Exploring Generalized Hardy-Type Inequalities via Nabla Calculus on Time Scales

Abstract

In this research, we aim to explore generalizations of Hardy-type inequalities using nabla Hölder’s inequality, nabla Jensen’s inequality, chain rule on nabla calculus and leveraging the properties of convex and submultiplicative functions. Nabla calculus on time scales provides a unified framework that unifies continuous and discrete calculus, making it a powerful tool for studying various mathematical problems on time scales. By utilizing this approach, we seek to extend Hardy-type inequalities beyond their classical continuous or discrete settings to a more general time scale domain. As specific instances of our discoveries, we have the integral inequalities previously established in the existing literature.

1. Introduction

In the work [1], Hardy introduced a discrete inequality:

$$\sum _{\tau =1}^{\infty }{\left(\frac{{\sum }_{j=1}^{\tau }\eta \left(j\right)}{\tau }\right)}^{\gamma }\le {\left(\frac{\gamma }{\gamma -1}\right)}^{\gamma }\sum _{\tau =1}^{\infty }{\eta }^{\gamma }\left(\tau \right),\gamma >1,$$

(1)

where $\eta \left(\tau \right)\ge 0$ for $\tau \ge 1,$ ${\sum }_{\tau =1}^{\infty }{\eta }^{\gamma }\left(\tau \right)<\infty$ and ${\left(\gamma /\left(\gamma -1\right)\right)}^{\gamma }$ is sharp.

In [2], the author used the calculus of variations to show the integral form of (1). Specifically, he showed that if $\gamma >1,$ $\kappa \ge 0$ and integrable over $(0,t)$ and ${\int }_0^{\infty }{\kappa }^{\gamma }\left(t\right)dt<\infty ,$ then

$${\int }_0^{\infty }{\left(\frac{{\int }_0^t\kappa \left(z\right)dz}{t}\right)}^{\gamma }dt\le {\left(\frac{\gamma }{\gamma -1}\right)}^{\gamma }{\int }_0^{\infty }{\kappa }^{\gamma }\left(t\right)dt.$$

(2)

Since the first Hardy-type inequalities were presented, more and more researchers have expanded upon these inequalities and used them in a variety of fields, see the papers [3,4,5] and the books [6,7,8].

In [9], the authors demonstrated that, for $0<\gamma <1,$ the reversed sign applies to (2). They explicitly mentioned that if $\kappa \left(\eta \right)\ge 0$ such that ${\int }_0^{\infty }{\kappa }^{\gamma }\left(\eta \right)d\eta <\infty ,$ then

$${\int }_0^{\infty }{\left(\frac{{\int }_t^{\infty }\kappa \left(z\right)dz}{\eta }\right)}^{\gamma }d\eta \ge {\left(\frac{\gamma }{1-\gamma }\right)}^{\gamma }{\int }_0^{\infty }{\kappa }^{\gamma }\left(\eta \right)d\eta ,$$

where ${\left(\gamma /\left(1-\gamma \right)\right)}^{\gamma }$ is sharp.

In [10], Levinson proved that (2) still holds for parameters $\beta$ and $ϵ.$ That is, the inequality

$${\int }_{\beta }^{ϵ}{\left(\frac{1}{t}{\int }_0^t\kappa \left(z\right)dz\right)}^{\gamma }d\lambda \le {\left(\frac{\gamma }{\gamma -1}\right)}^{\gamma }{\int }_{\beta }^{ϵ}{\kappa }^{\gamma }\left(t\right)dt,$$

(3)

is valid for $0<\beta <ϵ<\infty .$

In [11], G. H. Hardy showcased an overarching generalization of (2) as, if $\gamma >1$ and $\kappa >0$, then

$${\int }_0^{\infty }\frac{1}{t^q}{\left({\int }_0^t\kappa \left(z\right)dz\right)}^{\gamma }dt\le {\left(\frac{\gamma }{q-1}\right)}^{\gamma }{\int }_0^{\infty }\frac{1}{t^{q-\gamma }}{\kappa }^{\gamma }\left(t\right)dt,\mathrm{for}q>1,$$

(4)

and

$${\int }_0^{\infty }\frac{1}{t^q}{\left({\int }_t^{\infty }\kappa \left(z\right)dz\right)}^{\gamma }dt\le {\left(\frac{\gamma }{1-q}\right)}^{\gamma }{\int }_0^{\infty }\frac{1}{t^{q-\gamma }}{\kappa }^{\gamma }\left(t\right)dt,\mathrm{for}q<1.$$

(5)

In [12], the author found that

$${\int }_0^{\infty }exp\left(\frac{1}{\nu }{\int }_0^{\nu }ln\kappa \left(z\right)dz\right)d\nu \le e{\int }_0^{\infty }\kappa \left(\nu \right)d\nu .$$

(6)

This formulation is recognized as the Knopp inequality in mathematical literature, while its discrete counterpart is referred to as the Carleman inequality:

$$\sum _{\nu =1}^{\infty }{\left(\prod _{j=1}^{\nu }\eta \left(j\right)\right)}^{\frac{1}{\nu }}\le e\sum _{\nu =1}^{\infty }\eta \left(\nu \right).$$

In [13], S. Kaijser et al. showed that

$${\int }_0^{\infty }Ϝ\left(\frac{1}{\nu }{\int }_0^{\nu }\kappa \left(z\right)dz\right)\frac{d\nu }{\nu }\le {\int }_0^{\infty }Ϝ\left(\kappa \left(\nu \right)\right)\frac{d\nu }{\nu },$$

(7)

where $Ϝ$ is a convex function and $0<\kappa :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is a locally integrable function.

Later, in [14] authors generalized (7) as follows: Let $0\le \omega :(0,c)\to \mathbb{R}$ s.t. $s\to \omega \left(s\right)/s^2$ be locally integrable in $(0,c)$, $Ϝ$ be convex on $\left(\beta ,\delta \right);$ $-\infty <\beta \le \delta <\infty .$ Then,

$${\int }_0^c\omega \left(s\right)Ϝ\left(\frac{1}{s}{\int }_0^s\kappa \left(z\right)dz\right)\frac{ds}{s}\le {\int }_0^c\vartheta \left(s\right)Ϝ\left(\kappa \left(s\right)\right)\frac{ds}{s},$$

(8)

holds $\forall \kappa :(0,c)\to \mathbb{R}$ s.t. $\kappa \left(s\right)\in \left(\beta ,\delta \right)$ and

$$\vartheta \left(s\right)=s{\int }_s^b\frac{\omega \left(z\right)}{z^2}dz.$$

In [15], Sulaiman proved that if $\kappa ,\psi \ge 0$ are nondecreasing functions and $0<c<\infty ,$ then

$${\int }_0^c\psi \left(\frac{\Lambda \left(s\right)}{s}\right)ds\le {\int }_0^c\psi \left(\kappa \left(s\right)\right)ds.$$

(9)

In the same paper [15], he proved that if $\kappa \ge 0$ and $\xi >0$ s.t. $s/\xi \left(s\right)$ is non-increasing, $\gamma >1,$ $0<\alpha <1,$ then

$$\begin{array}{ccc}& & {\int }_0^{\infty }{\left(\frac{\Lambda \left(s\right)}{\xi \left(s\right)}\right)}^{\gamma }ds\hfill \\ & \le & \frac{1}{\alpha \left(\gamma -1\right){\left(1-\alpha \right)}^{\gamma -1}}{\int }_0^{\infty }{\left(\frac{s\kappa \left(s\right)}{\xi \left(s\right)}\right)}^{\gamma }ds.\hfill \end{array}$$

(10)

Also, he established if $\gamma >1$ and $\kappa ,\psi \ge 0$ s.t. $\phi$ is convex, then

$$\begin{array}{ccc}& & {\int }_0^{\infty }{\psi }^{\gamma }\left(\frac{\Lambda \left(s\right)}{s}\right)ds\hfill \\ & \le & {\left(\frac{\gamma }{\gamma -1}\right)}^{\gamma }{\int }_0^{\infty }{\psi }^{\gamma }\left(\kappa \left(s\right)\right)ds,\hfill \end{array}$$

(11)

where

$$\Lambda \left(s\right)={\int }_0^s\kappa \left(\varsigma \right)d\varsigma .$$

In the last few years, there has been significant attention given to the study of inequalities (1) and (2) on a time scale $\mathbb{T}$, which is referred to as an arbitrary nonempty closed subset of $\mathbb{R}$. The emergence of this novel idea has motivated scholars to investigate Hardy-type inequalities on $\mathbb{T}$. The initial exploration of this direction is credited to P. Řehák [16]. In fact, he demonstrated that

$${\int }_d^{\infty }{\left(\frac{1}{\sigma \left(s\right)-d}{\int }_d^{\sigma \left(s\right)}\kappa \left(\nu \right)\Delta \nu \right)}^h\Delta s\le {\left(\frac{h}{h-1}\right)}^h{\int }_d^{\infty }{\kappa }^h\left(s\right)\Delta s,s\in {[d,\infty )}_{\mathbb{T}}.$$

Given that $d>0,$ $h>1$ and $\kappa \ge 0$. If, furthermore, $\mu \left(s\right)/s\to 0$ as $s\to \infty ,$ then ${\left(h/(h-1)\right)}^h$ is sharp. Refer to Section 2 for the notations used here and in the upcoming content.

In [17], the authors derived the subsequent Jensen’s inequality over $\mathbb{T}:$

$$Ϝ\left(\frac{1}{d-e}{\int }_d^e\kappa \left(s\right)\Delta s\right)\le \frac{1}{d-e}{\int }_d^eϜ\left(\kappa \left(s\right)\right)\Delta s,$$

where $Ϝ$ is a convex function.

Furthermore, the authors ([18], Theorem 7.1.3) utilized the aforementioned outcome to establish that

$${\int }_d^{e}v\left(s\right)Ϝ\left(\frac{1}{\sigma \left(s\right)-d}{\int }_d^{\sigma \left(s\right)}\kappa \left(z\right)\Delta z\right)\frac{\Delta s}{s-d}\le {\int }_d^{e}w\left(s\right)Ϝ\left(\kappa \left(s\right)\right)\frac{\Delta s}{s-d},$$

(12)

where $0\le v\in C_{rd}({[d,e)}_{\mathbb{T}},\mathbb{R}),$ $Ϝ:(i,j)\to \mathbb{R}$ is continuous and convex such that $i,j\in \mathbb{R},$ $\kappa \in C_{rd}({[d,e)}_{\mathbb{T}},\mathbb{R})$ is a $\Delta$ integrable and

$$w\left(s\right)=(s-d){\int }_s^e\frac{v\left(\varsigma \right)\Delta \varsigma }{(\varsigma -d)\left(\sigma \right(\varsigma )-d)},s\in {[d,e)}_{\mathbb{T}}.$$

In [19], the authors demonstrated an extension of the inequality described in (12) as follows:

$${\int }_{{\Theta }_1}v\left(s\right)Ϝ\left(\frac{1}{\Re \left(s\right)}{\int }_{{\Theta }_2}\lambda (s,\varsigma )¸\left(\varsigma \right)\Delta \varsigma \right)\Delta s\le {\int }_{{\Theta }_2}w\left(\varsigma \right)Ϝ(¸\left(\varsigma \right))\Delta \varsigma .$$

Here $({\Theta }_1,\pounds ,{\Sigma }_{\Delta }),$ $({\Theta }_2,\yen ,{\Xi }_{\Delta })$ represent two time scale measure spaces, $\Re \left(s\right)={\int }_{{\Theta }_2}\lambda (s,\varsigma )\Delta \varsigma <\infty ,$ $s\in {\Theta }_1,$ $\lambda (s,\varsigma )\ge 0$ is a kernel, $v:{\Theta }_1\to \mathbb{R}$,

$$w\left(\varsigma \right)={\int }_{{\Theta }_1}\left(\frac{\lambda (s,\varsigma )v\left(s\right)}{\Re \left(s\right)}\right)\Delta ,\varsigma \in {\Theta }_2,$$

$Ϝ$ is a convex function and $¸:{\Theta }_2\to {\mathbb{R}}^n$ s.t. $¸\left({\Theta }_2\right)\subset {\mathbb{R}}^n$ are ${\Xi }_{\Delta }—$integrable functions.

Recently, the authors [20] presented a time scale form of (9)–(11) by using delta calculus, respectively, as follows: If $\kappa ,\psi \ge 0$ are nondecreasing functions, then

$${\int }_c^d\psi \left(\frac{{\int }_0^{\sigma \left(s\right)}\kappa \left(\varsigma \right)\Delta \varsigma }{\sigma \left(s\right)-c}\right)\Delta s\le {\int }_c^d\psi \left(\kappa \left(s\right)\right)\Delta s,$$

(13)

and if $\kappa \ge 0,$ $\xi >0$ s.t. $\left(\sigma \right(s)-c)/\xi \left(s\right)$ is non-increasing, $\gamma >1,$ $0<\alpha <1$, then

$$\begin{array}{ccc}& & {\int }_c^{\infty }{\left(\frac{{\int }_0^{\sigma \left(s\right)}\kappa \left(\varsigma \right)\Delta \varsigma }{\xi \left(s\right)}\right)}^{\gamma }\Delta s\hfill \\ & \le & \frac{L^{\alpha \left(\gamma -1\right)}}{\alpha \left(\gamma -1\right){\left(1-\alpha \right)}^{\gamma -1}}{\int }_c^{\infty }{\left((\sigma \left(s\right)-c)\frac{\kappa \left(s\right)}{\xi \left(s\right)}\right)}^{\gamma }\Delta s.\hfill \end{array}$$

(14)

Also, if $\gamma >1,$ $\kappa ,\psi \ge 0$ s.t. $\phi$ is convex, then

$$\begin{array}{ccc}& & {\int }_c^{\infty }{\psi }^{\gamma }\left(\frac{{\int }_0^{\sigma \left(s\right)}\kappa \left(j\right)\Delta j}{\sigma \left(s\right)-c}\right)\Delta s\hfill \\ & \le & {\left(\frac{h}{h-1}\right)}^h{L}^{\frac{h}{h-1}}{\int }_c^{\infty }{\psi }^{\gamma }\left(\kappa \left(s\right)\right)\Delta s,\hfill \end{array}$$

(15)

where $L>0$ is a constant s.t. $\sigma \left(s\right)-c\le L(s-c).$

For more information about Hardy’s inequality on time scales delta calculus, we suggest reading [21,22,23,24,25,26,27,28,29,30].

In continuation of this growing trend and with the aim of advancing the exploration of dynamic inequalities over $\mathbb{T}$, we shall present numerous novel of Hardy inequalities within this mathematical framework. The research findings will contribute to the understanding of mathematical inequalities and their applications on time scales, presenting a valuable addition to the existing body of knowledge in this field. Furthermore, these results may have potential implications in various mathematical and applied areas where time scales play a crucial role, such as difference equations, dynamic systems, and more.

While numerous outcomes exist in the realm of $\mathbb{T}$ calculus involving the delta derivative, there remains a notable scarcity of research concerning the nabla derivative. As a result, the primary goal of this study is to expand (9)–(11) for nabla time scale calculus. The foundational principles of these principal theorems draw inspiration from the work presented in the paper [20], wherein analogous findings were delineated within the domain of time scale delta calculus.

Our discussions are organized as follows. Basic ideas and a number of lemmas related to time scale calculus are presented in the section that follows. The main findings are presented in the final section.

2. Preliminaries

Before delving into the main results, we will introduce some preliminary concepts and background information to set the foundation for our study.

In [31], the authors introduced a definition for the backward jump operator $\rho$ and the backward graininess function v as follows: $\rho \left(\tau \right):=sup\{\varsigma \in \mathbb{T}:\varsigma <\tau \},$ and $v\left(\tau \right):=\tau -\rho \left(\tau \right)\ge 0$, respectively.

In the context of any function $\Gamma :\mathbb{T}\to \mathbb{R}$, ${\Gamma }^{\rho }\left(\varsigma \right)$ signifies the value of $\Gamma$ at $\rho \left(\varsigma \right)$. Additionally, we define $J_{\mathbb{T}}$ as $J_{\mathbb{T}}:=J\cap \mathbb{T},$ $J\subseteq \mathbb{R}$.

Definition 1

([32]). Γ is left-dense continuous ($ld—$continuous), if it exhibits continuity at each left-dense point in $\mathbb{T}$ and possesses well-defined right-sided limits at right-dense points within $\mathbb{T}$. The set encompassing all such ld-continuous functions is denoted as ${\mathrm{C}}_{ld}(\mathbb{T},\mathbb{R})$.

Definition 2

([32]). Γ is $\nabla —$differentiable at ς, if it is exists within a neighborhood V of ς and $\exists {\Gamma }^{\nabla }\left(\varsigma \right)\in \mathbb{R}$ is unique (which is referred to as the nabla derivative of Γ at ς), s.t. $\forall ϵ>0,$ $\exists N$ is a neighborhood of ς s.t. $N\subseteq V$ and

$$\left|{\Gamma }^{\rho }\left(\varsigma \right)-\Gamma \left(\vartheta \right)-{\Gamma }^{\nabla }\left(\varsigma \right)[\rho \left(\varsigma \right)-\vartheta ]\right|\le ϵ\left|\rho \left(\varsigma \right)-\vartheta \right|,\forall \vartheta \in N.$$

Theorem 1

([32]). Assuming κ and χ are $\nabla —$differentiable at $s\in \mathbb{T}$, then

 1.

The product $\kappa \chi$ is $\nabla —$differentiable at s and

$${\left(\kappa \chi \right)}^{\nabla }\left(s\right)={\kappa }^{\nabla }\left(s\right)\chi \left(s\right)+{\kappa }^{\rho }\left(s\right){\chi }^{\nabla }\left(s\right)=\kappa \left(s\right){\chi }^{\nabla }\left(s\right)+{\kappa }^{\nabla }\left(s\right){\chi }^{\rho }\left(s\right),$$

(16)

holds.

 2.

If $\chi \left(s\right){\chi }^{\rho }\left(s\right)\ne 0$, then $\kappa /\chi$ is $\nabla —$differentiable at s and

$${\left(\frac{\kappa }{\chi }\right)}^{\nabla }\left(s\right)=\frac{{\kappa }^{\nabla }\left(s\right)\chi \left(s\right)-\kappa \left(s\right){\chi }^{\nabla }\left(s\right)}{\chi \left(s\right){\chi }^{\rho }\left(s\right)},$$

(17)

holds.

Lemma 1

([33]). Assuming $\kappa :\mathbb{R}\to \mathbb{R}$ is continuously differentiable, $\chi :\mathbb{T}\to \mathbb{R}$ is continuous and $\nabla —$differentiable, then

$${\left(\kappa \circ \chi \right)}^{\nabla }\left(s\right)={\kappa }^{\prime }\left(\chi \left(d\right)\right){\chi }^{\nabla }\left(s\right),d\in [\rho \left(s\right),s].$$

(18)

Definition 3

([32]). Λ is considered a $\nabla —$antiderivative of κ if ${\Lambda }^{\nabla }\left(s\right)=\kappa \left(s\right)$ true for $s\in \mathbb{T}$. The nabla integral of κ is defined as

$${\int }_{\beta }^s\kappa \left(z\right)\nabla z=\Lambda \left(s\right)-\Lambda \left(\beta \right),\forall s\in \mathbb{T}.$$

Theorem 2

([32]). Let ${\delta }_1,{\delta }_2,{\delta }_3\in \mathbb{T},$ $ϵ,\epsilon \in \mathbb{R}$ and $\kappa ,\chi \in {\mathrm{C}}_{ld}(\mathbb{T},\mathbb{R})$. Then

 1.

${\int }_{{\delta }_1}^{{\delta }_2}\left[ϵ\kappa \left(s\right)+\epsilon \chi \left(s\right)\right]\nabla s=ϵ{\int }_{{\delta }_1}^{{\delta }_2}\kappa \left(s\right)\nabla s+\epsilon {\int }_{{\delta }_1}^{{\delta }_2}\chi \left(s\right)\nabla s$,

 2.

${\int }_{{\delta }_1}^{{\delta }_3}\kappa \left(s\right)\nabla s={\int }_{{\delta }_1}^{{\delta }_2}\kappa \left(s\right)\nabla s+{\int }_{{\delta }_2}^{{\delta }_3}\kappa \left(s\right)\nabla s$,

 3.

If $\kappa \left(s\right)\ge 0$$\forall s\in {[{\delta }_1,{\delta }_2]}_{\mathbb{T}}$, then ${\int }_{{\delta }_1}^{{\delta }_2}\kappa \left(s\right)\nabla s\ge 0,$

 4.

If $\kappa \left(s\right)\ge \chi \left(s\right)$$\forall s\in {[{\delta }_1,{\delta }_2]}_{\mathbb{T}}$, then ${\int }_{{\delta }_1}^{{\delta }_2}\kappa \left(s\right)\nabla s\ge {\int }_{{\delta }_1}^{{\delta }_2}\chi \left(s\right)\nabla s.$

Lemma 2

(The integration by parts [32]). If ${\delta }_0,\delta \in \mathbb{T}$ and $\kappa ,\chi \in {\mathrm{C}}_{ld}(\mathbb{T},\mathbb{R})$, then

$${\int }_{{\delta }_0}^{\delta }\kappa \left(s\right){\chi }^{\nabla }\left(s\right)\nabla s={\left.\kappa \left(s\right)\chi \left(s\right)\right|}_{{\delta }_0}^{\delta }-{\int }_{{\delta }_0}^{\delta }{\kappa }^{\nabla }\left(s\right){\chi }^{\rho }\left(s\right)\nabla s.$$

(19)

Lemma 3

(Hölder’s inequality [34]). Let ${\delta }_0,d\in \mathbb{T},$ $\kappa ,\xi \in {\mathrm{C}}_{ld}(\mathbb{T},\mathbb{R})$ and $ϵ,\epsilon >1$ such that $1/ϵ+1/\epsilon =1.$ Then,

$${\int }_{{\delta }_0}^d\left|\kappa \left(s\right)\xi \left(s\right)\right|\nabla s\le {\left[{\int }_{{\delta }_0}^d{\left|\kappa \left(s\right)\right|}^{ϵ}\nabla s\right]}^{\frac{1}{ϵ}}{\left[{\int }_{{\delta }_0}^d{\left|\xi \left(s\right)\right|}^{\epsilon }\nabla s\right]}^{\frac{1}{\epsilon }}.$$

(20)

Definition 4

([35]). $\psi :J_{\mathbb{T}}\to \mathbb{R}$ is termed convex if

$$\psi \left(\alpha \varsigma +\left(1-\alpha \right)s\right)\le \alpha \psi \left(\varsigma \right)+\left(1-\alpha \right)\psi \left(s\right),$$

$\forall \varsigma ,s\in J_{\mathbb{T}}$ and $\alpha \in \left[0,1\right]$ s.t. $\alpha \varsigma +\left(1-\alpha \right)s\in J_{\mathbb{T}}.$

Definition 5

([36]). $\psi :J_{\mathbb{T}}\to {\mathbb{R}}^{+}$ is submultiplicative if

$$\psi \left(\varsigma s\right)\le \psi \left(\varsigma \right)\psi \left(s\right),\forall \varsigma ,s\in J_{\mathbb{T}}\subset \mathbb{R}.$$

Lemma 4

(Jensen’s inequality [34]). Assume ${\varsigma }_0,\varsigma \in \mathbb{T}$ and $\delta ,r\in \mathbb{R}$. If $\Psi :({\delta }_0,\delta )\to \mathbb{R}$ is continuous and convex, $\kappa \in {\mathrm{C}}_{ld}({[{\varsigma }_0,\varsigma ]}_{\mathbb{T}},\mathbb{R})$ and $\phi \in {\mathrm{C}}_{ld}({[{\varsigma }_0,\varsigma ]}_{\mathbb{T}},({\delta }_0,\delta )),$ then

$$\Psi \left(\frac{{\int }_{{\varsigma }_0}^{\varsigma }\kappa \left(s\right)\phi \left(s\right)\nabla s}{{\int }_{{\varsigma }_0}^{\varsigma }\kappa \left(s\right)\nabla s}\right)\le \frac{{\int }_{{\varsigma }_0}^{\varsigma }\kappa \left(s\right)\Psi \left(\phi \left(s\right)\right)\nabla s}{{\int }_{{\varsigma }_0}^{\varsigma }\kappa \left(s\right)\nabla s}.$$

(21)

Lemma 5

([35]). Suppose $\psi :J_{\mathbb{T}}\to \mathbb{R}$ is a continuous function. If ${\psi }^{\nabla \nabla }$ exists on $J_{\mathbb{T}}$ and ${\psi }^{\nabla \nabla }\ge 0,$ then ψ is convex.

For more information about some properties about convexity, concavity and submultiplicative functions, see [37,38,39,40,41].

3. Principal Findings

Throughout our study, we will operate under the assumption that the functions involved are ld-continuous functions and we will consider integrals that exist within the context.

Theorem 3.

Given $\beta ,r\in \mathbb{T}$ and considering non-decreasing functions $f,\psi \ge 0$, we have

$${\int }_{\beta }^r\psi \left(\frac{F\left(s\right)}{s-\delta }\right)\nabla s\le {\int }_{\beta }^r\psi \left(f\left(s\right)\right)\nabla s,$$

(22)

where

$$F\left(s\right)={\int }_{\beta }^{s}f\left(\varsigma \right)\nabla \varsigma .$$

(23)

Proof. 

From (23), we find that

$${\int }_{\beta }^r\psi \left(\frac{F\left(s\right)}{s-\beta }\right)\nabla s={\int }_{\beta }^r\psi \left(\frac{{\int }_{\beta }^{s}f\left(\varsigma \right)\nabla \varsigma }{s-\beta }\right)\nabla s.$$

(24)

Let $\varsigma \le s.$ Then, $f\left(\varsigma \right)\le f\left(s\right)$ (as f is a nonnegative and nondecreasing), thus

$$\begin{array}{ccc}\hfill {\int }_{\beta }^{s}f\left(\varsigma \right)\nabla \varsigma & \le & {\int }_{\beta }^{s}f\left(s\right)\nabla \varsigma =f\left(s\right){\int }_{\beta }^s\nabla \varsigma \hfill \\ & =& \left(s-\beta \right)f\left(s\right).\hfill \end{array}$$

(25)

As $\psi \ge 0$ is a nondecreasing, it follows that

$$\psi \left(\frac{{\int }_{\beta }^{s}f\left(\varsigma \right)\nabla \varsigma }{s-\beta }\right)\le \psi \left(f\left(s\right)\right).$$

(26)

Using (26) into (24), we get

$${\int }_{\beta }^r\psi \left(\frac{F\left(s\right)}{s-\beta }\right)\nabla s\le {\int }_{\beta }^r\psi \left(f\left(s\right)\right)\nabla s.$$

This result corresponds to (22). □

Remark 1.

When $\mathbb{T}=\mathbb{R}$ and $\beta =0,$ we get (9), which is established in [15].

Theorem 4.

Assume $\beta \in \mathbb{T},$ $f\ge 0$ and $\xi >0$ such that $\left(\varsigma -\beta \right)/\xi \left(\varsigma \right)$ is non-increasing, $h>1$ and $\alpha \in (0,1).$ Consider the definition of F as provided in (23). Additionally, suppose $\exists G>0$ be a constant s.t.

$$s-\beta \le G\left(\rho \left(s\right)-\beta \right),$$

(27)

then,

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\left(\frac{F\left(s\right)}{\xi \left(s\right)}\right)}^h\nabla s\hfill \\ & \le & \frac{G^{\alpha \left(h-1\right)}}{\lambda \left(h-1\right){\left(1-\lambda \right)}^{h-1}}{\int }_{\beta }^{\infty }{\left(\left(\rho \left(s\right)-\beta \right)\frac{f\left(s\right)}{{\xi }^{\rho }\left(s\right)}\right)}^h\nabla s.\hfill \end{array}$$

(28)

Proof. 

From (23), we see that

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\left(\frac{F\left(s\right)}{\xi \left(s\right)}\right)}^h\nabla s\hfill \\ & =& {\int }_{\beta }^{\infty }{\xi }^{-h}\left(s\right){\left({\int }_{\beta }^{s}f\left(\varsigma \right)\nabla \varsigma \right)}^h\nabla s\hfill \\ & =& {\int }_{\beta }^{\infty }{\xi }^{-h}\left(s\right){\left({\int }_{\beta }^s\left[{\left(\varsigma -\beta \right)}^{{}^{\frac{\lambda \left(h-1\right)}{h}}}f\left(\varsigma \right)\right]{\left(\varsigma -\beta \right)}^{{}^{\frac{-\lambda \left(h-1\right)}{h}}}\nabla \varsigma \right)}^h\nabla s.\hfill \end{array}$$

(29)

Applying (20) on ${\int }_{\beta }^s\left[{\left(\varsigma -\beta \right)}^{\lambda \left(h-1\right)/h}f\left(\varsigma \right)\right]{\left(\varsigma -\beta \right)}^{-\lambda \left(h-1\right)/h}\nabla \varsigma ,$ with $h>1,$ $h/\left(h-1\right),$ we obtain

$$\begin{array}{ccc}& & {\int }_{\beta }^s\left[{\left(\varsigma -\beta \right)}^{{}^{\frac{\lambda \left(h-1\right)}{h}}}f\left(\varsigma \right)\right]{\left(\varsigma -\beta \right)}^{{}^{\frac{-\lambda \left(h-1\right)}{h}}}\nabla \varsigma \hfill \\ & \le & {\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\lambda \left(h-1\right)}{f}^h\left(\varsigma \right)\nabla \varsigma \right)}^{{}^{\frac{1}{h}}}{\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{-\lambda }\nabla s\right)}^{{}^{\frac{h-1}{h}}}.\hfill \end{array}$$

(30)

Substituting (30) into (29), we obtain

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\left(\frac{F\left(s\right)}{\xi \left(s\right)}\right)}^h\nabla s\hfill \\ & \le & {\int }_{\beta }^{\infty }{\xi }^{-h}\left(s\right)\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\lambda \left(h-1\right)}{f}^h\left(\varsigma \right)\nabla \varsigma \right){\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{-\lambda }\nabla \varsigma \right)}^{h-1}\nabla s.\hfill \end{array}$$

(31)

Applying (18) on ${\left(\varsigma -\beta \right)}^{1-\lambda }$ for $0<\lambda <1,$ we see that

$${\left({\left(\varsigma -\beta \right)}^{1-\lambda }\right)}^{\nabla }=\left(1-\lambda \right){\left(d-\beta \right)}^{-\lambda },d\in [\rho \left(\varsigma \right),\varsigma ].$$

Since $d\le \varsigma$, we see that

$${\left({\left(\varsigma -\beta \right)}^{1-\lambda }\right)}^{\nabla }\ge \left(1-\lambda \right){\left(\varsigma -\beta \right)}^{-\lambda },$$

and then

$$\begin{array}{ccc}\hfill {\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{-\lambda }\nabla \varsigma & \le & \frac{1}{1-\lambda }{\int }_{\beta }^s{\left({\left(\varsigma -\beta \right)}^{1-\lambda }\right)}^{\nabla }\nabla \varsigma \hfill \\ & =& \frac{1}{1-\lambda }{\left(s-\beta \right)}^{1-\lambda }.\hfill \end{array}$$

(32)

Using(32) in (31), we get

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\left(\frac{F\left(s\right)}{\xi \left(s\right)}\right)}^h\nabla s\hfill \\ & \le & \frac{1}{{\left(1-\lambda \right)}^{h-1}}{\int }_{\beta }^{\infty }{\left(s-\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)}{\xi }^{-h}\left(s\right)\hfill \\ & & \times \left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\lambda \left(h-1\right)}{f}^h\left(\varsigma \right)\nabla \varsigma \right)\nabla s.\hfill \end{array}$$

(33)

Applying (19) on ${\int }_{\beta }^{\infty }{\left(s-\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)}{\xi }^{-h}\left(s\right)\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\lambda \left(h-1\right)}{f}^h\left(\varsigma \right)\nabla \varsigma \right)\nabla s,$ with ${\chi }^{\nabla }\left(s\right)={\left(s-\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)}{\xi }^{-h}\left(s\right)$ and $\kappa \left(s\right)={\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\lambda \left(h-1\right)}{f}^h\left(\varsigma \right)\nabla \varsigma ,$ we observe that

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\left(s-\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)}{\xi }^{-h}\left(s\right)\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\lambda \left(h-1\right)}{f}^h\left(\varsigma \right)\nabla \varsigma \right)\nabla s\hfill \\ & =& {\left.\chi \left(s\right)\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\lambda \left(h-1\right)}{f}^h\left(\varsigma \right)\nabla \varsigma \right)\right|}_{\beta }^{\infty }-{\int }_{\beta }^{\infty }{\chi }^{\rho }\left(s\right){\left(s-\beta \right)}^{\lambda \left(h-1\right)}{f}^h\left(s\right)\nabla s,\hfill \end{array}$$

(34)

where

$$\chi \left(s\right)=-{\int }_s^{\infty }{\left(\varsigma -\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)}{\xi }^{-h}\left(\varsigma \right)\nabla \varsigma .$$

Since ${lim}_{s\to \infty }\chi \left(s\right)=0,$ we have

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\left(s-\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)}{\xi }^{-h}\left(s\right)\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\lambda \left(h-1\right)}{f}^h\left(\varsigma \right)\nabla \varsigma \right)\nabla s\hfill \\ & =& {\int }_{\beta }^{\infty }{\left(s-\beta \right)}^{\lambda \left(h-1\right)}{f}^h\left(s\right)\left({\int }_{\rho \left(s\right)}^{\infty }{\left(\varsigma -\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)}{\xi }^{-h}\left(\varsigma \right)\nabla \varsigma \right)\nabla s.\hfill \end{array}$$

(35)

Since the function $\left(\varsigma -\beta \right)/\xi \left(\varsigma \right)$ is non-increasing, we get

$$\begin{array}{ccc}& & {\int }_{\rho \left(s\right)}^{\infty }{\left(\varsigma -\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)}{\xi }^{-h}\left(\varsigma \right)\nabla \varsigma \hfill \\ & =& {\int }_{\rho \left(s\right)}^{\infty }{\left(\varsigma -\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)-h}{\left(\varsigma -\beta \right)}^h{\xi }^{-h}\left(\varsigma \right)\nabla \varsigma \hfill \\ & \le & {\left(\rho \left(s\right)-\beta \right)}^h{\xi }^{-h}\left(\rho \left(s\right)\right){\int }_{\rho \left(s\right)}^{\infty }{\left(\varsigma -\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)-h}\nabla \varsigma .\hfill \end{array}$$

(36)

Using (36) in (35), we find that

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\left(s-\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)}{\xi }^{-h}\left(s\right)\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\lambda \left(h-1\right)}{f}^h\left(\varsigma \right)\nabla \varsigma \right)\nabla s\hfill \\ & \le & {\int }_{\beta }^{\infty }{\left(s-\beta \right)}^{\lambda \left(h-1\right)}{f}^h\left(s\right){\left(\frac{\rho \left(s\right)-\beta }{{\xi }^{\rho }\left(s\right)}\right)}^h\hfill \\ & & \times \left({\int }_{\rho \left(s\right)}^{\infty }{\left(\varsigma -\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)-h}\nabla \varsigma \right)\nabla s.\hfill \end{array}$$

(37)

Applying (18) on ${\left(\varsigma -\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)-h+1},$ we see that

$$\begin{array}{ccc}& & \left(\frac{-1}{\lambda \left(h-1\right)}\right){\left[{\left(\varsigma -\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)-h+1}\right]}^{\nabla }\hfill \\ & =& \left(\frac{-1}{\lambda \left(h-1\right)}\right){\left[{\left(\varsigma -\beta \right)}^{-\lambda \left(h-1\right)}\right]}^{\nabla }={\left(d-\beta \right)}^{-\left[\lambda \left(h-1\right)+1\right]},\hfill \end{array}$$

(38)

where $d\in [\rho \left(\varsigma \right),\varsigma ].$ Since $d\le \varsigma ,$ $h>1$ and $-\left[\lambda \left(h-1\right)+1\right]<0,$ it is evident that

$${\left(d-\beta \right)}^{-\left[\lambda \left(h-1\right)+1\right]}\ge {\left(\varsigma -\beta \right)}^{-\left[\lambda \left(h-1\right)+1\right]}={\left(\varsigma -\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)-h},$$

and then we get

$${\left(\varsigma -\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)-h}\le \left(\frac{-1}{\lambda \left(h-1\right)}\right){\left[{\left(\varsigma -\beta \right)}^{-\lambda \left(h-1\right)}\right]}^{\nabla }.$$

By integrating the previous inequality over $\varsigma$ from $\rho \left(s\right)$ to $\infty ,$ (taking into account that $h>1,$ $0<\lambda <1$ and $\left(1-\lambda \right)\left(h-1\right)-h+1=-\lambda \left(h-1\right)<0$), we derive

$$\begin{array}{ccc}& & {\int }_{\rho \left(s\right)}^{\infty }{\left(\varsigma -\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)-h}\nabla \varsigma \hfill \\ & \le & \left(\frac{-1}{\lambda \left(h-1\right)}\right){\int }_{\rho \left(s\right)}^{\infty }{\left[{\left(\varsigma -\beta \right)}^{-\lambda \left(h-1\right)}\right]}^{\nabla }\nabla \varsigma \hfill \\ & =& \frac{1}{\lambda \left(h-1\right)}{\left(\rho \left(s\right)-\beta \right)}^{-\lambda \left(h-1\right)}.\hfill \end{array}$$

(39)

Using (39) in (37), we obtain

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\left(s-\beta \right)}^{\left(1-\lambda \right)\left(h-1\right)}{\xi }^{-h}\left(s\right)\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\lambda \left(h-1\right)}{f}^h\left(\varsigma \right)\nabla \varsigma \right)\nabla s\hfill \\ & \le & \frac{1}{\lambda \left(h-1\right)}{\int }_{\beta }^{\infty }{\left(\frac{s-\beta }{\rho \left(s\right)-\beta }\right)}^{\lambda \left(h-1\right)}{f}^h\left(s\right){\left(\frac{\rho \left(s\right)-\beta }{{\xi }^{\rho }\left(s\right)}\right)}^h\nabla s.\hfill \end{array}$$

(40)

Using (40) in (33), we observe that

$$\begin{array}{cc}\hfill & {\int }_{\beta }^{\infty }{\left(\frac{F\left(s\right)}{\xi \left(s\right)}\right)}^h\nabla s\hfill \\ \hfill & \le \frac{1}{\lambda \left(h-1\right){\left(1-\lambda \right)}^{h-1}}{\int }_{\beta }^{\infty }{\left(\frac{s-\beta }{\rho \left(s\right)-\beta }\right)}^{\lambda \left(h-1\right)}{f}^h\left(s\right){\left(\frac{\rho \left(s\right)-\beta }{{\xi }^{\rho }\left(s\right)}\right)}^h\nabla s.\hfill \end{array}$$

(41)

Using (27) in (41), we see that

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\left(\frac{F\left(s\right)}{\xi \left(s\right)}\right)}^h\nabla s\hfill \\ & \le & \frac{G^{\lambda \left(h-1\right)}}{\lambda \left(h-1\right){\left(1-\lambda \right)}^{h-1}}{\int }_{\beta }^{\infty }{\left(\left(\rho \left(s\right)-\beta \right)\frac{f\left(s\right)}{{\xi }^{\rho }\left(s\right)}\right)}^h\nabla s,\hfill \end{array}$$

which is (28). □

Remark 2.

When $\mathbb{T}=\mathbb{R}$ and $\beta =0$, then $\rho \left(s\right)=s,$ (27) holds with equality for $G=1$ and we get (10), which is proved in [15].

Theorem 5.

Assume $\beta \in \mathbb{T},$ $h>1$ and $f,\psi \ge 0$ s.t. ψ is convex. Consider the definition of F as provided in (23). Additionally, suppose $\exists G>0$ be a constant s.t.

$$s-\beta \le G\left(\rho \left(s\right)-\beta \right).$$

(42)

then,

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\psi }^h\left(\frac{F\left(s\right)}{s-\beta }\right)\nabla s\hfill \\ & \le & {\left(\frac{h}{h-1}\right)}^h{G}^{\frac{h-1}{h}}{\int }_{\beta }^{\infty }{\psi }^h\left(f\left(s\right)\right)\nabla s.\hfill \end{array}$$

(43)

Proof. 

Note that

$${\int }_{\beta }^{\infty }{\psi }^h\left(\frac{F\left(s\right)}{s-\beta }\right)\nabla s={\int }_{\beta }^{\infty }{\left[\psi \left(\frac{{\int }_{\beta }^{s}f\left(\varsigma \right)\nabla \varsigma }{s-\beta }\right)\right]}^h\nabla s.$$

(44)

Applying (21), we find that

$$\psi \left(\frac{{\int }_{\beta }^{s}f\left(\varsigma \right)\nabla \varsigma }{s-\beta }\right)\le \frac{{\int }_{\beta }^s\psi \left(f\left(\varsigma \right)\right)\nabla \varsigma }{s-\beta }.$$

(45)

Substituting (45) into (44), we have that

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\psi }^h\left(\frac{F\left(s\right)}{s-\beta }\right)\nabla s\hfill \\ & \le & {\int }_{\beta }^{\infty }{\left(s-\beta \right)}^{-h}{\left({\int }_{\beta }^s\psi \left(f\left(\varsigma \right)\right)\nabla \varsigma \right)}^h\nabla s\hfill \\ & =& {\int }_{\beta }^{\infty }{\left(s-\beta \right)}^{-h}{\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{-1}{h}}{\left(\varsigma -\beta \right)}^{\frac{1}{h}}\psi \left(f\left(\varsigma \right)\right)\nabla \varsigma \right)}^h\nabla s.\hfill \end{array}$$

(46)

Applying (30) on ${\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{-1}{h}}{\left(\varsigma -\beta \right)}^{\frac{1}{h}}\psi \left(f\left(\varsigma \right)\right)\nabla \varsigma ,$ with $h>1,$ $h/\left(h-1\right),$ we find that

$$\begin{array}{ccc}& & {\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{-1}{h}}{\left(\varsigma -\beta \right)}^{\frac{1}{h}}\psi \left(f\left(\varsigma \right)\right)\nabla \varsigma \hfill \\ & \le & {\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{-1}{h}}\nabla \varsigma \right)}^{\frac{h-1}{h}}{\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{-1}{h}}\left(\varsigma -\beta \right){\psi }^h\left(f\left(\varsigma \right)\right)\nabla \varsigma \right)}^{\frac{1}{h}}\hfill \\ & =& {\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{-1}{h}}\nabla \varsigma \right)}^{\frac{h-1}{h}}{\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{h-1}{h}}{\psi }^h\left(f\left(\varsigma \right)\right)\nabla \varsigma \right)}^{\frac{1}{h}}.\hfill \end{array}$$

(47)

Substituting (47) into (46), we find that

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\psi }^h\left(\frac{F\left(s\right)}{s-\beta }\right)\nabla s\hfill \\ & \le & {\int }_{\beta }^{\infty }{\left(s-\beta \right)}^{-h}{\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{-1}{h}}\nabla \varsigma \right)}^{h-1}\hfill \\ & & \times \left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{h-1}{h}}{\psi }^h\left(f\left(\varsigma \right)\right)\nabla \varsigma \right)\nabla s.\hfill \end{array}$$

(48)

Applying (18) on ${\left(\varsigma -\beta \right)}^{\frac{-1}{h}+1},$ we find that

$$\frac{h}{h-1}{\left[{\left(\varsigma -\beta \right)}^{\frac{-1}{h}+1}\right]}^{\nabla }={\left(d-\beta \right)}^{\frac{-1}{h}},$$

(49)

where $d\in \left[\rho \left(\varsigma \right),\varsigma \right].$ Since $d\le \varsigma$ and $h>1,$ we observe that

$${\left(d-\beta \right)}^{\frac{-1}{h}}\ge {\left(\varsigma -\beta \right)}^{\frac{-1}{h}},$$

and then we have from (49) that

$${\left(\varsigma -\beta \right)}^{\frac{-1}{h}}\le \frac{h}{h-1}{\left[{\left(\varsigma -\beta \right)}^{\frac{-1}{h}+1}\right]}^{\nabla }.$$

(50)

Integrating (50) over $\varsigma$ from $\beta$ to s, we see that

$$\begin{array}{ccc}& & {\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{-1}{h}}\nabla \varsigma \hfill \\ & \le & \frac{h}{h-1}{\int }_{\beta }^s{\left[{\left(\varsigma -\beta \right)}^{\frac{-1}{h}+1}\right]}^{\nabla }\nabla \varsigma \hfill \\ & =& \frac{h}{h-1}{\int }_{\beta }^s{\left[{\left(\varsigma -\beta \right)}^{\frac{h-1}{h}}\right]}^{\nabla }\nabla \varsigma =\frac{h}{h-1}{\left(s-\beta \right)}^{\frac{h-1}{h}}.\hfill \end{array}$$

(51)

Substituting (51) into (48), we observe that

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\psi }^h\left(\frac{F\left(s\right)}{s-\beta }\right)\nabla s\hfill \\ & \le & {\left(\frac{h}{h-1}\right)}^{h-1}{\int }_{\beta }^{\infty }{\left(s-\beta \right)}^{\frac{1}{h}-2}\hfill \\ & & \times \left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{h-1}{h}}{\psi }^h\left(f\left(\varsigma \right)\right)\nabla \varsigma \right)\nabla s.\hfill \end{array}$$

(52)

Applying (19) on ${\int }_{\beta }^{\infty }{\left(s-\beta \right)}^{\frac{1}{h}-2}\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{h-1}{h}}{\psi }^h\left(f\left(\varsigma \right)\right)\nabla \varsigma \right)\nabla s,$ with ${\chi }^{\nabla }\left(s\right)={\left(s-\beta \right)}^{\frac{1}{h}-2},$ and $\kappa \left(s\right)={\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{h-1}{h}}{\psi }^h\left(f\left(\varsigma \right)\right)\nabla \varsigma ,$ we see that

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\left(s-\beta \right)}^{\frac{1}{h}-2}\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{h-1}{h}}{\psi }^h\left(f\left(\varsigma \right)\right)\nabla \varsigma \right)\nabla s\hfill \\ & =& {\left.\chi \left(s\right)\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{h-1}{h}}{\psi }^h\left(f\left(\varsigma \right)\right)\nabla \varsigma \right)\right|}_{\beta }^{\infty }\hfill \\ & & -{\int }_{\beta }^{\infty }{\chi }^{\rho }\left(s\right){\left(s-\beta \right)}^{\frac{h-1}{h}}{\psi }^h\left(f\left(s\right)\right)\nabla s,\hfill \end{array}$$

(53)

where

$$\chi \left(s\right)=-{\int }_s^{\infty }{\left(\varsigma -\beta \right)}^{\frac{1}{h}-2}\nabla \varsigma .$$

Since ${lim}_{s\to \infty }v\left(s\right)=0,$ we have

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\left(s-\beta \right)}^{\frac{1}{h}-2}\left({\int }_{\beta }^s{\left(\varsigma -\beta \right)}^{\frac{h-1}{h}}{\psi }^h\left(f\left(\varsigma \right)\right)\nabla \varsigma \right)\nabla s\hfill \\ & =& {\int }_{\beta }^{\infty }\left({\int }_{\rho \left(s\right)}^{\infty }{\left(\varsigma -\beta \right)}^{\frac{1}{h}-2}\nabla \varsigma \right){\left(s-\beta \right)}^{\frac{h-1}{h}}{\psi }^h\left(f\left(s\right)\right)\nabla s.\hfill \end{array}$$

(54)

Substituting (54) into (52), we obtain

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\psi }^h\left(\frac{F\left(s\right)}{s-\beta }\right)\nabla s\hfill \\ & \le & {\left(\frac{h}{h-1}\right)}^{h-1}{\int }_{\beta }^{\infty }\left({\int }_{\rho \left(s\right)}^{\infty }{\left(\varsigma -\beta \right)}^{\frac{1}{h}-2}\nabla \varsigma \right)\hfill \\ & & \times {\left(s-\beta \right)}^{\frac{h-1}{h}}{\psi }^h\left(f\left(s\right)\right)\nabla s.\hfill \end{array}$$

(55)

Applying (18) on ${\left(\varsigma -\beta \right)}^{\frac{1}{h}-1},$ we get

$$\frac{h}{1-h}{\left[{\left(\varsigma -\beta \right)}^{\frac{1}{h}-1}\right]}^{\nabla }={\left(d-\beta \right)}^{\frac{1}{h}-2},$$

(56)

where $d\in \left[\rho \left(\varsigma \right),\varsigma \right].$ Since $d\le \varsigma$ and $(1/h)-2<0,$ we see that

$${\left(d-\beta \right)}^{\frac{1}{h}-2}\ge {\left(\varsigma -\beta \right)}^{\frac{1}{h}-2},$$

and then we have from (56) that

$${\left(\varsigma -\beta \right)}^{\frac{1}{h}-2}\le \frac{h}{1-h}{\left[{\left(\varsigma -\beta \right)}^{\frac{1}{h}-1}\right]}^{\nabla }.$$

(57)

Integrating (57) over $\varsigma$ from $\rho \left(s\right)$ to $\infty ,$ considering that $(1/h)-1<0,$ yields

$$\begin{array}{ccc}\hfill {\int }_{\rho \left(s\right)}^{\infty }{\left(\varsigma -\beta \right)}^{\frac{1}{h}-2}\nabla \varsigma & \le & {\int }_{\rho \left(s\right)}^{\infty }\frac{h}{1-h}{\left[{\left(\varsigma -\beta \right)}^{\frac{1}{h}-1}\right]}^{\nabla }\nabla \varsigma \hfill \\ & =& \left(\frac{h}{h-1}\right){\left(\rho \left(s\right)-\beta \right)}^{\frac{1}{h}-1}.\hfill \end{array}$$

(58)

Using (58) in (55), we have

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\psi }^h\left(\frac{F\left(s\right)}{s-\beta }\right)\nabla s\hfill \\ & \le & {\left(\frac{h}{h-1}\right)}^h{\int }_{\beta }^{\infty }{\left(\frac{s-\beta }{\rho \left(s\right)-\beta }\right)}^{\frac{h-1}{h}}{\psi }^h\left(f\left(s\right)\right)\nabla s.\hfill \end{array}$$

Using (42), the last inequality becomes

$$\begin{array}{ccc}& & {\int }_{\beta }^{\infty }{\psi }^h\left(\frac{F\left(s\right)}{s-\beta }\right)\nabla s\hfill \\ & \le & {\left(\frac{h}{h-1}\right)}^h{G}^{\frac{h-1}{h}}{\int }_{\beta }^{\infty }{\psi }^h\left(f\left(s\right)\right)\nabla s,\hfill \end{array}$$

which is (43). □

Remark 3.

When $\mathbb{T}=\mathbb{R}$ and $\beta =0$, then $\rho \left(s\right)=s,$ (42) holds with equality for $G=1$ and we get (11), which is established in [15].

Now, the upcoming lemma is both novel and crucial for demonstrating our fundamental outcomes.

Lemma 6.

Let $\beta \in \mathbb{T}$ and $\psi \ge 0$ be convex and submultiplicative on ${[\beta ,r]}_{\mathbb{T}}$ such that $\psi \left(\beta \right)=0.$ Then, $\psi \left(s\right)/\left(s-\beta \right)$ is non-decreasing.

Proof. 

Utilizing (17) on $\psi \left(s\right)/\left(s-\beta \right),$ it becomes evident that

$$\begin{array}{ccc}\hfill {\left(\frac{\psi \left(s\right)}{s-\beta }\right)}^{\nabla }& =& \frac{\left(s-\beta \right){\psi }^{\nabla }\left(s\right)-\psi \left(s\right)}{\left(s-\beta \right)\left(\rho \left(s\right)-\beta \right)}\hfill \\ & =& \frac{J\left(s\right)}{\left(s-\beta \right)\left(\rho \left(s\right)-\beta \right)},\hfill \end{array}$$

(59)

where

$$J\left(s\right)=\left(s-\beta \right){\psi }^{\nabla }\left(s\right)-\psi \left(s\right).$$

(60)

Applying (16) on $\left(s-\beta \right){\psi }^{\nabla }\left(s\right),$ we get

$${\left[\left(s-\beta \right){\psi }^{\nabla }\left(s\right)\right]}^{\nabla }={\psi }^{\nabla }\left(s\right)+\left(\rho \left(s\right)-\beta \right){\psi }^{\nabla \nabla }\left(s\right).$$

(61)

From (60), we have

$$J^{\nabla }\left(s\right)={\left[\left(s-\beta \right){\psi }^{\nabla }\left(s\right)\right]}^{\nabla }-{\psi }^{\nabla }\left(s\right).$$

(62)

Using (61) in (62), we obtain

$$J^{\nabla }\left(s\right)=\left(\rho \left(s\right)-\beta \right){\psi }^{\nabla \nabla }\left(s\right).$$

(63)

As $\psi$ is convex on ${[\beta ,r]}_{\mathbb{T}},$ we can deduce from (63) that

$$J^{\nabla }\left(s\right)\ge 0.$$

This implies that $J\left(s\right)$ is non-decreasing. As $s\ge \beta$, it follows that $J\left(s\right)\ge J\left(\beta \right).$ Considering $\psi \left(\beta \right)=0,$ we can infer from (60), that $J\left(\beta \right)=0$ and consequently

$$J\left(s\right)\ge 0.$$

(64)

Form (64) and (59), we have

$${\left(\frac{\psi \left(s\right)}{s-\beta }\right)}^{\nabla }\ge 0.$$

The proof is now concluded. □

Theorem 6.

Assume $\beta ,r\in \mathbb{T}$ and $h>1$. Additionally, consider $f,\psi \ge 0$, where ψ is both convex and submultiplicative, and it satisfies $\psi \left(\beta \right)=0,$ we can establish that

$$\begin{array}{ccc}& & {\int }_{\beta }^r\frac{{\left(s-\beta \right)}^{1-h}}{\psi \left(s\right)\psi \left(s-\beta \right)}\psi \left(F\left(s\right)\right)\nabla s\hfill \\ & \le & \frac{1}{h-1}{\int }_{\beta }^r\frac{1}{{\left(\rho \left(s\right)-\beta \right)}^{h-1}}\frac{\psi \left(f\left(s\right)\right)}{{\psi }^{\rho }\left(s\right)}\nabla s,\hfill \end{array}$$

(65)

where F is given in (23).

Proof. 

It’s important to note that

$$\begin{array}{ccc}& & {\int }_{\beta }^r\frac{{\left(s-\beta \right)}^{1-h}}{\psi \left(s\right)\psi \left(s-\beta \right)}\psi \left(F\left(s\right)\right)\nabla s\hfill \\ & =& {\int }_{\beta }^r\frac{{\left(s-\beta \right)}^{1-h}}{\psi \left(s\right)\psi \left(s-\beta \right)}\psi \left(\left(s-\beta \right)\frac{{\int }_{\beta }^{s}f\left(\varsigma \right)\nabla \varsigma }{s-\beta }\right)\nabla s.\hfill \end{array}$$

(66)

As $\psi$ is a submultiplicative, we can deduce that

$$\begin{array}{ccc}& & {\int }_{\beta }^r\frac{{\left(s-\beta \right)}^{1-h}}{\psi \left(s\right)\psi \left(s-\beta \right)}\psi \left(F\left(s\right)\right)\nabla s\hfill \\ & \le & {\int }_{\beta }^r\frac{{\left(s-\beta \right)}^{1-h}}{\psi \left(s\right)}\psi \left(\frac{{\int }_{\beta }^{s}f\left(\varsigma \right)\nabla \varsigma }{s-\beta }\right)\nabla s.\hfill \end{array}$$

(67)

Applying (21), we obtain

$$\psi \left(\frac{{\int }_{\beta }^{s}f\left(\varsigma \right)\nabla \varsigma }{s-\beta }\right)\le \frac{1}{s-\beta }{\int }_{\beta }^s\psi \left(f\left(\varsigma \right)\right)\nabla \varsigma .$$

(68)

Using (68) in (67), we get

$$\begin{array}{ccc}& & {\int }_{\beta }^r\frac{{\left(s-\beta \right)}^{1-h}}{\psi \left(s\right)\psi \left(s-\beta \right)}\psi \left(F\left(s\right)\right)\nabla s\hfill \\ & \le & {\int }_{\beta }^r\frac{{\left(s-\beta \right)}^{-h}}{\psi \left(s\right)}\left({\int }_{\beta }^s\psi \left(f\left(\varsigma \right)\right)\nabla \varsigma \right)\nabla s.\hfill \end{array}$$

(69)

Applying (18) on ${\int }_{\beta }^r\frac{{\left(s-\beta \right)}^{-h}}{\psi \left(s\right)}\left({\int }_{\beta }^s\psi \left(f\left(\varsigma \right)\right)\nabla \varsigma \right)\nabla s,$ with ${\chi }^{\nabla }\left(s\right)=\frac{{\left(s-\beta \right)}^{-h}}{\psi \left(s\right)}$ and $\kappa \left(s\right)={\int }_{\beta }^s\psi$$\left(f\left(\varsigma \right)\right)\nabla \varsigma ,$ then

$$\begin{array}{ccc}& & {\int }_{\beta }^r\frac{{\left(s-\beta \right)}^{-h}}{\psi \left(s\right)}\left({\int }_{\beta }^s\psi \left(f\left(\varsigma \right)\right)\nabla \varsigma \right)\nabla s\hfill \\ & =& {\left.\chi \left(s\right)\left({\int }_{\beta }^s\psi \left(f\left(\varsigma \right)\right)\nabla \varsigma \right)\right|}_{\beta }^r-{\int }_{\beta }^r{\chi }^{\rho }\left(s\right)\psi \left(f\left(s\right)\right)\nabla s,\hfill \end{array}$$

(70)

where

$$\chi \left(s\right)=-{\int }_s^r\frac{{\left(\varsigma -\beta \right)}^{-h}}{\psi \left(\varsigma \right)}\nabla \varsigma .$$

Since $\chi \left(r\right)=0,$ we have from (70) that

$$\begin{array}{ccc}& & {\int }_{\beta }^r\frac{{\left(s-\beta \right)}^{-h}}{\psi \left(s\right)}\left({\int }_{\beta }^s\psi \left(f\left(\varsigma \right)\right)\nabla \varsigma \right)\nabla s\hfill \\ & =& {\int }_{\beta }^r\psi \left(f\left(s\right)\right)\left({\int }_{\rho \left(s\right)}^r\frac{{\left(\varsigma -\beta \right)}^{-h}}{\psi \left(\varsigma \right)}\nabla \varsigma \right)\nabla s\hfill \\ & =& {\int }_{\beta }^r\psi \left(f\left(s\right)\right)\left({\int }_{\rho \left(s\right)}^r\frac{1}{\left(\varsigma -\beta \right)}\frac{{\left(\varsigma -\beta \right)}^{-h}}{\left(\frac{\psi \left(\varsigma \right)}{\varsigma -\beta }\right)}\nabla \varsigma \right)\nabla s.\hfill \end{array}$$

(71)

Since $\varsigma \ge \rho \left(s\right),$ we have $\varsigma -\beta \ge \rho \left(s\right)-\beta$ and

$$\frac{1}{\varsigma -\beta }\le \frac{1}{\rho \left(s\right)-\beta }.$$

Subsequently, we can conclude from (71) that

$$\begin{array}{ccc}& & {\int }_{\beta }^r\frac{{\left(s-\beta \right)}^{-h}}{\psi \left(s\right)}\left({\int }_{\beta }^s\psi \left(f\left(\varsigma \right)\right)\nabla \varsigma \right)\nabla s\hfill \\ & \le & {\int }_{\beta }^r\frac{\psi \left(f\left(s\right)\right)}{\rho \left(s\right)-\beta }\left({\int }_{\rho \left(s\right)}^r\frac{{\left(\varsigma -\beta \right)}^{-h}}{\left(\frac{\psi \left(\varsigma \right)}{\varsigma -\beta }\right)}\nabla \varsigma \right)\nabla s.\hfill \end{array}$$

(72)

By employingLemma 6, taking into account the non-decreasing nature of $\psi \left(\varsigma \right)/\left(\varsigma -\beta \right)$, we can deduce $\varsigma \ge \rho \left(s\right)$ and thus

$$\frac{\psi \left(\varsigma \right)}{\varsigma -\beta }\ge \frac{{\psi }^{\rho }\left(s\right)}{\rho \left(s\right)-\beta }.$$

(73)

Form (73) and (72), we get

$$\begin{array}{ccc}& & {\int }_{\beta }^r\frac{{\left(s-\beta \right)}^{-h}}{\psi \left(s\right)}\left({\int }_{\beta }^s\psi \left(f\left(\varsigma \right)\right)\nabla \varsigma \right)\nabla s\hfill \\ & \le & {\int }_{\beta }^r\frac{\psi \left(f\left(s\right)\right)}{{\psi }^{\rho }\left(s\right)}\left({\int }_{\rho \left(s\right)}^r{\left(\varsigma -\beta \right)}^{-h}\nabla \varsigma \right)\nabla s.\hfill \end{array}$$

(74)

Applying (18), on ${\left(\varsigma -\beta \right)}^{1-h},$ we have that

$$\frac{1}{1-h}{\left({\left(\varsigma -\beta \right)}^{-h+1}\right)}^{\nabla }={\left(d-\beta \right)}^{-h},$$

where $d\in \left[\rho \left(\varsigma \right),\varsigma \right].$ Since $d\le \varsigma$ and $h>1,$ we get

$$\frac{1}{1-h}{\left({\left(\varsigma -\beta \right)}^{-h+1}\right)}^{\nabla }\ge {\left(\varsigma -\beta \right)}^{-h}.$$

(75)

Integrating (75) over $\varsigma$ from $\rho \left(s\right)$ to $r,$ we obtain

$$\begin{array}{ccc}\hfill {\int }_{\rho \left(s\right)}^r{\left(\varsigma -\beta \right)}^{-h}\nabla \varsigma & \le & \frac{1}{1-h}{\int }_{\rho \left(s\right)}^r{\left({\left(\varsigma -\beta \right)}^{-h+1}\right)}^{\nabla }\nabla \varsigma \hfill \\ & =& \frac{1}{h-1}\left[{\left(\rho \left(s\right)-\beta \right)}^{-h+1}-{\left(r-\beta \right)}^{-h+1}\right]\hfill \\ & \le & \frac{1}{h-1}{\left(\rho \left(s\right)-\beta \right)}^{-h+1}.\hfill \end{array}$$

(76)

Using (76) in (74), we get

$$\begin{array}{ccc}& & {\int }_{\beta }^r\frac{{\left(s-\beta \right)}^{-h}}{\psi \left(s\right)}\left({\int }_{\beta }^s\psi \left(f\left(\varsigma \right)\right)\nabla \varsigma \right)\nabla s\hfill \\ & \le & \frac{1}{h-1}{\int }_{\beta }^r\frac{1}{{\left(\rho \left(s\right)-\beta \right)}^{h-1}}\frac{\psi \left(f\left(s\right)\right)}{{\psi }^{\rho }\left(s\right)}\nabla s.\hfill \end{array}$$

(77)

Using (77) in (69), we obtain

$$\begin{array}{ccc}& & {\int }_{\beta }^r\frac{{\left(s-\beta \right)}^{1-h}}{\psi \left(s\right)\psi \left(s-\beta \right)}\psi \left(F\left(s\right)\right)\nabla s\hfill \\ & \le & \frac{1}{h-1}{\int }_{\beta }^r\frac{1}{{\left(\rho \left(s\right)-\beta \right)}^{h-1}}\frac{\psi \left(f\left(s\right)\right)}{{\psi }^{\rho }\left(s\right)}\nabla s,\hfill \end{array}$$

which is (65). □

4. Conclusions

In this research, we have successfully demonstrated several dynamic inequalities of Hardy type by utilizing nabla calculus, specifically for convex, submultiplicative functions, and monotone functions. Looking ahead, we intend to extend our exploration by presenting similar inequalities using diamond-$\alpha$ calculus for $\alpha \in (0,1)$ as well as quantum calculus. It is intriguing to consider the prospect of introducing analogous inequalities on time scales through Riemann–Liouville-type fractional integrals. Moreover, there’s potential for us to generalize the dynamic inequalities discussed in this article to two or more dimensions, incorporating symmetry in both the functions and variables. The concept of symmetry has various implications for convex functions, submultiplicative functions and Hardy-type inequalities, impacting their properties, behavior, and generalization. Recognizing and utilizing symmetry can aid in proving inequalities and understanding their solutions and applications. By addressing the suggested future research directions, scholars can continue to deepen their understanding of Hardy-type inequalities and their broader implications. This research is a stepping stone towards further advancements in the field.