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Integral inequalities for s-convex functions via generalized conformable fractional integral operators

Abstract

We introduce new operators, the so-called left and right generalized conformable fractional integral operators. By using these operators we establish new Hermite–Hadamard inequalities for s-convex functions and products of two s-convex functions in the second sense. Also, we obtain two interesting identities for a differentiable function involving a generalized conformable fractional integral operator. By applying these identities we give Hermite–Hadamard and midpoint-type integral inequalities for s-convex functions. Different special cases have been identified and some known results are recovered from our general results. These results may motivate further research in different areas of pure and applied sciences.

Introduction

The theory of inequalities is known to play an important role in almost all areas of pure and applied sciences. Richard Bellman stated succinctly, at the Second International Conference on Mathematical Inequalities, Oberwolfach, Germany, July 30–August 5, 1978, that “there are three reasons for the study of inequalities: practical, theoretical, and aesthetic”. In the last few decades the theory of inequalities has attracted the attention of great number of researchers.

We use I to denote an interval in the real line \(\mathbb{R}=(-\infty, +\infty )\), and \(L[\lambda _{1},\lambda _{2}]\), where \(\lambda _{1}<\lambda _{2}\), the set of integrable functions on \([\lambda _{1},\lambda _{2}]\).

Definition 1

A function \(\mathfrak{h}:I\subset \mathbb{R}\longrightarrow \mathbb{R}\) is said to be convex if

$$ \mathfrak{h}\bigl(x \lambda _{1}+(1-x)\lambda _{2}\bigr)\leq x \mathfrak{h}( \lambda _{1})+(1-x) \mathfrak{h}(\lambda _{2}) $$
(1)

for all \(\lambda _{1}, \lambda _{2}\in I\) and \(x\in [0,1]\).

The interesting mean type inequality, known as the Hermite–Hadamard inequality for convex functions, is given by the following theorem.

Theorem 1

Let\(\mathfrak{h}:I\subset \mathbb{R}\longrightarrow \mathbb{R}\)be a convex function, and let\(\lambda _{1},\lambda _{2}\in I\)with\(\lambda _{1}<\lambda _{2}\). Then

$$ \mathfrak{h} \biggl(\frac{\lambda _{1}+\lambda _{2}}{2} \biggr)\leq \frac{1}{\lambda _{2}-\lambda _{1}} \int _{\lambda _{1}}^{\lambda _{2}} \mathfrak{h}(x)\,dx\leq \frac{\mathfrak{h}(\lambda _{1})+\mathfrak{h}(\lambda _{2})}{2}. $$
(2)

Inequality (2) is also acknowledged as the trapezium inequality.

The trapezium inequality has an extraordinary interest due to its wide applications in the field of mathematical analysis. Authors of recent decades have studied (2) in the premises of newly invented definitions due to the motivation of convex functions. The interested readers can see the references [27, 9, 1215, 18, 19, 21, 22, 2428, 3032, 3437].

Definition 2

([13])

A function \(\mathfrak{h}:[0,+\infty )\longrightarrow \mathbb{R}\) is said to be s-convex \((s\in (0,1])\) in the second sense if

$$ \mathfrak{h}\bigl(a \lambda _{1}+(1-a)\lambda _{2}\bigr)\leq a^{s}\mathfrak{h}( \lambda _{1})+(1-a)^{s}\mathfrak{h}(\lambda _{2}) $$
(3)

for all \(\lambda _{1}, \lambda _{2}\geq 0\) and \(a\in [0,1]\).

The s-convex functions in the second sense are presented in [13]. Also, researchers started to study conformable fractional integrals; see [1, 8, 12, 18, 19, 30, 32]. Khalil et al. [17] defined the fractional integral only of order \(0<\alpha \leq 1\), whereas Abdeljawad [1] generalized the definition of left and right conformable fractional integrals to any order \(\alpha > 0\). In 2017, Khan et al. implemented this definition by providing a class of Hermite-type inequalities.

Definition 3

Let \(f:[\lambda _{1}, +\infty )\longrightarrow \mathbb{R}\), \(\xi \in (n,n+1]\), set \(\zeta =\xi -n\). Then the left conformable fractional integral starting at \(\lambda _{1}\) is defined by

$$ \bigl(I_{\xi }^{\lambda _{1}}f \bigr) (x)=\frac{1}{n!} \int _{ \lambda _{1}}^{x}(x-\theta )^{n}(\theta - \lambda _{1})^{\zeta -1}f( \theta )\,d\theta,\quad x>\lambda _{1}. $$

Analogously, if \(f:(-\infty,\lambda _{2}]\longrightarrow \mathbb{R}\), \(\xi \in (n,n+1]\), and \(\zeta =\xi -n\), then the right conformable fractional integral at \(\lambda _{2}\) is defined by

$$ \bigl(^{\lambda _{2}}I_{\xi }f \bigr) (x)=\frac{1}{n!} \int _{x}^{ \lambda _{2}}(\theta -x)^{n}(\lambda _{2}-\theta )^{\zeta -1}f( \theta )\,d\theta,\quad x< \lambda _{2}. $$

Note that if \(\xi =n+1\), then \(\zeta =\xi -n=n+1-n=1\), where \(n=0,1,2,\ldots \) .

Set et al. [30] obtained a new generalized class of Hermite–Hadamard-type inequalities for s-convex functions by applying conformable fractional integrals.

Theorem 2

([30])

Let\(f:[\lambda _{1},\lambda _{2}]\longrightarrow \mathbb{R}\)be a function with\(0\leq \lambda _{1} < \lambda _{2}, s\in (0,1]\), and\(f\in L[\lambda _{1},\lambda _{2}]\). Iffis ans-convex function on\([\lambda _{1}, \lambda _{2}]\), then we have the following inequalities for conformable fractional integrals:

$$\begin{aligned} &\frac{\varGamma (\xi -n)}{\varGamma (\xi +1)}f \biggl( \frac{\lambda _{1}+\lambda _{2}}{2} \biggr)\\ &\quad \leq \frac{1}{2^{s}(\lambda _{2}-\lambda _{1})^{\xi }} \bigl[ \bigl(I_{\xi }^{ \lambda _{1}}f \bigr) (\lambda _{2})+ \bigl(^{\lambda _{2}}I_{\xi }f \bigr) (\lambda _{1}) \bigr] \\ &\quad \leq \biggl[\frac{\zeta (n+s+1,\xi -n)+\zeta (n+1,\xi -n+s)}{n!} \biggr]\frac{[f(\lambda _{1})+f(\lambda _{2})]}{2^{s}} \end{aligned}$$

with\(\xi \in (n,n+1], n=0,1,2,\ldots \) , whereΓis the Euler gamma function.

Sarikaya et al. [28] defined a function \(\varPhi:(0,+\infty )\longrightarrow [0,+\infty )\) satisfying the following conditions:

$$\begin{aligned} & \int _{0}^{1}\frac{\varPhi (\theta )}{\theta }\,d\theta < +\infty, \end{aligned}$$
(4)
$$\begin{aligned} & \frac{1}{\mathfrak{A}}\leq \frac{\varPhi (\theta _{1})}{\varPhi (\theta _{2})}\leq \mathfrak{A}\quad \text{for } \frac{1}{2}\leq \frac{\theta _{1}}{\theta _{2}}\leq 2, \end{aligned}$$
(5)
$$\begin{aligned} & \frac{\varPhi (\theta _{2})}{\theta _{2}^{2}}\leq \mathfrak{B} \frac{\varPhi (\theta _{1})}{\theta _{1}^{2}} \quad\text{for } \theta _{1} \leq \theta _{2}, \end{aligned}$$
(6)
$$\begin{aligned} & \biggl\vert \frac{\varPhi (\theta _{2})}{\theta _{2}^{2}}- \frac{\varPhi (\theta _{1})}{\theta _{1}^{2}} \biggr\vert \leq \mathfrak{C} \vert \theta _{2}-\theta _{1} \vert \frac{\varPhi (\theta _{2})}{\theta _{2}^{2}} \quad\text{for } \frac{1}{2}\leq \frac{\theta _{1}}{\theta _{2}}\leq 2, \end{aligned}$$
(7)

where \(\mathfrak{A}, \mathfrak{B}, \mathfrak{C}>0\) are independent of \(\theta _{2}, \theta _{1}>0\). If \(\varPhi (\theta _{2})\theta _{2}^{\xi }\) is increasing for some \(\xi \geq 0\) and \(\frac{\varPhi (\theta _{2})}{\theta _{2}^{\zeta }}\) is decreasing for some \(\zeta \geq 0\), then Φ satisfies (4)–(7); see [29]. Therefore the left–sided and right–sided generalized integral operators are defined as follows:

$$\begin{aligned} &{} _{\lambda _{1}^{+}}I_{\varPhi }f(x)= \int _{\lambda _{1}}^{x} \frac{\varPhi (x-\theta )}{x-\theta }f(\theta )\,d\theta,\quad x> \lambda _{1}, \end{aligned}$$
(8)
$$\begin{aligned} &{} _{\lambda _{2}^{-}}I_{\varPhi }f(x)= \int _{x}^{\lambda _{2}} \frac{\varPhi (\theta -x)}{\theta -x}f(\theta )\,d\theta,\quad x< \lambda _{2}. \end{aligned}$$
(9)

The most significant element of generalized integrals is that they produce Riemann–Liouville fractional integrals, k–Riemann–Liouville fractional integrals, Katugampola fractional integrals, and so on; see [10, 11, 16, 20, 23, 28, 29, 31, 33].

Now we are in position to introduce the following definitions of left and right generalized conformable fractional integral operators of any order \(\xi >0\), where \(\varPhi:[0,+\infty )\longrightarrow [0,+\infty )\) satisfies conditions (4)–(7).

Definition 4

Let \(\xi \in (n,n+1]\) and \(\zeta =\xi -n\), where \(n=0,1,2,3,\ldots \) . The left generalized conformable fractional integral operator starting at \(\lambda _{1}\) of order \(\xi >0\) is defined by

$$ _{\lambda _{1}^{+}}T_{\varPhi }f(x)= \int _{\lambda _{1}}^{x} \frac{\varPhi (x-\theta )}{x-\theta }(\theta -\lambda _{1})^{\zeta -1}f( \theta )\,d\theta,\quad x>\lambda _{1}. $$
(10)

Analogously, the right generalized conformable fractional integral operator of order \(\xi >0\) is defined by

$$ _{\lambda _{2}^{-}}T_{\varPhi }f(x)= \int _{x}^{\lambda _{2}} \frac{\varPhi (\theta -x)}{\theta -x}(\lambda _{2}-\theta )^{\zeta -1}f( \theta )\,d\theta,\quad x< \lambda _{2}. $$
(11)

Remark 1

Taking \(\zeta =1\) in Definition 4, we obtain the generalized fractional integral operators given from (8) and (9). Also, choosing \(\varPhi (t)=\frac{t^{n+1}}{n!}\) in Definition 4, we get Definition 3.

Motivated by the literature cited, our paper is organized as follows: In Sect. 2, using the new operators, we establish the so-called left and right generalized conformable fractional integral operators, new Hermite–Hadamard inequalities for s-convex functions and products of two s-convex functions in the second sense. In Sect. 3, we obtain two interesting identities for differentiable function involving generalized conformable fractional integral operator. By applying these identities we give Hermite–Hadamard and midpoint-type integral inequalities for s-convex functions. Various particular cases will be identified, and some known results will be recaptured from our general results. In Sect. 4, we give a brief conclusion.

Hermite–Hadamard inequalities

Throughout this study, let \(\xi \in (n,n+1]\) and \(\zeta =\xi -n\), where \(n=0,1,2,3,\ldots \) . Also, for all \(\theta \in [0,1]\), we define

$$\begin{aligned} & \varOmega _{\varPhi }^{\zeta }(\theta )= \int _{0}^{\theta } \frac{\varPhi (x(\lambda _{2}-\lambda _{1}))}{x}(1-x)^{\zeta -1} \,dx< + \infty, \quad \lambda _{1}< \lambda _{2}, \end{aligned}$$
(12)
$$\begin{aligned} & \varSigma _{\varPhi,1}^{\zeta }(x,\theta )= \int _{0}^{\theta } \frac{\varPhi (u(x-\lambda _{1}))}{u}(1-u)^{\zeta -1} \,du< +\infty, \quad x> \lambda _{1}, \end{aligned}$$
(13)
$$\begin{aligned} &\varSigma _{\varPhi,2}^{\zeta }(x,\theta )= \int _{0}^{\theta } \frac{\varPhi (u(\lambda _{2}-x))}{u}(1-u)^{\zeta -1} \,du< +\infty,\quad x< \lambda _{2}, \end{aligned}$$
(14)

and

$$ \varPsi _{\varPhi }^{\zeta }= \int _{\lambda _{1}}^{\lambda _{2}} \frac{\varPhi (x-\lambda _{1})}{x-\lambda _{1}}(\lambda _{2}-x)^{\zeta -1}\,dx= \int _{\lambda _{1}}^{\lambda _{2}} \frac{\varPhi (\lambda _{2}-x)}{\lambda _{2}-x}(x-\lambda _{1})^{\zeta -1}\,dx. $$
(15)

Let represent Hermite–Hadamard inequalities for s-convex functions in the second sense via general conformable fractional integral operators as follows.

Theorem 3

Let\(f:[\lambda _{1},\lambda _{2}]\longrightarrow \mathbb{R}\)be a positive function with\(0\leq \lambda _{1}<\lambda _{2}\)such that\(f\in L[\lambda _{1},\lambda _{2}]\). Iffiss-convex in the second sense on\([\lambda _{1},\lambda _{2}]\), then for any\(\xi >0\)and\(s\in (0,1]\), we have the following inequalities for generalized conformable fractional integral operators:

$$ f \biggl(\frac{\lambda _{1}+\lambda _{2}}{2} \biggr)\leq \frac{1}{2^{s} \varPsi _{\varPhi }^{\zeta }} \bigl[ {}_{\lambda _{1}^{+}}T_{ \varPhi }f(\lambda _{2}) + _{\lambda _{2}^{-}}T_{\varPhi }f( \lambda _{1}) \bigr]\leq \biggl[\frac{f(\lambda _{1})+f(\lambda _{2})}{2^{s}} \biggr] \Delta _{\varPhi }^{s,\zeta }, $$
(16)

where

$$ \Delta _{\varPhi }^{s,\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{s} \varPsi _{\varPhi }^{\zeta }} \int _{\lambda _{1}}^{\lambda _{2}} \frac{\varPhi (x-\lambda _{1})}{x-\lambda _{1}}(\lambda _{2}-x)^{\zeta -1} \bigl[(x-\lambda _{1})^{s}+( \lambda _{2}-x)^{s} \bigr]\,dx, $$
(17)

and\(\varPsi _{\varPhi }^{\zeta }\)is defined by (15).

Proof

Let \(x,y\in [\lambda _{1},\lambda _{2}]\). Since f is s-convex in the second sense on \([\lambda _{1},\lambda _{2}]\), we have

$$ f \biggl(\frac{x+y}{2} \biggr)\leq \frac{f(x)+f(y)}{2^{s}}. $$

Taking \(x=\theta \lambda _{1}+(1-\theta )\lambda _{2}\) and \(y=(1-\theta )\lambda _{1}+\theta \lambda _{2}\), we get

$$ 2^{s}f \biggl(\frac{\lambda _{1}+\lambda _{2}}{2} \biggr)\leq f\bigl( \theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)+f\bigl((1- \theta )\lambda _{1}+ \theta \lambda _{2}\bigr). $$
(18)

Multiplying both sides of inequality (18) by \({ \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}}\) and integrating the resulting inequality with respect to θ over \([0,1]\), we obtain

$$\begin{aligned} &2^{s}f \biggl(\frac{\lambda _{1}+\lambda _{2}}{2} \biggr) \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}\,d\theta \\ &\quad \leq \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}f\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)\,d\theta \\ &\qquad{}+ \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}f\bigl((1-\theta )\lambda _{1}+\theta \lambda _{2}\bigr)\,d\theta. \end{aligned}$$

So, we have

$$ 2^{s} \varPsi _{\varPhi }^{\zeta } f \biggl( \frac{\lambda _{1}+\lambda _{2}}{2} \biggr)\leq \bigl[{} _{\lambda _{1}^{+}}T_{ \varPhi }f(\lambda _{2}) + _{\lambda _{2}^{-}}T_{\varPhi }f(\lambda _{1}) \bigr], $$
(19)

which means that the left side of (16) is proved. To prove the right side of (16), since f is s-convex in the second sense on \([\lambda _{1},\lambda _{2}]\), we have the inequalities

$$ f\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)\leq \theta ^{s}f( \lambda _{1})+(1- \theta )^{s}f(\lambda _{2}) $$
(20)

and

$$ f\bigl((1-\theta )\lambda _{1}+\theta \lambda _{2}\bigr)\leq (1-\theta )^{s}f( \lambda _{1})+ \theta ^{s}f(\lambda _{2}). $$
(21)

Adding (20) and (21), we get

$$ f\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)+f\bigl((1-\theta )\lambda _{1}+ \theta \lambda _{2}\bigr)\leq \bigl[\theta ^{s}+(1-\theta )^{s} \bigr] \bigl[f( \lambda _{1})+f(\lambda _{2}) \bigr]. $$
(22)

Multiplying both sides of inequality (22) by \({ \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}}\) and integrating the resulting inequality with respect to t over \([0,1]\), we obtain

$$\begin{aligned} & \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}f\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)\,d\theta \\ &\qquad{}+ \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}f\bigl((1-\theta )\lambda _{1}+\theta \lambda _{2}\bigr)\,d\theta \\ &\quad\leq \bigl[f(\lambda _{1})+f(\lambda _{2}) \bigr] \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1} \bigl[\theta ^{s}+(1-\theta )^{s} \bigr]\,d \theta. \end{aligned}$$

So, we have

$$ \bigl[ {}_{\lambda _{1}^{+}}T_{\varPhi }f(\lambda _{2}) + _{\lambda _{2}^{-}}T_{ \varPhi }f(\lambda _{1}) \bigr]\leq \bigl[f( \lambda _{1})+f(\lambda _{2}) \bigr]\varPsi _{\varPhi }^{\zeta } \Delta _{\varPhi }^{s,\zeta }, $$
(23)

which means that the right side of (16) is proved. The proof of Theorem 3 is completed. □

Corollary 1

Taking\(s=1\)in Theorem 3, we get the following inequalities for convex functions via generalized conformable fractional integral operators:

$$ f \biggl(\frac{\lambda _{1}+\lambda _{2}}{2} \biggr)\leq \frac{1}{2 \varPsi _{\varPhi }^{\zeta }} \bigl[ {}_{\lambda _{1}^{+}}T_{\varPhi }f( \lambda _{2}) + _{\lambda _{2}^{-}}T_{\varPhi }f( \lambda _{1}) \bigr] \leq \frac{f(\lambda _{1})+f(\lambda _{2})}{2}. $$
(24)

Corollary 2

Taking\({\varPhi (\theta )=\frac{\theta ^{n+1}}{n!}}\)in Theorem 3, we get Theorem 2.

Remark 2

Taking \(\xi =s=1\) in Corollary 2, we get the well–known Hermite–Hadamard inequality (2).

Let us represent now Hermite–Hadamard inequalities for the product of two s-convex functions in the second sense via general conformable fractional integral operators.

Theorem 4

Let\(f, g:[\lambda _{1},\lambda _{2}]\longrightarrow \mathbb{R}\)be two positive functions with\(0\leq \lambda _{1}<\lambda _{2}\)and\(f\in L[\lambda _{1},\lambda _{2}]\). Iffandgares-convex in the second sense on\([\lambda _{1},\lambda _{2}]\), then for any\(\xi >0\)and\(s\in (0,1]\), we have the following inequalities for generalized conformable fractional integral operators:

$$\begin{aligned} &f \biggl(\frac{\lambda _{1}+\lambda _{2}}{2} \biggr)g \biggl( \frac{\lambda _{1}+\lambda _{2}}{2} \biggr) - \frac{1}{4^{s} \varPsi _{\varPhi }^{\zeta }} \bigl[ {}_{\lambda _{1}^{+}}T_{ \varPhi }f(\lambda _{2})g(\lambda _{2}) + _{\lambda _{2}^{-}}T_{\varPhi }f( \lambda _{1})g(\lambda _{1}) \bigr] \\ &\quad\leq \frac{1}{4^{s} \varPsi _{\varPhi }^{\zeta }} \bigl[ {}_{\lambda _{1}^{+}}T_{ \varPhi }f(\lambda _{2})g(\lambda _{1}) + _{\lambda _{2}^{-}}T_{\varPhi }f( \lambda _{1})g(\lambda _{2}) \bigr] \\ &\quad\leq \frac{1}{4^{s}} \bigl[2M(\lambda _{1},\lambda _{2})\varTheta _{ \varPhi }^{s,\zeta }+N(\lambda _{1},\lambda _{2}) \Delta _{\varPhi }^{2s, \zeta } \bigr], \end{aligned}$$
(25)

where

$$\begin{aligned} &M(\lambda _{1},\lambda _{2})=f(\lambda _{1})g(\lambda _{1})+f( \lambda _{2})g(\lambda _{2}), \end{aligned}$$
(26)
$$\begin{aligned} &N(\lambda _{1},\lambda _{2})=f(\lambda _{1})g(\lambda _{2})+f( \lambda _{2})g(\lambda _{1}), \end{aligned}$$
(27)
$$\begin{aligned} &\varTheta _{\varPhi }^{s,\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{2s} \varPsi _{\varPhi }^{\zeta }} \int _{\lambda _{1}}^{\lambda _{2}} \frac{\varPhi (x-\lambda _{1})}{x-\lambda _{1}}(\lambda _{2}-x)^{\zeta -1}(x- \lambda _{1})^{s}( \lambda _{2}-x)^{s}\,dx, \end{aligned}$$
(28)

and\(\varPsi _{\varPhi }^{\zeta }\)and\(\Delta _{\varPhi }^{2s,\zeta }\)are defined by (15) and (17).

Proof

Let \(x,y\in [\lambda _{1},\lambda _{2}]\). Since f and g are s-convex in the second sense on \([\lambda _{1},\lambda _{2}]\), we have

$$ f \biggl(\frac{x+y}{2} \biggr)\leq \frac{f(x)+f(y)}{2^{s}},\qquad g \biggl( \frac{x+y}{2} \biggr)\leq \frac{g(x)+g(y)}{2^{s}}. $$

Taking \(x=\theta \lambda _{1}+(1-\theta )\lambda _{2}\) and \(y=(1-\theta )\lambda _{1}+\theta \lambda _{2}\), we get

$$ 2^{s}f \biggl(\frac{\lambda _{1}+\lambda _{2}}{2} \biggr)\leq f\bigl( \theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)+f\bigl((1- \theta )\lambda _{1}+ \theta \lambda _{2}\bigr) $$
(29)

and

$$ 2^{s}g \biggl(\frac{\lambda _{1}+\lambda _{2}}{2} \biggr)\leq g\bigl( \theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)+g\bigl((1- \theta )\lambda _{1}+ \theta \lambda _{2}\bigr). $$
(30)

Multiplying both sides of inequalities (29) and (30), we obtain

$$\begin{aligned} & 4^{s}f \biggl(\frac{\lambda _{1}+\lambda _{2}}{2} \biggr)g \biggl( \frac{\lambda _{1}+\lambda _{2}}{2} \biggr) \\ &\quad\leq f\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr)g\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr) \\ &\qquad{}+f\bigl((1-\theta )\lambda _{1}+ \theta \lambda _{2} \bigr)g\bigl((1-\theta )\lambda _{1}+\theta \lambda _{2} \bigr) \\ &\qquad{}+f\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr)g\bigl((1-\theta ) \lambda _{1}+\theta \lambda _{2} \bigr) \\ &\qquad{}+f\bigl((1-\theta )\lambda _{1}+\theta \lambda _{2} \bigr)g\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr). \end{aligned}$$
(31)

Multiplying both sides of inequality (31) by \({ \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}}\) and integrating the resulting inequality with respect to θ over \([0,1]\), we have

$$\begin{aligned} &4^{s}f \biggl(\frac{\lambda _{1}+\lambda _{2}}{2} \biggr)g \biggl( \frac{\lambda _{1}+\lambda _{2}}{2} \biggr) \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}\,d\theta \\ &\quad\leq \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}f\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)g\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)\,d\theta \\ &\qquad{}+ \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}f\bigl((1-\theta )\lambda _{1}+\theta \lambda _{2}\bigr)g\bigl((1-\theta ) \lambda _{1}+\theta \lambda _{2}\bigr)\,d\theta \\ &\qquad{}+ \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}f\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)g\bigl((1-\theta ) \lambda _{1}+\theta \lambda _{2}\bigr)\,d\theta \\ &\qquad{}+ \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}f\bigl((1-\theta )\lambda _{1}+\theta \lambda _{2}\bigr)g\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)\,d\theta. \end{aligned}$$

So, we get

$$\begin{aligned} 4^{s} \varPsi _{\varPhi }^{\zeta } f \biggl( \frac{\lambda _{1}+\lambda _{2}}{2} \biggr)g \biggl( \frac{\lambda _{1}+\lambda _{2}}{2} \biggr)\leq{}& \bigl[ {}_{\lambda _{1}^{+}}T_{ \varPhi }f(\lambda _{2})g(\lambda _{1}) + _{\lambda _{2}^{-}}T_{\varPhi }f( \lambda _{1})g( \lambda _{2}) \bigr] \\ &{}+ \bigl[ {}_{\lambda _{1}^{+}}T_{\varPhi }f(\lambda _{2})g(\lambda _{2}) + _{\lambda _{2}^{-}}T_{\varPhi }f(\lambda _{1})g( \lambda _{1}) \bigr], \end{aligned}$$
(32)

which means that the left side of (25) is proved. To prove the right side of (25), since f and g are s-convex in the second sense on \([\lambda _{1},\lambda _{2}]\), we have the inequalities

$$\begin{aligned} &f\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)\leq \theta ^{s}f( \lambda _{1})+(1- \theta )^{s}f(\lambda _{2}), \end{aligned}$$
(33)
$$\begin{aligned} & f\bigl((1-\theta )\lambda _{1}+\theta \lambda _{2}\bigr)\leq (1-\theta )^{s}f( \lambda _{1})+ \theta ^{s}f(\lambda _{2}), \end{aligned}$$
(34)
$$\begin{aligned} &g\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)\leq \theta ^{s}g( \lambda _{1})+(1- \theta )^{s}g(\lambda _{2}) \end{aligned}$$
(35)

and

$$ g\bigl((1-\theta )\lambda _{1}+\theta \lambda _{2}\bigr)\leq (1-\theta )^{s}g( \lambda _{1})+ \theta ^{s}g(\lambda _{2}). $$
(36)

Applying inequalities (33) to (36), we have

$$\begin{aligned} &f\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr)g\bigl(\theta \lambda _{1}+(1- \theta )\lambda _{2} \bigr)+f\bigl((1-\theta )\lambda _{1}+\theta \lambda _{2} \bigr)g\bigl((1- \theta )\lambda _{1}+\theta \lambda _{2} \bigr) \\ &\qquad{}+f\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr)g\bigl((1-\theta ) \lambda _{1}+\theta \lambda _{2} \bigr)+f\bigl((1-\theta )\lambda _{1}+\theta \lambda _{2} \bigr)g\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr) \\ &\quad \leq f\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr)g\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr)+f\bigl((1-\theta )\lambda _{1}+ \theta \lambda _{2} \bigr)g\bigl((1-\theta )\lambda _{1}+\theta \lambda _{2} \bigr) \\ &\qquad{}+ \bigl[\theta ^{s}f(\lambda _{1})+(1-\theta )^{s}f(\lambda _{2}) \bigr]\cdot \bigl[(1-\theta )^{s}g(\lambda _{1})+\theta ^{s}g(\lambda _{2}) \bigr] \\ &\qquad{}+ \bigl[(1-\theta )^{s}f(\lambda _{1})+\theta ^{s}f(\lambda _{2}) \bigr]\cdot \bigl[\theta ^{s}g(\lambda _{1})+(1-\theta )^{s}g(\lambda _{2}) \bigr] \\ &\quad = f\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr)g\bigl(\theta \lambda _{1}+(1- \theta )\lambda _{2} \bigr)+f\bigl((1-\theta )\lambda _{1}+\theta \lambda _{2} \bigr)g\bigl((1- \theta )\lambda _{1}+\theta \lambda _{2} \bigr) \\ &\qquad{}+2\theta ^{s}(1-\theta )^{s}M(\lambda _{1}, \lambda _{2})+ \bigl[ \theta ^{2s}+(1-\theta )^{2s} \bigr]N(\lambda _{1},\lambda _{2}). \end{aligned}$$
(37)

Multiplying both sides of inequality (37) by \({ \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}}\) and integrating the resulting inequality with respect to θ over \([0,1]\), we obtain

$$\begin{aligned} & \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}f\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)g\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)\,d\theta \\ &\qquad{}+ \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}f\bigl((1-\theta )\lambda _{1}+\theta \lambda _{2}\bigr)g\bigl((1-\theta ) \lambda _{1}+\theta \lambda _{2}\bigr)\,d\theta \\ &\qquad{}+ \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}f\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)g\bigl((1-\theta ) \lambda _{1}+\theta \lambda _{2}\bigr)\,d\theta \\ &\qquad{}+ \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}f\bigl((1-\theta )\lambda _{1}+\theta \lambda _{2}\bigr)g\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)\,d\theta \\ &\quad\leq \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}f\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)g\bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2}\bigr)\,d\theta \\ &\qquad{}+ \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}f\bigl((1-\theta )\lambda _{1}+\theta \lambda _{2}\bigr)g\bigl((1-\theta ) \lambda _{1}+\theta \lambda _{2}\bigr)\,d\theta \\ &\qquad{}+2M(\lambda _{1},\lambda _{2}) \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1}\theta ^{s}(1-\theta )^{s}\,d\theta \\ &\qquad{}+N(\lambda _{1},\lambda _{2}) \int _{0}^{1} \frac{\varPhi (\theta (\lambda _{2}-\lambda _{1}))}{\theta }(1-\theta )^{ \zeta -1} \bigl[\theta ^{2s}+(1-\theta )^{2s} \bigr] \,d\theta. \end{aligned}$$

So, we get

$$\begin{aligned} & \bigl[ {}_{\lambda _{1}^{+}}T_{\varPhi }f(\lambda _{2})g(\lambda _{2}) + _{\lambda _{2}^{-}}T_{\varPhi }f(\lambda _{1})g( \lambda _{1}) \bigr]+ \bigl[ {}_{\lambda _{1}^{+}}T_{\varPhi }f(\lambda _{2})g(\lambda _{1}) + _{\lambda _{2}^{-}}T_{\varPhi }f( \lambda _{1})g(\lambda _{2}) \bigr] \\ &\quad\leq \bigl[ {}_{\lambda _{1}^{+}}T_{\varPhi }f(\lambda _{2})g( \lambda _{2}) + _{\lambda _{2}^{-}}T_{\varPhi }f(\lambda _{1})g(\lambda _{1}) \bigr]+2M( \lambda _{1}, \lambda _{2})\varTheta _{\varPhi }^{s,\zeta }+N(\lambda _{1}, \lambda _{2}) \varPsi _{\varPhi }^{\zeta } \Delta _{\varPhi }^{2s,\zeta }, \end{aligned}$$
(38)

which means that the right side of (25) is proved. The proof of Theorem 4 is completed. □

Corollary 3

Taking\(s=1\)in Theorem 4, we get the following inequalities for products of two convex functions via generalized conformable fractional integral operators:

$$\begin{aligned} &f \biggl(\frac{\lambda _{1}+\lambda _{2}}{2} \biggr)g \biggl( \frac{\lambda _{1}+\lambda _{2}}{2} \biggr) - \frac{1}{4 \varPsi _{\varPhi }^{\zeta }} \bigl[ {}_{\lambda _{1}^{+}}T_{\varPhi }f( \lambda _{2})g(\lambda _{2}) + _{\lambda _{2}^{-}}T_{\varPhi }f( \lambda _{1})g(\lambda _{1}) \bigr] \\ &\quad\leq \frac{1}{4 \varPsi _{\varPhi }^{\zeta }} \bigl[ {}_{\lambda _{1}^{+}}T_{ \varPhi }f( \lambda _{2})g(\lambda _{1}) + _{\lambda _{2}^{-}}T_{\varPhi }f( \lambda _{1})g(\lambda _{2}) \bigr] \\ &\quad\leq \frac{1}{4} \bigl[2M(\lambda _{1},\lambda _{2})\varTheta _{\varPhi }^{ \zeta }+N(\lambda _{1},\lambda _{2}) \Delta _{\varPhi }^{2,\zeta } \bigr], \end{aligned}$$
(39)

where

$$\begin{aligned} \varTheta _{\varPhi }^{\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{2} \varPsi _{\varPhi }^{\zeta }} \int _{\lambda _{1}}^{\lambda _{2}}\varPhi (x-\lambda _{1}) ( \lambda _{2}-x)^{ \zeta }\,dx \end{aligned}$$

and

$$\begin{aligned} \Delta _{\varPhi }^{2,\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{2} \varPsi _{\varPhi }^{\zeta }} \int _{\lambda _{1}}^{\lambda _{2}} \frac{\varPhi (x-\lambda _{1})}{x-\lambda _{1}}(\lambda _{2}-x)^{\zeta -1} \bigl[(x-\lambda _{1})^{2}+( \lambda _{2}-x)^{2} \bigr]\,dx. \end{aligned}$$

Corollary 4

Taking\(f=g\)in Theorem 4, we get

$$\begin{aligned} &f^{2} \biggl(\frac{\lambda _{1}+\lambda _{2}}{2} \biggr) - \frac{1}{4^{s} \varPsi _{\varPhi }^{\zeta }} \bigl[ {}_{\lambda _{1}^{+}}T_{ \varPhi }f^{2}(\lambda _{2}) + _{\lambda _{2}^{-}}T_{\varPhi }f^{2}( \lambda _{1}) \bigr] \\ &\quad \leq \frac{1}{4^{s} \varPsi _{\varPhi }^{\zeta }} \bigl[ {}_{\lambda _{1}^{+}}T_{ \varPhi }f( \lambda _{1})f(\lambda _{2}) + _{\lambda _{2}^{-}}T_{\varPhi }f( \lambda _{1})f(\lambda _{2}) \bigr] \\ &\quad\leq \frac{1}{4^{s}} \bigl[2P(\lambda _{1},\lambda _{2})\varTheta _{ \varPhi }^{s,\zeta }+Q(\lambda _{1},\lambda _{2}) \Delta _{\varPhi }^{2s, \zeta } \bigr], \end{aligned}$$
(40)

where

$$ P(\lambda _{1},\lambda _{2})=f^{2}(\lambda _{1})+f^{2}(\lambda _{2}),\qquad Q(\lambda _{1},\lambda _{2})=2f(\lambda _{1})f(\lambda _{2}). $$

Some other results

To establish the results of this section regarding general conformable fractional integral operators, we first prove the following two lemmas.

Lemma 3.1

Let\(f:[\lambda _{1},\lambda _{2}]\longrightarrow \mathbb{R}\)be a differentiable function on\((\lambda _{1},\lambda _{2})\). If\(f'\in L[\lambda _{1},\lambda _{2}]\), then we have the following identity for generalized conformable fractional integrals:

$$\begin{aligned} &\biggl[\frac{f(\lambda _{1})+f(\lambda _{2})}{2} \biggr]\varPsi _{\varPhi }^{ \zeta } - \frac{1}{2} \bigl[ {}_{\lambda _{1}^{+}}T_{\varPhi }f(\lambda _{2}) + _{\lambda _{2}^{-}}T_{\varPhi }f(\lambda _{1}) \bigr] \\ &\quad =\frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2} \int _{0}^{1} \bigl[ \varOmega _{\varPhi }^{\zeta }(1- \theta )-\varOmega _{\varPhi }^{\zeta }(\theta ) \bigr]f' \bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr)\,d \theta, \end{aligned}$$
(41)

where\(\varOmega _{\varPhi }^{\zeta }(\theta )\)is defined by (12).

Proof

We denote

$$ I_{f,\varOmega _{\varPhi }^{\zeta }}(\lambda _{1},\lambda _{2})= \frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2} \int _{0}^{1} \bigl[ \varOmega _{\varPhi }^{\zeta }(1- \theta )-\varOmega _{\varPhi }^{\zeta }(\theta ) \bigr]f' \bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr)\,d \theta. $$
(42)

We write (42) in the form

$$\begin{aligned} I_{f,\varOmega _{\varPhi }^{\zeta }}(\lambda _{1},\lambda _{2})= \frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2} \bigl[I_{f,\varOmega _{ \varPhi }^{\zeta }}^{(1)}( \lambda _{1},\lambda _{2})-I_{f,\varOmega _{\varPhi }^{ \zeta }}^{(2)}( \lambda _{1},\lambda _{2}) \bigr], \end{aligned}$$
(43)

where

$$\begin{aligned} I_{f,\varOmega _{\varPhi }^{\zeta }}^{(1)}(\lambda _{1},\lambda _{2})= \int _{0}^{1}\varOmega _{\varPhi }^{\zeta }(1- \theta )f' \bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr)\,d\theta \end{aligned}$$
(44)

and

$$\begin{aligned} I_{f,\varOmega _{\varPhi }^{\zeta }}^{(2)}(\lambda _{1},\lambda _{2})= \int _{0}^{1}\varOmega _{\varPhi }^{\zeta }( \theta )f' \bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr)\,d\theta. \end{aligned}$$
(45)

Integrating by parts (44) and changing the variable of integration \(x=\theta \lambda _{1}+(1-\theta )\lambda _{2}\), we get

$$\begin{aligned} &I_{f,\varOmega _{\varPhi }^{\zeta }}^{(1)}(\lambda _{1},\lambda _{2}) \\ &\quad = \varOmega _{\varPhi }^{\zeta }(1-\theta ) \frac{f (\theta \lambda _{1}+(1-\theta )\lambda _{2} )}{\lambda _{1}-\lambda _{2}}\bigg|_{0}^{1} \end{aligned}$$
(46)
$$\begin{aligned} &\qquad{}+\frac{1}{\lambda _{1}-\lambda _{2}} \int _{0}^{1} \frac{\varPhi ((1-\theta )(\lambda _{2}-\lambda _{1}))}{1-\theta }\bigl(1-(1- \theta )\bigr)^{\zeta -1}f \bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr)\,d\theta \end{aligned}$$
(47)
$$\begin{aligned} &\quad =\varPsi _{\varPhi }^{\zeta } \frac{f(\lambda _{2})}{(\lambda _{2}-\lambda _{1})^{\zeta }}- \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta }} \times{} _{\lambda _{2}^{-}}T_{ \varPhi }f( \lambda _{1}). \end{aligned}$$
(48)

Similarly, using (45), we obtain

$$\begin{aligned} I_{f,\varOmega _{\varPhi }^{\zeta }}^{(2)}(\lambda _{1},\lambda _{2}) = - \varPsi _{\varPhi }^{\zeta } \frac{f(\lambda _{1})}{(\lambda _{2}-\lambda _{1})^{\zeta }}+ \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta }} \times {}_{\lambda _{1}^{+}}T_{ \varPhi }f( \lambda _{2}). \end{aligned}$$
(49)

Substituting (48) and (49) into (43), we obtain the desired equality (41). □

Remark 3

Taking \({\varPhi (\theta )=\frac{\theta ^{n+1}}{n!}}\) in Lemma 3.1, we get ([30], Lemma 3.1).

Lemma 3.2

Let\(f:[\lambda _{1},\lambda _{2}]\longrightarrow \mathbb{R}\)be a differentiable function on\((\lambda _{1},\lambda _{2})\). If\(f'\in L[\lambda _{1},\lambda _{2}]\), then we have the following identity for generalized conformable fractional integrals:

$$\begin{aligned} &f(x) \varPsi _{\varPhi }^{\zeta } - \frac{1}{2} \bigl[ {}_{x^{+}}T_{\varPhi }f( \lambda _{2}) + _{x^{-}}T_{\varPhi }f(\lambda _{1}) \bigr] \\ &\quad=\frac{(x-\lambda _{1})^{\zeta }}{2} \int _{0}^{1}\varSigma _{\varPhi,1}^{ \zeta }(x, \theta )f' \bigl(\theta x+(1-\theta )\lambda _{1} \bigr) \,d \theta \\ &\qquad{}-\frac{(\lambda _{2}-x)^{\zeta }}{2} \int _{0}^{1}\varSigma _{\varPhi,2}^{ \zeta }(x, \theta )f' \bigl(\theta x+(1-\theta )\lambda _{2} \bigr) \,d \theta, \end{aligned}$$
(50)

where\(\varSigma _{\varPhi,1}^{\zeta }(x,\theta )\)and\(\varSigma _{\varPhi,2}^{\zeta }(x,\theta )\)are defined by (13) and (14), respectively. We denote

$$\begin{aligned} &I_{f,\varSigma _{\varPhi,1}^{\zeta },\varSigma _{\varPhi,2}^{\zeta }}(x; \lambda _{1},\lambda _{2}) \\ &\quad =\frac{(x-\lambda _{1})^{\zeta }}{2} \int _{0}^{1}\varSigma _{\varPhi,1}^{ \zeta }(x, \theta )f' \bigl(\theta x+(1-\theta )\lambda _{1} \bigr) \,d \theta \\ &\qquad{}-\frac{(\lambda _{2}-x)^{\zeta }}{2} \int _{0}^{1}\varSigma _{\varPhi,2}^{ \zeta }(x, \theta )f' \bigl(\theta x+(1-\theta )\lambda _{2} \bigr) \,d \theta. \end{aligned}$$
(51)

Proof

See the proof of Lemma 3.1. □

Theorem 5

Let\(f:[\lambda _{1},\lambda _{2}]\longrightarrow \mathbb{R}\)be a differentiable function on\((\lambda _{1},\lambda _{2})\). If\(f'\in L[\lambda _{1},\lambda _{2}]\)and\(|f'|^{q}\)iss-convex in the second sense with\(s\in (0,1]\), then for\(q>1\)and\(\frac{1}{p}+\frac{1}{q}=1\), wehave the following inequality for generalized conformable fractional integrals:

$$ \bigl\vert I_{f,\varOmega _{\varPhi }^{\zeta }}(\lambda _{1},\lambda _{2}) \bigr\vert \leq \frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2} \sqrt[p]{\varPi _{\varPhi }^{\zeta }(p)} \sqrt[q]{\frac{ \vert f'(\lambda _{1}) \vert ^{q}+ \vert f'(\lambda _{2}) \vert ^{q}}{s+1}}, $$
(52)

where

$$ \varPi _{\varPhi }^{\zeta }(p)=2 \int _{0}^{\frac{1}{2}} \biggl( \int _{ \theta }^{1-\theta }\frac{\varPhi (x(\lambda _{2}-\lambda _{1}))}{x}(1-x)^{ \zeta -1} \,dx \biggr)^{p}\,d\theta. $$

Proof

By Lemma 3.1, the s-convexity in the second sense of \(|f'|^{q}\), the Hölder inequality, and properties of the modulus we have

$$\begin{aligned} &\bigl\vert I_{f,\varOmega _{\varPhi }^{\zeta }}(\lambda _{1},\lambda _{2}) \bigr\vert \\ &\quad \leq \frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2} \int _{0}^{1} \bigl\vert \varOmega _{\varPhi }^{\zeta }(1-\theta )-\varOmega _{\varPhi }^{\zeta }( \theta ) \bigr\vert \bigl\vert f' \bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr) \bigr\vert \,d\theta \\ &\quad\leq \frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2} \biggl( \int _{0}^{1} \bigl\vert \varOmega _{\varPhi }^{\zeta }(1-\theta )-\varOmega _{\varPhi }^{\zeta }( \theta ) \bigr\vert ^{p}\,d\theta \biggr)^{\frac{1}{p}} \biggl( \int _{0}^{1} \bigl\vert f' \bigl( \theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr) \bigr\vert ^{q}\,d \theta \biggr)^{\frac{1}{q}} \\ &\quad\leq \frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2} \sqrt[p]{\varPi _{\varPhi }^{\zeta }(p)} \biggl( \int _{0}^{1} \bigl[\theta ^{s} \bigl\vert f'( \lambda _{1}) \bigr\vert ^{q}+(1- \theta )^{s} \bigl\vert f'(\lambda _{2}) \bigr\vert ^{q} \bigr]\,d\theta \biggr)^{\frac{1}{q}} \\ &\quad=\frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2} \sqrt[p]{\varPi _{\varPhi }^{\zeta }(p)} \sqrt[q]{\frac{ \vert f'(\lambda _{1}) \vert ^{q}+ \vert f'(\lambda _{2}) \vert ^{q}}{s+1}}. \end{aligned}$$

The proof of Theorem 5 is completed. □

We point out some particular cases of Theorem 5.

Corollary 5

Taking\({\varPhi (\theta )=\frac{\theta ^{n+1}}{n!}}\)in Theorem 5, we get ([30], Theorem 3.2).

Corollary 6

Taking\(s=1\)in Theorem 5, we have the following inequality for convex function via generalized conformable fractional integral operators:

$$ \bigl\vert I_{f,\varOmega _{\varPhi }^{\zeta }}(\lambda _{1},\lambda _{2}) \bigr\vert \leq \frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2} \sqrt[p]{\varPi _{\varPhi }^{\zeta }(p)} \sqrt[q]{\frac{ \vert f'(\lambda _{1}) \vert ^{q}+ \vert f'(\lambda _{2}) \vert ^{q}}{2}}. $$
(53)

Corollary 7

Taking\(|f'|\leq K\)in Theorem 5, we obtain

$$ \bigl\vert I_{f,\varOmega _{\varPhi }^{\zeta }}(\lambda _{1},\lambda _{2}) \bigr\vert \leq K\frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2} \sqrt[q]{ \frac{2}{s+1}}\sqrt[p]{\varPi _{\varPhi }^{\zeta }(p)}. $$
(54)

Theorem 6

Let\(f:[\lambda _{1},\lambda _{2}]\longrightarrow \mathbb{R}\)be a differentiable function on\((\lambda _{1},\lambda _{2})\). If\(f'\in L[\lambda _{1},\lambda _{2}]\)and\(|f'|^{q}\)iss-convex in the second sense with\(s\in (0,1]\), then for\(q\geq 1\), we have the following inequality for generalized conformable fractional integrals:

$$\begin{aligned} & \bigl\vert I_{f,\varOmega _{\varPhi }^{\zeta }}(\lambda _{1},\lambda _{2}) \bigr\vert \\ &\quad \leq \frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2\sqrt[q]{s+1}} \bigl\{ \bigl[L_{\varPhi,1}^{\zeta } \bigr]^{1-\frac{1}{q}} \sqrt[q]{ \bigl(L_{\varPhi,2}^{s,\zeta }+L_{\varPhi,3}^{s,\zeta } \bigr) \bigl\vert f'(\lambda _{1}) \bigr\vert ^{q}+ \bigl(\varPsi _{\varPhi }^{\zeta }-L_{\varPhi,4}^{s,\zeta }-L_{\varPhi,5}^{s,\zeta } \bigr) \bigl\vert f'(\lambda _{2}) \bigr\vert ^{q}} \\ &\qquad{}+ \bigl[L_{\varPhi,6}^{\zeta } \bigr]^{1-\frac{1}{q}} \sqrt[q]{ \bigl( \varPsi _{\varPhi }^{\zeta }-L_{\varPhi,7}^{s,\zeta }-L_{\varPhi,8}^{s,\zeta } \bigr) \bigl\vert f'(\lambda _{1}) \bigr\vert ^{q}+ \bigl(L_{\varPhi,9}^{s,\zeta }+L_{\varPhi,10}^{s,\zeta } \bigr) \bigl\vert f'(\lambda _{2}) \bigr\vert ^{q}} \bigr\} , \end{aligned}$$
(55)

where

$$\begin{aligned} &L_{\varPhi,1}^{\zeta }=\frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta }} \biggl[ \int _{\frac{\lambda _{1}+\lambda _{2}}{2}}^{\lambda _{2}} \frac{\varPhi (\theta -\lambda _{1})}{\theta -\lambda _{1}}(\lambda _{2}- \theta )^{\zeta }\,d\theta + \int _{\frac{\lambda _{1}+\lambda _{2}}{2}}^{ \lambda _{2}}\varPhi (\lambda _{2}-\theta ) (\theta -\lambda _{1})^{ \zeta -1}\,d\theta \biggr], \\ &L_{\varPhi,2}^{s,\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta +s}} \int _{ \frac{\lambda _{1}+\lambda _{2}}{2}}^{\lambda _{2}} \frac{\varPhi (\theta -\lambda _{1})}{\theta -\lambda _{1}}(\lambda _{2}- \theta )^{\zeta +s}\,d\theta, \\ &L_{\varPhi,3}^{s,\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta +s}} \int _{ \frac{\lambda _{1}+\lambda _{2}}{2}}^{\lambda _{2}}\varPhi (\lambda _{2}- \theta ) (\theta -\lambda _{1})^{\zeta -1}(\lambda _{2}- \theta )^{s}\,d \theta, \\ &L_{\varPhi,4}^{s,\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta +s}} \int _{ \frac{\lambda _{1}+\lambda _{2}}{2}}^{\lambda _{2}}\varPhi (\theta - \lambda _{1}) (\lambda _{2}-\theta )^{\zeta -1}(\theta -\lambda _{1})^{s}\,d \theta, \\ &L_{\varPhi,5}^{s,\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta +s}} \int _{ \frac{\lambda _{1}+\lambda _{2}}{2}}^{\lambda _{2}} \frac{\varPhi (\lambda _{2}-\theta )}{\lambda _{2}-\theta }(\theta - \lambda _{1})^{\zeta +s}\,d\theta, \\ &L_{\varPhi,6}^{\zeta }=\varPsi _{\varPhi }^{\zeta }- \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta }} \biggl[ \int _{\lambda _{1}}^{ \frac{\lambda _{1}+\lambda _{2}}{2}} \frac{\varPhi (\theta -\lambda _{1})}{\theta -\lambda _{1}}(\lambda _{2}- \theta )^{\zeta }\,d\theta \\ &\phantom{L_{\varPhi,6}^{\zeta }=}{}+ \int _{\lambda _{1}}^{ \frac{\lambda _{1}+\lambda _{2}}{2}}\varPhi (\lambda _{2}-\theta ) ( \theta -\lambda _{1})^{\zeta -1}\,d\theta \biggr], \\ &L_{\varPhi,7}^{s,\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta +s}} \int _{\lambda _{1}}^{ \frac{\lambda _{1}+\lambda _{2}}{2}}\varPhi (\lambda _{2}-\theta ) ( \theta -\lambda _{1})^{\zeta -1}(\lambda _{2}-\theta )^{s}\,d\theta, \\ &L_{\varPhi,8}^{s,\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta +s}} \int _{\lambda _{1}}^{ \frac{\lambda _{1}+\lambda _{2}}{2}} \frac{\varPhi (\theta -\lambda _{1})}{\theta -\lambda _{1}}(\lambda _{2}- \theta )^{\zeta +s}\,d\theta, \\ &L_{\varPhi,9}^{s,\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta +s}} \int _{\lambda _{1}}^{ \frac{\lambda _{1}+\lambda _{2}}{2}} \frac{\varPhi (\lambda _{2}-\theta )}{\lambda _{2}-\theta }(\theta - \lambda _{1})^{\zeta +s}\,d\theta, \\ &L_{\varPhi,10}^{s,\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta +s}} \int _{\lambda _{1}}^{ \frac{\lambda _{1}+\lambda _{2}}{2}}\varPhi (\theta -\lambda _{1}) ( \lambda _{2}-\theta )^{\zeta -1}(\theta -\lambda _{1})^{s}\,d\theta, \end{aligned}$$

and\(\varPsi _{\varPhi }^{\zeta }\)is defined by (15).

Proof

By Lemma 3.1, the s-convexity in the second sense of \(|f'|^{q}\), the well-known power mean inequality, and properties of the modulus we have

$$\begin{aligned} & \bigl\vert I_{f,\varOmega _{\varPhi }^{\zeta }}(\lambda _{1},\lambda _{2}) \bigr\vert \\ &\quad \leq \frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2} \int _{0}^{1} \bigl\vert \varOmega _{\varPhi }^{\zeta }(1-\theta )-\varOmega _{\varPhi }^{\zeta }( \theta ) \bigr\vert \bigl\vert f' \bigl(\theta \lambda _{1}+(1-\theta )\lambda _{2} \bigr) \bigr\vert \,d\theta \\ &\quad \leq \frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2} \biggl( \int _{0}^{1} \bigl\vert \varOmega _{\varPhi }^{\zeta }(1-\theta )-\varOmega _{\varPhi }^{\zeta }( \theta ) \bigr\vert \,d\theta \biggr)^{1-\frac{1}{q}} \\ &\qquad{}\times \biggl( \int _{0}^{1} \bigl\vert \varOmega _{\varPhi }^{\zeta }(1-\theta )- \varOmega _{\varPhi }^{\zeta }( \theta ) \bigr\vert \bigl\vert f' \bigl(\theta \lambda _{1}+(1- \theta )\lambda _{2} \bigr) \bigr\vert ^{q}\,d\theta \biggr)^{\frac{1}{q}} \\ &\quad\leq \frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2} \biggl\{ \biggl( \int _{0}^{\frac{1}{2}} \bigl[\varOmega _{\varPhi }^{\zeta }(1- \theta )- \varOmega _{\varPhi }^{\zeta }(\theta ) \bigr]\,d\theta \biggr)^{1-\frac{1}{q}} \\ &\qquad{}\times \biggl( \int _{0}^{\frac{1}{2}} \bigl[\varOmega _{\varPhi }^{\zeta }(1- \theta )-\varOmega _{\varPhi }^{\zeta }(\theta ) \bigr] \bigl[\theta ^{s} \bigl\vert f'( \lambda _{1}) \bigr\vert ^{q}+(1-\theta )^{s} \bigl\vert f'( \lambda _{2}) \bigr\vert ^{q} \bigr]\,d\theta \biggr)^{\frac{1}{q}} \\ &\qquad{}+ \biggl( \int _{\frac{1}{2}}^{1} \bigl[\varOmega _{\varPhi }^{\zeta }( \theta )-\varOmega _{\varPhi }^{\zeta }(1-\theta ) \bigr]\,d\theta \biggr)^{1- \frac{1}{q}} \\ &\qquad{}\times \biggl( \int _{\frac{1}{2}}^{1} \bigl[\varOmega _{\varPhi }^{\zeta }( \theta )-\varOmega _{\varPhi }^{\zeta }(1-\theta ) \bigr] \bigl[\theta ^{s} \bigl\vert f'( \lambda _{1}) \bigr\vert ^{q}+(1-\theta )^{s} \bigl\vert f'( \lambda _{2}) \bigr\vert ^{q} \bigr]\,d\theta \biggr)^{\frac{1}{q}} \biggr\} \\ &\quad =\frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2\sqrt[q]{s+1}} \bigl\{ \bigl[L_{\varPhi,1}^{\zeta } \bigr]^{1-\frac{1}{q}} \sqrt[q]{ \bigl(L_{\varPhi,2}^{s,\zeta }+L_{\varPhi,3}^{s,\zeta } \bigr) \bigl\vert f'(\lambda _{1}) \bigr\vert ^{q}+ \bigl(\varPsi _{\varPhi }^{\zeta }-L_{\varPhi,4}^{s,\zeta }-L_{\varPhi,5}^{s,\zeta } \bigr) \bigl\vert f'(\lambda _{2}) \bigr\vert ^{q}} \\ &\qquad{}+ \bigl[L_{\varPhi,6}^{\zeta } \bigr]^{1-\frac{1}{q}} \sqrt[q]{ \bigl( \varPsi _{\varPhi }^{\zeta }-L_{\varPhi,7}^{s,\zeta }-L_{\varPhi,8}^{s,\zeta } \bigr) \bigl\vert f'(\lambda _{1}) \bigr\vert ^{q}+ \bigl(L_{\varPhi,9}^{s,\zeta }+L_{\varPhi,10}^{s,\zeta } \bigr) \bigl\vert f'(\lambda _{2}) \bigr\vert ^{q}} \bigr\} . \end{aligned}$$

The proof of Theorem 6 is completed. □

We point out some particular cases of Theorem 6.

Corollary 8

Taking\({\varPhi (\theta )=\frac{\theta ^{n+1}}{n!}}\)and\(q=1\)in Theorem 6, we get ([30], Theorem 3.1).

Corollary 9

Taking\(s=1\)in Theorem 6, we get the following inequality for convex function via generalized conformable fractional integral operators:

$$\begin{aligned} & \bigl\vert I_{f,\varOmega _{\varPhi }^{\zeta }}(\lambda _{1},\lambda _{2}) \bigr\vert \\ &\quad \leq \frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2\sqrt[q]{2}} \bigl\{ \bigl[L_{\varPhi,1}^{\zeta } \bigr]^{1-\frac{1}{q}} \sqrt[q]{ \bigl(L_{\varPhi,2}^{\zeta }+L_{\varPhi,3}^{\zeta } \bigr) \bigl\vert f'(\lambda _{1}) \bigr\vert ^{q}+ \bigl(\varPsi _{\varPhi }^{\zeta }-L_{\varPhi,4}^{\zeta }-L_{\varPhi,5}^{\zeta } \bigr) \bigl\vert f'(\lambda _{2}) \bigr\vert ^{q}} \\ &\qquad+{} \bigl[L_{\varPhi,6}^{\zeta } \bigr]^{1-\frac{1}{q}} \sqrt[q]{ \bigl( \varPsi _{\varPhi }^{\zeta }-L_{\varPhi,7}^{\zeta }-L_{\varPhi,8}^{\zeta } \bigr) \bigl\vert f'(\lambda _{1}) \bigr\vert ^{q}+ \bigl(L_{\varPhi,9}^{\zeta }+L_{\varPhi,10}^{\zeta } \bigr) \bigl\vert f'(\lambda _{2}) \bigr\vert ^{q}} \bigr\} , \end{aligned}$$
(56)

where

$$\begin{aligned} &L_{\varPhi,2}^{\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta +1}} \int _{ \frac{\lambda _{1}+\lambda _{2}}{2}}^{\lambda _{2}} \frac{\varPhi (\theta -\lambda _{1})}{\theta -\lambda _{1}}(\lambda _{2}- \theta )^{\zeta +1}\,d\theta, \\ &L_{\varPhi,3}^{\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta +1}} \int _{ \frac{\lambda _{1}+\lambda _{2}}{2}}^{\lambda _{2}}\varPhi (\lambda _{2}- \theta ) (\theta -\lambda _{1})^{\zeta -1}(\lambda _{2}- \theta )\,d \theta, \\ &L_{\varPhi,4}^{\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta +1}} \int _{ \frac{\lambda _{1}+\lambda _{2}}{2}}^{\lambda _{2}}\varPhi (\theta - \lambda _{1}) (\lambda _{2}-\theta )^{\zeta -1}(\theta -\lambda _{1})\,d \theta, \\ &L_{\varPhi,5}^{\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta +1}} \int _{ \frac{\lambda _{1}+\lambda _{2}}{2}}^{\lambda _{2}} \frac{\varPhi (\lambda _{2}-\theta )}{\lambda _{2}-\theta }(\theta - \lambda _{1})^{\zeta +1}\,d\theta, \\ &L_{\varPhi,7}^{\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta +1}} \int _{\lambda _{1}}^{ \frac{\lambda _{1}+\lambda _{2}}{2}}\varPhi (\lambda _{2}-\theta ) ( \theta -\lambda _{1})^{\zeta -1}(\lambda _{2}-\theta )\,d\theta, \\ &L_{\varPhi,8}^{\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta +1}} \int _{\lambda _{1}}^{ \frac{\lambda _{1}+\lambda _{2}}{2}} \frac{\varPhi (\theta -\lambda _{1})}{\theta -\lambda _{1}}(\lambda _{2}- \theta )^{\zeta +1}\,d\theta, \\ &L_{\varPhi,9}^{\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta +1}} \int _{\lambda _{1}}^{ \frac{\lambda _{1}+\lambda _{2}}{2}} \frac{\varPhi (\lambda _{2}-\theta )}{\lambda _{2}-\theta }(\theta - \lambda _{1})^{\zeta +1}\,d\theta, \end{aligned}$$

and

$$\begin{aligned} &L_{\varPhi,10}^{\zeta }= \frac{1}{(\lambda _{2}-\lambda _{1})^{\zeta +1}} \int _{\lambda _{1}}^{ \frac{\lambda _{1}+\lambda _{2}}{2}}\varPhi (\theta -\lambda _{1}) ( \lambda _{2}-\theta )^{\zeta -1}(\theta -\lambda _{1})\,d\theta. \end{aligned}$$

Corollary 10

Taking\(|f'|\leq K\)in Theorem 6, we obtain

$$\begin{aligned} & \bigl\vert I_{f,\varOmega _{\varPhi }^{\zeta }}(\lambda _{1},\lambda _{2}) \bigr\vert \\ &\quad \leq K\frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2\sqrt[q]{s+1}} \bigl\{ \bigl[L_{\varPhi,1}^{\zeta } \bigr]^{1-\frac{1}{q}} \sqrt[q]{ \bigl(L_{\varPhi,2}^{s,\zeta }+L_{\varPhi,3}^{s,\zeta } \bigr)+ \bigl(\varPsi _{\varPhi }^{\zeta }-L_{\varPhi,4}^{s,\zeta }-L_{\varPhi,5}^{s,\zeta } \bigr)} \\ &\qquad{}+ \bigl[L_{\varPhi,6}^{\zeta } \bigr]^{1-\frac{1}{q}} \sqrt[q]{ \bigl( \varPsi _{\varPhi }^{\zeta }-L_{\varPhi,7}^{s,\zeta }-L_{\varPhi,8}^{s,\zeta } \bigr)+ \bigl(L_{\varPhi,9}^{s,\zeta }+L_{\varPhi,10}^{s,\zeta } \bigr)} \bigr\} . \end{aligned}$$
(57)

Theorem 7

Let\(f:[\lambda _{1},\lambda _{2}]\longrightarrow \mathbb{R}\)be a differentiable function on\((\lambda _{1},\lambda _{2})\). If\(f'\in L[\lambda _{1},\lambda _{2}]\)and\(|f'|^{q}\)iss-convex in the second sense with\(s\in (0,1]\), then for\(q>1\)and\(\frac{1}{p}+\frac{1}{q}=1\), we have the following inequality for generalized conformable fractional integrals:

$$\begin{aligned} & \bigl\vert I_{f,\varSigma _{\varPhi,1}^{\zeta },\varSigma _{\varPhi,2}^{\zeta }}(x; \lambda _{1},\lambda _{2}) \bigr\vert \\ &\quad \leq \frac{(x-\lambda _{1})^{\zeta }}{2} \sqrt[p]{\varXi _{\varPhi, 1}^{\zeta }(x,p)} \sqrt[q]{\frac{ \vert f'(x) \vert ^{q}+ \vert f'(\lambda _{1}) \vert ^{q}}{s+1}} \\ &\qquad{}+\frac{(\lambda _{2}-x)^{\zeta }}{2} \sqrt[p]{\varXi _{\varPhi, 2}^{\zeta }(x,p)} \sqrt[q]{\frac{ \vert f'(x) \vert ^{q}+ \vert f'(\lambda _{2}) \vert ^{q}}{s+1}}, \end{aligned}$$
(58)

where

$$ \varXi _{\varPhi, 1}^{\zeta }(x,p)= \int _{0}^{1} \bigl[\varSigma _{\varPhi,1}^{ \zeta }(x, \theta ) \bigr]^{p}\,d\theta, \qquad \varXi _{\varPhi, 2}^{\zeta }(x,p)= \int _{0}^{1} \bigl[\varSigma _{\varPhi,2}^{\zeta }(x, \theta ) \bigr]^{p}\,d \theta. $$

Proof

By Lemma 3.2, the s-convexity in the second sense of \(|f'|^{q}\), the Hölder inequality, and properties of the modulus we have

$$\begin{aligned} & \bigl\vert I_{f,\varSigma _{\varPhi,1}^{\zeta },\varSigma _{\varPhi,2}^{\zeta }}(x; \lambda _{1},\lambda _{2}) \bigr\vert \\ &\quad \leq \frac{(x-\lambda _{1})^{\zeta }}{2} \int _{0}^{1}\varSigma _{\varPhi,1}^{ \zeta }(x, \theta ) \bigl\vert f' \bigl(\theta x+(1-\theta )\lambda _{1} \bigr) \bigr\vert \,d \theta \\ &\qquad{}+\frac{(\lambda _{2}-x)^{\zeta }}{2} \int _{0}^{1}\varSigma _{\varPhi,2}^{ \zeta }(x, \theta ) \bigl\vert f' \bigl(\theta x+(1-\theta )\lambda _{2} \bigr) \bigr\vert \,d \theta \\ &\quad \leq \frac{(x-\lambda _{1})^{\zeta }}{2} \biggl( \int _{0}^{1} \bigl[ \varSigma _{\varPhi,1}^{\zeta }(x, \theta ) \bigr]^{p}\,d\theta \biggr)^{ \frac{1}{p}} \biggl( \int _{0}^{1} \bigl\vert f' \bigl( \theta x+(1-\theta ) \lambda _{1} \bigr) \bigr\vert ^{q} \,d\theta \biggr)^{\frac{1}{q}} \\ &\qquad{}+\frac{(\lambda _{2}-x)^{\zeta }}{2} \biggl( \int _{0}^{1} \bigl[ \varSigma _{\varPhi,2}^{\zeta }(x, \theta ) \bigr]^{p}\,d\theta \biggr)^{ \frac{1}{p}} \biggl( \int _{0}^{1} \bigl\vert f' \bigl( \theta x+(1-\theta ) \lambda _{2} \bigr) \bigr\vert ^{q} \,d\theta \biggr)^{\frac{1}{q}} \\ &\quad\leq\frac{(x-\lambda _{1})^{\zeta }}{2} \sqrt[p]{\varXi _{\varPhi, 1}^{\zeta }(x,p)} \biggl( \int _{0}^{1} \bigl[ \theta ^{s} \bigl\vert f'(x) \bigr\vert ^{q}+(1-\theta )^{s} \bigl\vert f'(\lambda _{1}) \bigr\vert ^{q} \bigr]\,d \theta \biggr)^{\frac{1}{q}} \\ &\qquad{}+\frac{(\lambda _{2}-x)^{\zeta }}{2} \sqrt[p]{\varXi _{\varPhi, 2}^{\zeta }(x,p)} \biggl( \int _{0}^{1} \bigl[ \theta ^{s} \bigl\vert f'(x) \bigr\vert ^{q}+(1-\theta )^{s} \bigl\vert f'(\lambda _{2}) \bigr\vert ^{q} \bigr]\,d \theta \biggr)^{\frac{1}{q}} \\ &\quad=\frac{(x-\lambda _{1})^{\zeta }}{2} \sqrt[p]{\varXi _{\varPhi, 1}^{\zeta }(x,p)} \sqrt[q]{\frac{ \vert f'(x) \vert ^{q}+ \vert f'(\lambda _{1}) \vert ^{q}}{s+1}} \\ &\qquad{}+\frac{(\lambda _{2}-x)^{\zeta }}{2} \sqrt[p]{\varXi _{\varPhi, 2}^{\zeta }(x,p)} \sqrt[q]{\frac{ \vert f'(x) \vert ^{q}+ \vert f'(\lambda _{2}) \vert ^{q}}{s+1}}. \end{aligned}$$

The proof of Theorem 7 is completed. □

We point out some particular cases of Theorem 7.

Corollary 11

Taking\({x=\frac{\lambda _{1}+\lambda _{2}}{2}}\)in Theorem 7, we get the following midpoint inequality via generalized conformable fractional integral operators:

$$\begin{aligned} & \biggl\vert I_{f,\varSigma _{\varPhi,1}^{\zeta },\varSigma _{\varPhi,2}^{\zeta }} \biggl( \frac{\lambda _{1}+\lambda _{2}}{2};\lambda _{1},\lambda _{2} \biggr) \biggr\vert \\ &\quad \leq \frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2^{\zeta +1}\sqrt[q]{s+1}} \sqrt[p]{\varXi _{\varPhi }^{\zeta }(p)} \\ &\qquad\times \biggl\{ \sqrt[q]{ \bigl\vert f'(\lambda _{1}) \bigr\vert ^{q}+ \biggl\vert f' \biggl(\frac{\lambda _{1}+\lambda _{2}}{2} \biggr) \biggr\vert ^{q}}+ \sqrt[q]{ \biggl\vert f' \biggl(\frac{\lambda _{1}+\lambda _{2}}{2} \biggr) \biggr\vert ^{q}+ \bigl\vert f'(\lambda _{2}) \bigr\vert ^{q}} \biggr\} , \end{aligned}$$
(59)

where

$$ \varXi _{\varPhi }^{\zeta }(p)= \int _{0}^{1} \biggl( \int _{0}^{\theta } \frac{\varPhi (u (\frac{\lambda _{2}-\lambda _{1}}{2} ) )}{u}(1-u)^{ \zeta -1} \,du \biggr)^{p}\,d\theta. $$

Corollary 12

Taking\(s=1\)in Theorem 7, we have the following inequality for convex function via generalized conformable fractional integral operators:

$$\begin{aligned} & \bigl\vert I_{f,\varSigma _{\varPhi,1}^{\zeta },\varSigma _{\varPhi,2}^{\zeta }}(x; \lambda _{1},\lambda _{2}) \bigr\vert \\ &\quad \leq \frac{(x-\lambda _{1})^{\zeta }}{2} \sqrt[p]{\varXi _{\varPhi, 1}^{\zeta }(x,p)} \sqrt[q]{\frac{ \vert f'(x) \vert ^{q}+ \vert f'(\lambda _{1}) \vert ^{q}}{2}} \\ &\qquad{} +\frac{(\lambda _{2}-x)^{\zeta }}{2} \sqrt[p]{\varXi _{\varPhi, 2}^{\zeta }(x,p)} \sqrt[q]{\frac{ \vert f'(x) \vert ^{q}+ \vert f'(\lambda _{2}) \vert ^{q}}{2}}. \end{aligned}$$
(60)

Corollary 13

Taking\(|f'|\leq K\)in Theorem 7, we obtain

$$ \bigl\vert I_{f,\varSigma _{\varPhi,1}^{\zeta },\varSigma _{\varPhi,2}^{\zeta }}(x; \lambda _{1},\lambda _{2}) \bigr\vert \leq \frac{K}{2} \sqrt[q]{ \frac{2}{s+1}} \bigl\{ (x-\lambda _{1})^{\zeta } \sqrt[p]{ \varXi _{\varPhi, 1}^{\zeta }(x,p)}+(\lambda _{2}-x)^{\zeta } \sqrt[p]{\varXi _{\varPhi, 2}^{\zeta }(x,p)} \bigr\} . $$
(61)

Theorem 8

Let\(f:[\lambda _{1},\lambda _{2}]\longrightarrow \mathbb{R}\)be a differentiable function on\((\lambda _{1},\lambda _{2})\). If\(f'\in L[\lambda _{1},\lambda _{2}]\)and\(|f'|^{q}\)iss-convex in the second sense with\(s\in (0,1]\), then for\(q\geq 1\), we have following inequality for generalized conformable fractional integrals:

$$\begin{aligned} & \bigl\vert I_{f,\varSigma _{\varPhi,1}^{\zeta },\varSigma _{\varPhi,2}^{\zeta }}(x; \lambda _{1},\lambda _{2}) \bigr\vert \\ &\quad \leq \frac{(x-\lambda _{1})^{\zeta }}{2}\bigl[M_{\varPhi,1}^{\zeta }(x) \bigr]^{1-\frac{1}{q}} \sqrt[q]{M_{\varPhi,2}^{s,\zeta }(x) \bigl\vert f'(x) \bigr\vert ^{q}+M_{\varPhi,3}^{s,\zeta }(x) \bigl\vert f'(\lambda _{1}) \bigr\vert ^{q}} \\ &\qquad{}+\frac{(\lambda _{2}-x)^{\zeta }}{2} \bigl[M_{\varPhi,4}^{\zeta }(x) \bigr]^{1-\frac{1}{q}} \sqrt[q]{M_{\varPhi,5}^{s,\zeta }(x) \bigl\vert f'(x) \bigr\vert ^{q}+M_{\varPhi,6}^{s,\zeta }(x) \bigl\vert f'(\lambda _{2}) \bigr\vert ^{q}}, \end{aligned}$$
(62)

where

$$\begin{aligned} &M_{\varPhi,1}^{\zeta }(x)= \frac{\varPsi _{\varPhi }^{\zeta }}{(x-\lambda _{1})^{\zeta -1}}- \frac{1}{(x-\lambda _{1})^{\zeta }} \int _{\lambda _{1}}^{x}\varPhi ( \theta -\lambda _{1}) (x-\theta )^{\zeta -1}\,d\theta, \\ &M_{\varPhi,2}^{s,\zeta }(x)= \frac{\varPsi _{\varPhi }^{\zeta }}{(s+1)(x-\lambda _{1})^{\zeta -1}}- \frac{1}{(s+1)(x-\lambda _{1})^{\zeta +s}} \int _{\lambda _{1}}^{x} \varPhi (\theta -\lambda _{1}) (\theta -\lambda _{1})^{s}(x-\theta )^{ \zeta -1}\,d\theta, \\ &M_{\varPhi,3}^{s,\zeta }(x)= \frac{1}{(s+1)(x-\lambda _{1})^{\zeta +s}} \int _{\lambda _{1}}^{x} \frac{\varPhi (\theta -\lambda _{1})}{\theta -\lambda _{1}}(x-\theta )^{ \zeta +s}\,d\theta, \\ &M_{\varPhi,4}^{\zeta }(x)= \frac{\varPsi _{\varPhi }^{\zeta }}{(\lambda _{2}-x)^{\zeta -1}}- \frac{1}{(\lambda _{2}-x)^{\zeta }} \int _{x}^{\lambda _{2}}\varPhi ( \lambda _{2}-\theta ) (\theta -x)^{\zeta -1}\,d\theta, \\ &M_{\varPhi,5}^{s,\zeta }(x)= \frac{\varPsi _{\varPhi }^{\zeta }}{(s+1)(\lambda _{2}-x)^{\zeta -1}}- \frac{1}{(s+1)(\lambda _{2}-x)^{\zeta +s}} \int _{x}^{\lambda _{2}} \varPhi (\theta -x) (\theta -x)^{s}(\lambda _{2}-\theta )^{\zeta -1}\,d \theta, \\ &M_{\varPhi,6}^{s,\zeta }(x)= \frac{1}{(s+1)(\lambda _{2}-x)^{\zeta +s}} \int _{x}^{\lambda _{2}} \frac{\varPhi (\theta -x)}{\theta -x}(\lambda _{2}-\theta )^{\zeta +s}\,d \theta, \end{aligned}$$

and\(\varPsi _{\varPhi }^{\zeta }\)is defined from (15).

Proof

By Lemma 3.2, the s-convexity in the second sense of \(|f'|^{q}\), the well–known power mean inequality, and properties of the modulus we have

$$\begin{aligned} & \bigl\vert I_{f,\varSigma _{\varPhi,1}^{\zeta },\varSigma _{\varPhi,2}^{\zeta }}(x; \lambda _{1},\lambda _{2}) \bigr\vert \\ &\quad \leq \frac{(x-\lambda _{1})^{\zeta }}{2} \int _{0}^{1}\varSigma _{\varPhi,1}^{ \zeta }(x, \theta ) \bigl\vert f' \bigl(\theta x+(1-\theta )\lambda _{1} \bigr) \bigr\vert \,d \theta \\ &\qquad{}+\frac{(\lambda _{2}-x)^{\zeta }}{2} \int _{0}^{1}\varSigma _{\varPhi,2}^{ \zeta }(x, \theta ) \bigl\vert f' \bigl(\theta x+(1-\theta )\lambda _{2} \bigr) \bigr\vert \,d \theta \\ &\quad \leq \frac{(x-\lambda _{1})^{\zeta }}{2} \biggl( \int _{0}^{1} \varSigma _{\varPhi,1}^{\zeta }(x, \theta )\,d\theta \biggr)^{1-\frac{1}{q}} \biggl( \int _{0}^{1}\varSigma _{\varPhi,1}^{\zeta }(x, \theta ) \bigl\vert f' \bigl( \theta x+(1-\theta )\lambda _{1} \bigr) \bigr\vert ^{q}\,d\theta \biggr)^{ \frac{1}{q}} \\ &\qquad{}+\frac{(\lambda _{2}-x)^{\zeta }}{2} \biggl( \int _{0}^{1}\varSigma _{ \varPhi,2}^{\zeta }(x, \theta )\,d\theta \biggr)^{1-\frac{1}{q}} \biggl( \int _{0}^{1}\varSigma _{\varPhi,2}^{\zeta }(x, \theta ) \bigl\vert f' \bigl(\theta x+(1- \theta )\lambda _{2} \bigr) \bigr\vert ^{q}\,d\theta \biggr)^{\frac{1}{q}} \\ &\quad \leq \frac{(x-\lambda _{1})^{\zeta }}{2} \bigl[M_{\varPhi,1}^{\zeta }(x) \bigr]^{1-\frac{1}{q}} \biggl( \int _{0}^{1}\varSigma _{\varPhi,1}^{\zeta }(x, \theta ) \bigl[\theta ^{s} \bigl\vert f'(x) \bigr\vert ^{q}+(1-\theta )^{s} \bigl\vert f'(\lambda _{1}) \bigr\vert ^{q} \bigr]\,d\theta \biggr)^{\frac{1}{q}} \\ &\qquad{}+\frac{(\lambda _{2}-x)^{\zeta }}{2} \bigl[M_{\varPhi,4}^{\zeta }(x) \bigr]^{1-\frac{1}{q}} \biggl( \int _{0}^{1}\varSigma _{\varPhi,2}^{\zeta }(x, \theta ) \bigl[\theta ^{s} \bigl\vert f'(x) \bigr\vert ^{q}+(1-\theta )^{s} \bigl\vert f'(\lambda _{2}) \bigr\vert ^{q} \bigr]\,d\theta \biggr)^{\frac{1}{q}} \\ &\quad =\frac{(x-\lambda _{1})^{\zeta }}{2} \bigl[M_{\varPhi,1}^{\zeta }(x) \bigr]^{1-\frac{1}{q}} \sqrt[q]{M_{\varPhi,2}^{s,\zeta }(x) \bigl\vert f'(x) \bigr\vert ^{q}+M_{\varPhi,3}^{s,\zeta }(x) \bigl\vert f'(\lambda _{1}) \bigr\vert ^{q}} \\ &\qquad{}+\frac{(\lambda _{2}-x)^{\zeta }}{2} \bigl[M_{\varPhi,4}^{\zeta }(x) \bigr]^{1-\frac{1}{q}} \sqrt[q]{M_{\varPhi,5}^{s,\zeta }(x) \bigl\vert f'(x) \bigr\vert ^{q}+M_{\varPhi,6}^{s,\zeta }(x) \bigl\vert f'(\lambda _{2}) \bigr\vert ^{q}}. \end{aligned}$$

The proof of Theorem 8 is completed. □

We point out some particular cases of Theorem 8.

Corollary 14

Taking\({x=\frac{\lambda _{1}+\lambda _{2}}{2}}\)in Theorem 8, we get the following midpoint inequality via generalized conformable fractional integral operators:

$$\begin{aligned} & \biggl\vert I_{f,\varSigma _{\varPhi,1}^{\zeta },\varSigma _{\varPhi,2}^{\zeta }} \biggl( \frac{\lambda _{1}+\lambda _{2}}{2};\lambda _{1},\lambda _{2} \biggr) \biggr\vert \\ &\quad \leq \frac{(\lambda _{2}-\lambda _{1})^{\zeta }}{2^{\zeta +1}} \\ &\qquad{}\times \biggl\{ \biggl[M_{\varPhi,1}^{\zeta } \biggl( \frac{\lambda _{1}+\lambda _{2}}{2} \biggr) \biggr]^{1-\frac{1}{q}} \\ &\qquad{}\times \sqrt[q]{M_{\varPhi,2}^{s,\zeta } \biggl(\frac{\lambda _{1}+\lambda _{2}}{2} \biggr) \biggl\vert f' \biggl( \frac{\lambda _{1}+\lambda _{2}}{2} \biggr) \biggr\vert ^{q}+M_{\varPhi,3}^{s,\zeta } \biggl(\frac{\lambda _{1}+\lambda _{2}}{2} \biggr) \bigl\vert f'(\lambda _{1}) \bigr\vert ^{q}} \\ &\qquad{}+ \biggl[M_{\varPhi,4}^{\zeta } \biggl( \frac{\lambda _{1}+\lambda _{2}}{2} \biggr) \biggr]^{1-\frac{1}{q}} \\ &\qquad{}\times \sqrt[q]{M_{\varPhi,5}^{s,\zeta } \biggl( \frac{\lambda _{1}+\lambda _{2}}{2} \biggr) \biggl\vert f' \biggl(\frac{\lambda _{1}+\lambda _{2}}{2} \biggr) \biggr\vert ^{q}+M_{\varPhi,6}^{s,\zeta } \biggl( \frac{\lambda _{1}+\lambda _{2}}{2} \biggr) \bigl\vert f'(\lambda _{2}) \bigr\vert ^{q}} \biggr\} . \end{aligned}$$
(63)

Corollary 15

Taking\(s=1\)in Theorem 8, we have the following inequality for convex functions via generalized conformable fractional integral operators:

$$\begin{aligned} & \bigl\vert I_{f,\varSigma _{\varPhi,1}^{\zeta },\varSigma _{\varPhi,2}^{\zeta }}(x; \lambda _{1},\lambda _{2}) \bigr\vert \\ &\quad \leq \frac{(x-\lambda _{1})^{\zeta }}{2} \bigl[M_{\varPhi,1}^{\zeta }(x) \bigr]^{1-\frac{1}{q}} \sqrt[q]{M_{\varPhi,2}^{\zeta }(x) \bigl\vert f'(x) \bigr\vert ^{q}+M_{\varPhi,3}^{\zeta }(x) \bigl\vert f'(\lambda _{1}) \bigr\vert ^{q}} \\ &\qquad{}+\frac{(\lambda _{2}-x)^{\zeta }}{2} \bigl[M_{\varPhi,4}^{\zeta }(x) \bigr]^{1-\frac{1}{q}} \sqrt[q]{M_{\varPhi,5}^{\zeta }(x) \bigl\vert f'(x) \bigr\vert ^{q}+M_{\varPhi,6}^{\zeta }(x) \bigl\vert f'(\lambda _{2}) \bigr\vert ^{q}}, \end{aligned}$$
(64)

where

$$\begin{aligned} &M_{\varPhi,2}^{\zeta }(x)= \frac{\varPsi _{\varPhi }^{\zeta }}{2(x-\lambda _{1})^{\zeta -1}}- \frac{1}{2(x-\lambda _{1})^{\zeta +1}} \int _{\lambda _{1}}^{x}\varPhi ( \theta -\lambda _{1}) (\theta -\lambda _{1}) (x-\theta )^{\zeta -1}\,d \theta, \\ &M_{\varPhi,3}^{\zeta }(x)=\frac{1}{2(x-\lambda _{1})^{\zeta +1}} \int _{ \lambda _{1}}^{x} \frac{\varPhi (\theta -\lambda _{1})}{\theta -\lambda _{1}}(x-\theta )^{ \zeta +1}\,d\theta, \\ &M_{\varPhi,5}^{\zeta }(x)= \frac{\varPsi _{\varPhi }^{\zeta }}{2(\lambda _{2}-x)^{\zeta -1}}- \frac{1}{2(\lambda _{2}-x)^{\zeta +1}} \int _{x}^{\lambda _{2}}\varPhi ( \theta -x) (\theta -x) ( \lambda _{2}-\theta )^{\zeta -1}\,d\theta, \end{aligned}$$

and

$$\begin{aligned} &M_{\varPhi,6}^{\zeta }(x)=\frac{1}{2(\lambda _{2}-x)^{\zeta +1}} \int _{x}^{ \lambda _{2}}\frac{\varPhi (\theta -x)}{\theta -x}(\lambda _{2}-\theta )^{ \zeta +1}\,d\theta. \end{aligned}$$

Corollary 16

Taking\(|f'|\leq K\)in Theorem 8, we obtain

$$\begin{aligned} & \bigl\vert I_{f,\varSigma _{\varPhi,1}^{\zeta },\varSigma _{\varPhi,2}^{\zeta }}(x; \lambda _{1},\lambda _{2}) \bigr\vert \\ &\quad \leq \frac{K}{2} \bigl\{ (x-\lambda _{1})^{\zeta } \bigl[M_{\varPhi,1}^{\zeta }(x) \bigr]^{1-\frac{1}{q}} \sqrt[q]{M_{\varPhi,2}^{s,\zeta }(x)+M_{\varPhi,3}^{s,\zeta }(x)} \\ &\qquad{}+(\lambda _{2}-x)^{\zeta } \bigl[M_{\varPhi,4}^{\zeta }(x) \bigr]^{1- \frac{1}{q}} \sqrt[q]{M_{\varPhi,5}^{s,\zeta }(x)+M_{\varPhi,6}^{s,\zeta }(x)} \bigr\} . \end{aligned}$$
(65)

Remark 4

Applying our Theorems 5, 6, 7, and 8 to suitable functions \(\varPhi (\theta )=\theta, \frac{\theta ^{\xi }}{\varGamma (\xi )}, \frac{\theta ^{\frac{\xi }{k}}}{k\varGamma _{k}(\xi )}\), \(\varPhi (\theta )=\theta (\lambda _{2}-\theta )^{\xi -1}\) for \(\xi \in (0,1), \varPhi (\theta )=\frac{\theta }{\xi }\exp (-A\theta )\), where \(A=\frac{1-\xi }{\xi }\) for \(\xi \in (0,1)\), and f such that \(|f'|^{q}\) is an s-convex function (\(f(x)=x^{2}\), etc.), we can construct some new generalized conformable fractional integral inequalities. Also, we can obtain several new general fractional integral inequalities using special means (arithmetic, geometric, logarithmic, etc.). Finally, some new bounds for the midpoint and trapezium quadrature formula using our results can be provided as well. We omit their proofs, and the details are left to the interested readers.

Conclusion

Trapezium-type integral inequalities for functions of divers natures are useful in numerical computations. Using the generalized conformable fractional integral operators defined in our paper, the interested reader can obtain in a similar way new results for different operators, such as k-Riemann–Liouville fractional integral, Katugampola fractional integrals, the conformable fractional integral, \((p,q)\)-quantum calculus, Hadamard fractional integrals, and so on. These results can be applied in convex analysis, optimization, probability, and also different areas of pure and applied sciences. The ideas and techniques of this paper may stimulate further research in the fascinating field of integral inequalities.

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Acknowledgements

We thank the reviewers for their great comments to improve the quality of our paper.

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This work was supported by National Natural Science Foundation of China (11761083).

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Correspondence to Wei Gao.

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Kashuri, A., Iqbal, S., Liko, R. et al. Integral inequalities for s-convex functions via generalized conformable fractional integral operators. Adv Differ Equ 2020, 217 (2020). https://doi.org/10.1186/s13662-020-02671-4

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MSC

  • 26A51
  • 26A33
  • 26D07
  • 26D10
  • 26D15

Keywords

  • Hermite–Hadamard inequality
  • Hölder inequality
  • Power mean inequality
  • Convexity
  • Conformable fractional integral
  • General fractional integral operators