A New Proof of Erdos's Theorem on Monotone Multiplicative Functions
Author(s): Everett Howe
Source: The American Mathematical Monthly, Vol. 93, No. 8 (Oct., 1986), pp. 593-595
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/2322314
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A NEW PROOF OF ERDOS'S THEOREM ON MONOTONE
MULTIPLICATIVE FUNCTIONS
EVERETTHOWE*
Pasadena, CA 91125
of Techn0ology,
CaliforniaInstitute
Department
ofMathematics,
zero,is said to be multiplicativeif
f, notidentically
function
An arithmetical
1. Introduction.
f( mn) = f( m)f( n)
whenever ( m, n) = 1,
if
and completely
multiplicative
forall m andn.
f(mn) =f(m)f(n)
functions
is due to Erdo's
multiplicative
increasing
theorem
concerning
remarkable
The following
[2].
a suchthatf( n) = na for
thenthereis a constant
andmultiplicative,
THEOREM.Iff is increasing
all n > 1.
and simplerproofshavebeengivenby Moserand
Erdo's'soriginalproofis rathercomplicated,
or notwell
[4].All of theseproofsare eitherlengthy
[1],and Schoenberg
Lambek[3],Besicovitch
multiplicative
motivated.This papershowsthattheresultis fairlyeasy to proveforcompletely
is completely
function
liesin showingthateveryincreasing
multiplicative
functions;
thedifficulty
intotwopartsas follows:
we splitErdo's'stheorem
Consequently,
multiplicative.
a suchthat
thenthereis a constant
multiplicative,
THEOREM
and completely
A. Iff is increasing
f(n) = na foralln >1.
is completely
multiplicative.
THEOREM
B. Everyincreasing
function
multiplicative
We proveTheorem
and completely
multiplicative.
2. ProofofTheoremA. Let f be increasing
Assumethereis no a suchthatf( n) = na forall n > 1. Thenlogf( n)/log n
A by contradiction.
integersm > 1, n > 1, suchthat
is notconstant,so thereexistdistinct
logf(m)
logf(n)
log n
log m
Takingx to be thelargerand y thesmallerof thesetworatios,we have
f(m)=mx
and
f(n)=n-Y,
withx > y.
A > 1 and B > 1 suchthat
Because y/x < 1 thereexistintegers
logn
y logn
x log m
log m
In fact,
B=
g) j +1
[(x- y)log n]
[
and A= [Blogn
login]
But thenwe haveboth
satisfytheabove inequalities.
EverettHowe: I am currentlycompletingmy senior year at the California Instituteof Technology.My
mathematicalinterestslie chieflyin thearea of numbertheory,
analyticnumbertheory,but I have been
particularly
spendinga lot of timelatelystudyingJuliasets.I am a 1985 PutnamFellow,and I was recentlyawardeda National
Science FoundationGraduateFellowship.
*Supportedby a CaltechSummerUndergraduateResearchFellowship.
593
594
[October
EVERETT HOWE
A logm < Blog n and Axlogm > Bylog n,
or
m
and
< nB
mAx > nBy
multiplicative,
However,sincef is completely
f(m A) = f(m)A
and
= mAx
f(nB)
=f(n
)
B=
nBy
thefactthatf is increasing.
This proves
so that mA < nB whilef(mA) > f(n B), contradicting
Theorem A.
lemma.
3. Proofof TheoremB. The proofof TheoremB is based on thefollowing
LEMMA
function
multiplicative
1. Givenan increasing
f and a primep, let
L=
inf
x#Omodp
f(x + p)
f(x)
Then L = 1.
We use Lemma1 to deduceTheoremB, and thenwe proveLemma1 in thenextsection.
it
multiplicative
To show thatf is completely
Assumef is increasingand multiplicative.
sufficesto showthatforeveryprimep and everyintegern > 1 we have
= f(p)fnl
f(pfl)
that
Fix a primep and let yn= f(pn)* We willshowthatyn= ynlor,equivalently,
Yn+ 1
Yn
forall n.
Consideranyintegerx notdivisiblebyp. Thenforany n we have
(px
Now each of px
-
-
1)pn < xp,,+l < (px + 1)pn.
1, x, and px + 1 is prime to p so
f(px
-
l)yn
Af(pX
<f(X)yn+1
+ l)Yn,
Therefore,
and multiplicative.
becausef is increasing
in+
<l
+ 1)
f(px + p2)
f
f(px
I,:,inf
K.,nf
f (x)
f f(x)
=f (p)inf
f(x +p)
f (x)
-y L
1
we find
wheretheinfis takenoverall x # 0 modp. Similarly,
Yn
y+
Yn
where
U=
sup
x#Omodp
x>p
f(x -p)
f_(x)
-
f(x)
sup
xOmod p f (x + p)
-.
1
L
Thus we have
~ ,Ly1.-
Yn? 1
UylK.
Yn
implies(1), which,in turn,proves
But L = 1 by Lemma1, so U = 1 also,and thislastinequality
TheoremB.
1986]
GRAPHICAL CONSTRUCTIONS
OF MEANS
595
of L, if
and multiplicative.
Fromthedefinition
4. Proofof Lemma1. Assumef is increasing
an integerx is notdivisiblebyp we havef(x + p) >? Lf(x), and hence
'f(x + kp) > Lkf(x)
foreveryintegerk > 0.
Now, givenany k > 0, we can findan integerx > kp whichis primeto bothp and 2. Then
2x > x + kp, and
f(2)f (x) = f(2x) > f(x + kp) > Lkf(x),
we also
we musthave L < 1. But sincef is increasing,
so thatLk < f (2). Sincek was arbitrary,
showthatL = 1 and proveLemma1.
have L > 1. Thesetwoinequalities
resultfor decreasingmultiplicative
5. Comments.Erdo's'stheoremimpliesa corresponding
withone restriction.
It is easy to showthatif f is decreasingand multiplicative,
then
functions
or elsef(1) = 1 and f (n) = 0 whenn > 2, with0 < f(2) < 1. Now,if
eitherf is alwayspositive,
so Erdo's'stheorem
impliesl/f(n) = no for
then1/f is increasing,
f is positiveand decreasing,
and iff(3) *A0, thenthere
some a, hencef (n) = n- . Thus,iff is monotoneand multiplicative
existsa constanta suchthatf(n) = no forn > 1.
The author wishes to thankProfessorTom M. Apostol for suggestingthis researchand for helpingin the
preparationof the manuscriptforpublication.
References
1. A. S. Besicovitch,On AdditiveFunctionsof a PositiveInteger,Studiesin MathematicalAnalysisand Related
Topics, pp. 38-41, StanfordUniversity
Press,Stanford,CA, 1962.
functionof additivefunctions,
Ann. of Math.,(2) 47 (1946) 1-20.
2. P. Erdos, On thedistribution
3. L. Moser and J. Lambek,On monotonemultiplicative
Proc. Amer.Math. Soc., 4 (1953) 544-545.
functions,
4. I. J. Schoenberg,On two theoremsof P. Erdo'sand A. Renyi,IllinoisJ.Math.,6 (1962) 53-58.
GRAPHICAL CONSTRUCTIONS OF MEANS
RICHARD P. SAVAGE, JR.
University,
Science,TennesseeTechnological
and Computer
ofMathematics
Department
Cookeville,TN 38505
and Polya,it is shown
In chapterthreeof Inequalitiesby Hardy,Littlewood,
1. Introduction.
can be usedto assign
definedon an interval
function
monotonic
and strictly
thateverycontinuous
The meanvalueassignedis onlya meanin
in theinterval.
a meanvalue to each setof n numbers
thanthelargest.
norgreater
a generalsense:themeanis notless thanthesmallestofthenumbers
mean
definedby an
for
the
hold
not
thatthefamiliarmeanspossessneed
Moreover,properties
a meanof
for
determining
technique
the
graphical
that
we
show
In
paper
function. this
arbitrary
a meanof n variables
to determining
twovariablesgivenby Moskovitzin [4] can be generalized
in Rn.Thisprovides
a graphicaltechnique
and Polya)through
(in thesenseof Hardy,Littlewood,
meansin R3 and more
ofmanywell-known
representations
geometric
someespeciallyinteresting
in R"1.
generally
RichlardP. Savage, Jr.: I receivedmy Ph. D. in 1981 fromthe Universityof Utah forworkin differential
geometry.My thesisadvisorwas Domingo Toledo. I was a facultymemberfirstat Moorhead State Universityin
wheremyfatheris also
Minnesota.Since 1982 I have been a facultymemberat TennesseeTechnologicalUniversity
interestsincludegenealogicalresearch,keepingup with
faculty.My nonmathematical
a memberof the mathematics
baseball, and readingtheclassics.