# On the Stability of Cubic and Quadratic Mapping in Random Normed Spaces under ArbitraryT-Norms

Authors: J. Vahidi; S. J. Lee; F. Fallah; R. Ahmadi
Journal: Journal of Function Spaces
 (2014)

---

## Abstract

We prove the stability of some functional equations in the random normed spaces under arbitraryt-norms.


---

## Body

## 1. Introduction and Preliminaries

The study of stability problems for functional equations is related to a question of Ulam [1] concerning the stability of group homomorphisms and affirmatively answered for Banach spaces by Hyers [2]. Subsequently, the result of Hyers was generalized by Aoki [3] for additive mappings and by Rassias [4] for linear mappings by considering an unbounded Cauchy difference. The paper [4] of Rassias has provided a lot of influence in the development of what we now call Hyers-Ulam-Rassias stability of functional equations. We refer the interested readers for more information on such problems to the papers [5–9]. In addition, some authors investigated the stability in the settings of fuzzy, probabilistic, and random normed spaces, and for more details see [10–22].The functional equation(1)
f
(
x
+
2
y
)
+
f
(
x
-
2
y
)
=
4
f
(
x
+
y
)
+
4
f
(
x
-
y
)
-
6
f
(
x
)
is said to be the cubic functional equation since the function f
(
x
)
=
c
x
3 is its solution. Every solution of the cubic functional equation is said to be a cubic mapping. The stability problem for the cubic functional equation was solved by Skof [23] for mappings f
:
X
→
Y, where X is a normed space and Y is a Banach space.The functional equation(2)
f
(
x
+
2
y
)
+
f
(
x
-
2
y
)
=
4
f
(
x
+
y
)
+
4
f
(
x
-
y
)
-
6
f
(
x
)
+
2
f
(
2
y
)
-
8
f
(
y
)
is said to be the quadratic functional equation since the function f
(
x
)
=
c
x
2 is its solution. Every solution of the quadratic functional equation is said to be a quadratic mapping (see [8, 9]).Throughout this paper, the space of all probability distribution functions is denoted byΔ
+
=
{
F
:
R
∪
{
-
∞
,
+
∞
}
→
[
0,1
]
:
F
is
left-continuous
and
non
decreasing
on
⁡
R
and
F
(
0
)
=
0
,
F
(
+
∞
)
=
1
} and the subset D
+
⊆
Δ
+ is the set D
+
=
{
F
∈
Δ
+
:
l
-
F
(
+
∞
)
=
1
}, where l
-
f
(
x
) denotes the left limit of the function f at the point x. The space is partially ordered by the usual pointwise ordering of functions; that is, F
⩽
G if and only if F
(
t
)
⩽
G
(
t
) for all t in R. The maximal element for Δ
+ in this order is the distribution function given by
(3)
ɛ
0
(
t
)
=
{
0
t
≤
0
1
t
>
0
.Definition 1 (see [24]).
A mappingT
:
[
0,1
]
×
[
0,1
]
→
[
0,1
] is a continuous triangular norm (briefly, a t-norm) if T satisfies the following conditions:(i)
Tis commutative and associative;
(ii)
T is continuous;
(iii)
T
(
a
,
1
)
=
a for all a
∈
[
0,1
];
(iv)
T
(
a
,
b
)
≤
T
(
c
,
d
) whenever a
≤
c and b
≤
d (a
,
b
,
c
,
d
∈
[
0,1
]).
Recall that, ifT is a t-norm and {
a
n
} is a given sequence of numbers in [
0,1
], T
i
=
1
n
a
i is defined recursively by T
i
=
1
1
a
i
=
a
1 and T
i
=
1
n
a
i
=
T
(
T
i
=
1
n
-
1
a
i
,
a
n
) for n
≥
2.Definition 2 (see [25]).
Arandom normed space (briefly, RN-space) is a triple (
X
,
μ
,
T
), where X is a vector space, T is a continuous t-norm, and μ is a mapping from X into D
+ such that, the following conditions hold:(RN1)μ
x
(
t
)
=
ɛ
0
(
t
) for all t
>
0 if and only if x
=
0;
(RN2)μ
α
x
(
t
)
=
μ
x
(
t
/
|
α
|
) for all x
∈
X, α
≠
0;
(RN3)μ
x
+
y
(
t
+
s
)
≥
T
(
μ
x
(
t
)
,
μ
y
(
s
)
) for all x
,
y
∈
X and t
,
s
≥
0.Definition 3.
Let(
X
,
μ
,
T
) be an RN-space. Consider the following.(1)
A sequence{
x
n
} in X is said to beconvergent to x in X if, for every ϵ
>
0 and λ
>
0, there exists a positive integer N such that μ
x
n
-
x
(
ϵ
)
>
1
-
λ whenever n
≥
N.
(2)
A sequence{
x
n
} in X is calledCauchy if, for every ϵ
>
0 and λ
>
0, there exists a positive integer N such that μ
x
n
-
x
m
(
ϵ
)
>
1
-
λ whenever n
≥
m
≥
N.
(3)
An RN-space(
X
,
μ
,
T
) is said to becomplete if every Cauchy sequence in X is convergent to a point in X.Theorem 4 (see [24]).
If(
X
,
μ
,
T
) is an RN-space and {
x
n
} is a sequence such that x
n
→
x, then lim
⁡
n
→
∞
μ
x
n
(
t
)
=
μ
x
(
t
) almost everywhere.


## 2. On the Stability of Cubic Mappings in RN-Spaces

Theorem 5.
LetX be a normed space, let (
R
+
,
μ
′
,
T
) be an RN-space, and let φ
:
X
×
X
→
[
0
,
+
∞
) be a mapping such that for some α
<
1,
(4)
μ
φ
(
x
,
y
)
′
(
t
)
≥
μ
8
α
φ
(
x
/
2
,
y
/
2
)
′
(
t
)
(
x
,
y
∈
X
,
t
>
0
)
,
(5)
lim
⁡
m
,
n
→
∞
T
k
=
1
n
+
1
(
μ
4
φ
(
x
,
x
)
′
(
8
t
2
∑
k
=
m
+
1
n
+
m
α
k
-
1
)
,
m
i
i
μ
φ
(
2
x
,
x
)
′
(
8
t
2
∑
k
=
m
+
1
n
+
m
α
k
-
1
)
)
=
1
(
x
,
y
∈
X
,
t
>
0
)
,
and let (
Y
,
μ
,
T
) be a complete RN-space. If f
:
X
→
Y is an odd mapping such that f
(
0
)
=
0 and
(6)
μ
f
(
x
+
2
y
)
+
f
(
x
-
2
y
)
-
4
f
(
x
+
y
)
-
4
f
(
x
-
y
)
+
6
f
(
x
)
(
t
)
≥
μ
φ
(
x
,
y
)
′
(
t
)
ssssssssssssssssssssssssssssssssssi
(
x
,
y
∈
X
,
t
>
0
)
then there exists a unique cubic mapping C
:
X
→
Y such that
(7)
μ
f
(
2
x
)
-
2
f
(
x
)
-
C
(
x
)
(
t
)
≥
T
k
=
1
∞
(
μ
(
1
/
2
)
φ
(
x
,
x
)
′
(
(
1
-
α
)
t
2
)
,
m
m
m
m
m
m
m
m
m
m
m
m
i
m
m
μ
(
1
/
8
)
φ
(
2
x
,
x
)
′
(
(
1
-
α
)
t
2
)
)
.Proof.
Lettingx
=
y in (6) we get
(8)
μ
f
(
y
+
2
y
)
+
f
(
y
-
2
y
)
-
4
f
(
y
+
y
)
-
4
f
(
y
-
y
)
+
6
f
(
y
)
(
t
)
≥
μ
φ
(
y
,
y
)
′
(
t
)
thus
(9)
μ
f
(
3
y
)
+
f
(
-
y
)
-
4
f
(
2
y
)
-
4
f
(
0
)
+
6
f
(
y
)
(
t
)
≥
μ
φ
(
y
,
y
)
′
(
t
)
since f is an odd mapping and f
(
0
)
=
0 then we have
(10)
μ
f
(
3
y
)
-
4
f
(
2
y
)
+
5
f
(
y
)
(
t
)
≥
μ
φ
(
y
,
y
)
′
(
t
)
(
y
∈
X
,
t
>
0
)
.
Replacing x by 2
y in (6)
(11)
μ
f
(
2
y
+
2
y
)
+
f
(
2
y
-
2
y
)
-
4
f
(
2
y
+
y
)
-
4
f
(
2
y
-
y
)
+
6
f
(
2
y
)
(
t
)
≥
μ
φ
(
2
y
,
y
)
′
(
t
)
we get
(12)
μ
f
(
4
y
)
-
4
f
(
3
y
)
+
6
f
(
2
y
)
-
4
f
(
y
)
(
t
)
≥
μ
φ
(
2
y
,
y
)
′
(
t
)
from Definition 2 we have
(13)
μ
4
f
(
3
y
)
-
16
f
(
2
y
)
+
20
f
(
y
)
+
f
(
4
y
)
-
4
f
(
3
y
)
+
6
f
(
2
y
)
-
4
f
(
y
)
(
t
)
≥
T
(
μ
4
f
(
3
y
)
-
16
f
(
2
y
)
+
20
f
(
y
)
(
t
2
)
,
μ
f
(
4
y
)
-
4
f
(
3
y
)
+
6
f
(
2
y
)
-
4
f
(
y
)
(
t
2
)
)
=
T
(
μ
4
(
f
(
3
y
)
-
4
f
(
2
y
)
+
5
f
(
y
)
)
(
t
2
)
,
μ
f
(
4
y
)
-
4
f
(
3
y
)
+
6
f
(
2
y
)
-
4
f
(
y
)
(
t
2
)
)
.
Therefore by (10) and (12), we get
(14)
μ
f
(
4
y
)
-
10
f
(
2
y
)
+
16
f
(
y
)
(
t
)
≥
T
(
μ
4
φ
(
y
,
y
)
′
(
t
2
)
,
μ
φ
(
2
y
,
y
)
′
(
t
2
)
)
(
y
∈
X
,
t
>
0
)
.
Replacing y by x and letting g
(
x
)
=
f
(
2
x
)
-
2
f
(
x
) in (14) we get
(15)
μ
f
(
4
x
)
-
10
f
(
2
x
)
+
16
f
(
x
)
(
t
)
=
μ
f
(
4
x
)
-
2
f
(
2
x
)
-
8
f
(
2
x
)
+
16
f
(
x
)
(
t
)
=
μ
g
(
2
x
)
-
8
g
(
x
)
(
t
)
.
Then
(16)
μ
(
1
/
8
)
g
(
2
x
)
-
g
(
x
)
(
t
)
≥
T
(
μ
4
φ
(
x
,
x
)
′
(
8
t
2
)
,
μ
φ
(
2
x
,
x
)
′
(
8
t
2
)
)
.
Replacing x by 2
n
x in (16) we obtain
(17)
μ
g
(
2
n
+
1
x
)
/
8
-
g
(
2
n
x
)
(
t
)
≥
T
(
(
8
t
2
)
μ
4
φ
(
2
n
x
,
2
n
x
)
′
(
8
t
2
)
,
m
m
m
m
m
m
m
m
m
m
μ
φ
(
2
n
+
1
x
,
2
n
x
)
′
(
8
t
2
)
)
,
then
(18)
μ
g
(
2
n
+
1
x
)
/
8
n
+
1
-
g
(
2
n
x
)
/
8
n
(
t
8
n
)
≥
T
(
μ
4
φ
(
2
n
x
,
2
n
x
)
′
(
8
t
2
)
,
μ
φ
(
2
n
+
1
x
,
2
n
x
)
′
(
8
t
2
)
)
.
Whence
(19)
μ
g
(
2
n
+
1
x
)
/
8
n
+
1
-
g
(
2
n
x
)
/
8
n
(
t
)
≥
T
(
(
8
n
+
1
t
2
)
μ
4
φ
(
2
n
x
,
2
n
x
)
′
(
8
n
+
1
t
2
)
,
m
m
m
m
m
m
m
m
m
i
m
m
m
m
m
m
μ
φ
(
2
n
+
1
x
,
2
n
x
)
′
(
8
n
+
1
t
2
)
)
and using (4) we have
(20)
μ
g
(
2
n
+
1
x
)
/
8
n
+
1
-
g
(
2
n
x
)
/
8
n
(
t
)
≥
T
(
μ
8
n
α
n
4
φ
(
x
,
x
)
′
(
8
n
+
1
t
2
)
,
μ
8
n
α
n
φ
(
2
x
,
x
)
′
(
8
n
+
1
t
2
)
)
≥
T
(
μ
4
φ
(
x
,
x
)
′
(
8
n
+
1
t
2
(
8
n
α
n
)
)
,
μ
φ
(
2
x
,
x
)
′
(
8
n
+
1
t
2
(
8
n
α
n
)
)
)
therefore
(21)
μ
g
(
2
n
+
1
x
)
/
8
n
+
1
-
g
(
2
n
x
)
/
8
n
(
t
)
≥
T
(
μ
4
φ
(
x
,
x
)
′
(
8
t
2
α
n
)
,
μ
φ
(
2
x
,
x
)
′
(
8
t
2
α
n
)
)
,
and from (21) we have
(22)
μ
g
(
2
n
+
1
x
)
/
8
n
+
1
-
g
(
2
n
x
)
/
8
n
(
α
n
t
8
)
≥
T
(
μ
4
φ
(
x
,
x
)
′
(
t
2
)
,
μ
φ
(
2
x
,
x
)
′
(
t
2
)
)
,
also, from above inequality we have
(23)
μ
Σ
k
=
1
n
g
(
2
k
x
)
/
8
k
-
g
(
2
k
-
1
x
)
/
8
k
-
1
(
∑
k
=
1
n
‍
α
k
-
1
t
8
)
=
T
(
(
α
n
-
1
t
8
)
μ
g
(
2
x
)
/
8
-
g
(
x
)
(
t
8
)
,
μ
g
(
2
2
x
)
/
8
2
-
g
(
2
x
)
/
8
(
α
t
8
)
,
i
i
i
i
i
…
,
μ
g
(
2
n
x
)
/
8
n
-
g
(
2
n
-
1
x
)
/
8
n
-
1
(
α
n
-
1
t
8
)
)
=
T
k
=
1
n
(
μ
g
(
2
k
x
)
/
8
k
-
g
(
2
k
-
1
x
)
/
8
k
-
1
(
α
k
-
1
t
8
)
)
≥
T
k
=
1
n
(
T
(
μ
4
φ
(
x
,
x
)
′
(
t
2
)
,
μ
φ
(
2
x
,
x
)
′
(
t
2
)
)
)
=
T
k
=
1
n
+
1
(
μ
4
φ
(
x
,
x
)
′
(
t
2
)
,
μ
φ
(
2
x
,
x
)
′
(
t
2
)
)
.
Then
(24)
μ
g
(
2
n
x
)
/
8
n
-
g
(
x
)
(
∑
k
=
1
n
‍
α
k
-
1
t
8
)
≥
T
k
=
1
n
+
1
(
μ
4
φ
(
x
,
x
)
′
(
t
2
)
,
μ
φ
(
2
x
,
x
)
′
(
t
2
)
)
,
(25)
μ
g
(
2
n
x
)
/
8
n
-
g
(
x
)
(
t
)
≥
T
k
=
1
n
+
1
(
μ
4
φ
(
x
,
x
)
′
(
8
t
2
∑
k
=
1
n
α
k
-
1
)
,
m
m
m
m
m
m
m
m
m
m
m
m
m
m
μ
φ
(
2
x
,
x
)
′
(
8
t
2
∑
k
=
1
n
α
k
-
1
)
)
.
By replacing x by 2
m
x in (25) we observe that
(26)
μ
g
(
2
n
+
m
x
)
/
8
n
-
g
(
2
m
x
)
(
t
)
≥
T
k
=
1
n
+
1
(
μ
4
φ
(
2
m
x
,
2
m
x
)
′
(
8
t
2
∑
k
=
1
n
α
k
-
1
)
,
μ
φ
(
2
m
+
1
x
,
2
m
x
)
′
(
8
t
2
∑
k
=
1
n
α
k
-
1
)
)
,
then
(27)
μ
g
(
2
n
+
m
x
)
/
8
n
+
m
-
g
(
2
m
x
)
/
8
m
(
t
8
m
)
≥
T
n
+
1
(
μ
4
φ
(
2
m
x
,
2
m
x
)
′
(
8
t
2
∑
k
=
1
n
α
k
-
1
)
,
m
m
m
m
m
m
μ
φ
(
2
m
+
1
x
,
2
m
x
)
′
(
8
t
2
∑
k
=
1
n
α
k
-
1
)
)
,
μ
g
(
2
n
+
m
x
)
/
8
n
+
m
-
g
(
2
m
x
)
/
8
m
(
t
)
≥
T
k
=
1
n
+
1
(
μ
4
φ
(
2
m
x
,
2
m
x
)
′
(
8
m
+
1
t
2
∑
k
=
1
n
α
k
-
1
)
,
m
m
m
m
m
m
m
m
μ
φ
(
2
m
+
1
x
,
2
m
x
)
′
(
8
m
+
1
t
2
∑
k
=
1
n
α
k
-
1
)
)
.
Using (4) and Definition 2 we have
(28)
μ
g
(
2
n
+
m
x
)
/
8
n
+
m
-
g
(
2
m
x
)
/
8
m
(
t
)
≥
T
k
=
1
n
+
1
(
μ
8
m
α
m
4
φ
(
x
,
x
)
′
(
8
m
+
1
t
2
∑
k
=
1
n
α
k
-
1
)
,
μ
8
m
α
m
φ
(
2
x
,
x
)
′
(
8
m
+
1
t
2
∑
k
=
1
n
α
k
-
1
)
)
≥
T
k
=
1
n
+
1
(
μ
4
φ
(
x
,
x
)
′
(
8
m
+
1
t
8
m
α
m
2
∑
k
=
1
n
α
k
-
1
)
,
μ
φ
(
2
x
,
x
)
′
(
8
m
+
1
t
8
m
α
m
2
∑
k
=
1
n
α
k
-
1
)
)
=
T
k
=
1
n
+
1
(
μ
4
φ
(
x
,
x
)
′
(
8
t
2
∑
k
=
m
+
n
+
m
α
k
-
1
)
,
μ
φ
(
2
x
,
x
)
′
(
8
t
2
∑
k
=
m
+
n
+
m
α
k
-
1
)
)
⟶
1
,
when m, n tend to ∞. Then {
g
(
2
n
x
)
/
8
n
} is a Cauchy sequence in (
Y
,
μ
,
T
). Since (
Y
,
μ
,
T
) is a complete RN-space this sequence converges to some point C
(
x
)
∈
Y. Fix x
∈
X and put m
=
0 in (28). Then
(29)
μ
g
(
2
n
x
)
/
8
n
-
g
(
x
)
(
t
)
≥
T
k
=
1
n
+
1
(
μ
4
φ
(
x
,
x
)
′
(
8
t
2
∑
k
=
1
n
α
k
-
1
)
,
μ
φ
(
2
x
,
x
)
′
(
8
t
2
∑
k
=
1
n
α
k
-
1
)
)
and from (29) for every t, δ
>
0,  x
∈
X, we have
(30)
μ
C
(
x
)
-
g
(
x
)
(
t
+
δ
)
=
μ
C
(
x
)
-
g
(
2
n
x
)
/
8
n
+
g
(
2
n
x
)
/
8
n
-
g
(
x
)
(
t
+
δ
)
≥
T
(
μ
C
(
x
)
-
g
(
2
n
x
)
/
8
n
(
δ
)
,
μ
g
(
2
n
x
)
/
8
n
-
g
(
x
)
(
t
)
)
≥
T
(
μ
C
(
x
)
-
g
(
2
n
x
)
/
8
n
(
δ
)
μ
′
φ
(
2
x
,
x
)
(
8
t
2
Σ
k
=
0
n
-
1
α
k
)
)
,
m
m
m
m
m
m
T
k
=
1
n
+
1
(
μ
4
φ
(
x
,
x
)
′
(
8
t
2
∑
k
=
0
n
-
1
α
k
)
,
m
m
m
m
m
m
m
m
m
m
μ
φ
(
2
x
,
x
)
′
(
8
t
2
∑
k
=
0
n
-
1
α
k
)
)
)
.
We know when n tend ∞, then g
(
2
n
x
)
/
8
n
→
C
(
x
), and ∑
k
=
1
n
‍
α
k
-
1
=
1
+
α
+
α
2
+
⋯
+
α
n
-
1
=
(
1
-
α
n
)
/
(
1
-
α
)
→
1
/
(
1
-
α
). Therefore taking the lim
⁡
n
→
∞ and using (30), and also by using Definition 1(iii) and Definition 2 and (
RN
2
) we get
(31)
μ
C
(
x
)
-
g
(
x
)
(
t
+
δ
)
≥
T
(
(
8
(
1
-
α
)
t
2
)
1
,
T
k
=
1
∞
(
μ
4
φ
(
x
,
x
)
′
(
8
(
1
-
α
)
t
2
)
,
m
m
m
m
m
m
μ
φ
(
2
x
,
x
)
′
(
8
(
1
-
α
)
t
2
)
)
)
=
T
k
=
1
∞
(
μ
(
1
/
2
)
φ
(
x
,
x
)
′
(
(
1
-
α
)
t
2
)
,
m
m
m
m
m
μ
(
1
/
8
)
φ
(
2
x
,
x
)
′
(
(
1
-
α
)
t
2
)
)
and since δ was arbitrary, by taking δ
→
0 in (31), we have
(32)
μ
g
(
x
)
-
C
(
x
)
(
t
)
≥
T
k
=
1
∞
(
μ
(
1
/
2
)
φ
(
x
,
x
)
′
(
(
1
-
α
)
t
2
)
,
μ
(
1
/
8
)
φ
(
2
x
,
x
)
′
(
(
1
-
α
)
t
2
)
)
which implies that the Inequality (7) holds. Replacing x and y by 2
n
x and 2
n
y in (6), respectively, we get
(33)
μ
f
(
2
n
x
+
2
n
+
1
y
)
+
f
(
2
n
x
-
2
n
+
1
y
)
-
4
f
(
2
n
x
+
2
n
y
)
-
4
f
(
2
n
x
-
2
n
y
)
+
16
f
(
2
n
x
)
(
t
)
≥
μ
φ
(
2
n
x
,
2
n
y
)
′
(
t
)
,
that by division by the 8
n we have
(34)
μ
f
(
2
n
(
x
+
2
y
)
)
/
8
n
+
f
(
2
n
(
x
-
2
y
)
)
/
8
n

-
4
f
(
2
n
(
x
+
y
)
)
/
8
n

-
4
f
(
2
n
(
x
-
y
)
)
/
8
n

+
16
f
(
2
n
x
)
/
8
n
×
(
t
8
n
)
≥
μ
φ
(
2
n
x
,
2
n
y
)
′
(
t
)
.
then(35)
μ
f
(
2
n
(
x
+
2
y
)
)
/
8
n

+
f
(
2
n
(
x
-
2
y
)
)
/
8
n
-
4
f
(
2
n
(
x
+
y
)
)
/
8
n

-
4
f
(
2
n
(
x
-
y
)
)
/
8
n
+

16
f
(
2
n
x
)
/
8
n
×
(
t
)
≥
μ
φ
(
2
n
x
,
2
n
y
)
′
(
8
n
t
)
≥
μ
8
n
α
n
φ
(
x
,
y
)
′
(
8
n
t
)
≥
μ
φ
(
x
,
y
)
′
(
t
α
n
)
for all x, y
∈
X and for all t
>
0. Since lim
⁡
n
→
∞
μ
φ
(
x
,
y
)
′
(
t
/
α
n
)
=
1, we conclude that C fulfills (1). To prove the uniqueness of the cubic function C, let us assume that there exists a quadratic function D
:
X
→
Y which satisfies (7). Obviously we have C
(
2
n
x
)
=
8
n
C
(
x
) and D
(
2
n
x
)
=
8
n
D
(
x
) for all x
∈
X and n
∈
ℕ. It follows from (7) and (32) that for each x
∈
X,
(36)
μ
C
(
x
)
-
D
(
x
)
(
t
)
≥
μ
C
(
2
n
x
)
-
D
(
2
n
x
)
(
8
n
t
)
≥
T
(
μ
C
(
2
n
x
)
-
g
(
2
n
x
)
(
8
n
t
2
)
,
m
i
μ
g
(
2
n
x
)
-
D
(
2
n
x
)
(
8
n
t
2
)
)
≥
T
(
μ
(
1
/
8
)
φ
(
2
n
+
1
x
,
2
n
x
)
′
(
(
1
-
α
)
8
n
t
2
)
)
1
,
T
k
=
1
∞
(
μ
(
1
/
8
)
φ
(
2
n
+
1
x
,
2
n
x
)
′
(
(
1
-
α
)
8
n
t
2
)
)
μ
(
1
/
2
)
φ
(
2
n
x
,
2
n
x
)
′
(
(
1
-
α
)
8
n
t
2
)
,
m
m
m
m
m
m
m
m
m
μ
(
1
/
8
)
φ
(
2
n
+
1
x
,
2
n
x
)
′
(
(
1
-
α
)
8
n
t
2
)
)
)
=
T
k
=
1
∞
(
μ
(
8
n
α
n
/
2
)
φ
(
x
,
x
)
′
(
(
1
-
α
)
8
n
t
2
)
,
m
i
μ
(
8
n
α
n
/
8
)
φ
(
2
x
,
x
)
′
(
(
1
-
α
)
8
n
t
2
)
)
≥
T
k
=
1
∞
(
μ
(
1
/
2
)
φ
(
x
,
x
)
′
(
(
1
-
α
)
t
2
α
n
)
,
m
m
m
m
m
m
i
μ
(
1
/
8
)
φ
(
2
x
,
x
)
′
(
(
1
-
α
)
t
2
α
n
)
)
⟶
1
,
then μ
C
(
x
)
-
D
(
x
)
(
t
)
=
1, for all t
>
0. And so C
(
x
)
=
D
(
x
). This completes the proof.


## 3. On the Stability of Quadratic Mappings in RN-Spaces

Theorem 6.
LetX be a normed space, let (
R
+
,
μ
′
,
T
) be an RN-space, and let φ
:
X
×
X
→
[
0
,
+
∞
) be a mapping such that for some α
<
1,
(37)
μ
φ
(
x
,
y
)
′
(
t
)
≥
μ
4
α
φ
(
x
/
2
,
y
/
2
)
′
(
t
)
(
x
,
y
∈
X
,
t
>
0
)
,
lim
⁡
m
,
n
→
∞
T
k
=
1
n
+
1
(
μ
4
φ
(
x
,
x
)
′
(
4
t
2
Σ
j
=
0
n
+
m
-
1
α
j
)
,
m
m
m
i
m
m
i
m
i
μ
φ
(
2
x
,
x
)
′
(
4
t
2
Σ
j
=
0
n
+
m
-
1
α
j
)
)
=
1
mmmmmmmmmmmmi
(
x
,
y
∈
X
,
t
>
0
)
,
and let (
Y
,
μ
,
T
) be a complete RN-space. If f
:
X
→
Y is an even mapping such that f
(
0
)
=
0 and(38)
μ
f
(
x
+
2
y
)
+
f
(
x
-
2
y
)
-
4
f
(
x
+
y
)
-
4
f
(
x
-
y
)
+
6
f
(
x
)
-
2
f
(
2
y
)
+
8
f
(
y
)
(
t
)
≥
μ
φ
(
x
,
y
)
′
(
t
)
(
x
,
y
∈
X
,
t
>
0
)
,then there exists a unique quadratic mapping Q
:
X
→
Y such that
(39)
μ
f
(
2
x
)
-
16
f
(
x
)
-
Q
(
x
)
(
t
)
≥
T
k
=
1
∞

(
μ
(
1
/
2
)
φ
(
x
,
x
)
′

(
(
1
-
α
)
t
2
)
,
μ
(
1
/
4
)
φ
(
2
x
,
x
)
′

(
(
1
-
α
)
t
2
)
)
.Proof.
Lettingx
=
y in (38) we get
(40)
μ
f
(
3
y
)
+
f
(
-
y
)
-
4
f
(
2
y
)
-
4
f
(
0
)
+
6
f
(
y
)
-
2
f
(
2
y
)
+
8
f
(
y
)
(
t
)
≥
μ
φ
(
y
,
y
)
′
(
t
)
.
Since f is an even mapping and f
(
0
)
=
0 then we have
(41)
μ
f
(
3
y
)
-
6
f
(
2
y
)
+
15
f
(
y
)
(
t
)
≥
μ
φ
(
y
,
y
)
′
(
t
)
(
y
∈
X
,
t
>
0
)
.
Replacing x by 2
y in (38) we get
(42)
μ
f
(
4
y
)
-
4
f
(
3
y
)
+
4
f
(
2
y
)
+
4
f
(
y
)
(
t
)
≥
μ
φ
(
2
y
,
y
)
′
(
t
)
and by (41) and (42), we get
(43)
μ
f
(
4
y
)
-
20
f
(
2
y
)
+
64
f
(
y
)
(
t
)
≥
T
(
μ
4
f
(
3
y
)
-
24
f
(
2
y
)
+
60
f
(
y
)
(
t
2
)
,
μ
f
(
4
y
)
-
4
f
(
3
y
)
+
4
f
(
2
y
)
+
4
f
(
y
)
(
t
2
)
)
=
T
(
(
t
2
)
μ
4
(
f
(
3
y
)
-
6
f
(
2
y
)
+
15
f
(
y
)
)
(
t
2
)
,
μ
f
(
4
y
)
-
4
f
(
3
y
)
+
4
f
(
2
y
)
+
4
f
(
y
)
(
t
2
)
)
.
Therefore
(44)
μ
f
(
4
y
)
-
20
f
(
2
y
)
+
64
f
(
y
)
(
t
)
≥
T
(
μ
4
φ
(
y
,
y
)
′
(
t
2
)
,
μ
φ
(
2
y
,
y
)
′
(
t
2
)
)
mmmmmimmimm
(
y
∈
X
,
t
>
0
)
.
Replacing y by x and letting g
(
x
)
=
f
(
2
x
)
-
16
f
(
x
) in (44), we get
(45)
μ
f
(
4
x
)
-
20
f
(
2
x
)
+
64
f
(
x
)
(
t
)
≥
T
(
μ
4
φ
(
x
,
x
)
′
(
t
2
)
,
μ
φ
(
2
x
,
x
)
′
(
t
2
)
)
,
μ
f
(
4
x
)
-
20
f
(
2
x
)
+
64
f
(
x
)
(
t
)
=
μ
f
(
4
x
)
-
16
f
(
2
x
)
-
4
f
(
2
x
)
+
64
f
(
x
)
(
t
)
=
μ
g
(
2
x
)
-
4
g
(
x
)
(
t
)
.
Then
(46)
μ
g
(
2
x
)
-
4
g
(
x
)
(
t
)
≥
T
(
μ
4
φ
(
x
,
x
)
′
(
t
2
)
,
μ
φ
(
2
x
,
x
)
′
(
t
2
)
)
that is,
(47)
μ
(
1
/
4
)
g
(
2
x
)
-
g
(
x
)
(
t
)
≥
T
(
μ
4
φ
(
x
,
x
)
′
(
4
t
2
)
,
μ
φ
(
2
x
,
x
)
′
(
4
t
2
)
)
.
Replacing x by 2
n
x in (46) we obtain
(48)
μ
g
(
2
n
+
1
x
)
/
4
-
g
(
2
n
x
)
(
t
)
≥
T
(
μ
4
φ
(
2
n
x
,
2
n
x
)
′
(
4
t
2
)
,
μ
φ
(
2
n
+
1
x
,
2
n
x
)
′
(
4
t
2
)
)
,
then
(49)
μ
g
(
2
n
+
1
x
)
/
4
n
+
1
-
g
(
2
n
x
)
/
4
n
(
t
4
n
)
≥
T
(
μ
4
φ
(
2
n
x
,
2
n
x
)
′
(
4
t
2
)
,
μ
φ
(
2
n
+
1
x
,
2
n
x
)
′
(
4
t
2
)
)
.
By using (2) and (37) we have
(50)
μ
g
(
2
n
+
1
x
)
/
4
n
+
1
-
g
(
2
n
x
)
/
4
n
(
t
)
≥
T
(
μ
4
φ
(
2
n
x
,
2
n
x
)
′
(
4
n
+
1
t
2
)
,
m
m
m
m
m
μ
φ
(
2
n
+
1
x
,
2
n
x
)
′
(
4
n
+
1
t
2
)
)
,
μ
g
(
2
n
+
1
x
)
/
4
n
+
1
-
g
(
2
n
x
)
/
4
n
(
t
)
≥
T
(
μ
4
n
α
n
4
φ
(
x
,
x
)
′
(
4
n
+
1
t
2
)
,
m
m
m
m
m
μ
4
n
α
n
φ
(
2
x
,
x
)
′
(
4
n
+
1
t
2
)
)
≥
T
(
μ
4
φ
(
x
,
x
)
′
(
4
n
+
1
t
2
(
4
n
α
n
)
)
,
m
m
m
μ
φ
(
2
x
,
x
)
′
(
4
n
+
1
t
2
(
4
n
α
n
)
)
)
,
so
(51)
μ
g
(
2
n
+
1
x
)
/
4
n
+
1
-
g
(
2
n
x
)
/
4
n
(
t
)
≥
T
(
μ
4
φ
(
x
,
x
)
′
(
4
t
2
α
n
)
,
m
m
i
m
m
μ
φ
(
2
x
,
x
)
′
(
4
t
2
α
n
)
)
,
and from (51) we have
(52)
μ
g
(
2
n
+
1
x
)
/
4
n
+
1
-
g
(
2
n
x
)
/
4
n
(
α
n
t
4
)
≥
T
(
μ
4
φ
(
x
,
x
)
′
(
t
2
)
,
μ
φ
(
2
x
,
x
)
′
(
t
2
)
)
,
and also, from the above inequality we have,
(53)
μ
∑
k
=
1
n
g
(
2
k
x
)
/
4
k
-
g
(
2
k
-
1
x
)
/
4
k
-
1
(
∑
k
=
1
n
‍
α
k
-
1
t
4
)
=
T
k
=
1
n
(
(
α
n
-
1
t
4
)
…
,
μ
g
(
2
n
x
)
/
4
n
-
g
(
2
n
-
1
x
)
/
4
n
-
1
(
α
n
-
1
t
4
)
μ
g
(
2
x
)
/
4
-
g
(
x
)
(
t
4
)
,
μ
g
(
2
2
x
)
/
4
2
-
g
(
2
x
)
/
4
(
α
t
4
)
,
…
,
μ
g
(
2
n
x
)
/
4
n
-
g
(
2
n
-
1
x
)
/
4
n
-
1
(
α
n
-
1
t
4
)
)
≥
T
k
=
1
n
(
μ
g
(
2
k
x
)
/
4
k
-
g
(
2
k
-
1
x
)
/
4
k
-
1
(
α
k
-
1
t
4
)
)
≥
T
k
=
1
n
(
T
(
μ
4
φ
(
x
,
x
)
′
(
t
2
)
,
μ
φ
(
2
x
,
x
)
′
(
t
2
)
)
)
=
T
k
=
1
n
+
1
(
μ
4
φ
(
x
,
x
)
′
(
t
2
)
,
μ
φ
(
2
x
,
x
)
′
(
t
2
)
)
.
Hence
(54)
μ
g
(
2
n
x
)
/
4
n
-
g
(
x
)
(
t
)
≥
T
k
=
1
n
+
1
(
μ
4
φ
(
x
,
x
)
′
(
4
t
2
∑
k
=
1
n
α
k
-
1
)
,
m
m
m
i
m
i
m
μ
φ
(
2
x
,
x
)
′
(
4
t
2
∑
k
=
1
n
α
k
-
1
)
)
.
By replacing x by 2
m
x in (54) we obtain
(55)
μ
g
(
2
n
+
m
x
)
/
8
n
-
g
(
2
m
x
)
(
t
)
≥
T
k
=
1
n
+
1
(
μ
4
φ
(
2
m
x
,
2
m
x
)
′
(
4
t
2
∑
j
=
0
n
-
1
α
j
)
,
m
m
m
m
i
m
m
μ
φ
(
2
m
+
1
x
,
2
m
x
)
′
(
4
t
2
∑
j
=
0
n
-
1
α
j
)
)
,
then
(56)
μ
g
(
2
n
+
m
x
)
/
4
n
+
m
-
g
(
2
m
x
)
/
4
m
(
t
4
m
)
≥
T
k
=
1
n
+
1
(
μ
4
φ
(
2
m
x
,
2
m
x
)
′
(
4
t
2
∑
j
=
0
n
-
1
α
j
)
,
m
m
m
m
i
i
i
i
i
μ
φ
(
2
m
+
1
x
,
2
m
x
)
′
(
4
t
2
∑
j
=
0
n
-
1
α
j
)
)
,
μ
g
(
2
n
+
m
x
)
/
4
n
+
m
-
g
(
2
m
x
)
/
4
m
(
t
)
≥
T
k
=
1
n
+
1
(
μ
4
φ
(
2
m
x
,
2
m
x
)
′
(
4
m
+
1
t
2
∑
j
=
0
n
-
1
α
j
)
,
m
m
m
m
i
i
i
i
i
μ
φ
(
2
m
+
1
x
,
2
m
x
)
′
(
4
m
+
1
t
2
∑
j
=
0
n
-
1
α
j
)
)
.
Using (37) and Definition 2 we observe that
(57)
μ
g
(
2
n
+
m
x
)
/
4
n
+
m
-
g
(
2
m
x
)
/
4
m
(
t
)
≥
T
k
=
1
n
+
1
(
μ
4
m
α
m
4
φ
(
x
,
x
)
′
(
4
m
+
1
t
2
∑
j
=
0
n
-
1
α
j
)
,
m
m
m
m
i
m
i
μ
4
m
α
m
φ
(
2
x
,
x
)
′
(
4
m
+
1
t
2
∑
j
=
0
n
-
1
α
j
)
)
≥
T
k
=
1
n
+
1
(
μ
4
φ
(
x
,
x
)
′
(
4
m
+
1
t
4
m
α
m
2
∑
j
=
0
n
-
1
α
j
)
,
m
m
m
i
m
m
i
μ
φ
(
2
x
,
x
)
′
(
4
m
+
1
t
4
m
α
m
2
∑
j
=
0
n
-
1
α
j
)
)
=
T
k
=
1
n
+
1
(
μ
4
φ
(
x
,
x
)
′
(
4
t
2
∑
j
=
0
n
+
m
-
1
α
j
)
,
m
m
m
m
i
m
i
μ
φ
(
2
x
,
x
)
′
(
4
t
2
∑
j
=
0
n
+
m
-
1
α
j
)
)
⟶
1
,
when m, n tend to ∞. Then {
g
(
2
n
x
)
/
4
n
} is a Cauchy sequence in (
Y
,
μ
,
T
). Since (
Y
,
μ
,
T
) is a complete RN-space this sequence converges to some point Q
(
x
)
∈
Y. Fix x
∈
X and put m
=
0 in (57). Thus
(58)
μ
g
(
2
n
x
)
/
4
n
-
g
(
x
)
(
t
)
≥
T
k
=
1
n
+
1
(
μ
4
φ
(
x
,
x
)
′
(
4
t
2
∑
j
=
0
n
-
1
α
j
)
,
m
m
m
m
i
i
i
i
m
μ
φ
(
2
x
,
x
)
′
(
4
t
2
∑
j
=
0
n
-
1
α
j
)
)
,
for every δ, t
>
0, x
∈
X, and from (58) we have
(59)
μ
Q
(
x
)
-
g
(
x
)
(
t
+
δ
)
=
μ
Q
(
x
)
-
g
(
2
n
x
)
/
4
n
+
g
(
2
n
x
)
/
4
n
-
g
(
x
)
(
t
+
δ
)
≥
T
(
μ
Q
(
x
)
-
g
(
2
n
x
)
/
4
n
(
δ
)
,
m
m
m
m
μ
g
(
2
n
x
)
/
4
n
-
g
(
x
)
(
t
)
)
≥
T
(
μ
′
φ
(
2
x
,
x
)
(
4
t
2
Σ
j
=
0
n
-
1
α
j
)
)
μ
Q
(
x
)
-
g
(
2
n
x
)
/
4
n
(
δ
)
,
m
m
m
m
m
T
k
=
1
n
+
1
(
μ
4
φ
(
x
,
x
)
′
(
4
t
2
∑
j
=
0
n
-
1
α
j
)
,
m
m
m
m
m
m
m
m
μ
φ
(
2
x
,
x
)
′
(
4
t
2
∑
j
=
0
n
-
1
α
j
)
)
)
,
we know when n tend ∞, then g
(
2
n
x
)
/
4
n
→
Q
(
x
), and ∑
j
=
0
n
-
1
‍
α
j
→
(
1
/
(
1
-
α
)
). Therefore taking the lim
⁡
n
→
∞ and using (59), also by using Definitions 1
(
iii
) and 2
(
RN
2
) we get
(60)
μ
Q
(
x
)
-
g
(
x
)
(
t
+
δ
)
≥
T
(
1
,
T
k
=
1
∞
(
μ
4
φ
(
x
,
x
)
′
(
4
(
1
-
α
)
t
2
)
,
m
m
m
m
m
m
m
m
m
μ
φ
(
2
x
,
x
)
′
(
4
(
1
-
α
)
t
2
)
)
)
≥
T
k
=
1
∞
(
μ
(
1
/
2
)
φ
(
x
,
x
)
′
(
(
1
-
α
)
t
2
)
,
m
m
m
m
m
μ
(
1
/
4
)
φ
(
2
x
,
x
)
′
(
(
1
-
α
)
t
2
)
)
,
since δ was arbitrary, and by taking δ
→
0 in (60), we get
(61)
μ
g
(
x
)
-
Q
(
x
)
(
t
)
≥
T
k
=
1
∞
(
μ
(
1
/
2
)
φ
(
x
,
x
)
′
(
(
1
-
α
)
t
2
)
,
m
m
m
i
μ
(
1
/
4
)
φ
(
2
x
,
x
)
′
(
(
1
-
α
)
t
2
)
)
,
which implies that the Inequality (39) holds. Replacing x and y by 2
n
x and 2
n
y in (38), respectively, we get
(62)
μ
f
(
2
n
x
+
2
n
+
1
y
)
+
f
(
2
n
x
-
2
n
+
1
y
)
-
4
f
(
2
n
x
+
2
n
y
)
-
4
f
(
2
n
x
-
2
n
y
)
+
6
f
(
2
n
x
)
-
2
f
(
2
n
+
1
y
)
+
8
f
(
2
n
y
)
(
t
)
≥
μ
φ
(
2
n
x
,
2
n
y
)
′
(
t
)
,that by division by the 4
n we have(63)
μ
f
(
2
n
(
x
+
2
y
)
)
/
4
n
+
f
(
2
n
(
x
-
2
y
)
)
/
4
n
-
4
f
(
2
n
(
x
+
y
)
)
/
4
n
-
4
f
(
2
n
(
x
-
y
)
)
/
4
n
+
6
f
(
2
n
x
)
/
4
n
-
2
f
(
2
n
(
2
y
)
)
/
4
n
+
8
f
(
2
n
y
)
/
4
n
(
t
)
≥
μ
φ
(
2
n
x
,
2
n
y
)
′
(
4
n
t
)
≥
μ
8
n
α
n
φ
(
x
,
y
)
′
(
4
n
t
)
≥
μ
φ
(
x
,
y
)
′
(
t
α
n
)
⟶
1
,for all x
,
y
∈
X and for all t
>
0, then Q fulfills (1). To prove the uniqueness of the quadratic function Q, let us assume that there exists a quadratic function H
:
X
→
Y which satisfies (39). Obviously we have Q
(
2
n
x
)
=
4
n
Q
(
x
) and H
(
2
n
x
)
=
4
n
H
(
x
) for all x
∈
X. From (39) and (61) we have
(64)
μ
Q
(
x
)
-
H
(
x
)
(
t
)
≥
T

(
μ
Q
(
2
n
x
)
-
g
(
2
n
x
)

(
4
n
t
2
)
,
μ
g
(
2
n
x
)
-
Q
(
2
n
x
)

(
4
n
t
2
)
)
≥
T
(
1
,
T
k
=
1
∞
(
μ
φ
(
2
n
x
,
2
n
x
)
′
(
(
1
-
α
)
4
n
t
2
)
,
m
i
μ
(
1
/
4
)
φ
(
2
n
+
1
x
,
2
n
x
)
′

(
(
1
-
α
)
4
n
t
2
)

)

)
=
T
k
=
1
∞
(
μ
4
n
α
n
φ
(
x
,
x
)
′
(
(
1
-
α
)
4
n
t
2
)
,
m
m
m
m
i
μ
(
4
n
α
n
/
4
)
φ
(
2
x
,
x
)
′
(
(
1
-
α
)
4
n
t
2
)
)
≥
T
k
=
1
∞
(
μ
φ
(
x
,
x
)
′
(
(
1
-
α
)
t
2
α
n
)
,
m
m
m
m
m
μ
(
1
/
4
)
φ
(
2
x
,
x
)
′
(
(
1
-
α
)
t
2
α
n
)
)
.
Since lim
⁡
n
→
∞
(
(
1
-
α
)
(
t
/
2
α
n
)
)
=
∞, we get lim
⁡
n
→
∞
μ
φ
(
x
,
x
)
′
(
(
1
-
α
)
(
t
/
2
α
n
)
)
=
1, and μ
n
→
∞
′
μ
(
1
/
4
)
φ
(
2
x
,
x
)
(
(
1
-
α
)
(
t
/
2
α
n
)
)
=
1, then μ
Q
(
x
)
-
H
(
x
)
(
t
)
=
1, for all t
>
0. And so Q
(
x
)
=
H
(
x
). This completes the proof.



---

*Source: 349281-2014-04-30.xml*