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Volume 2011, Article ID 734567,18pages doi:10.1155/2011/734567

Research Article

Approximate Quartic and Quadratic Mappings in Quasi-Banach Spaces

M. Eshaghi Gordji,

1

H. Khodaei,

1

and Hark-Mahn Kim

2

1Department of Mathematics, Semnan University, P. O. Box 35195-363, Semnan, Iran

2Department of Mathematics, Chungnam National University, 220 Yuseong-Gu, Daejeon 305-764, Republic of Korea

Correspondence should be addressed to Hark-Mahn Kim,[email protected] Received 17 March 2011; Accepted 13 May 2011

Academic Editor: Petru Jebelean

Copyrightq2011 M. Eshaghi Gordji 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 establish the general solution for a mixed type functional equation of aquartic and a quadratic mapping in linear spaces. In addition, we investigate the generalized Hyers-Ulam stability inp- Banach spaces.

1. Introduction and Preliminaries

The stability problem of functional equations originated from a question of Ulam1in 1940, concerning the stability of group homomorphisms. LetG1,·be a group, and letG2,∗be a metric group with the metricd·,·. Given > 0, does there exist aδ > 0 such that if a mappingh: G1G2 satisfies the inequalitydhx·y, hxhy < δfor allx, yG1, then there exists a homomorphismH :G1G2withdhx, Hx < for allxG1? 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. In 1941, Hyers 2 gave a first affirmative answer to the question of Ulam for Banach spaces. Let f:XXbe a mapping between Banach spaces such that

f xy

fxf

yδ, 1.1

(2)

for allx, yX, and for someδ >0. Then, there exists a unique additive mappingT :XX such that

fxTx≤δ, 1.2 for allxX.

The result of Hyers was generalized by Aoki3for approximate additive function and by Rassias4for approximate linear function by allowing the difference Cauchy equation fxyfxfy to be controlled by εxp yp. Taking into consideration a lot of influence of Ulam, Hyers and Rassias on the development of stability problems of functional equations, the stability phenomenon that was proved by Rassias may be called the Hyers-Ulam-Rassias stabilitysee5,6. In 1994, a generalization of Rassias theorem was obtained by G˘avruta7, who replacedεxpypby a general control functionϕx, y.

The functional equation f

xy f

xy

2fx 2f y

1.3 is related to a symmetric biadditive function8–10. It is natural that this equation is called a quadratic functional equation. In particular, every solution of the quadratic equation1.3is said to be a quadratic function. It is well known that a functionf between real vector spaces is quadratic if and only if there exists a unique symmetric biadditive functionB1 such that fx B1x, xfor allxin the vector space. The biadditive functionB1is given by

B1

x, y 1 4

f xy

f xy

. 1.4

A Hyers-Ulam stability problem for the quadratic functional equation1.3was proved by Skof for functionsf:XY, whereXis normed space andY is Banach spacesee11. In the paper12, Czerwik proved the Hyers-Ulam-Rassias stability of1.3.

Lee et al.13considered the following functional equation:

f 2xy

f 2x−y

4f xy

4f xy

24fx−6f y

. 1.5

In fact, they proved that a functionfbetween two real vector spacesXandYis a solution of 1.5if and only if there exists a unique symmetric biquadratic functionB2:X×XYsuch thatfx B2x, xfor allxX. The biquadratic functionB2is given by

B2

x, y 1 12

f xy

f xy

−2fx−2f y

. 1.6

It is easy to show that the functionfx ax4satisfies the functional equation1.5, which is called the quartic functional equationsee also14.

Jun and Kim15have obtained the generalized Hyers-Ulam stability for a mixed type of cubic and additive functional equation. In addition, the generalized Hyers-Ulam stability for a mixed type of cubic, quadratic, and additive functional equation has been investigated by Gordji and Khodaei16 see also17,18. The stability problems for several mixed types

(3)

of functional equations have been extensively investigated by a number of authors and there are many interesting results concerning this problem19–27.

In this paper, we deal with the following functional equation derived from quartic and quadratic functions:

f kxy

f kxy k2f

xy k2f

xy k2

k2−1 6

f2x−4fx

−2 k2−1

f

y 1.7

for fixed integersk /0,±1. It is easy to see that the functionfx ax4bx2is a solution of the functional equation1.7. In the sequel, we investigate the general solution of functional equation1.7whenfis a function between vector spaces, and then we prove the generalized Hyers-Ulam stability of 1.7 in the spirit of Hyers, Ulam, and Rassias using the direct method.

We recall some basic facts concerning quasi-Banach spaces and some preliminary results.

Definition 1.1see28,29. LetXbe a real linear space. A quasinorm is a real-valued function onXsatisfying the following:

1x ≥0 for allxXandx 0 if and only ifx 0, 2λ·x |λ| · xfor allλÊand allxX,

3there is a constantM≥1 such thatxy ≤Mxyfor allx, yX.

The pairX, · is called a quasinormed space if · is a quasinorm onX.

The smallest possibleMis called the modulus of concavity of · . A quasi-Banach space is a complete quasinormed space. A quasinorm · is called ap-norm0< p≤1if

xyp≤ xpyp, 1.8 for allx, yX. In this case, a quasi-Banach space is called ap-Banach space.

Given ap-norm, the formuladx, y: x−ypgives us a translation invariant metric onX. By the Aoki-Rolewicz Theorem29, each quasinorm is equivalent to somep-normsee also28. Since it is much easier to work withp-norms, henceforth we restrict our attention mainly top-norms.

Lemma 1.2see17. Letx1, x2, . . . , xnbe nonnegative real numbers. Then, one has n

i 1

xi

p

n

i 1

xip, 1.9

for a positive real numberpwithp1.

2. General Solution

We here present the general solution of1.7.

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Theorem 2.1. Let bothXandY be real vector spaces. A functionf :XY satisfies1.7for all x, yXif and only if there exists a unique symmetric biquadratic functionB2 :X×XY and a unique symmetric biadditive functionB1:X×XY such that

fx B2x, x B1x, x, 2.1

for allxX.

Proof. Letfsatisfy1.7and letg, h:XYbe functions defined by

gx: f2x−16fx, hx: f2x−4fx, 2.2

for allxX. We claim that the functionsgandhare quadratic and quartic, respectively.

Lettingx y 0 in1.7, we havef0 0. By puttingx 0 in1.7, one leads to the evennessf−y fyoff. Replacingybyxyin1.7, we have

f

k1xy f

k−1x−y

k2f 2xy

k2f

−y k2

k2−1 6

f2x−4fx 2

1−k2 f

xy ,

2.3

for allx, yX. Replacingyby−yin2.3, we obtain

f

k1x−y f

k−1xy

k2f 2x−y

k2f y

k2 k2−1

6

f2x−4fx 2

1−k2 f

xy ,

2.4

for allx, yX. Adding2.3to2.4, we get by evenness off,

f

k1xy f

k1x−y f

k−1xy f

k−1x−y

k2 f

2xy f

2x−y 2k2

k2−1 6

f2x−4fx

2

1−k2 f

xy f

xy

2k2f y

,

2.5

for allx, yX. From the substitutiony kxyin1.7, we have by evenness off,

f

2kxy f

y k2f

k1xy k2f

k−1xy k2

k2−1 6

f2x−4fx 2

1−k2 f

kxy ,

2.6

(5)

for allx, yX. Replacingyby−yin2.6, we get

f

2kx−y f

−y k2f

k1x−y k2f

k−1x−y

k2 k2−1

6

f2x−4fx 2

1−k2 f

kxy ,

2.7

for allx, yX. Adding2.6to2.7, we get by evenness off,

f

2kxy f

2kx−y k2

f

k1xy f

k1x−y f

k−1xy f

k−1x−y

2k2 k2−1

6

f2x−4fx 2

1−k2 f

kxy f

kxy

−2f y

, 2.8

for allx, yX. By using1.7and2.5, it follows from2.8that

f

2kxy f

2kx−y k2

k2

f 2xy

f

2x−y 2k2

k2−1 6

f2x−4fx

2

1−k2 f

xy f

xy

2k2f y

2

1−k2 k2f

xy k2f

xy k2

k2−1 6

f2x−4fx 2

1−k2 f

y 2k2

k2−1 6

f2x−4fx

−2f y

,

2.9

for allx, yX. If we replacexby 2xin1.7, then we get that

f

2kxy f

2kx−y k2f

2xy k2f

2x−y k2

k2−1 6

f4x−4f2x 2

1−k2 f

y ,

2.10

(6)

for allx, yX. It follows from2.9and2.10that

k2

k2 f

2xy f

2x−y 2k2

k2−1 6

f2x−4fx

2

1−k2 f

xy f

xy

2k2f y

2

1−k2 k2f

xy k2f

xy k2

k2−1 6

f2x−4fx 2

1−k2 f

y 2k2

k2−1 6

f2x−4fx

−2f y k2f

2xy k2f

2x−y k2

k2−1 6

f4x−4f2x 2

1−k2 f

y ,

2.11

for allx, yX. On the other hand, puttingy 0 in1.7, we get

fkx k2fx k2 k2−1

12

f2x−4fx

, 2.12

for allxX. Puttingy xin1.7, we get

fk1x fk−1x k2f2x k2 k2−1

6

f2x−4fx 2

1−k2 fx,

2.13

for allxX. Puttingy kxin1.7and using the evenness off, we obtain

f2kx k2

fk1x fk−1x k2

k2−1 6

f2x−4fx 2

1−k2 fkx,

2.14

for allxX. Lettingy 0 in2.10, we have

f2kx k2f2x k2 k2−1

12

f4x−4f2x

, 2.15

for allxX. It follows from2.14and2.15that k2

k2−1 12

f4x−4f2x k2

fk1x fk−1x k2

k2−1 6

f2x−4fx

2 1−k2

fkxk2f2x,

2.16

(7)

for allxX. Now, by using2.12,2.13and2.16, we lead to k2

k2−1 12

f4x−4f2x k2

k2f2x k2 k2−1

6

f2x−4fx 2

1−k2 fx

2

1−k2

k2fx k2 k2−1

12

f2x−4fx

k2 k2−1

6

f2x−4fx

k2f2x,

2.17

for allxX. Finally, comparing2.11with2.17, then we conclude that f

2xy f

2x−y 4f

xy 4f

xy 2

f2x−4fx

−6f y

, 2.18

for allx, yX. Replacingyby 2yin2.18, we get f

2x2y f

2x−2y 4f

x2y 4f

x−2y 2

f2x−4fx

−6f 2y

,

2.19

for allx, yX. Interchangingxwithyin2.18, one gets f

x2y f

x−2y 4f

xy 4f

xy 2

f 2y

−4f y

−6fx, 2.20

for allx, yX. It follows from2.19and2.20that f

2 xy

−16f xy

f 2

xy

−16f xy 2

f2x−16fx 2

f 2y

−16f y

, 2.21

for allx, yX. This means that g

xy g

xy

2gx 2g y

, 2.22

for allx, yX. So the functiong:XYdefined bygx: f2x−16fxis quadratic.

To prove thath:XY defined byhx: f2x−4fxis quartic, we need to show that

h 2xy

h 2x−y

4h xy

4h xy

24hx−6h y

, 2.23

for allx, yX. Replacingxandyby 2xand 2yin2.18, respectively, we obtain f

2

2xy f

2

2x−y 4f

2 xy

4f 2

xy 2

f4x−4f2x

−6f 2y

, 2.24

(8)

for allx, yX. But, since g2x 4gxfor allxX, where g : XY is a quadratic function defined above, we see that

f4x 20f2x−64fx, 2.25

for allxX. Hence, according to2.24and2.25, we get f

2

2xy f

2

2x−y 4f

2 xy

4f 2

xy 32

f2x−4fx

−6f 2y

, 2.26

for allx, yX. By multiplying 4 on both sides of2.18, we get that 4f

2xy 4f

2x−y 16f

xy 16f

xy 8

f2x−4fx

−24f y

,

2.27

for allx, yX. If we subtract the last equation from2.26, then we arrive at f

2

2xy

−4f 2xy

f 2

2x−y

−4f 2x−y 4

f 2

xy

−4f xy

4 f

2 xy

−4f xy 24

f2x−4fx

−6 f

2y

−4f y

,

2.28

for allx, yX. This means that hsatisfies2.23and, therefore, the functionh : XY is quartic. Thus, there exists a unique symmetric biquadratic functionB2 :X×XY and a unique symmetric biadditive functionB1 : X ×XY such thathx 12B2x, xand gx −12B1x, xfor allxXsee8,13. Therefore, we obtain from2.2that

fx 1

12hx− 1

12gx B2x, x B1x, x, 2.29 for allxX.

The proof of the converse is trivial.

3. Generalized Hyers-Ulam Stability

From this point on, assume thatX is a quasinormed space with quasinorm · Xand thatY is ap-Banach space withp-norm · Y. LetMbe the modulus of concavity of · Y.

Before taking up the main subject, given a mapping f : XY, we define the difference operatorDf :X×XY by

Df

x, y : f

kxy f

kxy

k2f xy

k2f xy

k2 k2−1

6

f2x−4fx 2

k2−1 f

y ,

3.1

for allx, yX. Letϕpx, y: ϕx, ypfor notational convenience.

(9)

Theorem 3.1. Letj ∈ {−1,1}be fixed and letϕq:X×X → 0,∞be a function such that

nlim→ ∞4njϕq x

2nj, y 2nj

0, 3.2

for allx, yXand

i 1j/2

4ipjϕpq u

2ij, y 2ij

<∞, 3.3

for allu, y∈ {x,0,2x,0,x, x,x, kx:xX}. Suppose that an even functionf :XY withf0 0 satisfies the inequality

Df

x, y

Yϕq

x, y

, 3.4

for allx, yX. Then, there exists a unique quadratic functionQ:XYsuch that f2x−16fx−Qx

YM2 4

ψqx1/p

, 3.5

for allxX, where

ψqx:

i 1j/2

4ipj k2pk2−1p

12k2p ϕpq

x 2ij, x

2ij

12

k2−1p ϕpq

x 2ij,0

6pϕpq 2x

2ij,0

12pϕpq x

2ij,kx 2ij

.

3.6

Proof. Letj 1. Settingy 0 in3.4, we have

2fkx−2k2fxk2 k2−1

6

f2x−4fx

Y

ϕqx,0, 3.7

for allxX. Puttingy xin3.4, we obtain

fk1x fk−1x−k2f2xk2 k2−1

6

f2x−4fx 2

k2−1 fx

Y

ϕqx, x,

3.8 for allxX. Replacingxby 2xin3.7, we see that

2f2kx−2k2f2xk2 k2−1

6

f4x−4f2x

Y

ϕq2x,0, 3.9

(10)

for allxX. Settingybykxin3.4and using the evenness off, we get

f2kxk2fk1x−k2fk−1x 2 k2−1

fkxk2 k2−1

6

f2x−4fx

Y

ϕqx, kx,

3.10

for allxX. It follows from3.9and3.10that

k2f2x k2 k2−1

12

f4x−4f2x

k2fk1x−k2fk−1x

2 k2−1

fkx− k2 k2−1

6

f2x−4fx

Y

M 1

2ϕq2x,0 ϕqx, kx

,

3.11

for allxX. Also, it follows from3.7and3.8that

k2fk1x k2fk−1x−2 k2−1

fkxk4f2x

k2 k2−1

6 f2x−4fx 4k2

k2−1 fx

Y

M

k2ϕqx, x k2−1

ϕqx,0 , 3.12

for allxX. Finally, using3.11and3.12, we obtain that f4x−20f2x 64fx

Y

M2 k2k2−1

12k2ϕqx, x 12 k2−1

ϕqx,0 6ϕq2x,0 12ϕqx, kx M2ψqx,

3.13

where

ψqx: 1 k2k2−1

12k2ϕqx, x 12 k2−1

ϕqx,0 6ϕq2x,0 12ϕqx, kx ,

3.14 for allxX. Letg :XY be a function defined bygx: f2x−16fxfor allxX.

From3.13, we conclude that

g2x−4gxYM2ψqx, 3.15

(11)

for allxX. If we replacexin3.15byx/2n1and multiply both sides of3.15by 4n, then we get

4n1g x

2n1

−4ngx 2n

Y

M24nψq x

2n1

, 3.16

for allxXand all non-negative integersn. SinceYis ap-Banach space, the inequality3.16 gives

4n1g x

2n1

−4mgx 2m

p

Y

n

i m

4i1g x

2i1

−4ig x

2i p

Y

M2p n i m

4ipψqp x

2i1

, 3.17 for all nonnegative integersnandmwithnmand allxX. Since 0< p≤1, by Lemma1.2 and3.14, we conclude that

ψpqx≤ 1 k2pk2−1p

12k2p

ϕpqx, x 12

k2−1p

ϕpqx,0 6pϕpq2x,0 12pϕpqx, kx , 3.18

for allxX. Therefore, it follows from3.3and3.18that

i 1

4ipψqp x

2i

<∞, 3.19

for allxX. It follows from3.17and3.19that the sequence{4ngx/2n}is a Cauchy for allxX. SinceY is complete, the sequence{4ngx/2n}converges for allxX. So one can define a functionQ:XY by

Qx lim

n→ ∞4ngx 2n

, 3.20

for allxX. Lettingm 0 and passing the limitn → ∞in3.17, we get gx−QxpYM2p

i 0

4ipψqp x

2i1

M2p 4p

i 1

4ipψpq x

2i

, 3.21

for allxX. Thus3.5follows from3.18and3.21. Now we show thatQis quadratic. It follows from3.16,3.19and3.20that

Q2x−4QxY lim

n→ ∞

4ng x

2n−1

−4n1gx 2n

Y

4 lim

n→ ∞

4n−1g x

2n−1

−4ngx 2n

Y

M2lim

n→ ∞4nψq

x 2n

0,

3.22

(12)

for allxX. So,

Q2x 4Qx, 3.23

for allxX. On the other hand, it follows from3.2,3.4and3.20that DQ

x, y

Y lim

n→ ∞4nDg

x 2n, y

2n

Y lim

n→ ∞4n Df

x 2n−1, y

2n−1

−16Df

x 2n, y

2n

Y

Mlim

n→ ∞4n Df

x 2n−1, y

2n−1

Y

16Df

x 2n, y

2n

Y

Mlim

n→ ∞4n

ϕq

x 2n−1, y

2n−1

16ϕq

x 2n, y

2n

0,

3.24 for all x, yX. Hence the function Q satisfies1.7. Thus, by Theorem 2.1, the function x Q2x−16Qxis quadratic. Therefore,3.23implies that the functionQis quadratic.

Now, to prove the uniqueness property ofQ, let Q : XY be another quadratic function satisfying3.5. It follows from3.3that

nlim→ ∞4np

i 1

4ipϕpq u

2ni, y 2ni

nlim→ ∞

i n1

4ipϕpq u

2i,y 2i

0, 3.25

for allu, y∈ {x,0,2x,0,x, x,x, kx:xX}. Hence,

nlim→ ∞4npψq

x 2n

0, 3.26

for allxX. It follows from3.5,3.20and3.26that QxQxp

Y lim

n→ ∞4npgx 2n

Qx 2n

p

YM2p 4p lim

n→ ∞4npψq

x 2n

0, 3.27

for allxX. SoQ Q.

Forj −1, we can prove the theorem by a similar argument.

Corollary 3.2. Letθ, r, sbe nonnegative real numbers such thatr,s >2 orr,s <2. Suppose that an even functionf:XY withf0 0 satisfies the inequality

Df

x, y

Yθ

xrXys

X

, 3.28

for allx, yX. Then there exists a unique quadratic functionQ:XYsatisfying f2x−16fx−Qx

YM2θ

k2k2−1γqx, 3.29

(13)

for allxX, where

γqx

⎜⎝12p

k2p k2−1p

2r−1p1

|4p−2rp| xrpX 12p

k2pksp

|4p−2sp| xspX

⎟⎠

1/p

. 3.30

Proof. In Theorem3.1, puttingϕqx, y: θxrXysXfor allx, yX, we get the desired result.

Corollary 3.3. Letθ0 andr, s >0 be real numbers such thatλ: rs /2. Suppose that an even functionf:XY withf0 0 satisfies the inequality

Df x, y

YθxrXys

X, 3.31

for allx, yX. Then there exists a unique quadratic functionQ:XYsatisfying f2x−16fx−Qx

YM2θ

k2k2−1 12p

k2pksp 4p−2λp

1/p

xλX, 3.32

for allxX.

Proof. In Theorem3.1, takingϕqx, y: θxrXysX, for allx, yX, we arrive at the desired result.

Theorem 3.4. Letj ∈ {−1,1}be fixed and letϕv:X×X → 0,∞be a function such that

nlim→ ∞16njϕv

x 2nj, y

2nj

0, 3.33

for allx, yXand

i 1j/2

16ipjϕpv u

2ij, y 2ij

<∞, 3.34

for allu, y∈ {x,0,2x,0,x, x,x, kx:xX}. Suppose that an even functionf :XY withf0 0 satisfies the inequality

Dfx,y

Yϕv

x, y

, 3.35

for allx, yX. Then there exists a unique quartic functionV :XY such that f2x−4fx−Vx

YM2 16

ψvx1/p

, 3.36

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for allxX, where

ψvx:

i 1j/2

16ipj k2pk2−1p

12k2p ϕpv

x 2ij, x

2ij

12

k2−1p ϕpv

x 2ij,0

6pϕpv

2x 2ij,0

12pϕpv

x 2ij,kx

2ij

.

3.37

Proof. Being similar to the proof of Theorem3.1, we omit its proof.

Corollary 3.5. Letθ, r, sbe nonnegative real numbers such thatr, s >4 orr, s <4. Suppose that an even functionf :XY withf0 0 satisfies the inequality3.28for allx, yX. Then there exists a unique quartic functionV :XY satisfying

f2x−4fx−Vx

YM2θ

k2k2−1γvx, 3.38 for allxX, where

γvx

⎜⎝12p k2p

k2−1p

2r−1p 1

|16p−2rp| xrpX 12p

k2pksp

|16p−2sp| xspX

⎟⎠

1/p

, 3.39

for allxX.

Corollary 3.6. Letθ0 andr, s >0 be real numbers such thatλ: rs /4. Suppose that an even functionf :XY withf0 0 satisfies the inequality3.31for allx, yX. Then, there exists a unique quartic functionV :XYsatisfying

f2x−4fx−Vx

YM2θ

k2k2−1 12p

k2pksp 16p−2λp

1/p

xλX, 3.40

for allxX.

Now, we are ready to prove the main theorem concerning the stability problem for 1.7.

Theorem 3.7. Letj ∈ {−1,1}be fixed and letϕ:X×X → 0,∞be a function such that

nlim→ ∞

1−j 2

4njϕ

x 2nj, y

2nj

1j

2

16njϕ x

2nj, y 2nj

0, 3.41

for allx, yXand i 1j/2

1−j 2

4ipjϕp

u 2ij, y

2ij

1j

2

16ipjϕp u

2ij, y 2ij

<∞, 3.42

(15)

for allu, y∈ {x,0,2x,0,x, x,x, kx:xX}. Suppose that an even functionf :XY withf0 0 satisfies the inequality

Dfx, y

Yϕ x, y

, 3.43

for allx, yX. Then, there exists a unique quadratic functionQ : XY and a unique quartic functionV :XY such that

fxQxVx

YM3 192

4

ψqx1/p

ψvx1/p

, 3.44

for allxX, where

ψqx:

i 1j/2

4ipj k2pk2−1p

12k2p

ϕp x

2ij, x 2ij

12

k2−1p

ϕp x

2ij,0

6pϕp 2x

2ij,0

12pϕp x

2ij,kx 2ij

,

ψvx:

i 1j/2

16ipj k2pk2−1p

12k2p ϕp

x 2ij, x

2ij

12

k2−1p ϕp

x 2ij,0

6pϕp 2x

2ij,0

12pϕp x

2ij,kx 2ij

.

3.45

Proof. By Theorems3.1and3.4, there exists a quadratic functionQ0 :XY and a quartic functionV0:XY such that

f2x−16fx−Q0x

YM2 4

ψqx1/p

, f2x−4fx−V0x

YM2 16

ψvx1/p , 3.46

for allxX. Therefore, it follows from3.46that fx 1

12Q0x− 1

12V0x Y

M3 192

4

ψqx1/p

ψvx1/p

, 3.47

for allxX. Thus we obtain3.44by lettingQx −1/12Q0xandVx 1/12V0x for allxX.

To prove the uniqueness property ofQandV, letQ, V:XYbe another quadratic and quartic functions satisfying3.44. LetQ QQandV VV. Hence,

Qx Vx

YM fxQxVx

Y fxQx−Vx

Y

!

M4 96

4

ψqx1/p

ψvx1/p ,

3.48

(16)

for allxX. Since limn→ ∞4npjψqx/2n limn→ ∞16npjψvx/2n 0, for allxX, we figure out that

nlim→ ∞16nQx 2n

Vx 2n

Y 0, 3.49

for allxX. Therefore, we getV 0 and thenQ 0.

Corollary 3.8. Letθ, r, sbe nonnegative real numbers such thatr, s >4 or 2< r,s <4 orr, s <2.

Suppose that an even functionf : XY with f0 0 satisfies the inequality3.28, for all x, yX. Then, there exists a unique quadratic functionQ:XY and a unique quartic function V :XY such that

fxQxVx

YM3θ

12k2k2−1

γqx γvx

, 3.50

for allxX, whereγqxandγvxare defined as in Corollaries3.2and3.5.

Corollary 3.9. Letθ0 andr, s >0 be non-negative real numbers such thatλ : rs∈0,2∪ 2,4∪4,∞. Suppose that an even function f : XY withf0 0 satisfies the inequality 3.31for allx, yX. Then there exist a unique quadratic functionQ:XYand a unique quartic functionV :XY such that

fxQxVx

Y

M3θ 12k2k2−1

⎧⎨

⎩ 12p

k2pksp 4p−2λp

1/p

12p

k2pksp 16p−2λp

1/p

⎭xλX, 3.51

for allxX.

Corollary 3.10. Suppose that an even functionf:XYwithf0 0 satisfies the inequality Df

x, y

Xε, 3.52

for allx, yXwhereε >0. Then there exist a unique quadratic functionQ:XY and a unique quartic functionV :XYsuch that

fxQxVx

Y

M3ε k2k2−1

⎧⎨

k2p

k2−1p

2−p1 4p−1

1/p

k2p k2−1p2−p1 16p−1

1/p

, 3.53 for allxX.

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Acknowledgment

Hark-Mahn Kim was supported by Basic Research Program through the National Research Foundation of Korea funded by the Ministry of Education, Science and Technologyno. 2011- 0002614.

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