Stability of a 2 Dimensional Functional Equation in a Class of Vector Variable Functions

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Volume 2010, Article ID 167042,12pages doi:10.1155/2010/167042

Research Article

Stability of a 2-Dimensional Functional Equation in

a Class of Vector Variable Functions

Won-Gil Park

1

_{and Jae-Hyeong Bae}

2

1_{Department of Mathematics Education, College of Education, Mokwon University,}

Daejeon 302-729, Republic of Korea

2_{College of Liberal Arts, Kyung Hee University, Yongin 449-701, Republic of Korea}

Correspondence should be addressed to Jae-Hyeong Bae,[email protected]

Received 16 March 2010; Accepted 24 June 2010

Academic Editor: Patricia J. Y. Wong

Copyrightq2010 W.-G. Park and J.-H. Bae. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We prove the Hyers-Ulam stability of a 2-dimensional quadratic functional equation in a class of vector variable functions in Banach modules over a unitalC_-algebra.

1. Introduction

In 1940, Ulam proposed the stability problemsee1:

LetG1be a group and letG2be a metric group with the metricd·,·. Givenε >0, does there exist aδ >0such that if a mappingh:G1 → G2 satisfies the inequalitydhxy, hxhy < δ for all x, y ∈ G1 then there is a homomorphism H : G1 → G2 with dhx, Hx < ε for allx∈G1?

In 1941, this problem was solved by Hyers2in the case of Banach space. Thereafter, we call that type the Hyers-Ulam stability. The authors investigated various functional equations and their Hyers-Ulam stability3–8. This Hyers-Ulam stability is a classical type of stability, but there is another kind of stability introduced by Risteski 9 for functional equations spanned over ann-dimensional complex vector space too.

LetXandY be real or complex vector spaces. For a mappingg:X → Y, consider the quadratic functional equation

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In 1989, Acz´el and Dhombres10obtained the solution of1.1for the case thatY acts on

X. The result also holds when X and Y are arbitrary real or complex vector spaces. For a mappingf:X×X → Y, consider the 2-dimensional quadratic functional equation:

fxy, z−wfx−y, zw2fx, z 2fy, w. 1.2

The quadratic formf:R×R → Rgiven byfx, y:ax2_by2_{is a solution of}_1.2_{. In 2008,}

the authors of8acquired the general solution and proved the stability of the 2-dimensional quadratic functional equation1.2for the case thatXandY are real vector spaces as follows. The results of8, Theorems 3 and 4also hold for complex vector spacesX andY. In this paper, we investigate the stability of1.2with two module actions in Banach modules over a unitalC_-algebra.

2. Preliminaries

LetAbe a unitalC-algebra with a norm| · |, and let AMandANbe left BanachA-modules

with norms · and · , respectively. Put A1 : {a ∈ A | |a| 1}, Ain : {a ∈ A |

ais invertible inA}, Asa : {a ∈ A | a a}, UA : {a ∈ A | aa aa 1},

A:{a∈Asa|Spa⊂0,∞},andA₁ :A1∩A.

Definition 2.1. A 2-dimensional vector variable quadratic mappingF : _AM ×AM → AN

satisfying1.2is calledA-quadraticifFax, ay a2_F_{x, y}_{for all}_a_∈_A_{and all}_{x, y}_∈ AM.

Definition 2.2. A unital C_-algebra _A_{is said to have} _{real rank} ₀ _see₁₁_{if the invertible}

self-adjoint elements are dense inAsa.

For any elementa∈A,a a1ia2, wherea1 : aa/2 anda2 : a−a/2iare

self-adjoint elements; furthermore,aa₁−a−₁ia₂−ia−₂, wherea₁, a−₁, a₂anda−₂are positive elementssee12, Lemma 38.8.

3. Results

Theorem 3.1. Letψ:R×R → Rbe a function satisfying

ψst, u−v ψs−t, uv 2ψs, u 2ψt, v 3.1

for alls, t, u, v∈R. If the functionψis a Borel function, then it also satisfies

ψs, t s2ψ1,0 t2ψ0,1 3.2

for alls, t∈R.

Proof . By8, Theorem 3, there exist two symmetric biadditive mappingsS, T :R×R → R such thatψs, t Ss, s Tt, tfor alls, t∈R. By the proof of Theorem 3 in8, we gain

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for allp, q∈Qand allu, v∈R. Lettingpv1 in the equality3.3, we get

ψu, qψu,0 q2ψ0,1 3.4

for allu∈Rand allq∈Q. Puttinguv1 in the equality3.3again, we have

ψp, qp2ψ1,0 q2ψ0,1 3.5

for allp, q∈Q. Since the functionv → ψu, vis measurable and satisfies1.1, by13, it is continuous. By the same reasoning,u → ψu, vis also continuous. Lets, t∈Rbe fixed. Since

ψis measurable, by14, Theorem 7.14.26, for everym∈Nthere is a closed setFm⊂s, s1

such thatμs, s1\Fm<1/mandψ|Fm×Ris continuous. SinceμFm → 1, one can choose

um∈Fmsatisfyingum → s. Take a sequence{qn}inQconverging tot. By the equality3.4,

we get

ψum, t ψ

um,_nlim_{→ ∞}qn

lim

n→ ∞ψ

um, qn

lim

n→ ∞

ψum,0 qn2ψ0,1

ψum,0 t2ψ0,1

3.6

for allm∈N. For each fixedm∈N, take a sequence{pn}inQconverging toum. By3.5and

the above equality, we have

ψum, t ψ

lim

n→ ∞pn,0

t2ψ0,1 lim

n→ ∞ψ

pn,0

t2ψ0,1

lim

n→ ∞p 2

nψ1,0 t2ψ0,1 um2ψ1,0 t2ψ0,1.

3.7

Hence we see that

ψs, t ψ

lim

m→ ∞um, t

lim

m→ ∞ψum, t mlim→ ∞

u2_mψ1,0 t2ψ0,1

s2ψ1,0 t2ψ0,1,

3.8

as desired.

Lemma 3.2. LetX andY be normed spaces andr /2a real number, and letf :X×X → Y be a mapping such that

_f

xy, z−wfx−y, zw−2fx, z−2fy, w≤xryrzrwr

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for allx, y, z, w ∈ X. Supposef0,0 0 forr > 2. IfY is complete, then there exists a unique

2-variable quadratic mappingF :X×X → Ysuch that

fx, y−Fx, y≤

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

1 2−2r−1

2xr 3yr1

3f0,0 r <2, 21−r

1−22−r

2xr 3yr r >2

3.10

for allx, y∈X. The mappingFis given by

Fx, y:

⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

lim

j→ ∞

1 4jf

2j_x,₂j_y _{r <}₂_,

lim

m→ ∞4 j_f

x

2j,

y

2j

r >2

3.11

for allx, y∈X.

Proof. Lettingyxandw−zin3.9, we gain

fx, z fx,−z−1

2

f0,0 f2x,2z≤ xrzr 3.12

for allx, z∈X. Puttingx0 in3.12, we get

f0, z f0,−z−1

2

f0,0 f0,2z≤ zr 3.13

for allz∈X. Replacingzby−zin the above inequality, we have

f0,−z f0, z−1

2

f0,0 f0,−2z≤ zr 3.14

for allz∈X. By the above two inequalities, we see that

f0,2z−f0,−2z≤4zr 3.15

for allz∈X. Settingyxandwzin3.9, we obtain that

f2x,0 f0,2z−4fx, z≤2xrzr 3.16

for allx, z∈X. Replacingzby−zin the above inequality, we see that

_f₂_x,₀ _f₀_,₋₂_z₋₄_f_x,₋_z_≤₂

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for allx, z∈X. By the last two inequalities, we know that

fx, z−fx,−z−1

4

f0,2z−f0,−2z≤ xrzr 3.18

for allx, z∈X. By3.12and3.18, we obtain that

fx, z−1

8

f0,2z−f0,−2z−1

4

f0,0 f2x,2z≤ xrzr 3.19

for allx, z∈X. By3.15and the above inequality, we have

fx, z−1

4

f0,0 f2x,2z≤ xr3

2z

r _3.20

for allx, z∈X. Thus we obtain that

₄1jf

2jx,2jz− 1

4j1

f0,0 f2j1x,2j1z≤2jr−2

xr 3

2z

r _3.21

for allx, z∈Xand allj. Replacingzbyyin the above inequality, we see that

₄1jf

2jx,2jy− 1

4j1

f0,0 f2j1x,2j1y≤2jr−2xr3

2y

r

3.22

for allx, y∈Xand allj. For given integersl, m0≤l < m, we obtain that

₄1mf

2mx,2my− 1

4lf

2lx,2ly1

3

1 4l −

1 4m

f0,0≤ 2

lr−2₋₂mr−2

1−2r−2

xr3

2y

r

3.23

for allx, y∈X.

Consider the case r < 2. By 3.23, the sequence {1/4j_f₂j_x,₂j_y_} _{is a Cauchy}

sequence for allx, y ∈ X. SinceY is complete, the sequence{1/4j_f₂j_x,₂j_y_} _converges

for allx, y∈X. DefineF:X×X → YbyFx, y:limj→ ∞1/4jf2jx,2jy for allx, y∈X.

By3.9, we have

₄1jf

2jxy,2jz−w 1

4jf

2jx−y,2jzw

−2

4jf

2jx,2jz− 2

4jf

2jy,2jw≤2r−2jxryrzrwr

3.24

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are four symmetric biadditive mappingsS, T, U, V :X×X → Y such thatFx, y Sx, x Ty, yandGx, y Ux, x Vy, yfor allx, y∈X. Thus we obtain that

Fx, y−Gx, ySx, x Ty, y−Ux, x−Vy, y

1

4nS2

n_x,₂n_x _T₂n_y,₂n_y₋_U₂n_x_,₂n_x₋_V₂n_y,₂n_y

1

4nF

2nx,2ny−G2nx,2ny

≤ 1

4nF

2n_x,₂n_y₋_f₂n_x,₂n_y 1

4nf

2n_x,₂n_y₋_G₂n_x,₂n_y

≤ 2nr−2

1−2r−2

2xr3yr21−2n

3 f0,0

−→0, asn−→ ∞

3.25

for allx, y ∈X. Hence the mappingFis the unique 2-dimensional vector variable quadratic mapping, as desired.

Next, consider the caser >2. Sincef0,0 0, by inequality3.20, we gain

4fx

2,

z

2

−fx, z≤ 1

2r−1

2xr3zr 3.26

for allx, z∈X. Thus we get

4j1_f

x

2j1,

z

2j1

−4j_f

x

2j,

z

2j

≤2j2−r1−r₂_xr₃_zr _3.27

for allx, z∈Xand allj. Replacingzbyyin the above inequality, we have

4j1f

x

2j1,

y

2j1

−4jf

x

2j,

y

2j

≤2j2−r1−r2xr3yr 3.28

for allx, y∈Xand allj. For given integersl, m0≤l < m, we obtain that

4mfx

2m,

y

2m

−4lf

x

2l,

y

2l

≤ 22−r−22−rm1

2−23−r

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for allx, y ∈X. By3.29, the sequence{4j_f_x/₂j_{, y/}₂j_}_{is a Cauchy sequence for all}_{x, y}_∈

X. Since Y is complete, the sequence {4j_f_x/₂j_{, y/}₂j_} _{converges for all}_{x, y} _∈ _X_{. Define}

F:X×X → Y byFx, y:limj→ ∞4jfx/2j, y/2jfor allx, y∈X. By3.9, we have

4jf

_x_y

2j ,

z−w

2j

4jf

_x₋_y

2j ,

zw

2j

−2·4jf

x

2j,

z 2j

−2·4jf

_y

2j,

w

2j

≤22−rjxryrzr 3wr

3.30

for allx, y, z, w ∈Xand allj. Lettingj → ∞, we see thatF satisfies1.2. Settingl 0 and takingm → ∞in3.29, one can obtain inequality3.10. IfG : X×X → Y is another 2-dimensional vector variable quadratic mapping satisfying3.10, by in8, Theorem 3, there are four symmetric biadditive mappingsS, T, U, V :X×X → Y such thatFx, y Sx, x Ty, yandGx, y Ux, x Vy, yfor allx, y∈X. Thus we obtain that

Fx, y−Gx, ySx, x Ty, y−Ux, x−Vy, y

4nSx

2n,

x

2n

Ty

2n,

y

2n

−Ux

2n,

x

2n

−Vy

2n,

y

2n

4nFx

2n,

y

2n

−Gx

2n,

y

2n

≤4n_Fx

2n,

y

2n

−fx

2n,

y

2n

₄n_fx

2n,

y

2n

−Gx

2n,

y

2n

≤ 22−rn1

1−22−r

2xr3yr

−→0, asn−→ ∞

3.31

for allx, y ∈X. Hence the mappingFis the unique 2-dimensional vector variable quadratic mapping, as desired.

Theorem 3.3. Letr /2be a real number, and letf : _AM ×AM → ANbe a mapping such that

faxay, az−awfax−ay, azaw−2a2fx, z−2a2fy, w

≤xryrzrwr

3.32

for alla∈A1and allx, y, z, w∈ AM. Ifftx, tyis continuous int∈Rfor each fixedx, y∈ AM, then there exists a unique2-dimensional vector variableA-quadratic mappingF: AM ×AM → AN satisfying1.2and3.10for allx, y∈ _AM.

Proof. Supposer <2. By Lemma3.2, it follows from the inequality of the statement fora1 that there exists a unique 2-dimensional vector variable quadratic mappingF: AM ×AM →

ANsatisfying1.2and inequality3.10for allx, y∈ AM.

Letx0, y0 ∈ AMbe fixed. And letL: AN → Rbe any continuous linear functional,

that is, L is an arbitrary element of the dual space of AN. For n ∈ N, consider two

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ξnt : 1/4nLf0,2nty0for all t ∈ R. By the assumption that ftx, tyis continuous

int∈Rfor each fixedx, y∈ _AM, the functionsζnandξnare continuous for alln∈N. Note

thatζnt 1/4nLf2ntx0,0 L1/4nf2ntx0,0andξnt 1/4nLf0,2nty0

L1/4n_f₀_,₂n_ty

0 for alln ∈ Nand all t ∈ R. By8, the sequences {ζnt} and{ξnt}

are Cauchy sequences for allt ∈ R. Define two functions ζ : R → Rand ξ : R → Rby

ζt: limn→ ∞ζntandξt: limn→ ∞ξntfor allt∈ R. Note thatζt LFtx0,0and

ξt LF0, ty0for allt∈R. SinceFsatisfies1.2, we get

ζst ζs−t LFstx0,0 LFs−tx0,0

LFstx0,0 Fs−tx0,0 LFsx0tx0,0 Fsx0−tx0,0 L2Fsx0,0 2Ftx0,0 2LFsx0,0 2LFtx0,0 2ζs 2ζt,

ξst ξs−t LF0,sty0

LF0,s−ty0

LF0,sty0

F0,s−ty0

LF0, sy0ty0

F0, sy0−ty0

L2F0, sy0

2F0, ty0

2LF0, sy0

2LF0, ty0

2ξs 2ξt

3.33

for alls, t∈R. Sinceζandξare the pointwise limits of continuous functions, they are Borel functions. Thus the functionsζandξas measurable quadratic functions are continuoussee

13, so have the formsζt t2_ζ₁_and_ξ_t _t2_ξ₁_{for all}_t_∈_R_{. Since}_F _satisfies_1.2_{, by} 8, Theorem 3, there exist two symmetric biadditive mappingsS, T : X ×X → Y such as

Fx, y Sx, x Ty, yfor allx, y∈X. Hence we have

LFtx0, ty0

LFtx0,0 F

0, ty0

LFtx0,0 L

F0, ty0

ζt ξt

t2ζ1 t2ξ1 t2LFx0,0 t2L

F0, y0

t2LFx0,0 F

0, y0

t2LSx0, x0 T

y0, y0

t2LFx0, y0

Lt2Fx0, y0

3.34

for allt∈R. SinceLis any continuous linear functional, the 2-dimensional quadratic mapping

F: AM ×AM → ANsatisfiesFtx0, ty0 t2Fx0, y0for allt∈R. Therefore we obtain

Ftx, tyt2Fx, y 3.35

for allt∈Rand allx, y∈ AM. Letjbe an arbitrary positive integer. Replacingxandzby 2jx

and 2j_z_{, respectively, and letting}_y_w_{0 in inequality}_3.32_{, we gain}

f2j_ax,₂j_az₋_a2_f₂j_x,₂j_z₋_a2_f₀_,₀_≤₂jr−1_xr_zr _3.36

for alla∈A1and allx, z∈ AM. Note that there is a constantK >0 such that the condition

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for eacha∈Aand eachv∈ ANsee12, Definition 12. For alla∈A1and allx, y∈ AM,

we get

1 4j

f2jax,2jay−a2f2jx,2jy≤2jr−2−1xrwrK|a|

2

4j f0,0−→0 3.38

asj → ∞. Hence we have

Fax, ay lim

j→ ∞

1 4jf

2jax,2jaya2lim

j→ ∞

1 4jf

2jx,2jya2Fx, y 3.39

for alla∈A1and allx, y ∈ AM. SinceFax, ay a2Fx, yfor eacha∈A1, by3.35, we

obtain

Fax, ayF

|a| a

|a|x,|a| a

|a|y

|a|2F

a

|a|x, a

|a|y

a2Fx, y 3.40

for all nonzeroa ∈ A and all x, y ∈ AM. By3.35, we getF0x,0y 02Fx, y for all

x, y∈ _AM. Therefore the mappingF : _AM ×AM → ANis the unique 2-dimensional vector

variableA-quadratic mapping satisfying1.2and3.10.

The proof of the caser >2 is similar to that of the caser <2.

Theorem 3.4. Letr /2be a real number andAof real rank0, and letf : _AM ×AM → ANbe a mapping such that

_f

axay, bz−bwfax−ay, bzbw−2abfx, z−2aby, w

<xryrzrwr 3.41

for alla, b ∈A₁ ∩Ain∪ {i}and allx, y, z, w ∈ AM. For each fixedx, y ∈ AM, let the sequence {1/4j_f₂j_ax,₂j_by_}_{converge uniformly on}_A

1×A1. Iffax, byis continuous ina, b∈A1∪ R2 _{for each fixed} _{x, y} _∈

AM, then there exists a unique 2-dimensional vector variable mapping

F : _AM ×AM → ANsatisfying1.2and3.10such thatFax, by abFx, yfor alla, b ∈

A₁∪ {i}and allx, y∈ AM.

Proof. Suppose r < 2. By 8, Theorem 4, there exists a unique 2-dimensional quadratic mapping F : _AM ×AM → AN satisfying 1.2 and inequality 3.10 on AM ×AM. Let

x0, y0 ∈ AMbe fixed. And letLbe an arbitrary element of the dual space of AN. Forn∈N,

consider the functions ψn : R×R → Rdefined by ψns, t : 1/4nLf2nsx0,2nty0

for all s, t ∈ R. By the assumption that fax, by is continuous ina, b ∈ A1 ∪R2 for

each fixed x, y ∈ AM, the function ψn is continuous for all n ∈ N. Note thatψns, t 1/4n_L_f₂n_sx

0,2nty0 L1/4nf2nsx0,2nty0for alln ∈ Nand alls, t ∈ R. By 8,

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ψs, t: limn→ ∞ψns, tfor alls, t ∈R. Note thatψs, t LFsx0, ty0for allt ∈R. Thus

we have

ψs1s2, t1−t2 ψs1−s2, t1t2 LFs1s2x0,t1−t2y0

LFs1−s2x0,t1t2y0

LFs1s2x0,t1−t2y0

Fs1−s2x0,t1t2y0

LFs1x0s2x0, t1y0−t2y0

Fs1x0−s2x0, t1y0t2y0

L2Fs1x0, t1y0

2Fs2x0, t2y0

2LFs1x0, t1y0

2LFs2x0, t2y0

2ψs1, t1 2ψs2, t2

3.42

for alls1, s2, t1, t2 ∈ R. Since ψ is the pointwise limit of continuous functions, it is a Borel

function. By Theorem3.1, we gainψs, t s2_ψ₁_,₀ _t2_ψ₀_,₁_{for all}_{s, t}_∈_R_{. Hence we get}

LFsx0, ty0

ψs, t s2_ψ₁_,₀ _t2_ψ₀_,₁ _s2_L_F_x

0,0 t2L

F0, y0

Ls2Fx0,0 t2F

0, y0

3.43

for all s, t ∈ R. Since L is any continuous linear functional, the 2-dimensional quadratic mappingF : AM ×AM → ANsatisfiesFsx0, ty0 s2Fx0,0 t2F0, y0for alls, t ∈R.

Therefore we obtain

Fsx, tys2Fx,0 t2F0, y 3.44

for alls, t∈Rand allx, y∈ _AM. Letjbe an arbitrary positive integer. Replacingxandzby 2j_x_{and 2}j_z_{, respectively, and letting}_y_w_{0 in the inequality}_3.41_{, we get}

f2jax,2jbz−abf2jx,2jz−abf0,0≤2jr−1xrzr 3.45

for alla, b∈A₁∩Ain∪{i}and allx, z∈ AM. By condition3.37, for alla, b∈A1∩Ain∪{i}

and allx, y∈ AM, we have

1 4j

f2jax,2jby−abf2jx,2jy≤2jr−2−1xrzrK|a||b|

4j f0,0 −→0, asj−→ ∞.

3.46

Hence we obtain that

Fax, by lim

j→ ∞

1 4jf

2jax,2jbyablim

j→ ∞

1 4jf

2jx,2jyabFx, y 3.47

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Let c, d ∈ A₁ \Ain. SinceAin ∩Asa is dense in Asa, there exist two sequences {cj}

and {dj}inAin ∩Asa such thatcj → c anddj → dasj → ∞. Put pj : 1/|cj|cj and

qj: 1/|dj|dj. Thenpj → candqj → dasj → ∞. Setaj :

pjpjandbj:

qjqj. Then

aj → candbj → dasj → ∞andaj, bj ∈A₁∩Ain. Since{1/4jf2jax,2jby}is uniformly

converges onA1×A1 andfax, byis continuous ina, b ∈A1, we see thatFax, byis also

continuous ina, b∈A1for eachx, y∈ AM. In fact, we gain

lim

a,b→c,dF

ax, by lim

a,b→c,djlim→ ∞

1 4jf

2jax,2jby lim

j→ ∞a,blim→c,d

1 4jf

2jax,2jby

lim

j→ ∞

1 4jf

2jcx,2jdyFcx, dy

3.48

for allx, y∈ _AM. Thus we get

lim

j→ ∞F

ajx, bjy

F

lim

j→ ∞ajx,jlim→ ∞bjy

Fcx, dy 3.49

for allx, y∈ AM. By equality3.47, we have

Fajx, bjy

−cdFx, yajbjF

x, y−cdFx, y−→cdFx, y−cdFx, y0

3.50

asj → ∞for allx, y∈ _AM.By equality3.49and the above convergence, we see that

Fcx, dy−cdFx, y≤Fcx, dy−Fajx, bjyF

ajx, bjy

−cdFx, y

−→0 asj −→ ∞

3.51

for all x, y ∈ AM. By equality 3.47 and the above convergence, we obtain Fax, by

abFx, yfor alla, b∈A₁∪ {i}and allx, y∈ AM.

The proof of the caser >2 is similar to that of the caser <2.

References

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