Special Issue on Space Dynamics

RESTRICTED COMPLEX DOMAINS

WATCHARAPON PIMSERT, VICHIAN LAOHAKOSOL, AND SAJEE PIANSKOOL Received 15 September 2004; Revised 22 March 2006; Accepted 4 April 2006

Using a method modified from that used by Pisot and Schoenberg in 1964-1965, a Cauchy’s functional equation with restricted domains in the complex field is solved for uniformly continuous solutions.

Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.

1. Introduction

Consider the Cauchy’s functional equation, hereby called CFE, f

⎛

⎝^A

j=1

ujαj

⎞

⎠=^A

j=1

fujαj

. (1.1)

Under various assumptions onAandα_js, Pisot and Schoenberg [8] solved for monotone solutions of (1.1). In a subsequent paper [9], they treated the case where the domain of solutions is a subset ofRⁿwith the following main result. Letα1,α2,...,αAbe elements of Rⁿ(n < A) satisfying the following conditions:

(1) every set ofnelements among theαiis linearly independent overR,

(2) the elementsα1,α2,...,αA are rationally independent, that is, linearly indepen- dent overQ.

LetS= { ^A_j=1ujαj|uj∈N0, the set of nonnegative integers},Ba Banach space and f : S→B. Iff is a uniformly continuous solution of the CFE (1.1), thenf(x) admits a unique representation of the form f(x)=λ(x) + ^A_m=1ϕ_m(x), whereλis a linear function from RⁿintoB, and eachϕ_m(m=1, 2,...,A) is a function fromR_m= {u_mα_m+ ^A_j=1,j=mk_jα_j| um∈N0,kj∈Z}intoBsatisfying

(1)ϕm(0)=0,

(2)ϕ_m(x+α_j)=ϕ_m(x) (j=m,x∈R_m),

(3)ϕmis a uniformly continuous function onRmintoB.

Studying both of Pisot’s and Schoenberg’s works, we observe two significant features, first, that their method of proof in [9], both beautiful and powerful, can be extended to

Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 69368, Pages1–12

DOI10.1155/IJMMS/2006/69368

wider classes of functional equations and second, that the rational independence can be replaced by denseness, even though the two conditions are not equivalent. We illustrate these observations by solving similar, but different, CFEs in the complex field together with relevant examples and several remarks relating them with the classical complex CFE (see [1]) and in particular with an old theorem of Erd¨os about monotone additive func- tions (see [4]). Our main results say roughly that solutions of complex CFEs, uniformly continuous over some dense subsets and additive with respect to certain real or complex base elements, consist generally of two parts, one linear and the other periodic, and the periodic part disappears if each of its elements has a natural dense subdomain.

2. Case of real base elements

In this section, we solve the CFE (1.1) with real base elementsαj and Gaussian integer coefficients.

Theorem 2.1. LetA∈N,A≥2, andα1,α2,...,α_A∈Rbe such that the set S⁺=

_A

m=1

u_m+iv_mα_m|u_m,v_m∈N0

(2.1) is dense inC. If f :S⁺→Cis a uniformly continuous solution of the functional equation

f _A

m=1

u_m+iv_mα_m

= ^A

m=1

fu_mα_m+fiv_mα_m, (2.2)

then f can be uniquely written as

f(x)=λ(x) + A m=1

ϕ_m(x), (2.3)

whereλ:C→Cis anR-linear function andϕm(m=1, 2,...,A) is a complex-valued func- tion defined on

Sm=

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

uma+iumb

αm+ A j=1 j=m

kja+ikjb

αjuma,umb∈N0;kja,kjb∈Z

⎫⎪

⎪⎪

⎬

⎪⎪

⎪⎭

⊇S⁺ (2.4)

satisfying

(1)ϕ_m(0)=0,

(2)ϕm(x+αj)=ϕm(x+iαj)=ϕm(x) (j=mandx∈Sm), (3)ϕmis uniformly continuous onSm.

Proof. Let f be a uniformly continuous solution of (2.2). There are three major steps in the proof. First, we show that the two limits lim_N→∞f(Nα_m)/N and lim_N→∞f(iNα_m)/iN exist. Then we define anR-linear functionλ:C→Cand show thatλis uniformly con- tinuous onS⁺. Finally, the periodic functionsϕmare constructed.

Step 1. Letε >0. By the uniform continuity of f, there existsδ >0 such that

∀z,w∈S⁺, |z−w|< δ=⇒f(z)−f(w)< ε. (2.5) SinceS⁺is dense inC, there arex,y∈S⁺such that|x−y|< δ. Write

x= A m=1

cma+icmb

αm, y= A m=1

dma+idmb

αm, (2.6)

where allc_ma,c_mb,d_ma,d_mb∈N0. Setq_mτ=c_mτ−d_mτ(m=1, 2,...,A),τ∈ {a,b}, and split {1, 2,...,A}into any two disjoint nonempty setsIandJwithI∪J= {1, 2,...,A}.

We claim that for eachM∈N, we have

t∈I

fcta+ictb αt

−fdta+idtb αt + 1

M

j∈J

ηja

fwja+Mqjaαj

−fwjaαj

+ 1 M

j∈J

η_jbfiw_jb+Mq_jbα_j−fiw_jbα_j< ε,

(2.7)

where, for j∈J,w_ja,w_jb are any nonnegative integers,η_jτ=1 if q_jτ≥0, η_jτ= −1 if q_jτ<0 (τ∈ {a,b}).

To prove this inequality, definec^(k)_jτ,d^(k)_jτ forτ∈ {a,b},j∈J, andk∈Nas follows:

ifqjτ≥0, setc⁽_jτ^k)=wjτ+kqjτ, d⁽_jτ^k)=wjτ+ (k−1)qjτ;

ifqjτ<0, setc⁽_jτ^k)=wjτ+ (k−1)qjτ, d⁽_jτ^k)=wjτ+kqjτ. (2.8) We see thatc⁽_jτ^k),d⁽_jτ^k)∈N0andq_jτ=c⁽_jτ^k)−d⁽_jτ^k). Now

t∈I

cta+ictb

αt+

j∈J

c⁽_ja^k)+ic⁽_jb^k)αj

−

t∈I

dta+idtb

αt+

j∈J

d⁽_ja^k)+id⁽_jb^k)αj

=

A m=1

qma+iqmb

αm

=

A m=1

cma+icmb

αm− A m=1

dma+idmb

αm

= |x−y|< δ, (2.9) which, by the uniform continuity condition (2.5) and the CFE (2.2), yields

t∈I

fcta+ictb

αt

−fdta+idtb

αt

+

j∈J

fc^(k)_jaαj

−fd^(k)_jaαj

+

j∈J

fic^(k)_jbαj

−fid^(k)_jbαj < ε.

(2.10)

In the last two sums, splitJinto two pairs of disjoint subsets

J=Ja⁺∪Ja⁻, J=Jb⁺∪Jb⁻, (2.11) whereJ_a⁺= {j∈J|q_ja≥0},J_a⁻ = {j∈J|q_ja<0},J_b⁺= {j∈J|q_jb≥0}, andJ_b⁻ = {j∈J|qjb<0}. Thus

j∈J

fc^(k)_jaαj

−fd^(k)_jaαj

=

j∈Ja+

fc^(k)_jaαj

−fd^(k)_jaαj

+

j∈Ja⁻

fc⁽_ja^k)α_j−fd⁽_ja^k)α_j,

(2.12)

and we also have a similar expression forbin place ofabut with purely imginary argu- ments. Note that

M k=1

j∈Ja+

fc⁽_ja^k)αj

−fd⁽_ja^k)αj

=

j∈Ja+

M k=1

fw_ja+kq_jaα_j−fw_ja+ (k−1)q_jaα_j

=

j∈Ja+

fwja+Mqja αj

−fwjaαj ,

(2.13)

and similarly, M k=1

j∈Ja⁻

fc⁽_ja^k)αj

−fd⁽_ja^k)αj

=

j∈Ja⁻

fwjaαj

−fwja+Mqjaαj

,

M k=1

j∈Jb+

fic^(k)_jbαj

−fid^(k)_jbαj

=

j∈Jb+

fiwjb+Mqjb

αj

−fiwjbαj

,

M k=1

j∈Jb⁻

fic⁽_jb^k)αj

−fid⁽_jb^k)αj

=

j∈Jb⁻

fiwjbαj

−fiwjb+Mqjbαj .

(2.14) Using these equations, summing over k=1, 2,...,M in (2.10) and dividing by M, the claim (2.7) follows. For eachm, by the denseness ofS⁺, there exist qka,qkb∈N0 (k= 1, 2,...,A;k=m) andq_ma,q_mb∈Nsuch that|(q1a+iq1b)α1+···+ (q_Aa+iq_Ab)α_A|< δ, and so componentwise|q1aα1+···+qAaαA|< δand|iq1bα1+···+iqAbαA|< δ. From (2.2), (2.7) with x=q1aα1+···+qAaαA, y=0, J= {m}, and I= {1, 2,...,A}\{m},

we deduce

t∈I

fqtaαt

+ 1 M

fwma+Mqma

αm

−fwmaαm

< ε, (2.15)

and similarly, withx=iq1bα1+···+iqAbαA, we deduce

t∈I

fiqtbαt + 1

M

fi(wmb+Mqmb αm

−fiwmbαm

< ε. (2.16)

For eachu,v∈N, by the division algorithm, we can writeu=wa+Maqmaandv=wb+ Mbqmb, where 0≤wa< qmaand 0≤wb< qmb. We see that ifu,v→ ∞, thenMaandMb→

∞. This allows us to consider in (2.15) and (2.16) the values ofw_maandw_mbin the same ranges aswaandwb, respectively. LettingM→ ∞in (2.15), we have

t∈I

fq_taα_t+q_maL≤ε, (2.17) whereLis any one of the possible limits of the sequence{f(uαm)/u}.

We now show thatLis unique. Suppose thatL1,L2are any two limits of the sequence.

From (2.17), we get|qma||L1−L2|≤2ε, and soL1=L2. This implies that lim_N→∞f(Nαm)/

N=λmR exists. The existence of lim_N→∞f(iNαm)/iN =λmI is similarly derived from (2.16).

Step 2. For each x∈C, if x = ^A_m=1(x_ma+ix_mb)α_m (x_ma,x_mb∈R), define λ(x)=

Am=1(xmaλmR+ixmbλmI). The representation of anyx∈Cwith respect toα1,α2,...,αA

above is clearly always possible but certainly not unique. We first verify that λ is in- deed well defined. To do so, it suffices to show that if 0= ^A_m=1(xma+ixmb)αm, then

Am=1(x_maλ_mR+ix_mbλ_mI)=0. There are two possible cases.

Case 1. x_mτ=0 for allm∈ {1, 2,...,A}and allτ∈ {a,b}. This case is trivial.

Case 2. There existsx_mτ=0 for somem∈ {1, 2,...,A}andτ∈ {a,b}.

From Dirichlet’s diophantine approximation theorem (see [5, Chapter I]), for each ν∈N, there aret^(ν),kmτ^(ν)∈Zwitht^(ν)>0 such that

t^(ν)xmτ−k^(ν)_mτ<1

ν (m=1, 2,...,A;τ=a,b), (2.18) withkmτ^(ν):=0 ifx_mτ=0. Note that we may chooset^(ν)→ ∞asν→ ∞, so that forx_mτ=0, we must have|kmτ^(ν)| → ∞asν→ ∞. From the representation of 0, we get

A m=1

k_ma^(ν)+ik_mb^(ν)αm

=

A m=1

k_ma^(ν)+ik⁽_mb^ν)αm−t^(ν) A m=1

xma+ixmb

αm

=

A m=1

k_ma^(ν)−t^(ν)xma

+ik^(ν)_mb−t^(ν)xmb

αm

≤2

ν A m=1

αm. (2.19)

Thus

νlim→∞

A m=1

k_ma^(ν)+ik^(ν)_mbαm

=0. (2.20)

From (2.18), using the fact thatxmτ=0, we arrive at

νlim→∞

kmτ^(ν)

k^(ν)_mτ ⁼ x_mτ

xmτ. (2.21)

Clearly, for eachm=1, 2,...,A, we have sgnk^(ν)mτ=sgnxmτwhenνis sufficiently large. Let U_a⁺= {m|xma>0},U_a⁻= {m|xma<0},U_b⁺= {m|xmb>0}, andU_b⁻= {m|xmb<0}. Rewriting (2.20) as

νlim→∞

⎛

⎜⎝

t∈Ua⁺

k_ta^(ν)α_t+

t∈Ub⁺

ik_tb^(ν)α_t

⎞

⎟⎠−

⎛

⎜⎝

s∈Ua⁻

k_sa^(ν)α_s+

s∈U_b⁻

ik_sb^(ν)α_s

⎞

⎟⎠

=0, (2.22)

using uniform continuity, (2.2), and f(0)=0, we get asν→ ∞,

t∈Ua⁺

fk⁽_ta^ν)α_t+

t∈U_b⁺

fik⁽_tb^ν)α_t−

s∈Ua⁻

fk_sa^(ν)α_s+

s∈U_b⁻

fik⁽_sb^ν)α_s−→0, (2.23) and so

t∈Ua⁺

fk_ta^(ν)α_t k^(ν)_mτ ^·

k_ta^(ν) k_ta^(ν)+

t∈Ub⁺

fik_tb^(ν)α_t k^(ν)_mτ ^·

ik_tb^(ν) ik_tb^(ν)

−

s∈Ua⁻

fk⁽sa^ν)α_s k_m^(ν)_τ ^·

ksa^(ν) ksa^(ν)−

s∈U⁻_b

fik⁽_sb^ν)α_s k_m^(ν)_τ ^·

ik⁽_sb^ν) ik⁽_sb^ν) −→0,

(2.24)

which, by (2.21) andStep 1, yields

t∈Ua⁺

λtR xta

xmτ +

t∈Ub⁺

λtI ixtb

xmτ−

s∈Ua⁻

λsR

xsa xmτ −

s∈Ub⁻

λsIixsb xmτ

=0. (2.25)

Thus ^A_m=1(xmaλmR+ixmbλmI)=0, that is,λ(0)=0, which shows thatλis a function.

Thatλis linear overR, and so uniformly continuous onS⁺, is easily checked.

Step 3. Define the functionω:S⁺→Cbyω(x)=f(x)−λ(x).

Thenωis uniformly continuous and satisfies (2.2) onS⁺as f andλare. Moreover, by Step 1and the linearity ofλ, we have

Nlim→∞

ωNαm

N ⁼0= lim

N→∞

ωiNαm

iN . (2.26)

For eachm=1, 2,...,A, defineϕm:Sm(⊇S⁺)→Cby (1)ϕm(0)=0,

(2)ϕ_m(x+α_j)=ϕ_m(x+iα_j)=ϕ_m(x) for allj=m,x∈S_m, (3)ϕm

(uma+iumb)αm

=ω(uma+iumb)αm .

To confirm the shape of solution, note that forx= ^A_m=1(xma+ixmb)αm∈S⁺, we have f(x)=λ(x) +ω(x)=λ(x) +

A m=1

ωx_ma+ix_mbα_m

=λ(x) + A m=1

ϕm

xma+ixmb

αm

=λ(x) + A m=1

ϕm(x).

(2.27)

Finally, to complete the proof, we are left only to show thatϕ_mis uniformly continuous onSm. Fixmand letε >0. Sinceωis uniformly continuous onS⁺, there existsδ^∗>0 such that for allx,y∈S⁺,|x−y|< δ^∗⇒ |ω(x)−ω(y)|< ε. Let

ζ=

u_ma+iu_mbα_m+ A j=1 j=m

k_ja+ik_jbα_j, η=

v_ma+iv_mbα_m+ A j=1 j=m

l_ja+il_jbα_j

(2.28) be elements ofSm with|ζ−η|< δ^∗. For each j=m, rewriteqja+iqjb=(kja+ikjb)− (lja+iljb). Chooseuja,ujb,vja,vjb∈N0so thatqja+iqjb=(uja+iujb)−(vja+ivjb). Let x=(u_ma+iu_mb)α_m+ ^A_j=1,j=m(u_ja+iu_jb)α_j, and y=(v_ma+iv_mb)α_m+ ^A_j=1,j=m(v_ja+ ivjb)αj. Then

|x−y| =

uma+iumb αm−

vma+ivmb αm

+ A j=1 j=m

qja+iqjb αj

=

uma+iumb

αm−

vma+ivmb

αm

+ A j=1 j=m

kja+ikjb

− lja+iljb

αj

= |ζ−η|< δ^∗.

(2.29) Applying (2.7) with f =ω,I= {m},J= {1, 2,...,A}\{m},w_ja=w_jb=0, andq_jτ=u_jτ− vjτ(τ=a,b), we get

ωu_ma+iu_mbα_m−ωv_ma+iv_mbα_m + 1

M

j∈J

ηja

ωMqjaαj

−0+ 1 M

j∈J

ηjb

ωiMqjbαj

−0 < ε.

(2.30)

Since the sequencesω(M|qja|αj)/Mandω(iM|qjb|αj)/Mare subsequences ofω(Nαj)/N andω(iNαj)/N, respectively, by (2.26), we have

Mlim→∞

ωMq_jaα_j

M ⁼0= lim

M→∞

ωiMq_jbα_j

M (j∈J). (2.31)

LettingM→ ∞in (2.30) and using (2.31), we get ωuma+iumb

αm

−ωvma+ivmb

αm≤ε. (2.32)

As ϕ_m(ζ)=ϕ_m((u_ma+iu_mb)α_m)=ω((u_ma+iu_mb)α_m) and similarly ϕ_m(η)=ω((v_ma+ ivmb)αm), we deduce that|ϕm(ζ)−ϕm(η)| ≤ε, that is,ϕm is uniformly continuous on

S_m, which completes the proof.

Corollary 2.2. LetA∈N,A≥2, andα1,α2,...,αA∈Rbe such that (1)S⁺= { ^A_m=1(u_m+iv_m)α_m|u_m,v_m∈N0}is dense inC,

(2)Tm= { ^A_j=1,j=m(uj+ivj)αj|uj,vj∈Z}(m=1,...,A) is dense inSm.

If f :S⁺→Cis a uniformly continuous solution of the CFE (2.2), then f is anR-linear function; in particular f(z)=az+bz, where¯ a,b are arbitrary complex constants and ¯z denotes the complex conjugate ofz.

Proof. The first part will follow fromTheorem 2.1if we show that eachϕmvanishes iden- tically. This is immediate from the facts that elements of S_m can be approximated ar- bitrarily closely by elements ofTm, whileϕm is continuous onSm and vanishes identi- cally onTm. The second part follows from the linearity ofλ, viz, ifz=x+iy∈C, then

f(z)=λ(z)=xλ(1) +yλ(i).

Corollary 2.2may be regarded as an extension of a special case of the classical complex CFE: f(z1+z2)= f(z1) +f(z2), whose general solution is of the form f(z)=az+bz; see¯ [1, Proposition 2 in Chapter 5].

The next two propositions provide examples forTheorem 2.1andCorollary 2.2. For convenience, we make use of the following notation:Z[i] denotes the ring of Gaussian integers, whileN0[i] denotes the subset of Gaussian integers both of whose real and imag- inary parts are nonnegative.

Proposition 2.3. LetA∈N,A≥2, andα1,α2,...,α_A∈R. If there are two rationally in- dependentα_jswhose ratio is a negative real number, thenS⁺:=α1N0[i] +α2N0[i] +···+ αAN0[i] is dense inC.

Proof. Suppose thatαmandαnare rationally independent andαm/αn<0. Then both are nonzero and at least one of them is irrational. Consider here the case whereαm is irra- tional>0 andα_n<0; other cases can be similarly handled. Letr1+ir2∈Cand>0. By Kronecker’s theorem, see [5, Chapter II], there are infinitely many natural numbersu1,v1

and integersu2,v2such that|u1(αm/|αn|)−u2−r1/|αn||< ε/2|αn|,|v1(αm/|αn|)−v2− r2/|α_n||< ε/2|α_n|. We can choose these integers so thatu1α_m−r1>0,v1α_m−r2>0, and sou2,v2are also positive. Thus|(u1+iv1)αm+ (u2+iv2)αn−(r1+ir2)| ≤ |u1αm+u2αn− r1|+|v1αm+v2αn−r2|< ε, yielding the denseness ofS⁺inC.