2. Bounded variation sequence spaces defined by Orlicz function

In this section, we define and study the concepts of I-convergence for double sequences defined by Orlicz function and present some basic results on the above definitions [8, 20].

$$\mathbb{E}\_2 \text{BV}\_{\sigma}^I(M) = \left\{ (\mathbf{x}\_{\not\equiv}) \in \_2\boldsymbol{w} : I - \lim M \left( \frac{|\phi\_{m \not\equiv i}(\boldsymbol{\chi}) - L|}{\rho} \right) = \mathbf{0}, \text{ for some} \ L \in \mathbb{C}, \rho > \mathbf{0} \right\}. \tag{6}$$

$$\mathbb{P}\_2\left(\_{0}BV^I\_{\sigma}(M)\right) = \left\{ \left( \mathbb{x}\_{\vec{\eta}} \right) \in \_2\omega : I - \lim M \left( \frac{|\phi\_{m\vec{\eta}j}(\mathbf{x})|}{\rho} \right) = \mathbf{0}, \rho > \mathbf{0} \right\},\tag{7}$$

$$\mathcal{I}\_2\left(\_{\infty}BV\_{\sigma}^{I}(\mathcal{M})\right) = \left\{ \left( \mathbf{x}\_{\vec{\eta}} \right) \in \_2\mathcal{W} : \begin{cases} (i,j) : \exists \ k > 0 \,\,\text{s.t.} \,\mathcal{M} \left( \frac{|\phi\_{\text{mnp}}(\mathbf{x})|}{\rho} \right) \ge k \right\} \in I, \rho > 0 \right\} \tag{8}$$

$$\mathbf{r}\_2(\_{\mathfrak{w}}BV\_{\sigma}(M)) = \left\{ (\mathbf{x}\_{\vec{\eta}}) \in \_{\mathfrak{z}} \mathbf{w} \, : \, \text{supp}M\left(\frac{|\phi\_{m \vec{\eta}j}(\mathbf{x})|}{\rho}\right) < \infty, \rho > 0 \right\}.\tag{9}$$

Now, we read some theorems based on these sequence spaces. These theorems are of general importance as indispensable tools in various theoretical and practical problems.

Theorem 2.1 Let M1, M<sup>2</sup> be two Orlicz functions with Δ<sup>2</sup> condition, then

$$\begin{aligned} \text{(a)}\ \chi(\mathsf{M}\_{2}) &\subseteq \chi(\mathsf{M}\_{1}\mathsf{M}\_{2})\\ \text{(b)}\ \chi(\mathsf{M}\_{1}) &\cap \chi(\mathsf{M}\_{2}) \subseteq \chi(\mathsf{M}\_{1} + \mathsf{M}\_{2}) \text{ for } \chi = \_{2}\mathsf{B}V^{I}\_{\sigma^{\*}2}(\_{0}\mathsf{B}V^{I}\_{\sigma}). \end{aligned}$$

˜ ° ˜ ° Proof. (a) Let <sup>x</sup> <sup>¼</sup> xij <sup>∈</sup> 2 0BV<sup>I</sup>ðM2<sup>Þ</sup> be an arbitrary element, so there exists <sup>σ</sup> ρ > 0 such that

$$I - \lim M\_2 \left( \frac{|\phi\_{m\nu ij}(\mathbf{x})|}{\rho} \right) = \mathbf{0}. \tag{10}$$

Let <sup>ϵ</sup> <sup>&</sup>gt; <sup>0</sup> and choose <sup>δ</sup> with 0 < <sup>δ</sup> <sup>&</sup>lt; <sup>1</sup> such that <sup>M</sup>1ð Þ<sup>t</sup> <sup>&</sup>lt; <sup>ε</sup> for 0 < <sup>t</sup> <sup>≤</sup> <sup>δ</sup>. ˇ ˘ Write <sup>∣</sup>ϕmnij yij <sup>¼</sup> <sup>M</sup><sup>2</sup> ð Þ <sup>x</sup> <sup>∣</sup> . Consider, <sup>ρ</sup>

$$\lim\_{\vec{\boldsymbol{\eta}}\vec{\boldsymbol{\eta}}} \boldsymbol{M}\_{1}(\boldsymbol{\mathcal{y}}\_{\vec{\boldsymbol{\eta}}}) = \lim\_{\boldsymbol{\mathcal{y}}\_{\vec{\boldsymbol{\eta}}} \leq \delta\_{2}} \boldsymbol{M}\_{1}(\boldsymbol{\mathcal{y}}\_{\vec{\boldsymbol{\eta}}}) + \lim\_{\boldsymbol{\mathcal{y}}\_{\vec{\boldsymbol{\eta}}} \rhd \boldsymbol{\delta}\_{2} \boldsymbol{i}\_{\vec{\boldsymbol{\eta}}} \in \mathbb{N}} \boldsymbol{M}\_{1}(\boldsymbol{\mathcal{y}}\_{\vec{\boldsymbol{\eta}}}).\tag{11}$$

Now, since M<sup>1</sup> is an Orlicz function so we have M1ð Þ λx ≤ λM1ð Þ x , 0 < λ < 1. Therefore, we have

$$\lim\_{\mathcal{Y}\_{ij}\leq\delta\_{2}i,j\in\mathbb{N}}M\_{1}(\mathcal{Y}\_{ij})\leq M\_{1}(\mathcal{Z})\lim\_{\mathcal{Y}\_{ij}\leq\delta\_{2}i,j\in\mathbb{N}}\left(\mathcal{Y}\_{ij}\right).\tag{12}$$

yij yij For yij>δ, we have yij <sup>&</sup>lt; < 1 <sup>þ</sup> <sup>δ</sup> . Now, since <sup>M</sup><sup>1</sup> is non-decreasing and convex, it <sup>δ</sup> follows that,

$$M\_1(\mathcal{Y}\_{\vec{\eta}}) \prec M\_1\left(\mathbf{1} + \frac{\mathcal{Y}\_{\vec{\eta}}}{\delta}\right) \prec \frac{1}{2}M\_1(\mathcal{Z}) + \frac{1}{2}M\_1\left(\frac{2\mathcal{Y}\_{\vec{\eta}}}{\delta}\right). \tag{13}$$

Since M<sup>1</sup> satisfies the Δ2-condition, so we have

$$\mathcal{M}\_1\left(\boldsymbol{y}\_{ij}\right) < \frac{1}{2}K\frac{\mathcal{Y}\_{ij}}{\delta}\mathcal{M}\_1(2) + \frac{1}{2}K\mathcal{M}\_1\left(\frac{2\mathcal{Y}\_{ij}}{\delta}\right)$$

$$\quad < \frac{1}{2}K\frac{\mathcal{Y}\_{ij}}{\delta}\mathcal{M}\_1(2) + \frac{1}{2}K\frac{\mathcal{Y}\_{ij}}{\delta}\mathcal{M}\_1(2) \tag{14}$$

$$= K\frac{\mathcal{Y}\_{ij}}{\delta}\mathcal{M}\_1(2).$$

This implies that,

$$\mathcal{M}\_1\left(\mathcal{Y}\_{\vec{\eta}}\right) \prec K \frac{\mathcal{Y}\_{\vec{\eta}}}{\delta} \mathcal{M}\_1(\mathbf{2}).\tag{15}$$

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

Hence, we have

$$\lim\_{y\_{ij}\succ \delta\_{\flat}} M\_1(y\_{ij}) \le \max\left\{ 1, K\delta^{-1} M\_1(2) \lim\_{\substack{y\_{ij}\succ \delta\_{\flat} i\_i j \in \mathbb{N}}} \left( y\_{ij} \right) \right\}.\tag{16}$$

Therefore from (12) and (16), we have

$$I - \lim\_{\vec{\eta}} M\_1 \left( \mathcal{Y}\_{\vec{\eta}} \right) = \mathbf{0}.$$

$$\Rightarrow I - \lim\_{\vec{\eta}} M\_1 M\_2 \left( \frac{|\phi\_{m m \vec{\eta}}(\mathbf{x})|}{\rho} \right) = \mathbf{0}.$$

� � � � This implies that x ¼ xij ð Þ . Hence χðM<sup>2</sup> ð Þ for � � ∈2 0BV<sup>I</sup> <sup>σ</sup> M1M<sup>2</sup> Þ ⊆ χ M1M<sup>2</sup> χ = 2 0BV<sup>I</sup> : The other cases can be proved in similar way. <sup>σ</sup>

� � � � � � (b) Let <sup>x</sup> <sup>¼</sup> xij <sup>∈</sup>2 0BV<sup>I</sup>ðM1<sup>Þ</sup> <sup>∩</sup> 2 0BV<sup>I</sup>ðM2<sup>Þ</sup> . Let <sup>ϵ</sup>>0 be given. Then there <sup>σ</sup> <sup>σ</sup> exist ρ>0, such that

$$I - \lim\_{\vec{\eta}} M\_1(\frac{|\phi\_{miij}(\infty)|}{\rho}) = 0,\tag{17}$$

and

$$I - \lim\_{\neq} M\_2\left(\frac{|\phi\_{m m \circ j}(\mathbf{x})|}{\rho}\right) = \mathbf{0}.\tag{18}$$

Therefore

$$I - \lim\_{\vec{\eta}\to 0} (M\_1 + M\_2) \left(\frac{|\phi\_{m\vec{\eta}}(\mathbf{x})|}{\rho}\right) = I - \lim\_{\vec{\eta}} M\_1 \left(\frac{|\phi\_{m\vec{\eta}}(\mathbf{x})|}{\rho}\right) + I - \lim\_{\vec{\eta}} M\_2 \left(\frac{|\phi\_{m\vec{\eta}}(\mathbf{x})|}{\rho}\right),$$

from Eqs. (17) and (18), we get

$$I - \lim\_{\vec{\eta}} \left( M\_1 + M\_2 \right) \left( \frac{|\phi\_{mnij}(\mathbf{x})|}{\rho} \right) = \mathbf{0}.$$

� � � � so we have <sup>x</sup> <sup>¼</sup> xij <sup>∈</sup>2 0BV<sup>I</sup>ðM<sup>1</sup> <sup>þ</sup> <sup>M</sup>2<sup>Þ</sup> : <sup>σ</sup>

� � � � � � Hence, 2 0BV<sup>I</sup>ðM1<sup>Þ</sup> <sup>∩</sup> 2 0BV<sup>I</sup>ðM2<sup>Þ</sup> <sup>⊆</sup> 2 0BV<sup>I</sup>ðM<sup>1</sup> <sup>þ</sup> <sup>M</sup>2<sup>Þ</sup> : For <sup>χ</sup> <sup>=</sup><sup>2</sup> BV<sup>σ</sup> <sup>I</sup> the <sup>σ</sup> <sup>σ</sup> <sup>σ</sup> inclusion are similar.

Corollary χ ⊆ χðMÞ for χ ¼<sup>2</sup> � BV<sup>I</sup> σ � and <sup>2</sup>BV<sup>I</sup> σ.

� � � � � � Proof. For this let Mð Þ¼ x x, for all x ¼ xij ∈X. Let us suppose that x ¼ xij ∈2 <sup>0</sup>BV<sup>σ</sup> <sup>I</sup> . Then for any given ϵ>0, we have

$$\left\{ (i, j) : |\phi\_{m n \ddot{j}}(\mathbf{x})| \ge \epsilon \right\} \in I.$$

n o Now let <sup>A</sup><sup>1</sup> <sup>¼</sup> <sup>i</sup>; <sup>j</sup> : <sup>j</sup>ϕmnijð Þj <sup>∈</sup> I, be such that <sup>A</sup><sup>c</sup> ð Þ <sup>x</sup> <sup>&</sup>lt; <sup>ϵ</sup> <sup>1</sup> <sup>∈</sup>I. Consider for <sup>ρ</sup> <sup>&</sup>gt; 0,

$$M\left(\frac{|\phi\_{m\nu\dot{\jmath}}(\infty)|}{\rho}\right) = \frac{|\phi\_{m\nu\dot{\jmath}}(\infty)|}{\rho} < \frac{\epsilon}{\rho} < \epsilon.$$

� � � � � � <sup>∣</sup>ϕmnijð Þ <sup>x</sup> <sup>∣</sup> This implies that <sup>I</sup> � lim <sup>M</sup> <sup>¼</sup> 0, which shows that <sup>ρ</sup> x ¼ xij ∈2 0BV<sup>σ</sup> <sup>I</sup>ð Þ <sup>M</sup> .

Hence, we have

$$\begin{aligned} \,\_2\left( \,\_0BV\_{\sigma}^{I} \right) & \subseteq\_2 \left( \,\_0BV\_{\sigma}^{I}(M) \right) . \\ & \Rightarrow \chi \subseteq \chi(M) . \end{aligned}$$

Using the definition of convergence free sequence space, let us give another theorem which will be of particular importance in our future work:

˜ ° Theorem 2.2 The spaces 2 0BV<sup>I</sup> and <sup>2</sup>BV<sup>I</sup> <sup>σ</sup> <sup>M</sup> <sup>σ</sup> ð Þ ð Þ <sup>M</sup> are not convergence free. Example 2.1 To show this let I ¼ If and MðxÞ ¼ x, for all x ¼ ½0; ∞Þ. Now ˜ ° ˛ ˝ consider the double sequence xij , yij which defined as follows:

$$\mathcal{X}\_{\vec{\eta}} = \frac{1}{i+j} \quad \text{and} \quad \mathcal{Y}\_{\vec{\eta}} = i+j, \forall i, j \in \mathbb{N} \dots$$

Then we have ˜ ° xij ° ˜ belong to both 2 0BV<sup>I</sup> <sup>σ</sup>ð Þ <sup>M</sup> ° <sup>σ</sup> and <sup>2</sup>BV<sup>I</sup>ð Þ <sup>M</sup> , but ° ˛ ˝ yij ˜ belong to 2 0BV<sup>I</sup> <sup>σ</sup>ð Þ <sup>M</sup> and <sup>2</sup>BV<sup>I</sup> <sup>σ</sup>ð Þ M . Hence, the sp <sup>σ</sup> ˜ aces 2 0BV<sup>I</sup>ð Þ <sup>M</sup> <sup>σ</sup> and <sup>2</sup>BV<sup>I</sup> does not ð Þ M are not convergence free.

To gain a good understanding of these double sequence spaces and related concepts, let us finally look at this theorem on inclusions:

Theorem 2.3 Let M be an Orlicz function. Then

$$\_2(\_0BV^I\_\sigma(M)) \subseteq \, \_2BV^I\_\sigma(M) \subseteq \, \_2(\_\inftyBV^I\_\sigma(M)) \,.$$

σ ˜ ° ˜ ° <sup>∈</sup> 2 0BV<sup>I</sup> Proof. For this let us consider <sup>x</sup> <sup>¼</sup> xij belong to <sup>2</sup>BV<sup>I</sup> σ ð Þ M : It is obvious that it must ð Þ M : Now consider

$$M\left(\frac{|\phi\_{m\text{nij}}(\mathbf{x}) - L|}{\rho}\right) \le M\left(\frac{|\phi\_{m\text{nij}}(\mathbf{x})|}{\rho}\right) + M\left(\frac{|L|}{\rho}\right).$$

Now taking the limit on both sides we get

$$I - \lim\_{\vec{\eta}} M\left(\frac{|\phi\_{m\vec{m}\vec{j}}(\infty) - L|}{\rho}\right) = \mathbf{0}.$$

σ ˜ °<sup>∈</sup> <sup>2</sup>BV<sup>I</sup> Hence <sup>x</sup> <sup>¼</sup> xij ð Þ <sup>M</sup> : Now it remains to show that

$$\_2\left(BV^I\_{\sigma}(\mathcal{M})\right) \subseteq \_2\left(\_{\infty}BV^I\_{\sigma}(\mathcal{M})\right).$$

σ ∈2BV<sup>I</sup> ˜ ° For this let us consider <sup>x</sup> <sup>¼</sup> xij ð Þ <sup>M</sup> this implies that there exist <sup>ρ</sup> <sup>&</sup>gt; <sup>0</sup> s.t

$$I - \lim\_{\vec{\eta}} M\left(\frac{|\phi\_{m\vec{\eta}j}(\mathbf{x}) - L|}{\rho}\right) = \mathbf{0}.$$

Now consider,

$$M\left(\frac{|\phi\_{m\text{ii}j}(\mathbf{x})|}{\rho}\right) \le M\left(\frac{|\phi\_{m\text{ii}j}(\mathbf{x}) - L|}{\rho}\right) + M\left(\frac{|L|}{\rho}\right).$$

Now taking the supremum on both sides, we get

$$\sup\_{\vec{\eta}} M\left(\frac{|\phi\_{mni\vec{\eta}}(\infty)|}{\rho}\right) < \infty.$$

: ▪ <sup>σ</sup> ˜ ° ˜ ° <sup>∞</sup>BV<sup>I</sup> Hence, <sup>x</sup> <sup>¼</sup> xij <sup>∈</sup> <sup>2</sup> ð Þ <sup>M</sup>

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