




































©2024 Ada Academica https://adac.eeEur. J. Math. Anal. 4 (2024) 11doi: 10.28924/ada/ma.4.11
On the Stability of Hyers Orthogonality Functional Equations in Non-Archimedean Spaces

Wenhui Xu, Qi Liu, Jinyu Xia∗
School of Mathematics and Physics, Anqing Normal University, Anqing 246133, P. R. Chinaxuwenhuiwww@163.com, liuq67@aqnu.edu.cn, Y23060036@stu.aqnu.edu.cn

∗Correspondence: Y23060036@stu.aqnu.edu.cn
Abstract. In this paper, we investigate the stability of specially orthogonally functional equationsderiving from additive and quadratic functions
4f (x + y) + 4f (x − y) + 10f (x) + 14f (−x)− 3f (y)− 3f (−y) = f (2x + y) + f (2x − y)

and
f

(
x + y + z

2

)
+ f

(
x + y − z
2

)
+ f

(
x − y + z
2

)
+ f

(
y + z − x
2

)
= f (x) + f (y) + f (z)

where f is a mapping from Abelian group to a non-Archimedean space. By adopting a new method,we have made an attempt to prove the Hyers-Ulam stability in non-Archimedean spaces.

1. Introduction and preliminaries
The stability problem of functional equations originated from Ulam in 1940 when he posedthe group homomorphism problem "Given an approximately linear mapping f , when does a linearmapping T exist that approximates f ?" In 1941, Hyers [1] explored the scenario of approximatelyadditive mapping f : X → Y where X and Y are Banach spaces and f satisfies

‖f (x + y)− f (x)− f (y)‖ 6 ε

for all x, y ∈ X . Then there is a unique mapping additive L : X → Y satisfying
‖f (x)− L(x)‖ 6 ε

with the limit
L(x) = lim

n→∞

f (2nx)

2n
.

Rassias [14] weakened the bounded Cauchy difference proposed by Hyers in the map and ex-tended it to the unbounded Cauchy difference
‖f (x + y)− f (x)− f (y)‖ 6 ε(‖x‖p + ‖y‖p)

Received: 5 Mar 2024.Key words and phrases. Orthogonality; Stability; Non-Archimedean space; Functional equations.1

https://adac.ee
https://doi.org/10.28924/ada/ma.4.11


Eur. J. Math. Anal. 10.28924/ada/ma.4.11 2where ε > 0 and p ∈ [0, 1), Hyers’ theorem was extended to approximately linear maps. R. Gerand J. Sikorska [7] restricted the conditions with (x, y) = 0 and investigated the stability of theCauchy functional
f (x + y) = f (x) + f (y) (1.1)

Of course it is easy to spot that the function f (x) = ‖x‖2 satisfies the functional equations (1.1)by the Pythagorean theorem. They founded that there exists a orthogonality additive mapping
g : X → Y such that

‖f (x)− g(x)‖ 6
16

3
ε

for all x ∈ X with restriction on definition domain (1.1) was denoted as a additive equations.Similarly, the equation was called as a quadratic equation which satisfies
f (x + y) + f (x − y) = 2f (x) + 2f (y). (1.2)

During several decades, mathematicians have achieved various fruits in studying the stability offunctional equations based one these two equations in the spirit of Hyers-Ulam-Rassias.Now let us introduce the concept of orthogonality ⊥ defined by Rätz [16]. Suppose X is a realvector space with dimX > 2 and ⊥ is a binary relation on X are characterized by the followingproperties:(i) totality of ⊥ for zero: x ⊥ 0, 0 ⊥ x for all x ∈ X;(ii) homogeneity: if x, y ∈ X , x ⊥ y , then λx ⊥ µy for all λ, µ ∈ R ;(iii) independence: if x, y ∈ X \ {0}, x ⊥ y , if and only if x, y are linearly independent;(iv) for any two-dimensional subspace P of X and for every x ∈ P , there exists λ, y ∈ P such that
x ⊥ y and x + y ⊥ λx − y .The pair (X , ⊥) is called an orthogonality space, which means an orthogonality space havinga normed structure. Various notions of othogonlity on a real normed space such as Roberts,Pythagorean, Isosceles, Birkhoff-James, Carlsson, Hermite–Hadamard (HH) type orthogonalitieson the basis of the fundamental properties.

Definition 1.1. [16] A function ‖·‖ :X → [0,∞) on a vector space over X a scalar field K with anon-Archimedean valuation | · |, is classified as a non-Archimedean norm if it meets the followingconditions:(i) nonnegativity: ‖x‖ > 0 and ‖x‖ = 0 if and only if x = 0;(ii) homogeneity: ‖λx‖ = |λ| ‖x‖ ∀λ ∈ K,∀x, y ∈ X;(iii) the strong triangle inequality
‖x + y‖ 6 max {‖x‖ , ‖y‖} ∀x, y ∈ X

Then (X,‖·‖) is called a non-Archimedean normed space.

https://doi.org/10.28924/ada/ma.4.11


Eur. J. Math. Anal. 10.28924/ada/ma.4.11 3Gordji [9] investigated the stability of the traditionally functional equations
D(x, y) = f (x + y)− f (x)− f (y)

where f : X → Y X, Y are both non-Arohimedean Banach spaces. They established the existenceof functions ϕ,ψ : A× A→ [0,∞) such that
‖D(x, y)‖ 6 ϕ(x, y)
‖f (xy)− f (x)f (y)‖ 6 ψ(x, y)

for all x, y ∈ X , and they considered the case if there exists a constant 0 < L < 1 such that
ϕ(2x, 2y) 6 |2|Lϕ(x, y)
ϕ(2x, 2y) 6 |2|2Lψ(x, y)

Then there exist a unique ring homomorphis H : X → Y such that
‖f (x)−H(x)‖ 6

1

|2|(1− L)ϕ(x, x)

Kang [10] explored the stability of the orthogonally functional equation(1.3) through the classi-fication of the oddness and evenness of f within the same spaces
4f (x + y) + 4f (x − y) + 10f (x) + 14f (−x)− 3f (y)− 3f (−y) = f (2x + y) + f (2x − y) (1.3)

Park [12] investigated the stability of the orthogonally additive-additive and orthogonallyquadratic-quadratic functional equation(1.4) in non-Archimedean orthogonality spaces using con-ventional methods
f

(
x + y + z

2

)
+ f

(
x + y − z
2

)
+ f

(
x − y + z
2

)
+ f

(
y + z − x
2

)
= f (x) + f (y) + f (z)(1.4)

Drawing inspiration from [14], this paper we explore different spaces and employ new methodsto investigete the stability of the aforementioned equation(1.4) and (1.3).
2. Stability of the orthogonally additive-quadratic functional equation

In this section, we will use the following symbol
D1f (x, y) = f (2x + y) + f (2x − y)− 4f (x + y)− 4f (x − y)

−10f (x)− 14f (−x) + 3f (y) + 3f (−y)
(2.1)

we deal with the stability problem for the orthogonally additive-quartic functional equation for
D1f (x, y) = 0 by referring to the stability proof of [13, 14].

https://doi.org/10.28924/ada/ma.4.11


Eur. J. Math. Anal. 10.28924/ada/ma.4.11 4

Lemma 2.1. Assume f : G → X be a mapping with G be an Abelian group and (X, ‖ · ‖) be acomplete non -Archimedean normed space. For all x, y ∈ G and there is a constant C > 0 suchthat ∥∥∥∥f (2x)− 38 f (4x) + 18 f (−4x)
∥∥∥∥ 6 C (2.2)

Then we define
h(x, n) =

∥∥∥∥f (2x)− 2n + 12 · 4n f
(
2n+1x

)
+
2n − 1
2 · 4n f

(
−2n+1x

)∥∥∥∥
and

gn(x) =
2n + 1

2 · 4n f (2
nx)−

2n − 1
2 · 4n f (−2

nx) . n ∈ N

(1)Then we have
|h(x, n + 1)− h(x, n)| 6

2n + 1

2 · 4n C (2.3)
h(x, n) 6 C (2.4)

(2)and {gn(x)} is a Cauchy sequence, for every x ∈ G. Hence, the mapping g : G → X can bedefined as
g(x) = lim

n→∞
gn(x)

and then we get
‖f (2x)− g(2x)‖ 6 C (2.5)

Proof: Adding one and subtracting one with h(x, n + 1) for matching and then using the in-equality, we obtain∥∥∥∥f (2x)− 2n+1 + 12 · 4n+1 f
(
2n+2x

)
+
2n+1 − 1
2 · 4n+1 f

(
−2n+2x

)∥∥∥∥
6

∥∥∥∥f (2x)− 2n + 12 · 4n f
(
2n+1x

)
+
2n − 1
2 · 4n f

(
−2n+1x

)∥∥∥∥
+
2n + 1

2 · 4n

∥∥∥∥f (2n+1x)− 38 f (2n+2x) +18 f (−2n+2x)
∥∥∥∥

+
2n − 1
2 · 4n

∥∥∥∥f (−2n+1 · x)+ 18 f (2n+2x)− 38 f (−2n+2x)
∥∥∥∥

6

∥∥∥∥f (2x)− 2n + 12 · 4n f
(
2n+1x

)
+
2n − 1
2 · 4n f

(
−2n+1x

)∥∥∥∥+ C ·max{2n + 12 · 4n ,
2n − 1
2 · 4n

}
Next, it is easy to get

|h(x, n + 1)− h(x, n)| 6
2n + 1

2 · 4n C

https://doi.org/10.28924/ada/ma.4.11


Eur. J. Math. Anal. 10.28924/ada/ma.4.11 5Then
h(x, n) =

∥∥∥∥∥
(
n∑
i=2

h(x, i)− h(x, i − 1)

)
+ h(x, 1)

∥∥∥∥∥
6 C ·max

{
2 + 1

2 · 4 ,
22 + 1

2 · 42 , · · · ,
2n + 1

2 · 4n , 1
}

= CNext, we have to prove that for every x ∈ G, the sequence
gn(x) =

2n + 1

2 · 4n f (2
nx)−

2n − 1
2 · 4n f (−2

nx) n ∈ N

is convergent in G. Since X is complete, it is sufficient to show that (gn(x))n∈N is a Cauchysequence for all x ∈ G. By matching ‖gn+1(x)− gn(x)‖twice then we have
‖gn+1(x)− gn(x)‖ 6

2n + 1

2 · 4n

∥∥∥∥f (2nx)− 38 f (2n+1x)+ 18 f (−2n+1x)
∥∥∥∥

+
2n − 1
2 · 4n

∥∥∥∥f (−2nx)− 38 f (−2n+1x)+ 18 f (2n+1x)
∥∥∥∥

6C ·max
{
2n + 1

2 · 4n ,
2n − 1
2 · 4n

}
=
2n + 1

2 · 4n Cfor each n ∈ N . This easily implies that {gn(x)} is a Cauchy sequence. The mapping g : G → Xcan be defined as
g(x) = lim

n→∞
gn(x)Through the above results, we can obtain

‖f (2x)− g(2x)‖ = ‖h(x, n) + gn(2x)− g(2x)‖ 6 C

In this section, let G be an Abelian group and let ⊥ be a binary relation defined on G with theproperties:(i) x ⊥ 0, 0 ⊥ x , for all x ∈ X;(ii) if x, y ∈ X and x ⊥ y , then x2 ⊥ y
2 , 2x ⊥ 2y , 4x ⊥ 4y and −x ⊥ −y .

Theorem 2.1. Suppose f : G → X where f is a mapping from an Abelian group to a completenon-Archimedean normed space. For ε > 0, when x ⊥ y for all x, y ∈ G,we obtain
‖D1f (x, y)‖ 6 ε (2.6)

and
‖f (x) + f (−x)‖ 6 ε (2.7)Then there exists a unique mapping g : G → X such that x ⊥ y implies

4g(x + y) + 4g(x − y) + 10g(x) + 14g(−x)− 3g(y)− 3g(−y) = g(2x + y) + g(2x − y) (2.8)

https://doi.org/10.28924/ada/ma.4.11


Eur. J. Math. Anal. 10.28924/ada/ma.4.11 6and
‖f (x)− g(x)‖ 6

7

2
ε (2.9)

for all x ∈ 2G = {2x : x ∈ G}.
Proof. For all x ∈ X , since 0 ⊥ x , x ⊥ 0 and 0 ⊥ 0, setting x = 0, y = 0 in (2.6), we obtain
‖24f (0)‖ 6 ε, respectively, setting y = 0 in (2.6),we obtain the following inequality:

‖2f (2x)− 18f (x)− 14f (−x) + 6f (0)‖ 6 ε (2.10)
By using the strong triangle inequality, we obtain
‖2f (2x)−18f (x)−14f (−x)‖ 6 max{‖2f (2x)−18f (x)−14f (−x)+6f (0)‖, ‖6f (0)‖} 6 ε (2.11)

By replacing x with 4x in (2.7) and applying the triangle inequality twice, we obtain
‖2f (2x)− 4f (x)‖ 6 max{‖2f (2x)− 18f (x)− 14f (−x)‖, 14‖f (x) + f (−x)‖} = 14ε (2.12)

Applying (2.7) and (2.12) to ‖3f (4x)− 8f (2x)− f (−4x)‖, we can conclude that
‖3f (4x)− 8f (2x)− f (−4x)‖

= ‖4[f (4x)− 2f (2x)]− [f (4x) + f (−4x)]‖

6 max {28ε, ε} = 28ε (2.13)
This means that ∥∥∥∥f (2x)− 38 f (4x) + 18 f (−4x)

∥∥∥∥ 6 72ε (2.14)
The next step resembles Lemma2.1, let

gn(x) =
2n + 1

2 · 4n f (2
nx)−

2n − 1
2 · 4n f (−2

nx) (2.15)
then we can define a mapping g

g : G → X g(x) = lim
n→∞

gn(x).

According to Lemma2.1, we obtain
‖f (2x)− g(2x)‖ 6

7

2
ε (2.16)

we consider the following inequality
‖D1gn(x, y)‖

6

∥∥∥∥2n + 12 · 4n D1f (2
nx, 2ny) +

2n − 1
2 · 4n D1f (2

nx, 2ny)

∥∥∥∥
6
2n + 1

2 · 4n ε (2.17)

https://doi.org/10.28924/ada/ma.4.11


Eur. J. Math. Anal. 10.28924/ada/ma.4.11 7for all x, y ∈ G. Then we let n →∞, we get (2.8). Now, in order to prove g is unique, we assume
g′ as another mapping satisfying (2.8) and (2.9) that∥∥g(x)− g′(x)∥∥ = ∥∥g(x)− f (x) + f (x)− g′(x)∥∥

6 max
{
‖g(x)− f (x)‖,

∥∥f (x)− g′(x)∥∥}
= ε (2.18)

for all x ∈ 2G = {2x : x ∈ G}On the other hand, the mapping g − g′ satisfy (2.6) and(2.8)
g(2x)− g′(2x) =

2n + 1

2 · 4n
[
g
(
2n+1x

)
− g′

(
2n+1x

)]
−
2n − 1
2 · 4n

[
g
(
−2n+1x

)
− g′

(
−2n+1x

)] (2.19)
and therefore∥∥g(2x)− g′(2x)∥∥
6 max

{(
2n + 1

2 · 4n

)∥∥g (2n+xx)− g′ (2n+1 · x)∥∥ ,(2n − 1
2 · 4n

)∥∥g (−2n+1x)− g′ (−2n+1x)∥∥
6 max

{(
2n + 1

2 · 4n

)
ε,

(
2n − 1
2 · 4n

)
ε

}
=
2n + 1

2 · 4n ε (2.20)
for x ∈ G. By using the nonnegativity of norm and the forced convergence we can get that themapping g is unique on the set 2G.

�

3. Stability of additive-additive and orthogonally quadratic-quadratic functional equation
In this section, we substituted the equations with the orthogonally additive-additive and orthog-onally quadratic-quadratic functional equation concerning [12] in the same method and by referringto the stability proof of [13, 14], we define D2(x, y , z) as the followig

D2f (x, y , z) = f

(
x + y + z

2

)
+ f

(
x + y − z
2

)
+ f

(
x − y + z
2

)
+ f

(
y + z − x
2

)
− f (x)− f (y)− f (z)

Theorem 3.1. Suppose f : G → X where f is a mapping from an Abelian group to a completenon-Archimedean normed space. For ε > 0, when x ⊥ y for all x, y , z ∈ G,we obtain
‖D2f (x, y , z)‖ 6 ε (3.1)

and
‖f (x) + f (−x)‖ 6 ε. (3.2)

https://doi.org/10.28924/ada/ma.4.11


Eur. J. Math. Anal. 10.28924/ada/ma.4.11 8Then there exists a unique mapping g : X → Y such that x ⊥ y implies
g

(
x + y + z

2

)
+g

(
x + y − z
2

)
+g

(
x − y + z
2

)
+g

(
y + z − x
2

)
= g(x)+g(y)+g(z) (3.3)

and
‖f (x)− g(x)‖ 6 ε (3.4)for all x ∈ 2G = {2x : x ∈ G}.

Proof. For all x ∈ G, since 0 ⊥ x , x ⊥ 0, and 0 ⊥ 0, setting x = 0, y = 0, z = 0 in inquality (3.1),we obtain ‖f (0)‖ 6 ε, then similarily setting y = 0, z = 0 in inequality (3.1), we obtain∥∥∥∥3f (x2)+ f
(
−x
2

)
− f (x)− 2f (0)

∥∥∥∥ 6 ε (3.5)
Then, by using the strong triangle inequality, we obtain∥∥∥∥3f (x2)+ f

(
−x
2

)
− f (x)

∥∥∥∥
6 max

{
‖2f (0)‖,

∥∥∥3f (x
2

)
+ f

(
−
x

2

)
− f (x)− 2f (0)

∥∥∥}
= 2ε (3.6)

By replacing x witn 2x in (3.6), we obtain
‖3f (x) + f (−x)− f (2x)‖ 6 2ε (3.7)

By using the strong triangle inequality twice, we can easily obtain
‖2f (x)− f (2x)‖

6 max{‖f (x) + f (−x)‖, ‖3f (x) + f (−x)− f (2x)‖} = 2ε (3.8)
By replacing x witn 4x in (3.2),then combining the following with (3.2)and(3.8), we can concludethat

‖3f (4x)− 8f (2x)− f (−4x)‖

= ‖4[f (4x)− 2f (2x)]− [f (4ẋ) + f (−4x)]‖

6 max{8ε, ε} = 8ε (3.9)
Then dividing both side of the inequality by 8,we obtain∥∥∥∥f (2x)− 38 f (4x) + 18 f (−4x)

∥∥∥∥ 6 ε (3.10)
The next step resembles Lemma2.1,

gn(x) =
2n + 1

2 · 4n f (2
nx)−

2n − 1
2 · 4n f (−2

nx) n ∈ N.

Let
g : G → X g(x) = lim

n→∞
gn(x).

https://doi.org/10.28924/ada/ma.4.11


Eur. J. Math. Anal. 10.28924/ada/ma.4.11 9According to Lemma2.1, we obtain
‖f (2x)− g(2x)‖ = ‖h(x, n) + gn(2x)− g(2x)‖ 6 ε (3.11)

For the purpose of proving that g is orthogonally additive, firstiy, we apply the strong triangleinequality and the nonnegativity property for the following , we obtain
‖D2g(x, y , z)‖

=

∥∥∥∥2n + 12 · 4n D2f (2
nx, 2ny , 2nz) +

2n − 1
2 · 4n D2f (2

nx, 2ny , 2nz)

∥∥∥∥
6max

{
2n + 1

2 · 4n ε,
2n − 1
2 · 4n ε

}
=
2n + 1

2 · 4n ε (3.12)
for all x, y , z ∈ G with x ⊥ y and n ∈ N, n > 1. When we let n → ∞, we get (3.3). The rest ofproof resembles Theorem 2.1, according to (2.18)to (2.20) , we can get the mapping g is unique onthe set 2G similarly. �

Theorem 3.2. Suppose f : G → X where f is a mapping from an Abelian group to a completenon-Archimedean normed space. For ε > 0, when x ⊥ y for all x, y , z ∈ G,we obtain
‖D2f (x, y , z)‖ 6 ε (3.13)

and
‖f (x)− f (−x)‖ 6 ε. (3.14)

Then there exists a unique mapping g : X → Y such that x ⊥ y implies
g

(
x + y + z

2

)
+g

(
x + y − z
2

)
+g

(
x − y + z
2

)
+g

(
y + z − x
2

)
= g(x)+g(y)+g(z) (3.15)

and
‖f (x)− g(x)‖ 6

1

2
ε (3.16)

for all x ∈ 2G = {2x : x ∈ G}.
Proof. Our proof resembles Theorem3.1, the same step from (3.5) to (3.7), we get that

‖3f (x) + f (−x)− f (2x)‖ 6 2ε (3.17)
Adding (3.14) to (3.17) and using the triangle inequality, we can obtain

‖f (2x)− 4f (x)‖

6 max {‖3f (x) + f (−x)− f (2x)‖, ‖f (x)− f (−x)‖} = 2ε (3.18)

https://doi.org/10.28924/ada/ma.4.11


Eur. J. Math. Anal. 10.28924/ada/ma.4.11 10Hence, by using the result, there is
‖3f (4x)− 8f (2x)− f (−4x)‖

= ‖2[f (4x)− 4f (2x)] + f (4x)− f (−4x)]‖

6 max{4ε, ε} = 4ε (3.19)
The rest of proof is similar to the Theorem 3.1. �

Acknowledgments
Thanks to all the members of the Functional Analysis Research team of the College of Mathemat-ics and Physics of Anqing Normal University for their discussion and correction of the difficultiesand errors encountered in this paper. This research work was funded by Anhui Province HigherEducation Science Research Project (Natural Science), 2023AH050487.

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dema-2020-0009.

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https://doi.org/10.1073/pnas.27.4.222
https://doi.org/10.1073/pnas.27.4.222
https://doi.org/10.1090/s0002-9939-1978-0507327-1
https://doi.org/10.1515/GMJ.1999.33
https://doi.org/10.1016/j.jmaa.2005.05.052
https://doi.org/10.1016/j.na.2007.09.023
https://doi.org/10.1016/j.jmaa.2007.03.104
https://doi.org/10.1007/s00010-006-2868-0
https://doi.org/10.1007/s00010-006-2868-0
https://doi.org/10.4064/ba58-1-3
https://doi.org/10.1155/2011/123656
https://doi.org/10.5899/2012/jnaa-00123
https://doi.org/10.11650/twjm/1500406797
https://doi.org/10.1186/1029-242x-2012-139
https://doi.org/10.1515/dema-2020-0009
https://doi.org/10.1515/dema-2020-0009


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bf02189629.

https://doi.org/10.28924/ada/ma.4.11
https://doi.org/10.1016/j.jmaa.2022.126744
http://eudml.org/doc/144593
https://doi.org/10.1007/bf02189629
https://doi.org/10.1007/bf02189629

	1. Introduction and preliminaries
	2. Stability of the orthogonally additive-quadratic functional equation
	3. Stability of additive-additive and orthogonally quadratic-quadratic functional equation
	Acknowledgments
	References

