Journal of Applied Mathematics

Journal of Applied Mathematics / 2012 / Article

Research Article | Open Access

Volume 2012 |Article ID 691981 | 12 pages | https://doi.org/10.1155/2012/691981

Stability of Jensen-Type Functional Equations on Restricted Domains in a Group and Their Asymptotic Behaviors

Academic Editor: Junjie Wei
Received20 Jun 2011
Accepted05 Sep 2011
Published29 Oct 2011

Abstract

We consider the Hyers-Ulam stability problems for the Jensen-type functional equations in general restricted domains. The main purpose of this paper is to find the restricted domains for which the functional inequality satisfied in those domains extends to the inequality for whole domain. As consequences of the results we obtain asymptotic behavior of the equations.

1. Introduction

The Hyers-Ulam stability problems of functional equations was originated by Ulam in 1960 when he proposed the following question [1].

Let 𝑓 be a mapping from a group 𝐺1 to a metric group 𝐺2 with metric 𝑑(⋅,⋅) such that 𝑑(𝑓(𝑥𝑦),𝑓(𝑥)𝑓(𝑦))≤𝜀.(1.1)Then does there exist a group homomorphism ℎ and 𝛿𝜖>0 such that 𝑑(𝑓(𝑥),ℎ(𝑥))≤𝛿𝜖(1.2)for all 𝑥∈𝐺1?

One of the first assertions to be obtained is the following result, essentially due to Hyers [2], that gives an answer for the question of Ulam.

Theorem 1.1. Suppose that 𝑆 is an additive semigroup, 𝑌 is a Banach space, 𝜖≥0, and 𝑓∶𝑆→𝑌 satisfies the inequality ‖𝑓(𝑥+𝑦)−𝑓(𝑥)−𝑓(𝑦)‖≤𝜖(1.3) for all 𝑥,𝑦∈𝑆. Then there exists a unique function 𝐴∶𝑆→𝑌 satisfying 𝐴(𝑥+𝑦)=𝐴(𝑥)+𝐴(𝑦)(1.4) for which ‖𝑓(𝑥)−𝐴(𝑥)‖≤𝜖(1.5) for all 𝑥∈𝑆.

We call the functions satisfying (1.4) additive functions. Perhaps Aoki in 1950 was the first author treating the generalized version of Hyers’ theorem [3]. Generalizing Hyers’ result he proved that if a mapping 𝑓∶𝑋→𝑌 between two Banach spaces satisfies ‖𝑓(𝑥+𝑦)−𝑓(𝑥)−𝑓(𝑦)‖≤Φ(𝑥,𝑦)for𝑥,𝑦∈𝑋(1.6) with Φ(𝑥,𝑦)=𝜖(‖𝑥‖𝑝+‖𝑦‖𝑝)(𝜖≥0,0≤𝑝<1), then there exists a unique additive function 𝐴∶𝑋→𝑌 such that ‖𝑓(𝑥)−𝐴(𝑥)‖≤2𝜖‖𝑥‖𝑝/(2−2𝑝) for all 𝑥∈𝑋. In 1951 Bourgin [4, 5] stated that if Φ is symmetric in ‖𝑥‖ and ‖𝑦‖ with âˆ‘âˆžğ‘—=1Φ(2𝑗𝑥,2𝑗𝑥)/2𝑗<∞ for each 𝑥∈𝑋, then there exists a unique additive function 𝐴∶𝑋→𝑌 such that ∑‖𝑓(𝑥)−𝐴(𝑥)â€–â‰¤âˆžğ‘—=1Φ(2𝑗𝑥,2𝑗𝑥)/2𝑗 for all 𝑥∈𝑋. Unfortunately, there was no use of these results until 1978 when Rassias [6] dealt with the inequality of Aoki [3]. Following Rassias’ result, a great number of papers on the subject have been published concerning numerous functional equations in various directions [6–15]. Among the results, stability problem in a restricted domain was investigated by Skof, who proved the stability problem of inequality (1.3) in a restricted domain [16, 17]. Developing this result, Jung, Rassias, and M. J. Rassias considered the stability problems in restricted domains for some functional equations including the Jensen functional equation [9] and Jensen-type functional equations [13]. We also refer the reader to [18–27] for some related results on Hyers-Ulam stabilities in restricted conditions. The results can be summarized as follows. Let 𝑋 and 𝑌 be a real normed space and a real Banach space, respectively. For fixed 𝑑≥0, if 𝑓∶𝑋→𝑌 satisfies the functional inequalities (such as that of Cauchy, quadratic, Jensen, and Jensen type) for all 𝑥,𝑦∈𝑋 with ‖𝑥‖+‖𝑦‖≥𝑑 (which is the case where the inequalities are given by two indeterminate variables 𝑥 and 𝑦), the inequalities hold for all 𝑥,𝑦∈𝑋. Following the approach in [28] we consider the Jensen-type equation in various restricted domains in an Abelian group. As applications, we obtain the stability problems for the above equations in more general restricted domains than that of the form {(𝑥,𝑦)∈𝑋∶‖𝑥‖+‖𝑦‖≥𝑑}, which generalizes and refines the stability theorems in [13]. As an application we obtain asymptotic behaviors of the equations.

2. Stability of Jensen-Type Functional Equations

Throughout this section, we denote by 𝐺, 𝑋, and 𝑌, an Abelian group, a real normed space, and a Banach space, respectively. In this section we consider the Hyers-Ulam stability of the Jensen and Jensen-type functional inequalities for the functions 𝑓∶𝐺→𝑌‖𝑓(𝑥+𝑦)+𝑓(𝑥−𝑦)−2𝑓(𝑥)‖≤𝜖,(2.1)‖𝑓(𝑥+𝑦)−𝑓(𝑥−𝑦)−2𝑓(𝑦)‖≤𝜖(2.2) in restricted domains 𝑈⊂𝐺×𝐺.

Inequalities (2.1) and (2.2) were previously treated by J. M. Rassias and M. J. Rassias [13], who proved the Hyers-Ulam stability of the inequalities in the restricted domain 𝑈={(𝑥,𝑦)∶‖𝑥‖+‖𝑦‖≥𝑑},𝑑≥0, for the functions 𝑓∶𝑋→𝑌:

Theorem 2.1. Let 𝑑≥0 and 𝜖>0 be fixed. Suppose that 𝑓∶𝑋→𝑌 satisfies the inequality ‖𝑓(𝑥+𝑦)+𝑓(𝑥−𝑦)−2𝑓(𝑥)‖≤𝜖(2.3) for all 𝑥,𝑦∈𝑋, with ‖𝑥‖+‖𝑦‖≥𝑑. Then there exists a unique additive function 𝐴∶𝑋→𝑌 such that 5‖𝑓(𝑥)−𝐴(𝑥)−𝑓(0)‖≤2𝜖(2.4) for all 𝑥∈𝑋.

Theorem 2.2. Let 𝑑≥0 and 𝜖>0 be fixed. Suppose that 𝑓∶𝑋→𝑌 satisfies the inequality ‖𝑓(𝑥+𝑦)−𝑓(𝑥−𝑦)−2𝑓(𝑦)‖≤𝜖(2.5) for all 𝑥,𝑦∈𝑋, with ‖𝑥‖+‖𝑦‖≥𝑑 and ‖𝑓(𝑥)+𝑓(−𝑥)‖≤3𝜖(2.6) for all 𝑥∈𝑋, with ‖𝑥‖≥𝑑. Then there exists a unique additive function 𝐴∶𝑋→𝑌 such that ‖𝑓(𝑥)−𝐴(𝑥)‖≤332𝜖(2.7) for all 𝑥∈𝑋.

We use the following usual notations. We denote by 𝐺×𝐺={(ğ‘Ž1,ğ‘Ž2)âˆ¶ğ‘Ž1,ğ‘Ž2∈𝐺} the product group; that is, for ğ‘Ž=(ğ‘Ž1,ğ‘Ž2),𝑏=(𝑏1,𝑏2)∈𝐺×𝐺, we define ğ‘Ž+𝑏=(ğ‘Ž1+𝑏1,ğ‘Ž2+𝑏2),ğ‘Žâˆ’ğ‘=(ğ‘Ž1−𝑏1,ğ‘Ž2−𝑏2). For a subset 𝐻 of 𝐺×𝐺 and ğ‘Ž,𝑏∈𝐺×𝐺, we define ğ‘Ž+𝐻={ğ‘Ž+â„Žâˆ¶â„Žâˆˆğ»}.

For given 𝑥,𝑦∈𝐺 we denote by 𝑃𝑥,𝑦,𝑄𝑥,𝑦 the subsets of points of the forms (not necessarily distinct) in 𝐺×𝐺, respectively, 𝑃𝑥,𝑦𝑄={(0,0),(𝑥,−𝑥),(𝑦,𝑦),(𝑥+𝑦,−𝑥+𝑦)},𝑥,𝑦={(−𝑥,𝑥),(𝑦,𝑦),(−𝑥+𝑦,𝑥+𝑦)}.(2.8) The set 𝑃𝑥,𝑦 can be viewed as the vertices of rectangles in 𝐺×𝐺, and 𝑄𝑥,𝑦 can be viewed as a subset of the vertices of rectangles in 𝐺×𝐺.

Definition 2.3. Let 𝑈⊂𝐺×𝐺. One introduces the following conditions (𝐽1) and (𝐽2) on 𝑈. For any 𝑥,𝑦∈𝐺, there exists a 𝑧∈𝐺 such that (𝐽1)(0,𝑧)+𝑃𝑥,𝑦={(0,𝑧),(𝑥,−𝑥+𝑧),(𝑦,𝑦+𝑧),(𝑥+𝑦,−𝑥+𝑦+𝑧)}⊂𝑈,(𝐽2)(𝑧,0)+𝑄𝑥,𝑦={(−𝑥+𝑧,𝑥),(𝑦+𝑧,𝑦),(−𝑥+𝑦+𝑧,𝑥+𝑦)}⊂𝑈,(2.9) respectively.

The sets (0,𝑧)+𝑃𝑥,𝑦,(𝑧,0)+𝑄𝑥,𝑦 can be understood as the translations of 𝑃𝑥,𝑦 and 𝑄𝑥,𝑦 by (0,𝑧) and (𝑧,0), respectively.

There are many interesting examples of the sets 𝑈 satisfying some of the conditions (𝐽1) and (𝐽2). We start with some trivial examples.

Example 2.4. Let 𝐺 be a real normed space. For 𝑑≥0,𝑥0,𝑦0∈𝐺, let 𝑈={(𝑥,𝑦)∈𝐺×𝐺∶𝑘‖𝑥‖+𝑠‖𝑦‖≥𝑑},𝑉={(𝑥,𝑦)∈𝐺×𝐺∶‖𝑘𝑥+𝑠𝑦‖≥𝑑}.(2.10) Then 𝑈 satisfies (𝐽1) if 𝑠>0, (𝐽2) if 𝑘>0 and 𝑉 satisfies (𝐽1) if 𝑠≠0, (𝐽2) if 𝑘≠0.

Example 2.5. Let 𝐺 be a real inner product space. For 𝑑≥0,𝑥0,𝑦0∈𝐺𝑈=(𝑥,𝑦)∈𝐺×𝐺∶⟨𝑥0,𝑥⟩+⟨𝑦0,𝑦⟩≥𝑑.(2.11) Then 𝑈 satisfies (𝐽1) if 𝑦0≠0, (𝐽2) if 𝑥0≠0.

Example 2.6. Let 𝐺 be the group of nonsingular square matrices with the operation of matrix multiplication. For 𝑘,𝑠∈ℝ,𝛿,𝑑≥0, let 𝑃𝑈=1,𝑃2||∈𝐺×𝐺∶det𝑃1||𝑘||det𝑃2||𝑠,𝑃≤𝛿𝑉=1,𝑃2||∈𝐺×𝐺∶det𝑃1||𝑘||det𝑃2||𝑠.≥𝑑(2.12) Then both 𝑈 and 𝑉 satisfy (𝐽1) if 𝑠≠0, (𝐽2) if 𝑘≠0.

In the following one can see that if 𝑃𝑥,𝑦 and 𝑄𝑥,𝑦 are replaced by arbitrary subsets of four points (not necessarily distinct) in 𝐺×𝐺, respectively, then the conditions become stronger; that is, there are subsets 𝑈1 and 𝑈2 which satisfy the conditions (𝐽1) and (𝐽2), respectively, but 𝑈1 and 𝑈2 fail to fulfill the following conditions (2.13) and (2.14), respectively. For any subset {𝑋1,𝑋2,𝑋3,𝑋4} of points (not necessarily distinct) in 𝐺×𝐺, there exists a 𝑧∈𝐺 such that 𝑋(0,𝑧)+1,𝑋2,𝑋3,𝑋4⊂𝑈1(𝑋,(2.13)𝑧,0)+1,𝑋2,𝑋3,𝑋4⊂𝑈2,(2.14) respectively.

Now we give examples of 𝑈1 and 𝑈2 which satisfy (𝐽1) and (𝐽2), respectively, but not (2.13) and (2.14), respectively.

Example 2.7. Let 𝐺=ℤ be the group of integers. Enumerate ğ‘Žâ„¤Ã—â„¤=1,𝑏1,î€·ğ‘Ž2,𝑏2î€¸î€·ğ‘Ž,…,𝑛,𝑏𝑛,…⊂ℝ2(2.15) such that ||ğ‘Ž1||+||𝑏1||≤||ğ‘Ž2||+||𝑏2||||ğ‘Žâ‰¤â‹¯â‰¤ğ‘›||+||𝑏𝑛||≤⋯,(2.16) and let 𝑃𝑛={(0,0),(ğ‘Žğ‘›,âˆ’ğ‘Žğ‘›),(𝑏𝑛,𝑏𝑛),(ğ‘Žğ‘›+𝑏𝑛,âˆ’ğ‘Žğ‘›+𝑏𝑛)},𝑛=1,2,…. Then it is easy to see that ⋃𝑈=âˆžğ‘›=1((0,2𝑛)+𝑃𝑛) satisfies the condition (𝐽1). Now let 𝑃={(𝑝1,ğ‘ž1),(𝑝2,ğ‘ž2)}⊂ℤ×ℤ with |ğ‘ž2âˆ’ğ‘ž1|≤|𝑝2−𝑝1|,𝑝1𝑝2>0. Then (0,𝑧)+𝑃 is not contained in 𝑈 for all 𝑧∈ℤ. Indeed, for any choices of (𝑥𝑛,𝑦𝑛)∈𝑃𝑛+(0,2𝑛),𝑛=1,2,…, we have 𝑦𝑚−𝑦𝑛>|𝑥𝑚−𝑥𝑛| for all 𝑚>𝑛,𝑚,𝑛=1,2,…. Thus, if (0,𝑧)+𝑃⊂𝑈 for some 𝑧∈ℤ, then 𝑃+(0,𝑧)⊂(0,2𝑛)+𝑃𝑛 for some 𝑛∈ℕ. Thus, it follows from the condition ğ‘ž2âˆ’ğ‘ž1≤|𝑝2−𝑝1| that the line segment joining the points of 𝑃+(−𝑧,𝑧) intersects the line 𝑥=0 in ℝ2, which contradicts the condition 𝑝1𝑝2>0. Similarly, let 𝑄𝑛={(âˆ’ğ‘Žğ‘›,ğ‘Žğ‘›),(𝑏𝑛,𝑏𝑛),(âˆ’ğ‘Žğ‘›+𝑏𝑛,ğ‘Žğ‘›+𝑏𝑛)}. Then it is easy to see that ⋃𝑈=âˆžğ‘›=1((2𝑛,0)+𝑄𝑛) satisfies the condition (𝐽2) but not (2.14).

Theorem 2.8. Let 𝑈⊂𝐺×𝐺 satisfy the condition (𝐽1) and 𝜖≥0. Suppose that 𝑓∶𝐺→𝑌 satisfies (2.1) for all (𝑥,𝑦)∈𝑈. Then there exists an additive function 𝐴∶𝐺→𝑌 such that ‖𝑓(𝑥)−𝐴(𝑥)−𝑓(0)‖≤2𝜖(2.17) for all 𝑥∈𝐺.

Proof. For given 𝑥,𝑦∈𝐺, choose a 𝑧∈𝐺 such that (0,𝑧)+𝑃𝑥,𝑦⊂𝑈. Replacing 𝑥 by 𝑥+𝑦, 𝑦 by −𝑥+𝑦+𝑧; 𝑥 by 𝑥, 𝑦 by −𝑥+𝑧; 𝑥 by 𝑦, 𝑦 by 𝑦+𝑧; 𝑥 by 0, 𝑦 by 𝑧 in (2.1), respectively, we have ‖(‖𝑓(2𝑦+𝑧)+𝑓(2𝑥−𝑧)−2𝑓(𝑥+𝑦)‖≤𝜖,‖𝑓(𝑧)+𝑓(2𝑥−𝑧)−2𝑓(𝑥)‖≤𝜖,𝑓(2𝑦+𝑧)+𝑓(−𝑧)−2𝑓(𝑦)‖≤𝜖,‖𝑓𝑧)+𝑓(−𝑧)−2𝑓(0)‖≤𝜖.(2.18) From (2.18), using the triangle inequality and dividing the result by 2, we have ||||𝑓(𝑥+𝑦)−𝑓(𝑥)−𝑓(𝑦)+𝑓(0)≤2𝜖(2.19) for all 𝑥,𝑦∈𝐺. From (2.19), using Theorem 1.1, we get the result.

Let 𝑑≥0,𝑠∈ℝ, and let 𝑈={(𝑥,𝑦)∶‖𝑥‖+𝑠‖𝑦‖≥𝑑}. Then 𝑈 satisfies the condition (𝐽1). Thus, as a direct consequence of Theorem 2.8, we obtain the following (cf. Theorem 2.1).

Corollary 2.9. Let 𝑑≥0,𝑠∈ℝ. Suppose that 𝑓∶𝑋→𝑌 satisfies inequality (2.1) for all 𝑥,𝑦∈𝑋, with ‖𝑥‖+𝑠‖𝑦‖≥𝑑. Then there exists a unique additive function 𝐴∶𝑋→𝑌 such that ‖𝑓(𝑥)−𝐴(𝑥)−𝑓(0)‖≤2𝜖(2.20) for all 𝑥∈𝑋.

Theorem 2.10. Let 𝑈⊂𝐺×𝐺 satisfy the condition (𝐽2) and 𝜖≥0. Suppose that 𝑓∶𝐺→𝑌 satisfies (2.2) for all (𝑥,𝑦)∈𝑈. Then there exists a unique additive function 𝐴∶𝐺→𝑌 such that 3‖𝑓(𝑥)−𝐴(𝑥)‖≤2𝜖(2.21) for all 𝑥∈𝐺.

Proof. For given 𝑥,𝑦∈𝐺, choose a 𝑧∈𝐺 such that (𝑧,0)+𝑄𝑥,𝑦⊂𝑈. Replacing 𝑥 by −𝑥+𝑦+𝑧, 𝑦 by 𝑥+𝑦; 𝑥 by −𝑥+𝑧, 𝑦 by 𝑥; 𝑥 by 𝑦+𝑧, 𝑦 by 𝑦 in (2.2), respectively, we have ‖‖𝑓(2𝑦+𝑧)−𝑓(−2𝑥+𝑧)−2𝑓(𝑥+𝑦)‖≤𝜖,‖𝑓(𝑧)−𝑓(−2𝑥+𝑧)−2𝑓(𝑥)‖≤𝜖,𝑓(2𝑦+𝑧)−𝑓(𝑧)−2𝑓(𝑦)‖≤𝜖.(2.22) From (2.22), using the triangle inequality and dividing the result by 2, we have ||||≤3𝑓(𝑥+𝑦)−𝑓(𝑥)−𝑓(𝑦)2𝜖.(2.23) Now by Theorem 1.1, we get the result.

Let 𝑑≥0,𝑘∈ℝ, and let 𝑈={(𝑥,𝑦)∶𝑘‖𝑥‖+‖𝑦‖≥𝑑}. Then 𝑈 satisfies the condition (J2). Thus, as a direct consequence of Theorem 2.10, we generalize and refine Theorem 2.2 as follows.

Corollary 2.11. Let𝑑≥0,𝑘∈ℝ. Suppose that 𝑓∶𝑋→𝑌 satisfies inequality (2.2) for all 𝑥,𝑦, with 𝑘‖𝑥‖+‖𝑦‖≥𝑑. Then there exists a unique additive function 𝐴∶𝑋→𝑌 such that 3‖𝑓(𝑥)−𝐴(𝑥)‖≤2𝜖(2.24) for all 𝑥∈𝑋.

Remark 2.12. Corollary 2.11 refines Theorem 2.2 in both the bounds and the condition (2.6).

Now we discuss other possible restricted domains. We assume that 𝐺 is a 2-divisible Abelian group. For given 𝑥,𝑦∈𝐺, we denote by 𝑅𝑥,𝑦,𝑆𝑥,𝑦⊂𝐺×𝐺, 𝑅𝑥,𝑦=(𝑥,𝑥),(𝑥,𝑦),𝑥−𝑦2,𝑥−𝑦2,𝑥−𝑦2,−𝑥+𝑦2,𝑆𝑥,𝑦=(𝑥,𝑥),(𝑦,𝑥),𝑥−𝑦2,𝑥−𝑦2,−𝑥+𝑦2,𝑥−𝑦2.(2.25) One can see that 𝑅𝑥,𝑦 and 𝑆𝑥,𝑦 consist of the vertices of parallelograms in 𝐺×𝐺, respectively.

Definition 2.13. Let 𝑈⊂𝐺×𝐺. One introduces the following conditions (𝐽3),(𝐽4) on 𝑈. For any 𝑥,𝑦∈𝐺, there exists a 𝑧∈𝐺 such that (𝐽3)(𝑧,−𝑧)+𝑅𝑥,𝑦=(𝑥+𝑧,𝑥−𝑧),(𝑥+𝑧,𝑦−𝑧),𝑥−𝑦2+𝑧,𝑥−𝑦2,−𝑧𝑥−𝑦2+𝑧,−𝑥+𝑦2−𝑧⊂𝑈,(𝐽4)(𝑧,−𝑧)+𝑆𝑥,𝑦=(𝑥+𝑧,𝑥−𝑧),(𝑦+𝑧,𝑥−𝑧),𝑥−𝑦2+𝑧,𝑥−𝑦2,−𝑧−𝑥+𝑦2+𝑧,𝑥−𝑦2−𝑧⊂𝑈,(2.26) respectively.

Example 2.14. Let 𝐺 be a real normed space. For 𝑘,𝑠,𝑑∈ℝ, let 𝑈={(𝑥,𝑦)∈𝐺×𝐺∶𝑘‖𝑥‖+𝑠‖𝑦‖≥𝑑},𝑉={(𝑥,𝑦)∈𝐺×𝐺∶‖𝑘𝑥+𝑠𝑦‖≥𝑑}.(2.27) Then 𝑈 satisfies (𝐽3) and (𝐽4) if 𝑘+𝑠>0, and 𝑉 satisfies (𝐽3) and (𝐽4) if 𝑘≠𝑠.

Example 2.15. Let 𝐺 be a real inner product space. For 𝑑≥0,𝑥0,𝑦0∈𝐺, 𝑈=(𝑥,𝑦)∈𝐺×𝐺∶⟨𝑥0,𝑥⟩+⟨𝑦0,𝑦⟩≥𝑑.(2.28) Then 𝑈 satisfies (𝐽3),(𝐽4) if 𝑥0≠𝑦0.

Example 2.16. Let 𝐺 be the group of nonsingular square matrices with the operation of matrix multiplication. For 𝑘,𝑠∈ℝ,𝛿,𝑑≥0, let 𝑃𝑈=1,𝑃2||∈𝐺×𝐺∶det𝑃1||𝑘||det𝑃2||𝑠,𝑃≤𝛿𝑉=1,𝑃2||∈𝐺×𝐺∶det𝑃1||𝑘||det𝑃2||𝑠.≥𝑑(2.29) Then 𝑈 and 𝑉 satisfy both (𝐽3) and (𝐽4) if 𝑘≠𝑠.

From now on, we assume that 𝐺 is a 2-divisible Abelian group.

Theorem 2.17. Let 𝑈⊂𝐺×𝐺 satisfy the condition (𝐽3) and 𝜖≥0. Suppose that 𝑓∶𝐺→𝑌 satisfies (2.1) for all (𝑥,𝑦)∈𝑈. Then there exists a unique additive function 𝐴∶𝐺→𝑌 such that ‖𝑓(𝑥)−𝐴(𝑥)−𝑓(0)‖≤4𝜖(2.30) for all 𝑥∈𝐺.

Proof. For given 𝑥,𝑦∈𝐺, choose a 𝑧∈𝐺 such that (𝑧,−𝑧)+𝑅𝑥,𝑦⊂𝑈. Replacing 𝑥 by 𝑥+𝑧, 𝑦 by 𝑥−𝑧; 𝑥 by 𝑥+𝑧, 𝑦 by 𝑦−𝑧; 𝑥 by (𝑥−𝑦)/2+𝑧, 𝑦 by (𝑥−𝑦)/2−𝑧; 𝑥 by (𝑥−𝑦)/2+𝑧, 𝑦 by (−𝑥+𝑦)/2−𝑧 in (2.1), respectively, we have ‖‖‖‖𝑓(2𝑥)+𝑓(2𝑧)−2𝑓(𝑥+𝑧)‖≤𝜖,‖𝑓(𝑥+𝑦)+𝑓(𝑥−𝑦+2𝑧)−2𝑓(𝑥+𝑧)‖≤𝜖,𝑓(𝑥−𝑦)+𝑓(2𝑧)−2𝑓𝑥−𝑦2‖‖‖‖‖‖+𝑧≤𝜖,𝑓(0)+𝑓(𝑥−𝑦+2𝑧)−2𝑓𝑥−𝑦2‖‖‖+𝑧≤𝜖.(2.31) From (2.31), using the triangle inequality, we have ||||𝑓(2𝑥)−𝑓(𝑥+𝑦)−𝑓(𝑥−𝑦)+𝑓(0)≤4𝜖(2.32) for all 𝑥,𝑦∈𝐺. Replacing 𝑥 by (𝑥+𝑦/2), 𝑦 by (𝑥−𝑦/2) in (2.32), we have ||||𝑓(𝑥+𝑦)−𝑓(𝑥)−𝑓(𝑦)+𝑓(0)≤4𝜖(2.33) for all 𝑥,𝑦∈𝐺. From (2.33), using Theorem 1.1, we get the result.

Let 𝑑≥0,𝑘,𝑠∈ℝ with 𝑘+𝑠>0, and let 𝑈={(𝑥,𝑦)∶𝑘‖𝑥‖+𝑠‖𝑦‖≥𝑑}. Then 𝑈 satisfies the conditions (𝐽3) and (𝐽4). Thus, as a direct consequence of Theorem 2.17 we generalize Theorem 2.1 as follows.

Corollary 2.18. Let𝑑≥0,𝑘,𝑠∈ℝ with 𝑘+𝑠>0. Suppose that 𝑓∶𝑋→𝑌 satisfies the inequality (2.1) for all 𝑥,𝑦, with 𝑘‖𝑥‖+𝑠‖𝑦‖≥𝑑. Then there exists a unique additive function 𝐴∶𝑋→𝑌 such that ‖𝑓(𝑥)−𝐴(𝑥)−𝑓(0)‖≤4𝜖(2.34) for all 𝑥∈𝑋.

Theorem 2.19. Let 𝑈⊂𝐺×𝐺 satisfy the condition (𝐽4) and 𝜖≥0. Suppose that 𝑓∶𝐺→𝑌 satisfies (2.2) for all (𝑥,𝑦)∈𝑈. Then there exists a unique additive function 𝐴∶𝐺→𝑌 such that ‖𝑓(𝑥)−𝐴(𝑥)‖≤4𝜖(2.35) for all 𝑥∈𝐺.

Proof. For given 𝑥,𝑦∈𝐺, choose a 𝑧∈𝐺 such that (𝑧,−𝑧)+𝑆𝑥,𝑦⊂𝑈. Replacing 𝑥 by 𝑥+𝑧, 𝑦 by 𝑥−𝑧; 𝑥 by 𝑦+𝑧, 𝑦 by 𝑥−𝑧; 𝑥 by (𝑥−𝑦)/2+𝑧, 𝑦 by (𝑥−𝑦)/2−𝑧; 𝑥 by (−𝑥+𝑦)/2+𝑧, 𝑦 by (𝑥−𝑦)/2−𝑧 in (2.2), respectively, we have ‖‖‖𝑓‖𝑓(2𝑥)−𝑓(2𝑧)−2𝑓(𝑥−𝑧)‖≤𝜖,‖𝑓(𝑥+𝑦)−𝑓(−𝑥+𝑦+2𝑧)−2𝑓(𝑥−𝑧)‖≤𝜖,(𝑥−𝑦)−𝑓(2𝑧)−2𝑓𝑥−𝑦2‖‖‖‖‖‖+𝑧≤𝜖,𝑓(0)−𝑓(−𝑥+𝑦+2𝑧)−2𝑓𝑥−𝑦2‖‖‖+𝑧≤𝜖.(2.36) From (2.36), using the triangle inequality, we have ||||𝑓(2𝑥)−𝑓(𝑥+𝑦)−𝑓(𝑥−𝑦)+𝑓(0)≤4𝜖(2.37) for all 𝑥,𝑦∈𝐺. Replacing 𝑥 by (𝑥+𝑦)/2, 𝑦 by (𝑥−𝑦)/2 in (2.37) and using Theorem 1.1, we get the result.

As a direct consequence of Theorem 2.19, we have the following.

Corollary 2.20. Let 𝑑≥0,𝑘,𝑠∈ℝ with 𝑘+𝑠>0. Suppose that 𝑓∶𝑋→𝑌 satisfies the inequality (2.2) for all 𝑥,𝑦, with 𝑘‖𝑥‖+𝑠‖𝑦‖≥𝑑. Then there exists a unique additive function 𝐴∶𝑋→𝑌 such that ‖𝑓(𝑥)−𝐴(𝑥)‖≤4𝜖(2.38) for all 𝑥∈𝑋.

3. Asymptotic Behavior of the Equations

In this section we discuss asymptotic behaviors of the equations which gives refined versions of the results in [13].

Using Theorems 2.8 and 2.17, we have the following (cf. [13]).

Theorem 3.1. Let 𝑈 satisfy (𝐽1) or (𝐽3). Suppose that 𝑓∶𝑋→𝑌 satisfies the asymptotic condition ‖𝑓(𝑥+𝑦)+𝑓(𝑥−𝑦)−2𝑓(𝑥)‖⟶0(3.1) as ‖𝑥‖+â€–ğ‘¦â€–â†’âˆž,(𝑥,𝑦)∈𝑈. Then there exists a unique additive function 𝐴∶𝑋→𝑌 such that 𝑓(𝑥)=𝐴(𝑥)+𝑓(0)(3.2) for all 𝑥∈𝑋.

Proof. By the condition (3.1), for each 𝑛∈ℕ, there exists 𝑑𝑛>0 such that 1‖𝑓(𝑥+𝑦)+𝑓(𝑥−𝑦)−2𝑓(𝑥)‖≤𝑛(3.3) for all (𝑥,𝑦)∈𝑈 with ‖𝑥‖+‖𝑦‖≥𝑑𝑛. Let 𝑈0=𝑈∩{(𝑥,𝑦)∶‖𝑥‖+‖𝑦‖≥𝑑𝑛}. Then 𝑈0 satisfies both the conditions (𝐽1) and (𝐽3). By Theorems 2.8 and 2.17, there exists a unique additive function 𝐴𝑛∶𝑋→𝑌 such that ‖‖𝑓(𝑥)−𝐴𝑛‖‖≤2(𝑥)−𝑓(0)𝑛4or𝑛(3.4) for all 𝑥∈𝑋. Putting 𝑛=𝑚 in (3.4) and using the triangle inequality, we have ‖‖𝐴𝑛(𝑥)−𝐴𝑚‖‖(𝑥)≤8(3.5) for all 𝑥∈𝑋. Using the additivity of 𝐴𝑛,𝐴𝑚, we have 𝐴𝑛=𝐴𝑚 for all 𝑛,𝑚∈ℕ. Letting ğ‘›â†’âˆž in (3.4), we get the result.

Corollary 3.2. Let 𝑘,𝑠∈ℝ satisfy one of the conditions: 𝑠>0,𝑘+𝑠>0. Suppose that 𝑓∶𝑋→𝑌 satisfies the condition ‖𝑓(𝑥+𝑦)+𝑓(𝑥−𝑦)−2𝑓(𝑥)‖⟶0(3.6) as 𝑘‖𝑥‖+ğ‘ â€–ğ‘¦â€–â†’âˆž. Then there exists a unique additive function 𝐴∶𝑋→𝑌 such that 𝑓(𝑥)=𝐴(𝑥)+𝑓(0)(3.7) for all 𝑥∈𝑋.

Using Theorems 2.10 and 2.19, we have the following (cf. [13]).

Theorem 3.3. Let 𝑈 satisfy (𝐽2) or (𝐽4). Suppose that 𝑓∶𝑋→𝑌 satisfies the condition ‖𝑓(𝑥+𝑦)−𝑓(𝑥−𝑦)−2𝑓(𝑦)‖⟶0(3.8) as ‖𝑥‖+â€–ğ‘¦â€–â†’âˆž,(𝑥,𝑦)∈𝑈. Then 𝑓 is an additive function.

Corollary 3.4. Let 𝑘,𝑠∈ℝ satisfy one of the conditions: 𝑘>0,𝑘+𝑠>0. Suppose that 𝑓∶𝑋→𝑌 satisfies the condition ‖𝑓(𝑥+𝑦)−𝑓(𝑥−𝑦)−2𝑓(𝑦)‖⟶0(3.9) as 𝑘‖𝑥‖+ğ‘ â€–ğ‘¦â€–â†’âˆž. Then 𝑓 is an additive function.

Acknowledgments

The first author was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (MEST) (no. 2011-0003898), and the second author was partially supported by the Research Institute of Mathematics, Seoul National University.

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