Almost-all results on the p^\lambda problem

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)