# Stability of a Quartic Functional Equation in Restricted Domains

Authors: Jaeyoung Chung; Yu-Min Ju
Journal: Journal of Function Spaces
 (2016)

---

## Abstract

LetX 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.


---

## Body

## 1. Introduction

LetR, 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 calleda 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.
Letϵ
≥
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
≤
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.
Letd
>
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.
Letd
>
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
≤
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 subsetΓ
⊂
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 off
:
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
⟶
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 thatΩ
⊂
X
×
X satisfies the following condition:For given
x
,
y
∈
X
there exists
t
∈
X
such that
(C)
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.
Letϵ
≥
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.
SinceΩ satisfies condition (C), for given x
,
y
∈
X, there exists t
∈
X such that(15)Fu,v≤ϵfor 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
ϵfor 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
ϵfor all x
,
y
∈
X. Using Theorem 1 with (17), we get (10). This completes the proof.Remark 5.
Lettingx
=
0 in (10) and dividing the result by 232 we have f
(
0
)
≤
509
ϵ
/
232
×
36. Thus, inequality (10) implies(18)fx-Tx≤231f0+50936ϵ≤2356678352ϵfor all x
∈
X.Letd
≥
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.
Letϵ
,
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, ifϵ
=
0, we have the following.Corollary 7.
Suppose thatf
:
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 setΓ of Lebesgue measure zero can satisfy condition (C) when X
=
R. From now on, we identify R
2 with C.Lemma 8.
LetP
=
{
(
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 coefficientsa
j
′ and b
j
′ are given by (19)aj′=ajcos⁡θ-bjsin⁡θ,bj′=ajsin⁡θ-bjcos⁡θfor all j
=
1,2
,
…
,
r. Now, the equation (20)∏j=1rajcos⁡θ-bjsin⁡θajsin⁡θ-bjcos⁡θ=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 setK
⊂
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
=
{
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=1∞Km,nc⊉M,∞.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.
Letd
≥
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.
LetP
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,t⊂K×K,Px,y,t⊂x,y:x+y≥d.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,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
)
∈
P
x
,
y
,
t. Thus, Γ satisfies (C). This completes the proof.Now, as a direct consequence of Theorems4 and 10 we have the following.Corollary 11.
Letϵ
,
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 Corollary11 we obtain an asymptotic behavior of f satisfying condition (11) as x
+
y
→
∞ only for (
x
,
y
)
∈
Γ
0.Corollary 12.
Suppose thatf
:
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
nfor all (
x
,
y
)
∈
Γ
d
n. By Corollary 11, there exists a unique quartic mapping T
n
:
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 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 defineΓ
⊂
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 thatf
:
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.



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*Source: 1804206-2016-07-26.xml*