Some Remarks Concerning the General Octic Functional Equation

In this article, we study the stability of various forms for the general octic functional equation 􏽐 9 i � 09 C i (− 1 ) 9 − i f ( x + iy ) � 0 . We frst fnd a special way of representing a given mapping as the sum of eight mappings. And by using the above representation, we will investigate the hyperstability of the general octic functional equation. Furthermore, we will discuss the Hyers–Ulam–Rassias stability of the general octic functional equation.


Introduction
Let X be a real normed space and Y be a real Banach space throughout this article.Consider the functional equation Tis functional equation is called as Jensen, general quadratic, general cubic, general quartic, general quintic, general sextic, general septic, and general octic functional equation, respectively, for n � 2, 3, 4, 5, 6, 7, 8, 9.And a function f: R ⟶ R given by f(x) ≔  n−1 i�0 a i x i is a particular solution of equation (1), when a monomial mapping f(x) ≔ ax n is a particular solution of the n-monomial functional equation. ( Te n-monomial functional equation is called as Cauchy, quadratic, cubic, quartic, quintic, sextic, septic, and octic functional equation, respectively, for n � 1, 2, 3, 4, 5, 6, 7, 8.
We let for all x, y ∈ X.Since D n f(x, y) � C n−1 f(x, x + y)− C n−1 f(x, y) and D n f(x, y) � D n−1 f(x, x + y) − D n−1 f(x, y) for all x, y ∈ X, if f: X ⟶ Y satisfes the functional equation C n−1 f(x, y) � 0 or D n−1 f(x, y) � 0, then f satisfes the functional equation D n f(x, y) � 0. Terefore, we say that equation ( 1) is a more general functional equation than the n-monomial functional (2).For additional information one can refer to [1].Actually, the general octic functional equation is a generalization of Cauchy, Jensen, quadratic, general quadratic, cubic, general cubic, quartic, general quartic, quintic, general quintic, sextic, general sextic, septic, general septic, and octic functional equation.Te stability of equations is a very important topic in the feld of mathematical analysis.Te stability of various forms for functional equation has been developed by Ulam [2], Hyers [3], Rassias [4], and Gȃvruta [5].Specially, Gilányi [6] investigated the stability of the monomial functional equations in real normed spaces.And Cȃdariu and Radu [7], Kaiser [8], and Lee [9] generalized the results of Gilányi [6].Moreover, one can refer to [9][10][11] for the hyperstability of the monomial functional equation.Te given functional equation is called hyperstable if any mapping satisfying the given equation approximately (in some sense) must be actually a solution to it (refer to [12]).
In this article, we investigated the hyperstability and Hyers-Ulam-Rassias stability of the general octic functional equation.
In Section 2, we will show that any given mapping f is expressed as the sum of eight mappings f 1 , f 2 , f 3 , f 4 , f 5 , f 6 , f 7 , and f 8 derived from f. Te expression of the mapping f in this way is an essential preliminary step in the proof of the hyperstability and stability of the general octic functional equation, which will be shown in Sections 3 and 4.
In Section 3, we will show the hyperstability of the functional general octic equation as follows: Let p be a given negative real number.If f: X ⟶ Y satisfes the inequality for all x, y ∈ X/ 0 { }, then  9 i�0 9 C i (−1) 9− i f(ix + y) � 0 for all x, y ∈ X.
In Section 4, we will show the stability of the functional general octic equation as follows: Let p ≠ 1, 2, 3, 4, 5, 6, 7, 8 be a given non-negative real number.If f: X ⟶ Y satisfes inequality (4) for all x, y ∈ X, then there exists a unique mapping F satisfying for all x ∈ X, where M is some constant.
As an example, we will consider 2 < p < 3 and see that F(x) can only be obtained as the limit of

Representation of a Mapping
In this section, we fnd a way of representing a given mapping f as the sum of eight mappings f 1 , f 2 , f 3 , f 4 , f 5 , f 6 , f 7 , and f 8 derived from f.
For a given mapping f: X ⟶ Y, we denote the following equation: And for a given mapping f: X ⟶ Y, let us consider the following system of nonhomogeneous linear equations: for all x ∈ X. Ten, we obtain the following two lemmas by the uniqueness of solution stated in Cramer's rule.
Lemma 1.Let f: X ⟶ Y be a given mapping and Ten, we have the following equation: Ten, for all x ∈ X, we have the following equation: From Lemma 1 and Lemma 2, the mappings f 1 , f 2 , f 3 , f 4 , f 5 , f 6 , f 7 , and f 8 used to prove the main theorem in Sections 3 and 4 can be calculated and summarized as follows.
and the following equation: holds for all x ∈ X.
Next, we will defne the mappings needed to prove the main theorems.
Defnition 1.For a given mapping f: X ⟶ Y, we defne the following equation: for all x, y ∈ X.
Next, by tedious computations, one can obtain the following lemma: Lemma 4. For a given mapping f: X ⟶ Y, the following equalities: hold for all x, y ∈ X, where f i (x) is the mapping in Lemma 3.
Proof.We omit the proof. □

Hyperstability of the General Octic Functional Equation
In this section, we will prove the hyperstability of the general octic functional equation.For the proof, we will use the mappings , and f 8 given in Lemma 3 and Lemma 4.
Step 3. By letting n � 0 and passing the limit m ⟶ ∞ in equation (36), we get the following inequality: and DF(x, y) � 0 for all x, y ∈ X, we have the following equality for any n ∈ N: for all x ∈ X\ 0 { }.Terefore, by p < 0 and equations ( 22) and (41), we have the following equation:

Hyers-Ulam-Rassias Stability of the General Octic Functional Equation
In this section, we will investigate the Hyers-Ulam-Rassias stability of the general octic functional equation, as follows: Lemma .If g: X ⟶ Y is a mapping such that Dg(x, y) � 0 for all x, y ∈ X with g(0) � 0, then for each i ∈ 1, 2, . . ., 8 { }, g i (x) in Lemma 3 satisfes the following equality: for all x, y ∈ X.
Case 1.If 0 ≤ p < 1, then it follows from Lemma 4 and equation (49) that for all x ∈ X.So, we obtain that for all x ∈ X and n, m ∈ N ∪ 0 { }.It follows from 0 ≤ p < 1 and equation (52) that the sequence  f 1 (2 n x)/2 n   is a Cauchy sequence for all x ∈ X.Since Y is complete, the sequence  f 1 (2 n x)/2 n   converges for all x ∈ X.Hence, we can defne the mapping F 1 : X ⟶ Y by the following equation: Journal of Mathematics � 0 and passing the limit m ⟶ ∞ in equation (52), we get the following inequality: for all x ∈ X.From the defnition of F 1 , we easily get F 1 (2x) � 2F 1 (x) for all x ∈ X and for all x, y ∈ X. Case 2. If p > 1, then it follows from Lemma 4 and equation (49) that for all x ∈ X.So, we have the following equation: for all x ∈ X and n, m ∈ N ∪ 0 { }.It follows from p > 1 and equation (57) that the sequence converges for all x ∈ X.Hence, we can defne the mapping F 1 : X ⟶ Y by the following equation: Moreover, letting n � 0 and passing the limit m ⟶ ∞ in equation ( 57), we get the following inequality: for all x ∈ X.From the defnition of F 1 , we easily get F 1 (2x) � 2F 1 (x) for all x ∈ X and DF 1 (x, y) � 0 for all x, y ∈ X.Hence, depending on the value of p, we can choose the defnition of F 1 (x) as the mapping F 1 (x) in equation ( 53) or equation ( 58).
Case 3. If p < 2, then it follows from Lemma 4 and equation (50) that for all x ∈ X and n, m ∈ N ∪ 0 { }.It follows from p < 2 and equation (60) that the sequence  f 2 (2 n x)/4 n   is a Cauchy sequence for all x ∈ X.Since Y is complete, the sequence  f 2 (2 n x)/4 n   converges for all x ∈ X.Hence, we can defne the mapping F 2 : X ⟶ Y by Moreover, letting n � 0 and passing the limit m ⟶ ∞ in equation (60), we get the following inequality: for all x ∈ X.From the defnition of F 2 , we easily get F 2 (2x) � 4F 2 (x) for all x ∈ X and DF 2 (x, y) � 0 for all x, y ∈ X.
for all x ∈ X and n, m ∈ N ∪ 0 { }.It follows from p > 2 and equation ( 63) that the sequence 4 n  f 2 (2 − n x) is a Cauchy sequence for all x ∈ X.Since Y is complete, the sequence converges for all x ∈ X.Hence, we can defne the mapping F 2 : X ⟶ Y by the following equation: Moreover, letting n � 0 and passing the limit m ⟶ ∞ in equation ( 63), we get the following inequality: for all x ∈ X.From the defnition of F 2 , we easily get F 2 (2x) � 4F 2 (x) for all x ∈ X and DF 2 (x, y) � 0 for all x, y ∈ X.Hence, depending on the value of p, we can choose the defnition of F 2 (x) as the mapping F 2 (x) in equation (61) or equation (64).
Case 5.If p < 3, then it follows from Lemma 4 and equation (49) that for all x ∈ X and n, m ∈ N ∪ 0 { }.It follows from p < 3 and equation (66) that the sequence  f 3 (2 n x)/8 n   is a Cauchy sequence for all x ∈ X.Since Y is complete, the sequence  f 3 (2 n x)/8 n   converges for all x ∈ X.Hence, we can defne the mapping F 3 : X ⟶ Y by the following equation: Moreover, letting n � 0 and passing the limit m ⟶ ∞ in equation (66), we get the following inequality: for all x ∈ X.From the defnition of F 3 , we easily get F 3 (2x) � 8F 3 (x) for all x ∈ X and DF 3 (x, y) � 0 for all x, y ∈ X.
Case 6.If p > 3, then it follows from Lemma 4 and equation (49) that for all x ∈ X and n, m ∈ N ∪ 0 { }.It follows from p > 3 and equation (69) that the sequence converges for all x ∈ X.Hence, we can defne the mapping F 3 : X ⟶ Y by the following equation: Moreover, letting n � 0 and passing the limit m ⟶ ∞ in equation (69), we get the following inequality: for all x ∈ X.From the defnition of F 3 , we easily get F 3 (2x) � 8F 3 (x) for all x ∈ X and DF 3 (x, y) � 0 for all x, y ∈ X.Hence, depending on the value of p, we can choose the defnition of F 3 (x) as the mapping F 3 (x) in equation (67) or equation (70).
Case 7. If p < 4, then we can defne the mapping F 4 : X ⟶ Y by the following equation: 16 n for all x ∈ X. (72) Moreover, we get the following inequality: for all x ∈ X.From the defnition of F 4 , we easily get F 4 (2x) � 16F 4 (x) for all x ∈ X and DF 4 (x, y) � 0 for all x, y ∈ X.
Case 8.If p > 4, then we can defne the mapping F 4 : X ⟶ Y by the following equation: Moreover, we get the following inequality: for all x ∈ X.From the defnition of F 4 , we easily get F 4 (2x) � 16F 4 (x) for all x ∈ X and DF 4 (x, y) � 0 for all x, y ∈ X.Hence, depending on the value of p, we can choose the defnition of F 4 (x) as the mapping F 4 (x) in equation (72) or equation (74).Case 9.If p < 5, then we can defne the mapping F 5 : X ⟶ Y by the following equation: 32 n for all x ∈ X. (76) Moreover, we get the following inequality: for all x ∈ X.From the defnition of F 5 , we easily get F 5 (2x) � 32F 5 (x) for all x ∈ X and DF 5 (x, y) � 0 for all x, y ∈ X.

Journal of Mathematics
p > 5, then we can defne the mapping F 5 : X ⟶ Y by the following equation: Moreover, we get the following inequality: for all x ∈ X.From the defnition of F 5 , we easily get F 5 (2x) � 32F 5 (x) for all x ∈ X and DF 5 (x, y) � 0 for all x, y ∈ X.
Hence, depending on the value of p, we can choose the defnition of F 5 (x) as the mapping F 5 (x) in equation ( 76) or equation (78).
Case 11.If p < 6, then we can defne the mapping F 6 : X ⟶ Y by the following equation: 64 n for all x ∈ X. (80) Moreover, we get the following inequality: for all x ∈ X.From the defnition of F 6 , we easily get F 6 (2x) � 64F 6 (x) for all x ∈ X and DF 6 (x, y) � 0 for all x, y ∈ X.
Case 12.If p > 6, then we can defne the mapping F 6 : X ⟶ Y by the following equation: Moreover, we get the following inequality: for all x ∈ X.From the defnition of F 6 , we easily get F 6 (2x) � 64F 6 (x) for all x ∈ X and DF 6 (x, y) � 0 for all x, y ∈ X.Hence, depending on the value of p, we can choose the defnition of F 6 (x) as the mapping F 6 (x) in equation (80) or equation (82).
Case 13.If p < 7, then we can defne the mapping F 7 : X ⟶ Y by the following equation: 128 n for all x ∈ X. (84) Moreover, we get the following inequality: for all x ∈ X.From the defnition of F 7 , we easily get F 7 (2x) � 128F 7 (x) for all x ∈ X and DF 7 (x, y) � 0 for all x, y ∈ X.
Case 14.If p > 7, then we can defne the mapping F 7 : X ⟶ Y by the following equation: Moreover, we get the following inequality: for all x ∈ X.From the defnition of F 7 , we easily get F 7 (2x) � 128F 7 (x) for all x ∈ X and DF 7 (x, y) � 0 for all x, y ∈ X.
Hence, depending on the value of p, we can choose the defnition of F 7 (x) as the mapping F 7 (x) in equation (84) or equation (86).
Case 15.If p < 8, then we can defne the mapping F 8 : X ⟶ Y by the following equation: 256 n for all x ∈ X. (88) Moreover, we get the following inequality: for all x ∈ X.From the defnition of F 8 , we easily get F 8 (2x) � 256F 8 (x) for all x ∈ X and DF 8 (x, y) � 0 for all x, y ∈ X.
Case 16.If p > 8, then we can defne the mapping F 8 : X ⟶ Y by the following equation: Moreover, we get the following inequality: for all x ∈ X.From the defnition of F 8 , we easily get F 8 (2x) � 256F 8 (x) for all x ∈ X and DF 8 (x, y) � 0 for all x, y ∈ X.Hence, depending on the value of p, we can choose the defnition of F 8 (x) as the mapping F 8 (x) in equation (88) or equation (90).
At last, we need to prove the uniqueness of F. Suppose that F ′ : X ⟶ Y be another mapping satisfying DF ′ (x, y) � 0 for all x, y ∈ X and inequality (49) for all x ∈ X with F ′ (0) � 0. By Lemma 5, the mappings F 1 ′ , . . ., F 8 ′ : X ⟶ Y satisfy the following equation: To prove the uniqueness of F(x), we will assume 2 < p < 3 for a given p.For other cases, one can prove by a similar method.If 2 < p < 3, then we have the following equation: ( Taking the limit in the above inequalities as n ⟶ ∞, we obtain the equality which means that F 2 (x) � F 2 ′ (x) for all x ∈ X.Also, we obtain the following equation: for all x ∈ X and all positive integer n.Taking the limit in the above inequalities as n ⟶ ∞, we obtain the equality x)/8 n for all x ∈ X, which means that F 3 (x) � F 3 ′ (x) for all x ∈ X.
And we easily show that F 1 � F 1 ′ , F 4 � F 4 ′ , F 5 � F 5 ′ , F 6 � F 6 ′ , F 7 � F 7 ′ , and F 8 � F 8 ′ by the similar method.Since F(x) �  8 i�1 F i (x) �  8 i�1 F i ′ (x) � F ′ (x), we complete the proof of the uniqueness of F for the case 2 < p < 3.For the other cases, we can easily prove the uniqueness of F in much the same way as the proof for the case 2 < p < 3.

Conclusion
In this article, we discussed the hyperstability and Hyers-Ulam-Rassias stability of the general octic functional equation.To prove main theorems, we proved that any given mapping f can be expressed as the sum of eight mappings f 1 , f 2 , f 3 , f 4 , f 5 , f 6 , f 7 , and f 8 derived from f. Te expression of the mapping f in this way is an essential preliminary step in the proof of the stabilities of the general octic functional equation, which are shown in Sections 3 and 4. In particular, we proved that f i (2x) � 2 i f(x) holds for i � 1, 2, 3, 4, 5, 6, 7, 8 if the mapping f satisfes the general octic functional equation.Tis means that f 1 , f 2 , f 3 , f 4 , f 5 , f 6 , f 7 , and f 8 are additive, quadratic, cubic, quartic, quintic, sextic, septic, and octic mapping, respectively, if the mapping f satisfes the general octic functional equation.Not much research has been conducted in the general octic functional equation.Te big advantage of this article is that we proved the uniqueness of the solution in the Hyers-Ulam-Rassias stability of the general octic functional equation.Te uniqueness of the solution in the Hyers-Ulam-Rassias stability of the monomial functional equation has been discussed in many researches.But the uniqueness of the solution in the Hyers-Ulam-Rassias stability of the general functional equations is a more complicated problem.As an example, for 2 < p < 3, the general octic mapping F satisfying inequality (53) can be expressed only as the limit of the following function: as n ⟶ ∞, while one obtain F(x) � lim n ⟶ ∞ f(2 n x)/256 n in the case of stability of the octic functional equation.As a follow-up research, we studied the stability of the general nonic functional equation.Finally, we will try to get generalized results of the stability for the general n-th functional equation.

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Journal of Mathematics we have D  f(x, y) � DF(x, y) � 0 for all x, y ∈ X.And by the fact, D  f(x, y) � Df(x, y), we have Df(x, y) � D  f(x, y) � DF(x, y) � 0 to complete the proof.