Arithmeti al aspe ts of ertain fun tional equations
by
Lutz G. Lu ht (Clausthal)
The lassi alsystemof fun tionalequations
1
n n 1
X
=0 F
x+
n
=n s
F(x) (n2N)
withs2C, investigatedforinstan ebyArtin(1931), Yoder(1975), Kubert
(1979),and Milnor (1983),isextended to
1
n n 1
X
=0 F
x+
n
= 1
X
d=1
n
(d)F(dx) (n2N)
with omplexvaluedsequen es
n
. Thisleadstonewresultsontheperiodi
integrable and theaperiodi ontinuoussolutionsF :R
+
!C interrelating
thetheory offun tionalequationsand thetheory ofarithmeti fun tions.
1. Repli ativityequations. The lassi alsystemofrepli ativityequa-
tions
(1
n )
1
n n 1
X
=0 F
x+
n
=n s
F(x)
foralln2N withsome xedparameters2C wasstudiedfrequently. Usu-
allyx varies over theadditivegroupT=R=ZorasuitableintervalDR,
for instan e (0;1) or R
+
. For D = Q= Z see Kubert [3℄. The solutions
F of theabove system form a linear spa e. Sin e any two equations (1
m )
and (1
n
) imply(1
mn
), itwouldsuÆ e to assume the equations(1
n
) forall
n2P=fp2N :pprimeg. The ase s=1 wasstudied byArtin[1℄ in on-
ne tionwiththegammafun tion. He proved thatevery integrable solution
F :T!C ofthesystem(1
n
)isalinear ombinationoflog (1 e 2ix
)andits
onjugatealmosteverywhere. Forarbitrarys2C Milnor[8℄gavea omplete
1991 Mathemati sSubje tClassi ation: 11A25,39B62,39A10.
hara terizationofthespa eof ontinuoussolutionsF :(0;1)!C and dis-
ussedtheir ontinuousextensionsto Tandto R
+
. Inparti ular,thisleads
tolinearrelationsbetweenthepolylogarithmandtheHurwitzzetafun tion,
andthere aresigni antappli ationstoDiri hletL-fun tions, theRiemann
zeta fun tion, the gamma fun tion and related fun tions. Yoder [12℄ re-
pla edthefa torsn s
onthe rightsideof theequations(1
n
) bythevalues
g(n) of a generating sequen e g : N ! C and showed that the existen e
of non-trivialsolutions F for esg to be totally multipli ative, whi h means
g(1) =1 and g(mn) =g(m)g(n) for all m;n2N. Moreover, the existen e
ofaperiodi ontinuoussolutionsF :R
+
!C impliesg(n)=n s
withsome
s2C and hen eleadsba kto the lassi alsystem.
Theaim ofthispaperisto investigatethesystemS ofextendedrepli a-
tivity equations, previouslyintrodu edin[6℄,
(2
n )
1
n n 1
X
=0 F
x+
n
= 1
X
d=1
n
(d)F(dx)
with some xed sequen es
n
:N ! C forall n2N. This systemprovides
a natural extension of the lassi al repli ativity system as it removes the
restraint onthegeneratingsequen egtohavetheformg(n) =n s
ortobe
totallymultipli ativeat least. Namely, onsiderF :T!C asrestri tionof
a fun tionG, holomorphi on theopen unitdis U C and normalized by
G(0)=0and G 0
(0)=1,to the boundaryU. Set
G(z)= 1
X
a=1 g(a)z
a
and writeF(x)=G(e(x)), wheree(x)=e 2ix
forx2R. The identity
1
n n 1
X
=0 e
a
n
=
1 if nja,
0 otherwise,
yields
1
n n 1
X
=0 G
ze
n
= 1
X
a=1
nja g(a)z
a
(z2U):
Now, apart from a remainder termof order jzj (d+1)n
, theright side an be
writtenasalinear ombination
n (1)G(z
n
)+
n (2)G(z
2n
)+::: +
n (d)G(z
dn
)
ofG(z n
);:::;G(z dn
)withuniquelydetermined oeÆ ients
n
(1);:::;
n (d)
2C forevery d2N. Assume that theremainder termvanishes asd!1.
Then
1
n n 1
X
=0 G
ze
n
= 1
X
n (d)G(z
nd
) (z 2U);
andinpassingtoF withxrepla edbyx=nweformallyarriveat(2
n
). Noti e
that S redu es to Yoder'ssystem ifand onlyifg istotally multipli ative,
in whi h ase
n
(1) = g(n) and
n
(d) = 0 for all d 2 N, d 6= 1, and all
n2N.
2. Periodi solutions. LetL(T) denote the spa e of integrable fun -
tionsF :T!C modulonullfun tions underthenorm
kFk=
\
T
jF(x)jdx<1:
Theabove onstru tionofthesystemS suggests investigatingitssolutions
F 2L(T). EvidentlyS hasnon-trivial onstant solutions F ifand onlyif
(3
n )
1
X
d=1
n
(d)=1
holdsfor all n2 N. The existen e of non- onstant solutions F 2 L(T) de-
pendsonhowthesequen es
n
(n2N)areinterrelated,whi hrequiressome
arithmeti alpreparation. Denote byF the omplexalgebraof arithmeti al
sequen es g : N ! C endowed with the usual pointwise linear operations
andwiththeDiri hlet onvolution:F 2
!F asmultipli ation,de nedby
(f g)(m)= X
a;b2N
ab=m
f(a)g(b) (m2N):
Observe that F ontains the multipli ative identity " with "(1) = 1 and
"(m) = 0 for all m 2 N, m 6= 1. The multipli ative group of F is F
=
fg2F :g(1)6=0g. By g 1
we denotethe multipli ativeinverse of g2F
,
whi h means gg 1
=". For g2F and n2N we de ne thesubsequen e
g(n
)2F byg(n
)(a)=g(na) foralla2N.
Theorem 1. For arbitrary sequen es
n
2 F (n 2 N), let g 2 F be
de ned by g(n) =
n
(1) for all n 2 N. If the system S has non- onstant
solutions F 2L(T) then
(a) g(1)=1 and
n
=g 1
g(n
) for all n2N,
(b) the Fourier oeÆ ients of F are given by
b
F(m)=
sgnm
g(jmj) if m2Z
=Znf0g,
0
if m=0,
withparameters
0
;
1
;
1 2C,
( )
0
=0, unless (3
n
) is valid for all n2N.
Proof. Let
X
b
F(m)e(mx) (x2T)
betheFourierseriesof F 2L(T). The Fourier oeÆ ientsaregiven by
b
F(m)=
\
T
F(x)e( mx)dx (m2Z):
Denotingtheleftsideof(2
n )byF
n
(x), weseethatF
n
2L(T) and b
F
n (m)=
b
F(nm)forallm2Z. Denotingtherightsideofequation(2
n )byS
n
(x), we
obtain S
n
2 L(T) from the onvergen e of the seriesto F
n
2 L(T), whi h
legalizes termwiseintegration andshowsthat
b
S
n (m)=
X
djm
n (d)
b
F(m=d) (m2Z):
Comparing the oeÆ ients leadsto
(4)
b
F(nm)= X
djm
n (d)
b
F(m=d) (m2Z):
In parti ular, for m = 1 this implies b
F(n) = g(n) b
F(1). Sin e, by
assumption, b
F 60 onZ
,weobtaing(1)=1and g(nm)=(
n
g)(m) for
all m2N, whi h isequivalent to (a). Putting
1
= b
F(1),
1
= b
F( 1) and
0
= b
F(0)we obtain(b), and ( ) followsfrom (4)form=0.
In the sequel we only onsidersystems S forwhi h a sequen e g2F
a ording to Theorem1(a) exists. We allg thegenerating sequen e ofS.
ByTheorem1thedimensionoftheC-linearspa eP
g
ofsolutionsF 2L(T)
of S isat most3. In fa t, fora large lass of generatingsequen es we an
redu e thisboundby1.
Forq2N,q 6=1,denote byR
q
the setof allsequen es g2F satisfying
a re urren eequation oftheform
(5
q
) g(q
l+k
)+
k 1 g(q
l+k 1
)+:::+
0 g(q
l
)=0
with ertain integers k;l 0 and omplex oeÆ ients
0 6= 0;
1
;:::;
k 1 .
For every xed g 2 R
q
there exist minimal numbers k = k(q), l = l(q) 2
N
0
=N[f0gsu hthat (5
q
) holds. Forthis hoi e ofk,l we asso iate with
g2R
q
the ompanion polynomial
(6) f
q
(z)=z k
+
k 1 z
k 1
+:::+
0 2C[z℄
and theset Z
q
=f 2C :f
q
()=0g of itszeros. By k
2N we denote the
multipli ityof 2Z
q .
Lemma 1. Let g 2 R
q
for some q > 1. Then, for every a 2 N, the
solutions of the re urren e equation(5
q
) are given by
g(q
a)= X
2Zq P
()
( 2N
0
; l)
with polynomials P
(x)2 C[x℄, depending on a2N, of degree k
1 or
P
(x)0.
Proof. For xeda2N set h()=g(q
a). Then(5
q
) implies
h(+k)+
k 1
h(+k 1)+:::+
0
h()=0 ( l):
Asis wellknown (see,forinstan e, Lidland Niederreiter[5℄, Chapter6, or
Methfessel [7℄, Se tion 1), the solutions h 2 F of this re urren e equation
have theform
h()= X
2Z
q P
()
with ertain polynomialsP
(x)2C[x℄ , a ordingto thestatement of Lem-
ma1.
Theorem2.Letg2R
q
forsomeq >1andg(1)=1. Thenthe omplex
ve tor spa e P
g
has dimension 2. If b
F(0)6=0 for some F 2P
g
then P
g
onsists of all onstant fun tions, anddimP
g
=1.
Proof. InviewofTheorem1weonlyneedto onsiderthe ase(3
n )for
alln2N. Bytakingthe onvolutionprodu twithg 1
we mayrewrite(5
q )
as
q l+k+
k 1
q
l+k 1+:::+
0
q l =0:
Summation over d 2 N of the values at d and insertion of (3
n
) for n 2
fq l
;:::;q l+k
g leadsto f
q
(1)=0, whi h is equivalent to 12 Z
q
. Hen e, by
Lemma1 and theminimalityofk =k(q),
g(q
a)=P
1 ()+
X
2Zqnf1g P
()
( l)
with ertainpolynomials P
(x)2C[x℄ dependingon a2N, and P
1
(x)60
forsomea2N. Fromthiswe on ludethatg(n)6=O (1)asn!1. Finally,
Theorem1 and theRiemann{Lebesgue lemmayield
1
=
1
=0.
As to the onverse of Theorems1 and 2,we add thefollowingresult.
Theorem 3. Assume that the Fourier series of F 2L(T) has the form
X
m2Z
sgnm
g(jmj)e(mx)
with some sequen eg 2 F, g(1)= 1, and some onstants
1
;
1
2C, and
let the system S begenerated by g. If all series
1
X
d=1
n
(d)F(dx) (n2N)
onverge in L(T) then F 2P .
Proof. Re all that, byFejer'stheorem,theseries
X
m2Z
sgnm
g(jmj)e(mx) (x2T)
is Cesaro-summable to F 2L(T). With thenotation of theproofof Theo-
rem 1,F
n
2L(T) hastheFourier series
X
m2Z
sgnm
g(njmj)e(mx) (x2T):
By assumption, the series S
n
(x) onverges for almost all x 2T, and S
n 2
L(T). A short al ulation similar to that inthe proof of Theorem 1 shows
that
X
m2Z
X
djm
n (d)
b
F(m=d)e(mx) (x2T)
isthe Fourierseriesof S
n
. Now observe that
X
djm
n (d)
b
F(m=d)=
sgnm (
n
g)(jmj)=
sgnm
g(njmj) (m2Z
):
Hen e the Fourier series of F
n
and S
n
oin ide, whi h gives F
n
= S
n in
L(T), and theassertionof Theorem3 follows.
3. Multipli ative sequen es. A sequen e g : N ! C is alled mul-
tipli ative if g(1) = 1 and g(mn) = g(m)g(n) for all oprime m;n 2 N.
The set M of multipli ative sequen es forms a subgroup of F
under the
Diri hlet onvolution. By T we denote the subsetof totally multipli ative
sequen es, and we writehni and hTi for the set of all d 2 N whose prime
fa torsdividen2N, andforthemultipli ativesemigroupgeneratedbythe
set T P respe tively. First,werestate Theorem2 from [6℄as
Lemma 2. Let g 2 F, g(1) = 1, and
n
= g 1
g(n
) for all n 2 N.
Then g 2Mis equivalent to supp
n
hni for all n2N, and in this ase
mn
=
m
n
for all oprime m;n2N.
ByLemma2,everysystemS generatedbysomeg2M anequivalently
be repla ed by the partialsystem of equations (2
n
) for all n 2P
=fp
:
p2P; 2Ng.
With g2Mand p2P we asso iate theformalDiri hlet series
e g
p (s)=
1
X
k=0 g(p
k
)p ks
(s2C) ;
andwe denotebyM
p
thesetofallg2Mhavingthepropertythateg
p (s)is
absolutely onvergent and zero-free inthe losed half plane Res0. The
Theorem4.Letg2M
p
forsomep2P. Thenthe omplexve torspa e
P
g
has dimension 2. In parti ular, b
F(0)=0 for every F 2P
g .
Proof. It follows from Wiener's inversion theorem (see, for instan e,
Rudin [11℄, Chapter 18) applied to the power series G
p
(z) = eg
p
(s) with
z=p s
thatM
p
formsasubgroupof M,forevery p2P. Forn=p
2P
we inferfrom Lemma2 that
1
X
d=1 j
n (d)j=
1
X
k=0 j
n (p
k
)j
1
X
%=0 jg
1
(p
%
)j 1
X
=
jg(p
)j 1
X
=
jg(p
)j=O(1)
as ! 1. Hen e equation (3
n
) failsto hold forall n2N, and Theorem1
givesthedesired on lusion.
A lo al version of Theorem 4 is easily obtained by assuming g(1) = 1
and supp
n
hpi for all n 2 hpi instead of the global multipli ativity of
thesequen e g inthede nitionofM
p .
The proof of Theorem 4, similarto that of Theorem 2, onsists in ex-
luding ertain generating sequen es g of S satisfyingthe equations (3
n ),
n2N. It would therefore be desirable to know how to re onstru t g2 M
from anin nite systemof equationsof thetype
(7
n
) f(n)=
1
X
d=1
n
(d)h(d)
forall n2N, where h2T andf 2F aresuitably hosensequen es.
Lemma 3. For g2 F with g(1) =1 let
n
=g 1
g(n
) for all n 2N,
and let h 2 T. Suppose that all series (7
n
) for n 2 N are onvergent, so
that f 2 F is de ned and f(1) =1. If g 2M then f 2 M, and if g 2T
then f 2T. Conversely, iff 2T andf(p)h(p)6=1 for all p2Pthen there
exists a uniquely determined g 2 M su h that (7
n
) is valid for all n 2 N,
namelyg=f 2T. Thisimpli ationdoesnotholdfor f 2T iff(p)h(p)=1
for some p2P.
Proof. The Diri hlet series
1
X
d=1
n
(d)h(d)
d s
(n2N)
onverges fors=0 andhen eforRes>0. The onvergen eisabsolute for
equation
1
X
d=1
mn
(d)h(d)
d s
= 1
X
a=1
m
(a)h(a)
a s
1
X
b=1
n (b)h(b)
b s
;
whi h extends analyti ally to the half plane Res > 0. By letting s ! 0
along the positive real axis, Abel's limit theorem for Diri hlet series gives
f(mn) =f(m)f(n), i.e. f 2M. If, moreover, g 2 T then
n
= g(n)" for
all n2N, and f =g2T follows.
Next, Lemma2 impliestheidentity
(8)
p
(p k
)
p
+1(p k 1
)=g(p
)g 1
(p k
)
validforg 2M, p 2P, 2N
0
,k 2N. Multiplywith h(p k
) and sum over
k 2N to obtain
(9) f(p
) h(p)f(p
+1
)=g(p
) 1
X
k=0 g
1
(p k
)h(p k
) (p2P; 2N
0 ):
In parti ular,for=0,
(10) 1 h(p)f(p)=
1
X
k=0 g
1
(p k
)h(p k
) (p2P):
By inserting (10) into (9) and dividing by 1 h(p)f(p) 6= 0 we see that
f 2 T implies g(p
) = f(p
) for all p
2 P
and hen e g = f. The last
assertionof Lemma2 resultsfrom thefollowing onstru tion.
Example 1. Let f;h 2 T, T = fp 2 P : f(p)h(p) = 1g 6= ; and
q:T !N su h thatq(p)p forp2T. Theng2Mde nedby
g(p k
)= 8
<
: f(p
k
)( 1) k
1=q(p)
k
ifp2T,
f(p k
) ifp62T,
satis es (7
n
) for all n 2 N and, moreover, g(n) = O (jf(n)j) as n ! 1,
n2hTi.
Inordertoverifytheassertionswemayrestri tourattentionton2hTi.
First,observe that(gh) 1
(gh)(n
)=
n h(n
),andrepla eghbygandfh
byf. Then it suÆ es to show that the restri tion of thesequen e g 2 M
with
g(p k
)=( 1) k
1=q(p)
k
(p2T; k2N)
to hTi satis esbothg(p k
)=O(1)asp k
!1,and (3 ) foralln2hTi.
In fa t, we have theestimate
0<g(p k
)= 1
kq(p) k 1
Y
=1
1+ 1
q(p)
1
kq(p)
(k 1)q(p)
Y
%=0
1+ 1
%
1=q(p)
= 1
kq(p)
((k 1)q(p)+1) 1=q(p)
<(kq(p))
1+1=q(p)
;
whi h givestheO-relation. Next, onsiderforjzj<1 thepowerseries
G
p (z)=
1
X
k=0 g(p
k
)z k
=(1 z) 1=q(p)
6=0;
and invert to obtain
G 1
p (z)=
1
X
k=0 g
1
(p k
)z k
=(1 z) 1=q(p)
= 1
X
k=0 ( 1)
k
1=q(p)
k
z k
:
Hen e
g 1
(p k
)=( 1) k
1=q(p)
k
(p2T; k2N);
and,similarto theabove,
0< g 1
(p k
) 1
q(p) k
1 1=q(p)
(p2T; k2N):
This shows the absolute onvergen e of the power series G 1
p
(z) even for
jzj1,and Abel'slimittheorem forpowerseriesyields
1
X
k=0 g
1
(p k
)=0 (p2T):
Finally,forjzj<1and n=p
withp2T, 2N we ndtherepresentation
1
X
k=0
n (p
k
)z k
=G 1
p (z)
1
X
k=0 g(p
+k
)z k
=z
1 G
1
p (z)
1
X
k=0 g(p
k
)z k
;
theright sideof whi his an absolutely onvergent powerseriesfor jzj1.
By applyingAbel's limittheoremagain, we arriveat (3
n
)forall n2T.
Inparti ular,Example1showsthatthereexistin nitelymanysequen es
g2Msatisfying(3
n
) forall n2N, with g(n)=O(1) asn!1. We leave
theproblemopenwhether the orresponding series
1
X
n=1
g(n)e(nx) (x2T)
4. Multipli ativere urrentsequen es. A ordingtoTheorem1(a),
every sequen e g 2 F with g(1) = 1 uniquely determines a system S of
fun tionalequations. Apartfromthe lassi al aseg2T,where
n
=g(n)"
forall n2N, there existsno generalexpli itformulaforthe orresponding
sequen es
n
2 F. However, for generating sequen es g 2 M\R
q we
des ribe a onstru tive method based on Lemmas 1 and 2, whi h enables
usto determineS inmany ases. The followingtheorem hara terizes the
lo al polynomial-exponentialtype stru tureof sequen esg2M\R
q .
Theorem 5. For g2M the following assertionsareequivalent:
(a) g2R
p
for some p2P,
(b) g2R
q
for some q 2N, q 6=1,
( ) there exist integers p 2P, k;l2N
0
with k+l 6=0,and oeÆ ients
0
;:::;
k 1
2C,
0
6=0, su h that g satis es the linear re urren e equation
g(p
+k
)+
k 1 g(p
+k 1
)+:::+
0 g(p
)=0 ( 2N
0
; l);
(d) there exist integers p2P, k;l;r2N
0
with k+l6=0, k
1
;:::;k
r 2N
satisfyingk
1
+:::+k
r
=k,andpolynomials P
%
(x)2C[ x℄of degreek
% 1
or P
%
(x)0 (%=1;:::;r), su h that
g(p
)= X
1%r P
% ()
%
( 2N
0
; l):
Proof. Trivially,(a)implies(b). Ifg2R
q
andq6=1isapowerofp2P
then a re urren e equation of type ( ) follows from (b) by taking a = p
with 2N
0 in(5
q
). Now, let q =q
1 q
2
beany oprimede ompositionwith
q
1
;q
2
2 N nf1g and a prime power q
2
. If g(q
%
1
b) 6= 0 for in nitely many
exponents % 2 N and some b2 N with (b;q
2
) =1 then it follows from (b)
and g2Mbytaking a=q
%
1 q
2 b in(5
q ) that
g(q l+%+k
1
b)g(q l++k
2
)+
k 1 g(q
l+%+k 1
1
b)g(q
l++k 1
2
)+:::
+
0 g(q
l+%
1
b)g(q l+
2
)=0:
Fix% and bsu hthat g(q l+%
1
b)6=0. Then
d
k g(q
l++k
2
)+d
k 1 g(q
l++k 1
2
)+:::+d
0 g(q
l+
2
)=0
with ertain oeÆ ients d
0
;:::;d
k 1
;d
k
2 C, d
0
6= 0, whi h gives a re-
urren e equation a ording to ( ). If g(q
%
1
b) = 0 for all suÆ iently large
%2N and all b2N with (b;q
2
)=1 then, bythe multipli ativityof g, also
g(q
%
1
a) = 0 for all suÆ iently large % 2 N and all a 2 N. Hen e g 2 R
q
1
where q
1
has a redu ed number of prime divisors. In this ase pro eed
indu tivelyto obtain( ).
The equivalen e of ( ) and (d) follows from Lemma 1. It remains to
form a=p
l
b with2N
0
, l. Nowg2Mimplies
g(p l+k
a)+
k 1 g(p
l+k 1
a)+:::+
0 g(p
l
a)
=g(b)(g(p
+k
)+
k 1 g(p
+k 1
)+:::+
0 g(p
))=0;
whi h ompletestheproof.
With g2Mand p2P we asso iate thep- bre g
p
2Mde ned by
g
p (n)=
g(n) ifn2hpi,
0 otherwise,
(n2N)
and theformalpowerseries
G
p (z)=
1
X
=0 g(p
)z
(z2C);
whi h takes the formof the Diri hlet serieseg
p
(s) of g
p
bysubstitutingz =
p s
. Further we set N
p
= fg 2 M :g = "on hpig. Obviously N
p
R
p .
Thefollowingtheoremdes ribesthemultipli ativestru tureofM\R
p with
respe tto theDiri hlet onvolution.
Theorem 6.For every p2P, M\R
p
forms a group underthe Diri h-
let onvolution,and thereisa group homomorphism withkernel N
p from
M\R
p
onto the multipli ative group of all quotients P(z)=Q(z) of rela-
tively prime polynomials P(z);Q(z) 2C[z℄ satisfying P(0)=Q(0)=1. In
parti ular,
(g)=G
p
(z); Q(z)=z k
f
p
1
z
; P(z)=Q(z)G
p (z);
wherek2N
0
isthedegreeofthe ompanionpolynomialf
p
(z)ofg2M\R
p .
Proof. First,observethatforg2Mtherationalityofthepowerseries
G
p
(z)isequivalentto g2R
p
. Hen e,ifg;h2M\R
p
thengh2M,and
the orrespondingpowerseriesG
p (z);H
p
(z)aswellastheirCau hyprodu t
G
p (z)H
p (z)=
1
X
=0 X
%+=
g
p (p
%
)h
p (p
)z
are rationalfun tions. This gives gh2M\R
p
and shows that M\R
p
is losed undertheDiri hlet onvolution.
Further, the inverse g 1
of g 2 M belongs to M. From Theorem 5
we know that the power series G
p
(z), asso iated with g 2 M\R
p and
representing a rational fun tion, has positive radius of onvergen e, and
G
p
(0) = 1. By the Cau hy{Taylor theorem, the same is true for G 1
p (z),
and the Cau hy produ t with G
p
(z) obviously orresponds to g 1
g = "
on theset hpi. Therefore, g 1
2M\R .
Next, noti ethatN
p
is asubgroupof thegroup M\R
p
. Themapping
g 7! g
p
is learly a group homomorphism with kernel N
p
. The Diri hlet
serieseg
p
(s)ofg
p
2M\R
p
isabsolutely onvergent insomerighthalfplane
of the omplexplane,and g
p 7! G
p (p
s
)=eg
p
(s) representsan isomorphi
mappingfrom(M\R
p )=N
p
ontothemultipli ativegroupofrationalpower
seriesR (z) witha positiveradius of onvergen e andR (0) =1.
Finally, by multiplying the re urren e equation in Theorem 5( ) with
z
+k
andsummingoverall 2N
0
,we seethatG
p (z)z
k
f
p
(1=z)is apolyno-
mialP(z) ofdegree k+l 1 withP(0)=1, whi his relativelyprime to
Q(z) =z k
f
p
(1=z), dueto the minimalityof k = k(p). This ompletes the
proof ofTheorem6.
The divisorfun tion 2Misde nedby =11,where1 denotesthe
onstant sequen e withvalue1 on N.
Example 2. Fors2C let thesystemS be generated by thesequen e
g = ( (n)n s
) 2 M. Observe that g 2 R
p
for every p 2 P with f
p (z) =
(z p s
) 2
. Forn=p
2P
we obtain
1
X
k=0
n (p
k
)z k
=(1 p s
z) 2
1
X
%=0
(+%+1)p (+%)s
z
%
=p
s
(+1 p s
z);
whi h gives
p
(p k
)= (
(+1)p
s
ifk =0,
p
(+1)s
ifk =1,
0 ifk >1.
Hen e S onsistsofthe equations
1
n n 1
X
=0 F
x+
n
=n s
X
djn
(d)
n
d
d s
F(dx) (n2N);
and dimP
g
2. In parti ular, for Res 0, s 6= 0, the null fun tion is
the only solutionF 2 L(T) of S, and for s= 0 the spa e P
g
of solutions
F 2L(T) of S onsists of all onstant fun tions. For Res>0 that spa e
isspannedbythe osineand thesinepart oftheseries
T(x)= 1
X
n=1
(n)n s
e 2inx
;
ifthese aretheFourierseriesofL(T) fun tionsrespe tively. Thisisobvious
forRes>1only,dueto theabsolute onvergen e ofT(x).
In a similar way extended repli ativity systems S generated, for in-
s s
respe tively theEuler fun tionand the sum of divisorsfun tion, and their
solutionsF 2L(T) an be treated.
5. Aperiodi solutions. ForintervalsI R wedenotebyC(I)theset
of ontinuousfun tions F :I !C. We areinterested inaperiodi solutions
F 2 C(R
+
) of systems S, generated by some sequen e g 2 F, g(1) = 1.
One mayexpe tthatthey arerather ex eptional.
It is wellknown thattheHurwitz zeta fun tion de ned by
(s;x)= 1
X
=0
(x+) s
(Res>1; x2R
+ )
extends to a fun tionholomorphi fors2 C nf1g and x 2C
=C nfx 2
R :x 0g, with a simple pole at s=1 withresidue 1. The polylogarithm
fun tion de ned by
L
s (z)=
1
X
=1
s
z
(s2C; z2U)
extends to a fun tion holomorphi fors2C and z2C n[1;1). In parti -
ular, we have the following two lemmas, whi h are taken from Milnor [8℄,
Lemma1, Lemma2 and Theorem1.
Lemma 4.Thefun tion
B
s
(x)= s(1 s;x)
is holomorphi for s 2 C and x 2 C
, with B
0
(x) = 1. For ea h s 2 C
the fun tion B
s : R
+
! C solves both the lassi al repli ativity system S
generated by g(n)=n s
and the di eren eequation
B
s
(x+1) B
s
(x)=sx s 1
:
Moreover,if s2N
0
then B
s
(x)2Q[ x℄ isthe sth Bernoulli polynomial and
B
s
=B
s
(0)is the sth Bernoulli number.
Lemma 5.For every s2C the linear spa e onsisting of all ontinuous
solutions F :(0;1) !C of the lassi al systemS generated by g(n) =n s
hasdimension2. Fors62N
0
itisspanned byB
s
(x)andB
s
(1 x),for s=0
by B
0
(x) and ot (x), for s2N by B
s
(x) and ReL
s (e
2ix
).
Now we an formulate ourmainresulton aperiodi solutionsof S.
Theorem 7.Letg2M\R
q
for two di erent q2P, andlet the system
S begenerated by g. Then S has an aperiodi solutionF 2C(R
+
) if and
only ifS is the lassi al system generated by g(n) =n s
with somes6=0.
In this ase
(11) F(x)= B
s
(x)+(x) (x2R
+ )
withsome onstant 2C
and some 1-periodi solution 2C(R ) of S.
The important point to note here is that the existen e of aperiodi so-
lutionsF 2C(R
+
) of extended repli ativitysystems S generated by mul-
tipli ative re urrent sequen es shrinks S to the lassi al system. This ex-
ludes the o urren e of aperiodi solutions F 2 C(R
+
) in many ases.
Observe that the previous result of Yoder [12℄, Theorem 4, is ontained in
Theorem 7 above, sin e every g 2 T has the propertyg 2 M\R
q for all
q2P.
Proof (of Theorem 7). By Lemmas 4 and 5 we only need to show
that the existen e of an aperiodi solution F 2 C(R
+
) of S impliesthat
g(n)=n s
with some s2C. We set
(x)=F(x+1) F(x) (x2R
+ )
and denote for ! 2 R
+ by R
!
the ring of !-periodi fun tions 2 C(R) .
The rst step oftheproof onsistsin derivingfromg2R
q
forsome q2N,
q6=1,a fun tionalequation of thetype
(12) q k
x
q k
+
k 1 q
(k 1)
x
q k 1
+:::+
0
(x)=0 (x2R
+ )
with ertain oeÆ ients
k 1
;:::;
0
2 C,
0
6= 0, and from g 2 M the
systemof fun tionalequations
(13)
x
p
+1
p
g(p
)
x
p
= 1
p p 1
X
%=0
x+%
p
+1
p
g(p
)
x+%
p
(x2R
+ )
for all p 2 P and 2 N. In the se ond step of the proof, while assuming
(12) fortwo numbersq 2N havingQ-linearly independentlogarithms(e.g.
fordi erent primes),weobtainthe expli itformula
(14) (x)=
X
s2S P
s
(logx)x s 1
(x2R
+ )
with some niteset S C and ertain polynomials P
s
(x) 2C[x℄. Further
we put(13) into theform
(15)
x
p
+1
p
g(p
)
x
p
= X
2T
p h
;p;
(x)
x
;
where T
p
is theset of zeros of thepolynomialt
p
(z) =(z p
1)=(z 1) p
and h
;p;
2R
1
. Inthethirdstep we insert(14) into (15)to on ludefrom
60 that(x)=ax s 1
withsome onstants a;s2C, a6=0. Finally,we
derivethat g(n)=n s
forall n2N.
Inparti ular,supposethatthesystemS isgeneratedbysomesequen e
g2R ,g(1)=1,where16=q2N. Then,similartotheproofofTheorem2,
equation (5
q
)mayberewrittenas
q l+k+
k 1
q
l+k 1+:::+
0
q l =0;
from whi hit follows that
q
l+k(d)F(dx)+
k 1
q
l+k 1(d)F(dx)+:::+
0
q
l(d)F(dx)=0
foralld2N, x2R
+
. Summingoverd 2N andwritingagainF
n
(x)forthe
left sideof equation (2
n
), we inferthat
(16) F
q
l+k(x)+
k 1 F
q
l+k 1(x)+:::+
0 F
q
l(x)=0 (x2R
+ ):
Observe that F
n
(x+1) F
n
(x) =(1=n)(x=n) forall n2N and x2R
+ ,
and subtra tequation (16) at x fromthat at x+1. It follows that
1
q l+k
x
q l+k
+
k 1 1
q l+k 1
x
q l+k 1
+ :::+
0 1
q l
x
q l
=0 (x2R
+ );
whi h isequivalent to (12).
Next, assumethatthesystemS isgeneratedbysomesequen e g2M.
Observe thatS turnsinto the systemoffun tionalequations for,
1
n
x
n
= 1
X
d=1
n (d)
d 1
X
Æ=0
(dx+Æ) (n2N):
By multiplyingtheidentity(8)with
p k
1
X
Æ=0
(p k
x+Æ);
summingoverk 2N and applyingLemma 2,we obtain
1
p
x
p
g(p
)(x) 1
p
+1 p 1
X
%=0
px+%
p
+1
=g(p
)
p
;
forall p2Pand 2N
0
,where
p
= 1
X
k=1 g
1
(p k
) p
k
1
X
Æ=0
(p k
x+Æ):
In parti ular,for=0we see that
p
= 1
p p 1
X
%=0
px+%
p
;
from whi h(13) follows withpxinstead ofx.
The derivation of (14) and (15) from (12) and (13) requires two more
Combining(14)and (15) gives
X
s2S
x
p
s 1
(p
s
P
s
(logx logp
+1
) g(p
)P
s
(logx logp))
=p
X
2Tp h
;p;
(x)
x
:
Observe that 1 2 T
p
and all 2 T
p
, 6=1, satisfy jj> 1. Therefore the
rightsideof theaboveequationredu esto thetermp
h
1;p;
(x)=h
p;
(x),
say, andwith pxinsteadof xwe arrive at
X
s2S x
s 1
(p
s
P
s
(logx logp
) g(p
)P
s
(logx))=h
p;
(px);
where h
p;
2 R
1
. It follows from (x) 60 that (x)= ax s 1
with some
onstantsa6=0,s2S. Moreover,g(p
)=p
s
ors=1. Sin e g2M, the
rst equation is validforall p2P and 2N ifs6=1. If s=1 thenwe see
from Lemma4 thatF(x)=aB
1
(x)+(x)withsome fun tion2R
1 . On
insertingthisinto(2
n
) we obtain
a 1
X
d=1
n (d)B
1 (dx)
a
n B
1 (x)=
1
n n 1
X
=0
x+
n
1
X
d=1
n
(d)(dx)
wheretherightsiderepresentsafun tion
n 2R
1
,say. NowB
1
(x)=x 1=2
yields
1
X
d=1
n
(d)d=n 1
and
a
2
1
n
f(n)
=
n (x);
where, forabbreviation,
f(n)= 1
X
d=1
n (d):
By integrating overTwesee that b
n (0)=
b
(0)(1 f(n)), leadingto
(17)
a
2
1
n
f(n)
= b
(0)(1 f(n)):
Supposethat b
(0)6=0. Then
f(n)
b
(0) a
= b
(0) a
and b
(0)6=
a
;
sothat
f(n)= b
(0) a=(2n)
b
(0) a=2 :
ByLemma3,f ismultipli ative,and we on ludefrom f(mn)=f(m)f(n)
for oprime m;n 2 N that b
(0) = 0, ontrary to our assumption. Hen e
b
(0) = 0 and f(n) = n 1
by (17). Applying Lemma 3 again we obtain
g(n)=n 1
,whi histhe desired on lusion.
Example 3. A ording to Example 2, the sequen e g = ( (n)n s
)
belongs to M\ R
q
for every q 2 P. By Theorem 7 the orresponding
system S has no aperiodi solutions F 2 C(R
+
). The same holds for
theextended repli ativitysystems,generated forinstan e bythe sequen es
('(n)n s
)and ((n)n s
).
6. Di eren eequations. Inthisse tionwe losethegapinthepre ed-
ingproofofTheorem7byderivingtheequations(14)and(15)respe tively
from the fun tional equations (12) and (13). We begin with two lemmas
on erningthe ontinuoussolutionsof homogeneous di eren eequations.
For xed ! 2 R
+
we are interested in the solutions 2 C(R) of the
linear homogeneousequation
(18) (x+k!)+
k 1
(x+(k 1)!)+:::+
0
(x)=0 (x2R)
where k 2 N,
0
;:::;
k 1
2 C,
0
6= 0. We may onsider C(R) as an
R
!
-moduleandinterprettheindeterminate zinthe ompanionpolynomial
(19) f(z)=z
k
+
k 1 z
k 1
+:::+
0 2C[z℄
of equation (18) as the endomorphism of C(R) de ned by 7! z with
z(x) = (x+!) for all x 2 R. Further, we may write the omposition
of endomorphisms as produ t. Then (18) takes the form f(z)(x) = 0,
and the problemof solving (18) onsists in determining kerf(z) C(R) .
Observe that kerf(z) is a k-dimensionalR
!
-submodule of C(R). Namely,
if
1
;:::;
r 2C
arethedistin tzeros off(z)oforderk
1
;:::;k
r
2N; k
1 +
:::+k
r
=k,then
kerf(z)=ker (z
1 )
k1
:::ker (z
r )
kr
:
It remains to determine ker (z ) k
where 2 C
, k 2 N. Noti e that
thebran hesofthemulti-valuedfun tion x=!
=e
(x=!)log
onlydi er bya
fa tor of theforme(mx=!)2R
!
withm2Z. Now
(z ) k
x=!
P(x)= k
x=!
(z 1) k
P(x)
vanishesforall x2R ifandonlyif(z 1) k
P(x)0,whi hisequivalentto
P(x)2R
!
[x℄,degP(x)k 1,orP(x)0. Wethusarriveatthefollowing
Lemma6.Let
% 2C
bethedistin tzerosoforderk
%
2N (%=1;:::;r)
of the ompanion polynomial (19), asso iated with equation(18). Then 2
C(R) solves (18) if and onlyif
(x)= X
1%r P
% (x)
x=!
%
with polynomials P
%
(x) 2 R
!
[x℄, degP
%
(x) k
%
1, or P
%
(x) 0 (% =
1;:::;r).
Inthesequel,weindi atethedependen yon! 2R
+
oftheshiftoperator
z de ned above bywriting z
!
. The followinglemma extends a theorem of
Popovi iu [10℄ (seeKu zma [4℄,Theorem 13.5).
Lemma 7.Let !
1
;!
2 2R
+
su hthat !
1
=!
2
62Q. Let f
1 (z);f
2
(z) 2C[z℄
satisfy f
1 (0)f
2
(0)6=0,andlet Abethe ( nite)setof (uniquely determined)
solutions 2C of the system
f
1 (e
!
1
)=f
2 (e
!
2
)=0:
Then every 2kerf
1 (z
!1
)\kerf
2 (z
!2
) has the form
(x)= X
2A P
(x)e
x
with ertain polynomials P
(x)2C[ x℄.
Proof. Be auseofthedire tsumrepresentationofthekernelsoff
1 (z
!
1 )
and f
2 (z
!2
) it suÆ es to prove the followingspe ial ase: Let
1
;
2 2C
and k 2 N. Then the existen e of a non-trivial solution 2 C(R) of the
di eren eequations
(z
!
1
1 )
k
=0; (z
!
2
2 )
k
=0
impliesthat
1
=e !
1
,
2
=e !
2
withauniquelydeterminednumber 2C,
and (x)=P(x)e x
withsome polynomialP(x)2C[ x℄, degP(x)k 1.
Fix some values
1
of (log
1 )=!
1
and
2
of (log
2 )=!
2
. Then, by the
irrationalityof !
1
=!
2
,theequation
(20)
1 +
2m
1
i
!
1
=
2 +
2m
2
i
!
2
hasat mostone solution(m
1
;m
2 )2Z
2
.
Lemma 6 yieldstherepresentations
(x)=P
1 (x)e
1 x
=P
2 (x)e
2 x
(x2R)
with polynomials P
1
(x) 2 R
!1 [x℄, P
2
(x) 2 R
!2
[x℄ of degree k 1. By
omparingsummandswe obtain
(21) h (x)x
e
1 x
=h (x)x
e
2 x
(x2R; 0<k)
where h
1
2R
!
1 ,h
2
2R
!
2
. In parti ular,withh=h
1
and Æ =
2
1 ,
itfollows that h(x+!
1
)=h(x) and h(x+!
2 )=e
Æ!
2
h(x) forall x2R or,
equivalently,
(22) h(x+m!
1 +n!
2 )=e
nÆ!
2
h(x) (m;n2Z; x2R):
By Krone ker's theorem (see for instan e Hardy and Wright [2℄, Theo-
rem 444) the set fm!
1 +n!
2
: m;n 2 Zg is dense in R. Consequently,
for every y 2 R there exists a sequen e of pairs (m
;n
) 2 Z
2
depending
onlyon y su h that the orresponding sequen e (m
!
1 +n
!
2
) tends to y
as tendsto in nity. Sin eh is ontinuous, (22) implies
h(x+y)=h(x)H(y) (x;y 2R)
with some fun tionH :R !C. We may assume that h(x)60. Then h is
zerofreeonR andh(y)=h(0)H(y). Therefore,His ontinuousandzerofree
too,and additionallysatis esthefun tionalequation
H(x+y)=H(x)H(y) (x;y2R) :
Thesefun tions aregiven byH(x)=e x
,whi hleadsto h(x)= e x
,with
some onstants 2C, 2C
. Byinserting weseefrom (22)thate !
1
=1
and e !2
=e Æ!2
or, equivalently, !
1
=2%
1
iand !
2
=Æ!
2 +2%
2
iwith
ertain numbers %
1
;%
2
2Z. It follows that
(23) Æ=
2ki
!
1
2li
!
2
;
whi h showsthat (20)hasatleastone solution. We on ludethat is well
de ned.
A tually,theequations(20),(21)and(23)yieldthatthefun tionde ned
by
x7!h
1
(x)e
%
1
!
1 x
=h
2
(x)e
%
2
!
2 x
(x2R)
belongs to R
!
1
\R
!
2
and therefore must be a onstant. This means that
(x)=P(x)e x
witha polynomialP(x)2C[x℄ ofdegree k 1,whi his
thedesired on lusion.
In order to derive(14) from (12) we multiply (12)with x, repla e x by
e x
,put(x)=e x
(e x
) and !=logq. Thisyieldsthedi eren e equation
(24) (x k!)+
k 1
(x (k 1)!)+:::+
0
(x)=0 (x2R):
With the ompanion polynomial f
q
(z) of g 2 R
q
a ording to (6), the
ompanion polynomial of equation (24) equals z k
f
q
(1=z). Therefore the
zeros
% 2 Z
q
(% = 1;:::;r) of order k
%
2 N of f
q
(z) orrespond to the
zeros 1
2C
ofthe same orderof z k
f
q
(1=z). By applyingLemma6 and
returningto we obtain
(x)= X
1%r P
%
(logx)x s
% 1
(x2R
+ )
with polynomials P
%
(x)2 R
!
[x℄, degP
%
(x) k
%
1, or P
%
(x) 0, where
s
%
issome xed value of (log
%
)=! for%=1;:::;r.
Now we asso iate with g 2 R
q
the set S
q
= fs 2 C : f
q (q
s
) = 0g.
Observe that S
q
= ; if and only if k = 0, and ardS
q
= 1 for k 2 N,
sin eevery zero 2Z
q
of the ompanion polynomialf
q
(z) ontributes the
distin telements
log+2mi
logq
(m2Z)
to S
q
,where log is some xed logarithm of 6=0. Assume now that g2
R
q
1
\R
q
2
,g(1)=1,with !
1
=logq
1 and !
2
=logq
2
linearlyindependent
over Q. Due to the irrationality of !
1
=!
2
, the interse tion S = S
q
1
\S
q
2
is always a niteset. Noti ethat for every s2S thenumbers
1
=e
!
1 s
,
2
=e
!2s
are zeros of f
q
1
(z) and f
q
2
(z), respe tively. Hen e, byapplying
Lemma 7 to the (transformed) fun tional equations (12) for q = q
1
;q
2 we
obtain(14).
In order to derive (15) from (13) denote the left hand side of (13) by
H(x). Then (13)takestheform ofthe di eren eequation
H(x)= 1
p p 1
X
%=0
H(x+%) (x2R
+ )
with the ompanion polynomial t
p
(z) = (z p
1)=(z 1) p having only
simplezeros 2T
p
. A ording to Lemma6, the solutionsH 2C(R
+ ) are
given by H(x) = P
2T
p h
(x)
x
with fun tions h
2R
1
, whi h gives (15)
and ompletesthe proofof Theorem7.
A loser inspe tionof the proof of Theorem 7 shows that the existen e
of aperiodi solutions F 2C(R
+
) of ertain extended repli ativitysystems
S an be ex luded by onsidering suitable nite subsystems only. This
requires lookingagain at the ompanionpolynomial
f
q
(z)=
k z
k
+
k 1 z
k 1
+:::+
0 2C[z℄
of g 2 R
q
, where
k
= 1 and
0
6= 0. We asso iate with g 2 R
q the
(non-empty) set
N
q
=fq l+
:0k;
6=0g
of powersof q,where k;l2N
0
are hosen minimalsu h that equation (5
q )
isvalid.
Theorem8.For di erentprimesq
1
;q
2
assumethat thesequen eg2F,
normalized by g(1)=1, satis es g2R and supp hqi for all n2hqi,
q 2 fq
1
;q
2
g. If the system S generated by g has aperiodi solutions F 2
C(R
+
)then thereexistssomes2C su h that g(n)=n s
for alln2hq
1 q
2 i.
Hen e,ifthe nitesubsystemofS onsistingoftheequations(2
n
)forall
n2N
q
1 [N
q
2
doesnot oin ide withthatofthe lassi alsystemgenerated
byg(n)=n s
thenS hasnoaperiodi solutionsF 2C(R
+
). Fortheproof
of Theorem8werefer to that ofTheorem 7.
Example4. A ordingtoExample2,thenon-existen eofaperiodi so-
lutionsF 2C(R
+
)ofthesystemS generatedby( (n)n s
)followsalready
from its equations(2
n
) forn=q;q 2
andtwodi erent primesq.
Referen es
[1℄ E.Artin,TheGammaFun tion,Holt,RinehartandWinston,NewYork,1964.
[2℄ G.H.HardyandE.M.Wright,AnIntrodu tiontotheTheory ofNumbers,3rd
ed.,OxfordUniv.Press,London,1960.
[3℄ D.Kubert,Theuniversalordinarydistribution,Bull.So .Math.Fran e107(1979),
179{202.
[4℄ M. Ku zma,Fun tional Equations in a Single Variable, PWN{Polish S ienti
Publishers,Warszawa,1968.
[5℄ R.LidlandH.Niederreiter,Introdu tiontoFiniteFieldsandtheirAppli ations,
CambridgeUniv.Press,London,1986.
[6℄ L. G.Lu ht,Arithmeti alsequen esandsystems offun tional equations,Aequa-
tionesMath.53(1997),73{90.
[7℄ C. Methfessel, Multipli ative and additive re urrent sequen es, Ar h. Math.
(Basel)63(1994),321{328.
[8℄ J. Milnor, On polylogarithms,Hurwitz zeta fun tions, and theKubertidentities,
Enseign.Math.29(1983),281{322.
[9℄ N.E.Norlund,VorlesungenuberDi erenzenre hnung,Springer,Berlin,1924.
[10℄ T.Popovi iu,Remarquessurlade nitionfon tionnelled'un polyn^omed'uneva-
riable reelle,Mathemati a(Cluj)12(1936),5{12.
[11℄ W.Rudin,RealandComplex Analysis,M Graw-Hill,London,1970.
[12℄ M.F.Yoder,Continuousrepli ativefun tions,AequationesMath.13(1975),251{
261.
InstitutfurMathematik
Te hnis heUniversitatClausthal
Erzstrasse1
38678Clausthal-Zellerfeld, Germany
E-mail:lu htmath.tu- lausthal.de
Re eivedon1.10.1996
andinrevisedformon4.12.1996 (3053)
!["Plundring, skatter och den fecdala statens framväxt. Organisatoriska tendenser i Sverige under övergången från vigingatid till tidig medeltid", Thomas Lindkvist, Opuscula Historica Upsaliensia 1, Uppsala 1988 : [recenzja]](data:image/gif;base64,R0lGODlhAQABAIAAAP///wAAACH5BAEAAAAALAAAAAABAAEAAAICRAEAOw==)