## 1 Introduction

The following definition is well known in the literature: a function $f:I\to \mathbb{R}$, $\mathrm{\varnothing }\ne I\subset \mathbb{R}$, is said to be convex on I if the inequality

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

holds for all $x,y\in I$ and $t\in \left[0,1\right]$. Geometrically, this means that if P, Q and R are three distinct points on the graph of f with Q between P and R, then Q is on or below the chord PR.

In their paper [1], Hudzik and Maligranda considered, among others, the class of functions which are s-convex in the second sense. This class is defined in the following way: a function $f:\left[0,\mathrm{\infty }\right)\to \mathbb{R}$ is said to be s-convex in the second sense if

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

holds for all $x,y\in \left[0,\mathrm{\infty }\right)$, $t\in \left[0,1\right]$ and for some fixed $s\in \left(0,1\right]$. The class of s-convex functions in the second sense is usually denoted by ${K}_{s}^{2}$.

It can be easily seen that for $s=1$ s-convexity reduces to the ordinary convexity of functions defined on $\left[0,\mathrm{\infty }\right)$.

In the same paper [1], Hudzik and Maligranda proved that if $s\in \left(0,1\right)$, $f\in {K}_{s}^{2}$ implies $f\left(\left[0,\mathrm{\infty }\right)\right)\subseteq \left[0,\mathrm{\infty }\right)$, i.e., they proved that all functions from ${K}_{s}^{2}$, $s\in \left(0,1\right)$, are nonnegative.

Example 1 [1]

Let $s\in \left(0,1\right)$ and $a,b,c\in \mathbb{R}$. We define the function $f:\left[0,\mathrm{\infty }\right)\to \mathbb{R}$ as

$f\left(t\right)=\left\{\begin{array}{cc}a,\hfill & t=0,\hfill \\ b{t}^{s}+c,\hfill & t>0.\hfill \end{array}$

It can be easily checked that

1. (i)

if $b\ge 0$ and $0\le c\le a$, then $f\in {K}_{s}^{2}$,

2. (ii)

if $b>0$ and $c<0$, then $f\notin {K}_{s}^{2}$.

Many important inequalities are established for the class of convex functions, but one of the most famous is the so-called Hermite-Hadamard’s inequality (or Hadamard’s inequality). This double inequality is stated as follows: Let f be a convex function on $\left[a,b\right]\subset \mathbb{R}$, where $a\ne b$. Then

$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\le \frac{f\left(a\right)+f\left(b\right)}{2}.$

For several recent results concerning Hadamard’s inequality, we refer the interested reader to [25].

In [6] Dragomir and Fitzpatrick proved a variant of Hadamard’s inequality which holds for s-convex functions in the second sense.

Theorem 1 Suppose that $f:\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ is an s-convex function in the second sense, where $s\in \left(0,1\right)$, and let $a,b\in \left[0,\mathrm{\infty }\right)$, $a. If $f\in L\left(\left[a,b\right]\right)$, then the following inequalities hold:

${2}^{s-1}f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\le \frac{f\left(a\right)+f\left(b\right)}{s+1}.$
(1.1)

The constant $k=1/\left(s+1\right)$ is best possible in the second inequality in [7].

The above inequalities are sharp. For recent results and generalizations concerning s-convex functions, see [813].

Along this paper, we consider a real interval $I\subset \mathbb{R}$, and we denote that ${I}^{\circ }$ is the interior of I.

The main aim of this paper is to establish new inequalities of Hermite-Hadamard type for the class of functions whose second derivatives at certain powers are s-convex functions in the second sense.

## 2 Main results

To prove our main results, we consider the following lemma.

Lemma 1 Let $f:I\subset \mathbb{R}\to \mathbb{R}$ be a differentiable mapping on ${I}^{\circ }$ where $a,b\in I$ with $a. If ${f}^{\mathrm{\prime }\mathrm{\prime }}\in L\left[a,b\right]$, then the following equality holds:

$\begin{array}{r}\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx-f\left(\frac{a+b}{2}\right)\\ \phantom{\rule{1em}{0ex}}=\frac{{\left(b-a\right)}^{2}}{16}\left[{\int }_{0}^{1}{t}^{2}{f}^{\mathrm{\prime }\mathrm{\prime }}\left(t\frac{a+b}{2}+\left(1-t\right)a\right)\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{2em}{0ex}}+{\int }_{0}^{1}{\left(t-1\right)}^{2}{f}^{\mathrm{\prime }\mathrm{\prime }}\left(tb+\left(1-t\right)\frac{a+b}{2}\right)\phantom{\rule{0.2em}{0ex}}dt\right].\end{array}$
(2.1)

Proof By integration by parts, we have the following identity:

$\begin{array}{rcl}{I}_{1}& =& {\int }_{0}^{1}{t}^{2}{f}^{\mathrm{\prime }\mathrm{\prime }}\left(t\frac{a+b}{2}+\left(1-t\right)a\right)\phantom{\rule{0.2em}{0ex}}dt\\ =& {t}^{2}\frac{2}{b-a}{f}^{\mathrm{\prime }}\left(t\frac{a+b}{2}+\left(1-t\right)a\right){|}_{0}^{1}-\frac{4}{b-a}{\int }_{0}^{1}t{f}^{\mathrm{\prime }}\left(t\frac{a+b}{2}+\left(1-t\right)a\right)\phantom{\rule{0.2em}{0ex}}dt\\ =& \frac{2}{b-a}{f}^{\mathrm{\prime }}\left(\frac{a+b}{2}\right)-\frac{4}{b-a}\left[t\frac{2}{b-a}f\left(t\frac{a+b}{2}+\left(1-t\right)a\right){|}_{0}^{1}\\ -\frac{2}{b-a}{\int }_{0}^{1}f\left(t\frac{a+b}{2}+\left(1-t\right)a\right)\phantom{\rule{0.2em}{0ex}}dt\right]\\ =& \frac{2}{b-a}{f}^{\mathrm{\prime }}\left(\frac{a+b}{2}\right)-\frac{8}{{\left(b-a\right)}^{2}}f\left(\frac{a+b}{2}\right)\\ +\frac{8}{{\left(b-a\right)}^{2}}{\int }_{0}^{1}f\left(t\frac{a+b}{2}+\left(1-t\right)a\right)\phantom{\rule{0.2em}{0ex}}dt.\end{array}$
(2.2)

Using the change of the variable $x=t\frac{a+b}{2}+\left(1-t\right)a$ for $t\in \left[0,1\right]$ and multiplying the both sides (2.2) by $\frac{{\left(b-a\right)}^{2}}{16}$, we obtain

$\begin{array}{r}\frac{{\left(b-a\right)}^{2}}{16}{\int }_{0}^{1}{t}^{2}{f}^{\mathrm{\prime }\mathrm{\prime }}\left(t\frac{a+b}{2}+\left(1-t\right)a\right)\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{1em}{0ex}}=\frac{b-a}{8}{f}^{\mathrm{\prime }}\left(\frac{a+b}{2}\right)-\frac{1}{2}f\left(\frac{a+b}{2}\right)+\frac{1}{b-a}{\int }_{a}^{\frac{a+b}{2}}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx.\end{array}$
(2.3)

Similarly, we observe that

$\begin{array}{r}\frac{{\left(b-a\right)}^{2}}{16}{\int }_{0}^{1}{\left(t-1\right)}^{2}{f}^{\mathrm{\prime }\mathrm{\prime }}\left(tb+\left(1-t\right)\frac{a+b}{2}\right)\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{1em}{0ex}}=-\frac{b-a}{8}{f}^{\mathrm{\prime }}\left(\frac{a+b}{2}\right)-\frac{1}{2}f\left(\frac{a+b}{2}\right)+\frac{1}{b-a}{\int }_{\frac{a+b}{2}}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx.\end{array}$
(2.4)

Thus, adding (2.3) and (2.4), we get the required identity (2.1). □

Theorem 2 Let $f:I\subset \left[0,\mathrm{\infty }\right)\to \mathbb{R}$ be a differentiable mapping on ${I}^{\circ }$ such that ${f}^{\mathrm{\prime }\mathrm{\prime }}\in L\left[a,b\right]$, where $a,b\in I$ with $a. If $|f|$ is s-convex on $\left[a,b\right]$, for some fixed $s\in \left(0,1\right]$, then the following inequality holds:

$\begin{array}{l}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{8\left(s+1\right)\left(s+2\right)\left(s+3\right)}\left\{|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right)|+\left(s+1\right)\left(s+2\right)|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right)|+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right)|\right\}\end{array}$
(2.5)
$\phantom{\rule{1em}{0ex}}\le \frac{\left[1+\left(s+2\right){2}^{1-s}\right]{\left(b-a\right)}^{2}}{8\left(s+1\right)\left(s+2\right)\left(s+3\right)}\left\{|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right)|+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right)|\right\}.$
(2.6)

Proof From Lemma 1, we have

$\begin{array}{l}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}\left[{\int }_{0}^{1}{t}^{2}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(t\frac{a+b}{2}+\left(1-t\right)a\right)|\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{2em}{0ex}}+{\int }_{0}^{1}{\left(t-1\right)}^{2}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(tb+\left(1-t\right)\frac{a+b}{2}\right)|\phantom{\rule{0.2em}{0ex}}dt\right]\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}{\int }_{0}^{1}{t}^{2}\left[{t}^{s}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right)|+{\left(1-t\right)}^{s}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right)|\right]\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{2em}{0ex}}+\frac{{\left(b-a\right)}^{2}}{16}{\int }_{0}^{1}{\left(t-1\right)}^{2}\left[{t}^{s}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right)|+{\left(1-t\right)}^{s}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right)|\right]\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{1em}{0ex}}=\frac{{\left(b-a\right)}^{2}}{16}\left[\frac{1}{s+3}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right)|+\frac{2}{\left(s+1\right)\left(s+2\right)\left(s+3\right)}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right)|\right]\\ \phantom{\rule{2em}{0ex}}+\frac{{\left(b-a\right)}^{2}}{16}\left[\frac{2}{\left(s+1\right)\left(s+2\right)\left(s+3\right)}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right)|+\frac{1}{s+3}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right)|\right]\\ \phantom{\rule{1em}{0ex}}=\frac{{\left(b-a\right)}^{2}}{8\left(s+1\right)\left(s+2\right)\left(s+3\right)}\left\{|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right)|+\left(s+1\right)\left(s+2\right)|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right)|+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right)|\right\},\end{array}$
(2.7)

where we have used the fact that

$\begin{array}{r}{\int }_{0}^{1}{t}^{2}{\left(1-t\right)}^{s}\phantom{\rule{0.2em}{0ex}}dt={\int }_{0}^{1}{\left(t-1\right)}^{2}{t}^{s}\phantom{\rule{0.2em}{0ex}}dt=\frac{2}{\left(s+1\right)\left(s+2\right)\left(s+3\right)},\\ {\int }_{0}^{1}{t}^{s+2}\phantom{\rule{0.2em}{0ex}}dt={\int }_{0}^{1}{\left(1-t\right)}^{s+2}\phantom{\rule{0.2em}{0ex}}dt=\frac{1}{s+3}.\end{array}$

This proves inequality (2.5). To prove (2.6), and since $|{f}^{\mathrm{\prime }\mathrm{\prime }}|$ is s-convex on $\left[a,b\right]$, for any $t\in \left[0,1\right]$, then by (1.1) we have

${2}^{s-1}f\left(\frac{a+b}{2}\right)\le \frac{f\left(a\right)+f\left(b\right)}{s+1}.$
(2.8)

Combining (2.7) and (2.8), we have

$\begin{array}{r}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{8\left(s+1\right)\left(s+2\right)\left(s+3\right)}\left\{|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right)|+\left(s+1\right)\left(s+2\right)|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right)|+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right)|\right\}\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{8\left(s+1\right)\left(s+2\right)\left(s+3\right)}\left\{|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right)|+\left(s+1\right)\left(s+2\right){2}^{1-s}\frac{f\left(a\right)+f\left(b\right)}{s+1}+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right)|\right\}\\ \phantom{\rule{1em}{0ex}}=\frac{\left[1+\left(s+2\right){2}^{1-s}\right]{\left(b-a\right)}^{2}}{8\left(s+1\right)\left(s+2\right)\left(s+3\right)}\left\{|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right)|+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right)|\right\},\end{array}$

which proves inequality (2.6), and thus the proof is completed. □

Corollary 1 In Theorem  2, if we choose $s=1$, we have

$\begin{array}{l}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{192}\left\{|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right)|+6|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right)|+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right)|\right\}\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{48}\left\{|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right)|+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right)|\right\}.\end{array}$
(2.9)

The next theorem gives a new upper bound of the left Hadamard inequality for s-convex mappings.

Theorem 3 Let $f:I\subset \left[0,\mathrm{\infty }\right)\to \mathbb{R}$ be a differentiable mapping on ${I}^{\circ }$ such that ${f}^{\mathrm{\prime }\mathrm{\prime }}\in L\left[a,b\right]$, where $a,b\in I$ with $a. If ${|{f}^{\mathrm{\prime }\mathrm{\prime }}|}^{q}$ is s-convex on $\left[a,b\right]$, for some fixed $s\in \left(0,1\right]$ and $q>1$ with $\frac{1}{p}+\frac{1}{q}=1$, then the following inequality holds:

$\begin{array}{l}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}{\left(\frac{1}{2p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{s+1}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}×\left[{\left(|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right){|}^{q}+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^{q}\right)}^{\frac{1}{q}}+{\left(|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^{q}+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right){|}^{q}\right)}^{\frac{1}{q}}\right].\end{array}$
(2.10)

Proof Suppose that $p>1$. From Lemma 1 and using the Hölder inequality, we have

$\begin{array}{r}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}\left[{\int }_{0}^{1}{t}^{2}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(t\frac{a+b}{2}+\left(1-t\right)a\right)|\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{2em}{0ex}}+{\int }_{0}^{1}{\left(t-1\right)}^{2}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(tb+\left(1-t\right)\frac{a+b}{2}\right)|\phantom{\rule{0.2em}{0ex}}dt\right]\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}{\left({\int }_{0}^{1}{t}^{2p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{p}}{\left({\int }_{0}^{1}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(t\frac{a+b}{2}+\left(1-t\right)a\right){|}^{q}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+\frac{{\left(b-a\right)}^{2}}{16}{\left({\int }_{0}^{1}{\left(t-1\right)}^{2p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{p}}{\left({\int }_{0}^{1}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(tb+\left(1-t\right)\frac{a+b}{2}\right){|}^{q}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}}.\end{array}$

Because ${|{f}^{\mathrm{\prime }\mathrm{\prime }}|}^{q}$ is s-convex, we have

${\int }_{0}^{1}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(t\frac{a+b}{2}+\left(1-t\right)a\right){|}^{q}\phantom{\rule{0.2em}{0ex}}dt\le \frac{1}{s+1}\left\{|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right){|}^{q}+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^{q}\right\}$

and

${\int }_{0}^{1}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(tb+\left(1-t\right)\frac{a+b}{2}\right){|}^{q}\phantom{\rule{0.2em}{0ex}}dt\le \frac{1}{s+1}\left\{|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^{q}+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right){|}^{q}\right\}.$

By a simple computation,

${\int }_{0}^{1}{t}^{2p}\phantom{\rule{0.2em}{0ex}}dt=\frac{1}{2p+1}$

and

${\int }_{0}^{1}{\left(t-1\right)}^{2p}\phantom{\rule{0.2em}{0ex}}dt={\int }_{0}^{1}{\left(1-t\right)}^{2p}\phantom{\rule{0.2em}{0ex}}dt=\frac{1}{2p+1}.$

Therefore, we have

$\begin{array}{r}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}{\left(\frac{1}{2p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{s+1}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}×\left[{\left(|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right){|}^{q}+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^{q}\right)}^{\frac{1}{q}}+{\left(|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^{q}+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right){|}^{q}\right)}^{\frac{1}{q}}\right].\end{array}$

This completes the proof. □

Corollary 2 Let $f:I\subset \left[0,\mathrm{\infty }\right)\to \mathbb{R}$ be a differentiable mapping on ${I}^{\circ }$ such that ${f}^{\mathrm{\prime }\mathrm{\prime }}\in L\left[a,b\right]$, where $a,b\in I$ with $a. If ${|{f}^{\mathrm{\prime }\mathrm{\prime }}|}^{q}$ is s-convex on $\left[a,b\right]$, for some fixed $s\in \left(0,1\right]$ and $q>1$ with $\frac{1}{p}+\frac{1}{q}=1$, then the following inequality holds:

$\begin{array}{r}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}{\left(\frac{1}{2p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{s+1}\right)}^{\frac{2}{q}}\\ \phantom{\rule{2em}{0ex}}×\left\{{2}^{\frac{1-s}{q}}+{\left({2}^{1-s}+s+1\right)}^{\frac{1}{q}}\right\}\left[|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right)|+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right)|\right].\end{array}$

Proof We consider inequality (2.10), and since ${|{f}^{\mathrm{\prime }\mathrm{\prime }}|}^{q}$ is s-convex on $\left[a,b\right]$, then by (1.1) we have

${2}^{s-1}f\left(\frac{a+b}{2}\right)\le \frac{f\left(a\right)+f\left(b\right)}{s+1}.$

Therefore

$\begin{array}{r}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}{\left(\frac{1}{2p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{s+1}\right)}^{\frac{2}{q}}\\ \phantom{\rule{2em}{0ex}}×\left[{\left(\left\{{2}^{1-s}+s+1\right\}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right){|}^{q}+{2}^{1-s}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right){|}^{q}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+{\left({2}^{1-s}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right){|}^{q}+\left\{{2}^{1-s}+s+1\right\}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right){|}^{q}\right)}^{\frac{1}{q}}\right].\end{array}$

We let ${a}_{1}=\left({2}^{1-s}+s+1\right)|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right){|}^{q}$, ${b}_{1}={2}^{1-s}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right){|}^{q}$, ${a}_{2}={2}^{1-s}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right){|}^{q}$ and ${b}_{2}=\left({2}^{1-s}+s+1\right)|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right){|}^{q}$. Here, $0<1/q<1$ for $q>1$. Using the fact

$\sum _{i=1}^{n}{\left({a}_{i}+{b}_{i}\right)}^{r}\le \sum _{i=1}^{n}{a}_{i}^{r}+\sum _{i=1}^{n}{b}_{i}^{r}$

for $0, ${a}_{1},{a}_{2},\dots ,{a}_{n}\ge 0$ and ${b}_{1},{b}_{2},\dots ,{b}_{n}\ge 0$, we obtain the inequalities

$\begin{array}{r}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}{\left(\frac{1}{2p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{s+1}\right)}^{\frac{2}{q}}\\ \phantom{\rule{2em}{0ex}}×\left[{\left(\left\{{2}^{1-s}+s+1\right\}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right){|}^{q}+{2}^{1-s}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right){|}^{q}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+{\left({2}^{1-s}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right){|}^{q}+\left\{{2}^{1-s}+s+1\right\}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right){|}^{q}\right)}^{\frac{1}{q}}\right]\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}{\left(\frac{1}{2p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{s+1}\right)}^{\frac{2}{q}}\\ \phantom{\rule{2em}{0ex}}×\left\{{2}^{\frac{1-s}{q}}+{\left({2}^{1-s}+s+1\right)}^{\frac{1}{q}}\right\}\left[|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right)|+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right)|\right].\end{array}$

□

Theorem 4 Let $f:I\subset \left[0,\mathrm{\infty }\right)\to \mathbb{R}$ be a differentiable mapping on ${I}^{\circ }$ such that ${f}^{\mathrm{\prime }\mathrm{\prime }}\in L\left[a,b\right]$, where $a,b\in I$ with $a. If ${|{f}^{\mathrm{\prime }\mathrm{\prime }}|}^{q}$, $q\ge 1$ is s-convex on $\left[a,b\right]$, for some fixed $s\in \left(0,1\right]$, then the following inequality holds:

$\begin{array}{r}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}{\left(\frac{1}{3}\right)}^{\frac{1}{p}}\left\{{\left(\frac{2}{\left(s+1\right)\left(s+2\right)\left(s+3\right)}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right){|}^{q}+\frac{1}{s+3}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^{q}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+{\left(\frac{1}{s+3}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^{q}+\frac{2}{\left(s+1\right)\left(s+2\right)\left(s+3\right)}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right){|}^{q}\right)}^{\frac{1}{q}}\right\}.\end{array}$

Proof Suppose that $p\ge 1$. From Lemma 1 and using the power mean inequality, we have

$\begin{array}{r}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}\left[{\int }_{0}^{1}{t}^{2}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(t\frac{a+b}{2}+\left(1-t\right)a\right)|\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{2em}{0ex}}+{\int }_{0}^{1}{\left(t-1\right)}^{2}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(tb+\left(1-t\right)\frac{a+b}{2}\right)|\phantom{\rule{0.2em}{0ex}}dt\right]\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}{\left({\int }_{0}^{1}{t}^{2}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{p}}{\left({\int }_{0}^{1}{t}^{2}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(t\frac{a+b}{2}+\left(1-t\right)a\right){|}^{q}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+\frac{{\left(b-a\right)}^{2}}{16}{\left({\int }_{0}^{1}{\left(t-1\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{p}}{\left({\int }_{0}^{1}{\left(t-1\right)}^{2}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(tb+\left(1-t\right)\frac{a+b}{2}\right){|}^{q}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}}.\end{array}$

Because ${|{f}^{\mathrm{\prime }\mathrm{\prime }}|}^{q}$ is s-convex, we have

${\int }_{0}^{1}{t}^{2}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(t\frac{a+b}{2}+\left(1-t\right)a\right){|}^{q}\phantom{\rule{0.2em}{0ex}}dt\le \frac{2}{\left(s+1\right)\left(s+2\right)\left(s+3\right)}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right){|}^{q}+\frac{1}{s+3}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^{q}$

and

$\begin{array}{c}{\int }_{0}^{1}{\left(t-1\right)}^{2}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(tb+\left(1-t\right)\frac{a+b}{2}\right){|}^{q}\phantom{\rule{0.2em}{0ex}}dt\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{s+3}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^{q}+\frac{2}{\left(s+1\right)\left(s+2\right)\left(s+3\right)}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right){|}^{q}.\hfill \end{array}$

Therefore, we have

$\begin{array}{r}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}{\left(\frac{1}{3}\right)}^{\frac{1}{p}}\left\{{\left(\frac{2}{\left(s+1\right)\left(s+2\right)\left(s+3\right)}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right){|}^{q}+\frac{1}{s+3}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^{q}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+{\left(\frac{1}{s+3}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^{q}+\frac{2}{\left(s+1\right)\left(s+2\right)\left(s+3\right)}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right){|}^{q}\right)}^{\frac{1}{q}}\right\}.\end{array}$

□

Corollary 3 In Theorem  4, if we choose $s=1$, we have

$\begin{array}{l}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{48}{\left(\frac{3}{4}\right)}^{\frac{1}{q}}\left\{{\left(\frac{|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(a\right){|}^{q}}{3}+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^{q}\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+{\left(|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^{q}+\frac{|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(b\right){|}^{q}}{3}\right)}^{\frac{1}{q}}\right\}.\end{array}$
(2.11)

Now, we give the following Hadamard-type inequality for s-concave mappings.

Theorem 5 Let $f:I\subset \left[0,\mathrm{\infty }\right)\to \mathbb{R}$ be a differentiable mapping on ${I}^{\circ }$ such that ${f}^{\mathrm{\prime }\mathrm{\prime }}\in L\left[a,b\right]$, where $a,b\in I$ with $a. If ${|{f}^{\mathrm{\prime }\mathrm{\prime }}|}^{q}$ is s-concave on $\left[a,b\right]$, for some fixed $s\in \left(0,1\right]$ and $q>1$ with $\frac{1}{p}+\frac{1}{q}=1$, then the following inequality holds:

$\begin{array}{r}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}\frac{{2}^{\frac{s-1}{q}}}{{\left(2p+1\right)}^{1/p}}\left\{|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{3a+b}{4}\right)|+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+3b}{4}\right)|\right\}.\end{array}$

Proof From Lemma 1 and using the Hölder inequality for $q>1$ and $\frac{1}{p}+\frac{1}{q}=1$, we obtain

$\begin{array}{l}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}\left[{\int }_{0}^{1}{t}^{2}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(t\frac{a+b}{2}+\left(1-t\right)a\right)|\phantom{\rule{0.2em}{0ex}}dt+{\int }_{0}^{1}{\left(t-1\right)}^{2}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(tb+\left(1-t\right)\frac{a+b}{2}\right)|\phantom{\rule{0.2em}{0ex}}dt\right]\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}{\left({\int }_{0}^{1}{t}^{2p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{p}}{\left({\int }_{0}^{1}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(t\frac{a+b}{2}+\left(1-t\right)a\right){|}^{q}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}}\\ \phantom{\rule{2em}{0ex}}+\frac{{\left(b-a\right)}^{2}}{16}{\left({\int }_{0}^{1}{\left(t-1\right)}^{2p}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{p}}{\left({\int }_{0}^{1}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(tb+\left(1-t\right)\frac{a+b}{2}\right){|}^{q}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{q}}.\end{array}$
(2.12)

Since ${|{f}^{\mathrm{\prime }\mathrm{\prime }}|}^{q}$ is s-concave, using inequality (1.1), we have

${\int }_{0}^{1}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(t\frac{a+b}{2}+\left(1-t\right)a\right){|}^{q}\phantom{\rule{0.2em}{0ex}}dt\le {2}^{s-1}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{3a+b}{4}\right){|}^{q}$
(2.13)

and

${\int }_{0}^{1}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(tb+\left(1-t\right)\frac{a+b}{2}\right){|}^{q}\phantom{\rule{0.2em}{0ex}}dt\le {2}^{s-1}|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+3b}{4}\right){|}^{q}.$
(2.14)

From (2.12)-(2.14), we get

$\begin{array}{r}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}\frac{{2}^{\frac{s-1}{q}}}{{\left(2p+1\right)}^{1/p}}\left\{|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{3a+b}{4}\right)|+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+3b}{4}\right)|\right\},\end{array}$

which completes the proof. □

Corollary 4 In Theorem  5, if we choose $s=1$ and $\frac{1}{3}<{\left(\frac{1}{2p+1}\right)}^{\frac{1}{p}}<1$, $p>1$, we have

$\begin{array}{l}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx|\\ \phantom{\rule{1em}{0ex}}\le \frac{{\left(b-a\right)}^{2}}{16}\left\{|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{3a+b}{4}\right)|+|{f}^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+3b}{4}\right)|\right\}.\end{array}$
(2.15)

## 3 Applications to special means

We now consider the means for arbitrary real numbers α, β ($\alpha \ne \beta$). We take:

1. (1)

Arithmetic mean:

$A\left(\alpha ,\beta \right)=\frac{\alpha +\beta }{2},\phantom{\rule{1em}{0ex}}\alpha ,\beta \in {\mathbb{R}}^{+};$
2. (2)

Logarithmic mean:

$L\left(\alpha ,\beta \right)=\frac{\alpha -\beta }{ln|\alpha |-ln|\beta |},\phantom{\rule{1em}{0ex}}|\alpha |\ne |\beta |,\alpha ,\beta \ne 0,\alpha ,\beta \in {\mathbb{R}}^{+};$

Generalized log-mean:

${L}_{n}\left(\alpha ,\beta \right)={\left[\frac{{\beta }^{n+1}-{\alpha }^{n+1}}{\left(n+1\right)\left(\beta -\alpha \right)}\right]}^{\frac{1}{n}},\phantom{\rule{1em}{0ex}}n\in \mathbb{Z}\mathrm{\setminus }\left\{-1,0\right\},\alpha ,\beta \in {\mathbb{R}}^{+}.$

Now, using the results of Section 2, we give some applications to special means of real numbers.

Proposition 1 Let $0 and $s\in \left(0,1\right)$. Then we have

$|{A}^{s}\left(a,b\right)-{L}_{s}^{s}\left(a,b\right)|\le \frac{|s\left(s-1\right)|{\left(b-a\right)}^{2}}{192}\left\{{a}^{s-2}+6{\left(\frac{a+b}{2}\right)}^{s-2}+{b}^{s-2}\right\}.$

Proof The assertion follows from (2.9) applied to the s-convex function in the second sense $f:\left[0,1\right]\to \left[0,1\right]$, $f\left(x\right)={x}^{s}$. □

Proposition 2 Let $0 and $s\in \left(0,1\right)$. Then we have

$\begin{array}{rcl}|{A}^{s}\left(a,b\right)-{L}_{s}^{s}\left(a,b\right)|& \le & \frac{|s\left(s-1\right)|{\left(b-a\right)}^{2}}{48}{\left(\frac{3}{4}\right)}^{q}\left\{{\left[\frac{{a}^{q\left(s-2\right)}}{3}+{\left(\frac{a+b}{2}\right)}^{q\left(s-2\right)}\right]}^{\frac{1}{q}}\\ +{\left[{\left(\frac{a+b}{2}\right)}^{q\left(s-2\right)}+\frac{{b}^{q\left(s-2\right)}}{3}\right]}^{\frac{1}{q}}\right\}.\end{array}$

Proof The assertion follows from (2.11) applied to the s-convex function in the second sense $f:\left[0,1\right]\to \left[0,1\right]$, $f\left(x\right)={x}^{s}$. □

Proposition 3 Let $0 and $p>1$. Then we have

$|{A}^{s}\left(a,b\right)-{L}_{s}^{s}\left(a,b\right)|\le {\left(b-a\right)}^{2}\left\{\frac{1}{{\left(3a+b\right)}^{2}}+\frac{1}{{\left(a+3b\right)}^{2}}\right\}.$

Proof The inequality follows from (2.15) applied to the concave function in the second sense $f:\left[a;b\right]\to \mathbb{R}$, $f\left(x\right)=lnx$. The details are omitted. □