# Density by Moduli and Statistical Boundedness

Authors: Vinod K. Bhardwaj; Shweta Dhawan; Sandeep Gupta
Journal: Abstract and Applied Analysis
 (2016)

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## Abstract

We have generalized the notion of statistical boundedness by introducing the concept off-statistical boundedness for scalar sequences where f is an unbounded modulus. It is shown that bounded sequences are precisely those sequences which are f-statistically bounded for every unbounded modulus f. A decomposition theorem for f-statistical convergence for vector valued sequences and a structure theorem for f-statistical boundedness have also been established.


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## Body

## 1. Introduction and Background

The idea of statistical convergence was given in the first edition (published in Warsaw in 1935) of the monograph of Zygmund [1]. Formally the concept of statistical convergence was introduced by Steinhaus [2] and Fast [3] and later reintroduced by Schoenberg [4]. Although statistical convergence was introduced over nearly last 60 years, it has become an active area of research in recent years. Statistical convergence has been studied most recently by several authors [5–15].The standard definition of “(xk) is convergent to L” requires that the set {k∈N:|xk-L|≥ɛ} should be finite for every ɛ>0, where N is the set of natural numbers.The number sequence(xk) is said to be statistically convergent to the number L provided that the set {k∈N:|xk-L|≥ɛ}, instead of being finite, has natural density 0, where the natural density of a subset K⊂N (see [16], chapter 11) is defined by(1)dK=limn→∞⁡1nk≤n:k∈K,where {k≤n:k∈K} denotes the number of elements of K not exceeding n. A set K is said to be statistically dense [17] if d(K)=1. A subsequence of a sequence is said to be statistically dense if the set of all indices of its elements is statistically dense. In other words, a subsequence of a sequence is said to be statistically dense if the complement of the set of all indices of its elements has natural density 0. Obviously, we have d(K)=0 provided that K is a finite set of positive integers.We will be particularly concerned with those subsets ofN which have natural density zero. To facilitate this, Fridy [18] introduced the following notation: if x=(xk) is a sequence such that xk satisfies property P for all k except a set of natural density zero, then we say that x=(xk) satisfies P for “almost all k” and we abbreviate this by “a.a. k.”Using this notation, we have the following.Definition 1.
The number sequencex=(xk) is said to be statistically convergent to the number L if, for each ɛ>0, (2)dAɛ=0,whereAɛ=k∈N:xk-L≥ɛ,that  is,limn→∞⁡1nk≤n:xk-L≥ɛ=0,that  is,xk-L<ɛa.a.k.Definition 2.
A number sequencex=(xk) is said to be statistically Cauchy if, for each ɛ>0, there exists a positive integer p=p(ɛ) such that(3)dk∈N:xk-xp≥ɛ=0,that  is,xk-xp<ɛa.a.k.In 1997, Fridy and Orhan [19] introduced the concept of statistical boundedness as follows.Definition 3.
The number sequencex=(xk) is said to be statistically bounded if there is a number M>0 such that d({k∈N:|xk|>M})=0; that is,(4)xk≤Ma.a.k.We denote the set of all statistically bounded sequences by S(b).In the same year, that is, 1997, Tripathy [20] proved a decomposition theorem for statistically bounded sequences and also established a necessary and sufficient condition for a sequence to be statistically bounded.Quite recently, Bhardwaj and Gupta [10] have introduced and studied the concepts of statistical boundedness of order α, λ-statistical boundedness, and λ-statistical boundedness of order α for scalar sequences. Bhardwaj et al. [11] have also introduced and studied a new concept of lacunary statistical boundedness as a lacunary analog of the concept of statistical boundedness.The idea of a modulus function was structured by Nakano [21] in 1953. Following Ruckle [22] and Maddox [23], we recall that a modulus f is a function from [0,∞) to [0,∞) such that (i) f(x)=0 if and only if x=0, (ii)f(x+y)≤f(x)+f(y) for x≥0,y≥0, (iii) f is increasing, and (iv) f is continuous from the right at 0. Hence f must be continuous everywhere on [0,∞). A modulus may be unbounded or bounded. For example, f(x)=xp, where 0<p<1, is unbounded, but f(x)=x/(1+x) is bounded.Connor [24], Pehlivan [25], Pehlivan and Fisher [26], Kolk [27], Ghosh and Srivastava [28], Bhardwaj and Singh [29, 30], Çolak [31], Altin and Et [32], and some others have used a modulus function to construct some sequence spaces.In the year 2014, Aizpuru et al. [6] have defined a new concept of density with the help of modulus function and consequently obtained a new concept of f-statistical convergence, which is in fact a generalization of the notion of statistical convergence. They proved that the ordinary convergence is equivalent to the f-statistical convergence for every unbounded modulus function f.Quite recently, Bhardwaj and Dhawan have introduced and studied the concepts off-statistical convergence of order α [9] and f-lacunary statistical convergence [33] by using the approach of Aizpuru et al. [6].Throughout the paper, unless otherwise specified,X will denote a real normed space.First we recall some definitions from [6].Definition 4.
Letf be an unbounded modulus function. The f-density of a set A⊂N is defined by(5)dfA=limn⁡fk≤n:k∈Afn,in case this limit exists. Clearly, finite sets have zero f-density and df(N-A)=1-df(A) does not hold, in general.Remark 5.
For any unbounded modulusf and A⊂N, df(A)=0 implies d(A)=0 (see [6]). But the converse need not be true in the sense that a set having natural density zero may not have f-density zero with respect to some unbounded modulus f.Definition 6.
A sequence(xk) in X is said to be f-statistically convergent to L∈X, if, for each ɛ>0, df({k∈N:xk-L≥ɛ})=0 and one writes it as f-stlimxk=L.In view of Definition6 and Remark 5, it follows that every f-statistically convergent sequence is statistically convergent but a statistically convergent sequence need not be f-statistically convergent for every unbounded modulus f.Definition 7.
A sequence(xk) in X is said to be f-statistically Cauchy if, for each ɛ>0, there exists a positive integer p=p(ɛ) such that(6)dfk∈N:xk-xp≥ɛ=0.We now introduce the following notation: ifx=(xk) is a sequence such that xk satisfies property P for all k except a set of f-density zero, then we say that x=(xk) satisfies P for “almost all k with respect to f,” where f is any unbounded modulus and we abbreviate this by “a.a. k w.r.t. f.”Using this notation, the definitions off-statistical convergence and f-statistically Cauchy can be reformulated as follows.Definition 8.
AnX-valued sequence (xk) is said to be f-statistically convergent to L∈X, if, for each ɛ>0, (7)xk-L<ɛa.a.kw.r.t.f.Definition 9.
AnX-valued sequence (xk) is said to be f-statistically Cauchy if, for each ɛ>0, there exists a positive integer p=p(ɛ) such that (8)xk-xp<ɛa.a.kw.r.t.f.The main object of this paper is to introduce and study a new concept off-statistical boundedness for scalar sequences defined as follows.Definition 10.
A number sequence(xk) is said to be f-statistically bounded if there exists M>0 such that df({k∈N:|xk|>M})=0; that is, |xk|≤Ma.a.kw.r.t.f. By sf(b), one will denote the space of all f-statistically bounded scalar sequences.In this paper, we establish a relation between statistical boundedness andf-statistical boundedness. It is shown that the concept of f-statistical boundedness is intermediate between the ordinary boundedness and the statistical boundedness. We also prove that bounded sequences are precisely those sequences which are f-statistically bounded for every unbounded modulus f. Apart from studying f-statistical boundedness we also propose to derive further properties concerning f-statistical convergence. We establish a decomposition theorem for f-statistical convergence. It is proved that the terms of an f-statistically convergent sequence (xk) are coincident with those of a convergent sequence for almost all k with respect to f. We also show that l∞f, the set of all bounded f-statistically convergent sequences of scalars, is a closed linear subspace of the normed linear space l∞ of all bounded sequences of scalars and hence is a nowhere dense set in l∞.


## 2. Some More Results onf-Statistical Convergence

Throughout this section, unless otherwise specified, we deal withX-valued sequences. By sf(X), we will denote the space of all X-valued f-statistically convergent sequences.We begin this section by establishing a decomposition theorem forf-statistical convergence. In fact, the decomposition theorem for statistically convergent sequences of scalars was given by Connor [34]. The following theorem extends the decomposition theorem of Connor [34] to f-statistical convergence.Theorem 11 (decomposition theorem).
Ifx=(xk) is f-statistically convergent to L, then there is a sequence y converging to L and an f-statistically null sequence z such that x=y+z. Moreover, if x is bounded then z is also bounded and z∞≤x∞+L.Proof.
Asx=(xk) is f-statistically convergent to L, there exists A⊂N with df(A)=0 such that limk∈N-Axk=L. For k∈N, let (9)yk=xk,ifk∈N-A;L,ifk∈A,zk=0,ifk∈N-A;xk-L,ifk∈A.Clearly x=y+z. As {k∈N:zk-0>ϵ}⊂A for every ϵ>0, we have df({k∈N:zk-0>ϵ})=0 for every ϵ>0. Thus z=(zk) is an f-statistically null sequence and z∞≤x∞+L, if x is bounded.
Fork∈N, we have(10)yk-L=xk-L,ifk∈N-A;0,ifk∈A;and so(11)k∈N:yk-L>ϵ⊂k∈N:xk-L>ϵ∩N-A.As limk∈N-Axk=L, the set on right hand side of (11) is finite for each ɛ>0 and hence limkyk=L.Remark 12.
If we denote the space of allX-valued convergent and f-statistically null sequences by c(X) and sf0(X), respectively, then in view of Theorem 11 we have sf(X)=c(X)+sf0(X). Moreover sf(X)≠c(X)⊕sf0(X) as c(X)∩sf0(X)⊃c0(X), the space of X-valued null sequences.Fridy [18] proved that, in case of scalar sequences, every statistically convergent sequence has entries coincident with those of a convergent sequence for almost all k. We establish a similar result for f-statistical convergence of X-valued sequences, which includes the above stated result of Fridy [18].Theorem 13.
A sequencex=(xk) is f-statistically convergent if and only if there exists a convergent sequence y=(yk) such that xk=yka.a.kw.r.t.f.Proof.
First supposex=(xk) is an f-statistically convergent sequence. Proceeding on the same lines as in Theorem 11, we get a convergent sequence y=(yk) with df({k∈N:xk≠yk})≤df(A)=0; that is, xk=yka.a.kw.r.t.f. Conversely, it is given that there exists a convergent sequence y=(yk) such that xk=yka.a.kw.r.t.f. Now(12)k∈N:xk-L>ϵ⊂k∈N:xk≠yk∪k∈N:yk-L>ϵ.Since limkyk=L, the latter set on the right hand side of (12) is finite. Therefore df({k∈N:xk-L>ϵ})=0 for every ϵ>0. Thus x=(xk) is an f-statistically convergent sequence.As an immediate consequence of Theorem13, we have the following.Corollary 14.
Ifx=(xk) is a sequence such that f-stlim⁡xk=L, then x has a subsequence y=(yk) such that limkyk=L.Aizpuru et al. [6] proved that, in a Banach space X, (xn) is f-statistically Cauchy if and only if (xn) is f-statistically convergent. Combining this result with Theorem 13, we get the following result which includes Theorem 7 of Fridy [18].Theorem 15.
LetX be a Banach space and f an unbounded modulus. Then the following are equivalent: (a)
(xk) is f-statistically convergent.(b)
(xk) is f-statistically Cauchy.(c)
There exists a convergent sequencey=(yk) such that xk=yka.a.kw.r.t.f.Remark 16.
We know that every subsequence of a convergent sequence is convergent but this is no longer true in case off-statistical convergence; that is, an f-statistically convergent sequence may have a subsequence which is not f-statistically convergent. This can be verified by the following example.Example 17.
ConsiderX=C, the space of complex numbers, and f(x)=xp with 0<p≤1. Let (xn)=(1,0,0,4,0,0,0,0,9,…). Now df({i∈N:|xi-0|>ϵ})=df(A) for every ϵ>0 where A={1,4,9,…}. Then |A(n)|=|{k≤n:k∈A}|≤n for every n∈N and so (13)fAnfn≤n1/2pnp⟶0asn⟶∞;that is, df(A)=0. Thus (xn) is f-statistically convergent, whereas (1,4,9,…) is a subsequence of (xn) which is not f-statistically convergent.Definition 18.
A subsequence of a sequence(xk) is said to be f-statistically dense if the complement of the set of all indices of its elements has f-density zero.Remark 19.
Everyf-statistically dense subsequence of a sequence (xk) is statistically dense.Burgin and Duman [17] proved that a number sequence (xk) is statistically convergent if and only if every statistically dense subsequence of it is statistically convergent. We extend this result to the following.Theorem 20.
A sequence(xk) is f-statistically convergent if and only if every f-statistically dense subsequence of it is f-statistically convergent.The proof is similar to Theorem2.1 of Burgin and Duman [17] and hence is omitted.Corollary 21.
f-statistically dense subsequences of an f-statistically convergent sequence are f-statistically convergent.The following theorem shows that continuous functions preserve thef-statistical convergence of sequences.Theorem 22.
Iff-stlim⁡xk=L and g(x) defined for all x∈X is continuous at L, then f-stlim⁡g(xk)=g(L).Proof.
Asf-stlim⁡xk=L, it follows by Theorem 3.1 of Aizpuru et al. [6] that there exists a set A⊂N such that df(A)=0 and limk∈N-Axk=L. As g is continuous at L, limk∈N-Ag(xk)=g(L). Using Theorem 3.1 of Aizpuru et al. [6], the result follows.Šalát [35] proved that the set m0 of all bounded statistically convergent sequences of real numbers is a closed linear subspace of the normed linear space m of all bounded sequences of real numbers. We establish a similar result for l∞f, the set of all bounded f-statistically convergent sequences of scalars.Theorem 23.
The setl∞f is a closed linear subspace of the normed linear space l∞.The proof is similar to Theorem2.1 of Šalát [35] and hence is omitted.The above theorem provides us with the following information related to the structure of the setl∞f.Theorem 24.
The setl∞f is a nowhere dense set in l∞.Since the sequence((-1)k)∈l∞ does not belong to l∞f, the proof follows from the fact that every proper closed linear subspace of an arbitrary normed linear space E is a nowhere dense set in E.


## 3.f-Statistical Boundedness

In this section we show that the concept off-statistical boundedness is intermediate between the ordinary boundedness and the statistical boundedness. Some of the results of this section include the corresponding earlier results of Bhardwaj and Gupta [10] on statistical boundedness. f-statistical analog of monotone convergence theorem is established in Theorem 38. Theorem 40 shows that a sequence which is f-statistically bounded for each modulus f is also bounded in the ordinary sense.Throughout this section, we deal with the sequences of scalars.Theorem 25.
Every bounded sequence isf-statistically bounded; however, the converse need not be true.Proof.
The result follows in view of the fact that empty set has zerof-density for every unbounded modulus f. For the converse part, the sequence x=(xk)=(1,0,0,4,0,0,0,0,9,…) of Example 17 serves the purpose.Theorem 26.
Everyf-statistically bounded sequence is statistically bounded.The proof follows in view of the fact thatA⊂N, df(A)=0 implies d(A)=0.Remark 27.
The converse of the above theorem need not be true which can be verified by the following example.Example 28.
Letf(x)=log⁡(x+1) and (xk)=(1,0,0,4,0,0,0,0,9,…). Let A={1,4,9,…}= the set of squares of natural numbers. For any M>0, (14)k∈N:xk>M=A-afinitesubsetofN.Since df(A)=1/2≠0 and d(A)=0, (xk)∉sf(b) and (xk)∈s(b). Consequently, sf(b)⊊s(b).Remark 29.
From Theorems25 and 26 and Example 28, we get l∞⊊sf(b)⊊s(b); that is, f-statistical boundedness is intermediate between the ordinary boundedness and the statistical boundedness and agrees with the statistical boundedness when f=I, the identity mapping.Aizpuru et al. [6] proved that f-stlimxk=L if and only if there exists A⊂N with df(A)=0 and limk∈N-Axk=L. We also establish a similar structure theorem for f-statistically bounded sequences; however, our proof is quite different from that of [6].Theorem 30 (structure theorem).
A sequencex=(xk) is f-statistically bounded if and only if there exists A⊂N such that df(A)=0 and (xk)k∈N-A∈l∞.Proof.
First, we suppose(xk) is f-statistically bounded. So there exists M>0 such that df({k∈N:|xk|>M})=0. Take A={k∈N:|xk|>M}. Then df(A)=0 and for k∈N-A, we have |xk|≤M; that is, (xk)k∈N-A∈l∞. Conversely, it is given that there exists A⊂N such that df(A)=0 and (xk)k∈N-A∈l∞. As (xk)k∈N-A∈l∞ there exists M>0 such that |xk|≤M for all k∈N-A. This implies that {k∈N:|xk|>M}⊂A and so df({k∈N:|xk|>M})=0. Hence (xk)k∈N is f-statistically bounded.Remark 31.
Let(xk)∈sf(b). Then there exists M>0 such that df({k∈N:|xk|>M})=0. Take A={k∈N:|xk|>M}. For k∈N, let(15)yk=xk,ifk∈N-A;0,ifk∈A,zk=0,ifk∈N-A;xk,ifk∈A.Clearly y=(yk)∈l∞ and z=(zk)∈sf0, the space of f-statistically null sequences. Here xk=yk+zk for k∈N. Thus sf(b)⊂l∞+sf0. As l∞,sf0⊂sf(b) and sf(b) is a linear space, we have sf(b)=l∞+sf0.It is easy to note thatsf(b)≠l∞⊕sf0 as l∞∩sf0≠{0}. In fact l∞∩sf0⊃c0, the space of null scalar sequences.Remark 32.
A subsequence of anf-statistically bounded sequence need not be f-statistically bounded. The sequence x=(xk)=(1,0,0,4,0,0,0,0,9,…) of Example 17 is f-statistically bounded whereas (1,4,9,…) is a subsequence of it which is not f-statistically bounded.We now characterizef-statistically bounded sequences in terms of their subsequences.Theorem 33.
A sequence isf-statistically bounded if and only if every f-statistically dense subsequence of it is f-statistically bounded.The proof is easy in view of Theorem2.1 of Burgin and Duman [17] and hence is omitted.Theorem 34.
Everyf-statistically convergent sequence is f-statistically bounded; however, the converse need not be true.Proof.
The proof follows from the fact that{k∈N:|xk|>|L|+ϵ}⊂{k∈N:|xk-L|>ϵ}. For the converse part, taking f=I, identity map, and xk=(-1)k, we get (xk)∈sf(b); however (xk)∉sf, the space of f-statistically convergent sequences of scalars.The following theorem shows that everyf-statistically bounded sequence has entries coincident with those of a bounded sequence for almost all k with respect to f.Theorem 35.
A sequencex=(xk) is f-statistically bounded if and only if there exists a bounded sequence y=(yk) such that xk=yka.a.kw.r.t.f.The proof is easy and hence omitted.Theorem 36.
Everyf-statistically Cauchy sequence is f-statistically bounded; however, the converse need not be true.Theorem 37.
(a)sf(b) is normal and hence monotone.
(b)sf(b) is a sequence algebra.
(c)sf(b) is not symmetric, in general.Theorem 38.
Every monotone andf-statistically bounded sequence is f-statistically convergent.Proof.
Let(xk) be a monotone and f-statistically bounded sequence. By Theorem 30, there exists A⊂N with df(A)=0 such that (xk)k∈N-A∈l∞. So there exists L∈C such that limk∈N-Axk=L. Using Theorem 3.1 of [6], we have (xk)∈sf.Lemma 39 (see [6]).
IfA⊂N is infinite, then there exists an unbounded modulus f such that df(A)=1.The following theorem shows that a sequence which isf-statistically bounded for each modulus f is also bounded in ordinary sense.Theorem 40.
If, for every unbounded modulusf, (xk)∈sf(b), then (xk)∈l∞.Proof.
Suppose, if possible,(xk)∉l∞. Then for every M>0, we have that A={k∈N:|xk|>M} is an infinite set and so by Lemma 39, there exists an unbounded modulus f such that df(A)=1 which contradicts the assumption that (xk)∈sf(b) for every modulus f.Remark 41.
From Theorem25, we have l∞⊂sf(b) for every unbounded modulus f. Using this and Theorem 40, we can say that bounded sequences are precisely those sequences which are f-statistically bounded for every unbounded modulus f.



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