In this paper, I study to solve functional inequalities and equations of type Cauchy-Jensen with 3k-variables in a general form. I first introduce the con-cept of the general Cauchy-Jensen equation and next, I use the direct method of proving the solutions of the Jensen-Cauchy functional inequalities relative to the general Cauchy-Jensen equations and then I show that their solutions are mappings that are additive mappings calculated and finally apply the de-rivative setup on fuzzy algebra also the results of the paper.
Functional Equation Functional Inequality Additivity Banach Space Derivation on Fuzzy-Algebras1. Introduction
Let G be an m-divisible group where m ∈ ℕ \ { 0 } and X , Y be a normed space on the same field K , and f : G → X ( f : G → Y ) be a mapping. I use the notation ‖ ⋅ ‖ X ( ‖ ⋅ ‖ Y ) for corresponding the norms on X and Y . In this paper, I investigate functional inequalities and equations when when G be an m-divisible group where m ∈ ℕ and X is a normed space with norm ‖ ⋅ ‖ X and that Y is a Banach space with norm ‖ ⋅ ‖ Y .
In fact, when G be an m-divisible group where m ∈ ℕ and X is a normed space with norm ‖ ⋅ ‖ X and that Y is a Banach space with norm ‖ ⋅ ‖ Y I solve and prove the Hyers-Ulam-Rassias type stability of following functional inequalities and equations.
‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + 2 k ∑ j = 1 k f ( z j ) ‖ Y ≤ ‖ 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) ‖ Y (1)
and
∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + 2 k ∑ j = 1 k f ( z j ) = 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) (2)
Where k is a positive integer.
The study of the functional equation stability originated from a question of S. M. Ulam [1] , concerning the stability of group homomorphisms. Let ( G , ∗ ) be a group and let ( G ′ , ∘ , d ) be a metric group with metric d ( ⋅ , ⋅ ) . Geven ε > 0 , does there exist a δ > 0 such that if f : G → G ′ satisfies: d ( f ( x ∗ y ) , f ( x ) ∘ f ( y ) ) < δ for all x , y ∈ G then there is a homomorphism h : G → G ′ with d ( f ( x ) , h ( x ) ) < ε , for all x ∈ G , if the answer, is affirmative, I would say that equation of homomophism h ( x ∗ y ) = h ( y ) ∘ h ( y ) is stable. The concept of stability for a functional equation arises when we replace a functional equation with an inequality which acts as a perturbation of the equation. Thus the stability question of functional equations is how the solutions of the inequality differ from those of the given function equation. Hyers gave a first affirmative answer the question Ulam as follows:
In 1941 D. H. Hyers [2] Let ε ≥ 0 and let f : E 1 → E 2 be a mapping between Banach space such that ‖ f ( x + y ) − f ( x ) − f ( y ) ‖ ≤ ε , for all x , y ∈ E 1 and some ε ≥ 0 . It was shown that the limit T ( x ) = lim n → ∞ f ( 2 n x ) 2 n exists for all x ∈ E 1 and that T : E 1 → E 2 is that unique additive mapping satisfying ‖ f ( x ) − T ( x ) ‖ ≤ ε , ∀ x ∈ E 1 .
Next in 1978 Th. M. Rassias [3] provided a generalization of Hyers’ Theorem which allows the Cauchy difference to be unbounded:
Consider E , E ′ to be two Banach spaces, and let f : E → E ′ be a mapping such that f ( t x ) is continous in t for each fixed x. Assume that there exist θ ≥ 0 and p ∈ [ 0,1 ) such that ‖ f ( x + y ) − f ( x ) − f ( y ) ‖ ≤ ε ( ‖ x ‖ p + ‖ y ‖ p ) , ∀ x , y ∈ E . then there exists a unique linear L : E → E ′ satifies ‖ f ( x ) − L ( x ) ‖ ≤ 2 θ 2 − 2 p ‖ x ‖ , x ∈ E .
Next J. M. Rassias [4] following the spirit of the innovative approach of Th. M. Rassias for the unbounded Cauchy difference proved a similar stability theorem in which he replaced the factor ‖ x ‖ p + ‖ y ‖ p by ‖ x ‖ p ‖ y ‖ p for p , q ∈ ℝ with p + q ≠ 1 .
Next in 1992, a generalized of Rassias’ Theorem was obtained by Găvruta [5] .
Let ( G , + ) be a group Abelian and E a Banach space.
Denote by ϕ : G × G → [ 0, ∞ ) a function such that ϕ ˜ ( x , y ) = ∑ n = 0 ∞ 2 − n ϕ ( 2 n x , 2 n y ) < ∞ for all x , y ∈ G . Suppose that f : G → E is a mapping satisfying ‖ f ( x + y ) − f ( x ) − f ( y ) ‖ ≤ ε , ∀ x , y ∈ G . There exists a unique additive mapping T : G → E such that ‖ f ( x ) − T ( x ) ‖ ≤ ϕ ˜ ( x , x ) , ∀ x , y ∈ G .
Generally speaking for a more specific problem, when considering this famous result, the additive Cauchy equation f ( x + y ) = f ( x ) + f ( y ) is said to have the Hyers-Ulam stability on ( E 1 , E 2 ) with E 1 and E 2 are Banach spaces if for each f : E 1 → E 2 satisfying ‖ f ( x + y ) − f ( x ) − f ( y ) ‖ ≤ ε for all x , y ∈ E 1 for some ε > 0 , there exists an additive h : E 1 → E 2 such that f − h is bounded on E 1 . The method which was provided by Hyers, and which produces the additive h, was called a direct method.
Afterward, Gilány showed that if satisfies the functional inequality
‖ 2 f ( x ) + 2 f ( y ) − f ( x y − 1 ) ‖ ≤ ‖ f ( x y ) ‖ (3)
Then f satisfies the Jordan-von Newman functional equation
2 f ( x ) + 2 f ( y ) = f ( x y ) + f ( x y − 1 ) (4)
Gilányi [6] and Fechner [7] proved the Hyers-Ulam stability of the functional inequality.
Recently, the authors studied the Hyers-Ulam stability for the following functional inequalities and equation
‖ f ( x ) + f ( y ) + 2 f ( y ) ‖ ≤ ‖ 2 f ( x + y 2 + z ) ‖ (5)
f ( x ) + f ( y ) + 2 f ( y ) = 2 f ( x + y 2 + z ) (6)
in Banach spaces.
In this paper, I solve and prove the Hyers-Ulam stability for inequality (1.1) is related to Equation (1.2), ie the functional inequalities and equation with 3k variables. Under suitable assumptions on spaces G and X or G and Y , I will prove that the mappings satisfy the (1.1) - (1.2). Thus, the results in this paper are generalization of those in [1] - [33] for inequality (1.1) is related to Equation (1.2) with 3k variables.
The paper is organized as follows:
In the section preliminary, I remind some basic notations such as:
Concept of the divisible group, definition of the stability of Cauchy-Jenen functional inequalities and functional equation, Solutions of the equation, functional inequalities and functional equation, the crucial problem when constructing solutions for Cauchy-Jensen inequalities.
Section 3: Establish a solution to the generalized Cauchy-Jensen functional inequalities (2.2) when I assume that G be a m-divisible abelian group and X is a normed space.
Section 4: Stability of functional inequalities (1.1) related to the Cauchy-Jensen equation when I assume that G be a m-divisible abelian group and Y is a Banach space.
Section 5: Establish solutions to functional inequalities (1.1) based on the definition when I assume that G be a m-divisible abelian group and Y is a Banach space.
Section 6: The stability of derivation on fuzzy-algebras.
2. Preliminaries2.1. Concept of Divisible Group
A group G is called divisible if for every x ∈ G and every positive integer n there is a y ∈ G so that n y = x , i.e., every element of G is divisible by every positive integer. A abelian group G is called divisible if for every x ∈ G and every n ∈ ℕ there is some y ∈ G so that x = n y . divisible by every positive integer. Let G be an n-divisible abelian group where n ∈ ℕ (i.e., a → n a : G → G is a surjection).
Denote by
M ( G , X ) = { f | f : G → X }
L ∞ ( G , X ) = { f : G → X | ‖ f ‖ ∞ : = sup x ∈ G ‖ f ‖ X < ∞ }
The sets M ( G , Y ) , M ( G r , X ) and M ( G r , ℝ + ) can be defined similarly where
G r = { ( x 1 , x 2 , ⋯ , x r ) : x j ∈ G , j = 1, ⋯ , k }
2.2. Definition of the Stability of Functional Inequalities and Functional Equation
Given mappings E : M ( G , X ) → M ( G r , ℝ + ) , φ : G r → ℝ and ψ : G → ℝ + . If E ( f ) ( x 1 , x 2 , ⋯ , x r ) ≤ φ ( x 1 , x 2 , ⋯ , x r ) for all x 1 , x 2 , ⋯ , x r ∈ G implies that there exists g ∈ M ( G , X ) such that E ( g ) ≤ 0 and ‖ f ( x ) − g ( x ) ‖ ∞ ≤ ψ ( x ) , for all x ∈ G , then we say that the inequality E ( f ) ≤ 0 is ( φ , ψ ) -stable in M ( G , X ) . In this case, we also say that the solutions of the inequality E ( f ) ≤ 0 is ( φ , ψ ) -stable in M ( G , X ) . Given mappings E : M ( G , X ) → M ( G r , ℝ + ) , φ : G r → ℝ and ψ : G → ℝ + if ‖ E ( f ) ( x 1 , x 2 , ⋯ , x r ) ‖ ∞ ≤ φ ( x 1 , x 2 , ⋯ , x r ) for all x 1 , x 2 , ⋯ , x r ∈ G , implies that there exists g ∈ M ( G , X ) such that E ( g ) = 0 and ‖ f ( x ) − g ( x ) ‖ ∞ ≤ ψ ( x ) , for all x ∈ G , then we say that the inequality E ( f ) ≤ 0 is ( φ , ψ ) -stable in M ( G , X ) . In this case, we also say that the solutions of the inequality E ( f ) = 0 is ( φ , ψ ) -stable in M ( G , X ) .
It is well known that if an additive function f : ℝ → ℝ satisfies one of the following conditions:
1) f is continuous at a point;
2) f is monotonic on an interval of positive length;
3) f is bounded on an interval of positive length;
4) f is integrable;
5) f is measurable;
then f is of the form f ( x ) = c x with a real constant c.
2.3. Solutions of the Equation
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 functional equation f ( x + y 2 ) = 1 2 f ( x ) + 1 2 f ( y ) is called the Jensen equation. In particular, every solution of the Jensen equation is said to be an Jensen additive mapping.
The functional equation f ( x ) + f ( y ) + 2 f ( z ) = 2 f ( x + y 2 + z ) is called the Cauchy-Jensen equation. In particular, every solution of the equation is said to be an additive mapping.
2.4. Solutions of the Functional Inequalities
The functional inequalities ‖ f ( x ) + f ( y ) + 2 f ( z ) ‖ ≤ ‖ 2 f ( x + y 2 + z ) ‖ is called the Cauchy-Jensen inequalities. In particular, every solution of the inequalities is said to be an additive mapping
2.5. The Crucial Problem When Constructing Solutions for Cauchy-Jensen Inequalities
Suppose a mapping f : G → X , the equation
∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + m ∑ j = 1 k f ( z j ) = m f ( ∑ j = 1 k x j + y j m + ∑ j = 1 k z j ) (7)
is said to a generalized Cauchy-Jensen equation.
And function inequalities
∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + m ∑ j = 1 k f ( z j ) ≤ m f ( ∑ j = 1 k x j + y j m + ∑ j = 1 k z j ) (8)
is said to a generalized Cauchy-Jensen function inequalitiess Note: case m = 2 and k = 1 so (7) it is called a classical Cauchy-Jensen equation, (8) it is called a Cauchy-Jensen function inequalities.
3. Establish a Solution to the Generalized Cauchy-Jensen Functional Inequality
Now, I first study the solutions of (8). Note that for inequalities, G be a m-divisible group where m ∈ ℕ \ { 0 } and X be a normed spaces. Under this setting, I can show that the mapping satisfying (8) is additive. These results are give in the following.
Lemma 1. Let f : G → X be a mapping such that satisfies
‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + m ∑ j = 1 k f ( z j ) ‖ X ≤ ‖ m f ( ∑ j = 1 k x j + y j m + ∑ j = 1 k z j ) ‖ X (9)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G if and only if f : G → X is additive.
Proof. Prerequisites
Assume that f : G → Y satisfies (9) Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (9), I get | 2 k + m | ‖ f ( 0 ) ‖ X ≤ | m | ‖ f ( 0 ) ‖ X ( | 2 k + m | − | m | ) ‖ f ( 0 ) ‖ X ≤ 0
So f ( 0 ) = 0 .
Next I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( − m z , ⋯ , − m z ,0, ⋯ ,0, z , ⋯ , z ) in (9), I get ‖ k f ( − m z ) + k m f ( z ) ‖ ≤ 0 and so
f ( − m z ) = − m f ( z ) (10)
for all z ∈ G .
Next I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , − x j + y j m , ⋯ , − x j + y j m ) in (9) and (10) I have
‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + m ∑ j = 1 k f ( z j ) ‖ X = ‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) − ∑ j = 1 k f ( x j + y j ) ‖ X ≤ ‖ m f ( ∑ j = 1 k x j + y j m − ∑ j = 1 k x j + y j m ) ‖ X = ‖ f ( 0 ) ‖ X = 0 (11)
Therefore
∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) = ∑ j = 1 k f ( x j + y j ) (12)
Finally we replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) by ( u , ⋯ , u , v , ⋯ , v ) in (12) so f ( u ) + f ( v ) = f ( u + v ) .
Sufficient conditions:
Suppose f : G → Y is additive. Then
f ( ∑ j = 1 k x j + ∑ j = 1 k y j ) = ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) (13)
and so f ( p ∑ j = 1 k x j ) = p ∑ j = 1 k f ( x j ) for all p ∈ ℚ and x 1 , x 2 , ⋯ , x r ∈ G .
Therefore
∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + m ∑ j = 1 k f ( z j ) = m f ( ∑ j = 1 k x j + y j m ) + m ∑ j = 1 k f ( z j ) = m f ( ∑ j = 1 k x j + y j m + ∑ j = 1 k z j ) (14)
So I have something to prove
‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + m ∑ j = 1 k f ( z j ) ‖ Y ≤ ‖ m f ( ∑ j = 1 k x j + y j m + ∑ j = 1 k z j ) ‖ Y (15)
From the proof of the lemma 2, I get the following corollary:
Corollary 1. Suppose a mapping f : G → X , The following clauses are equivalent
1) f is additive.
2) ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + m ∑ j = 1 k f ( z j ) = m f ( ∑ j = 1 k x j + y j m + ∑ j = 1 k z j ) ,
∀ x j , y j , z j ∈ G , j = 1, ⋯ , k .
3) ‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + m ∑ j = 1 k f ( z j ) ‖ ≤ ‖ m f ( ∑ j = 1 k x j + y j m + ∑ j = 1 k z j ) ‖
∀ x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G .
Note: Clearly, a vector space is a m-divisible abelian group, so Corollary 3.2 is right when G is a vector space.
Through the Lemma 2 proof, I have the remark:
Remark: When the letting m = 2k (means that m is always even) and G is an m-divisible abelian gourp then G must be a 2-divisible abelian gourp.
4. Stability of Functional Inequalities Related to the Cauchy-Jensen Equation
Now, I first study the solutions of (1.1). Note that for inequalities, G be a m-divisible group where m ∈ ℕ \ { 0 } and Y be a Banach spaces. Under this setting, I can show that the mapping satisfying (1.1) is additive. These results are give in the following.
Theorem 2. For ϕ : G 3 k → ℝ + be a function such that
lim n → ∞ 1 ( 2 k ) n ϕ ( ( 2 k ) n x 1 , ⋯ , ( 2 k ) n x k , ( 2 k ) n y 1 , ⋯ , ( 2 k ) n y k , ⋯ , ( 2 k ) n z 1 , ⋯ , ( 2 k ) n z k ) = 0 (16)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G .
And
ϕ ˜ ( x 1 , ⋯ , x k , z 1 , ⋯ , z k ) = ∑ n = 0 ∞ 1 ( 2 k ) n + 1 ϕ ( ( 2 k ) n + 1 x 1 , ⋯ , ( 2 k ) n + 1 x k , 0 , ⋯ , 0 , ( 2 k ) n z 1 , ⋯ , ( 2 k ) n z k ) < ∞ (17)
for all x 1 , ⋯ , x k , z 1 , ⋯ , z k , z j ∈ G . Suppose that an odd mapping f : G → Y satisfies
‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + 2 k ∑ j = 1 k f ( z j ) ‖ Y ≤ ‖ 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) ‖ Y + ϕ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) (18)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G .
Then there exists a unique additive mapping ψ : G → Y such that
‖ f ( x ) − ψ ( x ) ‖ Y ≤ ϕ ˜ ( x , ⋯ , x , x , ⋯ , x ) (19)
for all x ∈ G .
Proof. Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (18), we get
( | 2 k 2 + k | − | 2 k | ) ‖ f ( 0 ) ‖ Y ≤ 0. (20)
so f ( 0 ) = 0 .
Next I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 2 k x , ⋯ ,2 k x ,0, ⋯ ,0, − x , ⋯ , − x ) in (18), I get
‖ k f ( 2 k x ) − 2 k 2 f ( x ) ‖ Y ≤ ϕ ( 2 k x ,2 k x , ⋯ ,2 k x ,0,0, ⋯ ,0, − x , − x , ⋯ , − x ) (21)
‖ f ( x ) − 1 2 k f ( 2 k x ) ‖ Y ≤ 1 2 k 2 ϕ ( 2 k x ,2 k x , ⋯ ,2 k x ,0,0, ⋯ ,0, − x , − x , ⋯ , − x )
Hence
‖ 1 ( 2 k ) l f ( ( 2 k ) l x ) − 1 ( 2 k ) m f ( ( 2 k ) m x ) ‖ Y ≤ ∑ j = l m − 1 ‖ 1 ( 2 k ) j f ( ( 2 k ) j x ) − 1 ( 2 k ) j + 1 f ( ( 2 k ) j + 1 x ) ‖ Y ≤ 1 2 k 2 ∑ j = l + 1 m 1 ( 2 k ) j ϕ ( ( 2 k ) j + 1 x , ⋯ , ( 2 k ) j + 1 x , 0 , 0 , ⋯ , 0 , − ( 2 k ) j x , ⋯ , − ( 2 k ) j x ) = 0 (22)
for all nonnegative integers m and l with m > l and all x ∈ G . It follows from (22) that the sequence { 1 ( 2 k ) n f ( ( 2 k ) n x ) } is a cauchy sequence for all x ∈ G . Since Y is complete space, the sequence { 1 ( 2 k ) n f ( ( 2 k ) n x ) } coverges.
So one can define the mapping ψ : G → Y by ψ ( x ) : = lim n → ∞ 1 ( 2 k ) n f ( ( 2 k ) n x ) for all x ∈ G . Moreover, letting l = 0 and passing the limit m → ∞ in (22), I get (19).
Now, It follows from (18) I have
‖ ∑ j = 1 k ψ ( x j ) + ∑ j = 1 k ψ ( y j ) + 2 k ∑ j = 1 k ψ ( z j ) ‖ Y = lim n → ∞ ‖ 1 ( 2 k ) n ∑ j = 1 k f ( ( 2 k ) n x j ) + 1 ( 2 k ) n ∑ j = 1 k f ( ( 2 k ) n y j ) + 2 k 1 ( 2 k ) n ∑ j = 1 k f ( ( 2 k ) n z j ) ‖ Y = lim n → ∞ 1 ( 2 k ) n ‖ ∑ j = 1 k f ( ( 2 k ) n x j ) + ∑ j = 1 k f ( ( 2 k ) n y j ) + 2 k ∑ j = 1 k f ( ( 2 k ) n z j ) ‖ Y ≤ lim n → ∞ 1 ( 2 k ) n ( ‖ 2 k f ( ( 2 k ) n ∑ j = 1 k x j + y j 2 k + ( 2 k ) n ∑ j = 1 k z j ) ‖ Y + ϕ ( ( 2 k ) n x 1 , ⋯ , ( 2 k ) n x k , ( 2 k ) n y 1 , ⋯ , ( 2 k ) n y k , ( 2 k ) n z 1 , ⋯ , ( 2 k ) n z k ) ) = ‖ 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) ‖ Y (23)
So I have
‖ ∑ j = 1 k ψ ( x j ) + ∑ j = 1 k ψ ( y j ) + 2 k ∑ j = 1 k ψ ( z j ) ‖ Y ≤ ‖ 2 k ψ ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) ‖ Y (24)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G .
Hence from Lemma 1 and corollary 1 it follows that ψ is an additive mapping.
Finally I have to prove that ψ is a unique additive mapping.
Now, let ψ ′ : G → Y be another generalized Cauchy-Jensen additive mapping satisfying (19). Then I have
‖ ψ ( x ) − ψ ′ ( x ) ‖ Y = 1 ( 2 k ) n ‖ ψ ( ( 2 k ) n x ) − ψ ′ ( ( 2 k ) n x ) ‖ Y ≤ 1 ( 2 k ) n ( ‖ f ( ( 2 k ) n x ) − ψ ( ( 2 k ) n x ) ‖ Y + ‖ f ( ( 2 k ) n x ) − ψ ′ ( x 2 n ) ‖ Y ) ≤ 2 1 ( 2 k ) n ϕ ˜ ( ( 2 k ) n x , ⋯ , ( 2 k ) n x ,0, ⋯ ,0, ( 2 k ) n x , ⋯ , ( 2 k ) n x ) (25)
which tends to zero as n → ∞ for all x ∈ X . So we can conclude that ψ ( x ) = ψ ′ ( x ) for all x ∈ G . This proves the uniquence of ψ ′ .
From Theorem 2 I have the following corollarys.
Corollary 2. For G is a normed space and p , r ≠ 0 , q > 0 , θ > 0 . Suppose f : G → Y be a function such that
‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + 2 k ∑ j = 1 k f ( z j ) ‖ Y ≤ ‖ 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) ‖ Y + θ ⋅ ∏ j = 1 k ‖ x j ‖ p ⋅ ∏ j = 1 k ‖ y j ‖ q ⋅ ∏ j = 1 k ‖ z j ‖ r (26)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G then f í a additive mapping.
Corollary 3. For G is a normed space and 0 < p , r < 1 , q ≠ 0 , θ > 0 . Suppose f : G → Y be a function such that
‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + 2 k ∑ j = 1 k f ( z j ) ‖ Y ≤ ‖ 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) ‖ Y + θ ( ∑ j = 1 k ‖ x j ‖ p + ∑ j = 1 k ‖ y j ‖ q + ∑ j = 1 k ‖ z j ‖ r ) (27)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G . Then there exists a unique additive mapping ψ : G → Y such that
‖ f ( x ) − ψ ( x ) ‖ Y ≤ θ k ( ( 2 k ) p 2 k − ( 2 k ) p ‖ x ‖ p + 1 2 k − ( 2 k ) k ‖ x ‖ r ) (28)
for all x ∈ G .
Theorem 3. For ϕ : G 3 k → ℝ + be a function such that
lim n → ∞ ( 2 k ) n ϕ ( 1 ( 2 k ) n x 1 , ⋯ , 1 ( 2 k ) n x k , 1 ( 2 k ) n y 1 , ⋯ , 1 ( 2 k ) n y k , − 1 ( 2 k ) n z 1 , ⋯ , − 1 ( 2 k ) n z k ) = 0 (29)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G , and
ϕ ˜ ( x 1 , ⋯ , x k , z 1 , ⋯ , z k ) = ∑ n = 0 ∞ ϕ ( 2 k ) n ϕ ( 1 ( 2 k ) n x 1 , ⋯ , 1 ( 2 k ) n x k , 0 , 0 , ⋯ , 0 , ⋯ , 1 ( 2 k ) n + 1 z 1 , ⋯ , 1 ( 2 k ) n + 1 z k ) < ∞ (30)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G .
Suppose that be an odd mapping f : G → Y satisfies
‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + 2 k ∑ j = 1 k f ( z j ) ‖ Y ≤ ‖ 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) ‖ Y + ϕ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) (31)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G .
Then there exists a unique additive mapping ψ : G → Y such that
‖ f ( x ) − ψ ( x ) ‖ Y ≤ ϕ ˜ ( x , ⋯ , x , x , ⋯ , x ) (32)
for all x ∈ G .
Proof. Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (31), I get
( | 2 k 2 + k | − | 2 k | ) ‖ f ( 0 ) ‖ Y ≤ 0. (33)
so f ( 0 ) = 0 .
Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 2 k x , ⋯ ,2 k x ,0, ⋯ ,0, − x , ⋯ , − x ) in (31), I get
‖ k f ( 2 k x ) − 2 k 2 f ( x ) ‖ Y ≤ ϕ ( 2 k x ,2 k x , ⋯ ,2 k x ,0,0, ⋯ ,0, − x , − x , ⋯ , − x ) (34)
‖ f ( x ) − 2 k f ( x 2 k ) ‖ Y ≤ 1 k ϕ ( x , x , ⋯ , x ,0,0, ⋯ ,0, − x 2 k , − x 2 k , ⋯ , − x 2 k )
The remainder is similar to the proof of Theorem 2. This completes the proof.
From Theorem 2 andTheorem 2. I have the following corollarys.
Corollary 4. For G is a normed space and p , r ≠ 0 , q > 0 , θ > 0 . Suppose f : G → Y be a function such that
‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + 2 k ∑ j = 1 k f ( z j ) ‖ Y ≤ ‖ 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) ‖ Y + θ ⋅ ∏ j = 1 k ‖ x j ‖ p ⋅ ∏ j = 1 k ‖ y j ‖ q ⋅ ∏ j = 1 k ‖ z j ‖ r (35)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G , then f is a additive mapping.
Corollary 5. For G is a normed space and 0 < p , r < 1 , q ≠ 0 , θ > 0 . Suppose f : G → Y be a function such that
‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + 2 k ∑ j = 1 k f ( z j ) ‖ Y ≤ ‖ 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) ‖ Y + θ ( ∑ j = 1 k ‖ x j ‖ p + ∑ j = 1 k ‖ y j ‖ q + ∑ j = 1 k ‖ z j ‖ r ) (36)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G . Then there exists a unique additive mapping ψ : G → Y such that
‖ f ( x ) − ψ ( x ) ‖ Y ≤ θ k ( ( 2 k ) p ( 2 k ) p − 2 k ‖ x ‖ p + 1 ( 2 k ) k − 2 k ‖ x ‖ r ) (37)
for all x ∈ G .
5. Establish Solutions to Functional Inequalities Based on the Definition
Now, I first study the solutions of (1). We first consider the mapping E : M ( G , Y ) → M ( G r , ℝ * ) as E ( f ) ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) = ‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + 2 k ∑ j = 1 k f ( z j ) ‖ − ‖ 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) ‖ then the inequalities E f ≤ 0 is ( ϕ , ϕ ˜ ) -stable in M ( G , Y ) where ( ϕ , ϕ ˜ ) is as Theorem 2 and Theorem 3.
Note that for inequalities, G be a m-divisible group where m ∈ ℕ \ { 0 } and Y be a Banach spaces. Under this setting, we can show that the mapping satisfying (1) is additive. These results are give in the following.
Theorem 4. For ϕ : G 3 k → ℝ + be a function such that
lim n → ∞ 1 ( 2 k ) n ϕ ( ( 2 k ) n x 1 , ⋯ , ( 2 k ) n x k , ( 2 k ) n y 1 , ⋯ , ( 2 k ) n y k , ⋯ , ( 2 k ) n z 1 , ⋯ , ( 2 k ) n z k ) = 0 (38)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G , and
ϕ ˜ ( x 1 , ⋯ , x k , z 1 , ⋯ , z k ) = ∑ n = 0 ∞ 1 ( 2 k ) n + 1 ( ϕ ( ( 2 k ) n + 1 x 1 , ⋯ , ( 2 k ) n + 1 x k , 0 , ⋯ , 0 , − ( 2 k ) n z 1 , ⋯ , − ( 2 k ) n z k ) + ϕ ( − ( 2 k ) n + 1 x 1 , ⋯ , − ( 2 k ) n + 1 x k , 0 , ⋯ , 0 , ( 2 k ) n z 1 , ⋯ , ( 2 k ) n z k ) ) < ∞ (39)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G .
Suppose that a mapping f : G → Y satisfies f ( 0 ) = 0 for all x ∈ G , and
‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + 2 k ∑ j = 1 k f ( z j ) − 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) ‖ Y ≤ ϕ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) (40)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G .
Then there exists a unique additive mapping ψ : G → Y such that
‖ f ( x ) − ψ ( x ) ‖ Y ≤ ϕ ˜ ( x , ⋯ , x , x , ⋯ , x ) (41)
for all x ∈ G .
Proof. I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 2 k x , ⋯ ,2 k x ,0, ⋯ ,0, − x , ⋯ , − x ) in (40), I get
‖ k f ( 2 k x ) + 2 k 2 f ( − x ) ‖ Y ≤ ϕ ( 2 k x ,2 k x , ⋯ ,2 k x ,0,0, ⋯ ,0, − x , − x , ⋯ , − x ) (42)
continue I replace x by −x in (42), I have
‖ k f ( − 2 k x ) + 2 k 2 f ( x ) ‖ Y ≤ ϕ ( − 2 k x , − 2 k x , ⋯ , − 2 k x ,0,0, ⋯ ,0, x , x , ⋯ , x ) (43)
put
g ( x ) = f ( x ) − f ( − x ) 2 (44)
So since (45), (43) and (44), I have
‖ f ( x ) − 1 2 k f ( 2 k x ) ‖ Y ≤ 1 2 k 2 ( ϕ ( 2 k x ,2 k x , ⋯ ,2 k x ,0,0, ⋯ ,0, − x , − x , ⋯ , − x ) + ϕ ( − 2 k x , − 2 k x , ⋯ , − 2 k x ,0,0, ⋯ ,0, x , x , ⋯ , x ) ) (45)
Hence
‖ 1 ( 2 k ) l f ( ( 2 k ) l x ) − 1 ( 2 k ) m f ( ( 2 k ) m x ) ‖ Y ≤ ∑ j = l m − 1 ‖ 1 ( 2 k ) j f ( ( 2 k ) j x ) − 1 ( 2 k ) j + 1 f ( ( 2 k ) j + 1 x ) ‖ Y ≤ 1 2 k 2 ∑ j = l + 1 m 1 ( 2 k ) j ( ϕ ( ( 2 k ) j + 1 x , ⋯ , ( 2 k ) j + 1 x ,0,0, ⋯ ,0, − ( 2 k ) j x , ⋯ , − ( 2 k ) j x ) + ϕ ( − 2 k x , − 2 k x , ⋯ , − 2 k x ,0,0, ⋯ ,0, x , x , ⋯ , x ) ) = 0 (46)
for all nonnegative integers m and l with m > l and all x ∈ G . It follows from (46) that the sequence { 1 ( 2 k ) n f ( ( 2 k ) n x ) } is a cauchy sequence for all x ∈ G . Since Y is complete space, the sequence { 1 ( 2 k ) n f ( ( 2 k ) n x ) } coverges.
So one can define the mapping ψ : G → Y by ψ ( x ) : = lim n → ∞ 1 ( 2 k ) n f ( ( 2 k ) n x ) for all x ∈ G . Moreover, letting l = 0 and passing the limit m → ∞ in (46), I get (41).
Now, It follows from (40)we have
‖ ∑ j = 1 k ψ ( x j ) + ∑ j = 1 k ψ ( y j ) + 2 k ∑ j = 1 k ψ ( z j ) − 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) ‖ Y = lim n → ∞ ‖ 1 ( 2 k ) n ∑ j = 1 k f ( ( 2 k ) n x j ) + 1 ( 2 k ) n ∑ j = 1 k f ( ( 2 k ) n y j ) + 2 k 1 ( 2 k ) n ∑ j = 1 k f ( ( 2 k ) n z j ) − 2 k f ( ( 2 k ) n ∑ j = 1 k x j + y j 2 k + ( 2 k ) n ∑ j = 1 k z j ) ‖ Y = lim n → ∞ 1 ( 2 k ) n ‖ ∑ j = 1 k f ( ( 2 k ) n x j ) + ∑ j = 1 k f ( ( 2 k ) n y j ) + 2 k ∑ j = 1 k f ( ( 2 k ) n z j ) − 2 k f ( ( 2 k ) n ∑ j = 1 k x j + y j 2 k + ( 2 k ) n ∑ j = 1 k z j ) ‖ Y ≤ ϕ ( ( 2 k ) n x 1 , ⋯ , ( 2 k ) n x k , ( 2 k ) n y 1 , ⋯ , ( 2 k ) n y k , ( 2 k ) n z 1 , ⋯ , ( 2 k ) n z k ) = 0 (47)
So I have
∑ j = 1 k ψ ( x j ) + ∑ j = 1 k ψ ( y j ) + 2 k ∑ j = 1 k ψ ( z j ) = 2 k ψ ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) (48)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G .
Hence from Lemma 2 and corollary 1, it follows that ψ is an additive mapping.
Finally I have to prove that ψ is a unique additive mapping.
Now, let ψ ′ : G → Y be another generalized Cauchy-Jensen additive mapping satisfying (41). Then we have
‖ ψ ( x ) − ψ ′ ( x ) ‖ Y = 1 ( 2 k ) n ‖ ψ ( ( 2 k ) n x ) − ψ ′ ( ( 2 k ) n x ) ‖ Y ≤ 1 ( 2 k ) n ( ‖ f ( ( 2 k ) n x ) − ψ ( ( 2 k ) n x ) ‖ Y + ‖ f ( ( 2 k ) n x ) − ψ ′ ( x 2 n ) ‖ Y ) ≤ 2 1 ( 2 k ) n ϕ ˜ ( ( 2 k ) n x , ⋯ , ( 2 k ) n x ,0, ⋯ ,0, ( 2 k ) n x , ⋯ , ( 2 k ) n x ) = ∑ n = 0 ∞ 1 ( 2 k ) n + 1 ( ϕ ( ( 2 k ) n + 1 x 1 , ⋯ , ( 2 k ) n + 1 x k ,0, ⋯ ,0, − ( 2 k ) n z 1 , ⋯ , − ( 2 k ) n z k ) + ϕ ( − ( 2 k ) n + 1 x 1 , ⋯ , − ( 2 k ) n + 1 x k ,0, ⋯ ,0, ( 2 k ) n z 1 , ⋯ , ( 2 k ) n z k ) ) < ∞ (49)
which tends to zero as n → ∞ for all x ∈ X . So we can conclude that ψ ( x ) = ψ ′ ( x ) for all x ∈ X . This proves the uniquence of ψ ′ .
From Theorem 4 I have the following corollarys.
Corollary 6. For G is a normed space and p , r ≠ 0 , q > 0 , θ > 0 . Suppose f : G → Y be a function such that f ( 0 ) = 0 and
‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + 2 k ∑ j = 1 k f ( z j ) − 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) ‖ Y ≤ θ ⋅ ∏ j = 1 k ‖ x j ‖ p ⋅ ∏ j = 1 k ‖ y j ‖ q ⋅ ∏ j = 1 k ‖ z j ‖ r (50)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G then f is an additive mapping.
Corollary 7. For G is a normed space and 0 < p , r < 1 , q ≠ 0 , θ > 0 . Suppose f : G → Y be a function such that f ( 0 ) = 0 and
‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + 2 k ∑ j = 1 k f ( z j ) − 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) ‖ Y ≤ θ ( ∑ j = 1 k ‖ x j ‖ p + ∑ j = 1 k ‖ y j ‖ q + ∑ j = 1 k ‖ z j ‖ r ) (51)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G . Then there exists a unique additive mapping ψ : G → Y such that
‖ f ( x ) − ψ ( x ) ‖ Y ≤ θ k ( ( 2 k ) p 2 k − ( 2 k ) p ‖ x ‖ p + 1 2 k − ( 2 k ) k ‖ x ‖ r ) (52)
for all x ∈ G .
Theorem 5. For ϕ : G 3 k → ℝ + be a function such that
lim n → ∞ ( 2 k ) n ϕ ( 1 ( 2 k ) n x 1 , ⋯ , 1 ( 2 k ) n x k , 1 ( 2 k ) n y 1 , ⋯ , 1 ( 2 k ) n y k , ⋯ , 1 ( 2 k ) n z 1 , ⋯ , 1 ( 2 k ) n z k ) = 0 (53)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G .
And
ϕ ˜ ( x 1 , ⋯ , x k , z 1 , ⋯ , z k ) = ∑ n = 0 ∞ ( 2 k ) n − 1 ( ϕ ( ( 2 k ) − n x 1 , ⋯ , ( 2 k ) − ( n + 1 ) x k , 0 , ⋯ , 0 , − ( 2 k ) n + 1 z 1 , ⋯ , − ( 2 k ) n + 1 z k ) + ϕ ( − ( 2 k ) − n x 1 , ⋯ , − ( 2 k ) − n x k , 0 , ⋯ , 0 , ( 2 k ) n + 1 z 1 , ⋯ , ( 2 k ) n + 1 z k ) ) < ∞ (54)
for all x 1 , ⋯ , x k , z 1 , ⋯ , z k ∈ G .
Suppose that a mapping f : G → Y satisfies f ( 0 ) = 0 for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G .
And
‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + 2 k ∑ j = 1 k f ( z j ) − 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) ‖ Y ≤ ϕ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) (55)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G .
Then there exists a unique additive mapping ψ : G → Y such that
‖ f ( x ) − ψ ( x ) ‖ Y ≤ ϕ ˜ ( x , ⋯ , x , x , ⋯ , x ) (56)
for all x ∈ G .
The proof is similar to theorem 4.
Corollary 8. For G is a normed space and p , r ≠ 0 , q > 0 , θ > 0 . Suppose f : G → Y be a function such that f ( 0 ) = 0 and
‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + 2 k ∑ j = 1 k f ( z j ) − 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) ‖ Y ≤ θ ⋅ ∏ j = 1 k ‖ x j ‖ p ⋅ ∏ j = 1 k ‖ y j ‖ q ⋅ ∏ j = 1 k ‖ z j ‖ r (57)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G then f í a additive mapping.
Corollary 9. For G is a normed space and 0 < p , r < 1 , q ≠ 0 , θ > 0 . Suppose f : G → Y be a function such that f ( 0 ) = 0 and
‖ ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + 2 k ∑ j = 1 k f ( z j ) − 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) ‖ Y ≤ θ ( ∑ j = 1 k ‖ x j ‖ p + ∑ j = 1 k ‖ y j ‖ q + ∑ j = 1 k ‖ z j ‖ r ) (58)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ G . Then there exists a unique additive mapping ψ : G → Y such that
‖ f ( x ) − ψ ( x ) ‖ Y ≤ θ k ( ( 2 k ) p 2 k − ( 2 k ) p ‖ x ‖ p + 1 2 k − ( 2 k ) k ‖ x ‖ r ) (59)
for all x ∈ G .
6. The Stability of Derivation on Fuzzy-Algebras
Lemma 6. Let ( Y , ℕ ) be a fuzzy normed vector space and f : X → Y be a mapping such that
N ( ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( y j ) + 2 k ∑ j = 1 k f ( z j ) , t ) ≥ N ( 2 k f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k z j ) , t 2 k ) (60)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ Y and all t > 0 . Then f is Cauchy additive.
Proof. I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (60), I have
N ( ( 2 k 2 + 2 k ) f ( 0 ) , t ) = N ( f ( 0 ) , t 2 k 2 + 2 k ) ≥ N ( 2 k f ( 0 ) , t 2 k ) = 1 (61)
for all t > 0 . By N5 and N6, N ( f ( 0 ) , t 2 k ) = 1 . It follows N2 that f ( 0 ) = 0 .
Next I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( − y , ⋯ , − y , y , ⋯ , y ,0, ⋯ ,0 ) in (60), I have
N ( k f ( − y ) + k f ( y ) , t ) = N ( f ( − y ) + f ( y ) , t k ) ≥ N ( 2 k f ( 0 ) , t 2 k 2 + 2 k ) (62)
It follows N2 that f ( − y ) + f ( y ) = 0 .
So
f ( − y ) = − f ( y )
Next I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( − 2 z , ⋯ , − 2 z ,0, ⋯ ,0, z ,0 , ⋯ ,0 ) in (60), we have
N ( − k f ( 2 z ) + 2 k f ( z ) , t ) = N ( f ( − 2 z ) + 2 f ( z ) , t k ) ≥ N ( 2 k f ( 0 ) , t 2 k 2 + 2 k ) (63)
It follows N2 that f ( − 2 z ) + 2 f ( z ) = 0 .
So f ( 2 z ) = 2 f ( z ) for all t > 0 and for all z ∈ X .
Next I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x , ⋯ , x , y , ⋯ , y , z 1 = − x + y 2 , z 2 = 0 , ⋯ , 0 ) in (60), we have
N ( f ( x ) + f ( y ) − f ( x + y ) , t k ) = N ( f ( x ) + f ( y ) + 2 f ( − x + y 2 ) , t k ) ≥ N ( 2 k f ( 0 ) , t 2 k 2 + 2 k ) (64)
for all t > 0 . and for all x , y ∈ X Thus f ( x ) + f ( y ) = f ( x + y ) for all x , y ∈ X , as desired.
Theorem 7. Let ψ : X 3 k → [ 0, ∞ ) be a function such that there exists an L < 1 2 k
ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) ≤ L 2 k ψ ( 2 k x 1 , ⋯ ,2 k x k ,2 k y 1 , ⋯ ,2 k y k ,2 k z 1 , ⋯ ,2 k z k ) (65)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X and f ( 0 ) = 0 .
Let f : X → X be a mapping sattisfying
N ( 2 k f ( ∑ j = 1 k q x j + q y j 2 k + ∑ j = 1 k q z j ) − ∑ j = 1 k q f ( x j ) − ∑ j = 1 k q f ( y j ) − 2 k ∑ j = 1 k q f ( z j ) , t ) ≥ t t + ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) (66)
N ( f ( ∏ j = 1 k x j ⋅ y j ) − ∏ j = 1 k f ( x j ) ⋅ ∏ j = 1 k y j − ∏ j = 1 k x j ⋅ ∏ j = 1 k f ( y j ) , t ) ≥ t t + ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ,0, ⋯ ,0 ) (67)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k ∈ X , for all t > 0 and for all q > 0 . Then
H ( x ) = N − lim n → ∞ ( 2 k ) n f ( x ( 2 k ) n ) (68)
exists each x ∈ X and defines a fuzzy derivation H : X → X , such that
N ( f ( x ) − H ( x ) , t ) ≥ ( 1 − L ) t ( 1 − L ) + L ψ ( x 1 , ⋯ , x k ,0, ⋯ ,0 ) (69)
for all t > 0 and for all q > 0 .
Proof. Letting q = 1 and I replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x , 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (84), I get
N ( 2 k f ( x 2 k ) − f ( x ) , t ) ≥ t 1 + φ ( x , ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) (70)
for all x ∈ X . Now I consider the set M : = { h : X → Y } and introduce the generalized metric on S as follows:
d ( g , h ) : = inf { β ∈ ℝ + : N ( g ( x ) − h ( x ) , β t ) ≥ t t + φ ( x , 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) , ∀ x ∈ X , ∀ t > 0 } , (71)
where, as usual, inf ϕ = + ∞ . That has been proven by mathematicians ( M , d ) is complete (see [32] ).
Now I cosider the linear mapping T : M → M such that T g ( x ) : = 2 k g ( x 2 k ) for all x ∈ X . Let g , h ∈ M be given such that d ( g , h ) = ε then N ( g ( x ) − h ( x ) , ε t ) ≥ t t + φ ( x , 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) , ∀ x ∈ X , ∀ t > 0.
Hence
N ( g ( x ) − h ( x ) , ε t ) = N ( 2 k g ( x 2 k ) − 2 k h ( x 2 k ) , L ε t ) = N ( g ( x 2 k x ) − h ( x 2 k x ) , L 2 k ε t ) ≥ L t 2 k L t 2 k + φ ( x 2 k , ⋯ , 0,0, ⋯ ,0,0, ⋯ ,0 ) ≥ L t 2 k L t 2 k + L 2 k φ ( x ,0 , ⋯ ,0, ⋯ ,0,0, ⋯ ,0 ) = t t + φ ( x , x , ⋯ , x , x , ⋯ , x ) , ∀ x ∈ X , ∀ t > 0. (72)
So d ( g , h ) = ε implies that d ( T g , T h ) ≤ L ⋅ ε . This means that d ( T g , T h ) ≤ L d ( g , h ) for all g , h ∈ M . It folows from (70) that I have.
For all x ∈ X . So d ( f , T f ) ≤ 1 . By Theorem 1.2, there exists a mapping H : X → Y satisfying the fllowing:
1) H is a fixed point of T, i.e.,
H ( x 2 k ) = 1 2 k H ( x ) (73)
for all x ∈ X . The mapping H is a unique fixed point T in the set ℚ = { g ∈ M : d ( f , g ) < ∞ } .
This implies that H is a unique mapping satisfying (73) such that there exists a β ∈ ( 0, ∞ ) satisfying N ( f ( x ) − H ( x ) , β t ) ≥ t t + φ ( x , 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) , ∀ x ∈ X .
2) d ( T l f , H ) → 0 as l → ∞ . This implies equality N − lim l → ∞ ( 2 k ) l f ( x ( 2 k ) l ) = H ( x ) for all x ∈ X .
3) d ( f , H ) ≤ 1 1 − L d ( f , T f ) . which implies the inequality.
4) d ( f , H ) ≤ 1 1 − L .
This follows that the inequality (70) is satisfied.
By (85)
N ( ( 2 k ) p + 1 f ( ∑ j = 1 k q x j + q y j ( 2 k ) p + 1 + ∑ j = 1 k q z j ( 2 k ) p ) − ( 2 k ) p ∑ j = 1 k q f ( x j ( 2 k ) p ) − ( 2 k ) p ∑ j = 1 k q f ( y j ( 2 k ) p ) − ( 2 k ) p 2 k ∑ j = 1 k q f ( z j ( 2 k ) p ) , t ) ≥ t t + ψ ( x 1 ( 2 k ) p , ⋯ , x k ( 2 k ) p , y 1 ( 2 k ) p , ⋯ , y k ( 2 k ) p , z 1 ( 2 k ) p , ⋯ , z k ( 2 k ) p ) (74)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X , for all t > 0 and for all q ∈ ℝ . So
N ( ( 2 k ) p + 1 f ( ∑ j = 1 k q x j + q y j ( 2 k ) p + 1 + ∑ j = 1 k q z j ( 2 k ) p ) − ( 2 k ) p ∑ j = 1 k q f ( x j ( 2 k ) p ) − ( 2 k ) p ∑ j = 1 k q f ( y j ( 2 k ) p ) − ( 2 k ) p 2 k ∑ j = 1 k q f ( z j ( 2 k ) p ) , t ) ≥ t ( 2 k ) p t ( 2 k ) p + L p ( 2 k ) p ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) (75)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X , for all t > 0 and for all q ∈ ℝ .
Since lim n → ∞ t ( 2 k ) p t ( 2 k ) p + L p ( 2 k ) p ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) = 1 for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X , ∀ t > 0 , q ∈ ℝ . So
N ( 2 k H ( ∑ j = 1 k q x j + q y j 2 k + ∑ j = 1 k q z j ) − ∑ j = 1 k q H ( x j ) − ∑ j = 1 k q H ( y j ) − 2 k ∑ j = 1 k q H ( z j ) , t ) = 1 (76)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X , ∀ t > 0 , q ∈ ℝ . So
2 k H ( ∑ j = 1 k q x j + q y j 2 k + ∑ j = 1 k q z j ) − ∑ j = 1 k q H ( x j ) − ∑ j = 1 k q H ( y j ) − 2 k ∑ j = 1 k q H ( z j ) = 0 (77)
Thus the mapping H : X → X is additive and R -linear by (85) I have
N ( ( 2 k ) 2 p f ( ∏ j = 1 k x j ⋅ y j ( 2 k ) 2 p ) − ( 2 k ) p ∏ j = 1 k f ( x j ( 2 k ) p ) ⋅ ∏ j = 1 k y j − ∏ j = 1 k x j ⋅ ( 2 k ) p ∏ j = 1 k f ( y j ( 2 k ) p ) , t ) ≥ t t + ψ ( x 1 ( 2 k ) p , ⋯ , x k ( 2 k ) p , y 1 ( 2 k ) p , ⋯ , y k ( 2 k ) p ,0 , ⋯ ,0 ) (78)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k ∈ X , for all t > 0 .
N ( ( 2 k ) 2 p f ( ∏ j = 1 k x j ⋅ y j ( 2 k ) 2 p ) − ( 2 k ) p ∏ j = 1 k f ( x j ( 2 k ) p ) ⋅ ∏ j = 1 k y j − ∏ j = 1 k x j ⋅ ( 2 k ) p ∏ j = 1 k f ( y j ( 2 k ) p ) , t ) ≥ t ( 2 k ) 2 p t ( 2 k ) 2 p + L p ( 2 k ) p ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , 0 , ⋯ , 0 ) (79)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k ∈ X , for all t > 0 Since
lim p → ∞ t ( 2 k ) 2 p t ( 2 k ) 2 p + L p ( 2 k ) p ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ,0 , ⋯ ,0 ) = 1 (80)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k ∈ X , for all t > 0 Thus
N ( f ( ∏ j = 1 k x j ⋅ y j ) − ∏ j = 1 k f ( x j ) ⋅ ∏ j = 1 k y j − ∏ j = 1 k x j ⋅ ∏ j = 1 k f ( y j ) , t ) = 1 (81)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k ∈ X , for all t > 0 Thus
f ( ∏ j = 1 k x j ⋅ y j ) − ∏ j = 1 k f ( x j ) ⋅ ∏ j = 1 k y j − ∏ j = 1 k x j ⋅ ∏ j = 1 k f ( y j ) = 0 (82)
So the mapping H : X → X is a fuzzy derivation, as desired.
Theorem 8. Let ψ : X 3 k → [ 0, ∞ ) be a function such that there exists an L < 1
ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) ≤ 2 k ψ ( x 1 2 k , ⋯ , x k 2 k , y 1 2 k , ⋯ , y k 2 k , z 1 2 k , ⋯ , z k 2 k ) (83)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X and f ( 0 ) = 0 .
Let f : X → X be a mapping sattisfying
N ( 2 k f ( ∑ j = 1 k q x j + q y j 2 k + ∑ j = 1 k q z j ) − ∑ j = 1 k q f ( x j ) − ∑ j = 1 k q f ( y j ) − 2 k ∑ j = 1 k q f ( z j ) , t ) ≥ t t + ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) (84)
N ( f ( ∏ j = 1 k x j ⋅ y j ) − ∏ j = 1 k f ( x j ) ⋅ ∏ j = 1 k y j − ∏ j = 1 k x j ⋅ ∏ j = 1 k f ( y j ) , t ) ≥ t t + ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , 0 , ⋯ , 0 ) (85)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k ∈ X , for all t > 0 and for all q > 0 . Then
β ( x ) = N − lim n → ∞ 1 ( 2 k ) n f ( ( 2 k ) n x ) (86)
exists each x ∈ X and defines a fuzzy derivation H : X → X .
Such that
N ( f ( x ) − H ( x ) , t ) ≥ ( 1 − L ) t ( 1 − L ) + L ψ ( x 1 , ⋯ , x k ,0, ⋯ ,0 ) (87)
for all t > 0 and for all q > 0 .
7. Conclusion
In this article, I introduced the concept of the general Jensen Cauchy functional equation, then I used a direct method to show that the solutions of the Jensen-Cauchy functional inequality are additive maps related to the functional equation, Jensen-Cauchy. Then apply the derivative setup on fuzzy algebra.
Conflicts of Interest
The author declares no conflicts of interest.
Cite this paper
An, L.V. (2024) Tremendous Development of Functional Inequalities and Cauchy-Jensen Functional Equations with 3k-Variables on Banach Space and Stability Derivation on Fuzzy-Algebras. Open Access Library Journal, 11: e11241. https://doi.org/10.4236/oalib.1111241
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