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Journal of Applied Mathematics
Volume 2012, Article ID 610714, 14 pages
http://dx.doi.org/10.1155/2012/610714
Research Article

On a Generalized Hyers-Ulam Stability of Trigonometric Functional Equations

1Department of Mathematics, Kunsan National University, Kunsan 573-701, Republic of Korea
2Department of Mathematics Education, Dankook University, Yongin 448-701, Republic of Korea

Received 4 July 2011; Revised 3 January 2012; Accepted 3 January 2012

Academic Editor: Yongkun Li

Copyright © 2012 Jaeyoung Chung and Jeongwook Chang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Let 𝐺 be an Abelian group, let be the field of complex numbers, and let 𝑓,𝑔𝐺. We consider the generalized Hyers-Ulam stability for a class of trigonometric functional inequalities, |𝑓(𝑥𝑦)𝑓(𝑥)𝑔(𝑦)+𝑔(𝑥)𝑓(𝑦)|𝜓(𝑦),|𝑔(𝑥𝑦)𝑔(𝑥)𝑔(𝑦)𝑓(𝑥)𝑓(𝑦)|𝜓(𝑦), where 𝜓𝐺 is an arbitrary nonnegative function.

1. Introduction

The Hyers-Ulam stability problems of functional equations go back to 1940 when S. M. Ulam proposed a question concerning the approximate homomorphisms from a group to a metric group (see [1]). A partial answer was given by Hyers et al. [2, 3] under the assumption that the target space of the involved mappings is a Banach space. After the result of Hyers, Aoki [4], and Bourgin [5, 6] dealt with this problem, however, there were no other results on this problem until 1978 when Rassias [7] dealt again with the inequality of Aoki [4]. Following the Rassias' result, a great number of papers on the subject have been published concerning numerous functional equations in various directions [2, 721]. The following four functional equations are called trigonometric functional equations.𝑓(𝑥+𝑦)𝑓(𝑥)𝑔(𝑦)𝑔(𝑥)𝑓(𝑦)=0(1.1)𝑔(𝑥+𝑦)𝑔(𝑥)𝑔(𝑦)+𝑓(𝑥)𝑓(𝑦)=0(1.2)𝑓(𝑥𝑦)𝑓(𝑥)𝑔(𝑦)+𝑔(𝑥)𝑓(𝑦)=0(1.3)𝑔(𝑥𝑦)𝑔(𝑥)𝑔(𝑦)𝑓(𝑥)𝑓(𝑦)=0(1.4)The four functional equations have been investigated separately. The general solutions and regular solutions of the above equations are introduced [22, 23]. In particular, the last equation (1.4) is most interesting in the sense that (1.4) alone characterizes the two trigonometric functions 𝑓(𝑥)=cos(𝑎𝑥), 𝑔(𝑥)=sin(𝑎𝑥) under some regularities of 𝑔, which none of the remaining equations are able to do.

In [19], Székelyhidi developed his idea of using invariant subspaces of functions defined on a group or semigroup to obtain the Hyers-Ulam stability of the trigonometric functional equations (1.1) and (1.2). As results, he obtained the Hyers-Ulam stability when for each fixed 𝑦 the difference𝑇1(𝑥)=𝑓(𝑥+𝑦)𝑓(𝑥)𝑔(𝑦)𝑔(𝑥)𝑓(𝑦)(1.5) is a bounded function of 𝑥 and the Hyers-Ulam stability when for each fixed 𝑦 the difference𝑇2(𝑥)=𝑔(𝑥+𝑦)𝑔(𝑥)𝑔(𝑦)+𝑓(𝑥)𝑓(𝑦)(1.6) is a bounded function of 𝑥, where 𝑓,𝑔 are mappings from an Abelian (amenable) group 𝐺 to the field of complex numbers.

In this paper, we complete the parallel Hyers-Ulam stability to that of [19] for the functional equations (1.3) and (1.4). As results, we obtained the Hyers-Ulam stability when for each fixed 𝑦 the difference𝑇3(𝑥)=𝑓(𝑥𝑦)𝑓(𝑥)𝑔(𝑦)+𝑔(𝑥)𝑓(𝑦)(1.7) is a bounded function of 𝑥 and the Hyers-Ulam stability when for each fixed 𝑦 the difference𝑇4(𝑥)=𝑔(𝑥𝑦)𝑔(𝑥)𝑔(𝑦)𝑓(𝑥)𝑓(𝑦)(1.8) is a bounded function of 𝑥.

In fact, the authors [10] obtained weaker versions of the Hyers-Ulam stability for the functional equations (1.3) and (1.4), that is, we proved the Hyers-Ulam stability of (1.3) when the difference𝑇3(𝑥,𝑦)=𝑓(𝑥𝑦)𝑓(𝑥)𝑔(𝑦)+𝑔(𝑥)𝑓(𝑦)(1.9) is uniformly bounded for all 𝑥 and 𝑦, and we proved the Hyers-Ulam stability of (1.4) when the difference𝑇4(𝑥,𝑦)=𝑔(𝑥𝑦)𝑔(𝑥)𝑔(𝑦)𝑓(𝑥)𝑓(𝑦)(1.10) is uniformly bounded for all 𝑥 and 𝑦.

So, the results in this paper would be generalizations of those in [10]. We refer the reader to [9, 15, 16, 20, 21] for some related Hyers-Ulam stability of functional equations of trigonometric type.

2. Main Theorems

A function 𝑎 from a semigroup 𝑆,+ to the field of complex numbers is said to be an additive function provided that 𝑎(𝑥+𝑦)=𝑎(𝑥)+𝑎(𝑦) and 𝑚𝑆 is said to be an exponential function provided that 𝑚(𝑥+𝑦)=𝑚(𝑥)𝑚(𝑦). Throughout this paper, we denote by 𝐺 an Abelian group, the set of complex numbers, and 𝜓𝐺 a fixed nonnegative function. For the proof of stabilities of (1.3) and (1.4), we need the following.

Lemma 2.1 (see [2]). Let 𝑆 be a semigroup. Assume that 𝑓,𝑔𝑆 satisfy the inequality; for each 𝑦𝑆, there exists a positive constant 𝑀𝑦 such that ||||𝑓(𝑥+𝑦)𝑓(𝑥)𝑔(𝑦)𝑀𝑦,(2.1) for all 𝑥𝑆, then either 𝑓 is a bounded function or 𝑔 is an exponential function.

Proof. Suppose that 𝑔 is not exponential, then there are 𝑦,𝑧𝑆 such that 𝑔(𝑦+𝑧)𝑔(𝑦)𝑔(𝑧). Now we have 𝑓(𝑥+𝑦+𝑧)𝑓(𝑥+𝑦)𝑔(𝑧)=(𝑓(𝑥+𝑦+𝑧)𝑓(𝑥)𝑔(𝑦+𝑧))𝑔(𝑧)(𝑓(𝑥+𝑦)𝑓(𝑥)𝑔(𝑦))+𝑓(𝑥)(𝑔(𝑦+𝑧)𝑔(𝑦)𝑔(𝑧)),(2.2) and hence, 𝑓(𝑥)=(𝑔(𝑦+𝑧)𝑔(𝑦)𝑔(𝑧))1×((𝑓(𝑥+𝑦+𝑧)𝑓(𝑥+𝑦)𝑔(𝑧))(𝑓(𝑥+𝑦+𝑧)𝑓(𝑥)𝑔(𝑦+𝑧))+𝑔(𝑧)(𝑓(𝑥+𝑦)𝑓(𝑥)𝑔(𝑦))).(2.3)
In view of (2.1), the right hand side of (2.3) is bounded as a function of 𝑥. Consequently, 𝑓 is bounded.

We discuss the general solutions 𝑓,𝑔𝐺 of the corresponding trigonometric functional equations 𝑓(𝑥𝑦)𝑓(𝑥)𝑔(𝑦)+𝑔(𝑥)𝑓(𝑦)=0,(2.4)𝑔(𝑥𝑦)𝑔(𝑥)𝑔(𝑦)𝑓(𝑥)𝑓(𝑦)=0.(2.5)

Lemma 2.2 (see [22, 23]). The general solutions (𝑓,𝑔) of the functional equation (2.4) are given by one of the following:(i)𝑓=0 and 𝑔 is arbitrary,(ii)𝑓(𝑥)=𝜆1(𝑚(𝑥)𝑚(𝑥)) and 𝑔(𝑥)=𝜆2𝑓(𝑥)+(1/2)(𝑚(𝑥)+𝑚(𝑥)) for some 𝜆1,𝜆2, where 𝑚 is an exponential function,(iii)𝑓(𝑥)=𝑎(𝑥)𝑚(𝑥),𝑔(𝑥)=(1+𝜆𝑎(𝑥))𝑚(𝑥) for some 𝜆, where 𝑎 is an additive function and 𝑚 is an exponential function satisfying 𝑚21.
Also, the general solutions (𝑔,𝑓) of the functional equation (2.5) are given by one of the following:(i)𝑔(𝑥)=𝜆 and 𝑓(𝑥)=±𝜆𝜆2 for some 𝜆,(ii)𝑔(𝑥)=(1/2)(𝑚(𝑥)+𝑚(𝑥)) and 𝑓(𝑥)=(1/2𝑖)(𝑚(𝑥)𝑚(𝑥)), where 𝑚 is an exponential function.

Proof. The solutions of the functional equation (2.4) are given in [23, p. 217, Theorem  11]. For the functional equation (2.5), combining the result of L. Vietoris [22, p. 177] and that of J. A. Baker [23, p. 220], we obtain that every nonconstant function 𝑔 satisfying (2.5) has the form 1𝑔(𝑥)=2(𝑚(𝑥)+𝑚(𝑥)),(2.6) for some exponential function 𝑚. Thus, using (2.5), we have 1𝑓(𝑥)=2𝑖(𝑚(𝑥)𝑚(𝑥)).(2.7) This completes the proof.

For the proof of the stability of (1.1), we need the following. Throughout this paper, we denote by 𝜓 an arbitrary nonnegative function on 𝐺.

Lemma 2.3. Let 𝑓,𝑔𝐺 satisfy the inequality ||||𝑓(𝑥𝑦)𝑓(𝑥)𝑔(𝑦)+𝑔(𝑥)𝑓(𝑦)𝜓(𝑦),(2.8) for all 𝑥,𝑦𝐺, then either there exist 𝜆1,𝜆2, not both zero, and 𝑀>0 such that ||𝜆1𝑓(𝑥)𝜆2||𝑔(𝑥)𝑀,(2.9) or else 𝑓(𝑥𝑦)𝑓(𝑥)𝑔(𝑦)+𝑔(𝑥)𝑓(𝑦)=0,(2.10) for all 𝑥,𝑦𝐺.

Proof. Suppose that the inequality (2.9) holds only when 𝜆1=𝜆2=0. Let 𝑘(𝑥,𝑦)=𝑓(𝑥+𝑦)𝑓(𝑥)𝑔(𝑦)+𝑔(𝑥)𝑓(𝑦),(2.11) and choose 𝑦1 satisfying 𝑓(𝑦1)0. Now it can be easily calculated that 𝑔(𝑥)=𝜆0𝑓(𝑥)+𝜆1𝑓𝑥+𝑦1𝜆1𝑘𝑥,𝑦1,(2.12) where 𝜆0=𝑔(𝑦1)/𝑓(𝑦1) and 𝜆1=1/𝑓(𝑦1). By (2.11), we have 𝑓(𝑥+(𝑦+𝑧))=𝑓(𝑥)𝑔(𝑦𝑧)𝑔(𝑥)𝑓(𝑦𝑧)+𝑘(𝑥,𝑦+𝑧).(2.13) Also by (2.11) and (2.12), we have =𝜆𝑓((𝑥+𝑦)+𝑧)=𝑓(𝑥+𝑦)𝑔(𝑧)𝑔(𝑥+𝑦)𝑓(𝑧)+𝑘(𝑥+𝑦,𝑧)(𝑓(𝑥)𝑔(𝑦)𝑔(𝑥)𝑓(𝑦)+𝑘(𝑥,𝑦))𝑔(𝑧)0𝑓(𝑥+𝑦)+𝜆1𝑓𝑥+𝑦+𝑦1𝜆1𝑘𝑥+𝑦,𝑦1𝑓(𝑧)+𝑘(𝑥+𝑦,𝑧)=(𝑓(𝑥)𝑔(𝑦)𝑔(𝑥)𝑓(𝑦)+𝑘(𝑥,𝑦))𝑔(𝑧)𝜆0(𝑓(𝑥)𝑔(𝑦)𝑔(𝑥)𝑓(𝑦)+𝑘(𝑥,𝑦))𝑓(𝑧)𝜆1𝑓(𝑥)𝑔𝑦𝑦1𝑔(𝑥)𝑓𝑦𝑦1+𝑘𝑥,𝑦+𝑦1𝑓(𝑧)+𝜆1𝑘𝑥+𝑦,𝑦1𝑓(𝑧)+𝑘(𝑥+𝑦,𝑧).(2.14)
From (2.13) and (2.14), we have 𝑔(𝑦)𝑔(𝑧)𝜆0𝑔(𝑦)𝑓(𝑧)𝜆1𝑔𝑦𝑦1𝑓𝑓+(𝑧)𝑔(𝑦𝑧)(𝑥)𝑔(𝑦)𝑔(𝑧)+𝜆0𝑓(𝑦)𝑔(𝑧)+𝜆1𝑓𝑦𝑦1𝑓(𝑧)+𝑓(𝑦𝑧)𝑔(𝑥)=𝑘(𝑥,𝑦)𝑔(𝑧)+𝜆0𝑘(𝑥,𝑦)𝑓(𝑧)+𝜆1𝑘𝑥,𝑦+𝑦1𝑓(𝑧)𝜆1𝑘𝑥+𝑦,𝑦1𝑓(𝑧)𝑘(𝑥+𝑦,𝑧)+𝑘(𝑥,𝑦+𝑧).(2.15)
Since 𝑘(𝑥,𝑦) is bounded by 𝜓(𝑦), if we fix 𝑦, 𝑧, the right hand side of (2.15) is bounded by a constant 𝑀, where ||𝑔||||𝜆𝑀=𝜓(𝑦)(𝑧)+𝜓(𝑦)0𝑓||(𝑧)+𝜓𝑦𝑦1||𝜆1𝑓||(𝑧)+𝜓𝑦1||𝜆1||𝑓(𝑧)+𝜓(𝑧)+𝜓(𝑦𝑧).(2.16)
So by our assumption, the left hand side of (2.15) vanishes, so does the right hand side. Thus, we have 𝜆0𝑘(𝑥,𝑦)𝜆1𝑘𝑥,𝑦+𝑦1+𝜆1𝑘𝑥+𝑦,𝑦1𝑓(𝑧)+𝑘(𝑥,𝑦)𝑔(𝑧)=𝑘(𝑥,𝑦+𝑧)𝑘(𝑥+𝑦,𝑧).(2.17) Now by the definition of 𝑘, we have 𝑘(𝑥+𝑦,𝑧)𝑘(𝑥,𝑦+𝑧)=𝑓(𝑥+𝑦+𝑧)𝑓(𝑥+𝑦)𝑔(𝑧)+𝑔(𝑥+𝑦)𝑓(𝑧)𝑓(𝑥+𝑦+𝑧)+𝑓(𝑥)𝑔(𝑦𝑧)𝑔(𝑥)𝑓(𝑦𝑧)=𝑓(𝑦𝑧𝑥)𝑓(𝑦𝑧)𝑔(𝑥)+𝑔(𝑦𝑧)𝑓(𝑥)𝑓(𝑧𝑥𝑦)+𝑓(𝑧)𝑔(𝑥+𝑦)𝑔(𝑧)𝑓(𝑥+𝑦)=𝑘(𝑦𝑧,𝑥)𝑘(𝑧,𝑥𝑦).(2.18) Hence, the right hand side of (2.17) is bounded by 𝜓(𝑥)+𝜓(𝑥+𝑦). So if we fix 𝑥, 𝑦 in (2.17), the left hand side of (2.17) is a bounded function of 𝑧. Thus, by our assumption, we conclude that 𝑘(𝑥,𝑦)0. This completes the proof.

In the following theorem, we assume that Φ1(𝑥)=𝑘=02𝑘𝜓2𝑘𝑥<,(2.19) or Φ2(𝑥)=𝑘=12𝑘𝜓2𝑘𝑥<.(2.20)

For the proof, we discuss the following property.

Lemma 2.4. Let 𝑚𝐺 be a bounded exponential function satisfying 𝑚(𝑥)𝑚(𝑥) for some 𝑥𝐺, then there exists 𝑦𝐺 such that ||||𝑚(𝑦)𝑚(𝑦)3.(2.21) Furthermore, the constant 3 is the best one.

Proof. Since 𝑚 is a bounded exponential, there exists 𝐶>0 such that |𝑚(𝑥)|𝑘=|𝑚(𝑘𝑥)|𝐶 for all 𝑘 and 𝑥𝐺, which implies |𝑚(𝑥)|=1 for all 𝑥𝐺. Assume that 𝑚(𝑥0)𝑚(𝑥0), then we have 𝑚(𝑥0)±1, and we may assume that 𝑚(𝑥0)=𝑒𝑖𝜃,0<𝜃<𝜋. If 𝜃[𝜋/3,2𝜋/3], we have |𝑚(𝑥0)𝑚(𝑥0)|=|𝑒𝑖𝜃𝑒𝑖𝜃|3. If 𝜃[0,𝜋/3], there exists a positive integer 𝑘 such that 𝑘𝜃[𝜋/3,2𝜋/3], and we have |𝑚(𝑘𝑥0)𝑚(𝑘𝑥0)|=|𝑒𝑖𝑘0𝜃𝑒𝑖𝑘0𝜃|3. If 𝜃[2𝜋/3,5𝜋/6], then 2𝜃[4𝜋/3,5𝜋/3], and we have |𝑚(2𝑥0)𝑚(2𝑥0)|=|𝑒𝑖2𝜃𝑒𝑖2𝜃|3. Finally, if 𝜃[5𝜋/6,𝜋], there exists a positive integer 𝑘 such that 2𝑘𝜃[𝜋/3,2𝜋/3], and we have |𝑚(2𝑘𝑥0)𝑚(2𝑘𝑥0)|=|𝑒𝑖2𝑘𝜃𝑒𝑖2𝑘𝜃|3. Now define 𝑚 by 𝑚(𝑘)=𝑒𝑖𝑘𝜋/3. Then we have |𝑚(3𝑘+1)𝑚(3𝑘1)|=3 for all 𝑘. Thus, 3 is the biggest one. This completes the proof.

Theorem 2.5. Let 𝑓,𝑔𝐺 satisfy the inequality ||||𝑓(𝑥𝑦)𝑓(𝑥)𝑔(𝑦)+𝑔(𝑥)𝑓(𝑦)𝜓(𝑦),(2.22) for all 𝑥,𝑦𝐺, then (𝑓,𝑔) satisfies one of the following:(i)𝑓=0, 𝑔 is arbitrary,(ii)𝑓 and 𝑔 are bounded functions,(iii)𝑓(𝑥)=𝜆1(𝑚(𝑥)𝑚(𝑥)) and 𝑔(𝑥)=𝜆2𝑓(𝑥)+(1/2)(𝑚(𝑥)+𝑚(𝑥)) for some 𝜆1,𝜆2, where 𝑚 is an exponential function,(iv)there exist 𝜆 and a bounded exponential function 𝑚 such that 𝑔(𝑥)=𝜆𝑓(𝑥)+𝑚(𝑥),(2.23) for all 𝑥𝐺, and 𝑓 satisfies the condition; there exists 𝑑0 satisfying ||||2𝑓(𝑥)3(𝜓(𝑥)+𝑑),(2.24) for all 𝑥𝐺,(v)there exist 𝜆 and a bounded exponential function 𝑚 satisfying 𝑚21 such that 𝑔(𝑥)=𝜆𝑓(𝑥)+𝑚(𝑥),(2.25) for all 𝑥𝐺, and 𝑓 satisfies one of the following conditions; there exists an additive function 𝑎1𝐺 such that ||𝑓𝑎(𝑥)1𝑚||(𝑥)+𝑓(0)(𝑥)Φ1(𝑥),(2.26) for all 𝑥𝐺, or there exists an additive function 𝑎2𝐺 such that ||𝑓𝑎(𝑥)1𝑚||(𝑥)+𝑓(0)(𝑥)Φ2(𝑥),(2.27) for all 𝑥𝐺, where Φ1 and Φ2 are the functions given in (2.19) and (2.20).

Proof. In view of Lemma 2.3, we first consider the case when 𝑓,𝑔 satisfy (2.9). If 𝑓=0, 𝑔 is arbitrary which is the case (i). If 𝑓 is a nontrivial bounded function, in view of (2.22), 𝑔 is also bounded which gives the case (ii). If 𝑓 is unbounded, it follows from (2.9) that 𝜆20 and 𝑔(𝑥)=𝜆𝑓(𝑥)+𝑚(𝑥),(2.28) for some 𝜆 and a bounded function 𝑚. Putting (2.28) in (2.22), we have ||||𝑓(𝑥𝑦)𝑓(𝑥)𝑚(𝑦)+𝑚(𝑥)𝑓(𝑦)𝜓(𝑦),(2.29) for all 𝑥,𝑦𝐺. Replacing 𝑦 by 𝑦 and using the triangle inequality, we have, for some 𝐶>0, ||||||||𝑓(𝑥+𝑦)𝑓(𝑥)𝑚(𝑦)𝐶𝑓(𝑦)+𝜓(𝑦),(2.30) for all 𝑥,𝑦𝐺. By Lemma 2.1, 𝑚 is an exponential function. If 𝑚=0, putting 𝑦=0 in (2.29), we have ||||𝑓(𝑥)𝜓(0).(2.31) Thus, we have 𝑚0 since 𝑓 is unbounded. Since 𝑚 is a nonzero bounded exponential function, it follows from the equalities 𝑚(𝑥)=𝑚(𝑥𝑦)𝑚(𝑦),𝑥,𝑦𝐺(2.32) that 𝑚(0)=1 and 𝑚(𝑥)0, for all 𝑥𝐺. Putting 𝑥=0 in (2.29) and replacing 𝑦 by 𝑦 multiplying |𝑚(𝑥)| in the result, we have ||||𝑚(𝑥)𝑓(𝑦)+𝑚(𝑥)𝑓(𝑦)𝑓(0)𝑚(𝑥)𝑚(𝑦)𝜓(𝑦),(2.33) for all 𝑦𝐺. Replacing 𝑦 by 𝑦 in (2.29) and using (2.33), we have ||||𝑓(𝑥+𝑦)𝑓(𝑥)𝑚(𝑦)𝑚(𝑥)𝑓(𝑦)+𝑓(0)𝑚(𝑥)𝑚(𝑦)2𝜓(𝑦).(2.34)
First we consider the case 𝑚(𝑥)𝑚(𝑥) for some 𝑥𝐺. Replacing 𝑥 by 𝑦 and 𝑦 by 𝑥 in (2.34), we have ||||𝑓(𝑦+𝑥)𝑓(𝑦)𝑚(𝑥)𝑚(𝑦)𝑓(𝑥)+𝑓(0)𝑚(𝑦)𝑚(𝑥)2𝜓(𝑥),(2.35) for all 𝑥,𝑦𝐺. From (2.34) and (2.35), using the triangle inequality, putting 𝑦=𝑦0 such that |𝑚(𝑦0)𝑚(𝑦0)|3 and dividing |𝑚(𝑦0)𝑚(𝑦0)|, we have ||||2𝑓(𝑥)3(𝜓(𝑥)+𝑑),(2.36) for all 𝑥𝐺, where 𝑑=𝜓(𝑦0)+|𝑓(𝑦0)|+|𝑓(0)|, which gives (iv). Now we consider the case 𝑚(𝑥)=𝑚(𝑥), for all 𝑥𝐺. Dividing both the sides of (2.34) by 𝑚(𝑥)𝑚(𝑦), we have ||||𝐹(𝑥+𝑦)𝐹(𝑥)𝐹(𝑦)2𝜓(𝑦),(2.37) for all 𝑥,𝑦𝐺, where 𝐹(𝑥)=𝑓(𝑥)/𝑚(𝑥)𝑓(0). By the well-known results in [4], there exists a unique additive function 𝑎1(𝑥) given by 𝑎1(𝑥)=lim𝑛2𝑛𝑓(2𝑛𝑥)(2.38) such that ||𝐹(𝑥)𝑎1||(𝑥)Φ1(𝑥)(2.39) if Φ1(𝑥)=𝑘=02𝑘𝜓(2𝑘𝑥)<, and there exists a unique additive function 𝑎2(𝑥) given by 𝑎2(𝑥)=lim𝑛2𝑛𝑓(2𝑛𝑥)(2.40) such that ||𝐹(𝑥)𝑎2||(𝑥)Φ2(𝑥)(2.41) if Φ2(𝑥)=𝑘=12𝑘𝜓(2𝑘𝑥)<. Multiplying |𝑚(𝑥)| in both sides of (2.39) and (2.41), we get (v). Now we consider the case when 𝑓,𝑔 satisfy (2.10). In view of Lemma 2.2, the solutions of (2.10) are given by (i), (iii), or contained in the case (v). This completes the proof.

Let 𝑋 be a real normed space, and let 𝜓𝑋 be given by 𝜓(𝑥)=𝜖𝑥𝑝,𝑝0,𝑝1, then 𝜓 satisfies the conditions assumed in Theorem 2.5. In view of (2.19) and (2.20), we have Φ1(𝑥)=2𝜖𝑥𝑝22𝑝(2.42) if 0<𝑝<1, Φ2(𝑥)=2𝜖𝑥𝑝2𝑝2(2.43) if 𝑝>1. Thus, as a direct consequence of Theorem 2.5, we have the following.

Corollary 2.6. Let 𝑓,𝑔𝑋 satisfy the inequality ||||𝑓(𝑥𝑦)𝑓(𝑥)𝑔(𝑦)+𝑔(𝑥)𝑓(𝑦)𝜖𝑦𝑝,𝑝1,𝑝0,(2.44) for all 𝑥,𝑦𝑋, then (𝑓,𝑔) satisfies one of the following:(i)𝑓=0, 𝑔 is arbitrary,(ii)𝑓 and 𝑔 are bounded functions,(iii)𝑓(𝑥)=𝜆1(𝑚(𝑥)𝑚(𝑥)) and 𝑔(𝑥)=𝜆2𝑓(𝑥)+(1/2)(𝑚(𝑥)+𝑚(𝑥)) for some 𝜆1,𝜆2, where 𝑚 is an exponential function,(iv)there exist 𝜆 and a bounded exponential function 𝑚 such that 𝑔(𝑥)=𝜆𝑓(𝑥)+𝑚(𝑥),(2.45) for all 𝑥𝑋, and 𝑓 satisfies the condition; there exists 𝑑0 satisfying ||||2𝑓(𝑥)3(𝜓(𝑥)+𝑑),(2.46) for all 𝑥𝑋,(v)there exist 𝜆 and a bounded exponential function 𝑚 satisfying 𝑚21 such that 𝑔(𝑥)=𝜆𝑓(𝑥)+𝑚(𝑥),(2.47) for all 𝑥𝑋, and 𝑓 satisfies one of the following conditions; there exists an additive function 𝑎𝑋 such that ||||𝑓(𝑥)(𝑎(𝑥)+𝑓(0))𝑚(𝑥)2𝜖𝑥𝑝||22𝑝||,(2.48) for all 𝑥𝑋.

Now we prove the stability of (1.2). For the proof, we need the following.

Lemma 2.7. Let 𝑓,𝑔𝐺 satisfy the inequality ||||𝑔(𝑥𝑦)𝑔(𝑥)𝑔(𝑦)𝑓(𝑥)𝑓(𝑦)𝜓(𝑦),(2.49) for all 𝑥,𝑦𝐺, then either there exist 𝜆1,𝜆2, not both zero, and 𝑀>0 such that ||𝜆1𝑓(𝑥)𝜆2||𝑔(𝑥)𝑀,(2.50) or else 𝑔(𝑥𝑦)𝑔(𝑥)𝑔(𝑦)𝑓(𝑥)𝑓(𝑦)=0,(2.51) for all 𝑥,𝑦𝐺.

Proof. Suppose that 𝜆1𝑓(𝑥)𝜆2𝑔(𝑥) is bounded only when 𝜆1=𝜆2=0, and let 𝑙(𝑥,𝑦)=𝑔(𝑥+𝑦)𝑔(𝑥)𝑔(𝑦)𝑓(𝑥)𝑓(𝑦).(2.52) Since we may assume that 𝑓 is nonconstant, we can choose 𝑦1 satisfying 𝑓(𝑦1)0. Now it can be easily get that 𝑓(𝑥)=𝜆0𝑔(𝑥)+𝜆1𝑔𝑥+𝑦1𝜆1𝑙𝑥,𝑦1,(2.53) where 𝜆0=𝑔(𝑦1)/𝑓(𝑦1) and 𝜆1=1/𝑓(𝑦1). From the definition of 𝑙 and the use of (2.53), we have the following two equations: =+𝜆𝑔((𝑥+𝑦)+𝑧)=𝑔(𝑥+𝑦)𝑔(𝑧)+𝑓(𝑥+𝑦)𝑓(𝑧)+𝑙(𝑥+𝑦,𝑧)(𝑔(𝑥)𝑔(𝑦)+𝑓(𝑥)𝑓(𝑦)+𝑙(𝑥,𝑦))𝑔(𝑧)0𝑔(𝑥+𝑦)+𝜆1𝑔𝑥+𝑦+𝑦1𝜆1𝑙𝑥+𝑦,𝑦1𝑓(𝑧)+𝑙(𝑥+𝑦,𝑧)=(𝑔(𝑥)𝑔(𝑦)+𝑓(𝑥)𝑓(𝑦)+𝑙(𝑥,𝑦))𝑔(𝑧)+𝜆0(𝑔(𝑥)𝑔(𝑦)+𝑓(𝑥)𝑓(𝑦)+𝑙(𝑥,𝑦))𝑓(𝑧)+𝜆1𝑔(𝑥)𝑔𝑦𝑦1+𝑓(𝑥)𝑓𝑦𝑦1+𝑙𝑥,𝑦+𝑦1𝑓(𝑧)𝜆1𝑙𝑥+𝑦,𝑦1𝑓(𝑧)+𝑙(𝑥+𝑦,𝑧),(2.54)𝑔(𝑥+(𝑦+𝑧))=𝑔(𝑥)𝑔(𝑦𝑧)+𝑓(𝑥)𝑓(𝑦𝑧)+𝑙(𝑥,𝑦+𝑧).(2.55) Equating (2.54) and (2.55), we have 𝑔𝑔(𝑥)(𝑦)𝑔(𝑧)+𝜆0𝑔(𝑦)𝑓(𝑧)+𝜆1𝑔𝑦𝑦1𝑓(𝑧)𝑔(𝑦𝑧)+𝑓(𝑥)𝑓(𝑦)𝑔(𝑧)+𝜆0𝑓(𝑦)𝑓(𝑧)+𝜆1𝑓𝑦𝑦1𝑓(𝑧)𝑓(𝑦𝑧)=𝑙(𝑥,𝑦)𝑔(𝑧)𝜆0𝑙(𝑥,𝑦)𝑓(𝑧)𝜆1𝑙𝑥,𝑦+𝑦1𝑓(𝑧)+𝜆1𝑙𝑥+𝑦,𝑦1𝑓(𝑧)𝑙(𝑥+𝑦,𝑧)+𝑙(𝑥,𝑦+𝑧).(2.56) In (2.56), when 𝑦,𝑧 are fixed, the right hand side is bounded, so by our assumption, we have 𝜆𝑙(𝑥,𝑦)𝑔(𝑧)+0𝑙(𝑥,𝑦)+𝜆1𝑙𝑥,𝑦+𝑦1𝜆1𝑙𝑥+𝑦,𝑦1𝑓(𝑧)=𝑙(𝑥,𝑦+𝑧)𝑙(𝑥+𝑦,𝑧).(2.57) Also we can write 𝑙(𝑥,𝑦+𝑧)𝑙(𝑥+𝑦,𝑧)=𝑔(𝑥+𝑦+𝑧)𝑔(𝑥)𝑔(𝑦+𝑧)𝑓(𝑥)𝑓(𝑦+𝑧)𝑔(𝑥+𝑦+𝑧)+𝑔(𝑥+𝑦)𝑔(𝑧)+𝑓(𝑥+𝑦)𝑓(𝑧)=𝑙(𝑦+𝑧,𝑥)𝑙(𝑧,𝑥+𝑦)𝜓(𝑥)+𝜓(𝑥𝑦).(2.58) Thus, if we fix 𝑥,𝑦 in (2.57), the right hand side of (2.57) is bounded. By our assumption, we have 𝑙(𝑥,𝑦)0. This completes the proof.

Theorem 2.8. Let 𝑓,𝑔𝐺 satisfy the inequality ||||𝑔(𝑥𝑦)𝑔(𝑥)𝑔(𝑦)𝑓(𝑥)𝑓(𝑦)𝜓(𝑦),(2.59) for all 𝑥,𝑦𝐺, then (𝑓,𝑔) satisfies one of the following:(i)𝑓 and 𝑔 are bounded functions,(ii)𝑔(𝑥)=(1/2)(𝑚(𝑥)+𝑚(𝑥)) and 𝑓(𝑥)=(1/2)(𝑚(𝑥)𝑚(𝑥)), where 𝑚 is an exponential function,(iii)𝑓=±𝑖(𝑔𝑚) for a bounded exponential function 𝑚, and 𝑔 satisfies |||1𝑔(𝑥)2|||1(𝑔(0)𝑚(𝑥)+𝑚(𝑥))2𝜓(𝑥),(2.60) for all 𝑥𝐺. In particular if 𝜓(0)=0, one has 𝑔(0)=1,𝑓(0)=0.

Proof. In view of Lemma 2.7, we first consider the case when 𝑓,𝑔 satisfy (2.51). If 𝑓 is bounded, then in view of the inequality (2.59), for each 𝑦, 𝑔(𝑥+𝑦)𝑔(𝑥)𝑔(𝑦), is also bounded. It follows from Lemma 2.1 that 𝑔 is bounded or a nonzero exponential function. If 𝑔 is bounded, the case (i) follows. If 𝑔 is a nonzero exponential function, from (2.59), using the triangle inequality, we have for some 𝑑0, ||||𝑔(𝑥)(𝑔(𝑦)𝑔(𝑦))𝜓(𝑦)+𝑑,(2.61) for all 𝑥,𝑦𝐺. Thus, it follows that 𝑔(𝑦)=𝑔(𝑦),(2.62) for all 𝑦𝐺, or else 𝑔 is bounded, and equality (2.62) implies 𝑔21, which gives the case (i).
If 𝑓 is unbounded, then in view of (2.59), 𝑔 is also unbounded, and hence, 𝜆1𝜆20 and 𝑓(𝑥)=𝜆𝑔(𝑥)+𝑟(𝑥),(2.63) for some 𝜆0 and a bounded function 𝑟. Putting (2.63) in (2.59), replacing 𝑦 by 𝑦, and using the triangle inequality, we have ||𝑔𝜆(𝑥+𝑦)𝑔(𝑥)2𝑔||||||+1(𝑦)+𝜆𝑟(𝑦)(𝜆𝑔(𝑦)+𝑟(𝑦))𝑟(𝑥)+𝜓(𝑦).(2.64) From Lemma 2.1, we have 𝜆2𝑔+1(𝑦)+𝜆𝑟(𝑦)=𝑚(𝑦),(2.65) for some exponential function 𝑚. If 𝜆21, we have 𝑓(𝑥)=𝜆𝑚(𝑥)+𝑟(𝑥)𝜆2+1,𝑔(𝑥)=𝑚(𝑥)𝜆𝑟(𝑥)𝜆2.+1(2.66) Putting (2.66) in (2.59), multiplying |𝜆2+1| in the result, and using the triangle inequality, we have for some 𝑑0, ||𝑚||||𝜆(𝑥)(𝑚(𝑦)𝑚(𝑦))2||𝜓+1(𝑦)+𝑑,(2.67) for all 𝑥,𝑦𝐺. Since 𝑚 is unbounded, we have 𝑚(𝑦)=𝑚(𝑦),(2.68) for all 𝑦𝐺, which implies 𝑚21, contradicting to the fact that 𝑚 is unbounded. Thus, it follows that 𝜆2=1, and we have 𝑓=±𝑖(𝑔𝑚),(2.69) where 𝑚 is a bounded exponential function. Putting (2.69) in (2.59), we have ||||𝑔(𝑥𝑦)𝑔(𝑥)𝑚(𝑦)𝑔(𝑦)𝑚(𝑥)+𝑚(𝑥)𝑚(𝑦)𝜓(𝑦),(2.70) for all 𝑥,𝑦𝐺. Replacing 𝑦 by 𝑥 in (2.70) and dividing the result by 2𝑚(𝑥), we have |||1𝑔(𝑥)2|||1(𝑔(0)𝑚(𝑥)+𝑚(𝑥))2𝜓(𝑥),(2.71) for all 𝑥𝐺. From (2.69), (2.71), we get (iii). Now we consider the case when 𝑓,𝑔 satisfy (2.51). In view of Lemma 2.2, the solutions of (2.51) are contained in (i) or given by (ii). Furthermore, if 𝜓(0)=0, then putting 𝑥=𝑦=0 in (2.70), we have 𝑔(0)=1, and from (2.69), we also have 𝑓(0)=0. This completes the proof.

In particular, if 𝑓,𝑔𝑛 is a continuous function and 𝜓(𝑥)=𝜖|𝑥|𝑝,𝑝>0,𝑝1, then Theorem 2.8 is reduced as follows.

Corollary 2.9. Let 𝑓,𝑔𝑛 be a continuous function satisfying (2.59) for 𝜓(𝑥)=𝜖|𝑥|𝑝, then (𝑓,𝑔) satisfies one of the following:(i)𝑓 and 𝑔 are bounded functions,(ii)𝑔(𝑥)=cos(𝑐𝑥) and 𝑓(𝑥)=sin(𝑐𝑥) for some 𝑐𝑛,(iii)there exists 𝑎𝑛 such that||𝑓||𝜖(𝑥)sin(𝑎𝑥)2|𝑥|𝑝,||||𝜖𝑔(𝑥)cos(𝑎𝑥)2|𝑥|𝑝,(2.72) for all 𝑥𝑛.

Acknowledgment

The second author was supported by the research fund of Dankook University in 2010.

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