Now, based on Tian’s [18, 19] research results, we will give the following generalizations and refinements of the three-tuple diamond-α integral and n-tuple diamond-α integral Hölder inequality on time scales.
Theorem 3.1
Let
\(\mathbb{T}\)be a time scale
\(a, b \in \mathbb{T}\)with
\(a< b\)and
\(\alpha _{kj}\in \mathbb{R}\)
\((j=1, 2, \ldots , m, k=1,2, \ldots , s), \sum_{k=1}^{s}\frac{1}{p_{k}}=1, \sum_{k=1}^{s}\alpha _{kj}=0\). If
\(f_{j}(\delta )>0\), and
\(f_{j}\ (j=1,2,\ldots , m)\)is a continuous real-valued function on
\([\xi , \sigma ]_{\mathbb{T}}\), then
(1) for
\(p_{k}>1\), we have
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \prod _{j=1}^{m} f_{j}(\delta _{1},\delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2} \Diamond _{\alpha } \delta _{3} \\ &\quad\leq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}}\prod _{j=1}^{m} f_{j}^{1+p_{k}\alpha _{kj}}( \delta _{1},\delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2} \Diamond _{\alpha } \delta _{3} \Biggr)^{\frac{1}{p_{k}}}, \end{aligned} $$
(9)
(2) for
\(0< p_{s}<1, p_{k}<0\ (k=1,2,\ldots ,s-1)\), we have
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \prod _{j=1}^{m} f_{j}(\delta _{1},\delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2} \Diamond _{\alpha } \delta _{3} \\ &\quad\geq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}}\prod _{j=1}^{m} f_{j}^{1+p_{k}\alpha _{kj}}( \delta _{1},\delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2} \Diamond _{\alpha } \delta _{3} \Biggr)^{\frac{1}{p_{k}}}. \end{aligned} $$
(10)
Proof
(1) Set
$$ \begin{aligned} g_{k}(\delta _{1}, \delta _{2}, \delta _{3})= \Biggl(\prod _{j=1}^{m} f_{j} ^{1+p_{k} \alpha _{k j}}(\delta _{1}, \delta _{2}, \delta _{3}) \Biggr)^{1 / p_{k}}. \end{aligned} $$
Applying the assumptions \(\sum_{k=1}^{s}\frac{1}{p_{k}}=1\) and \(\sum_{k=1}^{s}\alpha _{kj}=0\), by computing, we can observe that
$$\begin{aligned} & \prod_{k=1}^{s} g_{k}(\delta _{1}, \delta _{2}, \delta _{3}) \\ &\quad=g_{1} g _{2} \cdots g_{s} \\ &\quad= \Biggl(\prod_{j=1}^{m} f_{j}^{1+a_{1} \alpha _{1 j}}( \delta _{1}, \delta _{2}, \delta _{3}) \Biggr)^{1 / a_{1}} \Biggl( \prod_{j=1} ^{m} f_{j}^{1+a_{2} \alpha _{2 j}}( \delta _{1}, \delta _{2}, \delta _{3}) \Biggr)^{1 / a_{2}}\cdots \\ &\qquad{}\times \Biggl(\prod_{j=1}^{m} f_{j}^{1+a_{s} \alpha _{s_{j}}}(\delta _{1}, \delta _{2}, \delta _{3}) \Biggr)^{1 / a_{s}} \\ &\quad=\prod_{j=1}^{m} f_{j}^{1 / a_{1}+\alpha _{1 j}}(\delta _{1}, \delta _{2}, \delta _{3}) \prod _{j=1}^{m} f_{j}^{1 / a_{2}+\alpha _{2 j}}( \delta _{1}, \delta _{2}, \delta _{3}) \cdots \\ &\qquad{}\times \prod_{j=1}^{m} f_{j} ^{1 / a_{s}+\alpha _{s j}}(\delta _{1}, \delta _{2}, \delta _{3}) \\ &\quad= \prod_{j=1}^{m} f_{j}^{1 / a_{1}+1 / a_{2}+\cdots +1 / a_{s}+\alpha _{1 j}+\alpha _{2 j}+\cdots +\alpha _{s j}}(\delta _{1}, \delta _{2}, \delta _{3})=\prod _{j=1}^{m} f_{j}(\delta _{1}, \delta _{2}, \delta _{3}). \end{aligned}$$
Hence, we obtain
$$ \begin{aligned}[b] &\int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \prod _{j=1}^{m} f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2} \Diamond _{\alpha } \delta _{3} \\ &\quad = \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \prod _{k=1} ^{s} g_{k}(\delta _{1}, \delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2}\Diamond _{\alpha } \delta _{3}. \end{aligned} $$
(11)
By the Hölder inequality (3), we find
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \prod _{k=1}^{s} g_{k}(\delta _{1}, \delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2} \Diamond _{\alpha } \delta _{3} \\ &\quad\leq \prod_{k=1}^{s} \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} g_{k}^{p_{k}}( \delta _{1}, \delta _{2}, \delta _{3}) \Diamond _{ \alpha } \delta _{1}\Diamond _{\alpha } \delta _{2}\Diamond _{\alpha } \delta _{3} \biggr)^{1 / p_{k}}. \end{aligned} $$
(12)
Substituting \(g_{k}(\delta _{1},\delta _{2},\delta _{3})\) into the inequality (12) can be obtained
$$ \begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \prod _{j=1}^{m} f_{j}(\delta _{1},\delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2} \Diamond _{\alpha } \delta _{3} \\ &\quad\leq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \prod _{j=1}^{m} f_{j}^{1+p_{k} \alpha _{k j}}( \delta _{1}, \delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1} \Diamond _{\alpha } \delta _{2} \Diamond _{\alpha } \delta _{3} \Biggr)^{\frac{1}{p_{k}}}. \end{aligned} $$
(2) After the same proof as inequality (9), we get
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \prod _{k=1}^{s} g_{k}(\delta _{1}, \delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2} \Diamond _{\alpha } \delta _{3} \\ &\quad\geq \prod_{k=1}^{s} \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} g_{k}^{p_{k}}( \delta _{1}, \delta _{2}, \delta _{3}) \Diamond _{ \alpha } \delta _{1}\Diamond _{\alpha } \delta _{2}\Diamond _{\alpha } \delta _{3} \biggr)^{1 / p_{k}}. \end{aligned} $$
(13)
Substituting \(g_{k}(\delta _{1},\delta _{2}.\delta _{3})\) into the (12) can be obtained
$$\begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \prod _{j=1}^{m} f_{j}(\delta _{1},\delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2} \Diamond _{\alpha } \delta _{3} \\ &\quad\geq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \prod _{j=1}^{m} f_{j}^{1+p_{k} \alpha _{k j}}( \delta _{1}, \delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1} \Diamond _{\alpha } \delta _{2} \Diamond _{\alpha } \delta _{3} \Biggr)^{\frac{1}{p_{k}}}. \end{aligned}$$
The proof of Theorem 3.1 is accomplished. □
Corollary 3.2
Under the conditions of Theorem
3.1, let
\(s=m, \alpha _{kj}=-t/p_{k}\)for
\(k \neq j\)and
\(\alpha _{jj}=t(1-1/p_{j})\)with
\(t \in \mathbb{R}\), then
(1) for
\(p_{k}>1\), we have the following inequality:
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \prod _{j=1}^{m} f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2} \Diamond _{\alpha } \delta _{3} \\ &\quad\leq \prod_{k=1}^{m} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \Biggl(\prod _{j=1}^{m} f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \Biggr)^{1-t} \\ &\qquad{}\times \bigl(f_{k}^{p_{k}}(\delta _{1}, \delta _{2}, \delta _{3}) \bigr)^{t} \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2}\Diamond _{\alpha } \delta _{3} \Biggr)^{1 / p_{k}}, \end{aligned} $$
(14)
(2) \(0< p_{m}<1, p_{k}<0\ (k=1,2,\ldots , m-1)\), we have the following reverse inequality:
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \prod _{j=1}^{m} f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2} \Diamond _{\alpha } \delta _{3} \\ &\quad\geq \prod_{k=1}^{m} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \Biggl(\prod _{j=1}^{m} f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \Biggr)^{1-t} \\ &\qquad{}\times \bigl(f_{k}^{p_{k}}(\delta _{1}, \delta _{2}, \delta _{3}) \bigr)^{t} \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2}\Diamond _{\alpha } \delta _{3} \Biggr)^{1 / p_{k}}. \end{aligned} $$
(15)
On the basis of Theorem 3.1, we give the n-tuple diamond-α integral Hölder’s inequality on time scales.
Theorem 3.3
Let
\(\mathbb{T}\)be a time scale
\(\xi ,\sigma \in \mathbb{T}\)with
\(\xi <\sigma \)and
\(\alpha _{kj}\in \mathbb{R}\)
\((j=1, 2, \ldots , m, k=1,2, \ldots , s), \sum_{k=1}^{s}\frac{1}{p_{k}}=1, \sum_{k=1}^{s} \alpha _{kj}=0\). If
\(f_{j}(\delta )>0\), and
\(f_{j}\ (j=1,2,\ldots , m)\)is a continuous real-valued function on
\([\xi , \sigma ]_{\mathbb{T}}\), then
(1) for
\(p_{k}>1\), we have the following inequality:
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \prod _{j=1}^{m} f_{j}(\delta _{1},\delta _{2},\ldots , \delta _{n}) \Diamond _{\alpha } \delta _{1}\Diamond _{ \alpha } \delta _{2}\cdots \Diamond _{\alpha } \delta _{n} \\ &\quad\leq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}}\prod _{j=1}^{m} f_{j}^{1+p_{k} \alpha _{kj}}( \delta _{1},\delta _{2},\ldots , \delta _{n})\Diamond _{ \alpha } \delta _{1}\Diamond _{\alpha } \delta _{2}\cdots \Diamond _{ \alpha } \delta _{n} \Biggr)^{\frac{1}{p_{k}}}, \end{aligned} $$
(16)
(2) for
\(0< p_{s}<1, p_{k}<0\ (k=1,2,\ldots ,s-1)\), we have the following reverse inequality:
$$\begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \prod _{j=1}^{m} f_{j}(\delta _{1},\delta _{2},\ldots , \delta _{n}) \Diamond _{\alpha } \delta _{1}\Diamond _{ \alpha } \delta _{2}\cdots \Diamond _{\alpha } \delta _{n} \\ &\quad\geq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}}\prod _{j=1}^{m} f_{j}^{1+p_{k} \alpha _{kj}}( \delta _{1},\delta _{2},\ldots , \delta _{n})\Diamond _{ \alpha } \delta _{1}\Diamond _{\alpha } \delta _{2}\cdots \Diamond _{ \alpha } \delta _{n} \Biggr)^{\frac{1}{p_{k}}}. \end{aligned}$$
(17)
Proof
Similar to the proof of Theorem 3.1, we get the result of Theorem 3.3. □
Remark 3.4
The three-tuple diamond-\(alpha\) inequalities in Theorem 3.1 and the n-tuple diamond-α inequalities in Theorem 3.3 are generalizations to Theorem 3.3 in Ref. [10].
Theorem 3.5
Let
\(\mathbb{T}\)be a time scale, \(\xi , \sigma \in \mathbb{T}\)with
\(\xi <\sigma \)and
\(r\in \mathbb{R}, \alpha _{kj}\in \mathbb{R}\ (j=1,2, \ldots ,m, k=1,2,\ldots ,s)\), \(\sum_{k=1}^{s}\frac{1}{p_{k}}=r, \sum_{k=1}^{s}\alpha _{kj}=0\). If
\(f_{j}(\delta )>0\), and
\(f_{j}\ (j=1,2, \ldots ,m)\)is a continuous real-valued function on
\([\xi , \sigma ]_{ \mathbb{T}}\), then
(1) for
\(rp_{k}>1\), we have the following inequality:
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \prod _{j=1}^{m} f_{j}(\delta _{1},\delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2} \Diamond _{\alpha } \delta _{3} \\ &\quad\leq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}}\prod _{j=1}^{m} f_{j}^{1+rp_{k}\alpha _{kj}}( \delta _{1}, \delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2}\Diamond _{\alpha } \delta _{3} \Biggr)^{\frac{1}{rp_{k}}}, \end{aligned} $$
(18)
(2) for
\(0< rp_{k}<1, rp_{k}<0\ (k=1, 2,\ldots , s-1)\), we have the following reverse inequality:
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \prod _{j=1}^{m} f_{j}(\delta _{1},\delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2} \Diamond _{\alpha } \delta _{3} \\ &\quad\geq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}}\prod _{j=1}^{m} f_{j}^{1+rp_{k}\alpha _{kj}}( \delta _{1}, \delta _{2}, \delta _{3}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2}\Diamond _{\alpha } \delta _{3} \Biggr)^{\frac{1}{rp_{k}}}. \end{aligned} $$
(19)
Proof
(1) According to \(rp_{k}>1\) and \(\sum_{k=1}^{s}\frac{1}{p_{k}}=r\), we get \(\sum_{k=1}^{s}\frac{1}{rp_{k}}=1\). Then, by inequality (9), we immediately obtain the inequality (18).
(2) According to \(0< rp_{k}<1, rp_{k}<0\ (k=1,2,\ldots ,s-1)\) and \(\sum_{k=1}^{s}\frac{1}{p_{k}}=r\), we have \(\sum_{k=1}^{s}\frac{1}{rp _{k}}=1\), by inequality (10), we immediately have the inequality (19). This completes the proof. □
Similarly, on the basis of Theorem 3.5, we give the n-tuple diamond-α integral Hölder’s inequality on time scales.
Theorem 3.6
Let
\(\mathbb{T}\)be a time scale, \(\xi , \sigma \in \mathbb{T}\)with
\(\xi <\sigma \)and
\(r\in \mathbb{R}, \alpha _{kj}\in \mathbb{R}\ (j=1,2, \ldots ,m, k=1,2,\ldots ,s)\), \(\sum_{k=1}^{s}\frac{1}{p_{k}}=r, \sum_{k=1}^{s}\alpha _{kj}=0\). If
\(f_{j}(\delta )>0\), and
\(f_{j}\ (j=1,2, \ldots ,m)\)is a continuous real-valued function on
\([\xi , \sigma ]_{ \mathbb{T}}\), then
(1) for
\(rp_{k}>1\), we have the following inequality:
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \prod _{j=1}^{m} f_{j}(\delta _{1},\delta _{2},\ldots , \delta _{n}) \Diamond _{\alpha } \delta _{1}\Diamond _{ \alpha } \delta _{2}\cdots \Diamond _{\alpha } \delta _{n} \\ &\quad\leq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}}\prod _{j=1}^{m} f_{j}^{1+rp _{k}\alpha _{kj}}( \delta _{1},\delta _{2},\ldots , \delta _{n}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2}\cdots \Diamond _{\alpha } \delta _{n} \Biggr)^{\frac{1}{rp_{k}}}, \end{aligned} $$
(20)
(2) for
\(0< rp_{k}<1, rp_{k}<0\ (k=1, 2,\ldots , s-1)\), we have the following reverse inequality:
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \prod _{j=1}^{m} f_{j}(\delta _{1},\delta _{2},\ldots , \delta _{n}) \Diamond _{\alpha } \delta _{1}\Diamond _{ \alpha } \delta _{2}\cdots \Diamond _{\alpha } \delta _{n} \\ &\quad\geq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}}\prod _{j=1}^{m} f_{j}^{1+rp _{k}\alpha _{kj}}( \delta _{1},\delta _{2},\ldots , \delta _{n}) \Diamond _{\alpha } \delta _{1}\Diamond _{\alpha } \delta _{2}\cdots \Diamond _{\alpha } \delta _{n} \Biggr)^{\frac{1}{rp_{k}}}. \end{aligned} $$
(21)
Proof
Similar to the proof of Theorem 3.5, we get the result of Theorem 3.6. □
Remark 3.7
For the inequality of Theorem 3.4 in the Reference [10], we put forward Theorem 3.5 and Theorem 3.6 as the generalization results.
Theorem 3.8
Assume that
\(\mathbb{T}\)is a time scale, \(\xi , \sigma \in \mathbb{T}\)with
\(\xi <\sigma \)and
\(p_{k}>0, \alpha _{kj}\in \mathbb{R}\ (j=1,2,\ldots ,m, k=1,2,\ldots , s), \sum_{k=1}^{s}\frac{1}{p _{k}}=1, \sum_{k=1}^{s}\alpha _{kj}=0, f_{j},h:\mathbb{T}\to \mathbb{R}\). Ifhand
\(f_{j}\)are ◊-integrable on
\([\xi , \sigma ]_{\mathbb{T}}\), then the following assertions hold true.
(1) For
\(p_{k}>1\), one has
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Biggl\vert \prod_{j=1}^{m} f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \Biggr\vert \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad \leq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \prod_{j=1}^{m} \bigl\vert f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{1+p_{k} \alpha _{k j}} \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \Biggr)^{1 / p_{k}}. \end{aligned} $$
(22)
(2) For
\(0< p_{s}<1, p_{k}<0\ (k=1,2,\ldots ,s-1), f_{j}^{1+p_{k}\alpha _{kj}}\)is ◊-integrable on
\([\xi , \sigma ]_{\mathbb{T}}\), one has
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Biggl\vert \prod_{j=1}^{m} f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \Biggr\vert \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad \geq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \prod_{j=1}^{m} \bigl\vert f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{1+p_{k} \alpha _{k j}} \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \Biggr)^{1 / p_{k}}. \end{aligned} $$
(23)
Proof
(1) Let
$$ \begin{aligned} g_{k}(\delta _{1}, \delta _{2}, \delta _{3})= \Biggl(\prod _{j=1}^{m} f_{j} ^{1+p_{k}\alpha _{kj}(\delta _{1}, \delta _{2}, \delta _{3})} \Biggr)^{1/p _{k}}. \end{aligned} $$
Based on the assumptions \(\sum_{k=1}^{s}\frac{1}{p_{k}}=1\) and \(\sum_{k=1}^{s}\alpha _{kj}=0\), from a direct computation, it is obvious to show that
$$\begin{aligned} &\prod_{k=1}^{s} g_{k}(\delta _{1}, \delta _{2}, \delta _{3})\\ &\quad=g_{1} g _{2} \cdots g_{s} \\ &\quad= \Biggl(\prod_{j=1}^{m} f_{j}^{1+a_{1} \alpha _{1 j}}( \delta _{1}, \delta _{2}, \delta _{3}) \Biggr)^{1 / a_{1}} \Biggl( \prod_{j=1} ^{m} f_{j}^{1+a_{2} \alpha _{2 j}}( \delta _{1}, \delta _{2}, \delta _{3}) \Biggr)^{1 / a_{2}}\cdots \\ &\qquad{}\times\Biggl(\prod_{j=1}^{m} f_{j}^{1+a_{s} \alpha _{s_{j}}}(\delta _{1}, \delta _{2}, \delta _{3}) \Biggr)^{1 / a_{s}} \\ &\quad=\prod_{j=1}^{m} f_{j}^{1 / a_{1}+\alpha _{1 j}}(\delta _{1}, \delta _{2}, \delta _{3}) \prod _{j=1}^{m} f_{j}^{1 / a_{2}+\alpha _{2 j}}( \delta _{1}, \delta _{2}, \delta _{3}) \cdots \\ &\qquad{}\times \prod_{j=1}^{m} f_{j} ^{1 / a_{s}+\alpha _{s j}}(\delta _{1}, \delta _{2}, \delta _{3}) \\ &\quad= \prod_{j=1}^{m} f_{j}^{1 / a_{1}+1 / a_{2}+\cdots +1 / a_{s}+\alpha _{1 j}+\alpha _{2 j}+\cdots +\alpha _{s j}}(\delta _{1}, \delta _{2}, \delta _{3})=\prod _{j=1}^{m} f_{j}(\delta _{1}, \delta _{2}, \delta _{3}). \end{aligned}$$
From the above result, we can obtain
$$ \begin{aligned} \prod_{k=1}^{s} g_{k}(\delta _{1}, \delta _{2}, \delta _{3})=\prod_{j=1} ^{m}f_{j}(\delta _{1}, \delta _{2}, \delta _{3}). \end{aligned} $$
Hence, we have
$$ \begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Biggl\vert \prod_{j=1}^{m} f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \Biggr\vert \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad= \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Biggl\vert \prod_{j=1}^{m} g_{k}(\delta _{1}, \delta _{2}, \delta _{3}) \Biggr\vert \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3}. \end{aligned} $$
It follows from Hölder’s inequality (5) that
$$ \begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Biggl\vert \prod_{j=1}^{m} g_{k}(\delta _{1}, \delta _{2}, \delta _{3}) \Biggr\vert \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad\leq \prod_{k=1}^{s} \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert g_{k}(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{p _{k}} \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{1/p _{k}}. \end{aligned} $$
Thus, we have
$$ \begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Biggl\vert \prod_{j=1}^{m} f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \Biggr\vert \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad \leq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \prod_{j=1}^{m} \bigl\vert f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{1+p_{k} \alpha _{k j}} \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \Biggr)^{1 / p_{k}}. \end{aligned} $$
(2) The proof of inequality (23) is similar to the proof of inequality (22), we have
$$ \begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Biggl\vert \prod_{j=1}^{m} g_{k}(\delta _{1}, \delta _{2}, \delta _{3}) \Biggr\vert \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad\geq \prod_{k=1}^{s} \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert g_{k}(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{p _{k}} \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{1/p _{k}}. \end{aligned} $$
Thus, we have
$$ \begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Biggl\vert \prod_{j=1}^{m} f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \Biggr\vert \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ & \quad\geq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \prod_{j=1}^{m} \bigl\vert f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{1+p_{k} \alpha _{k j}} \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \Biggr)^{1 / p_{k}}. \end{aligned} $$
Thus, the proof of Theorem 3.8 is completed. □
Corollary 3.9
Under the assumptions of Theorem
3.8, taking
\(s=m, \alpha _{kj}=-t/p_{k}\)for
\(j \neq k\)and
\(\alpha _{kk}=t(1-1/p_{k})\)with
\(t \in \mathbb{R}\), the following assertions hold true.
(1) For
\(p_{k}>1\), one has
$$ \begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Biggl\vert \prod_{j=1}^{m} f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \Biggr\vert \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad\leq \prod_{k=1}^{m} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Biggl(\prod_{j=1}^{m} \bigl\vert f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Biggr)^{1-t} \\ &\qquad{}\times \bigl( \bigl\vert f_{k}(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{p_{k}}\bigr)^{t}\Diamond \delta _{1} \Diamond \delta _{2}\Diamond \delta _{3} \Biggr)^{1/p_{k}}. \end{aligned} $$
(2) For
\(0< p_{m}<1, p_{k}<0\ (k=1,2,\ldots ,m-1)\), one has
$$ \begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Biggl\vert \prod_{j=1}^{m} f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \Biggr\vert \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad\geq \prod_{k=1}^{m} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Biggl(\prod_{j=1}^{m} \bigl\vert f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Biggr)^{1-t} \\ &\qquad{}\times \bigl( \bigl\vert f_{k}(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{p_{k}}\bigr)^{t}\Diamond \delta _{1} \Diamond \delta _{2}\Diamond \delta _{3} \Biggr)^{1/p_{k}}. \end{aligned} $$
Theorem 3.10
Assume that
\(\mathbb{T}\)is a time scale, \(\xi , \sigma \in \mathbb{T}\)with
\(\xi <\sigma \)and
\(p_{k}>0, \alpha _{kj}\in \mathbb{R}\ (j=1,2,\ldots ,m, k=1,2,\ldots , s), \sum_{k=1}^{s}\frac{1}{p _{k}}=1, \sum_{k=1}^{s}\alpha _{kj}=0, f_{j},h:\mathbb{T}\to \mathbb{R}\). Ifhand
\(f_{j}\)are ◊-integrable on
\([\xi , \sigma ]_{\mathbb{T}}\), then the following assertions hold true.
(1) For
\(p_{k}>1\), one has
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \bigl\vert h(\delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \Biggl\vert \prod_{j=1}^{m} f_{j}(\delta _{1},\delta _{2}, \ldots , \delta _{n}) \Biggr\vert \Diamond \delta _{1} \Diamond \delta _{2} \cdots \Diamond \delta _{n} \\ &\quad \leq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}} ^{\sigma _{n}} \bigl\vert h(\delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \\ &\qquad{}\times\prod_{j=1}^{m} \bigl\vert f_{j}(\delta _{1},\delta _{2}, \ldots , \delta _{n}) \bigr\vert ^{1+p_{k} \alpha _{k j}} \Diamond \delta _{1}\Diamond \delta _{2} \cdots \Diamond \delta _{n} \Biggr)^{1 / p_{k}}. \end{aligned} $$
(24)
(2) For
\(0< p_{s}<1, p_{k}<0\ (k=1,2,\ldots ,s-1), f_{j}^{1+p_{k}\alpha _{kj}}\)is ◊-integrable on
\([\xi , \sigma ]_{\mathbb{T}}\), one has
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \bigl\vert h(\delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \Biggl\vert \prod_{j=1}^{m} f_{j}(\delta _{1},\delta _{2}, \ldots , \delta _{n}) \Biggr\vert \Diamond \delta _{1} \Diamond \delta _{2} \cdots \Diamond \delta _{n} \\ & \quad\geq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}} ^{\sigma _{n}} \bigl\vert h(\delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \\ &\qquad{}\times\prod_{j=1}^{m} \bigl\vert f_{j}(\delta _{1},\delta _{2}, \ldots , \delta _{n}) \bigr\vert ^{1+p_{k} \alpha _{k j}} \Diamond \delta _{1}\Diamond \delta _{2} \cdots \Diamond \delta _{n} \Biggr)^{1 / p_{k}}. \end{aligned} $$
(25)
Proof
Similar to the proof of Theorem 3.8, we get the result of Theorem 3.10. □
Remark 3.11
The inequalities in Theorem 3.8 and Theorem 3.10 are the result of generalization of Theorem 4.1 in Ref. [17].
Theorem 3.12
Assume that
\(\mathbb{T}\)is a time scale, \(\xi , \sigma \in \mathbb{T}\)with
\(\xi <\sigma \)and
\(p_{k}>0, r\in \mathbb{R}, \alpha _{kj}\in \mathbb{R}\ (j=1,2,\ldots ,m, k=1,2,\ldots ,s), \sum_{k=1} ^{s}\frac{1}{p_{k}}=r, \sum_{k=1}^{s}\alpha _{kj}=0, f_{j},h: \mathbb{T}\to \mathbb{R}\). If
\(f_{j}\)and h are ◊-integrable on
\([\xi , \sigma ]_{T}\), then the following assertions hold true.
(1) For
\(rp_{k}>1\), one has
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Biggl\vert \prod_{j=1}^{m} f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \Biggr\vert \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad\leq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \prod_{j=1}^{m} \bigl\vert f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{1+rp _{k}\alpha _{kj}}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \Biggr)^{1/rp_{k}}. \end{aligned} $$
(26)
(2) For
\(0< rp_{k}<1, rp_{k}<0\ (k=1,2,\ldots , s-1), f_{j}^{1+rp_{k} \alpha _{kj}}\)is ◊-integrable on
\([\xi , \sigma ]_{ \mathbb{T}}\), one has
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Biggl\vert \prod_{j=1}^{m} f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \Biggr\vert \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad\geq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \prod_{j=1}^{m} \bigl\vert f_{j}(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{1+rp _{k}\alpha _{kj}}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \Biggr)^{1/rp_{k}}. \end{aligned} $$
(27)
Proof
(1) Since \(rp_{k}>1\) and \(\sum_{k=1}^{s}\frac{1}{rp_{k}}=1\). Then by inequality (22) we can obtain inequality (26).
(2) Since \(0< rp_{s}<1, rp_{k}<0\) and \(\sum_{k=1}^{s}\frac{1}{rp_{k}}=1\), by inequality (23), we can obtain inequality (27).
The proof of Theorem 3.12 is completed. □
Theorem 3.13
Assume that
\(\mathbb{T}\)is a time scale, \(\xi , \sigma \in \mathbb{T}\)with
\(\xi <\sigma \)and
\(p_{k}>0, r\in \mathbb{R}, \alpha _{kj}\in \mathbb{R}\ (j=1,2,\ldots ,m, k=1,2,\ldots ,s), \sum_{k=1} ^{s}\frac{1}{p_{k}}=r, \sum_{k=1}^{s}\alpha _{kj}=0, f_{j},h: \mathbb{T}\to \mathbb{R}\). If
\(f_{j}\)and h are ◊-integrable on
\([\xi , \sigma ]_{T}\), then the following assertions hold true.
(1) For
\(rp_{k}>1\), one has
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \bigl\vert h(\delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \Biggl\vert \prod_{j=1}^{m} f_{j}(\delta _{1},\delta _{2}, \ldots , \delta _{n}) \Biggr\vert \Diamond \delta _{1} \Diamond \delta _{2} \cdots \Diamond \delta _{n} \\ &\quad\leq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}} ^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \bigl\vert h(\delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \\ &\qquad{}\times\prod_{j=1}^{m} \bigl\vert f_{j}(\delta _{1},\delta _{2}, \ldots , \delta _{n}) \bigr\vert ^{1+rp_{k}\alpha _{kj}}\Diamond \delta _{1}\Diamond \delta _{2} \cdots \Diamond \delta _{n} \Biggr)^{1/rp_{k}}. \end{aligned} $$
(28)
(2) For
\(0< rp_{k}<1, rp_{k}<0\ (k=1,2,\ldots , s-1), f_{j}^{1+rp_{k} \alpha _{kj}}\)is ◊-integrable on
\([\xi , \sigma ]_{ \mathbb{T}}\), one has
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \bigl\vert h(\delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \Biggl\vert \prod_{j=1}^{m} f_{j}(\delta _{1},\delta _{2}, \ldots , \delta _{n}) \Biggr\vert \Diamond \delta _{1} \Diamond \delta _{2} \cdots \Diamond \delta _{n} \\ &\quad\geq \prod_{k=1}^{s} \Biggl( \int _{\xi _{1}} ^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \bigl\vert h(\delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \\ &\qquad{}\times\prod_{j=1}^{m} \bigl\vert f_{j}(\delta _{1},\delta _{2}, \ldots , \delta _{n}) \bigr\vert ^{1+rp_{k}\alpha _{kj}}\Diamond \delta _{1}\Diamond \delta _{2} \cdots \Diamond \delta _{n} \Biggr)^{1/rp_{k}}. \end{aligned} $$
(29)
Proof
Similar to the proof of Theorem 3.12, we get the result of Theorem 3.13. □
Theorem 3.14
Let
\(f, g, h:\mathbb{T}\to \mathbb{R}\)be ◊-integrable on
\([\xi , \sigma ]_{T}\), and
\(s, t \in \mathbb{R}\), and let
\(p=(s-t)/(1-t), q=(s-t)/(s-1)\).
(1) If
\(s<1<t\)or
\(s>1>t\), then
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert f( \delta _{1}, \delta _{2}, \delta _{3})g(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad\leq \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert f(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{sp}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{1/p^{2}} \\ &\qquad{} \times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert g(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{tq}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{1/q^{2}} \\ &\qquad{} \times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert f(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{tp}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\qquad{}\times \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert g(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{sq}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{1/pq}. \end{aligned} $$
(30)
(2) If
\(s>t>1\)or
\(s< t<1\); \(t>s>1\)or
\(t< s<1\), then
$$\begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert f( \delta _{1}, \delta _{2}, \delta _{3})g(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad\geq \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert f(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{sp}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{1/p^{2}} \\ & \qquad{}\times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert g(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{tq}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{1/q^{2}} \\ &\qquad{} \times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert f(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{tp}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\qquad{}\times \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert g(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{sq}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{1/pq}. \end{aligned}$$
(31)
Proof
(1) Let \(p=\frac{s-t}{1-t}\) and in view of \(s<1<t\) or \(s>1>t\), we have
$$ \begin{aligned} p=\frac{s-t}{1-t}>1, \end{aligned} $$
by Hölder’s inequality (5) with indices \(\frac{s-t}{1-t}\) and \(\frac{s-t}{s-1}\), we have
$$ \begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \vert h \vert \vert fg \vert \Diamond \delta _{1}\Diamond \delta _{2} \Diamond \delta _{3} \\ &\quad= \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \vert h \vert \vert fg \vert ^{s(1-t)/(s-t)} \vert fg \vert ^{t(s-1)/(s-t)} \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad\leq \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \vert h \vert \vert fg \vert ^{s}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{(1-t)/(s-t)} \\ &\qquad{}\times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \vert h \vert \vert fg \vert ^{t}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{(s-1)/(s-t)}. \end{aligned} $$
On the other hand, from Hölder’s inequality (5) again for \(p=\frac{s-t}{1-t}>1\), it follows that the following two inequalities are true:
$$ \begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \vert h \vert \vert fg \vert ^{s}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad\leq \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}} ^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \vert h \vert \vert f \vert ^{s(s-t)/(1-t)}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{(1-t)/(s-t)} \\ &\qquad{}\times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \vert h \vert \vert g \vert ^{s(s-t)/(s-1)}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{(s-1)/(s-t)} \end{aligned} $$
and
$$\begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \vert h \vert \vert fg \vert ^{t}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad\leq \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}} ^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \vert h \vert \vert f \vert ^{t(s-t)/(1-t)}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{(1-t)/(s-t)} \\ &\qquad{}\times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \vert h \vert \vert g \vert ^{t(s-t)/(s-1)}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{(s-1)/(s-t)}. \end{aligned}$$
Thus, we have
$$\begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert f( \delta _{1}, \delta _{2}, \delta _{3})g(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad\leq \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert f(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{sp}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{1/p^{2}} \\ & \qquad{}\times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert g(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{tq}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{1/q^{2}} \\ &\qquad{} \times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert f(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{tp}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\qquad{}\times \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert g(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{sq}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{1/pq}. \end{aligned}$$
(2) Let \(p=\frac{s-t}{1-t}\) and in view of \(s>t>1\) or \(s< t<1\), we have
$$ \begin{aligned} p=\frac{s-t}{1-t}< 0 \end{aligned} $$
and \(t>s>1\) or \(t< s<1\), we have \(0<\frac{s-t}{1-t}<1\), by the reverse Hölder inequality (6) with indices \(\frac{s-t}{1-t}\) and \(\frac{s-t}{s-1}\), we have
$$ \begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \vert h \vert \vert fg \vert \Diamond \delta _{1}\Diamond \delta _{2} \Diamond \delta _{3} \\ &\quad= \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \vert h \vert \vert fg \vert ^{s(1-t)/(s-t)} \vert fg \vert ^{t(s-1)/(s-t)} \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad\geq \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \vert h \vert \vert fg \vert ^{s}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{(1-t)/(s-t)} \\ &\qquad{}\times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \vert h \vert \vert fg \vert ^{t}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{(s-1)/(s-t)}. \end{aligned} $$
On the other hand, from Hölder’s inequality (6) again for \(0< p=\frac{s-t}{1-t}<1\) or \(p=\frac{s-t}{1-t}<0\), it follows that the following two inequalities are true:
$$ \begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \vert h \vert \vert fg \vert ^{s}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad\geq \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}} ^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \vert h \vert \vert f \vert ^{s(s-t)/(1-t)}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{(1-t)/(s-t)} \\ &\qquad{}\times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \vert h \vert \vert g \vert ^{s(s-t)/(s-1)}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{(s-1)/(s-t)} \end{aligned} $$
and
$$ \begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \vert h \vert \vert fg \vert ^{t}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad\geq \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}} ^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \vert h \vert \vert f \vert ^{t(s-t)/(1-t)}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{(1-t)/(s-t)} \\ &\qquad{}\times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \vert h \vert \vert g \vert ^{t(s-t)/(s-1)}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{(s-1)/(s-t)}. \end{aligned} $$
Thus, we have
$$ \begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}} ^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert f( \delta _{1}, \delta _{2}, \delta _{3})g(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\quad\geq \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert f(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{sp}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{1/p^{2}} \\ &\qquad{} \times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert g(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{tq}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{1/q^{2}} \\ & \qquad{}\times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert f(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{tp}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \\ &\qquad{}\times \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}} \int _{\xi _{3}}^{\sigma _{3}} \bigl\vert h(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert \bigl\vert g(\delta _{1}, \delta _{2}, \delta _{3}) \bigr\vert ^{sq}\Diamond \delta _{1}\Diamond \delta _{2}\Diamond \delta _{3} \biggr)^{1/pq}. \end{aligned} $$
Thus, the proof of Theorem 3.14 is completed. □
Theorem 3.15
Let
\(f, g, h:\mathbb{T}\to \mathbb{R}\)be ◊-integrable on
\([\xi , \sigma ]_{T}\), and
\(s, t \in \mathbb{R}\), and let
\(p=(s-t)/(1-t), q=(s-t)/(s-1)\).
(1) If
\(s<1<t\)or
\(s>1>t\), then
$$ \begin{aligned}[b] & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \bigl\vert h(\delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \bigl\vert f(\delta _{1},\delta _{2}, \ldots , \delta _{n})g( \delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \Diamond \delta _{1} \Diamond \delta _{2}\cdots \Diamond \delta _{n} \\ &\quad\leq \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \bigl\vert h(\delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \bigl\vert f(\delta _{1},\delta _{2}, \ldots , \delta _{n}) \bigr\vert ^{sp}\Diamond \delta _{1}\Diamond \delta _{2}\cdots \Diamond \delta _{n} \biggr)^{1/p^{2}} \\ &\qquad{}\times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \bigl\vert h( \delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \bigl\vert g(\delta _{1}, \delta _{2}, \ldots , \delta _{n}) \bigr\vert ^{tq}\Diamond \delta _{1}\Diamond \delta _{2}\cdots \Diamond \delta _{n} \biggr)^{1/q^{2}} \\ &\qquad{}\times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \bigl\vert h(\delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \bigl\vert f(\delta _{1},\delta _{2}, \ldots , \delta _{n}) \bigr\vert ^{tp}\Diamond \delta _{1}\Diamond \delta _{2}\cdots \Diamond \delta _{n} \\ &\qquad{}\times \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \bigl\vert h(\delta _{1},\delta _{2}, \ldots , \delta _{n}) \bigr\vert \bigl\vert g(\delta _{1},\delta _{2}, \ldots , \delta _{n}) \bigr\vert ^{sq}\Diamond \delta _{1}\Diamond \delta _{2}\cdots \Diamond \delta _{n} \biggr)^{1/pq}. \end{aligned} $$
(32)
(2) If
\(s>t>1\)or
\(s< t<1\); \(t>s>1\)or
\(t< s<1\), then
$$\begin{aligned} & \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \bigl\vert h(\delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \bigl\vert f(\delta _{1},\delta _{2}, \ldots , \delta _{n})g( \delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \Diamond \delta _{1} \Diamond \delta _{2}\cdots \Diamond \delta _{n} \\ &\quad\geq \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \bigl\vert h(\delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \bigl\vert f(\delta _{1},\delta _{2}, \ldots , \delta _{n}) \bigr\vert ^{sp}\Diamond \delta _{1}\Diamond \delta _{2}\cdots \Diamond \delta _{n} \biggr)^{1/p^{2}} \\ &\qquad{}\times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \bigl\vert h( \delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \bigl\vert g(\delta _{1}, \delta _{2}, \ldots , \delta _{n}) \bigr\vert ^{tq}\Diamond \delta _{1}\Diamond \delta _{2}\cdots \Diamond \delta _{n} \biggr)^{1/q^{2}} \\ &\qquad{}\times \biggl( \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \bigl\vert h(\delta _{1},\delta _{2},\ldots , \delta _{n}) \bigr\vert \bigl\vert f(\delta _{1},\delta _{2}, \ldots , \delta _{n}) \bigr\vert ^{tp}\Diamond \delta _{1}\Diamond \delta _{2}\cdots \Diamond \delta _{n} \\ &\qquad{}\times \int _{\xi _{1}}^{\sigma _{1}} \int _{\xi _{2}}^{\sigma _{2}}\cdots \int _{\xi _{n}}^{\sigma _{n}} \bigl\vert h(\delta _{1},\delta _{2}, \ldots , \delta _{n}) \bigr\vert \bigl\vert g(\delta _{1},\delta _{2}, \ldots , \delta _{n}) \bigr\vert ^{sq}\Diamond \delta _{1}\Diamond \delta _{2}\cdots \Diamond \delta _{n} \biggr)^{1/pq}. \end{aligned}$$
(33)
Proof
Similar to the proof of Theorem 3.14, we get the result of Theorem 3.15. □
Remark 3.16
For the inequality of Theorem 5.1 in the Reference [17], we put forward Theorem 3.14 and Theorem 3.15 as the generalization results.