Relative uniform convergence of difference double sequence of positive linear functions

Abstract

In this article we introduce the notion of relative uniform convergence of difference double sequence of positive linear functions. We define the difference double sequence spaces  $_2\ell _\infty (\varDelta , ru),~ _2c(\varDelta , ru),~ _2c_0(\varDelta , ru), _2{c_0}^B(\varDelta , ru),~_2c^B(\varDelta , ru), _2c^R(\varDelta , ru), ~_2{c_0}^R(\varDelta , ru)$ and study their topological properties.

1 Introduction

Throughout the article $_2\omega , ~_2\ell _\infty (\varDelta , ru),~ _2c(\varDelta , ru), ~_2c_0(\varDelta , ru),~_2c^R(\varDelta ,ru),~_2{c_0}^B(\varDelta ,ru),~_2{c_0}^R(\varDelta ,ru)$ and $_2c^B(\varDelta , ru)$ denote all sequence space, relative uniform bounded, relative uniform convergence, relative uniform null, regular relative uniform convergence, relative uniform null bounded, regular relative uniform null , relative uniform bounded convergence difference double sequence space respectively.

A double sequence is a double infinite array of numbers by $(x_{nk})$. The notion of double sequence was introduced by Pringsheim [15]. Some earlier works on double sequence spaces are found in Bromwich [2]. Hardy [11] introduced the notion of regular convergence of double sequence. The double sequence has been investigated from different aspects by Basarir and Sonalcan [1], Das et al. [6], Datta and Tripathy [7, 8], Tripathy and Sarma [19] and many others.

The notion of uniform convergence of sequence of functions relative to a scale function was introduced by E. H. Moore. Chittenden [3,4,5] gave a formulation of the definition given by Moore as follows:

Definition 1.1

A sequence $(f_n )$ of real, single-valued functions $f_n$ of a real variable x, ranging over a compact subset D of real numbers, converges relatively uniformly on D in case there exist functions g and $\sigma$, defined on D, and for every $\varepsilon >0$, there exists an integer $n_o$ (dependent on $\varepsilon )$ such that for every $n\ge n_o$, the inequality

$$\begin{aligned} \mid g(x)-f_n (x)\mid <\varepsilon \mid \sigma (x)\mid , \end{aligned}$$

holds for every element x of D.

The function $\sigma$ of the above definition is called a scale function.

The notion was further studied by many others [9, 10, 16].

Kizmaz [13] defined the difference sequence spaces $\ell _\infty (\varDelta ), c(\varDelta ), c_0(\varDelta )$ as follows:

$$\begin{aligned} Z(\varDelta )= \left\{ {x=(x_k):(\varDelta x_k) \in Z}\right\} , \end{aligned}$$

for $Z = \ell _\infty ,~ c,~ c_0$ where $\varDelta x_k= x_k-x_{k+1}, k\in N$.

These sequence spaces are Banach space under the norm

$$\begin{aligned} ||(x_k)||_\varDelta =\mid x_1\mid +\begin{array}{c} {sup}\\ {k\in N} \end{array}|\varDelta x_k|. \end{aligned}$$

Tripathy and Sarma [20, 21] studied difference double sequence spaces and their topological properties.

The notion was further studied from different aspects by Tripathy [17], Tripathy and Goswami [18], Jena et al. [12], Paikray et al. [14] and many others.

A double sequence $(x_{nk})$ is said to be convergent in Pringshiem’s sense if

$$\begin{aligned} \lim _{{n,k}\rightarrow \infty } x_{nk}=M, ~\text{ exists } \text{ where }~ n,k\in N. \end{aligned}$$

A double sequence $(x_{nk})$ is said to converge regularly if it is convergent in Pringsheim’s sense to limit M and the following limit exists:

$$\begin{aligned}&\lim _{{n}\rightarrow \infty } x_{nk}=P_k, ~\text{ exists } \text{ for } \text{ each }~ k\in N; \\&\lim _{{k}\rightarrow \infty } x_{nk}=L_n, ~\text{ exists } \text{ for } \text{ each }~ n\in N. \end{aligned}$$

2 Preliminaries

Definition 2.1

A subset E of the set of all double sequence $_2w$ is said to be solid or normal if $(f_{nk}(x))\in E \Rightarrow (\alpha _{nk}f_{nk}(x))\in E,$ for all $(\alpha _{nk})$ of sequence of scalars with $\mid \alpha _{nk}\mid \le 1$, for all $n,k \in N$.

Definition 2.2

Let

$$\begin{aligned} K= & {} \Bigl \{(n_i,k_j):i, j\in N;n_1<n_2<n_3<\cdots ~\text{ and }~\\&k_1<k_2<k_3<\cdots \Bigr \}\subseteq N\times N \end{aligned}$$

and E be a subset of the set of all double sequence $_2w$. A K-step space of E is a sequence space

$$\begin{aligned} \lambda _K^E=\{(f_{n_ik_j}(x))\in _2\omega :(f_{nk}(x))\in E\}. \end{aligned}$$

A canonical pre-image of a sequence of functions $(f_{n_ik_j}(x))\in E$ is a sequence of functions $(g_{nk}(x))\in E$ defined by

$$\begin{aligned} g_{nk}(x)= {\left\{ \begin{array}{ll} f_{nk}(x), &{} \text{ if }~ (n,k)\in K ;\\ \theta , &{} \text{ otherwise }. \end{array}\right. } \end{aligned}$$

Definition 2.3

A double sequence space E is said to be monotone if it contains the canonical pre-images of all its step spaces.

Remark 2.1

From the above notions, it follows that if a sequence space E is solid then, E is monotone.

Definition 2.4

A double sequence space E is said to be symmetric if

$(f_{nk}(x))\in E\Rightarrow (f_{\pi (n, k)}(x))\in E$, where $\pi$ is a permutation of N.

Definition 2.5

A difference double sequence of functions $(\varDelta f_{nk}(x))$ defined on a compact domain D is said to be relatively uniformly convergent if there exists a function $\sigma (x)$ defined on D and for every $\varepsilon >0$, there exists an integer $n_0=n_0(\varepsilon )$ such that

$$\begin{aligned} \mid \varDelta f_{nk}(x)-f(x)\mid <\varepsilon \mid \sigma (x)\mid , \end{aligned}$$

for all $n,k\ge n_0$ holds for every element x of D. The difference operator $\varDelta$ is defined by $\varDelta f_{nk}(x) = \varDelta f_{nk}(x) - \varDelta f_{nk}(x) - \varDelta f_{nk}(x)$, for all $n, k \in N$.

Remark 2.2

When $f=\theta$, the zero function, we get the definition of null relative uniform from the above definition.

Definition 2.6

A difference double sequence of functions $(\varDelta f_{nk}(x))$ defined on a compact domain D is said to be regular relative uniform convergent if there exist functions $g(x), g_k(x), f_n(x), \sigma (x), \xi _n(x), \eta _k(x)$ defined on D, for every $\varepsilon >0,$ there exists an integer $n_0=n_0(\varepsilon )$ such that for all $x\in D$,

$$\begin{aligned}&\mid \varDelta f_{nk}(x)-g(x)\mid<\varepsilon \mid \sigma (x)\mid , ~\text {for all}~ n,k\ge n_0; \\&\mid \varDelta f_{nk}(x)-g_k(x)\mid<\varepsilon \mid \eta _k(x)\mid , ~\text {for each} ~k\in N ~\text {and for all}~ n\ge n_0; \\&\mid \varDelta f_{nk}(x)-f_n(x)\mid <\varepsilon \mid \xi _n(x)\mid , ~\text {for each} ~n\in N~\text { and for all}~ k\ge n_0. \end{aligned}$$

Remark 2.3

When $g=g_k=f_n=\theta$, the zero function, we get the definition of regular null relative uniform from the above definition.

We introduce the following difference double sequence spaces defined over the normed space $(D, ||.||_{(\varDelta , \sigma )})$.

$$\begin{aligned} Z(\varDelta , ru)=\left\{ (f_{nk}(x)) :(\varDelta f_{nk}(x))\in Z ~\text {relative uniformly}~\text { w.r.t.}~ \sigma (x)\right\} , \end{aligned}$$

where $Z=_2\ell _\infty , ~_2{c_0}^B,~_2c^B,~ _2c^R ~_2{c_0}^R, _2c$ and $_2c_0$.

The above sequence spaces are normed by the norm defined by

$$\begin{aligned} ||f(x)||_{(\varDelta , \sigma )}= & {} \begin{array}{c} {\sup }\\ {n\ge 1} \end{array}\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||f_{n1}(x)||~ ||\sigma (x)||}{||x||}\\&+\begin{array}{c} {\sup }\\ {k\ge 1} \end{array}\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||f_{1k}(x)||~ ||\sigma (x)||}{||x||}+\begin{array}{c} {\sup }\\ {n\ge 1;k\ge 1} \end{array}~\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||\varDelta f_{nk}(x)||~ ||\sigma (x)||}{||x||} \end{aligned}$$

3 Main Results

In this section we establish the results of this article.

Theorem 3.1

The sequence spaces $Z(\varDelta , ru)$ where, $Z=_2\ell _\infty , ~_2{c_0}^B,~_2c^B,~ _2c^R$ and $~_2{c_0}^R$ are normed linear spaces.

Proof

We established the theorem for the case of $_2\ell _\infty (\varDelta , ru)$.

Let $\alpha$ and $\beta$ be the scalars and $(f_{nk}(x)) \in _2\ell _\infty (\varDelta , ru) , (g_{nk}(x))\in _2\ell _\infty (\varDelta , ru)$.

Then, we have

$$\begin{aligned}&||(\varDelta f_{nk}(x)\sigma (x))||\le M_1||x||, \quad \text {for all}~x\in D ~\text {and} \\&||(\varDelta g_{nk}(x)\sigma (x))|| \le M_2||x||,\quad \text {for all}~x\in D. \end{aligned}$$

Without the loss of generality, we can consider the same scale function for the sequence of functions $(f_{nk}(x))$ and $(g_{nk}(x))$.

We have,

$$\begin{aligned}&||(\varDelta (\alpha f_{nk}(x))+\beta g_{nk}(x) )\sigma (x))||\le \mid \alpha \mid ||(\varDelta f_{nk}(x)\sigma (x))||\\&\qquad +\mid \beta \mid ||(\varDelta g_{nk}(x) \sigma (x))||\\&\quad \le |\alpha | M_1||x||+|\beta |M_2||x||\\&\quad \le (|\alpha | M_1+|\beta | M_2)||x||,~ \alpha , \beta \hbox { are absolutely scalable}. \end{aligned}$$

Hence, $_2\ell _\infty (\varDelta , ru)$ is a linear space.

Next we prove that $_2\ell _\infty ( \varDelta , ru)$ is a normed space:

(i) Clearly $||f(x)||_{(\varDelta , \sigma )}\ge 0$, for all $x\in D$.

$$\begin{aligned}&||f(x)||_{(\varDelta ,\sigma )}= \Biggl \{\begin{array}{c} {\sup }\\ {n\ge 1} \end{array}\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||(f_{n1}(x)\sigma (x))||}{||x||}\\&\quad +\begin{array}{c} {\sup }\\ {k\ge 1} \end{array}\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||(f_{1k}(x)\sigma (x))||}{||x||}+ \begin{array}{c} {\sup }\\ {n\ge 1;k\ge 1} \end{array}~\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||(\varDelta f_{nk}(x)\sigma (x))||}{||x||}\Biggr \}=0 \end{aligned}$$

$\Rightarrow f_{n1}(x)\sigma (x)=0, f_{1k}(x)\sigma (x)=0~ \text {and}~ \varDelta f_{nk}(x)\sigma (x)=0$.

Therefore, $f(x)=(f_{nk}(x))=0$, since $\sigma (x)\ne 0$, for $x\in D$.

Conversely, let $f(x)=0$.

Then, we have $||f(x)||_{(\varDelta , ru)}=0$.

(ii)

$$\begin{aligned} ||\alpha f(x)||_{(\varDelta , \sigma )}= & {} \Biggl \{\begin{array}{c} {\sup }\\ {n\ge 1} \end{array}\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||(\alpha f_{n1}(x)\sigma (x))||}{||x||}\\&+\begin{array}{c} {\sup }\\ {k\ge 1} \end{array}\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||(\alpha f_{1k}(x)\sigma (x))||}{||x||}+ \begin{array}{c} {\sup }\\ {n\ge 1;k\ge 1} \end{array}~\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||(\alpha \varDelta f_{nk}(x)\sigma (x))||}{||x||}\Biggr \} \end{aligned}$$

$\Rightarrow ||\alpha f(x)||_{(\varDelta , \sigma )}=\mid \alpha \mid ||f(x)||_{(\varDelta ,\sigma )}.$

(iii) It can be easily verified that

$$\begin{aligned} ||f(x)+g(x)||_{(\varDelta , \sigma )}\le ||f(x)||_{(\varDelta ,\sigma )}+|| g(x)||_{(\varDelta ,\sigma )}. \end{aligned}$$

Since the three conditions of norms are satisfied, $_2\ell _\infty (\varDelta , ru)$ is a normed space.

Similarly, we can show that the other sequence spaces are also normed linear space.

$\square$

Theorem 3.2

Let $(D, ||.||_{(\varDelta , \sigma )})$ be a complete normed space. The sequence spaces $Z(\varDelta , ru)$ where, $Z=_2\ell _\infty , ~_2{c_0}^B,~_2c^B,~ _2c^R$ and $~_2{c_0}^R$ are complete.

Proof

Let $(f^i(x))=(f^i_{nk}(x))$ be a Cauchy sequence in $_2\ell _\infty (\varDelta , ru)$.

For a given $\varepsilon >0$, there exists $n_0\in N$ such that for all $x\in D$,

$||f^i(x)-f^j(x)||_{(\varDelta , \sigma )}\le \frac{\varepsilon }{3}$, for all $i,j\ge n_0$.

Then,

$$\begin{aligned}&\Biggl \{\begin{array}{c} {\sup }\\ {n\ge 1} \end{array}\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||(f^i_{n1}(x)-f^j_{n1}(x))\sigma (x)||}{||x||}\nonumber \\&\quad +\begin{array}{c} {\sup }\\ {k\ge 1} \end{array}\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||(f^i_{1k}(x)-f^j_{1k}(x))\sigma (x)||}{||x||}\nonumber \\&\quad +\begin{array}{c} {\sup }\\ {n\ge 1;k\ge 1} \end{array}~\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||(\varDelta f^i_{nk}(x)-\varDelta f^j_{nk}(x))\sigma (x)||}{||x||}\Biggr \}\le \frac{\varepsilon }{3}, \end{aligned}$$

(1)

for all $i,j \ge n_0$.

$\Rightarrow (f^i_{n1}(x))$ is a is relative uniform Cauchy in D w.r.t. $\sigma (x)$, for all $x\in D$ and for each $n\in N$.

$\Rightarrow (f^i_{n1}(x))$ converges relatively uniformly in D w.r.t. $\sigma (x)$, for all $x\in D$ and for each $n\in N$.

Let

$$\begin{aligned} \begin{array}{c} {\lim }\\ {i\rightarrow \infty } \end{array}f^i_{n1}(x)=f_{n1}(x), x\in D \end{aligned}$$

(2)

Similarly,

$$\begin{aligned}&\begin{array}{c} {\lim }\\ {i\rightarrow \infty } \end{array}f^i_{1k}(x)=f_{1k}(x), x\in D \end{aligned}$$

(3)

$$\begin{aligned}&\begin{array}{c} {\lim }\\ {i\rightarrow \infty } \end{array} \varDelta f^i_{nk}(x)=\varDelta f_{nk}(x), x\in D \end{aligned}$$

(4)

From the above three equations we get,

$$\begin{aligned} \begin{array}{c} {\lim }\\ {i\rightarrow \infty } \end{array}f^i_{nk}(x)=f_{nk}(x), ~\text {for all}~ x \in D \end{aligned}$$

(5)

From Eq. (1), we have for all $i, j\ge n_0$ and for all $x\in D$,

$$\begin{aligned} \begin{array}{c} {\lim }\\ {j\rightarrow \infty } \end{array}\frac{||(f^i_{n1}(x)-f^j_{n1}(x))\sigma (x)||}{||x||}=\frac{||(f^i_{n1}(x)-f_{n1}(x))\sigma (x)||}{||x||}\le \frac{\varepsilon }{3}, \end{aligned}$$

$\text {for each}~ n\in N$.

Similarly,

$$\begin{aligned} \begin{array}{c} {\lim }\\ {j\rightarrow \infty } \end{array}\frac{||(f^i_{1k}(x)-f^j_{1k}(x))\sigma (x)||}{||x||}=\frac{||(f^i_{1k}(x)-f_{1k}(x))\sigma (x)||}{||x||}\le \frac{\varepsilon }{3}, \end{aligned}$$

$\text {for each}~ k\in N$ and

$$\begin{aligned} \begin{array}{c} {\lim }\\ {j\rightarrow \infty } \end{array}\frac{||(\varDelta f^i_{nk}(x)-\varDelta f^j_{nk}(x))\sigma (x)||}{||x||}=\frac{||(\varDelta f^i_{nk}(x)-\varDelta f_{nk}(x))\sigma (x)||}{||x||}\le \frac{\varepsilon }{3}, \end{aligned}$$

$\text {for all}~ n,k\in N.$

Since $\frac{\varepsilon }{3}$ is not dependent on n and k we have,

$$\begin{aligned}&\begin{array}{c} {\sup }\\ {n\ge 1} \end{array}\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||(f^i_{n1}(x)-f_{n1}(x))\sigma (x)||}{||x||}\le \frac{\varepsilon }{3}; \\&\begin{array}{c} {\sup }\\ {k\ge 1} \end{array}\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||(f^i_{1k}(x)-f_{1k}(x))\sigma (x)||}{||x||}\le \frac{\varepsilon }{3}; \\&\begin{array}{c} {\sup }\\ {n\ge 1;k\ge 1} \end{array}~\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||(\varDelta f^i_{nk}(x)-\varDelta f_{nk}(x))\sigma (x)||}{||x||}\le \frac{\varepsilon }{3}. \\&\text{ Evidently, }~||f^i_{nk}(x)-f_{nk}(x)||_{(\varDelta , \sigma )}=\Biggl \{\begin{array}{c} {\sup }\\ {n\ge 1} \end{array}\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||(f^i_{n1}(x)-f_{n1}(x))\sigma (x)||}{||x||} \nonumber \\&\quad +\begin{array}{c} {\sup }\\ {k\ge 1} \end{array}\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||(f^i_{1k}(x)-f_{1k}(x))\sigma (x)||}{||x||} \\&\quad +\begin{array}{c} {\sup }\\ {n\ge 1;k\ge 1} \end{array}~\begin{array}{c} {\sup }\\ {||x||\le 1} \end{array}\frac{||(\varDelta f^i_{nk}(x)-\varDelta f_{nk}(x))\sigma (x)||}{||x||}\Biggr \}\le \varepsilon . \end{aligned}$$

$\Rightarrow ||f^i(x)-f(x)||_{(\varDelta , \sigma )}\le \varepsilon$, for all $i\ge n_0$, for all $x\in D$.

Therefore, $(f^i(x)-f(x))\in _2\ell _\infty (\varDelta , ru)$, for all $i\ge n_0$, for all $x\in D$.

Then, $f(x)=f^i(x)-(f^i(x)-f(x))\in _2\ell _\infty (\varDelta , ru)$ since, $_2\ell _\infty (\varDelta , ru)$ is a linear space.

Hence, $_2\ell _\infty (\varDelta , ru)$ is complete.

We can show that the other sequence spaces are also complete in the same process.$\square$

Result 3.1

The sequence spaces $Z(\varDelta , ru)$ where, $Z=_2\ell _\infty , ~_2{c_0}^B,~_2c^B,~ _2c^R, ~_2{c_0}^R, _2c$ and $_2c_0$ are not monotone.

Proof

The proof follows from the following example:

Example 3.1

Consider the sequence of functions $(f_{nk}(x)), f_{nk}:[0,1]\rightarrow R$ defined by

$$\begin{aligned} f_{nk}(x)= & {} {\left\{ \begin{array}{ll} \frac{1}{x}, &{} \text {when}~ x\in (0,1] ;\\ 0, &{} \text {when}~x=0. \end{array}\right. } \\ \varDelta f_{nk}(x)= & {} f_{nk}(x)-f_{n+1,k}(x)-f_{n,k+1}(x)+f_{n+1,k+1}(x)=0. \end{aligned}$$

$(\varDelta f_{nk}(x))$ converge uniformly to zero function on [0, 1] w.r.t. the constant scale function $\sigma (x)=1$.

Then, $(\varDelta f_{nk}(x))\in Z(\varDelta , ru)$ where, $Z=_2\ell _\infty , ~_2{c_0}^B,~_2c^B,~ _2c^R, ~_2{c_0}^R, _2c$ and $_2c_0$.

Let us consider the sequence of functions $(g_{nk}(x))$, the pre-image of the sequence of functions $(f_{nk}(x))$ defined by

$$\begin{aligned}&g_{nk}(x) = {\left\{ \begin{array}{ll} f_{nk}(x),&{} \text {if}~ n ~\text {and}~ k ~\text {both are odd};\\ 0, &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

The difference sequence of functions $(\varDelta g_{nk}(x))$ of the sequence of functions $(g_{nk}(x))$ is given by

$$\begin{aligned} \varDelta g_{nk}(x)= {\left\{ \begin{array}{ll} \frac{1}{x},&{} \text {when}~ n+k ~\text {is even}, ~\text {for all}~x\in (0,1];\\ -\frac{1}{x},&{} \text {when}~ n+k ~\text {is odd}, ~\text {for all}~x\in (0,1]. \end{array}\right. } \end{aligned}$$

$(\varDelta g_{nk}(x))$ is not null uniformly w.r.t. any scale function $\sigma (x)$.

Then, $(\varDelta g_{nk}(x))\notin _2c_0(\varDelta , ru)$.

Hence, the spaces $_2\ell _\infty (\varDelta , ru), ~_2{c_0}^B(\varDelta , ru), ~_2c^B(\varDelta ,ru),~ _2c^R(\varDelta , ru),~ _2c(\varDelta , ru), _2c_0(\varDelta , ru), ~_2{c_0}^R(\varDelta , ru)$ are not monotone.

Remark 3.1

Since soild implies monotone and from the Result 3.1. it follows that the spaces $_2\ell _\infty (\varDelta , ru), ~_2{c_0}^B(\varDelta , ru), ~_2c^B(\varDelta ,ru),~ _2c^R(\varDelta , ru),~ _2c(\varDelta , ru), ~ _2c_0(\varDelta , ru),_2{c_0}^R(\varDelta , ru)$ are not solid in general.

Result 3.2

The sequence spaces $Z(\varDelta , ru)$ where, $Z=_2\ell _\infty , ~_2{c_0}^B,~_2c^B,~ _2c^R, ~_2{c_0}^R, _2c$ and $_2c_0$ are not symmetric. $\square$

Proof

The proof of the result follows from the following example:

Example 3.2

Consider the sequence of functions $(f_{nk}(x)), f_{nk}:[0,1]\rightarrow R$ defined by

$$\begin{aligned} f_{nk}(x)= {\left\{ \begin{array}{ll} x,&{} \text {when}~ n=1,k\in N;\\ 0&{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

The difference sequence of functions $(\varDelta f_{nk}(x))$ of the sequence of functions $(f_{nk}(x))$ is given by

$$\begin{aligned} \varDelta f_{nk}(x)=f_{nk}(x)-f_{n,k+1}(x)-f_{n+1,k}(x)+f_{n+1,k+1}(x)=0. \end{aligned}$$

$(\varDelta f_{nk}(x))$ is relative uniform convergent w.r.t. the constant scale function 1.

Then, $(\varDelta f_{nk}(x))\in Z(\varDelta , ru)$ where $Z=_2\ell _\infty , ~_2{c_0}^B,~_2c^B,~ _2c^R, ~_2{c_0}^R, _2c$ and $_2c_0$ .

Let $(g_{nk}(x))$ be the rearranged sequence of functions of $(f_{nk}(x))$ defined by

$$\begin{aligned} g_{nk}(x)= {\left\{ \begin{array}{ll} x,&{} \text {when}~ n=k=i^2, i\in N;\\ 0,&{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

Then,

$$\begin{aligned} \varDelta g_{nk}(x)= {\left\{ \begin{array}{ll} x,&{} \text {for}~ n=k=i^2~\text {and}~n=k=i^2-1 ;i\in N;\\ -x,&{} \text {for}~ n=i^2-1, k=i^2 ~\text {and}~n=i^2, k=i^2-1;i\in N;\\ 0,&{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

One cannot get a scale function such that the sequence of functions $(\varDelta g_{nk}(x))$, a null relative uniform.

Then, $(\varDelta g_{nk}(x))\notin _2c_o(\varDelta , ru)$.

Hence, the sequence spaces $_2\ell _\infty (\varDelta , ru), ~_2{c_0}^B(\varDelta , ru),~_2c^B(\varDelta , ru),~ _2c^R(\varDelta , ru), _2c(\varDelta , ru),~ _2c_0(\varDelta , ru)$ and $~_2{c_0}^R(\varDelta , ru)$ are not symmetric. $\square$

Theorem 3.3

(i) $Z(ru)\subset Z(\varDelta , ru)$, for $Z=_2\ell _\infty , ~_2{c_0}^B,~_2c^B,~ _2c^R, ~_2{c_0}^R, ~ _2c$ and $_2c_0$ and the inclusions are strict.

(ii) $Z_0(\varDelta ,ru)\subset Z(\varDelta , ru)$, for $Z=_2\ell _\infty , ~_2{c_0}^B,~_2c^B,~ _2c^R, ~_2{c_0}^R, ~ _2c$ and $_2c_0$ and the inclusions are strict.

Proof

(i)

$$\begin{aligned} \text{ Let }~ (f_{nk}(x))\in _2c(ru) \end{aligned}$$

(6)

Then, for all $\varepsilon >0$, there exists an integer $n_0=n_0(\varepsilon )$ such that

$$\begin{aligned} \mid f_{nk}(x)-f(x)\mid <\frac{\varepsilon }{4}\mid \sigma (x)\mid , \end{aligned}$$

for all $n\ge n_0$ and $k\ge n_0.$

For all $n\ge n_0$ and $k\ge n_0$ we have,

$$\begin{aligned}&\mid \varDelta f_{nk}(x)\mid \le \mid f_{nk}(x)-f(x)\mid +\mid f_{n,k+1}(x)-f(x)\mid +\mid f_{n,k+1}(x)-f(x)\mid \nonumber \\&\qquad +\mid f_{n+1,k+1}(x)-f(x)\mid \nonumber \\&\quad \le \frac{\varepsilon }{4}\mid \sigma (x)\mid +\frac{\varepsilon }{4}\mid \sigma (x)\mid +\frac{\varepsilon }{4}\mid \sigma (x)\mid +\frac{\varepsilon }{4}\mid \sigma (x)\mid +\frac{\varepsilon }{4}\mid \sigma (x)\mid \nonumber \\&\quad \le \varepsilon \mid \sigma (x)\mid \nonumber \\&\mid \varDelta f_{nk}(x)\mid \le \varepsilon \mid \sigma (x)\mid .\nonumber \\&(f_{nk}(x))\in _2c_0(\varDelta , ru) \end{aligned}$$

(7)

From Eqs. (6) and (7) we get, $_2c(ru)\subset _2c_0(\varDelta , ru)$.

Similarly, we can prove for the other sequence spaces.

The following example shows that the inclusions are strict. $\square$

Example 3.3

Consider the sequence of functions $(f_{nk}(x)), f_{nk}(x):[0,1]\rightarrow R$ defined by

$$\begin{aligned} f_{nk}(x)= {\left\{ \begin{array}{ll} x,&{} \text {when}~ n~ \text {is odd},k\in N;\\ 0&{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

The difference sequence of functions $(\varDelta f_{nk}(x))$ of the sequence of functions $(f_{nk}(x))$ is given by

$$\begin{aligned} \varDelta f_{nk}(x)=f_{nk}(x)-f_{n,k+1}(x)-f_{n+1,k}(x)+f_{n+1,k+1}(x)=0. \end{aligned}$$

One cannot get a scale function such that the ordinary sequence of functions $(f_{nk}(x))$, a null relative uniform. However its difference sequence of functions, $(\varDelta f_{nk}(x))$ is null relative uniform w.r.t. the constant scale function 1.

Hence, the inclusion are strict.

(ii) It is obvious that $Z_0(\varDelta ,ru)\subset Z(\varDelta , ru)$, for $Z=_2\ell _\infty , ~_2{c_0}^B,~_2c^B,~ _2c^R, ~_2{c_0}^R, _2c$ and $_2c_0$, hence we omitted the prove.

The following example shows that inclusions are strict.

Example 3.4

Consider the sequence of functions $(f_{nk}(x)), f_{nk}(x):[0,1]\rightarrow R$ defined by

$$\begin{aligned} f_{nk}(x)= {\left\{ \begin{array}{ll} (n+k)x,&{} \text {for}~ n=1, k\in N;\\ nkx&{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

We have $\varDelta f_{nk}(x)=x$.

$(\varDelta f_{nk}(x))$ is relative uniform convergent to 1 on [0,1] w.r.t. the scale function $\sigma (x)$ defined by

$$\begin{aligned} \sigma (x)= {\left\{ \begin{array}{ll} \frac{1}{x},&{} \text {for}~ x\in (0,1];\\ 0&{} \text {for}~x=0. \end{array}\right. } \end{aligned}$$

Hence the inclusions are strict.

Result 3.3

A diference double sequence of functions $(\varDelta f_{nk}(x))$ is regular relative uniform convergent over a compact subset D w.r.t. a scale function $\sigma (x)$ then, $(\varDelta f_{nk}(x))$ is also relative uniform convergent w.r.t. a scale function $\sigma (x)$ but not conversely.

Example 3.5

Consider the sequence of functions $(f_{nk}(x)), f_{nk}(x):[0,1]\rightarrow R$ defined by

$$\begin{aligned} f_{nk}(x)= {\left\{ \begin{array}{ll} x,&{} \text {when}~ n=1~ \text {and}~ k~\text {is odd},k\in N;\\ 0&{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

$(\varDelta f_{nk}(x))$ is given by

$$\begin{aligned} \varDelta f_{nk}(x)= {\left\{ \begin{array}{ll} x,&{} \text {when}~ n=1;~ k~\text {is odd},k\in N;\\ -x,&{} \text {when}~ n=1;~k~ \text {is even},k\in N;\\ 0&{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

We have seen that $(\varDelta f_{nk}(x))$ is relative uniform convergent to the zero function $\theta$ on [0, 1] w.r.t. the constant scale function 1. However, in case of regular relative uniform convergence, the first row of the sequence of functions $(\varDelta f_{nk}(x))$ fails to converge relatively uniformly w.r.t. a scale function $\sigma (x)$.

Hence, the above result is justified.

Remark 3.2

All the results for the case of regularly convergent sequences will be same as that of the Pringsheim’s sense convergence.

4 Conclusions

In this article we have studied difference double sequences of positive linear operators from the point of view of relative uniform convergence. This is the first article on this topic and it is expected that it will attract researcher for further investigation and applications.