University of Paderborn
Faculty
of
Electrical Engineering, Computer Science and
Mathematics
Mean values of multiplicative
functions over multiplicative
arithmetical semigroups
Dissertation
by
László Germán
Supervisor
Prof. Dr. Dr. h.c. mult. Karl-Heinz Indlekofer
2010.
Acknowledgments
I would like to thank Prof. Dr. Dr. h.c. mult. Karl-Heinz Indlekofer for
his support, for his useful advices and for introducing me to the topic of
arithmetical semigroups. I would also like to thank Prof. Dr. Imre Kátai and
Prof. Dr. Oleg Klesov for their constructive remarks. Most of all, I thank my
family. Without them, I could not have completed this work. Special thanks
to colleagues and friends for their assistance.
Chapter 1
Introduction
Prime numbers play a central role in mathematics, due to their atomic na-
ture. The justication for this is the Fundamental Theorem of Arithmetic,
which says that each positive integer can be factorized into a product of
prime numbers and this factorization is unique up to the order of the terms.
This fact is not as obvious as it seems, which can be seen through examples
given by particular ideals of number elds. Since prime numbers serve as a
basic concept, it is natural to ask how many of them there are. More than
2000 years ago, Euclid proved that there are innitely many primes among
the naturals and Eratosthenes gave a method for recognizing prime numbers.
Since the ancient Greeks worked with complicated objects such as "propor-
tion" instead of numbers, the revolutionary work of Fibonacci (Liber Abaci,
1202) concerning the introduction of the Arabic Numeral System in Europe
was principal to the theory of prime numbers as well as to the whole math-
ematics. The next important step in the investigation of the distribution of
prime numbers was the work of Euler (early 18th century) who proved that
X
p
prime
1
p
diverges and who introduced the equation
X
n∈N
1
nσ=Y
p
prime
(1 −1
pσ)−1,
(1.1)
which is valid for
σ > 1
and turned out to be essential for today's research.
At the end of that century, Gauss and Legendre conjectured that the number
of primes up to a positive integer
n
is asymptotically
n
log n.
1
Moreover, Gauss later proposed the expression
Rn
2
1
log udu
instead of
n
log n
. The
conjecture of Gauss and Legendre was proved independently by Hadamard
and de la Vallée Poussin in 1896. As a consequence of their works, we have
for
x > 2
(π(x) :=) X
p≤x
p
prime
1 = x
log x+R(x),
(1.2)
where
R(x) = o(x
log x) (x→ ∞).
This result is now known as the Prime Number Theorem (PNT). In fact,
Hadamard and de la Vallée Poussin proved more than what was conjectured
by Legendre and Gauss, namely that
π(x) = Li(x) + R(x)
(1.3)
holds where
Li(x) = Zx
2
1
log udu,
and with some xed
c > 0
R(x)xexp(−cplog x).
The implied constant in the Vinogradov symbol
is uniform for all
x > 2
,
but not for
c
.
Their result was based on the work of Riemann, who combined equation (1.1)
with Fourier theoretic methods to obtain information on
π(x)
. He introduced
the function
ζ(s) = X
n∈N
1
ns,
(1.4)
which is absolutely convergent for all complex numbers
s
with
<s > 1
, and he
extended Euler's equation (1.1) for those
s
. Riemann showed how
ζ(s)
could
be analytically continued to a meromorphic function on the whole complex
plane, with a pole of order 1 at
s= 1
with residue 1, i.e. that
ζ(s) = 1
s−1+h(s)
(1.5)
2
holds for all
C\{1}
where
h(s)
is an entire function. This uniquely dened
meromorphic function is one of the most famous functions in mathematics,
and it is called Riemann's zeta function. He established a functional equation
for this function, from which one can see that the zeta function has zeros
which occur for all negative even integers
s=−2,−4,−6. . .
. Those zeros are
called "trivial" zeros while the other zeros are called "non-trivial" zeros. It
was recognized by Riemann that the size of the function
R(x)
in (1.3) depends
on the zeros of this function and he made some conjectures about it, most
of which seem to be unreachable even with today's tools of mathematics.
One of these conjectures states that all of the "non-trivial" zeros of the zeta
function lie on the vertical line with real part 1/2. It is equivalent to
R(x)x1/2log x
in (1.3) (See for example [38]). Many of the properties of the zeta function
depend on the fact that
(N(x) :=) X
n≤x
1 = x+O(1).
(1.6)
It is known that the Fundamental Theorem of Arithmetic is not valid in
general for the algebraic integers in algebraic number elds (as for example in
Q(√−6)
), but the uniqueness can be obtained for the multiplicative structure
of ideals. That is, every proper ideal in the ring of algebraic integers over
an algebraic number eld can be uniquely factorized into a product of prime
ideals. In 1897, Weber proved that
N(x)
, the number of integral ideals with
norm not exceeding
x
satises
N(x) = Ax +O(xθ),
where
A > 0
and
0< θ < 1
are constants and depend on the base eld. Lan-
dau proved in [34] that the prime ideals satisfy the Prime Number Theorem,
i.e. the number of prime ideals with norm not exceeding
x
satises formula
(1.2).
Knopfmacher, partially motivated by this and other results, introduced the
terminology of the so-called arithmetical semigroups in [31, 32]. Let
P
be a
nonempty set, and let
G= (G,·)
be the free commutative monoid generated
by
P
. Suppose that the mapping
|.|:G → R≥1
satises
1.
|m·n|=|m|·|n|
for all
m, n ∈ G
,
2.
#{n∈ G :|n| ≤ x}
is nite for each
x≥1
. (niteness property)
3
In this case we say that
G
forms a multiplicative arithmetical semigroup
(or simply multiplicative semigroup) with norm function
|.|
. The connec-
tion between the distribution of elements and prime elements of arithmetical
semigroups is a central question in that topic, and was intensively studied
by many authors. Let
(NG(x) =)N(x) := X
n∈G
|n|≤x
1,
and let
(πG(x) =)π(x) := X
p∈P
|p|≤x
1.
The assumptions of Knopfmacher, which he called Axiom A, considers all
multiplicative semigroups satisfying
N(x) = Axδ+O(xη),
where
A, δ > 0
,
0≤η < δ
. The result of Landau concerning the distribution
of Prime Ideals ts in this context and can be seen as an example of the fol-
lowing Abstract Prime Number Theorem, which was proved in its generality
by Knopfmacher in [31]
Theorem (Knopfmacher, Landau).
Let
G
be a multiplicative semigroup
which satises
N(x) = Axδ+O(xθ)
with some
θ < δ
, and
A > 0
for all
x≥1
. Then
π(x) = xδ
δlog x+o(xδ
log x) (x→ ∞).
Under Axiom A Knopfmacher developed a broad of arithmetical semigroups
paying attention to arithmetical functions which are of special interest. Sub-
sequent authors investigated the stability of the error term in the theorem
above, i.e. in which way the error term in the PNT depends on the error
term of
N(x)
. (See [3, 5, 6, 2, 45]) Moreover Fainleb investigates in [13] the
properties of
N(x)
using asymptotics for
ψ(x)
, where
ψ(x) = X
|p|α≤x
log |p|.
4
One could call his result the "inverse PNT".
The set of natural numbers is the standard example for a multiplicative semi-
group. Together with the above mentioned case with the integral ideals of
algebraic number elds it would be an example for a multiplicative semigroup
for which the norm function is integer valued. In [3] Beurling suggested the
following general structure. Let
P
be a sequence of positive real numbers
(p1, p2, . . .)
such that
1< p1≤p2≤ ··· , pj→ ∞
as
j→ ∞.
He called the elements of
P
the (generalized)g-primes. The g-integers are
then the real numbers
(n0, n1(= p1), n2, . . .)
of the form
Q∞
j=1 pνj
j
where each
νj
is allowed to range over all nonnegative integers. Note that the values of
g-integers are not necessarily distinct. In the case
ni−1< ni=ni+1 =··· =ni+m−1< ni+m
we say that
ni
has multiplicity
m
, but if we take them as if they were dis-
tinct (as it is the case in the literature) then the set of g-integers is just a
multiplicative arithmetical semigroup. Further examples and applications of
arithmetical semigroups can be found in [32, 14, 1, 19].
A large number of scientic papers investigate the connection between
N(x)
,
ζ(s)
and
π(x)
. In order to understand better the behaviour of primes of
natural numbers, there was a need to prove the PNT by elementary methods,
i.e. without using the theory of complex functions. Another motivation was
described by Hardy as quoted in [17]:
No elementary proof of the prime number theorem is known, and one
may ask whether it is reasonable to expect one. Now we know that the theorem
is roughly equivalent to a theorem about an analytic function, the theorem that
Riemann's zeta function has no roots on a certain line. A proof of such a
theorem, not fundamentally dependent on the theory of functions, seems to
me extraordinarily unlikely. It is rash to assert that a mathematical theorem
cannot be proved in a particular way; but one thing seems quite clear. We
have certain views about the logic of the theory; we think that some theorems,
as we say `lie deep' and others nearer to the surface. If anyone produces
an elementary proof of the prime number theorem, he will show that these
views are wrong, that the subject does not hang together in the way we have
supposed, and that it is time for the books to be cast aside and for the theory
to be rewritten.
More than a half century after the result of Hadamard and de la Valée
Poussin, A. Selberg was able to make the rst step into this direction by
5
developing his famous Symmetry Formula. Using this formula Erd®s and
Selberg gave the rst elementary proofs of the PNT [12, 41] independently.
Landau proved in [35] pp. 567-574 that
X
n≤x
µ(n) = o(x) (x→ ∞)
is equivalent to the PNT, where
µ
is the Möbius function which is dened by
µ(n) =
1
if
n= 1
0
if
p2|n
for some prime
p
(−1)r
if
n=p1···pr, p1<··· < pr.
It is clear that the Möbius function is uniquely determined by the values
on prime powers, and
µ(nm) = µ(n)µ(m)
whenever
m
and
n
do not have
common prime divisors. In general, an arithmetical function
f:N→C
is
said to be multiplicative if
f(nm) = f(n)f(m)
for all
(m, n) = 1
.
The result of Landau was of special interest in producing alternative proofs
of the PNT. In general, the asymptotic of the partial sums of
f
played an
important role, where
f
is an arithmetical function, i.e. what can we say
about the limit behaviour of
L(x)
as
x
tends to innity where
L(x) = 1
xX
n≤x
f(n) ?
The limit of
L(x)
as
(x→ ∞)
, if it exists, is called the
mean value
of
f
and
is denoted by
M(f)
. This question was intensively investigated by several
authors, especially for multiplicative functions with values in the unit disc.
For further references as well as for investigations of arithmetical functions
on subsets of the natural numbers see [28, 29, 30, 36, 37, 42, 43].
Then the result of Landau says that the PNT is equivalent to the fact that
the Möbius function possesses the mean value zero.
For multiplicative
f
with
|f(n)| ≤ 1
Delange [9] proved that the mean value
M(f)
exists and is dierent from zero if and only if the series
X
p
1−f(p)
p
(1.7)
converges, and for some positive
r
,
f(2r)6=−1
.
Assuming further that
f
is real-valued and the above series diverges, Wirsing
[49] proved that
f
has mean-value
M(f) = 0
. In particular, this means that
6
the mean value
M(f)
always exists for real-valued multiplicative functions
of modulus
≤1
, and that the PNT holds.
The result of Delange and Wirsing was extended by Halász in [20] to complex
valued functions.
Theorem (Halász).
Let
f
be a multiplicative function with
|f(n)| ≤ 1
,
satisfying
X
p
1−Ref(p)p−iτ
p<∞
for some real
τ
. Then
1
xX
n≤x
f(n) = xiτ
1 + iτ Y
p≤x1−1
pY
p≤x 1 + X
m≥1
f(pm)
pm(1+iτ)!+o(1)
as
x
tends to innity. On the other hand, if there is no such
τ
then
1
xX
n≤x
f(n) = o(1) (x→ ∞).
In [8], Daboussi and Indlekofer proved this result by elementary methods,
thus giving an alternative elementary proof of the PNT (compare [23],[43]).
Indlekofer developed in [24, 25] ,[26] a simpler way to investigate the asymp-
totic behaviour of
L(x)
. His method turns out to have applications in gener-
alizations of the questions mentioned just above as we will see in subsequent
chapters (compare [27]).
Let
f:N→C
be a multiplicative function with values in the unit disc, and
let the corresponding Dirichlet generating function formally dened by
F(s) = X
n
f(n)
ns
which is absolutely convergent for
<s > 1
. Let furthermore
M(x) := X
n≤x
f(n)
for
x > 1
.
Indlekofer's method could be described briey as follows:
By using a convolution identity after applying twice the "dierential like"
logarithm operator, we obtain with a suitable weight function
h
log2xM(x) = X
n≤x
M(x
n)h(n) + Error1(x).
7
Then partial summation allows us to use the special behaviour of
h
(as for
example Selberg's Symmetry Formula) to deduce
X
n≤x|M(x
n)h(n)| ≤ cx log xZx
1
|M(u)|
u2du +Error2(x),
from which an application of Parseval's identity leads to
log2xM(x)≤cx log3/2x(Z∞
−∞ |F(1 + 1
log x+iτ)
1 + 1
log x+iτ |2dτ)1/2+Error3(x).
Indlekofer obtained two other strong versions of the last inequality, one of
which includes the square integral of the derivative of the generating function
over the abscissa at
1/log x
, and the other averages this last square integral
over the abscissas between
1/log x
and 1. (See [26]) After some computation
we can ensure that the contribution of
Error3(x)
is negligible and after using
some elementary properties of
F(s)
on the half plane right to 1 we arrive to
an estimation of
M(x)
. This estimation lies at least as deep as the PNT,
since applying for
f=µ
,
X
n≤x
µ(n) = o(x) (x→ ∞)
would follow. Moreover, this method is suitable to give quantitative estima-
tions for the partial sums, as it was shown by Germán, Indlekofer and Klesov
in [16] for multiplicative functions
f
for which
|f(n)| ≤ 1
does not necessarily
hold.
A generalization of Halász's result for arithmetical semigroups under Axiom
A was done in [37] by Lucht and Reifenrat. They also show that their result
implies the PNT for such semigroups. It was pointed out by Zhang in [45, 46]
that in general arithmetical semigroups the Landau result concerning the
equivalence of the PNT and the mean value zero for the Möbius function
does not necessarily hold, thus disproving a conjecture of Hall in [21].
The aim of this work is to generalize the Halász theorem to arithmetical
semigroups
G
under quite general conditions by using the Indlekofer method
described above, to prove a quantitative version of Halász's theorem and to
characterize the limit distribution of additive arithmetical functions dened
on
G
.
A general assumption on multiplicative semigroups used throughout in the
present work, is that
N(x) = Axδlogβx+R1(x) (x→ ∞)
(1.8)
8
with
A, δ > 0
,
β≥0
and
R1(x) = o(xδlogβx) (x→ ∞).
In general, relation (1.8) does not imply the PNT. The possible zeros of the
generating zeta function play an important role here (See for example [3, 7]
and for the stability of the Prime Number Theorem [3, 18]). Beurling has
shown in [3] that if
c0, . . . , cr
are constants,
c > 0
and
N(x) = cx logβx+x
m
X
r=0
crlogβrx+O(xlog−ηx)
(1.9)
with
β0< β1<··· < βm< β
such that
β≥0
, then the corresponding zeta
function may have zeros on the vertical line with real part 1. Denoting the
imaginary part of these possible zeros by
tn,(n∈N)
, that is
ζ(1 + itn)=0 n∈N,
then
X
|tn|>0
α(tn)≤(β+ 1)/2
where
α(t)
- the degree of the corresponding zero - is dened by
lim inf
σ→1+
log |ζ(σ+it)|
log(σ−1) .
Hence there are only nitely many zeros. Denoting them by
t1, t2, . . . , tl
Beurling obtained that if
η > 1 + (β+ 1)/2
then the prime counting function
π(x)
is such that
π(x) = (β+ 1 −2
l
X
k=1
α(tk)
1 + t2
k
(tksin(tklog x) + cos(tklog x))) x
log x+o(x
log x)
(x→ ∞).
Thus in addition to (1.8) we shall suppose
π(x)=(β+ 1
δ−2
m
X
r=1
αr
pδ2+t2
r
cos(trlog x−θr)) xδ
log x+R2(x)
(1.10)
with some positive integer
m
, where
αr∈N0
,
tr>0
,
r= 1, . . . , m
such that
m
X
r=1
αr≤β+ 1
2,
9
and
θr
,
r= 1, . . . , m
are the angles such that
sin θr=tr
pδ2+t2
r
and
cos θr=δ
pδ2+t2
r
and
R2(x) = o(xδ) (x→ ∞).
In Chapter 3 we prove that under these conditions Halász theorem remains
valid, i.e. if
f
is a multiplicative function with
|f(n)| ≤ 1
such that the series
X
p
1−<f(p)|p|−ia
|p|δ
(1.11)
converges for some real
a
, then
X
|n|≤x
f(n) = N(x)xiaδ
δ+ia Y
|p|≤x
(1 −1
|p|δ)(1 + X
α
f(pα)
|p|α(δ+ia)) + o(N(x)) (x→ ∞).
(1.12)
On the other hand, if there is no such
a
then
X
|n|≤x
f(n) = o(N(x)) (x→ ∞).
Note that it would be enough to consider only the case
δ= 1
and then
by adjusting the value of
a
in (1.11) to deduce the case of an arbitrary
δ
.
Nevertheless, for more transparency of the results we develop the general case.
For a larger class of functions with
δ= 1
Zhang proved in [47] supposing (1.9)
with
η > η0
and supposing (1.10) together with
R2(x) = O(xlog−Mx)
where
0< η0≤M
, that
(F(σ+iτ) =)F(s) = c
(s−1−iα)ξL(1
σ−1) + o(|s|
σ−1) (σ→1+)
(1.13)
implies
X
|n|≤x
f(n) = cx1+iα log xξ−1
Γ(ξ)(1 + iα)L(log x) + o(xlogτ−1x) (x→ ∞),
10
where
ξ≥1
and
L(u)
is a slowly oscillating function with
|L(u)|= 1
and
Γ
is
the Gamma function. If
|f(n)| ≤ 1
, then
F(s)
satises (1.13), thus implying
the results of Chapter 3 but under stronger conditions. Unfortunately, his
method is not suitable to give quantitative results, which would have appli-
cation to the distributions of additive functions (compare [11]). We deal with
such estimations in Chapter 4. Assuming (1.9) we show how the error term
in (1.12) depends on the values of
f
at prime powers, if
f
is in some sense
close to a real
κ > 1
2(β+1)
. In Chapter 5 we show that under the conditions
of Chapter 4 for an additive arithmetical function
f:G → R
there exists a
distribution function
G
for which
lim
x→∞
1
N(x)#{n∈ G,|n| ≤ x:f(n)≤z}=G(z)
holds in all continuity points
z
of
G
if and only if the three series
X
|f(p)|≤1
f(p)
|p|δ,X
|f(p)|≤1
f2(p)
|p|δ,X
|f(p)|>1
1
|p|δ
converge. If the limit law
G
exists then the characteristic function
ψ(t)
of
G
is given by
ψ(t) = Y
p
(1 −1
|p|δ)(1 + X
α≥1
eitf(pα)
|p|δα )
and
G
is continuous if and only if
X
f(p)6=0
1
|p|δ
diverge. This result could be compared with Kolmogorov's three series the-
orem for the limit distribution of sums of independent random variables.
11
Chapter 2
Dirichlet convolution over
arithmetical semigroups and
generating zeta functions
The Dirichlet algebra
FG
of
G
consists of all arithmetical functions
f:G → C
with the usual linear operations and the Dirichlet convolution
∗:FG×FG→
FG
as multiplication, which for
f, g ∈ G
is dened by
f∗g(a) = X
mn=a
f(m)g(n) (a∈ G).
The convolution is associative and commutative, thus
FG
is a commutative
C
-algebra.
Let
f, g :G → C
be arithmetical functions on
G
. The logarithm function on
G
will be denoted by
L
, i.e.
L(n) = log |n|
,
n∈ G
.
It is clear that
L
is a dierential operator in some sense, that is
L·(f∗g) = (L·f)∗g+f∗(L·g).
(2.1)
Let
(n) = (1
if |n|=1
0
otherwise,
such that
f∗(n) = X
d|n
f(d)(n
d) = f(n).
13
If
f(1) 6= 0
then the inverse of
f
is dened by
f−1(1) = 1
f(1)
and
f−1(n) = −Pd|n
|d|<|n|f−1(d)f(n
d)
f(1)
for all
|n|>1
. It is easy to see that with this denition
f−1∗f=
holds. The corresponding von Mangoldt function
Λf
is dened by
Λf=Lf ∗f−1.
The above dened
f
is called multiplicative if
f(nm) = f(n)f(m)
if
(n, m) =
1
and is called completely multiplicative if
f(nm) = f(n)f(m)
holds for all
m, n ∈ G
. Here
(n, m) = 1
indicates that
n, m
does not have a common
prime divisor. Since for such functions
f(1) = 1
, the inverse
f−1
exists and
it is easy to see that for completely multiplicative
f
we have
f−1(n) = µ(n)f(n) (n∈ G).
Here
µ
is the inverse to the function
10
, where with an
a∈R
1a(n) = |n|ia (n∈ G).
For completely multiplicative
f
we clearly have
Λf=fΛ,
where
Λ = Λ10
. Furthermore,
Λ
is supported only on prime powers, and an
easy computation shows
Λ(pα) = log |p|(p∈ P).
Note, that if
f, g
are multiplicative then
f∗g, f−1
, is multiplicative as well.
It is convenient to associate the summatory function to an arbitrary
f:G →
C
(M(x) =)Mf(x) = X
n∈G
|n|≤x
f(n).
For further references on Dirichlet convolution see [31]. Let the generating
Riemann Zeta function of
G
be dened formally by
(ζG(s) =)ζ(s) := X
n∈G
1
|n|s.
14
It may happen that the series on the right hand side converges for some
s∈C
. Suppose that the abscissa of absolute convergence for
ζ
is nite, and
denote it by
δ
. It is well known that the Euler product formula
ζ(s) = Y
p∈P
(1 −|p|−s)−1
holds for all
s
with
<s>δ
and that
ζ(s)6= 0
for such
s
. In this region we
write
ζ(s) = H(s)
(s−δ)β+1
(2.2)
for
β∈R+
0
and appropriate
(Hβ(s) =)H(s)
. Suppose that for
c1>0
there
exist
c2, c3
positive constants such that
c2<|H(s)|< c3
holds for
σ > δ
and
|τ| ≤ c1
. Then
−ζ0
ζ(s) = −X
n∈G
Λ(n)
|n|s
=Ilog H(s)
(s−z)2dz +αIlog(z−δ)
(s−z)2dz
1
|s−δ|
(2.3)
holds in the same neighbourhood of
δ
to the right. Let
ψ(x) := X
|n|≤x
Λ(n).
The two counting functions
ψ(x)
and
π(x)
are connected by the partial sum-
mation formula such that it is easy to compute the one from the other. By
the absolute convergence of the generating zeta function we obtain by partial
summation that
−ζ0
ζ(s) = lim
x→∞x−sψ(x) + sZx
1
ψ(u)
us+1 du.
Since the series represented by the left hand side converges absolutely, and
both terms on the right hand side are positive for real
δ < s
, we obtain that
ψ(x)xδ+
15
for all
> 0
. It follows immediately that
−ζ0
ζ(s) = sZ∞
1
ψ(u)
us+1 du.
This relation plays an important role in the theory of prime numbers. A
common regularity assumption on
ψ(x)
is
ψ(x)xδ
which Chebyshev originated. An example of Hall in [21] shows that this does
not necessary holds, i.e. there are semigroups for which with
lim inf
x→∞
ψ(x)
N(x)=a
and
lim sup
x→∞
ψ(x)
N(x)=A
either
a= 0
or
A=∞
. It is an open question (see [48]) what values
a
and
A
can take.
Let
N(x) := X
|n|≤x
1.
Suppose that
N(x) = cxδlogβx+o(xδlogβx) (x→ ∞)
(2.4)
where
β≥0
,
c, δ > 0
. Then (2.2) holds. Generally, we have
Lemma 1.
Suppose that
N(x)∼cxδlogβx(x→ ∞)
where
c, δ > 0
,
β≥0
. Then
ζ(s) = cδΓ(β+ 1)
(s−δ)β+1 +o(1
(σ−δ)β+1 ) (σ→δ+)
uniformly for all
|τ| 1
.
16
Proof.
Using partial summation we obtain for all complex
s
with
σ > δ
X
|n|≤y
1
|n|s=y−sN(y) + sZy
1
N(u)
us+1 du.
(2.5)
Since
Z∞
1
N(u)
us+1 du
converges absolutely, letting
y
to innity, the right hand side of (2.5) tends
to a nite value. Thus
X
n
1
|n|s
converges absolutely for
σ > δ
and uniformly for
σ≥δ1> δ
. Let
> 0
be
xed. Then there exists
1< l()
, such that
Zy
1
N(u)
us+1 du =cZy
1
logβu
u(s−δ+1) du
+O(Zl()
1
logβu
u(σ−δ+1) du +Zy
l()
logβu
u(σ−δ+1) du)
(2.6)
holds if
y
is large enough. We compute
Zy
1
logβu
us−δ+1 du =Zy
1
exp(−(s−δ+ 1) log u+βlog log u)du
=Zlog y
0
exp(−(s−δ)t+βlog t)dt
=1
s−δZ(s−δ) log y
0
exp(−u)( u
s−δ)βdu
=1
(s−δ)β+1 Γ(β+ 1) + o(1
|s−δ|β+1 ) (y→ ∞).
(2.7)
Here we used that
exp(−u)uβ
is holomorphic for
<u > 0
, such that by Cauchy's theorem
Iexp(−u)uβdu = 0
17
holds for every closed path contained in
{z∈C:<z > 0}
. Since
Zlog y
(s−δ) log y
exp(−u)uβdu log yexp(−(σ−δ) log y) logβy
for all large enough
y
, (2.7) follows. Substituting into (2.6), then letting
y
to
innity, we deduce
|ζ(s)−scΓ(β+ 1)
(s−δ)β+1 | ≤ |s|
(σ−δ)β+1 (σ→δ+),
which - since
was arbitrary - proves the assertion.
By supposing a substantially more precise error term we obtain the following
Lemma 2.
Suppose that
N(x) = cxδlogβx+O(xδlog−ηx)
where
c, δ > 0
,
β≥0
and
η > 1
. Then there exists a function
g(s)
which is
continuous and bounded for
<s≥δ
, such that for
<s>δ
ζ(s) = cδΓ(β+ 1)
(s−δ)β+1 +cΓ(β+ 1)
(s−δ)β+sg(s).
Proof.
As in the proof of the previous lemma
ζ(s) =sZ∞
1
N(x)
xs+1 dx
=sZ∞
1
cxδlogβx
xs+1 dx +sZ∞
1
N(x)−cxδlogβx
xs+1 dx.
Putting
g(s) := Z∞
1
N(x)−cxδlogβx
xs+1 dx,
we obtain that by the conditions
g
is uniformly bounded and continuous for
<s≥1
.
Suppose that
c0, . . . , cr
are constants,
c > 0
and
N(x) = cx logβx+x
m
X
r=0
crlogβrx+R1(x)
18
with
−2< β0< β1··· < β
real numbers such that
β > −1
where
R1(x) = O(xlog−ηx)
with
η > max(3/2,1 + (β+ 1)/2)
. In [3] Beurling has shown that there are
certain non-zero real numbers
t1, t2...,tl
, such that
ψ(x) = (β+ 1 −2
l
X
k=1
1
p1 + t2
k
(cos(tklog x−arctan tk)))x+R2(x)
where
R2(x) = o(x) (x→ ∞).
In the following chapter we suppose
N(x) = cxδlogβx+R3(x),
ψ(x) = (β+ 1 −2
l
X
k=1
1
p1 + t2
k
(cos(tklog x−arctan tk)))xδ+R4(x),
where
c, δ > 0
,
β≥0
and
R3(x) = o(xδlogβx) (x→ ∞)
further,
t1, t2, . . . , tk
are non-zero real numbers and
R4(x) = o(xδ) (x→ ∞).
Under these conditons we show for multiplicative functions with values in
the unit disc that if
X
p
1−<f(p)|p|−ia
|p|δ
converges for some real
a
("convergent" case) then
X
|n|≤x
f(n) = N(x)xiaδ
δ+ia Y
|p|≤x
(1 −1
|p|δ)(1 + X
α
f(pα)
|p|α(δ+ia)) + o(N(x)) (x→ ∞).
On the other side, if there is no such
a
("divergent" case) then we deduce
X
|n|≤x
f(n) = o(N(x)) (x→ ∞).
19
Zhang proved a general result in [47], which as a consequence includes the
above result for
δ= 1
with stronger conditions. His condition was that with
appropriate positive
η0, M0
in the above formulation
N(x)
and
ψ(x)
satises
β≥0
and
R1(x) = O(xlog−ηx)
with
η > η0>0
and
R2(x) = O(xlog−Mx) (x→ ∞)
where
M > M0>0
. Furthermore, under the conditions
Z∞
1
|N(x)−Ax|
x2dx < ∞
and either
Zx
1
(N(t)−At) log t
tdt x
or
Z∞
1
|N(x)−Ax|2log x
x3dx < ∞
he proved the "divergent" case in [46].
20
Chapter 3
An analogue of the Theorem of
Halász
We are interested in the limit behaviour of
P
n∈G
|n|≤x
f(n)
as
x→ ∞
. In this
chapter we generalize the Theorem of Halász mentioned in the introduction
to arithmetical semigroups satisfying
N(x) = cxδlogβx+R(x),
ψ(x) = (β+ 1
δ−2
m
X
r=1
αr
pδ2+t2
r
cos(trlog x−θr))xδ+o(xδ) (x→ ∞)
with
c, δ > 0
,
β≥0
, and
R(x) = o(xδlogβx) (x→ ∞).
Here
αr∈N0
and
tr
,
r= 1, . . . , m
are real positive numbers such that
m
X
r=1
αr≤β+ 1
2,
and
θr
,
r= 1, . . . , m
are the angles which satisfy
sin θr=tr
pδ2+t2
r
and
cos θr=δ
pδ2+t2
r
.
The "divergent case" is entiled in Corollary 2 while the "convergent case" in
Corollary 1.
The partial sums of
f
will be estimated rst via convolution identity from
21
which we obtain a sum over the weighted partial sums of
f
. Then using the
regularity properties of the weights (Selberg Symmetry Formula), this leads
to an integral which avarages the partial sums of
f
. Using Parseval's identity
we relate this to the
L2
norm of
F(s)
s
where
F
is the Dirichlet generating
function of
f
. Then using elementary properties of the zeta function we
estimate the Fourier transform.
3.1
A convolution identity
The following result was proved for
G=N
in [25], and it remains true for
general arithmetical semigroups.
Theorem 1 (Indlekofer).
Let
f:G → C
be a multiplicative function, and
put
(M(x) =)Mf−A1a(x) = P|n|≤xf(n)−A1a(n)
, where
A∈C
and
a∈R
.
Dene
˜
f
to be a completely multiplicative function with
˜
f(p) = f(p)
, and let
g
be dened by
f=˜
f∗g
. Then for
x > 1
log2xM(x) = X
|n|≤x
M(x
|n|)˜
f(n){Λ∗Λ(n) + L(n)Λ(n)}
+X
|n|≤x{R1+R2+R3}(x
|n|)˜
f(n)Λ(n)
+ (R1+R2+R3)(x) log x,
where
R1(x) = X
|n|≤x
log x
|n|(f(n)−A1a(n)),
R2(x) = X
|n|≤x
(X
|m|≤ x
|n|
˜
f(m))g(n) log |n|,
and
R3(x) = −X
|n|≤x
(X
|m|≤ x
|n|
A1a(m))Λ(n)(1a(n)−˜
f(n)).
Proof.
We have
log xM(x) = X
|n|≤x
L(n)(f−A1a)(n) + R1(x),
22
where
R1(x) = X
|n|≤x
(f−A1a)(n) log x
|n|.
Since by (2.1)
Lf =L(˜
f∗g) = L˜
f∗g+˜
f∗Lg =Λ˜
f∗˜
f∗g+˜
f∗Lg
=Λ˜
f∗f+˜
f∗Lg
we obtain
X
|n|≤x
L(f−A1a)(n) = X
|n|≤x
Λ˜
f∗f(n)−AX
|n|≤x
Λ1a∗1a(n) + R2(x),
where
R2(x) = X
|n|≤x
˜
f∗Lg(n).
By rearranging the terms, it follows
log xM(x) = X
|n|≤x
M(x
|n|)Λ˜
f(n)+(R1+R2+R3)(x),
(3.1)
where
R3(x) = −AX
|n|≤x
1a∗(Λ1a−Λ˜
f)(n).
Thus,
log2xM(x) = X
|n|≤x
log x
|n|M(x
|n|)Λ˜
f(n)
+X
|n|≤x
M(x
|n|)Λ˜
fL(n)+(R1+R2+R3)(x) log x.
By (3.1) the right hand side equals
X
|n|≤xX
|m|≤ x
|n|
M(x
|mn|)Λ˜
f(m)Λ˜
f(n) + X
|n|≤x
(R1+R2+R3)( x
|n|)Λ˜
f(n)
+X
|n|≤x
M(x
|n|)Λ˜
fL(n)+(R1+R2+R3)(x) log x.
By rearranging the summation in the rst term, the assertion follows.
23
The set
|G| := {|n|:n∈ G}
is discrete. Its elements will be represented
by
nν
with
ν∈N
in a non-decreasing way. For an
x∈R≥1
let
([x]G=
)[x] := max {|n| ≤ x:n∈ G}
and let
λ(x)
be the uniquely dened integer
with
nλ(x)= [x]
.
Lemma 3 (Partial Summation).
Let
a, b :N→C
, and
y > 1
. Then
X
ν≤y
a(ν)b(ν) = X
ν≤y−1
A(ν){b(ν)−b(ν+ 1)}+A(y)b([y]N),
where
A(u) = X
µ≤u
a(µ).
Lemma 4.
Let
N(x)∼cxδlogβx(x→ ∞)
where
c, δ > 0
,
β≥0
. Then
nν=nν−1(1 + o(1))
as
ν→ ∞
.
Proof.
There exists a sequence
(ν)→0
as
ν→ ∞
with
N(nν−1) =N(nν−(ν))
=(c+o(1))(nν−(ν))δlogβ(nν−(ν)) (ν→ ∞).
On the other side
N(nν−1)=(c+o(1))nδ
ν−1logβnν−1
as
ν→ ∞
. Therefore,
(c+o(1))nδ
ν−1logβnν−1= (c+o(1))(nν−(ν))δlogβ(nν−(ν)) (ν→ ∞).
Since
(nν−(ν))δlogβ(nν−(ν)) = nδ
νlogβnν(1 + o(1))
as
ν→ ∞
, a simple computation shows that
nν= (1 + o(1))nν−1(ν→ ∞).
24
Lemma 5.
Let
η > 0
. Suppose that
N(x)∼cxδlogβx(x→ ∞)
where
c, δ > 0
,
β≥0
and that
X
|n|≤x
h(n) = (t(x) + δ(x))xδlog x(x→ ∞)
(3.2)
for some
h∈ FG
. Suppose further that
η < t(x)1
and that the derivative
satises
|t0(x)| x−1
for
x > 1
and that
δ(x)1
holds. Let
f∈ FG
. Then for all
x≥2
,
X
|n|≤x|M(x
|n|)|h(n)Zx
1|M(x
t)|log tdtδt(t)
(3.3)
+ ( max
1<u<x |δ(u)|log u+ 1)xδX
|n|≤x
|f(n)|
|n|δ
+N(x) log x.
Proof.
At rst note that under condition (3.2) we have
h(1) = 0
. For
u > 1
,
let
T(u) = 1
log uX
|n|≤u
h(n).
(3.4)
Let
˜
h(nν) = X
|n|=nν
n∈G
h(n)
and let
r(nν) := ˜
h(nν)−Znν
nν−1
log udT(u)
for
nν>1.
By the denition of
G
and condition (3.2) there exists a xed
> 0
such
that
P
|n|≤u
h(n)=0
for
1≤u < 1 +
. Since
log u
is of bounded variation in
25
[a, b]
if
1 + ≤a<b<∞
, it follows that
T(u)
is of bounded variation here
as well (See [39] Kap. VIII.3. Satz 4.). Furthermore, dening
T(u) = 0
for
1≤u < 1 +
we obtain that
Zu
1
log tdT(t), u > 1
exists. Therefore using (3.2)
X
1<nν≤u
r(nν) = X
|n|≤u
h(n)−Z[u]
1
log tdT(t) = Z[u]
1
T(t)dlog t
Zu
1
tδ
tdu
uδ.
(3.5)
Setting
a(ν) = r◦nν
and
b(ν) = |M|◦ x
nν
,
y=λ(x)
then using Lemma 3 and
(3.5) we obtain that
|X
1<nν≤x|M(x
nν
)|˜
h(nν)−X
1<nν≤x|M(x
nν
)|Znν
nν−1
log udT(u)|
=|X
1<nν≤x|M(x
nν
)|r(nν)|
does not exceed
X
nν≤nλ(x)−1
nδ
ν(||M(x
nν
)|−|M(x
nν+1
)||) + O(|M(x
nλ(x)
)|xδ).
Since
||a|−|b|| ≤ |a−b|
for all real
a
and
b
, an application of Lemma 3 again
leads to
X
nν≤nλ(x)−1
nδ
ν(||M(x
nν
)|−|M(x
nν+1
)||)X
nν≤x
H(x
nν
)(nδ
ν−nδ
ν−1)
+H(x
nλ(x)
)xδ,
(3.6)
where
H(x) = X
|n|≤x|f(n)|.
26
Furthermore, since
X
nν≤x
H(x
nν
)(nδ
ν−nδ
ν−1) = X
nν≤xX
|n|≤ x
nν
|f(n)|(nδ
ν−nδ
ν−1)
=X
|n|≤x|f(n)|X
nν≤x/|n|
(nδ
ν−nδ
ν−1)
X
|n|≤x|f(n)|(x/|n|)δ
xδX
|n|≤x
|f(n)|
|n|δ,
we deduce
X
nν≤x|M(x
nν
)|˜
h(nν) = X
1<nν≤x|M(x
nν
)|Znν
nν−1
log udT(u)
+O(xδX
|n|≤x
|f(n)|
|n|δ).
Since
N(x) = (c+o(1))xδlogβx
, varying
x
at most by an amount of
O(1
xδlogβ+1 x)
we can ensure that
Zx
1|M(x
t)|log tdT(t)
exists. The contribution of the mistake we make in (3.3) is at most
xδP
|n|≤x
|f(n)|
|n|δ
,
which is acceptable. Due to the change in
x
,
Znν
nν−1||M(x
nν
)|−|M(x
u)||log udT(u) = Znν−
nν−1||M(x
nν
)|−|M(x
u)||log udT(u).
(3.7)
The right hand side of (3.7) equals zero if
nν−1= 1
or otherwise it equals
the limit of
m+1
X
k=1 ||M(x
nν
)|−|M(x
ξk
)||log ξk(T(xk)−T(xk−1)),
(3.8)
as we take
max
k=0,...,m+1(xk−xk−1)→0
, where
x0=nν−1< x1< . . . < xm<
nν=xm+1
and
ξk∈[xk−1, xk]
,
k= 1, . . . , m + 1
. Using (3.4), by the mean
27
value theorem of calculus the above sum equals
−X
|n|≤nν−1
h(n)
m+1
X
k=1 ||M(x
nν
)|−|M(x
ξk
)||log ξk(1
%klog2%k
)(xk−xk−1)
(3.9)
with appropriate
%k∈[xk, xk−1]
. Since
M(x)
is continuous except of nitely
many points, the limit in (3.9) does not depend on the choice of
ξk
. Therefore
taking
ξk=%k
the integral equals
−X
|n|≤nν−1
h(n)Znν
nν−1||M(x
nν
)|−|M(x
u)|| 1
ulog udu.
Furthermore, we obtain
X
1<nν≤x|M(x
nν
)|Znν
nν−1
log udT(u) = X
1<nν≤xZnν
nν−1|M(x
u)|) log udT(u)
+Error,
where
Error =O(X
e<nν≤x
(X
|n|≤nν−1
h(n)){H(x
nν−1
)−H(x
nν
)}(log log nν−log log nν−1)
+xδX
|n|≤x
|f(n)|
|n|δ).
Then in view of condition (3.2)
Error =O(X
e<nν≤x
(H(x
nν−1
)−H(x
nν
))nδ
ν−1).
Here we used that by Lemma 4
log log nν−log log nν−11
log nν
uniformly for
nν> e
. A similar computation as in (3.6) above shows that
Error
is
O(xδX
|n|≤x
|f(n)|
|n|δ).
28
We arrive at
X
|n|≤x|M(x/|n|)|h(n)−Zx
1|M(x/u)|log udT(u)xδX
|n|≤x
|f(n)|
|n|δ.
Setting
L(u) := T(u)−t(u)uδ,
we obtain
Zx
1|M(x/u)|log udT(u) = Zx
1|M(x/u)|log udt(u)uδ
+|M(1)|log xL(x)−Zx
1
L(u)d|M(x/u)|log u.
We compute the second integral on the right hand side. It equals to the limit
of
m−1
X
k=0
L(ξk){|M(x
xk+1
)|log xk+1 −|M(x
xk
)|log xk}
as
max
0≤k≤m−1|xk+1 −xk| → 0
, where
x0= 1 < x1< . . . < xm=x
and
ξk∈[xk, xk+1]
. Furthermore, it is
O
m−1
X
k=0 |L(ξk)|log xk+1||M(x
xk+1
)|−|M(x
xk
)||
+
m−1
X
k=0 |L(ξk)||M(x
xk
)|(log xk+1 −log xk).
We denote the rst sum by
Σ1
and the second by
Σ2
. Then
Σ1
does not
exceed
m−1
X
k=0 |L(ξk)|log xk+1(H(x
xk
)−H(x
xk+1
))
max
1<u<x|δ(u)|log uX
xk<x
ξδ
k(H(x
xk
)−H(x
xk+1
)).
Since
H(x
u)
is a step function with jumps at
nν
, we obtain by taking
max
0≤k≤m−1|xk+1 −xk| → 0
that
Σ1
is at most
cmax
1<u<x |δ(u)|log uX
nν≤x
xδ
nδ
ν
(H(nν)−H(nν−1)).
29
Thus,
Σ1=O( max
1<u<x |δ(u)|log u)xδX
|n|≤x
|f(n)|
|n|δ.
To compute
Σ2
we note that by taking
max
0≤k≤m−1|xk+1 −xk| → 0
we have
|L(ξk)|H(x
xk
)N(x),
such that applying the mean value theorem to
log xk+1 −log xk
we obtain
Σ2N(x)Zx
1
1
udu
O(N(x) log x).
The proof is nished.
3.2
An application of Parseval's identity
Analogous to [39] Kap.VIII Satz 2 we have the following
Lemma 6.
Let
f: [a, b]→R
be an almost everywhere continuous function,
and let
g: [a, b]→R
be a function with an almost everywhere continuous
and bounded derivative on
[a, b]
such that
f
and
g0
does not have a common
point of discontinuity. Then
Zb
a
f(x)dg(x) = Zb
a
f(x)g0(x)dx.
Proof.
Obvious.
Lemma 7.
Suppose that
N(x)∼cxδlogβx(x→ ∞)
where
c, δ > 0
,
β≥0
and that
t(x)
has an almost everywhere continuous
derivative with
|t0(x)| 1
x
and
t(x)1
. Let
f∈ FG
with
|f| 1
. Then
30
i
Zx
1|M(x
t)|log tdtδt(t)xδlog3/2xH(1/log x),
(3.10)
ii
Zx
1|M(x
t)|log tdtδt(t)xδlog xT(1/log x),
(3.11)
iii
Zx
1|M(x
t)|log tdtδt(t)xδlog x{Z1/2
1/log x
T(u)
u1/2du +1
log1/2xT(1/log x)
+ 1},
(3.12)
where
H(u) := (Z∞
−∞ |F(δ+u+iτ)
δ+u+iτ |2dτ)1/2
and
T(u) := (Z∞
−∞ |F0(δ+u+iτ)
δ+u+iτ |2dτ)1/2
and
F
is the Dirichlet generating function of
f
.
Proof.
Using Lemma 6 we have
Zx
1|M(x
t)|log tdtδt(t)log xZx
1|M(x
t)|tδ−1dt
=xδlog xZx
1
|M(u)|
uδ+1 du.
To prove (3.10) we note that using the Cauchy-Schwarz inequality
(Zx
1
|M(u)|
uδ+1 du)2≤Zx
1
|M(u)|2
u2δ+1 du Zx
1
1
udu
= log xZlog x
0
|M(eω)|2
e2δω dω.
Since
1≤exp(2ω/ log x)≤e2
for
0≤ω≤log x,
31
we deduce
Zlog x
0
|M(eω)|2
e2δω dω ≤e2Z∞
0|M(eω)
e(δ+1/log x)ω|2dω.
(3.13)
Note that for
σ > δ
X
n
f(n)
|n|s=Z∞
1−
1
usdM(u)
=sZ∞
1
M(u)
us+1 du
=sZ∞
0
M(eω)
esω dω
=sZ∞
0
M(eω)
eσω e−iτωdω.
Hence, the Fourier transform of
H(ω) := M(eω)e−σω
equals
ˆ
H(τ) = (σ+iτ)−1F(σ+iτ).
Since
M(eω)eωωβ
and since by Lemma 1
F(σ+iτ)1
(σ−δ)β+1 ,
both functions
H, ˆ
H
belong to
L2(−∞,∞)
. Thus, by Parseval's identity
kHk2=kˆ
Hk2
such that with
σ=δ+ 1/log x
we obtain
Z∞
0|M(eω)
eσω |2dω =1
2πZ∞
−∞ |F(σ+iτ)
σ+iτ |2dτ.
(3.14)
To prove (3.11) let
K(u) = X
|n|≤u
f(n) log |n|.
32
Integration by parts shows that for
u > 2
,
M(u)−1 = Zu
1+
1
log tdK(t)
=K(u)
log u+Zu
1+
K(t)
tlog2tdt.
We deduce
Zx
2
|M(u)|
uδ+1 du Zx
2
|K(u)|
uδ+1 log udu +Zx
2
1
uδ+1 Zu
1+
|K(t)|
tlog2tdtdu + 1.
Further,
Zx
2
1
uδ+1 Zu
1+
|K(t)|
tlog2tdtdu =Zx
1+
|K(t)|
tlog2tZx
t
1
uδ+1 dudt
−Z2
1+
K(t)
tlog2tZ2
t
1
uδ+1 dudt
Zx
2
|K(t)|
tδ+1 log2tdt + 1.
It follows
Zx
2
|M(u)|
uδ+1 du Zx
2
|K(t)|
tδ+1 log tdt + 1.
Applying the Cauchy-Schwarz inequality, we obtain
(Zx
2
|K(t)|
tδ+1 log tdt)2Zx
1
|K(t)|2
t2δ+1 dt Zx
2
1
tlog2tdt
Zlog x
0
|K(eω)|2
e2δω dω.
Since if
u > 1
, then
1≤exp(2ω/ log u)≤e2
for
0≤ω≤log u
such that,
Zlog u
0
|K(eω)|2
e2δω dω ≤e2Z∞
0|K(eω)
e(δ+1/log u)ω|2dω.
(3.15)
33
Note that for
σ > δ
we have
X
n
f(n) log n
|n|s=Z∞
1−
1
usdK(u)
=sZ∞
1
K(u)
us+1 du
=sZ∞
0
K(eω)
esω dω
=sZ∞
0
K(eω)
eσω e−iτωdω.
Hence, the Fourier transform of
L(ω) := K(eω)e−σω
equals
ˆ
L(τ) = −(σ+iτ)−1F0(σ+iτ).
Since
K(eω)eωωβ+1
and since by Lemma 1 using Cauchy's theorem
F0(σ+iτ)1
(σ−δ)β+2 ,
both functions
L, ˆ
L
belong to
L2(−∞,∞)
. Thus, by Parseval's identity
kLk2=kˆ
Lk2
such that with
σ > δ
we obtain
Z∞
0|K(eω)
eσω |2dω =1
2πZ∞
−∞ |F0(σ+iτ)
σ+iτ |2dτ.
(3.16)
To prove (3.12) we note that
Zx
e2
|K(u)|
uδ+1 log udu Zx
e2
|K(u)|
uδ+1 log uZu
u1/2
1
tlog tdtdu
=Ze2
e
1
tlog tZt2
e2
|K(u)|
u2log ududt
+Zx1/2
e2
1
tlog tZt2
t
|K(u)|
uδ+1 log ududt
+Zx
x1/2
1
tlog tZx
t
|K(u)|
uδ+1 log ududt.
34
Furthermore, using (3.15) and (3.16) we obtain
Zx
1
|K(u)|
uδ+1 du log1/2xT(1/log x).
Hence,
Zx
x1/2
1
tlog tZx
t
|K(u)|
uδ+1 log ududt 1
log xZx
1
|K(u)|
uδ+1 du
1
log1/2xT(1/log x).
Thus,
Zx
2
|K(u)|
uδ+1 log udu Zx1/2
e
1
tlog tZt2
t
|K(u)|
uδ+1 log ududt +1
log1/2xT(1/log x)+1
Zx1/2
e
1
tlog2tZt2
1
|K(u)|
uδ+1 dudt +1
log1/2xT(1/log x)+1
Zx1/2
e
T(1/(2 log t))
tlog3/2tdt +1
log1/2xT(1/log x)+1
Z2
1/log x
T(u)
u1/2du +1
log1/2xT(1/log x)+1,
which proves (3.12).
3.3
Estimation of the Fourier transform
Lemma 8.
Suppose that
N(x)∼cxδlogβx(x→ ∞)
where
c, δ > 0
,
β≥0
. Let
f:G → C
be a multiplicative function such that
|f(n)| ≤ 1
. Then
F(s) = Y
|p|δ≤2
(1 + h(p, s)) exp( X
|p|δ>2
f(p)
|p|s)F1(s)
holds for
σ > δ
, where
h(p, s) = ∞
X
α=1
f(pα)
|p|αs
and
F1(s)
is analytic for
σ > δ/2
, and
|F1(s)| 1
for
σ≥δ
.
35
Proof.
Since
X
pX
α
f(pα)
|p|αs
converges absolutely for
σ > δ
, the Euler product representation
F(s) = Y
p
(1 + h(p, s))
holds for
σ > δ
. Dening
F1(s) := Y
p
|p|δ>2
(1 + h(p, s)) exp(−f(p)
ps),
we have to show that
F1(s)
has the asserted properties. We have
|h(p, s)| ≤|p|−σ1
1−1
|p|σ
(3.17)
for all
σ > 0
. Since for
0< < 1
|log(1 + z)−z| ≤ c()|z|2
holds for all
|z| ≤ 1−
, where
log z
is the principal value of the complex
logarithm function,
|log(1 + h(p, s)) −f(p)
|p|s| |p|−2σ
(3.18)
holds uniformly for all large enough primes. Since
X
p
1
|p|2σX
n
1
|n|2σ
converges absolutely and uniformly for all
σ≥σ0> δ/2
,
F1(s)
is dened
by a product, which converges uniformly and absolutely for such
σ
. Hence,
F1(s)
is analytic for
σ > δ/2
. If
σ≥δ
then by (3.17) with some
0>0
|h(p, s)| ≤ 1−0
holds uniformly for all primes
p
with
|p|δ>2
. Therefore, (3.18) holds for all
such primes. Consequently,
log |F1(s)| 1,
which proves the assertion. Here we note that by the denition of
G
, if
|p|δ>2
, then there exists an
> 0
such that
|p|δ>2 +
.
36
Remark.
The lemma gives for the Zeta function
ζ(s) = Y
|p|δ≤2
(1 + h(p, s)) exp( X
|p|δ>2
1
|p|s)ζ1(s), σ > δ,
where for
|p|δ≤2
|1 + h(p, s)|=|(1 −1
|p|s)−1|
≤1/0σ > σ0>0
is valid for some
0>0
. This is the result of the fact that
1 + < p
for all
primes
p
holds for some
> 0
by the denition of
G
. Furthermore,
|1 + h(p, s)| ≥ 1/2, σ > 0
for all
p
. Thus,
ζ(s)exp(X
p
1
|p|s)
uniformly for
σ > δ
.
The following lemma appeared for Dirichlet series in [38] (Lemma II.3.) but
we need it for Laplace-Stieltjes integrals.
Lemma 9 (Montgomery).
Let
A(ω)
be a function of bounded variation,
and let
B(ω)
be a non-decreasing function such that
A(0) = B(0) = 0
and
|A(ω1)−A(ω2)| ≤ B(ω1)−B(ω2)
(3.19)
for all
ω1> ω2
. Suppose that the Laplace-Stieltjes transform
Z∞
0
e−σtdB(t)
converges for some
σ > 0
. Let
F1(s) = Z∞
0
e−stdA(t),
and
F2(s) = Z∞
0
e−stdB(t).
Then
ZT
−T|F1(s)|2dτ ≤2Z2T
−2T|F2(s)|2dτ
holds for all
T > 0
.
37
Proof.
First we note that the Fejér kernel is non-negative, that is
Z1
−1
(1 −|t|)eitydt = (sin(y/2)
y/2)2≥0
(3.20)
for all
y > 0
. Therefore,
ZT
−T|F1(s)|2dτ ≤2Z2T
−2T
(1 −|τ|
2T)|F1(s)|2dτ
=2 Z2T
−2T
(1 −|τ|
2T)Z∞
0Z∞
0
e−sω1e−sω2dA(ω1)dA(ω2)dτ
=2 Z∞
0Z∞
0
e−σω1e−σω2Z2T
−2T
(1 −|τ|
2T)ei(−ω1+ω2)τdτdA(ω1)dA(ω2).
By (3.19) and (3.20), this last integral is at most
Z∞
0Z∞
0
e−σω1e−σω2Z2T
−2T
(1 −|τ|
2T)ei(−ω1+ω2)τdτdB(ω1)dB(ω2)
=Z2T
−2T
(1 −|τ|
2T)Z∞
0Z∞
0
e−sω1e−sω2dB(ω1)dB(ω2)dτ
≤Z2T
−2T|F2(s)|2dτ.
The following well known lemma (see for example [25]) is useful in many
cases and remains true for arithmetical semigroups.
Lemma 10.
Suppose that
N(x)∼cxδlogβx(x→ ∞)
where
c, δ > 0
,
β≥0
and that
ψ(x)xδ.
Then if
f∈ FG
is a multiplicative function with
|f| 1
, then
X
|n|≤x
f(n)ψ(x) log−1xX
|n|≤x
|f(n)|
|n|δ.
Proof.
At rst note that
X
|n|≤x|f(n)|log x
|n|xδX
|n|≤x
|f(n)|
nδ.
(3.21)
38
Since
log xX
|n|≤x
f(n) = X
|n|≤x
f(n) log |n|+O(X
|n|≤x|f(n)|log x
|n|)
=X
|n|≤x
f(n)X
pα||n
log |p|α+O(xδX
|n|≤x
|f(n)|
nδ)
=X
|pα|≤x
f(pα) log |p|αX
|n|≤ x
|p|α
p-n
f(n) + O(xδX
|n|≤x
|f(n)|
nδ)
X
|n|≤x|f(n)|X
|p|α≤x
|n|
|f(pα)|log |p|α+xδX
|n|≤x
|f(n)|
nδ
ψ(x)X
|n|≤x
|f(n)|
|n|δ,
the assertion follows.
Remark.
1. Since
X
|n|≤x
|f(n)|
|n|δY
|p|≤x
(1 + ∞
X
α=1
|f(pα)|
|p|δα )
exp(X
|p|≤x
|f(p)|
|p|δ),
it follows, that
X
|n|≤x|f(n)| ψ(x)
log xexp(X
|p|≤x
|f(p)|
|p|δ)
holds as well.
2. Since multiplicative functions are determined by their values on prime
powers, it is reasonable to nd an estimate for the partial sums of mul-
tiplicative functions by partial sums over their values on prime powers.
Contrary to its simplicity, this lemma allows us to estimate partial sums
quiet eectively even in non-trivial cases. For example, let
G=N
, fur-
thermore, let
0< z < 1
and let
f(pα) = z
. The Sathe-Selberg method
(see for example [44] Theorem II.6.1) gives us
X
n≤x
f(n) = xlogz−1(cz+O(1
log x))
39
while using the above lemma we obtain
X
n≤x
f(n)xlogz−1.
3.3.1
The "divergent" case
We will use (3.10) to obtain the following
Lemma 11.
Let
η > 0
. Let
N(x)∼cxδlogβx(x→ ∞)
where
c, δ > 0
,
β≥0
and let
η < t(x)1
which has an almost everywhere
continuous derivative with
t0(x)1
x
. Suppose further that for
x≥2
X
|p|≤x
1
|p|δ= (β+ 1) log log x+O(1).
(3.22)
If
f∈ FG
is multiplicative with
|f| ≤ 1
and
X
p
1−<f(p)|p|iτ
|p|δ
diverges for all real
τ
, then
Zx
1|M(x
t)|log tdtδt(t) = o(xδlogβ+2 x) (x→ ∞).
Proof.
In the case of rational integers it is possible to obtain the analogous
assertion with one of the inequalities (3.10), (3.11), (3.12). Using inequalities
(3.11) and (3.12) the problem reduces to the investigation of the logarithmic
derivative of the generating function
F
. Lemma 8 usually allows us to extract
some knowledge about the logarithmic derivative, which is closely connected
to prime number sums. Although we have some knowledge about prime
number sums in our case, the existence of the logarithmic derivative is not
always fullled. Therefore we proceed as in [25]. We use inequality (3.10)
with
A= 0
and
a= 0
, i.e.
Zx
1|M(x
t)|log tdtδt(t)xδlog3/2xH(1/log x),
(3.23)
40
where
H(u) := (Z∞
−∞ |F(δ+u+iτ)
δ+u+iτ |2dτ)1/2.
Let
M > 0
be a large xed number. Suppose that
X
p
1−<f(p)|p|iτ
|p|δ
diverges for all real
τ
. Then by Lemma 8
F(s)
ζ(σ)exp(−X
p
1−<f(p)|p|iτ
|p|σ)
holds uniformly for all real
τ
. Therefore by a theorem of Dini
F(s) = o((σ−δ)−β−1)σ→δ+
(3.24)
uniformly for all
|τ|< M
. Further
Z|τ|≤M|F(s)
s|2dτ max
|τ|≤M|F(s)|1/2Z|τ|≤M
|F(s)|3/2
|s|2dτ.
Let
h(n)
be a multiplicative function dened on prime powers by
h(pα) = (3f(p)
4
if
α= 1
0
otherwise
.
Note that by the denition of
h
,
Y
|p|δ≤2
(1 + X
α
h(pα)
|p|sα )1
uniformly for
σ > δ
. Thus using Lemma 8 and the remark concerning the
Zeta function after that lemma we obtain
|F(s)|3/4|exp(X
p
3f(p)
4|p|s)|
|X
n
h(n)
|n|s|
41
for
σ > δ
. Thus
Z|τ|≤M
|F(s)|3/2
|s|2dτ Z∞
−∞
|exp(Pp
h(p)
|p|s)|2
|s|2dτ.
Then applying Lemma 10 and Parseval's identity to
h
we deduce
Z∞
−∞
|exp(Pp
h(p)
|p|s)|2
|s|2dτ =Z∞
0|X
|n|≤eω
h(n)e(δ−σ)ω|2dω
Z∞
e|exp( X
|p|≤eω
|h(p)|
|p|δ)ω−1e(δ−σ)ω|2dω + 1.
Using condition (3.22) it follows that the last integral is at most
O((σ−δ)−3
2β−1/2),
and using (3.24) we arrive at
Z|τ|≤M|F(s)
s|2d=o((σ−δ)−2β−1).
For
|τ|> M
we have
Z|τ|>M |F(s)
s|2dτ X
|m|>M
1
m2Z|τ−m|≤1|F(s)|2dτ
M−1(σ−δ)−2β−1.
Here we used that since
˜
K(eω) := X
|n|≤eω
f(n)|n|−im
N(eω),
by (3.14)
Z1
−1|F(σ+i(τ+m))|2dτ Z∞
−∞ |F(σ+i(τ+m))
σ+iτ |2dτ
=2πZ∞
0|˜
K(eω)e−σω|2dτ
(σ−δ)−2β−1.
42
Putting everything together,
Z∞
−∞ |F(δ+u+iτ)
δ+u+iτ |2dτ =o(u−2β−1)
as
u→0
. Substituting into (3.23) the assertion follows.
Theorem 2.
Let
N(x)∼cxδlogβx(x→ ∞)
where
c, δ > 0
,
β≥0
and
ψ(x)xδ.
Suppose further that for
x≥2
X
|p|≤x
1
|p|δ= (β+ 1) log log x+O(1)
and that the Selberg Symmetry Formula holds, i.e. that
X
|n|≤x
Λ∗Λ(n) + Λ(n) log |n|=t(x)xδlog x+o(ψ(x) log x) (x→ ∞)
where
t(x)
is as in Lemma 5. Assume that
f∈ FG
is multiplicative with
|f| ≤ 1
and
X
p
1−<f(p)|p|iτ
|p|δ
diverges for all real
τ
. Then
X
|n|≤x
f(n) = o(N(x)) (x→ ∞).
Proof.
Choosing
A= 0
,
a= 0
in Theorem 1 we obtain
log2xX
|n|≤x
f(n)X
|n|≤x|M(x
n)|{Λ∗Λ(n)+ΛL(n)}
+ (R1+R2)(x) log x+X
|n|≤x
(R1+R2)( x
|n|)Λ(n),
43
where
R1(x)X
|n|≤x
log x
|n|
and
R2(x)N(x)X
|n|≤x
|g(n)|log |n|
|n|δ.
Here
g
is dened by
f=˜
f∗g
, where
˜
f
is completely multiplicative with
˜
f(p) = f
. By (3.21) we have
R1(x)N(x) log x−Zx
1
log udN(u)
=Zx
1
N(u)
u
N(u).
Further since
g=f∗˜
f−1=f∗µf,
we have
g(pα) = f(pα)−f(p)f(pα−1).
Therefore
g
is zero on primes and
|g(pα)| ≤ 2
. It follows that
R2(x)N(x)X
|n|≤x
|g(n)|log |n|
|n|δ
N(x).
Here we used that
|g(n2)| ≤ 2ω(n)
, such that
X
|n|≤x
|g(n)|
|n|3δ
4Y
|p|≤x
(1 + 2 X
α≥2
1
|p|α3δ
4
)
1,
where
ω(n)
counts the distinct prime divisors of
n
. Further since
X
|n|≤x
Λ(n)
|n|δ=Zx
1
1
uδdψ(u)
=ψ(x)
xδ+δZx
1
ψ(u)
uδ+1 du
log x,
44
we obtain
X
|n|≤x
(R1+R2)( x
|n|)Λ(n)N(x)X
|n|≤x
Λ(n)
|n|δ
N(x) log x.
By Lemma 5 we have
log2xM(x)Zx
1|M(x
t)|log tdtδt(t) + o(N(x) log2x) (x→ ∞)
and by Lemma 11
Zx
1|M(x
t)|log tdtδt(t) = o(N(x) log2x) (x→ ∞),
so that the proof is nished.
Corollary 1.
Let
N(x)∼cxδlogβx(x→ ∞)
where
c, δ > 0
,
β≥0
and let
ψ(x)=(β+ 1
δ−2
m
X
r=1
αr
pδ2+t2
r
cos(trlog x−θr))xδ+o(xδ) (x→ ∞)
where
αr∈N0
,
tr>0
,
r= 1, . . . , m
such that
m
X
r=1
αr≤β+ 1
2,
and
θr
,
r= 1, . . . , m
are the angles which satisfy
sin θr=tr
pδ2+t2
r
and
cos θr=δ
pδ2+t2
r
.
Assume further that
f∈ FG
is multiplicative with
|f| ≤ 1
and
X
p
1−<f(p)|p|iτ
|p|δ
diverges for all real
τ
. Then
X
|n|≤x
f(n) = o(N(x)) (x→ ∞).
45
Proof.
By Theorem 2 it is enough to prove that
X
|p|≤x
1
|p|δ= (β+ 1) log log x+O(1)
(3.25)
and that the Selberg Symmetry Formula holds. First note that by the con-
ditions using partial summation
X
|p|≤x
log |p|
|p|δlog x.
Then by Lemma 1 and by the remark concerning the zeta function after
Lemma 8
(β+ 1) log(σ−δ) = X
p
1
|p|σ+O(1)
for all
δ+ 1 > σ > δ
. Choosing
σ=δ+1
log x
we obtain
X
|p|≤x
1
|p|σ−1
|p|δX
|p|≤x
1
|p|δ(exp((σ−δ) log |p|)−1)
1
log xX
|p|≤x
log |p|
|p|δ
1.
Therefore
X
|p|≤x
1
|p|σ=X
|p|≤x
1
|p|δ+O(1),
for
σ=δ+1
log x
. Further, since
X
|p|>x
1
|p|σ=Z∞
x
1
uσlog udψ(u)
xδ−σ
log x+Z∞
x
ψ(u)
uσ+1 log udu
1 + 1
log xZ∞
x
1
uσ−δ+1 du
1,
46
(3.25) follows.
Concerning the Selberg Symmetry Formula we have to compute
Σ := X
|n|≤x
Λ∗Λ(n) + X
|n|≤x
Λ(n) log |n|.
(3.26)
We have
X
|n|≤x
Λ(n) log |n|=Zx
1
log udψ(u)
=ψ(x) log x−Zx
1
ψ(u)
udu
=ψ(x) log x+O(ψ(x)),
and
X
|n|≤x
Λ∗Λ(n) = X
|n|≤xX
d|n
Λ(d)Λ(n
d)
=X
|d|≤x
Λ(d)X
|n|≤ x
|d|
Λ(d)
=X
|d|≤x
Λ(d)ψ(x
|d|).
We obtain that (3.26) equals
[X
|d|≤x
Λ(d)
|d|δ(β+ 1
δ)−2
m
X
r=1
αr
pδ2+t2
rX
|d|≤x
Λ(d)
|d|δcos(trlog x
|d|−θr)
+X
|d|≤x
Λ(d)
|d|δox
|d|(1)]xδ+ψ(x) log x+O(ψ(x)) (x→ ∞).
(3.27)
We have to compute
Σ1:= X
|d|≤x
Λ(d)
|d|δ
and
Σ2:= X
|d|≤x
Λ(d)
|d|δcos(trlog x
|d|−θr)
47
and
Σ3:= X
|d|≤x
Λ(d)
|d|δox
|d|(1) (x→ ∞).
Concerning
Σ1
we have
X
|d|≤x
Λ(d)
|d|δ=Zx
1
1
uδdψ(u)
=ψ(x)
xδ+δZx
1
ψ(u)
uδ+1 du.
The integral on the right hand side equals
β+ 1
δZx
1
1
udu −2
m
X
r=1
αr
pδ2+t2
rZx
1
cos(trlog u−θr)
udu +Zx
1
ou(1)
udu
=β+ 1
δlog x+o(log x) (x→ ∞).
Further using the estimation above we have
Σ3=ox(1) X
|d|≤ x
log1/(2δ)x
Λ(d)
|d|δ+X
|d|>x
log1/(2δ)x
Λ(d)
|d|δ
=o(log x) + O(√log x
xδψ(x))
=o(log x) (x→ ∞).
It remains to estimate
Σ2
. We have
X
|d|≤x
Λ(d)
|d|δcos(trlog x
|d|−θr) = Zx
1
cos(trlog x
u−θr)
uδdψ(u)
=Zx
1
ψ(u)(δcos(trlog x
u−θr)−trsin(trlog x
u−θr))
uδ+1 du + cos(θr)ψ(x)
xδ
=pδ2+t2
rZx
1
ψ(u) cos(trlog x
u)
uδ+1 du +O(1).
In the view of the above computations it is enough to estimate
Zx
1
cos(tjlog u) cos(trlog x
u)
udu.
(3.28)
48
Since
cos αcos β=cos(α+β) + cos(α−β)
2,
the integral in (3.28) above is
O(1)
except for those
tr, tj,
for which
tr=tj
,
in which case it equals
cos(trlog x)
2log x+O(1).
Putting it all together we deduce that
Σ
equals
[β+ 1
δ(β+ 2) −2
m
X
r=1
α2
r
pδ2+t2
r
cos(trlog x)−2
m
X
r=1
αr
pδ2+t2
r
cos(trlog x−θr)
+o(1)]xδlog x(x→ ∞),
which, noting that with an appropriate
0< η
η < β+ 1
δ(β+ 2) −2
m
X
r=1
α2
r
pδ2+t2
r−2
m
X
r=1
αr
pδ2+t2
r
,
proves the assertion.
3.3.2
The "convergent" case
Before we continue with the rest we need some lemmas. For a real a let
Fa(s) := F(s+ia).
Note that by Lemma 8 we have
Fa(s)
ζ(s)=Q(s) exp(−X
|p|δ>2
1−f(p)|p|−ia
|p|s),
for
σ > δ
, where
Q(s) = Y
|p|δ≤2
(1 + X
α
f(pα)
|p|α(s+ia))Y
|p|δ>2
(1 + X
α
f(pα)
|p|α(s+ia)) exp(−f(p)
|p|s+ia )
×Y
|p|δ>2
(1 −1
|p|s) exp( 1
|p|s)Y
|p|δ<2
(1 −1
|p|s).
(3.29)
49
Since for all large enough primes
|log(1 + X
α
f(pα)
|p|α(s+ia))−f(p)
|p|s+ia | 1
|p|2σ
and since
|Y
|p|δ≤2
(1 + X
α
f(pα)
|p|α(s+ia))| 1
for
σ > 0
, we deduce that
|Q(s)| 1
uniformly for
σ > σ0> δ/2
. Further-
more each term in the product in (3.29) is continuous, therefore, by the same
argument as above,
Q(s)
is continuous as well in each rectangle
δ≤σ≤δ+1
,
|τ| ≤ M
. Setting
(Ax=)A=Q(1) exp(−X
|p|δ>2
|p|≤x
1−f(p)|p|−ia
|p|δ),
(3.30)
we obtain the following
Lemma 12.
Let
N(x)∼cxδlogβx(x→ ∞)
where
c, δ > 0
,
β≥0
and suppose that
ψ(x)xδ.
Let
M > 0
be an arbitrary real number. Further assume that
f∈ FG
is
multiplicative with
|f| ≤ 1
and that
X
p
1−<f(p)|p|−ia
|p|δ
converges for some
a∈R
. Then by setting
σ=δ+1
log x
,
Fa(s)−Aζ(s) = o(|s−δ|−β−1) (σ→δ+)
uniformly for
|τ| ≤ M(σ−δ)
.
Proof.
We have
X
|p|≤y|1
|p|s−1
|p|δ|=X
|p|≤y
1
|p|δ|exp((s−δ) log |p|)−1|
|s−δ|log y,
50
for
y≤x
. Furthermore, with
η≥δ
X
y<|p|<t
1
|p|η≤Zt
y
1
uηlog udψ(u)
ψ(y)
yηlog y+Zt
y
ψ(u)
uη+1 log udu.
The integral
Zt
y
ψ(u)
uη+1 log udu
is at most a constant times
1
η−δ
1
yη−δlog y
if
η > δ
and does not exceed
O(log log t
log y)
if
η=δ
. Thus, using the Cauchy-
Schwarz inequality
(X
y<|p|
1−f(p)|p|−ia
|p|σ)2X
y<|p|
|1−f(p)|p|−ia|2
|p|σX
y<|p|
1
|p|σ.
Further, since
|1−f(p)|p|−ia|2≤2(1 −Ref(p)p−ia),
we obtain that
X
|p|δ>2
|p|≤x
(1 −f(p)|p|−ia)( 1
|p|s−1
|p|δ) + X
|p|δ>2
x<|p|
|1−f(p)|p|−ia|1
|p|σ
does not exceed
X
|p|δ>2
|p|≤y
|1
|p|s−1
|p|δ|+X
|p|δ>2
y<|p|≤x
|1−f(p)|p|−ia|1
|p|δ+X
|p|δ>2
x<|p|
|1−f(p)|p|−ia|1
|p|σ,
which is at most
c{log y(1
log x+|τ|) + δ1/2(y) log1/2log x
log y+δ1/2(x)}
51
where
δ(z) = X
z<|p|
1−<f(p)|p|−ia
|p|δ.
(3.31)
Choosing
y= max{xδ(x), x1/√log x}
, we deduce that
Fa(s)
equals
Q(s) exp(−X
|p|δ>2
|p|≤x
1−f(p)|p|−ia
|p|δ)ζ(s)
×exp(−X
|p|δ>2
|p|≤x
(1 −f(p)|p|−ia)( 1
|p|s−1
|p|δ)−X
x<|p|
1−f(p)|p|−ia
|p|s)
=(Q(1) + o(1)) exp(−X
|p|δ>2
|p|≤x
1−f(p)|p|−ia
|p|δ)ζ(s)(1 + o(1))
uniformly for
|τ| ≤ M(σ−δ)
as
σ→δ+
. Hence, in the view of Lemma 1
the assertion follows.
Lemma 13.
Let
N(x)∼cxδlogβx(x→ ∞)
where
c, δ > 0
,
β≥0
and suppose that
f∈ FG
is multiplicative with
|f| ≤ 1
and that
X
p
1−<f(p)|p|−ia
|p|δ
converges for some
a∈R
. Then
Fa(s)1
K(β+1)/2(σ−δ)β+1 +o(1
(σ−δ)β+1 ) (σ→δ+)
holds uniformly for all
K(σ−δ)≤ |τ| ≤ K
. The constant implied by the
Vinogradov symbol does not depend on
K
.
Proof.
We have
2(1 −<|p|it) =|1−|p|it|2
≤2|1−f(p)|p|−ia|2+ 2|f(p)|p|−ia −|p|it|2
≤2|1−f(p)|p|−ia|2+ 4(1 −<f(p)|p|−ia|p|−it)
52
thus
exp(4 X
p
<f(p)|p|−it|p|−ia
|p|σ)≤exp(2 X
p
|1−f(p)|p|−ia|2
|p|σ) exp(2 X
p
1 + <|p|it
|p|σ).
Since
X
p
|1−f(p)|p|−ia|2
|p|σ1
uniformly for
σ≥1
, using Lemma 8 we deduce
|Fa(s+it)|2ζ(σ)|ζ(s−it)|.
Then using Lemma 1 the proof is nished.
Lemma 14.
Let
η > 0
,
N(x)∼cxδlogβx(x→ ∞)
where
c, δ > 0
,
β≥0
and
t(x)
is a real function with
η < t(x)1
which has
an almost everywhere continuous derivative with
t0(x)1
x
. Suppose further
that for
x≥2
X
|p|≤x
1
|p|δ= (β+ 1) log log x+O(1).
(3.32)
Assume that
f∈ FG
is multiplicative with
|f| ≤ 1
and that
X
p
1−<f(p)|p|−ia
|p|δ
converges for some
a∈R
. Then
Zx
1|Mf−A1a(x
t)|log tdtδt(t) = o(xδlogβ+2 x) (x→ ∞)
where
A
is dened by
(3.30)
.
Proof.
We use inequality (3.10), i.e. that
Zx
1|M(x
t)|log tdtδt(t)xδlog3/2xH(1/log x)
53
where
H(u) := (Z∞
−∞ |F(δ+u+iτ)−Aζ(δ+u+i(τ−a))
δ+u+iτ |2dτ)1/2.
We split the range of integration in
H(u)
into three parts
I1, I2, I3
. Let
M
be a xed large number. In
I1
we integrate over
|τ| ≤ Mu
. Since
Z∞
−∞ |F(δ+u+iτ)−Aζ(δ+u+i(τ−a))
δ+u+iτ |2dτ
Z∞
−∞ |F(δ+u+i(τ+a)) −Aζ(δ+u+iτ)
δ+u+iτ |2dτ =Z∞
−∞ |Fa(s)−Aζ(s)
s|2dτ,
by using Lemma 12 we obtain
Z|τ|≤M(σ−δ)|Fa(s)−Aζ(s)
s|2dτ =o(Z|τ|≤M(σ−δ)
1
|s−δ|2β+2|s|2dτ)
=o((σ−δ)−2β−1) (σ→δ+).
In
I2
we integrate over
M(σ−δ)≤ |τ| ≤ M
. By Lemma 13 and by Lemma
1
ZM(σ−δ)≤|τ|≤M|Fa(s)−Aζ(s)
s|2dτ max
M(σ−δ)≤|τ|≤M|Fa(s)|1/2Z∞
−∞
|Fa(s)|3/2
|s|2dτ
+|A|max
M(σ−δ)≤|τ|≤M|ζ(s)|1/2Z∞
−∞
|ζ(s)|3/2
|s|2dτ.
As we have seen in the proof of Lemma 11 here the two integrals do not
exceed
c(σ−δ)3β+1
2.
Thus,
I2(1
M(β+1)/4+o(1))(σ−δ)−2β−1.
It remains to estimate
I3
. In exactly the same way as in the proof of Lemma
11 we obtain that
I31
M(σ−δ)−2β−1
which - since
M
was arbitrary - proves the assertion.
54
Theorem 3.
Let
N(x)∼cxδlogβx(x→ ∞)
where
c, δ > 0
,
β≥0
and
ψ(x)xδ.
Suppose further that for
x≥2
X
|p|≤x
1
|p|δ= (β+ 1) log log x+O(1).
Assume that
f∈ FG
is multiplicative with
|f| ≤ 1
and that
X
p
1−<f(p)|p|−ia
|p|δ
converges for some real
a
. Suppose further that the Selberg Symmetry For-
mula holds, i.e. that
X
|n|≤x
Λ∗Λ(n) + Λ(n) log |n|=t(x)xδlog x+o(ψ(x) log x) (x→ ∞)
where
t(x)
is as in Lemma 5. Then
X
|n|≤x
f(n) = N(x)xiaδ
δ+ia Y
|p|≤x
(1 −1
|p|δ)(1 + X
α
f(pα)
|p|α(δ+ia)) + o(N(x)) (x→ ∞).
Proof.
We proceed as in the proof of Theorem 2. By Theorem 1 we obtain
log2x|X
|n|≤x
f(n)−A|n|ia| X
|n|≤x|M(x
n)|{Λ∗Λ(n)+ΛL(n)}
+ (R1+R2+R3)(x) log x
+X
|n|≤x
(R1+R2+R3)( x
|n|)Λ(n),
(3.33)
where
R1(x)X
|n|≤x
log x
|n|
55
and
R2(x)N(x)X
|n|≤x
|g(n)|log |n|
|n|δ.
Here
g
is dened by
f=˜
f∗g
where
˜
f
is completely multiplicative with
˜
f(p) = f
and
A
is dened by (3.30). Further
R3(x) = X
|n|≤x
(X
|m|≤ x
|n|
A1a(m))Λ(n)|1a(n)−˜
f(n)|.
Using Selberg's formula by Lemma 5 and Lemma 14 we have
X
|n|≤x|M(x
|n|)|{Λ∗Λ(n)+ΛL(n)}=o(N(x) log2x) (x→ ∞),
and in the proof of Theorem 2 we have seen that the error terms concerning
R1
,
R2
are
O(N(x) log2x)
. Therefore, the right hand side of (3.33) equals
R3(x) log x+X
|n|≤x
R3(x
|n|)Λ(n) + o(N(x) log2x) (x→ ∞).
Since
X
|n|≤x
log |p||f(p)−|p|ia|
|p|δ=X
|n|≤y
log |p||f(p)−|p|ia|
|p|δ+X
y<|n|≤x
log |p||f(p)−|p|ia|
|p|δ
log y+plog x(X
|p|≤x
log |p|
|p|δ)1/2(X
y<|p|
|f(p)|p|−ia −1|2
|p|δ)1/2
log y+ log x(X
y<|p|
1−<f(p)|p|−ia
|p|δ)1/2,
choosing
y= max{xδ(x), x1/√log x}
where
δ(x)
is dened by (3.31) we obtain
X
|n|≤x
Λ(n)|˜
f(n)−|n|ia|
|n|δ=o(log x) (x→ ∞).
Thus,
R3(x) |A|N(x)X
|n|≤x
Λ(n)|˜
f(n)−|n|ia|
|n|δ=o(N(x) log x) (x→ ∞).
56
Furthermore, similarly
X
|n|≤x
R3(x
|n|)Λ˜
f(n)AX
|n|≤xX
|m|≤ x
|n|
1a∗|Λ˜
f−Λ1a|(m)Λ˜
f(n)
=AX
|n|≤x
1a∗|Λ˜
f−Λ1a|∗Λ˜
f(n)
=AX
|n|≤x
L∗|Λ˜
f−Λ1a|(n)
=AX
|n|≤x
(X
|m|≤ x
|n|
log |m|)Λ(n)|˜
f(n)−|n|ia|
=o(N(x) log2x) (x→ ∞).
We have
X
|n|≤x|n|ia =Zx
1−
uiadN(u)
=N(x)xia +O(1) −ia Zx
1
N(u)uia−1du.
We distinguish two cases. For
β= 0
we have
Zx
1
uδ+ia−1logβudu =xδ+ia
δ+ia +O(1),
while if
β > 0
then
Zx
1
uδ+ia−1logβudu =xδ+ia logβx
δ+ia −β
δ+ia Zx
1
uδ+ia−1logβ−1udu.
Since
Zx
1
uδ+ia−1logβ−1udu =Z√x
1
uδ+ia−1logβ−1udu +Zx
√x
uδ+ia−1logβ−1udu
xδ/2log x+xδlogβ−1x
=o(N(x) (x→ ∞),
it remains to prove that
A=Y
|p|≤x
(1 −1
|p|δ)(1 + X
α
f(p)
|p|α(δ+ia)) + o(1) (x→ ∞).
57
But since
X
x<|p|
log(1 + X
α
f(p)
|p|α(δ+ia))−f(p)
|p|δ+ia +X
x<|p|
log(1 −1
|p|δ) + 1
|p|δ
X
x<|p|
1
|p|2δ,
the assertion follows.
In the view of the proof of Corollary 1 the following corollary is immediate.
Corollary 2.
Let
N(x)∼cxδlogβx(x→ ∞),
where
c, δ > 0
,
β≥0
and let
ψ(x)=(β+ 1
δ−2
m
X
r=1
αr
pδ2+t2
r
cos(trlog x−θr))xδ+o(xδ) (x→ ∞)
where
αr∈N0
,
tr>0
,
r= 1, . . . , m
such that
m
X
r=1
αr≤β+ 1
2,
and
θr
,
r= 1, . . . , m
are the angles which satisfy
sin θr=tr
pδ2+t2
r
and
cos θr=δ
pδ2+t2
r
.
Assume further that
f∈ FG
is multiplicative with
|f| ≤ 1
and that
X
p
1−<f(p)|p|−ia
|p|δ
converges for some real
a
. Then
X
|n|≤x
f(n) = N(x)xiaδ
δ+ia Y
|p|≤x
(1 −1
|p|δ)(1 + X
α
f(pα)
|p|α(δ+ia)) + o(N(x)) (x→ ∞).
58
Chapter 4
Quantitative estimations
4.1
Introduction
Now we investigate a quantitative version of the results obtained in Chapter
3. Let
f
be a multiplicative function over the natural numbers.
Germán, Indlekofer and Klesov assuming that
f
is in some sense close to
some positive real number larger then
1/2
, developed quantitative results for
the limit behavior of
1
xX
n≤x
f(n)
in [16]. Their theorem is as follows:
Theorem.
Let
f6= 0
be multiplicative. Let
κ > 1/2
and
0≤η0< κ
,
0< λ0≤2
. Let
˜
f
be dened by the equation
˜
fΛ=Λf
(for the notation see
Chapter 2) and let
τκ
be dened by
ζκ(s) = X
n
τκ(n)
ns(σ > 1).
Assume that
|˜
f(pα)−κ| ≤ ηα(2 −λ)α−1
for all primes
p
and all
α∈N,
(4.1)
where
0≤η≤η0
, and
λ0≤λ≤2
. Put
Ax= exp(X
p≤x
f(p)−κ
p).
(4.2)
59
Then there exist positive constants
c1, c2
which depend at most on
κ, λ0, η0
such that, for
x≥2
,
|X
n≤x
f(n)−AxX
n≤x
τκ(n)| ≤c1ηx logκ−1x|Ax|
(4.3)
+c1xlogκ−1xexp(X
p≤x
|f(p)|−κ
p){exp(−c2
η) + log−c2x}.
This theorem is a generalization of a result of Halász and Elliott [11] (The-
orem 19.2). In this part of the work we release the conditions of this last
theorem and we allow
f
to be a multiplicative function over a multiplicative
semigroup the values at primes of which are close to an arbitrary
κ > 1
2(β+1)
.
For the sake of simplicity we assume that
G
is a multiplicative semigroup
which satises the conditions mentioned at the beginning of Chapter 3 with
R(x)xδlog−ηx
where
0< η0< η
. Remember that it follows that
X
|p|≤x
1
|p|δ= (β+ 1) log log x+O(1)
and that the Selberg Symmetry formula
X
|n|≤x
Λ∗Λ(n) + Λ(n) log |n|=t(x)xδlog x+δ(x)ψ(x) log x
(4.4)
is valid, where
t(x)
is as in Lemma 5 and
δ(x)→0
as
x
tends to innity (for
the proof see the proof of Corollary 1).
Lemma 15.
Let
f:G → C
be a multiplicative function. The corresponding
generating Dirichlet function
F
is formally dened by
F(s) := X
n
f(n)
|n|s.
(4.5)
Let
κ > η > 0
and suppose
|Λf(n)−κΛ(n)| ≤ ηΛ(n).
Then
F
is absolutely convergent in the halfplane
σ > δ
and
F(s)6= 0
there
and for all such complex values the equation
F(s) = exp(X
|n|>1
˜
f(n)Λ(n)
|n|slog |n|)
(4.6)
holds with some multiplicative function
˜
f
, such that
Λf=˜
fΛ
.
60
Proof.
Let
g
be the positive real valued multiplicative function, for which
the generating Dirichlet function equals
ζ(s)κ+η
. Since
ζ(s)6= 0
for
σ > δ
, it
is easy to see that
ζκ+η(s) = exp(X
|n|>1
(κ+η)Λ(n)
|n|slog |n|),
and that
Λg= (κ+η)Λ.
Then by induction we have
|Lf|=|Λf∗f|
≤Λg∗g
=Lg.
Thus,
F
converges absolutely for
σ > δ
. By the absolute convergence we
have
F(s) = 1 + u(s)
where
|u(s)| → 0 (σ→ ∞).
Therefore, there is a halfplane where
log F(s)
is a holomorphic function,
where
log
is the principal value of the complex logarithm function. On the
other hand, since
F−1(s) =(1 + u(s))−1
=X
l
(−1)lu(s)l,
by rearranging the terms we obtain that
F−1(s) = X
n
h(n)
|n|s
converges absolutely in the same halfplane. Using the unicity of the Dirichlet
generating function,
h=f−1
. Consequently,
log0F(s) = F0
F(s) = −X
|n|>1
Lf ∗f−1(n)
|n|s
61
in the same halfplane, and the right hand side converges absolutely. Inte-
grating the left and right hand side of this equation we deduce that (4.6)
holds by analytic continuation for all
σ > δ
. From this exponential equation
follows, that
F(s)
is non-zero for
σ > δ
.
Remark.
A similar argument shows that the result remains true for multi-
plicative functions
f
wich satises
|Λf(pα)−κΛ(pα)|≤ ηαΛ(pα)(n2−λ)α
for all primes
p∈ P
and
α≥1
for some
0< λ ≤n2
. Remember that
n2
was
the minimum of the norms of the primes (See the notations before Lemma
3).
Let
κ > 1
2(β+1)
, and let
τκ
be the multiplicative function dened by
∞
X
n
τκ(n)
|n|s=ζκ(s),<(s)> δ,
(4.7)
where
ζ(s)
is the Riemann's zeta function belonging to
G
.
Remark.
It is easy to see that Lemma 10 remains true if
f
satises
X
|p|α≤x|f(pα)|log |p|αψ(x)
instead of
|f| 1
. Since
exp(log |n|
log x)1
uniformly for all
1≤n≤x
, using
Lemma 1
X
|n|≤x
τκ(n)xδ
log xζ(δ+1
log x)κ
xδlog(β+1)κ−1x.
(4.8)
Under some strong assumptions on the analytic behaviour of
ζ(s)
it is possible
to compute the right asymptotic properties of
X
|n|≤x
τκ(n).
We have the following
62
Theorem 4.
Let
x > 2
, and let
f:G → C
be multiplicative. Let
˜
f
be
dened by
(4.6)
and
τκ
by
(4.7)
. Assume that
κ > 1
2(β+1)
and
0≤η0< κ
,
0< λ0≤n2
. Suppose that
˜
f(pα) = κ
for all
p∈ P
and
α≥1
such that
|p|α> x
, and that
|˜
f(pα)−κ| ≤ ηα(n2−λ)α−1
for all
p∈ P, α ≥1,
(4.9)
where
0≤η≤η0
, and
λ0≤λ≤n2
. Put
A= exp(X
|p|≤x
f(p)−κ
|p|δ).
Then there exist positive constants
c1, c2
which depend at most on
κ, λ0, η0,G
such that
|X
|n|≤x
f(n)−AX
|n|≤x
τκ(n)| ≤c1xδlogκ(β+1)−1x|A|η
+c1xδlogκ(β+1)−1xexp(X
|p|≤x
|f(p)|−κ
|p|δ)
×{exp(−c2
η) + log−c2x}
+c1
xδ
log2xX
|n|≤x
|f(n)|
|n|δ( max
1<u<x |δ(u)|log u+ 1)
where
δ(u)
is dened by
(4.4)
.
Remark.
Since
Lf = Λf∗f
, by using the method of Lemma 10 it is easy to
see that under the conditions of the theorem above
|X
|n|≤u
f(n)| uδ
log uX
|n|≤u
|f(n)|
|n|δ
(4.10)
holds uniformly for
u≥2
.
We deduce the above theorem using
Theorem 5.
Let
x > 2
, and let
f:G → C
be multiplicative. Let
˜
f
be
dened by
(4.6)
and
τκ
by
(4.7)
. Assume that
κ > 1
2(β+1)
and
0≤η0< κ
,
0< λ0≤n2
. Suppose that
˜
f(pα) = κ
for all
p∈ P
and
α≥1
such that
|p|α> x
, and that
(4.9)
is satised. Putting
M(x) = X
|n|≤x
(f−Aτκ) (n)
63
we have
log2x|M(x)| xδlog x
x
Z
1
|M(u)|
u2du
+xδX
|n|≤x
|f(n)|
|n|δ( max
1<u<x |δ(u)|log u+ 1)
+ log x(η+1
log x)|A|xδlog(β+1)κx
(4.11)
uniformly for all
A∈C
where
δ(u)
is dened by
(4.4)
. The implied constant
depends at most on
κ, λ0, η0,G
.
Remark.
These theorems are not uniform in
G
. To be more precise, they are
uniform for all multiplicative semigroups with
n2>1 +
for some
> 0
.
The integral appearing in the above theorems can be estimated with the help
of Lemma 7.
4.2
A convolution identity
The quantitative estimation depends on a variant of a Theorem of Indlekofer
in [25]. For the sake of completeness we give the proof of
Lemma 16.
Let the arithmetical function
f, g ∈ FG
satisfy
f(1) 6= 0
and
g(1) 6= 0
. Putting
M(x) = P|n|≤x(f−g)(n)
we have
log2(x)M(x) = X
|n|≤x
M(x
|n|)(Λg∗Λg(n) + log |n|Λg(n))
+X
|n|≤x
(R1+R2)( x
|n|)Λg(n)
+ log x(R1+R2)(x),
where
R1=X
|n|≤x
log x
|n|(f−g)(n)
and
R2=X
|n|≤x
f∗(Λf−Λg)(n).
64
Proof.
We have
log xM(x) = X
|n|≤x
log x
|n|(f−g)(n) + X
|n|≤x
log |n|(f−g)(n)
and, putting
R1=P|n|≤xlog x
|n|(f−g)(n)
,
log xM(x) = X
|n|≤x
(f∗Λf)(n)−X
|n|≤x
(g∗Λg)(n) + R1(x)
=X
|n|≤x
(f−g)∗Λg(n) + X
|n|≤x
f∗(Λf−Λg)(n) + R1(x)
=X
|n|≤x
M(x
|n|)Λg(n) + R1(x) + R2(x),
(4.12)
where
R2=X
|n|≤x
f∗(Λf−Λg)(n).
We multiply (4.12) with
log x
and obtain
log2xM(x) = X
|n|≤x
log x
|n|M(x
|n|)Λg(n) + X
|n|≤x
M(x
|n|) log |n|Λg(n)
+ log xR1(x) + log xR2(x).
(4.13)
Then, by substituting (4.12) into (4.13) we arrive at
log2xM(x) = X
|n|≤x
M(x
|n|)(Λg∗Λg(n) + log |n|Λg(n))
+X
|n|≤x
(R1+R2)( x
|n|)Λg(n)
+ log x(R1+R2)(x)
which leads immediately to Lemma 16.
4.3
Proof of the theorems
Proof of Theorem 5.
We apply Lemma 16 and show rst that
R1(x) = O(xδ
log x(X
|n|≤x
|f(n)|
|n|δ+|A|log(β+1)κx))
(4.14)
65
and that
R2(x)xδ−1
log xZx
1|M(x
t)|log tdt + (η+ log−1x)|A|xδlog(β+1)κx
+ ( max
1<u<x |δ(u)|log u+ 1) xδ
log xX
|n|≤x
|f(n)|
|n|δ.
(4.15)
Then we deduce
X
|n|≤x
R1(x
|n|)Λτκ(n) = O
xδ(X
|n|≤x
|f(n)|
|n|δ+|A|log(β+1)κx)
,
and
X
|n|≤x
R2(x
|n|)Λτκ(n) = Oxδ−1Zx
1|M(x
t)|log tdt +η|A|xδlog(β+1)κ+1 x
+xδX
|n|≤x
|f(n)|
|n|δ.
Now using (4.10) we obtain
R1(x) = X
|n|≤x
log x
|n|(f−Aτκ)(n)
=X
|n|≤x
(f−Aτκ)(n)Zx
|n|
1
udu
=
x
Z
1
M(u)
udu
=
x
Z
1
Mf(u)−AMτκ(u)
udu
X
|n|≤x
|f(n)|
|n|δ
x
Z
1+
uδ−1
log udu +|A|log(β+1)κx
x
Z
1+
uδ−1
log udu
66
for some
> 0
. This proves (4.14) since the estimation in (4.8) holds. Since
X
|p|α≤u
α≥2
α(n2−λ)α−1log |p| X
|p|≤√u
log |p|X
α≤log u
log |p|
αexp(αlog(n2−λ0))
log uX
|p|≤n2δ
2
exp(log(n2−λ0)
log |p|log u)
+uδ/2−log2uX
|p|≤√u
log |p|
uδ−
(4.16)
for some appropriate
> 0
,
X
|n|≤u|Λf(n)−Λτκ(n)| ≤ηX
|p|α≤u
α(n2−λ)α−1log |p|
=ηX
|p|≤u
log |p|+ηX
|p|α≤u
α≥2
α(n2−λ)α−1log |p|
ηuδ
(4.17)
holds. This implies
X
|n|≤y
log x
|n|(Λf−Λτκ)(n) = Zy
1P|n|≤u(Λf−Λτκ)(n)
udu
yδ.
(4.18)
Thus rearranging the terms in the summation,
X
|n|≤x
log x
|n|f∗(Λf−Λτκ)(n)xδX
|n|≤x
|f(n)|
|n|δ.
Observe that
Lf = Λf∗f
and that
|Λf−Λτκ| ≤ ηΛ + c˜
Λ
where
˜
Λ(n) = (α(n2−λ0)αlog |p|
if
n=pα, α > 1
0
otherwise
.
67
This leads to
LR2(x) = X
|n|≤x
f∗(Λf∗(Λf−Λτκ)(n) + L(Λf−Λτκ)(n)) + O(xδX
|n|≤x
|f(n)|
|n|δ)
X
|n|≤x|(f−Aτk)( x
|n|)|[(ηΛ + ˜
Λ) ∗(Λ + ˜
Λ) + L(ηΛ + ˜
Λ)](n)
+xδX
|n|≤x
|f(n)|
|n|δ+|A|X
|n|≤x
τk∗(|Λf|∗|Λf−Λτκ|+L|Λf−Λτκ|)(n).
(4.19)
Note that by (4.16)
X
|n|≤x
˜
Λ∗˜
Λ(n)xδ
and
X
|n|≤x
Λ∗˜
Λ(n)xδ.
Thus, by Selberg's formula, Lemma 5 is applicable to the rst term on the
right hand side of (4.19) and we arrive at
R2(x)xδ−1
log xZx
1|M(x
t)|log tdt + (η+ log−1x)|A|xδlog(β+1)κx
+ ( max
1<u<x |δ(u)|log u+ 1) xδ
log xX
|n|≤x
|f(n)|
|n|δ,
which proves (4.15). Here in the last step we used the inequality
X
|n|≤x
τk∗(|Λf|∗|Λf−Λτκ|+L|Λf−Λτκ|)(n)X
|n|≤x|τκ(n)|xδ
|n|δ(ηlog x
|n|+ 1),
which is nothing else but
ηxδZx
1P|n|≤u|τκ(n)|
|n|δ
udu +xδX
|n|≤x
τk(n)
|n|δ(η+1
log x)xδlog(β+1)κ+1 x.
Concerning
P|n|≤xR1(x
|n|)Λτκ(n)
we have by partial summation that
X
|n|≤y
log y
|n|Λτκ(n) = Zy
1P|n|≤uΛτκ(n)
udu
yδ.
68
Therefore,
X
|n|≤x
R1(x
|n|)Λτκ(n) = X
|n|≤x
log x
|n|(f−Aτκ)∗Λτκ(n)
=X
|n|≤x
(f−Aτκ)(n)X
|m|≤ x
|n|
log x
|nm|Λτκ(m)
xδ(X
|n|≤x
|f(n)|
|n|δ+|A|log(β+1)κx).
We have to estimate
X
|n|≤x
R2(x
|n|)Λτκ(n) = X
|n|≤x
f∗(Λf−Λτκ)∗Λτκ(n).
Similarly as above,
X
|n|≤x|M(x
|n|)|(Λf−Λτκ)∗Λτκ(n)X
|n|≤x|M(x
|n|)|(Λ + ˜
Λ) ∗Λ(n).
Further
X
|n|≤x
˜
Λ∗Λ(n)xδX
|p|α≤x
α>1
α(n2−λ0)α
|p|α
xδ.
Therefore, Lemma 5 is applicable and we deduce that
X
|n|≤x
R2(x
|n|)Λτκ(n)xδ−1Zx
1|M(x
t)|log tdt
+η|A|xδlog(β+1)κ+1 x
+xδX
|n|≤x
|f(n)|
|n|δ( max
1<u<x |δ(u)|log u+ 1),
as asserted. Here in the last step we used that
X
|n|≤x
τκ∗(Λf−Λτκ)∗Λτκ(n) = X
|n|≤x
(Λf−Λτκ)∗Λτκ∗τκ(n)
ηX
|n|≤x
(x
|n|)δlog |n|τκ(n),
69
and that
X
|n|≤x
τκ(n)
|n|δ≤ζ(δ+1
log x)κ.
Since
Λτκ∗Λτκ(n)+Λτκ(n) log |n| Λ∗Λ(n) + Λ(n) log |n|
for each
n∈ G
, therefore, using Lemma 5 again
|X
|n|≤x
Mx
|n|(Λτκ∗Λτκ(n)+Λτκ(n) log |n|)|
xδ−1Zx
1|Mx
t|(log t)dt +xδX
|n|≤x
|f(n)|
|n|δ( max
1<u<x |δ(u)|log x+ 1).
Observing
Zx
1|Mx
t|(log t)dt ≤xlog x
x
Z
1
|M(u)|
u2du,
the proof is nished.
Proof of Theorem 4.
We use the estimation
ζκ(s) = O(1
|s−δ|(β+1)κ),
(4.20)
which is valid uniformly for all
|τ| 1
,
δ+ 1 > σ > δ
.
Lemma 17.
Let
x > 2
and let
f:G → C
be multiplicative. Assume that
κ > 1
2(β+1)
and
0≤η0< κ
,
0< λ0≤n2
,
c0>0
. Suppose that
˜
f(pα) = κ
for all
p∈ P
and
α≥1
such that
|p|α> x
and that
(4.9)
and
(4.10)
are
satised. We have
F0(s)−A(ζκ(s))0|A|
|s−δ|(β+1)κ{ηlog(2 + |s−δ|log x)}1
σ−δ
(4.21)
uniformly for all
τ1
,
δ < σ ≤δ+ 1
,
2< x
, as long as
ηlog(2 + |s−
δ|log x)1
.
70
Proof.
Since for
σ > δ
ζκ(s) = exp(X
|n|>1
κΛ(n)
|n|slog |n|)
and
F(s) = exp(X
|n|>1
˜
f(n)Λ(n)
|n|slog |n|),
we have
F(s)−Aζκ(s) =ζκ(s)(exp(X
|n|>1
Λ(n)( ˜
f(n)−κ)
|n|slog |n|)−A)
|ζκ(s)A||exp(X
|n|>1
Λ(n)( ˜
f(n)−κ)
|n|slog |n|−X
|p|≤x
f(p)−κ
|p|δ)−1|
|ζκ(s)A||exp(X
|p|≤x
(f(p)−κ)( 1
|p|s−1
|p|δ) + X
|p|>x
f(p)−κ
|p|s
+X
pα
α≥2
˜
f(pα)−κ
α|p|αs )−1|.
(4.22)
We compute
X
|p|≤x|1
|p|s−1
|p|δ|
for
δ≤σ≤δ+ 1
,
x > 1
. As usual let
a= exp( 1
|s−δ|)
. Then
X
|p|≤a|1
|p|s−1
|p|δ|=X
|p|≤a
1
|p|δ|exp((s−δ) log |p|)−1|
|s−δ|X
|p|≤a
log |p|
|p|δ
1.
For the rest we have
X
a<|p|≤x|1
|p|s−1
|p|δ| ≤2X
a<|p|≤x
1
|p|δ
Zx
a
1
uδlog udψ(u)
log log x
log a+ 1.
71
Thus,
X
|p|≤x|1
|p|s−1
|p|δ| log(2 + |s−δ|log x)
(4.23)
uniformly for
δ≤σ≤δ+ 1
,
1< x
. Therefore substituting it into the above
inequality we have for all
s
with
ηlog(2 + |s−δ|log x)1
that
F(s)−Aζκ(s)|ζκ(s)A|{ηlog(2 + |s−δ|log x)+Σ1}
×exp{ηlog(2 + |s−δ|log x)+Σ1}
|ζκ(s)A|{ηlog(2 + |s−δ|log x)+Σ1}exp{Σ1},
(4.24)
where by the conditions
(Σ1(σ) =)Σ1= sup
τ|X
|p|>x
f(p)−κ
|p|s+X
pα
α≥2
˜
f(pα)−κ
α|p|αs |
η.
Let
Γ
be the circular path surrounding
s
with radius
(σ−δ)/2
. It is easy to
check that the conditions for the above inequality are satised for the points
of
Γ
, therefore using Cauchy's theorem and (4.20) we obtain
F0(s)−A(ζκ(s))0=ZΓ
F(z)−Aζκ(z)
(z−s)2dz
|A|{ηlog(2 + |s−δ|log x)}
(σ−δ)2ZΓ
1
|z−δ|(β+1)κdz
|A|{ηlog(2 + |s−δ|log x)}
σ−δ
1
|s−δ|(β+1)κ
(4.25)
uniformly for
|τ| 1
,
δ < σ ≤δ+1
,
1< x
,
ηlog(2+|s−δ|log x)1
. Here
we used that
|s−δ|/2≤ |s−δ|−|z−s|≤|s−δ+z−s|=|z−δ|,
and that similarly
|z−δ| ≤ 3/2|s−δ|
hold for the points of
Γ
.
Let
F0(s) = exp(X
p
|f(p)|
|p|s).
72
Lemma 18.
Let
x > 2
and let
f:G → C
be multiplicative. Assume that
κ > 1
2(β+1)
and
0≤η0< κ
,
0< λ0≤n2
,
c0>0
. Suppose that
˜
f(pα) = κ
for all
p∈ P
and
α≥1
such that
|p|α> x
and that
(4.9)
and
(4.10)
are
satised. Let
θp= arg f(p)
with
−π < arg z≤π
for all complex numbers
z
.
Assume that there are real numbers
θ0
, and
1> ξ > 0
such that
|eiθ0−eiθp| ≥ ξ
is satised. Let
A > 1
be an arbitrary large number. Then there are positive
constants
τ0
,
K
so that the following inequalities are satised for
δ < σ ≤
δ+ 1
:
|F(s)|
F0(σ)≤Kexp −ξ3(κ−η0)
64πlog 1
σ−δ
(4.26)
if
τ0<|τ|<(σ−δ)−A, δ < σ ≤δ+ 1
and
|F(s)|
F0(σ)≤Kexp −ξ3(κ−η0)
32πlog 1 + |τ|
σ−δ
(4.27)
if
|τ| ≤ τ0
,
δ < σ ≤δ+ 1
.
Proof.
One can follow the proof of [11] Lemma 19.6. By the conditions we
have
|F(s)|
F0(σ)=|exp(X
pα
α≥2
˜
f(pα)
α|p|sα )|exp(X
p
(<{eiθp|p|−it}−1)|f(p)|
|p|σ)
≤Σ2exp(X
p
(<{eiθp|p|−it}−1)|f(p)|
|p|σ),
where by the conditions
(Σ2(σ) =)Σ2= sup
τ
exp(|X
pα
α≥2
˜
f(pα)
α|p|sα |)
1.
Let
0< % ≤1
be a xed parameter to be determined later. For
a, b ∈R
we
use the notation
|a−b|(mod 2π) := min
k∈Z|a−b+ 2kπ|.
73
Let
ψ(eiθ)∈C2π(R)
such that it is zero at
θ0±ξ/2
,
%ξ2/8
at
θ0
, and linear
on the intervals between these three points,
( mod 2π)
, and zero otherwise.
The Fourier series expansion of
ψ
is
ψ(eiθ) = X
l∈Z
aleilθ
(4.28)
where
al=1
2πZπ
−π
ψ(eiθ)e−iθldθ.
One has
a0=%ξ3
32π
, and integrating by parts gives
|al| ≤ %8
πl2
for all
l6= 0
. Since
1−<eiθp|p|−iτ ≥ξ2
8
if
|θ0−τlog p|(mod 2π)≤ξ/2
and since it is non-negative, it is at least as
large as
ψ(|p|iτ )
. Here we used that
|1−<eiθp|p|−iτ |=|1−eiθp|p|−iτ |2
2,
and that
|e−iτ log |p|−eiθ0| ≤ ξ/2.
It follows that
X
p
(1 −<{eiθp|p|−iτ })|f(p)||p|−σ≥X
p
(κ−η0)ψ(|p|iτ )|p|−σ.
(4.29)
Substituting (4.28) into (4.29) and rearranging the terms we obtain that the
sum here on the right side of (4.29) is
(κ−η0)X
l∈Z
alX
p
1
|p|σ+ilτ .
Since using Lemma 15 and the remark after this lemma we have
log ζ(s) = X
p
1
|p|s+O(1),
74
using that
X
l|al|=|a0|+X
l6=0 |al| ≤ %
32π+ 2 ∞
X
l=1
1
πl2.
we obtain that the right side of (4.29) equals
X
l∈Z
(κ−η0)allog ζ(σ−ilτ) + O(1).
Let rst
|τ|> τ0>0
. If
l6= 0
then by Lemma 2 we have
|log ζ(σ−ilτ)| ≤ log(2 + |τ|) + c1log(2 + |l|)
with an appropriate
c1>0
constant. By choosing
%≤3ξ3
64π2A
we deduce
|X
l6=0
allogζ(σ−ilτ)| ≤%π
3log(2 + |τ|) + c2
≤ξ3
64πlog 1
σ−δ+c2
where
c2
is a constant. Here we used that
∞
X
l=1
log(2 + l)
l2≤ ∞
and that
∞
X
l=1
1
l2≤π2
6
and that
|τ| ≤ (σ−δ)−A
. Since
a0log ζ(σ)≥a0log 1
σ−δ−c3δ < σ ≤δ+ 1
for some
c2>0
, (4.26) follows. Now suppose that
|τ| ≤ τ0
. If
l|τ|> τ0
then
as above
log ζ(σ−ilτ) =O(log(2 + |l||τ|))
=−(β+ 1) log(σ−δ+|τ|) + O(log(2 + |l|)).
(4.30)
If
l|τ| ≤ τ0
and
τ0
is small enough, then with an appropriate
c4>0
ζ(σ−ilτ) = 1
(σ−ilτ −δ)β+1 (c4+O(1)),
75
such that (4.30) remains valid. Thus,
X
l∈Z
allog ζ(σ−ilτ) =a0log ζ(σ)−(β+ 1) log(σ−δ+|τ|)X
l6=0
al+O(1)
=a0(β+ 1) log 1
σ−δ−(ψ(1) −a0)(β+ 1) log(σ−δ+|τ|)
+O(1)
≥a0log(1 + |τ|
σ−δ)−c5
for some
c5>0
. Since
ψ(1) ≥0
, this proves (4.27).
Lemma 19.
Under the conditions of Lemma 18. we have that
|F(s)|
F0(σ)≤Kexp −ξ3(κ−η0)
32π(A+ 2) log 1 + |τ|
σ−δ
(4.31)
uniformly for all
|τ| ≤ (σ−δ)−A
.
Proof of Lemma 19.
Since
log(1 + |τ|
σ−δ)≤(A+ 2) log( 1
σ−δ) + c
holds uniformly for all
|τ| ≤ (σ−δ)−A
, substituting this last inequality into
(4.26) we obtain that
|F(s)|
F0(σ)≤Kexp −ξ3(κ−η0)
32π(A+ 2) log 1 + |τ|
σ−δ
holds for all
τ0≤ |τ| ≤ (σ−δ)−A
. But the same inequality holds by (4.27)
for
|τ| ≤ τ0
, and we deduce that (4.31) is valid.
Dene
βy= exp(r)y
, and
v= exp(r)
with
2r=1
η+1/log log x
. Let
H2(1 + y) = Z∞
−∞ |F0(δ+y+it)−A(ζκ(δ+y+it))0
δ+y+it |2dt.
(4.32)
In the range
1/log−1x≤y≤vlog−1x
we treat the integral on the right side
for
|t| ≤ βy
,
βy<|t| ≤ T
and
T < |t|
separately where
T=y−D
with an
arbitrary large positive constant
D
. The integral over these three ranges will
76
be denoted by
I11, I12, I13
respectively. With
s:= δ+u+it
, considering
I11
we have that
ηlog(2 + |s−δ|log x)≤ηlog(2 + ylog x+yv log x)
ηlog v2
1,
and
y≤βy≤v2/log x≤1.
Using (4.21) it follows that
I11 η2|A|2
y2Z|t|≤βy
log2(2 + ylog x+tlog x)
|y+it|2κ(β+1) dt
η2|A|2
y2Z|t|≤y
log2(2 + ylog x+tlog x)
|y+it|2κ(β+1) dt
+η2|A|2
y2Zy<|t|≤βy
log2(2 + ylog x+tlog x)
|y+it|2κ(β+1) dt.
The rst term on the most right hand side above does not exceed
2η2|A|2log2(2 + 2ylog x)
y2κ(β+1)+1 ,
while the integral in the second term for a
κ > 1
2(β+1)
is at most
2Z∞
y
log2(2 + 2tlog x)
t2κ(β+1) dt log2κ(β+1)−1xZ∞
ylog x
log2(2 + 2u)
u2κ(β+1) du
log2(2 + 2ylog x)y−2κ(β+1)+1.
Thus
I11 η2|A|2log2(2 + 2ylog x)y−2κ(β+1)−1.
Concerning
I12
, using the Cauchy-Schwarz inequality, we have
I12 Zβy≤|t|≤T|F0(δ+y+it)
δ+y+it |2dt +Zβy≤|t|≤T|A(ζκ(δ+y+it))0
δ+y+it |2dt
=I121 +I122.
77
Keeping in mind that
F(s)6= 0
for
σ > δ
and using the factorization
F0(s) = F(s)F0(s)
F(s),
which is valid for all
<s>δ
, we obtain
I121 sup
βy≤|t|≤T|F(δ+y+it)|2Z∞
−∞ |F0(δ+y+iu)
F(δ+y+iu)(δ+y+iu)|2du.
Since using (4.17)
L(u) := X
|n|≤u
˜
f(n)Λ(n)
uδ,
by an application of Parseval's identity we deduce,
Z∞
−∞ |F0(δ+y+iu)
F(δ+y+iu)(δ+y+iu)|2du =2πZ∞
0|L(ew)|2e−2(δ+y)wdw
y−1.
Furthermore, the conditions of Lemma 18 are fullled. Under these circum-
stances we can choose
θ0=π
, and
v≤1−2η0
2−η0
in that Lemma, and using
Lemma 19 we have that
sup
βy≤|t|≤T|F(δ+y+it)|2F2
0(δ+y) exp(−2clog(1 + y−1βy))
with some appropriate positive constant
(cD,κ,η0,G=)c
. With
=κ−η0
F0(δ+y)exp( X
|p|≤exp(y−1)
|f(p)|−
|p|δ+y)y−
exp(X
|p|≤x
|f(p)|−
|p|δ+y)y−
exp(X
|p|≤x
|f(p)|−
|p|δ)y−
uniformly for
1/log x < y < 1
. Here we used that
X
|p|>exp(y−1)
1
|p|δ+y1,
78
uniformly for
0< y ≤1
which is a direct consequence of (4.23) and the
asymptotic estimations
X
|p|≤u
1
|p|δ=(β+ 1) log log u+O(1) (u > 2),
ζ(δ+y+it) = c1
(y+it)β+1 +c2
(y+it)β+O(1) (0 < y ≤1,|t| 1).
Using Lemma 2 we have that for some
τ0>0
ζ0(δ+y+it)|y+it|−(β+2)
uniformly for
0< y ≤1,|t| ≤ τ0,
and
ζ0(δ+y+it)|t|2
uniformly for
0< y ≤1,|t| ≥ τ0
and
ζ(δ+y+it)y−(β+1)
uniformly for all
t
. Applying these estimations in this order we obtain with
an appropriate
u1>0
I122 |A|2Zβy≤|t|≤1
1
|y+it|2κ(β+1)+2 dt +|A|2Z1≤|t|≤u1
t2dt
+|A|2y−2κ(β+1) Zu1≤|t||ζκ(δ+y+it)0
ζκ(δ+y+it)(δ+y+it)|2dt
|A|2(β−2κ(β+1)−1
y+u3
1+y−2κ(β+1)−1u−1
1).
Here we used that a similar argument as in the proof of Lemma 11 shows
that
Zu1≤|t||ζκ(δ+y+it)0
ζκ(δ+y+it)(δ+y+it)|2dt u−1
1.
Choosing
u1=y−u2
where
u2>0
is to be determined later we deduce that
I122 |A|2(v−2κ(β+1)−1y−2κ(β+1)−1+y−3u2+y−2κ(β+1)−1u−1
1)
=: E122.
We arrive at
I12 exp(2 X
|p|≤x
|f(p)|−
|p|δ)y−2−1exp(−2clog(1 + y−1βy)) + E122.
79
For
I13
we use that since
|F(δ+y+it)| y−B0< y ≤1, t ∈R
for some
0< B
, using Cauchy's theorem
|F0(δ+y+it)| y−B−10< y ≤1, t ∈R.
Choosing the constant
D
large enough we obtain
sup
t|F0(δ+y+it)|2ZT≤|t|
1
|δ+y+it|2dt yB.
Similarly, we deduce
I13 (|A|2+ 1)yB.
It remains to estimate
H2(δ+y)
for
v/ log x < y ≤1
. In this range we split
the integral appearing on the right hand side of (4.32) into two pieces, which
we denote by
I21
and
I22
. First we estimate the contribution of
|t| ≤ T
, then
that of
T < |t|
respectively. A similar computation as by
I12
shows that
I21 exp(2 X
|p|≤x
|f(p)|−
|p|δ)y−2−1+E122,
and like by
I13
we obtain that
I22 (|A|2+ 1)yB.
Putting it all together we deduce that
R1
1/log xH(δ+y)y−1/2dy
is at most
Z1
1/log x
[η|A|log(2 + 2ylog x)y−κ(β+1)−1
+ exp(X
|p|≤x
|f(p)|−
|p|δ)y−−1exp(−clog(1 + y−1βy))
+|A|v−κ(β+1)−1y−κ(β+1)−1+|A|y−3u2/2−1/2
+|A|y−κ(β+1)−1+u2/2+ (|A|+ 1)yB]dy
+Z1
v/ log x
[exp(X
|p|≤x
|f(p)|−
|p|δ)y−−1+|A|y−κ(β+1)−1
+|A|y−3u2/2−1/2+|A|y−κ(β+1)−1+u2/2]dy,
80
Chapter 5
The Theorem of Erd®s and
Wintner
In this chapter we always suppose that
G
satises the conditions mentioned
at the beginning of Chapter 3 with
R(x)xδlog−ηx
where
0< η0< η
. Let
f:G → R
be an arithmetic function. Let
Fx(y) := N(x)−1#{n∈ G,|n| ≤ x:f(n)≤y}.
For
x > 1Fx(y)
is a distribution function. We say that
f
possesses a limit
law if there exists a distribution function
F
such that
Fx(y)→F(y) (x→ ∞)
holds for all continuity points of
F
. In notation
Fx=⇒F(x→ ∞).
f
is said to be additive if
f(mn) = f(m) + f(n)
for all
(n, m) = 1
. Erd®s
and Wintner proved that in the case of
Nf
has a limit law if and only if the
three series
X
|f(p)|≤1
f(p)
p,X
|f(p)|≤1
f2(p)
p,X
|f(p)|>1
1
p
converge (See for example in [10]). Their work was pioneering in the topic
initiated by Hardy and Ramanujan in [22] which is known today as proba-
bilistic number theory. Probabilistic properties of primes and arithmetical
functions have been intensively investigated by several authors. For further
references see [4, 15, 33]. In this chapter we show that the Erd®s Wintner
Theorem remains valid for quite general arithmetical semigroups. We have
83
Theorem 6.
An additive arithmetical function
f:G → R
possesses a limit
law if and only if the three series
X
|f(p)|≤1
f(p)
|p|δ,X
|f(p)|≤1
f2(p)
|p|δ,X
|f(p)|>1
1
|p|δ
(5.1)
converge. The characteristic function
ψ(t)
of the limit law is given by the
convergent product
ψ(t) = Y
p
(1 −1
|p|δ)(1 + X
α≥1
eitf(pα)
|p|δα ).
The limit law is continuous if and only if
X
f(p)6=0
1
|p|δ
diverge.
To sketch the proof of the theorem we need some lemmas.
Lemma 20 (Delange).
Let
g:G → C
be a multiplicative function with
values in the unit disc. Then
g
possesses a non-zero mean value if and only
if
X
p
1−g(p)
|p|δ
converges, and
∞
X
α=1
g(pα)
|p|αδ 6=−1
for all primes p with
|p|δ≤2
. The non-zero mean value is given by the
convergent product
Y
p
(1 −1
|p|δ)(1 + X
α≥1
g(pα)
|p|δα )
Proof.
The lemma is an easy consequence of the generalization of the Theo-
rem of Halász in Chapter 3 and the fact that
X
x<|p|≤ax
1
|p|δ=o(1) (x→ ∞)
holds for all
a > 1
.
84
Scatch of the proof of Theorem 6.
The characteristic function of
Fx
equals
ψx(t) = N(x)−1X
|n|≤x
eiτf(n).
Suppose rst that
Fx
has a limit law. By the continuity theorem of Lévy
[44] (Theorem 2.4)
ψx(t)
converges uniformly for all bounded values of
t
to a
characteristic function, say
ψ
. Since
ψ
is continuous and
ψ(0) = 1
, with an
appropriate
T
we have
|ψ(t)|>1/2
for all
|t|< T
. Thus using Lemma 20
we obtain that
X
p
1−eitf(p)
|p|δ
converges. Therefore
ψ(t) = Y
p
(1 −1
|p|δ)(1 + X
α≥1
eitf(pα)
|p|δα ).
From this representation we easily obtain that the three series in (5.1) con-
verge. Conversely suppose that the three series converge. The uniform con-
vergence of
X
p
1−eitf(p)
|p|δ
for
|t| ≤ T
is immediate. Thus using Lemma 20 and then the Theorem of
Lévy again, we obtain that the limit law exists. By another theorem of Lévy
we have that the limit law is continuous if and only if
X
f(p)6=0
1
|p|δ
diverges.
85
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