AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation41747310.1155/2009/417473417473Research ArticleSolution and Stability of a Mixed Type Cubic and Quartic Functional Equation in Quasi-Banach SpacesEshaghi GordjiM.1ZolfaghariS.1RassiasJ. M.2SavadkouhiM. B.1LitsynElena1Department of MathematicsSemnan UniversityP. O. Box 35195-363SemnanIransemnan.ac.ir2Section of Mathematics and InformaticsPedagogical DepartmentNational and Capodistrian University of Athens4, Agamemnonos StreetAghia Paraskevi15342 AthensGreeceuoa.gr200919082009200924012009060820092009Copyright © 2009This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We obtain the general solution and the generalized Ulam-Hyers stability of the mixed type cubic and quartic functional equation f(x+2y)+f(x2y)=4(f(x+y)+f(xy))24f(y)6f(x)+3f(2y) in quasi-Banach spaces.

1. Introduction

We recall some basic facts concerning quasiBanach space. A quasinorm is a real-valued function on X satisfying the following.

x0 for all xX and x=0 if and only if x=0.

λ·x=|λ|·x for all λ and all xX.

There is a constant K1 such that x+yK(x+y) for all x,yX.

The pair (X,·) is called a quasinormed space if · is a quasinorm on X. A quasiBanach space is a complete quasinormed space. A quasinorm · is called a p-norm (0<p1) if x+ypxp+yp for all x,yX. In this case, a quasiBanach space is called a p-Banach space. Given a p-norm, the formula d(x,y):=x-yp gives us a translation invariant metric on X. By the Aoki-Rolewicz theorem  (see also ), each quasinorm is equivalent to some p-norm. Since it is much easier to work with p-norms, henceforth we restrict our attention mainly to p-norms. The stability problem of functional equations originated from a question of Ulam  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:G1G2 satisfies the inequality d(h(x·y),h(x)*h(y))<δ for all x,yG1, then there exists a homomorphism H:G1G2 with d(h(x),H(x))<ϵ for all xG1? In the 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  gave the first affirmative answer to the question of Ulam for Banach spaces. Let f:EE be a mapping between Banach spaces such that f(x+y)-f(x)-f(y)δ for all x,yE, and for some δ>0. Then there exists a unique additive mapping T:EE such that f(x)-T(x)δ   for all xE. Moreover if f(tx) is continuous in t for each fixed xE, then T is linear. Rassias  succeeded in extending the result of Hyers' Theorem by weakening the condition for the Cauchy difference controlled by (xp+yp), p[0,1) to be unbounded. This condition has been assumed further till now, through the complete Hyers direct method, in order to prove linearity for generalized Hyers-Ulam stability problem forms. A number of mathematicians were attracted to the pertinent stability results of Rassias , and stimulated to investigate the stability problems of functional equations. The stability phenomenon that was introduced and proved by Rassias is called Hyers-Ulam-Rassias stability. And then 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 [4, 5, 718]).

The following cubic functional equation, which is the oldest cubic functional equation, was introduced by the third author of this paper, Rassias  (in 2001): f(x+2y)+3f(x)=3f(x+y)+f(x-y)+6f(y). Jun and Kim  introduced the following cubic functional equation: 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×XY such that f(x)=C(x,x,x) for all xX, and C is symmetric for each fixed one variable and is additive for fixed two variables (see also ).

The quartic functional equation (1.6) was introduced by Rassias  (in 2000) and then (in 2005) was employed by Park and Bae  and others, such that: 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 multiadditive function Q:X×X×X×XY such that f(x)=Q(x,x,x,x) for all x (see also ). It is easy to show 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. In this paper we deal with the following functional equation: f(x+2y)+f(x-2y)=4(f(x+y)+f(x-y))-24f(y)-6f(x)+3f(2y) in quasiBanach spaces. It is easy to see that the function f(x)=ax3+bx4 is a solution of the functional equation (1.7). In the present paper we investigate the general solution of functional equation (1.7) when f is a mapping between vector spaces, and we establish the generalized Hyers-Ulam-Rassias stability of the functional equation (1.7) whenever f is a mapping between two quasiBanach spaces. We only mention here the papers [30, 31] concerning the stability of the mixed type functional equations.

2. General Solution

Throughout this section, X and Y will be real vector spaces. Before proceeding to the proof of Theorem 2.3 which is the main result in this section, we shall need the following two lemmas.

Lemma 2.1.

If an even function f:XY satisfies (1.7), then f is quartic.

Proof.

Putting x=y=0 in (1.7), we get f(0)=0. Setting x=0 in (1.7), by evenness of f we obtain f(2y)=16f(y) for all yX. Hence (1.7) can be written as f(x+2y)+f(x-2y)=4(f(x+y)+f(x-y))+24f(y)-6f(x). This means that f is quartic function, which completes the proof of the lemma.

Lemma 2.2.

If an odd function f:XY satisfies (1.7), then f is a cubic function.

Proof.

Setting x=y=0 in (1.7) gives f(0)=0. Putting x=0 in (1.7), then by oddness of f, we have f(2y)=8f(y). Hence (1.7) can be written as f(x+2y)+f(x-2y)=4f(x+y)+4f(x-y)-6f(x). Replacing x by x+y in (2.4), we obtain f(x+3y)+f(x-y)=4f(x+2y)-6f(x+y)+4f(x). Substituting -y for y in (2.5) gives f(x-3y)+f(x+y)=4f(x-2y)-6f(x-y)+4f(x). If we subtract (2.5) from (2.6), we obtain f(x+3y)-f(x-3y)=4f(x+2y)-4f(x-2y)-5f(x+y)+5f(x-y). Let us interchange x and y in (2.7). Then we see that f(3x+y)+f(3x-y)=4f(2x+y)+4f(2x-y)-5f(x+y)-5f(x-y). With the substitution y:=x+y in (2.4), we have f(3x+2y)-f(x+2y)=4f(2x+y)-4f(y)-6f(x). From the substitution y:=-y in (2.9) it follows that f(3x-2y)-f(x-2y)=4f(2x-y)+4f(y)-6f(x). If we add (2.9) to (2.10), we have f(3x+2y)+f(3x-2y)=4f(2x+y)+4f(2x-y)+f(x+2y)+f(x-2y)-12f(x). Replacing x by 2x in (2.7) and using (2.3), we obtain f(2x+3y)-f(2x-3y)=32f(x+y)-32f(x-y)-5f(2x+y)+5f(2x-y). Interchanging x with y in (2.12) gives the equation f(3x+2y)+f(3x-2y)=32f(x+y)+32f(x-y)-5f(x+2y)-5f(x-2y). If we compare (2.11) and (2.13) and employ (2.4), we conclude that f(2x+y)+f(2x-y)=2f(x+y)+2f(x-y)+12f(x).         This means that f is cubic function. This completes the proof of Lemma.

Theorem 2.3.

A function f:XY satisfies (1.7) for all x,yX if and only if there exists a unique function C:X×X×XY and a unique symmetric multiadditive function Q:X×X×X×XY such that f(x)=C(x,x,x)+Q(x,x,x,x) for all xX, and that C is symmetric for each fixed one variable and is additive for fixed two variables.

Proof.

Let f satisfy (1.7). We decompose f into the even part and odd part by setting fe(x)=12(f(x)+f(-x)),fo(x)=12(f(x)-f(-x)) for all xX. By (1.7), we have fe(x+2y)+fe(x-2y)=12[f(x+2y)+f(-x-2y)+f(x-2y)+f(-x+2y)]=4(fe(x+y)+fe(x-y))-24fe(y)-6fe(x)+3fe(2y) for all x,yX. This means that fe satisfies in (1.7). Similarly we can show that fo satisfies (1.7). By Lemmas 2.1 and 2.2, fe and fo are quartic and cubic, respectively. Thus there exists a unique function C:X×X×XY and a unique symmetric multiadditive function Q:X×X×X×XY such that fe(x)=Q(x,x,x,x) and that fo(x)=C(x,x,x) for all xX, and C is symmetric for each fixed one variable and is additive for fixed two variables. Thus f(x)=C(x,x,x)+Q(x,x,x,x) for all xX. The proof of the converse is trivial.

3. Stability

Throughout this section, X and Y will be a uniquely two-divisible abelian group and a quasiBanach spaces respectively, and p will be a fixed real number in [0,1]. We need the following lemma in the main theorems. Now before taking up the main subject, given f:XY, we define the difference operator Df:X×XY by Df(x,y)=f(x+2y)+f(x-2y)-4[f(x+y)+f(x-y)]-3f(2y)+24f(y)+6f(x) for all x,yX. We consider the following functional inequality: Df(x,y)ϕ(x,y) for an upper bound ϕ:X×X[0,).

Lemma 3.1.

Let x1,x2,,xn be nonnegative real numbers. Then (i=1nxi)pi=1nxip.

Theorem 3.2.

Let l{1,-1} be fixed and let φ:X×X+ be a function such that limn16lnφ(x2ln,y2ln)=0 for all x,yX and i=116ilpφp(0,y2li)< for all yX. Suppose that an even function f:XY with f(0)=0 satisfies the inequality Df(x,y)Yφ(x,y), for all x,yX. Then the limit Q(x):=limn16lnf(x2ln) exists for all xX and Q:XY is a unique quartic function satisfying f(x)-Q(x)YK16[ψ̃e(x)]1/p, where ψ̃e(x):=i=|l+1|/216ilpφp(0,x2li) for all xX.

Proof.

Let l=1. By putting x=0 in (3.6), we get f(2y)-16f(y)Yφ(0,y) for all yX. Replacing y by x in (3.10) yields f(2x)-16f(x)Yφ(0,x) for all xX. Let ψ(x)=φ(0,x) for all xX, then by (3.11), we get f(2x)-16f(x)Yψ(x) for all xX. Interchanging x with x/2n+1 in (3.12), and multiplying by 16n it follows that 16n+1f(x2n+1)-16nf(x2n)YK16nψ(x2n+1) for all xX and all nonnegative integers n. Since Y is p-Banach space, then by (3.13) we have 16n+1f(x2n+1)-16mf(x2m)Ypi=mn16i+1f(x2i+1)-16if(x2i)YpKpi=mn16ipψp(x2i+1) for all nonnegative integers n and m with nm and all xX. Since ψp(x)=φp(0,x) for all xX. Therefore by (3.5) we have i=116ipψp(x2i)< for all xX. Therefore we conclude from (3.14) and (3.15) that the sequence {16nf(x/2n)} is a Cauchy sequence for all xX. Since Y is complete, it follows that the sequence {16nf(x/2n)} converges for all xX. We define the mapping Q:XY by (3.7) for all xX. Letting m=0 and passing the limit n in (3.14), we get f(x)-Q(x)YpKpi=016ipψp(x2i+1)=Kp16pi=116ipψp(x2i) for all xX. Therefore (3.8) follows from (3.9) and (3.16). Now we show that Q is quartic. It follows from (3.4), (3.6) and (3.7) DQ(x,y)Y=limn16nDf(x2n,y2n)Ylimn16nφ(x2n,y2n)=0 for all x,yX. Therefore the mapping Q:XY satisfies (1.7). Since Q(0)=0, then by Lemma 2.1 we get that the mapping Q:XY is quartic. To prove the uniqueness of Q, let T:XY be another quartic mapping satisfies (3.8). Since limn16npi=116ipφp(x2i+n,y2i+n)=limni=n+116ipφp(x2i,y2i)=0 for all yX and all x{0}, then limn16npψ̃e(x2n)=0 for all xX. It follows from (3.8), (3.19) Q(x)-T(x)Yp=limn16npf(x2n)-T(x2n)YpKp16plimn16npψ̃e(x2n)=0 for all xX. Hence Q=T. For l=-1, we obtain f(2nx)16n-f(x)YpKp16pi=0φp(0,2ix)16ip, from which one can prove the result by a similar technique.

Corollary 3.3.

Let θ,r,s,u,v be nonnegative real numbers such that s4u+v. Suppose that an even function f:XY with f(0)=0 satisfies the inequality Df(x,y)Yθ(xXuyXv+xXr+yXs) for all x,yX. Then there exists a unique quartic function Q:XY satisfying f(x)-Q(x)YKθ{1|16p-2sp|}1/pxXs for all xX.

Proof.

It follows from Theorem 3.2thatφ(x,y):=θ(xXuyXv+xXr+yXs) for all x,yX.

Theorem 3.4.

Let l{1,-1} be fixed and let φ:X×X[0,) be a function such that limn8lnφ(x2ln,y2ln)=0 for all x,yX and i=18ilpφp(0,y2il)< for all yX. Suppose that an odd function f:XY satisfies the inequality Df(x,y)Yφ(x,y), for all x,yX. Then the limit C(x):=limn8lnf(x2ln) exists for all xX and C:XY is a unique cubic function satisfying f(x)-C(x)YK24[ϕ̃o(x)]1/p for all xX, where ϕ̃o(x):=i=|l+1|/28ilpφp(0,x2il).

Proof.

Let l=1. Setting x=0 in (3.26), we get 3f(2y)-24f(y)Yφ(0,y) for all yX. If we replace y in (3.30) by x and divide both sides of (3.30) by 3, we get f(2x)-8f(x)Y13φ(0,x) for all xX. Let ϕ(x)=(1/3)φ(0,x) for all xX, then by (3.31), we get f(2x)-8f(x)Yϕ(x) for all xX. Multiply (3.32) by 8n and replace x by x/2n+1, we obtain that 8n+1f(x2n+1)-8nf(x2n)YK8nϕ(x2n+1) for all xX and all nonnegative integers n. Since Y is a p-Banach space, (3.33) follows that 8n+1f(x2n+1)-8mf(x2m)Ypi=mn8i+1f(x2i+1)-8if(x2i)YpKpi=mn8ipϕp(x2i+1) for all nonnegative integers n and m with nm and all xX. Since ϕp(x)=(1/3p)φp(0,x) for all xX. Therefore it follows from (3.25) that i=18ipϕp(x2i)< for all xX, therefore we conclude from (3.34) and (3.35) that the sequence {8nf(x/2n)} is a Cauchy sequence for all xX. Since Y is complete, the sequence {8nf(x/2n)} converges for all xX. So one can define the mapping C:XY by (3.27) for all xX. Letting m=0 and passing the limit n in (3.34), we get f(x)-C(x)YpKpi=08ipϕp(x2i+1)=Kp8pi=18ipϕp(x2i) for all xX. Therefore (3.28) follows from (3.29) and (3.36). Now we show that C is cubic. It follows from (3.24), (3.26) and (3.27) DC(x,y)Y=limn8nDf(x2n,y2n)Ylimn8nφ(x2n,y2n)=0 for all x,yX. Therefore the mapping C:XY satisfies (1.7). Since f is an odd function, then (3.27) implies that the mapping odd. Therefore by Lemma 2.2 we get that the mapping C:XY is cubic. The rest of proof is similar to the proof of Theorem 3.2.

Corollary 3.5.

Let θ be a nonnegative real number and r,s be real numbers such that s3u+v. Suppose that an odd function f:XY satisfies the inequality Df(x,y)Yθ(xXuyXv+xXr+yXs) for all x,yX. Then there exists a unique cubic function C:XY satisfying f(x)-C(x)YKθ3{1|8p-2sp|}1/pxXs for all xX.

Proof.

It follows from (3.38) and Theorem 3.4thatφ(x,y):=θ(xXuyXv+xXr+yXs) for all x,yX.

Theorem 3.6.

Let l{1,-1} be fixed and let φ:X×X[0,) be a function which satisfies limn{(|l|+l2)16lnφ(x2ln,y2ln)+(|l|-l2)8lnφ(x2ln,y2ln)}=0 for all x,yX and i=|l+1|/2{(|l|+l2)16ilpφp(0,y2li)+(|l|-l2)8ilpφp(0,y2li)}< for all yX. Suppose that a function f:XY with f(0)=0 satisfies the inequality Df(x,y)Yφ(x,y) for all x,yX. Then there exists a unique quartic function Q:XY and a unique cubic function C:XY satisfying (1.7) and f(x)-Q(x)-C(x)YK332[ψ̃e(x)+ψ̃e(-x)]1/p+K348[ϕ̃o(x)+ϕ̃o(-x)]1/p for all xX, where ψ̃e(x) and ϕ̃o(x) that have been defined in (3.9) and (3.29), respectively.

Proof.

Let fe(x)=(1/2)(f(x)+f(-x)) for all xX. Then fe(0)=0,fe(-x)=fe(x), and Dfe(x,y)(K/2)[φ(x,y)+φ(-x,-y)] for all x,yX. Let Φ(x,y)=K2[φ(x,y)+φ(-x,-y)] for all x,yX. So limn16lnΦ(x2ln,y2ln)=0 for all x,yX. Since Φp(x,y)Kp2p[φp(x,y)+φp(-x,-y)] for all x,yX, then i=116ilpΦp(x2il,y2il)< for all yX and all x{0}. Hence, in view of Theorem 3.2, there exists a unique quartic function Q:XY satisfying fe(x)-Q(x)YK16[Ψ̃e(x)]1/p for all xX, where Ψ̃e(x):=i=116ilpΦp(0,x2il). We have Ψ̃e(x)Kp2p[ψ̃e(x)+ψ̃e(-x)] for all xX. Therefore it follows from (3.48) that, fe(x)-Q(x)YK232[ψ̃e(x)+ψ̃e(-x)]1/p for all xX. Let fo(x)=(1/2)(f(x)-f(-x)) for all xX. Then fo(0)=0,fo(-x)=-fo(x), and Dfo(x,y)Φ(x,y) for all x,yX. From Theorem 3.4, it follows that there exists a unique cubic function C:XY satisfying fo(x)-C(x)YK24[Φ̃o(x)]1/p for all xX, where Φ̃o(x):=i=18ipΦp(0,x2i). Since Φ̃o(x)Kp2p[ϕ̃o(x)+ϕ̃o(-x)] for all xX, it follows from (3.52) that, fo(x)-C(x)YK248[ϕ̃o(x)+ϕ̃o(-x)]1/p for all xX. Hence (3.43) follows from (3.51) and (3.55).

Corollary 3.7.

Let θ,r,s be nonnegative real numbers such that u+v,s(4,)(-,3). Suppose that a function f:XY with f(0)=0 satisfies the inequality Df(x,y)Yθ(xXuyXv+xXr+yXs) for all x,yX. Then there exists a unique quartic function Q:XY and a unique cubic function C:XY satisfying (1.7) and f(x)-Q(x)-C(x)YK3θ3{3[1|16p-2sp|]1/p+[1|8p-2sp|]1/p}xXs for all xX.

Proof.

It follows from Theorem 3.6that φ(x,y):=θ(xXuyXv+xXr+yXs) for all x,yX.

Acknowledgment

The second and fourth authors would like to thank the office of gifted students at Semnan University for its financial support.

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