ADDITIVE ρ-FUNCTIONAL INEQUALITIES

Abstract

In this paper, we solve the additive ρ-functional inequalities (0.1)${\parallel}f(x+y)+f(x-y)-2f(x){\parallel}$ $\leq$ ${\parallel}{\rho}(2f(\frac{x+y}{2})+f(x-y)-2f(x)){\parallel}$, where ρ is a fixed complex number with |ρ| < 1, and (0.2) ${\parallel}2f(\frac{x+y}{2})+f(x-y)-2f(x)){\parallel}$ $\leq$ ${\parallel}{\rho}f(x+y)+f(x-y)-2f(x){\parallel}$, where ρ is a fixed complex number with |ρ| < 1. Furthermore, we prove the Hyers-Ulam stability of the additive ρ-functional inequalities (0.1) and (0.2) in complex Banach spaces.

1. INTRODUCTION AND PRELIMINARIES

The stability problem of functional equations originated from a question of Ulam [11] concerning the stability of group homomorphisms.

The functional equation f(x+y) = f(x) + f(y) is called the Cauchy equation. In particular, every solution of the Cauchy equation is said to be an additive mapping. Hyers [6] gave a first affirmative partial answer to the question of Ulam for Banach spaces. Hyers’ Theorem was generalized by Aoki [1] for additive mappings and by Rassias [8] for linear mappings by considering an unbounded Cauchy difference. A generalization of the Rassias theorem was obtained by Gǎvruta [5] by replacing the unbounded Cauchy difference by a general control function in the spirit of Rassias’ approach.

The stability of quadratic functional equation was proved by Skof [10] for mappings f : E1 → E2, where E1 is a normed space and E2 is a Banach space. Cholewa [3] noticed that the theorem of Skof is still true if the relevant domain E1 is replaced by an Abelian group. See [2, 4, 7, 9, 12] for more information on the stability problems of functional equations.

In Section 2, we solve the additive ρ-functional inequality (0.1) and prove the Hyers-Ulam stability of the additive ρ-functional inequality (0.1) in complex Banach spaces.

In Section 3, we solve the additive ρ-functional inequality (0.2) and prove the Hyers-Ulam stability of the additive ρ-functional inequality (0.2) in complex Banach spaces.

Throughout this paper, let G be a 2-divisible abelian group. Assume that X is a real or complex normed space with norm ∥ · ∥ and that Y is a complex Banach space with norm ∥ · ∥.

 

2. ADDITIVE ρ-FUNCTIONAL INEQUALITY (0.1)

Throughout this section, assume that ρ is a fixed complex number with |ρ| < 1.

In this section, we solve and investigate the additive ρ-functional inequality (0.1) in complex Banach spaces.

Lemma 2.1. If a mapping f : G → Y satisfies f(0) = 0 and

for all x, y ∈ G, then f : G → Y is additive.

Proof. Assume that f : G → Y satisfies (2.1).

Letting y = x in (2.1), we get ∥f(2x) − 2f(x)∥ ≤ 0 and so f(2x) = 2f(x) for all x ∈ G. Thus

for all x ∈ G.

It follows from (2.1) and (2.2) that

and so f(x + y) + f(x − y) = 2f(x) for all x, y ∈ G. It is easy to show that f is additive.                                      □

We prove the Hyers-Ulam stability of the additive ρ-functional inequality (2.1) in complex Banach spaces.

Theorem 2.2. Let r > 1 and θ be nonnegative real numbers, and let f : X → Y be a mapping satisfying f(0) = 0 and

for all x, y ∈ X. Then there exists a unique additive mapping h : X → Y such that

for all x ∈ X.

Proof. Letting y = x in (2.3), we get

for all x ∈ X. So

for all x ∈ X. Hence

for all nonnegative integers m and l with m > l and all x ∈ X. It follows from (2.6) that the sequence is a Cauchy sequence for all x ∈ X. Since Y is complete, the sequence converges. So one can define the mapping h : X → Y by

for all x ∈ X. Moreover, letting l = 0 and passing the limit m → ∞ in (2.6), we get (2.4).

It follows from (2.3) that

for all x, y ∈ X. So

for all x, y ∈ X. By Lemma 2.1, the mapping h : X → Y is additive.

Now, let T : X → Y be another additive mapping satisfying (2.4). Then we have

which tends to zero as n → ∞ for all x ∈ X. So we can conclude that h(x) = T(x) for all x ∈ X. This proves the uniqueness of h. Thus the mapping h : X → Y is a unique additive mapping satisfying (2.4).                                      □

Theorem 2.3. Let r < 1 and θ be positive real numbers, and let f : X → Y be a mapping satisfying f(0) = 0 and (2.3). Then there exists a unique additive mapping h : X → Y such that

for all x ∈ X.

Proof. It follows from (2.5) that

for all x ∈ X. Hence

for all nonnegative integers m and l with m > l and all x ∈ X. It follows from (2.8) that the sequence is a Cauchy sequence for all x ∈ X. Since Y is complete, the sequence is a Cauchy sequence for all x converges. So one can define the mapping h : X → Y by

for all x ∈ X. Moreover, letting l = 0 and passing the limit m → ∞ in (2.8), we get (2.7).

The rest of the proof is similar to the proof of Theorem 2.2.                                       □

Remark 2.4. If ρ is a real number such that −1 < ρ < 1 and Y is a real Banach space, then all the assertions in this section remain valid.

 

3. ADDITIVE ρ-FUNCTIONAL INEQUALITY (0.2)

Throughout this section, assume that ρ is a fixed complex number with |ρ| < 1.

In this section, we solve and investigate the additive ρ-functional inequality (0.2) in complex Banach spaces.

Lemma 3.1. If a mapping f : G → Y satisfies

for all x, y ∈ G, then f : G → Y is additive.

Proof. Assume that f : G → Y satisfies (3.1).

Letting x = y = 0 in (3.1), we get ∥f(0)∥ ≤ 0. So f(0) = 0.

Letting y = 0 in (3.1), we get and so

for all x ∈ G.

It follows from (3.1) and (3.2) that

and so f(x + y) + f(x − y) = 2f(x) for all x, y ∈ G. . It is easy to show that f is additive.                                       □

We prove the Hyers-Ulam stability of the additive ρ-functional inequality (3.1) in complex Banach spaces.

Theorem 3.2. Let r > 1 and θ be nonnegative real numbers, and let f : X → Y be a mapping such that

for all x, y ∈ X. Then there exists a unique additive mapping h : X → Y such that

for all x ∈ X.

Proof. Letting x = y = 0 in (3.3), we get ∥f(0)∥ ≤ 0. So f(0) = 0.

Letting y = 0 in (3.3), we get

for all x ∈ X. So

for all nonnegative integers m and l with m > l and all x ∈ X. It follows from (3.6) that the sequence is a Cauchy sequence for all x ∈ X. Since Y is complete, the sequence converges. So one can define the mapping h : X → Y by

for all x ∈ X. Moreover, letting l = 0 and passing the limit m → ∞ in (3.6), we get (3.4).

It follows from (3.3) that

for all x, y ∈ X. So

for all x, y ∈ X. By Lemma 3.1, the mapping h : X → Y is additive.

Now, let T : X → Y be another additive mapping satisfying (3.4). Then we have

which tends to zero as n → ∞ for all x ∈ X. So we can conclude that h(x) = T(x) for all x ∈ X. This proves the uniqueness of h. Thus the mapping h : X → Y is a unique additive mapping satisfying (3.4).                                       □

Theorem 3.3. Let r < 1 and θ be positive real numbers, and let f : X → Y be a mapping satisfying (3.3). Then there exists a unique additive mapping h : X → Y such that

for all x ∈ X.

Proof. It follows from (3.5) that

for all x ∈ X. Hence

for all nonnegative integers m and l with m > l and all x ∈ X. It follows from (3.8) that the sequence is a Cauchy sequence for all x ∈ X. Since Y is complete, the sequence converges. So one can define the mapping h : X → Y by

for all x ∈ X. Moreover, letting l = 0 and passing the limit m → ∞ in (3.8), we get (3.7).

The rest of the proof is similar to the proof of Theorem 3.2.                                      □

Remark 3.4. If ρ is a real number such that −1 < ρ < 1 and Y is a real Banach space, then all the assertions in this section remain valid.