Discrete Dynamics in Nature and Society

Volume 2012 (2012), Article ID 835893, 12 pages

http://dx.doi.org/10.1155/2012/835893

## Stability of a Bi-Additive Functional Equation in Banach Modules Over a -Algebra

^{1}Department of Mathematics Education, College of Education, Mokwon University, Daejeon 302-729, Republic of Korea^{2}Graduate School of Education, Kyung Hee University, Yongin 446-701, Republic of Korea

Received 6 April 2012; Accepted 30 May 2012

Academic Editor: Baodong Zheng

Copyright © 2012 Won-Gil Park and Jae-Hyeong Bae. 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

We solve the bi-additive functional equation and prove that every bi-additive Borel function is bilinear. And we investigate the stability of a bi-additive functional equation in Banach modules over a unital -algebra.

#### 1. Introduction

In 1940, Ulam proposed the stability problem (see [1]).

Let be a group, and let be a metric group with the metric . Given , does there exist a such that if a mapping satisfies the inequality for all then there is a homomorphism with for all ?

In 1941, this problem was solved by Hyers [2] in the case of Banach space. Thereafter, many authors investigated solutions or stability of various functional equations (see [3–21]).

Let and be real or complex vector spaces. In 1989, Aczél and Dhombres [22] proved that a mapping satisfies the quadratic functional equation if and only if there exists a symmetric bi-additive mapping such that , where for all . For a mapping , consider the bi-additive functional equation: For a mapping satisfying (1.1), the Aczél's bi-additive mapping given by (1.2) is a solution of (1.3).

In this paper, we find out the general solution of the bi-additive functional equation (1.3) and investigate the linearity of bi-additive Borel functions. And we investigate the stability of (1.3) in Banach modules over a unital -algebra.

#### 2. Solution of the Bi-Additive Functional Equation (1.3)

The general solution of the bi-additive functional equation (1.3) is as follows.

Theorem 2.1. * A mapping satisfies (1.3) if and only if the mapping is bi-additive.*

*Proof. * Assume that the mapping satisfies (1.3). Letting in (1.3), we gain . Putting in (1.3), we get
for all . Setting in (2.1), we have
for all . Taking (resp., ) in the above equation, we obtain
for all (resp., for all ). Letting in (1.3) and using (2.3), we gain
for all . Putting in (2.1) and using (2.3), we get
for all . Replacing by in (2.1) and using (2.3), (2.4), and (2.5) and the above equation, we see that for all .

On the other hand, letting in (1.3) and using (2.3), we gain
for all . Putting in (1.3) and using (2.3), we get
for all . Setting in (2.6) and using (2.3), we have
for all . Replacing by in (2.6) and using (2.7) and (2.8), we obtain that for all .

The converse is trivial.

The bi-additive functional equation (1.3) is related to the quadratic functional equation (1.1).

If is a mapping satisfying (1.3) and is the mapping given by for all , then one can easily obtain that satisfies (1.1).

Let and be a mapping satisfying (1.1). If is the mapping given by for all , then one can easily prove that satisfies (1.3). Furthermore, holds for all if .

The following is a result on bi-additive Borel functions.

Theorem 2.2. * Let be a bi-additive Borel function; then it is bilinear, that is, it satisfies for all .*

*Proof. * Since the function is bi-additive, we gain
for all and all . Letting in equality (2.9), we get
for all and all . Putting in equality (2.9) again, we have
for all . Note that the function is measurable for each fixed (see [23, Proposition 2.34]). Since the function is additive for each fixed , by [24], it is continuous for each fixed . By the same reasoning, the function is also continuous for each fixed . Let be fixed. Since is measurable, by [25, Theorem 7.14.26], for every there is a closed set such that and is continuous. Since , one can choose satisfying . Take a sequence in converging to . For each fixed , take a sequence in converging to . By equalities (2.10) and (2.11), we see that
for all . Hence we obtain that
as desired.

#### 3. Stability of the Bi-Additive Functional Equation (1.3)

From now on, let be a normed space, a complete normed space, and a nonnegative real number. In this section, we investigate the stability of the bi-additive functional equation (1.3).

Lemma 3.1. * Let be a mapping such that
**
for all . Then there exists a unique bi-additive mapping satisfying (1.3) such that
**
for all . The mapping is given by
**
for all .*

*Proof. * Consider the case . Letting and in (3.1), we gain
for all . Putting in (3.4), we get . Putting in (3.1), we get
for all . Replacing by and by in the above inequality, we have
for all . Setting and in (3.1), we obtain
for all . By (3.4) and (3.6), we gain
for all . By (3.4) and (3.7), we get
for all . By (3.4), (3.6), and (3.7), we have
for all . Replacing by and by and dividing , we obtain that
for all and all . For given integers , we obtain that
for all . By (3.12), the sequence is a Cauchy sequence for all . Since is complete, the sequence converges for all . Define by for all . By (3.1), we have
for all and all . Letting in the above inequality, we see that satisfies (1.3). Setting and taking in (3.12), one can obtain inequality (3.2). If is another mapping satisfying (1.3) and (3.2), by Theorem 2.1, we obtain that
for all . Hence the mapping is the unique bi-additive mapping satisfying (1.3), as desired.

The proof of the case is similar to that of the case .

From now on, let be a unital -algebra with a norm , and let and be left Banach -modules with norms and , respectively. Put .

A bi-additive mapping satisfying (1.3) is called *-quadratic* if for all and all .

Theorem 3.2. * Let be a mapping such that
**
for all and all . If is continuous in for each fixed , then there exists a unique bi-additive -quadratic mapping satisfying (1.3) and inequality (3.2).*

*Proof. * Consider the case . By Lemma 3.1, it follows from the inequality of the statement for that there exists a unique bi-additive mapping satisfying (1.3) and inequality (3.2). Let be fixed. And let be any real continuous linear functional, that is, is an arbitrary real functional element of the dual space of restricted to the scalar field . For , consider the functions defined by for all . By the assumption that is continuous in for each fixed , the function is continuous for all . Note that for all and all . By the proof of Lemma 3.1, the sequence is a Cauchy sequence for all . Define a function by for all . Note that for all . Since is bi-additive, we get
for all . Since is the pointwise limit of continuous functions, it is a Borel function. Thus the function as a measurable quadratic function is continuous (see [26]) so has the form for all . Hence we have
for all . Since is any continuous linear functional, the bi-additive mapping satisfies for all . Therefore we obtain
for all and all . Let be an arbitrary positive integer. Replacing and by and , respectively, and letting in inequality (3.15), we gain
for all and all . Note that there is a constant such that the condition
for each and each (see [27, Definition 12]). For all and all , we get
as . Hence we have
for all and all . Since for each , by (3.18), we obtain
for all nonzero and all . By (3.18), we get for all . Therefore the bi-additive mapping is the unique -quadratic mapping satisfying the inequality (3.2).

The proof of the case is similar to that of the case .

We obtain the Hyers-Ulam stability of (1.3) as a corollary of Theorem 3.2.

Corollary 3.3. * Let be a complex normed space and a function such that
**
for all and all . If is continuous in for each fixed , then there exists a unique bi-additive -quadratic mapping satisfying (1.3) such that for all .*

Put , , , and .

A unital -algebra is said to have *real rank * (see [28]) if the invertible self-adjoint elements are dense in .

For any element , , where and are self-adjoint elements, furthermore, , where , and are positive elements (see [27, Lemma 38.8]).

Theorem 3.4. * Let be of real rank , and let be a mapping such that
**
for all and all . For each fixed , let the sequence converge uniformly on . If is continuous in for each fixed , then there exists a unique bi-additive -quadratic mapping satisfying (1.3) and inequality (3.2) such that for all and all .*

*Proof. * Consider the case . By Lemma 3.1, there exists a unique bi-additive mapping satisfying (1.3) and inequality (3.2) on . Let be fixed. And let be an arbitrary real functional element of the dual space of restricted to the scalar field . For , consider the functions defined by for all . By the assumption that is continuous in for each fixed , the function is continuous for all . Note that for all and all . By the proof of Lemma 3.1, the sequence is a Cauchy sequence for all . Define a function by for all . Note that for all . Since the mapping is bi-additive, we have
for all . Since is the pointwise limit of continuous functions, it is a Borel function. By Theorem 2.2, we gain for all . Hence we get
for all . Since is any continuous linear functional, the bi-additive mapping satisfies for all . Therefore we obtain
for all and all . Let be an arbitrary positive integer. Replacing and by and , respectively, and letting in inequality (3.25), we get
for all and all . By inequality (3.20) and the above inequality, for all and all , we have
Hence we obtain that
for all and all . Let . Since is dense in , there exists two sequences and in such that and as . Put and for all . Then and as . Set and for all . Then and as and . Since is uniformly converges on for each and is continuous in for each , we see that is also continuous in for each . In fact, we gain
for all . Thus we get
for all . By equality (3.31), we have
as for all . By equality (3.33) and the above convergence, we see that
as for all . By equality (3.31) and the above convergence, we obtain
for all and all . Since the mapping is bi-additive, we see that
for all and all . By (3.28) and equality (3.36), we have
for all and all . Note that . Hence we obtain that
for all and all .

The proof of the case is similar to that of the case .

#### Acknowledgment

This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant no. 2012003499).

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