Abstract

Let 𝑋,𝑌 be vector spaces and 𝑘 a fixed positive integer. It is shown that a mapping 𝑓(𝑘𝑥+𝑦)+𝑓(𝑘𝑥𝑦)=2𝑘2𝑓(𝑥)+2𝑓(𝑦) for all 𝑥,𝑦𝑋 if and only if the mapping 𝑓𝑋𝑌 satisfies 𝑓(𝑥+𝑦)+𝑓(𝑥𝑦)=2𝑓(𝑥)+2𝑓(𝑦) for all 𝑥,𝑦𝑋. Furthermore, the Hyers-Ulam-Rassias stability of the above functional equation in Banach spaces is proven.

1. Introduction

The stability problem of functional equations originated from a question of Ulam [1] concerning the stability of group homomorphisms. Hyers [2] gave a first affirmative answer to the question of Ulam for Banach spaces. Hyers' theorem was generalized by Aoki [3] for additive mapping and by Th. M. Rassias [4] for linear mappings by considering an unbounded Cauchy difference. The paper of Th. M. Rassias [4] has provided a lot of influence in the development of what we now call Hyers-Ulam-Rassias stability of functional equations. Th. M. Rassias [5] during the 27th International Symposium on Functional Equations asked the question whether such a theorem can also be proved for 𝑝1. Gajda [6], following the same approach as in [4], gave an affirmative solution to this question for 𝑝>1. It was shown by Gajda [6] as well as by Rassias and Šemrl [7] that one cannot prove a Th.M. Rassias' type theorem when 𝑝=1. J. M. Rassias [8], following the spirit of the innovative approach of Th. M. Rassias [4] for the unbounded Cauchy difference, proved a similar stability theorem in which he replaced the factor 𝑥𝑝+𝑦𝑝 by 𝑥𝑝𝑦𝑞 for 𝑝,𝑞 with 𝑝+𝑞1.

The functional equation𝑓(𝑥+𝑦)+𝑓(𝑥𝑦)=2𝑓(𝑥)+2𝑓(𝑦)(1.1)is called a quadratic functional equation. In particular, every solution of the quadratic functional equation is said to be a quadratic function. A Hyers-Ulam-Rassias stability problem for the quadratic functional equation was proved by Skof [9] for mappings 𝑓𝑋𝑌, where 𝑋 is a normed space and 𝑌 is a Banach space. Cholewa [10] noticed that the theorem of Skof is still true if the relevant domain 𝑋 is replaced by an Abelian group. In [11], Czerwik proved the Hyers-Ulam-Rassias stability of the quadratic functional equation. Several functional equations have been investigated in [1217].

Throughout this paper, assume that 𝑘 is a fixed positive integer.

In this paper, we solve the functional equation𝑓(𝑘𝑥+𝑦)+𝑓(𝑘𝑥𝑦)=2𝑘2𝑓(𝑥)+2𝑓(𝑦)(1.2)and prove the Hyers-Ulam-Rassias stability of the functional equation (1.2) in Banach spaces.

2. Hyers-Ulam-Rassias Stability of the Quadratic Functional Equation

Proposition 2.1. Let 𝑋 and 𝑌 be vector spaces. A mapping 𝑓𝑋𝑌 satisfies 𝑓(𝑘𝑥+𝑦)+𝑓(𝑘𝑥𝑦)=2𝑘2𝑓(𝑥)+2𝑓(𝑦)(2.1)for all 𝑥,𝑦𝑋 if and only if the mapping 𝑓𝑋𝑌 satisfies 𝑓(𝑥+𝑦)+𝑓(𝑥𝑦)=2𝑓(𝑥)+2𝑓(𝑦)(2.2)for all 𝑥,𝑦𝑋.

Proof. Assume that 𝑓𝑋𝑌 satisfies (2.1).
Letting 𝑥=𝑦=0 in (2.1), we get 𝑓(0)=0.
Letting 𝑦=0 in (2.1), we get 𝑓(𝑘𝑥)=𝑘2𝑓(𝑥) for all 𝑥𝑋.
Letting 𝑥=0 in (2.1), we get 𝑓(𝑦)=𝑓(𝑦) for all 𝑦𝑋.
It follows from (2.1) that𝑓(𝑘𝑥+𝑦)+𝑓(𝑘𝑥𝑦)=2𝑘2𝑓(𝑥)+2𝑓(𝑦)=2𝑓(𝑘𝑥)+2𝑓(𝑦)(2.3)for all 𝑥,𝑦𝑋. So the mapping 𝑓𝑋𝑌 satisfies𝑓(𝑥+𝑦)+𝑓(𝑥𝑦)=2𝑓(𝑥)+2𝑓(𝑦)(2.4)for all 𝑥,𝑦𝑋.
Assume that 𝑓𝑋𝑌 satisfies 𝑓(𝑥+𝑦)+𝑓(𝑥𝑦)=2𝑓(𝑥)+2𝑓(𝑦) for all 𝑥,𝑦𝑋.
We prove (2.1) for 𝑘=𝑗 by induction on 𝑗.
For the case 𝑗=1, (2.1) holds by the assumption.
For the case 𝑗=2, since𝑓(2𝑥+𝑦)+𝑓(2𝑥𝑦)=𝑓(𝑥+𝑦+𝑥)+𝑓(𝑥𝑦+𝑥)=2𝑓(𝑥+𝑦)+2𝑓(𝑥)𝑓(𝑦)+2𝑓(𝑥𝑦)+2𝑓(𝑥)𝑓(𝑦)=2𝑓(𝑥+𝑦)+2𝑓(𝑥𝑦)+4𝑓(𝑥)2𝑓(𝑦)=4𝑓(𝑥)+4𝑓(𝑦)+4𝑓(𝑥)2𝑓(𝑦)=8𝑓(𝑥)+2𝑓(𝑦)(2.5)for all 𝑥,𝑦𝑋, then (2.1) holds.
Assume that (2.1) holds for 𝑗=𝑛2 and 𝑗=𝑛1 (2<𝑛𝑘). By the assumption,𝑓(𝑛𝑥+𝑦)+𝑓(𝑛𝑥𝑦)=𝑓(𝑛1)𝑥+𝑦+𝑥+𝑓(𝑛1)𝑥𝑦+𝑥=2𝑓(𝑛1)𝑥+𝑦+2𝑓(𝑥)𝑓(𝑛2)𝑥+𝑦+2𝑓(𝑛1)𝑥𝑦+2𝑓(𝑥)𝑓(𝑛2)𝑥𝑦=4(𝑛1)2𝑓(𝑥)+4𝑓(𝑦)+4𝑓(𝑥)2(𝑛2)2𝑓(𝑥)2𝑓(𝑦)=2𝑛2𝑓(𝑥)+2𝑓(𝑦)(2.6)for all 𝑥,𝑦𝑋, (2.1) holds for 𝑗=𝑛. Hence the mapping 𝑓𝑋𝑌 satisfies (2.1) for 𝑗=𝑘.

From now on, assume that 𝑋 is a normed vector space with norm and that 𝑌 is a Banach space with norm .

For a given mapping 𝑓𝑋𝑌, we define𝐷𝑓(𝑥,𝑦)=𝑓(𝑘𝑥+𝑦)+𝑓(𝑘𝑥𝑦)2𝑘2𝑓(𝑥)2𝑓(𝑦)(2.7)for all 𝑥,𝑦𝑋.

Now we prove the Hyers-Ulam-Rassias stability of the quadratic functional equation 𝐷𝑓(𝑥,𝑦)=0.

Theorem 2.2. Let 𝑓𝑋𝑌 be a mapping with 𝑓(0)=0 for which there exists a function 𝜑𝑋2[0,) such that 𝜑(𝑥,𝑦)=𝑗=01𝑘2𝑗𝜑𝑘𝑗𝑥,𝑘𝑗𝑦<,(2.8)𝐷𝑓(𝑥,𝑦)𝜑(𝑥,𝑦)(2.9)𝑥,𝑦𝑋for all 𝑄𝑋𝑌. Then there exists a unique quadratic mapping 1𝑓(𝑥)𝑄(𝑥)2𝑘2𝜑(𝑥,0)(2.10) such that 𝑥𝑋for all 𝑦=0.

Proof. Letting 2𝑓(𝑘𝑥)2𝑘2𝑓(𝑥)𝜑(𝑥,0)(2.11) in (2.9), we get𝑥𝑋for all 1𝑓(𝑥)𝑘21𝑓(𝑘𝑥)2𝑘2𝜑(𝑥,0)(2.12). So𝑥𝑋for all 1𝑘2𝑙𝑓𝑘𝑙𝑥1𝑘2𝑚𝑓𝑘𝑚𝑥𝑚1𝑗=𝑙12𝑘2𝑗+2𝜑𝑘𝑗𝑥,0(2.13). Hence𝑚for all nonnegative integers 𝑙 and 𝑚>𝑙 with 𝑥𝑋 and all {(1/𝑘2𝑛)𝑓(𝑘𝑛𝑥)}. It follows from (2.13) that the sequence 𝑥𝑋 is Cauchy for all 𝑌. Since {(1/𝑘2𝑛)𝑓(𝑘𝑛𝑥)} is complete, the sequence 𝑄𝑋𝑌 converges. So one can define the mapping 𝑄(𝑥)=lim𝑛1𝑘2𝑛𝑓𝑘𝑛𝑥(2.14) by𝑥𝑋for all 𝐷𝑄(𝑥,𝑦)=lim𝑛1𝑘2𝑛𝑘𝐷𝑓𝑛𝑥,𝑘𝑛𝑦lim𝑛1𝑘2𝑛𝜑𝑘𝑛𝑥,𝑘𝑛𝑦=0(2.15).
By (2.8),𝑥,𝑦𝑋for all 𝐷𝑄(𝑥,𝑦)=0. So 𝑄𝑋𝑌. By Proposition 2.1, the mapping 𝑙=0 is quadratic. Moreover, letting 𝑚 and passing the limit 𝑇𝑋𝑌 in (2.13), we get (2.10).
Now, let =1𝑄(𝑥)𝑇(𝑥)𝑘2𝑛𝑄𝑘𝑛𝑥𝑘𝑇𝑛𝑥1𝑘2𝑛𝑄𝑘𝑛𝑥𝑘𝑓𝑛𝑥+𝑇𝑘𝑛𝑥𝑘𝑓𝑛𝑥1𝑘2𝑛+2𝑘𝜑𝑛,𝑥,0(2.16) be another quadratic mapping satisfying (2.1) and (2.10). Then we have𝑛which tends to zero as 𝑥𝑋 for all 𝑄(𝑥)=𝑇(𝑥). So we can conclude that 𝑥𝑋 for all 𝑄. This proves the uniqueness of 𝑄𝑋𝑌. So there exists a unique quadratic mapping 𝑝<2 satisfying (2.10).

Corollary 2.3. Let 𝜃 and 𝑓𝑋𝑌 be positive real numbers, and let 𝐷𝑓(𝑥,𝑦)𝜃𝑥𝑝+𝑦𝑝(2.17) be a mapping such that 𝑥,𝑦𝑋for all 𝑄𝑋𝑌. Then there exists a unique quadratic mapping 𝜃𝑓(𝑥)𝑄(𝑥)82𝑝+1𝑥𝑝(2.18) such that 𝑥𝑋for all 𝜑(𝑥,𝑦)=𝜃𝑥𝑝+𝑦𝑝(2.19).

Proof. The proof follows from Theorem 2.2 by taking𝑥,𝑦𝐴for all 𝑓𝑋𝑌.

Theorem 2.4. Let 𝑓(0)=0 be a mapping with 𝜑𝑋2[0,) for which there exists a function 𝜑(𝑥,𝑦)=𝑗=0𝑘2𝑗𝑥𝜑(𝑘𝑗,𝑦𝑘𝑗)<(2.20) satisfying (2.9) such that 𝑥,𝑦𝑋for all 𝑄𝑋𝑌. Then there exists a unique quadratic mapping 1𝑓(𝑥)𝑄(𝑥)2𝑥𝜑(𝑘,0)(2.21) such that 𝑥𝑋for all 𝑓(𝑥)𝑘2𝑥𝑓(𝑘1)2𝑥𝜑(𝑘,0)(2.22).

Proof. It follows from (2.11) that𝑥𝑋for all 𝑘2𝑙𝑥𝑓(𝑘𝑙)𝑘2𝑚𝑥𝑓(𝑘𝑚)𝑚1𝑗=𝑙𝑘2𝑗2𝑥𝜑(𝑘𝑗+1,0)(2.23). Hence𝑚for all nonnegative integers 𝑙 and 𝑚>𝑙 with 𝑥𝑋 and all {𝑘2𝑛𝑓(𝑥/𝑘𝑛)}. It follows from (2.23) that the sequence 𝑥𝑋 is Cauchy for all 𝑌. Since {𝑘2𝑛𝑓(𝑥/𝑘𝑛)} is complete, the sequence 𝑄𝑋𝑌 converges. So one can define the mapping 𝑄(𝑥)=lim𝑛𝑘2𝑛𝑥𝑓(𝑘𝑛)(2.24) by𝑥𝑋for all 𝐷𝑄(𝑥,𝑦)=lim𝑛𝑘2𝑛𝑥𝐷𝑓(𝑘𝑛,𝑦𝑘𝑛)lim𝑛𝑘2𝑛𝑥𝜑(𝑘𝑛,𝑦𝑘𝑛)=0(2.25).
By (2.20),𝑥,𝑦𝑋for all 𝐷𝑄(𝑥,𝑦)=0. So 𝑄𝑋𝑌. By Proposition 2.1, the mapping 𝑙=0 is quadratic. Moreover, letting 𝑚 and passing the limit 𝑝>2 in (2.23), we get (2.21).
The rest of the proof is similar to the proof of Theorem 2.2.

Corollary 2.5. Let 𝜃 and 𝑓𝑋𝑌 be positive real numbers, and let 𝑄𝑋𝑌 be a mapping satisfying (2.17). Then there exists a unique quadratic mapping 𝜃𝑓(𝑥)𝑄(𝑥)2𝑝+18𝑥𝑝(2.26) such that 𝑥𝑋for all 𝜑(𝑥,𝑦)=𝜃𝑥𝑝+𝑦𝑝(2.27).

Proof. The proof follows from Theorem 2.4 by taking𝑥,𝑦𝐴for all 𝑘=2.

From now on, assume that 𝑓𝑋𝑌.

Theorem 2.6. Let 𝑓(0)=0 be a mapping with 𝜑𝑋2[0,) for which there exists a function 𝜑(𝑥,𝑦)=𝑗=019𝑗𝜑3𝑗𝑥,3𝑗𝑦<(2.28) satisfying (2.9) such that 𝑥,𝑦𝑋for all 𝑄𝑋𝑌. Then there exists a unique quadratic mapping 1𝑓(𝑥)𝑄(𝑥)9𝜑(𝑥,𝑥)(2.29) such that 𝑥𝑋for all 𝑦=𝑥.

Proof. Letting 𝑓(3𝑥)9𝑓(𝑥)𝜑(𝑥,𝑥)(2.30) in (2.9), we get𝑥𝑋for all 1𝑓(𝑥)91𝑓(3𝑥)9𝜑(𝑥,𝑥)(2.31). So𝑥𝑋for all 19𝑙𝑓(3𝑙1𝑥)9𝑚𝑓3𝑚𝑥𝑚1𝑗=𝑙19𝑗+1𝜑3𝑗𝑥,3𝑗𝑥(2.32). Hence𝑚for all nonnegative integers 𝑙 and 𝑚>𝑙 with 𝑥𝑋 and all {(1/9𝑛)𝑓(3𝑛𝑥)}. It follows from (2.32) that the sequence 𝑥𝑋 is Cauchy for all 𝑌. Since {(1/9𝑛)𝑓(3𝑛𝑥)} is complete, the sequence 𝑄𝑋𝑌 converges. So one can define the mapping 𝑄(𝑥)=lim𝑛19𝑛𝑓3𝑛𝑥(2.33) by𝑥𝑋for all 𝐷𝑄(𝑥,𝑦)=lim𝑛19𝑛3𝐷𝑓𝑛𝑥,3𝑛𝑦lim𝑛19𝑛𝜑3𝑛𝑥,3𝑛𝑦=0(2.34).
By (2.28),𝑥,𝑦𝑋for all 𝐷𝑄(𝑥,𝑦)=0. So 𝑄𝑋𝑌. By Proposition 2.1, the mapping 𝑙=0 is quadratic. Moreover, letting 𝑚 and passing the limit 𝑝<1 in (2.32), we get (2.29).
The rest of the proof is similar to the proof of Theorem 2.2.

Corollary 2.7. Let 𝜃 and 𝑓𝑋𝑌 be positive real numbers, and let 𝐷𝑓(𝑥,𝑦)𝜃𝑥𝑝𝑦𝑝(2.35) be a mapping such that 𝑥,𝑦𝑋for all 𝑄𝑋𝑌. Then there exists a unique quadratic mapping 𝜃𝑓(𝑥)𝑄(𝑥)99𝑝𝑥2𝑝(2.36) such that 𝑥𝑋for all 𝜑(𝑥,𝑦)=𝜃𝑥𝑝𝑦𝑝(2.37).

Proof. The proof follows from Theorem 2.6 by taking𝑥,𝑦𝐴for all 𝑓𝑋𝑌.

Theorem 2.8. Let 𝑓(0)=0 be a mapping with 𝜑𝑋2[0,) for which there exists a function 𝜑(𝑥,𝑦)=𝑗=09𝑗𝑥𝜑(3𝑗,𝑦3𝑗)<(2.38) satisfying (2.9) such that 𝑥,𝑦𝑋for all 𝑄𝑋𝑌. Then there exists a unique quadratic mapping 𝑥𝑓(𝑥)𝑄(𝑥)𝜑(3,𝑥3)(2.39) such that 𝑥𝑋for all 𝑥𝑓(𝑥)9𝑓(3𝑥)𝜑(3,𝑥3)(2.40).

Proof. It follows from (2.30) that𝑥𝑋for all 9𝑙𝑥𝑓(3𝑙)9𝑚𝑥𝑓(3𝑚)𝑚1𝑗=𝑙9𝑗𝑥𝜑(3𝑗+1,𝑥3𝑗+1)(2.41). Hence𝑚for all nonnegative integers 𝑙 and 𝑚>𝑙 with 𝑥𝑋 and all {9𝑛𝑓(𝑥/3𝑛)}. It follows from (2.41) that the sequence 𝑥𝑋 is Cauchy for all 𝑌. Since {9𝑛𝑓(𝑥/3𝑛)} is complete, the sequence 𝑄𝑋𝑌 converges. So one can define the mapping 𝑄(𝑥)=lim𝑛9𝑛𝑥𝑓(3𝑛)(2.42) by𝑥𝑋for all 𝐷𝑄(𝑥,𝑦)=lim𝑛19𝑛3𝐷𝑓𝑛𝑥,3𝑛𝑦lim𝑛19𝑛𝜑3𝑛𝑥,3𝑛𝑦=0(2.43).
By (2.38),𝑥,𝑦𝑋for all 𝐷𝑄(𝑥,𝑦)=0. So 𝑄𝑋𝑌. By Proposition 2.1, the mapping 𝑙=0 is quadratic. Moreover, letting 𝑚 and passing the limit 𝑝>1 in (2.41), we get (2.39).
The rest of the proof is similar to the proof of Theorem 2.2.

Corollary 2.9. Let 𝜃 and 𝑓𝑋𝑌 be positive real numbers, and let 𝑄𝑋𝑌 be a mapping satisfying (2.35). Then there exists a unique quadratic mapping 𝜃𝑓(𝑥)𝑄(𝑥)9𝑝9𝑥2𝑝(2.44) such that 𝑥𝑋for all 𝜑(𝑥,𝑦)=𝜃𝑥𝑝𝑦𝑝(2.45).

Proof. The proof follows from Theorem 2.8 by taking𝑥,𝑦𝐴for all .

Acknowledgments

Jung Rye Lee was supported by Daejin University grants in 2007. The authors would like to thank the referees for a number of valuable suggestions regarding a previous version of this paper.