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

Authors: Xuemei Xue; Jian Tao
Journal: Journal of Function Spaces
 (2018)

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## Abstract

A new concept of statisticallye-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 e-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.


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## Body

## 1. Introduction

The first idea of statistical convergence goes back to Zygmund’s monograph [1], where Zygmund called italmost 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 u-uniformly convergent sequences in Riesz spaces, which is a statistical analogue of u-uniformly convergent sequences.The aim of our present paper is to continue to develop the theory of statistical order convergence and statisticallyu-uniform convergence. The present paper is organized as follows.In Section2, we first establish a useful characterization of statistically e-uniform convergence, which is a statistical analogue of an equivalent definition of e-uniform convergence. Secondly, we prove that a statistically e-uniformly convergent sequence almost coincides with an e-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 e-uniform pre-Cauchy sequence and proved that each statistically e-uniform pre-Cauchy sequence admits an e-uniform Cauchy subsequence. In this section, we introduce the notion of statistically e-uniform Cauchy sequence and prove that a sequence (xn)n∈N is statistically e-uniform Cauchy if and only if there exists a subset K={kj:j∈N} of N with δ(K)=1 such that (xkj)j∈N is an e-uniform Cauchy sequence. Using this result, we obtain that every monotone statistically e-uniform Cauchy sequence is e-uniform Cauchy. Furthermore, we show that every statistically e-uniform Cauchy sequence is statistically e-uniformly pre-Cauchy. Finally, we prove that every monotone statistically e-uniform Cauchy sequence in a Dedekind σ-complete Riesz space E is statistically e-uniformly convergent.Section3 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,N will denote the set of all positive integers and R will denote the set of all real numbers. Let E be a Riesz space. The set of all positive elements of E is denoted by E+. If K is a subset of N, then Kn denotes the set {k∈K:k≤n} and Kn denotes the number of elements in Kn. The natural density of K is defined by δ(K)=limn→∞⁡Kn/n. Let (xn)n∈N be a sequence in a Riesz space E. If (xn)n∈N is increasing (decreasing), we shall write xn↑ (xn↓). If, moreover, supn∈N⁡xn=x (infn∈N⁡xn=x) exists, then we denote this by xn↑x (xn↓x). Given a sequence (xn)n∈N in a normed Riesz space E and an infinite subset A⊆N, we say that (xn)n∈N converges in norm to x∈E along A if, for every ϵ>0, there exists a natural number N such that for each n>N,n∈A we have xn-x<ϵ. We denote this by limn∈A⁡xn=x. If a sequence xnn∈N in a Riesz space E satisfies property P for all n except a natural density zero set, then we say that xnn∈N satisfies property P for almost all n and we abbreviate this bya.a.n. [17]. A Riesz space E is calledDedekindσ-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

LetE be a Riesz space and let e∈E+. A sequence (xn)n∈N in E is said to converge e-uniformly to x∈E [13, Definition 39.1] whenever, for every ϵ>0, there exists a natural number N such that |xn-x|≤ϵe holds for all n≥N. We denote this by xn→rux(e). It is said that the sequence (xn)n∈N in E converges relatively uniformly to x whenever (xn)n∈N converges e-uniformly to x for some e∈E+. We shall write xn→x(ru).It should be noted that, in an Archimedean spaceE, a sequence xn→rux(e) if and only if there exists a sequence (ϵn)n∈N of positive reals with ϵn↓0 such that |xn-x|≤ϵne for all n∈N. See [18, Definition 2.23, p. 24] for this equivalent definition.Ercan [16] modified the definition of e-uniformly convergent sequences in Riesz spaces as follows.Definition 2 (see [16]).
LetE be a Riesz space and e∈E+. A sequence (xn)n∈N in E is said to be statistically e-uniformly convergent to x∈E if (1)limn→∞⁡1nk:k≤n,xk-x-ϵe+>0=0for each ϵ>0. Equivalently, |xn-x|≤ϵe,a.a.n. for each ϵ>0. We write xn→stx(e). We say that a sequence (xn)n∈N in E converges statistically relatively uniformly to x provided that xn→stx(e) for some e∈E+. We write xn→srux.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{Ai}i∈I be a countable collection of subsets of N such that δ(Ai)=1 for each i∈I. Then there is a subset A⊆N such that δ(A)=1 and |A\Ai|<∞ for all i∈I.The following characterization of statisticallye-uniform convergence is useful and will be used in the sequel.Theorem 4.
LetE be an Archimedean Riesz space and e∈E+. The following are equivalent for a sequence (xn)n∈N in E and x∈E:(1)
xn→stx(e).(2)
There exists a sequence(ϵn)n∈N of positive reals with ϵn↓0 such that (2)xn-x≤ϵne,a.a.n.Proof.
(1) ⇒ (2). Suppose that xn→stx(e). For each k∈N, we set(3)Ak=n∈N:xn-x≤1ke.Then δAk = 1 and Ak⊇Ak+1 for each k∈N.
LetEe be the principal ideal generated by the singleton {e}. We define (4)xe=inf⁡λ>0:x≤λefor each x∈Ee. Then ·e is a lattice norm on Ee and x≤xee for all x∈Ee.
By Lemma3, we get a subset A⊆N such that δ(A)=1 and |A\Ak|<∞ for all k∈N.
Claim 4. limn∈A∩A1⁡xn-xe=0.
Indeed, for everyϵ>0, we choose k0∈N with 1/k0<ϵ. We set N=maxn∈A\Ak0⁡n. Then, for each n>N,n∈A, we have (5)xn-xe≤1k0<ϵ.
LetA∩A1=kj:j∈N. Then δ(A∩A1)=1. We let ϵn=supj≥n⁡xkj-xe for each n∈N. Then ϵn↓0 and (6)xkn-x≤xkn-xee≤ϵne,n=1,2,….Let ηn=ϵj for kj-1<n≤kj(j=1,2,…), where k0=0. Then ηn↓0 and (7)xkj-x≤ϵje=ηkje,j=1,2,….This implies(8)A∩A1⊆n∈N:xn-x≤ηne.Hence(9)δn∈N:xn-x≤ϵne=1.
(2)⇒(1). Let (ϵn)n∈N be the sequence of positive reals as stated in (2). Let K=n∈N:xn-x≤ϵne and ϵ>0. Choose k0 with ϵk0<ϵ. Then, for n∈K and n>k0, we have (10)xn-x≤ϵne≤ϵk0e≤ϵe.Since δn∈K:n>k0=1, we get (11)δn∈N:xn-x≤ϵe=1.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(xn)n∈N be a sequence in R defined by(12)xn=n,n=k2k∈N1n,otherwise.Then, xn=1/n,a.a.n. and hence, by Theorem 4, we have xn→st0(1). But, it is obvious that the subsequence (xn)n∈K of (xn)n∈N, where K={k2:k∈N}, is not statistically relatively uniform convergent. Since the sequence (xn)n∈N is not order bounded, (xn)n∈N is not relatively uniform convergent.Remark 6.
Statistically relatively uniform convergence is stable; that is, ifxn→sru0, then there exists a sequence (λn)n∈N of reals with 0≤λn↑∞ such that λnxn→sru0. Indeed, there exist e∈E+ and a sequence (ϵn)n∈N of reals with ϵn↓0 such that (13)xn≤ϵne,a.a.n.That is, (14)1ϵnxn≤ϵne,a.a.n.Hence,(15)1ϵnxn→sru0.We give another useful characterization of statisticallye-uniform convergence.Theorem 7.
LetE be an Archimedean Riesz space and e∈E+. The following are equivalent for a sequence (xn)n∈N in E and x∈E: (1)
xn→stx(e).(2)
There exists a subsetK=kj:j∈N of N with δ(K)=1 such that xkj→rux(e).(3)
There exists a sequence(yn)n∈N in E such that xn=yna.a.n. and yn→rux(e).Proof.
The implication(1)⇒(2) follows from Theorem 4.
(2) ⇒ (3). Assume that there exists a subset K={kj:j∈N} of N with δ(K)=1 such that xkj→rux(e). Then there exists a sequence (ϵj)j∈N of positive reals with ϵj↓0 such that xkj-x≤ϵje for all j∈N.
We set(16)yn=xn,n∈Kx,n∉Kand ηn=ϵj for kj-1<n≤kj(j=1,2,…), where k0=0.
Then,xn=yna.a.n., ηn↓0, and |yn-x|≤ηne for all n∈N. Hence yn→rux(e).
( 3 ) ⇒ ( 1 ) is trivial.Corollary 8.
LetE be an Archimedean Riesz space. Then every monotone statistically relatively uniformly convergent sequence in E is relatively uniformly convergent.Proof.
Suppose that(xn)n∈N is an increasing sequence in E and xn→stx(e) for some x∈X and e∈E+. It follows from [16, Corollary 5] that supn∈N⁡xn exists and equals x. By Theorem 7, there exists a subset K={kj:j∈N} of N with δ(K)=1 such that xkj→rux(e). Thus, we get a sequence (ϵj)j∈N of positive reals with ϵj↓0 such that |xkj-x|≤ϵje for all j∈N. We set(17)ηn=ϵ1,1≤n≤k1ϵj-1,kj-1<n≤kjj=2,3,….
Thenηn↓0. We set u=(x-x1)∨e. Then, for 1≤n≤k1, we have (18)x-xn≤x-x1≤η1u.For kj-1<n≤kj(j=2,3,…), we have (19)x-xn≤x-xkj-1≤ηne≤ηnu.Hence, xn→rux(u). This completes the proof.LetE be a Riesz space and let e∈E+. Following [13], recall that a sequence (xn)n∈N in E is called an e-uniform Cauchy sequence whenever, for every ϵ>0, there exists a natural number N such that |xm-xn|≤ϵe holds for all m,n≥N. We say that a sequence (xn)n∈N isuniform Cauchy if it is e-uniform Cauchy for some e∈E+.Definition 9.
LetE be a Riesz space and let e∈E+. We say that a sequence (xn)n∈N in E is astatisticallye-uniform Cauchy sequence whenever, for every ϵ>0, there exists a natural number N such that |xn-xN|≤ϵe,  a.a.n. We say that a sequence (xn)n∈N isstatistically uniform Cauchy if it is statistically e-uniform Cauchy for some e∈E+.Following [15], a sequence (xn)n∈N in a Riesz space E is said to bestatistically order bounded if there exist y,z∈E such that y≤xn≤z,  a.a.n. It is easy to see that every statistically e-uniform Cauchy sequence is statistically order bounded for all e∈E+. 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.
LetK={k2:k∈N}. For each n∈N, define fn∈C[0,1] by(20)fn=n,n∈Ken,n∉K,where e(t)=t,t∈[0,1]. Then fn=e/n,a.a.n. and hence fn→st0(e). Thus, we see that the sequence (fn)n∈N is statistically e-uniform Cauchy, but not uniform Cauchy since supn∈N⁡fn=∞.Theorem 11.
LetE be a Riesz space and e∈E+. The following are equivalent for a sequence (xn)n∈N in E: (1)
(xn)n∈N is a statistically e-uniform Cauchy sequence.(2)
There exists a subsetK={kj:j∈N} of N with δ(K)=1 such that (xkj)j∈N is an e-uniform Cauchy sequence.Proof.
(1)⇒(2). For each k∈N, by (1), we get a natural number Nk such that (21)δn∈N:xn-xNk≤1ke=1.Let Ak=n∈N:xn-xNk≤1/ke(k=1,2,…). It follows from Lemma 3 that there exists a subset K⊆N with δ(K)=1 such that |K\Ak|<∞ for all k∈N. Let ϵ>0. Choose k0 such that 1/k0<ϵ/2. Since the set K\Ak0 is finite, there exists a natural number N such that, for all m,n>N,m,n∈K, we have (22)xm-xNk0≤1k0e≤ϵ2e,xn-xNk0≤1k0e≤ϵ2e.Hence, we get (23)xm-xn≤ϵe.This means that the sequence (xn)n∈N is an e-uniform Cauchy sequence along K.
(2) ⇒ (1). Let K be as stated in (2). Let ϵ>0. There exists a natural number j0 such that (24)xki-xkj≤ϵe,∀i,j≥j0.This implies that (25)kj:j≥j0⊆n∈N:xn-xkj0≤ϵe.Hence, we get that xn-xkj0≤ϵe,  a.a.n. This completes the proof.Corollary 12.
LetE be a Riesz space. Then every monotone statistically e-uniform Cauchy sequence is e-uniform Cauchy for all e∈E+.Proof.
Suppose that(xn)n∈N is an increasing statistically e-uniform Cauchy sequence in E. It follows from Theorem 11 that there exists a strictly increasing sequence (kj)j∈N of positive integers such that (xkj)j∈N is an e-uniform Cauchy sequence. Let ϵ>0. Then there exists a natural number j0 such that xki-xkj≤ϵe for all i,j≥j0. For all m>n≥kj0, if kj≤n<m<kj+1 for some j≥j0, then we have (26)0≤xm-xn≤xkj+1-xkj≤ϵe.If kj≤n<kj+1 for some j≥j0 and ki≤m<ki+1 for some i>j, then we have (27)0≤xm-xn≤xki+1-xkj≤ϵe.Hence, (xn)n∈N is e-uniform Cauchy.LetE be a Riesz space and 0<e∈E. Recall in [16] that a sequence (xn)n∈N in E is said to bestatisticallye-uniform pre-Cauchy if (28)limn→∞⁡1n2⁡i,j:i,j≤n,xi-xj-ϵe+>0=0for every ϵ>0. We say that a sequence (xn)n∈N isstatistically uniform pre-Cauchy if it is statistically e-uniform pre-Cauchy for some 0<e∈E.Corollary 13.
Every statisticallye-uniform Cauchy sequence in a Riesz space E is statistically e-uniform pre-Cauchy for all 0<e∈E.Proof.
Let(xn)n∈N be a statistically e-uniform Cauchy sequence in E. By Theorem 11, there exists a subset K⊆N with δ(K)=1 such that (xn)n∈K is e-uniform Cauchy along K. Let ϵ>0. There exists a subset A⊆K such that K\A is finite and (29)A×A⊆j,k:xj-xk≤ϵe.Note that, for each n∈N, we have (30)An2=j,k∈A×A:j,k≤n.Hence, (31)1n2An2≤1n2j,k:j,k≤n,xj-xk≤ϵe.Since δ(A)=1, we get, by letting n→∞, (32)limn→∞⁡1n2j,k:j,k≤n,xj-xk≤ϵe=1.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].
Let(33)xk=∑i=1m1i,m!≤k<m+1!,k,m=1,2,….Then (xk)k∈N is increasing and tends to ∞. Hence, the sequence (xk)k∈N has no convergent subsequences and so is not statistically convergent. It follows from [17, Theorem 1] that (xk)k∈N is not statistically Cauchy and hence is not statistically uniform Cauchy. It is proved in [19, Example 8] that (xk)k∈N is statistically pre-Cauchy. This means that (xk)k∈N is statistically 1-uniform pre-Cauchy.Theorem 15.
LetE be an Archimedean Riesz space and e∈E+. Then
(1) every statisticallye-uniformly convergent sequence in E is statistically e-uniform Cauchy;
(2) if, moreover,E is Dedekind σ-complete, then every monotone statistically e-uniform Cauchy sequence in E is statistically e-uniformly convergent.Proof.
(1) Suppose thatxn→stx(e). Let ϵ>0. Then δ({n∈N:|xn-x|≤ϵ/2e})=1. Let K=n∈N:xn-x≤ϵ/2e and N=minn∈Kn. Then, for every n∈K, we have (34)xn-xN≤xn-x+xN-x≤ϵ2e+ϵ2e=ϵe.This implies that K⊆{n∈N:xn-xN≤ϵe} and hence δ({n∈N:|xn-xN|≤ϵe})=1.
(2) Let(xn)n∈N be an increasing, statistically e-uniform Cauchy sequence in E. Then (xn)n∈N is statistically order bounded. Since (xn)n∈N is increasing, we see that (xn)n∈N is order bounded. Since E is Dedekind σ-complete, then supn∈Nxn exists and is denoted by x.
Letϵ>0. Then there is a natural number N such that |xn-xN|≤ϵe,  a.a.n. Let K=n∈N:xn-xN≤ϵe and A=n∈K:n≥N. Then δ(K)=δ(A)=1. Let us fix any n∈A. Then, for any m≥n,m∈A, we have (35)xm-xn≤xm-xN≤ϵe.That is, (36)xm≤xn+ϵe,∀m∈k∈A:k≥n.Since (xn)n∈N is increasing, we get x=supm∈{k∈A:k≥n}xm. Hence, we have (37)x≤xn+ϵe,∀n∈A.This means (38)0≤x-xn≤ϵe,a.a.n.Thus, (39)xn→stxe.It follows from Theorem15 and Corollary 13 that every statistically e-uniformly convergent sequence is statistically e-uniform pre-Cauchy for all e∈E+. The next result suggests that, under certain circumstances, the converse is also true. This result is an analogue of [19, Theorem 7].Theorem 16.
LetE be an Archimedean Riesz space and e∈E+. Assume that (xn)n∈N is a statistically e-uniform pre-Cauchy sequence in E. If (xn)n∈N has a subsequence (xn)n∈K which converges e-uniformly to x∈E and lim⁡infn→∞⁡|Kn|/n>0. Then xn→stx(e).Proof.
Letϵ>0. Since (xn)n∈K converges e-uniformly to x, there exists a subset A⊆K such that K\A is finite and (40)xj-x≤ϵ2e,∀j∈A.We set B={k∈N:xk-x-ϵe+≠0}. Then, for (j,k)∈A×B, we have (41)xj-xk-ϵ2e+≠0.That is, (42)A×B⊆j,k:xj-xk-ϵ2e+≠0.This implies(43)AnnBnn≤j,k:xj-xk-ϵ/2e+≠0,j,k≤nn2.Letting n→∞, we get δ(B)=0. This completes the proof.


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

Let us recall in [13] that a sequence (xn)n∈N in a Riesz space E isorder convergent to x∈E whenever there exists a sequence pn↓0 in E such that |xn-x|≤pn holds for all n.Definition 17 (see [15]).
Let(xn)n∈N be a sequence in a Riesz space E. If there exists a set K={kj}j∈N with δK=1 such that (xkj)j∈N is increasing and supj∈N⁡xkj=x for some x∈E, then one writes xn↑stx. Similarly, if (xkj)j∈N is decreasing and infj∈N⁡xkj=x for some x∈E, then one writes xn↓stx. If xn↑stx or xn↓stx, then one said that (xn)n∈N isstatistically monotone convergent tox.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]).
LetE be a Riesz space. A sequence (xn)n∈N in E isstatistically order convergent to x∈E provided that there exist a sequence (pn)n∈N such that pn↓st0 and a subset K of N with δ(K)=1 such that |xn-x|≤pn for every n∈K. One writes xn→st-ordx.Remark 19.
It is easy to see thatxn→st-ordx if and only if there exists a sequence (pn)n∈N in E with pn↓0 such that |xn-x|≤pn,a.a.n.Theorem 20.
Let(xn)n∈N be a sequence in E, x∈E and e∈E+. Then xn→rux(e) if and only if xn→st-ordx and (xn)n∈N is an e-uniform Cauchy sequence.Proof.
The necessary part is trivial. We only prove the sufficient part.
Suppose thatxn→st-ordx and (xn)n∈N is e-uniform Cauchy. By Remark 19, there exists a sequence (pn)n∈N in E with pn↓0 such that |xn-x|≤pn,  a.a.n. Let K=n∈N:xn-x≤pn. Then δ(K)=1. Let ϵ>0. Then there exists a natural number N such that |xm-xn|≤ϵe for all m,n≥N. Let us fix any n≥N. Then, for any m∈{j∈K:j≥n}, we have (44)xn-x≤xn-xm+xm-x≤ϵe+pm.Note that infm∈K,m≥n⁡pm=0. Hence, we get |xn-x|≤ϵe for all n≥N. This finishes the proof.Theorem 21.
Let(xn)n∈N be a sequence in E, x∈E and e∈E+. Then xn→stx(e) if and only if xn→st-ordx and (xn)n∈N is a statistically e-uniform Cauchy sequence.Proof.
It suffices to prove the sufficient part. Assume thatxn→st-ordx and (xn)n∈N is a statistically e-uniform Cauchy sequence. By [15, Theorem 5] and Theorem 7, we obtain a subset K=(kj)j∈N of N with δ(K)=1 such that the sequence (xkj)j∈N converges in order to x and is e-uniform Cauchy. Let ϵ>0. There exists a natural number j0 such that (45)xki-xkj≤ϵe,∀i,j≥j0.Let us fix any j≥j0. Letting i→∞, we get (46)x-xkj≤ϵe.This means that xkj→rux(e). It follows from Theorem 7 that xn→stx(e).Theorem 22.
LetE 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 xn→st-ord0. Then, by (1), there exists a sequence (λn)n∈N of reals with 0≤λn↑∞ such that λnxn→st-ord0. By Remark 19, there exists a sequence (pn)n∈N with pn↓0 such that (47)λnxn≤pn,a.a.n.This means(48)xn≤1λnp1,a.a.n.
Hence,xn→st0(p1) and then xn→sru0.
(2) ⇒ (1). Assume that xn→st-ord0. By (2), we see that xn→sru0. Then, there exist e∈E+ and a sequence ϵn↓0 such that (49)xn≤ϵne,a.a.n.Thus, we get 1/ϵnxn→st-ord0. 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 16.18, p. 90]. For each n∈N, we define fn∈C[0,1] by fn(t)=1 on 0,1/n and fn(t)=0 on 2/n,1 and fn(t) is linear on [1/n,2/n]. Clearly, fn↓0. Moreover, there is no sequence 0≤λn↑∞ such that λnfn→st-ord0. Otherwise, it follows from [15, Theorem 5] that there exists a subset K={kj:j∈N} of N with δ(K)=1 such that the sequence (λkjfkj)j∈N is order convergent 0. Hence, there exists a sequence (pj)j∈N in E with pj↓0 so that λkjfkj≤pj for all j∈N. This implies (50)1=fkj0≤1λkjpj0≤1λkjp10⟶0j⟶∞.This contradiction suggests that statistical order convergence in C[0,1] is not stable. Then Theorem 22 gives the conclusion.Following [12], a sequence (xn)n∈N in a Banach space E is said to benorm statistically convergent to x∈E provided that δn∈N:xn-x≥ϵ=0 for all ϵ>0. It is clear that a sequence (xn)n∈N in a Banach lattice E is norm statistically convergent whenever it is statistically relatively uniformly convergent. The sequence (xn)n∈N is said to beweakly statistically convergent to x provided that the sequence (〈x∗,xn-x〉)n∈N converges statistically to 0 for each x∗∈X∗. It is noted in [12] that if a sequence (xn)n∈N in a Banach space E is norm statistically convergent to x∈E, then there exists a subset K=(kj)j∈N⊆N with δ(K)=1 such that (xkj)j∈N converges in norm to x. Combining this observation, Theorems 11 and 7 with [14, Theorem 100.4], we obtain the following result.Theorem 24.
Let(xn)n∈N be a sequence in a Banach lattice E, x∈E and e∈E+. Then xn→stx(e) if and only if (xn)n∈N is norm statistically convergent to x and (xn)n∈N is a statistically e-uniform Cauchy sequence.It is known [18] that, in C(K)-spaces, the relatively uniform convergence coincides with the norm convergence. It follows from Theorem 7 that, in C(K)-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.
LetK=k2:k∈N. For each n∈N, we define fn∈C[0,1] by(51)fn=1,n∈Kgn,n∉K,where gn(t)=tn,t∈[0,1],n=1,2,….
Obviously,gn↓0 and hence fn→st-ord0. Since gn=1 for all n∈N, we see that the sequence (fn)n∈N 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(fn)n∈N in [14, Exercise 100.13] as follows.
LetE be the normed Riesz space of all real continuous functions on [0.1] with f=∫01|f|dt. We set(52)f112=1,f1=12,0≤f1≤1,f213=f223=1,f2=122,0≤f2≤1,and generally(53)fn1n+1=fn2n+1=⋯=fnnn+1=1,fn=12n,0≤fn≤1,for n=1,2,….
It is proved in [14, Exercise 100.13] that the sequence (fn)n∈N converges to 0 in norm and no subsequence of it converges to 0 in order.
We setK={k2:k∈N} and let(54)gn=1,n∈Kfn,n∉K.Clearly, (gn)n∈N is norm statistically convergent to 0. But (gn)n∈N is not statistically order convergent. Indeed, if (gn)n∈N is statistically order convergent to 0, it follows from [15, Theorem 5] that there exists a subset A of N with δ(A)=1 such that (gn)n∈A is order convergent to 0. Hence, the subsequence (fn)n∈A∩(N\K)=(gn)n∈A∩(N\K) is order convergent to 0. This is a contradiction.Example 27.
We take the sequence(x(r))r∈N in [20, Example 5] as follows.
Let(55)E=xnn∈N∈∑n⊕l∞∞:xn∈c0except  for  finitely  manyn.Then E is ideal in ∑n‍⊕l∞∞. Define x(r)=(xn(r))n∈N∈E for r=1,2,… by(56)xnrk=1rk=1,2,…,forn≤r,xnr=0,forn>r.
Thenx(r)=1/r→0(r→∞), but no subsequence of (x(r))r∈N is relatively uniformly convergent.
Then, as in Example26, we set K={k2:k∈N} and let(57)yr=0,r∈Kxr,r∉K.Clearly, (y(r))r∈N is norm statistically convergent to 0. An argument similar to Example 26 shows that (y(r))r∈N 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].
Pickg∈L1[0,1] such that g is continuous on (0,1],g(t)>0 for all t∈(0,1], and limt→0+⁡g(t)=∞. Let E be the Riesz subspace of L1[0,1] consisting of all f such that, for some δ(f)>0 and γ(f)∈R, we have f(t)=γ(f)g(t) for all 0≤t<δ(f). Define a Riesz norm · on E by setting (58)f=fL1+γf.Moore Jr. exhibited a sequence (fn)n∈N in E such that fn↓0 and limn→∞⁡fn=0, but fn↛0(ru), as follows.
Letf0=g. Define fn by induction so that(a)
fn is continuous on (0,1],(b)
0≤fn≤fn-1,(c)
fn1/k=1/kf0(1/k),k=1,2,…,n,(d)
fn(t)=1/n+1g(t),t∈(0,1/n+1],(e)
fnL1≤1/n+1gL1.
This implies that no subsequence of(fn)n∈N converges relatively uniformly to 0.
Similarly, we setK={k2:k∈N} and let(59)gn=0,n∈Kfn,n∉K.Clearly, (gn)n∈N is norm statistically convergent to 0 and gn→st-ord0. But Theorem 7 yields that (gn)n∈N is not statistically relatively uniformly convergent to 0. This is because no subsequence of (fn)n∈N converges relatively uniformly to 0.Theorem 29.
Let(xn)n∈N be a monotone sequence in a Banach lattice E. Then (1)
(xn)n∈N is norm statistically convergent if and only if it is norm convergent;(2)
(xn)n∈N is weakly statistically convergent if and only if it is weakly convergent;(3)
(xn)n∈N is norm statistically convergent if and only if it is weakly statistically convergent.In order to prove Theorem29, we need two simple lemmas. The first lemma is an immediate consequence of [21, Proposition 1.1.6].Lemma 30.
If a sequence(xn)n∈N in a Banach lattice E is increasing and converges norm statistically to x∈E, then x=supn∈N⁡xn.Actually, Lemma30 also holds for weakly statistical convergence as shown in the following.Lemma 31.
If a sequence(xn)n∈N in a Banach lattice E is increasing and converges weakly statistically to x∈E, then x=supn∈Nxn.Proof.
Letx∗∈(E∗)+. Then there exists a strictly increasing sequence (kj)j∈N of natural numbers so that limj→∞⁡x∗,xkj=〈x∗,x〉. For each n∈N, we choose kj0 with kj0>n. Then, we have (60)x∗,xn≤x∗,xkj,∀j>j0.This means that (61)x∗,xn≤x∗,x.Hence, x is an upper bound of (xn)n∈N.
Assume thatu is an arbitrary upper bound of (xn)n∈N. Then, for every x∗∈(E∗)+, we have (62)x∗,xn≤x∗,u,∀n∈N.
Since the sequence(〈x∗,xn〉)n∈N converges statistically to (〈x∗,x〉)n∈N, we have (63)x∗,x≤x∗,u.This implies that x≤u and we are done.Proof of Theorem29.
(1) Suppose that(xn)n∈N is increasing and norm statistically convergent to x∈E. By Lemma 30, we get x=supn∈N⁡xn. Let (kj)j∈N be a strictly increasing sequence of natural numbers so that limj→∞⁡xkj=x in norm. Let ϵ>0. Choose a natural number j0 such that xkj0-x<ϵ. Then, for any n>kj0, we have(64)0≤x-xn≤x-xkj0.This yields (65)x-xn≤x-xkj0<ϵ.Hence, limn→∞⁡xn=x in norm.
(2) Let(xn)n∈N be increasing and weakly statistically convergent to x∈E. By Lemma 31, we get x=supn∈Nxn. Let x∗∈(E∗)+. Since (xn)n∈N converges weakly statistically to x, (〈x∗,xn〉)n∈N converges statistically to 〈x∗,x〉. There exists a subset K⊆N with δ(K)=1 such that limn∈K⁡〈x∗,xn〉=〈x∗,x〉. Let ϵ>0. Choose N∈K such that |〈x∗,xN〉-〈x∗,x〉|<ϵ. Then, for all n>N, we have (66)0≤x-xn≤x-xN.This implies that(67)0≤x∗,x-xn≤x∗,x-xN<ϵ.Hence, (xn)n∈N converges weakly to x.
(3) follows from (1) and (2).



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