Some new generalizations of reversed Minkowski's inequality for several functions via time scales

1.   Introduction

In 1889, Hölder ^[1] proved that

where $\left\{ x_{k}\right\} _{k = 1}^{\xi }$ and $\left\{ y_{k}\right\} _{k = 1}^{\xi }$ are positive sequences and $p > 1$ and $1/p+1/q = 1.$ The inequality (1.1) is reversed if $p < 0$ or $q < 0$. The integral form of (1.1) is

where $\epsilon,$ $\iota \in \mathbb{R}$, $\gamma > 1, \ 1/\gamma +1/\nu = 1,$ and $\lambda,$ $\omega \in C\left([a, b], \mathbb{R}\right).$ If $0 < \gamma < 1,$ then (1.2) is reversed. For more informations about the applications of Hölder's inequality, see ^{[2,3,4,5,6,7,8]}.

In particular, Minkowski's inequality is considered as an application of H ölder's inequality, which states that, for $\delta \geq 1,$ if $\mathcal{G},$ $\mathcal{W}$ are nonnegative continuous functions on $[\check{c}, \breve{a }]$ such that

then

Sulaiman ^[9] introduced the following outcome pertaining to the reversed Minkowski inequality: If $\mathcal{G}, \mathcal{W} > 0,$ $\delta \geq 1,$ and

for all $\zeta \in \lbrack \check{c}, \breve{a}],$ then

Sroysang ^[10] showed that, if $\delta \geq 1$ and $\mathcal{G}, \mathcal{ W} > 0$ with

for all $\zeta \in \lbrack \check{c}, \breve{a}],$ then

Benaissa ^[11] introduced a novel finding concerning the inverse Minkowski inequality, proposing the following: If $\mathcal{G}, \mathcal{W} > 0,$ $\alpha > 0,$and $\delta \geq 1$ such that

for all $\zeta \in \lbrack \check{c}, \breve{a}],$ then

In ^[12], Benaissa showed that, if $\alpha > 0, \ 0 < p\leq \eta,$ $\mathcal{G},$ $\mathcal{W} > 0,$ $\mathcal{U}$ is a weight function, and

then

Also, the author of ^[12] proved that, if $1 < p\leq \eta,$ $\alpha > 0,$ $\mathcal{G},$ $\mathcal{W} > 0,$ and (1.6) holds, then

Time scale calculus is a unification of continuous calculus and discrete calculus. Many authors have proved inequalities on time scales. For example, in 2001, Bohner and Peterson ^[13] presented Hölder's inequality on time scales, stating that if $\mathbb{T}$ is a time scale, $\check{c},$ $\breve{a}\in \mathbb{T}$, $\lambda,$ $\omega \in \mathrm{C}_{rd}(\mathbb{[}\check{c}, \breve{a}\mathbb{]}_{\mathbb{T}},$ $\mathbb{R}^{+})$ and $\gamma > 1$ with $1/\gamma +1/\nu = 1$, then

The inequality (1.9) is reversed for $0 < \gamma < 1$ or $\gamma < 0.$ In addition the authors of ^[13] presented Minkowski's inequality (1.3) on time scales, stating that if $a,$ $b\in \mathbb{T},$ $\psi,$ $\varpi \in C_{rd}\left([a, b]_{\mathbb{T}}, \mathbb{R}^{+}\right)$ and $\alpha > 1$, then

The inequality (1.10) is reversed with the reversed sign when $\alpha < 0$ or $0 < \alpha < 1.$ In ^[14], Lütfi proved that if $g, h:\mathbb{I} \rightarrow \mathbb{R}$ are $\Delta$-integrable functions on $\mathbb{I = [} a, b]\in \mathbb{T}$ with $1 < l\leq g^{p},$ $h^{p}\leq L < \infty$ and $p > 1,$ then

Here, we aim to extend the inequality (1.7) and rectify (1.8) in time scale calculus which is defined in Section 2. Also, we can get some new inequalities in (continuous, discrete, and quantum) calculus.

2.   Preliminaries and basic lemmas

In 2001, Bohner and Peterson introduced the backward jump and forward jump operators, denoted by $\sigma (\zeta): = \inf \{\varrho \in \mathbb{T} :\varrho > \zeta \}$ and $\rho (\zeta): = \sup \{\varrho \in \mathbb{T} :\varrho < \zeta \},$ respectively ^[13]. For any function $\mathcal{G}: \mathbb{T}\rightarrow \mathbb{R}$, the notations $\mathcal{G}^{\sigma }(\zeta)$ and $\mathcal{G}^{\rho }(\zeta)$ denote $\mathcal{G}(\sigma (\zeta))$ and $\mathcal{G}(\rho (\zeta)),$ respectively. The time scale interval $[\check{c}, \breve{a}]_{\mathbb{T}}$ is denoted by $[\check{c}, \breve{a}]\cap \mathbb{T}.$ For more information about dynamic inequalities, see for instance ^{[15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35]}.

Theorem 2.1. ^[13] Let $\mathcal{G},$ $\mathcal{W}:\mathbb{T}\rightarrow \mathbb{R}$ be $\Delta$-differentiable at $\tau \in \mathbb{T}$. Then we have the following at $\tau$:

(1) The sum $\mathcal{G}+\mathcal{W}:\mathbb{T}\rightarrow \mathbb{R}$ is differentiable with

(2) $\alpha \mathcal{G}:\mathbb{T}\rightarrow \mathbb{R}$ is differentiable for any constant $\alpha$ with

(3) The product $\mathcal{GW}$ $:\mathbb{T}\rightarrow \mathbb{R}$ is differentiable and the product rule is defined by

(4) If $\mathcal{G}(\tau)\mathcal{G}(\sigma (\tau))\neq 0$, then $1/ \mathcal{G}:\mathbb{T}\rightarrow \mathbb{R}$ is differentiable and

(5) If $\mathcal{W}(\tau)\mathcal{W}(\sigma (\tau))\neq 0$, then the quotient $\mathcal{G}/\mathcal{W}:\mathbb{T}\rightarrow \mathbb{R}$ is differentiable and the quotient rule is defined as

Definition 2.1. ^[13] A function $F:\mathbb{T}\rightarrow \mathbb{R}$ is said to an antiderivative of $\mathcal{G}:\mathbb{T}\rightarrow \mathbb{R}$ if

In this case, the Cauchy integral of $\mathcal{G}$ is defined as

Definition 2.2. ^[13] A function $f:\mathbb{T}\rightarrow \mathbb{R}$ is called $rd$-continuous provided that it is continuous at right-dense points in $\mathbb{T}$ and its left-sided limits exist (finite) at left-dense points in $\mathbb{T}$. The set of $rd$-continuous functions $f:\mathbb{T}\rightarrow \mathbb{R}$ is denoted by $\mathrm{C}_{rd}({\mathbb{T}},$ ${\mathbb{R)}}$.

Theorem 2.2. ^[36] Assume that $\check{c},$ $\breve{a}\in \mathbb{T}$ and $\mathcal{G}\in C_{rd}(\mathbb{T}$, $\mathbb{R}).$ Then the next features satisfy the following conditions:

$(1)$ If $\mathbb{T = R}$, then

$(2)$ If $[\check{c}, \breve{a}]$ consists of only isolated points, then

$(3)$ If ${\mathbb{T}} = {\mathbb{Z}}$, then

$(4)$ If ${\mathbb{T = }}\delta {\mathbb{Z}}$, $\delta > 0,$ then

The AM-GM inequality yields that

where $\psi _{i}\left(\xi \right),$ $i = 1, 2, ..., \varrho$ are nonnegative functions.

Lemma 2.1. Assume that $\check{c},$ $\breve{a}\in \mathbb{T},$ $\breve{a} > \check{c},$ $0 < p\leq \delta$, and $h,$ $\lambda \in C_{rd}(\mathbb{[} \check{c}, \breve{a}\mathbb{]}_{\mathbb{T}},$ $\mathbb{R}^{+}).$ Then,

Proof. Applying (1.9) with $\gamma = \delta /p > 1$ and $\nu = \delta /\left(\delta -p\right),$ then

Thus,

which is (2.2). □

Lemma 2.2. (Jensen's inequality ^[36]) Let $\check{c},$ $\breve{a}\in \mathbb{T}$ with $\check{c} < \breve{a}$ and $c,$ $d\in \mathbb{R}$. Assuming that $\phi \in C_{rd}([\check{c}, \breve{a}]_{\mathbb{T}}, (c, d)),$ $\varkappa \in C_{rd}([\check{c}, \breve{a}]_{\mathbb{T}}, \mathbb{R}^{+}).$

If $F\in C\left((c, d), \mathbb{R}\right)$ is convex, then

The inequality (2.3) is reversed if $F$ is concave.

If $F(\varsigma) = \varsigma ^{\lambda },$ then

and

3.   Main results

During this study, we assume the existence of the integrals under consideration.

Theorem 3.1. Assume that $\check{c}, \breve{a}\in \mathbb{T},$ $\breve{a} > \check{c},$ $0 < p\leq \delta < \infty,$ $\alpha > 0$, and $\mathcal{G}_{x},$ $\mathcal{W}_{x},$ $\mathcal{U}_{x}\in \mathrm{C}_{rd}(\left[\check{c}, \breve{a}\right] _{\mathbb{T}}, \mathbb{R}^{+}),$ $x = 1, 2, ..., \iota$, with

Then,

Proof. From (3.1), we see that

Then,

Since $0 < p\leq \delta,$ we have from (3.3) that

and

Integrating (3.4) and (3.5) over $\varsigma$ from $\check{c}$ to $\breve{a},$ we see that

and

By applying (2.2) with $h\left(\varsigma \right) = \prod\limits_{x = 1}^{\iota }\mathcal{G}_{x}^{\frac{1}{\iota }}\left(\varsigma \right)$ and $\lambda (\varsigma) = \prod\limits_{x = 1}^{\iota }\left(\mathcal{U}_{x}\left(\varsigma \right) \right) ^{\frac{1}{\iota }},$ we see that

Then, from (3.6), we have

Using (3.1), we deduce that

Therefore

Since $0 < p\leq \delta,$ we have from (3.9) that

and

By applying (2.2) with $h(\varsigma) = \prod\limits_{x = 1}^{\iota } \mathcal{W}_{x}^{\frac{1}{\iota }}(\varsigma)$ and $\lambda (\varsigma) = \prod\limits_{x = 1}^{\iota }\left(\mathcal{U}_{x}\left(\varsigma \right) \right) ^{\frac{1}{\iota }},$ we have

Thus, (3.10) gives

From (3.8) and (3.12), we deduce that

Applying (2.1), we see that

and

Thus, (3.13) becomes

From (3.7) and (3.11), we have

Combining (3.14) and (3.15), we obtain

which is (3.2). □

Corollary 3.1. Taking $\mathbb{T = R}$ and $\iota = 1,$ then we get (1.7).

Corollary 3.2. Taking $\mathbb{T = N},$ $\check{c},$ $\breve{a}\in \mathbb{N},$ $0 < p\leq \delta < \infty,$ $\alpha > 0,$ and $\left\{ \mathcal{G}_{x}\right\} _{x = 1}^{\iota },$ $\left\{ \mathcal{W}_{x}\right\} _{x = 1}^{\iota },$ $\left\{ \mathcal{U}_{x}\right\} _{x = 1}^{\iota }$ are positive sequences such that

Then, $\sigma \left(\varsigma \right) = \varsigma +1,$ $\mu \left(\varsigma \right) = 1,$ $\rho \left(\breve{a}\right) = \breve{a}-1$, and

Remark 3.1. If $\mathbb{T = }$ $q^{\mathbb{N}}$ for $q > 1, \ \check{c},$ $\breve{a}\in \mathbb{T},$ $0 < p\leq \delta < \infty,$ $\alpha > 0,$ and $\left\{ \mathcal{G}_{x}\right\} _{x = 1}^{\iota },$ $\left\{ \mathcal{W}_{x}\right\} _{x = 1}^{\iota },$ $\left\{ \mathcal{U}_{x}\right\} _{x = 1}^{\iota }$ are positive sequences such that, for $\varsigma \in \mathbb{T},$

Then, $\sigma \left(\varsigma \right) = q\varsigma,$ $\mu \left(\varsigma \right) = \sigma \left(\varsigma \right) -\varsigma = \left(q-1\right) \varsigma,$ $\rho \left(\breve{a}\right) = \breve{a}/q,$ and

Example 3.1. In Theorem 3.1, assume that $\mathbb{T = R}$, $\check{c} = 0, \breve{a} \in \mathbb{R},$ $\iota,$ $p = 1,$ $\delta = 2,$ $B = 2, \ E = 1$, and $\alpha = 1.$ In addition, if $\mathcal{G},$ $\mathcal{W},$ $\mathcal{U}\in C(\left[\check{c}, \breve{a}\right] _{\mathbb{T}}, \mathbb{R}^{+})$ such that $\mathcal{U}\left(\varsigma \right) = \varsigma,$ $\mathcal{W}\left(\varsigma \right) = \varsigma$, $\mathcal{G}\left(\varsigma \right) = 3\varsigma,$ and $\ell = 5,$ then

Proof. Using the hypothesis, the left-hand side of (3.16) can be written as follows:

Also, we see that

Furthermore,

From (3.17)–(3.19), we see that the inequality (3.16) holds. The proof is complete. □

Theorem 3.2. Assume that $\check{c}, \breve{a}\in \mathbb{T}$, $\breve{a} > \check{c},$ $0 < \delta < \infty,$ $\alpha > 0,$ $\mathcal{G}_{x},$ $\mathcal{W }_{x},$ $\mathcal{U}_{x}\in C_{rd}(\left[\check{c}, \breve{a}\right] _{ \mathbb{T}}, \mathbb{R}^{+}),$ $x = 1, 2, ..., \iota,$ and (3.1) holds. If $p > 1,$ then

and, if $0 < p < 1,$ then

Proof. To prove this theorem, we have two cases:

Case 1: $p > 1.$ From (3.1), we deduce that

Then,

and

Integrating (3.22) and (3.23) over $\varsigma$ from $\check{c}$ to $\breve{a},$ we see that

and

Applying (2.5) with $\lambda = 1/p < 1,$ $\varkappa (\varsigma) = \prod\limits_{x = 1}^{\iota }\left(\mathcal{U}_{x}\left(\varsigma \right) \right) ^{\frac{1}{\iota }}$, and $\phi (\varsigma) = \prod\limits_{x = 1}^{\iota }\mathcal{G}_{x}^{\frac{\delta }{\iota } }(\varsigma),$ we get

Then, we have from (3.24) that

Using (3.1), we deduce that

Therefore,

Then, we have that

and

Applying (2.5) with $\lambda = 1/p < 1,$ $\phi (\varsigma) = \prod\limits_{x = 1}^{\iota }\mathcal{W}_{x}^{\frac{\delta }{\iota }}\left(\varsigma \right)$, and $\varkappa (\varsigma) = \prod\limits_{x = 1}^{\iota }\left(\mathcal{U}_{x}\left(\varsigma \right) \right) ^{\frac{1}{\iota }}$, we see that

Thus, (3.27) becomes

Adding (3.26) and (3.29), we see that

Applying (2.1) to the terms $\left(\int_{\check{c}}^{\breve{a} }\prod\limits_{x = 1}^{\iota }\left(\mathcal{U}_{x}\left(\varsigma \right) \right) ^{\frac{1}{\iota }}\mathcal{G}_{x}^{\frac{\delta }{\iota } }(\varsigma)\Delta \varsigma \right) ^{\frac{1}{\delta }}$ and $\left(\int_{\check{c}}^{\breve{a}}\prod\limits_{x = 1}^{\iota }\mathcal{W}_{x}^{ \frac{\delta }{\iota }}\left(\varsigma \right) \left(\mathcal{U} _{x}(\varsigma)\right) ^{\frac{1}{\iota }}\Delta \varsigma \right) ^{\frac{1 }{\delta }},$ then (3.30) becomes

From (3.25) and (3.28), we have

Combining (3.31) and (3.32), we get

which is (3.20).

Case 2: $0 < p < 1.$ Again, by using (3.1), we see that

Then,

and

Integrating (3.33) and (3.34), we have for $p,$ $\delta > 0,$ that

and

Applying (2.4) with $\lambda = 1/p > 1,$ $\varkappa (\varsigma) = \prod\limits_{x = 1}^{\iota }\left(\mathcal{U}_{x}\left(\varsigma \right) \right) ^{\frac{1}{\iota }}$, and $\phi \left(\varsigma \right) = \prod\limits_{x = 1}^{\iota }\mathcal{G}_{x}^{\frac{\delta }{\iota }}\left(\varsigma \right),$ we get

So, the inequality (3.35) can be written as follows

From (3.1), we deduce that

Thus,

Then, we have for $\delta > 0$ that

and

Applying (2.4) with $\lambda = 1/p > 1,$ $\phi \left(\varsigma \right) = \prod\limits_{x = 1}^{\iota }\mathcal{W}_{x}^{\frac{\delta }{\iota }}\left(\varsigma \right)$, and $\varkappa \left(\varsigma \right) = \prod\limits_{x = 1}^{\iota }\left(\mathcal{U}_{x}\left(\varsigma \right) \right) ^{\frac{1}{\iota }},$ we get

Then, the inequality (3.38) becomes

Adding (3.37) and (3.40), we get

Applying the inequality (2.1) on the right-hand side of (3.41), we have

From (3.36) and (3.39), we get

Combining (3.42) and (3.43), we see that

which is (3.21). □

Remark 3.2. Assume that $\mathbb{T = R}$; $\check{c}, \breve{a}\in \mathbb{R}$ with $\breve{a} > \check{c},$ $0 < \delta < \infty,$ $\alpha > 0$; and $\mathcal{G}_{x},$ $\mathcal{W}_{x},$ $\mathcal{U}_{x}\in C$ $(\left[\check{c}, \breve{ a}\right], \mathbb{R}^{+})$ with

In addition, if $p > 1,$ then

If $0 < p < 1,$ then

Remark 3.3. If $\mathbb{T = R}$, $\iota = 1,$ $p > 1,$ and $0 < \delta < \infty,$ then we obtain the correction and the generalization of (1.8) by replacing $\breve{a} -\check{c}$ with $\int_{\check{c}}^{\breve{a}}\mathcal{U}\left(\varsigma \right) d\varsigma.$

Remark 3.4. Suppose that $\mathbb{T = N};$ $\check{c},$ $\breve{a}\in \mathbb{N},$ $0 < \delta < \infty,$ $\alpha > 0$, and $\left\{ \mathcal{G}_{x}\right\} _{x = 1}^{\iota },$ $\left\{ \mathcal{W}_{x}\right\} _{x = 1}^{\iota },$ $\left\{ \mathcal{U} _{x}\right\} _{x = 1}^{\iota }$ are positive sequences such that

In addition, if $p > 1,$ then

and, if $0 < p < 1,$ then

Remark 3.5. Assume that $\mathbb{T = }q^{\mathbb{N}}$ for $q > 1;$ $\check{c},$ $\breve{a}\in \mathbb{T},$ $0 < \delta < \infty,$ $\alpha > 0;$ and $\left\{ \mathcal{G} _{x}\right\} _{x = 1}^{\iota },$ $\left\{ \mathcal{W}_{x}\right\} _{x = 1}^{\iota },$ $\left\{ \mathcal{U}_{x}\right\} _{x = 1}^{\iota }$ are positive sequences with

In addition, if $p > 1,$ then $\rho \left(\breve{a}\right) = \breve{a}/q,$ $\sigma \left(\varsigma \right) = q\varsigma,$ $\mu \left(\varsigma \right) = \sigma \left(\varsigma \right) -\varsigma = \left(q-1\right) \varsigma,$ and

and, if $0 < p < 1,$ then

4.   Conclusions

In this paper, we have established a novel extensions of the reversed Minkowski's inequality for various functions on delta calculus time scales by applying Jensen's and Hölder's inequalities on time scales. In addition, we have presented some new inequalities in different cases, like $\mathbb{T = N}$ and $q^{\mathbb{N}}$ for $q > 1$.

In the future, we will apply the reversed Minkowski inequality for various functions to diamond-alpha calculus and conformable calculus time scales.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The authors extend their appreciation to the Deputyship for Research & Innovation, Ministry of Education in Saudi Arabia for funding this research work through the project number ISP23-86.

Conflict of interest

The authors declare that they have no conflict of interest.