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Kyungpook Mathematical Journal 2019; 59(1): 47-63

Published online March 23, 2019

### A Note on Spliced Sequences and A-density of Points with respect to a Non-negative Matrix

Kumardipta Bose*, and Sayan Sengupta

Department of Mathematics, Jadavpur University, Kolkata 700032, India
e-mail: gabuaktafaltu6ele@hotmail.com and mathematics.sayan1729@gmail.com

Received: October 12, 2017; Revised: October 3, 2018; Accepted: October 10, 2018

For y ∈ ℝ, a sequence x = (xn) ∈ ℓ, and a non-negative regular matrix A, Bartoszewicz et. al., in 2015, defined the notion of the A-density δA(y) of the indices of those xn that are close to y. Their main result states that if the set of limit points of (xn) is countable and density δA(y) exists for any y ∈ ℝ where A is a non-negative regular matrix, then limn→∞(Ax)n = ∑y∈ℝδA(y) · y. In this note we first show that the result can be extended to a more general class of matrices and then consider a conjecture which naturally arises from our investigations.

Keywords: non-negative matrix, A-density, A-limit,ideal convergence, in nite splice, conjecture (*).

### 1. Introduction

We start by recalling the definition of natural density. For n, m ∈ ℕ with n < m, let [n, m] denote the set {n, n + 1, n + 2, …, m}. Let A ⊂ ℕ. Define

d¯(A)=limsupnA[1,n]n         and         d_(A)=liminfnA[1,n]n.

The numbers (A) and d(A) are called the upper natural density and the lower natural density of A, respectively. If (A) = d(A), then this common value is called the natural density of A and we denote it by d(A). Let ℐd be the family of all subsets of ℕ which have natural density 0. This ℐd is a proper nontrivial admissible ideal of subsets of ℕ. The notion of natural density was used by Fast [5] and Scoenberg [18] to define the notion of statistical convergence. Details of statistical convergence and later on, ideal convergence are thoroughly described in [1, 3, 4, 6, 7, 8, 10, 11, 12, 14, 15, 17, 19].

In [16] Osikiewicz had developed the ideas of finite and infinite splices. Let E1, E2, E3, …, Ek, … be a partition of ℕ into countable number of sequences. Let y1, y2, y3, …, yk, … be distinct real numbers. Let (xn) be such that

limn,nEixn=yi.

Then (xn) is called an infinite-splice (In the same way Osikiewicz defined an finite splice taking finite number of sequences and finite number of distinct real numbers). He proved the following:

### Theorem 1.1. ([16, Simplified version of Osikiewicz Theorem])

Assume that (xn) is a splice over a partition {Ei}. Letyi=limn,nEixn. Assume that d(Ei) exists for each i and

id(Ei)=1.

Then

limn1nk=1nxk=iyi·d(Ei).

In fact Osikiewicz considered a more general case, namely a regular matrix summability method A and A-density the details of which are presented in the preliminaries. Very recently in [2] a new approach was made to study the general version of Osikiewicz Theorem by defining the notion of the A-density of a point and an alternative version of the same result was established. In fact it was shown that the assumptions of Osikiewicz Theorem imply those of the following Theorem:

### Theorem 1.2

Suppose that x = (xn) is a bounded sequence, δA(y) exists for every y ∈ ℝ andyDδA(y)=1. Then

limn(Ax)n=yDδA(y)·y.

and consequently Osikiewicz result is a consequence of Theorem 1.2.

A natural question arises whether the result is only true for regular matrices. Note that one of the main condition of regularity of a non-negative matrix A = (an, k) is that limnk=1an,k=1. In this note we first show that actually the result can be extended to a larger class of matrices A = (an, k) satisfying limnk=1an,k< using almost the same arguments used in [2]. The main observation which leads to Theorem 2 is that if (xn) ∈ ℓ and δA(y) exists for all y ∈ ℝ, then D:= {y ∈ ℝ: δA(y) > 0} is countable. We produce an example of a matrix A = (an, k) with limnk=1an,k= and a bounded sequence (xn) for which δA(y) exists for all y ∈ ℝ but D:= {y ∈ ℝ: δA(y) > 0} is uncountable. This surprising example naturally gives rise to the following conjecture:

### Conjecture (★)

For any matrix A = (an, k) with limnk=1an,k= there exists a bounded sequence (xn) for which δA(y) exists for all y ∈ ℝ and D:= {y ∈ ℝ: δA(y) > 0} is uncountable.

In the last section of the note we deal with this conjecture essentially showing that it is false however showing that with imposition of certain conditions on the matrix the conjecture becomes true.

### 2. Preliminaries

We first present the necessary definitions and notations which will form the background of this article.

If x = (xn) is a sequence and A = (an, k) is a summability matrix, then by Ax we denote the sequence ((Ax)1, (Ax)2, (Ax)3, …) where (Ax)n=k=1an,kxk. The matrix A is called regular if limnxn=L implies limn(Ax)n=L. The well-known Silverman-Toeplitz theorem characterizes regular matrices in the following way. A matrix A is regular if and only if

• limnan,k=0,

• limnk=1an,k=1,

• supnk=1an,k<.

For a non-negative regular matrix A and E ⊂ ℕ, following Freedman and Sember [9], the A-density of E, denoted by δA(E), is defined as follows

δA¯(E)=limsupnkEan,k=limsupnk=1an,k1E(k)=limsupn(A1E)n,δA_(E)=liminfnkEan,k=liminfnk=1an,k1E(k)=liminfn(A1E)n

where 1E is a 0–1 sequence such that 1E(k) = 1 ⇔ kE. If δA¯(E)=δA_(E) then we say that the A-density of E exists and it is denoted by δA(E). Clearly, if A is the Cesàro matrix i.e.

an,k={1nif nk0otherwise

then δA coincides with the natural density.

Throughout by ℓ we denote the set of all bounded sequences of reals.

In [2] in a new approach, the authors had defined for a sequence (xn) a density δA(y) of indices of those xn which are close to y which was not dealt with till then in the literature. This was a more general approach than that of Osikiewicz [16].

Fix (xn) ∈ ℓ. For y ∈ ℝ let

δA¯(y)=limɛ0+δA¯({n:xn-yɛ})

and

δA_(y)=limɛ0+δA_({n:xn-yɛ}).

If δA¯(y)=δA_(y), then the common value is denoted by δA(y).

The main result of [2] was the following.

### Theorem 2.1

Let x = (xn) ∈ ℓ. Suppose that the set of limit points of (xn) is countable and δA(y) exists for any y ∈ ℝ where A is a non-negative regular matrix. Then

limn(Ax)n=yδA(y)·y.

Now recall that a non-empty family ℐ of subsets of ℕ is an ideal in ℕ if for A, B ⊂ ℕ, (i) A, B ∈ ℐ ⇒ AB ∈ ℐ; (ii)A ∈ ℐ, BAB ∈ ℐ. Further if AJA= ie. {n} ∈ ℐ ∀ n ∈ ℕ, then ℐ is called admissible or free. An ideal ℐ is called a P-ideal if for any sequence of sets (Dn) from ℐ, there is another sequence of sets (Cn) in ℐ such that DnCn is finite for all n and nCnJ. Equivalently if for each sequence (An) of sets from ℐ there exists A ∈ ℐ such that An A is finite for all n ∈ ℕ then ℐ becomes a P-ideal.

For a bounded sequence (xn), we now recall the following definitions (see [13]):

• (xn) is ℐ convergent to y if for any ɛ > 0, {n: |xny| ≥ ɛ} ∈ ℐ.

• A point y is called an ℐ-cluster point of (xn) if {n: |xny| ≤ ɛ} ∉ ℐ for any ɛ > 0.

• y is called an ℐ-limit point of (xn) if there is a set B ⊂ ℕ, B ∉ ℐ, such that limnBxn=y.

In this note we primarily consider non-negative matrices A = (an, k) satisfying

• an, k ≥ 0 for all n, k;

• limnan,k=0, for all k;

• limnk=1an,k<.

One should note that one can not replace the limn in (iii) above by supn for the simple fact that in that case if one considers a matrix A having infinitely many zero rows, like the matrix A = (an, k) where

an,k={1if n=kand nis even0otherwise

then even δA(ℕ) does not exist.

Throughout the next section by a non-negative matrix A we will always mean a matrix satisfying the above three conditions. It should be noted that the notion A-density can be similarly defined as in the case of regular matrices and all the axiomatic conditions, as stated in [9] are satisfied here except for the fact that the density of the set of natural numbers ℕ is now a finite real number, not necessarily 1. Further it is easy to check that {E ⊂ ℕ: δA(E) = 0} forms a P-ideal of ℕ. The proof being very similar to the case of regular matrices (see [2]) is omitted here.

### 3. Results for Non-negative Matrices

The main result which we are going to establish in this paper is the following.

### Theorem 3.1

Let x = (xn) ∈ ℓ, and the set of limit points of (xn) is countable. Let A be a non-negative matrix as defined before (necessarily withsupnk=1an,k<). Suppose δA(y) exists for all y ∈ ℝ. Then

limn(Ax)n=yδA(y)·y

### Lemma 3.2

Let (xn) ∈ ℓand δA(y) exists for all y ∈ ℝ. Then D:= {y ∈ ℝ: δA(y) > 0} is countable and

yDδA(y)M.

wheresupnk=1an,k=M.

Proof

Let (rn) be a strictly decreasing sequence converging to M. For m ∈ ℕ let

Dm:={yδA(y)1m}.

Note that D1D2 ⊂ … ⊂ Dm ⊂ … and D=mDm. Now if y1, y2, …, ylDm be distinct, let us choose ɛ=minijyi-yj3>0. Consequently the sets Ei = {n: xn ∈ (yiɛ, yi + ɛ)} are pairwise disjoint. Moreover

δA_(Ei)δA(yi)1mliminfkEian,k1m.

Then for any τ > 0, ∃ n1 ∈ ℕ such that kEian,k>1m-τnn1. Again limsupnk=1an,k<rnn. So for any fixed rp, we get n2 ∈ ℕ such that k=1an,k<M+δ<rpnn2 and for a suitably chosen δ. Let n0 = max{n1, n2}. As Ei’s are disjoint, we have

ki=1lEian,k=i=1,kEilan,klm-lτnn0.

But

ki=1lEian,kk=1an,k<rp.

Now note that lm-lτrplmrp1-τm. Hence choosing τ so that 1 − τm > 0 we observe that l must be finite. Thus Dm is finite for each m which implies that D=mDm can be at most countable.

Again

yDmδA(y)i=1lδA_(Ei)=i=1lliminfnkEian,ki=1l(kEian,k+ɛ0l)         nN(for some N)

where ɛ0 is arbitrary. So

yDmδA(y)ki=1lEian,k+ɛ0k=1an,k+ɛ0rp

for suitably chosen ɛ0. Finally in view of the fact that D=mDm we get

yDδA(y)=limmyDmδA(y)rp.

Since this is true for any rp, so letting p→∞ we get yDδA(y)M.

In [2] it was observed that generally it is not true that D = {y ∈ ℝ: δA(y) > 0} should be nonempty as also D¯:={y:δA¯(y)>0} need not be countable.

The next result extends Theorem 6 [2].

### Theorem 3.3

Let (xn) be a bounded sequence and A be a non-negative matrix such that δA(y) exists for every y ∈ ℝ and moreoveryDδA(y)=M. Then

limn(Ax)n=yDδA(y)·y.
Proof

Since (xn) is bounded, there exists a K > 0 such that |xn| ≤ K for every n ∈ ℕ. Let D = {yi: i = 1, 2, …} where yi’s are distinct. Let ɛ > 0 be given and let r ∈ ℕ be such that

i=1rδA(yi)>M-ɛand i=r+1yi·δA(yi)<ɛ.

Let N ∈ ℕ be such that

13minijyi-yj>1N

i, j ∈ 1, 2, …, r and such that the sets Ei={j:xj-yi<1N} have the following property

δA(yi)-ɛr(k+1)δA_(Ei)δA¯(Ei)δA(yi)+ɛr(k+1)

for i = 1, 2, …, r. Obviously E1, …, Er are pairwise disjoint. Now we can choose a m0 ∈ ℕ such that

δA_(Ei)-1N<kEian,k<δA¯(Ei)+1NkEian,k-δA(yi)<1N+ɛr(k+1)

for every nm0 and i = 1, 2, …, r. Then for nm0 we have

(Axn)=k=1an,kxkkE1an,k(y1+1N)+kE2an,k(y2+1N)++kEran,k(yr+1N)+K.k(E1Er)can,k.

Now we can choose m1 > m0 such that ∀nm1

k=1an,k<M+ɛ.

Then

M+ɛ>k=1an,k=ki=1rEian,k+k(i=1rEi)can,k

and consequently

ki=1rEian,k=i=1rkEian,k>i=1rδA(yi)-rN-ɛK+1.

Therefore for nm1 we have

k(i=1rEi)can,k<(M+ɛ)-(M-rN-(1+1K+1)ɛ)=rN+(2+1K+1)ɛ.

Subsequently we get for nm1,

(Ax)nkE1an,k(y1+1N)+kE2an,k(y2+1N)++kEran,k(yr+1N)+Krn+(2+1K+1)ɛK.

Analogously

(Ax)nkE1an,k(y1-1N)+kE2an,k(y2-1N)++kEran,k(yr-1N)-Krn-(2+1K+1)ɛK.

Thus

(Ax)n-i=1rkEian,k(yi+1N)Krn+(2+1K+1)ɛK

and

(Ax)n-i=1rkEian,k(yi-1N)-Krn-(2+1K+1)ɛK.

Hence using (1) and (2) we obtain

(Ax)n-i=1δA(yi).yi=(Ax)n-i=1rδA(yi).yi-i=r+1δA(yi).yi(Ax)n-i=1rδA(yi).yi+i=r+1δA(yi).yi(Ax)n-i=1rδA(yi).yi+ɛ=[(Ax)n-i=1rkEian,k(yi+1N)]+i=1r[kEian,k(yi+1N)-δA(yi).yi]+ɛi=1r[(kEian,k-δA(yi))(yi+1N)]+1Ni+1rδA(yi)+KrN+(2K+KK+1+1)ɛi=1r[(kEian,k-δA(yi))(yi+1N)]+Mn+KrN+(2K+KK+1+1)ɛr(1N+ɛr(K+1)).(K+1+1N)+Mn+KrN+(2K+KK+1+1)ɛ.

Analogously from (1) and (3) we get

(Ax)n-i=1δA(yi).yi-r(1N+ɛr(K+1)).(K+1+1N)-Mn-KrN-(2K+KK+1+1)ɛ.

Since N can be chosen arbitrarily large, we obtain

(Ax)n-i=1δA(yi).yi(2K+KK+1+1)ɛ

for every ɛ > 0. Therefore we can conclude that

limn(Ax)n=i=1δA(yi)·yi.

Next we prove the following result which is a variant of the corresponding result of [2] forming the basis of a necessary condition for the existence of a limit of {(Ax)n}.

### Proposition 3.4

Suppose x = (xn) ∈ l. If δA(y) = M, then My is a limit point of the sequence {(Ax)n}.

Proof

Since (xn) is bounded, there is K > 0 such that |xn| ≤ KN ∈ ℕ. Let y be such that δA¯(y)=M. Let N ∈ ℕ and let EN={j:xj-y<1N}. Then there exists kNN such that

kENakN,k>δA¯(EN)-1N=M-1N.

Again as we have limsupnk=1an,k=M, we get

k=1akN,k<M+1N.

Since y-1N<xk<y+1NxkEN and −KxkKxkEN, so we have

kENakN,k(y-1N)-k(EN)cakN,k·K(Ax)kNkENakN,k(y+1N)+k(EN)cakN,k·K.

Thus

y(k=1akN,k-M)-1NkENakN,k-k(EN)cakN,k(K+y)(Ax)kN-Myy(k=1akN,k-M)+1NkENakN,k+k(EN)cakN,k(K-y)

and consequently

(Ax)kN-MykENakN,k.1N+k(EN)cakN,k(K+y)+1Ny.

Since

k(EN)cakN,k=k=1akN,k-kENakN,k<M+1N-(M-1N)=2N

so

(Ax)kN-My(MN+1N2)+k(EN)cakN,k(K+y)+1Ny(MN+1N2)+2N(K+y)+1Ny.

Therefore

limN(Ax)kN=My.

### Corollary 3.5

Let (xn) be a bounded sequence. Suppose that there are y and z (yz) with δA(y) = δA(z) = M. Then the limitlimn(Ax)ndoes not exist.

Now we recall some important results from [2] which will be useful in the sequel.

### Lemma 3.6.([2])

Letbe an ideal of subsets of. Assume that X:= {n: xn ∈ [a, b]} ∉ ℐ. Suppose that

{n:axnt-ɛ}Jor{n:t+ɛxnb}J

for any t ∈ (a, b) and any ɛ > 0 such that ɛ < min{ta, bt}. Then there is y ∈ [a, b] such that {n: |xny| ≥ ɛ} ∈ ℐ for every ɛ > 0.

### Proposition 3.7.([2])

Letbe a P-ideal. Assume that (xn) ∈ ℓdoes not have any-limit points. Then the set of limit points of (xn), i.e. the set

{y:xnky   forsomeincreasingsequence(nk)ofnaturalnumbers},

is uncountable and closed.

### Corollary 3.8.([2])

Let [a, b] be a fixed interval andbe a P-ideal. Assume that {n: xn ∈ [a, b]} ∉ ℐ and any point y ∈ (a, b) is not an-limit point of (xn). Then the set of limit points of (xn) in [a, b], i.e. the set

{y(a,b):xnky   forsomeincreasingsequence(nk)ofnaturalnumbers},

is uncountable and closed.

### Corollary 3.9.([2])

Let (xn) ∈ ℓ. Assume that the set of limit points of (xn) is countable. Then the sequence (xn) has at least one-limit point for every P-ideal ℐ.

We can easily prove the following results analogous to the results of [2] which will help us to reach our final goal.

### Lemma 3.10

Let r ∈ (0, 1), r1r2r3 ≥ …, limnrn=rand let (En) be a decreasing sequence of subsets of ℕ.

• IfδA(En) = rn, n ∈ ℕ, then there is a subset E ofwithδA(E) = r and such that E*En, n ∈ ℕ, i.e. E En is finite for every n ∈ ℕ. Moreover, ifδA¯(En)r, then δA(E) = r.

• IfδA¯(En)=rn, n ∈ ℕ, then there is a subset E ofwithδA¯(E)=rand such that E*En, n ∈ ℕ.

### Theorem 3.11

Let (xn) ∈ ℓ. A point y ∈ ℝ is an A-statistical limit point of (xn) if and only ifδA¯(y)>0. Moreover if δA(y) > 0, then there is E ⊂ ℕ with δA(E) = δA(y) andlimnExn=y.

### Corollary 3.12

Let (xn) ∈ ℓ. A point y ∈ ℝ is an A-statistical cluster point of (xn) and it is not an A-statistical limit point if and only if

• δA¯({j:xj-y1/n})>0for every n;

• δA(y)=limnδA¯({j:xj-y1/n})=0.

### Proposition 3.13

Let (xn) ∈ ℓ. Assume that y1, y2, … are the only distinct real numbers such that δA(yi) > 0 ∀i. Then there exists a partition E1, E2, … such that δA(Ei) = δA(yi), i = 1, 2, … andlimnEixn=yi.

### Lemma 3.14

Assume that {En: n = 1, 2, … } is a partition ofsuch thatn=1δA(En)<M. Then there is a partition {Fn: n = 0, 1, 2, … } ofsuch that

• FnEn;

• δA(Fn) = δA(En);

• δA(F0)=M-n=1δA(En).

Finally we prove a sufficient condition for a bounded sequence (xn) to have the property that yδA(y)=M.

### Theorem 3.15

Let (xn) be a bounded sequence. Suppose that the set of limit points of (xn) is countable and δA(y) exists for all yD. ThenyDδA(y)=M.

Proof

Suppose that, on the contrary,

yDδA(y)<M.

Then by Corollary 14 in [2] the set D is non-empty. By Lemma 3.2 the set D is countable. Enumerate D as {y1, y2, … }. By Proposition 3.13 there is a partition {Ek: k = 1, 2, … } of ℕ such that δA(Ek) = δA(yk) and limn,nEkxn=yk. By Lemma 3.14 there is a partition {Fk: k = 0, 1, 2, … } such that FkEk, δA(Fk) = δA(Ek) for every k ≥ 1, F0=i=1Fi so that δA(F0)=M-k=1δA(Fk). Thus δA(F0) > 0. Consider the sequence (xn)nF0 and the ideal ℐA|F0 = {EF0: E ∈ ℐA}. Since δA(y) = 0 for every yD, so by Theorem 3.11 y can not be an A-statistical limit point of (xn) and so can not be an ℐA|F0-limit point of (xn)nF0 for any yDc. If yi was ℐA|F0-limit point of (xn)nF0, then there would be a set BF0 such that B ∉ ℐA|F0, and limnBxn=yi. But then B ∉ ℐA and since BFi = ∅︀ and limnBFixn=yi, we would have δA(y)=δA¯(y)δA¯(BFi)>δA¯(Fi)=δA(y), which gives a contradiction. Therefore the sequence (xn)nF0 has no ℐA|F0-limit points.

Note that ℐA|F0 is a P-ideal. To see this assume that A1, A2, · · · ∈ ℐA|F0. Then A1, A2, · · · ∈ ℐA and as ℐA is a P-ideal, we can find A ∈ ℐA such that An A is finite for every n. Since AF0 ∈ ℐA|F0 and each AnF0, then An (AF0) is finite for every n. Now by Proposition 14 in [2] applied to the sequence (xn)nF0 and P-ideal ℐA|F0, we obtain that the sequence (xn)nF0 has uncountably many limit points which are, in turn, limit points of (xn). This contradicts the assumption.

Finally combining Theorem 3.3 with Theorem 3.15 we get the desired proof of our main result.

### 4. Conjecture (★)

Throughout the paper we have considered the matrices with the following properties.

• an, k ≥ 0 for all n, k;

• limnan,k=0, for all k;

• limnk=1an,k<.

Now the natural question arises whether the results discussed above remain true if any one of the conditions for the matrix can be relaxed. The first condition is clearly essential.

Coming to the second condition, if limnan,k>0 for some k, then note that the axiomatic condition in [9] namely the condition ’for E1, E2 ⊂ ℕ such that E1E2 is finite, δA(E1) = δA(E2)’, which is very crucial in defining a density function does not hold anymore. For example, let

A=[11001121200113131300......11n1n1n0....]

Here an,1 → 1 as n→∞. We take E1 = ℕ, E2 = ℕ {1}. Then E1E2 = {1}, but δA(E1) = 2 whereas δA(E2) = 1.

Finally let us relax condition (iii) and assume that limnk=1an,k=. Observe that Lemma 3.6 forms the backbone of our main result as only then the sequence can be partitioned into countably infinitely many splices. Below we produce an example of a matrix A with limnk=1an,k= and a bounded real sequence (xn) for which {y ∈ ℝ: δA(y) > 0} is uncountable, i.e. Lemma 3.2 is not true when M = ∞.

### Proposition 4.1

Define the matrix A = (an, k) in the following way

an,k={1g(n)ifnk0otherwise

where g: ℕ → (0, ∞) is defined as g(k) = n + 1 if k ∈ [2n, 2n+1). Then there is a bounded sequence (xn) such that {y ∈ ℝ: δA(y) > 0} is uncountable.

Proof

Let (xn) be defined as follows:

x1=12,x2=14,x3=34,x4=18,x5=38,x6=58,x7=78,         etc

In general

x2n+k=2k+12n+1         for         k[0,2n).

Now consider a dyadic interval I=[k2p,k+12p). Then I contains 2np elements from {x2n+k: k ∈ [0, 2n)}. Therefore

δA{m:xmI}liminfm1g(m){l[0,2n):x2n+lI}=limn1n+22n-p=

where n is the largest natural number such that 2n+1m. Let y ∈ [0, 1]. Note that for every ɛ > 0 there is a dyadic interval I=[k2p,k+12p) contained in (yɛ, y + ɛ). Thus δA(y) = ∞ for every y ∈ [0, 1] (consequently δA(y) exists and it equals ∞). As for each point y ∈ ℝ [0, 1], δA(y) = 0 so we can conclude that δA(y) exists for all y ∈ ℝ and {y: δA(y) > 0} = [0, 1] is uncountable.

In view of the above example one can naturally think of the following conjecture.

### Conjecture (★)

For any matrix A = (an, k) with

• an, k ≥ 0 for all n, k,

• limnan,k=0 for all k,

• limnk=1an,k=.

there exists a bounded sequence (xn) for which δA(y) exists for all y ∈ ℝ and D:= {y ∈ ℝ: δA(y) > 0} is uncountable.

### Proposition 4.2

There is a non-negative matrix A = (an, k) satisfying all the above properties (as prescribed in Conjecture (★) such that for any bounded sequence (xn) for which δA(y) exists for all y ∈ ℝ the set D = {y: δA(y) > 0} is a singleton.

Proof

Let the matrix A = (an, k) be such that

an,k={nif n=k0otherwise

Clearly limnk=1an,k=. Let (xn) be a bounded sequence. First note that isolated points of the sequence have density 0 with respect to matrix A. Now if (xn) has more than one limit point, let y be one of them. Choose another limit point z of the sequence (xn). Let we choose 0<ɛ<y-z2 and consider the set Bɛ = {k: xk ∈ (yɛ, y +ɛ)}. Observe that the set Bɛc={k:xk(y-ɛ,y+ɛ)} must be infinite, or else z would not be a limit point of (xn). Let Bɛc={n1,n2,}. So kBɛani,k=0 for all i = 1, 2, …. This implies that liminfnkBɛan,k=0. But it is easy to see that limsupnkBɛan,k=. So δA(y) does not exist. This is true for all limit points of (xn). Finally if (xn) has only one limit point y, say, i.e. (xn) is convergent to y0 then obviously δA(y0) > 0 and D = {y: δA(y) > 0} = {y0}.

Finally one can ask for which matrices the Conjecture (★) is true ? The following result shows that the Conjecture is valid for matrices A satisfying certain conditions.

### Proposition 4.3

Let A = (an, k) be a non-negative matrix satisfying the conditions of Conjecture (★). In addition let there exist a sequence of positive real numbers(δk)k=1with the following properties.

• For any k, min{a11, a12, …, a1k} ≥ δk;

• limnδn.2n-p>0, for any p < n.

Then there exists a bounded sequence (xn) such that the set {y: δA(y) > 0} is uncountable.

Proof

Let (xn) be defined as in Proposition 4.1. Now consider a dyadic interval I=[k2p,k+12p). Then I contains 2np elements from {x2n+k: k ∈ [0, 2n)}. Therefore

δA{m:xmI}liminfmδm.{l[0,2n):x2n+lI}=limnδn2n-p=

where n is the largest natural number such that 2n+1m. Let y ∈ [0, 1]. Note that for every ɛ > 0 there is a dyadic interval I=[k2p,k+12p) contained in (yɛ, y + ɛ). Thus δA(y) = ∞ for every y ∈ [0, 1] (consequently δA(y) exists and it equals ∞). As for each point y ∈ ℝ [0, 1], δA(y) = 0 so we can conclude that δA(y) exists for all y ∈ ℝ and {y: δA(y) > 0} = [0, 1] is uncountable.

### Problem 1

Find the necessary and sufficient conditions or some other condition for a matrix A for the Conjecture (★) to be true.

### Acknowledgements

The authors are thankful to Prof. Pratulananda Das for his guidance during the preparation of this paper and of course to the referee for his/her valuable remarks.

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