C OMPOSITIO M ATHEMATICA
H. G. M EIJER R. T IJDEMAN
On additive functions
Compositio Mathematica, tome 29, n
o3 (1974), p. 265-271
<http://www.numdam.org/item?id=CM_1974__29_3_265_0>
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265
ON ADDITIVE FUNCTIONS
H. G.
Meijer
and R.Tijdeman
N oordhoft’ International Publishing
Printed in the Netherlands
1. Introduction
A function F defined on the set of
positive integers
is said to be additiveif
whenever
(n, m)
=1 ;
F is calledcompletely
additiveif (1)
holds for everypair
ofpositive integers n
and m.Obviously
acompletely
additivefunction is determined
by
its values on the set ofprimes.
The functionF(n)
= clog n (c constant)
is anexample
of acompletely
additive function.On the other hand
Erdôs [1] proved
that if an additive function F satisfies one of thefollowing
two conditionsor
then F is
necessarily
of the formF(n)
= clog n
for some constant c.The same results have been found
by
several other authors.Pisot and
Schoenberg [2]
considered thefollowing
situation. Let p,, - - -, pr denote r differentprimes,
let A be themultiplicative semigroup generated by
p 1, ’ ’ ’, p, and let F be an additive function defined on A.Pisot and
Schoenberg proved
that if F is nondecreasing
onA,
thenF(n)
= clog n
on Aprovided
that r ~3.(For
r = 2 monotonic additive functions exist which are not of the form clog n). Obviously
this resultis a
generalisation
of the first mentioned result of Erdôs. Now one may ask if a condition similar to condition(3)
wouldimply
thatF(n)
= clog n
on A. The present paper deals with thisquestion.
We write
A = {1 = n1 n2 n3 ···}
and restrict ourselves to266
functions F on A which are
completely
additive. In section 2 wegive
some results on the structure of A. For the
proofs
we refer to[3], [4]
and
[5].
In section 3 we prove thatimplies
thatF(n)
= clog n
on A.Subsequently
we prove in section 4 that everycompletely
additive function F on A satisfiesWe
conjecture
that the result of section 3 can beimproved considerably
and that the condition
is
already
sufficient toimply
thatF(n)
= clog n on
A. We are able to prove theconjecture
for r ~5;
see section 5.2. The structure of A
In the
sequel
we shall use thefollowing
notations.By
p1, ···, pr we denote r differentprimes (r
~2).
Themultiplicative semigroup generated by
them will be denotedby A;
1 = nl n2 n3 ... are the elements of A inincreasing
order.LEMMA 1: There exist
positive
constantsCl, C2
and N such thatPROOF: The first
inequality
is thecorollary
of[3]
Theorem 1. Thesecond can be found in
[4].
We shall use the
following
easy consequences of lemma 1 :LEMMA 2: Let p be one
of the primes {p1, ···, pr}.
Then there exists aninfinite
numberof pairs
ni,ni + 1 such that ni is a pure powerol
p and ni+1 is not divisibleby
p.PROOF : See
[5]
Theorem 2.LEMMA 3: Let p and q be two
different primes from {p1, ···, Prl.
Thereexist
infinitely
manypairs
ni, ni+1 such that oneof
the numbers ni, ni+1 iscomposed of
p and q and the other is neither divisibleby
p norby
q.PROOF : See
[5]
Theorem 3.3
In this section we prove the
following
theorem.THEOREM 1: Let F be a
completely
additivefunction defined
on Asatisfying
then
F(n)
= clog n
on Afor
some constant c.PROOF : Put
F(p03C1)
= cplog
p03C1(p
=1,..., r).
Without loss ofgenerality
we may assume that the p1, ···, pr are
arranged
in such a way thatc1 ~ c2 ~ ··· ~
cr. We shall prove c1 = ’ ’ ’ = cr, whichobviously implies
the assertion of the theorem.Suppose
Ci = c2 = ··· = cs >cs+1 ~ ··· ~ cr for some s ~ {1, ···, r-1}.
Let ni, ni+
1 be
apair
of consecutive elements of A such that ni iscomposed
of
primes
p1, ···, psonly
and ni+1 contains at least oneprime
from{ps+1, ···, pr}.
It is evident that an infinite sequence of suchpairs
exists.Put ni+1 =
pk11 s s ··· pkrr,
thenwhere
268
Since
F(nJ
= c1 log ni
we obtainIn view of
(7)
and(9)
this is a contradiction with(8).
4
Using
the estimates of ni+1 - ni we caneasily
deduce thefollowing
theorem.
THEOREM 2 : Let F be a
completely
additivefunction
on A. ThenPROOF : Put
F(p03C1)
= cplog pP ( p
=1,..., r)
and M =max03C1 |c03C1|.
Then weobtain
for ni
=pt11 ··· prr
thatHence
by (5),
Therefore
using (4)
we obtainfrom which the theorem follows.
5
For r ~ 5 we can
improve
theorem 1. To that purpose we prove thefollowing
theorem.THEOREM 3: Let F be a
completely
additivefunction
on A. PutF(pp)
= cplog
p,(p
=1,..., r).
We may suppose that p1, ···, Pr arearranged
in such a way thatc1 ~ c2 ~ ··· ~
cr .If
then it
follows
that cl = C2if
r = 2 and cl = c2 = C3 and cr- 2 = Cr-l = C,if
r ~ 3.COROLLARY :
If
r =2, 3, 4
or 5 then condition(10)
is asufficient
onefor
acompletely
additivefunction
on A toimply F(n)
= clog
n on A.Obviously
this is an improvementof
theorem 1.PROOF OF THEOREM 3:
Suppose
cl > c2. Choose 8 > 0 such thatWe obtain from
(10)
thatF(ni) ~ F(ni+1) + 03B5 log ni+1, if i is sufficiently large.
By
lemma 2 there exists an infinite number ofpairs
ni,ni+1 such that ni is a pure powerof p and
ni+1 is not divisibleby
p1. ThenF(ni)
= c1 log
ni andc2 ~ ··· ~
crimplies F(ni+1) ~
c2log
ni+1. Hencefrom which it follows
by (6),
This is a contradiction with
(11).
Thus c = c2 .Applying
this result tothe function - F we obtain c, = Cr - 1 .
Suppose
r~ 3
and cl = C2 > c3 . Choose e > 0 such thatWe obtain from
(10)
thatand
if i is
sufficiently large. By
lemma 3 there existinfinitely
manypairs
ni, ni+ 1 such that one of them is
composed of pl
and p, and the other iscomposed
of p3, ···, pr .270
Suppose
firstthat ni
iscomposed of p1,
p2 . ThenF(ni)
= clog
ni and sincec3 ~ ··· ~
Cr, we haveF(ni+1) ~
c3log
ni+1. Hence,by (13),
Suppose
on the other hand that ni+1 iscomposed of pl ,
p2 . ThenTherefore
by (14)
Since there is an infinite number
of integers
i for which(15)
or(16)
holdswe obtain by (6)
This is a contradiction with
(12).
Therefore cl = c2 = c3.Applying
thisresult to the function - F we obtain Cr = cr-1 Cr - 2 REMARK: In
[5]
thefollowing conjecture
is made:Let pl , ’ ’ ’, pr be different
primes
and{ni}
theincreasing
sequence ofintegers composed
of theseprimes.
Let t befixed,
1 ~ t ~ r -1. Then there existinfinitely
manypairs
n; , ni+1 such that one of the numbers ni, ni +1 is composed of p1, ···,
pt and the other iscomposed of Pt
+ 1, ···, pr.Under the
assumption
that thisconjecture
is true, we can prove, forall r,
that the conditionimplies
thatF(n)
= clog n
on A for some constant c.REFERENCES
[1] P. ERDÖS: On the distribution function of additive functions. Ann. of Math. 47 (1946) 1-20.
[2] C. PISOT and I. J. SCHOENBERG: Arithmetic problems concerning Cauchy’s functional equation. Illinois J. Math. 8 (1964) 40-56.
[3] R. TIJDEMAN: On integers with many small prime factors. Compositio Math. 26
(1973) 319-330.
[4] R. TIJDEMAN: On the maximal distance between integers composed of small primes.
Compositio Math. 28 (1974) 159-162.
[5] R. TIJDEMAN and H. G. MEIJER: On integers generated by a finite number of fixed
primes. Compositio Math. 29 (1974) 273-286.
(Oblatum 19-111-1974) University of Technology Department of Mathematics Delft, Netherlands
University of Leiden, Mathematical Institute
Leiden, Netherlands