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(2x+y)+f(2x-y)-4f(x+y)-4f(x-y)-24f(x)+6f(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 Education2015R1D1A3A010195731. 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)f2x+y+f2x-y-4fx+y-4fx-y-24fx+6fy≤ϵfor all x,y∈X. Then there exists a unique quartic mapping T:X→Y such that(3)fx-Tx≤ϵ24for all x∈X.

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 d>0. Suppose that f:X→Y satisfies the inequality(4)fx+y+fx-y-2fx-2fy≤δfor 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-qx≤32ϵfor all x∈X.

Also, Chung-Ju-Rassias [21] 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)f2x+y+f2x-y-2fx+y-2fx-y-12fx≤ϵfor 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-48f0≤7914ϵfor 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+231f0≤50936ϵfor 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⊂R2 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)∈R2: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)f2x+y+f2x-y-4fx+y-4fx-y-24fx+6fy⟶0as 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:

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,v≤ϵfor all (u,v)∈Px,y,t. Thus, dividing (14) by 6 and using the triangle inequality and (15) we have(16)f2x+y+f2x-y-4fx+y-4fx-y-24fx+6fy-5544f0≤10183ϵfor all x,y∈X. Let g(x)=f(x)+231f(0) for all x∈X. Then from (16) we have(17)g2x+y+g2x-y-4gx+y-4gx-y-24gx+6gy≤10183ϵfor 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)≤509ϵ/232×36. Thus, inequality (10) implies(18)fx-Tx≤231f0+50936ϵ≤2356678352ϵfor 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 Px,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 R2 with C.

Lemma 8.

Let P={(pj+ajt,qj+bjt):j=1,2,…,r}, where pj,qj,aj,bj∈R with aj2+bj2≠0 for all j=1,2,…,r. Then there exists θ∈0,2π such that eiθP≔{(pj′+aj′t,qj′+bj′t):j=1,2,…,r} satisfies aj′bj′≠0 for all j=1,2,…,r.

Proof.

The coefficients aj′ and bj′ 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 0,2π. Thus, we can choose θ∈0,2π such that ∏j=1raj′bj′≠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:u∈U,v∈V⊂K.

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={u1,u2,u3,…} and V={v1,v2,v3,…} and Km,nc=vm-1(Kc-un),m,n=1,2,3,…. Then, since Kc is of the first Baire category, Km,nc are also of the first Baire category for all m,n=1,2,3,…. Since each Km,nc 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=1∞Km,nc⊉M,∞.Thus, there exists λ≥M such that λ∉Km,nc for all m,n=1,2,3,…. This means that un+vmλ∈K for all m,n=1,2,3,…. This completes the proof.

Theorem 10.

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

Proof.

Let Px,y,t be the set in condition (C). Then by Lemma 8 we can choose θ∈0,2π such that eiθPx,y,t≔{(pj′+aj′t,qj′+bj′t):j=1,2,…,r} satisfies aj′bj′≠0 for all j=1,2,…,r. Let K be the set in Lemma 9. Then Γd=e-iθ(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,t⊂K×K,Px,y,t⊂x,y:x+y≥d.Let eiθPx,y,t≔{(pj′+aj′t,qj′+bj′t):j=1,2,…,r}, U={pj′,qj′:j=1,2,…,r}, and V={aj′,bj′:j=1,2,…}. Then we have(24)u,v:u,v∈eiθPx,y,t⊂U+tV.Now, by Lemma 9, for given x,y∈R there exists t≥x+y+d such that(25)U+tV⊂K.From (24) and (25) we have(26)eiθPx,y,t⊂K×K.Since t≥x+y+d, using the triangle inequality we have u+v≥d for all (u,v)∈Px,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)∈Γd. 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)∈Γ0.

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 dn>0 such that(27)f2x+y+f2x-y-4fx+y-4fx-y-24fx+6fy≤1nfor all (x,y)∈Γdn. By Corollary 11, there exists a unique quartic mapping Tn:X→Y such that(28)fx-Tnx+231f0≤50936nfor all x∈X. Replacing n by m in (28) and using the triangle inequality we have(29)Tmx-Tnx≤50936n+50936m≤50918for all x∈X. Let Tm,n(x)=Tm(x)-Tn(x) for all x∈X. Then by (29), Tm,n is a bounded quartic mapping. Thus, we have Tm,n=0 and hence Tm=Tn≔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⊂R2n as an appropriate rotation of 2n-product K2n of K, then Γd has 2n-dimensional measure 0 and satisfies condition (C). Consequently, we obtain the following.

Theorem 14.

Suppose that f:Rn→Y satisfies inequality (2) for all (x,y)∈Γd. Then there exists a unique quartic mapping T:Rn→Y satisfying (10) for all x∈Rn.

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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