Stability of a 2-Dimensional Functional Equation in a Class of Vector Variable Functions

Abstract

We prove the Hyers-Ulam stability of a 2-dimensional quadratic functional equation in a class of vector variable functions 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, we call that type the Hyers-Ulam stability. The authors investigated various functional equations and their Hyers-Ulam stability [3–8]. This Hyers-Ulam stability is a classical type of stability, but there is another kind of stability introduced by Risteski [9] for functional equations spanned over an -dimensional complex vector space too.

Let and be real or complex vector spaces. For a mapping , consider the quadratic functional equation

(11)

In 1989, Aczél and Dhombres [10] obtained the solution of (1.1) for the case that acts on . The result also holds when and are arbitrary real or complex vector spaces. For a mapping , consider the -dimensional quadratic functional equation:

(12)

The quadratic form given by is a solution of (1.2). In 2008, the authors of [8] acquired the general solution and proved the stability of the -dimensional quadratic functional equation (1.2) for the case that and are real vector spaces as follows.

The results of [8, Theorms 3 and 4] also hold for complex vector spaces and . In this paper, we investigate the stability of (1.2) with two module actions in Banach modules over a unital -algebra.

2. Preliminaries

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

Definition 2.1.

A -dimensional vector variable quadratic mapping satisfying (1.2) is called -quadratic if for all and all .

Definition 2.2.

A unital -algebra is said to have real rank (see [11]) 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 [12, Lemma ]).

3. Results

Theorem 3.1.

Let be a function satisfying

(31)

for all . If the function is a Borel function, then it also satisfies

(32)

for all .

Proof.

By [8, Theoem 3], there exist two symmetric biadditive mappings such that for all . By the proof of Therem 3 in [8], we gain

(33)

for all and all . Letting in the equality (3.3), we get

(34)

for all and all . Putting in the equality (3.3) again, we have

(35)

for all . Since the function is measurable and satisfies (1.1), by [13], it is continuous. By the same reasoning, is also continuous. Let be fixed. Since is measurable, by [14, Theore 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 . By the equality (3.4), we get

(36)

for all . For each fixed , take a sequence in converging to . By (3.5) and the above equality, we have

(37)

Hence we see that

(38)

as desired.

Lemma 3.2.

Let and be normed spaces and a real number, and let be a mapping such that

(39)

for all . Suppose for . If is complete, then there exists a unique -variable quadratic mapping such that

(310)

for all . The mapping is given by

(311)

for all .

Proof.

Letting and in (3.9), we gain

(312)

for all . Putting in (3.12), we get

(313)

for all . Replacing by in the above inequality, we have

(314)

for all . By the above two inequalities, we see that

(315)

for all . Setting and in (3.9), we obtain that

(316)

for all . Replacing by in the above inequality, we see that

(317)

for all . By the last two inequalities, we know that

(318)

for all . By (3.12) and (3.18), we obtain that

(319)

for all . By (3.15) and the above inequality, we have

(320)

for all . Thus we obtain that

(321)

for all and all . Replacing by in the above inequality, we see that

(322)

for all and all . For given integers , we obtain that

(323)

for all .

Consider the case . By (3.23), the sequence is a Cauchy sequence for all . Since is complete, the sequence converges for all . Define by for all . By (3.9), we have

(324)

for all and all . Letting , we see that satisfies (1.2). Setting and taking in (3.23), one can obtain inequality (3.10). If is another 2-dimensional vector variable quadratic mapping satisfying (3.10), by [8, Therem 3], there are four symmetric biadditive mappings such that and for all . Thus we obtain that

(325)

for all . Hence the mapping is the unique 2-dimensional vector variable quadratic mapping, as desired.

Next, consider the case . Since , by inequality (3.20), we gain

(326)

for all . Thus we get

(327)

for all and all . Replacing by in the above inequality, we have

(328)

for all and all . For given integers , we obtain that

(329)

for all . By (3.29), the sequence is a Cauchy sequence for all . Since is complete, the sequence converges for all . Define by for all . By (3.9), we have

(330)

for all and all . Letting , we see that satisfies (1.2). Setting and taking in (3.29), one can obtain inequality (3.10). If is another 2-dimensional vector variable quadratic mapping satisfying (3.10), by in [8, Thorem 3], there are four symmetric biadditive mappings such that and for all . Thus we obtain that

(331)

for all . Hence the mapping is the unique 2-dimensional vector variable quadratic mapping, as desired.

Theorem 3.3.

Let be a real number, and let be a mapping such that

(332)

for all and all . If is continuous in for each fixed , then there exists a unique -dimensional vector variable -quadratic mapping satisfying (1.2) and (3.10) for all .

Proof.

Suppose . By Lemma 3.2, it follows from the inequality of the statement for that there exists a unique -dimensional vector variable quadratic mapping satisfying (1.2) and inequality (3.10) for all .

Let be fixed. And let be any continuous linear functional, that is, is an arbitrary element of the dual space of . For , consider two functions and defined by and for all . By the assumption that is continuous in for each fixed , the functions and are continuous for all . Note that and for all and all . By [8], the sequences and are Cauchy sequences for all . Define two functions and by and for all . Note that and for all . Since satisfies (1.2), we get

(333)

for all . Since and are the pointwise limits of continuous functions, they are Borel functions. Thus the functions and as measurable quadratic functions are continuous (see [13]), so have the forms and for all . Since satisfies (1.2), by [8, Thorem 3], there exist two symmetric biadditive mappings such as for all . Hence we have

(334)

for all . Since is any continuous linear functional, the -dimensional quadratic mapping satisfies for all . Therefore we obtain

(335)

for all and all . Let be an arbitrary positive integer. Replacing and by and , respectively, and letting in inequality (3.32), we gain

(336)

for all and all . Note that there is a constant such that the condition

(337)

for each and each (see [12, Definitin 12]). For all and all , we get

(338)

as . Hence we have

(339)

for all and all . Since for each , by (3.35), we obtain

(340)

for all nonzero and all . By (3.35), we get for all . Therefore the mapping is the unique -dimensional vector variable -quadratic mapping satisfying (1.2) and (3.10).

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

Theorem 3.4.

Let be a real number and of real rank , and let be a mapping such that

(341)

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 -dimensional vector variable mapping satisfying (1.2) and (3.10) such that for all and all .

Proof.

Suppose . By [8, Teorem 4], there exists a unique -dimensional quadratic mapping satisfying (1.2) and inequality (3.10) on . Let be fixed. And let be an arbitrary element of the dual space of . 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 [8], the sequence is a Cauchy sequence for all . Define a function by for all . Note that for all . Thus we have

(342)

for all . Since is the pointwise limit of continuous functions, it is a Borel function. By Theorem 3.1, we gain for all . Hence we get

(343)

for all . Since is any continuous linear functional, the -dimensional quadratic mapping satisfies for all . Therefore we obtain

(344)

for all and all . Let be an arbitrary positive integer. Replacing and by and , respectively, and letting in the inequality (3.41), we get

(345)

for all and all . By condition (3.37), for all and all , we have

(346)

Hence we obtain that

(347)

for all and all .

Let . Since is dense in , there exist two sequences and in such that and as . Put and . Then and as . Set and . Then and as and . Since is uniformly converges on and is continuous in , we see that is also continuous in for each . In fact, we gain

(348)

for all . Thus we get

(349)

for all . By equality (3.47), we have

(350)

as for all By equality (3.49) and the above convergence, we see that

(351)

for all . By equality (3.47) and the above convergence, we obtain for all and all .

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