STABILITY OF EULER-LAGRANGE-JENSEN ’ S ( a , b )-SEXTIC FUNCTIONAL EQUATION IN MULTI-BANACH SPACES

In this paper, we prove the Hyers-Ulam Stability of Euler-Lagrange-Jensen’s (a, b)-Sextic Functional Equation in Multi-Banach Spaces.


Introduction and Preliminaries
The theory of stability is an important branch of the qualitative theory of functional equations.The concept of stability for a functional equation arises when one replaces a functional equation by an inequality which acts as a perturbation of the equation.The first stability problem of functional equation was raised by S.M. Ulam [17] about seventy seven years ago.Since then, this question has attracted the attention of many of authors, and there are many interesting results concerning this problem ( [3, 6, 7, 9, 11-13, 15, 16, 18, 19]).
The Hyers-Ulam stability of functional equation is investigated and the investigation is following.Here, we establish the Hyers-Ulam Stability of Euler-Lagrange-Jensen's (a, b)-Sextic Functional Equation is of the form where a = b such that k ∈ R, h = a + b = 0, ±1 in Multi-Banach Spaces by using direct and fixed point method.
x for each x ∈ A, and the following axioms are satisfied for each k ∈ N with k ≥ 2 : (1) In this case, we say that (A k , .k ) : k ∈ N is a multi -normed space.
Suppose that (A k , .k ) : k ∈ N is a multi -normed space, and take k ∈ N. We need the following two properties of multi -norms.They can be found in [2].
In this case, (A k , .k ) : k ∈ N is a multi -Banach space.

Stability of Functional Equation (1.1) in Multi-Banach Spaces: Direct Method
Theorem 2.1.Let X be a linear space and Then there exists a unique sextic mapping S : Proof.Letting y i = x i where i = 1, 2, ...k in (2.1), we arrive at Now, Replacing x i by 2x i where i = 1, 2, .., k and dividing by 2 in above equation, we get By using induction for a positive integer n, we obtain (2.5) Now, we have to show that the sequence f (2 n hx) 2 n h 6 is a Cauchy sequence, by fixing x ∈ X and replacing Using the definition of Multi-norm, we arrive at (2.6) Hence the above inequality (2.6), shows that f (2 n hx) Therefore, as n → ∞, the inequality (2.5) implies the inequality (2.2).Obviously, one can find the uniqueness of the mapping S : X → Y, using the definition of multi-norm.That is, we can prove S = S .
Corollary 2.1.Let X be a linear space and ((Y n , .n ) : n ∈ N ) be a multi-Banach space.Let f : X → Y be a mapping satisfying f (0) = 0 such that for all x 1 , .., x k , y 1 , .., y k ∈ X.Then there exists a unique sextic mapping S : X → Y such that for all x 1 , .., x k ∈ X.
Proof.Proof is similar to that of Theorem 2.1 by replacing the condition φ(x 1 , y 1 , ..., x k , y k ) in place of .
Corollary 2.2.Let X be a linear space and ((Y n , .n ) : n ∈ N ) be a multi-Banach space.Let 0 < p < 6 , θ ≥ 0 and f : X → Y be a mapping satisfying f (0) = 0 such that for all x 1 , .., x k , y 1 , .., y k ∈ X.Then there exists a unique sextic mapping S : X → Y such that Proof.Proof is similar to that of Theorem 2.1 by replacing the condition Proof.Letting y i = x i where i = 1, 2, ...k in (2.1), we arrive at Let Ψ = {l : X → Y |l(0) = 0} and introduce the generalized metric d defined on Ψ by Then it is easy to show that Ψ, d is a generalized complete metric space, See [8].
We define an operator J : Ψ → Ψ by We assert that J is a strictly contractive operator.Given l, m ∈ Ψ, let Ψ ∈ [0, ∞] be an arbitary constant with d(l, m) ≤ Ψ.From the definition if follows that sup Therefore, sup k∈N (J l(x 1 ) − J m(x 1 ), ..., J l( This Means that J is strictly contractive operator on Ψ with the Lipschitz constant L = 1 h 6 .By (3.3), we have d(J f, f ) ≤ 2h 6 .Applying the Theorem 2.2 in [10], we deduce the existence of a fixed point of J that is the existence of mapping S : X → Y such that Moreover, we have d (J n f, S) → 0, which implies S(x) = lim Hence the proof.

n→∞J 1 h
n f (x) = lim n→∞ f (h n x) h 6nfor all x ∈ X.Also, d(f, S) ≤ 1 1 − L d(J f, f ) implies the inequality ≤ 2(h 6 − 1) .Doing x 1 =, ..., = x k = h n x,and y 1 =, ..., = y k = h n y in (1.1) and dividing by h 6n .Now, applying the property (a) of multi-norms, we have DS(x, y) = lim n→∞ 6n Df (h n x, h n y) x, y ∈ X.The uniqueness of S follows from the fact that S is the unique fixed point of J with the property that there exists ∈ (0, ∞) such that sup k∈N (f (x 1 ) − S(x 1 ), ..., f (x k ) − S(x k )) k ≤ for all x 1 , ..., x k ∈ X.