# Hyers-Ulam-Rassias RNS Approximation of Euler-Lagrange-Type Additive Mappings

Authors: H. Azadi Kenary; H. Rezaei; A. Ebadian; A. R. Zohdi
Journal: Mathematical Problems in Engineering
 (2012)

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## Abstract

Recently the generalized Hyers-Ulam (or Hyers-Ulam-Rassias) stability of the following functional equation∑j=1mf(-rjxj+∑1≤i≤m,i≠jrixi)+2∑i=1mrif(xi)=mf(∑i=1mrixi) where r1,…,rm∈ℝ, proved in Banach modules over a unital C*-algebra. It was shown that if ∑i=1mri≠0, ri,rj≠0 for some 1≤i<j≤m and a mapping f:X→Y satisfies the above mentioned functional equation then the mapping f:X→Y is Cauchy additive. In this paper we prove the Hyers-Ulam-Rassias stability of the above mentioned functional equation in random normed spaces (briefly RNS).


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## Body

## 1. Introduction and Preliminaries

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 has provided a lot of influence in the development of what we call thegeneralized Hyers-Ulam stability of functional equations. In 1994, a generalization of Rassias’ theorem was obtained by Găvruţa [5] by replacing the bound  ϵ(∥x∥p+∥y∥p) by a general control function φ(x,y).The functional equation:(1.1)f(x+y)+f(x-y)=2f(x)+2f(y),
is called a quadratic functional equation. In particular, every solution of the quadratic functional equation is said to be a quadratic mapping. The generalized Hyers-Ulam stability problem for the quadratic functional equation was proved by Skof [6] for mappings f:X→Y, where X is a normed space and Y is a Banach space. Cholewa [7] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group. Czerwik [8] 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 [2, 4, 5, 9–28]).In the sequel, we will adopt the usual terminology, notions, and conventions of the theory of random normed spaces as in [29]. Throughout this paper, the spaces of all probability distribution functions are denoted by Δ+. Elements of Δ+ are functions F:ℝ∪{-∞,+∞}→[0,1], such that F is left continuous and nondecreasing on ℝ, F(0)=0 and F(+∞)=1. It’s clear that the subset D+={F∈Δ+:l-F(+∞)=1}, where l-f(x)=limt→x-f(t), is a subset of Δ+. The space Δ+ is partially ordered by the usual point-wise ordering of functions, that is, for all t∈R, F≤G if and only if F(t)≤G(t). For every a≥0, Ha(t) is the element of D+ defined by(1.2)Ha(t)={0ift≤a1ift>a.
One can easily show that the maximal element for Δ+ in this order is the distribution function H0(t).Definition 1.1.
A functionT:[0,1]2→[0,1] is a continuous triangular norm (briefly a t-norm) if T satisfies the following conditions:(i)
T is commutative and associative;(ii)
T is continuous;(iii)
T(x,1)=x for all x∈[0,1];(iv)
T(x,y)≤T(z,w) whenever x≤z and y≤w for all x,y,z,w∈[0,1].Three typical examples of continuoust-norms are TP(x,y)=xy,Tmax(x,y)=max{a+b-1,0}, and TM(x,y)=min(a,b). Recall that, if T is a t-norm and {xn} is a given of numbers in [0,1], Ti=1nxi is defined recursively by Ti=11x1 and Ti=1nxi=T(Ti=1n-1xi,xn) for n≥2.Definition 1.2.
A random normed space (brieflyRNS) is a triple (X,μ′,T), where X is a vector space, T is a continuous t-norm, and μ′:X→D+ is a mapping such that the following conditions hold.(i)
μx′(t)=H0(t) for all t>0 if and only if x=0.(ii)
μαx′(t)=μx′(t/|α|) for all α∈ℝ, α≠0, x∈X and t≥0.(iii)
μx+y′(t+s)≥T(μx′(t),μy′(s)), for all x,y∈X and t,s≥0.Definition 1.3.
Let(X,μ′,T) be an RNS.
(i)
A sequence{xn} in X is said to be convergent to x∈X in X if for all t>0, limn→∞μxn-x′(t)=1(ii)
A sequence{xn} in X is said to be Cauchy sequence in X if for all t>0,limn→∞μxn-xm′(t)=1.(iii)
TheRN-space (X,μ′,T) is said to be complete if every Cauchy sequence in X is convergent.Theorem 1.4.
If(X,μ′,T) is RNS and {xn} is a sequence such that xn→x, then limn→∞μxn′(t)=μx′(t).In this paper, we investigate the generalized Hyers-Ulam stability of the following additive functional equation of Euler-Lagrange type:(1.3)∑j=1mf(-rjxj+∑1≤i≤m,i≠jrixi)+2∑i=1mrif(xi)=mf(∑i=1mrixi),
where r1,…,rn∈ℝ,  ∑i=1mri≠0, and ri,rj≠0 for some 1≤i<j≤m, in random normed spaces.Every solution of the functional equation (1.3) is said to be a generalized Euler-Lagrange type additive mapping.


## 2. RNS Approximation of Functional Equation (1.3)

Remark 2.1.
Throughout this paper,r1,…,rm will be real numbers such that ri,rj≠0 for fixed 1≤i<j≤m.Theorem 2.2.
LetX be a real linear space, (Z,μ′,min) be an RN space, φ:Xn→Z be a function such that for some 0<α<2,
(2.1)μφ(2x1,…,2xm)′(t)≥μαφ(x1,…,xm)′(t)∀xi∈X,t>0.f(0)=0 and for all xi∈X and t>0(2.2)lim⁡n→∞⁡μ(φ(2nx1,…,2nxm)/2n)′(t)=1.
Let(Y,μ,min)  be a complete RN space. If f:x→Y is a mapping such that for all xi,xj∈X and t>0(2.3)μ∑j=1mf(-rjxj+∑1≤i≤m,i≠jrixi)+2∑i=1mrif(xi)-mf(∑i=1mrixi)(t)≥μ′φ(x1,…,xm)(t)
then there is a unique generalized Euler-Lagrange-type additive mapping EL:X→Y such that, for all x∈X and all t>0(2.4)μEL(x)-f(x)(t)≥TM(TM(μ′φi,j(x/2ri,-(x/2rj))((2-α)t6),μ′φi,j(x/2ri,0)((2-α)t6),μ′φi,j(0,-(x/2rj))((2-α)t6)),TM(μ′φi,j(x/ri,x/rj)((2-α)t3),μ′φi,j(x/ri,0)((2-α)t3),μ′φi,j(0,x/rj)((2-α)t3))).Proof.
For each1≤k≤m with k≠i,j, let xk=0 in (2.3). Then we get the following inequality:
(2.5)μλ(xi,xj)(t)≥μ′φi,j(xi,xj)(t),
for all xi,xj∈X, where
(2.6)φi,j(x,y):=φ(0,…,0,x︸ith,0,…,0,y︸jth,0,…,0),
for all x,y∈X and all 1≤i<j≤m, and
(2.7)λ(xi,xj)=f(-rixi+rjxj)+f(rixi-rjxj)-2f(rixi+rjxj)+2rif(xi)+2rjf(xj).
Lettingxi=0 in (2.5), we get
(2.8)μf(-rjxj)-f(rjxj)+2rjf(xj)(t)≥μ′φi,j(0,xj)(t),
for all xj∈X. Similarly, letting xj=0 in (2.5), we get
(2.9)μf(-rixi)-f(rixi)+2rif(xi)(t)≥μ′φi,j(xi,0)(t),
for all xi∈X. It follows from (2.5), (2.8), and (2.9) that for all xi,xj∈X(2.10)μλ(xi,xj)-(f(-rixi)-f(rixi)+2rif(xi))-(f(-rjxj)-f(rjxj)+2rjf(xj))(t)≥TM(μ′φi,j(xi,xj)(t3),μ′φi,j(xi,0)(t3),μ′φi,j(0,xj)(t3)).
Replacingxi and xj by (x/ri) and (y/rj) in (2.10), we get that
(2.11)μf(-x+y)+f(x-y)-2f(x+y)+f(x)+f(y)-f(-x)-f(-y)(t)≥TM(μ′φi,j(x/ri,y/rj)(t3),μ′φi,j(x/ri,0)(t3),μ′φi,j(0,y/rj)(t3)),
for all x,y∈X. Putting y=x in (2.11), we get
(2.12)μ2f(x)-2f(-x)-2f(2x)(t)≥TM(μ′φi,j(x/ri,x/rj)(t3),μ′φi,j(x/ri,0)(t3),μ′φi,j(0,x/rj)(t3)),
for all x∈X. Replacing x and y by (x/2) and -(x/2) in (2.11), respectively, we get
(2.13)μf(x)+f(-x)(t)≥TM(μ′φi,j(x/2ri,-(x/2rj))(t3),μ′φi,j(x/2ri,0)(t3),μ′φi,j(0,-(x/2rj))(t3)),
for all x∈X. It follows from (2.12) and (2.13) that
(2.14)μf(2x)-2f(x)(t)=μf(x)+f(-x)+((2f(x)-2f(-x)-2f(2x))/2)(t)≥TM(μf(x)+f(-x)(t2),μ2f(x)-2f(-x)-2f(2x)(t))≥TM(TM(μ′φi,j(x/2ri,-(x/2rj))(t6),μ′φi,j(x/2ri,0)(t6),μ′φi,j(0,-(x/2rj))(t6)),TM(μ′φi,j(x/ri,x/rj)(t3),μ′φi,j(x/ri,0)(t3),μ′φi,j(0,x/rj)(t3))),
for all x∈X. So
(2.15)μ(f(2x)/2)-f(x)(t)≥TM(μ′φi,j(x/ri,x/rj)(2t3),μ′φi,j(x/ri,0)(2t3),μ′φi,j(0,x/rj)(2t3)(2t3)TM(μ′φi,j(x/2ri,-(x/2rj))(t3),μ′φi,j(x/2ri,0)(t3),(2t3)(μ′φi,j(xri,xrj)(2t3),μ′φi,j(xri,0)(2t3),μ′φi,j(0,xrj)(2t3))μ′φi,j(0,-(x/2rj))(t3)(t3)),TM(μ′φi,j(x/ri,x/rj)(2t3),μ′φi,j(x/ri,0)(2t3),μ′φi,j(0,x/rj)(2t3))).
Replacingx by 2nx in (2.15) and using (2.1), we get
(2.16)μ(f(2n+1x)/2n+1)-(f(2nx)/2n)(t)≥TM(TM(μ′φi,j((2nx/2ri),-(2nx/2rj))(2nt3),μ′φi,j((2nx/2ri),0)(2nt3),μ′φi,j(0,-(2nx/2rj))(2nt3)),TM(μ′φi,j((2nx/ri),(2nx/rj))(2n+1t3),μ′φi,j((2nx/ri),0)(2n+1t3),μ′φi,j(0,(2nx/rj))(2n+1t3)))≥TM(TM(μ′φi,j(x/2ri,-(x/2rj))(2nt3αn),μ′φi,j(x/2ri,0)(2nt3αn),μ′φi,j(0,-(x/2rj))(2nt3αn)),TM(μ′φi,j(x/ri,x/rj)(2n+1t3αn),μ′φi,j(x/ri,0)(2n+1t3αn),μ′φi,j(0,x/rj)(2n+1t3αn))),
for all x∈X and all n∈ℕ. Therefore, we have
(2.17)μ(f(2nx)/2n)-f(x)(∑k=0n-1αkt2k)=μ∑k=0n-1((f(2k+1x)/2k+1)-(f(2kx)/2k))(∑k=0n-1αkt2k)≥Tk=0n-1(μ((f(2k+1x)/2k+1)-(f(2kx)/2k))(αkt2k))≥Tk=0n-1(TM(TM(μ′φi,j(x/2ri,-(x/2rj))(t3),μ′φi,j(x/2ri,0)(t3),μ′φi,j(0,-(x/2rj))(t3)),TM(μ′φi,j(x/ri,x/rj)(2t3),μ′φi,j(x/ri,0)(2t3),μ′φi,j(0,x/rj)(2t3))))=TM(TM(μ′φi,j(x/2ri,-(x/2rj))(t3),μ′φi,j(x/2ri,0)(t3),μ′φi,j(0,-(x/2rj))(t3)),TM(μ′φi,j(x/ri,x/rj)(2t3),μ′φi,j(x/ri,0)(2t3),μ′φi,j(0,x/rj)(2t3))),
for all x∈X. This implies that
(2.18)μ(f(2nx)/2n)-f(x)(t)≥TM(TM(μ′φi,j(x/2ri,-(x/2rj))(t3∑k=0n-1(αk/2k)),μ′φi,j(x/2ri,0)(t3∑k=0n-1αk/2k),μ′φi,j(0,-(x/2rj))(t3∑k=0n-1(αk/2k))),TM(μ′φi,j(x/ri,x/rj)(2t3∑k=0n-1(αk/2k)),μ′φi,j(x/ri,0)(2t3∑k=0n-1(αk/2k)),μ′φi,j(0,x/rj)(2t3∑k=0n-1(αk/2k)))).
Replacingx by 2px in (2.18), we obtain
(2.19)μ(f(2n+px)/2n+p)-(f(2px)/2p)(t)≥TM(TM(μ′φi,j(x/2ri,-(x/2rj))(t3∑k=pp+n-1(αk/2k)),μ′φi,j(x/2ri,0)(t3∑k=pp+n-1(αk/2k)),μ′φi,j(0,-(x/2rj))(t3∑k=pp+n-1(αk/2k))),TM(μ′φi,j(x/ri,x/rj)(2t3∑k=pp+n-1(αk/2k)),μ′φi,j(x/ri,0)(2t3∑k=pp+n-1(αk/2k))μ′φi,j(0,x/rj)(2t3∑k=pp+n-1(αk/2k))))
Since the right-hand side of the above inequality tends to 1, whenp,n→∞, then the sequence {f(2kx)/2k}n=1+∞ is a Cauchy sequence in complete RN space (Y,μ,min), so there exists some point EL(x)∈Y such that
(2.20)EL(x)=lim⁡n→∞f(2kx)2k,
for all x∈X.
Fixx∈X and put P=0 in (2.19). Then we obtain
(2.21)μ(f(2nx)/2n)-f(x)(t)≥TM(TM(μ′φi,j(x/2ri,-(x/2rj))(t3∑k=0n-1(αk/2k)),μ′φi,j(x/2ri,0)(t3∑k=0n-1(αk/2k)),μ′φi,j(0,-(x/2rj))(t3∑k=0n-1(αk/2k))),TM(μ′φi,j(x/ri,x/rj)(2t3∑k=0n-1(αk/2k)),μ′φi,j(x/ri,0)(2t3∑k=0n-1(αk/2k)),μ′φi,j(0,x/rj)(2t3∑k=0n-1(αk/2k)))),
and so, for every ϵ>0, we have
(2.22)μEL(x)-f(x)(t+ϵ)≥T(μEL(x)-(f(2nx)/2n)(ϵ),μ(f(2nx)/2n)-f(x)(t))≥T(μEL(x)-(f(2nx)/2n)(ϵ),TM(TM(μ′φi,j(x/2ri,-(x/2rj))(t3∑k=0n-1(αk/2k)),μ′φi,j(x/2ri,0)(t3∑k=0n-1(αk/2k))μ′φi,j(0,-(x/2rj))(t3∑k=0n-1(αk/2k))),TM(μ′φi,j(x/ri,x/rj)(2t3∑k=0n-1(αk/2k)),μ′φi,j(x/ri,0)(2t3∑k=0n-1(αk/2k)),μ′φi,j(0,x/rj)(2t3∑k=0n-1(αk/2k))))).
Taking the limit asn→∞ and using (2.22), we get
(2.23)μEL(x)-f(x)(t+ϵ)≥TM(TM(μ′φi,j(x/2ri,-(x/2rj))((2-α)t6),μ′φi,j(x/2ri,0)((2-α)t6),μ′φi,j(0,-(x/2rj))((2-α)t6)),TM(μ′φi,j(x/ri,x/rj)((2-α)t3),μ′φi,j(x/ri,0)((2-α)t3),μ′φi,j(0,x/rj)((2-α)t3))).
Sinceϵ was arbitrary by taking ϵ→0 in (2.23), we get
(2.24)μEL(x)-f(x)(t)≥TM(TM(μ′φi,j(x/2ri,-(x/2rj))((2-α)t6),μ′φi,j(x/2ri,0)((2-α)t6),μ′φi,j(0,-(x/2rj))((2-α)t6)),TM(μ′φi,j(x/ri,x/rj)((2-α)t3),μ′φi,j(x/ri,0)((2-α)t3),μ′φi,j(0,x/rj)((2-α)t3))).
Replacingxi by 2nxi for all 1≤i≤m, in (2.3), we get for all xi,xj∈X and for all t>0,
(2.25)μ∑j=1mf(-2nrjxj+∑1≤i≤m,i≠j2nrixi)+2∑i=1mrif(2nxi)-mf(∑i=1m2nrixi)/2n(t)≥μ′φ(2nx1,…,2nxm)/2n(t).
since(2.26)lim⁡n→∞⁡μφ(2nx1,…,2nxm)/2n′(t)=1,
We conclude that
(2.27)∑j=1mEL(-rjxj+∑1≤i≤m,i≠jrixi)+2∑i=1mriEL(xi)-mEL(∑i=1mrixi)=0.
To prove the uniqueness of mappingEL, assume that there exists another mapping A:X→Y which satisfies (2.4). Fix x∈X, clearly EL(2nx)=2nEL(x) and A(2nx)=2nA(x), for all n∈N. Since μEL(x)-A(x)(t)=limn→∞μ(EL(2nx)/2n)-(A(2nx)/2n)(t), so
(2.28)μ(EL(2nx)/2n)-(A(2nx)/2n)(t)≥min⁡{μ(EL(2nx)/2n)-(f(2nx)/2n)(t)(t2),μ(f(2nx)/2n)-(A(2nx)/2n)(t)(t2)}≥TM(TM(μ′φi,j(x/2ri,-(x/2rj))(2n(2-α)t12αn),μ′φi,j(x/2ri,0)(2n(2-α)t12αn),μ′φi,j(0,-(x/2rj))(2n(2-α)t12αn)),TM(μ′φi,j(x/ri,x/rj)(2n(2-α)t6αn),μ′φi,j(x/ri,0)(2n(2-α)t6αn),μ′φi,j(0,x/rj)(2n(2-α)t6αn))).
Since the right-hand side of the above inequality tends to 1, whenn→∞, therefore, it follows that for all t>0, μEL(x)-A(x)(t)=1 and so EL(x)=A(x). This completes the proof.Corollary 2.3.
LetX be a real linear space, (Z,μ′,min) be an RN space, and (Y,μ,min) a complete RN space. Let 0<p<1, z0∈Z and f:X→Y be a mapping with f(0)=0 and satisfying
(2.29)μ∑j=1mf(-rjxj+∑1≤i≤m,i≠jrixi)+2∑i=1mrif(xi)-mf(∑i=1mrixi)(t)≥μ′(∑k=1m‖xk‖p)z0(t),
for all xi,xj∈X and t>0. Then the limit EL(x)=limn→∞f(2nx)/2n  exists for all x∈X and defines a unique Euler-Lagrange additive mapping EL:X→Y such that
(2.30)μEL(x)-f(x)(t)≥TM(TM(μ′‖x‖pz0(2p|rirj|p(2-2p)t6(|ri|p+|rj|p)),μ′‖x‖pz0(|2ri|p(2-2p)t6),μ′‖x‖pz0(|2rj|p(2-2p)t6)),TM(μ′‖x‖pz0(|rirj|p(2-2p)t3(|ri|p+|rj|p)),μ′‖x‖pz0(|ri|p(2-2p)t3),μ′‖x‖pz0(|rj|p(2-2p)t3))),
for all x∈X and t>0.Proof.
Letα=2p and φ:Xm→Z be defined as φ(x1,…,xm)=(∑k=1m∥xi∥p)z0.Corollary 2.4.
LetX be a real linear space, (Z,μ′,min) be an RN space, and (Y,μ,min) a complete RN space. Let z0∈Z and f:X→Y be a mapping with f(0)=0 and satisfying
(2.31)μ∑j=1mf(-rjxj+∑1≤i≤m,i≠jrixi)+2∑i=1mrif(xi)-mf(∑i=1mrixi)(t)≥μ′δz0(t),
for all xi∈X for all 1≤i≤m and all t>0. Then, the limit C(x)=limn→∞(f(2nx)/2n) exists for all x∈X and defines a unique Euler-Lagrange additive mapping EL:X→Y such that
(2.32)μEL(x)-f(x)(t)≥TM(μ′δz0(t6),μ′δz0(t3)),
for all x∈X and t>0.Proof.
Letα=1 and φ:Xm→Z be defined as φ(x1,…,xm)=δz0.Theorem 2.5.
LetX be a real linear space, (Z,μ′,min) be an RN space, φ:Xm→Z be a function such that for some 0<α<1/2,
(2.33)μ′φ(x1/2,…,xm/2)(t)≥μ′αφ(x1,…,xm)(t)∀xi∈X,t>0,f(0)=0 and for all xi∈X and t>0, limn→∞μ′2nφ(x1/2n,…,xm/2n)(t)=1. Let (Y,μ,min) be a complete RN space. If f:X→Y is a mapping satisfying (2.3), then there is a unique generalized Euler-Lagrange-type additive mapping EL:X→Y such that, for all x∈X(2.34)μEL(x)-f(x)(t)≥TM(TM(μ′φi,j(x/ri,-x/rj)((1-2α)t6α),μ′φi,j(x/rj,0)((1-2α)t6α),μ′φi,j(0,-(x/rj))((1-2α)t6α)),TM(μ′φi,j(x/ri,x/rj)((1-2α)t3α),μ′φi,j(x/ri,0)((1-2α)t3α),μ′φi,j(0,x/rj)((1-2α)t3α))),
for all x∈X and all t>0.Proof.
Replacingx by x/2n+1 in (2.14) and using (2.33), we obtain
(2.35)μ2nf(x/2n)-2n+1f(x/2n+1)(t)≥TM(TM(μ′φi,j(x/2n+2ri,-(x/2n+2rj))(t2n⋅6),μ′φi,j(x/2n+2ri,0)(t2n⋅6),μ′φi,j(0,-(x/2n+2rj))(t2n⋅6)),TM(μ′φi,j(x/2n+1ri,x/2n+1rj)(t2n⋅3),μ′φi,j(x/2n+1ri,0)(t2n⋅3),μ′φi,j(0,x/2n+1rj)(t2n⋅3)))≥TM(TM(μ′φi,j(x/2ri,-(x/2rj))(tαn+12n⋅6),μ′φi,j(x/2ri,0)(tαn+12n⋅6),μ′φi,j(0,-(x/2rj))(tαn+12n⋅6)),TM(μ′φi,j(x/ri,x/j)(tαn+12n⋅3),μ′φi,j(x/ri,0)(tαn+12n⋅3),μ′φi,j(0,x/rj)(tαn+12n⋅3))).
So(2.36)μ2nf(x/2n)-f(x)(∑i=1n-12kαk+1t)≥TM(TM(μ′φi,j(x/ri,-(x/rj))(t6),μ′φi,j(x/ri,0)(t6),μ′φi,j(0,-(x/rj))(t6)),TM(μ′φi,j(x/ri,x/rj)(t3),μ′φi,j(x/ri,0)(t3),μ′φi,j(0,x/rj)(t3))),
for all x∈X. This implies that
(2.37)μ2nf(x/2n)-f(x)(t)≥TM(TM(μ′φi,j(x/ri,-(x/rj))(t6α∑k=0n-12kαk),μ′φi,j(x/ri,0)(t6α∑k=0n-12kαk),μ′φi,j(0,-(x/rj))(t6α∑k=0n-12kαk)),TM(μ′φi,j(x/ri,x/rj)(t3α∑k=0n-12kαk),μ′φi,j(x/ri,0)(t3α∑k=0n-12kαk),μ′φi,j(0,x/rj)(t3α∑k=0n-12kαk))).
Proceeding as in the proof of Theorem2.2, one can easily show that the sequence {2nf(x/2n)}n=1+∞ is a Cauchy sequence in complete RN space (Y,μ,min), so there exists some point EL(x)∈Y such that
(2.38)EL(x)=lim⁡n→∞2nf(x2n),
for all x∈X.
Taking the limitn→∞ from both sides of the above inequality, we obtain (2.34).
The rest of the proof is similar to the proof of Theorem2.2.Corollary 2.6.
LetX be a real linear space, (Z,μ′,min) be an RN space and (Y,μ,min) a complete RN space. Let p>1, z0∈Z and f:X→Y be a mapping with f(0)=0 and satisfying
(2.39)μ∑j=1mf(-rjxj+∑1≤i≤m,i≠jrixi)+2∑i=1mrif(xi)-mf(∑i=1mrixi)(t)≥μ′(∑k=1m‖xi‖p)z0(t),
for all xi∈X for all 1≤i≤m and all t>0. Then the limit EL(x)=limn→∞2nf(x/2n) exists for all x∈X and defines a unique Euler-Lagrange additive mapping EL:X→Y such that
(2.40)μEL(x)-f(x)(t)≥TM(TM(μ′‖x‖pz0(2p|rirj|p(2p-2)t6(|ri|p+|rj|p)),μ′‖x‖pz0(|2ri|p(2p-2)t6),μ′‖x‖pz0(|2rj|p(2p-2)t6)),TM(μ′‖x‖pz0(|rirj|p(2p-2)t3(|ri|p+|rj|p)),μ′‖x‖pz0(|ri|p(2p-2)t3),μ′‖x‖pz0(|rj|p(2p-2)t3))),
for all x∈X and t>0.Proof.
Letα=2-p and φ:Xm→Z be defined as φ(x1,…,xm)=(∑k=1m∥xi∥p)z0.Corollary 2.7.
LetX be a real linear space, (Z,μ′,min) be an RN space and (Y,μ,min) a complete RN space. Let z0∈Z and f:X→Y be a mapping with f(0)=0 and satisfying
(2.41)μ∑j=1mf(-rjxj+∑1≤i≤m,i≠jrixi)+2∑i=1mrif(xi)-mf(∑i=1mrixi)(t)≥μδz0′(t),
for all xi,xj∈X and t>0. Then, the limit EL(x)=limn→∞2nf(x/2n) exists for all x∈X and defines a unique Euler-Lagrange additive mapping EL:X→Y such that
(2.42)μEL(x)-f(x)(t)≥TM(μ′δz0(4t3),μ′δz0(2t3)),
for all x∈X and t>0.Proof.
Letα=1/4 and φ:Xm→Z be defined as φ(x1,…,xm)=δz0.



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*Source: 672531-2012-06-04.xml*