HYERS-ULAM-RASSIAS STABILITY OF THE K -QUADRATIC FUNCTIONAL EQUATION

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Hyers-Ulam-Rassias Stability M. Ait Sibaha, B. Bouikhalene

and E. Elqorachi vol. 8, iss. 3, art. 89, 2007

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HYERS-ULAM-RASSIAS STABILITY OF THE K -QUADRATIC FUNCTIONAL EQUATION

MOHAMED AIT SIBAHA, BELAID BOUIKHALENE

University of Ibn Tofail, Faculty of Sciences Department of Mathematics

Kenitra, Morocco

EMail:mohaait@yahoo.fr bbouikhalene@yahoo.fr

ELHOUCIEN ELQORACHI

Laboratory LAMA , Harmonic Analysis and Functional Equations Team Department of Mathematics, Faculty of Sciences

University Ibn Zohr, Agadir, Morocco EMail:elqorachi@hotmail.com

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 equation 1

|K|

X

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 by automorphisms ofG. As a consequence we can derive the Hyers-Ulam-Rassias stability of the quadratic and the additive functional equations.

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1. Introduction

In the book, A Collection of Mathematical Problems [32], S.M. Ulam posed the question of the stability of the Cauchy functional equation. Ulam asked: if we replace a given functional equation by a functional inequality, when can we assert that the solutions of the inequality lie near to the solutions of the strict equation?

Originally, he had proposed the following more specific question during a lecture given before the University of Wisconsin’s Mathematics Club in 1940.

Given a groupG₁, a metric group (G₂, d), a numberε >0and a mapping f : G₁ −→G₂ which satisfies the inequalityd(f(xy), f(x)f(y))< εfor all x, y ∈ G₁, does there exist an homomorphism h : G₁ −→ G₂ and a constantk >0, depending only onG₁ andG₂ such thatd(f(x), h(x))≤ kεfor allxinG₁?

A partial and significant affirmative answer was given by D.H. Hyers [9] under the condition thatG₁andG₂ are Banach spaces.

In 1978, Th. M. Rassias [18] provided a generalization of Hyers’s stability theorem which allows the Cauchy difference to be unbounded, as follows:

Theorem 1.1. Letf :V −→Xbe a mapping between Banach spaces and letp < 1 be fixed. Iff satisfies the inequality

kf(x+y)−f(x)−f(y)k ≤θ(kxk^p+kyk^p)

for someθ ≥ 0and for allx, y ∈ V (x, y ∈ V \ {0}ifp < 0), then there exists a unique additive mappingT :V −→X such that

kf(x)−T(x)k ≤ 2θ

|2−2^p|kxk^p for allx∈V (x∈V \ {0}ifp <0).

If, in addition,f(tx)is continuous intfor each fixedx, thenT is linear.

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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 to Rassias’ question for p > 1. However, it was showed that a similar result for the casep= 1does not hold.

In 1994, P. Gavrutˇa [8] provided a generalization of the above theorem by replacing the function(x, y) 7−→ θ(kxk^p +kyk^p)with a mappingϕ(x, y)which satisfies the following condition:

∞

X

n=0

2⁻ⁿϕ(2ⁿx,2ⁿy)<∞ or

∞

X

n=0

2ⁿϕ x 2ⁿ⁺¹, y

2ⁿ⁺¹

<∞ for everyx, yin a Banach spaceV.

Since then, a number of stability results have been obtained for functional equations 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,

whereGis an abelian group. In particular, for the classical equations of Cauchy and Jensen, 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 the action by automorphisms on a groupGof a compact transformation groupK. Writ- ing the action ofk ∈K onx∈Gask·xand lettingdkdenote the normalized Haar measure onK,the functional equations (1.1) and (1.2) have the form

(1.3)

Z

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|

X

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, andGis an abelian group.

The set up allows us to give a unified treatment of the stability of the additive functional 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 solutions formulas.

The stability problem for the quadratic equation (1.6) was first solved by Skof in [23]. In [4] Cholewa extended Skof’s result in the following way, whereGis an abelian group andEis a Banach space.

Theorem 1.2. Letη >0be 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 ∈ Gthe limitq(x) = limn−→+∞f(2ⁿx)

2²ⁿ exists and q : G −→ E is the 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. Letp6= 2,θ > 0, δ > 0be real numbers. Suppose that the function f :E₁ −→E₂ satisfies the inequality

kf(x+y) +f(x−y)−2f(x)−2f(y)k ≤δ+θ(kxk^p+kyk^p) for all x, y ∈E₁. Then there exists exactly one quadratic functionq :E1 −→E2such that

kf(x)−q(x)k ≤c+kθkxk^p for allx∈E₁ifp≥0and for allx∈E₁\ {0}ifp≤0, where

• c= ^kf^(0)k₃ ,k = ₄₋₂²p andq(x) = limn−→+∞f(2ⁿx)

4ⁿ , forp < 2.

• c= 0,k = ₂p²−4 andq(x) = limn−→+∞f(2ⁿx)

4ⁿ , forp >2.

Recently, B. Bouikhalene, E. Elqorachi and Th. M. Rassias [1], [2] and [3] proved the Hyers-Ulam-Rassias stability of the functional equation (1.4) with K = {I, σ}

(σis an automorphism ofGsuch thatσ◦σ=I).

The results obtained in the present paper encompass results from [2] and [18]

given in Corollaries2.5and2.6below.

General Set-Up. LetKbe a compact transformation group of an abelian topological group(G,+), acting by automorphisms ofG. We letdkdenote the normalized Haar measure onK, and the action ofk ∈ K onx ∈ Gis denoted byk·x. We assume that the functionk 7−→k·yis continuous for ally∈G.

A continuous mappingq : G −→ C is said to be K-quadratical if it satisfies the functional equation

(1.9)

Z

K

q(x+k·y)dk =q(x) +q(y), x, y ∈G.

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WhenK is finite, the normalized Haar measuredk onK is given by Z

K

h(k)dk = 1

|K|

X

k∈K

h(k)

for anyh : K −→ C, where|K|denotes the order of K. So equation (1.9) can in this case be written

(1.10) 1

|K|

X

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 following condition

(2.1) ψ(x, y)

=

∞

X

n=1

2⁻ⁿ Z

K

Z

K

· · · Z

K

ϕ



x+ X

ij<ij+1, k_ij∈{k₁,k2,...,kn−1}

(k_i₁k_i₂· · ·k_i_p)·x,

y+ X

ij<ij+1, k_ij∈{k₁,k2,...,kn−1}

(k_i₁k_i₂· · ·k_i_p)·y



dk₁dk₂. . . dkn−1 <∞, for allpsuch that1 ≤p ≤ n−1,for allx, y ∈ G(uniform convergence). In what follows, we setϕ(x) = ϕ(x, x)andψ(x) = ψ(x, x)for allx∈G.

The main results of the present paper are based on the following proposition.

Proposition 2.1. Let G be an abelian group and let ϕ : G × G −→ R⁺ be a continuous control mapping which satisfies (2.1). Suppose that f : G −→ C is continuous and satisfies the inequality

(2.2)

Z

K

f(x+k·y)dk−f(x)−f(y)

≤ϕ(x, y) for allx, y ∈G.Then, the formulaq(x) = lim

n−→∞

fn(x) 2ⁿ , with

(2.3) f₀(x) =f(x) and f_n(x) = Z

K

fn−1(x+k·x)dk for all n ≥1,

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defines a continuous function which satisfies

(2.4) |f(x)−q(x)| ≤ψ(x) and Z

K

q(x+k·x)dk = 2q(x) for all x∈G.

Furthermore, the continuous functionqwith the condition (2.4) is unique.

Proof. Replacingybyxin (2.2) gives (2.5) |f₁(x)−2f(x)|=

Z

K

f(x+k·x)dk−2f(x)

≤ϕ(x) and consequently

|f₂(x)−2f₁(x)|= Z

K

f₁(x+k₁ ·x)dk₁−2 Z

K

f(x+k₁·x)dk₁ (2.6)

≤ Z

K

|f₁(x+k1·x)−2f(x+k1 ·x)|dk₁

≤ Z

K

ϕ(x+k₁·x)dk₁. Next, we prove that

(2.7)

f_n(x)

2ⁿ − f_n−1(x) 2ⁿ⁻¹

≤2⁻ⁿ Z

K

Z

K

· · · Z

K

ϕ



x+ X

ij<ij+1, k_ij∈{k1,k2,...,kn−1}

(k_i₁k_i₂· · ·k_i_p)·x



dk₁dk₂. . . dk_n−1

for alln ∈ 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

|f_n+1(x)−2f_n(x)|= Z

K

f_n(x+k_n·x)dk_n−2 Z

K

fn−1(x+k_n·x)dk_n (2.8)

≤ Z

K

|f_n(x+k_n·x)−2f_n−1(x+k_n·x)|dk_n. Then

f_n+1(x)

2ⁿ⁺¹ − f_n(x) 2ⁿ

(2.9)

≤ 1 2

Z

K

f_n(x+k_n·x)

2ⁿ −fn−1(x+k_n·x) 2ⁿ⁻¹

dk_n

≤2⁻⁽ⁿ⁺¹⁾ Z

K

Z

K

· · · Z

K

ϕ



x+ X

ij<ij+1, k_ij∈{k1,k2,...,kn}

(ki1ki2· · ·kip)·x



dk1dk2. . . dkn,

so that the inductive assumption (2.7) is indeed true for all positive integers. Hence, forr > swe get

f_r(x)

2^r − f_s(x) 2^s

(2.10)

≤

r−1

X

n=s

f_n+1(x)

2ⁿ⁺¹ − f_n(x) 2ⁿ

≤

r−1

X

n=s

2⁻⁽ⁿ⁺¹⁾ Z

K

· · · Z

K

ϕ



x+ X

ij<ij+1, k_ij∈{k1,k2,...,kn}

(ki1ki2· · ·kir)·x



dk1. . . dkn,

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which by assumption (2.1) converges to zero (uniformly) asrandstend to infinity.

Thus the sequence of complex functions ^fⁿ₂^(x)n is a Cauchy sequence for each fixed x ∈ G and then this sequence converges for each fixedx ∈ Gto some limit inC, which is continuous onG. We call this limitq(x). Next, we prove that

(2.11)

f_n(x)

2ⁿ −f(x)

≤

n

X

l=1

2^−l Z

K

· · · Z

K

ϕ



x+ X

ij<ij+1, k_ij∈{k1,k2,...,kl−1}

(k_i₁k_i₂· · ·k_i_p)·x



dk₁. . . dkl−1

for alln∈N\ {0}. We get the case ofn = 1by (2.5) (2.12) |f₁(x)−2f(x)|=

Z

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 for n∈N\ {0}. By using (2.9), we obtain

|fn+1(x)−2ⁿ⁺¹f(x)|

(2.13)

≤ |f_n+1(x)−2f_n(x)|+ 2|f_n(x)−2ⁿf(x)|

≤ Z

K

· · · Z

K

ϕ



x+ X

ij<ij+1, k_ij∈{k₁,k2,...,kn}

(k_i₁k_i₂· · ·k_i_p)·x



dk₁dk₂. . . dk_n

+2ⁿ⁺¹

n

X

l=1

2^−l Z

K

· · · Z

K

ϕ



x+ X

ij<ij+1, k_ij∈{k1,k2,...,kl−1}

(k_i₁k_i₂· · ·k_i_p)·x



dk₁dk₂. . . dkl−1

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so (2.11) is true for all n ∈ N\ {0}. By letting n −→ +∞, we obtain the first assertion of (2.4). We now shall show thatq satisfies the second assertion of (2.4).

By using (2.2) we get

Z

K

f1(x+k1·x)dk1−f1(x)−f1(x) (2.14)

= Z

K

Z

K

f(x+k1·x+k2·(x+k1·x))dk1dk2

− Z

K

f(x+k1·x)dk1− Z

K

f(x+k1·x)dk1

≤ Z

K

Z

K

f(x+k1·x+k2·(x+k1·x))dk2

−f(x+k1·x)−f(x+k1·x)

dk1

≤ Z

K

ϕ(x+k₁·x)dk₁. Make the induction assumption (2.15)

Z

K

fn(x+k·x)dk−2fn(x)

≤ Z

K

· · · Z

K

ϕ



x+ X

ij<ij+1, k_ij∈{k1,k2,...,kn}

(k_i₁k_i₂· · ·k_i_p)·x



dk₁. . . dk_n,

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which is true forn = 1by (2.14). Forn+ 1we have

Z

K

f_n+1(x+k_n+1·x)dk_n+1−2f_n+1(x)

= Z

K

Z

K

f_n(x+k_n+1·x+k·(x+k_n+1·x)dk_n+1dk−2 Z

K

f_n(x+k_n+1·x))dk_n+1

≤ Z

K

Z

K

f_n(x+k_n+1·x+k·(x+k_n+1·x))dk−2f_n(x+k_n+1·x)

dk_n+1

≤ Z

K

(Z

K

· · · Z

K

ϕ x

+k_n+1·x+ X

ij<ij+1, k_ij∈{k₁,k2,...,kn}

(k_i₁k_i₂· · ·k_i_p)·(x+k_n+1·x)

!

dk₁. . . dk_n )

dk_n+1

= Z

K

· · · Z

K

ϕ



x+ X

ij<ij+1, k_ij∈{k1,k2,...,kn+1}

(k_i₁k_i₂· · ·k_i_p)·x



dk₁. . . dk_n+1.

Thus (2.15) is true for alln∈N\{0}. Now, in view of the condition (2.1),qsatisfies the second assertion of (2.4).

To demonstrate the uniqueness of the mappingq subject to (2.4), let us assume on the contrary that there is another mappingq⁰ :G−→Csuch that

|f(x)−q⁰(x)| ≤ψ(x) and Z

K

q⁰(x+k·x)dk = 2q⁰(x) for all x∈G.

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First, we prove by induction the following relation (2.16)

f_n(x)

2ⁿ −q⁰(x)

≤ 1 2ⁿ

Z

K

· · · Z

K

ψ



x+ X

ij<ij+1, k_ij∈{k₁,k2,...,kn}

(k_i₁k_i₂· · ·k_i_p)·x



dk₁. . . dk_n.

Forn= 1, we have

|f1(x)−2q⁰(x)|= Z

K

f(x+k·x)dk− Z

K

q⁰(x+k·x)dk (2.17)

≤ Z

K

ψ(x+k·x)dk so (2.16) is true forn = 1. By using the following

|f_n+1(x)−2ⁿ⁺¹q⁰(x)|= Z

K

f_n(x+k·x)dk−2ⁿ Z

K

q⁰(x+k·x)dk (2.18)

≤ Z

K

|fn(x+k·x)−2ⁿq⁰(x+k·x)|dk we get the rest of the proof by proving that

1 2ⁿ

Z

K

· · · Z

K

ψ



x+ X

ij<ij+1, k_ij∈{k1,k2,...,kn}

(k_i₁k_i₂· · ·k_i_p)·x



dk₁. . . dk_n

converges to zero. In fact by setting

X =x+ X

ij<ij+1, k_ij∈{k₁,k2,...,kn}

(k_i₁k_i₂· · ·k_i_p)·x

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it follows that 1

2ⁿ Z

K

· · · Z

K

ψ



x+ X

ij<ij+1, k_ij∈{k1,k2,...,kn}

(k_i₁k_i₂· · ·k_i_p)·x



dk₁. . . dk_n

= 1 2ⁿ

Z

K

· · · Z

K







+∞

X

r=1

2^−r Z

K

· · · Z

K

ϕ



X+ X

ij<ij+1, k_ij∈{k_n+1,kn+2,...,kn+r−1}

(k_i₁k_i₂· · ·k_i_q)·X





dkn+1. . . dkn+r−1

)

dk1. . . .dkn

=

+∞

X

r=1

2^−(n+r) Z

K

· · · Z

K

ϕ



x+ X

ij<ij+1, k_ij∈{k1,k2,...,kn+r−1}

(k_i₁k_i₂· · ·k_i_l)·x



dk₁. . . dkn+r−1

=

+∞

X

m=n+1

2^−m Z

K

· · · Z

K

ϕ



x+ X

ij<ij+1, k_ij∈{k1,k2,...,km−1}

(ki1ki2· · ·ki_l)·x



dk1. . . dkm−1.

In view of (2.1), this converges to zero, soq=q⁰. This ends the proof.

Our main result reads as follows.

Theorem 2.2. LetK be a finite cyclic subgroup of the group of automorphisms of the abelian group(G,+). Letϕ :G×G−→R⁺be a mapping such that

(2.19) ψ(x, y)

=

∞

X

n=1

|K| (2|K|)ⁿ

X

k1,...,kn−1∈K

ϕ



x+ X

ij<ij+1;k_ij∈{k₁,...,kn−1}

(k_i₁k_i₂· · ·k_i_p)·x,

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y+ X

ij<ij+1;k_ij∈{k1,...,kn−1}

(k_i₁k_i₂· · ·k_i_p)·y



<∞, for allx, y ∈G. Suppose thatf :G−→Csatisfies the inequality

(2.20)

1

|K|

X

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|

X

k∈K

fn−1(x+k·x) for all n ≥1, exists for allx ∈ G,andq : G −→ Cis the uniqueK-quadratical mapping which satisfies

(2.22) |f(x)−q(x)| ≤ψ(x) for allx∈G.

Proof. In this case, the induction relations corresponding to (2.7) and (2.11) can be written as follows

(2.23)

fn(x)

2ⁿ − fn−1(x) 2ⁿ⁻¹

≤ |K| (2|K|)ⁿ

X

k1,...,kn−1∈K

ϕ



x+ X

ij<ij+1;k_ij∈{k1,...,kn−1}





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for anyn ∈N\ {0}.

(2.24)

f_n(x)

2ⁿ −f(x)

≤

n

X

l=1

|K|

(2|K|)^l

X

k1,...,kl−1∈K

ϕ



x+ X

ij<ij+1;k_ij∈{k1,...,kl−1}



,

for all integersn ∈ N\ {0}. So, we can easily deduce thatq(x) = limn−→+∞fn(x) 2ⁿ

exists for allx∈Gandqsatisfies the inequality (2.22). Now, we will show thatqis aK-quadratical function. For allx,y∈G, we have

1

|K|

X

k∈K

f₁(x+k·y)−f₁(x)−f₁(y) (2.25)

=

1

|K|

X

k∈K

1

|K| X

k1∈K

f((x+k·y) +k₁·(x+k·y))

− 1

|K|

X

k1∈K

f(x+k₁·x)− 1

|K|

X

k1∈K

f(y+k₁·y)

=

1

|K|

X

k∈K

1

|K| X

k1∈K

f((x+k₁·x) +k·(y+k₁·y))

− 1

|K|

X

k1∈K

f(x+k₁·x)− 1

|K|

X

k1∈K

f(y+k₁·y)

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≤ 1

|K|

X

k1∈K

1

|K|

X

k∈K

f((x+k₁·x) +k·(y+k₁·y))

−f(x+k₁·x)−f(y+k₁·y)

≤ 1

|K|

X

k1∈K

ϕ(x+k₁·x, y+k₁·y).

Make the induction assumption (2.26)

1

|K|

X

k∈K

f_n(x+k·y)

2ⁿ −f_n(x)

2ⁿ − f_n(y) 2ⁿ

≤ 1

(2|K|)ⁿ X

k1,...,kn∈K

ϕ



x+ X

ij<ij+1;k_ij∈{k1,...,kn}

(k_i₁k_i₂· · ·k_i_p)·x, y

+ X

ij<ij+1;k_ij∈{k1,...,kn}

(k_i₁k_i₂· · ·k_i_p)·y





which is true forn = 1by (2.26). By using

1

|K|

X

k∈K

f_n(x+k·y)−f_n(x)−f_n(y)

=

1

|K|

X

k⁰∈K

1

|K| X

k⁰∈K

fn−1(x+k·y+k⁰ ·(x+k·y))

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− 1

|K| X

k⁰∈K

fn−1(x+k⁰·x)− 1

|K| X

k∈K

fn−1(y+k⁰·y)

=

1

|K|

X

k⁰∈K

1

|K| X

k∈K

fn−1(x+k⁰·x+k·(y+k⁰·y))

− 1

|K| X

k⁰∈K

fn−1(x+k⁰·x)− 1

|K| X

k⁰∈K

fn−1(y+k⁰·y)

≤ 1

|K|

X

k⁰∈K

1

|K|

X

k∈K

fn−1(x+k⁰·x+k·(y+k⁰·y))

−f_n−1(x+k⁰·x)−f_n−1(y+k⁰·y)

we get the result (2.26) for alln∈ N\ {0}. Now, in view of the condition (2.19),q is aK-quadratical function. This completes the proof.

Corollary 2.3. LetK be a finite cyclic subgroup of the group of automorphisms of G, letδ >0. Suppose thatf :G−→Csatisfies the inequality

(2.27)

X

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 f_n(x) = 1

|K|

X

k∈K

fn−1(x+k·x) for n≥1

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exists for allx ∈ G,andq : G −→ Cis the uniqueK-quadratical mapping which satisfies

(2.29) |f(x)−q(x)| ≤ δ

|K| for allx∈G.

Corollary 2.4. LetK be a finite cyclic subgroup of the group of automorphisms of the normed space(G,k·k), letθ ≥0andp <1. Suppose thatf :G−→Csatisfies the inequality

(2.30)

1

|K| X

k∈K

f(x+k·y)−f(x)−f(y)

≤θ(kxk^p+kyk^p) for all x, y ∈G.Then, the limitq(x) = limn−→∞ fn(x)

2ⁿ , with (2.31) f₀(x) = f(x) and f_n(x) = 1

|K|

X

k∈K

fn−1(x+k·x) for n≥1 exists for allx ∈ G,andq : G −→ Cis the uniqueK-quadratical mapping which satisfies

(2.32) |f(x)−q(x)|

≤

∞

X

n=1

|K|

(2|K|)ⁿ

X

k1,...,kn−1∈K

2θ

x+ X

ij<ij+1;k_ij∈{k₁,...,kn−1}

(k_i₁k_i₂· · ·k_i_p)·x

p

for allx∈G.

Corollary 2.5 ([18]). LetK = {I}, θ ≥ 0andp < 1. Suppose thatf : G −→ C satisfies the inequality

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(2.33) |f(x+y)−f(x)−f(y)| ≤θ(kxk^p+kyk^p) for allx, y ∈G.Then, the limitq(x) = lim_n−→∞ ^fⁿ₂^(x)n , with (2.34) f_n(x) = f(2ⁿx) for n ∈N\ {0}

exists for allx∈G,andq:G−→Cis the unique additive mapping which satisfies (2.35) |f(x)−q(x)| ≤ 2θkxk^p

2−2^p for allx∈G.

Corollary 2.6 ([2]). LetK ={I, σ}, whereσ :G−→Gis an involution ofG, and letϕ :G×G−→[0,∞)be a mapping satisfying the condition

(2.36) ψ(x, y) =

∞

X

n=0

2⁻²ⁿ⁻¹[ϕ(2ⁿx,2ⁿy)

+ (2ⁿ−1)ϕ(2ⁿ⁻¹x+ 2ⁿ⁻¹σ(x),2ⁿ⁻¹y+ 2ⁿ⁻¹σ(y))]<∞ for allx, y ∈G. Letf :G−→Csatisfy

(2.37) |f(x+y) +f(x+σ(y))−2f(x)−2f(y)| ≤ϕ(x, y)

for allx, y ∈G. Then, there exists a unique solutionq:G−→Cof the equation (2.38) q(x+y) +q(x+σ(y)) = 2q(x) + 2q(y) x, y ∈G

given by

(2.39) q(x) = lim

n−→+∞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 allx∈G.

Remark 1. We can replace in Theorem 2.2 the condition that K is a finite cyclic subgroup by the condition thatK is a compact commutative subgroup ofAut(G).

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