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If $p$ is a prime, the discussion of the congruence $x^2 \equiv a(p)$ is fairly easy. It is solvable iff $a^{(p-1) / 2} \equiv 1(p)$. With this fact in hand a complete analysis is a simple matter. However, if the question is turned around, the problem is much more difficult. Suppose that $a$ is an integer. For which primes $p$ is the congruence $x^2 \equiv a(p)$ solvable? The answer is provided by the law of quadratic reciprocity. This law was formulated by Euler and A. M. Legendre but Gauss was the first to provide a complete proof. Gauss was extremely proud of this result. He called it the Theorema Aureum, the golden theorem.
20 世纪 20 年代，范德尔科普特提出了一种估算三角和的方法，这种方法在数论中应用广泛。这一理论通常与狄利克雷的除数问题和圆问题有关–即对$(m, n) \in\left(\mathbb{Z}^{+}\right)^2$ 这样的$m^2+n^2 \leqslant x$ 的对数的评估。它更普遍地适用于给定平面轮廓内积分坐标点的计数。
早在 1922 年，van der Corput 就证明了 Dirichlet 除数问题中的余项有阶 $\ll_{\varepsilon} x^{33 / 100+\varepsilon}$ 。这个问题以及其他类似性质的问题后来引发了工作。其中很大一部分是基于van der Corput的观点。
We shall make use of the Poisson summation formula, which undoubtedly constitutes the most natural approach to the study of trigonometric sums. We define the Fourier transform by
$$
\widehat{f}(\vartheta):=\int_{-\infty}^{+\infty} f(t) \mathrm{e}(-\vartheta t) \mathrm{d} t \quad\left(f \in L^1(\mathbb{R})\right) .
$$
The following statement provides a precise framework for our usage of Poisson’s formula.
$$
\varphi(t)=\sum_{n \in \mathbb{Z}} f(n+t)
$$
converges for all $t$, and that its sum defines a function continuous at 0 and of bounded variation on $[0,1]$. Then we have
$$
\lim {N \rightarrow \infty} \sum{|\nu| \leqslant N} \widehat{f}(\nu)=\sum_{n \in \mathbb{Z}} f(n) .
$$
Proof. Jordan’s theorem on Fourier series asserts that any periodic function of bounded variation over a period is the sum of its Fourier series at each point of continuity. This implies $\lim {N \rightarrow \infty} \sum{|\nu| \leqslant N} \widehat{f}(\nu)=\sum_{n \in \mathbb{Z}} f(n)$. since the Fourier coefficient of order $\nu$ of $\varphi$ is $\widehat{f}(\nu)$.
bounding trigonometric integrals
Let $f \in \mathcal{C}^1(] a, b[)$ be such that $f^{\prime}(t)$ is monotonic, of constant sign on $] a, b\left[\right.$, and such that $m:=\inf _{a0}$. Then we have
$$
\left|\int_a^b \mathrm{e}(f(t)) \mathrm{d} t\right| \leqslant \frac{1}{\pi m} \text {. }
$$
Proof. Assuming without loss of generality that $f^{\prime}$ is non-increasing on ]$a, b[$, we can write
$$
\begin{aligned}
\left|2 \pi \int_a^b \mathrm{e}(f(t)) \mathrm{d} t\right| & =\left|\int_a^b \frac{\mathrm{d}{\mathrm{e}(f(t))}}{f^{\prime}(t)}\right|=\left|\left[\frac{\mathrm{e}(f(t))}{f^{\prime}(t)}\right]_a^b-\int_a^b \mathrm{e}(f(t)) \mathrm{d}\left{\frac{1}{f^{\prime}(t)}\right}\right| \
& \leqslant \frac{1}{\left|f^{\prime}(a)\right|}+\frac{1}{\left|f^{\prime}(b)\right|}+\frac{1}{\left|f^{\prime}(b)\right|}-\frac{1}{\left|f^{\prime}(a)\right|} \leqslant \frac{2}{m} .
\end{aligned}
$$
Let $f \in \mathcal{C}^2(] a, b[)$ be such that $f^{\prime \prime}(t)$ has constant sign on ]$a, b\left[\right.$ and set $r:=\inf {a0$. Then we have $$ \left|\int_a^b \mathrm{e}(f(t)) \mathrm{d} t\right| \leqslant \frac{4}{\sqrt{\pi r}} . $
Proof. Suppose, for example, that $f^{\prime \prime}(t) \leqslant-r<0$ for $a{c-\delta}^{c+\delta}+\int_{c+\delta}^b=I_1+I_2+I_3$
say, where the positive parameter $\delta$ satisfies $a+\delta \leqslant c \leqslant b-\delta$. We have
$$
\left|f^{\prime}(t)\right|=\left|\int_c^t f^{\prime \prime}(v) \mathrm{d} v\right| \geqslant r|t-c| \geqslant r \delta
$$
when $t$ is in $[a, c-\delta] \cup[c+\delta, b]$. it follows that
$$
\left|I_1\right|+\left|I_3\right| \leqslant 2 / \pi r \delta \text {. }
$$
Since, trivially, $\left|I_2\right| \leqslant 2 \delta$, we get
$$
|I| \leqslant 2 \delta+2 / \pi r \delta
$$
whence the required result, by selecting $\delta=\sqrt{1 / \pi r}$. It is clear that this upper bound remains valid if, with this choice of $\delta$, we have either $cb-\delta$.
van der Corput’s inequality
(van der Corput’s inequality). Let $a, b \in \mathbb{R}$ be such that $a0 \quad(a<t<b) .
$
Then we have
$$
\sum_{a<n \leqslant b} \mathrm{e}(f(n)) \ll(b-a+1) \lambda^{1 / 2}+\lambda^{-1 / 2} .
$$
Proof. We can assume $\lambda \leqslant 1$, since otherwise it is trivially satisfied. With the notation of above Theorem, the left-hand side of it is
$$
\begin{aligned}
& \ll(\beta-\alpha+1) \max _\nu\left|\int_a^b \mathrm{e}(f(t)-\nu t) \mathrm{d} t\right|+\ln (\beta-\alpha+2) \
& \ll(\beta-\alpha+1) \lambda^{-1 / 2}+\ln (\beta-\alpha+2),
\end{aligned}
$$
where the second upper bound follows from above thm Now we have
$$
\beta-\alpha=\left|\int_a^b f^{\prime \prime}(t) \mathrm{d} t\right| \asymp \lambda(b-a) .
$$
The previous bound is therefore
$$
\lambda^{1 / 2}(b-a)+\lambda^{-1 / 2}+1+\lambda(b-a) \ll \lambda^{1 / 2}(b-a)+\lambda^{-1 / 2} .
$$
(Weyl-van der Corput). For all integers $N \geqslant 1, Q \geqslant 1$ and any complex sequence $\left{z_n\right}_{n=1}^N$, we have
$$
\left|\sum_{1 \leqslant n \leqslant N} z_n\right|^2 \leqslant\left(1+\frac{N-1}{Q}\right) \sum_{|q|<Q}\left(1-\frac{|q|}{Q}\right) \sum_{1 \leqslant n, n+q \leqslant N} z_{n+q} \overline{z_n} .
$$
Proof. Let us put $z_n=0$ for $n \notin[1, N]$. With this convention, we have
$$
Q \sum_{n \in \mathbb{Z}} z_n=\sum_{1 \leqslant q \leqslant Q} \sum_{n \in \mathbb{Z}} z_{n+q}=\sum_{n \in \mathbb{Z}} \sum_{1 \leqslant q \leqslant Q} z_{n+q} .
$$
Only integers $n$ satisfying $1-Q \leqslant n \leqslant N-1$ may contribute to the last inner sum. Hence there are at most $N-1+Q$ such integers.
By the Cauchy-Schwarz inequality, we obtain
$$
\begin{aligned}
Q^2\left|\sum_{n \in \mathbb{Z}} z_n\right|^2 & \leqslant(N-1+Q) \sum_{n \in \mathbb{Z}}\left|\sum_{1 \leqslant q \leqslant Q} z_{n+q}\right|^2 \
& \leqslant(N-1+Q) \sum_{1 \leqslant q_1 \leqslant Q} \sum_{1 \leqslant q_2 \leqslant Q} \sum_{n \in \mathbb{Z}} z_{n+q_1} \overline{z_{n+q_2}} \
& \leqslant(N-1+Q) \sum_{1 \leqslant q_1 \leqslant Q} \sum_{1 \leqslant q_2 \leqslant Q} \sum_{m \in \mathbb{Z}} z_{m+q_1-q_2} \overline{z_m} \
& \leqslant(N-1+Q) \sum_{-Q \leqslant q \leqslant Q} r(q) \sum_{m \in \mathbb{Z}} z_{m+q} \overline{z_m}
\end{aligned}
$$
where $r(q)=\left|\left{\left(q_1, q_2\right): 1 \leqslant q_1 \leqslant Q, 1 \leqslant q_2 \leqslant Q, q_1-q_2=q\right}\right|$. We plainly have
$$
r(-q)=r(q)=\left|\left{q_1: \max (1, q+1) \leqslant q_1 \leqslant \min (Q, Q+q)\right}\right|=Q-|q| .
$$
This implies on dividing by $Q^2$.
下面是一些经典的解析数论代考van der Corput method的题目
$$
F_H(t):=\sum_{|h| \leqslant H}\left(1-\frac{|h|}{H}\right) \cos (2 \pi h t)=\frac{1}{H}\left(\frac{\sin \pi H t}{\sin \pi t}\right)^2
$$
the Fejér kernel of order $H \in \mathbb{N}^*$. Define the Fourier coefficient of index $h \in \mathbb{Z}$ of a function $f \in L^1(\mathbb{T})$ by
$$
c_h(f):=\int_{\mathrm{T}} \mathrm{e}(-h x) f(x) \mathrm{d} x,
$$
with $\mathrm{e}(u):=\exp (2 \pi i u)$. For $f, g \in L^1(\mathrm{~T})$ define the convolution product
$$
f * g(x):=\int_{\mathbf{T}} f(t) g(x-t) \mathrm{d} t .
$$
Finally, denote by $\langle x\rangle$ the fractional part of a real number $x$.
(a) Let $f \in L^1(\mathrm{~T})$ be a real-valued function satisfying, for a suitable integer $H \geqslant 4$,
$$
c_h(f)=0 \quad(|h| \leqslant H) .
$$
Show that $f * F_H=0$. Deduce that, for all $x \in \mathbb{T}$, we have
$$
\begin{aligned}
f(x)(1-I)+\int_{|t| \leqslant 2 / H} & {f(x-t \pm 2 / H)-f(x)} F_H(t) \mathrm{d} t \
& +\int_{2 / H<|t| \leqslant 1 / 2} f(x-t \pm 2 / H) F_H(t) \mathrm{d} t=0, \end{aligned} $$ where $I:=\int_{|t|>2 / H} F_H(t) \mathrm{d} t$.
(b) Show that $I \leqslant \frac{1}{4}$.
(c) Assume that $f$ is bounded on $\mathrm{T}$ and set
$$
K:=\sup {\substack{0 \leqslant y \leqslant 1 / H \ x \in \mathbf{T}}}{f(x+y)-f(x)} . $$ Show that, under the assumption, we have $$ |f|{\infty} \leqslant 4 K(1-I) /(1-2 I) \leqslant 6 K .
$$
(d) Show that, without assuming , we have
$$
|f|_{\infty} \leqslant 6 K+13 \sum_{|h| \leqslant H}\left|c_h(f)\right| .
$$
(e) Let $\left{u_n\right}_{n=1}^N$ be a finite sequence of real numbers. Define
$$
F(x):=\frac{1}{N} \sum_{1 \leqslant n \leqslant N} \mathbf{1}_{[0, x]}\left(\left\langle u_n\right\rangle\right) .
$$
By applying the previous results to the function
$$
f(x):=\langle x\rangle-\frac{1}{2}+c_0(F)-F(x),
$$
show that we have, for every integer $H \geqslant 1$,
$$
\sup {I \subset \mathbf{T}}\left|\frac{1}{N} \sum{1 \leqslant n \leqslant N} \mathbf{1}I\left(\left\langle u_n\right\rangle\right)-\right| I|| \leqslant \frac{12}{H}+9 \sum{0<h \leqslant H} \frac{\left|\sigma_N(h)\right|}{h},
$$
where $\sigma_N(h):=(1 / N) \sum_{1 \leqslant n \leqslant N} \mathrm{e}\left(h u_n\right)$.
With the $k$-th derivatives. Let $k$ be an integer $\geqslant 3$, let $I$ be an interval containing $N$ integers and let $f \in \mathrm{C}^k(I, \mathbb{R})$. Suppose there exist $\lambda>0$ and $C>1$ such that $\lambda \leqslant\left|f^{(k)}(x)\right| \leqslant C \lambda$ for $x \in I$. Show by induction on $k$ that
$$
\left|\sum_{n \in I} \mathrm{e}(f(n))\right| \ll N \lambda^{\alpha_k}+N^{1-\alpha_k} \lambda^{-\alpha_k}
$$
with $\alpha_k:=1 /\left{2^k-2\right}$.
