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

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 𝛿_𝜖>0 such that 𝑑(𝑓(𝑥),ℎ(𝑥))≤𝛿_𝜖(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/(2−2^𝑝) for all 𝑥∈𝑋. In 1951 Bourgin [4, 5] stated that if Φ is symmetric in ‖𝑥‖ and ‖𝑦‖ with ∑^∞𝑗=1Φ(2^𝑗𝑥,2^𝑗𝑥)/2^𝑗<∞ for each 𝑥∈𝑋, then there exists a unique additive function 𝐴∶𝑋→𝑌 such that ‖𝑓(𝑥)−𝐴(𝑥)‖≤∑^∞𝑗=1Φ(2^𝑗𝑥,2^𝑗𝑥)/2^𝑗 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‖‖𝑓𝑓((𝑥𝑥++𝑦𝑦))+−𝑓𝑓((𝑥𝑥−−𝑦𝑦))−−22𝑓𝑓((𝑥𝑦))‖‖≤≤𝜖𝜖,((22..12)) 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 ‖𝑓(𝑥)−𝐴(𝑥)−𝑓(0)‖≤52𝜖(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={(𝑎₁,𝑎₂)∶𝑎₁,𝑎₂∈𝐺} the product group; that is, for 𝑎=(𝑎₁,𝑎₂),𝑏=(𝑏₁,𝑏₂)∈𝐺×𝐺, we define 𝑎+𝑏=(𝑎₁+𝑏₁,𝑎₂+𝑏₂),𝑎−𝑏=(𝑎₁−𝑏₁,𝑎₂−𝑏₂). For a subset 𝐻 of 𝐺×𝐺 and 𝑎,𝑏∈𝐺×𝐺, we define 𝑎+𝐻={𝑎+ℎ∶ℎ∈𝐻}.

For given 𝑥,𝑦∈𝐺 we denote by 𝑃_𝑥,𝑦,𝑄_𝑥,𝑦 the subsets of points of the forms (not necessarily distinct) in 𝐺×𝐺, respectively, 𝑃_𝑥,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 𝑃_𝑥,𝑦 can be viewed as the vertices of rectangles in 𝐺×𝐺, and 𝑄_𝑥,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()𝐽2)(0,(𝑧𝑧),+0)𝑃+_𝑥,𝑦𝑄=_𝑥,𝑦{(0,𝑧),(𝑥,−𝑥+𝑧),(𝑦,𝑦+𝑧),(𝑥+𝑦,−𝑥+𝑦+𝑧)}⊂𝑈,={(−𝑥+𝑧,𝑥),(𝑦+𝑧,𝑦),(−𝑥+𝑦+𝑧,𝑥+𝑦)}⊂𝑈,(2.9) respectively.

The sets (0,𝑧)+𝑃_𝑥,𝑦,(𝑧,0)+𝑄_𝑥,𝑦 can be understood as the translations of 𝑃_𝑥,𝑦 and 𝑄_𝑥,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 𝑈𝑉=={({𝑥(,𝑥𝑦,)𝑦∈)∈𝐺𝐺××𝐺𝐺∶∶𝑘‖‖𝑘𝑥𝑥‖++𝑠𝑠𝑦‖‖𝑦‖≥≥𝑑}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∈𝐺𝑈=(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 𝑈𝑉==𝑃𝑃₁₁,𝑃₂∈𝐺×𝐺∶||det𝑃₁||^𝑘||det𝑃₂||^𝑠≤𝛿,,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 𝑃_𝑥,𝑦 and 𝑄_𝑥,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 ((0𝑧,,0𝑧))++𝑋𝑋₁₁,𝑋₂,𝑋₃,𝑋₄⊂𝑈₁,(2.13),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 ℤ×ℤ=𝑎₁,𝑏₁,𝑎₂,𝑏₂,…,𝑎_𝑛,𝑏_𝑛,…⊂ℝ²(2.15) such that ||𝑎₁||+||𝑏₁||≤||𝑎₂||+||𝑏₂||≤⋯≤||𝑎_𝑛||+||𝑏_𝑛||≤⋯,(2.16) and let 𝑃_𝑛={(0,0),(𝑎_𝑛,−𝑎_𝑛),(𝑏_𝑛,𝑏_𝑛),(𝑎_𝑛+𝑏_𝑛,−𝑎_𝑛+𝑏_𝑛)},𝑛=1,2,…. Then it is easy to see that 𝑈=⋃^∞𝑛=1((0,2^𝑛)+𝑃_𝑛) satisfies the condition (𝐽1). Now let 𝑃={(𝑝₁,𝑞₁),(𝑝₂,𝑞₂)}⊂ℤ×ℤ with |𝑞₂−𝑞₁|≤|𝑝₂−𝑝₁|,𝑝₁𝑝₂>0. Then (0,𝑧)+𝑃 is not contained in 𝑈 for all 𝑧∈ℤ. Indeed, for any choices of (𝑥_𝑛,𝑦_𝑛)∈𝑃_𝑛+(0,2^𝑛),𝑛=1,2,…, we have 𝑦_𝑚−𝑦_𝑛>|𝑥_𝑚−𝑥_𝑛| for all 𝑚>𝑛,𝑚,𝑛=1,2,…. Thus, if (0,𝑧)+𝑃⊂𝑈 for some 𝑧∈ℤ, then 𝑃+(0,𝑧)⊂(0,2^𝑛)+𝑃_𝑛 for some 𝑛∈ℕ. Thus, it follows from the condition 𝑞₂−𝑞₁≤|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 𝑄_𝑛={(−𝑎_𝑛,𝑎_𝑛),(𝑏_𝑛,𝑏_𝑛),(−𝑎_𝑛+𝑏_𝑛,𝑎_𝑛+𝑏_𝑛)}. Then it is easy to see that 𝑈=⋃^∞𝑛=1((2^𝑛,0)+𝑄_𝑛) 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)+𝑃_𝑥,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+‖𝑓‖𝑧‖𝑓)(𝑓2+(𝑦𝑧(𝑓𝑧)+)+(2+𝑧𝑓𝑥)𝑓+−(2(𝑓−𝑥𝑧)𝑧(−−−)𝑧−𝑧2))𝑓2−−𝑓(2𝑥2(𝑓𝑓0+)((‖𝑥𝑦𝑦)))≤‖‖‖𝜖≤≤≤.𝜖𝜖𝜖,,,(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 ‖𝑓(𝑥)−𝐴(𝑥)‖≤32𝜖(2.21) for all 𝑥∈𝐺.

Proof. For given 𝑥,𝑦∈𝐺, choose a 𝑧∈𝐺 such that (𝑧,0)+𝑄_𝑥,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(2‖𝑦‖𝑓𝑓+(𝑧(𝑧2))𝑦−−+𝑓𝑓𝑧()(−−−22𝑥𝑓𝑥+(+𝑧𝑧)𝑧)−)−−22𝑓2𝑓𝑓(𝑦((𝑥)𝑥‖)‖+≤≤𝑦𝜖)‖𝜖.,≤𝜖,(2.22) From (2.22), using the triangle inequality and dividing the result by 2, we have ||𝑓(𝑥+𝑦)−𝑓(𝑥)−𝑓(𝑦)||≤32𝜖.(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 ‖𝑓(𝑥)−𝐴(𝑥)‖≤32𝜖(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 𝑅_𝑥,𝑦,𝑆_𝑥,𝑦⊂𝐺×𝐺, 𝑅_𝑥,𝑦=(𝑥,𝑥),(𝑥,𝑦),𝑥−𝑦𝑆2,𝑥−2𝑦,𝑥−2𝑦,−𝑥2+𝑦_𝑥,𝑦,=(𝑥,𝑥),(𝑦,𝑥),𝑥−𝑦2,𝑥−2𝑦,−𝑥2+𝑦,𝑥−2𝑦.(2.25) One can see that 𝑅_𝑥,𝑦 and 𝑆_𝑥,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)+𝑅_𝑥,𝑦=(𝑥+𝑧,𝑥−𝑧),(𝑥+𝑧,𝑦−𝑧),𝑥−𝑦𝑥−𝑦2+𝑧,𝑥−2𝑦−𝑧,(𝐽4)(𝑧,−𝑧)+𝑆_𝑥,𝑦2+𝑧,−𝑥2+𝑦−𝑧⊂𝑈,=(𝑥+𝑧,𝑥−𝑧),(𝑦+𝑧,𝑥−𝑧),𝑥−𝑦−𝑥+𝑦2+𝑧,𝑥−2𝑦−𝑧,2+𝑧,𝑥−2𝑦−𝑧⊂𝑈,(2.26) respectively.

Example 2.14. Let 𝐺 be a real normed space. For 𝑘,𝑠,𝑑∈ℝ, let 𝑈𝑉=={({𝑥(,𝑥𝑦,)𝑦∈)∈𝐺𝐺××𝐺𝐺∶∶𝑘‖‖𝑘𝑥𝑥‖++𝑠𝑠𝑦‖‖𝑦‖≥≥𝑑}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 𝑈𝑉==𝑃𝑃₁₁,𝑃₂∈𝐺×𝐺∶||det𝑃₁||^𝑘||det𝑃₂||^𝑠≤𝛿,,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)+𝑅_𝑥,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 ‖‖‖‖𝑓𝑓(𝑥(𝑥‖+𝑓−𝑦(𝑦)2)𝑥++)𝑓+𝑓(𝑓𝑥(2(−𝑧2)𝑧𝑦−)+−22𝑓2𝑧𝑓)−(𝑥𝑥−2+𝑓𝑦𝑧()𝑥‖+≤𝑧)𝜖‖,≤𝜖,‖‖‖𝑓(0)+𝑓(𝑥−𝑦+2𝑧)−2𝑓𝑥2−+𝑦𝑧‖‖‖≤𝜖,2+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)+𝑆_𝑥,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(𝑥(+𝑥‖𝑓−𝑦)(𝑦2−)𝑥−)𝑓−𝑓(−𝑓(𝑥2(𝑧2+)𝑧𝑦−)−+2𝑓22𝑓𝑧)(𝑥−𝑥−−2𝑓𝑦𝑧)(‖𝑥−≤𝑧𝜖),‖≤𝜖,‖‖‖𝑓(0)−𝑓(−𝑥+𝑦+2𝑧)−2𝑓2𝑥−+𝑦𝑧‖‖‖≤𝜖,2+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‖+‖𝑦‖→∞,(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 𝑑_𝑛>0 such that ‖𝑓(𝑥+𝑦)+𝑓(𝑥−𝑦)−2𝑓(𝑥)‖≤1𝑛(3.3) for all (𝑥,𝑦)∈𝑈 with ‖𝑥‖+‖𝑦‖≥𝑑_𝑛. Let 𝑈₀=𝑈∩{(𝑥,𝑦)∶‖𝑥‖+‖𝑦‖≥𝑑_𝑛}. Then 𝑈₀ satisfies both the conditions (𝐽1) and (𝐽3). By Theorems 2.8 and 2.17, there exists a unique additive function 𝐴_𝑛∶𝑋→𝑌 such that ‖‖𝑓(𝑥)−𝐴_𝑛(𝑥)−𝑓(0)‖‖≤2𝑛or4𝑛(3.4) for all 𝑥∈𝑋. Putting 𝑛=𝑚 in (3.4) and using the triangle inequality, we have ‖‖𝐴_𝑛(𝑥)−𝐴_𝑚(𝑥)‖‖≤8(3.5) for all 𝑥∈𝑋. Using the additivity of 𝐴_𝑛,𝐴_𝑚, we have 𝐴_𝑛=𝐴_𝑚 for all 𝑛,𝑚∈ℕ. Letting 𝑛→∞ 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‖+𝑠‖𝑦‖→∞. 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‖+‖𝑦‖→∞,(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‖+𝑠‖𝑦‖→∞. 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.