In this paper, I establish homomorphisms, isomorphisms, and derivatives of quasi-algebras based on the general additive equation Cauchy-Jensen with 3k variables. First, I establish the homomorphisms for Equation (1.1); second, I establish the isomorphisms for Equation (1.2); and finally, I develop the derivative for Equation (1.3). These are the main results of this paper.
Let X and Y are two linear spaces on the same field K , and f : X → Y be a linear mapping. I use the notation ‖ ⋅ ‖ X ( ‖ ⋅ ‖ Y ) for corresponding the norms on X and Y . In this paper, I investigate the stability of generalized homomorphisms-isomorphism and derivatives when X is a quasi-normed algebras with quasi-norm ‖ ⋅ ‖ X and that Y is a p-Banach algebras with p-norm ‖ ⋅ ‖ Y .
In fact, when X is a quasi-normed algebras with quasi-norm ‖ ⋅ ‖ X and that Y is a p-Banach algebras with p-norm ‖ ⋅ ‖ Y , I solve and prove the Hyers-Ulam-Rassias type stability of generalized Homomorphisms-isomorphism and derivatives on quasi-Banach algebra, associated to the following generalized Cauchy-Jensen additive functional equations
k ϕ ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) + k ϕ ( ∑ i = 1 k x i − y i 2 k + ∑ i = 1 k z i ) = ∑ i = 1 k ϕ ( x i ) + 2 k ∑ i = 1 k ϕ ( z i ) (1)
k ϕ ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) − k ϕ ( ∑ i = 1 k x i − y i 2 k + ∑ i = 1 k z i ) = ∑ i = 1 k ϕ ( y i ) (2)
2 k ϕ ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) = ∑ i = 1 k ϕ ( x i ) + ∑ i = 1 k ϕ ( y i ) + 2 k ∑ i = 1 k ϕ ( z i ) (3)
The study the stability of generalized homomorphisms-isomorphism and derivatives in quasi-Banach algebras originated from a question of S. M. Ulam [
Let ( G , ∗ ) be a group and let ( G ′ , ∘ , d ) be a metric group with metric d ( ⋅ , ⋅ ) . Given ε > 0 , does there exist a δ > 0 such that if f : G → G ′ satisfies inequality
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
If the answer is affirmative, I would say that the equation of homomorphism
H ( x ∗ y ) = H ( x ) ∘ H ( y ) , ∀ x ∈ G
is stable. The concept of stability for a functional equation arises when I replace the functional equation by an inequality which acts as a perturbation of the equation. Thus the stability question of functional equations is that how do the solutions of the inequality differ from those of the given functional equation? In 1941, Hyers [
Let X and Y be Banach space. Assume that f : X → Y satisfies
f ( x + y ) − f ( x ) − f ( y ) ≤ ε (4)
for all x , y ∈ X and some ε > 0 . Then there exists a unique additive mapping T : X → Y such that
‖ f ( x ) − T ( x ) ‖ ≤ ε (5)
Next 1978 Th. M. Rassias [
J. M. Rassias [
‖ f ( x + y ) − f ( x ) − f ( y ) ‖ = θ ⋅ ‖ x ‖ p ⋅ ‖ y ‖ q (6)
∀ x , y ∈ E ; involving a product of different powers of norms, where θ > 0 and real p , q such that r = p + q ≠ 1 , and retained the condition of continuity f ( t x ) in t for fixed x. Analogous results could be investigated with additive type equations involving a product of powers of norms.
Next 1994 Găvruta [
Recently, the authors studied the Hyers-Ulam-Rassias type stability for the following functional equations (see [
f ( x + y 2 + z ) + f ( x − y 2 + z ) = f ( x ) + 2 f ( z ) (7)
f ( x + y 2 + z ) − f ( x − y 2 + z ) = f ( y ) (8)
2 f ( x + y 2 + z ) + f ( x − y 2 + z ) = f ( x ) + f ( y ) + 2 f ( z ) (9)
Next
f ( ∑ j = 1 k x j + 1 k ∑ j = 1 k x k + j ) = ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( x k + j k ) (10)
and
2 k f ( 1 2 k ∑ j = 1 k x j + 1 2 k 2 ∑ j = 1 k x k + j ) = ∑ j = 1 k f ( x j ) + ∑ j = 1 k f ( x k + j k ) (11)
Final
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 ) (12)
In this paper, I have built a general problem about homomorphisms-isomorphism and derivatives on quasi-Banach algebra based on the general additive Cauchy-Jensen equation. To write these problems, I follow the ideas of mathematicians around the world see [
In this paper, I solve and proved the Hyers-Ulam-Rassias type stability for functional Equations (1.1), (1.2) and (1.3), i.e., the functional equations with 3k variables. Under suitable assumptions on spaces X and Y , I will prove that the mappings satisfying the functional Equation (1.1), (1.2) and (1.3).
Thus, the results in this paper are generalization of those in [
In section preliminarier I remind some basic notations in such as Quasi-normed space―Quasi-Banach algebras. Some theorems ℝ -linear mapping and Solutions of the equations see ( [
Section 3: Constructing homomorphisms on quasi-Banach algebras for (1.1).
Section 4: Constructing isomorphisms on quasi-Banach algebras for (1.2).
Section 5: Constructing derivatives on quasi-Banach algebras for (1.3).
Let X be a real linear space. A quasi-norm is a real-valued function on X satisfying the following:
1) ‖ x ‖ ≥ 0 for all x ∈ X and ‖ x ‖ = 0 if and only if x = 0 .
2) ‖ λ x ‖ = | λ | ‖ x ‖ for all λ ∈ R and all x ∈ X .
3) There is a constant K ≥ 1 such that
‖ x + y ‖ ≤ K ( ‖ x ‖ + ‖ y ‖ ) , ∀ x , y ∈ X .
The pair ( X , ‖ ⋅ ‖ ) is called a quasi-normed space if ‖ ⋅ ‖ is a quasi-norm on X .
The smallest possible K is called the modulus of concavity of ‖ ⋅ ‖ .
A quasi-Banach space is a complete quasi-normed space.
A quasi-norm ‖ ⋅ ‖ is called a p-norm ( 0 < p ≤ 1 ) if
‖ x + y ‖ p ≤ ‖ x ‖ p + ‖ y ‖ p ∀ x , y ∈ X .
In this case, a quasi-Banach space is called a p-Banach space.
Note: Given a p-norm, the formula d ( x , y ) : = ‖ x − y ‖ p gives us a translation invariant metric on X. By the Aoki-Rolewicz Theorem [
Let ( X , ‖ ⋅ ‖ ) be a quasi-normed space. The quasi-normed space ( X , ‖ ⋅ ‖ ) is called a quasi-normed algebras if X is an algebras and there is a constant K > 0 such that
‖ x ⋅ y ‖ ≤ K ‖ x ‖ ‖ y ‖
A quasi-Banach algebras is a complete quasi-normed algebras.
If the quasi-norm ‖ ⋅ ‖ is a p-norm, quasi-Banach is called p-Banach algebras.
Theorem 1. Th. M. Rassias: Let f : E 1 → E 2 be a mapping from a normed vector space E 1 into a Banach space E 2 subject to the inequality
‖ f ( x + y ) − f ( x ) − f ( y ) ‖ ≤ ε ( ‖ x ‖ p + ‖ y ‖ p ) , (13)
for all x , y ∈ E 1 , where ε and p are constants with p < 1 and ε ≥ 0 . then the limit
T ( x ) = l i m n → ∞ f ( 2 n x ) 2 n (14)
exists for all x ∈ E 1 and that T : E 1 → E 2 is the unique additive mapping satisfying
‖ f ( x ) − T ( x ) ‖ ≤ 2 θ 2 − 2 p ‖ x ‖ p , ∀ x ∈ E 1 . (15)
If p < 0 then (2.1) holds for x , y ≠ 0 and (2.2) for x ≠ 0 . Also, if for each x ∈ E 1 the function If f(tx) is continuous in t ∈ ℝ , then T is linear.
Theorem 2. Let E be real normed linear space and E ′ a real complete normed linear space. Assume that f : E → E ′ is an approximately additive mapping for which there exists constants θ ≥ 0 and p ∈ ℝ − { 1 } such that f ( x ) satisfy inequality
‖ f ( x + y ) − f ( x ) − f ( y ) ‖ ≤ θ ‖ x ‖ p 2 ‖ y ‖ p 2 , ∀ x , y ∈ E . (16)
then there exists a unique additive mapping linear T : E → E ′ satisfies
‖ f ( x ) − L ( x ) ‖ ≤ θ 2 p − 2 ‖ x ‖ p , x ∈ E . (17)
If, in addition f : E → E ′ is a transformation t → f ( t x ) is continuous in t ∈ ℝ for each fixed x ∈ E , then T is an R -linear mapping.
Theorem 3. Let E be real normed linear space and E ′ a real complete normed linear space. Assume that f : E → E ′ is an approximately additive mapping for which there exists constants θ ≥ 0 such that f ( x ) satisfy inequality
‖ f ( ∑ i = 1 n x i ) − ∑ i = 1 n f ( x i ) ‖ ≤ θ K ( x 1 , x 2 , ⋯ , x n ) . (18)
( x 1 , x 2 , ⋯ , x n ) ∈ E and K : E n → R + − { 0 } is a non-negative real-valued function such that
R n ( x ) = ∑ i = 1 n 1 n j K ( n j x 1 , n j x 2 , ⋯ , n j x n ) < ∞ (19)
is a non-negative function of x, and the condition
lim m → ∞ 1 n m K ( n m x 1 , n m x 2 , ⋯ , n m x n ) = 0 (20)
holds then there exists a unique additive mapping T n : E → E ′ satisfies
‖ f ( x ) − T n ( x ) ‖ ≤ θ n R n ( x ) , x ∈ E . (21)
If, in addition f : E → E ′ is a transformation t → f ( t x ) is continuous in t ∈ ℝ for each fixed x ∈ E , then T is an ℝ -linear mapping.
Theorem 4. Let E be real normed linear space and E ′ a real complete normed linear space. Assume that f : E → E ′ is an approximately additive mapping for which there exists constants θ ≥ 0 and p , q ∈ ℝ such that p + q ≠ 1 and f satisfy inequality
‖ f ( x + y ) − f ( x ) − f ( y ) ‖ ≤ θ ‖ x ‖ p ‖ y ‖ q , ∀ x , y ∈ E . (22)
then there exists a unique additive mapping linear T : E → E ′ satisfies
‖ f ( x ) − L ( x ) ‖ ≤ θ 2 p − 2 ‖ x ‖ p , x ∈ E . (23)
If, in addition f : E → E ′ is a transformation t → f ( t x ) is continuous in t ∈ ℝ for each fixed x ∈ E , then T is an ℝ -linear mapping.
f ( x + y ) = f ( x ) + f ( y ) (24)
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 ) (25)
is called the Jensen equation. In particular, every solution of the Jensen equation is said to be a Jensen additive mapping.
The functional equation
2 f ( x + y 2 + z ) = f ( x ) + f ( y ) + 2 f ( z ) (26)
is called the Cauchy-Jensen equation. In particular, every solution of the Cauchy-Jensen equation is said to be a Jensen-Cauchy additive mapping.
Now I construct a homomorphism for (1.1) Note that: (1.2) and (1.3) are also built exactly the same.
Here I assume that, A is a quasi-normed with norm ‖ ⋅ ‖ A and that B is a p-Banach algebra with norm ‖ ⋅ ‖ B . Let K be the modulus of concavity of ‖ ⋅ ‖ B . Under this setting, I can show that the mappings satisfying (1.1) is homomorphisms.
Theorem 5. Let r > q with q ≥ 2 and θ be positive real numbers, and f : X → Y be a mapping such that
‖ k f ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) + k f ( ∑ i = 1 k x i − y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k f ( x i ) − 2 k ∑ i = 1 k f ( z i ) ‖ Y ≤ θ ( ∑ i = 1 k ‖ x i ‖ X r + ∑ i = 1 k ‖ y i ‖ X r + ∑ i = 1 k ‖ z i ‖ X r ) (27)
‖ f ( ∏ i = 1 n x i y i ) − ∏ i = 1 n f ( x i ) ∏ i = 1 n f ( y i ) ‖ Y ≤ θ ( ∑ i = 1 k ‖ x i ‖ X r + ∑ i = 1 k ‖ y i ‖ X r ) (28)
for all x i , y i , z i ∈ X , for all i = 1 → k . If f ( t x ) is continuous in t ∈ ℝ each fixed x ∈ X , then there exists a unique homomorphism H : X → Y such that
‖ f ( x ) − H ( x ) ‖ Y ≤ ( 2 + 1 k ) θ ( 2 p r − 2 ) 1 p ‖ x ‖ X r , ∀ x ∈ X . (29)
Proof. I replace ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x , ⋯ , x , x , ⋯ , x , x , ⋯ ,0 ) in (27), I have
‖ k f ( 2 x ) − 2 k f ( x ) ‖ Y ≤ ( 2 k + 1 ) θ ‖ x ‖ X r (30)
for all x ∈ X . So
‖ f ( x ) − 2 f ( x 2 ) ‖ Y ≤ ( 2 + 1 k ) θ 2 r ‖ x ‖ X r (31)
for all x ∈ X . From Y is p-Banach algebra so I have
‖ 2 l f ( x 2 l ) − 2 m f ( x 2 m ) ‖ Y p ≤ ∑ j = l m − 1 ‖ 2 j f ( x 2 j ) − 2 j + 1 f ( x 2 j + 1 ) ‖ Y p ≤ ( 2 + 1 k ) p θ p 2 p r ∑ j = l + 1 m − 1 2 p j 2 p r j ‖ x ‖ X p r (32)
for all nonnegative integers m and l with m > l and for all x ∈ X . It follows (32) that the sequence { 2 n f ( x 2 n ) } is a Cauchy sequence for all x ∈ X . Since Y is complete, the sequence { 2 n f ( x 2 n ) } converges. So one can define the mapping H : X → Y by
H ( x ) = l i m n → ∞ 2 n f ( x 2 n ) , (33)
for all x ∈ X . By (28) and (27),
‖ k H ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) + k H ( ∑ i = 1 k x i − y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k H ( x i ) − 2 k ∑ i = 1 k H ( z i ) ‖ Y = l i m n → ∞ 2 n ‖ k f ( ∑ i = 1 k x i + y i 2 n + 1 k + ∑ i = 1 k z i 2 n ) + k f ( ∑ i = 1 k x i − y i 2 n + 1 k + ∑ i = 1 k z i 2 n ) − ∑ i = 1 k f ( x i 2 n ) − 2 k ∑ i = 1 k f ( z i 2 n ) ‖ Y ≤ l i m n → ∞ 2 n θ 2 n r ( ∑ i = 1 k ‖ x i ‖ X r + ∑ i = 1 k ‖ y i ‖ X r + ∑ i = 1 k ‖ z i ‖ X r ) = 0 (34)
for all x i , y i , z i ∈ X , for all i = 1 → k .
So
k H ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) + k H ( ∑ i = 1 k x i − y i 2 k + ∑ i = 1 k z i ) = ∑ i = 1 k H ( x i ) + 2 k ∑ i = 1 k H ( z i ) (35)
for all x i , y i , z i ∈ X , for all i = 1 → k . By lemma 5 (see 24]), the mapping H : X → Y is Cauchy additive. (see the theorem of [
Then mapping H : X → Y is ℝ -linear. It follows from (28) that
‖ f ( ∏ i = 1 k x i y i ) − ∏ i = 1 k f ( x i ) ∏ i = 1 k f ( y i ) ‖ Y = l i m n → ∞ 2 2 n k ‖ f ( ∏ i = 1 k x i 2 n ⋅ y i 2 n ) − ∏ i = 1 k f ( x i 2 n ) ⋅ ∏ i = 1 k f ( y i 2 n ) ‖ Y ≤ l i m n → ∞ 2 2 n k θ 2 n r ( ∑ i = 1 k ‖ x i ‖ X r + ∑ i = 1 k ‖ y i ‖ X r ) = 0 (36)
∀ x , y ∈ X .
So
H ( ∏ i = 1 k x i ⋅ y i ) = ∏ i = 1 k H ( x i ) ∏ i = 1 k H ( y i ) (37)
∀ x , y ∈ X .
Now I prove the uniqueness of H. Assume that H 1 : X → Y is a Cauchy-Jensen additive mapping satisfying (29). Then I have
‖ H ( x ) − H 1 ( x ) ‖ Y = 2 n ‖ H ( 1 2 n x ) + H 1 ( 1 2 n x ) ‖ Y ≤ 2 n K ( ‖ H ( 1 2 n x ) − f ( 1 2 n x ) ‖ Y + ‖ f ( 1 2 n x ) + H 1 ( 1 2 n x ) ‖ Y ) ≤ 2 ( 2 + 1 k ) K θ ( 2 p r − 2 p ) 1 p ⋅ 2 n 2 n r ‖ x ‖ X r (38)
which tends to zero as n → ∞ for all x ∈ X . So I can conclude that H ( x ) = H 1 ( x ) for all x ∈ X . This proves the uniqueness of H. Thus the mapping H 1 : X → Y is a unique homomorphism satisfying (29). □
Theorem 6. Let r < q with q ≤ 1 and θ be positive real numbers, and f : X → Y be a mapping such that
‖ k f ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) + k f ( ∑ i = 1 k x i − y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k f ( x i ) − 2 k ∑ i = 1 k f ( z i ) ‖ Y ≤ θ ( ∑ i = 1 k ‖ x i ‖ X r + ∑ i = 1 k ‖ y i ‖ X r + ∑ i = 1 k ‖ z i ‖ X r ) (39)
‖ f ( ∏ i = 1 k x i y i ) − ∏ i = 1 k f ( x i ) ∏ i = 1 k f ( y i ) ‖ Y ≤ θ ( ∑ i = 1 k ‖ x i ‖ X r + ∑ i = 1 k ‖ y i ‖ X r ) (40)
for all x i , y i , z i ∈ X , for all i = 1 → k . If f ( t x ) is continuous in t ∈ ℝ each fixed x ∈ X , then there exists a unique homomorphism H : X → Y such that
‖ f ( x ) − H ( x ) ‖ Y ≤ ( 1 + 1 2 k ) θ ( 2 − 2 p r ) 1 p ‖ x ‖ X r , ∀ x ∈ X . (41)
Proof. I replace ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x , ⋯ , x , x , ⋯ , x , x , ⋯ ,0 ) in (39), I have
‖ k f ( 2 x ) − 2 k f ( x ) ‖ Y ≤ ( 2 k + 1 ) θ ‖ x ‖ X r (42)
for all x ∈ X . So
‖ f ( x ) − 1 2 f ( 2 x ) ‖ Y ≤ ( 1 + 1 2 k ) θ ‖ x ‖ X r (43)
for all x ∈ X . Since Y is a p-Banach algebra,
‖ 1 2 l f ( 2 l x ) − 1 2 m f ( 2 m x ) ‖ Y p ≤ ∑ j = l m − 1 ‖ 1 2 j f ( 2 j x ) − 1 2 j + 1 f ( 2 j + 1 x ) ‖ Y p ≤ ( 1 + 1 2 k ) p θ p ∑ j = l + 1 m − 1 2 p r j 2 p j ‖ x ‖ X p r (44)
for all nonnegative integers m and l with m > l and for all x ∈ X . It follows (44) that the sequence { 1 2 n f ( 2 n x ) } is a Cauchy sequence for all x ∈ X . Since Y is complete, the sequence { 1 2 n f ( 2 n x ) } converges. So one can define the mapping H : X → Y by
H ( x ) = l i m n → ∞ 1 2 n f ( 2 n x )
for all x ∈ X .
Moreover, letting l = 0 and passing the limit m → ∞ in (44) I get (41). The rest of the proof is similar to the proof of Theorem 5. □
Theorem 7. Let r > k , with k ≥ 1 and θ be positive real numbers, and f : X → Y be a mapping such that
‖ k f ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) + k f ( ∑ i = 1 k x i − y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k f ( x i ) − 2 k ∑ i = 1 k f ( z i ) ‖ Y ≤ θ ⋅ ∏ i = 1 k ‖ x i ‖ X r ⋅ ∏ i = 1 k ‖ y i ‖ X r ⋅ ‖ z 1 ‖ X r k ⋅ ( 1 + ∏ i = 2 k ‖ z i ‖ X r ) (45)
‖ f ( ∏ i = 1 k x i y i ) − ∏ i = 1 k f ( x i ) ∏ i = 1 k f ( y i ) ‖ Y ≤ θ ⋅ ∏ i = 1 k ‖ x i ‖ X r ⋅ ∏ i = 1 k ‖ y i ‖ X r (46)
for all x i , y i , z i ∈ X , for all i = 1 → k . If f ( t x ) is continuous in t ∈ ℝ each fixed x ∈ X , then there exists a unique homomorphism H : X → Y such that
‖ f ( x ) − H ( x ) ‖ Y ≤ θ k ( 2 3 p k r − 2 p ) 1 p ‖ x ‖ X 3 k r , ∀ x ∈ X . (47)
Proof. I replace ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x , ⋯ , x , x , ⋯ , x , x , ⋯ ,0 ) in (45), I have
‖ k f ( 2 x ) − 2 k f ( x ) ‖ Y ≤ θ ‖ x ‖ X 3 k r (48)
for all x ∈ X . So
‖ f ( x ) − 2 f ( x 2 ) ‖ Y ≤ 1 k θ 2 3 k r ‖ x ‖ X 3 k r (49)
for all x ∈ X . Since Y is a p-Banach algebras,
‖ 2 l f ( x 2 l ) − 2 m f ( x 2 m ) ‖ Y p ≤ ∑ j = l m − 1 ‖ 2 j f ( x 2 j ) − 2 j + 1 f ( x 2 j + 1 ) ‖ Y p ≤ 1 k p θ p 2 3 p k r ∑ j = l + 1 m − 1 2 p j 2 3 p k r j ‖ x ‖ X 3 p k r (50)
for all nonnegative integers m and l with m > l and for all x ∈ X . It follows (50) that the sequence { 2 n f ( x 2 n ) } is a Cauchy sequence for all x ∈ X . Since Y is complete, the sequence { 2 n f ( x 2 n ) } converges. So one can define the mapping H : X → Y by
H ( x ) = l i m n → ∞ 2 n f ( x 2 n ) , (51)
for all x ∈ X . By (46) and (45),
‖ k H ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) + k H ( ∑ i = 1 k x i − y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k H ( x i ) − 2 k ∑ i = 1 k H ( z i ) ‖ Y = l i m n → ∞ 2 n ‖ k f ( ∑ i = 1 k x i + y i 2 n + 1 k + ∑ i = 1 k z i 2 n ) + k f ( ∑ i = 1 k x i − y i 2 n + 1 k + ∑ i = 1 k z i 2 n ) − ∑ i = 1 k f ( x i 2 n ) − 2 k ∑ i = 1 k f ( z i 2 n ) ‖ Y ≤ l i m n → ∞ 2 n θ 2 3 n k r ∏ i = 1 k ‖ x i ‖ X r ⋅ ∏ i = 1 k ‖ y i ‖ X r ⋅ ‖ z 1 ‖ X r k ⋅ ( 1 + ∏ i = 2 k ‖ z i ‖ X r ) = 0 (52)
for all x i , y i , z i ∈ X , for all i = 1 → k .
So
k H ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) + k H ( ∑ i = 1 k x i − y i 2 k + ∑ i = 1 k z i ) = ∑ i = 1 k H ( x i ) + 2 k ∑ i = 1 k H ( z i ) (53)
for all x 1 , x 2 , ⋯ , x 3 k ∈ X . By lemma 5 (see [
It follows from (46) that
‖ f ( ∏ i = 1 k x i y i ) − ∏ i = 1 k f ( x i ) ∏ i = 1 k f ( y i ) ‖ Y = l i m n → ∞ 2 2 n k ‖ f ( ∏ i = 1 k x i 2 n ⋅ y i 2 n ) − ∏ i = 1 k f ( x i 2 n ) ⋅ ∏ i = 1 k f ( y i 2 n ) ‖ Y ≤ l i m n → ∞ 2 2 n k θ 2 2 n r ∏ i = 1 k ‖ x i ‖ X r ⋅ ∏ i = 1 k ‖ y i ‖ X r = 0 (54)
∀ x , y ∈ X .
So
H ( ∏ i = 1 k x i ⋅ y i ) = ∏ i = 1 k H ( x i ) ∏ i = 1 k H ( y i ) (55)
∀ x , y ∈ X .
The rest of the proof is similar to the proof of Theorem 5 □
Theorem 8. Let r < 1 q , with q ≥ 3 and θ be positive real numbers, and f : X → Y be a mapping such that
‖ k f ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) + k f ( ∑ i = 1 k x i − y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k f ( x i ) − 2 k ∑ i = 1 k f ( z i ) ‖ Y ≤ θ ⋅ ∏ i = 1 k ‖ x i ‖ X r ⋅ ∏ i = 1 k ‖ y i ‖ X r ⋅ ‖ z 1 ‖ X r k ⋅ ( 1 + ∏ i = 2 k ‖ z i ‖ X r ) (56)
‖ f ( ∏ i = 1 k x i y i ) − ∏ i = 1 k f ( x i ) ∏ i = 1 k f ( y i ) ‖ Y ≤ θ ⋅ ∏ i = 1 k ‖ x i ‖ X r ⋅ ∏ i = 1 k ‖ y i ‖ X r (57)
for all x i , y i , z i ∈ X , for all i = 1 → k . If f ( t x ) is continuous in t ∈ ℝ each fixed x ∈ X , then there exists a unique homomorphism H : X → Y such that
‖ f ( x ) − H ( x ) ‖ Y ≤ θ k ( 2 p − 2 3 p k r ) 1 p ‖ x ‖ X 3 k r , ∀ x ∈ X . (58)
Proof. I replace ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x , ⋯ , x , x , ⋯ , x , x , ⋯ ,0 ) in (56), I have
‖ k f ( 2 x ) − 2 k f ( x ) ‖ Y ≤ θ ‖ x ‖ X 3 k r (59)
for all x ∈ X . So
‖ f ( x ) − 1 2 f ( 2 x ) ‖ Y ≤ 1 2 k θ ‖ x ‖ X 3 k r
for all x ∈ X . Since Y is a p-Banach algebras,
‖ 1 2 l f ( 2 l x ) − 1 2 m f ( 2 m x ) ‖ Y p ≤ ∑ j = l m − 1 ‖ 1 2 j f ( 2 j x ) − 1 2 j + 1 f ( 2 j + 1 x ) ‖ Y p ≤ θ ( 2 k ) p ∑ j = l + 1 m − 1 2 3 p k r j 2 p j ‖ x ‖ X 3 p k r (60)
for all nonnegative integers m and l with m > l and for all x ∈ X . It follows (60) that the sequence { 1 2 n f ( 2 n x ) } is a Cauchy sequence for all x ∈ X . Since Y is complete, the sequence { 1 2 n f ( 2 n x ) } converges. So one can define the mapping H : X → Y by
H ( x ) = l i m n → ∞ 1 2 n f ( 2 n x )
for all x ∈ X . Moreover, letting l = 0 and passing the lim m → ∞ in (60) I have (56),
It follows from (57) that
‖ f ( ∏ i = 1 k x i y i ) − ∏ i = 1 k f ( x i ) ∏ i = 1 k f ( y i ) ‖ Y = l i m n → ∞ 1 2 2 n k ‖ f ( ∏ i = 1 k 2 n x i ⋅ 2 n y i ) − ∏ i = 1 k f ( 2 n x i ) ⋅ ∏ i = 1 k f ( 2 n y i ) ‖ Y ≤ l i m n → ∞ θ 2 2 n k r ∏ i = 1 k ‖ x i ‖ X r ⋅ ∏ i = 1 k ‖ y i ‖ X r = 0
∀ x , y ∈ X . So
H ( ∏ i = 1 k x i ⋅ y i ) = ∏ i = 1 k H ( x i ) ∏ i = 1 k H ( y i )
∀ x , y ∈ X .
The rest of the proof is similar to the proof of Theorem 5. □
Now I construct isomorphisms for (1.2). Note that: (1.1) and (1.3) are also built exactly the same.
Here I assume that A is a quasi-Banach with norm ‖ ⋅ ‖ A and unit e and that B is a p-Banach algebra with norm ‖ ⋅ ‖ B and unit e ′ . Let K be the modulus of concavity of ‖ ⋅ ‖ B . Under this setting, I can show that the mappings satisfying (1.2) is isomorphisms.
Theorem 9. Let r > q with q ≥ 1 and θ be positive real numbers, and f : X → Y be a mapping such that
‖ k f ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) − k f ( ∑ i = 1 k x i − y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k f ( x i ) ‖ Y ≤ θ ( ∑ i = 1 k ‖ x i ‖ X r + ∑ i = 1 k ‖ y i ‖ X r + ∑ i = 1 k ‖ z i ‖ X r ) (61)
f ( ∏ i = 1 k x i y i ) = ∏ i = 1 k f ( x i ) ∏ i = 1 k f ( y i ) (62)
for all x i , y i , z i ∈ X , for all i = 1 → k . If f ( t x ) is continuous in t ∈ ℝ each fixed x ∈ X and
l i m n → ∞ 2 n f ( e 2 n ) = e ′ (63)
then the mapping f : X → Y is an isomorphism.
Proof. I replace ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x , ⋯ , x , x , ⋯ , x , x , ⋯ ,0 ) in (61), I have
‖ k f ( 2 x ) − 2 k f ( x ) ‖ Y ≤ ( 2 k + 1 ) θ ‖ x ‖ X r (64)
for all x ∈ X . So
‖ f ( x ) − 2 f ( x 2 ) ‖ Y ≤ ( 2 + 1 k ) θ 2 r ‖ x ‖ X r (65)
for all x ∈ X . Since Y is a p-Banach algebra,
‖ 2 l f ( x 2 l ) − 2 m f ( x 2 m ) ‖ Y p ≤ ∑ j = l m − 1 ‖ 2 j f ( x 2 j ) − 2 j + 1 f ( x 2 j + 1 ) ‖ Y p ≤ ( 2 + 1 k ) p θ p 2 p r ∑ j = l + 1 m − 1 2 p j 2 p r j ‖ x ‖ X p r (66)
for all nonnegative integers m and l with m > l and for all x ∈ X . It follows (66) that the sequence { 2 n f ( x 2 n ) } is a Cauchy sequence for all x ∈ X . Since Y is complete, the sequence { 2 n f ( x 2 n ) } converges. So one can define the mapping H : X → Y by
H ( x ) = l i m n → ∞ 2 n f ( x 2 n ) , (67)
for all x ∈ X . Moreover, letting l = 0 and passing the limit m → ∞ in (66), I get
‖ f ( x ) − H ( x ) ‖ Y ≤ ( 2 + 1 k ) θ ( 2 p r − 2 p ) 1 p ‖ x ‖ X r , ∀ x ∈ X . (68)
It follows from (62) that
‖ k H ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) − k H ( ∑ i = 1 k x i − y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k H ( x i ) ‖ Y = l i m n → ∞ 2 n ‖ k f ( ∑ i = 1 k x i + y i 2 n + 1 k + ∑ i = 1 k z i 2 n ) + k f ( ∑ i = 1 k x i − y i 2 n + 1 k + ∑ i = 1 k z i 2 n ) − ∑ i = 1 k f ( x i 2 n ) ‖ Y ≤ l i m n → ∞ 2 n θ 2 n r ( ∑ i = 1 k ‖ x i ‖ X r + ∑ i = 1 k ‖ y i ‖ X r + ∑ i = 1 k ‖ z i ‖ X r ) = 0 (69)
for all x i , y i , z i ∈ X , for all i = 1 → k .
So
k H ( ∑ i = 1 k x i + x k + i 2 k + ∑ i = 1 k x 2 k + i ) − k H ( ∑ i = 1 k x i − x k + i 2 k + ∑ i = 1 k x 2 k + i ) = ∑ i = 1 k H ( x i ) (70)
for all x i , y i , z i ∈ X , for all i = 1 → k . By lemma 5 (see [
The mapping H : X → Y is ℝ -linear. Since
f ( ∏ i = 1 k x i y i ) = ∏ i = 1 k f ( x i ) ∏ i = 1 k f ( y i ) (71)
for all x i , y i , z i ∈ X , for all i = 1 → k .
H ( ∏ i = 1 k x i y i ) = lim n → ∞ 2 2 n k f ( ∏ i = 1 k x i 2 n ⋅ ∏ i = 1 k y i 2 n ) = lim n → ∞ 2 n k f ( ∏ i = 1 k x i 2 n ) ⋅ 2 n k f ( ∏ i = 1 k y i 2 n ) = H ( ∏ i = 1 k x i ) ⋅ H ( ∏ i = 1 k y i ) (72)
for all x i , y i , z i ∈ X , for all i = 1 → k . So the mapping H : X → Y is homomorphism.
It follows from (62) that
H ( x ) = H ( e x ) = lim n → ∞ 2 n f ( e x 2 n ) = lim n → ∞ 2 n f ( e 2 n ⋅ x ) = lim n → ∞ 2 n f ( e 2 n ) ⋅ f ( x ) = e ′ ⋅ f ( x ) = f ( x ) , ∀ x ∈ X (73)
∀ x ∈ X . So the mapping f : X → Y is an isomorphism. □
Theorem 10. Let r < q with q ≤ 1 and θ be positive real numbers, and f : X → Y be a mapping such that
‖ k f ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) − k f ( ∑ i = 1 k x i − y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k f ( x i ) ‖ Y ≤ θ ( ∑ i = 1 k ‖ x i ‖ X r + ∑ i = 1 k ‖ y i ‖ X r + ∑ i = 1 k ‖ z i ‖ X r ) (74)
f ( ∏ i = 1 k x i y i ) = ∏ i = 1 k f ( x i ) ∏ i = 1 k f ( y i ) (75)
for all x i , y i , z i ∈ X , for all i = 1 → k . If f ( t x ) is continuous in t ∈ ℝ each fixed x ∈ X and
l i m n → ∞ 2 n f ( e 2 n ) = e ′ (76)
then the mapping f : X → Y is an isomorphism.
The rest of the proof is similar to the proof of Theorem 9.
Theorem 11. Let r > 1 q with q ≥ 3 and θ be positive real numbers, and f : X → Y be a mapping such that
‖ k f ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) − k f ( ∑ i = 1 k x i − y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k f ( x i ) ‖ Y ≤ θ ⋅ ∏ i = 1 k ‖ x i ‖ X r ⋅ ∏ i = 1 k ‖ y i ‖ X r ⋅ ‖ z 1 ‖ X r k ⋅ ( 1 + ∏ i = 2 k ‖ z i ‖ X r ) (77)
f ( ∏ i = 1 k x i y i ) = ∏ i = 1 k f ( x i ) ∏ i = 1 k f ( y i ) (78)
for all x i , y i , z i ∈ X , for all i = 1 → k . If f ( t x ) is continuous in t ∈ ℝ each fixed x ∈ X and
l i m n → ∞ 2 n f ( e 2 n ) = e ′ (79)
then the mapping f : X → Y is an isomorphism.
Proof. I replace ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x , ⋯ , x , x , ⋯ , x , x , ⋯ ,0 ) in (77), I have
‖ k f ( 2 x ) − 2 k f ( x ) ‖ Y ≤ θ ‖ x ‖ X 3 k r (80)
for all x ∈ X . So
‖ f ( x ) − 2 f ( x 2 ) ‖ Y ≤ 1 k θ 2 3 r k ‖ x ‖ X 3 k r (81)
for all x ∈ X . Hence
‖ 2 l f ( x 2 l ) − 2 m f ( x 2 m ) ‖ Y p ≤ ∑ j = l m − 1 ‖ 2 j f ( x 2 j ) − 2 j + 1 f ( x 2 j + 1 ) ‖ Y p ≤ 1 k p θ p 2 3 k r p ∑ j = l + 1 m − 1 2 p j 2 p k r j ‖ x ‖ X 3 p k r (82)
for all nonnegative integers m and l with m > l and for all x ∈ X . It follows (82) that the sequence { 2 n f ( x 2 n ) } is a Cauchy sequence for all x ∈ X . Since Y is complete, the sequence { 2 n f ( x 2 n ) } converges. So one can define the mapping H : X → Y by
H ( x ) = l i m n → ∞ 2 n f ( x 2 n ) , (83)
for all x ∈ X .
Moreover, letting l = 0 and passing the limit m → ∞ in (82), I get
‖ f ( x ) − H ( x ) ‖ Y ≤ θ k ( 2 3 p k r − 2 p ) 1 p ‖ x ‖ X 3 k r , ∀ x ∈ X . (84)
The rest of the proof is similar to the proof of Theorems 7 and 9. □
Theorem 12. Let r > 1 q with q ≥ 3 and θ be positive real numbers, and f : X → Y be a mapping such that
‖ k f ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) − k f ( ∑ i = 1 k x i − y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k f ( x i ) ‖ Y ≤ θ ⋅ ∏ i = 1 k ‖ x i ‖ X r ⋅ ∏ i = 1 k ‖ y i ‖ X r ⋅ ‖ z 1 ‖ X r k ⋅ ( 1 + ∏ i = 2 k ‖ z i ‖ X r ) (85)
f ( ∏ i = 1 k x i y i ) = ∏ i = 1 k f ( x i ) ∏ i = 1 k f ( y i ) (86)
for all x i , y i , z i ∈ X , for all i = 1 → k . If f ( t x ) is continuous in t ∈ ℝ each fixed x ∈ X and
l i m n → ∞ 2 n f ( e 2 n ) = e ′ (87)
then the mapping f : X → Y is an isomorphism.
The rest of the proof is similar to the proof of Theorems 8 and 9.
Now I construct derivatives for (1.3). Note that: (1.1) and (1.2) are also built exactly the same.
Here I assume that, A is a p-Banach algebras with norm ‖ ⋅ ‖ A Let K be the modulus of concavity of ‖ ⋅ ‖ A . Under this setting, I can show that the mappings satisfying (1.3) is generalized derivation.
A generalized derivations β : X → X is linear and fulfills the generalized Leibniz rue
β ( ∏ i = 1 k x i y i z i ) = β ( ∏ i = 1 k x i y i ) ∏ i = 1 k z i − ∏ i = 1 k x i β ( ∏ i = 1 k y i ) ∏ i = 1 k z i + ∏ i = 1 k x i β ( ∏ i = 1 k y i z i ) (88)
for all x 1 , x 2 , ⋯ , x 3 k ∈ X .
Theorem 13. Let r > q with q ≥ 3 and θ be positive real numbers, and f : X → X be a mapping such that
‖ 2 k f ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k f ( x i ) − ∑ i = 1 k f ( y i ) − 2 k ∑ i = 1 k f ( z i ) ‖ Y ≤ θ ( ∑ i = 1 k ‖ x i ‖ X r + ∑ i = 1 k ‖ y i ‖ X r + ∑ i = 1 k ‖ z i ‖ X r ) (89)
‖ β ( ∏ i = 1 k x i y i z i ) − β ( ∏ i = 1 k x i y i ) ∏ i = 1 k z i + ∏ i = 1 k x i β ( ∏ i = 1 k y i ) ∏ i = 1 k z i − ∏ i = 1 k x i β ( ∏ i = 1 k y i z i ) ‖ X ≤ θ ( ∑ i = 1 k ‖ x i ‖ X r + ∑ i = 1 k ‖ y i ‖ X r + ∑ i = 1 k ‖ z i ‖ X r ) (90)
for all x i , y i , z i ∈ X , for all i = 1 → k . If f ( t x ) is continuous in t ∈ ℝ each fixed x ∈ X , then there exists a unique generalized derivation β : X → X such that
‖ f ( x ) − β ( x ) ‖ X ≤ ( 1 + 1 2 k ) θ ( 2 p r − 2 p ) 1 p ‖ x ‖ X r , ∀ x ∈ X . (91)
Proof. I replace ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x , ⋯ , x , x , ⋯ , x , x , ⋯ ,0 ) in (89), I have
‖ 2 k f ( 2 x ) − 4 k f ( x ) ‖ X ≤ ( 2 k + 1 ) θ ‖ x ‖ X r (92)
for all x ∈ X . So
‖ f ( x ) − 2 f ( x 2 ) ‖ X ≤ ( 1 + 1 2 k ) ⋅ θ 2 r ‖ x ‖ X r
for all x ∈ X . Since X is a p-Banach algebra,
‖ 2 l f ( x 2 l ) − 2 m f ( x 2 m ) ‖ Y p ≤ ∑ j = l m − 1 ‖ 2 j f ( x 2 j ) − 2 j + 1 f ( x 2 j + 1 ) ‖ Y p ≤ ( 1 + 1 2 k ) p ⋅ θ p 2 p r ∑ j = l + 1 m − 1 2 p j 2 p r j ‖ x ‖ X p r (93)
for all nonnegative integers m and l with m > l and for all x ∈ X . It follows (93) that the sequence { 2 n f ( x 2 n ) } is a Cauchy sequence for all x ∈ X . Since X is complete, the sequence { 2 n f ( x 2 n ) } converges. So one can define the mapping β : X → X by
β ( x ) = l i m n → ∞ 2 n f ( x 2 n )
for all x ∈ X . Moreover, letting l = 0 and passing the limit m → ∞ in (93), I get (91). It follows from (89) that
‖ 2 k β ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k β ( x i ) − ∑ i = 1 k β ( y i ) − 2 k ∑ i = 1 k β ( z i ) ‖ X = l i m n → ∞ 2 n ‖ 2 k f ( ∑ i = 1 k x i + y i 2 n + 1 k + ∑ i = 1 k z i 2 n ) − ∑ i = 1 k f ( x i 2 n ) − ∑ i = 1 k f ( y i 2 n ) − 2 k ∑ i = 1 k f ( z i 2 n ) ‖ X ≤ l i m n → ∞ 2 n 2 n r θ ( ∑ i = 1 k ‖ x i ‖ X r + ∑ i = 1 k ‖ y i ‖ X r + ∑ i = 1 k ‖ z i ‖ X r ) = 0
for all x i , y i , z i ∈ X , for all i = 1 → k .
So
2 k β ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k β ( x i ) − ∑ i = 1 k β ( y i ) − 2 k ∑ i = 1 k β ( z i ) = 0
for all x i , y i , z i ∈ X , for all i = 1 → k . By lemma 5 (see [
It follows from (90) that
‖ β ( ∏ i = 1 k x i y i z i ) − β ( ∏ i = 1 k x i y i ) ∏ i = 1 k z i + ∏ i = 1 k x i β ( ∏ i = 1 k y i ) ∏ i = 1 k z i − ∏ i = 1 k x i β ( ∏ i = 1 k y i z i ) ‖ X = l i m n → ∞ 2 3 n k ‖ f ( ∏ i = 1 k x i y i z i 2 3 n k ) − f ( ∏ i = 1 k x i y i 2 2 n k ) ∏ i = 1 k z i 2 n k + ∏ i = 1 k x i 2 n k f ( ∏ i = 1 k y i 2 n k ) ∏ i = 1 k z i 2 n k − ∏ i = 1 k x i 2 n k f ( ∏ i = 1 k x k + i z i 2 2 n k ) ‖ X ≤ l i m n → ∞ θ 2 3 n k 2 n k r ( ∑ i = 1 k ‖ x i ‖ X r + ∑ i = 1 k ‖ y i ‖ X r + ∑ i = 1 k ‖ z i ‖ X r ) = 0 (94)
for all x i , y i , z i ∈ X , for all i = 1 → k . So
β ( ∏ i = 1 k x i y i z i ) = β ( ∏ i = 1 k x i y i ) ∏ i = 1 k z i − ∏ i = 1 k x i β ( ∏ i = 1 k y i ) ∏ i = 1 k z i + ∏ i = 1 k x i β ( ∏ i = 1 k y i z i ) (95)
for all x i , y i , z i ∈ X , for all i = 1 → k .
Now I prove the uniqueness of β . Assume that β 1 : X → X is a Cauchy-Jensen additive mapping satisfying (91). Then I have
‖ β ( x ) − β 1 ( x ) ‖ X = 2 n ‖ β ( 1 2 n x ) + β 1 ( 1 2 n x ) ‖ X ≤ 2 n K ( ‖ β ( 1 2 n x ) − f ( 1 2 n x ) ‖ X + ‖ f ( 1 2 n x ) + β 1 ( 1 2 n x ) ‖ X ) ≤ 2 ( 1 + 1 2 k ) K θ ( 2 p r − 2 p ) 1 p ⋅ 2 n 2 n r ‖ x ‖ X r
which tends to zero as n → ∞ for all x ∈ X . So I can conclude that β ( x ) = β 1 ( x ) for all x ∈ X . This proves the uniqueness of β . Thus the mapping β 1 : X → X is a unique generalized derivation satisfying (91). □
Theorem 14. Let r < q with q ≤ 1 and θ be positive real numbers, and f : X → X be a mapping such that
‖ 2 k f ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k f ( x i ) − ∑ i = 1 k f ( y i ) − 2 k ∑ i = 1 k f ( z i ) ‖ Y ≤ θ ( ∑ i = 1 k ‖ x i ‖ X r + ∑ i = 1 k ‖ y i ‖ X r + ∑ i = 1 k ‖ z i ‖ X r ) (96)
‖ β ( ∏ i = 1 k x i y i z i ) − β ( ∏ i = 1 k x i y i ) ∏ i = 1 k z i + ∏ i = 1 k x i β ( ∏ i = 1 k y i ) ∏ i = 1 k z i − ∏ i = 1 k x i β ( ∏ i = 1 k y i z i ) ‖ X ≤ θ ( ∑ i = 1 k ‖ x i ‖ X r + ∑ i = 1 k ‖ y i ‖ X r + ∑ i = 1 k ‖ z i ‖ X r ) (97)
for all x i , y i , z i ∈ X , for all i = 1 → k . If f ( t x ) is continuous in t ∈ ℝ each fixed x ∈ X , then there exists a unique generalized derivation β : X → X such that
‖ f ( x ) − β ( x ) ‖ X ≤ ( 1 + 1 2 k ) θ ( 2 p − 2 p r ) 1 p ‖ x ‖ X r , ∀ x ∈ X . (98)
The rest of the proof is similar to the proof of Theorem 13.
Theorem 15. Let r > q with q > 1 and θ be positive real numbers, and f : X → X be a mapping such that
‖ 2 k f ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k f ( x i ) − ∑ i = 1 k f ( y i ) − 2 k ∑ i = 1 k f ( z i ) ‖ Y ≤ θ ⋅ ∏ i = 1 k ‖ x i ‖ X r ⋅ ∏ i = 1 k ‖ y i ‖ X r ⋅ ‖ z 1 ‖ X r k ⋅ ( 1 + ∏ i = 2 k ‖ z i ‖ X r ) (99)
‖ β ( ∏ i = 1 k x i y i z i ) − β ( ∏ i = 1 k x i y i ) ∏ i = 1 k z i + ∏ i = 1 k x i β ( ∏ i = 1 k y i ) ∏ i = 1 k z i − ∏ i = 1 k x i β ( ∏ i = 1 k y i z i ) ‖ X ≤ θ ⋅ ∏ i = 1 k ‖ x i ‖ X r ⋅ ∏ i = 1 k ‖ y i ‖ X r ⋅ ‖ z 1 ‖ X r k ⋅ ( 1 + ∏ i = 2 k ‖ z i ‖ X r ) (100)
for all x i , y i , z i ∈ X , for all i = 1 → k . If f ( t x ) is continuous in t ∈ ℝ each fixed x ∈ X , then there exists a unique generalized derivation β : X → X such that
‖ f ( x ) − β ( x ) ‖ X ≤ ( 2 + 1 k k r ) θ ( 2 3 p r − 2 p ) 1 p ‖ x ‖ X k r , ∀ x ∈ X . (101)
Proof. I replace ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x , ⋯ , x , x , ⋯ , x , x , ⋯ ,0 ) in (99), I have
‖ 2 k f ( 2 x ) − 4 k f ( x ) ‖ X ≤ θ ‖ x ‖ X 3 k r (102)
for all x ∈ X . So
‖ f ( x ) − 2 f ( x 2 ) ‖ X ≤ 1 2 k ⋅ θ 2 3 k r ‖ x ‖ X 3 k r
for all x ∈ X . Since X is a p-Banach algebra,
‖ 2 l f ( x 2 l ) − 2 m f ( x 2 m ) ‖ Y p ≤ ∑ j = l m − 1 ‖ 2 j f ( x 2 j ) − 2 j + 1 f ( x 2 j + 1 ) ‖ Y p ≤ 1 ( 2 k ) p ⋅ θ p 2 3 p k r ∑ j = l + 1 m − 1 2 p j 2 3 k p r j ‖ x ‖ X 3 p k r (103)
for all nonnegative integers m and l with m > l and for all x ∈ X . It follows (103) that the sequence { 2 n f ( x 2 n ) } is a Cauchy sequence for all x ∈ X . Since X is complete, the sequence { 2 n f ( x 2 n ) } converges. So one can define the mapping β : X → X by
β ( x ) = l i m n → ∞ 2 n f ( x 2 n )
for all x ∈ X . Moreover, letting l = 0 and passing the limit m → ∞ in (103), I get (101). It follows from (99) that
‖ 2 k β ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k β ( x i ) − ∑ i = 1 k β ( y i ) − 2 k ∑ i = 1 k β ( z i ) ‖ X = l i m n → ∞ 2 n ‖ 2 k f ( ∑ i = 1 k x i + y i 2 n + 1 k + ∑ i = 1 k z i 2 n ) − ∑ i = 1 k f ( x i 2 n ) − ∑ i = 1 k f ( y i 2 n ) − 2 k ∑ i = 1 k f ( z i 2 n ) ‖ X ≤ l i m n → ∞ θ ⋅ ∏ i = 1 k ‖ x i ‖ X r ⋅ ∏ i = 1 k ‖ y i ‖ X r ⋅ ‖ z 1 ‖ X r k ⋅ ( 1 + ∏ i = 2 k ‖ z i ‖ X r ) = 0 (104)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X .
So
2 k β ( ∑ i = 1 k x i + y i 2 k + ∑ i = 1 k z i ) − ∑ i = 1 k β ( x i ) − ∑ i = 1 k β ( y i ) − 2 k ∑ i = 1 k β ( z i ) = 0 (105)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X . By lemma 5 (see [
It follows from (100) that
‖ β ( ∏ i = 1 k x i y i z i ) − β ( ∏ i = 1 k x i y i ) ∏ i = 1 k z i + ∏ i = 1 k x i β ( ∏ i = 1 k y i ) ∏ i = 1 k z i − ∏ i = 1 k x i β ( ∏ i = 1 k y i z i ) ‖ X = lim n → ∞ 2 3 n k ‖ f ( ∏ i = 1 k x i x k + i x 2 k + i 2 3 n k ) − f ( ∏ i = 1 k x i x k + i 2 2 n k ) ∏ i = 1 k z i 2 n k + ∏ i = 1 k x i 2 n k f ( ∏ i = 1 k y i 2 n k ) ∏ i = 1 k z i 2 n k − ∏ i = 1 k x i 2 n k f ( ∏ i = 1 k x k + i x 2 k + i 2 2 n k ) ‖ ∥ X ≤ l i m n → ∞ 3 k θ 2 3 n k 2 n k r ( ∑ i = 1 k ‖ x i ‖ X r + ∑ i = 1 k ‖ y i ‖ X r + ∑ i = 1 k ‖ z i ‖ X r ) = 0 (106)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X . So
β ( ∏ i = 1 k x i y i z i ) = β ( ∏ i = 1 k x i y i ) ∏ i = 1 k z i − ∏ i = 1 k x i β ( ∏ i = 1 k y i ) ∏ i = 1 k z i + ∏ i = 1 k x i β ( ∏ i = 1 k y i z i ) (107)
for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ∈ X .
Now I prove the uniqueness of β . Assume that β 1 : X → X is a Cauchy-Jensen additive mapping satisfying (101). Then I have
‖ β ( x ) − β 1 ( x ) ‖ X = 2 n ‖ β ( 1 2 n x ) + β 1 ( 1 2 n x ) ‖ X ≤ 2 n K ( ‖ β ( 1 2 n x ) − f ( 1 2 n x ) ‖ X + ‖ f ( 1 2 n x ) + β 1 ( 1 2 n x ) ‖ X ) ≤ 2 ( 2 k + 1 k r − 1 ) K θ ( 2 p r − 2 p ) 1 p ⋅ 2 n 2 n r ‖ x ‖ X r (108)
which tends to zero as n → ∞ for all x ∈ X . So I can conclude that β ( x ) = β 1 ( x ) for all x ∈ X . This proves the uniqueness of β . Thus the mapping β 1 : X → X is a unique generalized derivation satisfying (101). □
In this paper, I construct extensions of homomorphisms, isomorphisms, and derivatives based on Banach algebra. The fundamental contribution here is the development of a general Cauchy-Jensen equation, which serves as the cornerstone for establishing mathematical links across various research areas in mathematics, without any restrictions on generality.
The author declares no conflicts of interest.
An, L.V. (2023) Extension of Homomorphisms-Isomorphisms and Derivatives on Quasi-Banach Algebra Based on the General Additive Cauchy-Jensen Equation. Open Access Library Journal, 10: e10095. https://doi.org/10.4236/oalib.1110095