# Remarks on some inequalities for s-convex functions and applications

## Abstract

Some new inequalities for functions whose second derivatives in absolute value at certain powers are s-convex in the second sense are established. Two mistakes in a recently published paper are pointed out and corrected.

## 1 Introduction

It is well known that a function $$f:I\to\mathbf{R}$$, $$\emptyset\neq I\subset\mathbf{R}$$ is called convex if

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

holds for all $$x,y\in I$$ and $$\lambda\in[0,1]$$. If $$-f:I\to\mathbf{R}$$ is convex, then we say that $$f:I\to\mathbf{R}$$ is concave.

In [1], the class of s-convex function in the second sense is defined in the following way: a function $$f:[0,\infty)\to\mathbf{R}$$ is said to be s-convex in the second sense if

$$f\bigl(\lambda x+(1-\lambda)y\bigr)\le\lambda^{s}f(x)+(1- \lambda)^{s}f(y)$$

holds for all $$x,y\in[0,\infty)$$, $$\lambda\in[0,1]$$ 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 ordinary convexity of functions defined on $$[0,\infty)$$. It is proved in [1] that all functions from $$K_{s}^{2}, s\in(0,1)$$ are nonnegative. Similarly, a function $$f:[0,\infty)\to\mathbf{R}$$ is said to be s-concave in the second sense for some fixed $$s\in(0,1]$$ if $$-f\in K_{s}^{2}$$. Thus we can conclude that an s-concave function is always nonpositive for any $$s\in(0,1)$$.

### Example 1

[1]

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

$$f(t)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} a, &t=0,\\ bt^{s}+c,&t>0. \end{array}\displaystyle \right .$$

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 $$f\notin K_{s}^{2}$$.

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

In a recent paper [2], Özdemir et al. proved the following inequalities for functions whose second derivatives in absolute value at certain powers are s-convex in the second sense.

### Theorem 1

([2], Theorem 2)

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

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl(\frac{a+b}{2}\biggr)\biggr| \\ &\quad\le\frac{(b-a)^{2}}{8(s+1)(s+2)(s+3)}\biggl[\bigl|f''(a)\bigr|+(s+1) (s+2) \biggl|f''\biggl(\frac{a+b}{2}\biggr)\biggr|+\bigl|f''(b)\bigr| \biggr] \\ &\quad\le\frac{[1+(s+2)2^{1-s}](b-a)^{2}}{8(s+1)(s+2)(s+3)} \bigl[\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr| \bigr]. \end{aligned}
(1)

### Theorem 2

([2], Theorem 4)

Let $$I\subset[0,\infty)$$, $$f:I\to\mathbf{R}$$ be a twice differentiable function on $$I^{\circ}$$ such that $$f''\in L^{1}[a,b]$$, where $$a,b\in I$$ with $$a< b$$. If $$|f''|^{q}$$ is s-convex in the second sense on $$[a,b]$$ for some fixed $$s\in(0,1]$$ and $$q\ge 1$$, then the following inequality holds:

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl(\frac{a+b}{2}\biggr)\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl(\frac{1}{3}\biggr)^{1-\frac{1}{q}}\biggl\{ \biggl(\frac{2}{(s+1)(s+2)(s+3)}\bigl|f''(a)\bigr|^{q}+ \frac{1}{s+3} \biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}\biggr)^{\frac{1}{q}} \\ &\qquad{}+\biggl(\frac{1}{s+3}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q}+\frac{2}{(s+1)(s+2)(s+3)} \bigl|f''(b)\bigr|^{q} \biggr)^{\frac{1}{q}}\biggr\} . \end{aligned}
(2)

However, it is a pity that Theorem 5 in [2] is not valid since nonnegative $$|f''|^{q}$$ could not be an s-concave function for any fixed $$s\in(0,1)$$ which has been mentioned in [3], and it is the Hölder inequality but not the power mean inequality that has been used in proving Theorem 4 of [2].

In this work, we will first derive a new general inequality for functions whose second derivatives in absolute value at certain powers are s-convex in the second sense, which not only provides generalization of Theorem 1 and Theorem 2 but also gives some other interesting special results. Then we establish another new general inequality for functions whose second derivatives in absolute value at certain powers are s-convex in the second sense, which also gives some interesting special results. Finally, applications to some special means of real numbers are considered.

## 2 Main results

We first provide a new general inequality for functions whose second derivatives in absolute value at certain powers are s-convex in the second sense, and so we need the following lemma.

### Lemma 1

Let $$I\subset\mathbf{R}$$, $$f:I\to\mathbf{R}$$ be a twice differentiable function on $$I^{\circ}$$ such that $$f''\in L^{1}[a,b]$$, where $$a,b\in I$$ with $$a< b$$. Then, for any $$\theta\in[0,1]$$, the following equality holds:

\begin{aligned} &\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-(1-\theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2} \\ &\quad=\frac{(b-a)^{2}}{16}\biggl[\int_{0}^{1} \bigl(t^{2}-2\theta t\bigr)f''\biggl(t \frac{a+b}{2}+(1-t)a\biggr)\,dt \\ &\qquad{}+\int_{0}^{1} \bigl(t^{2}-2\theta t\bigr)f''\biggl(t \frac{a+b}{2}+(1-t)b\biggr)\,dt\biggr]. \end{aligned}
(3)

### Proof

Integrating by parts, we have the following identity:

\begin{aligned} I_{1} =&\int_{0}^{1} \bigl(t^{2}-2\theta t\bigr)f''\biggl(t \frac{a+b}{2}+(1-t)a\biggr)\,dt \\ =&\bigl(t^{2}-2\theta t\bigr)\frac{2}{b-a}f' \biggl(t\frac{a+b}{2}+(1-t)a\biggr)\bigg|_{0}^{1}- \frac{4}{b-a} \int_{0}^{1}(t- \theta)f'\biggl(t\frac{a+b}{2}+(1-t)a\biggr)\,dt \\ =&\frac{2(1-2\theta)}{b-a}f'\biggl(\frac{a+b}{2}\biggr) -\frac{4}{b-a}\biggl[\frac{2(t-\theta)}{b-a}f\biggl(t\frac{a+b}{2}+(1-t)a \biggr)\bigg|_{0}^{1} \\ &{}-\frac{2}{b-a}\int_{0}^{1}f\biggl(t \frac{a+b}{2}+(1-t)a\biggr)\,dt\biggr] \\ =&\frac{2(1-2\theta)}{b-a}f'\biggl(\frac{a+b}{2}\biggr)- \frac{8(1-\theta)}{ (b-a)^{2}}f\biggl(\frac{a+b}{2}\biggr)-\frac{8\theta}{(b-a)^{2}}f(a) \\ &{}+\frac{8}{(b-a)^{2}}\int_{0}^{1}f\biggl(t \frac{a+b}{2}+(1-t)a\biggr)\,dt. \end{aligned}
(4)

Using the change of variable $$x=t\frac{a+b}{2}+(1-t)a$$ for $$t\in[0,1]$$ and multiplying both sides of (4) by $$\frac{(b-a)^{2}}{16}$$, we obtain

\begin{aligned} &\frac{(b-a)^{2}}{16}\int_{0}^{1} \bigl(t^{2}-2\theta t\bigr)f''\biggl(t \frac{a+b}{2}+(1-t)a\biggr)\,dt \\ &\quad=\frac{b-a}{8}(1-2\theta)f'\biggl(\frac{a+b}{2} \biggr)-\frac{1-\theta}{2} f\biggl(\frac{a+b}{2}\biggr)-\frac{\theta}{2}f(a)+ \frac{1}{b-a}\int_{a}^{\frac{a+b}{2}}f(x)\,dx. \end{aligned}
(5)

Similarly, we observe that

\begin{aligned} &\frac{(b-a)^{2}}{16}\int_{0}^{1} \bigl(t^{2}-2\theta t\bigr)f''\biggl(t \frac{a+b}{2}+(1-t)b\biggr)\,dt \\ &\quad=-\frac{b-a}{8}(1-2\theta)f'\biggl(\frac{a+b}{2} \biggr) -\frac{1-\theta}{2}f\biggl(\frac{a+b}{2}\biggr)-\frac{\theta}{2}f(b)+ \frac{1}{b-a}\int_{\frac{a+b}{2}}^{b}f(x)\,dx. \end{aligned}
(6)

Thus, adding (5) and (6), we get the required identity (3). □

### Theorem 3

Let $$I\subset[0,\infty)$$, $$f:I\to\mathbf{R}$$ be a twice differentiable function on $$I^{\circ}$$ such that $$f''\in L^{1}[a,b]$$, where $$a,b\in I$$ with $$a< b$$. If $$|f''|^{q}$$ is s-convex in the second sense on $$[a,b]$$ for some fixed $$s\in(0,1]$$ and $$q\ge 1$$, then the following inequalities hold:

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(x)\,dx-(1-\theta)f\biggl(\frac{a+b}{2}\biggr) -\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl(\frac{8\theta^{3}-3\theta+1}{3}\biggr)^{1-\frac{1}{q}}\biggl\{ \biggl[\frac{2(2\theta)^{s+3}-2(s+3)\theta+s+2}{(s+2)(s+3)}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\frac{4(1-2\theta)^{s+2}[(s+1)\theta+1]+2(s+3)\theta-2}{(s+1)(s+2)(s+3)} \bigl|f''(a)\bigr|^{q} \biggr]^{\frac{1}{q}} \\ &\qquad{}+\biggl[\frac{2(2\theta)^{s+3}-2(s+3)\theta+s+2}{(s+2)(s+3)}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\frac{4(1-2\theta)^{s+2}[(s+1)\theta+1] +2(s+3)\theta-2}{(s+1)(s+2)(s+3)}\bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}\biggr\} \end{aligned}
(7)

for $$0\le\theta\le\frac{1}{2}$$ and

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(x)\,dx-(1-\theta) f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl(\theta-\frac{1}{3}\biggr)^{1-\frac{1}{q}}\biggl\{ \biggl[\frac{2(s+3)\theta-s-2}{(s+2)(s+3)}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\frac{2(s+3)\theta-2}{(s+1)(s+2)(s+3)}\bigl|f''(a)\bigr|^{q} \biggr]^{\frac{1}{q}} +\biggl[\frac{2(s+3)\theta-s-2}{(s+2)(s+3)}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\frac{2(s+3)\theta-2}{(s+1)(s+2)(s+3)}\bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}\biggr\} \end{aligned}
(8)

for $$\frac{1}{2}\le\theta\le 1$$.

### Proof

In case $$0\le\theta\le\frac{1}{2}$$, by Lemma 1 and using the Hölder inequality, we have

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-(1- \theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl[\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\biggl|f''\biggl(t\frac{a+b}{2}+(1-t)a\biggr)\biggr|\,dt \\ &\qquad{}+\int_{0}^{1}\bigl|t^{2}-2\theta t\bigr|\biggl|f''\biggl(t\frac{a+b}{2}+(1-t)b\biggr)\biggr|\,dt \biggr] \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl[\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\,dt\biggr)^{1-\frac{1}{q}}\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\biggl|f''\biggl(t\frac{a+b}{2}+(1-t)a \biggr)\biggr|^{q} \,dt\biggr)^{\frac{1}{q}} \\ &\qquad{}+\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\,dt\biggr)^{1-\frac{1}{q}}\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\biggl|f''\biggl(t\frac{a+b}{2}+(1-t)b \biggr)\biggr|^{q} \,dt\biggr)^{\frac{1}{q}}\biggr] \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl[\biggl(\frac{8\theta^{3}-3\theta+1}{3}\biggr) ^{1-\frac{1}{q}} \biggl(\int_{0}^{1}\bigl|t^{2}-2\theta t\bigr|\biggl(t^{s}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q}\\ &\biggl.\qquad{}+(1-t)^{s}\bigl|f''(a)\bigr|^{q} \biggr)\,dt\biggr)^{\frac{1}{q}} \\ &\qquad{}+\biggl(\frac{8\theta^{3}-3\theta+1}{3}\biggr)^{1-\frac{1}{q}}\biggl(\int _{0}^{1}\bigl|t^{2}-2\theta t\bigr|\biggl(t^{s}\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}+(1-t)^{s}\bigl|f''(b)\bigr|^{q} \biggr)\,dt\biggr)^{\frac{1}{q}}\biggr] \\ &\quad=\frac{(b-a)^{2}}{16}\biggl[\biggl(\frac{8\theta^{3}-3\theta+1}{3}\biggr)^{1-\frac{1}{q}}\biggl(\int_{0}^{2\theta}t(2\theta-t) \biggl(t^{s}\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}\\ &\qquad{}+(1-t)^{s}\bigl|f''(a)\bigr|^{q} \biggr)\,dt +\int_{2\theta}^{1}t(t-2\theta) \biggl(t^{s}\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}+(1-t)^{s}\bigl|f''(a)\bigr|^{q} \biggr)\,dt\biggr)^{\frac{1}{q}} \\ &\qquad{}+\biggl(\frac{8\theta^{3}-3\theta+1}{3}\biggr)^{1-\frac{1}{q}}\biggl(\int _{0}^{2\theta}t(2\theta-t) \biggl(t^{s}\biggl|f'' \biggl(\frac{a+b}{2}\biggr)\biggr|^{q}+(1-t)^{s}\bigl|f''(b)\bigr|^{q} \biggr)\,dt \\ &\qquad{}+\int_{2\theta}^{1}t(t-2\theta) \biggl(t^{s}\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}+(1-t)^{s}\bigl|f''(b)\bigr|^{q} \biggr)\,dt\biggr)^{\frac{1}{q}}\biggr] \\ &\quad=\frac{(b-a)^{2}}{16}\biggl(\frac{8\theta^{3}-3\theta+1}{3}\biggr)^{1-\frac{1}{q}} \biggl\{ \biggl[\frac{2(2\theta)^{s+3}-2(s+3)\theta+s+2}{(s+2)(s+3)} \biggl|f''\biggl(\frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\frac{4(1-2\theta)^{s+2}[(s+1)\theta+1]+2(s+3) \theta-2}{(s+1)(s+2)(s+3)}\bigl|f''(a)\bigr|^{q} \biggr]^{\frac{1}{q}} \\ &\qquad{}+\biggl[\frac{2(2\theta)^{s+3}-2(s+3)\theta+s+2}{(s+2)(s+3)}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\frac{4(1-2\theta)^{s+2}[(s+1)\theta+1]+2(s+3) \theta-2}{(s+1)(s+2)(s+3)}\bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}\biggr\} , \end{aligned}

where

\begin{aligned}& \int_{0}^{1}\bigl|t^{2}-2\theta t\bigr|\,dt=\int _{0}^{2\theta}t(2\theta-t)\,dt+\int_{2\theta}^{1}t(t-2 \theta)\,dt =\frac{8\theta^{3}-3\theta+1}{3}, \\& \int_{0}^{2\theta}t^{s+1}(2\theta-t)\,dt= \frac{(2\theta)^{s+3}}{(s+2)(s+3)}, \\& \int_{2\theta}^{1}t^{s+1}(t-2\theta)\,dt= \frac{(2\theta)^{s+3}-2(s+3)\theta+s+2}{(s+2)(s+3)}, \\& \int_{0}^{2\theta}t(1-t)^{s}(2\theta-t)\,dt= \frac{(1-2\theta)^{s+2}[2(s+1)\theta+2]+2(s+3)\theta-2}{(s+1)(s+2)(s+3)}, \end{aligned}

and

$$\int_{2\theta}^{1}t(1-t)^{s}(t-2\theta)\,dt= \frac{(1-2\theta)^{s+2}[2(s+1)\theta+2]}{(s+1)(s+2)(s+3)}.$$

In case $$\frac{1}{2}\le\theta\le 1$$, by Lemma 1 and using the Hölder inequality, we have

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-(1- \theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl[\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\biggl|f''\biggl(t\frac{a+b}{2}+(1-t)a\biggr)\biggr|\,dt \\ &\qquad{}+\int_{0}^{1}\bigl|t^{2}-2\theta t\bigr|\biggl|f''\biggl(t\frac{a+b}{2}+(1-t)b\biggr)\biggr|\,dt \biggr] \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl[\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\,dt\biggr)^{1-\frac{1}{q}}\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\biggl|f''\biggl(t\frac{a+b}{2}+(1-t)a \biggr)\biggr|^{q} \,dt\biggr)^{\frac{1}{q}} \\ &\qquad{}+\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\,dt\biggr)^{1-\frac{1}{q}}\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\biggl|f''\biggl(t\frac{a+b}{2}+(1-t)b \biggr)\biggr|^{q} \,dt\biggr)^{\frac{1}{q}}\biggr] \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl\{ \biggl(\theta-\frac{1}{3} \biggr)^{1-\frac{1}{q}}\biggl(\int_{0}^{1}\bigl|t^{2}-2 \theta t\bigr|\biggl[t^{s}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q}+(1-t)^{s}\bigl|f''(a)\bigr|^{q} \biggr]\,dt\biggr)^{\frac{1}{q}} \\ &\qquad{}+\biggl(\theta-\frac{1}{3}\biggr)^{1-\frac{1}{q}}\biggl(\int _{0}^{1}\bigl|t^{2}-2\theta t\bigr|\biggl[t^{s}\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}+(1-t)^{s}\bigl|f''(b)\bigr|^{q} \biggr]\,dt\biggr)^{\frac{1}{q}}\biggr\} \\ &\quad=\frac{(b-a)^{2}}{16}\biggl(\theta-\frac{1}{3}\biggr)^{1-\frac{1}{q}}\biggl\{ \biggl[\biggl(\int_{0}^{1}t^{s+1}(2 \theta-t)\,dt\biggr)\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}\\ &\qquad{}+\biggl(\int_{0}^{1}t(1-t)^{s}(2 \theta-t)\,dt\biggr)\bigl|f''(a)\bigr|^{q} \biggr]^{\frac{1}{q}} \\ &\qquad{}+\biggl[\biggl(\int_{0}^{1}t^{s+1}(2 \theta-t)\,dt\biggr)\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}+\biggl(\int_{0}^{1}t(1-t)^{s}(2 \theta-t)\,dt\biggr)\bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}\biggr\} \\ &\quad=\frac{(b-a)^{2}}{16}\biggl(\theta-\frac{1}{3}\biggr)^{1-\frac{1}{q}}\biggl\{ \biggl[\frac{2(s+3)\theta-s-2}{(s+2)(s+3)}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\frac{2(s+3)\theta-2}{(s+1)(s+2)(s+3)}\bigl|f''(a)\bigr|^{q} \biggr]^{\frac{1}{q}} +\biggl[\frac{2(s+3)\theta-s-2}{(s+2)(s+3)}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\frac{2(s+3)\theta-2}{(s+1)(s+2)(s+3)}\bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}\biggr\} , \end{aligned}

where

\begin{aligned}& \int_{0}^{1}\bigl|t^{2}-2\theta t\bigr|\,dt=\int _{0}^{1}t(2\theta-t)\,dt=\theta-\frac{1}{3}, \\& \int_{0}^{1}t^{s+1}(2\theta-t)\,dt= \frac{2(s+3)\theta-s-2}{(s+2)(s+3)}, \end{aligned}

and

$$\int_{0}^{1}t(1-t)^{s}(2\theta-t)\,dt= \frac{2(s+3)\theta-2}{(s+1)(s+2)(s+3)}.$$

The proof is thus completed. □

### Remark 1

If we take $$\theta=0$$ in (7), then we get a midpoint type inequality

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl(\frac{a+b}{2}\biggr)\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl(\frac{1}{3}\biggr)^{1-\frac{1}{q}} \biggl\{ \biggl[\frac{1}{s+3}\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q} +\frac{2}{(s+1)(s+2)(s+3)}\bigl|f''(a)\bigr|^{q} \biggr]^{\frac{1}{q}} \\ &\qquad{}+\biggl[\frac{1}{s+3}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q}+\frac{2}{(s+1)(s+2)(s+3)} \bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}\biggr\} . \end{aligned}
(9)

If we take $$\theta=1$$ in (8), then we get a trapezoid type inequality

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl(\frac{2}{3}\biggr)^{1-\frac{1}{q}} \biggl\{ \biggl[\frac{s+4}{(s+2)(s+3)}\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|^{q}+\frac{2}{(s+1)(s+3)} \bigl|f''(a)\bigr|^{q} \biggr]^{\frac{1}{q}} \\ &\qquad{}+\biggl[\frac{s+4}{(s+2)(s+3)}\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q}+\frac{2}{(s+1)(s+3)} \bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}\biggr\} . \end{aligned}
(10)

If we take $$\theta=\frac{1}{3}$$ in (7), then we get a Simpson type inequality

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{6}\biggl[f(a)+4f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl(\frac{8}{81}\biggr)^{1-\frac{1}{q}} \biggl\{ \biggl[\biggl(\frac{s}{3(s+2)(s+3)}+\frac{2}{(s+2)(s+3)}\biggl(\frac{2}{s+3} \biggr)^{s+3}\biggr)\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+\biggl(\frac{4s+16}{(s+1)(s+2)(s+3)}\biggl(\frac{1}{3}\biggr)^{s+3}+ \frac{2s}{3(s+1)(s+2)(s+3)}\biggr)\bigl|f''(a)\bigr|^{q} \biggr]^{\frac{1}{q}} \\ &\qquad{}\times\biggl[\biggl(\frac{s}{3(s+2)(s+3)}+\frac{2}{(s+2)(s+3)}\biggl( \frac{2}{s+3}\biggr)^{s+3}\biggr)\biggl|f'' \biggl(\frac{a+b}{2}\biggr)\biggr|^{q} \\ &\qquad{}+ \biggl(\frac{4s+16}{(s+1)(s+2)(s+3)}\biggl(\frac{1}{3} \biggr)^{s+3}+\frac{2s}{3(s+1) (s+2)(s+3)}\biggr)\bigl|f''(b)\bigr|^{q} \biggr]^{\frac{1}{q}}\biggr\} . \end{aligned}
(11)

If we take $$\theta=\frac{1}{2}$$ in (7) or (8), then we get an averaged midpoint-trapezoid type inequality

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{4}\biggl[f(a)+2f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le\frac{(b-a)^{2}}{16}\biggl(\frac{1}{6}\biggr)^{1-\frac{1}{q}} \biggl\{ \biggl[\frac{1}{(s+2)(s+3)}\biggl(\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q}+\bigl|f''(a)\bigr|^{q} \biggr)\biggr]^{\frac{1}{q}} \\ &\qquad{}+\biggl[\frac{1}{(s+2)(s+3)}\biggl(\biggl|f''\biggl( \frac{a+b}{2}\biggr)\biggr|^{q}+\bigl|f''(b)\bigr|^{q} \biggr)\biggr]^{\frac{1}{q}}\biggr\} . \end{aligned}
(12)

### Corollary 1

Let $$I\subset[0,\infty)$$, $$f:I\to\mathbf{R}$$ be a twice differentiable function on $$I^{\circ}$$ such that $$f''\in L^{1}[a,b]$$, where $$a,b\in I$$ with $$a< b$$. If $$|f''|$$ is s-convex in the second sense on $$[a,b]$$ for some fixed $$s\in(0,1]$$, then the following inequalities hold:

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(x)\,dx-(1-\theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le \frac{(b-a)^{2}}{8}\biggl[\frac{2(2\theta)^{s+3}-2(s+3)\theta+s+2}{(s+2)(s+3)}\biggl|f'' \biggl(\frac{a+b}{2}\biggr)\biggr| \\ &\qquad{}+\frac{2(1-2\theta)^{s+2}[(s+1)\theta+1] +(s+3)\theta-1}{(s+1)(s+2)(s+3)}\bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr| \bigr)\biggr] \end{aligned}
(13)

for $$0\le\theta\le\frac{1}{2}$$ and

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(x)\,dx-(1-\theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{8}\biggl[\frac{2(s+3)\theta-s-2}{(s+2)(s+3)}\biggl|f'' \biggl(\frac{a+b}{2}\biggr)\biggr| \\ &\qquad{}+\frac{(s+3)\theta-1}{(s+1)(s+2)(s+3)}\bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr| \bigr)\biggr] \end{aligned}
(14)

for $$\frac{1}{2}\le\theta\le 1$$.

### Proof

Inequalities (13) and (14) are immediate by setting $$q=1$$ in (7) and (8) of Theorem 3. □

### Remark 2

If we take $$\theta=0$$ in (13), then we get a midpoint type inequality

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl(\frac{a+b}{2}\biggr)\biggr| \\ &\quad\le\frac{(b-a)^{2}}{8}\biggl[ \frac{1}{s+3}\biggl|f''\biggl(\frac{a+b}{2} \biggr)\biggr|+\frac{1}{(s+1)(s+2)(s+3)}\bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr| \bigr)\biggr]. \end{aligned}
(15)

If we take $$\theta=1$$ in (14), then we get a trapezoid type inequality

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{8}\biggl[\frac{s+4}{(s+2)(s+3)}\biggl|f'' \biggl(\frac{a+b}{2}\biggr)\biggr|+\frac{1}{(s+1)(s+3)} \bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr|\bigr) \biggr]. \end{aligned}
(16)

If we take $$\theta=\frac{1}{3}$$ in (13), then we get a Simpson type inequality

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{6}\biggl[f(a)+4f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le\frac{(b-a)^{2}}{8}\biggl\{ \biggl[\frac{s}{3(s+2)(s+3)}+\frac{2}{(s+2)(s+3)} \biggl(\frac{2}{3}\biggr)^{s+3}\biggr]\biggl|f'' \biggl(\frac{a+b}{2}\biggr)\biggr| \\ &\qquad{}+\biggl[\frac{2s+8}{(s+1)(s+2)(s+3)}\biggl(\frac{1}{3}\biggr)^{s+3} +\frac{s}{3(s+1)(s+2)(s+3)}\biggr]\bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr| \bigr)\biggr\} . \end{aligned}
(17)

If we take $$\theta=\frac{1}{2}$$ in (13) or (14), then we get an averaged midpoint-trapezoid type inequality

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{4}\biggl[f(a)+2f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le\frac{(b-a)^{2}}{8}\biggl[\frac{1}{(s+2)(s+3)} \biggl|f'' \biggl(\frac{a+b}{2}\biggr)\biggr|+\frac{1}{2(s+2)(s+3)}\biggr]\bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr| \bigr). \end{aligned}
(18)

### Remark 3

If we put $$M=\sup_{x\in[a,b]}|f''|$$ in (15)-(18), then we have

\begin{aligned}& \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl( \frac{a+b}{2}\biggr)\biggr|\le \frac{M(s^{2}+3s+4)(b-a)^{2}}{8(s+1)(s+2)(s+3)}, \end{aligned}
(19)
\begin{aligned}& \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt- \frac{f(a)+f(b)}{2}\biggr|\le \frac{M(s^{2}+7s+8)(b-a)^{2}}{8(s+1)(s+2)(s+3)}, \end{aligned}
(20)
\begin{aligned}& \begin{aligned}[b] &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{6}\biggl[f(a)+4f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le \frac{M(b-a)^{2}}{24(s+1)(s+2)(s+3)}\biggl[s^{2}+3s+(4s+16) \biggl( \frac{1}{3}\biggr)^{s+2}+6(s+1) \biggl(\frac{2}{3} \biggr)^{s+3}\biggr], \end{aligned} \end{aligned}
(21)

and

$$\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{4}\biggl[f(a)+2f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \le \frac{M(b-a)^{2}}{4(s+2)(s+3)}.$$
(22)

### Remark 4

If we further take $$s=1$$ in (19)-(22), i.e., for functions f with convex $$|f''|$$, we have

\begin{aligned}& \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl( \frac{a+b}{2}\biggr)\biggr|\le\frac{M(b-a)^{2}}{24}, \end{aligned}
(23)
\begin{aligned}& \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt- \frac{f(a)+f(b)}{2}\biggr|\le\frac{M(b-a)^{2}}{12}, \end{aligned}
(24)
\begin{aligned}& \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt- \frac{1}{6}\biggl[f(a)+4f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr|\le \frac{M(b-a)^{2}}{81}, \end{aligned}
(25)
\begin{aligned}& \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt- \frac{1}{4}\biggl[f(a)+2f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr|\le \frac{M(b-a)^{2}}{48}. \end{aligned}
(26)

Obviously, (23)-(26) indicate that the Simpson type inequality has the best error estimation for functions f with convex $$|f''|$$.

Now we turn to establish another new general inequality for functions whose second derivatives in absolute value at certain powers are s-convex in the second sense, we need the following lemma.

### Lemma 2

Let $$I\subset\mathbf{R}$$, $$f:I\to\mathbf{R}$$ be a twice differentiable function on $$I^{\circ}$$ such that $$f''\in L^{1}[a,b]$$, where $$a,b\in I$$ with $$a< b$$. Then, for any $$\theta\in[0,1]$$, the following equality holds:

\begin{aligned} &\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-(1-\theta) f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2} \\ &\quad=(b-a)^{2}\int_{0}^{1}k(t)f'' \bigl(ta+(1-t)b\bigr)\,dt, \end{aligned}
(27)

where

$$k(t)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{1}{2}t(t-\theta), & 0\le t\le\frac{1}{2},\\ \frac{1}{2}(1-t)(1-\theta-t), & \frac{1}{2}\le t\le 1. \end{array}\displaystyle \right .$$

### Proof

See, e.g., Lemma 2 in [4]. □

### Theorem 4

Let $$I\subset[0,\infty)$$, $$f:I\to\mathbf{R}$$ be a twice differentiable function on $$I^{\circ}$$ such that $$f''\in L^{1}[a,b]$$, where $$a,b\in I$$ with $$a< b$$. If $$|f''|^{q}$$ is s-convex in the second sense on $$[a,b]$$ for some fixed $$s\in(0,1]$$ and $$q\ge 1$$, then the following inequalities hold:

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(x)\,dx-(1-\theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{2}\biggl(\frac{8\theta^{3}-3\theta+1}{24}\biggr)^{1-\frac{1}{q}}\biggl\{ \biggl[\frac{2\theta^{s+3}}{(s+2)(s+3)} \\ &\qquad{}+\frac{2(1-\theta)^{s+2}[(s+1)\theta+2]+(s+3)\theta-2}{(s+1)(s+2)(s+3)}-\frac{1-\theta}{2^{s+1}(s+1)(s+2)}\biggr] \\ &\qquad{}\times\bigl(\bigl|f''(a)\bigr|^{q}+\bigl|f''(b)\bigr|^{q} \bigr)\biggr\} ^{\frac{1}{q}} \end{aligned}
(28)

for $$0\le\theta\le\frac{1}{2}$$ and

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(x)\,dx-(1-\theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{2}\biggl(\frac{3\theta-1}{24}\biggr)^{1-\frac{1}{q}}\biggl\{ \biggl[\frac{(s+3)\theta-2}{(s+1)(s+2)(s+3)}+\frac{1-\theta}{2^{s+1}(s+1)(s+2)}\biggr] \\ &\qquad{}\times\bigl(\bigl|f''(a)\bigr|^{q}+\bigl|f''(b)\bigr|^{q} \bigr)\biggr\} ^{\frac{1}{q}} \end{aligned}
(29)

for $$\frac{1}{2}\le\theta\le 1$$.

### Proof

In case $$0\le\theta\le\frac{1}{2}$$, by Lemma 2 and using the Hölder inequality, we have

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-(1- \theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le(b-a)^{2}\int_{0}^{1}\bigl|k(t)\bigr|\bigl|f'' \bigl(ta+(1-t)b\bigr)\bigr|\,dt \\ &\quad\le(b-a)^{2}\biggl[\int_{0}^{1}\bigl|k(t)\bigr|\,dt\biggr]^{1-\frac{1}{q}} \biggl[\int_{0}^{1}\bigl|k(t)\bigr|\bigl|f'' \bigl(ta+(1-t)b\bigr)\bigr|^{q} \,dt\biggr]^{\frac{1}{q}} \\ &\quad\le(b-a)^{2}\biggl(\frac{8\theta^{3}-3\theta+1}{24}\biggr)^{1-\frac{1}{q}}\biggl(\int_{0}^{1}\bigl|k(t)\bigr|\bigl[t^{s}\bigl|f''(a)\bigr|^{q}+(1-t)^{s}\bigl|f''(b)\bigr|^{q} \bigr]\,dt\biggr)^{\frac{1}{q}} \\ &\quad=\frac{(b-a)}{2}\biggl(\frac{8\theta^{3}-3\theta+1}{24}\biggr)^{1-\frac{1}{q}} \\ &\qquad{}\times\biggl[\biggl(\int_{0}^{\frac{1}{2}}t^{s+1}|t-\theta|\,dt+\int_{\frac{1}{2}}^{1}t^{s}(1-t)|1-\theta-t|\,dt\biggr)\bigl|f''(a)\bigr|^{q} \\ &\qquad{}+\biggl(\int_{0}^{\frac{1}{2}}t(1-t)^{s}|t-\theta|\,dt+\int_{\frac{1}{2}}^{1}(1-t)^{s+1}|1-\theta-t|\,dt\biggr)\bigl|f''(b)\bigr|^{q} \biggr] \\ &\quad=\frac{(b-a)}{2}\biggl(\frac{8\theta^{3}-3\theta+1}{24}\biggr) ^{1-\frac{1}{q}}\biggl[ \biggl(\int_{0}^{\theta}t^{s+1}(\theta-t)\,dt+ \int_{\theta}^{\frac{1}{2}}t^{s+1}(t-\theta)\,dt \\ &\qquad{}+\int_{\frac{1}{2}}^{1-\theta}t^{s}(1-t) (1-\theta-t)\,dt+\int_{1-\theta}^{1}t^{s}(1-t) ( \theta+t-1)\,dt\biggr)\bigl|f''(a)\bigr|^{q} \\ &\qquad{}+\biggl(\int_{0}^{\theta}t(1-t)^{s}(\theta-t)\,dt+\int_{\theta}^{\frac{1}{2}}t(1-t)^{s}(t-\theta)\,dt \\ &\qquad{}+\int_{\frac{1}{2}}^{1-\theta}(1-t)^{s+1}(1- \theta-t)\,dt+\int_{1-\theta}^{1}(1-t)^{s+1}( \theta+t-1)\,dt\biggr)\bigl|f''(b)\bigr|^{q}\biggr] \\ &\quad=\frac{(b-a)^{2}}{2}\biggl(\frac{8\theta^{3}-3\theta+1}{24}\biggr)^{1-\frac{1}{q}}\biggl\{ \biggl[\frac{2\theta^{s+3}}{(s+2)(s+3)} \\ &\qquad{}+\frac{2(1-\theta)^{s+2}[(s+1)\theta+2]+(s+3)\theta-2}{(s+1)(s+2)(s+3)}-\frac{1-\theta}{2^{s+1}(s+1)(s+2)}\biggr]\\ &\qquad{}\times\bigl(\bigl|f''(a)\bigr|^{q}+\bigl|f''(b)\bigr|^{q} \bigr)\biggr\} ^{\frac{1}{q}}, \end{aligned}

where

\begin{aligned}& \begin{aligned}[b] \int_{0}^{1}\bigl|k(t)\bigr|\,dt={}&\int _{0}^{\frac{1}{2}}\biggl|\frac{1}{2}t(t-\theta)\biggr|\,dt+ \int_{\frac{1}{2}}^{1}\biggl|\frac{1}{2}(1-t) (1- \theta-t)\biggr|\,dt \\ ={}&\frac{1}{2}\biggl[\int_{0}^{\theta}t(\theta-t)\,dt+\int_{\theta}^{\frac{1}{2}} t(t-\theta)\,dt\\ &{}+\int _{\frac{1}{2}}^{1-\theta}(1-t) (1-\theta-t)\,dt+ \int_{1-\theta}^{1}(1-t) (t-1+\theta)\,dt\biggr] \\ ={}&\frac{8\theta^{3}-3\theta+1}{24}, \end{aligned} \\& \int_{0}^{\theta}t^{s+1}(\theta-t)\,dt=\int _{1-\theta}^{1} (1-t)^{s+1}(\theta+t-1)\,dt= \frac{\theta^{s+3}}{(s+2)(s+3)}, \\& \begin{aligned}[b] \int_{0}^{\theta}t(1-t)^{s}(\theta-t)\,dt&=\int_{1-\theta}^{1}(1-t) (\theta+t-1)t^{s} \,dt \\ &=\frac{(1-\theta)^{s+2}[(s+1)\theta+2]+(s+3)\theta-2}{(s+1)(s+2)(s+3)}, \end{aligned} \\& \begin{aligned}[b] \int_{\theta}^{\frac{1}{2}}t^{s+1}(t-\theta)\,dt&=\int_{\frac{1}{2}}^{1-\theta} (1-t)^{s+1}(1- \theta-t)\,dt\\ &=\frac{(2\theta)^{s+3}-2(s+3)\theta+s+2}{2^{s+3}(s+2)(s+3)}, \end{aligned} \end{aligned}

and

\begin{aligned} \int_{\theta}^{\frac{1}{2}}t(1-t)^{s}(t- \theta)\,dt =&\int_{\frac{1}{2}}^{1-\theta}(1-t) (1- \theta-t)t^{s} \,dt \\ =&\frac{(1-\theta)^{s+2}[(s+1)\theta+2]}{(s+1)(s+2)(s+3)}+\frac{2(s+3)^{2}\theta-s^{2}-7s-14}{2^{s+3}(s+1)(s+2)(s+3)}. \end{aligned}

In case $$0\le\theta\le\frac{1}{2}$$, by Lemma 2 and using the Hölder inequality, we have

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-(1- \theta)f\biggl(\frac{a+b}{2}\biggr) -\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le(b-a)^{2}\int_{0}^{1}\bigl|k(t)\bigr|\bigl|f'' \bigl(ta+(1-t)b\bigr)\bigr|\,dt \\ &\quad\le(b-a)^{2}\biggl[\int_{0}^{1}\bigl|k(t)\bigr|\,dt\biggr]^{1-\frac{1}{q}} \biggl[\int_{0}^{1}\bigl|k(t)\bigr| \bigl|f''\bigl(ta+(1-t)b\bigr)\bigr|^{q} \,dt \biggr]^{\frac{1}{q}} \\ &\quad\le(b-a)^{2}\biggl(\frac{3\theta-1}{24}\biggr)^{1-\frac{1}{q}}\biggl(\int_{0}^{1}\bigl|k(t)\bigr|\bigl[t^{s}\bigl|f''(a)\bigr|^{q}++(1-t)^{s}\bigl|f''(b)\bigr|^{q} \bigr]\,dt\biggr)^{\frac{1}{q}} \\ &\quad=\frac{(b-a)}{2}\biggl(\frac{3\theta-1}{24}\biggr)^{1-\frac{1}{q}} \biggl[\biggl(\int_{0}^{\frac{1}{2}}t^{s+1}|t-\theta|\,dt+\int_{\frac{1}{2}}^{1} t^{s}(1-t)|1-\theta-t|\,dt\biggr) \bigl|f''(a)\bigr|^{q} \\ &\qquad{}+\biggl(\int_{0}^{\frac{1}{2}}t(1-t)^{s}|t-\theta|\,dt+\int_{\frac{1}{2}}^{1} (1-t)^{s+1}|1-\theta-t|\,dt\biggr) \bigl|f''(b)\bigr|^{q} \biggr] \\ &\quad=\frac{(b-a)}{2}\biggl(\frac{3\theta-1}{24}\biggr)^{1-\frac{1}{q}} \biggl[\biggl(\int_{0}^{\frac{1}{2}}t^{s+1}(\theta-t)\,dt+\int_{\frac{1}{2}}^{1}t^{s}(1-t) (\theta+t-1)\,dt\biggr) \bigl|f''(a)\bigr|^{q} \\ &\qquad{}+\biggl(\int_{0}^{\frac{1}{2}}t(1-t)^{s}(\theta-t)\,dt+\int_{\frac{1}{2}}^{1}(1-t)^{s+1} (\theta+t-1)\,dt\biggr) \bigl|f''(b)\bigr|^{q} \biggr] \\ &\quad=\frac{(b-a)^{2}}{2}\biggl(\frac{3\theta-1}{24}\biggr)^{1-\frac{1}{q}} \biggl\{ \biggl[\frac{(s+3)\theta-2}{(s+1)(s+2)(s+3)}+\frac{1-\theta}{2^{s+1}(s+1)(s+2)}\biggr]\\ &\qquad{}\times \bigl(\bigl|f''(a)\bigr|^{q}+\bigl|f''(b)\bigr|^{q} \bigr)\biggr\} ^{\frac{1}{q}}, \end{aligned}

where

\begin{aligned}& \begin{aligned}[b] \int_{0}^{1}\bigl|k(t)\bigr|\,dt&=\int _{0}^{\frac{1}{2}}\biggl|\frac{1}{2}t(t-\theta)\biggr|\,dt+ \int_{\frac{1}{2}}^{1}\biggl|\frac{1}{2}(1-t) (1- \theta-t)\biggr|\,dt \\ &=\frac{1}{2}\biggl[\int_{0}^{\frac{1}{2}}t( \theta-t)\,dt+\int_{\frac{1}{2}}^{1}(1-t) (\theta+t-1)\,dt\biggr] \\ &=\frac{3\theta-1}{24}, \end{aligned} \\& \int_{0}^{\frac{1}{2}}t^{s+1}(\theta-t)\,dt=\int_{\frac{1}{2}}^{1} (1-t)^{s+1}( \theta+t-1)\,dt=\frac{(2s+6)\theta-s-2}{2^{s+3}(s+2)(s+3)}, \end{aligned}

and

\begin{aligned} \int_{0}^{\frac{1}{2}}t(1-t)^{s}( \theta-t)\,dt =&\int_{\frac{1}{2}}^{1}(1-t) ( \theta+t-1)t^{s} \,dt \\ =&\frac{2^{s+3}[(s+3)\theta-2]-2(s+3)^{2}\theta+s^{2}+7s+14}{2^{s+3}(s+1)(s+2)(s+3)}. \end{aligned}

The proof is thus completed. □

### Remark 5

If we take $$\theta=0$$ in (28), then we get a midpoint type inequality

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl(\frac{a+b}{2}\biggr)\biggr| \\ &\quad\le\frac{(b-a)^{2}}{2}\biggl(\frac{1}{24}\biggr)^{1-\frac{1}{q}} \biggl[\frac{2^{s+2}-s-3}{2^{s+1}(s+1)(s+2)(s+3)}\bigl(\bigl|f''(a)\bigr|^{q}+\bigl|f''(b)\bigr|^{q} \bigr)\biggr]^{\frac{1}{q}}. \end{aligned}
(30)

If we take $$\theta=1$$ in (29), then we get a trapezoid type inequality

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{2}\biggl(\frac{1}{4}\biggr)^{1-\frac{1}{q}} \biggl[\frac{|f''(a)|^{q}+|f''(b)|^{q}}{(s+2)(s+3)}\biggr]^{\frac{1}{q}}. \end{aligned}
(31)

If we take $$\theta=\frac{1}{3}$$ in (28), then we get a Simpson type inequality

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{f(a)+4f(\frac{a+b}{2})+f(b)}{6}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{2}\biggl(\frac{1}{81}\biggr)^{1-\frac{1}{q}} \\ &\qquad{}\times\biggl[\frac{2^{s+4}(s+1)+2^{2s+6}(s+7)-6^{s+2}(6-2s)-8(s+3)3^{s+2}}{6^{s+3}(s+1)(s+2)(s+3)}\bigl(\bigl|f''(a)\bigr|^{q}+\bigl|f''(b)\bigr|^{q} \bigr)\biggr]^{\frac{1}{q}}. \end{aligned}
(32)

If we take $$\theta=\frac{1}{2}$$ in (28) or (29), then we get an averaged midpoint-trapezoid type inequality

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{4}\biggl[f(a)+2f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le\frac{(b-a)^{2}}{2}\biggl(\frac{1}{48}\biggr)^{1-\frac{1}{q}} \biggl[\frac{2^{s+1}(s-1)+s+3}{2^{s+2}(s+1)(s+2)(s+3)}\bigl(\bigl|f''(a)\bigr|^{q}+\bigl|f''(b)\bigr|^{q} \bigr)\biggr]^{\frac{1}{q}}. \end{aligned}
(33)

### Corollary 2

Let $$I\subset[0,\infty)$$, $$f:I\to\mathbf{R}$$ be a twice differentiable function on $$I^{\circ}$$ such that $$f''\in L^{1}[a,b]$$, where $$a,b\in I$$ with $$a< b$$. If $$|f''|$$ is s-convex in the second sense on $$[a,b]$$ for some fixed $$s\in(0,1]$$, then the following inequalities hold:

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(x)\,dx-(1-\theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{2}\biggl[\frac{2\theta^{s+3}}{(s+2)(s+3)} \\ &\qquad{}+\frac{2(1-\theta)^{s+2}((s+1)\theta+2)+(s+3)\theta-2}{(s+1)(s+2)(s+3)}-\frac{1-\theta}{2^{s+1}(s+1)(s+2)}\biggr] \bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr|\bigr) \end{aligned}
(34)

for $$0\le\theta\le\frac{1}{2}$$ and

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(x)\,dx-(1-\theta)f\biggl(\frac{a+b}{2}\biggr)-\theta\frac{f(a)+f(b)}{2}\biggr| \\ &\quad\le\frac{(b-a)^{2}}{2}\biggl[\frac{(s+3)\theta-2}{ (s+1)(s+2)(s+3)}+\frac{1-\theta}{2^{s+1}(s+1)(s+2)}\biggr] \bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr|\bigr) \end{aligned}
(35)

for $$\frac{1}{2}\le\theta\le 1$$.

### Proof

Inequalities (34) and (35) are immediate by setting $$q=1$$ in (28) and (29) of Theorem 4. □

### Remark 6

If we take $$\theta=0$$ in (34), then we get a midpoint type inequality

$$\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl(\frac{a+b}{2}\biggr)\biggr|\le\frac{2^{s+2}-s-3}{2^{s+2}(s+1)(s+2)(s+3)}(b-a)^{2} \bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr|\bigr).$$
(36)

If we take $$\theta=1$$ in (35), then we get a trapezoid type inequality

$$\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{f(a)+f(b)}{2}\biggr|\le\frac{1}{2(s+2)(s+3)}(b-a)^{2} \bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr|\bigr).$$
(37)

If we take $$\theta=\frac{1}{3}$$ in (34), then we get a Simpson type inequality

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{6}\biggl[f(a)+4f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le \frac{(s+1)2^{s+3}+(s+7)2^{2s+5}-(3-s)6^{s+2}-4(s+3) 3^{s+2}}{6^{s+3}(s+1)(s+2)(s+3)} \\ &\qquad{}\times(b-a)^{2}\bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr| \bigr). \end{aligned}
(38)

If we take $$\theta=\frac{1}{2}$$ in (34) or (35), then we get an averaged midpoint-trapezoid type inequality

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{4}\biggl[f(a)+2f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le\frac{s+3-2^{s+1}(1-s)}{2^{s+3}(s+1)(s+2)(s+3)}(b-a)^{2}\bigl(\bigl|f''(a)\bigr|+\bigl|f''(b)\bigr| \bigr). \end{aligned}
(39)

### Remark 7

If we put $$M=\sup_{x\in[a,b]}|f''|$$ in (36)-(39), then we have

\begin{aligned}& \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-f\biggl( \frac{a+b}{2}\biggr)\biggr|\le \frac{M(2^{s+2}-s-3)(b-a)^{2}}{2^{s+1}(s+1)(s+2)(s+3)}, \end{aligned}
(40)
\begin{aligned}& \biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt- \frac{f(a)+f(b)}{2}\biggr|\le \frac{M(b-a)^{2}}{(s+2)(s+3)}, \end{aligned}
(41)
\begin{aligned}& \begin{aligned}[b] &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{6}\biggl[f(a)+4f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le\frac{(s+1)2^{s+3}+(s+7)2^{2s+5}-(3-s)6^{s+2} -4(s+3)3^{s+2}}{3(s+1)(s+2)(s+3)6^{s+2}}M(b-a)^{2} \end{aligned} \end{aligned}
(42)

and

\begin{aligned} &\biggl|\frac{1}{b-a}\int_{a}^{b}f(t)\,dt-\frac{1}{4}\biggl[f(a)+2f\biggl(\frac{a+b}{2}\biggr)+f(b)\biggr]\biggr| \\ &\quad\le\frac{s+3-2^{s+1}(1-s)}{2^{s+2}(s+1)(s+2)(s+3)}M(b-a)^{2}. \end{aligned}
(43)

### Remark 8

If we further take $$s=1$$ in (40)-(43), i.e., for functions f with convex $$|f''|$$, then we recapture inequalities (23)-(26).

## 3 Applications to special means

We consider the means for arbitrary positive numbers α, β ($$\alpha\neq\beta$$) as follows:

1. (1)

The arithmetic mean

$$A(\alpha, \beta)=\frac{\alpha+\beta}{2}.$$
2. (2)

The geometric mean

$$G(\alpha, \beta)=\sqrt{\alpha\beta}.$$
3. (3)

The harmonic mean

$$H(\alpha, \beta)=\frac{2\alpha\beta}{\alpha+\beta}.$$
4. (4)

The logarithmic mean

$$L(\alpha, \beta)=\frac{\beta-\alpha}{\ln\beta-\ln\alpha}.$$
5. (5)

The generalized log-mean

$$L_{p}(\alpha,\beta)=\biggl[\frac{\beta^{p+1}-\alpha^{p+1}}{(p+1)(\beta-\alpha)}\biggr] ^{\frac{1}{p}},\quad p\neq -1,0.$$
6. (6)

The identric mean

$$I(\alpha,\beta)=\frac{1}{e}\biggl(\frac{\beta^{\beta}}{\alpha^{\alpha}}\biggr) \frac{1}{\beta-\alpha}.$$

### Proposition 1

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

\begin{aligned}& \bigl|A^{s}(a,b)-L_{s}^{s}(a,b)\bigr|\le\frac{s(1-s)(b-a)^{2}}{24a^{2-s}}, \\& \bigl|A\bigl(a^{s},b^{s}\bigr)-L_{s}^{s}(a,b)\bigr| \le\frac{s(1-s)(b-a)^{2}}{12a^{2-s}}, \\& \biggl|\frac{2A^{s}(a,b)+A(a^{s},b^{s})}{3}-L_{s}^{s}(a,b)\biggr|\le\frac{s(1-s)(b-a)^{2}}{81a^{2-s}}, \end{aligned}

and

$$\biggl|\frac{A^{s}(a,b)+A(a^{s},b^{s})}{2}-L_{s}^{s}(a,b)\biggr|\le\frac{s(1-s)(b-a)^{2}}{48a^{2-s}}.$$

### Proof

The assertion follows from applying inequalities (23)-(26) to the mapping $$f(x)=x^{s}$$, $$x\in[a,b]$$, which implies that $$|f''(x)|=s(1-s)x^{s-2}$$ is convex on $$[a,b]$$, and we may take $$M=\frac{s(1-s)}{a^{2-s}}$$. □

### Proposition 2

Let $$0< a< b$$. Then we have

\begin{aligned}& \bigl|A^{-1}(a,b)-L^{-1}(a,b)\bigr|\le\frac{(b-a)^{2}}{12a^{3}}, \\& \bigl|H^{-1}(a,b)-L^{-1}(a,b)\bigr|\le\frac{(b-a)^{2}}{6a^{3}}, \\& \biggl|\frac{2A^{-1}(a,b)+H^{-1}(a,b)}{3}-L^{-1}(a,b)\biggr|\le\frac{2(b-a)^{2}}{81a^{3}} \end{aligned}

and

$$\biggl|\frac{A^{-1}(a,b)+H^{-1}(a,b)}{2}-L^{-1}(a,b)\biggr|\le\frac{(b-a)^{2}}{24a^{3}}.$$

### Proof

The assertion follows from applying inequalities (23)-(26) to the mapping $$f(x)=\frac{1}{x}$$, $$x\in[a,b]$$, which implies that $$|f''(x)|=\frac{2}{x^{3}}$$ is convex on $$[a,b]$$, and we may take $$M=\frac{2}{a^{3}}$$. □

### Proposition 3

Let $$a,b\in\mathbf{R}$$, $$0< a< b$$. Then we have

\begin{aligned}& \bigl|\ln A(a,b)-\ln I(a,b)\bigr|\le\frac{(b-a)^{2}}{24a^{2}}, \\& \bigl|\ln G(a,b)-\ln I(a,b)\bigr|\le\frac{(b-a)^{2}}{12a^{2}}, \\& \biggl|\frac{2\ln A(a,b)+\ln G(a,b)}{3}-\ln I(a,b)\biggr|\le\frac{(b-a)^{2}}{81a^{2}} \end{aligned}

and

$$\biggl|\frac{\ln A(a,b)+\ln G(a,b)}{2}-\ln I(a,b)\biggr|\le\frac{(b-a)^{2}}{48a^{2}}.$$

### Proof

The assertion follows from applying inequalities (23)-(26) to the mapping $$f(x)=\ln x$$, $$x\in[a,b]$$, which implies that $$|f''(x)|=\frac{1}{x^{2}}$$ is convex on $$[a,b]$$, and we may take $$M=\frac{1}{a^{2}}$$. □

## References

1. Hudzik, H, Maligranda, L: Some remarks on s-convex functions. Aequ. Math. 17(2), 100-111 (1994)

2. Özdemir, ME, Yildiz, Ç, Akdemir, AO, Set, E: On some inequalities for s-convex functions and applications. J. Inequal. Appl. 2013, 333 (2013)

3. Liu, Z: A note on Ostrowski type inequalities related to some s-convex functions in the second sense. Bull. Korean Math. Soc. 49(4), 775-785 (2012)

4. Sarikaya, MZ, Aktan, N: On the generalization of some integral inequalities and their applications. arXiv:1005.2879v1 [math.CA] (17 May 2010)

## Acknowledgements

The author wishes to thank the referees for their helpful comments and suggestions.

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Correspondence to Zheng Liu.

### Competing interests

The author declares that he has no competing interests.

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Liu, Z. Remarks on some inequalities for s-convex functions and applications. J Inequal Appl 2015, 333 (2015). https://doi.org/10.1186/s13660-015-0851-9