Solutions and Stability of Generalized Mixed Type QC Functional Equations in Random Normed Spaces

Volume 2010, Article ID 403101,16pages doi:10.1155/2010/403101

Research Article

Solutions and Stability of

Generalized Mixed Type QC Functional Equations

in Random Normed Spaces

Yeol Je Cho,

_{Madjid Eshaghi Gordji,}

_{and Somaye Zolfaghari}

1_{Department of Mathematics Education and the RINS, Gyeongsang National University,} Chinju 660-701, Republic of Korea

2_{Department of Mathematics, Semnan University, P.O. Box 35195-363, Semnan, Iran} 3_{Department of Mathematics, Urmia University, Urmia, Iran}

Correspondence should be addressed to Madjid Eshaghi Gordji,[email protected]

Received 29 July 2010; Accepted 31 August 2010

Academic Editor: Vijay Gupta

Copyrightq2010 Yeol Je Cho et al. 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 achieve the general solution and the generalized stability result for the following functional equation in random normed spacesin the sense of Sherstnevunder arbitraryt-norms:fx ky fx−ky k2_{fxy fx−y 2k}2_−1/k2_k−2fkx−k3₋

k2_−k1/2k−

2f2x f2y−8fy, wherefy:fy−f−yfor any fixed integerkwith_{k /}0, ±1,2.

1. Introduction

In 1940, the stability problem of functional equations originated from a question given by Ulam1, that is, the stability of group homomorphisms.

Let G1,· be a group and G2,∗, d be a metric group with the metric d. For any

> 0, does there existδ > 0, such that, if a mapping h : G1 → G2 satisfies inequality

dhx·y, hx∗hy< δ.

For all x, y ∈ G1, then there exists a homomorphism H : G1 → G2 with

dhx, Hx< for allx∈G1?

In other words, under what condition, does there exists a homomorphism near an approximate homomorphism? The concept of stability for functional equation arises when we replace the functional equation by an inequality which acts as a perturbation of the equation.

Letfbe a mapping between Banach spacesEandEsuch that, for someδ >0,

_f

xy−fx−fy≤δ, ∀x, y∈E. 1.1

Then there exists a unique additive mappingT :E → Esuch that

_{fx−Tx≤δ, ∀x, y∈E. 1.2}

Moreover, ifftxis continuous int∈Rfor any fixedx∈E,thenTis linear.

In 1978, Rassias3 provided a generalization of Hyers’ Theorem which allows the Cauchy difference to be unbounded. In 1991, Gajda4answered the question for the case

p > 1, which was raised by Rassias. This new concept is known as theHyers-Ulam-Rassias stability of functional equationssee5–17.

The functional equation

fxyfx−y2fx 2fy 1.3

is related to symmetric biadditive mapping. It is natural that this equation is called aquadratic functional equation. In particular, every solution of the quadratic equation1.3is said to be a

quadratic mapping. It is well known that a mapping f between real vector spacesE andE

is quadratic if and only if there exits a unique symmetric biadditive mappingB such that

fx Bx, xfor allx∈Esee5,18. The biadditive mappingBis given by

Bx, y 1

4

fxy−fx−y, ∀x, y∈E. 1.4

Hyers-Ulam-Rassias stability problem for the quadratic functional equation1.3was proved by Skof19for a mappingf:E → E, whereEis a normed space andEis a Banach space. Cholewa20noticed that the theorem of Skof is still true if the relevant domainAis replaced by an abelian group. In21, Czerwik proved the Hyers-Ulam-Rassias stability of1.3and Grabiec22generalized these results mentioned above.

Recently, Jun and Kim23introduced the following cubic functional equation:

f2xyf2x−y2fxy2fx−y12fx, ∀x, y∈X, 1.5

where f is a mapping from a real vector space X into a real vector space Y and they established the general solution and the generalized Hyers-Ulam-Rassias stability for the functional equation1.5. The functionfx x3_{satisfies the functional equation1.5,which} is thus called acubic functional equation. Every solution of the cubic functional equation is said to be acubic function. Also, Jun and Kim proved that a functionfbetween real vector spaces

XandY is a solution of1.5if and only if there exits a unique functionC:X×X×X → Y

such thatfx Cx, x, xfor allx∈XandCis symmetric for each fixed one variable and is additive for fixed two variables.

functions, that is, the space of all mappingsF : R∪ {−∞,∞} → 0,1such thatF is left-continuous and nondecreasing onR,F0 0 andF∞ 1.Dis a subset ofΔconsisting of all functionsF ∈ Δ for which l−F∞ 1, wherel−fxdenotes the left limit of the functionfat the pointx, that is,l−fx limt→x−ft. The spaceΔis partially ordered by

the usual pointwise ordering of functions, that is,F≤Gif and only ifFt≤Gtfor allt∈R. The maximal element forΔin this order is the distribution functionε0given by

ε0t ⎧ ⎨ ⎩

0, ift≤0,

1, ift >0. 1.6

Definition 1.1see27. A mappingT:0,1×0,1 → 0,1is called acontinuous triangular normbriefly, acontinuoust-normifTsatisfies the following conditions:

aT is commutative and associative; bT is continuous;

cTa,1 afor alla∈0,1;

dTa, b≤Tc, dwhenevera≤candb≤dfor alla, b, c, d∈0,1.

Typical examples of continuoust-norms areTPa, b ab,TMa, b mina, band TLa, b maxab−1,0 the Lukasiewiczt-norm.

Recallsee29,30that, ifT is at-norm and{xn}is a given sequence of numbers in

0,1, thenT_in₁xiis defined recurrently by

T_i1₁xix1, Tin1xiT Tin−11xi, xn

, ∀n≥2, 1.7

andT_i∞_nxiis defined asT_i∞₁xni.

It is known30that, for the Lukasiewiczt-norm, the following implication holds:

lim

n→ ∞TL ∞

i1xni1⇐⇒

∞

n1

1−xn<∞. 1.8

Definition 1.2see28. Arandom normed spacebriefly,RN-spaceis a tripleX, μ, T, where Xis a vector space,T is a continuoust-norm andμis a mapping fromXintoD satisfying the following conditions:

RN1μxt ε0tfor allt >0 if and only ifx0;

RN2μαxt μxt/|α|for allx∈X, andα∈Rwithα /0;

RN3μxyts≥Tμxt, μysfor allx, y∈Xandt, s≥0.

Every normed spacesX, · defines a random normed spaceX, μ, TM, where

μxt t

tx, ∀t >0, 1.9

Definition 1.3. LetX, μ, Tbe a RN-space.

1A sequence{xn}inXis said to beconvergentto a pointx∈X if, for any >0 and λ >0, there exists a positive integerNsuch thatμxn−x>1−λfor alln≥N. 2A sequence{xn}inX is called aCauchy sequenceif, for any >0 andλ >0, there

exists a positive integerNsuch thatμxn−xm>1−λfor alln≥m≥N.

3A RN-space X, μ, T is said to be complete if every Cauchy sequence in X is convergent to a point inX.

Theorem 1.4see27. IfX, μ, Tis a RN-space and{xn}is a sequence inXsuch thatxn → x, thenlimn→ ∞μxnt μxtalmost everywhere.

The stability of different functional equations in fuzzy normed spaces and random normed spaces has been studied in13,31–42.

In this paper, we deal with the following functional equation for fixed integerskwith

k /0, ±1,2:

fxkyfx−kyk2fxyfx−y2

k2₋₁

k2_k−2fkx

− k3−₂k2−k1

k−2 f2x f

2y−8fy,

1.10

wherefy : fy−f−yon random normed spaces. It is easy to see that the mapping

fx ax2_bx3_{cis a solution of the functional equation1.10. InSection 2, we investigate} the general solution of functional equation1.10whenfis a mapping between vector spaces and, inSection 3, we establish the stability of the functional equation1.10in RN-spaces.

2. General Solutions

Before proceeding to the proof of Theorem 2.3which is the main result in this section, we need the following two lemmas.

Lemma 2.1. If an even functionf:X → Ywithf0 0satisfies1.10, thenfis quadratic.

Proof. Settingx0 in1.10, by the evenness off, we obtain

fkyk2fy, ∀y∈X. 2.1

Interchangingxwithyin2.1, we have

fkx k2fx, ∀x∈X. 2.2

Lettingy0 in1.10, we have

fx −1

k2_k−2fkx

k−1

It follows from2.2and2.3that

f2x 4fx, ∀x∈X. 2.4

According to2.2,2.4and using the evenness off,1.10can be written as

fxkyfx−kyk2_fxyk2_{fx−y2 1−k}2_{fx, ∀x, y∈X. 2.5}

Replacingxbykxin2.5and then using2.2, we obtain

fkxyfkx−yfxyfx−y2 k2−1fx, ∀x, y∈X. 2.6

Interchangingxwithyin2.5, by the evenness off, we have

fkxyfkx−yk2_fxyk2_{fx−y2 1−k}2_{fy, ∀x, y∈X. 2.7}

But, since_{k /}0, ±1,2, it follows from2.6and2.7that

fxyfx−y2fx 2fy, ∀x, y∈X, 2.8

which shows thatfis quadratic. This completes the proof.

Lemma 2.2. If an odd functionf:X → Ysatisfies1.10, thenfis cubic. Proof. Lettingx0 in1.10, by the oddness off, we have

f2y8fy, ∀y∈X. 2.9

Settingy0 in1.10and then using2.9, we obtain

fkx k3fx, ∀x∈X. 2.10

According to2.9,2.10and using the oddness off,1.10can be written as

fxkyfx−kyk2_fxyk2_{fx−y2 1−k}2_{fx, ∀x, y∈X. 2.11}

Lettingyxin2.11and using2.9, by the oddness off,it follows that

Replacingxbyk−1xin2.11, we have

fk−1xkyfk−1x−ky

k2fk−1xyk2fk−1x−y

2 1−k2fk−1x, ∀x, y∈X.

2.13

Now, replacingxbyk1xin2.11and using2.12, we have

fk1xkyfk1x−ky

k2fk1xyk2fk1x−y

2 1−k2fk−1x 4 1−k2 3k21fx, ∀x, y∈X.

2.14

Substitutingxywithxin2.11and thenx−ywithxin2.11, we obtain

fx k1yfx−k−1y

k2fx2y2 1−k2fxyk2fx, ∀x, y∈X,

2.15

fx−k1yfx k−1y

k2fx−2y2 1−k2fx−yk2fx, ∀x, y∈X. 2.16

If we subtract2.16from2.15, we have

fx k1y−fx−k1y

k2fx2y−k2fx−2yfx k−1y−fx−k−1y

2 1−k2fxy−2 1−k2fx−y, ∀x, y∈X.

2.17

Interchangingxwithyin2.17and using the oddness off, we get

fk1xyfk1x−y

k2f2xyk2f2x−yfk−1xyfk−1x−y

2 1−k2_{fxy2 1−k}2_{fx−y, ∀x, y∈X.}

Thus it follows from2.14and2.18that

fk1xkyfk1x−ky

k2fk−1xyk2fk−1x−yk4f2xyk4f2x−y

2k2 1−k2fxy2k2 1−k2fx−y

2 1−k2fk−1x 4 1−k2 3k21fx, ∀x, y∈X.

2.19

Again, substitutingxy with y in2.11 and thenx−y with y in2.11, we get, by the oddness off,

fk1xky−fk−1xky

k2_f2xyk2_{f−y2 1−k}2_{fx, ∀x, y∈X,} 2.20

fk1x−ky−fk−1x−ky

k2f2x−yk2fy2 1−k2fx, ∀x, y∈X.

2.21

Then, by adding2.20to2.21and then using2.13, we have

fk1xkyfk1x−ky

k2fk−1xyk2fk−1x−y2 1−k2fk−1x

k2f2xyk2f2x−y4 1−k2fx, ∀x, y∈X.

2.22

Finally, if we compare2.19with2.22, then we conclude that

f2xyf2x−y2fxy2fx−y12fx, ∀x, y∈X. 2.23

Therefore,fis a cubic function. This completes the proof.

Theorem 2.3. A functionf :X → Y withf0 0satisfies1.10for allx, y ∈Xif and only if

there exist functionsQ:X×X → Y andC:X×X×X → Ysuch thatfx Qx, x Cx, x, x for allx ∈X,where the functionCis symmetric for each fixed one variable and is additive for fixed two variables andQis symmetric biadditive.

Proof. Letf be a mapping withf0 0 satisfies1.10. We decomposefinto the even part and odd part by putting

fex

1 2

fx f−x, fox

1 2

It is clear thatfx fex foxfor allx ∈ X and it is easy to show that the functions fe andfo satisfy1.10. Hence, by Lemmas2.1and2.2, we know that the functionsfeand fo are quadratic and cubic, respectively. Thus there exist a symmetric biadditive function Q:X×X → Y such thatfex Qx, xfor allx∈Xand the functionC:X×X×X → Y

such thatfox Cx, x, xfor allx∈X,where the functionCis symmetric for each fixed one

variable and is additive for fixed two variables. Therefore, we getfx Cx, x, x Qx, x

for allx∈X.

Conversely, let fx Cx, x, x Qx, x for all x ∈ X, where the function C is symmetric for each fixed one variable and is additive for fixed two variables and Q is biadditive. By a simple computation, one can show that the functionsx → Cx, x, xand

x → Qx, xsatisfy the functional equation1.10. Thus the functionf satisfies1.10. This completes the proof.

3. Stability Problems

From now on, we suppose thatXis a real linear space,Y, μ, Tis a complete RN-space and

f : X → Y is a function withf0 0 for which there exists a mappingρ : X×X → D

ρx, yis denoted byρx,ywith the property:

μ_{fxkyfx−ky−k}2_{fxyfx−y−2k}2_−1/k2_k−2fkxk3_−k2_{−k1/2k−2f2x−f2y8fy}t

≥ρx,yt, ∀x, y∈X, t >0,

3.1

wherefy:fy−f−yfor ally∈X.

Theorem 3.1. Let

lim

m→ ∞T ∞

j1 ρ0,kmj−1_x k2mjt

1

lim

m→ ∞ρkmx,kmy k

2m_{t, ∀x, y∈X, t >0.}

3.2

Suppose that an even functionf:X → Y withf0 0satisfies inequality

μ_{fxkyfx−ky−k}2_{fxyfx−y−2k}2_−1/k2_k−2fkxk3_−k2_{−k1/2k−2f2x−f2y8fy}t

≥ρx,yt, ∀x, y∈X, t >0,

3.3

wherefy:fy−f−yfor ally∈X. Then there exists a unique quadratic mappingQ:X → Y such that

μQx−fxt≥Tj∞1 ρ0,kj−1_x kjt

Proof. It follows from3.3and the evenness offthat

μfxkyfx−ky−k2_{fxyfx−y−2k}2_−1/k2_k−2fkxk3_−k2_{−k1/2k−2f2x}t

≥ρx,yt, ∀x, y∈X, t >0.

3.5

Settingx0 in3.5, we get

μ2fky−2k2_fyt≥ρ_0,yt, ∀y∈X, t >0. 3.6

If we replaceybyxin3.6and divide both sides of3.6by 2, we get

μfkx−k2_fxt≥ρ_0,x2t≥ρ_0,xt, ∀x∈X, t >0. 3.7

In other words, we have

μfkx/k2_−fxt≥ρ_0,x k2t

, ∀x∈X, t >0. 3.8

Therefore, it follows that

μfkn1_x/k2n1_−fkn_x/k2n

t k2n

≥ρ0,kn_x k2t

, ∀x∈X, t >0, n∈N. 3.9

Hence we have

μfkn1_x/k2n1_−fkn_x/k2nt≥ρ_0,kn_x k2n1t

, ∀x∈X, t >0, n∈N. 3.10

This means that

μfkn1x/k2n1−fkn_x/k2n

t kn1

≥ρ0,kn_x kn1t

, ∀x∈X, t >0, n∈N. 3.11

Since 1>1/21/22_{· · ·1/2}m_{,by the triangle inequality, it follows that}

μfkm_x/k2m_−fxt≥T_nm−₀1

μfkn1x/k2n1−fkn_x/k2n _m−1

n0

t kn1

≥T_nm−₀1 ρ0,kn_x kn1t

T_jm₁ ρ0,kj−1_x kjt

, ∀x∈X, t >0.

3.12

In order to prove the convergence of the sequence{fkm_x/k2m_{}, we replacexwithk}m_xin

3.12to find that

μfkmm_x/k2mm_−fkm_x/k2mt≥T_jm₁ ρ_0,kjm−1_x kj2m

Since the right hand side of inequality 3.13 tends to 1 as m, m → ∞, the sequence

{fkm_x/k2m_{}is a Cauchy sequence. Therefore, we may define}

Qx lim

m→ ∞

fkm_x

k2m , ∀x∈X. 3.14

Now, we show thatQ is a quadratic mapping. In fact, replacingx, y with km_xand km_{y, respectively, in3.3and then taking the limit asm → ∞, we find thatQsatisfies1.10}

for allx, y∈X.Therefore, the mappingQ:X → Y is quadratic. To prove3.4,taking the limit asm → ∞in3.12, we get3.4.

Finally, to prove the uniqueness of the quadratic functionQsubject to3.4, assume that there exists a quadratic functionQwhich satisfies3.4. SinceQkm_{x k}2m_Qxand Qkm_{x k}2m_{Qxfor allx∈Xandm∈N,it follows from3.4that}

μQx−Qxt μQkm_x−Qkm_x k2mt

≥T μQkm_x−fkm_x k2m−1t

, μfkm_x−Qkm_x k2m−1t

≥T T_j∞₁ ρ0,kjm−1_x k2mjt

, T_j∞₁ ρ0,kjm−1_x k2mjt

3.15

for allx∈Xandt >0. Therefore, by lettingm → ∞in inequality3.15, we find thatQQ. This completes the proof.

Theorem 3.2. Letf :X → Ybe an odd mapping withf0 0satisfies inequality

μ_{fxkyfx−ky−k}2_{fxyfx−y−2k}2_−1/k2_k−2fkxk3_−k2_{−k1/2k−2f2x−f2y8fy}t

≥ρx,yt, ∀x, y∈X, t >0,

3.16

wherefy:fy−f−yfor ally∈X. If

lim

m→ ∞T

∞

j1

T

ρ0,kmj−1_x

2k2j3m k2_k−1t

, ρkmj−1_x,0

k3m2j_1−k2

k2_k−2 t

1, 3.17

lim

m→ ∞ρkmx,kmy k

3m_{t1, ∀x, y∈X, t >0, 3.18}

then there exists a unique cubic mappingC:X → Y such that

μCx−fxt≥Tj∞1

T

ρ0,kj−1_x

2k2j k2_k−1t

, ρkj−1_x,0

k2j_1−k2

k2_k−2 t

3.19

Proof. It follows from3.15and the oddness offthat

μfxkyfx−ky−k2_{fxyfx−y−2k}2_−1/k2_k−2fkxk3_−k2_{−k1/2k−2f2x−2f2y16fy}t

≥ρx,yt, ∀x, y∈X, t >0.

3.20

By puttingx0 in3.20and usingRN2,we obtain

μ2f2y−16fyt≥ρ0,yt, ∀y∈X, t >0. 3.21

If we replaceybyxin3.21and divide both sides of3.21by 2,we get

μf2x−8fxt≥ρ0,x2t≥ρ0,xt, ∀x∈X, t >0. 3.22

Lettingy0 in3.20, we get

μ21−k2_fx−2k2_−1/k2_k−2fkxk3_−k2_{−k1/2k−2f2x}t

≥ρx,0t, ∀x∈X, t >0.

3.23

Therefore, we have

μk2_{k−2fxfkx−k}2_k−1/4f2xt≥ρ_x,0

21−k2

k2_k−2t

, ∀x∈X, t >0. 3.24

It follows from3.22and3.24that

μfkx−k3_fxt ≥T

ρ0,x

2

k2_k−₁t

, ρx,0

1−k2

k2_k−₂t

∀x∈X, t >0, k∈N.

3.25

Let

ψx,xt T

ρ0,x

2

k2_k−1t

, ρx,0

1−k2 k2_k−2t

, ∀x∈X, t >0. 3.26

Then we get

μfkx−k3_fxt≥ψ_x,xt, ∀x∈X, t >0. 3.27

It follows that

μfkx/k3_−fxt≥ψ_x,x k3t

Hence we have

μfkn1x/k3n1−fkn_x/k3n

t k3n

≥ψkn_x,kn_x k3t

, ∀x∈X, t >0, 3.29

which implies that

μfkn1_x/k3n1_−fkn_x/k3nt≥ψ_kn_x,kn_x k3n1t

, ∀x∈X, t >0, n∈N. 3.30

Thus we have

μfkn1_x/k3n1_−fkn_x/k3n

t kn1

≥ψkn_x,kn_x k2n1t

, ∀x∈X, t >0, n∈N. 3.31

Since 1>1/21/22_{· · ·1/2}m_{,by the triangle inequality, it follows that}

μfkm_x/k3m_−fxt≥T_nm−₀1

μfkn1_x/k3n1_−fkn_x/k3n _m−1

n0

t kn1

≥T_nm−₀1 ψkn_x,kn_x k2n1t

T_jm₁ ψkj−1_x,kj−1_x k2jt

, ∀x∈X, t >0.

3.32

In order to prove the convergence of the sequence{fkm_x/k3m_{}, we replacexwithk}m_xin

3.32to find that

μfkmm_x/k3mm_−fkm_x/k3mt≥T_jm₁ ψ_kjm−1_x,kjm−1_x k2j3m

t. 3.33

Since the right hand side of inequality 3.33 tends to 1 as m, m → ∞, the sequence

{fkm_x/k3m_{}is a Cauchy sequence. Therefore, we can define}

Cx lim

m→ ∞

fkm_x

k3m , ∀x∈X. 3.34

Now, we show thatCis a cubic mapping. In fact, replacingx, ywithkm_xandkm_yin

3.15, respectively, and then taking the limitm → ∞, we find thatCsatisfies1.10for all

x, y∈X.Therefore, the mappingC:X → Y is cubic.

Finally, to prove the uniqueness of the cubic functionCsubject to inequality3.19, assume that there exists a cubic functionCwhich satisfies inequality3.19. SinceCkm_x k3m_CxandCkm_{x k}3m_{Cxfor allx∈Xandm∈N,it follows from3.19that}

μCx−C_xt

μCkm_x−Ckm_x k3mt

≥T μCkm_x−fkm_x k3m−1t

, μfkm_x−Ckm_x k3m−1t

≥T

T_j∞₁

T

ρ0,kmj−1_x

2k2j3m k2_k−1t

, ρkmj−1_x,0

k3m2j_1−k2

k2_k−2 t

,

T_j∞₁

T

ρ0,kmj−1_x

2k2j3m k2_k−1t

, ρkmj−1_x,0

k3m2j_1−k2

k2_k−2 t

3.35

for allx∈ X andt >0. By lettingm → ∞in inequality3.35, we know thatC C. This completes the proof.

Theorem 3.3. If

lim

m→ ∞T

∞

j1

T ρ0,kmj−1_x kj2mt

, ρ0,−kmj−1_x kj2mt

1

lim

m→ ∞T ∞

j1

T

ρ0,kmj−1_x

2k2j3m k2_k−1t

, ρkmj−1_x,0

k2j3m_1−k2

k2_k−2 t

,

T

ρ0,−kmj−1_x

2k2j3m k2_k−1t

, ρ−kmj−1_x,0

k2j3m_1−k2

k2_k−2 t

3.36

for allx∈Xandt >0and

lim

m→ ∞T ρkmx,kmy k

2n−1_{t, ρ}

km_x,km_y k2n−1t

1 lim

m→ ∞T ρkmx,kmy k

3n−1_{t, ρ}

km_x,km_y k3n−1t

, ∀x, y∈X, t >0,

3.37

μfx−Cx−Qxt

≥T

T_j∞₁

T

ρ0,kj−1_x

kj

2 t

, ρ0,−kj−1_x

kj

2t

,

T_j∞₁

T

ρ0,kj−1_x

k2j k2_k−1t

, ρkj−1_x,0

k2j_1−k2 2k2_k−2t

,

T

ρ0,−kj−1_x

k2j k2_k−1t

, ρ−kj−1_x,0

k2j_1−k2 2k2_k−2t

, ∀x∈X, t >0.

3.38

Proof. Iffex 1/2fx f−xfor allx∈X.Thenfe0 0, fe−x fex, and μ_{fexkyfex−ky−k}2_f

exyfex−y−2k2−1/k2k−2fekxk3−k2−k1/2k−2fe2x−fe2y8feyt

≥Tρx,y2t, ρ−x,−y2t

≥Tρx,yt, ρ−x,−yt

, ∀x, y∈X, t >0,

3.39

wherefey:fey−fe−yfor allx, y ∈X.Hence, in view ofTheorem 3.1, there exists a

unique quadratic functionQ:X → Y such that

μQx−fext≥T ∞

j1

T ρ0,kj−1_x kjt

, ρ0,−kj−1_x kjt

, ∀x∈X, t >0. 3.40

Letfox 1/2fx−f−xfor allx∈X.Thenfo0 0, fo−x −foxand

μ_{foxkyfox−ky−k}2_{foxyfox−y−2k}2_−1/k2_k−2fokxk3_−k2_{−k1/2k−2fo2x−fo2y8foy}t

≥Tρx,y2t, ρ−x,−y2t

≥Tρx,yt, ρ−x,−yt

, ∀x, y∈X, t >0,

3.41

wherefoy:foy−fo−yfor allx, y∈X.FromTheorem 3.2, it follows that there exists a

unique cubic mappingC:X → Y such that

μCx−foxt≥T ∞

j1

T

ρ0,kj−1_x

2k2j k2_k−1t

, ρkj−1_x,0

k2j_1−k2

k2_k−2 t

,

T

ρ0,−kj−1_x

2k2j k2_k−1t

, ρ−kj−1_x,0

k2j_1−k2

k2_k−2 t

, ∀x∈X, t >0.

3.42

Acknowledgment

This paper was supported by the Korea Research Foundation Grant funded by the Korean GovernmentKRF-2008-313-C00050.

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