hyperstability of cubic functional equation in ultrametric spaces...hyperstability of cubic...

Hyperstability of cubic functional equation inultrametric spaces

Youssef AribouUniversity of Ibn Tofail, Morocco

Muaadh AlmahalebiUniversity of Ibn Tofail, Morocco

andSamir kabbaj

University of Ibn Tofail, MoroccoReceived : December 2016. Accepted : March 2017

Proyecciones Journal of MathematicsVol. 36, No 3, pp. 461-484, September 2017.Universidad Catolica del NorteAntofagasta - Chile

Abstract

In this paper, we present the hyperstability results of cubic func-tional equations in ultrametric Banach spaces.

Keywords : Stability, hyperstability, ultrametric space, cubic func-tional equation.

Mathematics Subject Classification : Primary 39B82; Secondary39B52, 47H10.

462 Youssef Aribou, Muaadh Almahalebi and Samir kabbaj

1. Introduction

The starting point of studying the stability of functional equations seemsto be the famous talk of Ulam [29] in 1940, in which he discussed a numberof important unsolved problems. Among those was the question concerningthe stability of group homomorphisms.

Let G1 be a group and let G2 be a metric group with a metric d(., .).Given ε > 0, does there exists a δ > 0 such that if a mapping h : G1 → G2satisfies the inequality d

³h(xy), h(x)h(y)

´< δ for all x, y ∈ G1, then there

exists a homomorphism H : G1 → G2 with d³h(x),H(x)

´< ε for all

x ∈ G1.

The first partial answer, in the case of Cauchy equation in Banachspaces, to Ulam question was given by Hyers [22]. Later, the result of Hy-ers was first generalized by Aoki [?]. And only much later by Rassias [27]and Gavruta [20]. Since then, the stability problems of several functionalequations have been extensively investigated.

We say a functional equation is hyperstable if any function f satisfyingthe equation approximately (in some sense) must be actually a solution toit. It seems that the first hyperstability result was published in [11] andconcerned the ring homomorphisms. However, the term hyperstability hasbeen used for the first time in [25]. Quite often the hyperstability is con-fused with superstability, which admits also bounded functions. Numerouspapers on this subject have been published and we refer to [1]-[5], [8], [15]-[18], [21], [25], [26], [28].

Throughout this paper, N stands for the set of all positive integers,N0 :=N∪{0}, Nm0 the set of integers ≥ m0, R+ := [0,∞) and we use thenotation X0 for the set X \ {0}.

Let us recall (see, for instance, [24] some basic definitions and factsconcerning non-Archimedean normed spaces.

Definition 1.1. By a non-Archimedean field we mean a field K equippedwith a function (valuation) | · | : K→ [0,∞) such that for all r, s ∈ K, thefollowing conditions hold:

1. |r| = 0 if and only if r = 0,

Hyperstability of cubic functional equation in ultrametric spaces 463

2. |rs| = |r||s|,

3. |r + s| ≤ maxn|r|, |s|

o.

The pair (K, |.|) is called a valued field.

In any non-Archimedean field we have |1| = | − 1| = 1 and |n| ≤ 1 forn ∈N0. In any field K the function | · | : K→ R+ given by

|x| :=(0, x = 0,1, x 6= 0,

is a valuation which is called trivial, but the most important examples ofnon-Archimedean fields are p-adic numbers which have gained the interestof physicists for their research in some problems coming from quantumphysics, p-adic strings and superstrings.

Definition 1.2. Let X be a vector space over a scalar field K with a non-Archimedean non-trivial valuation | · |. A function || · ||∗ : X → R is anon-Archimedean norm (valuation) if it satisfies the following conditions:

1. kxk∗ = 0 if and only if x = 0,

2. krxk∗ = |r| kxk∗ (r ∈ K, x ∈ X),

3. The strong triangle inequality (ultrametric); namely

kx+ yk∗ ≤ maxnkxk∗, kyk∗

ox, y ∈ X.

Then (X, k · k∗) is called a non-Archimedean normed space or an ultra-metric normed space.

Definition 1.3. Let {xn} be a sequence in a non-Archimedean normedspace X.

1. A sequence{xn}∞n=1 in a non-Archimedean space is a Cauchy sequenceiff the sequence {xn+1 − xn}∞n=1 converges to zero;

2. The sequence {xn} is said to be convergent if, there exists x ∈ Xsuch that, for any ε > 0, there is a positive integer N such thatkxn − xk∗ ≤ ε, for all n ≥ N . Then the point x ∈ X is called thelimit of the sequence {xn}, which is denoted by limn→∞xn = x;


3. If every Cauchy sequence in X converges, then the non-Archimedeannormed space X is called a non-Archimedean Banach space or anultrametric Banach space.

Let X, Y be normed spaces. A function f : X → Y is Cubic providedit satisfies the functional equation

f (2x+ y) + f (2x− y) = 2f(x+ y) + 2f(x− y) + 12f(x) for all x, y ∈ X,

(1.1)

and we can say that f : X → Y is Cubic on X0 if it satisfies (1.1) for allx, y ∈ X0 such that x+ y 6= 0.

In 2013, A. Bahyrycz and al. [7] used the fixed point theorem from [12,Theorem 1] to prove the stability results for a generalization of p-Wrightaffine equation in ultrametric spaces. Recently, corresponding results formore general functional equations (in classical spaces) have been proved in[9], [10], [30] and [31].

In this paper, by using the fixed point method derived from [8], [15]and [14], we present some hyperstability results for the equation (1.1) inultrametric Banach spaces. Before proceeding to the main results, we stateTheorem 1.4 which is useful for our purpose. To present it, we introducethe following three hypotheses:

(H1) X is a nonempty set, Y is an ultrametric Banach space over a non-Archimedean field, f1, ..., fk : X −→ X and L1, ..., Lk : X −→ R+are given.

(H2) T : Y X −→ Y X is an operator satisfying the inequality°°°°T ξ(x)− T µ(x)°°°°∗≤ max1≤i≤k

½Li(x)

°°°°ξµfi(x)¶− µ

µfi(x)

¶°°°°∗

¾,

ξ, µ ∈ Y X , x ∈ X.

(H3) Λ : RX+ −→ RX

+ is a linear operator defined by

Λδ(x) := max1≤i≤k

½Li(x)δ

µfi(x)

¶¾, δ ∈ RX

+ , x ∈ X.


Thanks to a result due to J. Brzdek and K. Cieplinski [13, Remark 2],we state a slightly modified version of the fixed point theorem [12, Theorem1] in ultrametric spaces. We use it to assert the existence of a unique fixedpoint of operator T : Y X −→ Y X .

Theorem 1.4. Let hypotheses (H1)-(H3) be valid and functions ε : X −→R+ and ϕ : X −→ Y fulfil the following two conditions

kT ϕ(x)− ϕ(x)k∗ ≤ ε(x), x ∈ X,

limn→∞

Λnε(x) = 0, x ∈ X.

Then there exists a unique fixed point ψ ∈ Y X of T with

kϕ(x)− ψ(x)k∗ ≤ supn∈N0

Λnε(x), x ∈ X.

Moreoverψ(x) := lim

n→∞T nϕ(x), x ∈ X.

2. Main results

In this section, using Theorem 1.4 as a basic tool to prove the hyperstabilityresults of the cubic functional equation in ultrametric Banach spaces.

Theorem 2.1. Let (X, k·k) and (Y, k·k∗) be normed space and ultrametricBanach space respectively, c ≥ 0, p, q ∈ R, p + q < 0 and let f : X → Ysatisfy

kf (2x+ y) + f(2x− y)− 2f(x+ y)− 2f(x− y)− 12f(x)k∗ ≤ c kxkp kykq,(2.1)for all x, y ∈ X0. Then f is cubic on X0.

Proof. Take m ∈ N such that

αm :=

µm− 12

¶p+q< 1 and m ≥ 4.

Since p + q < 0, one of p, q must be negative. Assume that q < 0 andreplace y by mx and x by m+1

2 x in (2.1). Thus


°°°°12f µm+ 1

2x

¶+ 2f(

3m+ 1

2x) + 2f(

1−m

2x)− f((2m+ 1)x)− f(x)

°°°°∗

≤ c mq(m+ 1

2)p kxkp+q,

Define operators Tm : Y X0 → Y X0 and Λm : RX0+ → RX0

+ by

Tmξ(x) := 12ξµµ

m+ 1

2

¶x

¶+ 2ξ

µµ3m+ 1

2

¶x

¶+ 2ξ

µµ1−m

2

¶x

¶− ξ ((2m+ 1)x) , ξ ∈ Y X0 , x ∈ X0,(2.2)

Λmδ(x) := max

½δ

µµm+ 1

2

¶x

¶, δ

µµ3m+ 1

2

¶x

¶, δ

µµ1−m

2

¶x

¶δ(2m+ 1), δ ∈ RX0

+ , x ∈ X0(2.3)

and write

εm(x) := c mq(m+ 1

2)pkxkp+q, x ∈ X0.(2.4)

It is easily seen that Λm has the form described in (H3) with k = 4,

f1(x) =³m+12

´x, f2(x) =

³3m+12

´x, f3(x) =

³1−m2

´x, f4(x) = (2m+ 1)x

and L1(x) = L2(x) = L3(x) = L4(x) = 1. Further, (2.1) can be written inthe following way

kTmf(x)− f(x)k∗ ≤ εm(x), x ∈ X0.

Moreover, for every ξ, µ ∈ Y X0 , x ∈ X0°°°°Tmξ(x)− Tmµ(x)°°°°∗=

°°°°12ξ ³³m+12 ´x´+ 2ξ

³³3m+12

´x´+ 2ξ

³³1−m2

´x´

−ξ ((2m+ 1)x)− 12µ³³

m+12

´x´− 2µ

³³3m+12

´x´− 2µ

³³1−m2

´x´

+ξ ((2m+ 1)x)

°°°°∗

≤ maxn °°°ξ ³³m+12 ´

x´− µ

³³m+12

´x´°°°∗,°°°ξ(3m+12 x)− µ(3m+12 x)

°°°∗,°°°ξ(1−m2 x)− µ(1−m2 x)

°°°∗,°°°ξ(2m+ 1)x)− µ((2m+ 1)x)

°°°∗}.


So, (H2) is valid.

By using mathematical induction, we will show that for each x ∈ X0

we have

Λnmεm(x) = c mq(m+ 1

2)p kxkp+qαnm(2.5)

where αm =³m−12

´p+q. From (2.4), we obtain that (2.5) holds for n = 0.

Next, we will assume that (2.5) holds for n = k, where k ∈ N. Then wehave

Λk+1m εm(x) = Λm³Λkmεm(x)

´= max

nΛkmεm

³³m+12

´x´, Λkmεm

³³3m+12

´x´,

Λkmεm³³

m−12

´x´, Λkmεm((2m+ 1)x)

= max

½c mq(m+12 )pkxkp+q αkm

³m+12

´p+q, c mq(m+12 )pkxkp+q αkm

³3m+12

´p+q,

c mq(m+12 )pkxkp+q αkm³m−12

´p+qc mqkxkp+q αkm(2m+ 1)p+q

= c mqkxkp+q αkmmax½ ³

m+12

´p+q,³3m+12

´p+q,³m−12

´p+q, (2m+ 1)p+q

¾= c mq(m+12 )pkxkp+q αk+1m , x ∈ X0.

This shows that (2.5) holds for n = k + 1. Now we can conclude thatthe inequality (2.5) holds for all n ∈ N0. From (2.5), we obtain

limn→∞

Λnεm(x) = 0,

for all x ∈ X0. Hence, according to Theorem 1.4, there exists a uniquesolution Cm : X0 → Y of the equation

Cm(x) = 12Cm

µµm+ 1

2

¶x

¶+2Cm

µµ3m+ 1

2

¶x

¶+2Cm

µµ1−m

2

¶x

¶

− Cm((2m+ 1)x),(2.6)

x ∈ X0(2.7)

such that

kf(x)− Cm(x)k∗ ≤ supn∈N0

½c mq(

m+ 1

2)pkxkp+q αnm

¾, x ∈ X0.(2.8)


Moreover,

Cm(x) := limn→∞

T nmf(x)

for all x ∈ X0. Now we show that

kT nmf (2x+ y) + T n

mf (2x− y)− 2T nmf (x+ y)− T n

mf (x− y)− 12T nmf(x)k∗

≤ c αnmkxkp kykq,(2.9)

for every x, y ∈ X0 such that x+ y 6= 0. Since the case n = 0 is just (2.1),take k ∈ N and assume that (2.9) holds for n = k and every x, y ∈ X0 suchthat x+ y 6= 0. Then

°°°°T k+1m f (2x+ y)+T k+1

m f (2x− y)−2T k+1m f (x+ y)−2T k+1

m f (x− y)−12T k+1m f(x)

°°°°∗

=

°°°°12Tkmf

³³m+12

´(2x+ y)

´+2T k

mf³³

3m+12

´(2x+ y)

´+2T k

mf³³

1−m2

´(2x+ y)

´-Tk

mf ((2m+ 1) (2x+ y))+12T kmf

³³m+12

´(2x− y)

´+2T k

mf³³

3m+12

´(2x− y)

´+2Tk

mf³³

1−m2

´(2x− y)

´−T k

mf ((2m+ 1) (2x− y))−24T kmf

³³m+12

´(x+ y)

´-4Tk

mf³³

3m+12

´(x+ y)

´−4T k

mf³³

1−m2

´(x+ y)

´+2T k

mf ((2m+ 1) (x+ y))

-24Tkmf

³³m+12

´(x− y)

´− 4T k

mf³³

3m+12

´(x− y)

´−4T k

mf³³

1−m2

´(x− y)

´+2T k

mf ((2m+ 1) (x− y))− 144T kmf

³³m+12

´(x)´− 24T k

mf³³

3m+12

´(x)´

-24Tkmf

³³1−m2

´(x)´+ 12T k

mf ((2m+ 1) (x))

°°°°∗

(2.10)


≤ max½°°°°T n

mf³m+12 (2x+ y)

´+ T n

mf³m+12 (2x− y)

´− 2T n

mf³m+12 (x+ y)

´

−T nmf

³m+12 (x− y)

´− 12T n

mf(m+12 x)

°°°°∗,°°°°T n

mf³3m+12 (2x+ y)

´+ T n

mf³3m+12 (2x− y)

´− 2T n

mf³3m+12 (x+ y)

´

−T nmf

³3m+12 (x− y)

´− 12T n

mf(3m+12 x)

°°°°∗,°°°°T n

mf³1−m2 (2x+ y)

´+ T n

mf³1−m2 (2x− y)

´− 2T n

mf³1−m2 (x+ y)

´

−T nmf

³1−m2 (x− y)

´− 12T n

mf(1−m2 x)

°°°°∗,°°°°T n

mf ((2m+ 1)(2x+ y)) + T nmf ((2m+ 1)(2x− y))− 2T n

mf ((2m+ 1)(x+ y))

−T nmf ((2m+ 1)(x− y))− 12T n

mf((2m+ 1)x)

°°°°∗

¾

≤ max½c αkmkxkp kykq

³m+12

´p+q, c αkmkxkpkykq

³3m+12

´p+q,

c αkmkxkpkykq³1−m2

´p+q, c αkmkxkpkykq (2m+ 1)p+q

¾

= cαkmkxkpkykqmax½³

m+12

´p+q,³3m+12

´p+q,³m−12

´p+q, (3m+ 1)p+q

¾

≤ cαk+1m kxkpkykq

(2.11)


for all x, y ∈ X0 such that x + y 6= 0. Thus, by induction we have shownthat (2.9) holds for every n ∈ N0. Letting n→∞ in (2.9), we obtain that

Cm (2x+ y) + Cm (2x− y) = 2Cm (x+ y) + 2Cm (x− y) + 12Cm (x)

for all x, y ∈ X0 such that x + y 6= 0. In this way we obtain a sequence{Cm}m≥m0 of cubic functions on X0 such that


½c mq(

m+ 1

2)pkxkp+q αnm

¾, x0,

this implies that

kf(x)− Cm(x)k∗ ≤ c mq(m+ 1

2)pkxkp+q, x ∈ X0,

It follows, with m→∞, that f is Cubic on X0. 2In a similar way we can prove the following theorem.

Theorem 2.2. Let (X, k·k) and (Y, k·k∗) be normed space and ultrametricBanach space respectively, c ≥ 0, p, q ∈ R, p + q > 0 and let f : X → Ysatisfy

kf (2x+ y) + f(2x− y)− 2f(x+ y)− 2f(x− y)− 12f(x)k∗ ≤ c kxkp kykq,(2.12)for all x, y ∈ X0. Then f is Cubic on X0.

Proof. Take m ∈ N such that

αm =

µm− 4m

¶p+q< 1 and 6 < m.

Since p+ q > 0, one of p, q must be positive; let q > 0 and replace y by−2m y and x by m−2

2m y in (2.12). Thus

°°°°12f ³m−22m y´+ 2f(m+22m y) + 2f(m−62m y)− f(m−4m y)− f(y)

°°°°∗

≤ c³2m

´q ³m−22m

´pkykp+q, x ∈ X0.

(2.13)


Write

Tmξ(x) := 12ξ³³

m−22m

´x´+ 2ξ

³³m+22m

´x´+ 2ξ

³³m−62m

´x´− ξ

³³m−4m

´x´,

ξ ∈ Y X0 , x ∈ X0,(2.14)and

εm(x) := c

µ2

m

¶q µm− 12m

¶pkxkp+q, x ∈ X0,(2.15)

then (2.13) takes form


Define

Λmδ(x) := maxnδ³³

m−22m

´x´, δ³³

m+22m

´x´, δ³³

m−62m

´x´, δ³³

m−4m

´x´},

δ ∈ RX0+ , x ∈ X0.

(2.16)

Then it is easily seen that Λm has the form described in (H3) with k = 4,

f1(x) =³m−22m

´x, f2(x) =

³m+22m

´x,f3(x) =

³m−62m

´x,f2(x) =

³m−4m

´x and

L1(x) = 1, L2(x) = 1, L3(x) = 1, L4(x) = 1.

Moreover, for every ξ, µ ∈ Y X0 , x ∈ X0°°°°Tmξ(x)− Tmµ(x)°°°°∗=

°°°°12ξ ³³m−22m

´x´+ 2ξ

³³m+22m

´x´+ 2ξ

³³m−62m

´x´

−ξ³³

m−4m

´x´− 12µ

³³m−22m

´x´− 2µ

³³m+22m

´x´− 2µ

³³m−62m

´x´

+ξ³³

m−4m

´x´ °°°°

∗≤ max

n °°°ξ ³³m−22m

´x´− µ

³³m−22m

´x´°°°∗,°°°ξ(m+22m x)− µ(m+12m x)

°°°∗,°°°ξ(m−62m x)− µ(m−62m x)

°°°∗,°°°ξ(m−42m )x)− µ((m−42m )x)

°°°∗}.

So, (H2) is valid.By using mathematical induction, we will show that for each x ∈ X0

we have

Λnmεm(x) = c

µ2

m

¶q µm− 22m

¶pkxkp+q αnm(2.17)


where αm =³m−4m

´p+q. From (2.15), we obtain that (2.17) holds for n = 0.

Next, we will assume that (2.17) holds for n = k, where k ∈ N. Then wehaveΛk+1m εm(x) = Λm

³Λkmεm(x)

´= max

nΛkmεm

³³m−22m

´x´, Λkmεm

³³m+22m

´x´

Λkmεm³³

m−62m

´x´, Λkmεm

³³m−4m

´xo

= c³2m

´q ³m−22m

´pkxkp+qαkmmax

½³m−22m

´p+q,³m+22m

´p+q,³m−62m

´p+q,³m−4m

´p+q¾

= c³2m

´q ³m−22m

´pkxkp+q αk+1m , x ∈ X0.

This shows that (2.17) holds for n = k + 1. Now we can conclude thatthe inequality (2.17) holds for all n ∈ N0. From (2.17), we obtain

limn→∞

Λnεm(x) = 0,

for all x ∈ X0. Hence, according to Theorem 1.4, there exists a uniquesolution Cm : X0 → Y of the equation

Cm(x) = 12Cm

µµm− 22m

¶x

¶+ 2Cm

µµm+ 2

2m

¶x

¶+ 2Cm

µµm− 62m

¶x

¶

− Cm

µµm− 4m

¶x

¶, x ∈ X0(2.18)

such that


½c

µ2

m

¶q µm− 22m

¶pkxkp+q αnm

¾, x0.

(2.19)

Moreover,

Cm(x) := limn→∞

T nmf(x)

for all x ∈ X0. We show that



mf (2x− y)− 2T nmf (x+ y)− T n

mf (x− y)− 12T nmf(x)k∗

≤ c αnmkxkp kykq,(2.20)

for every x, y ∈ X0 such that x+ y 6= 0. Since the case n = 0 is just (2.12),take k ∈ N and assume that (2.20) holds for n = k and every x, y ∈ X0

such that x+ y 6= 0. Then

°°°T k+1m f (2x+ y) + T k+1

m f (2x− y)− 2T k+1m f (x+ y)− 2T k+1

m f (x− y)− 12T k+1m f(x)

°°°∗°°°= 12T k

mf³³

m−22m

´(2x+ y)

´+ 2T k

mf³³

m+22m

´(2x+ y)

´+ 2T k

mf³³

m−62m

´(2x+ y)

´

−T kmf

³(m−4m )(2x+ y)

´+ 12T k

mf³(m−22m )(2x− y)

´+ 2T k

mf³(m+22m )(2x− y)

´

+2T kmf

³(m−62m )(2x− y)

´− T k

mf³(m−4m )(2x− y)

´− 24T k

mf³(m−22m )(x+ y)

´

−4T kmf

³³m+22m

´(x+ y)

´− 4T k

mf³³

m−62m

´(x+ y)

´+ 2T k

mf³³

m−4m

´(x+ y)

´

−24T kmf

³³m−22m

´(x− y)

´− 4T k

mf³³

m+22m

´(x− y)

´− 4T k

mf³³

m−62m

´(x− y)

´

+2T kmf

³³m−4m

´(x− y)

´− 144T k

mf³³

m−22m

´x´− 24T k

mf³³

m+22m

´x´

−24T kmf

³³m−62m

´(x)´+ 12T k

mf³³

m−4m

´(x)´°°°∗

(2.21)


≤ maxn°°°T n

mf³m−22m (2x+ y)

´+ T n

mf³m−22m (2x− y)

´− 2T n

mf³m−22m (x+ y)

´

−T nmf

³m−22m (x− y)

´− 12T n

mf(m−22m x)

°°°∗,

°°°T nmf

³m+22m (2x+ y)

´+ T n

mf³m+22m (2x− y)

´− 2T n

mf³m+22m (x+ y)

´

−T nmf

³m+22m (x− y)

´− 12T n

mf(m+22m x)

°°°∗,

°°°T nmf

³m−62m (2x+ y)

´+ T n

mf³m−62m (2x− y)

´− 2T n

mf³m−62m (x+ y)

´

−T nmf

³m−62m (x− y)

´− 12T n

mf(m−62m x)

°°°∗,

°°°T nmf

³(m−4m )(2x+ y)

´+ T n

mf³(m−4m )(2x− y)

´− 2T n

mf³(m−4m )(x+ y)

´

−T nmf

³(m−4m )(x− y)

´− 12T n

mf((m−4m )x)

°°°∗

o

≤ maxncαkmkxkpkykq

³m−22m

´p+q, c αkmkxkpkykq

³m+12m

´p+q,

cαkmkxkpkykq³m−62m

´p+q, αkmkxkpkykq(m−4m )p+q

o= c αkmkxkp kykqmax

½ ³m−22m

´p+q,³m+22m

´p+q,³m−62m

´p+q, (m−4m )p+q

¾

≤ c αk+1m kxkp kykq


for all x, y ∈ X0 such that x + y 6= 0. Thus, by induction we have shownthat (2.20) holds for every n ∈ N0. Letting n → ∞ in (2.20), we obtainthat


for all x, y ∈ X0 such that x + y 6= 0. In this way we obtain a sequence{Cm}m≥m0 of Cubic functions on X0 such that


½c

µ2

m

¶q µm− 22m

¶pkxkp+q αnm

¾, x0,

this implies that

kf(x)−Cm(x)k∗ ≤ c

µ2

m

¶q µm− 22m

¶p|xkp+q, x ∈ X0.

It follows, with m→∞, that f is Cubic on X0. 2It easy to show the hyperstability of Cubic equation on the set contain-

ing 0. We present the following theorem and we refer to see [8, Theorem5].

Theorem 2.3. Let (X, k·k) and (Y, k·k∗) be normed space and ultrametricBanach space respectively, c ≥ 0, p, q ∈ R, p, q > 0 and let f : X → Ysatisfy

kf (2x+ y)) + f(2x+ y)− 2f(x+ y)− 2f(x− y)− 12f(x)k∗ ≤ c kxkp kykq,(2.22)for all x, y ∈ X0. Then f is Cubic on X0.

Proof. Same proof of last theoremThe above theorems imply in particular the following corollary, which

shows their simple application.

Corollary 2.4. Let (X, k·k) and (Y, k·k∗) be normed space and ultrametricBanach space respectively, G : X2 → Y and G(x, y) 6= 0 for some x, y ∈ Xand °°°G(x, y)°°°

∗≤ c kxkp kykq, x, y ∈ X(2.23)

where c ≥ 0, p, q ∈ R. Assume that the numbers p, q satisfy one of thefollowing conditions:


1. p+ q < 0, and (2.1) holds for all x, y ∈ X0,

2. p+ q > 0, and (2.12) holds for all x, y ∈ X0.

Then the functional equation

f (2x+ y) + f (2x+ y) = 2f(x+ y) + 2f(x− y) + 12f(x) +G(x, y), x, y ∈ X

(2.24)

has no solution in the class of functions g : X → Y .

In the following theorem, we present a general hyperstability for theCubic equation where the control function is ϕ(x)+ϕ(y), which correspondsto the approach introduced in [14]

Theorem 2.5. Let (X, k·k) be a normed space, (Y, k·k∗) be an ultrametricBanach space over a field K, and ϕ : X → R+ be a function such that

U := {n ∈ N : αn := max {λ(n) , λ(3n− 1) , λ(−n+ 1) , λ(4n− 1)} < 1}(2.25)is an infinite set, where λ(a) := inf{t ∈ R+ : ϕ(ax) ≤ tϕ(x) for all x ∈ X}for all a ∈ K0. Suppose that

lima→∞

λ(a) = 0 and lima→∞

λ(−a) = 0.

And f : X → Y satisfies the inequality

kf (2x+ y)) + f(2x+ y)− 2f(x+ y)− 2f(x− y)− 12f(x)k∗ ≤ ϕ(x)+ϕ(y),(2.26)for all x, y ∈ X0. Then f is Cubic on X0.

Proof. Replacing x by mx and y by (−2m + 1)x for m ∈ N in (2.27)we get

kf (x)) + f((4m− 1)x)− 2f((−m+ 1)x)− 2f((3m− 1)x)− 12f(mx)k∗


≤ ϕ³(−2m+ 1)x

´+ ϕ(mx)(2.27)

for all x ∈ X0. For each m ∈ U , we define the operator Tm : Y X0 → Y X0

by

Tmξ(x) := 12ξ(mx)+2ξ³(3m−1)x

´+2ξ

³(−m+1)x

´−ξ³(4m−1)x

´ξ ∈ Y X0 ,

x ∈ X0.

Further put

εm(x) := ϕ³(−2m+1)x

´+ϕ(mx) ≤

µλ(−2m+1)+λ(mx)

¶ϕ(x), x ∈ X0.

(2.28)

Then the inequality (2.28) takes the form


For each m ∈ U , the operator Λm : RX0+ → RX0

+ which is defined by

Λmδ(x) := max

½δ(mx) , δ

µ(3m− 1)x

¶¾, δ

µ(−m+ 1)x

¶,

δ

µ(4m− 1)x

¶, δ ∈ RX0

+ , x ∈ X0

has the form described in (H3) with k = 4 and

f1(x) = mx, f2(x) = (3m−1)x, f3(x) = (−m+1)x, f4(x) = (4m−1)x,

L1(x) = L2(x) = L2(x) = L4(x) = 1

for all x ∈ X0. Moreover, for every ξ, µ ∈ Y X0 , x ∈ X0°°°°Tmξ(x)− Tmµ(x)°°°°∗=

°°°°12ξ ((m)x) + 2ξ ((3m− 1)x) + 2ξ ((−m+ 1)x)

−ξ ((4m− 1)x)− 12µ ((m)x)− 2µ ((3m− 1)x)− 2µ ((−m+ 1)x)

+ξ ((4m− 1)x)°°°°∗

≤ maxnkξ ((m)x)− µ ((m)x)k∗ ,

°°°ξ((3m− 1)x)− µ((3m− 1)x)°°°∗,°°°ξ((−m+ 1)x)− µ((−m+ 1)x)

°°°∗,°°°ξ(4m− 1)x)− µ((4m− 1)x)

°°°∗}.


So, (H2) is valid. By using mathematical induction, we will show that foreach x ∈ X0 we have

Λnmεm(x) ≤µλ(−2m+ 1) + λ(m)

¶αnmϕ(x)(2.29)

From (2.29), we obtain that the inequality (2.30) holds for n = 0. Next,we will assume that (2.30) holds for n = k, where k ∈ N. Then we have

Λk+1m εm(x) = Λm³Λkmεm(x)

´= max

nΛkmεm(mx); , Λkmεm

³(3m− 1)x

´,

Λkmεm³(−m+ 1)x

´,Λkmεm

³(4m− 1)x

´≤³λ(m) + λ(−2m+ 1)

´αkm max

nϕ³mx

´, ϕ

³(3m− 1)x

´,

ϕ³(−m+ 1)x

´, ϕ³(4m− 1)x

´ o≤µλ(m) + λ(−2m+ 1)

¶αk+1m ϕ(x), x ∈ X0.

This shows that (2.30) holds for n = k + 1. Now we can conclude thatthe inequality (2.30) holds for all n ∈ N. From (2.30), we obtain

limn→∞

Λnεm(x) = 0,

for all x ∈ X0 and all m ∈ U . Hence, according to Theorem 1.4, thereexists, for each m ∈ U , a unique solution Cm : X0 → Y of the equation

Cm(x) = 12Cm(mx)+2Cm

³(3m−1)x)+2Cm

³(−m+1)x)−Cm

³(4m−1)x),

x ∈ X0(2.30)

such that


½µλ(m) + λ(−2m+ 1)

¶αnmϕ(x)

¾, x ∈ X0.

(2.31)

Moreover,

Cm(x) := limn→∞

³T nmf´(x)

for all x ∈ X0. Now we show that



mf (2x− y)− 2T nmf(x+ y)− 2T n

mf(x+ y)− 12T nmf(x)k∗

≤ αnm

µϕ(x) + ϕ(y)

¶(2.32)

for every x, y ∈ X0 such that x + y 6= 0 and n ∈ N. Since the case n = 0is just (2.27), take k ∈ N and assume that (2.33) holds for n = k, wherek ∈ N and every x, y ∈ X0 such that x+ y 6= 0. Then°°°T k+1

m f (2x+ y) + T k+1m f (2x− y)− 2T k+1

m f (x+ y)− 2T k+1m f (x− y)− 12T k+1

m f(x)°°°∗

=°°°12T k

mf (m(2x+ y)) + 2T kmf ((3m− 1) (2x+ y)) + 2T k

mf ((−m+ 1) (2x+ y))

−T kmf ((4m− 1) (2x+ y)) + 12T k

mf ((m) (2x− y)) + 2T kmf ((3m− 1) (2x− y))

+2T kmf ((−m+ 1)(2x− y))− T k

mf ((4m− 1)(2x− y))− 24T kmf ((m) (x+ y))

−4T kmf ((3m− 1) (x+ y))− 4T k

mf ((−m+ 1) (x+ y)) + 2T kmf ((4m− 1) (x+ y))

−24T kmf ((m) (x− y))− 4T k

mf³³

3m−12

´(x− y)

´− 4T k

mf ((−m+ 1) (x− y))

+ 2T kmf ((4m− 1) (x− y))− 144T k

mf ((m) (x))− 24T kmf

³³3m−12

´(x)´−

24T kmf ((−m+ 1) (x)) + 12T k

mf ((4m− 1) (x))¯∗

≤ maxn °°°T n

mf (m(2x+ y)) + T nmf (m(2x− y))− 2T n

mf (m(x+ y))

−T nmf (m(x− y))− 12T n

mf(mx)°°°∗,°°°T n

mf ((3m− 1)(2x+ y))+

T nmf ((3m− 1)(2x− y))− 2T n

mf ((3m− 1)(x+ y))− T nmf ((3m− 1)(x− y))−

12T nmf((3m− 1)x)

°°°∗,°°°T n

mf ((−m+ 1)(2x+ y)) + T nmf ((−m+ 1)(2x− y))−

2T nmf ((−m+ 1)(x+ y))− T n

mf ((−m+ 1)(x− y))− 12T nmf((−m+ 1)x)

°°°∗,

kT nmf ((4m− 1)(2x+ y)) + T n

mf ((4m− 1)(2x− y))− 2T nmf ((4m− 1)(x+ y))−

T nmf ((4m− 1)(x− y))− 12T n

mf((4m− 11)x)∗o

(2.33)


≤ max½αkm

µϕ³(mx

´+ ϕ

³(my

´¶, αkm

µϕ(3m− 1)x) + ϕ((3m− 1)y)

¶,

αkm

µϕ(−m+ 1)x) + ϕ((−m+ 1)y)

¶¶,

αkm

µϕ(4m− 1)x) + ϕ((4m− 1)y)

¶¾

≤ αkmmax

½λ(m) , λ(3m− 1) , λ(−m+ 1) , λ(4m− 1)

¾µϕ(x) + ϕ(y)

¶

= αk+1m

µϕ(x) + ϕ(y)

¶

.

Thus, by induction we have shown that (2.33) holds for every n ∈ N.Letting n→∞ in (2.33), we obtain that


for all x, y ∈ X0 such that x + y 6= 0. In this way we obtain a sequence{Cm}m∈U of Cubic functions on X0 such that


½µλ(m+ 2) + λ(−m)

¶αnmϕ(x)

¾, x ∈ X0,

this implies that

kf(x)−Cm(x)k∗ ≤µλ(m) + λ(−2m+ 1)

¶ϕ(x), x ∈ X0.

Because the precedent inequality holds for over n = 0 and αm < 1


It follows, with m→∞, that f is Cubic on X0. 2The following corollary is a particular case of Theorem 2.5 where ϕ(x) :=

c kxkp with c ≥ 0 and p < 0.

Corollary 2.6. Let (X, k·k) and (Y, k·k∗) be normed space and ultrametricBanach space respectively, c ≥ 0, p < 0 and let f : X → Y satisfy

kf (2x+ y)) + f(2x+ y)− 2f(x+ y)− 2f(x− y)− 12f(x)k∗

≤ c

µkxkp + kykp

¶,(2.34)

for all x, y ∈ X0. Then f is Cubic on X0.

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Youssef AribouDepartment of Mathematics,Faculty of Sciences,University of IbnTofail, B. P. 133 Kenitra,Moroccoe-mail : [email protected]


Muaadh AlmahalebiDepartment of Mathematics,Faculty of Sciences,University of IbnTofail, B. P. 133 Kenitra,Moroccoe-mail : [email protected]

and

Samir kabbajDepartment of Mathematics,Faculty of Sciences,University of IbnTofail, B. P. 133 Kenitra,Moroccoe-mail : [email protected]

hyperstability of cubic functional equation in ultrametric spaces...hyperstability of cubic...

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