Using Vinogradov's theorem for finding prime solutions to a linear equation (an exercise from Vaughan's book)

Question

I'm trying to solve an exercise from Vaughan's book, "The Hardy-Littlewood Method" (ex. 3 in chapter 3: Goldbach's problems, p.36), because I want to use the result stated in it. It is a variation of Vinogradov's theorem on every large enough odd integer being the sum of 3 primes.

Here I need to show that there are "many" triples of prime numbers less than or equal to $N$ that solve the equation with integer coefficients: $b_1p_1+b_2p_2+b_3p_3-b_4=0$. More specifically, I need to show that if $R(N)=\sum_{(p_1,p_2,p_3)\mid p_i\leq n, b_1p_1+b_2p_2+b_3p_3-b_4=0} (\log p_1)(\log p_2)(\log p_3)$ then $R(N)=J(N)\mathfrak S +O(N^2/\log ^AN)$ where $J(N)$ is the number of integer solutions to $b_1m_1+b_2m_2+b_3m_3-b_4=0$ satisfying $m_i\leq N$.

Following the Proof as given in Vaughan and Nathanzon, I have defined $F_i(x)=\sum _{p\leq N}\log p\cdot e(b_ipx)$ so that $R(N)=\int _0 ^1 F_1(x)F_2(x)F_3(x)e(-b_4x)dx$.

When integrating over the major arcs $\mathcal M$, I guess I should approximate $F_i(x)$ by $G_i(x)=\frac{c_q(b_i)}{\phi (q)}u_i \left(x-\frac{a}{q}\right)$ where $u_i(y)=\sum _{m\leq N}e(mb_iy)$ and evaluate the integral $\int _\mathcal M G_1(x)G_2(x)G_3(x)e(-b_4x)dx$.

So, I start by integrating $G_1(x)G_2(x)G_3(x)e(-b_4x)$ over a singel major arc $(\frac {a}{q}-\frac {Q}{N}, \frac {a}{q}+\frac {Q}{N})$. After a change of variables I end up with the integral $\int _\frac {-Q}{N}^\frac {Q}{N}u_1(y)u_2(y)u_3(y)e(-b_4y)dy$, and I want to bound the difference between the last integral to: $J(N)=\int _\frac {-1}{2}^\frac {1}{2}u_1(y)u_2(y)u_3(y)e(-b_4y)dy$. I guess this is how I get the $J(N)$ in the main term of the desired expression for $R(N)$.

If so, I want to bound $\int _\frac {-1}{2}^\frac {-Q}{N}u_1(y)u_2(y)u_3(y)e(-b_4y)dy$ and $\int _\frac {Q}{N}^\frac {1}{2}u_1(y)u_2(y)u_3(y)e(-b_4y)dy$. In the proof of Vinogradov's Theorem, this difference is $O(Q^2/N^2)$ where $Q=\log ^BN$. But now, in the case dealt with in the exercise, this difference seems to be huge: $O(N^3)$. Since, when $b_iy$ is an integer $e(b_iy)=1$ which gives $u_i(y)=N$ and $|u_1(y)u_2(y)u_3(y)|=O(N^3)$.

To summarize, my questions are: How do I bound $\int _\frac {Q}{N}^\frac {1}{2}u_1(y)u_2(y)u_3(y)e(-b_4y)dy$? Does anyone know of a place where this claim is proved?

Thanks!

Answer 1

It's a long time since this was posted. I don't know why you didn't ask me directly. There is an oversight in exercise 2. The main term on the RHS in question 2 should be $\gcd(a_1,a_2,a_3)^{-1}J(n)\frak S$. One has to use intervals of length $1/\gcd(a_1,a_2,a_3)$.

Answer 2

Hello,

I too would be very curious about this bound and will think further about this. The nice thing about the $u_i(y)$, as I recall my advisor having told me, is that the estimates for geometric sums can be independent of their length. So, similar to an estimate for geometric sums found in Chapter 25 of H. Davenport, Multiplicative Number Theory, Third Edition, one has that

$$ u_i(y) \ll \min\left(N, \frac{1}{\|b_iy\|}\right). $$

Following the perspective of this enlightening post, Montgomery's uncertainty principle, from the blog of Professor Tao, another nice way to think of it is that the $u_i(y)$ is the Fourier transform of a function $f: \mathbb{Z} \rightarrow \{0,1\}$ that avoids $p-1$ residue classes modulo $p$ for each prime $p$ dividing $b_i$.

A technique for the estimate you require which initially occurred to me was one I learnt from:

S. Baier, L. Zhao, Primes in Quadratic Progressions on Average, Math. Ann., Vol. 338, 2007, No. 4, pp. 963-982.

The analogous situation to the bound you need is treated there at equations (4.2) and (4.3). But there is a difference between that problem and this one. Over there, one only has the case $b_i=1$ and the maximum value of $\sum_{m\leq N}e(m y)$ can be avoided by avoiding $(-Q/N,Q/N)$. But here, when $b_i>1$, the interval $(Q/N, 1/2)$ still contains values where $u_i(y)$ attains its maximum.

So one observes that by naively applying Hölder's inequality with $\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}=1$ to obtain

$$ \int_{Q/N}^{1/2}\left|u_1(y)u_2(y)u_3(y)\right|dy \ll \prod_{i=1}^3\left(\int_{Q/N}^{1/2}\left|u_i(y)\right|^{n_i}dy\right)^{1/n_i}, $$

where, for each $u_i(y)$, if one ignores that one is integrating from $Q/N$ and instead integrates from $0$, one has that

$$ \int_{0}^{1/2}\left|u_i(y)\right|^{n_i}dy \ll b_i\left(\int_{0}^{1/(b_iN)}N^{n_i} dy + \int_{1/(b_iN)}^{1/2b_i}\frac{1}{(b_iy)^{n_i}}dy\right) \ll N^{n_i-1} $$

which only gives $O(N^2)$ for the term you need to estimate. It can be observed that throwing away the part from $(0,Q/N)$ will not save much like this. Letting $b=\max(b_1,b_2,b_3)$ and applying Hölder's inequality to the interval $(Q/N,1/(2b))$ gives

$$ \int_{Q/N}^{1/(2b)}u_1(y)u_2(y)u_3(y)dy \ll \prod_{i=1}^3\left(\int_{Q/N}^{1/(2b)}\left|\frac{1}{b_iy}\right|^{n_i}dy\right)^{1/n_i} \ll \frac{1}{b_1b_2b_3}\left(\frac{N}{Q}\right)^2, $$

but this is only the easy part of the interval to treat like this.

Here are some special cases:

Case 1 Amongst, $b_1,b_2,b_3$, at least one $b_i$=1. Without loss of generality, suppose $b_1=1$. Then by Cauchy's inequality and Parserval's identity,

$$ \int_{Q/N}^{1/2}u_1(y)u_2(y)u_3(y)dy \ll \sup_{y\in (Q/N,1/2)}u_1(y)\left(\int_0^1\left|u_2(y)\right|^2dy\right)^{1/2} \left(\int_0^1\left|u_3(y)\right|^2dy\right)^{1/2} $$

which is $O\left(N^2/Q\right)$.

Case 2 $b_1,b_2,b_3$ are pairwise relatively prime and $|b_i|<R$, say, for $R$ sufficiently small. The interval $(Q/N,1/(2b))$ was dealt with, so here we consider $(1/(2b),1/2)$. For $i=1,2,3$, let

$$ \mathfrak{M}_{b_i} = \bigcup_{k=0}^{b_i-1}\left[\frac{k}{b_i}-\frac{1}{b_iN},\frac{k}{b_i}+\frac{1}{b_iN}\right]. $$

Then the conditions for this case imply that the $\mathfrak{M}_{b_i}$ only intersect each other at an interval centered at zero that is contained in $(-1/(2b),1/(2b))$. The idea is that the $u_i(y)$ are not simultaneously large on $(1/(2b),1-1/(2b))$. Without loss of generality, consider $u_1(y)$ and $u_2(y)$ on $(1/(2b),1-1/(2b))$ and suppose that $u_1(y) \gg N^{1/2}$. Then $y$ is of the form

$$ y= \frac{k \pm \beta}{b_1}, $$

$k \not=0$, $\beta>0$ and $1/\beta > N^{1/2}$. But then,

$$ \frac{1}{\|b_2y\|} = \frac{b_1}{(kb_2 \bmod b_1) + \beta} = O(b_1) $$

since $k\not=0$. Therefore, this implies that on $(1/(2b),1/2)$, one has that $|u_1u_2u_3|\ll NR^2 + N^{3/2}$.

References:

There are some papers that deal with this kind of problem using the circle method. They are:

A. Balog, Linear Equations in Primes, Mathematika (1992), 39 : pp 367-378

M.C. Liu, K.M. Tsang, Small prime solutions of linear equations, Théorie des nombres (Quebec, PQ, 1987), 595–624, de Gruyter, Berlin, 1989.

M.C. Liu, K.M. Tsang, Small prime solutions of some additive equations, Monatsh. Math. 111 (1991), no. 2, 147–169.

Answer 3

First, we may assume $(b_1,b_2,b_3)=1$. That gives that for each $y\in (0,1)$ at most two of $\{b_1y,b_2y,b_3y \}$ can be integers. Now, take $$J^{(i)}_k=\left[\frac {k}{b_i}-\frac {1}{b_iN^\frac{1}{3}},\frac {k}{b_i}+\frac {1}{b_iN^\frac{1}{3}}\right]$$ for all $1\leq k\leq b_i-1$. Assume $N$ to be large enough such that no two of these intervals (for all $i=1,2,3$ and all $k$) intersect each other, unless when they are both centered around the same point $y_0$, i.e both $b_iy_0$ and $b_jy_0$ are integers. If so, at most two of these intervals can intersect each other.

Also, define: $$I^{(i)}_k=\left[\frac {k}{b_i}-\frac {1}{b_iN^\frac{1}{2}},\frac {k}{b_i}+\frac {1}{b_iN^\frac{1}{2}}\right]\subset J^{(i)}_k.$$

On $I^{(i)}_k$ we have that $|u_i(y)|\sim N$, so if $y\in I^{(i)}_{k_1}\cap I^{(j)}_{k_2}$ then $|u_i(y)u_j(y)|\sim N^2$.But, for $l\neq i,j$ we have that $|u_l(y)|\ll N^\frac {1}{3}$ as $y\notin J^{(l)}_k$ for all $k$. Over a certain $I^{(i)}_k$ we can therefore bound $|u_1(y)u_2(y)u_3(y)| $ by $$|u_1(y)u_2(y)u_3(y)|\ll N^2N^\frac {1}{3}.$$ As $|I^{(i)}_k|=\frac{2}{b_iN^\frac{1}{2}}$, we have that: $$\int _{I^{(i)}_k} |u_1(y)u_2(y)u_3(y)|dy \ll N^2N^\frac {1}{3}\frac{2}{b_iN^\frac{1}{2}}=\frac{2}{b_i}N^2N^\frac {-1}{6}$$

There are at most $|b_i|$ such $I^{(i)}_k$'s, so the integral over all of them is bounded by $2N^2N^\frac {-1}{6}$.

Now, say we're in $J^{(i)}_k\setminus I^{(i)}_k$. Then at least one of the $u_i$'s is bounded by $N^\frac{1}{3}$, and the other two are at bounded by $N^\frac{1}{2}$. The length of each $J^{(i)}_k$ is $\frac{2}{b_iN^\frac{1}{3}}$, so: $$\int _{J^{(i)}_k\setminus I^{(i)}_k} |u_1(y)u_2(y)u_3(y)|dy \ll N^\frac {1}{2}N^\frac {1}{2}N^\frac {1}{3}\frac{2}{b_iN^\frac{1}{3}}=\frac{2}{b_i}N.$$ Again, the integral over the total of intervals $J^{(i)}_k\setminus I^{(i)}_k$ is bounded by $2N$.

Outside of all the intervals $J^{(i)}_k$, for all $i=1,2,3$ and $1\leq k\leq b_i-1$, we know that $|u_i(y)|\ll N^\frac {1}{3}$ and therefore $|u_1(y)u_2(y)u_3(y)|\ll N$. Everything is inside $(0,1)$, so the total of the integral "outside of the $J^{(i)}_k$'s" is bounded by $N$.

To summarize: $$\int _\frac {Q}{N}^\frac{1}{2}|u_1(y)u_2(y)u_3(y)|dy\ll 2N^2N^\frac {-1}{6}+2N+N=O(N^{1\frac {5}{6}}).$$

But, this is true only when we are far enough from 0. At 0, all three functions $u_i$ obtain their maximum simultaneously, so the main contribution to the integral will be near 0. take $b$ such that $\frac{1}{bN^{\frac{1}{3}}}\leq\frac{1}{2b_{i}}$ for all $i=1,2,3$. Write: $$\mathcal S= \left[\frac {Q}{N},\frac{1}{bN^{\frac{1}{3}}} \right].$$ We estimate the integral on $\mathcal S$. Note that if $y\in \mathcal S$ then $b_{i}y\leq\frac{1}{2}$ and therefore $\left\Vert b_{i}y\right\Vert =\left|b_{i}y\right|$ for all $i=1,2,3$. Thus, $u_{i}\left(y\right)\ll\frac{1}{\left\Vert b_{i}y\right\Vert }=\frac{1}{\left|b_{i}y\right|}$ for $y\in\mathcal S$. Now: $$\int_{\mathcal S}|u_1(y)u_2(y)u_3(y)|dy \ll \frac{1}{\left|b_{1}b_{2}b_{3}\right|} \int_{\mathcal S}\frac{1}{y^{3}}dy=O\left(\frac{N^{2}}{Q^{2}}\right).$$

We get that The integral of $u_1(y)u_2(y)u_3(y)$ on $\left[\frac{Q}{N},\frac{1}{2}\right]$ is $O\left(\frac{N^{2}}{Q^{2}}\right)$, and this bound is tight.

Answer 4

Hello,

I wanted to suggest some simplifications/improvements.

First of all you can avoid doing case-analysis by using the elementary estimate

$v_1v_2v_3 \leq \min(v_1,v_2,v_3) \max(v_1,v_2,v_3)^2$,

which is valid for all positive real numbers $v_1,v_2,v_3$. In your problem you end up with the bound

$\int_{Q/N}^{1/2} |u_1(y)u_2(y)u_3(y)|dy$

$\leq \left(\sup_{y \in [Q/N,1/2]}\min_{i=1,2,3}|u_i(y)| \right) \int_{Q/N}^{1/2}\max_{i=1,2,3}(|u_i(y)|^2)dy$

$\leq \left(\sup_{y \in [Q/N,1/2]}\min_{i=1,2,3}|u_i(y)|\right) \int_{-1/2}^{1/2}\max_{i=1,2,3}(|u_i(y)|^2)dy.$

And by Parseval's identity

$\leq \left(\sup_{y \in [Q/N,1/2]}\min_{i=1,2,3}|u_i(y)|\right) N$

If $b_1,b_2,b_3$ are coprime (as we have to assume for this estimate), most values of $y$ are at least $(2b_1b_2b_3)^{-1}$ away from a point of the form $a/b_i$ for some $i \in \{1,2,3\}$. For those $y$ we have the estimate

$\sup_{y \in [1/(b_1b_2b_3),1/2]}\min_{i=1,2,3}|u_i(y)| \ll b_1b_2b_3$.

For the remaining values of $y$, which are close to $0$, we have to use the weaker estimate

$\sup_{y \in [Q/N,1/(b_1b_2b_3)]}\min_{i=1,2,3}|u_i(y)| \ll N/Q$.

Altogether we obtain

$\int_{Q/N}^{1/2} |u_1(y)u_2(y)u_3(y)|dy \ll N^2/Q$.

I hope I didn't screw things up somewhere...