# Compact Operators Defined on2-Normed and 2-Probabilistic Normed Spaces

Authors: Fatemeh Lael; Kourosh Nourouzi
Journal: Mathematical Problems in Engineering
 (2009)

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## Abstract

The compact operators defined on2-normed
spaces are investigated, and then the main ideas are generalized to operators
defined on 2-probabilistic normed spaces.


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## Body

## 1. Introduction

In 1963, Gähler [1] introducedthe notion of a 2-metric, real-valued function of pointtriples on a set X, whose abstract properties were suggested by the area function for a triangle determined by a triple in Euclidean space. A related concept in the category of linear spaces, the theory of 2-norm on a linear space, was also investigated by Gähler in [2]. Since these were studied in many papers, we mention [3–5].Also, due to vagueness about the distance between points in a metric space, probabilistic metric spaces were introduced by Menger [6] as a generalization of metric spaces. From the vantage point of a sixty-year history, it is safe to say that the probabilistic approach on deterministic results of linear normed spaces is playing an important role in applied mathematics.In this paper, we first investigate compact operators between2-normed spaces. Then, according to Menger's probabilistic approach, we discuss on 2-probabilistic normed spaces and extend the main ideas given in first section to operators defined between 2-probabilistic normed spaces.


## 2.2-Normed Spaces

In this section, after providing the required preliminaries, we discuss on compact operators between2-normed spaces.In the sequel of this paper, it is always assumed that all vector spaces are real with the dimension greater than one.Definition 2.1 ([7]).
LetX be a real linear space. A function ∥·,·∥:X2→ℝ is called a 2-norm on X if it satisfies the following conditions, for every α∈ℝ and x,y,z∈X: (a)
∥x,y∥=0⇔x and y are linearly dependent,(b)
∥x,y∥=∥y,x∥,(c)
∥αx,y∥=|α|∥x,y∥,(d)
∥x,y+z∥≤∥x,y∥+∥x,z∥.Then the pair (X,∥·,·∥) is said to be a linear2-normed space.
A most standard example of a2-normed space is X=ℝ2 equipped with the following 2-norm (the absolute value of the determinant):
(2.1)∥x1,x2∥E=abs|x11x12x21x22|,
where xi=(xi1,xi2) for i=1,2.Definition 2.2.
LetX and Y be two 2-normed spaces, and let T:X→Y be a linear operator. For any e∈X, we say that the operator T is e-bounded if there exists Me>0 such that ∥T(x),T(e)∥≤Me∥x,e∥ for all x∈X. An e-bounded operator T, for every e, will be called bounded.
For example, the operatorT(x)=cx, where c∈ℝ defined on any 2-normed space X is a bounded operator. More examples are the followings.Example 2.3.
The operatorT:(ℝ2,∥·,·∥E)→(ℝ2,∥·,·∥E) defined by
(2.2)T(x1,x2)=(x1,-x1-x2)
is a bounded linear operator. In fact, for each e,x∈X, we have
(2.3)∥T(x),T(e)∥E=abs|x1-x1-x2e1-e1-e2|=abs|x1x2e1e2|=∥x,e∥E.Example 2.4.
Consider the real vector spaceP of all real polynomials on the interval [0,1]. Define
(2.4)∥f,g∥=sup0≤t≤1|fg′-f′g|+∫01|fg′-f′g|2dt,
for all f,g∈P, where the prime denotes differentiation with respect to t. The operator T:P→P defined by
(2.5)Tf(t)=tf(t)
is a bounded operator. Indeed,
(2.6)∥Tf,Tg∥=sup0≤t≤1|(tf)(tg)′-(tf)′(tg)|+∫01|(tf)(tg)′-(tf)′(tg)|2dt=sup0≤t≤1t2|fg′-f′g|+∫01t2|fg′-f′g|2dt≤sup0≤t≤1|fg′-f′g|+∫01|fg′-f′g|2dt=∥f,g∥,
for all f,g∈P.Example 2.5.
Let(X,∥·∥) be a normed space. Whereas any normed space may be realized as a function space on the closed unit ball of the dual space X*, one can define a 2-norm on X by
(2.7)∥x,y∥=sup{|f(x)g(y)-g(x)f(y)|:f,g∈Ball(X*)},(x,y∈X).
Now suppose that T is a bounded linear operator on (X,∥·∥)  in the usual sense. It can be easily seen that T is bounded on (X,∥·,·∥).
We are interested in calling the2-norm given in Example 2.5 the 2-norm induced by (ordinary) norm.Definition 2.6.
A sequence{xn} of X is said to be convergent if there exists an element a∈X such that limn→∞∥xn-a,x∥=0, for all x∈X.
Evidently the limit of any convergent sequence is unique.Definition 2.7.
LetX and Y be two 2-normed spaces, and let  T:X→Y be a linear operator. The operator T is said to be sequentially continuous at x∈X if for any sequence {xn} of X converging to x we have T(xn)→T(x).Definition 2.8.
Theclosure of a subset E of a 2-normed space X is denoted by E¯ and defined by the set of all x∈X such that there is a sequence {xn} of E converging to x. We say that E is closed if E=E¯.
For a2-normed space X, consider the subsets,
(2.8)Be(a,r)={x:∥x-a,e∥<r},Be[a,r]={x:∥x-a,e∥≤r},
of X. It is clear that Be[a,r] is closed.Definition 2.9.
A subsetA of a 2-normed space X is said to be locally bounded if there exist e∈X∖{0}, and r>0 such that A⊆Be(0,r).Example 2.10.
Every bounded set inℝ2 is a locally bounded set in (ℝ2,∥·,·∥E). In fact, assume that A is a bounded set in ℝ2. There exists an M>0 such that for every (x,y)∈A, ∥(x,y)∥<M. Putting e=(1,0), we obtain A⊆Be(0,M).Example 2.11.
For a normed space(X,∥·∥) consider the 2-norm induced by its norm as given in Example 2.5. Suppose that A is a bounded set in (X,∥·∥) and e∈X∖{0}. It can be easily seen that A lies in Be(0,r), for some r>0.Definition 2.12.
A subsetB of a 2-normed space X is said to be compact if every sequence {xn} of B has a convergent subsequence in B.
It is clear that every compact set of a normed spaceX is also compact in its 2-norm induced by norm.Lemma 2.13.
Every compact subsetM of a 2-normed space is closed and locally bounded.Proof.
The proof of closedness is trivial. IfM were not locally bounded, it would contain a sequence {yn} such that ∥yn,e∥>n, for any nonzero fixed element e. Now this sequence could not have a convergent subsequence because if {ynk} were a convergent subsequence to y0, then ∥ynk-y0,e∥→0. And for ϵ there would exist a positive integer N such that ∥ynk,e∥-∥y0,e∥≤∥ynk-y0,e∥<ϵ, for each k>N which is a contradiction.The following example shows that the converse of Lemma2.13 is false in general.Example 2.14.
The subsetB(1,0)[0,1] of (ℝ2,∥·,·∥E) is not a compact set. Because the sequence {(n,0)} of  B(1,0)[0,1] has no convergent subsequence. Suppose on the contrary that (nk,0)→(a,b). Hence, for e=(0,1), we have ∥(nk,0)-(a,b),(0,1)∥E→0. That is, |nk-a|→0 which is impossible.Lemma 2.15.
LetX be a 2-normed space. Then X is of finite dimension if Be[a,r] is a compact set in X, for some a,e∈X, and r>0.Proof.
Suppose thatBe[a,r] is compact. Consider the normed space X/〈e〉 equipped with the norm:
(2.9)∥x+〈e〉∥=∥x,e∥∥e,e′∥,
where {e,e′} is a linearly independent set. The subset,
(2.10)A={x+〈e〉:∥x-a+〈e〉∥≤r∥e,e′∥},
is a closed ball in the usual sense of the normed space X/〈e〉. We aim to show that A is a compact set in the normed space X/〈e〉. Choose the sequence {xn+〈e〉} of A. Since, for every n,
(2.11)∥xn+〈e〉-(a+〈e〉)∥=∥xn-a,e∥∥e,e′∥≤r∥e,e′∥,
and therefore {xn} is a sequence in Be[a,r]. Hence {xn} has a convergent subsequence {xnk} to a point x0. We have
(2.12)limn→∞∥xnk+〈e〉-(x0+〈e〉)∥=limn→∞∥xnk-x0,e∥∥e,e′∥=0.
Hence {xnk+〈e〉}k=1∞ is a convergent subsequence of {xn+〈e〉}. This implies that A is compact. Therefore X is of finite dimension.In the rest of this section, the spaceX/〈e〉 will denote the normed space given in the proof of Lemma 2.15.It is well known that if{x1,…,xn} is a linear independent set of vectors in a normed space X (of any dimension), then there is a number c>0 such that for all scalars α1,…,αn we have
(2.13)∥α1x1+⋯+αnxn∥≥c(|α1|+⋯+|αn|).
The next lemma gives a similar assertion in 2-normed spaces.Lemma 2.16.
Let{x1,…,xn,e} be a linearly independent set of vectors in a 2-normed space X (of any dimension). Then, there is a positive number c such that for any choice of scalars α1,…,αn we have
(2.14)∥α1x1+⋯+αnxn,e∥≥c(|α1|+⋯+|αn|).Proof.
Consider the normed spaceX/〈e〉 and put λ=∥e,e′∥>0. Since {x1,…,xn,e} is linearly independent in X, so is {λx1+〈e〉,…,λxn+〈e〉} in X/〈e〉. Thus, there exists c>0 such that for every choice of scalar α1,…,αn we have
(2.15)∥α1(λx1+〈e〉)+⋯+αn(λxn+〈e〉)∥≥c(|α1|+⋯+|αn|).
Therefore ∥α1x1+…+αnxn,e∥≥c(|α1|+…+|αn|). This completes the proof.Definition 2.17.
LetX and Y be two 2-normed spaces. A linear operator T:X→Y is called a compact operator if it maps every locally bounded sequence {xn} in X onto a sequence {T(xn)} in Y which has a convergent subsequence.Lemma 2.18.
LetX and Y be two 2-normed spaces, and let T:X→Y be a compact operator. Then for every e∈X, T induces the ordinary compact operator T′:X/〈e〉→Y/〈T(e)〉 defined by T′(x+〈e〉)=T(x)+〈T(e)〉, for all x∈X.Proof.
Supposee∈X, and {xn+〈e〉} is a bounded sequence in the normed space X/〈e〉. There exists M>0 such that every ∥xn+〈e〉∥<M and so ∥xn,e∥<M∥e,e′∥, for all n. Since T is compact, the sequence {T(xn)} has a convergent subsequence {T(xnk)} to a point y0. Thus, limn→∞∥T(xnk)-y0,y∥=0, for all y∈Yorlimn→∞∥T(xnk)-y0+〈T(e)〉∥=0. This showsthat  T′ is a compact operator.Lemma 2.19.
LetX and Y be two 2-normed spaces. If T:X→Y is a surjective bounded linear operator, then it is sequentially continuous.Proof.
Ifxn→a, then ∥xn-a,e∥→0, for each e∈X. Since T is bounded for every e∈X, there exists Me such that ∥T(xn)-T(a),T(e)∥≤Me∥xn-a,e∥ for all n. Thus T(xn)→T(a).Corollary 2.20.
LetX and Y be two 2-normed spaces. Then(a)
every compact operatorT:X→Y is bounded;(b)
ifdimX=∞, then the identity operator I:X→X is not a compact operator.Proof.
(a) Choosee∈X. Let T′ be the compact operator induced by T(as explained in Lemma 2.18). Since T′ is a compact operator, there exists Me>0 such that
(2.16)∥T(x)+〈T(e)〉∥≤Me∥x+〈e〉∥,
for all x∈X. That is, for all x∈X(2.17)∥T(x),T(e)∥∥T(e),y0∥≤Me∥x,e∥∥e,e′∥,
where {y0,T(e)} and {e,e′} are linearly independent sets. This implies that T is bounded.
(b) Choosee∈X. The identity operator I maps Be[0,1] to itself. Suppose on the contrary that I is a compact operator. Let {xn} be a sequence of Be[0,1]. Because {xn} is a locally bounded sequence, it has a convergent subsequence. Hence Be[0,1] is compact and therefore X is of finite dimension by Lemma 2.15, which is a contradiction.Remark 2.21.
SupposeX and Yaretwo 2-normed spaces, T1 and T2 arecompact operators from X into Y, and c∈ℝ. Then cT1+T2 is a compact operator. To see this, let {xn} be any locally bounded sequence in X. The sequence {T1(xn)} has a convergent subsequence {T1(xnk)}. The sequence {T2(xnk)} has a convergent subsequence T2(zn). Let T1(zn)→u, and let T2(zn)→v. If y∈Y, c∈ℝ, we have
(2.18)limn→∞∥(cT1+T2)(zn)-cu-v,y∥≤limn→∞|c|∥T1(zn)-u,y∥+limn→∞∥T2(zn)-v,y∥.
Thus limn→∞∥cT1+T2(zn)-cu-v,y∥=0, for all y∈Y. This implies that cT1+T2 is a compact operator.Theorem 2.22.
LetX be a 2-normed space, let T:X→X be a compact operator, and let S:X→X be a bijective bounded operator. Then ST and TS are compact operators.Proof.
Let{xn} be any locally bounded sequence in X. Then {T(xn)} has a convergent subsequence {T(xnk)}. Put limn→∞T(xnk)=y0. Since S is bijective and bounded, by Lemma 2.19, we have S(T(xnk))→S(y0). Hence S(T(xn)) has a convergent subsequence. This proves ST is compact. Now, to show that TS is compact, for any locally bounded sequence {xn}, there exist e∈X and M>0 such that xn∈Be(0,M) for all n, that is, ∥xn,e∥<M, for all n≥1. Since S is bounded, the sequence {S(xn)} is a locally bounded sequence in X. Because T is compact, {T(S(xn))} has a convergent subsequence. This completes the proof.Theorem 2.23.
LetX and Y be two 2-normed spaces. If T:X→Y is a linear operator where dimX<∞, then T is bounded.Proof.
Choosee∈X. Since  dimX<∞, so dimX/〈e〉<∞. Therefore the operator T′:X/〈e〉→Y/〈T(e)〉 definedbyT′(x+〈e〉)=T(x)+〈Te〉, for all x∈X, is a bounded operator. Thus, for every e∈X there exists Me>0 suchthat ∥T(x)+〈T(e)〉∥≤Me∥x+〈e〉∥, for all x∈X. Therefore
(2.19)∥T(x),T(e)∥∥T(e),y0∥≤Me∥x,e∥∥e,e′∥,
where {y0,T(e)} and {e,e′} are linearly independent subsets. Thus T is bounded.Theorem 2.24.
LetT:X→X be a compact operator on a 2-normed space X. Then for every λ≠0, the null space N(Tλ) of Tλ=T-λI is of finite dimension.Proof.
Consider the subsetM=Be[a,r] of N(Tλ). We show that M is compact, then apply Lemma 2.15. If {xn} is a sequence in M, then {xn} is locally bounded and {T(xn)} has a convergent subsequence {T(xnk)}. Now xn∈M⊂N(Tλ) implies Tλ(xn)=T(xn)-λxn=0, so that xn=λ-1T(xn) because λ≠0. Consequently, {λ-1T(xnk)} will be a convergent subsequence of {xn} in M. Hence M is compact, because {xn} was arbitrary in M. This shows that dimN(Tλ)<∞.Definition 2.25.
A sequence{xn} of 2-normed space X is called a Cauchy sequence if limm,n→∞∥xn-xm,x∥=0, for all x∈X.
We will say that the2-normed space X is a 2-Banach space if every Cauchy sequence in X is a convergent sequence in X.Theorem 2.26.
LetX,Y, and Z be 2-normed spaces, let T:Z⊂X→Y be a surjective bounded operator, and let Y is a 2-Banach space. Then T has an extension T¯:Z¯→Y, where T¯ is an e-bounded operator for each e∈Z.Proof.
We consider anyx∈Z¯. There is a sequence {xn} in Z such that xn→x. Since T is linear and bounded for every e∈Z, there exists Me>0 such that
(2.20)∥T(xn)-T(xm),T(e)∥≤Me∥xn-xm,e∥,
for all n,m. This shows that {T(xn)} is Cauchy in Y, because {xn} is convergent. By assumption, Y is a 2-Banach space, so that {T(xn)} converges in Y say T(xn)→y. We define T¯ by T¯(x)=y. This definition is independent of the particular choice of a sequence in Z converging to x. Because suppose that xn→x and zn→x. Then vm→x, where {vm} is the sequence {x1,z1,x2,z2,…}. Hence {T(vm)} is convergent and the two subsequences {T(xn)} and {T(zn)} of {T(vm)} must have the same limit. This proves that T¯ is uniquely defined at every x∈Z¯. Clearly T¯ is linear and T¯(x)=T(x) for every x∈Z, so that T¯ is an extension of T. On the other hand,
(2.21)∥T(x),T(e)∥≤Me∥x,e∥
for all x. Thus,
(2.22)∥T¯(x),T(e)∥≤∥T¯(x)-T(xn),T(e)∥+∥T(xn),T(e)∥.
When n→∞, ∥T¯(x),T(e)∥≤Me∥x,e∥. Therefore T¯ is an e-bounded linear operator for each e∈Z.


## 3.2-Probabilistic Normed Spaces

In this section, we aim to consider compact operators between2-probabilistic normed spaces. We need some preliminaries which are given first.Definition 3.1.
A functionf:ℝ→[0,∞) is called a distribution function if it is nondecreasing and right-continuous with inft∈ℝf(t)=0, and supt∈ℝf(t)=1.
We will denote the set of all distribution functions by𝒟.Definition 3.2.
A pair(X,N) is called a 2-probabilistic normed space (briefly, a 2PN-space) if X is a real vector space with dimX>1, N is a mapping from X×X into 𝒟 (for x∈X, the distribution function N(x,y) is denoted by Nx,y, and Nx,y(t) is the value Nx,y at t∈ℝ) satisfyingthe followingconditions:(2PN-I)
Nx,y(0)=0, for all x,y∈X,(2PN-II)
Nx,y(t)=1 for all t>0 if and only if x and y are linearly dependent,(2PN-III)
Nx,y(t)=Ny,x(t), for all x,y∈X,(2PN-IV)
Nαx,y(t)=Nx,y(t/|α|), for all α∈ℝ∖{0}, and for all x,y∈X,(2PN-V)
Nx+y,z(s+t)≥min{Nx,z(s),Ny,z(t)} for all x,y,z∈X, and  s,t∈ℝ.We call the mapping (x,y)→Nx,y a 2-probabilistic norm(2P-norm) on X.Example 3.3.
Let(X,∥·,·∥) be a 2-normed space. Every 2-norm induces a 2P-norm on X as follows:
(3.1)Nx,y(t)={tt+∥x,y∥,t>0,0,t≤0.
This 2-probabilistic norm is called the standard2P-norm.Theorem 3.4 ([8]).
Let(X,N) be a 2PN-space. Assume that the condition (2PN-VI)Nx,y(t)>0, for all t∈(0,∞) implies that {x,y} is linearly dependent. For α∈(0,1), define
(3.2)∥x,y∥α=inf{t:Nx,y(t)≥α}.
Then {∥·,·∥α:α∈(0,1)} is an ascending family of 2-norms on X. These 2-norms are called α-2-norms on X corresponding to (or induced by) the 2-probabilistic norm N on X.The following example gives us a2PN-space satisfying condition (2PN-VI).Example 3.5.
Suppose that(X,∥·,·∥) is a 2-normed space. Define
(3.3)Nx,y(t)={0,t≤∥x,y∥,1,t>∥x,y∥,
where x,y∈X, and t∈ℝ. Then the 2PN-space (X,N) satisfies (2PN-VI).Definition 3.6.
Let(X,N) be a 2PN-space, and let {xn} be a sequence of X. Then the sequence {xn} is said to be convergent to x0∈X and denoted by xn→x0 if limn→∞Nxn-x0,x(t)=1 for all x∈X and t>0.Definition 3.7.
LetT:(X,N)→(Y,N′) be a linear operator, where (X,N) and (Y,N′) are 2PN-spaces. For an element e∈X, ( 1 )
the operatorT is called e-2-probabilistic continuous (e-2-PC) at z∈X if for any ϵ>0, and α∈(0,1) there exists δ>0 such that
(3.4)Nx-z,e(δ)≥α⇒NT(x)-T(z),T(e)′(ϵ)≥α,
for all x∈X.If T is e-2-PC at each point of X, then T is said to be e-2-PC on X. If T is e-2-PC on X for each e∈X, then T is said to be 2-probabilistic continuous(2-PC) on X.( 2 )
the linear operatorT is called e-2-probabilistic bounded(e-2-PB) for e∈X on X if for every α∈(0,1) there exists me,α>0 such that
(3.5)Nx,e(tme,α)≥α⇒NT(x),T(e)′(t)≥α,
for all x∈X and t∈ℝ.If T is e-2-PB on X for each e∈X, then T is said to be 2-probabilistic bounded(2-PB) on X.Example 3.8.
Suppose thatX is a 2PN-space and that T:X→X is a linear operator defined by T(x)=cx, c∈ℝ. Then T is a 2-PC operator. Because, for any e∈X, ϵ>0, and α∈(0,1) it suffices to choose δ=ϵ/c2. Now, for x∈X if Nx-z,e(δ)≥α, then NT(x)-T(z),T(e)′(ϵ)≥α.Theorem 3.9.
Let(X,N) and (Y,N′) be two 2PN-spaces, and let T:(X,N)→(Y,N′) be a linear operator. (a)
IfT is e-2-PC for e∈X at x0∈X, then T is e-2-PC on X.(b)
T is 2-PC on X if and only if T is 2-PB on X.Proof.
(a) SinceT is e-2-PC at x0, for each ϵ>0, and α∈(0,1), there exists δ>0 such that
(3.6)Nx-x0,e(δ)≥α⇒NT(x)-T(x0),T(e)′(ϵ)≥α,
for all x∈X. Taking y∈X, and x∈X such that Nx-y,e(δ)≥α we get
(3.7)NT(x+x0-y)-T(x0),T(e)′(ϵ)≥α,
or
(3.8)NT(x)-T(y),T(e)′(ϵ)≥α.
Since y is arbitrary, it follows that T is e-2-PC on X.
(b) First we suppose thatT is 2-PB on X. Choose e∈X, α∈(0,1), and ϵ>0 arbitrarily. There exists me,α>0 such that
(3.9)Nx,e(ϵme,α)≥α⇒NT(x),T(e)′(ϵ)≥α,
for all x∈X. This shows that T is e-2-PC at zero and by part (a) it is e-2-PC on X.
Conversely, suppose thatT is 2-PC at 0. Using e-2P continuity of T at 0 and taking ϵ=1, and α∈(0,1), there exists δ>0 such that
(3.10)Nx-0,e(δ)≥α⇒NT(x)-T(0),T(e)′(1)≥α,
or
(3.11)Nx,e(δ)≥α⇒NT(x),T(e)′(1)≥α,
for all x∈X. Choose me,α=1/δ. Then
(3.12)Nx,e(tme,α)=Nx/t,e(δ)≥α⇒NT(x/t),T(e)′(1)=NT(x),T(e)′(t)≥α,
for all x≠0 and t>0. This implies that T is e-2-PB on X. Because e was arbitrary, T is 2-PB.Theorem 3.10.
Let(X,N) and (Y,N′) be two 2PN-spaces satisfying (2PN-VI). If the linear operator T:(X,∥·,·∥α)→(Y,∥·,·∥α′) is bounded with respect to α-2-norms corresponding to N and N′ for each α∈(0,1), then T:(X,N)→(Y,N′) is 2-PB.Proof.
Fixe∈X. For any α∈(0,1), there exists me,α such that for all x∈X,
(3.13)∥T(x),T(e)∥α≤me,α∥x,e∥α.
Then for x≠0, and t>0,
(3.14)∥me,αx,e∥α≤t⇒∥T(x),T(e)∥α≤t.
On the other hand,
(3.15)inf{s:Nme,αx,e(s)≥α}≤t⇒inf{s:NT(x),T(e)′(s)≥α}≤t.
It is clear that
(3.16)inf{s:Nme,αx,e(s)≥α}≤t⟺Nme,αx,e(t)≥α,inf{s:NT(x),T(e)′(s)≥α}≤t⟺NT(x),T(e)′(t)≥α.
Thus, for any α∈(0,1), there exists me,α>0 such that for all t∈ℝ, x∈X,
(3.17)Nx,e(tme,α)≥α⇒NT(x),T(e)′(t)≥α,
that is, T is 2-PB.Theorem 3.11.
LetT:(X,N)→(Y,N′) be a linear surjective operator, where (X,N) and (Y,N′) are 2PN-spaces. If T is 2-PC on X, then T is sequentially continuous, that is, for any sequence {xn} converging to x, T(xn)→T(x).Proof.
Ifxn→x, then limn→∞Nxn-x,e(t)=1, for each e∈X, and t>0. Since T is 2-PC, it is 2-PB by Theorem 3.9. Thus, for each α∈(0,1) there exists me,α such that if Nxn-x,e(t/me,α)≥α, then NT(xn)-T(x),T(e)′(t)≥α, for all n∈ℕ, and t∈ℝ. Hence, T(xn)→T(x).Definition 3.12.
A subsetB in a 2PN-space (X,N) is called compact if each sequence of B has a convergent subsequence in B.Definition 3.13.
Let(X,N) be a 2PN-space. For e,x∈X, α∈(0,1), and r>0 we define the locally ballBe,α[x,r] by {y∈X:Nx-y,e(r)≥α}.
It is clear that every locally ball is a closed set.Definition 3.14.
A subsetB of a 2PN-space (X,N) is said to be 2-probabilistic locally bounded (2-PLB), if there are t>0,  e∈X∖{0}, and 0<r<1 such that Nx,e(t)>1-r,  for all x∈B.Example 3.15.
The subsetC={(x,y):y=arcsinx} is a 2-PLB set in (ℝ2,N), where N is the standard 2-probabilistic norm. In fact, C⊆B(1,0),1/2((0,0),1). Since, if (x,y)∈C, then ∥(x,y),(1,0)∥E=|y|<1. That is, (x,y)∈B(1,0)((0,0),1).Definition 3.16.
Theclosure of a subset E of a 2PN-space (X,N) is denoted by E¯ and defined by the set of all x∈X such that there is a sequence {xn} of E converging to x. We say that E is closed if E=E¯.Definition 3.17.
Let(X,N) and (Y,N′) be 2PN-spaces. A linear operator T:(X,N)→(Y,N′) is called a compact operator if it maps every 2-PLB sequence {xn} in X onto a sequence {T(xn)} in Y which has a convergent subsequence.Example 3.18.
Let∥·,·∥1 and ∥·,·∥2 be two 2-norms, and let T:(X,∥·,·∥1)→(Y,∥·,·∥2) be a compact operator. Then T:(X,N1)→(Y,N2) is a compact operator, where N1 and N2 are 2-probabilistic norms defined by
(3.18)Nix,y(t)={tt+∥x,y∥i,t>0,0,t≤0,
for i=1,2. To see this, let {xn} be a 2-PLB-sequence in (X,N1). There exist t0>0, e∈X, and α∈(0,1) such that for all n∈ℕ,
(3.19)N1xn,e(t0)>α.
Therefore t0/(t0+∥xn,e∥1)>α, and this implies that {xn} is a locally bounded sequence in (X,∥·,·∥1). Now, the compactness of T:(X,∥·,·∥1)→(Y,∥·,·∥2) implies that the sequence {T(xn)} has a convergent subsequence {T(xnk)}; that is, there exists a b∈Y such that
(3.20)limnk→∞∥T(xnk)-b,v∥2=0,
for all v∈Y. Hence,
(3.21)limnk→∞N2T(xnk)-b,v(t)=1,
for all v∈Y and t>0. Thus T:(X,N1)→(Y,N2) is a compact operator.Lemma 3.19.
Let(X,N) be a 2PN-space satisfying (2PN-VI), and let {xn} be a sequence in X. Then xn→x0 in (X,N) if and only if xn→x0 in (X,∥·,·∥α) for each α∈(0,1).Proof.
Suppose thatxn→x0 in (X,N). Choose α∈(0,1), x∈X, and t>0. There exists k∈ℕ such that Nxn-x0,x(t)>1-α, for all n≥k. It follows that ∥xn-x0,x∥1-α≤t, for all n≥k. Thus ∥xn-x0,x∥1-α→0. Conversely, choose x∈X. Let ∥xn-x0,x∥α→0, for every α∈(0,1). Fix α∈(0,1), and t>0. There exists k∈ℕ such that
(3.22)⋀{r>0:Nxn-x0,x(r)≥1-α}<t,
for all n≥k. Hence, for every n≥k there is 0<tn<t such that
(3.23)Nxn-x0,x(tn)≥1-α.
It implies that
(3.24)Nxn-x0,x(t)≥1-α,
for all n≥k, that is, xn→x0 in (X,N).Lemma 3.20.
Let(X,N) be a 2PN-space satisfying (2PN-VI). Then X is of finite dimension if the locally ball Be,α[x,r] is a compact set in (X,N).Proof.
Let∥·,·∥α be the α-2-norm induced by N. To show that X is of finite dimension, it suffices to prove that by Lemma 2.15, the subset
(3.25)Be[x,r]={y∈X:∥x-y,e∥α≤r}
is a compact set in (X,∥·,·∥α). It is clear that Be[x,r]=Be,α[x,r]. Choose a sequence {xn} of Be[x,r]. Since Be,α[x,r] is compact, it has a convergent subsequence {xnk}. Lemma 3.19 implies that {xnk} is convergent in ∥·,·∥α. Thus Be[x,r] is compact in (X,∥·,·∥α), and consequently X is of finite dimension.Remark 3.21.
The converse of Lemma3.20 generally is not true. For example, consider B(1,0),1/2[0,1] in (ℝ2,N), where N is a standard 2PN-norm. Clearly,
(3.26)B(1,0),1/2[0,1]=B(1,0)[0,1],
where B(1,0)[0,1] is the subset of standard 2-normed space ℝ2. On the contrary, if B(1,0),1/2[0,1] were a compact set, then for each {xn}∈B(1,0),1/2[0,1]=B(1,0)[0,1] there would exist a converging subsequence {xnk}. Say xnk→a, where a∈B(1,0),1/2[0,1]. Thus
(3.27)limn→∞Nxnk-a,e(t)=limn→∞tt+∥xnk-a,e∥=1,
for all e∈X, t>0. This implies that
(3.28)limn→∞∥xnk-a,e∥=0,
for all e. Therefore B(1,0)[0,1] is a compact set which is a contradiction by Example 2.14.Lemma 3.22.
LetT:(X,N)→(Y,N′) be a compact operator, where (X,N) and (Y,N′) are 2PN-spaces satisfying (2PN-VI). If the 2-norms ∥·,·∥α, ∥·,·∥α′ are α-2-norms induced by N and N′, respectively, then T:(X,∥·,·∥α)→(Y,∥·,·∥α′) is a compact operator for all α∈(0,1).Proof.
Letα∈(0,1). We show that for each locally bounded sequence {xn} in (X,∥·,·∥α), the sequence {T(xn)} has a convergent subsequence in (Y,∥·,·∥α′). Let {xn} be a locally bounded sequence in (X,∥·,·∥α). There exist e∈X and M>0 such that
(3.29)∥xn,e∥α<M,
for all n∈ℕ. By the definition of ∥·,·∥α, for every n≥1 there exists tn>0 such that tn<M and Nxn,e(tn)≥α for all n. Because N is nondecreasing, α≤Nxn,e(tn)≤Nxn,e(M). Hence
(3.30)Nxn,e(M)≥α,
for all n. That is, {xn} is 2-PLB in (X,N). Thus {T(xn)} has a convergent subsequence {T(xnk)} in (Y,N′). By Lemma 3.19, {T(xnk)} is convergent in ∥·,·∥α′.Theorem 3.23.
Let(X,N) and (Y,N′) be two 2PN-spaces satisfying (2PN-VI). Then(a)
every compact operatorT:(X,N)→(Y,N′) is 2-PC;(b)
ifdimX=∞, then the identity operator I:(X,N)→(X,N) is not a compact operator.Proof.
(a) Chooseα∈(0,1) and e∈X. By Lemma 3.22, T:(X,∥·,·∥α)→(Y,∥·,·∥α′) is a compact operator between 2-α-normed spaces (X,∥·,·∥α) and (Y,∥·,·∥α′), where ∥·,·∥α and ∥·,·∥α′ are induced 2-norms. Therefore T is bounded by Corollary 2.20. There exists me,α>0 such that
(3.31)∥T(x),T(e)∥α′≤me,α∥x,e∥α,
for all x∈X. Hence T is 2-PB by Theorem 3.10. Now Theorem 3.9 implies that T is 2-PC. (b) Choose e∈X and α∈(0,1). The identity operator I maps the locally ball Be,α[0,1] to itself. Suppose on the contrary that I is a compact operator. Let {xn} be a sequence in Be,α[0,1]. Because I is a compact operator, the 2-PLB sequence {xn} has a convergent subsequence. Hence Be,α[0,1] is compact. Thus X is of finite dimension by Lemma 3.20, which is a contradiction.Remark 3.24.
Let(X,N) and (Y,N′) be two 2PN-spaces. If T1 and T2 are compact operators from X into Y, and α∈ℝ, then αT1+T2 is a compact operator. Because, for each {xn} that is a 2-PLB sequence in X, the sequence {T1(xn)} has a convergent subsequence {T1(xnk)}, and the sequence {T2(xnk)} has a convergent subsequence {T2(zn)}. Hence, {T1(zn)} and {T2(zn)} are convergent sequences. Let T1(zn)→u and T2(zn)→v, where u,v∈Y. We have
(3.32)limn→∞N(T1+T2)(zn)-u-v,y′(t)≥limn→∞min{NT1(zn)-u,y′(t2),NT2(zn)-v,y′(t2)},
for all y∈Y and t>0. Thus
(3.33)limn→∞NT1+T2(zn)-u-v,y′(t)=1,
for all y∈Y, and t>0. This implies that T1+T2 is a compact operator. Now for all α∈ℝ∖{0} if T(xnk)→y0, then
(3.34)limn→∞NαT1(xnk)-αy0,y′(t)=limn→∞NT1(xnk)-y0,y′(t|α|)=1,
for all y∈Y and t>0. Hence αT1 is also a compact operator.Theorem 3.25.
Let(X,N) be a 2PN-space, let T:(X,N)→(X,N) be a compact operator, let and that S:(X,N)→(X,N) be a bijective 2-PC operator. Then ST and TS are compact operators.Proof.
Let{xn} be a 2-PLB sequence in X. Then {T(xn)} has a convergent subsequence {T(xnk)}. Let limn→∞T(xnk)=y. Since S is 2-PC, by Theorem 3.11 we have S(T(xnk))→S(y). Hence ST(xn) has a convergent subsequence and this shows that ST is compact. Now, we show that TS is compact. There are t0>0,e∈X, and r0∈(0,1) such that Nxn,e(t0)>1-r0 for all n≥1 since {xn} is a 2-PLB sequence. The operator S is 2-PB by Theorem 3.9 and so there is me,1-r0>0 such that
(3.35)Nxn,e(t0me,1-r0)≥1-r0⇒NS(xn),S(e)′(t0)≥1-r0,
for all n. It follows that {S(xn)} is a 2-PLB sequence in X. Because T is a compact operator, {T(S(xn))} has a convergent subsequence. This completes the proof.Theorem 3.26.
LetT:X→X be a compact operator on a 2PN-space X. Then for every λ≠0, the null space N(Tλ) of Tλ=T-λI is of finite dimension.Proof.
We choose a locally ballM in N(Tλ) and show that it is compact, then apply Lemma 3.20. Let {xn} be a sequence in M. Then {xn} is locally bounded, and {T(xn)} has a convergent subsequence {T(xnk)}. Now xn∈M⊂N(Tλ) implies Tλ(xn)=T(xn)-λxn=0, so that xn=λ-1T(xn). Consequently, xnk=λ-1T(xnk) is convergent. Now, the closedness of M implies that the limit of {xnk} belongs to M. This proves that dimN(Tλ)<∞.Definition 3.27.
Let(X,N) be a 2PN-space. A sequence {xn} in X is called a Cauchy sequence if limn→∞Nxn+p-xn,x(t)=1 for all x∈X, t>0, and p∈ℕ.
We say that a2PN-space (X,N) is complete if every Cauchy sequence in X is convergent to a point of X.Theorem 3.28.
LetX,Y, and Z be two 2PN-spaces, let T:Z⊂X→Y be a surjective 2-PB operator, and let Y be a complete space. Then T has an extension
(3.36)T¯:Z¯→Y,
where T¯ is an e-2-PB operator for each e∈Z.Proof.
We consider anyx∈Z¯. There is a sequence {xn} in Z such that xn→x. Since T is linear and 2-PB, for every α∈(0,1) and e∈Z there exists me,α>0 such that
(3.37)Nxm-xn,e(tme,α)≥α⇒NT(xm)-T(xn),T(e)′(t)≥α,
for all t>0 and m,n∈ℕ. But the sequence {xn} is Cauchy, thus for all t>0, there exists k∈ℕ such that for all m,n>k we have
(3.38)Nxm-xn,e(tme,α)≥α.
Therefore, for m,n>k,
(3.39)NT(xm)-T(xn),T(e)′(t)≥α.
This shows that {T(xn)} is Cauchy in Y. Thus {T(xn)} is convergent to an element y∈Y. Now, we define T¯ by T¯(x)=y. In exactly the same way as presented in the proof of Theorem 2.26 we see that this definition is independent of the particular choice of a sequence in Z converging to x. Clearly T¯ is linear and T¯(x)=T(x) for every x∈Z, so that T¯ is an extension of T. We now use the 2-probabilistic boundedness of T on Z. Let α∈(0,1) and e∈Z. There exists me,α>0 such that
(*)Nx,e(tme,α)≥α⇒NT(x),T(e)′(t)≥α,
for all t>0 and x∈Z. Choose t>0 and x∈Z¯ such that
(3.40)Nx,e(t4me,α)≥α.
Now, we show that NT¯(x),T(e)′(t)≥α. Because x∈Z¯, there exists {xn}⊆Z such that xn→x. Therefore, for n∈ℕ sufficiently large, we have
(3.41)Nxn,e(t2me,α)≥min{Nxn-x,e(t4me,α),Nx,e(t4me,α)}≥α.
By (*), NT(xn),T(e)′(t/2)≥α, and since limn→∞T(xn)=T¯(x), we obtain
(3.42)NT(xn)-T¯(x),T(e)′(t2)≥α.
Hence
(3.43)NT¯(x),T(e)′(t)≥min{NT¯(x)-T(xn),T(e)′(t2),NT(xn),T(e)′(t2)}≥α.
Therefore T¯ is a 2-PB linear operator on Z.



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*Source: 950234-2009-07-08.xml*