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A Note on the Stability of Functional Equations via a Celebrated Direct Method

Dongwen Zhang1, John Michael Rassias2, Qi Liu3, Yongjin Li4,∗
1School of Mathematics (Zhuhai), Sun Yat-sen University, Zhuhai 519082, P.R. China

zhangdw25@mail2.sysu.edu.cn
2National and Kapodistrian University of Athens, Department of Mathematics and Informatics,

Attikis 15342, Greece
jrassias@primedu.uoa.gr

3School of Mathematics and Physics, Anqing Normal University, Anqing 246133, P.R. China
liuq325@mail2.sysu.edu.cn

4Department of Mathematics, Sun Yat-sen University, Guangzhou, 510275, P.R. China
stslyj@mail.sysu.edu.cn

∗Correspondence: stslyj@mail.sysu.edu.cn

Abstract. More than ten years after Justyna Sikorska [8] attempted to solve the Heyers-Ulam sta-bility of a single variable equation by using direct method. In this paper, we will improve the resultsof Justyna Sikorska by using a more efficient approach. Relations between the generalized functionalequation, the dependence of their different parameters and several properties are also further ex-plored. To achieve the problem, we try to develop some new techniques to overcome the fundamentaldifficulties caused by the different properties of the function and the presence of several variables inthe equation. Furthermore, we continue to construct and study a couple of functional equations bymaking a new direct method.

1. Introduction
The core idea of the Hyers-Ulam stability for functional equations has been dated back to awell-known problem concerning about group homomorphisms solved by S.M. Ulam and D.H. Hyers(see [1–3]). In the last decades, a great number of papers treating the stability problem aboutfunctional equations has already been achieved and a great deal of important problems about thisfield has been studied ( [4–7]). It follows that the most efficient methods have been stated in manypapers ( [10, 18–24, 27]) such as the direct approach, the shadowing approach, and invariant meanapproach and so on. In particular, the direct method is always the main studying tool on theinvestigation of functional equations of different types.
Received: 15 Sep. 2022.
Key words and phrases. Stability; Several functional equations; Approximations; Odd function; Even function.1

https://adac.ee
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Eur. J. Math. Anal. 10.28924/ada/ma.3.7 2
The stability problems for an appropriate simple variable functional equations have earlier beeninvestigated by direct method. The direct method is familiar with many readers to derive thesolutions of the equations. The author in [8] have made full use of quite a general way to solve theHyers-Ulam stability problems on the functional equations under which many excellent outcomeshave been achieved without reduplicating the similar procedure in the whole process of computation.However, her results can only be used to derive the solutions of the equation where the mediatefunction is odd. This is exactly our contribution to the paper. In fact, a straightforward observationis that the inequality

‖f (x) −uf (e(x)) −vf (−e(x))‖6 δ(x)

can be solved if the function h is even. Next, the present studying approach calls us to investigatethe following functional inequality, by using a direct method, under which the result can not becovered by earlier works
‖f (x + y + z) + f (x) + f (y) + f (z) − f (x + y) − f (z + y) − f (x + z)‖
6 K (‖x‖r + ‖y‖r + ‖z‖r ) .

(1.1)
In fact, the functional inequality (1.1) comes from some equivalent characterizations of Hilbert spacein [15]. The investigator described several properties of an inner product space and applies theseresults to solve many interesting functional inequalities such as: Zarantone’s inequality, Hayashi’sinequality and so on. However, the more far reaching work can be done M. Fréchet in [16] underwhich he ascertained that the corresponding equation is a necessary prerequisite condition whencomplex or real normed completed spaces become Hilbert spaces. Investigator in [17] studied thestability of Fréchet functional equation from which a characterization of inner product spaces hadbeen achieved by using a stationary point theorem in Banach spaces. Compared with the beforestudying approaches, we further explored solutions of the equation (1.1) in this literature. Of course,to the best of our knowledge, it has also already been solved by [8] under which a direct methodwas to derive solutions of the equation (1.1) and to look for some improvement approximations.However, in this literature we make a new direct method to achieve the solution of equation (1.1)must be close to the approximate solution, approximately satisfying the corresponding equation.Besides this, we will consider that the functions on the functional equation of different typeshave been defined in a more general domain. For instance, the papers [11, 12] have defined anadditive ρ-functional inequalities in nonArchimedean normed spaces and Banach spaces. However,this phenomenon can not attract enough attention to the study of functional equations in themore general and complex nonlinear structure of F-spaces (see the definition in [13, 14]). But,the nonlinear structure of space has always stood in a very important position of leadership infunctional analysis. Based on the above analysis, it is of great significance that the functionalinequality is considered in β-homogeneous F-space.

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Eur. J. Math. Anal. 10.28924/ada/ma.3.7 3
In section 2, the counterpart of Theorem 2.1 from [8] where the mediate function is odd willbe considered. In the subsequent part, a new direct method for solving (1.1) in F-space will bedescribed and some new extended results of Theorem 2.1 from [8] will be presented. With it, twonew different applications of the results will be described in the final part.

2. A simple variable of abstract equation
In Theorem 2.1 from [8], Sikorska solved the equation (2.1) where the mediate function e is oddand the related parameters u, v are restricted on the real field. For simplicity in notation, weprovide traditionally our first result with the studying mapping defined in Banach space. By makinguse of small conjectures the more general form of the results will be provided in β-homogeneous

F-space in section 3. Therefore, our first result is simply considered in Banach space.
Theorem 2.1 Let (X, +) be a group, and (Y,‖ ·‖) be a Banach space, and assume the mapping

f : X → Y satisfying the inequality
‖f (x) −uf (e(x)) −vf (−e(x))‖6 δ(x), x ∈ X, (2.1)

where u, v ∈ (−∞, +∞), and the mappings e : X → X, δ : X → [0,∞) satisfy that e is even ( i.e.,
e(−x) = e(x) for every x ∈ X). Let the infinite progression ∑∞n=0 [|un|δ (en(x)) + |vn|δ (−en(x))]with

u0 := 1, un :=
[
u(u + v)n−1

]
,

v0 := 0, vn :=
[
v(u + v)n−1

]
, n ∈N(em states the m-th composition of function e ), be assumed convergence for every x ∈ X. Thenthere has a unique even mapping g : X → Y satisfying

g(x) = ung(e
n(x)) + vng(−en(x)),

and
‖f (x) −g(x)‖6

∞∑
i=0

[
|ui|δ

(
ei (x)

)
+ |vi|δ

(
−ei (x)

)]
, x ∈ X and n ∈N. (2.2)

Proof. We will prove that
‖f (x) −unf (en(x)) −vnf (−en(x))‖6 γn(x), x ∈ X, (2.3)

where
γn(x) :=

n−1∑
i=0

[
|ui|δ

(
ei (x)

)
+ |vi|δ

(
−ei (x)

)]
, x ∈ X,n ∈N.

First of all, consider with every m,n ∈N and it is easy to observe that
un+1 = uun + uvn, vn+1 = vvn + vun,

and
un+m = umun + vmun, vn+m = umvn + vmvn. (2.4)

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Eur. J. Math. Anal. 10.28924/ada/ma.3.7 4
From the definition of sequences (un) and (vn) we also have

uvn = vun, vmun = umvn.

First, (2.1) gives (2.3) with setting n = 1, and by mathematical induction, later we suppose that
(2.3) establishes for some n ∈ N. We prove that in the case for n + 1 by virtue of (2.1)

‖f (x) −un+1f
(
en+1(x)

)
−vn+1f

(
−en+1(x)

)
‖

6‖f (x) −unf (en(x)) −vnf (−en(x))‖

+ |un|
∥∥f (en(x)) −uf (en+1(x))−vf (−en+1(x))∥∥

+ |vn|
∥∥f (−en(x)) −uf (en+1(x))−vf (−en+1(x))∥∥

6
n−1∑
i=0

[
|ui|δ

(
ei (x)

)
+ |vi|δ

(
−ei (x)

)]
+ |un|δ (en(x)) + |vn|δ (−en(x))

=

n∑
i=0

[
|ui|δ

(
ei (x)

)
+ |vi|δ

(
−ei (x)

)]
.

Since the series ∑∞i=0 [|ui|δ(ei (x)) + |vi|δ(−ei (x))] is convergent for every x ∈ X, combinedwith (2.3) and by virtue of the completeness of Y , the mapping can be well defined as in thefollowing:
g(x) := lim

n→∞
[unf (e

n(x)) + vnf (−en(x))] , x ∈ X, (2.5)
and we prove the following properties of the function g.An easy computation is to prove that
ug(e(x)) + vg(−e(x))

= u lim
n→∞

[
unf

(
en+1(x)

)
+ vnf

(
−en+1(x)

)]
+ v lim

n→∞

[
unf

(
en+1(x)

)
+ vnf

(
−en+1(x)

)]
= lim
n→∞

[
(uun + vun) f

(
en+1(x)

)
+ (uvn + vvn) f

(
−en+1(x)

)]
= g(x).

Furthermore, we will prove the more general property of g
g(x) = ung (e

n(x)) + vng (−en(x)) , f or all x ∈ X and n ∈ N. (2.6)
By induction, we assume that the equation is true for all natural number k with k ≤ n for some

n ∈ N. Let us calculate with k = n + 1
g(x) = ung (e

n(x)) + vng (−en(x))

= un(ug
(
en+1(x)

)
+ vg

(
−en+1(x)

)
) + vn(ug

(
en+1(x)

)
+ vg

(
−en+1(x)

)
)

= un+1g
(
en+1(x)

)
+ vn+1g

(
−en+1(x)

)
).

In particular, we also have that the function g is even. An easy computation is to state that
g(−x) = ung (en(x)) + vng (−en(x)) = g(x), f or every x ∈ X and n ∈N.

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Eur. J. Math. Anal. 10.28924/ada/ma.3.7 5
In order to achieve the uniqueness of g, suppose further that ḡ : X → Y is the another mappingsuch that (2.2) and (2.6) hold. Then

‖g(x) − ḡ(x)‖6 2
∞∑
i=0

[
|ui|δ

(
ei (x)

)
+ |vi|δ

(
−ei (x)

)]
, x ∈ X.

Moreover, we have
g(x) − ḡ(x) = un [g (en(x)) − ḡ (en(x))] + vn [g (−en(x)) − ḡ (−en(x))] , x ∈ X,

and on account of (2.4) and (2.6) we can rewrite
‖g(x) − ḡ(x)‖6 |un + vn|‖g (en(x)) − ḡ (en(x))‖

6 2|un + vn|
∞∑
i=0

[
|ui|δ

(
ei+n(x)

)
+ |vi|δ

(
−ei+n(x)

)]
= 2

∞∑
i=0

[(|ui (un + vn)|) δ
(
ei+n(x)

)
+ |vi (un + vn)|δ

(
−ei+n(x)

)
]

= 2

∞∑
i=0

[
|ui+n|δ

(
ei+n(x)

)
+ |vi+n|δ

(
−ei+n(x)

)]
= 2

∞∑
j=n

[∣∣uj∣∣δ(ej(x)) + ∣∣vj∣∣δ(−ej(x))]
for every x ∈ X and n ∈N, where it states that g = ḡ as n →∞. This proves the theorem. �

The purpose of stating and proving this results is of particular interest and give out a solutionof a simple variable functional equation (2.1) at least. In section 3, we will extend the results ofTheorem 2.1 form [8] to a more general setting. In particular, the related parameters u, v can beextended to complex numbers.According to the above analysis, we give out a corollary of Theorem 2.1 (still quite general).First of all, we must state that the absolute of an element x ∈ X can be given out in the real fieldconsidering that the function h is even for the meaningful of the results, for example h(x) = a|x|.As a matter of fact, we can also present the absolute of x = (x1,x2, · · · ,xn) ∈ Rn by |x| =
(|x1|, |x2|, · · · , |xn|). Thus, it worth stating the results. In particular, we can present the followingresults in the Euclidean space if the more general setting can not be judged.

Corollary 2.1 Assume that (X, +) is a real or complex normed linear space and set (Y,‖ · ‖) isa Banach space. Suppose further that the mapping f : X → Y fulfils the inequality∥∥∥∥f (x) − a + 12a2 f (a|x|) + a− 12a2 f (−a|x|)
∥∥∥∥ 6 δ(x), x ∈ X, (2.7)

where a ∈ R with a > 1 and mappings e : X → X, δ : X → [0,∞) make that e is an even function
( i.e., e(−x) = e(x) for every x ∈ X). The infinite progression ∑∞i=0 1ai δ(ai|x|) is convergence for

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Eur. J. Math. Anal. 10.28924/ada/ma.3.7 6
every x ∈ X. Then there has a unique mapping g : X → Y fulfilling the following equations for all
x ∈ X

g(x) =
a + 1

2a2
g(a|x|) −

a− 1
2a2

g(−a|x|),

and
‖f (x) −g(x)‖6 ∆(x) + Λ(x),

where
∆(x) := 1

2

∑∞
i=0

1
ai

[
δ
(
ai|x|

)
+ δ

(
−ai|x|

)]
,

Λ(x) := 1
2

∑∞
i=0

1
a2i

[
δ
(
ai|x|

)
−δ

(
−ai|x|

)]
, x ∈ X.

Furthermore, g can be obtained in the following limiting equality
g(x) := lim

n→∞

(
an + 1

2a2n
f (an|x|) −

an − 1
2a2n

f (−an|x|)
)
, x ∈ X.

Proof. By using the results of Theorem 2.1, u := 1+a
2a2

, v := 1−a
2a2

and together e(x) := a|x|, for all
x ∈ X, a computation is to prove that

un :=
1 + a

2a2n
, vn :=

1 −a
2a2n

, n ∈N.

For the convergent series ∑∞i=0 1ai δ(ai|x|) with x ∈ X, therefore ∑∞i=0 1a2i δ(ai|x|) is convergence.Applying Theorem 2.1, there has a unique limiting function g : X → Y fulfilling
‖f (x) −g(x)‖6

∞∑
i=0

[∣∣∣∣1 + ai2a2i
∣∣∣∣δ(ai|x|) + ∣∣∣∣1 −ai2a2i

∣∣∣∣δ(−ai|x|)]
=

∞∑
i=0

1

a2i
[
δ
(
ai|x|

)
−δ

(
−ai|x|

)]
+

∞∑
i=0

1

ai
[
δ
(
ai|x|

)
+ δ

(
−ai|x|

)]
= Λ(x) + ∆(x).

Function g has been dated back to derived from (2.5). We complete the proof. �
Remark 2.2 If u = 1, v = 0 and e(x) = |x|, the above results may be trivial and meaningless.In the above results, suppose that a ∈ (−∞,∞) which is not equal to −1, 0, 1. Exchanging a with

−a, this transformation may not be different from primary inequality (2.7). This is a basic factleaving to the reader to check it. Assume that the convergent series ∑∞i=0 1|a|i δ(|a|i|x|) establishesfor every x ∈ X. In fact, the assertions with |a| exchanging for a has also been achieved by asimilar way.
Corollary 2.2 Assume that (X, +) is a group and set (Y,‖ · ‖) is a Banach space. Supposefurther that the mapping f : X → Y fulfils the inequality∥∥∥∥f (x) − a + 12a2 f (a|x|) + a− 12a2 f (−a|x|)

∥∥∥∥ 6 δ, x ∈ X,

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Eur. J. Math. Anal. 10.28924/ada/ma.3.7 7
where a ∈ (−∞,∞) with |a| > 1 and δ > 0 is constant. Then there is a unique limiting evenfunction g : X → Y fulfilling

‖f (x) −g(x)‖6
|a|δ
|a|− 1

.

Proof. Since the function δ is a positive constant, thus ∆(x) = |a||a|−1δ and Λ(x) = 0 for every
x ∈ X. The mapping g has been stated in the following shape:

g(x) := lim
n→∞

(
|a|n + 1

2a2n
f (|a|n|x|) −

|a|n − 1
2a2n

f (−|a|n|x|)
)
, x ∈ X.

This proves the proof. �
Remark 2.3 The above corollaries 2.1 and 2.2 will still establish in β-homogeneous F-spacewith a ∈ (−∞,∞) and |a| > 1. If we exchange a for 1

a
in the equation

f (x) −
a + 1

2a2
f (a|x|) +

a− 1
2a2

f (−a|x|)

from (2.7), the second group of results will also be obtained with a is a positive constant stated inthe following results.
Corollary 2.3 Assume that (X, +) is a group divisible by a with a ∈ (−∞,∞) and |a| > 1 andset (Y,‖·‖) is a Banach space. Suppose further that the mapping f : X → Y fulfils the inequality∥∥∥∥f (x) − a2 + a2 f

(
1

a
|x|
)
−
a2 −a

2
f

(
−

1

a
|x|
)∥∥∥∥ 6 δ(x), x ∈ X,

with δ : X → [0,∞) is such that the convergent series ∑∞i=0a2iδ( 1ai |x|) holds for every x ∈ X.Then there has a unique even limiting mapping g : X → Y fulfilling for every x ∈ X,
g(x) =

a2 + a

2
g

(
1

a
|x|
)

+
a2 −a

2
g

(
−

1

a
|x|
)
,

and
‖f (x) −g(x)‖6 ∆̃(x) + Λ̃(x).

Furthermore, the mapping g can be stated in the following shape:
g(x) := lim

n→∞

[
a2n + an

2
f

(
1

an
|x|
)

+
a2n −an

2
f

(
−

1

an
|x|
)]

, x ∈ X.

Proof. Applying for Theorem 2.1 for u := a2+a
2

, v := a2−a
2

and e(x) := 1
a
|x|, for all x ∈ X, an easycomputation is to show that

un :=
a2n + a2n−1

2
, vn :=

a2n −a2n−1

2
, n ∈N.

According to the convergent series ∑∞i=0a2iδ( 1ai |x|) for all x ∈ X, hence the series∑∞
i=0a

2i−1δ
(
1
ai
|x|
) is convergence, and there has a unique even limiting mapping g : X → Y

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Eur. J. Math. Anal. 10.28924/ada/ma.3.7 8
fulfilling
‖f (x) −g(x)‖6

∞∑
i=0

[∣∣∣∣a2i + ai2
∣∣∣∣δ( 1ai |x|

)
+

∣∣∣∣a2i −ai2
∣∣∣∣δ(− 1ai |x|

)]
=

∞∑
i=0

a2i

2

[
δ

(
1

ai
|x|
)

+ δ

(
−

1

ai
|x|
)]

+

∞∑
i=0

ai

2

[
δ

(
1

ai
|x|
)
−δ

(
−

1

ai
|x|
)]

= ∆̃(x) + Λ̃(x).

The definition of g is derived from (2.5). We complete the proof. �
Remark 2.4 Corollary 2.3 can be used to investigate the function from which it could be splitinto even and odd parts. There is a good point of the approach achieved here where the functionssplit into two two parts of odd and even functions can give more concise approximations than thebefore approximations in Theorem 2.1. The above results is the counterpart of the correspondingresults of Sikorska’s paper. However, it is not copied word by word. It is the counterpart of evenfunction.

3. The stability of functional equations in F-space
An F-space is called β-homogeneous if it satisfies ‖tx‖ = |t|β‖x‖ for every x ∈ X, t ∈ C. Inthis section of the first two theorems, β1, β2 are to be 0 < β1 ≤ 1 and 0 < β2 ≤ 1. Furthermore,we suppose X is β1-homogeneous F-space and Y is β2-homogeneous F-space. Before applyingTheorem 2.1 we would like to make an answer that all roads lead to Rome. Therefore anotherapproach to prove the following functional inequality has been stated in the following. In fact,there is also a similar solution about functional equation being stated in [8].
Theorem 3.1 Assume the mapping f : X → Y fulfilling for some K ≥ 0 and r < β2

β1

‖f (x + y + z) + f (x) + f (y) + f (z) − f (x + y) − f (z + y) − f (x + z)‖
6 K (‖x‖r + ‖y‖r + ‖z‖r )

(3.1)
for x,y,z ∈ X. Then there has a unique limiting mapping ψ1 : X → Y such that

‖f (x) −ψ1(x)‖6
(2 + 2rβ1 + 3 · 2β2)K

(2β1r − 22β2)(2β1r − 2β2)
‖x‖r

for x ∈ X. Moreover, ψ1 satisfying the above inequality is also satisfying the following equation
ψ1(x + y + z) + ψ1(x) + ψ1(z) + ψ1(y) = ψ1(x + y) + ψ1(z + y) + ψ1(x + z) (3.2)

for all x,y,z ∈ X.
Proof. From (x,x,x) in place of (x,y,z) in (3.1) we have

‖f (3x) + 3f (x) − 3f (2x)‖6 3K (‖x‖r ) .

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Eur. J. Math. Anal. 10.28924/ada/ma.3.7 9
Hence

‖2f (3x) + 6f (x) − 6f (2x)‖6 3 · 2β2K (‖x‖r ) .

Substitute (x,x, 2x) in place of (x,y,z) in (3.1), yielding that
‖f (4x) + 2f (x) − 2f (3x)‖6 (2 + 2rβ1)K (‖x‖r ) .

And combining the above two inequalities, we get
‖f (4x) + 8f (x) − 6f (2x)‖6 (2 + 2rβ1 + 3 · 2β2)K (‖x‖r ) . (3.3)

Let us define g(x) = f (2x) − 4f (x) for all x ∈ X. Hence
‖g(2x)/2 −g(x)‖6 (2 + 2rβ1 + 3 · 2β2)K (‖x‖r ) /2β2 (3.4)

for all x ∈ X. Therefore
‖g(2nx)/2n −g(2mx)/2m‖6

n−1∑
j=m

(2 + 2rβ1 + 3 · 2β2)K
2jβ1r

2β22jβ2
(‖x‖r ) (3.5)

for m, n ∈ N with n > m and all x ∈ X. Since the sequence {g(2nx)/2n} is a Cauchy sequencein Y for all x ∈ X and Y is complete, the mapping can be well defined as:
φ(x) = lim

n→∞
g(2nx)/2n

for all x ∈ X. In particular, letting m = 0 and setting n →∞ in (3.4), we have
‖φ(x) −g(x)‖6 (2+2

rβ1+3·2β2)K
2β1r−2β2 (‖x‖

r ) . (3.6)
Now, we prove the mapping φ is additive and is unique. From (x,y,y + x) in (3.1) yields that

‖f (2x + 2y) + f (x) + f (y) − f (x + 2y) − f (2x + y)‖6 K (‖x‖r + ‖y‖r + ‖x + y‖r ) .

From (x,x,y) in equation (3.1) yields that
‖f (2x + y) + 2f (x) + f (y) − f (2x) − 2f (x + y)‖6 K (2‖x‖r + ‖y‖r ) .

From (x,y,y) in (3.1) we have
‖f (x + 2y) + f (x) + 2f (y) − f (2y) − 2f (x + y)‖6 K (‖x‖r + 2‖y‖r ) .

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Eur. J. Math. Anal. 10.28924/ada/ma.3.7 10
Combining the above three inequalities, we have that for x,y,z ∈ X
‖φ(x + y) −φ(x) −φ(y)‖

= lim
n→∞

1

2β2n
‖f (2n+1x + 2n+1y) + 4f (2nx) + 4f (2ny) − f (2n+1x) − f (2n+1y) − 4f (2nx + 2ny)‖

6 lim
n→∞

1

2β2n

∥∥f (2n+1x + 2n+1y) + f (2nx) + f (2ny) − f (2n+1x + 2ny) − f (2n+1y + 2nx)∥∥
+ lim
n→∞

1

2β2n

∥∥f (2n+1x + 2ny) + 2f (2nx) + f (2ny) − f (2n+1x) − 2f (2nx + 2ny)∥∥
+ lim
n→∞

1

2β2n

∥∥f (2nx + 2n+1y) + f (2nx) + 2f (2ny) − f (2n+1y) − 2f (2nx + 2ny)∥∥
6 lim
n→∞

2β1rn

2β2n
K (4‖x‖r + 4‖y‖r + ‖x + y‖r ) .

So we have φ(x + y) = φ(x) + φ(y) for all x,y ∈ X.Next, the uniqueness of the mapping φ will be proved. Let u(x) be another additive mappingsuch that for some K2 ≥ 0 and r < β2β1 , ‖g(x) −u(x)‖6 K2‖x‖r2. Hence
‖φ(x) −u(x)‖ =‖φ(nx) −u(nx)‖/nβ2

6‖φ(nx) −g(nx)‖/nβ2 + ‖g(nx) −u(nx)‖/nβ2

6
(2 + 2rβ1 + 3 · 2β2)K

2β1r − 2β2
‖x‖rnrβ1−β2 + K2‖x‖r2nr2β1−β2

for all x ∈ X. Therefore φ(x) = u(x) for all x ∈ X. by the condition r < β2
β1

. So there has a uniqueadditive limiting mapping φ fulfilling
‖(f (x) −

1

2
φ(x)) − (f (2x) −

1

2
φ(2x))/4‖≤

(2 + 2rβ1 + 3 · 2β2)K
2β1r − 2β2

‖x‖r/22β2.

Hence
‖(f (x) −

1

2
φ(x)) − (f (2nx) −

1

2
φ(2nx))/4n‖≤

n−1∑
j=0

2β1rj

22β2j
(2 + 2rβ1 + 3 · 2β2)K

22β2(2β1r − 2β2)
‖x‖r.

Then the mapping can be well defined as
ψ(x) = lim

n→∞
(f (2nx) −

1

2
φ(2nx))/4n

for all x ∈ X, by the completeness of the space Y . Thus
‖ψ(x) − f (x) + φ(x)/2‖≤

(2 + 2rβ1 + 3 · 2β2)K
(2β1r − 2β2)(2β1r − 22β2)

‖x‖r.

Let u(x) be another limiting mapping which has the same property to the function ψ(x) such that,
‖u(x) −φ(x)‖6 ‖u(x) − (f (2nx) −

1

2
φ(2nx))/4n‖ + ‖(f (2nx) −

1

2
φ(2nx))/4n −φ(x)‖

6 2
∞∑
j=n

2β1rj

22β2j
(2 + 2rβ1 + 3 · 2β2)K

22β2(2β1r − 2β2)
‖x‖r

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Eur. J. Math. Anal. 10.28924/ada/ma.3.7 11
which shows that the approximation function φ(x) is unique. Finally, it remains to prove that φ(x)satisfies (3.2) and we obtain

1
4n
‖f (2nx + 2ny + 2nz) + f (2nx) + f (2ny) + f (2nz)−

f (2nx + 2ny) − f (2nz + 2ny) − f (2nx + 2nz)‖
6 2

β1n

4n
K (‖x‖r + ‖y‖r + ‖z‖r ) .

(3.7)
Letting n →∞, and we get our assertion by using the additivity of φ(x). �

In another direction, we will describe the similar stability results of the above Theorem 3.1.
Theorem 3.2 Let r > β2

β1
and assume that f : X → Y is a mapping satisfying the equation (3.1).Then there has a unique limiting mapping ψ1 : X → Y satisfying

‖f (x) −ψ1(x)‖6
(2 + 2rβ1 + 3 · 2β2)K

(2β1r − 22β2)(2β1r − 2β2)
‖x‖r

for all x ∈ X. Moreover, ψ1 solves also the following equation
ψ1(x + y + z) + ψ1(x) + ψ1(z) + ψ1(y) = ψ1(x + y) + ψ1(z + y) + ψ1(x + z) (3.8)

for all x,y,z ∈ X.
Proof. According to the equation (3.3), we obtain

‖g(x) − 2g(
x

2
)‖6 (2 + 2rβ1 + 3 · 2β2)K (‖x‖r ) /2β1r.

Therefore
‖2ng(

x

2n
) − 2mg(

x

2m
)‖6

n−1∑
j=m

(2 + 2rβ1 + 3 · 2β2)K
2jβ2

2jβ1r 2β1r
(‖x‖r )

for m,n ∈ N with n > m and x ∈ X. Since the sequence {2ng( x
2n

)} is a Cauchy sequence in Yfor all x ∈ X and Y is complete, the mapping can be well defined as:
φ(x) = lim

n→∞
2ng(

x

2n
)

for all x ∈ X. Using a similar manner, we can complete the rest part. �
If f (x) is odd, then (x,y,−x −y) in (3.2) can give a precise condition to ascertain the additiveproperty of the function f (x) (See [9]). Obviously, the additive property is stronger than theproperty of the equation (3.2), but vice versa is not true. In contrast with the subadditive property,we can not get obvious strong or weak property temporarily. By using another approach to solvethe Theorem 3.1, according to (3.6), we have

‖f (2nx)/22n − f (x) −
n−1∑
j=0

φ(2jx)/22(j+1)‖≤
n−1∑
j=0

2β1rj

22β2j
(2 + 2rβ1 + 3 · 2β2)K

22β2(2β1r − 2β2)
‖x‖r.

Then the mapping can be well defined as
ψ(x) = lim

n→∞
f (2nx)/22n

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Eur. J. Math. Anal. 10.28924/ada/ma.3.7 12
for all x ∈ X, by the completeness of the space Y . Thus

‖ψ(x) − f (x) −φ(x)/2‖≤
(2 + 2rβ1 + 3 · 2β2)K

(2β1r − 2β2)(2β1r − 22β2)
‖x‖r.

In a similar way, we can use two steps to prove that the mapping ψ(x) is unique. The first stepwe show that the mapping satisfies the property: ψ(kx) = k2ψ(x) for all k ∈ N, x ∈ X. We provethis by mathematical induction, for a fixed element x ∈ X. We will prove that the property is truefor k = 2. From (x,−x,x) in equation (3.1), we can get that
‖3f (x) + f (−x) − f (2x)‖6 3K (‖x‖r )

for all x ∈ X.Thus
‖f (−x) − f (x) −φ(x)‖≤‖f (2x) − 4f (x) −φ(x)‖ + ‖3f (x) + f (−x) − f (2x)‖

6 (
(2 + 2rβ1 + 3 · 2β2)K

2β1r − 2β2
+ 3K) (‖x‖r )

for all x ∈ X. Using the similar above argumentation together the above inequality and equation
(3.1), yields

ψ(−x) = ψ(x) + lim
n→∞

φ(x)

2nand
ψ(x + y + z) + ψ(x) + ψ(z) + ψ(y) = ψ(x + y) + ψ(z + y) + ψ(x + z) (3.9)

for all x,y,z ∈ X. From (x,−x,x) in equation (3.7), we achieve
ψ(2x) = 3ψ(x) + ψ(−x) = 4ψ(x).

Fixed x ∈ X, we prove this by induction. We have already proved that the property is true for
n = 2. Supposing that ψ(nx) = n2ψ(x) for all natural n ≤ 2k, with k ≥ 1, let us calculate
ψ((2k + 1)x). From (kx,kx,x) in (3.7), we know

ψ((2k + 1)x) = ψ(2kx) + 2ψ((k + 1)x) − 2ψ(kx) −ψ(x)

= (4k2 + 2(k + 1)2 − 2k2 − 1)ψ(x)

= (2k + 1)2ψ(x).

Now, we show the mapping ψ satisfies the property ψ(kx) = k2ψ(x) for all k ∈ N, x ∈ X. Thesecond step, we claim that the mapping φ is unique. Let u(x) be another limiting mapping suchthat for some K2 ≥ 0 and r < β2β1 ,
‖u(x) − f (x) −φ(x)/2‖6 K2‖x‖r2

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Eur. J. Math. Anal. 10.28924/ada/ma.3.7 13
which satisfies the property u(kx) = k2u(x) for all k ∈ N and x ∈ X. Therefore

‖ψ(x) −u(x)‖

=‖ψ1(kx) −u(kx)‖/k2β2

6‖u(xk) − f (kx) −φ(kx)/2‖/k2β2 + |ψ(xk) − f (kx) −φ(kx)/2‖/k2β2

6
(2 + 2rβ1 + 3 · 2β2)K

(2β1r − 22β2)(2β1r − 2β2)
‖x‖rkrβ1−2β2 + |K2‖x‖r2kr2β1−2β2.

Hence φ(x) = u(x) for all x ∈ X. This shows that ψ is unique. Let ψ1(x) = ψ(x) −φ(x)/2. Thiscompletes the uniqueness of ψ1(x). We have
‖ψ1(x) − f (x)‖6

(2 + 2rβ1 + 3 · 2β2)K
(2β1r − 22β2)(2β1r − 2β2)

‖x‖r

for all x ∈ X and also the equation (3.2) holds by using the additive property of φ and equation
(3.7). We complete the proof. We may also assume that limn→∞ φ(x)2n = limn→∞ g(2nx)/2n2n = 0.Otherwise, this limit may not be convergence to zero. Conversely, we may add some similar smalladditional assumptions to guarantee the convergence in Theorem 3.2.

4. The stability of functional equations in Banach space
In this section, we will prove the counterpart of the results of Theorem 2.1 from [8] to moregeneral case. We generalize the results of Sikorska in 2010. In particular, the related parameters

u, v can be extended to complex numbers by using a more efficient approach. Beyond that, westate that the first results in section 2 are presented and combined the first results in [8]. Ourcontribution to the parameters u,v are complex numbers. The results is stated in this section inmore detail.
Theorem 4.1 Suppose that (X, +) is a group, and (Y,‖ · ‖) is a Banach space, and let themapping f : X → Y satisfy the inequality

‖f (x) −uf (e(x)) −vf (−e(x))‖6 δ(x), x ∈ X,

where u,v ∈ C (C denotes the complex field.), and e : X → X, δ : X → [0,∞) are arbitrary givenfunctions.(1): If e is a even function ( i.e., e(−x) = e(x) for x ∈ X) and the convergent series
∞∑
n=0

[|un|δ (en(x)) + |vn|δ (−en(x))]

with
u0 := 1, un :=

[
u(u + v)n−1

]
, n ∈N,

v0 := 0, vn :=
[
v(u + v)n−1

]
, n ∈N

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Eur. J. Math. Anal. 10.28924/ada/ma.3.7 14
(and where en states the n-th composition of the function e ), establishes for every x ∈ X. Thenthere has a unique even limiting function g : X → Y fulfilling

g(x) = ung(e
n(x)) + vng(−en(x)), x ∈ X and n ∈ N, (4.1)

and
‖f (x) −g(x)‖6

∞∑
i=0

[
|ui|δ

(
ei (x)

)
+ |vi|δ

(
−ei (x)

)]
, x ∈ X. (4.2)

(2): If e is odd ( i·e., e(−x) = −e(x) for all x ∈ X) and the convergent series
∞∑
n=0

[|un|δ (en(x)) + |vn|δ (−en(x))] .

with
u0 := 1, un :=

1
2

[(u + v)n + (u −v)n] , n ∈N,
v0 := 0, vn :=

1
2

[(u + v)n − (u −v)n] , n ∈Nestablishes for all x ∈ X. Then there has a unique limiting mapping g : X → Y fulfilling (3.10)and (3.11).
Proof. We only need to prove the uniqueness of the approximation function. (1): Let us supposethat g̃ : X → Y is another approximating mapping. So let’s first prove the inequality together withthe equation (2.5) and g(−x) = g(x)

‖f (em(x)) −um(unf
(
en+m(x)

)
+ vnf

(
−em+n(x)

)
)

−vm(unf
(
en+m(x)

)
+ vnf

(
−em+n(x)

)
)‖

=‖f (em(x)) −un+mf
(
en+m(x)

)
−vn+mf

(
−en+m(x)

)
‖

6
n+m−1∑
j=m

[∣∣uj∣∣δ(ej(x)) + ∣∣vj∣∣δ(−ej(x))] ,
and letting n →∞ we have for any m ∈N

‖f (em(x)) −umg (em(x)) −vmg (−em(x))‖6
∞∑
j=m

[∣∣uj∣∣δ(ej(x)) + ∣∣vj∣∣δ(−ej(x))] ,
and we can rewrite

‖g(x) − g̃(x)‖6‖f (km(x)) −umg (em(x)) −vmg (−em(x))‖

+ ‖f (km(x)) −umg̃ (em(x)) −vmg̃ (em(x))‖

62
∞∑
j=m

[∣∣uj∣∣δ(ej(x)) + ∣∣vj∣∣δ(−ej(x))]
for any x ∈ X and m ∈N, which yields g = g̃ in X as m →∞.(2): Combined with the results of Theorem 2.1 from [8] where e is odd, we only need to prove theuniqueness of the approximation function. Let us suppose that g̃ : X → Y is another approximating

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Eur. J. Math. Anal. 10.28924/ada/ma.3.7 15
mapping. So let’s first prove the inequality together with the equation the results in Theorem 2.1in [8]

‖f (em(x)) −um(unf
(
en+m(x)

)
+ vnf

(
−em+n(x)

)
)

−vm(unf
(
−en+m(x)

)
+ vnf

(
em+n(x)

)
)‖

=‖f (em(x)) −un+mf
(
en+m(x)

)
−vn+mf

(
−en+m(x)

)
‖

6
n+m−1∑
j=m

[∣∣uj∣∣δ(ej(x)) + ∣∣vj∣∣δ(−ej(x))] ,
and letting n →∞ we have for any m ∈N

‖f (em(x)) −umg (em(x)) −vmg (−em(x))‖6
∞∑
j=m

[∣∣uj∣∣δ(ej(x)) + ∣∣vj∣∣δ(−ej(x))] ,
and we can rewrite

‖g(x) − g̃(x)‖6‖f (km(x)) −umg (em(x)) −vmg (−em(x))‖

+ ‖f (km(x)) −umg̃ (em(x)) −vmg̃ (em(x))‖

62
∞∑
j=m

[∣∣uj∣∣δ(ej(x)) + ∣∣vj∣∣δ(−ej(x))]
for any x ∈ X and m ∈N, which yields g = g̃ in X as m →∞. This completes the proof. �

For the Euler-Lagrange equation, we provide another method to solve it in contrast with [10].
Theorem 4.2 Suppose that (X, +) is a group, and (Y,‖ · ‖) is a Banach space and let themapping f : X → Y satisfy the inequality for all x,y,z ∈ X and some ε > 0
‖f (x + y + z) + f (x −y + z) + f (x + y −z) + f (x −y −z) − 4f (x) − 4f (y) − 4f (z)‖6 ε. (4.3)

Then there has a unique limiting function g : X → Y such that
g(x) =

2

9
g(3x) −

1

9
g(−3x), x ∈ X

and
‖f (x) −g(x)‖6

3ε

8
x ∈ X.

In particular, if X is Abelian, then g is a solution of the equation in the following
f (x + y + z) + f (x −y + z) + f (x + y −z) + f (x −y −z) = 4f (x) + 4f (y) + 4f (y), (4.4)

for all x,y ∈ X.
Proof. From (x,x,−x) in (4.3), we obtain

‖6f (x) + 3f (−x) − f (3x)‖6 ε, x ∈ X.

Replacing x by −x in the above inequality we obtain
‖6f (−x) + 3f (x) − f (−3x)‖6 ε, x ∈ X.

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Eur. J. Math. Anal. 10.28924/ada/ma.3.7 16
Consequently, combining the above two inequalities yield that

‖9f (x) + f (−3x) − 2f (3x)‖6 3ε, x ∈ X.

By using the results second part of Theorem 4.1, a computation is to prove that
un :=

3n + 1

2 · 9n
, vn :=

1 − 3n

2 · 9n
, n ∈N.

and the convergent series can be described as
∞∑
n=0

[|un|δ (en(x)) + |vn|δ (−en(x))] =
3ε

8
.

And we show that if X is commutative, by using (x,y,z) = (3nx, 3ny, 3nz), then
‖un[f (3n(x + y + z)) + f (3n(x −y + z)) + f (3n(x + y −z)) + f (3n(x −y −z))

− 4f (3nx) − 4f (3ny) − 4f (3ny)] + vn[f (3n(x + y + z)) + f (3n(x −y + z))

+ f (3n(x + y −z)) + f (3n(x −y −z)) − 4f (3nx) − 4f (3ny) − 4f (3ny)]‖

6
ε

9nwhich we achieve our result (3.13) by letting n →∞.
�

Theorem 4.3 Suppose that X is a group, and (Y,‖ · ‖) is a Banach space and let the mapping
f : X → Y satisfy the inequality for all x,y ∈ X and some ε > 0

‖f (x + y) + f (x −y) − 2f (x) − f (y) − f (−y)‖6 ε, x,y ∈ X. (4.5)
Then there has a unique limiting function g : X → Y fulfilling

g(x) =
3

8
g(2x) −

1

8
g(−2x), x ∈ X

and
‖f (x) −g(x)‖6

2ε

3
x ∈ X.

In particular, if X is commutative, then g also fulfils
g(x + y) + g(x −y) = 2g(x) + g(y) + g(−y), x,y ∈ X.

Proof. Substituting in the sequel (x,x) in (4.5), we obtain
‖f (2x) + f (0) − 3f (x) − f (−x)‖6 ε, x ∈ X. (4.6)

Replacing x by −x in (4.6) we have
‖f (−2x) + f (0) − 3f (−x) − f (x)‖6 ε, x ∈ X. (4.7)

Consequently, (4.6) and (4.7) yield that
‖8f (x) + f (−2x) − 3f (2x)‖6 4ε, x ∈ X.

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Eur. J. Math. Anal. 10.28924/ada/ma.3.7 17
By using the results of Theorem 3.3, a computation is to prove that

un :=
2n + 1

2 · 4n
, vn :=

1 − 2n

2 · 4n
, n ∈N.

and the convergent series
∞∑
n=0

[|un|δ (en(x)) + |vn|δ (−en(x))] =
2ε

3
.

And we show that if X is commutative, by using (x,y) = (2nx, 2ny), then
‖un[f (2nx + 2ny) + f (2nx − 2ny) − 2f (2nx) − f (2ny) − f (−2ny)]

+ vn[f (2
nx + 2ny) + f (2nx − 2ny) − 2f (2nx) − f (2ny) − f (−2ny)]‖

6
ε

4nwhich we achieve our result (∗) by letting n →∞. �
If we can not set f (0) = 0, then the approximate constat is 5

6
ε.

Acknowledgments
The authors express their gratitude to the anonymous reviewers and editor for their carefulreading the manuscript and for many valuable remarks and suggestions.

Conflict of Interest
The author(s) declare(s) that there is no conflict of interest regarding this manuscript.

Data Availability
Data sharing not applicable to this article as no datasets were generated or analysed duringthe current study.

Funding Statement
This work was supported by the National Natural Science Foundation of China (11971493) and(12071491).

References
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https://doi.org/10.1073/pnas.27.4.222.[4] D.H. Hyers, G. Isac, T.M. Rassias, Stability of Functional Equations in Several Variables, Birkhäuser Boston,Boston, MA, 1998. https://doi.org/10.1007/978-1-4612-1790-9.

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https://doi.org/10.1073/pnas.27.4.222
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https://doi.org/10.4134/BKMS.2005.42.1.057
https://doi.org/10.3390/math10071188

	1. Introduction
	2. A simple variable of abstract equation
	3. The stability of functional equations in F-space
	4. The stability of functional equations in Banach space
	Acknowledgments
	Conflict of Interest
	Data  Availability
	Funding  Statement
	References

