# Approximate Quartic and Quadratic Mappings in Quasi-Banach Spaces

Authors: M. Eshaghi Gordji; H. Khodaei; Hark-Mahn Kim
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

we establish the general solution for a mixed type functional equation of aquartic and a quadratic mapping in linear spaces. In addition, we investigate the generalized Hyers-Ulam stability inp-Banach spaces.


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## Body

## 1. Introduction and Preliminaries

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 a first affirmative answer to the question of Ulam for Banach spaces. Let f:X→X′ be a mapping between Banach spaces such that (1.1)‖f(x+y)-f(x)-f(y)‖≤δ,
for all x,y∈X, and for some δ>0. Then, there exists a unique additive mapping T:X→X′ such that (1.2)‖f(x)-T(x)‖≤δ,
for all x∈X.The result of Hyers was generalized by Aoki [3] for approximate additive function and by Rassias [4] for approximate linear function by allowing the difference Cauchy equation ∥f(x+y)-f(x)-f(y)∥ to be controlled by ɛ(∥x∥p+∥y∥p). Taking into consideration a lot of influence of Ulam, Hyers and Rassias on the development of stability problems of functional equations, the stability phenomenon that was proved by Rassias may be called the Hyers-Ulam-Rassias stability (see [5, 6]). In 1994, a generalization of Rassias theorem was obtained by Găvruta [7], who replaced ɛ(∥x∥p+∥y∥p) by a general control function φ(x,y). The functional equation(1.3)f(x+y)+f(x-y)=2f(x)+2f(y)
is related to a symmetric biadditive function [8–10]. It is natural that this equation is called a quadratic functional equation. In particular, every solution of the quadratic equation (1.3) is said to be a quadratic function. It is well known that a function f between real vector spaces is quadratic if and only if there exists a unique symmetric biadditive function B1 such that f(x)=B1(x,x) for all x in the vector space. The biadditive function B1 is given by(1.4)B1(x,y)=14(f(x+y)-f(x-y)).
A Hyers-Ulam stability problem for the quadratic functional equation (1.3) was proved by Skof for functions f:X→Y, where X is normed space and Y is Banach space (see [11]). In the paper [12], Czerwik proved the Hyers-Ulam-Rassias stability of (1.3).Lee et al. [13] considered the following functional equation:(1.5)f(2x+y)+f(2x-y)=4f(x+y)+4f(x-y)+24f(x)-6f(y).
In fact, they proved that a function f between two real vector spaces X and Y is a solution of (1.5) if and only if there exists a unique symmetric biquadratic function B2:X×X→Y such that f(x)=B2(x,x) for all x∈X. The biquadratic function B2 is given by(1.6)B2(x,y)=112(f(x+y)+f(x-y)-2f(x)-2f(y)).
It is easy to show that the function f(x)=ax4 satisfies the functional equation (1.5), which is called the quartic functional equation (see also [14]).Jun and Kim [15] have obtained the generalized Hyers-Ulam stability for a mixed type of cubic and additive functional equation. In addition, the generalized Hyers-Ulam stability for a mixed type of cubic, quadratic, and additive functional equation has been investigated by Gordji and Khodaei [16] (see also [17, 18]). The stability problems for several mixed types of functional equations have been extensively investigated by a number of authors and there are many interesting results concerning this problem [19–27].In this paper, we deal with the following functional equation derived from quartic and quadratic functions:(1.7)f(kx+y)+f(kx-y)=k2f(x+y)+k2f(x-y)+k2(k2-1)6(f(2x)-4f(x))-2(k2-1)f(y)
for fixed integers k≠0, ±1. It is easy to see that the function f(x)=ax4+bx2 is a solution of the functional equation (1.7). In the sequel, we investigate the general solution of functional equation (1.7) when f is a function between vector spaces, and then we prove the generalized Hyers-Ulam stability of (1.7) in the spirit of Hyers, Ulam, and Rassias using the direct method.We recall some basic facts concerning quasi-Banach spaces and some preliminary results.Definition 1.1 (see [28, 29]).
LetX be a real linear space. A quasinorm is a real-valued function on X satisfying the following:(1)
∥x∥≥0 for all x∈X and ∥x∥=0 if and only if x=0,(2)
∥λ·x∥=|λ|·∥x∥ for all λ∈ℝ and all x∈X,(3)
there is a constantM≥1 such that ∥x+y∥≤M(∥x∥+∥y∥) for all x,y∈X.
The pair(X,∥·∥) is called a quasinormed space if ∥·∥ is a quasinorm on X.The smallest possibleM is called the modulus of concavity of ∥·∥. A quasi-Banach space is a complete quasinormed space. A quasinorm ∥·∥ is called a p-norm (0<p≤1) if
(1.8)‖x+y‖p≤‖x‖p+‖y‖p,
for all x,y∈X. In this case, a quasi-Banach space is called a p-Banach space.Given ap-norm, the formula d(x,y):=∥x-y∥p gives us a translation invariant metric on X. By the Aoki-Rolewicz Theorem [29], each quasinorm is equivalent to some p-norm (see also [28]). Since it is much easier to work with p-norms, henceforth we restrict our attention mainly to p-norms.Lemma 1.2 (see [17]).
Letx1,x2,…,xn be nonnegative real numbers. Then, one has
(1.9)(∑i=1nxi)p≤∑i=1nxip,
for a positive real number p with p≤1.


## 2. General Solution

We here present the general solution of (1.7).Theorem 2.1.
Let bothX and Y be real vector spaces. A function f:X→Y satisfies (1.7) for all x,y∈X if and only if there exists a unique symmetric biquadratic function B2:X×X→Y and a unique symmetric biadditive function B1:X×X→Y such that
(2.1)f(x)=B2(x,x)+B1(x,x),
for all x∈X.Proof.
Letf satisfy (1.7) and let g,h:X→Y be functions defined by
(2.2)g(x):=f(2x)-16f(x),h(x):=f(2x)-4f(x),
for all x∈X. We claim that the functions g and h are quadratic and quartic, respectively.
Lettingx=y=0 in (1.7), we have f(0)=0. By putting x=0 in (1.7), one leads to the evenness f(-y)=f(y) of f. Replacing y by x+y in (1.7), we have(2.3)f((k+1)x+y)+f((k-1)x-y)=k2f(2x+y)+k2f(-y)+k2(k2-1)6(f(2x)-4f(x))+2(1-k2)f(x+y),
for all x,y∈X. Replacing y by -y in (2.3), we obtain
(2.4)f((k+1)x-y)+f((k-1)x+y)=k2f(2x-y)+k2f(y)+k2(k2-1)6(f(2x)-4f(x))+2(1-k2)f(x-y),
for all x,y∈X. Adding (2.3) to (2.4), we get by evenness of f,
(2.5)f((k+1)x+y)+f((k+1)x-y)+f((k-1)x+y)+f((k-1)x-y)=k2(f(2x+y)+f(2x-y))+2k2(k2-1)6(f(2x)-4f(x))+2(1-k2)(f(x+y)+f(x-y))+2k2f(y),
for all x,y∈X. From the substitution y=kx+y in (1.7), we have by evenness of f,
(2.6)f(2kx+y)+f(y)=k2f((k+1)x+y)+k2f((k-1)x+y)+k2(k2-1)6(f(2x)-4f(x))+2(1-k2)f(kx+y),
for all x,y∈X. Replacing y by -y in (2.6), we get
(2.7)f(2kx-y)+f(-y)=k2f((k+1)x-y)+k2f((k-1)x-y)+k2(k2-1)6(f(2x)-4f(x))+2(1-k2)f(kx-y),
for all x,y∈X. Adding (2.6) to (2.7), we get by evenness of f,
(2.8)f(2kx+y)+f(2kx-y)=k2(f((k+1)x+y)+f((k+1)x-y)+f((k-1)x+y)+f((k-1)x-y))+2k2(k2-1)6(f(2x)-4f(x))+2(1-k2)(f(kx+y)+f(kx-y))-2f(y),
for all x,y∈X. By using (1.7) and (2.5), it follows from (2.8) that
(2.9)f(2kx+y)+f(2kx-y)=k2[k2(f(2x+y)+f(2x-y))+2k2(k2-1)6(f(2x)-4f(x))+2(1-k2)(f(x+y)+f(x-y))+2k2f(y)2k2(k2-1)6]+2(1-k2)[k2f(x+y)+k2f(x-y)+k2(k2-1)6(f(2x)-4f(x))+2(1-k2)f(y)]+2k2(k2-1)6(f(2x)-4f(x))-2f(y),
for all x,y∈X. If we replace x by 2x in (1.7), then we get that
(2.10)f(2kx+y)+f(2kx-y)=k2f(2x+y)+k2f(2x-y)+k2(k2-1)6(f(4x)-4f(2x))+2(1-k2)f(y),
for all x,y∈X. It follows from (2.9) and (2.10) that
(2.11)k2[k2(f(2x+y)+f(2x-y))+2k2(k2-1)6(f(2x)-4f(x))+2(1-k2)(f(x+y)+f(x-y))+2k2f(y)2k2(k2-1)6]+2(1-k2)[k2f(x+y)+k2f(x-y)+k2(k2-1)6(f(2x)-4f(x))+2(1-k2)f(y)]+2k2(k2-1)6(f(2x)-4f(x))-2f(y)=k2f(2x+y)+k2f(2x-y)+k2(k2-1)6(f(4x)-4f(2x))+2(1-k2)f(y),
for all x,y∈X. On the other hand, putting y=0 in (1.7), we get
(2.12)f(kx)=k2f(x)+k2(k2-1)12(f(2x)-4f(x)),
for all x∈X. Putting y=x in (1.7), we get
(2.13)f((k+1)x)+f((k-1)x)=k2f(2x)+k2(k2-1)6(f(2x)-4f(x))+2(1-k2)f(x),
for all x∈X. Putting y=kx in (1.7) and using the evenness of f, we obtain
(2.14)f(2kx)=k2(f((k+1)x)+f((k-1)x))+k2(k2-1)6(f(2x)-4f(x))+2(1-k2)f(kx),
for all x∈X. Letting y=0 in (2.10), we have
(2.15)f(2kx)=k2f(2x)+k2(k2-1)12(f(4x)-4f(2x)),
for all x∈X. It follows from (2.14) and (2.15) that
(2.16)k2(k2-1)12(f(4x)-4f(2x))=k2(f((k+1)x)+f((k-1)x))+k2(k2-1)6(f(2x)-4f(x))+2(1-k2)f(kx)-k2f(2x),
for all x∈X. Now, by using (2.12), (2.13) and (2.16), we lead to
(2.17)k2(k2-1)12(f(4x)-4f(2x))=k2[k2f(2x)+k2(k2-1)6(f(2x)-4f(x))+2(1-k2)f(x)]+2(1-k2)[k2f(x)+k2(k2-1)12(f(2x)-4f(x))]+k2(k2-1)6(f(2x)-4f(x))-k2f(2x),
for all x∈X. Finally, comparing (2.11) with (2.17), then we conclude that
(2.18)f(2x+y)+f(2x-y)=4f(x+y)+4f(x-y)+2(f(2x)-4f(x))-6f(y),
for all x,y∈X. Replacing y by 2y in (2.18), we get
(2.19)f(2x+2y)+f(2x-2y)=4f(x+2y)+4f(x-2y)+2(f(2x)-4f(x))-6f(2y),
for all x,y∈X. Interchanging x with y in (2.18), one gets
(2.20)f(x+2y)+f(x-2y)=4f(x+y)+4f(x-y)+2(f(2y)-4f(y))-6f(x),
for all x,y∈X. It follows from (2.19) and (2.20) that
(2.21)f(2(x+y))-16f(x+y)+f(2(x-y))-16f(x-y)=2(f(2x)-16f(x))+2(f(2y)-16f(y)),
for all x,y∈X. This means that
(2.22)g(x+y)+g(x-y)=2g(x)+2g(y),
for all x,y∈X. So the function g:X→Y defined by g(x):=f(2x)-16f(x) is quadratic.
To prove thath:X→Y defined by h(x):=f(2x)-4f(x) is quartic, we need to show that(2.23)h(2x+y)+h(2x-y)=4h(x+y)+4h(x-y)+24h(x)-6h(y),
for all x,y∈X. Replacing x and y by 2x and 2y in (2.18), respectively, we obtain
(2.24)f(2(2x+y))+f(2(2x-y))=4f(2(x+y))+4f(2(x-y))+2(f(4x)-4f(2x))-6f(2y),
for all x,y∈X. But, since g(2x)=4g(x) for all x∈X, where g:X→Y is a quadratic function defined above, we see that
(2.25)f(4x)=20f(2x)-64f(x),
for all x∈X. Hence, according to (2.24) and (2.25), we get
(2.26)f(2(2x+y))+f(2(2x-y))=4f(2(x+y))+4f(2(x-y))+32(f(2x)-4f(x))-6f(2y),
for all x,y∈X. By multiplying 4 on both sides of (2.18), we get that
(2.27)4f(2x+y)+4f(2x-y)=16f(x+y)+16f(x-y)+8(f(2x)-4f(x))-24f(y),
for all x,y∈X. If we subtract the last equation from (2.26), then we arrive at
(2.28)f(2(2x+y))-4f(2x+y)+f(2(2x-y))-4f(2x-y)=4(f(2(x+y))-4f(x+y))+4(f(2(x-y))-4f(x-y))+24(f(2x)-4f(x))-6(f(2y)-4f(y)),
for all x,y∈X. This means that h satisfies (2.23) and, therefore, the function h:X→Y is quartic. Thus, there exists a unique symmetric biquadratic function B2:X×X→Y and a unique symmetric biadditive function B1:X×X→Y such that h(x)=12B2(x,x) and g(x)=-12B1(x,x) for all x∈X (see [8, 13]). Therefore, we obtain from (2.2) that
(2.29)f(x)=112h(x)-112g(x)=B2(x,x)+B1(x,x),
for all x∈X.
The proof of the converse is trivial.


## 3. Generalized Hyers-Ulam Stability

From this point on, assume thatX is a quasinormed space with quasinorm ∥·∥X and that Y is a p-Banach space with p-norm ∥·∥Y. Let M be the modulus of concavity of ∥·∥Y.Before taking up the main subject, given a mappingf:X→Y, we define the difference operator Df:X×X→Y by (3.1)Df(x,y):=f(kx+y)+f(kx-y)-k2f(x+y)-k2f(x-y)-k2(k2-1)6(f(2x)-4f(x))+2(k2-1)f(y),
for all x,y∈X. Let φp(x,y):=(φ(x,y))p for notational convenience.Theorem 3.1.
Letj∈{-1,1} be fixed and let φq:X×X→[0,∞) be a function such that
(3.2)lim⁡n→∞4njφq(x2nj,y2nj)=0,
for all x,y∈X and
(3.3)∑i=(1+j)/2∞4ipjφqp(u2ij,y2ij)<∞,
for all (u,y)∈{(x,0),(2x,0),(x,x),(x,kx):x∈X}. Suppose that an even function f:X→Y with f(0)=0 satisfies the inequality
(3.4)‖Df(x,y)‖Y≤φq(x,y),
for all x,y∈X. Then, there exists a unique quadratic function Q:X→Y such that
(3.5)‖f(2x)-16f(x)-Q(x)‖Y≤M24(ψ̃q(x))1/p,
for all x∈X, where
(3.6)ψ̃q(x):=∑i=(1+j)/2∞4ipjk2p(k2-1)p[(12k2)pφqp(x2ij,x2ij)+(12(k2-1))pφqp(x2ij,0)+6pφqp(2x2ij,0)+12pφqp(x2ij,kx2ij)].Proof.
Letj=1. Setting y=0 in (3.4), we have
(3.7)‖2f(kx)-2k2f(x)-k2(k2-1)6(f(2x)-4f(x))‖Y≤φq(x,0),
for all x∈X. Putting y=x in (3.4), we obtain
(3.8)‖f((k+1)x)+f((k-1)x)-k2f(2x)-k2(k2-1)6(f(2x)-4f(x))+2(k2-1)f(x)‖Y≤φq(x,x),
for all x∈X. Replacing x by 2x in (3.7), we see that
(3.9)‖2f(2kx)-2k2f(2x)-k2(k2-1)6(f(4x)-4f(2x))‖Y≤φq(2x,0),
for all x∈X. Setting y by kx in (3.4) and using the evenness of f, we get
(3.10)‖f(2kx)-k2f((k+1)x)-k2f((k-1)x)+2(k2-1)f(kx)-k2(k2-1)6(f(2x)-4f(x))‖Y≤φq(x,kx),
for all x∈X. It follows from (3.9) and (3.10) that
(3.11)‖k2f(2x)+k2(k2-1)12(f(4x)-4f(2x))-k2f((k+1)x)-k2f((k-1)x)+2(k2-1)f(kx)-k2(k2-1)6(f(2x)-4f(x))‖Y≤M[12φq(2x,0)+φq(x,kx)],
for all x∈X. Also, it follows from (3.7) and (3.8) that
(3.12)‖k2(k2-1)6k2f((k+1)x)+k2f((k-1)x)-2(k2-1)f(kx)-k4f(2x)-k2(k2-1)6(f(2x)-4f(x))+4k2(k2-1)f(x)‖Y≤M[k2φq(x,x)+(k2-1)φq(x,0)],
for all x∈X. Finally, using (3.11) and (3.12), we obtain that
(3.13)‖f(4x)-20f(2x)+64f(x)‖Y≤M2k2(k2-1)[12k2φq(x,x)+12(k2-1)φq(x,0)+6φq(2x,0)+12φq(x,kx)]=M2ψq(x),
where
(3.14)ψq(x):=1k2(k2-1)[12k2φq(x,x)+12(k2-1)φq(x,0)+6φq(2x,0)+12φq(x,kx)],
for all x∈X. Let g:X→Y be a function defined by g(x):=f(2x)-16f(x) for all x∈X. From (3.13), we conclude that
(3.15)∥g(2x)-4g(x)∥Y≤M2ψq(x),
for all x∈X. If we replace x in (3.15) by x/2n+1 and multiply both sides of (3.15) by 4n, then we get
(3.16)‖4n+1g(x2n+1)-4ng(x2n)‖Y≤M24nψq(x2n+1),
for all x∈X and all non-negative integers n. Since Y is a p-Banach space, the inequality (3.16) gives
(3.17)‖4n+1g(x2n+1)-4mg(x2m)‖Yp≤∑i=mn‖4i+1g(x2i+1)-4ig(x2i)‖Yp≤M2p∑i=mn4ipψqp(x2i+1),
for all nonnegative integers n and m with n≥m and all x∈X. Since 0<p≤1, by Lemma 1.2 and (3.14), we conclude that
(3.18)ψqp(x)≤1k2p(k2-1)p[(12k2)pφqp(x,x)+(12(k2-1))pφqp(x,0)+6pφqp(2x,0)+12pφqp(x,kx)],
for all x∈X. Therefore, it follows from (3.3) and (3.18) that
(3.19)∑i=1∞4ipψqp(x2i)<∞,
for all x∈X. It follows from (3.17) and (3.19) that the sequence {4ng(x/2n)} is a Cauchy for all x∈X. Since Y is complete, the sequence {4ng(x/2n)} converges for all x∈X. So one can define a function Q:X→Y by
(3.20)Q(x)=lim⁡n→∞4ng(x2n),
for all x∈X. Letting m=0 and passing the limit n→∞ in (3.17), we get
(3.21)‖g(x)-Q(x)‖Yp≤M2p∑i=0∞4ipψqp(x2i+1)=M2p4p∑i=1∞4ipψqp(x2i),
for all x∈X. Thus (3.5) follows from (3.18) and (3.21). Now we show that Q is quadratic. It follows from (3.16), (3.19) and (3.20) that
(3.22)‖Q(2x)-4Q(x)‖Y=lim⁡n→∞‖4ng(x2n-1)-4n+1g(x2n)‖Y=4lim⁡n→∞‖4n-1g(x2n-1)-4ng(x2n)‖Y≤M2lim⁡n→∞4nψq(x2n)=0,
for all x∈X. So,
(3.23)Q(2x)=4Q(x),
for all x∈X. On the other hand, it follows from (3.2), (3.4) and (3.20) that
(3.24)‖DQ(x,y)‖Y=lim⁡n→∞4n‖Dg(x2n,y2n)‖Y=lim⁡n→∞4n‖Df(x2n-1,y2n-1)-16Df(x2n,y2n)‖Y≤Mlim⁡n→∞4n{‖Df(x2n-1,y2n-1)‖Y+16‖Df(x2n,y2n)‖Y}≤Mlim⁡n→∞4n{φq(x2n-1,y2n-1)+16φq(x2n,y2n)}=0,
for all x,y∈X. Hence the function Q satisfies (1.7). Thus, by Theorem 2.1, the function x⇝Q(2x)-16Q(x) is quadratic. Therefore, (3.23) implies that the function Q is quadratic.
Now, to prove the uniqueness property ofQ, let Q′:X→Y be another quadratic function satisfying (3.5). It follows from (3.3) that(3.25)lim⁡n→∞4np∑i=1∞4ipφqp(u2n+i,y2n+i)=lim⁡n→∞∑i=n+1∞4ipφqp(u2i,y2i)=0,
for all (u,y)∈{(x,0),(2x,0),(x,x),(x,kx):x∈X}. Hence,
(3.26)lim⁡n→∞4npψ̃q(x2n)=0,
for all x∈X. It follows from (3.5), (3.20) and (3.26) that
(3.27)‖Q(x)-Q′(x)‖Yp=lim⁡n→∞4np‖g(x2n)-Q′(x2n)‖Yp≤M2p4plim⁡n→∞4npψ̃q(x2n)=0,
for all x∈X. So Q=Q′.
Forj=-1, we can prove the theorem by a similar argument.Corollary 3.2.
Letθ,r,s be nonnegative real numbers such that r, s>2 or r, s<2. Suppose that an even function f:X→Y with f(0)=0 satisfies the inequality
(3.28)‖Df(x,y)‖Y≤θ(‖x‖Xr+‖y‖Xs),
for all x,y∈X. Then there exists a unique quadratic function Q:X→Y satisfying
(3.29)‖f(2x)-16f(x)-Q(x)‖Y≤M2θk2(k2-1)γq(x),
for all x∈X, where
(3.30)γq(x)=(12p[k2p+(k2-1)p+(2r-1)p+1]|4p-2rp|‖x‖Xrp+12p(k2p+ksp)|4p-2sp|‖x‖Xsp)1/p.Proof.
In Theorem3.1, putting φq(x,y):=θ(∥x∥Xr+∥y∥Xs) for all x,y∈X, we get the desired result.Corollary 3.3.
Letθ≥0 and r,s>0 be real numbers such that λ:=r+s≠2. Suppose that an even function f:X→Y with f(0)=0 satisfies the inequality
(3.31)‖Df(x,y)‖Y≤θ‖x‖Xr‖y‖Xs,
for all x,y∈X. Then there exists a unique quadratic function Q:X→Y satisfying
(3.32)‖f(2x)-16f(x)-Q(x)‖Y≤M2θk2(k2-1)(12p(k2p+ksp)|4p-2λp|)1/p‖x‖Xλ,
for all x∈X.Proof.
In Theorem3.1, taking φq(x,y):=θ∥x∥Xr∥y∥Xs, for all x,y∈X, we arrive at the desired result.Theorem 3.4.
Letj∈{-1,1} be fixed and let φv:X×X→[0,∞) be a function such that
(3.33)lim⁡n→∞16njφv(x2nj,y2nj)=0,
for all x,y∈X and
(3.34)∑i=(1+j)/2∞16ipjφvp(u2ij,y2ij)<∞,
for all (u,y)∈{(x,0),(2x,0),(x,x),(x,kx):x∈X}. Suppose that an even function f:X→Y with f(0)=0 satisfies the inequality
(3.35)‖Df(x,y)‖Y≤φv(x,y),
for all x,y∈X. Then there exists a unique quartic function V:X→Y such that
(3.36)‖f(2x)-4f(x)-V(x)‖Y≤M216(ψ̃v(x))1/p,
for all x∈X, where
(3.37)ψ̃v(x):=∑i=(1+j)/2∞16ipjk2p(k2-1)p{(12k2)pφvp(x2ij,x2ij)+(12(k2-1))pφvp(x2ij,0)+6pφvp(2x2ij,0)+12pφvp(x2ij,kx2ij)}.Proof.
Being similar to the proof of Theorem3.1, we omit its proof.Corollary 3.5.
Letθ,r,s be nonnegative real numbers such that r,s>4 or r,s<4. Suppose that an even function f:X→Y with f(0)=0 satisfies the inequality (3.28) for all x,y∈X. Then there exists a unique quartic function V:X→Y satisfying
(3.38)‖f(2x)-4f(x)-V(x)‖Y≤M2θk2(k2-1)γv(x),
for all x∈X, where
(3.39)γv(x)=(12p[k2p+(k2-1)p+(2r-1)p+1]|16p-2rp|‖x‖Xrp+12p(k2p+ksp)|16p-2sp|‖x‖Xsp)1/p,
for all x∈X.Corollary 3.6.
Letθ≥0 and r,s>0 be real numbers such that λ:=r+s≠4. Suppose that an even function f:X→Y with f(0)=0 satisfies the inequality (3.31) for all x,y∈X. Then, there exists a unique quartic function V:X→Y satisfying
(3.40)‖f(2x)-4f(x)-V(x)‖Y≤M2θk2(k2-1)(12p(k2p+ksp)|16p-2λp|)1/p‖x‖Xλ,
for all x∈X.Now, we are ready to prove the main theorem concerning the stability problem for (1.7).Theorem 3.7.
Letj∈{-1,1} be fixed and let φ:X×X→[0,∞) be a function such that
(3.41)lim⁡n→∞((1-j2)4njφ(x2nj,y2nj)+(1+j2)16njφ(x2nj,y2nj))=0,
for all x,y∈X and
(3.42)∑i=(1+j)/2∞((1-j2)4ipjφp(u2ij,y2ij)+(1+j2)16ipjφp(u2ij,y2ij))<∞,
for all (u,y)∈{(x,0),(2x,0),(x,x),(x,kx):x∈X}. Suppose that an even function f:X→Y with f(0)=0 satisfies the inequality
(3.43)‖Df(x,y)‖Y≤φ(x,y),
for all x,y∈X. Then, there exists a unique quadratic function Q:X→Y and a unique quartic function V:X→Y such that
(3.44)‖f(x)-Q(x)-V(x)‖Y≤M3192{4(ψ̃q(x))1/p+(ψ̃v(x))1/p},
for all x∈X, where
(3.45)ψ̃q(x):=∑i=(1+j)/2∞4ipjk2p(k2-1)p{(12k2)pφp(x2ij,x2ij)+(12(k2-1))pφp(x2ij,0)+6pφp(2x2ij,0)+12pφp(x2ij,kx2ij)},ψ̃v(x):=∑i=(1+j)/2∞16ipjk2p(k2-1)p{(12k2)pφp(x2ij,x2ij)+(12(k2-1))pφp(x2ij,0)+6pφp(2x2ij,0)+12pφp(x2ij,kx2ij)}.Proof.
By Theorems3.1 and 3.4, there exists a quadratic function Q0:X→Y and a quartic function V0:X→Y such that
(3.46)‖f(2x)-16f(x)-Q0(x)‖Y≤M24[ψ̃q(x)]1/p,‖f(2x)-4f(x)-V0(x)‖Y≤M216[ψ̃v(x)]1/p,
for all x∈X. Therefore, it follows from (3.46) that
(3.47)‖f(x)+112Q0(x)-112V0(x)‖Y≤M3192(4[ψ̃q(x)]1/p+[ψ̃v(x)]1/p),
for all x∈X. Thus we obtain (3.44) by letting Q(x)=-(1/12)Q0(x) and V(x)=(1/12)V0(x) for all x∈X.
To prove the uniqueness property ofQ and V, let Q′,V′:X→Y be another quadratic and quartic functions satisfying (3.44). Let Q¯=Q-Q′ and V¯=V-V′. Hence,(3.48)‖Q¯(x)+V¯(x)‖Y≤M{‖f(x)-Q(x)-V(x)‖Y+‖f(x)-Q′(x)-V′(x)‖Y}≤M496(4[ψ̃q(x)]1/p+[ψ̃v(x)]1/p),
for all x∈X. Since limn→∞4npjψ̃q(x/2n)=limn→∞16npjψ̃v(x/2n)=0, for all x∈X, we figure out that
(3.49)lim⁡n→∞16n‖Q¯(x2n)+V¯(x2n)‖Y=0,
for all x∈X. Therefore, we get V¯=0 and then Q¯=0.Corollary 3.8.
Letθ,r,s be nonnegative real numbers such that r,s>4 or 2<r, s<4 or r,s<2. Suppose that an even function f:X→Y with f(0)=0 satisfies the inequality (3.28), for all x,y∈X. Then, there exists a unique quadratic function Q:X→Y and a unique quartic function V:X→Y such that
(3.50)‖f(x)-Q(x)-V(x)‖Y≤M3θ12k2(k2-1)(γq(x)+γv(x)),
for all x∈X, where γq(x) and γv(x) are defined as in Corollaries 3.2 and 3.5.Corollary 3.9.
Letθ≥0 and r,s>0 be non-negative real numbers such that λ:=r+s∈(0,2)∪(2,4)∪(4,∞). Suppose that an even function f:X→Y with f(0)=0 satisfies the inequality (3.31) for all x,y∈X. Then there exist a unique quadratic function Q:X→Y and a unique quartic function V:X→Y such that
(3.51)‖f(x)-Q(x)-V(x)‖Y≤M3θ12k2(k2-1){(12p(k2p+ksp)|4p-2λp|)1/p+(12p(k2p+ksp)|16p-2λp|)1/p}‖x‖Xλ,
for all x∈X.Corollary 3.10.
Suppose that an even functionf:X→Y with f(0)=0 satisfies the inequality
(3.52)‖Df(x,y)‖X≤ɛ,
for all x,y∈X where ɛ>0. Then there exist a unique quadratic function Q:X→Y and a unique quartic function V:X→Y such that
(3.53)‖f(x)-Q(x)-V(x)‖Y≤M3ɛk2(k2-1){(k2p+(k2-1)p+2-p+14p-1)1/p+(k2p+(k2-1)p+2-p+116p-1)1/p},
for all x∈X.



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*Source: 734567-2011-07-06.xml*