On Solution and Stability of a Two-Variable Functional Equations

Abstract

The main purpose of this paper is to investigate the stability of the functional equation 𝑓(𝑥+𝑦,𝑦+𝑧)=2𝑓(𝑥/2,𝑦/2)+2𝑓(𝑦/2,𝑧/2) in normed spaces. The solutions of such functional equations are considered.

1. Introduction

The stability of functional equations was originated from a question of Ulam in 1940, concerning the stability of group homomorphism [1]. Since then, the stability problems of various functional equations have been extensively investigated by a number of authors and there are many investigating results concerning this problem (see [2–12]).

Throughout this paper, we study the stability of the functional equation 𝑓(𝑥+𝑦,𝑦+𝑧)=2𝑓𝑥2,2𝑦+2𝑓2𝑦,𝑧2,(1.1) where 𝑋 is a normed space, 𝑌 is a Banach space, and 𝑓∶𝑋×𝑋→𝑌 is a mapping. The solutions of such functions are considered.

We recall that a mapping 𝑓∶𝑋×𝑋→𝑌 is called additive if 𝑓(𝑥+𝑧,𝑦+𝑤)=𝑓(𝑥,𝑦)+𝑓(𝑧,𝑤).(1.2)

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

2. The General Solution of the Functional Equation (1.1)

Before proceeding the proof of the main result, we shall need the following two lemmas.

Lemma 2.1. Let 𝑓∶𝑋×𝑋→𝑌 be a mapping satisfying (1.1) and 𝑓(−𝑥,−𝑦)=𝑓(𝑥,𝑦) for all 𝑥,𝑦∈𝑋. Then 𝑓 is zero.

Proof. Suppose that 𝑓∶𝑋×𝑋→𝑌 is a mapping satisfying (1.1) and 𝑓(−𝑥,−𝑦)=𝑓(𝑥,𝑦). It is clear that 𝑓(0,0)=0. Letting 𝑦=𝑧=0 and replacing 𝑥 by 2𝑥 in (1.1), we obtain 𝑓(2𝑥,0)=2𝑓(𝑥,0).(2.1)
Similarly, letting 𝑥=𝑦=0 and replacing 𝑧 by 2𝑥 in (1.1), we obtain 𝑓(0,2𝑥)=2𝑓(0,𝑥).(2.2) Putting 𝑦=0 and replacing 𝑧 by 𝑦 in (1.1), we get 𝑓(𝑥,𝑦)=2𝑓𝑥2,0+2𝑓0,2𝑦=𝑓(𝑥,0)+𝑓(0,𝑦),(2.3) for all 𝑥,𝑦∈𝑋, where the last equality follows from (2.1) and (2.2). It follows from (2.3) and (1.1) that 𝑓(𝑥+𝑦,0)+𝑓(0,𝑥−𝑦)==𝑓2𝑓(𝑥+𝑦𝑦,𝑥−𝑦)=2𝑓2𝑦,𝑥2+2𝑓𝑥2,−2𝑦=𝑓(𝑦,20),0+𝑓+(02,𝑓𝑥)0+,𝑓𝑥2(𝑥+,02)𝑓+𝑓𝑥2(,00,−+𝑦)2.𝑓0,−2𝑦(2.4) Replacing 𝑥 and 𝑦 by 𝑥/2 and 𝑥/2 in (2.4), respectively, we obtain 𝑓(𝑥,0)=2𝑓0,𝑥2+2𝑓𝑥2,0.(2.5) Also, replacing 𝑦 and 𝑥 by −𝑥/2 and 𝑥/2 in (2.4), respectively, we get 𝑓(0,𝑥)=2𝑓𝑥2,0+2𝑓0,𝑥2.(2.6) It follows from (2.5) and (2.6) that 𝑓(𝑥,0)=𝑓(0,𝑥).(2.7) Consequently, by (2.4), we obtain 𝑓(𝑥+𝑦,0)+𝑓(𝑥−𝑦,0)=2𝑓(𝑥,0)+2𝑓(𝑦,0).(2.8) Replacing 𝑦 by 𝑥 in (2.8), we get 𝑓(2𝑥,0)=4𝑓(𝑥,0).(2.9) By (2.1) and (2.9), we have 𝑓(𝑥,0)=0.(2.10) The equalities (2.7) and (2.10) imply that 𝑓(0,𝑥)=0.(2.11) Now, it follows from (2.3), (2.10), and (2.11) that 𝑓(𝑥,𝑦)=0.(2.12) This completes the proof.

Lemma 2.2. Let 𝑓∶𝑋×𝑋→𝑌 be a mapping satisfying (1.1) and 𝑓(−𝑥,−𝑦)=−𝑓(𝑥,𝑦) for all 𝑥,𝑦∈𝑋. Then 𝑓 is additive.

Proof. It is easy to show that 𝑓(0,0)=0. Putting 𝑦=𝑧=0 and replacing 𝑥 by 2𝑥 in (1.1), we obtain 𝑓(2𝑥,0)=2𝑓(𝑥,0).(2.13) Again, putting 𝑥=𝑦=0 and replacing 𝑧 by 2𝑥 in (1.1), we obtain 𝑓(0,2𝑥)=2𝑓(0,𝑥).(2.14) Putting 𝑦=0 and replacing 𝑧 by 𝑦 in (1.1), we get 𝑓(𝑥,𝑦)=2𝑓𝑥2,0+2𝑓0,2𝑦=𝑓(𝑥,0)+𝑓(0,𝑦),(2.15) for all 𝑥,𝑦∈𝑋, where the last equality follows from (2.13) and (2.14). It follows from (2.13), (2.14), and (2.15) that 𝑓(𝑥+𝑦,0)−𝑓(0,𝑥−𝑦)===𝑓𝑓𝑓(((𝑥𝑥𝑥,++𝑦)𝑦𝑦+,,𝑦0𝑓)−+(𝑦𝑥𝑓,)−(𝑥0,)𝑦.−𝑥)(2.16) Replacing 𝑦 by −𝑦 in (2.16), we obtain 𝑓(𝑥−𝑦,0)−𝑓(0,𝑥+𝑦)=𝑓(𝑥,−𝑦)+𝑓(−𝑦,−𝑥).(2.17) Also, we have 𝑓(𝑥+𝑦,0)+𝑓(0,𝑥−𝑦)==𝑓𝑓((𝑥𝑥,+−𝑦𝑦),𝑥+−𝑓𝑦()𝑦,𝑥).(2.18) Replacing 𝑦 by −𝑦 in (2.18), we get 𝑓(𝑥−𝑦,0)+𝑓(0,𝑥+𝑦)=𝑓(𝑥,𝑦)+𝑓(−𝑦,𝑥).(2.19)
Now, by (2.16), (2.17), (2.18), and (2.19), we have 𝑓(𝑥+𝑦,0)−𝑓(𝑥−𝑦,0)=2𝑓(𝑦,0).(2.20) Replacing 𝑥 with 𝑥+𝑦 in the above equality, we get 𝑓(𝑥+2𝑦,0)−𝑓(𝑥,0)=2𝑓(𝑦,0).(2.21) Now, replacing 𝑦 by 𝑦/2 in the previous equality, we obtain 𝑓(𝑥+𝑦,0)=𝑓(𝑥,0)+𝑓(𝑦,0).(2.22)
Similarly, one can prove that 𝑓(0,𝑧+𝑤)=𝑓(0,𝑧)+𝑓(0,𝑤).(2.23) By (2.22) and (2.23), we conclude that 𝑓(𝑥+𝑦,𝑧+𝑤)=𝑓(𝑥,𝑧)+𝑓(𝑦,𝑤),(2.24) which shows that 𝑓 is additive. This completes the proof.

Now, we are ready to present the general solution of (1.1).

Theorem 2.3. Every mapping 𝑓∶𝑋×𝑋→𝑌 satisfying (1.1) is additive.

Proof. One can write 𝑓(𝑥,𝑦)=𝑓^𝑒(𝑥,𝑦)+𝑓^𝑜(𝑥,𝑦),(2.25) where 𝑓^𝑒(𝑥,𝑦)∶=(𝑓(𝑥,𝑦)+𝑓(−𝑥,−𝑦))/2 and 𝑓^𝑜∶=(𝑓(𝑥,𝑦)−𝑓(−𝑥,−𝑦))/2. Since 𝑓 satisfies (1.1), 𝑓^𝑒 and 𝑓^𝑜 satisfy (1.1). Further, we have 𝑓^𝑒(−𝑥,−𝑦)=𝑓^𝑒(𝑥,𝑦) and 𝑓^𝑜(−𝑥,−𝑦)=−𝑓^𝑜(𝑥,𝑦) for all 𝑥,𝑦∈𝑋. By Lemmas 2.1 and 2.2, 𝑓^𝑒(𝑥,𝑦) is zero and 𝑓^𝑜(𝑥,𝑦) is additive, respectively. It follows that 𝑓 is additive. This completes the proof.

3. The Stability of the Functional Equation (1.1)

Throughout this section, we prove the stability of the functional equation (1.1).

Theorem 3.1. Let 𝑓∶𝑋×𝑋→𝑌 be a mapping such that ‖‖‖𝑓(𝑥+𝑦,𝑦+𝑧)−2𝑓𝑥2,2𝑦−2𝑓2𝑦,2𝑧‖‖‖≤𝜙(𝑥,𝑦,𝑧),(3.1) where 𝜙∶𝑋×𝑋×𝑋→[0,∞) satisfies ∑^∞𝑖=02^𝑖𝜙(𝑥/2^𝑖,𝑦/2^𝑖,𝑧/2^𝑖)<∞ for all 𝑥,𝑦,𝑧∈𝑋. Then there exists a mapping 𝐹∶𝑋×𝑋→𝑌 satisfying (1.1), ‖𝑓(𝑥,𝑦)−𝐹(𝑥,𝑦)‖≤^∞_𝑖=12^𝑖𝜙2𝑥^𝑖,0,0+𝜙0,0,𝑦lim_𝑛→∞2^𝑖+𝜙(𝑥,0,𝑦),2^𝑛𝑓𝑥2^𝑛,𝑦2^𝑛=𝐹(𝑥,𝑦),(3.2) for all 𝑥,𝑦∈𝑋.

Proof. Letting 𝑥=𝑦=𝑧=0, we have 𝑓(0,0)=0. Putting 𝑦=𝑧=0, we obtain ‖‖‖𝑓(𝑥,0)−2𝑓𝑥2,0‖‖‖≤𝜙(𝑥,0,0).(3.3) By induction, we conclude that ‖‖‖𝑓(𝑥,0)−2_𝑛𝑓𝑥2^𝑛,0‖‖‖≤^𝑛−1_𝑖=02^𝑖𝜙𝑥2^𝑖,0,0.(3.4) Put 𝐾_𝑚,𝑛=‖2^𝑚𝑓(𝑥/2^𝑚,0)−2^𝑛𝑓(𝑥/2^𝑛,0)‖. Then 𝐾_𝑚,𝑛≤𝐾_𝑚,𝑛+1+𝐾_𝑛+1,𝑛=𝐾_𝑚,𝑛+1+2^𝑛‖‖‖2𝑓𝑥2^𝑛+1,0−𝑓𝑥2^𝑛≤𝐾_𝑚,𝑛+1,0‖‖‖+2^𝑛𝜙𝑥2^𝑛,0,0,(3.5) for all 𝑚,𝑛∈ℕ with 𝑚>𝑛. By induction, we get ‖‖‖2_𝑚𝑓𝑥2^𝑚,0−2^𝑛𝑓𝑥2^𝑛,0‖‖‖≤^𝑚−1_𝑖=𝑛2^𝑖𝜙𝑥2^𝑖,0,0<𝜖,(3.6) which implies that {2^𝑛𝑓(𝑥/2^𝑛,0)} is a Cauchy sequence in 𝑌 for all 𝑥∈𝑋. Since 𝑌 is complete, there exists ℎ(𝑥) such that ‖‖‖2_𝑛𝑓𝑥2^𝑛,0−ℎ(𝑥)‖‖‖⟶0,(3.7) as 𝑛→∞ for all 𝑥∈𝑋. Since ‖‖‖ℎ(𝑥)−2ℎ𝑥2‖‖‖≤‖‖‖ℎ(𝑥)−2^𝑛𝑓𝑥2^𝑛,0‖‖‖+‖‖‖2^𝑛𝑓𝑥2^𝑛,0−2ℎ𝑥2‖‖‖⟶0,(3.8) as 𝑛→∞, ℎ(𝑥)=2ℎ(𝑥/2) for all 𝑥∈𝑋.
Similarly, we can prove that there exists a mapping 𝑔∶𝑋→𝑌 such that ‖‖‖2_𝑛𝑓0,𝑥2^𝑛−𝑔(𝑥)‖‖‖⟶0,(3.9) as 𝑛→∞ and 𝑔(𝑥)=2𝑔(𝑥/2) for all 𝑥∈𝑋.
Now, if 𝑑_𝑛(𝑥,𝑦,𝑧)=𝑓𝑥+𝑦2^𝑛,𝑦+𝑧2^𝑛−2𝑓2𝑥_𝑛+1,𝑦2^𝑛+1−2𝑓2𝑦^𝑛+1,𝑧2^𝑛+1,(3.10) and 𝐹_𝑛=‖2^𝑛𝑓((𝑥+𝑦)/2^𝑛,𝑦/2^𝑛)−ℎ(𝑥+𝑦)−𝑔(𝑦)‖, then we get 𝐹_𝑛=‖‖‖2^𝑛𝑑_𝑛(𝑥+𝑦,0,𝑦)+2^𝑛+1𝑓𝑥+𝑦2^𝑛+1,0+2^𝑛+1𝑓0,𝑦≤‖‖22^𝑛+1−ℎ(𝑥+𝑦)−𝑔(𝑦)_𝑛‖‖‖𝑑_𝑛(𝑥+𝑦,0,𝑦)‖‖+‖‖‖2^𝑛+1𝑓𝑥+𝑦2^𝑛+1,0−ℎ(𝑥+𝑦)‖‖‖+‖‖‖2^𝑛+1𝑓0,𝑦≤2_𝑛2^𝑛+1−𝑔(𝑦)‖‖‖‖‖𝑑_𝑛(𝑥+𝑦,0,𝑦)‖‖+‖‖‖2^𝑛+1𝑓𝑥+𝑦2^𝑛+1,0−ℎ(𝑥+𝑦)‖‖‖+‖‖‖2^𝑛+1𝑓0,𝑦≤2_𝑛2^𝑛+1−𝑔(𝑦)‖‖‖𝜙𝑥+𝑦2^𝑛,0,𝑦2^𝑛+‖‖‖2^𝑛+1𝑓𝑥+𝑦2^𝑛+1,0−ℎ(𝑥+𝑦)‖‖‖+‖‖‖2^𝑛+1𝑓0,𝑦2^𝑛+1−𝑔(𝑦)‖‖‖,(3.11) for all 𝑥,𝑦∈𝑋. Thus we obtain lim_𝑛→∞2^𝑛𝑓𝑥+𝑦2^𝑛,𝑦2^𝑛=ℎ(𝑥+𝑦)+𝑔(𝑦),(3.12) for all 𝑥,𝑦∈𝑋. Let 𝐺_𝑛=‖2^𝑛𝑓(𝑥/2^𝑛,𝑦/2^𝑛)−ℎ(𝑥+𝑦)−𝑔(𝑦)+ℎ(𝑦)‖. Then we conclude that 𝐺_𝑛=‖‖‖−2^𝑛−1𝑑_𝑛−1(𝑥,𝑦,0)+2^𝑛−1𝑓𝑥+𝑦2^𝑛−1,𝑦2^𝑛−1−2^𝑛𝑓𝑦2^𝑛≤‖‖2,0−ℎ(𝑥+𝑦)−𝑔(𝑦)+ℎ(𝑦)_𝑛−1‖‖‖𝑑_𝑛−1(𝑥,𝑦,0)‖‖+‖‖‖2^𝑛−1𝑓𝑥+𝑦2^𝑛−1,𝑦+‖‖‖22^𝑛−1−ℎ(𝑥+𝑦)−𝑔(𝑦)_𝑛‖‖‖𝑓𝑦2^𝑛≤2_𝑛−1,0−ℎ(𝑦)‖‖‖𝜙𝑥2^𝑛−1,𝑦2^𝑛−1,0+‖‖‖2^𝑛−1𝑓𝑥+𝑦2^𝑛−1,𝑦+‖‖‖22^𝑛−1−ℎ(𝑥+𝑦)−𝑔(𝑦)_𝑛‖‖‖𝑓𝑦2^𝑛,0−ℎ(𝑦)‖‖‖,(3.13) for all 𝑥,𝑦∈𝑋. It follows from (3.12) that lim_𝑛→∞2^𝑛𝑓𝑥2^𝑛,𝑦2^𝑛=ℎ(𝑥+𝑦)+𝑔(𝑦)−ℎ(𝑦),(3.14) for all 𝑥,𝑦∈𝑋. Put 𝐻_𝑛=‖2^𝑛𝑓(𝑥/2^𝑛,𝑦/2^𝑛)−ℎ(𝑥)−𝑔(𝑦)‖. Since 𝐻_𝑛=‖‖‖2^𝑛𝑑_𝑛(𝑥,0,𝑦)+2^𝑛+1𝑓𝑥2^𝑛+1,0+2^𝑛+1𝑓0,𝑦≤‖‖22^𝑛+1−ℎ(𝑥)−𝑔(𝑦)_𝑛‖‖‖𝑑_𝑛(𝑥,0,𝑦)‖‖+‖‖‖2^𝑛+1𝑓𝑥2^𝑛+1,0−ℎ(𝑥)‖‖‖+‖‖‖2^𝑛+1𝑓0,𝑦≤2_𝑛2^𝑛+1−𝑔(𝑦)‖‖‖𝜙𝑥2^𝑛,0,𝑦2^𝑛+‖‖‖2^𝑛+1𝑓𝑥2^𝑛+1,0−ℎ(𝑥)‖‖‖+‖‖‖2^𝑛+1𝑓𝑥2^𝑛+1⟶0,,0−ℎ(𝑥)‖‖‖(3.15) as 𝑛→∞, lim_𝑛→∞2^𝑛𝑓𝑥2^𝑛,𝑦2^𝑛=ℎ(𝑥)+𝑔(𝑦),(3.16) for all 𝑥,𝑦∈𝑋. It follows from (3.14) and (3.16) that ℎ(𝑥+𝑦)+𝑔(𝑦)−ℎ(𝑦)=ℎ(𝑥)+𝑔(𝑦),(3.17) for all 𝑥,𝑦∈𝑋. So ℎ(𝑥+𝑦)=ℎ(𝑥)+ℎ(𝑦) for all 𝑥,𝑦∈𝑋.
Similarly, we can show that 𝑔(𝑥+𝑦)=𝑔(𝑥)+𝑔(𝑦),(3.18) for all 𝑥,𝑦∈𝑋. Let 𝐹(𝑥,𝑦)=ℎ(𝑥)+𝑔(𝑦). Then we conclude that 𝐹(𝑥+𝑦,𝑦+𝑧)==ℎℎ(𝑥𝑥)+ℎ(𝑦)+𝑔(𝑦)+𝑔(𝑧)=2𝐹2𝑥+2ℎ2𝑦+2𝑔2𝑦+2𝑔2𝑧2,2𝑦+2𝐹2𝑦,2𝑧,(3.19) for all 𝑥,𝑦,𝑧∈𝑋.
Now, letting 𝐿_𝑛=‖𝑓(𝑥,𝑦)−2^𝑛𝑓(𝑥/2^𝑛,0)−2^𝑛𝑓(0,𝑦/2^𝑛)‖, we obtain ‖𝑓(𝑥,𝑦)−𝐹(𝑥,𝑦)‖≤‖‖‖𝑓(𝑥,𝑦)−2^𝑛𝑓𝑥2^𝑛,0−2^𝑛𝑓0,𝑦2^𝑛‖‖‖+𝐿^𝑛.(3.20) Since ‖‖‖2_𝑛𝑓𝑥2^𝑛,0+2^𝑛𝑓0,𝑦2^𝑛−𝐹(𝑥,𝑦)‖‖‖⟶0,(3.21) as 𝑛→∞, it is sufficient to show that 𝐿_𝑛≤^𝑛−1_𝑖=12^𝑖𝜙2𝑥^𝑖,0,0+𝜙0,0,𝑦2^𝑖+𝜙(𝑥,0,𝑦).(3.22) We have 𝐿_𝑛≤‖‖‖𝑓(𝑥,𝑦)−2𝑓𝑥2,0−2𝑓0,2𝑦‖‖‖+‖‖‖2𝑓𝑥2,0−2^𝑛𝑓𝑥2^𝑛+‖‖‖2𝑓0,𝑦,0‖‖‖2−2^𝑛𝑓0,𝑦≤_𝑛−12^𝑛‖‖‖_𝑖=12^𝑖𝜙2𝑥^𝑖,0,0+𝜙0,0,𝑦2^𝑖+𝜙(𝑥,0,𝑦).(3.23) As 𝑛→∞, we have ‖𝑓(𝑥,𝑦)−𝐹(𝑥,𝑦)‖≤^∞_𝑖=12^𝑖𝜙2𝑥^𝑖,0,0+𝜙0,0,𝑦2^𝑖+𝜙(𝑥,0,𝑦).(3.24) This completes the proof.

Corollary 3.2. Suppose that 𝑓∶𝑋×𝑋→𝑌 is a mapping such that ‖‖‖𝑓(𝑥+𝑦,𝑦+𝑧)−2𝑓𝑥2,2𝑦−2𝑓2𝑦,2𝑧‖‖‖≤𝜖‖𝑥‖^𝑝+‖𝑦‖^𝑝+‖𝑧‖^𝑝,(3.25) where 𝑝>1 and 𝜖>0. Then there exists a mapping 𝐹∶𝑋×𝑋→𝑌 satisfying (1.1), ‖𝑓(𝑥,𝑦)−𝐹(𝑥,𝑦)‖≤2^𝑝2^𝑝−2𝜖‖𝑥‖^𝑝lim_𝑛→∞+‖𝑦‖^𝑝,2^𝑛𝑓𝑥2^𝑛,𝑦2^𝑛=𝐹(𝑥,𝑦),(3.26) for all 𝑥,𝑦∈𝑋.

Corollary 3.3. Suppose that 𝑓∶𝑋×𝑋→𝑌 is a mapping such that ‖‖‖𝑓(𝑥+𝑦,𝑦+𝑧)−2𝑓𝑥2,2𝑦−2𝑓2𝑦,2𝑧‖‖‖≤𝜖‖𝑥‖^𝑝‖𝑦‖^𝑝+‖𝑦‖^𝑝‖𝑧‖^𝑝+‖𝑧‖^𝑝‖𝑥‖^𝑝,(3.27) where 𝑝>1/2 and 𝜖>0. Then there exists a mapping 𝐹∶𝑋×𝑋→𝑌 satisfying (1.1), ‖𝑓(l𝑥im,_𝑛→∞𝑦)−𝐹(𝑥,𝑦)‖≤𝜖‖𝑥‖^𝑝‖𝑦‖^𝑝,2^𝑛𝑓𝑥2^𝑛,𝑦2^𝑛=𝐹(𝑥,𝑦),(3.28) for all 𝑥,𝑦∈𝑋.

Example 3.4. Let 𝑋=𝑌=𝐶[𝑎,𝑏] with the norm ‖𝑥‖=sup_{𝑡∈[𝑎,𝑏]}|𝑥(𝑡)|, where 𝑏>𝑎>0. Define 𝑓∶𝐶[𝑎,𝑏]×𝐶[𝑎,𝑏]→𝐶[𝑎,𝑏] by 𝑓(𝑥,𝑦)(𝑡)=𝑥(𝑡)+𝑡𝑦(𝑡)²,(3.29) for all 𝑡∈[𝑎,𝑏], and 𝜙∶𝑋×𝑋×𝑋→ℝ by 𝜙(𝑥,𝑦,𝑧)=𝑏(‖𝑥‖+‖𝑦‖)².(3.30) It is not difficult to show that 𝜙 satisfies Theorem 3.1 since ‖‖‖𝑓(𝑥+𝑦,𝑦+𝑧)−2𝑓𝑥=sup_{𝑡∈[𝑎,𝑏]}2,2𝑦−2𝑓2𝑦,2𝑧‖‖‖|||𝑓(𝑥+𝑦,𝑦+𝑧)(𝑡)−2𝑓𝑥=sup_{𝑡∈[𝑎,𝑏]}2,2𝑦(𝑡)−2𝑓2𝑦,2𝑧(𝑡)||||||𝑡2𝑥(𝑡)²+𝑡2𝑦(𝑡)²≤𝑏‖𝑥‖₂+2𝑡𝑥(𝑡)𝑦(𝑡)|||+𝑏‖𝑦‖²+2𝑏‖𝑥‖‖𝑦‖=𝑏(‖𝑥‖+‖𝑦‖)²=𝜙(𝑥,𝑦,𝑧),(3.31) for all 𝑥,𝑦,𝑧∈𝑋. Theorem 3.1 implies that there exists a mapping 𝐹∶𝑋×𝑋→𝑌 satisfying (1.1) and ‖𝑓(𝑥,𝑦)−𝐹(𝑥,𝑦)‖≤2𝑏‖𝑥‖²,(3.32) for all 𝑥,𝑦∈𝑋.

Example 3.5. Let 𝑋 be a normed space. Define the function 𝜙₀∶𝑋×𝑋→[0,∞) by 𝜙₀(𝑥,𝑦)=⎛⎜⎜⎜⎝sup_{𝑓,𝑔∈𝐵𝑋′}||||||𝑓𝑔((𝑥𝑥))𝑓𝑔((𝑦𝑦))||||||⎞⎟⎟⎟⎠^𝑝,(3.33) where 𝐵_𝑋^′ is the closed unit ball of the dual space 𝑋^ and 𝑝>1. Then one can show that the function 𝜙∶𝑋×𝑋×𝑋→[0,∞), defined by 𝜙(𝑥,𝑦,𝑧)=𝜙₀(𝑥,𝑦)+𝜙₀(𝑦,𝑧),(3.34) satisfies Theorem 3.1.

Example 3.6. Define 𝜙(𝑥,𝑦,𝑧)=ℓ^𝑝×ℓ^𝑝×ℓ^𝑝→[0,∞) by 𝜙(𝑥,𝑦,𝑧)=^∞_𝑖=1||𝜉_𝑖||^𝑝+^∞_𝑖=1||𝜂_𝑖||^𝑝+^∞_𝑖=1||𝜁_𝑖||^𝑝,(3.35) where 𝑥=(𝜉₁,𝜉₂,…), 𝑦=(𝜂₁,𝜂₂,…), and 𝑧=(𝜁₁,𝜁₂,…). It is obvious that 𝜙 satisfies Theorem 3.1.

Theorem 3.7. Suppose that 𝑓∶𝑋×𝑋→𝑌 is a mapping and 𝜙∶𝑋×𝑋×𝑋→[0,∞) is a function such that ‖‖‖𝑓(𝑥+𝑦,𝑦+𝑧)−2𝑓𝑥2,2𝑦−2𝑓2𝑦,2𝑧‖‖‖≤𝜙(𝑥,𝑦,𝑧),(3.36)𝜙(0,0,0)=0 and ∑^∞𝑖=1(1/2^𝑖)𝜙(2^𝑖𝑥,2^𝑖𝑦,2^𝑖𝑧)<∞ for all 𝑥,𝑦,𝑧∈𝑋. Then there exists a mapping 𝐹∶𝑋×𝑋→𝑌 satisfying (1.1), ‖𝑓(𝑥,𝑦)−𝐹(𝑥,𝑦)‖≤^∞_𝑖=212^𝑖lim_𝑛→∞𝜙2^𝑖𝑥,0,0+𝜙0,0,2^𝑖𝑦+𝜙(𝑥,0,𝑦),12^𝑛𝑓(2^𝑛𝑥,2^𝑛𝑦)=𝐹(𝑥,𝑦),(3.37) for all 𝑥,𝑦∈𝑋.

Proof. It is clear that 𝑓(0,0)=0. Now, letting 𝑦=𝑧=0, we obtain ‖‖‖12𝑓(2𝑥,0)−𝑓(𝑥,0)‖‖‖≤12𝜙(2𝑥,0,0).(3.38) By induction, we have ‖‖‖12^𝑛𝑓(2^𝑛𝑥,0)−𝑓(𝑥,0)‖‖‖≤^𝑛_𝑖=112^𝑖𝜙2^𝑖𝑥,0,0.(3.39) It follows that, for any 𝜖>0, there exists 𝑁∈ℕ such that, for all 𝑛,𝑚∈ℕ with 𝑚>𝑛>𝑁, ‖‖‖12^𝑚𝑓(2^𝑚𝑥,0)−12^𝑛𝑓(2^𝑛𝑥,0)‖‖‖≤^𝑚_𝑖=_𝑛+112^𝑖𝜙2^𝑖𝑥,0,0<𝜖.(3.40) Thus {(1/2^𝑛)𝑓(2^𝑛𝑥,0)} is a Cauchy sequence in 𝑌. The completeness of 𝑌 implies that there exists ℎ(𝑥) such that lim_𝑛→∞‖‖‖12^𝑛𝑓(2^𝑛𝑥,0)−ℎ(𝑥)‖‖‖=0.(3.41) Since ‖‖‖ℎ(𝑥)−12ℎ(2𝑥)‖‖‖≤‖‖‖ℎ(𝑥)−21^𝑛𝑓(2^𝑛𝑥,0)‖‖‖+‖‖‖12^𝑛𝑓(2^𝑛𝑥,0)−12ℎ(2𝑥)‖‖‖⟶0,(3.42) as 𝑛→∞, ℎ(𝑥)=(1/2)ℎ(2𝑥).
Similarly, we conclude that there exists a mapping 𝑔∶𝑋→𝑌 such that lim_𝑛→∞‖‖‖12^𝑛𝑓(0,2^𝑛𝑧)−𝑔(𝑧)‖‖‖=0,(3.43) and 𝑔(𝑧)=(1/2)𝑔(2𝑧). Let 𝑑_𝑛(𝑥,𝑦,𝑧)=12𝑓2^𝑛+1(𝑥+𝑦),2^𝑛+1(𝑦+𝑧)−𝑓(2^𝑛𝑥,2^𝑛𝑦)−𝑓(2^𝑛𝑦,2^𝑛𝑧),(3.44) and 𝐹_𝑛=‖(1/2^𝑛+1)𝑓(2^𝑛+1(𝑥+𝑦),2^𝑛+1𝑦)−ℎ(𝑥+𝑦)−𝑔(𝑦)‖. Then we have 𝐹_𝑛=‖‖‖12^𝑛𝑑_𝑛(𝑥+𝑦,0,𝑦)+12^𝑛𝑓(2^𝑛(𝑥+𝑦),0)+12^𝑛≤‖‖‖1𝑓(0,2^𝑛𝑦)−ℎ(𝑥+𝑦)−𝑔(𝑦)‖‖‖2^𝑛𝑑_𝑛(𝑥+𝑦,0,𝑦)‖‖‖+‖‖‖12^𝑛𝑓(2^𝑛(𝑥+𝑦),0)−ℎ(𝑥+𝑦)‖‖‖+‖‖‖12^𝑛≤1𝑓(0,2^𝑛𝑦)−𝑔(𝑦)‖‖‖2^𝑛‖‖𝑑_𝑛(𝑥+𝑦,0,𝑦)‖‖+‖‖‖12^𝑛𝑓(2^𝑛(𝑥+𝑦),0)−ℎ(𝑥+𝑦)‖‖‖+‖‖‖12^𝑛≤1𝑓(0,2^𝑛𝑦)−𝑔(𝑦)‖‖‖2^𝑛𝜙(2^𝑛(𝑥+𝑦),0,2^𝑛𝑦)+‖‖‖12^𝑛𝑓(2^𝑛(𝑥+𝑦),0)−ℎ(𝑥+𝑦)‖‖‖+‖‖‖12^𝑛𝑓0,2^𝑛+1𝑦−𝑔(𝑦)‖‖‖,(3.45) for all 𝑥,𝑦∈𝑋. By hypothesis, (3.41) and (3.43), we get lim_𝑛→∞12^𝑛+1𝑓2^𝑛+1(𝑥+𝑦),2^𝑛+1𝑦=ℎ(𝑥+𝑦)+𝑔(𝑦),(3.46) for all 𝑥,𝑦∈𝑋.
Now, let 𝐿_𝑛=‖(1/2^𝑛−1)𝑓(2^𝑛−1𝑥,2^𝑛−1𝑦)−ℎ(𝑥+𝑦)−𝑔(𝑦)+ℎ(𝑦)‖. Then we obtain 𝐿_𝑛=‖‖‖−12^𝑛−1𝑑_𝑛−1(𝑥,𝑦,0)+12^𝑛𝑓(2^𝑛(𝑥+𝑦),2^𝑛𝑦)−12^𝑛≤‖‖‖1𝑓(2^𝑛𝑦,0)−ℎ(𝑥+𝑦)−𝑔(𝑦)+ℎ(𝑦)‖‖‖2^𝑛−1𝑑_𝑛−1(𝑥,𝑦,0)‖‖‖+‖‖‖12^𝑛+‖‖‖1𝑓(2^𝑛(𝑥+𝑦),2^𝑛𝑦)−ℎ(𝑥+𝑦)−𝑔(𝑦)‖‖‖2^𝑛−1≤1𝑓2^𝑛−1𝑦,0−ℎ(𝑦)‖‖‖2^𝑛𝜙(2^𝑛𝑥,2^𝑛𝑦,0)+‖‖‖12^𝑛+‖‖‖1𝑓2^𝑛(𝑥+𝑦),2^𝑛−1𝑦−ℎ(𝑥+𝑦)−𝑔(𝑦)‖‖‖2^𝑛−1𝑓2^𝑛−1𝑦,0−ℎ(𝑦)‖‖‖,(3.47) where the last inequality follows from (3.46). By (3.41) and (3.43), we conclude lim_𝑛→∞12^𝑛−1𝑓2^𝑛−1𝑥,2^𝑛−1𝑦=ℎ(𝑥+𝑦)+𝑔(𝑦)−ℎ(𝑦),(3.48) for all 𝑥,𝑦∈𝑋. Put 𝑀_𝑛=‖(1/2^𝑛+1)𝑓(2^𝑛+1𝑥,2^𝑛+1𝑦)−ℎ(𝑥)−𝑔(𝑦)‖. Since 𝑀_𝑛=‖‖‖12^𝑛𝑑_𝑛(𝑥,0,𝑦)+12^𝑛𝑓(2^𝑛𝑥,0)+12^𝑛≤‖‖‖1𝑓(0,2^𝑛𝑦)−ℎ(𝑥)−𝑔(𝑦)‖‖‖2^𝑛𝑑_𝑛(𝑥,0,𝑦)‖‖‖+‖‖‖12^𝑛𝑓(2^𝑛𝑥,0)−ℎ(𝑥)‖‖‖+‖‖‖12^𝑛≤1𝑓(0,2^𝑛𝑦)−𝑔(𝑦)‖‖‖2^𝑛+1𝜙2^𝑛+1𝑥,0,2^𝑛+1𝑦+‖‖‖12^𝑛+‖‖‖1𝑓(2^𝑛𝑥,0)−ℎ(𝑥)‖‖‖2^𝑛⟶0,𝑓(0,2^𝑛𝑦)−𝑔(𝑦)‖‖‖(3.49) as 𝑛→∞, where the last inequality follows from (3.41) and (3.43). So we have lim_𝑛→∞12^𝑛+1𝑓2^𝑛+1𝑥,2^𝑛+1𝑦=ℎ(𝑥)+𝑔(𝑦),(3.50) for all 𝑥,𝑦∈𝑋. By (3.48) and (3.50), we have ℎ(𝑥+𝑦)+𝑔(𝑦)−ℎ(𝑦)=ℎ(𝑥)+𝑔(𝑦),(3.51) for all 𝑥,𝑦∈𝑋. It follows that ℎ(𝑥+𝑦)=ℎ(𝑥)+ℎ(𝑦).
Similarly, we obtain 𝑔(𝑥+𝑦)=𝑔(𝑥)+𝑔(𝑦).
Now, define the mapping 𝐹∶𝑋×𝑋→𝑌 by 𝐹(𝑥,𝑦)=ℎ(𝑥)+𝑔(𝑦). Then we have 𝐹(𝑥+𝑦,𝑦+𝑧)==ℎ2ℎ(𝑥)𝑥+ℎ(𝑦)+𝑔(𝑦)+𝑔(𝑧)=2𝐹2𝑥+2ℎ2𝑦+2𝑔2𝑦+2𝑔2𝑧2,2𝑦+2𝐹2𝑦,2𝑧.(3.52) Now, we have ‖𝑓(𝑥,𝑦)−𝐹(𝑥,𝑦)‖≤‖‖‖𝑓(𝑥,𝑦)−12^𝑛𝑓(2^𝑛𝑥,0)−12^𝑛+‖‖‖1𝑓(0,2^𝑛𝑦)‖‖‖2^𝑛𝑓(2^𝑛𝑥,0)+12^𝑛𝑓(0,2^𝑛𝑦)−𝐹(𝑥,𝑦)‖‖‖.(3.53) Let 𝐻_𝑛=‖𝑓(𝑥,𝑦)−(1/2^𝑛)𝑓(2^𝑛𝑥,0)−(1/2^𝑛)𝑓(0,2^𝑛𝑦)‖. Since we know that lim_𝑛→∞‖‖‖12^𝑛𝑓(2^𝑛𝑥,0)+12^𝑛𝑓(0,2^𝑛𝑦)−𝐹(𝑥,𝑦)‖‖‖=0,(3.54) to prove (3.37), it is sufficient to show that 𝐻_𝑛≤^𝑛_𝑖=212^𝑖𝜙2^𝑖𝑥,0,0+𝜙0,0,2^𝑖𝑦+𝜙(𝑥,0,𝑦),(3.55) for all 𝑛∈ℕ. We have 𝐻_𝑛≤‖‖‖𝑓(𝑥,𝑦)−12𝑓(2𝑥,0)−12𝑓(0,2𝑦)‖‖‖+‖‖‖21𝑓(2𝑥,0)−21^𝑛+‖‖‖1𝑓(2^𝑛𝑥,0)‖‖‖2𝑓(0,2𝑦)−21^𝑛≤𝑓(0,2^𝑛𝑦)_𝑛‖‖‖_𝑖=212^𝑖𝜙2^𝑖𝑥,0,0+𝜙0,0,2^𝑖𝑦+𝜙(𝑥,0,𝑦),(3.56) for all 𝑥,𝑦∈𝑋. As 𝑛→∞, we conclude ‖𝑓(𝑥,𝑦)−𝐹(𝑥,𝑦)‖≤^∞_𝑖=212^𝑖𝜙2^𝑖𝑥,0,0+𝜙0,0,2^𝑖𝑦+𝜙(𝑥,0,𝑦),(3.57) for all 𝑥,𝑦∈𝑋. This completes the proof.

Corollary 3.8. Suppose that 𝜖>0 and 𝑓∶𝑋×𝑋→𝑌 is a mapping satisfying ‖‖‖𝑓(𝑥+𝑦,𝑦+𝑧)−2𝑓𝑥2,2𝑦−2𝑓2𝑦,2𝑧‖‖‖≤𝜖‖𝑥‖^𝑝+‖𝑦‖^𝑝+‖𝑧‖^𝑝,(3.58) for some 𝑝<1. Then there exists a mapping 𝐹∶𝑋×𝑋→𝑌 satisfying (1.1), ‖𝑓(𝑥,𝑦)−𝐹(𝑥,𝑦)‖≤2−2^𝑝+2^2𝑝−12−2^𝑝lim_𝑛→∞𝜖‖𝑥‖^𝑝+‖𝑦‖^𝑝,12^𝑛𝑓(2^𝑛𝑥,2^𝑛𝑦)=𝐹(𝑥,𝑦),(3.59) for all 𝑥,𝑦∈𝑋.

Example 3.9. (1) Let (𝑋,⟨⋅,⋅⟩) be an inner product space. Then one can show that the function 𝜙∶𝑋×𝑋×𝑋→ℝ, defined by 𝜙(𝑥,𝑦,𝑧)=||⟨𝑥,𝑦⟩||^𝑝+||⟨𝑦,𝑧⟩||^𝑝(𝑝<1),(3.60) satisfies Theorem 3.7.
(2) Suppose that (𝑋,‖⋅‖) is a normed space. Then the function 𝜙∶𝑋×𝑋×𝑋→ℝ, defined by 𝜙(𝑥,𝑦,𝑧)=‖𝑥‖‖𝑦‖‖𝑧‖, satisfies Theorem 3.7.

4. A Fixed Point Approach to the Stability

In this section, we apply the fixed point method to prove the stability of the functional equation (1.1).

Let 𝑋 be a set. A function 𝑑∶𝑋×𝑋→[0,∞] is called a generalized metric on 𝑋 if 𝑑 satisfies the following conditions:

(1) 𝑑(𝑥,𝑦)=0 for all 𝑥,𝑦∈𝑋 if and only if 𝑥=𝑦;

(2) 𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥) for all 𝑥,𝑦∈𝑋;

(3) 𝑑(𝑥,𝑧)≤𝑑(𝑥,𝑦)+𝑑(𝑦,𝑧) for all 𝑥,𝑦,𝑧∈𝑋.

We now introduce one of the fundamental results of fixed point theory.

Theorem 4.1 (see [13]). Let (𝑋,𝑑) be a complete generalized metric space and let 𝐽∶𝑋→𝑋 be a contractive mapping with constant 𝐿. Then for each 𝑥∈𝑋, either 𝑑𝐽^𝑛𝑥,𝐽^𝑛+1𝑥=∞,(4.1) for all 𝑛∈ℕ or there exists an 𝑛₀∈ℕ such that(a)𝑑(𝐽^𝑛𝑥,𝐽^𝑛+1𝑥)<∞ for all 𝑛≥𝑛₀;(b) the sequence {𝐽^𝑛𝑥} converges to fixed point 𝑥₀ of 𝐽;(c)𝑥₀ is the unique fixed point of 𝐽 in the set 𝑌={𝑦∈𝑋∶𝑑(𝐽^𝑛0𝑥,𝑦)<∞};(d)𝑑(𝑦,𝑦₀)≤(1/(1−𝐿))𝑑(𝑦,𝐽𝑦) for all 𝑦∈𝑌.

In 1996, Isac and Rassias [14] were the first to provide applications of stability theory of functional equations for the proof of new fixed point theorems with applications. By using fixed point methods, the stability problems of several functional equations have been extensively investigated by a number of authors (see [15–22]).

Theorem 4.2. Suppose that 𝑋 is a vector space and 𝑌 is a Banach space. Let 𝑓∶𝑋×𝑋→𝑌 be a mapping and let 𝜙∶𝑋×𝑋×𝑋→[0,∞) be a function satisfying the following conditions:(a)‖𝑓(𝑥+𝑦,𝑦+𝑧)−2𝑓(𝑥/2,𝑦/2)−2𝑓(𝑦/2,𝑧/2)‖≤𝜙(𝑥,𝑦,𝑧);(b)𝜙(0,0,0)=0;(c) for all 𝑥,𝑦∈𝑋, lim_𝑛→∞12^𝑛𝜙(2^𝑛𝑥,2^𝑛𝑦,2^𝑛𝑧)=0;(4.2)(d) there exists a number 𝐿∈[0,1) such that 𝜙(𝑥,𝑦,𝜎(𝑥,𝑦))≤2𝐿𝜙𝑥2,2𝑦,𝜎𝑥2,2𝑦,(4.3) where 𝜎∶𝑋×𝑋→𝑋 is a mapping such that 𝜎(𝑥,0)=0 for all 𝑥∈𝑋.Then there exists a unique mapping 𝐹∶𝑋×𝑋→𝑌 satisfying (1.1) and ‖𝑓(𝑥,𝑦)−𝐹(𝑥,𝑦)‖≤12−2𝐿𝜙(𝑥,𝑦,𝜎(𝑥,𝑦)),(4.4) for all 𝑥,𝑦∈𝑋.

Proof. Consider the set 𝒮={𝑔∶𝑋×𝑋⟶𝑌},(4.5) and define the generalized metric on 𝒮 by 𝑑(𝑔,ℎ)=inf{𝐶∶‖𝑔(𝑥,𝑦)−ℎ(𝑥,𝑦)‖≤𝐶𝜙(𝑥,𝑦,𝜎(𝑥,𝑦)),∀(𝑥,𝑦)∈𝑋×𝑋}.(4.6) It is easy to show that (𝒮,𝑑) is complete.
Now, we consider the linear mapping 𝐽∶𝒮→𝒮 such that 𝐽𝑔(𝑥,𝑦)=12𝑔(2𝑥,2𝑦),(4.7) for all 𝑥,𝑦∈𝑋. The definition of 𝐽 implies that 𝑑(𝐽𝑔,𝐽ℎ)≤𝐿𝑑(𝑔,ℎ),(4.8) for all 𝑔,ℎ∈𝒮. Replacing 𝑥 by 2𝑥 and putting 𝑦=𝑧=0 in the first statement, we get ‖𝑓(2𝑥,0)−2𝑓(𝑥,0)‖≤𝜙(𝑥,0,0)=𝜙(2𝑥,0,𝜎(2𝑥,0)),(4.9) for all 𝑥∈𝑋. So ‖‖‖𝑓(𝑥,0)−12𝑓(2𝑥,0)‖‖‖≤21𝜙(2𝑥,0,𝜎(2𝑥,0)),(4.10) for all 𝑥∈𝑋. Hence 𝑑(𝑓,𝐽𝑓)≤1/2. By Theorem 4.1, there exists a mapping 𝐹∶𝑋×𝑋→𝑌 such that 𝐹 is a fixed point of 𝐽, that is, 𝐹(2𝑥,2𝑦)=2𝐹(𝑥,𝑦),(4.11) for all 𝑥,𝑦∈𝑋. The mapping 𝐹 is a unique fixed point of 𝐽 in the set 𝒮′={𝑔∈𝒮∶𝑑(𝑓,𝑔)<∞}.(4.12) This implies that there exists a 𝐶<∞ such that ‖𝐹(𝑥,𝑦)−𝑓(𝑥,𝑦)‖≤𝐶𝜙(𝑥,𝑦,𝜎(𝑥,𝑦)),(4.13) for all 𝑥,𝑦∈𝑋. Also, we have 𝑑(𝐽^𝑛𝑓,𝐹)⟶0,(4.14) as 𝑛→∞. It follows that lim_𝑛→∞12^𝑛𝑓(2^𝑛𝑥,2^𝑛𝑦)=𝐹(𝑥,𝑦),(4.15) for all 𝑥,𝑦∈𝑋. By the third statement of Theorem 4.1, we have 𝑑(𝑓,𝐹)≤(1/(1−𝐿))𝑑(𝑓,𝐽𝑓). This implies that 𝑑(𝑓,𝐹)≤12−2𝐿.(4.16) Thus, by the definition of 𝑑(⋅,⋅), we conclude ‖𝑓(𝑥,𝑦)−𝐹(𝑥,𝑦)‖≤12−2𝐿𝜙(𝑥,𝑦,𝜎(𝑥,𝑦)),(4.17) for all 𝑥,𝑦∈𝑋. Let 𝐷_𝐹=𝐹(𝑥+𝑦,𝑦+𝑧)−2𝐹𝑥2,2𝑦−2𝐹2𝑦,𝑧2.(4.18) It follows from (4.15) that ‖‖𝐷_𝐹‖‖=lim_𝑛→∞12^𝑛‖‖‖𝑓(2_𝑛(𝑥+𝑦),2^𝑛(𝑦+𝑧))−2𝑓2^𝑛𝑥2,2^𝑛𝑦2−2𝑓2^𝑛𝑦2,2^𝑛𝑧≤lim_𝑛→∞2‖‖‖12^𝑛𝜙(2^𝑛𝑥,2^𝑛𝑦,2^𝑛𝑧)=0.(4.19) So we have 𝐹(𝑥+𝑦,𝑦+𝑧)=2𝐹𝑥2,2𝑦+2𝐹2𝑦,2𝑧,(4.20) for all 𝑥,𝑦,𝑧∈𝑋. So 𝐹 satisfies (1.1). This completes the proof.

Acknowledgments

Y. J. Cho was supported by the Korea Research Foundation Grant funded by the Korean Government KRF-2008-313-C00050.