Solution and Stability of a General Mixed Type Cubic and Quartic Functional Equation

Abstract

We consider the following mixed type cubic and quartic functional equation 𝜆[𝑓(𝑥+𝜆𝑦)+𝑓(𝑥−𝜆𝑦)] = 𝜆³[𝑓(𝑥+𝑦)+𝑓(𝑥−𝑦)]−2𝜆³(𝜆+1)𝑓(𝑦)−2𝜆(𝜆²−1)𝑓(𝑥)+2(𝜆+1)𝑓(𝜆𝑦), where 𝜆 is a fixed integer. We establish the general solution of the functional equation when the integer 𝜆≠0,±1, and then, by using the fixed point alternative, we investigate the generalized Hyers-Ulam-Rassias stability for this functional equation when the integer 𝜆≥2.

1. Introduction

In 1940, Ulam [1] asked the fundamental question for the stability for the group homomorphisms.

Let (𝐺₁,∗) be a group, and let (𝐺₂,⋄,𝑑) be a metric group with the metric 𝑑(⋅,⋅). Given 𝜀>0, does there exist a 𝛿(𝜀)>0 such that if a mapping ℎ∶𝐺₁→𝐺₂ satisfies the inequality 𝑑ℎ𝑥∗𝑦,ℎ(𝑥)⋄ℎ𝑦<𝛿 (1) for all 𝑥,𝑦∈𝐺₁, then there is a homomorphism 𝐻∶𝐺₁→𝐺₂ with 𝑑(ℎ(𝑥),𝐻(𝑥))<𝜀 (2) for all 𝑥∈𝐺₁?

In other words, under what conditions, does there exist a homomorphism near an approximately homomorphism? In the next year, Hyers [2] gave the first affirmative answer to the question of Ulam for Cauchy equation in the Banach spaces. Then, Rassias [3] generalized Hyers’ result by considering an unbounded Cauchy difference, and this stability phenomenon is known as generalized Hyers-Ulam-Rassias stability or Hyers-Ulam-Rassias stability. During the last three decades, the stability problem for several functional equations has been extensively investigated by many mathematicians; see, for example, [4–9] and the references therein. We also refer the readers to the books [10–13].

In [14], Jun and Kim introduced the following functional equation 𝑓2𝑥+𝑦+𝑓2𝑥−𝑦=2𝑓𝑥+𝑦+2𝑓𝑥−𝑦+12𝑓(𝑥). (3) It is easy to see that the function 𝑓(𝑥)=𝑥³ satisfies the functional equation (3). Thus, it is natural that (3) is called a cubic functional equation and every solution of the cubic functional equation is said to be a cubic function. In [14], Jun and Kim established the general solution and the generalized Hyers-Ulam-Rassias stability for (3). They proved that a function 𝑓∶𝑋→𝑌 between real vector spaces is a solution of the functional equation (3) if and only if there exists a function 𝐺∶𝑋×𝑋×𝑋→𝑌 such that 𝑓(𝑥)=𝐺(𝑥,𝑥,𝑥) for all 𝑥∈𝑋 and 𝐺 is symmetric for each fixed one variable and is additive for fixed two variables.

In [15], Lee et al. considered the following quartic functional equation 𝑓𝑓(𝑥)𝑦.2𝑥+𝑦+𝑓2𝑥−𝑦=4𝑥+𝑦+𝑓𝑥−𝑦+24𝑓−6𝑓 (4) Since the function 𝑓(𝑥)=𝑥⁴ satisfies the functional equation (4), the functional equation (4) is called a quartic functional equation and every solution of the quartic functional equation is said to be a quartic function. In [15], the authors solved the functional equation (4) and proved the stability for it. Actually, they obtained that a function 𝑓∶𝑋→𝑌 between real vector spaces satisfies the functional equation (4) if and only if there exists a symmetric biquadratic function 𝐵∶𝑋×𝑋→𝑌 such that 𝑓(𝑥)=𝐵(𝑥,𝑥) for all 𝑥∈𝑋. A function 𝑓∶𝑋→𝑌 between real vector spaces is said to be quadratic if it satisfies the following functional equation 𝑓𝑥+𝑦+𝑓𝑥−𝑦=2𝑓(𝑥)+2𝑓𝑦 (5) for all 𝑥,𝑦∈𝑋, and a function 𝐵∶𝑋×𝑋→𝑌 is said to be bi-quadratic if 𝐵 is quadratic for each fixed one variable (see [8]). The following mixed type cubic and quartic functional equation was introduced by Eshaghi Gordji et al. [16] 𝑓𝑓𝑦(𝑥).𝑥+2𝑦+𝑓𝑥−2𝑦=4𝑥+𝑦+𝑓𝑥−𝑦−24𝑓−6𝑓+3𝑓2𝑦 (6) It can be verified that the function 𝑓(𝑥)=𝑥³+𝑥⁴ satisfies the functional equation (6). In [16], the authors obtained the general solution and the generalized Hyers-Ulam-Rassias stability of (6) in quasi-Banach space. The literature on the stability of the mixed type functional equations is very rich; see [17–22].

In the present paper, we extend (6) and consider the following functional equation 𝜆𝑓𝑥+𝜆𝑦+𝑓𝑥−𝜆𝑦=𝜆³𝑓𝑥+𝑦+𝑓𝑥−𝑦−2𝜆³()𝑓𝑦𝜆𝜆+1−2𝜆²−1𝑓(𝑥)+2(𝜆+1)𝑓𝜆𝑦, (7) where 𝜆 is a fixed integer. One can see that the functional equation (6) is a special case of (7) when we take the integer 𝜆=2.

In 2003, Radu [23] noticed that the fixed point theorem, which was established by Diaz and Margolis [24], plays an important part in solving the stability problem of functional equations. Subsequently, this method has been successfully used by many mathematicians to investigate the stability of several functional equations; see, for example, [21, 25, 26] and the references therein.

In this paper, we first establish the general solution of functional equation (7) when the integer 𝜆≠0,±1 and 𝑓 is a mapping between vector spaces. Then, by using the fixed point method, we prove the generalized Hyers-Ulam-Rassias stability of the functional equation (7) when the integer 𝜆≥2 and 𝑓 is a mapping from the normed space to the Banach space.

2. Solution of the Functional Equation (7)

Recall form [14, 15] that every solution of the cubic functional equation (3) and the quartic functional equation (4) is said to be a cubic function and a quartic function, respectively. In this section, we investigate the general solution of the mixed type cubic and quartic functional equation (7). Throughout this section, let 𝑋 and 𝑌 be two real vector spaces, and we always assume that the integer 𝜆 in the functional equation (7) is different from 0, −1, and 1. Before proving our main theorem, we first give the following two lemmas.

Lemma 1. If an odd mapping 𝑓∶𝑋→𝑌 satisfies (7), then 𝑓 is cubic.

Proof. Note that, in view of the oddness of 𝑓, we have 𝑓(0)=0 and 𝑓(−𝑥)=−𝑓(𝑥) for all 𝑥∈𝑋. Letting 𝑥=0 in (7), we get 𝑓𝜆𝑦=𝜆³𝑓𝑦 (8) for all 𝑦∈𝑋. Applying (8) to (7), we obtain 𝑓𝑥+𝜆𝑦+𝑓𝑥−𝜆𝑦=𝜆²𝑓𝑥+𝑦+𝑓𝑥−𝑦−2𝜆²−1𝑓(𝑥) (9) for all 𝑥,𝑦∈𝑋. Replacing 𝑦 by 𝑥+𝑦 in (9), we get 𝑓(𝜆+1)𝑥+𝜆𝑦−𝑓(𝜆−1)𝑥+𝜆𝑦=𝜆²𝑓2𝑥+𝑦−𝑓𝑦−2𝜆²−1𝑓(𝑥) (10) for all 𝑥,𝑦∈𝑋. Now, if we replace 𝑥 by 𝜆𝑥+𝑦 in (9) and use (8), we see that 𝑓𝜆𝑥+(𝜆+1)𝑦+𝑓𝜆𝑥−(𝜆−1)𝑦=𝜆²𝑓𝜆𝑥+2𝑦+𝜆³𝑓(𝑥)−2𝜆²−1𝑓𝜆𝑥+𝑦 (11) for all 𝑥,𝑦∈𝑋. Interchanging 𝑥 with 𝑦 in (11) and using the oddness of 𝑓, we get the relation 𝑓(𝜆+1)𝑥+𝜆𝑦−𝑓(𝜆−1)𝑥−𝜆𝑦=𝜆²𝑓2𝑥+𝜆𝑦+𝜆³𝑓𝑦−2𝜆²−1𝑓𝑥+𝜆𝑦 (12) for all 𝑥,𝑦∈𝑋. Then, subtracting (12) from (10), one has 𝑓(𝜆−1)𝑥−𝜆𝑦−𝑓(𝜆−1)𝑥+𝜆𝑦=𝜆²𝑓−𝜆2𝑥+𝑦−𝑓2𝑥+𝜆𝑦₂+𝜆⁵𝑓𝑦+2𝜆²−1𝑓𝑥+𝜆𝑦−𝑓(𝑥) (13) for all 𝑥,𝑦∈𝑋. Now, replacing 𝑦 by −𝑦 in (13) and noting that 𝑓 is odd, we have 𝑓(𝜆−1)𝑥+𝜆𝑦−𝑓(𝜆−1)𝑥−𝜆𝑦=𝜆²𝑓2𝑥−𝑦−𝑓2𝑥−𝜆𝑦+𝜆²+𝜆⁵𝑓𝑦+2𝜆²−1𝑓𝑥−𝜆𝑦−𝑓(𝑥) (14) for all 𝑥,𝑦∈𝑋. Adding (13) to (14) gives 𝜆²𝑓𝜆2𝑥+𝑦+𝑓2𝑥−𝑦+2²−1𝑓𝑥+𝜆𝑦+𝑓𝑥−𝜆𝑦=𝜆²𝑓2𝑥+𝜆𝑦+𝑓2𝑥−𝜆𝑦+4𝜆²−1𝑓(𝑥) (15) for all 𝑥,𝑦∈𝑋. Replacing 𝑥 by 2𝑥 in (9) and using (8), we get 𝑓2𝑥+𝜆𝑦+𝑓2𝑥−𝜆𝑦=𝜆²𝑓2𝑥+𝑦+𝑓2𝑥−𝑦−16𝜆²−1𝑓(𝑥). (16) Applying (9) and (16) to (15) yields that 𝑓𝑓(𝑥)2𝑥+𝑦+𝑓2𝑥−𝑦=2𝑥+𝑦+𝑓𝑥−𝑦+12𝑓 (17) for all 𝑥,𝑦∈𝑋. Thus, the mapping 𝑓∶𝑋→𝑌 is cubic. This completes the proof.

Lemma 2. If an even mapping 𝑓∶𝑋→𝑌 satisfies (7) for all 𝑥,𝑦∈𝑋, then 𝑓 is quartic.

Proof. In view of the evenness of 𝑓, we have 𝑓(−𝑥)=𝑓(𝑥) for all 𝑥∈𝑋. Putting 𝑥=𝑦=0 in (7), we get 𝑓(0)=0. Then, let 𝑥=0 in (7), we obtain 𝑓𝜆𝑦=𝜆⁴𝑓𝑦 (18) for all 𝑦∈𝑋. Combing (7) and (18) implies the following equation 𝑓𝑥+𝜆𝑦+𝑓𝑥−𝜆𝑦=𝜆²𝑓𝜆𝑥+𝑦+𝑓𝑥−𝑦−2²−1𝑓(𝑥)+2𝜆²𝜆²−1𝑓𝑦 (19) for all 𝑥,𝑦∈𝑋. Replacing 𝑦 by 𝑥+𝑦 in (19) and note that 𝑓 is even, we get 𝑓(𝜆+1)𝑥+𝜆𝑦+𝑓(𝜆−1)𝑥+𝜆𝑦=𝜆²𝑓2𝑥+𝑦+𝑓𝑦−2𝜆²−1𝑓(𝑥)+2𝜆²𝜆²−1𝑓𝑥+𝑦 (20) for all 𝑥,𝑦∈𝑋. Replacing 𝑥 by 𝜆𝑥+𝑦 in (19) and using (18), we get 𝑓𝜆𝑥+(𝜆+1)𝑦+𝑓𝜆𝑥−(𝜆−1)𝑦=𝜆²𝑓𝜆𝑥+2𝑦+𝜆⁴𝑓(𝑥𝜆)−2²−1𝑓𝜆𝑥+𝑦+2𝜆²𝜆²−1𝑓𝑦 (21) for all 𝑥,𝑦∈𝑋. Interchanging the roles of 𝑥 and 𝑦 in (21), we obtain 𝑓(𝜆+1)𝑥+𝜆𝑦+𝑓(𝜆−1)𝑥−𝜆𝑦=𝜆²𝑓2𝑥+𝜆𝑦+𝜆⁴𝑓𝑦−2𝜆²−1𝑓𝑥+𝜆𝑦+2𝜆²𝜆²−1𝑓(𝑥) (22) for all 𝑥,𝑦∈𝑋. If we subtract (22) from (20), we obtain 𝑓(𝜆−1)𝑥+𝜆𝑦−𝑓(𝜆−1)𝑥−𝜆𝑦=𝜆²𝑓𝜆2𝑥+𝑦−𝑓2𝑥+𝜆𝑦+2²−1𝑓𝑥+𝜆𝑦+𝜆²−𝜆⁶𝑓𝑦+2𝜆²𝜆²−1𝑓𝑥+𝑦−2𝜆²+1𝜆²−1𝑓(𝑥) (23) for all 𝑥,𝑦∈𝑋. Replacing 𝑦 by −𝑦 in (23), we get 𝑓(𝜆−1)𝑥−𝜆𝑦−𝑓(𝜆−1)𝑥+𝜆𝑦=𝜆²𝑓𝜆2𝑥−𝑦−𝑓2𝑥−𝜆𝑦+2²−1𝑓𝑥−𝜆𝑦+𝜆²−𝜆⁶𝑓𝑦+2𝜆²𝜆²−1𝑓𝑥−𝑦−2𝜆²+1𝜆²−1𝑓(𝑥) (24) for all 𝑥,𝑦∈𝑋. If we add (23) to (24), we have 𝜆²𝑓2𝑥+𝑦+𝑓2𝑥−𝑦−𝜆²𝑓𝜆2𝑥+𝜆𝑦+𝑓2𝑥−𝜆𝑦+2²−1𝑓𝑥+𝜆𝑦+𝑓𝑥−𝜆𝑦+2𝜆²𝜆²𝑓𝜆−1𝑥+𝑦+𝑓𝑥−𝑦=4²+1𝜆²−1𝑓(𝑥)+2𝜆⁶−𝜆²𝑓𝑦 (25) for all 𝑥,𝑦∈𝑋. Replacing 𝑥 by 2𝑥 in (19) and using (18), we get 𝑓2𝑥+𝜆𝑦+𝑓2𝑥−𝜆𝑦=𝜆²𝑓𝜆2𝑥+𝑦+𝑓2𝑥−𝑦−32²−1𝑓(𝑥)+2𝜆²𝜆²−1𝑓𝑦 (26) for all 𝑥,𝑦∈𝑋. Applying (19) and (26) to (25), we obtain that 𝑓𝑓(𝑥)𝑦2𝑥+𝑦+𝑓2𝑥−𝑦=4𝑥+𝑦+𝑓𝑥−𝑦+24𝑓−6𝑓 (27) for all 𝑥,𝑦∈𝑋. Therefore, the mapping 𝑓∶𝑋→𝑌 is quartic and the proof is complete.

Now, we are ready to find out the general solution of (7).

Theorem 3. A mapping 𝑓∶𝑋→𝑌 satisfies (7) for all 𝑥,𝑦∈𝑋 if and only if there exist a symmetric multiadditive mapping 𝐺∶𝑋×𝑋×𝑋→𝑌 and a symmetric bi-quadratic mapping 𝐵∶𝑋×𝑋→𝑌 such that 𝑓(𝑥)=𝐺(𝑥,𝑥,𝑥)+𝐵(𝑥,𝑥) for all 𝑥∈𝑋.

Proof. First, we assume that there exist a symmetric multi-additive mapping 𝐺∶𝑋×𝑋×𝑋→𝑌 and a symmetric bi-quadratic mapping 𝐵∶𝑋×𝑋→𝑌 such that 𝑓(𝑥)=𝐺(𝑥,𝑥,𝑥)+𝐵(𝑥,𝑥) for all 𝑥∈𝑋. We need to show that the function 𝑓 satisfies (7). By a simple computation, one can obtain that the function 𝑥↦𝐺(𝑥,𝑥,𝑥) satisfies (7). Thus, to show that the function 𝑓 satisfies (7), we only need to show that the function 𝑥↦𝐵(𝑥,𝑥) also satisfies (7); namely, 𝜆𝐵𝑥+𝜆𝑦,𝑥+𝜆𝑦+𝐵𝑥−𝜆𝑦,𝑥−𝜆𝑦=𝜆³𝐵𝑥+𝑦,𝑥+𝑦+𝐵𝑥−𝑦,𝑥−𝑦−2𝜆³()𝐵𝜆𝜆+1𝑦,𝑦−2𝜆²−1𝐵(𝑥,𝑥)+2(𝜆+1)𝐵𝜆𝑦,𝜆𝑦 (28) for all 𝑥,𝑦∈𝑋. Since 𝐵∶𝑋×𝑋→𝑌 is a bi-quadratic mapping, it can be verified that 𝐵𝜆𝑥,𝜆𝑦=𝜆⁴𝐵𝑥,𝑦 (29) for all integer 𝜆 and all 𝑥,𝑦∈𝑋. Then, (28) becomes 𝐵𝑥+𝜆𝑦,𝑥+𝜆𝑦+𝐵𝑥−𝜆𝑦,𝑥−𝜆𝑦=𝜆²𝐵𝜆𝑥+𝑦,𝑥+𝑦+𝐵𝑥−𝑦,𝑥−𝑦−2²−1𝐵(𝑥,𝑥)+2𝜆²𝜆²−1𝐵𝑦,𝑦. (30) To establish (28), it suffices to show (30). Note that if (30) holds for some integer 𝜆, then so does −𝜆. Thus, in the following, we will show that (30) holds for all positive integers 𝜆 with 𝜆≥2 and all 𝑥,𝑦∈𝑋. To do this, we use induction on 𝜆. Fix any 𝑥,𝑦∈𝑋. In the case when 𝜆=2, we have 𝐵=𝐵+𝐵−𝐵𝐵𝐵−()𝐵𝐵𝐵𝐵−()𝐵𝐵𝐵()𝐵−()𝐵𝐵()𝐵𝐵()𝐵().𝑥+2𝑦,𝑥+2𝑦+𝐵𝑥−2𝑦,𝑥−2𝑦𝑥+𝑦+𝑦,𝑥+2𝑦+𝐵𝑥+𝑦−𝑦,𝑥+2𝑦𝑥−𝑦+𝑦,𝑥−2𝑦+𝐵𝑥−𝑦−𝑦,𝑥−2𝑦𝑥,𝑥+2𝑦+𝐵𝑥,𝑥−2𝑦=2𝑥+𝑦,𝑥+2𝑦+𝐵𝑦,𝑥+2𝑦+2𝑥−𝑦,𝑥−2𝑦+𝐵𝑦,𝑥−2𝑦2𝐵𝑥,𝑥+2𝐵𝑥,2𝑦=2𝑥+𝑦,𝑥+𝑦+𝑦+𝐵𝑥+𝑦,𝑥+𝑦−𝑦+2𝑥−𝑦,𝑥−𝑦−𝑦+𝐵𝑥−𝑦,𝑥−𝑦+𝑦−2𝑥+𝑦,𝑥+𝐵𝑥−𝑦,𝑥+2𝑦,𝑥+2𝑦+𝐵𝑦,𝑥−2𝑦2𝐵𝑥,𝑥+8𝐵𝑥,𝑦=4𝑥+𝑦,𝑥+𝑦+𝐵𝑥+𝑦,𝑦+4𝑥−𝑦,𝑥−𝑦+𝐵𝑥−𝑦,𝑦−4𝑥,𝑥+𝐵𝑦,𝑥+4𝑦,𝑥+𝐵𝑦,2𝑦2𝐵𝑥,𝑥+8𝐵𝑥,𝑦=4𝑥+𝑦,𝑥+𝑦+𝐵𝑥−𝑦,𝑥−𝑦+4𝑥+𝑦,𝑦+𝐵𝑥−𝑦,𝑦−6𝐵𝑥,𝑥−8𝐵𝑥,𝑦+16𝐵𝑦,𝑦=4𝑥+𝑦,𝑥+𝑦+𝐵𝑥−𝑦,𝑥−𝑦+8𝑥,𝑦+𝐵𝑦,𝑦−6𝐵𝑥,𝑥−8𝐵𝑥,𝑦+16𝐵𝑦,𝑦=4𝑥+𝑦,𝑥+𝑦+𝐵𝑥−𝑦,𝑥−𝑦−6𝐵𝑥,𝑥+24𝐵𝑦,𝑦 (31) So (30) is true for 𝜆=2. Here, we have used the bi-quadratic property of the function 𝐵 and (29). Now, assume that (30) is true for all positive integers that are less than or equal to some integer 𝜆(>2). Then, 𝐵()()𝑦()()𝑦=𝐵+𝐵−𝐵()()𝑦()()𝑦𝑥+𝜆+1𝑦,𝑥+𝜆+1+𝐵𝑥−𝜆+1𝑦,𝑥−𝜆+1𝑥+𝑦+𝜆𝑦,𝑥+𝑦+𝜆𝑦+𝐵𝑥+𝑦−𝜆𝑦,𝑥+𝑦−𝜆𝑦𝑥−𝑦+𝜆𝑦,𝑥−𝑦+𝜆𝑦+𝐵𝑥−𝑦−𝜆𝑦,𝑥−𝑦−𝜆𝑦𝑥+𝜆−1𝑦,𝑥+𝜆−1+𝐵𝑥−𝜆−1𝑦,𝑥−𝜆−1=𝜆²𝐵(𝜆𝑥+2𝑦,𝑥+2𝑦+𝐵𝑥,𝑥)−2²−1𝐵𝑥+𝑦,𝑥+𝑦+2𝜆²𝜆²−1𝐵𝑦,𝑦+𝜆²𝐵()𝑛𝑥,𝑥+𝐵𝑥−2𝑦,𝑥−2𝑦−2²−1𝐵𝑥−𝑦,𝑥−𝑦+2𝜆²𝜆²𝐵−)−1𝑦,𝑦(𝜆−1²𝐵()𝑥+𝑦,𝑥+𝑦+𝐵𝑥−𝑦,𝑥−𝑦−2𝜆−1²𝐵()()−1𝑥,𝑥+2𝜆−1²(𝜆−1)²−1𝐵𝑦,𝑦. (32) Applying (31) to (32), one can obtain that 𝐵()()𝑦()()𝑦=()𝑥+𝜆+1𝑦,𝑥+𝜆+1+𝐵𝑥−𝜆+1𝑦,𝑥−𝜆+1𝜆+1²𝐵()𝑥+𝑦,𝑥+𝑦+𝐵𝑥−𝑦,𝑥−𝑦−2𝜆+1²𝐵()()−1𝑥,𝑥+2𝜆+1²(𝜆+1)²−1𝐵𝑦,𝑦. (33) This means that (30) is true for 𝜆+1, and we have showed that the function 𝑥↦𝐵(𝑥,𝑥) satisfies (7). Therefore, the mapping 𝑓∶𝑋→𝑌 satisfies (7).
Conversely, we decompose 𝑓 into the odd part and the even part by putting 𝑓_𝑜(𝑥)=𝑓(𝑥)−𝑓(−𝑥)2,𝑓_𝑒(𝑥)=𝑓(𝑥)+𝑓(−𝑥)2 (34) for all 𝑥∈𝑋. Then, 𝑓(𝑥)=𝑓_𝑜(𝑥)+𝑓_𝑒(𝑥) for all 𝑥∈𝑋. It is easy to show that the mappings 𝑓_𝑜 and 𝑓_𝑒 satisfy (7). Hence, it follows from Lemmas 1 and 2 that the function 𝑓_𝑜 is cubic and 𝑓_𝑒 is quartic, respectively. Therefore, there exist a symmetric multi-additive mapping 𝐺∶𝑋×𝑋×𝑋→𝑌 such that 𝑓_𝑜(𝑥)=𝐺(𝑥,𝑥,𝑥) for all 𝑥∈𝑋 (see [14, Theorem  2.1]) and a symmetric bi-quadratic mapping 𝐵∶𝑋×𝑋→𝑌 such that 𝑓_𝑒(𝑥)=𝐵(𝑥,𝑥) for all 𝑥∈𝑋 (see [15, Theorem  2.1]). Hence, we get 𝑓(𝑥)=𝐺(𝑥,𝑥,𝑥)+𝐵(𝑥,𝑥) for all 𝑥∈𝑋. The proof is complete.

3. Generalized Hyers-Ulam-Rassias Stability of the Functional Equation (7)

In this section, we will investigate the stability of the functional equation (7) by using the fixed point alternative. Throughout this section, let 𝑋 be a real normed space and 𝑌 be a real Banach space, and we always assume that the integer 𝜆 used in the section is greater than or equal to 2. For convenience, we use the following abbreviation for a given function 𝑓∶𝑋→𝑌∶𝐷𝑓𝑥,𝑦∶=𝜆𝑓𝑥+𝜆𝑦+𝑓𝑥−𝜆𝑦−𝜆³𝑓𝑥+𝑦+𝑓𝑥−𝑦+2𝜆³(𝜆+1)𝑓𝑦+2𝜆𝜆²𝑓(𝑥)()𝑓−1−2𝜆+1𝜆𝑦 (35) for all 𝑥,𝑦∈𝑋.

Let us recall the following result by Diaz and Margolis.

Proposition 4 (see [24]). Let (𝐸,𝑑) be a complete generalized metric space (i.e., one for which 𝑑 may assume infinite value), and let 𝐽∶𝐸→𝐸 be a strictly contractive mapping with Lipschitz constant 𝐿<1; that is, 𝑑𝐽𝑥,𝐽𝑦≤𝐿𝑑𝑥,𝑦∀𝑥,𝑦∈𝑋. (36) Then, for each fixed element 𝑥₀∈𝐸, either 𝑑𝐽^𝑛𝑥₀,𝐽^𝑛+1𝑥₀=+∞ (37) for all nonnegative integers 𝑛 or there exists a non-negative integer 𝑛₀ such that(a)𝑑(𝐽^𝑛𝑥₀,𝐽^𝑛+1𝑥₀)<+∞ for all 𝑛≥𝑛₀;(b)the sequence {𝐽^𝑛𝑥₀} converges to a fixed point 𝑥^∗ of 𝐽;(c)𝑥^∗ is the unique fixed point of 𝐽 in the set ℱ∶={𝑥∈𝐸∣𝑑(𝐽^𝑛0𝑥₀,𝑥)<+∞};(d)𝑑(𝑥,𝑥^∗)≤(1/(1−𝐿))𝑑(𝑥,𝐽𝑥) for all 𝑥∈ℱ.

Lemma 5. Let 𝑓∶𝑋→𝑌 be an odd function for which there exists a function 𝜑∶𝑋×𝑋→[0,+∞) such that 𝐷𝑓𝑥,𝑦≤𝜑𝑥,𝑦 (38) for all 𝑥,𝑦∈𝑋. If there exists a constant 0<𝐿<1 such that 𝜑𝜆𝑥,𝜆𝑦≤𝜆³𝐿𝜑𝑥,𝑦 (39) for all 𝑥,𝑦∈𝑋, then there exists a unique cubic mapping 𝐶∶𝑋→𝑌 such that 𝑓(𝑥)(𝑥≤𝐿−𝐶)𝜙(𝑥)1−𝐿 (40) for all 𝑥∈𝑋, where 𝜙(𝑥)∶=(1/2(𝜆+1))𝜑(0,𝑥/𝜆).

Proof. It follows from (39) that lim_𝑟→∞𝜑𝜆^𝑟𝑥,𝜆^𝑟𝑦𝜆^3𝑟=0 (41) for all 𝑥,𝑦∈𝑋. Letting 𝑥=0 in (38) and replacing 𝑦 by 𝑥, we have 𝜆³𝑓(𝑥)−𝑓(𝜆𝑥)≤𝜙(𝜆𝑥) (42) for all 𝑥∈𝑋. By (39), we have 𝜙(𝜆𝑥)≤𝜆³𝐿𝜙(𝑥) for all 𝑥∈𝑋. This, together with (42), implies that 𝑓(𝑥)−1𝜆³𝑓(𝜆𝑥)≤𝐿𝜙(𝑥) (43) for all 𝑥∈𝑋. Let Ω be the set of all odd mappings 𝑔∶𝑋→𝑌. We introduce the generalized metric on Ω: 𝑑[)∣𝑔(𝑥)(𝑥(𝑥).𝑔,ℎ∶=inf𝐾∈0,∞−ℎ)≤𝐾𝜙,𝑥∈𝑋 (44) It is easy to show that (Ω,𝑑) is complete. Now, we define a function 𝑇∶Ω→Ω by (𝑥)1𝑇𝑔∶=𝜆³𝑔(𝜆𝑥) (45) for all 𝑔∈Ω and all 𝑥∈𝑋. Note that, for all 𝑔,ℎ∈Ω, 𝑑⟹𝑔(𝑥)(𝑥(𝑥)⟹1𝑔,ℎ≤𝐾−ℎ)≤𝐾𝜙,𝑥∈𝑋𝜆³𝑔()−1𝜆𝑥𝜆³ℎ(≤1𝜆𝑥)𝜆³()⟹1𝐾𝜙𝜆𝑥,𝑥∈𝑋𝜆³𝑔()−1𝜆𝑥𝜆³ℎ(𝜆𝑥)≤𝐿𝐾𝜙(𝑥),𝑥∈𝑋⟹𝑑𝑇𝑔,𝑇ℎ≤𝐿𝐾. (46) Hence, we obtain that 𝑑𝑇𝑔,𝑇ℎ≤𝐿𝑑𝑔,ℎ (47) for all 𝑔,ℎ∈Ω; that is, 𝑇 is a strictly contractive mapping of Ω with Lipschitz constant 𝐿. It follows from (43) that 𝑑(𝑇𝑓,𝑓)≤𝐿. Therefore, according to Proposition 4, the sequence {𝑇^𝑟𝑓} converges to a fixed point 𝐶 of 𝑇; that is, 𝐶∶𝑋⟶𝑌,𝐶(𝑥)=lim_𝑟→∞𝑇^𝑟𝑓(𝑥)=lim_𝑟→∞𝑓𝜆^𝑟𝑥𝜆^3𝑟 (48) and 𝐶(𝜆𝑥)=𝜆³𝐶(𝑥) for all 𝑥∈𝑋. Also 𝐶 is the unique fixed point of 𝑇 in the set Δ={𝑔∈Ω∣𝑑(𝑓,𝑔)<∞} and 𝑑≤1𝑓,𝐶𝑑≤𝐿1−𝐿𝑇𝑓,𝑓,1−𝐿 (49) which yields the inequality (40). It follows from the definition of 𝐶, (38), and (41) that 𝐷𝐶𝑥,𝑦=lim_𝑟→∞𝐷𝑓𝜆^𝑟𝑥,𝜆^𝑟𝑦𝜆^3𝑟≤lim_𝑟→∞𝜑𝜆^𝑟𝑥,𝜆^𝑟𝑦𝜆^3𝑟=0 (50) for all 𝑥,𝑦∈𝑋; that is, the mapping 𝐶∶𝑋→𝑌 satisfies (7). Since 𝑓 is odd, 𝐶 is odd. Therefore, Lemma 1 guarantees that 𝐶 is cubic. Finally, it remains to prove the uniqueness of 𝐶. Let 𝑆∶𝑋→𝑌 be another cubic function satisfying (40). Since 𝑑(𝑓,𝑆)≤𝐿/(1−𝐿) and 𝑆 is cubic, we get 𝑆∈Δ and 𝑇𝑆(𝑥)=(1/𝜆³)𝑆(𝜆𝑥)=𝑆(𝑥) for all 𝑥∈𝑋; that is, 𝑆 is a fixed point of 𝑇. Since 𝐶 is the unique fixed point of 𝑇 in Δ, it follows that 𝑆=𝐶.

Lemma 6. Let 𝑓∶𝑋→𝑌 be an odd function for which there exists a function 𝜑∶𝑋×𝑋→[0,+∞) such that 𝐷𝑓𝑥,𝑦≤𝜑𝑥,𝑦 (51) for all 𝑥,𝑦∈𝑋. If there exists a constant 0<𝐿<1 such that 𝜆³𝜑𝑥,𝑦≤𝐿𝜑𝜆𝑥,𝜆𝑦 (52) for all 𝑥,𝑦∈𝑋, then there exists a unique cubic mapping 𝐶∶𝑋→𝑌 such that 𝑓(𝑥)(𝑥≤1−𝐶)𝜓(𝑥)1−𝐿 (53) for all 𝑥∈𝑋, where 𝜓(𝑥)∶=(1/2(𝜆+1))𝜑(0,𝑥/𝜆).

Proof. It follows from (52) that lim_𝑟→∞𝜆^3𝑟𝜑𝑥𝜆^𝑟,𝑦𝜆^𝑟=0 (54) for all 𝑥,𝑦∈𝑋. Letting 𝑥=0 in (51) and replacing 𝑦 by 𝑥, we have 𝜆³𝑓𝑥𝜆−𝑓(𝑥)≤𝜓(𝑥) (55) for all 𝑥∈𝑋. We introduce the same definitions for Ω and 𝑑 as in the proof of Lemma 5 (by replacing 𝜙 by 𝜓) such that (Ω,𝑑) becomes a generalized complete metric space. Let 𝑇∶Ω→Ω be the mapping defined by 𝑇𝑔(𝑥)∶=𝜆³𝑔𝑥𝜆 (56) for all 𝑔∈Ω and all 𝑥∈𝑋. One can show that 𝑑(𝑇𝑔,𝑇ℎ)≤𝐿𝑑(𝑔,ℎ) for all 𝑔,ℎ∈Ω. It follows from (55) that 𝑑(𝑇𝑓,𝑓)≤1. Due to Proposition 4, the sequence {𝑇^𝑟𝑓} converges to a fixed point 𝐶 of 𝑇; that is, 𝐶∶𝑋⟶𝑌,𝐶(𝑥)=lim_𝑟→∞𝑇^𝑟𝑓(𝑥)=lim_𝑟→∞𝜆^3𝑟𝑓𝑥𝜆^𝑟 (57) and 𝐶(𝜆𝑥)=𝜆³𝐶(𝑥) for all 𝑥∈𝑋. Also, 𝑑≤1𝑓,𝐶𝑑≤11−𝐿𝑇𝑓,𝑓,1−𝐿 (58) which yields the inequality (53). The rest of the proof is similar to the proof of Lemma 5, and we omit the details.

Similarly, we can prove the following two lemmas on even functions.

Lemma 7. Let 𝑓∶𝑋→𝑌 be an even function with 𝑓(0)=0 for which there exists a function 𝜑∶𝑋×𝑋→[0,+∞) such that 𝐷𝑓𝑥,𝑦≤𝜑𝑥,𝑦 (59) for all 𝑥,𝑦∈𝑋. If there exists a constant 0<𝐿<1 such that 𝜑𝜆𝑥,𝜆𝑦≤𝜆⁴𝐿𝜑𝑥,𝑦 (60) for all 𝑥,𝑦∈𝑋, then there exists a unique quartic mapping 𝑄∶𝑋→𝑌 such that 𝑓(𝑥)(𝑥≤𝐿−𝑄)𝜙(𝑥)1−𝐿 (61) for all 𝑥∈𝑋, where 𝜙(𝑥)∶=(1/2)𝜑(0,𝑥/𝜆).

Lemma 8. Let 𝑓∶𝑋→𝑌 be an even function with 𝑓(0)=0 for which there exists a function 𝜑∶𝑋×𝑋→[0,+∞) such that 𝐷𝑓𝑥,𝑦≤𝜑𝑥,𝑦 (62) for all 𝑥,𝑦∈𝑋. If there exists a constant 0<𝐿<1 such that 𝜆⁴𝜑𝑥,𝑦≤𝐿𝜑𝜆𝑥,𝜆𝑦 (63) for all 𝑥,𝑦∈𝑋, then there exists a unique quartic mapping 𝑄∶𝑋→𝑌 such that 𝑓(𝑥)(𝑥≤1−𝑄)𝜓(𝑥)1−𝐿 (64) for all 𝑥∈𝑋, where 𝜓(𝑥)∶=(1/2)𝜑(0,𝑥/𝜆).

Now, we are ready to give our main theorems in this section.

Theorem 9. Let 𝑓∶𝑋→𝑌 be a function with 𝑓(0)=0 for which there exists a function 𝜑∶𝑋×𝑋→[0,+∞) such that 𝐷𝑓𝑥,𝑦≤𝜑𝑥,𝑦 (65) for all 𝑥,𝑦∈𝑋. If there exists a constant 0<𝐿<1 such that 𝜑𝜆𝑥,𝜆𝑦≤𝜆³𝐿𝜑𝑥,𝑦 (66) for all 𝑥,𝑦∈𝑋, then there exist a unique cubic mapping 𝐶∶𝑋→𝑌 and a unique quartic mapping 𝑄∶𝑋→𝑌 such that 𝑓(𝑥)(𝑥)(𝑥≤𝐿−𝐶−𝑄)𝜙1−𝐿^𝑜(𝑥)+𝜙_𝑒(𝑥) (67) for all 𝑥∈𝑋, where 𝜙_𝑜(𝑥)1∶=2()Φ𝑥𝜆+10,𝜆,𝜙_𝑒(𝑥)1∶=2Φ𝑥0,𝜆,Φ1𝑥,𝑦∶=2𝜑𝑥,𝑦+𝜑−𝑥,−𝑦. (68)

Proof. Let 𝑓_𝑜 and 𝑓_𝑒 denote the odd and the even part of 𝑓, respectively. Then, it can be verified from (65) that 𝐷𝑓_𝑜𝑥,𝑦≤Φ𝑥,𝑦,𝐷𝑓_𝑒𝑥,𝑦≤Φ𝑥,𝑦 (69) for all 𝑥,𝑦∈𝑋. Moreover, by (66), it is easy to compute that Φ𝜆𝑥,𝜆𝑦≤𝜆³𝐿Φ𝑥,𝑦 (70) for all 𝑥,𝑦∈𝑋. Thus, by applying Lemmas 5 and 7, one can obtain that there exist a unique cubic mapping 𝐶∶𝑋→𝑌 and a unique quartic mapping 𝑄∶𝑋→𝑌 such that 𝑓_𝑜(𝑥)(𝑥≤𝐿−𝐶)𝜙1−𝐿^𝑜(𝑥),𝑓_𝑒(𝑥)(𝑥≤𝐿−𝑄)𝜙1−𝐿^𝑒(𝑥) (71)(72) for all 𝑥∈𝑋, where 𝜙_𝑜(𝑥)∶=(1/2(𝜆+1))Φ(0,𝑥/𝜆) and 𝜙_𝑒(𝑥)∶=(1/2)Φ(0,𝑥/𝜆). Moreover, combining (71) and (72) yields the inequality (67). The proof is complete.

Theorem 10. Let 𝑓∶𝑋→𝑌 be a function with 𝑓(0)=0 for which there exists a function 𝜑∶𝑋×𝑋→[0,+∞) such that 𝐷𝑓𝑥,𝑦≤𝜑𝑥,𝑦 (73) for all 𝑥,𝑦∈𝑋. If there exists a constant 0<𝐿<1 such that 𝜆⁴𝜑𝑥,𝑦≤𝐿𝜑𝜆𝑥,𝜆𝑦 (74) for all 𝑥,𝑦∈𝑋, then there exist a unique cubic mapping 𝐶∶𝑋→𝑌 and a unique quartic mapping 𝑄∶𝑋→𝑌 such that 𝑓(𝑥)(𝑥)(𝑥≤1−𝐶−𝑄)𝜓1−𝐿^𝑜(𝑥)+𝜓_𝑒(𝑥) (75) for all 𝑥∈𝑋, where 𝜓_𝑜(𝑥)1∶=2()Ψ𝑥𝜆+10,𝜆,𝜓_𝑒(𝑥)1∶=2Ψ𝑥0,𝜆,Ψ1𝑥,𝑦∶=2𝜑𝑥,𝑦+𝜑−𝑥,−𝑦. (76)

Proof. Similar to the proof of Theorem 9, the result follows from Lemmas 6 and 8.

Conflict of Interests

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