# General Cubic-Quartic Functional Equation

Authors: M. Eshaghi Gordji; M. Kamyar; Th. M. Rassias
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We obtain the general solution and the generalized Hyers-Ulam stability of the general cubic-quartic functional equation for fixed integersk with k≠0,±1: f(x+ky)+f(x−ky)=k2(f(x+y)+f(x−y))+2(1−k2)f(x)+((k4−k2)/4)(f(2y)−8f(y))+f˜(2x)−16f˜(x), where f˜(x):=f(x)+f(−x).


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## Body

## 1. Introduction

The stability problem of functional equations originated from a question of Ulam [1] in 1940, concerning the stability of group homomorphisms. Let (G1,·) be a group and let (G2,*) be a metric group with the metric d(·,·). Given ϵ>0, does there exist a δ>0, such that if a mapping h:G1→G2 satisfies the inequality d(h(x·y),h(x)*h(y))<δ for all x,y∈G1, then there exists a homomorphism H:G1→G2 with d(h(x),H(x))<ϵ for all x∈G1? In other words, under what condition does there exists a homomorphism near an approximate homomorphism? The concept of stability for functional equation arises when we replace the functional equation by an inequality which acts as a perturbation of the equation. In 1941, Hyers [2] gave the first affirmative answer to the question of Ulam for Banach spaces. Let f:E→E′ be a mapping between Banach spaces such that
(1.1)‖f(x+y)-f(x)-f(y)‖≤δ
for all x,y∈E and for some δ>0. Then there exists a unique additive mapping T:E→E′ such that (1.2)‖f(x)-T(x)‖≤δ
for all x∈E. Moreover, if f(tx) is continuous in t∈ℝ for each fixed x∈E, then T is linear. In 1978, Rassias [3] proved the following theorem.Theorem 1.1.
Letf:E→E′ be a mapping from a normed vector space  E into a Banach space  E′ subject to the inequality
(1.3)‖f(x+y)-f(x)-f(y)‖≤ϵ(‖x‖p+‖y‖p)
for all x,y∈E, where ϵ and p are constants with ϵ>0 and p<1. Then there exists a unique additive mapping T:E→E′ such that
(1.4)‖f(x)-T(x)‖≤2ϵ2-2p‖x‖p
for all x∈E. If p<0 then inequality (1.3) holds for all x,y≠0 and (1.4) forx≠0. Also, if the function t↦f(tx) from ℝ into E′ is continuous in real t for each fixed x∈E, then T is linear.In 1990, Rassias during the 27th International Symposium on Functional Equations asked the question whether such a Theorem can also be proved for all real values ofp that are greater or equal to one. In 1991, Gajda [4], following the same approach as that of Rassias, provided an affirmative solution to this question for all real values of p that are strictly greater than one. The new concept of stability of the linear mapping that was inspired by Rassias' stability theorem is called Hyers-Ulam-Rassias stability of functional equations.Jun and Kim [5] introduced the following cubic functional equation: (1.5)f(2x+y)+f(2x-y)=2f(x+y)+2f(x-y)+12f(x),
and they established the general solution and the generalized Hyers-Ulam-Rassias stability for the functional equation (1.5). The function f(x)=x3 satisfies the functional equation (1.5), which is thus called a cubic functional equation. Every solution of the cubic functional equation is said to be a cubic function. Jun and Kim proved that a function f between real vector spaces X and Y is a solution of (1.5) if and only if there exists a unique function C:X×X×X→Y such that f(x)=C(x,x,x) for all x∈X and C is symmetric for each fixed one variable and is additive for fixed two variables. The stability of the quartic functional equations was studied by Park and Bae [6], when (1.6)f(x+2y)+f(x-2y)=4(f(x+y)+f(x-y))+24f(y)-6f(x).
In fact, they proved that a function f between real vector spaces X and Y is a solution of (1.6) if and only if there exists a unique symmetric multi-additive function Q:X×X×X×X→Y such that f(x)=Q(x,x,x,x) for all x∈X (see also [7, 8]). It is straightforward to verify that the function f(x)=x4 satisfies the functional equation (1.6), which is called a quartic functional equation and every solution of the quartic functional equation is said to be a quartic function.The stability problems of several functional equations have been extensively investigated by a number of authors, and there are many interesting results concerning this problem. (see [9–45]).In 2008, Gordji et al. [17] provided the solution as well as the stability of a mixed type cubic-quartic functional equation. We only mention here the papers [19, 32, 33] concerning the stability of the mixed type functional equations.In this paper, we deal with the following general cubic-quartic functional equation:(1.7)f(x+ky)+f(x-ky)=k2(f(x+y)+f(x-y))+2(1-k2)f(x)+k4-k24×(f(2y)-8f(y))+f̃(2x)-16f̃(x),wheref̃(x):=f(x)+f(-x).
Then it follows easily that the function f(x)=ax4+bx3 satisfies (1.7). We investigate the general solution and the generalized Hyers-Ulam-Rassias stability of the functional equation (1.7).


## 2. General Solution

In this section, we establish the general solution of functional equation (1.7).Theorem 2.1.
LetX,  Y be vector spaces and let  f:X→Y be a function. Then  f satisfies  (1.7) if and only if there exists a unique symmetric multiadditive function  Q:X×X×X×X→Y and a unique function  C:X×X×X→Y such that  f(x)=Q(x,x,x,x)+C(x,x,x) for all  x∈X, where the function  C is symmetric for each fixed one variable and is additive for fixed two variables.Proof.
Letf satisfies (1.7). We decompose f into the even part and odd part by setting
(2.1)fe(x)=12(f(x)+f(-x)),fo(x)=12(f(x)-f(-x))
for all x∈X. By (1.7), we have
(2.2)fe(x+ky)+fe(x-ky)=12[f(x+ky)+f(-x-ky)+f(x-ky)+f(-x+ky)]=12[f(x+ky)+f(x-ky)]+12[f((-x)+(-ky))+f((-x)-(-ky))]=12[k4-k24k2(f(x+y)+f(x-y))+2(1-k2)f(x)+k4-k24(f(2y)-8f(y))+f̃(2x)-16f̃(x)]+12[k2(f(-x-y)+f(-x+y))+2(1-k2)f(-x)+k4-k24×(f(-2y)-8f(-y))+f̃(-2x)-16f̃(-x)k4-k24]=k2[12(f(x+y)+f(-(x+y)))]+k2[12(f(x-y)+f(-(x-y)))]+2(1-k2)[12(f(x)+f(-x))]+k4-k24[12(f(2y)+f(-2y))]-k4-k24[12(8f(y)+8f(-y))]+[12(f̃(2x)+f̃(-2x))]-16[12(f̃(x)+f̃(-x))]=k2(fe(x+y)+fe(x-y))+2(1-k2)fe(x)+k4-k24(fe(2y)-8fe(y))+f̃e(2x)-16f̃e(x)
for all x,y∈X. This means that fe satisfies (1.7), or
(1.5(e))fe(x+ky)+fe(x-ky)=k2(fe(x+y)+fe(x-y))+2(1-k2)fe(x)+k4-k24(fe(2y)-8fe(y))+f̃e(2x)-16f̃e(x)
for all x,y∈X. Applying the fact that the function fe is even for all x,y∈X, 1.5(e) can be written in the form
(2.3)fe(x+ky)+fe(x-ky)=k2(fe(x+y)+fe(x-y))+2(1-k2)fe(x)+k4-k24(fe(2y)-8fe(y))+2fe(2x)-32fe(x)
for all x,y∈X. Now be setting x=y=0 in (2.3), we get fe(0)=0. Similarly, by setting y=0 in (2.3), we obtain
(2.4)fe(2x)=16fe(x)
for all x∈X. Hence (2.3) can be written as
(2.5)fe(x+ky)+fe(x-ky)=k2(fe(x+y)+fe(x-y))+2(1-k2)fe(x)+2(k4-k2)fe(y)
for all x,y∈X. By substituting x by x+y in (2.5), we have
(2.6)fe(x+(1+k)y)+fe(x+(1-k)y)=k2(fe(x+2y)+fe(x))+2(1-k2)fe(x+y)+2(k4-k2)fe(y)
for all x,y∈X. Substituting -y for y in (2.6), we get by evenness of f(2.7)fe(x-(1+k)y)+fe(x-(1-k)y)=k2(fe(x-2y)+fe(x))+2(1-k2)fe(x-y)+2(k4-k2)fe(y)
for all x,y∈X. Adding (2.6) to (2.7), we obtain
(2.8)fe(x+(1+k)y)+fe(x+(1-k)y)+fe(x-(1+k)y)+fe(x-(1-k)y)=k2(fe(x+2y)+fe(x-2y))+2k2fe(x)+2(1-k2)(fe(x+y)+fe(x-y))+4(k4-k2)fe(y)
for all x,y∈X. By substituting x by x-ky in (2.5), we have
(2.9)fe(x)+fe(x-2ky)=k2(fe(x+(1-k)y)+fe(x-(k+1)y))+2(1-k2)fe(x-ky)+2(k4-k2)fe(y)
for all x,y∈X. Substituting -x for x in (2.9), we get by evenness of fe(2.10)fe(x)+fe(x+2ky)=k2(fe(x+(k-1)y)+fe(x+(k+1)y))+2(1-k2)fe(x+ky)+2(k4-k2)fe(y)
for all x,y∈X. Adding (2.9) to (2.10), we obtain
(2.11)fe(x+2ky)+fe(x-2ky)=k2(fe(x+(1-k)y)+fe(x-(k+1)y)+fe(x+(k-1)y)+fe(x+(k+1)y))+2(1-k2)(fe(x-ky)+fe(x+ky))+4(k4-k2)fe(y)-2fe(x)
for all x,y∈X. Now, by using (2.5), (2.8), and (2.11), we lead to
(2.12)fe(x+2ky)+fe(x-2ky)=k4(fe(x+2y)+fe(x-2y))+4k2(1-k2)(fe(x+y)+fe(x-y))+8(k4-k2)fe(y)+(6k4-8k2+2)fe(x)
for all x,y∈X. If we replace y by 2y in (2.5), we get
(2.13)fe(x+2ky)+fe(x-2ky)=k2(fe(x+2y)+fe(x-2y))+2(1-k2)fe(x)+2(k4-k2)fe(2y)
for all x,y∈X. It follows from (2.12) and (2.13) that
(2.14)k4(fe(x+2y)+fe(x-2y))+4k2(1-k2)(fe(x+y)+fe(x-y))+8(k4-k2)fe(y)+(6k4-8k2+2)fe(x)=k2(fe(x+2y)+fe(x-2y))+2(1-k2)fe(x)+2(k4-k2)fe(2y)
for all x,y∈X. So we have
(2.15)fe(x+2y)+fe(x-2y)=4(fe(x+y)+fe(x-y))+24fe(y)-6fe(x)
for all x,y∈X. This means that fe is a quartic function. Thus there exists a unique symmetric multiadditive function Q:X×X×X×X→Y such that fe(x)=Q(x,x,x,x) for all x∈X. On the other hand, we can show that fo satisfies (1.7), or
(1.5(o))fo(x+ky)+fo(x-ky)=k2(fo(x+y)+fo(x-y))+2(1-k2)fo(x)+k4-k24(fo(2y)-8fo(y))+f̃o(2x)-16f̃o(x)
for all x,y∈X. By oddness of fo for all x,y∈X, 1.5(o) can be written as
(2.16)fo(x+ky)+fo(x-ky)=k2(fo(x+y)+fo(x-y))+2(1-k2)fo(x)+k4-k24(fo(2y)-8fo(y))
for all x,y∈X. Now by setting x=y=0 in (3.2), we get fo(0)=0, and by setting x=0 in (2.16), we obtain
(2.17)fo(2y)=8fo(y)
for all y∈X. Hence (2.16) can be written as
(2.18)fo(x+ky)+fo(x-ky)=k2(fo(x+y)+fo(x-y))+2(1-k2)fo(x)
for all x,y∈X. Replacing x by x-y in (2.18), we obtain
(2.19)fo(x+(k-1)y)+fo(x-(k+1)y)=k2(fo(x-2y)+fo(x))+2(1-k2)fo(x-y)
for all x,y∈X. Substituting -x for x in (2.19), we get by oddness of fo(2.20)-fo(x+(1-k)y)-fo(x+(k+1)y)=k2(-fo(x+2y)-fo(x))-2(1-k2)fo(x+y)
for all x,y∈X. If we subtract (2.19) from (2.20), we obtain
(2.21)fo(x+(k-1)y)+fo(x-(k+1)y)+fo(x+(1-k)y)+fo(x+(k+1)y)=k2(fo(x+2y)+fo(x-2y))+2k2fo(x)+2(1-k2)(fo(x+y)+fo(x-y))
for all x,y∈X. By substituting x by x+ky in (2.18), we have
(2.22)fo(x)+fo(x+2ky)=k2(fo(x+(k+1)y)+fo(x+(k-1)y))+2(1-k2)fo(x+ky)
for all x,y∈X. Substituting -y for y in (2.22), we get
(2.23)fo(x)+fo(x-2ky)=k2(fo(x-(k+1)y)+fo(x-(k-1)y))+2(1-k2)fo(x-ky)
for all x,y∈X. Adding (2.22) to (2.23), we obtain
(2.24)fo(x+2ky)+fo(x-2ky)=k2(fo(x+(k+1)y)+fo(x+(k-1)y)+fo(x-(k+1)y)+fo(x-(k-1)y))+2(1-k2)(fo(x+ky)+fo(x-ky))-2fo(x)
for all x,y∈X. Now, by using (2.18), (2.21), and (2.24), we lead to
(2.25)fo(x+2ky)+fo(x-2ky)=4k2(1-k2)(fo(x+y)+fo(x-y))+(6k4-8k2+2)fo(x)+k4(fo(x+2y)+fo(x-2y))
for all x,y∈X. If we replace y by 2y in (2.18), we get
(2.26)fo(x+2ky)+fo(x-2ky)=k2(fo(x+2y)+fo(x-2y))+2(1-k2)fo(x)
for all x,y∈X. If we compare (2.25) with (2.26), then we conclude that
(2.27)fo(x+2y)+fo(x-2y)=4(fo(x+y)+fo(x-y))-6fo(x)
for all x,y∈X. Replacing x by 2x in (2.27), we get
(2.28)fo(2(x+y))+fo(2(x-y))=4(fo(2x+y)+fo(2x-y))-6fo(2x)
for all x,y∈X. Finally, it follows from (2.17) and (2.28) that
(2.29)8(fo(x+y)+fo(x-y))=4(fo(2x+y)+fo(2x-y))-48fo(x)
for all x,y∈X. By multiplying both sides of (2.29) by 1/4, we get
(2.30)2(fo(x+y)+fo(x-y))=(fo(2x+y)+fo(2x-y))-12fo(x)
for all x,y∈X. This means that fo is a cubic function and that there exits a unique function C:X×X×X→Y such that fo(x)=C(x,x,x) for all x∈X and C is symmetric for each fixed one variable and is additive for fixed two variables. Thus for all x∈X, we have
(2.31)f(x)=fe(x)+fo(x)=C(x,x,x)+Q(x,x,x,x).
The proof of the converse is trivially.The following corollary is an alternative result of above Theorem2.1.Corollary 2.2.
LetX,  Y be vector spaces, and let  f:X→Y be a function satisfying  (1.7). Then the following assertions hold. (a)
Iff is even function, then f is quartic.(b)
Iff is odd function, then f is cubic.


## 3. Stability

We will investigate the generalized Hyers-Ulam-Rassias stability problem for the functional equation (1.7). In the following, let X be a real vector space and let Y be a Banach space. Given f:X→Y, we define the difference operator Df:X×X→Y by (3.1)Df(x,y)=f(x+ky)+f(x-ky)-k2(f(x+y)+f(x-y))-2(1-k2)f(x)-k4-k24(f(2y)-8f(y))-f̃(2x)+16f̃(x)
for all x,y∈X.Theorem 3.1.
Letj∈{-1,1} be fixed and let  φ:X×X→[0,∞) be a function such that
(3.2)∑i=(1+j)/2∞k4ijφ(xkij,ykij)<∞
for all x,y∈X. Suppose that an even function f:X→Y with f(0)=0 satisfies the inequality
(3.3)‖Df(x,y)‖≤φ(x,y)
for all x,y∈X. Then the limit
(3.4)Q(x):=lim⁡n→∞k4njf(xknj)
exists for all x∈X and Q:X→Y is a unique quartic function satisfying
(3.5)‖f(x)-Q(x)‖≤1k4ψ̃e(x)
for all x∈X, where
(3.6)ψ̃e(x)=∑i=(1+j)/2∞k4ij[12φ(0,xkij)+k4-k216φ(xkij,0)].Proof.
Letj=1. It follows from (3.3) and using evenness of f that
(3.7)‖k4-k24f(x+ky)+f(x-ky)-k2(f(x+y)+f(x-y))-2(1-k2)f(x)-k4-k24(f(2y)-8f(y))-2f(2x)+32f(x)‖≤φ(x,y)
for all x,y∈X. Replacing x and y by 0 and x in (3.7), respectively, we see that
(3.8)‖2f(kx)+(2k4-4k2)f(x)+k2-k44f(2x)‖≤φ(0,x)
for all x∈X. If we divide both sides of (3.8) by 2, we get
(3.9)‖f(kx)+(k4-2k2)f(x)+k2-k48f(2x)‖≤12φ(0,x)
for all x∈X. Putting y=0 in (3.7), we obtain
(3.10)‖2f(2x)-32f(x)‖≤φ(x,0)
for all x∈X. If we multiply both sides of (3.10) by (k4-k2)/16, then we have
(3.11)‖k4-k28f(2x)-2(k4-k2)f(x)‖≤k4-k216φ(x,0)
for all x∈X. It follows from (3.9) and (3.11) that
(3.12)‖f(kx)-k4f(x)‖≤12φ(0,x)+k4-k216φ(x,0)
for all x∈X. Let
(3.13)ψe(x)=12φ(0,x)+k4-k216φ(x,0)
for all x∈X. Thus by (3.12), we get
(3.14)‖f(kx)-k4f(x)‖≤ψe(x)
for all x∈X. If we replace x in (3.14) by x/kn+1 and multiply both sides of (3.14) by k4n, we see that
(3.15)‖k4(n+1)f(xkn+1)-k4nf(xkn)‖≤k4nψe(xkn+1)
for all x∈X and all nonnegative integers n. So
(3.16)‖k4(n+1)f(xkn+1)-k4mf(xkm)‖≤∑i=mn‖k4(i+1)f(xki+1)-k4if(xki)‖≤∑i=mnk4iψe(xki+1)
for all nonnegative integers n and m with n≥m and all x∈X. By (3.2), we infer that
(3.17)∑i=mnk4iψe(xki+1)<∞,lim⁡n→∞k4nψe(xkn+1)=0
for all x∈X. It follows from (3.16) and (3.17) that the sequence {k4nf(x/kn)} is a Cauchy sequence for all x∈X. Since Y is complete, the sequence {k4nf(x/kn)} converges for all x∈X. So one can define a mapping Q:X→Y by (3.4) for all x∈X. Letting m=0 and passing the limit n→∞ in (3.16), we obtain (3.5). It follows from (3.4), (3.15), and (3.17) that
(3.18)‖Q(x)-k4Q(xk)‖=lim⁡n→∞‖k4nf(xkn)-k4(n+1)f(xk(n+1))‖≤lim⁡n→∞k4nψe(xkn+1)=0
for all x∈X. So
(3.19)Q(kx)=k4Q(x)
for all x∈X. On the other hand, it follows from (3.2), (3.3), and (3.4) that
(3.20)‖DQ(x,y)‖=lim⁡n→∞k4n‖Df(xkn,ykn)‖≤lim⁡n→∞k4nφ(xkn,ykn)=0
for all x,y∈X. Therefore, by Corollary 2.2, the function Q:X→Y is quartic.
To prove the uniqueness ofQ, let Q′:X→Y be a another quartic function satisfying (3.5). Since
(3.21)lim⁡n→∞k4n∑i=1∞k4iφ(xkn+i,ykn+i)=lim⁡n→∞∑i=n+1∞k4iφ(xki,yki)=0
for all x,y∈X, hence
(3.22)lim⁡n→∞k4nψ̃e(xkn)=0
for all x∈X. So it follows from (3.5) and (3.22) that
(3.23)‖Q(x)-Q′(x)‖=lim⁡n→∞k4n‖f(xkn)-Q′(xkn)‖≤lim⁡n→∞k4nk4ψ̃e(xkn)=0
for all x∈X. Hence Q=Q′.
Forj=-1, the proof of the theorem is similar.Theorem 3.2.
Letj∈{-1,1} be fixed, and let  φ:X×X→[0,∞) be a function such that
(3.24)ψ̃c(x,y)=∑i=(1+j)/2∞23ijφ(x2ij,y2ij)<∞
for all x,y∈X. Suppose that an odd function f:X→Y with f(0)=0 satisfies the inequality (3.3). Then the limit
(3.25)C(x):=lim⁡n→∞23njf(x2nj)
exists for all x∈X and C:X→Y is a unique cubic function satisfying
(3.26)‖f(x)-C(x)‖≤12(k4-k2)ψ̃c(0,x)
for all x∈X.Proof.
Letj=1. It follows from (3.3) and using oddness of f that
(3.27)‖k4-k24f(x+ky)+f(x-ky)-k2(f(x+y)+f(x-y))-2(1-k2)f(x)-k4-k24(f(2y)-8f(y))‖≤φ(x,y)
for all x,y∈X. Replacing x and y by 0 and x in (3.27), respectively, we see that
(3.28)‖k4-k24(f(2x)-8f(x))‖≤φ(0,x)
for all x∈X. If we multiply both sides of (3.28) by 4/(k4-k2), we get
(3.29)‖f(2x)-8f(x)‖≤4k4-k2φ(0,x)
for all x∈X. If we replace x in (3.29) by x/2n+1 and multiply both sides of (3.29) by 23n, we see that
(3.30)‖23(n+1)f(x2n+1)-23nf(x2n)‖≤23n4k4-k2φ(0,x2n+1)
for all x∈X and all nonnegative integers n. So
(3.31)‖23(n+1)f(x2n+1)-23mf(x2m)‖≤∑i=mn‖23(i+1)f(x2i+1)-23if(x2i)‖≤4k4-k2∑i=mn23iφ(0,x2i+1)
for all nonnegative integers n and m with n≥m and all x∈X. By (3.24), we infer that
(3.32)∑i=mn23iφ(x2i+1,y2i+1)<∞,lim⁡n→∞23nφ(x2n+1,y2n+1)=0
for all x,y∈X. It follows from (3.31) and (3.32) that the sequence {23nf(x/2n)} is a Cauchy sequence for all x∈X. Since Y is complete, the sequence {23nf(x/2n)} converges for all x∈X. So one can define a mapping C:X→Y by (3.25) for all x∈X. Letting m=0 and passing the limit n→∞ in (3.31), we obtain (3.26). It follows from (3.25), (3.30), and (3.32) that
(3.33)‖C(x)-23C(x2)‖=lim⁡n→∞‖23nf(x2n)-23(n+1)f(x2(n+1))‖≤lim⁡n→∞4k4-k223nφ(0,x2n+1)=0
for all x∈X. So
(3.34)C(2x)=23C(x)
for all x∈X. On the other hand, it follows from (3.3), (3.24), and (3.25) that
(3.35)‖Dc(x,y)‖=lim⁡n→∞23n‖Df(x2n,y2n)‖≤lim⁡n→∞23nφ(x2n,y2n)=0
for all x,y∈X. Therefore by Corollary 2.2, the function C:X→Y is cubic.
To prove the uniqueness ofC, let C′:X→Y be a another cubic function satisfying (3.26). Since
(3.36)lim⁡n→∞23n∑i=1∞23iφ(x2n+i,y2n+i)=lim⁡n→∞∑i=n+1∞23iφ(x2i,y2i)=0
for all x,y∈X, hence
(3.37)lim⁡n→∞23nψ̃c(x2n,y2n)=0
for all x,y∈X. So it follows from (3.26) and (3.37) that
(3.38)‖C(x)-C′(x)‖=lim⁡n→∞23n‖f(x2n)-C′(x2n)‖≤lim⁡n→∞12(k4-k2)23nψ̃c(0,x2n)=0
for all x∈X. Hence C=C′.
Forj=-1, the proof of the theorem is similar.Theorem 3.3.
Letj∈{1,-1} be fixed. Suppose that a function  f:X→Y with  f(0)=0 satisfies the inequality  (3.3). If the upper bound  ϕ:X×X→[0,∞) is a mapping such that
(3.39)∑i=(1+j)/2∞[(1+j2)k4ijφ(xkij,ykij)+(1-j2)23ijφ(x2ij,y2ij)]<∞,
for all x,y∈X, then there exists a unique quartic function Q:X→Y and a unique cubic function C:X→Y satisfying
(3.40)‖f(x)-Q(x)-C(x)‖≤12k4[ψ̃e(x)+ψ̃e(-x)]+14(k4-k2)[ψ̃c(0,x)+ψ̃c(0,-x)]
for all x∈X, where
(3.41)ψ̃e(x)=∑i=(1+j)/2∞k4ij[12φ(0,xkij)+k4-k216φ(xkij,0)],ψ̃c(x,y)=∑i=(1+j)/2∞23njφ(x2ij,y2ij).Proof.
Letfe(x)=(1/2)(f(x)+f(-x)) for all x∈X. Then fe(0)=0 and fe is even function satisfying  ∥Dfe(x,y)∥≤(1/2)[ϕ(x,y)+ϕ(-x,-y)]  for all x,y∈X. By Theorem 3.1, there exists a unique quartic function Q:X→Y satisfying
(3.42)‖fe(x)-Q(x)‖≤12k4[ψ̃e(x)+ψ̃e(-x)]
for all x∈X, where
(3.43)ψ̃e(x)=∑i=(1+j)/2∞k4ij[12φ(0,xkij)+k4-k216φ(xkij,0)]
for all x∈X. Let now fo(x)=(1/2)(f(x)-f(-x)) for all x∈X. Then fe(0)=0 and fo is an odd function satisfying  ∥Dfo(x,y)∥≤(1/2)[ϕ(x,y)+ϕ(-x,-y)]  for all x,y∈X. Hence, in view of Theorem 3.2, there exists a unique cubic function C:X→Y such that
(3.44)‖fo(x)-Q(x)‖≤14(k4-k2)[ψ̃c(0,x)+ψ̃c(0,-x)]
for all x∈X, where
(3.45)ψ̃c(x,y)=∑i=(1+j)/2∞23njφ(x2ij,y2ij)
for all x,y∈X. On the other hand, we have f(x)=fe(x)+fo(x) for all x∈X. Then by combining (3.42) and (3.44), it follows that
(3.46)‖f(x)-C(x)-Q(x)‖≤‖fe(x)-Q(x)‖+‖fo(x)-C(x)‖≤12k4[ψ̃e(x)+ψ̃e(-x)]+14(k4-k2)[ψ̃c(0,x)+ψ̃c(0,-x)]
for all x∈X.We are going to investigate the Hyers-Ulam-Rassias stability problem for functional equation (1.7).Corollary 3.4.
Letp∈(-∞,3)∪(4,+∞),  θ>0. Suppose  f:X→Y satisfies  f(0)=0 and inequality
(3.47)‖Df(x,y)‖≤θ(‖x‖p+‖y‖p)
for all x,y∈X. Then there exist a unique quartic function Q:X→Y and a unique cubic function C:X→Y satisfying
(3.48)‖f(x)-Q(x)-C(x)‖≤{θ‖x‖p(1k4((12+k4-k216)(1kp-4-1))+12(k4-k2)(12p-3-1)),p>4,θ‖x‖p(1k4((12+k4-k216)(11-kp-4))+12(k4-k2)(11-2p-3)),p<3,
for all x∈X.Proof.
In Theorem3.3, put ϕ(x,y)=θ(∥x∥p+∥y∥p) for all x,y∈X.Similarly, one can solve Ulam stability problem for functional equation (1.7) when the norm of the Cauchy difference is controlled by the mixed type product-sum function
(3.49)(x,y)⟼θ(‖x‖Xu‖y‖Xv+‖x‖p+‖y‖p).Corollary 3.5.
Letu,v,p be real numbers such that  u+v,p∈(-∞,3)∪(4,+∞) and  θ>0. Suppose  f:X→Y satisfies  f(0)=0 and inequality
(3.50)‖Df(x,y)‖≤θ(‖x‖Xu‖y‖Xv+‖x‖p+‖y‖p)
for all x,y∈X. Then there exist a unique quartic function Q:X→Y and a unique cubic function C:X→Y satisfying
(3.51)‖f(x)-Q(x)-C(x)‖≤{θ‖x‖p(1k4((12+k4-k216)(1kp-4-1))+12(k4-k2)(12p-3-1)),p>4,θ‖x‖p(1k4((12+k4-k216)(11-kp-4))+12(k4-k2)(11-2p-3)),p<3,
for all x∈X.Applying Corollary3.4, one can obtain the stability of the functional equation (1.7) in the following form.Corollary 3.6.
Letϵ be a positive real number. Suppose  f:X→Y satisfies  f(0)=0 and  ∥Df(x,y)∥≤ϵ for all  x,y∈X. Then there exists a unique quartic function  Q:X→Y and a unique cubic function  C:X→Y satisfying
(3.52)‖f(x)-Q(x)-C(x)‖≤ϵ(1k4((12+k4-k216)(k4k4-1))+12(k4-k2)(88-1))
for all x∈X.



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*Source: 463164-2011-10-20.xml*