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

• (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 $f\notin K_{s}^{2}$.

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) The arithmetic mean

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

• (2) The geometric mean

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

• (3) The harmonic mean

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

• (4) The logarithmic mean

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

• (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) 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 $M=\frac{1}{a^{2}}$. □