# The Intuitionistic Fuzzy Normed Space of Coefficients

Authors: B. T. Bilalov; S. M. Farahani; F. A. Guliyeva
Journal: Abstract and Applied Analysis
 (2012)

---

## Abstract

Intuitionistic fuzzy normed space is defined using  concepts oft-norm and t-conorm. The concepts of fuzzy completeness, fuzzy minimality, fuzzy biorthogonality, fuzzy basicity, and fuzzy space of coefficients are introduced. Strong completeness of fuzzy space of coefficients with regard to fuzzy norm and strong basicity of canonical system in this space are proved. Strong basicity criterion in fuzzy Banach space is presented in terms of coefficient operator.


---

## Body

## 1. Introduction

The fuzzy theory, dating back to Zadeh [1], has emerged as the most active area of research in many branches of mathematics and engineering. Fuzzy set theory is a powerful handset for modeling uncertainty and vagueness in various problems arising in the field of science and engineering. The concept of fuzzy topology may have very important applications in quantum particle physics, particularly in connection with both string and ε(∞) theories introduced and studied by El Naschie [2–4] and further developed in [5]. So, further development in E-Infinity may lead to a set transitional resolution of quantum entanglement [6]. A large number of research works are appearing these days which deal with the concept of fuzzy set numbers, and the fuzzification of many classical theories has also been made. The concept of Schauder basis in intuitionistic fuzzy normed space and some results related to this concept have recently been studied in [7–9]. These works introduced the concepts of strongly and weakly intuitionistic fuzzy (Schauder) basis in intuitionistic fuzzy Banach  spaces (IFBS in short). Some of their properties are revealed. The concepts of strongly and weakly intuitionistic fuzzy approximation properties (sif-AP and wif-APin short, resp.) are also introduced in these works. It is proved that if the intuitionistic fuzzy space has a sif-basis, then it has a sif-AP. All the results in these works are obtained on condition that IFBS admits equivalent topology using the family of norms generated by t-norm and t-conorm (we will define them later).In our work, we define the basic concepts of classical basis theory inintuitionistic fuzzy normed spaces (IFNS in short). Concepts of weakly and strongly fuzzy spaces of coefficientsare introduced. Strong completenessof these spaces with regard to fuzzy norm and strong basicity of canonical system in them is proved. Strong basicitycriterion in fuzzy Banach space is presented in terms of coefficient operator.In Section2, we recall some notations and concepts. In Section 3, we state our main results. We first define the fuzzy space of coefficients and then introduce the corresponding fuzzy norms. We prove that for nondegenerate system the corresponding fuzzy space of coefficients is strongly fuzzy complete. Moreover, we show that the canonical system forms a strong basis for this space.


## 2. Some Preliminary Notations and Concepts

We will use the usual notations:N will denote the set of all positive integers, R will be the set of all real numbers, C will be the set of complex numbers, and K will denote a field of scalars (K≡R, or K≡C), R+≡(0,+∞). We state some concepts and facts from IFNS theory to be used later.One of the most important problems in fuzzy topology is to obtain an appropriate concept of intuitionistic fuzzy normed space. This problem has been investigated by Park [10]. He has introduced and studied a notion of intuitionistic fuzzy metric space. We recall it.Definition 2.1.
Abinary operation *:[0,1]2→[0,1]  is a continuous t-norm if it satisfies the following conditions:(a)
* is associative and commutative,(b)
* is continuous,(c)
a*1=a, ∀a∈[0,1],(d)
a*b≤c*d whenever a≤c and b≤d, ∀a, b, c, d∈[0,1].Example 2.2.
Two typical examples of continuoust-norm are a*b=ab and a*b=min{a;b}.Definition 2.3.
A binary operation⋄:[0,1]2→[0,1] is a continuous t-conorm if it satisfies the following conditions:(α)
⋄ is associative and commutative,(β)
⋄ is continuous,(γ)
a⋄0=a,
∀a∈[0,1],(η)
a⋄b≤c⋄d whenever a≤c and b≤d, ∀a,b,c,d∈[0,1].Example 2.4.
Two typical examples of continuoust-conorm are a⋄b=min{a+b;1} and a⋄b=max{a;b}.Definition 2.5.
LetXbe a linear space over a field K.Functions μ;ν:X×R→[0,1] are called fuzzy norms on Xif they hold the following conditions:(1)
μ(x;t)=0, ∀t≤0, ∀x∈X,(2)
μ(x;t)=1, ∀t>0⇒x=0,(3)
μ(cx;t)=μ(x;t/|c|), ∀c≠0,(4)
μ(x;·):R→[0,1] is a nondecreasing function of t for ∀x∈X and limt→∞μ(x;t)=1, ∀x∈X,(5)
μ(x;s)*μ(y;t)≤μ(x+y;s+t), ∀x,y∈X, ∀s,t∈R,(6)
ν(x;t)=1, ∀t≤0, ∀x∈X,(7)
ν(x;t)=0, ∀t<0⇒x=0,(8)
ν(cx;t)=ν(x;t/|c|), ∀c≠0,(9)
ν(x;·):R→[0,1] is a nonincreasing function of t for ∀x∈X and limt→∞ν(x;t)=0, ∀x∈X,(10)
ν(x;s)⋄ν(y;t)≥ν(x+y;s+t), ∀x,y∈X, ∀s,t∈R,(11)
μ(x;t)+ν(x;t)≤1, ∀x∈X, ∀t∈R.Then the 5-tuple (X;μ;ν;*;⋄) is said to be an intuitionistic fuzzy normed space (shortly IFNS).Example 2.6.
Let(X;∥·∥) be a normed space. Denote a*b=ab and a⋄b=min{a+b;1}, for∀a,b∈[0,1], and define μ and ν as follows:
(2.1)μ(x;t)={tt+‖x‖,t>0,0,t≤0,ν(x;t)={tt+‖x‖,t>0,1,t≤0.
Then (X;μ;ν;*;⋄) is an IFNS.
The above concepts allow to introduce the following kinds of convergence (or topology) in IFNS.Definition 2.7.
Let(X;μ;ν) be a fuzzy normed space, and let {xn}n∈N⊂X be some sequence, then it is said to be strongly intuitionistic fuzzy convergent to x∈X (denoted by xn→sx, n→∞ or s-limn→∞xn=x in short) if and only if for ∀ε>0, ∃n0=n0(ε):μ(xn-x;t)≥1-ε, ν(xn-x;t)≤ε, ∀n≥n0, ∀t∈R.Definition 2.8.
Let(X;μ;ν) be a fuzzy normed space, and let{xn}n∈N⊂X be some sequence, then it is said to be weakly intuitionistic fuzzy convergent to x∈X (denoted by xn→wx, n→∞,or w-limn→∞xn=x in short) if and only if for ∀t∈R+, ∀ε>0, ∃n0=n0(ε;t):μ(xn-x;t)≥1-ε, ν(xn-x;t)≤ε, ∀n≥n0. More details on these concepts can be found in [10–19].Let(X;μ;ν) be an IFNS, and let M⊂X be some set. By L[M], we denote the linear span of M in X. The weakly (strongly) intuitionistic fuzzy convergent closure of L[M] will be denoted by Ls[M]¯ (Lw[M]¯). If X is complete with respect to the weakly (strongly) intuitionistic fuzzy convergence, then we will call it intuitionistic fuzzy weakly (strongly) Banach space (IFBwS or Xw (IFBsS or Xs) in short). Let X be an IFBsS (IFBwS). We denote by Xs* (Xw*) the linear space of linear and continuous in IFBsS (IFBwS) functionals over the same field K.Now, we define the corresponding concepts of basis theory for IFNS. Let{xn}n∈N⊂X be some system.Definition 2.9.
System{xn}n∈N is called s-complete (w-complete) in Xs (in Xw) if Ls[{xn}n∈N]¯≡Xs(Lw[{xn}n∈N]¯≡Xw).Definition 2.10.
System{xn*}n∈N⊂Xs*({xn*}n∈N⊂Xw*) is called s-biorthogonal (w-biorthogonal) to the system {xn}n∈N if xn*(xk)=δnk, ∀n,k∈N, where δnk is the Kronecker symbol.Definition 2.11.
System{xn}n∈N⊂Xs({xn}n∈N⊂Xw) is called s-linearly (w-linearly) independent in X if ∑n=1∞λnxn=0 in Xs (in Xw) implies λn=0, ∀n∈N.Definition 2.12.
System{xn}n∈N⊂Xs({xn}n∈N⊂Xw) is called s-basis (w-basis)for Xs (for Xw) if ∀x∈X, ∃!{λn}n∈N⊂K:∑n=1∞λnxn=x in Xs (in Xw).We will also need the following concept.Definition 2.13.
System{xn}n∈N⊂X is called nondegenerate if xn≠0, ∀n∈N.


## 3. Main Results

### 3.1. Space of Coefficients

LetX be an IFNS, and let {xn}n∈N⊂X be some system.Assume that(3.1)Kx¯w≡{{λn}n∈N⊂C:∑n=1∞λnxnconvergesinXw},Kx¯s≡{{λn}n∈N⊂C:∑n=1∞λnxnconvergesinXs}.It is not difficult to see that𝒦x¯w and 𝒦x¯s are linear spaces with regard to component-specific summation and component-specific multiplication by a scalar. Take ∀λ≡{λn}n∈N∈𝒦x¯w, and assume that(3.2)μK(λ¯;t)=inf⁡mμ(∑n=1mλnxn;t);νK(λ¯;t)=sup⁡mν(∑n=1mλnxn;t).Let us show thatμK and νK satisfy the conditions (1)–(11).(1)
It is clear thatμK(λ¯;t)=0, ∀t≤0.(2)
LetμK(λ¯;t)=1, ∀t>0. Hence, μ(∑n=1mλnxn;t)=1, ∀m∈N, ∀t>0. Suppose that the system {xn}n∈N is nondegenerate. It follows from the above-stated relations that, for m=1, we have μ(λ1x1;t)=1, ∀t>0. Hence,  λ1x1=0⇒λ1=0. Continuing this way, we get at the end of this process thatλn=0, ∀n∈N, that is, λ¯=0.(3)
The validity of relationμK(Aλ¯;t)=μK(λ¯;t/|c|), ∀c≠0 is beyond any doubt.(4)
Asμ(x;·) is a nondecreasing function on R it is not difficult to see that μK(λ¯;·) has the same property. Let us show that limt→∞μK(λ¯;t)=1. Take ∀ε>0. Let Sm=∑n=1mλnxn and w-limm→∞Sm=S∈Xw. It is clear that ∃t0>0:μ(S;t0)≥1-ε. Then it follows from the definition of w-limm that∃m0(ε;t0):μ(Sm-S;t0)≥1-ε,∀m≥m0(ε;t0). Property (4) implies
(3.3)μ(Sm;2t0)=μ(Sm-S+S;t0+t0)≥μ(Sm-S;t0)*μ(S;t0).
As a result, we get
(3.4)μ(Sm;t0)≥1-ε,∀m≥m0(ε;t0).
Asμ(x;·) is a nondecreasing function oft, it follows from (3.4) that
(3.5)μ(Sm;t)≥1-ε,∀m≥m0(ε;t0),∀t≥t0.
We have
(3.6)μK(λ¯;t)=inf⁡mμ(Sm;t)=min⁡{μ(S1;t);…;μ(Sm0-1;t);inf⁡m≥m0μ(Sm;t)},
wherem0=m0(ε;t0). Aslimt→∞μ(Sk;t)=1 for∀k∈N, we have∃tk(ε);∀t≥tk(ε):μ(Sk;t)≥1-ε,k=1,m0-1¯. Lettε0=max{tk(ε),k=1,m0-1¯}, then it is clear that
(3.7)μ(Sk;t)≥1-ε,∀t≥tε0.
It follows from (3.5) and (3.6) that
(3.8)inf⁡m≥m0⁡μ(Sm;t)≥1-ε,∀t≥t0.
Lettε=max{t0;tε0}. Hence, we obtain from (3.6) and (3.7) that
(3.9)μK(λ¯;t)≥1-ε,∀t≥tε.
Thus,limt→∞μK(λ¯;t)=1,∀λ¯∈𝒦x¯w.(5)
Letλ¯,μ¯∈𝒦x¯w(λ¯≡{λn}n∈N;μ¯≡{μn}n∈N) and s,t∈R. We have
(3.10)μK(λ¯+μ¯;s+t)=inf⁡mμ(∑n=1m(λn+μn)xn;s+t)=inf⁡mμ(∑n=1mλnxn+∑n=1mμnxn;s+t)≥inf⁡m[μ(∑n=1mλnxn;s)*μ(∑n=1mμnxn;t)]=[inf⁡mμ(∑n=1mλnxn;s)]*[inf⁡mμ(∑n=1mμnxn;t)]=μ(λ¯;s)*μ(μ¯;t).(6)
Asν(x;t)=1, ∀t≤0, it is clear that νK(λ¯;t)=1, ∀t≤0, ∀λ¯∈𝒦x¯w.(7)
Let the system{xn}n∈N be nondegenerate. Assume that νK(λ¯;t)=0, ∀t>0, thenν(∑n=1mλnxn;t)=0,∀t>0,∀m∈N. Form=1, we haveν(λ1x1;t)=0,∀t>0⇒λ1x1=0⇒λ1=0. Continuing this process, we getλn=0, ∀n∈N⇒λ¯=0.(8)
Clearly,νK(cλ¯;t)=νK(λ¯;t/|c|), ∀c≠0.(9)
It follows from the property (9) thatν(x;·) is a nonincreasing function on R. Therefore, νK(λ¯;·) is a nonincreasing function on R. Let us show that limt→∞νK(λ¯;t)=0. Let Sm=∑n=1mλnxn and w-limm→∞Sm=S∈X. Take ∀ε>0. It is clear that ∃t0>0:ν(S;t0)≤ε. Then it follows from the definition of w-limm that ∃m0=m0(ε;t0):ν(Sm-S;t0)≤ε, ∀m≥m0. We have
(3.11)ν(Sm;t0)=ν(Sm-S+S;t0+t0)≤ν(Sm-S;t0)⋄ν(S;t0)≤ε,∀m≥m0
As ν(x;·) is a nonincreasing function, it is clear that
(3.12)ν(Sm;t)≤ε,∀m≥m0,∀t≥t0.
We have
(3.13)νK(λ¯;t)=sup⁡mν(Sm;t)=max⁡{ν(S1;t);…;ν(Sm0-1;t);sup⁡m≥m0ν(Sm;t)}.
As limt→∞ν(Sk;t)=0 for ∀k∈N, we have ∃tk(ε);∀t≥tk(ε):ν(Sk;t)≤ε, k=1,m0-1¯. Let tε0=max{tk(ε),k=1,m0-1¯}. It is clear that ν(Sk;t)≤ε, ∀t≥tε0. It follows from (3.12) that supm≥m0ν(Sm;t)≤ε, ∀t≥t0. Let tε=max{t0;tε0}, then it is clear that νK(λ¯;t)≤ε, ∀t≥tε⇒limt→∞νK(λ¯;t)=0.(10)
Letλ¯,μ¯∈𝒦x¯w(λ¯≡{λn}n∈N;μ¯≡{μn}n∈N) and s,t∈R. We have
(3.14)νK(λ¯+μ¯;s+t)=sup⁡mν(∑n=1m(λn+μn)xn;s+t)≤sup⁡m[ν(∑n=1mλnxn;s)⋄ν(∑n=1mμnxn;t)]=[sup⁡mν(∑n=1mλnxn;s)]⋄[sup⁡mν(∑n=1mμnxn;t)]=νK(λ¯;s)⋄νK(μ¯;t).(11)
Consider the following:(3.15)μK(λ¯;t)+νK(λ¯;t)=inf⁡mμ(∑n=1mλnxn;t)+sup⁡mν(∑n=1mλnxn;t)≤sup⁡m[μ(∑n=1mλnxn;t)+ν(∑n=1mλnxn;t)]≤1,∀λ¯∈Kx¯w,∀λ∈R.
Thus, we have proved the validity of the following.Theorem 3.1.
Let(X;μ;ν) be a fuzzy normed space, and let  {xn}n∈N⊂X be a nondegenerate system, then the space of coefficients  (𝒦x¯s;μK;νK) is also strongly fuzzy normed space.The following theorem is proved in absolutely the same way.Theorem 3.2.
Let(X;μ;ν) be a fuzzy normed space, and let  {xn}n∈N⊂X be a nondegenerate system, then the space of coefficients  (𝒦x¯w;μK;νK) is also weakly fuzzy normed space.


### 3.2. Completeness of the Space of Coefficients

Subsequently, we assume that(X;μ;ν) is IFBS. Let us show that (𝒦x¯s;μK;νK) is a strongly fuzzy complete normed space. First, we prove the following.Lemma 3.3.
Letx0≠0, x0∈X, and let  {λn}n∈N⊂R be some sequence. If  s-limn→∞(λnx0)=0, that is, ∀ε>0, ∃n0=n0(ε):μ(λnx0;t)>1-ε,  ν(λnx0;t)<ε, ∀t∈R+, and ∀n≥n0, then  λn→0,  n→∞.Proof.
Asx0≠0, it is clear that ∃t0>0:μ(x0;t0)<1. We have μ(λnx0;t)=μ(x0;(t)/|λn|) for λn≠0. Assume that the relation limn→∞λn=0 is not true, then ∃{λnk}k∈N and ∃δ>0:|λnk|≥δ, ∀k∈N. It is clear that limk→∞μ(λnkx0;t)=1 uniformly in t. On the other hand, for tk=|λnk|t0, we have μ(λnkx0;tk)=μ(x0;t0)<1. So we came upon a contradiction which proves the lemma.Further, we assume that the following condition is also fulfilled.(12)
The functionsμ(x;·),ν(x;·):R→[0,1] are continuous for ∀x∈X.Takes-fundamental sequence{λ¯n}n∈N⊂𝒦x¯s,λ¯n≡{λk(n)}k∈N. Thenlimn,m→∞μK(λ¯n-λ¯m;t)=1 uniformly int∈R, that is,
(3.16)lim⁡n,m→∞⁡inf⁡rμ(∑k=1r(λk(n)-λk(m))xk)=1,uniformly int∈R. Take ∀k0∈N and fix it. We have(3.17)(λk0(n)-λk0(m))xk0=∑k=1k0(λk(n)-λk(m))xk-∑k=1k0-1(λk(n)-λk(m))xk.Then from property (5), we get(3.18)μ((λk0(n)-λk0(m))xk0;t)≥μ(∑k=1k0(λk(n)-λk(m))xk;t2)*μ(∑k=1k0-1(λk(n)-λk(m))xk;t2).It follows directly from this relation thatlimn,m→∞μ((λk0(n)-λk0(m))xk0;t)=1 uniformly in t. As xk0≠0, Lemma 3.3 implies limn,m→∞|λk0(n)-λk0(m)|=0, that is, the sequence {λk0(n)}n∈N is fundamental in R. Let λk0(n)→λk0, as n→∞. Denote λ¯≡{λn}n∈N. Let us show that limn→∞μK(λ¯n-λ¯;t)=1 uniformly in t. Take ∀ε>0. It is clear that ∃n0, ∀n≥n0, ∀p∈N:μK(λ¯n-λ¯n+p;t)>1-ε, ∀t∈R. Consequently,(3.19)inf⁡rμ(∑k=1r(λk(n)-λk(n+p))xk;t)>1-ε,∀n≥n0,∀p∈N,∀t∈R+.Hence,(3.20)μ(∑k=1r(λk(n)-λk(n+p))xk;t)>1-ε,∀n≥n0,∀r,p∈N,∀t∈R+.
As shown above, limn,m→∞μ((λk(n)-λk(m))xk;t)=1 uniformly in t∈R+. Now let us take into account the fact that limm→∞μ(λk(m)xk;t)=μ(λkxk;t), ∀t∈R+. Indeed, if λk=0, then μ(0;t)=1, ∀t∈R+, and clearly, limm→∞μ(λk(m)xk;t)=1 for ∀t∈R+. If λk≠0, then for sufficiently large values of m we have λk(m)≠0, and as a result,(3.21)μ(λk(m)xk;t)=μ(xk;t|λk(m)|)⟶m→∞μ(xk;t|λk|)=μ(λkxk;t),∀t∈R+.Passage to the limit in the inequality (3.20) as p→∞ yields(3.22)μ(∑k=1r(λk(n)-λk)xk;t)≥1-ε,∀n≥n0,∀r∈N,∀t∈R+.
We have(3.23)μ(∑k=rr+p(λk(n)-λk)xk;t)=μ(∑k=1r+p(λk(n)-λk)xk-∑k=1r-1(λk(n)-λk)xk;t)≥μ(∑k=1r+p(λk(n)-λk)xk;t2)*μ(∑k=1r-1(λk(n)-λk)xk;t2)≥1-ε,∀n≥n0,∀r,p∈N,∀t∈R+.Asλ¯n∈𝒦x̅s, it is clear that ∃m0(n):∀m≥m0(n), ∀p∈N:(3.24)μ(∑k=mm+pλk(n)xk;t)>1-ε,∀t∈R+.
We have
(3.25)μ(∑k=mm+pλkxk;t)=μ(∑k=mm+p(λk-λk(n))xk+∑k=mm+pλk(n)xk;t)≥μ(∑k=mm+p(λk-λk(n))xk;t2)*μ(∑k=mm+pλk(n)xk;t2)≥1-ε,∀m≥m0(n),∀p∈N,∀t∈R+.It follows that the series∑k=1∞λkxk is strongly fuzzy convergent, that is, ∃s-limm→∞∑k=1mλkxk. Consequently,λ¯∈𝒦x̅s, and the relation (3.22) implies that limn→∞μK(λ̅n-λ̅;t)=1 uniformly in, ∀t∈R+. It can be proved in a similar way that limn→∞νK(λ̅n-λ̅;t)=0 uniformly in ∀t∈R+. As a result, we obtain that the space (𝒦x̅s;μK;νK) is strongly fuzzy complete. Thus, we have proved the following.Theorem 3.4.
Let(X;μ;ν) be a fuzzy Banach space with condition (12), and let  {xn}n∈N⊂X be a nondegenerate system, then the space of coefficients  (𝒦x̅s;μK;νK) is a strongly fuzzy complete normed space.Consider operatorT:𝒦x̅s→X defined by(3.26)Tλ¯=∑n=1∞λnxn,λ̅≡{λn}n∈N∈Kx̅s.Lets-limn→∞λ¯n=λ¯ in 𝒦x¯s,where λ¯n≡{λk(n)}k∈N∈𝒦xs. We have(3.27)μ(Tλ¯n-Tλ¯;t)=μ(∑k=1∞(λk(n)-λk)xk;t)≥inf⁡mμ(∑k=1m(λk(n)-λk)xk;t)=μ(λ¯n-λ¯;t).It follows directly thats-limn→∞Tλ¯n=Tλ¯, that is, the operator T is strongly fuzzy continuous. Let λ¯∈KerT, that is, Tλ¯=0⇒∑n=1∞λnxn=0, where λ¯≡{λn}n∈N∈𝒦x̅s. It is clear that if the system{xn}n∈N is s-linearly independent, then λn=0, ∀n∈N, and as a result, KerT={0}. In this case,∃T-1: ImT→𝒦x¯s. If, in addition, ImT is s-closedin X, thenT-1 is also continuous.Denote by{e̅n}n∈N⊂𝒦x¯s  a canonical system in𝒦x¯s, where e̅n={δnk}k∈N∈𝒦x¯s. Obviously, Te̅n=xn, ∀n∈N. Let us prove that {e̅n}n∈N forms an s-basis for 𝒦x¯s. Take ∀λ̅≡{λn}n∈N∈𝒦x¯s and show that the series ∑n=1∞λne̅n is strongly fuzzy convergent in 𝒦x¯s. In fact, the existence of s-limm→∞∑n=1mλnxn in Xs implies that ∀ε>0, ∃m0∈N,(3.28)μ(∑n=mm+pλnxn;t)>1-ε,∀m≥m0,∀p∈N,∀t∈R+.We have(3.29)μK(∑n=mm+pλne¯n;t)=inf⁡r(∑n=mrλnxn;t)≥1-ε,∀m≥m0,∀p∈N,∀t∈R+.It follows that the series∑n=1∞λne¯n is strongly fuzzy convergent in 𝒦x¯s. Moreover,(3.30)μK(λ¯-∑n=1mλne¯n;t)=μK({…;0;λm+1;…};t)=inf⁡rμ(∑n=m+1rλnxn;t)≥1-ε,∀m≥m0,∀t∈R+.Consequently,s-limm→∞∑n=1mλne¯n=λ¯, that is, λ¯=s∑n=1mλne¯n. Consider the functionals en*(λ¯)=λn, ∀n∈N. Let us show that they are s-continuous. Let s-limn→∞λ¯n=λ¯, where λ¯n≡{λk(n)}k∈N∈𝒦x¯s. As established in the proof of Theorem 3.4, we have λk(n)→λk as n→∞, that is, ek*(λ¯n)→ek*(λ¯) as n→∞ for ∀k∈N. Thus, ek* is s-continuous in 𝒦x¯s for ∀k∈N. On the other hand, it is easy to see that en*(e¯k)=δnk, ∀n,k∈N, that is, {en*}n∈N is s-biorthogonal to {e¯n}n∈N. As a result, we obtain that the system {e¯n}n∈N forms an s-basis for 𝒦x¯s. So we get the validity of the following.Theorem 3.5.
Let(X;μ;ν) be a fuzzy Banach space with condition (12), and let {xn}n∈N⊂X be a nondegenerate system. Then the corresponding space of coefficients  (𝒦x¯s;μK;νK) is strongly fuzzy complete with canonical  s-basis {e¯n}n∈N.Suppose that the system{xn}n∈N iss-linearly independentand ImT is closed, then it is easily seen that {xn}n∈N forms an s-basis for ImT, and in case of its s-completeness in Xs, it forms an s-basis for Xs. In this case, 𝒦x¯s and Xs are isomorphic, and T is an isomorphism between them. The opposite of it is also true, that is, if the above-defined operator T is an isomorphism between 𝒦x¯s and Xs, then the system {xn}n∈N forms an s-basisfor Xs. We will call T a coefficient operator. Thus, the following theorem holds.Theorem 3.6.
Let(X;μ;ν) be a fuzzy Banach space with condition (12), let  {xn}n∈N⊂X be a nondegenerate system, let  (𝒦x¯s;μK;νK) be a corresponding strongly fuzzy complete normed space, and let T:𝒦x̅s→Xs be a coefficient operator. System  {xn}n∈N forms an  s-basis for  Xs if and only if the operator  T is an isomorphism between  𝒦x̅s and  Xs.


## 3.1. Space of Coefficients

LetX be an IFNS, and let {xn}n∈N⊂X be some system.Assume that(3.1)Kx¯w≡{{λn}n∈N⊂C:∑n=1∞λnxnconvergesinXw},Kx¯s≡{{λn}n∈N⊂C:∑n=1∞λnxnconvergesinXs}.It is not difficult to see that𝒦x¯w and 𝒦x¯s are linear spaces with regard to component-specific summation and component-specific multiplication by a scalar. Take ∀λ≡{λn}n∈N∈𝒦x¯w, and assume that(3.2)μK(λ¯;t)=inf⁡mμ(∑n=1mλnxn;t);νK(λ¯;t)=sup⁡mν(∑n=1mλnxn;t).Let us show thatμK and νK satisfy the conditions (1)–(11).(1)
It is clear thatμK(λ¯;t)=0, ∀t≤0.(2)
LetμK(λ¯;t)=1, ∀t>0. Hence, μ(∑n=1mλnxn;t)=1, ∀m∈N, ∀t>0. Suppose that the system {xn}n∈N is nondegenerate. It follows from the above-stated relations that, for m=1, we have μ(λ1x1;t)=1, ∀t>0. Hence,  λ1x1=0⇒λ1=0. Continuing this way, we get at the end of this process thatλn=0, ∀n∈N, that is, λ¯=0.(3)
The validity of relationμK(Aλ¯;t)=μK(λ¯;t/|c|), ∀c≠0 is beyond any doubt.(4)
Asμ(x;·) is a nondecreasing function on R it is not difficult to see that μK(λ¯;·) has the same property. Let us show that limt→∞μK(λ¯;t)=1. Take ∀ε>0. Let Sm=∑n=1mλnxn and w-limm→∞Sm=S∈Xw. It is clear that ∃t0>0:μ(S;t0)≥1-ε. Then it follows from the definition of w-limm that∃m0(ε;t0):μ(Sm-S;t0)≥1-ε,∀m≥m0(ε;t0). Property (4) implies
(3.3)μ(Sm;2t0)=μ(Sm-S+S;t0+t0)≥μ(Sm-S;t0)*μ(S;t0).
As a result, we get
(3.4)μ(Sm;t0)≥1-ε,∀m≥m0(ε;t0).
Asμ(x;·) is a nondecreasing function oft, it follows from (3.4) that
(3.5)μ(Sm;t)≥1-ε,∀m≥m0(ε;t0),∀t≥t0.
We have
(3.6)μK(λ¯;t)=inf⁡mμ(Sm;t)=min⁡{μ(S1;t);…;μ(Sm0-1;t);inf⁡m≥m0μ(Sm;t)},
wherem0=m0(ε;t0). Aslimt→∞μ(Sk;t)=1 for∀k∈N, we have∃tk(ε);∀t≥tk(ε):μ(Sk;t)≥1-ε,k=1,m0-1¯. Lettε0=max{tk(ε),k=1,m0-1¯}, then it is clear that
(3.7)μ(Sk;t)≥1-ε,∀t≥tε0.
It follows from (3.5) and (3.6) that
(3.8)inf⁡m≥m0⁡μ(Sm;t)≥1-ε,∀t≥t0.
Lettε=max{t0;tε0}. Hence, we obtain from (3.6) and (3.7) that
(3.9)μK(λ¯;t)≥1-ε,∀t≥tε.
Thus,limt→∞μK(λ¯;t)=1,∀λ¯∈𝒦x¯w.(5)
Letλ¯,μ¯∈𝒦x¯w(λ¯≡{λn}n∈N;μ¯≡{μn}n∈N) and s,t∈R. We have
(3.10)μK(λ¯+μ¯;s+t)=inf⁡mμ(∑n=1m(λn+μn)xn;s+t)=inf⁡mμ(∑n=1mλnxn+∑n=1mμnxn;s+t)≥inf⁡m[μ(∑n=1mλnxn;s)*μ(∑n=1mμnxn;t)]=[inf⁡mμ(∑n=1mλnxn;s)]*[inf⁡mμ(∑n=1mμnxn;t)]=μ(λ¯;s)*μ(μ¯;t).(6)
Asν(x;t)=1, ∀t≤0, it is clear that νK(λ¯;t)=1, ∀t≤0, ∀λ¯∈𝒦x¯w.(7)
Let the system{xn}n∈N be nondegenerate. Assume that νK(λ¯;t)=0, ∀t>0, thenν(∑n=1mλnxn;t)=0,∀t>0,∀m∈N. Form=1, we haveν(λ1x1;t)=0,∀t>0⇒λ1x1=0⇒λ1=0. Continuing this process, we getλn=0, ∀n∈N⇒λ¯=0.(8)
Clearly,νK(cλ¯;t)=νK(λ¯;t/|c|), ∀c≠0.(9)
It follows from the property (9) thatν(x;·) is a nonincreasing function on R. Therefore, νK(λ¯;·) is a nonincreasing function on R. Let us show that limt→∞νK(λ¯;t)=0. Let Sm=∑n=1mλnxn and w-limm→∞Sm=S∈X. Take ∀ε>0. It is clear that ∃t0>0:ν(S;t0)≤ε. Then it follows from the definition of w-limm that ∃m0=m0(ε;t0):ν(Sm-S;t0)≤ε, ∀m≥m0. We have
(3.11)ν(Sm;t0)=ν(Sm-S+S;t0+t0)≤ν(Sm-S;t0)⋄ν(S;t0)≤ε,∀m≥m0
As ν(x;·) is a nonincreasing function, it is clear that
(3.12)ν(Sm;t)≤ε,∀m≥m0,∀t≥t0.
We have
(3.13)νK(λ¯;t)=sup⁡mν(Sm;t)=max⁡{ν(S1;t);…;ν(Sm0-1;t);sup⁡m≥m0ν(Sm;t)}.
As limt→∞ν(Sk;t)=0 for ∀k∈N, we have ∃tk(ε);∀t≥tk(ε):ν(Sk;t)≤ε, k=1,m0-1¯. Let tε0=max{tk(ε),k=1,m0-1¯}. It is clear that ν(Sk;t)≤ε, ∀t≥tε0. It follows from (3.12) that supm≥m0ν(Sm;t)≤ε, ∀t≥t0. Let tε=max{t0;tε0}, then it is clear that νK(λ¯;t)≤ε, ∀t≥tε⇒limt→∞νK(λ¯;t)=0.(10)
Letλ¯,μ¯∈𝒦x¯w(λ¯≡{λn}n∈N;μ¯≡{μn}n∈N) and s,t∈R. We have
(3.14)νK(λ¯+μ¯;s+t)=sup⁡mν(∑n=1m(λn+μn)xn;s+t)≤sup⁡m[ν(∑n=1mλnxn;s)⋄ν(∑n=1mμnxn;t)]=[sup⁡mν(∑n=1mλnxn;s)]⋄[sup⁡mν(∑n=1mμnxn;t)]=νK(λ¯;s)⋄νK(μ¯;t).(11)
Consider the following:(3.15)μK(λ¯;t)+νK(λ¯;t)=inf⁡mμ(∑n=1mλnxn;t)+sup⁡mν(∑n=1mλnxn;t)≤sup⁡m[μ(∑n=1mλnxn;t)+ν(∑n=1mλnxn;t)]≤1,∀λ¯∈Kx¯w,∀λ∈R.
Thus, we have proved the validity of the following.Theorem 3.1.
Let(X;μ;ν) be a fuzzy normed space, and let  {xn}n∈N⊂X be a nondegenerate system, then the space of coefficients  (𝒦x¯s;μK;νK) is also strongly fuzzy normed space.The following theorem is proved in absolutely the same way.Theorem 3.2.
Let(X;μ;ν) be a fuzzy normed space, and let  {xn}n∈N⊂X be a nondegenerate system, then the space of coefficients  (𝒦x¯w;μK;νK) is also weakly fuzzy normed space.


## 3.2. Completeness of the Space of Coefficients

Subsequently, we assume that(X;μ;ν) is IFBS. Let us show that (𝒦x¯s;μK;νK) is a strongly fuzzy complete normed space. First, we prove the following.Lemma 3.3.
Letx0≠0, x0∈X, and let  {λn}n∈N⊂R be some sequence. If  s-limn→∞(λnx0)=0, that is, ∀ε>0, ∃n0=n0(ε):μ(λnx0;t)>1-ε,  ν(λnx0;t)<ε, ∀t∈R+, and ∀n≥n0, then  λn→0,  n→∞.Proof.
Asx0≠0, it is clear that ∃t0>0:μ(x0;t0)<1. We have μ(λnx0;t)=μ(x0;(t)/|λn|) for λn≠0. Assume that the relation limn→∞λn=0 is not true, then ∃{λnk}k∈N and ∃δ>0:|λnk|≥δ, ∀k∈N. It is clear that limk→∞μ(λnkx0;t)=1 uniformly in t. On the other hand, for tk=|λnk|t0, we have μ(λnkx0;tk)=μ(x0;t0)<1. So we came upon a contradiction which proves the lemma.Further, we assume that the following condition is also fulfilled.(12)
The functionsμ(x;·),ν(x;·):R→[0,1] are continuous for ∀x∈X.Takes-fundamental sequence{λ¯n}n∈N⊂𝒦x¯s,λ¯n≡{λk(n)}k∈N. Thenlimn,m→∞μK(λ¯n-λ¯m;t)=1 uniformly int∈R, that is,
(3.16)lim⁡n,m→∞⁡inf⁡rμ(∑k=1r(λk(n)-λk(m))xk)=1,uniformly int∈R. Take ∀k0∈N and fix it. We have(3.17)(λk0(n)-λk0(m))xk0=∑k=1k0(λk(n)-λk(m))xk-∑k=1k0-1(λk(n)-λk(m))xk.Then from property (5), we get(3.18)μ((λk0(n)-λk0(m))xk0;t)≥μ(∑k=1k0(λk(n)-λk(m))xk;t2)*μ(∑k=1k0-1(λk(n)-λk(m))xk;t2).It follows directly from this relation thatlimn,m→∞μ((λk0(n)-λk0(m))xk0;t)=1 uniformly in t. As xk0≠0, Lemma 3.3 implies limn,m→∞|λk0(n)-λk0(m)|=0, that is, the sequence {λk0(n)}n∈N is fundamental in R. Let λk0(n)→λk0, as n→∞. Denote λ¯≡{λn}n∈N. Let us show that limn→∞μK(λ¯n-λ¯;t)=1 uniformly in t. Take ∀ε>0. It is clear that ∃n0, ∀n≥n0, ∀p∈N:μK(λ¯n-λ¯n+p;t)>1-ε, ∀t∈R. Consequently,(3.19)inf⁡rμ(∑k=1r(λk(n)-λk(n+p))xk;t)>1-ε,∀n≥n0,∀p∈N,∀t∈R+.Hence,(3.20)μ(∑k=1r(λk(n)-λk(n+p))xk;t)>1-ε,∀n≥n0,∀r,p∈N,∀t∈R+.
As shown above, limn,m→∞μ((λk(n)-λk(m))xk;t)=1 uniformly in t∈R+. Now let us take into account the fact that limm→∞μ(λk(m)xk;t)=μ(λkxk;t), ∀t∈R+. Indeed, if λk=0, then μ(0;t)=1, ∀t∈R+, and clearly, limm→∞μ(λk(m)xk;t)=1 for ∀t∈R+. If λk≠0, then for sufficiently large values of m we have λk(m)≠0, and as a result,(3.21)μ(λk(m)xk;t)=μ(xk;t|λk(m)|)⟶m→∞μ(xk;t|λk|)=μ(λkxk;t),∀t∈R+.Passage to the limit in the inequality (3.20) as p→∞ yields(3.22)μ(∑k=1r(λk(n)-λk)xk;t)≥1-ε,∀n≥n0,∀r∈N,∀t∈R+.
We have(3.23)μ(∑k=rr+p(λk(n)-λk)xk;t)=μ(∑k=1r+p(λk(n)-λk)xk-∑k=1r-1(λk(n)-λk)xk;t)≥μ(∑k=1r+p(λk(n)-λk)xk;t2)*μ(∑k=1r-1(λk(n)-λk)xk;t2)≥1-ε,∀n≥n0,∀r,p∈N,∀t∈R+.Asλ¯n∈𝒦x̅s, it is clear that ∃m0(n):∀m≥m0(n), ∀p∈N:(3.24)μ(∑k=mm+pλk(n)xk;t)>1-ε,∀t∈R+.
We have
(3.25)μ(∑k=mm+pλkxk;t)=μ(∑k=mm+p(λk-λk(n))xk+∑k=mm+pλk(n)xk;t)≥μ(∑k=mm+p(λk-λk(n))xk;t2)*μ(∑k=mm+pλk(n)xk;t2)≥1-ε,∀m≥m0(n),∀p∈N,∀t∈R+.It follows that the series∑k=1∞λkxk is strongly fuzzy convergent, that is, ∃s-limm→∞∑k=1mλkxk. Consequently,λ¯∈𝒦x̅s, and the relation (3.22) implies that limn→∞μK(λ̅n-λ̅;t)=1 uniformly in, ∀t∈R+. It can be proved in a similar way that limn→∞νK(λ̅n-λ̅;t)=0 uniformly in ∀t∈R+. As a result, we obtain that the space (𝒦x̅s;μK;νK) is strongly fuzzy complete. Thus, we have proved the following.Theorem 3.4.
Let(X;μ;ν) be a fuzzy Banach space with condition (12), and let  {xn}n∈N⊂X be a nondegenerate system, then the space of coefficients  (𝒦x̅s;μK;νK) is a strongly fuzzy complete normed space.Consider operatorT:𝒦x̅s→X defined by(3.26)Tλ¯=∑n=1∞λnxn,λ̅≡{λn}n∈N∈Kx̅s.Lets-limn→∞λ¯n=λ¯ in 𝒦x¯s,where λ¯n≡{λk(n)}k∈N∈𝒦xs. We have(3.27)μ(Tλ¯n-Tλ¯;t)=μ(∑k=1∞(λk(n)-λk)xk;t)≥inf⁡mμ(∑k=1m(λk(n)-λk)xk;t)=μ(λ¯n-λ¯;t).It follows directly thats-limn→∞Tλ¯n=Tλ¯, that is, the operator T is strongly fuzzy continuous. Let λ¯∈KerT, that is, Tλ¯=0⇒∑n=1∞λnxn=0, where λ¯≡{λn}n∈N∈𝒦x̅s. It is clear that if the system{xn}n∈N is s-linearly independent, then λn=0, ∀n∈N, and as a result, KerT={0}. In this case,∃T-1: ImT→𝒦x¯s. If, in addition, ImT is s-closedin X, thenT-1 is also continuous.Denote by{e̅n}n∈N⊂𝒦x¯s  a canonical system in𝒦x¯s, where e̅n={δnk}k∈N∈𝒦x¯s. Obviously, Te̅n=xn, ∀n∈N. Let us prove that {e̅n}n∈N forms an s-basis for 𝒦x¯s. Take ∀λ̅≡{λn}n∈N∈𝒦x¯s and show that the series ∑n=1∞λne̅n is strongly fuzzy convergent in 𝒦x¯s. In fact, the existence of s-limm→∞∑n=1mλnxn in Xs implies that ∀ε>0, ∃m0∈N,(3.28)μ(∑n=mm+pλnxn;t)>1-ε,∀m≥m0,∀p∈N,∀t∈R+.We have(3.29)μK(∑n=mm+pλne¯n;t)=inf⁡r(∑n=mrλnxn;t)≥1-ε,∀m≥m0,∀p∈N,∀t∈R+.It follows that the series∑n=1∞λne¯n is strongly fuzzy convergent in 𝒦x¯s. Moreover,(3.30)μK(λ¯-∑n=1mλne¯n;t)=μK({…;0;λm+1;…};t)=inf⁡rμ(∑n=m+1rλnxn;t)≥1-ε,∀m≥m0,∀t∈R+.Consequently,s-limm→∞∑n=1mλne¯n=λ¯, that is, λ¯=s∑n=1mλne¯n. Consider the functionals en*(λ¯)=λn, ∀n∈N. Let us show that they are s-continuous. Let s-limn→∞λ¯n=λ¯, where λ¯n≡{λk(n)}k∈N∈𝒦x¯s. As established in the proof of Theorem 3.4, we have λk(n)→λk as n→∞, that is, ek*(λ¯n)→ek*(λ¯) as n→∞ for ∀k∈N. Thus, ek* is s-continuous in 𝒦x¯s for ∀k∈N. On the other hand, it is easy to see that en*(e¯k)=δnk, ∀n,k∈N, that is, {en*}n∈N is s-biorthogonal to {e¯n}n∈N. As a result, we obtain that the system {e¯n}n∈N forms an s-basis for 𝒦x¯s. So we get the validity of the following.Theorem 3.5.
Let(X;μ;ν) be a fuzzy Banach space with condition (12), and let {xn}n∈N⊂X be a nondegenerate system. Then the corresponding space of coefficients  (𝒦x¯s;μK;νK) is strongly fuzzy complete with canonical  s-basis {e¯n}n∈N.Suppose that the system{xn}n∈N iss-linearly independentand ImT is closed, then it is easily seen that {xn}n∈N forms an s-basis for ImT, and in case of its s-completeness in Xs, it forms an s-basis for Xs. In this case, 𝒦x¯s and Xs are isomorphic, and T is an isomorphism between them. The opposite of it is also true, that is, if the above-defined operator T is an isomorphism between 𝒦x¯s and Xs, then the system {xn}n∈N forms an s-basisfor Xs. We will call T a coefficient operator. Thus, the following theorem holds.Theorem 3.6.
Let(X;μ;ν) be a fuzzy Banach space with condition (12), let  {xn}n∈N⊂X be a nondegenerate system, let  (𝒦x¯s;μK;νK) be a corresponding strongly fuzzy complete normed space, and let T:𝒦x̅s→Xs be a coefficient operator. System  {xn}n∈N forms an  s-basis for  Xs if and only if the operator  T is an isomorphism between  𝒦x̅s and  Xs.



---

*Source: 969313-2012-05-28.xml*