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In this paper, we investigate the Hyers–Ulam–Rassias stability of generalized Jensen type quadratic functional equations in Banach spaces

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Volume 20, No. 4, December 2007

STABILITY OF A GENERALIZED JENSEN TYPE QUADRATIC FUNCTIONAL EQUATIONS

Kil-Woung Jun* and Young-Sun Cho**

Abstract. In this paper, we investigate the Hyers–Ulam–Rassias stability of generalized Jensen type quadratic functional equations in Banach spaces.

1. Introduction

In 1940, S. M. Ulam [1] gave a talk before the Mathematics Club of the University of Wisconsin in which he discussed a number of un- solved problems. Among these was the following question concerning the stability of homomorphisms.

Let G be a group and let G0 be a metric group with metric ρ(·, ·).

Given ² > 0, does there exist a δ > 0 such that if f : G → G0 satisfies ρ(f (xy), f (x)f (y)) < δ for all x, y ∈ G, then a homomorphism h : G → G0 exists with ρ(f (x), h(x)) < ² for all x ∈ G?

In 1941, D. H. Hyers [2] considered the case of an approximately additive mapping f : E → E0, where E and E0 are Banach spaces and f satisfies Hyers inequality

kf (x + y) − f (x) − f (y)k ≤ ²

for all x, y ∈ E. It was shown that the limit L(x) = limn→∞f (22nnx)

exists for all x ∈ E and that L : E → E0 is the unique additive mapping satisfying

kf (x) − L(x)k ≤ ².

Let E1 and E2 be real vector spaces. A function f : E1 → E2, there exists a quadratic function if and only if f is a solution function of the

Received October 30, 2007.

2000 Mathematics Subject Classification: Primary 39B62; Secondary 39B82.

Key words and phrases: Jensen type quadratic mapping, generalized Hyers–

Ulam–Rassias stability, functional inequality.

This work was supported by the second Brain Korea 21 Project in 2006.

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quadratic functional equation

f (x + y) − f (x − y) = 2f (x) + 2f (y).

(1.1)

A stability problem for the quadratic functional equation (1.1) was solved by F. Skof [3] for a mapping f : E1 → E2, where E1 is a normed space and E2 is a Banach space.

In 1978, Th. M. Rassias [4] provided a generalization of Hyers’ The- orem which allows the Cauchy difference to be unbounded.

S. Czerwik [5] proved the Hyers-Ulam-Rassias stability of quadratic functional equation (1.1). Let E1 and E2 be a real normed space and a real Banach space, respectively, and let p 6= 2 be a positive constant. If a function f : E1→ E2 satisfies the inequality

kf (x + y) − f (x − y) − 2f (x) − 2f (y)k ≤ ²(kxkp+ kykp)

for some ² > 0 and for all x, y ∈ E1, then there exists a unique quadratic function q : E1 → E2 such that

kf (x) − q(x)k ≤

|4 − 2p|kxkp for all x ∈ G.

A mapping g : X → Y is called a Jensen mapping if g satisfies the functional equation

2g(x + y

2 ) = g(x) + g(y) for all x, y ∈ X.

Jun and Lee [6] proved the following : Let X and Y be Banach spaces.

Denote by ϕ : X \ {0} × X \ {0} → [0, ∞) function such that ψ(x, y) =

X j=0

3−kϕ(3kx, 3ky) < ∞

for all x, y ∈ X \ {0}. Suppose that f : X → Y is a mapping satisfying k2f (x + y

2 ) − f (x) − f (y)k ≤ ϕ(x, y)

for all x, y ∈ X \ {0}. Then there exists a unique additive mapping T : X → Y such that

kf (x) − f (0) − T (x)k ≤ 1

3(ψ(x, −x) + ψ(−x, 3x))

for all x ∈ X \ {0}. Recently C. Park and W. Park [7] applied the Jun and Lee’s result to the Jensen’s equation in Banach modules over a C-algebra.

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Now, we consider the following functional equation f (x + y

a + bz) + f (x + y

a − bz) + f (x − y

a + bz) + f (x − y a − bz)

= 4

a2f (x) + 4

a2f (y) + 4b2f (z), (1.2)

where a, b 6= 0 are real numbers.

In this paper, we will establish the general solution and the gener- alized Hyers-Ulam-Rassias stability problem for the equation (1.2) in Banach spaces.

2. Jesen type quadratic mapping in Banach spaces

Lemma 2.1. Let X and Y be vector spaces. If a mapping f : X → Y satisfies f (0) = 0 and

f (x + y

a + bz) + f (x + y

a − bz) + f (x − y

a + bz) + f (x − y a − bz)

= 4

a2f (x) + 4

a2f (y) + 4b2f (z) (2.1)

for all x, y, z ∈ X, then the mapping f is quadratic.

Proof. Letting x = y in (2.1), we get f (2x

a + bz) + f (2x

a − bz) + f (bz) + f (−bz)

= 8

a2f (x) + 4b2f (z) (2.2)

for all x, z ∈ X. Letting x = 0 in (2.2), we get 2f (bz) + 2f (−bz) = 4b2f (z). Setting y = −x in (2.1), we obtain

f (2x

a + bz) + f (2x

a − bz) + f (bz) + f (−bz)

= 4

a2f (x) + 4

a2f (−x) + 4b2f (z).

(2.3)

By (2.2) and (2.3), we conclude that f is even. And by setting z = 0 in (2.2), we get f (2xa) = a42f (x) for all x ∈ X. So, we get

f (2x

a + bz) + f (2x

a − bz) = 2f (2x

a ) + 2f (bz) for all x, z ∈ X. Hence f is quadratic.

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The mapping f : X → Y given in the statement of Lemma 2.1 is called a generalized Jensen type quadratic mapping. Putting z = 0 in (2.1) with a = 2, we get the Jensen type quadratic mapping 2f (x+y2 ) + 2f (x−y2 ) = f (x) + f (y), and putting x = y in (2.1) with a = 2 and b = 1, we get the quadratic mapping f (x + z) + f (x − z) = 2f (x) + 2f (z).

From now on, let X and Y be a normed vector space and a Banach space ,respectively.

For a given mapping f : X → Y , we define Df (x, y, z) := f (x + y

a + bz) + f (x + y

a − bz) + f (x − y a + bz) +f (x − y

a − bz) − 4

a2f (x) − 4

a2f (y) − 4b2f (z) for all x, y, z ∈ X

Theorem 2.2. Let f : X → Y be a mapping satisfying f (0) = 0 for which there exists a function φ : X3 → [0, ∞) such that

Φ(x, y, z) :=

X j=1

1 2

¡ 4 a2

¢j φ¡¡ a

2

¢j x,¡ a

2

¢j y,¡ a

2

¢j z¢

< ∞, (2.4)

kDf (x, y, z)k ≤ φ(x, y, z) (2.5)

for all x, y, z ∈ X. Then there exists a unique quadratic mapping Q : X → Y such that DQ(x, y, z) = 0 and

kQ(x) − f (x)k ≤ a2

4 Φ(x, x, 0) (2.6)

for all x ∈ X.

Proof. Letting x = y and z = 0 in (2.5), we get kf¡ 2

ax¢

4

a2f (x)k ≤ 1

2φ(x, x, 0) for all x ∈ X. So

kf (x) − 4 a2f¡ a

2x¢ k ≤ 1

2φ¡ a 2x,a

2x, 0¢

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for all x ∈ X. Hence k¡ 4

a2

¢l f¡¡ a

2

¢l x¢

¡ 4 a2

¢m f¡¡ a

2

¢m x¢

k

Xm j=l+1

k¡ 4 a2

¢j−1 f¡¡ a

2

¢j−1 x¢

¡ 4 a2

¢j f¡¡ a

2

¢j x¢

k

Xm j=l+1

¡ 4 a2

¢j−11 2φ¡¡ a

2

¢j x,¡ a

2

¢j x, 0¢

for all x ∈ X. It means that a sequence {(a42)nf ((a2)nx)} is Cauchy for all x ∈ X. Since Y is complete, the sequence {(a42)nf ((a2)nx)} converges. So one can define a mapping Q : X → Y by Q(x) := limn→∞(a42)nf ((a2)nx) for all x ∈ X.

By (2.4) and (2.5), kDQ(x, y, z)k = lim

n→∞

¡ 4 a2

¢n

kDf¡¡ a 2

¢n x,¡ a

2

¢n y,¡ a

2

¢n z¢

k (2.7)

≤ lim

n→∞

¡ 4 a2

¢n φ¡¡ a

2

¢n x,¡ a

2

¢n y,¡ a

2

¢n z¢

= 0 (2.8)

for all x, y, z ∈ X. So DQ(x, y, z) = 0. By Lemma 2.1, the mapping Q : X → Y is a quadratic mapping.

Moreover, letting l = 0 and passing the limit m → ∞ in (2.7), we get the approximation (2.6) of f by Q.

Now, let Q0 : X −→ Y be another quadratic mapping satisfying (2.6).

Then we obtain

kQ(x) − Q0(x)k =¡ 4 a2

¢n kQ¡¡ a

2

¢n x¢

− Q¡¡ a 2

¢n x¢

k

¡ 4 a2

¢n£ kQ¡¡ a

2

¢n x¢

− f¡¡ a 2)nx¢

k + kQ0¡¡ a 2

¢n x¢

− f¡¡ a 2

¢n x¢¢

k¤

≤ 2(4

a2)n−1Φ¡¡ a 2

¢n x,¡ a

2

¢n x, 0¢

,

which tends to zero as n → ∞. So we can conclude that Q(x) = Q0(x) for all x ∈ X. This proves the uniqueness of Q. Hence the mapping Q : X → Y is a unique quadratic mapping satisfying (2.6).

Corollary 2.3. Let p and θ be positive real numbers with p > 2 and 0 < |a| < 2. Let f : X → Y be a mapping satisfying f (0) = 0 and

kDf (x, y, z)k ≤ θ(kxkp+ kykp+ kzkp) (2.9)

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for all x, y, z ∈ X. Then there exists a unique quadratic mapping Q : X → Y such that

kf (x) − Q(x)k ≤ |a|p· θ

4(2p−2− |a|p−2)kxkp (2.10)

for all x ∈ X.

Proof. Define φ(x, y, z) = θ(kxkp+ kykp+ kzkp), and apply Theorem 2.2.

Corollary 2.4. Let p and θ be positive real numbers with p < 2 and |a| > 2. Let f : X → Y be a mapping satisfying f (0) = 0 and

kDf (x, y, z)k ≤ θ(kxkp+ kykp+ kzkp) (2.11)

for all x, y, z ∈ X. Then there exists a unique quadratic mapping Q : X → Y such that

kf (x) − Q(x)k ≤ |a|2· θ

2p(|a|2−p− 22−p)kxkp (2.12)

for all x ∈ X.

Proof. Define φ(x, y, z) = θ(kxkp+ kykp+ kzkp), and apply Theorem 2.2.

Theorem 2.5. Let f : X → Y be a mapping satisfying f (0) = 0 for which there exists a function φ : X3 → [0, ∞) such that

Φ(x, y, z) :=

X j=1

1 2

¡ a2 4

¢j φ¡¡ 2

a

¢j−1 x,¡ 2

a

¢j−1 y,¡ 2

a

¢j−1 z¢

< ∞, (2.13)

kDf (x, y, z)k ≤ φ(x, y, z) (2.14)

for all x, y, z ∈ X. Then there exists a unique quadratic mapping Q : X → Y such that DQ(x, y, z) = 0 and

kQ(x) − f (x)k ≤ Φ(x, x, 0) (2.15)

for all x ∈ X.

Proof. Letting x = y and z = 0 in (2.5), we get kf¡ 2

ax¢

4

a2f (x)k ≤ 1

2φ(x, x, 0) (2.16)

for all x ∈ X. So

kf (x) −a2 4 f¡ 2

ax¢

k ≤¡ a2 4

¢ 1

2φ(x, x, 0) (2.17)

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for all x ∈ X.

Hence k¡ a2

4

¢l f¡¡ 2

a

¢l x¢

−¡ a2 4

¢m f¡¡ 2

a

¢m x¢

k

Xm j=l+1

¡ a2 4

¢j1 2φ¡¡ 2

a

¢j−1 x,¡ 2

a

¢j−1 x, 0¢ (2.18)

for all x ∈ X. It means that a sequence {(a42)nf ((2a)nx)} is Cauchy for all x ∈ X. Since Y is complete, the sequence {(a42)nf ((2a)nx)} converges. So one can define a mapping Q : X → Y by Q(x) := limn→∞(a42)nf ((a2)nx) for all x ∈ X.

By (2.13) and (2.14), kDQ(x, y, z)k = lim

n→∞

¡ a2 4

¢n

kDf¡¡ 2 a

¢n x,¡ 2

a

¢n y,¡ 2

a)nz¢ (2.19)

≤ lim

n→∞

¡ a2 4

¢n φ¡¡ 2

a

¢n x,¡ 2

a

¢n y,¡ 2

a

¢n z¢

= 0 (2.20)

for all x, y, z ∈ X. So DQ(x, y, z) = 0. By Lemma 2.1, the mapping Q : X → Y is a quadratic mapping.

Moreover, letting l = 0 and passing the limit m → ∞ in (2.18), we get the approximation (2.15) of f by Q.

Now, let Q0 : X −→ Y be another quadratic mapping satisfying (2.15) . Then we obtain

kQ(x) − Q0(x)k =¡ a2 4

¢n kQ¡¡ 2

a

¢n x¢

− Q¡¡ 2 a

¢n x¢

k

≤¡ a2 4

¢n£ kQ¡¡ 2

a

¢n x¢

− f¡¡ 2 a

¢n x¢

k + kQ0¡¡ 2 a

¢n x¢

− f¡¡ 2 a

¢n x¢¢

k¤

≤ 2¡ a2 4

¢n Φ¡¡ 2

a

¢n x,¡ 2

a

¢n x, 0¢

,

which tends to zero as n → ∞. So we can conclude that Q(x) = Q0(x) for all x ∈ X. This proves the uniqueness of Q. Hence the mapping Q : X → Y is a unique quadratic mapping satisfying (2.15).

Corollary 2.6. Let p and θ be positive real numbers with p > 2 and |a| > 2. Let f : X → Y be a mapping satisfying f (0) = 0 and

kDf (x, y, z)k ≤ θ(kxkp+ kykp+ kzkp) (2.21)

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for all x, y, z ∈ X. Then there exists a unique quadratic mapping Q : X → Y such that

kf (x) − Q(x)k ≤ |a|p· θ

4(|a|p−2− 2p−2)kxkp (2.22)

for all x ∈ X.

Proof. Define φ(x, y, z) = θ(kxkp+ kykp+ kzkp), and apply Theorem 2.5.

Corollary 2.7. Let p and θ be positive real numbers with p < 2 and |a| < 2. Let f : X → Y be a mapping satisfying f (0) = 0 and

kDf (x, y, z)k ≤ θ(kxkp+ kykp+ kzkp) (2.23)

for all x, y, z ∈ X. Then there exists a unique Jensen type quadratic- quadratic mapping Q : X → Y such that

kf (x) − Q(x)k ≤ |a|2· θ

2p(22−p− |a|2−p)kxkp (2.24)

for all x ∈ X.

Proof. Define φ(x, y, z) = θ(kxkp+ kykp+ kzkp), and apply Theorem 2.5.

References

[1] S. M. Ulam, A Collection of the Mathematical Problems, Interscience Publ.

New York, 1960.

[2] D. H. Hyers, On the stability of the linear functional equation, Proc. Nat. Acad.

Sci. U.S.A. 27 (1941), 222–224.

[3] F. Skof, Local properties and approximations of operators(Italian), Rend. Sem.

Mat. Fis. Milano, 53 (1983), 113–129.

[4] Th. M. Rassias, On the stability of the linear mapping in Banach spaces, Proc.

Amer. Math. Soc. 72 (1978), 297–300.

[5] S. Czerwik, On the stability of the mapping in normed spaces, Abh. Math. Sem.

Univ. Hamburg, 62(1992), 59–64.

[6] K. Jun and Y. Lee, A generalization of the Hyers–Ulam–Rassias stability of Jensen’s equation, J. Math. Anal. Appl. 238 (1999), 305–315.

[7] C. Park, W. Park, On the Jensen’s equation in Banach modules, Taiqanese J.

Math. 6 (2002), 523–300.

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*

Department of Mathematics Chungnam National University Daejeon 305–764, Republic of Korea E-mail: kwjun@@math.cnu.ac.kr

**

Department of Mathematics Chungnam National University Daejeon 305–764, Republic of Korea E-mail: yscho@@cnu.ac.kr

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