Certain Hermite-Hadamard type inequalities via generalized k-fractional integrals

Abstract

Some Hermite-Hadamard type inequalities for generalized k-fractional integrals (which are also named $(k,s)$-Riemann-Liouville fractional integrals) are obtained for a fractional integral, and an important identity is established. Also, by using the obtained identity, we get a Hermite-Hadamard type inequality.

Introduction

Let $f:I\subseteq \mathbb{R}\to \mathbb{R}$ be a convex function defined on the interval I of real numbers and $a,b\in I$ with $a<b$. The following inequality

$$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}f(x)dx\le \frac{f(a)+f(b)}{2}$$ 1.1

holds. This double inequality is known in the literature as a Hermite-Hadamard integral inequality for convex functions [1].

Sarikaya et al. established the following results for Riemann-Liouville fractional integrals.

Theorem 1.1

see Theorem 2 in [2]

Let $f:[a,b]\to \mathbb{R}$ be a positive function with $0\le a<b$ and $f\in L_1[a,b]$. If f is a convex function on $[a,b]$, then the following inequality for fractional integrals holds:

$$f\left(\frac{a+b}{2}\right)\le \frac{\mathrm{\Gamma }(1+\alpha )}{2{(b-a)}^{\alpha }}[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a)]\le \frac{f(a)+f(b)}{2}$$ 1.2

with $\alpha >0$, where the symbols $J_{a^{+}}^{\alpha }$ and $J_{b^{-}}^{\alpha }$ denote the left-sided and right-sided Riemann-Liouville fractional integrals of the order $\alpha \in {\mathbb{R}}^{+}$ that are defined by

$$J_{a^{+}}^{\alpha }f(x)=\frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_a^{x}f(t){(x-t)}^{\alpha -1}dt(0\le a<x\le b)$$

and

$$J_{b^{-}}^{\alpha }f(x)=\frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_x^{b}f(t){(t-x)}^{\alpha -1}dt(0\le a\le x<b)$$

respectively. Here $\mathrm{\Gamma }(\cdot )$ denotes the classical gamma function [3], Chapter 6.

Theorem 1.2

see Theorem 3 in [2]

Let $f:[a,b]\to \mathbb{R}$ be a differentiable mapping on $(a,b)$ with $a<b$. If $f^{\prime }\in L[a,b]$, then the following inequality for Riemann-Liouville fractional integrals holds:

$$\begin{array}{c}|\frac{f(a)+f(b)}{2}-\frac{\mathrm{\Gamma }(\alpha +1)}{2{(b-a)}^{\alpha }}[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a)]|\hfill \\ \le \frac{b-a}{2(\alpha +1)}(1-\frac{1}{2^{\alpha }})\left(\right|f^{\prime }(a)|+|f^{\prime }(b)\left|\right)\hfill \end{array}$$ 1.3

with $\alpha >0$.

The Pochhammer k-symbol ${(x)}_{n,k}$ and the k-gamma function ${\mathrm{\Gamma }}_k$ are defined as follows (see [4]):

$${(x)}_{n,k}:=x(x+k)(x+2k)\cdots (x+(n-1)k)(n\in \mathbb{N};k>0)$$ 1.4

and

$${\mathrm{\Gamma }}_k(x):=\underset{n\to \mathrm{\infty }}{lim}\frac{n!k^n{(nk)}^{\frac{x}{k}-1}}{{(x)}_{n,k}}(k>0;x\in \mathbb{C}\setminus k{\mathbb{Z}}_0^{-}),$$ 1.5

where $k{\mathbb{Z}}_0^{-}:=\{kn:n\in {\mathbb{Z}}_0^{-}\}$. It is noted that the case $k=1$ of (1.4) and (1.5) reduces to the familiar Pochhammer symbol ${(x)}_n$ and the gamma function Γ. The function ${\mathrm{\Gamma }}_k$ is given by the following integral:

$${\mathrm{\Gamma }}_k(x)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-\frac{t^k}{k}}dt(\mathrm{\Re }(x)>0).$$ 1.6

The function ${\mathrm{\Gamma }}_k$ defined on ${\mathbb{R}}^{+}$ is characterized by the following three properties: (i) ${\mathrm{\Gamma }}_k(x+k)=x{\mathrm{\Gamma }}_k(x)$; (ii) ${\mathrm{\Gamma }}_k(k)=1$; (iii) ${\mathrm{\Gamma }}_k(x)$ is logarithmically convex. It is easy to see that

$${\mathrm{\Gamma }}_k(x)=k^{\frac{x}{k}-1}\mathrm{\Gamma }\left(\frac{x}{k}\right)(\mathrm{\Re }(x)>0;k>0).$$ 1.7

We want to recall the preliminaries and notations of some well-known fractional integral operators that will be used to obtain some remarks and corollaries.

The $(k,s)$-Riemann-Liouville fractional integral operator ${}_k{}^s\mathcal{J}_a^{\alpha }$ of order $\alpha >0$ for a real-valued continuous function $f(t)$ is defined as (see [5], p.79, 2.1. Definition):

$${}_k{}^s\mathcal{J}_a^{\alpha }f(x)=\frac{{(s+1)}^{1-\frac{\alpha }{k}}}{k{\mathrm{\Gamma }}_k(\alpha )}{\int }_a^x{(x^{s+1}-t^{s+1})}^{\frac{\alpha }{k}-1}{t}^{s}f(t)dt,$$ 1.8

where $k>0$, $\beta >0$ and $s\in \mathbb{R}\setminus \{-1\}$.

The most important feature of $(k,s)$-fractional integrals is that they generalize some types of fractional integrals (Riemann-Liouville fractional integral, k-Riemann-Liouville fractional integral, generalized fractional integral and Hadamard fractional integral). These important special cases of the integral operator ${}_k{}^s\mathcal{J}_a^{\alpha }$ are mentioned below.

1. For $k=1$, the operator in (1.8) yields the following generalized fractional integrals defined by Katugompola in [6]:

   $${}_a{}^r\mathcal{J}_t^{\alpha }f(x)=\frac{{(r+1)}^{1-\alpha }}{\mathrm{\Gamma }(\alpha )}{\int }_a^x{(x^{r+1}-t^{r+1})}^{\alpha -1}{t}^{r}f(t)dt.$$ 1.9
2. Firstly by taking $k=1$, after that by taking limit $r\to -1^{+}$ and using L’Hôpital’s rule, the operator in (1.8) leads to the Hadamard fractional integral operator [1, 7]. That is,

   $$\begin{array}{rcl}\underset{r\to -1^{+}}{lim}{}_a^r{\mathcal{J}}_t^{\alpha }f(x)& =& \underset{r\to -1^{+}}{lim}\frac{{(r+1)}^{1-\alpha }}{\mathrm{\Gamma }(\alpha )}{\int }_a^x\frac{f(t)t^r}{{(x^{r+1}-t^{r+1})}^{1-\alpha }}dt\\ & =& \frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_a^x\underset{r\to -1^{+}}{lim}f(t)t^r{\left(\frac{r+1}{x^{r+1}-t^{r+1}}\right)}^{1-\alpha }dt\\ & =& \frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_a^{x}f(t)\underset{r\to -1^{+}}{lim}{\left(\frac{r+1}{x^{r+1}-t^{r+1}}\right)}^{1-\alpha }\frac{dt}{t}\\ & =& \frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_a^{x}f(t){\left(\underset{r\to -1^{+}}{lim}\frac{r+1}{x^{r+1}-t^{r+1}}\right)}^{1-\alpha }\frac{dt}{t}\\ & =& \frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_a^x(log\frac{x}{t})f(t)\frac{dt}{t}\\ & =& {}_H\mathcal{J}^{\alpha }[f(t)]\end{array}$$ 1.10

   (see [8], p.569, eq. (3.13)).
3. If we take $s=0$ in (1.8), operator (1.8), reduces to the k-Riemann-Liouville fractional integral operator, which has been firstly defined by Mubeen and Habibullah in [9]. This relation is as follows:

   $${\mathcal{J}}_{a,k}^{\alpha }f(x)=\frac{1}{k{\mathrm{\Gamma }}_k(\alpha )}{\int }_a^x{(x-t)}^{\frac{\alpha }{k}-1}f(t)dt.$$ 1.11
4. Again, taking $s=0$ and $k=1$, operator (1.8) gives us the Riemann-Liouville fractional integration operator

   $$J_{a^{+}}^{\alpha }f(x)=\frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_a^x{(x-t)}^{\alpha -1}f(t)dt.$$ 1.12

In recent years, these fractional operators have been studied and used to extend especially Grüss, Chebychev-Grüss and Pólya-Szegö type inequalities. For more details, one may refer to the recent works and books [7, 10–21].

Main results

Let $f:I^{\circ }\to \mathbb{R}$ be a given function, where $a,b\in I^{\circ }$ and $0<a<b<\mathrm{\infty }$. We suppose that $f\in L_{\mathrm{\infty }}(a,b)$ such that ${}_k{}^{s}J_{a^{+}}^{\alpha }f(x)$ and ${}_k{}^{s}J_{b^{-}}^{\alpha }f(x)$ are well defined. We define functions

$$\tilde{f}(x):=f(a+b-x),x\in [a,b]$$

and

$$F(x):=f(x)+\tilde{f}(x),x\in [a,b].$$

Hermite-Hadamard’s inequality for convex functions can be represented in a $(k,s)$-fractional integral form as follows by using the change of variables $u=\frac{t-a}{x-a}$; we have from (1.8)

$$\begin{array}{rcl}{}_k{}^s\mathcal{J}_a^{\alpha }f(x)& =& (x-a)\frac{{(s+1)}^{1-\frac{\alpha }{k}}}{k{\mathrm{\Gamma }}_k(\alpha )}{\int }_0^1\frac{{(ux+(1-u)a)}^s}{{({(ux+(1-u)a)}^{s+1}-t^{s+1})}^{\frac{\alpha }{k}-1}}\\ & & \times f(ux+(1-u)a)ds,\end{array}$$ 2.1

where $x>a$.

Theorem 2.1

Let $\alpha ,k>0$ and $s\in \mathbb{R}\setminus \{-1\}$. If f is a convex function on $[a,b]$, then we have

$$\begin{array}{rcl}f\left(\frac{a+b}{2}\right)& \le & \frac{{(s+1)}^{\frac{\alpha }{k}}{\mathrm{\Gamma }}_k(\alpha +k)}{4{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}[{}_k^s{J}_{a^{+}}^{\alpha }F(b)+{}_k^s{J}_{b^{-}}^{\alpha }F(a)]\\ & \le & \frac{f(a)+f(b)}{2}.\end{array}$$ 2.2

Proof

For $u\in [0,1]$, let $\xi =au+(1-u)b$ and $\eta =(1-u)a+bu$. Using the convexity of f, we get

$$f\left(\frac{a+b}{2}\right)=f\left(\frac{\xi +\eta }{2}\right)\le \frac{1}{2}f(\xi )+\frac{1}{2}f(\eta ).$$

That is,

$$f\left(\frac{a+b}{2}\right)\le \frac{1}{2}f(au+(1-u)b)+\frac{1}{2}f((1-u)a+bu).$$ 2.3

Now, multiplying both sides of (2.3) by

$$(b-a)\frac{{(s+1)}^{1-\frac{\alpha }{k}}}{k{\mathrm{\Gamma }}_k(\alpha )}\frac{{(ub+(1-u)a)}^s}{{[b^{s+1}-{(ub+(1-u)a)}^{s+1}]}^{1-\frac{\alpha }{k}}}$$

and integrating over $(0,1)$ with respect to u, we get

$$\begin{array}{c}(b-a)\frac{{(s+1)}^{1-\frac{\alpha }{k}}}{k{\mathrm{\Gamma }}_k(\alpha )}f\left(\frac{a+b}{2}\right){\int }_0^1\frac{{(ub+(1-u)a)}^{s}du}{{[b^{s+1}-{(ub+(1-u)a)}^{s+1}]}^{1-\frac{\alpha }{k}}}\hfill \\ \le \frac{1}{2}(b-a)\frac{{(s+1)}^{1-\frac{\alpha }{k}}}{k{\mathrm{\Gamma }}_k(\alpha )}{\int }_0^1\frac{{(ub+(1-u)a)}^{s}f(au+(1-u)b)du}{{[b^{s+1}-{(ub+(1-u)a)}^{s+1}]}^{1-\frac{\alpha }{k}}}\hfill \\ +\frac{1}{2}(b-a)\frac{{(s+1)}^{1-\frac{\alpha }{k}}}{k{\mathrm{\Gamma }}_k(\alpha )}{\int }_0^1\frac{{(ub+(1-u)a)}^{s}f((1-u)a+bu)du}{{[b^{s+1}-{(ub+(1-u)a)}^{s+1}]}^{1-\frac{\alpha }{k}}}.\hfill \end{array}$$

Note that we have

$${\int }_0^1\frac{{(ub+(1-u)a)}^{s}du}{{[b^{s+1}-{(ub+(1-u)a)}^{s+1}]}^{1-\frac{\alpha }{k}}}=\frac{k{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}{\alpha (s+1)(b-a)}.$$

Using the identity

$$\tilde{f}((1-u)a+bu)=f(au+(1-u)b),$$

and from (2.1), we obtain

$$(b-a)\frac{{(s+1)}^{1-\frac{\alpha }{k}}}{k{\mathrm{\Gamma }}_k(\alpha )}{\int }_0^1\frac{{(ub+(1-u)a)}^{s}f(au+(1-u)b)du}{{[b^{s+1}-{(ub+(1-u)a)}^{s+1}]}^{1-\frac{\alpha }{k}}}={}_k^s{J}_{a^{+}}^{\alpha }\tilde{f}(b)$$

and

$$(b-a)\frac{{(s+1)}^{1-\frac{\alpha }{k}}}{k{\mathrm{\Gamma }}_k(\alpha )}{\int }_0^1\frac{{(ub+(1-u)a)}^{s}f((1-u)a+bu)du}{{[b^{s+1}-{(ub+(1-u)a)}^{s+1}]}^{1-\frac{\alpha }{k}}}={}_k^s{J}_{a^{+}}^{\alpha }f(b).$$

Accordingly, we have

$$\frac{{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}{{(s+1)}^{\frac{\alpha }{k}}{\mathrm{\Gamma }}_k(\alpha +k)}f\left(\frac{a+b}{2}\right)\le \frac{{}_k{}^{s}J_{a^{+}}^{\alpha }F(b)}{2}.$$ 2.4

Similarly, multiplying both sides of (2.3) by

$$(b-a)\frac{{(s+1)}^{1-\frac{\alpha }{k}}}{k{\mathrm{\Gamma }}_k(\alpha )}\frac{{(ub+(1-u)a)}^s}{{[{(bu+(1-u)a)}^{s+1}-a^{s+1}]}^{1-\frac{\alpha }{k}}},$$

integrating over $(0,1)$ with respect to u, and from (2.1), we also get

$$\frac{{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}{{(s+1)}^{\frac{\alpha }{k}}{\mathrm{\Gamma }}_k(\alpha +k)}f\left(\frac{a+b}{2}\right)\le \frac{{}_k{}^{s}J_{b^{-}}^{\alpha }F(a)}{2}.$$ 2.5

By adding inequalities (2.4) and (2.5), we get

$$f\left(\frac{a+b}{2}\right)\le \frac{{(s+1)}^{\frac{\alpha }{k}}{\mathrm{\Gamma }}_k(\alpha +k)}{4{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}[{}_k^s{J}_{a^{+}}^{\alpha }F(b)+{}_k^s{J}_{b^{-}}^{\alpha }F(a)],$$

which is the left-hand side of inequality (2.2).

Since f is convex, for $u\in [0,1]$, we have

$$f(au+(1-u)b)+f((1-u)a+bu)\le f(a)+f(b).$$ 2.6

Multiplying both sides of (2.6) by

$$(b-a)\frac{{(s+1)}^{1-\frac{\alpha }{k}}}{k{\mathrm{\Gamma }}_k(\alpha )}\frac{{(ub+(1-u)a)}^s}{{[b^{s+1}-{(ub+(1-u)a)}^{s+1}]}^{1-\frac{\alpha }{k}}}$$

and integrating over $(0,1)$ with respect to u, we get

$$\begin{array}{rl}& (b-a)\frac{{(s+1)}^{1-\frac{\alpha }{k}}}{k{\mathrm{\Gamma }}_k(\alpha )}{\int }_0^1\frac{{(ub+(1-u)a)}^{s}f(au+(1-u)b)du}{{[b^{s+1}-{(ub+(1-u)a)}^{s+1}]}^{1-\frac{\alpha }{k}}}\\ & +(b-a)\frac{{(s+1)}^{1-\frac{\alpha }{k}}}{k{\mathrm{\Gamma }}_k(\alpha )}{\int }_0^1\frac{{(ub+(1-u)a)}^{s}f((1-u)a+bu)du}{{[b^{s+1}-{(ub+(1-u)a)}^{s+1}]}^{1-\frac{\alpha }{k}}}\\ & \le (b-a)\frac{{(s+1)}^{1-\frac{\alpha }{k}}}{k{\mathrm{\Gamma }}_k(\alpha )}[f(a)+f(b)]{\int }_0^1\frac{{(ub+(1-u)a)}^{s}du}{{[b^{s+1}-{(ub+(1-u)a)}^{s+1}]}^{1-\frac{\alpha }{k}}}.\end{array}$$

That is,

$${}_k{}^{s}J_{a^{+}}^{\alpha }F(b)\le \frac{{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}{{(s+1)}^{\frac{\alpha }{k}}{\mathrm{\Gamma }}_k(\alpha +k)}[f(a)+f(b)].$$ 2.7

Similarly, multiplying both sides of (2.6) by

$$(b-a)\frac{{(s+1)}^{1-\frac{\alpha }{k}}}{k{\mathrm{\Gamma }}_k(\alpha )}\frac{{(ub+(1-u)a)}^s}{{[{(ub+(1-u)a)}^{s+1}-a^{s+1}]}^{1-\frac{\alpha }{k}}}$$

and integrating over $(0,1)$ with respect to u, we also get

$${}_k{}^{s}J_{b^{-}}^{\alpha }F(a)\le \frac{{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}{{(s+1)}^{\frac{\alpha }{k}}{\mathrm{\Gamma }}_k(\alpha +k)}[f(a)+f(b)].$$ 2.8

Adding inequalities (2.7) and (2.8), we obtain

$$\frac{{(s+1)}^{\frac{\alpha }{k}}{\mathrm{\Gamma }}_k(\alpha +k)}{4{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}[{}_k^s{J}_{a^{+}}^{\alpha }F(b)+{}_k^s{J}_{b^{-}}^{\alpha }F(a)]\le \frac{f(a)+f(b)}{2},$$

which is the right-hand side of inequality (2.2). So the proof is complete. □

We want to give the following function that we will use later: For $\alpha ,k>0$ and $s\in \mathbb{R}\setminus \{-1\}$, let ${\mathrm{\nabla }}_{\alpha ,s}:[0,1]\to \mathbb{R}$ be the function defined by

$$\begin{array}{rcl}{\mathrm{\nabla }}_{\alpha ,s}(t):& =& {({(ta+(1-t)b)}^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}-{({(bt+(1-t)a)}^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}\\ & & +{(b^{s+1}-{(tb+(1-t)a)}^{s+1})}^{\frac{\alpha }{k}}-{(b^{s+1}-{(ta+(1-t)b)}^{s+1})}^{\frac{\alpha }{k}}.\end{array}$$

In order to prove our main result, we need the following identity.

Lemma 2.1

Let $\alpha ,k>0$ and $s\in \mathbb{R}{I}^{\circ }$. If f is a differentiable function on $I^{\circ }$ such that $f^{\prime }\in L[a,b]$ with $a<b$, then we have the following identity:

$$\begin{array}{c}\frac{f(a)+f(b)}{2}-\frac{{(s+1)}^{\frac{\alpha }{k}}{\mathrm{\Gamma }}_k(\alpha +k)}{4{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}[{}_k^s{J}_{a^{+}}^{\alpha }F(b)+{}_k^s{J}_{b^{-}}^{\alpha }F(a)]\hfill \\ =\frac{(b-a)}{4{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}{\int }_0^1{\mathrm{\nabla }}_{\alpha ,s}(t)f^{\prime }(ta+(1-t)b)dt.\hfill \end{array}$$ 2.9

Proof

Using integration by parts, we obtain

$$\begin{array}{rcl}{}_k{}^{s}J_{a^{+}}^{\alpha }F(b)& =& \frac{{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}{{(s+1)}^{\frac{\alpha }{k}}{\mathrm{\Gamma }}_k(\alpha +k)}F(a)+\frac{(b-a)}{{(s+1)}^{\frac{\alpha }{k}}{\mathrm{\Gamma }}_k(\alpha +k)}\\ & & \times {\int }_0^1{\left[(b^{s+1}-{(bu+(1-u)a)}^{s+1})\right]}^{\frac{\alpha }{k}}{F}^{\prime }(bu+(1-u)a)du.\end{array}$$ 2.10

Similarly, we get

$$\begin{array}{rcl}{}_k{}^{s}J_{b^{-}}^{\alpha }F(a)& =& \frac{{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}{{(s+1)}^{\frac{\alpha }{k}}{\mathrm{\Gamma }}_k(\alpha +k)}F(b)-\frac{(b-a)}{{(s+1)}^{\frac{\alpha }{k}}{\mathrm{\Gamma }}_k(\alpha +k)}\\ & & \times {\int }_0^1{[{(bu+(1-u)a)}^{s+1}-a^{s+1}]}^{\frac{\alpha }{k}}{F}^{\prime }(bu+(1-u)a)du.\end{array}$$ 2.11

Using the fact that $F(x)=f(x)+\tilde{f}(x)$ and by simple computation, from equalities (2.10) and (2.11), we get

$$\begin{array}{c}\frac{4{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}{(b-a)}(\frac{f(a)+f(b)}{2}-\frac{{(s+1)}^{\frac{\alpha }{k}}{\mathrm{\Gamma }}_k(\alpha +k)}{4{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}[{}_k^s{J}_{a^{+}}^{\alpha }F(b)+{}_k^s{J}_{b^{-}}^{\alpha }F(a)])\hfill \\ ={\int }_0^1[{({(bu+(1-u)a)}^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}-{(b^{s+1}-{(bu+(1-u)a)}^{s+1})}^{\frac{\alpha }{k}}]\hfill \\ \times F^{\prime }(bu+(1-u)a)du.\hfill \end{array}$$ 2.12

Note that we have

$$F^{\prime }(bu+(1-u)a)=f^{\prime }(bu+(1-u)a)-f^{\prime }(au+(1-u)b),u\in [0,1].$$

Then we can easily obtain

$$\begin{array}{c}{\int }_0^1{({(bu+(1-u)a)}^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}{F}^{\prime }(bu+(1-u)a)du\hfill \\ ={\int }_0^1{({(ta+(1-t)b)}^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}{f}^{\prime }(ta+(1-t)b)dt\hfill \\ -{\int }_0^1{({(bt+(1-t)a)}^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}{f}^{\prime }(ta+(1-t)b)dt\hfill \end{array}$$ 2.13

and

$$\begin{array}{c}{\int }_0^1{(b^{s+1}-{(bu+(1-u)a)}^{s+1})}^{\frac{\alpha }{k}}{F}^{\prime }(bu+(1-u)a)du\hfill \\ ={\int }_0^1{(b^{s+1}-{(ta+(1-t)b)}^{s+1})}^{\frac{\alpha }{k}}{f}^{\prime }(ta+(1-t)b)dt\hfill \\ -{\int }_0^1{(b^{s+1}-{(bt+(1-t)a)}^{s+1})}^{\frac{\alpha }{k}}{f}^{\prime }(ta+(1-t)b)dt.\hfill \end{array}$$ 2.14

Thus, the desired inequality (2.9) follows from inequalities (2.12), (2.13) and (2.14). □

For $\alpha ,k>0$, we introduce the following operator:

$$\mathrm{\Im }(s,x,y):={\int }_a^{\frac{a+b}{2}}|x-u||y^{s+1}-u^{s+1}{|}^{\frac{\alpha }{k}}du-{\int }_{\frac{a+b}{2}}^b|x-u||y^{s+1}-u^{s+1}{|}^{\frac{\alpha }{k}}du,$$

$s\in \mathbb{R}\setminus \{-1\}$, $x,y\in [a,b]$.

Using Lemma 2.1, we can obtain the following $(k,s)$-fractional integral inequality.

Theorem 2.2

Let $\alpha ,k>0$ and $s\in \mathbb{R}\setminus \{-1\}$. If f is a differentiable function on $I^{\circ }$ such that $f^{\prime }\in L[a,b]$ with $a<b$ and $|f^{\prime }|$ is convex on $[a,b]$, then

$$\begin{array}{c}|\frac{f(a)+f(b)}{2}-\frac{{(s+1)}^{\frac{\alpha }{k}}{\mathrm{\Gamma }}_k(\alpha +k)}{4{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}[{}_k^s{J}_{a^{+}}^{\alpha }F(b)+{}_k^s{J}_{b^{-}}^{\alpha }F(a)]|\hfill \\ \le \frac{\mathrm{\Psi }(s,\alpha ,a,b)}{4{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}(b-a)}\left(\right|f^{\prime }(a)|+|f^{\prime }(b)\left|\right),\hfill \end{array}$$ 2.15

where

$$\mathrm{\Psi }(s,\alpha ,a,b)=\mathrm{\Im }(s,b,b)+\mathrm{\Im }(s,a,b)-\mathrm{\Im }(s,b,a)-\mathrm{\Im }(s,a,a).$$

Proof

Using Lemma 2.1 and the convexity of $|f^{\prime }|$, we obtain

$$\begin{array}{c}|\frac{f(a)+f(b)}{2}-\frac{{(s+1)}^{\frac{\alpha }{k}}{\mathrm{\Gamma }}_k(\alpha +k)}{4{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}[{}_k^s{J}_{a^{+}}^{\alpha }F(b)+{}_k^s{J}_{b^{-}}^{\alpha }F(a)]|\hfill \\ \le \frac{(b-a)}{4{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}{\int }_0^1|{\mathrm{\nabla }}_{\alpha ,s}(t)|\left|f^{\prime }(ta+(1-t)b)\right|dt\hfill \\ \le \frac{(b-a)}{4{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}}\left(\right|f^{\prime }(a)\left|{\int }_0^{1}t\right|{\mathrm{\nabla }}_{\alpha ,s}(t)|dt+|f^{\prime }(b)|{\int }_0^1(1-t)|{\mathrm{\nabla }}_{\alpha ,s}(t)\left|dt\right).\hfill \end{array}$$ 2.16

Note that

$${\int }_0^{1}t|{\mathrm{\nabla }}_{\alpha ,s}(t)|dt=\frac{1}{{(b-a)}^2}{\int }_a^b|\mathrm{\wp }(u)|(b-u)du,$$

where

$$\begin{array}{rcl}\mathrm{\wp }(u)& =& {(u^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}-{({(b+a-u)}^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}\\ & & +{(b^{s+1}-{(b+a-u)}^{s+1})}^{\frac{\alpha }{k}}-{(b^{s+1}-u^{s+1})}^{\frac{\alpha }{k}},u\in [a,b].\end{array}$$

Observe that ℘ is a non-decreasing function on $[a,b]$. Moreover, we have $\mathrm{\wp }(a)=-2{(b^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}<0$ and $\mathrm{\wp }(\frac{a+b}{2})=0$. Thus, we have

$$\{\begin{array}{cc}\mathrm{\wp }(u)\le 0\hfill & \text{if }a\le u\le \frac{a+b}{2},\hfill \\ \mathrm{\wp }(u)>0\hfill & \text{if }\frac{a+b}{2}<u\le b.\hfill \end{array}$$

So, we obtain

$${(b-a)}^2{\int }_0^{1}t|{\mathrm{\nabla }}_{\alpha ,s}(t)|dt={\zeta }_1+{\zeta }_2+{\zeta }_3+{\zeta }_4,$$

where

$$\begin{array}{c}{\zeta }_1={\int }_a^{\frac{a+b}{2}}(b-u){(b^{s+1}-u^{s+1})}^{\frac{\alpha }{k}}du-{\int }_{\frac{a+b}{2}}^b(b-u){(b^{s+1}-u^{s+1})}^{\frac{\alpha }{k}}du,\hfill \\ {\zeta }_2=-{\int }_a^{\frac{a+b}{2}}(b-u){(u^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}du+{\int }_{\frac{a+b}{2}}^b(b-u){(u^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}du,\hfill \\ {\zeta }_3={\int }_a^{\frac{a+b}{2}}(b-u){({(b+a-u)}^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}du-{\int }_{\frac{a+b}{2}}^b(b-u){({(b+a-u)}^{s+1}-a^{s+1})}^{\frac{\alpha }{k}}du,\hfill \\ {\zeta }_4=-{\int }_a^{\frac{a+b}{2}}(b-u){(b^{s+1}-{(b+a-u)}^{s+1})}^{\frac{\alpha }{k}}du+{\int }_{\frac{a+b}{2}}^b(b-u){(b^{s+1}-{(b+a-u)}^{s+1})}^{\frac{\alpha }{k}}du.\hfill \end{array}$$

Observe that ${\zeta }_1=\mathrm{\Im }(s,b,b)$ and ${\zeta }_2=-\mathrm{\Im }(s,b,a)$. Using the change of variable $v=a+b-u$, we get ${\zeta }_3=-\mathrm{\Im }(s,a,a)$ and ${\zeta }_4=\mathrm{\Im }(s,a,b)$. Thus, we obtain

$${\int }_0^{1}t|{\mathrm{\nabla }}_{\alpha ,s}(t)|dt=\frac{\mathrm{\Im }(s,b,b)+\mathrm{\Im }(s,a,b)-\mathrm{\Im }(s,b,a)-\mathrm{\Im }(s,a,a)}{{(b-a)}^2}.$$ 2.17

Similarly,

$${\int }_0^1(1-t)|{\mathrm{\nabla }}_{\alpha ,s}(t)|dt=\frac{\mathrm{\Im }(s,b,b)+\mathrm{\Im }(s,a,b)-\mathrm{\Im }(s,b,a)-\mathrm{\Im }(s,a,a)}{{(b-a)}^2}.$$ 2.18

So, the desired inequality (2.15) follows from inequalities (2.16), (2.17) and (2.18). □

Conclusions

Lastly, we conclude this paper by remarking that we have obtained a Hermite-Hadamard inequality, an identity and a Hermite-Hadamard type inequality for a generalized k-fractional integral operator. Therefore, by suitably choosing the parameters, one can further easily obtain additional integral inequalities involving the various types of fractional integral operators from our main results.