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.

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.