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# The joint distribution of multiplicative functions.

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Том 10 Выпуск 1 (2009)
УДК 519.14
THE JOINT DISTRIBUTION OF MULTIPLICATIVE
FUNCTIONS
A. Laurinikas
Аннотация
In the paper, the existene of a joint limit distribution for a real multipliative and a omplexvalued multipliative funtions is onsidered. For one
lass of multipliative funtions, the neessary and suient onditions for
the existene of a limit distribution are obtained.
Introdution. Denote by N, N0 , Z, R and C the sets of all positive integers,
nonnegative integers, integers, real and omplex numbers, respetively. A funtion
g : N ? C is alled multipliative if g 6? 0, m ? N, and g(mn) = g(m)g(n) for
all m, n ? N, (m, n) = 1. Clearly, we have that g(1) = 1. A funtion f : N ? C
is said to be additive if f (mn) = f (m) + f (n) for all m, n ? N, (m, n) = 1. This
denition implies f (1) = 0. The main problem of the probabilisti number theory
are asymptoti distribution laws for additive and multipliative funtions. We reall
some results on multipliative funtions. Let, for n ? N,
1
#{1 ? m ? n : ...},
n
where in plae of dots a ondition satised by m is to be written. As usual, by B(S)
denote the lass of Borel sets of the spae S , and by p a prime number. The rst
probabilisti result for multipliative funtions was obtained by P. Erd
os in [6?. Let
Pn and P be two probability measures on (R, B(R)). We say that Pn onverges m
weakly to P as n ? ? if Pn onverges weakly to P and limn?? Pn ({0}) = P ({0}).
In the ase P ({0}) = 1, the latter ondition is ommited. We use the notation
(
u if |u| ? 1,
||u|| =
1 if |u| > 1.
?n (...) =
Теорема
ty measure
1. ([6?). Let g(m) ? 0 be a multipliative funtion. Then the probabili-
?n (g(m) ? A),
A ? B(R),
(1)
onverges mweakly to a ertain probability measure P on (R, B(R)), P ({0}) 6= 1, as
n ? ?, if and only if the series
X ||g(p) ? 1||
X ||g(p) ? 1||2
(2)
,
p
p
p
p
onverge.
A. LAURINCIKAS
42
The rst result on real multipliative funtions of an arbitrary sign belongs to A.
Bakstys [1?. We reall that a probability measure P on (R, B(R)) is alled symmetri
if, for some a ? R,
P (??, a) = 1 ? P (??, a].
2. ([1?). Let g(m) be a real multipliative funtion. Then the probability measure (1) onverges mweakly to a ertain nonsymmetri probability measure
on (R, B(R)), as n ? ?, if and only if the series (2) and
Теорема
X 1
p
g(p)<0
onverge, and there exists ? ? N suh that g(2?) 6= ?1.
The problem of the existene of limit distribution for real multipliative funtions
was ompletely solved in [14?. Let, for a ? R and A ? B(R),
(
1 if a ? A,
Pa (A) =
0 if a ?
/ A.
Теорема 3. ([14?). Let g(m) be a real multipliative funtion. Then the probability measure (1) onverges mweakly to a ertain probability measure P on (R, B(R)),
P 6= Pa for every a ? R, as n ? ?, if and only if the series
X
g(p)6=0
| log |g(p)||<1
log |g(p)|
,
p
X 1 log2 |g(p)|
,
p 1 + log2 |g(p)|
g(p)6=0
X 1
p
g(p)=0
onverge.
Now let g(m) be a omplexvalued multipliative funtion. Dene
(
g(p)
if g(p) 6= 0,
|g(p)|
ug (p) =
0
if g(p) = 0,
and
(
log |g(p)| if e?1 ? |g(p)| ? e,
vg (p) =
if |g(p)| < e?1 or |g(p)| > e.
1
Let Pn and P be probability measures on (C, B(C)). We say that Pn onverges
weakly in the sense of C to P as n ? ? if Pn onverges weakly to P as n ? ?,
and, additionally, limn?? Pn ({0}) = P ({0}).
4. ([4?). Let g(m) be a omplexvalued multipliative funtion. Then
the probability measure
?n (g(m) ? A), A ? B(C),
Теорема
THE JOINT DISTRIBUTION OF MULTIPLICATIVE FUNCTIONS
43
onverges weakly to a ertain probability measure P on (C, B(C)), P ({0}) 6= 1, as
n ? ?, if and only if the following two hypotheses hold:
10 The series
X vg2 (p)
X vg (p)
,
p
p
p
p
onverge;
20 Either, for all m ? N and all t ? R,
?it
X 1 ? Re um
g (p)p
= +?,
p
p
or there exists at least one m ? N suh that the series
X 1 ? um
g (p)
p
onverges.
p
The joint distribution of real multipliative funtions was disussed in [7?. We
will state the result for two real multipliative funtions g1 (m) and g2 (m). In [7? limit
theorems are stated in terms of distribution funtions, however, it is not diult to
use for them a language of probability measures. Let Pn and P be a probability
measures on (R2 , B(R2 )). We say that Pn onverges mweakly to P as n ? ?, if
Pn onverges weakly to P and
lim Pn (R Ч {0}) = P (R Ч {0}),
n??
lim Pn ({0} Ч R) = P ({0} Ч R),
n??
lim Pn ({0} Ч {0}) = P ({0} Ч {0}).
n??
Теорема 5. ([7?). Let g1 (m) and g2 (m) be real multipliative funtions. Then
the probability measure
?n ((g1 (m), g2 (m)) ? A),
A ? B(R2 ),
onverges mweakly to a ertain probability measure P on (R2 , B(R2 )), P (R Ч A) 6=
Pa (A), P (A Ч R) 6= Pb (A), A ? B(R2 ), for every a, b ? R, as n ? ?, if and only if
the series
X
gj (p)6=0
| log |gj (p)||<1
onverge.
log |gj (p)|
,
p
j = 1, 2,
2
X
X 1 log2 |gj (p)|
,
2
p
1
+
log
|g
(p)|
j
j=1
g(p)6=0
2
X
X 1
p
j=1
gj (p)=0
A. LAURINCIKAS
44
A twodimensional limit theorem for omplexvalued multipliative funtion was
proved in[9?.
The aim of this paper is to prove a twodimensional limit theorem for a real and
a omplexvalued multipliative funtions. Let, for brevity, X = R Ч C. Let Pn and
P be probability measures on (X, B(X)). We say that Pn onverges mweakly in the
sense of X to P as n ? ? if Pn onverges weakly to P as n ? ?, and
lim Pn (R Ч {0}) = P (R Ч {0}),
n??
lim Pn ({0} Ч C) = P ({0} Ч C),
n??
lim Pn ({0} Ч {0}) = P ({0} Ч {0}).
n??
Dene PR (A) = P (A Ч C), A ? B(R), and PC (A) = P (R Ч A), A ? B(C).
6. . Let g1 (m) be a real and g2 (m) be a omplexvalued multipliative
funtions suh that the series
X 1
X 1
and
p
p
Теорема
g1 (p)>0
g1 (p)<0
do not diverge simultaneously. Then the probability measure
def
Pn (A) = ?n ((g1 (m), g2 (m)) ? A),
A ? B(X),
onverges mweakly in the sense of X to a ertain probability measure P on
(X, B(X)), PR 6= Pa for every a ? R and PC ({0}) 6= 1, as n ? ?, if and only if the
following hypotheses are satised:
10
X
X 1 log2 |g1 (p)|
X 1
log |g1 (p)|
,
,
p
p 1 + log2 |g1(p)|
p
g1 (p)6=0
| log |g1 (p)||<1
onverge;
20 The series
g1 (p)=0
g1 (p)6=0
X vg (p)
2
,
p
p
onverge;
30 Either, for all k ? N and all u ? R,
X vg2 (p)
2
p
p
X 1 ? Re ukg (p)p?iu
2
= +?,
p
p
or there exists at least one k ? N suh that the series
X 1 ? ukg (p)
2
p
onverges.
p
THE JOINT DISTRIBUTION OF MULTIPLICATIVE FUNCTIONS
45
Charateristi transforms. For the proof of Theorem 6 we will apply
the method of harateristi transforms. Therefore, we will reall denitions of
harateristi transforms of probability measures on (R, B(R)), (C, B(C)) and
(X, B(X)) as well as the orrespondene between probability measures and their
harateristi transforms.
Let P be a probability measure on (R, B(R)). The funtions
Z
vl (t) =
|x|it sgnl xdP,
t ? R, l = 0, 1,
R\{0}
are alled the harateristi transforms of P . The measure P is uniquely determined
by its harateristi transforms.
1. . Let Pn be a probability measure on (R, B(R)) with its harateristi
transforms vln (t), l = 0, 1. Suppose that
Лемма
lim vln (t) = vl (t),
t ? R, l = 0, 1,
n??
where the funtions vl (t), l = 0, 1, are ontinuous at t = 0. Then on (R, B(R)) there
exists a probability measure P suh that Pn onverges mweakly to P as n ? ?. In
this ase, vl (t), l = 0, 1, are the harateristi transforms of the measure P.
Now, onversely, suppose that the probability measure Pn onverges mweakly to
some probability measure P on (R, B(R)) as n ? ?. Then
lim vln (t) = vl (t),
n??
t ? R,
l = 0, 1,
where vl (t), l = 0, 1, are the harateristi transforms of the measure P.
Proof of the lemma is given in [11?.
Now let Pn and P be probability measures on (C, B(C)). The funtion
Z
v(t, k) =
|z|it eik arg z dP, t ? R, k ? Z,
C\{0}
is alled the harateristi transforms of P . The measure P is uniquely determined
by its harateristi transform.
2. . Let Pn be a probability measure on (C, B(C)) with its harateristi
transform wn (t, k). Suppose that
Лемма
lim wn (t, k) = w(t, k),
n??
t ? R,
k ? Z,
where the funtion v(t, 0) is ontinuous at t=0. Then on (C, B(C)) there exists a
probability measure P suh that Pn onverges weakly in the sense of C to P as
n ? ?. In this ase, v(t, k) is the harateristi transform of the measure P.
A. LAURINCIKAS
46
If Pn onverges weakly in the sense of C to some probability measure P on
(C, B(C)) as n ? ?, then
lim vn (t, k) = v(t, k),
t ? R,
n??
k ? Z,
where v(t, k) is the harateristi transform of P.
Proof of the lemma is given in [8?, see also [10?.
Finally, let P be a probability measure on (X, B(X)). Then the funtions
vl (t) =
Z
|x|it sgnl xdPR ,
t ? R, l = 0, 1,
R\{0}
vl (t, k) =
Z
r it eik? dPC ,
t ? R,
k ? Z,
C\{0}
and
vm (t1 , t2 , k) =
Z
|x|it1 sgnm xr it2 eik? dP,
t1 , t2 ? R,
k ? Z,
m = 0, 1,
X
where the last integrand is zero if x = 0 or r = 0, are alled the harateristi
transforms of P .
3. . A probability measure P on (X, B(X)) is uniquely determined by
its harateristi transforms (vl (t), v(t, k), vm (t1 , t2 , k), l = 0, 1, m = 0, 1).
Лемма
Лемма 4. . Let Pn be a probability measure on (X, B(X)) with its harateristi
transforms (vln (t), vn (t, k), vmn (t1 , t2 , k), l = 0, 1, m = 0, 1), n ? N. Suppose that
lim vln (t) = vl (t),
n??
lim vn (t, k) = v(t, k),
n??
l = 0, 1,
t ? R,
t ? R,
k ? Z,
and
lim vmn (t1 , t2 , k) = vm (t1 , t2 , k),
n??
m = 0, 1,
t1 , t2 ? R,
k ? Z,
where the funtions vl (t), l = 0, 1, v(t, 0), vm(0, t2 , 0) and vm (t1 , 0, 0), m = 0, 1, are
ontinuous at t = 0, t2 = 0 and t1 = 0, respetively. Then on (X, B(X)) there
exists a probability measure P suh that Pn onverges mweakly in the sense of X
to P as n ? ?. In this ase, (vl (t), v(t, k), vm(t1 , t2 , k), l = 0, 1, m = 0, 1) are the
harateristi transforms of the measure P.
THE JOINT DISTRIBUTION OF MULTIPLICATIVE FUNCTIONS
47
Лемма 5. . Let Pn and (vln (t), vn (t, k), vmn (t1 , t2 , k), l = 0, 1, m = 0, 1) be the
same as in Lemma 4. Suppose that Pn onverges mweakly in the sense of X to some
probability measure P on (X, B(X)) as n ? ?. Then
lim vln (t) = vl (t),
l = 0, 1,
n??
lim vn (t, k) = v(t, k),
t ? R,
n??
t ? R,
k ? Z,
and
lim vmn (t1 , t2 , k) = vm (t1 , t2 , k),
m = 0, 1,
n??
t1 , t2 ? R,
k ? Z,
where (vl (t), v(t, k), vm (t1 , t2 , k), l = 0, 1, m = 0, 1) are the harateristi transforms
of the measure P.
Proofs of Lemmas 35 are given in [12?.
Let g(m) be a multipliative
funtion. We say that the funtion g(m) has the mean value M(g) if
Mean values of multipliative funtions.
1X
g(m) = M(g).
x?? x
m?x
lim
In this setion, we will reall some known results on mean values of multipliative
funtions.
6. . In order that the mean value of the multipliative funtion g(m),
|g(m)| ? 1, exist and be zero, it is neessary and suient that one of the following
hypotheses should be satised:
10 For every u ? R,
X 1 ? Reg(p)p?iu
= +?.
p
p
Лемма
20 There exists a number u0 ? R suh that the series
X 1 ? Reg(p)p?iu0
p
p
= +?
onverges, and 2?riu0 g(2r ) = ?1 for all r ? N.
The lemma is a orollary of the result of [4?, see also [5?.
Лемма 7. . Let g(m) = g(m; t1 , ..., tr ) be a multipliative funtion, |g(m)| ? 1.
Suppose that there exists a funtion a(t1 , ..., tr ) suh that the series
X 1 ? Reg(p)e?ia(t1 ,...,tr )
p
p
A. LAURINCIKAS
48
onverges uniformly in tj , |tj | ? T, j = 1, ..., r . Then, as x ? ?,
xia(t1 ,...,tr )
1X
g(m) =
Ч
x m?x
1 + ia(t1 , ..., tr )
?
Y
X
1
g(p? )
Ч
1?
1+
+ o(1)
?(1+ia(t1 ,...,tr ))
p
p
?=1
p?x
uniformly in tj , |tj | ? T, j = 1, ..., r .
The assertion of the lemma is a partiular ase of a result from [13?.
z
Proof of Theorem 6. Suieny. We suppose, for onveniene, that 0 = 0
for all z ? C.
Denote by (vln (t), vn (t, k), vmn (t1 , t2 , k), l = 0, 1, m = 0, 1) the harateristi
transforms of the measure Pn in Theorem 6. Then we have that
n
1X
vln (t) =
|g1 (m)|it sgnl g1 (m),
n m=1
and
t ? R, l = 0, 1,
n
1X
vn (t, k) =
|g2 (m)|it eik arg g2 (m) ,
n m=1
t ? R,
n
1X
vrn (t1 , t2 , k) =
|g1(m)|it1 sgnr g1 (m)|g2 (m)|it2 eik arg g2 (m) ,
n m=1
k ? Z,
t1 , t2 ? R, k ? Z,
r = 0, 1.
Clearly, it sues to onsider wrn (t1 , t2 , k) for k ? N0 .
In view of Theorem 3 we have that the probability measure
?n (g1 (m) ? A),
A ? B(R),
onverges mweakly to a ertain probability measure P on (R, B(R)), P 6= Pa for
every a ? R, as n ? ?. Therefore, by the seond part of Lemma 1, we obtain that
lim vln (t) = vl (t),
n??
t ? R,
l = 0, 1,
(3)
where vl (t), l = 0, 1, are the harateristi transforms of P . Sine the harateristi
transforms are ontinuous funtions, vl (t), l = 0, 1, are ontinuous at t = 0.
Similarly, in view of Theorem 4 we have that the probability measure
?n (g2 (m) ? A),
A ? B(C),
onverges weakly to a ertain probability measure P on (C, B(C)), P ({0}) 6= 1, as
n ? ?. Therefore, by the seond part of Lemma 2 we have that
lim vn (t, k) = v(t, k),
n??
t ? R,
k ? Z,
(4)
THE JOINT DISTRIBUTION OF MULTIPLICATIVE FUNCTIONS
49
where w(t, k) is the harateristi transform of the limit measure P . Moreover, the
funtion v(t, 0) is ontinuous at t = 0.
It remains to onsider the funtion vrn (t1 , t2 , k).
From the hypothesis of the theorem we have that
X
g1 (p)g2 (p)=0
Let, for brevity,
P?
1
< ?.
p
(5)
means that the summation runs over those p for whih
p
g1 (p)g2 (p) 6= 0. Consider the series
def
Sr (t1 , t2 , k) =
X ? 1 ? Re sgnr g1 (p)|g1 (p)|it1 |g2(p)|it2 eik arg g2 (p)
.
p
p
First we study the ase r = 0. The hypotheses 10 and 20 show that the series
X ? 1 ? Re|gj (p)|itj
X ? 1 ? cos(tj log |gj (p)|)
X ? sin2 ((tj /2) log |gj (p)|)
=
=2
p
p
p
p
p
p
(6)
onverges uniformly in tj , |tj | ? t0 , for every xed t0 > 0, j = 1, 2. Therefore, in
virtue of the inequality
1 ? Rez1 z2 ? 2(1 ? Rez1 ) + 2(1 ? Rez2 )
(7)
valid for |zj | ? 1, j = 1, 2, see [2?, we have that the series S0 (t1 , t2 , 0) onverges
uniformly in tj , |tj | ? t0 . This, (7) and Lemma 7 show that uniformly in tj , |tj | ? t0 ,
as n ? ?,
?
Y
X
1
|g1 (p? )|it1 |g2 (p? )|it2
v0n (t1 , t2 , 0) =
1?
+ o(1).
(8)
1+
?
p
p
?=1
p?n
Now suppose that there exists k ? N suh that the series
X 1 ? ukg (p)
2
p
p
(9)
onverges. Then using the hypothesis 20 and reasoning as in [3?, p. 224230, we an
prove that there exists q ? N suh that the series (9) onverges if and only if q|k .
Then, for q|k , in view of (6), (7) and onvergene of series (9) we obtain that the
series S0 (t1 , t2 , k) onverges uniformly in tj , |tj | ? t0 . Thus, by Lemma 7, uniformly
in tj , |tj | ? t0 , as n ? ?,
X
? ?
Y
|g1 (p? )|it1 |g2 (p? )|it2 eik arg g2 (p )
v0n (t1 , t2 , k) =
1? ? 1p 1+
+o(1). (10)
p?
?=1
p?n
A. LAURINCIKAS
50
Now let g ? k . Then the method of [3? allows to prove that, for all u ? R,
X ? 1 ? Reukg (p)p?iu
2
= +?.
p
p
(11)
It is not diult to see that in view of the identity
1 ? z1 z2 z3 = 1 ? z1 + z1 (1 ? z2 ) + z1 z2 (1 ? z3 )
we have that by (11) and (4)
X ? 1 ? Re|g1 (p)|it1 |g2(p)|it2 eik arg g2 (p) p?iu
?
p
p?n
X ? 1 ? Re eik arg g2 (p) p?iu X ? 1 ? Re|g1 (p)|it1 X ? 1 ? Re|g2 (p)|it2
?
?
?
?
p
p
p
p?n
p?n
p?n
X
1/2 X
1/2
? 1 ? Re|g1 (p)|it1
? 1 ? Re|g2 (p)|it2
?2
?
p
p
p?n
p?n
1/2 X
1/2
X
? 1 ? Reeik arg g2 (p) p?iu
? 1 ? Re|g1 (p)|it1
?
?2
p
p
p?n
p?n
X
1/2 X
1/2
? 1 ? Re|g2 (p)|it2
? 1 ? Reeik arg g2 (p) p?iu
?2
? +?
p
p
p?n
p?n
as n ? ? uniformly in tj , |tj | ? t0 , and all u ? R. Therefore, by (5)
X 1 ? Re|g1 (p)|it1 |g2 (p)|it2 eik arg g2 (p) p?iu
p
p
= +?
for all t1 , t2 , u ? R. Thus, in this ase Lemma 6 shows that
lim v0n (t1 , t2 , k) = 0.
n??
(12)
If the series (9) does not onverge for any k ? N, then we use the ondition
X 1 ? Reukg (p)p?iu
2
= +?,
p
p
k ? N, u ? R, and, similarly to the ase q ? k , we obtain that, for all k ? N, t1 , t2 ? R,
lim v0n (t1 , t2 , k) = 0.
n??
Clearly,
? ?
X
1
|g1 (p? )|it1 |g2 (p? )|it2 eik arg g2 (p )
1?
1 +
=
?
p
p
?=1
(13)
THE JOINT DISTRIBUTION OF MULTIPLICATIVE FUNCTIONS
1
|g1 (p)|it1 |g2 (p)|it2 eik arg g2 (p)
=1?
+O 2
2
p
p
51
(14)
uniformly in tj , j = 1, 2. We learly have seen that the series S0 (t1 , t2 , k), for q|k ,
onverges uniformly in tj , |tj | ? t0 , j = 1, 2. So, in view of (12), for the onvergene
of w0n (t1 , t2 , k), q|k , it remains to onsider the series
def
S(t1 , t2 , k) =
Denote by
P ??
X ? Im|g1(p)|it1 |g2 (p)|it2 eik arg g2 (p)
.
p
p
P?
over those primes p for whih at least one of
P ???
inequalities | log |g1 (p)|| > 1 or | log |g2 (p)|| > 1 is satised, and let
mean the
p
P?
summation in
over primes p satisfying | log |g1 (p)|| ? 1 and | log |g2(p)|| ? 1.
p
the summation in
p
p
Then we have that
S(t1 , t2 , k) =
X ?? Im|g1 (p)|it1 |g2 (p)|it2 eik arg g2 (p)
+
p
p
X ??? Im|g1 (p)|it1 Re|g2 (p)|it2 Re eik arg g2 (p)
+
+
p
p
X ??? Re|g1 (p)|it1 Im|g2 (p)|it2 Im eik arg g2 (p)
+
+
p
p
X ??? Re|g1 (p)|it1 Re|g2 (p)|it2 Im eik arg g2 (p)
+
+
p
p
X
??? Im|g1 (p)|it1 Im|g2 (p)|it2 Im eik arg g2 (p)
def
+ ?
=
p
p
def
=
5 X
X
j=1
j
.
(15)
The hypothesis
of theorem imply the uniform onvergene in tj , j = 1, 2, for the
P
series 1 . Furthemore, for q|k , the series
X
2
=
X ??? (t1 log |g1 (p)| + O(t3 log2 |g1 (p)|))(1 + O(t2 log2 |g2 (p)|))
1
2
?
p
p
X ??? Im |g1(p)|it1 Re |g2 (p)|it2 (1 ? Re eik arg g2 (p) )
?
p
p
onverges uniformly in tj , |tj | ? t0 , j = 1, 2. Similarly, we nd that the series
X
3
=
X ??? (1 + O(t2 log2 |g1 (p)|))((t2 log |g2 (p)| + O(t3 log2 |g1 (p)|))
1
2
2
?
p
p
A. LAURINCIKAS
52
?
X
4
=
X ??? Re |g1 (p)|it1 Im |g2 (p)|it2 (1 ? Re eik arg g2 (p) )
,
p
p
X ??? Im eik arg g2 (p) X ??? (1 ? Re |g1(p)|it1 )Re |g2 (p)|it2 Im eik arg g2 (p)
?
?
p
p
p
p
X ??? (1 ? Re |g2 (p)|it2 )Im eik arg g2 (p)
?
,
p
p
X X
??? (1 ? Re 2 |g1 (p)|it1 )1/2 (1 ? Re 2 |g2 (p)|it2 )1/2 Im eik arg g2 (p) = ?
5
p
p
X
1/2 X
1/2
??? 1 ? Re |g1 (p)|it1
??? 1 ? Re |g2 (p)|it2
?2
,
p
p
p
p
for q|k , onverge uniformly in tj , |tj | ? t0 , j = 1, 2. Therefore, from this, (8), (10)
and (14) we dedue that, for q|k ,
lim v0n (t1 , t2 , k) = v0 (t1 , t2 , k),
n??
where
v0 (t1 , t2 , k) =
Y
p
1
1?
p
? ?
X
|g1 (p? )|it1 |g2 (p? )|it2 eik arg g2 (p )
.
1+
p?
?=1
Hene, and from (12) and (13) we have that
lim v0n (t1 , t2 , k) = v0 (t1 , t2 , k),
n??
t1 , t2 ? R,
k ? Z,
and the funtions v0 (0, t2 , 0) and v0 (t1 , 0, 0) are ontinuous at t2 = 0 and t1 = 0,
respetively.
It remains to study the harateristi transform w1n (t1 , t2 , k). We begin with
w1n (t1 , t2 , 0). First suppose that
X 1
< ?.
p
(16)
g(p)<0
Then, similarly to the ase r = 0, we obtain that the series S1 (t1 , t2 , 0) onverges
uniformly in tj , |tj | ? t0 , j = 1, 2, and thus, by Lemma 7,
v1n (t1 , t2 , k) =
Y
p?n
1
1?
p
1+
?
X
|g1 (p? )|it1 sgng1 (p? )|g2 (p? )|it2
?=1
p?
+ o(1)
THE JOINT DISTRIBUTION OF MULTIPLICATIVE FUNCTIONS
53
uniformly in tj , |tj | ? t0 , j = 1, 2, as n ? ?. From this, using the above arguments,
we obtain that
lim v1n (t1 , t2 , 0) = v1 (t1 , t2 , 0),
(17)
n??
where
v1 (t1 , t2 , 0) =
Y
p
1
1?
p
?
X
|g1 (p? )|it1 sgng1 (p? )|g2 (p? )|it2
1+
,
?
p
?=1
and the funtions v1 (0, t2 , 0) and v1 (t1 , 0, 0) are ontinuous at t2 = 0 and t1 = 0,
respetively.
Now let
X 1
(18)
= +?
p
g1 (p)<0
This ase is more ompliated. We onsider the series
X ? 1 ? sgn g1 (p)Re |g1 (p)|it1 |g2 (p)|it2 p?iu
,
p
p
u ? R.
If u = 0, then in view of (6) and (18)
X ? 1 ? sgn g1 (p)Re |g1(p)|it1 |g2 (p)|it2
X ? 1 ? sgn g1 (p)Re |g1 (p)|it1
?
?
p
p
p?n
p?n
X
1/2
X ? 1 ? Re |g2 (p)|it2
? 1 ? sgn g1 (p)Re |g1 (p)|it1
?
?2
Ч
p
p
p?n
p?n
X
1/2
X ? 2 ? (1 ? Re |g1 (p)|it1 )
? 1 ? Re |g2 (p)|it2
Ч
=
+ O(1)+
p
p
p?n
p?n
g1 (p)<0
X
1/2
? 2 ? (1 ? Re |g1 (p)|it1 )
+O
??
p
p?n
(19)
g2 (p)<0
as n ? ?.
Now suppose that u 6= 0. Then, taking into aount (6), we nd that, for ? > 1,
X ? 1 ? sgn g1 (p)Re |g1 (p)|it1 |g2 (p)|it2 p?iu
=
p?
p
X
X ? 1 ? sgn g1 (p)Re |g1 (p)|it1 p?iu
? 1 ? Re |g2 (p)|it2
=
+O
+
?
p
p
p
p
X
1/2 X
1/2 it
?iu
1
? 1 ? sgng1 (p)Re |g1 (p)| p
? 1 ? Re |g2 (p)|it2
+O
=
p?
p
p
p
A. LAURINCIKAS
54
X ? 1 ? sgn g1 (p)Re |g1 (p)|it1 p?iu
+
p?
p
X
1/2 ? 1 ? sgn g1 (p)Re |g1 (p)|it1 p?iu
+O
+ O(1).
?
p
p
=
(20)
It is not diult to see that, for g1 (p) 6= 0,
1 ? sgn g1 (p)Re |g1 (p)|it1 p?iu =
= 1 ? sgn g1 (p)Re p?iu + sgn g1 (p)Re (p?iu ? |g1 (p)|it1 p?iu ) =
= 1 ? sgn g1 (p) cos(u log p) + sgn g1 (p) cos(u log p)(1 ? cos(t1 log |g1 (p)|))+
+sgn g1 (p) sin(u log p) cos(t1 log |g1 (p)|).
Therefore, in virtue of (6)
X ? 1 ? sgn g1 (p)Re |g1 (p)|it1 p?iu X ? 1 ? sgn g1 (p) cos(u log p)
=
+
p?
p?
p
p
X
X
1/2
? 1 ? cos(t1 log |g1 (p)|)
? 1 ? sgng1 (p) cos(u log p)
+O
+O
Ч
?
p
p
p
p
1/2 X
X
? 1 ? sgn g1 (p) cos(u log p)
? 1 ? cos(t1 log |g1 (p)|)
+
Ч
=
p
p?
p
p
X
1/2 ? 1 ? sgn g1 (p) cos(u log p)
+O
+ O(1).
(21)
p?
p
?
Now let ? > 0 is a small xed number, and a = aros(1 ? ?) < 2 ?. Then in view
of (5)
X ? 1 ? sgn g1 (p) cos(u log p)
X ? 1 ? cos(u log p)
=
+
p?
p?
p
g1 (p)>0
X ? 1 + cos(u log p)
X
X
1
+
?
?
?
?
p?
p?
g1 (p)<0
g1 (p)>0
1?cos(u log p)<?
g1 (p)<0
1+cos(u log p)<?
?
X 1
X
X
1
?
?
?
?
?
p
p
l??a
l?+a
p
l=0
exp{ u }<p?exp{ u }
?
1
?
? c?3/2 log
? +?
2
??1
1
?
p?
??
as ? ? 1 + 0. Thus, (19)(22) and (5) show that, for all t1 , t2 , u ? R,
X 1 ? sgn g1 (p)Re |g1 (p)|it1 |g2(p)|it2 p?iu
? +?
p?
p
(22)
THE JOINT DISTRIBUTION OF MULTIPLICATIVE FUNCTIONS
55
as ? ? 1+0. Moreover, the last sum monotonially inreases as ? ? 1+0. Therefore,
for all t1 , t2 , u ? R
X 1 ? sgn g1 (p)Re g1 (p)Re |g1 (p)|it1 |g2 (p)|it2 p?iu
= +?.
p
p
Therefore, by Lemma 6
lim v1 (t1 , t2 , 0) = 0.
n??
(23)
Now suppose that there exists at least one k ? N suh that the series (9)
onverges. If (16) is true, then similarly to the ases r = 0 and w1n (t1 , t2 , 0) we
nd that, for q|k , the series S1 (t1 , t2 , k) onverges uniformly in tj , |tj | ? t0 , j = 1, 2,
and from this we dedue that
lim v1n (t1 , t2 , k) = v1 (t1 , t2 , k),
n??
(24)
where
v1 (t1 , t2 , k) =
Y
p
1
1?
p
? ?
X
|g1 (p? )|it1 sgng1 (p? )|g2 (p? )|it2 eik arg g2 (p )
1+
.
?
p
?=1
Now suppose that (18) is valid. Then, the onvergene of the series (9) shows
that, for q|k ,
X ? 1 ? sgn g1 (p)Re eik arg g2 (p)
X ? 1 ? Re eik arg g2 (p)
=
+
p
p
p
g1 (p)<0
X ? 2 ? (1 ? Re eik arg g2 (p) )
= +?.
+
p
(25)
g1 (p)>0
Now let u 6= 0. Then, for q|k and ? > 1, by (22)
X ? 1 ? sgn g1 (p)Re eik arg g2 (p? ) p?iu X ? 1 ? sgn g1 (p) cos(u log p)
=
+
?
p
p
p
p
X
X
1/2
? 1 ? cos(k arg g2 (p))
? 1 ? sgn g1 (p) cos(u log p)
+O
+O
Ч
p
p?
p
p
X
1/2 X
? 1 ? cos(k arg g2 (p))
? 1 ? sgn g1 (p) cos(u log p)
Ч
=
+
p
p?
p
p
X
1/2
? 1 ? sgn g1 (p) cos(u log p)
+O
? +?
?
p
p
as ? ? 1 + 0. Now this, (5), (25) and onvergene of the series (6) allow to prove
that, for all t1 , t2 , u ? R and q|k ,
X 1 ? sgn g1 (p)|g1 (p)|it1 |g2 (p)|it2 eik arg g2 (p) p?iu
? +?
p?
p
A. LAURINCIKAS
56
as ? ? 1 + 0. Hene, we have that, for all t1 , t2 , u ? R,
X 1 ? sgn g1 (p)|g1 (p)|it1 |g2(p)|it2 eik arg g2 (p) p?iu
= +?,
p
p
therefore, Lemma 6 implies that, for q|k ,
lim v1 (t1 , t2 , k) = 0.
n??
(26)
If q ? k , then, for all u ? R,
X ? 1 ? Re eik arg g2 (p) p?iu
= +?.
p
p
(27)
Therefore, in the ase of (16), for q ? k and u ? R,
X ? 1 ? Re eik arg g2 (p) p?iu
= +?.
p
g1 (p)>0
Hene, for q ? k and u ? R,
X ? 1 ? sgn g1 (p)Reeik arg g2 (p) p?iu
X ? 1 ? Reeik arg g2 (p) p?iu
=
+
p
p
p
p
X ? 1 + Reeik arg g2 (p) p?iu
+
= +?.
p
p
Thus, the above arguments show that, for g ? k ,
lim v1 (t1 , t2 , k) = 0.
n??
Now suppose that (18) and (27) take plae. Then we have that
X 1
< ?.
p
g1 (p)>0
Hene we have that, for all u ? R,
X ? 1 ? Reeik arg g2 (p) p?iu
= +?.
p
g1 (p)<0
Therefore, for all u ? R,
X ? 1 ? sgn g1 (p)Reeik arg g2 (p) p?iu
X ? 1 ? Reeik arg g2 (p) p?iu
=
+
p
p
p
g1 (p)>0
(28)
THE JOINT DISTRIBUTION OF MULTIPLICATIVE FUNCTIONS
57
X ? 2 ? (1 ? Reeik arg g2 (p? ) p?iu )
+
= +?.
p
g1 (p)<0
Hene we obtain that, for q ? k ,
lim v1 (t1 , t2 , k) = 0.
n??
(29)
If, for all k ? N and u ? R,
X 1 ? Reukg (p)p?iu
2
= +?,
p
p
the reasoning similarly to the ase q ? k , we obtain that, for all k ? N, also
lim v1 (t1 , t2 , k) = 0.
n??
(30)
Now (3), (4), (15), (17), (23), (24), (26), (28)(30) and Lemma 4 omplete the
proof of the suieny.
Neessity. Suppose that the measure Pn onverges mweakly in the sense of
X to a ertain probability measure P on (X, B(X)), PR 6= Pa for every a ? R and
PC ({0}) 6= 1, as n ? ?. Hene we nd that the probability measure
?n (g1 (m) ? A),
A ? B(R),
onverges mweakly to a ertain measure P on (R, B(R)), P 6= Pa for every a ? R,
as n ? ?. Therefore, by Theorem 3 we obtain hypothesis 10 of the theorem.
Similarly, we have that the probability measure
?n (g2 (m) ? A),
A ? B(C),
onverges weakly to a ertain measure P on (C, B(C)), P ({0}) 6= 1, as n ? ?.
Hene, by Theorem 4, we obtain hypotheses 20 and 30 of the theorem.
СПИСОК ЦИТИОВАННОЙ ЛИТЕАТУЫ
[1? Bakstys A. On limit distribution laws for arithmetial multipliative funtions
// Liet. matem. rink. 1968. No 8, P. 520 (in Russian).
[2? Delange H. A remark on multipliative funtions // Bull. London Math. So.
1970. No 2. P. 183185.
[3? Delange H. On the distribution modulo 1 of additive funtions // J. Indian
Math. So. 1970. No 34. P. 215235.
[4? Delange H. Sur la distribution des valeurs des fontions multipliative omplexes // C. R. Aad. S. Paris. 1973. No 276. Serie A. P. 161164.
58
A. LAURINCIKAS
[5? Elliott P. D. T. A. Probabilisti Number Theory I // SpringerVerlang. 1979.
[6? Erdos P. Some remarks about additive and multipliative funtions // Bull.
Amer. Math. So. 1946. No 52. P. 527537.
[7? Laurinikas A. The multidimensional distribution of values of multipliative
funtions // Liet. mat. rink. 1975. 25 No 2. P. 1324 (in Russian).
[8? Laurinikas A. On the distribution of values of omplex funtions // Liet.
matem. rink. 1975. 15 No 2. P. 2539.
[9? Laurinikas A. On the distribution of omplexvalued multipliative funtions
// J. Theorie Nombres Bordeaux. 1996. No 8. P. 183203.
[10? Laurinikas A. Limit Theorems for the Riemann Zeta - Funtion, Kluwer.
Dordreht. 1996.
[11? Laurinikas A. The harateristi transforms of probability measures // Siauliai.
Math. J. (to appear).
[12? Laurinikas A., Maaitien
e R. The harateristi transforms on RЧC // Integral
Transforms and Speial Funtions (to appear).
[13? Levin B. V., Timofeev N. M. Analyti method in probabilisti theory // Uh.
zap. Vladim. gos. ped. inst. 1971. No 57 (2). P. 57150 (In Russian).
[14? Levin B. V., Timofeev N. M., Tuliaganov S. T. Distribution of values of
multipliative funtions // Liet. matem. rink. 1973. No 13 (1). P. 87100 (In
Russian).
Department of Mathematis and Informatis
Vilnius University
Naugarduko 24
03225 Vilnius
Lithuania
E- mail: antanas.laurinikasmaf.vu.lt
Получено 13.05.2008
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