#### Abstract

Let be a real normed space and a Banach space and . We prove the Ulam-Hyers stability theorem for the quartic functional equation in restricted domains. As a consequence we consider a measure zero stability problem of the above inequality when .

#### 1. Introduction

Let , , and be the set of real numbers, a real normed space, and a real Banach space, respectively. A mapping is called a quartic mapping if satisfies the functional equationfor all . It is known in [1, Theorem 2.1] that the general solutions of (1) are given by for all , where is a symmetric function which is additive for each variable when the other three variables are fixed. The following is a modified version of [1, Theorem 3.1]. We refer the reader to [2, 3] for the stability of generalized quartic mappings.

Theorem 1. Let be fixed. Suppose that satisfies the cubic functional inequalityfor all . Then there exists a unique quartic mapping such thatfor all .

It is a very natural subject to study functional equations or inequalities satisfied on restricted domains or satisfied under restricted conditions . Among the results, Jung  and Rassias  proved the Ulam-Hyers stability of the quadratic functional equations in a restricted domain. As a refined version of the results in [14, 17] we state the result in .

Theorem 2. Let . Suppose that satisfies the inequalityfor all . Then there exists a unique mapping satisfyingfor all such thatfor all .

Also, Chung-Ju-Rassias  proved the following Ulam-Hyers stability of cubic functional equation in restricted domains.

Theorem 3. Let . Suppose that satisfies the inequalityfor all . Then there exists a unique mapping satisfyingfor all such thatfor all .

In this paper, we consider the Ulam-Hyers stability of quartic functional equation (1) in some restricted domains . First, imposing condition (C) on (see Section 2) we prove that if satisfies inequality (2) for all , then there exists a unique quartic mapping such thatfor all . Since satisfies condition (C), we obtain the parallel result for quartic functional equation as Theorem 2 for a quadratic functional equation and Theorem 3 for a cubic functional equation.

Secondly, constructing a subset of measure 0 satisfying condition (C) we consider a measure zero stability problem of quartic functional equation (1) for all , where and has 2-dimensional Lebesgue measure 0.

As an application we consider an asymptotic behavior of satisfying the weak conditionas only for , where has 2-dimensional Lebesgue measure 0.

#### 2. Stability of the Quartic Functional Equation in Restricted Domain

Throughout this section we assume that satisfies the following condition:For given there exists such that(C), , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , .

Theorem 4. Let be fixed. Suppose that satisfies inequality (2) for all . Then there exists a unique quartic mapping satisfying (10) for all .

Proof. Letfor all . ThenThus, we get the functional identitySince satisfies condition (C), for given , there exists such thatfor all . Thus, dividing (14) by 6 and using the triangle inequality and (15) we havefor all . Let for all . Then from (16) we havefor all . Using Theorem 1 with (17), we get (10). This completes the proof.

Remark 5. Letting in (10) and dividing the result by 232 we have . Thus, inequality (10) impliesfor all .

Let . It is easy to see that satisfies condition (C). Indeed, for given if we choose , then . Thus, as a direct consequence of Theorem 4 we obtain the following result (see [14, 16, 17] for similar results).

Corollary 6. Let be fixed. Suppose that satisfies inequality (2) for all such that . Then there exists a unique quartic mapping satisfying (10) for all .

In particular, if , we have the following.

Corollary 7. Suppose that (1) for all . Then, (1) holds for all .

#### 3. Stability Problem in a Set of Lebesgue Measure Zero

In this section, we show that even a set of Lebesgue measure zero can satisfy condition (C) when . From now on, we identify with .

Lemma 8. Let , where with for all . Then there exists such that satisfies for all .

Proof. The coefficients and are given by for all . Now, the equation has only a finite number of zeros in . Thus, we can choose such that . This completes the proof.

Lemma 9. One can find a set of Lebesgue measure 0 such that, for any countable subsets , and , there exists satisfying

Proof. It is shown in [22, Theorem 1.6] that there exists a set of Lebesgue measure 0 such that is of the first Baire category; that is, is a countable union of nowhere dense subsets of . Let and and . Then, since is of the first Baire category, are also of the first Baire category for all . Since each consists of a countable union of nowhere dense subsets, by the Baire category theorem, countably many of them cannot cover ; that is, Thus, there exists such that for all . This means that for all . This completes the proof.

Theorem 10. Let be fixed. Then there exists a set of 2-dimensional Lebesgue measure 0 which satisfies condition (C).

Proof. Let be the set in condition (C). Then by Lemma 8 we can choose such that satisfies for all . Let be the set in Lemma 9. Then has 2-dimensional Lebesgue measure 0. Now, we show that satisfies condition (C); that is, for given , there exists satisfying the conditionsLet , , and . Then we haveNow, by Lemma 9, for given there exists such thatFrom (24) and (25) we haveSince , using the triangle inequality we have for all . Thus, satisfies (C). This completes the proof.

Now, as a direct consequence of Theorems 4 and 10 we have the following.

Corollary 11. Let be fixed. Suppose that satisfies inequality (2) for all . Then there exists a unique quartic mapping satisfying (10) for all .

As a consequence of Corollary 11 we obtain an asymptotic behavior of satisfying condition (11) as only for .

Corollary 12. Suppose that satisfies condition (11). Then is a quartic mapping.

Proof. Condition (11) implies that, for each , there exists such thatfor all . By Corollary 11, there exists a unique quartic mapping such thatfor all . Replacing by in (28) and using the triangle inequality we havefor all . Let for all . Then by (29), is a bounded quartic mapping. Thus, we have and hence for all . Letting in (28) we have and hence for all . This completes the proof.

Remark 13. If we define as an appropriate rotation of -product of , then has -dimensional measure 0 and satisfies condition (C). Consequently, we obtain the following.

Theorem 14. Suppose that satisfies inequality (2) for all . Then there exists a unique quartic mapping satisfying (10) for all .

#### Competing Interests

The authors declare that there are no competing interests regarding the publication of this paper.

#### Acknowledgments

This research was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (no. 2015R1D1A3A01019573).