Some Inequalities of Čebyšev Type for Conformable k-Fractional Integral Operators

Abstract

In the article, the authors present several inequalities of the Čebyšev type for conformable k-fractional integral operators.

1. Introduction

The Čebyšev inequality [1] reads that

$$\frac{1}{b-a}{\int }_a^{b}f\left(x\right)g\left(x\right)dx\ge \left[\frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx\right]\left[\frac{1}{b-a}{\int }_a^{b}g\left(x\right)dx\right],$$

(1)

where f and g are two integrable and synchronous functions on $[a,b]$ and two functions f and g are called synchronous on $[a,b]$ if

$$\left[f\right(x)-f(y\left)\right]\left[g\right(x)-g(y\left)\right]\ge 0,x,y\in [a,b].$$

The inequality (1) has many applications in diverse research subjects such as numerical quadrature, transform theory, probability, existence of solutions of differential equations, and statistical problems (see ([2], Chapter IX) and the paper [3]). Many authors have investigated, generalized, and applied the Čebyšev inequality (1). For detailed information, please refer to [4,5] and closely related references.

In [6,7], the Riemann–Liouville fractional integrals ${\mathfrak{I}}_{a^{+}}^{\alpha }$ and ${\mathfrak{I}}_{b^{-}}^{\alpha }$ of order $\alpha >0$ are defined respectively by

$${\mathfrak{I}}_{a^{+}}^{\alpha }f\left(x\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_a^x{(x-t)}^{\alpha -1}f\left(t\right)dt,x>a,\Re \left(\alpha \right)>0$$

(2)

and

$${\mathfrak{I}}_{b^{-}}^{\alpha }f\left(x\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_x^b{(t-x)}^{\alpha -1}f\left(t\right)dt,x<b,\Re \left(\alpha \right)>0,$$

(3)

where $\mathrm{\Gamma }$ is the classical Euler gamma function [8,9,10].

In [11], Belarbi and Dahmani presented the following theorems related to the Čebyšev inequality (1) for the Riemann–Liouville fractional integral operators [12,13,14].

Theorem 1

([11], Theorem 3.1). Let f and g be two synchronous functions on $[0,\infty )$. Then, for $t,\alpha >0$, we have

$$J^{\alpha }\left(fg\right)\ge \frac{\mathrm{\Gamma }(\alpha +1)}{t^{\alpha }}{J}^{\alpha }f\left(t\right)J^{\alpha }g\left(t\right).$$

Theorem 2

([11], Theorem 3.2). Let f and g be two synchronous functions on $[0,\infty )$. Then, for all $t,\alpha ,\beta >0$, we have

$$\frac{t^{\alpha }}{\mathrm{\Gamma }(\alpha +1)}{J}^{\beta }\left(fg\right)\left(t\right)+\frac{t^{\beta }}{\mathrm{\Gamma }(\beta +1)}{J}^{\alpha }\left(fg\right)\left(t\right)\ge J^{\alpha }f\left(t\right)J^{\beta }g\left(t\right)+J^{\beta }f\left(t\right)J^{\alpha }g\left(t\right).$$

Theorem 3

([11], Theorem 3.3). Let $f_i$ for $1\le i\le n$ be n positive and increasing functions on $[0,\infty )$. Then, for $t,\alpha >0$, we have

$$J^{\alpha }\left(\prod _{i=1}^n{f}_i\right)\left(t\right)\ge {\left[J^{\alpha }\left(1\right)\right]}^{1-n}\prod _{i=1}^n{J}^{\alpha }{f}_i\left(t\right).$$

Theorem 4

([11], Theorem 3.4). Let f and g be two functions defined on $[0,\infty )$, such that f is increasing, g is differentiable, and there exists a real number $m={inf}_{t\ge 0}{g}^{\prime }\left(t\right)$. Then, the inequality

$$J^{\alpha }\left(fg\right)\left(t\right)\ge \frac{1}{J^{\alpha }\left(1\right)}{J}^{\alpha }f\left(t\right)J^{\alpha }g\left(t\right)-\frac{mt}{\alpha +1}{J}^{\alpha }f\left(t\right)+m{J}^{\alpha }\left(tf\left(t\right)\right)$$

is valid for $t,\alpha >0$.

In [15], the Riemann–Liouville k-fractional integrals are respectively defined by

$${\mathfrak{I}}_{k,a^{+}}^{\alpha }f\left(x\right)=\frac{1}{k{\mathrm{\Gamma }}_k\left(\alpha \right)}{\int }_a^x{(x-t)}^{\alpha /k-1}f\left(t\right)dt,x>a,\Re \left(\alpha \right)>0$$

and

$${\mathfrak{I}}_{k,b^{-}}^{\alpha }f\left(x\right)=\frac{1}{k{\mathrm{\Gamma }}_k\left(\alpha \right)}{\int }_x^b{(t-x)}^{\alpha /k-1}f\left(t\right)dt,x<b,\Re \left(\alpha \right)>0,$$

where ${\mathrm{\Gamma }}_k$ is the gamma k-function [16,17].

In [18], the left and right sided fractional conformable integral operators are respectively defined by

$${}^{\beta }{\mathfrak{I}}_{a^{+}}^{\alpha }f\left(x\right)=\frac{1}{\mathrm{\Gamma }\left(\beta \right)}{\int }_a^x{\left[\frac{{(x-a)}^{\alpha }-{(\tau -a)}^{\alpha }}{\alpha }\right]}^{\beta -1}\frac{f\left(\tau \right)}{{(\tau -a)}^{1-\alpha }}d\tau$$

(4)

and

$${}^{\beta }{\mathfrak{I}}_{b^{-}}^{\alpha }f\left(x\right)=\frac{1}{\mathrm{\Gamma }\left(\beta \right)}{\int }_a^x{\left[\frac{{(b-x)}^{\alpha }-{(b-\tau )}^{\alpha }}{\alpha }\right]}^{\beta -1}\frac{f\left(\tau \right)}{{(b-\tau )}^{1-\alpha }}d\tau ,$$

(5)

where $\Re \left(\beta \right)>0$. Obviously, if taking $a=0$ and $\alpha =1$, then the Equations (4) and (5) reduce to the Riemann–Liouville fractional integrals (2) and (3), respectively.

In [19], one sided conformable fractional integral operator was defined as

$${}^{\beta }{\mathfrak{I}}^{\alpha }f\left(x\right)=\frac{1}{\mathrm{\Gamma }\left(\beta \right)}{\int }_0^x{\left(\frac{x^{\alpha }-{\tau }^{\alpha }}{\alpha }\right)}^{\beta -1}\frac{f\left(\tau \right)}{{\tau }^{1-\alpha }}d\tau .$$

(6)

Recently, conformable k-fractional integrals were defined [20] by

$${}_k^{\beta }{\mathfrak{I}}_{a^{+}}^{\alpha }f\left(x\right)=\frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)}{\int }_a^x{\left[\frac{{(x-a)}^{\alpha }-{(\tau -a)}^{\alpha }}{\alpha }\right]}^{\beta /k-1}\frac{f\left(\tau \right)}{{(\tau -a)}^{1-\alpha }}d\tau$$

(7)

and

$${}_k^{\beta }{\mathfrak{I}}_{b^{-}}^{\alpha }f\left(x\right)=\frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)}{\int }_a^x{\left[\frac{{(b-x)}^{\alpha }-{(b-\tau )}^{\alpha }}{\alpha }\right]}^{\beta /k-1}\frac{f\left(\tau \right)}{{(b-\tau )}^{1-\alpha }}d\tau ,$$

where $\Re \left(\beta \right)>0$.

In this paper, we introduce the conformable k-fractional integral operator

$${}^{\beta }{\mathfrak{I}}_k^{\alpha }f\left(x\right)=\frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)}{\int }_0^x{\left(\frac{x^{\alpha }-{\tau }^{\alpha }}{\alpha }\right)}^{\beta /k-1}\frac{f\left(\tau \right)}{{\tau }^{1-\alpha }}d\tau .$$

(8)

When $k=1$, the Equations (7) to (8) reduces to the Equations (4) to (6), respectively.

2. Main Results

In this section, we present several Čebyšev type inequalities for conformable k-fractional integral operators defined in the Equation (8).

Theorem 5.

Let f and g be two integrable functions which are synchronous on $[0,\infty )$. Then,

$$\left({}^{\beta }{J}_k^{\alpha }fg\right)\left(x\right)\ge \frac{{\mathrm{\Gamma }}_k(\beta +k){\alpha }^{\beta /k}}{x^{\alpha \beta /k}}\left({}^{\beta }{J}_k^{\alpha }f\right)\left(x\right)\left({}^{\beta }{J}_k^{\alpha }g\right)\left(x\right),$$

where $\alpha ,\beta >0$.

Proof. 

Since f and g are synchronous on $[0,\infty )$, we have

$$f\left(u\right)g\left(u\right)+f\left(v\right)g\left(v\right)\ge f\left(u\right)g\left(v\right)+f\left(v\right)g\left(u\right).$$

(9)

Multiplying both sides of the Equation (9) by

$$\frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)u^{1-\alpha }}{\left(\frac{x^{\alpha }-u^{\alpha }}{\alpha }\right)}^{\beta /k-1},x\in \mathbb{R},0<u<x$$

results in

$$\begin{array}{c}\frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)u^{1-\alpha }}{\left(\frac{x^{\alpha }-u^{\alpha }}{\alpha }\right)}^{\beta /k-1}f\left(u\right)g\left(u\right)+\frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)u^{1-\alpha }}{\left(\frac{x^{\alpha }-u^{\alpha }}{\alpha }\right)}^{\beta /k-1}f\left(v\right)g\left(v\right)\hfill \\ \hfill \ge \frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)u^{1-\alpha }}{\left(\frac{x^{\alpha }-u^{\alpha }}{\alpha }\right)}^{\beta /k-1}f\left(u\right)g\left(v\right)+\frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)u^{1-\alpha }}{\left(\frac{x^{\alpha }-u^{\alpha }}{\alpha }\right)}^{\beta /k-1}f\left(v\right)g\left(u\right).\end{array}$$

Further integrating both sides with respect to u over $(0,x)$ gives

$$\begin{array}{c}\frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)}{\int }_0^x{\left(\frac{x^{\alpha }-u^{\alpha }}{\alpha }\right)}^{\beta /k-1}\frac{f\left(u\right)g\left(u\right)}{u^{1-\alpha }}du+\frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)}{\int }_0^x{\left(\frac{x^{\alpha }-u^{\alpha }}{\alpha }\right)}^{\beta /k-1}\frac{f\left(v\right)g\left(v\right)}{u^{1-\alpha }}du\hfill \\ \hfill \ge \frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)}{\int }_0^x{\left(\frac{x^{\alpha }-u^{\alpha }}{\alpha }\right)}^{\beta /k-1}\frac{f\left(u\right)g\left(v\right)}{u^{1-\alpha }}du+\frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)}{\int }_0^x{\left(\frac{x^{\alpha }-u^{\alpha }}{\alpha }\right)}^{\beta /k-1}\frac{f\left(v\right)g\left(u\right)}{u^{1-\alpha }}du.\end{array}$$

Consequently, it follows that

$$\left({}^{\beta }{J}_k^{\alpha }fg\right)\left(x\right)+f\left(v\right)g\left(v\right)\frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)}{\int }_0^x{\left(\frac{x^{\alpha }-u^{\alpha }}{\alpha }\right)}^{\beta /k-1}\frac{du}{u^{1-\alpha }}\ge g\left(v\right)\left({}^{\beta }{J}_k^{\alpha }f\right)\left(x\right)+f\left(v\right)\left({}^{\beta }{J}_k^{\alpha }g\right)\left(x\right)$$

and

$$\left({}^{\beta }{J}_k^{\alpha }fg\right)\left(x\right)+\frac{x^{\alpha \beta /k}}{{\mathrm{\Gamma }}_k(\beta +k){\alpha }^{\beta /k}}f\left(v\right)g\left(v\right)\ge g\left(v\right)\left({}^{\beta }{J}_k^{\alpha }f\right)\left(x\right)+f\left(v\right)\left({}^{\beta }{J}_k^{\alpha }g\right)\left(x\right),$$

(10)

where

$${\int }_0^x{\left(\frac{x^{\alpha }-u^{\alpha }}{\alpha }\right)}^{\beta /k-1}\frac{du}{u^{1-\alpha }}=\frac{k{x}^{\alpha \beta /k}}{\beta {\alpha }^{\beta /k}}.$$

Multiplying both sides of the Equation (10) by

$$\begin{array}{c}\hfill \frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)v^{1-\alpha }}{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\beta /k-1}\end{array}$$

arrives at

$$\begin{array}{c}\frac{\left({}^{\beta }{J}_k^{\alpha }fg\right)\left(x\right)}{k{\mathrm{\Gamma }}_k\left(\beta \right)v^{1-\alpha }}{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\beta /k-1}+\frac{f\left(v\right)g\left(v\right)}{k{\mathrm{\Gamma }}_k\left(\beta \right)v^{1-\alpha }}{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\beta /k-1}\frac{x^{\alpha \beta /k}}{{\mathrm{\Gamma }}_k(\beta +k){\alpha }^{\beta /k}}\hfill \\ \hfill \ge \frac{g\left(v\right)\left({}^{\beta }{J}_k^{\alpha }f\right)\left(x\right)}{k{\mathrm{\Gamma }}_k\left(\beta \right)v^{1-\alpha }}{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\beta /k-1}+\frac{f\left(v\right)\left({}^{\beta }{J}_k^{\alpha }g\right)\left(x\right)}{k{\mathrm{\Gamma }}_k\left(\beta \right)v^{1-\alpha }}{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\beta /k-1}.\end{array}$$

Now, integrating over $(0,x)$ reveals

$$\begin{array}{c}\left({}^{\beta }{J}_k^{\alpha }fg\right)\left(x\right)\frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)}{\int }_0^x{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\beta /k-1}\frac{dv}{v^{1-\alpha }}\hfill \\ +\frac{x^{\alpha \beta /k}}{{\mathrm{\Gamma }}_k(\beta +k){\alpha }^{\beta /k}}\frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)}{\int }_0^x{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\beta /k-1}\frac{f\left(v\right)g\left(v\right)}{v^{1-\alpha }}dv\\ \ge \left({}^{\beta }{J}_k^{\alpha }f\right)\left(x\right)\frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)}{\int }_0^x{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\beta /k-1}\frac{g\left(v\right)}{v^{1-\alpha }}dv\\ \hfill +\left({}^{\beta }{J}_k^{\alpha }g\right)\left(x\right)\frac{1}{k{\mathrm{\Gamma }}_k\left(\beta \right)}{\int }_0^x{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\beta /k-1}\frac{f\left(v\right)}{v^{1-\alpha }}dv.\end{array}$$

Therefore, we have

$$\frac{x^{\alpha \beta /k}}{{\mathrm{\Gamma }}_k(\beta +k){\alpha }^{\beta /k}}\left({}^{\beta }{J}_k^{\alpha }fg\right)\left(x\right)+\frac{x^{\alpha \beta /k}}{{\mathrm{\Gamma }}_k(\beta +k){\alpha }^{\beta /k}}\left({}^{\beta }{J}_k^{\alpha }fg\right)\left(x\right)\ge \left({}^{\beta }{J}_k^{\alpha }f\right)\left(x\right)\left({}^{\beta }{J}_k^{\alpha }g\right)\left(x\right)+\left({}^{\beta }{J}_k^{\alpha }f\right)\left(x\right)\left({}^{\beta }{J}_k^{\alpha }g\right)\left(x\right).$$

The proof of Theorem 5 is complete. □

Corollary 1.

Let f and g be two integrable functions which are synchronous on $[0,\infty )$. Then,

$$\left({}^{\beta }{J}_{k}fg\right)\left(x\right)\ge \frac{{\mathrm{\Gamma }}_k(\beta +k)}{x^{\beta /k}}\left({}^{\beta }{J}_{k}f\right)\left(x\right)\left({}^{\beta }{J}_{k}g\right)\left(x\right),\alpha ,\beta >0.$$

Proof. 

This follows from taking $\alpha =1$ in Theorem 5. □

Theorem 6.

Let f and g be two integrable functions which are synchronous on $[0,\infty )$. Then,

$$\frac{x^{\alpha \tau /k}}{{\mathrm{\Gamma }}_k(\tau +k){\alpha }^{\tau /k}}\left({}^{\beta }{J}_k^{\alpha }fg\right)\left(x\right)+\frac{x^{\alpha \beta /k}}{{\mathrm{\Gamma }}_k(\beta +k){\alpha }^{\beta /k}}\left({}^{\tau }{J}_k^{\alpha }fg\right)\left(x\right)\ge \left({}^{\beta }{J}_k^{\alpha }f\right)\left(x\right)\left({}^{\tau }{J}_k^{\alpha }g\right)\left(x\right)+\left({}^{\tau }{J}_k^{\alpha }f\right)\left(x\right)\left({}^{\beta }{J}_k^{\alpha }g\right)\left(x\right)$$

for $\alpha ,\beta ,\tau >0$.

Proof. 

Multiplying both sides of the equality (10) by

$$\frac{1}{k{\mathrm{\Gamma }}_k\left(\tau \right)v^{1-\alpha }}{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\tau /k-1}$$

yields

$$\begin{array}{c}\frac{1}{k{\mathrm{\Gamma }}_k\left(\tau \right)v^{1-\alpha }}{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\tau /k-1}\left({}^{\beta }{J}_k^{\alpha }fg\right)\left(x\right)+\frac{f\left(v\right)g\left(v\right)}{k{\mathrm{\Gamma }}_k\left(\tau \right)v^{1-\alpha }}{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\tau /k-1}\frac{x^{\alpha \beta /k}}{{\mathrm{\Gamma }}_k(\beta +k){\alpha }^{\beta /k}}\hfill \\ \hfill \ge \frac{1}{k{\mathrm{\Gamma }}_k\left(\tau \right)v^{1-\alpha }}{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\tau /k-1}g\left(v\right)\left({}^{\beta }{J}_k^{\alpha }f\right)\left(x\right)+\frac{1}{k{\mathrm{\Gamma }}_k\left(\tau \right)v^{1-\alpha }}{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\tau /k-1}f\left(v\right)\left({}^{\beta }{J}_k^{\alpha }g\right)\left(x\right).\end{array}$$

Further integrating both sides with respect to v over $(0,x)$ leads to

$$\begin{array}{c}\frac{\left({}^{\beta }{J}_k^{\alpha }fg\right)\left(x\right)}{k{\mathrm{\Gamma }}_k\left(\tau \right)}{\int }_0^x{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\tau /k-1}\frac{dv}{v^{1-\alpha }}+\frac{x^{\alpha \beta /k}}{{\mathrm{\Gamma }}_k(\beta +k){\alpha }^{\beta /k}}\frac{1}{k{\mathrm{\Gamma }}_k\left(\tau \right)}{\int }_0^x{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\tau /k-1}\frac{f\left(v\right)g\left(v\right)}{v^{1-\alpha }}dv\hfill \\ \hfill \ge \frac{\left({}^{\beta }{J}_k^{\alpha }f\right)\left(x\right)}{k{\mathrm{\Gamma }}_k\left(\tau \right)}{\int }_0^x{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\tau /k-1}\frac{g\left(v\right)}{v^{1-\alpha }}dv+\frac{\left({}^{\beta }{J}_k^{\alpha }g\right)\left(x\right)}{k{\mathrm{\Gamma }}_k\left(\tau \right)}{\int }_0^x{\left(\frac{x^{\alpha }-v^{\alpha }}{\alpha }\right)}^{\tau /k-1}\frac{f\left(v\right)}{v^{1-\alpha }}dv.\end{array}$$

Therefore, we have

$$\frac{x^{\alpha \tau /k}}{{\mathrm{\Gamma }}_k(\tau +k){\alpha }^{\tau /k}}\left({}^{\beta }{J}_k^{\alpha }fg\right)\left(x\right)+\frac{x^{\alpha \beta /k}}{{\mathrm{\Gamma }}_k(\beta +k){\alpha }^{\beta /k}}\left({}^{\tau }{J}_k^{\alpha }fg\right)\left(x\right)\ge \left({}^{\beta }{J}_k^{\alpha }f\right)\left(x\right)\left({}^{\tau }{J}_k^{\alpha }g\right)\left(x\right)+\left({}^{\tau }{J}_k^{\alpha }f\right)\left(x\right)\left({}^{\beta }{J}_k^{\alpha }g\right)\left(x\right).$$

Further integrating with respect to v over $(0,x)$, as did in the proof of Theorem 5, concludes Theorem 6. □

Remark 1.

Applying Theorem 6 to $\tau =\beta$ results in Theorem 5.

Corollary 2.

Let f and g be two integrable functions which are synchronoms on $[0,\infty )$. Then

$$\frac{x^{\tau /k}}{{\mathrm{\Gamma }}_k(\tau +k)}\left({}^{\beta }{J}_{k}fg\right)\left(x\right)+\frac{x^{\beta /k}}{{\mathrm{\Gamma }}_k(\beta +k)}\left({}^{\tau }{J}_{k}fg\right)\left(x\right)\ge \left({}^{\beta }{J}_{k}f\right)\left(x\right)\left({}^{\tau }{J}_{k}g\right)\left(x\right)+\left({}^{\tau }{J}_{k}f\right)\left(x\right)\left({}^{\beta }{J}_{k}g\right)\left(x\right)$$

for $\alpha ,\beta ,\tau >0$.

Proof. 

This follows from taking $\alpha =1$ in Theorem 6. □

Theorem 7.

Let $f_i$ for $1\le i\le n$ be positive and increasing functions on $[a,b]$. For $\alpha ,\beta >0$, we have

$$\left({}^{\beta }{J}_k^{\alpha }\prod _{i=1}^n{f}_i\right)\left(x\right)\ge {\left[\frac{{\mathrm{\Gamma }}_k(\beta +k){\alpha }^{\beta /k}}{x^{\alpha \beta /k}}\right]}^{n-1}\prod _{i=1}^n\left({}^{\beta }{J}_k^{\alpha }{f}_i\right)\left(x\right).$$

(11)

Proof. 

We prove this theorem by induction on $n\in \mathbb{N}$. Obviously, the case $n=1$ of (11) holds.

For $n=2$, since $f_1$ and $f_2$ are increasing, we have

$$[f_1\left(x\right)-f_1\left(y\right)][f_2\left(x\right)-f_2\left(y\right)]\ge 0.$$

Now, the left proof of the inequality (11) for $n=2$ is the same as that of Theorem 5.

Assume that the inequality (11) is true for some $n\ge 3$. We observe that, since $f_i$ is increasing, $f={\prod }_{i=1}^n{f}_i$ is increasing. Let $g=f_{n+1}$. Then, applying the case $n=2$ to the functions f and g yields

$$\begin{array}{cc}\hfill \left({}^{\beta }{J}_k^{\alpha }\prod _{i=1}^n{f}_i{f}_{n+1}\right)\left(x\right)& \ge \left[\frac{{\mathrm{\Gamma }}_k(\beta +k){\alpha }^{\beta /k}}{x^{\alpha \beta /k}}\right]\left({}^{\beta }{J}_k^{\alpha }\prod _{i=1}^n{f}_i\right)\left({}^{\beta }{J}_k^{\alpha }{f}_{n+1}\right)\left(x\right)\hfill \\ & \ge {\left(\frac{{\mathrm{\Gamma }}_k(\beta +k){\alpha }^{\beta /k}}{x^{\alpha \beta /k}}\right)}^n\prod _{i=1}^{n+1}\left({}^{\beta }{J}_k^{\alpha }{f}_i\right)\left(x\right),\hfill \end{array}$$

where the induction hypothesis for n is used in the deduction of the second inequality. The proof of Theorem 7 is complete. □

Corollary 3.

Let $f_i$ for $1\le i\le n$ be positive and increasing functions on $[a,b]$. For $\alpha ,\beta >0$, we have

$$\left({}^{\beta }{J}_k\prod _{i=1}^n{f}_i\right)\left(x\right)\ge {\left[\frac{{\mathrm{\Gamma }}_k(\beta +k)}{x^{\beta /k}}\right]}^{n-1}\prod _{i=1}^n\left({}^{\beta }{J}_k{f}_i\right)\left(x\right).$$

Proof. 

This follows from taking $\alpha =1$ in Theorem 7. □

Theorem 8.

Let $\alpha ,\beta >0$ and the functions $f,g:[0,\infty )\to \mathbb{R}$ be such that f is increasing, g is differentiable, and $g^{\prime }$ has a lower bound $m={inf}_{t\in [0,\infty )}{g}^{\prime }\left(t\right)$. Then,

$$\left({}^{\beta }{J}_k^{\alpha }fg\right)\left(x\right)\ge \frac{{\mathrm{\Gamma }}_k(\beta +k){\alpha }^{\beta /k}}{x^{\alpha \beta /k}}\left({}^{\beta }{J}_k^{\alpha }f\right)\left(x\right)\left({}^{\beta }{J}_k^{\alpha }g\right)\left(x\right)-\frac{kmx}{(\beta +k)}\left({}^{\beta }{J}_k^{\alpha }f\right)\left(x\right)+m\left({}^{\beta }{J}_k^{\alpha }if\right)\left(x\right),$$

where $i\left(x\right)$ is the identity function.

Proof. 

Let $h\left(x\right)=g\left(x\right)-mx$. We find that h is differentiable and increasing on $[0,\infty )$. As in the proof of Theorem 7, for clarity, let $p\left(x\right)=mx$, we obtain

$$\begin{array}{c}\left({}^{\beta }{J}_k^{\alpha }f(g-p)\right)\left(x\right)\ge \frac{{\mathrm{\Gamma }}_k(\beta +k){\alpha }^{\beta /k}}{x^{\alpha \beta /k}}\left({}^{\beta }{J}_k^{\alpha }f\right)\left(x\right)\left({}^{\beta }{J}_k^{\alpha }(g-p)\right)\left(x\right)\hfill \\ \hfill =\frac{{\mathrm{\Gamma }}_k(\beta +k){\alpha }^{\beta /k}}{x^{\alpha \beta /k}}\left({}^{\beta }{J}_k^{\alpha }f\right)\left(x\right)\left({}^{\beta }{J}_k^{\alpha }g\right)\left(x\right)-\frac{{\mathrm{\Gamma }}_k(\beta +k){\alpha }^{\beta /k}}{x^{\alpha \beta /k}}\left({}^{\beta }{J}_k^{\alpha }f\right)\left(x\right)\left({}^{\beta }{J}_k^{\alpha }p\right)\left(x\right),\end{array}$$

(12)

where

$$\left({}^{\beta }{J}_k^{\alpha }f(g-p)\right)\left(x\right)=\left({}^{\beta }{J}_k^{\alpha }fg\right)\left(x\right)-m\left({}^{\beta }{J}_k^{\alpha }if\right)\left(x\right)$$

(13)

and

$$\left({}^{\beta }{J}_k^{\alpha }p\right)\left(x\right)=\frac{m{x}^{\alpha \beta /k+1}{\mathrm{\Gamma }}_k\left(2k\right)}{{\mathrm{\Gamma }}_k(\beta +2k){\alpha }^{\beta /k}}.$$

Since ${\mathrm{\Gamma }}_k\left(k\right)=1$, see ([16], p. 183), then ${\mathrm{\Gamma }}_k\left(2k\right)=k$. Therefore, we derive

$$\left({}^{\beta }{J}_k^{\alpha }p\right)\left(x\right)=\frac{km{x}^{\alpha \beta /k+1}}{{\mathrm{\Gamma }}_k(\beta +2k){\alpha }^{\beta /k}}.$$

(14)

Substituting the Equations (13) and (14) into the Equation (12) leads to the desired result. □

Corollary 4.

Under conditions of Theorem 8, we have

$$\left({}^{\beta }{J}_{k}fg\right)\left(x\right)\ge \frac{{\mathrm{\Gamma }}_k(\beta +k)}{x^{\beta /k}}\left({}^{\beta }{J}_{k}f\right)\left(x\right)\left({}^{\beta }{J}_{k}g\right)\left(x\right)-\frac{kmx}{(\beta +k)}\left({}^{\beta }{J}_{k}f\right)\left(x\right)+m\left({}^{\beta }{J}_{k}if\right)\left(x\right),$$

where $i\left(x\right)$ is the identity function.

Proof. 

This follows from taking $\alpha =1$ in Theorem 8. □

3. Conclusions

In this paper, we established several Čebyšev type inequalities for conformable k-fractional integral operators. We observed that, if allowing $k=1$, inequalities obtained in this paper will reduce to those inequalities in [21]. Similarly, if letting $\alpha =k=1$, inequalities obtained in this paper will reduce to those inequalities in [11].