Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2010, Article ID 476249, 15 pages
doi:10.1155/2010/476249
Research Article
Hyers-Ulam Stability of a Bi-Jensen Functional
Equation on a Punctured Domain
Gwang Hui Kim1 and Yang-Hi Lee2
1
2
Department of Mathematics, Kangnam University, Yongin, Gyeonggi 446-702, South Korea
Department of Mathematics Education, Gongju National University of Education,
Gongju 314-711, South Korea
Correspondence should be addressed to Gwang Hui Kim, ghkim@kangnam.ac.kr
Received 16 November 2009; Revised 8 February 2010; Accepted 15 February 2010
Academic Editor: Yong Zhou
Copyright q 2010 G. H. Kim and Y.-H. Lee. 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.
We obtain the Hyers-Ulam stability of a bi-Jensen functional equation: 2fx y/2, z − fx, z −
fy, z 0 and simultaneously 2fx, y z/2 − fx, y − fx, z 0. And we get its stability on
the punctured domain.
1. Introduction
In 1940, Ulam 1 raised a question concerning the stability of homomorphisms: let G1 be a
group and let G2 be a metric group with the metric d·, ·. Given ε > 0, does there exist a δ > 0
such that if a mapping h : G1 → G2 satisfies the inequality
d h xy , hxh y < δ
1.1
for all x, y ∈ G1 , then there is a homomorphism H : G1 → G2 with
dhx, Hx < ε
1.2
for all x ∈ G1 ? The case of approximately additive mappings was solved by Hyers 2 under
the assumption that G1 and G2 are Banach spaces. In 1949, 1950, and 1978, Bourgin 3, Aoki
4, and Rassias 5 gave a generalization of it under the conditions bounded by variables.
Since then, the further generalization has been extensively investigated by a number of
mathematicians, such as Găvruta, Rassias, and so forth, 6–25.
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Journal of Inequalities and Applications
Throughout this paper, let X be a normed space and Y a Banach space. A mapping
g : X → Y is called a Jensen mapping if g satisfies the functional equation 2gx y/2 gx gy. For a given mapping f : X × X → Y , we define
xy
, z − fx, z − f y, z ,
J1 f x, y, z : 2f
2
yz
J2 f x, y, z : 2f x,
− f x, y − fx, z
2
1.3
for all x, y, z ∈ X. A mapping f : X × X → Y is called a bi-Jensen mapping if f satisfies the
functional equations J1 f 0 and J2 f 0.
In 2006, Bae and Park 26 obtained the generalized Hyers-Ulam stability of a bi-Jensen
mapping. The following result is a special case of Theorem 6 in 26.
Theorem A. Let ε > 0 and let f : X × X → Y be a mapping such that
J1 f x, y, z ≤ ε,
J2 f x, y, z ≤ ε
1.4
for all x, y, z ∈ X. Then there exist two bi-Jensen mappings F, F0 : X × X → Y such that
f x, y − f 0, y − F x, y ≤ ε,
f x, y − fx, 0 − F0 x, y ≤ ε
1.5
for all x, y ∈ X.
In Theorem A, they did not show that there exist a k ∈ R and a unique bi-Jensen
mapping F : X × X → Y such that fx, y − Fx, y ≤ kε for all x, y ∈ X. In 2008, Jun et al.
7, 8 improved Bae and Park’s results.
In Section 2, we show that there exists a unique bi-Jensen mapping F : X × X → Y
such that fx, y − Fx, y ≤ 4ε for all x, y ∈ X. In Section 3, we investigate the Hyers-Ulam
stability of a bi-Jensen functional equation on the punctured domain.
2. Stability of a Bi-Jensen Functional Equation
From Lemma 1 in 8, we get the following lemma.
Lemma 2.1. Let f : X × X → Y be a bi-Jensen mapping. Then
1 f x, y n f 2n x, 2n y 4
n 1
1 1 2
n
−
f2 x, 0 f 0, 2 y 1 − n f0, 0
2n 4n
2
for all x, y ∈ X and n ∈ N.
Now we will give the Hyers-Ulam stability for a bi-Jensen mapping.
2.1
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3
Theorem 2.2. Let ε > 0 and let f : X × X → Y be a mapping satisfying 1.4 for all x, y, z ∈ X.
Then there exists a unique bi-Jensen mapping F : X × X → Y such that
f x, y − F x, y ≤ 4ε
2.2
for all x, y ∈ X with F0, 0 f0, 0. In particular, the mapping F : X × X → Y is given by
1
1 j
1 j
F x, y : lim j f 2j x, 2j y −
x,
0
f
0,
2
y
f0, 0
f
2
j →∞ 4
2j 4j
2.3
for all x, y ∈ X.
Proof. Let fj be the map defined by
f 2j x, 2j y
1
1 j
1
1
j
fj x, y −
x,
0
f
0,
2
y
1
−
f
2
f0, 0 2.4
4j
2j 4j
2j−1 4j
for all x, y ∈ X and j ∈ N. By 1.4, we get
J1 f 2j1 x, 0, 0
J1 f 2j1 x, 0, 2j1 y
J1 f 2j1 x, 0, 2j y
fj x, y − fj1 x, y 2j1
2 · 4j1
4j1
−
3J1 f 2j1 x, 0, 0
2 · 4j1
J2 f 0, 0, 2j1 y
J2 f 2j1 x, 0, 2j1 y
2j1
J2 f 2j x, 0, 2j1 y
3J2 f 0, 0, 2j1 y −
j1
j1
4
2·4
≤
2 · 4j1
2.5
1
3
ε
2j 2 · 4j
for all x, y ∈ X and j ∈ N. For given integers l, m with 0 ≤ l < m, we obtain
1
m−1
3
fl x, y − fm x, y ≤
ε
2j 2 · 4j
jl
2.6
for all x, y ∈ X. By the above inequality, the sequence {fj x, y} is a Cauchy sequence for
all x, y ∈ X. Since Y is complete, the sequence {fj x, y} converges for all x, y ∈ X. Define
F : X × X → Y by
F x, y : lim fj x, y
j →∞
2.7
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Journal of Inequalities and Applications
for all x, y ∈ X. Putting l 0 and taking m → ∞ in 2.6, we obtain the inequality
f x, y − F x, y ≤ 4ε
2.8
for all x, y ∈ X. By 1.4 and the definition of F, we get
1
1
1 j
j
j
j
J1 F x, y, z lim j J1 f 2 x, 2 y −
f
2
x,
0
J
f
0,
2
y
0,
J
1
1
j →∞4
2j 4j
1
1
1 j
j
j
j
J2 F3 x, y, z lim j J2 f 2 x, 2 y −
f
2
x,
0
J
f
0,
2
y
0
J
2
2
j →∞4
2j 4j
2.9
for all x, y, z ∈ X. So F is a bi-Jensen mapping satisfying 2.2. Now, let F : X × X → Y be
another bi-Jensen mapping satisfying 2.2 with F 0, 0 f0, 0. By Lemma 2.1, we have
F x, y − F x, y F − F 2n x, 2n y 1
n
n 1 −
F − F 2 x, 0 F − F 0, 2 y 4n
2n 4n
F − f 2n x, 2n y F − f 0, 2n y F − f 2n x, 0 ≤
4n
2n
2n
f − F 2n x, 2n y f − F 0, 2n y f − F 2n x, 0 4n
2n
2n
1
1
≤
ε
2n−2 4n−1
2.10
for all x, y ∈ X and n ∈ N. As n → ∞, we may conclude that Fx, y F x, y for all x, y ∈ X.
Thus the bi-Jensen mapping F : X × X → Y is unique.
Example 2.3. Let f, F, F : R × R → R be the bi-Jensen mappings defined by
f x, y : 0,
F x, y : ε,
F x, y : −ε
2.11
for all x, y ∈ R. Then f, F, F satisfy 1.4 for all x, y, z ∈ R. In addition, f, F satisfy 2.2 for
F. Hence the condition
all x, y ∈ R and f, F also satisfy 2.2 for all x, y ∈ R. But we get F /
F0, 0 f0, 0 is necessary to show that the mapping F is unique.
Journal of Inequalities and Applications
5
3. Stability of a Bi-Jensen Functional Equation on
the Punctured Domain
Let A be a subset of X. X \ A and X × X \ A × A are punctured domain on the spaces X
and X × X, respectively.
Throughout this paper, for a given mapping f : X × X → Y , let f1 , A1 , A2 : X × X → Y
be the mappings defined by
f x, y − f −x, y − f x, −y f −x, −y
,
f1 x, y :
4
1
1 A1 x, y :
−1m1 J1 f −1m x, −1n · 3x, y ,
m0 n0
3.1
1 1
A2 x, y :
−1n1 J2 f x, −1m · 3y, −1n y
m0 n0
for all x, y ∈ X.
Lemma 3.1. Let A be a subset of X satisfying the following condition: for every x /
0, there exists
a positive integer nx such that kx /
∈ A for all integer k with |k| ≥ nx , and such that kx ∈ A for all
integer k with |k| < nx . Let f : X × X → Y be a mapping such that
J1 f x, y, z 0,
J2 f x, y, z 0
3.2
for all x, y, z ∈ X \ A. Then there exists a unique bi-Jensen mapping F : X × X → Y such that
F x, y f x, y
3.3
for all x, y ∈ X \ A. Moreover, the equality
F x, y f x, y
3.4
holds for all x, y ∈ X × X \ A × A.
Proof. Note that J1 fx, y, z 0, J2 fx, y, z 0, A1 x, y 0, and A2 x, y 0 for all
x, y ∈ X \ A. Let f0, y f0, −y/2 c ∈ Y for any y ∈ X \ A. From 3.2, we get the
equality
1 f 0, y f 0, −y fx, 0 f−x, 0 J1 f x, −x, y J1 f x, −x, −y − J2 f x, y, −y
2
−J2 f −x, y, −y
3.5
for all x, y ∈ X \ A, and we know that the equality
f 0, y f 0, −y
fx, 0 f−x, 0
c
2
2
3.6
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Journal of Inequalities and Applications
holds for all x, y ∈ X \ A. From 3.2, we have
A1 x, y
A1 x, −y
f1 2x, y
−
,
f1 x, y 2
16
16
f1 2x, 2y
A2 2x, y
A2 −2x, y
f1 2x, y
−
,
2
4
32
32
f 0, y f 0, −y
f 0, y − f 0, −y
fx, 0 − f−x, 0
f x, y 2
2
2
1
f1 x, y − 2J1 f x, −x, y J2 f x, y, −y − J2 f −x, y, −y ,
4
fx, 0 − f−x, 0 f2x, 0 − f−2x, 0
2
3.7
1
4J2 f x, y, −y − 4J1 f −x, y, −y − 2J1 f 2x, y, −y
8
2J1 f −2x, y, −y A1 x, y A1 x, −y ,
f 0, 2y − f 0, −2y
f 0, y − f 0, −y 2
1
4J1 f x, −x, y − 4J1 f x, −x, −y − 2J1 f x, −x, 2y
8
2J1 f x, −x, −2y A2 x, y A2 −x, y
for all x, y ∈ X \ A. From the above equalities, we obtain the equalities
f1 2x, y
,
f1 x, y 2
f2x, 0 − f−2x, 0
,
2
f 0, 2y − f 0, −2y
,
f 0, y − f 0, −y 2
f1 2n x, 2n y
f1 x, y ,
4n
fx, 0 − f−x, 0 fx, 0 − f−x, 0 fx, 0 f−x, 0 f2n x, 0 − f−2n x, 0
c,
2
2
2n1
f 0, y f 0, −y
f 0, 2n y − f 0, −2n y
f 0, y − f 0, −y
f 0, y c,
2
2
2n1
f 0, 2n y − f 0, −2n y f2n x, 0 − f−2n x, 0
f1 2n x, 2n y
f x, y c
4n
2n1
3.8
3.9
3.10
fx, 0 for all x, y ∈ X \ A and n ∈ N.
3.11
Journal of Inequalities and Applications
7
∈ A} for each x /
0. From the above
Let Ax be the set defined by Ax {n ∈ N | nx /
equalities, we can define F : X × X → Y by
⎧ k
f 0, 2k y − f 0, −2k y f 2k x, 0 − f −2k x, 0
f1 2 x, 2k y
⎪
⎪
⎪
c,
⎪
⎪
4k
2k1
⎪
⎪
⎪
⎪
⎪
for some 2k ∈ Ax ∩ Ay if x, y / 0,
⎪
⎪
⎪ ⎨
k
k
F x, y : f 2 x, 0 − f −2 x, 0 c, for some 2k ∈ Ax if x /
0, y 0,
⎪
k1
⎪
⎪
k 2 ⎪
⎪
⎪
f 0, 2 y − f 0, −2k y
⎪
⎪
⎪
c for some 2k ∈ Ay if x 0, y / 0,
⎪
k1
⎪
2
⎪
⎪
⎩
c
if x, y 0.
3.12
From the definition of F, we get the equalities
F x, y f x, y ,
F 0, y f 0, y ,
Fx, 0 fx, 0
3.13
for all x, y ∈ X \ A. By 3.10, we get the equality
1 f x, y − F x, y J2 f x, 2k 2 y, −2k y − J2 F x, 2k 2 y, −2k y 0
2
3.14
for all x ∈ X \ A and y /
0, where 2k ∈ Ay . And also we get the equality
1 f x, y − F x, y J1 f 2k 2 x, −2k x, y − J1 F 2k 2 x, −2k x, y 0
2
3.15
for all x /
0 and y ∈ X \ A, where 2k ∈ Ax . Hence the equality
f x, y F x, y
3.16
holds for all x, y ∈ X × X \ A × A. From 3.8, 3.9, 3.10, and the definition of F, we
easily get
J1 F x, −x, y 0,
J1 F x, 0, y 0,
J1 Fx, −x, 0 0,
J1 Fx, 0, 0 0,
J1 F 0, 0, y 0,
J1 F0, 0, 0 0
3.17
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Journal of Inequalities and Applications
for all x, y /
0. And we obtain
J2 f 2k−1 x y , z, −z − J2 f −2k−1 x y , z, −z
J1 F x, y, 0 2k1
−J2 f 2k x, z, −z J2 f−2x x, z, −z −J2 f 2k y, z, −z
2k2
2k2
k
k
J2 f −2 y, z, −z J1 f 2 x, 2k y, z − J1 f −2k x, −2k y, z
2k2
k
J1 f 2 x, 2k y, −z − J1 f −2k x, −2k y, −z
0
2k2
3.18
for all x, y /
0 with x y /
0, where 2k ∈ Ax ∩ Ay ∩ Axy and z /
∈ A. From this, we have
J1 f 2k x, 2k y, 2k z − J1 f −2k x, −2k y, 2k z
J1 F x, y, z 4k1
k
J1 f −2 x, −2k y, −2k z − J1 f 2k x, 2k y, −2k z
J1 F x, y, 0 0
4k1
3.19
for all x, y, z /
0 with x y /
0, where 2k ∈ Ax ∩ Ay ∩ Az . From the above equalities, we get
J1 F x, y, z 0
3.20
for all x, y, z ∈ X. By the similar method, we have
J2 F x, y, z 0
3.21
for all x, y, z ∈ X. Hence F is a bi-Jensen mapping. Let F be another bi-Jensen mapping
satisfying
F x, y f x, y F x, y
3.22
for all x, y ∈ X × X \ A × A. Using the above equality, we show that the equalities
1
F x, y − F x, y J1 F k 2x, −kx, y − J1 F k 2x, −kx, y 0,
2
1
F 0, y − F 0, y J1 F kx, −kx, y − J1 F kx, −kx, y 0
2
hold for all x /
0 and y ∈ X as we desired, where k ∈ Ax .
3.23
Journal of Inequalities and Applications
9
Corollary 3.2. Let f : X × X → Y be a mapping such that
J1 f x, y, z 0,
J2 f x, y, z 0
3.24
for all x, y, z ∈ X \ {0}. Then there exists a unique bi-Jensen mapping F : X × X → Y such that
F x, y f x, y
3.25
for all x, y /
0, 0.
Example 3.3. Let f : R × R → R be the mapping defined by
f x, y :
⎧
⎨ x 3 y 4
⎩1
x, y / 0, 0,
for x, y 0, 0,
for
3.26
and let F be the mapping defined by Fx, y : x 3y 4 for all x, y ∈ R. Then the
mappings f, F satisfy the conditions of Corollary 3.2 with f0, 0 /
F0, 0.
Now, we prove the Hyers-Ulam stability of a bi-Jensen functional equation on the
punctured domain X \ A.
Theorem 3.4. Let ε > 0 and x0 ∈ X \ A. Let f : X × X → Y be a mapping such that
J2 f x, y, z ≤ ε
J1 f x, y, z ≤ ε,
3.27
for all x, y, z ∈ X \ A. Then there exists a unique bi-Jensen mapping F : X × X → Y such that
f x, y − F x, y ≤ 17 ε
2
3.28
holds for all x, y ∈ X × X \ A × A with F0, 0 fx0 , 0 f−x0 , 0/2. The mapping
F : X × X → Y is given by
F x, y : lim
j →∞
for all x, y ∈ X.
f1 2j x, 2j y
4j
f 0, 2j y f 2j x, 0
2j1
fx0 , 0 f−x0 , 0
2
3.29
10
Journal of Inequalities and Applications
Proof. By 3.27, we get
f1 2j x, 2j y f1 2j1 x, 2j1 y −
4j
4j1
1 1 1 3ε
j
j
j
j
j1
j
j1
j
j2 A1 2 x, 2 y − A1 2 x, −2 y A2 2 x, 2 y − A2 −2 x, 2 y ≤ j1 ,
2
2
4
4
j
j
j1
j1
f 0, 2 y − f 0, −2 y
f 0, 2 y − f 0, −2 y −
j1
2j2
2
1 j
j
j1
f
x,
−x,
2
y
−
4J
f
x,
−x,
−2
y
−
2J
f
x,
−x,
2
y
4J
1
1
1
2j4
5ε
2J1 f x, −x, −2j1 y A2 x, 2j y A2 −x, 2j y ≤ j2 ,
2
f 0, y f 0, −y fx, 0 f−x, 0 −
2
2
1
J1 f x, −x, y J1 f x, −x, −y − J2 f x, y, −y − J2 f −x, y, −y ≤ ε
4
3.30
for all x, y ∈ X \ A and j ∈ N. For given integers l, m 0 ≤ l < m, we have
f1 2l x, 2l y f1 2m x, 2m y m−1
3ε
−
,
≤
m
l
jl 4j1
4
4
3.31
f 0, 2l y − f 0, −2l y f 0, 2m y − f 0, −2m y m−1
5ε
−
,
≤
m1
l1
jl 2j2
2
2
3.32
f 2l x, 0 − f −2l x, 0 f2m x, 0 − f−2m x, 0 m−1
5ε
−
,
≤
m1
l1
jl 2j2
2
2
3.33
fx, 0 f−x, 0 f 0, 2m y f 0, −2m y −
≤ ε,
2
2
f 0, y f 0, −y f2m x, 0 f−2m x, 0 −
≤ε
2
2
3.34
3.35
for all x, y ∈ X \ A. The sequences {f1 2j x, 2j y/4j }, {f0, 2j y − f0, −2j y/2j1 }, and
{f2j x, 0 − f−2j x, 0/2j1 } are Cauchy sequences for all x, y ∈ X \ A. Since Y is complete,
the above sequences converge for all x, y ∈ X \ A. From 3.34 and 3.35, we have
lim
j →∞
f 0, 2j y f 0, −2j y
2j1
lim
j →∞
f 2j x, 0 f −2j x, 0
2j1
0
3.36
Journal of Inequalities and Applications
11
for all x, y ∈ X. Using the inequalities 3.31–3.35 and the above equality, we can define the
mappings F1 , F2 , F3 : X × X → Y by
F1 x, y : lim
f1 2j x, 2j y
,
4j
f 0, 2j y − f 0, −2j y
j →∞
f 0, 2j y
F2 x, y : lim
lim
,
j →∞
j →∞
2j
2j1
f 2j x, 0
f 2j x, 0 − f −2j x, 0
F3 x, y : lim
lim
j →∞
j →∞
2j
2j1
3.37
for all x, y ∈ X. By 3.27 and the definition of F1 , we obtain
J1 f 2j x, 2j y, 2j z − J1 f −2j x, −2j y, 2j z
J1 F1 x, y, z lim
j →∞
4j1
J1 f 2j x, 2j y, −2j z − J1 f −2j x, −2j y, −2j z
0,
−
4j1
J2 f 2j x, 2j y, 2j z − J2 f −2j x, 2j y, 2j z
J2 F1 x, y, z lim
j →∞
4j1
J2 f 2j x, −2j y, −2j z − J2 f −2j x, −2j y, −2j z
0
−
4j1
3.38
for all x, y, z / 0. Since J2 F2 x, y, −y 0 and
J2 F2 x, y, z lim
J1 f w, −w, 2j−1 y z
2j
j →∞
−
J1 f w, −w, 2j z
2j1
−
J1 f w, −w, 2j y
2j1
J2 f w, 2j y, 2j z
2j1
J2 f −w, 2j y, 2j z
2j1
0
3.39
for all x, y, z /
0 with y z /
0, where w /
∈ A, we have
J1 F2 x, y, z 0,
J2 F2 x, y, z 0
3.40
J2 F3 x, y, z 0
3.41
for all x, y, z /
0. Similarly, the equalities
J1 F3 x, y, z 0,
12
Journal of Inequalities and Applications
hold for all x, y, z /
0. By Lemma 3.1, There exist bi-Jensen mappings F1 , F2 , F3 : X × X → Y
such that
F1 x, y F1 x, y ,
F2 x, y F2 x, y ,
F3 x, y F3 x, y
3.42
for all x, y /
0, 0. Since the equalities
F1 0, 0 F1 x, 0 F1 −x, 0 F1 x, 0 F1 −x, 0
F1 0, 0,
2
2
F2 0, 0 F2 x, 0 F2 −x, 0 F2 x, 0 F2 −x, 0
F2 0, 0,
2
2
F3 0, 0 F3 x, 0 F3 −x, 0 F3 x, 0 F3 −x, 0
F3 0, 0
2
2
3.43
hold, F1 , F2 , F3 are bi-Jensen mappings. Putting l 0 and taking m → ∞ in 3.31, 3.32, and
3.33, one can obtain the inequalities
1 f 0, y − f 0, −y − F2 x, y ≤ 5ε ,
2
2
1
fx, 0 − f−x, 0 − F3 x, y ≤ 5ε
2
2
f1 x, y − F1 x, y ≤ ε,
3.44
for all x, y ∈ X \ A. By 3.30 and the above equalities, we get
f x, y − F x, y ≤ f x, y − f1 x, y − f 0, y − fx, 0 − f−x, 0 2
f 0, y f 0, −y fx , 0 f−x , 0 0
0
−
f1 x, y − F1 x, y 2
2
f 0, y − f 0, −y
fx, 0 − f−x, 0
− F2 x, y −
F
x,
y
3
2
2
1
1
1
≤ − J1 f x, −x, y − J2 f x, y, −y J2 f −x, y, −y 7ε
2
4
4
≤ 8ε
3.45
for all x, y ∈ X \ A, where F is given by
fx0 , 0 f−x0 , 0
F x, y F1 x, y F2 x, y F3 x, y 2
3.46
Journal of Inequalities and Applications
13
and z /
∈ A. By 3.45, we get the inequalities
f x, y − F x, y 1 J1 f k 2x, −kx, y f k 2x, y − F k 2x, y
2
17
f −kx, y − F −kx, y ≤
ε,
2
f 0, y − F 0, y 1 J1 f kx, −kx, y f kx, y − F kx, y f −kx, y − F −kx, y 2
≤
17
ε
2
3.47
for all x ∈ A, where k ∈ Ax , and the inequalities
/ 0 and y /
f x, y − F x, y ≤ 17 ε,
2
fx, 0 − Fx, 0 ≤ 17 ε
2
3.48
for all y /
0 and x /
∈ A. Hence F is a bi-Jensen mapping satisfying 3.28.
Now, let F : X × X → Y be another bi-Jensen mapping satisfying 3.28 with F 0, 0 F0, 0. By Lemma 2.1, we have
F x, y − F x, y 1
n n
n
1 1
n
≤
x,
2
y
−
x,
0
F
−
f
0,
2
y
F
−
f
F
−
f
2
2
4n
2n 4n
1
n
n
n 17ε
1
1 n
≤
x,
2
y
−
x,
0
f
−
F
y
f
−
F
f
−
F
2
0,
2
2
4n
2n−1
2n 4n
3.49
for all x, y ∈ X \ A and n ∈ N. As n → ∞, we may conclude that Fx, y F x, y for all
x, y ∈ X \ A. By Lemma 3.1, F F as we desired.
Example 3.5. Let f : R × R → R be the mapping defined by
⎧ε
⎨
f x, y : 2
⎩0
if x, y 0, 0,
if x, y /
0, 0.
3.50
Let F : R × R → R be the mapping defined by Fx, y 1 for all x, y ∈ X. Then f satisfies the
conditions in Theorem 3.4, and F is a bi-Jensen mapping satisfying 3.28 but F0, 0 /
f0, 0.
14
Journal of Inequalities and Applications
Corollary 3.6. Let f : X ×X → Y be a mapping satisfying 3.13 and 3.27 for all x, y, z ∈ X \{0}.
Then there exists a bi-Jensen mapping F : X × X → Y such that
f x, y − F x, y ≤ 8ε
3.51
for all x, y /
0, 0.
Proof. Let F, F2 , F3 be as in the proof of Theorem 3.4. By 3.30, we obtain
f 0, y f 0, −y
fx
,
0
f−x
,
0
0
0
f 0, y − F 0, y ≤ −
2
2
f 0, y − f 0, −y
7ε
− F2 x, y ≤ ,
2
2
fx, 0 f−x, 0 f 0, y f 0, −y fx, 0 − Fx, 0 ≤ −
2
2
f 0, y f 0, −y fx , 0 f−x , 0 0
0
−
2
2
3.52
fx, 0 − f−x, 0
9ε
≤
−
F
x,
y
3
2
2
for x, y /
0. From the above inequalities and 3.45, we get the inequality
f x, y − F x, y ≤ 8ε
3.53
for all x, y / 0, 0.
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