ON THE CENTRAL LIMIT THEOREM FOR SOME GAP SERIES

Katusi FUKUYAMA0. Introduction.
P In this paper, we shall be concerned with the central limit theorem for gap series
ak f (βk ω) where f is an α-Lipschitz continuous function (0 < α ≤ 1) with period 2πsatisfying
ZZ
2π2πf (x) dx = 0f 2 (x) dx = 2π,

and00

and the sequence {βk } of positive numbers has large gaps, i.e.
βk+1 /βk → ∞ask → ∞.

As to this problem, Kac [5] noticed the following result.
Theorem A. Let Ω = [0, 2π] and P be the normalized Lebesgue measure on Ω. If asequence {nk } of positive integers has large gaps and a sequence {ak } of real numbers
satisﬁes(0.1)

A2n = a21 + · · · + a2n → ∞then the distribution of A−1
ndistribution.

andan = o(An )asn → ∞,

Pn

k=1 ak f (nk ω) converges weakly to the standard normal

There are various extensions of this result. For example, Takahashi [8] extended
the class of functions of f and Berkes [1] studied the case in which {nk } have smaller
gaps. Our aim is to extend Theorem A to the case in which Ω = R and the probability
measure P on Ω is not necessarily absolutely continuous with respect to the Lebesgue
measure but may be singular measure satisfying one of the following conditions:(0.2)|Pb(u)| = O(|u|−ρ/2 )(0.3)

P {[ω, ω + h]} ≤ M hasρu → ∞,(ω ∈ Ω, h > 0).

Here M and ρ is positive constants and Pb is the characteristic function of P . One canﬁnd some important and interesting examples of probability measures satisfying above
conditions in [7], [11] and [12].
Let us now state our results.

Typeset by AMS-TEX

Theorem 1. Let P satisfy (0.2), {ak } satisfy
Pn (0.1), {γk } be arbitrary and {βk } have−1large gaps. Then the distribution of An
k=1 ak f (βk ω + γk ) converges weakly to thestandard normal distribution.
Theorem 2. Let P satisfy (0.3), {ak } satisfy (0.1), {γk } be arbitrary and {βk } havelarge gaps. Then,Pn for almost all t ∈ R with respect to the Lebesgue measure, the distribu−1
tion of An
k=1 ak f (tβk ω + γk ) converges weakly to the standard normal distribution.
Theorem 3. Let P satisfy (0.3), {ak } satisfy (0.1), {γk } be arbitrary and {φ(k)}satisfy(0.4)

φ(1) > 0andφ(k + 1) − φ(k) → ∞ask → ∞.

Then,
almost all x > 1 with respect to the Lebesgue measure, the distribution ofPfor
nφ(k)−1Anω + γk ) converges weakly to the standard normal distribution.
k=1 ak f (xRemark. Under the condition (0.4), the sequence {xφ(k) } has large gaps if x > 1.
√In the special case f (x) = 2 cos x, the gap series is called a lacunary trigonometricseries. As to the lacunary trigonometric series, the corresponding results to our theorems were ﬁrst proved by Takahashi [10], [9] and Kaufman [6] respectively, assuming
Hadamard’s gap condition:
βk+1 /βk ≥ q > 1.
Recently the author [2] extended these results to the case of weaker gap conditions andthe author [4] also proved functional law of the iterated logarithm both for lacunary
trigonometric and gap series assuming the same conditions on P .
1. Approximation by trigonometric polynomials.
We shall ﬁrst prove Theorem 1. In this section we shall approximate the gap series
by some series of trigonometric polynomials, by modifying the method of Takahashi.
Let L denote the absolute
constant which may changePline by √line.
√P∞
0Let us put f (t) = j=1 dj 2 cos(jt + γj ) and σn (t) = j≤n dj 2 cos(jt + γj0 ). Since
f is α-Lipschitz continuous, classical results in Fourier analysis claim
kf − σn k∞ ≤ Ln−α log n,(1.1)∞X(1.2)d2j ≤ L2 n−2α ,j=nkf k∞ , kσn k∞ ≤ L(1.3)(n ∈ N).

Using (1.1) we can easily prove¯ n¯¯X¯¯¯ak (f (βk ω + γk ) − σk1/α (βk ω + γk ))¯ = o(An )¯¯¯asn → ∞.

k=1

Thus it is suﬃcient to prove the central limit theorem for {σk1/α (βk ω + γk )}. Without
loss of generality, we may assume
β1 ≥ 1,

βk+1 /βk ≥ 4ψ(k),2 ≥ ψ(k + 1)/ψ(k) ≥ 1

andlim ψ(k) = ∞,k→∞for some sequence {ψ(k)}. We shall here prove the following convergence in probability:n¢1 X ¡(1.4)
ak σk1/α (βk ω + γk ) − σψρ/4 (k) (βk ω + γk ) → 0 as n → ∞.
An
k=100
By expanding into Fourier series and putting γk,j= jγk + γj0 , we have
Ã n!2X ¡¢E
ak σk1/α (βk ω + γk ) − σψρ/4 (k) (βk ω + γk )k=1≤Xk≤n+2Xa2kd2jψ ρ/4 (k)<j≤k1/α

XXX|ak1 ak2 |²=±1 1≤k1 ≤k2 ≤nψρ/4|dj1 dj2 |(kq )<jq ≤kq1/α(q=1,2)|βk2 j2 +²βk1 j1 |≥βk2¯¡¢¯× ¯E cos {βk2 j2 + ²βk1 j1 }ω + {γk002 ,j2 + ²γk001 ,j1 } ¯XX|ak1 ak2 ||dj1 dj2 |+21≤k1 ≤k2 ≤nψ ρ/4 (kq )<jq ≤kq1/α(q=1,2)1/αβk2 ≥2βk1 k1¯¡¢¯× ¯E cos {βk2 j2 − βk1 j1 }ω + {γk002 ,j2 − γk001 ,j1 } ¯XX|ak1 ak2 ||dj1 dj2 |+2

1≤k1 ≤k2 ≤nψ ρ/4 (kq )<jq ≤kq1/α(q=1,2)1/αβk2 <2βk1 k1|βk2 j2 +²βk1 j1 |≥βk1 /2¯¡¢¯
× ¯E cos {βk2 j2 − βk1 j1 }ω + {γk002 ,j2 − γk001 ,j1 } ¯XX|ak1 ak2 ||dj1 dj2 |+2

1≤k1 <k2 ≤nψ ρ/4 (kq )<jq ≤kq1/α(q=1,2)|βk2 j2 +²βk1 j1 |<βk1 /2¯¡¢¯× ¯E cos {βk2 j2 − βk1 j1 }ω + {γk002 ,j2 − γk001 ,j1 } ¯XXXXX(say).
+2+2+2+2=54231P2Since ψ(k) diverges to inﬁnity, trivially we have
1 = o(An ). Using the estimatep|E cos(βω + γ)| ≤ |Pb(β)|, (0.2) and βk2 ≥ βk1 βk2 , we obtain!1/2
ÃXXX
a2k1 a2k2≤L
d2j1 d2j2
2

1≤k1 ≤k2 ≤nÃ×ψ ρ/4 (kq )<jq ≤kq1/α(q=1,2)

XX

1≤k1 ≤k2 ≤n ψ ρ/4 (kq )<jq ≤k1/αq(q=1,2)!1/2(βk1 βk2 )−ρ/2≤ L

Xa2k1≤k≤n= o(A2n )X

Xd2j  X

X−ρ/2 βk1≤k≤n ψ ρ/4 (k)<j≤k1/αψ ρ/4 (k)<j≤k1/αk 1/α 2−kρ/2 = o(A2n ).

k≤np1/α,wehave
βj−βj
≥
β−βk≥
β/2
≥βk1P
βk2 . Thus we can prove
k2k1kkk2
1
21 123P
22=o(A)inthesameway
as
the
proof
of=o(A).
Innn
324 , we have

InPP−1/αβk2 j2 − βk1 j1 ≥ βk1 /2 ≥ k1−1/αβk2 /4 ≥ k1pβk1 βk2 /2.

Thus we getX4= o(A2n )XXX

k 1/α (βk )−ρ/2 = o(A2n )k≤n ψ ρ/4 (k)<j≤k1/αk 2/α 2−kρ/2 = o(A2n ).

k≤nP
In 5 we have |βk2 j2 /βk1 − j1 | < 1/2. From this, introducing the notation [x]∗ = {n ∈N : |x − n| < 1/2}, we haveX5≤X|ak1 ak2 | 1≤k1 <k2 ≤n≤X

1/2 Xd2j2 |ak1 ak2 | 1≤k1 <k2 ≤n11/αψ(k2 )<j2 ≤k2

ψ(k2 )<j2 ≤k2

1/2 

X

d2[βk j2 /βk ]∗ 21/α1/2Xd2j 

1/2X

d2j .

βk2 /βk1 ≤jψ(k1 )<jHere, using (1.2) and βk2 /βk1 ≥ 2k2 −k1 , we get
X5



X

≤L

|ak1 ak2 | 1≤k1 <k2 ≤n=Ln−1X

2−iαi=1

≤L

n−1
X

2i=1

≤ L

nX
k=1

n−iX
k=1a2kX
ψ(k)<j2−α(k2 −k1 )1/2X|ak ai+k | 

k=1−iα d2j ψ(k1 )<jn−iX1/2X

d2j 

ψ(k)<ja2kX

1/2 Ã

d2j 

d2j !1/2a2i+k

k=1

ψ(k)<j1/2 Ãn−i
XnX

!1/2
a2k= o(A2n ).

k=1

Thus (1.4) is veriﬁed.
By (1.4), it is suﬃcient to prove the central limit theorem for {σψρ/4 (k) (βk ω + γk )}.

2. Method of weakly multiplicative systems.
To prove the central limit theorem, we shall use the method of weakly multiplicativesystems.
For the sequence {ξk } of random variables, putXbk1 ,...,kr = Eξk1 . . . ξkr
andkBr k =|bk1 ,...,kr |.
k1 <···<krIf supr∈N kBr k1/r < ∞, we say that {ξk } is a weakly multiplicative system.
We use the following theorem in [3].
Theorem B. Let {ξn } be a weakly multiplicative system in the above sense satisfyingkξn k∞ ≤ B(n ∈ N),¢¡ 2lim E (ξi − 1)(ξj2 − 1) = 0,(2.1)
(2.2)

i+j→∞
i6=jand {an } satisfy (0.1). Then the distributions of {A−1nthe standard normal distribution.

Pn

k=1 ak ξk } converges weakly toPut ξk = σψρ/4 (k) (βk ω + γk ). We get (2.1) by (1.3). Now let us verify the bound ofkBr k and (2.2). Usual calculation yields
X √ r|bk1 ,...,kr | =2 |E cos(βir jr ω + γi00r ,jr ) . . . cos(βi1 jr ω + γi001 ,j1 )|jq ≤ψ(kq )(q=1,...,r)≤LX¡βkr jr − βkr−1 jr−1 − · · · − βk1 j1jq ≤ψ ρ/4 (kq )(q=1,...,r)¢−ρ/2ρ/4 ¡¢−ρ/2βkr − βkr−1 ψ(kr−1 ) − · · · − βk1 ψ(k1 ).

≤ L (ψ(kr ) . . . ψ(k1 ))Using the condition on {βk } and {ψk }, we have
βkr − βkr−1 ψ(kr−1 ) − · · · − βk1 ψ(k1 ) ≥ βkr (1 − 4−1 − · · · − 4−kr )≥ L4kr ψ(kr − 1) . . . ψ(1).
Thus we have |bk1 ,...,kr | ≤ L2−kr ρ , and this yields
kBr k ≤ LX
k1 <···<kr−kr ρ2∞X1k r−1 2−kρ ≤ Lr .
≤L(r − 1)!
k=r

It implies that {ξk } is a weakly multiplicative system.
Next let us verify (2.2). We have¯ ¯¯¯XX¯¯−ρ/42 ¯¯E ξk2 −d≤2βk≤ L2−ρk ,j ¯¯¯¯j≤ψ ρ/4 (k)j1 ≤j2 ≤ψ ρ/4 (k)and in case k1 < k2 we obtain¯ µX¯¯E ξk2 −
1¯¶µ
d2j1

j1 ≤ψ ρ/4 (k1 )X≤4X

ξk22 −d2j2

j2 ≤ψ ρ/4 (k2 )X|dj1 di1 dj2 di2 |jq ≤iq ≤ψ ρ/4 (kq )(q=1,2)¶¯¯¯¯²q =±1 (q=1,2)²=±1²6=−1 if jq =iq|E cos ({βk2 (i2 + ²2 j2 ) + ²βk1 (i1 + ²1 j1 )}X≤L¢¯+{γk002 ,i2 + ²2 γk002 ,j2 + ²γk001 ,i1 + ²²1 γk001 ,j1 } ¯(βk2 )−ρ/2 ≤ L2−ρk2 .

jq ≤iq ≤ψ ρ/4 (kq )(q=1,2)

By these estimates we have¯¯¯E(ξk2 − 1)(ξk2 − 1)¯12¯¯¯ µ¶µ¶¯XX¯¯≤ ¯¯E ξk21 −d2j1ξk22 −d2j2 ¯¯¯¯
j1 ≤ψ ρ/4 (k1 )j2 ≤ψ ρ/4 (k2 )
¯¯¯
¶¯
µXX¯¯
222+ ¯¯
dj2 ¯¯
dj1 E ξk2 −¯j1 ≤ψρ/4 (k1 )¯j2 ≤ψ ρ/4 (k2 )
¯
¯¯µ¶¯
XXX
X¯¯
222+ ¯¯
dj2 E ξk1 −dj1 ¯¯ +d2j2
d2j1¯j2 ≤ψρ/4 (k2 )¯ j1 ≤ψρ/4 (k1 )j1 ≤ψ ρ/4 (k1 )j2 ≤ψ ρ/4 (k2 )³´≤ L 2−ρ(k1 ∨k2 ) + ψ −ρα/4 (k1 )2−ρk2 + ψ −ρα/4 (k2 )2−ρk1 + ψ −ρα/4 (k1 )ψ −ρα/4 (k2 )−→ 0ask1 + k2 → ∞.

This completes the proof.
3. Lemmas used in the proofs of Theorems 2 and 3.
In the proofs of Theorems 2 and 3 we use the following lemmas due to Takahashi [9]and Kaufman [6], respectively.
Lemma C. Let P satisfy (0.3). Then there exists a constant D depending only on ρand M satisfying
Zv+1v|Pb(ut)| dt ≤ D|u|−ρ/2(u, v ∈ R).

Lemma D. Let P satisfy (0.3). There exists a constant D depending only on ρ andM , such that for any g ∈ C 2 [v, v + 1] satisfying min g 00 (x) > 0,Z v+1x∈[v,v+1]|Pb(g(x))|dx ≤ Dv

³min´−ρ/(2+4ρ)g 00 (x).

x∈[v,v+1]

Using these lemmas and replacing the all estimates of |Pb(β)| appearing in the proof
Z v+1
Z v+1
|Pb(p(x))| dx where p(x) is polynomials, we|Pb(tβ)| dt orof Theorem 1 byvv

can prove Theorems 2 and 3. For details, see [3].
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Institute of Mathematics, University of Tsukuba, Tsukuba-shi, 305, Japan