# On the Generalized Hyers-Ulam-Rassias Stability of Quadratic Functional Equations

Authors: M. Eshaghi Gordji; H. Khodaei
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

We achieve the general solution and the generalized Hyers-Ulam-Rassias and Ulam-Gavruta-Rassias stabilities for quadratic functional equationsf(ax+by)+f(ax−by)=(b(a+b)/2)f(x+y)+(b(a+b)/2)f(x−y)+(2a2−ab−b2)f(x)+(b2−ab)f(y) where a, b are nonzero fixed integers with b≠±a,−3a, and f(ax+by)+f(ax−by)=2a2f(x)+2b2f(y) for fixed integers a, b with a,b≠0 and a±b≠0.


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## Body

## 1. Introduction

In 1940,  Ulam [1]  proposed the stability problem for functional equations in the following question regarding to the stability of group homomorphism.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 conditions does a homomorphism exist near an approximately homomorphism? Generally, the concept of stability for a functional equation comes up when we the functional equation is replaced by an inequality which acts as a perturbation of that equation. Hyers [2] answered to the question affirmatively in 1941 so if f:E→E' such that
(1.1)∥f(x+y)-f(x)-f(y)∥≤δ,
for all x,y∈E, and for some δ>0 where E, E' are Banach spaces; then there exists a unique additive mapping T:E→E' such that
(1.2)∥f(x)-T(x)∥≤δ,
for all x∈E. However, if f(tx) is a continuous mapping at t∈ℝ for each fixed x∈E then T is linear. In 1950, Hyers's theorem was generalized by Aoki [3] for additive mappings and independently, in 1978, by Rassias [4] for linear mappings considering the Cauchy difference controlled by sum of powers of norms. This stability phenomenon is called the Hyers-Ulam-Rassias stability.On the other hand, Rassias [5–10] considered the Cauchy difference controlled by a product of different powers of norm. However, there was a singular case; for this singularity a counterexample was given by Găvruţa [11]. This stability phenomenon is called the Ulam-Găvruţa-Rassias stability (see also [12, 13]). In addition, J. M. Rassias considered the mixed product-sum of powers of norms control function [14]. This stability is called JMRassias mixed product-sum stability (see also [15–22]).The functional equation(1.3)f(x+y)+f(x-y)=2f(x)+2f(y),
is related to symmetric biadditive function and is called a quadratic functional equation naturally, and every solution of the quadratic equation (1.3) is said to be a quadratic function. It is well known that a function f between two real vector spaces is quadratic if and only if there exists a unique symmetric biadditive function B such that f(x)=B(x,x) for all x where
(1.4)B(x,y)=14(f(x+y)-f(x-y))
(see [23, 24]). Skof proved Hyers-Ulam-Rassias stability problem for quadratic functional equation (1.3) for a class of functions f:A→B, where A is normed space and B is a Banach space, (see [25]). Cholewa [26] noticed that Skof's theorem is still true if relevant domain A alters to an abelian group. In 1992, Czerwik proved the Hyers-Ulam-Rassias stability of (1.3) (see [27]) and four years later, Grabiec [28] generalized the result mentioned above.Throughout this paper, assume thata, b are fixed integers with a,b≠0, we introduce the following functional equations, which are different from (1.3):
(1.5)f(ax+by)+f(ax-by)=b(a+b)2f(x+y)+b(a+b)2f(x-y)+(2a2-ab-b2)f(x)+(b2-ab)f(y),
where b≠±a,-3a, and
(1.6)f(ax+by)+f(ax-by)=2a2f(x)+2b2f(y),
where b≠±a.In this paper, we establish the general solution and the generalized Hyers-Ulam-Rassias and Ulam-Găvruţa-Rassias stabilities problem for (1.5), (1.6) which are equivalent to (1.3).


## 2. Solution of (1.5), (1.6)

LetX and Y be real vector spaces. We here present the general solution of (1.5), (1.6).Theorem 2.1.
A functionf:X→Y satisfies the functional equation (1.3) if and only if f:X→Y satisfies the functional equation (1.5). Therefore, every solution of functional equation (1.5) is also a quadratic function.Proof.
Letf satisfy the functional equation (1.3). Putting x=y=0 in (1.3), we get f(0)=0. Set x=0 in (1.3) to get f(-y)=f(y). Letting y=x and y=2x in (1.3), respectively, we obtain that f(2x)=4f(x) and f(3x)=9f(x) for all x∈X. By induction, we lead to f(kx)=k2f(x) for all positive integers k. Replacing x and y by 2x+y and 2x-y in (1.3), respectively, gives
(2.1)f(2x+y)+f(2x-y)=8f(x)+2f(y)
for all x,y∈X. Using (1.3) and (2.1), we lead to
(2.2)f(2x+y)+f(2x-y)=2f(x+y)+2f(x-y)+4f(x)-2f(y)
for all x,y∈X. Suppose that k≠0 is a fixed integer by using (1.3), we get
(2.3)kf(x+y)+kf(x-y)-2kf(x)-2kf(y)=0
for all x,y∈X. Using (2.2) and (2.3), we obtain
(2.4)f(2x+y)+f(2x-y)=(2+k)f(x+y)+(2+k)f(x-y)+2(2-k)f(x)-2(1+k)f(y)
for all x,y∈X. Replacing x and y by 3x+y and 3x-y in (1.3), respectively, then using (1.3) and (2.3), we have
(2.5)f(3x+y)+f(3x-y)=(3+k)f(x+y)+(3+k)f(x-y)+2(6-k)f(x)-2(2+k)f(y)
for all x,y∈X. By using the above method, by induction, we infer that
(2.6)f(ax+y)+f(ax-y)=(a+k)f(x+y)+(a+k)f(x-y)+2(a2-a-k)f(x)-2(a+k-1)f(y)
for all x,y∈X and each positive integer a≥1. For a negative integer a≤-1, replacing a by -a one can easily prove the validity of (2.6). Therefore (1.3) implies (2.6) for any integer a≠0. First, it is noted that (2.6) also implies the following equation
(2.7)f(bx+y)+f(bx-y)=(b+k)f(x+y)+(b+k)f(x-y)+2(b2-b-k)f(x)-2(b+k-1)f(y)
for all integers b≠0. Setting y=0 in (2.7) gives f(bx)=b2f(x). Substituting y with by into (2.7), one gets
(2.8)(b+k)f(x+by)+(b+k)f(x-by)=b2f(x+y)+b2f(x-y)-2(b2-b-k)f(x)+2b2(b+k-1)f(y)
for all x,y∈X. Replacing y by by in (2.6), we observe that
(2.9)f(ax+by)+f(ax-by)=(a+k)f(x+by)+(a+k)f(x-by)+2(a2-a-k)f(x)-2(a+k-1)f(by)
for all x,y∈X. Hence, according to (2.8) and (2.9), we get
(2.10)(b+k)f(ax+by)+(b+k)f(ax-by)=b2(a+k)f(x+y)+b2(a+k)f(x-y)+2(a2(b+k)-b2(a+k))f(x)-2b2(a-b)f(y)
for all x,y∈X. In particular, if we substitute k:=b in (2.10) and dividing it by 2b, we conclude that f satisfies (1.5).
Letf satisfy the functional equation (1.5), for nonzero fixed integers a,b with b≠±a,-3a. Putting x=y=0 in (1.5), we get
(2.11)(2a2-ba+b2-2)f(0)=0,
so
(2.12)(2a-b+16-7b22)(a-b-16-7b24)f(0)=0,
but since a,b≠0 and b≠±a,-3a, therefore f(0)=0. Setting y=0 in (1.5) gives f(ax)=a2f(x) for all x∈X. Letting y=-y in (1.5), we get
(2.13)f(ax-by)+f(ax+by)=b(a+b)2f(x-y)+b(a+b)2f(x+y)+(2a2-ab-b2)f(x)+(b2-ab)f(-y)
for all x,y∈X. If we compare (1.5) with (2.13), then since a,b≠0 and b≠±a,-3a, we conclude that f(-y)=f(y) for all y∈X. Letting x=0 in (1.5) and using the evenness of f give f(by)=b2f(y) for all y∈X. Therefore for all x∈X, we get f(abx)=a2b2f(x). Replacing x and y by bx and ay in (1.5), respectively, we have
(2.14)a2b2f(x+y)+a2b2f(x-y)=b(a+b)2f(bx+ay)+b(a+b)2f(bx-ay)+b2(2a2-ab-b2)f(x)+a2(b2-ab)f(y)
for all x,y∈X. On the other hand, if we interchange x with y in (1.5), we obtain
(2.15)f(ay+bx)+f(ay-bx)=b(a+b)2f(y+x)+b(a+b)2f(y-x)+(2a2-ab-b2)f(y)+(b2-ab)f(x)
for all x,y∈X. But since f is even, it follows from (2.15) that
(2.16)f(bx+ay)+f(bx-ay)=b(a+b)2f(x+y)+b(a+b)2f(x-y)+(b2-ab)f(x)+(2a2-ab-b2)f(y)
for all x,y∈X. Hence, according to (2.14) and (2.16), we obtain that
(2.17)a2b2f(x+y)+a2b2f(x-y)=b(a+b)2[b(a+b)2(f(x+y)+f(x-y))+(b2-ab)f(x)+(2a2-ab-b2)f(y)b(a+b)2b(a+b)2b(a+b)2]+b2(2a2-ab-b2)f(x)+a2(b2-ab)f(y)
for all x,y∈X. So from (2.17), we have
(2.18)b24(4a2-(a+b)2)(f(x+y)+f(x-y))=b22(3a2-2ab-b2)f(x)+b22(3a2-2ab-b2)f(y)
for all x,y∈X. But since a,b≠0 and b≠±a,-3a, we conclude that
(2.19)f(x+y)+f(x-y)=2f(x)+2f(y)
for all x,y∈X. Therefore, f satisfies (1.3).Theorem 2.2.
A functionf:X→Y satisfies the functional equation (1.3) if and only if f:X→Y satisfies the functional equation (1.6). Therefore, every solution of functional equation (1.6) is also a quadratic function.Proof.
Iff satisfies the functional equation (1.3), then f satisfies the functional equation (1.5). Now combining (1.3) with (1.5), we have
(2.20)f(ax+by)+f(ax-by)=b(a+b)2(2f(x)+2f(y))+(2a2-ab-b2)f(x)+(b2-ab)f(y)
for all x,y∈X. So from (2.20), we conclude that f satisfies (1.6).Let f satisfy the functional equation (1.6) for fixed integers a, b with a≠0, b≠0 and a±b≠0. Putting x=y=0 in (1.6), we get (2(a2+b2)-2)f(0)=0, and since a≠0, b≠0, therefore f(0)=0. Setting y=0 in (1.6) gives f(ax)=a2f(x) for all x∈X. Letting y:=-y in (1.6), we have
(2.21)f(ax-by)+f(ax+by)=2a2f(x)+2b2f(-y)
for all x,y∈X. If we compare (1.6) with (2.21), then since a,b≠0 and a±b≠0, we obtain that f(-y)=f(y) for all y∈X. Letting x=0 in (1.6) and using the evenness of f gives f(by)=b2f(y) for all y∈X. Therefore for all x∈X, we get f(abx)=a2b2f(x). Replacing x and y by bx and ay in (1.6), respectively, we have
(2.22)f(abx-aby)+f(abx+aby)=2a2f(bx)+2b2f(ay)
for all x,y∈X. Now, by using f(ax)=a2f(x),f(bx)=b2f(x) and f(abx)=a2b2f(x), it follows from (2.22) that
(2.23)f(x+y)+f(x-y)=2f(x)+2f(y)
for all x,y∈X. Which completes the proof of the theorem.Corollary 2.3 ([29, Proposition 2.1]).
A functionf:X→Y satisfies the following functional equation:
(2.24)f(ax+y)+f(ax-y)=2a2f(x)+2f(y)
for all x,y∈X if and only if f:X→Y satisfies the functional equation (1.3) for all x,y∈X.Proof.
Assume thatb=1 in functional equation (1.6) and apply Theorem 2.2.


## 3. Stability

We now investigate the generalized Hyers-Ulam-Rassias and Ulam-Gavruta-Rassias stabilities problem for functional equations (1.5), (1.6). From this point on, let X be a real vector space and let Y be a Banach space. Before taking up the main subject, we define the difference operator Δf:X×X→Y by
(3.1)Δf(x,y)=f(ax+by)+f(ax-by)-b(a+b)2f(x+y)-b(a+b)2f(x-y)-(2a2-ab-b2)f(x)-(b2-ab)f(y)
for all x,y∈X and a,b fixed integers such that a,b≠0 and a±b≠0 where f:X→Y is a given function.Theorem 3.1.
Letj∈{-1,1} be fixed, and let φ:X×X→[0,∞) be a function such that
(3.2)φ̃(x):=∑i=(1-j)/2∞‍1a2ijφ(aijx,0)<∞(3.3)limn→∞1a2njφ(anjx,anjy)=0
for all x,y∈X. Suppose that f:X→Y be a function satisfies
(3.4)∥Δf(x,y)∥≤φ(x,y)
for all x,y∈X. Furthermore, assume that f(0)=0 in (3.4) for the case j=1. Then there exists a unique quadratic function Q:X→Y such that
(3.5)∥f(x)-Q(x)∥≤12a1+jφ̃(xa(1-j)/2),
for all x∈X.Proof.
Forj=1, putting y=0 in (3.4), we have
(3.6)∥2f(ax)-2a2f(x)∥≤φ(x,0)
for all x∈X. So
(3.7)∥f(x)-1a2f(ax)∥≤12a2φ(x,0)
for all x∈X. Replacing x by ax in (3.7) and dividing by a2 and summing the resulting inequality with (3.7), we get
(3.8)∥f(x)-1a4f(a2x)∥≤12a2(φ(x,0)+φ(ax,o)a2)
for all x∈X. Hence
(3.9)∥1a2kf(akx)-1a2mf(amx)∥≤12a2∑i=km-1‍1a2iφ(aix,0)
for all nonnegative integers m and k with m>k and for all x∈X. It follows from (3.2) and (3.9) that the sequence {(1/a2n)f(anx)} is a Cauchy sequence for all x∈X. Since Y is complete, the sequence {(1/a2n)f(anx)} converges. So one can define the function Q:X→Y by
(3.10)Q(x):=limn→∞1a2nf(anx)
for all x∈X. By (3.3) for j=1 and (3.4),
(3.11)∥ΔQ(x,y)∥=limn→∞1a2n∥Δf(anx,any)∥≤limn→∞1a2nφ(anx,any)=0
for all x,y∈X. So ΔQ(x,y)=0. By Theorem 2.1, the function Q:X→Y is quadratic. Moreover, letting k=0 and passing the limit m→∞ in (3.9), we get the inequality (3.5) for j=1.
Now, letQ':X→Y be another quadratic function satisfying (1.5) and (3.5). Then we have
(3.12)∥Q(x)-Q'(x)∥=1a2n∥Q(anx)-Q'(anx)∥≤1a2n(∥Q(anx)-f(anx)∥+∥Q'(anx)-f(anx)∥)≤1a2a2nφ̃(anx,0),
which tends to zero as n→∞ for all x∈X. So we can conclude that Q(x)=Q'(x) for all x∈X. This proves the uniqueness of Q.
Also, forj=-1, it follows from (3.6) that
(3.13)∥f(x)-a2f(xa)∥≤12φ(xa,0)
for all x∈X. Hence
(3.14)∥a2kf(xak)-a2mf(xam)∥≤12∑i=km-1‍a2iφ(xai+1,0)
for all nonnegative integers m and k with m>k and for all x∈X. It follows from (3.14) that the sequence {a2nf(x/an)} is a Cauchy sequence for all x∈X. Since Y is complete, the sequence {a2nf(x/an)} converges. So one can define the function Q:X→Y by
(3.15)Q(x):=limn→∞a2nf(xan)
for all x∈X. By (3.3) for j=-1 and (3.4),
(3.16)∥ΔQ(x,y)∥=limn→∞a2n∥Δf(xan,yan)∥≤limn→∞a2nφ(xanyan)=0,
for all x,y∈X. So ΔQ(x,y)=0. By Theorem 2.1, the function Q:X→Y is quadratic. Moreover, letting k=0 and passing the limit m→∞ in (3.14), we get the inequality (3.5) for j=-1. The rest of the proof is similar to the proof of previous section.From Theorem3.1, we obtain the following corollaries concerning the JMRassias mixed product-sum stability of the functional equation (1.5).Corollary 3.2.
Letε,p,q≥0 and r,s>0 be real numbers such that p,q<2 and r+s≠2. Suppose that a function f:X→Y satisfies
(3.17)∥Δf(x,y)∥≤ε(∥x∥p+∥y∥q+∥x∥r∥y∥s)
for all x,y∈X. Then there exists a unique quadratic function Q:X→Y such that
(3.18)∥f(x)-Q(x)∥≤ε2(a2-ap)∥x∥p
for all x∈X.Proof.
In Theorem3.1, put j:=1 and φ(x,y):=ε(∥x∥p+∥y∥q+∥x∥r∥y∥s).Corollary 3.3.
Letε,p,q≥0 and r,s>0 be real numbers such that p,q>2 and r+s≠2. Suppose that a function f:X→Y with f(0)=0 satisfies (3.17) for all x,y∈X. Then there exists a unique quadratic function Q:X→Y such that
(3.19)∥f(x)-Q(x)∥≤ε2(ap-a2)∥x∥p
for all x∈X.Proof.
In Theorem3.1, put j:=-1 and φ(x,y):=ε(∥x∥p+∥y∥q+∥x∥r∥y∥s).Theorem 3.4.
Letj∈{-1,1} be fixed, and let φ:X×X→[0,∞) be a function such that
(3.20)φ̃(x):=∑i=(1-j)/2∞‍1a2ijφ(aijx,0)<∞,limn→∞1a2njφ(anjx,anjy)=0
for all x,y∈X. Suppose that f:X→Y be a function satisfies
(3.21)∥f(ax+by)+f(ax-by)-2a2f(x)-2b2f(y)∥≤φ(x,y)
for all x,y∈X. Furthermore, assume that f(0)=0 in (3.21) for the case j=1. Then there exists a unique quadratic function Q:X→Y such that
(3.22)∥f(x)-Q(x)∥≤12a1+jφ̃(xa1-j/2),
for all x∈X.Proof.
The proof is similar to the proof of Theorem3.1.



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