International Journal of Mathematical Analysis

Vol. 14, 2020, no. 5, 219 - 239

HIKARI Ltd, www.m-hikari.com

https://doi.org/10.12988/ijma.2020.91169

Hyers-Ulam Stability Additive β-Functional

Inequalities with Three Variables in

Non-Archimedean Banach Space and

Complex Banach Spaces

Ly Van An

Faculty of Mathematics Teacher Education

Tay Ninh University

Ninh Trung, Ninh Son, Tay Ninh Province, Vietnam

This article is distributed under the Creative Commons by-nc-nd Attribution License.

Copyright c© 2020 Hikari Ltd.

Abstract

In this paper we solve additive β-functional inequalities with three variables and their

Hyers-Ulam stability in non- Archimedean Banach spaces as well as in complex Banach

spaces. It is shown that the solutions of first and second inequalities are additive map-

pings. Then Hyers-Ulam stability of these inequalities is studied and proven.

Keywords: additive β-functional equation; additive β-functional inequality; non-

Archimedean normed space; complex Banach space; Hyers-Ulam stability

1. Introduction

Let X and Y be a normed spaces on the same field K, and f : X → Y . We usethe notation

∥∥ · ∥∥ for all the norm on both X and Y . In this paper, we investigatesome additive β-functional inequalities when X is non-Archimedean normed spaceand Y is non-Archimedean Banach space, or X is complex normed space and Y iscomplex Banach space.

220 Ly Van An

In fact, when X is non-Archimedean normed space and Y is non-Archimedean Ba-nach space we solve and prove the Hyers-Ulam stability of following additive β-functional inequality. Let

∣∣2∣∣ 6= 1 and let β be a non-Archimedean number∣∣β∣∣ < 1.∥∥∥∥f(x+ y

2+ z

)− f

(x+ y

2

)− f(z)

∥∥∥∥ ≤ ∥∥∥∥β(2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f(z)

)∥∥∥∥(1.1)

∥∥∥∥2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f(z)

∥∥∥∥≤∥∥∥∥β(f(x+ y

2+ z

)− f

(x+ y

2

)− f(z)

)∥∥∥∥(1.2)

when X is complex normed space and Y is complex Banach spaces we solve andprove the Hyers-Ulam stability of following additive β-functional inequality. Let βbe a complex number with

∣∣β∣∣ < 1.∥∥∥∥f(x+ y

2+ z

)− f

(x+ y

2

)− f(z)

∥∥∥∥ ≤ ∥∥∥∥β(2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f(z)

)∥∥∥∥(1.3)

∥∥∥∥2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f(z)

∥∥∥∥≤∥∥∥∥β(f(x+ y

2+ z

)− f

(x+ y

2

)− f(z)

)∥∥∥∥(1.4)

The notions of non-Archimedean normed space and complex normed spaces will

be reminded in the next section. The Hyers-Ulam stability was first investigated for

functional equation of Ulam in [6] concerning the stability of group homomorphisms.

The functional equation

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

is called the Cauchy equation. In particular, every solution of the Cauchy equation

is said to be an additive mapping.

The Hyers [7] gave first affirmative partial answer to the equation of Ulam in

Banach spaces. After that, Hyers’ Theorem was generalized by Aoki [1] for additive

mappings and by Rassias [8] for linear mappings considering an unbouned Cauchy

diffrence. A generalization of the Rassias theorem was obtained by Gavruta [4] by

replacing the unbounded Cauchy difference by a general control function in the spirit

Hyers-Ulam stability additive β-functional inequalities 221

of Rassias’ approach.

f

(x+ y

2

)=

1

2f(x)

+1

2f(y)

is called the Jensen equation. See [2, 3, 9, 10] for more information on functional

equations.

The Hyers-Ulam stability for functional inequalities have been investigated such

as in [14, 15]. Gilany showed that is if satisfies the functional inequality∥∥∥∥2f(x)

+ 2f(y)− f

(xy−1

)∥∥∥∥ ≤ ∥∥f(xy)∥∥Then f satisfies the Jordan-von Newman functional equation

2f(x)

+ 2f(y)− f

(xy−1

)(1.5)

See also [15,16]. Gilanyi [13] and Fechner [9] proved the Hyers-Ulam stability of the

functional inequality.

Choonkil Park [17] proved the Hyers-Ulam stability of additive β-functional inequal-

ities. Recently, in [18, 19, 20] the authors studied the Hyers-Ulam stability for the

following functional inequalities∥∥∥∥f(x+ y)− f

(x)− f(y)

∥∥∥∥ ≤ ∥∥∥∥ρ(2f

(x+ y

2

)− f

(x)− f

(y))∥∥∥∥ (1.6)

∥∥∥∥2f

(x+ y

2

)− f

(x)− f

(y)∥∥∥∥ ≤ ∥∥∥∥ρ(f(x+ y

)− f

(x)− f

(y))∥∥∥∥ (1.7)

in Non-Archimedean Banach spaces and complex Banach spaces

In this paper, we solve and prove the Hyers-Ulam stability for two β-functional

inequalities (1.1)-(1.2), i.e. the β -functional inequalities with three variables [4,

13,14]. Under suitable assumptions on spaces X and Y , we will prove that the

mappings satisfying the β-functional inequatilies (1.1) or (1.2). Thus, the results in

this paper are generalization of those in [17] for β-functional inequatilies with three

variables.

The paper is organized as follows: In section preliminaries we remind some basic

notations in [18, 19, 20] such as Non-Archimedean field, Non-Archimedean normed

space and Non-Archimedean Banach space.

Section 3 is devoted to prove the Hyers-Ulam stability of the addive β- functional

inequalities (1.1) and (1.2) when X Non-Archimedean normed space and Y Non-

Archimedean Banach space.

Section 4 is devoted to prove the Hyers-Ulam stability of the additive β- functional

inequalities (1.1) and (1.2) when X is complex normed space and Y is complex

Banach space.

222 Ly Van An

2. preliminaries

2.1. Non-Archimedean normed and Banach spaces. In this subsection we re-

call some basic notations [19, 20] such as Non-Archimedean fields, Non-Archimedean

normed spaces and Non-Archimedean normed spaces.

A valuation is a function∣∣ · ∣∣ from a field K into

[0,∞

)such that 0 is the unique

element having the 0 valuation,∣∣r.s∣∣ =∣∣r∣∣.∣∣s∣∣,∀r, s ∈ K

and the triangle inequality holds, ie;∣∣r + s∣∣ ≤ ∣∣r∣∣+

∣∣s∣∣,∀r, s ∈ K

A field K is called a valued filed if K carries a valuation. The usual absolute values

of R and C are examples of valuation. Let us consider a vavluation which satisfies

a stronger condition than the triangle inaquality. If the tri triangle inequality is

replaced by

∣∣r + s∣∣ ≤ max

{∣∣r∣∣, ∣∣s∣∣},∀r, s ∈ K

then the function∣∣ · ∣∣ is called a norm -Archimedean valuational, and filed. Clearly∣∣1∣∣ =

∣∣ − 1∣∣ = 1 and

∣∣n∣∣ ≤ 1, ∀n ∈ N . A trivial expamle of a non- Archimedean

valuation is the function∣∣ · ∣∣ talking everything except for 0 into 1 and

∣∣0∣∣ = 0 this

paper, we assume that the base field is a non-Archimedean filed, hence call it simply

a filed.

Definition 2.1.

Let be a vector space over a filed K with a non-Archimedean∣∣ · ∣∣. A function∥∥ · ∥∥ : X → [0,∞) is said a non-Archimedean norm if it satisfies the following

conditions:

(1)∥∥x∥∥ = 0 if and only if x = 0;

(2)∥∥rx∥∥ =

∣∣r∣∣∥∥x∥∥(r ∈ K, x ∈ X);

(3) the strong triangle inequlity

∥∥x+ y∥∥ ≤ max

{∥∥x∥∥, ∥∥y∥∥}, x, y ∈ Xhold. Then (X,

∥∥ · ∥∥) is called a norm -Archimedean norm space.

Hyers-Ulam stability additive β-functional inequalities 223

Definition 2.2.

(1) Let{xn}

be a sequence in a norm-Archimedean normed space X. Then

sequence{xn}

is called Cauchy if for a given ε > 0 there a positive integer

N such that ∥∥xn − x∥∥ ≤ ε

for all n,m ≥ N

(2) Let{xn}

be a sequence in a norm-Archimedean normed space X. Then

sequence{xn}

is called cauchy if for a given ε > 0 there a positive integer N

such that

∥∥xn − x∥∥ ≤ ε

for all n,m ≥ N. The we call x∈ X a limit of sequence xn and denote

limn→∞ xn = x.

(3) If every sequence Cauchy inX converger, then the norm-Archimedean normed

space X is called a norm-Archimedean Bnanch space.

2.2. Solutions of the inequalities. The functional equation

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

is called the cauchuy equation. In particular, every solution of the cauchuy equation

is said to be an additive mapping.

3. Additive β-functional inequality in Non-Archimedean Banach

space

Now, we first study the solutions of (1.1) and (1.2). Note that for these in-

equalities, X is non-Archimedean normed space and Y is non-Archimedean Banach

spaces. Under this setting, we can show that the mapping satisfying (1.1) and (1.2)

is additive. These results are given in the following.

Lemma 3.1. A mapping f:X→Y satilies∥∥∥∥f(x+ y

2+z

)−f(x+ y

2

)−f(z)∥∥∥∥ ≤ ∥∥∥∥β(2f

(x+ y

22+z

2

)−f(x+ y

2

)−f(z))∥∥∥∥

(3.1)

for all x,y,z∈ X if and only if f:X→Y is additive

224 Ly Van An

Proof. Assume that f : X →Y satisfies (3.1)

Letting x = y = z = 0 in (3.1), we get∥∥f(0)

∥∥ ≤ 0. So f(0) = 0

Letting x = y = z in (3.1), we get∥∥f(2x) − 2f

(x)∥∥ ≤ 0 and so f

(2x)

= 2f(x)

for

all x ∈X.

Thus

f(x

2) =

1

2f(x) (3.2)

for all x ∈ X It follows from (3.1) and (3.2) that:∥∥∥∥f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z)∥∥∥∥ ≤ ∥∥∥∥β(2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f

(z))∥∥∥∥

=

∥∥∥∥β(2f

( x+y2

+ z

2

)− f

(x+ y

2

)− f

(z))∥∥∥∥

≤∣∣β∣∣∥∥∥∥(f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z))∥∥∥∥

and so

f

(x+ y

2+ z

)= f

(x+ y

2

)+ f(z)

. for all x, y, z ∈ X The coverse is obviously true. �

Lemma 3.2. A mapping f : X → Y satisfy f(0) = 0 and∥∥∥∥2f

(x+ y

22+z

2

)−f(x+ y

2

)−f(z)∥∥∥∥ ≤ ∥∥∥∥β(f(x+ y

2+z

)−f(x+ y

2

)−f(z))∥∥∥∥(3.3)

for all x, y ,z∈ X if and if f : X → Y is additive.

Proof. Assume that f : X → Y (3.3).

Letting x = y = 0 in (3.3), we get∥∥∥∥2f

(z

2

)− f

(z)∥∥∥∥ ≤ 0 (3.4)

and so

f

(z

2

)− 1

2f(z)

It follows from (3.3) and (3.4) that∥∥∥∥f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z)∥∥∥∥ =

∥∥∥∥2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f

(z)∥∥∥∥

=

∥∥∥∥2f

( x+y2

+ z

2

)− f

(x+ y

2

)− f

(z)∥∥∥∥

≤∣∣β∣∣∥∥∥∥f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z))∥∥∥∥

Hyers-Ulam stability additive β-functional inequalities 225

and so.

f

(x+ y

2+ z

)= f

(x+ y

2

)+ f(z)

for all x, y, z ∈ X The converse is obvoiusly true.

�

Theorem 3.3. Let ϕ : X3 →[0,∞

)be a function and let f : X → Y be a mapping

such that

ψ(x, y, z) :=∞∑j=1

∣∣2∣∣jψ(x2,y

2,z

2

)<∞ (3.5)

∥∥∥∥f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z)∥∥∥∥

≤∥∥∥∥β(2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f

(z))∥∥∥∥+ ϕ

(x, y, z

)(3.6)

for all x, y, z ∈X.Then there exists a unique mapping h: X → Y such that∥∥f(x)− h(x)∥∥ ≤ 1∣∣2∣∣ψ(x, x, x) (3.7)

for all x ∈ X

Proof. Letting x = y = z in (3.6), we get∥∥f(2x)− 2f(x)∥∥ ≤ ϕ

(x, x, x

)(3.8)

for all x ∈ X.

So ∥∥∥∥f(x)− 2f

(x

2

)∥∥∥∥ ≤ ϕ

(x

2,x

2,x

2

)for all x ∈ X. Hence∥∥∥∥2lf

( x2k

)− 2mf

( x2m

)∥∥≤ max

{∥∥2lf

(x

2l

)− 2l+1f

(x

2l+1

)∥∥∥∥, ...,∥∥∥∥2m−1f

(x

2m−1

)− 2mf

(x

2m

)∥∥∥∥}= max

{∣∣2∣∣l∥∥∥∥f( x2l

)− 2f

(x

2l+1

)∥∥∥∥, ..., |2m−1|∥∥∥∥f( x

2m−1

)− 2f

(x

2m

)∥∥∥∥}≤

∞∑j=l

∣∣2∣∣jϕ( x

2j+1,x

2j+1,x

2j+1

)(3.9)

for all nonnegative integers m and l with m > l and all x ∈ X. It follows from

(3.9) that the sequence{

2nf(

x2n

)}is a cauchy sequence for all x ∈ X. Since Y is a

226 Ly Van An

Non-Archimedean Banach space, the sequence{

2nf(

x2n

)}coverges.

So one can define the mapping h : X → Y by

h(x)

:= limn→∞

2nf

(x

2n

)for all x ∈ X. Moreover, letting l = 0 and passing the limit m→∞ in (3.9), we get

(3.7).

Now, let T : X → Y be another additive mapping satistify (3.7) then we have

∥∥h(x)− T(x)∥∥ =

∥∥∥∥2qh

(x

2q

)− 2qT

(x

2q

)∥∥∥∥≤ max

{∥∥∥∥2qh

(x

2q

)− 2qf

(x

2q

)∥∥∥∥,∥∥∥∥2qT

(x

2q

)− 2qf

(x

2q

)∥∥∥∥}≤∣∣2∣∣q−1ψ( x

2q,x

2q,x

2q

)(3.10)

which tends to zero as q→∞ for all x ∈ X. So we can conclude that h(x)

= T(x)

for all x ∈ X. The proves the uniqueness of h. It follows from (3.5) and (3.6) that∥∥∥∥h(x+ y

2+ z

)− h(x+ y

2

)− h(z)∥∥∥∥ = lim

n→∞

∥∥2n(f

(x+ y

2n+1+

z

2n

)− f

(x+ y

2n+1

)− f

(z

2n

))∥∥∥∥≤ lim

n→∞

∥∥∥∥2nβ

(f

(x+ y

2n+2+

z

2n+1

)− f

(x+ y

2n+1

)− f

(z

2n

)∥∥∥∥+ lim

n→∞|2|nψ(

x

2n,y

2n,z

2n

))=

∥∥∥∥β(2h

(x+ y

2+ z

)− h(x+ y

2

)− h(z))∥∥∥∥ (3.11)

for all x,y,z ∈X

∥∥∥∥h(x+ y

2+ z

)− h(x+ y

2

)− h(z)∥∥∥∥ ≤ ∥∥∥∥β(2h

(x+ y

2+ z

)− h(x+ y

2

)− h(z))∥∥∥∥

for all x,y∈ X. By lemma 3.1, the mapping h : X → Y is additive. �

Theorem 3.4.

Let ϕ : X →[0,∞

)be a function and let f : X → Y be a mapping satisfying

ψ(x, y, z

):=

∞∑j=0

1∣∣2∣∣jϕ(2jx, 2jy, 2jz)<∞ (3.12)

Hyers-Ulam stability additive β-functional inequalities 227

and∥∥∥∥f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z)∥∥∥∥

≤∥∥∥∥β(2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f

(z))∥∥∥∥+ ϕ

(x, y, z

)(3.13)

for all x,y,z∈ X. Then there exists a unique additive mapping h : X → Y such

that ∥∥f(x)− h(x)∥∥ ≤ 1∣∣2∣∣ψ(x, x, x) (3.14)

for all x ∈ X

Proof. . Letting x = y = z in (3.13), we get∥∥∥∥f(x)− 1

2f(2x)∥∥∥∥ ≤ 1∣∣2∣∣ψ(x, x, x)

for all x ∈ X.

Hence∥∥∥∥ 1

2lf(2lx)− 1

2mf(2mx

)∥∥∥∥≤ max

{∥∥∥∥ 1

2lf(2lx

)− 1

2l+1f(2l+1x

)∥∥∥∥, ...,∥∥∥∥ 1

2m−1f(2m−1x

)− 1

2mf(2mx

)∥∥∥∥}= max

{1∣∣2∣∣l∥∥∥∥f(2lx

)− 1

2f(2l+1x

)∥∥∥∥, ..., 1∣∣2∣∣m−1∥∥∥∥f(2m−1x

)− 1

2f(2mx

)∥∥∥∥}

≤∞∑j=1

1∣∣2∣∣j+1ϕ(2jx, 2jy, 2jz

)(3.15)

for all nonnegative integers m and l with m > l and all x ∈ X . It folows (3.15)

that the sequence

{12nf(2nx)}

is sequence for all x ∈ X. Since Y is complete, the

sequence

{12nf(2nx)}

converger so one can define the mapping h : X → Y by

h(x)

:= limn→∞

1

2nf(2nx)

for all x ∈ X. Moreover, letting l = 0 and passing the limit m → ∞ in (3.15), we

get (3.14). the rest of the proof is similar to the proof of the Theorem 3.3.

�

228 Ly Van An

Theorem 3.5.

Let ϕ : X3 →[0,∞

)be a function and let f : X → Y be a mapping satisfying

f(0) = 0 and

ψ(x, y, z

):=

∞∑j=1

∣∣2∣∣jψ(x2,y

2,z

2

)<∞ (3.16)

∥∥∥∥2f

(x+ y

22+z

2

)−f(x+ y

2

)− f

(z)∥∥∥∥

≤∥∥∥∥β(f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z))∥∥∥∥ +ϕ

(x, y, z

)(3.17)

for all x, y, z ∈ X. Then there exists a unique mapping h : X → Y such that∥∥f(z)− h(z)∥∥ ≤ ψ(0, 0, z

)(3.18)

for all z ∈ X

Proof. Letting x = y = 0 in (3.17)∥∥∥∥2f

(z

2

)− f

(z)∥∥∥∥ ≤ ϕ

(0, 0, z

)(3.19)

for all x ∈ X .

So∥∥∥∥2lf

(z

2l

)− 2mf

(z

2m

)∥∥∥∥≤ max

{∥∥∥∥2lf

(z

2l

)− 2l+1f

(z

2l+1

)∥∥∥∥, ...,∥∥∥∥2m−1f

(z

2m−1

)− 2mf

(z

2m

)∥∥∥∥}= max

{∣∣2∣∣l∥∥∥∥f( z2l

)− 2f

(z

2l+1

)∥∥∥∥, ..., ∣∣2∣∣m−1∥∥∥∥f( z

2m−1

)− 2f

(z

2m

)∥∥∥∥}≤

∞∑j=l+1

∣∣2∣∣j+1ϕ(0, 0,

z

2j

)<∞ (3.20)

for all nonnegativers integers m and l with m > l and all x ∈ X. It follows from

(3.20) that the sequence{

2kf(

z2k

)}is Cauchy for all x ∈ X. Since Y is a non -

Archimedean Banach space, the sequence,{

2kf(

z2k

)}converges. So one can define

the mapping h : X → Y by

h(z)

:= limk→∞

2kf

(z

2k

)for all z ∈ X. Moreover, letting l = 0 and passing the limit m → ∞ in (3.20), we

get (3.18).

Hyers-Ulam stability additive β-functional inequalities 229

The rest of the proof is similar to the proof of the theorem (3.3).

�

Theorem 3.6.

Let ϕ : X3 →[0,∞

)be a function and let f : X → Y be a mapping satisfying

f(0) = 0 and

ψ(x, y, z) :=∞∑j=1

1

|2|jψ(2jx, 2jy2jz) <∞; (3.21)

∥∥∥∥2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f

(z)∥∥∥∥

≤∥∥∥∥β(f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z))∥∥∥∥+ ϕ

(x, y, z

)(3.22)

for all x, y, z ∈X. Then there exists a unique mapping h: X → Y such that

∥∥f(x)− h(x)∥∥ ≤ ψ(0, 0, z

)(3.23)

for all z ∈ X

Proof. . Letting x = y = 0 in (3.22)∥∥∥∥2f

(z

2

)− f

(z)∥∥∥∥ ≤ ϕ

(0, 0, z

)(3.24)

for all z ∈ X .

So ∥∥∥∥f(z)− 1

2f(2z)∥∥∥∥ ≤ 1∣∣2∣∣ϕ(0, 0, 2z)

for all z ∈ X. Hence

∥∥∥∥ 1

2lf(2lz)− 1

2mf(2mz

)∥∥∥∥≤ max

{∥∥∥∥ 1

2lf(2lz)− 1

2l+1f(2l+1z

)∥∥∥∥, ...,∥∥∥∥ 1

2m−1f(2m−1z

)− 1

2mf(2mz

)∥∥∥∥}= max

{1∣∣2∣∣l∥∥∥∥f(2lz

)− 1

2f(2l+1z

)∥∥∥∥, ..., 1∣∣2∣∣m−1∥∥∥∥f(2m−1z

)− 1

2f(2mz

)∥∥∥∥}

≤∞∑

j=l+1

1∣∣2∣∣jϕ(0, 0, 2jz) <∞ (3.25)

230 Ly Van An

for all nonnegative integers m and l with m > l and all z ∈ X. It follows from (3.25)

that the sequence{

2nf(

z2n

)}is a cauchy sequence for all z ∈ X. Since is complete,

the sequence{

2nf(

z2n

)}coverges. So one can define the mapping h : X → Y by

h(z)

:= limn→∞

1

2nf(2nz)

for all z ∈ X. Moreover, letting l = 0 and passing the limit m → ∞ in (3.25), we

get (3.24).

The rest of the proof is similar to the proof of theorem 3.3.

�

Corollary 3.7.

let r < 1 and θ be nonnegative real numbers, and let f:X → Y be a mapping such

that∥∥∥∥f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z)∥∥∥∥

≤∥∥∥∥β(2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f

(z))∥∥∥∥+ θ(‖x‖r + ‖y‖r + ‖z‖r)

for all x, y,z∈ X.

Then there exits a unique additive maaping h : X → Y such that∥∥∥∥f(x)− h(x)∥∥∥∥ ≤ 2θ∣∣2∣∣r − ∣∣2∣∣∥∥x∥∥rfor all x ∈ X

Corollary 3.8.

let r > 1 and θ be nonnegative real numbers, and let f:X → Y be a mapping such

that∥∥∥∥f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z)∥∥∥∥

≤∥∥∥∥β(2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f

(z))∥∥∥∥+ θ(‖x‖r + ‖y‖r + ‖z‖r)

for all x, y,z∈ X.

Then there exits a unique additive maaping h : X → Y such that∥∥∥∥f(x)− h(x)∥∥∥∥ ≤ 2θ∣∣2∣∣− ∣∣2∣∣r∥∥x∥∥rfor all x ∈ X.

Hyers-Ulam stability additive β-functional inequalities 231

Corollary 3.9.

Let r < 1 and θ be a nonnegativers real number and let f : X → Y be a mapping

satisfying f(0)

= 0 and∥∥∥∥2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f

(z)∥∥∥∥

≤∥∥∥∥β(f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z))∥∥∥∥+ θ

(∥∥x∥∥r +∥∥y∥∥r +

∥∥z∥∥r)for all x, y ,z∈ X. The there exists a unique additive mapping h : X → Y such that∥∥f(x)− h(x)‖ ≤ ∣∣2∣∣rθ∣∣2∣∣r − ∣∣2∣∣∥∥x∥∥rfor all x ∈ X.

Corollary 3.10.

Let r > 1 and θ be a nonnegativers real number and let f : X → Y be a mapping

satisfying f(0)

= 0 and∥∥∥∥2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f

(z)∥∥∥∥

≤∥∥∥∥β(f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z))∥∥∥∥+ θ

(∥∥x∥∥r +∥∥y∥∥r +

∥∥z∥∥r)for all x, y ,z∈ X. The there exists a unique additive mapping h : X → Y such that∥∥f(x)− h(x)‖ ≤ ∣∣2∣∣rθ∣∣2∣∣− ∣∣2∣∣r∥∥x∥∥rfor all x ∈ X.

4. Additive β-functional inequality in complex Banach space

Now, we study the solutions of (1.1) and (1.2). Note that for these inequalities,

X is complex normed space and Y is complex Banach spaces. Under this setting,

we can show that the mapping satisfying (1.1) and (1.2) is additive. These results

are give in the following.

Lemma 4.1.

Amapping f : X → Y satisfies∥∥∥∥f(x+ y

2+z

)−f(x+ y

2

)−f(z)∥∥∥∥ ≤ ∥∥∥∥β(2f

(x+ y

22+z

2

)−f(x+ y

2

)−f(z))∥∥∥∥(4.1)

for all x, y, z ∈ X if and only if f : X → Y is additive

Proof. The proof is simlar to the proof of lemma 3.1

232 Ly Van An

Lemma 4.2.

Amapping f : X → Y satisfies f(0) = 0 and∥∥∥∥2f

(x+ y

22+z

2

)−f(x+ y

2

)−f(z)∥∥∥∥ ≤ ∥∥∥∥β(f(x+ y

2+z

)−f(x+ y

2

)−f(z))∥∥∥∥(4.2)

for all x,y,z∈ X if and only if f:X→Y is additive

Proof. The proof is simlar to the proof of lemma 3.2

Theorem 4.3.

Let ϕ : X3 →[0,∞

)be a function and let f : X → Y be a mapping such that

ψ(x, y, z

):=

∞∑j=1

2jϕ

(x

2j,y

2j,z

2j

)<∞ (4.3)

∥∥∥∥f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z)∥∥∥∥

≤∥∥∥∥β(2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f

(z))∥∥∥∥+ ϕ

(x, y, z

)(4.4)

for all x, y, z ∈X. Then there exists a unique mapping h : X → Y such that∥∥f(x)− h(x)∥∥ ≤ 1

2ψ(x, x, x

)(4.5)

for all x ∈ X

Proof. Let x = y = z in (4.4), we get∥∥f(2x)− 2f(x)∥∥ ≤ ϕ

(x, x, x

)(4.6)

for all x ∈ X. So ∥∥∥∥f(x)− 2f

(x

2

)∥∥∥∥ ≤ ϕ

(x

2,y

2,z

2

)for all x ∈ X. Hence∥∥∥∥2lf

(x

2l

)− 2mf

(x

2m

)∥∥∥∥≤

m−1∑j=1

∥∥∥∥2jf

(x

2j

)− 2j+1f

(x

2j+1

)∥∥∥∥ ≤ m−1∑j=1

‖2jϕ

(x

2j+1,x

2j+1,x

2j+1

)∥∥∥∥(4.7)

for all nonnegative integers m and l with m > l and all x ∈ X. It follows from

(4.7) that the sequence

{2nf

(x2n

)}is a cauchy sequence for all x ∈ X. Since

Hyers-Ulam stability additive β-functional inequalities 233

is complete, the sequence

{2nf

(x2n

)}coverges. So one can define the mapping

h : X → Y by

h(x)

:= limn→∞

1

2nf(2nx)

for all x ∈ X. Moreover, letting l = 0 and passing the limit m→∞ in (4.7), we get

(4.5).

Now, let T : X → Y be another additive mapping satisfying (4.5). Then we have∥∥∥∥h(x)− T(x)∥∥∥∥ =

∥∥∥∥2qh

(x

2q

)− 2qT

(x

2q

)∥∥∥∥≤∥∥∥∥2qh

(x

2q

)− 2qf

(x

2q

)∥∥∥∥+

∥∥∥∥2qT

(x

2q

)− 2qf

(x

2q

)∥∥∥∥≤ 2qψ

(x

2q,x

2q,x

2q

)which tends to zero as q →∞ for all x ∈ X.So we can conclude that h

(x)

= T(x)

for all x ∈ X. This proves the uniqueness of h. It follows from (4.3) and (4.4) that∥∥∥∥h(x+ y

2+ z

)− h(x+ y

2

)− h(z)∥∥∥∥

= limn→∞

∥∥∥∥2n

(f

(x+ y

2n+2+

z

2n+1

)− f

(x+ y

2n+1

)− f

(z

2n

))∥∥∥∥≤ lim

n→∞

∥∥∥∥2nβ

(2f

(x+ y

2n+3+

z

2n+2

)− f

(x+ y

2n+1

)− f

(z

2n

))∥∥∥∥+ lim

n→∞2nϕ

(x

2n,y

2n,z

2n

)=

∥∥∥∥β(2h

(x+ y

22+z

2

)− h(x+ y

2

)− h(z))∥∥∥∥ (4.8)

for all x, y, z ∈ X.

So∥∥∥∥h(x+ y

2+ z

)−h(x+ y

2

)−h(z)∥∥∥∥ ≤ ∥∥∥∥β(2h

(x+ y

22+z

2

)−h(x+ y

2

)−h(z))∥∥∥∥

for all x, y, z ∈ X. By Lemma 4.1, the mapping h : X → Y is additive. �

Theorem 4.4.

Let ϕ : X3 →[0,∞

)be a function and let f : X → Y be a mapping such that

ψ(x, y, z

):=

∞∑j=1

1

2jϕ(2jx, 2jy, 2jz) <∞ (4.9)

234 Ly Van An

and∥∥∥∥f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z)∥∥∥∥

≤∥∥∥∥β(2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f

(z))∥∥∥∥+ ϕ

(x, y, z

)(4.10)

for all x, y, z ∈X. Then there exists a unique mapping h : X → Y such that∥∥f(x)− h(x)∥∥ ≤ 1

2ψ(x, x, x

)(4.11)

for all x ∈ X

Proof. . It follows from (4.10) that∥∥∥∥f(x)− 1

2f(2x)∥∥∥∥ ≤ 1

2ϕ(x, x, x

)for all x ∈ X. Hence

∥∥∥∥ 1

2lf(2lx)− 1

2mf(2mx

)∥∥∥∥ ≤ m−1∑j=1

∥∥∥∥ 1

2jf(2jx

)− 1

2j+1f(2j+1x

)∥∥∥∥≤

m−1∑j=1

1

2jϕ(2jx, 2jx, 2jx

)∥∥ (4.12)

for all nonnegative integers m and l with m > l and all x ∈ X. It follows from (4.12)

that the sequence { 12nf(2nx)} is a cauchy sequence for all x ∈ X. Since is complete,

the sequence { 12nf(2nx)} coverges. So one can define the mapping h : X → Y by

h(x)

:= limn→∞

1

2nf(2nx)

for all x ∈ X. Moreover, letting l = 0 and passing the limit m → ∞ in (4.12), we

get (4.11).

The rest of the proof is similar to the proof of theorem 4.3.

�

Theorem 4.5.

Let ϕ : X3 →[0,∞

)be a function and let f : X → Y be a mapping such that

ψ(x, y, z

):=

∞∑j=1

2jϕ( x

2j,y

2j,z

2j

)<∞ (4.13)

Hyers-Ulam stability additive β-functional inequalities 235∥∥∥∥2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f

(z)∥∥∥∥

≤∥∥∥∥β(f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z))∥∥∥∥+ ϕ

(x, y, z

)(4.14)

for all x, y, z ∈X. Then there exists a unique mapping h: X → Y such that∥∥f(x)− h(x)∥∥ ≤ 1

2ψ(x, x, x

)(4.15)

for all x ∈ X

Proof. . Letting y = x = 0 in (4.14), we get∥∥∥∥f(z)− 2f(z

2

)∥∥∥∥ =

∥∥∥∥2f(z

2

)− f

(z)∥∥∥∥ ≤ ϕ

(0, 0, z

)(4.16)

for all x ∈ X. So

∥∥∥∥2lf( z

2l

)− 2mf

( z2m

)∥∥≤

m−1∑j=1

∥∥∥∥2jf

(z

2j

)− 2j+1f

(z

2j+1

)∥∥∥∥ ≤ m−1∑j=1

2jϕ(0, 0,

z

2j

)(4.17)

for all nonnegative integers m and l with m > l and all x ∈ X. It follows from (4.17)

that the sequence{

2nf(

z2n

)}is a cauchy sequence for all z ∈ X. Since is complete,

the sequence{

2nf(

z2n

)}coverges. So one can define the mapping h : X → Y by

h(z)

:= limn→∞

2nf( z

2n

)(4.18)

for all z ∈ X. Moreover, letting l = 0 and passing the limit m → ∞ in (4.17), we

get (4.15). The rest of the proof is similar to the proof of theorem 4.3. �

Theorem 4.6.

Let ϕ : X3 →[0,∞

)be a function and let f : X → Y be a mapping satisfying

f(0) = 0, and

ψ(x, y, z) :=∞∑j=1

1

2jϕ(2jx, 2jy, 2jz

)<∞ (4.19)

∥∥∥∥2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f

(z)∥∥∥∥

≤∥∥∥∥β(f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z))∥∥∥∥+ ϕ

(x, y, z

)(4.20)

236 Ly Van An

for all x, y, z ∈X. Then there exists a unique mappingh : X → Y such that∥∥f(z)− h(z)∥∥ ≤ 1

2ψ(0, 0, z

)(4.21)

for all z ∈ X

Proof. Letting y = x = 0 in (4.20), we get∥∥∥∥f(z)− 2f(z

2

)∥∥∥∥ =

∥∥∥∥2f(z

2

)− f

(z)∥∥∥∥ ≤ ϕ

(0, 0, z

)(4.22)

. It follows from (4.20) that∥∥∥∥f(z)− 1

2f(2z)∥∥∥∥ ≤ 1

2ϕ(0, 0, 2z

)for all z ∈ X. Hence∥∥∥∥ 1

2lf(2lz)− 1

2mf(2mz

)∥∥∥∥≤

m−1∑j=1

∥∥∥∥ 1

2jf(2jz)− 1

2j+1f(2j+1z

)∥∥∥∥ ≤ m−1∑j=1

1

2jϕ(0, 0, 2jx

)(4.23)

for all nonnegative integers m and l with m > l and all z ∈ X. It follows from

(4.23) that the sequence{

12nf(2nz)}

is a cauchy sequence for all z ∈ X. Since

is complete, the sequence{

12nf(2nz)}

coverges. So one can define the mapping

h : X → Y by

h(z)

:= limn→∞

1

2nf(2nz)

(4.24)

for all z ∈ X. Moreover, letting l = 0 and passing the limit m → ∞ in (4.23), we

get (4.22). The rest of the proof is similar to the proof of theorem 4.3.

�

Corollary 4.7.

Let r > 1 and θ be positive real numbers, and let f : X → Y be a mapping such

that∥∥∥∥f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z))∥∥∥∥

≤∥∥∥∥β(2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f

(z))∥∥∥∥+ θ(‖x‖r + ‖y‖r + ‖z‖r)

for all x, y, z ∈ X. Then there exists a unique additive mapping h : X → Y such that

‖f(x)− h(x)‖ ≤ 2θ

2r − 2‖x‖r

Hyers-Ulam stability additive β-functional inequalities 237

for all x ∈ X

Corollary 4.8.

Let r < 1 and θ be positive real numbers, and let f : X → Y be a mapping such

that∥∥∥∥(x+ y

2+ z

)− f

(x+ y

2

)− f

(z))∥∥∥∥

≤∥∥∥∥β(2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f

(z))∥∥∥∥+ θ

(∥∥x∥∥r +∥∥y∥∥r +

∥∥z∥∥r)for all x, y, z ∈ X. Then there exists a unique additive mapping h : X → Y such

that ∥∥f(x)− h(x)∥∥ ≤ 2θ

2r − 2

∥∥x∥∥rfor all x ∈ X

Corollary 4.9.

Let ϕ : X3 →[0,∞) be a function and let f : X → Y be a mapping satisfying

f(0) = 0,

ψ(x, y, z

):=

∞∑j=1

2jϕ( x

2j,y

2j,z

2j

)<∞

∥∥∥∥2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f

(z)− β

(f

(x+ y

2+ z

)− f

(x+ y

2

)− f

(z))∥∥∥∥

≤ ϕ(x, y, z

)for all x, y, z ∈X. Then there exists a unique mapping h: X → Y such that

∥∥f(z)− h(z)∥∥ ≤ 1

2ψ(0, 0, z

)for all z ∈ X

Corollary 4.10.

Let r > 1 and θ be positive real numbers, and let f : X → Y be a mapping such

that

238 Ly Van An

∥∥∥∥2f

(x+ y

22+z

2

)− f

(x+ y

2

)− f

(z))∥∥∥∥

≤∥∥∥∥β(f(x+ y

2+ z

)− f

(x+ y

2

)− f

(z))∥∥∥∥+ θ

(∥∥x∥∥r +∥∥y∥∥r +

∥∥z∥∥r)for all x, y, z ∈ X. Then there exists a unique additive mapping h : X → Y such that

∥∥f(x)− h(x)∥∥ ≤ 2θ

2r − 2

∥∥x∥∥rfor all x ∈ XRemark: If β is a real number such that −1 < β < 1 and is Y is a real Banach

space, then all the assertions in this sections remain valid.

5. Conclusion

In this paper, it is shown that the solutions of the first and second β-functional

inequalities are additive mappings and the Hyers-Ulam stability is proved. These

are the main results of the paper, which are a generalization of the results in [17,

20].

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Received: December 11, 2019; Published: June 12, 2020