Several New Integral Inequalities of Hermite–Hadamard Type for Extended ϕ_h-s-Convex Functions

1. Introduction

In this paper, let $I\subseteq (-\infty ,\infty )$ stand for an interval.

In the 2016 paper [7], the authors introduced the following concept of ${\varphi }_{h\text{-}s}$-convex functions and studied some properties of ${\varphi }_{h\text{-}s}$-convex functions. See also [6].

Definition 1

(Essentially modified version of [7, Definition 2.1]). Let $h:(0,1)\to {\mathbb{R}}_{+}=(0,\infty )$ and $\varphi :I\subseteq \mathbb{R}\to \mathbb{R}$ such that $\varphi \left(I\right)=\left\{\varphi \right(x),x\in I\}$ is an interval with inner points. For $s\in [0,1]$, a function $f:\varphi \left(I\right)\to \mathbb{R}$ is said to be ${\varphi }_{h\text{-}s}$-convex on $\varphi \left(I\right)$ if the inequality

$$f(t\varphi \left(x\right)+(1-t)\varphi \left(y\right))\le {\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^{s}f\left(\varphi \left(x\right)\right)+{\left[{\displaystyle \frac{1-t}{h(1-t)}}\right]}^{s}f\left(\varphi \left(y\right)\right)$$

(1)

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

Remark 1.

Under the condition of Definition 1,

• if $s\in (0,1]$, $\varphi \left(x\right)=x$ for $x\in I$, and $h\left(t\right)=1$ for $t\in (0,1)$, then the ${\varphi }_{h\text{-}s}$-convex function becomes an s-convex function, see [1,4];
• if $s\in (0,1]$, $\varphi \left(x\right)=x$ for $x\in I$, and $h\left(t\right)={\displaystyle \frac{1}{t^2}}$ for $t\in (0,1)$, then the ${\varphi }_{h\text{-}s}$-convex function is an s-Godunova-Levin convex function, see [2];
• if $s=1$, $\varphi \left(x\right)=x$ for $x\in I$, and $h\left(t\right)=2\sqrt{t(1-t)}$ for $t\in (0,1)$, then the ${\varphi }_{h\text{-}s}$-convex function is an MT-convex function, see [8];
• if $s=1$, $\varphi \left(x\right)=x$ for $x\in I$, and $h\left(t\right)={\displaystyle \frac{t}{h_1\left(t\right)}}$ for $t\in (0,1)$, then the ${\varphi }_{h\text{-}s}$-convex function is an h-convex function defined by [9, p. 304, Definition 4].

In [10], the concept of extended s-convex functions was innovated below.

Definition 2

([10, Definition 2.1]). A function $f:I\subseteq \mathbb{R}\to \mathbb{R}$ is said to be extended s-convex on an interval I if

$$f(tx+(1-t)y)\le t^{s}f\left(x\right)+{(1-t)}^{s}f\left(y\right)$$

(2)

holds for all $x,y\in I$ and $t\in (0,1)$ and for some fixed $s\in [-1,1]$.

In [11], the authors extended $s\in [0,1]$ in Definition 1 to $s\in [-1,1]$ and improved Definition 1 by introducing the following extended ${\varphi }_{h\text{-}s}$-convex functions.

Definition 3

(Slightly modified version of [11, Definition 2.1]). Let $h:(0,1)\to {\mathbb{R}}_{+}$ and $\varphi :I\subseteq \mathbb{R}\to \mathbb{R}$ such that $\varphi \left(I\right)=\left\{\varphi \right(x),x\in I\}$ is an interval with inner points. For some $s\in [-1,1]$, a function $f:\varphi \left(I\right)\to \mathbb{R}$ is said to be extended ${\varphi }_{h\text{-}s}$-convex on $\varphi \left(I\right)$ if the inequality (1) holds for all $x,y\in I$ and $t\in (0,1)$.

Remark 2.

Under the condition of Definition 3,

• if $s\in [0,1]$, then the extended ${\varphi }_{h\text{-}s}$-convex function is a ${\varphi }_{h\text{-}s}$-convex function, see [7];
• if $\varphi \left(x\right)=x$ for $x\in I$ and $h\left(t\right)=1$ for $t\in (0,1)$, then the extended ${\varphi }_{h\text{-}s}$-function is an extended s-convex function, see [10].

Example 1.

For some fixed $s\in [-1,1]\setminus \left\{0\right\}$ and $0<p<1$, define $f\left(x\right)=x^{p/(p+1)}$ for $x\in (0,1]$, $h\left(t\right)=t^{1-p/s(p+1)}$ for $t\in (0,1)$, and

$$\varphi \left(x\right)=\{\begin{array}{cc}{x}^{p+1},\hfill & x\in (0,1)\setminus \left\{{\displaystyle \frac{1}{2}}\right\};\hfill \\ 1,\hfill & x={\displaystyle \frac{1}{2}};\hfill \\ {\left({\displaystyle \frac{1}{2}}\right)}^{p+1},\hfill & x=1.\hfill \end{array}$$

Then $f\left(\varphi \left(x\right)\right)=x^p$ and ${\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s=t^{p/(p+1)}$ for $x\in (0,1)\setminus \left\{{\displaystyle \frac{1}{2}}\right\}$ and $t\in (0,1)$, with $f\left(\varphi \left({\displaystyle \frac{1}{2}}\right)\right)=1$ and $f\left(\varphi \left(1\right)\right)={\left({\displaystyle \frac{1}{2}}\right)}^p$.

Using the inequality ${(u+1)}^r\le u^r+1$ for $u>0$ and $0<r<1$, for all $x,y\in (0,1]$ and $t\in (0,1)$, we obtain

$${\left[t{x}^{p+1}+(1-t)y^{p+1}\right]}^{p/(p+1)}\le t^{p/(p+1)}{x}^p+{(1-t)}^{p/(p+1)}{y}^p.$$

This means that, although ϕ is not continuous, but $\varphi \left(I\right)=(0,1]$ is an interval with inner points, the function $f\left(x\right)=x^{p/(p+1)}$ is still extended ${\varphi }_{h\text{-}s}$-convex on $(0,1]$.

Example 2.

When $\varphi \left(I\right)=\{2,5\}$, which is not an interval with inner points, the function f in (1) is defined on the set $\varphi \left(I\right)=\{2,5\}$ with two points. But the left hand side in (1) requires f to take values in the interval $(2,5)=\{2t+5(1-t),t\in (0,1\left)\right\}=\{5-3t,t\in (0,1\left)\right\}$. This is the reason why we required $\varphi \left(I\right)$ in Definitions 1 and 3 to be an interval with inner points.

Example 3.

For $0<p<1$, let

$${\varphi }_1\left(x\right)=2x^{p+1},x\in I_1=\left(0,2^{-1/(p+1)}\right]$$

and

$${\varphi }_2\left(x\right)=x^{p+1},x\in I_2=(0,1].$$

It is easy to see that ${\varphi }_1\left(I_1\right)={\varphi }_2\left(I_2\right)=(0,1]$. If a function f were extended ${{\varphi }_1}_{h\text{-}s}$-convex on ${\varphi }_1\left(I_1\right)=(0,1]$ and extended ${{\varphi }_2}_{h\text{-}s}$-convex on ${\varphi }_2\left(I_2\right)=(0,1]$, we can still differentiate them naturally.

In [11], some integral inequalities for extended ${\varphi }_{h\text{-}s}$-convex functions were established. We now modify and restate Theorems 3.1 to 3.4 without any change of their proofs in [11] as follows.

Theorem 1

([11, Theorem 3.1]). Let $h:(0,1)\to {\mathbb{R}}_{+}$, let $\varphi :I\subseteq \mathbb{R}\to \mathbb{R}$ and $f:\varphi \left(I\right)\to \mathbb{R}$ be differentiable, and let $|{\varphi }^{\prime }|$ be convex. For some $s\in [-1,1]$ and $q\ge 1$, if $|f^{\prime }{|}^q$ is an increasing extended ${\varphi }_{h\text{-}s}$-convex function on $\varphi \left(I\right)$, $f^{\prime }\left(\varphi \right){\varphi }^{\prime }\in L_1\left(I\right)$, and $(1-2x){\left[{\displaystyle \frac{x}{h\left(x\right)}}\right]}^s\in L_1\left([0,1]\right)$, then for $a,b\in I^{\circ }$ with $a<b$, we have

$$\begin{array}{cc}\left|{\displaystyle \frac{f\left(\varphi \right(a\left)\right)+f\left(\varphi \right(b\left)\right)}{2}}-{\displaystyle \frac{1}{b-a}}{\int }_a^{b}f\left(\varphi \left(x\right)\right)\mathrm{d}x\right|\hfill & \\ & \hfill \le {\displaystyle \frac{(b-a)\parallel {\varphi }^{\prime }{\parallel }_{\infty }}{2^{2-1/q}}}\left({\left(|f^{\prime }\left(\varphi \left(a\right)\right){|}^q+|f^{\prime }\left(\varphi \left(b\right)\right)\right|}^q\right){\int }_0^1|1-2t|{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t{)}^{1/q},\end{array}$$

where $\parallel {\varphi }^{\prime }{\parallel }_{\infty }={sup}_{x\in [a,b]}\left|{\varphi }^{\prime }\left(x\right)\right|$.

Theorem 2

([11, Theorem 3.2]). Let $h:(0,1)\to {\mathbb{R}}_{+}$, let $\varphi :I\subseteq \mathbb{R}\to \mathbb{R}$ and $f:\varphi \left(I\right)\to \mathbb{R}$ be differentiable, and let $|{\varphi }^{\prime }|$ be convex. For some $s\in [-1,1]$ and $q>1$, if $|f^{\prime }{|}^q$ is an increasing extended ${\varphi }_{h\text{-}s}$-convex function on $\varphi \left(I\right)$, $f^{\prime }\left(\varphi \right){\varphi }^{\prime }\in L_1\left(I\right)$, and ${\left[{\displaystyle \frac{x}{h\left(x\right)}}\right]}^s\in L_1\left([0,1]\right)$, then for $a,b\in I^{\circ }$ with $a<b$, we have

$$\begin{array}{cc}\left|{\displaystyle \frac{f\left(\varphi \right(a\left)\right)+f\left(\varphi \right(b\left)\right)}{2}}-{\displaystyle \frac{1}{b-a}}{\int }_a^{b}f\left(\varphi \left(x\right)\right)\mathrm{d}x\right|\hfill & \\ & \hfill \le {\displaystyle \frac{(b-a)\parallel {\varphi }^{\prime }{\parallel }_{\infty }}{2}}{\left({\displaystyle \frac{q-1}{2q-1}}\right)}^{1-1/q}{\left(\left[|f^{\prime }{\left(\varphi \left(a\right)\right)|}^q+{\left|f^{\prime }\left(\varphi \left(b\right)\right)\right|}^q\right]{\int }_0^1{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t\right)}^{1/q},\end{array}$$

where $\parallel {\varphi }^{\prime }{\parallel }_{\infty }={sup}_{x\in [a,b]}\left|{\varphi }^{\prime }\left(x\right)\right|$.

Theorem 3

([11, Theorem 3.3]). Let $h:(0,1)\to {\mathbb{R}}_{+}$, let $\varphi :I\subseteq \mathbb{R}\to \mathbb{R}$ and $f:\varphi \left(I\right)\to \mathbb{R}$ be differentiable, and let $|{\varphi }^{\prime }|$ be convex. For some $s\in [-1,1]$ and $q\ge 1$, if $|f^{\prime }{|}^q$ is an increasing extended ${\varphi }_{h\text{-}s}$-convex function on $\varphi \left(I\right)$, $f^{\prime }\left(\varphi \right){\varphi }^{\prime }\in L_1\left(I\right)$, and ${\displaystyle \frac{x^{s+1}}{h^s\left(x\right)}},{\displaystyle \frac{(1-x)x^s}{h^s\left(x\right)}}\in L_1\left([0,1]\right)$, then for $a,b\in I^{\circ }$ with $a<b$, we have

$$\begin{array}{c}\left|{\displaystyle \frac{1}{b-a}}{\int }_a^{b}f\left(\varphi \left(x\right)\right)\mathrm{d}x-f\left(\varphi \left({\displaystyle \frac{a+b}{2}}\right)\right)\right|\hfill \\ \le {\displaystyle \frac{(b-a)\parallel {\varphi }^{\prime }{\parallel }_{\infty }}{2^{3(1-1/q)}}}{\left[{\left(|f^{\prime }\left(\varphi \left(a\right)\right){|}^q{\int }_0^{1/2}t{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t+|f^{\prime }\left(\varphi \left(b\right)\right)\right|}^q{\int }_{1/2}^1(1-t){\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t\right)}^{1/q}\\ \hfill +{\left(|f^{\prime }\left(\varphi \left(a\right)\right){|}^q{\int }_{1/2}^1(1-t){\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t+|f^{\prime }\left(\varphi \left(b\right)\right)\right|}^q{\int }_0^{1/2}t{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t{)}^{1/q}],\end{array}$$

where $\parallel {\varphi }^{\prime }{\parallel }_{\infty }={sup}_{x\in [a,b]}\left|{\varphi }^{\prime }\left(x\right)\right|$.

Theorem 4

([11, Theorem 3.4]). Let $h:(0,1)\to {\mathbb{R}}_{+}$, let $\varphi :I\subseteq \mathbb{R}\to \mathbb{R}$ and $f:\varphi \left(I\right)\to \mathbb{R}$ be differentiable, and let $|{\varphi }^{\prime }|$ be convex. For some $s\in [-1,1]$ and $q>1$, if $|f^{\prime }{|}^q$ is an increasing extended ${\varphi }_{h\text{-}s}$-convex function on $\varphi \left(I\right)$, $f^{\prime }\left(\varphi \right){\varphi }^{\prime }\in L_1\left(I\right)$, and ${\displaystyle \frac{x^{s+1}}{h^s\left(x\right)}},{\displaystyle \frac{(1-x)x^s}{h^s\left(x\right)}}\in L_1\left([0,1]\right)$, then for $a,b\in I^{\circ }$ with $a<b$, we have

$$\begin{array}{c}\left|{\displaystyle \frac{1}{b-a}}{\int }_a^{b}f\left(\varphi \left(x\right)\right)\mathrm{d}x-f\left(\varphi \left({\displaystyle \frac{a+b}{2}}\right)\right)\right|\le {\displaystyle \frac{(b-a)\parallel {\varphi }^{\prime }{\parallel }_{\infty }}{2^{(2q-1)/q}}}{\left({\displaystyle \frac{q-1}{2q-1}}\right)}^{1-1/q}\hfill \\ \times {\left\{{\left(|f^{\prime }\left(\varphi \left(a\right)\right){|}^q{\int }_0^{1/2}{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t+|f^{\prime }\left(\varphi \left(b\right)\right)\right|}^q{\int }_{1/2}^1{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t\right)}^{1/q}\\ \hfill +{\left(|f^{\prime }\left(\varphi \left(a\right)\right){|}^q{\int }_{1/2}^1{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t+|f^{\prime }\left(\varphi \left(b\right)\right)\right|}^q{\int }_0^{1/2}{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t{)}^{1/q}\},\end{array}$$

where $\parallel {\varphi }^{\prime }{\parallel }_{\infty }={sup}_{x\in [a,b]}\left|{\varphi }^{\prime }\left(x\right)\right|$.

In this paper, we first establish two new integral identities under the condition that the range of the function $\varphi$ is an interval, and then, by virtue of these two new integral identities and Hölder type integral inequalities, present some integral inequalities of the Hermite–Hadamard type for extended ${\varphi }_{h\text{-}s}$-convex functions.

2. Two New Integral Identities

In this section, we establish two new integral identities.

Lemma 1.

Let $\varphi :I\subseteq \mathbb{R}\to \mathbb{R}$, let $\varphi \left(I\right)$ be an interval with inner points, and let $f:\varphi \left(I\right)\to {\mathbb{R}}_0$ be a differentiable function. If $f^{\prime }\in L_1\left(\varphi \left(I\right)\right)$, then

$${\displaystyle \frac{f\left(\varphi \right(a\left)\right)+f\left(\varphi \right(b\left)\right)}{2}}-{\displaystyle \frac{1}{\varphi \left(b\right)-\varphi \left(a\right)}}{\int }_{\varphi \left(a\right)}^{\varphi \left(b\right)}f\left(u\right)\mathrm{d}u={\displaystyle \frac{\varphi \left(b\right)-\varphi \left(a\right)}{2}}{\int }_0^1(1-2t)f^{\prime }(t\varphi \left(a\right)+(1-t)\varphi \left(b\right))\mathrm{d}t$$

(3)

and

$$\begin{array}{c}{\displaystyle \frac{1}{\varphi \left(b\right)-\varphi \left(a\right)}}{\int }_{\varphi \left(a\right)}^{\varphi \left(b\right)}f\left(u\right)\mathrm{d}u-f\left({\displaystyle \frac{\varphi \left(a\right)+\varphi \left(b\right)}{2}}\right)\hfill \\ \hfill =[\varphi \left(b\right)-\varphi \left(a\right)]\left[{\int }_0^{1/2}t{f}^{\prime }(t\varphi \left(a\right)+(1-t)\varphi \left(b\right))\mathrm{d}t+{\int }_{1/2}^1(t-1)f^{\prime }(t\varphi \left(a\right)+(1-t)\varphi \left(b\right))\mathrm{d}t\right]\end{array}$$

(4)

for $a,b\in I^{\circ }$ such that $\varphi \left(a\right)\ne \varphi \left(b\right)$

Proof. 

The identity (3) can be proved by using [3, Lemma 2.1].

The identity (4) follows from [5, Lemma 2.1]. □

3. New Integral Inequalities of Hermite–Hadamard Type

In this section, we present several new integral inequalities of the Hermite–Hadamard type for extended ${\varphi }_{h\text{-}s}$-convex functions.

Theorem 5.

Let $h:(0,1)\to {\mathbb{R}}_{+}$, let $\varphi :I\subseteq \mathbb{R}\to \mathbb{R}$, and let $f:\varphi \left(I\right)\to {\mathbb{R}}_0$ is an extended ${\varphi }_{h\text{-}s}$-convex function on $\varphi \left(I\right)$ for some fixed $s\in [-1,1]$. If $f\in L_1\left(\varphi \left(I\right)\right)$ and ${\left[{\displaystyle \frac{x}{h\left(x\right)}}\right]}^s\in L_1\left([0,1]\right)$, then for $a,b\in I^{\circ }$ such that $\varphi \left(a\right)\ne \varphi \left(b\right)$, we have

$${\displaystyle \frac{{\left[2h(1/2)\right]}^s}{2}}f\left({\displaystyle \frac{\varphi \left(a\right)+\varphi \left(b\right)}{2}}\right)\le {\displaystyle \frac{1}{\varphi \left(b\right)-\varphi \left(a\right)}}{\int }_{\varphi \left(a\right)}^{\varphi \left(b\right)}f\left(u\right)\mathrm{d}u\le {\displaystyle \frac{{\left[2h(1/2)\right]}^s[f\left(\varphi \left(a\right)\right)+f\left(\varphi \left(b\right)\right)]}{2}}{\int }_0^1{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t.$$

Proof. 

For $t\in (0,1)$, using the extended ${\varphi }_{h\text{-}s}$-convexity of f, we have

$$\begin{array}{cc}\hfill f\left({\displaystyle \frac{\varphi \left(a\right)+\varphi \left(b\right)}{2}}\right)& =f\left({\displaystyle \frac{t\varphi \left(a\right)+(1-t)\varphi \left(b\right)+(1-t)\varphi \left(a\right)+t\varphi \left(b\right)}{2}}\right)\hfill \\ \hfill & \le {\left[{\displaystyle \frac{1/2}{h(1/2)}}\right]}^{s}f(t\varphi \left(a\right)+(1-t)\varphi \left(b\right))+{\left[{\displaystyle \frac{1/2}{h(1/2)}}\right]}^{s}f((1-t)\varphi \left(a\right)+t\varphi \left(b\right))\hfill \\ \hfill & ={\displaystyle \frac{f\left(t\varphi \right(a)+(1-t\left)\varphi \right(b\left)\right)+f\left(\right(1-t\left)\varphi \right(a)+t\varphi (b\left)\right)}{{\left[2h(1/2)\right]}^s}}.\hfill \end{array}$$

(5)

Integrating with respect to $t\in (0,1)$ on the very ends of the inequality (5) and making the variable transform $u=t\varphi \left(a\right)+(1-t)\varphi \left(b\right)$ for $t\in (0,1)$ result in

$$\begin{array}{cc}\hfill f\left({\displaystyle \frac{\varphi \left(a\right)+\varphi \left(b\right)}{2}}\right)& ={\int }_0^{1}f\left({\displaystyle \frac{t\varphi \left(a\right)+(1-t)\varphi \left(b\right)+(1-t)\varphi \left(a\right)+t\varphi \left(b\right)}{2}}\right)\mathrm{d}t\hfill \\ \hfill & \le {\displaystyle \frac{1}{{\left[2h(1/2)\right]}^s}}{\int }_0^1[f(t\varphi \left(a\right)+(1-t)\varphi \left(b\right))+f((1-t)\varphi \left(a\right)+t\varphi \left(b\right))]\mathrm{d}t\hfill \\ \hfill & ={\displaystyle \frac{2{\left[2h(1/2)\right]}^{-s}}{\varphi \left(b\right)-\varphi \left(a\right)}}{\int }_{\varphi \left(a\right)}^{\varphi \left(b\right)}f\left(u\right)\mathrm{d}u.\hfill \end{array}$$

Further, making the variable transform $u=t\varphi \left(a\right)+(1-t)\varphi \left(b\right)$ for $t\in (0,1)$ and using the extended ${\varphi }_{h\text{-}s}$-convexity of f lead to

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{\varphi \left(b\right)-\varphi \left(a\right)}}{\int }_{\varphi \left(a\right)}^{\varphi \left(b\right)}f\left(u\right)\mathrm{d}u& ={\int }_0^{1}f(t\varphi \left(a\right)+(1-t)\varphi \left(b\right))\mathrm{d}t\hfill \\ \hfill & \le {\int }_0^1\left({\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^{s}f\left(\varphi \left(a\right)\right)+{\left[{\displaystyle \frac{1-t}{h(1-t)}}\right]}^{s}f\left(\varphi \left(b\right)\right)\right)\mathrm{d}t\hfill \\ \hfill & =[f\left(\varphi \left(a\right)\right)+f\left(\varphi \left(b\right)\right)]{\int }_0^1{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t.\hfill \end{array}$$

Theorem 5 is thus proved. □

Corollary 1.

Under conditions of Theorem 5,

• if $s=1$, we have

  $$h\left({\displaystyle \frac{1}{2}}\right)f\left({\displaystyle \frac{\varphi \left(a\right)+\varphi \left(b\right)}{2}}\right)\le {\displaystyle \frac{1}{\varphi \left(b\right)-\varphi \left(a\right)}}{\int }_{\varphi \left(a\right)}^{\varphi \left(b\right)}f\left(u\right)\mathrm{d}u\le h\left({\displaystyle \frac{1}{2}}\right)[f\left(\varphi \left(a\right)\right)+f\left(\varphi \left(b\right)\right)]{\int }_0^1{\displaystyle \frac{t}{h\left(t\right)}}\mathrm{d}t;$$
• if $s=0$, we have

  $$\begin{array}{c}\hfill {\displaystyle \frac{1}{2}}f\left({\displaystyle \frac{\varphi \left(a\right)+\varphi \left(b\right)}{2}}\right)\le {\displaystyle \frac{1}{\varphi \left(b\right)-\varphi \left(a\right)}}{\int }_{\varphi \left(a\right)}^{\varphi \left(b\right)}f\left(u\right)\mathrm{d}u\le {\displaystyle \frac{f\left(\varphi \right(a\left)\right)+f\left(\varphi \right(b\left)\right)}{2}};\end{array}$$
• if $s=-1$, we have

  $${\displaystyle \frac{1}{4h(1/2)}}f\left({\displaystyle \frac{\varphi \left(a\right)+\varphi \left(b\right)}{2}}\right)\le {\displaystyle \frac{1}{\varphi \left(b\right)-\varphi \left(a\right)}}{\int }_{\varphi \left(a\right)}^{\varphi \left(b\right)}f\left(u\right)\mathrm{d}u\le {\displaystyle \frac{f\left(\varphi \right(a\left)\right)+f\left(\varphi \right(b\left)\right)}{4h(1/2)}}{\int }_0^1{\displaystyle \frac{h\left(t\right)}{t}}\mathrm{d}t.$$

Corollary 2.

Let $0<p<1$ and $a,b\in {\mathbb{R}}_{+}$ with $a<b$. Then

$${\displaystyle \frac{2^{1/(p+1)}}{4}}{\left({\displaystyle \frac{a^{p+1}+b^{p+1}}{2}}\right)}^{p/(p+1)}\le {\displaystyle \frac{(p+1)(b^{2p+1}-a^{2p+1})}{(2p+1)(b^{p+1}-a^{p+1})}}\le {\displaystyle \frac{2^{1/(p+1)}(p+1)}{4(2p+1)}}(a^p+b^p).$$

Proof. 

Let $\varphi \left(x\right)=x^{p+1}$, $f\left(x\right)=x^{p/(p+1)}$ for $x\in {\mathbb{R}}_{+}$, and $h\left(t\right)=t^{1-p/s(p+1)}$ for $t\in (0,1)$ and for some fixed $s\in [-1,1]\setminus \left\{0\right\}$. By virtue of [11, Example 2.1], we deduce that $f\left(x\right)=x^{p/(p+1)}$ is extended ${\varphi }_{h\text{-}s}$-convex on ${\mathbb{R}}_{+}$. Since $f\left(\varphi \left(x\right)\right)=x^p$ and ${\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s=t^{p/(p+1)}$ for $x\in {\mathbb{R}}_{+}$ and $t\in (0,1)$, with the help of Theorem 5, we arrive at

$$\begin{array}{cc}\hfill {\displaystyle \frac{{\left[2h(1/2)\right]}^s}{2}}f\left({\displaystyle \frac{\varphi \left(a\right)+\varphi \left(b\right)}{2}}\right)& ={\displaystyle \frac{1}{2^{1/(p+1)}}}{\left({\displaystyle \frac{a^{p+1}+b^{p+1}}{2}}\right)}^{p/(p+1)},\hfill \\ \hfill {\displaystyle \frac{1}{\varphi \left(b\right)-\varphi \left(a\right)}}{\int }_{\varphi \left(a\right)}^{\varphi \left(b\right)}f\left(u\right)\mathrm{d}u& ={\displaystyle \frac{(p+1)(b^{2p+1}-a^{2p+1})}{(2p+1)(b^{p+1}-a^{p+1})}},\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\displaystyle \frac{{\left[2h(1/2)\right]}^s[f\left(\varphi \left(a\right)\right)+f\left(\varphi \left(b\right)\right)]}{2}}{\int }_0^1{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t& ={\displaystyle \frac{(p+1)(a^p+b^p)}{2^{1/(p+1)}(2p+1)}}.\hfill \end{array}$$

The proof of Corollary 2 is completed. □

Theorem 6.

Let $h:(0,1)\to {\mathbb{R}}_{+}$, let $\varphi :I\subseteq \mathbb{R}\to \mathbb{R}$, and let $f:\varphi \left(I\right)\to \mathbb{R}$ be differentiable. For some fixed $s\in [-1,1]$, if $f^{\prime }\in L_1\left(\varphi \left(I\right)\right)$ and $|f^{\prime }|$ is an extended ${\varphi }_{h\text{-}s}$-convex function on $\varphi \left(I\right)$, then for $a,b\in I^{\circ }$ such that $\varphi \left(a\right)\ne \varphi \left(b\right)$,

• when ${\displaystyle \frac{(1-2x)x^s}{h^s\left(x\right)}}\in L_1\left([0,1]\right)$, we have

  $$\begin{array}{c}\left|{\displaystyle \frac{f\left(\varphi \right(a\left)\right)+f\left(\varphi \right(b\left)\right)}{2}}-{\displaystyle \frac{1}{\varphi \left(b\right)-\varphi \left(a\right)}}{\int }_{\varphi \left(a\right)}^{\varphi \left(b\right)}f\left(u\right)\mathrm{d}u\right|\hfill \\ \hfill \le {\displaystyle \frac{\left|\varphi \right(b)-\varphi (a\left)\right|}{2}}\left[\right|f^{\prime }\left(\varphi \left(a\right)\right)|+|f^{\prime }\left(\varphi \left(b\right)\right)\left|\right]{\int }_0^1|1-2t|{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t;\end{array}$$

  (6)
• when ${\displaystyle \frac{x^{s+1}}{h^s\left(x\right)}}\in L_1\left(\left[0,{\displaystyle \frac{1}{2}}\right]\right)$ and ${\displaystyle \frac{x^s(1-x)}{h^s\left(x\right)}}\in L_1\left(\left[{\displaystyle \frac{1}{2}},1\right]\right)$, we have

  $$\begin{array}{c}\left|{\displaystyle \frac{1}{\varphi \left(b\right)-\varphi \left(a\right)}}{\int }_{\varphi \left(a\right)}^{\varphi \left(b\right)}f\left(u\right)\mathrm{d}u-f\left({\displaystyle \frac{\varphi \left(a\right)+\varphi \left(b\right)}{2}}\right)\right|\hfill \\ \hfill \le |\varphi \left(b\right)-\varphi \left(a\right)|\left[\right|f^{\prime }\left(\varphi \left(a\right)\right)|+|f^{\prime }\left(\varphi \left(b\right)\right)\left|\right]\left({\int }_0^{1/2}t{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t+{\int }_{1/2}^1(1-t){\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t\right).\end{array}$$

  (7)

Proof. 

Since $|f^{\prime }|$ is a ${\varphi }_{h\text{-}s}$-convex function, we acquire

$$\begin{array}{cc}\hfill & \left|f^{\prime }(t\varphi \left(x\right)+(1-t)\varphi \left(y\right))\right|\le {\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s|f^{\prime }\left(\varphi \left(x\right)\right)|+{\left[{\displaystyle \frac{1-t}{h(1-t)}}\right]}^s|f^{\prime }\left(\varphi \left(y\right)\right)|\hfill \end{array}$$

(8)

for all $\varphi \left(x\right),\varphi \left(y\right)\in \varphi \left(I\right)$ and any $t\in (0,1)$.

From Lemma 1 and the inequality (8), it follows that

$$\begin{array}{cc}\hfill & \left|{\displaystyle \frac{f\left(\varphi \right(a\left)\right)+f\left(\varphi \right(b\left)\right)}{2}}-{\displaystyle \frac{1}{\varphi \left(b\right)-\varphi \left(a\right)}}{\int }_{\varphi \left(a\right)}^{\varphi \left(b\right)}f\left(u\right)\mathrm{d}u\right|\hfill \\ \hfill & \le {\displaystyle \frac{\left|\varphi \right(b)-\varphi (a\left)\right|}{2}}{\int }_0^1|1-2t||f^{\prime }(t\varphi \left(a\right)+(1-t)\varphi \left(b\right))|\mathrm{d}t\hfill \\ \hfill & \le {\displaystyle \frac{\left|\varphi \right(b)-\varphi (a\left)\right|}{2}}\left\{{\int }_0^1|1-2t|\left({\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s|f^{\prime }\left(\varphi \left(a\right)\right)|+{\left[{\displaystyle \frac{1-t}{h(1-t)}}\right]}^s|f^{\prime }\left(\varphi \left(b\right)\right)\right|\right)\mathrm{d}t\}\hfill \\ \hfill & ={\displaystyle \frac{\left|\varphi \right(b)-\varphi (a\left)\right|}{2}}\left[\right|f^{\prime }\left(\varphi \left(a\right)\right)|+|f^{\prime }\left(\varphi \left(b\right)\right)\left|\right]{\int }_0^1|1-2t|{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t.\hfill \end{array}$$

The inequality (6) is proved.

Utilizing Lemma 1 and the inequality (8), we gain

$$\begin{array}{cc}\hfill & \left|{\displaystyle \frac{1}{\varphi \left(b\right)-\varphi \left(a\right)}}{\int }_{\varphi \left(a\right)}^{\varphi \left(b\right)}f\left(u\right)\mathrm{d}u-f\left({\displaystyle \frac{\varphi \left(a\right)+\varphi \left(b\right)}{2}}\right)\right|\hfill \\ \hfill & \le |\varphi \left(b\right)-\varphi \left(a\right)|\left[{\int }_0^{1/2}t|f^{\prime }(t\varphi \left(a\right)+(1-t)\varphi \left(b\right))|\mathrm{d}t+{\int }_{1/2}^1(1-t)\left|f^{\prime }(t\varphi \left(a\right)+(1-t)\varphi \left(b\right))\right|\mathrm{d}t\right]\hfill \\ \hfill & \le |\varphi \left(b\right)-\varphi \left(a\right)|\left\{{\int }_0^{1/2}t\left({\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s|f^{\prime }\left(\varphi \left(a\right)\right)|+{\left[{\displaystyle \frac{1-t}{h(1-t)}}\right]}^s|f^{\prime }\left(\varphi \left(b\right)\right)\right|\right)\mathrm{d}t\hfill \\ \hfill & +{\int }_{1/2}^1(1-t)\left({\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s|f^{\prime }\left(\varphi \left(a\right)\right)|+{\left[{\displaystyle \frac{1-t}{h(1-t)}}\right]}^s|f^{\prime }\left(\varphi \left(b\right)\right)\right|\left)\mathrm{d}t\right\}\hfill \\ \hfill & =|\varphi \left(b\right)-\varphi \left(a\right)|\left[\right|f^{\prime }\left(\varphi \left(a\right)\right)|+|f^{\prime }\left(\varphi \left(b\right)\right)\left|\right]\left({\int }_0^{1/2}t{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t+{\int }_{1/2}^1(1-t){\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t\right).\hfill \end{array}$$

Thus, the inequality (7) is proved. Theorem 6 is thus proved. □

Theorem 7.

Let $h:(0,1)\to {\mathbb{R}}_{+}$, let $\varphi :I\subseteq \mathbb{R}\to \mathbb{R}$, and let $f:\varphi \left(I\right)\to \mathbb{R}$ be differentiable. For some fixed $s\in [-1,1]$, $q>1$, and $q>r\ge 0$, if $|f^{\prime }{|}^q$ is an extended ${\varphi }_{h\text{-}s}$-convex function on $\varphi \left(I\right)$, $f^{\prime }\in L_1\left(\varphi \left(I\right)\right)$, and ${\displaystyle \frac{{|1-2x|}^r{x}^s}{{\left[h\left(x\right)\right]}^s}}\in L_1\left([0,1]\right)$, then for $a,b\in I^{\circ }$ such that $\varphi \left(a\right)\ne \varphi \left(b\right)$, we have

$$\begin{array}{c}\left|{\displaystyle \frac{f\left(\varphi \right(a\left)\right)+f\left(\varphi \right(b\left)\right)}{2}}-{\displaystyle \frac{1}{\varphi \left(b\right)-\varphi \left(a\right)}}{\int }_{\varphi \left(a\right)}^{\varphi \left(b\right)}f\left(u\right)\mathrm{d}u\right|\hfill \\ \hfill \le {\displaystyle \frac{\left|\varphi \right(b)-\varphi (a\left)\right|}{2}}{\left({\displaystyle \frac{q-1}{2q-r-1}}\right)}^{1-1/q}\left({\left[|f^{\prime }\left(\varphi \left(a\right)\right){|}^q+|f^{\prime }\left(\varphi \left(b\right)\right)\right|}^q\right]{\int }_0^1{|1-2t|}^r{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t{)}^{1/q}.\end{array}$$

Proof. 

As in the proof of Theorem 6, using Lemma 1, the inequality (8), and Hölder’s integral inequality, we obtain

$$\begin{array}{cc}\hfill & \left|{\displaystyle \frac{f\left(\varphi \right(a\left)\right)+f\left(\varphi \right(b\left)\right)}{2}}-{\displaystyle \frac{1}{\varphi \left(b\right)-\varphi \left(a\right)}}{\int }_{\varphi \left(a\right)}^{\varphi \left(b\right)}f\left(u\right)\mathrm{d}u\right|\hfill \\ \hfill & \le {\displaystyle \frac{\left|\varphi \right(b)-\varphi (a\left)\right|}{2}}{\int }_0^1|1-2t||f^{\prime }(t\varphi \left(a\right)+(1-t)\varphi \left(b\right))|\mathrm{d}t\hfill \\ \hfill & \le {\displaystyle \frac{\left|\varphi \right(b)-\varphi (a\left)\right|}{2}}{\left({\int }_0^1{|1-2t|}^{(q-r)/(q-1)}\mathrm{d}t\right)}^{1-1/q}{\left({\int }_0^1{|1-2t|}^r{\left|f^{\prime }(t\varphi \left(a\right)+(1-t)\varphi \left(b\right))\right|}^q\mathrm{d}t\right)}^{1/q}\hfill \\ \hfill & \le {\displaystyle \frac{\left|\varphi \right(b)-\varphi (a\left)\right|}{2}}{\left({\displaystyle \frac{q-1}{2q-r-1}}\right)}^{1-1/q}\left\{{\int }_0^1{|1-2t|}^r{\left({\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s|f^{\prime }\left(\varphi \left(a\right)\right){|}^q+{\left[{\displaystyle \frac{1-t}{h(1-t)}}\right]}^s|f^{\prime }\left(\varphi \left(b\right)\right)\right|}^q\right)\mathrm{d}t{\}}^{1/q}\hfill \\ \hfill & ={\displaystyle \frac{\left|\varphi \right(b)-\varphi (a\left)\right|}{2}}{\left({\displaystyle \frac{q-1}{2q-r-1}}\right)}^{1-1/q}({\left[|f^{\prime }\left(\varphi \left(a\right)\right){|}^q+|f^{\prime }\left(\varphi \left(b\right)\right)\right|}^q]{\int }_0^1{|1-2t|}^r{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t{)}^{1/q}.\hfill \end{array}$$

The proof of Theorem 7 is completed. □

Theorem 8.

Let $h:(0,1)\to {\mathbb{R}}_{+}$, let $\varphi :I\subseteq \mathbb{R}\to \mathbb{R}$, and let $f:\varphi \left(I\right)\to \mathbb{R}$ be differentiable. For some fixed $s\in [-1,1]$, $q>1$, and $q>r\ge 0$, if $|f^{\prime }{|}^q$ is an extended ${\varphi }_{h\text{-}s}$-convex function on $\varphi \left(I\right)$, $f^{\prime }\in L_1\left(\varphi \left(I\right)\right)$, ${\displaystyle \frac{x^{s+r}}{h^s\left(x\right)}}\in L_1\left(\left[0,{\displaystyle \frac{1}{2}}\right]\right)$, and ${\displaystyle \frac{x^s{(1-x)}^r}{h^s\left(x\right)}}\in L_1\left(\left[{\displaystyle \frac{1}{2}},1\right]\right)$, then for $a,b\in I^{\circ }$ such that $\varphi \left(a\right)\ne \varphi \left(b\right)$, we have

$$\begin{array}{c}\left|{\displaystyle \frac{1}{\varphi \left(b\right)-\varphi \left(a\right)}}{\int }_{\varphi \left(a\right)}^{\varphi \left(b\right)}f\left(u\right)\mathrm{d}u-f\left({\displaystyle \frac{\varphi \left(a\right)+\varphi \left(b\right)}{2}}\right)\right|\le |\varphi \left(b\right)-\varphi \left(a\right)|{\left({\displaystyle \frac{q-1}{2q-r-1}}\right)}^{1-1/q}\hfill \\ \times {\left\{{\left(|f^{\prime }\left(\varphi \left(a\right)\right){|}^q{\int }_0^{1/2}{t}^r{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t+|f^{\prime }\left(\varphi \left(b\right)\right)\right|}^q{\int }_{1/2}^1{(1-t)}^r{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t\right)}^{1/q}\\ \hfill +{\left(|f^{\prime }\left(\varphi \left(a\right)\right){|}^q{\int }_{1/2}^1{(1-t)}^r{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t+|f^{\prime }\left(\varphi \left(b\right)\right)\right|}^q{\int }_0^{1/2}{t}^r{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t{)}^{1/q}\}.\end{array}$$

Proof. 

As argued in the proofs of Theorems 6 and 7, we deduce

$$\begin{array}{cc}\hfill & \left|{\displaystyle \frac{1}{\varphi \left(b\right)-\varphi \left(a\right)}}{\int }_{\varphi \left(a\right)}^{\varphi \left(b\right)}f\left(u\right)\mathrm{d}u-f\left({\displaystyle \frac{\varphi \left(a\right)+\varphi \left(b\right)}{2}}\right)\right|\hfill \\ \hfill & \le |\varphi \left(b\right)-\varphi \left(a\right)|[{\left({\int }_0^{1/2}{t}^{(q-r)/(q-1)}\mathrm{d}t\right)}^{1-1/q}{\left({\int }_0^{1/2}{t}^r{\left|f^{\prime }(t\varphi \left(a\right)+(1-t)\varphi \left(b\right))\right|}^q\mathrm{d}t\right)}^{1/q}\hfill \\ \hfill & +{\left({\int }_{1/2}^1{(1-t)}^{(q-r)/(q-1)}\mathrm{d}t\right)}^{1-1/q}{\left({\int }_{1/2}^1{(1-t)}^r{\left|f^{\prime }(t\varphi \left(a\right)+(1-t)\varphi \left(b\right))\right|}^q\mathrm{d}t\right)}^{1/q}]\hfill \\ \hfill & \le |\varphi \left(b\right)-\varphi \left(a\right)|{\left({\displaystyle \frac{q-1}{2q-r-1}}\right)}^{1-1/q}{\left\{\left[{\int }_0^{1/2}{t}^r{\left({\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s|f^{\prime }\left(\varphi \left(a\right)\right){|}^q+{\left[{\displaystyle \frac{1-t}{h(1-t)}}\right]}^s|f^{\prime }\left(\varphi \left(b\right)\right)\right|}^q\right)\mathrm{d}t\right]}^{1/q}\hfill \\ \hfill & +\left[{\int }_{1/2}^1{\left({(1-t)}^r{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s|f^{\prime }\left(\varphi \left(a\right)\right){|}^q+{\left[{\displaystyle \frac{1-t}{h(1-t)}}\right]}^s|f^{\prime }\left(\varphi \left(b\right)\right)\right|}^q\right)\mathrm{d}t{]}^{1/q}\}\hfill \\ \hfill & =|\varphi \left(b\right)-\varphi \left(a\right)|{\left({\displaystyle \frac{q-1}{2q-r-1}}\right)}^{1-1/q}{\left\{{\left(|f^{\prime }\left(\varphi \left(a\right)\right){|}^q{\int }_0^{1/2}{t}^r{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t+|f^{\prime }\left(\varphi \left(b\right)\right)\right|}^q{\int }_{1/2}^1{(1-t)}^r{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t\right)}^{1/q}\hfill \\ \hfill & +{\left(|f^{\prime }\left(\varphi \left(a\right)\right){|}^q{\int }_{1/2}^1{(1-t)}^r{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t+|f^{\prime }\left(\varphi \left(b\right)\right)\right|}^q{\int }_0^{1/2}{t}^r{\left[{\displaystyle \frac{t}{h\left(t\right)}}\right]}^s\mathrm{d}t{)}^{1/q}\}.\hfill \end{array}$$

Theorem 8 is thus proved. □

4. Remarks

Finally, we give several remarks about related stuffs.

Remark 3.

Definition 3 in this paper is a slightly modification of [11, Definition 2.1].

Remark 4.

Remark 3.1 in [11] should be corrected as follows:

• If ϕ has a derivative of the first order and $|{\varphi }^{\prime }|$ is concave, if for $q>1$ the function $|f^{\prime }{|}^q$ is decreasing extended ${\varphi }_{h\text{-}s}$-convex on $\varphi \left(I\right)$, then Theorems 1 to 4 in this paper still hold.