DISTRIBUTION OF THE TRADITIONALLY NORMALIZED ADDITIVE

Annales Univ. Sci. Budapest., Sect. Comp. 14(1994) 115–133
DISTRIBUTION OF THE TRADITIONALLY
NORMALIZED ADDITIVE FUNCTIONS
E. Manstavičius
1. Results.
Let h(m) be a real–valued additive function,
∑ h(pk )
A(x) =
,
pk
k
B(x) =
p ≤x
(
∑ h2 (pk )
pk
k
)1/2
,
p ≤x
where p denotes a prime number and k ≥ 1 . Put I(. . .) for the indicator of the set of
natural numbers m satisfying the conditions written in the parenthesis. In the present
paper we consider the conditions when there exist distribution functions ν(v) and µ(v)
such that
∑ (
)
νx (v) := [ x ]−1
I h(m) − A(x) < vB(x) ⇒ ν(v)
(1)
m≤x
or
(
µx (v) :=
∑ 1
m
)−1
m≤x
∑ 1 (
)
I h(m) − A(x) < vB(x) ⇒ µ(v) .
m
(2)
m≤x
Here and in what follows we suppose that B(x) → ∞ , the notation ⇒ denotes weak
convergence of bounded nondecreasing functions, and the notation of the limiting passage
x → ∞ is omitted.
In the first problem only the case of the improper limiting distribution concentrated at
v = a (denoted in the sequel by Ea (v) ) has been investigated completely (see Ruzsa [16]).
In the second problem, in addition to the law of large numbers (see Levin and Timofeev
[9]), we have solved (Manstavičius [14]) the case of the limiting distribution having the zero
mean and the unit variance. Now we shall summarize our results announced in several
contributions (Manstavičius [10–13 ]), the theorems from the papers (Manstavičius [14],
Timofeev [20]) will be generalized, and their proofs will be simplified considerably.
It seems likely that the conditions necessary for (1) or (2) with proper distribution
functions ν(v) , µ(v) should contain the existence of a nondecreasing bounded function
K(w) such that
∑
1
h2 (p)
⇒ K(w) ,
K(−∞) = 0 .
(3)
B 2 (x)
p
p≤x
h(p)<wB(x)
Moreover, on the contrary to the theory of distribution of independent random variables,
these conditions should involve (explicitly or not) the relations
A(xu ) − A(x)
ψx (u) :=
→ ψ(u) ,
B(x)
1
B(xu )
φx (u) :=
→ φ(u) ,
B(x)
(4)
valid uniformly in u belonging to any closed interval of the half–axis (0, ∞) , where
ψ(u) and φ(u) are continuous functions. Then we had (see, Levin and Timofeev [9]) their
expressions
ψ(u) = c (uρ − 1) ,
φ(u) = uρ ,
(5)
where c ∈ IR and ρ ≥ 0 ( c = 0 when ρ = 0 ).
Now the following question arises: what does the constants c and ρ control ? The role
of the limiting distribution should not to be overrated because we can have the standard
normal limit distribution for νx (v) , when ρ = 0 or ρ = 1/2 (see Timofeev [21]). As our
result in (Manstavičius [14]) shows, the situation with µx (v) is pretty different.
We have observed the quantities determining c and p . Denote
αx =
∑ h(pk ) log pk
1
,
B(x) log x k
pk
p ≤x
βx =
∑ h2 (pk ) log pk
1
.
B 2 (x) log x k
pk
p ≤x
Definition. An additive function h(m) belongs to the class H(α) if αx → α ; h(m)
belongs to the class H(α, β) if αx → α and βx → β .
{ √ √ }
We have |α| ≤ min 1/ 2 , β and 0 ≤ β ≤ 1 . The celebrated Kubilius H class
(Kubilius [5]) agrees with H(0, 0) . As we shall see, in the case of the standard normalized
additive functions the condition h(m) ∈ H(α, β) substitutes for the following relation
1
log x
∑
p≤x
h(p)<wB(x)
log p
⇒ F (w) ,
p
where F (u) is a distribution function, used in the paper (Levin and Timofeev [8]), or the
condition (3) in Theorem 18.1 (Elliott [1]).
In the paper (Levin and Timofeev [9]) the problem of joint convergence of νx (v) and
µx (v) has been solved. We modify this problem considering the weak convergence of either
of the sequences together with convergence of some sequences of moments.
For a distribution function F (v) put
∫
∫
a(F ) = v dF (v) ,
σ(F ) = v 2 dF (v) .
IR
IR
We have a(νx ) = o(1) , σ(νx ) ≤ 3/2+o(1) (see Kubilius [6], [7]), and a(µx ) = −αx +o(1) ,
σ(µx ) ≤ eγ + o(1) , where γ denotes the Euler constant (corollary of the estimate (13) in
(Manstavičius [14])). Therefore the limiting distributions ν(v) and µ(v) , when they do
exist, will have the first two finite moments satisfying the following relations
a(ν) = 0 ,
a(µ) = −α ,
σ(ν) ≤ lim inf σ(νx ) ≤ 3/2 ,
σ(µ) ≤ lim inf σ(µx ) ≤ eγ ,
2
a(µ)2 ≤ σ(µ) .
The first part of the following theorem will consider a special case of Theorem 1 of
the paper (Levin and Timofeev [9]).
Theorem 1. Let the conditions (3) and (4) be satisfied. Then
i1 ) νx (v) ⇒ ν(v) and µx (v) ⇒ µ(v) ;
i2 ) h ∈ H(α, β) , where
∫1
α=−
∫1
ψ(u) du = −a(µ) ,
β =1−
0
φ2 (u) du < 1 ;
0
(
)
i3 ) φ(u) = uβ/2(1−β) , ψ(u) = 0 if β = 0 and ψ(u) = α(2 − β)/β uβ/2(1−β) − 1 if
β ̸= 0 ;
i4 ) σ(µ) − (2 − β) a(µ)2 = (1 − β) σ(ν) , where σ(µ) ̸= 0 if and only if K(+∞) > 0 ;
i5 ) σ(νx ) = o(1) if and only if α2 = 1/2 .
Suppose h(m) ∈ H(α, β) and νx (v) ⇒ ν(v) ̸= E0 (v) . Then the conditions (3) and
(4) are satisfied and, in addition to the relations i1 ) – i4 ) , we have α2 = a(µ)2 < 1/2 and
σ(µ) ̸= (2 − β) a(µ)2 .
Thus, considering the necessity of the conditions (3) and (4) there is no need to
take µx (v) ⇒ µ(v) ̸= Ea (v) in advance. The influence of the moment convergence show
following results.
Theorem 2. In order that νx (v) ⇒ ν(v) ̸= E0 (v) or µx (v) ⇒ µ(v) ̸= Ea (v) jointly
with
σ(νx ) → σ1 > 0 ,
a(µx ) → a ,
σ(µx ) → σ2 > 0
(6)
it is necessary and sufficient that the condition (3), where K(+∞) > 0 , and
( the condition
)
cρ
2ρ
(4) be satisfied. Under these conditions the formulae (5) are true, h ∈ H ρ+1
, 2ρ+1
,
2 2
σ1 = 1 + 2c ρ
∞
r
∑
(−1)
−cρ
,
ρ+1
s−ρ
s+ρ
)2
(7)
σ1
2a2 (1 + ρ)
+
.
(1 + 2ρ)
(1 + 2ρ)
(8)
(r + ρ)
σ2 =
(
,
2
r=1
a=
r−1
∏
s=1
Moreover, the case a = 0 , σ1 = σ2 = 1 occurs if and only if ρ = 0 , c = 0 .
Remark. Considering µx (v) ⇒ µ(v) ̸= Ea (u) the second of the relations (6) is
superfluous.
Corollary 1. In order that νx (v) ⇒ ν(v) ̸= E0 (u) or µx (v) ⇒ µ(v) ̸= Ea (u) jointly
with
a(µx ) → 0 ,
σ(µx ) → σ > 0
(9)
it is necessary and sufficient that the conditions (3) with K(+∞) > 0 and (4) with
ψ(u) = 0 , φ(u) = u(1−σ)/2σ , 0 < σ ≤ 1 , be satisfied. Under these conditions σ(νx ) → 1
too.
3
Corollary 2. (Manstavičius [14]). In order that µx (v) ⇒ µ(v) with a(µ) = 0
and σ(µ) = 1 it is necessary and sufficient that h ∈ H(0, 0) and the condition (3) with
K(+∞) = 1 be satisfied.
Sometimes the behavior of moments is given indirectly. In part it is shown by the
following theorem.
Theorem 3. Let ε > 0 be arbitrary, the star ⋆ denote the condition h(p) < −εB(x),
and
Tx :=
1 ∑⋆ log p
→ 0.
log x
p
p≤x
In order that νx (v) ⇒ ν(v) with σ(ν) = 1 it is necessary and sufficient that
h ∈ H(0, 0) and the condition (3) with K(+∞) = 1 be satisfied.
In order that νx (v) ⇒ ν(v) ̸= Ea (v) jointly with σ(νx ) → 1 it is necessary and
sufficient that h ∈ H(0, 0) and the condition (3) with K(+∞) > 0 be satisfied. Under
this condition σ(ν) = K(+∞) .
This result in the case of the standard normal law has been announced by N. M. Timofeev ([20]). His rather complicated proof is presented in the thesis ([23]). Later a new
attempt has been done (Timofeev [22]). We follow the idea of P. D. T. A. Elliott given in
Supplement of the book [2]. Our criticism concerning the original calculations is expressed
in Concluding remark at the end of this paper.
2. Proofs.
Our approach is based upon relations of the sequences of distribution functions νx (v)
and µx (v) . On the contrary to the paper (Levin and Timofeev [9]), we do not use their joint
convergence in advance. We observe that either of them is relatively compact. Further,
adding other conditions we prove that the convergence of one of the sequences implies that
for the other one. Thus, then the result of (Levin and Timofeev [9]) can be applied.
To consider the relative compactness as well as features of the normalizing sequences
A(x) and B(x) , we shall apply Lemmas from our paper (Manstavičius [14]). While the
proof of Lemma 1 [14] has been then omitted, having this opportunity we present it in
more abstract setting influenced by the Ruzsa’s paper [18]. That is motivated by our future
plans too.
Let G be a commutative metric group with respect to additively written algebraic
operation. Put d ( . , .) for the metric and ∥g∥ = d (g, 0) . We consider the asymptotic
distribution of additive functions f : IN → G . In the set M(G) of probabilistic measures
defined on the Borel σ-algebra B(G) of the sets of G we introduce the Lévy–Prokhorov
metric
{
}
ε
ε
L(η1 , η2 ) = inf ε | η1 (A) ≤ η2 (A ) + ε, η2 (A) ≤ η1 (A ) + ε; A ∈ B(G) ,
where Aε denotes the ε-neighborhood of the set A , and η1 , η2 ∈ M(G) . Let further, Uε
denote the ε-neighborhood of the zero element,
{
}
Λ(η) = inf ε + η (Ūε ) | ε > 0 ,
η ∈ M(G) ,
Ūε = G \ Uε .
4
The total variation of the signed measure ∆ will be denoted by |∆| . Denote by δ(v) the
unit mass at a point v and δ = δ(0) . Put
Θy = [ y ]
−1
∑
δ(f (m) − g) ,
m≤y
where the function f = fx , g = gx ∈ G , and y = yx → ∞ as x → ∞ .
√
Lemma 1. Let z = y 1+u , |u| ≤ 1/5 , y1 = y |u| , y ≥ 3 ,
R=
∑
y1 <p≤y
1
,
p
τy = R
∑
−1
y1 <p≤y
(
)
δ h(p)
.
p
Then
L(Θy , Θz ) ≪ Λ(τy ) + R−1/2 ,
where the constant implied is absolute.
{
}
Let t⋆ = min | t | , 1 sgn t , where t ∈ IR . In the notations of Lemma 1 we obtain
Corollary. We have
(
L(Θy , Θz ) ≪ R−1/3
∑ ∥f (p)∥⋆ 2
p≤y
)1/3
p
+ R−1/2 .
In the case G = IR having in advance a limiting distribution for the values fx (m)−γx ,
γx ∈ IR , we can derive the boundiness of the sum in the last estimate, when f (p) is
replaced by f (p)−λx log p with some λx ∈ IR . Then applied for fx (m)−γx −λx log(m/x)
Corollary is quite useful. Lemma 2 below shows only one of its applications.
Proof of Lemma 1 is based on A. Hildebrand’s arguments (see Lemma 4 in (Hildebrand
[3])).
Observe that it is sufficient to deal with the case 0 < u ≤ 1/4 only. Later, in the case
−1/5 ≤ u < 0 , one can use the inversions z → y and y → z . For 0 ≤ u ≤ log−3/2 y our
estimate follows from the simple inequality
|Θz − Θy | ≤
(z − y + 1)
.
[y]
(10)
In that follows we suppose log−3/2 y < u ≤ 1/4 and y sufficiently large. Then
R−
(√
)−1
{
}
| log u|
≪
u log y
≪ min u1/6 , log−1/4 y .
2
5
(11)
We start with an averaging of Θy . For an arbitrary bm ∈ C, 1 ≤ m ≤ x , we have the
inequality (see Chapter 4 in Elliott [1])
2
∑ ∑
∑
1 ∑
2
p
bm −
bm ≪ y
|bm | .
m≤y
p
p≤y
m≡0modp
m≤y
m≤y
Applying it for bm = I{f (m) ∈ A + g} with arbitrary set A ⊂ G we obtain
(
)2
∑
(
)
p−1 Θy/p A − f (p) − Θy (A) ≪ 1 .
y1 <p≤y
Now the Cauchy inequality yields
Θy (A) = R−1
∑
(
)
p−1 Θy/p A − f (p) + BR−1/2 .
(12)
y1 <p≤y
Here and in what follows B denotes some quantity bounded by an absolute constant
C > 0 (in different places we shall not use indeces). Hence for the probabilistic measure
∑
Ξy := R−1
p−1 Θy/p
y1 <p≤y
we obtain
L(Θy , Ξy ) ≪ Λ(τy ) + R−1/2 .
(13)
√
u+u
Let us repeat our considerations with Θz taking z1 = y
instead of the previous
y1 . If τz is defined from τy , substituting y → z and y1 → z1 , via the estimate (10) and
∑ 1
B
1+u
√ +√
= R + Bu1/6
(14)
R1 :=
= log
p
u+ u
u log y
z1 <p≤z
we have |τy − τz | ≪ R−1 . If Ξz is obtained from Ξy after these substitutions, then the
same calculations as in the proof of (13) yield
L(Θz , Ξz ) ≪ Λ(τy ) + R−1/2 .
(15)
In the next step we shall obtain the integral representations of the measures Ξz and
Ξy . All the estimates will mean the estimates of the total variations of the signed measures.
√
Since the primes p ∈ ( y, y ] contribute to Ξy the term BR−1 only, we divide the interval
√
k
(y1 , y ] by the points w = wk := (1 + u) , k = 0, 1, . . . . For the typical interval (the
last one may be incomplete) in virtue of the prime number theorem with logarithmical
error term and (10) we have
( (
)
∑
(
) −1
)
−1
p Θy/p = Θy/w + Bu w (1 + Bu) π w(1 + u) − π(w) =
w<p≤w(1+u)
(Θy/w + Bu)u
=
=
log w
w(1+u)
∫
w
6
Θy/t + Bu
dt .
t log t
Therefore( summing
these estimates with respect to k and adding one with respect to the
)
√
interval
y , y we obtain
Ξy = R−1
∫y
Θy/t + Bu
dt + BR−1 = R−1
t log t
y1
y/y
∫ 1
Θt
dt + BR−1 .
t log y/t
1
Similarly, using the points z1 (1 + u)k , k ≥ 0 , we deduce
Ξy =
R1−1
z/z
∫ 1
Θt
dt + BR1−1 .
t log z/t
1
Hence and from the estimate (11), (14) in virtue of y/y1 = z/z1 we obtain
Ξy − Ξz ≪ R−1
y/y
∫ 1
log z/y dt
+ R−1 ≪ R−1 .
2
t
log (y/t)
1
Thus, in the case log−3/2 ≤ u ≤ 1/4 the assertion of Lemma 1 follows from (13), (15),
and the last estimate.
Now we return to G = IR . Proofs of Theorems are based on few elementary observations.
Lemma 2. Suppose the proposition (1) or (2) is valid with a proper limiting distribution. Then for each sequence y1 → ∞ there exist a subsequence y → ∞ , continuous
functions ψ(u) and φ(u) given on (0, +∞) and such that the condition (4) is satisfied
for the subsequence x = y .
Proof. In the case (2) the proposition has been prooved in (Levin and Timofeev
[9]). We shall consider the first case. The existence of a subsequence y → ∞ such that
φy (u) → φ(u) for u > 0 has been proved in Theorem 4 (Šiaulys [19]). The arguments
given on page 379 of this paper (in virtue of µ(n) = 0 there) yield, in fact, ψy (u) → ψ(u)
too. Observe that a possibility to choose y → ∞ such that φy (u) → φ(u) and ψy (u) →
ψ(u) is contained in Lemma 3 of the paper (Manstavičius [14]). Our idea can be seen in
the next step of the proof.
To prove continuity of ψ(u) and φ(u) , introduce
νx (u, v) = [xu ]
−1
∑
(
)
I h(m) − A(x) < vB(x) .
m≤xu
In what foolows L(. , .) denotes the Lévy metric in the distribution function space. We
have
(
(
))
L νy (u, .), ν . φ(u) + ψ(u) → 0
7
as y → ∞ . But then Corollary of Lemma 1 yields
( (
) (
))
L ν . φ(u + δ) + ψ(u + δ) , ν . φ(u) + ψ(u) → 0
as δ → 0 . Now by the well–known probabilistic lemma φ(u + δ) → φ(u) and ψ(u + δ) →
ψ(u) as δ → 0 . Lemma is proved.
Lemma 3. Let for a subsequence x = y → ∞ the relations (4) with some continuous
functions ψ(u) and φ(u) , u > 0 , hold. If h(m) ∈ H(α) , then almost averywhere in the
Lebesgue sense
(
)
φ(u)
ψ ′ (u) = α φ′ (u) +
.
(16)
u
If h(m) ∈ H(α, β) , then the functions ψ(u) and φ(u) do not depend on the subsequence
y → ∞ , the formulae (5) are valid with ρ = β/2(1 − β) , β < 1 ; c = 0 if β = 0 and
c = α(2 − β)/β if β ̸= 0 .
Proof. While φ(u) is continuous and nondecreasing, it is differentiable almost everywhere. Further, summing by parts we have
)
(
A y u+∆ − A(y u ) =
1
(u + ∆) log y
∑
y u <p≤y u+∆
h(p) log p
+
p
u+∆
∫
u
∑
y u <p≤y t
h(p) log p dt
,
p
t2 log y
where 0 < u − |∆| ≤ u + |∆| ≤ 1 . Dividing by B(y) and letting y → ∞ we derive
(
)
ψ(u + ∆) − ψ(u) =α φ(u + ∆) − φ(u)
u+∆
∫
+α
u
∆
+ φ(u + ∆)
+
u+∆
u+∆
φ(t) dt
∆
− αφ(u)
.
t
u+∆
u
Now dividing by ∆ and letting ∆ → 0 we obtain (16) almost everywhere in u .
Summation by parts (here and in what follows we do not indicate trivial estimates to
be done at first in order to avoid the influence of the zero neighborhood) applied in the
definition of βy yields
∫1
β = 1 − φ2 (u) du ,
(17)
0
which in virtue of φ(1) = 1 implies β < 1 . The same idea as earlier leads to the relation
(
)
(1 − β) φ2 (u + ∆) − φ2 (u) = β
u+∆
∫
u
8
φ2 (t)
dt ,
t
which gives the desired expression for φ(u) . Further, solving (16) we use the condition
ψ(1) = 0 . Lemma 3 is proved.
Lemma 4. Let for a subsequence x = y → ∞ the relations (4) with some continuous
functions ψ(u) and φ(u) , u > 0 , hold. If a(µx ) → a , σ(νx ) → σ1 , σ(µ) → σ2 > 0 ,
then the functions ψ(u) and φ(u) do not depend on the subsequence y → ∞ and the
formulae (5) are true. We have two possibilities:
i1 ) ρ = 0 , c = 0 if and only if a = 0 , σ1 = σ2 = 1 ;
i2 ) ρ =
2a(2 +σ1 −σ)2
2 σ2
−a2
1 +σ2
,
c = −a 2a2σ+σ
1 −σ2
> 0,
2a2 + σ1 − σ2 > 0 .
Proof. At first we observe that the condition a(µx ) → a implies h ∈ H(−a) and by
Lemma 3
(
)
φ(u)
′
′
ψ (u) = −a φ (u) +
.
u
almost everywhere.
Let ε , 0 < ε < 1 , be fixed and ε ≤ v ≤ ε−1 . In virtue of a(νx ) = o(1) and
1/2
ψx (u) ≪ | log u|
summing by parts we obtain
∫
∑
(
)
1
−1
v 2
−1
m
h(m) − A(x ) = O(ε) + v
σ(νxu )φ2x (u) du+
2
B (x)v log x
v
v
m≤x
+ v −1
∫v
ε
ψx2 (u) du − 2ψx (v)v −1
ε
∫v
(
)
ψx (u) du + ψx2 (v) + oε φ2x (ε−1 ) .
ε
Applying it for the subsequence y → ∞ we deduce the equality
∫v
2
∫v
σ2 vφ (v) = σ1
2
φ (u) du +
0
∫v
ψ (u) du − 2ψ(v)
2
0
ψ(u) du + vψ 2 (v) ,
(18)
0
which together with the differential equation above leads to the following system of equations for ψ = ψ(v) and φ = φ(v) :
{ ′
vψ = −avφ′ − aφ
∫v
2σ2 vφφ′ = (σ1 − σ2 )φ2 + 2vψψ ′ − 2ψ ′ 0 ψ(u) du ,
when φ(1) = 1 − ψ(1) = 1 . The system easily reduces to
{ ′
vψ = −avφ′ − aφ
2(σ2 − a2 )vφ′ = (σ1 − σ2 + 2a2 )φ .
Observe that the factors on the left side of the second equation are nonnegative. If σ2 = a2 ,
then σ1 = a2 = 0 , which contradicts to the condition σ2 > 0 . The equality σ1 −σ2 +2a2 =
0 yields the solution φ(v) = 1 . Further, having a = 0 we obtain ψ(v) = 0 . In this case
9
h ∈ H = H(0, 0) , hence σ1 = 1 = σ2 . When σ1 − σ2 + 2a2 > 0 , the only solution of the
system is given in Lemma. That ends the proof.
In the sequel we shall use the following corollary of the Hildebrand’s result [4].
Lemma 5. If h ∈ H(0) , then σ(νx ) = 1 + o(1) . If the conditions (4) and (5) are
satisfied, then σ(νx ) = σ1 + o(1) , where σ1 is given by the formula (7).
Proof. According to Theorem (Hildebrand [4]) we have
(
)
Θ
σ(νx ) = 1 + √ + ok (1) Vk (x) .
|Θ| ≤ 1 ,
k
where
( k
)
k
∑
∑
λ2x (r)(−1)r
2
Vk (x) = 1 +
+ ok (1) + ok
λx (r) ,
r
r=1
r=1
λx (r) =
∑ h(p)lrx (p)
,
pB(x)
p≤x
(
lrx (p) = qr
log p
log x
)
,
qr (t) = tgr−1 (t) , r ∈ IN .
Here gr−1 (t) , r ∈ IN , denote the sequence of orthogonal polynomials in the interval [0, 1]
with the weight t , that is,
)( )
r−1 (
∑
r−1 r
1/2
1/2 (1,0)
gr−1 (t) = (2r) Pr−1 (1−2t) = (2r)
(−1)r−1−s t r−1−s (1−t)s , r ≥ 1 .
s
s
s=0
While αx → 0 implies λx (r) → 0 , the first assertion evidently foolows from the
Hildebrand’s formulae above.
Further, since |λx (r)| ≤ 1 (see, [4]), the last term in the formula for Vk (x) tends to
zero. We shall prove convergence of the main term as x → ∞ and then convergence as
k → ∞ . All we need for this purpose is to prove that λx (r) → λ(r) , to find these limits,
(r) ≤ λ2 (r − 1) , r ≥ r0 . As in the proof of Lemma 3 summing by
and to verify that λ2√
parts we have λx (1)/ 2 = αx → α = cρ/(ρ + 1) and, further for r ≥ 2 ,
∫1
λx (r) = α
(
)′
gr−1 (t) φ(t)t dt + o(1) =
(19)
0
1/2
= cρ (2r)
)( )
∫1
r−1 (
∑
r−1 r
r−1−s
(−1)
tr−s−1+ρ (1 − t)s dt + o(1) =
s
s
s=0
0
r−1 (
∑
)( )
r−1 r
Γ(r + ρ − s) Γ(s + 1)
1/2
= cρ (2r)
+ o(1) =
(−1)r−1−s
(r + ρ) Γ(r + ρ)
s
s
s=0
(
)( )(
)−1
r−1
∑
r
r−1+ρ
1/2
−1
r−1−s r − 1
= cρ (2r) (r + ρ)
(−1)
+ o(1) .
s
s
s
s=0
10
Applying the summation formula 35.3 on page 633 of the book (Prudnikov et all [15]) we
have
(
)(
)−1
r−1+ρ
1/2
−1
r−1 ρ − 1
λx (r) = cρ (2r) (r + ρ) (−1)
+ o(1) .
r−1
r−1
If cρ = 0 , the assertion of Lemma 5 has been just proved. If cρ ̸= 0 , r > 1 + ρ , then
λ(r)
=
λ(r − 1)
(
r
r−1
)1/2
r−1−ρ
≤
r+ρ
(
r−1
r
)1/2
≤ 1.
Now since lim lim Vk (x) = σ1 , Lemma 5 is proved.
k→∞ x→∞
Proof of Theorem 1. The existence of limiting distributions is given by Theorem 1
of the paper (Levin and Timofeev [9]). As in the proof of (17), the assertion i2 ) follows
by partial summation. Then Lemma 2 implies i3 ) . To prove i4 ) , we shall consider the
relation between the characteristic functions f (t) and g(t) of distributions ν(v) and
µ(v)
∫1
(
) itψ(v)
(
)
vf tφ(v) e
= g tφ(u) eitψ(u) du ,
0
which easily can be obtained by partial summation (see, Levin and Timofeev [9], p. 340).
As we have remarked, the limit distributions have two finite moments, hence f (t) and
g(t) are twice differentiable. So we obtain
(
)
(
)
vf ′ tφ(v) φ(v)eitψ(v) + ivf tφ(v) ψ(v)eitψ(v) =
∫1 (
=
)
(
)
(
)
g ′ tφ(u) φ(v)eitψ(u) + ig tφ(u) ψ(v)eitψ(u) du .
0
When t = 0 , this is the same differential equation (16). Applying it in equality for the
second derivatives at the point t = 0 , we deduce
( 2 )′ (
)
(
)
φ (v) σ(µ) − a(µ)2 = φ2 (v) σ(ν) + 2a(µ)2 − σ(µ) .
Hence and from i3 ) we have the relation
(
)
(
)
β σ(µ) − a(µ)2 = (1 − β) σ(ν) + 2a(µ)2 − σ(µ) ,
which is the same as that in i4 ) . The condition σ(µ) > 0 has been observed in (Levin
and Timofeev [9]).
The following Ruzsa’s estimate (see, [17])

( k
)2 
h(p )


k


∑ B(x) − λ log p
2
σ(νx ) ≍ min λ +
= 1 − 2αx2 + o(1)
k

λ∈IR 
p


pk ≤x
11
implies i5 ) .
In order to prove the converse statement, we apply Lemmas 3 and 2. Further we can
use the proved part of Theorem.
Example of the function h(m) = log m shows that ν(v) = E0 (v) does not imply
µ(v) = Ea (v) for some a ∈ IR . For this function µ(v) exists, a(µ)2 = 1/2 , and σ(µ) =
2/3 .
Proof of Theorem 2. The existence of the limit distributions ν(v) and µ(v) follows
from Theorem 1. Formula (7) has been obtained in Lemma 5. While a(µx ) = −αx + o(1) ,
the first of the formulae (8) follows from i2 ) of Theorem 1. The desired expression for σ2
is presented by (18) when v = 1 .
Necessity follows from Lemmas 2 and 4.
Proof of Corollary 1. Only the necessity is not evident. By Lemma 2 for an arbitrary
sequence y1 → ∞ we have a subsequence y → ∞ such that ψy (u) → ψ(u) and φy (u) →
φ(u) , where ψ(u) and φ(u) are continuous functions. The first assumption of (9) implies
h ∈ H(0) . Hence ψ(u) = 0 independently of the sequence y1 → ∞ and, further by
Lemma 5, σ(νx ) → 1 . Thus, the desired result follows from Theorem 2, the exponent ρ
is determined by the formula (8).
Proof of Corollary 2. Behaving as in the proof of the first Corollary we obtain ψ(u) = 0
and σ(νx ) → 1 . To determine φ(u) , we apply the relation (18) but obtained using the
subsequence y → ∞ only. We have
∫1
1 = σ(µ) ≤ lim σ(µy ) =
φ2 (u) du ≤ 1 .
y→∞
0
Hence φ(u) = 1 , and Corollary is proved.
Proof of Theorem 3. Sufficiency is well–known (Kubilius [5]). To prove necessity, we
denote
∑ h(p)h(q)
Sx := B −2 (x)
.
pq
p, q≤x
pq>x
The Cauchy inequality yields the estimates
∑
B −2 (x)
p, q≤x, pq>x
|h(p)|<εB(x)
∑1
h(p)h(q)
≪ε
pq
p
p≤x
and
B −2 (x)

≪
∑
⋆
p, q≤x, pq>x
∑⋆
√
p≤ x
(
h(p)h(q)
≪
pq
(
∑⋆ 1
p
p≤x
∑
x/p<q≤x
∑
x/p<q≤x
1
q
1
q
)1/2
≪ε
)1/2
≪
1/2
∑
⋆ 1
log x

log
+
≪ Tx = o(1) ,
log x − log p
p
√
x<p≤x
12
where as earlier ⋆ stands for the condition h(p) < −εB(x) . Hence
Sx = B −2 (x)
∑
p, q≤x, pq>x
h(p)>0, h(q)>0
h(p)h(q)
+ o(1) .
pq
But (see Kubilius [6], [7] or Hildebrand [4]) σ(νx ) = 1 − Sx + o(1) , thus, observing that
either of the assertions of Theorem 3 yields σ(νx ) = 1 + o(1) , we obtain Sx = o(1) and,
further,

2
 ∑ h(p) 

 ≤ Sx + o(1) = o(1) .
pB(x)
√
x<p≤x
h(p)>0
Now repeatedly changing x to
obtain
∑
√
x and using the condition of Theorem once more we
xδ <p≤x
|h(p)|≥εB(x)
1
≪ δ −1 Tx + oε (1) = oε,δ (1)
p
for each ε > 0 and 0 < δ ≤ 1 . It means that the function h is of the Kubilius type (see
Ruzsa, [18]) and one can apply the Kubilius truncation procedure ([5]). Corresponding
limit theorem for the sum of independent random variables ξp , p ≤ x , normalized by
A(x) and B(x) , implies the condition (4). Theorem 3 is proved.
3. Concluding remark
Corollary 3 in (Hildebrand [4]) asserts that σ(νx ) = 1 + o(1) implies ψ(u) = 0 ,
0 < u ≤ 1 . That in the special case can be seen from our Lemma 5. The Hildebrand’s
proof is based upon the equality
Q
∫1 ∑
∑ h(p)h(q) ∫
h(p)h(q)
dt =
dt ,
pq
pq
p, q≤x
p, q≤xt
0
{
,
log q
log x
P
}
log p
log q
and Q = log
x + log x , which is false as the example
{
}
log p
log q
h(p) ≡ 1 shows. It should be Q = min log x + log x , 1 , but then the approach fails.
where P = max
log p
log x
pq>x
pq>xt
The mistake was repeated in Supplement of the book (Elliott [2]), therefore we could not
use these calculations directly.
In connection to the problems considered in the present paper it would be very desirable to obtain necessary and sufficient conditions for the relation σ(νx ) = σ + o(1) . Is it
true that sup lim sup σ(νx ) ≤ 1 ?
h̸=0
13
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14
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15