4. Bounded variation sequence spaces defined by modulus function

In this section, we study some new double sequence spaces of invariant means defined by ideal and modulus function. Furthermore, we also study several properties relevant to topological structures and inclusion relations between these spaces. The following classes of double sequence spaces are as follows:

$$\,\_2BV\_{\sigma}^I(f) = \left\{ (\mathbf{x}\_{\vec{\eta}}) \in \_2\omega : \left\{ (i, j) : \sum\_{m, n = 0}^{\infty} f\left( |\phi\_{m \vec{\eta} j}(\mathbf{x}) - L| \right) \ge \epsilon \right\} \in I; \text{for some } L \in \mathbb{C} \right\}; \tag{26}$$

$$\mathbb{P}\_2\left(\_{0}BV\_{\sigma}^{l}(f)\right) = \left\{ \left( \mathbf{x}\_{\vec{\eta}} \right) \in \_{2}\boldsymbol{\rho} : \left\{ (\dot{\boldsymbol{\iota}}, \boldsymbol{j}) : \sum\_{m\_{\mathfrak{p}} = 0}^{\infty} f\left( |\phi\_{m \vec{\eta}j}(\mathbf{x})| \right) \geq \mathbf{e} \right\} \in I \right\};\tag{27}$$

$$\pi\_2\left(\_{\mathfrak{so}}BV\_{\sigma}^I(f)\right) = \left\{ \left( \mathfrak{x}\_{\vec{\imath}} \right) \in \_2\omega : \left\{ (\mathfrak{i}, j) : \exists K \times \mathbf{0} : \sum\_{m\_1=0}^{\infty} f\left( \left| \phi\_{m:\vec{\imath}j}(\mathbf{x}) \right| \right) \geq K \right\} \in I \right\}; \tag{28}$$

$$\mathbb{P}\_2(\_{\infty}BV\_{\sigma}(f)) = \left\{ \left( \mathbb{x}\_{\vec{\eta}} \right) \in \_2\omega : \sup\_{i,j} \sum\_{m,n=0}^{\infty} f\left( \left| \phi\_{mnj}(\mathbf{x}) \right| \right) < \infty \right\}.\tag{29}$$

We also denote

$$\_2\mathcal{M}^l\_{BV\_\sigma}(f) = \_2BV^l\_\sigma(f) \cap \_2(\_\infty BV\_\sigma(f))$$

and

$$\_2\left(\_0\mathcal{M}^I\_{BV\_\sigma}(f)\right) = \_2\left(\_0BV^I\_\sigma(f)\right) \cap \_2\left(\_\inftyBV\_\sigma(f)\right).$$

We shall now consider important theorems of these double sequence spaces by using modulus function.

� � � � Theorem 4.1 For any modulus function f, the classes of double sequence 2 0BVIð Þ<sup>f</sup> , <sup>σ</sup> <sup>2</sup>BVIð Þ<sup>f</sup> ð Þ<sup>f</sup> and <sup>2</sup>M<sup>I</sup> ð Þ<sup>f</sup> are linear spaces. <sup>σ</sup> , 2 0M<sup>I</sup> BV<sup>σ</sup> BV<sup>σ</sup>

� � � � � � � � Proof. Suppose <sup>x</sup> <sup>¼</sup> xij and <sup>y</sup> <sup>¼</sup> yij <sup>∈</sup> <sup>2</sup>BVIð Þ<sup>f</sup> be any two arbitrary elements. <sup>σ</sup> Let <sup>α</sup>, <sup>β</sup> are scalars. Now, since xij , yij <sup>∈</sup> <sup>2</sup>BV<sup>I</sup>ð Þ<sup>f</sup> . Then this implies that there <sup>σ</sup> exists some positive numbers L1, L<sup>2</sup> ∈ C and such that the sets

$$\mathcal{A}\_1 = \left\{ (i, j) : \sum\_{m\_1 = 0}^{\infty} f\left( |\phi\_{m \lor j}(\infty) - L\_1| \right) \ge \frac{\epsilon}{2} \right\} \in I,\tag{30}$$

$$A\_2 = \left\{ (i, j) : \sum\_{m\_2, n=0}^{\infty} f\left( |\phi\_{mnj}(y) - L\_2| \right) \ge \frac{\epsilon}{2} \right\} \in I. \tag{31}$$

Now, assume

$$B\_1 = \left\{ (i, j) : \sum\_{m\_1 n = 0}^{\infty} f\left( |\phi\_{m n j}(\infty) - L\_1| \right) < \frac{\epsilon}{2} \right\} \in \mathcal{F}(I), \tag{32}$$

$$B\_2 = \left\{ (i, j) : \sum\_{m\_2=0}^{\infty} f\left( |\phi\_{mnij}(y) - L\_2| \right) < \frac{\varepsilon}{2} \right\} \in \mathcal{F}(I) \tag{33}$$

be such that B<sup>c</sup> 1, Bc <sup>2</sup> ∈ I: Since f is a modulus function, we have

$$\begin{split} &\sum\_{m\_{\mathfrak{m}}=0}^{\infty} \widehat{f}\left(|\phi\_{m\boldsymbol{m}\boldsymbol{j}}(\boldsymbol{\alpha}\boldsymbol{\omega}+\beta\boldsymbol{\gamma})-(a\boldsymbol{L}\_{1}+\beta\boldsymbol{L}\_{2})|\right) \\ &=\sum\_{m\_{\mathfrak{m}}=0}^{\infty} \widehat{f}\left(|\left(a\phi\_{m\boldsymbol{m}\boldsymbol{j}}(\boldsymbol{\alpha})+\beta\phi\_{m\boldsymbol{m}\boldsymbol{j}}(\boldsymbol{\gamma})\right)-(a\boldsymbol{L}\_{1}+\beta\boldsymbol{L}\_{2})|\right) \\ &=\sum\_{m\_{\mathfrak{m}}=0}^{\infty} \widehat{f}\left(|\alpha\left(\phi\_{m\boldsymbol{m}\boldsymbol{j}}(\boldsymbol{\alpha})-\boldsymbol{L}\_{1}\right)+\beta\left(\phi\_{m\boldsymbol{m}\boldsymbol{j}}(\boldsymbol{\gamma})-\boldsymbol{L}\_{2}\right)|\right) \\ &\leq\sum\_{m\_{\mathfrak{m}}=0}^{\infty} \widehat{f}\left(|\boldsymbol{\alpha}|\big|\phi\_{m\boldsymbol{m}\boldsymbol{j}}(\boldsymbol{\alpha})-\boldsymbol{L}\_{1}|\right)+\sum\_{m\_{\mathfrak{m}}=0}^{\infty} \widehat{f}\left(|\beta|\big|\phi\_{m\boldsymbol{m}\boldsymbol{j}}(\boldsymbol{\gamma})-\boldsymbol{L}\_{2}|\right) \\ &\leq\sum\_{m\_{\mathfrak{m}}=0}^{\infty} \widehat{f}\left(|\phi\_{m\boldsymbol{m}\boldsymbol{j}}(\boldsymbol{\alpha})-\boldsymbol{L}\_{1}|\right)+\sum\_{m\_{\mathfrak{m}}=0}^{\infty} \widehat{f}\left(|\phi\_{m\boldsymbol{m}\boldsymbol{j}}(\boldsymbol{\gamma})-\boldsymbol{L}\_{2}|\right) \\ &\leq\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon \end{split}$$

� � � � � � � � <sup>n</sup> � � o <sup>∞</sup> This implies that <sup>i</sup>; <sup>j</sup> m, <sup>n</sup>¼<sup>0</sup> <sup>f</sup> <sup>j</sup> mnij ð Þ : <sup>∑</sup> <sup>ϕ</sup> <sup>ð</sup>α<sup>x</sup> <sup>þ</sup> <sup>β</sup>yÞ � <sup>ð</sup>αL<sup>1</sup> <sup>þ</sup> <sup>β</sup>L2Þj ≥ ∈ <sup>ϵ</sup> <sup>I</sup>: Thus α xij þ β yij ∈ <sup>2</sup>BV<sup>σ</sup> <sup>I</sup>ð Þ<sup>f</sup> : As xij and yij are two arbitrary element then

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A Study of Bounded Variation Sequence Spaces DOI: http://dx.doi.org/10.5772/intechopen.81907

� � � � � � � � <sup>α</sup> xij <sup>þ</sup> <sup>β</sup> yij <sup>∈</sup> <sup>2</sup>BVIð Þ<sup>f</sup> for all xij , yij <sup>∈</sup> <sup>2</sup>BV<sup>σ</sup> <sup>I</sup>ð Þ<sup>f</sup> and for all scalars <sup>α</sup>, <sup>β</sup>. Hence <sup>σ</sup> <sup>2</sup>BVIð Þ<sup>f</sup> is linear space. The proof for other spaces will follow similarly. <sup>σ</sup>

▪

We may go a step further and define another theorem on ideal convergence which basically depends upon the set in the filter associated with the same ideal.

� � Theorem 4.2 <sup>A</sup> sequence <sup>x</sup> <sup>¼</sup> xij <sup>∈</sup> <sup>2</sup>M<sup>I</sup> ð Þf I-convergent if and only if for BV<sup>σ</sup> every ϵ>0, there exists Mε, N<sup>ε</sup> ∈ N such that

$$\left\{ (i, j) : \sum\_{m\_\bullet n = 0}^{\infty} f\left( |\phi\_{m n i j}(\mathbf{x}\_{ij}) - \phi\_{m n i j}(\mathbf{x}\_{M\_\bullet, N\_\epsilon})| \right) < \varepsilon \right\} \in \mathcal{F}(I).$$

� � <sup>2</sup>M<sup>I</sup> Proof. Let <sup>x</sup> <sup>¼</sup> xij <sup>∈</sup> ð Þ<sup>f</sup> . Suppose <sup>I</sup> � lim <sup>x</sup> <sup>¼</sup> <sup>L</sup>. Then, the set BV<sup>σ</sup>

$$B\_{\epsilon} = \left\{ (i, j) : \sum\_{m\_1 n = 0}^{\infty} f\left( |\phi\_{m n j}(\mathbf{x}\_{ij}) - L| \right) < \frac{\epsilon}{2} \right\} \in F(I), \quad \text{for all } \epsilon > 0. \right\}$$

Fix Mε, N<sup>ε</sup> ∈Bε: Then we have

$$\begin{aligned} \sum\_{m\_2}^{\infty} \bar{f}\left( |\phi\_{mnj}(\mathbf{x}\_{ij}) - \phi\_{mnj}(\mathbf{x}\_{M\_\epsilon, N\_\epsilon})| \right) &\leq \sum\_{m\_2}^{\infty} \bar{f}\left( |\phi\_{mnj}(\mathbf{x}\_{M\_\epsilon, N\_\epsilon}) - L| \right) \\ &+ \sum\_{m\_2, n=0}^{\infty} \bar{f}\left( |L - \phi\_{mnj}(\mathbf{x}\_{ij})| \right) \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{aligned}$$

which holds for all ð Þ i; j B ∈ <sup>ϵ</sup>: Hence

$$\left\{ (i, j) : \sum\_{m\_2, n = 0}^{\infty} f \left( |\phi\_{m n i j}(\mathbf{x}\_{ij}) - \phi\_{m n i j}(\mathbf{x}\_{M\_\epsilon, N\_\epsilon})| \right) < \epsilon \right\} \in \mathcal{F}(I). \epsilon$$

Conversely, suppose that

$$\left\{ (i, j) : \sum\_{m\_\bullet n = 0}^{\infty} f\left( |\phi\_{m n i j}(\mathbf{x}\_{ij}) - \phi\_{m n i j}(\mathbf{x}\_{M\_\bullet, N\_\epsilon})| \right) < \varepsilon \right\} \in \mathcal{F}(I).$$

Then, being f a modulus function and by using basic triangular inequality, we have

$$\left\{ (i,j) : |\sum\_{m\_1=0}^{\infty} f\left( |\phi\_{m m j}(\mathbf{x}\_{ij})| \right) - \sum\_{m\_2=0}^{\infty} f\left( |\phi\_{m m j}(\mathbf{x}\_{M\_\varepsilon, N\_\varepsilon})| \right)| < \varepsilon \right\} \in F(I), \text{ for all } \varepsilon > 0.$$

Then, the set

$$C\_{\varepsilon} = \left\{ (i, j) : \sum\_{m, n = 0}^{\infty} f\left( |\phi\_{mnij}(\mathbf{x}\_{ij})| \right) \in$$

$$\left[ \sum\_{m, n = 0}^{\infty} f\left( |\phi\_{mnij}(\mathbf{x}\_{M\_{\varepsilon}N\_{\varepsilon}})| \right) - \mathfrak{e}, \sum\_{m, n = 0}^{\infty} f\left( |\phi\_{mnij}(\mathbf{x}\_{M\_{\varepsilon}N\_{\varepsilon}})| \right) + \mathfrak{e} \right] \right\} \in \mathcal{F}(I).$$

$$\text{Let } f\_{\varepsilon} = \left[ \sum\_{m, n = 0}^{\infty} f\left( |\phi\_{mnij}(\mathbf{x}\_{M\_{\varepsilon}N\_{\varepsilon}})| \right) - \mathfrak{e}, \sum\_{m, n = 0}^{\infty} f\left( |\phi\_{mnij}(\mathbf{x}\_{M\_{\varepsilon}N\_{\varepsilon}})| \right) + \mathfrak{e} \right].$$

123

If we fix ϵ>0 then, we have C<sup>ϵ</sup> ∈F Ið Þ as well as C<sup>ϵ</sup> ∈ Fð ÞI . <sup>2</sup> Hence C<sup>ϵ</sup> ∩C<sup>ϵ</sup> ∈ Fð ÞI . This implies that <sup>2</sup>

$$J = J\_e \cap J\_\sharp \neq \Phi.$$

That is

$$\left\{ (i,j) : \sum\_{m\_1=0}^{\infty} f\left( |\phi\_{m m j}(\mathbf{x}\_{ij})| \right) \in I \right\} \in \mathcal{F}(I).$$

This shows that

diam J ≤ diam J<sup>ϵ</sup>

where the diam J denotes the length of interval J. In this way, by induction we get the sequence of closed intervals

$$J\_e = I\_0 \supseteq I\_1 \supseteq I\_2 \supseteq \dots \supseteq I\_k \supseteq \dots$$

� � with the property that diam Ik <sup>≤</sup> <sup>1</sup> diam Ik�<sup>1</sup> for <sup>ð</sup><sup>k</sup> <sup>¼</sup> <sup>2</sup>; <sup>3</sup>; <sup>4</sup>; …<sup>Þ</sup> and n � � o 2 <sup>∞</sup> <sup>i</sup>; <sup>j</sup> m, <sup>n</sup>¼<sup>0</sup> <sup>f</sup> <sup>j</sup>ϕmnij xij ð Þ : <sup>∑</sup> <sup>j</sup> <sup>∈</sup>Ik <sup>∈</sup> F Ið Þ for <sup>ð</sup><sup>k</sup> <sup>¼</sup> <sup>1</sup>; <sup>2</sup>; <sup>3</sup>; <sup>4</sup>; …Þ. Then there exists a ξ∈ ∩ Ik where k∈ N such that

$$\xi = I - \lim\_{i,j \quad m\_\bullet} \sum\_{n=0}^\infty f\left( |\phi\_{m n i j}\left(\mathbf{x}\_{i j}\right)| \right),$$

� � showing that <sup>x</sup> <sup>¼</sup> xij <sup>∈</sup> <sup>2</sup>MI ð Þ<sup>f</sup> is <sup>I</sup>-convergent. Hence the result holds. BV<sup>σ</sup>

As the reader knows about solid and monotone sequence space now turn to theorem on solid and monotone double sequence spaces of invariant mean defined by ideal and modulus function.

� � � � Theorem 4.3 For any modulus function f, the spaces 2 0BV<sup>I</sup> <sup>σ</sup>ð Þf and

2 0M<sup>I</sup> ð Þ<sup>f</sup> are solid and monotone. BV<sup>σ</sup>

� � � � � � � � Proof. We consider 2 0BV<sup>I</sup>ð Þ<sup>f</sup> and for 2 0M<sup>I</sup> ð Þ<sup>f</sup> the proof shall be similar. <sup>σ</sup> BV<sup>σ</sup> Let <sup>x</sup> <sup>¼</sup> xij <sup>∈</sup> 2 0BV<sup>I</sup>ð Þ<sup>f</sup> be an arbitrary element, then the set <sup>σ</sup>

$$\left\{ (i, j) : \sum\_{m\_1=0}^{\infty} f\left( |\phi\_{mnj}(\mathbf{x})| \right) \ge \epsilon \right\} \in I. \tag{34}$$

� � Let αij be a sequence of scalars with ∣αij∣ ≤ 1 for all i, j∈ N:

Now, since f is a modulus function. Then the result follows from (2.18) and the inequality

$$f\left(|a\_{\vec{\eta}}\phi\_{mn\vec{\eta}}(\mathbf{x})|\right) \le |a\_{\vec{\eta}}|f\left(|\phi\_{mn\vec{\eta}}(\mathbf{x})|\right) \le f\left(|\phi\_{mn\vec{\eta}}(\mathbf{x})|\right).$$

Therefore,

$$\left\{ (i,j) : \sum\_{m\_1 n = 0}^{\infty} f\left( |a\_{\overrightarrow{\eta}} \phi\_{m \overrightarrow{\eta}j}(\infty)| \right) \ge \epsilon \right\} \subseteq \left\{ (i,j) : \sum\_{m\_1 n = 0}^{\infty} f\left( |\phi\_{m \overrightarrow{\eta}j}(\infty)| \right) \ge \epsilon \right\} \in I^{\epsilon}$$

implies that

A Study of Bounded Variation Sequence Spaces DOI: http://dx.doi.org/10.5772/intechopen.81907

$$\left\{ (i,j) : \sum\_{m,n=0}^{\infty} f\left( |a\_{ij}\phi\_{mnj}(\mathbf{x})| \right) \ge \epsilon \right\} \in I.$$

� � � � � � � � <sup>∈</sup> 2 0BV<sup>I</sup> : Hence 2 0BV<sup>I</sup> Thus we have <sup>σ</sup> <sup>σ</sup> <sup>α</sup>ijxij ð Þ<sup>f</sup> ð Þ<sup>f</sup> is solid. Therefore 2 0BV<sup>I</sup> <sup>σ</sup>ð Þf is monotone. Since every solid sequence space is monotone.

� � Remark 4.1 The space <sup>2</sup>BV<sup>I</sup> σ MBVI <sup>σ</sup> ð Þf and <sup>2</sup> ð Þf are neither solid nor monotone in general.

Example 4.1 � Here � we give counter example for establishment of this result. Let <sup>2</sup> BV<sup>I</sup> � � MBVI <sup>σ</sup> X¼ <sup>σ</sup> and <sup>2</sup> . Let us consider I ¼ If and fð Þ¼ x x, for all x ¼ xij and <sup>x</sup> ½ Þ. Consider, the K-step space XKð Þ<sup>f</sup> of X fð Þ defined as follows: ij <sup>∈</sup> <sup>0</sup>; <sup>∞</sup> � � Let <sup>x</sup> <sup>¼</sup> xij <sup>∈</sup>X fð Þ and <sup>y</sup> <sup>¼</sup> yij <sup>∈</sup>XKð Þ<sup>f</sup> be such that

� �

$$\mathcal{Y}\_{ij} = \begin{cases} \mathcal{X}\_{ij}, & \text{if } i, \ j \text{ are even} \\ \mathbf{0}, & \text{otherwise}. \end{cases}$$

� � � �

σ Consider the sequence xij defined by xij <sup>¼</sup> <sup>1</sup> for all i, <sup>j</sup><sup>∈</sup> <sup>N</sup>. � �<sup>∈</sup> <sup>2</sup>BV<sup>I</sup> belong to BV<sup>I</sup> . Thus, <sup>2</sup>BV<sup>I</sup> σ σ Then, x ¼ xij <sup>σ</sup> ð Þf and <sup>2</sup>MBVI ð Þf , but K-step space preimage does not ð Þ<sup>f</sup> and <sup>2</sup>M<sup>I</sup> BV<sup>σ</sup> ð Þ<sup>f</sup> and <sup>2</sup>M<sup>I</sup> BV<sup>σ</sup> ð Þf ð Þf are not monotone and hence they are not solid.

After discussing about solid and monotone sequence space now we come to the concept of sequence algebra which will help to understand our further work.

� � σ σ Theorem 4.4 For any modulus function <sup>f</sup>, the spaces 2 0BV<sup>I</sup> and <sup>2</sup>BV<sup>I</sup> ð Þ<sup>f</sup> ð Þ<sup>f</sup> are sequence algebra.

� � � � σ � � <sup>∈</sup> 2 0BV<sup>I</sup> Proof. Let <sup>x</sup> <sup>¼</sup> xij , y <sup>¼</sup> <sup>y</sup> ð Þ<sup>f</sup> be any two arbitrary elements. ij Then, the sets

$$\left\{ (i,j) : \sum\_{m\_1=0}^{\infty} f\left( |\phi\_{m n j}(\mathbf{x})| \right) \ge \epsilon \right\} \in I$$

and

$$\left\{ (i,j) : \sum\_{m\_1=0}^{\infty} f\left( |\phi\_{m m j}(\wp)| \right) \ge \epsilon \right\} \in I.$$

Therefore,

$$\left\{ (i,j) : \sum\_{m\_\nu=0}^{\infty} f\left( |\phi\_{m n j}(\infty).\phi\_{m n i j}(\gg)| \right) \ge \epsilon \right\} \in I. $$

� � � � � � � � σ σ σ ∈ 2 0BV<sup>I</sup> : Hence 2 0BV<sup>I</sup> And for <sup>2</sup>BV<sup>I</sup> Thus, we have x ð Þf ð Þf is sequence algebra. ij : yij ð Þf the result can be proved similarly.

� � σ <sup>¼</sup> If then the spaces <sup>2</sup>BV<sup>I</sup> Remark 4.2 If I is not maximal and I 6ð Þf and 2 0BV<sup>σ</sup> <sup>I</sup>ð Þ<sup>f</sup> are not symmetric.

� � Example 4.2 Let A ∈ I be an infinite set and f xð Þ¼ x for all x ¼ xij and xij ∈ ½0; ∞Þ: If

$$\mathfrak{x}\_{\vec{\eta}} = \begin{cases} \mathbf{1}, & \text{if } (i, j) \in A \\ \mathbf{0}, & \text{otherwise} \end{cases}$$

° ˛ ° ˛ Then, it is clearly seen that xij ∈2 0BV<sup>I</sup> <sup>σ</sup>ðf<sup>Þ</sup> <sup>⊂</sup> <sup>2</sup>BVIðfÞ: <sup>σ</sup>

Let <sup>K</sup> <sup>⊆</sup> <sup>N</sup> � <sup>N</sup> be such that <sup>K</sup> <sup>∉</sup> <sup>I</sup> and <sup>K</sup><sup>c</sup> <sup>∉</sup>I: Let <sup>ϕ</sup> : <sup>K</sup> ! <sup>A</sup> and <sup>ψ</sup> : <sup>K</sup><sup>c</sup> ! <sup>A</sup><sup>c</sup> be <sup>a</sup> bijective maps (as all four sets are infinite). Then, the mapping π : N � N ! N � N defined by

$$
\pi(i,j) = \begin{cases}
\phi(i,j), & \text{if } \ (i,j) \in K \\
\psi(i,j), & \text{otherwise.}
\end{cases}
$$

is <sup>a</sup> permutation on <sup>N</sup> � <sup>N</sup>: ° ˛ ° ˛ ° ˛ But <sup>x</sup><sup>π</sup> <sup>i</sup>;<sup>j</sup> <sup>∉</sup>2BV<sup>I</sup> <sup>f</sup> and hence ð Þ <sup>0</sup>BV<sup>I</sup> <sup>f</sup> showing that <sup>2</sup>BV<sup>I</sup> ð Þ <sup>σ</sup>ð Þ x<sup>π</sup> <sup>i</sup>;<sup>j</sup> ∉<sup>2</sup> <sup>σ</sup>ð Þ <sup>σ</sup>ð Þf and ° ˛ 2 0BV<sup>I</sup> <sup>σ</sup>ðfÞ are not symmetric double sequence spaces.
