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Generalized fractional integral inequalities of Hermite–Hadamard type for \({(\alpha,m)}\)-convex functions

Abstract

In the paper, the authors establish some generalized fractional integral inequalities of the Hermite–Hadamard type for \((\alpha,m)\)-convex functions, show that one can find some Riemann–Liouville fractional integral inequalities and classical integral inequalities of the Hermite–Hadamard type, and generalize and extend some known results.

Introduction

Let \(h: I\subseteq \mathbb{R}\to \mathbb{R}\) be a convex function with \(a< b\) and \(a, b\in I\). Then

$$ h \biggl(\frac{a+b}{2} \biggr)\le \frac{1}{b-a} \int _{a}^{b} h(x)\,\operatorname {d}x \le \frac{h(a)+h(b)}{2}. $$
(1.1)

Inequality (1.1) is well known in the literature as the Hermite–Hadamard inequality. A number of mathematicians have devoted their efforts to generalize, refine, counterpart, and extend the Hermite–Hadamard inequality (1.1) for different classes of convex functions and mappings. For several recent results concerning inequality (1.1), we may refer the interested reader to [1, 10, 14, 27, 33, 34, 39].

Let us recall some definitions and known results concerning convexity.

Definition 1.1

([33])

A function \(h: I\subseteq \mathbb{R}\to \mathbb{R}\) is said to be convex on an interval I if the inequality

$$ h \bigl(\lambda x+(1-\lambda )y \bigr)\le \lambda h(x)+(1-\lambda )h(y) $$

holds for all \(x,y\in I\) and \(\lambda \in (0,1)\).

Definition 1.2

([1, 33])

A function: \(h: [0,b]\to \mathbb{R}\) is said to be m-convex if

$$ h \bigl(\lambda a+m(1-\lambda )b \bigr)\le \lambda h(a)+m(1-\lambda )h(b) $$

holds for all \(a,b\in [0,b]\) and \(\lambda \in [0,1]\) and for some \(m\in (0,1]\).

Definition 1.3

([1, 33])

Let \((\alpha ,m)\in (0,1]^{2}\). A function: \(h: [0,b]\to \mathbb{R}\) is said to be \((\alpha ,m)\)-convex if

$$ h \bigl(\lambda a+m(1-\lambda )b \bigr)\le \lambda ^{\alpha }h(a)+m \bigl(1- \lambda ^{\alpha } \bigr)h(b) $$

holds for all \(a,b\in [0,b]\) and \(\lambda \in [0,1]\) and for some \(m\in (0,1]\).

The Riemann–Liouville integrals \(J_{a+}^{\alpha }h(t)\) and \(J_{b-}^{\alpha }h(t)\) of order \(\alpha \ge 0\) are defined in [5] respectively by \(J_{a+}^{0} h(t)=J _{b-}^{0} h(t)=h(t)\),

$$ J_{a+}^{\alpha }h(t)=\frac{1}{\varGamma (\alpha )} \int _{a}^{t}(t-u)^{ \alpha -1}h(u)\,\operatorname {d}u, \quad t>a $$

and

$$ J_{b-}^{\alpha }h(t)=\frac{1}{\varGamma (\alpha )} \int _{t}^{b}(u-t)^{ \alpha -1}h(u)\,\operatorname {d}u, \quad t< b $$

for \(h \in L_{1}([a,b])\) and \(\alpha >0\), where Γ denotes the classical Euler gamma function which can be defined [17, 22] by

$$ \varGamma (w)=\lim_{n\to \infty }\frac{n!n^{w}}{\prod_{k=0}^{n}(w+k)}, \quad w\in \mathbb{C} \setminus \{0,-1,-2,\ldots \} $$

or by

$$ \varGamma (w)= \int ^{\infty }_{0}u^{w-1} e^{-u} \,\operatorname {d}u, \quad \Re (w)>0. $$

Recently, the following integral identity and the Riemann–Liouville fractional integral inequalities of the Hermite–Hadamard type for \((\alpha ,m)\)-convex functions were obtained.

Lemma 1.1

([26, Lemma 2.1])

Let \(h:[a,b]\subseteq \mathbb{R}\to \mathbb{R}\) be differentiable on an interval \((a,b)\) with \(a< b\) such that \(h'\in L_{1}([a,b])\). Then

$$\begin{aligned} Q_{\alpha }(a,b)&=\frac{b-a}{16} \biggl[ \int _{0}^{1} \bigl(1-u^{\alpha } \bigr)h' \biggl(\frac{3a+b}{4}u+\frac{a+b}{2}(1-u) \biggr) \,\operatorname {d}u \\ &\quad{} - \int _{0}^{1} u^{\alpha }h' \biggl(au+\frac{3a+b}{4}(1-u) \biggr)\,\operatorname {d}u \\ &\quad{}+ \int _{0} ^{1} \bigl(1-u^{\alpha } \bigr)h' \biggl(\frac{a+3b}{4}u+b(1-u) \biggr)\,\operatorname {d}u \\ &\quad{}- \int _{0}^{1} u^{\alpha }h' \biggl( \frac{a+b}{2}u+\frac{a+3b}{4}(1-u) \biggr)\,\operatorname {d}u \biggr] \end{aligned}$$
(1.2)

for \(\alpha >0\), where

$$\begin{aligned} Q_{\alpha }(a,b)&=\frac{1}{2} \biggl[\frac{h(a)+h(b)}{2}+h \biggl( \frac{a+b}{2} \biggr) \biggr] -\frac{4^{\alpha -1}\varGamma (\alpha +1)}{(b-a)^{ \alpha }} \biggl[J_{a+}^{\alpha }h \biggl(\frac{3a+b}{4} \biggr) \\ &\quad{} +J_{[(3a+b)/4]+}^{\alpha }h \biggl(\frac{a+b}{2} \biggr)+J_{[(a+b)/2]+} ^{\alpha }h \biggl(\frac{a+3b}{4} \biggr)+J_{[(a+3b)/4]+}^{\alpha }h(b) \biggr]. \end{aligned}$$

Theorem 1.1

([26, Theorem 3.1])

Let \(h: [0,\infty )\to \mathbb{R}\) be differentiable on \([0,\infty )\) and \(h'\in L_{1}([a,b])\) for \(0\le a< b\) and \(\alpha >0\). If \(\vert h'\vert ^{q}\) is \((\alpha _{1},m)\)-convex on \([0,\frac{b}{m} ]\) for some \((\alpha _{1},m)\in (0,1]^{2}\) and \(q\ge 1\), then

$$\begin{aligned} \bigl\vert Q_{\alpha }(a,b) \bigr\vert &\le \frac{b-a}{16(\alpha +1)} \biggl[ \frac{1}{( \alpha _{1}+1)(\alpha +\alpha _{1}+1)} \biggr]^{1/q} \biggl[ \biggl(( \alpha +1) (\alpha _{1}+1) \bigl\vert h'(a) \bigr\vert ^{q} \\ & \quad {} +m\alpha _{1}(\alpha _{1}+1) \biggl\vert h' \biggl(\frac{3a+b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q}+\alpha \biggl((\alpha +1) \biggl\vert h' \biggl(\frac{3a+b}{4} \biggr) \biggr\vert ^{q} \\ & \quad {} +m\alpha _{1}(\alpha _{1}+\alpha +2) \biggl\vert h' \biggl( \frac{a+b}{2m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\ & \quad {} + \biggl((\alpha +1) (\alpha _{1}+1) \biggl\vert h' \biggl( \frac{a+b}{2} \biggr) \biggr\vert ^{q} +m\alpha _{1}( \alpha _{1}+1) \biggl\vert h' \biggl(\frac{a+3b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\ & \quad {} +\alpha \biggl((\alpha +1) \biggl\vert h' \biggl( \frac{a+3b}{4} \biggr) \biggr\vert ^{q}+m\alpha _{1}( \alpha _{1}+\alpha +2) \biggl\vert h' \biggl( \frac{b}{m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \biggr]. \end{aligned}$$
(1.3)

Theorem 1.2

([26, Theorem 3.2])

Let \(h: [0,\infty )\to \mathbb{R}\) be differentiable on \([0,\infty )\) and \(h'\in L_{1}([a,b])\) for \(0\le a< b\) and \(\alpha >0\). If \(\vert h'\vert ^{q}\) is \((\alpha _{1},m)\)-convex on \([0,\frac{b}{m} ]\) for some \((\alpha _{1},m)\in (0,1]^{2}\) and for \(q>1\) and \(q\ge r\ge 0\), then

$$\begin{aligned} \bigl\vert Q_{\alpha }(a,b) \bigr\vert &\le \frac{b-a}{16} \biggl\{ \biggl(\frac{q-1}{ \alpha (q-r)+q-1} \biggr)^{1-1/q} \biggl[ \frac{1}{\alpha r+\alpha _{1}+1} \bigl\vert h'(a) \bigr\vert ^{q} \\ & \quad {} +\frac{m\alpha _{1}}{(\alpha r+1)(\alpha r+\alpha _{1}+1)} \biggl\vert h' \biggl( \frac{3a+b}{4m} \biggr) \biggr\vert ^{q} \biggr]^{1/q} \\ & \quad {} +\frac{1}{\alpha }B^{1-1/q} \biggl(\frac{2q-r-1}{q-1}, \frac{1}{\alpha } \biggr) \biggl[B \biggl(r+1,\frac{\alpha _{1}+1}{\alpha } \biggr) \biggl\vert h' \biggl(\frac{3a+b}{4} \biggr) \biggr\vert ^{q} \\ & \quad {} +m \biggl(B \biggl(r+1,\frac{1}{\alpha } \biggr) -B \biggl(r+1, \frac{ \alpha _{1}+1}{\alpha } \biggr) \biggr) \biggl\vert h' \biggl( \frac{a+b}{2m} \biggr) \biggr\vert ^{q} \biggr]^{1/q} \\ & \quad {} + \biggl(\frac{q-1}{\alpha (q-r)+q-1} \biggr)^{1-1/q} \biggl[ \frac{1}{ \alpha r+\alpha _{1}+1} \biggl\vert h' \biggl(\frac{a+b}{2} \biggr) \biggr\vert ^{q} \\ & \quad {} +\frac{m\alpha _{1}}{(\alpha r+1)(\alpha r+\alpha _{1}+1)} \biggl\vert h' \biggl( \frac{a+3b}{4m} \biggr) \biggr\vert ^{q} \biggr]^{1/q} \\ & \quad {} +\frac{1}{\alpha }B^{1-1/q} \biggl(\frac{2q-r-1}{q-1}, \frac{1}{\alpha } \biggr) \biggl[B \biggl(r+1,\frac{\alpha _{1}+1}{\alpha } \biggr) \biggl\vert h' \biggl(\frac{a+3b}{4} \biggr) \biggr\vert ^{q} \\ & \quad {} +m \biggl(B \biggl(r+1,\frac{1}{\alpha } \biggr)-B \biggl(r+1, \frac{ \alpha _{1}+1}{\alpha } \biggr) \biggr) \biggl\vert h' \biggl( \frac{b}{m} \biggr) \biggr\vert ^{q} \biggr]^{1/q} \biggr\} , \end{aligned}$$
(1.4)

where \(B(s,t)\) denotes the classical beta function which can be defined [18, 19] by

$$ B(s,t)= \int _{0}^{1}u^{s-1}(1-u)^{t-1} \,\operatorname {d}u,\quad s, t>0. $$

For more information about the Hermite–Hadamard type inequalities for \((\alpha ,m)\)-convex functions, please refer to the papers [2, 3, 6, 15, 21, 26, 28,29,30, 32, 35, 38] and closely related references therein.

A review for generalized fractional integral operators

Now we recall some necessary definitions and mathematical preliminaries of the generalized fractional integrals which are defined by Sarikaya and Ertuğral in [24].

Let \(\varphi : [0,\infty )\to [0,\infty )\) satisfy the condition \(\int _{0}^{1} \frac{\varphi (t)}{t}\,\operatorname {d}t<\infty \).

We now define the left-sided and right-sided generalized fractional integral operators \({}_{a^{+}}I_{\varphi }h(t)\) and \({}_{b^{-}}I_{ \varphi }h(t)\) by

$$ {}_{a^{+}}I_{\varphi }h(t)= \int _{a}^{t} \frac{\varphi (t-u)}{t-u}h(u)\,\operatorname {d}u, \quad t>a $$
(2.1)

and

$$ {}_{b^{-}}I_{\varphi }h(t)= \int _{t}^{b} \frac{\varphi (u-t)}{u-t}h(u)\,\operatorname {d}u, \quad t< b. $$
(2.2)

The most important feature of generalized fractional integrals is that they generalize some types of fractional integrals such as the Riemann–Liouville fractional integrals [25, 26, 31], the k-Riemann–Liouville fractional integrals [11, 36], the Katugampola fractional integrals [7, 8], conformable fractional integrals [23, 37], the Hadamard fractional integrals [16], and so on. These important special cases of the integral operators in (2.1) and (2.2) are mentioned below.

  1. 1.

    If we take \(\varphi (u)=u\), the operators in (2.1) and (2.2) reduce to the Riemann integrals

    $$ I_{a^{+}}h(t)= \int _{a}^{t} h(u)\,\operatorname {d}u, \quad t>a \quad \text{and} \quad I_{b^{-}}h(t)= \int _{t}^{b} h(u)\,\operatorname {d}u, \quad t< b. $$
  2. 2.

    If we take \(\varphi (u)=\frac{u^{\alpha }}{\varGamma (\alpha )}\), the operators in (2.1) and (2.2) become the Riemann–Liouville fractional integrals

    $$ I_{a^{+}}h(t)=\frac{1}{\varGamma (\alpha )} \int _{a}^{t} (t-u)^{\alpha -1}h(u)\,\operatorname {d}u, \quad t>a $$

    and

    $$ I_{b^{-}}h(t)=\frac{1}{\varGamma (\alpha )} \int _{t}^{b} (u-t)^{\alpha -1}h(u)\,\operatorname {d}u, \quad t< b. $$
  3. 3.

    If we take \(\varphi (u)=\frac{u^{\alpha /k}}{k\varGamma _{k}(\alpha )}\), the operators in (2.1) and (2.2) are the k-Riemann–Liouville fractional integrals

    $$ I_{a^{+},k}h(t)=\frac{1}{k\varGamma _{k}(\alpha )} \int _{a}^{t} (t-u)^{ \alpha /k-1}h(u)\,\operatorname {d}u, \quad t>a $$

    and

    $$ I_{b^{-},k}h(t)=\frac{1}{k\varGamma _{k}(\alpha )} \int _{t}^{b} (u-t)^{ \alpha /k-1}h(u)\,\operatorname {d}u, \quad t< b, $$

    where

    $$ \varGamma _{k}(\alpha )= \int _{0}^{\infty }u^{\alpha -1}e^{-u^{k}/k}\,\operatorname {d}u, \quad \mathbb{R}(\alpha )>0 $$

    and

    $$ \varGamma _{k}(\alpha )=k^{\alpha /k-1}\varGamma \biggl( \frac{\alpha }{k} \biggr),\quad \mathbb{R}(\alpha )>0, k>0 $$

    are given in [13, 20].

  4. 4.

    If we take \(\varphi (u)=\frac{u}{\alpha }\exp (-\frac{1-\alpha }{ \alpha }u )\), the operators in (2.1) and (2.2) reduce to the right-sided and left-sided fractional integral operators with exponential kernel for \(\alpha \in (0,1)\)

    $$ \mathcal{I}_{a^{+}}^{\alpha }h(t)=\frac{1}{\alpha } \int _{a}^{t} \exp \biggl(-\frac{1-\alpha }{\alpha }(t-u) \biggr)h(u)\,\operatorname {d}u, \quad t>a $$

    and

    $$ \mathcal{I}_{b^{-}}^{\alpha }h(t)=\frac{1}{\alpha } \int _{t}^{b} \exp \biggl(-\frac{1-\alpha }{\alpha }(u-t) \biggr)h(u)\,\operatorname {d}u, \quad t< b $$

    which are defined in [9].

Recently, Sarikaya and Ertuğral [24] established the following trapezoid inequalities for generalized fractional integrals.

Theorem 2.1

([24])

Let \(h: [a,b]\to \mathbb{R}\) be differentiable on \((a,b)\) with \(a< b\). If \(\vert h'\vert \) is convex on \([a,b]\), then

$$\begin{aligned}& \biggl\vert \frac{h(a)+h(b)}{2}-\frac{1}{2\varLambda (1)} \bigl[{}_{a^{+}}I _{\varphi }h(b)+{}_{b^{-}}I_{\varphi }h(a) \bigr] \biggr\vert \\& \quad \le \frac{ \vert h'(a) \vert + \vert h'(b) \vert }{2}\frac{b-a}{\varLambda (1)} \int _{0}^{1} u \bigl\vert \varLambda (1-u)- \varLambda (u) \bigr\vert \,\operatorname {d}u, \end{aligned}$$

where

$$ \varLambda (u)= \int _{0}^{u} \frac{\varphi ((b-a)t)}{t}\,\operatorname {d}t< \infty . $$

Theorem 2.2

([24])

Let \(h: [a,b]\to \mathbb{R}\) be differentiable on \((a,b)\) with \(a< b\). If \(\vert h'\vert ^{q}\) is convex on \([a,b]\) for \(p,q>1\) and \(\frac{1}{p}+ \frac{1}{q}=1\), then

$$\begin{aligned}& \biggl\vert \frac{h(a)+h(b)}{2}-\frac{{}_{a^{+}}I_{\varphi }h(b)+{}_{b ^{-}}I_{\varphi }h(a)}{2\varLambda (1)} \biggr\vert \\& \quad \le \frac{b-a}{2\varLambda (1)} \biggl[\frac{ \vert h'(a) \vert ^{q}+ \vert h'(b) \vert ^{q}}{2} \biggr]^{1/q} \biggl[ \int _{0}^{1} \bigl\vert \varLambda (1-u)- \varLambda (u) \bigr\vert ^{p}\,\operatorname {d}u \biggr]^{1/p}. \end{aligned}$$

In [4], Ertuğral and Sarikaya established the following trapezoid inequalities for generalized fractional integrals.

Theorem 2.3

([4])

Let \(h: [a,b]\to \mathbb{R}\) be absolutely continuous on \(I^{\circ }\) such that \(h'\in L_{1}([a,b])\) with \(a,b\in I^{\circ }\) with \(a< b\). If the mapping \(\vert h'\vert \) is convex on \([a,b]\), then

$$\begin{aligned}& \biggl\vert \frac{\nabla (0)h(b)+\Delta (0)h(a)}{b-a}-\frac{1}{b-a} \bigl[{}_{a^{+}}I_{\varphi }h(b)+{}_{b^{-}}I_{\varphi }h(a) \bigr] \biggr\vert \\& \quad \le \frac{b-x}{b-a} \bigl\vert h'(x) \bigr\vert \int _{0}^{1} \bigl\vert \nabla (u) \bigr\vert u \,\operatorname {d}u+ \frac{x-a}{b-a} \bigl\vert h'(x) \bigr\vert \int _{0}^{1} \bigl\vert \Delta (u) \bigr\vert u \,\operatorname {d}u \\ & \quad\quad{} +\frac{b-x}{b-a} \bigl\vert h'(b) \bigr\vert \int _{0}^{1} \bigl\vert \nabla (u) \bigr\vert (1-u)\,\operatorname {d}u+ \frac{x-a}{b-a} \bigl\vert h'(a) \bigr\vert \int _{0}^{1} \bigl\vert \Delta (u) \bigr\vert (1-u)\,\operatorname {d}u, \end{aligned}$$

where

$$ \Delta (u)= \int _{t}^{1} \frac{\varphi ((x-a)t)}{t}\,\operatorname {d}t< \infty \quad \textit{and}\quad \nabla (u)= \int _{t}^{1} \frac{\varphi ((b-x)t)}{t}\,\operatorname {d}t< \infty . $$

Theorem 2.4

([4])

Let \(h: [a,b]\to \mathbb{R}\) be differentiable on \((a,b)\) with \(a< b\). If \(\vert h'\vert ^{q}\) for \(q>1\) is convex on \([a,b]\), then

$$\begin{aligned}& \biggl\vert \frac{h(a)+h(b)}{2}-\frac{1}{2\varLambda (1)} \bigl[{}_{a^{+}}I _{\varphi }h(b)+{}_{b^{-}}I_{\varphi }h(a) \bigr] \biggr\vert \\ & \quad \le \frac{b-x}{b-a} \biggl[ \int _{0}^{1} \bigl\vert \nabla (u) \bigr\vert ^{p} \,\operatorname {d}u \biggr]^{1/p} \biggl[\frac{ \vert h'(a) \vert ^{q}+ \vert h'(b) \vert ^{q}}{2} \biggr]^{1/q} \\ & \quad\quad{} +\frac{b-x}{b-a} \biggl[ \int _{0}^{1} \bigl\vert \Delta (u) \bigr\vert ^{p} \,\operatorname {d}u \biggr]^{1/p} \biggl[\frac{ \vert h'(a) \vert ^{q}+ \vert h'(b) \vert ^{q}}{2} \biggr]^{1/q}, \end{aligned}$$

where \(\frac{1}{p}+\frac{1}{q}=1\).

Most recently, Mohammed and Sarikayain [12] established some generalized fractional integral inequalities of midpoint and trapezoid types for twice differential functions.

Theorem 2.5

([12])

Let \(h:I\subseteq \mathbb{R}\to \mathbb{R}\) be a twice differentiable function on \(I^{\circ }\) such that \(h''\in L_{1}([a,b])\) with \(a,b\in I^{\circ }\) and \(a< b\). If the function \(\vert h''\vert \) is convex on \([a,b]\), then

$$\begin{aligned}& \biggl\vert \bigl[{}_{ (\frac{a+b}{2} )^{+}}I_{\varphi }h(b)+ {}_{ (\frac{a+b}{2} )^{-}}I_{\varphi }h(a) \bigr] -2\nabla (1)h \biggl(\frac{a+b}{2} \biggr) \biggr\vert \\ & \quad \le \frac{(b-a)^{2}}{4} \bigl( \bigl\vert h''(a) \bigr\vert + \bigl\vert h''(b) \bigr\vert \bigr) \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert \,\operatorname {d}t, \end{aligned}$$

where

$$ \Delta (t)= \int _{0}^{t} \nabla (u)\,\operatorname {d}u< \infty \quad \textit{and} \quad \nabla (u)= \int _{0}^{u}\frac{\varphi ( (\frac{b-a}{2} )s )}{s}\,\operatorname {d}s< \infty . $$

Theorem 2.6

([12])

Let \(h:I\subseteq \mathbb{R}\to \mathbb{R}\) be twice differentiable on \(I^{\circ }\) such that \(h''\in L_{1}([a,b])\) with \(a,b\in I^{\circ }\) and \(a< b\). If \(\vert h''\vert ^{q}\) for \(q>1\) is convex on \([a,b]\), then

$$\begin{aligned} & \biggl\vert \bigl[{}_{ (\frac{a+b}{2} )^{+}}I_{\varphi }h(b)+ {}_{ (\frac{a+b}{2} )^{-}}I_{\varphi }h(a) \bigr] -2\nabla (1)h \biggl( \frac{a+b}{2} \biggr) \biggr\vert \\ &\quad \le \frac{(b-a)^{2}}{4} \biggl( \int _{0}^{1} \bigl\vert \Delta (u) \bigr\vert ^{p}\,\operatorname {d}u \biggr)^{1/p} \biggl\{ \biggl( \frac{ \vert h''(a) \vert ^{q}+3 \vert h''(b) \vert ^{q}}{4} \biggr)^{1/q} \\ &\quad \quad {} + \biggl(\frac{3 \vert h''(a) \vert ^{q}+ \vert h''(b) \vert ^{q}}{4} \biggr)^{1/q} \biggr\} \\ &\quad \le \frac{(b-a)^{2}}{2^{2/q}} \biggl( \int _{0}^{1} \bigl\vert \Delta (u) \bigr\vert ^{p}\,\operatorname {d}u \biggr)^{1/p} \bigl( \bigl\vert h''(a) \bigr\vert + \bigl\vert h''(b) \bigr\vert \bigr), \end{aligned}$$

where \(\frac{1}{p}+\frac{1}{q}=1\).

Theorem 2.7

([12])

Let \(h:I\subseteq \mathbb{R}\to \mathbb{R}\) be twice differentiable on \(I^{\circ }\) such that \(h''\in L_{1}([a,b])\) with \(a,b\in I^{\circ }\) and \(a< b\). If \(\vert h''\vert \) is convex on \([a,b]\), then

$$\begin{aligned}& \biggl\vert \frac{h(a)+h(b)}{2}-\frac{1}{2\varPhi (0)} \bigl[{}_{ ( \frac{a+b}{2} )^{+}}I_{\varphi }h(b) +{}_{ (\frac{a+b}{2} )^{-}}I_{\varphi }h(a) \bigr] \biggr\vert \\& \quad \le \frac{(b-a)^{2}}{8\varPhi (0)} \biggl[ \int _{0}^{1} \bigl\vert \Delta (u) \bigr\vert \,\operatorname {d}u \bigl( \bigl\vert h''(a) \bigr\vert + \bigl\vert h''(b) \bigr\vert \bigr) \biggr]. \end{aligned}$$

Theorem 2.8

([12])

Let \(h:I\subseteq \mathbb{R}\to \mathbb{R}\) be twice differentiable on \(I^{\circ }\) such that \(h''\in L_{1}([a,b])\) with \(a,b\in I^{\circ }\) and \(a< b\). If \(\vert h''\vert ^{q}\) for \(q>1\) is convex on \([a,b]\), then

$$\begin{aligned} & \biggl\vert \frac{h(a)+h(b)}{2}-\frac{1}{2\varPhi (0)} \bigl[{}_{ (\frac{a+b}{2} )^{+}}I_{\varphi }h(b) +{}_{ ( \frac{a+b}{2} )^{-}}I_{\varphi }h(a) \bigr] \biggr\vert \\ &\quad \le \frac{(b-a)^{2}}{8\varPhi (0)} \biggl( \int _{0}^{1} \bigl\vert \Delta (u) \bigr\vert ^{p}\,\operatorname {d}u \biggr)^{1/p} \biggl\{ \biggl( \frac{ \vert h''(a) \vert ^{q}+3 \vert h''(b) \vert ^{q}}{4} \biggr)^{1/q} \\ &\quad \quad {} + \biggl(\frac{3 \vert h''(a) \vert ^{q}+ \vert h''(b) \vert ^{q}}{4} \biggr)^{1/q} \biggr\} \\ &\quad \le \frac{(b-a)^{2}}{2^{3/q}\varPhi (0)} \biggl( \int _{0}^{1} \bigl\vert \Delta (u) \bigr\vert ^{p}\,\operatorname {d}u \biggr)^{1/p} \bigl( \bigl\vert h''(a) \bigr\vert + \bigl\vert h''(b) \bigr\vert \bigr), \end{aligned}$$

where \(\frac{1}{p}+\frac{1}{q}=1\).

A generalized fractional integral identity

Before stating and proving our main results, we formulate the following important fractional integral identity.

Lemma 3.1

Let \(f:[a,b]\to \mathbb{R}\) be a differentiable function on \((a,b)\) with \(a< b\) such that \(f\in L_{1}([a,b])\). Then

$$\begin{aligned}& \frac{f(a)+f(b)}{2}+2f \biggl(\frac{a+b}{2} \biggr)- \frac{2}{\Delta (1)} \biggl[{}_{a^{+}}I_{\varphi }f \biggl( \frac{3a+b}{4} \biggr) \\& \quad\quad {} +{}_{ (\frac{3a+b}{4} )^{+}}I_{\varphi }f \biggl( \frac{a+b}{2} \biggr) +{}_{ (\frac{a+b}{2} )^{+}}I_{\varphi }f \biggl(\frac{a+3b}{4} \biggr) +{}_{ (\frac{a+3b}{4} )^{-}}I _{\varphi }f(b) \biggr] \\& \quad =\frac{b-a}{8\Delta (1)} \biggl[ \int _{0}^{1}\nabla (t)f' \biggl( \frac{3a+b}{4}t+\frac{a+b}{2}(1-t) \biggr)\,\operatorname {d}t \\& \quad \quad {} - \int _{0} ^{1}\Delta (t)f' \biggl(at+ \frac{3a+b}{4}(1-t) \biggr)\,\operatorname {d}t + \int _{0}^{1}\nabla (t)f' \biggl( \frac{a+3b}{4}t+b(1-t) \biggr)\,\operatorname {d}t \\& \quad \quad {} - \int _{0}^{1}\Delta (t)f' \biggl( \frac{a+b}{2}t+\frac{a+3b}{4}(1-t) \biggr)\,\operatorname {d}t \biggr], \end{aligned}$$
(3.1)

where

$$ \Delta (t)= \int _{0}^{t}\frac{\varphi ( (\frac{b-a}{4} )u )}{u}\,\operatorname {d}u< \infty \quad \textit{and}\quad \nabla (t)= \int _{t} ^{1}\frac{\varphi ( (\frac{b-a}{4} )u )}{u}\,\operatorname {d}u< \infty . $$

Proof

Integrating by parts gives

$$\begin{aligned} I_{1}&=\frac{b-a}{8\Delta (1)} \int _{0}^{1}\nabla (t)f' \biggl( \frac{3a+b}{4}t+\frac{a+b}{2}(1-t) \biggr)\,\operatorname {d}t =\frac{1}{2 [ \Delta (1)+\nabla (0) ]} \\ &\quad{} \times \biggl[\nabla (0)f \biggl(\frac{a+b}{2} \biggr) - \int _{0}^{1} \frac{ \varphi ( (\frac{b-a}{4} )t )}{t} f \biggl( \frac{3a+b}{4}t+\frac{a+b}{2}(1-t) \biggr)\,\operatorname {d}t \biggr]. \end{aligned}$$

Changing the variable \(x=\frac{3a+b}{4}t+\frac{a+b}{2}(1-t)\) yields

$$\begin{aligned} I_{1} &=\frac{1}{2\Delta (1)} \biggl[\nabla (0)f \biggl(\frac{a+b}{2} \biggr) + \int _{\frac{a+b}{2}}^{\frac{3a+b}{4}}\frac{\varphi ( \frac{a+b}{2}-x )}{\frac{a+b}{2}-x}f(x)\,\operatorname {d}x \biggr] \\ &=\frac{1}{2\Delta (1)} \biggl[\nabla (0)f \biggl(\frac{a+b}{2} \biggr) -{}_{ (\frac{3a+b}{4} )^{+}}I_{\varphi }f \biggl(\frac{a+b}{2} \biggr) \biggr]. \end{aligned}$$

Similarly, we obtain

$$\begin{aligned}& \begin{aligned} I_{2} &=-\frac{b-a}{8\Delta (1)} \int _{0}^{1}\Delta (t)f' \biggl(at+ \frac{3a+b}{4}(1-t) \biggr)\,\operatorname {d}t \\ &=\frac{1}{2\Delta (1)} \biggl[\Delta (1)f(a)- \int _{0}^{1}\frac{ \varphi ( (\frac{b-a}{4} )t )}{t} f \biggl(at+ \frac{3a+b}{4}(1-t) \biggr)\,\operatorname {d}t \biggr] \\ &=\frac{1}{2\Delta (1)} \biggl[\Delta (1)f(a)-{}_{a^{+}}I_{\varphi }f \biggl(\frac{3a+b}{4} \biggr) \biggr], \end{aligned} \\& \begin{aligned} I_{3} &=\frac{b-a}{8\Delta (1)} \int _{0}^{1}\nabla (t)f' \biggl( \frac{a+3b}{4}t+b(1-t) \biggr)\,\operatorname {d}t \\ &=\frac{1}{2\Delta (1)} \biggl[\nabla (0)f(b)- \int _{0}^{1}\frac{ \varphi ( (\frac{b-a}{4} )t )}{t} f \biggl( \frac{a+3b}{4}t+b(1-t) \biggr)\,\operatorname {d}t \biggr] \\ &=\frac{1}{2\Delta (1)} \bigl[\nabla (0)f(b)-{}_{ ( \frac{a+3b}{4} )^{+}}I_{\varphi }f(b) \bigr] \end{aligned} \end{aligned}$$

and

$$\begin{aligned} I_{4} &=-\frac{b-a}{8\Delta (1)} \int _{0}^{1}\Delta (t)f' \biggl( \frac{3a+b}{4}t+\frac{a+b}{2}(1-t) \biggr)\,\operatorname {d}t \\ &=\frac{1}{2\Delta (1)} \biggl[\Delta (1)f \biggl(\frac{a+b}{2} \biggr)- \int _{0}^{1}\frac{\varphi ( (\frac{b-a}{4} )t )}{t} f \biggl( \frac{3a+b}{4}t+\frac{a+b}{2}(1-t) \biggr)\,\operatorname {d}t \biggr] \\ &=\frac{1}{2\Delta (1)} \biggl[\Delta (1)f \biggl(\frac{a+b}{2} \biggr) -{}_{ (\frac{a+b}{2} )^{+}}I_{\varphi }f \biggl(\frac{a+3b}{4} \biggr) \biggr], \end{aligned}$$

where we used the fact that \(\Delta (1)=\nabla (0)\). Adding \(I_{1}\), \(I_{2}\), \(I_{3}\), and \(I_{4}\) results in identity (3.1). The proof is thus completed. □

Remark 3.1

Since \(\Delta (1)=\nabla (0)\), we can write identity (3.1) in Lemma 3.1 as

$$\begin{aligned} & \frac{f(a)+f(b)}{2}+2f \biggl(\frac{a+b}{2} \biggr)-\frac{2}{ \nabla (0)} \biggl[{}_{a^{+}}I_{\varphi }f \biggl(\frac{3a+b}{4} \biggr) \\ &\quad \quad {} +{}_{ (\frac{3a+b}{4} )^{+}}I_{\varphi }f \biggl( \frac{a+b}{2} \biggr) +{}_{ (\frac{a+b}{2} )^{+}}I_{\varphi }f \biggl(\frac{a+3b}{4} \biggr) +{}_{ (\frac{a+3b}{4} )^{-}}I _{\varphi }f(b) \biggr] \\ &\quad =\frac{b-a}{8\nabla (0)} \biggl[ \int _{0}^{1}\nabla (t)f' \biggl( \frac{3a+b}{4}t+\frac{a+b}{2}(1-t) \biggr)\,\operatorname {d}t \\ &\quad\quad {} - \int _{0}^{1}\Delta (t)f' \biggl(at+ \frac{3a+b}{4}(1-t) \biggr)\,\operatorname {d}t + \int _{0}^{1}\nabla (t)f' \biggl( \frac{a+3b}{4}t \\ &\quad\quad {} +b(1-t) \biggr)\,\operatorname {d}t- \int _{0}^{1}\Delta (t)f' \biggl( \frac{a+b}{2}t+\frac{a+3b}{4}(1-t) \biggr)\,\operatorname {d}t \biggr]. \end{aligned}$$

Remark 3.2

Under assumptions of Lemma 3.1, if \(\varphi (t)=t\), then identity (3.1) reduces to

$$\begin{aligned} & \frac{1}{2} \biggl[\frac{f(a)+f(b)}{2}+f \biggl(\frac{a+b}{2} \biggr) \biggr]-\frac{1}{b-a} \int _{a}^{b} f(x)\,\operatorname {d}x \\ &\quad =\frac{b-a}{16} \biggl[ \int _{0}^{1}(1-t)f' \biggl( \frac{3a+b}{4}t+ \frac{a+b}{2}(1-t) \biggr)\,\operatorname {d}t \\ &\quad \quad {} - \int _{0}^{1}tf' \biggl(at+ \frac{3a+b}{4}(1-t) \biggr)\,\operatorname {d}t + \int _{0}^{1}(1-t)f' \biggl( \frac{a+3b}{4}t \\ &\quad \quad {} +b(1-t) \biggr)\,\operatorname {d}t- \int _{0}^{1}tf' \biggl( \frac{a+b}{2}t+ \frac{a+3b}{4}(1-t) \biggr)\,\operatorname {d}t \biggr], \end{aligned}$$

which has been proved in [26].

Remark 3.3

Under assumptions of Lemma 3.1, if \(\varphi (t)=\frac{t^{ \alpha }}{\varGamma (\alpha )}\), then identity (3.1) reduces to identity (1.2).

Remark 3.4

Under assumptions of Lemma 3.1, if \(\varphi (t)=\frac{t^{ \alpha /k}}{{k\varGamma _{k}(\alpha )}}\), then

$$\begin{aligned} &\frac{1}{2} \biggl[\frac{f(a)+f(b)}{2}+f \biggl(\frac{a+b}{2} \biggr) \biggr] -\frac{4^{\alpha /k-1}\varGamma _{k}(\alpha +1)}{(b-a)^{ \alpha /k}} \biggl[I^{\alpha }_{a^{+},k}f \biggl( \frac{3a+b}{4} \biggr) \\ &\quad \quad {} +I^{\alpha }_{ (\frac{3a+b}{4} )^{+},k}f \biggl( \frac{a+b}{2} \biggr) +I^{\alpha }_{ (\frac{a+b}{2} )^{+},k}f \biggl(\frac{a+3b}{4} \biggr) +I^{\alpha }_{ (\frac{a+3b}{4} )^{+},k}f(b) \biggr] \\ &\quad =\frac{b-a}{16} \biggl[ \int _{0}^{1} \bigl(1-t^{\alpha /k} \bigr)f' \biggl(\frac{3a+b}{4}t+\frac{a+b}{2}(1-t) \biggr) \,\operatorname {d}t \\ &\quad \quad {} - \int _{0}^{1} t^{\alpha /k} f' \biggl(at+\frac{3a+b}{4}(1-t) \biggr)\,\operatorname {d}t \\ &\quad \quad {} + \int _{0}^{1} \bigl(1-t^{\alpha /k} \bigr)f' \biggl( \frac{a+3b}{4}t+b(1-t) \biggr)\,\operatorname {d}t \\ &\quad \quad {} - \int _{0}^{1} t^{\alpha /k}f' \biggl( \frac{a+b}{2}t+ \frac{a+3b}{4}(1-t) \biggr)\,\operatorname {d}t \biggr]. \end{aligned}$$

Remark 3.5

Under assumptions of Lemma 3.1, applying \(\varphi (t)=\frac{t}{ \alpha }\exp (-\frac{1-\alpha }{\alpha }t )\) gives

$$\begin{aligned} &\frac{f(a)+f(b)}{2}+2f \biggl(\frac{a+b}{2} \biggr) - \frac{2(1- \alpha )}{1-\exp (-A)} \biggl[\mathcal{I}^{\alpha }{}_{a^{+}}f \biggl( \frac{3a+b}{4} \biggr) \\ &\quad \quad {} +\mathcal{I}^{\alpha }_{ (\frac{3a+b}{4} )^{+}}f \biggl( \frac{a+b}{2} \biggr) +\mathcal{I}^{\alpha }{}_{ ( \frac{a+b}{2} )^{+}}f \biggl( \frac{a+3b}{4} \biggr) +\mathcal{I} ^{\alpha }_{ (\frac{a+3b}{4} )^{-}}f(b) \biggr] \\ &\quad =\frac{b-a}{8[1-\exp (-A)]} \biggl[ \int _{0}^{1} \bigl[\exp (-At)-\exp (-A) \bigr]f' \biggl(\frac{3a+b}{4}t \\ &\quad \quad {} +\frac{a+b}{2}(1-t) \biggr)\,\operatorname {d}t - \int _{0}^{1} \bigl[1-\exp (-At) \bigr](t)f' \biggl(at+\frac{3a+b}{4}(1-t) \biggr)\,\operatorname {d}t \\ &\quad \quad {} + \int _{0}^{1} \bigl[\exp (-At)-\exp (-A) \bigr]f' \biggl(\frac{a+3b}{4}t+b(1-t) \biggr)\,\operatorname {d}t \\ &\quad \quad {} - \int _{0}^{1} \bigl[1-\exp (-At) \bigr]f' \biggl(\frac{a+b}{2}t+ \frac{a+3b}{4}(1-t) \biggr) \,\operatorname {d}t \biggr] \end{aligned}$$

for \(A=\frac{1-\alpha }{\alpha }\frac{b-a}{2}\).

Generalized fractional integral inequalities of Hermite–Hadamard type

Now we are in a position to state and prove our main results.

Theorem 4.1

Let \(f:[a,b]\to \mathbb{R}\) be a differentiable function on \((a,b)\) and \(f'\in L_{1}([a,b])\) for \(0\le a< b\) and \(\alpha >0\). If the mapping \(\vert f'\vert ^{q}\) for \(q\ge 1\) is \((\alpha _{1},m)\)-convex on \([0,\frac{b}{m} ]\) for some \((\alpha _{1},m)\in (0,1]^{2}\), then

$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}+2f \biggl(\frac{a+b}{2} \biggr) - \frac{2}{ \Delta (1)} \biggl[{}_{a^{+}}I_{\varphi }f \biggl( \frac{3a+b}{4} \biggr) \\ &\quad \quad {} +{}_{ (\frac{3a+b}{4} )^{+}}I_{\varphi }f \biggl( \frac{a+b}{2} \biggr) +{}_{ (\frac{a+b}{2} )^{+}}I_{\varphi }f \biggl(\frac{a+3b}{4} \biggr) +{}_{ (\frac{a+3b}{4} )^{-}}I _{\varphi }f(b) \biggr] \biggr\vert \\ &\quad \le \frac{b-a}{8\Delta (1)} \biggl[ \biggl( \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert \,\operatorname {d}t \biggr)^{1-1/q} \biggl(A_{1} \bigl\vert f'(a) \bigr\vert ^{q}+A_{2} \biggl\vert f' \biggl( \frac{3a+b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\ &\quad \quad {} + \biggl( \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert \,\operatorname {d}t \biggr)^{1-1/q} \biggl(B_{1} \biggl\vert f' \biggl(\frac{3a+b}{4} \biggr) \biggr\vert ^{q} +B_{2} \biggl\vert f' \biggl(\frac{a+b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\ &\quad \quad {} + \biggl( \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert \,\operatorname {d}t \biggr)^{1-1/q} \biggl(A_{1} \biggl\vert f' \biggl(\frac{a+b}{4} \biggr) \biggr\vert ^{q} +A_{2} \biggl\vert f' \biggl(\frac{a+3b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\ &\quad \quad {} + \biggl( \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert \,\operatorname {d}t \biggr)^{1-1/q} \biggl(B_{1} \biggl\vert f' \biggl(\frac{a+3b}{4} \biggr) \biggr\vert ^{q}+B_{2} \biggl\vert f' \biggl( \frac{b}{m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \biggr], \end{aligned}$$

where the constants \(A_{1}\), \(A_{2}\), \(B_{1}\), and \(B_{2}\) are defined by

$$\begin{aligned}& A_{1} = \int _{0}^{1} t^{\alpha _{1}} \bigl\vert \Delta (t) \bigr\vert \,\operatorname {d}t, \qquad A_{2}= \int _{0}^{1} m \bigl(1-t^{\alpha _{1}} \bigr) \bigl\vert \Delta (t) \bigr\vert \,\operatorname {d}t, \\& B_{1} = \int _{0}^{1} t^{\alpha _{1}} \bigl\vert \nabla (t) \bigr\vert \,\operatorname {d}t, \qquad B_{2}= \int _{0}^{1} m \bigl(1-t^{\alpha _{1}} \bigr) \bigl\vert \nabla (t) \bigr\vert \,\operatorname {d}t. \end{aligned}$$

Proof

Using Lemma 3.1, the well-known power mean inequality, and the \((\alpha _{1},m)\)-convexity of \(\vert f'\vert ^{q}\) on \([0,\frac{b}{m} ]\) gives

$$\begin{aligned}& \biggl\vert \frac{f(a)+f(b)}{2}+2f \biggl( \frac{a+b}{2} \biggr) -\frac{2}{ \Delta (1)} \biggl[{}_{a^{+}}I_{\varphi }f \biggl( \frac{3a+b}{4} \biggr) \\& \quad \quad {} +{}_{ (\frac{3a+b}{4} )^{+}}I_{\varphi }f \biggl( \frac{a+b}{2} \biggr) +{}_{ (\frac{a+b}{2} )^{+}}I_{\varphi }f \biggl(\frac{a+3b}{4} \biggr) +{}_{ (\frac{a+3b}{4} )^{-}}I _{\varphi }f(b) \biggr] \biggr\vert \\& \quad \le \frac{b-a}{8\Delta (1)} \biggl[ \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert \biggl\vert f' \biggl(at+\frac{3a+b}{4}(1-t) \biggr) \biggr\vert \,\operatorname {d}t \\& \quad\quad {} + \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert \biggl\vert f' \biggl(\frac{3a+b}{4}t+\frac{a+b}{2}(1-t) \biggr) \biggr\vert \,\operatorname {d}t \\& \quad \quad {} + \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert \biggl\vert f' \biggl(\frac{a+b}{2}t+\frac{a+3b}{4}(1-t) \biggr) \biggr\vert \,\operatorname {d}t \\& \quad \quad {} + \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert \biggl\vert f' \biggl(\frac{a+3b}{4}t+b(1-t) \biggr) \biggr\vert \,\operatorname {d}t \biggr] \\& \quad \le \frac{b-a}{8\Delta (1)} \biggl\{ \biggl( \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert \,\operatorname {d}t \biggr)^{1-1/q} \biggl[ \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert \biggl(t^{\alpha _{1}} \bigl\vert f'(a) \bigr\vert ^{q} \\& \quad \quad {} +m \bigl(1-t^{\alpha _{1}} \bigr) \biggl\vert f' \biggl( \frac{3a+b}{4m} \biggr) \biggr\vert ^{q} \biggr)\,\operatorname {d}t \biggr]^{1/q} \\& \quad \quad {} + \biggl( \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert \,\operatorname {d}t \biggr)^{1-1/q} \biggl[ \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert \biggl(t^{\alpha _{1}} \biggl\vert f' \biggl( \frac{3a+b}{4} \biggr) \biggr\vert ^{q} \\& \quad \quad {} +m \bigl(1-t^{\alpha _{1}} \bigr) \biggl\vert f' \biggl( \frac{a+b}{2m} \biggr) \biggr\vert ^{q} \biggr)\,\operatorname {d}t \biggr]^{1/q} \\& \quad \quad {} + \biggl( \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert \,\operatorname {d}t \biggr)^{1-1/q} \biggl[ \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert \biggl(t^{\alpha _{1}} \biggl\vert f' \biggl(\frac{a+b}{2} \biggr) \biggr\vert ^{q} \\& \quad \quad {} +m \bigl(1-t ^{\alpha _{1}} \bigr) \biggl\vert f' \biggl( \frac{a+3b}{4m} \biggr) \biggr\vert ^{q} \biggr)\,\operatorname {d}t \biggr]^{1/q} \\& \quad\quad {} + \biggl( \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert \,\operatorname {d}t \biggr)^{1-1/q} \biggl[ \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert \biggl(t^{\alpha _{1}} \biggl\vert f' \biggl(\frac{a+3b}{4} \biggr) \biggr\vert ^{q} \\& \quad \quad {} +m \bigl(1-t ^{\alpha _{1}} \bigr) \biggl\vert f' \biggl( \frac{b}{m} \biggr) \biggr\vert ^{q} \biggr)\,\operatorname {d}t \biggr]^{1/q} \biggr\} \\& \quad =\frac{b-a}{8\Delta (1)} \biggl[ \biggl( \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert \,\operatorname {d}t \biggr)^{1-1/q} \biggl(A_{1} \bigl\vert f'(a) \bigr\vert ^{q}+A_{2} \biggl\vert f' \biggl( \frac{3a+b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\& \quad \quad {} + \biggl( \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert \,\operatorname {d}t \biggr)^{1-1/q} \biggl(B_{1} \biggl\vert f' \biggl(\frac{3a+b}{4} \biggr) \biggr\vert ^{q}+B_{2} \biggl\vert f' \biggl( \frac{a+b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\& \quad \quad {} + \biggl( \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert \,\operatorname {d}t \biggr)^{1-1/q} \biggl(A_{1} \biggl\vert f' \biggl(\frac{a+b}{4} \biggr) \biggr\vert ^{q}+A_{2} \biggl\vert f' \biggl( \frac{a+3b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\& \quad \quad {} + \biggl( \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert \,\operatorname {d}t \biggr)^{1-1/q} \biggl(B_{1} \biggl\vert f' \biggl(\frac{a+3b}{4} \biggr) \biggr\vert ^{q}+B_{2} \biggl\vert f' \biggl( \frac{b}{m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \biggr]. \end{aligned}$$

This completes the proof. □

Remark 4.1

Under assumptions of Theorem 4.1, if \(\varphi (t)=t\) and \(m=\alpha _{1}=1\), then

$$\begin{aligned}& \biggl\vert \frac{1}{2} \biggl[\frac{f(a)+f(b)}{2}+f \biggl( \frac{a+b}{2} \biggr) \biggr]-\frac{1}{b-a} \int _{a}^{b} f(x)\,\operatorname {d}x \biggr\vert \\& \quad \le \frac{b-a}{32} \biggl(\frac{1}{3} \biggr)^{1/q} \biggl[ \biggl(2 \bigl\vert f'(a) \bigr\vert ^{q}+ \biggl\vert f' \biggl(\frac{3a+b}{4} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\& \quad\quad {} + \biggl( \biggl\vert f' \biggl(\frac{3a+b}{4} \biggr) \biggr\vert ^{q}+2 \biggl\vert f' \biggl( \frac{a+b}{2} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\& \quad\quad {} + \biggl(2 \biggl\vert f' \biggl(\frac{a+b}{2} \biggr) \biggr\vert ^{q}+ \biggl\vert f' \biggl( \frac{a+3b}{4} \biggr) \biggr\vert ^{q} \biggr)^{1/q} + \biggl( \biggl\vert f' \biggl(\frac{a+3b}{4} \biggr) \biggr\vert ^{q}+2 \bigl\vert f'(b) \bigr\vert ^{q} \biggr)^{1/q} \biggr], \end{aligned}$$

which was proved in [26].

Remark 4.2

Under assumptions of Theorem 4.1, if \(\varphi (t)=\frac{t^{ \alpha }}{\varGamma (\alpha )}\), then the inequality in Theorem 4.1 reduces to inequality (1.3).

Corollary 4.1

Under assumptions of Theorem 4.1, if \(\varphi (t)=\frac{t^{ \alpha /k}}{{k\varGamma _{k}(\alpha )}}\), then

$$\begin{aligned} & \biggl\vert \frac{1}{2} \biggl[\frac{f(a)+f(b)}{2}+f \biggl( \frac{a+b}{2} \biggr) \biggr] -\frac{4^{\alpha /k-1}\varGamma _{k}(\alpha +1)}{(b-a)^{ \alpha /k}} \biggl[I^{\alpha }_{a^{+},k}f \biggl(\frac{3a+b}{4} \biggr) \\ &\quad \quad {} +I^{\alpha }_{ (\frac{3a+b}{4} )^{+},k}f \biggl( \frac{a+b}{2} \biggr) +I^{\alpha }_{ (\frac{a+b}{2} )^{+},k}f \biggl(\frac{a+3b}{4} \biggr) +I^{\alpha }_{ (\frac{a+3b}{4} )^{+},k}f(b) \biggr] \biggr\vert \\ &\quad \le \frac{b-a}{16 (\frac{\alpha }{k}+1 )} \biggl[\frac{1}{( \alpha _{1}+1) (\frac{\alpha }{k}+\alpha _{1}+1 )} \biggr]^{1/q} \biggl\{ \biggl[ \biggl(\frac{\alpha }{k}+1 \biggr) (\alpha _{1}+1) \bigl\vert f'(a) \bigr\vert ^{q} \\ &\quad \quad {} +m\alpha _{1}(\alpha _{1}+1) \biggl\vert f' \biggl( \frac{3a+b}{4m} \biggr) \biggr\vert ^{q} \biggr]^{1/q} + \frac{\alpha }{k} \biggl[ \biggl(\frac{\alpha }{k}+1 \biggr) \biggl\vert f' \biggl(\frac{3a+b}{4} \biggr) \biggr\vert ^{q} \\ &\quad \quad {} +m\alpha _{1} \biggl(\alpha _{1}+ \frac{\alpha }{k}+2 \biggr) \biggl\vert f' \biggl( \frac{a+b}{2m} \biggr) \biggr\vert ^{q} \biggr]^{1/q} + \biggl[ \biggl( \frac{\alpha }{k}+1 \biggr) (\alpha _{1}+1) \\ &\quad \quad {} \times \biggl\vert f' \biggl(\frac{a+b}{2} \biggr) \biggr\vert ^{q}+m\alpha _{1}(\alpha _{1}+1) \biggl\vert f' \biggl(\frac{a+3b}{4m} \biggr) \biggr\vert ^{q} \biggr]^{1/q} \\ &\quad \quad {} +\alpha \biggl[ \biggl(\frac{\alpha }{k}+1 \biggr) \biggl\vert f' \biggl(\frac{a+b}{2} \biggr) \biggr\vert ^{q} +m \alpha _{1} \biggl( \alpha _{1}+ \frac{\alpha }{k}+2 \biggr) \biggl\vert f' \biggl( \frac{b}{m} \biggr) \biggr\vert ^{q} \biggr]^{1/q} \biggr\} . \end{aligned}$$

Corollary 4.2

Under assumptions of Theorem 4.1, if \(\alpha _{1}=1\) and \(\varphi (t)=\frac{t}{\alpha }\exp (-\frac{1-\alpha }{\alpha }t )\), then

$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}+2f \biggl(\frac{a+b}{2} \biggr) - \frac{2(1- \alpha )}{1-\exp (-A)} \biggl[\mathcal{I}^{\alpha }_{a^{+}}f \biggl( \frac{3a+b}{4} \biggr) \\ &\quad \quad {} +\mathcal{I}^{\alpha }_{ (\frac{3a+b}{4} )^{+}}f \biggl( \frac{a+b}{2} \biggr) +\mathcal{I}^{\alpha }_{ ( \frac{a+b}{2} )^{+}}f \biggl( \frac{a+3b}{4} \biggr) +\mathcal{I} ^{\alpha }_{ (\frac{a+3b}{4} )^{-}}f(b) \biggr] \biggr\vert \\ &\quad \le \frac{b-a}{8[1-\exp (-A)]} \biggl[ \biggl(\frac{A+\exp (-A)-1}{A} \biggr)^{1-1/q} \biggl(A_{3} \bigl\vert f'(a) \bigr\vert ^{q} \\ &\quad \quad {} +A_{4} \biggl\vert f' \biggl( \frac{3a+b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} + \biggl( \frac{A\exp (-A)+\exp (-A)-1}{A} \biggr)^{1-1/q} \\ &\quad \quad {} \times \biggl(B_{3} \biggl\vert f' \biggl( \frac{3a+b}{4} \biggr) \biggr\vert ^{q}+B_{4} \biggl\vert f' \biggl(\frac{a+b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\ &\quad \quad {} + \biggl(\frac{A+\exp (-A)-1}{A} \biggr)^{1-1/q} \biggl(A_{3} \biggl\vert f' \biggl(\frac{a+b}{4} \biggr) \biggr\vert ^{q} +A_{4} \biggl\vert f' \biggl( \frac{a+3b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\ &\quad \quad {} + \biggl(\frac{A\exp (-A)+\exp (-A)-1}{A} \biggr)^{1-1/q} \\ &\quad \quad {} \times \biggl(B_{3} \biggl\vert f' \biggl( \frac{a+3b}{4} \biggr) \biggr\vert ^{q}+B_{4} \biggl\vert f' \biggl(\frac{b}{m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \biggr], \end{aligned}$$

where

$$\begin{aligned}& A=\frac{1-\alpha }{\alpha }\frac{b-a}{2}, \quad \quad A_{3}= \frac{A^{2}+2A \exp (-A)+2\exp (-A)-2}{2A^{2}}, \\& A_{4}=\frac{m(A+\exp (-A)-1)}{A^{2}}, \\& B_{3}=\frac{A^{2}\exp (-A)+2A\exp (-A)+2\exp (-A)-2}{2A^{2}}, \\& B_{4}=\frac{m (A^{2}+2A+2\exp (-A)-A^{2}\exp (-A)-2 )}{2A ^{2}}. \end{aligned}$$

Theorem 4.2

Let \(f:[a,b]\to \mathbb{R}\) be a differentiable function on \((a,b)\) and \(f'\in L_{1}([a,b])\) for \(0\le a< b\). If the mapping \(\vert f'\vert ^{q}\) is \((\alpha _{1},m)\)-convex on \([0,\frac{b}{m} ]\) for some \((\alpha _{1},m)\in (0,1]^{2}\), \(q\ge 1\), and \(q\ge r\ge 0\), then

$$ \begin{aligned}[b] & \biggl\vert \frac{f(a)+f(b)}{2}+2f \biggl(\frac{a+b}{2} \biggr) -\frac{2}{ \Delta (1)} \biggl[{}_{a^{+}}I_{\varphi }f \biggl(\frac{3a+b}{4} \biggr) \\ &\quad \quad {} +{}_{ (\frac{3a+b}{4} )^{+}}I_{\varphi }f \biggl( \frac{a+b}{2} \biggr) +{}_{ (\frac{a+b}{2} )^{+}}I_{\varphi }f \biggl(\frac{a+3b}{4} \biggr) +{}_{ (\frac{a+3b}{4} )^{-}}I _{\varphi }f(b) \biggr] \biggr\vert \\ &\quad \le \frac{b-a}{8\Delta (1)} \biggl[ \biggl( \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert ^{\frac{q-r}{q-1}}\,\operatorname {d}t \biggr) ^{1-1/q} \biggl(C_{1} \bigl\vert f'(a) \bigr\vert ^{q}+C_{2} \biggl\vert f' \biggl(\frac{3a+b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\ &\quad \quad {} + \biggl( \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert ^{\frac{q-r}{q-1}}\,\operatorname {d}t \biggr)^{1-1/q} \biggl(D_{1} \biggl\vert f' \biggl(\frac{3a+b}{4} \biggr) \biggr\vert ^{q}+D_{2} \biggl\vert f' \biggl( \frac{a+b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\ &\quad \quad {} + \biggl( \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert ^{\frac{q-r}{q-1}}\,\operatorname {d}t \biggr)^{1-1/q} \biggl(C_{1} \biggl\vert f' \biggl(\frac{a+b}{4} \biggr) \biggr\vert ^{q}+C_{2} \biggl\vert f' \biggl( \frac{a+3b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\ &\quad \quad {} + \biggl( \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert ^{\frac{q-r}{q-1}}\,\operatorname {d}t \biggr)^{1-1/q} \biggl(D_{1} \biggl\vert f' \biggl(\frac{a+3b}{4} \biggr) \biggr\vert ^{q}+D_{2} \biggl\vert f' \biggl( \frac{b}{m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \biggr], \end{aligned} $$
(4.1)

where the constants \(C_{1}\), \(C_{2}\), \(D_{1}\), and \(D_{2}\) are defined by

$$\begin{aligned}& C_{1} = \int _{0}^{1} t^{\alpha _{1}} \bigl\vert \Delta (t) \bigr\vert ^{r}\,\operatorname {d}t, \quad\quad C_{2}= \int _{0}^{1} m \bigl(1-t^{\alpha _{1}} \bigr) \bigl\vert \Delta (t) \bigr\vert ^{r}\,\operatorname {d}t, \\& D_{1} = \int _{0}^{1} t^{\alpha _{1}} \bigl\vert \nabla (t) \bigr\vert ^{r}\,\operatorname {d}t, \quad\quad D_{2}= \int _{0}^{1} m \bigl(1-t^{\alpha _{1}} \bigr) \bigl\vert \nabla (t) \bigr\vert ^{r}\,\operatorname {d}t. \end{aligned}$$

Proof

By Lemma 3.1, the well-known Hölder inequality, and the \((\alpha _{1},m)\)-convexity of \(\vert f'\vert ^{q}\) on \([0,\frac{b}{m} ]\), we have

$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}+2f \biggl(\frac{a+b}{2} \biggr) - \frac{2}{ \Delta (1)} \biggl[{}_{a^{+}}I_{\varphi }f \biggl( \frac{3a+b}{4} \biggr) \\ &\quad \quad {} +{}_{ (\frac{3a+b}{4} )^{+}}I_{\varphi }f \biggl( \frac{a+b}{2} \biggr) +{}_{ (\frac{a+b}{2} )^{+}}I_{\varphi }f \biggl(\frac{a+3b}{4} \biggr) +{}_{ (\frac{a+3b}{4} )^{-}}I _{\varphi }f(b) \biggr] \biggr\vert \\ &\quad \le \frac{b-a}{8\Delta (1)} \biggl[ \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert \biggl\vert f' \biggl(at+\frac{3a+b}{4}(1-t) \biggr) \biggr\vert \,\operatorname {d}t + \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert \biggl\vert f' \biggl(\frac{3a+b}{4}t \\ &\quad \quad {} +\frac{a+b}{2}(1-t) \biggr) \biggr\vert \,\operatorname {d}t + \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert \biggl\vert f' \biggl(\frac{a+b}{2}t+\frac{a+3b}{4}(1-t) \biggr) \biggr\vert \,\operatorname {d}t \\ &\quad \quad {} + \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert \biggl\vert f' \biggl(\frac{a+3b}{4}t+b(1-t) \biggr) \biggr\vert \,\operatorname {d}t \biggr] \\ &\quad \le \frac{b-a}{8\Delta (1)} \biggl\{ \biggl( \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert ^{ \frac{q-r}{q-1}}\,\operatorname {d}t \biggr)^{1-1/q} \biggl[ \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert ^{r} \biggl(t^{\alpha _{1}} \bigl\vert f'(a) \bigr\vert ^{q} \\ &\quad \quad {} +m \bigl(1-t^{\alpha _{1}} \bigr) \biggl\vert f' \biggl( \frac{3a+b}{4m} \biggr) \biggr\vert ^{q} \biggr)\,\operatorname {d}t \biggr]^{1/q} + \biggl( \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert ^{\frac{q-r}{q-1}}\,\operatorname {d}t \biggr)^{1-1/q} \\ &\quad \quad {} \times \biggl[ \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert ^{r} \biggl(t^{\alpha _{1}} \biggl\vert f' \biggl( \frac{3a+b}{4} \biggr) \biggr\vert ^{q} +m \bigl(1-t^{ \alpha _{1}} \bigr) \biggl\vert f' \biggl(\frac{a+b}{2m} \biggr) \biggr\vert ^{q} \biggr)\,\operatorname {d}t \biggr]^{1/q} \\ &\quad \quad {} + \biggl( \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert ^{\frac{q-r}{q-1}}\,\operatorname {d}t \biggr)^{1-1/q} \biggl[ \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert ^{r} \biggl(t^{\alpha _{1}} \biggl\vert f' \biggl( \frac{a+b}{2} \biggr) \biggr\vert ^{q} \\ &\quad \quad {} +m \bigl(1-t^{\alpha _{1}} \bigr) \biggl\vert f' \biggl( \frac{a+3b}{4m} \biggr) \biggr\vert ^{q} \biggr)\,\operatorname {d}t \biggr]^{1/q}+ \biggl( \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert ^{\frac{q-r}{q-1}}\,\operatorname {d}t \biggr)^{1-1/q} \\ &\quad \quad {} \times \biggl[ \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert ^{r} \biggl(t^{\alpha _{1}} \biggl\vert f' \biggl( \frac{a+3b}{4} \biggr) \biggr\vert ^{q} +m \bigl(1-t^{ \alpha _{1}} \bigr) \biggl\vert f' \biggl(\frac{b}{m} \biggr) \biggr\vert ^{q} \biggr)\,\operatorname {d}t \biggr]^{1/q} \biggr\} \\ &\quad =\frac{b-a}{8\Delta (1)} \biggl[ \biggl( \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert ^{ \frac{q-r}{q-1}}\,\operatorname {d}t \biggr)^{1-1/q} \biggl(C_{1} \bigl\vert f'(a) \bigr\vert ^{q}+C _{2} \biggl\vert f' \biggl(\frac{3a+b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\ &\quad \quad {} + \biggl( \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert ^{\frac{q-r}{q-1}}\,\operatorname {d}t \biggr)^{1-1/q} \biggl(D_{1} \biggl\vert f' \biggl(\frac{3a+b}{4} \biggr) \biggr\vert ^{q}+D_{2} \biggl\vert f' \biggl( \frac{a+b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\ &\quad \quad {} + \biggl( \int _{0}^{1} \bigl\vert \Delta (t) \bigr\vert ^{\frac{q-r}{q-1}}\,\operatorname {d}t \biggr)^{1-1/q} \biggl(C_{1} \biggl\vert f' \biggl(\frac{a+b}{4} \biggr) \biggr\vert ^{q}+C_{2} \biggl\vert f' \biggl( \frac{a+3b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \\ &\quad \quad {} + \biggl( \int _{0}^{1} \bigl\vert \nabla (t) \bigr\vert ^{\frac{q-r}{q-1}}\,\operatorname {d}t \biggr)^{1-1/q} \biggl(D_{1} \biggl\vert f' \biggl(\frac{a+3b}{4} \biggr) \biggr\vert ^{q}+D_{2} \biggl\vert f' \biggl( \frac{b}{m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \biggr]. \end{aligned}$$

The required proof is complete. □

Remark 4.3

Under assumptions of Theorem 4.2, if \(\varphi (t)=t\), then inequality (4.1) reduces to

$$\begin{aligned} & \biggl\vert \frac{1}{2} \biggl[\frac{f(a)+f(b)}{2}+f \biggl( \frac{a+b}{2} \biggr) \biggr]-\frac{1}{b-a} \int _{a}^{b} f(x)\,\operatorname {d}x \biggr\vert \\ &\quad \le \frac{b-a}{16} \biggl(\frac{q-1}{2q-r-1} \biggr)^{1-1/q} \biggl[ \biggl(\frac{1}{r+\alpha _{1}+1} \bigl\vert f'(a) \bigr\vert ^{q} +\frac{m\alpha _{1}}{(r+1)(r+ \alpha _{1}+1)} \\ &\quad \quad {} \times \biggl\vert f' \biggl(\frac{3a+b}{4} \biggr) \biggr\vert ^{q} \biggr)^{1/q} + \biggl(B(r+1,\alpha _{1}+1) \biggl\vert f' \biggl(\frac{3a+b}{4} \biggr) \biggr\vert ^{q} +m \biggl(\frac{1}{r+1} \\ &\quad \quad {} -B(r+1,\alpha _{1}+1) \biggr) \biggl\vert f' \biggl(\frac{a+b}{2} \biggr) \biggr\vert ^{q} \biggr)^{1/q} + \biggl(\frac{1}{r+\alpha _{1}+1} \biggl\vert f' \biggl(\frac{a+b}{2} \biggr) \biggr\vert ^{q} \\ &\quad \quad {} +\frac{m\alpha _{1}}{(r+1)(r+\alpha _{1}+1)} \biggl\vert f' \biggl( \frac{a+3b}{4} \biggr) \biggr\vert ^{q} \biggr)^{1/q} +B(r+1,\alpha _{1}+1) \\ &\quad \quad {} \times \biggl( \biggl\vert f' \biggl( \frac{a+3b}{4} \biggr) \biggr\vert ^{q} +m \biggl( \frac{1}{r+1}-B(r+1,\alpha _{1}+1) \biggr) \biggl\vert f' \biggl( \frac{b}{m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \biggr], \end{aligned}$$

which was proved in [26].

Remark 4.4

Under assumptions of Theorem 4.2, if \(\varphi (t)=\frac{t^{ \alpha }}{\varGamma (\alpha )}\), then inequality (4.1) reduces to inequality (1.4).

Corollary 4.3

Under assumptions of Theorem 4.2, if \(\varphi (t)=\frac{t^{ \alpha /k}}{{k\varGamma _{k}(\alpha )}}\), then

$$\begin{aligned} & \biggl\vert \frac{1}{2} \biggl[\frac{f(a)+f(b)}{2}+f \biggl( \frac{a+b}{2} \biggr) \biggr] -\frac{4^{\alpha /k-1}\varGamma _{k}(\alpha +1)}{(b-a)^{ \alpha /k}} \biggl[I^{\alpha }_{a^{+},k}f \biggl(\frac{3a+b}{4} \biggr) \\ &\quad \quad {} +I^{\alpha }_{ (\frac{3a+b}{4} )^{+},k}f \biggl( \frac{a+b}{2} \biggr) +I^{\alpha }_{ (\frac{a+b}{2} )^{+},k}f \biggl(\frac{a+3b}{4} \biggr) +I^{\alpha }_{ (\frac{a+3b}{4} )^{+},k}f(b) \biggr] \biggr\vert \\ &\quad \le \frac{b-a}{16} \biggl\{ \biggl[ \frac{q-1}{\frac{\alpha }{k}(q-r)+q-1} \biggr]^{1-1/q} \biggl[\frac{1}{\frac{ \alpha r}{k}+\alpha _{1}+1} \bigl\vert f'(a) \bigr\vert ^{q} \\ &\quad \quad {} +\frac{m\alpha _{1}}{ (\frac{\alpha r}{k}+1 ) (\frac{ \alpha r}{k}+\alpha _{1}+1 )} \biggl\vert f \biggl(\frac{3a+b}{4m} \biggr) \biggr\vert ^{q} \biggr]^{1/q}+\frac{k}{\alpha } B^{1-1/q} \biggl(\frac{2q-r-1}{q-1},\frac{k}{\alpha } \biggr) \\ &\quad \quad {} \times \biggl[ \biggl(r+1,\frac{k(\alpha _{1}+1)}{\alpha } \biggr) \biggl\vert f \biggl(\frac{3a+b}{4} \biggr) \biggr\vert ^{q}+m \biggl[B \biggl(r+1,\frac{k}{\alpha } \biggr) -B \biggl(r+1, \\ &\quad \quad \frac{k(\alpha _{1}+1)}{\alpha } \biggr) \biggr] \biggl\vert f \biggl( \frac{a+b}{2m} \biggr) \biggr\vert ^{q} \biggr]^{1/q} + \biggl[\frac{q-1}{\frac{ \alpha }{k}(q-r)+q-1} \biggr]^{1-1/q} \biggl[\frac{1}{ \frac{\alpha r}{k} +\alpha _{1}+1} \\ &\quad \quad {} \times \bigl\vert f'(a) \bigr\vert ^{q}+ \frac{m\alpha _{1}}{ ( \frac{\alpha r}{k}+1 ) (\frac{\alpha r}{k}+\alpha _{1}+1 )} \biggl\vert f \biggl(\frac{a+3b}{4m} \biggr) \biggr\vert ^{q} \biggr]^{1/q} \\ &\quad \quad {} +\frac{k}{\alpha }B^{1-1/q} \biggl(\frac{2q-r-1}{q-1}, \frac{k}{ \alpha } \biggr) \biggl[ \biggl(r+1,\frac{k(\alpha _{1}+1)}{\alpha } \biggr) \biggl\vert f \biggl(\frac{a+3b}{4} \biggr) \biggr\vert ^{q} \\ &\quad \quad {} +m \biggl[B \biggl(r+1,\frac{k}{\alpha } \biggr)-B \biggl(r+1, \frac{k( \alpha _{1}+1)}{\alpha } \biggr) \biggr] \biggl\vert f \biggl( \frac{b}{m} \biggr) \biggr\vert ^{q} \biggr]^{1/q} \\ &\quad \quad {} + \biggl[\frac{q-1}{\frac{\alpha }{k}(q-r)+q-1} \biggr]^{1-1/q} \biggr\} . \end{aligned}$$

Corollary 4.4

Under assumptions of Theorem 4.2, if \(r=0\) and \(\varphi (t)=\frac{t}{ \alpha }\exp (-\frac{1-\alpha }{\alpha }t )\), then

$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}+2f \biggl(\frac{a+b}{2} \biggr) - \frac{2(1- \alpha )}{1-\exp (-A)} \biggl[\mathcal{I}^{\alpha }_{a^{+}}f \biggl( \frac{3a+b}{4} \biggr) \\ &\quad \quad {} +\mathcal{I}^{\alpha }_{ (\frac{3a+b}{4} )^{+}}f \biggl( \frac{a+b}{2} \biggr)+ \mathcal{I}^{\alpha }_{ ( \frac{a+b}{2} )^{+}}f \biggl( \frac{a+3b}{4} \biggr) +\mathcal{I} ^{\alpha }_{ (\frac{a+3b}{4} )^{-}}f(b) \biggr] \biggr\vert \\ &\quad \le \frac{b-a}{8[1-\exp (-A)]} \biggl[ \biggl( \int _{0}^{1} \bigl[1-\exp (-At) \bigr]^{p} \,\operatorname {d}t \biggr)^{1/p} \biggl( \bigl\vert f'(a) \bigr\vert ^{q} \\ &\quad \quad {} + \biggl\vert f' \biggl(\frac{3a+b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q}+ \biggl( \int _{0}^{1} \bigl[\exp (-At)-\exp (-A) \bigr]^{p}\,\operatorname {d}t \biggr)^{1/p} \\ &\quad \quad {} \times \biggl( \biggl\vert f' \biggl( \frac{3a+b}{4} \biggr) \biggr\vert ^{q} + \biggl\vert f' \biggl( \frac{a+b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q}+ \biggl( \int _{0}^{1} \bigl[1-\exp (-At) \bigr]^{p} \,\operatorname {d}t \biggr)^{1/p} \\ &\quad \quad {} \times \biggl( \biggl\vert f' \biggl( \frac{a+b}{4} \biggr) \biggr\vert ^{q} + \biggl\vert f' \biggl( \frac{a+3b}{4m} \biggr) \biggr\vert ^{q} \biggr)^{1/q}+ \biggl( \int _{0}^{1} \bigl[\exp (-At) \\ &\quad \quad {} -\exp (-A) \bigr]^{p}\,\operatorname {d}t \biggr)^{1/p} \biggl( \biggl\vert f' \biggl( \frac{a+3b}{4} \biggr) \biggr\vert ^{q}+ \biggl\vert f' \biggl(\frac{b}{m} \biggr) \biggr\vert ^{q} \biggr)^{1/q} \biggr], \end{aligned}$$

where \(\frac{1}{p}+\frac{1}{q}=1\) and \(A=\frac{1-\alpha }{\alpha } \frac{b-a}{2}\).

Remark 4.5

Under assumptions of Theorem 4.2, if \(A=\frac{1-\alpha }{ \alpha }\frac{b-a}{2}\), \(\alpha _{1}=r=1\), and \(\varphi (t)=\frac{t}{ \alpha }\exp (-\frac{1-\alpha }{\alpha }t )\), then Theorem 4.2 reduces to Corollary 4.2.

Conclusions

In this work, we establish generalized fractional integral inequalities, the Riemann–Liouville fractional integral inequalities, and some classical integral inequalities of the Hermite–Hadamard type for \((\alpha ,m)\)-convex functions. The results presented in this paper would provide generalizations and extensions of those given in earlier works.

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Acknowledgements

The authors appreciate the anonymous referees for their careful corrections to and valuable comments on the original version of this paper.

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Qi, F., Mohammed, P.O., Yao, JC. et al. Generalized fractional integral inequalities of Hermite–Hadamard type for \({(\alpha,m)}\)-convex functions. J Inequal Appl 2019, 135 (2019). https://doi.org/10.1186/s13660-019-2079-6

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  • DOI: https://doi.org/10.1186/s13660-019-2079-6

MSC

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

Keywords

  • Generalized fractional integral
  • Hermite–Hadamard type
  • Integral inequality
  • Convex function
  • Riemann–Liouville fractional integral