Volume 2008, Article ID 816963,22pages doi:10.1155/2008/816963
Research Article
On the Stability of a New Pexider-Type
Functional Equation
Kil-Woung Jun,1, 2 Yang-Hi Lee,3and Juri Lee2
1National Institute for Mathematical Sciences, Daejeon 305-340, South Korea
2Department of Mathematics, Chungnam National University, Daejeon 305-764, South Korea
3Department of Mathematics Education, Gongju National University of Education,
Gongju 314-711, South Korea
Correspondence should be addressed to Yang-Hi Lee,[email protected]
Received 1 October 2007; Accepted 31 January 2008
Recommended by Patricia Wong
We establish the generalized Hyers-Ulam stability of a Pexider-type functonal equationf1xy
z f2x−y f3x−z−f4x−y−z−f5xy−f6xz 0, which is mixed of a quadratic and
an additive functional equations. Also, we obtain its general solution from the stability results.
Copyrightq2008 Kil-Woung Jun et al. 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.
1. Introduction
In 1940, Ulam 1raised the following question. Under what conditions does there exist an additive mapping near an approximately additive mapping?
In 1941, Hyers2proved that iff :V→Xis a mapping satisfying
fxy−fx−fy≤δ 1.1
for allx, y∈V, whereV andXare Banach spaces andδis a given positive number, then there exists a unique additive mappingT :V→Xsuch that
fx−Tx≤δ 1.2
the so-called generalized Hyers-Ulam-Rassias stability see also 4, 7–10. Jun et al.11–13 also obtained the Hyers-Ulam-Rassias stability of the Pexider equation of fxy gx hy. Quadratic functional equation was used to characterize inner product spaces14. Several other functional equations were also to characterize inner product spaces. A square norm on an inner product space satisfies the important parallelogram equality
xy2x−y22x2y2. 1.3
The functional equation
fxy fx−y 2fx 2fy 1.4
is related to a symmetric biadditive function14. It is natural that each equation is called a quadratic functional equation. A stability problem for the quadratic functional equation was proved by Skof15for a functionf:V→X, whereV is a normed space andXa Banach space. Cholewa16noticed that the theorem of Skof is still true if the relevant domainV is replaced by an Abelian group. Czerwik17proved the Hyers-Ulam-Rassias stability of the quadratic functional equation. Jun and Lee13, 18–22proved the Hyers-Ulam-Rassias stability of the Pexiderized quadratic equation
fxy gx−y 2hx 2ky. 1.5
Now, we introduce the following new Pexider type functional equation:
f1xyz f2x−y f3z−x−f4x−y−z−f5xy−f6xz 0, 1.6
which is mixed of a quadratic and an additive functional equations. In this paper, we establish the generalized Hyers-Ulam-Rassias stability for1.6on the punctured domainV \ {0}and obtain its general solution from the stability results. Throughout this paper, letV andXbe a normed space and a Banach space, respectively. For convenience, we employ the operators as follows: for a given functionϕ:V\{0}×V\{0}×V\{0}→0,∞, letϕ, ϕe, ϕe :V\{0}3→0,∞, M, M, M
e, Me :V \ {0}→0,∞be functions defined by
ϕx, y, z: 1
2
ϕx, y, z ϕ−x, y, z,
ϕex, y, z: 1 2
ϕx, y, z ϕ−x,−y,−z,
ϕ
ex, y, z: 14ϕx, y, z ϕ−x, y, z ϕ−x,−y,−z ϕx,−y,−z,
Mx:ϕx
2, 3x
2 ,− x 2
2ϕ
x
2, x 2,−
x 2
ϕx
2, x 2,
x 2
,
Mx:ϕx
2, x 2,−
3x 2
2ϕ
x
2, x 2,−
x 2
ϕx
2, x 2,
x 2
,
Mx:ϕx
2, 3x
2 ,− x 2
2ϕ
x
2, x 2,−
x 2
ϕx
2, x 2,
x 2
,
M
ex:ϕe x
2, x 2,−
3x 2
2ϕe
x
2, x 2,−
x 2
ϕ
e x
2, x 2,
x 2
1.7
2. Generalized Hyers-Ulam-Rassias stability
We need the following lemma to prove our main results.
Lemma 2.1. Letabe a positive real number. LetΦ:V \ {0}→0,∞be a map such that
Φx:∞ l0
1 al1Φ
2lx<∞ ∀ x∈V \ {0} 2.1
or
Φx:∞ l0
alΦ x 2l1
<∞ ∀x∈V \ {0}. 2.2
Suppose that the functionf :V→Xsatisfies the inequality
fx−f2ax≤ Φax 2.3
for allx∈V \ {0}andf0 0. Then, there exists exactly one functionF:V→Xsatisfying
fx−Fx≤Φx ∀x∈V \ {0}, aFx F2x ∀x∈V. 2.4
Proof. First we assume thatΦsatisfies
∞
l0
Φ2lx
al1 <∞ 2.5
for allx∈V \ {0}. Replacingxby 2nxand dividing it byanin2.3, we have
fa2nnx −
f2n1x an1
≤ Φ
2nx
an1 2.6
for alln∈Nandx∈V \ {0}. Induction argument implies that
fx− f
2nx an
≤n−1
s0
Φ2sx
as1 2.7
for alln∈Nandx∈V \ {0}. Hence
fa2nnx −
f2mx am
≤m−1
sn
Φ2sx
as1 2.8
for all positive integers m > n and x ∈ V \ {0}. This shows that {f2nx/an} is a Cauchy sequence forx∈V \ {0}and thus converges. Therefore, we can defineF :V→Xsuch that
Fx ⎧ ⎪ ⎨ ⎪ ⎩
lim n→∞
f2nx
an , if x /0, 0, if x0
for allx∈V. From2.7and the definition ofF, we obtain
fx−Fx≤Φx, aFx F2x 2.10
for allx∈V \ {0}. Now, letF:V \ {0}→Xbe another mapping satisfying the above inequality and equality. Then, it follows that
Fx−Fx≤f
2mx am −
F2mx am
f
2mx am −
F2mx am
≤ Φ
2mx am
2.11
which tends to zero by the definition ofΦ as m→∞ for all x ∈ V. So we can conclude that Fx Fxfor allx∈V. This proves the uniqueness ofF.
Next we assume thatΦsatisfies
∞
l0
alΦ x 2l1
<∞ 2.12
for allx∈V \ {0}. Replacingxby 2−n−1xand multiplying it byan1in2.3, we have
anf2−nx−an1f2−n−1x≤anΦ2−n−1x 2.13
for alln∈Nandx∈V \ {0}. Induction argument implies that
fx−anf2−nx≤n−1 s0
asΦ2−s−1x 2.14
for alln∈Nandx∈V \ {0}. Hence
anf2−nx−amf2−mx≤m−1 sn
asΦ2−s−1x 2.15
for all positive integers m > n and x ∈ V \ {0}. This shows that {anf2−nx} is a Cauchy sequence forx∈V \ {0}and thus converges. Therefore we can defineF :V→Xsuch that
Fx ⎧ ⎨ ⎩
lim n→∞a
nf2−nx ifx /0,
0 ifx0 2.16
for allx∈V. From2.14and the definition ofF, we obtain
fx−Fx≤Φx, aFx F2x 2.17
for allx∈V \ {0}.
The uniqueness ofFis proved similarly as the first case. This completes the proof.
Theorem 2.2. Letϕ:V \ {0} ×V \ {0} ×V \ {0}→0,∞be a function such that
ϕx, y, z:∞ l0
1 4l1ϕ
2lx,2ly,2lz<∞ a
holds for allx, y, z∈V \ {0}. If the even functionsf1, f2, f3, f4, f5, f6:V→Xsatisfy the inequality
f1xyz f2x−y f3x−z−f4x−y−z−f5xy−f6xz≤ϕx, y, z 2.18
for all x, y, z ∈ V \ {0}, then there exists exactly one quadratic function Q : V→X satisfying the inequalities
f1x−f10−Qx≤
1 2
ϕx
2, 3 2x,−x
ϕx
2, x 2,−x
1
2M2x Mx ϕ
x
4, x 4,−
x 2
ϕx
4, x 4, x 2 , 2.19
f2x−f20−Qx≤Mx ϕx
4, 3x
4 ,− x 4 ϕ x 4, x 4, x 4 ,
f3x−f30−Qx≤Mx ϕ
x
4, x 4,−
3x 4
ϕx
4, x 4, x 4 ,
f4x−f40−Qx≤ 1 2 ϕ x 2, 3 2x,−x
ϕ
x
2, x 2,−x
1
2M2x Mx ϕ
x
4, x 4,−
x 2 ϕ x 4, x 4, x 2 ,
f5x−f50−Qx≤Mx ϕx
4, 3x
4 ,− x 4 ϕ x 4, x 4, x 4 ,
f6x−f60−Qx≤Mx ϕ
x
4, x 4,−
3x 4
ϕx
4, x 4, x 4 2.20
for allx∈V \ {0}, where
Mx:∞ l0
1 4l1M
2lx, Mx:
∞
l0
1 4l1M
2lx 2.21
for allx∈V \ {0}. Moreover, the functionQis given by
Qx lim n→∞
fk2nx
4n 2.22
Proof. Replacexby−xin2.18to obtain
f1x−y−z f2xy f3xz−f4xyz−f5x−y−f6x−z≤ϕ−x, y, z
2.23
for allx, y, z∈V \ {0}. From2.18and2.23, we get f1f4
xyz f2f5
x−y f3f6
x−z
−f1f4
x−y−z−f2f5
xy−f3f6
xz≤ϕ−x, y, z ϕx, y, z 2.24
for allx, y, z∈V \ {0}. Let the functionsF, G, H:V→Xbe defined by
Fx 1 2
f1x f4x−f10−f40
,
Gx 1 2
f2x f5x−f20−f50
,
Hx 1 2
f3x f6x−f30−f60
2.25
for allx, y, z∈V. Then, it follows from2.24that
Fxyz Gx−y Hx−z−Fx−y−z−Gxy−Hxz≤ϕx, y, z
2.26
for allx, y, z∈V \ {0}, whereϕx, y, z 1/2ϕx, y, z ϕ−x, y, z. Replaceyandzbyx and−xin2.26to get
H2x−G2x≤ϕx, x,−x 2.27
for allx∈V \ {0}.
Replacingy,zbyxin2.26and using2.27, we get
F3x−Fx−2G2x≤ϕx, x, x ϕx, x,−x 2.28
for allx∈V \ {0}. Replacingx,y,zbyx, 3x,−xin2.26and using2.27, one obtains
F3x−Fx−G4x 2G2x≤ϕx,3x,−x ϕx, x,−x 2.29
for allx∈V \ {0}. From2.28and the above inequality, we have
G4x−4G2x≤ϕx,3x,−x 2ϕx, x,−x ϕx, x, x 2.30
for allx∈V \ {0}. Replacingxbyx/2 and dividing it by 4 in the above inequality, we get
Gx−G2x
4
≤ Mx
for allx∈V \ {0}. ByLemma 2.1, there exists limn→∞G2nx/4nfor allx∈V satisfying
Gx−lim n→∞
G2nx 4n
≤Mx 2.32
for allx∈V \ {0}, where
Mx:∞ l0
1 4l1M
2lx. 2.33
By the similar method in obtaining inequality2.32, we get
Hx− lim n→∞
H2nx 4n
≤Mx 2.34
for allx∈V \ {0}, where
Mx:∞
l0
1 4l1M
2lx. 2.35
From2.27, we have
lim n→∞
G2nx 4n nlim→∞
H2nx
4n 2.36
for allx∈V. From2.36, we can define a mapQ :V→Xby
Qx lim n→∞
G2nx
4n 2.37
for allx∈V. It follows from2.26,2.32, and2.37that Fx−Qx≤ 1
2
Fx Gx H
3 2x
−G2x−H x
2
1
2Gx−Qx
Fx Gx H
1 2x
−H
3 2x
1
2G2x−Q2x
≤ 1
2ϕ
x
2, 3 2x,−x
1
2M2x 1 2ϕ
x
2, x 2,−x
Mx
2.38
for allx∈V \ {0}. Replacingxby 2nx, dividing it by 4nin the above inequality and taking the limit in the resulted inequality asn→∞, we have
lim n→∞
F2nx
4n Qx 2.39
for allx∈V. Using2.26,2.36,2.37, and2.39, we obtain
for allx, y, z∈V \ {0}. Replacingxandzbyx/2 in2.40and using the factQ0 0, we have
Qxy Q x
2 −y
−Q−y−Q x
2 y
−Qx 0 2.41
for allx, y ∈V. Replacexandzbyx/2 and−x/2 in2.40to have
Qy Q x
2 −y
Qx−Qx−y−Q x
2 y
0 2.42
for allx, y ∈V. Subtracting2.41from2.42and using the evenness ofQ, we lead to
Qxy Qx−y−2Qx−2Qy 0, 2.43
for allx, y ∈V.
On the other hand, it follows from2.18and2.23that f1−f4
xyz f2−f5
x−y f3−f6
x−z f1−f4
x−y−z f2−f5
xy f3−f6
−xz≤ϕ−x, y, z ϕx, y, z 2.44
for allx, y, z∈V \ {0}. Let the functionsF, G, H:V→Xbe defined by
Fx 1
2
f1x−f4x
, Gx 1
2
f2x−f5x
, Hx 1
2
f3x−f6x
2.45
for allx, y, z∈V.
From2.44, we have
Fxyz Gx−y Hx−z Fx−y−z Gxy Hxz≤ϕx, y, z
2.46
for allx, y, z∈V \ {0}. Replacey,zbyxin2.46to get
F3x Fx G2x G0 H2x H0≤ϕx, x, x 2.47
for allx, y, z∈V \ {0}. Replacex,y,zbyx, 3x,−xin2.46to get
F3x Fx G2x G4x H2x H0≤ϕx,3x,−x 2.48
for allx, y, z∈V \ {0}. From2.47and the above inequality, we have
G4x−G0≤ϕx,3x,−x ϕx, x, x 2.49
for allx∈V \ {0}.
Replacex,y,zbyx,x,−3xin2.46to get
for allx, y, z∈V \ {0}. From2.47and the above inequality, we get
H4x−H0≤ϕx, x,−3x ϕx, x, x 2.51
for allx∈V \ {0}. It follows from2.46that
F4x−F0≤F0 G0 H3x F2x G2x Hx
F4x G0 Hx F2x G2x H3x
≤ϕx, x,−2x ϕx, x,2x
2.52
for allx∈V \ {0}. By the definitions ofF, G, H, F, G, H, we have f1x−f10−Qx Fx Fx−F0−Qx,
f2x−f20−Qx Gx Gx−F0−Qx,
f3x−f30−Qx Hx Hx−H0−Qx,
f4x−f40−Qx Fx−Fx F0−Qx,
f5x−f50−Qx Gx−Gx G0−Qx,
f6x−f60−Qx Hx−Hx H0−Qx
2.53
for allx∈V \ {0}. Hence by using2.32,2.34,2.36,2.37,2.38,2.49,2.51, and2.52, the inequalities in2.19can be shown. The uniqueness ofQfollows fromLemma 2.1.
Theorem 2.3. Letϕ:V \ {0} ×V \ {0} ×V \ {0}→0,∞be a function such that
ϕx, y, z:∞ l0
4lϕ x
2l1,
y 2l1,
z 2l1
<∞ a
holds for all x, y, z ∈ V \ {0}. If the even functions f1, f2, f3, f4, f5, f6 : V→X satisfy inequality
2.18for allx, y, z ∈ V \ {0}, then there exists exactly one quadratic functionQ : V→Xsatisfying inequalities2.19for allx∈V \ {0}, where
Mx:∞ l0
4lM x
2l1
, Mx:∞ l0
4lM x
2l1
. 2.54
Moreover, the functionQis given by
Qx lim n→∞4
nf
k2−nx−fk0 2.55 for allx∈V and fork1,2,3,4,5,6.
Proof. The proof is similar to that ofTheorem 2.2.
Applying Theorems 2.2and2.3, we get the following corollary in the sense of Rassias inequality.
Corollary 2.4. Letp /2andε >0. If the even functionsfi:V→X,i1,2, . . . ,6, satisfy f1xyz f2x−y f3x−z−f4x−y−z−f5xy−f6xz
≤εxpypzp 2.56
Then there exist exactly one quadratic functionQ:V→Xsatisfying f1x−f10−Qx≤
1
3p112p2 2·2p2p−4
73p 2·2p
4 4p
·ε·xp,
f4x−f40−Qx≤13p112p2 2·2p2p−4
73p 2·2p
4 4p
·ε·xp,
fjx−fj0−Qx≤ 3p11 2p2p−4
3p5 4p
·ε·xp
2.57
for allx∈V \ {0}andj 2,3,5,6. Moreover, the functionQis given by
Qx ⎧ ⎪ ⎨ ⎪ ⎩ lim n→∞
fk2nx
4n if p <2,
lim n→∞4
nf
k2−nx−fk0 if p >2
2.58
for allx∈V \ {0}andk1,2,3,4,5,6
Proof. ApplyTheorem 2.2forp <2 andTheorem 2.3forp >2.
We establish Theorems2.5and2.6for the odd functions.
Theorem 2.5. Letϕ:V \ {0} ×V \ {0} ×V \ {0}→0,∞be a function such that
ϕx, y, z:∞ l0
1 2l1ϕ
2lx,2ly,2lz<∞ b
holds for allx, y, z∈V \ {0}. If the odd functionsf1, f2, f3, f4, f5, f6:V→Xsatisfy
f1xyz f2x−y f3x−z−f4x−y−z−f5xy−f6xz≤ϕx, y, z
2.59 for allx, y, z∈V \ {0}, then there exist exactly three additive functionsA, A1, A2:V→Xsatisfying
f1x−Ax A1x A2x≤ϕx
4, x 4,−
x 2
ϕx
4, x 4, x 2 2M x 2
ϕx
2, x 2, x
ϕx
2,− x 2, x
,
2.60
f2x−Ax−A1x≤Mx ϕ
x
2, 3x
2 ,− x 2
ϕx
2, x 2, x 2 ,
f3x−Ax−A2x≤Mx ϕx
2,− x 2,
3x 2
ϕx
2, x 2, x 2 ,
f4x−Ax−A1x−A2x≤ϕx
4, x 4,−
x 2
ϕx
4, x 4, x 2 2M x 2
ϕx
2, x 2, x
ϕx
2,− x 2, x
,
f5x−Ax A1x≤Mx ϕx
2, 3x
2 ,− x 2
ϕx
2, x 2, x 2 ,
f6x−Ax A2x≤Mx ϕx
2,− x 2,
3x 2
ϕx
for allx∈V \ {0}, where
Mx:∞ l0
1 2l1M
2lx, Mx:
∞
l0
1 2l1M
2lx,
ϕx, y, z:∞
l0
1 2l1ϕ
2lx,2ly,2lz. 2.62
Moreover, the functionsA, A1, A2are given by
Ax lim n→∞
f1
2nxf4
2nx 2n1 ,
A1x nlim→∞
f2
2nx−f5
2nx 2n1 ,
A2x nlim→∞
f3
2nx−f6
2nx 2n1
2.63
for allx∈V.
Proof. Replacexby−xin2.59to obtain
−f1x−y−z−f2xy−f3xz f4xyz f5x−y f6x−z≤ϕ−x, y, z 2.64
for allx, y, z∈V \ {0}. Let the functionsF, G, H:V→Xbe defined by
Fx 1 2
f1x f4x
, Gx 1
2
f2x f5x
, Hx 1
2
f3x f6x
2.65
for allx, y, z∈V. From2.59and2.64, we get
Fxyz Gx−y Hx−z−Fx−y−z−Gxy−Hxz≤ϕx, y, z
2.66
for allx, y, z∈V \ {0}. From2.66, we have
H2x−G2x≤ϕx, x,−x, 2.67
for allx∈V \ {0}. It follows from2.66and2.67that
G4x−2G2x−F3x−Fx G2x G4x−H2x
2H2x−2G2xF3x Fx−G2x−H2x
≤ϕx, x, x 2ϕx, x,−x ϕx,3x,−x
2.68
for allx∈V \ {0}. Replacingxbyx/2 and dividing it by 2 in the above inequality, we obtain
Gx−G2x
2
≤ Mx
for allx∈V \ {0}. ApplyingLemma 2.1, we obtain
Gx− lim n→∞
G2nx 2n
≤Mx 2.70
for allx∈V \ {0}. Similarly we have
Hx− lim n→∞
H2nx 2n
≤Mx 2.71
for allx∈V \ {0}. From2.67, we get
lim n→∞
G2nx 2n nlim→∞
H2nx
2n 2.72
for allx∈V and we can define a functionA:V→Xby
Ax lim n→∞
G2nx 2n nlim→∞
H2nx
2n 2.73
for allx∈V \ {0}. It follows from2.66and2.70that Fx−AxFx−Hx
4
Fx 2
−Gx 2
−H3x 4
H3x 4
−Fx 2
−Gx 2
Hx 4
2G
x
2
−2A x
2
≤ϕx
4, x 4,−
x 2
ϕx
4, x 4,
x 2
2M
x
2
2.74
for allx∈V \ {0}. Replacingxby 2nx, dividing it by 2nin the above inequality and taking the limit in the resulted inequality asn→∞, we obtain
lim n→∞
F2nx
2n Ax 2.75
for allx∈V \ {0}. From2.73and2.75, we have
Axyz Ax−y Ax−z−Ax−y−z−Axy−Axz 0 2.76
for allx, y, z∈V \ {0}. Replaceyandzby 2yandxin2.76to obtain
A2x2y Ax−2y A2y−Ax2y−A2x 0 2.77
for allx, y, z∈V \ {0}. Replaceyandzby−2yandxin2.76to get
for allx, y ∈ V \ {0}. Since A0 0 andA2x 2Ax, using the above two equalities, we have
Ax−y Axy−A2x 0 2.79
for allx, y ∈V. Hence,Ais an additive function. Let the functionsF, G, H:V→Xbe defined by
Fx 1
2
f1x−f4x
, Gx 1
2
f2x−f5x
, Hx 1
2
f3x−f6x
2.80
for allx, y, z∈V. From2.59and2.64, we have
Fxyz Gx−y Hx−z Fx−y−z Gxy Hxz≤ϕx, y, z
2.81
for allx, y, z∈V \ {0}. It follows from2.81that
Gx−G2x
2 ≤ 1
2 F3x
2
−Fx
2
−Gx G2x Hx
1 2
F3x
2
−Fx
2
Gx Hx
≤ 1
2ϕ
x
2, 3x
2 ,− x 2
1
2ϕ
x
2, x 2,
x 2
2.82
for allx∈V \ {0}. ApplyingLemma 2.1, we obtain an odd functionA1:V→Xdefined by
A1x nlim→∞
G2nx
2n ; 2.83
and the inequality
Gx−A
1x≤ϕ
x
2, 3x
2 ,− x 2
ϕx
2, x 2,
x 2
2.84
holds for allx∈V \ {0}. Similarly we have an odd functionA2:V→Xdefined by
A2x nlim→∞
H2nx
2n 2.85
for allx∈V and the inequality Hx−A
2x≤ϕ
x
2,− x 2,
3x 2
ϕx
2, x 2,
x 2
2.86
for allx∈V \ {0}. Replacex,y,zbyx,x,−xin2.81to get
for allx∈V\ {0}. Replacingxby 2n−1xand dividing it by 2nin the above inequality, we obtain
2F2n−1xG2n2nxH2nx ≤ ϕ
2nx,2nx,−2nx
2n 2.88
for allx∈V \ {0}. Taking the limit in the above inequality asn→∞, we have
lim n→∞
F2nx
2n −A1x−A2x 2.89
for allx∈V \ {0}. It follows from2.81that
Fx− F2x
2 ≤ 1
2
F2x G0−Hx
2
−Fx Gx H3x
2
1 2
Fx Gx−Hx
2
F0 G0 H3x
2
≤ 1
2
ϕx
2, x 2, x
ϕx
2, x 2,−x
2.90
for allx∈V \ {0}. ApplyingLemma 2.1and2.89, we have Fx A
1x A2x≤ϕ
x
2, x 2, x
ϕx
2, x 2,−x
2.91
for allx∈V \ {0}. From2.81,2.83,2.85, and2.89, we have
−A1xyz−A2xyz A1x−y A2x−z−A1x−y−z
−A2x−y−z A1xy A2xz 0
2.92
for allx, y, z∈V \ {0}. Replaceyandzby 2yandxin2.92to get
−A12x2y−A22x2y A1x−2y A12y A22y A22x A1x2y 0
2.93
for allx, y ∈V \ {0}. Replaceyandzbyxand 2yin2.92to get
−A12x2y−A22x2y A2x−2y A12y A22y A12x A2x2y 0
2.94
for allx, y ∈V \ {0}. From the above two equalities, we get
A1−A2
x−2y−A1−A2
2x A1−A2
x2y 0 2.95
for allx, y ∈V \ {0}. SinceA0 0, we have
A1−A2
x−2y−A1−A2
2x A1−A2
for allx, y ∈V. HenceA1−A2is additive, that is,
A1−A2
xy A1−A2
x A1−A2
y 2.97
for allx, y ∈V. Replacezby−yin2.92to obtain
−A1x−A2x A1x−y A2xy−A1x−A2x A1xy A2x−y 0 2.98
for allx, y ∈V \ {0}. SinceA1−A2is additive, we have
A22x−A2xy−A2x−y A12x−A1xy−A1x−y 2.99
for allx, y ∈V \ {0}. From this and2.98, we get
−A14x 2A1x−y 2A1xy 0 2.100
for allx, y ∈V \ {0}. From this andA10 0, we have
A1xy A1x A1y 2.101
for allx, y ∈V. SinceA1andA1−A2are additive,A2is additive.
From2.74,2.91, and the definitions ofF, F, we have f1x−Ax A1x A2x≤Fx−AxFx A
1x A2x
≤ϕx
4, x 4,−
x 2
ϕx
4, x 4, x 2 2M x 2
ϕx
2, x 2, x
ϕx
2, x 2,−x
2.102
for allx∈V \ {0}. The rest of inequalities in2.60can be shown by the similar method.
Theorem 2.6. Letϕ:V \ {0} ×V \ {0} ×V \ {0}→0,∞be a function such that
ϕx, y, z:∞ l0
2lϕ x
2l1,
y 2l1,
z 2l1
<∞ b
holds for allx, y, z∈V\{0}. If the odd functionsf1, f2, f3, f4, f5, f6:V→Xsatisfy inequalities2.59
for allx, y, z∈V \ {0}, then there exist exactly three additive functionsA, A1, A2 :V→Xsatisfying
the inequalities2.60for allx∈V \ {0}, where
Mx:∞ l0
2lM x
2l1
, Mx:∞
l0
2lM x
2l1
,
ϕx, y, z:∞
l0
2lϕ x
2l1,
y 2l1,
z 2l1
.
2.103
Moreover, the functionsA, A1, A2are given by
Ax lim n→∞2
n−2
f1 x 2n f4 x 2n
−f1
− x
2n
−f4
− x
2n
,
A1x nlim→∞2n−2
f2
x
2n
−f5
x
2n
−f2
− x 2n f5 − x 2n ,
A2x nlim→∞2n−2
f3
x
2n
−f6
x
2n
−f3
− x 2n f6 − x 2n 2.104
Proof. The proof is similar to that ofTheorem 2.5.
Applying Theorems 2.5and2.6, we get the following corollary in the sense of Rassias inequality.
Corollary 2.7. Letp /1. If the odd functionsfi:V→X,i1,2, . . . ,6, satisfy
f1xyz f2x−y f3x−z−f4x−y−z−f5xy−f6xz
≤εxpypzp 2.105
for allx, y, z∈V \ {0}.
Then there exist exactly three additive functionsA, A1, A2:V→Xsatisfying
f1x−f10−Ax A1x A2x≤ 2 2p
4 4p
23p114·2p2·4p 4p2p−2
·ε·xp,
f2x−f20−Ax−A1x≤
23p8 2p2p−2·ε·x
p,
f3x−f30−Ax−A2x≤ 23p8 2p2p−2·ε·x
p,
f4x−f40−Ax−A1x−A2x≤
2 2p
4 4p
23p114·2p2·4p 4p2p−2
·ε·xp,
f5x−f50−Ax A1x≤
23p8 2p2p−2·ε·x
p,
f6x−f60−Ax A2x≤ 23p8 2p2p−2·ε·x
p
2.106
for allx∈V \ {0}. Moreover, the functionsA, A1, A2are given by
Ax ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ lim n→∞ f1
2nxf4
2nx−f1
−2nx−f4
−2nx
2n2 , if p <1,
lim n→∞2
n−2
f1 x 2n f4 x 2n
−f1
− x
2n
−f4
− x
2n
, if p >1,
A1x
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ lim n→∞ f2
2nx−f5
2nx−f2
−2nxf5
−2nx
2n2 , if p <1,
lim n→∞2
n−2
f2
x
2n
−f5
x
2n
−f2
− x 2n f5 − x 2n
, if p >1,
A2x
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ lim n→∞ f3
2nx−f6
2nx−f3
−2nxf6
−2nx
2n2 , if p <1,
lim n→∞2
n−2
f3
x
2n
−f6
x
2n
−f3
− x 2n f6 − x 2n
, if p >1
2.107
Proof. ApplyTheorem 2.5forp <1 andTheorem 2.6forp >1.
We establish the following theorem for the general case from Theorems2.2and2.5.
Theorem 2.8. Letϕ:V \ {0} ×V \ {0} ×V \ {0}→0,∞be a function that satisfies conditionsa andb. Suppose that the functionsfi:V→X,i1,2, . . . ,6, satisfy the inequality
f1xyz f2x−y f3x−z−f4x−y−z−f5xy−f6xz≤ϕx, y, z
2.108
for allx, y, z∈ V \ {0}. Then there exist exactly one quadratic functionQ : V→Xand exactly three additive functionsA, A1, A2:V→Xsatisfying
f1x−f10−Qx−Ax A1x A2x
≤ 1 2 ϕ e x 2, 3 2x,−x
ϕ e x 2, x 2,−x
2ϕe
x
4, x 4,−
x 2
2ϕe
x 4, x 4, x 2 1
2Me2x Mex 2Me x 2 ϕ e x 2, x 2, x
ϕ
e x
2,− x 2, x
,
f2x−f20−Qx−Ax−A1x
≤Mex ϕ
e x
4, 3x
4 ,− x 4 ϕ e x 4, x 4, x 4
Mex ϕ
e x
2, 3x
2 ,− x 2 ϕ e x 2, x 2, x 2 ,
f3x−f30−Qx−Ax−A2x
≤M
ex ϕe x
4, x 4,−
3x 4 ϕ e x 4, x 4, x 4 M
ex ϕe x
2, x 2,−
3x 2 ϕ e x 2, x 2, x 2 ,
f4x−f40−Qx−Ax−A1x−A2x
≤ 1 2 ϕ e x 2, 3 2x,−x
ϕ e x 2, x 2,−x
2ϕe
x
4, x 4,−
x 2
2ϕe
x 4, x 4, x 2 1
2Me2x Mex 2Me x 2 ϕ e x 2, x 2, x
ϕ
e x
2,− x 2, x
,
f5x−f50−Qx−Ax A1x
≤Mex ϕ
e x
4, 3x
4 ,− x 4 ϕ e x 4, x 4, x 4
Mex ϕ
e x
2, 3x
2 ,− x 2 ϕ e x 2, x 2, x 2 ,
f6x−f60−Qx−Ax A2x
≤Mx ϕ
e x
4, x 4,−
3x 4 ϕ e x 4, x 4, x 4 M
ex ϕe x
2, x 2,−
for allx∈V \ {0}, where
Mex:∞ l0
1 4l1Me
2lx, Mex:
∞
l0
1 4l1M
e2lx,
Me:
∞
l0
1 2l1Me
2lx, Me:
∞
l0
1 2l1Me
2lx,
ϕex, y, z:∞
l0
1 2l1ϕ
e2lx,2ly,2lz
2.110
for allx∈V \ {0}. Moreover, the functionQis given by
Qx lim n→∞
fk2nxfk−2nx
2·4n 2.111
fori1,2,3,4,5,6and the functionsA, A1, A2are given by
Ax lim n→∞
f1
2nxf4
2nx−f1
−2nx−f4
−2nx 2n2 ,
A1x nlim→∞
f2
2nx−f5
2nx−f2
−2nxf5
−2nx 2n2 ,
A2x nlim→∞
f3
2nx−f6
2nx−f3
−2nxf6
−2nx 2n2
2.112
for all, x∈V.
Proof. From2.108, we obtain
f1−x−y−z f2−xy f3−xz−f4−xyz−f5−x−y−f6−x−z
≤ϕ−x,−y,−z 2.113
for allx, y, z∈V \ {0}. From2.108and this inequality, one gets
f1exyz f2ex−y f3ex−z−f4ex−y−z−f5exy−f6exz≤ϕex, y, z,
f1oxyz f2ox−y f3ox−z−f4ox−y−z−f5oxy−f6oxz≤ϕex, y, z
2.114
for all x, y, z ∈ V \ {0}, wherefkex fkx fk−x/2, fkox fkx−fk−x/2 for all x ∈ V \ {0}, k 1,2,3,4,5,6. Since fke is an even function, fko is an odd function, and fk fkefko, we can apply Theorems2.2and2.5to get the desired result.
We establish the following theorem for the general case from Theorems2.2and2.6.
Theorem 2.9. Letϕ : V \ {0} ×V \ {0} ×V \ {0}→0,∞be a function that satisfies conditions aandb. If the functionsf1, f2, f3, f4, f5, f6 : V→X satisfy inequalities2.108for allx, y, z ∈
A, A1, A2:V→Xsatisfying the inequalities inTheorem 2.8for allx∈V \ {0}, whereMe, Meare as inTheorem 2.8and
Mex:∞ l0
2lMe x
2l1
, M
ex:
∞
l0
2lMe x
2l1
,
ϕex, y, z:∞
l0
2lϕe x
2l1,
y 2l1,
z 2l1
2.115
for allx∈V\ {0}. Moreover, the functionQis given by2.111and the functionsA, A1, A2are given
by
Ax lim n→∞2
n−2
f1
x
2n
f4
x
2n
−f1
− x
2n
−f4
− x
2n
, 2.116
A1x nlim→∞2n−2
f2
x
2n
−f5
x
2n
−f2
− x
2n
f5
− x
2n
,
A2x nlim→∞2n−2
f3
x
2n
−f6
x
2n
−f3
− x
2n
f6
− x
2n
2.117
for allx∈V.
We establish the following theorem for the general case from Theorems2.3and2.6.
Theorem 2.10. Letϕ : V \ {0} ×V \ {0} ×V \ {0}→0,∞be a function that satisfies conditions aandb. If the functions f1, f2, f3, f4, f5, f6 : V→Xsatisfy inequalities 2.108for all x, y, z ∈
V \ {0}, then there exist exactly one quadratic functionQ:V→Xand exactly three additive functions A, A1, A2:V→Xsatisfying the inequalities inTheorem 2.8for allx∈V \ {0}, whereMe, Me,ϕeare as inTheorem 2.9and
Mex:∞ l0
4lMe x
2l1
, M
ex:
∞
l0
4lMe x
2l1
2.118
for allx∈V \ {0}. Moreover, the functionQis given by
Qx lim n→∞2·4
n−1f
k2−nxfk−2−nx−2fk0 2.119
fori1,2,3,4,5,6and the functionsA, A1, A2are given by2.116for allx∈V.
Corollary 2.11. Letp /1,2andε >0. Suppose that the functionsfi:V→X,i1,2, . . . ,6, satisfy f1xyz f2x−y f3x−z−f4x−y−z−f5xy−f6xz
≤εxpypzp 2.120
Then there exist exactly one quadratic function Q : V→X and three additive functions A, A1, A2:V→Xsatisfying
f1x−f10−Qx−Ax A1x A2x
≤
3p112p2
2·2p2p−4
113p 2·2p 1
8 4p
23p114·2p2·4p 4p2p−2
·ε·xp,
f2x−f20−Qx−Ax−A1x≤ 3p11 2p2p−4
3p5 4p
23p8 2p2p−2
·ε·xp,
f3x−f30−Qx−Ax−A2x≤ 3p11 2p2p−4
3p5 4p
23p8 2p2p−2
·ε·xp,
f4x−f40−Qx−Ax−A1x−A2x
≤
3p112p2
2·2p2p−4
113p 2·2p 1
8 4p
23p114·2p2·4p 4p2p−2
·ε·xp
f5x−f50−Qx−Ax A1x≤
3p11 2p2p−4
3p5 4p
23p8 2p2p−2
·ε·xp,
f6x−f60−Qx−Ax A2x≤ 3p11 2p2p−4
3p5 4p
23p8 2p2p−2
·ε·xp
2.121
for allx∈V \ {0}. Moreover, the functionQis given by2.111forp < 2and2.119forp > 2and the functionsA, A1, A2k1,2,3are given by2.112forp <1and2.116forp >1.
Corollary 2.12. Letε >0be a fixed real number. Suppose that the functionsfi:V→X,i1,2, . . . ,6, satisfy
f1xyz f2x−y f3x−z−f4x−y−z−f5xy−f6xz≤ε 2.122 for allx, y, z∈V \ {0}.
Then there exist exactly one quadratic function Q : V→X and three additive functions A, A1, A2:V→Xsatisfying
f1x−f10−Qx−Ax A1x A2x≤17ε,
f2x−f20−Qx−Ax−A1x≤ 28
3 ε, f3x−f30−Qx−Ax−A2x≤ 28
3 ε,
f4x−f40−Qx−Ax−A1x−A2x≤17ε,
f5x−f50−Qx−Ax A1x≤ 28
3 ε, f6x−f60−Qx−Ax A2x28
3 ε
