Metric discrepancy results for subsequences of geometricprogressions
K. Fukuyama1, *1

and K. Suzaki1, **Department of Mathematics, Kobe University
Abstract—In the previous work, we proved the law of the iterated logarithm for a subsequenceof geometric progression {θk x} and showed that the speed of convergence toward the uniform
distribution is faster than that of the original sequence. It is natural to ask if the speed becomes
faster again if we take a subsequence of the subsequence. In this note, we give a negative answer
to this question by giving a counterexample.
2010 Mathematical Subject Classification: 11K38, 42A55, 60F15
Keywords and phrases: discrepancy, lacunary sequence, law of the iterated logarithm

1. INTRODUCTION
The following discrepancies are frequently used to measure the speed of convergence of theempirical distribution of a sequence {xk } toward the uniform distribution:
1 ∑e1a′ ,a (xk ) ;DN ({xk }) = sup0≤a′ <a≤1 NN1 ∑e10,a (xk ) ;sup0≤a≤1 NN∗DN({xk }) =k=1

k=1ea′ ,a (x) = 1[a′ ,a) (⟨x⟩) − (a − a′ ), ⟨x⟩ denotes the fractional part x − [ x ] of x, and 1[a′ ,a)where 1denotes the indicator function of [ a′ , a).
For uniform distributed independent random variables {Uk }, Chung-Smirnov theorem [2, 14]claimsN D∗ ({Uk })N DN ({Uk })1lim √ N=√=N →∞2
2N log log N
2N log log Na.s.

Philipp [13] modified the method of Takahashi [15] and proved the bounded law of the iterated
logarithm√N D∗ ({nk x})
N DN ({nk x})
1/4 2 ≤ lim √ N≤ lim √≤ C(q) < ∞,N →∞N→∞
2N log log N
2N log log Na.e.

for {nk } satisfying Hadamard’s gap condition nk+1 /nk > q > 1. Aistleitner [1] proved the lower
bound can be replaced by 1/2 − 8q −1/4 and the upper bound by 1/2 + 6q −1/4 . It shows that thelimsup becomes nearer to 1/2 if the sequence {nk } grows faster.
***Electronic address: fukuyama@math.kobe-u.ac.jp
Electronic address: suzakik10@gmail.com

2If a sequence {nk } of real numbers grows very fast and satisfies a stronger gap condition
lim nk+1 /nk = ∞,

(1)

k→∞

we can prove (see Proposition in [5]) thatN D∗ ({nk x})
N DN ({nk x})
1
lim √ N= lim √=N →∞N→∞
2
2N log log N
2N log log Na.e.,(2)which coincides with Chung-Smirnov theorem and shows that {⟨nk x⟩} is almost independent.
For geometric progressions, it was proved in [3, 4] that the above limsups equal to a constantand the exact law of the iterated logarithm holds. Actually the following results were proved. For
all real numbers θ > 1, there exists a constant Σθ such thatN D∗ ({θk x})
N DN ({θk x})
lim √ N= lim √= Σθ
N →∞
N →∞
2N log log N
2N log log Na.e.

(3)

We have Σθ = 1/2 if θ satisfies the conditionθr ∈/ Q for allr ∈ N,(4)and Σθ > 1/2 otherwise. For concrete evaluation of Σθ in this case, see [3, 6–8, 11, 12].
In our previous work [9], we proved the following.

Denote by N the set of all strictlyincreasing sequences of natural numbers. For any θ > 1 and {m(k)} ∈ N , there exists a constantΣθ,{m(k)} ≥ 1/2 such that
∗ ({θ m(k) x})
N DN
N DN ({θm(k) x})√= lim √= Σθ,{m(k)}
N →∞
N →∞
2N log log N
2N log log Nlima.e.

(5)

We have{Σθ,{m(k)} {m(k)} ∈ N}

= [ 1/2, Σθ ].

(6)

By this result we see that exact law of the iterated logarithm holds for any subsequence ofgeometric progression, and see that the constant appearing there decreases if we take a subsequence.
Hence it is very natural to ask if the constant decreases again if we take a subsequence of thesubsequence. Surprisingly, the answer is negative. Now, we are in a position to state our result.
Theorem 1. We can construct a subsequence {nk } of {θk } satisfying the following properties: Itholds that
N D∗ ({nk x})
N DN ({nk x})
1lim √ N= lim √=N →∞N →∞2
2N log log N
2N log log Na.e.,(7)

and for any Σ ∈ [ 1/2, Σθ ], there exists a subsequence {νk } of {nk } such that
N D∗ ({νk x})
N DN ({νk x})
lim √ N= lim √=ΣN →∞N →∞2N log log N
2N log log Na.e.

(8)

By the same method of construction, we can prove a result concerning with subsequences ofErdős-Fortet sequence {2k − 1}. In [10], it was proved that
∗ ({(2k − 1)x})
N DN√= ΣEF (x)N →∞2N log log N

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where ΣEF (x) is a non-constant function.
Theorem 2. We can construct a subsequence {nk } of {2k − 1} satisfying the following properties:
It holds thatN D∗ ({nk x})
1lim √ N=N →∞22N log log Na.e.,

and there exists a subsequence {νk } of {nk } such that
N D∗ ({νk x})
lim √ N
= ΣEF (x)N →∞2N log log Na.e.

2. CONSTRUCTION OF THE SEQUENCE
In case of (4), by (6) we see Σθ,{m(k)} = 1/2 for all {m(k)} and there is nothing to be proved.
Hereafter, we assume that (4) does not hold. In this case we have Σθ > 1/2.
∗ can be done in the same
For simplicity we prove the results only for DN , since the proof for DNway.
Divide N into consecutive blocks ∆1 , ∆′1 , ∆2 , ∆′2 , . . . satisfying # ∆′i = i3 and # ∆i = i. Denote∆ = ∆1 ∪ ∆2 ∪ . . . and ∆′ = ∆′1 ∪ ∆′2 ∪ . . . . By∑√ea′ ,a (xk ) ≤ # (∆ ∩ [ 1, N ]) =: γ(N ) ∼ N ,1k≤N,k∈∆we see∑ea′ ,a (xk ) − γ(N ) ≤1k≤N,k∈∆′∑∑ea′ ,a (xk ) ≤1k≤Nea′ ,a (xk ) + γ(N ).
1k≤N,k∈∆′Hence by taking supremum for a′ and a, we haveδ(N )Dδ(N ) ({xk }k∈∆′ ) − γ(N ) ≤ N DN ({xk }) ≤ δ(N )Dδ(N ) ({xk }k∈∆′ ) + γ(N ),√where δ(N ) = # (∆′ ∩ [ 1, N ]) ∼ N .
By noting γ(N ) = O( δ(N )), and by dividing by√√2δ(N ) log log δ(N ) ∼ 2N log log N , we haveδ(N )Dδ(N ) ({xk }k∈∆′ )N DN ({xk })lim √= lim √.
N →∞N →∞
2N log log N
2δ(N ) log log δ(N )(9)Denote i− = min ∆i , i+ = max ∆i , and define m(k) in the following way.
m(0) = 0,m(k) = m(i− − 1) + k − i− + 1m(k) = m(i+ ) + k 2 − (i+ )2(k ∈ ∆i ),(k ∈ ∆′i ).

If k, k + 1 ∈ ∆′i , then m(k + 1) − m(k) = 2k + 1. If k = max ∆′i−1 = i− − 1 and l = min ∆′i =i+ + 1, by m(i+ ) = m(i− − 1) + i+ − i− + 1 = m(k) + i we have m(l) = m(i+ ) + (i+ + 1)2 − (i+ )2 =m(k) + 2i+ + i + 1. Therefore we see that the increment of the sequence {m(k)}k∈∆′ diverges toinfinity. Hence {θm(k) }k∈∆′ satisfies the condition (1) and (2) claims thatN DN ({θm(k) x}k∈∆′ )
1
√
=N →∞
2
2N log log Nlim

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a.e.

(10)

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By denoting {θm(k) }k∈N by {nk }, and by noting (10) and (9), we can prove (7).
To prove the existence of {νk }, we prepare a Bernoulli i.i.d. {Xk } such that P (Xk = 1) =1 − P (Xk = 0) = p, where p ∈ (0, 1 ].
We now denote {θm(k) | k ∈ ∆, Xk = 1} by {νk }. In the same way as the proof in [9] we see (8)holds for some Σ ∈ (1/2, Σθ ] with probability one. We also see that Σ moves continuously from
1/2 to Σθ if we vary p from 0 to 1.
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