Int. J. Anal. Appl. (2022), 20:43 Generalized Stability Additive λ-Functional Inequalities With 3k-Variable in α-Homogeneous F-Spaces Ly Van An∗ Faculty of Mathematics Teacher Education, Tay Ninh University, Ninh Trung, Ninh Son, Tay Ninh Province, Vietnam ∗Corresponding author: lyvanan145@gmail.com Abstract. In this paper, we study to solve two additive λ-functional inequalities with 3k-variables in α-homogeneous F spaces. Then we will show that the solutions of the first and second inequalities are additive mappings. 1. Introduction Let X and Y be a normed spaces on the same field K, and f : X → Y. We use the notation ∥∥ ·∥∥ for all the norm on both X and Y. In this paper, we investisgate some additive λ-functional inequality in α-homogeneous F-spaces. In fact, when X is a α1-homogeneous F-spaces and that Y is a α2-homogeneous F-spaces we solve and prove the Hyers-Ulam-Rassias type stability of two forllowing additive α-functional inequality. ∥∥∥∥∥f ( (m + 1) k∑ j=1 xj + yj 2k − k∑ j=1 zj ) − k∑ j=1 f ( m xj + yj 2k −zj ) − k∑ j=1 f (xj + yj 2k )∥∥∥∥∥ Y ≤ ∥∥∥∥∥λ ( f ( k∑ j=1 xj + yj 2k + k∑ j=1 zj ) − k∑ j=1 f (xj + yj 2k ) − k∑ j=1 f ( zj ))∥∥∥∥∥ Y (1.1) Received: Jun. 10, 2022. 2010 Mathematics Subject Classification. 39B62, 39B72, 39B52. Key words and phrases. additive β-functional inequality; α-homogeneous F-space; Hyers-Ulam stability. https://doi.org/10.28924/2291-8639-20-2022-43 ISSN: 2291-8639 © 2022 the author(s). https://doi.org/10.28924/2291-8639-20-2022-43 2 Int. J. Anal. Appl. (2022), 20:43 and∥∥∥∥∥f ( k∑ j=1 xj + yj 2k + k∑ j=1 zj ) − k∑ j=1 f (xj + yj 2k ) − k∑ j=1 f ( zj )∥∥∥∥∥ Y ≤ ∥∥∥∥∥λ ( f ( (m + 1) k∑ j=1 xj + yj 2k − k∑ j=1 zj ) − k∑ j=1 f ( m xj + yj 2k −zj ) − k∑ j=1 f (xj + yj 2k ))∥∥∥∥∥ Y (1.2) where λ is a fixed complex number with ∣∣λ∣∣ < 1, α1,α2 ∈ R+,α1,α2 ≤ 1 and m is a fixed integer with m > 1. The Hyers-Ulam stability was first investigated for functional equation of Ulam in [19] 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 [9] gave firts affirmative partial answer to the equation of Ulam in Banach spaces. After that, Hyers’Theorem was generalized by Aoki [1] additive mappings and by Rassias [18] for linear mappings considering an unbouned Cauchy diffrence. Ageneralization of the Rassias theorem was obtained by Gǎvruta [6] by replacing the unbounded Cauchy difference by a general control function in the spirit of Rassias’ approach. The Hyers-Ulam stability for functional inequalities have been investigated such as in [5], [10], [13], [16], [17], [18]. Gilány showed that is if satisfies the functional inequality∥∥2f(x) + 2f(y)− f(x −y)∥∥ ≤ ∥∥f(x + y)∥∥ (1.3) Then f satisfies the Jordan-von Newman functional equation 2f ( x ) + 2f ( y ) = f ( x + y ) + f ( x −y ) (1.4) . Gilányi [8] and Fechner [5] proved the Hyers-Ulam stability of the functional inequality (1.3). Next Chookil [16] proved the of additive β-functional inequalities in non-Archimedean Banach spaces and in complex Banach spaces, and Harin Leea [11] proved the Hyers-Ulam stability of additive β- functional inequalities in ρ-homogeneous F space. Recently, the author has studied the additive inequalities of mathematicians around the world, on spaces complex Banach spaces , non-Archimedan Banach spaces or additive β-functional inequalities in p-homogeneous F-space.. So in this paper, we solve and proved the Hyers-Ulam stability for two ff-functional inequalities (1.1)-(1.2), ie 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 (1.1) or (1.2). Thus, the results in this paper are generalization of those in [2], [11] for α-functional inequatilies with Int. J. Anal. Appl. (2022), 20:43 3 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 and F∗-space . Section 3: is devoted to prove the Hyers-Ulam stability of the addive λ- functional inequalities (1.1) when when X is a α1-homogeneous F-spaces and that Y is a α2-homogeneous F-spaces. Section 4: is devoted to prove the Hyers-Ulam stability of the addive λ- functional inequalities (1.2) when when X is a α1-homogeneous F-spaces and that Y is a α2-homogeneous F-spaces. 2. Preliminaries 2.1. F∗- spaces. Definition 2.1. Let X be a ( complex ) linear space. A nonnegative valued function ∥∥ ·∥∥ is an F-norm if it satisfies the following conditions: (1) ∥∥∥x∥∥∥ = 0 if and only if x = 0; (2) ∥∥∥λx∥∥∥ = ∥∥∥x∥∥∥ for all x ∈ X and all λ with ∣∣∣λ∣∣∣ = 1; (3) ∥∥∥x + y∥∥∥ ≤ ∥∥∥x∥∥∥ + ∥∥∥y∥∥∥ for all x,y ∈ X; (4) ∥∥∥λnx∥∥∥ → 0, λn → 0; (5) ∥∥∥λnx∥∥∥ → 0, xn → 0. Then ( X, ∥∥∥ · ∥∥∥ ) is called an F∗-space. An F-space is a complete F∗-space. An F-norm is called β-homgeneous ( β > 0 ) if ∥∥∥tx∥∥∥ = ∣∣∣t∣∣∣β∥∥∥x∥∥∥ for all x ∈ X and for all t ∈ C and (X,∥∥∥ ·∥∥∥) is called α-homogeneous F-space. 2.2. Solutions of the inequalities. 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. 4 Int. J. Anal. Appl. (2022), 20:43 3. Hyers-Ulam-Rassias stability Additive λ-functional inequalities (1.1) in α-homogeneous F-spaces Now, we first study the solutions of (1.1). Note that for these inequalities, when X is a α1- homogeneous F-spaces and that Y is a α2-homogeneous F-spaces. Under this setting, we can show that the mapping satisfying (1.1) is additive. These results are give in the following. Lemma 3.1. Let m ∈N and a mapping f : X → Y satilies ∥∥∥∥∥f ( (m + 1) k∑ j=1 xj + yj 2k − k∑ j=1 zj ) − k∑ j=1 f ( m xj + yj 2k −zj ) − k∑ j=1 f (xj + yj 2k )∥∥∥∥∥ Y ≤ ∥∥∥∥∥λ ( f ( k∑ j=1 xj + yj 2k + k∑ j=1 zj ) − k∑ j=1 f (xj + yj 2k ) − k∑ j=1 f ( zj ))∥∥∥∥∥ Y (3.1) for all xj,yj,zj ∈ X for j = 1 → n, then f : X → Y is additive Proof. Assume that f : G → Y satisfies (3.1). We replacing ( x1, ...,xk,y1, ...,yk,z1, ...,zk ) by ( 0, ..., 0, 0, ..., 0, 0, ..., 0 ) in (3.1), we have ∥∥∥2kf(0)∥∥∥ ≤ ∥∥∥λ(2k − 1)f (0)∥∥∥ Y ≤ 0 therefore (∣∣∣2k∣∣∣α2 − ∣∣∣λ(2k − 1)∣∣∣α2)∥∥∥f(0)∥∥∥ Y ≤ 0 So f ( 0 ) = 0. Replacing ( x1, ...,xk,y1, ...,yk,z1, ...,zk ) by ( 0, ..., 0, 0, ..., 0,z, 0, ..., 0) ) , in (3.1), we get ∥∥∥f(−z)− f(−z)∥∥∥ Y ≤ 0 and so f is an odd mapping. Replacing ( x1, ...,xk,y1, ...,yk,z1, ...,zk ) by ( x1, ...,xk,y1, ...,yk,m · x1+y12k −v1, ...,m · xk+yk 2k −vk ) in (3.1), we have ∥∥∥f( k∑ j=1 xj + yj 2k + k∑ j=1 vj ) − k∑ j=1 f (xj + yj 2k ) − k∑ j=1 f ( vj )∥∥∥ Y ≤ ∥∥∥∥∥λ ( f ( (m + 1) k∑ j=1 xj + yj 2k − k∑ j=1 vj ) − k∑ j=1 f ( m xj + yj 2k −vj ) − k∑ j=1 f (xj + yj 2k ))∥∥∥∥∥ Y (3.2) Int. J. Anal. Appl. (2022), 20:43 5 for all x1, ...,xk,y1, ...,yk,m x1+y1 2k −v1, ...,mxk+yk2k −vk ∈ G. From (3.1) and (3.2) we infer that∥∥∥∥∥f ( k∑ j=1 xj + yj 2k + k∑ j=1 zj ) − k∑ j=1 f (xj + yj 2k ) − k∑ j=1 f ( zj )∥∥∥∥∥ Y ≤ ∥∥∥∥∥λ ( f ( (m + 1) k∑ j=1 xj + yj 2k − k∑ j=1 zj ) − k∑ j=1 f ( m xj + yj 2k −zj ) − k∑ j=1 f (xj + yj 2k ))∥∥∥∥∥ Y ≤ ∥∥∥∥∥λ2 ( f ( k∑ j=1 xj + yj 2k + k∑ j=1 zj ) − k∑ j=1 f (xj + yj 2k ) − k∑ j=1 f ( zj ))∥∥∥∥∥ Y (3.3) and so f ( k∑ j=1 xj + yj 2k + k∑ j=1 zj ) = k∑ j=1 f (xj + yj 2k ) + k∑ j=1 f ( zj ) for all xj,yj,zj ∈ G for j = 1 → n, as we expected. � Theorem 3.2. Let r > α2 α1 ,m ∈ Z,m > 1, θ be nonngative real number, and let f : X → Y be a mapping such that ∥∥∥∥∥f ( (m + 1) k∑ j=1 xj + yj 2k − k∑ j=1 zj ) − k∑ j=1 f ( m xj + yj 2k −zj ) − k∑ j=1 f (xj + yj 2k )∥∥∥∥∥ Y ≤ ∥∥∥∥∥λ ( f ( k∑ j=1 xj + yj 2k + k∑ j=1 zj ) − k∑ j=1 f (xj + yj 2k ) − k∑ j=1 f ( zj ))∥∥∥∥∥ Y + θ ( k∑ j=1 ∥∥xj∥∥rX + k∑ j=1 ∥∥yj∥∥rX + k∑ j=1 ∥∥zj∥∥rX) (3.4) for all xj,yj,zj ∈ X for all j = 1 → n. Then there exists a unique mapping φ : X → Y such that∥∥∥f(x)−h(x)∥∥∥ Y ≤ ∑m−1 q=1 ( qα1r + 2kα1r )( 1 − ∣∣λ∣∣α2)(mα1r −mα2)θ∥∥x∥∥rX. (3.5) for all x ∈ X Proof. Assume that f : X → Y satisfies (3.4). Replacing ( x1, ...,xk,y1, ...,yk,z1, ...,zk ) by ( 0, ..., 0, 0, ..., 0, 0, ..., 0 ) in (3.4), we have ∥∥∥2kf(0)∥∥∥ Y ≤ ∥∥∥λ(2k − 1)f (0)∥∥∥ Y ≤ 0 therefore (∣∣∣2k∣∣∣α2 − ∣∣∣λ(2k − 1)∣∣∣α2)∥∥∥f(0)∥∥∥ Y ≤ 0 Sof ( 0 ) = 0. Next we replacing ( x1, ...,xk,y1, ...,yk,z1, ...,zk ) by ( kx, 0, ..., 0,kx, 0, ..., 0, 0, ..., 0 ) in (3.4), we get 6 Int. J. Anal. Appl. (2022), 20:43 ∥∥∥f((m + 1)x)− f(mx)− f(x)∥∥∥ Y ≤ 2kα1rθ ∥∥∥x∥∥∥r X (3.6) for all x ∈ X. Thus for q ∈N, we replacing ( x1, ...,xk,y1, ...,yk,z1, ...,zk ) by ( kx, 0, ..., 0,kx, 0, ..., 0,qx, 0, ..., 0 ) in (3.4), we have∥∥∥∥∥f((m−q + 1)x)− f((m−q)x)− f(x) ∥∥∥∥∥ Y ≤ ∥∥∥∥∥λ(f((q + 1)x)− f(qx)− f(x) ∥∥∥∥∥ Y + θ ( 2kα1r + qα1r )∥∥∥x∥∥∥r X (3.7) for all x ∈ X. For (3.6) and (3.7) m−1∑ q=1 ∥∥∥f((m−q + 1)x)− f((m−q)x)− f(x)∥∥∥ Y ≤ m−1∑ q=1 ∥∥∥λ(f((q + 1)x)− f(qx)− f(x)∥∥∥ Y + θ (m−1∑ q=1 ( 2kα1r + qα1r )∥∥∥x∥∥∥r X ) (3.8) for all x ∈ X. From (3.7) and (3.8) and triangle inequality, we have ( 1 − ∣∣λ∣∣α2)∥∥∥f(mx)−mf(x)∥∥∥ Y = ( 1 − ∣∣λ∣∣α2)m−1∑ q=1 ∥∥∥f((q + 1)x)− f(qx)− f(x)∥∥∥ Y ≤ m−1∑ q=1 ( 1 − ∣∣λ∣∣α2)∥∥∥(f((q + 1)x)− f(qx)− f(x)∥∥∥ Y ≤ m−1∑ q=1 ∥∥∥f((q + 1)x)− f(qx)− f(x)∥∥∥ Y − m−1∑ q=1 ∥∥∥λ(f((q + 1)x)− f(qx)− f(x))∥∥∥ Y ≤ θ (m−1∑ q=1 ( 2kα1r + qα1r )∥∥∥x∥∥∥r X ) (3.9) for all x ∈ X. from m−1∑ q=1 ∥∥∥f((m−q + 1)x)− f((m−q)x)− f(x)∥∥∥ Y = m−1∑ q=1 ∥∥∥(f((q + 1)x)− f(qx)− f(x)∥∥∥ Y Int. J. Anal. Appl. (2022), 20:43 7 Since ∣∣λ∣∣ < 1, the mapping f satisfies the inequalities ∥∥∥f(mx)−mf(x)∥∥∥ Y ≤ θ (∑m−1 q=1 ( 2kα1r + qα1r )∥∥∥x∥∥∥r X ) ( 1 − ∣∣λ∣∣α2) for all x ∈ X. Therefore ∥∥∥f(x)−mf( x m )∥∥∥ Y ≤ θ (∑m−1 q=1 ( 2kα1r + qα1r )∥∥∥x∥∥∥r X ) ( 1 − ∣∣λ∣∣α2)mα1r (3.10) for all x ∈ X. So ∥∥∥mlf( x ml ) −mpf ( x mp )∥∥∥ Y ≤ p−1∑ j=l ∥∥∥mjf( x mj ) −mj+1f ( x mj+1 )∥∥∥ Y ≤ θ (∑m−1 q=1 ( 2kα1r + qα1r )) ( 1 − ∣∣λ∣∣)mα1r p−1∑ j=l mα2j mα1rj ∥∥∥x∥∥∥r X (3.11) for all nonnegative integers p, l with p > l and all x ∈ X. It follows from (3.11) that the sequence{ mnf ( x mn )} is a cauchy sequence for all x ∈ X. Since Y is complete, the sequence {mnf ( x mn )} coverges. So one can define the mapping φ : X → Y by φ ( x ) := limn→∞m nf ( x mn ) for all x ∈ X. Moreover, letting l = 0 and passing the limit m →∞ in (3.11), we get (3.5). It follows from (3.4) that∥∥∥∥∥φ ( (m + 1) k∑ j=1 xj + yj 2k − k∑ j=1 zj ) − k∑ j=1 φ ( m xj + yj 2k −zj ) − k∑ j=1 φ (xj + yj 2k )∥∥∥∥∥ Y = lim n→∞ ∥∥∥∥∥mn ( f ((m + 1) mn k∑ j=1 xj + yj 2k − 1 mn k∑ j=1 zj ) − k∑ j=1 f ( m mn xj + yj 2k − 1 mn zj ) − k∑ j=1 f ( 1 mn xj + yj 2k ))∥∥∥∥∥ Y ≤ lim n→∞ ∥∥∥∥∥mnλ (( f ( 1 mn k∑ j=1 xj + yj 2k + 1 mn k∑ j=1 zj ) − k∑ j=1 f ( 1 mn xj + yj 2k ) − k∑ j=1 f ( 1 mn zj ))∥∥∥∥∥ Y + lim n→∞ mα2n mα1nr θ( k∑ j=1 ‖xj‖rX + k∑ j=1 ‖yj‖rX + k∑ j=1 ‖zj‖rX) = ∣∣λ∣∣α2 ∥∥∥∥∥φ ( k∑ j=1 xj + yj 2k + k∑ j=1 zj ) − k∑ j=1 φ (xj + yj 2k ) − k∑ j=1 φ ( zj )∥∥∥∥∥ Y (3.12) 8 Int. J. Anal. Appl. (2022), 20:43 for all xj,yj,zj ∈ X for all j = 1 → n.∥∥∥∥∥φ ( (m + 1) k∑ j=1 xj + yj 2k − k∑ j=1 zj ) − k∑ j=1 φ ( m xj + yj 2k −zj ) − k∑ j=1 φ (xj + yj 2k )∥∥∥∥∥ Y ≤ ∣∣λ∣∣α2 ∥∥∥∥∥ Y φ ( k∑ j=1 xj + yj 2k + k∑ j=1 zj ) − k∑ j=1 φ (xj + yj 2k ) − k∑ j=1 φ ( zj )∥∥∥∥∥ Y for all xj,yj,zj ∈ X for all j = 1 → n. So by lemma 3.1 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 (3.5). Then we have ∥∥∥φ(x)−φ′(x)∥∥∥ Y = mα2n‖φ ( x mn ) −φ′ ( x mn ) ‖Y ≤ mα2n (∥∥∥φ( x mn ) − f ( x mn )∥∥∥ Y + ‖φ′ ( x mn ) − f ( x mn )∥∥∥ Y ) ≤ 2.mα2n · ∑m−1 q=1 ( qα1r + 2kα1r )( 1 − ∣∣λ∣∣α2)mα1nr (mα1r −mα2)θ∥∥x∥∥rX (3.13) 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(3.5) as we expected. � Theorem 3.3. Let r < α2 α1 ,m ∈ Z,m > 1, θ be nonngative real number, and let f : X → Y be a mapping such that ∥∥∥∥∥f ( (m + 1) k∑ j=1 xj + yj 2k − k∑ j=1 zj ) − k∑ j=1 f ( m xj + yj 2k −zj ) − k∑ j=1 f (xj + yj 2k )∥∥∥∥∥ Y ≤ ∥∥∥∥∥λ ( f ( k∑ j=1 xj + yj 2k + k∑ j=1 zj ) − k∑ j=1 f (xj + yj 2k ) − k∑ j=1 f ( zj ))∥∥∥∥∥ Y + θ ( k∑ j=1 ∥∥xj∥∥rX + k∑ j=1 ∥∥yj∥∥rX + k∑ j=1 ∥∥zj∥∥rX) (3.14) for all xj,yj,zj ∈ X for all j = 1 → n. Then there exists a unique mapping φ : X → Y such that ∥∥∥f(x)−φ(x)∥∥∥ Y ≤ ∑m−1 q=1 ( qα1r + 2kα1r )( 1 − ∣∣λ∣∣α2)(mα2 −mα1r)θ∥∥x∥∥rX. (3.15) for all x ∈ X. The rest of the proof is similar to the proof of Theorem 3.2. Int. J. Anal. Appl. (2022), 20:43 9 4. Hyers-Ulam-Rassias stability Additive λ-functional inequalities (1.2) in α-homogeneous F-spaces Additive β-functional inequality in complex Banach space Now, we study the solutions of (1.2). Note that for these inequalities, when X is a α1-homogeneous F-spaces and that Y is a α2- homogeneous F-spaces . Under this setting, we can show that the mapping satisfying (1.2) is additive. These results are give in the following. Lemma 4.1. Let m ∈N and a mapping f : Y → Y satisfies ∥∥∥∥∥f ( k∑ j=1 xj + yj 2k + k∑ j=1 zj ) − k∑ j=1 f (xj + yj 2k ) − k∑ j=1 f ( zj )∥∥∥∥∥ Y ≤ ∥∥∥∥∥λ ( f ( (m + 1) k∑ j=1 xj + yj 2k − k∑ j=1 zj ) − k∑ j=1 f ( m xj + yj 2k −zj ) − k∑ j=1 f (xj + yj 2k ))∥∥∥∥∥ Y (4.1) for all xj,yj,zj ∈ X for j = 1 → n, then f : X → Y is additive Proof. Assume that f : X → Y satisfies (4.1). Replacing ( x1, ...,xk,y1, ...,yk,z1, ...,zk ) by ( 0, ..., 0, 0, ..., 0, 0, ..., 0 ) in (4.1), we have ∥∥∥(2k − 1)f(0)∥∥∥ Y ≤ ∥∥∥kλf (0)∥∥∥ Y ≤ 0 (∣∣∣2k − 1∣∣∣α2 − ∣∣∣λk∣∣∣α2)∥∥∥f(0)∥∥∥ Y ≤ 0 So f ( 0 ) = 0. Replacing ( x1, ...,xk,y1, ...,yk,z1, ...,zk ) by ( 0, ..., 0, 0, ..., 0,−z, 0, ..., 0) ) , in (4.1), we get ∥∥∥f(−z)− f(−z)∥∥∥ Y ≤ 0 and so f is an odd mapping. Replacing ( x1, ...,xk,y1, ...,yk,z1, ...,zk ) by ( x1, ...,xk,y1, ...,yk,m · x1+y12k −v1, ...,m · xk+yk 2k −vk ) in (4.1), we have ∥∥∥f((m + 1) k∑ j=1 xj + yj 2k − k∑ j=1 vj ) − k∑ j=1 f ( m xj + yj 2k −vj ) − k∑ j=1 f (xj + yj 2k )∥∥∥ Y ≤ ∥∥∥∥∥λ ( f ( k∑ j=1 xj + yj 2k + k∑ j=1 vj ) − k∑ j=1 f (xj + yj 2k ) − k∑ j=1 f ( vj ))∥∥∥∥∥ Y (4.2) 10 Int. J. Anal. Appl. (2022), 20:43 for all x1, ...,xk,y1, ...,yk,m x1+y1 2k −v1, ...,mxk+yk2k −vk ∈ G. From (4.1) and (4.2) we infer that∥∥∥∥∥f ( k∑ j=1 xj + yj 2k + k∑ j=1 vj ) − k∑ j=1 f (xj + yj 2k ) − k∑ j=1 f ( vj )∥∥∥∥∥ Y ≤ ∥∥∥∥∥λ ( f ( (m + 1) k∑ j=1 xj + yj 2k − k∑ j=1 vj ) − k∑ j=1 f ( m xj + yj 2k −vj ) − k∑ j=1 f (xj + yj 2k ))∥∥∥∥∥ Y ≤ ∥∥∥∥∥λ2 ( f ( k∑ j=1 xj + yj 2k + k∑ j=1 vj ) − k∑ j=1 f (xj + yj 2k ) − k∑ j=1 f ( vj ))∥∥∥∥∥ Y (4.3) and so f ( k∑ j=1 xj + yj 2k + k∑ j=1 zj ) = k∑ j=1 f (xj + yj 2k ) + k∑ j=1 f ( zj ) for all xj,yj,zj ∈ G for j = 1 → n, as we expected. � Theorem 4.2. Let r > α2 α1 ,m ∈ Z,m > 1, θ be nonngative real number, and let f : X → Y be a mapping such that ∥∥∥∥∥f ( k∑ j=1 xj + yj 2k + k∑ j=1 zj ) − k∑ j=1 f (xj + yj 2k ) − k∑ j=1 f ( zj )∥∥∥∥∥ Y ≤ ∥∥∥∥∥λ ( f ( (m + 1) k∑ j=1 xj + yj 2k − k∑ j=1 zj ) − k∑ j=1 f ( m xj + yj 2k −zj ) − k∑ j=1 f (xj + yj 2k ))∥∥∥∥∥ Y + θ ( k∑ j=1 ∥∥xj∥∥rX + k∑ j=1 ∥∥yj∥∥rX + k∑ j=1 ∥∥zj∥∥rX) (4.4) for all xj,yj,zj ∈ X for all j = 1 → n. Then there exists a unique mapping φ : X → Y such that∥∥∥f(x)−h(x)∥∥∥ Y ≤ ∑m−1 q=1 ( qα1r + 2kα1r )( 1 − ∣∣λ∣∣α2)(mα1r −mα2)θ∥∥x∥∥rX. (4.5) for all x ∈ X Proof. Assume that f : X → Y satisfies (4.4). Replacing ( x1, ...,xk,y1, ...,yk,z1, ...,zk ) by ( 0, ..., 0, 0, ..., 0, 0, ..., 0 ) in (4.4), we have e ∥∥∥2kf(0)∥∥∥ Y ≤ ∥∥∥λ(2k − 1)f (0)∥∥∥ Y ≤ 0 therefore (∣∣∣2k∣∣∣α2 − ∣∣∣λ(2k − 1)∣∣∣α2)∥∥∥f(0)∥∥∥ Y ≤ 0 Sof ( 0 ) = 0. Next we replacing ( x1, ...,xk,y1, ...,yk,z1, ...,zk ) by ( kx, 0, ..., 0,kx, 0, ..., 0, 0, ..., 0 ) in (4.4), we get Int. J. Anal. Appl. (2022), 20:43 11 ∥∥∥f((m + 1)x)− f(mx)− f(x)∥∥∥ Y ≤ 2kα1rθ ∥∥∥x∥∥∥r X (4.6) for all x ∈ X. Thus for q ∈N, we replacing ( x1, ...,xk,y1, ...,yk,z1, ...,zk ) by ( kx, 0, ..., 0,kx, 0, ..., 0,qx, 0, ..., 0 ) in (4.4), we have∥∥∥∥∥f((m−q + 1)x)− f((m−q)x)− f(x) ∥∥∥∥∥ Y ≤ ∥∥∥∥∥λ(f((q + 1)x)− f(qx)− f(x) ∥∥∥∥∥ Y + θ ( 2kα1r + qα1r )∥∥∥x∥∥∥r X (4.7) for all x ∈ X. For (4.6) and (4.7) m−1∑ q=1 ∥∥∥f((m−q + 1)x)− f((m−q)x)− f(x)∥∥∥ Y ≤ m−1∑ q=1 ∥∥∥λ(f((q + 1)x)− f(qx)− f(x)∥∥∥ Y + θ (m−1∑ q=1 ( 2kα1r + qα1r )∥∥∥x∥∥∥r X ) (4.8) for all x ∈ X. From (4.7) and (4.8) and triangle inequality, we have ( 1 − ∣∣λ∣∣α2)∥∥∥f(mx)−mf(x)∥∥∥ Y = ( 1 − ∣∣λ∣∣α2)m−1∑ q=1 ∥∥∥f((q + 1)x)− f(qx)− f(x)∥∥∥ Y ≤ m−1∑ q=1 ( 1 − ∣∣λ∣∣α2)∥∥∥(f((q + 1)x)− f(qx)− f(x)∥∥∥ Y ≤ m−1∑ q=1 ∥∥∥f((q + 1)x)− f(qx)− f(x)∥∥∥ Y − m−1∑ q=1 ∥∥∥λ(f((q + 1)x)− f(qx)− f(x))∥∥∥ Y ≤ θ (m−1∑ q=1 ( 2kα1r + qα1r )∥∥∥x∥∥∥r X ) (4.9) for all x ∈ X. from m−1∑ q=1 ∥∥∥f((m−q + 1)x)− f((m−q)x)− f(x)∥∥∥ Y = m−1∑ q=1 ∥∥∥(f((q + 1)x)− f(qx)− f(x)∥∥∥ Y 12 Int. J. Anal. Appl. (2022), 20:43 Since ∣∣λ∣∣ < 1, the mapping f satisfies the inequalities ∥∥∥f(mx)−mf(x)∥∥∥ ≤ θ (∑m−1 q=1 ( 2kα1r + qα1r )∥∥∥x∥∥∥r X ) ( 1 − ∣∣λ∣∣α2) for all x ∈ X. therefore ∥∥∥f(x)−mf( x m )∥∥∥ Y ≤ θ (∑m−1 q=1 ( 2kα1r + qα1r )∥∥∥x∥∥∥r X ) ( 1 − ∣∣λ∣∣α2)mα1r (4.10) for all x ∈ X. So ∥∥∥mlf( x ml ) −mpf ( x mp )∥∥∥ Y ≤ p−1∑ j=l ∥∥∥mjf( x mj ) −mj+1f ( x mj+1 )∥∥∥ Y ≤ θ (∑m−1 q=1 ( 2kr + qr )) ( 1 − ∣∣λ∣∣α2)mα1r p−1∑ j=l mα2j mα1rj ∥∥∥x∥∥∥r X (4.11) for all nonnegative integers p, l with p > l and all x ∈ X. It follows from (4.11) that the sequence{ mnf ( x mn )} is a cauchy sequence for all x ∈ X. Since Y is complete, the sequence {mnf ( x mn )} coverges. So one can define the mapping φ : X → Y by φ ( x ) := limn→∞m nf ( x mn ) for all x ∈ X. Moreover, letting l = 0 and passing the limit m →∞ in (4.11), we get (4.5). It follows from (4.4) that∥∥∥∥∥φ ( k∑ j=1 xj + yj 2k + k∑ j=1 zj ) −φ ( k∑ j=1 xj + yj 2k ) − k∑ j=1 φ ( zj )∥∥∥∥∥ Y = lim n→∞ ∥∥∥∥∥mn ( f ( 1 mn k∑ j=1 xj + yj 2k + 1 mn k∑ j=1 zj ) − f ( 1 mn k∑ j=1 xj + yj 2k ) − k∑ j=1 f ( 1 mn zj ))∥∥∥∥∥ Y + lim n→∞ mα2n mα1nr θ( k∑ j=1 ‖xj‖rX + k∑ j=1 ‖yj‖rX + k∑ j=1 ‖zj‖rX) ≤ lim n→∞ ∣∣λ∣∣α2 ∥∥∥∥∥mn ( f ((m + 1) mn k∑ j=1 xj + yj 2k − 1 mn k∑ j=1 zj ) − k∑ j=1 f ( m mn xj + yj 2k − 1 mn zj ) − k∑ j=1 f ( 1 mn zj ))∥∥∥∥∥ Y = ∣∣λ∣∣α2 ∥∥∥∥∥φ (( m + 1 ) k∑ j=1 xj + yj 2k + k∑ j=1 zj ) − k∑ j=1 φ ( m xj + yj 2k −zj ) − k∑ j=1 φ ( zj )∥∥∥∥∥ Y (4.12) Int. J. Anal. Appl. (2022), 20:43 13 for all xj,yj,zj ∈ X for all j = 1 → n. So∥∥∥∥∥φ ( k∑ j=1 xj + yj 2k + k∑ j=1 zj ) − k∑ j=1 φ (xj + yj 2k ) − k∑ j=1 φ ( zj )∥∥∥∥∥ Y ≤ ∣∣λ∣∣α2 ∥∥∥∥∥φ ( (m + 1) k∑ j=1 xj + yj 2k − k∑ j=1 zj ) − k∑ j=1 φ ( m xj + yj 2k −zj ) − k∑ j=1 φ (xj + yj 2k )∥∥∥∥∥ Y for all xj,yj,zj ∈ 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 (4.5) . Then we have ∥∥∥φ(x)−φ′(x)∥∥∥ Y = mα2n‖φ ( x mn ) −φ′ ( x mn ) ‖Y ≤ mα2n (∥∥∥φ( x mn ) − f ( x mn )∥∥∥ Y + ‖φ′ ( x mn ) − f ( x mn )∥∥∥ Y ) ≤ 2.mα2n · ∑m−1 q=1 ( qα1r + 2kα1r )( 1 − ∣∣λ∣∣α2)mnα1r (mα1r −mα2)θ∥∥x∥∥rX (4.13) 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(4.5) as we expected. � Theorem 4.3. Let r < α2 α1 ,m ∈ Z,m > 1, θ be nonngative real number, and let f : X → Y be a mapping such that ∥∥∥∥∥f ( k∑ j=1 xj + yj 2k + k∑ j=1 zj ) − k∑ j=1 f (xj + yj 2k ) − k∑ j=1 f ( zj )∥∥∥∥∥ Y ≤ ∥∥∥∥∥λ ( f ( (m + 1) k∑ j=1 xj + yj 2k − k∑ j=1 zj ) − k∑ j=1 f ( m xj + yj 2k −zj ) − k∑ j=1 f (xj + yj 2k ))∥∥∥∥∥ Y + θ ( k∑ j=1 ∥∥xj∥∥rX + k∑ j=1 ∥∥yj∥∥rX + k∑ j=1 ∥∥zj∥∥rX) (4.14) for all xj,yj,zj ∈ X for all j = 1 → n. Then there exists a unique mapping φ : X → Y such that ∥∥∥f(x)−h(x)∥∥∥ Y ≤ ∑m−1 q=1 ( qα1r + 2kα1r )( 1 − ∣∣λ∣∣α2)(mα2 −mα1r )θ∥∥x∥∥rX. (4.15) for all x ∈ X. The proof is similar to theorem 4.2. 14 Int. J. Anal. Appl. (2022), 20:43 5. Conclusion In this paper, I have shown that the solutions of the first and second k − variable β-functional inequalities are additive mappings. The Hyers-Ulam stability for these given from theorems. These are the main results of the paper , which are the generalization of the results [2], [11]. 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