Integral inequalities of Hermite-Hadamard type for functions whose derivatives are α-preinvex

Abstract

In the article, the authors introduce a new notion, ‘α-preinvex function’, establish an integral identity for the newly introduced function, and find some Hermite-Hadamard type integral inequalities for a function of which the power of the absolute value of the first derivative is α-preinvex.

MSC: 26D15, 26A51, 26B12, 41A55.

1 Introduction

Let us recall some definitions of various convex functions.

Definition 1 A function $f:I\subseteq \mathbb{R}=(-\mathrm{\infty },\mathrm{\infty })\to \mathbb{R}$ is said to be convex if

$$f(\lambda x+(1-\lambda )y)\le \lambda f(x)+(1-\lambda )f(y)$$

(1)

holds for all $x,y\in I$ and $\lambda \in [0,1]$.

Definition 2 ([1])

For $f:[0,b]\to \mathbb{R}$ and $m\in (0,1]$, if

$$f(tx+m(1-t)y)\le tf(x)+m(1-t)f(y)$$

(2)

is valid for all $x,y\in [0,b]$ and $t\in [0,1]$, then we say that $f(x)$ is an m-convex function on $[0,b]$.

Definition 3 ([2])

For $f:[0,b]\to \mathbb{R}$ and $(\alpha ,m)\in (0,1]\times (0,1]$, if

$$f(tx+m(1-t)y)\le t^{\alpha }f(x)+m(1-t^{\alpha })f(y)$$

(3)

is valid for all $x,y\in [0,b]$ and $t\in [0,1]$, then we say that $f(x)$ is an $(\alpha ,m)$-convex function on $[0,b]$.

Definition 4 ([3–5])

A set $S\subseteq {\mathbb{R}}^n$ is said to be invex with respect to the map $\eta :S\times S\to {\mathbb{R}}^n$ if for every $x,y\in S$ and $t\in [0,1]$

$$y+t\eta (x,y)\in S.$$

(4)

Definition 5 ([6])

Let $S\subseteq {\mathbb{R}}^n$ be an invex set with respect to $\eta :S\times S\to {\mathbb{R}}^n$. For every $x,y\in S$, the η-path $P_{xv}$ joining the points x and $v=x+\eta (y,x)$ is defined by

$$P_{xv}=\{z\mid z=x+t\eta (y,x),t\in [0,1]\}.$$

(5)

Definition 6 ([4])

Let $S\subseteq {\mathbb{R}}^n$ be an invex set with respect to $\eta :S\times S\to {\mathbb{R}}^n$. A function $f:S\to \mathbb{R}$ is said to be preinvex with respect to η, if for every $x,y\in S$ and $t\in [0,1]$,

$$f(y+t\eta (x,y))\le tf(x)+(1-t)f(y).$$

(6)

Let us reformulate some inequalities of Hermite-Hadamard type for the above mentioned convex functions.

Theorem 1 ([[7], Theorem 2.2])

Let $f:I^{\circ }\subseteq \mathbb{R}\to \mathbb{R}$ be a differentiable mapping and $a,b\in I^{\circ }$ with $a<b$. If $|f^{\prime }|$ is convex on $[a,b]$, then

$$|\frac{f(a)+f(b)}{2}-\frac{1}{b-a}{\int }_a^{b}f(x)\mathrm{d}x|\le \frac{b-a}{8}\left(\right|f^{\prime }(a)|+|f^{\prime }(b)\left|\right).$$

(7)

Theorem 2 ([8, 9])

Let $f:{\mathbb{R}}_0=[0,\mathrm{\infty })\to \mathbb{R}$ be m-convex and $m\in (0,1]$. If $f\in L([a,b])$ for $0\le a<b<\mathrm{\infty }$, then

$$\frac{1}{b-a}{\int }_a^{b}f(x)\mathrm{d}x\le min\{\frac{f(a)+mf(b/m)}{2},\frac{mf(a/m)+f(b)}{2}\}.$$

(8)

Theorem 3 ([[10], Theorem 3.1])

Let $I\supset {\mathbb{R}}_0$ be an open real interval and let $f:I\to \mathbb{R}$ be a differentiable function on I such that $f^{\prime }\in L([a,b])$ for $0\le a<b<\mathrm{\infty }$. If ${|f^{\prime }|}^q$ is $(\alpha ,m)$-convex on $[a,b]$ for some given numbers $m,\alpha \in (0,1]$ and $q\ge 1$, then

$$\begin{array}{c}|\frac{f(a)+f(b)}{2}-\frac{1}{b-a}{\int }_a^{b}f(x)\mathrm{d}x|\hfill \\ \le \frac{b-a}{2}{\left(\frac{1}{2}\right)}^{1-1/q}min\{{\left[v_1\right|f^{\prime }(a){|}^q+v_{2}m{\left|f^{\prime }\left(\frac{b}{m}\right)\right|}^q]}^{1/q},\hfill \\ {[v_{2}m{\left|f^{\prime }\left(\frac{a}{m}\right)\right|}^q+v_1|f^{\prime }(b){|}^q]}^{1/q}\},\hfill \end{array}$$

where

$$v_1=\frac{1}{(\alpha +1)(\alpha +2)}(\alpha +\frac{1}{2^{\alpha }})\mathit{\text{and}}{v}_2=\frac{1}{(\alpha +1)(\alpha +2)}(\frac{{\alpha }^2+\alpha +2}{2}-\frac{1}{2^{\alpha }}).$$

Theorem 4 ([[4], Theorem 2.1])

Let $A\subseteq \mathbb{R}$ be an open invex subset with respect to $\theta :A\times A\to \mathbb{R}$ and let $f:A\to \mathbb{R}$ be a differentiable function. If $|f^{\prime }|$ is preinvex on A, then for every $a,b\in A$ with $\theta (a,b)\ne 0$ we have

$$|\frac{f(b)+f(b+\theta (a,b))}{2}-\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x|\le \frac{|\theta (a,b)|}{8}\left[\right|f^{\prime }(a)|+|f^{\prime }(b)\left|\right].$$

(9)

For more information on Hermite-Hadamard type inequalities for various convex functions, please refer to recently published articles [11–21] and closely related references therein.

In this article, we will introduce a new notion ‘α-preinvex function’, establish an integral identity for such a kind of functions, and find some Hermite-Hadamard type integral inequalities for a function that the power of the absolute value of its first derivative is α-preinvex.

2 A new definition and a lemma

The so-called ‘α-preinvex function’ may be introduced as follows.

Definition 7 Let $S\subseteq {\mathbb{R}}^n$ be an invex set with respect to $\eta :S\times S\to {\mathbb{R}}^n$. A function $f:S\to \mathbb{R}$ is said to be α-preinvex with respect to η for $\alpha \in (0,1]$, if for every $x,y\in S$ and $t\in [0,1]$,

$$f(y+t\eta (x,y))\le t^{\alpha }f(x)+(1-t^{\alpha })f(y).$$

(10)

Remark 1 If $\alpha =1$ and $f(x)$ is an α-preinvex function, then $f(x)$ is a preinvex function.

For establishing our new integral inequalities of Hermite-Hadamard type for α-preinvex functions, we need the following integral identity.

Lemma 1 Let $A\subseteq \mathbb{R}$ be an open invex subset with respect to $\theta :A\times A\to \mathbb{R}$ and let $a,b\in A$ with $\theta (a,b)\ne 0$. If $f:A\to \mathbb{R}$ is a differentiable function and $f^{\prime }$ is integrable on the θ-path $P_{bc}$: $c=b+\theta (a,b)$, then

$$\begin{array}{c}\frac{1}{2}[\frac{f(b)+f(b+\theta (a,b))}{2}+f\left(\frac{2b+\theta (a,b)}{2}\right)]-\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x\hfill \\ =\frac{\theta (a,b)}{4}{\int }_0^1(\frac{1}{2}-t)[f^{\prime }(b+\frac{1-t}{2}\theta (a,b))+f^{\prime }(b+\frac{2-t}{2}\theta (a,b))]\mathrm{d}t.\hfill \end{array}$$

Proof Since $a,b\in A$ and A is an invex set with respect to θ, for every $t\in [0,1]$, we have $b+t\theta (a,b)\in A$. Integrating by parts gives

$$\begin{array}{c}{\int }_0^1(\frac{1}{2}-t)[f^{\prime }(b+\frac{1-t}{2}\theta (a,b))+f^{\prime }(b+\frac{2-t}{2}\theta (a,b))]\mathrm{d}t\hfill \\ =-\frac{2}{\theta (a,b)}[(\frac{1}{2}-t)f(b+\frac{1-t}{2}\theta (a,b)){|}_0^1+{\int }_0^{1}f(b+\frac{1-t}{2}\theta (a,b))\mathrm{d}t\hfill \\ +(\frac{1}{2}-t)f(b+\frac{2-t}{2}\theta (a,b)){|}_0^1+{\int }_0^{1}f(b+\frac{2-t}{2}\theta (a,b))\mathrm{d}t]\hfill \\ =\frac{2}{\theta (a,b)}[\frac{1}{2}f(b)+\frac{1}{2}f\left(\frac{2b+\theta (a,b)}{2}\right)-{\int }_0^{1}f(b+\frac{1-t}{2}\theta (a,b))\mathrm{d}t]\hfill \\ +\frac{2}{\theta (a,b)}[\frac{1}{2}f\left(\frac{2b+\theta (a,b)}{2}\right)+\frac{1}{2}f(b+\theta (a,b))-{\int }_0^{1}f(b+\frac{2-t}{2}\theta (a,b))\mathrm{d}t]\hfill \\ =\frac{2}{\theta (a,b)}[\frac{f(b)+f(b+\theta (a,b))}{2}+f\left(\frac{2b+\theta (a,b)}{2}\right)]-\frac{4}{{\theta }^2(a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x.\hfill \end{array}$$

The proof of Lemma 1 is completed. □

3 Some new integral inequalities of Hermite-Hadamard type

We are now in a position to establish some Hermite-Hadamard type integral inequalities for a function that the power of the absolute value of its first derivative is α-preinvex.

Theorem 5 Let $A\subseteq \mathbb{R}$ be an invex subset with respect to $\theta :A\times A\to \mathbb{R}$ and $a,b\in A$ with $\theta (a,b)\ne 0$. Suppose that $f:A\to \mathbb{R}$ is a differentiable function, $f^{\prime }$ is integrable on the θ-path $P_{bc}$: $c=b+\theta (a,b)$, and $\alpha \in (0,1]$. If ${|f^{\prime }|}^q$ is α-preinvex on A for $q\ge 1$, then

$$\begin{array}{c}|\frac{1}{2}[\frac{f(b)+f(b+\theta (a,b))}{2}+f\left(\frac{2b+\theta (a,b)}{2}\right)]-\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x|\hfill \\ \le \frac{|\theta (a,b)|}{16}{\left[\frac{1}{(\alpha +1)(\alpha +2)2^{2\alpha }}\right]}^{1/q}\left\{\right[2(1+\alpha 2^{\alpha })|f^{\prime }(a){|}^q\hfill \\ {+(2^{2\alpha +1}-2+\alpha (3\times 2^{\alpha }-2)2^{\alpha }+{\alpha }^2{2}^{2\alpha })|f^{\prime }(b){|}^q]}^{1/q}\hfill \\ +\left[(\alpha (2^{\alpha +1}-1)2^{\alpha +1}+2\times 3^{\alpha +2}-(2^{\alpha }+1)2^{\alpha +3})\right|f^{\prime }(a){|}^q\hfill \\ {+({\alpha }^2{2}^{2\alpha }+(\alpha (1-2^{\alpha -1})+4+5\times 2^{\alpha })2^{\alpha +1}-2\times 3^{\alpha +2})|f^{\prime }(b){|}^q]}^{1/q}\}.\hfill \end{array}$$

Proof Since $b+t\theta (a,b)\in A$ for every $t\in [0,1]$, by Lemma 1 and Hölder’s inequality, we have

$$\begin{array}{c}|\frac{1}{2}[\frac{f(b)+f(b+\theta (a,b))}{2}+f\left(\frac{2b+\theta (a,b)}{2}\right)]-\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x|\hfill \\ \le \frac{|\theta (a,b)|}{4}{\int }_0^1|\frac{1}{2}-t|[\left|f^{\prime }(b+\frac{1-t}{2}\theta (a,b))\right|+\left|f^{\prime }(b+\frac{2-t}{2}\theta (a,b))\right|]\mathrm{d}t\hfill \\ \le \frac{|\theta (a,b)|}{4}{\left({\int }_0^1|\frac{1}{2}-t|\mathrm{d}t\right)}^{1-1/q}\{{\left[{\int }_0^1|\frac{1}{2}-t|{\left|f^{\prime }(b+\frac{1-t}{2}\theta (a,b))\right|}^q\mathrm{d}t\right]}^{1/q}\hfill \\ +{\left[{\int }_0^1|\frac{1}{2}-t|{\left|f^{\prime }(b+\frac{2-t}{2}\theta (a,b))\right|}^q\mathrm{d}t\right]}^{1/q}\}.\hfill \end{array}$$

(11)

Using the α-preinvexity of ${|f^{\prime }|}^q$, we have

$$\begin{array}{c}{\int }_0^1|\frac{1}{2}-t|{\left|f^{\prime }(b+\frac{1-t}{2}\theta (a,b))\right|}^q\mathrm{d}t\hfill \\ \le {\int }_0^1|\frac{1}{2}-t|\left[{\left(\frac{1-t}{2}\right)}^{\alpha }\right|f^{\prime }(a){|}^q+(1-{\left(\frac{1-t}{2}\right)}^{\alpha })|f^{\prime }(b){|}^q]\mathrm{d}t\hfill \\ =\frac{1}{(\alpha +1)(\alpha +2)2^{2(\alpha +1)}}\left[2(1+\alpha 2^{\alpha })\right|f^{\prime }(a){|}^q\hfill \\ +(2^{2\alpha +1}-2+\alpha (3\times 2^{\alpha }-2)2^{\alpha }+{\alpha }^2{2}^{2\alpha })|f^{\prime }(b){|}^q]\hfill \end{array}$$

and

$$\begin{array}{c}{\int }_0^1|\frac{1}{2}-t|{\left|f^{\prime }(b+\frac{2-t}{2}\theta (a,b))\right|}^q\mathrm{d}t\hfill \\ \le {\int }_0^1|\frac{1}{2}-t|\left[{\left(\frac{2-t}{2}\right)}^{\alpha }\right|f^{\prime }(a){|}^q+(1-{\left(\frac{2-t}{2}\right)}^{\alpha })|f^{\prime }(b){|}^q]\mathrm{d}t\hfill \\ =\frac{1}{(\alpha +1)(\alpha +2)2^{2(\alpha +1)}}[(\alpha (2^{\alpha +1}-1)2^{\alpha +1}+2\times 3^{\alpha +2}-(2^{\alpha }+1)2^{\alpha +3})\hfill \\ \times |f^{\prime }(a){|}^q+({\alpha }^2{2}^{2\alpha }+(\alpha (1-2^{\alpha -1})+4+5\times 2^{\alpha })2^{\alpha +1}-2\times 3^{\alpha +2})|f^{\prime }(b){|}^q].\hfill \end{array}$$

Substituting the above two inequalities into (11) yields

$$\begin{array}{c}|\frac{1}{2}[\frac{f(b)+f(b+\theta (a,b))}{2}+f\left(\frac{2b+\theta (a,b)}{2}\right)]-\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x|\hfill \\ \le \frac{|\theta (a,b)|}{4}{\left({\int }_0^1|\frac{1}{2}-t|\mathrm{d}t\right)}^{1-1/q}\left\{\right[{\int }_0^1|\frac{1}{2}-t|\left({\left(\frac{1-t}{2}\right)}^{\alpha }\right|f^{\prime }(a){|}^q\hfill \\ {+(1-{\left(\frac{1-t}{2}\right)}^{\alpha })|f^{\prime }(b){|}^q)\mathrm{d}t]}^{1/q}\hfill \\ +{\left[{\int }_0^1|\frac{1}{2}-t|\left({\left(\frac{2-t}{2}\right)}^{\alpha }\right|f^{\prime }(a){|}^q+(1-{\left(\frac{2-t}{2}\right)}^{\alpha })|f^{\prime }(b){|}^q)\mathrm{d}t\right]}^{1/q}\}\hfill \\ =\frac{|\theta (a,b)|}{4}{\left(\frac{1}{4}\right)}^{1-1/q}{\left[\frac{1}{(\alpha +1)(\alpha +2)2^{2(\alpha +1)}}\right]}^{1/q}\left\{\right[2(1+\alpha 2^{\alpha })|f^{\prime }(a){|}^q\hfill \\ {+(2^{2\alpha +1}-2+\alpha (3\times 2^{\alpha }-2)2^{\alpha }+{\alpha }^2{2}^{2\alpha })|f^{\prime }(b){|}^q]}^{1/q}\hfill \\ +\left[(\alpha (2^{\alpha +1}-1)2^{\alpha +1}+2\times 3^{\alpha +2}-(2^{\alpha }+1)2^{\alpha +3})\right|f^{\prime }(a){|}^q\hfill \\ {+({\alpha }^2{2}^{2\alpha }+(\alpha (1-2^{\alpha -1})+4+5\times 2^{\alpha })2^{\alpha +1}-2\times 3^{\alpha +2})|f^{\prime }(b){|}^q]}^{1/q}\}.\hfill \end{array}$$

The proof of Theorem 5 is completed. □

Corollary 1 Under the conditions of Theorem  5, if $\alpha =1$, we have

$$\begin{array}{c}|\frac{1}{2}[\frac{f(b)+f(b+\theta (a,b))}{2}+f\left(\frac{2b+\theta (a,b)}{2}\right)]-\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x|\hfill \\ \le \frac{|\theta (a,b)|}{16}\{{\left[\frac{|f^{\prime }(a){|}^q+3|f^{\prime }(b){|}^q}{4}\right]}^{1/q}+{\left[\frac{3|f^{\prime }(a){|}^q+|f^{\prime }(b){|}^q}{4}\right]}^{1/q}\}.\hfill \end{array}$$

Theorem 6 Let $A\subseteq \mathbb{R}$ be an invex subset with respect to $\theta :A\times A\to \mathbb{R}$ and $a,b\in A$ with $\theta (a,b)\ne 0$. Suppose that $f:A\to \mathbb{R}$ is a differentiable function, $f^{\prime }$ is integrable on the θ-path $P_{bc}$: $c=b+\theta (a,b)$, and $\alpha \in (0,1]$. If ${|f^{\prime }|}^q$ is α-preinvex on A for $q>1$, then

$$\begin{array}{c}|\frac{1}{2}[\frac{f(b)+f(b+\theta (a,b))}{2}+f\left(\frac{2b+\theta (a,b)}{2}\right)]-\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x|\hfill \\ \le \frac{|\theta (a,b)|}{8}{\left(\frac{q-1}{2q-1}\right)}^{1-1/q}{\left[\frac{1}{(\alpha +1)2^{\alpha }}\right]}^{1/q}\{{\left[\right|f^{\prime }(a){|}^q+((\alpha +1)2^{\alpha }-1)|f^{\prime }(b){|}^q]}^{1/q}\hfill \\ +{\left[(2^{\alpha +1}-1)\right|f^{\prime }(a){|}^q+(1-(1-\alpha )2^{\alpha })|f^{\prime }(b){|}^q]}^{1/q}\}.\hfill \end{array}$$

Proof Since $b+t\theta (a,b)\in A$ for every $t\in [0,1]$, by Lemma 1, Hölder’s inequality, and the α-preinvexity of ${|f^{\prime }|}^q$, we have

$$\begin{array}{c}|\frac{1}{2}[\frac{f(b)+f(b+\theta (a,b))}{2}+f\left(\frac{2b+\theta (a,b)}{2}\right)]-\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x|\hfill \\ \le \frac{|\theta (a,b)|}{4}{\int }_0^1|\frac{1}{2}-t|[\left|f^{\prime }(b+\frac{1-t}{2}\theta (a,b))\right|+\left|f^{\prime }(b+\frac{2-t}{2}\theta (a,b))\right|]\mathrm{d}t\hfill \\ \le \frac{|\theta (a,b)|}{4}{\left({\int }_0^1{|\frac{1}{2}-t|}^{q/(q-1)}\mathrm{d}t\right)}^{1-1/q}\{{\left[{\int }_0^1{\left|f^{\prime }(b+\frac{1-t}{2}\theta (a,b))\right|}^q\mathrm{d}t\right]}^{1/q}\hfill \\ +{\left[{\int }_0^1{\left|f^{\prime }(b+\frac{2-t}{2}\theta (a,b))\right|}^q\mathrm{d}t\right]}^{1/q}\}\hfill \\ \le \frac{|\theta (a,b)|}{4}{\left({\int }_0^1{|\frac{1}{2}-t|}^{q/(q-1)}\mathrm{d}t\right)}^{1-1/q}\hfill \\ \times \{{\left[{\int }_0^1\left({\left(\frac{1-t}{2}\right)}^{\alpha }\right|f^{\prime }(a){|}^q+(1-{\left(\frac{1-t}{2}\right)}^{\alpha })|f^{\prime }(b){|}^q)\mathrm{d}t\right]}^{1/q}\hfill \\ +{\left[{\int }_0^1\left({\left(\frac{2-t}{2}\right)}^{\alpha }\right|f^{\prime }(a){|}^q+(1-{\left(\frac{2-t}{2}\right)}^{\alpha })|f^{\prime }(b){|}^q)\mathrm{d}t\right]}^{1/q}\}\hfill \\ =\frac{|\theta (a,b)|}{8}{\left(\frac{q-1}{2q-1}\right)}^{1-1/q}{\left[\frac{1}{(\alpha +1)2^{\alpha }}\right]}^{1/q}\{{\left[\right|f^{\prime }(a){|}^q+((\alpha +1)2^{\alpha }-1)|f^{\prime }(b){|}^q]}^{1/q}\hfill \\ +{\left[(2^{\alpha +1}-1)\right|f^{\prime }(a){|}^q+(1-(1-\alpha )2^{\alpha })|f^{\prime }(b){|}^q]}^{1/q}\}.\hfill \end{array}$$

The proof of Theorem 6 is complete. □

Corollary 2 Under the conditions of Theorem  6, if $\alpha =1$, we have

$$\begin{array}{c}|\frac{1}{2}[\frac{f(b)+f(b+\theta (a,b))}{2}+f\left(\frac{2b+\theta (a,b)}{2}\right)]-\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x|\hfill \\ \le \frac{|\theta (a,b)|}{8}{\left(\frac{q-1}{2q-1}\right)}^{1-1/q}\{{\left[\frac{|f^{\prime }(a){|}^q+3|f^{\prime }(b){|}^q}{4}\right]}^{1/q}+{\left[\frac{3|f^{\prime }(a){|}^q+|f^{\prime }(b){|}^q}{4}\right]}^{1/q}\}.\hfill \end{array}$$

Theorem 7 Let $A\subseteq \mathbb{R}$ be an invex subset with respect to $\theta :A\times A\to \mathbb{R}$ and $a,b\in A$ with $\theta (a,b)\ne 0$. Suppose that $f:A\to \mathbb{R}$ is a differentiable function, $f^{\prime }$ is integrable on the θ-path $P_{bc}$: $c=b+\theta (a,b)$, and $\alpha \in (0,1]$. If ${|f^{\prime }|}^q$ is α-preinvex on A for $q>1$ and $q\ge r>0$, then

$$\begin{array}{c}|\frac{1}{2}[\frac{f(b)+f(b+\theta (a,b))}{2}+f\left(\frac{2b+\theta (a,b)}{2}\right)]-\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x|\hfill \\ \le \frac{|\theta (a,b)|}{4}{\left(\frac{q-1}{2q-r-1}\right)}^{1-1/q}{\left(\frac{1}{2}\right)}^{(q-r)/q}\hfill \\ \times \{{\left[(\frac{1}{(2r+1)2^{2r+1}}+\frac{1}{(2\alpha +1)2^{2\alpha +1}})\right|f^{\prime }(a){|}^q+\frac{3}{(r+1)2^{r+2}}|f^{\prime }(b){|}^q]}^{1/q}\hfill \\ +{\left[(\frac{1}{(2r+1)2^{2r+1}}+\frac{2^{2\alpha +1}-1}{(2\alpha +1)2^{2\alpha +1}})\right|f^{\prime }(a){|}^q+\frac{1}{(r+1)2^{r+2}}|f^{\prime }(b){|}^q]}^{1/q}\}.\hfill \end{array}$$

(12)

Proof Since $b+t\theta (a,b)\in A$ for every $t\in [0,1]$, by Lemma 1, Hölder’s inequality, and the α-preinvexity of ${|f^{\prime }|}^q$, we have

$$\begin{array}{c}|\frac{1}{2}[\frac{f(b)+f(b+\theta (a,b))}{2}+f\left(\frac{2b+\theta (a,b)}{2}\right)]-\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x|\hfill \\ \le \frac{|\theta (a,b)|}{4}{\int }_0^1|\frac{1}{2}-t|[\left|f^{\prime }(b+\frac{1-t}{2}\theta (a,b))\right|+\left|f^{\prime }(b+\frac{2-t}{2}\theta (a,b))\right|]\mathrm{d}t\hfill \\ \le \frac{|\theta (a,b)|}{4}{\left({\int }_0^1{|\frac{1}{2}-t|}^{(q-r)/(q-1)}\mathrm{d}t\right)}^{1-1/q}\{{\left[{\int }_0^1{|\frac{1}{2}-t|}^r{\left|f^{\prime }(b+\frac{1-t}{2}\theta (a,b))\right|}^q\mathrm{d}t\right]}^{1/q}\hfill \\ +{\left[{\int }_0^1{|\frac{1}{2}-t|}^r{\left|f^{\prime }(b+\frac{2-t}{2}\theta (a,b))\right|}^q\mathrm{d}t\right]}^{1/q}\}\hfill \\ \le \frac{|\theta (a,b)|}{4}{\left(\frac{q-1}{2q-r-1}\right)}^{1-1/q}{\left(\frac{1}{2}\right)}^{(q-r)/q}\hfill \\ \times \{{\left[{\int }_0^1{|\frac{1}{2}-t|}^r\left({\left(\frac{1-t}{2}\right)}^{\alpha }\right|f^{\prime }(a){|}^q+(1-{\left(\frac{1-t}{2}\right)}^{\alpha })|f^{\prime }(b){|}^q)\mathrm{d}t\right]}^{1/q}\hfill \\ +{\left[{\int }_0^1{|\frac{1}{2}-t|}^r\left({\left(\frac{2-t}{2}\right)}^{\alpha }\right|f^{\prime }(a){|}^q+(1-{\left(\frac{2-t}{2}\right)}^{\alpha })|f^{\prime }(b){|}^q)\mathrm{d}t\right]}^{1/q}\}.\hfill \end{array}$$

(13)

Since $x^r{y}^{\alpha }\le \frac{x^{2r}+y^{2\alpha }}{2}$ and $z\le z^{\alpha }$ for $x,y\ge 0$ and $0\le z\le 1$, we obtain

$$\begin{array}{c}{\int }_0^1{|\frac{1}{2}-t|}^r\left[{\left(\frac{1-t}{2}\right)}^{\alpha }\right|f^{\prime }(a){|}^q+(1-{\left(\frac{1-t}{2}\right)}^{\alpha })|f^{\prime }(b){|}^q]\mathrm{d}t\hfill \\ \le {\int }_0^1\left[\frac{1}{2}({|\frac{1}{2}-t|}^{2r}+{\left(\frac{1-t}{2}\right)}^{2\alpha })\right|f^{\prime }(a){|}^q+\left(\frac{1+t}{2}\right){|\frac{1}{2}-t|}^r|f^{\prime }(b){|}^q]\mathrm{d}t\hfill \\ =[\frac{1}{(2r+1)2^{2r+1}}+\frac{1}{(2\alpha +1)2^{2\alpha +1}}]|f^{\prime }(a){|}^q+\frac{3}{(r+1)2^{r+2}}|f^{\prime }(b){|}^q\hfill \end{array}$$

(14)

and

$$\begin{array}{c}{\int }_0^1{|\frac{1}{2}-t|}^r\left[{\left(\frac{2-t}{2}\right)}^{\alpha }\right|f^{\prime }(a){|}^q+(1-{\left(\frac{2-t}{2}\right)}^{\alpha })|f^{\prime }(b){|}^q]\mathrm{d}t\hfill \\ \le {\int }_0^1\left[\frac{1}{2}({|\frac{1}{2}-t|}^{2r}+{\left(\frac{2-t}{2}\right)}^{2\alpha })\right|f^{\prime }(a){|}^q+\frac{t}{2}{|\frac{1}{2}-t|}^r|f^{\prime }(b){|}^q]\mathrm{d}t\hfill \\ =[\frac{1}{(2r+1)2^{2r+1}}+\frac{2^{2\alpha +1}-1}{(2\alpha +1)2^{2\alpha +1}}]|f^{\prime }(a){|}^q+\frac{1}{(r+1)2^{r+2}}|f^{\prime }(b){|}^q.\hfill \end{array}$$

(15)

Substituting (14) and (15) into (13) results in (12). The proof of Theorem 7 is complete. □

Corollary 3 Under the conditions of Theorem  7, if $\alpha =1$, we have

$$\begin{array}{c}|\frac{1}{2}[\frac{f(b)+f(b+\theta (a,b))}{2}+f\left(\frac{2b+\theta (a,b)}{2}\right)]-\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x|\hfill \\ \le \frac{|\theta (a,b)|}{4}{\left(\frac{q-1}{2q-r-1}\right)}^{1-1/q}{\left(\frac{1}{2}\right)}^{(q-r)/q}\hfill \\ \times \{{\left[(\frac{1}{(2r+1)2^{2r+1}}+\frac{1}{24})\right|f^{\prime }(a){|}^q+\frac{3}{(r+1)2^{r+2}}|f^{\prime }(b){|}^q]}^{1/q}\hfill \\ +{\left[(\frac{1}{(2r+1)2^{2r+1}}+\frac{7}{24})\right|f^{\prime }(a){|}^q+\frac{|f^{\prime }(b){|}^q}{(r+1)2^{r+2}}]}^{1/q}\}.\hfill \end{array}$$

Theorem 8 Let $A\subseteq \mathbb{R}$ be an invex subset with respect to $\theta :A\times A\to \mathbb{R}$ and $a,b\in A$ with $\theta (a,b)\ne 0$. Suppose that $f:A\to \mathbb{R}$ is a differentiable function and $f^{\prime }$ is integrable on the θ-path $P_{bc}$: $c=b+\theta (a,b)$. If ${|f^{\prime }|}^q$ is preinvex on A for $q>1$ and $q\ge r>0$, then

$$\begin{array}{c}|\frac{1}{2}[\frac{f(b)+f(b+\theta (a,b))}{2}+f\left(\frac{2b+\theta (a,b)}{2}\right)]-\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x|\hfill \\ \le \frac{|\theta (a,b)|}{8}{\left(\frac{q-1}{2q-r-1}\right)}^{1-1/q}{\left(\frac{1}{r+1}\right)}^{1/q}\hfill \\ \times \{{\left[\frac{|f^{\prime }(a){|}^q+3|f^{\prime }(b){|}^q}{4}\right]}^{1/q}+{\left[\frac{3|f^{\prime }(a){|}^q+|f^{\prime }(b){|}^q}{4}\right]}^{1/q}\}.\hfill \end{array}$$

Proof Since $b+t\theta (a,b)\in A$ for every $t\in [0,1]$, by Lemma 1, Hölder’s inequality, and the preinvexity of ${|f^{\prime }|}^q$, we have

$$\begin{array}{c}|\frac{1}{2}[\frac{f(b)+f(b+\theta (a,b))}{2}+f\left(\frac{2b+\theta (a,b)}{2}\right)]-\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x|\hfill \\ \le \frac{|\theta (a,b)|}{4}{\int }_0^1|\frac{1}{2}-t|[\left|f^{\prime }(b+\frac{1-t}{2}\theta (a,b))\right|+\left|f^{\prime }(b+\frac{2-t}{2}\theta (a,b))\right|]\mathrm{d}t\hfill \\ \le \frac{|\theta (a,b)|}{4}{\left({\int }_0^1{|\frac{1}{2}-t|}^{(q-r)/(q-1)}\mathrm{d}t\right)}^{1-1/q}\hfill \\ \times \{{\left[{\int }_0^1{|\frac{1}{2}-t|}^r{\left|f^{\prime }(b+\frac{1-t}{2}\theta (a,b))\right|}^q\mathrm{d}t\right]}^{1/q}\hfill \\ +{\left[{\int }_0^1{|\frac{1}{2}-t|}^r{\left|f^{\prime }(b+\frac{2-t}{2}\theta (a,b))\right|}^q\mathrm{d}t\right]}^{1/q}\}\hfill \\ \le \frac{|\theta (a,b)|}{4}{\left(\frac{q-1}{2q-r-1}\right)}^{1-1/q}{\left(\frac{1}{2}\right)}^{(q-r)/q}\hfill \\ \times \{{\left[{\int }_0^1{|\frac{1}{2}-t|}^r\left(\frac{1-t}{2}\right|f^{\prime }(a){|}^q+\frac{1+t}{2}|f^{\prime }(b){|}^q)\mathrm{d}t\right]}^{1/q}\hfill \\ +{\left[{\int }_0^1{|\frac{1}{2}-t|}^r\left(\frac{2-t}{2}\right|f^{\prime }(a){|}^q+\frac{t}{2}|f^{\prime }(b){|}^q)\mathrm{d}t\right]}^{1/q}\}\hfill \\ =\frac{|\theta (a,b)|}{4}{\left(\frac{q-1}{2q-r-1}\right)}^{1-1/q}{\left(\frac{1}{2}\right)}^{(q-r)/q}\hfill \\ \times \{{\left[\frac{1}{(r+1)2^{r+2}}\right|f^{\prime }(a){|}^q+\frac{3}{(r+1)2^{r+2}}|f^{\prime }(b){|}^q]}^{1/q}\hfill \\ +{\left[\frac{3}{(r+1)2^{r+2}}\right|f^{\prime }(a){|}^q+\frac{1}{(r+1)2^{r+2}}|f^{\prime }(b){|}^q]}^{1/q}\}.\hfill \end{array}$$

The proof of Theorem 8 is complete. □

Corollary 4 Under the conditions of Theorem  8, if $r=q$, we have

$$\begin{array}{c}|\frac{1}{2}[\frac{f(b)+f(b+\theta (a,b))}{2}+f\left(\frac{2b+\theta (a,b)}{2}\right)]-\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x|\hfill \\ \le \frac{|\theta (a,b)|}{8}{\left(\frac{1}{q+1}\right)}^{1/q}\{{\left[\frac{|f^{\prime }(a){|}^q+3|f^{\prime }(b){|}^q}{4}\right]}^{1/q}+{\left[\frac{3|f^{\prime }(a){|}^q+|f^{\prime }(b){|}^q}{4}\right]}^{1/q}\}.\hfill \end{array}$$

Theorem 9 Let $A\subseteq \mathbb{R}$ be an invex subset with respect to $\theta :A\times A\to \mathbb{R}$ and $a,b\in A$ with $\theta (a,b)\ne 0$. Suppose that $f:A\to \mathbb{R}$ is a differentiable function, $f^{\prime }$ is integrable on the θ-path $P_{bc}$: $c=b+\theta (a,b)$, and $\alpha \in (0,1]$. If f is α-preinvex on A, then

$$\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x\le min\{\frac{f(a)+\alpha f(b)}{\alpha +1},\frac{\alpha f(a)+f(b)}{\alpha +1}\}.$$

(16)

Proof Since $b+t\theta (a,b)\in A$ for $0\le t\le 1$, letting $x=(1-t)b+t(b+\theta (a,b))=b+t\theta (a,b)$ for $0\le t\le 1$ and using the α-preinvexity of f, we have

$$\begin{array}{rcl}\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x& =& {\int }_0^{1}f(b+t\theta (a,b))\mathrm{d}t\\ \le & {\int }_0^1[t^{\alpha }f(a)+(1-t^{\alpha })f(b)]\mathrm{d}t=\frac{f(a)+\alpha f(b)}{\alpha +1}.\end{array}$$

The proof of Theorem 9 is complete. □

Corollary 5 Under the conditions of Theorem  9, if $\alpha =1$, we have

$$\frac{1}{\theta (a,b)}{\int }_b^{b+\theta (a,b)}f(x)\mathrm{d}x\le \frac{f(a)+f(b)}{2}.$$