# Solution and Hyers-Ulam-Rassias Stability of

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

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## Abstract

By using fixed point methods and direct method, we establish the
generalized Hyers-Ulam stability of the following additive-quadratic functional equationf(x+ky)+f(x−ky)=f(x+y)+f(x−y)+(2(k+1)/k)f(ky)−2(k+1)f(y) for fixed integers k with k≠0,±1 in fuzzy Banach spaces.


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## Body

## 1. Introduction and Preliminaries

The stability problem of functional equations was originated from a question of Ulam [1] in 1940, concerning the stability of group homomorphisms. Let (G1,·) be a group and let (G2,*,d) be a metric group with the metric d(·,·). Given ϵ>0, does there exist a δ>0, such that if a mapping h:G1→G2 satisfies the inequality d(h(x·y),h(x)*h(y))<δ for all x,y∈G1, then there exists a homomorphism H:G1→G2 with d(h(x),H(x))<ϵ for all x∈G1? In other words, under what condition does there exist a homomorphism near an approximate homomorphism? The concept of stability for functional equation arises when we replace the functional equation by an inequality which acts as a perturbation of the equation. In 1941, Hyers [2] gave a first affirmative answer to the question of Ulam for Banach spaces. Let f:E→E′ be a mapping between Banach spaces such that(1.1)‖f(x+y)-f(x)-f(y)‖≤δ,
for all x,y∈E, and for some δ>0. Then there exists a unique additive mapping T:E→E′ such that(1.2)‖f(x)-T(x)‖≤δ,
for all x∈E. Moreover if f(tx) is continuous in t∈ℝ for each fixed x∈E, then T is linear. In 1978, Rassias [3] provided a generalization of Hyers’ Theorem which allows the Cauchy difference to be unbounded. In 1991, Gajda [4] answered the question for the case p>1, which was raised by Rassias. This new concept is known as Hyers-Ulam-Rassias stability of functional equations (see [5–17]).The functional equation(1.3)f(x+y)+f(x-y)=2f(x)+2f(y)
is related to a symmetric biadditive function. It is natural that this equation is called a quadratic functional equation. In particular, every solution of the quadratic equation (1.3) is said to be a quadratic function. It is well known that a function f between real vector spaces is quadratic if and only if there exists a unique symmetric biadditive function B such that f(x)=B(x,x) for all x (see [6, 18]). The biadditive function B is given by(1.4)B(x,y)=14(f(x+y)-f(x-y)).
A Hyers-Ulam-Rassias stability problem for the quadratic functional equation (1.3) was proved by Skof for functions f:A→B, where A is normed space and B Banach space (see [19–22]). Borelli and Forti [23] generalized the stability result of quadratic functional equations as follows (cf. [24, 25]): let G be an Abelian group, and X a Banach space. Assume that a mapping f:G→X satisfies the functional inequality:(1.5)‖f(x+y)+f(x-y)-2f(x)-2f(y)‖≤φ(x,y),
for all x,y∈G, and φ:G×G→[0,∞) is a function such that(1.6)Φ(x,y):=∑i=0∞14i+1φ(2ix,2iy)<∞,
for all x,y∈G. Then there exists a unique quadratic mapping Q:G→X with the property(1.7)‖f(x)-Q(x)‖≤Φ(x,x),
for all x∈G.Now, we introduce the following functional equation for fixed integersk with k≠0,±1:(1.8)f(x+ky)+f(x-ky)=f(x+y)+f(x-y)+2(k+1)kf(ky)-2(k+1)f(y),
with f(0)=0 in a non-Archimedean space. It is easy to see that the function f(x)=ax+bx2 is a solution of the functional equation (1.8), which explains why it is called additive-quadratic functional equation. For more detailed definitions of mixed type functional equations, we can refer to [26–47].Definition 1.1 (see [48]).
LetX be a real vector space. A function N:X×ℝ→[0,1] is called a fuzzy norm on X if for all x,y∈X and all s,t∈ℝ,(N1)
N(x,t)=0 for t≤0;(N2)
x=0 if and only if N(x,t)=1 for all t>0;(N3)
N(cx,t)=N(x,t/|c|) if c≠0;(N4)
N(x+y,s+t)≥min{N(x,s),N(y,t)};(N5)
N(x,·) is a nondecreasing function of ℝ and limt→∞N(x,t)=1;(N6)
forx≠0,N(x,·) is continuous on ℝ.The pair(X,N) is called a fuzzy normed vector space.Example 1.2.
Let(X,∥·∥) be a normed linear space and α,β>0.Then
(1.9)N(x,t)={αtαt+β‖x‖,t>0,x∈X,0,t≤0,x∈X,
is a fuzzy norm on X.Definition 1.3.
Let(X,N) be a fuzzy normed vector space. A sequence {xn} in X is said to be convergent or converge if there exists an x∈X such that limn→∞N(xn-x,t)=1 for all t>0.In this case,x is called the limit of the sequence {xn} in X and one denotes it by N-limn→∞xn=x.Definition 1.4.
Let(X,N) be a fuzzy normed vector space. A sequence {xn} in X is called Cauchy if for each ϵ>0 and each t>0 there exists an n0∈ℕ such that for all n≥n0 and all p>0, one has N(xn+p-xn,t)>1-ϵ.It is well known that every convergent sequence in a fuzzy normed vector space is Cauchy. If each Cauchy sequence is convergent, then the fuzzy norm is said to be complete and the fuzzy normed vector space is called a fuzzy Banach space.Example 1.5.
LetN:ℝ×ℝ→[0,1] be a fuzzy norm on ℝ defined by
(1.10)N(x,t)={tt+|x|,t>0,0,t≤0.
The (ℝ,N) is a fuzzy Banach space. Let {xn} be a Cauchy sequence in ℝ,δ>0, and ϵ=δ/(1+δ).Then there exist m∈ℕ such that for all n≥m and all p>0,one has
(1.11)11+|xn+p-xn|≥1-ϵ.
So |xn+p-xn|<δ for all n≥m and all p>0.Therefore {xn} is a Cauchy sequence in (ℝ,|·|).Let xn→x0∈ℝ as n→∞.Then limn→∞N(xn-x0,t)=1 for all t>0.We say that a mappingf:X→Y between fuzzy normed vector spaces X and Y is continuous at a point x∈X if for each sequence {xn} converging to x0∈X, the sequence {f(xn)} converges to f(x0). If f:X→Y is continuous at each x∈X, then f:X→Y is said to be continuous on X ([49]).Definition 1.6.
LetX be a set. A function d:X×X→[0,∞] is called a generalized metric on X if d satisfies the following conditions:(1)
d(x,y)=0 if and only if x=y for all x,y∈X;(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.Theorem 1.7.
Let (X,d) be a complete generalized metric space and letJ:X→X be a strictly contractive mapping with Lipschitz constant  L<1. Then, for all  x∈X, either
(1.12)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 n0≥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.We have the following theorem from [42], which investigates the solution of (1.8).Theorem 1.8.
A functionf:X→Y with  f(0)=0 satisfies (1.8) for all  x,y∈X if and only if there exist functions  A:X→Y and  Q:X×X→Y, such that  f(x)=A(x)+Q(x,x) for all  x∈X, where the function  Q is symmetric biadditive and  A is additive.


## 2. A Fixed Point Method

Using the fixed point methods, we prove the Hyers-Ulam stability of the additive-quadratic functional equation (1.8) in fuzzy Banach spaces. Throughout this paper, assume that X is a vector space and that (Y,N) is a fuzzy Banach space.Theorem 2.1.
Letφ:X2→[0,∞) be a mapping such that there exists an  α<1 with
(2.1)φ(x,y)≤|k|αφ(xk,yk),
for all x,y∈X. Let f:X→Y be an odd function satisfying f(0)=0 and
(2.2)N(f(k(x+y))+f(k(x-y))-f(kx+y)-f(kx-y)-2(k+1)kf(ky)+2(k+1)f(y),t)≥tt+φ(x,y),
for all x,y∈X and all t>0. Then A(x):=N-limn→∞(f(knx)/kn) exists for all x∈X and defines a unique additive mapping A:X→Y such that
(2.3)N(f(x)-A(x),t)≥(|2k+2|-|2k+2|α)t(|2k+2|-|2k+2|α)t+φ(0,x),
for all x∈X and t>0.Proof.
Note thatf(0)=0 and f(-x)=-f(x) for all x∈X since f is an odd function. Putting x=0 in (2.2), we get
(2.4)N(f(ky)k-f(y),t|2k+2|)≥tt+φ(0,y),
for all y∈X and all t>0. Replacing y by x in (2.4), we have
(2.5)N(f(kx)k-f(x),t|2k+2|)≥tt+φ(0,x),
for all x∈X and all t>0. Consider the set S:={h:X→Y;h(0)=0} and introduce the generalized metric on S:
(2.6)d(g,h)=inf⁡μ∈(0,+∞){N(g(x)-h(x),μt)≥tt+φ(0,x),∀x∈X},
where, as usual, infϕ=+∞. It is easy to show that (S,d) is complete (see [50]). We consider the mapping J:(S,d)→(S,d) as follows:
(2.7)Jg(x):=1kg(kx),
for all x∈X. Let g,h∈S be given such that d(g,h)=β. Then
(2.8)N(g(x)-h(x),βt)≥tt+φ(0,x),
for all x∈X and all t>0. Hence
(2.9)N(Jg(x)-Jh(x),αβt)=N(1kg(kx)-1kh(kx),αβt)=N(g(kx)-h(kx),|k|αβt)≥|k|αt|k|αt+φ(0,x)≥|k|αt|k|αt+|k|αφ(0,x)=tt+φ(0,x),
for all x∈X and all t>0. So d(g,h)=β implies that d(Jg,Jh)≤αβ. This means that d(Jg,Jh)≤αd(g,h) for all g,h∈S. It follows from (2.5) that
(2.10)d(f,Jf)≤1|2k+2|.
By Theorem 1.7, there exists a mapping A:X→Y satisfying the following.
(1)A is a fixed point of J, that is,
(2.11)kA(x)=A(kx),
for all x∈X. The mapping A is a unique fixed point of J in the set M={g∈S:d(h,g)<∞}. This implies that A is a unique mapping satisfying (2.11) such that there exists a μ∈(0,∞) satisfying
(2.12)N(f(x)-A(x),μt)≥tt+φ(0,x),
for all x∈X.
(2)d(Jnf,A)→0 as n→∞. This implies the equality limn→∞(f(knx)/kn)=A(x), for all x∈X.
(3)d(f,A)≤(1/(1-α))d(f,Jf), which implies the inequality
(2.13)d(f,A)≤1|2k+2|-|2k+2|α.
This implies that the inequality (2.3) holds.
It follows from (2.1) and (2.2) that
(2.14)N(f(kn(x+ky))kn+f(kn(x-ky))kn-f(kn(x+y))kn-f(kn(x-y))kn-2(k+1)kf(kn+1y)kn+2(k+1)f(kny)kn,tkn)≥tt+φ(knx,kny),
for all x,y∈X, all t>0, and all n∈ℕ. So
(2.15)N(f(kn(x+ky))kn+f(kn(x-ky))kn-f(kn(x+y))kn-f(kn(x-y))kn-2(k+1)kf(kn+1y)kn+2(k+1)f(kny)kn,t)≥|k|nt|k|nt+|k|nαnφ(x,y),
for all x,y∈X, all t>0, and all n∈ℕ. Since limn→∞(|k|nt/(|k|nt+|k|nαnφ(x,y)))=1 for all x,y∈X and all t>0, we obtain that
(2.16)N(2(k+1)kA(k(x+y))+A(k(x-y))-A(kx+y)-A(kx-y)-2(k+1)kA(ky)+2(k+1)A(y),t2(k+1)k)=1,
for all x,y,z∈X and all t>0. Hence the mapping A:X→Y is additive, as desired.Corollary 2.2.
Letθ≥0 and let  r be a real positive number with  r<1. Let  X be a normed vector space with norm  ∥·∥. Let  f:X→Y be an odd mapping satisfying
(2.17)N(f(k(x+y))+f(k(x-y))-f(kx+y)-f(kx-y)-2(k+1)kf(ky)+2(k+1)f(y),t)≥tt+θ(‖x‖r+‖y‖r),
for all x,y∈X and all t>0. Then the limit A(x):=N-limn→∞(f(knx)/kn) exists for each x∈X and defines a unique additive mapping A:X→Y such that
(2.18)N(f(x)-A(x),t)≥|2k+2|(|k|-|k|r)t|2k+2|(|k|-|k|r)t+|k|θ‖x‖r,
for all x∈X and all t>0.Proof.
The proof follows from Theorem2.1 by taking φ(x,y):=θ(∥x∥r+∥y∥r) for all x,y∈X. Then we can choose α=|k|r-1 and we get the desired result.Theorem 2.3.
Letφ:X2→[0,∞) be a mapping such that there exists an  α<1 with
(2.19)φ(xk,yk)≤α|k|φ(x,y),
for all x,y∈X. Let f:X→Y be an odd mapping satisfying f(0)=0 and (2.2). Then the limit A(x):=N-limn→∞knf(x/kn) exists for all x∈X and defines a unique additive mapping A:X→Y such that
(2.20)N(f(x)-A(x),t)≥(|2k+2|-|2k+2|α)t(|2k+2|-|2k+2|α)t+αφ(0,x),
for all x∈X and all t>0.Proof.
Let(S,d) be the generalized metric space defined as in the proof of Theorem 2.1.
Consider the mappingJ:S→S by
(2.21)Jg(x):=kg(xk),
for all g∈S. Let g,h∈S be given such that d(g,h)=β. Then
(2.22)N(g(x)-h(x),βt)≥tt+φ(0,x),
for all x∈X and all t>0. Hence
(2.23)N(Jg(x)-Jh(x),αβt)=N(kg(xk)-kh(xk),αβt)=N(g(xk)-h(xk),αβt|k|)≥(αt/|k|)αt/|k|+φ(0,x/k)≥tt+φ(0,x),
for all x∈X and all t>0. So d(g,h)=β implies that d(Jg,Jh)≤αβ. This means that d(Jg,Jh)≤αd(g,h) for all g,h∈S. It follows from (2.5) that
(2.24)N(kf(xk)-f(x),kt|2k+2|)≥tt+φ(0,x/k)≥tt+(α/|k|)φ(0,x),
for all x∈X and all t>0. Therefore
(2.25)N(kf(xk)-f(x),αt|2k+2|)≥tt+φ(0,x).
So d(f,Jf)≤α. By Theorem 1.7, there exists a mapping A:X→Y satisfying the following.
(1)A is a fixed point of J, that is,
(2.26)A(xk)=1kA(x),
for all x∈X. The mapping A is a unique fixed point of J in the set Ω={h∈S:d(g,h)<∞}. This implies that A is a unique mapping satisfying (2.26) such that there exists μ∈(0,∞) satisfying
(2.27)N(f(x)-A(x),μt)≥tt+φ(0,x),
for all x∈X and t>0.
(2)d(Jnf,A)→0 as n→∞. This implies the equality N-limn→∞knf(x/kn)=A(x) for all x∈X.
(3)d(f,A)≤d(f,Jf)/(1-L) with f∈Ω, which implies the inequality
(2.28)d(f,A)≤α|2k+2|-|2k+2|α.
This implies that the inequality (2.20) holds.
The rest of proof is similar to the proof of Theorem2.1.Corollary 2.4.
Letθ≥0 and let  r be a real number with  r>1. Let  X be a normed vector space with norm  ∥·∥. Let  f:X→Y be an odd mapping satisfying (2.17). Then  A(x):=N-limn→∞knf(x/kn) exists for each  x∈X and defines a unique additive mapping  A:X→Y such that
(2.29)N(f(x)-A(x),t)≥|2k+2|(|k|r-|k|)t|2k+2|(|k|r-|k|)t+|k|θ‖x‖r,
for all x∈X and all t>0.Proof.
The proof follows from Theorem2.3 by taking φ(x,y):=θ(∥x∥r+∥y∥r) for all x,y∈X. Then we can choose α=|k|1-r and we get the desired result.Theorem 2.5.
Letφ:X2→[0,∞) be a function such that there exists an  α<1 with
(2.30)φ(x,y)≤k2αφ(xk,yk),
for all x,y∈X. Let f:X→Y be an even mapping with f(0)=0 and satisfying (2.2). Then Q(x):=N-limn→∞(f(knx)/k2n) exists for all x∈X and defines a unique quadratic mapping Q:X→Y such that
(2.31)N(f(x)-Q(x),t)≥(2|k|-2|k|α)t(2|k|-2|k|α)t+φ(0,x),
for all x∈X and all t>0.Proof.
Replacingx by kx in (2.2), we get
(2.32)N(f(k(x+y))+f(k(x-y))-f(kx+y)-f(kx-y)-2(k+1)kf(ky)+2(k+1)f(y),t)≥tt+φ(kx,y),
for all x,y∈X and all t>0. Putting x=0 and replacing y by x in (2.32), we have
(2.33)N(f(kx)k-kf(x),t2)≥tt+φ(0,x),
for all x∈X and all t>0. By (2.33), (N3), and (N4), we get
(2.34)N(f(kx)k2-f(x),t2|k|)≥tt+φ(0,x),
for all x∈X and all t>0. Consider the set S*:={h:X→Y;h(0)=0} and introduce the generalized metric on S*:
(2.35)d(g,h)=inf⁡μ∈(0,+∞){N(g(x)-h(x),μt)≥tt+φ(0,x),∀x∈X},
where, as usual, infϕ=+∞. It is easy to show that (S*,d) is complete (see [50]). Now we consider the linear mapping J:(S*,d)→(S*,d) such that
(2.36)Jg(x):=1k2g(kx),
for all x∈X. Proceeding as in the proof of Theorem 2.1, we obtain that d(g,h)=β implies that d(Jg,Jh)≤αβ. This means that d(Jg,Jh)≤αd(g,h) for all g,h∈S. It follows from (2.34) that
(2.37)d(f,Jf)≤12|k|.
By Theorem 1.7, there exists a mapping Q:X→Y such that one has the folowing.
(1)Q is a fixed point of J, that is,
(2.38)k2Q(x)=Q(kx),
for all x∈X. The mapping Q is a unique fixed point of J in the set M={g∈S*:d(h,g)<∞}. This implies that Q is a unique mapping satisfying (2.38) such that there exists a μ∈(0,∞) satisfying N(f(x)-Q(x),μt)≥t/(t+φ(0,x)) for all x∈X.
(2)d(Jnf,Q)→0 as n→∞. This implies the equality limn→∞(f(knx)/k2n)=Q(x) for all x∈X.
(3)d(f,Q)≤(1/(1-α))d(f,Jf), which implies the inequality d(f,Q)≤1/(2|k|-2|k|α). This implies that the inequality (2.31) holds.
The rest of the proof is similar to the proof of Theorem2.1.Corollary 2.6.
Letθ≥0 and let  r be a real positive number with  r<1. Let  X be a normed vector space with norm  ∥·∥. Let  f:X→Y be an even mapping with  f(0)=0 and satisfying (2.17). Then the limit  Q(x):=N-limn→∞(f(knx)/k2n) exists for each  x∈X and defines a unique quadratic mapping  Q:X→Y such that
(2.39)N(f(x)-Q(x),t)≥(2k2-2k2r)t(2k2-2k2r)t+|k|θ‖x‖r,
for all x∈X and all t>0.Proof.
The proof follows from Theorem2.5 by taking φ(x,y):=θ(∥x∥r+∥y∥r) for all x,y∈X. Then we can choose α=k2r-2 and we get the desired result.Theorem 2.7.
Letφ:X2→[0,∞) be a function such that there exists an  α<1 with
(2.40)φ(xk,yk)≤αk2φ(x,y),
for all x,y∈X. Let f:X→Y be an even mapping with f(0)=0 and satisfying (2.2). Then the limit Q(x):=N-limn→∞k2nf(x/kn) exists for all x∈X and defines a unique quadratic mapping Q:X→Y such that
(2.41)N(f(x)-Q(x),t)≥(2|k|-2|k|α)t(2|k|-2|k|α)t+αφ(0,x),
for all x∈X and t>0.Proof.
Let(S*,d) be the generalized metric space defined as in the proof of Theorem 2.5. It follows from (2.34) that
(2.42)N(k2f(xk)-f(x),|k|t2)≥tt+φ(0,x/k)≥tt+(α/k2)φ(0,x),
for all x∈X and t>0. So
(2.43)N(f(x)-k2f(xk),αt2|k|)≥tt+φ(0,x).
The rest of the proof is similar to the proofs of Theorems 2.1 and 2.3.Corollary 2.8.
Letθ≥0 and let  r be a real number with  r>1. Let  X be a normed vector space with norm  ∥·∥. Let  f:X→Y be an even mapping with  f(0)=0 and satisfying (2.17). Then  Q(x):=N-limn→∞k2nf(x/kn) exists for each  x∈X and defines a unique quadratic mapping  Q:X→Y such that
(2.44)N(f(x)-Q(x),t)≥(2|k|2r+1-2|k|3)t(2|k|2r+1-2|k|3)t+k2θ‖x‖r,
for all x∈X and all t>0.Proof.
It follows from Theorem2.7 by taking φ(x,y):=θ(∥x∥r+∥y∥r) for all x,y∈X. Then we can choose α=k2-2r and we get the desired result.


## 3. Direct Method

In this section, using direct method, we prove the Hyers-Ulam stability of functional equation (1.8) in fuzzy Banach spaces. Throughout this section, we assume that X is a linear space, (Y,N) is a fuzzy Banach space, and (Z,N′) is a fuzzy normed space. Moreover, we assume that N(x,·) is a left continuous function on ℝ.Theorem 3.1.
Assume that a mappingf:X→Y is an odd mapping with  f(0)=0 satisfying the inequality
(3.1)N(f(k(x+y))+f(k(x-y))-f(kx+y)-f(kx-y)-2(k+1)kf(ky)+2(k+1)f(y),t)≥N′(φ(x,y),t),
for all x,y∈X, t>0, and φ:X2→Z is a mapping for which there is a constant r∈ℝ satisfying 0<|r|<1/|k| such that
(3.2)N′(φ(xk,yk),t)≥N′(φ(x,y),t|r|),
for all x,y∈X and all t>0. Then there exists a unique additive mapping A:X→Y satisfying (1.8) and the inequality
(3.3)N(f(x)-A(x),t)≥N′(φ(0,x),|2k+2|(1-|kr|)t|r|),
for all x∈X and all t>0.Proof.
It follows from (3.2) that
(3.4)N′(φ(xkj,ykj),t)≥N′(φ(x,y),t|r|j),
for all x,y∈X and all t>0. Putting x=0 in (3.1) and then replacing y by x/k, we get
(3.5)N(kf(xk)-f(x),|k|t|2k+2|)≥N′(φ(0,xk),t),
for all x∈X and all t>0. Replacing x by x/kj in (3.5), we have
(3.6)N(kj+1f(xkj+1)-kjf(xkj),|k|j+1t|2k+2|)≥N′(φ(0,xkj+1),t)≥N′(φ(0,x),t|r|j+1),
for all x∈X, all t>0, and all integer j≥0. So
(3.7)N(f(x)-knf(xkn),∑j=0n-1|k|j+1|r|j+1t|2k+2|)=N(∑j=0n-1kj+1f(xkj+1)-kjf(xkj),∑j=0n-1|k|j+1|r|j+1t|2k+2|)≥min⁡0≤j≤n-1⁡{N(kj+1f(xkj+1)-kjf(xkj),|k|j+1|r|j+1t|2k+2|)}≥min⁡0≤j≤n-1⁡{N′(φ(0,x),t)}=N′(φ(0,x),t),
which yields
(3.8)N(kn+pf(xkn+p)-kpf(xkp),∑j=0n-1|k|j+p+1|r|j+1t|2k+2|)≥N′(φ(0,x2p),t)≥N′(φ(0,x),t|r|p),
for all x∈X, t>0, and all integers n>0,  p≥0. So
(3.9)N(kn+pf(xkn+p)-kpf(xkp),∑j=0n-1|k|j+p+1|r|j+p+1t|2k+2|)≥N′(φ(0,x),t),
for all x∈X, t>0, and any integers n>0,  p≥0. Hence one can obtain
(3.10)N(kn+pf(xkn+p)-kpf(xkp),t)≥N′(φ(0,x),t∑j=0n-1(|k|j+p+1|r|j+p+1/|2k+2|)),
for all x∈X, t>0, and any integers n>0,  p≥0. Since the series ∑j=0+∞kj|r|j is a convergent series, we see by taking the limit p→∞ in the last inequality that the sequence {knf(x/kn)} is a Cauchy sequence in the fuzzy Banach space (Y,N) and so it converges in Y. Therefore a mapping A:X→Y defined by A(x):=N-limn→∞knf(x/kn) is well defined for all x∈X. This means that
(3.11)lim⁡n→∞N(A(x)-knf(xkn),t)=1,
for all x∈X and all t>0. In addition, it follows from (3.10) that
(3.12)N(f(x)-knf(xkn),t)≥N′(φ(0,x),t∑j=0n-1(|k|j+1|r|j+1/|2k+2|)),
for all x∈X and all t>0. So
(3.13)N(f(x)-A(x),t)≥min⁡{N(f(x)-knf(xkn),(1-ϵ)t),N(A(x)-knf(xkn),ϵt)}≥N′(φ(0,x),ϵt∑j=0n-1(|k|j+1|r|j+1/|2k+2|))≥N′(φ(0,x),|2k+2|(1-|k||r|)ϵt|kr|),
for sufficiently large n and for all x∈X, t>0, and ϵ with 0<ϵ<1. Since ε is arbitrary and N′ is left continuous, we obtain
(3.14)N(f(x)-A(x),t)≥N′(φ(0,x),|2k+2|(1-|k||r|)t|kr|),
for all x∈X and t>0. It follows from (3.1) that
(3.15)N(f(kn(x+ky))kn+f(kn(x-ky))kn-f(kn(x+y))kn-f(kn(x-y))kn-2(k+1)kf(kn+1y)kn+2(k+1)f(kny)kn,t)≥N′(φ(knx,kny),t|k|n)≥N′(φ(x,y),t|r|n|k|n)⟶1asn⟶+∞,
for all x,y∈X and all t>0. Therefore, we obtain in view of (3.11)
(3.16)N(2(k+1)kA(k(x+y))+A(k(x-y))-A(kx+y)-A(kx-y)-2(k+1)kA(ky)+2(k+1)A(y),t2(k+1)k)≥min⁡{N(f(kn+1y)knA(k(x+y))+A(k(x-y))-A(kx+y)-A(kx-y)-2(k+1)kA(ky)+2(k+1)A(y)-f(kn(x+ky))kn+f(kn(x-ky))kn-f(kn(x+y))kn-f(kn(x-y))kn-2(k+1)kf(kn+1y)kn+2(k+1)f(kny)kn,t2),N(f(kn(x+ky))kn+f(kn(x-ky))kn-f(kn(x+y))kn-f(kn(x-y))kn-2(k+1)kf(kn+1y)kn+2(k+1)f(kny)kn,t2)}=N(f(kn(x+ky))kn+f(kn(x-ky))kn-f(kn(x+y))kn-f(kn(x-y))kn-2(k+1)kf(kn+1y)kn+2(k+1)f(kny)kn,t2)(forsufficientlylargen)≥N′(φ(x,y),t2|k|n|r|n)⟶1asn⟶+∞,
for all x,y∈X and all t>0, which implies that
(3.17)A(k(x+y))+A(k(x-y))=A(kx+y)+A(kx-y)+2(k+1)kA(ky)-2(k+1)A(y).
Hence the mapping A:X→Y is additive, as desired.
To prove the uniqueness, let there be another mappingL:X→Y which satisfies the inequality (3.3). Since L(knx)=knL(x) for all x∈X, we have
(3.18)N(A(x)-L(x),t)=N(knA(xkn)-knL(xkn),t)≥min⁡{N(knA(xkn)-knf(xkn),t2),N(knf(xkn)-knL(xkn),t2)}≥N′(φ(0,xkn),|2k+2|(1-|k||r|)t2|k|n+1|r|)≥N′(φ(0,x),|2k+2|(1-|k||r|)t2|k|n+1|r|n+1)⟶1asn⟶∞,
for all t>0. Therefore A(x)=L(x) for all x∈X. This completes the proof.Corollary 3.2.
LetX be a normed space and let  (ℝ,N′) be a fuzzy Banach space. Assume that there exist real numbers  θ≥0 and  p>1 such that an odd mapping  f:X→Y with  f(0)=0 satisfies the following inequality:
(3.19)N(f(k(x+y))+f(k(x-y))-f(kx+y)-f(kx-y)-2(k+1)kf(ky)+2(k+1)f(y),t)≥N′(θ(‖x‖p+‖y‖p),t),
for all x,y∈X and t>0. Then there is a unique additive mapping A:X→Y satisfying (1.8) and the inequality
(3.20)N(f(x)-A(x),t)≥N′(θ‖x‖p|2k+2|,(|k|p-|k||k|)t).Proof.
Letφ(x,y):=θ(∥x∥p+∥y∥p) and |r|=|k|-p. Applying Theorem 3.1, we get desired results.Theorem 3.3.
Letf:X→Y be an odd mapping with  f(0)=0 satisfying the inequality (3.1) and let  φ:X2→Z be a mapping for which there exists a constant  r∈ℝ satisfying  0<|r|<|k| such that
(3.21)N′(φ(x,y),|r|t)≥N′(φ(xk,yk),t),
for all x,y∈X and all t>0. Then there exists a unique additive mapping A:X→Y satisfying (1.8) and the following inequality:
(3.22)N(f(x)-A(x),t)≥N′(φ(0,x),|2k+2|(|k|-|r|)t|k|),
for all x∈X and all t>0.Proof.
It follows from (3.5) that
(3.23)N(f(kx)k-f(x),t|2k+2|)≥N′(φ(0,x),t),
for all x∈X and all t>0. Replacing x by knx in (3.41), we obtain
(3.24)N(f(kn+1x)kn+1-f(knx)kn,t|2k+2|kn)≥N′(φ(0,knx),t)≥N′(φ(0,x),t|r|n).
So
(3.25)N(f(kn+1x)kn+1-f(knx)kn,|r|nt|2k+2||k|n)≥N′(φ(0,x),t),
for all x∈X and all t>0. Proceeding as in the proof of Theorem 3.1, we obtain that
(3.26)N(f(x)-f(knx)kn,∑j=0n-1|r|jt|2k+2||k|j)≥N′(φ(0,x),t),
for all x∈X, all t>0, and any integer n>0. So
(3.27)N(f(x)-f(knx)kn,t)≥N′(φ(0,x),t∑j=0n-1(|r|j/|2k+2||k|j)).
The rest of the proof is similar to the proof of Theorem 3.1.Corollary 3.4.
LetX be a normed space and let  (ℝ,N′) be a fuzzy Banach space. Assume that there exist real numbers  θ≥0 and  0<p<1 such that an odd mapping  f:X→Y with  f(0)=0 satisfies (3.19). Then there exists a unique additive mapping  A:X→Y satisfying (1.8) and the inequality
(3.28)N(f(x)-A(x),t)≥N′(φ(0,x),|2k+2|(|k|-|k|p)t|k|).Proof.
Letφ(x,y):=θ(∥x∥p+∥y∥p) and |r|=|k|p. Applying Theorem 3.3, we get the desired results.Theorem 3.5.
Letf:X→Y be an even mapping with  f(0)=0 satisfying the inequality (3.1) and let  φ:X2→Z be a mapping for which there exists a constant  r∈ℝ such that  0<|r|<1/k2 and that
(3.29)N′(φ(xk,yk),t)≥N′(φ(x,y),t|r|),
for all x,y∈X and all t>0. Then there exists a unique quadratic mapping Q:X→Y satisfying (1.8) and the inequality
(3.30)N(f(x)-Q(x),t)≥N′(φ(0,x),2(1-|k2r|)t|kr|),
for all x∈X and all t>0.Proof.
Replacingx by kx in (3.1), we get
(3.31)N(f(k(x+y))+f(k(x-y))-f(kx+y)-f(kx-y)-2(k+1)kf(ky)+2(k+1)f(y),t)≥N′(φ(kx,y),t),
for all x,y∈X and all t>0. Putting x=0 and replacing y by x in (3.31), we have
(3.32)N(f(kx)k2-f(x),t|2k|)≥N′(φ(0,x),t),
for all x∈X and all t>0. Replacing x by x/k in (3.32), we find
(3.33)N(k2f(xk)-f(x),|k|t2)≥N′(φ(0,xk),t),
for all x∈X and all t>0. Also, replacing x by x/kn in (3.33), we obtain
(3.34)N(k2n+2f(xkn)-k2nf(xkn),|k|2n+1t2)≥N′(φ(0,xkn+1),t)≥N′(φ(0,x),t|r|n+1).
So
(3.35)N(k2n+2f(xkn)-k2nf(xkn),|k|2n+1|r|n+1t2)≥N′(φ(0,x),t),
for all x∈X and all t>0. Proceeding as in the proof of Theorem 3.1, we obtain that
(3.36)N(f(x)-k2nf(xkn),∑j=0n-1|k|2j+1|r|j+1t2)≥N′(φ(0,x),t),
for all x∈X, all t>0, and any integer n>0. So
(3.37)N(f(x)-k2nf(xkn),t)≥N′(φ(0,x),t∑j=0n-1(|k|2j+1|r|j+1t/2)).
The rest of the proof is similar to the proof of Theorem 3.1.Corollary 3.6.
LetX be a normed space and let  (ℝ,N′) be a fuzzy Banach space. Assume that there exist real numbers  θ≥0 and  p>1 such that an even mapping  f:X→Y with  f(0)=0 satisfies the inequality (3.19). Then there exists a unique quadratic mapping  Q:X→Y satisfying (1.8) and the inequality
(3.38)N(f(x)-Q(x),t)≥N′(θ‖x‖p,2(k2p-k2)t|k|).Proof.
Letφ(x,y):=θ(∥x∥p+∥y∥p) and |r|=|k|-2p. Applying Theorem 3.5, we get the desired results.Theorem 3.7.
Assume that an even mappingf:X→Y with  f(0)=0 satisfies the inequality (3.1) and  φ:X2→Z is a mapping for which there is a constant  r∈ℝ satisfying  0<|r|<k2 such that
(3.39)N′(φ(x,y),|r|t)≥N′(φ(xk,yk),t),
for all x,y∈X and all t>0. Then there exists a unique quadratic mapping Q:X→Y satisfying (1.8) and the following inequality
(3.40)N(f(x)-Q(x),t)≥N′(φ(0,x),2(k2-|r|)t|k|),
for all x∈X and all t>0.Proof.
It follows from (3.32) that
(3.41)N(f(kx)k2-f(x),t|2k|)≥N′(φ(0,x),t),
for all x∈X and all t>0. Replacing x by knx in (3.41), we obtain
(3.42)N(f(kn+1x)k2n+2-f(knx)k2n,t2|k|2n+1)≥N′(φ(0,knx),t)≥N′(φ(0,x),t|r|n),
for all x∈X and all t>0. So
(3.43)N(f(kn+1x)k2n+2-f(knx)k2n,|r|nt2|k|2n+1)≥N′(φ(0,x),t),
for all x∈X and all t>0. So
(3.44)N(f(x)-f(knx)k2n,t)≥N′(φ(0,x),t∑j=0n-1(|r|jt/2|k|2j+1)).
The rest of the proof is similar to the proof of Theorem 3.1.Corollary 3.8.
LetX be a normed space and let  (ℝ,N′) be a fuzzy Banach space. Assume that there exist real numbers  θ≥0 and  0<p<1 such that an even mapping  f:X→Y with  f(0)=0 satisfies (3.19). Then there is a unique quadratic mapping  Q:X→Y satisfying (1.8) and the inequality
(3.45)N(f(x)-Q(x),t)≥N′(φ(0,x),2(k2-k2p)t|k|),
for all x∈X, all t>0.Proof.
Letφ(x,y):=θ(∥x∥p+∥y∥p) and |r|=k2p. Applying Theorem 3.7, we get the desired results.



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*Source: 953938-2012-04-12.xml*