New approach to Minkowski fractional inequalities using generalized k-fractional integral operator

TL;DR

This paper introduces new Minkowski fractional inequalities utilizing a generalized k-fractional integral operator expressed through the Gauss hypergeometric function, expanding the mathematical framework of fractional calculus.

Contribution

It presents novel Minkowski fractional inequalities based on a generalized k-fractional integral operator involving hypergeometric functions, advancing fractional calculus theory.

Findings

Derived new Minkowski fractional inequalities

Utilized hypergeometric functions in fractional integral operators

Extended existing fractional calculus frameworks

Abstract

In this paper, we obtain new results related to Minkowski fractional integral inequality using generalized k-fractional integral operator which is in terms of the Gauss hypergeometric function.

Equations

$$L_{p,k}[0,\infty )=\left\{f:\Vert f{\Vert }_{L_{p,k}[0,\infty )}={\left({\int }_0^{\infty }|f(x){|}^p{x}^{k}dx\right)}^{\frac{1}{p}}<\infty 1\le p<\infty k\ge 0\right\},$$

$$I^{\alpha ,k}f(x)=\frac{(k+1{)}^{1-\alpha }}{\Gamma (\alpha )}{\int }_0^x(x^{k+1}-t^{k+1}{)}^{\alpha -1}{t}^{k}f(t)dt.$$

$$\begin{array}{rl}{I}_{x,k}^{\alpha ,\beta ,\eta ,\mu }[f(x)]& =\frac{(k+1{)}^{\mu +\beta +1}{x}^{(k+1)(-\alpha -\beta -2\mu )}}{\Gamma (\alpha )}{\int }_0^x{\tau }^{(k+1)\mu }(x^{k+1}-{\tau }^{k+1}{)}^{\alpha -1}\times \\ & {}_2{F}_1(\alpha +\beta +\mu ,-\eta ;\alpha ;1-(\frac{\tau }{x}{)}^{k+1}){\tau }^{k}f(\tau )d\tau .\end{array}$$

$${}_2{F}_1(a,b;c;x)=\sum _{n=0}^{\infty }\frac{(a{)}_n(b{)}_n}{(c{)}_n}\frac{x^n}{n!},$$

$$(a{)}_n=a(a+1)...(a+n-1)=\frac{\Gamma (a+n)}{\Gamma (a)},(a{)}_0=1.$$

$$\begin{array}{rl}F(x,\tau )& =\frac{(k+1{)}^{\mu +\beta +1}{x}^{(k+1)(-\alpha -\beta -2\mu )}}{\Gamma (\alpha )}{\tau }^{(k+1)\mu }\\ & (x^{k+1}-{\tau }^{k+1}{)}^{\alpha -1}{\times }_2{F}_1(\alpha +\beta +\mu ,-\eta ;\alpha ;1-(\frac{\tau }{x}{)}^{k+1})\\ & =\sum _{n=0}^{\infty }\frac{(\alpha +\beta +\mu {)}_n(-n{)}_n}{\Gamma (\alpha +n)n!}{x}^{(k+1)(-\alpha -\beta -2\mu -\eta )}{\tau }^{(k+1)\mu }(x^{k+1}-{\tau }^{k+1}{)}^{\alpha -1+n}(k+1{)}^{\mu +\beta +1}\\ & =\frac{{\tau }^{(k+1)\mu }(x^{k+1}-{\tau }^{k+1}{)}^{\alpha -1}(k+1{)}^{\mu +\beta +1}}{x^{k+1}(\alpha +\beta +2\mu )\Gamma (\alpha )}+\\ & \frac{{\tau }^{(k+1)\mu }(x^{k+1}-{\tau }^{k+1}{)}^{\alpha }(k+1{)}^{\mu +\beta +1}(\alpha +\beta +\mu )(-n)}{x^{k+1}(\alpha +\beta +2\mu +1)\Gamma (\alpha +1)}+\\ & \frac{{\tau }^{(k+1)\mu }(x^{k+1}-{\tau }^{k+1}{)}^{\alpha +1}(k+1{)}^{\mu +\beta +1}(\alpha +\beta +\mu )(\alpha +\beta +\mu +1)(-n)(-n+1)}{x^{k+1}(\alpha +\beta +2\mu +1)\Gamma (\alpha +2)2!}+...\end{array}$$

$${\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[f^p(x)]\right]}^{\frac{1}{p}}+{\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[g^q(x)]\right]}^{\frac{1}{p}}\le \frac{1+M(m+2)}{(m+1)(M+1)}{\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[(f+g{)}^p(x)]\right]}^{\frac{1}{p}},$$

$$(M+1{)}^{p}f(\tau )\le M^p(f+g{)}^p(\tau ).$$

$$\begin{array}{rl}& (M+1{)}^p\frac{(k+1{)}^{\mu +\beta +1}{x}^{(k+1)(-\alpha -\beta -2\mu )}}{\Gamma (\alpha )}{\int }_0^x{\tau }^{(k+1)\mu }(x^{k+1}-{\tau }^{k+1}{)}^{\alpha -1}\times \\ & {}_2{F}_1(\alpha +\beta +\mu ,-\eta ;\alpha ;1-(\frac{\tau }{x}{)}^{k+1}){\tau }^k{f}^p(\tau )d\tau \\ & \le M^p\frac{(k+1{)}^{\mu +\beta +1}{x}^{(k+1)(-\alpha -\beta -2\mu )}}{\Gamma (\alpha )}{\int }_0^x{\tau }^{(k+1)\mu }(x^{k+1}-{\tau }^{k+1}{)}^{\alpha -1}\times \\ & {}_2{F}_1(\alpha +\beta +\mu ,-\eta ;\alpha ;1-(\frac{\tau }{x}{)}^{k+1}){\tau }^k(f+g{)}^p(\tau )d\tau ,\end{array}$$

$$I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[f^p(x)]\le \frac{M^p}{(M+1{)}^p}\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[(f+g{)}^p(x)]\right],$$

$${\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[f^p(x)]\right]}^{\frac{1}{p}}\le \frac{M}{(M+1)}{\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[(f+g{)}^p(x)]\right]}^{\frac{1}{p}}.$$

$$(1+\frac{1}{m})g(\tau )\le \frac{1}{m}(f(\tau )+g(\tau )),$$

$$(1+\frac{1}{m}{)}^p{g}^p(\tau )\le (\frac{1}{m}{)}^p(f(\tau )+g(\tau ){)}^p.$$

$${\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[g^p(x)]\right]}^{\frac{1}{p}}\le \frac{1}{(m+1)}{\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[(f+g{)}^p(x)]\right]}^{\frac{1}{p}}.$$

$$\begin{array}{rl}{\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[f^p(x)]\right]}^{\frac{2}{p}}+{\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[g^q(x)]\right]}^{\frac{2}{p}}\ge & (\frac{(M+1)(m+1)}{M}-2){\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[f^p(x)]\right]}^{\frac{1}{p}}+\\ & {\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[g^q(x)]\right]}^{\frac{1}{p}}.\end{array}$$

$$\frac{(M+1)(m+1)}{M}{\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[f^p(x)]\right]}^{\frac{1}{p}}\times {\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[g^q(x)]\right]}^{\frac{1}{p}}\le {\left([I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[(f(x)+g(x){)}^p]{]}^{\frac{1}{p}}\right)}^2.$$

$$({\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[(f(x)+g(x){)}^p]\right]}^{\frac{1}{p}}{)}^2\le ({\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[f^p(x)]\right]}^{\frac{1}{p}}+{\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[g^q(x)]\right]}^{\frac{1}{p}}{)}^2,$$

$$\begin{array}{rl}{\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[(f(x)+g(x){)}^p]\right]}^{\frac{2}{p}}\le & {\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[f^p(x)]\right]}^{\frac{2}{p}}+{\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[g^q(x)]\right]}^{\frac{2}{p}}\\ & +2{\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[f^p(x)]\right]}^{\frac{1}{p}}{\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[g^q(x)]\right]}^{\frac{1}{p}}.\end{array}$$

$${\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[f(x)]\right]}^{\frac{1}{p}}{\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[g(x)]\right]}^{\frac{1}{q}}\le (\frac{M}{m}{)}^{\frac{1}{pq}}\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }[[f(x){]}^{\frac{1}{p}}[g(x){]}^{\frac{1}{q}}]\right],$$

$$[g(\tau ){]}^{\frac{1}{p}}\ge M^{\frac{-1}{q}}[f(\tau ){]}^{\frac{1}{q}},$$

$$\begin{array}{rl}[f(\tau ){]}^{\frac{1}{p}}[g(\tau ){]}^{\frac{1}{q}}& \ge M^{\frac{-1}{q}}[f(\tau ){]}^{\frac{1}{q}}[f(\tau ){]}^{\frac{1}{p}}\\ & \ge M^{\frac{-1}{q}}[f(\tau ){]}^{\frac{1}{q}+\frac{1}{q}}\\ & \ge M^{\frac{-1}{q}}[f(\tau )].\end{array}$$

$$\begin{array}{rl}& \frac{(k+1{)}^{\mu +\beta +1}{x}^{(k+1)(-\alpha -\beta -2\mu )}}{\Gamma (\alpha )}{\int }_0^x{\tau }^{(k+1)\mu }(x^{k+1}-{\tau }^{k+1}{)}^{\alpha -1}\times \\ & {}_2{F}_1(\alpha +\beta +\mu ,-\eta ;\alpha ;1-(\frac{\tau }{x}{)}^{k+1}){\tau }^{k}f(\tau {)}^{\frac{1}{p}}g(\tau {)}^{\frac{1}{q}}d\tau \\ & \le M^{\frac{-1}{q}}\frac{(k+1{)}^{\mu +\beta +1}{x}^{(k+1)(-\alpha -\beta -2\mu )}}{\Gamma (\alpha )}{\int }_0^x{\tau }^{(k+1)\mu }(x^{k+1}-{\tau }^{k+1}{)}^{\alpha -1}\times \\ & {}_2{F}_1(\alpha +\beta +\mu ,-\eta ;\alpha ;1-(\frac{\tau }{x}{)}^{k+1}){\tau }^{k}f(\tau )d\tau ,\end{array}$$

$$I_{x,k}^{\alpha ,\beta ,\eta ,\mu }\left[[f(x){]}^{\frac{1}{p}}[g(x){]}^{\frac{1}{q}}\right]\le M^{\frac{-1}{q}}\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }f(x)\right].$$

$${\left(I_{x,k}^{\alpha ,\beta ,\eta ,\mu }\left[[f(x){]}^{\frac{1}{p}}[g(x){]}^{\frac{1}{q}}\right]\right)}^{\frac{1}{p}}\le M^{\frac{-1}{pq}}{\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }f(x)\right]}^{\frac{1}{p}},$$

$$[f(\tau ){]}^{\frac{1}{p}}\ge m^{\frac{1}{p}}[g(\tau ){]}^{\frac{1}{p}},$$

$$[f(\tau ){]}^{\frac{1}{p}}[g(\tau ){]}^{\frac{1}{q}}\ge m^{\frac{1}{p}}[g(\tau ){]}^{\frac{1}{q}}[g(\tau ){]}^{\frac{1}{p}}=m^{\frac{1}{p}}[g(\tau )].$$

$$\begin{array}{rl}& \frac{(k+1{)}^{\mu +\beta +1}{x}^{(k+1)(-\alpha -\beta -2\mu )}}{\Gamma (\alpha )}{\int }_0^x{\tau }^{(k+1)\mu }(x^{k+1}-{\tau }^{k+1}{)}^{\alpha -1}\times \\ & {}_2{F}_1(\alpha +\beta +\mu ,-\eta ;\alpha ;1-(\frac{\tau }{x}{)}^{k+1}){\tau }^{k}f(\tau {)}^{\frac{1}{p}}g(\tau {)}^{\frac{1}{q}}d\tau \\ & \le M^{\frac{1}{p}}\frac{(k+1{)}^{\mu +\beta +1}{x}^{(k+1)(-\alpha -\beta -2\mu )}}{\Gamma (\alpha )}{\int }_0^x{\tau }^{(k+1)\mu }(x^{k+1}-{\tau }^{k+1}{)}^{\alpha -1}\times \\ & {}_2{F}_1(\alpha +\beta +\mu ,-\eta ;\alpha ;1-(\frac{\tau }{x}{)}^{k+1}){\tau }^{k}g(\tau )d\tau ,\end{array}$$

$$I_{x,k}^{\alpha ,\beta ,\eta ,\mu }\left[[f(x){]}^{\frac{1}{p}}[g(x){]}^{\frac{1}{q}}\right]\le m^{\frac{1}{p}}\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }g(x)\right].$$

$${\left(I_{x,k}^{\alpha ,\beta ,\eta ,\mu }\left[[f(x){]}^{\frac{1}{p}}[g(x){]}^{\frac{1}{q}}\right]\right)}^{\frac{1}{q}}\le m^{\frac{1}{pq}}{\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }f(x)\right]}^{\frac{1}{q}},$$

$${\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }{f}^p(x)\right]}^{\frac{1}{p}}{\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }{g}^q(x)\right]}^{\frac{1}{q}}\le (\frac{M}{m}{)}^{\frac{1}{pq}}\left[I_{x,k}^{\alpha ,\beta ,\eta ,\mu }(f(x)g(x))\right]hold.$$

Full text

Keywords : Minkowski fractional integral inequality, generalized k-fractional integral operator and Gauss hypergeometric function.

Mathematics Subject Classification: 26D10, 26A33, 05A30.

1 Introduction

In the last decades many researchers have worked on fractional integral inequalities using Riemann-Liouville, generalized Riemann-Liouville, Hadamard and Siago, see [2, 3, 4, 7, 8, 9, 10]. W. Yang [23] proved the Chebyshev and Gr$\ddot{u}$ss-type integral inequalities for Saigo fractional integral operator. S. Mubeen and S. Iqbal [14] has proved the Gr$\ddot{u}$ss-type integral inequalities generalized k-fractional integral. In [1, 6, 13, 24] authors have studied some fractional integral inequalities using generalized k-fractional integral operator (in terms of the Gauss hypergeometric function). Recently many researchers have shown development of fractional integral inequalities associated with hypergeometric functions, see [11, 13, 15, 17, 18, 19, 21, 22, 24]. Also, in [3, 7] authors established reverse Minkowski fractional integral inequality using Hadamard and Riemann-Liouville integral operator respectively.

In literature few results have been obtained on some fractional integral inequalities using Saigo fractional integral operator, see [5, 12, 15, 16, 24]. Motivated from [2, 6, 7, 13], our purpose in this paper is to establish some new results using generalized k-fractional integral in terms of Gauss hypergeometric function. The paper has been organized as follows, in section 2, we define basic definitions and proposition related to generalized k-fractional integral. In section 3, we give the results about reverse Minkowski fractional integral inequality using fractional generalized k-fractional integral, In section 4, we give some other inequalities using fractional generalized k-fractional integral.

2 Preliminaries

In this section, we give some necessary definitions which will be used latter.

Definition 2.1

[13, 24]** The function $f(x)$, for all $x>0$ is said to be in the $L_{p,k}[0,\infty),$ if

[TABLE]

Definition 2.2

[13, 20, 24]** Let $f\in L_{1,k}[0,\infty),$. The generalized Riemann-Liouville fractional integral $I^{\alpha,k}f(x)$ of order $\alpha,k\geq 0$ is defined by

[TABLE]

Definition 2.3

[13, 24]** Let $k\geq 0,\alpha>0,\mu>-1$ and $\beta,\eta\in R$. The generalized k-fractional integral $I^{\alpha,\beta,\eta,\mu}_{x,k}$ (in terms of the Gauss hypergeometric function)of order $\alpha$ for real-valued continuous function $f(t)$ is defined by

[TABLE]

where, the function ${}_{2}F_{1}(-)$ in the right-hand side of (2.3) is the Gaussian hypergeometric function defined by

[TABLE]

and $(a)_{n}$ is the Pochhammer symbol

[TABLE]

Consider the function

[TABLE]

It is clear that $F(x,\tau)$ is positive because for all $\tau\in(0,x)$ , $(x>0)$ since each term of the (2.5) is positive.

3 Reverse Minkowski fractional integral inequality

In this section, we establish reverse Minkowski fractional integral inequality using generalized k-fractional integral operator (in terms of the Gauss hypergeometric function).

Theorem 3.1

Let $p\geq 1$ and let $f$, $g$ be two positive function on $[0,\infty)$, such that for all $x>0$, $I^{\alpha,\beta,\eta,\mu}_{x,k}[f^{p}(x)]<\infty$, $I^{\alpha,\beta,\eta,\mu}_{x,k}[g^{q}(x)]<\infty$. If $0<m\leq\frac{f(\tau)}{g(\tau)}\leq M$, $\tau\in(0,x)$ we have

[TABLE]

for all $k\geq 0,$ $\alpha>max\{0,-\beta-\mu\}$, $\beta<1,$ $\mu>-1,$ $\beta-1<\eta<0.$

Proof: Using the condition $\frac{f(\tau)}{g(\tau)}\leq M$, $\tau\in(0,x)$, $x>0$, we can write

[TABLE]

Multiplying both side of (3.2) by $F(x,\tau)$, then integrating resulting identity with respect to $\tau$ from [math] to $x$, we get

[TABLE]

which is equivalent to

[TABLE]

hence, we can write

[TABLE]

On other hand, using condition $m\leq\frac{f(\tau)}{g(\tau)}$, we obtain

[TABLE]

therefore,

[TABLE]

Now, multiplying both side of (3.7) by $F(x,\tau)$, ( $\tau\in(0,x)$, $x>0$), where $G(x,\tau)$ is defined by (2.5). Then integrating resulting identity with respect to $\tau$ from [math] to $x$, we have

[TABLE]

The inequalities (3.1) follows on adding the inequalities (3.5) and (3.8).

Our second result is as follows.

Theorem 3.2

Let $p\geq 1$ and $f$, $g$ be two positive function on $[0,\infty)$, such that for all $x>0$, $I^{\alpha,\beta,\eta,\mu}_{x,k}[f^{p}(x)]<\infty$, $I^{\alpha,\beta,\eta,\mu}_{x,k}[g^{q}(x)]<\infty$. If $0<m\leq\frac{f(\tau)}{g(\tau)}\leq M$, $\tau\in(0,x)$ then we have

[TABLE]

for all $k\geq 0,$ $\alpha>max\{0,-\beta-\mu\}$, $\beta<1,$ $\mu>-1,$ $\beta-1<\eta<0.$

Proof: Multiplying the inequalities (3.5) and (3.8), we obtain

[TABLE]

Applying Minkowski inequalities to the right hand side of (3.10), we have

[TABLE]

which implies that

[TABLE]

Hence, from (3.10) and (3.12), we obtain (3.9). Theorem 3.2 is thus proved.

4 Other fractional integral inequalities related to Minkowski inequality

In this section, we establish some new integral inequalities related to Minkowski inequality using generalized k-fractional integral operator (in terms of the Gauss hypergeometric function).

Theorem 4.1

Let $p>1$, $\frac{1}{p}+\frac{1}{q}=1$ and $f$, $g$ be two positive function on $[0,\infty)$, such that $I^{\alpha,\beta,\eta,\mu}_{x,k}[f(x)]<\infty$, $I^{\alpha,\beta,\eta,\mu}_{x,k}[g(x)]<\infty$. If $0<m\leq\frac{f(\tau)}{g(\tau)}\leq M<\infty$, $\tau\in[0,x]$ we have

[TABLE]

for all $k\geq 0,$ $\alpha>max\{0,-\beta-\mu\}$, $\beta<1,$ $\mu>-1,$ $\beta-1<\eta<0.$

Proof:- Since $\frac{f(\tau)}{g(\tau)}\leq M$, $\tau\in[0,x]$ $x>0$, therefore

[TABLE]

and also,

[TABLE]

Multiplying both side of (4.3) by $F(x,\tau)$, ( $\tau\in(0,x)$, $x>0$), where $F(x,\tau)$ is defined by (2.5). Then integrating resulting identity with respect to $\tau$ from [math] to $x$, we have

[TABLE]

which implies that

[TABLE]

Consequently,

[TABLE]

on other hand, since $mg(\tau)\leq f(\tau)$,   $\tau\in[0,x)$, $x>0$, then we have

[TABLE]

multiplying equation (4.7) by $[g(\tau)]^{\frac{1}{q}}$, we have

[TABLE]

Multiplying both side of (4.8) by $F(x,\tau)$, ( $\tau\in(0,x)$, $x>0$), where $F(x,\tau)$ is defined by (2.5). Then integrating resulting identity with respect to $\tau$ from [math] to $x$, we have

[TABLE]

which implies that

[TABLE]

Hence, we can write

[TABLE]

multiplying equation (4.6) and (4.11) we get the result (4.1).

Theorem 4.2

*Let $f$ and $g$ be two positive function on $[0,\infty[$, such that

$I^{\alpha,\beta,\eta,\mu}_{x,k}[f^{p}(x)]<\infty$, $I^{\alpha,\beta,\eta,\mu}_{x,k}[g^{q}(x)]<\infty$. $x>0$, If $0<m\leq\frac{f(\tau)^{p}}{g(\tau)^{q}}\leq M<\infty$, $\tau\in[0,x]$. Then we have*

[TABLE]

Where $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, for all $k\geq 0,$ $\alpha>max\{0,-\beta-\mu\}$, $\beta<1,$ $\mu>-1,$ $\beta-1<\eta<0.$

Proof:- Replacing $f(\tau)$ and $g(\tau)$ by $f(\tau)^{p}$ and $g(\tau)^{q}$, $\tau\in[0,x]$, $x>0$ in theorem 4.1, we obtain required inequality.

Now, here we present fractional integral inequality related to Minkowsky inequality as follows

Theorem 4.3

let $f$ and $g$ be two integrable functions on $[1,\infty]$ such that $\frac{1}{p}+\frac{1}{q}=1,p>1,$ and $0<m<\frac{f(\tau)}{g(\tau)}<M,\tau\in(0,x),x>0.$ Then for all $\alpha>0,$ we have

[TABLE]

for all $k\geq 0,$ $\alpha>max\{0,-\beta-\mu\}$, $\beta<1,$ $\mu>-1,$ $\beta-1<\eta<0.$

Proof:- Since, $\frac{f(\tau)}{g(\tau)}<M,\tau\in(0,x),x>0,$ we have

[TABLE]

Taking $p^{th}$ power on both side and multiplying resulting identity by $F(x,\tau)$, we obtain

[TABLE]

therefore,

[TABLE]

on other hand, $0<m<\frac{f(\tau)}{g(\tau)},\tau\in(0,x),x>0,$ we can write

[TABLE]

therefore,

[TABLE]

consequently, we have

[TABLE]

Now, using Young inequality

[TABLE]

Multiplying both side of (4.19) by $F(x,\tau)$, which is positive because $\tau\in(0,x)$, $x>0$, then integrate the resulting identity with respect to $\tau$ from [math] to $x$, we get

[TABLE]

from equation (4.15), (4.18) and (4.20) we get

[TABLE]

now using the inequality $(a+b)^{r}\leq 2^{r-1}(a^{r}+b^{r}),r>1,a,b\geq 0,$ we have

[TABLE]

and

[TABLE]

Injecting (4.22), (2.23) in (4.21) we get required inequality (4.12). This complete the proof.

Theorem 4.4

Let $f$, $g$ be two positive function on $[0,\infty)$, such that $f$ is non-decreasing and $g$ is non-increasing. Then

[TABLE]

for all $k\geq 0,$ $\alpha>max\{0,-\beta-\mu\}$, $\beta<1,$ $\mu>-1,$ $\beta-1<\eta<0.$

Proof:- let $\tau,\rho\in[0,x]$, $x>0$, for any $\delta>0$, $\gamma>0$, we have

[TABLE]

[TABLE]

therefore

[TABLE]

Now, multiplying both side of (4.27) by $F(x,\tau)$, ( $\tau\in(0,x)$, $x>0$), where $F(x,\tau)$ is defined by (2.5). Then integrating resulting identity with respect to $\tau$ from [math] to $x$, we have

[TABLE]

[TABLE]

Again, multiplying both side of (4.29) by $F(x,\rho)$, ( $\rho\in(0,x)$, $x>0$), where $F(x,\rho)$ is defined by (2.5). Then integrating resulting identity with respect to $\rho$ from [math] to $x$, we have

[TABLE]

then we can write

[TABLE]

This proves the result (4.24).

**Competing interests

**The authors declare that they have no competing interests.