# On the Stability Problem in Fuzzy Banach Space

Authors: G. Zamani Eskandani; P. Găvruţa; Gwang Hui Kim
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We investigate the generalized Ulam-Hyers stability of the Cauchy functional equation and pose two open problems in fuzzy Banach space.


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## Body

## 1. Introduction and Preliminaries

In 1940, Ulam [1] asked the first question on the stability problem. In 1941, Hyers [2] solved the problem of Ulam. This result was generalized by Aoki [3] for additive mappings and by Th. M. Rassias [4] for linear mappings by considering an unbounded Cauchy difference.Theorem 1.1 (Th. M. Rassias).
Letf:E→E′ be a mapping from a normed vector space E into a Banach space E′ subject to the inequality:
(1.1)‖f(x+y)-f(x)-f(y)‖≤ϵ(‖x‖p+‖y‖p)
for all x,y∈E, where ϵ and p are constants with ϵ>0 and 0≤p<1. Then, the limit L(x)=limn→∞(1/2n)f(2nx) exists for all x∈E and L:E→E′ is the unique additive mapping which satisfies
(1.2)‖f(x)-L(x)‖≤2ϵ2-2p‖x‖p
for all x∈E. Also, if for each x∈E the function f(tx) is continuous in t∈ℝ, then L is linear.In 1990, Th. M. Rassias [5] during the 27th International Symposium on Functional Equations asked the question whether such a theorem can also be proved for p≥1. In 1991, Gajda [6] gave an affirmative solution to this question for p>1. It was shown by Gajda [6], as well as by Th. M. Rassias and  Šemrl [7], that one cannot prove a Th. M. Rassias type theorem when p=1. Găvruţa [8] proved that the function f(x)=xln|x|, if x≠0 and f(0)=0 satisfies (1.1) with ϵ=p=1 but
(1.3)sup⁡x≠0|f(x)-A(x)||x|≥sup⁡n∈N|nln⁡n-A(n)|n=sup⁡n∈N|ln⁡n-A(1)|=∞
for any additive function A:ℝ→ℝ. J. M. Rassias [9] replaced the factor ∥x∥p+∥y∥p by ∥x∥p1∥y∥p2 for p1,p2∈ℝ with p1+p2≠1 (see also [10, 11]) and has obtained the following theorem.Theorem 1.2.
LetX be a real normed linear space and Y a real complete normed linear space. Assume that f:X→Y is an approximately additive mapping for which there exist constants θ≥0 and p=p1+p2≠1 such that f satisfies the inequality:
(1.4)‖f(x+y)-f(x)-f(y)‖≤θ‖x‖p1‖y‖p2
for all x,y∈X. Then, there exists a unique additive mapping L:X→Y satisfying
(1.5)‖f(x)-L(x)‖≤θ|2p-2|‖x‖p
for all x∈X. If, in addition, f:X→Y is a mapping such that the transformation t→f(tx) is continuous in t∈ℝ for each fixed x∈X, then L is an ℝ-linear mapping.In the casep=1, we do not have stability [12]. In 1994, a further generalization of Th. M. Rassias’ Theorem was obtained by Găvruţa [13], in which he replaced the bound ϵ(∥x∥p+∥y∥p) by a general control function φ(x,y). Isac and Th. M. Rassias [14] replaced the factor ∥x∥p+∥y∥p by ∥x∥p1+∥y∥p2 in Theorem 1.1 and solved stability problem when p2≤p1<1 or 1<p2≤p1, also they asked the question whether such a theorem can be proved for p2<1<p1. Găvruţa [8] gave a negative answer to this question. Isac and Th. M. Rassias [15] applied the Ulam-Hyers-Rassias stability theory to prove fixed point theorems and study some new applications in nonlinear analysis. During the last two decades, a number of papers and research monographs have been published on various generalizations and applications of the generalized Ulam-Hyers stability to a number of functional equations and mappings (see [16–40]). We also refer the readers to the books of Czerwik [41] and Hyers et al. [42].Th. M. Rassias [43] has obtained the following theorem and posed a problem.Theorem 1.3.
LetE1 and E2 be two Banach spaces, and let f:E1→E2 be a mapping such that f(tx) is continuous in t for each fixed x. Assume that there exist θ≥0 and p∈[0,1) such that
(1.6)‖f(x+y)-f(x)-f(y)‖≤θ(‖x‖p+‖y‖p)
for all x,y∈X. Let k be a positive integer k>2. Then, there exists a unique linear mapping T:E1→E2 such that
(1.7)‖f(x)-T(x)‖≤kθk-kp‖x‖ps(k,p)
for all x∈X, where
(1.8)s(k,p)=1+1k∑m=2k-1‍mp.Th. M. Rassias Problem
What is the best possible value ofk in Theorem 1.3?
Găvruţa et al. have given a generalization of [13] and have answered to Th. M. Rassias problem [44].
In [45], J. M. Rassias et al. have investigated the generalized Ulam-Hyers “product-sum” stability of  functional equations and have obtained the following theorem.Theorem 1.4 (see [45]).
Letf:E→F be a mapping which satisfies the inequality
(1.9)‖f(mx+y)+f(mx-y)-2f(x+y)-2f(x-y)-2(m2-2)f(x)+2f(y)‖F≤ϵ(‖x‖Ep‖y‖Ep+‖x‖E2p+‖y‖E2p)
for allx,y∈E with x⊥y, where ϵ and p are constants with ϵ,p>0 and either m>1, p<1 or m<1, p>1 with m≠0,m≠±1, m≠±2, and -1≠|m|p-1<1. Then, the limit limn→∞m-2nf(mnx) exists for all x∈E and Q:E→F is the unique orthogonally Euler-Lagrange quadratic mapping such that
(1.10)‖f(x)-Q(x)‖F≤ϵ2|m2-m2p|‖x‖E2p
for all x∈E.Note that the mixed “product-sum” function was introduced by J. M. Rassias in 2008-2009 [46–48].We recall some basic facts concerning fuzzy normed space.LetX be a real linear space. A function N:X×ℝ→[0,1] (so-called fuzzy subset) is said to be a fuzzy norm on X if for all x,y∈X and all c,t∈ℝ,(
N
1
)
N
(
x
,
c
)
=
0 for c≤0;(
N
2
)
x
=
0 if and only if N(x,c)=1 for all c>0;(
N
3
)
N
(
c
x
,
t
)
=
N
(
x
,
t
/
|
c
|
) if c≠0;(
N
4
)
N
(
x
+
y
,
t
)
≥
min
{
N
(
x
,
t
)
,
N
(
y
,
t
)
};(
N
5
)
N
(
x
,
·
) is a nondecreasing function of ℝ and
(1.11)lim⁡t→∞N(x,t)=1.The pair(X,N) is called a fuzzy normed linear space. The properties of fuzzy normed vector spaces and examples of fuzzy norms are given in [49–51].Let(X,N) be a fuzzy normed space and let  {xn} be a sequence in X. Then,  {xn} is said to be convergent if there exists x∈X such that limn→∞N(xn-x,t)=1 for all t>0. In that case, x is called the limit of the sequence {xn} and we denote it by limn→∞xn=x.A sequence{xn} in a fuzzy normed space (X,N) is called Cauchy if, for each ϵ>0 and δ>0, one can find some n0 such that
(1.12)N(xm-xn,δ)>1-ϵfor alln,m≥n0.It is known that every convergent sequence in a fuzzy normed space is Cauchy. If, in a fuzzy-normed space, each Cauchy sequence is convergent, then the fuzzy-norm is said to be complete and the fuzzy normed space is called a fuzzy Banach space.Stability of Cauchy, Jensen, quadratic, and cubic function equation in fuzzy normed spaces have first been investigated in [50–53].In this paper, we give a generalization of the results from [13] and pose two open problems in fuzzy Banach space. For convenience, we use the following abbreviation for a given mapping f:
(1.13)Df(x,y)=:f(x+y)-f(x)-f(y).


## 2. Stability of the Cauchy Functional Equation

Hereafter, unless otherwise stated, we will assume thatX is real vector space, (Y,N) is a complete fuzzy norm space and k is a fixed integer greater than 1.Theorem 2.1.
Let(Z,N′) be a fuzzy normed space and φ:X×X→Z be a mapping such that, φ(kx,ky)=αφ(x,y) for some α with 0<α<k. Suppose that f:X→Y be mapping such that
(2.1)N(Df(x,y),t)≥N′(φ(x,y),t)
for all x,y∈X and all positive real number t. Then, there is a unique additive mapping Tk:X→Y such that Tk(x)=limn→∞f(knx)/kn and
(2.2)N(Tk(x)-f(x),t)≥Mk(x,(k-α)t),
where  Mk(x,t):=min{N′(φ(x,ix),t):1≤i<k}.Proof.
By induction onk, we show that
(2.3)N(f(kx)-kf(x),t)≥Mk(x,t):=min⁡{N′(φ(x,ix),t):1≤i<k}
for all x∈X and all positive real number t. Letting y=x in (2.1), we get
(2.4)N(f(2x)-2f(x),t)≥N′(φ(x,x),t).
So we get (2.3) for k=2.
Assume that (2.3) holds for k with k>2. Letting y=kx in (2.1), we get
(2.5)N(f((k+1)x)-f(x)-f(kx),t)≥N′(φ(x,kx),t).
for all x∈X. By using (2.3) and (2.5), we get (2.3) for k+1 and this completes the induction argument. Replacing x by knx in (2.3), we get
(2.6)N(f(kn+1x)-kf(knx),t)≥Mk(knx,t).
Thus
(2.7)N(f(kn+1x)kn+1-f(knx)kn,tkn+1)≥Mk(x,tαn)
for all x∈X and all positive real number t. Hence,
(2.8)N(1kn+1f(kn+1x)-1kmf(kmx),∑i=mn‍αiki+1t)≥N(∑i=mn‍1ki+1f(ki+1x)-1kif(kix),∑i=mn‍αiki+1t)≥min⁡⋃i=mn‍{N(1ki+1f(ki+1x)-1kif(kix),αiki+1t)}≥Mk(x,t).
Let ϵ>0 and δ>0 be given. Since limt→∞Mk(x,t)=1, there is some t0>0 such that Mk(x,t0)>1-ϵ. Since ∑n=0∞‍(αn/kn)t0<∞, there is some n0∈N such that ∑i=mn‍(αi/ki)t0<kδ for all n>m≥n0. It follows that
(2.9)N(1kn+1f(kn+1x)-1kmf(kmx),δ)≥N(1kn+1f(kn+1x)-1kmf(kmx),∑i=mn‍αiki+1t0)≥Mk(x,t0)>1-ϵ
for allx∈X and all nonnegative integers n and m with n>m≥n0. Therefore, the sequence {(1/kn)f(knx)} is a Cauchy sequence in (Y,N) for all x∈X. Since (Y,N) is complete, the sequence {(1/kn)f(knx)} converges in Y for all x∈X. So one can define the mapping Tk:X→Y by
(2.10)Tk(x):=lim⁡n→∞1knf(knx)
for all x∈X. Now, we show that Tk is an additive mapping. It follows from (2.1) and (2.10) that
(2.11)N(DTk(x,y),t)=lim⁡n→∞N(Df(knx,kny)kn,t)≥lim⁡n→∞N′(φ(knx,kny)kn,t)=lim⁡n→∞N′(φ(x,y),knαnt)=1
for allx,y∈X and all positive real number t. Therefore, the mapping Tk is additive. Moreover, if we put m=0 in (2.8), we observe that
(2.12)N(1kn+1f(kn+1x)-f(x),∑i=0n‍αiki+1t)≥Mk(x,t).
Therefore,
(2.13)N(1kn+1f(kn+1x)-f(x),t)≥Mk(x,t∑i=0n‍(αi/ki+1)).
It follows from (2.13), for large enough n, that
(2.14)N(Tk(x)-f(x),t)≥min⁡{N(f(kn+1x)kn+1-f(x),t),N(Tk(x)-f(kn+1x)kn+1,t)}≥Mk(x,t∑i=0n‍(αi/ki+1))≥Mk(x,(k-α)t).
Now, we show that Tk is unique. Let T′ be another additive mapping from X into Y, which satisfies the required inequality. Then, for each x∈X and t>0, we have
(2.15)N(Tk(x)-T′(x),t)≥min⁡{N(Tk(x)-f(x),t),N(f(x)-T′(x),t)}≥Mk(x,(k-α)t).
So,
(2.16)N(Tk(x)-T′(x),t)=N(Tk(knx)kn-T′(knx)kn,t)=N(Tk(knx)-T′(knx),knt)≥Mk(knx,(k-α)knt)≥Mk(x,(k-α)knαnt).
Hence, the right-hand side of the above inequality tends to 1 as n→∞. It follows that Tk(x)=T′(x) for all x∈X.Theorem 2.2.
Let(Z,N′) be a fuzzy normed space and, Φ:X×X→Z be a mapping such that Φ(k-1x,k-1y)=α-1Φ(x,y) for some α with α>k. Suppose that f:X→Y be mapping such that
(2.17)N(Df(x,y),t)≥N′(Φ(x,y),t)
for all x,y∈X and all positive real number t. Then, there is a unique additive mapping Tk:X→Y such that Tk(x)=limn→∞knf(x/kn) and
(2.18)N(Tk(x)-f(x),t)≥Mk(x,(α-k)t),
where  Mk(x,t):=min{N′(Φ(x,ix),t):1≤i<k}.Proof.
Similarly to the proof of Theorem2.1, we have
(2.19)N(f(kx)-kf(x),t)≥Mk(x,t)
for all x∈X and all positive real number t. Replacing x by x/kn+1 in (2.19), we get
(2.20)N(f(xkn)-kf(xkn+1),t)≥Mk(xkn+1,t).
Thus,
(2.21)N(knf(xkn)-kn+1f(xkn+1),knt)≥Mk(x,αn+1t)
for all x∈X and all positive real number t. Hence,
(2.22)N(kn+1f(xkn+1)-kmf(xkm),∑i=mn‍kiαi+1t)≥N(∑i=mn‍ki+1f(xki+1)-kif(xki),∑i=mn‍kiαi+1t)≥min⁡⋃i=mn‍{N(ki+1f(xki+1)-kif(xki),kiαi+1t)}≥Mk(x,t).
Letϵ>0 and δ>0 be given. Since limt→∞Mk(x,t)=1, there is some t0>0 such that Mk(x,t0)>1-ϵ. Since ∑n=0∞‍(kn/αn)t0<∞, there is some n0∈N such that ∑i=mn‍(ki/αi)t0<αδ for all n>m≥n0. It follows from (2.22) that
(2.23)N(kn+1f(xkn+1)-kmf(xkm),δ)≥N(kn+1f(xkn+1)-kmf(xkm),∑i=mn‍kiαi+1t0)≥Mk(x,t0)>1-ϵ
for all x∈X and all nonnegative integers n and m with n>m≥n0. Therefore, the sequence {knf(x/kn)} is a Cauchy sequence in (Y,N) for all x∈X. Since (Y,N) is complete, the sequence {knf(x/kn)} converges in Y for all x∈X. So one can define the mapping Tk:X→Y by
(2.24)Tk(x):=lim⁡n→∞knf(xkn)
for all x∈X. The rest of the proof is similar to the proof of Theorem 2.1Theorem 2.3.
LetX be a normed space, let (Z,N′) be a fuzzy normed space, and let  ψ:[0,∞)→[0,∞) be a function such that
(1)
ψ
(
t
s
)
=
ψ
(
t
)
ψ
(
s
),
(2)
ψ
(
t
)
<
t for all t>1.
Suppose that a mappingf:X→Y satisfies the inequality:
(2.25)N(Df(x,y),t)≥N′((ψ(‖x‖)+ψ(‖y‖))z0,t)
for all x,y∈X and all positive real number t, where z0 is a fixed vector of Z. Then, there exists a unique additive mapping Tk:X→Y satisfying Tk(x):=limn→∞(f(knx)/kn) and
(2.26)N(Tk(x)-f(x),t)≥N′(ψ(‖x‖)z0,k-ψ(k)σk(ψ)t)
for all x∈X, where σk(ψ)=max{1+ψ(i):1≤i<k}. Moreover, Tk=T2 for all k≥2.Proof.
Let(2.27)φ(x,y)=(ψ(∥x∥)+ψ(∥y∥))z0
for all x,y∈X. So,
(2.28)φ(kx,ky)=ψ(k)φ(x,y).
where ψ(k)<k. By using Theorem 2.1, we can get (2.26). Now, we show that Tk=T2. It follows from (1) that ψ(kn)=(ψ(k))n. Replacing x by 2nx in (2.26), we get
(2.29)N(Tk(2nx)-f(2nx),t)≥N′(ψ(‖2nx‖)z0,k-ψ(k)σk(ψ)t)
for all x∈X. So we have
(2.30)N(Tk(x)-f(2nx)2n,t)≥N′(ψ(‖x‖)z0,k-ψ(k)σk(ψ)ψ(2n)2nt)
Using (2) and passing the limit n→∞ in (2.30), we get Tk=T2.Theorem 2.4.
LetX be a normed space, let (Z,N′) be a fuzzy normed space, and let  ψ:[0,∞)→[0,∞) be a function such that
(1)
ψ
(
t
s
)
=
ψ
(
t
)
ψ
(
s
),
(2)
ψ
(
t
)
>
t for all t>1.
Suppose that a mappingf:X→Y satisfies the inequality:
(2.31)N(Df(x,y),t)≥N′((ψ(‖x‖)+ψ(‖y‖))z0,t)
for all x,y∈X and all positive real number t, where z0 is a fixed vector of Z. Then, there exists a unique additive mapping Tk:X→Y satisfying Tk(x):=limn→∞knf(x/kn) and
(2.32)N(Tk(x)-f(x),t)≥N′(ψ(‖x‖)z0,ψ(k)-kσk(ψ)t)
for all x∈X, where
(2.33)σk(ψ)=max⁡{1+ψ(i):1≤i<k}.
Moreover,Tk=T2   for all k≥2.Proof.
Let(2.34)Φ(x,y)=(ψ(‖x‖)+ψ(‖y‖))z0
for all x,y∈X. So, we have
(2.35)Φ(k-1x,k-1y)=ψ(k-1)Φ(x,y),
where ψ(k-1)=ψ(k)-1<k-1. It follows from (1) that ψ(k-n)=(ψ(k))-n. By using Theorem 2.2, we can get (2.32). Now, we show that Tk=T2. Replacing x by x/2n in (2.32), we get
(2.36)N(Tk(x2n)-f(x2n),t)≥N′(ψ(‖(x2n)‖)z0,ψ(k)-kσk(ψ)t).
for all x∈X. So we have
(2.37)N(Tk(x)-2nf(x2n),t)≥N′(ψ(‖x‖)z0,ψ(k)-k2nσk(ψ)ψ(2-n)t).
Using (2) and passing the limit n→∞ in (2.37), we get Tk=T2.Theorem 2.5.
LetX be a normed space, let p be a nonnegative real number such that p≠1, and let H:[0,∞)×[0,∞)→[0,∞) be a homogeneous function of degree p. Suppose that (Z,N′) be a fuzzy normed space and let  f:X→Y be mapping such that
(2.38)N(Df(x,y),t)≥N′(H(‖x‖,‖y‖)z0,t)
for all x,y∈X and all positive real number t, where z0 is a fixed vector of Z. Then, there exists a unique additive mapping Tk:X→Y such that
(2.39)N(Tk(x)-f(x),t)≥Mk(x,|kp-k|t),
where  Mk(x,t):=min{N′(∥x∥pH(1,i)z0,t):1≤i<k}.Proof.
The proof follows from Theorems2.1 and 2.2.For the particular casesH(x,y)=θ(xp+yp),H(x,y)=xrys,H(x,y)=xrys+xr+s+yr+s(r+s=p), and H(x,y)=min{xp,yp}, we have the following corollaries.Corollary 2.6.
LetX be a normed space, let  p be a nonnegative real number such that p≠1. Suppose that (Z,N′) be a fuzzy normed space and f:X→Y be mapping such that
(2.40)N(Df(x,y),t)≥N′((‖x‖p+‖y‖p)θ,t)
for all x,y∈X and all positive real number t, where θ is a fixed vector of Z. Then, there exists a unique additive mapping Tk:X→Y such that
(2.41)N(Tk(x)-f(x),t)≥N′(‖x‖pθ,|kp-k|1+(k-1)pt).Corollary 2.7.
LetX be a normed space, r,s be non-negative real numbers such that p:=r+s≠1. Suppose that (Z,N′) be a fuzzy normed space and f:X→Y be mapping such that
(2.42)N(Df(x,y),t)≥N′(‖x‖r‖y‖sθ,t)
for all x,y∈X and all positive real number t, where θ is a fixed vector of Z. Then there exists a unique additive mapping Tk:X→Y such that
(2.43)N(Tk(x)-f(x),t)≥N′(‖x‖pθ,|kp-k|(k-1)st).Corollary 2.8.
LetX be a normed space, and let  r,s be nonnegative real numbers such that p:=r+s≠1. Suppose that (Z,N′) be a fuzzy normed space and let  f:X→Y be mapping such that
(2.44)N(Df(x,y),t)≥N′(θ‖x‖r‖y‖s+θ‖x‖r+s+θ‖y‖r+s,t)
for all x,y∈X and all positive real number t, where θ is a fixed vector of Z. Then, there exists a unique additive mapping Tk:X→Y such that
(2.45)N(Tk(x)-f(x),t)≥N′(‖x‖pθ,|kp-k|(k-1)s+(k-1)p+1t).Corollary 2.9.
LetX be a normed space, let p be a nonnegative real number such that p≠1. Suppose that (Z,N′) be a fuzzy normed space and let  f:X→Y be mapping such that
(2.46)N(Df(x,y),t)≥N′(min⁡{‖x‖p,‖y‖p}θ,t)
for all x,y∈X and all positive real number t, where θ is a fixed vector of Z. Then, there exists a unique additive mapping Tk:X→Y such that
(2.47)N(Tk(x)-f(x),t)≥N′(‖x‖pθ,|kp-k|t).Problem 1.
Whether Theorem2.5 and/or such Corollaries can be proved for p=1?Problem 2.
What is the best possible value ofk in Corollaries 2.6 and 2.7?



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*Source: 763728-2012-07-05.xml*