Stability of Cubic Functional Equation in the Spaces of Generalized Functions

Volume 2007, Article ID 79893,13pages doi:10.1155/2007/79893

Research Article

Stability of Cubic Functional Equation in

the Spaces of Generalized Functions

Young-Su Lee and Soon-Yeong Chung

Received 24 April 2007; Accepted 13 September 2007

Recommended by H. Bevan Thompson

In this paper, we reformulate and prove the Hyers-Ulam-Rassias stability theorem of the cubic functional equation f(ax+y) +f(ax−y)=a f(x+y) +a f(x−y) + 2a(a2₋ 1)f(x) for fixed integerawitha=0,±1 in the spaces of Schwartz tempered distributions and Fourier hyperfunctions.

Copyright © 2007 Y.-S. Lee and S.-Y. Chung. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, dis-tribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

In 1940, Ulam [1] raised a question concerning the stability of group homomorphisms: “Let f be a mapping from a groupG1to a metric groupG2with metricd(·,·)such that

df(xy),f(x)f(y)≤ε. (1.1)

Then does there exist a group homomorphismL:G1→G2andδ>0such that

df(x),L(x)≤δ (1.2)

for allx∈G1?”

Let bothE1 andE2 be real vector spaces. Jun and Kim [13] proved that a function

f :E1→E2satisfies the functional equation

f(2x+y) +f(2x−y)=2f(x+y) + 2f(x−y) + 12f(x) (1.3)

if and only if there exists a mappingB:E1×E1×E1→E2such that f(x)=B(x,x,x) for allx∈E1, whereBis symmetric for each fixed one variable and additive for each fixed two variables. The mappingBis given by

B(x,y,z)= 1

24

f(x+y+z) +f(x−y−z)−f(x+y−z)−f(x−y+z) (1.4)

for allx,y,z∈E1. It is natural that (1.3) is called a cubic functional equation because the mapping f(x)=ax3_{satisfies (1.3). Also Jun et al. generalized cubic functional equation,} which is equivalent to (1.3),

f(ax+y) +f(ax−y)=a f(x+y) +a f(x−y) + 2aa2_{−1f(x) (1.5)}

for fixed integerawitha=0,±1 (see [14]).

In this paper, we consider the general solution of (1.5) and prove the stability theorem of this equation in the space᏿(Rn_{) of Schwartz tempered distributions and the space}

Ᏺ_(Rn_{) of Fourier hyperfunctions. Following the notations as in [15,16] we reformulate}

(1.5) and related inequality as

u◦A1+u◦A2=au◦B1+au◦B2+ 2a

a2_{−1u◦P, (1.6) u◦A1+u◦A2−au◦B1−au◦B2−2a}

a2_{−1u◦P≤|x|}p_+|y|q_{, (1.7)}

respectively, whereA1,A2,B1,B2, andPare the functions defined by

A1(x,y)=ax+y, A2(x,y)=ax−y,

B1(x,y)=x+y, B2(x,y)=x−y, P(x,y)=x, (1.8)

and p, qare nonnegative real numbers with p,q=3. We note thatpneed not be equal toq. Hereu◦A1, u◦A2, u◦B1, u◦B2, andu◦P are the pullbacks ofuin᏿(Rn) or Ᏺ_(Rn_{) byA1,A2,B1,B2, andP, respectively. Also| · |denotes the Euclidean norm, and}

the inequalityv ≤ψ(x,y) in (1.7) means that| v,ϕ| ≤ ψϕL1 for all test functions

ϕ(x,y) defined onR2n_.

If p <0 orq <0, the right-hand side of (1.7) does not define a distribution and so inequality (1.7) makes no sense. If p,q=3, it is not guaranteed whether Hyers-Ulam-Rassias stability of (1.5) is hold even in classical case (see [13,14]). Thus we consider only the case 0≤p,q <3, orp,q >3.

We prove as results that every solutionuin᏿(Rn_{) orᏲ(R}n_{) of inequality (1.7) can}

be written uniquely in the form

u=

1≤i≤j≤k≤n

whereh(x) is a measurable function such that

_h(x)≤

2|a|3_{− |a|}p|x|p. (1.10)

2. Preliminaries

We first introduce briefly spaces of some generalized functions such as Schwartz tempered distributions and Fourier hyperfunctions. Here we use the multi-index notations,|α| = α1+···+αn,α!=α1!···αn!,xα=x1α1···xnαn, and∂α=∂α11···∂αnn forx=(x1,...,xn)∈

Rn_,α=(α

1,...,αn)∈N0n, whereN0is the set of nonnegative integers and∂j=∂/∂xj.

Definition 2.1[17,18]. Denote by᏿(Rn_{) the Schwartz space of all infinitely differentiable}

functionsϕinRn_satisfying

ϕα,β=sup x∈Rn

_xα_∂β_{ϕ(x)<∞ (2.1)}

for allα,β∈Nn

0, equipped with the topology defined by the seminorms · α,β. A linear

formuon᏿(Rn_{) is said to beSchwartz tempered distributionif there is a constantC≥0}

and a nonnegative integerNsuch that _u,ϕ≤C

|α|,|β|≤N

sup

x∈Rn

_xα_∂β_{ϕ (2.2)}

for allϕ∈᏿(Rn_{). The set of all Schwartz tempered distributions is denoted by᏿(R}n_).

Imposing growth conditions on · α,βin (2.1), Sato and Kawai introduced the space

Ᏺof test functions for the Fourier hyperfunctions.

Definition 2.2[19]. Denote byᏲ(Rn_{) the Sato space of all infinitely differentiable}

func-tionsϕinRn_{such that}

ϕA,B=sup x,α,β

_xα_∂β_ϕ(x)

A|α|_B|β|_α!β! <∞ (2.3)

for some positive constantsA,Bdepending only onϕ. We say thatϕj→0 as j→ ∞if

ϕjA,B→0 as j→ ∞for someA,B >0, and denote byᏲ(Rn) the strong dual ofᏲ(Rn)

and call its elementsFourier hyperfunctions.

It can be verified that the seminorms (2.3) are equivalent to

ϕh,k= sup x∈Rn_,α∈Nn

0

_∂α_ϕ(x)expk|x|

h|α|_α! <∞ (2.4)

for some constantsh,k >0. It is easy to see the following topological inclusion:

In order to solve (1.6), we employ then-dimensional heat kernel, that is, the fundamental solutionEt(x) of the heat operator∂t− xinRnx×R+t given by

Et(x)=

⎧ ⎪ ⎨ ⎪ ⎩

(4πt)−n/2_exp

−|x|2

4t

, t >0,

0, t≤0.

(2.6)

Since for eacht >0,Et(·) belongs to᏿(Rn), the convolution

u(x,t)=u∗Et

(x)=uy,Et(x−y)

, x∈Rn_{,t >0, (2.7)}

is well defined for eachu∈᏿_(Rn_{) andu∈Ᏺ(R}n_{), which is called theGauss transform}

ofu. Also we use the following result which is called theheat kernel method(see [20]). Letu∈᏿_(Rn_{). Then its Gauss transformu(x,t) is aC}∞_{-solution of the heat equation}

∂ ∂t−Δ

u(x,t)=0 (2.8)

satisfying the following.

(i) There exist positive constantsC,M, andNsuch that

u(x,t)≤Ct−M_{1 +|x|}N _inRn_{×(0,δ). (2.9)}

(ii)u(x,t)→uast→0+_{in the sense that for everyϕ∈᏿(R}n_),

u,ϕ =lim

t→0+

u(x,t)ϕ(x)dx. (2.10)

Conversely, everyC∞-solutionU(x,t) of the heat equation satisfying the growth condi-tion (2.9) can be uniquely expressed asU(x,t)=u(x,t) for someu∈᏿_(Rn_).

Similarly, we can represent Fourier hyperfunctions as initial values of solutions of the heat equation as a special case of the results (see [21]). In this case, the estimate (2.9) is replaced by the following.

For everyε >0 there exists a positive constantCεsuch that

u(x,t)≤Cεexp

ε

|x|+1

t

inRn_{×(0,δ). (2.11)}

We refer to [17, Chapter VI] for pullbacks and to [16,18,20] for more details of᏿(Rn₎

andᏲ(Rn_).

3. General solution in᏿(Rn_)andᏲ(Rn₎

Jun and Kim (see [22]) showed that every continuous solution of (1.5) inRis a cubic function f(x)=f(1)x3_{for allx∈R. Using induction argument on the dimensionn, it} is easy to see that every continuous solution of (1.5) inRn_{is a cubic form}

f(x)=

1≤i≤j≤k≤n

In this section, we consider the general solution of the cubic functional equation in the spaces of᏿(Rn_{) andᏲ(R}n_{). It is well known that thesemigroup propertyof the heat}

kernel

Et∗Es

(x)=Et+s(x) (3.2)

holds for convolution. Semigroup property will be useful to convert (1.6) into the classical functional equation defined on upper-half plane.

Convolving the tensor productEt(ξ)Es(η) ofn-dimensional heat kernels in both sides

of (1.6), we have

u◦A1

∗Et(ξ)Es(η)

(x,y)

=u◦A1,Et(x−ξ)Es(y−η)=

uξ,a−n

Et

x−ξ−η a

Es(y−η)dη

=uξ,a−n

Et

_{ax+y−ξ−η}

a

Es(η)dη

=uξ,

Ea2_t(ax+y−ξ−η)E_s(η)dη

=uξ,Ea2_t∗E_s(ax+y−ξ)=u_ξ,E_a2_t+s(ax+y−ξ)=uax+y,a2t+s,

(3.3)

and similarly we get

u◦A2

∗Et(ξ)Es(η)

(x,y)=uax−y,a2_t+s,

u◦B1

∗Et(ξ)Es(η)

(x,y)=u(x+y,t+s),

u◦B2

∗Et(ξ)Es(η)

(x,y)=u(x−y,t+s),

u◦P∗Et(ξ)Es(η)

(x,y)=u(x,t).

(3.4)

Thus (1.6) is converted into the classical functional equation

uax+y,a2t+s+uax−y,a2t+s

=au(x+y,t+s) +au(x−y,t+s) + 2aa2−1u(x,t) (3.5)

for allx,y∈Rn_{,t,s >0.}

Lemma3.1. Let f :Rn_{×(0,∞)→Cbe a continuous function satisfying}

fax+y,a2_{t+s+fax−y,a}2_t+s

=a f(x+y,t+s) +a f(x−y,t+s) + 2aa2−1f(x,t) (3.6)

for fixed integerawitha=0,±1. Then the solution is of the form

f(x,t)=

1≤i≤j≤k≤n

ai jkxixjxk+t

1≤i≤n

Proof. In view of (3.6) and given the continuity, f(x, 0+_{) :=lim}

t→0+f(x,t) exists. Define

h(x,t) :=f(x,t)−f(x, 0+_{), thenh(x, 0}+_{)=0 and}

hax+y,a2t+s+hax−y,a2t+s

=ah(x+y,t+s) +ah(x−y,t+s) + 2aa2−1h(x,t) (3.8)

for allx,y∈Rn_{,t,s >0. Settingy=0, s→0}+_{in (3.8), we have}

hax,a2_t=a3_{h(x,t). (3.9)}

Puttingy=0,s=a2_{sin (3.8), and using (3.9), we get}

a2h(x,t+s)=hx,t+a2s+a2−1h(x,t). (3.10)

Lettingt→0+_{in (3.10), we obtain}

a2_h(x,s)=hx,a2_{s. (3.11)}

Replacingtbya2_{tin (3.10) and using (3.11), we have}

hx,a2t+s=h(x,t+s) +a2−1h(x,t). (3.12)

Switchingtwithsin (3.12), we get

hx,t+a2_{s=h(x,t+s) +a}2_{−1h(x,s). (3.13)}

Adding (3.10) to (3.13), we obtain

h(x,t+s)=h(x,t) +h(x,s), (3.14)

which shows that

h(x,t)=h(x, 1)t. (3.15)

Lettingt→0+_{,s=1 in (3.8), we have}

h(ax+y, 1) +h(ax−y, 1)=ah(x+y, 1) +ah(x−y, 1). (3.16)

Also lettingt=1,s→0+_{in (3.8), and using (3.11), we get}

a2h(ax+y, 1) +a2h(ax−y, 1)=ah(x+y, 1) +ah(x−y, 1) + 2aa2−1h(x, 1). (3.17)

Now taking (3.16) into (3.17), we obtain

Replacingx, yby (x+y)/2, y=(x−y)/2 in (3.18), respectively, we see thath(x, 1) satis-fies Jensen functional equation

2h

_x+y

2 , 1

=h(x, 1) +h(y, 1). (3.19)

Puttingx=y=0 in (3.16), we geth(0, 1)=0. This shows thath(x, 1) is additive. On the other hand, lettingt=s→0+_{in (3.6), we can see that f(x, 0}+_{) satisfies (1.5).} Given the continuity, the solution f(x,t) is of the form

f(x,t)=

1≤i≤j≤k≤n

ai jkxixjxk+t

1≤i≤n

bixi, ai jk,bi∈C, (3.20)

which completes the proof.

As a direct consequence of the above lemma, we present the general solution of the cubic functional equation in the spaces of᏿(Rn_{) andᏲ(R}n_).

Theorem3.2. Suppose thatuin᏿(Rn_)orᏲ(Rn_{)satisfies the equation}

u◦A1+u◦A2=au◦B1+au◦B2+ 2aa2−1u◦P (3.21)

for fixed integerawitha=0,±1. Then the solution is the cubic form

u=

1≤i≤j≤k≤n

ai jkxixjxk, ai jk∈C. (3.22)

Proof. Convolving the tensor productEt(ξ)Es(η) ofn-dimensional heat kernels in both

sides of (3.21), we have the classical functional equation

uax+y,a2_{t+s+uax−y,a}2_t+s

=au(x+y,t+s) +au(x−y,t+s) + 2aa2_−1u(x,t) (3.23)

for allx,y∈Rn_{,t,s >0, whereuis the Gauss transform ofu. ByLemma 3.1, the solution}

uis of the form

u(x,t)=

1≤i≤j≤k≤n

ai jkxixjxk+t

1≤i≤n

bixi, ai jk,bi∈C. (3.24)

Thus we get

u,ϕ =

1≤i≤j≤k≤n

ai jkxixjxk+t

1≤i≤n

bixi,ϕ

(3.25)

for all test functionsϕ. Now lettingt→0+_{, it follows from the heat kernel method that}

u,ϕ =

1≤i≤j≤k≤n

ai jkxixjxk,ϕ

(3.26)

4. Stability in᏿(Rn_)andᏲ(Rn₎

We are going to prove the stability theorem of the cubic functional equation in the spaces of᏿(Rn_{) andᏲ(R}n_).

We note that the Gauss transform

ψp(x,t) :=

|ξ|p_E

t(x−ξ)dξ (4.1)

is well defined andψp(x,t)→ |x|plocally uniformly ast→0+. Alsoψp(x,t) satisfies

semi-homogeneous property

ψp

rx,r2t=rpψp(x,t) (4.2)

for allr≥0.

We are now in a position to state and prove the main result of this paper.

Theorem4.1. Letabe fixed integer witha=0,±1and let, p, qbe real numbers such that≥0and0≤p, q <3, or p,q >3. Suppose thatuin᏿(Rn_)orᏲ(Rn_{)satisfy the}

inequality

_{u◦A1−u◦A2−au◦B1−au◦B2−2a}

a2−1u◦P≤|x|p_+|y|q_{. (4.3)}

Then there exists a unique cubic form

c(x)=

1≤i≤j≤k≤n

ai jkxixjxk (4.4)

such that

_u−c(x)≤

2|a|3_{− |a|}p|x|p. (4.5)

Proof. Letv:=u◦A1−u◦A2−au◦B1−au◦B2−2a(a2−1)u◦P. Convolving the ten-sor productEt(ξ)Es(η) ofn-dimensional heat kernels inv, we have

_v∗

Et(ξ)Es(η)(x,y)=v,Et(x−ξ)Es(y−η)

≤|ξ|p_+|η|q_E

t(x−ξ)Es(y−η)L1

=ψp(x,t) +ψq(y,s)

.

(4.6)

Also we see that, as inTheorem 3.2,

v∗Et(ξ)Es(η)

(x,y)=uax+y,a2_{t+s+uax−y,a}2_t+s

whereuis the Gauss transform ofu. Thus inequality (4.3) is converted into the classical functional inequality

uax+y,a2_{t+s+uax−y,a}2_{t+s−au(x+y,t+s)−au(x−y,t+s)−2aa}2_−1u(x,t)

≤ψp(x,t) +ψq(y,s)

(4.8)

for allx,y∈Rn_{,t,s >0.}

We first prove for 0≤p,q <3. Lettingy=0,s→0+_{in (4.8) and dividing the result by} 2|a|3_{, we get}

uax_a,3a2t−u(x,t) ≤

2|a|3ψp(x,t). (4.9)

By virtue of the semihomogeneous property ofψp, substitutingx, tbyax,a2t,

respec-tively, in (4.9) and dividing the result by|a|3_{, we obtain}

ua2_ax6,a4t−

uax,a2_t

a3 ≤

2|a|3|a|p− 3

ψp(x,t). (4.10)

Using induction argument and triangle inequality, we have

uan_ax3,na2nt−u(x,t)

≤

2|a|3ψp(x,t)

n−1

j=0

|a|(p−3)j _(4.11)

for alln∈N, x∈Rn_{, t >0. Let us prove the sequence{a}−3n_u(an_x,a2n_{t)}is convergent}

for allx∈Rn_{,t >0. Replacingx,tbya}m_x,a2m_{t, respectively, in (4.11) and dividing the}

result by|a|3m,_{we see that}

uam+_an3(xm,a+2(n)m+n)t−

uam_x,a2m_t

a3m

≤

2|a|3ψp(x,t)

n−1

j=m

|a|(p−3)j_{. (4.12)}

Lettingm→ ∞, we have{a−3n_u(an_x,a2n_{t)}is a Cauchy sequence. Therefore, we may}

de-fine

G(x,t)=lim

n→∞a

−3n_uan_x,a2n_{t (4.13)}

for allx∈Rn_{,t >0.}

Now we verify that the given mappingGsatisfies (3.6). Replacingx, y,t,sbyan_x,an_y,

a2n_t,a2n_{sin (4.8), respectively, and then dividing the result by|a|}3n_{, we get}

|a|−3n_uan_(ax+y),a2n_a2_t+s+uan_(ax−y),a2n_a2_t+s

−auan(x+y),a2n(t+s)−auan(x+y),a2n(t+s)−2aa2−1uanx,a2nt ≤ |a|−3n_ψ

p

anx,a2nt+ψq

any,a2ns =|a|(p−3)n_ψ

p(x,t) +|a|(q−3)nψq(y,s)

.

Now lettingn→ ∞, we see by definition ofGthatGsatisfies

Gax+y,a2_{t+s+Gax−y,a}2_t+s

=aG(x+y,t+s) +aG(x−y,t+s) + 2aa2_−1G(x,t) (4.15)

for allx,y∈Rn_{,t,s >0. ByLemma 3.1,G(x,t) is of the form}

G(x,t)=

1≤i≤j≤k≤n

ai jkxixjxk+t

1≤i≤n

bixi, ai jk,bi∈C. (4.16)

Lettingn→ ∞in (4.11) yields

_{G(x,t)−u(x,t)≤}

2|a|3_{− |a|}pψp(x,t). (4.17)

To prove the uniqueness ofG(x,t), we assume thatH(x,t) is another function satisfying (4.15) and (4.17). Settingy=0 ands→0+_{in (4.15), we have}

Gax,a2_t=a3_{G(x,t). (4.18)}

Then it follows from (4.15), (4.17), and (4.18) that _{G(x,t)−H(x,t)}

= |a|−3n_Gan_x,a2n_t−Han_x,a2n_{t≤ |a|}−3n_Gan_x,a2n_t−uan_x,a2n_t

+|a|−3nuan_x,a2n_t−Han_x,a2n_t≤

|a|3n_|a|3_{− |a|}pψp(x,t)

(4.19)

for alln∈N,x∈Rn_{,t >0. Lettingn→ ∞, we haveG(x,t)=H(x,t) for allx∈R}n_{,t >0.}

This proves the uniqueness.

It follows from the inequality (4.17) that _{G(x,t)−u(x,t),ϕ≤}

2|a|3_{− |a|}p

ψp(x,t),ϕ

(4.20)

for all test functionsϕ. Lettingt→0+_{, we have the inequality}

u−

1≤i≤j≤k≤n

ai jkxixjxk

≤_2|a|3_{− |a|}p. (4.21)

Now we consider the case p,q >3. For this case, replacingx, y,tbyx/a, 0, t/a2 _in (4.8), respectively, and lettings→0+_{and then multiplying the result by|a|}3_{, we have}

u(x,t)−a3_u x a, t a2 ≤

2|a|3|a| 3−p_ψ

p(x,t). (4.22)

Substitutingx,tbyx/a,t/a2_{, respectively, in (4.22) and multiplying the result by|a|}3_we get

a3u

x a, t a2

−a6u

x a2,

t a4

≤

2|a|3|a| 2(3−p)

Using induction argument and triangle inequality, we obtain

u(x,t)−a3n_u

x an,

t a2n

≤

2|a|3ψp(x,t)

n

j=1

|a|(3−p)j _(4.24)

for alln∈N,x∈Rn_{,t >0. Following the same method as in the case 0≤p,q <3, we see}

that

G(x,t) :=lim

n→∞a 3n_ux

an,

t a2n

(4.25)

is the unique function satisfying (4.15). Lettingn→ ∞in (4.24), we get

u(x,t)−C(x,t)≤

2|a|p_{− |a|}3ψp(x,t). (4.26)

Now lettingt→0+_{in (4.26), we have the inequality}

u−

1≤i≤j≤k≤n

ai jkxixjxk

≤2|a|p_{− |a|}3. (4.27)

This completes the proof.

Remark 4.2. The above norm inequality

_u−c(x)≤

2|a|p_{− |a|}3|x|p (4.28)

implies thatu−c(x) is a measurable function. Thus all the solution u in᏿(Rn_{) orᏲ(R}n₎

can be written uniquely in the form

u=c(x) +h(x), (4.29)

where|h(x)| ≤(/(2||a|p_{− |a|}3_|))|x|p_.

Corollary4.3. Letabe fixed integer witha=0,±1and≥0. Suppose thatuin᏿(Rn₎

orᏲ(Rn_{)satisfy the inequality}

_{u◦A1−u◦A2−au◦B1−au◦B2−2a}

a2−1u◦P≤. (4.30)

Then there exists a unique cubic form

c(x)=

1≤i≤j≤k≤n

ai jkxixjxk (4.31)

such that

_u−c(x)≤

Acknowledgment

This work was supported by Korea Research Foundation Grant (KRF-2003-041-C00023).

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Young-Su Lee: Department of Mathematics, Sogang University, Seoul 121-741, South Korea

Email address:[email protected]

Soon-Yeong Chung: Department of Mathematics and Program of Integrated Biotechnology, Sogang University, Seoul 121-741, South Korea