Compact Operators Defined on 2-Normed and 2-Probabilistic Normed Spaces

Abstract

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

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 between 2-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 between 2-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]). Let 𝑋 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 linear 2-normed space.
A most standard example of a 2-normed space is 𝑋=ℝ² equipped with the following 2-norm (the absolute value of the determinant): ‖𝑥₁,𝑥₂‖_𝐸=abs𝑥₁₁𝑥₁₂𝑥₂₁𝑥₂₂, (2.1) where 𝑥_𝑖=(𝑥_𝑖1,𝑥_𝑖2) for 𝑖=1,2.

Definition 2.2. Let 𝑋 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 𝑀_𝑒>0 such that ‖𝑇(𝑥),𝑇(𝑒)‖≤𝑀_𝑒‖𝑥,𝑒‖ for all 𝑥∈𝑋. An 𝑒-bounded operator 𝑇, for every 𝑒, will be called bounded.
For example, the operator 𝑇(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 operator 𝑇∶(ℝ²,‖⋅,⋅‖_𝐸)→(ℝ²,‖⋅,⋅‖_𝐸) defined by 𝑇(𝑥₁,𝑥₂)=(𝑥₁,−𝑥₁−𝑥₂) (2.2) is a bounded linear operator. In fact, for each 𝑒,𝑥∈𝑋, we have ‖𝑇(𝑥),𝑇(𝑒)‖_𝐸=abs𝑥₁−𝑥₁−𝑥₂𝑒₁−𝑒₁−𝑒₂=abs𝑥₁𝑥₂𝑒₁𝑒₂=‖𝑥,𝑒‖_E. (2.3)

Example 2.4. Consider the real vector space 𝑃 of all real polynomials on the interval [0,1]. Define ‖𝑓,𝑔‖=sup_0≤𝑡≤1|𝑓𝑔^′−𝑓^′𝑔|+¹₀|𝑓𝑔^′−𝑓^′𝑔|²𝑑𝑡, (2.4) for all 𝑓,𝑔∈𝑃, where the prime denotes differentiation with respect to 𝑡. The operator 𝑇∶𝑃→𝑃 defined by 𝑇𝑓(𝑡)=𝑡𝑓(𝑡) (2.5) is a bounded operator. Indeed, ‖𝑇𝑓,𝑇𝑔‖=sup_0≤𝑡≤1|(𝑡𝑓)(𝑡𝑔)^′−(𝑡𝑓)^′(𝑡𝑔)|+¹₀|(𝑡𝑓)(𝑡𝑔)^′−(𝑡𝑓)^′(𝑡𝑔)|²𝑑𝑡=sup_0≤𝑡≤1𝑡²|𝑓𝑔^′−𝑓^′𝑔|+¹₀𝑡²|𝑓𝑔^′−𝑓^′𝑔|²𝑑𝑡≤sup_0≤𝑡≤1|𝑓𝑔^′−𝑓^′𝑔|+¹₀|𝑓𝑔^′−𝑓^′𝑔|²𝑑𝑡=‖𝑓,𝑔‖, (2.6) 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 ‖𝑥,𝑦‖=sup{|𝑓(𝑥)𝑔(𝑦)−𝑔(𝑥)𝑓(𝑦)|∶𝑓,𝑔∈Ball(𝑋^∗)},(𝑥,𝑦∈𝑋). (2.7) 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 the 2-norm given in Example 2.5 the 2-norm induced by (ordinary) norm.

Definition 2.6. A sequence {𝑥_𝑛} of 𝑋 is said to be convergent if there exists an element 𝑎∈𝑋 such that lim_𝑛→∞‖𝑥_𝑛−𝑎,𝑥‖=0, for all 𝑥∈𝑋.
Evidently the limit of any convergent sequence is unique.

Definition 2.7. Let 𝑋 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 {𝑥_𝑛} of 𝑋 converging to 𝑥 we have 𝑇(𝑥_𝑛)→𝑇(𝑥).

Definition 2.8. The closure 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 {𝑥_𝑛} of 𝐸 converging to 𝑥. We say that 𝐸 is closed if 𝐸=𝐸.
For a 2-normed space 𝑋, consider the subsets, 𝐵_𝑒𝐵(𝑎,𝑟)={𝑥∶‖𝑥−𝑎,𝑒‖<𝑟},_𝑒[𝑎,𝑟]={𝑥∶‖𝑥−𝑎,𝑒‖≤𝑟}, (2.8) of 𝑋. It is clear that 𝐵_𝑒[𝑎,𝑟] is closed.

Definition 2.9. A subset 𝐴 of a 2-normed space 𝑋 is said to be locally bounded if there exist 𝑒∈𝑋⧵{0}, and 𝑟>0 such that 𝐴⊆𝐵_𝑒(0,𝑟).

Example 2.10. Every bounded set in ℝ² 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⊆𝐵_𝑒(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 𝐵_𝑒(0,𝑟), for some 𝑟>0.

Definition 2.12. A subset 𝐵 of a 2-normed space 𝑋 is said to be compact if every sequence {𝑥_𝑛} of 𝐵 has a convergent subsequence in 𝐵.
It is clear that every compact set of a normed space 𝑋 is also compact in its 2-norm induced by norm.

Lemma 2.13. Every compact subset 𝑀 of a 2-normed space is closed and locally bounded.

Proof. The proof of closedness is trivial. If 𝑀 were not locally bounded, it would contain a sequence {𝑦_𝑛} such that ‖𝑦_𝑛,𝑒‖>𝑛, for any nonzero fixed element 𝑒. Now this sequence could not have a convergent subsequence because if {𝑦_𝑛𝑘} were a convergent subsequence to 𝑦₀, then ‖𝑦_𝑛𝑘−𝑦₀,𝑒‖→0. And for 𝜖 there would exist a positive integer 𝑁 such that ‖𝑦_𝑛𝑘,𝑒‖−‖𝑦₀,𝑒‖≤‖𝑦_𝑛𝑘−𝑦₀,𝑒‖<𝜖, for each 𝑘>𝑁 which is a contradiction.

The following example shows that the converse of Lemma 2.13 is false in general.

Example 2.14. The subset 𝐵_(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 (𝑛_𝑘,0)→(𝑎,𝑏). Hence, for 𝑒=(0,1), we have ‖(𝑛_𝑘,0)−(𝑎,𝑏),(0,1)‖_𝐸→0. That is, |𝑛_𝑘−𝑎|→0 which is impossible.

Lemma 2.15. Let 𝑋 be a 2-normed space. Then 𝑋 is of finite dimension if 𝐵_𝑒[𝑎,𝑟] is a compact set in 𝑋, for some 𝑎,𝑒∈𝑋, and 𝑟>0.

Proof. Suppose that 𝐵_𝑒[𝑎,𝑟] is compact. Consider the normed space 𝑋/⟨𝑒⟩ equipped with the norm: ‖𝑥+⟨𝑒⟩‖=‖𝑥,𝑒‖‖𝑒,𝑒^′‖, (2.9) where {𝑒,𝑒^′} is a linearly independent set. The subset, 𝑟𝐴={𝑥+⟨𝑒⟩∶‖𝑥−𝑎+⟨𝑒⟩‖≤‖𝑒,𝑒^′‖}, (2.10) 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 {𝑥_𝑛+⟨𝑒⟩} of 𝐴. Since, for every 𝑛, ‖𝑥_𝑛+⟨𝑒⟩−(𝑎+⟨𝑒⟩)‖=‖𝑥_𝑛−𝑎,𝑒‖‖𝑒,𝑒^′‖≤𝑟‖𝑒,𝑒^′‖, (2.11) and therefore {𝑥_𝑛} is a sequence in 𝐵_𝑒[𝑎,𝑟]. Hence {𝑥_𝑛} has a convergent subsequence {𝑥_𝑛𝑘} to a point 𝑥₀. We have lim_𝑛→∞‖𝑥_𝑛𝑘+⟨𝑒⟩−(𝑥₀+⟨𝑒⟩)‖=lim_𝑛→∞‖𝑥_𝑛𝑘−𝑥₀,𝑒‖‖𝑒,𝑒^′‖=0. (2.12) Hence {𝑥_𝑛𝑘+⟨𝑒⟩}^∞_𝑘=1 is a convergent subsequence of {𝑥_𝑛+⟨𝑒⟩}. This implies that 𝐴 is compact. Therefore 𝑋 is of finite dimension.

In the rest of this section, the space 𝑋/⟨𝑒⟩ will denote the normed space given in the proof of Lemma 2.15.

It is well known that if {𝑥₁,…,𝑥_𝑛} 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,…,𝛼_𝑛 we have ‖𝛼₁𝑥₁+⋯+𝛼_𝑛𝑥_𝑛‖≥𝑐(|𝛼₁|+⋯+|𝛼_𝑛|). (2.13) The next lemma gives a similar assertion in 2-normed spaces.

Lemma 2.16. Let {𝑥₁,…,𝑥_𝑛,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,…,𝛼_𝑛 we have ‖𝛼₁𝑥₁+⋯+𝛼_𝑛𝑥_𝑛,𝑒‖≥𝑐(|𝛼₁|+⋯+|𝛼_𝑛|). (2.14)

Proof. Consider the normed space 𝑋/⟨𝑒⟩ and put 𝜆=‖𝑒,𝑒^′‖>0. Since {𝑥₁,…,𝑥_𝑛,𝑒} is linearly independent in 𝑋, so is {𝜆𝑥₁+⟨𝑒⟩,…,𝜆𝑥_𝑛+⟨𝑒⟩} in 𝑋/⟨𝑒⟩. Thus, there exists 𝑐>0 such that for every choice of scalar 𝛼₁,…,𝛼_𝑛 we have ‖𝛼₁(𝜆𝑥₁+⟨𝑒⟩)+⋯+𝛼_𝑛(𝜆𝑥_𝑛+⟨𝑒⟩)‖≥𝑐(|𝛼₁|+⋯+|𝛼_𝑛|). (2.15) Therefore ‖𝛼₁𝑥₁+…+𝛼_𝑛𝑥_𝑛,𝑒‖≥𝑐(|𝛼₁|+…+|𝛼_𝑛|). This completes the proof.

Definition 2.17. Let 𝑋 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 {𝑥_𝑛} in 𝑋 onto a sequence {𝑇(𝑥_𝑛)} in 𝑌 which has a convergent subsequence.

Lemma 2.18. Let 𝑋 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. Suppose 𝑒∈𝑋, and {𝑥_𝑛+⟨𝑒⟩} is a bounded sequence in the normed space 𝑋/⟨𝑒⟩. There exists 𝑀>0 such that every ‖𝑥_𝑛+⟨𝑒⟩‖<𝑀 and so ‖𝑥_𝑛,𝑒‖<𝑀‖𝑒,𝑒^′‖, for all 𝑛. Since 𝑇 is compact, the sequence {𝑇(𝑥_𝑛)} has a convergent subsequence {𝑇(𝑥_𝑛𝑘)} to a point 𝑦₀. Thus, lim_𝑛→∞‖𝑇(𝑥_𝑛𝑘)−𝑦₀,𝑦‖=0, for all 𝑦∈𝑌orlim_𝑛→∞‖𝑇(𝑥_𝑛𝑘)−𝑦₀+⟨𝑇(𝑒)⟩‖=0. This showsthat 𝑇^′ is a compact operator.

Lemma 2.19. Let 𝑋 and 𝑌 be two 2-normed spaces. If 𝑇∶𝑋→𝑌 is a surjective bounded linear operator, then it is sequentially continuous.

Proof. If 𝑥_𝑛→𝑎, then ‖𝑥_𝑛−𝑎,𝑒‖→0, for each 𝑒∈𝑋. Since 𝑇 is bounded for every 𝑒∈𝑋, there exists 𝑀_𝑒 such that ‖𝑇(𝑥_𝑛)−𝑇(𝑎),𝑇(𝑒)‖≤𝑀_𝑒‖𝑥_𝑛−𝑎,𝑒‖ for all 𝑛. Thus 𝑇(𝑥_𝑛)→𝑇(𝑎).

Corollary 2.20. Let 𝑋 and 𝑌 be two 2-normed spaces. Then(a)every compact operator 𝑇∶𝑋→𝑌 is bounded; (b)if dim𝑋=∞, then the identity operator 𝐼∶𝑋→𝑋 is not a compact operator.

Proof. (a) Choose 𝑒∈𝑋. Let 𝑇^′ be the compact operator induced by 𝑇(as explained in Lemma 2.18). Since 𝑇^′ is a compact operator, there exists 𝑀_𝑒>0 such that ‖𝑇(𝑥)+⟨𝑇(𝑒)⟩‖≤𝑀_𝑒‖𝑥+⟨𝑒⟩‖, (2.16) for all 𝑥∈𝑋. That is, for all 𝑥∈𝑋‖𝑇(𝑥),𝑇(𝑒)‖‖𝑇(𝑒),𝑦₀‖≤𝑀_𝑒‖𝑥,𝑒‖‖𝑒,𝑒^′‖, (2.17) where {𝑦₀,𝑇(𝑒)} and {𝑒,𝑒^′} are linearly independent sets. This implies that 𝑇 is bounded.
(b) Choose 𝑒∈𝑋. The identity operator 𝐼 maps 𝐵_𝑒[0,1] to itself. Suppose on the contrary that 𝐼 is a compact operator. Let {𝑥_𝑛} be a sequence of 𝐵_𝑒[0,1]. Because {𝑥_𝑛} is a locally bounded sequence, it has a convergent subsequence. Hence 𝐵_𝑒[0,1] is compact and therefore 𝑋 is of finite dimension by Lemma 2.15, which is a contradiction.

Remark 2.21. Suppose 𝑋 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 {𝑥_𝑛} be any locally bounded sequence in 𝑋. The sequence {𝑇₁(𝑥_𝑛)} has a convergent subsequence {𝑇₁(𝑥_𝑛𝑘)}. The sequence {𝑇₂(𝑥_𝑛𝑘)} has a convergent subsequence 𝑇₂(𝑧_𝑛). Let 𝑇₁(𝑧_𝑛)→𝑢, and let 𝑇₂(𝑧_𝑛)→𝑣. If 𝑦∈𝑌, 𝑐∈ℝ, we have lim_𝑛→∞‖(𝑐𝑇₁+𝑇₂)(𝑧_𝑛)−𝑐𝑢−𝑣,𝑦‖≤lim_𝑛→∞|𝑐|‖𝑇₁(𝑧_𝑛)−𝑢,𝑦‖+lim_𝑛→∞‖𝑇₂(𝑧_𝑛)−𝑣,𝑦‖. (2.18) Thus lim_𝑛→∞‖𝑐𝑇₁+𝑇₂(𝑧_𝑛)−𝑐𝑢−𝑣,𝑦‖=0, for all 𝑦∈𝑌. This implies that 𝑐𝑇₁+𝑇₂ is a compact operator.

Theorem 2.22. Let 𝑋 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 {𝑥_𝑛} be any locally bounded sequence in 𝑋. Then {𝑇(𝑥_𝑛)} has a convergent subsequence {𝑇(𝑥_𝑛𝑘)}. Put lim_𝑛→∞𝑇(𝑥_𝑛𝑘)=𝑦₀. Since 𝑆 is bijective and bounded, by Lemma 2.19, we have 𝑆(𝑇(𝑥_𝑛𝑘))→𝑆(𝑦₀). Hence 𝑆(𝑇(𝑥_𝑛)) has a convergent subsequence. This proves 𝑆𝑇 is compact. Now, to show that 𝑇𝑆 is compact, for any locally bounded sequence {𝑥_𝑛}, there exist 𝑒∈𝑋 and 𝑀>0 such that 𝑥_𝑛∈𝐵_𝑒(0,𝑀) for all 𝑛, that is, ‖𝑥_𝑛,𝑒‖<𝑀, for all 𝑛≥1. Since 𝑆 is bounded, the sequence {𝑆(𝑥_𝑛)} is a locally bounded sequence in 𝑋. Because 𝑇 is compact, {𝑇(𝑆(𝑥_𝑛))} has a convergent subsequence. This completes the proof.

Theorem 2.23. Let 𝑋 and 𝑌 be two 2-normed spaces. If 𝑇∶𝑋→𝑌 is a linear operator where dim𝑋<∞, then 𝑇 is bounded.

Proof. Choose 𝑒∈𝑋. Since dim𝑋<∞, so dim𝑋/⟨𝑒⟩<∞. Therefore the operator 𝑇^′∶𝑋/⟨𝑒⟩→𝑌/⟨𝑇(𝑒)⟩ definedby𝑇^′(𝑥+⟨𝑒⟩)=𝑇(𝑥)+⟨𝑇_𝑒⟩, for all 𝑥∈𝑋, is a bounded operator. Thus, for every 𝑒∈𝑋 there exists 𝑀_𝑒>0 suchthat ‖𝑇(𝑥)+⟨𝑇(𝑒)⟩‖≤𝑀_𝑒‖𝑥+⟨𝑒⟩‖, for all 𝑥∈𝑋. Therefore ‖𝑇(𝑥),𝑇(𝑒)‖‖𝑇(𝑒),𝑦₀‖≤𝑀_𝑒‖𝑥,𝑒‖‖𝑒,𝑒^′‖, (2.19) where {𝑦₀,𝑇(𝑒)} and {𝑒,𝑒^′} are linearly independent subsets. Thus 𝑇 is bounded.

Theorem 2.24. Let 𝑇∶𝑋→𝑋 be a compact operator on a 2-normed space 𝑋. Then for every 𝜆≠0, the null space 𝑁(𝑇_𝜆) of 𝑇_𝜆=𝑇−𝜆𝐼 is of finite dimension.

Proof. Consider the subset 𝑀=𝐵_𝑒[𝑎,𝑟] of 𝑁(𝑇_𝜆). We show that 𝑀 is compact, then apply Lemma 2.15. If {𝑥_𝑛} is a sequence in 𝑀, then {𝑥_𝑛} is locally bounded and {𝑇(𝑥_𝑛)} has a convergent subsequence {𝑇(𝑥_𝑛𝑘)}. Now 𝑥_𝑛∈𝑀⊂𝑁(𝑇_𝜆) implies 𝑇_𝜆(𝑥_𝑛)=𝑇(𝑥_𝑛)−𝜆𝑥_𝑛=0, so that 𝑥_𝑛=𝜆⁻¹𝑇(𝑥_𝑛) because 𝜆≠0. Consequently, {𝜆⁻¹𝑇(𝑥_𝑛𝑘)} will be a convergent subsequence of {𝑥_𝑛} in 𝑀. Hence 𝑀 is compact, because {𝑥_𝑛} was arbitrary in 𝑀. This shows that dim𝑁(𝑇_𝜆)<∞.

Definition 2.25. A sequence {𝑥_𝑛} of 2-normed space 𝑋 is called a Cauchy sequence if lim_{𝑚,𝑛→∞}‖𝑥_𝑛−𝑥_𝑚,𝑥‖=0, for all 𝑥∈𝑋.
We will say that the 2-normed space 𝑋 is a 2-Banach space if every Cauchy sequence in 𝑋 is a convergent sequence in 𝑋.

Theorem 2.26. Let 𝑋,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 any 𝑥∈𝑍. There is a sequence {𝑥_𝑛} in 𝑍 such that 𝑥_𝑛→𝑥. Since 𝑇 is linear and bounded for every 𝑒∈𝑍, there exists 𝑀_𝑒>0 such that ‖𝑇(𝑥_𝑛)−𝑇(𝑥_𝑚),𝑇(𝑒)‖≤𝑀_𝑒‖𝑥_𝑛−𝑥_𝑚,𝑒‖, (2.20) for all 𝑛,𝑚. This shows that {𝑇(𝑥_𝑛)} is Cauchy in 𝑌, because {𝑥_𝑛} is convergent. By assumption, 𝑌 is a 2-Banach space, so that {𝑇(𝑥_𝑛)} converges in 𝑌 say 𝑇(𝑥_𝑛)→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 𝑥_𝑛→𝑥 and 𝑧_𝑛→𝑥. Then 𝑣_𝑚→𝑥, where {𝑣_𝑚} is the sequence {𝑥₁,𝑧₁,𝑥₂,𝑧₂,…}. Hence {𝑇(𝑣_𝑚)} is convergent and the two subsequences {𝑇(𝑥_𝑛)} and {𝑇(𝑧_𝑛)} of {𝑇(𝑣_𝑚)} 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, ‖𝑇(𝑥),𝑇(𝑒)‖≤𝑀_𝑒‖𝑥,𝑒‖ (2.21) for all 𝑥. Thus, ‖𝑇(𝑥),𝑇(𝑒)‖≤‖𝑇(𝑥)−𝑇(𝑥_𝑛),𝑇(𝑒)‖+‖𝑇(𝑥_𝑛),𝑇(𝑒)‖. (2.22) When 𝑛→∞, ‖𝑇(𝑥),𝑇(𝑒)‖≤𝑀_𝑒‖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 between 2-probabilistic normed spaces. We need some preliminaries which are given first.

Definition 3.1. A function 𝑓∶ℝ→[0,∞) is called a distribution function if it is nondecreasing and right-continuous with inf_𝑡∈ℝ𝑓(𝑡)=0, and sup_𝑡∈ℝ𝑓(𝑡)=1.
We will denote the set of all distribution functions by 𝒟.

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 𝑁_𝑥,𝑦, and 𝑁_𝑥,𝑦(𝑡) is the value 𝑁_𝑥,𝑦 at 𝑡∈ℝ) satisfyingthe followingconditions:(2PN-I)𝑁_𝑥,𝑦(0)=0, for all 𝑥,𝑦∈𝑋,(2PN-II)𝑁_𝑥,𝑦(𝑡)=1 for all 𝑡>0 if and only if 𝑥 and 𝑦 are linearly dependent, (2PN-III)𝑁_𝑥,𝑦(𝑡)=𝑁_𝑦,𝑥(𝑡), for all 𝑥,𝑦∈𝑋, (2PN-IV)𝑁_{𝛼𝑥,𝑦}(𝑡)=𝑁_𝑥,𝑦(𝑡/|𝛼|), for all 𝛼∈ℝ⧵{0}, and for all 𝑥,𝑦∈𝑋, (2PN-V)𝑁_{𝑥+𝑦,𝑧}(𝑠+𝑡)≥min{𝑁_𝑥,𝑧(𝑠),𝑁_𝑦,𝑧(𝑡)} for all 𝑥,𝑦,𝑧∈𝑋, and 𝑠,𝑡∈ℝ. We call the mapping (𝑥,𝑦)→𝑁_𝑥,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: 𝑁_𝑥,𝑦𝑡(𝑡)=𝑡+‖𝑥,𝑦‖,𝑡>0,0,𝑡≤0. (3.1) This 2-probabilistic norm is called the standard  2𝑃-norm.

Theorem 3.4 ([8]). Let (𝑋,𝑁) be a 2𝑃𝑁-space. Assume that the condition (2𝑃𝑁-𝑉𝐼)𝑁_𝑥,𝑦(𝑡)>0, for all 𝑡∈(0,∞) implies that {𝑥,𝑦} is linearly dependent. For 𝛼∈(0,1), define ‖𝑥,𝑦‖_𝛼=inf{𝑡∶𝑁_𝑥,𝑦(𝑡)≥𝛼}. (3.2) 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 a 2𝑃𝑁-space satisfying condition (2𝑃𝑁-𝑉𝐼).

Example 3.5. Suppose that (𝑋,‖⋅,⋅‖) is a 2-normed space. Define 𝑁_𝑥,𝑦(𝑡)=0,𝑡≤‖𝑥,𝑦‖,1,𝑡>‖𝑥,𝑦‖, (3.3) 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 {𝑥_𝑛} be a sequence of 𝑋. Then the sequence {𝑥_𝑛} is said to be convergent to 𝑥₀∈𝑋 and denoted by 𝑥_𝑛→𝑥₀ if lim_𝑛→∞𝑁_{𝑥𝑛−𝑥0,𝑥}(𝑡)=1 for all 𝑥∈𝑋 and 𝑡>0.

Definition 3.7. Let 𝑇∶(𝑋,𝑁)→(𝑌,𝑁^′) be a linear operator, where (𝑋,𝑁) and (𝑌,𝑁^′) are 2𝑃𝑁-spaces. For an element 𝑒∈𝑋, ( 1 )the operator 𝑇 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 𝑁_{𝑥−𝑧,𝑒}(𝛿)≥𝛼⇒𝑁^′_{𝑇(𝑥)−𝑇(𝑧),𝑇(𝑒)}(𝜖)≥𝛼, (3.4) 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 operator 𝑇 is called 𝑒- 2-probabilistic bounded  (𝑒-2-𝑃𝐵) for 𝑒∈𝑋 on 𝑋 if for every 𝛼∈(0,1) there exists 𝑚_𝑒,𝛼>0 such that 𝑁_𝑥,𝑒(𝑡𝑚_𝑒,𝛼)≥𝛼⇒𝑁^′_{𝑇(𝑥),𝑇(𝑒)}(𝑡)≥𝛼, (3.5) 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 that 𝑋 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 𝑁_{𝑥−𝑧,𝑒}(𝛿)≥𝛼, then 𝑁^′_{𝑇(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)If 𝑇 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) Since 𝑇 is 𝑒-2-𝑃𝐶 at 𝑥₀, for each 𝜖>0, and 𝛼∈(0,1), there exists 𝛿>0 such that 𝑁_{𝑥−𝑥0,𝑒}(𝛿)≥𝛼⇒𝑁^′_{𝑇(𝑥)−𝑇(𝑥0),𝑇(𝑒)}(𝜖)≥𝛼, (3.6) for all 𝑥∈𝑋. Taking 𝑦∈𝑋, and 𝑥∈𝑋 such that 𝑁_{𝑥−𝑦,𝑒}(𝛿)≥𝛼 we get 𝑁^′_{𝑇(𝑥+𝑥0−𝑦)−𝑇(𝑥0),𝑇(𝑒)}(𝜖)≥𝛼, (3.7) or 𝑁^′_{𝑇(𝑥)−𝑇(𝑦),𝑇(𝑒)}(𝜖)≥𝛼. (3.8) Since 𝑦 is arbitrary, it follows that 𝑇 is 𝑒-2-𝑃𝐶 on 𝑋.
(b) First we suppose that 𝑇 is 2-𝑃𝐵 on 𝑋. Choose 𝑒∈𝑋, 𝛼∈(0,1), and 𝜖>0 arbitrarily. There exists 𝑚_𝑒,𝛼>0 such that 𝑁_𝑥,𝑒(𝜖𝑚_𝑒,𝛼)≥𝛼⇒𝑁^′_{𝑇(𝑥),𝑇(𝑒)}(𝜖)≥𝛼, (3.9) 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 that 𝑇 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 𝑁_{𝑥−0,𝑒}(𝛿)≥𝛼⇒𝑁^′_{𝑇(𝑥)−𝑇(0),𝑇(𝑒)}(1)≥𝛼, (3.10) or 𝑁_𝑥,𝑒(𝛿)≥𝛼⇒𝑁^′_{𝑇(𝑥),𝑇(𝑒)}(1)≥𝛼, (3.11) for all 𝑥∈𝑋. Choose 𝑚_𝑒,𝛼=1/𝛿. Then 𝑁_𝑥,𝑒(𝑡𝑚_𝑒,𝛼)=𝑁_{𝑥/𝑡,𝑒}(𝛿)≥𝛼⇒𝑁^′_{𝑇(𝑥/𝑡),𝑇(𝑒)}(1)=𝑁^′_{𝑇(𝑥),𝑇(𝑒)}(𝑡)≥𝛼, (3.12) 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. Fix 𝑒∈𝑋. For any 𝛼∈(0,1), there exists 𝑚_𝑒,𝛼 such that for all 𝑥∈𝑋, ‖𝑇(𝑥),𝑇(𝑒)‖_𝛼≤𝑚_𝑒,𝛼‖𝑥,𝑒‖_𝛼. (3.13) Then for 𝑥≠0, and 𝑡>0, ‖𝑚_𝑒,𝛼𝑥,𝑒‖_𝛼≤𝑡⇒‖𝑇(𝑥),𝑇(𝑒)‖_𝛼≤𝑡. (3.14) On the other hand, inf{𝑠∶𝑁_{𝑚𝑒,𝛼𝑥,𝑒}(𝑠)≥𝛼}≤𝑡⇒inf{𝑠∶𝑁^′_{𝑇(𝑥),𝑇(𝑒)}(𝑠)≥𝛼}≤𝑡. (3.15) It is clear that inf{𝑠∶𝑁_{𝑚𝑒,𝛼𝑥,𝑒}(𝑠)≥𝛼}≤𝑡⟺𝑁_{𝑚𝑒,𝛼𝑥,𝑒}(𝑡)≥𝛼,inf{𝑠∶𝑁^′_{𝑇(𝑥),𝑇(𝑒)}(𝑠)≥𝛼}≤𝑡⟺𝑁^′_{𝑇(𝑥),𝑇(𝑒)}(𝑡)≥𝛼. (3.16) Thus, for any 𝛼∈(0,1), there exists 𝑚_𝑒,𝛼>0 such that for all 𝑡∈ℝ, 𝑥∈𝑋, 𝑁_𝑥,𝑒(𝑡𝑚_𝑒,𝛼)≥𝛼⇒𝑁^′_{𝑇(𝑥),𝑇(𝑒)}(𝑡)≥𝛼, (3.17) that is, 𝑇 is 2-𝑃𝐵.

Theorem 3.11. Let 𝑇∶(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 {𝑥_𝑛} converging to 𝑥, 𝑇(𝑥_𝑛)→𝑇(𝑥).

Proof. If 𝑥_𝑛→𝑥, then lim_𝑛→∞𝑁_{𝑥𝑛−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 𝑚_𝑒,𝛼 such that if 𝑁_{𝑥𝑛−𝑥,𝑒}(𝑡/𝑚_𝑒,𝛼)≥𝛼, then 𝑁^′_{𝑇(𝑥𝑛)−𝑇(𝑥),𝑇(𝑒)}(𝑡)≥𝛼, for all 𝑛∈ℕ, and 𝑡∈ℝ. Hence, 𝑇(𝑥_𝑛)→𝑇(𝑥).

Definition 3.12. A subset 𝐵 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 ball 𝐵_𝑒,𝛼[𝑥,𝑟] by {𝑦∈𝑋∶𝑁_{𝑥−𝑦,𝑒}(𝑟)≥𝛼}.
It is clear that every locally ball is a closed set.

Definition 3.14. A subset 𝐵 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 𝑁_𝑥,𝑒(𝑡)>1−𝑟, for all 𝑥∈𝐵.

Example 3.15. The subset 𝐶={(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. The closure 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 {𝑥_𝑛} 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 {𝑥_𝑛} in 𝑋 onto a sequence {𝑇(𝑥_𝑛)} 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 𝑁_{𝑖𝑥,𝑦}𝑡(𝑡)=𝑡+‖𝑥,𝑦‖_𝑖,𝑡>0,0,𝑡≤0, (3.18) for 𝑖=1,2. To see this, let {𝑥_𝑛} be a 2-𝑃𝐿𝐵-sequence in (𝑋,𝑁₁). There exist 𝑡₀>0, 𝑒∈𝑋, and 𝛼∈(0,1) such that for all 𝑛∈ℕ, 𝑁_{1𝑥𝑛,𝑒}(𝑡₀)>𝛼. (3.19) Therefore 𝑡₀/(𝑡₀+‖𝑥_𝑛,𝑒‖₁)>𝛼, and this implies that {𝑥_𝑛} is a locally bounded sequence in (𝑋,‖⋅,⋅‖₁). Now, the compactness of 𝑇∶(𝑋,‖⋅,⋅‖₁)→(𝑌,‖⋅,⋅‖₂) implies that the sequence {𝑇(𝑥_𝑛)} has a convergent subsequence {𝑇(𝑥_𝑛𝑘)}; that is, there exists a 𝑏∈𝑌 such that lim_{𝑛𝑘→∞}‖𝑇(𝑥_𝑛𝑘)−𝑏,𝑣‖₂=0, (3.20) for all 𝑣∈𝑌. Hence, lim_{𝑛𝑘→∞}𝑁_{2𝑛𝑘𝑇(𝑥)−𝑏,𝑣}(𝑡)=1, (3.21) 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 {𝑥_𝑛} be a sequence in 𝑋. Then 𝑥_𝑛→𝑥₀ in (𝑋,𝑁) if and only if 𝑥_𝑛→𝑥₀ in (𝑋,‖⋅,⋅‖_𝛼) for each 𝛼∈(0,1).

Proof. Suppose that 𝑥_𝑛→𝑥₀ in (𝑋,𝑁). Choose 𝛼∈(0,1), 𝑥∈𝑋, and 𝑡>0. There exists 𝑘∈ℕ such that 𝑁_{𝑥𝑛−𝑥0,𝑥}(𝑡)>1−𝛼, for all 𝑛≥𝑘. It follows that ‖𝑥_𝑛−𝑥₀,𝑥‖_1−𝛼≤𝑡, for all 𝑛≥𝑘. Thus ‖𝑥_𝑛−𝑥₀,𝑥‖_1−𝛼→0. Conversely, choose 𝑥∈𝑋. Let ‖𝑥_𝑛−𝑥₀,𝑥‖_𝛼→0, for every 𝛼∈(0,1). Fix 𝛼∈(0,1), and 𝑡>0. There exists 𝑘∈ℕ such that {𝑟>0∶𝑁_{𝑥𝑛−𝑥0,𝑥}(𝑟)≥1−𝛼}<𝑡, (3.22) for all 𝑛≥𝑘. Hence, for every 𝑛≥𝑘 there is 0<𝑡_𝑛<𝑡 such that 𝑁_{𝑥𝑛−𝑥0,𝑥}(𝑡_𝑛)≥1−𝛼. (3.23) It implies that 𝑁_{𝑥𝑛−𝑥0,𝑥}(𝑡)≥1−𝛼, (3.24) for all 𝑛≥𝑘, that is, 𝑥_𝑛→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 𝐵_𝑒,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 𝐵_𝑒[𝑥,𝑟]={𝑦∈𝑋∶‖𝑥−𝑦,𝑒‖_𝛼≤𝑟} (3.25) is a compact set in (𝑋,‖⋅,⋅‖_𝛼). It is clear that 𝐵_𝑒[𝑥,𝑟]=𝐵_𝑒,𝛼[𝑥,𝑟]. Choose a sequence {𝑥_𝑛} of 𝐵_𝑒[𝑥,𝑟]. Since 𝐵_𝑒,𝛼[𝑥,𝑟] is compact, it has a convergent subsequence {𝑥_𝑛𝑘}. Lemma 3.19 implies that {𝑥_𝑛𝑘} is convergent in ‖⋅,⋅‖_𝛼. Thus 𝐵_𝑒[𝑥,𝑟] is compact in (𝑋,‖⋅,⋅‖_𝛼), and consequently 𝑋 is of finite dimension.

Remark 3.21. The converse of Lemma 3.20 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, 𝐵_(1,0),1/2[0,1]=𝐵_(1,0)[0,1], (3.26) 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 {𝑥_𝑛}∈𝐵_(1,0),1/2[0,1]=𝐵_(1,0)[0,1] there would exist a converging subsequence {𝑥_𝑛𝑘}. Say 𝑥_𝑛𝑘→𝑎, where 𝑎∈𝐵_(1,0),1/2[0,1]. Thus lim_𝑛→∞𝑁_{𝑥𝑛𝑘−𝑎,𝑒}(𝑡)=lim_𝑛→∞𝑡𝑡+‖𝑥_𝑛𝑘−𝑎,𝑒‖=1, (3.27) for all 𝑒∈𝑋, 𝑡>0. This implies that lim_𝑛→∞‖𝑥_𝑛𝑘−𝑎,𝑒‖=0, (3.28) for all 𝑒. Therefore 𝐵_(1,0)[0,1] is a compact set which is a contradiction by Example 2.14.

Lemma 3.22. Let 𝑇∶(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. Let 𝛼∈(0,1). We show that for each locally bounded sequence {𝑥_𝑛} in (𝑋,‖⋅,⋅‖_𝛼), the sequence {𝑇(𝑥_𝑛)} has a convergent subsequence in (𝑌,‖⋅,⋅‖^′_𝛼). Let {𝑥_𝑛} be a locally bounded sequence in (𝑋,‖⋅,⋅‖_𝛼). There exist 𝑒∈𝑋 and 𝑀>0 such that ‖𝑥_𝑛,𝑒‖_𝛼<𝑀, (3.29) for all 𝑛∈ℕ. By the definition of ‖⋅,⋅‖_𝛼, for every 𝑛≥1 there exists 𝑡_𝑛>0 such that 𝑡_𝑛<𝑀 and 𝑁_{𝑥𝑛,𝑒}(𝑡_𝑛)≥𝛼 for all 𝑛. Because 𝑁 is nondecreasing, 𝛼≤𝑁_{𝑥𝑛,𝑒}(𝑡_𝑛)≤𝑁_{𝑥𝑛,𝑒}(𝑀). Hence 𝑁_{𝑥𝑛,𝑒}(𝑀)≥𝛼, (3.30) for all 𝑛. That is, {𝑥_𝑛} is 2-𝑃𝐿𝐵 in (𝑋,𝑁). Thus {𝑇(𝑥_𝑛)} has a convergent subsequence {𝑇(𝑥_𝑛𝑘)} in (𝑌,𝑁^′). By Lemma 3.19, {𝑇(𝑥_𝑛𝑘)} is convergent in ‖⋅,⋅‖^′_𝛼.

Theorem 3.23. Let (𝑋,𝑁) and (𝑌,𝑁^′) be two 2𝑃𝑁-spaces satisfying (2𝑃𝑁-𝑉𝐼). Then(a)every compact operator 𝑇∶(𝑋,𝑁)→(𝑌,𝑁^′) is 2-𝑃𝐶;(b)if dim𝑋=∞, then the identity operator 𝐼∶(𝑋,𝑁)→(𝑋,𝑁) is not a compact operator.

Proof. (a) Choose 𝛼∈(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 𝑚_𝑒,𝛼>0 such that ‖𝑇(𝑥),𝑇(𝑒)‖^′_𝛼≤𝑚_𝑒,𝛼‖𝑥,𝑒‖_𝛼, (3.31) 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 𝐵_𝑒,𝛼[0,1] to itself. Suppose on the contrary that 𝐼 is a compact operator. Let {𝑥_𝑛} be a sequence in 𝐵_𝑒,𝛼[0,1]. Because 𝐼 is a compact operator, the 2-𝑃𝐿𝐵 sequence {𝑥_𝑛} has a convergent subsequence. Hence 𝐵_𝑒,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 {𝑥_𝑛} that is a 2-𝑃𝐿𝐵 sequence in 𝑋, the sequence {𝑇₁(𝑥_𝑛)} has a convergent subsequence {𝑇₁(𝑥_𝑛𝑘)}, and the sequence {𝑇₂(𝑥_𝑛𝑘)} has a convergent subsequence {𝑇₂(𝑧_𝑛)}. Hence, {𝑇₁(𝑧_𝑛)} and {𝑇₂(𝑧_𝑛)} are convergent sequences. Let 𝑇₁(𝑧_𝑛)→𝑢 and 𝑇₂(𝑧_𝑛)→𝑣, where 𝑢,𝑣∈𝑌. We have lim_𝑛→∞𝑁^′_{(𝑇1+𝑇2)(𝑧𝑛)−𝑢−𝑣,𝑦}(𝑡)≥lim_𝑛→∞min{𝑁^′_{𝑇1(𝑧𝑛)−𝑢,𝑦}(𝑡2),𝑁^′_{𝑇2(𝑧𝑛)−𝑣,𝑦}(𝑡2)}, (3.32) for all 𝑦∈𝑌 and 𝑡>0. Thus lim_𝑛→∞𝑁^′_{𝑇1+𝑇2(𝑧𝑛)−𝑢−𝑣,𝑦}(𝑡)=1, (3.33) for all 𝑦∈𝑌, and 𝑡>0. This implies that 𝑇₁+𝑇₂ is a compact operator. Now for all 𝛼∈ℝ⧵{0} if 𝑇(𝑥_𝑛𝑘)→𝑦₀, then lim_𝑛→∞𝑁^′_{𝛼𝑇1(𝑥𝑛𝑘)−𝛼𝑦0,𝑦}(𝑡)=lim_𝑛→∞𝑁^′_{𝑇1(𝑥𝑛𝑘)−𝑦0,𝑦}(𝑡|𝛼|)=1, (3.34) 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 {𝑥_𝑛} be a 2-𝑃𝐿𝐵 sequence in 𝑋. Then {𝑇(𝑥_𝑛)} has a convergent subsequence {𝑇(𝑥_𝑛𝑘)}. Let lim_𝑛→∞𝑇(𝑥_𝑛𝑘)=𝑦. Since 𝑆 is 2-𝑃𝐶, by Theorem 3.11 we have 𝑆(𝑇(𝑥_𝑛𝑘))→𝑆(𝑦). Hence 𝑆𝑇(𝑥_𝑛) 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 𝑁_{𝑥𝑛,𝑒}(𝑡₀)>1−𝑟₀ for all 𝑛≥1 since {𝑥_𝑛} is a 2-𝑃𝐿𝐵 sequence. The operator 𝑆 is 2-𝑃𝐵 by Theorem 3.9 and so there is 𝑚_{𝑒,1−𝑟0}>0 such that 𝑁_{𝑥𝑛,𝑒}(𝑡₀𝑚_{𝑒,1−𝑟0})≥1−𝑟₀⇒𝑁^′_{𝑆(𝑥𝑛),𝑆(𝑒)}(𝑡₀)≥1−𝑟₀, (3.35) for all 𝑛. It follows that {𝑆(𝑥_𝑛)} is a 2-𝑃𝐿𝐵 sequence in 𝑋. Because 𝑇 is a compact operator, {𝑇(𝑆(𝑥_𝑛))} has a convergent subsequence. This completes the proof.

Theorem 3.26. Let 𝑇∶𝑋→𝑋 be a compact operator on a 2𝑃𝑁-space 𝑋. Then for every 𝜆≠0, the null space 𝑁(𝑇_𝜆) of 𝑇_𝜆=𝑇−𝜆𝐼 is of finite dimension.

Proof. We choose a locally ball 𝑀 in 𝑁(𝑇_𝜆) and show that it is compact, then apply Lemma 3.20. Let {𝑥_𝑛} be a sequence in 𝑀. Then {𝑥_𝑛} is locally bounded, and {𝑇(𝑥_𝑛)} has a convergent subsequence {𝑇(𝑥_𝑛𝑘)}. Now 𝑥_𝑛∈𝑀⊂𝑁(𝑇_𝜆) implies 𝑇_𝜆(𝑥_𝑛)=𝑇(𝑥_𝑛)−𝜆𝑥_𝑛=0, so that 𝑥_𝑛=𝜆⁻¹𝑇(𝑥_𝑛). Consequently, 𝑥_𝑛𝑘=𝜆⁻¹𝑇(𝑥_𝑛𝑘) is convergent. Now, the closedness of 𝑀 implies that the limit of {𝑥_𝑛𝑘} belongs to 𝑀. This proves that dim𝑁(𝑇_𝜆)<∞.

Definition 3.27. Let (𝑋,𝑁) be a 2𝑃𝑁-space. A sequence {𝑥_𝑛} in 𝑋 is called a Cauchy sequence if lim_𝑛→∞𝑁_{𝑥𝑛+𝑝−𝑥𝑛,𝑥}(𝑡)=1 for all 𝑥∈𝑋, 𝑡>0, and 𝑝∈ℕ.
We say that a 2𝑃𝑁-space (𝑋,𝑁) is complete if every Cauchy sequence in 𝑋 is convergent to a point of 𝑋.

Theorem 3.28. Let 𝑋,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 𝑇∶𝑍→𝑌, (3.36) where 𝑇 is an 𝑒-2-𝑃𝐵 operator for each 𝑒∈𝑍.

Proof. We consider any 𝑥∈𝑍. There is a sequence {𝑥_𝑛} in 𝑍 such that 𝑥_𝑛→𝑥. Since 𝑇 is linear and 2-𝑃𝐵, for every 𝛼∈(0,1) and 𝑒∈𝑍 there exists 𝑚_𝑒,𝛼>0 such that 𝑁_{𝑥𝑚−𝑥𝑛,𝑒}(𝑡𝑚_𝑒,𝛼)≥𝛼⇒𝑁^′_{𝑇(𝑥𝑚)−𝑇(𝑥𝑛),𝑇(𝑒)}(𝑡)≥𝛼, (3.37) for all 𝑡>0 and 𝑚,𝑛∈ℕ. But the sequence {𝑥_𝑛} is Cauchy, thus for all 𝑡>0, there exists 𝑘∈ℕ such that for all 𝑚,𝑛>𝑘 we have 𝑁_{𝑥𝑚−𝑥𝑛,𝑒}(𝑡𝑚_𝑒,𝛼)≥𝛼. (3.38) Therefore, for 𝑚,𝑛>𝑘, 𝑁^′_{𝑇(𝑥𝑚)−𝑇(𝑥𝑛),𝑇(𝑒)}(𝑡)≥𝛼. (3.39) This shows that {𝑇(𝑥_𝑛)} is Cauchy in 𝑌. Thus {𝑇(𝑥_𝑛)} 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 𝑚_𝑒,𝛼>0 such that 𝑁_𝑥,𝑒(𝑡𝑚_𝑒,𝛼)≥𝛼⇒𝑁^′_{𝑇(𝑥),𝑇(𝑒)}(𝑡)≥𝛼, (∗) for all 𝑡>0 and 𝑥∈𝑍. Choose 𝑡>0 and 𝑥∈𝑍 such that 𝑁_𝑥,𝑒(𝑡4𝑚_𝑒,𝛼)≥𝛼. (3.40) Now, we show that 𝑁^′_{𝑇(𝑥),𝑇(𝑒)}(𝑡)≥𝛼. Because 𝑥∈𝑍, there exists {𝑥_𝑛}⊆𝑍 such that 𝑥_𝑛→𝑥. Therefore, for 𝑛∈ℕ sufficiently large, we have 𝑁_{𝑥𝑛,𝑒}(𝑡2𝑚_𝑒,𝛼)≥min{𝑁_{𝑥𝑛−𝑥,𝑒}(𝑡4𝑚_𝑒,𝛼),𝑁_𝑥,𝑒(𝑡4𝑚_𝑒,𝛼)}≥𝛼. (3.41) By (*), 𝑁^′_{𝑇(𝑥𝑛),𝑇(𝑒)}(𝑡/2)≥𝛼, and since lim_𝑛→∞𝑇(𝑥_𝑛)=𝑇(𝑥), we obtain 𝑁^′_{𝑇(𝑥𝑛)−𝑇(𝑥),𝑇(𝑒)}(𝑡2)≥𝛼. (3.42) Hence 𝑁^′_{𝑇(𝑥),𝑇(𝑒)}(𝑡)≥min{𝑁^′_{𝑇(𝑥)−𝑇(𝑥𝑛),𝑇(𝑒)}(𝑡2),𝑁^′_{𝑇(𝑥𝑛),𝑇(𝑒)}(𝑡2)}≥𝛼. (3.43) Therefore 𝑇 is a 2-𝑃𝐵 linear operator on 𝑍.

Acknowledgment

This research was in part supported by a Grant from IPM (no. 87460021).