Let X and Y are two fuzzy normed vector spaces on the same field K , and map f : X → Y be continuously on X . We use the notation N X , N for corresponding the norms on X and Y . In this paper, we investigate the stability of generalized Jensen-type additive function equation with 2k-variables when X is a fuzzy normed-algebras with norm N X and Y is a fuzzy Banach algebras with norm N.

In fact, when X is a fuzzy normed algebras with norm N X and Y is a fuzzy Banach algebras with norm N, we solve and prove the Hyers-Ulam-Rassias type stability of generalized Jensen-type additive function equation in fuzzy Banach algebras, associated to the Jensen type additive functional equation

m f ( α ∑ j = 1 k x j + α ∑ j = 1 k y j m ) = ∑ j = 1 k     α f ( x j ) + ∑ j = 1 k     α f ( y j ) (1)

The study of the stability of generalized Jensen-type additive function equation in fuzzy Banach algebras is 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 ( ⋅ , ⋅ ) . Given ε > 0 , there exists a δ > 0 such that if f : G → G ′ satisfies

d ( f ( x ∗ y ) , f ( x ) ∘ f ( y ) ) < δ , ∀ x ∈ G

then there is a homomorphism h : G → G ′ with

d ( f ( x ) , h ( x ) ) < ε , ∀ x ∈ G

Since Hyers’ answer to Ulam’s question [2] , many ideas have arisen from mathematicians who have built theories about space such as the Theory of fuzzy space. It has much progressed in developing the theory of randomness. Some mathematicians have defined fuzzy norms on a vector space from various points of view. Following Bag and Samanta [3] and Cheng and Mordeson [4] gave an idea of a fuzzy norm in such a manner that the corresponding fuzzy metric was of Kramosil and Michalek type [5] and investigated some properties of fuzzy normed spaces. We use the definition of fuzzy normed spaces given in [3] [6] [7] [8] to investigate a fuzzy version of the Hyers-Ulam stability for the Jensen functional equation in the fuzzy normed algebra setting.

The functional equation f ( x + y ) + f ( x − y ) = 2 f ( x ) + 2 f ( y ) is called a quadratic functional equation. The Hyers-Ulam stability of the quadratic functional equation was proved by Skof [9] for mapping f : X → Y , where X is a normed space and Y is a Banach space. Cholewa [10] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group. Czerwik [11] proved the Hyers-Ulam stability of the quadratic functional equation.

The stability problems for several functional equations have been extensively investigated by a number of authors and there are many interesting results concerning this problem. Such as in 2008, Choonkil Park [12] have established and investigated the Hyers-Ulam-Rassias stability of homomorphisms in quasi-Banach algebras the following Jensen functional equation

2 f ( x + y 2 ) = f ( x ) + f ( y )

And next in 2009, M. Éhaghi Gordji and M. Bavand Savadkouhi [13] have established and investigated the approximation of generalized stability of homomorphisms in quasi-Banach algebras the following Jensen functional equation

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

Next, in 2022 Ly Van An [14] have established and investigated the approximation of generalized stability of homomorphisms in quasi-Banach algebras the following Jensen type functional equation

m f ( ∑ j = 1 k x j + ∑ j = 1 k x k + j m ) = ∑ j = 1 k     f ( x j ) + ∑ j = 1 k     f ( x k + j ) (2)

Recently, Ly Van An continued to conduct extensive research (1.2) in the Hyers-Ulam-Rassias type on fuzzy Banach algebras for the following equation

m f ( α ∑ j = 1 k x j + α ∑ j = 1 k y j m ) = ∑ j = 1 k     α f ( x j ) + ∑ j = 1 k     α f ( y j )

i.e., the functional equation with 2k-variables. Under suitable assumptions on spaces X and Y , we will prove that the mappings satisfying the functional (1). Thus, the results in this paper are generalization of those in [12] [13] [14] for functional equation with 2k-variables.

In this paper, I build a general homomorphism based on Jensen equation with 2k-variables on fuzzy Banach algebra. This is an extended problem for the field of homotopy research, exploiting unlimited problems of variables to build this problem based on the ideas of mathematicians around the world. See [1] - [30] . Allow me to express my deep gratitude to the mathematicians.

The paper is organized as follows:

In Section 2, we remind some basic notations in [3] [6] [7] [8] [16] [25] [30] such as Fuzzy normed spaces, extended metric space theorem and solutions of the Jensen function equation.

Section 3: Using the fixed point method, establish extended homomorphisms on fuzzy Banach algebra.

Section 4: Using the direct method, establish extended homomorphisms on fuzzy Banach algebra.

Now we study extended homomorphism by fixed point method.

When X is a fuzzy normed algebra with quasi-norm N X , Y is a fuzzy Banach algebras with norm N. Under this setting, we need to show that the mapping must satisfy (1). These results are given in the following.

Here we assume that m ≥ 2 is a positive integer and α ∈ ℝ .

Theorem 3. Suppose ψ : X 2 k → [ 0, ∞ ) be a function such that there exists an L < 1 m

ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) ≤ L m ( m x 1 , ⋯ , m x k , m y 1 , ⋯ , m y k ) (3)

for all x j , y j ∈ X for j = 1 → k . If f : X → Y be a mapping satisfying f ( 0 ) = 0 and

N ( m f ( ∑ j = 1 k α x j + ∑ j = 1 k α y j m ) − ∑ j = 1 k     α f ( x j ) − ∑ j = 1 k     α f ( y j ) , t ) ≥ t t + ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) (4)

N ( f ( ∏ j = 1 k     x j ⋅ y j ) − ∏ j = 1 k     f ( x j ) ⋅ ∏ j = 1 k     f ( y j ) , t ) ≥ t t + ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) (5)

for all x j , y j ∈ X for j = 1 → k , for all t > 0 and all α ∈ ℝ . Then

A ( x ) = N − lim n → ∞ m n f ( x m n )

exists for each x ∈ X and defines a fuzzy algebras generalized homomorphism A : X → Y such that

N ( f ( x ) − A ( x ) , t ) ≥ ( 1 − L ) t ( 1 − L ) t + ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) (6)

Proof. Putting α = 1 .

Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) by ( x , ⋯ ,0,0, ⋯ ,0 ) in hypothesis (4), we have

N ( m f ( x m ) − f ( x ) , t ) ≥ t t + ψ ( x ,0 , ⋯ ,0,0, ⋯ ,0 ) (7)

for all x ∈ X .

Now we consider the set

M : = { g : X → Y }

and introduce the generalized metric on M as follows:

d ( g , h ) : = inf { β ∈ ℝ + : N ( g ( x ) − h ( x ) , β t ) ≥ t t + φ ( x ,0 , ⋯ ,0,0, ⋯ ,0 ) , ∀ x ∈ X , ∀ t > 0 } ,

where, as usual, inf ϕ = + ∞ . That has been proven by mathematicians ( M , d ) is complete [18] Now we consider the linear mapping T : M → M such that

T g ( x ) : = m g ( x m )

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 ) , ∀ x ∈ X , ∀ t > 0.

Hence

N ( g ( x ) − h ( x ) , L ε t ) = N ( m g ( x m ) − m h ( x m ) , L ε t ) = N ( g ( x m ) − h ( x m ) , L m ε t ) ≥ L t m L t m + φ ( x m ,0 , ⋯ ,0,0, ⋯ ,0 ) ≥ L t m L t m + L m φ ( x ,0 , ⋯ ,0,0, ⋯ ,0 ) = t t + φ ( x ,0 , ⋯ ,0,0, ⋯ ,0 ) , ∀ x ∈ X , ∀ t > 0. (8)

So d ( g , h ) = ε implies d ( T g , T h ) ≤ L ⋅ ε . This means that

d ( T g , T h ) ≤ L d ( g , h )

for all g , h ∈ M . On ther hand, (6) implies that d ( f , T f ) ≤ 1 .

By Theorem 2.5, there exists a mapping A : X → Y satisfying the following:

(1) A is a fixed point of T, i.e.,

A ( x m ) = 1 m A ( x ) (9)

for all x ∈ X . The mapping A is a unique fixed point T in the set

ℚ = { g ∈ M : d ( f , g ) < ∞ }

This implies that A is a unique mapping satisfying (9) such that there exists a β ∈ ( 0, ∞ ) satisfying.

N ( f ( x ) − A ( x ) , β t ) ≥ t t + φ ( x ,0 , ⋯ ,0,0, ⋯ ,0 ) , ∀ x ∈ X .

(2) d ( T n f , A ) → 0 as n → ∞ . This implies equality

N − l i m n → ∞ m n f ( x m n ) = A ( x )

for all x ∈ X

(3) d ( f , A ) ≤ 1 1 − L d ( f , T f ) ,

which implies the inequality

d ( f , A ) ≤ 1 1 − L

This implies that the inequality (6)

By (4), I have

N ( m p + 1 f ( ∑ j = 1 k α x j + ∑ j = 1 k α y j m p + 1 ) − m n ∑ j = 1 k     α f ( x m p ) − m p ∑ j = 1 k     α f ( x m p ) , m p t ) ≥ t t + ψ ( x 1 m p , ⋯ , x k m p , y 1 m p , ⋯ , y k m p ) (10)

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X , ∀ t > 0 , α ∈ ℝ . So

N ( m p + 1 f ( ∑ j = 1 k α x j + ∑ j = 1 k α y j m p + 1 ) − m n ∑ j = 1 k     α f ( x m p ) − m p ∑ j = 1 k     α f ( x m p ) , m p t ) ≥ t m p t m p + L p m p ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) (11)

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X , ∀ t > 0 , α ∈ ℝ . So

Since

l i m p → ∞ t m p t m p + L p m p ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) = 1

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X , ∀ t > 0 , m ∈ ℝ . So

N ( m A ( ∑ j = 1 k α x j + ∑ j = 1 k α y j m ) − ∑ j = 1 k     α A ( x j ) − ∑ j = 1 k     α A ( y j ) , t ) = 1 (12)

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X , ∀ t > 0 , ∀ α ∈ ℝ . So we have

m A ( ∑ j = 1 k α x j + ∑ j = 1 k α y j m ) − ∑ j = 1 k     α A ( x j ) − ∑ j = 1 k     α A ( y j ) = 0

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X , ∀ t > 0 , α ∈ ℝ . So the mapping A : X → Y is additive and ℝ -linear. From (5)

N ( m 2 k f ( ∏ j = 1 k x j m k ⋅ y j m k ) − m k ∏ j = 1 k     f ( x j m k ) ⋅ m k ∏ j = 1 k     f ( y j m k ) , m 2 k t ) ≥ t t + ψ ( x 1 m p , ⋯ , x k m p , y 1 m p , ⋯ , y k m p ) (13)

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X , and all t > 0 .So

N ( m 2 k f ( ∏ j = 1 k x j ⋅ y j m 2 k ) − m k ∏ j = 1 k     f ( x j m k ) ⋅ m k ∏ j = 1 k     f ( y j m k ) , m 2 k t ) ≥ t m 2 p t m 2 p + L p m p ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) (14)

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X , and all t > 0 . Since

l i m p → ∞ t m 2 p t m 2 p + L p m p ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) = 1

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X , and all t > 0 ,

N ( A ( ∏ j = 1 k     x j ⋅ y j ) − ∏ j = 1 k     A ( x j ) ⋅ ∏ j = 1 k     A ( y j ) , t ) = 1. (15)

Thus

A ( ∏ j = 1 k     x j ⋅ y j ) − ∏ j = 1 k     A ( x j ) ⋅ ∏ j = 1 k     A ( y j ) = 0

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X , and all t > 0 . So that mapping A : X → Y is a fuzzy algebra generalized homomorphism, as desired.

□

Theorem 4. Suppose ψ : X 2 k → [ 0, ∞ ) be a function such that there exists an L < 1 m

ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) ≤ m L ( x 1 m , ⋯ , x k m , y 1 m , ⋯ , y k m ) (16)

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X , if f : X → Y be a mapping satisfying f ( 0 ) = 0 and

N ( m f ( ∑ j = 1 k α x j + ∑ j = 1 k α y j m ) − ∑ j = 1 k     α f ( x j ) − ∑ j = 1 k     α f ( y j ) , t ) ≥ t t + ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) (17)

N ( f ( ∏ j = 1 k     x j ⋅ y j ) − ∏ j = 1 k     f ( x j ) ⋅ ∏ j = 1 k     f ( y j ) , t ) ≥ t t + ψ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) (18)

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X , all t > 0 and all α ∈ ℝ . Then

A ( x ) = N − l i m p → ∞ m p f ( x m p )

exists for each x ∈ X and defines a fuzzy algebras generalized homomorphism A : X → Y such that

N ( f ( x ) − A ( x ) , t ) ≥ ( 1 − L ) t ( 1 − L ) t + L ψ ( x , ⋯ ,0,0, ⋯ ,0 ) (19)

for all x ∈ X and all t > 0 .

Proof. Let ( M , d ) be the generalized metric space defined on the proof of Theorem 3. Now we consider the linear mapping T : M → M such that

T g ( x ) : = 1 m m g ( 2 x )

for all x ∈ X .

Next putting α = 1 .

Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) by ( x , ⋯ ,0,0, ⋯ ,0 ) in hypothesis (17), we have

N ( m f ( x m ) − f ( x ) , t ) ≥ t t + ψ ( x , 0 , ⋯ ,0,0, ⋯ ,0 ) (20)

for all x ∈ X , all t > 0 . So

N ( f ( x ) − 1 m f ( m x ) , t m ) ≥ t t + ψ ( m x , 0 , ⋯ ,0,0, ⋯ ,0 ) ≥ t t + m L ψ ( x , 0 , ⋯ ,0,0, ⋯ ,0 ) (21)

for all x ∈ X , all t > 0 .

Thus

d ( f , T f ) ≤ L .

Hence

d ( f , A ) ≤ L 1 − L .

which implies that the inequality (19) holds. The rest of the proof is similar to the proof of Theorem 3.

□

Now we study extended homomorphism by direct method.

Where X is a fuzzy normed algebra with quasi-norm N X , Y is a fuzzy Banach algebras with norm N. Under this setting, we need to show that the mapping must satisfy (1). These results are given in the following.

Here we assume that m ≥ 2 is a positive integer and α ∈ ℝ .

Theorem 5. Suppose ψ : X 2 k → [ 0, ∞ ) be a function such that

∑ j = 0 ∞     m 2 k j ( x 1 m j , ⋯ , x k m j , y 1 m j , ⋯ , y k m j ) < ∞ (22)

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X , if f : X → Y be a mapping satisfying f ( 0 ) = 0 and

l i m t → ∞ N ( m f ( ∑ j = 1 k α x j + ∑ j = 1 k α y j m )         − ∑ j = 1 k     α f ( x j ) − ∑ j = 1 k     α f ( y j ) , t ψ ˜ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) ) = 1 (23)

uniformly on X 2 k for each α ∈ ℝ , and

lim t → ∞ N ( f ( ∏ j = 1 k     x j ⋅ y j ) − ∏ j = 1 k     f ( x j ) ⋅ ∏ j = 1 k     f ( y j ) , t ψ ˜ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) ) = 1 (24)

uniformly on X 2 k , where

ψ ˜ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) : = ∑ j = 0 ∞     m j ψ ( x 1 m j , ⋯ , x k m j , y 1 m j , ⋯ , y k m j ) < ∞ (25)

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X , then

A ( x ) = N − lim n → ∞ m n f ( x m n )

exists for each x ∈ X and defines a fuzzy algebras generalized homomorphism A : X → Y such that if for each θ > 0 , β > 0

N ( m f ( ∑ j = 1 k α x j + ∑ j = 1 k α y j m )         − ∑ j = 1 k     α f ( x j ) − ∑ j = 1 k     α f ( y j ) , θ ψ ˜ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) ) ≥ β (26)

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X , then

N ( f ( x ) − A ( x ) , θ ψ ˜ ( x , 0 , ⋯ ,0,0, ⋯ ,0 ) ) ≥ β (27)

for all x ∈ X .

Furthermore, the fuzzy algebra generalized homomorphism A : X → Y is a unique mapping such that

lim t → ∞ N ( f ( x ) − A ( x ) , t ψ ˜ ( x , 0 , ⋯ ,0,0, ⋯ ,0 ) ) = 1 (28)

uniformly on X .

Proof. We put α = 1 in (23).With ε > 0 , by (23), we can exist some t > 0 such that

N ( m f ( ∑ j = 1 k α x j + ∑ j = 1 k α y j m ) − ∑ j = 1 k     α f ( x j ) − ∑ j = 1 k     α f ( y j ) , t ) ≥ 1 − ε (29)

for all t ≥ t 0 . Next we replace ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) by ( x , ⋯ ,0,0, ⋯ ,0 ) in hypothesis (23), and we have

N ( m f ( x m ) − f ( x ) , t ψ ( x , 0 , ⋯ ,0,0, ⋯ ,0 ) ) ≥ 1 − ε (30)

for all x ∈ X . By induction on n, we will show that

N ( f ( x ) − m n f ( x m n ) , t ∑ p = 0 n − 1     m p ψ ( x m p , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) ) ≥ 1 − ε (31)

for all t ≥ t 0 , for all x ∈ X , all n ∈ ℕ . It follows from (30) and (31) holds for n = 1 We now assume that (31) satisfies all n ∈ ℕ . Then

N ( f ( x ) − m n f ( x m n ) , t ∑ p = 0 n − 1     m p ψ ( x m p , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) ) ≥ min { N ( f ( x ) − m n f ( x m n ) , t 0 ∑ p = 0 n − 1     m p ψ ( x m p , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) ) ,         N ( m n f ( x m n ) − m n + 1 f ( x m n + 1 ) , m n t 0 ψ ( x m p , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) ) } ≥ { 1 − ε , 1 − ε } = 1 − ε . (32)

This completes the induction argument. Letting t = t 0 and we replace n and x by q and x m n in (31), respectively, we get

N ( m n f ( x m n ) − m n + q f ( x m n + q ) , m n t 0 ∑ p = 0 q − 1     m p ψ ( x m q + p , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) ) ≥ 1 − ε (33)

for all n ≥ 0 , q > 0 . It follows from (22) and the equality

∑ p = 0 q − 1     m n + p ψ ( x m n + p , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) = ∑ p = n n + q − 1     m p ψ ( x m p , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) (34)

That for a given θ > 0 there is an n 0 ∈ ℕ such that

t 0 ∑ p = n n + q − 1     m p ψ ( x m p , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) < θ (35)

for all n ≥ n 0 and q > 0 . Now we deuce since (31) that

N ( m n f ( x m n ) − m n + q f ( x m n + q ) , θ ) ≥ N ( m n f ( x m n ) − m n + q f ( x m n + q ) , m n t 0 ∑ p = 0 q − 1     m p ψ ( x m n + p , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) ) ≥ 1 − ε . (36)

for all n ≥ n 0 , all q > 0 and all x ∈ X . It follows from (36) that the sequence { m n f ( x m n ) } is a Cauchy sequence for all x ∈ X . Since Y is a fuzzy complete (fuzzy Banach space), the sequence { m n f ( x m n ) } converges. So one can define the mapping A : X → Y by

A ( x ) : = N − lim n → ∞ m n f ( 1 m n x ) ∈ Y (37)

In other words, for each t ≥ 0 and ∀ x ∈ X

lim n → ∞ N ( m n f ( 1 m n x ) − A ( x ) , t ) = 1 (38)

Now we are fixed t > 0 and 0 < ε < 1 . Since

lim n → ∞ m n ψ ( x 1 m n , ⋯ , x k m n , y 1 m n , ⋯ , y k m n ) = 0 ,

there is an n ′ > n 0 such that

t 0 m n ψ ( x 1 m n , ⋯ , x k m n , y 1 m n , ⋯ , y k m n ) ≤ t m 2 k , ∀ n > n ′ .

Hence for each p > n ′ , we get

N ( f ( x ) − m n f ( x m n ) , t ∑ p = 0 n − 1     m p ψ ( x m p , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) ) ≥ 1 − ε (39)

for all t ≥ t 0 , for all x ∈ X and for all n ∈ ℕ . It follows from (30) and (31) holds for n = 1 We now assume that (31) satisfies all n ∈ ℕ . Then

N ( A ( ∑ j = 1 k x j + ∑ j = 1 k y j m ) − ∑ j = 1 k     A ( x j ) − ∑ j = 1 k     A ( y j ) , t ) ≥ min { N ( A ( ∑ j = 1 k x j + ∑ j = 1 k y j m ) − m p + 1 f ( ∑ j = 1 k x j + ∑ j = 1 k y j m p + 1 ) , t m 2 k ) ,       N ( A ( x 1 ) − m n f ( x m n ) , t m 2 k ) , ⋯ , N ( A ( x k ) − m n f ( x k m n ) , t m 2 k ) ,       N ( A ( y 1 ) − m n f ( y 1 m n ) , t m 2 k ) , ⋯ , N ( A ( y k ) − m n f ( y k m n ) , t m 2 k ) ,

      N ( m p + 1 f ( ∑ j = 1 k x j + ∑ j = 1 k y j m p + 1 ) − m p ∑ j = 1 k     f ( x m p ) − m p ∑ j = 1 k     f ( x m p ) , t m 2 k ) } ≥ N ( m p + 1 f ( ∑ j = 1 k x j + ∑ j = 1 k y j m p + 1 ) − m p ∑ j = 1 k     f ( x m p )             − m p ∑ j = 1 k     f ( x m p ) , t 0 m n ψ ( x 1 m n , ⋯ , x k m n , y 1 m n , ⋯ , y k m n ) ) ≥ 1 − ε . (40)

for all t ≥ t 0 and all x ∈ X .

Thus

N ( m A ( ∑ j = 1 k x j + ∑ j = 1 k y j m ) − ∑ j = 1 k     A ( x j ) − ∑ j = 1 k     A ( y j ) , t ) = 1

for all t > 0 , by N 2 ,

m A ( ∑ j = 1 k x j + ∑ j = 1 k y j m ) − ∑ j = 1 k     A ( x j ) − ∑ j = 1 k     A ( y j ) = 0, ∀ x ∈ X

Hence the mapping A : X → Y is additive.

Next we replace ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) by ( x , ⋯ ,0,0, ⋯ ,0 ) in hypothesis (23). ∀ ε > 0 , by (23), then exists t 0 > 0 such that

N ( m f ( ∑ j = 1 k α x j m ) − ∑ j = 1 k     α f ( x j ) , t ψ ( x , ⋯ , x , 0 , ⋯ , 0 ) ) ≥ 1 − ε , ∀ t ≥ t 0 . (41)

It follows from (41), we have

A ( ∑ j = 1 k     α x j ) = m A ( ∑ j = 1 k α x j m ) = α ∑ j = 1 k     A ( x j )

for all α ∈ ℝ and all x ∈ X .

Similarly, it follows from (24) that

f ( ∏ j = 1 k     x j ⋅ y j ) = ∏ j = 1 k     f ( x j ) ⋅ ∏ j = 1 k     f ( y j )

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k ∈ X . We now assume that ∀ θ > 0 and β > 0 satisfied (26). Put

ψ n ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) = ∑ j = 0 q − 1     m j ψ ( x 1 m j , ⋯ , x k m j , y 1 m j , ⋯ , y k m j ) (42)

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X .

Suppose by the same reasoning as in the beginning of the proof, one can deuce from (26) that

N ( f ( x ) − m n f ( x m n ) , θ ∑ p = 0 n − 1     m n − p , θ ψ ˜ ( x m p , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) ) ≥ β (43)

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X , then for all positive integer n. Suppose t > 0 we have

N ( f ( x ) − A ( x ) , θ ψ n ( x m n , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) + t ) ≥ min { N ( f ( x ) − m n f ( x m n ) , θ ψ n ( x m n , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) ) ,       N ( m n f ( x m n ) − A ( x ) , t ) } (44)

Combining (43) and (44). If f : X → Y be a mapping satisfying f ( 0 ) = 0 and the fact that lim n → ∞ N ( m n f ( x m n ) − A ( x ) , t ) = 1 , we observe that

N ( f ( x ) − A ( x ) , θ ψ n ( x ,0, ⋯ ,0,0, ⋯ ,0 ) + t ) ≥ α

For large enough n ∈ ℕ . Thanks to the continuity of the function

N ( f ( x ) − A ( x ) , ⋅ ) ,

we see that

N ( f ( x ) − A ( x ) , θ ψ ˜ n ( x , 0 , ⋯ ,0,0, ⋯ ,0 ) + t ) ≥ α

Now I give t → 0 , we conclude that

N ( f ( x ) − A ( x ) , θ ψ ˜ n ( x , 0 , ⋯ ,0,0, ⋯ ,0 ) + t ) ≥ α

In the end I still have to prove the uniqueness. Suppose A ′ be another additive mapping satisfying (27) and (28). Fix η > 0 . Given ε > 0 , follow (28) for A, and A ′ , then exist t 0 > 0 such that

N ( f ( x ) − A ( x ) , t ψ ˜ n ( x , 0 , ⋯ ,0,0, ⋯ ,0 ) ) ≥ 1 − ε

N ( f ( x ) − A ′ ( x ) , t ψ ˜ n ( x , 0 , ⋯ ,0,0, ⋯ ,0 ) ) ≥ 1 − ε

for all x ∈ X and ∀ t ≥ t 0 . With fixed x ∈ X then exists n 0 ∈ ℕ such that

t 0 ∑ j = 0 ∞     m p ψ ( x m p ,0, ⋯ ,0,0, ⋯ ,0 ) < η m

for all n ≥ n 0 . From

∑ j = 0 ∞     m p ψ ( x m p , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) = m n ∑ j = 0 ∞     m p − n ψ ( x m n − p , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) = m n ∑ j = 0 ∞     m i ψ ( 1 m i x m n , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) = m n ψ ˜ ( m − i x m n , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) (45)

N ( A ( x ) − A ′ ( x ) , η ) ≥ min { N ( m n f ( x m n ) − A ( x ) , η m ) , N ( A ′ ( x ) − m n f ( x m n ) , η m ) } = min { N ( f ( x m n ) − A ( x m n ) , η m n + 1 ) , N ( A ′ ( x m n + 1 ) − f ( x m n ) , η m ) } ≥ min { N ( f ( x m n ) − A ( x m n ) , m − n t 0 ∑ j = 0 ∞     m p ψ ( x m p , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) ) ,       N ( A ( x m n ) − f ( x m n ) , m − n t 0 ∑ j = 0 ∞     m p ψ ( x m p , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) ) } ,

= min { N ( f ( x m n ) − A ( x m n ) , t 0 ψ ˜ ( x m p , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) ) ,       N ( A ′ ( x m n + 1 ) − f ( x m n ) , t 0 ψ ˜ ( x m p , 0 , ⋯ , 0 , 0 , ⋯ , 0 ) ) } ≥ 1 − ε (46)

It follows that N ( A ( x ) − A ′ ( x ) , η ) = 1 for all η > 0 . So A ( x ) = A ′ ( x ) , ∀ x ∈ X . □

Theorem 6. Suppose ψ : X 2 k → [ 0, ∞ ) be a function such that

∑ j = 0 ∞     m − 2 k j ( x 1 m j x 1 , ⋯ , m j x k , m j y 1 , ⋯ , m j y k ) < ∞ (47)

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X . If f : X → Y be a mapping satisfying f ( 0 ) = 0 and

lim t → ∞ N ( m f ( ∑ j = 1 k α x j + ∑ j = 1 k α y j m )         − ∑ j = 1 k     α f ( x j ) − ∑ j = 1 k     α f ( y j ) , t ψ ˜ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) ) = 1 (48)

uniformly on X 2 k for each α ∈ ℝ , and

lim t → ∞ N ( f ( ∏ j = 1 k     x j ⋅ y j ) − ∏ j = 1 k     f ( x j ) ⋅ ∏ j = 1 k     f ( y j ) , t ψ ˜ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) ) = 1 (49)

uniformly on X 2 k , where

ψ ˜ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) : = ∑ j = 0 ∞     m − j ψ ( m j x 1 , ⋯ , m j x k , m j y 1 , ⋯ , m j y k ) < ∞ (50)

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X . Then

A ( x ) = N − lim n → ∞ m − n f ( m j x )

exists for each x ∈ X and defines a fuzzy algebras generalized homomorphism A : X → Y such that if for each θ > 0 , β > 0

N ( m f ( ∑ j = 1 k α x j + ∑ j = 1 k α y j m ) − ∑ j = 1 k     α f ( x j ) − ∑ j = 1 k     α f ( y j ) , θ ψ ˜ ( x 1 , ⋯ , x k , y 1 , ⋯ , y k ) ) ≥ β (51)

for all ( x 1 ,... , x k , y 1 , ... , y k ) ∈ X , then

N ( f ( x ) − A ( x ) , θ ψ ˜ ( x , 0 , ⋯ ,0,0, ⋯ ,0 ) ) ≥ β (52)

for all x ∈ X .

Furthermore, the fuzzy algebra generalized homomorphism A : X → Y is a unique mapping such that

lim t → ∞ N ( f ( x ) − A ( x ) , θ ψ ˜ ( x , 0 , ⋯ ,0,0, ⋯ ,0 ) ) = 1 (53)

uniformly on X .

In this paper, I built the existence of extended homomorphism on fuzzy Banach algebra based on Jensen equation 2k variables by two methods such as fixed point and direct method to check.

The author declares no conflicts of interest.

An, L.V. (2023) Building Extended Homomorphism on Fuzzy Banach Algebra Based on Jensen Equation with 2k-Variables by Fixed Point Methods and Direct Methods. Open Access Library Journal, 10: e10206. https://doi.org/10.4236/oalib.1110206