Hyers-Ulam Stability for a Class of Quadratic Functional Equations via a Typical Form

Abstract

We investigate the following typical form of a certain class of quadratic functional equations: 𝑓(𝑎𝑥+𝑏𝑦)+𝑓(𝑎𝑥−𝑏𝑦)+𝑐[𝑓(𝑥+𝑦)+𝑓(𝑥−𝑦)−2𝑓(𝑥)−2𝑓(𝑦)]=2𝑎²𝑓(𝑥)+2𝑏²𝑓(𝑦). Furthermore, we provide a systematic program to prove the generalized Hyers-Ulam stability for the class of functional equations via the stability for the typical form.

1. Introduction

In 1964, Ulam [1] proposed the following stability problem:

“let 𝐺₁ be a group and 𝐺₂ a metric group with the metric 𝑑. Given a constant 𝛿>0, does there exist a constant 𝑐>0 such that if a mapping 𝑓∶𝐺₁→𝐺₂ satisfies 𝑑(𝑓(𝑥𝑦),𝑓(𝑥)𝑓(𝑦))<𝑐 for all 𝑥,𝑦∈𝐺₁, then there exists a unique homomorphism ℎ∶𝐺₁→𝐺₂ with 𝑑(𝑓(𝑥),ℎ(𝑥))<𝛿 for all 𝑥∈𝐺₁?”

In 1941, Hyers [2] answered this problem under the assumption that the groups are Banach spaces. Aoki [3] and Rassias [4] generalized the result of Hyers. Rassias [4] solved the generalized Hyers-Ulam stability of the functional inequality 𝑓𝑥+𝑦−𝑓(𝑥)−𝑓𝑦≤𝜖‖𝑥‖^𝑝+𝑦^𝑝 (1) for some 𝜖≥0 and 𝑝 with 𝑝<1 and for all 𝑥,𝑦∈𝑋, where 𝑓∶𝑋→𝑌 is a function between Banach spaces. The paper of Rassias [4] has provided a lot of influence in the development of what we call the generalized Hyers-Ulam stability or the Hyers-Ulam-Rassias stability of functional equations. A generalization of the Rassias theorem was obtained by Găvruţa [5] by replacing the unbounded Cauchy difference by a general control function in the spirit of Rassias approach.

The functional equation 𝑓𝑥+𝑦+𝑓𝑥−𝑦=2𝑓(𝑥)+2𝑓𝑦 (2) is called a quadratic functional equation and a solution of a quadratic functional equation is called quadratic. A generalized Hyers-Ulam stability problem for the quadratic functional equation was proved by Skof [6] for mappings 𝑓∶𝑋→𝑌, where 𝑋 is a normed space and 𝑌 is a Banach space. Cholewa [7] noticed that the theorem of Skof is still true if the relevant domain 𝑋 is replaced by an Abelian group. Czerwik [8] proved the generalized Hyers-Ulam stability for the quadratic functional equation, and Park [9] proved the generalized Hyers-Ulam stability of the quadratic functional equation in Banach modules over a 𝐶^∗-algebra. Also, the stability problems of functional equations related to quadratic functions can be found in many papers (e.g., [10, 11], etc.).

Rassias [12] investigated the following Euler-Lagrange functional equation: 𝑓𝑎𝑥+𝑏𝑦+𝑓𝑏𝑥−𝑎𝑦=2𝑎²+𝑏²𝑓(𝑥)+𝑓𝑦, (3) and Gordji and Khodaei [13] investigated other Euler-Lagrange functional equations 𝑓=𝑏()𝑎𝑥+𝑏𝑦+𝑓𝑎𝑥−𝑏𝑦𝑎+𝑏2𝑓+𝑏()𝑥+𝑦𝑎+𝑏2𝑓+𝑥−𝑦2𝑎²−𝑎𝑏−𝑏²𝑓(𝑥)+𝑏²−𝑎𝑏𝑓𝑦 (4) for fixed integers 𝑎,𝑏 with 𝑏≠𝑎,−𝑎,−3𝑎, and 𝑓𝑎𝑥+𝑏𝑦+𝑓𝑎𝑥−𝑏𝑦=2𝑎²𝑓(𝑥)+2𝑏²𝑓𝑦 (5) for fixed integers 𝑎,𝑏 with 𝑎²≠𝑏² and 𝑎𝑏≠0.

In this paper, we consider the sum of two functional equations (2) and (5), that is, 𝑓𝑓(𝑥)𝑦𝑎𝑥+𝑏𝑦+𝑓𝑎𝑥−𝑏𝑦+𝑐𝑥+𝑦+𝑓𝑥−𝑦−2𝑓−2𝑓=2𝑎²𝑓(𝑥)+2𝑏²𝑓𝑦 (6) for fixed nonzero real numbers 𝑎,𝑏,𝑐 with 𝑎≠1,−1 and 𝑎²≠𝑏², and prove the generalized Hyers-Ulam stability for it.

As applications of theorems in Sections 2 and 3, we have a systematic program to prove the generalized Hyers-Ulam stability for functional inequalities which can be deformed into the following functional inequality: 𝑓𝑎𝑥+𝑏𝑦+𝑓𝑎𝑥−𝑏𝑦−2𝑎²𝑓(𝑥)−2𝑏²𝑓𝑦+𝑐𝑓𝑥+𝑦+𝑓𝑥−𝑦−2𝑓(𝑥)−2𝑓𝑦≤𝜙𝑥,𝑦. (7)

Throughout this paper, assume that 𝑋 is a normed space and 𝑌 is a Banach space.

2. Solutions of (6)

In this section, we investigate solutions of (6). In Corollary 5, it can be concluded that any solution of (6) is quadratic if 𝑎 is a rational number. We start with the following lemma.

Lemma 1. Let 𝑓∶𝑋→𝑌 be a mapping with 𝑓(0)=0. Suppose that 𝑓 satisfies (6); then the following equation holds: 𝑐𝑐+²𝑎²𝑓𝑏2𝑥+𝑦+𝑓2𝑥−𝑦−2²𝑐+2𝑐+²𝑎²𝑓+𝑥+𝑦+𝑓𝑥−𝑦4𝑏²𝑏+5𝑐+²𝑐𝑎²+2𝑐²𝑎²𝑓𝑦+𝑏𝑐−²𝑐𝑎²𝑓−𝑦+4𝑏²𝑐+2𝑐+²𝑎²𝑓(𝑥)𝑐−2𝑐+²𝑎²𝑓(2𝑥)=0 (8) for all 𝑥,𝑦∈𝑋.

Proof. Letting 𝑦=0 in (6), we have 𝑓(𝑎𝑥)=𝑎²𝑓(𝑥) (9) for all 𝑥∈𝑋. Setting 𝑥=0 and 𝑦=𝑥 in (6), we have 𝑓(𝑏𝑥)+𝑓(−𝑏𝑥)+𝑐𝑓(−𝑥)−𝑓(𝑥)=2𝑏²𝑓(𝑥) (10) for all 𝑥∈𝑋. Letting 𝑥=−𝑥 in (10) and adding the two equations, we have 𝑓(𝑏𝑥)+𝑓(−𝑏𝑥)=𝑏²𝑓(𝑥)+𝑓(−𝑥) (11) for all 𝑥∈𝑋.
Replacing 𝑦 by 𝑥+𝑦 in (6), we have 𝑓𝑓𝑎𝑎𝑥+𝑏𝑥+𝑦+𝑓𝑎𝑥−𝑏𝑥+𝑦+𝑐2𝑥+𝑦+𝑓−𝑦=2²+𝑐𝑓(𝑥)+2𝑏²+𝑐𝑓𝑥+𝑦 (12) for all 𝑥,𝑦∈𝑋, and letting 𝑦=−𝑦 in (12), we have 𝑓𝑓𝑦𝑎𝑎𝑥+𝑏𝑥−𝑦+𝑓𝑎𝑥−𝑏𝑥−𝑦+𝑐2𝑥−𝑦+𝑓=2²+𝑐𝑓(𝑥)+2𝑏²+𝑐𝑓𝑥−𝑦 (13) for all 𝑥,𝑦∈𝑋.
Replacing 𝑥 and 𝑦 by 𝑥+(𝑏/𝑎)𝑦 and 𝑥 in (6), respectively, we have 𝑓𝑓𝑏𝑎𝑥+𝑏𝑥+𝑦+𝑓𝑎𝑥−𝑏𝑥−𝑦+𝑐2𝑥+𝑎𝑦𝑏+𝑓𝑎𝑦=2𝑎²𝑓𝑏+𝑐𝑥+𝑎𝑦+2𝑏²+𝑐𝑓(𝑥) (14) for all 𝑥,𝑦∈𝑋, and letting 𝑦=−𝑦 in (14), we have 𝑓𝑓𝑏𝑎𝑥+𝑏𝑥−𝑦+𝑓𝑎𝑥−𝑏𝑥+𝑦+𝑐2𝑥−𝑎𝑦−𝑏+𝑓𝑎𝑦=2𝑎²𝑓𝑏+𝑐𝑥−𝑎𝑦+2𝑏²+𝑐𝑓(𝑥) (15) for all 𝑥,𝑦∈𝑋. By (12), (13), (14), and (15), we have 𝑐𝑓𝑦−𝑏2𝑥+𝑦+𝑓2𝑥−𝑦+𝑓+𝑓−𝑦₂𝑐𝑎²𝑓𝑦−𝑐+𝑓−𝑦𝑎²𝑓𝑎2𝑎𝑥+𝑏𝑦+𝑓2𝑎𝑥−𝑏𝑦=4²−𝑏²𝑓(𝑥)+2𝑏²𝑓𝑐+𝑐𝑥+𝑦+𝑓𝑥−𝑦−21+𝑎²𝑓𝑎𝑥+𝑏𝑦+𝑓𝑎𝑥−𝑏𝑦 (16) for all 𝑥,𝑦∈𝑋. By (6) and (16), we have 𝑐𝑓𝑦−𝑏2𝑥+𝑦+𝑓2𝑥−𝑦+𝑓+𝑓−𝑦₂𝑐𝑎²𝑓𝑦+𝑐+𝑓−𝑦₂𝑎²𝑓2𝑥+𝑦+𝑓2𝑥−𝑦−2𝑓(2𝑥)−2𝑓𝑦−2𝑎²𝑓(2𝑥)−2𝑏²𝑓𝑦=4𝑎²−𝑏²𝑓(𝑥)+2𝑏²𝑓𝑐+𝑐𝑥+𝑦+𝑓𝑥−𝑦+2𝑐1+𝑎²𝑓(𝑥)𝑦𝑥+𝑦+𝑓𝑥−𝑦−2𝑓−2𝑓−2𝑎²𝑓(𝑥)−2𝑏²𝑓𝑦 (17) for all 𝑥,𝑦∈𝑋. Now, just simplifying this equation, we can get the result.

Next three theorems deal with (6) for the different cases.

Theorem 2. Let 𝑓∶𝑋→𝑌 be a mapping with 𝑓(0)=0. Suppose that 𝑓 satisfies (6). If 𝑐≠−𝑏² and 𝑐²≠𝑎²𝑏², then 𝑓 is quadratic.

Proof. By (10) and (11) in the proof of Lemma 1, we have 𝑏²+𝑐𝑓(𝑥)=𝑏²+𝑐𝑓(−𝑥) (18) for all 𝑥∈𝑋. Since 𝑏²+𝑐≠0, we have 𝑓(−𝑥)=𝑓(𝑥) (19) for all 𝑥∈𝑋. By (10) and (19), we have 𝑓(𝑏𝑥)=𝑏²𝑓(𝑥) (20) for all 𝑥∈𝑋. Replacing 𝑥 and 𝑦 by 𝑏𝑥 and 𝑎𝑦 in (6), respectively, by (20), we have 𝑎²𝑏²𝑓𝑓𝑥+𝑦+𝑓𝑥−𝑦+𝑐𝑏𝑥+𝑎𝑦+𝑓𝑏𝑥−𝑎𝑦−2𝑏²𝑓(𝑥)−2𝑎²𝑓𝑦=2𝑎²𝑏²𝑓(𝑥)+𝑓𝑦 (21) for all 𝑥,𝑦∈𝑋. Replacing 𝑥 and 𝑦 by 𝑦 and 𝑥 in (6), respectively, by (19), we have 𝑓𝑓(𝑥)𝑦𝑏𝑥+𝑎𝑦+𝑓𝑏𝑥−𝑎𝑦+𝑐𝑥+𝑦+𝑓𝑥−𝑦−2𝑓−2𝑓=2𝑎²𝑓𝑦+2𝑏²𝑓(𝑥) (22) for all 𝑥,𝑦∈𝑋. By (21) and (22), we have 𝑎²𝑏²−𝑐²𝑓𝑥−𝑦+𝑓𝑥+𝑦−2𝑓(𝑥)−2𝑓𝑦=0 (23) for all 𝑥,𝑦∈𝑋. Since 𝑎²𝑏²−𝑐²≠0, then 𝑓𝑥+𝑦+𝑓𝑥−𝑦=2𝑓(𝑥)+2𝑓𝑦 (24) for all 𝑥,𝑦∈𝑋. Hence, 𝑓 is quadratic.

Theorem 3. Let 𝑓∶𝑋→𝑌 be a mapping with 𝑓(0)=0. Suppose that 𝑓 satisfies (6) and 𝑎 is a rational number. If 𝑐²=𝑎²𝑏², then 𝑓 is quadratic.

Proof. Since 𝑏²+𝑐=𝑏(𝑏±𝑎)≠0, by the first few lines in the proof of Theorem 2, 𝑓 is even. Hence, in this case, we can easily check that (8) can be reduced to 𝑏(𝑓(𝑦(𝑥)(𝑏±𝑎)2𝑥+𝑦+𝑓2𝑥−𝑦=𝑏𝑏±𝑎)4𝑓𝑥+𝑦+4𝑓𝑥−𝑦−6𝑓+8𝑓−2𝑓2𝑥) (25) for all 𝑥,𝑦∈𝑋. Since 𝑏(𝑏±𝑎)≠0, we have 𝑓𝑦(𝑥)()2𝑥+𝑦+𝑓2𝑥−𝑦=4𝑓𝑥+𝑦+4𝑓𝑥−𝑦−6𝑓+8𝑓−2𝑓2𝑥 (26) for all 𝑥,𝑦∈𝑋. By [14], a function 𝑓 satisfying (26) is quartic-quadratic. But in our case, 𝑓 also satisfies (26) and since 𝑓(𝑎𝑥)=𝑎²𝑓(𝑥), 𝑓 is quadratic.

Theorem 4. Let 𝑓∶𝑋→𝑌 be a mapping with 𝑓(0)=0. Suppose that 𝑓 satisfies (6) and 𝑎 is a rational number. If 𝑐=−𝑏², then 𝑓 is quadratic.

Proof. Suppose that 𝑐=−𝑏². By (8), we have 𝑏²𝑎²𝑏²−𝑎²𝑓=𝑏2𝑥+𝑦+𝑓2𝑥−𝑦₂𝑎²𝑏²−𝑎²(𝑥)()𝑦2𝑓𝑥+𝑦+2𝑓𝑥−𝑦−4𝑓+2𝑓2𝑥−𝑓−𝑓−𝑦 (27) for all 𝑥,𝑦∈𝑋. By [15], 𝑓 is quadratic-cubic and since 𝑓(𝑎𝑥)=𝑎²𝑓(𝑥), 𝑓 is quadratic.

Combining Theorems 2, 3, and 4 we can get the following corollary as the conclusion of this section.

Corollary 5. Let 𝑓∶𝑋→𝑌 be a mapping with 𝑓(0)=0. Suppose that 𝑓 satisfies (6) and 𝑎 is a rational number. Then 𝑓 is quadratic.

3. The Generalized Hyers-Ulam Stability for (6)

In this section, we will prove the generalized Hyers-Ulam stability for (6).

Theorem 6. Let 𝜙∶𝑋²→[0,∞) be a function such that ^∞_𝑛=0𝑎^−2𝑛𝜙𝑎^𝑛𝑥,𝑎^𝑛𝑦<∞ (28) for all 𝑥,𝑦∈𝑋. Let 𝑓∶𝑋→𝑌 be a mapping such that 𝑓𝑎𝑥+𝑏𝑦+𝑓𝑎𝑥−𝑏𝑦−2𝑎²𝑓(𝑥)−2𝑏²𝑓𝑦+𝑐𝑓𝑥+𝑦+𝑓𝑥−𝑦−2𝑓(𝑥)−2𝑓𝑦≤𝜙𝑥,𝑦 (29) for a fixed rational number 𝑎 and fixed nonzero real numbers 𝑏, 𝑐 with 𝑎≠1,0−1 and 𝑎²≠𝑏². Then there exists a unique quadratic mapping 𝑄∶𝑋→𝑌 satisfying (6) and 𝑄(𝑥)(𝑥)(0≤1−𝑓−𝑓)2^∞_𝑛=0𝑎^−2(𝑛+1)𝜙𝑎^𝑛𝑥,0+𝜙(0,0) (30) for all 𝑥∈𝑋.

Proof. Let 𝑔(𝑥)=𝑓(𝑥)−𝑓(0). Then 𝑔(0)=0 and 𝑔𝑎𝑥+𝑏𝑦+𝑔𝑎𝑥−𝑏𝑦−2𝑎²𝑔(𝑥)−2𝑏²𝑔𝑦𝑔(𝑥)𝑦()+𝑐𝑥+𝑦+𝑔𝑥−𝑦−2𝑔−2𝑔≤𝜙𝑥,𝑦+𝜙0,0 (31) for all 𝑥,𝑦∈𝑋. Setting 𝑦=0 in (31), we have 𝑎²𝑔(𝑥)(≤1−𝑔𝑎𝑥)2𝜙₀(𝑥,0) (32) for all 𝑥∈𝑋, where 𝜙₀(𝑥,𝑦)=𝜙(𝑥,𝑦)+𝜙(0,0). Replacing 𝑥 by 𝑎^𝑛𝑥 in (32) and dividing (32) by 𝑎^2(𝑛+1), we have 𝑎^−2𝑛𝑔𝑎^𝑛𝑥−𝑎^−2(𝑛+1)𝑔𝑎^𝑛+1𝑥≤12𝑎^−2(𝑛+1)𝜙₀𝑎^𝑛𝑥,0 (33) for all 𝑥∈𝑋 and all nonnegative integers 𝑛. For 𝑚,𝑛∈ℕ∪{0} with 𝑚<𝑛, 𝑎^−2𝑚𝑔𝑎^𝑚𝑥−𝑎^−2𝑛𝑔𝑎^𝑛𝑥≤12^𝑛−1_𝑘=𝑚𝑎^−2(𝑘+1)𝜙₀𝑎^𝑘=1𝑥,02𝑎^2𝑛−1_𝑘=𝑚𝑎^−2𝑘𝜙₀𝑎^𝑘𝑥,0 (34) for all 𝑥∈𝑋. Since (28) holds for 𝑦=0, ∑^𝑛−1_𝑘=𝑚𝑎^−2𝑘𝜙₀(𝑎^𝑘𝑥,0) goes to 0 as 𝑚→∞. So {𝑎^−2𝑛𝑔(𝑎^𝑛𝑥)} is a Cauchy sequence in 𝑌, and since 𝑌 is a Banach space, there exists a mapping 𝑄∶𝑋→𝑌 such that 𝑄(𝑥)=lim_𝑛→∞𝑎^−2𝑛𝑔𝑎^𝑛𝑥 (35) for all 𝑥∈𝑋 and 𝑄(𝑥)(𝑥≤1−𝑔)2^∞_𝑛=0𝑎^−2(𝑛+1)𝜙₀𝑎^𝑛𝑥,0 (36) for all 𝑥∈𝑋. Replacing 𝑥 and 𝑦 by 𝑎^𝑛𝑥 and 𝑎^𝑛𝑦 in (31), respectively, and dividing (31) by 𝑎^2𝑛, we have 𝑎^−2𝑛𝑔𝑎^𝑛𝑎𝑥+𝑏𝑦+𝑎^−2𝑛𝑔𝑎^𝑛𝑎𝑎𝑥−𝑏𝑦−𝑐^−2𝑛𝑔𝑎^𝑛𝑥+𝑦+𝑎^−2𝑛𝑔𝑎^𝑛𝑥−𝑦−2⋅𝑎^−2𝑛𝑔𝑎^𝑛𝑥−2⋅𝑎^−2𝑛𝑔𝑎^𝑛𝑦−2⋅𝑎²⋅𝑎^−2𝑛𝑔𝑎^𝑛𝑥−2⋅𝑏²⋅𝑎^−2𝑛𝑔𝑎^𝑛𝑦≤𝑎^−2𝑛𝜙₀𝑎^𝑛𝑥,𝑎^𝑛𝑦 (37) for all 𝑥,𝑦∈𝑋 and letting 𝑛→∞ in the above inequality, we can show that 𝑄 satisfies (6). By Corollary 5, 𝑄 is quadratic.
Now, we show the uniqueness of the quadratic mapping 𝑄. Suppose that 𝑄₀ is a quadratic mapping satisfying (6) and (30). Then we have 𝑄(𝑥)−𝑄₀(𝑥)=𝑎^−2𝑘𝑄𝑎^𝑘𝑥−𝑄₀𝑎^𝑘𝑥≤𝑎^−2𝑘𝑄𝑎^𝑘𝑥−𝑔𝑎^𝑘𝑥+𝑎^−2𝑘𝑄₀𝑎^𝑘𝑥−𝑔𝑎^𝑘𝑥≤𝑎^∞−2𝑘_𝑛=0𝑎^−2(𝑛+1)𝜙₀𝑎^𝑛+𝑘𝑥,0=𝑎^∞−2_𝑛=0𝑎^{−2(𝑛+𝑘)}𝜙₀𝑎^𝑛+𝑘𝑥,0 (38) for all 𝑥∈𝑋 and for all positive integers 𝑘. Hence, letting 𝑘→∞ in the above inequality by (28) the tail part ∑^∞_𝑛=0𝑎^{−2(𝑛+𝑘)}𝜙₀(𝑎^𝑛+𝑘𝑥,0)=∑^∞_𝑛=𝑘𝑎^−2𝑛𝜙₀(𝑎^𝑛𝑥,0) goes to 0. So we have 𝑄(𝑥)=𝑄₀(𝑥) (39) for all 𝑥∈𝑋.

We remark that if 𝑓(0)=0 in Theorem 6, inequality (30) can be replaced by 𝑄(𝑥)(𝑥≤1−𝑓)2^∞_𝑛=0𝑎^−2(𝑛+1)𝜙𝑎^𝑛𝑥,0. (40)

Related with Theorem 6, we can also have the following theorem. And the proof is similar to that of Theorem 6.

Theorem 7. Let 𝜙∶𝑋²→[0,∞) be a function such that ^∞_𝑛=0𝑎^2𝑛𝜙𝑎^−𝑛𝑥,𝑎^−𝑛𝑦<∞ (41) for all 𝑥,𝑦∈𝑋. Let 𝑓∶𝑋→𝑌 be a mapping such that 𝑓𝑎𝑥+𝑏𝑦+𝑓𝑎𝑥−𝑏𝑦−2𝑎²𝑓(𝑥)−2𝑏²𝑓𝑦+𝑐𝑓𝑥+𝑦+𝑓𝑥−𝑦−2𝑓(𝑥)−2𝑓𝑦≤𝜙𝑥,𝑦 (42) for a fixed rational number 𝑎 and fixed nonzero real numbers 𝑏, 𝑐 with 𝑎≠1,0−1 and 𝑎²≠𝑏². Then there exists a unique quadratic mapping 𝑄∶𝑋→𝑌 such that 𝑄(𝑥)(𝑥)(0≤1−𝑓−𝑓)2^∞_𝑛=0𝑎^2(𝑛+1)𝜙𝑎^−𝑛𝑥,0+𝜙(0,0) (43) for all 𝑥∈𝑋.

For the stability problem of quadratic functional equations, we can show that many quadratic functional equations turn out to be types of (6) or to be deformed into the type of (6). For example, Gordji and Khodaei [13] investigated the following functional equation: 𝑓=𝑏()𝑎𝑥+𝑏𝑦+𝑓𝑎𝑥−𝑏𝑦𝑎+𝑏2𝑓+𝑏()𝑥+𝑦𝑎+𝑏2𝑓+𝑥−𝑦2𝑎²−𝑎𝑏−𝑏²𝑓(𝑥)+𝑏²𝑓𝑦,𝑓−𝑎𝑏𝑎𝑥+𝑏𝑦+𝑓𝑎𝑥−𝑏𝑦=2𝑎²𝑓(𝑥)+𝑏²𝑓𝑦. (44)(45) Indeed, the functional equation (44) can be written as 𝑓𝑓(𝑥)𝑦𝑎𝑥+𝑏𝑦+𝑓𝑎𝑥−𝑏𝑦+𝑐𝑥+𝑦+𝑓𝑥−𝑦−2𝑓−2𝑓=2𝑎²𝑓(𝑥)+2𝑏²𝑓𝑦, (46) where 𝑐=−𝑏(𝑎+𝑏)/2. Hence the functional equations (44) and (45) are special cases of the functional equation (6).

As another example, Jun et al. [16] investigated the following functional equation: 𝑓𝑎𝑥+𝑦+𝑎𝑓𝑥−𝑦=(𝑎+1)𝑓𝑦+𝑎(𝑎+1)𝑓(𝑥), (47) where 𝑎 is an integer with 𝑎≠−1,0. Suppose that 𝑓 satisfies (47). Then clearly, 𝑓 is even, and hence the functional equation (47) can be deformed into 𝑓𝑓(𝑥)𝑦𝑎𝑥+𝑦+𝑓𝑎𝑥−𝑦+𝑎𝑥+𝑦+𝑓𝑥−𝑦−2𝑓−2𝑓=2𝑎²𝑓(𝑥)+2𝑓𝑦 (48) for all 𝑥,𝑦∈𝑋. That is, we can transform (47) into the type of (6).

As an example of 𝜙(𝑥,𝑦) in Theorems 6 and 7, we can take 𝜙(𝑥,𝑦)=𝜖(‖𝑥‖^𝑝‖𝑦‖^𝑝+‖𝑥‖^2𝑝+‖𝑦‖^2𝑝) which appeared in [17]. Then we can formulate the following corollary.

Corollary 8. Let 𝑝 be a real number with 𝑝≠1. Let 𝑓∶𝑋→𝑌 be a mapping such that 𝑓𝑎𝑥+𝑏𝑦+𝑓𝑎𝑥−𝑏𝑦−2𝑎²𝑓(𝑥)−2𝑏²𝑓𝑦𝑓(𝑥)𝑦‖𝑥‖+𝑐𝑥+𝑦+𝑓𝑥−𝑦−2𝑓−2𝑓≤𝜖^𝑝𝑦^𝑝+‖𝑥‖^2𝑝+𝑦^2𝑝 (49) for a fixed rational number 𝑎 and fixed nonzero real numbers 𝑏, 𝑐 with 𝑎≠1,0,−1 and 𝑎²≠𝑏². Then there exists a unique quadratic mapping 𝑄∶𝑋→𝑌 such that 𝑄(𝑥)(𝑥)(0≤𝜖‖𝑥‖−𝑓−𝑓)^2𝑝21−𝑎^2(1−𝑝)|𝑎||𝑎|,𝑄(𝑥)(𝑥)(0≤𝜖‖𝑥‖𝑝>1,>1,𝑜𝑟𝑝<1,<1−𝑓−𝑓)^2𝑝21−𝑎^2(𝑝−1)𝑝<1,|𝑎|<1,𝑜𝑟𝑝>1,|𝑎|>1 (50) for all 𝑥∈𝑋.

We remark that the functional equation (6) is not stable for 𝑝=1 in Corollary 8. The following example, which is a special case of the example in [18], shows that (6) is not stable for 𝑝=1 especially in the case of 𝑎=2, 𝑏=1, and 𝑐=−1. We give a proof for the reader’s convenience.

Example 9. Let 𝑡∶ℝ→ℝ be a mapping defined by 𝑡(𝑥)=𝑥²,if|𝑥|<1,1,otherwise, (51) and define a mapping 𝑓∶ℝ→ℝ by 𝑓(𝑥)=^∞_𝑛=0𝑡2^𝑛𝑥4^𝑛. (52)
We will show that 𝑓 satisfies the functional inequality 𝑓(𝑥)𝑦−𝑓(𝑥)𝑦≤2𝑥+𝑦+𝑓2𝑥−𝑦−8𝑓−2𝑓𝑥+𝑦+𝑓𝑥−𝑦−2𝑓−2𝑓6403|𝑥|𝑦+|𝑥|²+𝑦² (53) for all 𝑥,𝑦∈ℝ, but there do not exist a quadratic mapping 𝑄∶ℝ→ℝ and a positive constant 𝐾 such that 𝑄(𝑥)−𝑓(𝑥)≤𝐾‖𝑥‖² (54) for all 𝑥∈ℝ.

Note that |𝑓(𝑥)|≤4/3 for all 𝑥∈ℝ. For any mapping 𝑔∶ℝ→ℝ, let (𝑥)𝑦−𝑔(𝑥)𝑦𝐷𝑔𝑥,𝑦=𝑔2𝑥+𝑦+𝑔2𝑥−𝑦−8𝑔−2𝑔𝑥+𝑦+𝑔𝑥−𝑦−2𝑔−2𝑔 (55) for all 𝑥,𝑦∈ℝ.

First, suppose that 1/16≤|𝑥||𝑦|+|𝑥|²+|𝑦|². Then |𝐷𝑓(𝑥,𝑦)|≤(640/3)(|𝑥||𝑦|+|𝑥|²+|𝑦|²) for all 𝑥,𝑦∈ℝ.

Now suppose that 1/16>|𝑥||𝑦|+|𝑥|²+|𝑦|². Then there is a positive integer 𝑚 such that 12^2𝑚+3≤|𝑥|𝑦+|𝑥|²+𝑦²<12^2𝑚+2, (56) and so 2^𝑚|𝑥|<12,2^𝑚𝑦<12. (57) Hence, we have 2^𝑚−12𝑥±𝑦,2^𝑚−1𝑥±𝑦,2^𝑚−1𝑥,2^𝑚−1𝑦⊆(−1,1). (58) Hence for any 𝑛=0,1,2,…,𝑚−1, 𝐷𝑡2^𝑛𝑥,2^𝑛𝑦=0, (59) and so 𝐷𝑓𝑥,𝑦≤^∞_𝑛=014^𝑛𝐷𝑡2^𝑛𝑥,2^𝑛𝑦=^∞_𝑛=𝑚14^𝑛𝐷𝑡2^𝑛𝑥,2^𝑛𝑦≤403×2^2𝑚≤3203|𝑥|𝑦+|𝑥|²+𝑦². (60) Thus 𝑓 satisfies (53).

Suppose that there exist a quadratic mapping 𝑄∶ℝ→ℝ and a positive constant 𝐾 with (54). Since |𝑓(𝑥)|≤4/3, −𝐾𝑥²−43≤𝑄(𝑥)≤𝐾𝑥²+43 (61) for all 𝑥∈ℝ, and since 𝑄 is quadratic, −𝐾𝑥²−43𝑛²≤𝑄(𝑥)≤𝐾𝑥²+43𝑛² (62) for all 𝑥∈ℝ and all natural numbers 𝑛. Hence, we have |𝑄(𝑥)|≤𝐾𝑥² (63) for all 𝑥∈ℝ, and so, by (54), we have 𝑓(𝑥)≤2𝐾𝑥² (64) for all 𝑥∈ℝ.

Take a positive integer 𝑙 such that 𝑙>2𝐾, and pick 𝑥∈ℝ with 2^𝑙−1|𝑥|<1. Then 𝑓(𝑥)=^∞_𝑛=0𝑡2^𝑛𝑥4^𝑛>^𝑙−1_𝑛=0𝑡2^𝑛𝑥4^𝑛=^𝑙−1_𝑛=0𝑥²=𝑙𝑥²>2𝐾𝑥², (65) which contradicts (64).

4. Deforming Inequalities into the Type of (29)

It turns out that lots of functional inequalities can be deformed into inequality (29). So we can regard inequality (29) as a typical form of a certain class of functional inequalities. In this point of view, we have a following systematic program to prove the generalized Hyers-Ulam stability of certain functional inequalities.

Step 1. Deform a given inequality into the type of (29) and get a modified bound function.

Step 2. Apply Theorem 6 for the modified bound function.

It should be remarked that if a functional inequality can be deformed into the type of (29), then a solution of the original functional equation is quadratic. And, it can be easily checked that the resulting unique quadratic mapping 𝑄 in Step 2 also satisfies the original functional equation. So we don’t need to worry anything about the given functional equation in our program. In this section, we illustrate just two of them.

First, we consider the following functional equation: 𝑓+()𝑓=()𝑓(𝑥)+()𝑓𝑦𝑎𝑥+𝑦𝑎−1𝑥−𝑦−𝑓𝑥+𝑦𝑎+2)(𝑎−1𝑎−1 (66) for some rational number 𝑎 with 𝑎≠1,−1,0.

Theorem 10. Let 𝜙∶𝑋²→[0,∞) be a function with (28). Let 𝑓∶𝑋→𝑌 be a mapping satisfying 𝑓(0)=0 and 𝑓+()𝑓−()𝑓(𝑥)−()𝑓𝑦𝑎𝑥+𝑦𝑎−1𝑥−𝑦−𝑓𝑥+𝑦𝑎+2)(𝑎−1𝑎−1≤𝜙𝑥,𝑦 (67) for some rational number 𝑎 with 𝑎≠1,−1,0. Then there exists a unique quadratic mapping 𝑄∶𝑋→𝑌 such that 𝑄 satisfies (66) and 𝑄(𝑥)−𝑓(𝑥)≤^∞_𝑛=0𝑎^−2(𝑛+1)𝜙𝑎^𝑛+1𝑥,02𝜙(0,0) (68) for all 𝑥∈𝑋.

Proof. Setting 𝑦=−𝑦 in (67), we have 𝑓+()𝑓−()𝑓(𝑥)−()𝑓,𝑎𝑥−𝑦𝑎−1𝑥+𝑦−𝑓𝑥−𝑦𝑎+2)(𝑎−1𝑎−1−𝑦≤𝜙𝑥,−𝑦 (69) and by (67) and (69), we have 𝑓𝑎𝑥+𝑦+𝑓𝑎𝑥−𝑦−2𝑎²𝑓(𝑥)𝑦+(𝑓(𝑥)𝑦+(𝑓𝑦−2𝑓𝑎−2)𝑥+𝑦+𝑓𝑥−𝑦−2𝑓−2𝑓𝑎−1)−𝑓−𝑦≤𝜙𝑥,𝑦+𝜙𝑥,−𝑦 (70) for all 𝑥,𝑦∈𝑋. Letting 𝑥=0 in (67), we have (𝑎−1)𝑓𝑦−𝑓−𝑦≤𝜙0,𝑦 (71) for all 𝑦∈𝑋. Hence, by (70) and (71), we have 𝑓𝑎𝑥+𝑦+𝑓𝑎𝑥−𝑦−2𝑎²𝑓(𝑥)𝑦+(𝑓(𝑥)𝑦−2𝑓𝑎−2)𝑥+𝑦+𝑓𝑥−𝑦−2𝑓−2𝑓≤Φ𝑥,𝑦 (72) for all 𝑥,𝑦∈𝑋, where Φ(𝑥,𝑦)=𝜙(𝑥,𝑦)+𝜙(𝑥,−𝑦)+𝜙(0,𝑦). So by Theorem 6, we get the result.

Remark 11. It would be interesting to see how Theorem 10 works well for a simple case of 𝜙. Take 𝜙(𝑥,𝑦)=𝜖>0. Then the original inequality in Theorem 10 is 𝑓+()𝑓−()𝑓(𝑥)−()𝑓𝑦𝑎𝑥+𝑦𝑎−1𝑥−𝑦−𝑓𝑥+𝑦𝑎+2)(𝑎−1𝑎−1≤𝜖. (73) After the deforming process, inequality (73) turns into the following new inequality which is standard in our sense: 𝑓𝑎𝑥+𝑦+𝑓𝑎𝑥−𝑦−2𝑎²𝑓(𝑥)𝑦+(𝑓(𝑥)𝑦−2𝑓𝑎−2)𝑥+𝑦+𝑓𝑥−𝑦−2𝑓−2𝑓≤3𝜖. (74) With 𝑓(0)=0, apply Theorem 6 or Theorem 7 (𝜙(𝑥,𝑦)=3𝜖 in the theorems) to inequality (74); we get the following conclusion.
There exists a unique quadratic mapping 𝑄∶𝑋→𝑌 such that 𝑄 satisfies (66) and 𝑄(𝑥)−𝑓(𝑥)≤3𝜖2𝑎².−1 (75)

Now, we consider the following functional equation: 𝑓(𝑥)𝑦.3𝑥+2𝑦+2𝑓3𝑥−2𝑦−6𝑓𝑥−𝑦=21𝑓+6𝑓 (76)

Theorem 12. Let 𝜙∶𝑋²→[0,∞) be a function with (28) for 𝑎=3. Let 𝑓∶𝑋→𝑌 be a mapping satisfying 𝑓(0)=0 and 𝑓(𝑥)𝑦.3𝑥+2𝑦+2𝑓3𝑥−2𝑦−6𝑓𝑥−𝑦−21𝑓−6𝑓≤𝜙𝑥,𝑦 (77) Then there exists a unique quadratic mapping 𝑄∶𝑋→𝑌 such that 𝑄 satisfies (76) and 𝑄(𝑥)(𝑥≤1−𝑓)3^∞_𝑛=03^−2(𝑛+1)𝜙3^𝑛𝑥,0+6𝜙(0,0) (78) for all 𝑥∈𝑋.

Proof. Setting 𝑥=0 in (77), we have 𝑓2𝑦+2𝑓−2𝑦−6𝑓𝑦+𝑓−𝑦≤𝜙0,𝑦 (79) for all 𝑦∈𝑋. Letting 𝑦=−𝑦 in (79), we have 2𝑓2𝑦+𝑓−2𝑦−6𝑓𝑦+6𝑓−𝑦≤𝜙0,−𝑦 (80) for all 𝑦∈𝑋. By (79) and (80), we have 𝑓2𝑦−𝑓−2𝑦≤𝜙0,𝑦+𝜙0,−𝑦 (81) for all 𝑦∈𝑋. Hence by (81), we have 𝑓𝑦𝑦−𝑓−𝑦≤𝜙0,2+𝜙0,−𝑦2 (82) for all 𝑦∈𝑋.
Letting 𝑦=−𝑦 in (77), we get 𝑓(𝑥),3𝑥−2𝑦+2𝑓3𝑥+2𝑦−6𝑓𝑥+𝑦−21𝑓−6𝑓−𝑦≤𝜙𝑥,−𝑦 (83) and by (77) and (83), we have 3𝑓(𝑥)𝑦𝑓(𝑥)𝑦𝑓𝑦3𝑥−2𝑦+𝑓3𝑥+2𝑦−18𝑓−8𝑓−6𝑥+𝑦+𝑓𝑥−𝑦−2𝑓−2𝑓+6−𝑓−𝑦≤𝜙𝑥,𝑦+𝜙𝑥,−𝑦 (84) for all 𝑥,𝑦∈𝑋, and so by (82), we have 𝑓(𝑥)𝑦𝑓(𝑥)𝑦3𝑥−2𝑦+𝑓3𝑥+2𝑦−18𝑓−8𝑓−2𝑥+𝑦+𝑓𝑥−𝑦−2𝑓−2𝑓≤Φ𝑥,𝑦 (85) for all 𝑥,𝑦∈𝑋, where Φ(𝑥,𝑦)=(1/3)[𝜙(𝑥,𝑦)+𝜙(𝑥,−𝑦)]+2[𝜙(0,𝑦/2)+𝜙(0,−𝑦/2)]. So by Theorem 6, we get the result.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The first author was supported by the research fund of Dankook University in 2013.