# On the Stability of Quadratic Functional Equations

Authors: Jung Rye Lee; Jong Su An; Choonkil Park
Journal: Abstract and Applied Analysis
 (2008)

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## Abstract

LetX,Y be vector spaces and k a fixed positive integer. It is shown that a mapping f(kx+y)+f(kx-y)=2k2f(x)+2f(y) for all x,y∈X if and only if the mapping f:X→Y satisfies f(x+y)+f(x-y)=2f(x)+2f(y) for all
x,y∈X. Furthermore,  the Hyers-Ulam-Rassias stability of the above functional
equation in Banach spaces is proven.


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## Body

## 1. Introduction

The stability problem of functional equations originated
from a question of Ulam [1] concerning the stability of group homomorphisms.
Hyers [2] gave a first
affirmative answer to the question of Ulam for Banach spaces. Hyers' theorem
was generalized by Aoki [3]
for additive mapping and by Th. M. Rassias [4] for linear mappings by considering an unbounded Cauchy difference. The paper of
Th. M. Rassias [4] has
provided a lot of influence in the development of what we now call Hyers-Ulam-Rassias stability of functional
equations. Th. M. Rassias [5] during the 27th International Symposium on Functional
Equations asked the question whether such a theorem can also be proved for p≥1. Gajda [6], following the same approach as in
[4], gave an
affirmative solution to this question for p>1. It was shown by Gajda [6] as well as by Rassias
and Šemrl [7] that one cannot prove a Th.M. Rassias' type theorem
when p=1. J. M. Rassias [8], following the spirit of the innovative approach of
Th. M. Rassias [4] for
the unbounded Cauchy difference, proved a similar stability theorem in which he
replaced the factor ∥x∥p+∥y∥p by ∥x∥p⋅∥y∥q for p,q∈ℝ with p+q≠1.The functional equation(1.1)f(x+y)+f(x−y)=2f(x)+2f(y)is called a quadratic functional equation. In
particular, every solution of the quadratic functional equation is said to be a quadratic function. A
Hyers-Ulam-Rassias stability problem for the quadratic functional equation was
proved by Skof [9]
for mappings f:X→Y, where X is a normed
space and Y is a Banach
space. Cholewa [10]
noticed that the theorem of Skof is still true if the relevant domain X is replaced by
an Abelian group. In [11],
Czerwik proved the Hyers-Ulam-Rassias stability of the quadratic functional
equation. Several functional equations have been investigated in [12–17].Throughout this paper, assume thatk is a fixed
positive integer.In this paper, we solve the functional
equation(1.2)f(kx+y)+f(kx−y)=2k2f(x)+2f(y)and prove the Hyers-Ulam-Rassias
stability of the functional equation
(1.2) in Banach spaces.


## 2. Hyers-Ulam-Rassias Stability of the Quadratic Functional Equation

Proposition 2.1.
LetX and Y be vector
spaces. A mapping f:X→Y satisfies(2.1)f(kx+y)+f(kx−y)=2k2f(x)+2f(y)for all x,y∈X if and only if
the mapping f:X→Y satisfies(2.2)f(x+y)+f(x−y)=2f(x)+2f(y)for all x,y∈X.Proof.
Assume thatf:X→Y satisfies (2.1).
Lettingx=y=0 in  (2.1), we get f(0)=0.
Lettingy=0 in (2.1), we get f(kx)=k2f(x) for all x∈X.
Lettingx=0 in (2.1), we get f(−y)=f(y) for all y∈X.
It follows from
(2.1) that(2.3)f(kx+y)+f(kx−y)=2k2f(x)+2f(y)=2f(kx)+2f(y)for all x,y∈X. So the mapping f:X→Y satisfies(2.4)f(x+y)+f(x−y)=2f(x)+2f(y)for all x,y∈X.
Assume thatf:X→Y satisfies f(x+y)+f(x−y)=2f(x)+2f(y) for all x,y∈X.
We prove (2.1)
for k=j by induction on j.
For the casej=1, (2.1) holds
by the assumption.
For the casej=2, since(2.5)f(2x+y)+f(2x−y)=f(x+y+x)+f(x−y+x)=2f(x+y)+2f(x)−f(y)+2f(x−y)+2f(x)−f(−y)=2f(x+y)+2f(x−y)+4f(x)−2f(y)=4f(x)+4f(y)+4f(x)−2f(y)=8f(x)+2f(y)for all x,y∈X, then (2.1)
holds.
Assume that
(2.1) holds for j=n−2 and j=n−1 (2<n≤k). By the
assumption,(2.6)f(nx+y)+f(nx−y)=f((n−1)x+y+x)+f((n−1)x−y+x)=2f((n−1)x+y)+2f(x)−f((n−2)x+y)+2f((n−1)x−y)+2f(x)−f((n−2)x−y)=4(n−1)2f(x)+4f(y)+4f(x)−2(n−2)2f(x)−2f(y)=2n2f(x)+2f(y)for all x,y∈X, (2.1) holds
for j=n. Hence the mapping f:X→Y satisfies (2.1) for j=k.From now on, assume thatX is a normed
vector space with norm ∥⋅∥ and that Y is a Banach
space with norm ∥⋅∥.For a given mappingf:X→Y, we define(2.7)Df(x,y):=f(kx+y)+f(kx−y)−2k2f(x)−2f(y)for all x,y∈X.Now we prove the Hyers-Ulam-Rassias stability of the
quadratic functional equationDf(x,y)=0.Theorem 2.2.
Letf:X→Y be a mapping
with f(0)=0 for which there
exists a function φ:X2→[0,∞) such
that(2.8)φ˜(x,y):=∑j=0∞1k2jφ(kjx,kjy)<∞,(2.9)∥Df(x,y)∥≤φ(x,y)for all x,y∈X. Then there exists a unique quadratic mapping Q:X→Y such
that(2.10)∥f(x)−Q(x)∥≤12k2φ˜(x,0)for all x∈X.Proof.
Lettingy=0 in (2.9), we get(2.11)∥2f(kx)−2k2f(x)∥≤φ(x,0)for all x∈X. So(2.12)∥f(x)−1k2f(kx)∥≤12k2φ(x,0)for all x∈X. Hence(2.13)∥1k2lf(klx)−1k2mf(kmx)∥≤∑j=lm−112k2j+2φ(kjx,0)for all nonnegative integers m and l with m>l and all x∈X. It follows from
(2.13) that the sequence {(1/k2n)f(knx)} is Cauchy for
all x∈X. Since Y is complete,
the sequence {(1/k2n)f(knx)} converges. So
one can define the mapping Q:X→Y by(2.14)Q(x):=lim⁡n→∞1k2nf(knx)for all x∈X.
By
(2.8),(2.15)∥DQ(x,y)∥=lim⁡n→∞1k2n∥Df(knx,kny)∥≤lim⁡n→∞1k2nφ(knx,kny)=0for all x,y∈X. So DQ(x,y)=0. By Proposition 2.1, the mapping Q:X→Y is quadratic.
Moreover, letting l=0 and passing the
limit m→∞ in  (2.13), we get
(2.10).
Now, letT:X→Y be another
quadratic mapping satisfying (2.1)
and (2.10). Then we have(2.16)∥Q(x)−T(x)∥=1k2n∥Q(knx)−T(knx)∥≤1k2n(∥Q(knx)−f(knx)∥+∥T(knx)−f(knx)∥)≤1k2n+2φ˜(knx,0),which tends to zero as n→∞ for all x∈X. So we can conclude that Q(x)=T(x) for all x∈X. This proves the uniqueness of Q. So there exists a unique quadratic mapping Q:X→Y satisfying  (2.10).Corollary 2.3.
Letp<2 and θ be positive
real numbers, and let f:X→Y be a mapping
such that(2.17)∥Df(x,y)∥≤θ(||x||p+||y||p)for all x,y∈X. Then there exists a unique quadratic mapping Q:X→Y such
that(2.18)∥f(x)−Q(x)∥≤θ8−2p+1||x||pfor all x∈X.Proof.
The
proof follows from Theorem2.2 by taking(2.19)φ(x,y):=θ(||x||p+||y||p)for all x,y∈A.Theorem 2.4.
Letf:X→Y be a mapping
with f(0)=0 for which there
exists a function φ:X2→[0,∞) satisfying  (2.9) such that(2.20)φ˜(x,y):=∑j=0∞k2jφ(xkj,ykj)<∞for all x,y∈X. Then there exists a unique quadratic mapping Q:X→Y such
that(2.21)∥f(x)−Q(x)∥≤12φ˜(xk,0)for all x∈X.Proof.
It
follows from  (2.11) that(2.22)∥f(x)−k2f(xk)∥≤12φ(xk,0)for all x∈X. Hence(2.23)∥k2lf(xkl)−k2mf(xkm)∥≤∑j=lm−1k2j2φ(xkj+1,0)for all nonnegative integers m and l with m>l and all x∈X. It follows from
(2.23) that the sequence {k2nf(x/kn)} is Cauchy for
all x∈X. Since Y is complete,
the sequence {k2nf(x/kn)} converges. So
one can define the mapping Q:X→Y by(2.24)Q(x):=lim⁡n→∞k2nf(xkn)for all x∈X.
By
(2.20),(2.25)∥DQ(x,y)∥=lim⁡n→∞k2n∥Df(xkn,ykn)∥≤lim⁡n→∞k2nφ(xkn,ykn)=0for all x,y∈X. So DQ(x,y)=0. By Proposition 2.1, the mapping Q:X→Y is quadratic.
Moreover, letting l=0 and passing the
limit m→∞ in  (2.23), we get  (2.21).
The rest of the proof is similar to the proof of
Theorem2.2.Corollary 2.5.
Letp>2 and θ be positive
real numbers, and let f:X→Y be a mapping
satisfying  (2.17). Then there exists a
unique quadratic mapping Q:X→Y such
that(2.26)∥f(x)−Q(x)∥≤θ2p+1−8||x||pfor all x∈X.Proof.
The
proof follows from Theorem2.4 by taking(2.27)φ(x,y):=θ(||x||p+||y||p)for all x,y∈A.From now on, assume thatk=2.Theorem 2.6.
Letf:X→Y be a mapping
with f(0)=0 for which there
exists a function φ:X2→[0,∞) satisfying  (2.9) such that(2.28)φ˜(x,y):=∑j=0∞19jφ(3jx,3jy)<∞for all x,y∈X. Then there exists a unique quadratic mapping Q:X→Y such
that(2.29)∥f(x)−Q(x)∥≤19φ˜(x,x)for all x∈X.Proof.
Lettingy=x in  (2.9), we get(2.30)∥f(3x)−9f(x)∥≤φ(x,x)for all x∈X. So(2.31)∥f(x)−19f(3x)∥≤19φ(x,x)for all x∈X. Hence(2.32)∥19lf(3lx)−19mf(3mx)∥≤∑j=lm−119j+1φ(3jx,3jx)for all nonnegative integers m and l with m>l and all x∈X. It follows from
(2.32) that the sequence {(1/9n)f(3nx)} is Cauchy for
all x∈X. Since Y is complete,
the sequence {(1/9n)f(3nx)} converges. So
one can define the mapping Q:X→Y by(2.33)Q(x):=lim⁡n→∞19nf(3nx)for all x∈X.
By
(2.28),(2.34)∥DQ(x,y)∥=lim⁡n→∞19n∥Df(3nx,3ny)∥≤lim⁡n→∞19nφ(3nx,3ny)=0for all x,y∈X. So DQ(x,y)=0. By Proposition 2.1, the mapping Q:X→Y is quadratic.
Moreover, letting l=0 and passing the
limit m→∞ in  (2.32), we get  (2.29).
The rest of the proof is similar to the proof of
Theorem2.2.Corollary 2.7.
Letp<1 and θ be positive
real numbers, and let f:X→Y be a mapping
such that(2.35)∥Df(x,y)∥≤θ⋅||x||p⋅||y||pfor all x,y∈X. Then there exists a unique quadratic mapping Q:X→Y such
that(2.36)∥f(x)−Q(x)∥≤θ9−9p||x||2pfor all x∈X.Proof.
The
proof follows from Theorem2.6 by taking(2.37)φ(x,y):=θ⋅||x||p⋅||y||pfor all x,y∈A.Theorem 2.8.
Letf:X→Y be a mapping
with f(0)=0 for which there
exists a function φ:X2→[0,∞) satisfying  (2.9) such that(2.38)φ˜(x,y):=∑j=0∞9jφ(x3j,y3j)<∞for all x,y∈X. Then there exists a unique quadratic mapping Q:X→Y such
that(2.39)∥f(x)−Q(x)∥≤φ˜(x3,x3)for all x∈X.Proof.
It
follows from  (2.30) that(2.40)∥f(x)−9f(x3)∥≤φ(x3,x3)for all x∈X. Hence(2.41)∥9lf(x3l)−9mf(x3m)∥≤∑j=lm−19jφ(x3j+1,x3j+1)for all nonnegative integers m and l with m>l and all x∈X. It follows from
(2.41) that the sequence {9nf(x/3n)} is Cauchy for
all x∈X. Since Y is complete,
the sequence {9nf(x/3n)} converges. So
one can define the mapping Q:X→Y by(2.42)Q(x):=lim⁡n→∞9nf(x3n)for all x∈X.
By
(2.38),(2.43)∥DQ(x,y)∥=lim⁡n→∞19n∥Df(3nx,3ny)∥≤lim⁡n→∞19nφ(3nx,3ny)=0for all x,y∈X. So DQ(x,y)=0. By Proposition 2.1, the mapping Q:X→Y is quadratic.
Moreover, letting l=0 and passing the
limit m→∞ in  (2.41), we get  (2.39).
The rest of the proof is similar to the proof of
Theorem2.2.Corollary 2.9.
Letp>1 and θ be positive
real numbers, and let f:X→Y be a mapping
satisfying  (2.35). Then there exists a
unique quadratic mapping Q:X→Y such
that(2.44)∥f(x)−Q(x)∥≤θ9p−9||x||2pfor all x∈X.Proof.
The
proof follows from Theorem2.8 by taking(2.45)φ(x,y):=θ⋅||x||p⋅||y||pfor all x,y∈A.



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