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Box 35195-363, Semnan, Iran 2 Department of Mathematics, Payame Noor University of Mashhad, Mashhad, Iran 3 Section of Mathematics and Informatics, Pedagogical Department, National and C

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Volume 2009, Article ID 826130, 17 pages

doi:10.1155/2009/826130

Research Article

Solution and Stability of a Mixed Type Additive, Quadratic, and Cubic Functional Equation

M Eshaghi Gordji,1 S Kaboli Gharetapeh,2 J M Rassias,3

and S Zolfaghari1

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

2 Department of Mathematics, Payame Noor University of Mashhad, Mashhad, Iran

3 Section of Mathematics and Informatics, Pedagogical Department, National and Capodistrian

University of Athens, 4 Agamemnonos St., Aghia Paraskevi, Athens 15342, Greece

Correspondence should be addressed to M Eshaghi Gordji,madjid.eshaghi@gmail.com

Received 24 January 2009; Revised 13 April 2009; Accepted 26 June 2009

Recommended by Patricia J Y Wong

We obtain the general solution and the generalized Hyers-Ulam-Rassias stability of the mixed type

additive, quadratic, and cubic functional equation fx 2y − fx − 2y 2fx y − fx − y 2f 3y − 6f2y 6fy.

Copyrightq 2009 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

1 Introduction

The stability problem of functional equations originated from a question of Ulam1 in 1940, concerning the stability of group homomorphisms LetG1, · be a group, and let G2,∗ be

a metric group with the metric d·, · Given > 0, does there exist a δ > 0, such that if a mapping h : G1 → G2 satisfies the inequality dhx · y, hx ∗ hy < δ for all x, y ∈ G1,

then there exists a homomorphism H : G1 → G2with dhx, Hx < for all x ∈ G1? In other words, under what condition does there exist a homomorphism near an approximate homomorphism?

In 1941, Hyers2 gave a first aﬃrmative answer to the question of Ulam for Banach

spaces Let f : E → Ebe a mapping between Banach spaces such that

f

x y− fx − fy ≤ δ, 1.1

for all x, y ∈ E and for some δ > 0 Then there exists a unique additive mapping T : E → E

such that

fx − Tx ≤ δ, 1.2

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for all x ∈ E Moreover if ftx is continuous in t for each fixed x ∈ E, then T is linear see

also3 In 1950, Aoki 4 generalized Hyers’ theorem for approximately additive mappings

In 1978, Th M Rassias5 provided a generalization of Hyers’ theorem which allows the Cauchy diﬀerence to be unbounded This new concept is known as Hyers-Ulam-Rassias stability of functional equationssee 2 24

The functional equation

f

x y fx − y 2fx 2fy

1.3

is related to symmetric biadditive function In the real case it has f x x2 among its solutions Thus, it has been called quadratic functional equation, and each of its solutions

is said to be a quadratic function Hyers-Ulam-Rassias stability for the quadratic functional equation1.3 was proved by Skof for functions f : A → B, where A is normed space and B

Banach spacesee 25–28

The following cubic functional equation was introduced by the third author of this paper, J M Rassias29,30 in 2000-2001:

f

x 2y 3fx 3fx y fx − y 6fy

Jun and Kim13 introduced the following cubic functional equation:

f

2x y f2x − y 2fx y 2fx − y 12fx, 1.5

and they established the general solution and the generalized Hyers-Ulam-Rassias stability for the functional equation1.5

The function fx x3satisfies the functional equation1.5, which explains why it is called cubic functional equation

Jun and Kim proved that a function f between real vector spaces X and Y is a solution

of1.5 if and only if there exists a unique function C : X × X × X → Y such that fx Cx, x, x for all x ∈ X, and C is symmetric for each fixed one variable and is additive for

fixed two variablessee also 31–33

We deal with the following functional equation deriving from additive, cubic and quadratic functions:

f

x 2y− fx − 2y 2f

x y− fx − y 2f3y

− 6f2y

6fy

. 1.6

It is easy to see that the function fx ax3 bx2 cx is a solution of the functional

equation1.6 In the present paper we investigate the general solution and the generalized Hyers-Ulam-Rassias stability of the functional equation1.6

2 General Solution

In this section we establish the general solution of functional equation1.6

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Theorem 2.1 Let X,Y be vector spaces, and let f : X → Y be a function Then f satisfies 1.6 if and only if there exists a unique additive function A : X → Y, a unique symmetric and biadditive function Q : X × X → Y, and a unique symmetric and 3-additive function C : X × X × X → Y such that fx Ax Qx, x Cx, x, x for all x ∈ X.

Proof Suppose that fx Ax Qx, x Cx, x, x for all x ∈ X, where A : X → Y is additive, Q : X × X → Y is symmetric and biadditive, and C : X × X × X → Y is symmetric and 3-additive Then it is easy to see that f satisfies1.6 For the converse let f satisfy 1.6

We decompose f into the even part and odd part by setting

f e x  1

2

f x f−x, f o x  1

2

fx − f−x, 2.1

for all x ∈ X By 1.6, we have

f e

x 2y− f e

x − 2y

1

2

f

x 2y f−x − 2y− fx − 2y− f−x 2y

1

2

f

x 2y− fx − 2y1

2

f

−x −2y− f−x −−2y

1

2

2f

x y− 2fx − y 2f3y

− 6f2y

6fy

1

2

2f

−x − y− 2f−x y 2f−3y− 6f−2y 6f−y

2

1

2

f

x y f−x − y− 2

1 2

f

x − y f−x y

2

1

2

f

3y

f−3y− 6

1 2

f

2y

f−2y 6

1 2

f

y

f−y

2f e

x y− f e

x − y 2f e

3y

− 6f e

2y

6f e

y

,

2.2

for all x, y ∈ X This means that f esatisfies1.6, that is,

f e

x 2y− f e

x − 2y 2f e

x y− f e

x − y 2f e

3y

− 6f e

2y

6f e

y

. 2.3

Now putting x y 0 in 2.3, we get f e 0 0 Setting x 0 in 2.3, by evenness of f ewe obtain

3f e

2y

 f e

3y

3f e

y

Replacing x by y in2.3, we obtain

4f e

2y

 f e

3y

7f e

y

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Comparing2.4 with 2.5, we get

f e

3y

 9f e

y

By utilizing2.5 with 2.6, we obtain

f e

2y

 4f e

y

Hence, according to2.6 and 2.7, 2.3 can be written as

f e

x 2y− f e

x − 2y 2f e

x y− 2f e

x − y. 2.8

With the substitution x : x y, y : x − y in 2.8, we have

f e

3x − y− f e

x − 3y 8f e x − 8f e

y

Replacing y by −y in above relation, we obtain

f e

3x y− f e

x 3y 8f e x − 8f e

y

Setting x y instead of x in 2.8, we get

f e

x 3y− f e

x − y 2f e

x 2y− 2f e x. 2.11

Interchanging x and y in2.11, we get

f e

3x y− f e

x − y 2f e

2x y− 2f e

y

. 2.12

If we subtract2.12 from 2.11 and use 2.10, we obtain

f e

x 2y− f e

2x y 3f e

y

− 3f e x, 2.13

which, by putting y : 2y and using 2.7, leads to

f e

x 4y− 4f e

x y 12f e

y

− 3f e x. 2.14

Let us interchange x and y in2.14 Then we see that

f e

4x y− 4f e

x y 12f e x − 3f e

y

and by adding2.14 and 2.15, we arrive at

f e

x 4y f e

4x y 8f e

x y 9f e x 9f e

y

. 2.16

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Replacing y by x y in 2.8, we obtain

f e

3x 2y− f e

x 2y 2f e

2x y− 2f e

y

. 2.17

Let us Interchange x and y in2.17 Then we see that

f e

2x 3y− f e

2x y 2f e

x 2y− 2f e x. 2.18 Thus by adding2.17 and 2.18, we have

f e

2x 3y f e

3x 2y 3f e

x 2y 3f e

2x y− 2f e x − 2f e

y

. 2.19

Replacing x by 2x in2.11 and using 2.7 we have

f e

2x 3y− f e

2x − y 8f e

x y− 8f e x, 2.20

and interchanging x and y in2.20 yields

f e

3x 2y− f e

x − 2y 8f e

x y− 8f e

y

. 2.21

If we add2.20 to 2.21, we have

f e

2x 3y f e

3x 2y f e

2x − y f e

x − 2y 16f e

x y− 8f e x − 8f e

y

.

2.22

f e

2x y− f e

2x − y 2f e

x y− 2f e

x − y, 2.23 and by adding the last equation and2.8 with 2.19, we get

f e

2x 3y f e

3x 2y− f e

2x − y− f e

x − 2y

 2f e

x 2y 2f e

2x y 4f e

x y− 4f e

x − y− 2f e x − 2f e

y

. 2.24 Now according to2.22 and 2.24, it follows that

f e

x 2y f e

2x y 6f e

x y 2f e

x − y− 3f e x − 3f e

y

. 2.25

From the substitution y −y in 2.25 it follows that

f e

x − 2y f e

2x − y 6f e

x − y 2f e

x y− 3f e x − 3f e

y

. 2.26

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Replacing y by 2y in2.25 we have

f e

x 4y 4f e

x y 6f e

x 2y 2f e

x − 2y− 3f e x − 12f e

y

, 2.27

and interchanging x and y yields

f e

4x y 4f e

x y 6f e

2x y 2f e

2x − y− 12f e x − 3f e

y

. 2.28

By adding2.27 and 2.28 and then using 2.25 and 2.26, we lead to

f e

x 4y f e

4x y 32f e

x y 24f e

x − y− 39f e x − 39f e

y

. 2.29

If we compare2.16 and 2.29, we conclude that

f e

x y f e

x − y 2f e x 2f e

y

This means that f e is quadratic Thus there exists a unique quadratic function Q : X × X → Y such that f e x Qx, x, for all x ∈ X On the other hand we can show that f osatisfies1.6, that is,

f o

x 2y− f o

x − 2y 2f o

x y− f o

x − y 2f o

3y

− 6f o

2y

6f o

y

. 2.31

Now we show that the mapping g : X → Y defined by gx : f o 2x − 8f o x is additive and the mapping h : X → Y defined by hx : f o 2x − 2f o x is cubic Putting x 0 in

2.31, then by oddness of f o , we have

4f o

2y

 5f o

y

f o

3y

Hence2.31 can be written as

f o

x 2y− f o

x − 2y 2f o

x y− 2f o

x − y 2f o

2y

− 4f o

y

. 2.33

From the substitution y : −y in 2.33 it follows that

f o

x − 2y− f o

x 2y 2f o

x − y− 2f o

x y− 2f o

2y

4f o

y

. 2.34

Interchange x and y in2.33, and it follows that

f o

2x y f o

2x − y 2f o

x y 2f o

x − y 2f o 2x − 4f o x. 2.35

With the substitutions x : x − y and y : x y in 2.35, we have

f o

3x − y f o

x − 3y 2f o

2x − 2y− 4f o

x − y 2f o 2x − 2f o

2y

. 2.36

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Replace x by x − y in 2.34 Then we have

f o

x − 3y− f o

x y 2f o

x − 2y− 2f o x − 2f o

2y

4f o

y

. 2.37

Replacing y by −y in 2.37 gives

f o

x 3y− f o

x − y 2f o

x 2y− 2f o x 2f o

2y

− 4f o

y

. 2.38

f o

3x y f o

x − y 2f o

2x y− 2f o

y

2f o 2x − 4f o x. 2.39

f o

x 3y f o

3x y

 2f o

x 2y 2f o

2x y 2f o 2x 2f o

2y

− 6f o x − 6f o

y

. 2.40

f o

x 3y f o

3x y 2f o

2x 2y− 4f o

x y 2f o 2x 2f o

2y

. 2.41

By comparing2.40 with 2.41, we arrive at

f o

x 2y f o

2x y f o

2x 2y− 2f o

x y 3f o x 3f o

y

. 2.42

f o

x − 2y f o

2x − y f o

2x − 2y− 2f o

x − y 3f o x − 3f o

y

. 2.43

With the substitution y : x y in 2.43, we have

f o

x − y− f o

x 2y −f o

2y

− 3f o

x y 3f o x 2f o

y

, 2.44 and replacing−y by y gives

f o

x y− f o

x − 2y f o

2y

− 3f o

x − y 3f o x − 2f o

y

. 2.45

Let us interchange x and y in2.45 Then we see that

f o

x y f o

2x − y f o 2x 3f o

x − y− 2f o x 3f o

y

. 2.46

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f o

2x − y− f o

x − 2y f o 2x − 2f o

x y f o x f o

2y

f o

y

. 2.47 Adding2.42 to 2.47 and using 2.33 and 2.35, we obtain

f o

2

x y− 8f o

x yf o 2x − 8f o xf o

2y

− 8f o

y

, 2.48

for all x, y ∈ X The last equality means that

g

x y gx gy

for all x, y ∈ X Therefore the mapping g : X → Y is additive With the substitutions x : 2x and y : 2y in 2.35, we have

f o

4x 2y f o

4x − 2y 2f o

2x 2y 2f o

2x − 2y 2f o 4x − 4f o 2x. 2.50

Let g : X → Y be the additive mapping defined above It is easy to show that f ois

cubic-additive function Then there exists a unique function C : X × X × X → Y and a unique additive function A : X → Y such that f o x Cx, x, x Ax, for all x ∈ X, and C is symmetric and 3-additive Thus for all x ∈ X, we have

f x f e x f o x Qx, x Cx, x, x Ax. 2.51 This completes the proof of theorem

The following corollary is an alternative result ofTheorem 2.1

Corollary 2.2 Let X,Y be vector spaces, and let f : X → Y be a function satisfying 1.6 Then the following assertions hold.

a If f is even function, then f is quadratic.

b If f is odd function, then f is cubic-additive.

3 Stability

We now investigate the generalized Hyers-Ulam-Rassias stability problem for functional equation 1.6 From now on, let X be a real vector space, and let Y be a Banach space Now before taking up the main subject, given f : X → Y, we define the diﬀerence operator

D f : X × X → Y by

D f

x, y

fx2y−fx−2y−2f

xy−fx−y−2f3y

6f2y

− 6fy

, 3.1

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for all x, y ∈ X We consider the following functional inequality:

D f

x, y ≤ φx,y, 3.2

for an upper bound φ : X × X → 0, ∞.

Theorem 3.1 Let s ∈ {1, −1} be fixed Suppose that an even mapping f : X → Y satisfies f0 0,

and

D fx, y ≤ φx,y, 3.3

for all x, y ∈ X If the upper bound φ : X × X → 0, ∞ is a mapping such that

∞

i0

4si

φ 2−si x, 2 −si x

1

2φ 0, 2

−si x

<∞ 3.4

and that

lim

n 4sn φ

2−sn x, 2 −sn y

for all x, y ∈ X, then the limit

Qx : lim

n 4sn f

2−sn x

3.6

exists for all x ∈ X, and Q : X → Y is a unique quadratic function satisfying 1.6, and

fx − Qx ≤ 1

8

∞

is1/2

4si

φ 2−si x, 2 −si x

1

2φ 0, 2

−si x , 3.7

for all x ∈ X.

Proof Let s 1 Putting x 0 in 3.3, we get

2

f

3y

− 3f2y

3fy ≤ φ0,y, 3.8

for all y ∈ X On the other hand by replacing y by x in 3.3, it follows that

−f3y 4f2y − 7fy ≤ φy,y, 3.9

for all y ∈ X Combining 3.8 and 3.9, we lead to

2f

2y

− 8fy ≤ 2φy,y φ0,y, 3.10

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for all y ∈ X With the substitution y : x/2 in 3.10 and then dividing both sides of inequality by 2, we get

fx − 4f

2

≤ 1 2

2φ

2,

x

2

φ 0, x

2

Now, using methods similar as in 8, 34,35, we can easily show that the function

Q : X → Y defined by Qx lim n→ ∞4n fx/2 n for all x ∈ X is unique quadratic function

satisfying1.6 and 3.7 Let s −1 Then by 3.10 we have

f 2x4 − fx

≤ 182φx, x φ0, x, 3.12

for all x ∈ X And analogously, as in the case s 1, we can show that the function Q : X →

Y defined by Qx : lim n→ ∞4−n f2n x is unique quadratic function satisfying 1.6 and

3.7

Theorem 3.2 Let s ∈ {1, −1} be fixed Let φ : X × X → 0, ∞ is a mapping such that

∞

i1

2si

φ

x

2si , x

2si1 φ

0, x

2si1 <∞ 3.13

and that

lim

n→ ∞2sn φ

2sn , y

2sn

for all x, y ∈ X.

Suppose that an odd mapping f : X → Y satisfies

D fx, y ≤ φx,y, 3.15

for all x, y ∈ X.

Then the limit

Ax : lim

n→ ∞2sn

f

x

2sn−1 − 8f

2sn

3.16

exists, for all x ∈ X, and A : X → Y is a unique additive function satisfying 1.6, and

f2x − 8fx − Ax ≤ ∞

i|s−1|/2

2si φ

x

2si , x

2si1 2 ∞

i|s−1|/2

2si φ

0, x

2si1 , 3.17

for all x ∈ X.

for all y ∈ X With the substitution y : x/2 in 3.10 and then dividing both sides of inequality by 2, we get

...

for all x ∈ X And analogously, as in the case s 1, we can show that the function Q : X →

Y defined by Qx : lim n→ ∞4−n f2n...

2sn

3.16

exists, for all x ∈ X, and A : X → Y is a unique additive function satisfying 1.6, and< /i>

f2x − 8fx − A x ≤ ∞

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