# Generalized Hyers-Ulam Stability of a Mixed Type Functional Equation

Authors: Yang-Hi Lee; Soon-Mo Jung
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

We investigate the stability of a functional equationfx+y+z + fx−y+z + fx+y−z+f−x+y+z=3fx+f−x+3fy+f−y+3fz+f−z by applying the direct method in the sense of Hyers and Ulam.


---

## Body

## 1. Introduction

In 1940, Ulam [1] gave a wide ranging talk before the mathematics club of the University of Wisconsin in which he discussed a number of important unsolved problems. Among those was the question concerning the stability of group homomorphisms: LetG1 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 function h:G1→G2 satisfies the inequality d(h(xy),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?The Ulam’s problem for the Cauchy additive functional equation was solved by Hyers under the assumption thatG1 and G2 are Banach spaces. Indeed, Hyers [2] proved that every solution of the inequality ∥f(x+y)-f(x)-f(y)∥≤ε (for all x and y) can be approximated by an additive function. In this case, the Cauchy additive functional equation, f(x+y)=f(x)+f(y), is said to satisfy the Hyers-Ulam stability.Thereafter, Rassias [3] attempted to weaken the condition for the bound of norm of the Cauchy difference as follows:
(1)∥f(x+y)-f(x)-f(y)∥≤ε(∥x∥p+∥y∥p),
and he proved that Hyers’ theorem is also true for this case. Indeed, Rassias proved the generalized Hyers-Ulam stability (or the Hyers-Ulam-Rassias stability) of the Cauchy additive functional equation between Banach spaces. We here remark that a paper of Aoki [4] was published concerning the generalized Hyers-Ulam stability of the Cauchy functional equation earlier than Rassias’ paper.The stability concept that was introduced by Rassias’ theorem provided a large influence to a number of mathematicians to develop the notion of what is known today with the term generalized Hyers-Ulam stability of functional equations. Since then, the stability problems of several functional equations have been extensively investigated by several mathematicians (e.g., see [5–10] and the references therein).Almost all subsequent proofs in this very active area have used the Hyers’ method presented in [2]. Namely, starting from the given mapping f that approximately satisfies a given functional equation, a solution F of the functional equation is explicitly constructed by using the formula:
(2)F(x)≔limn→∞12nf(2nx)orF(x)≔limn→∞2nf(x2n),
which approximates the mapping f. This method of Hyers is called the direct method.We remark that another method for proving the Hyers-Ulam stability of various functional equations was introduced by Baker [11], which is called the fixed-point method. This method is very powerful technique of proving the stability of functional equations (see [12, 13]).Now we consider the following functional equation:(3)f(x+y+z)+f(x-y+z)+f(x+y-z)+f(-x+y+z)=3f(x)+f(-x)+3f(y)+f(-y)+3f(z)+f(-z),
which is called the mixed type functional equation. The mapping f(x)=ax2+bx is a solution of this functional equation, where a,b are real constants. Every solution of (3) will be called a quadratic-additive mapping.In 1998, Jung [14] proved the stability of (3) by decomposing f into the odd and even parts. In his proof, using the direct method, an additive mapping A and a quadratic mapping Q are separately constructed from the odd and even parts of f, and then A and Q are combined to provide a quadratic-additive mapping F which is close to f.In this paper, we will prove the generalized Hyers-Ulam stability of (3) by making use of the direct method. In particular, we will approximate the given mapping f by a solution F of (3) without decomposing f into its odd and even parts, while in the Jung’s paper [14] the mapping f was decomposed into the odd and even parts, and each of them was separately approximated by the corresponding part of a solution F of (3).


## 2. Main Results

Throughout this paper, letX be a (real or complex) normed space and Y a Banach space. For an arbitrary p∈ℝ, we define s≔sign(2-p) and t≔sign(1-p).For a given mappingf:X→Y, we use the following abbreviations:
(4)fo(x)≔f(x)-f(-x)2,fe(x)≔f(x)+f(-x)2,Jnf(x)≔9-sn2(f(3snx)+f(-3snx))+3-tn2(f(3tnx)-f(-3tnx)),Af(x,y)≔f(x+y)-f(x)-f(y),Qf(x,y)≔f(x+y)+f(x-y)-2f(x)-2f(y),Df(x,y,z)≔f(x+y+z)+f(x-y+z)+f(x+y-z)+f(-x+y+z)-3f(x)-f(-x)-3f(y)-f(-y)-3f(z)-f(-z)
for all x,y,z∈X.As we stated in the previous section,f is called a quadratic-additive mapping provided that f satisfies the functional equation Df(x,y,z)=0 for all x,y,z∈X.Proposition 1.
A mappingf:X→Y is a solution of (3) if and only if fe is a quadratic mapping and fo is an additive mapping.Proof.
Assume thatf:X→Y is a solution of (3). Then we have
(5)Qfe(x,y,z)=Dfe(x,y,0)2=0,Afo(x,y,z)=Dfo(x,y,0)2=0
for all x,y,z∈X, that is, fe is a quadratic mapping and fo is an additive mapping.
Conversely, assume thatfe is a quadratic mapping and fo is an additive mapping. Then we get
(6)Df(x,y,z)=Dfe(x,y,z)+Dfo(x,y,z)=Qfe(x+y,z)+Qfe(x-y,z)+2Qfe(x,y)+Afo(x+y,z)+Afo(x+y,-z)+Afo(x-y,z)+Afo(-x+y,z)+2Afo(x,y)=0
for all x,y,z∈X; that is, f is a solution of (3).We first prove the following lemma.Lemma 2.
If a mappingf:X→Y satisfies Df(x,y,z)=0 for all x,y,z∈X∖{0} and f(0)=0, then f is a quadratic-additive mapping.Proof.
Using the hypothesis, we have(7)f(2x)-3f(x)-f(-x)=11112(Df(4x,3x,x)-Df(4x,2x,2x)-Df(2x,2x,2x)+2Df(2x,x,x)+3Df(x,x,x)+Df(-x,-x,-x))-3112(Df(-4x,-3x,-x)-Df(-4x,-2x,-2x)-Df(-2x,-2x,-2x)+2Df(-2x,-x,-x)+3Df(-x,-x,-x)+Df(x,x,x))=0
for all x∈X∖{0}. Furthermore, by the last equality, we get
(8)Df(x,y,0)=Df(x,y2,y2)+Df(y,x2,x2)-2f(x)+6f(x2)+2f(-x2)-2f(y)+6f(y2)+2f(-y2)=0
for all x,y∈X∖{0}. Since Df(x,y,z) is invariant with respect to the permutation of (x,y,z), it holds that Df(0,y,z)=Df(x,0,z)=0 for all x,y,z∈X∖{0}. It is also easy to show that Df(x,0,0)=0, Df(0,y,0)=0, Df(0,0,z)=0, and Df(0,0,0)=0 for all x,y,z∈X∖{0} as we desired.In the following theorem, we can prove the generalized Hyers-Ulam stability of the functional equation (3) by making use of the direct method.Theorem 3.
If a mappingf:X→Y satisfies f(0)=0 and
(9)∥Df(x,y,z)∥≤∥x∥p+∥y∥p+∥z∥p
for all x,y,z∈X∖{0} with a real constant p∉{1,2}, then there exists a unique quadratic-additive mapping F:X→Y such that
(10)∥f(x)-F(x)∥≤{3∥x∥p3p-9(forp>2),(39-3p+33p-3)∥x∥p(for1<p<2),3∥x∥p3-3p(for0≤p<1)
for all x∈X∖{0}. Moreover, if p<0, then f itself is a quadratic-additive mapping.Proof.
Let us defineτs,n≔s(n+1/2)-1/2, where s∈{-1,1}. From the definitions of Jnf(x) and Df(x,y,z), we have
(11)Jnf(x)-Jn+1f(x)=-12(9τ-s,n(Df(3τs,nx,3τs,nx,3τs,nx)+Df(-3τs,nx,-3τs,nx,-3τs,nx))s+3τ-t,n(Df(3τt,nx,3τt,nx,3τt,nx)-Df(-3τt,nx,-3τt,nx,-3τt,nx))t)
for all x∈X∖{0} and n∈ℕ0. It follows from (9) and (11) that
(12)∥Jnf(x)-Jn+mf(x)∥=∑j=nn+m-1‍∥Jjf(x)-Jj+1f(x)∥≤12∑j=nn+m-1‍(∥9τ-s,jDf(3τs,jx,3τs,jx,3τs,jx)s+3τ-t,jDf(3τt,jx,3τt,jx,3τt,jx)t∥+∥9τ-s,jDf(-3τs,jx,-3τs,jx,-3τs,jx)s-3τ-t,jDf(-3τt,jx,-3τt,jx,-3τt,jx)t∥)≤{∑j=nn+m-13-j∥3jx∥p(forp<1),∑j=nn+m-1(3-2j-1∥3jx∥p+3j+1∥3-j-1x∥p)(for1<p<2),∑j=nn+m-1‍32j+1∥3-j-1x∥p(forp>2)≤{3np∥x∥p3n-1(3-3p)(forp<1),3np∥x∥p32n-1(9-3p)+3n+1∥x∥p3np(3p-3)(for1<p<2),32n+1∥x∥p3np(3p-9)(forp>2)
for all x∈X∖{0}. So, it is easy to show that the sequence {Jnf(x)} is a Cauchy sequence for all x∈X∖{0}.
SinceY is complete and f(0)=0, the sequence {Jnf(x)} converges for all x∈X. Hence, we can define a mapping F:X→Y by
(13)F(x)≔limn→∞Jnf(x)
for all x∈X. Moreover, if we put n=0 and let m→∞ in (12), we obtain the inequality (10).
From the definition ofF, we get
(14)DF(x,y,z)=limn→∞9-sn2(Df(3snx,3sny,3snz)+Df(-3snx,-3sny,-3snz))+limn→∞3-tn2(Df(3tnx,3tny,3tnz)-Df(-3tnx,-3tny,-3tnz))≤limn→∞(3-s(2-p)n+3-tn(1-p))(∥x∥p+∥y∥p+∥z∥p)=0
for all x,y,z∈X∖{0}. By Lemma 2, F is a quadratic-additive mapping.
Now, we will show thatF is uniquely determined. Let F′:X→Y be another quadratic-additive mapping satisfying (10). It is easy to show that F′(0)=0 for all quadratic-additive mapping F′. It follows from (11) that
(15)F′(x)-JnF′(x)=∑j=0n-1‍(JjF′(x)-Jj+1F′(x))=-12∑j=0n-1‍(9τ-s,j(DF′(3τs,jx,3τs,jx,3τs,jx)+DF′(-3τs,jx,-3τs,jx,-3τs,jx))s+3τ-t,j(DF′(3τt,jx,3τt,jx,3τt,jx)-DF′(-3τt,jx,-3τt,jx,-3τt,jx))t)=0
for all n∈ℕ and x∈X. Since F and F′ are quadratic-additive, if we replace x with 3nx in (10), then we have
(16)∥F(x)-F′(x)∥=∥JnF(x)-JnF′(x)∥≤9-sn2(∥(F-f)(3snx)∥+∥(f-F′)(3snx)∥+∥(F-f)(-3snx)∥+∥(f-F′)(-3snx)∥)+2-tn2(∥(F-f)(3tnx)∥+∥(F′-f)(3tnx)∥+∥(F-f)(-3tnx)∥+∥(F′-f)(-3tnx)∥)≤(6|9-3p|+6|3p-3|)(3-sn(2-p)+3-tn(1-p))∥x∥p
for all x∈X∖{0} and n∈ℕ. Taking the limit in the above inequality as n→∞, we can conclude that F(x)=F′(x) for all x∈X, which proves the uniqueness of F.
Since(17)∥f(x)-F(x)∥≤∥Df((2k-1)x,kx,kx)-DF((2k-1)x,kx,kx)∥+∥F((4k-1)x)-f((4k-1)x)∥+∥f((2k-1)x)-F((2k-1)x)∥+∥f((1-2k)x)-F((1-2k)x)∥+2∥f(-kx)-F(-kx)∥+6∥f(kx)-F(kx)∥≤((2k-1)p+2kp+3((4k-1)p+2(2k-1)p+8kp)3-3p)×∥x∥p
for all x∈X∖{0} and k∈ℕ, if p<0, then we conclude that f(x)=F(x) for all x∈X∖{0} by letting k→∞ in the previous inequality. From the fact that f(0)=0, f is a quadratic-additive mapping.Theorem 4.
If a mappingf:X→Y satisfies
(18)∥Df(x,y,z)∥≤∥x∥p+∥y∥p+∥z∥p
for all x,y,z∈X and for a nonnegative real constant p∉{1,2}, then there exists a unique quadratic-additive mapping F:X→Y such that
(19)∥f(x)-F(x)∥≤{∥x∥p2p-4(forp>2),(14-2p+12p-2)∥x∥p(for1<p<2),∥x∥p2-2p(for0<p<1),2516(forp=0)
for all x∈X.Proof.
Since(20)∥f(0)∥=∥18Df(0,0,0)∥≤3∥0∥p8,
we get f(0)=0 for p∉{0,1,2} and ∥f(0)∥≤3/8 for p=0. From the definitions of Jnf(x) and Df(x,y,z), we have
(21)Jnf(x)-Jn+1f(x)=-14(4τ-s,n(Df(2τs,nx,2τs,nx,0)+Df(-2τs,nx,-2τs,nx,0))s+2τ-t,n(Df(2τt,nx,2τt,nx,0)-Df(-2τt,nx,-2τt,nx,0))t)+124τ-s,nf(0)
for all x∈X and n∈ℕ0, where τs,n is defined by τs,n=s(n+1/2)-1/2 and s∈{-1,1}.
It follows from (18) and (21) that
(22)∥Jnf(x)-Jn+mf(x)∥=∑j=nn+m-1‍∥Jjf(x)-Jj+1f(x)∥≤14∑j=nn+m-1‍(∥4τ-s,jDf(2τs,jx,2τs,jx,0)s+2τ-t,jDf(2τt,jx,2τt,jx,0)t∥+∥4τ-s,jDf(-2τs,jx,-2τs,jx,0)s-2τ-t,jDf(-2τt,jx,-2τt,jx,0)t∥+2·4τ-s,j∥f(0)∥)≤{∑j=nn+m-1‍32j+2+3·2-2j-38(forp=0),∑j=nn+m-1‍2-j-1∥2jx∥p(for0<p<1),∑j=nn+m-1‍2-2j-2∥2jx∥p+2j∥2-j-1x∥p(for1<p<2),∑j=nn+m-1‍22j∥2-j-1x∥p(forp>2)≤{32n+1+122n+4(forp=0),2np∥x∥p2n(2-2p)(for0<p<1),2np∥x∥p4n(4-2p)+2n∥x∥p2np(2p-2)(for1<p<2),4n∥x∥p2np(2p-4)(forp>2)
for all x∈X. So, it is easy to show that the sequence {Jnf(x)} is a Cauchy sequence for all x∈X.
SinceY is complete, the sequence {Jnf(x)} converges for all x∈X. Hence, we can define a mapping F:X→Y by
(23)F(x)≔limn→∞Jnf(x)
for all x∈X. Moreover, putting n=0 and letting m→∞ in (22), we get the inequality (19).



---

*Source: 472531-2013-06-13.xml*