hyers–ulam stability of the quadratic functional equation
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Hyers-Ulam-RassiasStability
M. Ait Sibaha, B. Bouikhaleneand E. Elqorachi
vol. 8, iss. 3, art. 89, 2007
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HYERS-ULAM-RASSIAS STABILITY OF THEK-QUADRATIC FUNCTIONAL EQUATION
MOHAMED AIT SIBAHA, BELAID BOUIKHALENEUniversity of Ibn Tofail, Faculty of SciencesDepartment of MathematicsKenitra, MoroccoEMail: [email protected] [email protected]
ELHOUCIEN ELQORACHILaboratory LAMA , Harmonic Analysis and Functional Equations TeamDepartment of Mathematics, Faculty of SciencesUniversity Ibn Zohr, Agadir, MoroccoEMail: [email protected]
Received: 26 January, 2007
Accepted: 08 June, 2007
Communicated by: K. Nikodem
2000 AMS Sub. Class.: 39B82, 39B52, 39B32.
Key words: Group, Additive equation, Quadratic equation, Hyers-Ulam-Rassias stability.
Abstract: In this paper we obtain the Hyers-Ulam-Rassias stability for the functional equation1
|K|∑k∈K
f(x + k · y) = f(x) + f(y), x, y ∈ G,
whereK is a finite cyclic transformation group of the abelian group(G, +), acting byautomorphisms ofG. As a consequence we can derive the Hyers-Ulam-Rassias stability ofthe quadratic and the additive functional equations.
Hyers-Ulam-RassiasStability
M. Ait Sibaha, B. Bouikhalene
and E. Elqorachi
vol. 8, iss. 3, art. 89, 2007
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Contents
1 Introduction 3
2 Main Results 8
Hyers-Ulam-RassiasStability
M. Ait Sibaha, B. Bouikhalene
and E. Elqorachi
vol. 8, iss. 3, art. 89, 2007
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1. Introduction
In the book,A Collection of Mathematical Problems[32], S.M. Ulam posed thequestion of the stability of the Cauchy functional equation. Ulam asked: if we re-place a given functional equation by a functional inequality, when can we assert thatthe solutions of the inequality lie near to the solutions of the strict equation?
Originally, he had proposed the following more specific question during a lecturegiven before the University of Wisconsin’s Mathematics Club in 1940.
Given a groupG1, a metric group (G2, d), a numberε > 0 and a mappingf : G1 −→ G2 which satisfies the inequalityd(f(xy), f(x)f(y)) < ε forall x, y ∈ G1, does there exist an homomorphismh : G1 −→ G2 and aconstantk > 0, depending only onG1 andG2 such thatd(f(x), h(x)) ≤kε for all x in G1?
A partial and significant affirmative answer was given by D.H. Hyers [9] underthe condition thatG1 andG2 are Banach spaces.
In 1978, Th. M. Rassias [18] provided a generalization of Hyers’s stability theo-rem which allows the Cauchy difference to be unbounded, as follows:
Theorem 1.1.Letf : V −→ X be a mapping between Banach spaces and letp < 1be fixed. Iff satisfies the inequality
‖f(x+ y)− f(x)− f(y)‖ ≤ θ(‖x‖p + ‖y‖p)
for someθ ≥ 0 and for all x, y ∈ V (x, y ∈ V \ {0} if p < 0), then there exists aunique additive mappingT : V −→ X such that
‖f(x)− T (x)‖ ≤ 2θ
|2− 2p|‖x‖p
for all x ∈ V (x ∈ V \ {0} if p < 0).If, in addition,f(tx) is continuous int for each fixedx, thenT is linear.
Hyers-Ulam-RassiasStability
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During the 27th International Symposium on functional equations, Th. M. Ras-sias asked the question whether such a theorem can also be proved forp ≥ 1. Z.Gajda [7] following the same approach as in [18], gave an affirmative answer toRassias’ question forp > 1. However, it was showed that a similar result for thecasep = 1 does not hold.
In 1994, P. Gavruta [8] provided a generalization of the above theorem by replac-ing the function(x, y) 7−→ θ(‖x‖p + ‖y‖p) with a mappingϕ(x, y) which satisfiesthe following condition:
∞∑n=0
2−nϕ(2nx, 2ny) <∞ or∞∑
n=0
2nϕ( x
2n+1,y
2n+1
)<∞
for everyx, y in a Banach spaceV .Since then, a number of stability results have been obtained for functional equa-
tions of the forms
(1.1) f(x+ y) = g(x) + h(y), x, y ∈ G,and
(1.2) f(x+ y) + f(x− y) = g(x) + h(y), x, y ∈ G,whereG is an abelian group. In particular, for the classical equations of Cauchy andJensen, the quadratic and the Pexider equations, the reader can be referred to [4] –[22] for a comprehensive account of the subject.
In the papers [24] – [31], H. Stetkær studied functional equations related to theaction by automorphisms on a groupG of a compact transformation groupK. Writ-ing the action ofk ∈ K onx ∈ G ask · x and lettingdk denote the normalized Haarmeasure onK, the functional equations (1.1) and (1.2) have the form
(1.3)∫
K
f(x+ k · y)dk = g(x) + h(y), x, y ∈ G,
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whereK = {I} andK = {I,−I}, respectively,I denoting the identity.The purpose of this paper is to investigate the Hyers-Ulam-Rassias stability of
(1.4)1
|K|∑k∈K
f(x+ k · y) = f(x) + f(y), x, y ∈ G,
whereK is a finite cyclic subgroup ofAut(G) (the group of automorphisms ofG),|K| denotes the order ofK, andG is an abelian group.
The set up allows us to give a unified treatment of the stability of the additivefunctional equation
(1.5) f(x+ y) = f(x) + f(y), x, y ∈ G,
and the quadratic functional equation
(1.6) f(x+ y) + f(x− y) = 2f(x) + 2f(y), x, y ∈ G.
In particular, we want to see how the compact subgroupK enters into the solutionsformulas.
The stability problem for the quadratic equation (1.6) was first solved by Skofin [23]. In [4] Cholewa extended Skof’s result in the following way, whereG is anabelian group andE is a Banach space.
Theorem 1.2.Letη > 0 be a real number andf : G −→ E satisfies the inequality
(1.7) |f(x+ y) + f(x− y)− 2f(x)− 2f(y)| ≤ η for all x, y ∈ G.
Then for everyx ∈ G the limit q(x) = limn−→+∞f(2nx)
22n exists andq : G −→ E isthe unique quadratic function satisfying
(1.8) |f(x)− q(x)| ≤ η
2, x ∈ G.
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In [5] Czerwik obtained a generalization of the Skof-Chelewa result.
Theorem 1.3. Let p 6= 2, θ > 0, δ > 0 be real numbers. Suppose that the functionf : E1 −→ E2 satisfies the inequality
‖f(x+ y) + f(x− y)− 2f(x)− 2f(y)‖ ≤ δ+ θ(‖x‖p + ‖y‖p) for all x, y ∈ E1.
Then there exists exactly one quadratic functionq : E1 −→ E2 such that
‖f(x)− q(x)‖ ≤ c+ kθ‖x‖p
for all x ∈ E1 if p ≥ 0 and for allx ∈ E1 \ {0} if p ≤ 0, where
• c = ‖f(0)‖3
, k = 24−2p andq(x) = limn−→+∞
f(2nx)4n , for p < 2.
• c = 0, k = 22p−4
andq(x) = limn−→+∞f(2nx)
4n , for p > 2.
Recently, B. Bouikhalene, E. Elqorachi and Th. M. Rassias [1], [2] and [3] provedthe Hyers-Ulam-Rassias stability of the functional equation (1.4) with K = {I, σ}(σ is an automorphism ofG such thatσ ◦ σ = I).
The results obtained in the present paper encompass results from [2] and [18]given in Corollaries2.5and2.6below.
General Set-Up.LetK be a compact transformation group of an abelian topologicalgroup(G,+), acting by automorphisms ofG. We letdk denote the normalized Haarmeasure onK, and the action ofk ∈ K on x ∈ G is denoted byk · x. We assumethat the functionk 7−→ k · y is continuous for ally ∈ G.A continuous mappingq : G −→ C is said to beK-quadratical if it satisfies thefunctional equation
(1.9)∫
K
q(x+ k · y)dk = q(x) + q(y), x, y ∈ G.
Hyers-Ulam-RassiasStability
M. Ait Sibaha, B. Bouikhalene
and E. Elqorachi
vol. 8, iss. 3, art. 89, 2007
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WhenK is finite, the normalized Haar measuredk onK is given by∫K
h(k)dk =1
|K|∑k∈K
h(k)
for anyh : K −→ C, where|K| denotes the order ofK. So equation (1.9) can inthis case be written
(1.10)1
|K|∑k∈K
q(x+ k · y) = q(x) + q(y), x, y ∈ G.
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2. Main Results
Let ϕ : G × G −→ R+ be a continuous mapping which satisfies the followingcondition
(2.1) ψ(x, y)
=∞∑
n=1
2−n
∫K
∫K
· · ·∫
K
ϕ
x+∑
ij<ij+1, kij∈{k1,k2,...,kn−1}
(ki1ki2 · · · kip) · x,
y +∑
ij<ij+1, kij∈{k1,k2,...,kn−1}
(ki1ki2 · · · kip) · y
dk1dk2 . . . dkn−1 <∞,
for all p such that1 ≤ p ≤ n − 1, for all x, y ∈ G (uniform convergence). In whatfollows, we setϕ(x) = ϕ(x, x) andψ(x) = ψ(x, x) for all x ∈ G.
The main results of the present paper are based on the following proposition.
Proposition 2.1. Let G be an abelian group and letϕ : G × G −→ R+ be acontinuous control mapping which satisfies (2.1). Suppose thatf : G −→ C iscontinuous and satisfies the inequality
(2.2)
∣∣∣∣∫K
f(x+ k · y)dk − f(x)− f(y)
∣∣∣∣ ≤ ϕ(x, y)
for all x, y ∈ G. Then, the formulaq(x) = limn−→∞
fn(x)2n , with
(2.3) f0(x) = f(x) and fn(x) =
∫K
fn−1(x+ k · x)dk for all n ≥ 1,
Hyers-Ulam-RassiasStability
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defines a continuous function which satisfies
(2.4) |f(x)− q(x)| ≤ ψ(x) and∫
K
q(x+ k · x)dk = 2q(x) for all x ∈ G.
Furthermore, the continuous functionq with the condition (2.4) is unique.
Proof. Replacingy by x in (2.2) gives
(2.5) |f1(x)− 2f(x)| =∣∣∣∣∫
K
f(x+ k · x)dk − 2f(x)
∣∣∣∣ ≤ ϕ(x)
and consequently
|f2(x)− 2f1(x)| =∣∣∣∣∫
K
f1(x+ k1 · x)dk1 − 2
∫K
f(x+ k1 · x)dk1
∣∣∣∣(2.6)
≤∫
K
|f1(x+ k1 · x)− 2f(x+ k1 · x)|dk1
≤∫
K
ϕ(x+ k1 · x)dk1.
Next, we prove that
(2.7)
∣∣∣∣fn(x)
2n− fn−1(x)
2n−1
∣∣∣∣≤ 2−n
∫K
∫K
· · ·∫
K
ϕ
x+∑
ij<ij+1, kij∈{k1,k2,...,kn−1}
(ki1ki2 · · · kip) · x
dk1dk2 . . . dkn−1
for all n ∈ N \ {0}. Clearly (2.7) is true for the casen = 1, since settingn = 1 in(2.7) gives (2.5). Now, assume that the induction assumption is true forn ∈ N\{0},
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and consider
|fn+1(x)− 2fn(x)| =∣∣∣∣∫
K
fn(x+ kn · x)dkn − 2
∫K
fn−1(x+ kn · x)dkn
∣∣∣∣(2.8)
≤∫
K
|fn(x+ kn · x)− 2fn−1(x+ kn · x)|dkn.
Then ∣∣∣∣fn+1(x)
2n+1− fn(x)
2n
∣∣∣∣(2.9)
≤ 1
2
∫K
∣∣∣∣fn(x+ kn · x)2n
− fn−1(x+ kn · x)2n−1
∣∣∣∣ dkn
≤ 2−(n+1)
∫K
∫K
· · ·∫
K
ϕ
x+∑
ij<ij+1, kij∈{k1,k2,...,kn}
(ki1ki2 · · · kip) · x
dk1dk2 . . . dkn,
so that the inductive assumption (2.7) is indeed true for all positive integers. Hence,for r > s we get∣∣∣∣fr(x)
2r− fs(x)
2s
∣∣∣∣(2.10)
≤r−1∑n=s
∣∣∣∣fn+1(x)
2n+1− fn(x)
2n
∣∣∣∣≤
r−1∑n=s
2−(n+1)
∫K
· · ·∫
K
ϕ
x+∑
ij<ij+1 , kij∈{k1,k2,...,kn}
(ki1ki2 · · · kir) · x
dk1 . . . dkn,
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which by assumption (2.1) converges to zero (uniformly) asr ands tend to infinity.Thus the sequence of complex functionsfn(x)
2n is a Cauchy sequence for each fixedx ∈ G and then this sequence converges for each fixedx ∈ G to some limit inC,which is continuous onG. We call this limitq(x). Next, we prove that
(2.11)
∣∣∣∣fn(x)
2n− f(x)
∣∣∣∣≤
n∑l=1
2−l
∫K
· · ·∫
K
ϕ
x+∑
ij<ij+1 , kij∈{k1,k2,...,kl−1}
(ki1ki2 · · · kip) · x
dk1 . . . dkl−1
for all n ∈ N \ {0}. We get the case ofn = 1 by (2.5)
(2.12) |f1(x)− 2f(x)| =∣∣∣∣∫
K
f(x+ k · x)dk − 2f(x)
∣∣∣∣ ≤ ϕ(x),
so the induction assumption (2.11) is true forn = 1. Assume that (2.11) is true forn ∈ N \ {0}. By using (2.9), we obtain
|fn+1(x)− 2n+1f(x)|(2.13)
≤ |fn+1(x)− 2fn(x)|+ 2|fn(x)− 2nf(x)|
≤∫
K
· · ·∫
K
ϕ
x+∑
ij<ij+1, kij∈{k1,k2,...,kn}
(ki1ki2 · · · kip) · x
dk1dk2 . . . dkn
+2n+1
n∑l=1
2−l
∫K
· · ·∫
K
ϕ
x+∑
ij<ij+1, kij∈{k1,k2,...,kl−1}
(ki1ki2 · · · kip) · x
dk1dk2 . . . dkl−1
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so (2.11) is true for alln ∈ N \ {0}. By letting n −→ +∞, we obtain the firstassertion of (2.4). We now shall show thatq satisfies the second assertion of (2.4).By using (2.2) we get∣∣∣∣∫
K
f1(x+ k1 · x)dk1 − f1(x)− f1(x)
∣∣∣∣(2.14)
=
∣∣∣∣∫K
∫K
f(x+ k1 · x+ k2 · (x+ k1 · x))dk1dk2
−∫
K
f(x+ k1 · x)dk1 −∫
K
f(x+ k1 · x)dk1
∣∣∣∣≤∫
K
∣∣∣∣∫K
f(x+ k1 · x+ k2 · (x+ k1 · x))dk2
− f(x+ k1 · x)− f(x+ k1 · x)∣∣∣∣dk1
≤∫
K
ϕ(x+ k1 · x)dk1.
Make the induction assumption
(2.15)
∣∣∣∣∫K
fn(x+ k · x)dk − 2fn(x)
∣∣∣∣≤∫
K
· · ·∫
K
ϕ
x+∑
ij<ij+1, kij∈{k1,k2,...,kn}
(ki1ki2 · · · kip) · x
dk1 . . . dkn,
Hyers-Ulam-RassiasStability
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and E. Elqorachi
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which is true forn = 1 by (2.14). Forn+ 1 we have∣∣∣∣∫K
fn+1(x+ kn+1 · x)dkn+1 − 2fn+1(x)
∣∣∣∣=
∣∣∣∣∫K
∫K
fn(x+ kn+1 · x+ k · (x+ kn+1 · x)dkn+1dk − 2
∫K
fn(x+ kn+1 · x))dkn+1
∣∣∣∣≤∫
K
∣∣∣∣∫K
fn(x+ kn+1 · x+ k · (x+ kn+1 · x))dk − 2fn(x+ kn+1 · x)∣∣∣∣ dkn+1
≤∫
K
{∫K
· · ·∫
K
ϕ
(x
+ kn+1 · x+∑
ij<ij+1, kij∈{k1,k2,...,kn}
(ki1ki2 · · · kip) · (x+ kn+1 · x)
)dk1 . . . dkn
}dkn+1
=
∫K
· · ·∫
K
ϕ
x+∑
ij<ij+1, kij∈{k1,k2,...,kn+1}
(ki1ki2 · · · kip) · x
dk1 . . . dkn+1.
Thus (2.15) is true for alln ∈ N\{0}. Now, in view of the condition (2.1), q satisfiesthe second assertion of (2.4).
To demonstrate the uniqueness of the mappingq subject to (2.4), let us assumeon the contrary that there is another mappingq′ : G −→ C such that
|f(x)− q′(x)| ≤ ψ(x) and∫
K
q′(x+ k · x)dk = 2q′(x) for all x ∈ G.
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First, we prove by induction the following relation
(2.16)
∣∣∣∣fn(x)
2n− q′(x)
∣∣∣∣≤ 1
2n
∫K
· · ·∫
K
ψ
x+∑
ij<ij+1, kij∈{k1,k2,...,kn}
(ki1ki2 · · · kip) · x
dk1 . . . dkn.
Forn = 1, we have
|f1(x)− 2q′(x)| =∣∣∣∣∫
K
f(x+ k · x)dk −∫
K
q′(x+ k · x)dk∣∣∣∣(2.17)
≤∫
K
ψ(x+ k · x)dk
so (2.16) is true forn = 1. By using the following
|fn+1(x)− 2n+1q′(x)| =∣∣∣∣∫
K
fn(x+ k · x)dk − 2n
∫K
q′(x+ k · x)dk∣∣∣∣(2.18)
≤∫
K
|fn(x+ k · x)− 2nq′(x+ k · x)|dk
we get the rest of the proof by proving that
1
2n
∫K
· · ·∫
K
ψ
x+∑
ij<ij+1, kij∈{k1,k2,...,kn}
(ki1ki2 · · · kip) · x
dk1 . . . dkn
converges to zero. In fact by setting
X = x+∑
ij<ij+1, kij∈{k1,k2,...,kn}
(ki1ki2 · · · kip) · x
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it follows that
1
2n
∫K
· · ·∫
K
ψ
x+∑
ij<ij+1, kij∈{k1,k2,...,kn}
(ki1ki2 · · · kip) · x
dk1 . . . dkn
=1
2n
∫K
· · ·∫
K
+∞∑r=1
2−r
∫K
· · ·∫
K
ϕ
X +∑
ij<ij+1, kij∈{kn+1,kn+2,...,kn+r−1}
(ki1ki2 · · · kiq) ·X
dkn+1 . . . dkn+r−1
}dk1 . . . .dkn
=+∞∑r=1
2−(n+r)
∫K
· · ·∫
K
ϕ
x+∑
ij<ij+1, kij∈{k1,k2,...,kn+r−1}
(ki1ki2 · · · kil) · x
dk1 . . . dkn+r−1
=+∞∑
m=n+1
2−m
∫K
· · ·∫
K
ϕ
x+∑
ij<ij+1, kij∈{k1,k2,...,km−1}
(ki1ki2 · · · kil) · x
dk1 . . . dkm−1.
In view of (2.1), this converges to zero, soq = q′. This ends the proof.
Our main result reads as follows.
Theorem 2.2. LetK be a finite cyclic subgroup of the group of automorphisms ofthe abelian group(G,+). Letϕ : G×G −→ R+ be a mapping such that
(2.19) ψ(x, y)
=∞∑
n=1
|K|(2|K|)n
∑k1,...,kn−1∈K
ϕ
x+∑
ij<ij+1; kij∈{k1,...,kn−1}
(ki1ki2 · · · kip) · x,
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y +∑
ij<ij+1; kij∈{k1,...,kn−1}
(ki1ki2 · · · kip) · y
<∞,
for all x, y ∈ G. Suppose thatf : G −→ C satisfies the inequality
(2.20)
∣∣∣∣∣ 1
|K|∑k∈K
f(x+ k · y)− f(x)− f(y)
∣∣∣∣∣ ≤ ϕ(x, y)
for all x, y ∈ G. Then, the limitq(x) = limn−→∞fn(x)2n , with
(2.21) f0(x) = f(x) and fn(x) =1
|K|∑k∈K
fn−1(x+ k · x) for all n ≥ 1,
exists for allx ∈ G, andq : G −→ C is the uniqueK-quadratical mapping whichsatisfies
(2.22) |f(x)− q(x)| ≤ ψ(x) for all x ∈ G.
Proof. In this case, the induction relations corresponding to (2.7) and (2.11) can bewritten as follows
(2.23)
∣∣∣∣fn(x)
2n− fn−1(x)
2n−1
∣∣∣∣≤ |K|
(2|K|)n
∑k1,...,kn−1∈K
ϕ
x+∑
ij<ij+1; kij∈{k1,...,kn−1}
(ki1ki2 · · · kip) · x
Hyers-Ulam-RassiasStability
M. Ait Sibaha, B. Bouikhalene
and E. Elqorachi
vol. 8, iss. 3, art. 89, 2007
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for anyn ∈ N \ {0}.
(2.24)
∣∣∣∣fn(x)
2n− f(x)
∣∣∣∣≤
n∑l=1
|K|(2|K|)l
∑k1,...,kl−1∈K
ϕ
x+∑
ij<ij+1; kij∈{k1,...,kl−1}
(ki1ki2 · · · kip) · x
,
for all integersn ∈ N \ {0}. So, we can easily deduce thatq(x) = limn−→+∞fn(x)2n
exists for allx ∈ G andq satisfies the inequality (2.22). Now, we will show thatq isaK-quadratical function. For allx, y ∈ G, we have∣∣∣∣∣ 1
|K|∑k∈K
f1(x+ k · y)− f1(x)− f1(y)
∣∣∣∣∣(2.25)
=
∣∣∣∣∣ 1
|K|∑k∈K
1
|K|∑k1∈K
f((x+ k · y) + k1 · (x+ k · y))
− 1
|K|∑k1∈K
f(x+ k1 · x)−1
|K|∑k1∈K
f(y + k1 · y)
∣∣∣∣∣=
∣∣∣∣∣ 1
|K|∑k∈K
1
|K|∑k1∈K
f((x+ k1 · x) + k · (y + k1 · y))
− 1
|K|∑k1∈K
f(x+ k1 · x)−1
|K|∑k1∈K
f(y + k1 · y)
∣∣∣∣∣
Hyers-Ulam-RassiasStability
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≤ 1
|K|∑k1∈K
∣∣∣∣∣ 1
|K|∑k∈K
f((x+ k1 · x) + k · (y + k1 · y))
− f(x+ k1 · x)− f(y + k1 · y)
∣∣∣∣∣≤ 1
|K|∑k1∈K
ϕ(x+ k1 · x, y + k1 · y).
Make the induction assumption
(2.26)
∣∣∣∣∣ 1
|K|∑k∈K
fn(x+ k · y)2n
− fn(x)
2n− fn(y)
2n
∣∣∣∣∣≤ 1
(2|K|)n
∑k1,...,kn∈K
ϕ
x+∑
ij<ij+1; kij∈{k1,...,kn}
(ki1ki2 · · · kip) · x, y
+∑
ij<ij+1; kij∈{k1,...,kn}
(ki1ki2 · · · kip) · y
which is true forn = 1 by (2.26). By using∣∣∣∣∣ 1
|K|∑k∈K
fn(x+ k · y)− fn(x)− fn(y)
∣∣∣∣∣=
∣∣∣∣∣ 1
|K|∑k′∈K
1
|K|∑k′∈K
fn−1(x+ k · y + k′ · (x+ k · y))
Hyers-Ulam-RassiasStability
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− 1
|K|∑k′∈K
fn−1(x+ k′ · x)− 1
|K|∑k∈K
fn−1(y + k′ · y)
∣∣∣∣∣=
∣∣∣∣∣ 1
|K|∑k′∈K
1
|K|∑k∈K
fn−1(x+ k′ · x+ k · (y + k′ · y))
− 1
|K|∑k′∈K
fn−1(x+ k′ · x)− 1
|K|∑k′∈K
fn−1(y + k′ · y)
∣∣∣∣∣≤ 1
|K|∑k′∈K
∣∣∣∣∣ 1
|K|∑k∈K
fn−1(x+ k′ · x+ k · (y + k′ · y))
− fn−1(x+ k′ · x)− fn−1(y + k′ · y)
∣∣∣∣∣we get the result (2.26) for all n ∈ N \ {0}. Now, in view of the condition (2.19), qis aK-quadratical function. This completes the proof.
Corollary 2.3. LetK be a finite cyclic subgroup of the group of automorphisms ofG, let δ > 0. Suppose thatf : G −→ C satisfies the inequality
(2.27)
∣∣∣∣∣∑k∈K
f(x+ k · y)− |K|f(x)− |K|f(y)
∣∣∣∣∣ ≤ δ
for all x, y ∈ G. Then, the limitq(x) = limn−→∞fn(x)2n , with
(2.28) f0(x) = f(x) and fn(x) =1
|K|∑k∈K
fn−1(x+ k · x) for n ≥ 1
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exists for allx ∈ G, andq : G −→ C is the uniqueK-quadratical mapping whichsatisfies
(2.29) |f(x)− q(x)| ≤ δ
|K|for all x ∈ G.
Corollary 2.4. LetK be a finite cyclic subgroup of the group of automorphisms ofthe normed space(G, ‖·‖), let θ ≥ 0 andp < 1. Suppose thatf : G −→ C satisfiesthe inequality
(2.30)
∣∣∣∣∣ 1
|K|∑k∈K
f(x+ k · y)− f(x)− f(y)
∣∣∣∣∣ ≤ θ(‖x‖p + ‖y‖p)
for all x, y ∈ G. Then, the limitq(x) = limn−→∞fn(x)2n , with
(2.31) f0(x) = f(x) and fn(x) =1
|K|∑k∈K
fn−1(x+ k · x) for n ≥ 1
exists for allx ∈ G, andq : G −→ C is the uniqueK-quadratical mapping whichsatisfies
(2.32) |f(x)− q(x)|
≤∞∑
n=1
|K|(2|K|)n
∑k1,...,kn−1∈K
2θ
∥∥∥∥∥∥x+∑
ij<ij+1; kij∈{k1,...,kn−1}
(ki1ki2 · · · kip) · x
∥∥∥∥∥∥p
for all x ∈ G.
Corollary 2.5 ([18]). LetK = {I}, θ ≥ 0 andp < 1. Suppose thatf : G −→ Csatisfies the inequality
Hyers-Ulam-RassiasStability
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(2.33) |f(x+ y)− f(x)− f(y)| ≤ θ(‖x‖p + ‖y‖p)
for all x, y ∈ G. Then, the limitq(x) = limn−→∞fn(x)2n , with
(2.34) fn(x) = f(2nx) for n ∈ N \ {0}exists for allx ∈ G, andq : G −→ C is the unique additive mapping which satisfies
(2.35) |f(x)− q(x)| ≤ 2θ ‖x‖p
2− 2pfor all x ∈ G.
Corollary 2.6 ([2]). LetK = {I, σ}, whereσ : G −→ G is an involution ofG, andlet ϕ : G×G −→ [0,∞) be a mapping satisfying the condition
(2.36) ψ(x, y) =∞∑
n=0
2−2n−1[ϕ(2nx, 2ny)
+ (2n − 1)ϕ(2n−1x+ 2n−1σ(x), 2n−1y + 2n−1σ(y))] <∞for all x, y ∈ G. Letf : G −→ C satisfy
(2.37) |f(x+ y) + f(x+ σ(y))− 2f(x)− 2f(y)| ≤ ϕ(x, y)
for all x, y ∈ G. Then, there exists a unique solutionq : G −→ C of the equation
(2.38) q(x+ y) + q(x+ σ(y)) = 2q(x) + 2q(y) x, y ∈ Ggiven by
(2.39) q(x) = limn−→+∞
2−2n{f(2nx) + (2n − 1)f(2n−1x+ 2n−1σ(x))}
which satisfies the inequality
(2.40) |f(x)− q(x)| ≤ ψ(x, x)
for all x ∈ G.
Remark1. We can replace in Theorem2.2 the condition thatK is a finite cyclicsubgroup by the condition thatK is a compact commutative subgroup ofAut(G).
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