R E S E A R C H
Open Access
A fixed-point approach to the stability of a
functional equation on quadratic forms
Jae-Hyeong Bae
1and Won-Gil Park
2** Correspondence: [email protected]
2Department of Mathematics
Education, College of Education, Mokwon University, Daejeon, 302-729, Korea
Full list of author information is available at the end of the article
Abstract
Using the fixed-point method, we prove the generalized Hyers-Ulam stability of the functional equation
f(x+y,z+w) +f(x−y,z−w) = 2f(x,z) + 2f(y,w).
The quadratic formf:ℝ×ℝ®ℝgiven byf(x,y) =ax2 +bxy+cy2is a solution of the above functional equation.
Keywords:alternative of fixed point, functional equation, quadratic form, stability
1. Introduction
In 1940, S. M. 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 pro-blems. Among those was the question concerning the stability of group homomorphisms:
Let G1 be a group and let G2 be a metric group with the metric d(· ,·). Givenε>0,
does there exist aδ>0 such that if a function h:G1 ®G2satisfies the inequality d(h
(xy),h(x)h(y)) <δfor all x,y ÎG1, then there is a homomorphism H:G1®G2with d
(h(x),H(x)) <εfor all xÎG1?
The case of approximately additive mappings was solved by D. H. Hyers [2] under the assumption thatG1 andG2are Banach spaces. Thereafter, many authors
investi-gated solutions or stability of various functional equations (see [3-11]).
LetX be a set. A functiond:X ×X ®[0,∞] is called a generalized metric onXifd
satisfies
(1)d(x,y) = 0 if and only ifx=y; (2)d(x,y) =d(y,x) for allx,yÎX;
(3)d(x,z)≤d(x,y) +d(y,z) for allx,y,zÎX.
Note that the only substantial difference of the generalized metric from the metric is that the range of generalized metric includes the infinity.
Throughout this paper, letX andYbe two real vector spaces and let:X×X×X×
X® [0,∞) be a function. For a mappingf:X ×X ®Y, consider the functional equa-tion:
f(x+y,z+w) +f(x−y,z−w) = 2f(x,z) + 2f(y,w). (1:1)
The quadratic formf:ℝ×ℝ® ℝgiven byf(x,y) : =ax2+bxy+cy2 is a solution of the Equation 1.1.
The authors [12] acquired the general solution and proved the stability of the func-tional Equation 1.1 for the case that XandYare real vector spaces as follows.
Theorem A.A mapping f:X×X®Y satisfies the Equation1.1for all x,y,z,wÎX if and only if there exist two symmetric bi-additive mappings S,T :X ×X ® Y and a bi-additive mapping B:X×X®Y such that
f(x,y) =S(x,x) +B(x,y) +T(y,y)
for all x,y ÎX.
From now on, letYbe a complete normed space.
Theorem B.Assume thatsatisfies the condition ˜
ϕ(x,y,z,w) :=
∞
j=0
1 4j+1ϕ(2
jx, 2jy, 2jz, 2jw)<∞
for all x,y,z,wÎX. Let f:X×X®Y be a mapping such that
f(x+y,z+w) +f(x−y,z−w)−2f(x,z)−2f(y,w) ≤ϕ(x,y,z,w)
for all x,y, z,wÎ X. Then, there exists a unique mapping F: X×X ® Y satisfying the Equation 1.1such that
f(x,y)−F(x,y) ≤ ˜ϕ(x,x,y,y) (1:2)
for all x,y ÎX. The mapping F is given by
F(x,y) := lim
j→∞
1 4jf(2
jx, 2jy)
for all x,y ÎX.
In this paper, we prove the stability of the Equation 1.1 using the fixed-point method.
2. Stability using the alternative of fixed point
In this section, we investigate the stability of the functional Equation 1.1 using the alternative of fixed point. Before proceeding the proof, we will state the theorem, the alternative of fixed point.
Theorem 2.1. (The alternative of fixed point[13,14]). Suppose that we are given a complete generalized metric space (Ω, d)and a strictly contractive mapping T: Ω®
Ωwith Lipschitz constant L. Then, for each given x ÎΩ, either
d(Tnx,Tn+1x) =∞for all n≥0,
or
there exists a positive integer n0such that
• d(Tnx,Tn+1x)<∞for all n≥n0;
• the sequence(Tnx)is convergent to a fixed point y*of T;
• y*is the unique fixed point of T in the set ={y∈|d(Tn0x,y)<∞};
• d(y,y∗)≤ 1
Lemma 2.2.Letψ:X ×X® [0,∞)be a function given by
ψ(x,y) :=ϕ x
2, x 2,
y 2,
y 2
for all x, yÎX. Consider the setΩ: = {g|g:X×X ®Y,g(0, 0) = 0}and the gener-alized metric d onΩgiven by
d(g,h) =dψ(g,h) := infSψ(g,h),
where Sψ(g,h) : = {KÎ[0,∞] | ||g(x,y)- h(x,y) || ≤Kψ(x,y)for all x,y ÎX}for all g,hÎΩ. Then, (Ω,d)is complete.
Proof. Let {gn} be a Cauchy sequence in (Ω,d). Then, givenε>0, there existsNsuch
that d(gn, gk) <εifn, k≥ N. Let n, k≥ N. Sinced(gn, gk) = inf Sψ(gn, gk)< ε, there
existsKÎ [0,ε) such that
gn(x,y)−gk(x,y) ≤Kψ(x,y)≤εψ(x,y) (2:1)
for allx,yÎ X. So, for each x,yÎ X, {gn(x,y)} is a Cauchy sequence inY. SinceYis
complete, for eachx, yÎX, there existsg(x, y)ÎYsuch thatgn(x, y)®g(x,y) asn®
∞andg(0, 0) = 0. Thus, we havegÎΩ. By (2.1), we obtain that
n≥N⇒gn(x,y)−g(x,y) ≤εψ(x,y) for allx,y∈X ⇒ε∈Sψ(gn,g)
⇒d(gn,g) = infSψ(gn,g)≤ε.
Hence, gn® gÎΩasn®∞.
By using an idea of Cădariu and Radu (see [15]), we will prove the Hyers-Ulam stabi-lity of the functional equation related to quadratic forms.
Theorem 2.3.Assume thatsatisfies the condition
lim
n→∞
1 4nϕ(2
nx, 2ny, 2nz, 2nw) = 0
for all x,y,z, wÎ X. Suppose that a mapping f: X×X ®Y satisfies the functional inequality
f(x+y,z+w) +f(x−y,z−w)−2f(x,z)−2f(y,w) ≤ϕ(x,y,z,w) (2:2)
for all x, y, z, wÎ X and f(0, 0) = 0. If there exists L <1 such that the functionψ given in Lemma 2.2has the property
ψ(x,y)≤4Lψ x
2, y 2
(2:3)
for all x, yÎ X, then there exists a unique mapping F: X ×X ® Y satisfying(1.1)
such that the inequality
f(x,y)−F(x,y) ≤ L
1−Lψ(x,y) (2:4)
holds for all x,yÎ X.
Tg(x,y) := 1
4g(2x, 2y)
for all gÎΩand allx,y ÎX. Observe that, for allg,hÎΩ,
K∈Sψ(g,h) andK<K
⇒ g(x,y)−h(x,y) ≤Kψ(x,y)≤Kψ(x,y) for allx,y∈X
⇒ K∈Sψ(g,h).
Let g,hÎΩand εÎ (0,∞]. Then, there is aK’Î Sψ(g,h) such thatK’< d(g,h) +ε. By the above observation, we gaind(g,h) +εÎSψ(g,h). So we get ||g(x,y)- h(x, y)||≤ (d(g,h) +ε)ψ(x,y) for allx,yÎ X. Thus, we have
1
4g(2x, 2y)− 1
4h(2x, 2y) ≤ 1
4(d(g,h) +ε)ψ(2x, 2y)
for all x,yÎX. By (2.3), we obtain that
1
4g(2x, 2y)− 1
4h(2x, 2y)≤L(d(g,h) +ε)ψ(x,y)
for all x,yÎX. Hence,d(Tg,Th)≤L(d(g,h) +ε). Now we obtain that
d(Tg,Th)≤L(d(g,h) +ε)
for all εÎ (0,∞]. Taking the limit asε®0+ in the above inequality, we get
d(Tg,Th)≤Ld(g,h)
for all g,hÎΩ, that is,T is a strictly contractive mapping ofΩwith Lipschitz con-stantL.
Putting y=xand w=zin (2.2), by (2.3), we have the inequality
f(x,z)− 1
4f(2x, 2z) ≤ 1
4ϕ(x,x,z,z) = 1
4ψ(2x, 2z)≤Lψ(x,z) (2:5)
for all x,zÎX. Thus, we obtain that
d(f,Tf)≤L<∞. (2:6)
Applying the alternative of fixed point, we see that there exists a fixed point FofT
inΩsuch that
F(x,y) = lim
n→∞
1 4nf(2
nx, 2ny)
for all x,y Î X. Replacingx, y, z, wby 2nx, 2ny, 2nz, 2nwin (2.2), respectively, and dividing by 4n, we have
F(x+y,z+w) +F(x−y,z−w)−2F(x,z)−2F(y,w) = lim
n→∞
1 4n f(2
n(x+y), 2n(z+w)) +f(2n(x−y), 2n(z−w)) −2f(2nx, 2nz)−2f(2ny, 2nw)
≤ lim
n→∞
1 4nϕ(2
for all x, y, z, wÎ X. Thus, the mappingF satisfies the Equation 1.1. By (2.3) and (2.5), we obtain that
Tnf(x,y)−Tn+1f(x,y) = 1 4n f(2
nx, 2ny)−1
4f(2
n+1x, 2n+1y)
≤ L
4nψ(2
nx, 2ny)≤ · · · ≤ L
4n(4L) nψ(x,y)
=Ln+1ψ(x,y)
for allx,y ÎX and allnÎ N, that is,d(Tnf,Tn+1f)≤Ln+1<∞for allnÎN. By the fixed-point alternative, there exists a natural numbern0 such that the mappingFis the
unique fixed point of T in the set ={g∈|d(Tn0f,g)<∞}. So we have d(Tn0f,F)<∞. Since
d(f,Tn0f)≤d(f,Tf) +d(Tf,T2f) +· · ·+d(Tn0−1f,Tn0f)<∞,
we getfÎΔ. Thus, we have d(f,F)≤d(f,Tm0f) +d(Tm0f,F)<∞. Hence, we obtain
||f(x,y)−F(x,y)|| ≤Kψ(x,y)
for all xÎXand someKÎ[0,∞). Again using the fixed-point alternative, we have
d(f,F)≤ 1
1−Ld(f,Tf).
By (2.6), we may conclude that
d(f,F)≤ L 1−L,
which implies the inequality (2.4).
Lemma 2.4.Letψ:X ×X® [0,∞)be a function given by
ψ(x,y) :=ϕ(2x, 2x, 2y, 2y)
for all x, yÎX. Consider the setΩ: = {g|g:X×X ®Y,g(0, 0) = 0}and the gener-alized metric d onΩgiven by
d(g,h) =dψ(g,h) := infSψ(g,h),
where Sψ(g,h) : =KÎ[0,∞] | ||g(x,y)- h(x,y)||≤Kψ(x,y)for all x,yÎX}for all g,
hÎΩ. Then, (Ω,d)is complete.
Proof. The proof is similar to the proof of Lemma 2.2.
Theorem 2.5.Assume thatsatisfies the condition
lim
n→∞ 4 nϕx
2n,
y 2n,
z 2n,
w 2n
= 0
for all x,y,z, wÎ X. Suppose that a mapping f: X×X ®Y satisfies the functional inequality (2.2)for all x, y, z,wÎ X and f(0, 0) = 0. If there exists L <1such that the function ψgiven in Lemma 2.4 has the property
ψ(x,y)≤ L
for all x, yÎ X, then there exists a unique mapping F: X ×X ® Y satisfying(1.1)
such that the inequality
f(x,y)−F(x,y) ≤ L
2
16(1−L)ψ(x,y) (2:8)
holds for all x,yÎ X.
Proof. Consider the complete generalized metric space (Ω,d) given in Lemma 2.4. Now we define a mappingT :Ω®Ωby
Tg(x,y) := 4gx 2,
y 2
for allgÎΩand allx,y ÎX. By the same argument of the proof of Theorem 2.3,T
is a strictly contractive mapping of Ωwith Lipschitz constantL.
Replacing x, y, z, wby 2x,2x,z2,2z in (2.2), respectively, and using (2.7), we have the inequality
f(x,z)−4f
x 2,
z
2
≤ϕx 2, x 2, z 2, z 2 =ψ x 4, z 4 ≤ L 4ψ x 2, z 2
≤ L2
16ψ(x,y) (2:9)
for all x,zÎX. Thus, we obtain that
d(f,Tf)≤ L
2
16 <∞. (2:10)
Applying the alternative of fixed point, we see that there exists a fixed point FofT
inΩsuch that
F(x,y) = lim
n→∞4 nf x
2n,
y 2n
for all x,y ÎX. Replacingx,y,z, wby 2xn, y
2n,2zn,2wn in (2.2), respectively, and
multi-plying by 4n, we have
F(x+y,z+w) +F(x−y,z−w)−2F(x,z)−2F(y,w)
= lim
n→∞4
nfx+y
2n , z+w
2n
+f
x−y
2n , z−w
2n
−2f x
2n, z
2n
−2f y
2n, w
2n
≤ lim
n→∞4
nϕ x
2n, y
2n, z
2n, w
2n
= 0
for all x, y, z, wÎ X. Thus, the mappingF satisfies the Equation 1.1. By (2.7) and (2.9), we obtain that
||Tnf(x,y)−Tn+1f(x,y)||= 4nf x 2n,
y 2n
−4f x
2n+1,
y 2n+1
≤4n−2L2ψ x
2n,
y 2n
≤4n−3L3ψ x
2n−1,
y 2n−1
≤ · · · ≤Ln+2
16 ψ(x,y)
for all x,yÎX and allnÎN, that is, d(Tnf,Tn+1f)≤ Ln+2
16 <∞for all nÎ N. By the
d(f,F)≤ 1
1−Ld(f,Tf).
By (2.10), we may conclude that
d(f,F)≤ L
2
16(1−L),
which implies the inequality (2.8).
Acknowledgements
The authors would like to thank the referee for a number of valuable suggestions regarding a previous version of this paper.
Author details
1
Graduate School of Education, Kyung Hee University, Yongin, 446-701, Korea2Department of Mathematics Education, College of Education, Mokwon University, Daejeon, 302-729, Korea
Authors’contributions
Both authors contributed equally to this work. The authors read and approved the final manuscript.
Competing interests
The authors declare that they have no competing interests.
Received: 15 April 2011 Accepted: 10 October 2011 Published: 10 October 2011
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