VOL. LXIX 1995 FASC. 1
A MULTIFRACTAL ANALYSIS
OF AN INTERESTING CLASS OF MEASURES
BY
ANTONIS B I S B A S (IRAKLION)
1. Introduction. Let µn = pnδ(0) + (1 − pn)δ(1/2n), n = 1, 2, . . . , where pn ∈ [0, 1] and δ(x) denotes the probability atom at x. The infinite convolution product of the µn converges in the weak* sense to a probability measure µ on [0, 1] which is known as a coin tossing measure [9],
µ =n=1∗∞ µn.
Let x = P∞
n=1εn(x)/2n, where εn(x) ∈ {0, 1}, be the 2-adic expansion of x ∈ [0, 1]. It is not difficult to see that if
dνa,N =
N
Y
n=1
(1 + anrn(x))dλ, N = 1, 2, . . . ,
where a = (an)n≥1, λ denotes the Lebesgue measure, rn(x) = 1 − 2εn(x) is the nth Rademacher function and pn= (1 + an)/2, then
N →∞lim νa,N = µa
in the weak* sense and µ = µa (see also [12]). So we have two ways to describe the same measure. In this work we shall use the second way. The characterizations of the sequences (an)n≥1which give continuous or singular measures are given in [5], [6], [9], [11]. In a previous work [4] we have proved that
lim inf
n→∞
log µa(En,k(x))
−n log 2 = δa µa-a.e., where
δa = 1 − lim sup
N →∞
1 N log 4
N
X
n=1
log[(1 + an)1+an(1 − an)1−an]
1991 Mathematics Subject Classification: Primary 28A78.
Key words and phrases: Hausdorff dimension, multifractal, Rademacher Riesz pro- ducts.
[37]
and En,k(x) is the segment [k/2n, (k + 1)/2n) containing x, for some k = 0, 1, 2, . . . , 2n− 1. From this relation we deduce that µa is δa-dimensional [8] and dim µa = δa, where dim µa = inf{dim E : µa(E) = 1} and dim E denotes the Hausdorff dimension (HD) of the Borel set E (see [1]). If there are infinitely many c ∈ R such that dim Ec> 0, where
Ec=
x : lim inf
n→∞
log µa(En,k(x))
−n log 2 = c
,
then we say that µa is multifractal [8], [10]. We have seen [2], [3] that some special cases of Markov measures are multifractal. In Section 2 we shall give a necessary and sufficient condition for µa to be multifractal under the condition supn|an| < 1. In Section 3 we give an application which permits us to give a lower bound for the HD of a set Mβ(b), where
(1) Mβ(b) =
x : lim inf
N →∞
1 N
N
X
n=1
βnεn(x) ≤ b
,
b, βn∈ R, β = (βn)n≥1, |βn| ≤ M , M > 0. In some special cases our method gives equality.
2. A multifractal analysis. We need the following lemma, which can be deduced from [4]:
Lemma 1. Let γ = (γn)n≥1 and µγ be the corresponding coin tossing measure. If supn|an| < 1, then
lim sup
N →∞
1 N
N
X
n=1
log (1 + anrn(x))
= lim sup
N →∞
1 2N
N
X
n=1
log [(1 + an)1+γn(1 − an)1−γn] µγ-a.e.
Theorem 1. If the sequence a = (an)n≥1 is such that supn|an| < 1, then µa is multifractal if and only if
lim sup
N →∞
1 N
N
X
n=1
|an| > 0.
P r o o f. (i) Suppose that lim supN →∞(1/N )PN
n=1|an| = 0. Then by the Cauchy–Schwarz inequality we have equivalently limN →∞(1/N )PN
n=1|an|2
= 0. Since µa(EN,k(x)) = νa,N(EN,k(x)) = 2−NQN
n=1(1 + anrn(x)) and Ec =
x : 1 − lim sup
N →∞
1 N log 4
N
X
n=1
log(1 − a2n) + rn(x) log 1 + an 1 − an
= c
,
using the uniform convergence of the Taylor series for the function log(1+x),
|x| ≤ supn|an|, we see that Ec = ∅ if c 6= 1 and Ec= [0, 1] if c = 1 and so µa is not multifractal.
(ii) Suppose that
lim sup
N →∞
1 N
N
X
n=1
|an| > 0.
It is clear that c must be such that
(2) 1 − lim sup
N →∞
λN ≤ c ≤ 1 − lim sup
N →∞
κN, where
λN = 1 N log 2
N
X
n=1
log(1 + |an|) and κN = 1 N log 2
N
X
n=1
log(1 − |an|), otherwise the set Ec is empty. We define the function
f (y) = lim sup
N →∞
[κN + y(λN − κN)], y ∈ [0, 1].
If 0 ≤ y0< y ≤ 1, then using the properties of lim sup we obtain 0 ≤ f (y) − f (y0) ≤ (y − y0) lim sup
N →∞
(λN − κN).
This implies that f (y) is continuous on [0, 1]. Since f (0) = lim supN →∞κN
and f (1) = lim supN →∞λN, from the intermediate value theorem there is γ0∈ (−1, 1) such that f ((1 + γ0)/2) = 1 − c ∈ (f (0), f (1)), f (0) ≤ 0 < f (1).
We consider the measure µγ, where γ = (γn)n≥1 with γn = γ0sgn log 1 + an
1 − an
(sgn is the sign function, sgn 0 = 0). From Lemma 1 we have lim sup
N →∞
1 N log 4
N
X
n=1
log(1 − a2n) + rn(x) log 1 + an
1 − an
= lim sup
N →∞
1 N log 4
N
X
n=1
log (1 − a2n) + γ0log 1 + |an| 1 − |an|
= lim sup
N →∞
κN +1 + γ0
2 (λN − κN)
= 1 − c µγ-a.e.
From this we get µγ(Ec) = 1 and so dim Ec≥ dim µγ = δγ > 0, for infinitely many c ∈ R (f (1) > 0). This means that µa is multifractal.
3. Application. We consider the set of (1). It is clear that we can find a sequence a = (an)n≥1 such that
βn = log 1 + an 1 − an
, with supn|an| < 1. We also have
N →∞lim 1 N
N
X
n=1
|βn| = 0 ⇔ lim
N →∞
1 N
N
X
n=1
|an| = 0.
If limN →∞(1/N )PN
n=1|βn| = 0, then dim Mβ(b) is 0 if b < 0 and is 1 if b ≥ 0.
Suppose that lim supN →∞(1/N )PN
n=1|an| > 0. From (1) we see that b must be such that
lim inf
N →∞
1 N
N
X
n=1 βn<0
βn ≤ b ≤ lim inf
N →∞
1 N
N
X
n=1 βn>0
βn,
or equivalently, (3) lim inf
N →∞
1 N
N
X
n=1
log 1 + an
1 + |an|
≤ b ≤ lim inf
N →∞
1 N
N
X
n=1
log 1 + an
1 − |an|
, Otherwise dim Mβ(b) = 0 or 1.
Let b > 0 and
c = 1 + b
log 2 − lim sup
N →∞
1 N log 2
N
X
n=1
log(1 + an).
Using elementary properties of lim sup, lim inf and (3) we easily see that c satisfies (2). From the proof of Theorem 1 we have
dim Ec ≥ dim µγ, where
Ec=
x : 1 − lim sup
N →∞
1 N log 4
N
X
n=1
[log (1 − a2n) + (1 − 2εn(x))βn] = c
and γ = (γn)n≥1, γn= γ0sgn βn with lim sup
N →∞
κN +1 + γ0
2 (λN − κN)
= 1 − c.
If x ∈ Ec then
lim inf
N →∞
1 N
N
X
n=1
βnεn(x) ≤ b
and so Ec⊂ Mβ(b), which means that dim Mβ(b) ≥ dim µγ.
R e m a r k. If βn > β06= 0, n = 1, 2, . . . , b > 0, using the above method we get
β0= log 1 + a0
1 − a0
, c = 1 + b
log 2− 1
log 2log (1 + a0), γn= γ0sgn βn
and
1 − c = log (1 − a0) log 2 + β0
log 4(1 + γ0).
This gives
b β0
= 1 − γ0
2 . Since
Mβ(b) =
x : lim inf
N →∞
1 N
N
X
n=1
εn(x) ≤ b β0
and
dim µγ = 1 − 1 log 2
b β0
log 2b β0
+
1 − b
β0
log
2
1 − b
β0
, using Eggleston’s Theorem [7] we get dim Mβ(b) = dim µγ for b ≤ β0/2.
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DEPARTMENT OF MATHEMATICS UNIVERSITY OF CRETE
IRAKLION 71409, GREECE
E-mail: BISBAS@TALOS.CC.UCH.GR
Re¸cu par la R´edaction le 7.6.1994