# Intuitionistic Fuzzy Stability of Functional Equations Associated with Inner Product Spaces

Authors: Zhihua Wang; Themistocles M. Rassias
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

In intuitionistic fuzzy normed spaces, we investigate some stability results for the functional equation∑i=1nf(xi-(1/n)∑j=1nxj)=∑i=1nf(xi)-nf((1/n)∑i=1nxi) which is said to be a functional equation associated with inner products space.


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## Body

## 1. Introduction and Preliminaries

The aim of this article is to prove an intuitionistic fuzzy version of the Hyers-Ulam-Rassias stability for the functional equation:(1.1)∑i=1nf(xi-1n∑j=1nxj)=∑i=1nf(xi)-nf(1n∑i=1nxi),
which is said to be a functional equation associated with inner product spaces. It was shown by Rassias [1] that the norm defined over a real vector space X is induced by an inner product if and only if for a fixed integer n≥2 it follows
(1.2)∑i=1n‖xi-1n∑j=1nxj‖2=∑i=1n‖xi‖2-n‖1n∑i=1nxi‖2,
for all xi,…,xn∈X. Interesting new results concerning functional equations associated with inner product spaces have recently been obtained by Park et al. [2, 3] and Najati and Rassias [4] as well as for the fuzzy stability of a functional equation associated with inner product spaces [5].Stability problem of a functional equation was first posed by Ulam [6] which was answered by Hyers [7] on approximately additive mappings and then generalized by Aoki [8] and Rassias [9] for additive mappings and linear mappings, respectively. Later there have been proved several new results on stability of various classes of functional equations in the Hyers-Ulam sense (cf. the following books and papers [10–18] and the references cited therein), as well as various fuzzy stability results concerning Cauchy, Jensen, quadratic and cubic functional equations (cf. [19–22]). Furthermore some stability results concerning Jensen, cubic, mixed-type additive and cubic functional equations were investigated (cf. [23–26]) in the spirit of intuitionistic fuzzy normed spaces, while the idea of intuitionistic fuzzy normed space was introduced in [27] and further studied in [28–35].In this section, we recall some notations and basic definitions used in this paper as follows.Definition 1.1 (cf. [36]).
A binary operation*:[0,1]×[0,1]→[0,1] is said to be a continuous t-norm if it satisfies the following conditions:(a)
* is commutative and associative, (b) * is continuous,(c)
a*1=a for all a∈[0,1],(d)
a*b≤c*d whenever a≤c and b≤d for each a,b,c,d∈[0,1].Definition 1.2 (cf. [36]).
A binary operation⋄:[0,1]×[0,1]→[0,1] is said to be a continuous t-conorm if it satisfies the following conditions:(a’)
⋄ is commutative and associative, (b’) ⋄ is continuous,(c’)
a⋄0=a for all a∈[0,1],(d’)
a⋄b≤c⋄d whenever a≤c and b≤d for each a,b,c,d∈[0,1].
Using the notions of continuoust-norm and continuous t-conorm, Saadati and Park [27] have recently introduced the concept of intuitionistic fuzzy normed spaces as follows.Definition 1.3.
The five-tuple(X,μ,ν,*,⋄) is said to be an intuitionistic fuzzy normed space (for short, IFNS) if X is a vector space, * is a continuous t-norm, ⋄ is a continuous t-conorm, and μ,ν are fuzzy sets on X×(0,∞) satisfying the following conditions. For every x,y∈X and s,t>0,(i)
μ(x,t)+ν(x,t)≤1,(ii)
μ(x,t)>0,(iii)
μ(x,t)=1 if and only if x=0,(iv)
μ(αx,t)=μ(x,t/|α|) for each α≠0,(v)
μ(x,t)*μ(y,s)≤μ(x+y,t+s),(vi)
μ(x,·):(0,∞)→[0,1] is continuous,(vii)
limt→∞μ(x,t)=1 and limt→0μ(x,t)=0,(viii)
ν(x,t)<1, (ix) ν(x,t)=0 if and only if x=0,(x)
ν(αx,t)=ν(x,t/|α|) for each α≠0, (xi) ν(x,t)⋄ν(y,s)≥ν(x+y,t+s),(xii)
ν(x,·):(0,∞)→[0,1] is continuous,(xiii)
limt→∞ν(x,t)=0 and limt→0ν(x,t)=1.In this case (μ,ν) is called an intuitionistic fuzzy norm.Example 1.4 (cf. [37]).
Let(X,∥·∥)be a normed space, a*b=ab,anda⋄b=min(a+b,1)for all a,b∈[0,1].For all x∈Xand everyt>0andk=1,2,consider
(1.3)μk(x,t)={tt+k‖x‖,ift>0,0,ift≤0,νk(x,t)={k‖x‖t+k‖x‖,ift>0,0,ift≤0.
Then (X,μ,ν,*,⋄) is an IFNS.The concepts of convergence and Cauchy sequences in an intuitionistic fuzzy normed space are studied in [27].Let(X,μ,ν,*,⋄) be an IFNS. Then, a sequence {xk} is said to be intuitionistic fuzzy convergent to x∈X if, for every ɛ>0 and t>0, there exists k0∈ℕ such that μ(xk-x,t)>1-ɛ and ν(xk-x,t)<ɛ for all k≥k0. In this case we write (μ,ν)-limxk=x. The sequence {xk} is said to be intuitionistic fuzzy Cauchy sequence if, for every ɛ>0 and t>0, there exists k0∈ℕ such that μ(xk-xℓ,t)>1-ɛ and ν(xk-xℓ,t)<ɛ for all k,ℓ≥k0. (X,μ,ν,*,⋄) is said to be complete if every intuitionistic fuzzy Cauchy sequence in (X,μ,ν,*,⋄) is intuitionistic fuzzy convergent in (X,μ,ν,*,⋄).


## 2. Intuitionistic Fuzzy Stability

Throughout this section, assume thatX, (Z,μ′,ν′), and (Y,μ,ν) are linear space, IFNS, and intuitionistic fuzzy Banach space, respectively. For convenience, we use the following abbreviation for a given function f:X→Y: (2.1)Δf(x1,…,xn)=∑i=1nf(xi-1n∑j=1nxj)-∑i=1nf(xi)+nf(1n∑i=1nxi).We begin with the Hyers-Ulam-Rassias type theorem in IFNS for the functional (1.1) which is said to be a functional equation associated with inner product spaces.Theorem 2.1.
Letφ:X→Z be a function such that φ(2x)=αφ(x) for some real number α with 0<|α|<4. Suppose that an even function f:X→Y with f(0)=0 satisfies the inequality
(2.2)μ(Δf(x1,…,xn),t1+⋯+tn)≥μ′(φ(x1),t1)*⋯*μ′(φ(xn),tn),ν(Δf(x1,…,xn),t1+⋯+tn)≤ν′(φ(x1),t1)⋄⋯⋄ν′(φ(xn),tn),
for all x1,…,xn∈X and all t1,…,tn>0. Then there exists a unique quadratic function Q:X→Y such that
(2.3)μ(Q(x)-f(x),t)≥μ1′′(x,(4-|α|)t8),ν(Q(x)-f(x),t)≤ν1′′(x,(4-|α|)t8),
for all x∈X and t>0, where
(2.4)μ1′′(x,t):=μ′(φ(nx),8(n-1)2n2+9nt)*μ′(φ((n-1)x),8(n-1)2n2+9nt)*μ′(φ(x),8(n-1)2n2+9nt)*μ′(φ(0),8(n-1)2n2+9nt),ν1′′(x,t):=ν′(φ(nx),8(n-1)2n2+9nt)⋄ν′(φ((n-1)x),8(n-1)2n2+9nt)⋄ν′(φ(x),8(n-1)2n2+9nt)⋄ν′(φ(0),8(n-1)2n2+9nt).Proof.
Putx1=nx1,xi=nx2(i=2,…,n),ti=t(i=1,…,n) in (2.2), and, using the evenness of f, we obtain
(2.5)μ(nf(x1+(n-1)x2)+f((n-1)(x1-x2))+(n-1)f(x1-x2)-f(nx1)-(n-1)f(nx2),nt)≥μ′(φ(nx1),t)*μ′(φ(nx2),t),ν(nf(x1+(n-1)x2)+f((n-1)(x1-x2))+(n-1)f(x1-x2)-f(nx1)-(n-1)f(nx2),nt)≤ν′(φ(nx1),t)⋄ν′(φ(nx2),t),
for all x1,x2∈X and t>0. Interchanging x1 with x2 in (2.5) and using the evenness of f, we get
(2.6)μ(nf((n-1)x1+x2)+f((n-1)(x1-x2))+(n-1)f(x1-x2)-(n-1)f(nx1)-f(nx2),nt)≥μ′(φ(nx1),t)*μ′(φ(nx2),t),ν(nf((n-1)x1+x2)+f((n-1)(x1-x2))+(n-1)f(x1-x2)-(n-1)f(nx1)-f(nx2),nt)≤ν′(φ(nx1),t)⋄ν′(φ(nx2),t),
for all x1,x2∈X and t>0. It follows from (2.5) and (2.6) that
(2.7)μ(nf((n-1)x1+x2)+nf(x1+(n-1)x2)+2f((n-1)(x1-x2))+2(n-1)f(x1-x2)-nf(nx1)-nf(nx2),2nt)≥μ′(φ(nx1),t)*μ′(φ(nx2),t),ν(nf((n-1)x1+x2)+nf(x1+(n-1)x2)+2f((n-1)(x1-x2))+2(n-1)f(x1-x2)-nf(nx1)-nf(nx2),2nt)≤ν′(φ(nx1),t)⋄ν′(φ(nx2),t),
for all x1,x2∈X and t>0. Putting x1=nx1,x2=-nx2,xi=0(i=3,…,n),ti=t(i=1,…,n) in (2.2) and using the evenness of f, we obtain
(2.8)μ(f((n-1)x1+x2)+f(x1+(n-1)x2)+2(n-1)f(x1-x2)-f(nx1)-f(nx2),nt)≥μ′(φ(nx1),t)*μ′(φ(-nx2),t)*μ′(φ(0),t),ν(f((n-1)x1+x2)+f(x1+(n-1)x2)+2(n-1)f(x1-x2)-f(nx1)-f(nx2),nt)≤ν′(φ(nx1),t)⋄ν′(φ(-nx2),t)⋄ν′(φ(0),t),
for all x1,x2∈X and t>0. Hence, we obtain from (2.7) and (2.8) that
(2.9)μ(f((n-1)(x1-x2))-(n-1)2f(x1-x2),n2+2n2t)≥μ′(φ(nx1),t)*μ′(φ(nx2),t)*μ′(φ(-nx2),t)*μ′(φ(0),t),ν(f((n-1)(x1-x2))-(n-1)2f(x1-x2),n2+2n2t)≤ν′(φ(nx1),t)⋄ν′(φ(nx2),t)⋄ν′(φ(-nx2),t)⋄ν′(φ(0),t),
for all x1,x2∈X and t>0. So
(2.10)μ(f((n-1)x)-(n-1)2f(x),n2+2n2t)≥μ′(φ(nx),t)*μ′(φ(0),t),ν(f((n-1)x)-(n-1)2f(x),n2+2n2t)≤ν′(φ(nx),t)⋄ν′(φ(0),t),
for all x∈X and t>0. Putting x1=nx,xi=0(i=2,…,n),ti=t(i=1,…,n) in (2.2), we obtain
(2.11)μ(f(nx)-f((n-1)x)-(2n-1)f(x),nt)≥μ′(φ(nx),t)*μ′(φ(0),t),ν(f(nx)-f((n-1)x)-(2n-1)f(x),nt)≤ν′(φ(nx),t)⋄ν′(φ(0),t),
for all x∈X and t>0. It follows from (2.10) and (2.11) that
(2.12)μ(f(nx)-n2f(x),n2+4n2t)≥μ′(φ(nx),t)*μ′(φ(0),t),ν(f(nx)-n2f(x),n2+4n2t)≤ν′(φ(nx),t)⋄ν′(φ(0),t),
for all x∈X and t>0. Letting x2=-(n-1)x1 in (2.8) and replacing x1 by x/n in the obtained inequality, we get
(2.13)μ(f((n-1)x)-f((n-2)x)-(2n-3)f(x),nt)≥μ′(φ(x),t)*μ′(φ((n-1)x),t)*μ′(φ(0),t),ν(f((n-1)x)-f((n-2)x)-(2n-3)f(x),nt)≤ν′(φ(x),t)⋄ν′(φ((n-1)x),t)⋄ν′(φ(0),t),
for all x∈X and t>0. It follows from (2.10), (2.11), (2.12), and (2.13) that
(2.14)μ(f((n-2)x)-(n-1)2f(x),n2+4n2t)≥μ′(φ(nx),t)*μ′(φ((n-1)x),t)*μ′(φ(x),t)*μ′(φ(0),t),ν(f((n-2)x)-(n-1)2f(x),n2+4n2t)≤ν′(φ(nx),t)⋄ν′(φ((n-1)x),t)⋄ν′(φ(x),t)⋄ν′(φ(0),t),
for all x∈X and t>0. Applying (2.12) and (2.14), we obtain
(2.15)μ(f(nx)-f((n-2)x)-4(n-1)f(x),(n2+4n)t)≥μ′(φ(nx),t)*μ′(φ((n-1)x),t)*μ′(φ(x),t)*μ′(φ(0),t),ν(f(nx)-f((n-2)x)-4(n-1)f(x),(n2+4n)t)≤ν′(φ(nx),t)⋄ν′(φ((n-1)x),t)⋄ν′(φ(x),t)⋄ν′(φ(0),t),
for all x∈X and t>0. Setting x1=x2=nx,xi=0(i=3,…,n),ti=t(i=1,…,n) in (2.2), we obtain
(2.16)μ(f((n-2)x)+(n-1)f(2x)-f(nx),n2t)≥μ′(φ(nx),t)*μ′(φ(0),t),ν(f((n-2)x)+(n-1)f(2x)-f(nx),n2t)≤ν′(φ(nx),t)⋄ν′(φ(0),t),
for all x∈X and t>0. It follows from (2.15) and (2.16) that
(2.17)μ(f(2x)-4f(x),2n2+9n2n-2t)≥μ′(φ(nx),t)*μ′(φ((n-1)x),t)*μ′(φ(x),t)*μ′(φ(0),t),ν(f(2x)-4f(x),2n2+9n2n-2t)≤ν′(φ(nx),t)⋄ν′(φ((n-1)x),t)⋄ν′(φ(x),t)⋄ν′(φ(0),t).
It follows that
(2.18)μ(f(x)-4-1f(2x),t)≥μ′(φ(nx),8(n-1)2n2+9nt)*μ′(φ((n-1)x),8(n-1)2n2+9nt)*μ′(φ(x),8(n-1)2n2+9nt)*μ′(φ(0),8(n-1)2n2+9nt),ν(f(x)-4-1f(2x),t)≤ν′(φ(nx),8(n-1)2n2+9nt)⋄ν′(φ((n-1)x),8(n-1)2n2+9nt)⋄ν′(φ(x),8(n-1)2n2+9nt)⋄ν′(φ(0),8(n-1)2n2+9nt).
Define
(2.19)μ1′′(x,t):=μ′(φ(nx),8(n-1)2n2+9nt)*μ′(φ((n-1)x),8(n-1)2n2+9nt)*μ′(φ(x),8(n-1)2n2+9nt)*μ′(φ(0),8(n-1)2n2+9nt),ν1′′(x,t):=ν′(φ(nx),8(n-1)2n2+9nt)⋄ν′(φ((n-1)x),8(n-1)2n2+9nt)⋄ν′(φ(x),8(n-1)2n2+9nt)⋄ν′(φ(0),8(n-1)2n2+9nt).
Then, by our assumption,
(2.20)μ1′′(2x,t)=μ1′′(x,tα),ν1′′(2x,t)=ν1′′(x,tα).
Replacing x by 2nx in (2.18) and applying (2.20), we get
(2.21)μ(f(2nx)4n-f(2n+1x)4n+1,αnt4n)=μ(f(2nx)-f(2n+1x)4,αnt)≥μ1′′(2nx,αnt)=μ1′′(x,t),ν(f(2nx)4n-f(2n+1x)4n+1,αnt4n)=ν(f(2nx)-f(2n+1x)4,αnt)≤ν1′′(2nx,αnt)=ν1′′(x,t).
Thus for each n>m, we have
(2.22)μ(f(2mx)4m-f(2nx)4n,∑k=mn-1αkt4k)=μ(∑k=mn-1f(2kx)4k-f(2k+1x)4k+1,∑k=mn-1αkt4k)≥∏k=mn-1μ(f(2kx)4k-f(2k+1x)4k+1,αkt4k)≥μ1′′(x,t),ν(f(2mx)4m-f(2nx)4n,∑k=mn-1αkt4k)=ν(∑k=mn-1f(2kx)4k-f(2k+1x)4k+1,∑k=mn-1αkt4k)≤∐k=mn-1ν(f(2kx)4k-f(2k+1x)4k+1,αkt4k)≤ν1′′(x,t),
where ∏j=1naj=a1*a2*⋯*an,∐j=1naj=a1⋄a2⋄⋯⋄an. Let ɛ>0 and δ>0 be given. Since limt→∞μ1′′(x,t)=1 and limt→∞ν1′′(x,t)=0, there exists some t0>0 such that μ1′′(x,t0)>1-ɛ and ν1′′(x,t0)<ɛ. Since ∑k=0∞(αkt0/4k)<∞, there is some n0∈ℕ such that ∑k=mn-1(αkt0/4k)<δ for each n>m≥n0. It follows that
(2.23)μ(f(2mx)4m-f(2nx)4n,δ)≥μ(f(2mx)4m-f(2nx)4n,∑k=mn-1αkt04k)≥μ1′′(x,t0)>1-ε,ν(f(2mx)4m-f(2nx)4n,δ)≤ν(f(2mx)4m-f(2nx)4n,∑k=mn-1αkt04k)≤ν1′′(x,t0)<ε,
for all t>t0. This shows that the sequence {f(2nx)/4n} is Cauchy in (Y,μ,ν). Since (Y,μ,ν) is intuitionistic fuzzy Banach space, {f(2nx)/4n} converges to some point Q(x)∈Y. Thus, we can define a mapping Q(x):X→Y such that Q(x):=(μ,ν)-limn→∞(f(2nx)/4n). Moreover, if we put m=0 in (2.22), we get
(2.24)μ(f(2nx)4n-f(x),∑k=0n-1αkt4k)≥μ1′′(x,t),ν(f(2nx)4n-f(x),∑k=0n-1αkt4k)≤ν1′′(x,t).
Thus,
(2.25)μ(f(2nx)4n-f(x),t)≥μ1′′(x,t∑k=0n-1(α/4)k),ν(f(2nx)4n-f(x),t)≤ν1′′(x,t∑k=0n-1(α/4)k).
Now, we will show thatQ is quadratic. Setting xi=2mxi(i=1,…,n) and ti=(t/n)(i=1,…,n) in (2.2), we obtain
(2.26)μ(Δf(2mx1,…,2mxn)4m,t)≥μ′(φ(2mx1),4mtn)*⋯*μ′(φ(2mxn),4mtn)=μ′(φ(x1),(4α)mtn)*⋯*μ′(φ(xn),(4α)mtn),ν(Δf(2mx1,…,2mxn)4m,t)≤ν′(φ(2mx1),4mtn)⋄⋯⋄ν′(φ(2mxn),4mtn)=ν′(φ(x1),(4α)mtn)⋄⋯⋄ν′(φ(xn),(4α)mtn),
for all x1,…,xn∈X and t>0. Letting n→∞ in (2.26), we obtain
(2.27)μ(ΔQ(x1,…,xn),t)=1,ν(ΔQ(x1,…,xn),t)=0,
for all x1,…,xn∈X and all t>0. This means that Q satisfies the functional (1.1) and so it is quadratic (see Lemma 2.2 of [4]).
Next, we approximate the difference betweenf and Q in intuitionistic fuzzy sense. By (2.25), we have
(2.28)μ(Q(x)-f(x),t)≥μ(Q(x)-f(2nx)4n,t2)*μ(f(2nx)4n-f(x),t2)≥μ1′′(x,t2∑k=0∞(α/4)k)=μ1′′(x,(4-α)t8),ν(Q(x)-f(x),t)≤ν(Q(x)-f(2nx)4n,t2)⋄ν(f(2nx)4n-f(x),t2)≤ν1′′(x,t2∑k=0∞(α/4)k)=ν1′′(x,(4-α)t8),
for every x∈X,t>0 and large enough n. To prove the uniqueness of Q, assume that Q′ is another quadratic mapping from X to Y, which satisfies the required inequality. Then, for each x∈X and t>0,
(2.29)μ(Q(x)-Q′(x),t)≥μ(Q(x)-f(x),t2)*μ(Q′(x)-f(x),t2)≥μ1′′(x,(4-α)t16),ν(Q(x)-Q′(x),t)≤ν(Q(x)-f(x),t2)⋄ν(Q′(x)-f(x),t2)≤ν1′′(x,(4-α)t16).
Since Q and Q′ are quadratic, we have
(2.30)μ(Q(x)-Q′(x),t)=μ(Q(2nx)-Q′(2nx),4nt)≥μ1′′(x,(4/α)n(4-α)t16),ν(Q(x)-Q′(x),t)=ν(Q(2nx)-Q′(2nx),4nt)≤ν1′′(x,(4/α)n(4-α)t16),
for all x∈X,t>0 and n∈ℕ. Since 0<α<4 and limn→∞(4/α)n=∞, weget(2.31)lim⁡n→∞μ1′′(x,(4/α)n(4-α)t16)=1,lim⁡n→∞ν1′′(x,(4/α)n(4-α)t16)=0.
Therefore μ(Q(x)-Q′(x),t)=1 and ν(Q(x)-Q′(x),t)=0 for all x∈X and t>0. Hence, Q(x)=Q′(x) for all x∈X. This completes the proof of the theorem.Theorem 2.2.
Letφ:X→Z be a function such that φ(2x)=αφ(x) for some real number α with 0<|α|<2. Suppose that an odd function f:X→Y satisfies the inequality
(2.32)μ(Δf(x1,…,xn),t1+⋯+tn)≥μ′(φ(x1),t1)*⋯*μ′(φ(xn),tn),ν(Δf(x1,…,xn),t1+⋯+tn)≤ν′(φ(x1),t1)⋄⋯⋄ν′(φ(xn),tn),
for all x1,…,xn∈X and all t1,…,tn>0. Then there exists a unique additive function A:X→Y such that
(2.33)μ(A(x)-f(x),t)≥μ2′′(x,(2-|α|)t4),ν(A(x)-f(x),t)≤ν2′′(x,(2-|α|)t4),
for all x∈X and t>0, where
(2.34)μ2′′(x,t):=μ′(φ(2x),4n2+4nt)*μ′(φ(x),4n2+4nt)*μ′(φ(-x),4n2+4nt)*μ′(φ(0),4n2+4nt),ν2′′(x,t):=ν′(φ(2x),4n2+4nt)⋄ν′(φ(x),4n2+4nt)⋄ν′(φ(-x),4n2+4nt)⋄ν′(φ(0),4n2+4nt).Proof.
Putx1=nx1,xi=nx1′(i=2,…,n),ti=t(i=1,…,n) in (2.32) and using the oddness of f, we obtain
(2.35)μ(nf(x1+(n-1)x1′)+f((n-1)(x1-x1′))-(n-1)f(x1-x1′)-f(nx1)-(n-1)f(nx1′),nt)≥μ′(φ(nx1),t)*μ′(φ(nx1′),t),ν(nf(x1+(n-1)x1′)+f((n-1)(x1-x1′))-(n-1)f(x1-x1′)-f(nx1)-(n-1)f(nx1′),nt)≤ν′(φ(nx1),t)⋄ν′(φ(nx1′),t),
for all x1,x1′∈X and t>0. Interchanging x1 with x1′ in (2.35) and using the oddness of f, we get
(2.36)μ(nf((n-1)x1+x1′)-f((n-1)(x1-x1′))+(n-1)f(x1-x1′)-(n-1)f(nx1)-f(nx1′),nt)≥μ′(φ(nx1),t)*μ′(φ(nx1′),t),ν(nf((n-1)x1+x1′)-f((n-1)(x1-x1′))+(n-1)f(x1-x1′)-(n-1)f(nx1)-f(nx1′),nt)≤ν′(φ(nx1),t)⋄ν′(φ(nx1′),t),
for all x1,x1′∈X and t>0. It follows from (2.35) and (2.36) that
(2.37)μ(nf(x1+(n-1)x1′)-nf((n-1)x1+x1′)+2f((n-1)(x1-x1′))-2(n-1)f(x1-x1′)+(n-2)f(nx1)-(n-2)f(nx1′),2nt)≥μ′(φ(nx1),t)*μ′(φ(nx1′),t),ν(nf(x1+(n-1)x1′)-nf((n-1)x1+x1′)+2f((n-1)(x1-x1′))-2(n-1)f(x1-x1′)+(n-2)f(nx1)-(n-2)f(nx1′),2nt)≤ν′(φ(nx1),t)⋄ν′(φ(nx1′),t),
for all x1,x1′∈X and t>0. Setting x1=nx1,x2=-nx1′,xi=0(i=3,…,n),ti=t(i=1,…,n) in (2.32) and using the oddness of f, we get
(2.38)μ(f((n-1)x1+x1′)-f(x1+(n-1)x1′)+2f(x1-x1′)-f(nx1)+f(nx1′),nt)≥μ′(φ(nx1),t)*μ′(φ(-nx1′),t)*μ′(φ(0),t),ν(f((n-1)x1+x1′)-f(x1+(n-1)x1′)+2f(x1-x1′)-f(nx1)+f(nx1′),nt)≤ν′(φ(nx1),t)⋄ν′(φ(-nx1′),t)⋄ν′(φ(0),t),
for all x1,x1′∈X and t>0. It follows from (2.37) and (2.38) that
(2.39)μ(f((n-1)(x1-x1′))+f(x1-x1′)-f(nx1)+f(nx1′),n2+2n2t)≥μ′(φ(nx1),t)*μ′(φ(nx1′),t)*μ′(φ(-nx1′),t)*μ′(φ(0),t),ν(f((n-1)(x1-x1′))+f(x1-x1′)-f(nx1)+f(nx1′),n2+2n2t)≤ν′(φ(nx1),t)⋄ν′(φ(nx1′),t)⋄ν′(φ(-nx1′),t)⋄ν′(φ(0),t),
for all x1,x1′∈X and t>0. Putting x1=n(x1-x1′),xi=0(i=2,…,n),ti=t(i=1,…,n) in (2.32), we obtain
(2.40)μ(f(n(x1-x1′))-f((n-1)(x1-x1′))-f(x1-x1′),nt)≥μ′(φ(n(x1-x1′)),t)*μ′(φ(0),t),ν(f(n(x1-x1′))-f((n-1)(x1-x1′))-f(x1-x1′),nt)≤ν′(φ(n(x1-x1′)),t)⋄ν′(φ(0),t),
for all x1,x1′∈X and t>0. It follows from (2.39) and (2.40) that
(2.41)μ(f(n(x1-x1′))-f(nx1)+f(nx1′),n2+4n2t)≥μ′(φ(n(x1-x1′)),t)*μ′(φ(nx1),t)*μ′(φ(nx1′),t)*μ′(φ(-nx1′),t)*μ′(φ(0),t),ν(f(n(x1-x1′))-f(nx1)+f(nx1′),n2+4n2t)≤ν′(φ(n(x1-x1′)),t)⋄ν′(φ(nx1),t)⋄ν′(φ(nx1′),t)⋄ν′(φ(-nx1′),t)⋄ν′(φ(0),t),
for all x1,x1′∈X and t>0. Replacing x1 and x1′ by x/n and -x/n in (2.41); respectively, we have
(2.42)μ(f(2x)-2f(x),n2+4n2t)≥μ′(φ(2x),t)*μ′(φ(x),t)*μ′(φ(-x),t)*μ′(φ(0),t),ν(f(2x)-2f(x),n2+4n2t)≤ν′(φ(2x),t)⋄ν′(φ(x),t)⋄ν′(φ(-x),t)⋄ν′(φ(0),t).
It follows that
(2.43)μ(f(x)-2-1f(2x),t)≥μ′(φ(2x),4n2+4nt)*μ′(φ(x),4n2+4nt)*μ′(φ(-x),4n2+4nt)*μ′(φ(0),4n2+4nt),ν(f(x)-2-1f(2x),t)≤ν′(φ(2x),4n2+4nt)⋄ν′(φ(x),4n2+4nt)⋄ν′(φ(-x),4n2+4nt)⋄ν′(φ(0),4n2+4nt).
Define
(2.44)μ2′′(x,t):=μ′(φ(2x),4n2+4nt)*μ′(φ(x),4n2+4nt)*μ′(φ(-x),4n2+4nt)*μ′(φ(0),4n2+4nt),ν2′′(x,t):=ν′(φ(2x),4n2+4nt)⋄ν′(φ(x),4n2+4nt)⋄ν′(φ(-x),4n2+4nt)⋄ν′(φ(0),4n2+4nt).
Then by the assumption
(2.45)μ2′′(2x,t)=μ2′′(x,tα),ν2′′(2x,t)=ν2′′(x,tα).
Replacing x by 2nx in (2.43) and using (2.45), we obtain
(2.46)μ(f(2nx)2n-f(2n+1x)2n+1,αnt2n)=μ(f(2nx)-f(2n+1x)2,αnt)≥μ2′′(2nx,αnt)=μ2′′(x,t),ν(f(2nx)2n-f(2n+1x)2n+1,αnt2n)=ν(f(2nx)-f(2n+1x)2,αnt)≤ν2′′(2nx,αnt)=ν2′′(x,t).
Thus, for each n>m, we have
(2.47)μ(f(2mx)2m-f(2nx)2n,∑k=mn-1αkt2k)=μ(∑k=mn-1f(2kx)2k-f(2k+1x)2k+1,∑k=mn-1αkt2k)≥∏k=mn-1μ(f(2kx)2k-f(2k+1x)2k+1,αkt2k)≥μ2′′(x,t),ν(f(2mx)2m-f(2nx)2n,∑k=mn-1αkt2k)=ν(∑k=mn-1f(2kx)2k-f(2k+1x)2k+1,∑k=mn-1αkt2k)≤∐k=mn-1ν(f(2kx)2k-f(2k+1x)2k+1,αkt2k)≤ν2′′(x,t),
where ∏j=1naj=a1*a2*⋯*an, ∐j=1naj=a1⋄a2⋄⋯⋄an. Let ɛ>0 and δ>0 be given. Since limt→∞μ2′′(x,t)=1 and limt→∞ν2′′(x,t)=0, there exists some t0>0 such that μ2′′(x,t0)>1-ɛ and ν2′′(x,t0)<ɛ. Since ∑k=0∞αkt0/2k<∞, there is some n0∈ℕ such that ∑k=mn-1αkt0/2k<δ for each n>m≥n0. It follows that
(2.48)μ(f(2mx)2m-f(2nx)2n,δ)≥μ(f(2mx)2m-f(2nx)2n,∑k=mn-1αkt02k)≥μ2′′(x,t0)>1-ε,ν(f(2mx)2m-f(2nx)2n,δ)≤ν(f(2mx)2m-f(2nx)2n,∑k=mn-1αkt02k)≤ν2′′(x,t0)<ε,
for all t>t0. This shows that the sequence {f(2nx)/2n} is Cauchy in (Y,μ,ν). Since (Y,μ,ν) is intuitionistic fuzzy Banach space, {f(2nx)/4n} converges to some point A(x)∈Y. Thus, we can define a mapping A(x):X→Y such that A(x):=(μ,ν)-limn→∞f(2nx)/2n. Moreover, if we put m=0 in (2.47), we get
(2.49)μ(f(2nx)2n-f(x),∑k=0n-1αkt2k)≥μ2′′(x,t),ν(f(2nx)2n-f(x),∑k=0n-1αkt2k)≤ν2′′(x,t).
Thus,
(2.50)μ(f(2nx)2n-f(x),t)≥μ2′′(x,t∑k=0n-1(α/2)k),ν(f(2nx)2n-f(x),t)≤ν2′′(x,t∑k=0n-1(α/2)k).
Next we will show that A is additive. Putting xi=2mxi(i=1,…,n) and ti=(t/n)(i=1,…,n) in (2.32), we obtain
(2.51)μ(Δf(2mx1,…,2mxn)2m,t)≥μ′(φ(2mx1),2mtn)*⋯*μ′(φ(2mxn),2mtn)=μ′(φ(x2),(2α)mtn)*⋯*μ′(φ(xn),(2α)mtn),ν(Δf(2mx1,…,2mxn)2m,t)≤ν′(φ(2mx1),2mtn)⋄⋯⋄ν′(φ(2mxn),2mtn)=ν′(φ(x2),(2α)mtn)⋄⋯⋄ν′(φ(xn),(2α)mtn),
for all x1,…,xn∈X and t>0. Letting n→∞ in (2.51), we obtain
(2.52)μ(ΔA(x1,…,xn),t)=1,ν(ΔA(x1,…,xn),t)=0,
for all x1,…,xn∈X and all t>0. This means that A satisfies the functional (1.1), and so it is additive (see Lemma  2.1 of [4]).
Now, we approximate the difference betweenf and A in intuitionistic fuzzy sense. For every x∈X,t>0, and sufficiently large n, by (2.50), we have
(2.53)μ(A(x)-f(x),t)≥μ(A(x)-f(2nx)2n,t2)*μ(f(2nx)2n-f(x),t2)≥μ2′′(x,t2∑k=0∞(α/2)k)=μ2′′(x,(2-α)t4),ν(A(x)-f(x),t)≤ν(A(x)-f(2nx)2n,t2)⋄ν(f(2nx)2n-f(x),t2)≤ν2′′(x,t2∑k=0∞(α/2)k)=ν2′′(x,(2-α)t4).
To prove the uniqueness of A, assume that A′ is another additive mapping from X to Y, which satisfies the required inequality. Then, for each x∈X and t>0,
(2.54)μ(A(x)-A′(x),t)≥μ(A(x)-f(x),t2)*μ(A′(x)-f(x),t2)≥μ2′′(x,(2-α)t8),ν(A(x)-A′(x),t)≤ν(A(x)-f(x),t2)⋄ν(A′(x)-f(x),t2)≤ν2′′(x,(2-α)t8).
Therefore, by the additivity of A and A′, we have
(2.55)μ(A(x)-A′(x),t)=μ(A(2nx)-A′(2nx),2nt)≥μ2′′(x,(2/α)n(2-α)t8),ν(A(x)-A′(x),t)=ν(A(2nx)-A′(2nx),2nt)≤ν2′′(x,(2/α)n(2-α)t8),
for all x∈X,t>0 and n∈ℕ. Since 0<α<2 and limn→∞(2/α)n=∞, we get
(2.56)lim⁡n→∞μ2′′(x,(2/α)n(2-α)t8)=1,lim⁡n→∞ν2′′(x,(4/α)n(2-α)t8)=0.
Therefore, μ(A(x)-A′(x),t)=1 and ν(A(x)-A′(x),t)=0 for all x∈X and t>0. Hence, A(x)=A′(x) for all x∈X. This completes the proof of the theorem.Theorem 2.3.
Letφ:X→Z be a function such that φ(2x)=αφ(x) for some real number α with 0<|α|<2. Suppose that a function f:X→Y with f(0)=0 satisfies the inequality
(2.57)μ(Δf(x1,…,xn),t1+⋯+tn)≥μ′(φ(x1),t1)*⋯*μ′(φ(xn),tn),ν(Δf(x1,…,xn),t1+⋯+tn)≤ν′(φ(x1),t1)⋄⋯⋄ν′(φ(xn),tn),
for all x1,…,xn∈X and all t1,…,tn>0. Then there exists a unique quadratic function Q:X→Y and a unique additive function A:X→Y such that
(2.58)μ(Q(x)-A(x)-f(x),t)≥M1′′(x,(4-|α|)t16)*M̃1′′(x,(2-|α|)t8),ν(Q(x)-A(x)-f(x),t)≤M2′′(x,(4-|α|)t16)⋄M̃2′′(x,(2-|α|)t8),
for all x∈X and t>0, where
(2.59)M1′′(x,t):=μ′(φ(nx),8(n-1)2n2+9nt)*μ′(φ((n-1)x),8(n-1)2n2+9nt)*μ′(φ(x),8(n-1)2n2+9nt)*μ′(φ(-nx),8(n-1)2n2+9nt)*μ′(φ(-(n-1)x),8(n-1)2n2+9nt)*μ′(φ(-x),8(n-1)2n2+9nt)*μ′(φ(0),8(n-1)2n2+9nt),M̃1′′(x,t):=μ′(φ(2x),4n2+4nt)*μ′(φ(x),4n2+4nt)*μ′(φ(-x),4n2+4nt)*μ′(φ(-2x),4n2+4nt)*μ′(φ(0),4n2+4nt),M2′′(x,t):=ν′(φ(nx),8(n-1)2n2+9nt)⋄ν′(φ((n-1)x),8(n-1)2n2+9nt)⋄ν′(φ(x),8(n-1)2n2+9nt)⋄ν′(φ(-nx),8(n-1)2n2+9nt)⋄ν′(φ(-(n-1)x),8(n-1)2n2+9nt)⋄ν′(φ(-x),8(n-1)2n2+9nt)⋄ν′(φ(0),8(n-1)2n2+9nt),M̃2′′(x,t):=ν′(φ(2x),4n2+4nt)⋄ν′(φ(x),4n2+4nt)⋄ν′(φ(-x),4n2+4nt)⋄ν′(φ(-2x),4n2+4nt)⋄sν′(φ(0),4n2+4nt).Proof.
Passing to the even partfe and odd part fo of f, we deduce from (2.57) that
(2.60)μ(Δfe(x1,…,xn),t1+⋯+tn)≥μ′(φ(x1),t1)*μ′(φ(-x1),t1)*⋯*μ′(φ(xn),tn)*μ′(φ(-xn),tn),ν(Δfe(x1,…,xn),t1+⋯+tn)≤ν′(φ(x1),t1)⋄ν′(φ(-x1),t1)⋄⋯⋄ν′(φ(xn),tn)⋄ν′(φ(-xn),tn).
On the other hand,
(2.61)μ(Δfo(x1,…,xn),t1+⋯+tn)≥μ′(φ(x1),t1)*μ′(φ(-x1),t1)*⋯*μ′(φ(xn),tn)*μ′(φ(-xn),tn),ν(Δfo(x1,…,xn),t1+⋯+tn)≤ν′(φ(x1),t1)⋄ν′(φ(-x1),t1)⋄⋯⋄ν′(φ(xn),tn)⋄ν′(φ(-xn),tn).
Applying the proofs of Theorems 2.1 and 2.2, we get a unique quadratic function Q and a unique additive function A satisfying
(2.62)μ(Q(x)-fe(x),t)≥M1′′(x,(4-|α|)t8),ν(Q(x)-fe(x),t)≤M2′′(x,(4-|α|)t8).
Also,
(2.63)μ(A(x)-fo(x),t)≥M̃1′′(x,(2-|α|)t4),ν(A(x)-fo(x),t)≤M̃2′′(x,(2-|α|)t4).
Therefore,
(2.64)μ(Q(x)-A(x)-f(x),t)≥M1′′(x,(4-|α|)t16)*M̃1′′(x,(2-|α|)t8),ν(Q(x)-A(x)-f(x),t)≤M2′′(x,(4-|α|)t16)⋄M̃2′′(x,(2-|α|)t8).
This completes the proof of the theorem.



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*Source: 456182-2011-11-17.xml*