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 [4–19]. Among the results, Jung [14] and Rassias [17] 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 [20].
Theorem 2. Let . Suppose that satisfies the inequalityfor all . Then there exists a unique mapping satisfyingfor all such thatfor all .
Also, Chung-Ju-Rassias [21] 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).