Hyperstability of cubic functional equation in ultrametric spaces...Hyperstability of cubic...
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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;
464 Youssef Aribou, Muaadh Almahalebi and Samir kabbaj
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.
Hyperstability of cubic functional equation in ultrametric spaces 465
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
466 Youssef Aribou, Muaadh Almahalebi and Samir kabbaj
°°°°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)
°°°∗}.
Hyperstability of cubic functional equation in ultrametric spaces 467
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)
468 Youssef Aribou, Muaadh Almahalebi and Samir kabbaj
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)
Hyperstability of cubic functional equation in ultrametric spaces 469
≤ 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)
470 Youssef Aribou, Muaadh Almahalebi and Samir kabbaj
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
kf(x)− Cm(x)k∗ ≤ supn∈N0
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)
Hyperstability of cubic functional equation in ultrametric spaces 471
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
kTmf(x)− f(x)k∗ ≤ εm(x), x ∈ X0.
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)
472 Youssef Aribou, Muaadh Almahalebi and Samir kabbaj
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
kf(x)− Cm(x)k∗ ≤ supn∈N0
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
Hyperstability of cubic functional equation in ultrametric spaces 473
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,(2.20)
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)
474 Youssef Aribou, Muaadh Almahalebi and Samir kabbaj
≤ 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
Hyperstability of cubic functional equation in ultrametric spaces 475
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
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
kf(x)− Cm(x)k∗ ≤ supn∈N0
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:
476 Youssef Aribou, Muaadh Almahalebi and Samir kabbaj
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∗
Hyperstability of cubic functional equation in ultrametric spaces 477
≤ 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
kTmf(x)− f(x)k∗ ≤ εm(x), x ∈ X0.
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)
°°°∗}.
478 Youssef Aribou, Muaadh Almahalebi and Samir kabbaj
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
kf(x)− Cm(x)k∗ ≤ supn∈N0
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
Hyperstability of cubic functional equation in ultrametric spaces 479
kT nmf (2x+ y) + T n
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)
480 Youssef Aribou, Muaadh Almahalebi and Samir kabbaj
≤ 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
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∈U of Cubic functions on X0 such that
kf(x)− Cm(x)k∗ ≤ supn∈N0
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
Hyperstability of cubic functional equation in ultrametric spaces 481
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]
484 Youssef Aribou, Muaadh Almahalebi and Samir kabbaj
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]