Properties and Riemann-Liouville fractional Hermite-Hadamard inequalities for the generalized $(\alpha,m)$-preinvex functions

Abstract

The authors first introduce the concepts of generalized $(\alpha,m)$-preinvex function, generalized quasi m-preinvex function and explicitly $(\alpha, m)$-preinvex function, and then provide some interesting properties for the newly introduced functions. The more important point is that we give a necessary and sufficient condition respecting the relationship between the generalized $(\alpha, m)$-preinvex function and the generalized quasi m-preinvex function. Second, a new Riemann-Liouville fractional integral identity involving twice differentiable function on m-invex is found. By using this identity, we establish the right-sided new Hermite-Hadamard-type inequalities via Riemann-Liouville fractional integrals for generalized $(\alpha,m)$-preinvex mappings. These inequalities can be viewed as generalization of several previously known results.

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 $\eta: K \times K \times(0,1]\rightarrow\mathbb{R}^{n}$.

• (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.

• (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 $q>1$.