# Approximation of Mixed-Type Functional Equations in Menger PN-Spaces

Authors: M. Eshaghi Gordji; H. Khodaei; Y. W. Lee; G. H. Kim
Journal: Abstract and Applied Analysis
 (2012)

---

## Abstract

LetX and Y be vector spaces. We show that a function f:X→Y with f(0)=0 satisfies Δf(x1,…,xn)=0 for all x1,…,xn∈X, if and only if there exist functions C:X×X×X→Y, B:X×X→Y and A:X→Y such that f(x)=C(x,x,x)+B(x,x)+A(x) for all x∈X, where the function C is symmetric for each fixed one variable and is additive for fixed two variables, B is symmetric bi-additive, A is additive and Δf(x1,…,xn)=∑k=2n(∑i1=2k∑i2=i1+1k+1⋯∑in-k+1=in-k+1n)f(∑i=1,i≠i1,…,in-k+1nxi-∑r=1n-k+1xir)+f(∑i=1nxi)-2n-2∑i=2n(f(x1+xi)+f(x1-xi))+2n-1(n-2)f(x1) (n∈ℕ, n≥3) for all x1,…,xn∈X. Furthermore, we solve the stability problem for a given function f satisfying Δf(x1,…,xn)=0, in the Menger probabilistic normed spaces.


---

## Body

## 1. Introduction and Preliminaries

Menger [1] introduced the notion of a probabilistic metric space in 1942 and since then the theory of probabilistic metric spaces has developed in many directions [2]. The idea of Menger was to use distribution functions instead of nonnegative real numbers as values of the metric. The notion of a probabilistic metric space corresponds to situations when we do not know exactly the distance between two points, but we know probabilities of possible values of this distance. A probabilistic generalization of metric spaces appears to be interest in the investigation of physical quantities and physiological thresholds. It is also of fundamental importance in probabilistic functional analysis. Probabilistic normed spaces were introduced by Sherstnev in 1962 [3] by means of a definition that was closely modelled on the theory of (classical) normed spaces and used to study the problem of best approximation in statistics. In the sequel, we will adopt the usual terminology, notation and conventions of the theory of probabilistic normed spaces as in [2, 4–6, 6, 7, 7–18].Throughout this paper, letΔ+ be the space of distribution functions, that is,
(1.1)Δ+:={F(+∞)F:R∪{-∞,∞}⟶[0,1]:Fisleft-continuous,nondecreasingon⁡R,F(0)=0,F(+∞)=1},
and the subset D+⊆Δ+ is the set:
(1.2)D+={F∈Δ+:l-F(+∞)=1},
where, l-f(x) denotes the left limit of the function f at the point x. The space Δ+ is partially ordered by the usual point-wise ordering of functions, that is, F≤G if and only if F(t)≤G(t) for all t∈ℝ. The maximal element for Δ+ in this order is the distribution function given by
(1.3)ε0(t)={0,ift≤0,1,ift>0.Definition 1.1 (see [2]).
A mappingT:[0,1]×[0,1]→[0,1]   is a continuous triangular norm (briefly, a continuous  t-norm) if T satisfies the following conditions:(1)
T is commutative and associative;(2)
T is continuous;(3)
T(a,1)=a for all a∈[0,1];(4)
T(a,b)≤T(c,d) whenever a≤c and b≤d for all a,b,c,d∈[0,1].Two typical examples of continuoust-norms are TP(a,b)=ab, TM(a,b)=min(a,b). Recall (see [19, 20]) that if T is a t-norm and {xn} is a given sequence of numbers in [0,1], Ti=1nxi is defined recurrently by
(1.4)Ti=1nxi={x1,ifn=1,T(Ti=1n-1xi,xn),ifn≥2.
Ti=n+1∞xi is defined as Ti=1∞xn+i.Definition 1.2.
A Menger probabilistic normed space (briefly, Menger PN-space) is a triple(X,Λ,T) where,  X  is a vector space,  T  is a continuous  t-norm, and  Λ  is a mapping from  X  into  D+  such that the following conditions hold: (PN1)
Λx(0)=0 for all x∈X;(PN2)
Λx(t)=ε0(t) for all t>0 if and only if x=0;(PN3)
Λαx(t)=Λx(t/|α|) for all x∈X, α≠0 and all t>0;(PN4)
Λx+y(t+s)≥T(Λx(t),Λy(s)) for all x,y∈X and all t,s≥0.Clearly, every Menger PN-space is probabilistic metric space having a metrizable uniformity onX if supa<1T(a,a)=1.Definition 1.3.
Let(X,Λ,T) be a Menger PN-space.(i)
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 Λxn-x(ϵ)>1-λ whenever n≥N.(ii)
A sequence{xn} in X is called Cauchy sequence if, for every ϵ>0 and λ>0, there exists positive integer N such that Λxn-xm(ϵ)>1-λ whenever n≥m≥N.(iii)
A Menger PN-space(X,Λ,T) is said to be complete if and only if every Cauchy sequence in X is convergent to a point in X.Theorem 1.4.
If(X,Λ,T) is a Menger PN-space and  {xn} is a sequence such that  xn→x, then  limn→∞Λxn(t)=Λx(t).The concept of stability of a functional equation arises when one replaces a functional equation by an inequality which acts as a perturbation of the equation. The first stability problem concerning group homomorphisms was raised by Ulam [21] in 1940 and affirmatively solved by Hyers [22]. The result of Hyers was generalized by Aoki [23] for approximate additive function and by Rassias [24] for approximate linear functions by allowing the difference Cauchy equation ∥f(x+y)-f(x)-f(y)∥ to be controlled by ε(∥x∥p+∥y∥p). Taking into consideration a lot of influence of Ulam, Hyers, and Rassias on the development of stability problems of functional equations, the stability phenomenon that was proved by Rassias is called the Hyers-Ulam-Rassias stability. In 1994, a generalization of Rassias’ theorem was obtained by Găvruţa [25], who replaced ε(∥x∥p+∥y∥p) by a general control function φ(x,y). The functional equation,
(1.5)f(x1+x2)+f(x1-x2)=2f(x1)+2f(x2),
is related to symmetric biadditive function and is called a quadratic functional equation and every solution of the quadratic equation (1.5) is said to be a quadratic function. For more details about the results concerning such problems, the reader is referred to [26–28].The functional equation,(1.6)f(2x1+x2)+f(2x1-x2)=2f(x1+x2)+2f(x1-x2)+12f(x1),
is called the cubic functional equation, since the function f(x)=cx3 is its solution. Every solution of the cubic functional equation is said to be a cubic mapping. The stability results for the cubic functional equation were proved by Jun and Kim [29].Eshaghi Gordji and Khodaei [30] have established the general solution and investigated the Hyers-Ulam-Rassias stability for a mixed type of cubic, quadratic, and additive functional equations, with f(0)=0,
(1.7)f(x1+kx2)+f(x1-kx2)=k2f(x1+x2)+k2f(x1-x2)+2(1-k2)f(x1)
in quasi-Banach spaces, where k is nonzero integer numbers with k≠±1. It is easy to see that the function f(x)=ax+bx2+cx3 is a solution of the functional equation (1.7), see [31, 32]. The stability of different functional equations in probabilistic normed spaces, RN-spaces, and fuzzy normed spaces has been studied in [6, 7, 33–37].Now, we introduce the new mixed type of cubic, quadratic, and additive functional equation inn-variables as follows:
(1.8)∑k=2n(∑i1=2k∑i2=i1+1k+1⋯∑in-k+1=in-k+1n)f(∑i=1,i≠i1,…,in-k+1nxi-∑r=1n-k+1xir)+f(∑i=1nxi)=2n-2∑i=2n(f(x1+xi)+f(x1-xi))-2n-1(n-2)f(x1),
where n≥3 and f(0)=0. As a special case, if n=3 in (1.8), then (1.8) reduces to
(1.9)∑i1=22∑i2=i1+13f(∑i=1,i≠i1,i23xi-∑r=12xir)+∑i1=23f(∑i=1,i≠i13xi-xi1)+f(∑i=13xi)=2∑i=23(f(x1+xi)+f(x1-xi))-22f(x1),
that is,
(1.10)f(x1-x2-x3)+f(x1-x2+x3)+f(x1+x2-x3)+f(x1+x2+x3)=2(f(x1+x2)+f(x1-x2)+f(x1+x3)+f(x1-x3))-4f(x1).
The main purpose of this paper is to prove the stability for (1.8), in Menger probabilistic normed spaces.


## 2. Results in Menger Probabilistic Normed Spaces

We start our work with a general solution for the mixed functional equation (1.8) and then investigate the stability of this equation in Menger PN-space.Theorem 2.1.
LetX and  Y be vector spaces. A function  f:X→Y with  f(0)=0 satisfies  (1.8) for all  x1,…,xn∈X if and only if there exist functions  C:X×X×X→Y,  B:X×X→Y and  A:X→Y such that  f(x)=C(x,x,x)+B(x,x)+A(x) for all  x∈X, where the function  C is symmetric for each fixed one variable and is additive for fixed two variables,  B is symmetric biadditive and  A is additive.Proof.
If there exists a functionC that is symmetric for each fixed one variable and is additive for fixed two variables, B is biadditive, and A is additive, then by a simple computation one can show that the functions x↦C(x,x,x), x↦B(x,x), and x↦A(x) satisfy the functional equation (1.8). Therefore, the function f satisfies (1.8).
Conversely, letf with f(0)=0 satisfies (1.8). Hence, according to (1.8), we get
(2.1)∑i1=22∑i2=i1+13⋯∑in-1=in-2+1nf(∑i=1,i≠i1,…,in-1nxi-∑r=1n-1xir)+∑i1=23∑i2=i1+14⋯∑in-2=in-3+1nf(∑i=1,i≠i1,…,in-2nxi-∑r=1n-2xir)+⋯+∑i1=2nf(∑i=1,i≠i1nxi-xi1)+f(∑i=1nxi)=2n-2∑i=2n(f(x1+xi)+f(x1-xi))-2n-1(n-2)f(x1)
for all x1,…,xn∈X. Setting xi=0(i=4,…,n) in (2.1), we have
(2.2)f(x1-x2-x3)+((n-3)f(x1-x2-x3)+f(x1-x2+x3)+f(x1+x2-x3))+((n-32)f(x1-x2-x3)+(n-3n-4)f(x1-x2+x3)+(n-3n-4)f(x1+x2-x3)+(n-3n-3)f(x1+x2+x3))+((n-33)f(x1-x2-x3)+(n-3n-5)f(x1-x2+x3)+(n-3n-5)f(x1+x2-x3)+(n-3n-4)f(x1+x2+x3))+⋯+((n-3n-3)f(x1-x2-x3)+(n-31)f(x1-x2+x3)+(n-31)f(x1+x2-x3)+(n-32)f(x1+x2+x3))+(f(x1-x2+x3)+f(x1+x2-x3)+(n-3)f(x1+x2+x3))+f(x1+x2+x3)=2n-2∑i=23(f(x1+xi)+f(x1-xi))-2n-1f(x1),
that is,
(2.3)(1+∑l=1n-3(n-3l))(f(x1-x2-x3)+f(x1-x2+x3)+f(x1+x2-x3)+f(x1+x2+x3))=2n-2(f(x1+x2)+f(x1-x2)+f(x1+x3)+f(x1-x3))-2n-1f(x1)
for all x1,x2,x3∈X. On the other hand, we have the relation:
(2.4)1+∑l=1n-3(n-3l)=∑i=0n-3(n-3l)=2n-3
for all n≥3. Hence, we obtain from (2.3) and (2.4) that
(2.5)f(x1-x2-x3)+f(x1-x2+x3)+f(x1+x2-x3)+f(x1+x2+x3)=2(f(x1+x2)+f(x1-x2)+f(x1+x3)+f(x1-x3))-4f(x1)
for all x1,x2,x3∈X. Replacing x3 by x2 in (2.5), one gets
(2.6)f(x1+2x2)+f(x1-2x2)=4f(x1+x2)+4f(x1-x2)-6f(x1)
for all x1,x2∈X. Therefore, f satisfies (1.7) for k=2. By Theorem  2.3 of [30], there exist an additive function A:X→Y, symmetric biadditive function B:X×X→Y, and C:X×X×X→Y such that f(x)=C(x,x,x)+B(x,x)+A(x) for all x∈X, where the function C is symmetric for each fixed one variable and is additive for fixed two variables.From now on, letX be a real linear space and let (Y,Λ,T) be a complete Menger PN-space. For convenience, we use the following abbreviation for a given function f:X→Y:(2.7)Δf(x1,…,xn)=∑k=2n(∑i1=2k∑i2=i1+1k+1⋯∑in-k+1=in-k+1n)f(∑i=1,i≠i1,…,in-k+1nxi-∑r=1n-k+1xir)+f(∑i=1nxi)-2n-2∑i=2n(f(x1+xi)+f(x1-xi))+2n-1(n-2)f(x1)
for all x1,…,xn∈X.Theorem 2.2.
Letξ:Xn→D+(n∈ℕ,  n≥3 and  ξ(x1,…,xn) is denoted by  ξx1,…,xn) be a function such that
(2.8)lim⁡m→∞ξ2mx1,…,2mxn(22mt)=1
for all x1,…,xn∈X, t>0 and
(2.9)lim⁡m→∞Tl=1∞(ξ0,…,0,2m+l-1x,2m+l-1x(22m+l+n-2t))=1
for all x∈X and t>0. Suppose that an even function f:X→Y with f(0)=0 satisfies the inequality:
(2.10)ΛΔf(x1,…,xn)(t)≥ξx1,…,xn(t)
for all x1,…,xn∈X and t>0. Then, there exists a unique quadratic function Q:X→Y such that
(2.11)Λf(x)-Q(x)(t)≥Tl=1∞(ξ0,…,0,2l-1x,2l-1x(2l+n-2t))
for all x∈X and t>0.Proof.
Settingxi=0(i=1,…,n-2) and xn-1=xn=x in (2.10) and then use f(0)=0, we obtain that
(2.12)Λ(1+∑l=1n-3(n-3l)(n-3l))(f(2x)+f(-2x))-2n-1(f(x)+f(-x))(t)≥ξ0,…,0,x,x(t)
for all x∈X and t>0. By using evenness of f and the relation 1+∑ℓ=1n-3(n-3ℓ)=∑i=0n-3(n-3ℓ)=2n-3, we get
(2.13)Λ2n-2f(2x)-2nf(x)(t)≥ξ0,…,0,x,x(t)
for all x∈X and t>0. So,
(2.14)Λf(2x)/22-f(x)(t)≥ξ0,…,0,x,x(2nt)≥ξ0,…,0,x,x(2n-1t)
for all x∈X and t>0, which implies that
(2.15)Λf(2l+1x)/22(l+1)-f(2lx)/22l(t)≥ξ0,…,0,2lx,2lx(2n+2lt)
for all x∈X, t>0 and ℓ∈ℕ. It follows from (2.15) and (PN4) that
(2.16)Λf(22x)/24-f(x)(t)≥T(Λf(22x)/24-f(2x)/22(t2),Λf(2x)/22-f(x)(t2))≥T(ξ0,…,0,2x,2x(2n+1t),ξ0,…,0,x,x(2n-1t))≥T(ξ0,…,0,2x,2x(2nt),ξ0,…,0,x,x(2n-1t)),Λf(23x)/26-f(x)(t)≥T(Λf(23x)/26-f(2x)/22(t2),Λf(2x)/22-f(x)(t2))≥T(T(Λf(23x)/26-f(22x)/24(t4),Λf(22x)/24-f(2x)/22(t4)),Λf(2x)/22-f(x)(t2))≥T(T(ξ0,…,0,22x,22x(2n+2t),ξ0,…,0,2x,2x(2nt)),ξ0,…,0,x,x(2n-1t))≥T(T(ξ0,…,0,22x,22x(2n+1t),ξ0,…,0,2x,2x(2nt)),ξ0,…,0,x,x(2n-1t))=T(ξ0,…,0,x,x(2n-1t),T(ξ0,…,0,2x,2x(2nt),ξ0,…,0,22x,22x(2n+1t)))=T(T(ξ0,…,0,x,x(2n-1t),ξ0,…,0,2x,2x(2nt)),ξ0,…,0,22x,22x(2n+1t))
for all x∈X and t>0. Thus,
(2.17)Λf(2mx)/22m-f(x)(t)≥Tl=1m(ξ0,…,0,2l-1x,2l-1x(2l+n-2t))
for all x∈X and t>0. In order to prove the convergence of the sequence {f(2mx)/22m}, we replace x with 2m′x in (2.17) to find that
(2.18)Λf(2m+m′x)/22(m+m′)-f(2m′x)/22m′(t)≥Tl=1m(ξ0,…,0,2m′+l-1x,2m′+l-1x(22m′+l+n-2t))
for all x∈x and t>0. Since the right-hand side of the inequality tends to 1 as m′ and m tend to infinity, the sequence {f(2mx)/22m} is a Cauchy sequence. Therefore, one can define the function Q:X→Y by Q(x):=limm→∞(1/22m)f(2mx) for all x∈X. Now, if we replace x1,…,xn with 2mx1,…,2mxn in (2.10), respectively, it follows that
(2.19)ΛΔf(2mx1,…,2mxn)/22m(t)≥ξ2mx1,…,2mxn(22mt)
for all x1,…,xn∈x and t>0. By letting m→∞ in (2.19), we find that ΛΔQ(x1,…,xn)(t)=1 for all t>0, which implies that ΔQ(x1,…,xn)=0 thus Q satisfies (1.8). Hence, by Theorem 2.1 (see [30, Lemma  2.1]), the function Q:X→Y is quadratic.
To prove (2.11), take the limit as m→∞ in (2.17).
Finally, to prove the uniqueness of the quadratic functionQ subject to (2.11), let us assume that there exists a quadratic function Q′ which satisfies (2.11). Since Q(2mx)=22mQ(x) and Q′(2mx)=22mQ′(x) for all x∈X and m∈ℕ, from (2.11), it follows that
(2.20)ΛQ(x)-Q′(x)(t)=ΛQ(2mx)-Q′(2mx)(22mt)≥T(ΛQ(2mx)-f(2mx)(22m-1t),Λf(2mx)-Q′(2mx)(22m-1t))≥T(Tl=1∞(ξ0,…,0,2m+l-1x,2m+l-1x(22m+l+n-2t)),Tl=1∞(ξ0,…,0,2m+l-1x,2m+l-1x(22m+l+n-2t)))
for all x∈X and t>0. By letting m→∞ in (2.20), we find that Q=Q′.Theorem 2.3.
Letξ:Xn→D+ be a function such that
(2.21)lim⁡m→∞T(ξ2mx1,…,2mxn(2mt),ξ2mx1,…,2mxn(2m-4t))=1
for all x1,…,xn∈X, t>0 and
(2.22)lim⁡m→∞Tl=1∞(T(ξ2m+lx,2m+l-1x,2m+l-1x,0,…,0,(2m+n-5t),ξ2m+l-1x,2m+l-1x,2m+l-1x,0,…,0(2m+n-7t)))=1
for all x∈X and t>0. Suppose that an odd function f:X→Y satisfies (2.10) for all x1,…,xn∈X and t>0. Then, there exists a unique additive function A:X→Y such that
(2.23)Λf(2x)-8f(x)-A(x)(t)≥Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(2n-4t),ξ2l-1x,2l-1x,2l-1x,0,…,0(2n-6t)))
for all x∈X and t>0.Proof.
Settingx1=x2=x3=x and xi=0(i=4,…,n) in (2.10), we obtain
(2.24)Λ∑l=0n-3(n-3l)(f(3x)+2f(x)+f(-x))-2n-1f(2x)+2n-1f(x)(t)≥ξx,x,x,0,…,0(t)
for all x∈X and t>0. By using oddness of f and the relation ∑i=0n-3(n-3ℓ)=2n-3, we lead to
(2.25)Λf(3x)-4f(2x)+5f(x)(t)≥ξx,x,x,0,…,0(2n-3t)
for all x∈X and t>0. Putting x1=2x,x2=x3=x and xi=0(i=4,…,n) in (2.10), we have
(2.26)Λ∑l=0n-3(n-3l)(f(4x)+2f(2x))-2n-2(2f(3x)+2f(x))+2n-1f(2x)(t)≥ξ2x,x,x,0,…,0(t)
for all x∈X and t>0. So
(2.27)Λf(4x)-4f(3x)+6f(2x)-4f(x)(t)≥ξ2x,x,x,0,…,0(2n-3t)
for all x∈X and t>0. It follows from (2.25), (2.27), and (PN4) that
(2.28)Λf(4x)-10f(2x)+16f(x)(t)≥T(ξ2x,x,x,0,…,0(2n-4t),ξx,x,x,0,…,0(2n-6t))
for all x∈X and t>0. Let g:X→Y be a function defined by g(x):=f(2x)-8f(x) for all x∈X. From (2.28), we conclude that
(2.29)Λg(2x)/2-g(x)(t)≥T(ξ2x,x,x,0,…,0(2n-3t),ξx,x,x,0,…,0(2n-5t))≥T(ξ2x,x,x,0,…,0(2n-4t),ξx,x,x,0,…,0(2n-6t))
for all x∈X and t>0, which implies that
(2.30)Λg(2l+1x)/2l+1-g(2lx)/2l(t)≥T(ξ2l+1x,2lx,2lx,0,…,0(2l+n-3t),ξ2lx,2lx,2lx,0,…,0(2l+n-5t))
for all x∈X, t>0 and ℓ∈ℕ. It follows from (2.30) and (PN4) that
(2.31)Λg(22x)/22-g(x)(t)≥T(Λg(22x)/22-g(2x)/2(t2),Λg(2x)/2-g(x)(t2))≥T(T(ξ22x,2x,2x,0,…,0(2n-3t),ξ2x,2x,2x,0,…,0(2n-5t)),T(ξ2x,x,x,0,…,0(2n-4t),ξx,x,x,0,…,0(2n-6t)))≥T(T(ξ22x,2x,2x,0,…,0(2n-4t),ξ2x,2x,2x,0,…,0(2n-6t)),T(ξ2x,x,x,0,…,0(2n-4t),ξx,x,x,0,…,0(2n-6t))),Λg(23x)/23-g(x)(t)≥T(Λg(23x)/23-g(2x)/2(t2),Λg(2x)/2-g(x)(t2))≥T(T(Λg(23x)/23-g(22x)/22(t4),Λg(22x)/22-g(2x)/2(t4)),Λg(2x)/2-g(x)(t2))≥T(T(T(ξ23x,22x,22x,0,…,0(2n-3t),ξ22x,22x,22x,0,…,0(2n-5t)),T(ξ22x,2x,2x,0,…,0(2n-4t),ξ2x,2x,2x,0,…,0(2n-6t))),T(ξ2x,x,x,0,…,0(2n-4t),ξx,x,x,0,…,0(2n-6t)))≥T(T(T(ξ23x,22x,22x,0,…,0(2n-4t),ξ22x,22x,22x,0,…,0(2n-6t)),T(ξ22x,2x,2x,0,…,0(2n-4t),ξ2x,2x,2x,0,…,0(2n-6t))),T(ξ2x,x,x,0,…,0(2n-4t),ξx,x,x,0,…,0(2n-6t)))=T(T(T(ξ2x,x,x,0,…,0(2n-4t),ξx,x,x,0,…,0(2n-6t)),T(ξ22x,2x,2x,0,…,0(2n-4t),ξ2x,2x,2x,0,…,0(2n-6t))),T(ξ23x,22x,22x,0,…,0(2n-4t),ξ22x,22x,22x,0,…,0(2n-6t)))
for all x∈X and t>0. Thus,
(2.32)Λg(2mx)/2m-g(x)(t)≥Tl=1m(T(ξ2lx,2l-1x,2l-1x,0,…,0(2n-4t),ξ2l-1x,2l-1x,2l-1x,0,…,0(2n-6t)))
for all x∈X and t>0. In order to prove the convergence of the sequence {g(2mx)/2m}, we replace x with 2m′x in (2.32) to find that
(2.33)Λg(2m+m′x)/2m+m′-g(2m′x)/2m′(t)≥Tl=1m(T(ξ2m′+lx,2m′+l-1x,2m′+l-1x,0,…,0(2m′+n-4t),ξ2m′+l-1x,2m′+l-1x,2m′+l-1x,0,…,0(2m′+n-6t)))
for all x∈X and t>0. Since the right-hand side of the inequality tends to 1 as m′ and m tend to infinity, the sequence {g(2mx)/2m} is a Cauchy sequence. Therefore, one can define the function A:X→Y by A(x):=limm→∞(1/2m)g(2mx) for all x∈X. Now, if we replace x1,…,xn with 2mx1,…,2mxn in (2.10), respectively, it follows that
(2.34)ΛΔg(2mx1,…,2mxn)/2m(t)=ΛΔf(2m+1x1,…,2m+1xn)/2m-8(Δf(2mx1,…,2mxn)/2m)(t)≥T(ΛΔf(2m+1x1,…,2m+1xn)/2m(t2),ΛΔf(2mx1,…,2mxn)/2m-3(t2))≥T(ξ2m+1x1,…,2m+1xn(2m-1t),ξ2mx1,…,2mxn(2m-4t))
for all x1,…,xn∈X and t>0. By letting m→∞ in (2.34), we find that ΛΔA(x1,…,xn)(t)=1 for all t>0, which implies ΔA(x1,…,xn)=0, thus A satisfies (1.8). Hence, by Theorem 2.1 (see [30, Lemma  2.2]) the function A:X→Y is additive.
To prove (2.23), take the limit as m→∞ in (2.32).
Finally, to prove the uniqueness of the additive functionA subject to (2.23), let us assume that there exists a additive function A′ which satisfies (2.23). Since A(2mx)=2mA(x) and A′(2mx)=2mA′(x) for all x∈X and m∈ℕ, from (2.23), it follows that
(2.35)ΛA(x)-A′(x)(t)=ΛA(2mx)-A′(2mx)(2mt)≥T(ΛA(2mx)-g(2mx)(2m-1t),Λg(2mx)-A′(2mx)(2m-1t))≥T(Tl=1∞(T(ξ2m+lx,2m+l-1x,2m+l-1x,0,…,0,(2m+n-5t),ξ2m+l-1x,2m+l-1x,2m+l-1x,0,…,0(2m+n-7t))),Tl=1∞(T(ξ2m+lx,2m+l-1x,2m+l-1x,0,…,0,(2m+n-5t),ξ2m+l-1x,2m+l-1x,2m+l-1x,0,…,0(2m+n-7t))))
for all x∈X and t>0. By letting m→∞ in (2.35), we find that A=A′.Theorem 2.4.
Letξ:Xn→D+ be a function such that
(2.36)lim⁡m→∞T(ξ2mx1,…,2mxn(23mt),ξ2mx1,…,2mxn(23m-2t))=1
for all x1,…,xn∈X, t>0 and
(2.37)lim⁡m→∞Tl=1∞(T(ξ2m+lx,2m+l-1x,2m+l-1x,0,…,0,(23m+2l+n-5t),ξ2m+l-1x,2m+l-1x,2m+l-1x,0,…,0(23m+2l+n-7t)))=1
for all x∈X and t>0. Suppose that an odd function f:X→Y satisfies (2.10) for all x1,…,xn∈X and t>0. Then, there exists a unique cubic function C:X→Y such that
(2.38)Λf(2x)-2f(x)-C(x)(t)≥Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(22l+n-4t),ξ2l-1x,2l-1x,2l-1x,0,…,0(22l+n-6t)))
for all x∈X and t>0.Proof.
Similar to proof of Theorem2.3, we obtain (2.28) for all x∈X and t>0. Let h:X→Y be a function defined by h(x):=f(2x)-2f(x) for all x∈X. Therefore, (2.28) implies that
(2.39)Λh(2x)/23-h(x)(t)≥T(ξ2x,x,x,0,…,0(2n-1t),ξx,x,x,0,…,0(2n-3t))≥T(ξ2x,x,x,0,…,0(2n-2t),ξx,x,x,0,…,0(2n-4t))
for all x∈X and t>0, which implies that
(2.40)Λh(2l+1x)/23(l+1)-h(2lx)/23l(t)≥T(ξ2l+1x,2lx,2lx,0,…,0(23l+n-1t),ξ2lx,2lx,2lx,0,…,0(23l+n-3t))
for all x∈X, t>0, and ℓ∈ℕ. It follows from (2.40) and (PN4) that
(2.41)Λh(2mx)/23m-h(x)(t)≥Tl=1m(T(ξ2lx,2l-1x,2l-1x,0,…,0(22l+n-4t),ξ2l-1x,2l-1x,2l-1x,0,…,0(22l+n-6t)))
for all x∈X and t>0. In order to prove the convergence of the sequence {h(2mx)/23m}, we replace x with 2m′x in (2.41) to find that
(2.42)Λh(2m+m′x)/23(m+m′)-h(2m′x)/23m′(t)≥Tl=1m(T(ξ2m′+lx,2m′+l-1x,2m′+l-1x,0,…,0(23m′+2l+n-4t),ξ2m′+l-1x,2m′+l-1x,23m′+l-1x,0,…,0(23m′+2l+n-6t)))
for all x∈x and t>0. Since the right-hand side of the inequality tends to 1 as m′ and m tend to infinity, the sequence {h(2mx)/23m} is a Cauchy sequence. Therefore, one can define the function C:X→Y by C(x):=limm→∞(1/23m)h(2mx) for all x∈X. Now, if we replace x1,…,xn with 2mx1,…,2mxn in (2.10), respectively, it follows that
(2.43)ΛΔh(2mx1,…,2mxn)/23m(t)≥T(ξ2m+1x1,…,2m+1xn(23m-1t),ξ2mx1,…,2mxn(23m-2t))
for all x1,…,xn∈x and t>0. By letting m→∞ in (2.43), we find that ΛΔC(x1,…,xn)(t)=1 for all t>0, which implies ΔC(x1,…,xn)=0, thus C satisfies (1.8). Hence, by Theorem 2.1 (see [30, Lemma  2.2]), the function C:X→Y is cubic. The rest of the proof is similar to the proof of Theorem 2.3.Theorem 2.5.
Letξ:Xn→D+ be a function satisfies  (2.21) for all  x1,…,xn∈X,  t>0 and  (2.22) for all  x∈X and  t>0. Suppose that an odd function  f:X→Y satisfies  (2.10) for all  x1,…,xn∈X and  t>0. Then, there exists a unique additive function  A:X→Y and a unique cubic function  C:X→Y such that(2.44)Λf(x)-A(x)-C(x)(t)≥T(Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(3.2n-4t),ξ2l-1x,2l-1x,2l-1x,0,…,0(3.2n-6t))),Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(3.22l+n-4t),ξ2l-1x,2l-1x,2l-1x,0,…,0(3.22l+n-6t))))
for all x∈X and t>0.Proof.
By Theorems2.3 and 2.4, there exist an additive function A0:X→Y and a cubic function C0:X→Y such that
(2.45)Λf(2x)-8f(x)-A0(x)(t)≥Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(2n-4t),ξ2l-1x,2l-1x,2l-1x,0,…,0(2n-6t))),Λf(2x)-2f(x)-C0(x)(t)≥Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(22l+n-4t),ξ2l-1x,2l-1x,2l-1x,0,…,0(22l+n-6t)))
for all x∈X and t>0. It follows from (2.45) that
(2.46)Λf(x)+(1/6)A0(x)-(1/6)C0(x)(t)≥T(Λf(2x)-8f(x)-A0(x)(3t),Λf(2x)-2f(x)-C0(x)(3t))
for all x∈X and t>0. Thus, we obtain (2.44) by letting A(x)=-(1/6)A0(x) and C(x)=(1/6)C0(x) for all x∈X.
To prove the uniqueness property ofA and C, let A′,C′:X→Y be another additive and cubic functions satisfying (2.44). Let A¯=A-A′ and C¯=C-C′. So,
(2.47)ΛA¯(x)+C¯(x)(t)≥T(Λf(x)-A(x)-C(x)(t2),Λf(x)-A′(x)-C′(x)(t2))≥T(T(Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(3.2n-5t),ξ2l-1x,2l-1x,2l-1x,0,…,0(3.2n-7t))),Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(3.22l+n-5t),ξ2l-1x,2l-1x,2l-1x,0,…,0(3.22l+n-7t)))),T(Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(3.2n-5t),ξ2l-1x,2l-1x,2l-1x,0,…,0(3.2n-7t))),Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(3.22l+n-5t),ξ2l-1x,2l-1x,2l-1x,0,…,0(3.22l+n-7t)))))
for all x∈X and t>0, then (2.47) implies that
(2.48)ΛA¯(2mx)/23m+C¯(2mx)/23m(t)≥T(T(Tl=1∞(T(ξ2m+lx,2m+l-1x,2m+l-1x,0,…,0(3.23m+n-5t),ξ2m+l-1x,2m+l-1x,2m+l-1x,0,…,0(3.23m+n-7t))),Tl=1∞(T(ξ2m+lx,2m+l-1x,2m+l-1x,0,…,0(3.23m+2l+n-5t),ξ2m+l-1x,2m+l-1x,2m+l-1x,0,…,0(3.23m+2l+n-7t)))),T(Tl=1∞(T(ξ2m+lx,2m+l-1x,2m+l-1x,0,…,0(3.23m+n-5t),ξ2m+l-1x,2m+l-1x,2m+l-1x,0,…,0(3.23m+n-7t))),Tl=1∞(T(ξ2m+lx,2m+l-1x,2m+l-1x,0,…,0(3.23m+2l+n-5t),ξ2m+l-1x,2m+l-1x,2m+l-1x,0,…,0(3.23m+2l+n-7t)))))
for all x∈X and t>0. Since the right-hand side of the inequality tends to 1 as m tends to infinity, thus limm→∞A¯(2mx)/23m+C¯(2mx)/23m=0 for all x∈X, which implies that C¯=0. So, from (2.47), we lead to A¯=0.Now, we are ready to prove the main theorem concerning the stability results for (1.8), in Menger PN-space.Theorem 2.6.
Letξ:Xn→D+ be a function satisfies  (2.8) and  (2.21) for all  x1,…,xn∈X,  t>0 and satisfies  (2.9) and  (2.22) for all  x∈X and  t>0. Suppose that a function  f:X→Y satisfies  (2.10) for all  x1,…,xn∈X and  t>0. Furthermore, assume that  f(0)=0 in  (2.10) for the case  f is even. Then, there exists a unique additive function  A:X→Y, a unique quadratic function  Q:X→Y, and a unique cubic function  C:X→Y satisfying
(2.49)Λf(x)-A(x)-Q(x)-C(x)(t)≥T[T(Tl=1∞(ξ0,…,0,2l-1x,2l-1x(2l+n-3t)),Tl=1∞(ξ0,…,0,2l-1x,2l-1x(2l+n-3t))),T(T(Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(3.2n-5t),ξ2l-1x,2l-1x,2l-1x,0,…,0(3.2n-7t))),Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(3.22l+n-5t),ξ2l-1x,2l-1x,2l-1x,0,…,0(3.22l+n-7t)))),T(Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(3.2n-5t),ξ2l-1x,2l-1x,2l-1x,0,…,0(3.2n-7t))),Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(3.22l+n-5t),ξ2l-1x,2l-1x,2l-1x,0,…,0(3.22l+n-7t)))))]
for all x∈X and t>0.Proof.
Letfe(x)=(1/2)(f(x)+f(-x)) for all x∈X. Then, fe(0)=0,fe(-x)=fe(x) and
(2.50)ΛΔfe(x1,…,xn)(t)=Λ(Δf(x1,…,xn)+Δf(-x1,…,-xn))/2(t)≥T(ΛΔf(x1,…,xn)(t),ΛΔf(-x1,…,-xn)(t))≥T(ξx1,…,xn(t),ξ-x1,…,-xn(t))=T(ξx1,…,xn(t),ξx1,…,xn(t))
for all x1,…,xn∈X and t>0. Hence, in view of Theorem 2.2, there exist a unique quadratic function Q:X→Y such that
(2.51)Λfe(x)-Q(x)(t)≥T(Tl=1∞(ξ0,…,0,2l-1x,2l-1x(2l+n-2t)),Tl=1∞(ξ0,…,0,2l-1x,2l-1x(2l+n-2t)))
for all x∈X and t>0. On the other hand, let fo(x)=(1/2)(f(x)-f(-x)) for all x∈X. Then, fo(0)=0, fo(-x)=-fo(x) and by using the above method, from Theorem 2.5, there exist a unique additive function A:X→Y and a unique cubic function C:X→Y such that
(2.52)Λfo(x)-A(x)-C(x)(t)≥T(T(Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(3.2n-4t),ξ2l-1x,2l-1x,2l-1x,0,…,0(3.2n-6t))),Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(3.22l+n-4t),ξ2l-1x,2l-1x,2l-1x,0,…,0(3.22l+n-6t)))),T(Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(3.2n-4t),ξ2l-1x,2l-1x,2l-1x,0,…,0(3.2n-6t))),Tl=1∞(T(ξ2lx,2l-1x,2l-1x,0,…,0(3.22l+n-4t),ξ2l-1x,2l-1x,2l-1x,0,…,0(3.22l+n-6t)))))
for all x∈X and t>0. Hence, (2.49) follows from (2.51) and (2.52).



---

*Source: 392179-2012-05-07.xml*