ACTA ARITHMETICA XCII.4 (2000)
A problem of Galambos on Engel expansions
by
Jun Wu (Wuhan)
1. Introduction. Given x in (0, 1], let x = [d
1(x), d
2(x), . . .] denote the Engel expansion of x, that is,
(1) x = 1
d
1(x) + 1
d
1(x)d
2(x) + . . . + 1
d
1(x)d
2(x) . . . d
n(x) + . . . ,
where {d
j(x), j ≥ 1} is a sequence of positive integers satisfying d
1(x) ≥ 2 and d
j+1(x) ≥ d
j(x) for j ≥ 1. (See [3].) In [3], J´anos Galambos proved that for almost all x ∈ (0, 1],
(2) lim
n→∞
d
1/nn(x) = e.
He conjectured ([3], P132) that the Hausdorff dimension of the set where (2) fails is one. In this paper, we prove this conjecture:
Theorem. dim
H{x ∈ (0, 1] : (2) fails} = 1.
We use L
1to denote the one-dimensional Lebesgue measure on (0, 1] and dim
Hto denote the Hausdorff dimension.
2. Proof of Theorem. The aim of this section is to prove the main result of this paper.
By Egoroff’s Theorem, there exists a Borel set A ⊂ (0, 1] with L
1(A)
≥ 1/2 such that {d
1/nn(x), n ≥ 1} converges to e uniformly on A. In partic- ular, there exists a positive number N such that
(3) 2 ≤ d
1/nn(x) ≤ 3 for all n ≥ N and x ∈ A.
Choose a positive integer M satisfying M ≥ N . For any x = [d
1, d
2, . . .] = [d
1, d
2, . . . , d
M, d
M +1, d
M +2, . . . , d
kM +1, d
kM +2, . . . , d
(k+1)M, . . .], we cons- truct a new point x ∈ (0, 1] as follows:
x = [d
1, d
2, . . .],
2000 Mathematics Subject Classification: Primary 11K55; Secondary 28A80.
Project supported by National Natural Science Foundation of China.
[383]
384 J. Wu
where d
k(M +1)+l= d
kM +lfor all k ≥ 0 and 0 ≤ l ≤ M . That is, (4) x = [d
1, d
2, . . . , d
M, d
M, d
M +1, d
M +2, . . . ,
d
kM +1, d
kM +2, . . . , d
(k+1)M, d
(k+1)M, . . .].
Lemma 1. {x : x ∈ A} ⊂ {x ∈ (0, 1] : (2) fails}.
P r o o f. Note that for any k ≥ 1, d
k(M +1)(x) = d
kM(x). We have (5) lim
k→∞
d
1/(k(M +1))k(M +1)
(x) = lim
k→∞
(d
1/(kM )kM(x))
kM/(k(M +1))= e
M/(M +1), and this proves the assertion.
For any x = [d
1(x), d
2(x), . . .] ∈ (0, 1], y = [d
1(y), d
2(y), . . .] ∈ (0, 1], define
%(x, y) = inf{j : d
j(x) 6= d
j(y)} (inf ∅ = ∞).
For any x, y ∈ (0, 1], x 6= y, suppose %(x, y) = k. Without loss of gen- erality, assume d
k(x) < d
k(y). Then x > y and x ∈ (B, C], y ∈ (D, E]
with
B = 1
d
1(x) + 1
d
1(x)d
2(x) + . . . + 1
d
1(x)d
2(x) . . . d
k−1(x)d
k(x)
+ 1
d
1(x)d
2(x) . . . d
k(x)d
k+1(x) ,
C = 1
d
1(x) + 1
d
1(x)d
2(x) + . . . + 1
d
1(x)d
2(x) . . . d
k−1(x)(d
k(x) − 1) ,
D = 1
d
1(y) + 1
d
1(y)d
2(y) + . . . + 1
d
1(y)d
2(y) . . . d
k−1(y)d
k(y) ,
E = 1
d
1(y) + 1
d
1(y)d
2(y) + . . . + 1
d
1(y)d
2(y) . . . d
k−1(y)d
k(y)
+ 1
d
1(y)d
2(y) . . . d
k(y)(d
k+1(y) − 1) , hence
(6) 1
d
1(x)d
2(x) . . . d
k(x)d
k+1(x) ≤ |x − y| ≤ 1
d
1(x)d
2(x) . . . d
k−1(x) , where d
0(x) ≡ 1.
Let
ε = 6 log 3
M log 2 , c = 1 3
4M (M +1). Lemma 2. For any x, y ∈ A,
(7) |x − y| ≥ c|x − y|
1+2ε.
P r o o f. Without loss of generality, assume x > y. Suppose %(x, y) = k.
Galambos on Engel expansions 385
(a) If k ≤ 2M , then by (3), (4) and (6), we have
|x−y| ≥ 1
d
1(x)d
2(x) . . . d
k(x)d
k+1(x)d
k+2(x) ≥
1 3
2M 2M +2≥ c|x−y|
1+2ε. (b) If pM < k ≤ (p + 1)M for some p ≥ 2, then by (4) and (6), we have (8) |x − y| ≥
p−1
Y
j=0
Y
Ml=1
1 d
jM +l(x)
1
d
jM +M(x)
k+1Y
j=pM +1
1 d
j(x) . For 1 ≤ j ≤ p − 1, by (3), we have
(9) d
jM +M(x) ≤ 3
jM +M≤ 3
2jM≤
Y
Ml=1
2
jM +l ε≤
Y
Ml=1
d
jM +l(x)
ε, thus
(10)
Y
Ml=1
1 d
jM +l(x)
1
d
jM +M(x) ≥
Y
Ml=1
1 d
jM +l(x)
1+ε, 1 ≤ j ≤ p − 1.
For j = 0, (11)
Y
Ml=1
1 d
l(x)
1
d
M(x) ≥ 1 3
M (M +1). On the other hand,
d
k(x)d
k+1(x) ≤ 3
2k+1≤ 3
3k≤ (2
M2
M +1. . . 2
k−1)
ε(12)
≤ (d
M(x) . . . d
k−1(x))
ε, hence
(13) 1
d
k(x)d
k+1(x) ≥
1
d
1(x)d
2(x) . . . d
k−1(x)
ε. Combining (10), (11) and (13), we have
|x − y|
≥ 1
3
M (M +1) p−1Y
j=1
Y
Ml=1
1 d
jM +l(x)
1+ε k−1Y
j=pM +1
1 d
j(x)
1
d
k(x) · 1 d
k+1(x)
≥ 1
3
M (M +1) k−1Y
i=1