JFS Journal of Function Spaces 2314-8888 2314-8896 Hindawi Publishing Corporation 10.1155/2016/1804206 1804206 Research Article Stability of a Quartic Functional Equation in Restricted Domains http://orcid.org/0000-0003-2043-5135 Chung Jaeyoung 1 http://orcid.org/0000-0003-4125-582X Ju Yu-Min 1 Leśniak Zbigniew Department of Mathematics Kunsan National University Kunsan 573-701 Republic of Korea kunsan.ac.kr 2016 2672016 2016 24 03 2016 11 05 2016 16 05 2016 2672016 2016 Copyright © 2016 Jaeyoung Chung and Yu-Min Ju. 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.

Let X be a real normed space and Y a Banach space and f : X Y . We prove the Ulam-Hyers stability theorem for the quartic functional equation f ( 2 x + y ) + f ( 2 x - y ) - 4 f ( x + y ) - 4 f ( x - y ) - 24 f ( x ) + 6 f ( y ) = 0 in restricted domains. As a consequence we consider a measure zero stability problem of the above inequality when f : R Y .

Ministry of Education 2015R1D1A3A01019573
1. Introduction

Let R , X , and Y be the set of real numbers, a real normed space, and a real Banach space, respectively. A mapping f : X Y is called a quartic mapping if f satisfies the functional equation(1)f2x+y+f2x-y-4fx+y-4fx-y-24fx+6fy=0for all x , y X . It is known in [1, Theorem 2.1] that the general solutions f of (1) are given by f ( x ) = B ( x , x , x , x ) for all x X , where B : X × X × X × X Y 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 ϵ 0 be fixed. Suppose that f : X Y satisfies the cubic functional inequality (2) f 2 x + y + f 2 x - y - 4 f x + y - 4 f x - y - 24 f x + 6 f y ϵ for all x , y X . Then there exists a unique quartic mapping T : X Y such that(3)fx-Txϵ24for all x X .

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 d > 0 . Suppose that f : X Y satisfies the inequality(4)fx+y+fx-y-2fx-2fyδfor all x , y D { ( x , y ) X × X : x + y d } . Then there exists a unique mapping q : X Y satisfying(5)fx+y+fx-y=2fx+2fyfor all x , y X such that(6)fx-qx32ϵfor all x X .

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

Theorem 3.

Let d > 0 . Suppose that f : X Y satisfies the inequality (7) f 2 x + y + f 2 x - y - 2 f x + y - 2 f x - y - 12 f x ϵ for all x , y D { ( x , y ) X × X : x + y d } . Then there exists a unique mapping c : X Y satisfying(8)c2x+y+c2x-y-2cx+y-2cx-y-12cx=0for all x , y X such that(9)fx-cx-48f07914ϵfor all x X .

In this paper, we consider the Ulam-Hyers stability of quartic functional equation (1) in some restricted domains Ω X × X . First, imposing condition (C) on Ω (see Section 2) we prove that if f : X Y satisfies inequality (2) for all ( x , y ) Ω , then there exists a unique quartic mapping T : X Y such that(10)fx-Tx+231f050936ϵfor all x X . Since Ω = { ( x , y ) X × X : x + y d } 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 Γ R 2 of measure 0 satisfying condition (C) we consider a measure zero stability problem of quartic functional equation (1) for all x , y Γ , where f : R Y and Γ { ( x , y ) R 2 : x + y d } has 2-dimensional Lebesgue measure 0.

As an application we consider an asymptotic behavior of f : R Y satisfying the weak condition (11) f 2 x + y + f 2 x - y - 4 f x + y - 4 f x - y - 24 f x + 6 f y 0 as x + y only for ( x , y ) Γ , where Γ has 2-dimensional Lebesgue measure 0.

2. Stability of the Quartic Functional Equation in Restricted Domain

Throughout this section we assume that Ω X × X satisfies the following condition:

For given x , y X there exists t X such that

P x , y , t { ( x - t , y + 2 t ) , ( x + t , y + 2 t ) , ( x - t , - y + 2 t ) , ( x + t , - y + 2 t ) , ( x , y + 4 t ) , ( x , - y + 4 t ) , ( - t , x + y + 2 t ) , ( t , x - y - 2 t ) , ( - t , x - y + 2 t ) , ( t , x + y - 2 t ) , ( - t , x + y + t ) , ( t , x - y - t ) , ( - t , x - y + t ) , ( t , x + y - t ) , ( t , x + y ) , ( - t , x - y ) , ( t , x - y ) , ( - t , x + y ) , ( x + t , y ) , ( x - t , y ) , ( x , y + 2 t ) , ( x , - y + 2 t ) , ( 0 , x + y + 4 t ) , ( 0 , x + y - 4 t ) , ( 0 , x - y + 4 t ) , ( 0 , x - y - 4 t ) , ( 0 , x + y + 3 t ) , ( 0 , x + y - 3 t ) , ( 0 , x - y + 3 t ) , ( 0 , x - y - 3 t ) , ( 0 , x + y + 2 t ) , ( 0 , x + y - 2 t ) , ( 0 , x - y + 2 t ) , ( 0 , x - y - 2 t ) , ( 0 , x + y - t ) , ( 0 , x + y + t ) , ( 0 , x - y - t ) , ( 0 , x - y + t ) , ( 2 t , y ) , ( t , 0 ) , ( 0 , t ) } Ω .

Theorem 4.

Let ϵ 0 be fixed. Suppose that f : X Y satisfies inequality (2) for all ( x , y ) Ω . Then there exists a unique quartic mapping T : X Y satisfying (10) for all x X .

Proof.

Let(12)Fx,y=f2x+y+f2x-y-4fx+y-4fx-y-24fx+6fyfor all x , y X . Then(13)Fx-t,y+2t=f2x+y+f2x-y-4t-4fx+y+t-4fx-y-3t-24fx-t+6fy+2t,Fx+t,y+2t=f2x+y+4t+f2x-y-4fx+y+3t-4fx-y-t-24fx+t+6fy+2t,Fx-t,-y+2t=f2x-y+f2x+y-4t-4fx-y+t-4fx+y-3t-24fx-t+6f-y+2t,Fx+t,-y+2t=f2x-y+4t+f2x+y-4fx-y+3t-4fx+y-t-24fx+t+6f-y+2t,Fx,y+4t=f2x+y+4t+f2x-y-4t-4fx+y+4t-4fx-y-4t-24fx+6fy+4t,Fx,-y+4t=f2x-y+4t+f2x+y-4t-4fx-y+4t-4fx+y-4t-24fx+6f-y+4t,F-t,x+y+2t=fx+y+f-x-y-4t-4fx+y+t-4f-x-y-3t-24f-t+6fx+y+2t,Ft,x-y-2t=fx-y+f-x+y+4t-4fx-y-t-4f-x+y+3t-24ft+6fx-y-2t,F-t,x-y+2t=fx-y+f-x+y-4t-4fx-y+t-4f-x+y-3t-24f-t+6fx-y+2t,Ft,x+y-2t=fx+y+f-x-y+4t-4fx+y-t-4f-x-y+3t-24ft+6fx+y-2t,F-t,x+y+t=fx+y-t+f-x-y-3t-4fx+y-4f-x-y-2t-24f-t+6fx+y+t,Ft,x-y-t=fx-y+t+f-x+y+3t-4fx-y-4f-x+y+2t-24ft+6fx-y-t,F-t,x-y+t=fx-y-t+f-x+y-3t-4fx-y-4f-x+y-2t-24f-t+6fx-y+t,Ft,x+y-t=fx+y+t+f-x-y+3t-4fx+y-4f-x-y+2t-24ft+6fx+y-t,Ft,x+y=fx+y+2t+f-x-y+2t-4fx+y+t-4f-x-y+t-24ft+6fx+y,F-t,x-y=fx-y-2t+f-x+y-2t-4fx-y-t-4f-x+y-t-24f-t+6fx-y,Ft,x-y=fx-y+2t+f-x+y+2t-4fx-y+t-4f-x+y+t-24ft+6fx-y,F-t,x+y=fx+y-2t+f-x-y-2t-4fx+y-t-4f-x-y-t-24f-t+6fx+y,Fx+t,y=f2x+y+2t+f2x-y+2t-4fx+y+t-4fx-y+t-24fx+t+6fy,Fx-t,y=f2x+y-2t+f2x-y-2t-4fx+y-t-4fx-y-t-24fx-t+6fy,Fx,y+2t=f2x+y+2t+f2x-y-2t-4fx+y+2t-4fx-y-2t-24fx+6fy+2t,Fx,-y+2t=f2x-y+2t+f2x+y-2t-4fx-y+2t-4fx+y-2t-24fx+6f-y+2t,F0,x+y+4t=3fx+y+4t-3f-x-y-4t-24f0,F0,x+y-4t=3fx+y-4t-3f-x-y+4t-24f0,F0,x-y+4t=3fx-y+4t-3f-x+y-4t-24f0,F0,x-y-4t=3fx-y-4t-3f-x+y+4t-24f0,F0,x+y+3t=3fx+y+3t-3f-x-y-3t-24f0,F0,x+y-3t=3fx+y-3t-3f-x-y+3t-24f0,F0,x-y+3t=3fx-y+3t-3f-x+y-3t-24f0,F0,x-y-3t=3fx-y-3t-3f-x+y+3t-24f0,F0,x+y+2t=3fx+y+2t-3f-x-y-2t-24f0,F0,x+y-2t=3fx+y-2t-3f-x-y+2t-24f0,F0,x-y+2t=3fx-y+2t-3f-x+y-2t-24f0,F0,x-y-2t=3fx-y-2t-3f-x+y+2t-24f0,F0,x+y+t=3fx+y+t-3f-x-y-t-24f0,F0,x+y-t=3fx+y-t-3f-x-y+t-24f0,F0,x-y+t=3fx-y+t-3f-x+y-t-24f0,F0,x-y-t=3fx-y-t-3f-x+y+t-24f0,F2t,y=fy+4t+f-y+4t-4fy+2t-4f-y+2t-24f2t+6fy,Ft,0=2f2t-32ft+6f0,F0,t=3ft-3f-t-24f0.

Thus, we get the functional identity(14)3Fx-t,y+2t+3Fx+t,y+2t+3Fx-t,-y+2t+3Fx+t,-y+2t-3Fx,y+4t-3Fx,-y+4t-12F-t,x+y+2t-12Ft,x-y-2t-12F-t,x-y+2t-12Ft,x+y-2t-36F-t,x+y+t-36Ft,x-y-t-36F-t,x-y+t-36Ft,x+y-t-24Ft,x+y-24F-t,x-y-24Ft,x-y-24F-t,x+y-6Fx+t,y-6Fx-t,y+6Fx,y+2t+6Fx,-y+2t-4F0,x+y+4t-4F0,x+y-4t-4F0,x-y+4t-4F0,x-y-4t+4F0,x+y+3t+4F0,x+y-3t+4F0,x-y+3t+4F0,x-y-3t+40F0,x+y+2t+40F0,x+y-2t+40F0,x-y+2t+40F0,x-y-2t+32F0,x+y+t+32F0,x+y-t+32F0,x-y+t+32F0,x-y-t+18F2t,y+216Ft,0+1152F0,t=6f2x+y+6f2x-y-24fx+y-24fx-y-144fx+36fy-33264f0.

Since Ω satisfies condition (C), for given x , y X , there exists t X such that(15)Fu,vϵfor all ( u , v ) P x , y , t . Thus, dividing (14) by 6 and using the triangle inequality and (15) we have (16) f 2 x + y + f 2 x - y - 4 f x + y - 4 f x - y - 24 f x + 6 f y - 5544 f 0 1018 3 ϵ for all x , y X . Let g ( x ) = f ( x ) + 231 f ( 0 ) for all x X . Then from (16) we have (17) g 2 x + y + g 2 x - y - 4 g x + y - 4 g x - y - 24 g x + 6 g y 1018 3 ϵ for all x , y X . Using Theorem 1 with (17), we get (10). This completes the proof.

Remark 5.

Letting x = 0 in (10) and dividing the result by 232 we have f ( 0 ) 509 ϵ / 232 × 36 . Thus, inequality (10) implies(18)fx-Tx231f0+50936ϵ2356678352ϵfor all x X .

Let d 0 . It is easy to see that { ( x , y ) X × X : x + y d } satisfies condition (C). Indeed, for given x , y X if we choose t x + y + d , then P x , y , t { ( x , y ) X × X : x + y d } . Thus, as a direct consequence of Theorem 4 we obtain the following result (see [14, 16, 17] for similar results).

Corollary 6.

Let ϵ , d 0 be fixed. Suppose that f : X Y satisfies inequality (2) for all x , y X such that x + y d . Then there exists a unique quartic mapping T : X Y satisfying (10) for all x X .

In particular, if ϵ = 0 , we have the following.

Corollary 7.

Suppose that f : X Y (1) for all ( x , y ) Ω . Then, (1) holds for all x , y X .

3. Stability Problem in a Set of Lebesgue Measure Zero

In this section, we show that even a set Γ of Lebesgue measure zero can satisfy condition (C) when X = R . From now on, we identify R 2 with C .

Lemma 8.

Let P = { ( p j + a j t , q j + b j t ) : j = 1,2 , , r } , where p j , q j , a j , b j R with a j 2 + b j 2 0 for all j = 1,2 , , r . Then there exists θ 0,2 π such that e i θ P { ( p j + a j t , q j + b j t ) : j = 1,2 , , r } satisfies a j b j 0 for all j = 1,2 , , r .

Proof.

The coefficients a j and b j are given by (19)aj=ajcosθ-bjsinθ,bj=ajsinθ-bjcosθfor all j = 1,2 , , r . Now, the equation (20)j=1rajcosθ-bjsinθajsinθ-bjcosθ=0has only a finite number of zeros in 0,2 π . Thus, we can choose θ 0,2 π such that j = 1 r a j b j 0 . This completes the proof.

Lemma 9.

One can find a set K R of Lebesgue measure 0 such that, for any countable subsets U R , V R { 0 } , and M > 0 , there exists λ M satisfying(21)U+λV=u+λv:uU,vVK.

Proof.

It is shown in [22, Theorem 1.6] that there exists a set K R of Lebesgue measure 0 such that R K is of the first Baire category; that is, R K is a countable union of nowhere dense subsets of R . Let U = { u 1 , u 2 , u 3 , } and V = { v 1 , v 2 , v 3 , } and K m , n c = v m - 1 ( K c - u n ) , m , n = 1,2 , 3 , . Then, since K c is of the first Baire category, K m , n c are also of the first Baire category for all m , n = 1,2 , 3 , . Since each K m , n c consists of a countable union of nowhere dense subsets, by the Baire category theorem, countably many of them cannot cover M , ; that is, (22)m,n=1Km,ncM,.Thus, there exists λ M such that λ K m , n c for all m , n = 1,2 , 3 , . This means that u n + v m λ K for all m , n = 1,2 , 3 , . This completes the proof.

Theorem 10.

Let d 0 be fixed. Then there exists a set Γ d { ( x , y ) : x + y d } of 2-dimensional Lebesgue measure 0 which satisfies condition (C).

Proof.

Let P x , y , t be the set in condition (C). Then by Lemma 8 we can choose θ 0,2 π such that e i θ P x , y , t { ( p j + a j t , q j + b j t ) : j = 1,2 , , r } satisfies a j b j 0 for all j = 1,2 , , r . Let K be the set in Lemma 9. Then Γ d = e - i θ ( K × K ) { ( x , y ) : x + y d } has 2-dimensional Lebesgue measure 0. Now, we show that Γ satisfies condition (C); that is, for given x , y R , there exists t R satisfying the conditions(23)eiθPx,y,tK×K,Px,y,tx,y:x+yd.Let e i θ P x , y , t { ( p j + a j t , q j + b j t ) : j = 1,2 , , r } , U = { p j , q j : j = 1,2 , , r } , and V = { a j , b j : j = 1,2 , } . Then we have(24)u,v:u,veiθPx,y,tU+tV.Now, by Lemma 9, for given x , y R there exists t x + y + d such that(25)U+tVK.From (24) and (25) we have(26)eiθPx,y,tK×K.Since t x + y + d , using the triangle inequality we have u + v d for all ( u , v ) P x , y , t . Thus, Γ satisfies (C). This completes the proof.

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

Corollary 11.

Let ϵ , d 0 be fixed. Suppose that f : R Y satisfies inequality (2) for all ( x , y ) Γ d . Then there exists a unique quartic mapping T : X Y satisfying (10) for all x R .

As a consequence of Corollary 11 we obtain an asymptotic behavior of f satisfying condition (11) as x + y only for ( x , y ) Γ 0 .

Corollary 12.

Suppose that f : R Y satisfies condition (11). Then f is a quartic mapping.

Proof.

Condition (11) implies that, for each n N , there exists d n > 0 such that (27) f 2 x + y + f 2 x - y - 4 f x + y - 4 f x - y - 24 f x + 6 f y 1 n for all ( x , y ) Γ d n . By Corollary 11, there exists a unique quartic mapping T n : X Y such that(28)fx-Tnx+231f050936nfor all x X . Replacing n by m in (28) and using the triangle inequality we have(29)Tmx-Tnx50936n+50936m50918for all x X . Let T m , n ( x ) = T m ( x ) - T n ( x ) for all x X . Then by (29), T m , n is a bounded quartic mapping. Thus, we have T m , n = 0 and hence T m = T n T for all m , n N . Letting n in (28) we have f ( 0 ) = 0 and hence f ( x ) = T ( x ) for all x X . This completes the proof.

Remark 13.

If we define Γ R 2 n as an appropriate rotation of 2 n -product K 2 n of K , then Γ d has 2 n -dimensional measure 0 and satisfies condition (C). Consequently, we obtain the following.

Theorem 14.

Suppose that f : R n Y satisfies inequality (2) for all ( x , y ) Γ d . Then there exists a unique quartic mapping T : R n Y satisfying (10) for all x R n .

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).

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