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Journal of Function Spaces
Volume 2018, Article ID 9092136, 9 pages
https://doi.org/10.1155/2018/9092136
Research Article

## Statistical Order Convergence and Statistically Relatively Uniform Convergence in Riesz Spaces

1School of Mathematics and Statistics, Northeast Normal University, Changchun, Jilin 130024, China
2School of Mathematical Sciences, Xiamen University, Xiamen 361005, China

Correspondence should be addressed to Jian Tao; nc.ude.unen@joat

Received 25 January 2018; Revised 25 February 2018; Accepted 6 March 2018; Published 15 April 2018

Copyright © 2018 Xuemei Xue and Jian Tao. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

A new concept of statistically -uniform Cauchy sequences is introduced to study statistical order convergence, statistically relatively uniform convergence, and norm statistical convergence in Riesz spaces. We prove that, for statistically -uniform Cauchy sequences, these three kinds of convergence for sequences coincide. Moreover, we show that the statistical order convergence and the statistically relatively uniform convergence need not be equivalent. Finally, we prove that, for monotone sequences in Banach lattices, the norm statistical convergence coincides with the weak statistical convergence.

#### 1. Introduction

The first idea of statistical convergence goes back to Zygmund’s monograph [1], where Zygmund called it almost convergence. The concept of statistical convergence was formally introduced by Steinhaus [2] and Fast [3] and later reintroduced by Schoenberg [4]. Statistical convergence, as a generalization of the usual notion of convergence, has attracted much attention since it was introduced over sixty years ago. This concept has been applied in various areas such as number theory [5], trigonometric series [1], summability theory [6], measure theory [7], and probability theory [8]. Furthermore, generalizations have appeared in Hausdorff topological spaces and function spaces [9], locally convex spaces [10, 11], and Banach spaces [12].

The concepts of order convergence and relatively uniform convergence are two fundamental concepts and basic tools in the study of Riesz spaces. The two big books [13, 14] present a good and detailed investigation of these two concepts. Sençimen and Pehlivan [15] introduced the concept of statistical order convergence, which is a natural generalization of order convergence. Meanwhile, some basic definitions and results are established in [15]. Ercan [16] introduced the notion of statistically -uniformly convergent sequences in Riesz spaces, which is a statistical analogue of -uniformly convergent sequences.

The aim of our present paper is to continue to develop the theory of statistical order convergence and statistically -uniform convergence. The present paper is organized as follows.

In Section 2, we first establish a useful characterization of statistically -uniform convergence, which is a statistical analogue of an equivalent definition of -uniform convergence. Secondly, we prove that a statistically -uniformly convergent sequence almost coincides with an -uniformly convergent sequence. As a corollary, we show that every monotone statistically relatively uniformly convergent sequence is relatively uniformly convergent. In [16], Ercan introduced the notion of statistically -uniform pre-Cauchy sequence and proved that each statistically -uniform pre-Cauchy sequence admits an -uniform Cauchy subsequence. In this section, we introduce the notion of statistically -uniform Cauchy sequence and prove that a sequence is statistically -uniform Cauchy if and only if there exists a subset of with such that is an -uniform Cauchy sequence. Using this result, we obtain that every monotone statistically -uniform Cauchy sequence is -uniform Cauchy. Furthermore, we show that every statistically -uniform Cauchy sequence is statistically -uniformly pre-Cauchy. Finally, we prove that every monotone statistically -uniform Cauchy sequence in a Dedekind -complete Riesz space is statistically -uniformly convergent.

Section 3 is concerned with close relationship among statistically relatively uniform convergence, statistical order convergence, and norm statistical convergence. In this section, we prove that, for uniform Cauchy sequences, the relatively uniform convergence coincides with the statistical order convergence. Similarly, we also show that the statistically relatively uniform convergence coincides with the statistical order convergence for statistically uniform Cauchy sequences. Moreover, we show that the statistical order convergence and the statistically relatively uniform convergence need not be equivalent. The rest of this section is devoted to norm statistical convergence. We prove that, for statistically uniform Cauchy sequences, the statistically relatively uniform convergence also coincides with the norm statistical convergence. Finally, we show that the norm statistical convergence and the weakly statistical convergence are equivalent for monotone sequences.

Notation 1. Throughout the paper, will denote the set of all positive integers and will denote the set of all real numbers. Let be a Riesz space. The set of all positive elements of is denoted by . If is a subset of , then denotes the set and denotes the number of elements in . The natural density of is defined by . Let be a sequence in a Riesz space . If is increasing (decreasing), we shall write (). If, moreover, () exists, then we denote this by (). Given a sequence in a normed Riesz space and an infinite subset , we say that converges in norm to along if, for every , there exists a natural number such that for each we have We denote this by If a sequence in a Riesz space satisfies property for all except a natural density zero set, then we say that satisfies property for almost all and we abbreviate this by a.a.n. [17]. A Riesz space is called Dedekind -complete if every nonempty finite or countable subset which is order bounded from above has a supremum and every nonempty finite or countable subset which is order bounded from below has an infimum.
The reader is referred to [13, 14] for more details about Riesz spaces.

#### 2. Statistically Relatively Uniform Convergence

Let be a Riesz space and let . A sequence in is said to converge -uniformly to [13, Definition 39.1] whenever, for every , there exists a natural number such that holds for all . We denote this by . It is said that the sequence in converges relatively uniformly to whenever converges -uniformly to for some . We shall write .

It should be noted that, in an Archimedean space , a sequence if and only if there exists a sequence of positive reals with such that for all . See [18, Definition 2.23, p. 24] for this equivalent definition.

Ercan [16] modified the definition of -uniformly convergent sequences in Riesz spaces as follows.

Definition 2 (see [16]). Let be a Riesz space and . A sequence in is said to be statistically -uniformly convergent to if for each . Equivalently, . for each . We write We say that a sequence in converges statistically relatively uniformly to provided that for some . We write

The following well-known lemma about sets of natural density is the key to establishing many results regarding statistically convergent sequences.

Lemma 3 (see [6]). Let be a countable collection of subsets of such that for each . Then there is a subset such that and for all .

The following characterization of statistically -uniform convergence is useful and will be used in the sequel.

Theorem 4. Let be an Archimedean Riesz space and . The following are equivalent for a sequence in and :(1).(2)There exists a sequence of positive reals with such that

Proof.
(1) ⇒ (2). Suppose that . For each , we setThen = 1 and for each .
Let be the principal ideal generated by the singleton . We define for each . Then is a lattice norm on and for all .
By Lemma 3, we get a subset such that and for all .
Claim 4. .
Indeed, for every , we choose with . We set . Then, for each , we have Let . Then . We let for each . Then and Let for , where . Then and This impliesHence(2) (1). Let be the sequence of positive reals as stated in (2). Let and . Choose with . Then, for and , we have Since , we get This completes the proof.

Note that a subsequence of a statistically relatively uniform convergent sequence need not be statistically relatively uniform convergent and a statistically relatively uniform convergent sequence need not be relatively uniform convergent. These facts can be seen in the following easy example.

Example 5. Let be a sequence in defined byThen, . and hence, by Theorem 4, we have . But, it is obvious that the subsequence of , where , is not statistically relatively uniform convergent. Since the sequence is not order bounded, is not relatively uniform convergent.

Remark 6. Statistically relatively uniform convergence is stable; that is, if , then there exists a sequence of reals with such that . Indeed, there exist and a sequence of reals with such that That is, Hence,

We give another useful characterization of statistically -uniform convergence.

Theorem 7. Let be an Archimedean Riesz space and . The following are equivalent for a sequence in and : (1).(2)There exists a subset of with such that .(3)There exists a sequence in such that . and .

Proof. The implication follows from Theorem 4.
(2) ⇒ (3). Assume that there exists a subset of with such that . Then there exists a sequence of positive reals with such that for all .
We setand for , where .
Then, ., , and for all . Hence .
is trivial.

Corollary 8. Let be an Archimedean Riesz space. Then every monotone statistically relatively uniformly convergent sequence in is relatively uniformly convergent.

Proof. Suppose that is an increasing sequence in and for some and . It follows from [16, Corollary 5] that exists and equals . By Theorem 7, there exists a subset of with such that . Thus, we get a sequence of positive reals with such that for all . We setThen . We set . Then, for , we have For , we have Hence, . This completes the proof.

Let be a Riesz space and let . Following [13], recall that a sequence in is called an -uniform Cauchy sequence whenever, for every , there exists a natural number such that holds for all . We say that a sequence is uniform Cauchy if it is -uniform Cauchy for some .

Definition 9. Let be a Riesz space and let . We say that a sequence in is a statistically -uniform Cauchy sequence whenever, for every , there exists a natural number such that ,   We say that a sequence is statistically uniform Cauchy if it is statistically -uniform Cauchy for some .

Following [15], a sequence in a Riesz space is said to be statistically order bounded if there exist such that ,  . It is easy to see that every statistically -uniform Cauchy sequence is statistically order bounded for all . It should be noted that a statistically uniform Cauchy sequence need not be uniform Cauchy as can be seen in the following example.

Example 10. Let . For each , define bywhere . Then . and hence . Thus, we see that the sequence is statistically -uniform Cauchy, but not uniform Cauchy since .

Theorem 11. Let be a Riesz space and . The following are equivalent for a sequence in : (1) is a statistically -uniform Cauchy sequence.(2)There exists a subset of with such that is an -uniform Cauchy sequence.

Proof.
(1) (2). For each , by (1), we get a natural number such that Let . It follows from Lemma 3 that there exists a subset with such that for all . Let . Choose such that . Since the set is finite, there exists a natural number such that, for all , we have Hence, we get This means that the sequence is an -uniform Cauchy sequence along .
(2) ⇒ (1). Let be as stated in (2). Let . There exists a natural number such that This implies that Hence, we get that ,   This completes the proof.

Corollary 12. Let be a Riesz space. Then every monotone statistically -uniform Cauchy sequence is -uniform Cauchy for all .

Proof. Suppose that is an increasing statistically -uniform Cauchy sequence in . It follows from Theorem 11 that there exists a strictly increasing sequence of positive integers such that is an -uniform Cauchy sequence. Let . Then there exists a natural number such that for all . For all , if for some , then we have If for some and for some , then we have Hence, is -uniform Cauchy.

Let be a Riesz space and . Recall in [16] that a sequence in is said to be statistically -uniform pre-Cauchy if for every . We say that a sequence is statistically uniform pre-Cauchy if it is statistically -uniform pre-Cauchy for some .

Corollary 13. Every statistically -uniform Cauchy sequence in a Riesz space is statistically -uniform pre-Cauchy for all .

Proof. Let be a statistically -uniform Cauchy sequence in . By Theorem 11, there exists a subset with such that is -uniform Cauchy along . Let . There exists a subset such that is finite and Note that, for each , we have Hence, Since , we get, by letting , We are done.

The following example shows that a statistically uniform pre-Cauchy sequence need not be statistically uniform Cauchy.

Example 14. We take [19, Example 8].
LetThen is increasing and tends to . Hence, the sequence has no convergent subsequences and so is not statistically convergent. It follows from [17, Theorem 1] that is not statistically Cauchy and hence is not statistically uniform Cauchy. It is proved in [19, Example 8] that is statistically pre-Cauchy. This means that is statistically -uniform pre-Cauchy.

Theorem 15. Let be an Archimedean Riesz space and . Then
(1) every statistically -uniformly convergent sequence in is statistically -uniform Cauchy;
(2) if, moreover, is Dedekind -complete, then every monotone statistically -uniform Cauchy sequence in is statistically -uniformly convergent.

Proof. (1) Suppose that . Let . Then Let and . Then, for every , we have This implies that and hence .
(2) Let be an increasing, statistically -uniform Cauchy sequence in . Then is statistically order bounded. Since is increasing, we see that is order bounded. Since is Dedekind -complete, then exists and is denoted by .
Let . Then there is a natural number such that ,   Let and . Then . Let us fix any . Then, for any , we have That is, Since is increasing, we get Hence, we have This means Thus,

It follows from Theorem 15 and Corollary 13 that every statistically -uniformly convergent sequence is statistically -uniform pre-Cauchy for all . The next result suggests that, under certain circumstances, the converse is also true. This result is an analogue of [19, Theorem 7].

Theorem 16. Let be an Archimedean Riesz space and . Assume that is a statistically -uniform pre-Cauchy sequence in . If has a subsequence which converges -uniformly to and . Then .

Proof. Let . Since converges -uniformly to , there exists a subset such that is finite and We set . Then, for , we have That is, This impliesLetting , we get This completes the proof.

#### 3. Relationship to Statistical Order Convergence and Norm Statistical Convergence

Let us recall in [13] that a sequence in a Riesz space is order convergent to whenever there exists a sequence in such that holds for all .

Definition 17 (see [15]). Let be a sequence in a Riesz space . If there exists a set with such that is increasing and for some , then one writes . Similarly, if is decreasing and for some , then one writes . If or , then one said that is statistically monotone convergent to .

Sençimen and Pehlivan [15] modified the definition of the usual order convergence and introduced the notion of statistical order convergence as follows.

Definition 18 (see [15]). Let be a Riesz space. A sequence in is statistically order convergent to provided that there exist a sequence such that and a subset of with such that for every . One writes

Remark 19. It is easy to see that if and only if there exists a sequence in with such that

Theorem 20. Let be a sequence in , and . Then if and only if and is an -uniform Cauchy sequence.

Proof. The necessary part is trivial. We only prove the sufficient part.
Suppose that and is -uniform Cauchy. By Remark 19, there exists a sequence in with such that ,   Let . Then . Let . Then there exists a natural number such that for all . Let us fix any . Then, for any , we have Note that Hence, we get for all . This finishes the proof.

Theorem 21. Let be a sequence in , and . Then if and only if and is a statistically -uniform Cauchy sequence.

Proof. It suffices to prove the sufficient part. Assume that and is a statistically -uniform Cauchy sequence. By [15, Theorem 5] and Theorem 7, we obtain a subset of with such that the sequence converges in order to and is -uniform Cauchy. Let . There exists a natural number such that Let us fix any . Letting , we get This means that . It follows from Theorem 7 that .

Theorem 22. Let be an Archimedean Riesz space. The following statements are equivalent:
(1) Statistical order convergence is stable.
(2) Statistical order convergence and statistically relatively uniform convergence are equivalent.

Proof.
(1) ⇒ (2). Suppose that Then, by (1), there exists a sequence of reals with such that By Remark 19, there exists a sequence with such that This meansHence, and then .
(2) ⇒ (1). Assume that By (2), we see that . Then, there exist and a sequence such that Thus, we get And we are done.

Note that statistical order convergence and statistically relatively uniform convergence are not equivalent in general, as can be seen in the following example.

Example 23. Let us take a counterexample in [13, Exercise , p. ]. For each , we define by on and on and is linear on . Clearly, . Moreover, there is no sequence such that Otherwise, it follows from [15, Theorem 5] that there exists a subset of with such that the sequence is order convergent . Hence, there exists a sequence in with so that for all . This implies This contradiction suggests that statistical order convergence in is not stable. Then Theorem 22 gives the conclusion.

Following [12], a sequence in a Banach space is said to be norm statistically convergent to provided that for all . It is clear that a sequence in a Banach lattice is norm statistically convergent whenever it is statistically relatively uniformly convergent. The sequence is said to be weakly statistically convergent to provided that the sequence converges statistically to 0 for each . It is noted in [12] that if a sequence in a Banach space is norm statistically convergent to , then there exists a subset with such that converges in norm to . Combining this observation, Theorems 11 and 7 with [14, Theorem 100.4], we obtain the following result.

Theorem 24. Let be a sequence in a Banach lattice , and . Then if and only if is norm statistically convergent to and is a statistically -uniform Cauchy sequence.

It is known [18] that, in -spaces, the relatively uniform convergence coincides with the norm convergence. It follows from Theorem 7 that, in -spaces, the statistically relatively uniform convergence coincides with the norm statistical convergence. In Example 23, utilizing Theorem 22, we show that statistical order convergence and statistically relatively uniform convergence are not equivalent in general. Now we construct a statistical order convergent sequence that is not statistically relatively uniformly convergent as follows.

Example 25. Let . For each , we define bywhere
Obviously, and hence . Since for all , we see that the sequence is not norm statistically convergent and hence not statistically relatively uniformly convergent.

Example  5 in [15] suggests that the statistical order convergence need not be norm statistically convergent. The following example suggests that the norm statistical convergence need not be statistically order convergent and hence not be statistically relatively uniformly convergent.

Example 26. We take the sequence in [14, Exercise 100.13] as follows.
Let be the normed Riesz space of all real continuous functions on with . We setand generallyfor
It is proved in [14, Exercise ] that the sequence converges to in norm and no subsequence of it converges to in order.
We set and letClearly, is norm statistically convergent to . But is not statistically order convergent. Indeed, if is statistically order convergent to , it follows from [15, Theorem 5] that there exists a subset of with such that is order convergent to . Hence, the subsequence is order convergent to . This is a contradiction.

Example 27. We take the sequence in [20, Example 5] as follows.
LetThen is ideal in . Define for byThen , but no subsequence of is relatively uniformly convergent.
Then, as in Example 26, we set and letClearly, is norm statistically convergent to . An argument similar to Example 26 shows that is not relatively uniformly convergent.

The following example illustrates that even a combination of statistical order convergence and norm statistical convergence does not imply the statistically relatively uniform convergence.

Example 28. Let us take Example 2 in [20].
Pick such that is continuous on for all , and . Let be the Riesz subspace of consisting of all such that, for some and , we have for all . Define a Riesz norm on by setting Moore Jr. exhibited a sequence in such that and , but , as follows.
Let . Define by induction so that(a) is continuous on ,(b),(c),(d),(e)This implies that no subsequence of converges relatively uniformly to .
Similarly, we set and letClearly, is norm statistically convergent to and But Theorem 7 yields that is not statistically relatively uniformly convergent to . This is because no subsequence of converges relatively uniformly to .

Theorem 29. Let be a monotone sequence in a Banach lattice . Then (1) is norm statistically convergent if and only if it is norm convergent;(2) is weakly statistically convergent if and only if it is weakly convergent;(3) is norm statistically convergent if and only if it is weakly statistically convergent.

In order to prove Theorem 29, we need two simple lemmas. The first lemma is an immediate consequence of [21, Proposition 1.1.6].

Lemma 30. If a sequence in a Banach lattice is increasing and converges norm statistically to , then

Actually, Lemma 30 also holds for weakly statistical convergence as shown in the following.

Lemma 31. If a sequence in a Banach lattice is increasing and converges weakly statistically to , then

Proof. Let . Then there exists a strictly increasing sequence of natural numbers so that For each , we choose with . Then, we have This means that Hence, is an upper bound of .
Assume that is an arbitrary upper bound of . Then, for every , we have Since the sequence converges statistically to , we have This implies that and we are done.

Proof of Theorem 29. (1) Suppose that is increasing and norm statistically convergent to . By Lemma 30, we get Let be a strictly increasing sequence of natural numbers so that in norm. Let . Choose a natural number such that Then, for any , we haveThis yields Hence, in norm.
(2) Let be increasing and weakly statistically convergent to . By Lemma 31, we get . Let . Since converges weakly statistically to , converges statistically to . There exists a subset with such that Let . Choose such that Then, for all , we have This implies thatHence, converges weakly to .
(3) follows from (1) and (2).

#### Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

#### Acknowledgments

This research is supported by the National Natural Science Foundation of China (Grant no. 11571069).

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