Let X and Y be normed spaces on the same field K , and f : X → Y . We use the notation ‖   ⋅   ‖ for all the norms on both X and Y . In this paper, we investigate some additive α-functional inequality when X is a real or complex normed space and Y is a complex Banach space.

In fact, when X is a real or complex normed space and Y is a complex Banach space, we solve and prove the Hyers-Ulam stability of following additive α-functional inequality.

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

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

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

So (1) and (2) are equivalent propositions.

Where α is a fixed complex number with | α | < 1 and m be a fixed integer with m > 1 .

The Hyers-Ulam stability was first investigated for functional equation of Ulam in [1] concerning the stability of group homomorphisms.

The functional equation

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

is called the Cauchy equation. In particular, every solution of the Cauchy equation is said to be an additive mapping.

The Hyers [2] gave first affirmative partial answer to the equation of Ulam in Banach spaces. After that, Hyers’ Theorem was generalized by Aoki [3] additive mappings and by Rassias [4] for linear mappings considering an unbouned Cauchy diffrence. Ageneralization 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 Hyers-Ulam stability for functional inequalities has been investigated such as in [5] [6] [7]. Gilány showed that if it satisfies the functional inequality

‖ 2 f ( x ) + 2 f ( y ) − f ( x − y ) ‖ ≤ ‖ f ( x + y ) ‖ (3)

Then f satisfies the Jordan-von Newman functional equation

2 f ( x ) + 2 f ( y ) = f ( x + y ) + f ( x − y ) (4)

Gilányi [5] and Fechner [8] proved the Hyers-Ulam stability of the functional inequality (3).

Next Choonkil Park [9] proved the Hyers-Ulam stability of additive β-functional inequalities. Recently, the author has studied the addition inequalities of mathematicians in the world as [5] [8] [10] - [24] and I have introduced two general additive function inequalities (1) and (2) based on the ( β 1 , β 2 ) -function inequality result, see [25]. When inserting the parameter m this is the opening for modern functional equations. That is, it demonstrates the superiority of the field of functional equations and is also a bright horizon for the special development of functional equations. So in this paper, we solve and proved the Hyers-Ulam stability for two α-functional inequalities (1)-(2), i.e. the α-functional inequalities with 3k-variables. Under suitable assumptions on spaces X and Y , we will prove that the mappings satisfying the α-functional inequatilies (1) or (2). Thus, the results in this paper are generalization of those in [7] [9] [17] [25] [26] [27] for α-functional inequatilies with 3k-variables. The paper is organized as followns: In section preliminarier we remind a basic property such as We only redefine the solution definition of the equation of the additive function.

Notice here that we make the general assumption that: G be a k-divisible abelian group.

Section 3: is devoted to prove the Hyers-Ulam stability of the addive α-functional inequalities (1) when X is a real or complex normed space and Y complex Banach space.

Section 4: is devoted to prove the Hyers-Ulam stability of the addive α-functional inequalities (2) when X is a real or complex normed space and Y complex Banach space.

The functional equation

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

is called the cauchuy equation. In particular, every solution of the cauchuy equation is said to be an additive mapping.

Now, we first study the solutions of (1). Note that for these inequalities, G be a k-divisible abelian group, X is a real or complex normed space and Y is a complex Banach spaces. Under this setting, we can show that the mapping satisfying (1.1) is additive. These results are give in the following.

Lemma 1. Let m ∈ ℕ and a mapping f : G → Y satilies

‖ f ( ( m + 1 ) ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − ∑ j = 1 k     f ( m x j + y j 2 k − z j ) − ∑ j = 1 k     f ( x j + y j 2 k ) ‖ Y ≤ ‖ α ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) − ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) ) ‖ Y (5)

for all x j , y j , z j ∈ G for j = 1 → n , then f : G → Y is additive

Proof. Assume that f : G → Y satisfies (5).

We replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (5), we have

‖ ( 2 k − 1 ) f ( 0 ) ‖ Y ≤ ‖ α ( 2 k − 1 ) f ( 0 ) ‖ Y ≤ 0

therefore

( | 2 k − 1 | − | α ( 2 k − 1 ) | ) ‖ f ( 0 ) ‖ Y ≤ 0

So f ( 0 ) = 0 .

Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , m ⋅ x 1 + y 1 2 k − v 1 , ⋯ , m ⋅ x k + y k 2 k − v k ) in (5), we have

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     v j ) − ∑ j = 1 k f ( x j + y j 2 k ) − ∑ j = 1 k     f ( v j ) ‖ Y ≤ ‖ α ( f ( ( m + 1 ) ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     v j ) − ∑ j = 1 k     f ( m x j + y j 2 k − v j ) − ∑ j = 1 k     f ( x j + y j 2 k ) ) ‖ Y (6)

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , x 1 + y 1 2 k − v 1 , ⋯ , x k + y k 2 k − v k ∈ G . From (5) and (6) we infer that

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) − ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) ‖ Y ≤ ‖ α ( f ( ( m + 1 ) ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − ∑ j = 1 k     f ( m x j + y j 2 k − z j ) − ∑ j = 1 k     f ( x j + y j 2 k ) ) ‖ Y ≤ ‖ α 2 ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) − ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) ) ‖ Y (7)

and so

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

for all x j , y j , z j ∈ G for j = 1 → n , as we expected.

Theorem 2. Let r > 1 , m ∈ ℤ , m > 1 , θ be nonngative real number, and let f : X → Y be a mapping such that

‖ f ( ( m + 1 ) ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − ∑ j = 1 k     f ( m x j + y j 2 k − z j ) − ∑ j = 1 k     f ( x j + y j 2 k ) ‖ Y ≤ ‖ α ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) − ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) ) ‖ Y         + θ ( ∑ j = 1 k ‖ x j ‖ X r + ∑ j = 1 k ‖ y j ‖ X r + ∑ j = 1 k ‖ z j ‖ X r ) (8)

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 ) − h ( x ) ‖ Y ≤ ∑ q = 1 m − 1 ( q r + 2 k r ) ( 1 − | α | ) ( m r − m ) θ ‖ x ‖ X r . (9)

for all x ∈ X .

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

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

‖ ( 2 k − 1 ) f ( 0 ) ‖ Y ≤ ‖ α ( 2 k − 1 ) f ( 0 ) ‖ Y ≤ 0

therefore

( | 2 k − 1 | − | α ( 2 k − 1 ) | ) ‖ f ( 0 ) ‖ Y ≤ 0

So f ( 0 ) = 0 .

Next we:

Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( k x , 0, ⋯ ,0, k x , 0, ⋯ ,0,0, ⋯ ,0 ) in (8), we get

‖ f ( ( m + 1 ) x ) − f ( m x ) − f ( x ) ‖ Y ≤ 2 k r θ ‖ x ‖ X r (10)

for all x ∈ X . Thus for q ∈ ℕ .

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

‖ f ( ( m − q + 1 ) x ) − f ( ( m − q ) x ) − f ( x ) ‖ Y ≤ ‖ α ( f ( ( q + 1 ) x ) − f ( q x ) − f ( x ) ) ‖ Y + θ ( 2 k r + q r ) ‖ x ‖ Y r (11)

for all x ∈ X .

For (10) and (11)

∑ q = 1 m − 1 ‖ f ( ( m − q + 1 ) x ) − f ( ( m − q ) x ) − f ( x ) ‖ Y ≤ ∑ q = 1 m − 1 ‖ α ( f ( ( q + 1 ) x ) − f ( q x ) − f ( x ) ) ‖ Y + θ ( ∑ q = 1 m − 1 ( 2 k r + q r ) ‖ x ‖ X r ) (12)

for all x ∈ X .

From (11) and (12) and triangle inequality, we have

( 1 − | α | ) ‖ f ( m x ) − m f ( x ) ‖ Y = ( 1 − | α | ) ∑ q = 1 m − 1 ‖ f ( ( q + 1 ) x ) − f ( q x ) − f ( x ) ‖ Y ≤ ∑ q = 1 m − 1 ( 1 − | α | ) ‖ f ( ( q + 1 ) x ) − f ( q x ) − f ( x ) ‖ Y ≤ ∑ q = 1 m − 1 ‖ f ( ( q + 1 ) x ) − f ( q x ) − f ( x ) ‖ Y − ∑ q = 1 m − 1 ‖ α ( f ( ( q + 1 ) x ) − f ( q x ) − f ( x ) ) ‖ Y ≤ θ ( ∑ q = 1 m − 1 ( 2 k r + q r ) ‖ x ‖ X r ) (13)

for all x ∈ X . from

∑ q = 1 m − 1 ‖ f ( ( m − q + 1 ) x ) − f ( ( m − q ) x ) − f ( x ) ‖ Y = ∑ q = 1 m − 1 ‖ f ( ( q + 1 ) x ) − f ( q x ) − f ( x ) ‖ Y

Since | α | < 1 , the mapping f satisfies the inequalities

‖ f ( m x ) − m f ( x ) ‖ Y ≤ θ ( ∑ q = 1 m − 1 ( 2 k r + q r ) ‖ x ‖ X r ) 1 − | α |

for all x ∈ X .

Therefore

‖ f ( x ) − m f ( x m ) ‖ Y ≤ θ ( ∑ q = 1 m − 1 ( 2 k r + q r ) ‖ x ‖ X r ) ( 1 − | α | ) m r (14)

for all x ∈ X . So

‖ m l f ( x m n ) − m p f ( x m h ) ‖ Y ≤ ∑ j = l p − 1 ‖ m j f ( x m j ) − m j + 1 f ( x m j + 1 ) ‖ Y ≤ θ ( ∑ q = 1 m − 1 ( 2 k r + q r ) ) ( 1 − | α | ) m r ∑ j = l p − 1 m j m r j ‖ x ‖ X r (15)

for all nonnegative integers p , l with p > l and all x ∈ X . It follows from (15) that the sequence { m n f ( x m n ) } is a cauchy sequence for all x ∈ X . Since Y is complete, the sequence { m n f ( x m n ) } coverges.

So one can define the mapping ϕ : X → Y by ϕ ( x ) : = lim n → ∞ m n f ( x m n ) for all x ∈ X . Moreover, letting l = 0 and passing the limit m → ∞ in (15), we get (9).

It follows from (8) that

‖ ϕ ( ( m + 1 ) ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − ∑ j = 1 k     ϕ ( m x j + y j 2 k − z j ) − ∑ j = 1 k     ϕ ( x j + y j 2 k ) ‖ Y = l i m n → ∞ m n ‖ f ( m + 1 m n ∑ j = 1 k x j + y j 2 k − 1 m n ∑ j = 1 k     z j ) − ∑ j = 1 k     f ( m m n x j + y j 2 k − 1 m n z j )       − ∑ j = 1 k     f ( 1 m n x j + y j 2 k ) ‖ Y ≤ l i m n → ∞ m n ‖ α ( f ( 1 m n ∑ j = 1 k x j + y j 2 k + 1 m n ∑ j = 1 k     z j ) − ∑ j = 1 k     f ( 1 m n x j + y j 2 k )         − ∑ j = 1 k     f ( 1 m n z j ) ) ‖ Y + l i m n → ∞ m n m n r θ ( ∑ j = 1 k ‖ x j ‖ X r + ∑ j = 1 k ‖ y j ‖ X r + ∑ j = 1 k ‖ z j ‖ X r ) ≤ | α | ‖ ϕ ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) − ∑ j = 1 k     ϕ ( x j + y j 2 k ) − ∑ j = 1 k     ϕ ( z j ) ‖ Y (16)

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

‖ ϕ ( ( m + 1 ) ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − ∑ j = 1 k     ϕ ( m x j + y j 2 k − z j ) − ∑ j = 1 k     ϕ ( x j + y j 2 k ) ‖ Y ≤ | α | ‖ ϕ ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) − ∑ j = 1 k     ϕ ( x j + y j 2 k ) − ∑ j = 1 k     ϕ ( z j ) ‖ Y

for all x j , y j , z j ∈ X for all j = 1 → n . So by lemma 21 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 (9). Then we have

‖ ϕ ( x ) − ϕ ′ ( x ) ‖ Y = m n ‖ ϕ ( x m n ) − ϕ ′ ( x m n ) ‖ Y ≤ m n ( ‖ ϕ ( x m n ) − f ( x m n ) ‖ Y + ‖ ϕ ′ ( x m n ) − f ( x m n ) ‖ Y ) ≤ 2 ⋅ m n ⋅ ∑ q = 1 m − 1 ( q r + 2 k r ) ( 1 − | α | ) m n r ( m r − m ) θ ‖ x ‖ X r (17)

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 (9) as we expected.

Theorem 3. Let r > 1 , m ∈ ℤ , m > 1 , θ be nonngative real number, and let f : X → Y be a mapping such that

‖ f ( ( m + 1 ) ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − ∑ j = 1 k     f ( m x j + y j 2 k − z j ) − ∑ j = 1 k     f ( x j + y j 2 k ) ‖ Y ≤ ‖ α ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) − ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( z j ) ) ‖ Y         + θ ( ∑ j = 1 k ‖ x j ‖ X r + ∑ j = 1 k ‖ y j ‖ X r + ∑ j = 1 k ‖ z j ‖ X r ) (18)

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

‖ f ( x ) − ϕ ( x ) ‖ Y ≤ m n ⋅ ∑ q = 1 m − 1 ( q r + 2 k r ) ( 1 − | α | ) ( m − m r ) θ ‖ x ‖ X r . (19)

for all x ∈ X .

The rest of the proof is similar to the proof of Theorem 2.2.

Next, we study the solutions of (2). Note that for these inequalities, when X be a real or complete normed space and Y complex Banach space. Now, we study the solutions of (2). Note that for these inequalities, G be a k-divisible abelian group, X is a real or complex normed space and Y is complex Banach spaces. Under this setting, we can show that the mapping satisfying (2) is additive. These results are give in the following.

Lemma 4. Let m ∈ ℕ and a mapping f : G → Y satilies

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

≤ ‖ α ( f ( ( m + 1 ) ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − ∑ j = 1 k     f ( m x j + y j 2 k − z j ) − ∑ j = 1 k     f ( x j + y j 2 k ) ) ‖ Y (20)

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

Proof. Assume that f : G → Y satisfies (20).

Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (20), we have

‖ ( 2 k − 1 ) f ( 0 ) ‖ Y ≤ ‖ ( 2 k − 1 ) α f ( 0 ) ‖ Y ≤ 0

therefore

( | 2 k − 1 | − | α ( 2 k − 1 ) | ) ‖ f ( 0 ) ‖ Y ≤ 0

So f ( 0 ) = 0 .

Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , m ⋅ x 1 + y 1 2 k − v 1 , ⋯ , m ⋅ x k + y k 2 k − v k ) in (20), we have

‖ f ( ( m + 1 ) ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     v j ) − ∑ j = 1 k     f ( m x j + y j 2 k − v j ) − ∑ j = 1 k     f ( x j + y j 2 k ) ‖ Y ≤ ‖ α ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     v j ) − ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( v j ) ) ‖ Y (21)

for all x 1 , ⋯ , x k , y 1 , ⋯ , y k , x 1 + y 1 2 k − v 1 , ⋯ , x k + y k 2 k − v k ∈ G . From (20) and (21) we infer that

‖ f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     v j ) − ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( v j ) ‖ Y ≤ ‖ α ( f ( ( m + 1 ) ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     v j ) − ∑ j = 1 k     f ( m x j + y j 2 k − v j ) − ∑ j = 1 k     f ( x j + y j 2 k ) ) ‖ Y ≤ ‖ α 2 ( f ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     v j ) − ∑ j = 1 k     f ( x j + y j 2 k ) − ∑ j = 1 k     f ( v j ) ) ‖ Y (22)

and so

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

for all x j , y j , z j ∈ G for j = 1 → n , as we expected.

Theorem 5. Let r > 1 , m ∈ ℤ , m > 1 , θ be nonngative real number, and let f : X → Y be a mapping such that

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

  + θ ( ∑ j = 1 k ‖ x j ‖ r + ∑ j = 1 k ‖ y j ‖ r + ∑ j = 1 k ‖ z j ‖ r ) (23)

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

‖ f ( x ) − h ( x ) ‖ Y ≤ ∑ q = 1 m − 1 ( q r + 2 k r ) ( 1 − | α | ) ( m − m r ) θ ‖ x ‖ X r . (24)

for all x ∈ X .

Proof. Assume that f : X → Y satisfies (23).

Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( 0, ⋯ ,0,0, ⋯ ,0,0, ⋯ ,0 ) in (23), we have

‖ 2 k f ( 0 ) ‖ ≤ ‖ α ( 2 k − 1 ) f ( 0 ) ‖ Y ≤ 0

therefore

( | 2 k − 1 | − | α ( 2 k − 1 ) | ) ‖ f ( 0 ) ‖ Y ≤ 0

So f ( 0 ) = 0 .

Next we:

Replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( k x , 0, ⋯ ,0, k x , 0, ⋯ ,0 , 0, ⋯ ,0 ) in (23), we get

‖ f ( ( m + 1 ) x ) − f ( m x ) − f ( x ) ‖ Y ≤ 2 k r θ ‖ x ‖ X r (25)

for all x ∈ X . Thus for q ∈ ℕ .

We replacing ( x 1 , ⋯ , x k , y 1 , ⋯ , y k , z 1 , ⋯ , z k ) by ( k x , 0, ⋯ ,0, k x , 0, ⋯ ,0, q x , 0, ⋯ ,0 ) in (23), we have

‖ f ( ( m − q + 1 ) x ) − f ( ( m − q ) x ) − f ( x ) ‖ Y ≤ ‖ α ( f ( ( q + 1 ) x ) − f ( q x ) − f ( x ) ) ‖ + θ ( 2 k r + q r ) ‖ x ‖ Y r (26)

for all x ∈ X .

For (25) and (26)

∑ q = 1 m − 1 ‖ f ( ( m − q + 1 ) x ) − f ( ( m − q ) x ) − f ( x ) ‖ Y ≤ ∑ q = 1 m − 1 ‖ α ( f ( ( q + 1 ) x ) − f ( q x ) − f ( x ) ) ‖ Y + θ ( ∑ q = 1 m − 1 ( 2 k r + q r ) ‖ x ‖ r ) (27)

for all x ∈ X .

From (26) and (27) and triangle inequality, we have

( 1 − | α | ) ‖ f ( m x ) − m f ( x ) ‖ Y = ( 1 − | α | ) ∑ q = 1 m − 1 ‖ f ( ( q + 1 ) x ) − f ( q x ) − f ( x ) ‖ Y ≤ ∑ q = 1 m − 1 ( 1 − | α | ) ‖ f ( ( q + 1 ) x ) − f ( q x ) − f ( x ) ‖ Y

≤ ∑ q = 1 m − 1 ‖ f ( ( q + 1 ) x ) − f ( q x ) − f ( x ) ‖ − ∑ q = 1 m − 1 ‖ α ( f ( ( q + 1 ) x ) − f ( q x ) − f ( x ) ) ‖ Y ≤ θ ( ∑ q = 1 m − 1 ( 2 k r + q r ) ‖ x ‖ X r ) (28)

for all x ∈ X . from

∑ q = 1 m − 1 ‖ f ( ( m − q + 1 ) x ) − f ( ( m − q ) x ) − f ( x ) ‖ Y = ∑ q = 1 m − 1 ‖ f ( ( q + 1 ) x ) − f ( q x ) − f ( x ) ‖ Y

Since | α | < 1 , the mapping f satisfies the inequalities

‖ f ( m x ) − m f ( x ) ‖ Y ≤ θ ( ∑ q = 1 m − 1 ( 2 k r + q r ) ‖ x ‖ X r ) 1 − | α |

for all x ∈ X .

Therefore

‖ f ( x ) − m f ( x m ) ‖ Y ≤ θ ( ∑ q = 1 m − 1 ( 2 k r + q r ) ‖ x ‖ X r ) ( 1 − | α | ) m r (29)

for all x ∈ X . So

‖ m l f ( x m n ) − m p f ( x m h ) ‖ Y ≤ ∑ j = l p − 1 ‖ m j f ( x m j ) − m j + 1 f ( x m j + 1 ) ‖ Y ≤ θ ( ∑ q = 1 m − 1 ( 2 k r + q r ) ) ( 1 − | α | ) m r ∑ j = l p − 1 m j m r j ‖ x ‖ X r (30)

for all nonnegative integers p , l with p > l and all x ∈ X . It follows from (30) that the sequence { m n f ( x m n ) } is a Cauchy sequence for all x ∈ X . Since Y is complete, the sequence { m n f ( x m n ) } coverges.

So one can define the mapping ϕ : X → Y by ϕ ( x ) : = lim n → ∞ m n f ( x m n ) for all x ∈ X . Moreover, letting l = 0 and passing the limit m → ∞ in (30), we get (24).

It follows from (23) that

‖ ϕ ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) − ∑ j = 1 k     ϕ ( x j + y j 2 k ) − ∑ j = 1 k     ϕ ( z j ) ‖ Y = lim n → ∞ m n ‖ f ( 1 m n ∑ j = 1 k x j + y j 2 k + 1 m n ∑ j = 1 k     z j ) − ∑ j = 1 k     f ( 1 m n ∑ j = 1 k x j + y j 2 k )         − ∑ j = 1 k     f ( 1 m n z j ) ‖ Y + lim n → ∞ m n m n r θ ( ∑ j = 1 k ‖ x j ‖ X r + ∑ j = 1 k ‖ y j ‖ X r + ∑ j = 1 k ‖ z j ‖ X r ) ≤ lim n → ∞ m n | α | ‖ f ( m + 1 m n ∑ j = 1 k x j + y j 2 k − 1 m n ∑ j = 1 k     z j )

        − ∑ j = 1 k     f ( m m n ( x j + y j 2 k ) − 1 m n z j ) − ∑ j = 1 k     f ( 1 m n z j ) ‖ Y ≤ | α | ‖ ϕ ( ( m + 1 ) ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) − ∑ j = 1 k     ϕ ( m x j + y j 2 k − z j ) − ∑ j = 1 k     ϕ ( z j ) ‖ Y (31)

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

‖ ϕ ( ∑ j = 1 k x j + y j 2 k + ∑ j = 1 k     z j ) − ∑ j = 1 k     ϕ ( x j + y j 2 k ) − ∑ j = 1 k     ϕ ( z j ) ‖ Y ≤ | α | ‖ ϕ ( ( m + 1 ) ∑ j = 1 k x j + y j 2 k − ∑ j = 1 k     z j ) − ∑ j = 1 k     ϕ ( m x j + y j 2 k − z j ) − ∑ j = 1 k     ϕ ( x j + y j 2 k ) ‖ Y

for all x j , y j , z j ∈ X for all 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 an additive mapping that satisfies (24). Then we have

‖ ϕ ( x ) − ϕ ′ ( x ) ‖ = m n ‖ ϕ ( x m n ) − ϕ ′ ( x m n ) ‖ ≤ m n ( ‖ ϕ ( x m n ) − f ( x m n ) ‖ + ‖ ϕ ′ ( x m n ) − f ( x m n ) ‖ ) ≤ 2 ⋅ m n ⋅ ∑ q = 1 m − 1 ( q r + 2 k r ) ( 1 − | α | ) m n r ( m r − m ) θ ‖ x ‖ r (32)

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 (24) as we expected.

In this article, I have solved two problems posed as establishing the solution of the additive α-function inequality (1) and (2) in complex Banach spaces with 3k variable. So when I develop this result, I rely on the inequality ( β 1 , β 2 ) -function.

The author declares no conflicts of interest.

An, L.V. (2022) Generalized Hyers-Ulam-Rassias Type Stability Additive α-Functional Inequalities with 3k-Variable in Complex Banach Spaces. Open Access Library Journal, 9: e9373. https://doi.org/10.4236/oalib.1109373