We prove in this post an estimate for the prime exponential sums \[\begin{eqnarray}S(N;\alpha):=\sum_{p\leq N}e(\alpha p)\end{eqnarray}\] using an identity of Vaughan that splits the von Magoldt function $\Lambda(n)$, which is a natural weighted version of the characteristic function of the primes, into pieces that are suited to efficient estimation of oscillating sums $\sum_{n\leq N}f(n)$ over primes. Vinogradov was the first to prove nontrivial cancellation in prime exponential sums in 1937, but the proof remained long and technical until Vaughan discovered in 1977 his formula that simplifies the estimation considerably, giving at the same time a much better bound of $N^{\frac{4}{5}}$ (times logarithmic factors) when $\alpha$ has bounded continued fraction coefficients. This estimate was also required in this post on the ternary Goldbach problem. We will also apply the derived bounds to show that the fractional parts $\{\alpha p\}$ are uniformly distributed for any irrational $\alpha$ as $p$ ranges through the primes.
Vaughan's identity
We follow Davenport's Multiplicative Number Theory and Harman's Prime Detecting Sieves. The following is an elementary combinatorial splitting of the von Mangoldt function, and yet it will be seen to yield very powerful results.
Theorem 1 (Vaughan's identity). Let $U,V\geq 1$. For all positive integers $n$ we have
Theorem 1 (Vaughan's identity). Let $U,V\geq 1$. For all positive integers $n$ we have
\[\begin{eqnarray}\Lambda(n)=\Lambda(n)1_{[1,U]}(n)+\sum_{ab=n\atop{a\leq V}}\mu(a)\log b-\sum_{ab=n\atop a\leq UV}\sum_{cd=a\atop c\leq V, d\leq U}\mu(c)\Lambda(d)-\sum_{ab=n\atop a>U,b>V}\Lambda(a)\sum_{c\mid b\atop c\leq V}\mu(c).\end{eqnarray}\]
Proof. We may assume $n>U$, because otherwise the last sum is empty, while the first and the second sum contribute
\[\begin{eqnarray}\sum_{ab=n\atop a\leq V}\mu(a)\log b-\sum_{ab=n}\sum_{cd=a\atop c\leq V}\mu(c)\Lambda(d)=\sum_{ab=n\atop a\leq V}\sum_{cd=b}\mu(a)\Lambda(d)-\sum_{ab=n}\sum_{cd=a\atop c\leq V}\mu(c)\Lambda(d)=0,\end{eqnarray}\]
since $\log m=\sum_{d\mid m}\Lambda(d)$ by the fundamental theorem of arithmetic. Now let $n>U$ and split the simple identity $\Lambda(n)=\sum_{d\mid n}\mu(d)\log(\frac{n}{d})$ into two parts:
\[\begin{eqnarray}\Lambda(n)&=&\sum_{ab=n}\mu(a)\log b\\&=&\sum_{ab=n\atop a\leq V}\mu(a)\log b+\sum_{ab=n\atop a>V}\mu(a)\log b.\end{eqnarray}\]
The second sum equals
\[\begin{eqnarray}\sum_{ab=n\atop a>V}\mu(a)\sum_{d\mid b}\Lambda(d)&=&\sum_{m\mid n}\Lambda(m)\sum_{a\mid \frac{n}{m}\atop a>V}\mu(a)\\&=&-\sum_{m\mid n}\Lambda(m)\sum_{a\mid \frac{n}{m}\atop a\leq V}\mu(a)\end{eqnarray}\]
since $\sum_{h\mid k}\mu(k)=0$ unless $k=1,$ and in the situation above $k=1$ cannot happen. We do another splitting of the sum to find that it equals
\[\begin{eqnarray}&&-\sum_{m\mid n\atop m\leq U}\Lambda(m)\sum_{a\mid \frac{n}{m}\atop a\leq V}\mu(a)-\sum_{m\mid n\atop m> U}\Lambda(m)\sum_{a\mid \frac{n}{m}\atop a\leq V}\mu(a)\\&=&\sum_{km=n\atop m\leq U}\Lambda(m)\sum_{a\mid k\atop a\leq V}\mu(a)-\sum_{km=n\atop m>U}\Lambda(m)\sum_{a\mid k\atop a\leq V}\mu(a).\end{eqnarray}\]
The former sum is the last sum in Vaughan's identity, if we add the condition $k>V$ (this is clearly allowed), while the latter is
\[\begin{eqnarray}\sum_{a\ell m=n\atop m\leq U,a\leq V}\Lambda(m)\mu(a)&=&\sum_{\ell r=n\atop r\leq UV}\sum_{am=r\atop a\leq V, m\leq U}\Lambda(m)\mu(a),\end{eqnarray}\]
which is the missing term in the identity. ■
We remark that there is a very short proof of Vaughan's identity, based on comparing the Dirichlet coefficients in the identity
\[\begin{eqnarray}-\frac{\zeta'}{\zeta}(s)=F(s)-G(s)\zeta'(s)-F(s)G(s)\zeta(s)+\left(-\frac{\zeta'}{\zeta}(s)-F(s)\right)(1-\zeta(s)G(s)),\end{eqnarray}\]
where $\zeta(s)$ is the Riemann zeta function and
\[\begin{eqnarray}F(s)=\sum_{n\leq U}\frac{\Lambda(n)}{n^s},\quad G(s)=\sum_{n\leq V}\frac{\mu(n)}{n^s}\end{eqnarray}\]
are truncated Dirichlet series for $\frac{\zeta'(s)}{\zeta(s)}$ and $\frac{1}{\zeta(s)}$. However, it is perhaps easier to see how the identity arises in the proof given above.
Sums of oscillating functions over primes
Now we apply Vaughan's identity to estimate sums of the form
\[\begin{eqnarray}\sum_{n\leq N}f(n)\Lambda(n),\end{eqnarray}\]
where $f$ could be an arbitrary complex-valued function with $|f(n)|\leq 1$, but $f$ should be oscillating for the following estimates to be useful. We use Vaughan's identity to split $\Lambda(n)$ into four parts, and $\sum_{n\leq N}f(n)\lambda(n)$ then splits correspondingly into four parts, say $S_1,S_2,S_3,S_4$ with
\[\begin{eqnarray}S_1=\sum_{n\leq U}f(n)\Lambda(n)\ll U, \quad (1)\end{eqnarray}\]
and
\[\begin{eqnarray}S_2&=&\sum_{n\leq N}\sum_{ab=n\atop a\leq V}\mu(a)\log bf(ab)\\&=&\sum_{a\leq V}\mu(a)\sum_{b\leq \frac{N}{a}}f(ab)\log b,\end{eqnarray}\]
and
\[\begin{eqnarray}S_3&=&-\sum_{n\leq N}\sum_{ab=n\atop a\leq UV}\sum_{cd=a\atop c\leq V,d\leq U}\mu(c)\Lambda(d)f(ab)\\&=&\sum_{a\leq UV}\left(\sum_{cd=a\atop c\leq V,d\leq U}\mu(c)\Lambda(d)\right)\sum_{b\leq \frac{N}{UV}}f(ab),\end{eqnarray}\]
and lastly
\[\begin{eqnarray}S_4=\sum_{n\leq N}\sum_{ab=n\atop a>U,b>V}\Lambda(a)\sum_{c\mid b\atop c\leq V}\mu(c)f(ab)=\sum_{U\leq a\leq \frac{N}{V}}\Lambda(a)\sum_{V<b\leq \frac{N}{a}}\left(\sum_{c\mid b\atop c\leq V}\mu(c)\right)f(ab).\end{eqnarray}\]
The sum $S_1$ is harmless in estimations since it is $\ll U$ and we want to choose $U$ as well as $V$ to be considerably smaller than $N$ (in the next section we end up with $U=V=N^{\frac{2}{5}}$). The sum $S_3$ is also in a desirable form, since the inner sum is over values of $f$ in an arithmetic progression, so we expect a lot of cancellation, while the coefficients in front of these sums are small; they are at most $\Lambda(a)$. For that reason, the simple bound
\[\begin{eqnarray}S_3\ll (\log N)\sum_{a\leq UV}\left|\sum_{b\leq \frac{N}{UV}}f(ab)\right|\quad (2)\end{eqnarray}\]
should be significantly smaller than $N$ (which would be asymptotically the bound produced for the sum on the right if the absolute values were taken inside the inner sum).
The sum $S_2$ can be transformed into a form similar to $S_3$ by writing
\[\begin{eqnarray}S_2&=&\sum_{a\leq V}\mu(a)\sum_{b\leq \frac{N}{a}}f(ab)\int_{1}^{b}\frac{dz}{z}\\&=&\int_{1}^{N}\sum_{a\leq V}\mu(a)\sum_{b\leq \frac{N}{a}}f(ab)1_{[1,b]}(z)\frac{dz}{z}\\&\ll& \int_{1}^{N}\frac{dz}{z}\cdot \sum_{a\leq V}\max_{z}\left|\sum_{z\leq b\leq \frac{N}{a}}f(ab)\right|\\&\ll& (\log N)\sum_{a\leq V}\max_{z}\left|\sum_{z\leq b\leq \frac{N}{a}}f(ab)\right|\quad (3)\end{eqnarray}\]
We see that both $S_2$ and $S_3$ are dominated by this expression, and again putting absolute values inside everything would give a trivial estimate of $N\log^2 N$, but keeping the sum over the oscillating function intact, we expect to win a lot compared to trivial estimates.
The sum $S_4$ is of different type compared to $S_2$ and $S_3$ as we cannot immediately remove the coefficients $\sum_{c\mid b\atop c\leq V}\mu(c)$ that multiply the values of $f$. For such a sum, Cauchy-Schwarz offers a way to proceed with estimations. To simplify notations, let $\beta(m)=\sum_{c\mid m\atop c\leq V}\mu(c)$. We partition the interval $[U,\frac{N}{V}]$ into dyadic intervals $[U,2U),[2U,4U),...$ (the last interval is shorter than twice the previous one). For any interval $[M,2M]$, we use Cauchy-Schwarz to obtain
\[\begin{eqnarray}S_4&=&\sum_{M\leq a\leq 2M}\Lambda(a)\sum_{V<b\leq \frac{N}{a}}\beta(b)f(ab)\\&\ll& \left(\sum_{M\leq a\leq 2M}\Lambda(a)^2\right)^{\frac{1}{2}} \left(\sum_{M\leq a \leq 2M}\left|\sum_{V<b\leq \frac{N}{a}}\beta(b)f(ab)\right|^2\right)^{\frac{1}{2}}.\end{eqnarray}\]
In the second term we apply the trivial but useful identity $|z|^2=z\bar{z}$ to reveal possible cancellation:
\[\begin{eqnarray}\sum_{M\leq a \leq 2M}\left|\sum_{V<b\leq \frac{N}{a}}\beta(b)f(ab)\right|^2&=&\sum_{V<j\leq \frac{N}{M}}\sum_{V<k<\frac{N}{M}}\beta(j)\overline{\beta(k)}\sum_{\substack{M\leq a \leq 2M\\a\leq \frac{N}{j}\\a\leq\frac{N}{k}}}f(aj)\overline{f(ak)}.\end{eqnarray}\]
Notice that the dyadic partition implies that $\frac{a}{M}$ is almost constant, and this enabled changing the order of summation nicely (so that the resulting sums have length much less than $N$). The easy inequality $|\beta(j)\overline{\beta(k)}|\leq \frac{1}{2}(|\beta(j)|^2+|\beta(k)|^2)$ yields
\[\begin{eqnarray}S_4&\ll& \sum_{V<j\leq \frac{N}{M}}|\beta(j)|^2\sum_{V<k\leq \frac{N}{M}}\Big|\sum_{\substack{M\leq a\leq 2M\\a\leq \frac{N}{j}\\a\leq\frac{N}{k}}}f(aj)\overline{f(ak)}\Big|\\&\ll& \left(\sum_{j\leq \frac{N}{M}}|\beta(j)|^2\right)\Big(\max_{V\leq j\leq \frac{N}{M}}\sum_{V< k\leq \frac{N}{M}}\Big|\sum_{\substack{M\leq a\leq 2M\\a\leq \frac{N}{j}\\a\leq\frac{N}{k}}}f(aj)\overline{f(ak)}\Big|\Big).\end{eqnarray}\]
In principle, there should be another term involving $|\beta(k)|^2$ on the first line, but by symmetry it contributes an equal amount and can hence be forgotten. We finally sum over the choices $M=U,2U,4U,...,2^{r}U$ (where $r$ is the largest number with $2^rU\leq \frac{N}{2V}$ so $r\ll \log N$) and $M=\frac{N}{2V}$ to arrive at
\[\begin{eqnarray}S_4\ll N^{\frac{1}{2}}(\log^3 N)\max_{U\leq M\leq \frac{N}{V}}\max_{V\leq j\leq \frac{N}{M}}\Big(\sum_{V<k\leq \frac{N}{M}}\Big|\sum_{\substack{M\leq a\leq 2M\\a\leq\frac{N}{j}\\a\leq\frac{N}{k}}}f(aj)\overline{f(ak)}\Big|\Big)^{\frac{1}{2}}\quad (4).\end{eqnarray}\]
We applied the bounds
\[\begin{eqnarray}\sum_{a\leq N}\Lambda(a)^2\leq (\log N)\sum_{a\leq N}\Lambda(a)\ll N\log N\end{eqnarray}\]
and (denoting by $\tau$ is the number of divisors function)
\[\begin{eqnarray}\sum_{n\leq N}\beta(n)^2&\leq& N\sum_{n\leq N}\frac{\tau(n)^2}{n}\\&\leq& N\prod_{p\leq N}\left(1+\frac{2^2}{p}+\frac{3^2}{p^2}+\frac{4^2}{p^3}...\right)\\&\leq& N\prod_{p\leq N}\left(1+\frac{4}{p}+\frac{C}{p^2}\right)\\&\ll& N\prod_{p\leq N}\left(1+\frac{4}{p}\right)\\&\leq& N\prod_{p\leq N}\left(1+\frac{1}{p}+\frac{1}{p^2}+...\right)^4\\&\leq& N\log^4 N\end{eqnarray}\]
by Mertens' bound. We record the bound that follows for $\sum_{n\leq N}f(n)\Lambda(n)$ from $(1),(2),(3)$ and $(4)$ as a theorem.
Theorem 2 (Vaughan). For any arithmetic function $f$ with $|f(n)|\leq 1$ and any $U,V\geq 1$ we have
\[\begin{eqnarray}&&\sum_{n\leq N}f(n)\Lambda(n)\\&\ll& U+(\log N)\sum_{a\leq UV}\max_{z}\left|\sum_{z\leq b\leq \frac{N}{a}}f(ab)\right|\\&+& N^{\frac{1}{2}}(\log^3 N)\max_{U\leq M\leq \frac{N}{V}}\max_{V\leq j\leq \frac{N}{M}}\Big(\sum_{V<k\leq \frac{N}{M}}\Big|\sum_{\substack{M\leq a\leq 2M\\a\leq \frac{N}{j}\\a\leq\frac{N}{k}}}f(aj)\overline{f(ak)}\Big|\Big)^{\frac{1}{2}}.\end{eqnarray}\]
It will soon be seen that this inequality leads to a whole power saving of $N^{\frac{1}{5}}$ when $f(n)=e(\alpha n)$ with $\alpha$ being an irrational number with bounded continued fraction coefficients, say (weaker assumptions would suffice). We note, however, that the inequality above is not appropriate for estimating sums of multiplicative functions of unit modulus; in particular character sums. This is because for $f$ completely multiplicative we get
\[\begin{eqnarray}\max_{V\leq j\leq \frac{N}{M}}\sum_{V<k\leq \frac{N}{M}}\Big|\sum_{\substack{M\leq a \leq 2M\\a\leq \frac{N}{j}\\a\leq \frac{N}{k}}}f(aj)\overline{f(ak)}\Big|&=&\max_{V\leq j\leq \frac{N}{M}}\sum_{V<k\leq \frac{N}{M}}\sum_{\substack{M\leq a \leq 2M\\a\leq \frac{N}{j}\\a\leq \frac{N}{k}}}1\\&=&\sum_{V<k\leq \frac{N}{M}}\sum_{\substack{M\leq a \leq 2M\\a\leq \frac{N}{V}}}1\\&=&\sum_{V<k\leq \frac{N}{M}}M=N-VM\end{eqnarray}\]
for $M\in[U,\frac{N}{2V}]$. Taking the maximum over $M$, we get a bound of size $ N^{\frac{1}{2}}(\log^3 N)\sqrt{N-UV}$ for the last term in the theorem above. This loses to the trivial estimate unless $UV\geq cN$ for some constant $c.$ In that case
\[\begin{eqnarray}\sum_{n\leq UV}\left|\sum_{b\leq \frac{N}{a}} f(ab)\right|\geq \sum_{n\leq cN}\left|\sum_{b\leq \frac{N}{a}} f(b)\right|\geq \sum_{\frac{N}{k_0}<n<\frac{N}{k_0-1}}\left|\sum_{b\leq k_0-1}f(b)\right|\gg N,\end{eqnarray}\]
where $k_0$ is any constant such that $\sum_{b\leq k_0-1}f(b)\neq 0$ (it cannot equal to zero for two consecutive values of $k_0$). In conclusion, the bound of Theorem 2 loses to the trivial bound for completely multiplicative functions of modulus $1$. Nevertheless, for many other functions the theorem is very valuable.
Bounding prime exponential sums
We now turn to proving a bound for the prime exponential sum $\sum_{p\leq N}e(\alpha p)$ using Theorem 2. As we already noticed in the post on Goldbach's ternary problem, the size of the exponential sum depends crucially on approximations to $\alpha$ by rationals. We assume the Dirichlet condition $|\alpha-\frac{a}{q}|\leq \frac{1}{q^2}$ with $a$ and $q$ coprime (for any irrational $\alpha$, there are infinitely many solutions to this inequality). When applying Theorem 2, we make use of the simple but good estimate
\[\begin{eqnarray}\left|\sum_{M_1\leq m\leq M_2}e(\beta m)\right|\leq \min\left\{M_2-M_1,\frac{1}{2\|\beta\|}\right\},\end{eqnarray}\]
repeatedly. To bound the resulting sums, we need the following lemma.
Lemma 3. Suppose $|\alpha-\frac{a}{q}|\leq \frac{1}{q^2}$ with $a$ and $q$ coprime. It holds that
\[\begin{eqnarray}\sum_{t\leq T}\min\left\{\frac{N}{t},\frac{1}{\|t\alpha\|}\right\}\ll \left(\frac{N}{q}+q+T\right)\log (qT).\end{eqnarray}\]
Proof. Let $\alpha=\frac{a}{q}+z$ with $|z| \leq \frac{1}{q^2}$. We write the sum under consideration as
\[\begin{eqnarray}&&\sum_{0\leq k\leq \frac{T}{q}}\sum_{1\leq r\leq \frac{q}{2}}\min\left\{\frac{N}{qk+r},\frac{1}{\|\frac{ar}{q}+rz+qkz\|}\right\}\\&+&\sum_{0\leq k\leq \frac{T}{q}}\sum_{\frac{q}{2}<r\leq q}\min\left\{\frac{N}{qk+r},\frac{1}{\|\frac{ar}{q}+rz+qkz\|}\right\}. \quad (5)\\&:=&\Sigma_1+\Sigma_2.\end{eqnarray}\]
We estimate $\Sigma_2$ first. Consider the numbers $x_r=\frac{ar}{q}+rz+qkz,\frac{q}{2}<r\leq q$. Assume that two of them, say $x_r$ and $x_s$ with $r>s$, lie on the same inerval of length $\frac{1}{3q}$ modulo $1$. Then $\frac{a(r-s)}{q}+(r-s)z\mod 1\in [0,\frac{1}{3q}]$, so $\frac{a(r-s)}{q}\mod 1\in [-\frac{1}{q},\frac{4}{3q}]$. This shows that given $r$, there are at most two three such values of $s$. Therefore, each interval of length $\frac{1}{3q}$ contains at most six numbers $\|x_r\|,\frac{q}{2}< r<q$ (because the function $\|x\|$ attains every value at most twice on an interval of length $1$).
If $\frac{1}{\|x_r\|}\in [-\frac{1}{q},\frac{1}{q}]$, we choose the term $\frac{N}{qk+r}$ in the minimum in $\Sigma_2$, and otherwise we choose the term $\frac{1}{\|x_r\|}$ in $\Sigma_2$. Hence $\Sigma_2$ is dominated by
\[\begin{eqnarray}\Sigma_2&\ll& \sum_{0\leq k\leq \frac{T}{q}}\left(\frac{N}{qk+\frac{q}{2}}+\sum_{1\leq r \leq \frac{q}{2}}\frac{1}{\frac{r}{3q}}\right)\\&\ll& N\log \frac{T}{q}+T\log q.\end{eqnarray}\]
For the sum $\Sigma_1$ with the terms $k=0$ excluded, we get the same bound by the same argument. The essential difference in the case $k=0$ is that then $\frac{N}{qk+r}=\frac{N}{r}$, which can be of size $\gg N$. To estimate
\[\begin{eqnarray}\sum_{1\leq r<\frac{q}{2}}\frac{1}{\|\frac{ar}{q}+rz\|},\end{eqnarray}\]
we observe that $\frac{ar}{q}+rz$ differs from any integer by at least $\frac{1}{2q}$ and, as before, at most a constant number of them can lie on any interval of length $\frac{1}{3q}$. Thus
\[\begin{eqnarray}\sum_{1\leq r<\frac{q}{2}}\frac{1}{\|\frac{ar}{q}+rz\|}\ll \sum_{1\leq r<\frac{q}{2}}\frac{1}{\frac{r}{3q}}\ll q \log q.\end{eqnarray}\]
Combining the estimates gives the lemma.■
Now we can formulate and prove Vaughan's bound for prime exponential sums.
Theorem 4 (Vaughan). Let $|\alpha-\frac{a}{q}|\leq \frac{1}{q^2}$ with $1\leq q \leq N$ and $(a,q)=1$. Then
\[\begin{eqnarray}S(N,\alpha):=\sum_{p\leq N}e(\alpha p)\ll \left(\frac{N}{\sqrt{q}}+\sqrt{Nq}+N^{\frac{4}{5}}\right)\log^3 N.\end{eqnarray}\]
Proof. Let
\[\begin{eqnarray}S^{*}(N;\alpha):=\sum_{n\leq N}e(\alpha n)\Lambda(n).\end{eqnarray}\]
By partial summation we see that if $S^{*}(N;\alpha)\ll X$, then $S(N;\alpha)\ll \frac{X}{\log N}$, and vice versa. Therefore we can consider $S^{*}(N;\alpha)$. For any $U$ and $V$, Theorem 2 gives the bound
\[\begin{eqnarray}S^{*}(N,\alpha)&\ll& U+\log N\left(\sum_{t\leq UV}\min\left\{\frac{N}{t},\frac{1}{\|r\alpha\|}\right\}\right)\\&+&N^{\frac{1}{2}}(\log^3 N)\max_{U\leq M\leq \frac{N}{V}}\max_{V\leq j\leq \frac{N}{M}}\Big(M+\sum_{V<k\leq \frac{N}{M}\atop k\neq j}\min\left\{M,\frac{1}{\|(j-k)\alpha\|}\right\}\Big)^{\frac{1}{2}}.\end{eqnarray}\]
The first sum is $\ll \left(\frac{N}{q}+q+UV\right)\log N$ by Lemma 3. Also by that lemma, we have
\[\begin{eqnarray}\sum_{V<k\leq \frac{N}{M}\atop k>j}\min\left\{M,\frac{1}{\|(j-k)\alpha\|}\right\}&\ll& \sum_{V<k\leq \frac{N}{M}\atop k> j}\min\left\{\frac{N}{k-j},\frac{1}{\|(j-k)\alpha\|}\right\}\\&\ll& \left(\frac{N}{q}+q+\frac{N}{M}\right)\log N,\end{eqnarray}\]
and the same estimate holds for the terms $k<j$. Putting it all together,
\[\begin{eqnarray}&&S^{*}(N,\alpha)\\&\ll& U+(\log^2 N)\left(\frac{N}{q}+q+UV\right)\\&+&N^{\frac{1}{2}}(\log^4 N) \max_{U\leq M\leq \frac{N}{V}}\left(\sqrt{M}+\sqrt{\frac{N}{q}}+\sqrt{q}+\sqrt{\frac{N}{M}}\right)\\&\ll& U+(\log^2 N)UV+N^{\frac{1}{2}}(\log^4 N)\left(\sqrt{\frac{N}{U}}+\sqrt{\frac{N}{V}}+\sqrt{q}\right).\end{eqnarray}\]
The last thing to be done is optimization of $U$ and $V$. When we set $UV=\frac{N}{\sqrt{U}}=\frac{N}{\sqrt{V}}$, the errors have the same size, and we solve $U=V=N^{\frac{2}{5}}$. Now the upper bound simplifies to
\[\begin{eqnarray}\ll \left(\frac{N}{\sqrt{q}}+\sqrt{Nq}+N^{\frac{4}{5}}\right)\log^4 N,\end{eqnarray}\]
as wanted. ■
Uniform distribution of $\alpha p$ modulo one
The prime exponential sum bound that we proved completes the proof of the ternary Goldbach problem for large numbers from an earlier post. Moreover, we are going to show that it proves the uniform distribution of the fractional parts $\{\alpha p\}$ on $[0,1]$, where $\alpha$ is any irrational number and $p$ ranges through the primes. The usual method for proving uniform distribution results is the following theorem.
Theorem 5 (Weyl's criterion). Let $a_1,a_2,...$ be a sequence of real numbers on $[0,1]$. The following are equivalent.
$(i)$ The sequence $(a_n)$ is uniformly distributed in the sense that for any $a,b\in (0,1),a<b$ we have
\[\begin{eqnarray}\frac{|\{n\leq x:a_n\in (a,b)\}|}{x}\xrightarrow{x\to \infty} b-a.\end{eqnarray}\]
$(ii)$ For any Riemann integrable function $f:[0,1]\to \mathbb{C}$ we have
\[\begin{eqnarray}\frac{1}{N}\sum_{k=1}^{N}f(a_k)\xrightarrow{N\to \infty} \int_{0}^1 f(x)dx.\end{eqnarray}\]
$(iii)$ The estimate
\[\begin{eqnarray}\sum_{n\leq x}e(ka_n)=o(x)\end{eqnarray}\]
holds for every fixed $k\in \mathbb{Z}\setminus\{0\}$.
Proof. The implication $(i)\Rightarrow (ii)$ follows by choosing Riemann integrable functions $f_n$ such that $f_n(x)=1$ for $x\in (a,b)$, $f_n(x)=0$ for $x<a-\frac{1}{n}$ and $x>b+\frac{1}{n}$ and $0\leq f_n(x)\leq 1$, and eventually letting $n\to \infty$. The implication $(ii)\Rightarrow (i)$ follows by using the fact that for $N>N_{\varepsilon}$ each interval $[\frac{k}{N},\frac{k+1}{N}],k=0,..,,N-1$ contains $N(\frac{1}{k}+\theta\varepsilon)$ elements $a_1,a_2,...,a_N$ ($\theta$ depends on the variables but is bounded by $1$ in modulus). Therefore, $\frac{1}{N}\sum_{k=1}{N}f(a_k)$ approaches the limit defining the Riemann integral.
Plainly $(ii)$ implies $(iii)$, so it remains to show that $(iii)$ implies $(ii)$. From $(iii)$ the condition $(ii)$ readily follows for trigonometric polynomials. By the Weierstrass approximation theorem, trigonometric polynomials are dense in the space of continuous functions $f:[0,1]\to \mathbb{C}$, and these in turn are dense in the space of Riemann integrable functions. ■
We remark that $(ii)$ was useful mainly in proving Weyl's criterion, as $(iii)$ is always easier check. An immediate corollary of the theorem is that the fractional parts $\{\alpha n\},n\in \mathbb{Z_+}$ are uniformly distributed for irrational $\alpha$ (for rational $\alpha$ they cannot be). Indeed, for any fixed $k$ and any integer $N\geq 1$,
\[\begin{eqnarray}\sum_{n\leq N}e(k\alpha n)\ll \frac{1}{\|k\alpha\|},\end{eqnarray}\]
and this is a bound not depending on $N$, so it is certainly $o(N)$.
Showing that $\{\alpha p\}$ is equidistributed is much more difficult, but Theorem 3 enables showing it.
Theorem 6. Let $\alpha$ be irrational. Then the sequence $\{\alpha p\},p\in \mathbb{P}$ is uniformly distributed.
Proof. Let $N$ be a large positive integer. It suffices to consider the case $k=1$, since the following estimates hold if $\alpha$ is replaced by $k\alpha$. Let $\frac{a_1}{q_1},\frac{a_2}{q_2},...$ be the continued fraction convergents of $\alpha$, with $(a_i,q_i)=1$ and $q_1<q_2<...$. Then $|\alpha-\frac{a_i}{q_i}|\leq \frac{1}{q_i^2}$ holds. Suppose $\log^{12}N\leq q_k\leq \frac{N}{\log^{12}N}$ for some $k$. Then Theorem 3 yields
\[\begin{eqnarray}\sum_{p\leq N}e(\alpha p)\ll \frac{N}{\log^2 N}=o(\pi(N)).\end{eqnarray}\]
Now suppose that there is no $k$ with the desired property. Choose $r$ so that $q_r<\log^{12}N,q_{r+1}>\frac{N}{\log^{12}N}$. It is well known that the denominators of the continued fraction convergents satisfy $|\alpha-\frac{p_r}{q_r}|<\frac{1}{q_r q_{r+1}}$, so $|\alpha-\frac{p_r}{q_r}|<\frac{\log^{12}N}{N}$. This means that $\alpha$ belongs to a major arc in the sense of this post (we can take $A=B=12$ in the definition of the major arcs). A result of that post gives
\[\begin{eqnarray}S(N,\alpha)=\frac{\mu(q_r)}{\varphi(q_r)}\int_{2}^{N}\frac{e(zx)}{\log x}dx+O\left(\frac{N}{\log^{12} N}\right),\end{eqnarray}\]
where $z=\alpha-\frac{a_r}{q_r}$. Hence
\[\begin{eqnarray}S(N,\alpha)\ll\frac{1}{\varphi(q_r)}\frac{N}{\log N}+O\left(\frac{N}{\log^{12}N}\right).\end{eqnarray}\]
As $N\to \infty$, also $r\to \infty$, so $\varphi(q_r)\to \infty$. We get $S(N,\alpha)=o(\pi(N))$, which completes the proof. ■
