In this paper, we obtain bounds on the L¹-norm of the sum ∑_(n≤x)τ(n)e(αn) where τ(n) is the divisor function

Goldston, D. A.

Pandey, M.

Caltech Authors - Main

On the L¹ norm of an exponential sum involving the divisor function

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ON THE L1 NORM OF AN EXPONENTIAL SUM INVOLVING
THE DIVISOR FUNCTION
D. A. GOLDSTON AND M. PANDEY
1. Introduction
Let τ(n) =
∑
d|n 1 be the divisor function, and
S(α) =
∑
n≤x
τ(n)e(nα), e(α) = e2πiα.
In 2001 Brudern [1] considered the L1 norm of S(α) and claimed to prove
(1)
√
x ≪
∫ 1
0
|S(α)|dα ≪
√
x.
However there is a mistake in the proof given there which depends on a lemma
which is false. In this note we prove the following result.
Theorem 1. We have
(2)
√
x ≪
∫ 1
0
|S(α)|dα ≪
√
x log x.
The upper bound here is obtained by following Brudern’s proof with corrections.
The lower bound is based on the method Vaughan introduced to study the L1 norm
for exponential sums over primes [3], and also makes use of a more recent result of
Pongsriiam and Vaughan [2] on the divisor sum in arithmetic progressions. We do
not know whether the upper bound or the lower bound reflects the actual size of
the L1 norm here.
2. Proof of the upper bound
Let u and v always be positive integers. Following Brüdern, we have
S(α) =
∑
n≤x
(
∑
uv=n
1
)
e(nα)
=
∑
uv≤x
e(uvα)
= 2
∑
u≤√x
∑
u<v≤x/u
e(uvα) +
∑
u≤√x
e(u2α)
:= 2T (α) + V (α).
Date: April 21, 2017.
∗ The first author was in residence at the Mathematical Sciences Research Institute in Berkeley,
California (supported by the National Science Foundation under Grant No. DMS-1440140), during
the Spring 2017 semester.
1
2 D. A. GOLDSTON AND M. PANDEY
By Cauchy-Schwarz and Parseval
∫ 1
0
|V (α)| dα ≤
(
∫ 1
0
|V (α)|2 dα
)
1
2
=
√
⌊
√
x⌋ ≤ x 14 ,
and therefore by the triangle inequality
∫ 1
0
|S(α)| dα = 2
∫ 1
0
|T (α)| dα+O(x 14 ).
Thus to prove the upper bound in Theorem 1 we need to establish
(3)
∫ 1
0
|T (α)|dα ≪
√
x log x.
We proceed as in the circle method. Clearly in (3) we can replace the integration
range [0, 1] by [1/Q, 1 + 1/Q]. By Dirichlet’s theorem for any α ∈ [1/Q, 1 + 1/Q]
we can find a fraction aq , 1 ≤ q ≤ Q, 1 ≤ a ≤ q, (a, q) = 1, with |α − aq | ≤ 1/(qQ).
Thus the intervals [aq − 1qQ , aq + 1qQ ] cover the interval [1/Q, 1 + 1/Q]. Taking
(4) 2
√
x ≤ Q ≪
√
x,
we obtain
∫ 1
0
|T (α)|dα ≤
∑
q≤Q
q
∑
a=1
(a,q)=1
∫ a
q
+1/(2q
√
x)
a
q
−1/(2q√x)
|T (α)| dα.
On each interval [aq − 12q√x ,
a
q +
1
2q
√
x
] we decompose T (α) into
T (α) = Fq(α) +Gq(α)
where
Fq(α) =
∑
u≤√x
q|u
∑
u<v≤x/u
e(uvα)
and
Gq(α) =
∑
u≤√x
q∤u
∑
u<v≤x/u
e(uvα),
and have
∫ 1
0
|T (α)|dα ≤
∑
q≤Q
q
∑
a=1
(a,q)=1
∫ a
q
+1/(2q
√
x)
a
q
−1/(2q√x)
|Fq(α)| dα+
∑
q≤Q
q
∑
a=1
(a,q)=1
∫ a
q
+1/(2q
√
x)
a
q
−1/(2q√x)
|Gq(α)| dα
:= IF + IG.
The upper bound in Theorem 1 follows from the following two lemmas.
Lemma 2 (Brüdern). We have
IF ≪
√
x.
Lemma 3. We have
IG ≪
√
x log x.
In what follows we always assume (a, q) = 1, and define the new variable β by
(5) α =
a
q
+ β.
ON THE L1 NORM OF AN EXPONENTIAL SUM INVOLVING THE DIVISOR FUNCTION 3
Proof of Lemma 2. The proof follows from the estimate
(6) Fq(α) ≪
{
min
(
x, 1|β|
)
log 2
√
x
q
q , if q ≤
√
x, |β| ≤ 1
2q
√
x
;
0, if q >
√
x;
since this implies
IF ≪
∑
q≤√x
q
∫ 1/(2q
√
x)
0
min
(
x,
1
|β|
)
log 2
√
x
q
q
dβ
≪
∑
q≤√x
log
2
√
x
q
(
∫ 1/(2x)
0
x dβ +
∫ 1/(2q
√
x)
1/(2x)
1
β
dβ
)
≪
∑
q≤√x
(
log
2
√
x
q
)2
≪
√
x.
To prove (6), we first note that the conditions q|u and u ≤ √x force Fq(α) = 0
when q >
√
x. Next, when q ≤ √x we write u = jq and have
Fq(α) =
∑
j≤
√
x
q
∑
jq≤v≤ x
jq
e(jqvβ).
Making use of the estimate
(7)
∑
N1<n≤N2
e(nα) ≪ min
(
N2 −N1,
1
‖α‖
)
we have
Fq(α) ≪
∑
j≤
√
x
q
min
(
x
jq
,
1
‖jqβ‖
)
.
In this sum jq ≤ √x so that |jqβ| ≤ |β|√x, and hence the condition |β| ≤ 1
2q
√
x
implies |jqβ| ≤ 12q ≤ 12 . Hence ‖jqβ‖ = jq|β|, and we have
Fq(α) ≪
∑
j≤
√
x
q
1
jq
min
(
x,
1
|β|
)
≪ min(x, 1|β| )
log 2
√
x
q
q
.

Proof of Lemma 3. The proof follows from the estimate,
(8) Gq(α) ≪ (
√
x+ q) log q, for α =
a
q
+ β, |β| ≤ 1
2q
√
x
,
since this implies
IG ≪
∑
q≤Q
q
∫ 1/(2q
√
x)
0
(
√
x+ q) log q dβ
≪ 1√
x
(Q(
√
x+Q) logQ ≪
√
x log x
4 D. A. GOLDSTON AND M. PANDEY
by (4). To prove (8), we apply (7) to the sum over v in Gq(α) and obtain
Gq(α) ≪
∑
u≤√x
q∤u
min
(
x
u
,
1
‖uα‖
)
.
Recalling ‖x‖ = ‖ − x‖ and the triangle inequality ‖x+ y‖ ≤ ‖x‖+ ‖y‖, and using
the conditions 1 ≤ u ≤ √x, q ∤ u, |β| ≤ 1
2q
√
x
, we have
‖uα‖ ≥
∥
∥
∥
∥
au
q
∥
∥
∥
∥
− ‖uβ‖
≥
∥
∥
∥
∥
au
q
∥
∥
∥
∥
− u|β|
≥
∥
∥
∥
∥
au
q
∥
∥
∥
∥
− u
2q
√
x
≥
∥
∥
∥
∥
au
q
∥
∥
∥
∥
− 1
2q
≥ 1
2
∥
∥
∥
∥
au
q
∥
∥
∥
∥
,
and therefore
Gq(α) ≪
∑
u≤√x
q∤u
1
∥
∥
∥
au
q
∥
∥
∥
.
Here
∥
∥
∥
au
q
∥
∥
∥
= bq for some integer 1 ≤ b ≤
q
2 and since the integers {au : 1 ≤ u ≤
q
2}
are distinct modulo q since (a, q) = 1, we see
∑
1≤u≤ q
2
1
∥
∥
∥
au
q
∥
∥
∥
=
∑
1≤b≤ q
2
q
b
≪ q log q.
If q >
√
x then
∑
u≤√x
q∤u
1
∥
∥
∥
au
q
∥
∥
∥
≤ 2
∑
1≤u≤ q
2
1
∥
∥
∥
au
q
∥
∥
∥
≪ q log q,
while if q ≤ √x then the sum bounding Gq(α) can be split into ≪
√
x
q sums of this
type and
Gq(α) ≪
√
x
q
(q log q) ≪
√
x log q.

3. Proof of the lower bound
Following Brüdern, consider the intervals |α − aq | ≤ 1/(4x) for 1 ≤ a ≤ q ≤ Q,
where we take 12
√
x ≤ Q ≤ √x. These intervals are pairwise disjoint because for
two distinct fractions |a/q − a′/q′| ≥ 1/(qq′) ≥ 1/x. (We will see later why these
intervals have been chosen shorter than required to be disjoint.) Hence, using (5)
∫ 1
0
|S(α)| dα =
∫ 1+1/Q
1/Q
|S(α)| dα ≥
∑
q≤ 1
2
√
x
q
∑
a=1
(a,q)=1
∫ 1/(4x)
−1/(4x)
∣
∣
∣
∣
S
(
a
q
+ β
)∣
∣
∣
∣
dβ.
ON THE L1 NORM OF AN EXPONENTIAL SUM INVOLVING THE DIVISOR FUNCTION 5
Next we follow Vaughan’s method [3] and apply the triangle inequality to obtain
the lower bound
∫ 1
0
|S(α)| dα ≥
∑
q≤ 1
2
√
x
∫ 1/(4x)
−1/(4x)
∣
∣
∣
∣
∣
∣
∣
q
∑
a=1
(a,q)=1
S
(
a
q
+ β
)
∣
∣
∣
∣
∣
∣
∣
dβ.
Letting
(9) Uq(x;β) :=
q
∑
a=1
(a,q)=1
S
(
a
q
+ β
)
=
∑
n≤x
τ(n)cq(n)e(nβ),
where
cq(n) =
q
∑
a=1
(a,q)=1
e
(
an
q
)
is the Ramanujan sum, our lower bound may now be written as
(10)
∫ 1
0
|S(α)| dα ≥
∑
q≤ 1
2
√
x
∫ 1/(4x)
−1/(4x)
|Uq(x;β)| dβ.
To complete the proof of the lower bound we need the following lemma, which we
prove at the end of this section.
Lemma 4. For q ≥ 1 we have
Uq(x; 0) =
ϕ(q)
q
x(log(x/q2) + 2γ − 1) +O(qτ(q)(x 13 + q 12 )xǫ),(11)
where γ is Euler’s constant.
Proof of the lower bound in Theorem 1. For any exponential sum T (x;β) =
∑
n≤x ane(nβ)
we have by partial summation or direct verification
T (x;β) = e(βx)T (x; 0)− 2πiβ
∫ x
1
e(βy)T (y; 0) dy.
Taking T (x;β) = Uq(x;β) we thus obtain from (10) and the triangle inequality
(12)
∫ 1
0
|S(α)| dα ≥
∑
q≤ 1
2
√
x
∫ 1/(4x)
−1/(4x)
(
|Uq(x; 0)| − 2π|β|
∫ x
1
|Uq(y; 0)| dy
)
dβ.
By Lemma 4, with q ≤ 12
√
x,
∫ x
1
|Uq(y; 0)| dy ≤
ϕ(q)
q
(
∫ x
1
y| log(y/q2)|+ (2γ − 1)y dy
)
+O(xqτ(q)(x
1
3 + q
1
2 )xǫ)
≤ ϕ(q)
q
(
∫ q2
1
y log(q2/y) dy +
∫ x
q2
y log(y/q2) dy +
1
2
x2(2γ − 1)
)
+O(xqτ(q)(x
1
3 + q
1
2 )xǫ)
=
x
2
(
ϕ(q)
q
(
x(log(x/q2) + 2γ − 1)− x
2
+
q4
x
)
+O(qτ(q)(x
1
3 + q
1
2 )xǫ)
)
≤ x
2
Uq(x, 0) +O(qτ(q)(x
1
3 + q
1
2 )xǫ).
6 D. A. GOLDSTON AND M. PANDEY
Using |β| ≤ 1/(4x), we have
|Uq(x; 0)| − 2πβ
∫ x
1
|Uq(y; 0)| dy ≥
(
1− π
4
)
|Uq(x; 0)| −O(qτ(q)(x
1
3 + q
1
2 )xǫ).
We conclude, returning to (12) and making use of Lemma 4 again,
∫ 1
0
|S(α)| dα ≥ 4− π
8x
∑
q≤ 1
2
√
x
(
|Uq(x; 0)| −O(qτ(q)(x
1
3 + q
1
2 )xǫ)
)
≥ 4− π
8
∑
q≤ 1
2
√
x
ϕ(q)
q
(log(x/q2) + 2γ − 1)−O(x 13+ǫ).
It is easy to see that the sum above is ≫ √x which suffices to proves the lower
bound. More precisely, using
ϕ(n)
n
=
∑
d|n
µ(d)
d
a simple calculation gives the well-known result
∑
n≤x
ϕ(n)
n
=
6
π2
x+O(log x),
and then by partial summation we find
∑
q≤ 1
2
√
x
ϕ(q)
q
(log(x/q2) + 2γ − 1) ∼ 6
π2
(log 2 + γ − 1)
√
x.

Proof of Lemma 4. Pongsriiam and Vaughan [2] recently proved the following very
useful result on the divisor function in arithmetic progressions. For inteqer a and
d ≥ 1 and real x ≥ 1 we have
∑
n≤x
n≡a(mod d)
τ(n) =
x
d
∑
r|d
cr(a)
r
(
log
x
r2
+ 2γ − 1
)
+O((x
1
3 + d
1
2 )xǫ),
where γ is Euler’s constant and cr(a) is the Ramanujan sum. We need the special
case when a = 0 which along with the situation (a, d) > 1 is explicitly allowed in
this formula. Hence we have
(13)
∑
n≤x
d|n
τ(n) =
x
d
fx(d) +O((x
1
3 + d
1
2 )xǫ)
where
(14) fx(d) =
∑
r|d
gx(r), gx(r) =
ϕ(r)
r
(log(x/r2) + 2γ − 1).
Making use of
cq(n) =
∑
d|(n,q)
dµ
( q
d
)
,
ON THE L1 NORM OF AN EXPONENTIAL SUM INVOLVING THE DIVISOR FUNCTION 7
and (13) we have
Uq(x; 0) =
∑
n≤x
τ(n)cq(n)
=
∑
d|q
dµ
( q
d
)
∑
n≤x
d|n
τ(n)
= x
∑
d|q
µ
( q
d
)
fx(d) +O(qτ(q)(x
1
3 + d
1
2 )xǫ).
We evaluate the sum above using Dirichlet convolution and the identity 1 ∗ µ = δ
where δ(n) is the identity for Dirichlet convolution defined to be 1 if n = 1 and
zero otherwise. Hence
∑
d|q
µ
( q
d
)
fx(d) = (fx ∗ µ)(q) = ((gx ∗ 1) ∗ µ)(q) = (gx ∗ δ)(q) = gx(q),
and Lemma 4 is proved. 
References
[1] Jörg Brüdern, Exponential sums over products and their L1-norm, Arch. Math. 76 (2001),
196–201.
[2] Prapanpong Pongsriiam and Robert C. Vaughan, The divisor function on residue classes I,
Acta Arithmetica 168(4) (2015), 369–381.
[3] R.C. Vaughan, The L1 mean of exponential sums over primes, Bull. London Math. Soc.
20(1988), 121-123.
D. A. Goldston (daniel.goldston@sjsu.edu)
Department of Mathematics and Statistics
San Jose State University
San Jose, CA 95192
USA
M. Pandey (mayankpandey9973@gmail.com)
12861 Regan Lane
Saratoga, CA 95070
USA


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https://authors.library.caltech.edu/90280/2/1704.05953

On the L¹ norm of an exponential sum involving the divisor function

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