q^{1/2})\\ &\ll(\log n)^4(nq^{-1/2}+n^{4/5}+n^{1/2}q^{1/2}) \end{align*}$$

as required.

To estimate

$$\int_{\mathfrak{m}}f(\alpha)^3e(-\alpha n)\text{d}\alpha$$

it is now only necessary to make two observations. First that Parseval's identity and elementary prime number theory together give

$$\int_0^1|f(\alpha)|^2\text{d}\alpha=\sum_{p\le n}(\log p)^2\ll n \log n$$

Second that, by Theorem 3.1 (cf. the proof of Theorem 2.1),

$$\sup_{\alpha \in \mathfrak{m}}|f(\alpha)|\ll n(\log n)^{4-B/2}$$

Thus

Theorem 3.2\textbf{Theorem 3.2}\quad Suppose that AA is a positive constant and B≥2A+10B\ge 2A+10.

Then

$$\int_{\mathfrak{m}}|f(\alpha)|^3\text{d}\alpha\ll n^2(\log n)^{-A}$$

The treatment of the major arcs, although straightforward, requires an appeal to the theory of the distribution of primes in arithmetic progressions.

Lemma 3.1\textbf{Lemma 3.1}\quadLet

v(β)=∑m=1ne(βm)v(\beta)=\sum_{m=1}^ne(\beta m)

(3.6)

Then there is a positive constant CC such that whenever $1\le a\le q\le P,(a,q)=1,\alpha\in\mathfrak{M}(q,a)$ one has

$$f(\alpha)=\frac{\mu(q)}{\phi(q)}v(\alpha-a/q)+O(n\exp(-C(\log n)^{1/2}))$$

ProofProof\quadLet

fX(α)=∑p≤X(log⁡p)e(αp)f_X(\alpha)=\sum_{p\le X}(\log p)e(\alpha p)

Then

$$f_X(a/q)=\sum_{\substack{r=1\\(r,q)=1}}^qe(ar/q)\vartheta(X,q,r)+O((\log X)(\log q))$$

where

$$\vartheta(X,q,r)=\sum_{\substack{p\le X\\\\p\equiv r (\text{mod }q)}}\log p$$

By theorem 53 and (40) of Estermann (1952) it follows that whenever n<X≤n\sqrt{n}<X\le n one has

$$f_X(a/q)=\frac{X}{\phi(q)}\sum_{\substack{r=1\\(r,q)=1}}^qe(ar/q)+O(n\exp(-C_1(\log n)^{1/2}))$$

Observe that this is trivial when X≤nX\le\sqrt{n}. Also, by Theorem 271 of Hardy & Wright (1979)

∑r=1(r,q)=1qe(ar/q)=μ(q)\sum_{\substack{r=1\\(r,q)=1}}^qe(ar/q)=\mu(q)

Hence by (3.1), (3.6), (3.7) and Lemma 2.6 with X=n.F(m)=e(βm),β=α−a/qX=n.F(m)=e(\beta m),\beta=\alpha-a/q,

$$c_m= \begin{cases} e(am/q)\log m-\mu(q)/\phi(q)&\text{when }m\text{ is prime}\\ -\mu(q)/\phi(q)&\text{otherwise} \end{cases}$$

one has

$$f(\alpha)-\frac{\mu(q)}{\phi(q)}v(\alpha-a/q)\ll(1+n|\alpha-a/q|)n\exp(-C_1(\log n)^{1/2})$$

With (3.3) and (3.2) this establishes the lemma.

Let α∈M(q,a)\alpha\in\mathfrak{M}(q,a). Then, by the above lemma,

$$f(\alpha)^3-\frac{\mu(q)}{\phi(q)^3}v(\alpha-a/q)^3\ll n^3\exp(-C(\log n)^{1/2})$$

Now integrating over M\mathfrak{M} gives

$$\sum_{q\le P}\sum_{\substack{a=1\\(a,q)=1}}^q\int_{\mathfrak{M}(q,a)}(f(\alpha)^3-\frac{\mu(q)}{\phi(q)^3}v(\alpha-a/q)^3)e(-\alpha n)\text{d}\alpha\\ \ll P^3n^2\exp(-C(\log n)^{1/2})$$

Therefore, by (3.3),

$$\begin{align*} \int_{\mathfrak{M}}f(\alpha)^3e(-\alpha n)\text{d}\alpha=&\mathfrak{S}(n,P)\int_{-P/n}^{P/n}v(\beta)^3e(-\beta n)\text{d}\beta\\ &+O(P^3n^2\exp(-C(\log n)^{1/2}) \end{align*}$$

(3.8)

where

$$\mathfrak{S}(n,P)=\sum_{q\le P}\sum_{\substack{a=1\\(a,q)=1}}^q\frac{\mu(q)}{\phi(q)^3}e(-an/q)$$

(3.9)

By (3.6), when β\beta is not an integer,

v(β)≪∣∣β∣∣−1v(\beta)\ll ||\beta||^{-1}

(3.10)

Hence the interval of integration [−P/n,P/n]\lbrack-P/n,P/n\rbrack can be replaced by [−12,12]\lbrack-\frac{1}{2},\frac{1}{2}\rbrack with a total error

≪∑q≤Pϕ(q)−2n2P−2\ll\sum_{q\le P}\phi(q)^{-2}n^2P^{-2}

Therefore, by (3.2),

$$\int_{\mathfrak{M}}f(\alpha)^3e(-\alpha n)\text{d}\alpha=\mathfrak{S}(n,p)J(n)+O(n^2(\log n)^{-2B})$$

(3.11)

where

$$J(n)=\int_{-\frac{1}{2}}^{\frac{1}{2}}v(\beta)^3e(-\beta n)\text{d}\beta$$

By (3.6), J(n)J(n) is the number of solutions of m1+m2+m3=nm_1+m_2+m_3=n with 1≤mj≤n1\le m_j\le n. Thus

J(n)=12(n−1)(n−2)J(n)=\frac{1}{2}(n-1)(n-2)

(3.12)

Also, by (3.9),

$$\mathfrak{S}(n,p)=\mathfrak{S}(n)+O(\sum_{q>P}\phi(q)^{-2})$$

where

$$\mathfrak{S}(n)=\sum_{q=1}^\infty\frac{\mu(q)}{\phi(q)^3}\sum_{\substack{a=1\\(a,q)=1}}^qe(-an/q)$$

(3.13)

Hence, by (3.1), (3.11) and Theorem 327 of Hardy & Wright (1979) Ramanujan's sum,

cq(n)=∑a=1(a,q)=1qe(−an/q)c_q(n)=\sum_{\substack{a=1\\(a,q)=1}}^qe(-an/q)

is a multiplicative function of qq and satisfies

cq(n)=μ(q/(q,n))ϕ(q)ϕ(q/(q,n))c_q(n)=\frac{\mu(q/(q,n))\phi(q)}{\phi(q/(q,n))}

(3.14)

Hence, by (3.13),

$$\mathfrak{S}(n)=(\prod_{p|n}(1+(p-1)^{-3}))\prod_{p|n}(1-(p-1)^{-2})$$

(3.15)

This establishes

Theorem 3.3\textbf{Theorem 3.3}\quadSuppose that AA is a positive constant and B≥2AB\ge 2A. Then

$$\int_{\mathfrak{M}}f(\alpha)^3e(-\alpha n)\text{d}\alpha=\frac{1}{2}n^2\mathfrak{S}(n)+O(n^2(\log n)^{-A})$$

where S(n)\mathfrak{S}(n) satisfies (3.15).

Note that S(n)≫1\mathfrak{S}(n)\gg 1 when nn is odd and S(n)=0\mathfrak{S}(n)=0 when nn is even. When coupled with Theorem 3.2 and (3.4), Theorem 3.3 yields

Theorem 3.4\textbf{Theorem 3.4}\quadSuppose that AA is a positive constant and R(n)R(n) satisfies (3.5). Then

$$R(n)=\frac{1}{2}n^2\mathfrak{S}(n)+O(n^2(\log n)^{-A})$$

where S(n)\mathfrak{S}(n) satisfies (3.15).

Corollary\textbf{Corollary}\quadEvery sufficiently large odd number is the sum of three primes.

3.2 The binary Goldbach problem

In the binary Goldbach problem it is not possible to obtain an asymptotic formula in the same manner as in §\text{\sect} 3.1. However, a non-trivial estimate can be obtained for

∑m=1n(R1(m)−mS1(m))2\sum_{m=1}^n(R_1(m)-m\mathfrak{S}_1(m))^2

where

$$R_1(m)=\sum_{\substack{p_1,p_2\\\\p_1+p_2=m}}(\log p_1)(\log p_2)$$

and S1(m)\mathfrak{S}_1(m) is the corresponding singular series. This is because the above expression corresponds to a quaternary problem, rather than to a binary problem. It leads to less precise conclusion that almost every even number is a sum of two primes.

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