# Stability of a Functional Equation Deriving from Cubic and Quartic Functions

Authors: M. Eshaghi Gordji; A. Ebadian; S. Zolfaghari
Journal: Abstract and Applied Analysis
 (2008)

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## Abstract

We obtain the general solution and the generalized
Ulam-Hyers  stability of the cubic and quartic functional equation4(f(3x+y)+f(3x−y))=−12(f(x+y)+f(x−y))+12(f(2x+y)+f(2x−y))−8f(y)−192f(x)+f(2y)+30f(2x).


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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 the other words, under what condition does there exist 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)∥≤δfor 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)∥≤δfor all x∈E. Moreover, if f(tx) is continuous in t∈ℝ for each fixed x∈E, then T is linear. Finally, in 1978, Th. M. 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∥pfor all x∈E. If p<0,
then inequality (1.3) holds for all x,y≠0,
and (1.4) for x≠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 1991, Gajda [4] answered the question for the case p>1,
which was raised by Rassias. This new concept is known as Hyers-Ulam-Rassias
stability of functional equations (see [2, 4–13]). On the other
hand, J. M. Rassias [14–16] generalized the Hyers stability result by
presenting a weaker condition controlled by a product of different powers of
norms. According to J. M. Rassias theorem.Theorem 1.2.
If it is assumed that there exist
constantsΘ≥0 and p1,p2∈ℝ such that p=p1+p2≠1, and f:E→E′ is a map from a norm space E into a Banach space E′ such that the inequality(1.3p)∥f(x+y)−f(x)−f(y)∥≤ϵ∥x∥p1∥y∥p2for all x,y∈E, then there exists a unique additive mapping T:E→E′ such that(1.5)∥f(x)−T(x)∥≤Θ2−2p∥x∥p,for all x∈E. If in addition for every x∈E,f(tx) is continuous in real t for each fixed x, then T is linear (see [14, 15, 17–22]).The oldest cubic functional equation, and was
introduced by J. M. Rassias [23, 24] is as follows:(1.6)f(x+2y)+3f(x)=3f(x+y)+f(x−y)+6f(y).Jun and Kim [25] introduced the
following cubic functional equation:(1.7)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 (1.7). The function f(x)=x3 satisfies the functional equation (1.7), 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 (1.7) 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 oldest quartic functional equation,
and was introduced by J. M. Rassias [16, 26], and then was employed by Park and Bae [27], such that
(1.8)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 (1.8) if and only if there exists a unique symmetric multiadditive
function Q:X×X×X×X→Y such that f(x)=Q(x,x,x,x) for all x (see also [27–33]). It is easy
to show that the function f(x)=x4 satisfies the functional equation (1.8), which
is called a quartic functional equation and every solution of the quartic
functional equation is said to be a quartic function.We deal with the following functional equation
deriving from quartic and cubic functions:(1.9)4(f(3x+y)+f(3x−y))=−12(f(x+y)+f(x−y))+12(f(2x+y)+f(2x−y))−8f(y)−192f(x)+f(2y)+30f(2x).It is easy to see that the
function f(x)=ax4+bx3 is a solution of the functional equation
(1.9). In the present paper, we investigate the general solution and the
generalized Hyers-Ulam-Rassias stability of the functional equation (1.9).


## 2. General solution

In this
section, we establish the general solution of functional equation (1.9).Theorem 2.1.
LetX,Y be vector spaces, and let f:X→Y be a function. Then f satisfies (1.9) 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 C is symmetric for each fixed one variable and
is additive for fixed two variables, and that f(x)=Q(x,x,x,x)+C(x,x,x) for all x∈X.Proof.
Suppose
there exists a symmetric multiadditive functionQ:X×X×X×X→Y and a function C:X×X×X→Y such that C is symmetric for each fixed one variable and
is additive for fixed two variables, and that f(x)=Q(x,x,x,x)+C(x,x,x) for all x∈X. Then it is easy to see that f satisfies (1.9). For the convlet f satisfy (1.9). 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 (1.9), we have(2.2)4fe(3x+y)+4fe(3x−y)=12[4f(3x+y)+4f(−3x−y)+4f(3x−y)+4f(−3x+y)]=12[4f(3x+y)+4f(3x−y)]+12[4f((−3x)+(−y))+4f((−3x)−(−y))]=12[12f(2x+y)+12f(2x−y)−12f(x+y)−12f(x−y)−8f(y)−192f(x)+f(2y)+30f(2x)]+12[12f(−2x−y)+12f((−2x)+y))−12f(−x−y)−12f(−x+y)−8f(−y)−192f(−x)+f(−2y)+30f(−2x)]=12[12(f(2x+y)+f(−(2x+y)))]+12[12(f(2x−y)+f(−(2x−y)))]−12[12(f(x+y)+f(−(x+y)))]−12[12(f(x−y)+f(−(x−y)))]−8[12(f(y)+f(−y))]−192[12(f(x)+f(−x))]+12[f(2y)+f(−2y)]+30[12(f(2x)+f(−2x))]=12(fe(2x+y)+fe(2x−y))−12(fe(x+y)+fe(x−y))−8fe(y)−192fe(x)+fe(2y)+30fe(2x)for all x,y∈X. This means that fe satisfies (1.9), or(1.9e)4(fe(3x+y)+fe(3x−y))=−12(fe(x+y)+fe(x−y))+12(fe(2x+y)+fe(2x−y))−8fe(y)−192fe(x)+fe(2y)+30fe(2x).Now, putting x=y=0 in (1.9e), we get fe(0)=0.
Setting x=0 in (1.9e), by evenness of fe we obtain
(2.3)fe(2y)=16fe(y)for all y∈X. Hence, (1.9e) can be written
as(2.4)fe(3x+y)+fe(3x−y)+3(fe(x+y)+fe(x−y))=3(fe(2x+y)+fe(2x−y))+72fe(x)+2fe(y)for all x,y∈X. With the substitution y:=2y in (2.4), we have(2.5)fe(3x+2y)+fe(3x−2y)+3fe(x+2y)+3fe(x−2y)=48fe(x+y)+48fe(x−y)+72fe(x)+32fe(y).Replacing y by x+2y in (2.4), we obtain(2.6)16fe(2x+y)+16fe(x−y)+48fe(x+y)+48fe(y)=3fe(3x+2y)+3fe(x−2y)+2fe(x+2y)+72fe(x).Substituting −y for y in (2.6) gives(2.7)16fe(2x−y)+16fe(x+y)+48fe(x−y)+48fe(y)=3fe(3x−2y)+3fe(x+2y)+2fe(x−2y)+72fe(x).By utilizing (2.5), (2.6), and
(2.7), we obtain(2.8)4fe(2x+y)+4fe(2x−y)+fe(x+2y)+fe(x−2y)=20fe(x+y)+20fe(x−y)+90fe(x).Interchanging x and y in (2.5), we get(2.9)fe(2x+3y)+fe(2x−3y)+3fe(2x+y)+3fe(2x−y)=48fe(x+y)+48fe(x−y)+32fe(x)+72fe(y).If we add (2.5) to (2.9), we
have(2.10)fe(2x+3y)+fe(3x+2y)+fe(2x−3y)+fe(3x−2y)+3fe(2x+y)+3fe(x+2y)+3fe(2x−y)+3fe(x−2y)=96fe(x+y)+96fe(x−y)+104fe(x)+104fe(y).And by utilizing (2.6), (2.7),
and (2.10), we arrive at(2.11)3fe(2x+3y)+3fe(2x−3y)=−25fe(2x+y)−25fe(2x−y)−4fe(x−2y)−4fe(x+2y)+224fe(x+y)+224fe(x−y)+456fe(x)+216fe(y).Let us interchange x and y in (2.11). Then we see that(2.12)3fe(3x+2y)+3fe(3x−2y)=−25fe(x+2y)−25fe(x−2y)−4fe(2x−y)−4fe(2x+y)+224fe(x+y)+224fe(x−y)+456fe(y)+216fe(x).Comparing (2.12) with (2.5), we
get(2.13)4fe(2x−y)+4fe(2x+y)=−16fe(x+2y)−16fe(x−2y)+80fe(x+y)+80fe(x−y)+360fe(y).If we compare (2.13) and (2.8),
we conclude that(2.14)fe(x+2y)+fe(x−2y)+6fe(x)=4fe(x+y)+4fe(x−y)+24fe(y).This means that fe is 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 (1.9), or(1.9o)4(fo(3x+y)+fo(3x−y))=−12(fo(x+y)+fo(x−y))+12(fo(2x+y)+fo(2x−y))−8fo(y)−192fo(x)+fo(2y)+30fo(2x).Now setting x=y=0 in (1.9o) gives fo(0)=0. Putting x=0 in (1.9o), then by oddness of fo,
we have(2.15)fo(2y)=8fo(y).Hence, (1.9o) can be written
as(2.16)fo(3x+y)+fo(3x−y)+3fo(x+y)+3fo(x−y)=3fo(2x+y)+3fo(2x−y)+12fo(x)for all x,y∈X. Replacing x by x+y,
and y by x−y in (2.16) we have(2.17)8fo(2x+y)+8fo(x+2y)+24fo(x)+24fo(y)=3fo(3x+y)+3fo(x+3y)+12fo(x+y)and interchanging x and y in (2.16) yields(2.18)fo(x+3y)−fo(x−3y)+3fo(x+y)−3fo(x−y)=3fo(x+2y)−3fo(x−2y)+12fo(y).Which on substitution of −y for y in (2.16) gives(2.19)fo(3x−y)+fo(3x+y)+3fo(x−y)+3fo(x+y)=3fo(2x−y)+3fo(2x+y)+12fo(x).Replace y by x+2y in (2.16). Then we have(2.20)8fo(2x+y)+8fo(x−y)+24fo(x+y)−24fo(y)=3fo(3x+2y)+3fo(x−2y)+12fo(x).From the substitution y:=−y in (2.20) it follows that(2.21)8fo(2x−y)+8fo(x+y)+24fo(x−y)+24fo(y)=3fo(3x−2y)+3fo(x+2y)+12fo(x).If we add (2.20) to (2.21), we
have(2.22)3fo(3x−2y)+3fo(3x+2y)=8fo(2x+y)+8fo(2x−y)−3fo(x+2y)−3fo(x−2y)+32fo(x−y)+32fo(x+y)−24fo(x).Let us interchange x and y in (2.22). Then we see that(2.23)3fo(2x+3y)−3fo(2x−3y)=8fo(x+2y)−8fo(x−2y)−3fo(2x+y)+3fo(2x−y)+32fo(x+y)−32fo(x−y)−24fo(y).With the substitution y:=x+y in (2.16), we have(2.24)fo(4x+y)+fo(2x−y)+3fo(2x+y)−3fo(y)=3fo(3x+y)+3fo(x−y)+12fo(x),and replacing −y by y gives(2.25)fo(4x−y)+fo(2x+y)+3fo(2x−y)+3fo(y)=3fo(3x−y)+3fo(x+y)+12fo(x).If we add (2.24) to (2.25), we
have(2.26)fo(4x+y)+fo(4x−y)=3fo(3x+y)+3fo(3x−y)−4fo(2x−y)−4fo(2x+y)+3fo(x−y)+3fo(x+y)+24fo(x).By comparing (2.19) with (2.26),
we arrive at(2.27)fo(4x+y)+fo(4x−y)=5fo(2x+y)+5fo(2x−y)−6fo(x+y)−6fo(x−y)+60fo(x)and replacing y by 2y in (2.16) gives(2.28)fo(3x+2y)+fo(3x−2y)=24fo(x+y)+24fo(x−y)−3fo(x+2y)−3fo(x−2y)+12fo(x).By comparing (2.28) with (2.22),
we arrive at(2.29)3fo(x+2y)+3fo(x−2y)=20fo(x+y)+20fo(x−y)−4fo(2x+y)−4fo(2x−y)+30fo(x).Let us interchange x and y in (2.28). Then we see that(2.30)fo(2x+3y)−fo(2x−3y)=24fo(x+y)−24fo(x−y)−3fo(2x+y)+3fo(2x−y)+12fo(y).Thus combining (2.30) with (2.23)
yields(2.31)4fo(x+2y)−4fo(x−2y)=3fo(2x−y)−3fo(2x+y)+20fo(x+y)−20fo(x−y)+30fo(y).By comparing (2.31) with (2.18),
we arrive at(2.32)4fo(x+3y)−4fo(x−3y)=9fo(2x−y)−9fo(2x+y)+48fo(x+y)−48fo(x−y)+138fo(y).Which, by putting y:=2y in (2.17), leads to(2.33)64fo(x+y)+8fo(x+4y)+24fo(x)+192fo(y)=3fo(3x+2y)+3fo(x+6y)+12fo(x+2y).Replacing y by −y in (2.33) gives(2.34)64fo(x−y)+8fo(x−4y)+24fo(x)−192fo(y)=3fo(3x−2y)+3fo(x−6y)+12fo(x−2y).If we subtract (2.33) from
(2.34), we obtain(2.35)8fo(x+4y)−8fo(x−4y)=3fo(3x+2y)−3fo(3x−2y)+3fo(x+6y)−3fo(x−6y)+12fo(x+2y)−12fo(x−2y)+64fo(x−y)−64fo(x+y)−384fo(y).Setting x instead of y and y instead of x in (2.27), we get(2.36)fo(x+4y)−fo(x−4y)=5fo(x+2y)−5fo(x−2y)+6fo(x−y)−6fo(x+y)+60fo(y).Combining (2.35) and (2.36)
yields(2.37)3fo(3x+2y)−3fo(3x−2y)=28fo(x+2y)−28fo(x−2y)+3fo(x−6y)−3fo(x+6y)+16fo(x+y)−16fo(x−y)+864fo(y)and subtracting (2.21) from
(2.20), we obtain(2.38)3fo(3x+2y)−3fo(3x−2y)=3fo(x+2y)−3fo(x−2y)+8fo(2x+y)−8fo(2x−y)+16fo(x+y)−16fo(x−y)−48fo(y).By comparing (2.37) with (2.38),
we arrive at(2.39)3fo(x+6y)−3fo(x−6y)=25fo(x+2y)−25fo(x−2y)+8fo(2x−y)−8fo(2x+y)+912fo(y).Interchanging y with 2y in (2.32) gives the equation(2.40)4fo(x+6y)−4fo(x−6y)=48fo(x+2y)−48fo(x−2y)+72fo(x−y)−72fo(x+y)+1104fo(y).We obtain from (2.39) and
(2.40)(2.41)44fo(x+2y)−44fo(x−2y)=32fo(2x−y)−32fo(2x+y)+216fo(x+y)−216fo(x−y)+336fo(y).By using (2.31) and (2.41), we
lead to(2.42)fo(2x+y)−fo(2x−y)=4fo(x+y)−4fo(x−y)−6fo(y).And interchanging x with y in (2.42) gives(2.43)fo(x+2y)+fo(x−2y)=4fo(x+y)+4fo(x−y)−6fo(x).If we compare (2.43) and (2.29),
we conclude that(2.44)8fo(x+y)+8fo(x−y)+48fo(x)=4fo(2x+y)+4fo(2x−y).This means that fo is 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(2.45)f(x)=fe(x)+fo(x)=C(x,x,x)+Q(x,x,x,x).This completes the proof of
theorem.The following corollary is an
alternative result of Theorem2.1.Corollary 2.2.
LetX,Y be vector spaces, and let f:X→Y be a function satisfying (1.9). 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 now investigate the generalized Hyers-Ulam-Rassias
stability problem for functional equation (1.9). From now on, let X be a real
vector space and let Y be a Banach space. Now before taking up the main
subject, given f:X→Y,
we define the difference operator Df:X×X→Y by(3.1)Df(x,y)=4[f(3x+y)+f(3x−y)]−12[f(2x+y)+f(2x−y)]+12[f(x+y)+f(x−y)]−f(2y)+8f(y)−30f(2x)+192f(x)for all x,y∈X. We consider the following functional
inequality: (3.2)∥Df(x,y)∥≤ϕ(x,y)for an upper bound ϕ:X×X→[0,∞).Theorem 3.1.
Lets∈{1,−1} be fixed. Suppose that an even mapping f:X→Y satisfies f(0)=0, and(3.3)∥Df(x,y)∥≤ϕ(x,y)for all x,y∈X. If the upper bound ϕ:X×X→[0,∞) is a mapping such that the series ∑i=0∞24siϕ(0,x/2si) converges, and that limn→∞24snϕ(x/2sn,y/2sn)=0 for all x,y∈X, then the limit Q(x)=limn→∞24snf(x/2sn) exists for all x∈X, and Q:X→Y is a unique quartic function satisfying (1.9),
and(3.4)∥f(x)−Q(x)∥≤116∑i=(s−1)/2∞24s(i+1)ϕ(0,x2s(i+1))for all x∈X.Proof.
Lets=1. Putting x=0 in (3.3), we get
(3.5)∥f(2y)−16f(y)∥≤ϕ(0,y).Replacing y by x/2 in (3.5), yields
(3.6)∥f(x)−16f(x2)∥≤ϕ(0,x2).Interchanging x with x/2 in (3.6), and multiplying by 16 it follows that
(3.7)∥16f(x2)−162f(x4)∥≤16ϕ(0,x4).Combining (3.6) and (3.7), we lead to
(3.8)∥162f(x4)−f(x)∥≤ϕ(0,x2)+16ϕ(0,x4).From the inequality (3.6) we use
iterative methods and induction on n to prove our next relation:
(3.9)∥16nf(x2n)−f(x)∥≤116∑i=0n−116i+1ϕ(0,x2i+1).We multiply (3.9) by 16m and replace x by x/2m to obtain that(3.10)∥16m+nf(x2m+n)−16mf(x2m)∥≤∑i=0n−116m+iϕ(0,x2i+m+1).This shows that {16nf(x/2n)} is a Cauchy sequence in Y by taking the limit m→∞.
Since Y is a Banach space, it follows that the sequence {16nf(x/2n)} converges. We define Q:X→Y by Q(x)=limn→∞24nf(x/2n) for all x∈X. It is clear that Q(−x)=Q(x) for all x∈X,
and it follows from (3.3) that(3.11)∥DQ(x,y)∥=limn→∞16n∥Df(x2n,y2n)∥≤limn→∞16nϕ(x2n,y2n)=0for all x,y∈X. Hence, by Corollary 2.2, Q is quartic. It
remains to show that Q is unique. Suppose that there exists another quartic
function Q′:X→Y which satisfies (1.9) and (3.4). Since Q(2nx)=16nQ(x),
and Q′(2nx)=16nQ′(x) for all x∈X, we conclude that(3.12)∥Q(x)−Q′(x)∥=16n∥Q(x2n)−Q′(x2n)∥≤16n∥Q(x2n)−f(x2n)∥+16n∥Q′(x2n)−f(x2n)∥≤2∑i=0∞16n+iϕ(0,x2n+i+1)for all x∈X. By letting n→∞ in this inequality, it follows that Q(x)=Q′(x) for all x∈X, which gives the conclusion. For s=−1, we obtain
(3.13)∥f(2mx)16m−f(x)∥≤116∑i=−1n−2ϕ(0,2i+1x)16i+1,from which one can prove the
result by a similar technique.Theorem 3.2.
Lets∈{1,−1} be fixed. Suppose that an odd mapping f:X→Y satisfies
(3.14)∥Df(x,y)∥≤ϕ(x,y)for all x,y∈X. If the upper bound ϕ:X×X→[0,∞) is a mapping such that ∑i=0∞23siϕ(0,x/2si) converges, and that limn→∞23siϕ(x/2si,y/2si)=0 for all x,y∈X, then the limit C(x)=limn→∞23snf(x/2sn) exists for all x∈X, and C:X→Y is a unique cubic function satisfying (1.9),
and(3.15)∥f(x)−C(x)∥≤18∑i=(s−1)/2∞23s(i+1)ϕ(0,x2s(i+1))for all x∈X.Proof.
Lets=1. Set x=0 in (3.14). We obtain
(3.16)∥8f(y)−f(2y)∥≤ϕ(0,y).Replacing y by x/2 in (3.16) to get
(3.17)∥8f(x2)−f(x)∥≤ϕ(0,x2).An induction argument now implies
(3.18)∥8nf(x2n)−f(x)∥≤18∑i=0n−18i+1ϕ(0,x2i+1).Multiply (3.18) by 8m and replace x by x/2m, we obtain that(3.19)∥8m+nf(x2m+n)−8mf(x2m)∥≤∑i=0n−18m+iϕ(0,x2m+i+1).The right hand side of the
inequality (3.19) tends to 0 as m→∞ because of(3.20)∑i=0∞8iϕ(0,x2i+1)<∞by assumption, and thus the
sequence {23nf(x/2n)} is Cauchy in Y, as desired. Therefore we may
define a mapping C:X→Y as C(x)=limn→∞23nf(x/2n). The rest of proof is similar to the proof of
Theorem 3.1.Theorem 3.3.
Lets∈{1,−1} be fixed. Suppose a mapping f:X→Y satisfies f(0)=0, and ∥Df(x,y)∥≤ϕ(x,y) for all x,y∈X. If the upper bound ϕ:X×X→[0,∞) is a mapping such that(3.21)∑i=0∞[(|s|+s)24siϕ(0,x2si−1)+(|s|−s)23siϕ(0,x2si−1)]<∞,limn→∞[(|s|+s)2(4sn−1)ϕ(x2sn,y2sn)+(|s|+s)23snϕ(x2sn,y2sn)]=0for all x,y∈X. Then there exists a unique quartic function Q:X→Y and a unique cubic function C:X→Y satisfying (3.22)∥f(x)−Q(x)−C(x)∥≤∑i=(s−1)/2∞{(24s(i+1)32+23s(i+1)16)[ϕ(0,x2s(i+1))+ϕ(0,−x2s(i+1))]}for all x∈X.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. From Theorem 3.1, it follows that there exists
a unique quartic function Q:X→Y satisfies
(3.23)∥fe(x)−Q(x)∥≤132∑i=(s−1)/2∞{24s(i+1)ϕ(0,x2s(i+1))+24s(i+1)ϕ(0,−x2s(i+1))}for all x∈X. Let now fo(x)=(1/2)(f(x)−f(−x)) for all x∈X. Then fo is odd function satisfying
(3.24)∥Dfo(x,y)∥≤12[ϕ(x,y)+ϕ(−x,−y)]for all x,y∈X. Hence, in view of Theorem 3.2, it follows that
there exists a unique cubic function C:X→Y such that
(3.25)∥fo(x)−C(x)∥≤116∑i=(s−1)/2∞{23s(i+1)ϕ(0,x2s(i+1))+23s(i+1)ϕ(0,−x2s(i+1))}for all x∈X. On the other hand, we have f(x)=fe(x)+fo(x) for all x∈X. Then by combining (3.23) and (3.25), it
follows that(3.26)∥f(x)−C(x)−Q(x)∥≤∥fe(x)−Q(x)∥+∥fo(x)−C(x)∥≤∑i=(s−1)/2∞{(24s(i+1)32+23s(i+1)16)[ϕ(0,x2s(i+1))+ϕ(0,−x2s(i+1))]}for all x∈X, and the proof of theorem is complete.We are going to investigate the
Hyers-Ulam-Rassias stability problem for functional equation
(1.9).Corollary 3.4.
Letp∈(−∞,3)∪(4,+∞),θ>0. Suppose f:X→Y satisfies f(0)=0, and inequality(3.27)∥Df(x,y)∥≤θ(∥x∥p+∥y∥p),for all x,y∈X.
Then there exists a unique quartic function Q:X→Y,
and a unique cubic function C:X→Y satisfying
(3.28)∥f(x)−Q(x)−C(x)∥≤{θ∥x∥p(12p−24+12p−23),p>4,θ∥x∥p(124−2p+123−2p),p<3for all x∈X.Proof.
Lets=1 in Theorem 3.3. Then by taking ϕ(x,y)=θ(∥x∥p+∥y∥p) for all x,y∈X, the relations (3.21) hold for p>4. Then there exists a unique quartic function Q:X→Y and a unique cubic function C:X→Y satisfying
(3.29)∥f(x)−Q(x)−C(x)∥≤θ∥x2∥p(11−24−p+11−23−p)for all x∈X. Let now s=−1 in Theorem 3.3 and put ϕ(x,y)=θ(∥x∥p+∥y∥p) for all x,y∈X. Then the relations (3.21) hold for p<3. Then there exists a unique quartic function Q:X→Y and a unique cubic function C:X→Y satisfying
(3.30)∥f(x)−Q(x)−C(x)∥≤θ∥x∥p(124−2p+123−2p)for all x∈X.Similarly, we can prove the following Ulam stability
problem for functional equation (1.9) controlled by the mixed type product-sum
function(3.31)(x,y)↦θ(∥x∥Xu∥y∥Xv+∥x∥p+∥y∥p)introduced by J. M. Rassias
(e.g., [34]).Corollary 3.5.
Letu,v,p be real numbers such that u+v,p∈(−∞,3)∪(4,+∞), and let θ>0. Suppose f:X→Y satisfies f(0)=0, and inequality(3.32)∥Df(x,y)∥≤θ(∥x∥Xu∥y∥Xv+∥x∥p+∥y∥p),for all x,y∈X.
Then there exists a unique quartic function Q:X→Y,
and a unique cubic function C:X→Y satisfying(3.33)∥f(x)−Q(x)−C(x)∥≤{θ∥x∥p(12p−24+12p−23),p>4,θ∥x∥p(124−2p+123−2p),p<3for all x∈X.By Corollary3.4, we solve the following Hyers-Ulam
stability problem for functional equation (1.9).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 (3.34)∥f(x)−Q(x)−C(x)∥≤22105ϵfor all x∈X.



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*Source: 801904-2009-02-02.xml*