Some hyperstability and stability results for the Cauchy and Jensen equations

In this paper we give some hyperstability and stability results for the Cauchy and Jensen functional equations on restricted domains. We provide a simple and short proof for Brzdȩk’s result concerning a hyperstability result for the Cauchy equation.


Introduction
Let V and W be linear spaces. A function f : V → W is called • additive if f (x + y) = f (x) + f (y) for all x, y ∈ V ; • Jensen if 2f ( x+y 2 ) = f (x) + f (y) for all x, y ∈ V . The main motivation for the investigation of the stability of functional equations originated from a question of Ulam [21] concerning the stability of group homomorphisms. Hyers [9] gave an affirmative answer to the question of Ulam. The stability and hyperstability problems for various functional equations have been investigated by numerous mathematicians. For more information on this area of research and further references, see [1,2,4,7,10,11,[13][14][15]20].
Let us state the following theorem that is one of the classical results concerning the stability problem for the Cauchy functional equation f (x + y) = f (x) + f (y). Theorem 1.1 ([3, 5, 8, 9, 18, 19]) Let ε ≥ 0 and f : X → Y , where X is a normed space and Y is a Banach space. Let p = 1 be a real number and Then, there exists a unique additive function A : X → Y such that Rassias [16,17] considered the case f (x + y)f (x)f (y) ≤ ε x p y q , where p, q are real numbers with p + q ∈ [0, 1). Brzdȩk [6,Theprem 1.3] provided a complement for this result in the case p + q < 0. His proof was based on a fixed-point theorem. We provide a simple and short proof for Brzdȩk's result. In addition, some further results on the hyperstability of the Cauchy and Jensen functional equations are investigated.

Superstability
Denote by N the set of positive integers. A version of the following theorem is introduced by Brzdȩk [6, Theorem 1.3] and its proof is based on a fixed-point theorem. A simple and brief proof is given here. Theorem 2.1 Let X and Y be normed spaces, and E ⊆ X \ {0} be a nonempty set. Take ε ≥ 0 and let p, q be real numbers with p + q < 0. Assume that for each x ∈ E there exists a positive integer m x such that nx ∈ E for all n ∈ N with n ≥ m x . Then, every function f : X → Y satisfying the inequality Proof Without loss of generality, we may assume that q < 0. Let x, y ∈ E with x + y ∈ E. By assumption, there exists a positive integer m such that nx, ny, n(x + y) ∈ E for all n ≥ m. Then, (2.1) yields Letting n → ∞ in the above inequalities, we obtain Then, ≤ lim sup n→∞ f (n + 1)(x + y)f (n + 1)xf (n + 1)y ≤ lim sup n→∞ ε (n + 1) p+q + n p+q x p y q = 0.
Hence, f (x + y) = f (x) + f (y) for all x, y ∈ E with x + y ∈ E. This completes the proof.

Remark 2.2
The assumption p + q < 0 is necessary in Theorem 2.1. For example, the func- However, f is not additive.
The following theorem states a hyperstability result for the Jensen functional equation on a restricted domain.

Theorem 2.3
Let X and Y be normed spaces, and E ⊆ X \ {0} be a nonempty set. Take ε ≥ 0 and let p, q be real numbers with p + q < 0. Assume that for each x ∈ E there exists a positive integer m x such that nx 2 ∈ E for all n ∈ N with n ≥ m x . Then, every function f : X → Y satisfying the inequality 2f x is Jensen on E, that is Proof Without loss of generality, we may assume that q < 0. Let x, y ∈ E with x+y 2 ∈ E. By assumption, there exists a positive integer m such that { nx 2 , nx 2 , n(x+y) Letting n → ∞ in the above inequalities, we obtain Then, This ends the proof.
Then, f satisfies (2.2) with p + q > 0. However, f is not Jensen on E.
In the following, we obtain other hyperstability results for the Cauchy and Jensen functional equations.
Theorem 2.5 Let X and Y be normed spaces, and E ⊆ X \ {0} be a nonempty set. Take θ , ε ≥ 0 and let p, q, r be real numbers with p + q + r < 0 and p + q + 2r < 0. Assume that for each x ∈ E there exists a positive integer m x such that nx ∈ E for all n ∈ N with n ≥ m x . Then, every function f : X → Y satisfying the inequality Since p + q + 2r < 0, we may assume that q + r < 0 without loss of generality. Let x, y ∈ E with x + y ∈ E. By assumption, there exists a positive integer m such that nx, ny, n(x + y) ∈ E for all n ≥ m. By a similar argument as in the proof of Theorem 2.1, we obtain Then, This completes the proof.
Example 2.6 Let E = [1, +∞) and f be a function defined by f (x) = x 3 . It is clear that Then, f satisfies (2.3) with p = q = r = 1. However, f is not additive on E.
Theorem 2.7 Let X and Y be normed spaces, and E ⊆ X \ {0} be a nonempty set. Take θ , ε ≥ 0 and let p, q, r be real numbers with p + q + r < 0 and p + q + 2r < 0. Assume that for each x ∈ E there exists a positive integer m x such that nx 2 ∈ E for all n ∈ N with n ≥ m x . Suppose that a function f : X → Y satisfies the inequality 2f x for all x, y ∈ E with x+y 2 ∈ E. Then, f is Jensen on E.
Without loss of generality we may assume that q + r < 0. Let x, y ∈ E with x+y 2 ∈ E. By assumption, there exists a positive integer m such that { nx 2 , ny 2 , n(x+y) 4 } ⊆ E for all n ≥ m. By a similar argument as in the proof of Theorem 2.3, we obtain Then, Then, f satisfies (2.4) with p = q = 2 and r = 1. However, f is not Jensen on E.

Theorem 2.9
Assume that X is a linear space over the field F, and Y is a normed space over the field K. Let a, b ∈ F \ {0} and ϕ : X × X → [0, +∞) be a function such that for all x, y ∈ E d = {z ∈ X : z ≥ d} for some d > 0. Then, f satisfies

7)
for all x, y ∈ X. Moreover, for all x ∈ X.
In the following corollaries X and Y are normed spaces. (i) p + q + r < 0 and f (ax + by) -Af (x) -Bf (y) -C ≤ x p y q ε x + y r + θ xy r ; (ii) p + r < 0, q + r < 0 and (iii) p, q < 0 and for all x, y ∈ E d = {z ∈ X : z ≥ d} for some d > 0.

Corollary 2.11
Every function f : X → Y satisfies one of the following assertions: -C x r y s = +∞ for all real numbers r, s with r + s > 0.

Corollary 2.12
Every function f : X → Y satisfies one of the following assertions: x r y s x r + y s f (ax + by) -Af (x) -Bf (y) -C = +∞ for all real nonnegative numbers r, s.

Stability on restricted domains
Jung [12] proved the stability of Jensen's functional equation on a restricted and unbounded domain. In the following theorem, we improve the bound and thus the result of Jung [12] by obtaining sharper estimates. Theorem 3.1 Let X be a normed space and Y a Banach space. Take ε ≥ 0 and let a function f : X → Y satisfy the inequality 2f x for all x, y ∈ E d = {z ∈ X : z ≥ d} for some d > 0. Then, there exists a unique additive function T : Proof Letting y = -x in (3.1), we obtain

2)
Letting y = -3x in (3.1), we obtain Now, adding (3.2) and (3.3), we have Then, It is easy to see that It is clear that T(0) = 0 and T(3x) = 3T(x) for all x ∈ X . In view of the definition of T, (3.1) yields Putting y = 3x in (3.6) and using T(3x) = 3T(x), we infer that T(2x) = 2T(x) for all x ∈ X . Hence, (3.6) implies that T is Jensen (additive) on X . Letting m = 0 and taking the limit as n → ∞ in (3.5), one obtains To extend (3.7) to the whole X , let z ∈ X \ {0} and choose a positive integer n such that nz ≥ d. Take x = 2(n + 1)z and y = -2nz. Then, (3.7) yields Using these inequalities together with (3.1), we obtain Since T is Jensen and z = x+y 2 , we obtain This inequality is valid for z = 0 because of T(0) = 0. The uniqueness of T follows easily from the last inequality.
Proof Letting y = 0 in (3.1), we obtain 2f It is easy to see that x ≥ d. Let T e and T o be the even part and the odd part of T. Then, T e and T o satisfy (3.9) for all x, y ∈ X with x + y = 0. Since T o is odd, (3.9) yields that T o is additive on X . It follows from (3.9) that 2T( x 2 ) = T(x) for all x = 0, and then T e (xy) = 2T e xy 2 = T e (x) + T e (y) = 2T e x + y 2 = T e (x + y), x ± y = 0.
Putting y = 3x and using T e (2x) = 2T e (x), we infer that T e (x) = 0 for all x ∈ X . Hence, (3.9) implies that T is Jensen (additive) on X . Letting m = 0 and taking the limit as n → ∞ in (3.8), one obtains