Some Fractional Dynamic Inequalities of Hardy’s Type via Conformable Calculus

Abstract

In this article, we prove some new fractional dynamic inequalities on time scales via conformable calculus. By using chain rule and Hölder’s inequality on timescales we establish the main results. When $\alpha =1$ we obtain some well-known time-scale inequalities due to Hardy, Copson, Bennett and Leindler inequalities.

1. Preliminaries and Basic Lemmas

We present the fundamental results about the fractional timescales calculus. The results are adapted from [16,17,31,32]. A time-scale $\mathbb{T}$ is non-empty closed subset of $\mathbb{R}$ ($\mathbb{R}$ is the real numbers). The operators of backward jump and forward jump express of the closest point $t\in \mathbb{T}$ on the right and left of t is defined by, respectively:

$$\rho \left(t\right):=sup\{s\in \mathbb{T}:s<t\},$$

(16)

$$\sigma \left(t\right):=inf\{s\in \mathbb{T}:s>t\},$$

(17)

where $sup\varphi =inf\mathbb{T}$ and $inf\varphi =sup\mathbb{T}$ ($\varphi$ denotes the empty set), for any $t\in$$\mathbb{T}$ the notation $f^{\sigma }\left(t\right)$ refer to $f\left(\sigma \right(t\left)\right),$ i.e., $f^{\sigma }=f\circ \sigma$. The graininess function $\mu$$:\mathbb{T}\to [0$, $\infty )$, defined by $\mu \left(t\right):=\sigma \left(t\right)-t$.

Definition 1.

The number$T_{\alpha }^{\Delta }\left(f\right)\left(t\right)$(provided it exists) of the function$f:\mathbb{T}\to \mathbb{R}$, for$t>0$and$\alpha \in (0,1]$is the number which has the property that for any$ϵ>0,$there exists a neighborhood U of t S. T.

$$|[f^{\sigma }\left(t\right)-f\left(s\right)]t^{1-\alpha }-T_{\alpha }^{\Delta }\left(f\left(t\right)\right)(\sigma \left(t\right)-s)|\le ϵ\left|\sigma \right(t)-s)|,forallt\in U$$

$T_{\alpha }^{\Delta }\left(f\left(t\right)\right)$is called the conformable$\alpha —$fractional derivative of function f of order α at t, for conformable fractional derivative on$\mathbb{T}$at 0, we define it with$T_{\alpha }^{\Delta }(f\left(0\right)={lim}_{t\to 0^{+}}{T}_{\alpha }^{\Delta }\left(f\left(t\right)\right).$

The conformable fractional derivative has the following properties.

Theorem 1.

Let$v,$$u:\mathbb{T}\to \mathbb{R}$are conformable fractional derivative from order$\alpha \in (0,1]$, then the following properties are hold:

(i) 

The$v+u:\mathbb{T}\to \mathbb{R}$is conformable fractional derivative and

$$T_{\alpha }^{\Delta }(v+u)=T_{\alpha }^{\Delta }\left(v\right)+T_{\alpha }^{\Delta }\left(u\right).$$

(ii) 

Fora all$k\in \mathbb{R}$, then$kv:\mathbb{T}\to \mathbb{R}$is$\alpha —$fractional differentiable and

$$T_{\alpha }^{\Delta }\left(kv\right)=k{T}_{\alpha }^{\Delta }\left(v\right).$$

(iii) 

If v and u are$\alpha —$fractional differentiable, we have$vu:\mathbb{T}\to \mathbb{R}$is$\alpha —$fractional differentiable and

$$T_{\alpha }^{\Delta }\left(vu\right)=T_{\alpha }^{\Delta }\left(v\right)u+(v\circ \sigma )T_{\alpha }^{\Delta }\left(u\right)=T_{\alpha }^{\Delta }\left(v\right)(u\circ \sigma )+v{T}_{\alpha }^{\Delta }\left(u\right).$$

(iv) 

If v is$\alpha —$fractional differentiable, then$1/v$is$\alpha —$fractional differentiable with

$$T\left(\frac{1}{v}\right)=-\frac{T_{\alpha }^{\Delta }\left(v\right)}{v(v\circ \sigma )}.$$

(v) 

If v and u are$\alpha —$fractional differentiable, then$v/u$ is $\alpha —$fractional differentiable with

$$T_{\alpha }^{\Delta }(v/u)=\frac{T_{\alpha }^{\Delta }\left(v\right)u-v{T}_{\alpha }^{\Delta }\left(u\right)}{u(u\circ \sigma )},$$

valid $\forall t\in {\mathbb{T}}^k$, where $u\left(t\right)\left(u\right(\sigma \left(t\right))\ne 0.$

Lemma 1.

Let$v:\mathbb{T}\to \mathbb{R}$is continuous and$\alpha —$fractional differentiable at$t\in \mathbb{T}$for$\alpha \in (0,1]$, and$u:\mathbb{R}\to \mathbb{R}$is continuous and differentiable. Then there exists d$\in [t,\sigma (t\left)\right]$with

$$T_{\alpha }^{\Delta }(u\circ v)\left(t\right)=u^{{}^{\prime }}\left(v\left(d\right)\right)T_{\alpha }^{\Delta }\left(v\left(t\right)\right).$$

(18)

Lemma 2.

Let$u:\mathbb{R}\to \mathbb{R}$is continuously differentiable,$\alpha \in (0,1]$, and$v:\mathbb{T}\to \mathbb{R}$be$\alpha —$fractional differentiable. Then$(u\circ v):\mathbb{T}\to \mathbb{R}$is$\alpha —$fractional differentiable and we have

$$T_{\alpha }^{\Delta }(u\circ v)\left(s\right)=\left(\underset{0}{\overset{1}{\int }}{u}^{{}^{\prime }}\right(v\left(s\right)+h\mu \left(s\right)s^{\alpha -1}{T}_{\alpha }^{\Delta }\left(v\left(s\right)\right)\left)dh\right)T_{\alpha }^{\Delta }\left(v\left(s\right)\right).$$

(19)

Definition 2.

Let$0<\alpha \le 1,$the$\alpha —$fractional integral of$f$, is defined as

$$\int f\left(s\right){\Delta }_{\alpha }s=\int f\left(s\right)s^{\alpha -1}\Delta s.$$

The conformable fractional integral satisfying the next properties.

Theorem 2.

Assume a, b,$c\in \mathbb{T}$,$\lambda \in \mathbb{R}$. Let u,$v:\mathbb{T}\to \mathbb{R}$. Then

(i) 

${\int }_a^b[v\left(s\right)+u\left(s\right)]{\Delta }_{\alpha }s=$${\int }_a^{b}v\left(s\right){\Delta }_{\alpha }s+{\int }_a^{b}u\left(s\right){\Delta }_{\alpha }s.$

(ii) 

${\int }_a^b\lambda v\left(s\right){\Delta }_{\alpha }s$$=\lambda {\int }_a^{b}v\left(s\right){\Delta }_{\alpha }s.$

(iii) 

${\int }_a^{b}v\left(s\right){\Delta }_{\alpha }s$$=-{\int }_b^{a}v\left(s\right){\Delta }_{\alpha }s.$

(iv) 

${\int }_a^{b}v\left(s\right){\Delta }_{\alpha }s$$={\int }_a^{c}v\left(s\right){\Delta }_{\alpha }s+{\int }_c^{b}v\left(s\right){\Delta }_{\alpha }s.$

(v) 

${\int }_a^{a}v\left(s\right){\Delta }_{\alpha }s=0.$

Lemma 3.

Assume$\mathbb{T}$be a time-scale, a,$b\in \mathbb{T}$where$b>a$. Let u, v are conformable$\alpha$—fractional differentiable,$\alpha \in (0,1]$. Then the formula of integration by parts is given by

$${\int }_a^{b}v\left(s\right)T_{\alpha }^{\Delta }u\left(s\right){\Delta }_{\alpha }s={\left[v\right(s\left)u\right(s\left)\right]}_a^b-{\int }_a^b{u}^{\sigma }\left(s\right)T_{\alpha }^{\Delta }v\left(s\right){\Delta }_{\alpha }s.$$

(20)

Lemma 4.

Assume$\mathbb{T}$be a time-scale, a,$b\in \mathbb{T}$and$\alpha \in (0$,$1]$. Let u,$v:\mathbb{T}\to \mathbb{R}$. Then

$${\int }_a^b\left|v\left(s\right)u\left(s\right)\right|{\Delta }_{\alpha }s\le {\left[{\int }_a^b{\left|v\left(s\right)\right|}^k{\Delta }_{\alpha }s\right]}^{\frac{1}{k}}{\left[{\int }_a^b{\left|u\left(s\right)\right|}^l{\Delta }_{\alpha }s\right]}^{\frac{1}{l}},$$

(21)

where$k>1$and$1/k+1/l=1.$

2. Main Results

Throughout the paper, we will assume that the functions are nonnegative on ${[a,\infty )}_{\mathbb{T}}$ and its integrals exist and are finite. We start with the fractional time-scale inequality of Copson’s type.

Theorem 3.

Assume$1<c<$k, define

$$\Phi \left(x\right):={\int }_a^x\lambda \left(s\right){\Delta }_{\alpha }s\mathrm{and}\Omega \left(x\right):={\int }_a^x\lambda \left(s\right)g\left(s\right){\Delta }_{\alpha }s.$$

If

$$\Omega (\infty )<\infty ,\mathrm{and}{\int }_a^{\infty }\frac{\lambda \left(s\right)}{{\left({\Phi }^{\sigma }\left(s\right)\right)}^{c-\alpha +1}}{\Delta }_{\alpha }s<\infty ,$$

then

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k{\Delta }_{\alpha }x\le {\left(\frac{k}{c-\alpha }\right)}^k{\int }_a^{\infty }\frac{\lambda \left(x\right){\Phi }^{k(\alpha -c)}\left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{(c-\alpha +1)(1-k)}}{g}^k\left(x\right){\Delta }_{\alpha }x.$$

(22)

Proof. 

By employing the formula of integration by parts (20) on the term

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k{\Delta }_{\alpha }x,$$

with $u^{\sigma }\left(x\right)={\left({\Omega }^{\sigma }\left(x\right)\right)}^k$ and $x_{\alpha }^{\Delta }v\left(x\right)=\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}$, we have that

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k{\Delta }_{\alpha }x={v\left(x\right){\Omega }^k\left(x\right)|}_a^{\infty }+{\int }_a^{\infty }-v\left(x\right)x_{\alpha }^{\Delta }\left({\Omega }^k\left(x\right)\right){\Delta }_{\alpha }x,$$

(23)

where

$$-v\left(x\right)={\int }_x^{\infty }\frac{\lambda \left(s\right)}{{\left({\Phi }^{\sigma }\left(s\right)\right)}^{c-\alpha +1}}{\Delta }_{\alpha }s={\int }_x^{\infty }{x}_{\alpha }^{\Delta }\Phi \left(s\right){\left({\Phi }^{\sigma }\left(s\right)\right)}^{\alpha -c-1}{\Delta }_{\alpha }s.$$

By using the chain rule (18), we obtain that

$$\begin{array}{ccc}\hfill -x_{\alpha }^{\Delta }\left({\Phi }^{\alpha -c}\left(x\right)\right)& =& -(\alpha -c){\Phi }^{\alpha -c-1}\left(d\right)x_{\alpha }^{\Delta }\Phi \left(x\right),\mathrm{where}d\in [x,\sigma \left(x\right)]\hfill \\ & =& \frac{(c-\alpha )x_{\alpha }^{\Delta }\Phi \left(x\right)}{{\Phi }^{c-\alpha +1}\left(d\right)}\hfill \\ & \ge & \frac{(c-\alpha )x_{\alpha }^{\Delta }(\Phi (x\left)\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}.\hfill \end{array}$$

Then we have

$$x_{\alpha }^{\Delta }(\Phi (x)){\left({\Phi }^{\sigma }\left(x\right)\right)}^{\alpha -c-1}\le \frac{-1}{c-\alpha }{x}_{\alpha }^{\Delta }{\Phi }^{\alpha -c}\left(x\right),$$

and thus

$$-v\left(x\right)={\int }_x^{\infty }\frac{\lambda \left(s\right)}{{\left({\Phi }^{\sigma }\left(s\right)\right)}^{c-\alpha +1}}{\Delta }_{\alpha }s\le \frac{-1}{c-\alpha }{\int }_x^{\infty }{x}_{\alpha }^{\Delta }{\Phi }^{\alpha -c-1}\left(s\right){\Delta }_{\alpha }s\le \frac{{\Phi }^{\alpha -c}\left(x\right)}{c-\alpha }.$$

(24)

Again, by using the chain rule (18) to calculate

$$x_{\alpha }^{\Delta }\left({\Omega }^k\left(x\right)\right)=k{\Omega }^{k-1}\left(d\right)x_{\alpha }^{\Delta }(\Omega (x\left)\right),\mathrm{where}d\in [x,\sigma \left(x\right)],$$

at $x_{\alpha }^{\Delta }(\Omega (x\left)\right)=\lambda \left(x\right)g\left(x\right)\ge 0$ and $d\le \sigma \left(x\right)$, we get that

$$x_{\alpha }^{\Delta }\left({\Omega }^k\left(x\right)\right)\le k\lambda \left(x\right)g\left(x\right){\left({\Omega }^{\sigma }\left(x\right)\right)}^{k-1}.$$

(25)

Since $\Omega \left(a\right)=0$, $v(\infty )=0$ and from (24), (25) and (23) we have

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k{\Delta }_{\alpha }x\le \frac{k}{c-\alpha }{\int }_a^{\infty }{\Phi }^{\alpha -c}\left(x\right)\lambda \left(x\right)g\left(x\right){\left({\Omega }^{\sigma }\left(x\right)\right)}^{k-1}{\Delta }_{\alpha }x,$$

which reformulated as

$$\begin{array}{c}{\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k{\Delta }_{\alpha }x\\ =\frac{k}{c-\alpha }{\int }_a^{\infty }\frac{\lambda \left(x\right){\Phi }^{\alpha -c}\left(x\right)g\left(x\right)}{{\left(\lambda \left(x\right){\left({\Phi }^{\sigma }\left(x\right)\right)}^{\alpha -c-1}\right)}^{{}^{\frac{k-1}{k}}}}{\left(\frac{\lambda \left(x\right){\left({\Omega }^{\sigma }\left(x\right)\right)}^k}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}\right)}^{{}^{\frac{k-1}{k}}}{\Delta }_{\alpha }x.\end{array}$$

Using Hölder’s inequality (21) on

$${\int }_a^{\infty }\frac{\lambda \left(x\right){\Phi }^{\alpha -c}\left(x\right)g\left(x\right)}{{\left(\lambda \left(x\right){\left({\Phi }^{\sigma }\left(x\right)\right)}^{\alpha -c-1}\right)}^{{}^{\frac{k-1}{k}}}}{\left(\frac{\lambda \left(x\right){\left({\Omega }^{\sigma }\left(x\right)\right)}^k}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}\right)}^{{}^{\frac{k-1}{k}}}{\Delta }_{\alpha }x$$

with indices k and $k/(k-1)$, we have

$$\begin{array}{c}{\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k{\Delta }_{\alpha }x\le \\ \frac{k}{c-\alpha }{\left[{\int }_a^{\infty }{\left[\frac{\lambda \left(x\right){\Phi }^{\alpha -c}\left(x\right)g\left(x\right)}{{\left[\lambda \left(x\right){\left({\Phi }^{\sigma }\left(x\right)\right)}^{\alpha -c-1}\right]}^{{}^{\frac{k-1}{k}}}}\right]}^k{\Delta }_{\alpha }x\right]}^{\frac{1}{k}}\\ \times {\left[{\int }_a^{\infty }{\left[{\left[\frac{\lambda \left(x\right){\left({\Omega }^{\sigma }\left(x\right)\right)}^k}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}\right]}^{{}^{\frac{k-1}{k}}}\right]}^{\frac{k}{k-1}}{\Delta }_{\alpha }x\right]}^{\frac{k-1}{k}},\end{array}$$

then

$${\left[{\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k{\Delta }_{\alpha }x\right]}^{\frac{1}{k}}\le \frac{k}{c-\alpha }{\left[{\int }_a^{\infty }\frac{\lambda \left(x\right){\Phi }^{k(\alpha -c)}\left(x\right)g^k\left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{(c-\alpha +1)(1-k)}}{\Delta }_{\alpha }x\right]}^{\frac{1}{k}}.$$

This leads to

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k{\Delta }_{\alpha }x\le {\left(\frac{k}{c-\alpha }\right)}^k{\int }_a^{\infty }\frac{\lambda \left(x\right){\Phi }^{k(\alpha -c)}\left(x\right)g^k\left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{(c-\alpha +1)(1-k)}}{\Delta }_{\alpha }x,$$

that is the desired inequality (22). The proof is complete. □

Corollary 1.

At$\alpha =1$in Theorem 3, we obtain the inequality

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^c}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k\Delta x\le {\left(\frac{k}{c-1}\right)}^k{\int }_a^{\infty }\frac{{\Phi }^{k(1-c)}\left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c(1-k)}}\lambda \left(x\right)g^k\left(x\right)\Delta x.$$

that is the timescales version of inequality (2.8) in [33].

Corollary 2.

At$\alpha =1$, and$\mathbb{T}=\mathbb{R}$$({\Phi }^{\sigma }\left(x\right)=\Phi \left(x\right))$in Theorem 3, we obtain the integral inequality

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^c\left(x\right)}{\left({\int }_a^x\lambda \left(s\right)g\left(s\right)ds\right)}^{k}dx\le {\left(\frac{k}{c-1}\right)}^k{\int }_a^{\infty }{\Phi }^{k-c}\left(x\right)\lambda \left(x\right)g^k\left(x\right)dx,$$

which is of Copson type.

Corollary 3.

At$\alpha =1$, T$=\mathbb{R}$,$\lambda \left(x\right)=1$and$a=0$,$(\Phi \left(x\right)={\int }_0^x\lambda \left(s\right)ds=x)$in Theorem 3, we have Hardy-Littlewood integral inequality (3)

$${\int }_0^{\infty }\frac{1}{x^c}{\left({\int }_0^{x}g\left(s\right)ds\right)}^{k}dx\le {\left(\frac{k}{c-1}\right)}^k{\int }_0^{\infty }\frac{1}{x^{c-k}}{g}^k\left(x\right)dx.$$

Also, if$c=k$, we obtain the standard Hardy inequality (2)

$${\int }_0^{\infty }\frac{1}{x^k}{\left({\int }_0^{x}g\left(s\right)ds\right)}^{k}dx\le {\left(\frac{k}{k-1}\right)}^k{\int }_0^{\infty }{g}^k\left(x\right)dx.$$

Theorem 4.

Let$0\le c<1$and$k>1$. Define

$$\Phi \left(x\right):={\int }_a^x\lambda \left(s\right){\Delta }_{\alpha }s\mathrm{and}\Omega \left(x\right):={\int }_x^{\infty }\lambda \left(s\right)g\left(s\right){\Delta }_{\alpha }s.$$

If

$$\Omega \left(a\right)<\infty ,\mathrm{and}{\int }_a^{\infty }\frac{\lambda \left(s\right)}{{\left({\Phi }^{\sigma }\left(s\right)\right)}^{c-\alpha +1}}{\Delta }_{\alpha }s<\infty ,$$

then

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}{\Omega }^k\left(x\right){\Delta }_{\alpha }x\le {\left(\frac{k}{\alpha -c}\right)}^k{\int }_a^{\infty }{\left({\Phi }^{\sigma }\left(x\right)\right)}^{k-c+\alpha -1}\lambda \left(x\right)g^k\left(x\right){\Delta }_{\alpha }x.$$

(26)

Proof. 

By using the formula of integration by parts (20) on

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}{\Omega }^k\left(x\right){\Delta }_{\alpha }x,$$

with $v\left(x\right)={\Omega }^k\left(x\right)$ and $x_{\alpha }^{\Delta }u\left(x\right)=\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}$, we have

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}{\Omega }^k\left(x\right){\Delta }_{\alpha }x={u\left(x\right){\Omega }^k\left(x\right)|}_a^{\infty }+{\int }_a^{\infty }{u}^{\sigma }\left(x\right)x_{\alpha }^{\Delta }(-{\Omega }^k\left(x\right)){\Delta }_{\alpha }x,$$

(27)

where

$$u\left(x\right)={\int }_a^x\frac{\lambda \left(s\right)}{{\left({\Phi }^{\sigma }\left(s\right)\right)}^{c-\alpha +1}}{\Delta }_{\alpha }s={\int }_a^x{x}_{\alpha }^{\Delta }\Phi \left(s\right){\left({\Phi }^{\sigma }\left(s\right)\right)}^{\alpha -c-1}{\Delta }_{\alpha }s.$$

By using chain rule (18), then for $d\in [x,\sigma (x\left)\right],$ we get that

$$\begin{array}{ccc}\hfill x_{\alpha }^{\Delta }\left({\Phi }^{\alpha -c}\left(x\right)\right)& =& (\alpha -c){\Phi }^{\alpha -c-1}\left(d\right)x_{\alpha }^{\Delta }(\Phi (x))=\frac{(\alpha -c)x_{\alpha }^{\Delta }(\Phi (x\left)\right)}{{\Phi }^{c-\alpha +1}\left(d\right)}\hfill \\ & \ge & \frac{(\alpha -c)x_{\alpha }^{\Delta }(\Phi (x\left)\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}.\hfill \end{array}$$

So

$$x_{\alpha }^{\Delta }(\Phi (x)){\left({\Phi }^{\sigma }\left(x\right)\right)}^{\alpha -c-1}\le \frac{1}{\alpha -c}{x}_{\alpha }^{\Delta }({\Phi }^{\alpha -c}\left(x\right)),$$

and then,

$$\begin{array}{ccc}\hfill u^{\sigma }\left(x\right)& =& {\int }_a^{\sigma \left(x\right)}{x}_{\alpha }^{\Delta }(\Phi (s)){\left({\Phi }^{\sigma }\left(s\right)\right)}^{\alpha -c-1}{\Delta }_{\alpha }s\hfill \\ & \le & \frac{1}{\alpha -c}{\int }_a^{\sigma \left(x\right)}{x}_{\alpha }^{\Delta }({\Phi }^{\alpha -c}\left(s\right)){\Delta }_{\alpha }s\le \frac{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{\alpha -c}}{\alpha -c}.\hfill \end{array}$$

(28)

Again, by using chain rule (18), we obtain

$$-x_{\alpha }^{\Delta }\left({\Omega }^k\left(x\right)\right)=-k{\Omega }^{k-1}\left(d\right)x_{\alpha }^{\Delta }(\Omega (x\left)\right),\mathrm{where}d\in [x,\sigma \left(x\right)],$$

since $x_{\alpha }^{\Delta }(\Omega (x\left)\right)=-\lambda \left(x\right)g\left(x\right)\ge 0$ and $d\ge x,$ then

$$-x_{\alpha }^{\Delta }\left({\Omega }^k\left(x\right)\right)\le k\lambda \left(x\right)g\left(x\right){\Omega }^{k-1}\left(x\right).$$

(29)

Using $\Phi \left(a\right)=0$, $\Omega (\infty )=0$ and (28), (29) and (27), we have that

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}{\Omega }^k\left(x\right){\Delta }_{\alpha }x\le \frac{k}{\alpha -c}{\int }_a^{\infty }{\left({\Phi }^{\sigma }\left(x\right)\right)}^{\alpha -c}\lambda \left(x\right)g\left(x\right){\Omega }^{k-1}\left(x\right){\Delta }_{\alpha }x,$$

which reformulated as

$$\begin{array}{c}{\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}{\Omega }^k\left(x\right){\Delta }_{\alpha }x=\\ \frac{k}{\alpha -c}{\int }_a^{\infty }\frac{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{\alpha -c}\lambda \left(x\right)g\left(x\right)}{{\left(\lambda \left(x\right){\left({\Phi }^{\sigma }\left(x\right)\right)}^{\alpha -c-1}\right)}^{{}^{\frac{k-1}{k}}}}{\left(\frac{\lambda \left(x\right){\Omega }^k\left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}\right)}^{{}^{\frac{k-1}{k}}}{\Delta }_{\alpha }x.\end{array}$$

By employing Hölder’s inequality (21) on

$${\int }_a^{\infty }\frac{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{\alpha -c}\lambda \left(x\right)g\left(x\right)}{{\left(\lambda \left(x\right){\left({\Phi }^{\sigma }\left(x\right)\right)}^{\alpha -c-1}\right)}^{{}^{\frac{k-1}{k}}}}{\left(\frac{\lambda \left(x\right){\Omega }^k\left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}\right)}^{{}^{\frac{k-1}{k}}}{\Delta }_{\alpha }x$$

with indices k and $k/(k-1)$, we have

$$\begin{array}{c}{\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}{\Omega }^k\left(x\right){\Delta }_{\alpha }x\le \\ \frac{k}{\alpha -c}{\left[{\int }_a^{\infty }{\left[\frac{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{1-c}\lambda \left(x\right)g\left(x\right)}{{\left[\lambda \left(x\right){\left({\Phi }^{\sigma }\left(x\right)\right)}^{\alpha -c-1}\right]}^{{}^{\frac{k-1}{k}}}}\right]}^k{\Delta }_{\alpha }x\right]}^{\frac{1}{k}}\\ \times {\left[{\int }_a^{\infty }{\left[{\left[\frac{\lambda \left(x\right){\Omega }^k\left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}\right]}^{{}^{\frac{k-1}{k}}}\right]}^{\frac{k}{k-1}}{\Delta }_{\alpha }x\right]}^{\frac{k-1}{k}},\end{array}$$

then we have

$${\left[{\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-\alpha +1}}{\Omega }^k\left(x\right){\Delta }_{\alpha }x\right]}^{\frac{1}{k}}\le \frac{k}{\alpha -c}{\left[{\int }_a^{\infty }\frac{\lambda \left(x\right)g^k\left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{c-k-\alpha +1}}{\Delta }_{\alpha }x\right]}^{\frac{1}{k}}.$$

This leads to

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^c}{\Omega }^k\left(x\right){\Delta }_{\alpha }x\le {\left(\frac{k}{\alpha -c}\right)}^k{\int }_a^{\infty }{\left({\Phi }^{\sigma }\left(x\right)\right)}^{k-c+\alpha -1}\lambda \left(x\right)g^k\left(x\right){\Delta }_{\alpha }x,$$

that is the desired inequality (26). The proof is complete. □

Corollary 4.

At$\alpha =1$in Theorem 4, then

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\left({\Phi }^{\sigma }\left(x\right)\right)}^c}{\Omega }^k\left(x\right)\Delta x\le {\left(\frac{k}{1-c}\right)}^k{\int }_a^{\infty }{\left({\Phi }^{\sigma }\left(x\right)\right)}^{k-c}\lambda \left(x\right)g^k\left(x\right)\Delta x.$$

which is the timescales version inequality (2.22) in [33].

Corollary 5.

At$\alpha =1$, and T$=\mathbb{R}$in Theorem 4, we obtain the next integral inequality

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^c\left(x\right)}{\left({\int }_x^{\infty }\lambda \left(s\right)g\left(s\right)ds\right)}^{k}dx\le {\left(\frac{k}{1-c}\right)}^k{\int }_a^{\infty }{\Phi }^{k-c}\left(x\right)\lambda \left(x\right)g^k\left(x\right)dx,$$

which considered an extension of Hardy’s inequality (4) as in the following corollary.

Corollary 6.

At$\alpha =1$, T$=\mathbb{R}$,$\lambda \left(x\right)=1$and$a=0$in Theorem 4, we have Hardy-Littlewood integral inequality (4)

$${\int }_0^{\infty }\frac{1}{x^c}{\left({\int }_x^{\infty }g\left(s\right)ds\right)}^{k}dx\le {\left(\frac{k}{1-c}\right)}^k{\int }_0^{\infty }\frac{1}{x^{c-k}}{g}^k\left(x\right)dx.$$

A generalization of Leindler’s inequality (14) on fractional time scales will be proved in the next theorem.

Theorem 5.

Assume$0\le c<$$1<k$, define

$$\Phi \left(x\right):={\int }_x^{\infty }\lambda \left(s\right){\Delta }_{\alpha }s\mathrm{and}\Omega \left(x\right):={\int }_a^x\lambda \left(s\right)g\left(s\right){\Delta }_{\alpha }s.$$

If

$$\Omega (\infty )<\infty ,\mathrm{and}{\int }_a^{\infty }\frac{\lambda \left(s\right)}{{\Phi }^{c-\alpha +1}\left(s\right)}{\Delta }_{\alpha }s<\infty ,$$

then

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k{\Delta }_{\alpha }x\le {\left(\frac{k}{\alpha -c}\right)}^k{\int }_a^{\infty }{\Phi }^{k-c+\alpha -1}\left(x\right)\lambda \left(x\right)g^k\left(x\right){\Delta }_{\alpha }x.$$

(30)

Proof. 

Using the formula of integration by parts (20) on

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}{\Omega }^k\left(x\right){\Delta }_{\alpha }x,$$

with $u^{\sigma }\left(x\right)={\left({\Omega }^{\sigma }\left(x\right)\right)}^k$ and $x_{\alpha }^{\Delta }v\left(x\right)=\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}$, we have

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k{\Delta }_{\alpha }x={v\left(x\right){\Omega }^k\left(x\right)|}_a^{\infty }+{\int }_a^{\infty }-v\left(x\right)x_{\alpha }^{\Delta }\left({\Omega }^k\left(x\right)\right){\Delta }_{\alpha }x,$$

(31)

where

$$v\left(x\right)=-{\int }_x^{\infty }\frac{\lambda \left(s\right)}{{\Phi }^{c-\alpha +1}\left(s\right)}{\Delta }_{\alpha }s.$$

By chain rule (18), we see for $d\in [x,\sigma (x\left)\right]$ that

$$\begin{array}{ccc}\hfill -x_{\alpha }^{\Delta }{\Phi }^{\alpha -c}\left(x\right)& =& -(\alpha -c){\Phi }^{\alpha -c-1}\left(d\right)x_{\alpha }^{\Delta }(\Phi (x))=\frac{-(\alpha -c)(-\lambda (x\left)\right)}{{\Phi }^{c-\alpha +1}\left(d\right)}\hfill \\ & \ge & \frac{(\alpha -c)\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}.\hfill \end{array}$$

Hence

$$-v\left(x\right)={\int }_x^{\infty }\frac{\lambda \left(s\right)}{{\Phi }^{c-\alpha +1}\left(s\right)}{\Delta }_{\alpha }s\le \frac{-1}{\alpha -c}{\int }_x^{\infty }{x}_{\alpha }^{\Delta }{\Phi }^{\alpha -c}\left(s\right){\Delta }_{\alpha }s\le \frac{{\Phi }^{\alpha -c}\left(x\right)}{\alpha -c},$$

(32)

from chain rule (18), we obtain

$$x_{\alpha }^{\Delta }\left({\Omega }^k\left(x\right)\right)=k{\Omega }^{k-1}\left(d\right)x_{\alpha }^{\Delta }(\Omega (x\left)\right),\mathrm{where}d\in [x,\sigma \left(x\right)],$$

since

$$x_{\alpha }^{\Delta }(\Omega (x\left)\right)=\lambda \left(x\right)g\left(x\right)\ge 0\mathrm{and}d\le \sigma \left(x\right),$$

we get

$$x_{\alpha }^{\Delta }\left({\Omega }^k\left(x\right)\right)\le k\lambda \left(x\right)g\left(x\right){\left({\Omega }^{\sigma }\left(x\right)\right)}^{k-1}.$$

(33)

Using $\Omega \left(a\right)=0$, $v(\infty )=0$ and (32), (33) and (31 ), we get that

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k{\Delta }_{\alpha }x\le \frac{k}{\alpha -c}{\int }_a^{\infty }{\Phi }^{\alpha -c}\left(x\right)\lambda \left(x\right)g\left(x\right){\left({\Omega }^{\sigma }\left(x\right)\right)}^{k-1}{\Delta }_{\alpha }x,$$

which reformulated as

$$\begin{array}{c}{\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k{\Delta }_{\alpha }x\le \\ \frac{k}{\alpha -c}{\int }_a^{\infty }\frac{{\Phi }^{(c-\alpha +1)\left(\frac{k-1}{k}\right)}\left(x\right)}{{\lambda }^{\frac{k-1}{k}}\left(x\right){\Phi }^{c-\alpha }\left(x\right)}\lambda \left(x\right)g\left(x\right)\frac{{\lambda }^{\frac{k-1}{k}}\left(x\right){\left({\Omega }^{\sigma }\left(x\right)\right)}^{k-1}}{{\Phi }^{(c-\alpha +1)\left(\frac{k-1}{k}\right)}\left(x\right)}{\Delta }_{\alpha }x.\end{array}$$

Using Hölder’s inequality (21) on

$${\int }_a^{\infty }\frac{{\Phi }^{(c-\alpha +1)\left(\frac{k-1}{k}\right)}\left(x\right)}{{\lambda }^{\frac{k-1}{k}}\left(x\right){\Phi }^{c-\alpha }\left(x\right)}\lambda \left(x\right)g\left(x\right)\frac{{\lambda }^{\frac{k-1}{k}}\left(x\right){\left({\Omega }^{\sigma }\left(x\right)\right)}^{k-1}}{{\Phi }^{(c-\alpha +1)\left(\frac{k-1}{k}\right)}\left(x\right)}{\Delta }_{\alpha }x$$

with indices k and $k/(k-1)$, we have

$$\begin{array}{c}{\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k{\Delta }_{\alpha }x\le \\ \frac{k}{\alpha -c}{\left[{\int }_a^{\infty }{\left[\frac{{\Phi }^{(c-\alpha +1)\left(\frac{k-1}{k}\right)}\left(x\right)}{{\lambda }^{\frac{k-1}{k}}\left(x\right){\Phi }^{c-\alpha }\left(x\right)}\lambda \left(x\right)g\left(x\right)\right]}^k{\Delta }_{\alpha }x\right]}^{\frac{1}{k}}\times \\ {\left[{\int }_a^{\infty }{\left[{\lambda }^{\frac{k-1}{k}}\left(x\right)\frac{{\left({\Omega }^{\sigma }\left(x\right)\right)}^{k-1}}{{\Phi }^{(c-\alpha +1))(\frac{k-1}{k})}\left(x\right)}\right]}^{\frac{k}{k-1}}{\Delta }_{\alpha }x\right]}^{\frac{k-1}{k}},\end{array}$$

then

$${\left[{\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k{\Delta }_{\alpha }x\right]}^{\frac{1}{k}}\le \frac{k}{\alpha -c}{\left[{\int }_a^{\infty }{\Phi }^{k-c+\alpha -1}\left(x\right)\lambda \left(x\right)g^k\left(x\right){\Delta }_{\alpha }x\right]}^{\frac{1}{k}}.$$

This leads to

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k{\Delta }_{\alpha }x\le {\left(\frac{k}{\alpha -c}\right)}^k{\int }_a^{\infty }{\Phi }^{k-c+\alpha -1}\left(x\right)\lambda \left(x\right)g^k\left(x\right){\Delta }_{\alpha }x,$$

that is the desired inequality (30). The proof is complete. □

Corollary 7.

At$\alpha =1$in Theorem 5, we get

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^c\left(x\right)}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k\Delta x\le {\left(\frac{k}{1-c}\right)}^k{\int }_a^{\infty }{\Phi }^{k-c}\left(x\right)\lambda \left(x\right)g^k\left(x\right)\Delta x,$$

which inequality (2.36) in [33].

A generalization of Bennett’s inequality (15) on fractional timescales will be proved in the next theorem.

Theorem 6.

Assume$0<\alpha \le 1$,$1<c\le k$, and define

$$\Phi \left(x\right):={\int }_x^{\infty }\lambda \left(s\right){\Delta }_{\alpha }s\mathrm{and}\Omega \left(x\right):={\int }_x^{\infty }\lambda \left(s\right)g\left(s\right){\Delta }_{\alpha }s.$$

If

$$\Omega \left(a\right)<\infty \mathrm{and}{\int }_a^{\infty }\frac{\lambda \left(s\right)}{{\Phi }^{c-\alpha +1}\left(s\right)}{\Delta }_{\alpha }s<\infty ,$$

then

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}{\Omega }^k\left(x\right){\Delta }_{\alpha }x\le {\left(\frac{k}{c-\alpha }\right)}^k{\int }_a^{\infty }{\Phi }^{k-c+\alpha -1}\left(x\right)\lambda \left(x\right)g^k\left(x\right){\Delta }_{\alpha }x.$$

(34)

Proof. 

Using the formula of integration by parts (20) on

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}{\Omega }^k\left(x\right){\Delta }_{\alpha }x,$$

with $v\left(x\right)={\Omega }^k\left(x\right)$ and $x_{\alpha }^{\Delta }u\left(x\right)=\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}$, then

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}{\Omega }^k\left(x\right){\Delta }_{\alpha }x={u\left(x\right){\Omega }^k\left(x\right)|}_a^{\infty }+{\int }_a^{\infty }{u}^{\sigma }\left(x\right)x_{\alpha }^{\Delta }(-{\Omega }^k\left(x\right)){\Delta }_{\alpha }x,$$

(35)

where

$$u\left(x\right)={\int }_a^x\frac{\lambda \left(s\right)}{{\Phi }^{c-\alpha +1}\left(s\right)}{\Delta }_{\alpha }s.$$

By using chain rule (18), we see for $d\in [x,\sigma (x\left)\right]$ that

$$\begin{array}{ccc}\hfill x_{\alpha }^{\Delta }\left({\Phi }^{\alpha -c}\left(x\right)\right)& =& (\alpha -c){\Phi }^{\alpha -c-1}\left(d\right)x_{\alpha }^{\Delta }(\Phi (x))=\frac{(\alpha -c)(-\lambda (x\left)\right)}{{\Phi }^{c-\alpha +1}\left(d\right)}\hfill \\ & \ge & \frac{(c-\alpha )\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}.\hfill \end{array}$$

we get,

$$\begin{array}{ccc}\hfill u^{\sigma }\left(x\right)& =& {\int }_a^{\sigma \left(x\right)}\frac{\lambda \left(s\right)}{{\Phi }^{c-\alpha +1}\left(s\right)}{\Delta }_{\alpha }s\le \frac{1}{c-\alpha }{\int }_a^{\sigma \left(x\right)}{x}_{\alpha }^{\Delta }{\Phi }^{\alpha -c}\left(s\right){\Delta }_{\alpha }s\hfill \\ & =& \frac{{\left({\Phi }^{\sigma }\left(x\right)\right)}^{\alpha -c}}{c-\alpha }-\frac{{\Phi }^{\alpha -c}\left(a\right)}{c-\alpha }\le \frac{{\Phi }^{\alpha -c}\left(x\right)}{c-\alpha },\hfill \end{array}$$

(36)

from chain rule (18), we find that

$$x_{\alpha }^{\Delta }\left({\Omega }^k\left(x\right)\right)=k{\Omega }^{k-1}\left(d\right)x_{\alpha }^{\Delta }\Omega \left(x\right),\mathrm{where}d\in [x,\sigma \left(x\right)],$$

since

$$x_{\alpha }^{\Delta }(\Omega (x\left)\right)=-\lambda \left(x\right)g\left(x\right)\le 0\mathrm{and}x\le d,$$

then

$$-x_{\alpha }^{\Delta }\left({\Omega }^k\left(x\right)\right)\le k\lambda \left(x\right)g\left(x\right){\Omega }^{k-1}\left(x\right).$$

(37)

Using $v\left(a\right)=0$, $\Omega (\infty )=0$ and (36), (37) and (35), we get that

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k{\Delta }_{\alpha }x\le \frac{k}{c-\alpha }{\int }_a^{\infty }{\Phi }^{\alpha -c}\left(x\right)\lambda \left(x\right)g\left(x\right){\Omega }^{k-1}\left(x\right){\Delta }_{\alpha }x,$$

which reformulated as

$$\begin{array}{c}{\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}{\Omega }^k\left(x\right){\Delta }_{\alpha }x\le \\ \frac{k}{c-\alpha }{\int }_a^{\infty }\frac{{\Phi }^{(c-\alpha +1)\left(\frac{k-1}{k}\right)}\left(x\right)}{{\lambda }^{\frac{k-1}{k}}\left(x\right){\Phi }^{c-\alpha }\left(x\right)}\lambda \left(x\right)g\left(x\right){\lambda }^{\frac{k-1}{k}}\left(x\right)\frac{{\Omega }^{k-1}\left(x\right)}{{\Phi }^{(c-\alpha +1)\left(\frac{k-1}{k}\right)}\left(x\right)}{\Delta }_{\alpha }x,\end{array}$$

Using Hölder’s inequality (21) on

$${\int }_a^{\infty }\lambda \left(x\right)g\left(x\right)\frac{{\Phi }^{(c-\alpha +1)\left(\frac{k-1}{k}\right)}\left(x\right)}{{\lambda }^{\frac{k-1}{k}}\left(x\right){\Phi }^{c-\alpha }\left(x\right)}\frac{{\lambda }^{\frac{k-1}{k}}\left(x\right){\Omega }^{k-1}\left(x\right)}{{\Phi }^{(c-\alpha +1)\left(\frac{k-1}{k}\right)}\left(x\right)}{\Delta }_{\alpha }x$$

with indices k and $k/(k-1)$, we have

$$\begin{array}{c}{\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}{\Omega }^k\left(x\right){\Delta }_{\alpha }x\le \\ \frac{k}{c-\alpha }{\left[{\int }_a^{\infty }{\left[\lambda \left(x\right)g\left(x\right)\frac{{\Phi }^{(c-\alpha +1)\left(\frac{k-1}{k}\right)}\left(x\right)}{{\lambda }^{\frac{k-1}{k}}\left(x\right){\Phi }^{c-\alpha }\left(x\right)}\right]}^k{\Delta }_{\alpha }x\right]}^{\frac{1}{k}}\times \\ {\left[{\int }_a^{\infty }{\left[{\lambda }^{\frac{k-1}{k}}\left(x\right)\frac{{\Omega }^{k-1}\left(x\right)}{{\Phi }^{(c-\alpha +1)\left(\frac{k-1}{k}\right)}\left(x\right)}\right]}^{\frac{k}{k-1}}{\Delta }_{\alpha }x\right]}^{\frac{k-1}{k}},\end{array}$$

then

$${\left[{\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}{(\Omega (x\left)\right)}^k{\Delta }_{\alpha }x\right]}^{\frac{1}{k}}\le \frac{k}{c-\alpha }{\left[{\int }_a^{\infty }{\Phi }^{k-c+\alpha -1}\left(x\right)\lambda \left(x\right)g^k\left(x\right){\Delta }_{\alpha }x\right]}^{\frac{1}{k}}.$$

This leads to

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^{c-\alpha +1}\left(x\right)}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k{\Delta }_{\alpha }x\le {\left(\frac{k}{c-\alpha }\right)}^k{\int }_a^{\infty }{\Phi }^{k-c+\alpha -1}\left(x\right)\lambda \left(x\right)g^k\left(x\right){\Delta }_{\alpha }x,$$

that is the desired inequality (34). The proof is complete. □

Corollary 8.

At$\alpha =1$in Theorem 5, we have the inequality

$${\int }_a^{\infty }\frac{\lambda \left(x\right)}{{\Phi }^c\left(x\right)}{\left({\Omega }^{\sigma }\left(x\right)\right)}^k\Delta x\le {\left(\frac{k}{c-1}\right)}^k{\int }_a^{\infty }{\Phi }^{k-c}\left(x\right)\lambda \left(x\right)g^k\left(x\right)\Delta x,$$

which the inequality (2.49) in [33].

3. Conclusions

The new fractional calculus on timescales is presented with applications to some new fractional inequalities on timescales like Hardy, Bennett, Copson and Leindler types. Inequalities are considered in rather general forms and contain several special integral and discrete inequalities. The technique is based on the applications of well-known inequalities and new tools from fractional calculus.