# Nonlinear Random Stability via Fixed-Point Method

Authors: Yeol Je Cho; Shin Min Kang; Reza Saadati
Journal: Journal of Applied Mathematics
 (2012)

---

## Abstract

We prove the generalized Hyers-Ulam stability of the following additive-quadratic-cubic-quartic functional equationf(x+2y)+f(x−2y)=4f(x+y)+4f(x−y)−6f(x)+f(2y)+f(−2y)−4f(y)−4f(−y) in various complete random normed spaces.


---

## Body

## 1. Introduction

The stability problem of functional equations originated from a question of Ulam [1] concerning the stability of group homomorphisms. Hyers [2] gave a first affirmative partial answer to the question of Ulam for Banach spaces. Hyers' theorem was generalized by Aoki [3] for additive mappings and by Rassias [4] for linear mappings by considering an unbounded Cauchy difference. The paper of Rassias [4] has provided a lot of influence in the development of what we call generalized Hyers-Ulam stability or as Hyers-Ulam-Rassias stability of functional equations. A generalization of the Rassias theorem was obtained by Găvruţa [5] by replacing the unbounded Cauchy difference by a general control function in the spirit of Rassias approach.The functional equation(1.1)f(x+y)+f(x-y)=2f(x)+2f(y)is called aquadratic functional equation. In particular, every solution of the quadratic functional equation is said to be a quadratic mapping. A generalized Hyers-Ulam stability problem for the quadratic functional equation was proved by Cholewa [6] for mappings f:X→Y, where X is a normed space and Y is a Banach space. Czerwik [7] proved the generalized Hyers-Ulam stability of the quadratic functional equation. The stability problems of several functional equations have been extensively investigated by a number of authors, and there are many interesting results concerning this problem (see [8–12]).In [13], Jun and Kim consider the following cubic functional equation:(1.2)f(2x+y)+f(2x-y)=2f(x+y)+2f(x-y)+12f(x).It is easy to show that the functionf(x)=x3 satisfies the functional equation (1.2), which is called a cubic functional equation, and every solution of the cubic functional equation is said to be a cubic mapping.Considered the following quartic functional equation(1.3)f(2x+y)+f(2x-y)=4f(x+y)+4f(x-y)+24f(x)-6f(y).
It is easy to show that the function f(x)=x4 satisfies the functional equation, which is called a quartic functional equation, and every solution of the quartic functional equation is said to be a quartic mapping. One can easily show that an odd mapping f:X→Y satisfies the additive-quadratic-cubic-quadratic functional equation
(1.4)f(x+2y)+f(x-2y)=4f(x+y)+4f(x-y)-6f(x)+f(2y)+f(-2y)-4f(y)-4f(-y)
if and only if it is an additive-cubic mapping, that is, (1.5)f(x+2y)+f(x-2y)=4f(x+y)+4f(x-y)-6f(x).It was shown in Lemma  2.2 of [14] that  g(x):=f(2x)-2f(x) and h(x):=f(2x)-8f(x) are cubic and additive, respectively, and that f(x)=(1/6)g(x)-(1/6)h(x).One can easily show that an even mappingf:X→Y satisfies  (1.4) if and only if it is a quadratic-quartic mapping, that is, (1.6)f(x+2y)+f(x-2y)=4f(x+y)+4f(x-y)-6f(x)+2f(2y)-8f(y).Alsog(x):=f(2x)-4f(x) and h(x):=f(2x)-16f(x) are quartic and quadratic, respectively, and f(x)=(1/12)g(x)-(1/12)h(x).For a given mappingf:X→Y, we define (1.7)Df(x,y)∶=f(x+2y)+f(x-2y)-4f(x+y)-4f(x-y)+6f(x)-f(2y)-f(-2y)+4f(y)+4f(-y)for allx,y∈X.LetX be a set. A function d:X×X→[0,∞] is called a generalized metric on X if d satisfies(1)
d(x,y)=0 if and only if x=y,(2)
d(x,y)=d(y,x) for all x,y∈X,(3)
d(x,z)≤d(x,y)+d(y,z) for all x,y,z∈X.We recall the fixed-point alternative of Diaz and Margolis.Theorem 1.1 (see [15, 16]).
Let(X,d) be a complete generalized metric space and let  J:X→X be a strictly contractive mapping with Lipschitz constant  L<1, then for each given element x∈X, either
(1.8)d(Jnx,Jn+1x)=∞
for all nonnegative integers n or there exists a positive integer n0 such that(1)
d(Jnx,Jn+1x)<∞ for all n≥n0,(2)
the sequence{Jnx} converges to a fixed point y* of J,(3)
y* is the unique fixed point of J in the set Y={y∈X∣d(Jn0x,y)<∞},(4)
d(y,y*)≤(1/(1-L))d(y,Jy) for all y∈Y.In 1996, Isac and Rassias [17] were the first to provide applications of stability theory of functional equations for the proof of new fixed-point theorems with applications. By using fixed point methods, the stability problems of several functional equations have been extensively investigated by a number of authors (see [18–21]).


## 2. Preliminaries

In the sequel, we adopt the usual terminology, notations, and conventions of the theory of random normed spaces, as in [22–26]. Throughout this paper, Δ+ is the space of all probability distribution functions, that is, the space of all mappings F:ℝ∪{-∞,+∞}→[0,1], such taht F is left continuous, nondecreasing on ℝ, F(0)=0 and {F(+∞)=1}. D+ is a subset of Δ+ consisting of all functions F∈Δ+ for which l-F(+∞)=1, where l-f(x) denotes the left limit of the function f at the point x, that is, l-f(x)=limt→x-f(t). The space Δ+ is partially ordered by the usual pointwise ordering of functions, that is, F≤G if and only if F(t)≤G(t) for all t in ℝ. The maximal element for Δ+ in this order is the distribution function ɛ0 given by(2.1)ɛ0(t)={0,ift≤0,1,ift>0.Atriangular norm (shortly t-norm) is a binary operation on the unit interval [0,1], that is, a function T:[0,1]×[0,1]→[0,1], such that for all a,b,c∈[0,1] the following four axioms satisfied:(T1)
T(a,b)=T(b,a)     (commutativity),(T2)
T(a,(T(b,c)))=T(T(a,b),c)     (associativity),(T3)
T(a,1)=a     (boundary condition),(T4)
T(a,b)≤T(a,c) whenever b≤c     (monotonicity).Basic examples are the Łukasiewiczt-norm  TL,TL(a,b)=max(a+b-1,0) for all a,b∈[0,1] and the t-norms  TP,TM,TD, where    TP(a,b):=ab, TM(a,b):=min{a,b}, (2.2)TD(a,b):={min⁡(a,b),ifmax⁡(a,b)=1,0,otherwise.IfT is a t-norm, then xT(n) is defined for every x∈[0,1] and n∈N∪{0} by 1, if n=0 and T(xT(n-1),x) if n≥1. A t-normT is said to be of Hadžić type (we denote by T∈ℋ) if the family (xT(n))n∈N is equicontinuous at x=1 (cf. [27]).Other important triangular norms are the following (see [28]):(1)
TheSugeno-Weber family{TλSW}λ∈[-1,∞] is defined by T-1SW=TD, T∞SW=TP and
(2.3)TλSW(x,y)=max⁡(0,x+y-1+λxy1+λ)
if  λ∈(-1,∞).(2)
TheDomby family{TλD}λ∈[0,∞] is defined by TD if λ=0, TM if λ=∞, and
(2.4)TλD(x,y)=11+(((1-x)/x)λ+((1-y)/y)λ)1/λ
if λ∈(0,∞).(3)
TheAczel-Alsina family{TλAA}λ∈[0,∞] is defined by TD if λ=0, TM if λ=∞ and
(2.5)TλAA(x,y)=e-(|log⁡x|λ+|log⁡y|λ)1/λ
if λ∈(0,∞).At-norm    T can be extended (by associativity) in a unique way to an n-array operation taking for (x1,…,xn)∈[0,1]n the value T(x1,…,xn) defined by (2.6)Ti=10xi=1,Ti=1nxi=T(Ti=1n-1xi,xn)=T(x1,…,xn).T can also be extended to a countable operation taking for any sequence (xn)n∈N in [0,1] the value(2.7)Ti=1∞xi=lim⁡n→∞Ti=1nxi.
The limit on the right side of (6.4) exists since the sequence (Ti=1nxi)n∈ℕ is nonincreasing and bounded from below.Proposition 2.1 (see [28]).
We have the following.(1)
ForT≥TL, the following implication holds:
(2.8)lim⁡n→∞Ti=1∞xn+i=1⟺∑n=1∞(1-xn)<∞.(2)
IfT is of Hadžić type, then
(2.9)lim⁡n→∞Ti=1∞xn+i=1
for every sequence (xn)n∈N in [0,1] such that limn→∞xn=1.(3)
IfT∈{TλAA}λ∈(0,∞)∪{TλD}λ∈(0,∞), then
(2.10)lim⁡n→∞Ti=1∞xn+i=1⟺∑n=1∞(1-xn)α<∞.(4)
IfT∈{TλSW}λ∈[-1,∞), then
(2.11)lim⁡n→∞Ti=1∞xn+i=1⟺∑n=1∞(1-xn)<∞.Definition 2.2 (see [26]).
ARandom normed space (briefly, RN-space) is a triple (X,μ,T), where X is a vector space, T is a continuous t-norm, and μ is a mapping from X into D+ such that, the following conditions hold: (RN1)μx(t)=ɛ0(t) for all t>0 if and only if x=0,(RN2)μαx(t)=μx(t/|α|) for all x∈X, and α≠0,(RN3)μx+y(t+s)≥T(μx(t),μy(s)) for all x,y∈X and t,s≥0.Definition 2.3.
Let(X,μ,T) be an RN-space.(1)
A sequence{xn} in X is said to be convergent to x in X if, for every ϵ>0 and λ>0, there exists positive integer N such that μxn-x(ϵ)>1-λ whenever n≥N.(2)
A sequence{xn} in X is called a Cauchy sequence if, for every ϵ>0 and λ>0, there exists positive integer N such that μxn-xm(ϵ)>1-λ whenever n≥m≥N.(3)
An RN-space(X,μ,T) is said to be complete if and only if every Cauchy sequence in X is convergent to a point in X. A complete RN-space is said to be random Banach space.Theorem 2.4 (see [25]).
If(X,μ,T) is an RN-space and {xn} is a sequence such that xn→x, then limn→∞μxn(t)=μx(t) almost everywhere.The theory of random normed spaces (RN-spaces) is important as a generalization of deterministic result of linear normed spaces and also in the study of random operator equations. The RN-spaces may also provide us with the appropriate tools to study the geometry of nuclear physics and have important application in quantum particle physics. The generalized Hyers-Ulam stability of different functional equations in random normed spaces, RN-spaces, and fuzzy normed spaces has been recently studied [20, 24, 29–39].


## 3. Non-Archimedean Random Normed Space

By anon-Archimedean field, we mean a field 𝒦 equipped with a function (valuation) |·| from K into [0,∞) such that |r|=0 if and only if r=0, |rs|=|r||s|, and |r+s|≤max{|r|,|s|} for all r,s∈𝒦. Clearly, |1|=|-1|=1 and |n|≤1 for all n∈ℕ. By the trivial valuation, we mean the mapping |·| taking everything but 0 into 1 and |0|=0.  Let X be a vector space over a field 𝒦 with a non-Archimedean nontrivial valuation |·|. A function ∥·∥:X→[0,∞) is called a non-Archimedean norm if it satisfies the following conditions:(NAN1)∥x∥=0 if and only if x=0,(NAN2) for anyr∈𝒦 and x∈X, ∥rx∥=|r|∥x∥,(NAN3) the strong triangle inequality (ultrametric), namely,(3.1)‖x+y‖≤max⁡{‖x‖,‖y‖}(x,y∈X),then(X,∥·∥) is called a non-Archimedean normed space. Due to the fact that (3.2)‖xn-xm‖≤max⁡{‖xj+1-xj‖:m≤j≤n-1}(n>m),a sequence{xn} is a Cauchy sequence if and only if {xn+1-xn} converges to zero in a non-Archimedean normed space. By a complete non-Archimedean normed space, we mean one in which every Cauchy sequence is convergent.In 1897, Hensel [40] discovered the p-adic numbers of as a number theoretical analogues of power series in complex analysis. Fix a prime number p. For any nonzero rational number x, there exists a unique integer nx∈ℤ such that x=(a/b)pnx, where a and b are integers not divisible by p. Then |x|p:=p-nx defines a non-Archimedean norm on ℚ. The completion of ℚ with respect to the metric d(x,y)=|x-y|p is denoted by ℚp, which is called the p-adic number field.Throughout the paper, we assume thatX is a vector space and Y is a complete non-Archimedean normed space.Definition 3.1.
A non-Archimedean random normed space (briefly, non-Archimedean RN-space) is a triple (X,μ,T), where X is a linear space over a non-Archimedean field 𝒦, T is a continuous t-norm, and μ is a mapping from X into D+ such that the following conditions hold:(NA-RN1)μx(t)=ɛ0(t) for all t>0 if and only if x=0,(NA-RN2)μαx(t)=μx(t/|α|) for all x∈X,  t>0, and α≠0,(NA-RN3)μx+y(max{t,s})≥T(μx(t),μy(s)) for all x,y,z∈X and t,s≥0.
It is easy to see that if (NA-RN3) holds, then so is(RN3)μx+y(t+s)≥T(μx(t),μy(s)).As a classical example, if(X,∥.∥) is a non-Archimedean normed linear space, then the triple (X,μ,TM), where (3.3)μx(t)={0,t≤‖x‖,1,t>‖x‖,is a non-Archimedean RN-space.Example 3.2.
Let(X,∥·∥) be a non-Archimedean normed linear space. Define
(3.4)μx(t)=tt+‖x‖(x∈X,t>0),
then (X,μ,TM) is a non-Archimedean RN-space.Definition 3.3.
Let(X,μ,T) be a non-Archimedean RN-space. Let {xn} be a sequence in X, then {xn} is said to be convergent if there exists x∈X such that
(3.5)lim⁡n→∞μxn-x(t)=1
for all t>0. In that case, x is called the limit of the sequence {xn}.A sequence{xn} in X is called a Cauchy sequenceif for each ɛ>0 and each t>0 there exists n0 such that for all n≥n0 and all p>0, we have μxn+p-xn(t)>1-ɛ.If each Cauchy sequence is convergent, then the random norm is said to becomplete and the non-Archimedean RN-space is called a non-Archimedean random Banach  space.Remark 3.4 (see [41]).
Let(X,μ,TM) be a non-Archimedean RN-space, then
(3.6)μxn+p-xn(t)≥min⁡{μxn+j+1-xn+j(t):j=0,1,2,…,p-1}.
So, the sequence {xn} is a Cauchy sequence if for each ɛ>0 and t>0 there exists n0 such that for all n≥n0,
(3.7)μxn+1-xn(t)>1-ɛ.


## 4. Generalized Ulam-Hyers Stability for a Quartic Functional Equation in Non-Archimedean RN-Spaces of Functional Equation (1.4): An Odd Case

Let𝒦 be a non-Archimedean field, let X be a vector space over 𝒦, and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦.Next, we define a random approximately AQCQ mapping. LetΨ be a distribution function on X×X×[0,∞) such that Ψ(x,y,·) is nondecreasing and (4.1)Ψ(cx,cx,t)≥Ψ(x,x,t|c|)(x∈X,c≠0).Definition 4.1.
A mappingf:X→Y is said to be Ψ-approximately AQCQ if
(4.2)μDf(x,y)(t)≥Ψ(x,y,t)(x,y∈X,t>0).In this section, we assume that2≠0 in 𝒦 (i.e., characteristic of 𝒦 is not 2). Our main result, in this section, is the following.We prove the generalized Hyers-Ulam stability of the functional equationDf(x,y)=0 in non-Archimedean random spaces, an odd case.Theorem 4.2.
Let𝒦 be a non-Archimedean field, let X be a vector space over 𝒦 and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦. Let f:X→Y be an odd mapping and Ψ-approximately AQCQ mapping. If for some α∈ℝ, α>0, and some integer k, k>3 with |2k|<α,
(4.3)Ψ(2-kx,2-ky,t)≥Ψ(x,y,αt)(x∈X,t>0),(4.4)lim⁡n→∞Tj=n∞M(2x,αjt|8|kj)=1(x∈X,t>0),
then there exists a unique cubic mapping C:X→Y such that
(4.5)μf(x)-2f(x/2)-C(x/2)(t)≥Ti=1∞M(x,αi+1t|8|ki)
for all x∈X and t>0, where
(4.6)M(x,t)∶=Tk-1[Ψ(x2,x2,t|4|),Ψ(x,x2,t),…,Ψ(2k-1x2,2k-1x2,t|4|),Ψ(2k-1x,2k-1x2,t)](x∈X,t>0).Proof.
Lettingx=y in (4.2), we get
(4.7)μf(3y)-4f(2y)+5f(y)(t)≥Ψ(y,y,t)
for all y∈X and t>0. Replacing x by 2y in (4.2), we get
(4.8)μf(4y)-4f(3y)+6f(2y)-4f(y)(t)≥Ψ(2y,y,t)
for all y∈X and t>0. By (4.7) and (4.8), we have
(4.9)μf(4y)-10f(2y)+16f(y)(t)≥T(μ4(f(3y)-4f(2y)+5f(y))(t),μf(4y)-4f(3y)+6f(2y)-4f(y)(t))=T(μf(3y)-4f(2y)+5f(y)(t|4|),μf(4y)-4f(3y)+6f(2y)-4f(y)(t))≥T(Ψ(y,y,t|4|),Ψ(2y,y,t))
for all y∈X and t>0. Letting y:=x/2 and g(x):=f(2x)-2f(x) for all x∈X in (4.9), we get
(4.10)μg(x)-8g(x/2)(t)≥T(Ψ(x2,x2,t|4|),Ψ(x,x2,t))
for all x∈X and t>0. Now, we show by induction on j that for all x∈X, t>0 and j≥1,
(4.11)μg(2j-1x)-8jg(x/2)(t)≥Mj(x,t)∶=T2j-1[Ψ(x2,x2,t|4|),Ψ(x,x2,t),…,Ψ(2j-1x2,2j-1x2,t|4|),Ψ(2j-1x,2j-1x2,t)].
Putting j=1 in (4.11), we obtain (4.10). Assume that (4.11) holds for some j≥1. Replacing x by 2jx in (4.10), we get
(4.12)μg(2jx)-8g(2j-1x)(t)≥T(Ψ(2j-1x,2j-1x,t|4|),Ψ(2jx,2j-1x,t)).
Since |8|≤1,
(4.13)μg(2jx)-8j+1g(x/2)(t)≥T(μg(2jx)-8g(2j-1x)(t),μ8g(2j-1x)-8j+1g(x/2)(t))=T(μg(2jx)-8g(2j-1x)(t),μg(2j-1x)-8jg(x/2)(t|8|))≥T2(Ψ(2j-1x,2j-1x,t|4|),Ψ(2jx,2j-1x,t),Mj(x,t))=Mj+1(x,t)for all x∈X and t>0. Thus, (4.11) holds for all j≥2. In particular,
(4.14)μg(2k-1x)-8kg(x/2)(t)≥M(x,t)(x∈X,t>0).
Replacing x by 2-(kn+k-1)x in (4.14) and using inequality (4.3), we obtain
(4.15)μg(x/2kn)-8kg(x/2k(n+1))(t)≥M(2x2k(n+1),t)(x∈X,t>0,n=0,1,2,…).
Then
(4.16)μ8kng(x/2kn)-8k(n+1)g(x/2k(n+1))(t)≥M(2x,αn+1|8k(n+1)|t)(x∈X,t>0,n=0,1,2,…).
Hence
(4.17)μ8kng(x/2kn)-8k(n+p)g(x/2k(n+p))(t)≥Tj=nn+p(μ8kjg(x/2kj)-8k(j+p)g(x/2k(j+p))(t))≥Tj=nn+pM(2x,αj+1|(8k)j+1|t)≥Tj=nn+pM(2x,αj+1|(8k)j+1|t)(x∈X,t>0,n=0,1,2,…).
Since
(4.18)lim⁡n→∞Tj=n∞M(2x,αj+1|(8k)j+1|t)=1(x∈X,t>0),
then
(4.19){8kng(x2kn)}n∈Nis a Cauchy sequence in the non-Archimedean random Banach space (Y,μ,T). Hence we can define a mapping C:X→Y such that
(4.20)lim⁡n→∞μ(88k)ng(x/2kn)-C(x)(t)=1(x∈X,t>0).
Next for eachn≥1, x∈X and t>0,
(4.21)μg(x)-(88k)ng(x/2kn)(t)=μ∑i=0n-1(88k)ig(x/2ki)-(88k)i+1g(x/2k(i+1))(t)≥Ti=0n-1(μ(88k)ig(x/2ki)-(88k)i+1g(x/2k(i+1))(t))≥Ti=0n-1M(2x,αi+1t|8k|i+1).Therefore,
(4.22)μg(x)-C(x)(t)≥T(μg(x)-(88k)ng(x/2kn)(t),μ(88k)ng(x/2kn)-C(x)(t))≥T(Ti=0n-1M(2x,αi+1t|8k|i+1),μ(88k)ng(x/2kn)-C(x)(t)).
By letting n→∞, we obtain
(4.23)μg(x)-C(x)(t)≥Ti=1∞M(2x,αi+1t|8k|i+1).
So,
(4.24)μf(x)-2f(x/2)-C(x/2)(t)≥Ti=1∞M(x,αi+1t|8k|i+1).
This proves (4.5). From Dg(x,y)=Df(2x,2y)-2Df(x,y), by (4.2), we deduce that
(4.25)μDf(2x,2y)(t)≥Ψ(2x,2y,t),μ-2Df(x,y)(t)=μDf(x,y)(t|2|)≥μDf(x,y)(t)≥Ψ(x,y,t),
and so, by (NA-RN3) and (4.2), we obtain
(4.26)μDg(x,y)(t)≥T(μDf(2x,2y)(t),μ-2Df(x,y)(t))≥T(Ψ(2x,2y,t),Ψ(x,y,t))∶=N(x,y,t).
It follows that
(4.27)μ8knDg(x/2kn,y/2kn)(t)=μDg(x/2kn,y/2kn)(t|8|kn)≥N(x2kn,y2kn,t|8|kn)≥⋯≥N(x,y,αn-1t|8|k(n-1))for all x,y∈X, t>0, and n∈ℕ. Since
(4.28)lim⁡n→∞N(x,y,αn-1t|8|k(n-1))=1for all x,y∈X and t>0, by Theorem 2.4, we deduce that
(4.29)μDC(x,y)(t)=1for all x,y∈X and t>0. Thus, the mapping C:X→Y satisfies (1.4).
Now, we have(4.30)C(2x)-8C(x)=lim⁡n→∞[8ng(x2n-1)-8n+1g(x2n)]=8lim⁡n→∞[8n-1g(x2n-1)-8ng(x2n)]=0for all x∈X. Since the mapping x→C(2x)-2C(x) is cubic (see Lemma  2.2 of [14]), from the equality C(2x)=8C(x), we deduce that the mapping C:X→Y is cubic.Corollary 4.3.
Let𝒦 be a non-Archimedean field, let X be a vector space over 𝒦, and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦 under a t-norm T∈ℋ. Let f:X→Y be an odd and Ψ-approximately AQCQ mapping. If, for some α∈ℝ, α>0, and some integer k, k>3, with |2k|<α,
(4.31)Ψ(2-kx,2-ky,t)≥Ψ(x,y,αt)(x∈X,t>0),
then there exists a unique cubic mapping C:X→Y such that
(4.32)μf(x)-2f(x/2)-C(x/2)(t)≥Ti=1∞M(x,αi+1t|8|ki)
for all x∈X and t>0.Proof.
Since(4.33)lim⁡n→∞M(x,αjt|8|kj)=1(x∈X,t>0)
and T is of Hadžić type, from Proposition 2.1, it follows that
(4.34)lim⁡n→∞Tj=n∞M(x,αjt|8|kj)=1(x∈X,t>0).Now, we can apply Theorem 4.2 to obtain the result.Example 4.4.
Let(X,μ,TM) be non-Archimedean random normed space in which
(4.35)μx(t)=tt+‖x‖(x∈X,t>0).
And let (Y,μ,TM) be a complete non-Archimedean random normed space (see Example 3.2). Define
(4.36)Ψ(x,y,t)=t1+t.
It is easy to see that (4.3) holds for α=1. Also, since
(4.37)M(x,t)=t1+t,
we have
(4.38)lim⁡n→∞TM,j=n∞M(x,αjt|8|kj)=lim⁡n→∞(lim⁡m→∞TM,j=nmM(x,t|8|kj))=lim⁡n→∞lim⁡m→∞(tt+|8k|n)=1(x∈X,t>0).
Let f:X→Y be an odd and Ψ-approximately AQCQ mapping. Thus, all the conditions of Theorem 4.2 hold, and so there exists a unique cubic mapping C:X→Y such that
(4.39)μf(x)-2f(x/2)-C(x/2)(t)≥tt+|8k|.Theorem 4.5.
Let𝒦 be a non-Archimedean field, let X be a vector space over 𝒦, and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦. Let f:X→Y be an odd mapping and Ψ-approximately AQCQ mapping. If for some α∈ℝ, α>0, and some integer k, k>1 with |2k|<α,
(4.40)Ψ(2-kx,2-ky,t)≥Ψ(x,y,αt)(x∈X,t>0),lim⁡n→∞⁡Tj=n∞M(2x,αjt|2|kj)=1(x∈X,t>0),
then there exists a unique additive mapping A:X→Y such that
(4.41)μf(x)-8f(x/2)-A(x/2)(t)≥Ti=1∞M(x,αi+1t|2|ki)
for all x∈X and t>0, where
(4.42)M(x,t)∶=Tk-1[Ψ(x2,x2,t|4|),Ψ(x,x2,t),…,Ψ(2k-1x2,2k-1x2,t|4|),Ψ(2k-1x,2k-1x2,t)](x∈X,t>0)Proof.
Lettingy:=x/2 and g(x)∶=f(2x)-8f(x) for all x∈X in (4.9), we get
(4.43)μg(x)-2g(x/2)(t)≥T(Ψ(x2,x2,t|4|),Ψ(x,x2,t))
for all x∈X and t>0.
The rest of the proof is similar to the proof of Theorem4.2.Corollary 4.6.
Let𝒦 be a non-Archimedean field, let X be a vector space over 𝒦, and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦 under a t-norm T∈ℋ. Let f:X→Y be an odd and Ψ-approximately AQCQ mapping. If, for some α∈ℝ,α>0, and some integer k,k>1, with |2k|<α,
(4.44)Ψ(2-kx,2-ky,t)≥Ψ(x,y,αt)(x∈X,t>0),
then there exists a unique additive mapping A:X→Y such that
(4.45)μf(x)-8f(x/2)-A(x/2)(t)≥Ti=1∞M(x,αi+1t|2|ki)
for all x∈X and t>0.Proof.
Since(4.46)lim⁡n→∞M(x,αjt|2|kj)=1(x∈X,t>0)
and T is of Hadžić type, from Proposition 2.1, it follows that
(4.47)lim⁡n→∞Tj=n∞M(x,αjt|2|kj)=1(x∈X,t>0).Now, we can apply Theorem 4.5 to obtain the result.Example 4.7.
Let(X,μ,TM) non-Archimedean random normed space in which
(4.48)μx(t)=tt+‖x‖(x∈X,t>0),and let (Y,μ,TM) be a complete non-Archimedean random normed space (see Example 3.2). Define
(4.49)Ψ(x,y,t)=t1+t.
It is easy to see that (4.3) holds for α=1. Also, since
(4.50)M(x,t)=t1+t,
we have
(4.51)lim⁡n→∞TM,j=n∞M(x,αjt|2|kj)=lim⁡n→∞(lim⁡m→∞TM,j=nmM(x,t|2|kj))=lim⁡n→∞lim⁡m→∞(tt+|2k|n)=1(x∈X,t>0).
Let f:X→Y be an odd and Ψ-approximately AQCQ mapping. Thus, all the conditions of Theorem 4.2 hold, and so there exists a unique additive mapping A:X→Y such that
(4.52)μf(x)-8f(x/2)-A(x/2)(t)≥tt+|2k|.


## 5. Generalized Hyers-Ulam Stability of the Functional Equation (1.4) in Non-Archimedean Random Normed Spaces: An Even Case

Now, we prove the generalized Hyers-Ulam stability of the functional equationDf(x,y)=0 in non-Archimedean Banach spaces, an even case.Theorem 5.1.
Let𝒦 be a non-Archimedean field, let X be a vector space over 𝒦, and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦. Let f:X→Y be an even mapping, f(0)=0, and Ψ-approximately AQCQ mapping. If for some α∈ℝ, α>0, and some integer k, k>4 with |2k|<α,
(5.1)Ψ(2-kx,2-ky,t)≥Ψ(x,y,αt)(x∈X,t>0),lim⁡n→∞Tj=n∞M(2x,αjt|16|kj)=1(x∈X,t>0),
then there exists a unique quartic mapping Q:X→Y such that
(5.2)μf(x)-4f(x/2)-Q(x/2)(t)≥Ti=1∞M(x,αi+1t|16|ki)
for all x∈X and t>0, where
(5.3)M(x,t)∶=Tk-1[Ψ(x2,x2,t|4|),Ψ(x,x2,t),…,Ψ(2k-1x2,2k-1x2,t|4|),Ψ(2k-1x,2k-1x2,t)](x∈X,t>0).Proof.
Lettingx=y in (4.2), we get
(5.4)μf(3y)-6f(2y)+15f(y)(t)≥Ψ(y,y,t)
for all y∈X and t>0. Replacing x by 2y in (4.2), we get
(5.5)μf(4y)-4f(3y)+4f(2y)+4f(y)(t)≥Ψ(2y,y,t)
for all y∈X and t>0. By (5.4) and (5.5), we have
(5.6)μf(4y)-20f(2y)+64f(y)(t)≥T(μ4(f(3y)-4f(2y)+5f(y))(t),μf(4y)-4f(3y)+6f(2y)-4f(y)(t))=T(μf(3y)-4f(2y)+5f(y)(t|4|),μf(4y)-4f(3y)+6f(2y)-4f(y)(t))≥T(Ψ(y,y,t|4|),Ψ(2y,y,t))
for all y∈X and t>0. Letting y∶=x/2 and g(x)∶=f(2x)-4f(x) for all x∈X in (5.6), we get
(5.7)μg(x)-16g(x/2)(t)≥T(Ψ(x2,x2,t|4|),Ψ(x,x2,t))
for all x∈X and t>0.
The rest of the proof is similar to the proof of Theorem4.2.Corollary 5.2.
Let𝒦 be a non-Archimedean field, let X be a vector space over 𝒦, and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦 under a t-norm T∈ℋ. Let f:X→Y be an even, f(0)=0, and Ψ-approximately AQCQ mapping. If,   for some α∈ℝ, α>0, and some integer k, k>4, with |2k|<α,
(5.8)Ψ(2-kx,2-ky,t)≥Ψ(x,y,αt)(x∈X,t>0),
then there exists a unique quartic mapping Q:X→Y such that
(5.9)μf(x)-4f(x/2)-Q(x/2)(t)≥Ti=1∞M(x,αi+1t|16|ki)
for all x∈X and t>0.Proof.
Since(5.10)lim⁡n→∞M(x,αjt|16|kj)=1(x∈X,t>0)
and T is of Hadžić type, from Proposition 2.1, it follows that
(5.11)lim⁡n→∞Tj=n∞M(x,αjt|16|kj)=1(x∈X,t>0).
Now, we can apply Theorem 5.1 to obtain the result.Example 5.3.
Let(X,μ,TM) be non-Archimedean random normed space in which
(5.12)μx(t)=tt+‖x‖(x∈X,t>0).
And let (Y,μ,TM) be a complete non-Archimedean random normed space (see Example 3.2). Define
(5.13)Ψ(x,y,t)=t1+t.
It is easy to see that (4.3) holds for α=1. Also, since
(5.14)M(x,t)=t1+t,
we have
(5.15)lim⁡n→∞TM,j=n∞M(x,αjt|16|kj)=lim⁡n→∞(lim⁡m→∞TM,j=nmM(x,t|16|kj))=lim⁡n→∞lim⁡m→∞(tt+|16k|n)=1(x∈X,t>0).
Let f:X→Y be an even, f(0)=0, and Ψ-approximately AQCQ mapping. Thus all the conditions of Theorem 5.1 hold, and so there exists a unique quartic mapping Q:X→Y such that
(5.16)μf(x)-4f(x/2)-Q(x/2)(t)≥tt+|16k|.Theorem 5.4.
Let𝒦 be a non-Archimedean field, let X be a vector space over 𝒦 and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦. Let f:X→Y be an even mapping, f(0)=0 and Ψ-approximately AQCQ mapping. If for some α∈ℝ, α>0, and some integer k, k>2 with |2k|<α,
(5.17)Ψ(2-kx,2-ky,t)≥Ψ(x,y,αt)(x∈X,t>0),lim⁡n→∞Tj=n∞M(2x,αjt|4|kj)=1(x∈X,t>0),
then there exists a unique quadratic mapping Q:X→Y such that
(5.18)μf(x)-16f(x/2)-Q(x/2)(t)≥Ti=1∞M(x,αi+1t|4|ki)
for all x∈X and t>0, where
(5.19)M(x,t)∶=Tk-1[Ψ(x2,x2,t|4|),Ψ(x,x2,t),…,Ψ(2k-1x2,2k-1x2,t|4|),Ψ(2k-1x,2k-1x2,t)](x∈X,t>0).Proof.
Lettingy:=x/2 and g(x)∶=f(2x)-16f(x) for all x∈X in (5.6), we get
(5.20)μg(x)-4g(x/2)(t)≥T(Ψ(x2,x2,t|4|),Ψ(x,x2,t))
for all x∈X and t>0.
The rest of the proof is similar to the proof of Theorem5.1.Corollary 5.5.
Let𝒦 be a non-Archimedean field, let X be a vector space over 𝒦, and let (Y,μ,T) be a non-Archimedean random Banach space over 𝒦 under a t-norm T∈ℋ. Let f:X→Y be an even, f(0)=0, and Ψ-approximately AQCQ mapping. If,   for some α∈ℝ, α>0, and some integer k, k>2, with |2k|<α,
(5.21)Ψ(2-kx,2-ky,t)≥Ψ(x,y,αt)(x∈X,t>0),
then there exists a unique quadratic mapping Q:X→Y such that
(5.22)μf(x)-16f(x/2)-Q(x/2)(t)≥Ti=1∞M(x,αi+1t|4|ki)
for all x∈X and t>0.Proof.
Since(5.23)lim⁡n→∞M(x,αjt|4|kj)=1(x∈X,t>0)
and T is of Hadžić type, from Proposition 2.1, it follows that
(5.24)lim⁡n→∞Tj=n∞M(x,αjt|4|kj)=1(x∈X,t>0).Now, we can apply Theorem 5.4 to obtain the result.Example 5.6.
Let(X,μ,TM) be a non-Archimedean random normed space in which
(5.25)μx(t)=tt+‖x‖(x∈X,t>0).
And let (Y,μ,TM) be a complete non-Archimedean random normed space (see Example 3.2). Define
(5.26)Ψ(x,y,t)=t1+t.
It is easy to see that (4.3) holds for α=1. Also, since
(5.27)M(x,t)=t1+t,
we have
(5.28)lim⁡n→∞TM,j=n∞M(x,αjt|4|kj)=lim⁡n→∞(lim⁡m→∞TM,j=nmM(x,t|4|kj))=lim⁡n→∞lim⁡m→∞(tt+|4k|n)=1(x∈X,t>0).
Let f:X→Y be an even, f(0)=0, and Ψ-approximately AQCQ mapping. Thus, all the conditions of Theorem 5.4 hold, and so there exists a unique quadratic mapping Q:X→Y such that
(5.29)μf(x)-16f(x/2)-Q(x/2)(t)≥tt+|4k|.


## 6. Latticetic Random Normed Space

Letℒ=(L,≥L) be a complete lattice, that is, a partially ordered set in which every nonempty subset admits supremum and infimum, and 0ℒ=infL, 1ℒ=supL. The space of latticetic random distribution functions, denoted by ΔL+, is defined as the set of all mappings F:ℝ∪{-∞,+∞}→L such that F is left continuous and nondecreasing on ℝ, F(0)=0ℒ, F(+∞)=1ℒ.DL+⊆ΔL+ is defined as DL+={F∈ΔL+:l-F(+∞)=1ℒ}, where l-f(x) denotes the left limit of the function f at the point x. The space ΔL+ is partially ordered by the usual pointwise ordering of functions, that is, F≥G if and only if F(t)≥LG(t) for all t in ℝ. The maximal element for ΔL+ in this order is the distribution function given by(6.1)ɛ0(t)={0L,ift≤0,1L,ift>0.In Section2, we defined t-norms on [0,1], and now we extend t-norms on a complete lattice.Definition 6.1 (see [42]).
Atriangular norm (t-norm) on L is a mapping 𝒯:(L)2→L satisfying the following conditions:(a)
(forallx∈L)(𝒯(x,1ℒ)=x)     (boundary condition);(b)
(forall(x,y)∈(L)2)(𝒯(x,y)=𝒯(y,x))   (commutativity);(c)
(forall(x,y,z)∈(L)3)(𝒯(x,𝒯(y,z))=𝒯(𝒯(x,y),z))   (associativity);(d)
(forall(x,x′,y,y′)∈(L)4)(x≤Lx′andy≤Ly′⇒𝒯(x,y)≤L𝒯(x′,y′))   (monotonicity).Let{xn} be a sequence in L converges to x∈L (equipped order topology). The t-norm 𝒯 is said to be a continuoust-norm if
(6.2)lim⁡n→∞T(xn,y)=T(x,y)
for all y∈L.At-norm 𝒯 can be extended (by associativity) in a unique way to an n-array operation taking for (x1,…,xn)∈Ln the value 𝒯(x1,…,xn) defined by (6.3)Ti=10xi=1,Ti=1nxi=T(Ti=1n-1xi,xn)=T(x1,…,xn).𝒯 can also be extended to a countable operation taking for any sequence (xn)n∈N in L the value(6.4)Ti=1∞xi=lim⁡n→∞Ti=1nxi.
The limit on the right side of (6.4) exists since the sequence (𝒯i=1nxi)n∈ℕ is nonincreasing and bounded from below.Note that we put𝒯=T whenever L=[0,1]. If T is a t-norm, then xT(n) is defined for every x∈[0,1] and n∈N∪{0} by 1 if n=0 and T(xT(n-1),x) if n≥1. A t-norm T is said to be of Hadžić type, (we denote by T∈ℋ) if the family (xT(n))n∈N is equicontinuous at x=1 (cf. [27]).Definition 6.2 (see [42]).
A continuoust-norm 𝒯 on L=[0,1]2 is said to be continuoust–representable if there exist a continuous t-norm * and a continuous t-conorm ⋄ on [0,1] such that, for all x=(x1,x2), y=(y1,y2)∈L,
(6.5)T(x,y)=(x1*y1,x2⋄y2).For example,(6.6)T(a,b)=(a1b1,min⁡{a2+b2,1}),M(a,b)=(min⁡{a1,b1},max⁡{a2,b2})
for all a=(a1,a2), b=(b1,b2)∈[0,1]2 are continuous t-representable. Define the mapping 𝒯∧ from L2 to L by
(6.7)T∧(x,y)={x,ify≥Lx,y,ifx≥Ly.Recall (see [27, 28]) that if {xn} is a given sequence in L, (𝒯∧)i=1nxi is defined recurrently by (𝒯∧)i=11xi=x1 and (𝒯∧)i=1nxi=𝒯∧((𝒯∧)i=1n-1xi,xn) for all n≥2.A negation onℒ is any decreasing mapping 𝒩:L→L satisfying 𝒩(0ℒ)=1ℒ and 𝒩(1ℒ)=0ℒ. If 𝒩(𝒩(x))=x, for all x∈L, then 𝒩 is called an involutive negation. In the following, ℒ is endowed with a (fixed) negation 𝒩.Definition 6.3.
Alatticetic random normed space (in short LRN-space) is a triple (X,μ,𝒯∧), where X is a vector space and μ is a mapping from X into DL+ such that the following conditions hold: (LRN1)μx(t)=ɛ0(t) for all t>0 if and only if x=0,(LRN2)μαx(t)=μx(t/|α|) for all x in X, α≠0 and t≥0,(LRN3)μx+y(t+s)≥L𝒯∧(μx(t),μy(s)) for all x,y∈X and t,s≥0.We note that from (LPN2) it follows thatμ-x(t)=μx(t) for all x∈X and t≥0.Example 6.4.
LetL=[0,1]×[0,1] and operation ≤L be defined by
(6.8)L={(a1,a2):(a1,a2)∈[0,1]×[0,1],a1+a2≤1},(a1,a2)≤L(b1,b2)⟺a1≤b1,a2≥b2,∀a=(a1,a2),b=(b1,b2)∈L.
then (L,≤L) is a complete lattice (see [42]). In this complete lattice, we denote its units by 0L=(0,1) and 1L=(1,0). Let (X,∥·∥) be a normed space. Let 𝒯(a,b)=(min{a1,b1},max{a2,b2}) for all a=(a1,a2), b=(b1,b2)∈[0,1]×[0,1] and μ be a mapping defined by
(6.9)μx(t)=(tt+‖x‖,‖x‖t+‖x‖)(t∈R+),
then (X,μ,𝒯) is a latticetic random normed spaces.
If(X,μ,𝒯∧) is a latticetic random normed space, then
(6.10)V={V(ɛ,λ):ɛ>L0L,λ∈L∖{0L,1L}},V(ɛ,λ)={x∈X:Fx(ɛ)>LN(λ)},
is a complete system of neighborhoods of null vector for a linear topology on X generated by the norm F.Definition 6.5.
Let(X,μ,𝒯∧) be a latticetic random normed spaces.(1)
A sequence{xn} in X is said to be convergent to x in X if, for every t>0 and ε∈L∖{0ℒ}, there exists a positive integer N such that μxn-x(t)>L𝒩(ε) whenever n≥N.(2)
A sequence{xn} in X is called a Cauchy sequence if, for every t>0 and ε∈L∖{0ℒ}, there exists a positive integer N such that μxn-xm(t)>L𝒩(ε) whenever n≥m≥N.(3)
A latticetic random normed spaces(X,μ,𝒯∧) is said to be complete if and only if every Cauchy sequence in X is convergent to a point in X.Theorem 6.6.
If(X,μ,𝒯∧) is a latticetic random normed space and {xn} is a sequence such that xn→x, then limn→∞μxn(t)=μx(t).Proof.
The proof is the same as classical random normed spaces, see [25].


## 7. Generalized Hyers-Ulam Stability of the Functional Equation (1.4): An Odd Case via Fixed-Point Method

Using the fixed point method, we prove the generalized Hyers-Ulam stability of the functional equationDf(x,y)=0 in random Banach spaces: an odd case.Theorem 7.1.
LetX be a linear space, let (Y,μ,𝒯∧) be a complete LRN-space, and Φ let be a mapping from X2 to DL+  (Φ(x,y) is denoted by Φx,y) such that, for some 0<α<1/8,
(7.1)Φ2x,2y(t)≤LΦx,y(αt)(x,y∈X,t>0).
Let f:X→Y be an odd mapping satisfying
(7.2)μDf(x,y)(t)≥LΦx,y(t)
for all x,y∈X and t>0. Then
(7.3)C(x)∶=lim⁡n→∞8n(f(x2n-1)-2f(x2n))
exists for each x∈X and defines a cubic mapping C:X→Y such that
(7.4)μf(2x)-2f(x)-C(x)(t)≥LT∧(Φx,x(1-8α5αt),Φ2x,x(1-8α5αt))
for all x∈X and t>0.Proof.
Lettingx=y in (7.2), we get
(7.5)μf(3y)-4f(2y)+5f(y)(t)≥LΦy,y(t)
for all y∈X and t>0. Replacing x by 2y in (7.2), we get
(7.6)μf(4y)-4f(3y)+6f(2y)-4f(y)(t)≥LΦ2y,y(t)
for all y∈X and t>0. By (7.5) and (7.6),
(7.7)μf(4y)-10f(2y)+16f(y)(5t)≥LT∧(μ4(f(3y)-4f(2y)+5f(y))(4t),μf(4y)-4f(3y)+6f(2y)-4f(y)(t))=T∧(μf(3y)-4f(2y)+5f(y)(t),μf(4y)-4f(3y)+6f(2y)-4f(y)(t))≥LT∧(Φy,y(t),Φ2y,y(t))
for all y∈X and t>0. Letting y:=x/2 and g(x)∶=f(2x)-2f(x) for all x∈X, we get
(7.8)μg(x)-8g(x/2)(5t)≥LT∧(Φx/2,x/2(t),Φx,x/2(t))
for all x∈X and t>0.
Consider the set(7.9)S∶={h:X⟶Y,h(0)=0}
and introduce the generalized metric on S:
(7.10)d(h,k)=inf⁡{u∈R+:μh(x)-k(x)(ut)≥LT∧(Φx,x(t),Φ2x,x(t)),∀x∈X,∀t>0}
where, as usual, inf∅=+∞. It is easy to show that (S,d) is complete (see the proof of Lemma  2.1 of [24]).
Now, we consider the linear mappingJ:S→S such that
(7.11)Jh(x)∶=8h(x2)
for all x∈X, and we prove that J is a strictly contractive mapping with the Lipschitz constant 8α.
Leth,k∈S be given such that d(h,k)<ɛ. Then
(7.12)μh(x)-k(x)(ɛt)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. Hence
(7.13)μJh(x)-Jk(x)(8αɛt)=μ8h(x/2)-8k(x/2)(8αɛt)=μh(x/2)-k(x/2)(αɛt)≥T∧(Φx/2,x/2(αt),Φx,x/2(αt))≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(h,k)<ɛ implies that
(7.14)d(Jh,Jk)≤α8ɛ.
This means that
(7.15)d(Jh,Jk)≤α8d(h,k)
for all h,k∈S.  It follows from (7.8) that
(7.16)μg(x)-8g(x/2)(5αt)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(g,Jg)≤5α≤5/8.
By Theorem1.1, there exists a mapping C:X→Y satisfying the following:(1)
C is a fixed point of J, that is,
(7.17)C(x2)=18C(x)
for all x∈X. Since g:X→Y is odd, C:X→Y is an odd mapping. The                                     mapping C is a unique fixed point of J in the set
(7.18)M={h∈S:d(h,g)<∞}.
This implies that C is a unique mapping satisfying (7.17) such that there exists a u∈(0,∞) satisfying
(7.19)μg(x)-C(x)(ut)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0.(2)
d(Jng,C)→0 as n→∞. This implies the equality
(7.20)lim⁡n→∞8ng(x2n)=C(x)
for all x∈X.(3)
d(h,C)≤(1/(1-8α))d(h,Jh) with h∈M, which implies the inequality
(7.21)d(g,C)≤5α1-8α,
from which it follows that
(7.22)μg(x)-C(x)(5α1-8αt)≥LT∧(Φx,x(t),Φ2x,x(t)).
This implies that the inequality (7.4) holds. From Dg(x,y)=Df(2x,2y)-2Df(x,y), by (7.2), we deduce that
(7.23)μDf(2x,2y)(t)≥LΦ2x,2y(t),μ-2Df(x,y)(t)=μDf(x,y)(t2)≥LΦx,y(t2)
and so, by (LRN3) and (7.1), we obtain
(7.24)μDg(x,y)(3t)≥LT∧(μDf2x,2y(t),μ-2Df(x,y)(2t))≥LT∧(Φ2x,2y(t),Φx,y(t))≥LΦ2x,2y(t).
It follows that
(7.25)μ8nDg(x/2n,y/2n)(3t)=μDg(x/2n,y/2n)(3t8n)≥Φx/2n-1,y/2n-1(t8n)≥L⋯≥LΦx,y(18t(8α)n-1)
for all x,y∈X, t>0 and n∈ℕ.
Sincelimn→∞Φx,y((3/8)(t/(8α)n-1))=1 for all x,y∈X and t>0, by Theorem 2.4, we deduce that
(7.26)μDC(x,y)(3t)=1L
for all x,y∈X and t>0. Thus the mapping C:X→Y satisfies (1.4).
Now, we have(7.27)C(2x)-8C(x)=lim⁡n→∞⁡[8ng(x2n-1)-8n+1g(x2n)]=8lim⁡n→∞⁡[8n-1g(x2n-1)-8ng(x2n)]=0
for all x∈X. Since the mapping x→C(2x)-2C(x) is cubic (see Lemma  2.2 of [14]), from the equality C(2x)=8C(x), we deduce that the mapping C:X→Y is cubic.Corollary 7.2.
Letθ≥0 and let p be a real number with p>3. Let X be a normed vector space with norm ∥·∥. Let f:X→Y be an odd mapping satisfying
(7.28)μDf(x,y)(t)≥tt+θ(‖x‖p+‖y‖p)
for all x,y∈X and t>0. Note that (X,μ,TM) is a complete LRN-space, in which L=[0,1], then
(7.29)C(x)∶=lim⁡n→∞⁡8n(f(x2n-1)-2f(x2n))
exists for each x∈X and defines a cubic mapping C:X→Y such that
(7.30)μf(2x)-2f(x)-C(x)(t)≥(2p-8)t(2p-8)t+5(1+2p)θ‖x‖p
for all x∈X and t>0.Proof.
The proof follows from Theorem7.1 by taking
(7.31)Φx,y(t)∶=tt+θ(‖x‖p+‖y‖p)
for all x,y∈X and t>0. Then we can choose α=2-p, and we get
(7.32)μf(2x)-2f(x)-C(x)(t)≥min⁡((1-23-p)t(1-23-p)t+5⋅2-pθ(2‖x‖p),(1-23-p)t(1-23-p)t+5⋅2-pθ(‖2x‖p+‖x‖p))≥(1-23-p)t(1-23-p)t+5⋅2-pθ(‖2x‖p+‖x‖p)=(2p-8)t(2p-8)t+5⋅(2p+1)θ‖x‖p,
which is the desired result.Theorem 7.3.
LetX be a linear space, let (Y,μ,𝒯∧) be a complete LRN-space, and let Φ be a mapping from X2 to DL+  (Φ(x,y) is denoted by Φx,y) such that, for some 0<α<8,
(7.33)Φx/2,y/2(t)≤LΦx,y(αt)(x,y∈X,t>0).
Let f:X→Y be an odd mapping satisfying (7.2), then
(7.34)C(x)∶=lim⁡n→∞⁡18n(f(2n+1x)-2f(2nx))
exists for each x∈X and defines a cubic mapping C:X→Y such that
(7.35)μf(2x)-2f(x)-C(x)(t)≥LT∧(Φx,x(8-α5t),Φ2x,x(8-α5t))
for all x∈X and t>0.Proof.
Let(S,d) be the generalized metric space defined in the proof of Theorem 7.1.
Consider the linear mappingJ:S→S such that
(7.36)Jh(x)∶=18h(2x)
for all x∈X, and we prove that J is a strictly contractive mapping with the Lipschitz constant α/8.
Leth,k∈S be given such that d(h,k)<ɛ, then
(7.37)μh(x)-k(x)(ɛt)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. Hence
(7.38)μJh(x)-Jk(x)(α8ɛt)=μ(1/8)h(2x)-(1/8)k(2x)(α8ɛt)=μh(2x)-k(2x)(αɛt)≥LT∧(Φ2x,2x(αt),Φ4x,2x(αt))≥T∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(h,k)<ɛ implies that
(7.39)d(Jh,Jk)≤α8ɛ.
This means that(7.40)d(Jh,Jk)≤α8d(h,k)
for all g,h∈S. Letting g(x)∶=f(2x)-2f(x) for all x∈X, from (7.8), we get that
(7.41)μg(x)-(1/8)g(2x)(58t)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(g,Jg)≤5/8.
By Theorem1.1, there exists a mapping C:X→Y satisfying the following:(1)
C is a fixed point of J, that is,
(7.42)C(2x)=8C(x)
for all x∈X. Since g:X→Y is odd, C:X→Y is an odd mapping. The mapping C is a unique fixed point of J in the set
(7.43)M={h∈S:d(h,g)<∞}.
This implies that C is a unique mapping satisfying (7.42) such that there exists a u∈(0,∞) satisfying
(7.44)μg(x)-C(x)(ut)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0.(2)
d(Jng,C)→0 as n→∞. This implies the equalit
(7.45)lim⁡n→∞⁡18ng(2nx)=C(x)
for all x∈X.(3)
d(h,C)≤(1/(1-α/8))d(h,Jh) for every h∈M, which implies the inequality
(7.46)d(g,C)≤58-α,
from which it follows that
(7.47)μg(x)-C(x)(58-αt)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. This implies that the inequality (7.35) holds.
From(7.48)μDg(x,y)(3t)≥LT∧(Φ2x,2y(t),Φx,y(t))≥LT∧(Φ2x,2y(t),Φx,y(t8)),
by (7.33), we deduce that
(7.49)μ8-nDg(2nx,2ny)(3t)=μDg(2nx,2ny)(3⋅8nt)≥LΦ2nx,2ny(8n-1t)≥L⋯≥Φx,y((8α)n-1tα)
for all x,y∈X, t>0, and n∈ℕ. As n→∞, we deduce that
(7.50)μDC(x,y)(3t)=1L
for all x,y∈X and t>0. Thus the mapping C:X→Y satisfies (1.4).
Now, we have(7.51)C(2x)-8C(x)=lim⁡n→∞⁡[18ng(2n+1x)-18n-1g(2nx)]=8lim⁡n→∞⁡[18n+1g(2n+1x)-18ng(2nx)]=0
for all x∈X. Since the mapping x→C(2x)-2C(x) is cubic (see Lemma  2.2 of [14]), from the equality C(2x)=8C(x), we deduce that the mapping C:X→Y is cubic.Corollary 7.4.
Letθ≥0 and let p be a real number with 0<p<3. Let X be a normed vector space with norm ∥·∥. Let f:X→Y be an odd mapping satisfying  (7.28), then
(7.52)C(x)∶=lim⁡n→∞⁡18n(f(2n+1x)-2f(2nx))
exists for each x∈X and defines a cubic mapping C:X→Y such that
(7.53)μf(2x)-2f(x)-C(x)(t)≥(8-2p)t(8-2p)t+5(1+2p)θ‖x‖p
for all x∈X and t>0. Note that (X,μ,TM) is a complete LRN-space, in which L=[0,1].Proof.
The proof follows from Theorem7.3 by taking
(7.54)μDf(x,y)(t)≥tt+θ(‖x‖p+‖y‖p)
for all x,y∈X and t>0. Then we can choose α=2p, and we get the desired result.Theorem 7.5.
LetX be a linear space, let (Y,μ,𝒯∧) be a complete LRN-space, and let Φ be a mapping from X2 to DL+  (Φ(x,y) is denoted by Φx,y) such that, for some 0<α<1/2,
(7.55)Φ2x,2y(t)≤LΦx,y(αt)(x,y∈X,t>0).
Let f:X→Y be an odd mapping satisfying  (7.2), then
(7.56)A(x)∶=lim⁡n→∞⁡2n(f(x2n-1)-8f(x2n))
exists for each x∈X and defines an additive mapping A:X→Y such that
(7.57)μf(2x)-8f(x)-A(x)(t)≥LT∧(Φx,x(1-2α5αt),Φ2x,x(1-2α5αt))
for all x∈X and t>0.Proof.
Let(S,d) be the generalized metric space defined in the proof of Theorem 7.1.
Lettingy∶=x/2 and g(x)∶=f(2x)-8f(x) for all x∈X in (7.7), we get
(7.58)μg(x)-2g(x/2)(5t)≥LT∧(Φx/2,x/2(t),Φx,x/2(t))
for all x∈X and t>0.
Now, we consider the linear mappingJ:S→S such that
(7.59)Jh(x)∶=2h(x2)
for all x∈X. It is easy to see that J is a strictly contractive self-mapping on S with the Lipschitz constant 2α.
It follows from (7.58) and (7.55) that
(7.60)μg(x)-2g(x/2)(5αt)≥TM(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(g,Jg)≤5α<∞.
By Theorem1.1, there exists a mapping A:X→Y satisfying the following:(1)
A is a fixed point of J, that is,
(7.61)A(x2)=12A(x)
for allx∈X. Sinceg:X→Y is odd,A:X→Y is an odd mapping. The mappingA is a unique fixed point ofJ in the set
(7.62)M={h∈S:d(h,g)<∞}.
This implies that A is a unique mapping satisfying (7.61) such that there exists a u∈(0,∞) satisfying
(7.63)μg(x)-A(x)(ut)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0.(2)
d(Jng,A)→0 as n→∞. This implies the equality
(7.64)lim⁡n→∞⁡2ng(x2n)=A(x)
for all x∈X.(3)
d(h,A)≤(1/(1-2α))d(h,Jh) for each h∈M, which implies the inequality
(7.65)d(g,A)≤5α1-2α.
This implies that the inequality (7.57) holds. Since μDg(x,y)(3t)≥LΦ2x,2y(t), it follows that
(7.66)μ2nDg(x/2n,y/2n)(3t)=μDg(x/2n,y/2n)(3t2n)≥Φx/2n-1,y/2n-1(t2n)≥L⋯≥LΦx,y(12t(2α)n-1)
for all x,y∈X, t>0, and n∈ℕ. As n→∞, we deduce that
(7.67)μDA(x,y)(3t)=1L
for all x,y∈X and t>0. Thus, the mapping A:X→Y satisfies (1.4).
Now, we have(7.68)A(2x)-2A(x)=lim⁡n→∞⁡[2ng(x2n-1)-2n+1g(x2n)]=2lim⁡n→∞⁡[2n-1g(x2n-1)-2ng(x2n)]=0
for all x∈X. Since the mapping x→A(2x)-8A(x) is additive (see Lemma  2.2 of [14]), from the equality A(2x)=2A(x), we deduce that the mapping A:X→Y is additive.Corollary 7.6.
Letθ≥0 and let p be a real number with p>1. Let X be a normed vector space with norm ∥·∥. Let f:X→Y be an odd mapping satisfying  (7.28), then
(7.69)A(x)∶=lim⁡n→∞⁡2n(f(x2n-1)-8f(x2n))
exists for each x∈X and defines an additive mapping A:X→Y such that
(7.70)μf(2x)-8f(x)-A(x)(t)≥(2p-2)t(2p-2)t+5(1+2p)θ‖x‖p
for all x∈X and t>0, where (X,μ,TM) is a complete LRN-space in which L=[0,1].Proof.
The proof follows from Theorem7.5 by taking
(7.71)μDf(x,y)(t)≥tt+θ(‖x‖p+‖y‖p)
for all x,y∈X and t>0. Then we can choose α=2-p, and we get the desired result.Theorem 7.7.
LetX be a linear space, let (Y,μ,𝒯∧) be a complete LRN-space, and let Φ be a mapping from X2 to DL+  (Φ(x,y) is denoted by Φx,y) such that, for some 0<α<2,
(7.72)Φx,y(αt)≥LΦx/2,y/2(t)(x,y∈X,t>0).
Let f:X→Y be an odd mapping satisfying  (7.2), then
(7.73)A(x)∶=lim⁡n→∞⁡12n(f(2n+1x)-8f(2nx))
exists for each x∈X and defines an additive mapping A:X→Y such that
(7.74)μf(2x)-8f(x)-A(x)(t)≥LT∧(Φx,x(2-α5αt),Φ2x,x(2-α5αt))
for all x∈X and t>0.Proof.
Let(S,d) be the generalized metric space defined in the proof of Theorem 7.1.
Consider the linear mappingJ:S→S such that
(7.75)Jh(x)∶=12h(2x)
for all x∈X. It is easy to see that J is a strictly contractive self-mapping on S with the Lipschitz constant α/2.  Let g(x)=f(2x)-8f(x), from (7.58), it follows that
(7.76)μg(x)-1/2g(2x)(52t)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(g,Jg)≤5/2. By Theorem 1.1, there exists a mapping A:X→Y satisfying the following:(1)
A is a fixed point of J, that is,
(7.77)A(2x)=2A(x)
for all x∈X. Since h:X→Y is odd, A:X→Y is an odd mapping. The mapping A is a unique fixed point of J in the set
(7.78)M={h∈S:d(h,g)<∞}.
This implies that A is a unique mapping satisfying (7.77) such that there exists a u∈(0,∞) satisfying
(7.79)μg(x)-A(x)(ut)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0.(2)
d(Jng,A)→0 as n→∞. This implies the equality
(7.80)lim⁡n→∞⁡12ng(2nx)=A(x)
for all x∈X.(3)
d(h,A)≤(1/(1-α/2))d(h,Jh), which implies the inequality
(7.81)d(g,A)≤52-α.
This implies that the inequality (7.74) holds.
Proceeding as in the proof of Theorem7.5, we obtain that the mapping A:X→Y satisfies (1.4). Now, we have
(7.82)A(2x)-2A(x)=lim⁡n→∞⁡[12ng(2n+1x)-12n-1g(2nx)]=2lim⁡n→∞⁡[12n+1g(2n+1x)-12ng(2nx)]=0
for all x∈X. Since the mapping x→A(2x)-8A(x) is additive (see Lemma  2.2 of [14]), from the equality A(2x)=2A(x), we deduce that the mapping A:X→Y is additive.Corollary 7.8.
Letθ≥0 and let p be a real number with 0<p<1. Let X be a normed vector space with norm ∥·∥. Let f:X→Y be an odd mapping satisfying  (7.28), then
(7.83)A(x)∶=lim⁡n→∞⁡12n(f(2n+1x)-8f(2nx))
exists for each x∈X and defines an additive mapping A:X→Y such that
(7.84)μf(2x)-8f(x)-A(x)(t)≥(2-2p)t(2-2p)t+5(1+2p)θ‖x‖p
for all x∈X and t>0, where (X,μ,TM) is a complete LRN-space in which L=[0,1].Proof.
The proof follows from Theorem7.7 by taking
(7.85)μDf(x,y)(t)≥tt+θ(‖x‖p+‖y‖p)
for all x,y∈X and t>0. Then we can choose α=2p, and we get the desired result.


## 8. Generalized Hyers-Ulam Stability of the Functional Equation (1.4): An Even Case via Fixed-Point Method

Using the fixed point method, we prove the generalized Hyers-Ulam stability of the functional equationDf(x,y)=0 in random Banach spaces, an even case.Theorem 8.1.
LetX be a linear space, let (Y,μ,𝒯∧) be a complete LRN-space, and let Φ be a mapping from X2 to DL+  (Φ(x,y) is denoted by Φx,y) such that, for some 0<α<1/16,
(8.1)Φx,y(αt)≥LΦ2x,2y(t)(x,y∈X,t>0).
Let f:X→Y be an even mapping satisfying f(0)=0 and  (7.2), then
(8.2)Q(x)∶=lim⁡n→∞⁡16n(f(x2n-1)-4f(x2n))
exists for each x∈X and defines a quartic mapping Q:X→Y such that
(8.3)μf(2x)-4f(x)-Q(x)(t)≥LT∧(Φx,x(1-16α5αt),Φ2x,x(1-16α5αt))
for all x∈X and t>0.Proof.
Lettingx=y in (7.2), we get
(8.4)μf(3y)-6f(2y)+15f(y)(t)≥LΦy,y(t)
for all y∈X and t>0.  Replacing x by 2y in (7.2), we get
(8.5)μf(4y)-4f(3y)+4f(2y)+4f(y)(t)≥LΦ2y,y(t)
for all y∈X and t>0. By (8.4) and (8.5),
(8.6)μf(4x)-20f(2x)+64f(x)(5t)≥LT∧(μ4(f(3x)-6f(2x)+15f(x))(4t),μf(4x)-4f(3x)+4f(2x)+4f(x)(t))≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. Letting g(x)∶=f(2x)-4f(x) for all x∈X, we get
(8.7)μg(x)-16g(x/2)(5t)≥LT∧(Φx/2,x/2(t),Φx,x/2(t))
for all x∈X and t>0.   Let (S,d) be the generalized metric space defined in the proof of Theorem 7.1.
Now we consider the linear mappingJ:S→S such that Jh(x)∶=16h(x/2) for all x∈X. It is easy to see that J is a strictly contractive self-mapping on S with the Lipschitz constant 16α. It follows from (8.7) that
(8.8)μg(x)-16g(x/2)(5αt)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So,
(8.9)d(g,Jg)≤5α≤516<∞.
By Theorem1.1, there exists a mapping Q:X→Y satisfying the following:(1)
Q is a fixed point of J, that is,
(8.10)Q(x2)=116Q(x)
for allx∈X. Sinceg:X→Y is even withg(0)=0,Q:X→Y is an even mapping withQ(0)=0. The mappingQ is a unique fixed point ofJ in the set
(8.11)M={h∈S:d(h,g)<∞}.
This implies that Q is a unique mapping satisfying (8.10) such that there exists a u∈(0,∞) satisfying
(8.12)μg(x)-Q(x)(ut)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0.(2)
d(Jng,Q)→0 as n→∞. This implies the equality
(8.13)lim⁡n→∞⁡16ng(x2n)=Q(x)
for all x∈X.(3)
d(h,Q)≤(1/(1-16α))d(h,Jh) for every h∈M, which implies the inequality
(8.14)d(g,Q)≤5α1-16α.
This implies that the inequality (8.3) holds.
Proceeding as in the proof of Theorem7.1, we obtain that the mapping Q:X→Y satisfies (1.4). Now, we have
(8.15)Q(2x)-16Q(x)=lim⁡n→∞⁡[16ng(x2n-1)-16n+1g(x2n)]=16lim⁡n→∞⁡[16n-1g(x2n-1)-16ng(x2n)]=0
for all x∈X.  Since the mapping x→Q(2x)-4Q(x) is quartic, we get that the mapping Q:X→Y is quartic.Corollary 8.2.
Letθ≥0 and let p be a real number with p>4. Let X be a normed vector space with norm ∥·∥.  Let f:X→Y be an even mapping satisfying f(0)=0 and  (7.28), then
(8.16)Q(x)∶=lim⁡n→∞⁡16n(f(x2n-1)-4f(x2n))
exists for each x∈X and defines a quartic mapping Q:X→Y such that
(8.17)μf(2x)-4f(x)-Q(x)(t)≥(2p-16)t(2p-16)t+5(1+2p)θ‖x‖p
for all x∈X and t>0, where (X,μ,TM) is a complete LRN-space in which L=[0,1].Proof.
The proof follows from Theorem8.1 by taking
(8.18)μDf(x,y)(t)≥tt+θ(‖x‖p+‖y‖p)
for all x,y∈X and t>0. Then we can choose α=2-p, and we get the desired result.Theorem 8.3.
LetX be a linear space, let (Y,μ,𝒯∧) be a complete LRN-space, and let Φ be a mapping from X2 to DL+  (Φ(x,y) is denoted by Φx,y) such that, for some 0<α<16,
(8.19)Φx,y(αt)≥Φx/2,y/2(t)(x,y∈X,t>0).
Let f:X→Y be an even mapping satisfying f(0)=0 and  (7.2), then
(8.20)Q(x)∶=lim⁡n→∞⁡116n(f(2n+1x)-4f(2nx))
exists for each x∈X and defines a quartic mapping Q:X→Y such that
(8.21)μf(2x)-4f(x)-Q(x)(t)≥LT∧(Φx,x(16-α5t),Φ2x,x(16-α5t))
for all x∈X and t>0.Proof.
In the generalized metric space(S,d) defined in the proof of Theorem 7.1, we consider the linear mapping J:S→S such that
(8.22)Jh(x)∶=116h(2x)
for all x∈X. It is easy to see that J is a strictly contractive self-mapping on S with the Lipschitz constant α/16.
Lettingg(x)∶=f(2x)-4f(x) for all x∈X, by (8.7), we get
(8.23)μg(x)-(1/16)g(2x)(516t)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(g,Jg)≤5/16.
By Theorem1.1, there exists a mapping Q:X→Y satisfying the following:(1)
Q is a fixed point of J, that is,
(8.24)Q(2x)=16Q(x)
for all x∈X. Since g:X→Y is even with g(0)=0, Q:X→Y is an even mapping with Q(0)=0. The mapping Q is a unique fixed point of J in the set
(8.25)M={h∈S:d(h,g)<∞}.
This implies that Q is a unique mapping satisfying (8.24) such that there exists a u∈(0,∞) satisfying
(8.26)μg(x)-Q(x)(ut)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0.(2)
d(Jng,Q)→0 as n→∞. This implies the equality
(8.27)lim⁡n→∞⁡116ng(2nx)=Q(x)
for all x∈X.(3)
d(g,Q)≤(16/(16-α))d(g,Jg) for each h∈M, which implies the inequality
(8.28)d(g,Q)≤5/(16-α).
This implies that the inequality (8.21) holds.
Proceeding as in the proof of Theorem7.3, we obtain that the mapping Q:X→Y satisfies (1.4). Now, we have
(8.29)Q(2x)-16Q(x)=lim⁡n→∞⁡[116ng(2n+1x)-116n-1g(2nx)]=16lim⁡n→∞⁡[116n+1g(2n+1x)-116ng(2nx)]=0
for all x∈X. Since the mapping x→Q(2x)-4Q(x) is quartic, we get that the mapping Q:X→Y is quartic.Corollary 8.4.
Letθ≥0 and let p be a real number with 0<p<4. Let X be a normed vector space with norm ∥·∥. Let f:X→Y be an even mapping satisfying f(0)=0 and (7.28), then
(8.30)Q(x)∶=lim⁡n→∞⁡116n(f(2n+1x)-4f(2nx))exists for each x∈X  and defines a quartic mapping Q:X→Y such that
(8.31)μf(2x)-4f(x)-Q(x)(t)≥(16-2p)t(16-2p)t+5(1+2p)θ‖x‖p
for all x∈X and t>0, where (X,μ,TM) is a complete LRN-space in which L=[0,1].Proof.
The proof follows from Theorem8.3 by taking
(8.32)μDf(x,y)(t)≥tt+θ(‖x‖p+‖y‖p)
for all x,y∈X and t>0.  Then we can choose α=2p, and we get the desired result.Theorem 8.5.
LetXbe a linear space, let (Y,μ,𝒯∧) be a complete LRN-space,  and let Φ be a mapping from X2 to DL+  (Φ(x,y) is by denoted Φx,y) such that, for some 0<α<1/4,
(8.33)Φx,y(αt)≥LΦ2x,2y(t)(x,y∈X,t>0).
Let f:X→Y be an even mapping satisfying f(0)=0 and  (7.2), then
(8.34)T(x)∶=lim⁡n→∞⁡4n(f(x2n-1)-16f(x2n))
exists for each x∈X and defines a quadratic mapping T:X→Y such that
(8.35)μf(2x)-16f(x)-T(x)(t)≥LT∧(Φx,x(1-4α5αt),Φ2x,x(1-4α5αt))
for all x∈X and t>0.Proof.
Let(S,d) be the generalized metric space defined in the proof of Theorem 7.1.
Lettingg(x)∶=f(2x)-16f(x) for all x∈X in (8.6), we get
(8.36)μg(x)-4g(x/2)(5t)≥LT∧(Φx/2,x/2(t),Φx,x/2(t))
for all x∈X and t>0. It is easy to see that the linear mapping J:S→S such that
(8.37)Jh(x)∶=4h(x2)
for all x∈X, is a strictly contractive self-mapping with the Lipschitz constant 4α.
It follows from (8.36) that
(8.38)μg(x)-4g(x/2)(5αt)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(g,Jg)≤5α<∞.
By Theorem1.1, there exists a mapping T:X→Y satisfying the following:(1)
T is a fixed point of J, that is,
(8.39)T(x2)=14T(x)
for all x∈X. Since g:X→Y is even with g(0)=0, T:X→Y is an even mapping with T(0)=0. The mapping T is a unique fixed point of J in the set M={h∈S:d(h,g)<∞}. This implies that T is a unique mapping satisfying (8.39) such that there exists a u∈(0,∞) satisfying
(8.40)μg(x)-T(x)(ut)≥LT∧(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0.(2)
d(Jng,T)→0 as n→∞. This implies the equality
(8.41)lim⁡n→∞⁡4ng(x2n)=T(x)
for all x∈X.(3)
d(h,T)≤(1/(1-4α))d(h,Jh) for each h∈M, which implies the inequality
(8.42)d(g,T)≤5α1-4α.
This implies that the inequality (8.35) holds.
Proceeding as in the proof of Theorem7.1, we obtain that the mapping T:X→Y satisfies (1.4). Now, we have
(8.43)T(2x)-4T(x)=lim⁡n→∞⁡[4ng(x2n-1)-4n+1g(x2n)]=4lim⁡n→∞⁡[4n-1g(x2n-1)-4ng(x2n)]=0
for all x∈X. Since the mapping x→T(2x)-16T(x) is quadratic, we get that the mapping T:X→Y is quadratic.Corollary 8.6.
Letθ≥0 and let p be a real number with p>2. Let X be a normed vector space with norm ∥·∥. Let f:X→Y be an even mapping satisfying f(0)=0 and  (7.28), then
(8.44)T(x)∶=lim⁡n→∞⁡4n(f(x2n-1)-16f(x2n))
exists for each x∈X and defines a quadratic mapping T:X→Y such that
(8.45)μf(2x)-16f(x)-T(x)(t)≥(2p-4)t(2p-4)t+5(1+2p)θ‖x‖p
for all x∈X and t>0.Proof.
The proof follows from Theorem8.5 by taking
(8.46)Φx,y(t)∶=tt+θ(‖x‖p+‖y‖p)
for all x,y∈X. Then we can choose α=2-p, and we get the desired result.Theorem 8.7.
LetX be a linear space, let (Y,μ,TM) be a complete RN-space, and let Φ be a mapping from X2 to D+  (Φ(x,y) is denoted by Φx,y) such that, for some 0<α<4,
(8.47)Φx,y(αt)≥Φx/2,y/2(t)(x,y∈X,t>0).
Let f:X→Y be an even mapping satisfying f(0)=0 and  (7.2), then
(8.48)T(x)∶=lim⁡n→∞⁡14n(f(2n+1x)-16f(2nx))
exists for each x∈X and defines a quadratic mapping T:X→Y such that
(8.49)μf(2x)-16f(x)-T(x)(t)≥TM(Φx,x(4-α5t),Φ2x,x(4-α5t))
for all x∈X and t>0.Proof.
Let(S,d) be the generalized metric space defined in the proof of Theorem 7.1.
It is easy to see that the linear mappingJ:S→S such that
(8.50)Jh(x)∶=14h(2x)
for all x∈X is a strictly contractive self-mapping with the Lipschitz constant α/4.
Lettingg(x)∶=f(2x)-16f(x) for all x∈X, from (8.36), we get
(8.51)μg(x)-1/4g(2x)(54t)≥TM(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0. So, d(g,Jg)≤5/4.
By Theorem1.1, there exists a mapping T:X→Y satisfying the following:(1)T is a fixed point of J, that is,
(8.52)T(2x)=4T(x)
for all x∈X. Since g:X→Y is even with g(0)=0, T:X→Y is an even mapping with T(0)=0. The mapping T is a unique fixed point of J in the set
(8.53)M={h∈S:d(h,g)<∞}.
This implies that T is a unique mapping satisfying (8.52) such that there exists a u∈(0,∞) satisfying
(8.54)μg(x)-T(x)(ut)≥TM(Φx,x(t),Φ2x,x(t))
for all x∈X and t>0.(2)d(Jng,T)→0 as n→∞. This implies the equality
(8.55)lim⁡n→∞⁡14ng(2nx)=T(x)
for all x∈X.(3)d(h,T)≤(1/(1-α/4))d(h,Jh) for each h∈M, which implies the inequality
(8.56)d(g,T)≤5/(4-α).
This implies that the inequality (8.49) holds.
Proceeding as in the proof of Theorem  2.3, we obtain that the mappingQ:X→Y satisfies (1.4). Now, we have
(8.57)T(2x)-4T(x)=lim⁡n→∞⁡[14ng(2n+1x)-14n-1g(2nx)]=4lim⁡n→∞⁡[14n+1g(2n+1x)-14ng(2nx)]=0
for all x∈X. Since the mapping x→T(2x)-16T(x) is quadratic, we get that the mapping T:X→Y is quadratic.Corollary 8.8.
Letθ≥0 and let p be a real number with 0<p<2. Let X be a normed vector space with norm ∥·∥. Let f:X→Y be an even mapping satisfying f(0)=0 and  (7.28). Then
(8.58)T(x)∶=lim⁡n→∞⁡14n(f(2n+1x)-16f(2nx))
exists for each x∈X and defines a quadratic mapping T:X→Y such that
(8.59)μf(2x)-16f(x)-T(x)(t)≥(4-2p)t(4-2p)t+5(1+2p)θ‖x‖p
for all x∈X and t>0, where (X,μ,TM) is a complete LRN-space in which L=[0,1].Proof.
The proof follows from Theorem8.5 by taking
(8.60)Φx,y(t)∶=tt+θ(‖x‖p+‖y‖p)
for all x,y∈X and t>0. Then we can choose α=2p, and we get the desired result.



---

*Source: 902931-2012-04-05.xml*