Generalizations of Hermite–Hadamard like inequalities involving $\chi _{{\kappa }}$-Hilfer fractional integrals

Abstract

The main objective of this paper is to obtain a new κ-fractional analogue of Hermite–Hadamard’s inequality using the class of s-convex functions and $\chi _{{\kappa }}$-Hilfer fractional integrals. In order to obtain other main results of the paper we derive two new fractional integral identities using the definitions of $\chi _{{\kappa }}$-Hilfer fractional integrals. For the validity of these identities we also take some particular examples. Using these identities we then obtain some more new variants of Hermite–Hadamard’s inequality using s-convex functions.

1 Introduction

Theory of convexity has played significant role in the development of theory of inequalities. Many famous results known in theory of inequalities are direct consequences of the applications of convex functions. In this regard Hermite–Hadamard’s inequality which can be viewed as necessary and sufficient condition for a function to be convex is one of the most studied result. In recent years it has been observed that a number of new generalizations of classical Hermite–Hadamard’s inequality have been obtained in the literature. Dragomir and Pearce [3] has written a very informative monograph on some recent developments and applications of Hermite–Hadamard’s inequality.

Sarikaya et al. [16] utilized the concepts of fractional calculus and obtained fractional analogues of Hermite–Hadamard’s inequality. This particular article has opened a new venue of research and consequently several new fractional analogues of Hermite–Hadamard’s inequality have been obtained using different approaches. Since the birth of fractional calculus this subject has received special attention by the mathematicians and resultantly the classical concepts of fractional calculus have been extended and generalized in different directions according to the need of problem. This motivated inequalities experts and as a result they used new generalized concepts of fractional calculus in obtaining novel generalized fractional analogues of classical inequalities.

The main motivation of this paper is to derive some new fractional analogues of Hermite–Hadamard’s inequality using $\chi _{{\kappa }}$-Hilfer fractional integrals via s-convex functions of Breckner type. In order to obtain the main results of the paper we first derive new fractional integral identities. To check the validity of these new identities we take some particular examples. We hope that the ideas and techniques of this paper will inspire interested readers working in this field.

Before we proceed, let us recall some previously known concepts and results which will be used during the study of this paper.

Riemann–Liouville fractional integrals are defined as follows.

Definition 1.1

([8])

Let ${\Xi }\in L_{1}[a,b]$. Then Riemann–Liouville integrals $J_{a^{+}}^{\alpha }{\Xi }$ and $J_{b^{-}}^{\alpha }{\Xi }$ of order $\alpha >0$ with $a\geq 0$ are defined by

$$ J_{a^{+}}^{\alpha }{\Xi }(x)=\frac{1}{\Gamma (\alpha )} \int _{a}^{x}(x-{ \lambda })^{\alpha -1}{\Xi }({\lambda })\,\mathrm{d} {\lambda },\quad x>a, $$

and

$$ J_{b^{-}}^{\alpha }{\Xi }(x)=\frac{1}{\Gamma (\alpha )} \int _{x}^{b}({ \lambda }-x)^{\alpha -1}{\Xi }({\lambda })\,\mathrm{d} {\lambda },\quad x< b, $$

where

$$ \Gamma (\alpha )= \int _{0}^{\infty }e^{-x}x^{\alpha -1} \,\mathrm{d}x, $$

is the well-known Gamma function.

Mubeen and Habibullah [10] were the first to define the notion of κ-fractional integrals. Sarikaya et al. [15] introduced the κ-analogue of Riemann–Liouville fractional integrals and discussed some of its basic properties. They defined this concept in the following way: To be more precise let Ξ be piecewise continuous on $I^{*}=(0,\infty )$ and integrable on any finite subinterval of $I=[0,\infty ]$. Then, for ${\lambda }>0$, we consider κ-Riemann–Liouville fractional integral of Ξ of order α

$$ _{k}J_{a}^{\alpha }{\Xi }(x)= \frac{1}{{\kappa }\Gamma _{{\kappa }}(\alpha )} \int _{a}^{x}(x-{ \lambda })^{\frac{\alpha }{{\kappa }}-1}{ \Xi }({\lambda })\,\mathrm{d} { \lambda },\quad x>a,{\kappa }>0. $$

If ${\kappa }\rightarrow 1$, then κ-Riemann–Liouville fractional integrals reduces to classical Riemann–Liouville fractional integral.

Another important generalization of Riemann–Liouville fractional integrals the generalized R–L integrals with respect to another function χ (in the Hilfer sense [5]).

Definition 1.2

([8])

Let $(a,b)$ ($-\infty \leq a< b\leq \infty$) be a finite interval of the real line $\mathbb{R}$ and $\alpha >0$. Also let $\chi (x)$ be an increasing and positive monotone function on $(a,b]$, having a continuous derivative $\chi '(x)$ on $(a,b)$. The left and right-sided χ-fractional integrals in the Hilfer sense of a function Ξ with respect to another function χ on $[a,b]$ are defined as

$$\begin{aligned}& I_{a^{+}}^{\alpha ;\chi }{\Xi }(x) =\frac{1}{\Gamma (\alpha )} \int _{a}^{x}\chi '({\lambda }) \bigl(\chi (x)-\chi ({\lambda }) \bigr)^{\alpha -1}{ \Xi }({\lambda }) \, \mathrm{d} {\lambda }, \\& I_{b^{-}}^{\alpha ;\chi }{\Xi }(x) =\frac{1}{\Gamma (\alpha )} \int _{x}^{b}\chi '({\lambda }) \bigl(\chi ({\lambda })-\chi (x) \bigr)^{\alpha -1}{ \Xi }({\lambda }) \, \mathrm{d} {\lambda }, \end{aligned}$$

respectively; $\Gamma (\cdot)$ is the gamma function.

Liu et al. [9] and Zhao et al. [20] obtained some interesting results pertaining to Hermite–Hadamard’s inequality involving $\chi _{{\kappa }}$-Riemann–Liouville fractional integrals.

Recently Awan et al. [1] introduced the notion of $\chi _{{\kappa }}$-Hilfer fractional integrals and obtained some new variants of Hermite–Hadamard’s inequality.

Definition 1.3

([1])

Let $(a,b)$ ($-\infty \leq a< b\leq \infty$) be a finite interval of the real line $\mathbb{R}$ and $\alpha >0$. Also let $\chi (x)$ be an increasing and positive monotone function on $(a,b]$, having a continuous derivative $\chi '(x)$ on $(a,b)$. The left- and right-sided $\chi _{{\kappa }}$-fractional integrals in the Hilfer sense of a function Ξ with respect to another function $\chi _{{\kappa }}$ on $[a,b]$ and ${\kappa }>0$ are defined as

$$\begin{aligned}& {}_{{\kappa }}I_{a^{+}}^{\alpha ;\chi }{\Xi }(x) = \frac{1}{{\kappa }\Gamma _{{\kappa }}(\alpha )} \int _{a}^{x} \chi '({\lambda }) \bigl(\chi (x)-\chi ({\lambda }) \bigr)^{\frac{\alpha }{{\kappa }}-1}{ \Xi }({\lambda }) \, \mathrm{d} {\lambda }, \\& {}_{{\kappa }}I_{b^{-}}^{\alpha ;\chi }{\Xi }(x) = \frac{1}{{\kappa }\Gamma _{{\kappa }}(\alpha )} \int _{x}^{b} \chi '({\lambda }) \bigl(\chi ({\lambda })-\chi (x) \bigr)^{\frac{\alpha }{{\kappa }}-1}{ \Xi }({\lambda }) \, \mathrm{d} {\lambda }, \end{aligned}$$

respectively;

$$ \Gamma _{{\kappa }}(x)= \int _{0}^{\infty }{\lambda }^{x-1}e^{- \frac{{\lambda }^{{\kappa }}}{{\kappa }}} \,\mathrm{d} {\lambda },\quad \Re (x)>0, $$

is the κ-analogue of the gamma function.

By taking ${\kappa }\to 1$, Definition 1.3 reduces to Definition 1.2. This shows that $\chi _{{\kappa }}$-fractional integrals in the Hilfer sense is the significant generalization of χ-fractional integrals in the Hilfer sense.

The κ-analogues of beta function and incomplete beta function are, respectively, defined as

$$ \mathrm{B}_{{\kappa }}(x,y)=\frac{1}{{\kappa }} \int _{0}^{1}{ \lambda }^{\frac{x}{{\kappa }}-1}(1-{ \lambda })^{\frac{x}{{\kappa }}-1} \,\mathrm{d} {\lambda } $$

and

$$ \mathrm{B}_{{\kappa }}(z;x,y)=\frac{1}{{\kappa }} \int _{0}^{z}{ \lambda }^{\frac{x}{{\kappa }}-1}(1-{ \lambda })^{\frac{b}{y}-1} \,\mathrm{d} {\lambda }. $$

For some more interesting details and applications about some special functions and their generalizations, see [4, 6, 7, 11–14, 17–19].

Proposition 1.4

Let $a_{i}, b_{i}>0$ for $i=1,2,\ldots ,n$, then

$$ \sum_{i=1}^{n}(a_{i}+b_{i})^{r} \leq \sum_{i=1}^{n}a_{i}^{r}+ \sum_{i=1}^{n}b_{i}^{r} $$

(1.1)

for $0< r<1$.

We now define the class of s-convex functions of Breckner type.

Definition 1.5

([2])

Let $s\in (0,1]$. A function ${\Xi }:I\subset \mathbb{R}_{+}\to \mathbb{R}_{+}$ is said to be an s-convex function, if

$$ {\Xi } \bigl({\lambda }x_{1}+(1-{\lambda })x_{2} \bigr) \leq {\lambda }^{s}{\Xi }(x_{1})+(1-{ \lambda })^{s}{\Xi }(x_{2}),\quad \forall x_{1},x_{2} \in I, {\lambda } \in [0,1]. $$

2 Main results

We now discuss our main results. The first result is the fractional analogue of Hermite–Hadamard’s inequality using s-convexity property of the functions involving $\chi _{{\kappa }}$-Hilfer fractional integrals.

Theorem 2.1

Let $0\leq a_{1}< a_{2}$ and ${\Xi }:[a_{1},a_{2}]\rightarrow \mathbb{R}$ be a positive function and ${\Xi }\in L_{1}[a_{1},a_{2}]$, Also suppose that Ξ is an s-convex function on $[a_{1},a_{2}]$, $\chi (x)$ is an increasing and positive monotone function on $(a_{1},a_{2}]$, having a continuous derivative $\chi ^{\prime }(x)$ on $(a_{1},a_{2})$ and $\alpha \in (0,1)$, then, for ${\kappa }>0$, we have

$$\begin{aligned} &2^{s-1}{\Xi } \biggl(\frac{a_{1}+a_{2}}{2} \biggr) \\ &\quad \leq \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1}) ^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr] \\ &\quad \leq \biggl[\frac{3\alpha }{\alpha +ks}- \frac{\alpha }{(\alpha +sk)2^{\frac{\alpha +sk}{{\kappa }}}} \biggr] \frac{{\Xi }(a_{1})+{\Xi }(a_{2})}{2}. \end{aligned}$$

Proof

Let $x_{1},x_{2} \in [a_{1},a_{2}]$ and using the s-convexity of Ξ, we have

$$ {\Xi } \biggl(\frac{x_{1}+x_{2}}{2} \biggr)\leq \frac{{\Xi }(x_{1})}{2^{s}}+ \frac{{\Xi }(x_{2})}{2^{s}}. $$

Let $x_{1}={\lambda }a_{1}+(1-{\lambda })a_{2}$ and $x_{2}=(1-{\lambda })a_{1}+{\lambda }a_{2}$, we have

$$ 2^{s}{\Xi } \biggl(\frac{a_{1}+a_{2}}{2} \biggr)\leq {\Xi } \bigl({ \lambda }a_{1}+(1-{ \lambda })a_{2} \bigr)+{\Xi } \bigl((1-{\lambda })a_{1}+{\lambda }a_{2} \bigr). $$

Multiplying both sides by ${\lambda }^{\frac{\alpha }{{\kappa }}-1}$ and then integrating, we have

$$ \frac{2^{s}{\kappa }}{\alpha }{\Xi } \biggl(\frac{a_{1}+a_{2}}{2} \biggr) \leq \int _{0}^{1}{\lambda }^{\frac{\alpha }{{\kappa }}-1}{\Xi } \bigl({\lambda }a_{1}+(1-{ \lambda })a_{2} \bigr) \, \mathrm{d} {\lambda }+ \int _{0}^{1}{\lambda }^{ \frac{\alpha }{{\kappa }}-1}{\Xi } \bigl((1-{\lambda })a_{1}+{\lambda }a_{2} \bigr) \, \mathrm{d} {\lambda }. $$

Now

$$\begin{aligned} &\frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr] \\ &\quad = \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \frac{1}{{\kappa }\Gamma _{{\kappa }}(\alpha )} \biggl[ \int ^{\chi ^{-1}(a_{2})}_{ \chi ^{-1}(a_{1})} \bigl(a_{2}-\chi (v) \bigr)^{\frac{\alpha }{{\kappa }}}({\Xi } \circ \chi ) (v)\chi ^{\prime }(v) \, \mathrm{d}v \\ &\qquad {}+ \int ^{\chi ^{-1}(a_{2})}_{\chi ^{-1}(a_{1})} \bigl(\chi (v)-a_{1} \bigr)^{ \frac{\alpha }{{\kappa }}}({\Xi }\circ \chi ) (v)\chi ^{\prime }(v) \, \mathrm{d}v \biggr] \\ &\quad =\frac{\alpha }{2{\kappa }} \biggl[ \int _{0}^{1}{\lambda }^{ \frac{\alpha }{{\kappa }}-1}{\Xi } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \, \mathrm{d} {\lambda }+ \int _{0}^{1}{\lambda }^{\frac{\alpha }{{\kappa }}-1}{ \Xi } \bigl((1-{\lambda })a_{1}+{\lambda }a_{2} \bigr) \, \mathrm{d} {\lambda } \biggr] \\ &\quad \geq 2^{s-1}{\Xi } \biggl(\frac{a_{1}+a_{2}}{2} \biggr). \end{aligned}$$

To prove the right-hand side, we use the fact that Ξ is an s-convex function, then

$$ {\Xi } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \leq {\lambda }^{s}{\Xi }(a_{1})+(1-{ \lambda })^{s}{\Xi }(a_{2}) $$

and

$$ {\Xi } \bigl((1-{\lambda })a_{1}+{\lambda }a_{2} \bigr) \leq (1-{\lambda })^{s}{\Xi }(a_{1})+{ \lambda }^{s}{\Xi }(a_{2}). $$

Now

$$ {\Xi } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr)+{\Xi } \bigl((1-{\lambda })a_{1}+{ \lambda }a_{2} \bigr)\leq \bigl({\lambda }^{s}+(1-{\lambda })^{s} \bigr) \bigl({\Xi }(a_{1})+{ \Xi }(a_{2}) \bigr). $$

Multiplying both sides by ${\lambda }^{\frac{\alpha }{{\kappa }}-1}$ and then integrating with respect to λ on $[0,1]$, we obtain

$$\begin{aligned} &\int _{0}^{1}{\lambda }^{\frac{\alpha }{{\kappa }}-1}{\Xi } \bigl({\lambda }a_{1}+(1-{ \lambda })a_{2} \bigr) \, \mathrm{d} {\lambda }+ \int _{0}^{1}{\lambda }^{ \frac{\alpha }{{\kappa }}-1}{\Xi } \bigl((1-{\lambda })a_{1}+{\lambda }a_{2} \bigr) \, \mathrm{d} {\lambda } \\ &\quad \leq \biggl[\frac{3{\kappa }}{\alpha +ks}- \frac{{\kappa }}{(\alpha +sk) 2^{\frac{\alpha +sk}{{\kappa }}}} \biggr] \bigl[{ \Xi }(a_{1})+{\Xi }(a_{2}) \bigr]. \end{aligned}$$

This implies

$$\begin{aligned} &\frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr] \\ &\quad \leq \biggl[\frac{3\alpha }{\alpha +ks}- \frac{\alpha }{(\alpha +sk)2^{\frac{\alpha +sk}{{\kappa }}}} \biggr] \frac{{\Xi }(a_{1})+{\Xi }(a_{2})}{2}. \end{aligned}$$

The proof is complete. □

We now prove two new fractional integral identities which will be used as auxiliary results in the development of our next results.

Lemma 2.2

Let $a_{1}< a_{2}$ and ${\Xi }:[a_{1},a_{2}]\rightarrow \mathbb{R}$ be a differentiable mapping on $(a_{1},a_{2})$. Also suppose that ${\Xi }\in L[a_{1},a_{2}]$, then, for ${\kappa }>0$, we have

$$\begin{aligned} &\frac{{\Xi }(a_{1})+{\Xi }(a_{2})}{2}- \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+ {}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl( \chi ^{-1}(a_{1}) \bigr) \bigr] \\ &\quad =\frac{a_{2}-a_{1}}{2} \int ^{1}_{0} \bigl[(1-{\lambda })^{ \frac{\alpha }{{\kappa }}}-{\lambda }^{\frac{\alpha }{{\kappa }}} \bigr]{\Xi }^{ \prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \, \mathrm{d} {\lambda }. \end{aligned}$$

Proof

From [1], we have

$$\begin{aligned} &\frac{{\Xi }(a_{1})+{\Xi }(a_{2})}{2}- \frac{\Gamma _{{\kappa }} (\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+ {}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl( \chi ^{-1}(a_{1}) \bigr) \bigr] \\ &\quad =\frac{1}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \int ^{\chi ^{-1}(a_{2})}_{ \chi ^{-1}(a_{1})} \bigl[ \bigl(\chi (v)-a_{1} \bigr)^{\frac{\alpha }{{\kappa }}}- \bigl(a_{2}- \chi (v) \bigr)^{\frac{\alpha }{{\kappa }}} \bigr] \bigl({\Xi }^{\prime }\circ \chi \bigr) (v) \chi ^{\prime }(v)\,\mathrm{d}v \\ &\quad =\frac{1}{2} \int ^{\chi ^{-1}(a_{2})}_{\chi ^{-1}(a_{1})} \biggl[ \biggl( \frac{\chi (v)-a_{1}}{a_{2}-a_{1}} \biggr)^{ \frac{\alpha }{{\kappa }}}- \biggl(\frac{a_{2}-\chi (v)}{a_{2}-a_{1}} \biggr)^{\frac{\alpha }{{\kappa }}} \biggr] \bigl({\Xi }^{\prime }\circ \chi \bigr) (v) \chi ^{\prime }(v)\,\mathrm{d}v \\ &\quad =\frac{a_{2}-a_{1}}{2} \int ^{1}_{0} \bigl[(1-{\lambda })^{ \frac{\alpha }{{\kappa }}}-{\lambda }^{\frac{\alpha }{{\kappa }}} \bigr]{\Xi }^{ \prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \, \mathrm{d} {\lambda }. \end{aligned}$$

This completes the proof. □

Example 2.3

Let $a_{1}=2$, $a_{2}=3$, $\alpha =\frac{1}{2}$, ${\kappa }=2$, ${\Xi }(x)=x^{2}$, $\chi (x)=x$. Then all the assumptions in Lemma 2.2 are satisfied.

One can observe that $\frac{{\Xi }(a_{1})+{\Xi }(a_{2})}{2}=\frac{13}{2}$. We have

$$\begin{aligned} &\frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr] \\ &\quad =\frac{\Gamma _{2}(\frac{1}{2})}{8} \biggl[ \frac{1}{\Gamma _{2}(\frac{1}{2})} \int _{2}^{3}{\lambda }^{2}(3-{ \lambda })^{-\frac{3}{4}}\,\mathrm{d} {\lambda }+ \frac{1}{\Gamma _{2}(\frac{1}{2})} \int _{2}^{3}{\lambda }^{2}({ \lambda }-2)^{- \frac{3}{4}}\,\mathrm{d} {\lambda } \biggr]=\frac{577}{90}. \end{aligned}$$

It follows that

$$\begin{aligned}& \frac{{\Xi }(a_{1})+{\Xi }(a_{2})}{2}- \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }} I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr] \\& \quad = \frac{4}{45}. \end{aligned}$$

On the other hand

$$\begin{aligned} &\frac{a_{2}-a_{1}}{2} \int ^{1}_{0} \bigl[(1-{\lambda })^{ \frac{\alpha }{{\kappa }}}-{\lambda }^{\frac{\alpha }{{\kappa }}} \bigr]{\Xi }^{ \prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \, \mathrm{d} {\lambda } \\ &\quad =\frac{1}{2} \int _{0}^{1} \bigl[(1-{\lambda })^{\frac{1}{4}}-{\lambda }^{ \frac{1}{4}} \bigr](6-2{\lambda })\,\mathrm{d} { \lambda }=\frac{4}{45}. \end{aligned}$$

Example 2.4

Let $a_{1}=2$, $a_{2}=3$, $\alpha =\frac{1}{2}$, ${\kappa }=\frac{1}{2}$, ${\Xi }(x)=x^{2}$, $\chi (x)=x$. Then all the assumptions in Lemma 2.2 are satisfied.

One can observe that $\frac{{\Xi }(a_{1})+{\Xi }(a_{2})}{2}=\frac{13}{2}$. We have

$$\begin{aligned} &\frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr] \\ &\quad =\frac{\Gamma _{\frac{1}{2}}(\frac{1}{2})}{2} \biggl[ \frac{1}{\Gamma _{\frac{1}{2}}(\frac{1}{2})} \int _{2}^{3}{\lambda }^{2} \, \mathrm{d} {\lambda }+\frac{1}{\Gamma _{\frac{1}{2}}(\frac{1}{2})} \int _{2}^{3}{ \lambda }^{2} \, \mathrm{d} {\lambda } \biggr]=\frac{19}{3}. \end{aligned}$$

It follows that

$$\begin{aligned}& \frac{{\Xi }(a_{1})+{\Xi }(a_{2})}{2}- \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }} I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr] \\& \quad = \frac{1}{6}. \end{aligned}$$

On the other hand

$$\begin{aligned} &\frac{a_{2}-a_{1}}{2} \int ^{1}_{0} \bigl[(1-{\lambda })^{ \frac{\alpha }{{\kappa }}}-{\lambda }^{\frac{\alpha }{{\kappa }}} \bigr]{\Xi }^{ \prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \, \mathrm{d} {\lambda } \\ &\quad =\frac{1}{2} \int _{0}^{1}(1-2{\lambda }) (6-2{\lambda }) \, \mathrm{d} { \lambda }=\frac{1}{6}. \end{aligned}$$

Lemma 2.5

Let $a_{1}< a_{2}$ and ${\Xi }:[a_{1},a_{2}]\rightarrow \mathbb{R}$ be a differentiable mapping on $(a_{1},a_{2})$. Also suppose that ${\Xi }\in L[a_{1},a_{2}]$, then, for ${\kappa }>0$, we have

$$\begin{aligned} &\frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr]-{\Xi } \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \\ &\quad =\frac{a_{2}-a_{1}}{2} \int ^{1}_{0} \bigl[\mu +{\lambda }^{ \frac{\alpha }{{\kappa }}}-(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr]{\Xi }^{ \prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \, \mathrm{d} {\lambda }, \end{aligned}$$

where

$$ \mu = \textstyle\begin{cases} 1, & \textit{for } 0\leq {\lambda }< \frac{1}{2}, \\ -1 , & \textit{for } \frac{1}{2}\leq {\lambda }< 1. \end{cases} $$

Proof

It suffices to show that

$$\begin{aligned} &\frac{a_{2}-a_{1}}{2} \int ^{1}_{0}\mu {\Xi }^{\prime } \bigl({ \lambda }a_{1}+(1-{ \lambda })a_{2} \bigr)\, \mathrm{d} { \lambda } \\ &\quad =\frac{a_{2}-a_{1}}{2} \int ^{\frac{1}{2}}_{0}{\Xi }^{\prime } \bigl({ \lambda }a_{1}+(1-{ \lambda })a_{2} \bigr)\,\mathrm{d} { \lambda }-\frac{a_{2}-a_{1}}{2} \int ^{1}_{ \frac{1}{2}} {\Xi }^{\prime } \bigl({ \lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \,\mathrm{d} { \lambda } \\ &\quad =\frac{{\Xi }(a_{2})-{\Xi } (\frac{a_{1}+a_{2}}{2} )}{2}+ \frac{{\Xi }(a_{1})-{\Xi } (\frac{a_{1}+a_{2}}{2} )}{2} \\ &\quad =\frac{{\Xi }(a_{1})+{\Xi }(a_{2})}{2}-{\Xi } \biggl( \frac{a_{1}+a_{2}}{2} \biggr). \end{aligned}$$

By Lemma 2.2, we have

$$\begin{aligned} &\frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+ {}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl( \chi ^{-1}(a_{1}) \bigr) \bigr]-{\Xi } \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \\ &\quad = \biggl[\frac{{\Xi }(a_{1})+{\Xi }(a_{2})}{2}-{\Xi } \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \biggr] \\ &\qquad {}- \biggl\{ \frac{{\Xi }(a_{1})+{\Xi }(a_{2})}{2} \\ &\qquad {}- \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+ {}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl( \chi ^{-1}(a_{1}) \bigr) \bigr] \biggr\} \\ &\quad =\frac{a_{2}-a_{1}}{2} \int ^{1}_{0}\mu {\Xi }^{\prime } \bigl({ \lambda }a_{1}+(1-{ \lambda })a_{2} \bigr)\, \mathrm{d} { \lambda } \\ &\qquad {}-\frac{a_{2}-a_{1}}{2} \int ^{1}_{0} \bigl[(1-{ \lambda })^{\frac{\alpha }{{\kappa }}}-{\lambda }^{ \frac{\alpha }{{\kappa }}} \bigr] {\Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \, \mathrm{d} {\lambda } \\ &\quad =\frac{a_{2}-a_{1}}{2} \int ^{1}_{0} \bigl[\mu +{\lambda }^{ \frac{\alpha }{{\kappa }}}-(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr] { \Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \,\mathrm{d} {\lambda }. \end{aligned}$$

This completes the proof. □

Example 2.6

Let $a_{1}=2$, $a_{2}=3$, $\alpha =\frac{1}{2}$, ${\kappa }=2$, ${\Xi }(x)=x^{2}$, $\chi (x)=x$. Then all the assumptions in the Lemma 2.5 are satisfied.

One can observe that ${\Xi }(\frac{a_{1}+a_{2}}{2})=\frac{25}{4}$.

$$\begin{aligned} &\frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr] \\ &\quad =\frac{\Gamma _{2}(\frac{1}{2})}{8} \biggl[ \frac{1}{\Gamma _{2}(\frac{1}{2})} \int _{2}^{3}{\lambda }^{2}(3-{ \lambda })^{-\frac{3}{4}}\,\mathrm{d} {\lambda }+ \frac{1}{\Gamma _{2}(\frac{1}{2})} \int _{2}^{3}{\lambda }^{2}({ \lambda }-2)^{- \frac{3}{4}}\,\mathrm{d} {\lambda } \biggr] \\ &\quad =\frac{577}{90}. \end{aligned}$$

It follows that

$$\begin{aligned}& \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr]-{\Xi } \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \\& \quad =\frac{29}{180}. \end{aligned}$$

On the other hand

$$\begin{aligned} &\frac{a_{2}-a_{1}}{2} \int ^{1}_{0} \bigl[\mu +{\lambda }^{ \frac{\alpha }{{\kappa }}}-(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr]{\Xi }^{ \prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \, \mathrm{d} {\lambda } \\ &\quad =\frac{a_{2}-a_{1}}{2} \int ^{1}_{0}\mu {\Xi }^{\prime } \bigl({ \lambda }a_{1}+(1-{ \lambda })a_{2} \bigr)\, \mathrm{d} { \lambda } \\ &\qquad {}-\frac{a_{2}-a_{1}}{2} \int ^{1}_{0} \bigl[{ \lambda }^{\frac{\alpha }{{\kappa }}}-(1-{\lambda })^{ \frac{\alpha }{{\kappa }}} \bigr]{\Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \, \mathrm{d} {\lambda }=\frac{29}{180}. \end{aligned}$$

Example 2.7

Let $a_{1}=2$, $a_{2}=3$, $\alpha =\frac{1}{2}$, ${\kappa }=\frac{1}{2}$, ${\Xi }(x)=x^{2}$, $\chi (x)=x$. Then all the assumptions in Lemma 2.5 are satisfied.

One can observe that ${\Xi }(\frac{a_{1}+a_{2}}{2})=\frac{25}{4}$. We have

$$\begin{aligned} &\frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr] \\ &\quad =\frac{\Gamma _{\frac{1}{2}}(\frac{1}{2})}{2} \biggl[ \frac{1}{\Gamma _{\frac{1}{2}}(\frac{1}{2})} \int _{2}^{3}{\lambda }^{2} \, \mathrm{d} {\lambda }+\frac{1}{\Gamma _{2}(\frac{1}{2})} \int _{2}^{3}{ \lambda }^{2} \, \mathrm{d} {\lambda } \biggr]=\frac{19}{3}. \end{aligned}$$

It follows that

$$\begin{aligned}& \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr]-{\Xi } \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \\& \quad =\frac{1}{12}. \end{aligned}$$

On the other hand

$$\begin{aligned} &\frac{a_{2}-a_{1}}{2} \int ^{1}_{0} \bigl[\mu +{\lambda }^{ \frac{\alpha }{{\kappa }}}-(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr]{\Xi }^{ \prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \, \mathrm{d} {\lambda } \\ &\quad =\frac{a_{2}-a_{1}}{2} \int ^{1}_{0}\mu {\Xi }^{\prime } \bigl({ \lambda }a_{1}+(1-{ \lambda })a_{2} \bigr)\, \mathrm{d} { \lambda } \\ &\qquad {}-\frac{a_{2}-a_{1}}{2} \int ^{1}_{0} \bigl[{ \lambda }^{\frac{\alpha }{{\kappa }}}-(1-{\lambda })^{ \frac{\alpha }{{\kappa }}} \bigr]{\Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \, \mathrm{d} {\lambda }=\frac{1}{12}. \end{aligned}$$

Theorem 2.8

Let $a_{1}< a_{2}$ and ${\Xi }:[a_{1},a_{2}]\rightarrow \mathbb{R}$ be a differentiable mapping on $(a_{1},a_{2})$. Also suppose that $|{\Xi }^{\prime }|$ is s-convex on $[a_{1},a_{2}]$, $\chi (x)$ is an increasing and positive monotone function on $(a_{1},a_{2}]$, having a continuous derivative $\chi ^{\prime }(x)$ on $(a_{1},a_{2})$ and $\alpha \in (0,1)$, then, for ${\kappa }>0$, we have

$$\begin{aligned} & \biggl\vert \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr]-{\Xi } \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \biggr\vert \\ &\quad \leq \frac{a_{2}-a_{1}}{2(1+s)} \bigl( \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert + \bigl\vert {\Xi }^{ \prime }(a_{2}) \bigr\vert \bigr). \end{aligned}$$

Proof

Using Lemma 2.5 and the fact that $|{\Xi }^{\prime }|$ is s-convex, we have

$$\begin{aligned} & \biggl\vert \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr]-{\Xi } \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \biggr\vert \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \int ^{1}_{0} \bigl\vert \mu +{\lambda }^{ \frac{\alpha }{{\kappa }}}-(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr\vert \bigl|{ \Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr)\bigr|\,\mathrm{d} {\lambda } \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \biggl[ \int _{0}^{\frac{1}{2}} \bigl[1+{\lambda }^{ \frac{\alpha }{{\kappa }}}-(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr] \bigl[{ \lambda }^{s} \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert +(1-{\lambda })^{s}\bigl|{\Xi }^{\prime }(a_{2})\bigr| \bigr] \,\mathrm{d} {\lambda } \\ &\qquad {} + \int ^{1}_{\frac{1}{2}} \bigl[1-{\lambda }^{ \frac{\alpha }{{\kappa }}}+(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr] \bigl[{ \lambda }^{s} \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert +(1-{\lambda })^{s}\bigl|{\Xi }^{\prime }(a_{2})\bigr| \bigr] \,\mathrm{d} {\lambda } \biggr] \\ &\quad =\frac{a_{2}-a_{1}}{2} \biggl\{ \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert \biggl[ \int _{0}^{ \frac{1}{2}} \bigl[{\lambda }^{s}-{\lambda }^{s}(1-{\lambda })^{ \frac{\alpha }{{\kappa }}}+{ \lambda }^{\frac{\alpha }{{\kappa }}+s} \bigr] \,\mathrm{d} {\lambda } \\ &\qquad {} + \int _{\frac{1}{2}}^{1} \bigl[{\lambda }^{s}+{\lambda }^{s}(1-{ \lambda })^{\frac{\alpha }{{\kappa }}}-{ \lambda }^{ \frac{\alpha }{{\kappa }}+s} \bigr]\,\mathrm{d} {\lambda } \biggr] \\ &\qquad {} + \bigl\vert {\Xi }^{\prime }(a_{2}) \bigr\vert \biggl[ \int _{0}^{\frac{1}{2}} \bigl[(1-{ \lambda })^{s}+{\lambda }^{\frac{\alpha }{{\kappa }}}(1-{\lambda })^{s}-(1-{ \lambda })^{\frac{\alpha }{{\kappa }}+s} \bigr]\,\mathrm{d} {\lambda } \\ &\qquad {} + \int _{\frac{1}{2}}^{1} \bigl[(1-{\lambda })^{s}-{ \lambda }^{\frac{\alpha }{{\kappa }}}(1-{\lambda })^{s}+(1-{ \lambda })^{ \frac{\alpha }{{\kappa }}+s} \bigr]\,\mathrm{d} {\lambda } \biggr] \biggr\} \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \biggl\{ \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert \biggl[ \int _{0}^{\frac{1}{2}}{\lambda }^{s} \, \mathrm{d} {\lambda } + \int _{ \frac{1}{2}}^{1}{\lambda }^{s} \, \mathrm{d} {\lambda } \biggr] \\ &\qquad {}+ \bigl\vert {\Xi }^{ \prime }(a_{2}) \bigr\vert \biggl[ \int _{0}^{\frac{1}{2}}(1-{\lambda })^{s} \, \mathrm{d} {\lambda }+ \int _{\frac{1}{2}}^{1}(1-{\lambda })^{s} \, \mathrm{d} {\lambda } \biggr] \biggr\} \\ &\quad =\frac{a_{2}-a_{1}}{2} \biggl\{ \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert \int _{0}^{1}{ \lambda }^{s} \, \mathrm{d} {\lambda }+ \bigl\vert {\Xi }^{\prime }(a_{2}) \bigr\vert \int _{0}^{1}(1-{ \lambda })^{s} \,\mathrm{d} {\lambda } \biggr\} \\ &\quad =\frac{a_{2}-a_{1}}{2(1+s)} \bigl( \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert + \bigl\vert {\Xi }^{ \prime }(a_{2}) \bigr\vert \bigr). \end{aligned}$$

This completes the proof. □

Theorem 2.9

Let $a_{1}< a_{2}$ and ${\Xi }:[a_{1},a_{2}]\rightarrow \mathbb{R}$ be a differentiable mapping on $(a_{1},a_{2})$. Also suppose that $|{\Xi }^{\prime }|^{q}$ is s-convex on $[a_{1},a_{2}]$, $\chi (x)$ is an increasing and positive monotone function on $(a_{1},a_{2}]$, having a continuous derivative $\chi ^{\prime }(x)$ on $(a_{1},a_{2})$ and $\alpha \in (0,1)$, then, for ${\kappa }>0$, we have

$$\begin{aligned} & \biggl\vert \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr]-{\Xi } \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \biggr\vert \\ &\quad \leq (a_{2}-a_{1}) \biggl( \frac{{\kappa }}{(\alpha p+{\kappa })2^{\frac{\alpha p+{\kappa }}{{\kappa }}}} \biggr)^{\frac{1}{p}} \biggl(\frac{1}{(s+1)2^{s+1}} \biggr)^{ \frac{1}{q}} \bigl\{ \bigl( \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert ^{q}+ \bigl(2^{s+1}-1 \bigr) \bigl\vert {\Xi }^{\prime }(a_{2}) \bigr\vert ^{q} \bigr)^{\frac{1}{q}} \\ &\qquad {} + \bigl(2^{s+1}-1 \bigr) \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert ^{q}+ \bigl\vert {\Xi }^{ \prime }(a_{2}) \bigr\vert ^{q} )^{\frac{1}{q}} \bigr\} , \end{aligned}$$

(2.1)

where $p^{-1}+q^{-1}=1$

Proof

Using Lemma 2.5, Hölder’s integral inequality and the fact that $|{\Xi }^{\prime }|^{q}$ is s-convex, we have

$$\begin{aligned} & \biggl\vert \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr]-{\Xi } \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \biggr\vert \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \int ^{1}_{0} \bigl\vert \mu +{\lambda }^{ \frac{\alpha }{{\kappa }}}-(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr\vert \bigl|{ \Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr)\bigr|\,\mathrm{d} {\lambda } \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \biggl[ \int _{0}^{\frac{1}{2}} \bigl[1+{\lambda }^{ \frac{\alpha }{{\kappa }}}-(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr] \bigl\vert { \Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \bigr\vert \,\mathrm{d} {\lambda } \\ &\qquad {} + \int ^{1}_{\frac{1}{2}} \bigl[1-{\lambda }^{ \frac{\alpha }{{\kappa }}}+(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr] \bigl\vert { \Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \bigr\vert \,\mathrm{d} {\lambda } \biggr] \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \biggl[ \biggl( \int _{0}^{\frac{1}{2}} \bigl[1+{ \lambda }^{\frac{\alpha }{{\kappa }}}-(1-{\lambda })^{ \frac{\alpha }{{\kappa }}} \bigr]^{p} \, \mathrm{d} {\lambda } \biggr)^{\frac{1}{p}} \biggl( \int _{0}^{\frac{1}{2}} \bigl\vert {\Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{ \lambda })a_{2} \bigr) \bigr\vert ^{q}\,\mathrm{d} {\lambda } \biggr)^{\frac{1}{q}} \\ &\qquad {} + \biggl( \int ^{1}_{\frac{1}{2}} \bigl[1-{\lambda }^{ \frac{\alpha }{{\kappa }}}+(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr]^{p} \, \mathrm{d} {\lambda } \biggr)^{\frac{1}{p}} \biggl( \int ^{1}_{ \frac{1}{2}} \bigl\vert {\Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \bigr\vert ^{q} \,\mathrm{d} {\lambda } \biggr)^{\frac{1}{q}} \biggr] \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \biggl( \int _{0}^{\frac{1}{2}} \bigl[1+{\lambda }^{ \frac{\alpha }{{\kappa }}}-(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr]^{p} \, \mathrm{d} {\lambda } \biggr)^{\frac{1}{p}} \\ &\qquad {}\times\biggl\{ \biggl( \int _{0}^{ \frac{1}{2}} \bigl({\lambda }^{s} \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert ^{q}+(1-{\lambda })^{s} \bigl\vert { \Xi }^{\prime }(a_{2}) \bigr\vert ^{q} \bigr) \, \mathrm{d} {\lambda } \biggr)^{\frac{1}{q}} \\ &\qquad {} + \biggl( \int ^{1}_{\frac{1}{2}} \bigl({\lambda }^{s} \bigl\vert {\Xi }^{ \prime }(a_{1}) \bigr\vert ^{q}+(1-{\lambda })^{s} \bigl\vert {\Xi }^{\prime }(a_{2}) \bigr\vert ^{q} \bigr) \, \mathrm{d} {\lambda } \biggr)^{\frac{1}{q}} \biggr\} \\ &\quad =\frac{a_{2}-a_{1}}{2} \biggl( \int _{0}^{\frac{1}{2}} \bigl[1+{\lambda }^{ \frac{\alpha }{{\kappa }}}-(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr]^{p} \, \mathrm{d} {\lambda } \biggr)^{\frac{1}{p}} \\ &\qquad {}\times\biggl\{ \biggl( \frac{1}{(s+1)2^{s+1}} \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert ^{q}+\frac{1}{s+1} \biggl(1- \frac{1}{2^{s+1}} \biggr) \bigl\vert {\Xi }^{\prime }(a_{2}) \bigr\vert ^{q} \biggr)^{ \frac{1}{q}} \\ &\qquad {} + \biggl(\frac{1}{s+1} \biggl(1-\frac{1}{2^{s+1}} \biggr) \bigl\vert {\Xi }^{ \prime }(a_{1}) \bigr\vert ^{q}+\frac{1}{(s+1)2^{s+1}} \bigl\vert {\Xi }^{\prime }(a_{2}) \bigr\vert ^{q} \biggr)^{\frac{1}{q}} \biggr\} \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \biggl(2^{p} \int _{0}^{\frac{1}{2}}{ \lambda }^{\frac{\alpha p}{{\kappa }}} \, \mathrm{d} {\lambda } \biggr)^{ \frac{1}{p}} \biggl(\frac{1}{(s+1)2^{s+1}} \biggr)^{\frac{1}{q}} \bigl\{ \bigl( \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert ^{q}+ \bigl(2^{s+1}-1 \bigr) \bigl\vert { \Xi }^{ \prime }(a_{2}) \bigr\vert ^{q} \bigr)^{\frac{1}{q}} \\ &\qquad {} + \bigl(2^{s+1}-1 \bigr) \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert ^{q}+ \bigl\vert {\Xi }^{ \prime }(a_{2}) \bigr\vert ^{q} )^{\frac{1}{q}} \bigr\} \\ &\quad =(a_{2}-a_{1}) \biggl( \frac{{\kappa }}{(\alpha p+{\kappa })2^{\frac{\alpha p+{\kappa }}{{\kappa }}}} \biggr)^{\frac{1}{p}} \biggl(\frac{1}{(s+1)2^{s+1}} \biggr)^{ \frac{1}{q}} \bigl\{ \bigl( \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert ^{q}+ \bigl(2^{s+1}-1 \bigr) \bigl\vert {\Xi }^{\prime }(a_{2}) \bigr\vert ^{q} \bigr)^{\frac{1}{q}} \\ &\qquad {} + \bigl(2^{s+1}-1 \bigr) \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert ^{q}+ \bigl\vert {\Xi }^{ \prime }(a_{2}) \bigr\vert ^{q} )^{\frac{1}{q}} \bigr\} . \end{aligned}$$

This completes the proof. □

Corollary 2.10

Let $a_{1}< a_{2}$ and ${\Xi }:[a_{1},a_{2}]\rightarrow \mathbb{R}$ be a differentiable mapping on $(a_{1},a_{2})$. Also suppose that $|{\Xi }^{\prime }|^{q}, q>1$, is s-convex on $[a_{1},a_{2}]$, $\chi (x)$ is an increasing and positive monotone function on $(a_{1},a_{2}]$, having a continuous derivative $\chi ^{\prime }(x)$ on $(a_{1},a_{2})$ and $\alpha \in (0,1)$, then, for ${\kappa }>0$, we have

$$\begin{aligned} & \biggl\vert \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr]-{\Xi } \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \biggr\vert \\ &\quad \leq (a_{2}-a_{1}) \biggl( \frac{{\kappa }}{(\alpha p+{\kappa })2^{\frac{\alpha p+{\kappa }}{{\kappa }}}} \biggr)^{\frac{1}{p}} \biggl(\frac{1}{(s+1)2^{s+1}} \biggr)^{ \frac{1}{q}} \\ &\qquad {} \times \bigl(1+ \bigl(2^{s+1}-1 \bigr)^{\frac{1}{q}} \bigr) \bigl( \bigl\vert { \Xi }^{\prime }(a_{1}) \bigr\vert ^{q}+ \bigl\vert {\Xi }^{\prime }(a_{2}) \bigr\vert ^{q} \bigr), \end{aligned}$$

where $p^{-1}+q^{-1}=1$.

Proof

Use (2.1) and let $a_{1}=|{\Xi }^{\prime }(a_{1})|^{q}$, $b_{1}= (2^{s+1}-1 )|{ \Xi }^{\prime }(a_{2})|^{q}$, $a_{2}= (2^{s+1}-1 )|{\Xi }^{ \prime }(a_{1})|^{q}$, $b_{2}=|{\Xi }^{\prime }(a_{2})|^{q}$. Here $0<\frac{1}{q}<1$ for $q>1$. Then using (1.1), we obtain the required result. □

Theorem 2.11

Let $a_{1}< a_{2}$, $q>1$ and ${\Xi }:[a_{1},a_{2}]\rightarrow \mathbb{R}$ be a differentiable mapping on $(a_{1},a_{2})$. Also suppose that $|{\Xi }^{\prime }|^{q}$ is s-convex on $[a_{1},a_{2}]$, $\chi (x)$ is an increasing and positive monotone function on $(a_{1},a_{2}]$, having a continuous derivative $\chi ^{\prime }(x)$ on $(a_{1},a_{2})$ and $\alpha \in (0,1)$, then, for ${\kappa }>0$, we have

$$\begin{aligned} & \biggl\vert \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr]-{\Xi } \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \biggr\vert \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \biggl(\frac{{\kappa }}{\alpha +{\kappa }} \biggr)^{1-\frac{1}{q}} \biggl(\frac{\alpha -{\kappa }}{2{\kappa }}+ \frac{1}{2^{\frac{\alpha }{{\kappa }}}} \biggr)^{1-\frac{1}{q}} \biggl( \frac{1}{(s+1)2^{s+1}} \biggr)^{\frac{1}{q}} \\ &\qquad {} \times \bigl\{ \bigl( \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert ^{q}+ \bigl(2^{s+1}-1 \bigr) \bigl\vert { \Xi }^{\prime }(a_{2}) \bigr\vert ^{q} \bigr)^{\frac{1}{q}}+ \bigl(2^{s+1}-1 \bigr) \bigl\vert { \Xi }^{\prime }(a_{1}) \bigr\vert ^{q}+ \bigl\vert {\Xi }^{\prime }(a_{2}) \bigr\vert ^{q} )^{ \frac{1}{q}} \bigr\} . \end{aligned}$$

(2.2)

Proof

Using Lemma 2.5, the power mean integral inequality and the fact that $|{\Xi }^{\prime }|^{q}$ is s-convex, we have

$$\begin{aligned} & \biggl\vert \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr]-{\Xi } \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \biggr\vert \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \int ^{1}_{0} \bigl\vert \mu +{\lambda }^{ \frac{\alpha }{{\kappa }}}-(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr\vert \bigl|{ \Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr)\bigr|\,\mathrm{d} {\lambda } \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \biggl[ \int _{0}^{\frac{1}{2}} \bigl[1+{\lambda }^{ \frac{\alpha }{{\kappa }}}-(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr] \bigl\vert { \Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \bigr\vert \,\mathrm{d} {\lambda } \\ &\qquad {} + \int ^{1}_{\frac{1}{2}} \bigl[1-{\lambda }^{ \frac{\alpha }{{\kappa }}}+(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr] \bigl\vert { \Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \bigr\vert \,\mathrm{d} {\lambda } \biggr] \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \biggl[ \biggl( \int _{0}^{\frac{1}{2}} \bigl[1+{ \lambda }^{\frac{\alpha }{{\kappa }}}-(1-{\lambda })^{ \frac{\alpha }{{\kappa }}} \bigr]\,\mathrm{d} {\lambda } \biggr)^{1-\frac{1}{q}} \\ &\qquad {} \times \biggl( \int _{0}^{\frac{1}{2}} \bigl[1+{\lambda }^{ \frac{\alpha }{{\kappa }}}-(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr] \bigl\vert { \Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \bigr\vert ^{q}\,\mathrm{d} { \lambda } \biggr)^{\frac{1}{q}} \\ &\qquad {} + \biggl( \int ^{1}_{\frac{1}{2}} \bigl[1-{\lambda }^{ \frac{\alpha }{{\kappa }}}+(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr] \,\mathrm{d} {\lambda } \biggr)^{1-\frac{1}{q}} \\ &\qquad {} \times \biggl( \int ^{1}_{\frac{1}{2}} \bigl[1-{\lambda }^{ \frac{\alpha }{{\kappa }}}+(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr] \bigl\vert { \Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \bigr\vert ^{q}\,\mathrm{d} { \lambda } \biggr)^{\frac{1}{q}} \biggr] \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \biggl( \int _{0}^{\frac{1}{2}} \bigl[1+{\lambda }^{ \frac{\alpha }{{\kappa }}}-(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr] \,\mathrm{d} {\lambda } \biggr)^{1-\frac{1}{q}} \\ &\qquad {} \times \biggl\{ \biggl( \int _{0}^{\frac{1}{2}} \bigl[1-{\lambda }^{ \frac{\alpha }{{\kappa }}}+(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr] \bigl({ \lambda }^{s} \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert ^{q}+(1-{\lambda })^{s} \bigl\vert {\Xi }^{ \prime }(a_{2}) \bigr\vert ^{q} \bigr) \, \mathrm{d} {\lambda } \biggr)^{\frac{1}{q}} \\ &\qquad {} + \biggl( \int ^{1}_{\frac{1}{2}} \bigl[1-{\lambda }^{ \frac{\alpha }{{\kappa }}}+(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr] \bigl({ \lambda }^{s} \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert ^{q}+(1-{\lambda })^{s} \bigl\vert {\Xi }^{ \prime }(a_{2}) \bigr\vert ^{q} \bigr) \, \mathrm{d} {\lambda } \biggr)^{\frac{1}{q}} \biggr\} \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \biggl(\frac{{\kappa }}{\alpha +{\kappa }} \biggr)^{1-\frac{1}{q}} \biggl(\frac{\alpha -{\kappa }}{2{\kappa }}+ \frac{1}{2^{\frac{\alpha }{{\kappa }}}} \biggr)^{1-\frac{1}{q}} \\ &\qquad {} \times \biggl\{ \biggl( \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert ^{q} \int _{0}^{ \frac{1}{2}} \bigl[{\lambda }^{s}-{\lambda }^{s}(1-{\lambda })^{ \frac{\alpha }{{\kappa }}}+{ \lambda }^{\frac{\alpha }{{\kappa }}+s} \bigr] \,\mathrm{d} {\lambda } \\ &\qquad {} + \bigl\vert {\Xi }^{\prime }(a_{2}) \bigr\vert ^{q} \int _{0}^{\frac{1}{2}} \bigl[(1-{ \lambda })^{s}+{\lambda }^{\frac{\alpha }{{\kappa }}}(1-{\lambda })^{s}-(1-{ \lambda })^{\frac{\alpha }{{\kappa }}+s} \bigr]\,\mathrm{d} {\lambda } \biggr)^{ \frac{1}{q}} \\ &\qquad {} + \biggl( \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert ^{q} \int _{\frac{1}{2}}^{1} \bigl[{ \lambda }^{s}+{\lambda }^{s}(1-{\lambda })^{\frac{\alpha }{{\kappa }}}-{ \lambda }^{\frac{\alpha }{{\kappa }}+s} \bigr]\,\mathrm{d} {\lambda } \\ &\qquad {} + \bigl\vert {\Xi }^{\prime }(a_{2}) \bigr\vert ^{q} \int _{\frac{1}{2}}^{1} \bigl[(1-{ \lambda })^{s}-{\lambda }^{\frac{\alpha }{{\kappa }}}(1-{\lambda })^{s}+(1-{ \lambda })^{\frac{\alpha }{{\kappa }}+s} \bigr]\,\mathrm{d} {\lambda } \biggr)^{ \frac{1}{q}} \biggr\} \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \biggl(\frac{{\kappa }}{\alpha +{\kappa }} \biggr)^{1-\frac{1}{q}} \biggl(\frac{\alpha -{\kappa }}{2{\kappa }}+ \frac{1}{2^{\frac{\alpha }{{\kappa }}}} \biggr)^{1-\frac{1}{q}} \\ &\qquad {} \times \biggl\{ \biggl( \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert ^{q} \int _{0}^{ \frac{1}{2}}{\lambda }^{s} \, \mathrm{d} {\lambda }+ \bigl\vert {\Xi }^{\prime }(a_{2}) \bigr\vert ^{q} \int _{0}^{\frac{1}{2}}(1-{\lambda })^{s} \, \mathrm{d} {\lambda } \biggr)^{ \frac{1}{q}} \\ &\qquad {} + \biggl( \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert ^{q} \int _{\frac{1}{2}}^{1}{ \lambda }^{s} \, \mathrm{d} {\lambda }+ \bigl\vert {\Xi }^{\prime }(a_{2}) \bigr\vert ^{q} \int _{ \frac{1}{2}}^{1}(1-{\lambda })^{s} \, \mathrm{d} {\lambda } \biggr)^{ \frac{1}{q}} \biggr\} \\ &\quad =\frac{a_{2}-a_{1}}{2} \biggl(\frac{{\kappa }}{\alpha +{\kappa }} \biggr)^{1-\frac{1}{q}} \biggl(\frac{\alpha -{\kappa }}{2{\kappa }}+ \frac{1}{2^{\frac{\alpha }{{\kappa }}}} \biggr)^{1-\frac{1}{q}} \\ &\qquad {} \times \biggl\{ \biggl(\frac{1}{(s+1)2^{s+1}} \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert ^{q}+ \frac{1}{s+1} \biggl(1-\frac{1}{2^{s+1}} \biggr) \bigl\vert {\Xi }^{\prime }(a_{2}) \bigr\vert ^{q} \biggr)^{\frac{1}{q}} \\ &\qquad {} + \biggl(\frac{1}{s+1} \biggl(1-\frac{1}{2^{s+1}} \biggr) \bigl\vert {\Xi }^{ \prime }(a_{1}) \bigr\vert ^{q}+\frac{1}{(s+1)2^{s+1}} \bigl\vert {\Xi }^{\prime }(a_{2}) \bigr\vert ^{q} \biggr)^{\frac{1}{q}} \biggr\} \\ &\quad =\frac{a_{2}-a_{1}}{2} \biggl(\frac{{\kappa }}{\alpha +{\kappa }} \biggr)^{1-\frac{1}{q}} \biggl(\frac{\alpha -{\kappa }}{2{\kappa }}+ \frac{1}{2^{\frac{\alpha }{{\kappa }}}} \biggr)^{1-\frac{1}{q}} \biggl( \frac{1}{(s+1)2^{s+1}} \biggr)^{\frac{1}{q}} \\ &\qquad {} \times \bigl\{ \bigl( \bigl\vert {\Xi }^{\prime }(a_{1}) \bigr\vert ^{q}+ \bigl(2^{s+1}-1 \bigr) \bigl\vert { \Xi }^{\prime }(a_{2}) \bigr\vert ^{q} \bigr)^{\frac{1}{q}}+ \bigl(2^{s+1}-1 \bigr) \bigl\vert { \Xi }^{\prime }(a_{1}) \bigr\vert ^{q}+ \bigl\vert {\Xi }^{\prime }(a_{2}) \bigr\vert ^{q} )^{ \frac{1}{q}} \bigr\} . \end{aligned}$$

This completes the proof. □

Corollary 2.12

Let $a_{1}< a_{2}$ and ${\Xi }:[a_{1},a_{2}]\rightarrow \mathbb{R}$ be a differentiable mapping on $(a_{1},a_{2})$. Also suppose that $|{\Xi }^{\prime }|^{q}, q>1$ is s-convex on $[a_{1},a_{2}]$, $\chi (x)$ is an increasing and positive monotone function on $(a_{1},a_{2}]$, having a continuous derivative $\chi ^{\prime }(x)$ on $(a_{1},a_{2})$ and $\alpha \in (0,1)$, then, for ${\kappa }>0$, we have

$$\begin{aligned} & \biggl\vert \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr]-{\Xi } \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \biggr\vert \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \biggl(\frac{{\kappa }}{\alpha +{\kappa }} \biggr)^{1-\frac{1}{q}} \biggl(\frac{\alpha -{\kappa }}{2{\kappa }}+ \frac{1}{2^{\frac{\alpha }{{\kappa }}}} \biggr)^{1-\frac{1}{q}} \biggl( \frac{1}{(s+1)2^{s+1}} \biggr)^{\frac{1}{q}} \\ &\qquad {} \times \bigl(1+ \bigl(2^{s+1}-1 \bigr)^{\frac{1}{q}} \bigr) \bigl( \bigl\vert { \Xi }^{\prime }(a_{1}) \bigr\vert ^{q}+ \bigl\vert {\Xi }^{\prime }(a_{2}) \bigr\vert ^{q} \bigr). \end{aligned}$$

Proof

Using the same technique as in the proof of Corollary 2.10, the proof is complete. □

Theorem 2.13

Let $a_{1}< a_{2}$ and ${\Xi }:[a_{1},a_{2}]\rightarrow \mathbb{R}$ be a differentiable mapping on $(a_{1},a_{2})$. Also suppose that $|{\Xi }^{\prime }|^{q}$ is s-concave on $[a_{1},a_{2}]$, $\chi (x)$ is an increasing and positive monotone function on $(a_{1},a_{2}]$, having a continuous derivative $\chi ^{\prime }(x)$ on $(a_{1},a_{2})$ and $\alpha \in (0,1)$, then, for ${\kappa }>0$, we have

$$\begin{aligned} & \biggl\vert \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr]-{\Xi } \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \biggr\vert \\ &\quad \leq (a_{2}-a_{1}) \biggl( \frac{{\kappa }}{(\alpha p+{\kappa })2^{\alpha p+{\kappa }}} \biggr)^{ \frac{1}{p}} \biggl(\frac{1}{2} \biggr)^{\frac{1}{q}} \biggl( \biggl\vert { \Xi }^{\prime } \biggl(\frac{a_{1}+3a_{2}}{4} \biggr) \biggr\vert + \biggl\vert { \Xi }^{\prime } \biggl( \frac{3a_{1}+a_{2}}{4} \biggr) \biggr\vert \biggr), \end{aligned}$$

where $p^{-1}+q^{-1}=1$.

Proof

Using Lemma 2.5, Hölder’s integral inequality and the fact that $|{\Xi }^{\prime }|^{q}$ is s-concave, we have

$$\begin{aligned} & \biggl\vert \frac{\Gamma _{{\kappa }}(\alpha +{\kappa })}{2(a_{2}-a_{1})^{\frac{\alpha }{{\kappa }}}} \bigl[{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{1})^{+}}({ \Xi } \circ \chi ) \bigl(\chi ^{-1}(a_{2}) \bigr)+{}_{{\kappa }}I^{\alpha :\chi }_{\chi ^{-1}(a_{2})^{-}}({ \Xi }\circ \chi ) \bigl(\chi ^{-1}(a_{1}) \bigr) \bigr]-{\Xi } \biggl( \frac{a_{1}+a_{2}}{2} \biggr) \biggr\vert \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \int ^{1}_{0} \bigl\vert \mu +{\lambda }^{ \frac{\alpha }{{\kappa }}}-(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr\vert \bigl|{ \Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr)\bigr|\,\mathrm{d} {\lambda } \\ &\quad \leq \frac{a_{2}-a_{1}}{2} \biggl[ \int _{0}^{\frac{1}{2}} \bigl[1+{\lambda }^{ \frac{\alpha }{{\kappa }}}-(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr] \bigl\vert { \Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \bigr\vert \,\mathrm{d} {\lambda } \\ &\qquad {} + \int ^{1}_{\frac{1}{2}} \bigl[1-{\lambda }^{ \frac{\alpha }{{\kappa }}}+(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr] \bigl\vert { \Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \bigr\vert \,\mathrm{d} {\lambda } \biggr] \\ &\quad \leq \frac{a_{2}-a_{1}}{2}\biggl[ \biggl( \int _{0}^{\frac{1}{2}} \bigl[1+{ \lambda }^{\frac{\alpha }{{\kappa }}}-(1-{\lambda })^{ \frac{\alpha }{{\kappa }}} \bigr]^{p} \, \mathrm{d} {\lambda } \biggr)^{\frac{1}{p}} \biggl( \int _{0}^{\frac{1}{2}} \bigl\vert {\Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{ \lambda })a_{2} \bigr) \bigr\vert ^{q}\,\mathrm{d} {\lambda } \biggr)^{\frac{1}{q}} \\ &\qquad {} + \biggl( \int ^{1}_{\frac{1}{2}} \bigl[1-{\lambda }^{ \frac{\alpha }{{\kappa }}}+(1-{\lambda })^{\frac{\alpha }{{\kappa }}} \bigr]^{p} \, \mathrm{d} {\lambda } \biggr)^{\frac{1}{p}} \biggl( \int ^{1}_{ \frac{1}{2}} \bigl\vert {\Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{\lambda })a_{2} \bigr) \bigr\vert ^{q} \,\mathrm{d} {\lambda } \biggr)^{\frac{1}{q}}\biggr] \\ &\quad =(a_{2}-a_{1}) \biggl( \frac{{\kappa }}{(\alpha p+{\kappa })2^{\alpha p+{\kappa }}} \biggr)^{ \frac{1}{p}} \biggl( \int _{0}^{\frac{1}{2}} \bigl\vert {\Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{ \lambda })a_{2} \bigr) \bigr\vert ^{q}\,\mathrm{d} {\lambda } \biggr)^{\frac{1}{q}} \\ &\qquad {} + \biggl( \int ^{1}_{\frac{1}{2}} \bigl\vert {\Xi }^{\prime } \bigl({\lambda }a_{1}+(1-{ \lambda })a_{2} \bigr) \bigr\vert ^{q}\,\mathrm{d} {\lambda } \biggr)^{\frac{1}{q}} ]. \end{aligned}$$

Since $|{\Xi }^{\prime }|^{q}$ is a concave function on $[a_{1},a_{2}]$ and using the Jensen integral inequality, we have

$$\begin{aligned} \int _{0}^{\frac{1}{2}}\bigl|{\Xi }^{\prime } \bigl({ \lambda }a_{1}+(1-{\lambda })a_{2} \bigr)\bigr|^{q} \,\mathrm{d} {\lambda }&\leq \biggl( \int _{0}^{\frac{1}{2}}{\lambda }^{*} \, \mathrm{d} {\lambda } \biggr) \biggl\vert {\Xi }^{\prime } \biggl( \frac{ (\int _{0}^{\frac{1}{2}}({\lambda }a_{1}+(1-{\lambda })a_{2})\,\mathrm{d}{\lambda } )}{ (\int _{0}^{\frac{1}{2}}{\lambda }^{*}\,\mathrm{d}{\lambda } )} \biggr) \biggr\vert \\ &\leq \frac{1}{2} \biggl\vert {\Xi }^{\prime } \biggl( \frac{a_{1}+3a_{2}}{4} \biggr) \biggr\vert . \end{aligned}$$

Similarly

$$\begin{aligned} \int _{\frac{1}{2}}^{1}\bigl|{\Xi }^{\prime } \bigl({ \lambda }a_{1}+(1-{\lambda })a_{2} \bigr)\bigr|^{q} \,\mathrm{d} {\lambda }&\leq \biggl( \int _{\frac{1}{2}}^{1}{\lambda }^{*} \, \mathrm{d} {\lambda } \biggr) \biggl\vert {\Xi }^{\prime } \biggl( \frac{ (\int _{\frac{1}{2}}^{1}({\lambda }a_{1}+(1-{\lambda })a_{2})\,\mathrm{d}{\lambda } )}{ (\int _{0}^{\frac{1}{2}}{\lambda }^{*}\,\mathrm{d}{\lambda } )} \biggr) \biggr\vert \\ &\leq \frac{1}{2} \biggl\vert {\Xi }^{\prime } \biggl( \frac{3a_{1}+a_{2}}{4} \biggr) \biggr\vert . \end{aligned}$$

This completes the proof. □