# Hyperstability of thek-Cubic Functional Equation in Non-Archimedean Banach Spaces

Authors: Youssef Aribou; Mohamed Rossafi
Journal: Journal of Mathematics
 (2020)

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## Abstract

Using the fixed point approach, we investigate a general hyperstability results for the followingk-cubic functional equations fkx+y+fkx−y=kfx+y+kfx−y+2kk2−1fx, where k is a fixed positive integer ≥2, in ultrametric Banach spaces.


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## Body

## 1. Introduction

The starting point of studying the stability of functional equations seems to be the famous talk of Ulam [1] in 1940, in which he discussed a number of important unsolved problems. Among those was the question concerning the stability of group homomorphisms.Ulam’s problem: letG1 be a group and let G2 be a metric group with a metric d. Given ε>0, there exists δ>0 such that if a mapping h:G1⟶G2 satisfies the inequality(1)dhxy,hxhy<δ,for all x,y∈G1, then there exists a homomorphism H:G1⟶G2 with(2)dhx,Hx<ε,for all x∈G1.The first partial answer, in the case of Cauchy equation in Banach spaces, to Ulam question was given by Hyers [2]. Later, the result of Hyers was first generalized by Aoki [3], and only much later by Rassias [4] and Găvruţa [5]. Since then, the stability problems of several functional equations have been extensively investigated [6–10].We say a functional equation ishyperstable if any function f satisfying the equation approximately (in some sense) must be actually a solution to it. It seems that the first hyperstability result was published in [11] and concerned the ring homomorphisms. However, the term hyperstability has been used for the first time in [12]. Quite often, hyperstability is confused with superstability, which also admits bounded functions. Numerous papers on this subject have been published, and we refer, for example, to [3, 12–29].Throughout this paper,ℕ stands for the set of all positive integers and Nm0, the set of integers greater than or equal m0, ℝ+≔0,∞, and we use the notation X0 for the set X∖0.Let us recall (see, for instance, [30]) some basic definitions and facts concerning non-Archimedean normed spaces.Definition 1.
By a non-Archimedean field, we mean a fieldK equipped with a function (valuation) ⋅:K⟶0,∞ such that, for all r,s∈K, the following conditions hold:(1)
r=0 if and only if r=0(2)
rs=rs(3)
r+s≤maxr,s
The pairK,. is called a valued field.
In any non-Archimedean field, we have1=−1=1 and n≤1, for n∈ℕ0. In any field K, the function ⋅:K⟶ℝ+ given by(3)x≔0,x=0,1,x≠0,is a valuation which is called trivial, but the most important examples of non-Archimedean fields are p-adic numbers which have gained the interest of physicists for their research in some problems coming from quantum physics, p-adic strings, and superstrings.Definition 2.
LetX be a vector space over a scalar field K with a non-Archimedean nontrivial valuation ⋅. A function ⋅∗:X⟶ℝ is a non-Archimedean norm (valuation) if it satisfies the following conditions:(1)
x∗=0 if and only if x=0,(2)
rx∗=rx∗r∈K,x∈X,(3)
The strong triangle inequality (ultrametric), namely,(4)x+y∗≤maxx∗,y∗,x,y∈X.
Then,X,⋅∗ is called a non-Archimedean normed space or an ultrametric normed space.Definition 3.
Letxn be a sequence in a non-Archimedean normed space X.(1)
A sequencexnn=1∞ in a non-Archimedean space is a Cauchy sequence iff the sequence xn+1−xnn=1∞ converges to zero.(2)
The sequencexn is said to be convergent if there exists x∈X such that, for any ε>0, there is a positive integer N such that xn−x∗≤ε, for all n≥N. Then, the point x∈X is called the limit of the sequence xn, which is denoted by limn⟶∞xn=x.(3)
If every Cauchy sequence inX converges, then the non-Archimedean normed space X is called a non-Archimedean Banach space or an ultrametric Banach space.
LetX and Y be normed spaces. A function f:X⟶Y is called a k−cubic function provided it satisfies the functional equation:(5)fkx+y+fkx−y=kfx+y+kfx−y+2kk2−1fx,for allx,y∈X,and we can say that f:X⟶Y is k−cubic on X0 if it satisfies (5) for all x,y∈X0.
In 2013, Bahyrycz et al. [31] used the fixed point theorem from Theorem 1 in [24] to prove the stability results for the generalization of p-Wright affine equation in ultrametric spaces. Recently, corresponding results for more general functional equations (in classical spaces) have been proved in [32–35].
In this paper, by using the fixed point method derived from [20, 21, 36], we present some hyperstability results for equation (5) in ultrametric Banach spaces. Before proceeding to the main results, we state Theorem 1 which is useful for our purpose. To present it, we introduce the 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:YX⟶YX is an operator satisfying the inequality(6)Tξx−Tμx∗≤max1≤i≤kLixξfix−μfix∗,ξ,μ∈YX,x∈X.(H3):Λ:ℝ+X⟶ℝ+X is a linear operator defined by(7)Λδx≔max1≤i≤kLixδfix,δ∈ℝ+X,x∈X.
Thanks to a result due to Brzdçk and Ciepliński ([25], Remark 2), we state a slightly modified version of the fixed point theorem ([24], Theorem 1) in ultrametric spaces. We use it to assert the existence of a unique fixed point of operator T:YX⟶YX.Theorem 1.
Let hypotheses (H1)–(H3) be valid, and functions ε:X⟶ℝ+ and φ:X⟶Y fulfill the following two conditions:(8)Tφx−φx∗≤εx,x∈X,limn⟶∞Λnεx=0,x∈X.Then, there exists a unique fixed pointψ∈YX of T with(9)φx−ψx∗≤supn∈ℕ0Λnεx,x∈X.Moreover,(10)ψx≔limn⟶∞Tnφx,x∈X.


## 2. Main Results

In this section, we use Theorem1 as a basic tool to prove the hyperstability results of the k-cubic functional equation (5) in ultrametric Banach spaces.Theorem 2.
LetX,⋅ and Y,⋅∗ be normed space and ultrametric Banach space, respectively, c≥0, p,q∈ℝ, and p+q<0, and let f:X⟶Y satisfies(11)fkx+y+fkx−y−kfx+y−kfx−y−2k3−kfx∗≤cxpyq,for all x,y∈X0 such that kx+y≠0, kx−y≠0, x+y≠0, and x−y≠0. Then, f is k-cubic on X0.Proof.
Takem∈ℕ such that(12)αm≔m+1kp+q<1,m≥k.
Sincep+q<0, one of p,q must be negative. Assume that q<0 and replacing y by mx and x by m+1/kx in (11), we obtain(13)2k3−kfm+1kx+kfkm+m+1kx+kfm−km+1kx−f2m+1x−fx∗≤cmqm+1kpxp+q.
Define operatorsTm:YX0⟶YX0 and Λm:ℝ+X0⟶ℝ+X0 by(14)Tmξx≔2k3−kξm+1kx+kξkm+m+1kx+2ξm−km+1kx−ξ2m+1x,ξ∈YX0,x∈X0,Λmδx≔maxδm+1kx,δkm+m+1kx,δm−km+1kx,δ2m+1,δ∈ℝ+X0,x∈X0,and write(15)εmx≔cmqm+1kpxp+q,x∈X0.
It is easily seen thatΛm has the form described in (H3) with k=4, f1x=m+1/kx, f2x=km+m+1/kx, f3x=m−km+1/kx, f4x=2m+1x, and L1x=L2x=L3x=L4x=1. Furthermore, (11) can be written in the following way:(16)Tmfx−fx∗≤εmx,x∈X0.
Moreover, for everyξ,μ∈YX0, x∈X0,(17)Tmξx−Tmμx∗=2k3−kξm+1kx+kξkm+m+1kx+kξm−km+1kx−ξ2m+1x−2k3−kμm+1kx−kμkm+m+1kx−kμm−km+1kx+ξ2m+1x∗≤maxξm+1kx−μm+1kx∗,ξkm+m+1kx−μkm+m+1kx∗,ξm−km+1Kx−μm−km+1kx∗,ξ2m+1x−μ2m+1x∗.
So, (H2) is valid.
By using mathematical induction, we will show that, for eachx∈X0, we have(18)Λmnεmx=cmqm+1kpxp+qαmn,where αm=m−1/kp+q. We obtain that (18) holds for n=0. Next, we will assume that (18) holds for n=r, where r∈ℕ. Then, we have(19)Λmr+1εmx=ΛmΛmrεmx=maxΛmrεmm+1kx,Λmrεmkm+m+1kx,Λmrεmkm−m−1kx,Λmrεm2m+1x=maxcmqm+1kpxp+qαmrm+1kp+q,cmqm+1kpxp+qαmrkm+m+1kp+q,cmqm+1Kpxp+qαmrkm−m−1kp+q,cmqxp+qαmk2m+1p+q=cmqxp+qαmrmaxm+1kp+q,km+m+1kp+q,km−m−1kp+q,2m+1p+q=cmqm+1kpxp+qαmr+1,x∈X0.
This shows that (18) holds for n=r+1. Now, we can conclude that inequality (18) holds for all n∈ℕ0. From (18), we obtain(20)limn⟶∞Λnεmx=0,for all x∈X0.
Hence, according to Theorem1, there exists a unique solution Cm:X0⟶Y of the equation:(21)Cmx=2k3−kCmm+1kx+kCmkm+m+1kx+kCm1−km+mkx−Cm2m+1x,x∈X0,such that(22)fx−Cmx∗≤supn∈ℕ0cmqm+1kpxp+qαmn,x∈X0.
Moreover,(23)Cmx≔limn⟶∞Tmnfx,for all x∈X0.
Now, we show that(24)Tmnfkx+y+Tmnfkx−y−kTmnfx+y−kTmnfx−y−2k3−kTmnfx∗≤cαmnxpyq,for every x,y∈X0 such that x+y≠0, x−y≠0. Since the case n=0 is just (11), take r∈ℕ, and suppose the last inequality holds for n=r and every x,y∈X0 such that x+y≠0, x−y≠0. Then,(25)Tmr+1fkx+y+Tmr+1fkx−y−kTmr+1fx+y−kTmr+1fx−y−2k3−kTmr+1fx∗=2k3−kTmkfm+1kkx+y+kTmrfkm+m+1kkx+y+kTmrf1+m−kmkkx+y−Tmrf2m+1kx+y+2k3−kTmrfm+1kkx−y+kTmrfkm+m+1kkx−y+kTmrf1+m−kmkkx−y−Tmrf2m+1kx−y−2kk3−kTmrfm+1kx+y−k2Tmrfkm+m+1kx+y−k2Tmrf1+m−kmkx+y+kTmrf2m+1x+y−2kk3−kTmrfm+1kx−y−k2Tmrfkm+m+1kx−y−k2Tmrf1+m−kmkx−y+Tmrf2m+1x−y−2k3−k2Tmrfm+1kx−2kk3−kTmrfkm+m+1kx−2kk3−kTmrf1+m−kmkx+2k3−kTmkf2m+1x∗≤maxTmnfm+1kkx+y+Tmnfm+1kkx−y−kTmnfm+1kx+y−kTmnfm+1kx−y−2k3−kTmnfm+1kx∗,Tmnfkm+m+1kkx+y+Tmnfkm+m+1kkx−y−kTmnfkm+m+1kx+y−kTmnfkm+m+1kx−y−2k3−kTmnfkm+m+1kx∗,Tmnf1+m−kmkkx+y+Tmnf1+m−kmkkx−y−kTmnf1+m−kmkx+y−kTmnf1+m−kmkx−y−2k3−kTmnf1+m−kmkx∗,Tmnf2m+1kx+y+Tmnf2m+1kx−y−kTmnf2m+1x+y−Tmnf2m+1x−y−2k3−kTmnf2m+1x∗≤maxcαmkxpyqm+1kp+q,cαmkxpyqkm+m+1kp+q,cαmkxpyq1+m−kmkp+q,cαmkxpyq2m+1p+q=cαmkxpyqmaxm+1kp+q,km+m+1kp+q,km−m−1kp+q,km+m+1p+q≤cαmk+1xpyq,for all x,y∈X0 such that x+y≠0 and x−y≠0. Thus, by induction, we have shown that suppose the last inequality holds for every n∈ℕ0. Letting n⟶∞, we obtain that(26)Cmkx+y+Cmkx−y=kCmx+y+kCmx−y+2k3−kCmx,for all x,y∈X0 such that x+y≠0, x−y≠0. In this way, we obtain a sequence Cmm≥m0 of k−cubic functions on X0 such that(27)fx−Cmx∗≤supn∈ℕ0cmqm+1kpxp+qαmn,x∈X0.
This implies that(28)fx−Cmx∗≤cmqm+1kpxp+q,x∈X0.
It follows, withm⟶∞, that f is k−cubic on X0. In a similar way, we can prove the following theorem.Theorem 3.
LetX,⋅ and Y,⋅∗ be normed space and ultrametric Banach space, respectively, c≥0, p,q∈ℝ, and p+q>0, and let f:X⟶Y satisfies(29)fkx+y+fkx−y−kfx+y−kfx−y−2k3−kfx∗≤cxpyq,for all x,y∈X0 such that x+y≠0 and x−y≠0. Then, f is k−cubic on X0.Proof.
Takem∈ℕ such that(30)αm=m−4mp+q<1.
Sincep+q>0, one of p,q must be positive, and let q>0 and replace y by −2/my and x by m−2/kmy in (29). Thus,(31)2k3−kfm−2kmy+kfm+2kmy+kfm−6kmy−fm−4my−fy∗≤c2mqm−2kmpyp+q,x∈X0.
Writing(32)Tmξx≔2k3−kξm−2kmx+kξm+2kmx+kξm−6kmx−ξm−4mx,ξ∈YX0,x∈X0,(33)εmx≔c2mqm−1kmpxp+q,x∈X0,(31) takes form(34)Tmfx−fx∗≤εmx,x∈X0.
The rest of the proof is similar to the proof of the last theorem. It easy to show the hyperstability of cubic equation on the set containing 0. The above theorems imply, in particular, the following corollary, which shows their simple application.Corollary 1.
LetX,⋅ and Y,⋅∗ be normed space and ultrametric Banach space, respectively, G:X2⟶Y and Gx0,y0≠0 for some x0,y0∈X, and(35)Gx,y∗≤cxpyq,x,y∈X,where c≥0, p,q∈ℝ. Assume that the numbers pandq satisfy one of the following conditions:(1)
p+q<0, and (11) holds for all x,y∈X0(2)
p+q>0, and (36) holds for all x,y∈X0
Then, the functional equation(36)fkx+y+fkx+y=kfx+y+kfx−y+2k3−kfx+Gx,y,x,y∈X,has no solution in the class of functions g:X⟶Y.In the following theorem, we present a general hyperstability for thek−cubic equation where the control function is φx+φy, which corresponds to the approach introduced in [36].Theorem 4.
LetX,⋅ be a normed space, Y,⋅∗ be an ultrametric Banach space over a field K, and φ:X⟶ℝ+ be a function such that(37)U≔n∈ℕ:αn≔maxλn,λkn+n−1,λ−kn+n+1,λ2kn−1<1,is a nonempty set, where λa≔inft∈ℝ+:φax≤tφxforx∈X for all a∈K0. Suppose that(38)lima⟶∞λa=0,lima⟶∞λ−a=0,and f:X⟶Y satisfies the inequality(39)fkx+y+fkx+y−kfx+y−kfx−y−2k3−kfx∗≤φx+φy,for all x,y∈X0, such that x+y≠0 and x−y≠0. Then, f is cubic on X0.Proof.
Replacingx by mx and y by −km+1x for m∈ℕ in (39), we get(40)fx+f2km−1x−kf−km+m+1x−kfkm+m−1x−2k3−kfmx∗≤φ−km+1x+φmx,for all x∈X0. For each m∈U, we define the operator Tm:YX0⟶YX0 by(41)Tmξx≔2k3−kξmx+kξkm+m−1x+kξ−km+m+1x−ξ2km−1x,ξ∈YX0,x∈X0.
Furthermore, we put(42)εmx≔φ−km+1x+φmx≤λ−km+1+λmxφx,x∈X0.
Then, inequality (40) takes the form(43)Tmfx−fx∗≤εmx,x∈X0.
For eachm∈U, the operator Λm:ℝ+X0⟶ℝ+X0 which is defined by Λmδx≔maxδmx,δkm+m−1x,δ−km+m+1x,δ2km−1x,δ∈ℝ+X0,x∈X0, has the form described in (H3) with k=4 and(44)f1x=mx,f2x=km+m−1x,f3x=−km+m+1x,f4x=2km−1x,L1x=L2x=L2x=L4x=1,for all x∈X0. Moreover, for every ξ,μ∈YX0, x∈X0,(45)Tmξx−Tmμx∗=2k3−kξmx+kξkm+m−1x+kξ−km+m+1x−ξ2km−1x−2k3−kμmx−kμkm+m−1x−kμ−km+m+1x+ξ2km−1x∗≤maxξmx−μmx∗,ξkm+m−1x−μkm+m−1x∗,ξ−km+m+1x−μ−km+m+1x∗,ξ2km−1x−μ2km−1x∗.
So, (H2) is valid. By using mathematical induction, we will show that, for eachx∈X0, we have(46)Λmnεmx≤λ−km+1+λmαmnφx.
From (42), we obtain that inequality (46) holds for n=0. Next, we will assume that (46) holds for n=r, where r∈ℕ. Then, we have(47)Λmr+1εmx=ΛmΛmrεmx=maxΛmrεmmx,Λmrεmkm+m−1x,Λmrεm−km+m+1x,Λmrεm2km−1x≤λm+λ−km+1αmrmaxφmx,φkm+m−1x,φ−km+m+1x,φ2km−1x≤λm+λ−km+1αmr+1φxx∈X0.
This shows that (46) holds for n=r+1. Now, we can conclude that inequality (46) holds for all n∈ℕ. From (46), we obtain limn⟶∞Λnεmx=0, for all x∈X0 and all m∈U. Hence, according to Theorem 1, there exists, for each m∈U, a unique solution Cm:X0⟶Y of the equation:(48)Cmx=2k3−kCmmx+kCmkm+m−1x+kCm−km+m+1x−Cm2km−1x,x∈X0, such that(49)fx−Cmx∗≤supn∈ℕ0λm+λ−km+1αmnφx,x∈X0.
Moreover,Cmx≔limn⟶∞Tmnfx, for all x∈X0. Now, we show that(50)Tmnfkx+y+Tmnfkx−y−kTmnfx+y−kTmnfx+y−2k3−kTmnfx∗≤αmnφx+φy,for every x,y∈X0 such that x+y≠0, x−y≠0, and n∈ℕ. Since the case n=0 is just (39), take k∈ℕ and assume that (50) holds for n=r, where r∈ℕ and every x,y∈X0 such that x+y≠0. Then,(51)Tmr+1fkx+y+Tmk+1fkx−y−kTmr+1fx+y−kTmr+1fx−y−2k3−kTmr+1fx∗=2k3−kTmrfmkx+y+kTmrfkm+m−1kx+y+kTmrf−km+m+1kx+y−Tmrf2km−1kx+y+2k3−kTmrfmkx−y+kTmrfkm+m−1kx−y+kTmrf−km+m+1kx−y−Tmrfkkm−1kx−y−kk3−kTmrfmx+y−k2Tmrfkm+m−1x+y−k2Tmrf−km+m+1x+y+kTmrf2km−1x+y−kk3−kTmrfmx−y−k2Tmrfkm+m−1x−y−k2Tmrf−km+m+1x−ykTmrf2km−1x−y−2k3−k2Tmrfmx−2kk3−kTmrfkm+m−1x−2kk3−kTmrf−km+m+1x+2k3−kTmrf2km−1x|∗≤maxTmnfmkx+y+Tmnfmkx−y−kTmnfmx+y−kTmnfmx−y−2k3−kTmnfmx∗,Tmnfkm+m−1kx+y+Tmnfkm+m−1kx−y−kTmnfkm+m−1x+y−Tmnfkm+m−1x−y−2k3−kTmnfkm+m−1x∗,Tmnfm−km+1kx+y+Tmnfm−km+1kx−y−kTmnfm−km+1x+y−Tmnfm−km+1x−y−2k3−kTmnfm−km+1x∗,Tmnf2km−1kx+y+Tmnf2km−1kx−y−kTmnf2km−1x+y−Tmnf2km−1x−y−2k3−kTmnf2km−1x∗,≤maxαmkφmx+φmy,αmkφkm+m−1x+φKm+m−1yαmkφ−km+m+1x+φ−km+m+1y,αmkφ2km−1x+φ2km−1y≤αmkmaxλm,λkm+m−1,λ−km+m+1,λ2km−1φx+φy=αmk+1φx+φy.
Thus, by induction, we have shown that (50) holds for every n∈ℕ. Letting n⟶∞ in (50), we obtain that(52)Cmkx+y+Cmkx−y=kCmx+y+kCmx−y+2k3−kCmx,for all x,y∈X0 such that x+y≠0 and x−y≠0. In this way, we obtain a sequence Cmm∈U of k−cubic functions on X0 such that(53)fx−Cmx∗≤supn∈ℕ0λm+λ−km+1αmnφx,x∈X0.
This implies that(54)fx−Cmx∗≤λm+λ−km+1φx,x∈X0,because the precedent inequality holds for over n=0 and αm<1.
It follows, withm⟶∞, that f is cubic on X0. The following corollary is a particular case of Theorem 4 where φx≔cxp with c≥0 and p<0.Corollary 2.
LetX,⋅ and Y,⋅∗ be normed space and ultrametric Banach space, respectively, c≥0, and p<0, and let f:X⟶Y satisfies(55)fkx+y+fkx+y−kfx+y−kfx−y−2k3−kfx∗≤cxp+yp,for all x,y∈X0. Then, f is k−cubic on X0.



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