Definition 2.1
Let f be a modulus function. Two nonnegative sequences \(x=(x_{kj})\) and \(y=(y_{kj})\) are said to be f-asymptotically \(\mathcal{I}_{2}\)-equivalent of multiple L if for every \(\varepsilon >0\),
$$\begin{aligned} \biggl\{ (k,j)\in \mathbb{N}\times \mathbb{N} :f \biggl( \biggl\vert \frac{x _{kj}}{y_{kj}}-L \biggr\vert \biggr) \geq \varepsilon \biggr\} \in \mathcal{I}_{2} \end{aligned}$$
(denoted by \(x_{kj}\overset{{\mathcal{I}_{2}^{L}}(f)}{\sim }y_{kj}\)) and simply f-asymptotically \(\mathcal{I}_{2}\)-equivalent if \(L=1\).
Definition 2.2
Let f be a modulus function. The two nonnegative sequences \(x=(x_{kj})\) and \(y=(y_{kj})\) are said to be strongly f-asymptotically \(\mathcal{I}_{2}\)-equivalent of multiple L if for every \(\varepsilon >0\),
$$\begin{aligned} \Biggl\{ (m,n)\in \mathbb{N}\times \mathbb{N}: \frac{1}{mn}\sum _{k,j=1} ^{m,n}f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr)\geq \varepsilon \Biggr\} \in \mathcal{I}_{2} \end{aligned}$$
(denoted by \(x_{kj}\overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y_{kj}\)) and simply strongly f-asymptotically \(\mathcal{I}_{2}\)-equivalent if \(L=1\).
Theorem 2.1
Let
f
be a modulus function. Then
\(x_{kj} \overset{[{\mathcal{I}_{2}^{L}}]}{\sim }y_{kj}\Rightarrow x_{kj} \overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y_{kj}\).
Proof
Suppose that \(x_{kj}\overset{[{\mathcal{I}_{2}^{L}}]}{\sim }y_{kj}\), and let \(\varepsilon >0\) be given. Select \(0<\delta <1\) such that \(f(t)<\varepsilon \) for \(0\leq t\leq \delta \). We can write
$$\begin{aligned} \frac{1}{mn}\sum_{k,j=1}^{m,n}f \biggl( \biggl\vert \frac{x_{kj}}{y _{kj}}-L \biggr\vert \biggr) = & \frac{1}{mn} \sum_{\substack{k,j=1\\ \vert \frac{x _{kj}}{y_{kj}}-L \vert \leq \delta }}^{m,n} f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) \\ & {}+ \frac{1}{mn} {\sum_{\substack{k,j=1\\\vert \frac{x_{kj}}{y_{kj}}-L \vert > \delta}}^{m,n}}f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr), \end{aligned}$$
and so by Lemma 1
$$\begin{aligned} \frac{1}{mn}\sum_{k,j=1}^{m,n}f \biggl( \biggl\vert \frac{x_{kj}}{y _{kj}}-L \biggr\vert \biggr) < \varepsilon + \biggl(\frac{2f(1)}{\delta } \biggr) \frac{1}{mn} \sum _{k,j=1}^{m,n} \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert . \end{aligned}$$
Thus, for any \(\gamma >0\),
$$\begin{aligned} & \Biggl\{ (m,n)\in \mathbb{N}\times \mathbb{N}:\frac{1}{mn}\sum _{k,j=1}^{m,n}f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) \geq \gamma \Biggr\} \\ & \quad \subseteq \Biggl\{ (m,n)\in \mathbb{N}\times \mathbb{N}: \frac{1}{mn} \sum_{k,j=1}^{m,n} \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \geq \frac{(\gamma -\varepsilon )\delta }{2f(1)} \Biggr\} . \end{aligned}$$
Since \(x_{kj}\overset{[{\mathcal{I}_{2}^{L}}]}{\sim }y_{kj}\), it follows that the second set and thus the first set in the above expression belong to \(\mathcal{I}_{2}\). This proves that \(x_{kj} \overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y_{kj}\). □
Theorem 2.2
If
\(\lim_{t\rightarrow \infty }\frac{f(t)}{t}=\alpha >0\), then
\(x_{kj}\overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y_{kj}\Leftrightarrow x_{kj}\overset{[{\mathcal{I}_{2}^{L}}]}{\sim }y_{kj}\).
Proof
We showed that \(x_{kj}\overset{[{\mathcal{I}_{2}^{L}}]}{\sim }y_{kj} \Rightarrow x_{kj}\overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y_{kj}\) in Theorem 2.1. Now we must show that \(x_{kj} \overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y_{kj}\Rightarrow x_{kj} \overset{[{\mathcal{I}_{2}^{L}}]}{\sim }y_{kj}\).
Let \(\lim_{t\rightarrow \infty }\frac{f(t)}{t}=\alpha >0\). Then we have \(f(t)\geq \alpha t\) for all \(t\geq 0\). Assume that \(x_{kj} \overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y_{kj}\). Since
$$\begin{aligned} \frac{1}{mn}\sum_{k,j=1}^{m,n} f \biggl( \biggl\vert \frac{x_{kj}}{y _{kj}}-L \biggr\vert \biggr) \geq & \frac{1}{mn}{\sum_{k,j=1}^{m,n}} \alpha \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) =\alpha \Biggl(\frac{1}{mn}{\sum_{k,j=1}^{m,n}} \biggl\vert \frac{x _{kj}}{y_{kj}}-L \biggr\vert \Biggr), \end{aligned}$$
it follows that for each \(\varepsilon >0\), we have
$$\begin{aligned} & \Biggl\{ (m,n)\in \mathbb{N}\times \mathbb{N}:\frac{1}{mn}\sum _{k,j=1}^{m,n} \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \geq \varepsilon \Biggr\} \\ & \quad \subseteq \Biggl\{ (m,n)\in \mathbb{N}\times \mathbb{N}: \frac{1}{mn} \sum_{k,j=1}^{m,n}f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) \geq \alpha \varepsilon \Biggr\} . \end{aligned}$$
Since \(x_{kj}\overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y_{kj}\), it follows that the latter set and hence the former set in the above expression belong to \(\mathcal{I}_{2}\). This proves that \(x_{kj} \overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y_{kj}\Leftrightarrow x _{kj}\overset{[{\mathcal{I}_{2}^{L}}]}{\sim }y_{kj}\). □
Definition 2.3
Let f be a modulus function. Two nonnegative sequences \(x=(x_{kj})\) and \(y=(y_{kj})\) are said to be f-asymptotically lacunary \(\mathcal{I}_{2}\)-equivalent of multiple L if for every \(\varepsilon >0\),
$$\begin{aligned} \biggl\{ (k,j)\in I_{ru}:f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) \geq \varepsilon \biggr\} \in \mathcal{I}_{2} \end{aligned}$$
(denoted by \(x_{kj} \overset{{\mathcal{I}^{L}_{\theta _{2}}}(f)}{\sim }y_{kj}\))) and simply f-asymptotically lacunary \(\mathcal{I}_{2}\)-equivalent if \(L=1\).
Definition 2.4
Let f be a modulus function. Two nonnegative sequences \(x=(x_{kj})\) and \(y=(y_{kj})\) are said to be strongly f-asymptotically lacunary \(\mathcal{I}_{2}\)-equivalent of multiple L if for every \(\varepsilon >0\),
$$\begin{aligned} \biggl\{ (r,u)\in \mathbb{N}\times \mathbb{N}: \frac{1}{h_{ru}} \sum _{(k,j)\in I_{ru}}f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) \geq \varepsilon \biggr\} \in \mathcal{I}_{2} \end{aligned}$$
(denoted by \(x_{kj} \overset{[{\mathcal{I}^{L}_{\theta _{2}}}(f)]}{\sim }y_{kj}\))) and simply strongly f-asymptotically lacunary \(\mathcal{I}_{2}\)-equivalent if \(L=1\).
Theorem 2.3
Let
f
be a modulus function. Then, \(x_{kj} \overset{[{\mathcal{I}^{L}_{\theta _{2}}}]}{\sim }y_{kj}\Rightarrow x _{kj}\overset{[{\mathcal{I}^{L}_{\theta _{2}}}(f)]}{\sim }y_{kj}\).
Proof
Let \(x_{kj}\overset{[{\mathcal{I}^{L}_{\theta _{2}}}]}{\sim }y_{kj}\), and let \(\varepsilon >0\) be given. Choose \(0<\delta <1\) such that \(f(t)<\varepsilon \) for \(0\leq t\leq \delta \). We can write
$$\begin{aligned} \frac{1}{h_{ru}}\sum_{(k,j)\in I_{ru}}f \biggl( \biggl\vert \frac{x _{kj}}{y_{kj}}-L \biggr\vert \biggr) = &\frac{1}{h_{ru}} {\sum _{\substack{(k,j)\in I _{ru}\\\vert \frac{x _{kj}}{y_{kj}}-L \vert \leq \delta}}}f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) \\ &{} + \frac{1}{h_{ru}} {\sum_{\substack{(k,j)\in I_{ru}\\\vert \frac{x_{kj}}{y_{kj}}-L \vert > \delta}}}f \biggl( \biggl\vert \frac{x_{kj}}{y _{kj}}-L \biggr\vert \biggr), \end{aligned}$$
and so by Lemma 1
$$ \frac{1}{h_{ru}}\sum_{(k,j)\in I_{ru}}f \biggl( \biggl\vert \frac{x _{kj}}{y_{kj}}-L \biggr\vert \biggr)< \varepsilon + \biggl( \frac{2f(1)}{\delta } \biggr)\frac{1}{h_{ru}}\sum _{(k,j)\in I_{r}} \biggl\vert \frac{x _{kj}}{y_{kj}}-L \biggr\vert . $$
Thus, for each any \(\gamma >0\),
$$\begin{aligned} & \biggl\{ (r,u)\in \mathbb{N}\times \mathbb{N}: \frac{1}{h_{ru}} \sum _{(k,j)\in I_{ru}}f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) \geq \gamma \biggr\} \\ & \quad \subseteq \biggl\{ (r,u)\in \mathbb{N}\times \mathbb{N}: \frac{1}{h _{ru}}\sum_{(k,j)\in I_{ru}} \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \geq \frac{(\gamma -\varepsilon )\delta }{2f(1)} \biggr\} . \end{aligned}$$
Since \(x_{kj}\overset{[{\mathcal{I}^{L}_{\theta _{2}}}]}{\sim }y_{kj}\), it follows that the latter set and hence the former set in the above expression belong to \(\mathcal{I}_{2}\). This proves that \(x_{kj} \overset{[{\mathcal{I}^{L}_{\theta _{2}}}(f)]}{\sim }y_{kj}\). □
Theorem 2.4
If
\(\lim_{t\rightarrow \infty }\frac{f(t)}{t}=\alpha >0\), then
\(x_{kj}\overset{[{\mathcal{I}^{L}_{\theta _{2}}}(f)]}{\sim }y_{kj} \Leftrightarrow x_{kj} \overset{[{\mathcal{I}^{L}_{\theta _{2}}}]}{\sim }y_{kj} \).
Proof
We showed that \(x_{kj} \overset{[{\mathcal{I}^{L}_{\theta _{2}}}]}{\sim }y_{kj}\Rightarrow x _{kj}\overset{[{\mathcal{I}^{L}_{\theta _{2}}}(f)]}{\sim }y_{kj}\) in Theorem 2.3. Now we must show that \(x_{kj} \overset{[{\mathcal{I}^{L}_{\theta _{2}}}(f)]}{\sim }y_{kj}\Rightarrow x_{kj}\overset{[{\mathcal{I}^{L}_{\theta _{2}}}]}{\sim }y_{kj}\).
Let \(\lim_{t\rightarrow \infty }\frac{f(t)}{t}=\alpha >0\). Then we have \(f(t)\geq \alpha t\) for all \(t\geq 0\). Assume that \(x_{kj} \overset{[{\mathcal{I}^{L}_{\theta _{2}}}(f)]}{\sim }y_{kj}\). From
$$\begin{aligned} \frac{1}{h_{ru}}\sum_{(k,j)\in I_{ru}}f \biggl( \biggl\vert \frac{x _{kj}}{y_{kj}}-L \biggr\vert \biggr) \geq & \frac{1}{h_{ru}}{\sum_{(k,j)\in I_{ru}}}\alpha \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) \\ & \\ =& \alpha \biggl(\frac{1}{h_{ru}}{\sum_{(k,j)\in I_{ru}}} \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) \end{aligned}$$
it follows that for each \(\varepsilon >0\), we have
$$\begin{aligned} & \biggl\{ (r,u)\in \mathbb{N}\times \mathbb{N}:\frac{1}{h_{ru}} \sum _{(k,j)\in I_{ru}} \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \geq \varepsilon \biggr\} \\ & \quad \subseteq \biggl\{ (r,u)\in \mathbb{N}\times \mathbb{N}: \frac{1}{h _{ru}}\sum_{(k,j)\in I_{ru}}f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) \geq \alpha \varepsilon \biggr\} . \end{aligned}$$
Since \(x_{kj}\overset{[{\mathcal{I}^{L}_{\theta _{2}}}(f)]}{\sim }y _{kj}\), it follows that the latter set and hence the former set in the above expression belong to \(\mathcal{I}_{2}\). This proves that \(x_{kj}\overset{[{\mathcal{I}^{L}_{\theta _{2}}}]}{\sim }y_{kj}\Leftrightarrow x_{kj}\overset{[{\mathcal{I}^{L}_{\theta _{2}}}(f)]}{\sim }y_{kj}\). □
Theorem 2.5
Let
f
be a modulus function. If
\(\liminf_{r,u}q_{r,u}>1\), then
$$\begin{aligned} x_{kj}\overset{\bigl[{\mathcal{I}_{2}^{L}}(f) \bigr]}{\sim }y_{kj}\quad \Rightarrow\quad x _{kj}\overset{\bigl[{ \mathcal{I}_{\theta _{2}}^{L}}(f)\bigr]}{\sim }y_{kj}. \end{aligned}$$
Proof
Suppose that \(\liminf_{r,u}q_{r,u}>1\). Then there exists \(\eta >0\) such that \(q_{r,u}\geq 1+\eta \) for sufficiently large r, u. Then we have
$$\begin{aligned} \frac{h_{ru}}{k_{r}j_{u}}\geq \frac{\eta }{1+\eta }. \end{aligned}$$
Let \(x_{kj}\overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y_{kj}\). For sufficiently large r, u, we have
$$\begin{aligned} \frac{1}{k_{r}j_{u}}\sum_{k,j=1,1}^{k_{r}j_{u}}f \biggl( \biggl\vert \frac{x _{kj}}{y_{kj}}-L \biggr\vert \biggr) \geq & \frac{1}{k_{r}j_{u}}{\sum_{(k,j)\in I_{ru}}} f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) \\ = & \biggl(\frac{h_{ru}}{k_{r}j_{u}} \biggr)\frac{1}{h_{ru}}{\sum _{(k,j)\in I_{ru}}}f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) \\ \geq & \frac{\eta }{1 +\eta }\frac{1}{h_{ru}} {\sum _{(k,j)\in I_{ru}}}f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr), \end{aligned}$$
which gives, for any \(\varepsilon >0\),
$$\begin{aligned} & \biggl\{ (r,u)\in \mathbb{N}\times \mathbb{N}:\frac{1}{h_{ru}} \sum _{(k,j)\in I_{ru}} f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr)\geq \varepsilon \biggr\} \\ & \quad \subseteq \Biggl\{ (r,u)\in \mathbb{N}\times \mathbb{N}: \frac{1}{k _{r}j_{u}}\sum_{k,j=1,1}^{k_{r}j_{u}}f \biggl( \biggl\vert \frac{x_{kj}}{y _{kj}}-L \biggr\vert \biggr) \geq \frac{\varepsilon \eta }{1+\eta } \Biggr\} . \end{aligned}$$
Since \(x_{kj}\overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y_{kj}\), it follows that the latter set and hence the former set belong to \(\mathcal{I}_{2}\). This shows that \(x_{kj} \overset{[{\mathcal{I}_{\theta _{2}}^{L}}(f)]}{\sim }y_{kj}\). □
Theorem 2.6
Let
f
be a modulus function. Then, \(x_{kj} \overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y_{kj}\Rightarrow x_{kj} \overset{{\mathcal{I}_{2}(S)}}{\sim }y_{kj}\).
Proof
Assume that \(x_{kj}\overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y_{kj}\), and let \(\varepsilon >0\) be given. From
$$\begin{aligned} \frac{1}{mn}\sum_{k,j=1}^{m,n}f \biggl( \biggl\vert \frac{x_{kj}}{y _{kj}}-L \biggr\vert \biggr) \geq & \frac{1}{mn} {\sum_{\substack{k,j=1\\\vert \frac{x _{kj}}{y_{kj}}-L \vert \geq \varepsilon}}^{n}}f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) \\ \geq & f(\varepsilon )\cdot\frac{1}{mn} \biggl\vert \biggl\{ k\leq m, j \leq n: \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \geq \varepsilon \biggr\} \biggr\vert \end{aligned}$$
it follows that for any \(\gamma >0\), we have
$$\begin{aligned}& \biggl\{ (m,n)\in \mathbb{N}\times \mathbb{N}:\frac{1}{mn} \biggl\vert \biggl\{ k\leq m, j\leq n: \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \geq \varepsilon \biggr\} \biggr\vert \geq \frac{\gamma }{f(\varepsilon )} \biggr\} \\& \quad \subseteq \Biggl\{ (m,n)\in \mathbb{N}\times \mathbb{N}:\frac{1}{mn} \sum_{k,j=1}^{m,n} f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr)\geq \gamma \Biggr\} . \end{aligned}$$
Since \(x_{kj}\overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y_{kj}\), it follows the latter set and hence the former set in the above expression belong to \(\mathcal{I}_{2}\). Therefore \(x_{kj} \overset{{\mathcal{I}_{2}(S)}}{\sim }y_{kj}\). □
Theorem 2.7
Let
f
be a modulus function. If
f
is bounded, then
\(x_{kj} \overset{{\mathcal{I}_{2}(S)}}{\sim }y_{kj}\Leftrightarrow x_{kj} \overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y_{kj} \).
Proof
We showed that \(x_{kj}\overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y _{kj}\Rightarrow x_{kj}\overset{{\mathcal{I}_{2}(S)}}{\sim }y_{kj}\) in Theorem 2.6. Now we must show that \(x_{kj} \overset{{\mathcal{I}_{2}(S)}}{\sim }y_{kj}\Rightarrow x_{kj} \overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y_{kj}\).
Assume that f is bounded and let \(x_{kj} \overset{{\mathcal{I}_{2}(S)}}{\sim }y_{kj}\). Since f is bounded, there exists a positive real number M such that \(|f(x)|\leq M\) for all \(x\geq 0\). We have
$$\begin{aligned} \frac{1}{mn}\sum_{k,j=1}^{m,n}f \biggl( \biggl\vert \frac{x_{kj}}{y _{kj}}-L \biggr\vert \biggr) = & \frac{1}{mn} {\sum_{\substack{k,j=1\\\vert \frac{x_{kj}}{y _{kj}}-L \vert \geq \varepsilon}}^{m,n}} f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) \\ &{} +\frac{1}{mn} {\sum_{\substack{k,j=1\\\vert \frac{x_{kj}}{y_{kj}}-L \vert < \varepsilon}}^{m,n}} f \biggl( \biggl\vert \frac{x_{kj}}{y _{kj}}-L \biggr\vert \biggr) \\ \leq & \frac{M}{mn} \biggl\vert \biggl\{ k\leq m, j\leq n: \biggl\vert \frac{x _{kj}}{y_{kj}}-L \biggr\vert \geq \varepsilon \biggr\} \biggr\vert +f( \varepsilon ). \end{aligned}$$
This proves that \(x_{kj}\overset{[{\mathcal{I}_{2}^{L}}(f)]}{\sim }y _{kj}\). □
Theorem 2.8
Let
f
be a modulus function. Then
\(x_{kj} \overset{[{\mathcal{I}^{L}_{\theta _{2}}}(f)]}{\sim }y_{kj}\Rightarrow x_{kj}\overset{{\mathcal{I}(S_{\theta _{2})}}}{\sim }y_{kj}\).
Proof
Assume that \(x_{k} \overset{[{\mathcal{I}^{L}_{\theta _{2}}}(f)]}{\sim }y_{k}\), and let \(\varepsilon >0\) be given. From
$$\begin{aligned} \frac{1}{h_{ru}}\sum_{(k,j)\in I_{ru}}f \biggl( \biggl\vert \frac{x _{kj}}{y_{kj}}-L \biggr\vert \biggr) \geq & \frac{1}{h_{ru}} {\sum_{\substack{(k,j)\in I_{ru}\\\vert \frac{x_{kj}}{y_{kj}}-L \vert \geq \varepsilon}}}f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) \\ \geq & f(\varepsilon )\cdot\frac{1}{h_{ru}} \biggl\vert \biggl\{ (k,j) \in I_{ru}: \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \geq \varepsilon \biggr\} \biggr\vert \end{aligned}$$
it follows that for any \(\gamma >0\),
$$\begin{aligned}& \biggl\{ (r,u)\in \mathbb{N}\times \mathbb{N}:\frac{1}{h_{ru}} \biggl\vert \biggl\{ (k,j)\in I_{ru}: \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \geq \varepsilon \biggr\} \biggr\vert \geq \gamma \biggr\} \\& \quad \subseteq \biggl\{ (r,u)\in \mathbb{N}\times \mathbb{N}: \frac{1}{h _{ru}} \sum_{(k,j)\in I_{ru}}f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) \geq \gamma f(\varepsilon ) \biggr\} . \end{aligned}$$
Since \(x_{kj}\overset{[{\mathcal{I}^{L}_{\theta _{2}}}(f)]}{\sim }y _{kj}\), the last set belongs to \(\mathcal{I}_{2}\), and so by the definition of an ideal the first set belongs to \(\mathcal{I}_{2}\). Therefore \(x_{kj}\overset{{\mathcal{I}(S_{\theta _{2}})}}{\sim }y_{kj}\). □
Theorem 2.9
Let
f
be a modulus function. If
f
is bounded, then
\(x_{kj} \overset{{\mathcal{I}(S_{\theta _{2}})}}{\sim }y_{kj}\Leftrightarrow x_{kj}\overset{[{\mathcal{I}^{L}_{\theta _{2}}}(f)]}{\sim }y_{kj}\).
Proof
We showed that \(x_{kj} \overset{[{\mathcal{I}^{L}_{\theta _{2}}}(f)]}{\sim }y_{kj}\Rightarrow x_{kj}\overset{{\mathcal{I}(S_{\theta _{2}})}}{\sim }y_{kj}\) in Theorem 2.8. Now we must show that \(x_{kj} \overset{{\mathcal{I}(S_{\theta _{2}})}}{\sim }y_{kj}\Rightarrow x _{kj}\overset{[{\mathcal{I}^{L}_{\theta _{2}}}(f)]}{\sim }y_{kj}\).
Assume that f is bounded and let \(x_{kj} \overset{{\mathcal{I}(S_{\theta _{2}})}}{\sim }y_{kj}\). Since f is bounded, there exists a positive real number M such that \(|f(x)| \leq M\) for all \(x\geq 0\). We have
$$\begin{aligned} \frac{1}{h_{ru}}\sum_{(k,j)\in I_{ru}}f \biggl( \biggl\vert \frac{x _{kj}}{y_{kj}}-L \biggr\vert \biggr) = & \frac{1}{h_{ru}} {\sum_{\substack{(k,j)\in I_{ru}\\\vert \frac{x _{kj}}{y_{kj}}-L \vert \geq \varepsilon}}}f \biggl( \biggl\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert \biggr) \\ &{} +\frac{1}{h_{ru}} {\sum_{\substack{(k,j)\in I_{ru}\\\vert \frac{x_{kj}}{y_{kj}}-L \biggr\vert < \varepsilon}}}f \biggl( \biggl\vert \frac{x_{kj}}{y _{kj}}-L \biggr\vert \biggr) \\ \leq & \frac{M}{h_{ru}} \biggl\vert \biggl\{ (k,j)\in I_{ru}: \biggl\vert \frac{x _{kj}}{y_{kj}}-L \biggr\vert \geq \varepsilon \biggr\} \biggr\vert +f( \varepsilon ) . \end{aligned}$$
This proves that \(x_{kj} \overset{[{\mathcal{I}^{L}_{\theta _{2}}}(f)]}{\sim }y_{kj}\). □