VNU Journal of Science, Mathematics - Physics 25 (2009) 57-68
The extreme value of local dimension of convolution of the
cantor measure
Vu Thi Hong Thanh
1,∗
, Nguyen Ngoc Quynh
2
, Le Xuan Son
1
1
Department of Mathematics, Vinh University
2
Department of Fundamental Science, Vietnam academy of Traditional medicine
Received 5 March 2009
Abstract. Let µ be the m−fold convolution of the standard Cantor measure and α
m
be
the lower extreme value of the local dimension of the measure µ. The values of α
m
for
m = 2, 3, 4 were showed in [4] and [5]. In this paper, we show that
α
5
= |
log

2
3.2
5

√

m

j=1
p
j
= 1) be a set of
probability weights. Then, there exists a unique probability measure µ satisfying
µ(A) =
m

j=1
p
j
µ(S
−1
j
(A))
for all Borel measurable sets A (see [1]). We call µ a self-similar measure and {S
j
}
m
j=1
a system
iterat ed functions.
When S
1
, , S
m
are similarities with equal contraction ratio ρ ∈ (0, 1) on R, i.e., S
j

µ
ρ
(A) = P{ω : S(ω) ∈ A}
is called a fractal measure and µ
ρ
≡ µ (see [2]).
Let ν be the standard Cantor measure, then ν can be considered to be generated by the two
maps S
i
(x) =
1
3
x +
2
3
i, i = 0, 1 with weight
1
2
on each S
i
. Then the attractor of this system
∗
Corresponding author. E-mail: vu
hong
57
58 V.T.H. Thanh et al. / VNU Journal of Science, Mathematics - Physics 25 (2009) 57-68
iterated functions is the standard Cantor set C, i.e., C = S
0
(C) ∪ S
1

n

j=1
3
−j
X
j
and µ, µ
n
be the distribution
measure of S, S
n
respectively. It is well known that µ is either singular or absolutely continuous (see
[2]).
Recall that let µ be a probability measure on R. For s ∈ supp µ, the local dimension of µ at s
is denoted by α(s) and defined by
α(s) = lim
h→0
+
log µ(B
h
(s))
log h
if the limit exists. Otherwise, let
α(s) and α(s) denote the upper and lower dimension by taking the
upper and lower limits respectively. Let E = {α(s) : s ∈ supp µ} be the set of the attainable local
dimensions of the measure µ and for each m = 2, 3, , put
α
m
= inf{α

x
j
, x
j
= 0, , m, and the associated probability. After that, in [5], Pablo
Shmerkin showed the α
m
for m = 2, 3, 4 by the other way. He used the spectral radius of matrixes
to define his results. He said that the identifying formulae for α
m
for m ≥ 5 was a difficult problem,
and he only estimated the values of α
m
for 5  m  10.
Now, in this paper, we are interested in the identifying α
m
for m = 5 and we show that our
result coincides with Pablo Shmerkin’s estimate. We have
2. Main result
Main Theorem. Let µ be the 5−fold convolution o f the standard Cantor measure, then the lower
extreme value of the local dimension of µ is
α
5
= |
log

2
3.2
5
√

. Then its m−fold convolution µ = ν ∗ ∗ ν is generated
by S
i
(x) =
1
3
x +
2
3
i with weight
C
i
m
2
m
on with S
i
for i = 0, 1, , m.
Proposition 2 ([4]). Let m ≥ 2, then α(s) = lim
n→∞
|
log µ
n
(s
n
)
n log 3
| provided that the limit exists. Otherwise,
we can replace α(s) by α(s) and α(s) and consider th e up per and the lower limits.
Put D = {0, 1, , 5} and for each n ∈ N we denote

, , y
n
) ∈ D
n
:
n

i=1
3
−i
y
i
=
n

i=1
3
−i
x
i
}.
If (z
1
, , z
n
) ∈ (x
1
, , x
n
), then we denote (z

1
, , x
m
). (1)
We denote
(x
1
, , x
n
, x) = {(y
1
, , y
n
, x) : (y
1
, , y
n
) ∈ (x
1
, , x
n
)}.
The following lemma will be used frequently in this paper.
Lemma 1. Let s
n
=
n

j=1
3

, x
2
, ) = (2, 3, 2, 3, ) ∈ D
∞
, we ha ve
i) If n is even then (y
1
, , y
n
) ∈ (x
1
, , x
n
) = (2, 3, , 2, 3) iff
(y
1
, , y
n
) ∈ (x
1
, , x
n−1
, 3) or (y
1
, , y
n
) ∈ (x
1
, , x
n−2

, 5).
Proof.
i) The case n is even.
If (y
1
, , y
n
) ∈ (x
1
, , x
n
) = (2, 3, , 2, 3) then we have
(y
1
− 2)3
n−1
+ (y
2
−3)3
n−2
+ + (y
n−1
−2)3 + (y
n
− 3) = 0. (2)
Therefore, y
n
−3 ≡ 0 (mod 3). Since y
n
∈ D, we have y

1
, , x
n−1
). By (1) we have (y
1
, , y
n
) ∈ (x
1
, , x
n−1
, 3).
60 V.T.H. Thanh et al. / VNU Journal of Science, Mathematics - Physics 25 (2009) 57-68
b) If y
n
= 0 then y
n
−3 = −3. By (2) we have
(y
1
−2)3
n−2
+ (y
2
− 3)3
n−3
+ + (y
n−2
−3)3 + (y
n−1

1
, , x
n−1
, 3), then we have
(y
1
, , y
n
) ∈ (x
1
, , x
n
).
So we consider the case (y
1
, , y
n
) ∈ (x
1
, , x
n−2
, x
n−2
, 0). Then we have y
n
= 0 and (y
1
, , y
n−1
) ∈

+ (y
2
− 3)3
n−3
+ + (y
n−3
−3)3 + (y
n−2
−3) = 0.
Therefore, (y
1
, , y
n−2
) ∼ (2, 3, , 2, 3) = (x
1
, , x
n−2
) and (y
n−1
, y
n
) = (3, 0). Since (3, 0) ∼
(2, 3), by (1) we have (y
1
, , y
n
) ∈ (x
1
, , x
n

−2)3 + y
n−2
−4 = 0. (3)
Therefore, y
n−2
− 4 ≡ 0 (mod 3). Since y
n−2
∈ D, we have y
n−2
= 4 or y
n−2
= 1. We
consider the two following cases.
Case 1. y
n−2
= 4, then (y
n−2
, y
n−1
, y
n
) = (4, 0, 0) and y
n−2
− 4 = 0. By (3) we
have (y
1
, , y
n−3
) ∈  (2, 3, , 2, 3, 2). Since (4, 0, 0) ∼ (3, 2, 3), by (1) we have (y
1

n−3
+ + (y
n−4
−3)3 + y
n−3
−3 = 0. (4)
Therefore, (y
1
, , y
n−3
) ∈ (2, 3, . , 2, 3, 3). By similar argument, we get y
n−3
= 0 or y
n−3
= 3.
+) If y
n−3
= 3 then (y
n−3
, y
n−2
, y
n−1
, y
n
) = (3, 1, 0, 0) and from (4) we get (y
1
, , y
n−4
) ∈

n−2
+ + (y
n−1
− 3)3 + y
n
− 2 = 0. (5)
This implies y
n
= 2 or y
n
= 5.
a) If y
n
= 2 then from (5), we have
(y
1
, , y
n−1
) ∈ (2, 3, . , 2, 3)= (x
1
, , x
n−1
).
This means
(y
1
, , y
n
) ∈ (x
1

). This implies
(y
1
, , y
n
) ∈ (x
1
, , x
n−2
, x
n−2
, 5).
Conversely, if (y
1
, , y
n
) ∈ (x
1
, , x
n−1
, 2)then we have immediately that (y
1
, , y
n
) ∈ (x
1
, , x
n
).
So we consider the following case

1
, , x
n
).
In fact, since (y
1
, , y
n−1
) ∈ (2, 3, . , 2, 3, 2, 2), we have
(y
1
− 2)3
n−2
+ (y
2
−3)3
n−3
+ + (y
n−2
−2)3 + y
n−1
−2 = 0. (6)
Therefore, y
n−1
= 2 or y
n−1
= 5.
a) If y
n−1
= 2 then from (6), we have (y

n−3
−3)3 + y
n−2
−1 = 0. (7)
Therefore, y
n−2
= 1 or y
n−2
= 4.
b1) If y
n−2
= 1 then from (7), we have (y
1
, , y
n−3
) ∈ (2, 3, , 2, 3) and (y
n−2
, y
n−1
, y
n
) =
(1, 5, 5). Since (1, 5, 5) ∼ (2, 3, 2), by (1) we have
(y
1
, , y
n
) ∈ (2, 3, . , 2, 3, 2)= (x
1
, , x

n−2
, x
n−2
) = (2, 3, , 2, 3, 2, 2).
We have the assertion of the proposition.
From Proposition 3 we have the following corollary.
Corrolary 1. Let x = (x
1
, x
2
, ) = (2, 3, 2, 3, ) ∈ D
∞
. For each n ∈ N, put s
n
=
n

i=1
3
−i
x
i
and
s
′
n
=
n

i=1

) =
10
2
5
, µ
2
(s
2
) =
110
2
10
, µ
2
(s
′
2
) =
105
2
10
.
ii) µ
n
(s
n
) =
10
2
5

) = P (X
1
= 2) =
10
2
5
.
For n = 2 we have (x
1
, x
2
) = {(2, 3), (3, 0)} and (x
′
1
, x
′
2
) = {(2, 2), (1, 5)}. Therefore,
µ
2
(s
2
) =
10
2
5
.
10
2
5

.
1
2
5
=
105
2
10
.
ii) By Proposition 3, we have
a) If n is even then
(x
1
, , x
n
) = (x
1
, , x
n−1
, 3)∪ (x
′
1
, , x
′
n−1
, 0).
b) If n is odd then
(x
1
, , x

)
=
10
2
5
µ
n−1
(s
n−1
) +
1
2
5
µ
n−1
(s
′
n−1
).
The corollary is proved. 
To have the recurrence formula of µ
n
(s
n
), we need the following proposition.
Proposition 4. Let x = (x
1
, x
2
, ) = (2, 3, 2, 3, ) ∈ D

) ∈ (x
1
, , x
n−1
, 2)∪ (x
1
, , x
n−2
, 1, 5)∪ (x
′
1
, , x
′
n−2
, 4, 5).
ii) If n is odd then (y
1
, , y
n
) ∈ (x
′
1
, , x
′
n
) = (2, 3, , 2, 3, 3) iff
(y
1
, , y
n

) ∈ (x
1
, , x
n−1
). Therefore,
by (1) we have
(y
1
, , y
n
) ∈ (2, 3, . , 2, 3, 2, 2)= (x
′
1
, , x
′
n
).
b) If (y
1
, , y
n
) ∈ (x
1
, , x
n−2
, 1, 5) then y
n
= 5, y
n−1
= 1 and

(y
1
, , y
n
) ∈ (2, 3, . , 2, 3, 2, 2)= (x
′
1
, , x
′
n
)
since (2, 2, 4, 5) ∼ (2, 3, 2, 2).
V.T.H. Thanh et al. / VNU Journal of Science, Mathematics - Physics 25 (2009) 57-68 63
Conversely, if (y
1
, , y
n
) ∈ (2, 3, , 2, 3, 2, 2) then we have
(y
1
−2)3
n−1
+ (y
2
− 3)3
n−2
+ + (y
n−1
− 2)3 + y
n

n
= 5 then y
n
− 2 = 3. Hence, from (9), we get
(y
1
− 2)3
n−2
+ (y
2
−3)3
n−3
+ + (y
n−2
−3)3 + y
n−1
−1 = 0. (10)
This implies y
n−1
= 1 or y
n−1
= 4.
b1) If y
n−1
= 1 then from (10) we have
(y
1
, , y
n−2
) ∈ (2, 3, . , 2, 3)= (x

n
) ∈ (x
1
, , x
n−2
, 4, 5).
ii) The case n is odd.
a) Clearly that if (y
1
, , y
n
) ∈ (x
1
, , x
n−1
, 3) then
(y
1
, , y
n
) ∈ (2, 3, . , 2, 3, 3)= (x
′
1
, , x
′
n
).
b) If (y
1
, , y

1
, , y
n
) ∈ (2, 3, . , 2, 3, 3)= (x
′
1
, , x
′
n
),
since (3, 1, 0) ∼ (2, 3, 3).
Conversely, if (y
1
, , y
n
) ∈ (x
′
1
, , x
′
n
) = (2, 3, , 2, 3, 3), then we have
(y
1
−2)3
n−1
+ (y
2
− 3)3
n−2

, , x
n−1
, 3).
b) If y
n
= 0 then y
n
− 3 = −1. Hence, from (11) we have
(y
1
− 2)3
n−2
+ (y
2
−3)3
n−3
+ + (y
n−2
−2)3 + y
n−1
−4 = 0. (12)
This implies y
n−1
= 1 or y
n−1
= 4.
b1) If y
n−1
= 1 then y
n−1

1
, , y
n−2
) ∈ (2, 3, . , 2, 3, 2)= (x
1
, , x
n−2
).
This implies (y
1
, , y
n
) ∈ (x
1
, , x
n−2
, 4, 0). The proposition is proved. 
From Proposition 4, we have the following corollary, which will be used to establish the recur-
rence formula of µ
n
(s
n
) for each n ∈ N.
Corrolary 2. Let x = (x
1
, x
2
, ) = (2, 3, 2, 3, ) ∈ D
∞
. For each n ∈ N, put s

n
) = (x
1
, , x
n−1
, x
n−1
), we have
µ
n
(s
′
n
) =
10
2
5
µ
n−1
(s
n−1
) +
5
2
10

µ
n−2
(s
n−2

, 4, 5).
Therefore,
µ
n
(s
′
n
) =
10
2
5
µ
n−1
(s
n−1
) +
1
2
5
.
5
2
5
µ
n−2
(s
n−2
) +
1
2

′
n−2
)].
b) If n is odd then
(x
′
1
, , x
′
n
) = (x
1
, , x
n−1
, 3)∪ (x
1
, , x
n−2
, 4, 0)∪(x
′
1
, , x
′
n−2
, 1, 0).
Therefore,
µ
n
(s
′

n−2
(s
′
n−2
)
=
10
2
5
µ
n−1
(s
n−1
) +
5
2
10
[µ
n−2
(s
n−2
) + µ
n−2
(s
′
n−2
)].
Hence,
µ
n

Corrolary 3. Let x = (x
1
, x
2
, ) = (2, 3, 2, 3, ) ∈ D
∞
. For each n ∈ N, put s
n
=
n

i=1
3
−i
x
i
. Then
we have
µ
n
(s
n
) =
10
2
5
µ
n−1
(s
n−1

) +
1
2
5
µ
n−1
(s
′
n−1
) (13)
µ
n−1
(s
′
n−1
) =
10
2
5
µ
n−2
(s
n−2
) +
5
2
10

µ
n−3

n−3
). (15)
V.T.H. Thanh et al. / VNU Journal of Science, Mathematics - Physics 25 (2009) 57-68 65
From (13), (14) and (15), the assertion of the corollary follows.
2.2 The proof of the main theorem
Lemma 2. Let x = (x
1
, x
2
, ) = (2, 3, 2, 3, ) ∈ D
∞
. For each n ∈ N, put s
n
=
n

i=1
3
−i
x
i
. Then
we have µ
n
(s
n
) ≥ µ
n
(t
n

2
5
}
for all t
1
∈ supp µ
1
. Assume that the lemma is true for n = k, i.e.,
µ
k
(s
k
) ≥ µ
k
(t
k
) for all t
k
∈ supp µ
k
.
We will show that the lemma is true for n = k +1. For any y = (y
1
, y
2
, ) ∈ D
∞
, put t
n
=

−(k+1)
= t
′
k
+ 4.3
−(k+1)
.
Therefore, by induction hypothesis, we have
µ
k+1
(t
k+1
) = µ
k
(t
k
)P (X
k+1
= 1) + µ
k
(t
′
k
)P (X
k+1
= 4)
 µ
k
(t
k

) ≥ µ
k+1
(t
k+1
).
Case 2. If y
k+1
= 0 (or 3), then by Lemma 1, t
k+1
has at most two representations
t
k+1
= t
k
+ 0.3
−(k+1)
= t
′
k
+ 3.3
−(k+1)
.
a) If y
k
= 0 (or 3), then (y
k
, y
k+1
) ∈ {(0, 0), (0, 3)}. Therefore, by Lemma 1 we have (y
′

k+1
= 0) + P (X
k
= 3)P (X
k+1
= 0)
+P (X
k
= 2)P (X
k+1
= 3) + P(X
k
= 5)P (X
k+1
= 3)
+P (X
k
= 0)P (X
k+1
= 3) + P(X
k
= 3)P (X
k+1
= 3)
+P (X
k
= 1)P (X
k+1
= 0) + P(X
k

5
+
10
2
5
.
10
2
5
+
1
2
5
.
10
2
5
+
10
2
5
.
10
2
5
+
5
2
5
.

µ
k
(s
k
) ≥
241
2
10
µ
k−1
(s
k−1
) = µ
k+1
(t
k+1
).
b) If y
k
= 4 (or 1), then (y
k
, y
k+1
) ∈ {(4, 0), (4, 3)}. Therefore, by Lemma 1 we have (y
′
1
, , y
′
k+1
) ∈

+P (X
k
= 1)P (X
k+1
= 3) + P(X
k
= 4)P (X
k+1
= 3)
+P (X
k
= 0)P (X
k+1
= 3) + P(X
k
= 1)P (X
k+1
= 0)
+P (X
k
= 3)P (X
k+1
= 3) + P(X
k
= 4)P (X
k+1
= 0)]
= µ
k−1
(s

10
2
5
+
1
2
5
.
10
2
5
+
5
2
5
.
1
2
5
+
10
2
5
.
10
2
5
+
5
2

2
10
µ
k−1
(s
k−1
) ≥ µ
k+1
(t
k+1
).
c) If y
k
= 2 (or 5), then (y
k
, y
k+1
) ∈ {(2, 0), (2, 3)}. Therefore, by Lemma 1 we have (y
′
1
, , y
′
k+1
) ∈
(y
1
, , y
k+1
) iff
(y

k
= 5)P (X
k+1
= 3)
+P (X
k
= 2)P (X
k+1
= 0) + P(X
k
= 5)P (X
k+1
= 0)
+P (X
k
= 1)P (X
k+1
= 3) + P(X
k
= 4)P (X
k+1
= 3)]
= µ
k−1
(s
k−1
)(
1
2
5

2
5
.
10
2
5
+
5
2
5
.
10
2
5
+
10
2
5
.
1
2
5
+
1
2
5
.
1
2
5

k−1
) ≥ µ
k+1
(t
k+1
).
Case 3. If y
k+1
= 2 (or 5). This case is proved similarly to the Case 2.
Therefore, the lemma is proved.
By resolving Fibonacci recurrence formula of µ
n
(s
n
) in Corollary 3, we have the following
corollary.
Corrolary 4. Let x = (x
1
, x
2
, ) = (2, 3, 2, 3, ) ∈ D
∞
. For each n ∈ N, put s
n
=
n

i=1
3
−i

[
√
145 cos(
arccos
427
59
√
145
3
) + 5] ≃ 0, 3435055158
X
2
=
−2
3.2
5
[
√
145 cos(
arccos
427
59
√
145
3
+
π
3
) + 5] ≃ 0, 04959875748
X

1
X
1
+ a
2
X
2
+ a
3
X
3
µ
2
(s
2
) = a
1
X
2
1
+ a
2
X
2
2
+ a
3
X
2
3

(s
3
) are the values in Corollary 1.
From Lemma 2, Corollary 3 and Proposition 2, we have
Theorem. Let µ is the 5−fold convolution of the standard Cant or measure, then the lower extreme
value of the loca l dimension of µ is
α
5
= |
log

2
3.2
5

√
145 cos(
arccos
427
59
√
145
3
) + 5

log 3
| ≃ 0, 972638.
68 V.T.H. Thanh et al. / VNU Journal of Science, Mathematics - Physics 25 (2009) 57-68
References
[1] K.J. Falconer, Fractal Geometry-Mathematical Founda tions and Applications (1990), John Wiley &