Abstract

We prove the generalized Hyers-Ulam stability of the following additive-quadratic-cubic-quartic functional equation 𝑓(𝑥+2𝑦)+𝑓(𝑥2𝑦)=4𝑓(𝑥+𝑦)+4𝑓(𝑥𝑦)6𝑓(𝑥)+𝑓(2𝑦)+𝑓(2𝑦)4𝑓(𝑦)4𝑓(𝑦) in various complete random normed spaces.

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 𝑓(𝑥+𝑦)+𝑓(𝑥𝑦)=2𝑓(𝑥)+2𝑓(𝑦)(1.1)

is called a quadratic 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 𝑓𝑋𝑌, 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 [812]).

In [13], Jun and Kim consider the following cubic functional equation:𝑓(2𝑥+𝑦)+𝑓(2𝑥𝑦)=2𝑓(𝑥+𝑦)+2𝑓(𝑥𝑦)+12𝑓(𝑥).(1.2)

It is easy to show that the function 𝑓(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 equation𝑓(2𝑥+𝑦)+𝑓(2𝑥𝑦)=4𝑓(𝑥+𝑦)+4𝑓(𝑥𝑦)+24𝑓(𝑥)6𝑓(𝑦).(1.3) 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 𝑓𝑋𝑌 satisfies the additive-quadratic-cubic-quadratic functional equation 𝑓(𝑥+2𝑦)+𝑓(𝑥2𝑦)=4𝑓(𝑥+𝑦)+4𝑓(𝑥𝑦)6𝑓(𝑥)+𝑓(2𝑦)+𝑓(2𝑦)4𝑓(𝑦)4𝑓(𝑦)(1.4) if and only if it is an additive-cubic mapping, that is, 𝑓(𝑥+2𝑦)+𝑓(𝑥2𝑦)=4𝑓(𝑥+𝑦)+4𝑓(𝑥𝑦)6𝑓(𝑥).(1.5)

It was shown in Lemma  2.2 of [14] that 𝑔(𝑥)=𝑓(2𝑥)2𝑓(𝑥) and (𝑥)=𝑓(2𝑥)8𝑓(𝑥) are cubic and additive, respectively, and that 𝑓(𝑥)=(1/6)𝑔(𝑥)(1/6)(𝑥).

One can easily show that an even mapping 𝑓𝑋𝑌 satisfies (1.4) if and only if it is a quadratic-quartic mapping, that is, 𝑓(𝑥+2𝑦)+𝑓(𝑥2𝑦)=4𝑓(𝑥+𝑦)+4𝑓(𝑥𝑦)6𝑓(𝑥)+2𝑓(2𝑦)8𝑓(𝑦).(1.6)

Also 𝑔(𝑥)=𝑓(2𝑥)4𝑓(𝑥) and (𝑥)=𝑓(2𝑥)16𝑓(𝑥) are quartic and quadratic, respectively, and 𝑓(𝑥)=(1/12)𝑔(𝑥)(1/12)(𝑥).

For a given mapping 𝑓𝑋𝑌, we define 𝐷𝑓(𝑥,𝑦)=𝑓(𝑥+2𝑦)+𝑓(𝑥2𝑦)4𝑓(𝑥+𝑦)4𝑓(𝑥𝑦)+6𝑓(𝑥)𝑓(2𝑦)𝑓(2𝑦)+4𝑓(𝑦)+4𝑓(𝑦)(1.7)

for all 𝑥,𝑦𝑋.

Let 𝑋 be a set. A function 𝑑𝑋×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,𝑦𝑋,(3)𝑑(𝑥,𝑧)𝑑(𝑥,𝑦)+𝑑(𝑦,𝑧) for all 𝑥,𝑦,𝑧𝑋.

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 𝐽𝑋𝑋 be a strictly contractive mapping with Lipschitz constant 𝐿<1, then for each given element 𝑥𝑋, either 𝑑𝐽𝑛𝑥,𝐽𝑛+1𝑥=(1.8) for all nonnegative integers 𝑛 or there exists a positive integer 𝑛0 such that(1)𝑑(𝐽𝑛𝑥,𝐽𝑛+1𝑥)< for all 𝑛𝑛0,(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={𝑦𝑋𝑑(𝐽𝑛0𝑥,𝑦)<},(4)𝑑(𝑦,𝑦)(1/(1𝐿))𝑑(𝑦,𝐽𝑦) for all 𝑦𝑌.

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 [1821]).

2. Preliminaries

In the sequel, we adopt the usual terminology, notations, and conventions of the theory of random normed spaces, as in [2226]. Throughout this paper, Δ+ is the space of all probability distribution functions, that is, the space of all mappings 𝐹{,+}[0,1], such taht 𝐹 is left continuous, nondecreasing on , 𝐹(0)=0 and {𝐹(+)=1}. 𝐷+ is a subset of Δ+ consisting of all functions 𝐹Δ+ for which 𝑙𝐹(+)=1, where 𝑙𝑓(𝑥) denotes the left limit of the function 𝑓 at the point 𝑥, that is, 𝑙𝑓(𝑥)=lim𝑡𝑥𝑓(𝑡). The space Δ+ is partially ordered by the usual pointwise ordering of functions, that is, 𝐹𝐺 if and only if 𝐹(𝑡)𝐺(𝑡) for all 𝑡 in . The maximal element for Δ+ in this order is the distribution function 𝜀0 given by𝜀0(𝑡)=0,if𝑡0,1,if𝑡>0.(2.1)

A triangular 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 𝑏𝑐 (monotonicity).

Basic examples are the Łukasiewicz  𝑡-norm  𝑇𝐿,𝑇𝐿(𝑎,𝑏)=max(𝑎+𝑏1,0) for all 𝑎,𝑏[0,1] and the 𝑡-norms  𝑇𝑃,𝑇𝑀,𝑇𝐷, where   𝑇𝑃(𝑎,𝑏)=𝑎𝑏, 𝑇𝑀(𝑎,𝑏)=min{𝑎,𝑏}, 𝑇𝐷(𝑎,𝑏)=min(𝑎,𝑏),ifmax(𝑎,𝑏)=1,0,otherwise.(2.2)

If 𝑇 is a 𝑡-norm, then 𝑥𝑇(𝑛) is defined for every 𝑥[0,1] and 𝑛𝑁{0} by 1, if 𝑛=0 and 𝑇(𝑥𝑇(𝑛1),𝑥) if 𝑛1. A 𝑡-norm𝑇 is said to be of Hadžić type (we denote by 𝑇) if the family (𝑥𝑇(𝑛))𝑛𝑁 is equicontinuous at 𝑥=1 (cf. [27]).

Other important triangular norms are the following (see [28]):(1)The Sugeno-Weber family {𝑇SW𝜆}𝜆[1,] is defined by 𝑇SW1=𝑇𝐷, 𝑇SW=𝑇𝑃 and 𝑇SW𝜆(𝑥,𝑦)=max0,𝑥+𝑦1+𝜆𝑥𝑦1+𝜆(2.3) if  𝜆(1,).(2)The Domby family {𝑇𝐷𝜆}𝜆[0,] is defined by 𝑇𝐷 if 𝜆=0, 𝑇𝑀 if 𝜆=, and 𝑇𝐷𝜆1(𝑥,𝑦)=1+((1𝑥)/𝑥)𝜆+((1𝑦)/𝑦)𝜆1/𝜆(2.4) if 𝜆(0,).(3)The Aczel-Alsina family {𝑇AA𝜆}𝜆[0,] is defined by 𝑇𝐷 if 𝜆=0, 𝑇𝑀 if 𝜆= and 𝑇AA𝜆(𝑥,𝑦)=𝑒(|log𝑥|𝜆+|log𝑦|𝜆)1/𝜆(2.5) if 𝜆(0,).

A 𝑡-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 𝑇0𝑖=1𝑥𝑖=1,𝑇𝑛𝑖=1𝑥𝑖𝑇=𝑇𝑛1𝑖=1𝑥𝑖,𝑥𝑛𝑥=𝑇1,,𝑥𝑛.(2.6)

𝑇 can also be extended to a countable operation taking for any sequence (𝑥𝑛)𝑛𝑁 in [0,1] the value𝑇𝑖=1𝑥𝑖=lim𝑛𝑇𝑛𝑖=1𝑥𝑖.(2.7) The limit on the right side of (6.4) exists since the sequence (𝑇𝑛𝑖=1𝑥𝑖)𝑛 is nonincreasing and bounded from below.

Proposition 2.1 (see [28]). We have the following.(1)For 𝑇𝑇𝐿, the following implication holds: lim𝑛T𝑖=1𝑥𝑛+𝑖=1𝑛=11𝑥𝑛<.(2.8)(2)If 𝑇 is of Hadžić type, then lim𝑛T𝑖=1𝑥𝑛+𝑖=1(2.9) for every sequence (𝑥𝑛)𝑛𝑁 in [0,1] such that lim𝑛𝑥𝑛=1.(3)If 𝑇{𝑇AA𝜆}𝜆(0,){𝑇𝐷𝜆}𝜆(0,), then lim𝑛T𝑖=1𝑥𝑛+𝑖=1𝑛=11𝑥𝑛𝛼<.(2.10)(4)If 𝑇{𝑇SW𝜆}𝜆[1,), then lim𝑛T𝑖=1𝑥𝑛+𝑖=1𝑛=11𝑥𝑛<.(2.11)

Definition 2.2 (see [26]). A Random 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)=𝜀0(𝑡) for all 𝑡>0 if and only if 𝑥=0,(RN2) 𝜇𝛼𝑥(𝑡)=𝜇𝑥(𝑡/|𝛼|) for all 𝑥𝑋, and 𝛼0,(RN3) 𝜇𝑥+𝑦(𝑡+𝑠)𝑇(𝜇𝑥(𝑡),𝜇𝑦(𝑠)) for all 𝑥,𝑦𝑋 and 𝑡,𝑠0.

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 𝜇𝑥𝑛𝑥(𝜖)>1𝜆 whenever 𝑛𝑁.(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 𝜇𝑥𝑛𝑥𝑚(𝜖)>1𝜆 whenever 𝑛𝑚𝑁.(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 𝑥𝑛𝑥, then lim𝑛𝜇𝑥𝑛(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, 2939].

3. Non-Archimedean Random Normed Space

By a non-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,𝑠𝒦. Clearly, |1|=|1|=1 and |𝑛|1 for all 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 any 𝑟𝒦 and 𝑥𝑋, 𝑟𝑥=|𝑟|𝑥,(NAN3) the strong triangle inequality (ultrametric), namely, 𝑥+𝑦max{𝑥,𝑦}(𝑥,𝑦𝑋),(3.1)

then (𝑋,) is called a non-Archimedean normed space. Due to the fact that 𝑥𝑛𝑥𝑚𝑥max𝑗+1𝑥𝑗𝑚𝑗𝑛1(𝑛>𝑚),(3.2)

a sequence {𝑥𝑛} is a Cauchy sequence if and only if {𝑥𝑛+1𝑥𝑛} 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 such that 𝑥=(𝑎/𝑏)𝑝𝑛𝑥, where 𝑎 and 𝑏 are integers not divisible by 𝑝. Then |𝑥|𝑝=𝑝𝑛𝑥 defines a non-Archimedean norm on . The completion of with respect to the metric 𝑑(𝑥,𝑦)=|𝑥𝑦|𝑝 is denoted by 𝑝, which is called the 𝑝-adic number field.

Throughout the paper, we assume that 𝑋 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)=𝜀0(𝑡) for all 𝑡>0 if and only if 𝑥=0,(NA-RN2) 𝜇𝛼𝑥(𝑡)=𝜇𝑥(𝑡/|𝛼|) for all 𝑥𝑋, 𝑡>0, and 𝛼0,(NA-RN3) 𝜇𝑥+𝑦(max{𝑡,𝑠})𝑇(𝜇𝑥(𝑡),𝜇𝑦(𝑠)) for all 𝑥,𝑦,𝑧𝑋 and 𝑡,𝑠0.
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 𝜇𝑥(𝑡)=0,𝑡𝑥,1,𝑡>𝑥,(3.3)

is a non-Archimedean RN-space.

Example 3.2. Let (𝑋,) be a non-Archimedean normed linear space. Define 𝜇𝑥(𝑡𝑡)=(𝑡+𝑥𝑥𝑋,𝑡>0),(3.4) 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 𝑥𝑋 such that lim𝑛𝜇𝑥𝑛𝑥(𝑡)=1(3.5) 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 sequence if for each 𝜀>0 and each 𝑡>0 there exists 𝑛0 such that for all 𝑛𝑛0 and all 𝑝>0, we have 𝜇𝑥𝑛+𝑝𝑥𝑛(𝑡)>1𝜀.

If each Cauchy sequence is convergent, then the random norm is said to be complete 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 𝜇𝑥𝑛+𝑝𝑥𝑛𝜇(𝑡)min𝑥𝑛+𝑗+1𝑥𝑛+𝑗(𝑡)𝑗=0,1,2,,𝑝1.(3.6) So, the sequence {𝑥𝑛} is a Cauchy sequence if for each 𝜀>0 and 𝑡>0 there exists 𝑛0 such that for all 𝑛𝑛0, 𝜇𝑥𝑛+1𝑥𝑛(𝑡)>1𝜀.(3.7)

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

Let 𝒦 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. Let Ψ be a distribution function on 𝑋×𝑋×[0,) such that Ψ(𝑥,𝑦,) is nondecreasing and 𝑡Ψ(𝑐𝑥,𝑐𝑥,𝑡)Ψ𝑥,𝑥,|𝑐|(𝑥𝑋,𝑐0).(4.1)

Definition 4.1. A mapping 𝑓𝑋𝑌 is said to be Ψ-approximately AQCQ if 𝜇𝐷𝑓(𝑥,𝑦)(𝑡)Ψ(𝑥,𝑦,𝑡)(𝑥,𝑦𝑋,𝑡>0).(4.2)

In this section, we assume that 20 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 equation 𝐷𝑓(𝑥,𝑦)=0 in non-Archimedean random spaces, an odd case.

Theorem 4.2. Let 𝒦 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 𝑓𝑋𝑌 be an odd mapping and Ψ-approximately AQCQ mapping. If for some 𝛼, 𝛼>0, and some integer 𝑘, 𝑘>3 with |2𝑘|<𝛼, Ψ2𝑘𝑥,2𝑘𝑦,𝑡Ψ(𝑥,𝑦,𝛼𝑡)(𝑥𝑋,𝑡>0),(4.3)lim𝑛T𝑗=𝑛𝑀𝛼2𝑥,𝑗𝑡||8||𝑘𝑗=1(𝑥𝑋,𝑡>0),(4.4) then there exists a unique cubic mapping 𝐶𝑋𝑌 such that 𝜇𝑓(𝑥)2𝑓(𝑥/2)𝐶(𝑥/2)(𝑡)𝑇𝑖=1𝑀𝛼𝑥,𝑖+1𝑡||8||𝑘𝑖(4.5) for all 𝑥𝑋 and 𝑡>0, where 𝑀(𝑥,𝑡)=𝑇𝑘1Ψ𝑥2,𝑥2,𝑡||4||𝑥,Ψ𝑥,22,𝑡,,Ψ𝑘1𝑥2,2𝑘1𝑥2,𝑡||4||2,Ψ𝑘12𝑥,𝑘1𝑥2,𝑡(𝑥𝑋,𝑡>0).(4.6)

Proof. Letting 𝑥=𝑦 in (4.2), we get 𝜇𝑓(3𝑦)4𝑓(2𝑦)+5𝑓(𝑦)(𝑡)Ψ(𝑦,𝑦,𝑡)(4.7) for all 𝑦𝑋 and 𝑡>0. Replacing 𝑥 by 2𝑦 in (4.2), we get 𝜇𝑓(4𝑦)4𝑓(3𝑦)+6𝑓(2𝑦)4𝑓(𝑦)(𝑡)Ψ(2𝑦,𝑦,𝑡)(4.8) for all 𝑦𝑋 and 𝑡>0. By (4.7) and (4.8), we have 𝜇𝑓(4𝑦)10𝑓(2𝑦)+16𝑓(𝑦)𝜇(𝑡)𝑇4(𝑓(3𝑦)4𝑓(2𝑦)+5𝑓(𝑦))(𝑡),𝜇𝑓(4𝑦)4𝑓(3𝑦)+6𝑓(2𝑦)4𝑓(𝑦)𝜇(𝑡)=𝑇𝑓(3𝑦)4𝑓(2𝑦)+5𝑓(𝑦)𝑡||4||,𝜇𝑓(4𝑦)4𝑓(3𝑦)+6𝑓(2𝑦)4𝑓(𝑦)Ψ𝑡(𝑡)𝑇𝑦,𝑦,||4||,Ψ(2𝑦,𝑦,𝑡)(4.9) for all 𝑦𝑋 and 𝑡>0. Letting 𝑦=𝑥/2 and 𝑔(𝑥)=𝑓(2𝑥)2𝑓(𝑥) for all 𝑥𝑋 in (4.9), we get 𝜇𝑔(𝑥)8𝑔(𝑥/2)Ψ𝑥(𝑡)𝑇2,𝑥2,𝑡||4||𝑥,Ψ𝑥,2,𝑡(4.10) for all 𝑥𝑋 and 𝑡>0. Now, we show by induction on 𝑗 that for all 𝑥𝑋, 𝑡>0 and 𝑗1, 𝜇𝑔(2𝑗1𝑥)8𝑗𝑔(𝑥/2)(𝑡)𝑀𝑗(𝑥,𝑡)=𝑇2𝑗1Ψ𝑥2,𝑥2,𝑡||4||𝑥,Ψ𝑥,22,𝑡,,Ψ𝑗1𝑥2,2𝑗1𝑥2,𝑡||4||2,Ψ𝑗12𝑥,𝑗1𝑥2.,𝑡(4.11) Putting 𝑗=1 in (4.11), we obtain (4.10). Assume that (4.11) holds for some 𝑗1. Replacing 𝑥 by 2𝑗𝑥 in (4.10), we get 𝜇𝑔(2𝑗𝑥)8𝑔(2𝑗1𝑥)Ψ2(𝑡)𝑇𝑗1𝑥,2𝑗1𝑡𝑥,||4||2,Ψ𝑗𝑥,2𝑗1𝑥,𝑡.(4.12) Since |8|1, 𝜇𝑔(2𝑗𝑥)8𝑗+1𝑔(𝑥/2)𝜇(𝑡)𝑇𝑔(2𝑗𝑥)8𝑔(2𝑗1𝑥)(𝑡),𝜇8𝑔(2𝑗1𝑥)8𝑗+1𝑔(𝑥/2)𝜇(𝑡)=𝑇𝑔(2𝑗𝑥)8𝑔(2𝑗1𝑥)(𝑡),𝜇𝑔(2𝑗1𝑥)8𝑗𝑔(𝑥/2)𝑡||8||𝑇2Ψ2𝑗1𝑥,2𝑗1𝑡𝑥,||4||2,Ψ𝑗𝑥,2𝑗1𝑥,𝑡,𝑀𝑗(𝑥,𝑡)=𝑀𝑗+1(𝑥,𝑡)(4.13)for all 𝑥𝑋 and 𝑡>0. Thus, (4.11) holds for all 𝑗2. In particular, 𝜇𝑔(2𝑘1𝑥)8𝑘𝑔(𝑥/2)(𝑡)𝑀(𝑥,𝑡)(𝑥𝑋,𝑡>0).(4.14) Replacing 𝑥 by 2(𝑘𝑛+𝑘1)𝑥 in (4.14) and using inequality (4.3), we obtain 𝜇𝑔(𝑥/2𝑘𝑛)8𝑘𝑔(𝑥/2𝑘(𝑛+1))(𝑡)𝑀2𝑥2𝑘(𝑛+1),𝑡(𝑥𝑋,𝑡>0,𝑛=0,1,2,).(4.15) Then 𝜇8𝑘𝑛𝑔(𝑥/2𝑘𝑛)8𝑘(𝑛+1)𝑔(𝑥/2𝑘(𝑛+1))𝛼(𝑡)𝑀2𝑥,𝑛+1||8𝑘(𝑛+1)||𝑡(𝑥𝑋,𝑡>0,𝑛=0,1,2,).(4.16) Hence 𝜇8𝑘𝑛𝑔(𝑥/2𝑘𝑛)8𝑘(𝑛+𝑝)𝑔(𝑥/2𝑘(𝑛+𝑝))(𝑡)𝑇𝑛+𝑝𝑗=𝑛𝜇8𝑘𝑗𝑔(𝑥/2𝑘𝑗)8𝑘(𝑗+𝑝)𝑔(𝑥/2𝑘(𝑗+𝑝))(𝑡)𝑇𝑛+𝑝𝑗=𝑛𝑀𝛼2𝑥,𝑗+1|||8𝑘𝑗+1|||𝑡𝑇𝑛+𝑝𝑗=𝑛𝑀𝛼2𝑥,𝑗+1|||8𝑘𝑗+1|||𝑡(𝑥𝑋,𝑡>0,𝑛=0,1,2,).(4.17) Since lim𝑛𝑇𝑗=𝑛𝑀𝛼2𝑥,𝑗+1|||8𝑘𝑗+1|||𝑡=1(𝑥𝑋,𝑡>0),(4.18) then 8𝑘𝑛𝑔𝑥2𝑘𝑛𝑛(4.19)is a Cauchy sequence in the non-Archimedean random Banach space (𝑌,𝜇,𝑇). Hence we can define a mapping 𝐶𝑋𝑌 such that lim𝑛𝜇(88𝑘)𝑛𝑔(𝑥/2𝑘𝑛)𝐶(𝑥)(𝑡)=1(𝑥𝑋,𝑡>0).(4.20)
Next for each 𝑛1, 𝑥𝑋 and 𝑡>0, 𝜇𝑔(𝑥)(88𝑘)𝑛𝑔(𝑥/2𝑘𝑛)(𝑡)=𝜇𝑛1𝑖=0(88𝑘)𝑖𝑔(𝑥/2𝑘𝑖)(88𝑘)𝑖+1𝑔(𝑥/2𝑘(𝑖+1))(𝑡)𝑇𝑛1𝑖=0𝜇(88𝑘)𝑖𝑔(𝑥/2𝑘𝑖)(88𝑘)𝑖+1𝑔(𝑥/2𝑘(𝑖+1))(𝑡)𝑇𝑛1𝑖=0𝑀𝛼2𝑥,𝑖+1𝑡||8𝑘||𝑖+1.(4.21)Therefore, 𝜇𝑔(𝑥)𝐶(𝑥)𝜇(𝑡)𝑇𝑔(𝑥)(88𝑘)𝑛𝑔(𝑥/2𝑘𝑛)(𝑡),𝜇(88𝑘)𝑛𝑔(𝑥/2𝑘𝑛)𝐶(𝑥)𝑇(𝑡)𝑇𝑛1𝑖=0𝑀𝛼2𝑥,𝑖+1𝑡||8𝑘||𝑖+1,𝜇(88𝑘)𝑛𝑔(𝑥/2𝑘𝑛)𝐶(𝑥).(𝑡)(4.22) By letting 𝑛, we obtain 𝜇𝑔(𝑥)𝐶(𝑥)(𝑡)𝑇𝑖=1𝑀𝛼2𝑥,𝑖+1𝑡||8𝑘||𝑖+1.(4.23) So, 𝜇𝑓(𝑥)2𝑓(𝑥/2)𝐶(𝑥/2)(𝑡)𝑇𝑖=1𝑀𝛼𝑥,𝑖+1𝑡||8𝑘||𝑖+1.(4.24) This proves (4.5). From 𝐷𝑔(𝑥,𝑦)=𝐷𝑓(2𝑥,2𝑦)2𝐷𝑓(𝑥,𝑦), by (4.2), we deduce that 𝜇𝐷𝑓(2𝑥,2𝑦)𝜇(𝑡)Ψ(2𝑥,2𝑦,𝑡),2𝐷𝑓(𝑥,𝑦)(𝑡)=𝜇𝐷𝑓(𝑥,𝑦)𝑡||2||𝜇𝐷𝑓(𝑥,𝑦)(𝑡)Ψ(𝑥,𝑦,𝑡),(4.25) and so, by (NA-RN3) and (4.2), we obtain 𝜇𝐷𝑔(𝑥,𝑦)𝜇(𝑡)𝑇𝐷𝑓(2𝑥,2𝑦)(𝑡),𝜇2𝐷𝑓(𝑥,𝑦)(𝑡)𝑇(Ψ(2𝑥,2𝑦,𝑡),Ψ(𝑥,𝑦,𝑡))=𝑁(𝑥,𝑦,𝑡).(4.26) It follows that 𝜇8𝑘𝑛𝐷𝑔(𝑥/2𝑘𝑛,𝑦/2𝑘𝑛)(𝑡)=𝜇𝐷𝑔(𝑥/2𝑘𝑛,𝑦/2𝑘𝑛)𝑡||8||𝑘𝑛𝑥𝑁2𝑘𝑛,𝑦2𝑘𝑛,𝑡||8||𝑘𝑛𝛼𝑁𝑥,𝑦,𝑛1𝑡||8||𝑘(𝑛1)(4.27)for all 𝑥,𝑦𝑋, 𝑡>0, and 𝑛. Since lim𝑛𝑁𝛼𝑥,𝑦,𝑛1𝑡||8||𝑘(𝑛1)=1(4.28)for all 𝑥,𝑦𝑋 and 𝑡>0, by Theorem 2.4, we deduce that 𝜇𝐷𝐶(𝑥,𝑦)(𝑡)=1(4.29)for all 𝑥,𝑦𝑋 and 𝑡>0. Thus, the mapping 𝐶𝑋𝑌 satisfies (1.4).
Now, we have 𝐶(2𝑥)8𝐶(𝑥)=lim𝑛8𝑛𝑔𝑥2𝑛18𝑛+1𝑔𝑥2𝑛=8lim𝑛8𝑛1𝑔𝑥2𝑛18𝑛𝑔𝑥2𝑛=0(4.30)for all 𝑥𝑋. Since the mapping 𝑥𝐶(2𝑥)2𝐶(𝑥) is cubic (see Lemma  2.2 of [14]), from the equality 𝐶(2𝑥)=8𝐶(𝑥), we deduce that the mapping 𝐶𝑋𝑌 is cubic.

Corollary 4.3. Let 𝒦 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. Let 𝑓𝑋𝑌 be an odd and Ψ-approximately AQCQ mapping. If, for some 𝛼, 𝛼>0, and some integer 𝑘, 𝑘>3, with |2𝑘|<𝛼, Ψ2𝑘𝑥,2𝑘𝑦,𝑡Ψ(𝑥,𝑦,𝛼𝑡)(𝑥𝑋,𝑡>0),(4.31) then there exists a unique cubic mapping 𝐶𝑋𝑌 such that 𝜇𝑓(𝑥)2𝑓(𝑥/2)𝐶(𝑥/2)(𝑡)𝑇𝑖=1𝑀𝛼𝑥,𝑖+1𝑡||8||𝑘𝑖(4.32) for all 𝑥𝑋 and 𝑡>0.

Proof. Since lim𝑛𝑀𝛼𝑥,𝑗𝑡||8||𝑘𝑗=1(𝑥𝑋,𝑡>0)(4.33) and 𝑇 is of Hadžić type, from Proposition 2.1, it follows that lim𝑛𝑇𝑗=𝑛𝑀𝛼𝑥,𝑗𝑡||8||𝑘𝑗=1(𝑥𝑋,𝑡>0).(4.34)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 𝜇𝑥(𝑡𝑡)=(𝑡+𝑥𝑥𝑋,𝑡>0).(4.35) And let (𝑌,𝜇,𝑇𝑀) be a complete non-Archimedean random normed space (see Example 3.2). Define 𝑡Ψ(𝑥,𝑦,𝑡)=.1+𝑡(4.36) It is easy to see that (4.3) holds for 𝛼=1. Also, since 𝑡𝑀(𝑥,𝑡)=,1+𝑡(4.37) we have lim𝑛𝑇𝑀,𝑗=𝑛𝑀𝛼𝑥,𝑗𝑡||8||𝑘𝑗=lim𝑛lim𝑚𝑇𝑚𝑀,𝑗=𝑛𝑀𝑡𝑥,||8||𝑘𝑗=lim𝑛lim𝑚𝑡||8𝑡+𝑘||𝑛=1(𝑥𝑋,𝑡>0).(4.38) Let 𝑓𝑋𝑌 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 𝐶𝑋𝑌 such that 𝜇𝑓(𝑥)2𝑓(𝑥/2)𝐶(𝑥/2)(𝑡𝑡)||8𝑡+𝑘||.(4.39)

Theorem 4.5. Let 𝒦 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 𝑓𝑋𝑌 be an odd mapping and Ψ-approximately AQCQ mapping. If for some 𝛼, 𝛼>0, and some integer 𝑘, 𝑘>1 with |2𝑘|<𝛼, Ψ2𝑘𝑥,2𝑘𝑦,𝑡Ψ(𝑥,𝑦,𝛼𝑡)(𝑥𝑋,𝑡>0),lim𝑛𝑇𝑗=𝑛𝑀𝛼2𝑥,𝑗𝑡||2||𝑘𝑗=1(𝑥𝑋,𝑡>0),(4.40) then there exists a unique additive mapping 𝐴𝑋𝑌 such that 𝜇𝑓(𝑥)8𝑓(𝑥/2)𝐴(𝑥/2)(𝑡)𝑇𝑖=1𝑀𝛼𝑥,𝑖+1𝑡||2||𝑘𝑖(4.41) for all 𝑥𝑋 and 𝑡>0, where 𝑀(𝑥,𝑡)=𝑇𝑘1Ψ𝑥2,𝑥2,𝑡||4||𝑥,Ψ𝑥,22,𝑡,,Ψ𝑘1𝑥2,2𝑘1𝑥2,𝑡||4||2,Ψ𝑘12𝑥,𝑘1𝑥2,𝑡(𝑥𝑋,𝑡>0)(4.42)

Proof. Letting 𝑦=𝑥/2 and 𝑔(𝑥)=𝑓(2𝑥)8𝑓(𝑥) for all 𝑥𝑋 in (4.9), we get 𝜇𝑔(𝑥)2𝑔(𝑥/2)Ψ𝑥(𝑡)𝑇2,𝑥2,𝑡||4||𝑥,Ψ𝑥,2,𝑡(4.43) for all 𝑥𝑋 and 𝑡>0.
The rest of the proof is similar to the proof of Theorem 4.2.

Corollary 4.6. Let 𝒦 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. Let 𝑓𝑋𝑌 be an odd and Ψ-approximately AQCQ mapping. If, for some 𝛼,𝛼>0, and some integer 𝑘,𝑘>1, with |2𝑘|<𝛼, Ψ2𝑘𝑥,2𝑘𝑦,𝑡Ψ(𝑥,𝑦,𝛼𝑡)(𝑥𝑋,𝑡>0),(4.44) then there exists a unique additive mapping 𝐴𝑋𝑌 such that 𝜇𝑓(𝑥)8𝑓(𝑥/2)𝐴(𝑥/2)(𝑡)𝑇𝑖=1𝑀𝛼𝑥,𝑖+1𝑡||2||𝑘𝑖(4.45) for all 𝑥𝑋 and 𝑡>0.

Proof. Since lim𝑛𝑀𝛼𝑥,𝑗𝑡||2||𝑘𝑗=1(𝑥𝑋,𝑡>0)(4.46) and 𝑇 is of Hadžić type, from Proposition 2.1, it follows that lim𝑛𝑇𝑗=𝑛𝑀𝛼𝑥,𝑗𝑡||2||𝑘𝑗=1(𝑥𝑋,𝑡>0).(4.47)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 𝜇𝑥=𝑡(𝑡)𝑡+𝑥(𝑥𝑋,𝑡>0),(4.48)and let (𝑌,𝜇,𝑇𝑀) be a complete non-Archimedean random normed space (see Example 3.2). Define 𝑡Ψ(𝑥,𝑦,𝑡)=.1+𝑡(4.49) It is easy to see that (4.3) holds for 𝛼=1. Also, since 𝑡𝑀(𝑥,𝑡)=,1+𝑡(4.50) we have lim𝑛𝑇𝑀,𝑗=𝑛𝑀𝛼𝑥,𝑗𝑡||2||𝑘𝑗=lim𝑛lim𝑚𝑇𝑚𝑀,𝑗=𝑛𝑀𝑡𝑥,||2||𝑘𝑗=lim𝑛lim𝑚𝑡||2𝑡+𝑘||𝑛=1(𝑥𝑋,𝑡>0).(4.51) Let 𝑓𝑋𝑌 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 𝐴𝑋𝑌 such that 𝜇𝑓(𝑥)8𝑓(𝑥/2)𝐴(𝑥/2)(𝑡𝑡)||2𝑡+𝑘||.(4.52)

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 equation 𝐷𝑓(𝑥,𝑦)=0 in non-Archimedean Banach spaces, an even case.

Theorem 5.1. Let 𝒦 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 𝑓𝑋𝑌 be an even mapping, 𝑓(0)=0, and Ψ-approximately AQCQ mapping. If for some 𝛼, 𝛼>0, and some integer 𝑘, 𝑘>4 with |2𝑘|<𝛼, Ψ2𝑘𝑥,2𝑘𝑦,𝑡Ψ(𝑥,𝑦,𝛼𝑡)(𝑥𝑋,𝑡>0),lim𝑛T𝑗=𝑛𝑀𝛼2𝑥,𝑗𝑡||||16𝑘𝑗=1(𝑥𝑋,𝑡>0),(5.1) then there exists a unique quartic mapping 𝑄𝑋𝑌 such that 𝜇𝑓(𝑥)4𝑓(𝑥/2)𝑄(𝑥/2)(𝑡)T𝑖=1𝑀𝛼𝑥,𝑖+1𝑡||||16𝑘𝑖(5.2) for all 𝑥𝑋 and 𝑡>0, where 𝑀(𝑥,𝑡)=𝑇𝑘1Ψ𝑥2,𝑥2,𝑡||4||𝑥,Ψ𝑥,22,𝑡,,Ψ𝑘1𝑥2,2𝑘1𝑥2,𝑡||4||2,Ψ𝑘12𝑥,𝑘1𝑥2,𝑡(𝑥𝑋,𝑡>0).(5.3)

Proof. Letting 𝑥=𝑦 in (4.2), we get 𝜇𝑓(3𝑦)6𝑓(2𝑦)+15𝑓(𝑦)(𝑡)Ψ(𝑦,𝑦,𝑡)(5.4) for all 𝑦𝑋 and 𝑡>0. Replacing 𝑥 by 2𝑦 in (4.2), we get 𝜇𝑓(4𝑦)4𝑓(3𝑦)+4𝑓(2𝑦)+4𝑓(𝑦)(𝑡)Ψ(2𝑦,𝑦,𝑡)(5.5) for all 𝑦𝑋 and 𝑡>0. By (5.4) and (5.5), we have 𝜇𝑓(4𝑦)20𝑓(2𝑦)+64𝑓(𝑦)𝜇(𝑡)𝑇4(𝑓(3𝑦)4𝑓(2𝑦)+5𝑓(𝑦))(𝑡),𝜇𝑓(4𝑦)4𝑓(3𝑦)+6𝑓(2𝑦)4𝑓(𝑦)𝜇(𝑡)=𝑇𝑓(3𝑦)4𝑓(2𝑦)+5𝑓(𝑦)𝑡||4||,𝜇𝑓(4𝑦)4𝑓(3𝑦)+6𝑓(2𝑦)4𝑓(𝑦)Ψ𝑡(𝑡)𝑇𝑦,𝑦,||4||,Ψ(2𝑦,𝑦,𝑡)(5.6) for all 𝑦𝑋 and 𝑡>0. Letting 𝑦=𝑥/2 and 𝑔(𝑥)=𝑓(2𝑥)4𝑓(𝑥) for all 𝑥𝑋 in (5.6), we get 𝜇𝑔(𝑥)16𝑔(𝑥/2)Ψ𝑥(𝑡)𝑇2,𝑥2,𝑡||4||𝑥,Ψ𝑥,2,𝑡(5.7) for all 𝑥𝑋 and 𝑡>0.
The rest of the proof is similar to the proof of Theorem 4.2.

Corollary 5.2. Let 𝒦 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. Let 𝑓𝑋𝑌 be an even, 𝑓(0)=0, and Ψ-approximately AQCQ mapping. If,   for some 𝛼, 𝛼>0, and some integer 𝑘, 𝑘>4, with |2𝑘|<𝛼, Ψ2𝑘𝑥,2𝑘𝑦,𝑡Ψ(𝑥,𝑦,𝛼𝑡)(𝑥𝑋,𝑡>0),(5.8) then there exists a unique quartic mapping 𝑄𝑋𝑌 such that 𝜇𝑓(𝑥)4𝑓(𝑥/2)𝑄(𝑥/2)(𝑡)T𝑖=1𝑀𝛼𝑥,𝑖+1𝑡||||16𝑘𝑖(5.9) for all 𝑥𝑋 and 𝑡>0.

Proof. Since lim𝑛𝑀𝛼𝑥,𝑗𝑡||||16𝑘𝑗=1(𝑥𝑋,𝑡>0)(5.10) and 𝑇 is of Hadžić type, from Proposition 2.1, it follows that lim𝑛𝑇𝑗=𝑛𝑀𝛼𝑥,𝑗𝑡||||16𝑘𝑗=1(𝑥𝑋,𝑡>0).(5.11) 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 𝜇𝑥(𝑡𝑡)=(𝑡+𝑥𝑥𝑋,𝑡>0).(5.12) And let (𝑌,𝜇,𝑇𝑀) be a complete non-Archimedean random normed space (see Example 3.2). Define 𝑡Ψ(𝑥,𝑦,𝑡)=.1+𝑡(5.13) It is easy to see that (4.3) holds for 𝛼=1. Also, since 𝑡𝑀(𝑥,𝑡)=,1+𝑡(5.14) we have lim𝑛𝑇𝑀,𝑗=𝑛𝑀𝛼𝑥,𝑗𝑡||||16𝑘𝑗=lim𝑛lim𝑚𝑇𝑚𝑀,𝑗=𝑛𝑀𝑡𝑥,||||16𝑘𝑗=lim𝑛lim𝑚𝑡||𝑡+16𝑘||𝑛=1(𝑥𝑋,𝑡>0).(5.15) Let 𝑓𝑋𝑌 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 𝑄𝑋𝑌 such that 𝜇𝑓(𝑥)4𝑓(𝑥/2)𝑄(𝑥/2)(𝑡𝑡)||𝑡+16𝑘||.(5.16)

Theorem 5.4. Let 𝒦 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 𝑓𝑋𝑌 be an even mapping, 𝑓(0)=0 and Ψ-approximately AQCQ mapping. If for some 𝛼, 𝛼>0, and some integer 𝑘, 𝑘>2 with |2𝑘|<𝛼, Ψ2𝑘𝑥,2𝑘𝑦,𝑡Ψ(𝑥,𝑦,𝛼𝑡)(𝑥𝑋,𝑡>0),lim𝑛T𝑗=𝑛𝑀𝛼2𝑥,𝑗𝑡||4||𝑘𝑗=1(𝑥𝑋,𝑡>0),(5.17) then there exists a unique quadratic mapping 𝑄𝑋𝑌 such that 𝜇𝑓(𝑥)16𝑓(𝑥/2)𝑄(𝑥/2)(𝑡)T𝑖=1𝑀𝛼𝑥,𝑖+1𝑡||4||𝑘𝑖(5.18) for all 𝑥𝑋 and 𝑡>0, where 𝑀(𝑥,𝑡)=𝑇𝑘1Ψ𝑥2,𝑥2,𝑡||4||𝑥,Ψ𝑥,22,𝑡,,Ψ𝑘1𝑥2,2𝑘1𝑥2,𝑡||4||2,Ψ𝑘12𝑥,𝑘1𝑥2,𝑡(𝑥𝑋,𝑡>0).(5.19)

Proof. Letting 𝑦=𝑥/2 and 𝑔(𝑥)=𝑓(2𝑥)16𝑓(𝑥) for all 𝑥𝑋 in (5.6), we get 𝜇𝑔(𝑥)4𝑔(𝑥/2)Ψ𝑥(𝑡)𝑇2,𝑥2,𝑡||4||𝑥,Ψ𝑥,2,𝑡(5.20) for all 𝑥𝑋 and 𝑡>0.
The rest of the proof is similar to the proof of Theorem 5.1.

Corollary 5.5. Let 𝒦 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. Let 𝑓𝑋𝑌 be an even, 𝑓(0)=0, and Ψ-approximately AQCQ mapping. If,   for some 𝛼, 𝛼>0, and some integer 𝑘, 𝑘>2, with |2𝑘|<𝛼, Ψ2𝑘𝑥,2𝑘𝑦,𝑡Ψ(𝑥,𝑦,𝛼𝑡)(𝑥𝑋,𝑡>0),(5.21) then there exists a unique quadratic mapping 𝑄𝑋𝑌 such that 𝜇𝑓(𝑥)16𝑓(𝑥/2)𝑄(𝑥/2)(𝑡)T𝑖=1𝑀𝛼𝑥,𝑖+1𝑡||4||𝑘𝑖(5.22) for all 𝑥𝑋 and 𝑡>0.

Proof. Since lim𝑛𝑀𝛼𝑥,𝑗𝑡||4||𝑘𝑗=1(𝑥𝑋,𝑡>0)(5.23) and 𝑇 is of Hadžić type, from Proposition 2.1, it follows that lim𝑛𝑇𝑗=𝑛𝑀𝛼𝑥,𝑗𝑡||4||𝑘𝑗=1(𝑥𝑋,𝑡>0).(5.24)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 𝜇𝑥(𝑡𝑡)=(𝑡+𝑥𝑥𝑋,𝑡>0).(5.25) And let (𝑌,𝜇,𝑇𝑀) be a complete non-Archimedean random normed space (see Example 3.2). Define 𝑡Ψ(𝑥,𝑦,𝑡)=.1+𝑡(5.26) It is easy to see that (4.3) holds for 𝛼=1. Also, since 𝑡𝑀(𝑥,𝑡)=,1+𝑡(5.27) we have lim𝑛𝑇𝑀,𝑗=𝑛𝑀𝛼𝑥,𝑗𝑡||4||𝑘𝑗=lim𝑛lim𝑚𝑇𝑚𝑀,𝑗=𝑛𝑀𝑡𝑥,||4||𝑘𝑗=lim𝑛lim𝑚𝑡||4𝑡+𝑘||𝑛=1(𝑥𝑋,𝑡>0).(5.28) Let 𝑓𝑋𝑌 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 𝑄𝑋𝑌 such that 𝜇𝑓(𝑥)16𝑓(𝑥/2)𝑄(𝑥/2)(𝑡𝑡)||4𝑡+𝑘||.(5.29)

6. Latticetic Random Normed Space

Let =(𝐿,𝐿) be a complete lattice, that is, a partially ordered set in which every nonempty subset admits supremum and infimum, and 0=inf𝐿, 1=sup𝐿. The space of latticetic random distribution functions, denoted by Δ+𝐿, is defined as the set of all mappings 𝐹{,+}𝐿 such that 𝐹 is left continuous and nondecreasing on , 𝐹(0)=0, 𝐹(+)=1.

𝐷+𝐿Δ+𝐿 is defined as 𝐷+𝐿={𝐹Δ+𝐿𝑙𝐹(+)=1}, where 𝑙𝑓(𝑥) denotes the left limit of the function 𝑓 at the point 𝑥. The space Δ+𝐿 is partially ordered by the usual pointwise ordering of functions, that is, 𝐹𝐺 if and only if 𝐹(𝑡)𝐿𝐺(𝑡) for all 𝑡 in . The maximal element for Δ+𝐿 in this order is the distribution function given by𝜀00(𝑡)=1,if𝑡0,,if𝑡>0.(6.1)

In Section 2, we defined 𝑡-norms on [0,1], and now we extend 𝑡-norms on a complete lattice.

Definition 6.1 (see [42]). A triangular norm (𝑡-norm) on 𝐿 is a mapping 𝒯(𝐿)2𝐿 satisfying the following conditions:(a)(forall𝑥𝐿)(𝒯(𝑥,1)=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)(𝑥𝐿𝑥and𝑦𝐿𝑦𝒯(𝑥,𝑦)𝐿𝒯(𝑥,𝑦)) (monotonicity).

Let {𝑥𝑛} be a sequence in 𝐿 converges to 𝑥𝐿 (equipped order topology). The 𝑡-norm 𝒯 is said to be a continuous 𝑡-norm if lim𝑛𝒯𝑥𝑛,𝑦=𝒯(𝑥,𝑦)(6.2) for all 𝑦𝐿.

A 𝑡-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 𝒯0𝑖=1𝑥𝑖=1,𝒯𝑛𝑖=1𝑥𝑖𝒯=𝒯𝑛1i=1𝑥𝑖,𝑥𝑛𝑥=𝒯1,,𝑥𝑛.(6.3)

𝒯 can also be extended to a countable operation taking for any sequence (𝑥𝑛)𝑛𝑁 in 𝐿 the value𝒯𝑖=1𝑥𝑖=lim𝑛𝒯𝑛𝑖=1𝑥𝑖.(6.4) The limit on the right side of (6.4) exists since the sequence (𝒯𝑛𝑖=1𝑥𝑖)𝑛 is nonincreasing and bounded from below.

Note that we put 𝒯=𝑇 whenever 𝐿=[0,1]. If 𝑇 is a 𝑡-norm, then 𝑥𝑇(𝑛) is defined for every 𝑥[0,1] and 𝑛𝑁{0} by 1 if 𝑛=0 and 𝑇(𝑥𝑇(𝑛1),𝑥) if 𝑛1. A 𝑡-norm 𝑇 is said to be of Hadžić type, (we denote by 𝑇) if the family (𝑥𝑇(𝑛))𝑛𝑁 is equicontinuous at 𝑥=1 (cf. [27]).

Definition 6.2 (see [42]). A continuous 𝑡-norm 𝒯 on 𝐿=[0,1]2 is said to be continuous 𝑡–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, 𝒯𝑥(𝑥,𝑦)=1𝑦1,𝑥2𝑦2.(6.5)

For example, 𝑎𝒯(𝑎,𝑏)=1𝑏1𝑎,min2+𝑏2,𝑎,1𝐌(𝑎,𝑏)=min1,𝑏1𝑎,max2,𝑏2(6.6) for all 𝑎=(𝑎1,𝑎2), 𝑏=(𝑏1,𝑏2)[0,1]2 are continuous 𝑡-representable. Define the mapping 𝒯 from 𝐿2 to 𝐿 by 𝒯(𝑥,𝑦)=𝑥,if𝑦𝐿𝑥,𝑦,if𝑥𝐿𝑦.(6.7)

Recall (see [27, 28]) that if {𝑥𝑛} is a given sequence in 𝐿, (𝒯)𝑛𝑖=1𝑥𝑖 is defined recurrently by (𝒯)1𝑖=1𝑥𝑖=𝑥1 and (𝒯)𝑛𝑖=1𝑥𝑖=𝒯((𝒯)𝑛1𝑖=1𝑥𝑖,𝑥𝑛) for all 𝑛2.

A negation on is any decreasing mapping 𝒩𝐿𝐿 satisfying 𝒩(0)=1 and 𝒩(1)=0. If 𝒩(𝒩(𝑥))=𝑥, for all 𝑥𝐿, then 𝒩 is called an involutive negation. In the following, is endowed with a (fixed) negation 𝒩.

Definition 6.3. A latticetic 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 𝐷+𝐿 such that the following conditions hold: (LRN1) 𝜇𝑥(𝑡)=𝜀0(𝑡) for all 𝑡>0 if and only if 𝑥=0, (LRN2) 𝜇𝛼𝑥(𝑡)=𝜇𝑥(𝑡/|𝛼|) for all 𝑥 in 𝑋, 𝛼0 and 𝑡0, (LRN3) 𝜇𝑥+𝑦(𝑡+𝑠)𝐿𝒯(𝜇𝑥(𝑡),𝜇𝑦(𝑠)) for all 𝑥,𝑦𝑋 and 𝑡,𝑠0.

We note that from (LPN2) it follows that 𝜇𝑥(𝑡)=𝜇𝑥(𝑡) for all 𝑥𝑋 and 𝑡0.

Example 6.4. Let 𝐿=[0,1]×[0,1] and operation 𝐿 be defined by 𝑎𝐿=1,𝑎2𝑎1,𝑎2[]×[]0,10,1,𝑎1+𝑎2,𝑎11,𝑎2𝐿𝑏1,𝑏2𝑎1𝑏1,𝑎2𝑏2𝑎,𝑎=1,𝑎2𝑏,𝑏=1,𝑏2𝐿.(6.8) 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 𝜇𝑥𝑡(𝑡)=,𝑡+𝑥𝑥𝑡+𝑥𝑡+,(6.9) then (𝑋,𝜇,𝒯) is a latticetic random normed spaces.
If (𝑋,𝜇,𝒯) is a latticetic random normed space, then 𝑉𝒱=(𝜀,𝜆)𝜀>𝐿00,𝜆𝐿,1,𝑉(𝜀,𝜆)=𝑥𝑋𝐹𝑥(𝜀)>𝐿𝒩(𝜆),(6.10) 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 𝜀𝐿{0}, there exists a positive integer 𝑁 such that 𝜇𝑥𝑛𝑥(𝑡)>𝐿𝒩(𝜀) whenever 𝑛𝑁.(2)A sequence {𝑥𝑛} in 𝑋 is called a Cauchy sequence if, for every 𝑡>0 and 𝜀𝐿{0}, there exists a positive integer 𝑁 such that 𝜇𝑥𝑛𝑥𝑚(𝑡)>𝐿𝒩(𝜀) whenever 𝑛𝑚𝑁.(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 𝑥𝑛𝑥, then lim𝑛𝜇𝑥𝑛(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 equation 𝐷𝑓(𝑥,𝑦)=0 in random Banach spaces: an odd case.

Theorem 7.1. Let 𝑋 be a linear space, let (𝑌,𝜇,𝒯) be a complete LRN-space, and Φ let be a mapping from 𝑋2 to 𝐷+𝐿 (Φ(𝑥,𝑦) is denoted by Φ𝑥,𝑦) such that, for some 0<𝛼<1/8, Φ2𝑥,2𝑦(𝑡)𝐿Φ𝑥,𝑦(𝛼𝑡)(𝑥,𝑦𝑋,𝑡>0).(7.1) Let 𝑓𝑋𝑌 be an odd mapping satisfying 𝜇𝐷𝑓(𝑥,𝑦)(𝑡)𝐿Φ𝑥,𝑦(𝑡)(7.2) for all 𝑥,𝑦𝑋 and 𝑡>0. Then 𝐶(𝑥)=lim𝑛8𝑛𝑓𝑥2𝑛1𝑥2𝑓2𝑛(7.3) exists for each 𝑥𝑋 and defines a cubic mapping 𝐶𝑋𝑌 such that 𝜇𝑓(2𝑥)2𝑓(𝑥)𝐶(𝑥)(𝑡)𝐿𝒯Φ𝑥,𝑥18𝛼𝑡5𝛼,Φ2𝑥,𝑥18𝛼𝑡5𝛼(7.4) for all 𝑥𝑋 and 𝑡>0.

Proof. Letting 𝑥=𝑦 in (7.2), we get 𝜇𝑓(3𝑦)4𝑓(2𝑦)+5𝑓(𝑦)(𝑡)𝐿Φ𝑦,𝑦(𝑡)(7.5) for all 𝑦𝑋 and 𝑡>0. Replacing 𝑥 by 2𝑦 in (7.2), we get 𝜇𝑓(4𝑦)4𝑓(3𝑦)+6𝑓(2𝑦)4𝑓(𝑦)(𝑡)𝐿Φ2𝑦,𝑦(𝑡)(7.6) for all 𝑦𝑋 and 𝑡>0. By (7.5) and (7.6), 𝜇𝑓(4𝑦)10𝑓(2𝑦)+16𝑓(𝑦)(5𝑡)𝐿𝒯𝜇4(𝑓(3𝑦)4𝑓(2𝑦)+5𝑓(𝑦))(4𝑡),𝜇𝑓(4𝑦)4𝑓(3𝑦)+6𝑓(2𝑦)4𝑓(𝑦)(𝑡)=𝒯𝜇𝑓(3𝑦)4𝑓(2𝑦)+5𝑓(𝑦)(𝑡),𝜇𝑓(4𝑦)4𝑓(3𝑦)+6𝑓(2𝑦)4𝑓(𝑦)(𝑡)𝐿𝒯Φ𝑦,𝑦(𝑡),Φ2𝑦,𝑦(𝑡)(7.7) for all 𝑦𝑋 and 𝑡>0. Letting 𝑦=𝑥/2 and 𝑔(𝑥)=𝑓(2𝑥)2𝑓(𝑥) for all 𝑥𝑋, we get 𝜇𝑔(𝑥)8𝑔(𝑥/2)(5𝑡)𝐿𝒯Φ𝑥/2,𝑥/2(𝑡),Φ𝑥,𝑥/2(𝑡)(7.8) for all 𝑥𝑋 and 𝑡>0.  
Consider the set 𝑆={𝑋𝑌,(0)=0}(7.9) and introduce the generalized metric on 𝑆: 𝑑(,𝑘)=inf𝑢+𝜇(𝑥)𝑘(𝑥)(𝑢𝑡)𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡),𝑥𝑋,𝑡>0(7.10) 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 mapping 𝐽𝑆𝑆 such that 𝑥𝐽(𝑥)=82(7.11) for all 𝑥𝑋, and we prove that 𝐽 is a strictly contractive mapping with the Lipschitz constant 8𝛼.
Let ,𝑘𝑆 be given such that 𝑑(,𝑘)<𝜀. Then 𝜇(𝑥)𝑘(𝑥)(𝜀𝑡)𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(7.12) for all 𝑥𝑋 and 𝑡>0. Hence 𝜇𝐽(𝑥)𝐽𝑘(𝑥)(8𝛼𝜀𝑡)=𝜇8(𝑥/2)8𝑘(𝑥/2)(8𝛼𝜀𝑡)=𝜇(𝑥/2)𝑘(𝑥/2)(𝛼𝜀𝑡)𝒯Φ𝑥/2,𝑥/2(𝛼𝑡),Φ𝑥,𝑥/2(𝛼𝑡)𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(7.13) for all 𝑥𝑋 and 𝑡>0. So, 𝑑(,𝑘)<𝜀 implies that 𝑑𝛼(𝐽,𝐽𝑘)8𝜀.(7.14) This means that 𝑑𝛼(𝐽,𝐽𝑘)8𝑑(,𝑘)(7.15) for all ,𝑘𝑆. It follows from (7.8) that 𝜇𝑔(𝑥)8𝑔(𝑥/2)(5𝛼𝑡)𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(7.16) for all 𝑥𝑋 and 𝑡>0. So, 𝑑(𝑔,𝐽𝑔)5𝛼5/8.
By Theorem 1.1, there exists a mapping 𝐶𝑋𝑌 satisfying the following:(1)𝐶 is a fixed point of 𝐽, that is, 𝐶𝑥2=18𝐶(𝑥)(7.17) for all 𝑥𝑋. Since 𝑔𝑋𝑌 is odd, 𝐶𝑋𝑌 is an odd mapping. The mapping 𝐶 is a unique fixed point of 𝐽 in the set 𝑀={𝑆𝑑(,𝑔)<}.(7.18) This implies that 𝐶 is a unique mapping satisfying (7.17) such that there exists a 𝑢(0,) satisfying 𝜇𝑔(𝑥)𝐶(𝑥)(𝑢𝑡)𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(7.19) for all 𝑥𝑋 and 𝑡>0.(2)𝑑(𝐽𝑛𝑔,𝐶)0 as 𝑛. This implies the equality lim𝑛8𝑛𝑔𝑥2𝑛=𝐶(𝑥)(7.20) for all 𝑥𝑋.(3)𝑑(,𝐶)(1/(18𝛼))𝑑(,𝐽) with 𝑀, which implies the inequality 𝑑(𝑔,𝐶)5𝛼,18𝛼(7.21) from which it follows that 𝜇𝑔(𝑥)𝐶(𝑥)5𝛼𝑡18𝛼𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥.(𝑡)(7.22) This implies that the inequality (7.4) holds. From 𝐷𝑔(𝑥,𝑦)=𝐷𝑓(2𝑥,2𝑦)2𝐷𝑓(𝑥,𝑦), by (7.2), we deduce that 𝜇𝐷𝑓(2𝑥,2𝑦)(𝑡)𝐿Φ2𝑥,2𝑦𝜇(𝑡),2𝐷𝑓(𝑥,𝑦)(𝑡)=𝜇𝐷𝑓(𝑥,𝑦)𝑡2𝐿Φ𝑥,𝑦𝑡2(7.23) and so, by (LRN3) and (7.1), we obtain 𝜇𝐷𝑔(𝑥,𝑦)(3𝑡)𝐿𝒯𝜇𝐷𝑓2𝑥,2𝑦(𝑡),𝜇2𝐷𝑓(𝑥,𝑦)(2𝑡)𝐿𝒯Φ2𝑥,2𝑦(𝑡),Φ𝑥,𝑦(𝑡)𝐿Φ2𝑥,2𝑦(𝑡).(7.24) It follows that 𝜇8𝑛𝐷𝑔(𝑥/2𝑛,𝑦/2𝑛)(3𝑡)=𝜇𝐷𝑔(𝑥/2𝑛,𝑦/2𝑛)3𝑡8𝑛Φ𝑥/2𝑛1,𝑦/2𝑛1𝑡8𝑛𝐿𝐿Φ𝑥,𝑦18𝑡(8𝛼)𝑛1(7.25) for all 𝑥,𝑦𝑋, 𝑡>0 and 𝑛.
Since lim𝑛Φ𝑥,𝑦((3/8)(𝑡/(8𝛼)𝑛1))=1 for all 𝑥,𝑦𝑋 and 𝑡>0, by Theorem 2.4, we deduce that 𝜇𝐷𝐶(𝑥,𝑦)(3𝑡)=1(7.26) for all 𝑥,𝑦𝑋 and 𝑡>0. Thus the mapping 𝐶𝑋𝑌 satisfies (1.4).
Now, we have 𝐶(2𝑥)8𝐶(𝑥)=lim𝑛8𝑛𝑔𝑥2𝑛18𝑛+1𝑔𝑥2𝑛=8lim𝑛8𝑛1𝑔𝑥2𝑛18𝑛𝑔𝑥2𝑛=0(7.27) for all 𝑥𝑋. Since the mapping 𝑥𝐶(2𝑥)2𝐶(𝑥) is cubic (see Lemma  2.2 of [14]), from the equality 𝐶(2𝑥)=8𝐶(𝑥), we deduce that the mapping 𝐶𝑋𝑌 is cubic.

Corollary 7.2. Let 𝜃0 and let 𝑝 be a real number with 𝑝>3. Let 𝑋 be a normed vector space with norm . Let 𝑓𝑋𝑌 be an odd mapping satisfying 𝜇𝐷𝑓(𝑥,𝑦)(𝑡𝑡)𝑡+𝜃𝑥𝑝+𝑦𝑝(7.28) for all 𝑥,𝑦𝑋 and 𝑡>0. Note that (𝑋,𝜇,𝑇𝑀) is a complete LRN-space, in which 𝐿=[0,1], then 𝐶(𝑥)=lim𝑛8𝑛𝑓𝑥2𝑛1𝑥2𝑓2𝑛(7.29) exists for each 𝑥𝑋 and defines a cubic mapping 𝐶𝑋𝑌 such that 𝜇𝑓(2𝑥)2𝑓(𝑥)𝐶(𝑥)(𝑡)(2𝑝8)𝑡(2𝑝8)𝑡+5(1+2𝑝)𝜃𝑥𝑝(7.30) for all 𝑥𝑋 and 𝑡>0.

Proof. The proof follows from Theorem 7.1 by taking Φ𝑥,𝑦(𝑡𝑡)=𝑡+𝜃𝑥𝑝+𝑦𝑝(7.31) for all 𝑥,𝑦𝑋 and 𝑡>0. Then we can choose 𝛼=2𝑝, and we get 𝜇𝑓(2𝑥)2𝑓(𝑥)𝐶(𝑥)(𝑡)min123𝑝𝑡123𝑝𝑡+52𝑝𝜃2𝑥𝑝,123𝑝𝑡123𝑝𝑡+52𝑝𝜃2𝑥𝑝+𝑥𝑝123𝑝𝑡123𝑝𝑡+52𝑝𝜃2𝑥𝑝+𝑥𝑝=(2𝑝8)𝑡(2𝑝8)𝑡+5(2𝑝+1)𝜃𝑥𝑝,(7.32) which is the desired result.

Theorem 7.3. Let 𝑋 be a linear space, let (𝑌,𝜇,𝒯) be a complete LRN-space, and let Φ be a mapping from 𝑋2 to 𝐷+𝐿 (Φ(𝑥,𝑦) is denoted by Φ𝑥,𝑦) such that, for some 0<𝛼<8, Φ𝑥/2,𝑦/2(𝑡)𝐿Φ𝑥,𝑦(𝛼𝑡)(𝑥,𝑦𝑋,𝑡>0).(7.33) Let 𝑓𝑋𝑌 be an odd mapping satisfying (7.2), then 𝐶(𝑥)=lim𝑛18𝑛𝑓2𝑛+1𝑥2𝑓(2𝑛𝑥)(7.34) exists for each 𝑥𝑋 and defines a cubic mapping 𝐶𝑋𝑌 such that 𝜇𝑓(2𝑥)2𝑓(𝑥)𝐶(𝑥)(𝑡)𝐿𝒯Φ𝑥,𝑥8𝛼5𝑡,Φ2𝑥,𝑥8𝛼5𝑡(7.35) for all 𝑥𝑋 and 𝑡>0.

Proof. Let (𝑆,𝑑) be the generalized metric space defined in the proof of Theorem 7.1.  
Consider the linear mapping 𝐽𝑆𝑆 such that 1𝐽(𝑥)=8(2𝑥)(7.36) for all 𝑥𝑋, and we prove that 𝐽 is a strictly contractive mapping with the Lipschitz constant 𝛼/8.
Let ,𝑘𝑆 be given such that 𝑑(,𝑘)<𝜀, then 𝜇(𝑥)𝑘(𝑥)(𝜀𝑡)𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(7.37) for all 𝑥𝑋 and 𝑡>0. Hence 𝜇𝐽(𝑥)𝐽𝑘(𝑥)𝛼8𝜀𝑡=𝜇(1/8)(2𝑥)(1/8)𝑘(2𝑥)𝛼8𝜀𝑡=𝜇(2𝑥)𝑘(2𝑥)(𝛼𝜀𝑡)𝐿𝒯Φ2𝑥,2𝑥(𝛼𝑡),Φ4𝑥,2𝑥(𝛼𝑡)𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(7.38) for all 𝑥𝑋 and 𝑡>0. So, 𝑑(,𝑘)<𝜀 implies that 𝑑𝛼(𝐽,𝐽𝑘)8𝜀.(7.39)
This means that 𝑑𝛼(𝐽,𝐽𝑘)8𝑑(,𝑘)(7.40) for all 𝑔,𝑆. Letting 𝑔(𝑥)=𝑓(2𝑥)2𝑓(𝑥) for all 𝑥𝑋, from (7.8), we get that 𝜇𝑔(𝑥)(1/8)𝑔(2𝑥)58𝑡𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(7.41) for all 𝑥𝑋 and 𝑡>0. So, 𝑑(𝑔,𝐽𝑔)5/8.
By Theorem 1.1, there exists a mapping 𝐶𝑋𝑌 satisfying the following:(1)𝐶 is a fixed point of 𝐽, that is, 𝐶(2𝑥)=8𝐶(𝑥)(7.42) for all 𝑥𝑋. Since 𝑔𝑋𝑌 is odd, 𝐶𝑋𝑌 is an odd mapping. The mapping 𝐶 is a unique fixed point of 𝐽 in the set 𝑀={𝑆𝑑(,𝑔)<}.(7.43) This implies that 𝐶 is a unique mapping satisfying (7.42) such that there exists a 𝑢(0,) satisfying 𝜇𝑔(𝑥)𝐶(𝑥)(𝑢𝑡)𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(7.44) for all 𝑥𝑋 and 𝑡>0.(2)𝑑(𝐽𝑛𝑔,𝐶)0 as 𝑛. This implies the equalit lim𝑛18𝑛𝑔(2𝑛𝑥)=𝐶(𝑥)(7.45) for all 𝑥𝑋.(3)𝑑(,𝐶)(1/(1𝛼/8))𝑑(,𝐽) for every 𝑀, which implies the inequality 5𝑑(𝑔,𝐶),8𝛼(7.46) from which it follows that 𝜇𝑔(𝑥)𝐶(𝑥)5𝑡8𝛼𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(7.47) for all 𝑥𝑋 and 𝑡>0. This implies that the inequality (7.35) holds.
From 𝜇𝐷𝑔(𝑥,𝑦)(3𝑡)𝐿𝒯Φ2𝑥,2𝑦(𝑡),Φ𝑥,𝑦(𝑡)𝐿𝒯Φ2𝑥,2𝑦(𝑡),Φ𝑥,𝑦𝑡8,(7.48) by (7.33), we deduce that 𝜇8𝑛𝐷𝑔(2𝑛𝑥,2𝑛𝑦)(3𝑡)=𝜇𝐷𝑔(2𝑛𝑥,2𝑛𝑦)(38𝑛𝑡)𝐿Φ2𝑛𝑥,2𝑛𝑦8𝑛1𝑡𝐿Φ𝑥,𝑦8𝛼𝑛1𝑡𝛼(7.49) for all 𝑥,𝑦𝑋, 𝑡>0, and 𝑛. As 𝑛, we deduce that 𝜇𝐷𝐶(𝑥,y)(3𝑡)=1(7.50) for all 𝑥,𝑦𝑋 and 𝑡>0. Thus the mapping 𝐶𝑋𝑌 satisfies (1.4).
Now, we have 𝐶(2𝑥)8𝐶(𝑥)=lim𝑛18𝑛𝑔2𝑛+1𝑥18𝑛1𝑔(2𝑛𝑥)=8lim𝑛18𝑛+1𝑔2𝑛+1𝑥18𝑛𝑔(2𝑛𝑥)=0(7.51) for all 𝑥𝑋. Since the mapping 𝑥𝐶(2𝑥)2𝐶(𝑥) is cubic (see Lemma  2.2 of [14]), from the equality 𝐶(2𝑥)=8𝐶(𝑥), we deduce that the mapping 𝐶𝑋𝑌 is cubic.

Corollary 7.4. Let 𝜃0 and let 𝑝 be a real number with 0<𝑝<3. Let 𝑋 be a normed vector space with norm . Let 𝑓𝑋𝑌 be an odd mapping satisfying (7.28), then 𝐶(𝑥)=lim𝑛18𝑛𝑓2𝑛+1𝑥2𝑓(2𝑛𝑥)(7.52) exists for each 𝑥𝑋 and defines a cubic mapping 𝐶𝑋𝑌 such that 𝜇𝑓(2𝑥)2𝑓(𝑥)𝐶(𝑥)(𝑡)(82𝑝)𝑡(82𝑝)𝑡+5(1+2𝑝)𝜃𝑥𝑝(7.53) for all 𝑥𝑋 and 𝑡>0. Note that (𝑋,𝜇,𝑇𝑀) is a complete LRN-space, in which 𝐿=[0,1].

Proof. The proof follows from Theorem 7.3 by taking 𝜇𝐷𝑓(𝑥,𝑦)(𝑡𝑡)𝑡+𝜃𝑥𝑝+𝑦𝑝(7.54) for all 𝑥,𝑦𝑋 and 𝑡>0. Then we can choose 𝛼=2𝑝, and we get the desired result.

Theorem 7.5. Let 𝑋 be a linear space, let (𝑌,𝜇,𝒯) be a complete LRN-space, and let Φ be a mapping from 𝑋2 to 𝐷+𝐿 (Φ(𝑥,𝑦) is denoted by Φ𝑥,𝑦) such that, for some 0<𝛼<1/2, Φ2𝑥,2𝑦(𝑡)𝐿Φ𝑥,𝑦(𝛼𝑡)(𝑥,𝑦𝑋,𝑡>0).(7.55) Let 𝑓𝑋𝑌 be an odd mapping satisfying (7.2), then 𝐴(𝑥)=lim𝑛2𝑛𝑓𝑥2𝑛1𝑥8𝑓2𝑛(7.56) exists for each 𝑥𝑋 and defines an additive mapping 𝐴𝑋𝑌 such that 𝜇𝑓(2𝑥)8𝑓(𝑥)𝐴(𝑥)(𝑡)𝐿𝒯Φ𝑥,𝑥12𝛼𝑡5𝛼,Φ2𝑥,𝑥12𝛼𝑡5𝛼(7.57) for all 𝑥𝑋 and 𝑡>0.

Proof. Let (𝑆,𝑑) be the generalized metric space defined in the proof of Theorem 7.1.
Letting 𝑦=𝑥/2 and 𝑔(𝑥)=𝑓(2𝑥)8𝑓(𝑥) for all 𝑥𝑋 in (7.7), we get 𝜇𝑔(𝑥)2𝑔(𝑥/2)(5𝑡)𝐿𝒯Φ𝑥/2,𝑥/2(𝑡),Φ𝑥,𝑥/2(𝑡)(7.58) for all 𝑥𝑋 and 𝑡>0.
Now, we consider the linear mapping 𝐽𝑆𝑆 such that 𝑥𝐽(𝑥)=22(7.59) for all 𝑥𝑋. 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 𝜇𝑔(𝑥)2𝑔(𝑥/2)(5𝛼𝑡)𝑇𝑀Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(7.60) for all 𝑥𝑋 and 𝑡>0. So, 𝑑(𝑔,𝐽𝑔)5𝛼<.
By Theorem 1.1, there exists a mapping 𝐴𝑋𝑌 satisfying the following:(1)𝐴 is a fixed point of 𝐽, that is, 𝐴𝑥2=12𝐴(𝑥)(7.61) for all𝑥𝑋. Since𝑔𝑋𝑌 is odd,𝐴𝑋𝑌 is an odd mapping. The mapping𝐴 is a unique fixed point of𝐽 in the set 𝑀={𝑆𝑑(,𝑔)<}.(7.62) This implies that 𝐴 is a unique mapping satisfying (7.61) such that there exists a 𝑢(0,) satisfying 𝜇𝑔(𝑥)𝐴(𝑥)(𝑢𝑡)𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(7.63) for all 𝑥𝑋 and 𝑡>0.(2)𝑑(𝐽𝑛𝑔,𝐴)0 as 𝑛. This implies the equality lim𝑛2𝑛𝑔𝑥2𝑛=𝐴(𝑥)(7.64) for all 𝑥𝑋.(3)𝑑(,𝐴)(1/(12𝛼))𝑑(,𝐽) for each 𝑀, which implies the inequality 𝑑(𝑔,𝐴)5𝛼.12𝛼(7.65) This implies that the inequality (7.57) holds. Since 𝜇𝐷𝑔(𝑥,𝑦)(3𝑡)𝐿Φ2𝑥,2𝑦(𝑡), it follows that 𝜇2𝑛𝐷𝑔(𝑥/2𝑛,𝑦/2𝑛)(3𝑡)=𝜇𝐷𝑔(𝑥/2𝑛,𝑦/2𝑛)3𝑡2𝑛Φ𝑥/2𝑛1,𝑦/2𝑛1𝑡2𝑛𝐿𝐿Φ𝑥,𝑦12𝑡(2𝛼)𝑛1(7.66) for all 𝑥,𝑦𝑋, 𝑡>0, and 𝑛. As 𝑛, we deduce that 𝜇𝐷𝐴(𝑥,𝑦)(3𝑡)=1(7.67) for all 𝑥,𝑦𝑋 and 𝑡>0. Thus, the mapping 𝐴𝑋𝑌 satisfies (1.4).
Now, we have 𝐴(2𝑥)2𝐴(𝑥)=lim𝑛2𝑛𝑔𝑥2𝑛12𝑛+1𝑔𝑥2𝑛=2lim𝑛2𝑛1𝑔𝑥2𝑛12𝑛𝑔𝑥2𝑛=0(7.68) for all 𝑥𝑋. Since the mapping 𝑥𝐴(2𝑥)8𝐴(𝑥) is additive (see Lemma  2.2 of [14]), from the equality 𝐴(2𝑥)=2𝐴(𝑥), we deduce that the mapping 𝐴𝑋𝑌 is additive.

Corollary 7.6. Let 𝜃0 and let 𝑝 be a real number with 𝑝>1. Let 𝑋 be a normed vector space with norm . Let 𝑓𝑋𝑌 be an odd mapping satisfying (7.28), then 𝐴(𝑥)=lim𝑛2𝑛𝑓𝑥2𝑛1𝑥8𝑓2𝑛(7.69) exists for each 𝑥𝑋 and defines an additive mapping 𝐴𝑋𝑌 such that 𝜇𝑓(2𝑥)8𝑓(𝑥)𝐴(𝑥)(𝑡)(2𝑝2)𝑡(2𝑝2)𝑡+5(1+2𝑝)𝜃𝑥𝑝(7.70) for all 𝑥𝑋 and 𝑡>0, where (𝑋,𝜇,𝑇𝑀) is a complete LRN-space in which 𝐿=[0,1].

Proof. The proof follows from Theorem 7.5 by taking 𝜇𝐷𝑓(𝑥,𝑦)(𝑡𝑡)𝑡+𝜃𝑥𝑝+𝑦𝑝(7.71) for all 𝑥,𝑦𝑋 and 𝑡>0. Then we can choose 𝛼=2𝑝, and we get the desired result.

Theorem 7.7. Let 𝑋 be a linear space, let (𝑌,𝜇,𝒯) be a complete LRN-space, and let Φ be a mapping from 𝑋2 to 𝐷+𝐿 (Φ(𝑥,𝑦) is denoted by Φ𝑥,𝑦) such that, for some 0<𝛼<2, Φ𝑥,𝑦(𝛼𝑡)𝐿Φ𝑥/2,𝑦/2(𝑡)(𝑥,𝑦𝑋,𝑡>0).(7.72) Let 𝑓𝑋𝑌 be an odd mapping satisfying (7.2), then 𝐴(𝑥)=lim𝑛12𝑛𝑓2𝑛+1𝑥8𝑓(2𝑛𝑥)(7.73) exists for each 𝑥𝑋 and defines an additive mapping 𝐴𝑋𝑌 such that 𝜇𝑓(2𝑥)8𝑓(𝑥)𝐴(𝑥)(𝑡)𝐿𝒯Φ𝑥,𝑥2𝛼𝑡5𝛼,Φ2𝑥,𝑥2𝛼𝑡5𝛼(7.74) for all 𝑥𝑋 and 𝑡>0.

Proof. Let (𝑆,𝑑) be the generalized metric space defined in the proof of Theorem 7.1.
Consider the linear mapping 𝐽𝑆𝑆 such that 1𝐽(𝑥)=2(2𝑥)(7.75) for all 𝑥𝑋. 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 𝜇𝑔(𝑥)1/2𝑔(2𝑥)52𝑡𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(7.76) for all 𝑥𝑋 and 𝑡>0. So, 𝑑(𝑔,𝐽𝑔)5/2. By Theorem 1.1, there exists a mapping 𝐴𝑋𝑌 satisfying the following:(1)𝐴 is a fixed point of 𝐽, that is, 𝐴(2𝑥)=2𝐴(𝑥)(7.77) for all 𝑥𝑋. Since 𝑋𝑌 is odd, 𝐴𝑋𝑌 is an odd mapping. The mapping 𝐴 is a unique fixed point of 𝐽 in the set 𝑀={𝑆𝑑(,𝑔)<}.(7.78) This implies that 𝐴 is a unique mapping satisfying (7.77) such that there exists a 𝑢(0,) satisfying 𝜇𝑔(𝑥)𝐴(𝑥)(𝑢𝑡)𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(7.79) for all 𝑥𝑋 and 𝑡>0.(2)𝑑(𝐽𝑛𝑔,𝐴)0 as 𝑛. This implies the equality lim𝑛12𝑛𝑔(2𝑛𝑥)=𝐴(𝑥)(7.80) for all 𝑥𝑋.(3)𝑑(,𝐴)(1/(1𝛼/2))𝑑(,𝐽), which implies the inequality 5𝑑(𝑔,𝐴).2𝛼(7.81) This implies that the inequality (7.74) holds.
Proceeding as in the proof of Theorem 7.5, we obtain that the mapping 𝐴𝑋𝑌 satisfies (1.4). Now, we have 𝐴(2𝑥)2𝐴(𝑥)=lim𝑛12𝑛𝑔2𝑛+1𝑥12𝑛1𝑔(2𝑛𝑥)=2lim𝑛12𝑛+1𝑔2𝑛+1𝑥12𝑛𝑔(2𝑛𝑥)=0(7.82) for all 𝑥𝑋. Since the mapping 𝑥𝐴(2𝑥)8𝐴(𝑥) is additive (see Lemma  2.2 of [14]), from the equality 𝐴(2𝑥)=2𝐴(𝑥), we deduce that the mapping 𝐴𝑋𝑌 is additive.

Corollary 7.8. Let 𝜃0 and let 𝑝 be a real number with 0<𝑝<1. Let 𝑋 be a normed vector space with norm . Let 𝑓𝑋𝑌 be an odd mapping satisfying (7.28), then 𝐴(𝑥)=lim𝑛12𝑛𝑓2𝑛+1𝑥8𝑓(2𝑛𝑥)(7.83) exists for each 𝑥𝑋 and defines an additive mapping 𝐴𝑋𝑌 such that 𝜇𝑓(2𝑥)8𝑓(𝑥)𝐴(𝑥)(𝑡)(22𝑝)𝑡(22𝑝)𝑡+5(1+2𝑝)𝜃𝑥𝑝(7.84) for all 𝑥𝑋 and 𝑡>0, where (𝑋,𝜇,𝑇𝑀) is a complete LRN-space in which 𝐿=[0,1].

Proof. The proof follows from Theorem 7.7 by taking 𝜇𝐷𝑓(𝑥,𝑦)(𝑡𝑡)𝑡+𝜃𝑥𝑝+𝑦𝑝(7.85) for all 𝑥,𝑦𝑋 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 equation 𝐷𝑓(𝑥,𝑦)=0 in random Banach spaces, an even case.

Theorem 8.1. Let 𝑋 be a linear space, let (𝑌,𝜇,𝒯) be a complete LRN-space, and let Φ be a mapping from 𝑋2 to 𝐷+𝐿 (Φ(𝑥,𝑦) is denoted by Φ𝑥,𝑦) such that, for some 0<𝛼<1/16, Φ𝑥,𝑦(𝛼𝑡)𝐿Φ2𝑥,2𝑦(𝑡)(𝑥,𝑦𝑋,𝑡>0).(8.1) Let 𝑓𝑋𝑌 be an even mapping satisfying 𝑓(0)=0 and (7.2), then 𝑄(𝑥)=lim𝑛16𝑛𝑓𝑥2𝑛1𝑥4𝑓2𝑛(8.2) exists for each 𝑥𝑋 and defines a quartic mapping 𝑄𝑋𝑌 such that 𝜇𝑓(2𝑥)4𝑓(𝑥)𝑄(𝑥)(𝑡)𝐿𝒯Φ𝑥,𝑥116𝛼𝑡5𝛼,Φ2𝑥,𝑥116𝛼𝑡5𝛼(8.3) for all 𝑥𝑋 and 𝑡>0.

Proof. Letting 𝑥=𝑦 in (7.2), we get 𝜇𝑓(3𝑦)6𝑓(2𝑦)+15𝑓(𝑦)(𝑡)𝐿Φ𝑦,𝑦(𝑡)(8.4) for all 𝑦𝑋 and 𝑡>0. Replacing 𝑥 by 2𝑦 in (7.2), we get 𝜇𝑓(4𝑦)4𝑓(3𝑦)+4𝑓(2𝑦)+4𝑓(𝑦)(𝑡)𝐿Φ2𝑦,𝑦(𝑡)(8.5) for all 𝑦𝑋 and 𝑡>0. By (8.4) and (8.5), 𝜇𝑓(4𝑥)20𝑓(2𝑥)+64𝑓(𝑥)(5𝑡)𝐿𝒯𝜇4(𝑓(3𝑥)6𝑓(2𝑥)+15𝑓(𝑥))(4𝑡),𝜇𝑓(4𝑥)4𝑓(3𝑥)+4𝑓(2𝑥)+4𝑓(𝑥)(𝑡)𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(8.6) for all 𝑥𝑋 and 𝑡>0. Letting 𝑔(𝑥)=𝑓(2𝑥)4𝑓(𝑥) for all 𝑥𝑋, we get 𝜇𝑔(𝑥)16𝑔(𝑥/2)(5𝑡)𝐿𝒯Φ𝑥/2,𝑥/2(𝑡),Φ𝑥,𝑥/2(𝑡)(8.7) for all 𝑥𝑋 and 𝑡>0. Let (𝑆,𝑑) be the generalized metric space defined in the proof of Theorem 7.1.  
Now we consider the linear mapping 𝐽𝑆𝑆 such that 𝐽(𝑥)=16(𝑥/2) for all 𝑥𝑋. 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 𝜇𝑔(𝑥)16𝑔(𝑥/2)(5𝛼𝑡)𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(8.8) for all 𝑥𝑋 and 𝑡>0. So, 5𝑑(𝑔,𝐽𝑔)5𝛼16<.(8.9)
By Theorem 1.1, there exists a mapping 𝑄𝑋𝑌 satisfying the following:(1)𝑄 is a fixed point of 𝐽, that is, 𝑄𝑥2=116𝑄(𝑥)(8.10) for all𝑥𝑋. Since𝑔𝑋𝑌 is even with𝑔(0)=0,𝑄𝑋𝑌 is an even mapping with𝑄(0)=0. The mapping𝑄 is a unique fixed point of𝐽 in the set 𝑀={𝑆𝑑(,𝑔)<}.(8.11) This implies that 𝑄 is a unique mapping satisfying (8.10) such that there exists a 𝑢(0,) satisfying 𝜇𝑔(𝑥)𝑄(𝑥)(𝑢𝑡)𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(8.12) for all 𝑥𝑋 and 𝑡>0.(2)𝑑(𝐽𝑛𝑔,𝑄)0 as 𝑛. This implies the equality lim𝑛16𝑛𝑔𝑥2𝑛=𝑄(𝑥)(8.13) for all 𝑥𝑋.(3)𝑑(,𝑄)(1/(116𝛼))𝑑(,𝐽) for every 𝑀, which implies the inequality 𝑑(𝑔,𝑄)5𝛼.116𝛼(8.14) This implies that the inequality (8.3) holds.
Proceeding as in the proof of Theorem 7.1, we obtain that the mapping 𝑄𝑋𝑌 satisfies (1.4). Now, we have 𝑄(2𝑥)16𝑄(𝑥)=lim𝑛16𝑛𝑔𝑥2𝑛116𝑛+1𝑔𝑥2𝑛=16lim𝑛16𝑛1𝑔𝑥2𝑛116𝑛𝑔𝑥2𝑛=0(8.15) for all 𝑥𝑋.  Since the mapping 𝑥𝑄(2𝑥)4𝑄(𝑥) is quartic, we get that the mapping 𝑄𝑋𝑌 is quartic.

Corollary 8.2. Let 𝜃0 and let 𝑝 be a real number with 𝑝>4. Let 𝑋 be a normed vector space with norm .  Let 𝑓𝑋𝑌 be an even mapping satisfying 𝑓(0)=0 and (7.28), then 𝑄(𝑥)=lim𝑛16𝑛𝑓𝑥2𝑛1𝑥4𝑓2𝑛(8.16) exists for each 𝑥𝑋 and defines a quartic mapping 𝑄𝑋𝑌 such that 𝜇𝑓(2𝑥)4𝑓(𝑥)𝑄(𝑥)(𝑡)(2𝑝16)𝑡(2𝑝16)𝑡+5(1+2𝑝)𝜃𝑥𝑝(8.17) for all 𝑥𝑋 and 𝑡>0, where (𝑋,𝜇,𝑇𝑀) is a complete LRN-space in which 𝐿=[0,1].

Proof. The proof follows from Theorem 8.1 by taking 𝜇𝐷𝑓(𝑥,𝑦)(𝑡𝑡)𝑡+𝜃𝑥𝑝+𝑦𝑝(8.18) for all 𝑥,𝑦𝑋 and 𝑡>0. Then we can choose 𝛼=2𝑝, and we get the desired result.

Theorem 8.3. Let 𝑋 be a linear space, let (𝑌,𝜇,𝒯) be a complete LRN-space, and let Φ be a mapping from 𝑋2 to 𝐷+𝐿 (Φ(𝑥,𝑦) is denoted by Φ𝑥,𝑦) such that, for some 0<𝛼<16, Φ𝑥,𝑦(𝛼𝑡)Φ𝑥/2,𝑦/2(𝑡)(𝑥,𝑦𝑋,𝑡>0).(8.19) Let 𝑓𝑋𝑌 be an even mapping satisfying 𝑓(0)=0 and (7.2), then 𝑄(𝑥)=lim𝑛116𝑛𝑓2𝑛+1𝑥4𝑓(2𝑛𝑥)(8.20) exists for each 𝑥𝑋 and defines a quartic mapping 𝑄𝑋𝑌 such that 𝜇𝑓(2𝑥)4𝑓(𝑥)𝑄(𝑥)(𝑡)𝐿𝒯Φ𝑥,𝑥16𝛼5𝑡,Φ2𝑥,𝑥16𝛼5𝑡(8.21) for all 𝑥𝑋 and 𝑡>0.

Proof. In the generalized metric space (𝑆,𝑑) defined in the proof of Theorem 7.1, we consider the linear mapping 𝐽𝑆𝑆 such that 1𝐽(𝑥)=16(2𝑥)(8.22) for all 𝑥𝑋. It is easy to see that 𝐽 is a strictly contractive self-mapping on 𝑆 with the Lipschitz constant 𝛼/16.  
Letting 𝑔(𝑥)=𝑓(2𝑥)4𝑓(𝑥) for all 𝑥𝑋, by (8.7), we get 𝜇𝑔(𝑥)(1/16)𝑔(2𝑥)5𝑡16𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(8.23) for all 𝑥𝑋 and 𝑡>0. So, 𝑑(𝑔,𝐽𝑔)5/16.
By Theorem 1.1, there exists a mapping 𝑄𝑋𝑌 satisfying the following:(1)𝑄 is a fixed point of 𝐽, that is, 𝑄(2𝑥)=16𝑄(𝑥)(8.24) for all 𝑥𝑋. Since 𝑔𝑋𝑌 is even with 𝑔(0)=0, 𝑄𝑋𝑌 is an even mapping with 𝑄(0)=0. The mapping 𝑄 is a unique fixed point of 𝐽 in the set 𝑀={𝑆𝑑(,𝑔)<}.(8.25) This implies that 𝑄 is a unique mapping satisfying (8.24) such that there exists a 𝑢(0,) satisfying 𝜇𝑔(𝑥)𝑄(𝑥)(𝑢𝑡)𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(8.26) for all 𝑥𝑋 and 𝑡>0.(2)d(𝐽𝑛𝑔,𝑄)0 as 𝑛. This implies the equality lim𝑛116𝑛𝑔(2𝑛𝑥)=𝑄(𝑥)(8.27) for all 𝑥𝑋.(3)𝑑(𝑔,𝑄)(16/(16𝛼))𝑑(𝑔,𝐽𝑔) for each 𝑀, which implies the inequality 𝑑(𝑔,𝑄)5/(16𝛼).(8.28) This implies that the inequality (8.21) holds.
Proceeding as in the proof of Theorem 7.3, we obtain that the mapping 𝑄𝑋𝑌 satisfies (1.4). Now, we have 𝑄(2𝑥)16𝑄(𝑥)=lim𝑛116𝑛𝑔2𝑛+1𝑥116𝑛1𝑔(2𝑛𝑥)=16lim𝑛116𝑛+1𝑔2𝑛+1𝑥116𝑛𝑔(2𝑛𝑥)=0(8.29) for all 𝑥𝑋. Since the mapping 𝑥𝑄(2𝑥)4𝑄(𝑥) is quartic, we get that the mapping 𝑄𝑋𝑌 is quartic.

Corollary 8.4. Let 𝜃0 and let 𝑝 be a real number with 0<𝑝<4. Let 𝑋 be a normed vector space with norm . Let 𝑓𝑋𝑌 be an even mapping satisfying 𝑓(0)=0 and (7.28), then 𝑄(𝑥)=lim𝑛116𝑛𝑓2𝑛+1𝑥4𝑓(2𝑛𝑥)(8.30)exists for each 𝑥𝑋 and defines a quartic mapping 𝑄𝑋𝑌 such that 𝜇𝑓(2𝑥)4𝑓(𝑥)𝑄(𝑥)(𝑡)(162𝑝)𝑡(162𝑝)𝑡+5(1+2𝑝)𝜃𝑥𝑝(8.31) for all 𝑥𝑋 and 𝑡>0, where (𝑋,𝜇,𝑇𝑀) is a complete LRN-space in which 𝐿=[0,1].

Proof. The proof follows from Theorem 8.3 by taking 𝜇𝐷𝑓(𝑥,𝑦)(𝑡𝑡)𝑡+𝜃𝑥𝑝+𝑦𝑝(8.32) for all 𝑥,𝑦𝑋 and 𝑡>0.  Then we can choose 𝛼=2𝑝, and we get the desired result.

Theorem 8.5. Let 𝑋be a linear space, let (𝑌,𝜇,𝒯) be a complete LRN-space, and let Φ be a mapping from 𝑋2 to 𝐷+𝐿 (Φ(𝑥,𝑦) is by denoted Φ𝑥,𝑦) such that, for some 0<𝛼<1/4, Φ𝑥,𝑦(𝛼𝑡)𝐿Φ2𝑥,2𝑦(𝑡)(𝑥,𝑦𝑋,𝑡>0).(8.33) Let 𝑓𝑋𝑌 be an even mapping satisfying 𝑓(0)=0 and (7.2), then 𝑇(𝑥)=lim𝑛4𝑛𝑓𝑥2𝑛1𝑥16𝑓2𝑛(8.34) exists for each 𝑥𝑋 and defines a quadratic mapping 𝑇𝑋𝑌 such that 𝜇𝑓(2𝑥)16𝑓(𝑥)𝑇(𝑥)(𝑡)𝐿𝒯Φ𝑥,𝑥14𝛼𝑡5𝛼,Φ2𝑥,𝑥14𝛼𝑡5𝛼(8.35) for all 𝑥𝑋 and 𝑡>0.

Proof. Let (𝑆,𝑑) be the generalized metric space defined in the proof of Theorem 7.1.
Letting 𝑔(𝑥)=𝑓(2𝑥)16𝑓(𝑥) for all 𝑥𝑋 in (8.6), we get 𝜇𝑔(𝑥)4𝑔(𝑥/2)(5𝑡)𝐿𝒯Φ𝑥/2,𝑥/2(𝑡),Φ𝑥,𝑥/2(𝑡)(8.36) for all 𝑥𝑋 and 𝑡>0. It is easy to see that the linear mapping 𝐽𝑆𝑆 such that 𝑥𝐽(𝑥)=42(8.37) for all 𝑥𝑋, is a strictly contractive self-mapping with the Lipschitz constant 4𝛼.
It follows from (8.36) that 𝜇𝑔(𝑥)4𝑔(𝑥/2)(5𝛼𝑡)𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(8.38) for all 𝑥𝑋 and 𝑡>0. So, 𝑑(𝑔,𝐽𝑔)5𝛼<.
By Theorem 1.1, there exists a mapping 𝑇𝑋𝑌 satisfying the following:(1)𝑇 is a fixed point of 𝐽, that is, 𝑇𝑥2=14𝑇(𝑥)(8.39) for all 𝑥𝑋. Since 𝑔𝑋𝑌 is even with 𝑔(0)=0, 𝑇𝑋𝑌 is an even mapping with 𝑇(0)=0. The mapping 𝑇 is a unique fixed point of 𝐽 in the set 𝑀={𝑆𝑑(,𝑔)<}. This implies that 𝑇 is a unique mapping satisfying (8.39) such that there exists a 𝑢(0,) satisfying 𝜇𝑔(𝑥)𝑇(𝑥)(𝑢𝑡)𝐿𝒯Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(8.40) for all 𝑥𝑋 and 𝑡>0.(2)𝑑(𝐽𝑛𝑔,𝑇)0 as 𝑛. This implies the equality lim𝑛4𝑛𝑔𝑥2𝑛=𝑇(𝑥)(8.41) for all 𝑥𝑋.(3)𝑑(,𝑇)(1/(14𝛼))𝑑(,𝐽) for each 𝑀, which implies the inequality 𝑑(𝑔,𝑇)5𝛼.14𝛼(8.42) This implies that the inequality (8.35) holds.
Proceeding as in the proof of Theorem 7.1, we obtain that the mapping 𝑇𝑋𝑌 satisfies (1.4). Now, we have 𝑇(2𝑥)4𝑇(𝑥)=lim𝑛4𝑛𝑔𝑥2𝑛14𝑛+1𝑔𝑥2𝑛=4lim𝑛4𝑛1𝑔𝑥2𝑛14𝑛𝑔𝑥2𝑛=0(8.43) for all 𝑥𝑋. Since the mapping 𝑥𝑇(2𝑥)16𝑇(𝑥) is quadratic, we get that the mapping 𝑇𝑋𝑌 is quadratic.

Corollary 8.6. Let 𝜃0 and let 𝑝 be a real number with 𝑝>2. Let 𝑋 be a normed vector space with norm . Let 𝑓𝑋𝑌 be an even mapping satisfying 𝑓(0)=0 and (7.28), then 𝑇(𝑥)=lim𝑛4𝑛𝑓𝑥2𝑛1𝑥16𝑓2𝑛(8.44) exists for each 𝑥𝑋 and defines a quadratic mapping 𝑇𝑋𝑌 such that 𝜇𝑓(2𝑥)16𝑓(𝑥)𝑇(𝑥)(𝑡)(2𝑝4)𝑡(2𝑝4)𝑡+5(1+2𝑝)𝜃𝑥𝑝(8.45) for all 𝑥𝑋 and 𝑡>0.

Proof. The proof follows from Theorem 8.5 by taking Φ𝑥,𝑦(𝑡𝑡)=𝑡+𝜃𝑥𝑝+𝑦𝑝(8.46) for all 𝑥,𝑦𝑋. Then we can choose 𝛼=2𝑝, and we get the desired result.

Theorem 8.7. Let 𝑋 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, Φ𝑥,𝑦(𝛼𝑡)Φ𝑥/2,𝑦/2(𝑡)(𝑥,𝑦𝑋,𝑡>0).(8.47) Let 𝑓𝑋𝑌 be an even mapping satisfying 𝑓(0)=0 and (7.2), then 𝑇(𝑥)=lim𝑛14𝑛𝑓2𝑛+1𝑥16𝑓(2𝑛𝑥)(8.48) exists for each 𝑥𝑋 and defines a quadratic mapping 𝑇𝑋𝑌 such that 𝜇𝑓(2𝑥)16𝑓(𝑥)𝑇(𝑥)(𝑡)𝑇𝑀Φ𝑥,𝑥4𝛼5𝑡,Φ2𝑥,𝑥4𝛼5𝑡(8.49) for all 𝑥𝑋 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 mapping 𝐽𝑆𝑆 such that 1𝐽(𝑥)=4(2𝑥)(8.50) for all 𝑥𝑋 is a strictly contractive self-mapping with the Lipschitz constant 𝛼/4.
Letting 𝑔(𝑥)=𝑓(2𝑥)16𝑓(𝑥) for all 𝑥𝑋, from (8.36), we get 𝜇𝑔(𝑥)1/4𝑔(2𝑥)54𝑡𝑇𝑀Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(8.51) for all 𝑥𝑋 and 𝑡>0. So, 𝑑(𝑔,𝐽𝑔)5/4.  
By Theorem 1.1, there exists a mapping 𝑇𝑋𝑌 satisfying the following:(1) 𝑇 is a fixed point of 𝐽, that is, 𝑇(2𝑥)=4𝑇(𝑥)(8.52) for all 𝑥𝑋. Since 𝑔𝑋𝑌 is even with 𝑔(0)=0, 𝑇𝑋𝑌 is an even mapping with 𝑇(0)=0. The mapping 𝑇 is a unique fixed point of 𝐽 in the set 𝑀={𝑆𝑑(,𝑔)<}.(8.53) This implies that 𝑇 is a unique mapping satisfying (8.52) such that there exists a 𝑢(0,) satisfying 𝜇𝑔(𝑥)𝑇(𝑥)(𝑢t)𝑇𝑀Φ𝑥,𝑥(𝑡),Φ2𝑥,𝑥(𝑡)(8.54) for all 𝑥𝑋 and 𝑡>0.(2) 𝑑(𝐽𝑛𝑔,𝑇)0 as 𝑛. This implies the equality lim𝑛14𝑛𝑔(2𝑛𝑥)=𝑇(𝑥)(8.55) for all 𝑥𝑋.(3) 𝑑(,𝑇)(1/(1𝛼/4))𝑑(,𝐽) for each 𝑀, which implies the inequality 𝑑(𝑔,𝑇)5/(4𝛼).(8.56) This implies that the inequality (8.49) holds.
Proceeding as in the proof of Theorem  2.3, we obtain that the mapping 𝑄𝑋𝑌 satisfies (1.4). Now, we have 𝑇(2𝑥)4𝑇(𝑥)=lim𝑛14𝑛𝑔2𝑛+1𝑥14𝑛1𝑔(2𝑛𝑥)=4lim𝑛14𝑛+1𝑔2𝑛+1𝑥14𝑛𝑔(2𝑛𝑥)=0(8.57) for all 𝑥𝑋. Since the mapping 𝑥𝑇(2𝑥)16𝑇(𝑥) is quadratic, we get that the mapping 𝑇𝑋𝑌 is quadratic.

Corollary 8.8. Let 𝜃0 and let 𝑝 be a real number with 0<𝑝<2. Let 𝑋 be a normed vector space with norm . Let 𝑓𝑋𝑌 be an even mapping satisfying 𝑓(0)=0 and (7.28). Then 𝑇(𝑥)=lim𝑛14𝑛𝑓2𝑛+1𝑥16𝑓(2𝑛𝑥)(8.58) exists for each 𝑥𝑋 and defines a quadratic mapping 𝑇𝑋𝑌 such that 𝜇𝑓(2𝑥)16𝑓(𝑥)𝑇(𝑥)(𝑡)(42𝑝)𝑡(42𝑝)𝑡+5(1+2𝑝)𝜃𝑥𝑝(8.59) for all 𝑥𝑋 and 𝑡>0, where (𝑋,𝜇,𝑇𝑀) is a complete LRN-space in which 𝐿=[0,1].

Proof. The proof follows from Theorem 8.5 by taking Φ𝑥,𝑦(𝑡𝑡)=𝑡+𝜃𝑥𝑝+𝑦𝑝(8.60) for all 𝑥,𝑦𝑋 and 𝑡>0. Then we can choose 𝛼=2𝑝, and we get the desired result.

Acknowledgments

The authors are grateful to the area Editor Professor Yeong-Cheng Liou and the reviewer for their valuable comments and suggestions. Y. J. Cho was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant no. 2011-0021821).