Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2009, Article ID 153084, 26 pages
doi:10.1155/2009/153084
Research Article
On the Stability of a Generalized
Quadratic and Quartic Type Functional
Equation in Quasi-Banach Spaces
M. Eshaghi Gordji,1 S. Abbaszadeh,1 and Choonkil Park2
1
2
Department of Mathematics, Semnan University, P.O. Box 35195-363, Semnan, Iran
Department of Mathematics, Research Institute for Natural Sciences, Hanyang University,
Seoul 133-791, South Korea
Correspondence should be addressed to Choonkil Park, baak@hanyang.ac.kr
Received 31 May 2009; Accepted 9 September 2009
Recommended by Nikolaos Papageorgiou
We establish the general solution of the functional equation fnx y fnx − y n2 fx y n2 fx − y 2fnx − n2 fx − 2n2 − 1fy for fixed integers n with n /
0, ±1 and investigate the
generalized Hyers-Ulam stability of this equation in quasi-Banach spaces.
Copyright q 2009 M. Eshaghi Gordji 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
The stability problem of functional equations originated from a question of Ulam 1 in 1940,
concerning the stability of group 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 dhx · y, hx ∗ hy < δ for all x, y ∈ G1 ,
then there exists a homomorphism H : G1 → G2 with dhx, Hx < ε for all x ∈ G1 ? In
other words, under what condition does there exist a homomorphism near an approximate
homomorphism? The concept of stability for functional equation arises when we replace the
functional equation by an inequality which acts as a perturbation of the equation. In 1941,
Hyers 2 gave a first affirmative answer to the question of Ulam for Banach spaces. Let
f : E → E be a mapping between Banach spaces such that
f x y − fx − f y ≤ δ
1.1
for all x, y ∈ E, and for some δ > 0. Then there exists a unique additive mapping T : E → E
such that
fx − T x ≤ δ
1.2
2
Journal of Inequalities and Applications
for all x ∈ E. Moreover, if ftx is continuous in t ∈ R for each fixed x ∈ E, then T is R-linear.
In 1978, Th. M. Rassias 3 provided a generalization of Hyers’ theorem which allows the
Cauchy difference to be unbounded. The functional equation
f x y f x − y 2fx 2f y
1.3
is related to a symmetric biadditive mapping 4–7. It is natural that this functional equation is
called a quadratic functional equation. In particular, every solution of the quadratic equation
1.3 is said to be a quadratic mapping. It is well known that a mapping f between real vector
spaces is quadratic if and only if there exists a unique symmetric biadditive mapping B such
that fx Bx, x for all x see 4, 7. The biadditive mapping B is given by
1 f xy −f x−y .
B x, y 4
1.4
The generalized Hyers-Ulam stability problem for the quadratic functional equation 1.3 was
proved by Skof for mappings f : A → B, where A is a normed space and B is a Banach space
see 8. Cholewa 9 noticed that the theorem of Skof is still true if relevant domain A is
replaced by an abelian group. In 10 , Czerwik proved the generalized Hyers-Ulam stability
of the functional equation 1.3. Grabiec 11 has generalized these results mentioned above.
In 12, Park and Bae considered the following quartic functional equation:
f x 2y f x − 2y 4 f x y f x − y 6f y − 6fx.
1.5
In fact, they proved that a mapping f between two real vector spaces X and Y is a solution
of 1.5 if and only if there exists a unique symmetric multiadditive mapping D : X × X × X ×
X → Y such that fx Dx, x, x, x for all x. It is easy to show that fx x4 satisfies the
functional equation 1.5, which is called a quartic functional equation see also 13.
In addition, Kim 14 has obtained the generalized Hyers-Ulam stability for a mixed
type of quartic and quadratic functional equation between two real linear Banach spaces.
Najati and Zamani Eskandani 15 have established the general solution and the generalized
Hyers-Ulam stability for a mixed type of cubic and additive functional equation, whenever f
is a mapping between two quasi-Banach spaces see also 16, 17.
Now we introduce the following functional equation for fixed integers n with n /
0, ±1:
f nx y f nx − y n2 f x y n2 f x − y 2fnx − 2n2 fx − 2 n2 − 1 f y
1.6
in quasi-Banach spaces. It is easy to see that the function fx ax4 bx2 is a solution
of the functional equation 1.6. In the present paper we investigate the general solution of
the functional equation 1.6 when f is a mapping between vector spaces, and we establish
the generalized Hyers-Ulam stability of this functional equation whenever f is a mapping
between two quasi-Banach spaces.
We recall some basic facts concerning quasi-Banach space and some preliminary
results.
Journal of Inequalities and Applications
3
Definition 1.1 See 18, 19. Let X be a real linear space. A quasinorm is a real-valued function
on X satisfying the following.
1 x ≥ 0 for all x ∈ X and x 0 if and only if x 0.
2 λ · x |λ| · x for all λ ∈ R and all x ∈ X.
3 There is a constant M ≥ 1 such that x y ≤ Mx y for all x, y ∈ X.
It follows from the condition 3 that
2m 2m
xi ≤ Mm xi ,
i1 i1
2m1
2m1
xi ≤ Mm1
xi i1 i1
1.7
for all m ≥ 1 and all x1 , x2 , . . . , x2m1 ∈ X.
The pair X, · is called a quasinormed space if · is a quasinorm on X. The smallest
possible M is called the modulus of concavity of · . A quasi-Banach space is a complete
quasi-normed space.
A quasi-norm · is called a p-norm 0 < p ≤ 1 if
x yp ≤ xp yp
1.8
for all x, y ∈ X. In this case, a quasi-Banach space is called a p-Banach space.
Given a p-norm, the formula dx, y : x − yp gives us a translation invariant metric
on X. By the Aoki-Rolewicz theorem 19 see also 18, each quasi-norm is equivalent
to some p-norm. Since it is much easier to work with p-norms, henceforth we restrict our
attention mainly to p-norms. In 20, Tabor has investigated a version of Hyers-Rassias-Gajda
theorem see 3, 21 in quasi-Banach spaces.
2. General Solution
Throughout this section, X and Y will be real vector spaces. We here present the general
solution of 1.6.
Lemma 2.1. If a mapping f : X → Y satisfies the functional equation 1.6, then f is a quadratic
and quartic mapping.
Proof. Letting x y 0 in 1.6, we get f0 0. Setting x 0 in 1.6, we get fy f−y
for all y ∈ X. So the mapping f is even. Replacing x by x y in 1.6 and then x by x − y in
1.6, we get
f nx n 1y f nx n − 1y
n2 f x 2y n2 fx 2f nx ny − 2n2 f x y − 2 n2 − 1 f y ,
2.1
f nx − n − 1y f nx − n 1y
n2 fx n2 f x − 2y 2f nx − ny − 2n2 f x − y − 2 n2 − 1 f y
2.2
4
Journal of Inequalities and Applications
for all x, y ∈ X. Interchanging x and y in 1.6 and using the evenness of f, we get the relation
f x ny f x − ny
n2 f x y n2 f x − y 2f ny − 2n2 f y − 2 n2 − 1 fx
2.3
for all x, y ∈ X. Replacing y by ny in 1.6 and then using 2.3, we have
f nx ny f nx − ny
n4 f x y n4 f x − y 2f ny 2fnx − 2n4 fx − 2n4 f y
2.4
for all x, y ∈ X. If we add 2.1 to 2.2 and use 2.4, then we have
f nx n 1y f nx − n 1y f nx n − 1y f nx − n − 1y
n2 f x 2y n2 f x − 2y 2n2 n2 − 1 f x y 2n2 n2 − 1 f x − y
2.5
4f ny 4fnx −4n4 2n2 fx −4n4 − 4n2 4 f y
for all x, y ∈ X. Replacing y by x y in 1.6 and then y by x − y in 1.6 and using the
evenness of f, we obtain that
f n 1x y f n − 1x − y
n2 f 2x y n2 f y 2fnx − 2n2 fx − 2 n2 − 1 f x y ,
2.6
f n 1x − y f n − 1x y
n2 f 2x − y n2 f y 2fnx − 2n2 fx − 2 n2 − 1 f x − y
2.7
for all x, y ∈ X. Interchanging x with y in 2.6 and 2.7 and using the evenness of f, we get
the relations
f x n 1y f x − n − 1y
n2 f x 2y n2 fx 2f ny − 2n2 f y − 2 n2 − 1 f x y ,
2.8
f x − n 1y f x n − 1y
n2 f x − 2y n2 fx 2f ny − 2n2 f y − 2 n2 − 1 f x − y
2.9
Journal of Inequalities and Applications
5
for all x, y ∈ X. Replacing y by n 1y in 1.6 and then y by n − 1y in 1.6, we have
f nx n 1y f nx − n 1y
n2 f x n 1y n2 f x − n 1y 2fnx − 2n2 fx − 2 n2 − 1 f n 1y ,
2.10
f nx n − 1y f nx − n − 1y
n2 f x n − 1y n2 f x − n − 1y 2fnx − 2n2 fx − 2 n2 − 1 f n − 1y
2.11
for all x, y ∈ X. Replacing x by y in 1.6, we obtain
f n 1y f n − 1y n2 f 2y − 2 2n2 − 1 f y 2f ny
2.12
for all y ∈ X. Adding 2.10 to 2.11 and using 2.8, 2.9, and 2.12, we get
f nx n 1y f nx − n 1y f nx n − 1y f nx − n − 1y
n4 f x 2y n4 f x − 2y − 2n2 n2 − 1 f x y − 2n2 n2 − 1 f x − y
4f ny 4fnx − 2n2 n2 − 1 f 2y 2n4 − 4n2 fx 4n4 − 12n2 4 f y
2.13
for all x, y ∈ X. By 2.5 and 2.13, we obtain
f x 2y f x − 2y 4f x y 4f x − y 2f 2y − 8f y − 6fx
2.14
for all x, y ∈ X. Interchanging x and y in 2.14 and using the evenness of f, we get the
relation
f 2x y f 2x − y 4f x y 4f x − y 2f2x − 8fx − 6f y
2.15
for all x, y ∈ X.
Now we show that 2.15 is a quadratic and quartic functional equation. To get this,
we show that the mappings g : X → Y , defined by gx f2x − 16fx, and h : X → Y ,
defined by hx f2x − 4fx, are quadratic and quartic, respectively.
Replacing y by 2y in 2.15 and using the evenness of f, we have
f 2x 2y f 2x − 2y 4f 2y x 4f 2y − x 2f2x − 8fx − 6f 2y
2.16
6
Journal of Inequalities and Applications
for all x, y ∈ X. Interchanging x with y in 2.16 and then using 2.15, we obtain by the
evenness of f
f 2x 2y f 2x − 2y
4f 2x y 4f 2x − y 2f 2y − 8f y − 6f2x
16f x y 16f x − y 2f2x 2f 2y − 32fx − 32f y
2.17
for all x, y ∈ X. By 2.17, we have
f 2x2y − 16f xy f 2x−2y −16f x−y 2 f2x−16fx 2 f 2y −16f y
2.18
for all x, y ∈ X. This means that
g x y g x − y 2gx 2g y
2.19
for all x, y ∈ X. Thus the mapping g : X → Y is quadratic.
To prove that h : X → Y is quartic, we have to show that
h 2x y h 2x − y 4h x y 4h x − y 24hx − 6h y
2.20
for all x, y ∈ X. Replacing x and y by 2x and 2y in 2.15, respectively, we get
f 4x 2y f 4x − 2y 4f 2x 2y 4f 2x − 2y 2f4x − 8f2x − 6f 2y
2.21
for all x, y ∈ X. Since g2x 4gx for all x ∈ X and g : X → Y is a quadratic mapping, we
have
f4x 20f2x − 64fx
2.22
for all x ∈ X. So it follows from 2.15, 2.21, and 2.22 that
h 2x y h 2x − y
f 4x 2y − 4f 2x y f 4x − 2y − 4f 2x − y
4 f 2x 2y − 4f x y 4 f 2x − 2y − 4f x − y
24 f2x − 4fx − 6 f 2y − 4f y
4h x y 4h x − y 24hx − 6h y
for all x, y ∈ X. Thus h : X → Y is a quartic mapping.
2.23
Journal of Inequalities and Applications
7
Theorem 2.2. A mapping f : X → Y satisfies 1.6 if and only if there exist a unique symmetric
multiadditive mapping D : X × X × X × X → Y and a unique symmetric bi-additive mapping
B : X × X → Y such that
fx Dx, x, x, x Bx, x
2.24
for all x ∈ X.
Proof. We first assume that the mapping f : X → Y satisfies 1.6. Let g, h : X → Y be
mappings defined by
gx : f2x − 16fx,
hx : f2x − 4fx
2.25
for all x ∈ X. By Lemma 2.1, we achieve that the mappings g and h are quadratic and quartic,
respectively, and
fx :
1
1
hx − gx
12
12
2.26
for all x ∈ X. Thus there exist a unique symmetric multiadditive mapping D : X ×X ×X ×X →
Y and a unique symmetric bi-additive mapping B : X × X → Y such that Dx, x, x, x 1/12hx and Bx, x −1/12gx for all x ∈ X see citead, ki. So
fx Dx, x, x, x Bx, x
2.27
fx Dx, x, x, x Bx, x
2.28
for all x ∈ X.
Conversely assume that
for all x ∈ X, where the mapping D : X × X × X × X → Y is symmetric multi-additive and
B : X × X → Y is bi-additive. By a simple computation, one can show that the mappings D
and B satisfy the functional equation 1.6, so the mapping f satisfies 1.6.
3. Generalized Hyers-Ulam Stability of 1.6
From now on, let X and Y be a quasi-Banach space with quasi-norm · X and a p-Banach
space with p-norm · Y , respectively. Let M be the modulus of concavity of · Y . In this
section, using an idea of Găvruta 22, we prove the stability of 1.6 in the spirit of Hyers,
Ulam, and Rassias. For convenience we use the following abbreviation for a given mapping
8
Journal of Inequalities and Applications
f : X → Y:
Δf x, y f nx y f nx − y − n2 f x y − n2 f x − y
− 2fnx 2n2 fx 2 n2 − 1 f y
3.1
p
for all x, y ∈ X. Let ϕq x, y : ϕq x, yp . We will use the following lemma in this section.
Lemma 3.1 see 15. Let 0 < p ≤ 1 and let x1 , x2 , . . . , xn be nonnegative real numbers. Then
p
n
n
xi
≤
xi p .
i1
3.2
i1
Theorem 3.2. Let ϕq : X × X → 0, ∞ be a function such that
x y 0
,
2m 2m
3.3
∞
x y
4pi ϕq p i , i < ∞
2 2
i1
3.4
lim 4m ϕq
m→∞
for all x, y ∈ X and
for all x ∈ X and all y ∈ {x, 2x, 3x, nx, n 1x, n − 1x, n 2x, n − 2x, n − 3x}. Suppose
that a mapping f : X → Y with f0 0 satisfies the inequality
Δf x, y ≤ ϕq x, y
Y
3.5
x x
Qx : lim 4m f m−1 − 16f m
m→∞
2
2
3.6
for all x, y ∈ X. Then the limit
exists for all x ∈ X and Q : X → Y is a unique quadratic mapping satisfying
11 f2x − 16fx − Qx ≤ M ψq x 1/p
Y
4
3.7
Journal of Inequalities and Applications
9
for all x ∈ X, where
ψq x :
∞
i1
pi
4
1
p x n 2x
p x n − 2x
p p x n 1x
ϕ
,
,
ϕ
,
ϕ
4
q
q
q
2i
2i
2i
2i
2i
2i
n2p n2 − 1p
p 2x 2x
p p x n − 1x
p p x nx
p p 2x x
4 ϕq i ,
,
,
4 ϕq i , i ϕq
4 ϕq
2
2i
2 2
2i 2i
2i 2i
p p x 2x p p x x 2p p x 3x
p
2
2
n ϕq i , i 2 3n − 1 ϕq i , i 17n − 8 ϕq i , i
2 2
2 2
2 2
xn 1x
n2p
n − 3x
n − 1x
p
p
p p
ϕq 0,
10 ϕq 0,
2
p ϕq 0,
2i
2i
2i
n − 1
p 4
n 1
nx
2x
n − 2x
p
p p
p p
4 ϕq 0, i 4 ϕq 0,
2
p ϕq 0, i
2
2i
2
n − 1
p 4
2 3n − n2 2
x
p
ϕq 0, i
.
p
2
n2 − 1
3.8
Proof. Setting x 0 in 3.5 and then interchanging x and y, we get
2
n − 1 fx − n2 − 1 f−x ≤ ϕq 0, x
3.9
for all x ∈ X. Replacing y by x, 2x, nx, n 1x and n − 1x in 3.5, respectively, we get
fn 1x fn − 1x − n2 f2x − 2fnx 4n2 − 2 fx ≤ ϕq x, x,
fn 2x fn − 2x − n2 f3x − n2 f−x − 2fnx 2n2 fx
2 n2 − 1 f2x ≤ ϕq x, 2x,
f2nx − n2 fn 1x − n2 f1 − nx 2 n2 − 2 fnx 2n2 fx ≤ ϕq x, nx,
f2n 1x f−x − n2 fn 2x − n2 f−nx − 2fnx 2n2 fx
2 n2 − 1 fn 1x ≤ ϕq x, n 1x,
f2n − 1x fx − n2 f2 − nx − n2 2 fnx 2n2 fx
2 n2 − 1 fn − 1x ≤ ϕq x, n − 1x,
f2n 1x f−2x − n2 fn 3x − n2 f−n 1x − 2fnx
2n2 fx 2 n2 − 1 fn 2x ≤ ϕq x, n 2x,
3.10
3.11
3.12
3.13
3.14
3.15
10
Journal of Inequalities and Applications
f2n − 1x f2x − n2 fn − 1x − n2 f−n − 3x − 2fnx 2n2 fx
3.16
2
2 n − 1 fn − 2x ≤ ϕq x, n − 2x,
fn 3x fn − 3x − n2 f4x − n2 f−2x − 2fnx 2n2 fx
2 n2 − 1 f3x ≤ ϕq x, 3x
3.17
for all x ∈ X. Combining 3.9 and 3.11–3.17, respectively, yields the following inequalities:
fn 2x fn − 2x − n2 f3x − n2 fx − 2fnx 2n2 fx 2 n2 − 1 f2x
≤ ϕq x, 2x n2
ϕq 0, x,
−1
n2
f2nx − n2 fn 1x − n2 fn − 1x 2 n2 − 2 fnx 2n2 fx
≤ ϕq x, nx n2
ϕq 0, n − 1x,
2
n −1
3.18
3.19
f2n 1x fx − n2 fn 2x − n2 fnx − 2fnx 2n2 fx 2 n2 − 1 fn 1x
≤ ϕq x, n 1x n2
1
ϕq 0, nx 2
ϕq 0, x,
2
n −1
n −1
3.20
f2n − 1x fx − n2 fn − 2x − n2 2 fnx 2n2 fx 2 n2 − 1 fn − 1x
≤ ϕq x, n − 1x n2
ϕq 0, n − 2x,
n2 − 1
3.21
f2n1xf2x−n2 fn3x−n2 fn1x−2fnx2n2 fx2 n2 −1 fn2x
≤ ϕq x, n 2x n2
ϕq 0, n 1x ϕq 0, 2x,
n2 − 1
3.22
f2n−1xf2x−n2 fn−1x−n2 fn−3x−2fnx2n2 fx2 n2 −1 fn−2x
≤ ϕq x, n − 2x n2
ϕq 0, n − 3x,
−1
n2
fn 3x fn − 3x − n2 f4x − n2 f2x − 2fnx 2n2 fx 2 n2 − 1 f3x
≤ ϕq x, 3x for all x ∈ X.
n2
ϕq 0, 2x
n2 − 1
3.23
3.24
Journal of Inequalities and Applications
11
Replacing x and y by 2x and x in 3.5, respectively, we obtain
f2n 1x f2n − 1x − n2 f3x − 2f2nx 2n2 f2x n2 − 2 fx ≤ ϕq 2x, x
3.25
for all x ∈ X. Putting 2x and 2y instead of x and y in 3.5, respectively, we have
f2n 1x f2n − 1x − n2 f4x − 2f2nx 2 2n2 − 1 f2x ≤ ϕq 2x, 2x
3.26
for all x ∈ X. It follows from 3.10, 3.18, 3.19, 3.20, 3.21, and 3.25 that
f3x − 6f2x 15fx
M5
≤ 2 2
ϕq x, n 1x ϕq x, n − 1x ϕq 2x, x 2ϕq x, nx n2 ϕq x, 2x
n n − 1
4n2 − 2 ϕq x, x n4 1
ϕq 0, x
2
n −1
n2 2ϕq 0, n − 1x ϕq 0, nx ϕq 0, n − 2x
n2 − 1
3.27
for all x ∈ X. Also, from 3.10, 3.18, 3.19, 3.22, 3.23, 3.24, and 3.26, we conclude
f4x − 4f3x 4f2x 4fx
M6
≤ 2 2
ϕq x, n 2x ϕq x, n − 2x ϕq 2x, 2x 2ϕq x, nx
n n − 1
n2 ϕq x, 3x ϕq x, x 2 n2 − 1 ϕq x, 2x
n2 2ϕq 0, n − 1x ϕq 0, n − 3x ϕq 0, n 1x
−1
n4 1
2
ϕq 0, 2x 2n ϕq 0, x
2
n −1
n2
3.28
12
Journal of Inequalities and Applications
for all x ∈ X. Finally, combining 3.27 and 3.28 yields
f4x − 24f2x 64fx
M8
≤ 2 2
ϕq x, n 2x ϕq x, n − 2x 4ϕq x, n 1x
n n − 1
4ϕq x, n − 1x 10ϕq x, nx ϕq 2x, 2x 4ϕq 2x, x
n2 ϕq x, 3x 2 3n2 − 1 ϕq x, 2x 17n2 − 8 ϕq x, x
2
3.29
n
ϕq 0, n 1x ϕq 0, n − 3x 10ϕq 0, n − 1x
n2 − 1
n4 1
ϕq 0, 2x
4ϕq 0, nx4ϕq 0, n − 2x 2
n −1
2 3n4 − n2 2
ϕq 0, x
n2 − 1
for all x ∈ X. Let
ψq x :
1
n2 n2
− 1
ϕq x, n 2x ϕq x, n − 2x 4ϕq x, n 1x
4ϕq x, n − 1x 10ϕq x, nx ϕq 2x, 2x 4ϕq 2x, x
n2 ϕq x, 3x 2 3n2 − 1 ϕq x, 2x 17n2 − 8 ϕq x, x
n2 ϕq 0, n1xϕq 0, n−3x10ϕq 0, n−1x4ϕq 0, nx
−1
2 3n4 − n2 2
n4 1
4ϕq 0, n−2x 2
ϕq 0, 2x
ϕq 0, x .
n −1
n2 − 1
3.30
n2
Then the inequality 3.29 implies that
f4x − 20f2x 64fx ≤ M8 ψq x
3.31
for all x ∈ X.
Let g : X → Y be a mapping defined by gx : f2x − 16fx for all x ∈ X. From
3.31, we conclude that
g2x − 4gx ≤ M8 ψq x
3.32
Journal of Inequalities and Applications
13
for all x ∈ X. If we replace x in 3.32 by x/2m1 and multiply both sides of 3.32 by 4m , then
we get
x m1
x
x
m
≤ M8 4m ψq
4 g
g
−
4
2m Y
2m1
2m1
3.33
for all x ∈ X and all nonnegative integers m. Since Y is a p-Banach space, the inequality 3.33
gives
p p
m i1
m1
x
x x
x k
i
4 g
4 g i1 − 4 g i −4 g k ≤
2
2m1
2
2
Y
Y
ik
≤ M8p
m
4ip ψq p
ik
x
2i1
3.34
for all nonnegative integers m and k with m ≥ k and all x ∈ X. Since 0 < p ≤ 1, by Lemma 3.1
and 3.30, we conclude that
p
ψq x
1
p
p
p
ϕq x, n 2x ϕq x, n − 2x 4p ϕq x, n 1x
≤ 2p 2
n n − 1p
p
p
p
p
4p ϕq x, n − 1x10p ϕq x, nxϕq 2x, 2x 4p ϕq 2x, x
p p
p p
p
n2p ϕq x, 3x2p 3n2 − 1 ϕq x, 2x 17n2 − 8 ϕq x, x
n2p
p
n2 − 1
p
p
p
p
× ϕq 0, n1xϕq 0, n−3x10p ϕq 0, n−1x4p ϕq 0, nx
p
4
n4 1p p
2 3n − n2 2
p
2
ϕq 0, x
p ϕq 0, 2x
p
n − 1
n2 − 1
3.35
p
4p ϕq 0, n−2x
for all x ∈ X. Therefore, it follows from 3.4 and 3.35 that
∞
i1
4ip ψq p
x
2i
<∞
3.36
for all x ∈ X. It follows from 3.34 and 3.36 that the sequence {4m gx/2m } is Cauchy for
all x ∈ X. Since Y is complete, the sequence {4m gx/2m } converges for all x ∈ X. So one can
define the mapping Q : X → Y by
Qx lim 4m g
m→∞
x 2m
3.37
14
Journal of Inequalities and Applications
for all x ∈ X. Letting k 0 and passing the limit m → ∞ in 3.34, we get
∞
gx − Qxp ≤ M8p 4ip ψq p
Y
i0
x
2i1
∞
M8p p x
ip
4
ψ
q
4p i1
2i
3.38
for all x ∈ X. Thus 3.7 follows from 3.4 and 3.38.
Now we show that Q is quadratic. It follows from 3.3, 3.33 and 3.37 that
x m
x
m1
Q2x − 4QxY lim 4 g m−1 − 4 g m m→∞
2 Y
2
x m−1
x
m
4 lim 4 g m−1 − 4 g m m→∞
2 4
3.39
Y
≤ M11 lim 4m ψq
m→∞
x 0
2m
for all x ∈ X. So
Q2x 4Qx
3.40
for all x ∈ X. On the other hand, it follows from 3.3, 3.5, 3.6 and 3.37 that
x y ΔQx, y lim 4m Δg m , m Y
m→∞
2 2
Y
x y y
x
m
lim 4 Δf m−1 , m−1 − 16Δf m , m m→∞
2 2 Y
2
2
x y y
x
≤ M lim 4 Δf m−1 , m−1 16Δf m , m m→∞
2
2
2
2
Y
Y
3.41
m
x y y
x
≤ M lim 4 ϕq m−1 , m−1 16ϕq m , m
0
m→∞
2 2
2
2
m
for all x, y ∈ X. Hence the mapping Q satisfies 1.6. By Lemma 2.1, the mapping Q2x −
4Qx is quadratic. Hence 3.40 implies that the mapping Q is quadratic.
It remains to show that Q is unique. Suppose that there exists another quadratic
mapping Q : X → Y which satisfies 1.6 and 3.7. Since Q x/2m 1/4m Q x and
Qx/2m 1/4m Qx for all x ∈ X, we conclude from 3.7 that
x p M8p
x x mp
Qx − Q xp lim 4mp −
Q
≤
lim
4
ψ
g
q
Y
m→∞
2m
2m Y
4p m → ∞
2m
3.42
Journal of Inequalities and Applications
15
for all x ∈ X. On the other hand, since
lim 4mp
m→∞
∞
4ip ϕq p
i1
y
,
2mi 2mi
x
lim
∞
4ip ϕq p
m→∞
im1
x y
,
2i 2i
0
3.43
for all x ∈ X and all y ∈ {x, 2x, 3x, nx, n 1x, n − 1x, n 2x, n − 2x, n − 3x}, then
lim 4mp ψq
m→∞
x 0
2m
3.44
for all x ∈ X. Using 3.44 and 3.42, we get Q Q , as desired.
Theorem 3.3. Let ϕq : X × X → 0, ∞ be a function such that
lim
m→∞
1 m
ϕq 2 x, 2m y 0
4m
3.45
for all x, y ∈ X and
∞
1 p i
i
2
ϕ
x,
2
y
<∞
q
pi
i0 4
3.46
for all x ∈ X and all y ∈ {x, 2x, 3x, nx, n 1x, n − 1x, n 2x, n − 2x, n − 3x}. Suppose
that a mapping f : X → Y with f0 0 satisfies the inequality
Δf x, y ≤ ϕq x, y
Y
3.47
1 m1 m
f
2
x
−
16f2
x
m → ∞ 4m
3.48
for all x, y ∈ X. Then the limit
Qx : lim
exists for all x ∈ X and Q : X → Y is a unique quadratic mapping satisfying
8
f2x − 16fx − Qx ≤ M ψq x 1/p
Y
4
3.49
16
Journal of Inequalities and Applications
for all x ∈ X, where
∞
1
ψq x : pi
4
i0
1
p
p
i
i
i
i
p p
i
i
ϕ
2
2
2
x,
2
x,
2
ϕ
x,
2
ϕ
4
n2x
n−2x
n1x
q
q
q
p
n2p n2 − 1
p
p
p
p
4p ϕq 2i x, 2i n−1x 10p ϕq 2i x, 2i nx ϕq 2i 2x, 2i 2x 4p ϕq 2i 2x, 2i x
p p p p p
n2p ϕq 2i x, 2i 3x 2p 3n2 − 1 ϕq 2i x, 2i 2x 17n2 − 8 ϕq 2i x, 2i x
n2p p i
p
i
ϕ
0,
2
0,
2
1x
ϕ
−
3x
n
n
q
q
p
n2 − 1
p
p
p
10p ϕq 0, 2i n − 1x 4p ϕq 0, 2i nx 4p ϕq 0, 2i n − 2x
p
n4 1
n2 − 1
p
p ϕq
0, 2 2x i
p
4
2 3n − n2 2
n2 − 1
p
p
ϕq
i
0, 2 x
.
3.50
Proof. The proof is similar to the proof of Theorem 3.2.
Corollary 3.4. Let θ, r, s be nonnegative real numbers such that r, s > 2 or s < 2. Suppose that a
mapping f : X → Y with f0 0 satisfies the inequality
Δfx, y ≤
Y
⎧
⎪
θ,
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨θxrX ,
r s 0,
r > 0, s 0,
s
⎪
⎪
⎪
θyX ,
⎪
⎪
⎪
⎪
⎪
⎪
⎩θxr ys ,
X
X
r 0, s > 0,
3.51
r, s > 0
for all x, y ∈ X. Then there exists a unique quadratic mapping Q : X → Y satisfying
f2x − 16fx − Qx ≤
Y
⎧
⎪
δq ,
⎪
⎪
⎪
⎪
⎪
⎪
⎪α x,
⎪
⎨ q
r s 0,
r > 0, s 0,
M8 θ
n2 n2 − 1 ⎪
⎪
r 0, s > 0,
⎪βq x,
⎪
⎪
⎪
⎪
1/p
⎪
⎪
⎩ αpq x βqp x
, r, s > 0
3.52
Journal of Inequalities and Applications
17
for all x ∈ X, where
δq p p p
p
1
p
2
2
2
2
4
2
6n
n
−
2
−
1
17n
−
8
n
−
1
6n
−
2n
4
4p − 1n2 − 1p
p
p
p
n2p 2 10p 2 ∗ 4p n4 1 n2p n2 − 1 3 ∗ 4p n2 − 1
10
p
p
p 1/p
2
n −1 3 n −1
,
2
1/p
p p
4p 2 2rp 10p 6n2 − 2 17n2 − 8 2rp n2p
αq x xrX ,
|4p − 2rp |
p p
p
1
p
p
sp
2
2
2
2
4
2
6n
n
−
2
−
1
17n
−
8
n
−
1
6n
−
2n
4
2
βq x p
n2 − 1 |4p − 2sp |
n2p n 1sp n − 3sp 10p n − 1sp 4p nsp 4p n − 2sp
p
p
p
p
2sp n4 1 3sp n2p n2 − 1 4p n2 − 1 n 2sp n2 − 1
p
p
p
n − 2sp n2 − 1 4p n 1sp n2 − 1 4p n − 1sp n2 − 1
p 1/p
10p nsp n2 − 1
xsX .
3.53
Proof. In Theorem 3.2, putting ϕq x, y : θxrX ysX for all x, y ∈ X, we get the desired
result.
Corollary 3.5. Let θ ≥ 0 and r, s > 0 be nonnegative real numbers such that λ : r s / 2. Suppose
that a maping f : X → Y with f0 0 satisfies the inequality
Δf x, y ≤ θxr ys
X
X
Y
3.54
for all x, y ∈ X. Then there exists a unique quadratic mapping Q : X → Y satisfying
f2x − 16fx − Qx
Y
8
1
M θ
n 2sp n − 2sp 4p n 1sp
≤ 2 2
n n − 1 4p − 2λp 4p n − 1sp 10p nsp 2rsp 4p 2rp n2p 3sp
2
sp
for all x ∈ X.
p p
6n − 2 17n2 − 8
2
1/p
xλX
3.55
18
Journal of Inequalities and Applications
Proof. In Theorem 3.2, putting ϕq x, y : θxrX ysX for all x, y ∈ X, we get the desired
result.
Theorem 3.6. Let ϕt : X × X → 0, ∞ be a function such that
lim 16m ϕt
m→∞
x y 0
,
2m 2m
3.56
for all x, y ∈ X and
∞
16pi ϕt p
i1
x y
,
2i 2i
<∞
3.57
for all x ∈ X and all y ∈ {x, 2x, 3x, nx, n 1x, n − 1x, n 2x, n − 2x, n − 3x}. Suppose
that a mapping f : X → Y with f0 0 satisfies the inequality
Δf x, y ≤ ϕt x, y
Y
3.58
x x
T x : lim 16 f m−1 − 4f m
m→∞
2
2
3.59
for all x, y ∈ X. Then the limit
m
exists for all x ∈ X and T : X → Y is a unique quartic mapping satisfying
8
f2x − 4fx − T x ≤ M ψt x 1/p
Y
16
3.60
for all x ∈ X, where
ψt x :
∞
i1
16pi
1
p x n 2x
p x n − 2x
p p x n 1x
,
,
ϕ
,
ϕ
ϕ
4
t
t
2i
2i
2i
2i
2i
n2p n2 − 1p t 2i
p 2x 2x
p p x n − 1x
p p x nx
p p 2x x
4 ϕt
,
,
,
,
10 ϕt
ϕt
4 ϕt
2i
2i
2i 2i
2i 2i
2i 2i
p p x 2x p p x x 2p p x 3x
p
2
2
n ϕt
,
,
,
2 3n − 1 ϕt
17n − 8 ϕt
2i 2i
2i 2i
2i 2i
xn 1x
n2p
n − 3x
n − 1x
p
p
p p
ϕ
ϕ
0,
ϕ
0,
10
0,
2
p
t
t
t
2i
2i
2i
n − 1
p 4
n 1
nx
2x
n − 2x
p
p p
p p
4 ϕt 0, i 4 ϕt 0,
2
0, i
pϕ
2
2i
2
n − 1 t
4
p
2 3n − n2 2
x
p
.
ϕt 0, i
p
2
2
n − 1
3.61
Journal of Inequalities and Applications
19
Proof. Similar to the proof Theorem 3.2, we have
f4x − 20f2x 64fx ≤ M8 ψt x
3.62
for all x ∈ X, where
1
ψt x 2 2
ϕt x, n 2x ϕt x, n − 2x 4ϕt x, n 1x
n n − 1
4ϕt x, n − 1x 10ϕt x, nx ϕt 2x, 2x 4ϕt 2x, x
n2 ϕt x, 3x 2 3n2 − 1 ϕt x, 2x 17n2 − 8 ϕt x, x
n2 ϕt 0, n 1x ϕt 0, n − 3x 10ϕt 0, n − 1x 4ϕt 0, nx
−1
2 3n4 − n2 2
n4 1
4ϕt 0, n − 2x 2
ϕt 0, 2x ϕt 0, x .
n −1
n2 − 1
3.63
n2
Let h : X → Y be a mapping defined by hx : f2x − 4fx. Then we conclude that
3.64
h2x − 16hx ≤ M8 ψt x
for all x ∈ X. If we replace x in 3.65 by x/2m1 and multiply both sides of 3.65 by 16m ,
then we get
x m1
x
x
m
≤ M8 16m ψt
16 h
−
16
h
2m Y
2m1
2m1
3.65
for all x ∈ X and all nonnegative integers m. Since Y is a p-Banach space, the inequality 3.66
gives
p p
m i1
m1
x
x x
x k
i
16 h
16 h i1 − 16 h i − 16 h k ≤
m1
2 2
2
2
Y
Y
ik
≤M
8p
m
ik
pi
16 ψt
p
x
2i1
3.66
20
Journal of Inequalities and Applications
for all nonnegative integers m and k with m ≥ k and all x ∈ X. Since 0 < p ≤ 1, by Lemma 3.1,
we conclude from 3.64 that
p
ψt x
1
p
p
p
ϕ x, n 2x ϕt x, n − 2x 4p ϕt x, n 1x
≤ 2p 2
n n − 1p t
p
p
p
p
4p ϕt x, n − 1x 10p ϕt x, nx ϕt 2x, 2x 4p ϕt 2x, x
p p
p p
p
n2p ϕt x, 3x2p 3n2 −1 ϕt x, 2x 17n2 −8 ϕt x, x
n2p p
p
p p
p ϕt 0, n 1x ϕt 0, n − 3x 10 ϕt 0, n − 1x
n2 − 1
n4 1p p
p
p
4p ϕt 0, nx 4p ϕt 0, n − 2x 2
p ϕ 0, 2x
n − 1 t
p
4
2 3n − n2 2
p
ϕt 0, x
p
n2 − 1
3.67
for all x ∈ X. It follows from 3.57 and 3.67 that
∞
pi
16 ψt
p
i1
x
2i
<∞
3.68
for all x ∈ X. Thus we conclude from 3.67 and 3.69 that the sequence {16m hx/2m } is
Cauchy for all x ∈ X. Since Y is complete, the sequence {16m hx/2m } converges for all
x ∈ X. So one can define the mapping T : X → Y by
x T x lim 16m h m
m→∞
2
3.69
for all x ∈ X. Letting k 0 and passing the limit m → ∞ in 3.67, we get
hx −
p
T xY
≤M
8p
∞
i0
pi
16 ψt
p
x
2i1
∞
M11p pi p x
16 ψt
16p i1
2i
3.70
for all x ∈ X. Thus 3.60 follows from 3.58 and 3.70.
Now we show that T is quartic. From 3.57, 3.66, and 3.70, it follows that
x m
x
m1
T 2x − 16T xY lim 16 h m−1 − 16 h m m→∞
2
2
Y
m−1
x
x m
− 16 h m 16 lim 16 h
m→∞
2 Y
16m−1
x ≤ M8 lim 16m ψt m 0
m→∞
2
3.71
Journal of Inequalities and Applications
21
for all x ∈ X. So
T 2x 16T x
3.72
for all x ∈ X. On the other hand, by 3.59, 3.69, and 3.70, we have
x y ΔT x, y lim 16m Δh m , m Y
m→∞
2 2
Y
x y y
x
Δf
−
4Δf
lim 16m ,
, m m
m−1
m−1
m→∞
2 2 Y
2
2
x y y
x
4
Δf
≤ M lim 16m ,
Δf m , m m→∞
2 2
2m−1 2m−1 Y
Y
x y y
x
≤ M lim 16m ϕt m−1 , m−1 4ϕt m , m
0
m→∞
2 2
2
2
3.73
for all x, y ∈ X. Hence the mapping T satisfies 1.6. By Lemma 2.1, the mapping T 2x −
16T x is quartic. Therefore, 3.75 implies that the mapping T is quartic.
To prove the uniqueness property of T, let T : X → Y be another quartic mapping
satisfying 3.61. Since
lim 16mp
m→∞
∞
16pi ϕt p
i1
x
, mi
mi
2
2
x
lim
m→∞
∞
16pi ϕt p
im1
x x
,
2i 2i
0
3.74
for all x ∈ X and all y ∈ {x, 2x, 3x, nx, n 1x, n − 1x, n 2x, n − 2x, n − 3x}, then
lim 16mp ψt
m→∞
x 0
2m
3.75
for all x ∈ X. It follows from 3.61 and 3.86 that
x x p M8p
x T x − T x lim 16mp lim 16mp ψt m 0
h m − T m ≤
Y
p
m→∞
2
2
16 m → ∞
2
Y
3.76
for all x ∈ X. So T T , as desired.
Theorem 3.7. Let ϕt : X × X → 0, ∞ be a function such that
lim
1
ϕt 2m x, 2m y 0
m
m → ∞ 16
3.77
∞
1
p
i
i
2
ϕ
x,
2
y
<∞
t
pi
i0 16
3.78
for all x, y ∈ X and
22
Journal of Inequalities and Applications
for all x ∈ X and all y ∈ {x, 2x, 3x, nx, n 1x, n − 1x, n 2x, n − 2x, n − 3x}. Suppose
that a mapping f : X → Y with f0 0 satisfies the inequality
Δf x, y ≤ ϕt x, y
Y
3.79
1 m1 f 2 x − 4f2m x
m
m → ∞ 16
3.80
for all x, y ∈ X. Then the limit
T x : lim
exists for all x ∈ X and T : X → Y is a unique quartic mapping satisfying
8
f2x − 4fx − T x ≤ M ψt x 1/p
Y
16
3.81
for all x ∈ X, where
ψt x :
∞
p
p
1
1
i
i
i
i
p p
i
i
ϕ
2
2
2
x,
2
x,
2
ϕ
x,
2
ϕ
4
n2x
n−2x
n1x
p
t
t
t
pi n2p
n2 − 1
i0 16
p
p
p
p
4p ϕt 2i x, 2i n−1x 10p ϕt 2i x, 2i nx ϕt 2i 2x, 2i 2x 4p ϕt 2i 2x, 2i x
p p p p p
n2p ϕt 2i x, 2i 3x 2p 3n2 − 1 ϕt 2i x, 2i 2x 17n2 − 8 ϕt 2i x, 2i x
n2p p i
p
i
p p
i
p ϕt 0, 2 n 1x ϕt 0, 2 n − 3x 10 ϕt 0, 2 n − 1x
− 1
n4 1p p p p
i
p p
i
i
0,
2
ϕ
2x
4 ϕt 0, 2 nx 4 ϕt 0, 2 n − 2x 2
p
n − 1 t
p 4
2 3n − n2 2
p
i
ϕt 0, 2 x
.
p
n2 − 1
3.82
n2
Proof. The proof is similar to the proof of Theorem 3.6.
Corollary 3.8. Let θ, r, s be nonnegative real numbers such that r, s > 4 or 0 ≤ r, s < 4. Suppose that
a mapping f : X → Y with f0 0 satisfies the inequality3.51 for all x, y ∈ X. Then there exists
a unique quartic mapping T : X → Y satisfying
f2x − 4fx − T x ≤
Y
⎧
⎪
⎪
⎪δt ,
⎪
⎪
⎪
⎪
⎪
⎨αt x,
M8 θ
n2 n2 − 1 ⎪
βt x,
⎪
⎪
⎪
⎪
⎪
1/p
⎪
⎪
⎩ αp x βp x
,
t
t
r s 0,
r > 0, s 0,
r 0, s > 0,
r, s > 0
3.83
Journal of Inequalities and Applications
23
for all x ∈ X, where
δt p p p p p
1
2
2
2
2
4
2
n
n
−
2
−
1
17n
−
8
−
1
6n
−
2n
4
6n
p
16p − 1n2 − 1
p
p
p
n2p 2 10p 2 · 4p n4 1 n2p n2 − 1 3 · 4p n2 − 1
p
p 1/p
10p n2 − 1 3 n2 − 1
,
1/p
p p
4p 2 2rp 10p 6n2 − 2 17n2 − 8 2rp n2p
xrX ,
|16p − 2rp |
p p p p p
1
sp
2
2
2
2
4
2
6n
n
n
−2
−1
17n
−8
−1
6n
−2n
4
2
p
sp
n2 − 1 |16 − 2 |
n2p n 1sp n − 3sp 10p n − 1sp 10p nsp 4p n − 2sp
p
p
p
2sp n4 1 3sp n2p n2 − 1 4p n2 − 1
αt x βt x p
p
p
p
n 2sp n2 −1 n − 2sp n2 − 1 4p n 1sp n2 − 1
4 n − 1
p
sp
p
p 1/p
p sp
2
n −1 4 n n −1
xsX .
2
3.84
Proof. In Theorem 3.6, putting ϕt x, y : θxrX ysX for all x, y ∈ X, we get the desired
result.
Corollary 3.9. Let θ ≥ 0 and r, s > 0 be nonnegative real numbers such that λ : r s /
4. Suppose
that a mapping f : X → Y with f0 0 satisfies the inequality 3.56 for all x, y ∈ X. Then there
exists a unique quartic mapping T : X → Y satisfying
f2x − 4fx − T x
≤
M8 θ
n2 n2 − 1
Y
1
n 2sp n − 2sp 4p n 1sp
16p − 2λp sp
rsp
4 n − 1 10 n 2
p
p sp
3.85
4 2 n 3
p rp
2p sp
1/p
p
2sp 6n2 − 2 17n2 − 8p
xλX
for all x ∈ X.
Proof. In Theorem 3.6, putting ϕt x, y : θxrX ysX for all x, y ∈ X, we get the desired
result.
24
Journal of Inequalities and Applications
Theorem 3.10. Let ϕ : X × X → 0, ∞ be a function such that
x y 1 lim 4m ϕ m , m 0 lim m ϕ 2m x, 2m y
m→∞
m
→
∞
2 2
16
3.86
for all x, y ∈ X and
∞
pi p x y
4 ϕ
,
< ∞,
2i 2i
i1
∞
1 p i
i
2
ϕ
x,
2
y
<∞
pi
i0 16
3.87
for all x ∈ X and all y ∈ {x, 2x, 3x, nx, n 1x, n − 1x, n 2x, n − 2x, n − 3x}. Suppose
that a mapping f : X → Y with f0 0 satisfies the inequality
Δf x, y ≤ ϕ x, y
Y
3.88
for all x, y ∈ X. Then there exist a unique quadratic mapping Q : X → Y and a unique quartic
mapping T : X → Y such that
9 fx − Qx − T x ≤ M 4 ψq x 1/p ψt x 1/p
Y
192
3.89
for all x ∈ X, where ψq x and ψt x are defined in Theorems 3.2 and 3.7, respectively.
Proof. By Theorems 3.2 and 3.7, there exist a quadratic mapping Q0 : X → Y and a quartic
mapping T0 : X → Y such that
8
f2x − 16fx − Q0 x ≤ M ψq x 1/p ,
Y
4
8
f2x − 4fx − T0 x ≤ M ψt x 1/p
Y
16
3.90
for all x ∈ X. It follows from the last inequalities that
9 fx 1 Q0 x − 1 T0 x ≤ M 4 ψq x 1/p ψt x 1/p
12
12
192
Y
3.91
for all x ∈ X. So we obtain 3.92 by letting Qx −1/12Q0 x and T x 1/12T0 x for
all x ∈ X.
To prove the uniqueness property of Q and T, we first show the uniqueness property
for Q0 and T0 and then we conclude the uniqueness property of Q and T. Let Q1 , T1 : X → Y
Journal of Inequalities and Applications
25
be another quadratic and quartic mappings satisfying 3.92 and let Q2 1/12Q0 , T2 1/12T0 , Q3 Q2 − Q1 and T3 T2 − T1 . So
Q3 x − T3 xY ≤ M fx − Q2 x − T2 xY fx − Q1 x − T1 xY
≤
3.92
1/p 1/p M10 4 ψq x
ψt x
96
for all x ∈ x. Since
lim 4mp ψq
m→∞
x 1
lim mp ψt 2m x 0
m → ∞ 16
2m
3.93
for all x ∈ X, 3.62 implies that limm → ∞ 4m Q3 x/2m 1/16m T3 2m xY 0 for all x ∈ X.
Thus T3 Q3 . But T3 is only a quartic function and Q3 is only a quadratic function.
Therefore, we have T3 Q3 0 and this completes the uniqueness property of Q and
T. We can prove the other results similarly.
Corollary 3.11. Let θ, r, s be nonnegative real numbers such that r, s > 4 or 2 < r, s < 4 or 0 ≤ r, s <
2. Suppose that a mapping f : X → Y satisfies the inequality3.51 for all x, y ∈ X. Then there exist
a unique quadratic mapping Q : X → Y and a unique quartic mapping T : X → Y such that
fx − Qx − T x
Y
⎧
⎪
⎪
⎪δq δt ,
⎪
⎪
⎪
⎪
⎪
⎨αq x αt x,
9
M θ
≤
12n2 n2 − 1 ⎪
βq x βt x,
⎪
⎪
⎪
⎪
⎪
1/p p
1/p
⎪
p
⎪
⎩ αpq x βqp x
α x β x
,
t
t
r s 0,
r > 0, s 0,
3.94
r 0, s > 0,
r, s > 0
for all x ∈ X, where δq , δt , αq x, αt x, βq x, and βt x are defined as in Corollaries 3.4 and 3.8.
Corollary 3.12. Let θ ≥ 0 and r, s > 0 be nonnegative real numbers such that λ : r s ∈ 0, 2 ∪
2, 4 ∪ 4, ∞. Suppose that a mapping f : X → Y satisfies the inequality 3.56 for all x, y ∈ X.
Then there exist a unique quadratic mapping Q : X → Y and a unique quartic function T : X → Y
such that
fx − Qx − T x
M9 θ
≤
12n2 n2 − 1
Y
1
n 2sp n − 2sp 4p n 1sp
4p − 2λp 4p n − 1sp 10p nsp 2rsp 4p 2rp n2p 3sp
p p
sp
2
2
xλX
2 6n − 2 17n − 8
for all x ∈ X.
3.95
26
Journal of Inequalities and Applications
Acknowledgment
The third and corresponding author was supported by the Basic Science Research Program
through the National Research Foundation of Korea funded by the Ministry of Education,
Science and Technology NRF-2009-0070788. Also, the second author would like to thank
the office of gifted students at Semnan University for its financial support.
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