1 Introduction

The following notation is used throughout this paper. We use I to denote an interval on the real line \(\mathbb{R}=(-\infty,\infty)\). For any subset \(K\subseteq\mathbb{R}^{n} \), \(K^{\circ}\) is used to denote the interior of K and \(\mathbb{R}^{n}\) is used to denote a generic n-dimensional vector space. The set of integrable functions on the interval \([a,b]\) is denoted by \(L^{1}[a,b]\). The non-negative real numbers and the positive real numbers are denoted by \(\mathbb {R}_{0}=[0,\infty)\) and \(\mathbb{R}_{+}=(0,\infty)\), respectively.

Let \(f : I \subseteq\mathbb{R}\rightarrow\mathbb{R}\) be a convex mapping defined on the interval I of real numbers and \(a, b \in I\) with \(a < b\). The inequality

$$ f \biggl(\frac{a+b}{2} \biggr)\leq\frac{1}{b-a} \int^{b}_{a}f(x)\,\mathrm{d}x\leq\frac{f(a)+f(b)}{2}, $$
(1.1)

referred to as Hermite-Hadamard inequality, is one of the most famous results for convex mappings. A number of papers have been written on this inequality providing new proofs, noteworthy extensions, generalizations, refinements and new inequalities connected with the Hermite-Hadamard inequality. The reader may refer to [1–9] and the references therein.

We need, now, some necessary definitions and preliminary results as follows.

Definition 1.1

[10, 11]

A set \(S\subseteq\mathbb{R}^{n}\) is said to be invex set with respect to the mapping \(\eta: S \times S\rightarrow\mathbb{R}^{n}\) if \(x + t\eta(y, x) \in S\) for every \(x, y\in S\) and \(t\in[0, 1]\). The invex set S is also called an η-connected set.

Notice that every convex set is invex with respect to the mapping \(\eta (y,x)=y-x\), but the converse is not necessarily true. See [10], for example.

Definition 1.2

[12]

A set \(K\subseteq\mathbb{R}^{n}\) is said to be m-invex with respect to the mapping \(\eta:K \times K\times(0, 1]\rightarrow\mathbb{R}^{n}\) for some fixed \(m\in(0, 1]\), if \(mx+\lambda\eta(y, x, m) \in K\) holds for each \(x, y \in K\) and any \(\lambda\in[0, 1]\).

The Definition 1.2 essentially says that there is a path for some fixed \(m\in(0,1]\), starting from mx, which is contained in K. We do not require that y should be one of the end points of the path. However, if we demand that y should be an end point of the path for every pair \(x,y\), then \(\eta(y,x,m)=y-mx\) with \(m=1\), reducing to convexity.

It is noticed that every convex set is m-invex with respect to the mapping \(\eta(y,x,m)=y-mx\) with \(m=1\), but the converse is not necessarily true. See [12], for example.

Definition 1.3

[13]

The function \(f:[0,b] \rightarrow\mathbb{R}\), \(b > 0\), is said to be \((\alpha,m)\)-convex where \((\alpha,m)\in(0,1]\times(0,1]\), if we have

$$ f \bigl(tx + m(1 - t)y \bigr) \leq t^{\alpha}f(x) + m\bigl(1 - t^{\alpha}\bigr) f(y) $$
(1.2)

for all \(x, y \in[0, b]\) and \(t\in[0, 1]\).

Definition 1.4

[11]

The function f defined on the invex set \(K\subseteq\mathbb{R}^{n}\) is said to be preinvex with respect to η if for every \(x, y\in K\) and \(t\in[0, 1]\), we have

$$ f \bigl(x+t\eta{(y,x)} \bigr)\leq(1-t)f(x)+tf(y). $$
(1.3)

The function f is said to be preconcave if and only if −f is preinvex.

The concept of preinvexity is more general than convexity since every convex function is preinvex with respect to the mapping \(\eta(y, x)=y-x\). Further, there exist preinvex mappings which are not convex.

Theorem 1.1

[14]

Let \(f:K=[a, a+\eta(b, a)] \rightarrow(0, \infty)\) be a preinvex function on the interval of the real numbers \(K^{\circ}\) and \(a, b\in K^{\circ }\) with \(\eta(b, a)>0\). Then the following inequality holds:

$$ f \biggl(\frac{2a+\eta(b, a)}{2} \biggr) \leq\frac{1}{\eta(b, a)} \int ^{a+\eta(b, a)}_{a}f(x)\,\mathrm{d}x \leq \frac{f(a)+f(b)}{2}. $$
(1.4)

The inequality (1.4) is usually termed the Hermite-Hadamard-Noor-type inequality for preinvex mappings. This result is analogous to the original Hermite-Hadamard inequalities. If \(\eta(b, a)=b-a\), then the inequality (1.4) reduces to the remarkable Hermite-Hadamard’s inequality (1.1).

For recent results on some new generalizations, refinements of integral inequalities involved with the preinvex functions, one can see [12, 15–18] and the references therein.

In [19], Latif and Shoaib raised the so-called \((\alpha ,m)\)-preinvex function below.

Definition 1.5

[19]

The function f on the invex set \(K\subseteq[0,b^{*}]\), \(b^{*} > 0\), is said to be \((\alpha,m)\)-preinvex with respect to η if

$$ f \bigl(x+t\eta{(y,x)} \bigr) \leq\bigl(1-t^{\alpha}\bigr)f(x) + mt^{\alpha}f \biggl(\frac {y}{m} \biggr). $$
(1.5)

holds for all \(x, y \in K\), \(t\in[0,1]\) and \((\alpha,m) \in(0,1]\times (0,1]\). The function f is said to be \((\alpha,m)\)-preincave if and only if −f is \((\alpha ,m)\)-preinvex.

We also need the following fractional calculus background.

Definition 1.6

[20]

Let \(f\in L^{1}[a, b]\). The left-sided and right-sided Riemann-Liouville fractional integrals of order \(\alpha>0\) with \(a\geq0\) are defined by

$$ J^{\alpha}_{a^{+}}f(x) = \frac{1}{\Gamma(\alpha)} \int^{x}_{a}(x-t)^{\alpha -1}f(t)\,\mathrm{d}t, \quad a< x, $$

and

$$ J^{\alpha}_{b^{-}}f(x) = \frac{1}{\Gamma(\alpha)} \int^{b}_{x}(t-x)^{\alpha -1}f(t)\,\mathrm{d}t, \quad x< b, $$

respectively, where \(\Gamma(\cdot)\) is Gamma function and its definition is \(\Gamma(\alpha)=\int_{0}^{\infty}e^{-u}u^{\alpha-1}\,\mathrm{d}u\). It is to be noted that \(J^{0}_{a^{+}}f(x)=J^{0}_{b^{-}}f(x)=f(x)\).

In the case \(\alpha=1\), the Riemann-Liouville fractional integral becomes the classical integral.

In [21], Sarikaya et al. established the following interesting inequalities of Hermite-Hadamard-type involving Riemann-Liouville fractional integrals.

Theorem 1.2

Let \(f:[a,b]\rightarrow\mathbb{R}\) be a positive function with \(0\leq a < b\) and \(f\in L^{1}[a,b]\). If f is a convex function on \([a,b]\), then the following inequalities for fractional integrals hold:

$$ f \biggl(\frac{a+b}{2} \biggr) \leq\frac{\Gamma(\alpha+1)}{2(b-a)^{\alpha}} \bigl[J^{\alpha}_{a^{+}}f(b)+J^{\alpha}_{b^{-}}f(a) \bigr]\leq\frac{f(a)+f(b)}{2} $$
(1.6)

with \(\alpha>0\).

Observe that, for \(\alpha=1\), the inequalities (1.6) becomes the original Hermite-Hadamard inequality (1.1).

For some recent results associated with the fractional integral inequalities, one can consult [22–32].

In a very recently published paper [33] by Hussain and Qaisar, they found some Hermite-Hadamard integral inequalities for mapping whose absolute values of derivatives are \((\alpha, m)\)-preinvex, and in the article [34] by Qaisar et al., they also obtained Riemann-Liouville fractional Hadamard-type integral inequalities for mappings whose absolute value of first derivatives are preinvex.

Motivated by this idea and based on our previous work [2, 12, 17, 35, 36], in the present paper, the next section we are going to introduce new concepts, to be referred as the generalized \((\alpha, m)\)-preinvex function, the generalized quasi m-preinvex function and the explicitly \((\alpha, m)\)-preinvex function, respectively, and then we derive some interesting properties for the newly introduced functions. In this section, the more important point is that we give a necessary and sufficient condition with respect to the relationship between the generalized \((\alpha, m)\)-preinvex function and the generalized quasi m-preinvex function. In Section 3, we will discover a Riemann-Liouville fractional integral identity involving twice differentiable preinvex functions. By using this identity, we explore the right-sided new Hermite-Hadamard-type inequalities for mappings whose absolute value of second derivatives are generalized \((\alpha,m)\)-preinvex via Riemann-Liouville fractional integrals. These inequalities can be viewed as generalization of the results of [37, 38].

2 New definitions and properties

As one can see, the definitions of the preinvex, \((\alpha,m)\)-convex, and \((\alpha,m)\)-preinvex mappings have similar configurations. This observation leads us to generalize these varieties of convexity.

We next give new definitions, to be referred to as the generalized \((\alpha,m)\)-preinvex function, the generalized quasi m-preinvex function and the explicitly \((\alpha,m)\)-preinvex function, respectively.

Definition 2.1

Let \(K \subseteq\mathbb{R}^{n}\) be an open m-invex set with respect to \(\eta: K \times K \times(0,1]\rightarrow\mathbb{R}^{n}\).

  1. (i)

    For \(f:K\rightarrow\mathbb{R}\) and some fixed \(\alpha, m \in (0,1]\), if

    $$ f \bigl(m x+ \lambda\eta(y, x, m) \bigr)\leq m\bigl(1- \lambda^{\alpha}\bigr) f(x)+ \lambda^{\alpha}f(y) $$
    (2.1)

    is valid for all \(x, y \in K\), \(\lambda\in[0, 1]\), then we say that \(f(x)\) is a generalized \((\alpha,m)\)-preinvex function with respect to η.

  2. (ii)

    For \(f:K\rightarrow\mathbb{R}\) and some fixed \(m \in(0,1]\), if

    $$ f \bigl(m x+ \lambda\eta(y, x, m) \bigr)\leq\max\bigl\{ f(x), f(y) \bigr\} $$
    (2.2)

    is valid for all \(x, y \in K\), \(\lambda\in[0, 1]\), then we say that \(f(x)\) is a generalized quasi m-preinvex function with respect to η.

The function \(f(x)\) is said to be strictly generalized \((\alpha ,m)\)-preinvex function on K with respect to η, if a strict inequality holds on (2.1) for any \(x,y\in K\) and \(x\neq y\).

Remark 2.1

In Definition 2.1, it is worthwhile to note that generalized \((\alpha,m)\)-preinvex function is an \((\alpha,m)\)-convex function on K with respect to \(\eta(y,x,m)=y-mx\).

Definition 2.2

Let \(K \subseteq\mathbb{R}^{n}\) be an open m-invex set with respect to \(\eta: K \times K \times(0,1]\rightarrow\mathbb{R}^{n}\). For \(f:K\rightarrow\mathbb{R}\) and some fixed \(\alpha, m \in(0,1]\), if \(\forall\lambda\in(0,1)\), \(\forall x, y \in K\) and \(f(x)\neq f(y)\), we have

$$ f \bigl(mx+\lambda\eta(y,x,m) \bigr)< m\bigl(1- \lambda^{\alpha}\bigr) f(x)+ \lambda ^{\alpha}f(y), $$
(2.3)

then we say that \(f(x)\) is an explicitly \((\alpha,m)\)-preinvex function with respect to η.

Example 2.1

Let \(f(x)=\sin x\), \(\alpha=1\), and let

$$ \eta(y, x, m)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{\sin y-m\sin x}{m\cos x}, & y\geq x; \\ 0, & y< x. \end{array}\displaystyle \right . $$

Then \(f(x)\) is a generalized \((1,\frac{1}{2})\)-preinvex function with respect to \(\eta: \mathbb{R} \times\mathbb{R} \times(0, 1]\rightarrow \mathbb{R}\). However, it is obvious that \(f(x)=\sin x\) is not a convex function on \(\mathbb{R}\). By letting \(x>y=\frac{\pi}{2}, \lambda=\frac {1}{2}\), we have

$$ f \bigl(mx+\lambda\eta(y,x,m) \bigr)=f \biggl(\frac{1}{2}x+ \frac{1}{2}\eta \biggl(\frac{\pi}{2},x,m\biggr) \biggr)=\sin \biggl( \frac{1}{2}x \biggr) $$

and

$$ m\bigl(1-\lambda^{\alpha}\bigr) f(x)+ \lambda^{\alpha}f(y)= \frac{1}{4}\sin x+\frac{1}{2}. $$

Thus, there must exist an \(x_{0}>y=\frac{\pi}{2}\) such that \(f(x_{0})\neq f(y)=f(\frac{\pi}{2})=1\) and

$$ \sin \biggl(\frac{1}{2}x_{0} \biggr)=\frac{1}{4}\sin x_{0}+\frac{1}{2}. $$

Hence, f is not also an explicitly \((\alpha,m)\)-preinvex function on \(\mathbb{R}\) with respect to η for \(\alpha=1\) and \(m=\frac {1}{2}\).

The so-called ‘generalized \((\alpha,m)\)-logarithmically preinvexity’, may be introduced as follows.

Definition 2.3

Let \(K\subseteq\mathbb{R}^{n}\), be an open m-invex set with respect to \(\eta: K \times K \times(0,1]\rightarrow\mathbb{R}^{n}\). For \(f:K\rightarrow\mathbb{R_{+}}\) and some fixed \(\alpha, m \in(0,1]\), if \(\forall\lambda\in(0,1)\), \(\forall x, y \in K\), we have

$$ f \bigl(mx+\lambda\eta(y,x,m) \bigr)\leq \bigl[f(x) \bigr]^{m(1-\lambda^{\alpha})}\bigl[f(y)\bigr]^{\lambda^{\alpha}}, $$
(2.4)

then we say that \(f(x)\) is a generalized \((\alpha,m)\)-logarithmically preinvex function with respect to η.

Based on the above Definition 2.1 and Definition 2.2, we investigate, now, some interesting properties of the generalized \((\alpha,m)\)-preinvex function, generalized quasi m-preinvex function and explicitly \((\alpha,m)\)-preinvex function. The first observation is given as follows.

Proposition 2.1

If \(f: K\subseteq\mathbb{R}^{n}\rightarrow\mathbb{R}_{0}\) is a generalized \((\alpha,m)\)-preinvex function on m-invex set K with respect to η, then f is also a generalized quasi m-preinvex function on m-invex set K with respect to η.

Proof

Since f is a non-negative generalized \((\alpha,m)\)-preinvex function, we assume that \(f(x)\leq f(y)\), \(\forall x,y\in K\), for every \(\lambda\in[0,1]\), we have

$$ f \bigl(m x+ \lambda\eta(y, x, m) \bigr) \leq m\bigl(1-\lambda^{\alpha}\bigr) f(x)+ \lambda^{\alpha}f(y) \leq\bigl[m\bigl(1-\lambda^{\alpha}\bigr)+\lambda^{\alpha}\bigr] f(y) \leq f(y). $$

In the same way, let \(f(y)\leq f(x)\), \(\forall x,y\in K\), we can also get

$$ f \bigl(m x+ \lambda\eta(y, x, m) \bigr)\leq f(x). $$

Consequently,

$$ f \bigl(m x+ \lambda\eta(y, x, m) \bigr)\leq\max\bigl\{ f(x), f(y)\bigr\} . $$

That is, f is a generalized quasi m-preinvex function on m-invex set K with respect to η, the required result. □

The proofs of Propositions 2.2 and 2.3 are all easy to verify.

Proposition 2.2

If \(f_{i}: K\subseteq\mathbb{R}^{n} \rightarrow\mathbb{R}\) \((i=1,2,\ldots, n)\) are generalized \((\alpha,m)\)-preinvex (explicitly \((\alpha ,m)\)-preinvex) functions on m-invex set K with respect to the same \(\eta: K \times K \times(0, 1]\rightarrow\mathbb{R}\) for same fixed \(\alpha, m\in(0,1]\), then the function

$$ f=\sum_{i=1}^{n} a_{i}f_{i}, a_{i}\geq0\quad (i=1,2, \ldots, n) $$

is also a generalized \((\alpha,m)\)-preinvex (explicitly \((\alpha, m)\)-preinvex) functions on m-invex set K with respect to the same η for fixed \(\alpha, m\in(0,1]\).

Proposition 2.3

If \(f_{i}: K\subseteq\mathbb{R}^{n}\rightarrow\mathbb{R}\) \((i=1,2,\ldots, n)\) are generalized \((\alpha,m)\)-preinvex (explicitly \((\alpha ,m)\)-preinvex) functions on m-invex set K with respect to the same \(\eta: K \times K \times(0, 1]\rightarrow\mathbb{R}\) for same fixed \(\alpha, m\in(0,1]\), then the function

$$ f=\max\{f_{i}, i=1,2, \ldots, n\} $$

is also a generalized \((\alpha,m)\)-preinvex (an explicitly \((\alpha, m)\)-preinvex) function on m-invex set K with respect to the same η for fixed \(\alpha, m\in(0,1]\).

In Proposition 2.4 we prove that the combination of a generalized \((\alpha,m)\)-preinvex function with a sublinear and nondecreasing function is a generalized \((\alpha,m)\)-preinvex function.

Proposition 2.4

Let K be a nonempty m-invex set in \(\mathbb{R}^{n}\) with respect to \(\eta: K\times K\times(0, 1]\rightarrow\mathbb{R}^{n}\), \(f: K\rightarrow \mathbb{R}\) be a generalized \((\alpha,m)\)-preinvex function with respect to η for some fixed \(\alpha, m\in(0,1]\), and let \(g: W\rightarrow\mathbb{R}\) (\(W\subseteq\mathbb{R}\)) be a sublinear and nondecreasing function, where \(\operatorname{rang}(f)\subseteq W\). Then the composite function \(g(f)\) is a generalized \((\alpha,m)\)-preinvex function with respect to η on K for fixed \(\alpha, m\in(0,1]\).

Proof

Since f is a generalized \((\alpha,m)\)-preinvex function, for all \(x,y\in K\), we have

$$ f \bigl(mx+\lambda\eta(y,x,m) \bigr)\leq m\bigl(1-\lambda^{\alpha}\bigr)f(x)+\lambda ^{\alpha}f(y) $$

holds for any \(\lambda\in[0,1]\). Notice that g is a sublinear and nondecreasing function, it yields

$$\begin{aligned} g \bigl(f \bigl(mx+\lambda\eta(y,x,m) \bigr) \bigr) & \leq g \bigl(m\bigl(1- \lambda ^{\alpha}\bigr)f(x)+\lambda^{\alpha}f(y) \bigr) \leq m\bigl(1-\lambda^{\alpha}\bigr)g \bigl(f(x) \bigr)+ \lambda^{\alpha}g \bigl(f(y) \bigr), \end{aligned}$$

from which it follows that \(g(f)\) is a generalized \((\alpha ,m)\)-preinvex function with respect to η on K for some fixed \(\alpha,m\in(0,1]\). □

Proposition 2.5

Let K be a nonempty m-invex set in \(\mathbb{R}^{n}\) with respect to \(\eta: K\times K\times(0, 1]\rightarrow\mathbb{R}^{n}\), and \(f,g: K\rightarrow\mathbb{R}\) be generalized \((\alpha,m)\)-preinvex functions with respect to the same η for some fixed \(\alpha, m\in(0,1]\). Then their product fg is also a generalized \((\alpha,m)\)-preinvex function provided that f and g are similarly ordered functions with \(fg\geq0\).

Proof

Since f and g are two similarly ordered generalized \((\alpha,m)\)-preinvex functions, we have

$$\begin{aligned} & f \bigl(mx+\lambda\eta(y,x,m) \bigr)g \bigl(mx+\lambda\eta(y,x,m) \bigr) \\ &\quad \leq\bigl[m\bigl(1-\lambda^{\alpha}\bigr)f(x)+\lambda^{\alpha}f(y)\bigr] \bigl[m\bigl(1-\lambda^{\alpha}\bigr)g(x)+\lambda^{\alpha}g(y)\bigr] \\ &\quad = \bigl[m\bigl(1-\lambda^{\alpha}\bigr)\bigr]^{2}f(x)g(x)+ \bigl(\lambda^{\alpha}\bigr)^{2}f(y)g(y)+m\bigl(1- \lambda^{\alpha}\bigr)\lambda^{\alpha}\bigl[f(x)g(y)+f(y)g(x)\bigr] \\ &\quad \leq\bigl[m\bigl(1-\lambda^{\alpha}\bigr)\bigr]^{2}f(x)g(x)+ \bigl(\lambda^{\alpha}\bigr)^{2}f(y)g(y)+m\bigl(1- \lambda^{\alpha}\bigr)\lambda^{\alpha}\bigl[f(x)g(x)+f(y)g(y)\bigr] \\ &\quad = m\bigl(1-\lambda^{\alpha}\bigr)\bigl[m\bigl(1- \lambda^{\alpha}\bigr)+\lambda^{\alpha}\bigr]f(x)g(x)+ \lambda^{\alpha}\bigl[m\bigl(1-\lambda^{\alpha}\bigr)+ \lambda^{\alpha}\bigr]f(y)g(y) \\ &\quad \leq m\bigl(1-\lambda^{\alpha}\bigr)f(x)g(x)+\lambda^{\alpha}f(y)g(y), \end{aligned}$$

where we used the required condition \(fg\geq0\). This shows that the product of two generalized \((\alpha,m)\)-preinvex functions is also a generalized \((\alpha,m)\)-preinvex function. □

Proposition 2.6

If \(g_{i}: \mathbb{R}^{n}\rightarrow\mathbb{R} \) (\(i=1,2,\ldots, n\)) are generalized \((\alpha,m)\)-preinvex functions with respect to the same η for same fixed \(\alpha,m\in(0,1]\), then the set \(M=\{x\in\mathbb {R}^{n}: g_{i}(x)\leq0, i=1,2,\ldots, n\} \) is an m-invex set.

Proof

Since \(g_{i}(x)\) \((i=1,2,\ldots, n)\) are generalized \((\alpha ,m)\)-preinvex functions, for all \(x,y\in\mathbb{R}^{n}\), we have

$$ g_{i} \bigl(mx+\lambda\eta(y,x,m) \bigr)\leq m\bigl(1- \lambda^{\alpha}\bigr)g_{i}(y)+\lambda^{\alpha}g_{i}(x),\quad i=1,2,\ldots, n, $$

holds for any \(\lambda\in[0,1]\). When \(x,y\in M\), we know \(g_{i}(x)\leq0\) and \(g_{i}(y)\leq0\). From the above inequality, it yields

$$ g_{i} \bigl(mx+\lambda\eta(y,x,m) \bigr)\leq0, \quad i=1,2,\ldots, n. $$

That is, \(mx+\lambda\eta(y,x,m)\in M\). Hence, M is an m-invex set. □

Proposition 2.7

Let \(f: \mathbb{R}_{0}\rightarrow\mathbb{R}_{0}\) is a generalized \((\alpha ,m)\)-preinvex function with respect to \(\eta: \mathbb{R}_{0}\times \mathbb{R}_{0} \times(0, 1]\rightarrow\mathbb{R}_{0}\) for some fixed \(\alpha, m\in(0,1]\). Assume that f is monotone decreasing, η is monotone increasing regarding m for fixed \(x,y\in\mathbb{R}_{0}\), and \(m_{1}\leq m_{2}\) (\(m_{1},m_{2}\in(0,1]\)). If f is a generalized \((\alpha ,m_{1})\)-preinvex function on \(\mathbb{R}_{0}\) with respect to η, then f is also a generalized \((\alpha,m_{2})\)-preinvex function on \(\mathbb{R}_{0}\) with respect to η.

Proof

Since f is a generalized \((\alpha,m_{1})\)-preinvex function, for all \(x,y\in\mathbb{R}_{0}\), we have

$$ f \bigl(m_{1}x+\lambda\eta(y,x,m_{1}) \bigr)\leq m_{1}\bigl(1-\lambda^{\alpha}\bigr)f(x)+\lambda^{\alpha}f(y). $$

Combining the monotone decreasing of the function f with the monotone increasing of the mapping η regarding m for fixed \(x,y\in\mathbb{R}_{0}\), and \(m_{1}\leq m_{2}\), it follows that

$$ f \bigl(m_{2}x+\lambda\eta(y,x,m_{2}) \bigr) \leq f \bigl(m_{1}x+\lambda\eta (y,x,m_{1}) \bigr) $$

and

$$ m_{1}\bigl(1-\lambda^{\alpha}\bigr)f(x)+\lambda^{\alpha}f(y)\leq m_{2}\bigl(1-\lambda^{\alpha}\bigr)f(x)+ \lambda^{\alpha}f(y). $$

Following the above two inequalities, we have

$$ f \bigl(m_{2}x+\lambda\eta(y,x,m_{2}) \bigr)\leq m_{2}\bigl(1-\lambda^{\alpha}\bigr)f(x)+\lambda^{\alpha}f(y). $$

Hence, f is also a generalized \((\alpha,m_{2})\)-preinvex function on \(\mathbb{R}_{0}\) with respect to η for fixed \(\alpha\in (0,1]\), which ends the proof. □

Proposition 2.8

Let K be a nonempty m-invex set in \(\mathbb{R}^{n}\) with respect to \(\eta: K\times K\times(0, 1]\rightarrow\mathbb{R}^{n}\), and \(f_{i}: K\rightarrow\mathbb{R}\) (\(i\in I=\{1,2, \ldots, n\}\)) be a family of real-valued fucntions which are explicitly \((\alpha,m)\)-preinvex functions with respect to the same η for same fixed \(\alpha, m\in (0,1]\) and bounded from above on K. Then the function \(f(x)=\sup\{ f_{i}(x), i\in I\}\) is also an explicitly \((\alpha,m)\)-preinvex function on K with respect to the same η for fixed \(\alpha, m\in(0,1]\).

Proof

Since each \(f_{i}(x)\) (\(i\in I\)) is an explicitly \((\alpha ,m)\)-preinvex function with respect to the same η for same fixed \(\alpha, m\in(0,1]\), we have for each \(i\in I\)

$$ f_{i} \bigl(mx+ \lambda\eta(y, x, m) \bigr)< m\bigl(1- \lambda^{\alpha}\bigr) f_{i}(x)+ \lambda^{\alpha}f_{i}(y), \quad\forall x,y\in K, \lambda\in(0,1). $$

Therefore, for each \(i\in I\),

$$ f_{i} \bigl(mx+ \lambda\eta(y, x, m) \bigr)< m\bigl(1- \lambda^{\alpha}\bigr) \sup_{i\in I}f_{i}(x)+ \lambda^{\alpha}\sup_{i\in I}f_{i}(y),\quad \forall x,y\in K, \lambda \in(0,1). $$

Taking the sup of the left-hand side of the above inequality, we obtain

$$ \sup_{i\in I}f_{i} \bigl(mx+ \lambda\eta(y, x, m) \bigr)< m\bigl(1-\lambda^{\alpha}\bigr) \sup_{i\in I}f_{i}(x)+ \lambda^{\alpha}\sup_{i\in I}f_{i}(y),\quad \forall x,y\in K, \lambda\in(0,1). $$

That is, \(f(x)=\sup\{f_{i}(x), i\in I\}\) is also an explicitly \((\alpha,m)\)-preinvex function on K with respect to the same η for fixed \(\alpha, m\in(0,1]\). □

Proposition 2.9 below reveals that a local minimum of an explicitly \((\alpha,m)\)-preinvex function on an m-invex set is a global one under some conditions.

Proposition 2.9

Let K be a nonempty m-invex set in \(\mathbb{R}^{n}\) with respect to \(\eta: K\times K\times(0, 1]\rightarrow\mathbb{R}^{n}\), and \(f: K\rightarrow\mathbb{R}_{0}\) be an explicitly \((\alpha,m)\)-preinvex function with respect to η for some fixed \(\alpha, m\in(0,1]\). If \(\bar{x}\in K\) is a local minimum to the problem of minimizing \(f(x)\) subject to \(x\in K\), then x̄ is a global one.

Proof

Suppose that \(\bar{x}\in K\) is a local minimum to the problem of minimizing \(f(x)\) subject to \(x\in K\). Then there is an ε-neighborhood \(N_{\varepsilon}(\bar{x})\) around x̄ such that

$$ f(\bar{x})\leq f(x),\quad \forall x\in K\cap N_{\varepsilon}( \bar{x}). $$
(2.5)

If x̄ is not global minimum of \(f(x)\) on K, then there exists an \(x^{*}\in K\) such that

$$ f\bigl(x^{*}\bigr)< f(\bar{x}). $$

By the explicit \((\alpha,m)\)-preinvexity of \(f(x)\) and the fact that \(m(1-\lambda^{\alpha})+\lambda^{\alpha}\leq1\), we can deduce that

$$ f \bigl(m \bar{x}+ \lambda\eta\bigl(x^{*}, \bar{x}, m\bigr) \bigr)< m\bigl(1- \lambda^{\alpha}\bigr) f(\bar{x})+ \lambda^{\alpha}f\bigl(x^{*}\bigr)< \bigl[m\bigl(1-\lambda^{\alpha}\bigr)+\lambda^{\alpha}\bigr]f( \bar{x})< f(\bar{x}) $$

for all \(0<\lambda<1\). For a sufficiently small \(\lambda>0\), it follows that

$$ m \bar{x}+ \lambda\eta\bigl(x^{*}, \bar{x}, m\bigr)\in K\cap N_{\varepsilon}( \bar{x}), $$

which is a contradiction to (2.5). This completes the proof. □

By Proposition 2.9, we can conclude that explicitly \((\alpha ,m)\)-preinvex functions constitute an important class of generalized convex functions in mathematical programming. The function in Example 2.1 is not an explicitly \((\alpha,m)\)-preinvex function with respect to η based on Proposition 2.9.

For investigating the relationship between the generalized \((\alpha ,m)\)-preinvex function and the generalized quasi m-preinvex function, we will present the extended Condition C and Lemma 2.1.

Let us recall the Condition C introduced by Mohan and Neogy [39] as follows.

Condition C: Let \(\eta: \mathbb{R}^{n}\times\mathbb {R}^{n}\rightarrow\mathbb{R}^{n}\), we say that the mapping η satisfies the condition C if for any \(x,y\in\mathbb{R}^{n}\),

(C1):

\(\eta (x,x+\lambda\eta(y,x) )=-\lambda \eta(y,x)\),

(C2):

\(\eta (y,x+\lambda\eta(y,x) )=(1-\lambda )\eta(y,x)\),

for all \(\lambda\in[0,1]\), hold.

Similarly, we present here the so-called ‘extended Condition C’.

Extended Condition C: Let \(\eta: \mathbb {R}^{n}\times\mathbb{R}^{n}\times(0, 1]\rightarrow\mathbb{R}^{n}\), we say that the mapping η satisfies the extended condition C if for any \(x,y\in\mathbb{R}^{n}\),

(C1):

\(\eta (x,mx+\lambda\eta(y,x,m),m )=-\lambda\eta(y,x,m)\),

(C2):

\(\eta (y,mx+\lambda\eta(y,x,m),m )=(1-\lambda)\eta(y,x,m)\),

(C3):

\(\eta(y,x,m)=-\eta(x,y,m)\),

for all \(\lambda\in[0,1]\) and fixed \(m\in(0,1]\), hold.

Lemma 2.1

Let \(K\subseteq\mathbb{R}^{n}\) be a nonempty m-invex set with respect to the mapping \(\eta: \mathbb{R}^{n}\times\mathbb{R}^{n}\times(0, 1]\rightarrow\mathbb{R}^{n}\) and η satisfies the extended Condition C. If \(f:K\rightarrow\mathbb{R}_{0}\) satisfies \(f (mx+\eta(y,x,m) )\leq f(y)\), \(\forall x,y\in K\), and there exists a \(t\in(0,1)\) such that

$$ f \bigl(mx+t\eta(y,x,m) \bigr)\leq m\bigl(1-t^{\alpha}\bigr)f(x)+t^{\alpha}f(y), \quad \forall x,y\in K, $$
(2.6)

then the set \(A=\{\lambda\in[0,1]|f (mx+\lambda\eta(y,x,m) )\leq m(1-\lambda^{\alpha})f(x)+\lambda^{\alpha}f(y),\forall x,y\in K\}\) is dense in \([0,1]\).

The proof of Lemma 2.1 is much akin to that of given method for Lemma 3.2 in [40], p.232. The details are left to the interested reader. The next theorem shows the relationship between the generalized \((\alpha,m)\)-preinvex function and the generalized quasi m-preinvex function.

Theorem 2.1

Let K be a nonempty m-invex set in \(\mathbb{R}_{0}\) with respect to \(\eta: \mathbb{R}\times\mathbb{R}\times(0, 1]\rightarrow\mathbb{R}\), where η satisfies the extended Condition C. Then the real-value decrease function \(f:K\rightarrow\mathbb{R}_{0}\) is a generalized \((\alpha,m)\)-preinvex function if and only if it is a generalized quasi m-preinvex function on K and there exists a \(t\in(0,1)\) such that

$$ f \bigl(mx+t\eta(y,x,m) \bigr)\leq m\bigl(1-t^{\alpha}\bigr)f(x)+t^{\alpha}f(y), \quad \forall x,y\in K. $$
(2.7)

Proof

The necessity is proofed by Proposition 2.1. We only need prove the sufficiency.

For every \(x,y\in K\), let \(z_{\lambda}=mx+\lambda\eta(y,x,m)\), \(\lambda\in[0,1]\). Two different situations where \(mf(x)=f(y)\) or \(mf(x)\neq f(y)\) will be considered as follows, respectively.

(I) \(mf(x)=f(y)\). We need to prove that

$$ f \bigl(mx+\lambda\eta(y,x,m) \bigr)\leq m\bigl(1-\lambda^{\alpha}\bigr)f(x)+\lambda ^{\alpha}f(y), \quad\forall\lambda\in[0,1]. $$

By contradiction, assume that there exists \(\beta\in(0,1]\) such that

$$ f(z_{\beta})=f \bigl(mx+\beta\eta(y,x,m) \bigr)> m\bigl(1- \beta^{\alpha}\bigr)f(x)+\beta ^{\alpha}f(y)=mf(x)=f(y). $$
(2.8)

(i) Suppose that \(0<\gamma<\beta\leq1\). Let \(\mu=\frac{\beta-\gamma }{1-\gamma}\). From the extended Condition C, we have

$$ z_{\beta}=z_{\mu}+\gamma\eta(y,z_{\mu},m). $$

From (2.7) and (2.8) and the decrease of f on K, we deduce that

$$\begin{aligned} f(z_{\beta}) & =f \bigl(z_{\mu}+\gamma \eta(y,z_{\mu},m) \bigr) \\ & \leq f \bigl(mz_{\mu}+\gamma\eta(y,z_{\mu},m) \bigr) \\ & \leq m\bigl(1-\gamma^{\alpha}\bigr)f(z_{\mu})+ \gamma^{\alpha}f(y) \\ & < mf(z_{\mu}) \\ &< f(z_{\mu}). \end{aligned}$$
(2.9)

To prove the third inequality above, we used the fact that \(f(y)< mf(z_{\mu})\). Otherwise, this breeds a contradiction to (2.8). On the other hand, let \(\delta=\frac{\beta-\mu}{\beta}\) and from the extended Condition C, we get

$$z_{\mu}=z_{\beta}+\delta\eta(x,z_{\beta},m). $$

Consequently, from the decrease of f on K and the generalized quasi m-preinvexity of f, we derive that

$$ f(z_{\mu}) =f \bigl(z_{\beta}+\delta \eta(x,z_{\beta},m) \bigr) \leq f \bigl(mz_{\beta}+\delta \eta(x,z_{\beta},m) \bigr) \leq\max\bigl\{ f(z_{\beta}),f(x)\bigr\} . $$
(2.10)

(a) If \(f(x)\leq f(z_{\beta})\), from the inequality (2.10), we have \(f(z_{\mu})\leq f(z_{\beta})\), which contradicts the inequality (2.9).

(b) If \(f(x)> f(z_{\beta})\), from the inequality (2.10), we have \(f(z_{\mu})\leq f(x)\), which contradicts the fact that \(f(x)< f(z_{\mu})\).

(ii) Assume that \(0<\beta<\gamma\leq1\). Let \(\mu=\frac{\beta }{\gamma}>\beta\). From the extended Condition C, we obtain

$$ z_{\beta}=mx+\gamma\eta(z_{\mu},x,m). $$
(2.11)

From (2.7) and (2.11) as well as (2.8), we deduce that

$$ f(z_{\beta}) =f \bigl(mx+\gamma\eta(z_{\mu},x,m) \bigr) \leq m\bigl(1-\gamma^{\alpha}\bigr)f(x)+\gamma^{\alpha}f(z_{\mu}) < f(z_{\mu}). $$
(2.12)

Let \(\delta=\frac{\mu-\beta}{1-\beta}\), by the extended Condition C, we have

$$ z_{\mu}=z_{\beta}+\delta\eta(y,z_{\beta},m). $$
(2.13)

In the same way, from (2.7) and (2.13) as well as (2.8), we get

$$\begin{aligned} f(z_{\mu}) & =f \bigl(z_{\beta}+\delta\eta(y,z_{\beta},m) \bigr) \\ & \leq f \bigl(mz_{\beta}+\delta\eta(y,z_{\beta},m) \bigr) \\ & \leq m\bigl(1-\gamma^{\alpha}\bigr)f(z_{\beta})+ \gamma^{\alpha}f(y) \\ & < f(z_{\beta}). \end{aligned}$$

which contradicts the inequality (2.12).

(II) \(mf(x)\neq f(y)\). In this case, we also need to prove that

$$ f \bigl(mx+\lambda\eta(y,x,m) \bigr)\leq m\bigl(1-\lambda^{\alpha}\bigr)f(x)+\lambda ^{\alpha}f(y),\quad \forall\lambda\in[0,1]. $$

By contradiction, assume that there exists \(\beta\in(0,1)\) such that

$$ f(z_{\beta})=f \bigl(mx+\beta\eta(y,x,m) \bigr)> m\bigl(1- \beta^{\alpha}\bigr)f(x)+\beta ^{\alpha}f(y). $$
(2.14)

From Lemma 2.1, we know that, for A, defined in Lemma 2.1,

$$ f\bigl(mx+\lambda\eta(y,x,m)\bigr)\leq m\bigl(1-\lambda^{\alpha}\bigr)f(x)+ \lambda^{\alpha}f(y),\quad\forall\lambda\in A. $$

(1) Assume that \(mf(x)>f(y)\). Then from (2.14) and the density of A, there exists \(\mu\in A\) with \(\mu<\beta\) such that

$$\begin{aligned} f(z_{\mu}) & =f \bigl(mx+\mu\eta(y,x,m) \bigr) \\ & \leq m\bigl(1-\mu^{\alpha}\bigr)f(x)+\mu^{\alpha}f(y) \\ & \leq f \bigl(mx+\beta\eta(y,x,m) \bigr) \\ & =f(z_{\beta}). \end{aligned}$$
(2.15)

Let \(\delta=\frac{\beta-\mu}{1-\mu}\). Clearly \(0<\delta<1\) and from the extended Condition C, we have

$$ z_{\beta}=z_{\mu}+\delta\eta(y,z_{\mu},m). $$

(a) If \(f(y)\leq f(z_{\mu})\), from the decrease generalized quasi-m-preinvexity of f, we obtain

$$\begin{aligned} f(z_{\beta}) & =f \bigl(z_{\mu}+\delta\eta(y,z_{\mu},m) \bigr) \\ & \leq f \bigl(mz_{\mu}+\delta\eta(y,z_{\mu},m) \bigr) \\ & \leq\max\bigl\{ f(z_{\mu}),f(y)\bigr\} \\ & \leq f(z_{\mu}), \end{aligned}$$

which contradicts the inequality (2.15).

(b) If \(f(y)> f(z_{\mu})\), similarly, by the decrease generalized quasi-m-preinvexity of f and \(mf(x)>f(y)\) we obtain

$$\begin{aligned} f(z_{\beta}) & =f \bigl(z_{\mu}+\delta\eta(y,z_{\mu},m) \bigr) \\ & \leq f \bigl(mz_{\mu}+\delta\eta(y,z_{\mu},m) \bigr) \\ & \leq\max\bigl\{ f(z_{\mu}),f(y)\bigr\} \\ & \leq f(y) \\ & < m\bigl(1-\beta^{\alpha}\bigr)f(x)+\beta^{\alpha}f(y) \\ & < f(z_{\beta}), \end{aligned}$$

which is a contradiction.

(2) Assume that \(mf(x)< f(y)\). Then from (2.14) and the density of A, there exists \(\mu\in A\) with \(\mu>\beta\) such that

$$\begin{aligned} f(z_{\mu}) & =f \bigl(mx+\mu\eta(y,x,m) \bigr) \\ & \leq m\bigl(1-\mu^{\alpha}\bigr)f(x)+\mu^{\alpha}f(y) \\ & \leq f \bigl(mx+\beta\eta(y,x,m) \bigr) \\ & =f(z_{\beta}). \end{aligned}$$
(2.16)

Let \(\delta=\frac{\beta}{\mu}\). Obviously \(0<\delta<1\) and from the extended Condition C, we have

$$ z_{\beta}=mx+\delta\eta(z_{\mu},x,m). $$
(2.17)

(a) If \(mf(x)\leq f(z_{\mu})\), from (2.7) and (2.17), we obtain

$$ f(z_{\beta}) =f \bigl(mx+\delta\eta(z_{\mu},x,m) \bigr) \leq m \bigl(1-\delta^{\alpha}\bigr)f(x)+\delta^{\alpha}f(z_{\mu}) \leq f(z_{\mu}), $$

which contradicts the inequality (2.16).

(b) If \(mf(x)> f(z_{\mu})\), in the same way, and utilizing \(mf(x)< f(y)\), we obtain

$$\begin{aligned} f(z_{\beta}) & =f \bigl(mx+\delta\eta(z_{\mu},x,m) \bigr) \\ & \leq m\bigl(1-\delta^{\alpha}\bigr)f(x)+\delta^{\alpha}f(z_{\mu}) \\ & \leq mf(x) \\ & < m\bigl(1-\beta^{\alpha}\bigr)f(x)+\beta^{\alpha}f(y) \\ & < f(z_{\beta}), \end{aligned}$$

which is a contradiction. This completes the proof. □

The result established by Theorem 2.1 shows that under certain conditions the generalized \((\alpha,m)\)-preinvexity is equivalent to the generalized quasi-m-preinvexity when there exists a point to satisfy generalized \((\alpha,m)\)-preinvexity. The extended Condition C seems to be an indispensable hypothesis.

3 Riemann-Liouville fractional Hermite-Hadamard inequalities

Let f: \(K \rightarrow\mathbb{R}\) be a differentiable function, throughout this section we will take

$$\begin{aligned} R_{f}(\alpha;\eta,m,a,b):={}& \frac{f(ma)+f (ma+\eta(b,a,m) )}{2} \\ &{}-\frac{\Gamma(\alpha+1)}{2\eta^{\alpha}(b,a,m)} \bigl[J^{\alpha}_{ma^{+}}f \bigl(ma+ \eta(b,a,m) \bigr) +J^{\alpha}_{(ma+\eta(b,a,m))^{-}}f(ma) \bigr], \end{aligned}$$

where \(K\subseteq\mathbb{R}\) be an open m-invex subset with respect to \(\eta: K \times K \times(0, 1]\rightarrow\mathbb{R}\) for some fixed \(m\in(0, 1]\), \(a, b \in K\) with \(a< b\), \(\alpha>0\) and Γ is the Euler Gamma function.

We prove the following lemma to obtain our new results in this section.

Lemma 3.1

Let \(K\subseteq\mathbb{R}\) be an open m-invex subset with respect to \(\eta: K \times K \times(0, 1]\rightarrow\mathbb{R}\) for some fixed \(m\in(0, 1]\) and let \(a, b \in K\), \(a < b\) with \(\eta(b, a, m)>0\). Assume that \(f:K \rightarrow\mathbb{R}\) is a twice differentiable function, \(f''\) is integrable on \([ma, m a+ \eta(b, a, m)]\), then the following identity for the Riemann-Louville fractional integral with \(\alpha>0\) and x∈ \([ma,ma+\eta(b,a,m)]\) holds:

$$\begin{aligned} R_{f}(\alpha;\eta,m,a,b) =\frac{\eta^{2}(b,a,m)}{2} \int_{0}^{1}\frac{1-t^{\alpha+1}-(1-t)^{\alpha +1}}{\alpha+1}f'' \bigl(ma+t\eta(b,a,m) \bigr)\,\mathrm{d}t. \end{aligned}$$
(3.1)

Proof

Set

$$\begin{aligned} I = \frac{\eta^{2}(b,a,m)}{2} \int_{0}^{1}\frac{1-t^{\alpha+1}-(1-t)^{\alpha +1}}{\alpha+1}f'' \bigl(ma+t\eta(b,a,m) \bigr)\,\mathrm{d}t. \end{aligned}$$

Since \(a, b\in K \) and K is an m-invex subset with respect to η, for every \(t \in[0, 1]\) and some fixed \(m\in(0, 1]\), we have \(ma+t\eta(b,a,m)\in K \). Integrating by part yields

$$\begin{aligned} I ={}& \frac{\eta^{2}(b,a,m)}{2} \biggl[\frac{1-t^{\alpha+1}-(1-t)^{\alpha +1}}{(\alpha+1)\eta(b,a,m)} f' \bigl(ma+t \eta(b,a,m) \bigr) \bigg|_{0}^{1} \\ &{}- \int_{0}^{1} \frac{-(\alpha+1)t^{\alpha}+(1-t)^{\alpha}(\alpha+1)}{(\alpha +1)\eta(b,a,m)} f' \bigl(ma+t\eta(b,a,m) \bigr)\,\mathrm{d}t \biggr] \\ = {}& \frac{\eta^{2}(b,a,m)}{2} \biggl[\frac{f (ma+\eta(b,a,m) )}{\eta ^{2}(b,a,m)} + \frac{f(ma)}{\eta^{2}(b,a,m)} \\ &{}- \int_{0}^{1} \frac{\alpha t^{\alpha-1}+\alpha(1-t)^{\alpha-1}}{\eta ^{2}(b,a,m)} f \bigl(ma+t \eta(b,a,m) \bigr) \,\mathrm{d}t \biggr] \\ = {}& \frac{f(ma)+f (ma+\eta(b,a,m) )}{2} \\ &{}- \frac{\alpha}{2} \biggl[ \int_{0}^{1} \bigl(t^{\alpha-1}+(1-t)^{\alpha-1} \bigr) f \bigl(ma+t\eta(b,a,m) \bigr) \,\mathrm{d}t \biggr]. \end{aligned}$$

Let \(u = ma+t\eta(b,a,m)\), then \(\mathrm{d}u = \eta (b,a,m)\,\mathrm{d}t\), and using the reduction formula \(\Gamma(\alpha+1) = \alpha\Gamma(\alpha)\) \((\alpha>0)\) for Euler Gamma function, we have

$$\begin{aligned} \frac{\alpha}{2} \biggl[ \int_{0}^{1}t^{\alpha-1} f \bigl(ma+t\eta(b,a,m) \bigr) \,\mathrm{d}t\biggr] = \frac{\Gamma(\alpha+1)}{2\eta^{\alpha}(b,a,m)} J^{\alpha}_{(ma+\eta(b,a,m))^{-}}f(ma) \end{aligned}$$

and similarly we get

$$\begin{aligned} \frac{\alpha}{2} \biggl[ \int_{0}^{1} (1-t)^{\alpha-1} f \bigl(ma+t\eta (b,a,m) \bigr) \,\mathrm{d}t\biggr] = \frac{\Gamma(\alpha+1)}{2\eta^{\alpha}(b,a,m)} J^{\alpha}_{ma^{+}}f (ma+\eta(b,a,m). \end{aligned}$$

Thus, we have conclusion (3.1). □

Remark 3.1

If \(\eta(b,a,m) = b-ma\) with \(m=1\) in Lemma 3.1, then the identity (3.1) reduces to the following identity:

$$\begin{aligned} & \frac{f(a)+f(b)}{2} - \frac{\Gamma(\alpha+1)}{2(b-a)^{\alpha}} \bigl[J^{\alpha}_{a^{+}}f(b)+J^{\alpha}_{b^{-}}f(a) \bigr] \\ & \quad=\frac{(b-a)^{2}}{2} \int_{0}^{1}\frac{1-t^{\alpha+1}-(1-t)^{\alpha +1}}{\alpha+1} f'' \bigl(tb+(1-t)a \bigr)\,\mathrm{d}t. \end{aligned}$$
(3.2)

By using \(J^{\alpha}_{b^{+}}f(a)+J^{\alpha}_{a^{-}}f(b) = (-1)^{\alpha}[J^{\alpha}_{a^{+}}f(b)+J^{\alpha}_{b^{-}}f(a)]\) and exchanging a with b in (3.2), it follows that

$$\begin{aligned} & \frac{f(a)+f(b)}{2} - \frac{\Gamma(\alpha+1)}{2(b-a)^{\alpha}} \bigl[J^{\alpha}_{a^{+}}f(b)+J^{\alpha}_{b^{-}}f(a) \bigr] \\ &\quad =\frac{(b-a)^{2}}{2} \int_{0}^{1}\frac{1-t^{\alpha+1}-(1-t)^{\alpha +1}}{\alpha+1} f'' \bigl(ta+(1-t)b \bigr)\,\mathrm{d}t, \end{aligned}$$
(3.3)

which is proved by Wang et al. [30]. Based on this identity, they established some interesting Riemann-Liouville fractional integrals for m-convex and \((s,m)\)-convex mappings, respectively.

If we choose \(\alpha= 1\) in (3.3), it follows that

$$ \frac{f(a)+f(b)}{2} - \frac{1}{b-a} \int_{a}^{b} f(t)\,\mathrm{d}t =\frac{(b-a)^{2}}{2} \int_{0}^{1} t(1-t)f'' \bigl(ta+(1-t)b \bigr)\,\mathrm{d}t, $$

which is used by Ödemir, Avci and Set in [6] to establish many interesting Hermite-Hadamard-type inequalities for m-convexity.

With the help of Lemma 3.1, new upper bound for the right-hand side of (1.6) for generalized \((\alpha,m)\)-preinvex functions via the Riemann-Liouville fractional integral is presented in the following theorem.

Theorem 3.1

Let \(A\subseteq\mathbb{R}\) be an open m-invex subset with respect to \(\eta: A \times A \times(0, 1]\rightarrow\mathbb{R}\) for some fixed \(m\in(0, 1]\) and let \(a, b \in A\), \(a < b\) with \(\eta(b, a, m)>0\). Assume that \(f:A \rightarrow\mathbb{R}\) is a twice differentiable function, \(|f''|\) is a generalized \((\alpha,m)\)-preinvex function on A for some fixed \(\alpha,m \in(0,1]\) and x∈ \([ma,ma+\eta(b,a,m)]\), then the following inequality for the Riemann-Louville fractional integral with \(0<\alpha\leq1\) holds:

$$\begin{aligned} & \bigl|R_{f}(\alpha;\eta,m,a,b) \bigr| \\ &\quad\leq\frac{\eta^{2}(b,a,m)}{2(\alpha+1)} \biggl[m \biggl(\frac{2{\alpha }^{2}+\alpha-2}{(\alpha+2)(2\alpha+2)}+\beta(\alpha+1, \alpha+2) \biggr)\bigl|f''(a)\bigr| \\ &\qquad{} + \biggl(\frac{1}{2\alpha+2}-\beta(\alpha+1,\alpha+2) \biggr)\bigl|f''(b)\bigr| \biggr]. \end{aligned}$$
(3.4)

Proof

Since \(ma+t\eta(b,a,m)\in A\) for each \(t\in[0,1]\), by using the properties of modulus on Lemma 3.1, we can obtain

$$ \bigl|R_{f}(\alpha;\eta,m,a,b) \bigr| \leq\frac{\eta^{2}(b,a,m)}{2} \int_{0}^{1} \biggl|\frac{1-t^{\alpha +1}-(1-t)^{\alpha+1}}{\alpha+1} \biggr|\bigl|f'' \bigl(ma+t\eta(b,a,m) \bigr)\bigr|\,\mathrm{d}t. $$

Using the generalized \((\alpha,m)\)-preinvexity of \(|f''|\) on A, we have

$$\begin{aligned} & \int_{0}^{1} \biggl|\frac{1-t^{\alpha+1}-(1-t)^{\alpha+1}}{\alpha+1} \biggr|\bigl|f'' \bigl(ma+t\eta(b,a,m) \bigr)\bigr|\,\mathrm{d}t \\ &\quad\leq\frac{1}{\alpha+1} \int_{0}^{1} \bigl(1-t^{\alpha+1}-(1-t)^{\alpha +1} \bigr) \bigl(m\bigl(1-t^{\alpha}\bigr)\bigl|f''(a)\bigr|+t^{\alpha}\bigl|f''(b)\bigr| \bigr)\,\mathrm{d}t \\ &\quad\leq\frac{1}{\alpha+1} \biggl[m \biggl(\frac{2{\alpha}^{2}+\alpha-2}{(\alpha +2)(2\alpha+2)}+\beta(\alpha+1, \alpha+2) \biggr)\bigl|f''(a)\bigr| \\ &\qquad{}+ \biggl(\frac{1}{2\alpha+2}-\beta(\alpha+1,\alpha+2) \biggr)\bigl|f''(b)\bigr| \biggr]. \end{aligned}$$

To prove the second inequality above, we used the facts that

$$\begin{aligned} & \int_{0}^{1} \bigl(1-t^{\alpha+1}-(1-t)^{\alpha+1}-t^{\alpha}+t^{2\alpha +1} \bigr)\,\mathrm{d}t=\frac{2\alpha^{2}+\alpha-2}{(\alpha+2)(2\alpha+2)}, \\ & \int_{0}^{1}\bigl(t^{\alpha}-t^{2\alpha+1} \bigr)\,\mathrm{d}t=\frac{1}{2\alpha+2}, \end{aligned}$$

and

$$\int_{0}^{1}t^{\alpha}(1-t)^{\alpha+1}\, \mathrm{d}t=\beta(\alpha+1,\alpha+2), $$

where the Beta function,

$$\beta(x,y)= \int_{0}^{1}t^{x-1}(1-t)^{y-1}\, \mathrm{d}t,\quad \forall x, y>0, $$

which completes the proof. □

By means of elementary calculation, it is easy to deduce the following results.

Corollary 3.1

With the same assumptions given in Theorem 3.1, if \(\eta (b,a,m)=b-ma\), we obtain

$$\begin{aligned} & \biggl|\frac{f(ma)+f(b)}{2}-\frac{\Gamma(\alpha+1)}{2(b-ma)^{\alpha}} \bigl[J^{\alpha}_{ma^{+}}f(b) +J^{\alpha}_{b^{-}}f(ma) \bigr] \biggr| \\ &\quad\leq \frac{(b-ma)^{2}}{2(\alpha+1)} \biggl[m \biggl(\frac{2{\alpha }^{2}+\alpha-2}{(\alpha+2)(2\alpha+2)}+\beta(\alpha+1, \alpha+2) \biggr)\bigl|f''(a)\bigr| \\ &\qquad{} + \biggl(\frac{1}{2\alpha+2}-\beta(\alpha+1,\alpha+2) \biggr)\bigl|f''(b)\bigr| \biggr], \end{aligned}$$

specially for \(\alpha=m=1\), we get

$$ \biggl|\frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int^{b}_{a}f(x)\,\mathrm{d}x \biggr| \leq \frac{(b-a)^{2}}{12} \biggl[ \frac{|f''(a)| + |f''(b)|}{2} \biggr] . $$

This is one of the inequalities given in [38], Theorem 2.

Corollary 3.2

In Theorem 3.1, if the mapping \(\eta(b,a,m)\) with \(m=1\) degenerates into \(\eta(b,a)\) and we choose \(\alpha=1\), then (3.4) becomes

$$\begin{aligned} \biggl|\frac{f(a)+f (a+\eta(b,a) )}{2}-\frac{1}{\eta(b,a)} \int ^{a+\eta(b,a)}_{a}f(x)\,\mathrm{d}x \biggr| \leq \frac{\eta^{2}(b,a)}{24} \bigl[ \bigl|f''(a)\bigr| + \bigl|f''(b)\bigr| \bigr], \end{aligned}$$

which is the same as the inequality established in [37], Theorem 4.1.

Theorem 3.2

Let f be defined as in Theorem 3.1, If the function \(|f''|^{q}\) for \(q >1\) is a generalized \((\alpha,m)\)-preinvex function on A for some fixed \(\alpha,m \in(0,1]\) and x∈ \([ma,ma+\eta(b,a,m)]\), then the following inequality for the Riemann-Louville fractional integral with \(0<\alpha\leq1\) holds:

$$\begin{aligned} & \bigl|R_{f}(\alpha;\eta,m,a,b) \bigr| \\ &\quad\leq\frac{\eta^{2}(b,a,m)}{2(\alpha+1)} \biggl\{ m \biggl[\frac{q\alpha+\alpha+1}{(\alpha+1)(q+1)}- \frac {2}{q(\alpha+1)+1}+\beta \bigl(\alpha+1,q(\alpha+1)+1 \bigr) \biggr]\bigl|f''(a)\bigr|^{q} \\ &\qquad{}+ \biggl[\frac{q}{(\alpha+1)(q+1)}-\beta \bigl(\alpha+1,q(\alpha+1)+1 \bigr) \biggr]\bigl|f''(b)\bigr|^{q} \biggr\} ^{\frac{1}{q}}. \end{aligned}$$
(3.5)

Proof

Since \(ma+t\eta(b,a,m)\in A\) for every \(t\in[0,1]\), by using the properties of modulus on Lemma 3.1 and making use of Hölder’s integral inequality for \(q>1\), we can obtain

$$\begin{aligned} & \bigl|R_{f}(\alpha;\eta,m,a,b) \bigr| \\ &\quad\leq\frac{\eta^{2}(b,a,m)}{2} \int_{0}^{1} \biggl|\frac{1-t^{\alpha +1}-(1-t)^{\alpha+1}}{\alpha+1} \biggr|\bigl|f'' \bigl(ma+t\eta(b,a,m) \bigr)\bigr|\,\mathrm{d}t \\ &\quad\leq\frac{\eta^{2}(b,a,m)}{2(\alpha+1)} \biggl( \int_{0}^{1} 1\,\mathrm{d}t \biggr)^{1-\frac{1}{q}} \biggl[ \int_{0}^{1} \bigl(1-t^{\alpha+1}-(1-t)^{\alpha+1} \bigr)^{q}\bigl|f'' \bigl(ma+t\eta(b,a,m) \bigr)\bigr|^{q}\,\mathrm{d}t \biggr]^{\frac{1}{q}} \\ &\quad\leq\frac{\eta^{2}(b,a,m)}{2(\alpha+1)} \biggl[ \int_{0}^{1} \bigl(1-t^{q(\alpha+1)}-(1-t)^{q(\alpha+1)} \bigr)\bigl|f'' \bigl(ma+t\eta(b,a,m) \bigr)\bigr|^{q} \,\mathrm{d}t \biggr]^{\frac{1}{q}}. \end{aligned}$$

To prove the third inequality above, we used the following inequality:

$$ \bigl(1-t^{\alpha+1}-(1-t)^{\alpha+1} \bigr)^{q}\leq1-t^{q(\alpha +1)}-(1-t)^{q(\alpha+1)} $$
(3.6)

for any \(t\in[0,1]\), which follows from

$$ (A-B)^{q}\leq A^{q}-B^{q} $$

for any \(A>B\geq0\) and \(q\geq1\). □

Using the generalized \((\alpha,m)\)-preinvexity of \(|f''|^{q}\) on A, we have

$$\begin{aligned} & \int_{0}^{1} \bigl(1-t^{q(\alpha+1)}-(1-t)^{q(\alpha+1)} \bigr)\bigl|f'' \bigl(ma+t\eta(b,a,m) \bigr)\bigr|^{q} \,\mathrm{d}t \\ &\quad\leq \int_{0}^{1} \bigl(1-t^{(\alpha+1)q}-(1-t)^{(\alpha+1)q} \bigr) \bigl(m\bigl(1-t^{\alpha}\bigr)\bigl|f''(a)\bigr|^{q} + t^{\alpha}\bigl|f''(b)\bigr|^{q} \bigr)\, \mathrm{d}t \\ &\quad=m \biggl[\frac{\alpha q+\alpha+1}{(\alpha+1)(q+1)}-\frac{2}{q(\alpha +1)+1}+\beta \bigl(\alpha+1,q( \alpha+1)+1 \bigr) \biggr]\bigl|f''(a)\bigr|^{q} \\ &\qquad{}+ \biggl[\frac{q}{(\alpha+1)(q+1)}-\beta \bigl(\alpha+1,q(\alpha+1)+1 \bigr) \biggr]\bigl|f''(b)\bigr|^{q}. \end{aligned}$$

Thus, we can get the desired result.

Direct computation yields the following corollary.

Corollary 3.3

With the same assumptions given in Theorem 3.2, if \(\eta (b,a,m)=b-ma\), we obtain

$$\begin{aligned} & \biggl|\frac{f(ma)+f(b)}{2} -\frac{\Gamma(\alpha+1)}{2(b-ma)^{\alpha}} \bigl[J^{\alpha}_{ma^{+}}f(b) +J^{\alpha}_{b^{-}}f(ma) \bigr] \biggr| \\ &\quad\leq\frac{(b-ma)^{2}}{2(\alpha+1)} \biggl\{ m \biggl[\frac{q\alpha+\alpha+1}{(\alpha+1)(q+1)}- \frac {2}{q(\alpha+1)+1}+\beta \bigl(\alpha+1,q(\alpha+1)+1 \bigr) \biggr]\bigl|f''(a)\bigr|^{q} \\ &\qquad{}+ \biggl[\frac{q}{(\alpha+1)(q+1)}-\beta \bigl(\alpha+1,q(\alpha+1)+1 \bigr) \biggr]\bigl|f''(b)\bigr|^{q} \biggr\} ^{\frac{1}{q}}; \end{aligned}$$

specially for \(\alpha=m=1\) and \(|f''| \leq K\) on \([a,b]\), we get

$$\begin{aligned} \biggl|\frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int^{b}_{a}f(x)\,\mathrm{d}x \biggr| &\leq \frac{(b-a)^{2}}{4} \biggl( \frac{2q-1}{2q+1} \biggr)^{\frac{1}{q}} K \\ & \leq \frac{(b-a)^{2}}{4} K. \end{aligned}$$
(3.7)

For proving the second inequality of (3.7), we use the facts that

$$ \lim_{q \to1^{+}} \biggl(\frac{2q-1}{2q+1} \biggr)^{\frac{1}{q}} = \frac{1}{3} $$

and

$$ \lim_{q \to\infty} \biggl(\frac{2q-1}{2q+1} \biggr)^{\frac{1}{q}} = 1. $$

Therefore, we have

$$ \frac{1}{3} < \biggl(\frac{2q-1}{2q+1} \biggr)^{\frac{1}{q}} < 1, \quad q\in (1,\infty). $$
(3.8)

A similar result is presented in the following theorem.

Theorem 3.3

Let f be defined as in Theorem 3.1 with \(\frac{1}{p}+\frac {1}{q}=1\), \(q >1\). If \(|f''|^{q}\) is a generalized \((\alpha ,m)\)-preinvex function on A for some fixed \(\alpha,m \in(0,1]\) and x∈ \([ma,ma+\eta(b,a,m)]\), then the following inequality for the Riemann-Louville fractional integral with \(0<\alpha\leq1\) holds:

$$\begin{aligned} &\bigl|R_{f}(\alpha;\eta,m,a,b) \bigr| \\ &\quad\leq\frac{\eta^{2}(b,a,m)}{2(\alpha+1)} \biggl(\frac{p\alpha+p-1}{p\alpha +p+1} \biggr)^{\frac{1}{p}} \biggl(\frac{m\alpha}{\alpha+1}\bigl|f''(a)\bigr|^{q}+ \frac {1}{\alpha+1}\bigl|f''(b)\bigr|^{q} \biggr)^{\frac{1}{q}}. \end{aligned}$$
(3.9)

Proof

Since \(ma+t\eta(b,a,m)\in A\) for every \(t\in[0,1]\), by using the properties of modulus on Lemma 3.1 and Hölder’s integral inequality for \(q>1\), we can obtain

$$\begin{aligned} & \bigl|R_{f}(\alpha;\eta,m,a,b) \bigr| \\ &\quad\leq\frac{\eta^{2}(b,a,m)}{2} \int_{0}^{1} \biggl|\frac{1-t^{\alpha +1}-(1-t)^{\alpha+1}}{\alpha+1} \biggr|\bigl|f'' \bigl(ma+t\eta(b,a,m) \bigr)\bigr|\,\mathrm{d}t \\ &\quad\leq\frac{\eta^{2}(b,a,m)}{2(\alpha+1)} \biggl( \int_{0}^{1}\bigl|1-t^{\alpha +1}-(1-t)^{\alpha+1}\bigr|^{p} \,\mathrm{d}t \biggr)^{\frac{1}{p}} \biggl( \int_{0}^{1}\bigl|f'' \bigl(ma+t\eta(b,a,m) \bigr)\bigr|^{q}\,\mathrm{d}t \biggr)^{\frac {1}{q}}. \end{aligned}$$

Using the inequality (3.6) and the generalized \((\alpha,m)\)-preinvexity of \(|f''|^{q}\) on A, we have

$$\begin{aligned} & \bigl|R_{f}(\alpha;\eta,m,a,b) \bigr| \\ &\quad\leq\frac{\eta^{2}(b,a,m)}{2(\alpha+1)} \biggl( \int_{0}^{1} \bigl(1-t^{p(\alpha +1)}-(1-t)^{p(\alpha+1)} \bigr)\,\mathrm{d}t \biggr)^{\frac{1}{p}} \\ &\qquad{}\times \biggl( \int_{0}^{1} \bigl( m\bigl(1-t^{\alpha} \bigr)\bigl|f''(a)\bigr|^{q}+t^{\alpha }\bigl|f''(b)\bigr|^{q} \bigr)\,\mathrm{d}t \biggr)^{\frac{1}{q}} \\ &\quad=\frac{\eta^{2}(b,a,m)}{2(\alpha+1)} \biggl(\frac{p\alpha+p-1}{p\alpha +p+1} \biggr)^{\frac{1}{p}} \biggl(\frac{m\alpha}{\alpha+1}\bigl|f''(a)\bigr|^{q}+ \frac{1}{\alpha +1}\bigl|f''(b)\bigr|^{q} \biggr)^{\frac{1}{q}}. \end{aligned}$$

Therefore, we can get the required results. □

Elementary calculation provides the following corollaries.

Corollary 3.4

With the same assumptions given in Theorem 3.3, if \(\eta (b,a,m)=b-ma\), we obtain

$$\begin{aligned} & \biggl|\frac{f(ma)+f(b)}{2} -\frac{\Gamma(\alpha+1)}{2(b-ma)^{\alpha}} \bigl[J^{\alpha}_{ma^{+}}f(b) +J^{\alpha}_{b^{-}}f(ma) \bigr] \biggr| \\ &\quad\leq\frac{(b-ma)^{2}}{2(\alpha+1)} \biggl(\frac{p\alpha+p-1}{p\alpha +p+1} \biggr)^{\frac{1}{p}} \biggl(\frac{m\alpha}{\alpha+1}\bigl|f''(a)\bigr|^{q}+ \frac {1}{\alpha+1}\bigl|f''(b)\bigr|^{q} \biggr)^{\frac{1}{q}}, \end{aligned}$$

specially for \(\alpha=m=1\) and \(|f''| \leq K\) on \([a,b]\), we get

$$ \biggl|\frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int^{b}_{a}f(x)\,\mathrm{d}x \biggr| \leq \frac{(b-a)^{2}}{4} \biggl(\frac{2p-1}{2p+1} \biggr)^{\frac{1}{p}} K. $$

Corollary 3.5

In Theorem 3.3, if the mapping \(\eta(b,a,m)\) with \(m=1\) degenerates into \(\eta(b,a)\) and we choose \(\alpha=1\), then (3.9) becomes

$$\begin{aligned} & \biggl|\frac{f(a)+f (a+\eta(b,a) )}{2}-\frac{1}{\eta(b,a)} \int ^{a+\eta(b,a)}_{a}f(x)\,\mathrm{d}x \biggr| \\ &\quad\leq \frac{\eta^{2}(b-a)}{4} \biggl(\frac{2p-1}{2p+1} \biggr)^{\frac{1}{p}} \biggl(\frac{1}{2} \biggr)^{\frac{1}{q}} \bigl(\bigl|f''(a)\bigr|^{q}+\bigl|f''(b)\bigr|^{q} \bigr)^{\frac {1}{q}} \\ &\quad\leq \frac{\eta^{2}(b-a)}{4} \biggl(\frac{1}{2} \biggr)^{\frac{1}{q}} \bigl(\bigl|f''(a)\bigr|^{q}+\bigl|f''(b)\bigr|^{q} \bigr)^{\frac{1}{q}}, \end{aligned}$$
(3.10)

where we also use the inequality (3.8) for \(p>1\) and \(\frac {1}{p}+\frac{1}{q}=1\).

It is noted that the result of the second inequality (3.10) is the same as the one presented by Barani, Ghazanfari, and Dragomir in [37], Theorem 4.3. Clearly, the result of the first inequality (3.10) is better than the inequality established by Barani et al. in [37], Theorem 4.3.

A different approach leads to the following results.

Theorem 3.4

Suppose that all the assumptions of Theorem 3.2 are satisfied. Then the following inequality for the Riemann-Louville fractional integral with \(0<\alpha\leq1\) holds:

$$\begin{aligned} & \bigl|R_{f}(\alpha;\eta,m,a,b) \bigr| \\ &\quad\leq \frac{\eta^{2}(b,a,m)}{2(\alpha+1)} \biggl(\frac{\alpha}{\alpha +2} \biggr)^{1-\frac{1}{q}} \biggl[m \biggl(\frac{2{\alpha}^{2}+\alpha-2}{(\alpha+2)(2\alpha+2)} +\beta (\alpha+1,\alpha+2) \biggr)\bigl|f''(a)\bigr|^{q} \\ &\qquad{} + \biggl(\frac{1}{2\alpha+2}-\beta(\alpha+1,\alpha+2) \biggr)\bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}. \end{aligned}$$
(3.11)

Proof

Since \(ma+t\eta(b,a,m)\in A\) for every \(t\in[0,1]\), by utilizing the properties of modulus on Lemma 3.1 and using Hölder’s integral inequality for \(q>1\), we can obtain

$$\begin{aligned} & \bigl|R_{f}(\alpha;\eta,m,a,b) \bigr| \\ &\quad\leq \frac{\eta^{2}(b,a,m)}{2} \int_{0}^{1} \biggl|\frac{1-t^{\alpha +1}-(1-t)^{\alpha+1}}{\alpha+1} \biggr|\bigl|f'' \bigl(ma+t\eta(b,a,m) \bigr)\bigr|\,\mathrm{d}t \\ &\quad\leq \frac{\eta^{2}(b,a,m)}{2(\alpha+1)} \biggl[ \int_{0}^{1} \bigl(1-t^{\alpha +1}-(1-t)^{\alpha+1} \bigr)\,\mathrm{d}t \biggr]^{1-\frac{1}{q}} \\ &\qquad{}\times \biggl[ \int_{0}^{1} \bigl(1-t^{\alpha+1}-(1-t)^{\alpha+1} \bigr)\bigl|f'' \bigl(ma+t\eta(b,a,m) \bigr)\bigr|^{q} \,\mathrm{d}t \biggr]^{\frac{1}{q}} \\ &\quad= \frac{\eta^{2}(b,a,m)}{2(\alpha+1)} \biggl(\frac{\alpha}{\alpha+2} \biggr)^{1-\frac{1}{q}} \biggl[ \int_{0}^{1} \bigl(1-t^{\alpha+1}-(1-t)^{\alpha+1} \bigr)\bigl|f'' \bigl(ma+t\eta(b,a,m) \bigr)\bigr|^{q} \,\mathrm{d}t \biggr]^{\frac{1}{q}}. \end{aligned}$$

Using the generalized \((\alpha,m)\)-preinvexity of \(|f''|^{q}\) on A, we have

$$\begin{aligned} & \int_{0}^{1} \bigl(1-t^{\alpha+1}-(1-t)^{\alpha+1} \bigr)\bigl|f'' \bigl(ma+t\eta (b,a,m) \bigr)\bigr|^{q} \,\mathrm{d}t \\ &\quad\leq \int_{0}^{1} \bigl(1-t^{\alpha+1}-(1-t)^{\alpha+1} \bigr) \bigl(m\bigl(1-t^{\alpha}\bigr)\bigl|f''(a)\bigr|^{q} + t^{\alpha}\bigl|f''(b)\bigr|^{q} \bigr)\, \mathrm{d}t \\ &\quad= m \biggl(\frac{2\alpha^{2}+\alpha-2}{(\alpha+2)(2\alpha+2)} +\beta (\alpha+1,\alpha+2) \biggr)\bigl|f''(a)\bigr|^{q} \\ &\qquad{}+ \biggl(\frac{1}{2\alpha+2}-\beta(\alpha+1,\alpha+2) \biggr)\bigl|f''(b)\bigr|^{q}. \end{aligned}$$

Thus, we get the desired inequality (3.11). □

Simple calculation yields the following results.

Corollary 3.6

With the same assumptions given in Theorem 3.4, if \(\eta (b,a,m)=b-ma\), we obtain

$$\begin{aligned} & \biggl|\frac{f(ma)+f(b)}{2} -\frac{\Gamma(\alpha+1)}{2(b-ma)^{\alpha}} \bigl[J^{\alpha}_{ma^{+}}f(b) +J^{\alpha}_{b^{-}}f(ma) \bigr] \biggr| \\ &\quad\leq \frac{(b-ma)^{2}}{2(\alpha+1)} \biggl(\frac{\alpha}{\alpha+2} \biggr)^{1-\frac{1}{q}} \biggl[m \biggl(\frac{2{\alpha}^{2}+\alpha-2}{(\alpha+2)(2\alpha+2)}+\beta (\alpha+1,\alpha+2) \biggr)\bigl|f''(a)\bigr|^{q} \\ &\qquad{} + \biggl(\frac{1}{2\alpha+2}-\beta(\alpha+1,\alpha+2) \biggr)\bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}, \end{aligned}$$

specially for \(\alpha=m=1\) and \(|f''| \leq K\) on \([a,b]\), we get

$$ \biggl|\frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int^{b}_{a}f(x)\,\mathrm{d}x \biggr| \leq \frac{(b-a)^{2}}{12} K. $$
(3.12)

It is worthwhile to note that the inequality in (3.12) is better than the inequality in (3.7).

Corollary 3.7

In Theorem 3.4, if the mapping \(\eta(b,a,m)\) with \(m=1\) degenerates into \(\eta(b,a)\) and we choose \(\alpha=1\), then (3.11) becomes

$$\begin{aligned} & \biggl|\frac{f(a)+f (a+\eta(b,a) )}{2}-\frac{1}{\eta(b,a)} \int ^{a+\eta(b,a)}_{a}f(x)\,\mathrm{d}x \biggr| \\ &\quad\leq \frac{\eta^{2}(b,a)}{12} \biggl(\frac{1}{2} \biggr)^{\frac{1}{q}} \bigl[\bigl|f''(a)\bigr|^{q} + \bigl|f''(b)\bigr|^{q} \bigr] ^{\frac{1}{q}}, \end{aligned}$$

which is the inequality established by Barani et al. in [37], Theorem 4.3.

Finally we shall prove the following result.

Theorem 3.5

Suppose that all the assumptions of Theorem 3.3 are satisfied. Then the following inequality for the Riemann-Louville fractional integral with \(0<\alpha\leq1\) holds:

$$\begin{aligned} & \bigl|R_{f}(\alpha;\eta,m,a,b) \bigr| \\ &\quad\leq\frac{\eta^{2}(b,a,m)}{2(\alpha+1)} \biggl[\frac{(q-p)\alpha -p+1}{(q-p)\alpha+2q-p-1} \biggr]^{\frac{q-1}{q}} \\ &\qquad{}\times \biggl\{ m \biggl[\frac{\alpha p+\alpha+1}{(\alpha+1)(p+1)}-\frac {2}{p(\alpha+1)+1}+\beta \bigl(\alpha+1,p(\alpha+1)+1 \bigr) \biggr]\bigl|f''(a)\bigr|^{q} \\ &\qquad{}+ \biggl[\frac{p}{(\alpha+1)(p+1)}-\beta \bigl(\alpha+1,p(\alpha+1)+1 \bigr) \biggr]\bigl|f''(b)\bigr|^{q} \biggr\} ^{\frac{1}{q}}. \end{aligned}$$
(3.13)

Proof

Since \(ma+t\eta(b,a,m)\in A\) for every \(t\in[0,1]\), by using the properties of modulus on Lemma 3.1 and Hölder’s integral inequality for \(q>1\), we can obtain

$$\begin{aligned} & \bigl|R_{f}(\alpha;\eta,m,a,b) \bigr| \\ &\quad\leq\frac{\eta^{2}(b,a,m)}{2} \int_{0}^{1} \biggl|\frac{1-t^{\alpha +1}-(1-t)^{\alpha+1}}{\alpha+1} \biggr|\bigl|f'' \bigl(ma+t\eta(b,a,m) \bigr)\bigr|\,\mathrm{d}t \\ &\quad\leq\frac{\eta^{2}(b,a,m)}{2(\alpha+1)} \biggl[ \int_{0}^{1} \bigl(1-t^{{\frac{q-p}{q-1}} (\alpha+1)}-(1-t)^{{\frac{q-p}{q-1}}(\alpha+1)} \bigr)\,\mathrm{d}t \biggr]^{\frac{q-1}{q}} \\ &\qquad{}\times \biggl[ \int_{0}^{1} \bigl(1-t^{\alpha+1}-(1-t)^{\alpha+1} \bigr)^{p}\bigl|f'' \bigl(ma+t\eta(b,a,m) \bigr)\bigr|^{q}\,\mathrm{d}t \biggr]^{\frac{1}{q}} \\ &\quad= \frac{\eta^{2}(b,a,m)}{2(\alpha+1)} \biggl[\frac{(q-p)\alpha -p+1}{(q-p)\alpha+2q-p-1} \biggr]^{\frac{q-1}{q}} \\ &\qquad{}\times \biggl[ \int_{0}^{1} \bigl(1-t^{\alpha+1}-(1-t)^{\alpha+1} \bigr)^{p}\bigl|f'' \bigl(ma+t\eta(b,a,m) \bigr)\bigr|^{q}\,\mathrm{d}t \biggr]^{\frac{1}{q}}, \end{aligned}$$
(3.14)

where we used the inequality (3.6) and the fact that

$$ \int_{0}^{1} \bigl(1-t^{{\frac{q-p}{q-1}}(\alpha+1)}-(1-t)^{{\frac {q-p}{q-1}}(\alpha+1)} \bigr)\,\mathrm{d}t = \frac{(q-p)\alpha-p+1}{(q-p)\alpha+2q-p-1}. $$

Utilizing the inequality (3.6) again and the generalized \((\alpha,m)\)-preinvexity of \(|f''|^{q}\) on A, we have

$$\begin{aligned} & \int_{0}^{1} \bigl(1-t^{\alpha+1}-(1-t)^{\alpha+1} \bigr)^{p}\bigl|f'' \bigl(ma+t\eta (b,a,m) \bigr)\bigr|^{q}\,\mathrm{d}t \\ &\quad\leq \int_{0}^{1} \bigl(1-t^{(\alpha+1)p}-(1-t)^{(\alpha+1)p} \bigr) \bigl(m\bigl(1-t^{\alpha}\bigr)\bigl|f''(a)\bigr|^{q} + t^{\alpha}\bigl|f''(b)\bigr|^{q} \bigr)\, \mathrm{d}t \\ &\quad=m \biggl[\frac{\alpha p+\alpha+1}{(\alpha+1)(p+1)}-\frac{2}{p(\alpha +1)+1}+\beta \bigl(\alpha+1,p( \alpha+1)+1 \bigr) \biggr]\bigl|f''(a)\bigr|^{q} \\ &\qquad{}+ \biggl[\frac{p}{(\alpha+1)(p+1)}-\beta \bigl(\alpha+1,p(\alpha+1)+1 \bigr) \biggr]\bigl|f''(b)\bigr|^{q}. \end{aligned}$$
(3.15)

Using (3.15) in (3.14), we get the desired inequality (3.5). □

Corollary 3.8

With the same assumptions given in Theorem 3.5, if \(\eta (b,a,m)=b-ma\), we obtain

$$\begin{aligned} & \biggl|\frac{f(ma)+f(b)}{2}-\frac{\Gamma(\alpha+1)}{2(b-ma)^{\alpha}} \bigl[J^{\alpha}_{ma^{+}}f(b) +J^{\alpha}_{b^{-}}f(ma) \bigr] \biggr| \\ &\quad\leq \frac{(b-ma)^{2}}{2(\alpha+1)} \biggl[\frac{(q-p)\alpha -p+1}{(q-p)\alpha+2q-p-1} \biggr]^{\frac{q-1}{q}} \\ &\qquad{} \times \biggl\{ m \biggl[\frac{\alpha p+\alpha+1}{(\alpha+1)(p+1)}-\frac {2}{p(\alpha+1)+1}+\beta \bigl(\alpha+1,p(\alpha+1)+1 \bigr) \biggr]\bigl|f''(a)\bigr|^{q} \\ &\qquad{}+ \biggl[\frac{p}{(\alpha+1)(p+1)}-\beta \bigl(\alpha+1,p(\alpha+1)+1 \bigr) \biggr]\bigl|f''(b)\bigr|^{q} \biggr\} ^{\frac{1}{q}}, \end{aligned}$$

specially for \(\alpha=m=1\) and \(|f''| \leq K\) on \([a,b]\), we get

$$\begin{aligned} & \biggl|\frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int^{b}_{a}f(x)\,\mathrm{d}x \biggr| \\ &\quad\leq \frac{(b-a)^{2}}{4} \biggl(\frac{q-2p+1}{3q-2p-1} \biggr)^{\frac {q-1}{q}} \biggl(\frac{2p-1}{2p+1} \biggr)^{\frac{1}{q}} K, \end{aligned}$$

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