On some inequalities for s-convex functions and applications

Abstract

Some new results related to the left-hand side of the Hermite-Hadamard type inequalities for the class of functions whose second derivatives at certain powers are s-convex functions in the second sense are obtained. Also, some applications to special means of real numbers are provided.

MSC:26A51, 26D15.

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

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

holds for all $x,y\in I$ and $t\in [0,1]$. 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 $f:[0,\mathrm{\infty })\to \mathbb{R}$ is said to be s-convex in the second sense if

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

holds for all $x,y\in [0,\mathrm{\infty })$, $t\in [0,1]$ and for some fixed $s\in (0,1]$. 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 $[0,\mathrm{\infty })$.

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

Example 1 [1]

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

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

It can be easily checked that

• (i) if $b\ge 0$ and $0\le c\le a$, then $f\in K_s^2$,

• (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 $[a,b]\subset \mathbb{R}$, where $a\ne b$. Then

$$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}f(x)dx\le \frac{f(a)+f(b)}{2}.$$

For several recent results concerning Hadamard’s inequality, we refer the interested reader to [2–5].

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

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

(1.1)

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

The above inequalities are sharp. For recent results and generalizations concerning s-convex functions, see [8–13].

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 $a<b$. If $f^{\mathrm{\prime }\mathrm{\prime }}\in L[a,b]$, then the following equality holds:

$$\begin{array}{r}\frac{1}{b-a}{\int }_a^{b}f(x)dx-f\left(\frac{a+b}{2}\right)\\ =\frac{{(b-a)}^2}{16}[{\int }_0^1{t}^2{f}^{\mathrm{\prime }\mathrm{\prime }}(t\frac{a+b}{2}+(1-t)a)dt\\ +{\int }_0^1{(t-1)}^2{f}^{\mathrm{\prime }\mathrm{\prime }}(tb+(1-t)\frac{a+b}{2})dt].\end{array}$$

(2.1)

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

$$\begin{array}{rcl}{I}_1& =& {\int }_0^1{t}^2{f}^{\mathrm{\prime }\mathrm{\prime }}(t\frac{a+b}{2}+(1-t)a)dt\\ =& t^2\frac{2}{b-a}{f}^{\mathrm{\prime }}(t\frac{a+b}{2}+(1-t)a){|}_0^1-\frac{4}{b-a}{\int }_0^{1}t{f}^{\mathrm{\prime }}(t\frac{a+b}{2}+(1-t)a)dt\\ =& \frac{2}{b-a}{f}^{\mathrm{\prime }}\left(\frac{a+b}{2}\right)-\frac{4}{b-a}[t\frac{2}{b-a}f(t\frac{a+b}{2}+(1-t)a){|}_0^1\\ -\frac{2}{b-a}{\int }_0^{1}f(t\frac{a+b}{2}+(1-t)a)dt]\\ =& \frac{2}{b-a}{f}^{\mathrm{\prime }}\left(\frac{a+b}{2}\right)-\frac{8}{{(b-a)}^2}f\left(\frac{a+b}{2}\right)\\ +\frac{8}{{(b-a)}^2}{\int }_0^{1}f(t\frac{a+b}{2}+(1-t)a)dt.\end{array}$$

(2.2)

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

$$\begin{array}{r}\frac{{(b-a)}^2}{16}{\int }_0^1{t}^2{f}^{\mathrm{\prime }\mathrm{\prime }}(t\frac{a+b}{2}+(1-t)a)dt\\ =\frac{b-a}{8}{f}^{\mathrm{\prime }}\left(\frac{a+b}{2}\right)-\frac{1}{2}f\left(\frac{a+b}{2}\right)+\frac{1}{b-a}{\int }_a^{\frac{a+b}{2}}f(x)dx.\end{array}$$

(2.3)

Similarly, we observe that

$$\begin{array}{r}\frac{{(b-a)}^2}{16}{\int }_0^1{(t-1)}^2{f}^{\mathrm{\prime }\mathrm{\prime }}(tb+(1-t)\frac{a+b}{2})dt\\ =-\frac{b-a}{8}{f}^{\mathrm{\prime }}\left(\frac{a+b}{2}\right)-\frac{1}{2}f\left(\frac{a+b}{2}\right)+\frac{1}{b-a}{\int }_{\frac{a+b}{2}}^{b}f(x)dx.\end{array}$$

(2.4)

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

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

$$\begin{array}{l}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_a^{b}f(x)dx|\\ \le \frac{{(b-a)}^2}{8(s+1)(s+2)(s+3)}\left\{\right|f^{\mathrm{\prime }\mathrm{\prime }}(a)|+(s+1)(s+2)|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right)|+|f^{\mathrm{\prime }\mathrm{\prime }}(b)\left|\right\}\end{array}$$

(2.5)

$$\le \frac{[1+(s+2)2^{1-s}]{(b-a)}^2}{8(s+1)(s+2)(s+3)}\left\{\right|f^{\mathrm{\prime }\mathrm{\prime }}(a)|+|f^{\mathrm{\prime }\mathrm{\prime }}(b)\left|\right\}.$$

(2.6)

Proof From Lemma 1, we have

$$\begin{array}{l}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_a^{b}f(x)dx|\\ \le \frac{{(b-a)}^2}{16}\left[{\int }_0^1{t}^2\right|f^{\mathrm{\prime }\mathrm{\prime }}(t\frac{a+b}{2}+(1-t)a)|dt\\ +{\int }_0^1{(t-1)}^2\left|f^{\mathrm{\prime }\mathrm{\prime }}(tb+(1-t)\frac{a+b}{2})\right|dt]\\ \le \frac{{(b-a)}^2}{16}{\int }_0^1{t}^2\left[t^s\right|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right)|+{(1-t)}^s|f^{\mathrm{\prime }\mathrm{\prime }}(a)\left|\right]dt\\ +\frac{{(b-a)}^2}{16}{\int }_0^1{(t-1)}^2\left[t^s\right|f^{\mathrm{\prime }\mathrm{\prime }}(b)|+{(1-t)}^s|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right)\left|\right]dt\\ =\frac{{(b-a)}^2}{16}\left[\frac{1}{s+3}\right|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right)|+\frac{2}{(s+1)(s+2)(s+3)}|f^{\mathrm{\prime }\mathrm{\prime }}(a)\left|\right]\\ +\frac{{(b-a)}^2}{16}\left[\frac{2}{(s+1)(s+2)(s+3)}\right|f^{\mathrm{\prime }\mathrm{\prime }}(b)|+\frac{1}{s+3}|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right)\left|\right]\\ =\frac{{(b-a)}^2}{8(s+1)(s+2)(s+3)}\left\{\right|f^{\mathrm{\prime }\mathrm{\prime }}(a)|+(s+1)(s+2)|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right)|+|f^{\mathrm{\prime }\mathrm{\prime }}(b)\left|\right\},\end{array}$$

(2.7)

where we have used the fact that

$$\begin{array}{r}{\int }_0^1{t}^2{(1-t)}^{s}dt={\int }_0^1{(t-1)}^2{t}^{s}dt=\frac{2}{(s+1)(s+2)(s+3)},\\ {\int }_0^1{t}^{s+2}dt={\int }_0^1{(1-t)}^{s+2}dt=\frac{1}{s+3}.\end{array}$$

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

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

(2.8)

Combining (2.7) and (2.8), we have

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

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

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

$$\begin{array}{l}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_a^{b}f(x)dx|\\ \le \frac{{(b-a)}^2}{192}\left\{\right|f^{\mathrm{\prime }\mathrm{\prime }}(a)|+6|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right)|+|f^{\mathrm{\prime }\mathrm{\prime }}(b)\left|\right\}\\ \le \frac{{(b-a)}^2}{48}\left\{\right|f^{\mathrm{\prime }\mathrm{\prime }}(a)|+|f^{\mathrm{\prime }\mathrm{\prime }}(b)\left|\right\}.\end{array}$$

(2.9)

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

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

$$\begin{array}{l}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_a^{b}f(x)dx|\\ \le \frac{{(b-a)}^2}{16}{\left(\frac{1}{2p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{s+1}\right)}^{\frac{1}{q}}\\ \times [{\left(\right|f^{\mathrm{\prime }\mathrm{\prime }}(a){|}^q+\left|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^q\right)}^{\frac{1}{q}}+{\left(\right|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^q+|f^{\mathrm{\prime }\mathrm{\prime }}(b){|}^q)}^{\frac{1}{q}}].\end{array}$$

(2.10)

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

$$\begin{array}{r}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_a^{b}f(x)dx|\\ \le \frac{{(b-a)}^2}{16}\left[{\int }_0^1{t}^2\right|f^{\mathrm{\prime }\mathrm{\prime }}(t\frac{a+b}{2}+(1-t)a)|dt\\ +{\int }_0^1{(t-1)}^2\left|f^{\mathrm{\prime }\mathrm{\prime }}(tb+(1-t)\frac{a+b}{2})\right|dt]\\ \le \frac{{(b-a)}^2}{16}{\left({\int }_0^1{t}^{2p}dt\right)}^{\frac{1}{p}}{\left({\int }_0^1\right|f^{\mathrm{\prime }\mathrm{\prime }}(t\frac{a+b}{2}+(1-t)a){|}^{q}dt)}^{\frac{1}{q}}\\ +\frac{{(b-a)}^2}{16}{\left({\int }_0^1{(t-1)}^{2p}dt\right)}^{\frac{1}{p}}{\left({\int }_0^1\right|f^{\mathrm{\prime }\mathrm{\prime }}(tb+(1-t)\frac{a+b}{2}){|}^{q}dt)}^{\frac{1}{q}}.\end{array}$$

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

$${\int }_0^1|f^{\mathrm{\prime }\mathrm{\prime }}(t\frac{a+b}{2}+(1-t)a){|}^{q}dt\le \frac{1}{s+1}\left\{\right|f^{\mathrm{\prime }\mathrm{\prime }}(a){|}^q+\left|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^q\right\}$$

and

$${\int }_0^1|f^{\mathrm{\prime }\mathrm{\prime }}(tb+(1-t)\frac{a+b}{2}){|}^{q}dt\le \frac{1}{s+1}\left\{\right|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^q+|f^{\mathrm{\prime }\mathrm{\prime }}(b){|}^q\}.$$

By a simple computation,

$${\int }_0^1{t}^{2p}dt=\frac{1}{2p+1}$$

and

$${\int }_0^1{(t-1)}^{2p}dt={\int }_0^1{(1-t)}^{2p}dt=\frac{1}{2p+1}.$$

Therefore, we have

$$\begin{array}{r}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_a^{b}f(x)dx|\\ \le \frac{{(b-a)}^2}{16}{\left(\frac{1}{2p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{s+1}\right)}^{\frac{1}{q}}\\ \times [{\left(\right|f^{\mathrm{\prime }\mathrm{\prime }}(a){|}^q+\left|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^q\right)}^{\frac{1}{q}}+{\left(\right|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^q+|f^{\mathrm{\prime }\mathrm{\prime }}(b){|}^q)}^{\frac{1}{q}}].\end{array}$$

This completes the proof. □

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

$$\begin{array}{r}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_a^{b}f(x)dx|\\ \le \frac{{(b-a)}^2}{16}{\left(\frac{1}{2p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{s+1}\right)}^{\frac{2}{q}}\\ \times \{2^{\frac{1-s}{q}}+{(2^{1-s}+s+1)}^{\frac{1}{q}}\}\left[\right|f^{\mathrm{\prime }\mathrm{\prime }}(a)|+|f^{\mathrm{\prime }\mathrm{\prime }}(b)\left|\right].\end{array}$$

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

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

Therefore

$$\begin{array}{r}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_a^{b}f(x)dx|\\ \le \frac{{(b-a)}^2}{16}{\left(\frac{1}{2p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{s+1}\right)}^{\frac{2}{q}}\\ \times [{\left(\{2^{1-s}+s+1\}\right|f^{\mathrm{\prime }\mathrm{\prime }}(a){|}^q+2^{1-s}|f^{\mathrm{\prime }\mathrm{\prime }}(b){|}^q)}^{\frac{1}{q}}\\ +{\left(2^{1-s}\right|f^{\mathrm{\prime }\mathrm{\prime }}(a){|}^q+\{2^{1-s}+s+1\}|f^{\mathrm{\prime }\mathrm{\prime }}(b){|}^q)}^{\frac{1}{q}}].\end{array}$$

We let $a_1=(2^{1-s}+s+1)|f^{\mathrm{\prime }\mathrm{\prime }}(a){|}^q$, $b_1=2^{1-s}|f^{\mathrm{\prime }\mathrm{\prime }}(b){|}^q$, $a_2=2^{1-s}|f^{\mathrm{\prime }\mathrm{\prime }}(a){|}^q$ and $b_2=(2^{1-s}+s+1)|f^{\mathrm{\prime }\mathrm{\prime }}(b){|}^q$. Here, $0<1/q<1$ for $q>1$. Using the fact

$$\sum _{i=1}^n{(a_i+b_i)}^r\le \sum _{i=1}^n{a}_i^r+\sum _{i=1}^n{b}_i^r$$

for $0<r<1$, $a_1,a_2,\dots ,a_n\ge 0$ and $b_1,b_2,\dots ,b_n\ge 0$, we obtain the inequalities

$$\begin{array}{r}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_a^{b}f(x)dx|\\ \le \frac{{(b-a)}^2}{16}{\left(\frac{1}{2p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{s+1}\right)}^{\frac{2}{q}}\\ \times [{\left(\{2^{1-s}+s+1\}\right|f^{\mathrm{\prime }\mathrm{\prime }}(a){|}^q+2^{1-s}|f^{\mathrm{\prime }\mathrm{\prime }}(b){|}^q)}^{\frac{1}{q}}\\ +{\left(2^{1-s}\right|f^{\mathrm{\prime }\mathrm{\prime }}(a){|}^q+\{2^{1-s}+s+1\}|f^{\mathrm{\prime }\mathrm{\prime }}(b){|}^q)}^{\frac{1}{q}}]\\ \le \frac{{(b-a)}^2}{16}{\left(\frac{1}{2p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{s+1}\right)}^{\frac{2}{q}}\\ \times \{2^{\frac{1-s}{q}}+{(2^{1-s}+s+1)}^{\frac{1}{q}}\}\left[\right|f^{\mathrm{\prime }\mathrm{\prime }}(a)|+|f^{\mathrm{\prime }\mathrm{\prime }}(b)\left|\right].\end{array}$$

 □

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

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

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

$$\begin{array}{r}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_a^{b}f(x)dx|\\ \le \frac{{(b-a)}^2}{16}\left[{\int }_0^1{t}^2\right|f^{\mathrm{\prime }\mathrm{\prime }}(t\frac{a+b}{2}+(1-t)a)|dt\\ +{\int }_0^1{(t-1)}^2\left|f^{\mathrm{\prime }\mathrm{\prime }}(tb+(1-t)\frac{a+b}{2})\right|dt]\\ \le \frac{{(b-a)}^2}{16}{\left({\int }_0^1{t}^{2}dt\right)}^{\frac{1}{p}}{\left({\int }_0^1{t}^2\right|f^{\mathrm{\prime }\mathrm{\prime }}(t\frac{a+b}{2}+(1-t)a){|}^{q}dt)}^{\frac{1}{q}}\\ +\frac{{(b-a)}^2}{16}{\left({\int }_0^1{(t-1)}^{2}dt\right)}^{\frac{1}{p}}{\left({\int }_0^1{(t-1)}^2\right|f^{\mathrm{\prime }\mathrm{\prime }}(tb+(1-t)\frac{a+b}{2}){|}^{q}dt)}^{\frac{1}{q}}.\end{array}$$

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

$${\int }_0^1{t}^2|f^{\mathrm{\prime }\mathrm{\prime }}(t\frac{a+b}{2}+(1-t)a){|}^{q}dt\le \frac{2}{(s+1)(s+2)(s+3)}|f^{\mathrm{\prime }\mathrm{\prime }}(a){|}^q+\frac{1}{s+3}|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^q$$

and

$$\begin{array}{c}{\int }_0^1{(t-1)}^2|f^{\mathrm{\prime }\mathrm{\prime }}(tb+(1-t)\frac{a+b}{2}){|}^{q}dt\hfill \\ \le \frac{1}{s+3}|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^q+\frac{2}{(s+1)(s+2)(s+3)}|f^{\mathrm{\prime }\mathrm{\prime }}(b){|}^q.\hfill \end{array}$$

Therefore, we have

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

 □

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

$$\begin{array}{l}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_a^{b}f(x)dx|\\ \le \frac{{(b-a)}^2}{48}{\left(\frac{3}{4}\right)}^{\frac{1}{q}}\{{(\frac{|f^{\mathrm{\prime }\mathrm{\prime }}(a){|}^q}{3}+|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^q)}^{\frac{1}{q}}\\ +{\left(\right|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+b}{2}\right){|}^q+\frac{|f^{\mathrm{\prime }\mathrm{\prime }}(b){|}^q}{3})}^{\frac{1}{q}}\}.\end{array}$$

(2.11)

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

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

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

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

$$\begin{array}{l}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_a^{b}f(x)dx|\\ \le \frac{{(b-a)}^2}{16}\left[{\int }_0^1{t}^2\right|f^{\mathrm{\prime }\mathrm{\prime }}(t\frac{a+b}{2}+(1-t)a)|dt+{\int }_0^1{(t-1)}^2|f^{\mathrm{\prime }\mathrm{\prime }}(tb+(1-t)\frac{a+b}{2})\left|dt\right]\\ \le \frac{{(b-a)}^2}{16}{\left({\int }_0^1{t}^{2p}dt\right)}^{\frac{1}{p}}{\left({\int }_0^1\right|f^{\mathrm{\prime }\mathrm{\prime }}(t\frac{a+b}{2}+(1-t)a){|}^{q}dt)}^{\frac{1}{q}}\\ +\frac{{(b-a)}^2}{16}{\left({\int }_0^1{(t-1)}^{2p}dt\right)}^{\frac{1}{p}}{\left({\int }_0^1\right|f^{\mathrm{\prime }\mathrm{\prime }}(tb+(1-t)\frac{a+b}{2}){|}^{q}dt)}^{\frac{1}{q}}.\end{array}$$

(2.12)

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

$${\int }_0^1|f^{\mathrm{\prime }\mathrm{\prime }}(t\frac{a+b}{2}+(1-t)a){|}^{q}dt\le 2^{s-1}|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{3a+b}{4}\right){|}^q$$

(2.13)

and

$${\int }_0^1|f^{\mathrm{\prime }\mathrm{\prime }}(tb+(1-t)\frac{a+b}{2}){|}^{q}dt\le 2^{s-1}|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+3b}{4}\right){|}^q.$$

(2.14)

From (2.12)-(2.14), we get

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

which completes the proof. □

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

$$\begin{array}{l}|f\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_a^{b}f(x)dx|\\ \le \frac{{(b-a)}^2}{16}\left\{\right|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{3a+b}{4}\right)|+|f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{a+3b}{4}\right)\left|\right\}.\end{array}$$

(2.15)

3 Applications to special means

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

• (1) Arithmetic mean:

  $$A(\alpha ,\beta )=\frac{\alpha +\beta }{2},\alpha ,\beta \in {\mathbb{R}}^{+};$$

• (2) Logarithmic mean:

  $$L(\alpha ,\beta )=\frac{\alpha -\beta }{ln|\alpha |-ln|\beta |},|\alpha |\ne |\beta |,\alpha ,\beta \ne 0,\alpha ,\beta \in {\mathbb{R}}^{+};$$

Generalized log-mean:

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

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

Proposition 1 Let $0<a<b$ and $s\in (0,1)$. Then we have

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

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

Proposition 2 Let $0<a<b$ and $s\in (0,1)$. Then we have

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

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

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

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

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