Sums of two relatively prime k-th powers.
Roger C. Baker (Provo, UT)
§1 Introduction.
Let k be a natural number, k ≥ 3. Let Vk (x) be the number of solutions
(u, v) in Z2 of
|u|k + |v|k ≤ x, (u, v) = 1
and let
Ek (x) = Vk (x) − ck x2/k ,
2
(1/k)
where ck = π3Γ2 Γ(2/k)
, be the error term in the asymptotic formula for Vk (x).
Recent progress in estimating Ek (x) has been conditional on the Riemann
hypothesis. The best currently known result for E3 (x) under the Riemann
hypothesis is
E3 (x) = O(xθ3 + )
(1.1)
for every > 0, where θ3 = 9581/36864 = 0.2599 . . . (Baker [2]).
Although I cannot improve (1.1) at present, I shall show that it can be
proved without the full strength of the Riemann hypothesis.
Theorem 1 Suppose that ζ(s) has no zero with real part greater than
ρ3 =
123θ3 − 30
= 0.5802 . . . .
90θ3 − 20
Then (1.1) holds.
For E4 (x), we have the bound
(1.2)
E4 (x) = O(xβ+ ), β =
1
107
= 0.2089 . . .
512
under the Riemann hypothesis. This result is given in Zhai [18]. (Earlier
papers on Ek (x) are listed in [18].) It is claimed by Zhai and Cao [20] that
(1.2) holds with β replaced by 37/184 = 0.2010 . . ., but the proof contains
an error. On page 167 of [20], it is shown that
x− E4 (x) 7
X
xηj ,
j=1
where the ηj are given explicitly and η4 = 37/184. However, η7 = 0.2096 . . .,
so the result that ensues is weaker than (1.2).
In the present paper I shall obtain
E4 (x) = O(xθ4 + )
(1.3)
under the Riemann hypothesis, where
θ4 =
7801
= 0.2073 . . . .
37616
As above, I can reach the same result with a narrower zero-free strip.
Theorem 2 We have (1.3) for every > 0, provided that ζ(s) has no zero
with real part greater than
ρ4 =
32θ4 − 5
= 0.7058 . . . .
16θ4 − 1
It is of interest to examine the mean square of Ek (x). The objective here
is to prove a result of the form
Z X
2
2
Ek (x)2 dx = dk X 1+2/k−2/k + O(X 1+2/k−2/k −η )
(1.4)
0
for a positive constant η. Here
0
ck2 ek
dk =
2 1 + k2 −
2
k2
,
with
c0k =
8Γ(1/k)
πk
k
2π
1/k
, ek =
∞
X
k=1
2

2
X

µ(d)d2/k  n−2−2/k .
d|n
The asymptotic formula (1.4) was obtained by Zhai [19] for k ≥ 6, and
Zhai and Cao [20] for k = 5, under the Riemann hypothesis, with an explicitly
given η = η(k). In the present paper I fill in the missing cases k = 3, 4, and
as above, assume only a narrower zero-free strip.
Theorem 3 Suppose that ζ(s) has no zero with real part greater than χ,
where χ < 1 − 1/k. Then the asymptotic formula (1.4) holds with a positive
constant η = η(χ, k).
The proof permits the calculation of a value for η(χ, k). I leave some of
the details of this calculation to the interested reader. The improvement over
the earlier results stems from a relatively simple tool (Lemma 7 below).
Let rk (n) denote the number of representations of the positive integer n
in the form
n = |u|k + |v|k , (u, v) ∈ Z2 .
The Dirichlet series
∞
X
rk (n)
Zk (s) =
ns
n=1
is known to have an extension to a function analytic in
Re s > 1/k − 1/k 2 ,
except for a simple pole at s = 2/k; see, for example, Zhai [19]. To obtain
our theorems, we need to study the mean value
Z 2T
Mk (σ, T ) =
|Zk (σ + it)|2 dt.
T
I shall show that
Mk (σ, T ) T 2+
for σ ≥ 1/k − 1/k 2 + . This is used in the proof of Theorem 3. The stronger
estimate
(1.5)
Mk (σ, T ) T 1+
seems inaccessible without increasing σ substantially. For Theorems 1 and
2, we need σ as small as possible in (1.5) to narrow our zero-free strip. Zhai
3
− 2k12 .
[19] obtains (1.5) with σ = 2k
3
Theorem 4 The bound (1.5) holds provided that
σ ≥ 2/5 (k = 3),
∞
P
n=1
σ ≥ 3/2k − 1/k 2 (k = 4, 5, . . .).
We isolate as a theorem a result on the mean values of partial sums of
rk (n)
.
ns
Theorem 5 Let σ ≥ 2/5 (k = 3), σ ≥ (4k − 4)/k(3k − 2) (k ≥ 4). Let
4
α = max
− 2σ, 3 − 2σk .
k
Suppose that X ≥ 1 and X α ≤ T . Then
2
Z 2T X
r
(n)
k
dt T 1+ .
σ+it n
T
n≤X
This result is used in the proof of Theorem 4.
Most of the estimates for exponential sums and integrals used below can
be traced back to the ideas of van der Corput. However, the paper of Robert
and Sargos [12] not only plays an important role in a result from [2] re-used
here, but is used afresh. In particular, an exponential sum estimate based
on counting solutions of
q
(h1 + `q1 )1/q (hq2 + `q2 )1/q < ∆,
−
n1
n2
in the proof of Theorem 1, depends on [12].
Constants implicit in the ‘O’ and ‘’ notations may depend (unless otherwise stated) on k and ; other dependencies are made explicit where they
occur. Let C(k) be a sufficiently large positive constant depending on k. We
write A B for A B A. The notation ‘n − a ∼ N ’ (where n is an
integer variable and a is fixed) means N < n − a ≤ 2N . We write e(z) for
e2πiz .
I would like to acknowledge the friendly hospitality of the Department
of Mathematics, University of Florida, where much of the work was accomplished.
4
§2 Preliminary results
Let us write ψ(w) = w − [w] − 1/2, with [. . .] the integer part function. The
nice paper of Kühleitner [11] is a helpful source for the present topic. We
find there the formula
2
Tk (x) = Ak x2/k + c0k Φk (x2/k )x1/k−1/k + Pk (x2/k ) + Bk (x)
P
2 (1/k)
,
for the summatory function Tk (x) =
rk (n). Here Ak = 2Γ
kΓ(2/k)
(2.1)
n≤x
∞
X
Φk (u) =
m
−1−1/k
m=1
(2.2)
1
1
1/2
cos 2π mu −
1+
,
4
k
X
Pk (u) = −8
ψ((uk/2 − nk )1/k ),
2−1/k u1/2 ≤n≤u1/2
and Bk (x) = O(1).
Kuba [10] has shown that
Pk (u) = O(u23/73+ ).
(2.3)
Presumably this could be sharpened by a careful application of the recent
work of Huxley [7] within the argument of [10]. Kühleitner [11] gives an
asymptotic formula for the mean value of Pk (u),
Z X
(2.4)
Pk (u)2 du = Ck X 3/2 + O(X 3/2−δk )
0
where Ck and δk are positive numbers given explicitly.
For u in a range [U, 2U ], U large and positive, Kühleitner splits up the
interval of summation in (2.2) using subintervals [Nr , Nr+1 ], where
Nr = Nr (u) =
u1/2
, r = 0, 1, . . . , R.
(1 + 2−rq )1/k
Here q = k/(k − 1) and R is the least integer such that
√
u − NR < 1 for u ∈ [U, 2U ].
It is easy to see that
Nr+1 − Nr = O(U 1/2 2−rq ),
5
2qR U 1/2
(2.5)
, R = O(log U ),
and that
(2.6)
Pk (u) = −8
Nr+1
R
X
X
ψ (uk/2 − nk )1/k + O(log U ).
r=0 n=Nr
There are two well-known approximations to ψ. The first is elementary
(Jones [9]):
2
Z 1 X
e(hw)
(2.7)
dw H −1 .
ψ(w) +
h
0
0<|h|≤H
The second, due to Vaaler [15], is likewise important in the present paper:
X
(2.8)
ah e(hw) ≤ B(w),
ψ(w) −
0<|h|≤H
where B(w) =
P
bh e(hw) is a non-negative trigonometric polynomial, and
|h|≤H
1
1
, bh .
h
H
(The ah and bh are given explicitly by Vaaler. See also the appendix to
[3].) It is worth noting that (2.8), (2.9) are valid even when H < 1, since
|ψ(w)| ≤ 1/2.
Thus for U ≤ u ≤ 2U , and Hr ≥ 1 (0 ≤ r ≤ R)
Nr+1
R
X
X
X
k/2
k 1/k (2.10)
ah
e h(u − n )
Pk (u) + 8
r=0
ah (2.9)
0<|h|≤Hr
≤
R
X
X
r=0 |h|≤Hr
n=Nr
Nr+1
bh
X
e h(uk/2 − nk )1/k + C(k) log U.
n=Nr
Moreover, the van der Corput B-process yields
Nr+1
(2.11)
X00
e(−1/8)
h u1/4
(hm)−1+q/2 ×
e(h(uk/2 − nk )1/k ) = √
k
−
1
n=Nr
m∈[h2r ,h2r+1 ]
X
× |(h, m)|−q+1/2 e(−u1/2 |(h, m)|) + O(log(|h|U + 2)).
6
Here and subsequently,
|(h, m)| = (|h|q + |m|q )1/q ,
P
and 00 indicates that the first and last terms are weighted with a factor
1/2. See Kühleitner [11] for more details.
It is convenient to write ∆k (x) = Tk (x) − Ak x2/k .
For y > 1, let f (y, s) denote the meromorphic function
f (y, s) =
X
1
−
µ(n)−s .
ζ(s) n≤y
Lemma 1 Let X ≥ 1. The function Zk (s) has a meromorphic continuation
to the region
1
1
Re s > − 2
k k
given by
Z ∞
X rk (n) 2 Ak X 2/k−s
∆k (ω)
−s
Zk (s) =
+
−
X
∆
(X)
+
s
dω.
k
s
s+1
n
k
s
−
2/k
ω
X
n≤X
Proof. See, for example, the proof of Lemma 3.1 of Zhai [19].
Lemma 2 Let y > 1. For a suitable positive constant C = C(k), we have
Ek (x) =
X
d≤y
whenever
1
k
−
1
k2
µ(d)∆k
x
dk
+≤λ≤
2
k
1
+
2πi
Z
λ+ixC
λ−ixC
xs
f (y, ks)Zk (s) ds + O(1)
s
− .
Proof. This can easily be adapted from the proof of Lemma 19 of [2], for
example.
Lemma 3 Let > 0. Suppose that ζ(s) has no zero with Re s > θ, where
1
≤ θ < 1−. Then ζ(s) and ζ(s)−1 are O(t ) for s = σ +it, t ≥ 2, σ ≥ θ +.
2
Proof. In view of results in Titchmarsh [14], Chapter 5, we suppose that
σ < 1. Following Titchmarsh [14], §14.2, we apply the Borel-Carathéodory
theorem ([13], §5.5) to the function log ζ(z) and the circles with center 2 + it
7
and radii 2 − θ − 2δ , 2 − θ − δ, where 0 < δ < 1 − . On the larger circle,
writing B1 , B2 , . . . for absolute constants,
Re(log ζ(z)) = log |ζ(z)| < B1 log t.
Hence, on the smaller circle,
| log ζ(z)| ≤
4 − 2θ − 2δ
4 − 2θ − 3δ/2
B1 log t +
| log ζ(2 + it)|
δ/2
δ/2
< B2 δ −1 log t.
In particular, we find that
| log ζ(σ + it)| < B2 δ −1 log t.
(2.12)
Now let σ1 = −3 and δ = 4 , and apply Hadamard’s three-circles theorem
([13], §5.3) to circles of center σ1 + it and radii r1 < r2 < r3 ,
r1 = σ1 − 1 − δ, r2 = σ1 − σ, r3 = σ1 − θ − δ.
If the maxima of | log ζ(z)| on the respective circles are M1 , M2 , M3 , we
obtain
M2 ≤ M11−a M3a ,
where
a=
log 1 +
log r2 /r1
=
log r3 /r1
log 1 +
=
1+δ−σ
σ1 −1−δ
1−θ
σ1 −1−δ
1−σ+δ
+ O(σ1−1 )
1−θ
1 − σ + O(2 )
.
=
1−θ
The last two implied constants are absolute.
By (2.12), M3 < B2 δ −1 log t, and it is easy to show (see [14], §14.2) that
M1 < B3 δ −1 . Since σ + it is on the middle circle,
1−a a
B3
B2 log t
| log ζ(σ + it)| <
δ
δ
< C()(log t)(1−σ+/2)/(1−θ) .
This is stronger than the required bound.
8
Lemma 4 Suppose that ζ(s) has no zero with Re s > θ, where
Then
1
2
≤ θ < 1−.
f (y, s) = O(y θ−σ+ |t| )
(2.13)
for y > 1, s = σ + it, θ + ≤ σ ≤ k, |t| ≥ 2.
Proof. It suffices to prove (2.13) when y is half an odd integer. In Lemma
1
3.12 of [14], take an = µ(n), f (s) = ζ(s)
, c = 2. We obtain
X µ(n)
ns
n<y
1
=
2πi
Z
2+iT
2−iT
1
xw
dw + O
ζ(s + w) w
y2
T
for T > 0. Take T = y k+2 , so that
y2
= O(y θ−σ ).
T
We have
(2.14)
2+iT
1
yw
dw
2−iT ζ(s + w) w
Z θ+ −σ+iT Z 2−iT
2
=
−
Z
θ+ 2 −σ−iT
2+iT
Z
!
+
θ+ 2 −σ−iT
θ+ 2 −σ+iT
1
yw
dw.
ζ(s + w) w
We may now apply Lemma 3. The horizontal integrals on the right-hand
side of (2.14) are
!
Z
2
O T −1+
y u du
= O(T −1+ y 2 )
θ+ 2 −σ
= O(y θ−σ ).
The vertical integral is
Z
θ+ 2 −σ
O y
T
(1 + |t|)
−1+/4k
dt = O(y θ−σ+ ).
−T
The lemma follows on combining these estimates.
9
Lemma 5 Let A > 0, A < B ≤ 2A, C ≥ 2, C < D ≤ 2C. Let f be a
bounded measurable function on [A, B]. Then
Z
D
C
Z
B
2
Z
f (x)x dx dt A log C
B
it
A
|f (x)|2 dx.
A
Proof. We have
Z
D
C
Z
B
A
Z
2
f (x)xit dx dt
B
Z
B
=
Z
D
f (x1 )f (x2 )
A
A
C
x1
x2
it
dt dx1 dx2
(by Fubini’s theorem)
Z
B
Z
B
1
≤
(|f (x1 )| + |f (x2 )| ) min C,
| log x1 /x2 |
A
A
Z B
Z B
1
2
=2
|f (x1 )|
min C,
dx2 dx1 .
| log x1 /x2 |
A
A
2
2
dx1 dx2
In the inner integral, substitute v = x1 /x2 ; then | log v| |v − 1|, so that
Z 2
Z B
1
1
min C,
dx2 A
dv
min C,
1
| log x1 /x2 |
|v
−
1|
A
2
A log C.
The lemma follows at once.
Lemma 6 For 1/k − 1/k 2 + ≤ σ ≤ 1 and T ≥ 2,
Mk (σ, T ) T 2 log T.
Proof. By Lemma 1 with X = 1, s = σ + it, T ≤ t ≤ 2T ,
Z ∞
∆k (ω)
Zk (s) = s
dω + O(1).
ω σ+it+1
1
10
Hence Cauchy’s inequality yields
2
Z 2T X
∞ Z 2j
∆
(ω)
k
dω Mk (σ, T ) T + T 2
dt
σ+it+1 ω
j−1
2
T
j=1
Z
T + T2
2T
2 
! ∞
X Z 2j ∆k (ω)dω j −2 
j2  dt.
2j−1 ω σ+it+1 j=1
j=1
∞
X
T
Since
∆k (ω) ω 1/k−1/k
2
from (2.1), (2.3), we find that
Z
2j
2j−1
|∆k (ω)|2
dω 2−j(1+) .
2σ+2
ω
Applying Lemma 5,
2T
Z
T
Z j
2
2
∆k (ω) dt 2−j log T
2j−1 ω σ+it+1 and
Mk (σ, T ) T + T 2
∞
X
!
j 2 2−j log T T 2 log T.
j=1
Lemma 7 Let D > C ≥ 2, B > A > 1 and suppose that g(t) is a bounded
measurable function on [C, D]. Then
Z
B
A
Z
D
C
2
Z
g(t)x dt dx B log D
D
it
|g(t)|2 dt.
C
Proof. This is a slight variant of Harman [4], Lemma 9.1.
Lemma 8 Let F , G be real differentiable functions on [a, b] such that G/F 0
is monotonic and either F 0 /G ≥ M > 0, or F 0 /G ≤ −M < 0. Then
Z b
iF
(x)
≤ 4 .
G(x)e
dx
M
a
11
Proof. This is Lemma 4.3 of [14].
Lemma 9 Let F be a real differentiable function in [a, b], such that F 0 is
monotonic and 0 < M ≤ |F 0 | ≤ 1 − . Then
X
e(f (n)) = O(M −1 ).
a<n≤b
Proof. This result is known as the Kusmin-Landau theorem. It is a consequence of Lemma 8 in conjunction with Lemma 4.8 of [14]; there is a different
proof in [3].
For H ≥ 1, K ≥ 1, P ≥ 1, Q ≥ 1 and a given quadruple of real numbers
a = (a1 , a2 , a3 , a4 ), let us write
N (a, H, K, P, Q, ∆)
for the number of quadruples (m1 , m2 , m3 , m4 ) with m1 ∼ H, m2 ∼ P H,
m3 ∼ K, m4 ∼ QK,
|a1 mq1 + a2 mq2 + a3 mq3 + a4 mq4 | ≤ ∆(P H)q .
We write more succinctly
N (N, ∆) = N ((1, 1, −1 − 1), N, N, 1, 1, ∆).
Lemma 10 Suppose that 0 < c1 ≤ |aj | ≤ c2 (j = 1, . . . , 4) and c1 P H ≤
QK ≤ c2 P H. Then
N (a, H, K, P, Q, ∆) (P H) (P H 2 + ∆P 3 H 4 )1/2 (QK 2 + ∆Q3 K 4 )1/2 .
Here and in the proof, the implied constants depend on c1 and c2 .
Proof. Let M1 = H, M2 = P H, M3 = K, M4 = QK, and
X
Sj (u) =
e(umq ).
m∼Mj
By a slight variant of Lemma 2.1 of [16],
2
q
Z
N (a, H, K, P, Q, ∆) ≤ π ∆(P H)
0
12
1/2∆(P H)q
4
Y
j=1
|Sj (aj u)| du.
By Hölder’s inequality
∆(P H)q /2
Z
0
4
Y
|Sj (aj u)| du ≤
j=1
Z
4
Y
j=1
Z
4
Y
!1/4
1/2∆(P H)q
|Sj (aj u)|4 du
0
!1/4
aj /2∆(P H)q
|Sj (u)|4 du
,
0
j=1
so that
(2.15)
N (a, H, K, P, Q, ∆) 4
Y
∆(P H)q
Z
!1/4
c2 /2∆(P H)q
|Sj (u)|4 du
0
j=1
Again by Lemma 2.1 of [16],
Z c2 /2∆(P H)q
q
(2.16)
∆(P H)
|Sj (u)|4 du
0
N
∆(P H)q
Mj ,
c2 Mjq
.
We now apply the inequality
N (Mj , ηj ) Mj2+ + Mj4+ ηj ,
which is Theorem 2 of Robert and Sargos [12]. We take
ηj Mjq =
4∆(P H)q
.
c2
Thus, since P H QK,
∆(P H)q
N M1 ,
H 2+ + H 4+ (P q ∆),
q
c2 M1
∆(P H)q
N M2 ,
(P H)2+ + (P H)4+ ∆,
c2 M2q
∆(P H)q
N M3 ,
K 2+ + K 4+ (Qq ∆)
q
c2 M3
13
.
and
N
∆(P H)q
M4 ,
c2 M4q
(QK)2+ + (QK)4+ ∆.
Moreover,
1
((P H)2 + (P H)4 ∆)1/4 (H 2 + H 4 P q ∆) 4
1
(P 2 H 4 + P 4 H 6 ∆ + P 4+q H 8 ∆2 ) 4
(P 2 H 4 + P 6 H 8 ∆2 )1/4 (P H 2 + P 3 H 4 ∆)1/2 .
There is a similar bound
((QK)2 + (QK)4 ∆)1/4 (K 2 + K 4 Qq ∆)1/4
(QK 2 + Q3 K 4 ∆)1/2 ,
so that (2.16) gives the desired bound for the right-hand side of (2.15).
Lemma 11 Let 1 ≤ H ≤ P H, N ≥ 1. The number of solutions N of
|(h1 , `1 )| |(h2 , `2 )| ∆P H
<
−
(2.17)
n1
n2 N
with H ≤ hi ≤ 2H, P H ≤ `i ≤ 2P H, N ≤ ni < 2N is
O((P H) (P 3 H 4 N 2 ∆ + P 3/2 H 3 N )).
Proof. Let d be an integer in [1, 2N ). We count the number of solutions
Nd of (2.17) with (n1 , n2 ) = d. Write nj = kj d, (k1 , k2 ) = 1, k1 ≤ 2N/d,
k2 ≤ 2N/d.
First fix k1 , k2 . Then (2.17) implies
|(h1 , `1 )| −
(2.18)
k1
|(h2 , `2 )| ∆P H
k2
and indeed
(2.19)
hq1
+
`q1
−
k1
k2
q
(hq2 + `q2 ) ∆(P H)q .
14
By Lemma 10 the number of solutions h1 , h2 , `1 , `2 of (2.19) is
(P H)/2 (P H 2 + ∆P 3 H 4 ).
Hence
/2
Nd (P H)
N 2 P H 2 ∆N 2 P 3 H 4
+
d2
d2
.
On the other hand, if we fix h1 , `1 , h2 , `2 , then (2.18) implies
|(h1 , `1 )| k1 |(h2 , `2 )| − k2 ≤ 2∆.
Since the numbers k1 /k2 are spaced at least d2 /4N 2 apart, the number of
solutions of the last inequality is
∆N 2
+ 1.
d2
Hence
2
Nd P H
4
∆N 2
+1
d2
and indeed
Nd (P H)
(P H)
2
∆N 2 P 3 H 4
N P H2 2 4
+ min
,P H
d2
d2
!
2
1/2
N P H2
∆N 2 P 3 H 4
+
(P 2 H 4 )1/2 .
d2
d2
The lemma follows on summing this bound over d.
Lemma 12 Let f be a complex-valued function on [D, D0 ), where 2 ≤ D <
D0 ≤ 2D. Suppose that 0 < U ≤ D1/3 , B > 0, and
X
X
am
f (mn) B
m∼M
n∼N
D≤mn<D0
whenever M N D, N DU −1 and |am | ≤ 1. Suppose further that
X
X
am
bn f (mn) B
m∼M
n∼N
D≤mn<D0
15
whenever M N D, U N D1/2 and |am | ≤ 1, |bn | ≤ 1. Then
X
µ(d)f (d) BD .
D≤d<D0
Proof. This is essentially Lemma 2(ii) of [2]. (The idea is much older; see
[4] for a broader discussion.)
Lemma 13 Let (κ, λ) be an exponent pair. Let α, β be constants, α 6= 0,
α < 1, β < 0. Let X > 0, M ≥ 1/2, N ≥ 1/2, M N D, N0 = min(M, N ),
L = log(D + 2). Let |am | ≤ 1, |bn | ≤ 1, Im ⊆ (N, 2N ], and
X
X Xmβ nα S1 =
am
e
,
M βN α
m∼M
n∈I
m
S2 =
X
am
X
bn e
m∼M
n∼N
D<mn≤D0
Xmβ nα
M βN α
.
(i) We have
2(λ−κ) 1/(6+4κ)
S1 L2 {DN −1/2 + DX −1 + (D4+4κ X 1+2κ N −(1+2κ) N0
)
}.
(ii) If N M and X D, we have
S2 L7/4 (DN −1/2 + DM −1/4 + (D11+10κ X 1+2κ N 2(λ−κ) )1/(14+12κ) ).
The implied constants depend on α, β, κ and λ.
Proof. See [2], Theorems 4 and 5.
Lemma 14 Let α, β be real constants with αβ(α − 1)(β − 1) 6= 0. Let κ, λ,
X, M , N , L, S2 be as in Lemma 13. Then
S2 L3 {(X 2+4κ M 8+10κ N 9+11κ+λ )1/(12+16κ) + X 1/6 M 2/3 N 3/4+λ/(12+12κ)
+ (XM 3 N 4 )1/5 + (XM 7 N 10 )1/11 + M 2/3 N 11/12+λ/(12+12κ)
+ M N 1/2 + (X −1 M 14 N 23 )1/22 + X −1/2 M N }.
16
Proof. At the cost of a factor L, we can remove the condition D < mn ≤ D0
from the sum S2 . See [4], pp. 49–50. Now the result follows at once from
Theorem 2 of [17].
Rb
We recall some facts about Riemann-Stieltjes integrals a f (t)dα(t), as
presented in Apostol [1], Chapter 9. Sometimes these integrals do not receive
enough care in the number theory literature. The functions f and α are
assumed to be real-valued and bounded on [a, b]. We must be careful to
avoid both α and f being discontinuous from (e.g.) the left at any point,
Rb
since then a f (t)d α(t) may not exist (see [1], Theorem 9.28). If we begin
with f a function of bounded variation continuous from the left, and α the
sum of continuous
function and a step function continuous
from the right,
Rb
Rb
then I = a f (t)d α(t) does exist. Moreover, J = a α(t)d f (t) exists and
I + J = f (b)α(b) − f (a)α(a)
([1], Theorems 9.2, 9.6, 9.11 and 9.21). Moreover, if it happens that f is
continuously differentiable on [a, b], then
Z b
Z b
α(t) d f (t) =
α(t)f 0 (t) dt
a
a
([1], Theorem 9.8). We now derive some basic inequalities for the RiemannR 2X
Stieltjes integrals X f (t)d∆k (t) that we shall encounter. Here X ≥ 1. From
the definition, ∆k (t) = Tk (t) − Ak t2/k is the sum of a continuous function
and a step function continuous from the right.
Lemma 15 Let f (t), g(t) be real functions of bounded variation continuous
from the left on [X, 2X], |f (t)| ≤ g(t) (t ∈ [X, 2X]). Then
(i) We have
Z
2X
f (t) d∆k (t) kf k∞ X 2/k .
X
(ii) We have
Z 2X
Z
2X
f (t) d∆k (t) X
2/k−1
g(t) t
X
Z
dt + 2X
X
g(t) d∆k (t) .
(iii) If f is continuously differentiable on [X, 2X], then
Z 2X
Z 2X
1/k−1/k2
0
f (t) d∆k (t) kf k∞ X
+
f (t) ∆k (t)dt .
X
X
17
Here kf k∞ =
sup |f (t)|.
X≤t≤2X
Proof. (i) We have
Z
2X
2X
Z
2/k
f (t)d∆k (t) = −
(2.20)
f (t)d(Ak t
X
2X
Z
)+
f (t)dTk (t).
X
X
and ([1], Theorem 9.23)
Z 2X
2/k 2/k
2/k
f
(t)d(A
t
)
k
kf k∞ Ak ((2X) − X ),
X
Z
2X
X
f (t)d(Tk (t)) ≤ kf k∞ (Tk (2X) − Tk (X)).
Now (i) follows from simple bounds for the expressions used to bound the
two integrals.
(ii) From (2.20), and [1], Theorem 9.22,
Z
Z
Z 2X
2Ak 2X
2/k−1
+
≤
f
(t)t
dt
f
(t)d∆
(t)
k
k X
X
2Ak
≤
k
Z
4Ak
=
k
Z
2X
g(t)d Tk (t)
X
2X
2/k−1
g(t)t
Z
2X
dt +
g(t)d Tk (t)
X
X
2X
2/k−1
g(t)t
Z
2X
dt +
X
g(t)d∆k (t).
X
(iii) We have
2X
Z
X
Z
2X Z
2X
f (t)d∆k (t) = ∆k (t)f (t) −
∆k (t)f 0 (t)dt
X
X
Z
2X
f (t)d∆k (t) ≤ k∆k k∞ kf k∞ + X
2X
X
kf k∞ X
1/k−1/k2
∆k (t)f (t)dt
0
Z
+ 2X
X
18
∆k (t)f (t)dt .
0
§3 Proof of Theorem 5.
By a splitting-up argument and Minkowski’s inequality, it suffices to show
that
2
Z 2T X
rk (n) (3.1)
dt T 1+/2 .
σ+it n
T
n∼X
The left-hand side of (3.1) is
(3.2)
X X rk (n) rk (m) Z 2T
(m/n)it dt
σ
σ
n
m
T
n∼X m∼X
X rk (n)
≤4
nσ
n∼X
!
X
n≤m≤X
1
rk (m)
min T,
.
σ
m
log m/n
The contribution to the last double sum in (2) from m = n is
T
since rk (n) n and
P
n∼X
rk (n)
n2σ
X r2 (n)
k
T,
2σ
n
n∼X
X 2/k−2σ .
By a further splitting-up argument, it suffices to show that the contribution to the last double sum in (3.2) from n ∼ X, m − n ∼ Y is O(T 1+/3 ) for
1
≤ Y < X. Moreover, for m − n ∼ Y ,
2
log
m
m−n
Y
.
n
n
X
Thus we must show that
X
X X
−2σ
rk (n)
rk (m) T 1+/3 .
X
min T,
Y n∼X
m−n∼Y
Now
X
rk (m) = Ak ((n + 2Y )2/k − (n + Y )2/k )
m−n∼Y
+ ∆k (n + 2Y ) − ∆k (n + Y ),
19
and
X
−2σ
X X
min T,
rk (n)((n + 2Y )2/k − (n + Y )2/k )
Y n∼X
X −2σ+1 Y −1 · X 2/k · Y X 2/k−1 T,
since
X 4/k−2σ ≤ T.
Accordingly, it suffices to show that
(3.3)
X
rk (n)(G(n + 2Y ) − G(n + Y )) T
/3
X
2σ
n∼X
TY
1+
X
2
where G(ω) = c0k ω 1/k−1/k Φk (ω 2/k ) + Bk (ω), and that
X
(3.4)
rk (n)(Pk ((n + 2Y )2/k ) − Pk (n + Y )2/k )
n∼X
T
/3
X
2σ
TY
.
1+
X
Let L = Y −1 X 1−1/k . Then in [X, 3X],
2
G(ω) = H(ω) + O(X 1/k−1/k L−1/k ),
with
H(ω) =
2
c0k ω 1/k−1/k
X
−1−1/k
`
cos 2π `ω
`≤L
1/k
1
−
4
1
1+
.
k
(Possibly H(ω) = 0.) For ω ∈ [X, 2X],
X
2
2
H 0 (ω) X 2/k−1/k −1
`−1/k X 2/k−1/k −1 L1−1/k ,
`≤L
G(ω + 2Y ) − G(ω + Y ) X 2/k−1/k
Y 1/k .
Hence the left-hand side of (3.3) is
X 2/k Y 1/k .
20
2 −1
2
L1−1/k Y + X 1/k−1/k L−1/k
If Y ≤ X/T , then
X 2/k Y 1/k X 3/k T −1/k X 2σ ,
since X 3−2kσ ≤ T . If Y > X/T , then
X
2/k
Y
1/k
X
2/k
X
T
−(1−1/k)
Y = Y X 3/k−1 T 1−1/k Y T X 2σ−1
for the same reason. This proves (3.3).
Let ψ ∗ (u) = ψ(u) for u 6∈ Z, ψ ∗ (u) = 1/2 for u ∈ Z. Then ψ ∗ is of
bounded variation and continuous from the left, as is
X
Pk∗ (u) = −8
ψ ∗ ((uk/2 − nk )1/k ).
2−1/k u1/2 ≤n≤u1/2
We observe that
Pk∗ (ω 2/k ) − Pk (ω 2/k ) X (ω ∈ [X, 2X])
since ω − nk = mk , (m an integer) for O(X ) values of n in [2−1/k ω 1/k , ω 1/k ].
Since σ > 1/k, it suffices to prove a variant of (3.4) with Pk replaced by Pk∗ ;
that is, to prove
Z 2X
{Pk∗ ((ω + 2Y )2/k ) − Pk∗ ((ω + Y )2/k )}d Tk (ω)
X
TY
/3 2σ
.
T X
1+
X
21
Moreover,
Z 2X
{Pk∗ ((ω + 2Y )2/k ) − Pk∗ ((ω + Y )2/k )}d(Ak ω 2/k )
X
(Z
2X+2Y
2Ak
=
Pk∗ (ω 2/k )(ω − 2Y )2/k−1 dω
k
X+2Y
Z
)
2X+Y
Pk∗ (ω 2/k )(ω
−
−Y)
2/k−1
dω
X+Y
2Ak
=
k
(Z
2X+2Y
Pk∗ (ω 2/k ){(ω − 2Y )2/k−1 − (ω − Y )2/k−1 }dω
X+2Y
Z
X+2Y
Pk∗ (ω 2/k )(ω − Y )2/k−1 dω
−
X+Y
Z
X+2Y
+
Pk∗ (ω 2/k )(ω − Y )2/k−1 dω.
X+Y
In the last expression, the first integral is estimated using (2.3) as
X 2/3k Y X 2/k−1 X 2σ−1 T Y,
since
X 4/k−2σ T.
The last two integrals are also
X 2/3k Y X 2/k−1 X 2σ−1 T Y.
Thus it remains to prove
Z 2X
TY
∗
/3 2σ
(3.5)
Pk (ω + Y1 )d∆k (ω) T X
1+
X
X
for Y1 = Y , 2Y . We may suppose that
(3.6)
Y < X 1−4/3k .
22
For in the contrary case, the left-hand side of (3.5) can be estimated by
Lemma 15(i) as
X 8/3k X 2σ−1+4/k−2σ Y
X 2σ−1 T Y,
since X 4/k−2σ ≤ T .
Write ω1 = ω + Y1 and Hr = X 3/k−2σ 2−rq . We observe that, for ω ∈
[X, 2X],
R
X
X
X
2/k
ah
(3.7) Pk∗ (ω1 ) + 8
e(h(ω1 − nk )1/k )
2/k
2/k
r=0 0<|h|≤Hr
≤
R
X
X
Nr (ω1
)≤n≤Nr+1 (ω1
X
br
r=0 |h|≤Hr
)
2/k
Nr (ω1
e(h(ω1 − nk )1/k )
2/k
)≤n≤Nr+1 (ω1
)
3
+ C(k)(log 2T ) ,
2/k
with R = O(log 2T ). This follows from (2.10), (2.11) if ω1 is not an integer,
and by a limiting argument otherwise. Hence Lemma 15 (ii) yields
(3.8)
Z 2X
2/k
Pk∗ (ω1 )d∆k (ω)
X
= −8
R Z
X
r=0
2X
X
X
X
ah
2/k
0<|h|≤Hr
Nr (ω1
R Z 2X X
X
+O bh
X
r=0
|h|≤Hr
R Z 2X X
X
+
bh
X
r=0
|h|≤Hr
e(h(ω1 − nk )1/k )d∆k (ω)
)≤n≤Nr+1 (ω 2/k )
X
2/k
Nr (ω1
)≤n≤Nr+1 (ω 2/k )
X
2/k
Nr (ω1
2/k
)≤n≤Nr+1 (ω1
)
k 1/k
e(h(ω1 − n ) )d∆k (ω)
k 1/k
2/k−1
e(h(ω1 − n ) )ω
dω Z
Z
!
2X
2X
+
(log T )3 d∆k (ω) +
(log T )3 ω 2/k−1 dω .
X
X
23
The last two O-terms in (3.8) contribute
O(X 2/k (log 2T )3 ) = O(X 2σ )
by Lemma 15 (i). The contributions from b0 in the sums over h are both
!
R
3/k −rq
X
x 2
= O(X 2σ log 2T )
O
Hr
r=0
from the choice of Hr .
Fix a value of r, 0 ≤ r ≤ R, and write P = 2r . After a splitting-up
argument, we see that it suffices to prove
Z 2X X
X
−1
ch
(3.9)
H
e(h(ω1 − nk )1/k )d∆k (ω)
X
h∼H
2/k
Nr (ω1
T
)≤n≤Nr+1 (ω 2/k )
/4
X
2σ
TY
1+
X
and
(3.10)
H
−1
Z
2X
X
X
h∼H
X
ch
2/k
Nr (ω1
T
e(h(ω1 − nk )1/k )d(ω 2/k )
)≤n≤Nr+1 (ω 2/k )
/4
X
2σ
TY
1+
X
whenever
1
≤ H < X 3/k−2σ P −q , |ch | ≤ 1.
2
Let a = (2k − 1)/(2k − 2). Using (2.11), we write the integrands in (3.9),
(3.10) as
X
X
1/2k
P −a H −1/2 ω1
b(h, m)e(ω 1/k |(h, m)|) + O(log 2T ),
h∼H m∈[hP,2hP ]
with b(h, m) 1. We have already shown that the term O(log 2T ) gives rise
to an acceptable error. This reduces our task to showing that
Z 2X
X
X
1/2k
1/k
−a −3/2
(3.11) P H
b(h, m)
ω1 e(ω1 |(h, m)|)d∆k (ω)
X
h∼H m∈[hP,2hP ]
T
24
/4
X
2σ
TY
1+
X
and
(3.12) P
−a
H
−3/2
X
X
Z
2X
b(h, m)
1/2k
ω1
1/k
e(ω1 |(h, m)|)ω 2/k−1 dω
X
h∼H m∈[hP,2hP ]
T
/4
X
2σ
TY
1+
.
X
The bound (3.12) gives no trouble. The integrals are
O(X 5/2k−1 (|(h, m)|X 1/k−1 )−1 ) = O(X 3/2k (P H)−1 )
from Lemma 8. Thus the left-hand side of (3.12) is
P 1−a H 1/2 X 3/2k (P H)−1 X 3/2k X 2σ .
For the integrals in (3.11), we use Lemma 15 (iii):
Z 2X
1/k
1/2k
e(ω1 |(h, m)|)ω1 d∆k (ω)
X
Z
2X
(1/2k)−1
1/k
X
+
ω1
e(ω1 |(h, m)|)∆k (ω)dω X
Z
2X
(3/2k)−1
1/k
ω1
|(h, m)| e(ω1 |(h, m)|)∆k (ω)dω .
+
X
3/2k−1/k2
The contribution of the first two terms in this bound to the left-hand side of
(3.11) is
P 1−a H 1/2 X 3/2k−1/k
2
2
X 3/2k−σ+3/2k−1/k X 2σ .
Recalling (2.1) once more, it remains to show that
(3.13)
P 1−a H −1/2
X
h∼H
Z
2X
2
(3/2k)−1
1/k
ω1
e(ω1 |(h, m)|)ω 1/k−1/k Φk (ω 2/k )dω X
m∈[hP,2hP ]
TY
/4 2σ
T X
1+
X
X
25
and
(3.14) P 1−a H −1/2
X
h∼H
Z
2X
(3/2k)−1
1/k
ω1
e(ω1 |(h, m)|)Pk (ω 2/k )dω X
m∈[hP,2hP ]
TY
/4 2σ
1+
T X
.
X
X
For (3.13), we have the bound
Z 2X
(3/2k)−1 1/k−1/k2
1/k
ω1
ω
e(ω1 |(h, m)| ± `ω 1/k )dω
X
2
X 3/2k−1/k |` − |(h, m)||−1
unless
(3.15)
|` − |(h, m)|| < C(k)
Y
P H + 1.
X
Now
X
|`−|(h,m)||≥C(k)
`−1−1/k |` − |(h, m)||−1 (P H)−1 .
Y
X
P H+1
is clearly
For the contribution from ` − |(h, m)| > P H and ` < |(h,m)|
2
−1
O((P H) ). The remaining ` contribute
!
X
−1−1/k 0 −1
O
(P H)
(` )
= O((P H)−1 ).
1≤`0 ≤P H
We also observe that
X
|`−|(h,m)||≤C(k)
`−1−1/k
Y
X
P H+1
−1−1/k
(P H)
Y
PH + 1 .
X
Combining these estimates, we see that the integral in (3.13) is
Y
2
−1 3/2k−1/k2
−1−1/k
(P H) X
+ (P H)
P H + 1 X 5/2k−1/k .
X
26
The left-hand side of (3.13) is thus
2
H 1/2 X 3/2k−1/k + (P H)3/2−1/k X 5/2k−1/k
2 −1
2
Y + H 1/2−1/k X 5/2k−1/k .
Now
2
2
H 1/2 X 3/2k−1/k X 3/2k−σ+3/2k−1/k X 2σ
since σ > 1/k;
2
H 1/2−1/k X 5/2k−1/k X (1/2−1/k)(3/k−2σ)+5/2k−1/k
2
X 2σ
since σ ≥ (4k − 4)/(3k 2 − 2k). Finally,
(P H)3/2−1/k X 5/2k−1/k
2 −1
Y X (3/2−1/k)(3/k−2σ)+5/2k−1/k
2 −1
Y
X 2σ−1 T Y
because
X (3/2−1/k)(3/k−2σ)+5/2k−1/k
2 −2σ
= X 7/k−4/k
2 −σ(5−2/k)
≤ X 4/k−2σ ≤ T.
This is a consequence of the obvious inequality
2
3
2
σ 3−
> − 2.
k
k k
This establishes (3.13).
Turning to (3.14), another application of (2.10) gives
(3.16)
Z 2X
3/2k−1
1/k
ω1
Pk (ω 2/k )e(ω1 |(h, m)|)dω
X
= −8
R
X
X
Z
ah1
s=0 0<|h1 |≤Ks

+O
R Z
X
s=0
2X
Ns+1 (ω 2/k )
X
3/2k−1
ω1
X
3/2k−1
X
|h1 |≤Ks
1/k
e(h1 (ω − nk )1/k + ω1 |(h, m)|)dω
n=Ns (ω 2/k )
Ns+1 (ω 2/k )
ω1
X
2X
X
bh1
n=Ns (ω 2/k )
+ O(X 3/2k (log 2T )3 ).
27

e(h1 (ω − nk )1/k )dω 
Here
Ks = P 1−a 2−sq H 3/2 X 5/2k−2σ ,
so that
Z
2X
Ns+1
X
3/2k−1
ω1
b0
X
1 dω n=Ns
X 5/2k 2−sq
Ks
P a−1 H −3/2 X 2σ .
Thus the terms arising from b0 in (3.16) contribute to the left-hand side of
(3.14) an amount
P 1−a H 3/2 P a−1 H −3/2 X 2σ X 2σ .
The contribution arising from the term O(X 3/2k (log 2T )3 ) in (3.16) is
(3.17)
O(P H 3/2 X 3/2k (log 2T )3 ) = O(X 3/2(3/k−2σ)+3/2k (log 2T )3 )
= O(X 2σ (log 2T )3 ),
since
4k − 4
6
2
≥
(k ≥ 4), σ ≥
(k = 3).
2
3k − 2k
5k
5
For the remaining terms on the right-hand side of (3.16), select a particular s, write Q = 2s , and apply (2.11) to the sum over n, with s in place of r.
Since the term O(log(|h| U + 2)) leads to a further error O(X 3/2k (log 2T )2 ),
our task now reduces to showing that
σ≥
(3.18)
P 1−a H −1/2 Q−a K −3/2
X
X
X
X
|Iδ (h, m, h1 , m1 )|
h∼H P h≤m≤2P h h1 ∼K Qh1 ≤m1 ≤2Qh1
T
/5
X
2σ
TY
1+
.
X
Here 1 ≤ K ≤ Ks ,
Z
2X
Iδ (h, m, h1 , m1 ) =
3/2k−1
ω1
1/k
ω 1/2k e |(h1 , m1 )| ω 1/k + δ |(h, m)| ω1
dω,
X
28
and δ may be 0, 1 or −1.
We first consider (3.18) when either δ = 0 or 1, or QK > C(k)P H. In
this case
d
1/k
(|(h1 , m1 )| ω 1/k + δ |(h, m)| ω1 ) QKX 1/k−1 ,
dω
Iδ (h, m, h1 , m1 ) X 2/k−1 (QKX 1/k−1 )−1 = X 1/k (QK)−1
from Lemma 8. The left-hand side of (3.18) is
P 2−a H 3/2 Q1−a K 1/2 X 1/k (QK)−1
P −1/2 (P H)3/2 (QK)−1/2 X 1/k P HX 1/k X 2σ
as we saw in (3.17). Similarly, when P H > C(k)QK we have
I−1 (h, m, h1 , m1 ) X 1/k (P H)−1 ,
and the left-hand side of (3.18) is
P 2−a H 3/2 Q1−a K 1/2 X 1/k (P H)−1
(P H)1/2 (QK)1/2 X 1/k P HX 1/k X 2σ ,
as we saw in (3.17).
For the case δ = −1, QK P H, we observe that
(3.19)
d
1/k
(|h, m)| ω1 − |(h1 , m1 )| ω 1/k−1 )
dω
1 P HY 1/k−1
1/k−1 =
(|(h, m)| − |(h1 , m1 )|)ω
+O
ω
.
k
X
Consider the contribution to (3.18) from quadruples with
C(k)Y
P H < ||(h, m)| − |(h1 , m1 )|| ≤ 2∆P H.
(3.20)
∆−
X
where ∆ runs over the O(log 2T ) values
∆ = 2t
C(k)Y
, t = 0, 1, . . . , ∆ 1.
X
29
It suffices to show that for each ∆, these quadruples contribute O T /6 X 2σ 1 +
to the left-hand side of (3.18). From Lemma 10, the number of quadruples
satisfying (3.20) is
T /6 (P H 2 + ∆P 3 H 4 )1/2 (QK 2 + ∆Q3 K 4 )1/2 .
We now consider three cases.
Case 1. We have ∆ < (P H)−2 .
In this case the number of quadruples satisfying (3.20) is
P 1/2 HQ1/2 K (P H)2 (P Q)−1/2 .
Estimating the integral trivially, the contribution to the left-hand side of
(3.18) is
P 1−a Q−a H −1/2 K −3/2 (P H)2 (P Q)−1/2 X 2/k
(P H)3/2 (QK)−3/2 X 2/k X 2/k
X 2σ .
Case 2. We have ∆ ≥ (P H)−2 and t = 0, that is, ∆ = C(k)Y X −1 .
In this case the number of quadruples satisfying (3.20) is
T /6 (∆P 3 H 4 )1/2 (∆Q3 K 4 )1/2
T /6 ∆(P H)4 (P Q)−1/2
T /6 Y X −1 (P H)4 (P Q)−1/2 .
Estimating the integral trivially, the contribution to the left-hand side of
(3.18) is
T /6 P 1−a Q−a H −1/2 K −3/2 Y X −1 (P H)4 (P Q)−1/2 X 2/k
T /6 (P H)7/2 (QK)−3/2 Y X 2/k−1
T /6 (P H)2 Y X 2/k−1 T /6 X 8/k−4σ−1 Y
X 2σ−1 Y T 1−/6
since
X 8/k−6σ ≤ X 4/k−2σ ≤ T.
30
TY
X
Case 3. ∆ ≥ (P H)−2 and t ≥ 1, so that ∆ ≥ 2C(k)Y /X.
We can now infer from (3.19) that the quadruples satisfying (3.20) have
I−1 (h, m, h1 , m1 ) X 2/k−1 (∆P HX 1/k−1 )−1
X 1/k (P H)−1 ∆.
The number of quadruples is again T /6 ∆(P H)4 (P Q)−1/2 . Thus the contribution to the left-hand side of (3.18) is
T /6 P 1−a Q−a H −1/2 K −3/2 ∆(P H)4 (P Q)−1/2 X 1/k (P H)−1 ∆−1
T /6 P −1 (P H)5/2 (QK)−3/2 X 1/k
T /6 HX 1/k T X 2σ
from (3.17). This completes the proof of Theorem 5.
§4 Proof of Theorem 4.
3
3
Let σ0 be fixed, 52 ≤ σ0 ≤ 12 (k = 3), 2k
− k12 ≤ σ0 ≤ 2k
(k = 4, 5 . . .). Define
the positive number X by X 2σ0 = T . It is immediate from Lemma 1 that
Mk (σ0 , T ) W1 + T 2 W2 + T,
(4.1)
where
Z
2T
W1 =
T
Z
W2 =
T
2T
X r (n) 2
k
dt,
σ
+it
0
n
n≤X
Z
∞
X
2
∆k (ω)
dt.
dω
σ
+it+1
ω 0
We may apply Theorem 5 to W1 ; the conditions
2
(k = 3), σ0 ≥ (4k − 4)/(3k 2 − 2k),
5
4
max
− 2σ0 , 3 − 2σ0 k ≤ 2σ0
k
σ0 ≥
31
are easily seen to be satisfied.
It follows that W1 T 1+ . Recalling the decomposition of ∆k (x) in (2.1),
we have only to show that
2
Z 2T Z ∞
2/k
Φ
(ω
)
k
dt T −1+ ,
W3 =
dω
σ
+it
1
ω
T
X
where σ1 = σ0 + 1 − 1/k + 1/k 2 , and
2
Z 2T Z ∞
F
(ω)
k
dt T −1+ ,
W4 =
dω
σ
+it
2
T
X ω
where σ2 = σ0 + 1, Fk (ω) = Pk (ω) + Bk (ω k/2 ).
Crudely, we have
P −1−1/k
n
cos 2πnω 1/k −
Z 2T Z ∞
n>B
ω σ1 +it
T
X
1
4
1+
1
k
2
dω dt
T B −2/k X −2σ1 +2 T −1
if we choose
B = T k X k(−σ1 +1) .
After a splitting up of the sum
X
n
−1−1/k
cos 2πnω
1/k
n≤B
1
1+
,
k
1
2
≤ N < B, we have
2
Z 2T Z ∞
−it g(ω)ω
W3 T −1 + (log T )2
dt,
we find that for some N ,
(4.2)
1
−
4
T
X
where
1
1
1/k
g(ω) = ω
n
cos 2πnω −
1+
.
4
k
n∼N
Z ∞
∞ Z
X
We decompose the integral
as
, where
−σ1
X
−1−1/k
X
j=0
J(j)
J(j) = [X2j , X2j+1 ].
32
We have
(4.3)
Z
Z
X
t
log
ω
1
n−1−1/k
ω −σ1 e nω 1/k −
dω g(ω)ω −it dω ≤ 2
2π
J(j)
J(j)
n∼N
Z
X
t log ω
1
+ ω −σ1 e −nω 1/k −
n−1−1/k
dω .
2
2π
J(j)
n∼N
ω
and F1 (ω) = −nω 1/k − t log
. We have
2π
T
0
j 1/k−1
|F1 (ω)| max N (X2 )
,
X2j
Let F (ω) = nω 1/k −
t log ω
2π
in (4.3). If
k −1 N (X2j )1/k >
(4.4)
T
,
π
we have
|F 0 (ω)| N (X2j )1/k−1
in (4.3), while if
2k −1 N (X2j+1 )1/k <
(4.5)
T
,
4π
we have instead
|F 0 (ω)| T (X2j )−1
in (4.3).
We conclude from Lemma 8 that
Z
(4.6)
g(ω)ω −it dω N −1−1/k (2j X)−σ1 −1/k+1
J(j)
N −1/k T −1 (2j X)1−σ1
if (4.4) holds, while
Z
(4.7)
g(ω)ω −it dω N −1/k T −1 (2j X)1−σ1
J(j)
33
if (4.5) holds. There are only O(1) ‘exceptional’ values of j satisfying neither
(4.4) nor (4.5). For these j,
N (X2j )1/k T.
(Of course, there are no exceptional j unless N T X −1/k .)
If there are no exceptional j, then we can apply (4.6), (4.7) as follows:
2
Z 2T Z ∞
−it
(4.8)
g(ω)ω dω dt
T
X
2T
Z
≤
T
∞
X
∞
X
!
j
−2
j=1
2T
Z
T
∞
X
j=1
j=1
Z
j 2
Z
j 2
g(ω)ω
J(j)
g(ω)ω
−it
J(j)
−it
2 !
dω dt
2
dω dt
1
1
T −1 X 2−2σ1 = T −1 X −2(σ0 − k − k2 ) .
Recalling (4.2), we obtain the bound
W3 T −1
using only the lower bound σ0 > k1 − k12 .
Suppose now that there are exceptional j. For some fixed j0 satisfying
N (X2j0 )1/k T,
(4.9)
we can modify the above calculation to obtain
2
Z 2T Z ∞
−it
(4.10)
g(ω)ω dω dt
T
X
Z
T
2T
Z
2
−it
g(ω)ω dω dt
J
+ T −1 X 2−2σ1 .
Here J = J(j0 ).
34
We now appeal to Lemma 5. We have
2
Z 2T Z
−it
(4.11)
g(ω)ω dω dt
T
J
2
Z X
n−1−1/k e(nω 1/k ) dω.
(X2j0 )1−2σ1 log T
J n∼N
A change of variable shows that the integral on the right-hand side of (4.11)
is
2
Z (2j0 +1 X)1/k
X
k−1
−1−1/k
kv
n
e(nv) dv
(2j0 X)1/k
n∼N
(2j0 X)1−1/k N −(1+2/k) (2j0 X)1/k
by Parseval’s equality applied to subintervals of [(2j0 X)1/k , (2j0 +1 X)1/k ] having length 1. We conclude that
2
Z 2T Z
g(ω)ω −it dω dt
T
J
N −(1+2/k) (2j0 X)2−2σ1 log T
N −(1+2/k) (T k N −k )2−2σ1 log T
by (4.9). This bound is
N −(1+2/k) (T k N −k )−1/k log T T −1 log T
since σ1 ≥ 1 +
1
.
2k
Recalling (4.2), (4.10), we always have
W3 T −1 (log T )3 .
Now we have to show that
W4 T −1+ .
Arguing as in (4.8), it suffices to show that
2
Z 2T Z
Fk (ω) (4.12)
dω dt T −1+ j −4 .
σ2 +it
ω
T
J(j)
35
Lemma 5 yields, for any measurable function E(ω) such that
Z
2T
(4.13)
T
E(ω) T /8 on [X2j , X2j+1 ],
Z
2
E(ω)
dω
dt
J(j) ω σ2 +it (X2j )−2σ2 +2 (log T )T /4
(X2j )−2σ0 T /3 T −1+/3 2−2jσ0 .
Thus it suffices to show that
2
Z 2T Z
Pk (ω 2/k ) dω dt T −1+ j −4 .
σ2 +it
ω
T
J(j)
Define R as in Section 2 with U = (X2j )k/2 . In view of (4.13) and the
decomposition (2.6), it suffices to show for a fixed r, 0 ≤ r ≤ R, that
2
Z 2T Z
Nr+1 (ω 2/k )
X
ω −σ2 −it
ψ((ω − nk )1/k )dω dt T −1+/2 j −4 .
T
J(j)
2/k
n=Nr (ω
)
Let H = H(T, r) be a positive integer, to be chosen below. Let
Nr+1 (ω 2/k )
f (ω) =
X
ψ((ω − nk )1/k ),
n=Nr (ω 2/k )
1
g(ω) = −
2πi
Nr+1 (ω 2/k )
X e(h(ω − nk )1/k )
.
h
X
n=Nr (ω 2/k ) 0<|h|≤H
It will suffice to show that
2
Z 2T Z
−σ2 −it
(4.14)
ω
(f
(ω)
−
g(ω))dω
dt T −1+/2 j −4 .
T
J(j)
and
Z
(4.15)
T
2T
Z
2
ω −σ2 −it g(ω)dω dt T −1+/2 j −4 .
J(j)
36
We begin with (4.14). Let P = 2r . For n (X2j )1/k , we write fn (ω) for
the indicator function of the interval
I(n) = [nk (1 + (2P )−q ), nk (1 + P −q )].
Let
I1 (n) = [X2j , X2j+1 ] ∩ I(n).
Now
2
k 1/k X
1
e(h(ω − n ) ) ψ((ω − nk )1/k ) +
dω
2πi
h
I1 (n) Z
0<|h|≤H
2
X
1
e(hw)
=
kwk−1 ψ(w) +
dw
2πi
h
k
k
w +n ∈I1 (n)
Z
0<|h|≤H
X2j P −q
Z
0
1
2
1 X e(hw) ψ(w) +
dw
2πi
h 0<|h|≤H
X2j P −q H −1
by (2.7). (Note that the variable w introduced by the change of variable
satisfies w nP −q/k .) Hence
Z
|f (ω) − g(ω)|2 dω
J(j)
)2
X e(h(ω − nk )1/k ) 1
fn (ω) ψ((ω − nk )1/k ) +
dω
2πi
n
1/k
j
0<|h|≤H
n(X2 )
2
X Z
X e(h(ω − nk )1/k ) 1
fn (ω)ψ((ω − nk )1/k ) +
(X2j )1/k
dω
2πi
h
J(j)
j 1/k
=
J(j) Z
(
X
0<|h|≤H
n(X2 )
(by Cauchy’s inequality)
= (X2j )1/k
X
n(X2j )1/k
2
X e(h(ω k − nk )1/k ) 1
ψ((ω − nk )1/k ) +
dω
2πi
h
I1 (n) Z
0<|h|≤H
(X2j )1+2/k P −q H −1 .
37
In view of Lemma 5, the left-hand side of (4.14) is
Z
j 1−2σ2
log T
|f (ω) − g(ω)|2 dω
(X2 )
I1 (n)
(X2j )−2σ0 +2/k (log T )P −q H −1 T −1 log T 2−j(2σ0 −2/k)
if we now specify that
H = T X −2σ0 +2/k P −q = X 2/k P −q .
Turning to (4.15), it is sufficient to show for a fixed K, 21 ≤ K ≤ H, that
2
Z 2T Z
ω −σ2 +αit gK (ω)dω dt T −1+/3 j −4 ,
T
J(j)
where α = 1 or −1 and
Nr+1 (ω 2/k )
X
gK (ω) =
X
e(h(ω − nk )1/k ).
h∼K n=Nr (ω 2/k )
Recalling (4.13), we can reduce this to showing that
2
Z 2T Z
−a −3/2 −σ3 −αit
P K
ω
G(ω)dω dt T −1+/3 j −4 .
T
J(j)
Here
σ3 = σ2 − 1/2k = σ0 + 1 − 1/2k
and
G(ω) =
X
b(h, m)e(ω 1/k |(h, m)|)
(h,m)∈E
with b(h, m) 1,
E = {(h, m) : h ∼ K, P h ≤ m ≤ 2P h}.
Compare the reduction of (3.10) to (3.11).
Arguing as in the discussion of W3 , we find that after excluding O(1)
‘exceptional’ values of j for which
(4.16)
c2 (k)T
c1 (k)T
< PK <
,
j
1/k
(X2 )
(X2j )1/k
38
(with c1 (k) > 0) we have
Z
e(|(h, m)|ω 1/k )
dω
ω σ3 +αit
J(j)
min (X2j )1−σ3 −1/k (P K)−1 , (X2j )1−σ3 T −1
for all (h, m) ∈ E.
Since |E| P K 2 , the ‘non-exceptional’ j satisfy
2
Z 2T Z
G(ω)
−2a −3
(4.17) P K
dω dt
σ
+αit
3
J(j) ω
T
P −2a K −3 T (P K 2 )2 min((X2j )2−2/k−2σ3 P −2 K −2 ,
(X2j )2−2σ3 T −2 )
P −(2a−1) min((P K)−1 T (X2j )2−2/k−2σ3 , P KT −1 (X2j )2−2σ3 )
(X2j )2−2σ2 T −1 2−2σ0 j .
Now suppose that j satisfies (4.16). We have
(4.18)
Z
G(ω) 2
ω σ3 dω
J(j)
=
X
X
Z
b(h1 , m1 ) b(h2 , m2 )
J(j)
(h1 ,m1 )∈E (h2 ,m2 )∈E
X
X
(h1 ,m1 )∈E (h2 ,m2 )∈E
ω −2σ3 e (|(h1 , m1 )| − |(h2 , m2 )|)ω 1/k dω
(X2j )1−2σ2
|(h1 , m1 )| − |(h2 , m2 )|
j 1−2σ3 min (X2 )
!
,
by Lemma 8.
We consider the contribution to the last sum from h1 , m1 , h2 , m2 satisfying
(4.19)
∆P K − (X2j )−1/k ≤ |(h1 , m1 )| − |(h2 , m2 )| < 2∆P K.
Here ∆ runs over the numbers in (0, 3] of the form
∆ = (P K)−1 (X2j )−1/k 2h
39
(h = 0, 1, . . .).
Clearly (4.19) implies
|hq1 + mq1 − hq2 − mq2 | ∆(P K)q .
By Lemma 11, the number N of such quadruples satisfies
(4.20)
N (P K)/5 (P K 2 + ∆P 3 K 4 ).
The contribution of these quadruples to the last sum in (4.18) is
(X2j )1−2σ2
2
j 1−2σ3
/5
3 4
PK +
(X2 )
(P K)
∆P K
∆P K
= (P K)/5 {(X2j )1−2σ3 P K 2 + (X2j )1−2σ2 P 2 K 3 }.
Summing over O(j + log T ) values of ∆, we find that
Z
G(ω) 2
σ3 dω jT /4 {(X2j )1−2σ3 P K 2 + (X2j )1−2σ3 P 2 K 3 }.
J(j) ω
Applying Lemma 5, and recalling (4.16), we have
2
Z 2T Z
G(ω)
(4.21)
dω
P −2a K −3
dt
σ
+αit
3
J(j) ω
T
jT /3 {(X2j )2−2σ3 (P K)−1 + (X2j )2−2σ2 }
jT /3 {(X2j )2−2σ3 +1/k T −1 + (X2j )−2σ0 }.
We recall that X 2σ0 = T and that
2 − 2σ3 + 1/k = −2σ0 + 2/k < 0.
Thus the left-hand side of (4.21) is
jT −1+/3 2−(2σ0 −2/k)j .
Combining this with (4.17), we see that the proof of (4.15) is complete. As
already noted before (4.14), this finishes the proof of Theorem 4.
40
§5 Proof of Theorem 3.
It suffices to show that
Z 2X
2
2
Ek (x)2 dx = dk ((2X)1+2/k−2/k − X 1−2/k−2/k )
X
+ O(X 1+2/k−2/k
2 −η
)
for large X. We write (in this section only) k . . . k for the L2 norm on [X, 2X].
We note that
kF + Gk2 = kF k2 + O(kF k kGk)
if kGk = O(kF k). Accordingly it suffices to write Ek (x) in the form
(5.1)
Ek (x) = F (x) + G1 (x) + · · · + G4 (x),
and to show that
(5.2)
2
2
kF k2 = dk ((2X)1+2/k−2/k − X 1+2/k−2/k )
+ O(X 1+2/k−2/k
2 −η
),
and that each Gj satisfies
(5.3)
kGj k2 = O(X 1+2/k−2/k
2 −2η
).
Let λ = 1/k − 1/k 2 + . Let c = c(χ, k) be a small positive constant,
c < 1/(2k + 1), and let y = X c . By combining Lemma 2 with (2.1), we
obtain
2/k 2/k X
X
x
x
0 1/k−1/k2
+
µ(d)Pk
Ek (x) = ck x
µ(d)Φk
2
d
d2
d≤y
d≤y
1
+
2πi
Z
λ+ixC
f (y, ks)Zk (s)
λ−ixC
41
xs
ds + O(y).
s
Thus in (5.1), we may choose
F (x) =
2
c0k x1/k−1/k
X
−1−1/k
`
`≤X
G1 (x) =
2
c0k x1/k−1/k
X
−1−1/k
`
`>X
1/k
X µ(d)
1
1
`x
−
1+
,
cos 2π
d1−1/k
d
4
k
d≤y
1/k
X µ(d)
1
1
`x
−
1+
cos 2π
d1−1/k
d
4
k
d≤y
= O(1),
2/k X
x
G2 (x) =
µ(d)Pk
,
d2
d≤y
Z
λ+ixC
G3 (x) =
xs
ds,
s
f (y, ks)Zk (s)
λ−ixC
G4 (x) = O(y).
Obviously G1 , G4 satisfy (5.3) for η ≤ 1/k − 1/k 2 − c. It remains to prove
(5.2), and (5.3) for G2 , G3 .
We can dismiss G2 quickly. By Cauchy’s inequality, a change of variable,
and (2.4),
Z
2X
2
G2 (x) dx ≤ y
XZ
X
y
2X
Pk
d≤y
X
X
k
d≤y
d
X 2/k
d2
x2/k
d2
2
dx
3/2+(k/2−1)
= yX 1+1/k
X1
d≤y
d
X 1+1/k y log y.
Thus G2 satisfies (5.3) for 2η < 1/k − 2/k 2 − c.
For G3 , we note that, with T running over powers of 2, 2 ≤ T ≤ (2X)C ,
!
X Z 2T
2
(5.4)
G3 (x) x(k−1)/k + 1 +
g(t)xit dt .
T
Here
g(t) =
T
f (y, kλ + kit)Zk (λ + it)
.
λ + it
42
By Lemmas 7 and 4,
2
Z 2X Z 2T
Z
it
(5.5)
g(t)x dt dx X log T
X
T
2T
|g(t)|2 dt
T
X 1+ 2(χ−kλ)
y
T2
Z
2T
|Zk (λ + it)|2 dt.
T
Recalling Lemma 6, the right-hand side of (5.5) is
X 1+2 y 2(χ−kλ) .
Combining (5.4), (5.5),
Z 2X
2
G3 (x)2 dx X 1+2/k−2/k +2c(χ−kλ)+5 .
X
Since 2c(χ − kλ) = 2c χ − 1 + k1 − k , we see that (5.3) is valid for G3 if
η < c 1 − k1 − χ .
Our treatment of F (x) resembles that of Zhai [19], but we give the details
for the convenience of the reader. Using the identity
2 cos A cos B = cos(A − B) + cos(A + B),
we can write
0
2
2F (x)2 = ck2 x2/k−2/k (KX + F1 (x) + F2 (x)),
where
KX =
X
`1 ,`2 ≤X,d1 ,d2 ≤y
`1 d2 =`2 d1
F1 (x) =
X
`1 ,`2 ≤X,d1 ,d2 ≤y
`1 d2 6=`2 d1
µ(d1 )µ(d2 )
,
(`1 `2 )1+1/k (d1 d2 )1−1/k
µ(d1 )µ(d2 )
cos 2π
(`1 `2 )1+1/k (d1 d2 )1−1/k
µ(d1 )µ(d2 )
cos 2π
(`1 `2 )1+1/k (d1 d2 )1−1/k
`1
`2
−
d1 d2
`1
`2
+
d1 d2
x
1/k
and
F2 (x) =
X
`1 ,`2 ≤X,d1 ,d2 ≤y
43
x
1/k
1
−
2
1
1+
k
.
We now apply the particular case
Z 2X
2
2
x2/k−2/k cos 2π(∆x1/k + a)dx X 1+1/k−2/k |∆|−1
X
of Lemma 8. This yields
Z 2X
F (x)2 dx =
(5.6) 2
X
0
ck2 KX
1+2/k−2/k2
1+2/k−2/k2
(2X)
−
X
1 + 2/k − 2/k 2
1+1/k−2/k2
+O X
(S1 + S2 ) .
Here
S1 =
−1
`1
1
− `2 ,
1+1/k
1−1/k
(`1 `2 )
(d1 d2 )
d1 d2 X
`1 ,`2 ≤X
d1 ,d2 ≤y
`1 d2 6=`2 d1
S2 =
X
`1 ,`2 ≤X
d1 ,d2 ≤y
1
1+1/k
(`1 `2 )
(d1 d2 )1−1/k
`1
`2
+
d1 d2
−1
.
We evaluate KX as follows. We may write
X
KX =
b(n),
n≥1
where
X
b(n) =
n=`1 d2 =`2 d1
`1 ,`2 ≤X, d1 ,d2 ≤y
µ(d1 )µ(d2 )
.
(`1 `2 )1+1/k (d1 d2 )1−1/k
If n ≤ y, then clearly
b(n) =
X X
µ(d1 )µ(d2 )
d1 | n d2 | n
2

= n−2−2/k 
n2
d1 d2
X
µ(d)d2/k  .
d|n
44
−1−1/k
(d1 d2 )1/k−1
A similar calculation yields
2

|b(n)| ≤ n−2−2/k 
X
d2/k  ≤ n−2+2/k d2 (n)
d|n
for all n. Hence.
(5.7)
KX =
∞
X
2

n−2−2/k 
n=1
X
µ(d)d2/k  + O
!
X
n−2+2/k d2 (n)
n>y
d|n
= ek + O(y −1+2/k X ).
Turning to S1 , it is clear that
X
S1 ≤ y 2/k
(`1 `2 )−1−1/k |`1 d2 − `2 d1 |−1
`1 ,`2 ≤X
d1 ,d2 ≤y
`1 d2 6=`2 d1
y 2/k
X
(`1 `2 )−1−1/k r−1
`1 ,`2 ,d1 ,r
d1 ≤y, r≤Xy
y 2/k+1 log X.
Similar reasoning yields the same bound for S2 .
Recalling (5.6) and (5.7),
Z
2X
2
2
F (x)2 dx = dk (2X)1+2/k−2/k − X 1+2/k−2/k
X
+ O(X 1+2/k−2/k
2 +
y −1+2/k + X 1+1/k−2/k
Thus (5.2) is satisfied provided that
2
and
η <c 1−
k
η < 1/k − c(1 + 2/k).
As noted above, this completes the proof of Theorem 3.
45
2 +
y 1+2/k ).
§6 Proofs of Theorems 1 and 2.
Let k = 3 or 4, σ3 = 25 , σ4 =
5
.
16
1
Ek (x) =
2πi
(6.1)
+
Just as in §5, we find that
Z
σk +ixC
f (yk , s)Zk (s)
σk −ixC
X
µ(d)∆k
d≤yk
x
dk
xs
ds
s
+ O(1).
Here yk is to be specified below, yk > 1. Moreover, we can argue as in §5 to
obtain a large T , T ≤ xC with
Z σk +ixC
xs
(6.2)
f (yk , s)Zk (s) ds (log x)xσk max |f (yk , kσk + it)|
s
|t|≤xC
σk −ixC
Z 2T
−1
T
|Zk (σk + it)| dt + 1
T
xσk + ykρk −kσk .
We used Theorem 4 (together with the Cauchy-Schwarz inequality) and
Lemma 4 in the last step.
We take
y3 = x6θ3 −4/3 = x0.2260... , y4 = x4θ4 /3−1/12 = x0.1931... .
It is easily verified that in each case, (6.2) yields
Z σk +ixC
xs
(6.3)
f (yk , s)Zk (s) ds xθk + .
s
σk −ixC
From (2.1),
(6.4)
X
d≤yk
µ(d)∆k
x
dk
= Xk + Yk + O(yk ),
where
Xk =
2
c0k x1/k−1/k
∞
X µ(d) X
2π`x1/k 1
1
−1−1/k
`
cos
−
1+
1−1/k
d
d
4
k
`=1
d≤y
46
and
Yk = −8
X
d≤y
X
µ(d)
x
2dk
≤nk ≤ xk
d
ψ
x
dk
k
−n
1/k .
Clearly, in bounding Xk we need only show that for 21 ≤ D ≤ yk , and
` ≥ 1, we have
1/k X
`x
2
`1/k D1−1/k xθk −1/k+1/k + .
(6.5)
µ(d)e
d
d∼D
Since (6.5) is trivial for D ≤ xkθk −1+1/k , we assume that
(6.6)
D > xkθk −1+1/k
in proving (6.5). By appealing to [2], §7, we can suppose much more when
k = 3, namely
(6.7)
D > x0.221 .
For Yk , we follow the initial stages of the argument of Zhai and Cao, with
their (2.10) as the point of departure. Let P > 1 and write
H = max(x1/k−θk P −q , 1).
We find as in (5.7), (5.9) of [20] that in bounding Yk , it is sufficient to prove
(6.8)
1/2 X
1/k
X
D
x1/(2k)
x |(h, `)|
µ(d)
b(h, `)e −
P (1+q)/2 D1/2 K 3/2 d∼D
d
d
(h,`)∈E
xθk +
for 1 ≤ K ≤ H and 1 ≤ D ≤ yk . Here
E = E(K, P ) = {(h, `) : h ∼ K, P h ≤ ` ≤ 2P h}
as in Section 4, and |b(h, `)| ≤ 1. Strictly speaking, one also needs to prove
the analogue of (6.8) with µ(d) replaced by 1; this is easier and need not be
discussed separately.
47
Naturally we may apply Lemma 12. Thus we need only prove in place of
(6.5) that, for a suitable U , 0 < U ≤ D1/3 ,
1/k X X
`x
2
`1/k D1−1/k xθk −1/k+1/k +
(6.9)
SI =
am e
mn
m∼M n∼N
mn∼D
for ` ≥ 1, M N D, N DU −1 , |am | ≤ 1; and that
1/k X X
`x
2
(6.10)
SII =
am c n e
`1/k D1−1/k xθk −1/k+1/k +
mn
m∼M n∼N
mn∼D
for ` ≥ 1, M N D, U N D1/2 , |am | ≤ 1, |cn | ≤ 1.
It turns out that (6.8) requires no new work in the case k = 4. It is shown
in §5 of Zhai and Cao [20] that
9/28
Y4 x (y4 + x1/7 y4
5/12
+ x1/8 y4
1/9
+ x1/6 y4
+ x0.1875 ),
which is easily seen to be stronger than we need. In the case k = 3, we can
quote the result we need from [2], §6 when D < x2/9 . Thus we suppose that
x2/9 < D ≤ y3
(6.11)
in proving (6.8) for k = 3. Appealing to Lemma 12, we need only prove in
place of (6.8) that, for a suitable U , 0 < U ≤ D1/3 ,
1/3
X X D 1/2 X
−x |(h, `)|
x1/6
(6.12)
b(h, `)e
am
P 5/4 D1/2 K 3/2 m∼M n∼N
mn
mn
(h,`)∈E
mn∼D
xθ3 +
whenever M N D, N DU −1 and |am | ≤ 1; and that
1/3
X X
X
x1/6
−x |(h, `)|
(6.13)
am c n
b(h, `)e
P 5/4 D1/2 K 3/2 m∼M n∼N
mn
(h,`)∈E
mn∼D
xθ3 +
whenever M N D, U N D1/2 and |am | ≤ 1, |bn | ≤ 1.
48
1/3
We begin with (6.9), (6.10) for k = 3. We take
U = D . By Lemma 13
2 4
1/3 −1
with X `x D , N0 = M and (κ, λ) = 7 , 7 , the left-hand side of (6.9)
is
(log x)2 {DN −1/2 + D2 x−1/3 + (D3+2κ `1+2κ x(1+2κ)/3 N −(1+2κ) M 2(λ−κ) )1/(6+4κ) }
(log x)2 {D2/3 + (`11 D25 x11/3 N −11 M 4 )1/50 }.
Moreover,
(D25 x11/3 N −11 M 4 )1/50 < D19/50 x11/150 < D2/3 xθ3 −2/9
as a consequence of (6.7).
For (6.10), we appeal to Lemma 14 with X `x1/4 D−1 , (κ, λ) = (1/2, 1/2),
obtaining
(6.14)
SII (log x)3 `1/5 {x1/15 D9/20 N 1/10 + x1/18 D1/2 N 1/9 + x1/15 D2/5 N 1/5
+ x1/33 D6/11 N 3/11 + D2/3 N 5/18 + DN −1/2
+ x−1/66 D15/22 N 9/22 + D3/2 x−1/6 }
(log x)3 `1/5 {x1/15 D1/2 + x1/18 D5/9 + x1/33 D15/22
+ D5/6 + x−1/66 D39/44 + D3/2 x−1/6 }.
Now
D5/6 ≤ D2/3 xθ3 −2/9 ,
because D ≤ y3 = x6θ3 −4/3 . The remaining terms in the last expression
in (6.14) are easily seen to be of smaller order than `1/5 D2/3 xθ3 −2/9 . This
establishes (6.10) and completes the proof of (6.5) for k = 3.
Turning to (6.9), (6.10) for k = 4, we suppose that
x0.0795... = x4θ4 −3/4 < D ≤ y4 .
Let U = D1/2 x−2θ4 +3/8 . It is easily verified that 1 ≤ U ≤ D1/3 . According to
Lemma 13 (i) with (κ, λ) = (1/14, 11/14), X `x1/4 D−1 , N0 = M , we have
(log x)−2 SI DN −1/2 + D2 x−1/4 + (D64 `16 x4 N −36 )1/88 .
Now
DN −1/2 D2/3 D3/4 xθ4 −3/16 ,
D2 x−1/4 D3/4 xθ4 −3/16 .
49
Finally,
(D64 x4 N −36 )1/88 (D28 x4 U 36 )1/88
= (D46 x35/2−72θ4 )1/88 D3/4 xθ4 −3/16
since
D > x4θ4 −3/4 > x17/10−8θ4 .
This proves (6.9).
To obtain (6.10) in the range
x4θ4 −3/4 < D ≤ x0.125 ,
89 1
89
we apply Lemma 13 (ii) with X `x1/4 D−1 , (κ, λ) = 570
, 2 + 570
. This
exponent pair is due to Huxley [5], [6]; his significantly deeper work in [8]
hardly affects the value of θ4 . The condition X D follows from D ≤ x0.125 .
We have
(log x)−7/4 SII DN −1/2 + DM −1/4 + (D10+8κ `1+2κ x(1+2κ)/4 N )1/(14+12κ) .
Now
(6.15)
DN −1/2 DU −1/2 = D3/4 xθ4 −3/16 ,
DM −1/4 D7/8 D3/4 xθ4 −3/16
since D ≤ x0.125 < x8θ4 −3/2 . Finally
(D10+8κ x(1+2κ)/4 N )1/(14+12κ) = (D6412 x187 N 570 )1/9048
(D6697 x187 )1/9048 D3/4 xθ4 −3/16
since
(6.16)
D89 ≥ x89(4θ4 −3/4) = x3767/2−9048θ4 .
This is where the precise value of θ4 arises.
It remains to obtain (6.9) for
(6.17)
x0.125 < D ≤ y4 .
50
According to Lemma 14, with X `x1/4 D−1 , (κ, λ) = (1/2, 1/2), we have
(6.18)
(log x)−3 SII `1/5 {x1/20 D9/20 N 1/10 + x1/24 D1/2 N 1/9
+ x1/20 D2/5 N 1/5 + x1/44 D6/11 N 3/11 + D2/3 N 5/18
+ DN −1/2 + x−1/88 D15/22 N 9/22 + D3/2 x−1/8 }
`1/5 {x1/20 D1/2 + x1/24 D5/9 + x1/44 D15/22
+ x−1/88 D39/44 + D3/2 x−1/8 + D3/4 xθ4 −3/16 },
where we have applied (6.15) in the last step. Now since D ≤ y4 = x4θ4 /3−1/12 ,
we have
D3/2 x−1/8 ≤ D3/4 xθ4 −3/16 .
Moreover,
x1/24 D5/9 < D3/4 xθ4 −3/16 ,
since
D > x0.125 > x33/28−36θ4 /7 .
It is easily verified that the remaining three terms in the last expression in
(6.18) are smaller than `1/5 D3/4 xθ4 −3/16 . This completes the proof of (6.10),
and indeed (6.5), for k = 4. Since we already have (6.3), (6.5) and (6.8), we
have finished the proof of Theorem 2.
It remains only to prove (6.12) and (6.13) for the short range (6.11) of D.
If H = 1, then we can argue as on pp. 137–138 of Baker [2] to obtain (6.12),
(6.13). Thus we suppose that H ≥ K ≥ 1, and it follows that
KP 3/2 ≤ x1/3−θ3
from the choice of H.
We can dispose of the case
K ≥ DP −1/2 x−5/27
by repeating verbatim the argument in the last paragraph of [2], §6. We
suppose that
(6.19)
K < DP −1/2 x−5/27 .
We shall prove (6.12), (6.13) with
(6.20)
U = Dx−2θ3 +1/3 P −1/2 ;
51
obviously U < D1/3 .
We can easily dispose of (6.12) using the Kusmin-Landau theorem. For
if N DU −1 = P 1/2 x2θ3 −1/3 , then
x1/3 P K
d x1/3 |(h, `)|
dn
mn
DN
x−2θ3 +2/3 P 1/2 KD−1
x−2θ3 +13/27 = x−0.03...
from (6.19). A partial summation gives
X D 1/3 −x1/3 |(h, `)| DN
,
e
1/3
mn
D
x PK
n∼N
mn∼D
and the left-hand side of (6.12) is
x1/6
DN
· P K 2 M · 1/3
x−1/6 D3/2 xθ3 .
5/4
1/2
3/2
P D K
x PK
Turning to (6.13), we may remove the condition mn ∼ D from the sum
to be estimated at the cost of a factor log x, as noted earlier. Let us suppose
this has been done. Let
Q = max(64[x1/3 P KM −2 N −1 ], 1).
We divide the interval 0, 8PNH into Q equal subintervals I1 , . . . , IQ , and
bound
1/3
X
X
X
−x |(h, `)|
S=
am
cn
b(h, `)e
mn
m∼M
n∼N
(h,`)∈E
as follows:
Q X X X
x1/3 |(h, `)| |S| ≤
cn b(h, `)e −
.
mn
m∼M q=1 n∼N, (h,`)∈E
|(h,`)|/n∈Iq
52
Cauchy’s inequality yields
(6.21)
2
Q
1/3
X X X
−x
|(h,
`)|
2
cn b(h, `)e
|S| ≤ M Q
mn
q=1 m∼M n∼N, (h,`)∈E
|(h,`)|/n∈Iq
X x1/3 |(h , ` )| |(h , ` )| X
1 1
2 2
≤ MQ
e −
−
,
m
n
n
1
2
m∼M
n1 ,n2 ,(h1 ,`1 ),(h2 ,`2 )
(6.22)
where n1 , n2 , (h1 , `1 ), (h2 , `2 ) are restricted in the last summation by
|(h1 , `1 )| |(h2 , `2 )| 8P K
≤
(6.22)
nj ∼ N, (hj , `j ) ∈ E , −
.
n1
n2 NQ
A splitting-up argument yields
(6.23)
|S|2 M Q log x
X
n1 ,n2 ,(h1 ,`1 ),(h2 ,`2 )
(6.24)
X x1/3 |(h , ` )| |(h , ` )| 1 1
2 2
e −
−
,
m
n
n
1
2
m∼M
where the outer summation (6.23) is restricted by nj ∼ N , (hj , `j ) ∈ E and
|(h1 , `1 )| |(h2 , `2 )| 2∆P K
MN
PK
<
(6.24)
∆ − 1/3
≤ −
.
x PK
N
n1
n2 N
The positive number ∆ is of the form
∆=
2h M N
8
, ∆≤
, h ≥ 0.
1/3
x PK
Q
We can apply the Kusmin-Landau theorem again, since
1/3 d
16x1/3 P K
x
|(h
,
`
)|
|(h
,
`
)|
1
1
1
2
2
≤
−
≤
dm
2
m
n1
n2
M NQ
2
53
by the choice of Q. Thus the inner sum in (6.23) is
M 2N
.
min M, 1/3
x P K∆
According to Lemma 11, the number of solutions n1 , n2 , (h1 , `1 ), (h2 , `2 )
of (6.24) is
x (∆P 3 K 4 N 2 + P 3/2 K 3 N ).
Thus
2
3
4
2
|S| x M Q log x(∆P K N + P
3/2
M 2N
K N ) min M, 1/3
x P K∆
3
x M Q log x(P 2 K 3 N D2 x−1/3 + P 3/2 K 3 D),
S x Q1/2 P K 3/2 (D3/2 x−1/6 + DN −1/2 ).
The left-hand side of (6.12) is now seen to be
x1/6+ P −1/4 D−1/2 Q1/2 (D3/2 x−1/6 + DN −1/2 ).
To verify that this is xθ3 + reduces to showing that
(6.25)
Q min(x2θ3 P 1/2 D−2 , x2θ3 −1/3 P 1/2 D−1 N ).
If Q = 1, then (6.25) is a simple consequence of the lower bound N U .
Otherwise (6.25) reduces to the two assertions
x1/3 P KM −2 N −1 x2θ3 P 1/2 D−2
(6.26)
and
(6.27)
x1/3 P KM −2 N −1 x2θ3 −1/3 P 1/2 D−1 N.
Both assertions follow from (6.19). In the case of (6.26),
x1/3−2θ3 P 1/2 KM −2 N −1 D2 x4/27−2θ3 D3/2 1.
In the case of (6.27),
x2/3−2θ3 P 1/2 KM −2 N −2 D x13/27−2θ3 1.
This completes the proof of (6.13). All the required bounds are now in place,
and the proof of Theorem 1 is complete.
54
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Department of Mathematics
Brigham Young University
Provo, UT 84602
U.S.A.
[email protected]
56