Some dynamic Hardy-type inequalities with negative parameters on time scales nabla calculus

1.   Introduction

In 1920, Hardy ^[1] proved that

where $\epsilon > 1,$ $h(\tau)\geq 0$ for $\tau \geq 1$, and $\sum_{\tau = 1}^{\infty }h^{\epsilon }(\tau) < \infty.$ In 1925, Hardy [2, Theorem A] showed that if $\epsilon > 1,$ $\varpi \geq 0$ such that $\int_{0}^{\infty }\varpi ^{\epsilon }(\xi)d\xi < \infty$, then $\varpi$ is integrable over any finite interval $(0, \xi)$, $\xi \in (0, \infty)$ and

The constant $(\epsilon /(\epsilon -1))^{\epsilon }$ in (1.1) and (1.2) is the best possible.

In 1927, Hardy and Littlewood ^[3] proved that if $0 < \epsilon < 1,$ $\varpi (\xi)\geq 0$ for $\xi \in \left(0, \infty \right)$, and $\int_{0}^{\infty }\varpi ^{\epsilon }(\xi)d\xi < \infty$, then

where the constant $\left(\epsilon /\left(1-\epsilon \right) \right) ^{\epsilon }$ is the best possible.

The Hardy inequalities mentioned above are proved for a positive parameter $0 < \epsilon < 1$ and $\epsilon > 1.$ Also, these inequalities depend on the power rule inequality in case of the positive parameter. So, some authors discovered new inequalities of Hardy type with negative parameters and noted that they proved these inequalities with a different technique which depends on the power rule inequality.

For example, in 2007, Bicheng ^[4] established the integral inequality of Hardy type with negative parameter and proved that if $\epsilon < 0,$ $r > 1,$ $\varpi \left(\vartheta \right) \geq 0$, and$0 < \int_{0}^{\infty }\xi ^{-r}\left(\xi \varpi (\xi)\right) ^{\epsilon }d\xi < \infty,$ then

where the constant factor $\left(\epsilon /\left(1-r\right) \right) ^{\epsilon }$ is the best possible. Also, in ^[4] it was proved that if $\epsilon < 0,$ $r < 1,$ $\varpi \left(\vartheta \right) \geq 0$, and$0 < \int_{0}^{\infty }\xi ^{-r}\left(\xi \varpi (\xi)\right) ^{\epsilon }d\xi < \infty,$ then

where the constant factor $\left(\epsilon /\left(r-1\right) \right) ^{\epsilon }$ is the best possible.

In 2020, Benaissa and Sarikaya ^[5] generalized (1.3), and proved that if$\epsilon < 0,$ $r > 1$, and $\varpi,$ $\Xi > 0$ such that the function $\xi /\Xi (\xi)$ is nondecreasing, then

and if $\epsilon < 0,$ $0\leq r < 1$, and $\varpi,$ $\Xi > 0$ such that the function $\xi /\Xi (\xi)$ is nonincreasing, then

Also, the authors of ^[5] proved that if $\epsilon < 0,$ $r < 0$, and $\varpi,$ $\Xi > 0$ such that the function $\xi /\Xi (\xi)$ is nondecreasing, then

The time scale $\mathbb{T}$ is an arbitrary nonempty closed subset of the real numbers. As an application on time scales, we can get the continuous and discrete forms of any inequality, i.e., $\mathbb{T = R}$ and $\mathbb{T = N}$. For more details about the dynamic inequalities on time scales, refer to ^{[6,7,8,9,10,11,12,13,14,15,16]}, and for the applications of dynamic inequalities in the study of qualitative behavior of dynamic equations, refer to ^{[17,18,19,20,21,22,23,24]}. For including an exploration of advanced methodologies, we refer to the papers ^[25,26].

The objective of this paper is to introduce novel generalizations of the continuous the inequalities (1.5)–(1.7) on nabla time scales. The proofs of these results rely on employing the reverse Hö lder's inequality and the chain rule formula adapted to time scales.

This paper is divided into three sections: In Section 2, we present some lemmas on time scales needed in Section 3 where we prove our results. These results as special cases when $\mathbb{T = R}$ give the inequalities (1.5)–(1.7), while for $\mathbb{T = N}_{0}$, the results are fundamentally original.

2.   Preliminaries and basic lemmas

In 2001, Bohner and Peterson ^[27] introduced the time scale $\mathbb{T}$ as an arbitrary nonempty closed subset of the real numbers $\mathbb{R}$. Also, they defined the backward jump operator by $\rho (\tau): = \sup \{s\in \mathbb{T}:s < \tau \}.$ For any function $\varpi :\mathbb{T} \rightarrow \mathbb{R}$, the notation $\varpi ^{\rho }(\tau)$ signifies $\varpi (\rho (\tau)).$ The time scale interval $[\varrho, \delta]_{ \mathbb{T}}$ is defined as $[\varrho, \delta]_{\mathbb{T}}: = [\varrho, \delta]\cap \mathbb{T}.$

Definition 2.1. ^[28] A function $\lambda :\mathbb{T\rightarrow R}$ is left-dense continuous or $ld$-continuous provided that it is continuous at left-dense points in $\mathbb{T}$ and its right-sided limits exist at right-dense points in $\mathbb{T}$. The space of $ld$-continuous functions is denoted by $\mathrm{C} _{ld}({\mathbb{T}},$ ${\mathbb{R)}}$.

The set $\mathbb{T}^{\kappa }$ is derived from the time scale $\mathbb{T}$ as follows: If $\mathbb{T}$ has a left-scattered maximum $m$, then $\mathbb{T}^{\kappa } = \mathbb{T}-\left\{ m\right\}.$ Otherwise, $\mathbb{T} ^{\kappa } = \mathbb{T}$. In summary,

Definition 2.2. ^[28] A function $\psi :\mathbb{T\rightarrow R}$ is said to be $\nabla$-differentiable at $\vartheta \in \mathbb{T}^{\kappa }$ if $\psi$ is defined in a neighbourhood $U$ of $\vartheta$ and there exists a unique real number $\psi ^{\nabla }(\vartheta)$, called the nabla derivative of $\psi$ at $\vartheta$, such that for each $\epsilon > 0$, there exists a neighbourhood $N$ of $\vartheta$ with $N\subseteq U$ and

Theorem 2.1. ^[28] Assume $\psi,$ $\Theta :\mathbb{T\rightarrow R}$ are nabla differentiable at $\vartheta \in \mathbb{T}$. Then:

(1) The product $\psi \Theta :\mathbb{T\rightarrow R}$ is nabla differentiable at $\vartheta$, and we get the product rule

(2) If $\Theta (\vartheta)\Theta ^{\rho }(\vartheta)\neq 0$, then $\psi /\Theta$ is nabla differentiable at $\vartheta$, and we get the quotient rule

Lemma 2.1. ^[29] Let $\psi :\mathbb{R\rightarrow R}$ be continuously differentiable and suppose that $\Theta :\mathbb{T\rightarrow R}$ is continuous and nabla differentiable. Then, $\psi \circ \Theta :\mathbb{T\rightarrow R}$ is nabla differentiable and

Definition 2.3. ^[28] A function $\Lambda :\mathbb{T\rightarrow R}$ is called a nabla antiderivative of $\psi :\mathbb{T\rightarrow R},$ if $\Lambda ^{\nabla }(\vartheta) = \psi (\vartheta)$ $\forall \vartheta \in \mathbb{T}$. We then define the nabla integral of $\psi$ by

Lemma 2.2. ^[28] If $\varrho,$ $\delta \in \mathbb{T}$ and $\varpi,$ $\Xi :\mathbb{ T\rightarrow R}$ are $ld$-continuous, then

Lemma 2.3. ^[27] If $\varpi \in \mathrm{C}_{ld}(\mathbb{T},$ $\mathbb{R})$ and $\vartheta \in \mathbb{T},$ then

where $\nu \left(\vartheta \right) = \vartheta -\rho \left(\vartheta \right).$

Lemma 2.4. (Reverse Hölder's inequality ^[30]) If $\varrho,$ $\delta \in \mathbb{T}$, $\alpha < 0,$ $1/\alpha +1/\beta = 1,$ and $\phi,$ $\omega \in \mathrm{C}_{ld}(\mathbb{[}\varrho, \delta \mathbb{]}_{\mathbb{T}},$ $\mathbb{ R}^{+}),$ then

3.   Main results

In this document, we will make the assumption that the functions are ld-continuous on the interval $[\varrho, \infty)_{\mathbb{T}}$ and we also assume the existence of the integrals under consideration. Additionally, we posit the existence of a positive constant $K$ such that

Now, we can state and prove our results.

Theorem 3.1. Let $\varrho \in \mathbb{T}$, $\epsilon < 0,$ $\epsilon ^{\ast } = \epsilon /\left(\epsilon -1\right),$ $r > 1$, and $\varpi,$ $\Xi \in C_{ld}\left([\varrho, \infty)_{\mathbb{T}}, \mathbb{R}^{+}\right)$ where $\left(\xi -\varrho \right) /\Xi (\xi)$ is nondecreasing. If (3.1) holds, then

where $F(\xi) = \int_{\xi }^{\infty }\varpi (\vartheta)\nabla \vartheta$ and

Proof. Note that

Applying Lemma 2.4 to

with

we get

Applying (2.1) to $\left(\vartheta -\varrho \right) ^{\frac{r-1}{ \epsilon }},$ we see that

Since $\epsilon < 0,$ $r > 1,$and $d\geq \rho (\vartheta),$ we have that $\left(r-1-\epsilon \right) /\epsilon < 0$ and

Substituting (3.6) into (3.5), we see that

From (3.7)$,$ (note $\epsilon < 0$ and $\epsilon ^{\ast } = \epsilon /\left(\epsilon -1\right) > 0$)$,$ we observe that

and then

Substituting (3.8) into (3.4), we see that

From (3.3) and (3.9), we have for $\epsilon < 0$ that

Multiplying (3.10) by $\Xi ^{-r}(\xi)$ and then integrating over $\xi$ from $\varrho$ to $\infty,$ we get

Applying (2.2) to

with

we have that

where

Since $v_{1}(\varrho) = 0,$ we observe that

Note that

Since $\left(\vartheta -\varrho \right) /\Xi \left(\vartheta \right)$ is nondecreasing and $r > 1$, we have for $\vartheta \leq \xi$ that

Substituting (3.14) into (3.13), we observe that

Substituting (3.15) into (3.12), we have

To complete the proof, we have two cases:

Case 1: For $1-r\leq \epsilon.$

Applying (2.1) to the term $\left(\vartheta -\varrho \right) ^{\frac{ r-1}{\epsilon ^{\ast }}-r+1},$ we see that

where $d\in \lbrack \rho \left(\vartheta \right), \vartheta].$ Since $\epsilon < 0$ and $1-r\leq \epsilon$ (note $\left(1-r\right) /\epsilon -1\geq 0$), we have for $d\geq \rho \left(\vartheta \right)$ that

From (3.17) and (3.18), we observe that

Integrating (3.19) over $\vartheta$ from $\varrho$ to $\rho \left(\xi \right)$, since $r > 1$ and $\epsilon < 0,$ we have that

Using (3.1), since $\frac{r-1}{\epsilon ^{\ast }}-r = \frac{1-r}{\epsilon }-1 > 0,$ then (3.16) becomes

Substituting (3.20) into (3.21) and using (3.1), we see that

From (3.11) and (3.22), we see that

which is (3.2) with $C = \left(\epsilon /\left(1-r\right) \right) ^{\epsilon }K^{\frac{r-1}{\epsilon ^{\ast }}}.$

Case 2: For $1-r\geq \epsilon.$

Applying (2.1) to $\left(\vartheta -\varrho \right) ^{\frac{1-r}{ \epsilon }},$ we see that

Since $\epsilon < 0$, and $1-r\geq \epsilon$ (note that $\left(\left(1-r\right) /\epsilon \right) -1\leq 0$), we have for $d\leq \vartheta$ that

From (3.23) and (3.24), we observe that

Integrating (3.25) over $\vartheta$ from $\varrho$ to $\rho \left(\xi \right)$, we have that

Substituting (3.26) into (3.16), we observe that

Using (3.1), (3.27) then becomes

From (3.11) and (3.28), we see that

which is (3.2) with $C = \left(\epsilon /\left(1-r\right) \right) ^{\epsilon }K^{r}.$ □

Remark 3.1. In Theorem 3.1, if $\mathbb{T = R}$, and $\varrho = 0,$ then $\rho \left(\xi \right) = \xi$, and we observe that (3.1) holds with $K = 1$. Then, we get (1.5), and for $\Xi (\xi) = \xi,$ we get (1.3).

Corollary 3.1. If $\mathbb{T = N}_{0},$ $\varrho = 0$, and $\varpi,$ $\Xi$ are positive sequences such that $\tau /\Xi (\tau)$ is nondecreasing, then

where

Here, we used that for $\rho \left(\tau \right) > \varrho,$ we have for $\tau \geq 2,$ and

and thus inequality (3.1) holds with $K = 2.$

Theorem 3.2. Let $\varrho \in \mathbb{T}$, $\epsilon < 0,$ $0\leq r < 1$, and $\varpi,$ $\Xi \in C_{ld}\left([\varrho, \infty)_{\mathbb{T}}, \mathbb{ R}^{+}\right)$ where $\left(\vartheta -\varrho \right) /\Xi (\vartheta)$ is nonincreasing. If (3.1) holds, then

where $\Omega (\xi) = \int_{\varrho }^{\xi }\varpi (\vartheta)\nabla \vartheta$ and

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

Case 1: For $\left(r-1\right) /\epsilon \geq 1.$

Note that

where $\epsilon ^{\ast } = \epsilon /(\epsilon -1).$ Applying Lemma 2.4 to $\int_{\varrho }^{\xi }\left(\vartheta -\varrho \right) ^{\frac{ r-\epsilon -1}{\epsilon \epsilon ^{\ast }}}\left(\vartheta -\varrho \right) ^{\frac{1+\epsilon -r}{\epsilon \epsilon ^{\ast }}}\varpi (\vartheta)\nabla \vartheta,$ with

we get

From (3.30), and (3.31), we see that

Applying (2.1) to $\left(\vartheta -\varrho \right) ^{\frac{r-1}{ \epsilon }},$ we observe that

where $d\in \lbrack \rho \left(\vartheta \right), \vartheta].$ Since $d\leq \vartheta$, and $\left(r-1\right) /\epsilon \geq 1,$ we have from (3.33) that

By integrating (3.34) over $\vartheta$ from $\varrho$ to $\xi,$ we get

Substituting (3.35) into (3.32), since $\epsilon ^{\ast } > 0$, we observe that

and then we have for $\epsilon < 0,$ that

Multiplying (3.36) with $\Xi ^{-r}(\xi)$ and then integrating over $\xi$ from $\varrho$ to $\infty,$ we see that

Applying (2.2) to

with

we get

where

Since $\lim_{\xi \rightarrow \infty }v_{3}(\xi) = 0,$ we have from (3.38) that

Note that

Since $\left(\vartheta -\varrho \right) /\Xi (\vartheta)$ is nonincreasing, $0\leq r < 1$, and $\vartheta \geq \xi$, we observe that

and then we have from (3.40) that

Substituting (3.41) into (3.39), we observe that

Using (3.23), since $\left(1-r\right) /\epsilon < 0$ and $d\leq \vartheta,$ we have that

and then

Substituting (3.43) into (3.42), and then using (3.1), we observe that

Substituting (3.44) into (3.37), we see that

which is (3.29) with $D = \left(\epsilon /\left(r-1\right) \right) ^{\epsilon }K^{\frac{r-1}{\epsilon }}.$

Case 2: For $\left(r-1\right) /\epsilon \leq 1.$

Note that

Applying Lemma 2.4 to the term

with

we get

From (3.45), and (3.46), we see that

Using (3.5), since $d\geq \rho \left(\vartheta \right)$, and $0 < \left(r-1\right) /\epsilon \leq 1,$ we have that

and then

Substituting (3.48) into (3.47), we see that

and then (note $\epsilon < 0$)

Multiplying (3.49) with $\Xi ^{-r}(\xi)$ and then integrating over $\xi$ from $\varrho$ to $\infty,$ we observe that

Applying (2.2) to

with

we see that

where

Since $\lim_{\xi \rightarrow \infty }v_{4}(\xi) = 0,$ we have from (3.51) that

Using (3.41) and (3.52), we observe that

Using (3.23), since $\left(1-r\right) /\epsilon < 0$ and $d\leq \vartheta,$ we have that

and then

Substituting (3.54) into (3.53), and then using (3.1), we obtain

Substituting (3.55) into (3.50), we observe that

which is (3.29) with $D = \left(\epsilon /\left(r-1\right) \right) ^{\epsilon }K^{r}.$ □

Remark 3.2. In Theorem 3.2, if $\mathbb{T = R}$ and $\varrho = 0,$ then $\rho \left(\xi \right) = \xi$ and we see that (3.1) holds with $K = 1$. Then, we get (1.6), and for $\Xi \left(\xi \right) = \xi,$ we get (1.4).

Corollary 3.2. If $\mathbb{T = N}_{0}$, $\varrho = 0$, and $\varpi,$ $\Xi$are positive sequences such that $\tau /\Xi (\tau)$ is nonincreasing, then

where

Here, inequality (3.1) holds with $K = 2.$

Theorem 3.3. Assume that $\varrho \in \mathbb{T}$, $\epsilon < 0,$ $r < 0$, and $\varpi,$ $\Xi \in C_{ld}\left([\varrho, \infty)_{\mathbb{T}}, \mathbb{ R}^{+}\right)$ such that the function $\left(\vartheta -\varrho \right) /\Xi (\vartheta)$ is nondecreasing. Then,

where $\Omega (\xi) = \int_{\varrho }^{\xi }\varpi (\vartheta)\nabla \vartheta$ and

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

Case 1: For $\left(r-1\right) /\epsilon \leq 1.$

Note that

where $\epsilon ^{\ast } = \epsilon /(\epsilon -1)$. Applying Lemma 2.4 to $\int_{\varrho }^{\xi }\left(\rho \left(\vartheta \right) -\varrho \right) ^{\frac{r-\epsilon -1}{\epsilon \epsilon ^{\ast }}}\left(\rho \left(\vartheta \right) -\varrho \right) ^{\frac{1+\epsilon -r}{\epsilon \epsilon ^{\ast }}}\varpi (\vartheta)\nabla \vartheta$ with

we get

From (3.58) and (3.59), we see that

Using (3.5), since $d\geq \rho \left(\vartheta \right)$, and $0 < \left(r-1\right) /\epsilon \leq 1,$ we have that

and then

Substituting (3.61) into (3.60), since $\epsilon ^{\ast } > 0$, we see that

and then (note $\epsilon < 0$)

Multiplying (3.62) with $\Xi ^{-r}(\xi)$ and then integrating over $\xi$ from $\varrho$ to $\infty,$ we observe that

Applying (2.2) to $\int_{\varrho }^{\infty }\left(\xi -\varrho \right) ^{\frac{r-1}{\epsilon ^{\ast }}}\Xi ^{-r}(\xi)\left(\int_{\varrho }^{\xi }\left(\rho \left(\vartheta \right) -\varrho \right) ^{\frac{1+\epsilon -r }{\epsilon ^{\ast }}}\varpi ^{\epsilon }(\vartheta)\nabla \vartheta \right) \nabla \xi,$ with

we see that

where

Since $\lim_{\xi \rightarrow \infty }v_{5}(\xi) = 0,$ we have from (3.64) that

Since

Since $\left(\vartheta -\varrho \right) /\Xi (\vartheta)$ is nondecreasing, $r < 0$, and $\vartheta \geq \xi,$ we see that

Substituting the last inequality into (3.66), we get

Substituting (3.67) into (3.65), we observe that

Using (3.23), since $\left(1-r\right) /\epsilon < 0$, and $d\leq \vartheta,$ we have that

and then

Substituting (3.69) into (3.68), we obtain

Since $r < 0$, and $\xi \geq \rho \left(\xi\right),$ inequality (3.70) becomes

Substituting (3.71) into (3.63), we observe that

which is (3.57) with $E = \left(\epsilon /\left(r-1\right) \right) ^{\epsilon }.$

Case 2: For $\left(r-1\right) /\epsilon \geq 1.$

Note that

Applying Lemma 2.4 to $\int_{\varrho }^{\xi }\left(\vartheta -\varrho \right) ^{\frac{r-\epsilon -1}{\epsilon \epsilon ^{\ast }}}\left(\vartheta -\varrho \right) ^{\frac{1+\epsilon -r}{\epsilon \epsilon ^{\ast }}}\varpi (\vartheta)\nabla \vartheta,$ with

we get

From (3.72) and (3.73), we see that

Using (3.5), since $d\leq \vartheta$ and $\left(r-1\right) /\epsilon \geq 1,$ we have that

By integrating (3.75) over $\vartheta$ from $\varrho$ to $\xi,$ we get

Substituting (3.76) into (3.74), we observe that

and then we have for $\epsilon < 0$ that

Multiplying (3.77) with $\Xi ^{-r}(\xi)$ and then integrating over $\xi$ from $\varrho$ to $\infty,$ we see that

Applying (2.2) to $\int_{\varrho }^{\infty }\left(\xi -\varrho \right) ^{\frac{r-1}{\epsilon ^{\ast }}}\Xi ^{-r}(\xi)\left(\int_{\varrho }^{\xi }\left(\vartheta -\varrho \right) ^{\frac{1+\epsilon -r}{\epsilon ^{\ast }} }\varpi ^{\epsilon }(\vartheta)\nabla \vartheta \right) \nabla \xi,$ with

we get

where

Since $\lim_{\xi \rightarrow \infty }v_{6}(\xi) = 0,$ we have from (3.79) that

Since $\left(\vartheta -\varrho \right) /\Xi (\vartheta)$ is nondecreasing and $r < 0$, we have for $\vartheta \geq \xi$ that

and then

Substituting (3.81) into (3.80), we see that

Using (3.23), since $\left(1-r\right) /\epsilon < 0$, and $d\leq \vartheta,$ we have that

and then

Substituting (3.83) into (3.82), we observe that

and then we have from (3.1) that

Substituting (3.84) into (3.78), we see that

which is (3.57) with $E = \left(\epsilon /\left(r-1\right) \right) ^{\epsilon }K^{\frac{r-1}{\epsilon }}.$ □

Remark 3.3. In Theorem 3.3, if $\mathbb{T = R}$ and $\varrho = 0,$ then $\rho \left(\xi \right) = \xi$ and we see that (3.1) holds with $K = 1$, and thus we get (1.7). In addition, if $\Xi \left(\xi \right) = \xi,$ then we get (1.4).

Corollary 3.3. If $\mathbb{T = N}_{0},$ $\varrho = 0$, and $\varpi,$ $\Xi$are positive sequences such that $\tau /\Xi (\tau)$ is nondecreasing$,$ then we see that (3.1) holds with $K = 2$ and then

where

4.   Conclusions

In this paper, we established some new generalizations of the continuous inequalities on nabla calculus time scales. These inequalities were proved by employing the reverse Hölder's inequality and the chain rule formula adapted to 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 conflicts of interest.