Let X and Y be a normed spaces on the same field K , and f : X → Y . I use the notation ‖   ⋅   ‖ for all the norm on both X and Y . In this paper, I investisgate some additive-quadraic λ -functional inequality in ( α 1 ; α 2 ) -homogeneous F-spaces.

In fact, when X is a α 1 -homogeneous F-spaces and that Y is a α 2 -homogeneous F-spaces, I solve and prove the Hyers-Ulam-Rassias type stability of two forllowing additive-quadratic g ( λ ) -functional inequality.

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ g ( λ ) ( 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y (1)

and when I change the role of the function inequality (1), I continue to prove the following function inequality.

‖ 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ g ( λ ) ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y (2)

H = { h : ℂ \ { 0 } → ℂ , h ( λ ) = λ } (3)

where g ∈ H .

The stability problem of functional equations originated from a question of Ulam [1] concerning the stability of group homomorphisms.

The functional equation

f ( x + y ) = f ( x ) + f ( y ) (4)

is called the Cauchy equation.

In particular, every solution of the Cauchy equation is said to be an additive mapping. Hyers [2] gave a first affirmative partial answer to the question of Ulam for Banach spaces. Hyers’ Theorem was generalized by Aoki [3] for additive mappings and by Rassias [4] for linear mappings by considering an unbounded Cauchy difference. A generalization of the Rassias theorem was obtained by Găvruta [5] by replacing the unbounded Cauchy difference by a general control function in the spirit of Rassias’ approach.

The functional equation

f ( x + y ) + f ( x − y ) = 2 f ( x ) + 2 f ( y ) (5)

is called the quadratic functional equation. In particular, every solution of the quadratic functional equation is said to be a quadratic mapping. The stability of quadratic functional equation was proved by Skof [6] for mappings f : E 1 → E 2 , where E 1 is a normed space and E 2 is a Banach space. Cholewa [7] noticed that the theorem of Skof is still true if the relevant domain E₁ is replaced by an Abelian group.

Recently, the I has studied the additive function inequalities or quadratic function inequalities of mathematicians around the world see [1] - [24] , on spaces as complex Banach spaces, non-Archimedan Banach spaces or homogeneous F-space let me give two general additive-quadratic functional inequalities and show their solutions exist on ( α 1 , α 2 ) -homogeneous F-space.

In this article, I successfully built quadratic functional inequalities with the number of variables more than 3 on F-homogeneous space and I showed their solutions. This is a great step forward in the field of functional equations. Application to solve problems in many spaces with no limit on the number of variables.

The paper is organized as followns: In section preliminarier I remind a basic property such as I only redefine the solution definition of the equation of the additive function and F*-space.

Section 3: is devoted to prove the Hyers-Ulam stability of the addive g ( λ ) -functional inequalities (1) when when X is a α 1 -homogeneous F-spaces and that Y is a α 2 -homogeneous F-spaces.

Section 4: is devoted to prove the Hyers-Ulam stability of the addive g ( λ ) -functional inequalities (2) when when X is a α 1 -homogeneous F-spaces and that Y is a α 2 -homogeneous F-spaces.

Section 5: is devoted to prove the Hyers-Ulam stability of the quadratic g ( λ ) -functional inequalities (1) when when X is a α 1 -homogeneous F-spaces and that Y is a α 2 -homogeneous F-spaces.

Section 6: is devoted to prove the Hyers-Ulam stability of the quadratic g ( λ ) -functional inequalities (2) when when X is a α 1 -homogeneous F-spaces and that Y is a α 2 -homogeneous F-spaces.

Now, I first study the solutions of (1). Note that for these inequalities, when X is a α 1 -homogeneous F-spaces and that Y is a α 2 -homogeneous F-spaces. Under this setting, I can show that the mapping satisfying (1) is additive. These results are give in the following.

Where: α 1 , α 1 ∈ ℝ + and α 1 , α 1 ≤ 1 .

Lemma 1. Let f : X → Y be an odd mapping satilies

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ λ ( 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y (8)

for all x j , y j , z j ∈ X for j = 1 → n , if and only if f : X → Y is additive.

Proof. Assume that f : X → Y satisfies (8).

We replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (8), we have

‖ ( 4 k − 2 ) f ( 0 ) ‖ ≤ | g ( λ ) | α 2 ‖ 2 k f ( 0 ) ‖ ≤ 0

therefore

So f ( 0 ) = 0 .

Next replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( k x , ⋯ , k x , k x , ⋯ , k x , x , ⋯ , x ) in (8), we have

Thus

‖ f ( 2 k x ) − 2 k f ( x ) ‖ ≤ 0

f ( x 2 k ) = 1 2 k f ( x ) (9)

for all x ∈ X .

From (8) and (9) I infer that

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ g ( λ ) ( 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y = | g ( λ ) | α 2 ‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y (10)

for all x j , y j , z j ∈ X for j = 1 → n , and so

f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) = 2 ∑ j = 1 k     f ( x j + y j 2 k ) (11)

for all x j , y j , z j ∈ X for j = 1 → n .

Next we replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( k x , ⋯ , k x , k x , ⋯ , k x , z , ⋯ , z ) in (11), we have

f ( k x + k z ) + f ( k x − k z ) = 2 k f ( x ) (12)

for all x , z ∈ X .

Now letting p = k x + k z , q = k x − k z when that in (12), we get

f ( p ) + f ( q ) = 2 k f ( p + q 2 k ) = 2 k ⋅ 1 2 k f ( p + q ) = f ( p + q ) (13)

for all p , q ∈ X . So f is an additive mapping. as we expected. The couverse is obviously true.

Corollary 1. Let f : X → Y be an even mapping satilies

f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j )   − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) = g ( λ ) ( 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) (14)

for all x j , y j , z j ∈ X for j = 1 → n , if and only if f : X → Y is additive.

Note! The functional equation (14) is called an additive λ-functional equation.

Theorem 2. Assume for r > α 2 α 1 , θ be nonngative real number, and Suppose f : X → Y be a mapping such that

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ g ( λ ) ( 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y   + θ ( ∑ j = 1 k ‖ x j ‖ r + ∑ j = 1 k ‖ y j ‖ r + ∑ j = 1 k ‖ z j ‖ r ) (15)

for all x j , y j , z j ∈ X for all j = 1 → n . Then there exists a unique additive mapping ϕ : X → Y such that

‖ f ( x ) − ϕ ( x ) ‖ ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 1 r − ( 2 k ) α 2 θ ‖ x ‖ r . (16)

for all x ∈ X .

Proof. Assume that f : X → Y satisfies (15).

We replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (15), we have

‖ ( 4 k − 2 ) f ( 0 ) ‖ ≤ ‖ 2 k g ( λ ) f ( 0 ) ‖

therefore

( | 4 k − 2 | α 2 − | 2 k g ( λ ) | α 2 ) ‖ f ( 0 ) ‖

So f ( 0 ) = 0 .

Next replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( k x , ⋯ , k x , k x , ⋯ , k x , x , ⋯ , x ) in (15) we have

‖ f ( 2 k x ) − 2 k f ( x ) ‖ ≤ ( 2 k α 1 r + 1 + 1 ) θ ‖ x ‖ r (17)

for all x ∈ X . Thus

‖ f ( x ) − 2 k f ( x 2 k ) ‖ ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 1 r θ ‖ x ‖ r (18)

for all x ∈ X .

‖ ( 2 k ) l f ( x ( 2 k ) l ) − ( 2 k ) m f ( x ( 2 k ) m ) ‖ ≤ ∑ j = 1 m − 1 ‖ ( 2 k ) j f ( x ( 2 k ) j ) − ( 2 k ) j + 1 f ( x ( 2 k ) j + 1 ) ‖ ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 1 r θ ∑ j = 1 m − 1 ( 2 k ) α 2 j ( 2 k ) α 1 r j ‖ x ‖ r (19)

for all nonnegative integers p , l with p > l and all x ∈ X . It follows from (19) that the sequence { ( 2 k ) n f ( x ( 2 k ) n ) } is a cauchy sequence for all x ∈ X . Since Y is complete, the sequence { ( 2 k ) n f ( x ( 2 k ) n ) } coverges.

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

ϕ ( x ) : = lim n → ∞ ( 2 k ) n f ( x ( 2 k ) n )

for all x ∈ X . Moreover, letting l = 0 and passing the limit m → ∞ in (19), we get (16).

Form f : X → Y is even, the mapping ϕ : X → Y is even.

It follows from (15) that

‖ ϕ ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + ϕ ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     ϕ ( x j + y j 2 k ) − ∑ j = 1 k     ϕ ( z j ) − ∑ j = 1 k     ϕ ( − z j ) ‖ = lim n → ∞ ( 2 k ) α 2 n ‖ f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j 2 k + 1 ( 2 k ) n ∑ j = 1 k     z j )   + f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j 2 k − 1 ( 2 k ) n ∑ j = 1 k     z j ) − ∑ j = 1 k     f ( 1 ( 2 k ) n x j + y j 2 k ) − ∑ j = 1 k     f ( 1 ( 2 k ) n z j ) − ∑ j = 1 k     f ( − 1 ( 2 k ) n z j ) ‖

≤ lim n → ∞ ( 2 k ) α 2 n | g ( λ ) | α 2 ‖ 2 k f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 ( 2 k ) n + 1 ∑ j = 1 k     z j )   + 2 k f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 ( 2 k ) n + 1 ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( 1 ( 2 k ) n x j + y j 2 k ) − ∑ j = 1 k     f ( − 1 ( 2 k ) n x j + y j 2 k ) − ∑ j = 1 k     f ( 1 ( 2 k ) n z j ) − ∑ j = 1 k     f ( − 1 ( 2 k ) n z j ) ‖   + lim n → ∞ ( 2 k ) α 2 n ( 2 k ) α 1 n r θ ( ∑ j = 1 k ‖ x j ‖ r + ∑ j = 1 k ‖ y j ‖ r + ∑ j = 1 k ‖ z j ‖ r )

= | g ( λ ) | α 2 ‖ 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y (20)

for all   x j , y j , z j ∈ X for all j = 1 → n .

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ g ( λ ) ( 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y

for all x j , y j , z j ∈ X for j = 1 → n , So by Lemma 1 it follows that the mapping ϕ : X → Y is additive. Now we need to prove uniqueness, Suppose ϕ ′ : X → Y is also an additive mapping that satisfies (16). Then we have

‖ ϕ ( x ) − ϕ ′ ( x ) ‖ = ( 2 k ) α 2 n ‖ ϕ ( x ( 2 k ) n ) − ϕ ′ ( x ( 2 k ) n ) ‖ ≤ ( 2 k ) α 2 n ( ‖ ϕ ( x ( 2 k ) n ) − f ( x ( 2 k ) n ) ‖ + ‖ ϕ ′ ( x ( 2 k ) n ) − f ( x ( 2 k ) n ) ‖ ) ≤ 2 ⋅ ( 2 k ) α 2 n ⋅ ( 2 k α 1 r + 1 + 1 ) ( 2 k ) α 1 n r ( ( 2 k ) α 1 r − ( 2 k ) α 2 ) θ ‖ x ‖ r (21)

which tends to zero as n → ∞ for all x ∈ X . So we can conclude that ϕ ( x ) = ϕ ′ ( x ) for all x ∈ X . This proves thus the mapping ϕ : X → Y is a unique mapping satisfying (16) as we expected.

Theorem 3. Assume for r < α 2 α 1 , θ be nonngative real number, and Suppose f : X → Y be a mapping such that

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ g ( λ ) ( 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y   + θ ( ∑ j = 1 k ‖ x j ‖ r + ∑ j = 1 k ‖ y j ‖ r + ∑ j = 1 k ‖ z j ‖ r ) (22)

for all x j , y j , z j ∈ X for all j = 1 → n . Then there exists a unique additive mapping ϕ : X → Y such that

‖ f ( x ) − ϕ ( x ) ‖ ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 2 − ( 2 k ) α 1 r θ ‖ x ‖ r . (23)

for all x ∈ X .

Proof. Assume that f : X → Y satisfies (22).

We replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (22), we have

‖ ( 4 k − 2 ) f ( 0 ) ‖ ≤ ‖ 2 k g ( λ ) f ( 0 ) ‖

therefore

( | 4 k − 2 | α 2 − | 2 k g ( λ ) | α 2 ) ‖ f ( 0 ) ‖

So f ( 0 ) = 0 .

Next replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( k x , ⋯ , k x , k x , ⋯ , k x , x , ⋯ , x ) in (22) we have

‖ f ( 2 k x ) − 2 k f ( x ) ‖ ≤ ( 2 k α 1 r + 1 + 1 ) θ ‖ x ‖ r (24)

for all x ∈ X . Thus

‖ f ( x ) − 1 2 k f ( 2 k x ) ‖ ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 2 θ ‖ x ‖ r (25)

for all x ∈ X .

‖ 1 ( 2 k ) l f ( ( 2 k ) l x ) − 1 ( 2 k ) m f ( ( 2 k ) m x ) ‖ ≤ ∑ j = 1 m − 1 ‖ 1 ( 2 k ) j f ( ( 2 k ) j x ) − 1 ( 2 k ) j + 1 f ( ( 2 k ) j + 1 x ) ‖ ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 2 θ ∑ j = 1 m − 1 ( 2 k ) α 1 r j ( 2 k ) α 2 j ‖ x ‖ r (26)

for all nonnegative integers p , l with p > l and all x ∈ X . It follows from (26) that the sequence { 1 ( 2 k ) n f ( ( 2 k ) n x ) } is a cauchy sequence for all x ∈ X . Since Y is complete, the sequence { 1 ( 2 k ) n f ( ( 2 k ) n x ) } coverges.

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

ϕ ( x ) : = lim n → ∞ 1 ( 2 k ) n f ( ( 2 k ) n x )

for all x ∈ X . Moreover, letting l = 0 and passing the limit m → ∞ in (26), we get (23).

The rest of the proof is similar to the proof of Theorem 2.

Now, we study the solutions of (2). Note that for these inequalities, when X is a α 1 -homogeneous F-spaces and that Y is a α 2 -homogeneous F-spaces. Under this setting, I can show that the mapping satisfying (2) is additive. These results are give in the following.

Lemma 4. Let f : X → Y be an odd mapping satilies

‖ 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ g ( λ ) ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y (27)

for all x j , y j , z j ∈ X for j = 1 → n , if and only if f : X → Y is additive.

Proof. Assume that f : X → Y satisfies (27).

We replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (27), we have

‖ 2 k f ( 0 ) ‖ ≤ | g ( λ ) | α 2 ‖ ( 4 k − 2 ) f ( 0 ) ‖

So f ( 0 ) = 0 .

Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 2 k x , ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (27), we have

Thus

‖ 4 k f ( x 2 k ) − 2 f ( x ) ‖ ≤ 0

f ( x 2 k ) = 1 2 k f ( x ) (28)

for all x ∈ X .

From (27) and (28) we infer that

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j )

− 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ | g ( λ ) | α 2 ‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y (29)

for all x j , y j , z j ∈ X for j = 1 → n , and so

f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) = 2 ∑ j = 1 k     f ( x j + y j 2 k )

for all x j , y j , z j ∈ X for j = 1 → n , as we expected. The couverse is obviously true.

Corollary 2. Let f : X → Y be an even mapping satilies

2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j )   − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) = g ( λ ) ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) (30)

for all x j , y j , z j ∈ X for j = 1 → n , if and only if f : X → Y is additive.

Note! The functional equation (30) is called an additive λ-functional equation.

Theorem 5. Assume for r > α 2 α 1 , θ be nonngative real number, and Suppose f : X → Y be a mapping such that f ( 0 ) = 0 and

‖ 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ λ ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y   + θ ( ∑ j = 1 k ‖ x j ‖ r + ∑ j = 1 k ‖ y j ‖ r + ∑ j = 1 k ‖ z j ‖ r ) (31)

for all x j , y j , z j ∈ X for all j = 1 → n . Then there exists a unique additive mapping ϕ : X → Y such that

‖ f ( x ) − ϕ ( x ) ‖ ≤ ( 2 k ) α 1 r ( 2 k ) α 1 r − ( 4 k ) α 2 θ ‖ x ‖ r . (32)

for all x ∈ X .

Proof. Assume that f : X → Y satisfies (38).

We replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (38), we have

‖ 2 f ( 0 ) ‖ ≤ | λ | α 2 ‖ ( 4 k − 2 ) f ( 0 ) ‖

therefore

( | 4 k − 2 | α 2 − | 2 λ | α 2 ) ‖ f ( 0 ) ‖ ≤ 0

So f ( 0 ) = 0 .

Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 2 k x , ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (38) we have

‖ 4 f ( x 2 k ) − 1 k f ( x ) ‖ Y ≤ ( 2 k ) α 1 r θ ‖ x ‖ r (33)

for all x ∈ X . Thus

‖ 4 k f ( x 2 k ) − f ( x ) ‖ ≤ ( 2 k ) α 1 r k α 2 θ ‖ x ‖ r (34)

for all x ∈ X .

‖ ( 4 k ) l f ( x ( 2 k ) l ) − ( 4 k ) m f ( x ( 2 k ) m ) ‖ ≤ ∑ j = 1 m − 1 ‖ ( 4 k ) j f ( x ( 2 k ) j ) − ( 4 k ) j + 1 f ( x ( 2 k ) j + 1 ) ‖ ≤ ( 2 k ) α 1 r k α 2 θ ∑ j = 1 m − 1 ( 4 k ) α 2 j ( 2 k ) α 1 r j ‖ x ‖ r (35)

for all nonnegative integers p , l with p > l and all x ∈ X . It follows from (35) that the sequence { ( 4 k ) n f ( x ( 2 k ) n ) } is a cauchy sequence for all x ∈ X . Since Y is complete, the sequence { ( 4 k ) n f ( x ( 2 k ) n ) } coverges.

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

ϕ ( x ) : = lim n → ∞ ( 4 k ) n f ( x ( 2 k ) n )

for all x ∈ X . Moreover, letting l = 0 and passing the limit m → ∞ in (35), we get (39). Form f : X → Y is even, the mapping

ϕ : X → Y

is even. It follows from (38) that

‖ 2 ϕ ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 ϕ ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 2 k ∑ j = 1 k     ϕ ( x j + y j 2 k ) + 1 2 k ∑ j = 1 k     ϕ ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k     ϕ ( z j ) − 1 2 k ∑ j = 1 k     ϕ ( − z j ) ‖ = lim n → ∞ ( 4 k ) α 2 n ‖ 2 f ( ∑ j = 1 k x j + y j ( 2 k ) n + 2 + 1 ( 2 k ) n + 1 ∑ j = 1 k     z j )   + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) n + 2 − 1 ( 2 k ) n + 1 ∑ j = 1 k     z j ) − 3 2 k ∑ j = 1 k     f ( x j + y j ( 2 k ) n + 1 ) + 1 2 k ∑ j = 1 k     f ( − x j + y j ( 2 k ) n + 1 ) − 1 2 k ∑ j = 1 k     f ( z j ( 2 k ) n ) − 1 2 k ∑ j = 1 k     f ( − z j ( 2 k ) n ) ‖

≤ lim n → ∞ ( 4 k ) α 2 n | g ( λ ) | α 2 ‖ 2 f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j 2 k + 1 ( 2 k ) n ∑ j = 1 k     z j )   + 2 f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j 2 k − 1 ( 2 k ) n ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( 1 ( 2 k ) n x j + y j 2 k ) − 1 2 k ∑ j = 1 k     f ( 1 ( 2 k ) n z j ) − 1 2 k ∑ j = 1 k     f ( − 1 ( 2 k ) n z j ) ‖ + lim n → ∞ ( 4 k ) α 2 n ( 2 k ) α 1 n r θ ( ∑ j = 1 k ‖ x j ‖ r + ∑ j = 1 k ‖ y j ‖ r + ∑ j = 1 k ‖ z j ‖ r )

= ‖ ϕ ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + ϕ ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     ϕ ( x j + y j 2 k ) − ∑ j = 1 k     ϕ ( z j ) − ∑ j = 1 k     ϕ ( − z j ) ‖ Y (36)

for all x j , y j , z j ∈ X for all j = 1 → n .

‖ 2 ϕ ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 ϕ ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 2 k ∑ j = 1 k     ϕ ( x j + y j 2 k ) + 1 2 k ∑ j = 1 k     ϕ ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k     ϕ ( z j ) − 1 2 k ∑ j = 1 k     ϕ ( − z j ) ‖ Y ≤ ‖ g ( λ ) ( ϕ ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + ϕ ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     ϕ ( x j + y j 2 k ) − ∑ j = 1 k     ϕ ( z j ) − ∑ j = 1 k     ϕ ( − z j ) ) ‖ Y

for all x j , y j , z j ∈ X for j = 1 → n , So by Lemma 4.1 it follows that the mapping ϕ : X → Y is additive. Now we need to prove uniqueness, Suppose ϕ ′ : X → Y is also a quadratic mapping that satisfies (39). Then we have

‖ ϕ ( x ) − ϕ ′ ( x ) ‖ = ( 4 k ) α 2 n ‖ ϕ ( x ( 2 k ) n ) − ϕ ′ ( x ( 2 k ) n ) ‖ ≤ ( 4 k ) α 2 n ( ‖ ϕ ( x ( 2 k ) n ) − f ( x ( 2 k ) n ) ‖ + ‖ ϕ ′ ( x ( 2 k ) n ) − f ( x ( 2 k ) n ) ‖ ) ≤ 2 ⋅ ( 4 k ) α 2 n ⋅ ( 2 k ) α 1 r ( 2 k ) α 1 n r ( ( 2 k ) α 1 r − ( 4 k ) α 2 ) θ ‖ x ‖ r (37)

which tends to zero as n → ∞ for all x ∈ X . So we can conclude that ϕ ( x ) = ϕ ′ ( x ) for all x ∈ X . This proves thus the mapping ϕ : X → Y is a unique mapping satisfying (39) as we expected.

Theorem 6. Assume for r < 2 α 2 α 1 , θ be nonngative real number, f ( 0 ) = 0 and Suppose f : X → Y be a mapping such that

‖ 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 2 k ∑ j = 1 k     f ( x j + y j 2 k ) + 1 2 k ∑ j = 1 k     f ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k     f ( z j ) − 1 2 k ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ g ( λ ) ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y   + θ ( ∑ j = 1 k ‖ x j ‖ r + ∑ j = 1 k ‖ y j ‖ r + ∑ j = 1 k ‖ z j ‖ r ) (38)

for all x j , y j , z j ∈ X for all j = 1 → n . Then there exists a unique addtive mapping ϕ : X → Y such that

‖ f ( x ) − ϕ ( x ) ‖ ≤ ( 2 k ) α 1 r ( 4 k ) α 2 − ( 2 k ) α 1 r θ ‖ x ‖ r . (39)

for all x ∈ X .

The proof is similar to theorem 5.

Now, we first study the solutions of (1). Note that for these inequalities, when X is a α 1 -homogeneous F-spaces and that Y is a α 2 -homogeneous F-spaces. Under this setting, we can show that the mapping satisfying (1) is quadratic. These results are give in the following.

Lemma 7. Let f : X → Y be an even mapping satilies

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ g ( λ ) ( 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y (40)

for all x j , y j , z j ∈ X for j = 1 → n , if and only if f : X → Y is quadratic.

Proof. Assume that f : X → Y satisfies (40).

We replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (40), we have

‖ ( 4 k − 2 ) f ( 0 ) ‖ ≤ | λ | α 2 ‖ 2 k f ( 0 ) ‖ ≤ 0

therefore

So f ( 0 ) = 0 .

Next replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( k x , ⋯ , k x , k x , ⋯ , k x , x , ⋯ , x ) in (40), we have

Thus

‖ f ( 2 k x ) − 2 k f ( x ) ‖ ≤ 0

f ( x 2 k ) = 1 2 k f ( x ) (41)

for all x ∈ X .

From (40) and (41) we infer that

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ λ ( 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y = | λ | α 2 ‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y (42)

for all x j , y j , z j ∈ X for j = 1 → n , and so

f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) = 2 ∑ j = 1 k     f ( x j + y j 2 k ) + 2 ∑ j = 1 k     f ( z j ) (43)

for all x j , y j , z j ∈ X for j = 1 → n .

As we expected. The couverse is obviously true.

Corollary 3. Let f : X → Y be an even mapping satilies

f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j )   − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) = g ( λ ) ( 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) (44)

for all x j , y j , z j ∈ X for j = 1 → n , if and only if f : X → Y is quadratic.

Note! The functional equation (44) is called an quadratic g ( λ ) -functional equation.

Theorem 8. Assume for r > 2 α 2 α 1 , θ be nonngative real number, and Suppose f : X → Y be an even mapping such that

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ g ( λ ) ( 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y   + θ ( ∑ j = 1 k ‖ x j ‖ r + ∑ j = 1 k ‖ y j ‖ r + ∑ j = 1 k ‖ z j ‖ r ) (45)

for all x j , y j , z j ∈ X for all j = 1 → n . Then there exists a unique quadratic mapping ϕ : X → Y such that

‖ f ( x ) − ϕ ( x ) ‖ ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 1 r − ( 2 k ) α 2 θ ‖ x ‖ r (46)

for all x ∈ X .

Proof. Assume that f : X → Y satisfies (45).

We replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (45), we have

‖ ( 4 k − 2 ) f ( 0 ) ‖ ≤ ‖ 2 k g ( λ ) f ( 0 ) ‖

therefore

( | 4 k − 2 | α 2 − | 2 k g ( λ ) | α 2 ) ‖ f ( 0 ) ‖

So f ( 0 ) = 0 .

Next replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( k x , ⋯ , k x , k x , ⋯ , k x , x , ⋯ , x ) in (45) we have

‖ f ( 2 k x ) − 2 k f ( x ) ‖ ≤ ( 2 k α 1 r + 1 + 1 ) θ ‖ x ‖ r (47)

for all x ∈ X . Thus

‖ f ( x ) − 2 k f ( x 2 k ) ‖ ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 1 r θ ‖ x ‖ r (48)

for all x ∈ X .

‖ ( 2 k ) l f ( x ( 2 k ) l ) − ( 2 k ) m f ( x ( 2 k ) m ) ‖ ≤ ∑ j = 1 m − 1 ‖ ( 2 k ) j f ( x ( 2 k ) j ) − ( 2 k ) j + 1 f ( x ( 2 k ) j + 1 ) ‖ ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 1 r θ ∑ j = 1 m − 1 ( 2 k ) α 2 j ( 2 k ) α 1 r j ‖ x ‖ r (49)

for all nonnegative integers p , l with p > l and all x ∈ X . It follows from (49) that the sequence { ( 2 k ) n f ( x ( 2 k ) n ) } is a cauchy sequence for all x ∈ X . Since Y is complete, the sequence { ( 2 k ) n f ( x ( 2 k ) n ) } coverges.

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

ϕ ( x ) : = lim n → ∞ ( 2 k ) n f ( x ( 2 k ) n )

for all x ∈ X . Moreover, letting l = 0 and passing the limit m → ∞ in (49), we get (46).

Form f : X → Y is even, the mapping ϕ : X → Y is even.

It follows from (45) that

‖ ϕ ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + ϕ ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     ϕ ( x j + y j 2 k ) − ∑ j = 1 k     ϕ ( z j ) − ∑ j = 1 k     ϕ ( − z j ) ‖ = lim n → ∞ ( 2 k ) α 2 n ‖ f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j 2 k + 1 ( 2 k ) n ∑ j = 1 k     z j )

+ f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j 2 k − 1 ( 2 k ) n ∑ j = 1 k     z j ) − ∑ j = 1 k     f ( 1 ( 2 k ) n x j + y j 2 k ) − ∑ j = 1 k     f ( 1 ( 2 k ) n z j ) − ∑ j = 1 k     f ( − 1 ( 2 k ) n z j ) ‖ ≤ lim n → ∞ ( 2 k ) α 2 n | g ( λ ) | α 2 ‖ 2 k f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 ( 2 k ) n + 1 ∑ j = 1 k     z j )   + 2 k f ( 1 ( 2 k ) n ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 ( 2 k ) n + 1 ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( 1 ( 2 k ) n x j + y j 2 k ) − ∑ j = 1 k     f ( − 1 ( 2 k ) n x j + y j 2 k ) − ∑ j = 1 k     f ( 1 ( 2 k ) n z j ) − ∑ j = 1 k     f ( − 1 ( 2 k ) n z j ) ‖

+ lim n → ∞ ( 2 k ) α 2 n ( 2 k ) α 1 n r θ ( ∑ j = 1 k ‖ x j ‖ r + ∑ j = 1 k ‖ y j ‖ r + ∑ j = 1 k ‖ z j ‖ r ) = | g ( λ ) | α 2 ‖ 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y (50)

for all x j , y j , z j ∈ X for all j = 1 → n .

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ g ( λ ) ( 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y

for all x j , y j , z j ∈ X for j = 1 → n , So by Lemma 7 it follows that the mapping ϕ : X → Y is quadratc. Now we need to prove uniqueness, Suppose ϕ ′ : X → Y is also an additive mapping that satisfies (46). Then we have

‖ ϕ ( x ) − ϕ ′ ( x ) ‖ = ( 2 k ) α 2 n ‖ ϕ ( x ( 2 k ) n ) − ϕ ′ ( x ( 2 k ) n ) ‖ ≤ ( 2 k ) α 2 n ( ‖ ϕ ( x ( 2 k ) n ) − f ( x ( 2 k ) n ) ‖ + ‖ ϕ ′ ( x ( 2 k ) n ) − f ( x ( 2 k ) n ) ‖ ) ≤ 2 ⋅ ( 2 k ) α 2 n ⋅ ( 2 k α 1 r + 1 + 1 ) ( 2 k ) α 1 n r ( ( 2 k ) α 1 r − ( 2 k ) α 2 ) θ ‖ x ‖ r (51)

which tends to zero as n → ∞ for all x ∈ X . So we can conclude that ϕ ( x ) = ϕ ′ ( x ) for all x ∈ X . This proves thus the mapping ϕ : X → Y is a unique mapping satisfying (46) as we expected.

Theorem 9. Assume for r < α 2 α 1 , θ be nonngative real number, and Suppose f : X → Y be a mapping such that

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ g ( λ ) ( 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y   + θ ( ∑ j = 1 k ‖ x j ‖ r + ∑ j = 1 k ‖ y j ‖ r + ∑ j = 1 k ‖ z j ‖ r ) (52)

for all x j , y j , z j ∈ X for all j = 1 → n . Then there exists a unique quadratic mapping ϕ : X → Y such that

‖ f ( x ) − ϕ ( x ) ‖ ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 2 − ( 2 k ) α 1 r θ ‖ x ‖ r (53)

for all x ∈ X .

Proof. Assume that f : X → Y satisfies (52).

We replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (52), we have

‖ ( 4 k − 2 ) f ( 0 ) ‖ ≤ ‖ 2 k g ( λ ) f ( 0 ) ‖

therefore

( | 4 k − 2 | α 2 − | 2 k g ( λ ) | α 2 ) ‖ f ( 0 ) ‖

So f ( 0 ) = 0 .

Next replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( k x , ⋯ , k x , k x , ⋯ , k x , x , ⋯ , x ) in (52) we have

‖ f ( 2 k x ) − 2 k f ( x ) ‖ ≤ ( 2 k α 1 r + 1 + 1 ) θ ‖ x ‖ r (54)

for all x ∈ X . Thus

‖ f ( x ) − 1 2 k f ( 2 k x ) ‖ ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 2 θ ‖ x ‖ r (55)

for all x ∈ X .

‖ 1 ( 2 k ) l f ( ( 2 k ) l x ) − 1 ( 2 k ) m f ( ( 2 k ) m x ) ‖ ≤ ∑ j = 1 m − 1 ‖ 1 ( 2 k ) j f ( ( 2 k ) j x ) − 1 ( 2 k ) j + 1 f ( ( 2 k ) j + 1 x ) ‖ ≤ 2 k α 1 r + 1 + 1 ( 2 k ) α 2 θ ∑ j = 1 m − 1 ( 2 k ) α 1 r j ( 2 k ) α 2 j ‖ x ‖ r (56)

for all nonnegative integers p , l with p > l and all x ∈ X . It follows from (56) that the sequence { 1 ( 2 k ) n f ( ( 2 k ) n x ) } is a cauchy sequence for all x ∈ X . Since Y is complete, the sequence { 1 ( 2 k ) n f ( ( 2 k ) n x ) } coverges.

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

ϕ ( x ) : = lim n → ∞ 1 ( 2 k ) n f ( ( 2 k ) n x )

for all x ∈ X . Moreover, letting l = 0 and passing the limit m → ∞ in (56), we get (53).

The rest of the proof is similar to the proof of Theorem 5.

Now, we study the solutions of (2). Note that for these inequalities, when X is a α 1 -homogeneous F-spaces and that Y is a α 2 -homogeneous F-spaces. Under this setting, we can show that the mapping satisfying (2) is quadratic. These results are give in the following.

Lemma 10. Let f : X → Y be an even mapping satilies

‖ 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ g ( λ ) ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y (57)

for all x j , y j , z j ∈ X for j = 1 → n , if and only if f : X → Y is quadratic.

Proof. Assume that f : X → Y satisfies (57).

We replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (57), we have

‖ 2 k f ( 0 ) ‖ ≤ | g ( λ ) | α 2 ‖ ( 4 k − 2 ) f ( 0 ) ‖

So f ( 0 ) = 0 .

Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 2 k x , ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (57), we have

Thus

‖ 4 k f ( x 2 k ) − 2 f ( x ) ‖ ≤ 0

f ( x 2 k ) = 1 2 k f ( x ) (58)

for all x ∈ X .

From (57) and (58) we infer that

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y = ‖ 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ | g ( λ ) | α 2 ‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y (59)

for all x j , y j , z j ∈ X for j = 1 → n , and so

f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) = 2 ∑ j = 1 k     f ( x j + y j 2 k ) + 2 ∑ j = 1 k     f ( z j )

for all x j , y j , z j ∈ X for j = 1 → n , as we expected. The couverse is obviously true.

Corollary 4. Let f : X → Y be an even mapping satilies

2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) = g ( λ ) ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) (60)

for all x j , y j , z j ∈ X for j = 1 → n , if and only if f : X → Y is quadratic.

Note! The functional equation (60) is called a quadratic g ( λ ) -functional equation.

Theorem 11. Assume for r > 2 α 2 α 1 , θ be nonngative real number, and Suppose f : X → Y be a even mapping such that f ( 0 ) = 0 and

‖ 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 k f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z j ) − 3 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( − x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ g ( λ ) ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y   + θ ( ∑ j = 1 k ‖ x j ‖ r + ∑ j = 1 k ‖ y j ‖ r + ∑ j = 1 k ‖ z j ‖ r ) (61)

for all x j , y j , z j ∈ X for all j = 1 → n . Then there exists a unique quadratic mapping ϕ : X → Y such that

‖ f ( x ) − ϕ ( x ) ‖ ≤ ( 2 k ) α 1 r ( 2 k ) α 1 r − ( 4 k ) α 2 θ ‖ x ‖ r (62)

for all x ∈ X .

Proof. Assume that f : X → Y satisfies (61).

We replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (61), we have

‖ 2 f ( 0 ) ‖ ≤ | g ( λ ) | α 2 ‖ ( 4 k − 2 ) f ( 0 ) ‖

therefore

( | 4 k − 2 | α 2 − | 2 g ( λ ) | α 2 ) ‖ f ( 0 ) ‖ ≤ 0

So f ( 0 ) = 0 .

Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 2 k x , ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (61) we have

‖ 4 f ( x 2 k ) − 1 k f ( x ) ‖ Y ≤ ( 2 k ) α 1 r θ ‖ x ‖ r (63)

for all x ∈ X . Thus

‖ 4 k f ( x 2 k ) − f ( x ) ‖ ≤ ( 2 k ) α 1 r k α 2 θ ‖ x ‖ r (64)

for all x ∈ X .

‖ ( 4 k ) l f ( x ( 2 k ) l ) − ( 4 k ) m f ( x ( 2 k ) m ) ‖ ≤ ∑ j = 1 m − 1 ‖ ( 4 k ) j f ( x ( 2 k ) j ) − ( 4 k ) j + 1 f ( x ( 2 k ) j + 1 ) ‖ ≤ ( 2 k ) α 1 r k α 2 θ ∑ j = 1 m − 1 ( 4 k ) α 2 j ( 2 k ) α 1 r j ‖ x ‖ r (65)

for all nonnegative integers p , l with p > l and all x ∈ X . It follows from (65) that the sequence { ( 4 k ) n f ( x ( 2 k ) n ) } is a cauchy sequence for all x ∈ X . Since Y is complete, the sequence { ( 4 k ) n f ( x ( 2 k ) n ) } coverges.

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

ϕ ( x ) : = lim n → ∞ ( 4 k ) n f ( x ( 2 k ) n )

for all x ∈ X . Moreover, letting l = 0 and passing the limit m → ∞ in (65), we get (62). The rest of the proof is similar to the proof of Theorem 8.

Theorem 12. Assume for r < 2 α 2 α 1 , θ be nonngative real number, f ( 0 ) = 0 and Suppose f : X → Y be a mapping such that

‖ 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 + 1 2 k ∑ j = 1 k     z j ) + 2 f ( ∑ j = 1 k x j + y j ( 2 k ) 2 − 1 2 k ∑ j = 1 k     z ) j − 3 2 k ∑ j = 1 k     f ( x j + y j 2 k ) + 1 2 k ∑ j = 1 k     f ( − x j + y j 2 k ) − 1 2 k ∑ j = 1 k     f ( z j ) − 1 2 k ∑ j = 1 k     f ( − z j ) ‖ Y ≤ ‖ g ( λ ) ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) + f ( ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − 2 ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) − ∑ j = 1 k     f ( − z j ) ) ‖ Y   + θ ( ∑ j = 1 k ‖ x j ‖ r + ∑ j = 1 k ‖ y j ‖ r + ∑ j = 1 k ‖ z j ‖ r ) (66)

for all x j , y j , z j ∈ X for all j = 1 → n . Then there exists a unique quadratic mapping ϕ : X → Y such that

‖ f ( x ) − ϕ ( x ) ‖ ≤ ( 2 k ) α 1 r ( 4 k ) α 2 − ( 2 k ) α 1 r θ ‖ x ‖ r . (67)

for all x ∈ X .

The proof is similar to theorem 8 and 9.

In this article, I construct two general functional inequalities with multivariables on homogeneous space and show that their solutions are additive-quadratic maps.

The author declares no conflicts of interest.

An, L.V. (2023) Considerable Development of the Type Additive-Quadratic -Functional Inequalities with 3k-Variable in -Homogeneous F-Spaces. Open Access Library Journal, 10: e10970. https://doi.org/10.4236/oalib.1110970