Almost-all results on the p^\lambda problem

<strong>Almost</strong>-<strong>all</strong> <strong>results</strong> on the p λ problem II ∗
Stephan Baier
01.11.04
Abstract:
Suppose that 1/2 ≤ λ < 1. Balog and Harman proved that for any real θ there exist
infinitely many primes p satisfying �
pλ − θ �
< p−(1−λ)/2+ε (with an asymptotic result).
In the present paper we establish that for almost <strong>all</strong> θ in the interval 0 ≤ θ < 1 there
exist infinitely many primes p such that �
pλ − θ �
< p− min{(2−λ)/6,(14−9λ)/32}+ε . Thus we
obtain a better result for almost-<strong>all</strong> θ than for a single θ if λ > 1/2.
1 Introduction
In [1], [2], [3], [4], [5], [6], [8], [9], Balog, Harman and the author of the present paper
studied the distribution of fractional parts of p λ , p running over the prime numbers and
λ being a given real number lying in the interval 0 < λ < 1. In the sequel, we refer to
this question as the p λ problem. Balog’s and Harman’s result for the range 1/2 ≤ λ < 1
(see [4], [6, Theorem 2] and [8]) can be formulated as follows.
Theorem 1: Suppose that B > 0, λ ∈ [1/2, 1) and a real θ are given. Then for any
fixed τ in the range 0 ≤ τ < (1 − λ)/2 we have
as N → ∞.
�
n ≤ N,
� n λ − θ � < n −τ
Λ(n) =
N 1−τ
1 − τ
+ O
� N 1−τ
(log N) B
Here, as in the following, Λ(n) denotes the von Mangoldt function.
It is an interesting question whether sharper <strong>results</strong> than Theorem 1 hold true for
almost <strong>all</strong> θ. In [2] the author already established almost-<strong>all</strong> <strong>results</strong> on the p λ problem
for 0 < λ < 1/2. He obtained the following result by using density theorems for the
non-trivial zeta zeros.
∗ Mathematics Subject Classification (2000): 11N05, 11L20
1
�
(1)

Theorem 2: Let B > 0 and λ ∈ (0, 1/2) be given. Define
⎧
⎪⎨ 5/12 − λ/6 if λ ≤ 1/8,
H(λ) := � �
⎪⎩ 1 + λ/2 �2
/3 − λ if 1/8 < λ ≤ 1/2.
Then the asymptotic formula (1) holds true for almost <strong>all</strong> real θ with respect to the
Lebesgue measure if τ is any fixed real number lying in the range 0 ≤ τ < H(λ).
In the present paper we use a completely different approach to prove the following
almost-<strong>all</strong> result for the range 1/2 ≤ λ < 1.
Theorem 3: Let B > 0 and λ ∈ [1/2, 1) be given. Define
H(λ) := min{(2 − λ)/6, (14 − 9λ)/32}.
Then the asymptotic formula (1) holds true for almost <strong>all</strong> real θ with respect to the
Lebesgue measure if τ is any fixed real number lying in the range 0 ≤ τ < H(λ).
We note that the function H(λ) defined in Theorems 2 and 3 is continuous on the
interval (0, 1) and that for λ0 = 1/2 we have H(λ0) = 1/4 = (1 − λ0)/2. For λ0 = 1/2
Theorem 3 therefore does’nt provide any improvement beyond the result of Theorem 1.
However, for any λ ∈ (1/2, 1) the τ-range in Theorem 3 is wider than that in Theorem
1.
In [3] we improved the result of Theorem 3 for single θ’s in the range 59/85 =
0.694... < λ < 1. There we proved that in this range the exponent −(1 − λ)/2 + ε in
Theorem 1 may be replaced by −G(λ) + ε, where
G(λ) := min{max{(35 − 22λ)/129, 1/7}, 5/18 − λ/6}.
Even this result is exceeded by that of Theorem 3 for almost-<strong>all</strong> θ since in the range
59/85 < λ < 1 we have H(λ) > G(λ).
In a similar manner like [2, Theorem 2] was derived from [2, Proposition], we sh<strong>all</strong>
later derive the assertion of Theorem 3 from
Theorem 4: Let the conditions of Theorem 3 be kept, and suppose that C > 0.
Then we have
as N → ∞.
�1
0
�
�
�
�
�
�
�
�
�
�
�
n ≤ N,
� n λ − θ � < n −τ
Λ(n) −
N 1−τ
1 − τ
�
�2
�
�
�
�
�
�
�
�
dθ ≪ N 2(1−τ )
(log N) C
Acknowledgements. This research has been supported by a Marie Curie Fellowship
of the European Community programme “Improving the Human Research Potential
and the Socio-Economic Knowledge Base” under contract number HPMF-CT-2002-
02157.
2
(2)

2 Reduction to exponential sums
We sh<strong>all</strong> employ a similar method like in [3], where we linked the p λ problem with
Piatetski-Shapiro primes, that is, primes of the form [n c ] with fixed c > 1 and n ∈ �.
A standard procedure for attacking problems related to Piatetski-Shapiro primes is as
follows (see [11] for instance): Firstly, one reduces the sum of Λ(n) in question (in our
paper, this is the sum on the left-hand side of (1)) to exponential sums of the form
�
n∼N
Λ(n)e �
hn λ�
.
Secondly, one breaks up these exponential sums into certain bilinear exponential sums
(so-c<strong>all</strong>ed type I and type II sums) by using Vaughan-type identities. Fin<strong>all</strong>y, one estimates
these bilinear exponential sums in a non-trivial way. Here we also follow this
approach. However, since the first two steps were completely carried out in several papers
(see [11] for example), we omit some details in these steps.
Throughout the following, we suppose that λ, τ and t are given real numbers satisfying
1/2 ≤ λ < 1 and 0 ≤ τ < t < 1, and we put
H := N t .
The parameter t will later be choosen suitably subject to λ and τ. Furthermore, we
suppose that η is a (sm<strong>all</strong>) positive real number. By the notation k ∼ K we mean k to
lie in some interval K1 ≤ k ≤ K2 with K/2 ≤ K1 ≤ K2 ≤ 2K.
Using the Fourier series expansion of {x}, we obtain
Lemma 1: Suppose that (ah) is a sequence of complex numbers. Assume that we
uniformly have, as N → ∞,
�
ah
�
1≤h≤H N

�
1 − e � ��
−τ
hx ≪ hx −τ−1 ,
∂
∂x
and taking into consideration that the O-constant in (1) depends only on the O-constant
in (3).
Lemma 2: We have (2) if
for a suitable η > 0.
sup
N

We sh<strong>all</strong> prove the following lemma by adapting a method of Heath-Brown (see [11,
page 257]).
Lemma 4: Let Qh be any positive integer. Then
|Lh| 2 �
≪ QhX �
hX λ Y λ Q −1
�1/2 �
h Q −1
h Y 2 + Y �
Then
Proof: For q ∈ � we define the interval Iq by
Iq := �
Y λ (q − 1)/Qh, Y λ �
q/Qh .
|Lh| ≪ �
q≤2Qh
�
m∼X
| �
n ∼ Y,
n λ ∈ Iq,
mn ∼ N
�
2−λ
+ QhX h −1 Y 2−λ + Y X λ��
(log N).
bne �
hm λ n λ�
|.
From that, using the Cauchy-Schwarz inequality, we obtain
|Lh| 2 ≪ QhX � �
| �
bne �
hm λ n λ�
| 2
q≤2Qh
≪ QhX �
q≤2Qh
≪ QhX �
m∼X
�
n ∼ Y,
n λ ∈ Iq
n, r ∼ Y,
|t| ≤ 2Y λ /Qh
n ∼ Y,
nλ ∈ Iq,
mn ∼ N
�
r ∼ Y,
r λ ∈ Iq
| �
m ∼ X,
mn ∼ N,
mr ∼ N
| �
m ∼ X,
mn ∼ N,
mr ∼ N
e �
h(n λ − r λ )m λ�
|
e �
htm λ�
|, (5)
where we put t := n λ − r λ . Applying [11, Lemma 1] to the inner sum in the last line of
(5), we get
�
m ∼ X,
mn ∼ N,
mr ∼ N
e �
htm λ�
�
≪ min X, (ht) −1 X 1−λ + �
htX λ�1/2 �
≤ �
htX λ� 1/2
+ X min �
1, (ht) −1 X −λ�
. (6)
Combining (5) and (6), and using partial summation, we obtain
where
|Lh| 2 ≪ QhX �
hX λ Y λ Q −1
�1/2 h S �
2Y λ Q −1
�
h +
QhX 2 S �
h −1 X −λ�
+ h −1 QhX 2−λ (log N) max
∆≥h−1X −λ S(∆)∆−1 , (7)
5

Obviously, we have
S(∆) := �
n, r ∼ Y,
|t| ≤ ∆
Combining (7) and (8), we obtain the desired estimate. ✷
To optimize the estimate in Lemma 4, we choose
1.
S(∆) ≪ ∆Y 2−λ + Y. (8)
Qh := 1 + �
h 1/3 X (λ−2)/3 Y (λ+2)/3�
.
Then, by a short calculation using J ≤ H = N t and XY = N, we obtain
when
Lemma 5: We have
�
h≤H
This implies
|Lh| 2 ≪ �
N 1+λ/2+3t/2 + N 2−λ + N 2+t Y −1 + N 4/3+λ/3+4t/3 Y 1/3 +
N 2/3+2λ/3+5t/3 Y 2/3 + N 4/3−2λ/3+t/3 Y 4/3�
(log N).
Lemma 6: For every sufficiently sm<strong>all</strong> fixed η > 0 we have
provided the conditions
are satisfied.
4 Treatment of Kh
�
|Lh|
h≤H
2 ≪ N 2−2η
N t+100η ≤ Y ≤ N 2−λ−4t−100η ,
t < (2 − λ)/3, t > 2(1 − λ)/5 (9)
Heath-Brown established the following two estimates for Kh (see [11, pages 261-262]).
Lemma 7: (i) We have
6

�
2
Kh ≪ (log N) N 1−λ h −1 + Xh 1/2 N λ/2�
.
(ii) Let (p, q) be any exponent pair for which 0 < p ≤ 1/2 ≤ q ≤ 1. Let P be any
positive integer. Then
|Kh| 2 ≪ (log N) 8 (X + NP −1 ) �
N + N (p+1/2)λ Y q−2p−1/2 H p+1/2 P p+3/2 +
N 1−λ/2 Y 1/2 h −1/2 P 1/2 + Y P �
.
Taking the exponent pair (p, q) = (2/7, 4/7), and choosing
P := �
N (2−(1+2p)(λ+t))/(2p+3) Y (4p+1−2q)/(2p+3)�
= �
N (14−11(λ+t))/25 Y 7/25�
,
from Lemma 7, we obtain
Lemma 8: (i) We have
(ii) We have
�
h≤H
�
h≤H
|Kh| 2 �
4
≪ (log N) N 2(1−λ) + X 2 N λ+2t�
.
|Kh| 2 �
8
≪ (log N) N (36+11λ+36t)/25 Y −7/25 + N (43−7λ+18t)/25 Y 9/25 + N 1+t Y +
provided the condition
is satisfied.
when
N 2+t Y −1 + N (57−18λ+7t)/25 Y −9/25�
,
From the preceding lemma, we derive
t < 14/11 − λ (10)
Lemma 9: (i) For every sufficiently sm<strong>all</strong> fixed η > 0 we have
�
|Kh|
h≤H
2 ≪ N 2−2η
N (λ+2t)/2+100η ≤ Y.
(ii) For every sufficiently sm<strong>all</strong> fixed η > 0 we have (11) when
N max{(11λ+36t)/7−2,t}+100η ≤ Y ≤ N min{1−t,7(1+λ)/9−2t}−100η ,
7
(11)

provided the condition (10) is satisfied.
We are now in a position to prove
Lemma 10: For every sufficiently sm<strong>all</strong> fixed η > 0 we have (11) when
N t+100η ≤ Y,
provided the conditions in (9) and (10) as well as the conditions
are satisfied.
t < (2 − λ)/5, t < (14 + 5λ)/54, t < (14 − 9λ)/32 (12)
Proof: We first note that the sum �
same manner as �
h≤H
h≤H
|Kh| 2 can alternatively be estimated in the
|Lh| 2 was estimated in Lemma 6. Thus, we also have (11) when
N t+100η ≤ Y ≤ N 2−λ−4t−100η ,
provided the conditions in (9) are satisfied. By this observation and Lemma 9, for proving
Lemma 10 it suffices to show that
and
(λ + 2t)/2 < min{1 − t, 7(1 + λ)/9 − 2t} (13)
max{(11λ + 36t)/7 − 2, t} < 2 − λ − 4t. (14)
Indeed, (13) follows from the first and the second condition in (12), and (14) follows
from the first and the third condition in (12). This completes the proof. ✷
5 Proof of Theorem 4
We suppose that 1/2 ≤ λ < 1 and
We choose t in such a manner that
and we take
0 ≤ τ < min{(2 − λ)/6, (14 − 9λ)/32}.
max{τ, 2(1 − λ)/5} < t < min{(2 − λ)/6, (14 − 9λ)/32},
u := N t+100η ,
v := N 2−λ−4t−100η ,
z := N t+100η .
8

Then the conditions in (9), (10), (12) to λ and t are satisfied. Moreover, since t <
(2 − λ)/6, the conditions in Lemma 3 to u, v, z are satsified if η is sufficiently sm<strong>all</strong> and
N is sufficiently large.
Now, putting Lemmas 2, 3, 6 and 10 together, we obtain the result of Theorem 4. ✷
6 Proof of Theorem 3
Fin<strong>all</strong>y, we derive Theorem 3 from Theorem 4 by using the following lemma in metric
number theory which is essenti<strong>all</strong>y due to W.M.Schmidt.
Lemma 11: (Lemma 1.5 in [10]) Let X be a measure space with measure µ such
that 0 < µ(X) < ∞. Let fk(x) (k = 1, 2, ...) be a sequence of non-negative µ-measurable
functions, and let fk, ϕk be sequences of real numbers such that
Write
0 ≤ fk ≤ ϕk (k = 1, 2, ...). (15)
Φ(M) := �
and suppose that Φ(M) → ∞ as M → ∞. Suppose that for arbitrary integers m, n
(1 ≤ m < n) we have
�
X
k≤M
ϕk,
�
�
�
� �
�2
�
�
�
� (fk(x) − fk) �
� dµ ≤ D0 ϕk
�m≤k

k
(log rk) B+1 ≪ rk − rk−1 ≪
rk
(log rk) B+1
for <strong>all</strong> k ≥ 1, where B is the constant in Theorem 3. We note that
k ≪ (log rk) B+2
by this construction. We put X := [0, 1], define µ to be the Lebesgue measure, and
define
fk(θ) :=
fk := r1−τ k
�
rk−1 < n ≤ rk,
� n λ − θ � < n −τ
− r1−τ k−1
1 − τ
,
Λ(n),
r
ϕk := D1 ·
2(1−τ)
k
,
(log rk) 3B+6
where D1 is choosen in such a manner that (15) holds for <strong>all</strong> positive integers k. We
note that
fk ≪
r1−τ
k
(log rk) B+1
by (18) and Taylor’s formula.
Then the condition “Φ(M) → ∞ as M → ∞” in Lemma 11 is satisfied since we
have rk → ∞ as k → ∞. By Theorem 4, also the inequality (16) is satisfied for a
suitable constant D0 (to see this, we apply the Cauchy-Schwarz inequality). Hence,
the asymptotic estimate (17) indeed holds for almost <strong>all</strong> x ∈ [0, 1] with respect to the
Lebesgue measure. Moreover, by (19) we have
Φ(M) ≪
r2(1−τ)
M
(log rM) 2B+4
as M → ∞. Using (17), (20) and (21), we obtain, as M → ∞,
�
n ≤ rM ,
� n λ − θ � < n −τ
Λ(n) = r1−τ M
1 − τ
+ O �
r 1−τ
M (log rM) −(B+1)�
for almost <strong>all</strong> θ ∈ [0, 1].
Fin<strong>all</strong>y, we pick θ ∈ [0, 1] for which (22) holds, and we seek to show that then even
(1) holds for this θ. To do so, we proceed as follows: For a given N ≥ 2 let M be the
positive integer for which rM−1 < N ≤ rM. Then, by construction of the sequence (rk),
we have
rM − N ≪
rM
≪
(log rM) B+1
10
N
.
(log N) B+1
(18)
(19)
(20)
(21)
(22)

Using this inequality, (22) and Taylor’s formula, we obtain the asymptotic estimate (1).
This completes the proof of Theorem 3.
References
[1] S. Baier, On the p λ problem, Acta Arith. 113 (2004), 77-101.
[2] S. Baier, <strong>Almost</strong>-<strong>all</strong> <strong>results</strong> on the p λ problem, to appear in Acta Math. Hungarica.
[3] S. Baier, An extension of the Piatetski-Shapiro prime number theorem, preprint
(2004).
[4] A. Balog, On the fractional part of p θ , Arch. Math. (Basel) 40 (1983), 434-440.
[5] A. Balog, On the distribution of p θ mod 1, Acta Math. Hung. 45 (1985), 179-199.
[6] A. Balog, G. Harman, On mean values of Dirichlet polynomials, Arch. Math.
(Basel) 57 (1991), 581-587.
[7] S.W. Graham, G. Kolesnik, Van der Corput ′ s Method of Exponential Sums, Cambridge
University Press, 1991.
[8] G. Harman, On the distribution of √ p modulo one, Mathematika 30 (1983), 104-
116.
[9] G. Harman, Fractional and integral parts of p λ , Acta Arith. 58 (1991), 141-152.
[10] G. Harman, Metric Number Theory, Oxford Science Publications, Clarendon Press,
Oxford (1998).
[11] D.R. Heath-Brown, The Pjateckii-Shapiro Prime Number Theorem, J. Number
Theory 16 (1983), 242-266.
[12] R.C. Vaughan, On the Estimation of Trigonometrical Sums over Primes, and Related
Questions, Report No. 9, Institut Mittag-Leffler, 1977.
HARISH-CHANDRA RESEARCH INSTITUTE
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JHUSI
ALLAHABAD 211 019
INDIA
E-MAIL: SBAIER@MRI.ERNET.IN
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