Novel Hardy-Type Inequalities with Submultiplicative Functions on Time Scales Using Delta Calculus

Abstract

In this study, we apply Hölder’s inequality, Jensen’s inequality, chain rule and the properties of convex functions and submultiplicative functions to develop an innovative category of dynamic Hardy-type inequalities on time scales delta calculus. A time scale, denoted by $\mathbb{T},$ is any closed nonempty subset of $\mathbb{R}.$ In time scale calculus, results are unified and extended. As particular cases of our findings (when $\mathbb{T}=\mathbb{R}$), we have the continuous analogues of inequalities established in some the literature. Furthermore, we can find other inequalities in different time scales, such as $\mathbb{T}=\mathbb{N},$ which, to the best of the authors’ knowledge, is a largely novel conclusion.

In 1920, Hardy made the discovery of the discrete inequality [1]. He claimed that

$$\sum _{j=1}^{\infty }{\left(\frac{1}{j}\sum _{i=1}^j\zeta \left(i\right)\right)}^p\le {\left(\frac{p}{p-1}\right)}^p\sum _{j=1}^{\infty }{\zeta }^p\left(j\right),p>1,$$

(1)

where $\zeta \left(j\right)\ge 0$ for $j\ge 1,$ ${\sum }_{j=1}^{\infty }{\zeta }^p\left(j\right)<\infty$ and the constant ${\left(p/\left(p-1\right)\right)}^p$ is sharp.

In [2], Theorem A, Hardy demonstrated the integral form of (1) as follows: let $p>1,$ $\xi \ge 0$ and ${\xi }^p$ be integrable and convergent over $(0,\infty ),$ then $\xi$ be integrable over $(0,\lambda )$ for each $\lambda >0$ and

$${\int }_0^{\infty }{\left(\frac{1}{\lambda }{\int }_0^{\lambda }\xi \left(\tau \right)d\tau \right)}^{p}d\lambda \le {\left(\frac{p}{p-1}\right)}^p{\int }_0^{\infty }{\xi }^p\left(\lambda \right)d\lambda .$$

(2)

Hardy inequalities (1) and (2) are the typical forms seen in many analysis textbooks. Since the initial Hardy-type inequality was introduced, more and more researchers have further generalized and refined these inequalities and applied it in various fields, such as dynamical systems, classical real, complex analysis, numerical analysis, qualitative theory of differential equations and their applications; see [3,4,5,6] for more details. In the following, we will give some of these findings that support and guide the contents of this work.

In [7], the authors established that when $0<p<1,$ (2) holds with the sign reversed. They specifically demonstrated that if $\xi \left(\lambda \right)\ge 0$ and ${\int }_0^{\infty }{\xi }^p\left(\lambda \right)d\lambda <\infty ,$ then

$${\int }_0^{\infty }{\left(\frac{1}{\lambda }{\int }_{\lambda }^{\infty }\xi \left(\tau \right)d\tau \right)}^{p}d\lambda >{\left(\frac{p}{1-p}\right)}^p{\int }_0^{\infty }{\xi }^p\left(\lambda \right)d\lambda ,$$

(3)

unless $f\equiv 0$.

In [8], G. H. Hardy demonstrated the following generalization of (2) as

$${\int }_0^{\infty }{\left(\frac{1}{\lambda }{\int }_0^{\lambda }\xi \left(\tau \right)d\tau \right)}^p{\lambda }^{\delta }d\lambda >{\left(\frac{p}{p-1-\delta }\right)}^p{\int }_0^{\infty }{\xi }^p\left(\lambda \right){\lambda }^{\delta }d\lambda ,$$

(4)

where $p\ge 1,$ $\delta <p-1$ and ${\left(p/\left(p-1-\delta \right)\right)}^p$ is sharp. Additionally, in the same paper [8], he showed that if $p>1$ and $\xi \left(\tau \right)>0$ is an integrable function, then

$${\int }_0^{\infty }\frac{1}{{\lambda }^m}{\left({\int }_0^{\lambda }\xi \left(\tau \right)d\tau \right)}^{p}d\lambda \le {\left(\frac{p}{m-1}\right)}^p{\int }_0^{\infty }\frac{1}{{\lambda }^{m-p}}{\xi }^p\left(\lambda \right)d\lambda ,\mathrm{for}m>1,$$

(5)

and

$${\int }_0^{\infty }\frac{1}{{\lambda }^m}{\left({\int }_{\lambda }^{\infty }\xi \left(\tau \right)d\tau \right)}^{p}d\lambda \le {\left(\frac{p}{1-m}\right)}^p{\int }_0^{\infty }\frac{1}{{\lambda }^{m-p}}{\xi }^p\left(\lambda \right)d\lambda ,\mathrm{for}m<1.$$

(6)

Moreover, in [9], Knopp obtained that

$${\int }_0^{\infty }exp\left(\frac{1}{\lambda }{\int }_0^{\lambda }ln\xi \left(\tau \right)d\tau \right)d\lambda \le e{\int }_0^{\infty }\xi \left(\lambda \right)d\lambda ,$$

(7)

where $0<f\in L^1\left({\mathbb{R}}_{+}\right)$ and e is sharp. (7) is referred to as a Knopp-type inequality (also known as Pólya-Knopp inequality). (7) can be thought of as a limit of (2) for approaching to infinity; therefore, for the function $f^{1/p},$ we have

$${\int }_0^{\infty }{\left(\frac{1}{\lambda }{\int }_0^{\lambda }{\xi }^{\frac{1}{p}}\left(\tau \right)d\tau \right)}^{p}d\lambda \le {\left(\frac{p}{p-1}\right)}^p{\int }_0^{\infty }\xi \left(\lambda \right)d\lambda .$$

Indeed,

$$\underset{p\to \infty }{lim}{\left(\frac{1}{\lambda }{\int }_0^{\lambda }{\xi }^{\frac{1}{p}}\left(\tau \right)d\tau \right)}^p=exp\left(\frac{1}{\lambda }{\int }_0^{\lambda }ln\xi \left(\tau \right)d\tau \right),$$

while ${\left(p/\left(p-1\right)\right)}^p\to e$ as $p\to \infty .$ If we replace $\xi \left(\tau \right)$ by $\xi \left(\tau \right)/\tau$ in (7), then

$$\begin{array}{ccc}\hfill {\int }_0^{\infty }exp\left(\frac{1}{\lambda }{\int }_0^{\lambda }ln\frac{\xi \left(\tau \right)}{\tau }dt\right)d\lambda & =& {\int }_0^{\infty }exp\left(\frac{1}{\lambda }{\int }_0^{\lambda }ln\xi \left(\tau \right)d\tau -\frac{1}{\lambda }{\int }_0^{\lambda }\left(ln\tau \right)d\tau \right)d\lambda \hfill \\ & =& {\int }_0^{\infty }exp\left(\frac{1}{\lambda }{\int }_0^{\lambda }ln\xi \left(\tau \right)d\tau -ln\lambda +1\right)d\lambda \hfill \\ & =& e{\int }_0^{\infty }\frac{1}{\lambda }exp\left(\frac{1}{\lambda }{\int }_0^{\lambda }ln\xi \left(\tau \right)d\tau \right)d\lambda ,\hfill \end{array}$$

where ${\int }_0^{\lambda }\left(ln\tau \right)d\tau =\lambda ln\lambda -\lambda .$ Therefore, we have from (7) by replacing $\xi \left(\tau \right)$ with $\xi \left(\tau \right)/\tau$ that

$${\int }_0^{\infty }exp\left(\frac{1}{\lambda }{\int }_0^{\lambda }ln\xi \left(\tau \right)d\tau \right)\frac{d\lambda }{\lambda }\le {\int }_0^{\infty }\xi \left(\lambda \right)\frac{d\lambda }{\lambda }.$$

(8)

In [10], S. Kaijser et al. generalized (8) with a convex function $\mathsf{\Phi }$ on ${\mathbb{R}}^{+}$ and proved that

$${\int }_0^{\infty }\mathsf{\Phi }\left(\frac{1}{\lambda }{\int }_0^{\lambda }\xi \left(\tau \right)d\tau \right)\frac{d\lambda }{\lambda }\le {\int }_0^{\infty }\mathsf{\Phi }\left(\xi \left(\lambda \right)\right)\frac{d\lambda }{\lambda },$$

(9)

where $0<f:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is a locally integrable function.

In [11], the authors proved a generalization of (9) with two weight functions as follows: if $0<d<1,$ $0\le v:(0,d)\to \mathbb{R}$ s.t. $\lambda \to v\left(\lambda \right)/{\lambda }^2$ is locally integrable in $(0,d)$, $\mathsf{\Psi }$ is convex on $\left(\alpha ,\beta \right)$ where $-\infty <\alpha \le \beta <\infty ,$ then

$${\int }_0^{d}v\left(\lambda \right)\mathsf{\Psi }\left(\frac{1}{\lambda }{\int }_0^{\lambda }\xi \left(\tau \right)d\tau \right)\frac{d\lambda }{\lambda }\le {\int }_0^{d}z\left(\lambda \right)\mathsf{\Psi }\left(\xi \left(\lambda \right)\right)\frac{d\lambda }{\lambda },$$

(10)

holds for all integrable functions $\xi :(0,d)\to \mathbb{R}$ s.t. $\xi \left(\lambda \right)\in \left(\alpha ,\beta \right)$$\forall \lambda \in (0,d)$, and v is

$$z\left(\tau \right)=\tau {\int }_{\tau }^d\frac{v\left(\lambda \right)}{{\lambda }^2}d\lambda ,\tau (0,d).$$

The inequality (9) has been proven using convex functions, and this is not a special characteristic of these functions to produce this modern inequality. There is a question centered in our imagination, which is whether this inequality is expected to be proven for new conditions other than the previously mentioned or not.

Yes, this inequality has been proven for non-decreasing functions. In particular, in 2012, Sulaiman [12] established and generalized (9) and proved that if $\xi ,\varphi$ are nonnegative and nondecreasing functions, then

$${\int }_0^{\lambda }\varphi \left(\frac{\mathsf{\Omega }\left(z\right)}{z}\right)dz\le {\int }_0^{\lambda }\varphi \left(\xi \left(z\right)\right)dz,\mathrm{for}\lambda \in (0,\infty ],$$

(11)

where

$$\mathsf{\Omega }\left(z\right)={\int }_0^z\xi \left(t\right)dt.$$

(12)

In the same work [12], he demonstrated that if $0<\beta <1,$ $\xi \ge 0,$ $g>0$ such that $z/g\left(z\right)$ is non-increasing, $r>1$ and $\mathsf{\Omega }$ is defined as in (12), then

$$\begin{array}{ll}& {\int }_0^{\infty }{\left(\frac{\mathsf{\Omega }\left(z\right)}{g\left(z\right)}\right)}^{r}dz\\ \le & \frac{1}{\beta \left(r-1\right){\left(1-\beta \right)}^{r-1}}{\int }_0^{\infty }{\left(\frac{z\xi \left(z\right)}{g\left(z\right)}\right)}^{r}dz.\hfill \end{array}$$

(13)

Additionally, he proved that if $r>1$ and $\xi ,\phi \ge 0$ such that $\phi$ is convex, then

$$\begin{array}{ccc}& & {\int }_0^{\infty }{\phi }^r\left(\frac{\mathsf{\Omega }\left(z\right)}{z}\right)dz\hfill \\ & \le & {\left(\frac{r}{r-1}\right)}^r{\int }_0^{\infty }{\phi }^r\left(\xi \left(z\right)\right)dz.\hfill \end{array}$$

(14)

Recently, there are more new results about Hardy’s inequality via other kinds of functions that have been introduced to establish Hardy-type inequalities, such as convex functions [13], superquadratic functions [14,15,16], submultiplicative functions [17] and so on.

In 2005, P. Řehák presented the time scale form of Hardy’s inequality [18] in the following manner: assume $\mathbb{T}$ is a time scale and define

$$\mathsf{\Omega }\left(z\right)={\int }_d^z\xi \left(\delta \right)\Delta \delta ,\mathrm{for}z\in {[d,\infty )}_{\mathbb{T}}.$$

If $r>1,$ then

$${\int }_d^{\infty }{\left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{\sigma \left(z\right)-d}\right)}^r\Delta z<{\left(\frac{r}{r-1}\right)}^r{\int }_d^{\infty }{\xi }^r\left(z\right)\Delta z,$$

unless $\xi \equiv 0$. If, in addition, $\mu \left(z\right)/z\to 0$ as $z\to \infty ,$ then ${\left(\frac{r}{r-1}\right)}^r$ is sharp.

In ([19], Theorem 7.1.3), the authors mentioned the following Hardy-type inequality for a convex function on $\mathbb{T}:$

$${\int }_k^{l}v\left(\lambda \right)\mathsf{\Psi }\left(\frac{1}{\sigma \left(\lambda \right)-k}{\int }_k^{\sigma \left(\lambda \right)}\xi \left(\tau \right)\Delta \tau \right)\frac{\Delta \lambda }{\lambda -k}\le {\int }_k^{l}z\left(\lambda \right)\mathsf{\Psi }\left(\xi \left(\lambda \right)\right)\frac{\Delta \lambda }{\lambda -k},$$

(15)

where $0\le v\in C_{rd}([k,l),\mathbb{R}),$ $\mathsf{\Psi }:(m,n)\to \mathbb{R}$ is continuous and convex, $m,n\in \mathbb{R}$, $\xi \in C_{rd}([k,l),\mathbb{R})$ is a delta integrable function s.t. $\xi \left(\lambda \right)\in \left(m,n\right)$$\forall \lambda \in [k,l)$, and v is

$$z\left(\tau \right)=(\tau -k){\int }_{\tau }^l\frac{v\left(\lambda \right)\Delta \lambda }{(\lambda -k)\left(\sigma \right(\lambda )-k)},\tau \in [k,l).$$

In [20], T. Donchev et al. obtained the following Hardy-type inequality involving multivariate convex functions on $\mathbb{T}:$

$${\int }_{{\mathsf{\Omega }}_1}\xi \left(\lambda \right)\mathsf{\Psi }\left(\frac{1}{K\left(\lambda \right)}{\int }_{{\mathsf{\Omega }}_2}k(\lambda ,z)\mathit{f}\left(z\right)\Delta z\right)\Delta \lambda \le {\int }_{{\mathsf{\Omega }}_2}w\left(z\right)\mathsf{\Psi }\left(\mathit{f}\left(z\right)\right)\Delta z,$$

where $({\mathsf{\Omega }}_1,\mathsf{\Gamma },{\mu }_{\Delta }),$ $({\mathsf{\Omega }}_2,\mathrm{{\rm Y}},{\eta }_{\Delta })$ are two time scale measure spaces, $V\subset {\mathbb{R}}^n$ is a closed convex set, $K={\int }_{{\mathsf{\Omega }}_2}k(\lambda ,z)\Delta z<\infty ,$ $\lambda \in {\mathsf{\Omega }}_1,$ $\xi :{\mathsf{\Omega }}_1\to \mathbb{R}$,

$$w\left(z\right)={\int }_{{\mathsf{\Omega }}_1}\left(\frac{k(\lambda ,z)\xi \left(\lambda \right)}{K\left(\lambda \right)}\right)\Delta \lambda ,z\in {\mathsf{\Omega }}_2,$$

$\mathsf{\Psi }$ is a convex function and $\mathit{f}:{\mathsf{\Omega }}_1\to {\mathbb{R}}^n$ s.t. $\mathit{f}\left({\mathsf{\Omega }}_2\right)\subset V\subset {\mathbb{R}}^n$ is a ${\eta }_{\Delta }$-integrable function.

Since the first Hardy-type inequality on time scales was proposed, many researchers have further generalized and refined this inequality (see, [21,22,23,24]). Recently, there are more new results about the Hardy inequality via other kinds of time scale calculus, such as time scale delta integral [25] and time scale nabla integral [26,27].

The purpose of this paper is to advance the study of dynamic inequalities of Hardy’s type on $\mathbb{T}$. Specifically, we will prove and generalize the inequalities (11), (13) and (14) via delta integral on the theory of time scale calculus. Time scale calculus theory is used to combine discrete and continuous analysis into a single comprehensive form. This approach is also frequently applied to dynamic inequalities.

The paper is divided into two sections: the first section includes fundamentals and some lemmas on time scale calculus, the second section includes our main results. Ultimately, in the third section, we give a few examples of our fundamental findings obtained.

1. Preliminaries

In [28], Bohner and Peterson defined the forward jump operator $\sigma$ and the graininessfunction $\mu$ by $\sigma \left(z\right):=inf\{\delta \in \mathbb{T}:\delta >z\}$ and $\mu \left(z\right):=\sigma \left(z\right)-z\ge 0$, respectively.

In the following, we use the notations ${\xi }^{\sigma }\left(z\right)=\xi \left(\sigma \left(z\right)\right)$ for any function $\xi :\mathbb{T}\to \mathbb{R}$ and $I_{\mathbb{T}}:=I\cap \mathbb{T}$ for any interval on $\mathbb{T},$ where I is any interval on $\mathbb{R}.$

Definition 1

([28]). $\xi :\mathbb{T}\to \mathbb{R}$ is $rd$-continuous if 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 collection of $rd$-continuous functions is symbolized as ${\mathrm{C}}_{rd}(\mathbb{T},\mathbb{R})$.

Definition 2

([28]). Assume $\xi :\mathbb{T}\to \mathbb{R}$ and $z\in \mathbb{T}.$ We define the delta derivative ${\xi }^{\Delta }\left(z\right)$ to be the number if it exists as follows: for any $ϵ>0$, there is a neighborhood $U=(z-\delta ,$ $z+\delta )\cap \mathbb{T}$ for some $\delta >0$ of z, s.t.

$$|\xi \left(\sigma \left(z\right)\right)-\xi \left(t\right)-{\xi }^{\Delta }\left(z\right)(\sigma \left(z\right)-t)|\le ϵ|\sigma \left(z\right)-t|\forall t\in U,t\ne \sigma \left(z\right).$$

Theorem 1

([28]). Let $\xi ,\omega :\mathbb{T}\to \mathbb{R}$ be differentiable at $z\in \mathbb{T}$. Then,

1. 

$\xi \omega :\mathbb{T}\to \mathbb{R}$ is differentiable at z and the “product rule”

$${\left(\xi \omega \right)}^{\Delta }\left(z\right)={\xi }^{\Delta }\left(z\right)\omega \left(z\right)+\xi \left(\sigma \left(z\right)\right){\omega }^{\Delta }\left(z\right)=\xi \left(z\right){\omega }^{\Delta }\left(z\right)+{\xi }^{\Delta }\left(z\right)\omega \left(\sigma \left(z\right)\right)$$

holds.

2. 

If $\omega \left(z\right)\omega \left(\sigma \right(z\left)\right)\ne 0$, then $\xi /\omega :\mathbb{T}\to \mathbb{R}$ is differentiable at z and the “quotient rule”

$${\left(\frac{\xi }{\omega }\right)}^{\Delta }\left(z\right)=\frac{{\xi }^{\Delta }\left(z\right)\omega \left(z\right)-\xi \left(z\right){\omega }^{\Delta }\left(z\right)}{\omega \left(z\right)\omega \left(\sigma \right(z\left)\right)}$$

holds.

Theorem 2

(Chain Rule ([28], Theorem 1.90)). Assume $\eta :\mathbb{R}\to \mathbb{R}$ is continuous, $\eta :\mathbb{T}\to \mathbb{R}$ is delta differentiable on $\mathbb{T}$ and $\xi :\mathbb{R}\to \mathbb{R}$ is continuously differentiable; then,

$${\left(\xi \circ \eta \right)}^{\Delta }\left(z\right)={\xi }^{{}^{\prime }}\left(\eta \left(c\right)\right){\eta }^{\Delta }\left(z\right),c\in [z,\sigma \left(z\right)].$$

(16)

Definition 3

([28]). $\xi :\mathbb{T}\to \mathbb{R}$ is an antiderivative of $\eta :\mathbb{T}\to \mathbb{R}$ if

$${\xi }^{\Delta }\left(z\right)=\eta \left(z\right),\mathit{it}\mathit{holds}\mathit{that}\forall z\in {\mathbb{T}}^k.$$

In this case, the delta integral of η is

$${\int }_r^s\eta \left(z\right)\Delta z=\xi \left(s\right)-\xi \left(r\right),\forall r,s\in \mathbb{T}.$$

Theorem 3

([28]). Every rd-continuous function $\eta :\mathbb{T}\to \mathbb{R}$ has an antiderivative, and if $z_0\in \mathbb{T}$, then

$${\left({\int }_{z_0}^z\eta \left(t\right)\Delta t\right)}^{\Delta }=\eta \left(z\right),\forall z\in \mathbb{T}.$$

Theorem 4

([28]). If $r,z,z_0\in \mathbb{T},$ $\alpha ,\beta \in \mathbb{R}$ and $\xi ,\eta \in {\mathrm{C}}_{rd}{([r,z]}_{\mathbb{T}},$ $\mathbb{R})$, then

1. 

${\int }_r^z\left[\alpha \xi \left(\delta \right)+\beta \eta \left(\delta \right)\right]\Delta \delta =\alpha {\int }_r^z\xi \left(\delta \right)\Delta \delta +\beta {\int }_r^z\eta \left(\delta \right)\Delta \delta$.

2. 

${\int }_r^z\xi \left(\delta \right)\Delta \delta ={\int }_r^{z_0}\xi \left(\delta \right)\Delta \delta +{\int }_{z_0}^z\xi \left(\delta \right)\Delta \delta$.

3. 

$\left|{\int }_r^z\xi \left(\delta \right)\Delta \delta \right|\le {\int }_r^z\left|\xi \left(\delta \right)\right|\Delta \delta .$

4. 

If $\xi \left(\delta \right)\ge 0,$ $\forall \delta \in {[r,z]}_{\mathbb{T}},$ then ${\int }_r^z\xi \left(\delta \right)\Delta \delta \ge 0.$

Lemma 1

(Integration by Parts [29]). If $r,z\in \mathbb{T}$ and $u,v\in {\mathrm{C}}_{rd}{([r,z]}_{\mathbb{T}},$ $\mathbb{R}),$ then

$${\int }_r^{z}u\left(\delta \right)v^{\Delta }\left(\delta \right)\Delta \delta ={\left[u\left(\delta \right)v\left(\delta \right)\right]}_r^z-{\int }_r^z{u}^{\Delta }\left(\delta \right)v^{\sigma }\left(\delta \right)\Delta \delta .$$

(17)

Lemma 2

(Hölder’s Inequality [29]). If $r,z\in \mathbb{T}$ and $\lambda ,\omega \in {\mathrm{C}}_{rd}{([r,z]}_{\mathbb{T}},$ ${\mathbb{R}}^{+}),$ then

$${\int }_r^z\lambda \left(\delta \right)\omega \left(\delta \right)\Delta \delta \le {\left[{\int }_r^z{\lambda }^{\gamma }\left(\delta \right)\Delta \delta \right]}^{\frac{1}{\gamma }}{\left[{\int }_r^z{\omega }^{\nu }\left(\delta \right)\Delta \delta \right]}^{\frac{1}{\nu }},$$

(18)

where $\gamma >1$ and $1/\gamma +1/\nu =1.$ (18) is reversed if $0<\gamma <1$ or $\gamma <0.$

Lemma 3

(Jensen’s Inequality). Assume $z_0,z\in \mathbb{T}$ and $r_0,r\in \mathbb{R}$. If $\lambda \in {\mathrm{C}}_{rd}({[z_0,z]}_{\mathbb{T}},$ $\mathbb{R}),$ $\phi \in {\mathrm{C}}_{rd}({[z_0,z]}_{\mathbb{T}},(r_0,r))$ and $\mathsf{\Psi }:(r_0,r)\to \mathbb{R}$ is continuous and convex, then

$$\mathsf{\Psi }\left(\frac{1}{{\int }_{z_0}^z\lambda \left(\delta \right)\Delta \delta }{\int }_{z_0}^z\lambda \left(\delta \right)\phi \left(\delta \right)\Delta \delta \right)\le \frac{1}{{\int }_{z_0}^z\lambda \left(\delta \right)\Delta \delta }{\int }_{z_0}^z\lambda \left(\delta \right)\mathsf{\Psi }\left(\phi \left(\delta \right)\right)\Delta \delta .$$

(19)

Definition 4

([30]). $\phi :I_{\mathbb{T}}\to \mathbb{R}$ is convex on $I_{\mathbb{T}}$ if

$$\phi \left(\lambda \delta +\left(1-\lambda \right)z\right)\le \lambda \phi \left(\delta \right)+\left(1-\lambda \right)\phi \left(z\right),$$

$\forall \delta ,z\in I_{\mathbb{T}}$ and all $\lambda \in \left[0,1\right]$ such that $\lambda \delta +\left(1-\lambda \right)z\in I_{\mathbb{T}}.$

Lemma 4

([30]). Let $\phi :I_{\mathbb{T}}\to \mathbb{R}$ be a continuous function. If ${\phi }^{\Delta \Delta }$ exists on $I_{\mathbb{T}}$ and ${\phi }^{\Delta \Delta }\ge 0,$ then φ is convex.

Definition 5

([31]). $Z:I_{\mathbb{T}}\to {\mathbb{R}}^{+}$ is submultiplicative if

$$Z\left(\alpha \delta \right)\le Z\left(\alpha \right)Z\left(\delta \right),\forall \alpha ,\delta \in I_{\mathbb{T}}\subset \mathbb{R}.$$

The following lemma is new and needed to prove our essential results.

Lemma 5.

Let $d\in \mathbb{T}$ and $\phi \ge 0$ be a convex function on $I_{\mathbb{T}}$s.t. $\phi \left(d\right)=0.$ Then, $\phi \left(z\right)/\left(z-d\right)$ is non-decreasing.

Proof. 

Applying the derivative of the quotient rule on $\phi \left(z\right)/\left(z-d\right),$ we see that

$$\begin{array}{ccc}\hfill {\left(\frac{\phi \left(z\right)}{z-d}\right)}^{\Delta }& =& \frac{\left(z-d\right){\phi }^{\Delta }\left(z\right)-\phi \left(z\right)}{\left(z-d\right)\left(\sigma \left(z\right)-d\right)}\hfill \\ & =& \frac{L\left(z\right)}{\left(z-d\right)\left(\sigma \left(z\right)-d\right)},\hfill \end{array}$$

(20)

where

$$L\left(z\right)=\left(z-d\right){\phi }^{\Delta }\left(z\right)-\phi \left(z\right).$$

(21)

Applying the derivative of the product rule on $\left(z-d\right){\phi }^{\Delta }\left(z\right),$ we observe that

$${\left[\left(z-d\right){\phi }^{\Delta }\left(z\right)\right]}^{\Delta }={\phi }^{\Delta }\left(z\right)+\left(\sigma \left(z\right)-d\right){\phi }^{\Delta \Delta }\left(z\right).$$

(22)

From (21), we have that

$$L^{\Delta }\left(z\right)={\left[\left(z-d\right){\phi }^{\Delta }\left(z\right)\right]}^{\Delta }-{\phi }^{\Delta }\left(z\right).$$

(23)

Substituting (22) into (23), we obtain

$$L^{\Delta }\left(z\right)=\left(\sigma \left(z\right)-d\right){\phi }^{\Delta \Delta }\left(z\right).$$

(24)

Since $\phi$ is convex on $I_{\mathbb{T}},$ (i.e., ${\phi }^{\Delta \Delta }\left(z\right)\ge 0$), then we have from (24) that

$$L^{\Delta }\left(z\right)\ge 0,$$

which indicates that $L\left(z\right)$ is non-decreasing. Since $z\ge d$, we see that $L\left(z\right)\ge L\left(d\right).$ Since $\phi \left(d\right)=0,$ we have from (21) that $L\left(d\right)=0$ and then $L\left(z\right)\ge 0.$ Substituting the last inequality into (20), we can obtain

$${\left(\frac{\phi \left(z\right)}{z-d}\right)}^{\Delta }\ge 0,$$

i.e., the function $\phi \left(z\right)/\left(z-d\right)$ is non-decreasing. □

2. Main Results

During the work, we will make the assumption that the functions are rd-continuous functions and that the integrals under consideration exist.

Theorem 5.

Suppose $a,b\in \mathbb{T}$ and $\xi ,\varphi \ge 0$ are nondecreasing functions, then

$${\int }_a^b\varphi \left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{\sigma \left(z\right)-a}\right)\Delta z\le {\int }_a^b\varphi \left(\xi \left(z\right)\right)\Delta z,$$

(25)

where

$$\mathsf{\Omega }\left(z\right)={\int }_a^z\xi \left(\delta \right)\Delta \delta .$$

(26)

Proof. 

From (26), we observe that

$${\int }_a^b\varphi \left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{\sigma \left(z\right)-a}\right)\Delta z={\int }_a^b\varphi \left(\frac{{\int }_a^{\sigma \left(z\right)}\xi \left(\delta \right)\Delta \delta }{\sigma \left(z\right)-a}\right)\Delta z.$$

(27)

Note that

$$\begin{array}{ccc}\hfill {\int }_a^{\sigma \left(z\right)}\xi \left(\delta \right)\Delta \delta & =& {\int }_a^z\xi \left(\delta \right)\Delta \delta +{\int }_z^{\sigma \left(z\right)}\xi \left(\delta \right)\Delta \delta \hfill \\ & =& {\int }_a^z\xi \left(\delta \right)\Delta \delta +\mu \left(z\right)\xi \left(z\right).\hfill \end{array}$$

(28)

Let $t\le z.$ Then, $\xi \left(\delta \right)\le \xi \left(z\right)$ (where $\xi \ge 0$ is a nondecreasing function), and thus, (28) becomes

$$\begin{array}{ccc}\hfill {\int }_a^{\sigma \left(z\right)}\xi \left(\delta \right)\Delta \delta & \le & {\int }_a^z\xi \left(z\right)\Delta \delta +\mu \left(z\right)\xi \left(z\right)\hfill \\ & =& \xi \left(z\right){\int }_a^z\Delta \delta +\mu \left(z\right)\xi \left(z\right)\hfill \\ & =& \left(z-a\right)\xi \left(z\right)+\mu \left(z\right)\xi \left(z\right)\hfill \\ & =& \left(\sigma \left(z\right)-a\right)\xi \left(z\right).\hfill \end{array}$$

(29)

Since $\varphi$ is a nonnegative and nondecreasing function, we obtain from (29) that

$$\varphi \left(\frac{{\int }_a^{\sigma \left(z\right)}\xi \left(\delta \right)\Delta \delta }{\sigma \left(z\right)-a}\right)\le \varphi \left(\xi \left(z\right)\right).$$

(30)

Substituting (30) into (27), we obtain that

$${\int }_a^b\varphi \left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{\sigma \left(z\right)-a}\right)\Delta z\le {\int }_a^b\varphi \left(\xi \left(z\right)\right)\Delta z,$$

which is (25). □

Remark 1.

If $\mathbb{T}=\mathbb{R}$ and $a=0,$ we obtain (11), proved by Sulaiman [12].

Corollary 1.

Let $\mathbb{T}=\mathbb{N},a=1$ and $\xi ,\varphi \ge 0$ be nondecreasing sequences. Then, $\sigma \left(n\right)=n+1$ and

$$\sum _{n=1}^N\varphi \left(\frac{{\sum }_{k=1}^n\xi \left(k\right)}{n}\right)\le \sum _{n=1}^N\varphi \left(\xi \left(n\right)\right),\forall N\in \mathbb{N}.$$

Theorem 6.

Suppose $a\in \mathbb{T},$ $\xi \ge 0,$ $g>0$ s.t. $\left(\sigma \left(z\right)-a\right)/g\left(z\right)$ is non-increasing, $p>1$ and $0<\alpha <1.$ Let Ω be defined as in (26), and assume that there exists a constant $K>0$ s.t.

$$\sigma \left(z\right)-a\le K\left(z-a\right).$$

(31)

Then,

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{g\left(z\right)}\right)}^p\Delta z\hfill \\ & \le & \frac{K^{\alpha \left(p-1\right)}}{\alpha \left(p-1\right){\left(1-\alpha \right)}^{p-1}}{\int }_a^{\infty }{\left(\left(\sigma \left(z\right)-a\right)\frac{\xi \left(z\right)}{g\left(z\right)}\right)}^p\Delta z.\hfill \end{array}$$

(32)

Proof. 

From (26), we have

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{g\left(z\right)}\right)}^p\Delta z\hfill \\ & =& {\int }_a^{\infty }{g}^{-p}\left(z\right){\left({\int }_a^{\sigma \left(z\right)}\xi \left(\delta \right)\Delta \delta \right)}^p\Delta z\hfill \\ & =& {\int }_a^{\infty }{g}^{-p}\left(z\right){\left({\int }_a^{\sigma \left(z\right)}\left[{\left(\sigma \left(\delta \right)-a\right)}^{{}^{\frac{\alpha \left(p-1\right)}{p}}}\xi \left(\delta \right)\right]{\left(\sigma \left(\delta \right)-a\right)}^{{}^{\frac{-\alpha \left(p-1\right)}{p}}}\Delta \delta \right)}^p\Delta z.\hfill \end{array}$$

(33)

Using Hölder’s inequality on

$${\int }_a^{\sigma \left(z\right)}\left[{\left(\sigma \left(\delta \right)-a\right)}^{{}^{\frac{\alpha \left(p-1\right)}{p}}}\xi \left(\delta \right)\right]{\left(\sigma \left(\delta \right)-a\right)}^{{}^{\frac{-\alpha \left(p-1\right)}{p}}}\Delta \delta ,$$

with $p>1,$ $p/\left(p-1\right),$ $\lambda \left(\delta \right)={\left(\sigma \left(\delta \right)-a\right)}^{\alpha \left(p-1\right)/p}\xi \left(\delta \right)$ and $\omega \left(\delta \right)={\left(\sigma \left(\delta \right)-a\right)}^{-\alpha \left(p-1\right)/p},$ we have that

$$\begin{array}{ccc}& & {\int }_a^{\sigma \left(z\right)}\left[{\left(\sigma \left(\delta \right)-a\right)}^{{}^{\frac{\alpha \left(p-1\right)}{p}}}\xi \left(\delta \right)\right]{\left(\sigma \left(\delta \right)-a\right)}^{{}^{\frac{-\alpha \left(p-1\right)}{p}}}\Delta \delta \hfill \\ & \le & {\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\alpha \left(p-1\right)}{\xi }^p\left(\delta \right)\Delta \delta \right)}^{{}^{\frac{1}{p}}}\hfill \\ & & \times {\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{-\alpha }\Delta \delta \right)}^{{}^{\frac{p-1}{p}}}.\hfill \end{array}$$

(34)

Substituting (34) into (33), we obtain that

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{g\left(z\right)}\right)}^p\Delta z\hfill \\ & \le & {\int }_a^{\infty }{g}^{-p}\left(z\right)\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\alpha \left(p-1\right)}{\xi }^p\left(\delta \right)\Delta \delta \right)\hfill \\ & & \times {\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{-\alpha }\Delta \delta \right)}^{p-1}\Delta z.\hfill \end{array}$$

(35)

Applying the chain rule on ${\left(\delta -a\right)}^{1-\alpha }$for $0<\alpha <1,$ we see that

$${\left({\left(\delta -a\right)}^{1-\alpha }\right)}^{\Delta }=\left(1-\alpha \right){\left(\gamma -a\right)}^{-\alpha },\gamma \in [\delta ,\sigma \left(\delta \right)].$$

Since $\gamma \le \sigma \left(\delta \right)$, we have that

$${\left({\left(\delta -a\right)}^{1-\alpha }\right)}^{\Delta }\ge \left(1-\alpha \right){\left(\sigma \left(\delta \right)-a\right)}^{-\alpha },$$

and then

$$\begin{array}{ccc}\hfill {\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{-\alpha }\Delta \delta & \le & \frac{1}{1-\alpha }{\int }_a^{\sigma \left(z\right)}{\left({\left(\delta -a\right)}^{1-\alpha }\right)}^{\Delta }\Delta \delta \hfill \\ & =& \frac{1}{1-\alpha }{\left(\sigma \left(z\right)-a\right)}^{1-\alpha }.\hfill \end{array}$$

(36)

Substituting (36) into (35), we obtain

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{g\left(z\right)}\right)}^p\Delta z\hfill \\ & \le & \frac{1}{{\left(1-\alpha \right)}^{p-1}}{\int }_a^{\infty }{\left(\sigma \left(z\right)-a\right)}^{\left(1-\alpha \right)\left(p-1\right)}{g}^{-p}\left(z\right)\hfill \\ & & \times \left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\alpha \left(p-1\right)}{\xi }^p\left(\delta \right)\Delta \delta \right)\Delta z.\hfill \end{array}$$

(37)

Applying the integration by parts on

$${\int }_a^{\infty }{\left(\sigma \left(z\right)-a\right)}^{\left(1-\alpha \right)\left(p-1\right)}{g}^{-p}\left(z\right)\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\alpha \left(p-1\right)}{\xi }^p\left(\delta \right)\Delta \delta \right)\Delta z,$$

with $u^{\Delta }\left(z\right)={\left(\sigma \left(z\right)-a\right)}^{\left(1-\alpha \right)\left(p-1\right)}/g^p\left(z\right)$ and $v^{\sigma }\left(z\right)={\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\alpha \left(p-1\right)}{\xi }^p\left(\delta \right)\Delta \delta ,$ we see that

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\left(\sigma \left(z\right)-a\right)}^{\left(1-\alpha \right)\left(p-1\right)}{g}^{-p}\left(z\right)\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\alpha \left(p-1\right)}{\xi }^p\left(\delta \right)\Delta \delta \right)\Delta z\hfill \\ & =& {\left.u\left(z\right)\left({\int }_a^z{\left(\sigma \left(\delta \right)-a\right)}^{\alpha \left(p-1\right)}{\xi }^p\left(\delta \right)\Delta \delta \right)\right|}_a^{\infty }\hfill \\ & & -{\int }_a^{\infty }u\left(z\right){\left(\sigma \left(z\right)-a\right)}^{\alpha \left(p-1\right)}{\xi }^p\left(z\right)\Delta z,\hfill \end{array}$$

(38)

where

$$u\left(z\right)=-{\int }_z^{\infty }{\left(\sigma \left(\delta \right)-a\right)}^{\left(1-\alpha \right)\left(p-1\right)}{g}^{-p}\left(\delta \right)\Delta \delta .$$

Since ${lim}_{z\to \infty }u\left(z\right)=0,$ we have from (38) that

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\left(\sigma \left(z\right)-a\right)}^{\left(1-\alpha \right)\left(p-1\right)}{g}^{-p}\left(z\right)\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\alpha \left(p-1\right)}{\xi }^p\left(\delta \right)\Delta \delta \right)\Delta z\hfill \\ & =& {\int }_a^{\infty }{\left(\sigma \left(z\right)-a\right)}^{\alpha \left(p-1\right)}{\xi }^p\left(z\right)\left({\int }_z^{\infty }{\left(\sigma \left(\delta \right)-a\right)}^{\left(1-\alpha \right)\left(p-1\right)}{g}^{-p}\left(\delta \right)\Delta \delta \right)\Delta z.\hfill \end{array}$$

(39)

Since $\left(\sigma \left(z\right)-a\right)/g\left(z\right)$ is non-increasing, we see that

$$\begin{array}{ccc}& & {\int }_z^{\infty }{\left(\sigma \left(\delta \right)-a\right)}^{\left(1-\alpha \right)\left(p-1\right)}{g}^{-p}\left(\delta \right)\Delta \delta \hfill \\ & \le & {\int }_z^{\infty }{\left(\sigma \left(z\right)-a\right)}^p{g}^{-p}\left(z\right){\left(\sigma \left(\delta \right)-a\right)}^{\left(1-\alpha \right)\left(p-1\right)-p}\Delta \delta \hfill \\ & =& {\left(\sigma \left(z\right)-a\right)}^p{g}^{-p}\left(z\right){\int }_z^{\infty }{\left(\sigma \left(\delta \right)-a\right)}^{\left(1-\alpha \right)\left(p-1\right)-p}\Delta \delta .\hfill \end{array}$$

(40)

Substituting (40) into (39), we obtain that

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\left(\sigma \left(z\right)-a\right)}^{\left(1-\alpha \right)\left(p-1\right)}{g}^{-p}\left(z\right)\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\alpha \left(p-1\right)}{\xi }^p\left(\delta \right)\Delta \delta \right)\Delta z\hfill \\ & \le & {\int }_a^{\infty }{\left(\sigma \left(z\right)-a\right)}^{\alpha \left(p-1\right)}{\xi }^p\left(z\right){\left(\frac{\sigma \left(z\right)-a}{g\left(z\right)}\right)}^p\hfill \\ & & \times \left({\int }_z^{\infty }{\left(\sigma \left(\delta \right)-a\right)}^{\left(1-\alpha \right)\left(p-1\right)-p}\Delta \delta \right)\Delta z.\hfill \end{array}$$

(41)

Applying the chain rule on ${\left(\delta -a\right)}^{\left(1-\alpha \right)\left(p-1\right)-p+1},$ we see that

$$\begin{array}{ccc}& & \left(\frac{-1}{\alpha \left(p-1\right)}\right){\left[{\left(\delta -a\right)}^{\left(1-\alpha \right)\left(p-1\right)-p+1}\right]}^{\Delta }\hfill \\ & =& \left(\frac{-1}{\alpha \left(p-1\right)}\right){\left[{\left(\delta -a\right)}^{-\alpha \left(p-1\right)}\right]}^{\Delta }\hfill \\ & =& {\left(\gamma -a\right)}^{-\left[\alpha \left(p-1\right)+1\right]},\hfill \end{array}$$

(42)

where $\gamma \in [\delta ,\sigma (\delta \left)\right].$ Since $\gamma \le \sigma \left(\delta \right),$ $p>1$ and $-\left[\alpha \left(p-1\right)+1\right]<0,$ we observe that

$${\left(\gamma -a\right)}^{-\left[\alpha \left(p-1\right)+1\right]}\ge {\left(\sigma \left(\delta \right)-a\right)}^{-\left[\alpha \left(p-1\right)+1\right]}={\left(\sigma \left(\delta \right)-a\right)}^{\left(1-\alpha \right)\left(p-1\right)-p},$$

and then we have from (42) that

$${\left(\sigma \left(\delta \right)-a\right)}^{\left(1-\alpha \right)\left(p-1\right)-p}\le \left(\frac{-1}{\alpha \left(p-1\right)}\right){\left[{\left(\delta -a\right)}^{-\alpha \left(p-1\right)}\right]}^{\Delta }.$$

Integrating the last inequality over $\delta$ from z to $\infty ,$ (note that $p>1,$ $0<\alpha <1$ and $\left(1-\alpha \right)\left(p-1\right)-p+1=-\alpha \left(p-1\right)<0$), we obtain

$$\begin{array}{ccc}& & {\int }_z^{\infty }{\left(\sigma \left(\delta \right)-a\right)}^{\left(1-\alpha \right)\left(p-1\right)-p}\Delta \delta \hfill \\ & \le & \left(\frac{-1}{\alpha \left(p-1\right)}\right){\int }_z^{\infty }{\left[{\left(\delta -a\right)}^{-\alpha \left(p-1\right)}\right]}^{\Delta }\Delta \delta \hfill \\ & =& \frac{1}{\alpha \left(p-1\right)}{\left(z-a\right)}^{-\alpha \left(p-1\right)}.\hfill \end{array}$$

(43)

Substituting (43) into (41), we see that

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\left(\sigma \left(z\right)-a\right)}^{\left(1-\alpha \right)\left(p-1\right)}{g}^{-p}\left(z\right)\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\alpha \left(p-1\right)}{\xi }^p\left(\delta \right)\Delta \delta \right)\Delta z\hfill \\ & \le & \frac{1}{\alpha \left(p-1\right)}{\int }_a^{\infty }{\left(\frac{\sigma \left(z\right)-a}{z-a}\right)}^{\alpha \left(p-1\right)}{\xi }^p\left(z\right){\left(\frac{\sigma \left(z\right)-a}{g\left(z\right)}\right)}^p\Delta z.\hfill \end{array}$$

(44)

Substituting (44) into (37), we observe that

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{g\left(z\right)}\right)}^p\Delta z\hfill \\ & \le & \frac{1}{\alpha \left(p-1\right){\left(1-\alpha \right)}^{p-1}}{\int }_a^{\infty }{\left(\frac{\sigma \left(z\right)-a}{z-a}\right)}^{\alpha \left(p-1\right)}{\left(\left(\sigma \left(z\right)-a\right)\frac{\xi \left(z\right)}{g\left(z\right)}\right)}^p\Delta z.\hfill \end{array}$$

(45)

Substituting (31) into (45), we see that

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{g\left(z\right)}\right)}^p\Delta z\hfill \\ & \le & \frac{K^{\alpha \left(p-1\right)}}{\alpha \left(p-1\right){\left(1-\alpha \right)}^{p-1}}{\int }_a^{\infty }{\left(\left(\sigma \left(z\right)-a\right)\frac{\xi \left(z\right)}{g\left(z\right)}\right)}^p\Delta z,\hfill \end{array}$$

which is (32). □

Remark 2.

If $\mathbb{T}=\mathbb{R}$ and $a=0,$ then (31) holds with the equality for $K=1$ and obtain (13), proved by Sulaiman [12].

Corollary 2.

If $\mathbb{T}={\mathbb{N}}_0,a=0,$ $\xi \ge 0,$ $g>0$ s.t. $\left(n+1\right)/g\left(n\right)$ is nonincreasing, $p>1,$ $0<\alpha <1$, then $\sigma \left(n\right)=n+1$ and

$$\begin{array}{ccc}& & \sum _{n=1}^{\infty }{\left(\frac{\mathsf{\Omega }(n+1)}{g\left(n\right)}\right)}^p\hfill \\ & \le & \frac{2^{\alpha \left(p-1\right)}}{\alpha \left(p-1\right){\left(1-\alpha \right)}^{p-1}}\sum _{n=1}^{\infty }{\left(\left(n+1\right)\frac{\xi \left(n\right)}{g\left(n\right)}\right)}^p,\hfill \end{array}$$

where $\mathsf{\Omega }\left(n\right)={\sum }_{k=1}^{n-1}\xi \left(k\right).$ Here, we used that $\left(n-a\right)/\left(\sigma \left(n\right)-a\right)=n/\left(n+1\right)=1-\left[1/\left(n+1\right)\right],$ for $n\ge 1.$ So, $n+1\ge 2,$ $1/\left(n+1\right)\le 1/2$ and

$$\frac{n-a}{\sigma \left(n\right)-a}\ge 1-\frac{1}{2}=\frac{1}{2}.$$

This indicates that (31) holds with $K=2.$

Remark 3.

If $\mathbb{T}={\mathbb{N}}_0,a=0,$ $\xi \ge 0$, $g>0$ s.t. $g\left(n\right)=n+1$, $p>1,$ $0<\alpha <1$, then $\sigma \left(n\right)=n+1$ and

$$\begin{array}{ccc}& & \sum _{n=1}^{\infty }{\left(\frac{\mathsf{\Omega }(n+1)}{n+1}\right)}^p\hfill \\ & \le & \frac{2^{\alpha \left(p-1\right)}}{\alpha \left(p-1\right){\left(1-\alpha \right)}^{p-1}}\sum _{n=1}^{\infty }{\left(\xi \left(n\right)\right)}^p,\hfill \end{array}$$

where $\mathsf{\Omega }\left(n\right)={\sum }_{k=1}^{n-1}\xi \left(k\right).$

Theorem 7.

Assume $a\in \mathbb{T},$ $p>1$ and $\xi ,\phi \ge 0$ s.t. φ is convex. Suppose there exists a constant $K>0$ s.t.

$$\sigma \left(z\right)-a\le K\left(z-a\right).$$

(46)

Then,

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\phi }^p\left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{\sigma \left(z\right)-a}\right)\Delta z\hfill \\ & \le & {\left(\frac{p}{p-1}\right)}^p{K}^{\frac{p-1}{p}}{\int }_a^{\infty }{\phi }^p\left(\xi \left(z\right)\right)\Delta z,\hfill \end{array}$$

(47)

where Ω is defined as in (26).

Proof. 

Note that

$${\int }_a^{\infty }{\phi }^p\left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{\sigma \left(z\right)-a}\right)\Delta z={\int }_a^{\infty }{\left[\phi \left(\frac{{\int }_a^{\sigma \left(z\right)}\xi \left(t\right)\Delta t}{\sigma \left(z\right)-a}\right)\right]}^p\Delta z.$$

(48)

Applying Jensen’s inequality, we see that

$$\phi \left(\frac{{\int }_a^{\sigma \left(z\right)}\xi \left(\delta \right)\Delta \delta }{\sigma \left(z\right)-a}\right)\le \frac{{\int }_a^{\sigma \left(z\right)}\phi \left(\xi \left(\delta \right)\right)\Delta \delta }{\sigma \left(z\right)-a}.$$

(49)

Substituting (49) into (48), we obtain

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\phi }^p\left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{\sigma \left(z\right)-a}\right)\Delta z\hfill \\ & \le & {\int }_a^{\infty }{\left(\sigma \left(z\right)-a\right)}^{-p}{\left({\int }_a^{\sigma \left(z\right)}\phi \left(\xi \left(\delta \right)\right)\Delta \delta \right)}^p\Delta z\hfill \\ & =& {\int }_a^{\infty }{\left(\sigma \left(z\right)-a\right)}^{-p}{\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{-1}{p}}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{1}{p}}\phi \left(\xi \left(\delta \right)\right)\Delta \delta \right)}^p\Delta z.\hfill \end{array}$$

(50)

Applying Hölder’s inequality on

$${\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{-1}{p}}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{1}{p}}\phi \left(\xi \left(\delta \right)\right)\Delta \delta ,$$

with $p>1,$ $p/\left(p-1\right),$ $\lambda \left(\delta \right)={\left(\sigma \left(\delta \right)-a\right)}^{\frac{-1}{p}}$ and $\omega \left(\delta \right)={\left(\sigma \left(\delta \right)-a\right)}^{\frac{1}{p}}\phi \left(\xi \left(\delta \right)\right),$ we have that

$$\begin{array}{ccc}& & {\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{-1}{p}}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{1}{p}}\phi \left(\xi \left(\delta \right)\right)\Delta \delta \hfill \\ & \le & {\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{-1}{p}}\Delta \delta \right)}^{\frac{p-1}{p}}\hfill \\ & & \times {\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{-1}{p}}\left(\sigma \left(\delta \right)-a\right){\phi }^p\left(\xi \left(\delta \right)\right)\Delta \delta \right)}^{\frac{1}{p}}\hfill \\ & =& {\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{-1}{p}}\Delta \delta \right)}^{\frac{p-1}{p}}\hfill \\ & & \times {\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{p-1}{p}}{\phi }^p\left(\xi \left(\delta \right)\right)\Delta \delta \right)}^{\frac{1}{p}}.\hfill \end{array}$$

(51)

Substituting (51) into (50), we observe that

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\phi }^p\left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{\sigma \left(z\right)-a}\right)\Delta z\hfill \\ & \le & {\int }_a^{\infty }{\left(\sigma \left(z\right)-a\right)}^{-p}{\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{-1}{p}}\Delta \delta \right)}^{p-1}\hfill \\ & & \times \left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{p-1}{p}}{\phi }^p\left(\xi \left(\delta \right)\right)\Delta \delta \right)\Delta z.\hfill \end{array}$$

(52)

Applying the chain rule on ${\left(\delta -a\right)}^{\frac{-1}{p}+1},$ we see that

$$\frac{p}{p-1}{\left[{\left(\delta -a\right)}^{\frac{-1}{p}+1}\right]}^{\Delta }={\left(\gamma -a\right)}^{\frac{-1}{p}},$$

(53)

where $\gamma \in \left[\delta ,\sigma \left(\delta \right)\right].$ Since $\gamma \le \sigma \left(\delta \right)$ and $p>1,$ we observe that

$${\left(\gamma -a\right)}^{\frac{-1}{p}}\ge {\left(\sigma \left(\delta \right)-a\right)}^{\frac{-1}{p}},$$

and then we have from (53) that

$${\left(\sigma \left(\delta \right)-a\right)}^{\frac{-1}{p}}\le \frac{p}{p-1}{\left[{\left(\delta -a\right)}^{\frac{-1}{p}+1}\right]}^{\Delta }.$$

(54)

Integrating (54) over $\delta$ from a to $\sigma \left(z\right)$, we see that

$$\begin{array}{ccc}& & {\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{-1}{p}}\Delta \delta \hfill \\ & \le & \frac{p}{p-1}{\int }_a^{\sigma \left(z\right)}{\left[{\left(\delta -a\right)}^{\frac{-1}{p}+1}\right]}^{\Delta }\Delta \delta \hfill \\ & =& \frac{p}{p-1}{\left(\sigma \left(z\right)-a\right)}^{\frac{p-1}{p}}.\hfill \end{array}$$

(55)

Substituting (55) into (52), we observe that

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\phi }^p\left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{\sigma \left(z\right)-a}\right)\Delta z\hfill \\ & \le & {\left(\frac{p}{p-1}\right)}^{p-1}{\int }_a^{\infty }{\left(\sigma \left(z\right)-a\right)}^{\frac{1}{p}-2}\hfill \\ & & \times \left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{p-1}{p}}{\phi }^p\left(\xi \left(\delta \right)\right)\Delta \delta \right)\Delta z.\hfill \end{array}$$

(56)

Applying the integration by parts on

$${\int }_a^{\infty }{\left(\sigma \left(z\right)-a\right)}^{\frac{1}{p}-2}\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{p-1}{p}}{\phi }^p\left(\xi \left(\delta \right)\right)\Delta \delta \right)\Delta z,$$

with $u^{\Delta }\left(z\right)={\left(\sigma \left(z\right)-a\right)}^{\frac{1}{p}-2}$ and $v^{\sigma }\left(z\right)={\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{p-1}{p}}{\phi }^p\left(\xi \left(\delta \right)\right)\Delta \delta ,$ we see that

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\left(\sigma \left(z\right)-a\right)}^{\frac{1}{p}-2}\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{p-1}{p}}{\phi }^p\left(\xi \left(\delta \right)\right)\Delta \delta \right)\Delta z\hfill \\ & =& {\left.u\left(z\right)\left({\int }_a^z{\left(\sigma \left(\delta \right)-a\right)}^{\frac{p-1}{p}}{\phi }^p\left(\xi \left(\delta \right)\right)\Delta \delta \right)\right|}_a^{\infty }\hfill \\ & & -{\int }_a^{\infty }u\left(z\right){\left(\sigma \left(z\right)-a\right)}^{\frac{p-1}{p}}{\phi }^p\left(\xi \left(z\right)\right)\Delta z,\hfill \end{array}$$

(57)

where

$$u\left(z\right)=-{\int }_z^{\infty }{\left(\sigma \left(\delta \right)-a\right)}^{\frac{1}{p}-2}\Delta \delta .$$

Since ${lim}_{z\to \infty }u\left(z\right)=0,$ we have from (57) that

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\left(\sigma \left(z\right)-a\right)}^{\frac{1}{p}-2}\left({\int }_a^{\sigma \left(z\right)}{\left(\sigma \left(\delta \right)-a\right)}^{\frac{p-1}{p}}{\phi }^p\left(\xi \left(\delta \right)\right)\Delta \delta \right)\Delta z\hfill \\ & =& {\int }_a^{\infty }\left({\int }_z^{\infty }{\left(\sigma \left(\delta \right)-a\right)}^{\frac{1}{p}-2}\Delta t\right){\left(\sigma \left(z\right)-a\right)}^{\frac{p-1}{p}}{\phi }^p\left(\xi \left(z\right)\right)\Delta z.\hfill \end{array}$$

(58)

Substituting (58) into (56), we obtain

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\phi }^p\left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{\sigma \left(z\right)-a}\right)\Delta z\hfill \\ & \le & {\left(\frac{p}{p-1}\right)}^{p-1}{\int }_a^{\infty }\left({\int }_z^{\infty }{\left(\sigma \left(\delta \right)-a\right)}^{\frac{1}{p}-2}\Delta \delta \right)\hfill \\ & & \times {\left(\sigma \left(z\right)-a\right)}^{\frac{p-1}{p}}{\phi }^p\left(\xi \left(z\right)\right)\Delta z.\hfill \end{array}$$

(59)

Applying the chain rule on ${\left(\delta -a\right)}^{\frac{1}{p}-1},$ we observe that

$$\frac{p}{1-p}{\left[{\left(\delta -a\right)}^{\frac{1}{p}-1}\right]}^{\Delta }={\left(\gamma -a\right)}^{\frac{1}{p}-2},$$

(60)

where $\gamma \in \left[\delta ,\sigma \left(\delta \right)\right].$ Since $\gamma \le \sigma \left(\delta \right)$ and $1-2p<0,$ we have that

$${\left(\gamma -a\right)}^{\frac{1}{p}-2}\ge {\left(\sigma \left(\delta \right)-a\right)}^{\frac{1}{p}-2},$$

and then we obtain from (60) that

$${\left(\sigma \left(\delta \right)-a\right)}^{\frac{1}{p}-2}\le \frac{p}{1-p}{\left[{\left(\delta -a\right)}^{\frac{1}{p}-1}\right]}^{\Delta }.$$

Integrating the last inequality over $\delta$ from z to $\infty ,$ we obtain

$$\begin{array}{ccc}\hfill {\int }_z^{\infty }{\left(\sigma \left(\delta \right)-a\right)}^{\frac{1}{p}-2}\Delta \delta & \le & {\int }_z^{\infty }\frac{p}{1-p}{\left[{\left(\delta -a\right)}^{\frac{1}{p}-1}\right]}^{\Delta }\Delta \delta \hfill \\ & =& \left(\frac{p}{p-1}\right){\left(z-a\right)}^{\frac{1}{p}-1}.\hfill \end{array}$$

(61)

Substituting (61) into (59), we observe that

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\phi }^p\left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{\sigma \left(z\right)-a}\right)\Delta z\hfill \\ & \le & {\left(\frac{p}{p-1}\right)}^p{\int }_a^{\infty }{\left(z-a\right)}^{\frac{1}{p}-1}{\left(\sigma \left(z\right)-a\right)}^{\frac{p-1}{p}}{\phi }^p\left(\xi \left(z\right)\right)\Delta z.\hfill \end{array}$$

Using (46), the last inequality becomes

$$\begin{array}{ccc}& & {\int }_a^{\infty }{\phi }^p\left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{\sigma \left(z\right)-a}\right)\Delta z\hfill \\ & \le & {\left(\frac{p}{p-1}\right)}^p{K}^{\frac{p-1}{p}}{\int }_a^{\infty }{\phi }^p\left(\xi \left(z\right)\right)\Delta z,\hfill \end{array}$$

which is (47). □

Remark 4.

If $\mathbb{T}=\mathbb{R}$ and $a=0,$ then (46) holds with the equality for $K=1$ and obtain (14), proved by Sulaiman [12].

Corollary 3.

Let $\mathbb{T}={\mathbb{N}}_0,a=0,$ $\xi ,\phi \ge 0$, s.t. φ is convex and $p>1$. Then, $\sigma \left(n\right)=n+1$ and

$$\begin{array}{ccc}& & \sum _{n=0}^{\infty }{\phi }^p\left(\frac{\mathsf{\Omega }(n+1)}{n+1}\right)\hfill \\ & \le & {\left(\frac{p}{p-1}\right)}^p{2}^{\frac{p-1}{p}}\sum _{n=0}^{\infty }{\phi }^p\left(\xi \left(n\right)\right),\hfill \end{array}$$

where $\mathsf{\Omega }\left(n\right)={\sum }_{k=0}^{n-1}\xi \left(k\right).$ Here, we used that $\left(n-a\right)/\left(\sigma \left(n\right)-a\right)=n/\left(n+1\right)=1-\left[1/\left(n+1\right)\right],$ for $n\ge 1.$ So, $n+1\ge 2,$ $1/\left(n+1\right)\le 1/2$ and

$$\frac{n-a}{\sigma \left(n\right)-a}\ge 1-\frac{1}{2}=\frac{1}{2}.$$

This indicates that (46) holds with $K=2.$

Theorem 8.

Assume $a,b\in \mathbb{T}$ and $p>1.$ Let $\xi ,\phi \ge 0$ s.t. φ be a convex and submultiplicative function with $\phi \left(a\right)=0$. Then,

$$\begin{array}{ccc}& & {\int }_a^b\frac{{\left(\sigma \left(z\right)-a\right)}^{1-p}}{\phi \left(z\right)\phi \left(\sigma \left(z\right)-a\right)}\phi \left({\mathsf{\Omega }}^{\sigma }\left(z\right)\right)\Delta z\hfill \\ & \le & \frac{1}{p-1}{\int }_a^b\frac{\phi \left(\xi \left(z\right)\right)}{{\left(z-a\right)}^{p-1}\phi \left(z\right)}\Delta z,\hfill \end{array}$$

(62)

where Ω is defined as in (26).

Proof. 

Note that

$$\begin{array}{ccc}& & {\int }_a^b\frac{{\left(\sigma \left(z\right)-a\right)}^{1-p}}{\phi \left(z\right)\phi \left(\sigma \left(z\right)-a\right)}\phi \left({\mathsf{\Omega }}^{\sigma }\left(z\right)\right)\Delta z\hfill \\ & =& {\int }_a^b\frac{{\left(\sigma \left(z\right)-a\right)}^{1-p}}{\phi \left(z\right)\phi \left(\sigma \left(z\right)-a\right)}\phi \left(\left(\sigma \left(z\right)-a\right)\frac{{\int }_a^{\sigma \left(z\right)}\xi \left(\delta \right)\Delta \delta }{\sigma \left(z\right)-a}\right)\Delta z.\hfill \end{array}$$

(63)

Since $\phi$ is a submultiplicative function, we have from (63) that

$$\begin{array}{ccc}& & {\int }_a^b\frac{{\left(\sigma \left(z\right)-a\right)}^{1-p}}{\phi \left(z\right)\phi \left(\sigma \left(z\right)-a\right)}\phi \left({\mathsf{\Omega }}^{\sigma }\left(z\right)\right)\Delta z\hfill \\ & \le & {\int }_a^b\frac{{\left(\sigma \left(z\right)-a\right)}^{1-p}}{\phi \left(z\right)}\phi \left(\frac{{\int }_a^{\sigma \left(z\right)}\xi \left(\delta \right)\Delta \delta }{\sigma \left(z\right)-a}\right)\Delta z.\hfill \end{array}$$

(64)

Applying Jensen’s inequality, we see that

$$\phi \left(\frac{{\int }_a^{\sigma \left(z\right)}\xi \left(\delta \right)\Delta \delta }{\sigma \left(z\right)-a}\right)\le \frac{1}{\sigma \left(z\right)-a}{\int }_a^{\sigma \left(z\right)}\phi \left(\xi \left(\delta \right)\right)\Delta \delta .$$

(65)

Substituting (65) into (64), we obtain

$$\begin{array}{ccc}& & {\int }_a^b\frac{{\left(\sigma \left(z\right)-a\right)}^{1-p}}{\phi \left(z\right)\phi \left(\sigma \left(z\right)-a\right)}\phi \left({\mathsf{\Omega }}^{\sigma }\left(z\right)\right)\Delta z\hfill \\ & \le & {\int }_a^b\frac{{\left(\sigma \left(z\right)-a\right)}^{-p}}{\phi \left(z\right)}\left({\int }_a^{\sigma \left(z\right)}\phi \left(\xi \left(\delta \right)\right)\Delta \delta \right)\Delta z.\hfill \end{array}$$

(66)

Applying the integration by parts on

$${\int }_a^b\frac{{\left(\sigma \left(z\right)-a\right)}^{-p}}{\phi \left(z\right)}\left({\int }_a^{\sigma \left(z\right)}\phi \left(\xi \left(\delta \right)\right)\Delta \delta \right)\Delta z,$$

with $u^{\Delta }\left(z\right)=\frac{{\left(\sigma \left(z\right)-a\right)}^{-p}}{\phi \left(z\right)}$ and $v^{\sigma }\left(z\right)={\int }_a^{\sigma \left(z\right)}\phi \left(\xi \left(\delta \right)\right)\Delta \delta ,$ we see that

$$\begin{array}{ccc}& & {\int }_a^b\frac{{\left(\sigma \left(z\right)-a\right)}^{-p}}{\phi \left(z\right)}\left({\int }_a^{\sigma \left(z\right)}\phi \left(\xi \left(\delta \right)\right)\Delta \delta \right)\Delta z\hfill \\ & =& {\left.u\left(z\right)\left({\int }_a^z\phi \left(\xi \left(\delta \right)\right)\Delta \delta \right)\right|}_a^b-{\int }_a^{b}u\left(z\right)\phi \left(\xi \left(z\right)\right)\Delta z,\hfill \end{array}$$

(67)

where

$$u\left(z\right)=-{\int }_z^b\frac{{\left(\sigma \left(\delta \right)-a\right)}^{-p}}{\phi \left(\delta \right)}\Delta \delta .$$

Since $u\left(b\right)=0,$ we have from (67) that

$$\begin{array}{ccc}& & {\int }_a^b\frac{{\left(\sigma \left(z\right)-a\right)}^{-p}}{\phi \left(z\right)}\left({\int }_a^{\sigma \left(z\right)}\phi \left(\xi \left(\delta \right)\right)\Delta \delta \right)\Delta z\hfill \\ & =& {\int }_a^b\phi \left(\xi \left(z\right)\right)\left({\int }_z^b\frac{{\left(\sigma \left(\delta \right)-a\right)}^{-p}}{\phi \left(\delta \right)}\Delta \delta \right)\Delta z\hfill \\ & =& {\int }_a^b\phi \left(\xi \left(z\right)\right)\left({\int }_z^b\frac{1}{\left(\delta -a\right)}\frac{{\left(\sigma \left(\delta \right)-a\right)}^{-p}}{\left(\frac{\phi \left(\delta \right)}{\delta -a}\right)}\Delta \delta \right)\Delta z.\hfill \end{array}$$

(68)

Since $\delta \ge z,$ we observe that $\delta -a\ge z-a$ and

$$\frac{1}{\delta -a}\le \frac{1}{z-a},$$

and then we have from (68) that

$$\begin{array}{ccc}& & {\int }_a^b\frac{{\left(\sigma \left(z\right)-a\right)}^{-p}}{\phi \left(z\right)}\left({\int }_a^{\sigma \left(z\right)}\phi \left(\xi \left(\delta \right)\right)\Delta \delta \right)\Delta z\hfill \\ & \le & {\int }_a^b\frac{\phi \left(\xi \left(z\right)\right)}{z-a}\left({\int }_z^b\frac{{\left(\sigma \left(\delta \right)-a\right)}^{-p}}{\left(\frac{\phi \left(\delta \right)}{\delta -a}\right)}\Delta \delta \right)\Delta z.\hfill \end{array}$$

(69)

Applying Lemma 5 where $\phi \left(z\right)/\left(z-a\right)$ is non-decreasing, we have for $\delta \ge z$ that

$$\frac{\phi \left(\delta \right)}{\delta -a}\ge \frac{\phi \left(z\right)}{z-a}.$$

(70)

Substituting (70) into (69), we observe that

$$\begin{array}{ccc}& & {\int }_a^b\frac{{\left(\sigma \left(z\right)-a\right)}^{-p}}{\phi \left(z\right)}\left({\int }_a^{\sigma \left(z\right)}\phi \left(\xi \left(\delta \right)\right)\Delta \delta \right)\Delta z\hfill \\ & \le & {\int }_a^b\frac{\phi \left(\xi \left(z\right)\right)}{\phi \left(z\right)}\left({\int }_z^b{\left(\sigma \left(\delta \right)-a\right)}^{-p}\Delta \delta \right)\Delta z.\hfill \end{array}$$

(71)

Applying the chain rule on ${\left(\delta -a\right)}^{1-p},$ we see that

$$\frac{1}{1-p}{\left({\left(\delta -a\right)}^{-p+1}\right)}^{\Delta }={\left(\gamma -a\right)}^{-p},$$

where $\gamma \in \left[\delta ,\sigma \left(\delta \right)\right].$ Since $\gamma \le \sigma \left(\delta \right)$ and $p>1,$ we obtain

$$\frac{1}{1-p}{\left({\left(\delta -a\right)}^{-p+1}\right)}^{\Delta }\ge {\left(\sigma \left(\delta \right)-a\right)}^{-p}.$$

Integrating the last inequality over $\delta$ from z to $b,$ we obtain

$$\begin{array}{ccc}\hfill {\int }_z^b{\left(\sigma \left(\delta \right)-a\right)}^{-p}\Delta t& \le & \frac{1}{1-p}{\int }_z^b{\left({\left(\delta -a\right)}^{-p+1}\right)}^{\Delta }\Delta \delta \hfill \\ & =& \frac{1}{p-1}\left[{\left(z-a\right)}^{-p+1}-{\left(b-a\right)}^{-p+1}\right]\hfill \\ & \le & \frac{1}{p-1}{\left(z-a\right)}^{-p+1}.\hfill \end{array}$$

(72)

Substituting (72) into (71), we obtain that

$$\begin{array}{ccc}& & {\int }_a^b\frac{{\left(\sigma \left(z\right)-a\right)}^{-p}}{\phi \left(z\right)}\left({\int }_a^{\sigma \left(z\right)}\phi \left(\xi \left(\delta \right)\right)\Delta \delta \right)\Delta z\hfill \\ & \le & \frac{1}{p-1}{\int }_a^b\frac{\phi \left(\xi \left(z\right)\right)}{{\left(z-a\right)}^{p-1}\phi \left(z\right)}\Delta z.\hfill \end{array}$$

(73)

Substituting (73) into (66), we obtain

$$\begin{array}{ccc}& & {\int }_a^b\frac{{\left(\sigma \left(z\right)-a\right)}^{1-p}}{\phi \left(z\right)\phi \left(\sigma \left(z\right)-a\right)}\phi \left({\mathsf{\Omega }}^{\sigma }\left(z\right)\right)\Delta z\hfill \\ & \le & \frac{1}{p-1}{\int }_a^b\frac{\phi \left(\xi \left(z\right)\right)}{{\left(z-a\right)}^{p-1}\phi \left(z\right)}\Delta z,\hfill \end{array}$$

which is (62). □

3. Examples

In this section, we give some examples to demonstrate our findings obtained.

Example 1.

In Theorem 5, if $a>0$, $\xi \left(z\right)=1$ and $\varphi \left(x\right)=x,$ then R.H.S. of (25) becomes

$${\int }_a^b\varphi \left(\xi \left(z\right)\right)\Delta z={\int }_a^b\varphi \left(1\right)\Delta z={\int }_a^b\Delta z=b-a.$$

(74)

The L.H.S. of (25) becomes

$$\begin{array}{ccc}\hfill {\int }_a^b\varphi \left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{\sigma \left(z\right)-a}\right)\Delta z& =& {\int }_a^b\varphi \left(\frac{{\int }_a^{\sigma \left(z\right)}\xi \left(\delta \right)\Delta \delta }{\sigma \left(z\right)-a}\right)\Delta z\hfill \\ & =& {\int }_a^b\varphi \left(1\right)\Delta z={\int }_a^b\Delta z=b-a.\hfill \end{array}$$

(75)

From (74) and (75), we observe that (25) holds with the equality.

Example 2.

In Theorem 5, if $a>0,$ $\xi \left(z\right)=z+\sigma \left(z\right)$ and $\varphi \left(x\right)=x+2a,$ then

$$\varphi \left(\xi \left(z\right)\right)=\varphi \left(z+\sigma \left(z\right)\right)=z+a+\sigma \left(z\right)+a={\left[{\left(z+a\right)}^2\right]}^{\Delta },$$

and then the R.H.S. of (25) becomes

$$\begin{array}{ccc}\hfill {\int }_a^b\varphi \left(\xi \left(z\right)\right)\Delta z& =& {\int }_a^b{\left[{\left(z+a\right)}^2\right]}^{\Delta }\Delta z\hfill \\ & =& {\left(b+a\right)}^2-{\left(2a\right)}^2=\left(b-a\right)\left(b+3a\right).\hfill \end{array}$$

(76)

Since

$$\begin{array}{ccc}\hfill \frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{\sigma \left(z\right)-a}& =& \frac{{\int }_a^{\sigma \left(z\right)}(\delta +\sigma \left(\delta \right))\Delta \delta }{\sigma \left(z\right)-a}\hfill \\ & =& \frac{{\int }_a^{\sigma \left(z\right)}{\left({\delta }^2\right)}^{\Delta }\Delta \delta }{\sigma \left(z\right)-a}=\frac{{\left[\sigma \left(z\right)\right]}^2-a^2}{\sigma \left(z\right)-a}=\sigma \left(z\right)+a.\hfill \end{array}$$

Then, the L.H.S. of (25) becomes

$${\int }_a^b\varphi \left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{\sigma \left(z\right)-a}\right)\Delta z={\int }_a^b\varphi \left(\sigma \left(z\right)+a\right)\Delta z={\int }_a^b\left(\sigma \left(z\right)+3a\right)\Delta z,$$

and then we have for $a\le z$ that

$$\begin{array}{ccc}\hfill {\int }_a^b\varphi \left(\frac{{\mathsf{\Omega }}^{\sigma }\left(z\right)}{\sigma \left(z\right)-a}\right)\Delta z& \le & {\int }_a^b\left(z+a+\sigma \left(z\right)+a\right)\Delta z\hfill \\ & =& {\int }_a^b{\left[{\left(z+a\right)}^2\right]}^{\Delta }\Delta z=\left(b-a\right)\left(b+3a\right).\hfill \end{array}$$

(77)

From (76) and (77), we observe that

$${\int }_a^b\varphi \left(\frac{{\mathsf{\Omega }}^{\sigma }\sigma \left(z\right)}{\sigma \left(z\right)-a}\right)\Delta z\le {\int }_a^b\varphi \left(\xi \left(z\right)\right)\Delta z=\left(b-a\right)\left(b+3a\right),$$

which indicates that (25) holds.

4. Discussion and Conclusions

In this study, we demonstrate several dynamic inequalities of the Hardy-type for convex, submultiplicative functions and monotone functions using delta calculus. These findings are proven by the chain rule formula, Hölder’s inequality and Jensen’s inequality. In the future, we will be able to present such inequalities by employing nabla calculus, diamond-$\alpha$ calculus for $\alpha \in (0,1)$ and quantum calculus. It will be very fascinating to present similar inequalities on time scales using Riemann–Liouville type fractional integrals. Additionally, we will establish these inequalities for weighted functions on time scales and also for different spaces. In our next paper, we plan to present new results involving other well-known inequalities such as Hermite, Copson and Hilbert, among others. Furthermore, we can generalize the dynamic inequalities discussed in this article in two or more dimensions. In conclusion, this study has contributed significant findings to the field of dynamic inequalities on time scale calculus, but there is still much to explore. 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.