MATHEMATICS

N. M. KOROBOV

ESTIMATES OF TRIGONOMETRIC SUMS WITH COMPLETELY UNIFORMLY DISTRIBUTED FUNCTIONS

(Presented by Academician I. M. Vinogradov on 5 IV 1960)

A function (f(x)) is called completely uniformly distributed ((^{1})) if, for every (s \geqslant 1), for any choice of integers (m_1,\ldots,m_s) not all simultaneously equal to zero, the estimate

[
\sum_{x=1}^{P} e^{2\pi i F_s(x)} = O(P),
\tag{1}
]

holds, where the function (F_s(x)) is defined by the equality

[
F_s(x)=m_1 f(x+1)+\cdots+m_s f(x+s).
\tag{2}
]

In the paper ((^{2})), completely uniformly distributed functions are indicated for which the estimate

[
\sum_{x=1}^{P} e^{2\pi i F_s(x)} = O(P^{3/4}\ln P)
\tag{3}
]

is valid.

In the present paper completely uniformly distributed functions are constructed for which the estimates of the corresponding trigonometric sums are sharper than estimate (3), and the result obtained no longer admits any further substantial improvement. The proof is based on a combination of considerations contained in the papers ((^{3,4})).

Let, for (k=1,2,\ldots), primes (p_k) be given from the interval ((2k+1)^k < p_k < 2(k+1)^k). It is easy to show that there exist integers (a_{k,\nu}) ((\nu=1,2,\ldots,k)) such that, for (|x_\nu|\leqslant k), the congruence

[
a_{k,1}x_1+\cdots+a_{k,k}x_k \equiv 0 \pmod {p_k}
]

has no solutions except the trivial one*. Further, let (\psi(k)>k^2) be an arbitrary integer-valued function, (\tau_0=0) and (\tau_k=\tau_{k-1}+k\psi(k)p_k). Every integer (x\geqslant 1) can, evidently, be represented uniquely in the form

[
x=\tau_{k-1}+ky+z,
]

where (k\geqslant 1,\quad 0\leqslant y\leqslant \psi(k)p_k-1) and (1\leqslant z\leqslant k).

Theorem 1. The fractional parts of every function (f(x)) defined by the equalities

[
x=\tau_{k-1}+ky+z,\qquad f(x)=\frac{a_{k,z}}{p_k}\,y,
]

are completely uniformly distributed.

* One may, for example, choose (a_{k,\nu}=(2k+1)^{\nu-1}).

Proof. Suppose that for (\nu=1,2,\ldots,s) at least one of the quantities (m_\nu) is nonzero and (|m_\nu|\leq m). Consider the sum

[
S_k=\sum_{u=1}^{rkp_k} e^{2\pi i F_s(\tau_{k-1}+u)},
\tag{4}
]

where (r\leq \psi(k)-1) and the function (F_s) is defined by equality (2). We shall show that for (k\geq k_0), where (k_0=\max(m,s)), the sums (S_k) vanish. Indeed, noting that for (0\leq y\leq \psi(k)p_k-2) and (1\leq z\leq k), for (\nu=1,2,\ldots,s) the equalities

[
f(\tau_{k-1}+ky+z+\nu)=
\begin{cases}
\dfrac{a_{k,z+\nu}}{p_k}\,y, & \text{if } \nu\leq k-z,\[6pt]
\dfrac{a_{k,z+\nu-k}}{p_k}\,(y+1), & \text{if } \nu>k-z,
\end{cases}
]

hold, we obtain

[
F_s(\tau_{k-1}+ky+z)=
\tag{5}
]

[

\begin{cases}
\dfrac{(m_1a_{k,z+1}+\cdots+m_sa_{k,z+s})y}{p_k}, & \text{if } z\leq k-s,\[8pt]
\dfrac{(m_1a_{k,z+1}+\cdots+m_{k-z}a_{k,k})y+(m_{k-z+1}a_{k,1}+\cdots+m_sa_{k,z+s-k})(y+1)}{p_k}, & \text{if } z>k-s.
\end{cases}
]

Let us split the interval of summation of the sum (S_k) into parts:

[
|S_k|=\left|\sum_{z=1}^{k}\sum_{y=0}^{rp_k-1}
e^{2\pi i F_s(\tau_{k-1}+ky+z)}\right|\leq
]

[
\leq
\sum_{z=1}^{k-s}\left|\sum_{y=0}^{rp_k-1}
e^{2\pi i F_s(\tau_{k-1}+ky+z)}\right|
+
\sum_{z=k-s+1}^{k}\left|\sum_{y=0}^{rp_k-1}
e^{2\pi i F_s(\tau_{k-1}+ky+z)}\right|.
]

Using the choice of the quantities (a_{k,z}), from (5) we obtain

[
|S_k|\leq
\sum_{z=1}^{k-s}\left|\sum_{y=0}^{rp_k-1}
e^{2\pi i\frac{m_1a_{k,z+1}+\cdots+m_sa_{k,z+s}}{p_k}y}\right|
+
]

[
+
\sum_{z=k-s+1}^{k}\left|\sum_{y=0}^{rp_k-1}
e^{2\pi i\frac{m_1a_{k,z+1}+\cdots+m_{k-z}a_{k,k}+m_{k-z+1}a_{k,1}+\cdots+m_sa_{k,z+s-k}}{p_k}y}\right|
=0.
]

Now estimate the sum

[
S=\sum_{x=1}^{P} e^{2\pi i F_s(x)}.
]

Define (n) from the condition (\tau_{n-1}<P\leq \tau_n) and represent (P) in the form

[
P=\tau_{n-1}+r_1np_n+r_2
\qquad
(0\leq r_1\leq \psi(n)-1,\quad 1\leq r_2\leq np_n).
]

Choosing in (4) (r=\psi(k)-1) for (k<n) and (r=r_1) for (k=n), we obtain

[
|S|\leq \tau_{k_0-1}+
\sum_{k=k_0}^{n-1}\left|\sum_{u=1}^{\psi(k)kp_k}
e^{2\pi i F_s(\tau_{k-1}+u)}\right|
+
\left|\sum_{u=1}^{r_1np_n}
e^{2\pi i F_s(\tau_{n-1}+u)}\right|
+np_n\leq
]

[
\leq
\sum_{k=k_0}^{n}|S_k|+\tau_{k_0-1}+k_0p_{k_0}+\cdots+np_n.
\tag{6}
]

Using the choice of (p_k), it is easy to verify that (p_{k-1}\ll \dfrac1k p_k), and, consequently,

[
k_0p_{k_0}+\cdots+np_n\ll \frac{n(n-1)}2p_{n-1}+np_n\ll 2(n-1)p_n .
]

Further, in view of the fact that for (k\ge k_0), (S_k=0), from (6) we obtain

[
|S|\ll \tau_{k_0-1}+2(n-1)p_n . \tag{7}
]

On the other hand, using the fact that

[
p_n<2(2n+1)^n=2(2n+1)\left(1+\frac{2}{2n-1}\right)^{n-1}(2n-1)^{n-1}<
]

[
<30(n-1)p_{n-1}\le
30\,\frac{(n-1)\psi(n-1)p_{n-1}}{(n-1)^2}
<\frac{30}{(n-1)^2}P,
]

we obtain

[
|S|<\tau_{k_0-1}+\frac{60}{n-1}P=O(P),
]

which proves the theorem.

Theorem 2. Whatever monotone function (\varphi(P)), tending arbitrarily slowly to infinity as (P\to\infty), there exists a completely uniformly distributed function (f(x)) such that

[
S=\sum_{x=1}^{P} e^{2\pi i F_s(x)}=O(\varphi(P)).
]

For no completely uniformly distributed function can this estimate be improved to (O(1)).

Proof. Denote by (\widetilde{\varphi}(k)) the function inverse to the function (\varphi(k)), and choose (\psi(k)=[\widetilde{\varphi}(k^2p_{k+1})]+1). Obviously, defining (n), as before, by the condition (\tau_{n-1}<P\le \tau_n), we obtain

[
\widetilde{\varphi}(n-1)^2p_n<\psi(n-1)<P,\qquad (n-1)^2p_n<\varphi(P).
]

Hence, in view of (7), the first assertion of the theorem follows:

[
|S|\ll \tau_{k_0-1}+\frac{2}{n-1}\varphi(P)=O(\varphi(P)).
]

Suppose that the second assertion of the theorem is false. Then there exists a completely uniformly distributed function (f(x)) such that, for every (P\ge1), the estimate

[
\left|\sum_{x=1}^{P} e^{2\pi i F_s(x)}\right|<M, \tag{8}
]

holds, where (M) does not depend on (P).

Choose (n>2M). Since the function (f(x)) is completely uniformly distributed, for any natural numbers (s) and (n) the fractional parts of the system of functions (f(x+1),\ldots,f(x+n+s)) are uniformly distributed in the unit ((n+s))-dimensional cube. Consequently, there exists (x_0) such that the inequalities

[
{f(x_0+\nu)}<\frac{1}{2\pi(n+1)ms}\qquad (\nu=1,2,\ldots,n+s), \tag{9}
]

hold, where (m=\max_{1\le \nu\le s}|m_\nu|). Denote by (\sigma_n) the sum

[
\sigma_n=\sum_{x=x_0}^{x_0+n} e^{2\pi i F_s(x)} .
]

Using estimate (8), we obtain

[
|\sigma_n|=\left|\sum_{x=1}^{x_0+n} e^{2\pi i F_s(x)}-\sum_{x=1}^{x_0-1} e^{2\pi i F_s(x)}\right|<2M.
\tag{10}
]

By (9), for every (x) in the interval (x_0\leqslant x\leqslant x_0+n) it follows that

[
\left|1-e^{2\pi i F_s(x)}\right|
=
2\left|\sin \pi\bigl(m_1{f(x+1)}+\cdots+m_s{f(x+s)}\bigr)\right|
<
\frac{1}{n+1}.
]

But then

[
|\sigma_n|
=
\left|n+1-\sum_{x=x_0}^{x_0+n}\left(1-e^{2\pi i F_s(x)}\right)\right|
\geqslant
]

[
\geqslant
n+1-(n+1)\max_{x_0\leqslant x\leqslant x_0+n}
\left|1-e^{2\pi i F_s(x)}\right|>n.
]

Hence, by (10), the inequality (n<2M) follows, contradicting the choice of (n), and the theorem is thereby completely proved.

Mathematical Institute named after V. A. Steklov
Academy of Sciences of the USSR

Received
30 III 1960

CITED LITERATURE

1. N. M. Korobov, Izv. AN SSSR, ser. matem., 14, 215 (1950).
2. N. M. Korobov, Izv. AN SSSR, ser. matem., 20, 649 (1956).
3. L. P. Starchenko, DAN, 129, No. 3 (1959).
4. N. M. Korobov, DAN, 89, No. 4, 597 (1953).