A Functional Equation Originating from Elliptic Curves

Abstract

We obtain the general solution and the stability of the functional equation 𝑓(𝑥+𝑦+𝑧,𝑢+𝑣+𝑤)+𝑓(𝑥+𝑦−𝑧,𝑢+𝑣+𝑤)+2𝑓(𝑥,𝑢−𝑤)+2𝑓(𝑦,𝑣−𝑤)=𝑓(𝑥+𝑦,𝑢+𝑤)+𝑓(𝑥+𝑦,𝑣+𝑤)+𝑓(𝑥+𝑧,𝑢+𝑤)+𝑓(𝑥−𝑧,𝑢+𝑣−𝑤)+𝑓(𝑦+𝑧,𝑣+𝑤)+𝑓(𝑦−𝑧,𝑢+𝑣−𝑤). The function 𝑓(𝑥,𝑦)=𝑥³+𝑎𝑥+𝑏−𝑦² having level curves as elliptic curves is a solution of the above functional equation.

1. Introduction

A graph of an equation of the form 𝑦²=𝑥³+𝑎𝑥+𝑏 is called an elliptic curve [1], where 𝑎 and 𝑏 are constants. Since 𝑓(𝑥,𝑦)=𝑥³+𝑎𝑥+𝑏−𝑦² has level curves as elliptic curves, functional equations having the mapping 𝑓(𝑥,𝑦)=𝑥³+𝑎𝑥+𝑏−𝑦² as a solution are helpful to study cryptography and their applications.

Recently, Jun and Kim [2] solved the cubic functional equation𝑔(𝑥+𝑦+𝑧)+𝑔(𝑥+𝑦−𝑧)+2𝑔(𝑥)+2𝑔(𝑦)=2𝑔(𝑥+𝑦)+𝑔(𝑥+𝑧)+𝑔(𝑥−𝑧)+𝑔(𝑦+𝑧)+𝑔(𝑦−𝑧). (1.1)

We consider the quadratic functional equation𝑔(𝑥+𝑦+𝑧)+𝑔(𝑥−𝑧)+𝑔(𝑦−𝑧)=𝑔(𝑥+𝑦−𝑧)+𝑔(𝑥+𝑧)+𝑔(𝑦+𝑧), (1.2)and the 2-dimensional vector variable functional equation𝑓(𝑥+𝑦+𝑧,𝑢+𝑣+𝑤)+𝑓(𝑥+𝑦−𝑧,𝑢+𝑣+𝑤)+2𝑓(𝑥,𝑢−𝑤)+2𝑓(𝑦,𝑣−𝑤)=𝑓(𝑥+𝑦,𝑢+𝑤)+𝑓(𝑥+𝑦,𝑣+𝑤)+𝑓(𝑥+𝑧,𝑢+𝑤)+𝑓(𝑥−𝑧,𝑢+𝑣−𝑤)+𝑓(𝑦+𝑧,𝑣+𝑤)+𝑓(𝑦−𝑧,𝑢+𝑣−𝑤). (1.3)

The function 𝑔∶ℝ→ℝ given by 𝑔(𝑥)∶=𝑎𝑥³+𝑏𝑥+𝑐 is a particular solution of (1.1), and the function 𝑔∶ℝ→ℝ given by 𝑔(𝑥)∶=𝑎𝑥²+𝑏 is a particular solution of (1.2). The function 𝑓∶ℝ×ℝ→ℝ given by 𝑓(𝑥,𝑦)∶=𝑥³+𝑎𝑥+𝑏−𝑦² is a particular solution of (1.3). The functional (1.3) is a mixed type of (1.1) and (1.2).

Problems on solutions or stability of various functional equations have been extensively investigated by a number of mathematicians [3–12].

In this paper, let 𝑋, 𝑌, and 𝑍 be real vector spaces. We find out the general solution and investigate the stability of (1.2) and (1.3).

2. Solutions

We find out the general solution of the functional (1.2) as follows.

Lemma 2.1. A mapping 𝑔∶𝑋→𝑌 satisfies (1.2) if and only if there exists a symmetric biadditive mapping 𝑆∶𝑋×𝑋→𝑌 such that𝑔(𝑥)=𝑆(𝑥,𝑥)+𝑔(0) (2.1)for all 𝑥∈𝑋.

Proof. We first assume that 𝑔 is a solution of (1.2). Let ℎ∶𝑋→𝑌 be a mapping given by ℎ(𝑥)∶=𝑔(𝑥)−𝑔(0) for all 𝑥∈𝑋. Then, ℎ(0)=0 andℎ(𝑥+𝑦+𝑧)+ℎ(𝑥−𝑧)+ℎ(𝑦−𝑧)=ℎ(𝑥+𝑦−𝑧)+ℎ(𝑥+𝑧)+ℎ(𝑦+𝑧) (2.2)for all 𝑥,𝑦,𝑧∈𝑋. Putting 𝑥=𝑦=0 in (2.2), ℎ is even. Replacing 𝑧 by 𝑥+𝑦 in (2.2),ℎ(2𝑥+2𝑦)+ℎ(𝑦)+ℎ(𝑥)=ℎ(2𝑥+𝑦)+ℎ(𝑥+2𝑦) (2.3)for all 𝑥,𝑦∈𝑋. Taking 𝑧=𝑥 in (2.2), we get thatℎ(2𝑥+𝑦)+ℎ(𝑥−𝑦)=ℎ(𝑦)+ℎ(2𝑥)+ℎ(𝑥+𝑦) (2.4)for all 𝑥,𝑦∈𝑋. Interchanging 𝑥 and 𝑦 in (2.4), we see thatℎ(𝑥+2𝑦)+ℎ(𝑥−𝑦)=ℎ(𝑥)+ℎ(2𝑦)+ℎ(𝑥+𝑦) (2.5)for all 𝑥,𝑦∈𝑋. By (2.3), (2.4), and (2.5), we obtain that ℎ(2𝑥+2𝑦)=ℎ(2𝑥+𝑦)−ℎ(𝑦)+ℎ(𝑥+2𝑦)−ℎ(𝑥)=ℎ(2𝑥)+ℎ(𝑥+𝑦)−ℎ(𝑥−𝑦)+ℎ(2𝑦)+ℎ(𝑥+𝑦)−ℎ(𝑥−𝑦)=ℎ(2𝑥)+ℎ(2𝑦)+2ℎ(𝑥+𝑦)−2ℎ(𝑥−𝑦) (2.6) for all 𝑥,𝑦∈𝑋. Letting 𝑦=𝑥 in (2.6), we get thatℎ(2𝑥)=4ℎ(𝑥) (2.7)for all 𝑥∈𝑋. By (2.6) and (2.7), we obtain thatℎ(𝑥+𝑦)+ℎ(𝑥−𝑦)=2ℎ(𝑥)+2ℎ(𝑦) (2.8)for all 𝑥,𝑦∈𝑋. By [13], there exists a symmetric biadditive mapping 𝑆∶𝑋×𝑋→𝑌 such that ℎ(𝑥)=𝑆(𝑥,𝑥) for all 𝑥∈𝑋.
Conversely, we assume that there exists a symmetric biadditive mapping 𝑆∶𝑋×𝑋→𝑌 such that 𝑔(𝑥)=𝑆(𝑥,𝑥)+𝑔(0) for all 𝑥∈𝑋. Since 𝑆 is biadditive,𝑔(𝑥+𝑦+𝑧)+𝑔(𝑥−𝑧)+𝑔(𝑦−𝑧)=𝑆(𝑥+𝑦+𝑧,𝑥+𝑦+𝑧)+𝑆(𝑥−𝑧,𝑥−𝑧)+𝑆(𝑦−𝑧,𝑦−𝑧)+3𝑔(0)=2[𝑆(𝑥,𝑥)+𝑆(𝑥,𝑦)+𝑆(𝑦,𝑦)]+3𝑆(𝑧,𝑧)+3𝑔(0)=𝑆(𝑥+𝑦−𝑧,𝑥+𝑦−𝑧)+𝑆(𝑥+𝑧,𝑥+𝑧)+𝑆(𝑦+𝑧,𝑦+𝑧)+3𝑔(0)=𝑔(𝑥+𝑦−𝑧)+𝑔(𝑥+𝑧)+𝑔(𝑦+𝑧) (2.9)for all 𝑥,𝑦,𝑧∈𝑋.

Example 2.2. Let 𝑟/𝑠 be a rational number, where 𝑟 and 𝑠 are integers with gcd(𝑟,𝑠)=1. Define 𝐻(𝑟/𝑠)∶=max{|𝑟|,|𝑠|} and 𝑔(𝑟/𝑠)∶=log𝐻(𝑟/𝑠). Let ℚ be the set of rational numbers and let 𝐸(ℚ)∶={∞}∪{(𝑥,𝑦)∈ℚ×ℚ∣𝑦²=𝑥³+𝑎𝑥+𝑏} be an elliptic curve over ℚ. The addition in 𝐸(ℚ) is given in [1]. The height function ℎ∶𝐸(ℚ)→[0,∞) is defined by ℎ(𝑥,𝑦)=𝑔(𝑥),ℎ(∞)=0 (2.10)for all (𝑥,𝑦)∈𝐸(ℚ). The canonical height function ℎ∶𝐸(ℚ)→[0,∞) given by1ℎ(𝑃)=2lim_𝑛→∞14^𝑛ℎ(2^𝑛𝑃) (2.11)is a solution of (1.2).

We find out the general solution of the functional (1.3) as follows.

Theorem 2.3. A mapping 𝑓∶𝑋×𝑌→𝑍 satisfies (1.3) if and only if there exist a symmetric multiadditive mapping 𝑆₁∶𝑋×𝑋×𝑋→𝑍, an additive mapping 𝐴∶𝑋→𝑍, and a symmetric biadditive mapping 𝑆₂∶𝑌×𝑌→𝑍 such that𝑓(𝑥,𝑦)=𝑆₁(𝑥,𝑥,𝑥)+𝐴(𝑥)+𝑓(0,0)+𝑆₂(𝑦,𝑦) (2.12)for all 𝑥∈𝑋 and all 𝑦∈𝑌.

Proof. We first assume that 𝑓 satisfies (1.3). Define 𝑓₁∶𝑋→𝑍 and 𝑓₂∶𝑌→𝑍 by 𝑓₁(𝑥)∶=𝑓(𝑥,0) for all 𝑥∈𝑋 and 𝑓₂(𝑦)∶=𝑓(0,𝑦) for all 𝑦∈𝑌. Then, 𝑓₁ satisfies (1.1) and 𝑓₂ satisfies (1.2). By [2], there exist a symmetric multiadditive mapping 𝑆₁∶𝑋×𝑋×𝑋→𝑍 and an additive mapping 𝐴∶𝑋→𝑍 such that𝑓₁(𝑥)=𝑆₁(𝑥,𝑥,𝑥)+𝐴(𝑥)+𝑓₁(0) (2.13)for all 𝑥∈𝑋. By Lemma 2.1, there exists a symmetric biadditive mapping 𝑆₂∶𝑌×𝑌→𝑍 such that𝑓₂(𝑦)=𝑆₂(𝑦,𝑦)+𝑓₂(0) (2.14)for all 𝑦∈𝑌. Note that 𝑓₁(0)=𝑓₂(0)=𝑓(0,0) and that 𝑓(0,0)≠0 in general. Let 𝑘∶𝑋×𝑌→𝑍 be a mapping given by 𝑘(𝑥,𝑦)∶=𝑓(𝑥,𝑦)−𝑓(0,0) for all 𝑥∈𝑋 and all 𝑦∈𝑌. Then, 𝑘(0,0)=0 and𝑘(𝑥+𝑦+𝑧,𝑢+𝑣+𝑤)+𝑘(𝑥+𝑦−𝑧,𝑢+𝑣+𝑤)+2𝑘(𝑥,𝑢−𝑤)+2𝑘(𝑦,𝑣−𝑤)=𝑘(𝑥+𝑦,𝑢+𝑤)+𝑘(𝑥+𝑦,𝑣+𝑤)+𝑘(𝑥+𝑧,𝑢+𝑤)+𝑘(𝑥−𝑧,𝑢+𝑣−𝑤)+𝑘(𝑦+𝑧,𝑣+𝑤)+𝑘(𝑦−𝑧,𝑢+𝑣−𝑤) (2.15)for all 𝑥,𝑦,𝑧∈𝑋 and all 𝑢,𝑣,𝑤∈𝑌.
Since 𝑘(0,𝑢)=𝑆₂(𝑢,𝑢) for all 𝑢∈𝑌, 𝑘(0,−𝑢)=𝑘(0,𝑢) and 𝑘(0,2𝑢)=4𝑘(0,𝑢) for all 𝑢∈𝑌. Letting 𝑦=𝑧=𝑢=𝑣=0 in (2.15), we get𝑘(𝑥,𝑤)=𝑘(𝑥,−𝑤) (2.16)for all 𝑥∈𝑋 and all 𝑤∈𝑌. Putting 𝑦=𝑧=0 in (2.15), we get2𝑘(𝑥,𝑢+𝑣+𝑤)+2𝑘(𝑥,𝑢−𝑤)+2𝑘(0,𝑣−𝑤)=2𝑘(𝑥,𝑢+𝑤)+𝑘(𝑥,𝑣+𝑤)+𝑘(0,𝑣+𝑤)+𝑘(𝑥,𝑢+𝑣−𝑤)+𝑘(0,𝑢+𝑣−𝑤) (2.17)for all 𝑥∈𝑋 and all 𝑢,𝑣,𝑤∈𝑌. Setting 𝑣=𝑤=−𝑢 in (2.17) and using (2.16),𝑘(𝑥,𝑢)+𝑘(𝑥,2𝑢)=2𝑘(𝑥,0)+5𝑘(0,𝑢) (2.18)for all 𝑥∈𝑋 and all 𝑢∈𝑌. Taking 𝑣=−𝑤=−𝑢 in (2.17) and using (2.16),𝑘(𝑥,𝑢)+𝑘(𝑥,0)+7𝑘(0,𝑢)=2𝑘(𝑥,2𝑢) (2.19)for all 𝑥∈𝑋 and all 𝑢∈𝑌. By (2.18) and (2.19), 𝑘(𝑥,𝑢)=𝑘(𝑥,0)+𝑘(0,𝑢) for all 𝑥∈𝑋 and all 𝑢∈𝑌. Hence, we obtain that𝑓(𝑥,𝑦)=𝑘(𝑥,𝑦)+𝑓(0,0)=𝑘(𝑥,0)+𝑘(0,𝑦)+𝑓(0,0)=𝑓₁(𝑥)−𝑓₁(0)+𝑓₂(𝑦)−𝑓₂(0)+𝑓(0,0)=𝑆₁(𝑥,𝑥,𝑥)+𝐴(𝑥)+𝑓(0,0)+𝑆₂(𝑦,𝑦) (2.20)for all 𝑥∈𝑋 and all 𝑦∈𝑌.
Conversely, we assume that there exist a symmetric multiadditive mapping 𝑆₁∶𝑋×𝑋×𝑋→𝑍, an additive mapping 𝐴∶𝑋→𝑍, and a symmetric biadditive mapping 𝑆₂∶𝑌×𝑌→𝑍 such that𝑓(𝑥,𝑦)=𝑆₁(𝑥,𝑥,𝑥)+𝐴(𝑥)+𝑓(0,0)+𝑆₂(𝑦,𝑦) (2.21)for all 𝑥∈𝑋 and all 𝑦∈𝑌. Since 𝑆₁ is symmetric multiadditive and 𝑆₂ is symmetric biadditive,𝑓(𝑥+𝑦+𝑧,𝑢+𝑣+𝑤)+𝑓(𝑥+𝑦−𝑧,𝑢+𝑣+𝑤)+2𝑓(𝑥,𝑢−𝑤)+2𝑓(𝑦,𝑣−𝑤)=𝑆₁(𝑥+𝑦+𝑧,𝑥+𝑦+𝑧,𝑥+𝑦+𝑧)+𝑆₁(𝑥+𝑦−𝑧,𝑥+𝑦−𝑧,𝑥+𝑦−𝑧)+2𝑆₁(𝑥,𝑥,𝑥)+2𝑆₁(𝑦,𝑦,𝑦)+𝐴(𝑥+𝑦+𝑧)+𝐴(𝑥+𝑦−𝑧)+2𝐴(𝑥)+2𝐴(𝑦)+2𝑆₂(𝑢+𝑣+𝑤,𝑢+𝑣+𝑤)+2𝑆₂(𝑢−𝑤,𝑢−𝑤)+2𝑆₂(𝑣−𝑤,𝑣−𝑤)+6𝑓(0,0)=4𝑆₁(𝑥,𝑥,𝑥)+4𝑆₁(𝑦,𝑦,𝑦)+6𝑆₁(𝑥,𝑥,𝑦)+6𝑆₁(𝑥,𝑦,𝑦)+6𝑆₁(𝑥,𝑧,𝑧)+6𝑆₁(𝑦,𝑧,𝑧)+4𝐴(𝑥)+4𝐴(𝑦)+4𝑆₂(𝑢,𝑢)+4𝑆₂(𝑢,𝑣)+4𝑆₂(𝑣,𝑣)+6𝑆₂(𝑤,𝑤)+6𝑓(0,0)=2𝑆₁(𝑥+𝑦,𝑥+𝑦,𝑥+𝑦)+𝑆₁(𝑥+𝑧,𝑥+𝑧,𝑥+𝑧)+𝑆₁(𝑥−𝑧,𝑥−𝑧,𝑥−𝑧)+𝑆₁(𝑦+𝑧,𝑦+𝑧,𝑦+𝑧)+𝑆₁(𝑦−𝑧,𝑦−𝑧,𝑦−𝑧)+2𝐴(𝑥+𝑦)+𝐴(𝑥+𝑧)+𝐴(𝑥−𝑧)+𝐴(𝑦+𝑧)+𝐴(𝑦−𝑧)+2𝑆₂(𝑢+𝑤,𝑢+𝑤)+2𝑆₂(𝑣+𝑤,𝑣+𝑤)+2𝑆₂(𝑢+𝑣−𝑤,𝑢+𝑣−𝑤)+6𝑓(0,0)=𝑓(𝑥+𝑦,𝑢+𝑤)+𝑓(𝑥+𝑦,𝑣+𝑤)+𝑓(𝑥+𝑧,𝑢+𝑤)+𝑓(𝑥−𝑧,𝑢+𝑣−𝑤)+𝑓(𝑦+𝑧,𝑣+𝑤)+𝑓(𝑦−𝑧,𝑢+𝑣−𝑤) (2.22)for all 𝑥,𝑦,𝑧∈𝑋 and all 𝑢,𝑣,𝑤∈𝑌.

Example 2.4. Let 𝑋 and 𝑌 be the vector spaces 𝑀₃(ℂ) and 𝑀₂(ℂ), respectively. Consider a function 𝑓∶𝑋×𝑌→ℂ given by 𝑓(𝑥,𝑦)∶=det(𝑥)+det(𝑦) for all 𝑥∈𝑋 and all 𝑦∈𝑌. One can easily verify that 𝑓 satisfies (1.3). By Theorem 2.3, there exist a symmetric multiadditive mapping 𝑆₁∶𝑋×𝑋×𝑋→ℂ, an additive mapping 𝐴∶𝑋→ℂ, and a symmetric biadditive mapping 𝑆₂∶𝑌×𝑌→ℂ such that 𝑓(𝑥,𝑦)=𝑆₁(𝑥,𝑥,𝑥)+𝐴(𝑥)+𝑓(0,0)+𝑆₂(𝑦,𝑦) (2.23)for all 𝑥∈𝑋 and all 𝑦∈𝑌. In fact,𝑆₁1(𝑥,𝑦,𝑧)=𝑆24[det(𝑥+𝑦+𝑧)+det(𝑥−𝑦−𝑧)−det(𝑥+𝑦−𝑧)−det(𝑥−𝑦+𝑧)],𝐴(𝑥)=𝑓(0,0)=0,₂1(𝑢,𝑣)=4[det(𝑢+𝑣)−det(𝑢−𝑣)] (2.24)for all 𝑥,𝑦,𝑧∈𝑋 and all 𝑢,𝑣∈𝑌

3. Stability

From now on, let 𝑌 be a Banach space, and let 𝜑∶𝑋³→[0,∞) be a function satisfying^0𝑥0𝜑^02𝑑𝑐(𝑥,𝑦,𝑧)∶=^∞_𝑗=014^𝑗+1𝜑(2^𝑗𝑥,2^𝑗𝑦,2^𝑗𝑧)<∞ (3.1)for all 𝑥,𝑦,𝑧∈𝑋.

Lemma 3.1. Let 𝑔∶𝑋→𝑌 be a mapping such that
‖𝑔(𝑥+𝑦+𝑧)+𝑔(𝑥−𝑧)+𝑔(𝑦−𝑧)−𝑔(𝑥+𝑦−𝑧)−𝑔(𝑥+𝑧)−𝑔(𝑦+𝑧)‖≤𝜑(𝑥,𝑦,𝑧) (3.2) for all 𝑥,𝑦,𝑧∈𝑋. Then, there exists a unique quadratic mapping 𝐺∶𝑋→𝑌 satisfying (1.2) such that‖𝑔(𝑥)−𝐺(𝑥)‖≤^{0𝑥002𝑑𝑐}𝑥𝜑(2,𝑥2,𝑥)+2^{0𝑥002𝑑𝑐}𝑥𝜑(2,𝑥2,𝑥2)+2^{0𝑥002𝑑𝑐}𝑥𝜑(0,0,2)+13[𝜑(0,0,0)+‖𝑔(0)‖] (3.3)for all 𝑥∈𝑋. The mapping 𝐺 is given by 𝐺(𝑥)∶=lim_𝑗→∞14^𝑗𝑔(2^𝑗𝑥) for all 𝑥∈𝑋.

Proof. Letting 𝑥=𝑦=0 in (3.2), we have‖𝑔(−𝑧)−𝑔(𝑧)‖≤𝜑(0,0,𝑧) (3.4)for all 𝑧∈𝑋.
Replacing 𝑧 by 𝑥+𝑦 in (3.2), we get
‖𝑔(2𝑥+2𝑦)+𝑔(−𝑦)+𝑔(−𝑥)−𝑔(0)−𝑔(2𝑥+𝑦)−𝑔(𝑥+2𝑦)‖≤𝜑(𝑥,𝑦,𝑥+𝑦) (3.5)for all 𝑥,𝑦∈𝑋. Taking 𝑧=𝑥 in (3.2), we see that‖𝑔(2𝑥+𝑦)+𝑔(0)+𝑔(𝑦−𝑥)−𝑔(𝑦)−𝑔(2𝑥)−𝑔(𝑥+𝑦)‖≤𝜑(𝑥,𝑦,𝑥) (3.6)for all 𝑥,𝑦∈𝑋. Interchanging 𝑥 and 𝑦 in (3.6), we see that‖𝑔(𝑥+2𝑦)+𝑔(0)+𝑔(𝑥−𝑦)−𝑔(𝑥)−𝑔(2𝑦)−𝑔(𝑥+𝑦)‖≤𝜑(𝑦,𝑥,𝑥) (3.7)for all 𝑥,𝑦∈𝑋. By (3.5), (3.6), and (3.7),‖𝑔(2𝑥+2𝑦)−2𝑔(𝑥+𝑦)−𝑔(2𝑥)−𝑔(2𝑦)‖≤𝜑(𝑥,𝑦,𝑥+𝑦)+𝜑(𝑥,𝑦,𝑥)+𝜑(𝑦,𝑥,𝑥)+‖𝑔(−𝑥)+𝑔(𝑥)‖+‖𝑔(−𝑦)−𝑔(𝑦)‖+‖𝑔(𝑦−𝑥)−𝑔(𝑥−𝑦)‖+‖𝑔(0)‖≤𝜑(𝑥,𝑦,𝑥+𝑦)+𝜑(𝑥,𝑦,𝑥)+𝜑(𝑦,𝑥,𝑥)+𝜑(0,0,𝑥)+𝜑(0,0,𝑦)+𝜑(0,0,𝑥−𝑦)+‖𝑔(0)‖ (3.8)for all 𝑥,𝑦∈𝑋. Putting 𝑦=𝑥 in the above inequality, we get‖𝑔(4𝑥)−4𝑔(2𝑥)‖≤𝜑(𝑥,𝑥,2𝑥)+2𝜑(𝑥,𝑥,𝑥)+2𝜑(0,0,𝑥)+𝜑(0,0,0)+‖𝑔(0)‖ (3.9)for all 𝑥∈𝑋. So,1‖𝑔(𝑥)−41𝑔(2𝑥)‖≤4𝜑^∗(𝑥), (3.10)where𝜑^∗𝑥(𝑥)∶=𝜑(2,𝑥2𝑥,𝑥)+2𝜑(2,𝑥2,𝑥2𝑥)+2𝜑(0,0,2)+𝜑(0,0,0)+‖𝑔(0)‖ (3.11)for all 𝑥∈𝑋.
Thus, we obtain‖14^𝑗𝑔(2^𝑗1𝑥)−4^𝑗+1𝑔(2^𝑗+11𝑥)‖≤4^𝑗+1𝜑^∗(2^𝑗𝑥) (3.12)for all 𝑥∈𝑋 and all 𝑗. For given integers 𝑙,𝑚(0≤𝑙<𝑚), we get‖14^𝑙𝑔(2^𝑙1𝑥)−4^𝑚𝑔(2^𝑚𝑥)‖≤^𝑚−1_𝑗=𝑙14^𝑗+1𝜑^∗(2^𝑗𝑥) (3.13)for all 𝑥∈𝑋. By (3.13), the sequence {(1/4^𝑗)𝑔(2^𝑗𝑥)} is a Cauchy sequence for all 𝑥∈𝑋. Since 𝑌 is complete, the sequence {(1/4^𝑗)𝑔(2^𝑗𝑥)} converges for all 𝑥∈𝑋. Define 𝐺∶𝑋→𝑌 by𝐺(𝑥)∶=lim_𝑗→∞14^𝑗𝑔(2^𝑗𝑥) (3.14)for all 𝑥∈𝑋. By (3.2), we have‖14^𝑗[𝑔(2^𝑗(𝑥+𝑦+𝑧))+𝑔(2^𝑗(𝑥−𝑧))+𝑔(2^𝑗(𝑦−𝑧))−𝑔(2^𝑗(𝑥+𝑦−𝑧))−𝑔(2^𝑗(𝑥+𝑧))−𝑔(2^𝑗1(𝑦+𝑧))]‖≤4^𝑗𝜑(2^𝑗𝑥,2^𝑗𝑦,2^𝑗𝑧) (3.15)for all 𝑥,𝑦,𝑧∈𝑋 and all 𝑗. Letting 𝑗→∞ and using (3.1), we see that 𝐺 satisfies (1.2). Setting 𝑙=0 and taking 𝑚→∞ in (3.13), one can obtain the inequality (3.3). If 𝐻∶𝑋→𝑌 is another quadratic mapping satisfying (1.2) and (3.3), we obtain1‖𝐺(𝑥)−𝐻(𝑥)‖=4^𝑛‖𝐺(2^𝑛𝑥)−𝐻(2^𝑛≤1𝑥)‖4^𝑛‖𝐺(2^𝑛𝑥)−𝑔(2^𝑛1𝑥)‖+4^𝑛‖𝑔(2^𝑛𝑥)−𝐻(2^𝑛≤2𝑥)‖4^𝑛[𝜑^{0𝑥002𝑑𝑐}(2^𝑛−1𝑥,2^𝑛−1𝑥,2^𝑛𝑥)+2𝜑^{0𝑥002𝑑𝑐}(2^𝑛−1𝑥,2^𝑛−1𝑥,2^𝑛−1𝑥)+2𝜑^{0𝑥002𝑑𝑐}(0,0,2^𝑛−1+1𝑥)3[𝜑(0,0,0)+‖𝑔(0)‖]]⟶0as𝑛⟶∞ (3.16)for all 𝑥∈𝑋. Hence, 𝐺 is a unique quadratic mapping, as desired.

Example 3.2. If the height function ℎ in Example 2.2 satisfies ‖ℎ(𝑃+𝑄+𝑅)+ℎ(𝑃−𝑅)+ℎ(𝑄−𝑅)−ℎ(𝑃+𝑄−𝑅)−ℎ(𝑃+𝑅)−ℎ(𝑄+𝑅)‖≤𝑐 (3.17)for all 𝑃,𝑄,𝑅∈𝐸(ℚ), then there exists a unique quadratic function ^0𝑥0ℎ^02𝑑𝑐∶𝐸(ℚ)→[0,∞) satisfying (1.2) such that‖ℎ(𝑃)−ℎ^{0𝑥002𝑑𝑐}(𝑃)‖≤2𝑐 (3.18)for all 𝑃∈𝐸(ℚ).

Let 𝜓∶𝑋³×𝑌³→[0,∞) be a function satisfying ^0𝑥0𝜓^02𝑑𝑐(𝑥,𝑦,𝑧,𝑢,𝑣,𝑤)∶=^∞_𝑗=0[12^𝑗+1𝜓₁(2^𝑗𝑥,2^𝑗𝑦,2^𝑗1𝑧)+4^𝑗+1𝜓₂(2^𝑗𝑢,2^𝑗𝑣,2^𝑗𝑤)]<∞, (3.19) where𝜓₁(𝑥,𝑦,𝑧)∶=𝜓(𝑥,𝑦,𝑧,0,0,0),𝜓₂(𝑢,𝑣,𝑤)∶=𝜓(0,0,0,𝑢,𝑣,𝑤) (3.20)for all 𝑥,𝑦,𝑧∈𝑋 and all 𝑢,𝑣,𝑤∈𝑌. Also, let ^0𝑥0𝜓^02𝑑𝑐₁∶𝑋³→[0,∞), and ^0𝑥0𝜓^02𝑑𝑐₂∶𝑌³→[0,∞) be two functions given by^0𝑥0𝜓^02𝑑𝑐₁(𝑥,𝑦,𝑧)∶=𝜓^{0𝑥002𝑑𝑐}(𝑥,𝑦,𝑧,0,0,0),𝜓^{0𝑥002𝑑𝑐}₂(𝑢,𝑣,𝑤)∶=𝜓^{0𝑥002𝑑𝑐}(0,0,0,𝑢,𝑣,𝑤) (3.21)for all 𝑥,𝑦,𝑧∈𝑋 and all 𝑢,𝑣,𝑤∈𝑌.

Theorem 3.3. Let 𝑓∶𝑋×𝑌→𝑍 be a mapping such that‖𝑓(𝑥+𝑦+𝑧,𝑢+𝑣+𝑤)+𝑓(𝑥+𝑦−𝑧,𝑢+𝑣+𝑤)+2𝑓(𝑥,𝑢−𝑤)+2𝑓(𝑦,𝑣−𝑤)−𝑓(𝑥+𝑦,𝑢+𝑤)−𝑓(𝑥+𝑦,𝑣+𝑤)−𝑓(𝑥+𝑧,𝑢+𝑤)−𝑓(𝑥−𝑧,𝑢+𝑣−𝑤)−𝑓(𝑦+𝑧,𝑣+𝑤)−𝑓(𝑦−𝑧,𝑢+𝑣−𝑤)‖≤𝜓(𝑥,𝑦,𝑧,𝑢,𝑣,𝑤) (3.22) for all 𝑥,𝑦,𝑧∈𝑋 and all 𝑢,𝑣,𝑤∈𝑌. Then, there exists a mapping 𝐹∶𝑋×𝑌→𝑍 satisfying (1.3) such that ‖𝑓(𝑥,𝑦)−𝐹(𝑥,𝑦)‖≤𝜓(𝑥,0,0,𝑦,0,0)+𝜓^{0𝑥002𝑑𝑐}₁(𝑥2,𝑥2,𝑥21)+3^{0𝑥002𝑑𝑐}𝜓₁𝑥(𝑥,2,𝑥2)+𝜓^{0𝑥002𝑑𝑐}₂(𝑦2,𝑦2,𝑦)+2𝜓^{0𝑥002𝑑𝑐}₂(𝑦2,𝑦2,𝑦2)+2𝜓^{0𝑥002𝑑𝑐}₂𝑦(0,0,2)+13[𝜓₂(0,0,0)+‖𝑓(0,0)‖] (3.23) for all 𝑥∈𝑋 and all 𝑦∈𝑌. The mapping 𝐹 is given by𝐹(𝑥,𝑦)∶=lim_𝑗→∞[4^𝑗+1−13⋅8^𝑗𝑓(2^𝑗4𝑥,0)−^𝑗−16⋅8^𝑗𝑓(2^𝑗+11𝑥,0)+4^𝑗𝑓(0,2^𝑗𝑦)]−𝑓(0,0) (3.24)for all 𝑥∈𝑋 and all 𝑦∈𝑌.

Proof. Letting 𝑦=𝑧=𝑣=𝑤=0 in (3.22), we have‖𝑓(𝑥,𝑢)−𝑓(𝑥,0)−𝑓(0,𝑢)+𝑓(0,0)‖≤𝜓(𝑥,0,0,𝑢,0,0) (3.25)for all 𝑥∈𝑋 and all 𝑢∈𝑌.
Define 𝑓₁,𝑓₂∶𝑋→𝑌 by 𝑓₁(𝑥)∶=𝑓(𝑥,0) and 𝑓₂(𝑢)∶=𝑓(0,𝑢) for all 𝑥∈𝑋 and all 𝑢∈𝑌. Putting 𝑢=𝑣=𝑤=0 in (3.22), we have‖𝑓₁(𝑥+𝑦+𝑧)+𝑓₁(𝑥+𝑦−𝑧)+2𝑓₁(𝑥)+2𝑓₁(𝑦)−2𝑓₁(𝑥+𝑦)−𝑓₁(𝑥+𝑧)−𝑓₁(𝑥−𝑧)−𝑓₁(𝑦+𝑧)−𝑓₁(𝑦−𝑧)‖≤𝜓₁(𝑥,𝑦,𝑧) (3.26)for all 𝑥,𝑦,𝑧∈𝑋. Setting 𝑥=𝑦=𝑧=0 in (3.22), we have‖𝑓₂(𝑢+𝑣+𝑤)+𝑓₂(𝑢−𝑤)+𝑓₂(𝑣−𝑤)−𝑓₂(𝑢+𝑣−𝑤)−𝑓₂(𝑢+𝑤)−𝑓₂(𝑣+𝑤)‖≤𝜓₂(𝑢,𝑣,𝑤) (3.27)for all 𝑢,𝑣,𝑤∈𝑌.
By [2], there exists a cubic mapping 𝐹₁∶𝑋→𝑍 satisfying (1.1) such that‖𝑓₁(𝑥)−𝐹₁(𝑥)‖≤𝜓^{0𝑥002𝑑𝑐}₁(𝑥2,𝑥2,𝑥21)+3^{0𝑥002𝑑𝑐}𝜓₁𝑥(𝑥,2,𝑥2) (3.28)for all 𝑥∈𝑋. By Lemma 3.1, there exists a quadratic mapping 𝐹₂∶𝑌→𝑍 satisfying (1.2) such that‖𝑓₂(𝑦)−𝐹₂(𝑦)‖≤𝜓^{0𝑥002𝑑𝑐}₂(𝑦2,𝑦2,𝑦)+2𝜓^{0𝑥002𝑑𝑐}₂(𝑦2,𝑦2,𝑦2)+2𝜓^{0𝑥002𝑑𝑐}₂𝑦(0,0,21)+3[𝜓₂(0,0,0)+‖𝑓(0,0)‖] (3.29)for all 𝑦∈𝑌.
If we define𝐹(𝑥,𝑦)∶=𝐹₁(𝑥)+𝐹₂(𝑦)−𝑓(0,0) (3.30)for all 𝑥∈𝑋 and all 𝑦∈𝑌, we conclude that‖𝑓(𝑥,𝑦)−𝐹(𝑥,𝑦)‖≤𝜓(𝑥,0,0,𝑦,0,0)+𝜓^{0𝑥002𝑑𝑐}₁(𝑥2,𝑥2,𝑥21)+3^{0𝑥002𝑑𝑐}𝜓₁𝑥(𝑥,2,𝑥2)+𝜓^{0𝑥002𝑑𝑐}₂(𝑦2,𝑦2,𝑦)+2𝜓^{0𝑥002𝑑𝑐}₂(𝑦2,𝑦2,𝑦2)+2𝜓^{0𝑥002𝑑𝑐}₂𝑦(0,0,2)+13[𝜓₂(0,0,0)+‖𝑓(0,0)‖] (3.31)
for all 𝑥∈𝑋 and all 𝑦∈𝑌.

Let 𝜑^′∶𝑋³→[0,∞) be a function satisfying^∞_𝑗=04^𝑗+1𝜑^′(2^−𝑗𝑥,2^−𝑗𝑦,2^−𝑗𝑧)<∞ (3.32)for all 𝑥,𝑦,𝑧∈𝑋. For the function 𝜑^′, we can obtain similar results to Lemma 3.1 and Theorem 3.3.

As a corollary, one can obtain a result when the control mapping is a summation of terms ‖⋅‖^𝑝, in which 𝑝 is a suitable constant by referring to [14].